The Congruences of a Finite Lattice: A "Proof-by-Picture" Approach [3 ed.] 3031290623, 9783031290626

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George Grätzer

The Congruences of a Finite Lattice A “Proof-by-Picture” Approach Third Edition

George Grätzer

The Congruences of a Finite Lattice A “Proof-by-Picture” Approach Third Edition

George Grätzer Toronto, ON, Canada

ISBN 978-3-031-29062-6 ISBN 978-3-031-29063-3 (eBook) https://doi.org/10.1007/978-3-031-29063-3 Mathematics Subject Classification (2020): 06B10, 06D05, 06C05, 06C10 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2006, 2016, 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This book is published under the imprint Birkhäuser, www.birkhauser-science.com by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

o Fuchs, aszl´ To L´ my thesis advisor, my teacher, who taught me to set the bar high; and to my coauthors, who helped to raise the bar; abor Cz´edli, especially to G´ for his help in the last dozen years or so.

Short Contents

Glossary of Notation

xix

Picture Gallery

xxiii

Preface

xxv

Introduction

I

xxvii

A Brief Introduction to Lattices

1

1

Basic Concepts

2

Special Concepts

17

3

Congruences

35

4

Planar Semimodular Lattices

45

II

Some Special Techniques

67

5

Chopped Lattices

69

6

Boolean Triples

79

7

Cubic Extensions

93

III 8

3

RTs

99

Sectionally Complemented RT vii

101

viii

SHORT CONTENTS

9

Minimal RT

117

10

Semimodular RT

129

11

Rectangular RT

141

12

Modular RT

147

13

Uniform RT

161

IV

ETs

175

14

Sectionally Complemented ET

177

15

Semimodular ET

185

16

Isoform ET

195

17

Magic Wands

213

V

Congruence Lattices of Two Related Lattices

18

Sublattices

237

19

Ideals

249

20

Two Convex Sublattices

265

21

Tensor Extensions

275

VI

The Ordered Set of Principal Congruences

235

295

22

The RT for Principal Congruences

297

23

Minimal RTs

307

24

Principal Congruence Representable Sets

339

25

Isotone Maps

353

VII 26

Congruence Extensions and Prime Intervals The Prime-projectivity Lemma

359 361

SHORT CONTENTS

ix

27

The Swing Lemma

365

28

Fork Congruences

381

VIII 29

The Six Congruence Properties of SPS lattices 389

Six Major Properties

391

Bibliography

403

Index

421

Contents

Glossary of Notation

xix xxiii

Picture Gallery Preface

xxv xxvii xxvii xxix xxx

Introduction The Topics Proof-by-Picture Outline and notation

I 1

A Brief Introduction to Lattices Basic Concepts 1.1 Ordering 1.1.1 Ordered sets 1.1.2 Diagrams 1.1.3 Constructions of ordered sets 1.1.4 Partitions 1.2 Lattices and semilattices 1.2.1 Lattices 1.2.2 Semilattices and closure systems 1.3 Some algebraic concepts 1.3.1 Homomorphisms 1.3.2 Sublattices 1.3.3 Congruences xi

1 3 3 3 5 6 7 8 8 10 11 11 12 13

xii

Contents

2

Special Concepts 2.1 Elements and lattices 2.2 Direct and subdirect products 2.3 Terms and identities 2.4 Gluing and generalizations 2.4.1 Gluing 2.4.2 Generalizations 2.5 Modular and distributive lattices 2.5.1 The characterization theorems 2.5.2 Finite distributive lattices 2.5.3 Finite modular lattices

17 17 18 20 25 25 27 27 27 30 32

3

Congruences 3.1 Congruence spreading 3.2 Finite lattices and prime intervals 3.3 Congruence-preserving extensions and variants

35 35 38 40

4

Planar Semimodular Lattices 4.1 Planar lattices 4.2 Two acronyms: SPS and SR 4.3 SPS lattices 4.4 Forks 4.5 Rectangular lattices 4.6 Rectangular intervals 4.7 Special diagrams for SR lattices 4.8 Natural diagrams and C1 -diagrams 4.9 Discussion

45 45 47 48 50 52 57 60 62 64

II

Some Special Techniques

67

5

Chopped Lattices 5.1 Basic definitions 5.2 Compatible vectors of elements 5.3 Compatible vectors of congruences 5.4 From the chopped lattice to the ideal lattice 5.5 Sectional complementation

69 69 71 72 74 76

6

Boolean Triples 6.1 The general construction 6.2 The congruence-preserving extension property 6.3 The distributive case 6.4 Two interesting intervals 6.5 Discussion

79 79 82 84 85 92

Contents

7

III

Cubic Extensions 7.1 The construction 7.2 The basic property

RTs

xiii 93 93 95

99

8

Sectionally Complemented RT 8.1 The Basic RT 8.2 Proof-by-Picture 8.3 Computing 8.4 Sectionally complemented lattices 8.5 The N-relation 8.6 Discussion

101 101 102 104 106 108 113

9

Minimal RT 9.1 The results 9.2 Proof-by-Picture for the minimal construction 9.3 The formal construction 9.4 Proof-by-Picture for minimality 9.5 Computing minimality 9.6 Discussion

117 117 118 120 121 123 124

10

Semimodular RT 129 10.1 Semimodular lattices 129 10.2 Proof-by-Picture 130 10.3 Construction and proof 131 10.4 All congruences principal RT for planar semimodular lattices 138 10.5 Discussion 139

11

Rectangular RT 11.1 Results 11.2 Proof-by-Picture 11.3 All congruences principal RT 11.4 Discussion

141 141 142 143 143

12

Modular RT 12.1 Modular lattices 12.2 Proof-by-Picture 12.3 Construction and proof 12.4 Discussion

147 147 148 151 157

xiv 13

IV

Contents

Uniform RT 13.1 Uniform lattices 13.2 Proof-by-Picture 13.3 The lattice N (A, B) 13.4 Formal proof 13.5 Discussion

ETs

161 161 162 164 169 170

175

14

Sectionally Complemented ET 14.1 Sectionally complemented lattices 14.2 Proof-by-Picture 14.3 Easy extensions 14.4 Formal proof 14.5 Discussion

177 177 178 180 182 184

15

Semimodular ET 15.1 Semimodular lattices 15.2 Proof-by-Picture 15.3 The conduit 15.4 The construction 15.5 Formal proof 15.6 Rectangular ET 15.7 Discussion

185 185 185 188 189 190 191 194

16

Isoform ET 16.1 Isoform lattices 16.2 Proof-by-Picture 16.3 Formal construction 16.4 The congruences 16.5 The isoform property 16.6 Discussion 16.6.1 Variants 16.6.2 Problems 16.6.3 The Congruence Lattice and the Automorphism Group 16.6.4 More problems

195 195 195 199 205 206 207 207 209 210 211

17

Magic Wands 17.1 Constructing congruence lattices 17.2 Proof-by-Picture for bijective maps 17.3 Verification for bijective maps 17.4 2/3-Boolean triples 17.5 Proof-by-Picture for surjective maps

213 213 215 218 221 227

Contents

17.6 Verification for surjective maps 17.7 Discussion

xv 229 230

V

Congruence Lattices of Two Related Lattices

235

18

Sublattices 18.1 The results 18.2 Proof-by-Picture 18.3 Multi-coloring 18.4 Formal proof 18.5 Discussion

237 237 239 241 242 244

19

Ideals 19.1 The results 19.2 Proof-by-Picture for the main result 19.3 Formal proof 19.4 Proof-by-Picture for planar lattices 19.5 Discussion

249 249 250 252 261 263

20

Two Convex Sublattices 20.1 Introduction 20.2 Proof-by-Picture 20.3 Proof 20.4 Discussion

265 265 267 269 272

21

Tensor Extensions 21.1 The problem 21.2 Three unary functions 21.3 Defining tensor extensions 21.4 Computing 21.5 Congruences 21.6 The congruence isomorphism 21.7 Discussion

275 275 276 278 280 284 292 293

VI

The Ordered Set of Principal Congruences

295

22

The RT for Principal Congruences 22.1 Representing the ordered set of principal congruences 22.2 Proving the RT 22.3 An independence theorem 22.3.1 Frucht lattices 22.3.2 An independence result 22.4 Discussion

297 297 297 303 303 303 304

xvi

Contents

23

Minimal RTs 23.1 The Minimal RT 23.2 Three or more dual atoms 23.3 Exactly two dual atoms 23.3.1 Constructing the lattice L 23.3.2 Fusion of ordered sets 23.3.3 Splitting an element of an ordered set 23.3.4 Admissible congruences and extensions 23.3.5 The Bridge Theorem 23.3.6 Some technical results and proofs 23.4 Small distributive lattices 23.5 Full representability and planarity 23.6 Discussion

307 307 309 311 311 313 316 318 321 328 332 335 336

24

Principal Congruence Representable Sets 24.1 Chain representability 24.2 Proving the Necessity Theorem 24.3 Proof-by-Picture for the Sufficiency Theorem 24.3.1 A colored chain 24.3.2 The frame lattice Frame C and the lattice W (p, q) 24.3.3 Flag lattices 24.4 Construction for the Sufficiency Theorem 24.5 Proving the Sufficiency Theorem 24.5.1 Two preliminary lemmas 24.5.2 The congruences of a W (p, q) lattice 24.5.3 The congruences of flag lattices 24.5.4 The congruences of L 24.5.5 Principal congruences of L 24.6 Discussion

339 339 341 342 342 343 343 343 347 347 347 348 348 351 352

25

Isotone Maps 25.1 Two isotone maps 25.2 Sublattices, sketching the proof 25.3 Isotone surjective maps 25.4 Proving the Representation Theorem 25.5 Discussion

353 353 355 355 356 357

VII 26

Congruence Extensions and Prime Intervals The 26.1 26.2 26.3

Prime-projectivity Lemma Introduction Proof Discussion

359 361 361 363 364

Contents

27

The 27.1 27.2 27.3 27.4 27.5

28

Fork Congruences 28.1 The statements 28.2 Proofs 28.3 Discussion

VIII 29

Swing Lemma The statement Proving the Swing Lemma Some variants and consequences The Two-Cover Condition is not sufficient Applying the Swing Lemma to trajectories

xvii 365 365 367 375 377 378 381 381 382 387

The Six Congruence Properties of SPS lattices 389

Six Major Properties 29.1 Introduction 29.2 Cz´edli’s four properties 29.2.1 Proofs 29.3 The 3P3C property 29.3.1 Some relations 29.3.2 Proof 29.4 Discussion

391 391 391 392 395 395 399 401

Bibliography

403

Index

421

Glossary of Notation

Symbol

Explanation

Page

a∗ (a∗ ) 0I and 1I Atom(U ) Aut L Bn Cl (L), Cll (L), Cul (L) Cn con(a, b) con(H) con(p) Con L J(Con L) Cr (L), Clr (L), Cur (L) Cube K D Diag(K) Dn P ext : Con K → Con L fil(a) fil(H) Flag(c3 ) FreeD (3) FreeK (H) FreeM (3) Fuse(P, A) hom{∨,0} (X, Y ) id(a) id(H) Id L (Id)

the unique lower (upper) cover of a the zero and unit of the interval I set of atoms of the ideal U automorphism group of L Boolean lattice with n atoms left boundary chains of a planar lattice n-element chain smallest congruence under which a ≡ b smallest congruence collapsing H principal congruence generated by p congruence lattice of L join-irreducible congruences of L right boundary chains of a planar lattice L cubic extension of K class (variety) of distributive lattices diagonal embedding of K into Cube K down sets of the ordered set P for K ≤ L, extension map: α 7→ conL (α) filter generated by the element a filter generated by the set H flag lattice free distributive lattice on three generators free lattice generated by H in a variety K free modular lattice on three generators Fusion of ordered sets {∨, 0}-homomorphisms of X into Y ideal generated by the element a ideal generated by the set H ideal lattice of L condition to define ideals xix

17 12 106 11 4 47 4 15 15 38 14, 70 38 47 93 22 93 5, 9 41 13 13 343 23 23 25 313 283 13 13 13, 71 13, 71

xx

Table of Notation

Symbol

Explanation

J(D) J+ (D) J(a) ker(γ) L Lbottom , Ltop lc(L) Lleft , Lright M Max mcr(n) mcr(n, V) M3 M3 [L] M3 [L, a] M3 [L, a, b] M3 [a, b] M(D) N5 N5,5 N6 = N (p, q) N (A, B) Part A Pow X Princ L rc(L) Prime(L) re : Con L → Con K (SD∨ ), (SD∧ ) SecComp SemiMod Simp K (SP∨ ), (SP∧ ) Split(P, a) S(p, q), W (p, q) sub(H)

ordered set of join-irreducible elements of D J(D) ∪ {0, 1} set of join-irreducible elements below a congruence kernel of γ class (variety) of all lattices bottom and top of a rectangular lattice L left corner of a rectangular lattice L left and right boundary of L class (variety) of modular lattices maximal elements of an ordered set minimal congruence representation function mcr for a class V five-element modular nondistributive lattice ordered set of Boolean triples of L interval of M3 [L] interval of M3 [L] Boolean triples of [a, b] ordered set of meet-irreducible elements of D five-element nonmodular lattice seven-element nonmodular lattice six-element nonmodular lattice a lattice construction partition lattice of A power set lattice of X ordered set of principal congruences L right corner of a rectangular lattice set of prime intervals of L restriction map: α 7→ α⌉K semidistributive laws class of sectionally complemented lattices class of semimodular lattices simple extension of K substitution properties splitting an element small lattices used in lattice constructions sublattice generated by H

17 305 17 15 22 55 47 47 22 71 126 126 28 80 85 88 80 17 28 118 102 164 7, 9 5 37 47 38 40 50 18 139 93 13 316 298 12

p q, p swings to q set of all trajectories of L

365 48

,

in

,

ex

↶

↶ ↶ ↶

Swing, Traj L

Page

Glossary of Notation

Symbol

Explanation

Page

binary relations congruences zero of Part A and Con L unit of Part A and Con L a and b in the same block of π a and b in relation ϱ a and b in relation α block containing a blocks represented by H product of α and β reflexive product of α and β restriction of α to the sublattice K quotient lattice quotient congruence projection map: L1 × · · · × Ln → Li direct product of congruences

3 13 8 8 7 3 3 7, 13 7 18 26 13 15 16 18 18

ordering ordering, inverse notation K a sublattice of L ordering of P restricted to a subset Q a incomparable with b a is covered by b b covers a zero, least element of an ordered set unit, largest element of an ordered set join operation least upper bound of H meet operation greatest lower bound of H dual of the ordered set (lattice) P interval down set generated by H down set generated by {a} ordered set (lattice) P isomorphic to Q

3 3 12 4 3 5 5 4 4 8 3 8 4 4, 10 12 5 5 4, 11

Relations and Congruences A2 α, β, . . . 0 1 a ≡ b (mod π) aϱb a ≡ b (mod α) a/π H/π α◦β r α◦β α⌉K L/α β/α πi α×β Ordered sets ≤, < ≥, > K≤L ≤Q a∥b a≺b b≻a 0 1 a W∨ b H a ∧ V b H Pδ [a, b] ↓H ↓a P ∼ =Q

xxi

xxii

Table of Notation

Symbol

Explanation

Page

direct product of P and Q sum of P and Q glued sum of P and Q tensor extension of A by B tensor product of A and B modular lattice construction

6, 18 6 16 278 275 151

Constructions P ×Q P +Q P ∔Q A[B] A⊗B U ⊛V Prime intervals p, q, . . . con(p) Prime(L) Princ L

prime intervals principal congruence generated by p set of prime intervals of L the ordered set of principal congruences

38 38 37

[a, b] perspective to [c, d] [a, b] up-perspective to [c, d]

32 32

[a, b] ∼ [c, d] [a, b] ≈ [c, d] [a, b] ↠ [c, d]

[a, b] down-perspective to [c, d] [a, b] projective to [c, d] [a, b] congruence-perspective onto [c, d]

32 32 36

[a, b] ↠ [c, d]

[a, b] up congruence-perspective onto [c, d]

35

Perspectivities [a, b] ∼ [c, d] up [a, b] ∼ [c, d] dn

up dn

[a, b] ↠ [c, d] [a, b] ⇒ [c, d] p p −→ q p-up p −→ q

[a, b] down congruence-perspective onto [c, d] [a, b] congruence-projective onto [c, d] p prime-perspective to q p prime-perspective up to q

35 35 361 361

p prime-perspective down to q p prime-projective to q p swings to q

361 361 365

↶

q

internal swing

365

↶

q

external swing

365

p-dn

p −→ q p p =⇒ q q p ↶

p p

in ex

Miscellaneous x ∅ ⌉

closure of x empty set restriction

10 4 4

slim, planar, semimodular lattice slim rectangular lattice

47 47

Acronyms SPS lattice SR lattice

Picture Gallery

C3

C2

N6

N5,5

N5

S7

B2

S8

M3

xxiii

Preface

A book such as this is largely autobiographical; it references about a third of my (mathematical) publications from 1956 to 2022. Compared with the first edition (in 2006), this book grew from 281 to 430 pages, from 265 to 360 statements, and from 123 to 262 references. The manuscript was read for the publisher by Friedrich Wehrung; he offered many corrections. I received detailed reports from G´abor Cz´edli, Shriram K. Nimbhorkar, Sylvia Pulmannov´a, Andreja Tepavˇcevi´c, offering new evidence that Fred’s Law is true (a manuscript of n characters has 2n typos; this book has about 900,000 characters). And after all this, I received a 30-page report from Gregory L. Cherlin, full of corrections and suggestions. I’m not sure how I can thank him for his contribution. I updated the Introduction and the freely available Part I: arXiv:2104.06539 https://www.researchgate.net/publication/360184868

Toronto, Ontario February 2, 2023

George Gr¨atzer

xxv

Introduction

The topics This book is an introduction to congruences of finite lattices, which naturally splits into four fields of research: A. Congruence lattices of finite lattices. B. The ordered set of principal congruences of finite lattices. C. The congruence structure of finite lattices. D. Congruence properties of slim, planar, semimodular (SPS) lattices. These topics cover about 80 years of research and 250 papers. Some of the results extend naturally, for instance, to lattices of finite length and to some classes of universal algebras. To keep this book reasonably short, we do not include these results. Topic A. Congruence lattices of finite lattices The congruences of a finite lattice L form a lattice, called the congruence lattice of L, denoted by Con L. According to a 1942 result of N. Funayama and T. Nakayama [95], the lattice Con L is a finite distributive lattice. The converse is a result of R. P. Dilworth from around 1944 (see the book [22]). Basic Representation Theorem. Every finite distributive lattice D can be represented as the congruence lattice of a finite lattice L. We will refer to this result as the Basic RT (Representation Theorem). The Basic RT was not published until 1962 in my joint paper with E. T. Schmidt [172]. A large number of papers have strengthened and generalized the Basic RT. These papers form two distinct subfields: A1. RTs of finite distributive lattices as congruence lattices of finite lattices with special properties. xxvii

xxviii

Introduction

A2. The Congruence Lattice Problem (CLP): Can congruence lattices of general lattices be characterized as distributive algebraic lattices? A1. A finite distributive lattice D of more than one element is determined by the ordered set J(D) of join-irreducible elements. So a representation of D as the congruence lattice of a finite lattice L is the same as a representation of a finite ordered set P (= J(D)), as the ordered set, J(Con L), of join-irreducible congruences of a finite lattice L. A join-irreducible congruence of a finite lattice of more than one element L is exactly the same as a congruence of the form con(a, b), where a ≺ b in L; that is, the smallest congruence collapsing a prime interval [a, b]. Therefore, it is enough to concentrate on such congruences, and make sure that they are ordered as required by P . A2. The infinite case is much different. There are really only two general positive results. 1. The ideal lattice of a distributive lattice with zero is the congruence ak [230]). lattice of a lattice (see E. T. Schmidt [241] and also P. Pudl´ 2. Any distributive algebraic lattice with at most ℵ1 compact elements is the congruence lattice of a lattice (A. P. Huhn [214] and [215], see also H. Dobbertin [80]). The big breakthrough for negative results came in 2007 in F. Wehrung [258] (based on his paper [255]). Wehrung proved that there is a distributive algebraic lattice with ℵω+1 compact elements that is not representable as the congruence lattice of a lattice. P. R˚ uˇziˇcka [235] improved this result: there is a distributive algebraic lattice with ℵ2 compact elements that is not representable as the congruence lattice of a lattice. This is the sharp bound. This book deals almost exclusively with the finite case. F. Wehrung [259]– [261] (Chapters 7–9 of the book LTS1-[206]) provide a detailed review of the infinite case. The two types of RTs RTs for the finite case are all of the same general type. We represent a finite distributive lattice D as the congruence lattice of a “nice” finite lattice L. For instance, in my joint 1962 paper with E. T. Schmidt [172], we proved that the finite lattice L for the Basic RT can be constructed as a sectionally complemented lattice. To understand the second, more sophisticated, type of RT, we need the concept of a congruence-preserving extension. Let L be a lattice, and let K be a sublattice of L. In general, there is not much connection between the congruence lattices of L and K. If they happen to be naturally isomorphic, we call L a congruence-preserving extension of K. (More formally, see Section 3.3.) For sectionally complemented lattices, the congruence-preserving extension theorem, ET, was published in my 1999 paper with E. T. Schmidt [184], Every

Introduction

xxix

finite lattice K has a finite, sectionally complemented, congruence-preserving extension L. In particular this reduces the result of [172] to the Basic RT. Reading this statement for the first time, it is difficult to appreciate how much stronger this theorem is than the Basic RT. For a finite distributive lattice D, the 1962 theorem provides a finite sectionally complemented lattice L whose congruence lattice is isomorphic to D; the 1999 theorem starts with an arbitrary finite lattice K, and builds a sectionally complemented lattice L extending it with the “same” congruence structure. Topic B. The ordered set of principal congruences of finite lattices A large part of this book investigates the congruence lattice, Con L, of a finite lattice L. But Con L is not the only interesting congruence construct we can associate with a finite lattice L. A newer one, from a decade ago, is Princ L, the ordered set of principal congruences of L. We discuss this topic in Part VI. Topic C. The congruence structure of finite lattices The spreading of a congruence from a prime interval to another prime interval involves intervals of arbitrary size. Can we describe such a spreading with prime intervals only? We can indeed, by introducing the concept of prime-projectivity (see Chapter 26), and obtaining the Prime-projectivity Lemma (see my paper [112]). Then in Chapter 27, we develop much sharper forms of this result for SPS (slim, planar, semimodular) lattices. The main result is the Swing Lemma (see my paper [114]), from which we derive many of the known results of G. Cz´edli and myself concerning congruences of SPS lattices. Topic D. Congruence properties of slim, planar, semimodular (SPS) lattices. A finite ordered set P satisfies the Two-Cover Condition, if any element of P has at most two covers. The Two-Cover Theorem (Theorem 27.7) states that the ordered set of join-irreducible congruences of an SPS lattice L has the Two-Cover Condition (see my paper [117]). This theorem is the start of a new field covered in Part VIII.

Proof-by-Picture In 1960, trying with E. T. Schmidt to prove the Basic RT (unpublished at the time), we came up with the construction—more or less—as presented in Section 8.2. In 1960, we did not anticipate my 1968 result with H. Lakser [139], establishing that the construction of a chopped lattice solves the problem. So we translated the chopped lattice construction to a closure space, as in

xxx

Introduction

Section 8.4, proved that the closed sets form a sectionally complemented lattice L, and based on that, we verified that the congruence lattice of L represents the given finite distributive lattice. When we submitted the paper [172] for publication, it had a three-page section explaining the chopped lattice construction and its translation to a closure space. The referee was strict: “You cannot have a three-page explanation for a two-page proof.” I believe that in the 50 plus years since the publication of that article, few readers have developed an understanding of the idea behind the published proof. The referee’s dictum is quite in keeping with mathematical tradition and practice. When mathematicians discuss new results, they explain the constructions and the ideas with examples; when these same results are published, the motivation and the examples are largely gone. We publish definitions, constructions, and formal proofs (and conjectures, Paul Erd˝os would have added). After Gauss proved one of his famous results, he was not yet ready to publicize it because the proof gave away too much as to how the theorem was discovered. “I have had my results for a long time: but I do not yet know how I am to arrive at them,” Gauss is quoted in A. Arber [13]. I try to break with this tradition in this book. In many chapters, after stating the main result, I include a section: Proof-by-Picture. This is a misnomer. A Proof-by-Picture is not a proof. The Pythagorean Theorem has many well-known Proofs-by-Picture—sometimes called “Visual Proofs;” these are really proofs. My Proof-by-Picture is an attempt to convey the idea of the proof. I trust that if the idea is properly understood, the reader should be able to skip the formal proof, or should at least have less trouble reading it. Think of a Proof-by-Picture as a lecture to an informed audience, concluding with “its formal details now you can provide.”

Outline and notation Part I. A Brief Introduction to Lattices In the last paragraph, I call an audience “informed” if they are familiar with the basic concepts and techniques of lattice theory. Part I provides this. It can be downloaded at arXiv:submit/4276288 https://www.researchgate.net/publication/360184868 I am quite selective as to what to include. There are few proofs in this part (with a few exceptions) they are easy enough for the readers to work them out on their own. For proofs, lots of exercises, and a more detailed exposition, I refer the reader to my book LTF-[105]. This part has only two sections with novel results.

Introduction

xxxi

Part II. Some Special Techniques Most of the research in this book deals with RTs; lattices with certain properties are constructed with prescribed congruence structures. The constructions are ad hoc. Nevertheless, there are three basic techniques to verify them. • Chopped lattices, used in almost every chapter in Parts III–V. • Boolean triples, used in Chapters 12, 14, and 17, and generalized in Chapter 21. also used in some papers that did not make it in this book, for instance, my joint paper with E. T. Schmidt [186]. • Cubic extensions, used in most chapters of Part IV. These are presented in Part II with proofs. There are two more basic techniques. Multi-coloring is used in several papers; however, it appears in the book only in Chapter 18, so we introduce it there. Pruning is utilized in Chapters 13 and 16—it would seem to qualify for Part II; however, there are only concrete uses of pruning, there is no general theory to discuss. Part III. RTs This major part contains the RTs of congruence lattices of finite lattices, requiring only chopped lattices from Part II. I cover the following topics. • The Basic RT and the RT for sectionally complemented lattices in Chapter 8 (my joint paper with E. T. Schmidt [172], P. Crawley and R. P. Dilworth [24]; see also the book [22]). The closure relation of Section 8.4 is generalized in Section 8.5 using the N-relation (called D-relation or J-relation in the literature) due to J. B. Nation et al. • Minimal representations in Chapter 9; that is, for a given |J(D)|, we minimize the size of L representing the finite distributive lattice D (G. Gr¨atzer, H. Lakser, and E. T. Schmidt [156], G. Gr¨atzer, Rival, and N. Zaguia [168]). • The semimodular RT in Chapter 10 (G. Gr¨ atzer, H. Lakser, and E. T. Schmidt [159]). • The rectangular RT (my joint paper with E. Knapp [137]) in Chapter 11. • The RT for modular lattices in Chapter 12 (E. T. Schmidt [238] and my joint paper with E. T. Schmidt [189]); we are forced to represent with a countable lattice L, since the congruence lattice of a finite modular lattice is always Boolean. • The RT for uniform lattices (that is, lattices in which any two congruence classes of a congruence are of the same size) in Chapter 13 (G. Gr¨ atzer, E. T. Schmidt, and K. Thomsen [195]).

xxxii

Introduction

Part IV. ETs I present the ETs for the following classes of lattices. • Sectionally complemented lattices in Chapter 14 (my joint paper with E. T. Schmidt [184]). • Semimodular lattices in Chapter 15 (my joint paper with E. T. Schmidt [187]). • Isoform lattices (that is, lattices in which any two congruence classes of a congruence are isomorphic) in Chapter 16 (G. Gr¨atzer, R. W. Quackenbush, and E. T. Schmidt [167]). These three constructions are based on cubic extensions, introduced in Part II. Finally, in Chapter 17, I discuss two congruence “destroying” extensions, which we call “magic wands” (my joint paper with E. T. Schmidt [190], G. Gr¨atzer, M. Greenberg, and E. T. Schmidt [132]). Part V. Congruence Lattices of Two Related Lattices What happens if we consider the congruence lattices of two related lattices, such as a lattice and a sublattice? I take up some variants of this question in this part. Let L be a finite lattice, and let K be a sublattice of L. As we discuss in Section 3.3, there is an extension map ext : Con K → Con L: for a congruence α of K, let the image ext α be the congruence conL (α) of L generated by α. The map ext is a {0}-separating join-homomorphism. Chapter 18 proves the converse, a 1974 result of A. P. Huhn [213]. It is presented in a stronger form due to G. Gr¨atzer, H. Lakser, and E. T. Schmidt [157]. I deal with ideals in Chapter 19. Let K be an ideal of a lattice L. Then the restriction map re : Con L → Con K (which assigns to a congruence α of L, the restriction α⌉K of α to K) is a bounded homomorphism (that is, {0, 1}homomorphism). We prove the corresponding representation theorem for finite lattices, based on my joint paper with H. Lakser [140] (see Theorem 19.1). Let D and E be finite distributive lattices. Let φ be a bounded homomorphism of D into E. Then there exists a finite lattice L and an ideal I of L such that D ∼ = Con L, E ∼ = Con I, and φ is represented by re, the restriction map. This is an abstract/abstract result. The congruence lattices are given as D and E, they are abstract finite distributive lattices, whereas the finite lattices L and G are constructed.

Introduction

xxxiii

G. Cz´edli [28] improved on the abstract/abstract representation (see Theorem 20.2), obtaining a concrete/abstract result, where E is given as Con L. Earlier, E. T. Schmidt [246] proved this result for injective homomorphisms. In Chapter 19, we also prove two variants. The first is in my joint paper with H. Lakser [147], stating that this result also holds for sectionally complemented lattices. The second is in another joint paper with H. Lakser [145] stating that this result also holds for planar lattices. Chapter 20 contains sharper forms of Theorem 19.1. Now let L be a lattice and let F and G be convex sublattices of L. How should we map Con F into Con G? We could try to map a congruence α of the convex sublattice F to α, the minimal extension of α to L, and restrict it to the convex sublattice G as follows, σ : α → α⌉G,

α ∈ Con F.

Minimal extensions do not preserve meets, so σ is not a bounded homomorphism of Con F to Con G, in general. This problem does not arise if we assume that L is a congruence-preserving extension of F . This leads us to Theorem 20.3 (see my joint paper with H. Lakser [152]), which is a concrete/concrete result. The final Chapter 21 is a first contribution to the following class of problems. Let ⊛ be a construction for finite lattices (that is, if D and E are finite lattices, then so is D ⊛ E). Find a construction ⊚ of finite distributive lattices (that is, if K and L are finite distributive lattices, then so is K ⊚ L) satisfying Con(K ⊛ L) ∼ = Con K ⊚ Con L. If the lattice construction is the direct product, the answer is obvious since Con(K × L) ∼ = Con K × Con L. In Chapter 21, we take up the construction defined as the distributive lattice of all isotone maps from J(E) to D. In my joint paper with M. Greenberg [128], we introduced another construction: the tensor extension, A[B], for nontrivial finite lattices A and B. In Chapter 21, we prove that Con(A[B]) ∼ = (Con A)[Con B]. Part VI. The Ordered Set of Principal Congruences In 2013, I raised the question whether one can associate with a finite lattice L a structure of some of its congruences? We could take the ordered set of the principal congruences generated by prime intervals, but this is just J(Con L), which is “equivalent” to Con L (see Section 2.5.2 for an explanation). I proposed

xxxiv

Introduction

to consider the ordered set Princ K of principal congruences of a lattice K. In Part VI, we state and prove the two major results of this new field. Chapter 22 contains the first RT for Principal Congruences (Theorem 22.1), characterizing Princ K of a finite lattice K as a finite, bounded, ordered set (my paper [107]). This chapter also contains the Independence Theorem: for a finite lattice L, the two related structures Princ L and Aut L are independent (G. Cz´edli [35], see also my paper [119]). The second major result is the Minimal RT in Chapter 23, based on my joint paper with H. Lakser [150]. A finite lattice L has a minimal set of principal congruences, if all proper principal congruences are join-irreducible. Similarly, for a finite distributive lattice D, we call the finite lattice L a minimal representation of D, if D and Con L are isomorphic and L has a minimal set of principal congruences. The Minimal RT states that a finite distributive lattice D has a minimal representation L iff D has at most two dual atoms. Chapter 24 deals with a related result. Let D be a finite distributive lattice and let Q ⊆ D. We call the subset Q of D principal congruence representable, if there is a finite lattice L such that Con L is isomorphic to D and Princ L corresponds to Q under this isomorphism. We introduce a simple combinatorial condition, called chain representability (see Section 24.1), for a subset Q of a finite distributive lattice D and we prove two results. • The Necessity Theorem. Let D be a finite distributive lattice and let Q ⊆ D. If Q is representable, then it is chain representable. • The Sufficiency Theorem. Let D be a finite distributive lattice with a join-irreducible unit element. Then Q ⊆ D is representable iff it is chain representable. Part VII. Congruence Extensions and Prime Interval We discuss in Section 4.4, how an SPS lattice can be constructed by inserting forks and removing corners. In Chapter 28, we examine the extendibility of congruences to a fork extension. Since by the Structure Theorem for SR Lattices (Theorem 4.11), any SR lattice can be obtained from a grid by inserting forks, this gives us an insight into the congruence lattice of SR lattices. The results are mostly technical, except for the Two-cover Theorem we discussed in Topic D. The spreading of a congruence from a prime interval to another prime interval involves intervals of arbitrary size (as illustrated by Figure 3.2). We would like to describe such a spreading with prime intervals only. We do this in Chapter 26 with the Prime-projectivity Lemma (see my paper [112]). Chapter 27 sharpens the Prime-projectivity Lemma to the Swing Lemma, a very strong form of the Prime-projectivity Lemma, for slim, planar, and

Introduction

xxxv

semimodular lattices (see my paper [114]). Almost all results for congruences of slim, planar, and semimodular lattices can be derived from the Swing Lemma. Part VIII. Six Congruence Properties of SPS lattices We introduced this field in Topic D. We state and prove Cz´edli’s four properties (G. Cz´edli [46] and my paper [125]). The sixth major property is from my joint paper with G. Cz´edli [59], the proof is from my paper [127]. Notation Lattice-theoretic terminology and notation evolved from the three editions of G. Birkhoff’s Lattice Theory, [21], by way of my books, UA-[97], GLT-[99], UA2[100], GLT2-[102], CFL-[103], LTF-[105], CFL2-[115], and R. N. McKenzie, G. F. McNulty, and W. F. Taylor [224], changing quite a bit in the process. Birkhoff’s notation for the congruence lattice and ideal lattice of a lattice changed from Θ(L) and I(L) to Con L and Id L, respectively. The advent of LATEX promoted the use of operators for lattice constructions. I try to be consistent: I use an operator when a new structure is constructed; so I use Con L, Id L, and Aut L, and so on, without parentheses, unless required for readability, for instance, J(D) and Con(Id L). I use functional notation when sets are constructed, as in Atom(L) and J(a). “Generated by” uses the same letters as the corresponding lattice construction, but starting with a lower case letter: Con L is the congruence lattice of L and con(H) is the congruence generated by H, whereas Id L is the ideal lattice of L and id(H) is the ideal generated by H. New concepts introduced in more recent research papers exhibit the usual richness in notation and terminology. I use this opportunity, with the wisdom of hindsight, to make their use more consistent. The reader will often find different notation and terminology when reading the original papers. The detailed Table of Notation and Index may help. In combinatorial results, I use Landau’s big O notation: for the functions f and g, we write f = O(g) to mean that |f | ≤ C|g| for a suitable constant C. Natural numbers start at 1. In Section 4.2, we introduce acronyms for two classes of lattices: SPS (slim, planar, semimodular) and SR (slim rectangular). We also use acronyms for my books; for instance, LTF for “Lattice Theory: Foundation;” we use them in the form LTF-[105].

Part I

A Brief Introduction to Lattices

1

Chapter

1

Basic Concepts

In this chapter we introduce the most basic order theoretic concepts: ordered sets, lattices, diagrams, and the most basic algebraic concepts: sublattices, congruences, products.

1.1.

Ordering

1.1.1

Ordered sets

A binary relation ϱ on a nonempty set A is a subset of A2 , that is, a set of ordered pairs (a, b), with a, b ∈ A. For (a, b) ∈ ϱ, we will write a ϱ b or a ≡ b (mod ϱ). A binary relation ≤ on a set P is called an ordering if it is reflexive (a ≤ a for all a ∈ P ), antisymmetric (a ≤ b and b ≤ a imply that a = b for all a, b ∈ P ), and transitive (a ≤ b and b ≤ c imply that a ≤ c for all a, b, c ∈ P ). An ordered set (P, ≤) consists of a nonempty set P and an ordering ≤. a < b means that a ≤ b and a ̸= b. We also use the “inverse” relations, a ≥ b defined as b ≤ a and a > b for b < a. If more than one ordering is being considered, we write ≤P for the ordering of (P, ≤); on the other hand if the ordering is understood, we will say that P (rather than (P, ≤)) is an ordered set. An ordered set P is trivial if P has only one element. The elements a and b of the ordered set P are comparable if a ≤ b or b ≤ a. Otherwise, a and b are incomparable, in notation, a ∥ b. Let H ⊆ P and a ∈ P . Then a is an upper bound of H iff h ≤ a for all h ∈ H. An upper bound a of H is the least upper bound W of H iff a ≤ bWfor any upper bound b of H; in this case, we will write a = H. If a = H 3

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_1

4

1. Basic Concepts

W exists, then it is unique. By definition, ∅ exist (∅ is the empty set) iff P has a smallest element, zero, denoted by 0. The concepts of lower V bound and are similarly defined; the latter is denoted by greatest lower bound H. Note V that ∅ exists iff P has a largest element, unit, denoted by 1. A bounded ordered set hasWboth 0 and V 1. We often denote the 0 and 1 of P by 0P and 1P . The notation H and H will also be used for families of elements. The adverb “similarly” (in “similarly defined”) in the previous paragraph can be given concrete meaning. Let (P, ≤) be an ordered set. Then (P, ≥) is also an ordered set, called the dual of (P, ≤). The dual of the ordered set P will be denoted by P δ . Now if Φ is a “statement” about ordered sets, and if we replace all occurrences of ≤ by ≥ in Φ, then we get the dual of Φ. Duality Principle for Ordered Sets. If a statement Φ is true for all ordered sets, then its dual is also true for all ordered sets. For a, b ∈ P , if a is an upper bound of {b}, then a is an upper bound of b. If for all a, b ∈ P , the set {a, b} has an upper bound, then the ordered set P is directed. A chain (linearly ordered set, totally ordered set ) is an ordered set with no incomparable elements. An antichain is one in which a ∥ b for all a ̸= b. Let (P, ≤) be an ordered set and let Q be a nonempty subset of P . Then there is a natural ordering ≤Q on Q induced by ≤: for a, b ∈ Q, let a ≤Q b iff a ≤ b; we call (Q, ≤Q ) (or simply, (Q, ≤), or even simpler, Q) an ordered subset (or suborder ) of (P, ≤). We denote this ordered set (P, ≤)⌉Q, the restriction of (P, ≤) to Q. A chain C in an ordered set P is a nonempty subset, which, as a suborder, is a chain. An antichain C in an ordered set P is a nonempty subset which, as a suborder, is an antichain. The length of a finite chain C, length C, is |C| − 1. An ordered set P is said to be of length n (in symbols, length P = n), where n is a natural number iff there is a chain in P of length n and all chains in P are of length ≤ n. ∼ Q) and the map The ordered sets P and Q are isomorphic (written as P = ψ : P → Q is an isomorphism iff ψ is one-to-one and onto and a ≤ b in P

iff

ψa ≤ ψb in Q.

Let Cn denote the set {0, . . . , n − 1} ordered by 0 < 1 < 2 < · · · < n − 1. Then Cn is an n-element chain. Observe that length Cn = n − 1. If C = {x0 , . . . , xn−1 } is an n-element chain and x0 < x1 < · · · < xn−1 , then ψ : i 7→ xi is an isomorphism between Cn and C. Therefore, the n-element chain is unique up to isomorphism. Let Bn denote the set of all subsets of the set {0, . . . , n − 1} ordered by containment. Observe that the ordered set Bn has 2n elements and length Bn = n.

1.1. Ordering

5

In general, for a set X, we denote by Pow X the power set of X, that is, the set of all subsets of X ordered by set inclusion. A quasiordered set is a nonempty set Q with a binary relation ≤ that is reflexive and transitive. Let us define the binary relation a ≈ b on Q as a ≤ b and b ≤ a. Then ≈ is an equivalence relation. Define the set P as Q/≈, and on P define the binary relation ≤: a/≈ ≤ b/≈

iff

a ≤ b in Q.

It is easy to see that the definition of ≤ on P is well defined and that P is an ordered set. We will call P the ordered set associated with the quasiordered set Q. Starting with a binary relation ≺ on the set Q, we can define the reflexivetransitive closure ≤ of ≺ by the formula: for a, b ∈ Q, let a ≤ b iff a = b or if a = x0 ≺ x1 ≺ · · · ≺ xn = b for elements x1 , . . . , xn−1 ∈ Q. Then ≤ is a quasiordering on Q. A cycle on Q is a sequence x1 , . . . , xn ∈ Q satisfying x1 ≺ x2 ≺ · · · ≺ xn ≺ x1 (n > 1). The quasiordering ≤ is an ordering iff there are no cycles. For an ordered set P , call A ⊆ P a down set iff x ∈ A and y ≤ x in P , imply that y ∈ A. For H ⊆ P , there is a smallest down set containing H, namely, { x | x ≤ h, for some h ∈ H }; we use the notation ↓ H for this set. If H = {a}, we write ↓ a for ↓ {a}. Let Dn P denote the set of all down sets ordered by set inclusion. If P is an antichain, then Dn P ∼ = Bn , where n = |P |. The map ψ : P → Q is an isotone map (resp., antitone map) of the ordered set P into the ordered set Q iff a ≤ b in P implies that ψa ≤ ψb (resp., ψa ≥ ψb) in Q. Then ψP is a suborder of Q. Even if ψ is one-to-one, the ordered sets P and ψP need not be isomorphic. If both P and Q are bounded, then the map ψ : P → Q is bounded or a bounded map, if it preserves the bounds, that is, ψ0P = 0Q and ψ1P = 1Q . Most often, we talk about bounded isotone maps (and bounded homomorphisms, see Section 1.3.1). 1.1.2

Diagrams

In the ordered set P , the element a is covered by b or b covers a (written as a ≺ b or b ≻ a) iff a < b and a < x < b for no x ∈ P . The binary relation ≺ is called the covering relation. The covering relation determines the ordering. Let P be a finite ordered set. Then a ≤ b iff a = b or if there exists a finite sequence of elements x1 , x2 , . . . , xn such that a = x1 ≺ x2 ≺ · · · ≺ xn = b. A diagram of an ordered set P represents the elements with small circles ; the circles representing two elements x, y are connected by a line segment iff one covers the other; if x is covered by y, then the circle representing x is placed lower than the circle representing y.

6

1. Basic Concepts

The diagram of a finite ordered set determines the order up to isomorphism. In a diagram the intersection of two line segments does not indicate an element. A diagram is planar if no two line segments intersect. An ordered set P is planar if it has a diagram that is planar. Figure 1.1 shows three diagrams of the same ordered set P . Since the third diagram is planar, P is a planar ordered set. 1.1.3

Constructions of ordered sets

Given the ordered sets P and Q, we can form the direct product P × Q, consisting of all ordered pairs (x1 , x2 ), with x1 ∈ P and x2 ∈ Q, ordered componentwise, that is, (x1 , x2 ) ≤ (y1 , y2 ) iff x1 ≤ y1 and x2 ≤ y2 . Therefore, (x1 , x2 ) ≺ (y1 , y2 ) in P × Q iff x1 ≺ y1 , x2 = y2 or x1 = y1 , x2 ≺ y2 . If P = Q, then we write P 2 for P × Q. Similarly, we use the notation P n for P n−1 × P for n > 2. Figure 1.2 shows a diagram of C2 × P , where P is the ordered set with diagrams in Figure 1.1. Another often used construction is the (ordinal) sum P + Q of P and Q, defined on the (disjoint) union P ∪ Q and ordered as follows:   x ≤P y for x, y ∈ P ; x ≤ y iff x ≤Q y for x, y ∈ Q;   x ∈ P, y ∈ Q. Figure 1.3 shows diagrams of C2 + P and P + C2 , where P is the ordered

Figure 1.1: Three diagrams of the ordered set P

Figure 1.2: A diagram of C2 × P

1.1. Ordering

7

Figure 1.3: Diagrams of C2 + P , P + C2 , and P ∔ C2 set with diagrams in Figure 1.1. In both diagrams, the elements of C2 are black-filled. Figure 1.3 also shows the diagram of P ∔ C2 . A variant construction is the glued sum, P ∔ Q, applied to an ordered set P with largest element 1P and an ordered set Q with smallest element 0Q ; then P ∔ Q is P + Q in which 1P and 0Q are identified (that is, 1P = 0Q in P ∔ Q). 1.1.4

Partitions

We now give a nontrivial example of an ordered set. A partition of a nonempty set A is a set π of nonempty pairwise disjoint subsets of A whose union is A. The members of π are called the blocks of π. The block containing a ∈ A will be denoted by a/π. A singleton as a block is called trivial. If the elements a and b of A belong to the same block, we write a ≡ b (mod π) or a π b or a/π = b/π. In general, for H ⊆ A, H/π = { a/π | a ∈ H }, a collection of blocks. An equivalence relation ε on the set A is a reflexive, symmetric (a ε b implies that b ε a, for all a, b ∈ A), and transitive binary relation. Given a partition π, we can define an equivalence relation ε by (x, y) ∈ ε iff x/π = y/π. Conversely, if ε is an equivalence relation, then π = { a/ε | a ∈ A } is a partition of A. There is a one-to-one correspondence between partitions and equivalence relations; we will use the two terms interchangeably. Part A will denote the set of all partitions of A ordered by π1 ≤ π2

iff

x ≡ y (mod π1 ) implies that x ≡ y (mod π2 ).

We draw a picture of a partition by drawing the boundary lines of the (nontrivial) blocks. Then π1 ≤ π2 iff the boundary lines of π2 are also boundary lines of π1 (but π1 may have some more boundary lines). Equivalently, the blocks of π2 are unions of blocks of π1 (see Figure 1.4).

8

1. Basic Concepts

A

π1 ≤ π2 and

π1 : π2 :

Figure 1.4: Drawing a partition Part A has a zero and a unit, denoted by 0 and 1, respectively, defined by x≡y x≡y

(mod 0) (mod 1)

iff x = y; for all x, y ∈ A.

Figure 1.5 shows the diagrams of Part A for |A| ≤ 4. The partitions are labeled by listing the nontrivial blocks.

1.2.

Lattices and semilattices

1.2.1

Lattices

We need two basic concepts from Universal Algebra. An (n-ary) operation on a nonempty set A is a map from An to A. For n = 2, we call the operation binary. An algebra is a nonempty set W A with operations defined on A. V An ordered set (L, ≤) is a lattice if {a, b} and {a, b} exist for all a, b ∈ L. A lattice L is trivial if it has only one element; otherwise, it is nontrivial. We will use the notations _ a ∨ b = {a, b}, ^ a ∧ b = {a, b}, and call ∨ the join, and ∧ the meet. They are both binary operations that are idempotent (a ∨ a = a and a ∧ a = a), commutative (a ∨ b = b ∨ a and a ∧ b = b ∧ a), associative ((a ∨ b) ∨ c = a ∨ (b ∨ c) and (a ∧ b) ∧ c = a ∧ (b ∧ c)), and absorptive (a ∨ (a ∧ b) = a and a ∧ (a ∨ b) = a). These properties of the operations are also called the idempotent identities, commutative identities, associative identities, and absorption identities, respectively. (Identities, in general, are introduced in Section 2.3.) As always in algebra, associativity makes it possible to write a1 ∨ a2 ∨ · · · ∨ an without using parentheses (and the same for ∧).

1.2. Lattices and semilattices

9

1 = {1, 2, 3}

1 = {1, 2}

Part {1, 2}

Part {1}

0

{1, 2}

{1, 3}

Part {1, 2, 3}

{2, 3}

0

{1, 2, 3, 4} = 1 {1, 2, 3} {1, 2}, {3, 4}

{1, 2}

{2, 3, 4} {1, 3, 4}

{1, 3}

{1, 4}, {2, 3}

{1, 3}, {2, 4}

{1, 4}

{2, 3}

0

{2, 4}

{1, 2, 4}

{3, 4}

Part {1, 2, 3, 4}

Figure 1.5: Part A for |A| ≤ 4 For instance, for A, B ∈ Pow X, we have A ∨ B = A ∪ B and A ∧ B = A ∩ B. So Pow X is a lattice. For α, β ∈ Part A, if we regard α and β as equivalence relations, then the meet formula is trivial: α ∧ β = α ∩ β, but the formula for joins is a bit more complicated: x ≡ y (mod α ∨ β) iff there is a sequence x = z0 , z1 , . . . , zn = y of elements of A such that zi ≡ zi+1 (mod α) or zi ≡ zi+1 (mod β) for all 0 ≤ i < n. So Part A is a lattice; it is called the partition lattice on A. For an ordered set P , the order Dn P is a lattice: A ∨ B = A ∪ B and A ∧ B = A ∩ B for A, B ∈ Dn P . To treat lattices as algebras, define an algebra (L, ∨, ∧) a lattice iff L is a nonempty set, ∨ and ∧ are binary operations on L, both ∨ and ∧ are idempotent, commutative, and associative, and they satisfy the two absorption identities. A lattice as an algebra and a lattice as an ordered set are “equivalent” concepts: Let the order L = (L, ≤) be a lattice. Then the algebra

10

1. Basic Concepts

La = (L, ∨, ∧) is a lattice. Conversely, let the algebra L = (L, ∨, ∧) be a lattice. Define a ≤ b iff a ∨ b = b. Then Lp = (L, ≤) is an ordered set, and the ordered set Lp is a lattice. For an ordered set L that is a lattice, we have Lap = L; for an algebra L that is a lattice, we have Lpa = L. Note that for lattices as algebras, the Duality Principle takes on the following very simple form. Duality Principle for Lattices. Let Φ be a statement about lattices expressed in terms of ∨ and ∧. The dual of Φ is the statement we get from Φ by interchanging ∨ and ∧. If Φ is true for all lattices, then the dual of Φ is also true for all lattices. If the operations are understood, we will say that L (rather than (L, ∨, ∧)) is a lattice. The dual of the lattice L will be denoted by Lδ ; the ordered set Lδ is also a lattice. In this book, we deal almost exclusively with finite lattices. Some concepts, however, are more natural to introduce W in a more V general context. An ordered set (L, ≤) is a complete lattice if X and X exist for all X ⊆ L. All finite lattices are complete, of course. 1.2.2

Semilattices and closure systems

A semilattice (S, ◦) is an algebra: a nonempty set S with an idempotent, commutative, and associative binary operation ◦. A join-semilattice (S, ∨, ≤) is a structure, where (S, ∨) is a semilattice, (S,W≤) is an ordered set, and a ≤ b iff a∨b = b. In the ordered set (S, ≤), we have {a, b} = a∨b. As conventional, we write (S, ∨) for (S, ∨, ≤) or just S if the operation is understood. Similarly, a meet-semilattice (S, ∧, ≤) is a structure, where (S, ∧) is a semilattice, (S, ≤) isVan ordered set, and a ≤ b iff a ∧ b = a. In the ordered set (S, ≤), we have {a, b} = a ∧ b. As conventional, we write (S, ∧) for (S, ∧, ≤) or just S if the operation is understood. If (L, ∨, ∧) is a lattice, then (L, ∨) is a join-semilattice and (L, ∧) is a meet-semilattice; moreover, the orderings agree. The converse also holds. Let L be a lattice and let C be a nonempty subset of L with the property that for every x ∈ L, there is a smallest element x of C with x ≤ x. We call C a closure system in L, and x the closure of x in C. Obviously, C, as an ordered subset of L, is a lattice: For x, y ∈ C, the meet in C is the same as the meet in L, and the join is x ∨C y = x ∨L y.

V Let L be a complete lattice and V V let C be -closed subset of L, that is, if X ⊆ C, then X ∈ C. (Since ∅ = 1, such a subset is nonempty and contains the 1 of L.) Then C is a closure system in L, and for every x ∈ L, ^ x = ( y ∈ C | x ≤ y ).

1.3. Some algebraic concepts

1.3.

Some algebraic concepts

1.3.1

Homomorphisms

11

The lattices L1 = (L1 , ∨, ∧) and L2 = (L2 , ∨, ∧) are isomorphic as algebras (in symbols, L1 ∼ = L2 ), and the map φ : L1 → L2 is an isomorphism iff φ is one-to-one and onto and (1) (2)

φ(a ∨ b) = φa ∨ φb, φ(a ∧ b) = φa ∧ φb

for a, b ∈ L1 . A map, in general, and a homomorphism, in particular, is called injective if it is one-to-one, surjective if it is onto, and bijective if it is one-to-one and onto. An isomorphism of a lattice with itself is called an automorphism. The automorphisms of a lattice L form a group Aut L under composition. A lattice L is rigid if the identity map is the only automorphism of L, that is, if Aut L is the one-element group. It is easy to see that two lattices are isomorphic as ordered sets iff they are isomorphic as algebras. Let us define a homomorphism of the join-semilattice (S1 , ∨) into the join-semilattice (S2 , ∨) as a map φ : S1 → S2 satisfying (1); similarly, for meet-semilattices, we require (2). A lattice homomorphism (or simply, homomorphism) φ of the lattice L1 into the lattice L2 is a map of L1 into L2 satisfying both (1) and (2). A homomorphism of a lattice into itself is called an endomorphism. A one-to-one homomorphism is also called an embedding. Note that meet-homomorphisms, join-homomorphisms, and (lattice) homomorphisms are all isotone. Figure 1.6 shows three maps of the four-element lattice B2 into the threeelement chain C3 . The first map is isotone but it is neither a meet- nor a join-homomorphism. The second map is a join-homomorphism but is not a meet-homomorphism, thus not a homomorphism. The third map is a (lattice) homomorphism. Various versions of homomorphisms and embeddings will be used. For instance, for lattices and join-semilattices, there are also {∨, 0}-homomorphism,

Figure 1.6: Morphism

12

1. Basic Concepts

and so on, with obvious meanings. An onto homomorphism φ is also called surjective, whereas a one-to-one homomorphism is called injective; it is the same as an embedding. For bounded lattices, we often use bounded homomorphisms and bounded embeddings, that is, {0, 1}-homomorphisms and {0, 1}-embeddings. (In the literature, bounded homomorphisms sometimes have a different definition; this is unlikely to cause any confusion.) It should always be clear from the context what kind of homomorphism we are considering. If we say, “let φ be a homomorphism of K into L,” where K and L are lattices, then φ is a lattice homomorphism, unless otherwise stated. 1.3.2

Sublattices

A sublattice (K, ∨, ∧) of the lattice (L, ∨, ∧) is defined on a nonempty subset K of L with the property that a, b ∈ K implies that a ∨ b, a ∧ b ∈ K (where the operations ∨, ∧ are formed in (L, ∨, ∧)), and the ∨ and the ∧ of (K, ∨, ∧) are restrictions to K of the ∨ and the ∧ of (L, ∨, ∧), respectively. Instead of “(K, ∨, ∧) is a sublattice of (L, ∨, ∧),” we will simply say that “K is a sublattice of L”—in symbols, K ≤ L. Of course, a sublattice of a lattice is again a lattice. If K is a sublattice of L, then we call L an extension of K—in symbols, L ≥ K. ̸ ∅, there is a smallest sublattice sub(H) ⊆ L For every H ⊆ L, H = containing H called the sublattice of L generated by H. We say that H is a generating set of sub(H). For a bounded lattice L, the sublattice K is bounded (also called a bounded sublattice) if the 0 and 1 of L are in K. Similarly, we can define a {0}-sublattice, bounded extension, and so on. The subset K of the lattice L is called convex iff a, b ∈ K, c ∈ L, and a ≤ c ≤ b imply that c ∈ K. We can add the adjective “convex” to sublattices, extensions, and embeddings. A sublattice K of the lattice L is convex if it is a convex subset of L. Let L be an extension of K; then L is a convex extension if K is a convex sublattice. An embedding is convex if the image is a convex sublattice. For a, b ∈ L, a ≤ b, the interval I = [a, b] = { x | a ≤ x ≤ b } is an important example of a convex sublattice. We will use the notation 1 I for the largest element of I, that is, b, and 0I for the smallest element of I, that is, a. An interval [a, b] is trivial if a = b. The smallest nontrivial intervals are called prime; that is, [a, b] is prime iff a ≺ b. For planar lattices, the term prime interval is used interchangeably with edge. In Chapters 27 and 29, we use only edges. Another important example of a convex sublattice is an ideal. A nonempty subset I of L is an ideal iff it is a down set with the property:

1.3. Some algebraic concepts

13

(Id) a, b ∈ I implies that a ∨ b ∈ I. ̸ L. Since the intersection of any number of An ideal I of L is proper if I = ideals is an ideal, unless empty, we can define id(H), the ideal generated by a subset H of the lattice L, provided that H ̸= ∅. If H = {a}, we write id(a) for id({a}), and call it a principal ideal. Obviously, id(a) = { x | x ≤ a } = ↓ a. The set Id L of all ideals of L is an ordered set under set inclusion, and as an ordered set it is a lattice. In fact, for I, J ∈ Id L, the lattice operations in Id L are I ∨ J = id(I ∪ J) and I ∧ J = I ∩ J. So we obtain the formula for the ideal join: x ∈ I ∨ J iff x ≤ i ∨ j for some i ∈ I, j ∈ J. We call Id L the ideal lattice of L. Now observe the formulas: id(a) ∨ id(b) = id(a ∨ b), id(a) ∧ id(b) = id(a ∧ b). Since a ̸= b implies that id(a) ̸= id(b), these yield: The map a 7→ id(a) embeds L into Id L. Since the definition of an ideal uses only ∨ and ≤, it applies to any joinsemilattice S. The ordered set Id S is a join-semilattice and the same join formula holds as the one for lattices. Since the intersection of two ideals could be empty, Id S is not a lattice, in general. However, for a {∨, 0}-semilattice (a join-semilattice with zero), Id S is a lattice. For lattices (join-semilattices) S and T , let ε : S → T be an embedding. We call ε an ideal-embedding if εS is an ideal of T . Then, of course, for any ideal I of S, we have that εI is an ideal of T . Ideal-embeddings play a major role in Chapter 19. By dualizing, we get the concepts of a filter , fil(H), the filter generated by a subset H of the lattice L, provided that H ̸= ∅, principal filter fil(a), and so on. 1.3.3

Congruences

An equivalence relation α on a lattice L is called a congruence relation, or congruence, of L iff a ≡ b (mod α) and c ≡ d (mod α) imply that (SP∧ )

(SP∨ )

a∧c≡b∧d a∨c≡b∨d

(mod α), (mod α)

(Substitution Properties). Trivial examples are the relations 0 and 1 (introduced in Section 1.1.4). As in Section 1.1.4, for a ∈ L, we write a/α for the congruence class (congruence block ) containing a; observe that a/α is a convex sublattice. If L is a lattice, K ≤ L, and α a congruence on L, then α⌉K, the restriction of α to K, is a congruence of K. Formally, for x, y ∈ K, x ≡ y (mod α⌉K)

iff

x ≡ y (mod α) in L.

14

1. Basic Concepts

We call α discrete on K if α⌉K = 0. Sometimes it is tedious to compute that a binary relation is a congruence relation. Lemma 1.1, referred to as the Technical Lemma in the literature, often facilitates such computations (see my joint paper with E. T. Schmidt [170] and also the paper F. Maeda [222]). Lemma 1.1 (Technical Lemma). A reflexive binary relation α on a lattice L is a congruence relation iff the following three properties are satisfied for x, y, z, t ∈ L: (i) x ≡ y (mod α)

iff

x ∧ y ≡ x ∨ y (mod α).

(ii) x ≤ y ≤ z, x ≡ y (mod α), and y ≡ z (mod α) imply that x ≡ z (mod α). (iii) x ≤ y and x ≡ y (mod α) imply that x ∧ t ≡ y ∧ t (mod α) and x ∨ t ≡ y ∨ t (mod α). For finite lattices there is a stronger form (see my paper [111]). Lemma 1.2 (Technical Lemma for Finite Lattices). Let L be a finite lattice. Let α be an equivalence relation on L with intervals as equivalence classes. Then α is a congruence relation iff the following condition and its dual hold for L. (C∨ )

If x ≺ y, z ∈ L and x ≡ y

(mod α), then x ≡ y ∨ z

(mod α).

Let Con L denote the set of all congruence relations on L ordered by set inclusion (remember that we can view α ∈ Con L as a subset of L2 ). We use the Technical Lemma to prove the following result. Theorem 1.3. Con L is a lattice. For α, β ∈ Con L, α ∧ β = α ∩ β. The join, α ∨ β, can be described as follows. x ≡ y (mod α ∨ β) iff there is a sequence x ∧ y = z0 ≤ z1 ≤ · · · ≤ z n = x ∨ y of elements of L such that zi ≡ zi+1 (mod α) or zi ≡ zi+1 (mod β) for every i with 0 ≤ i < n. Remark. For the binary relations γ and δ on a set A, we define the binary relation γ ◦ δ, the product of γ and δ, as follows: for a, b ∈ A, the relation a (γ ◦ δ) b holds iff a γ x and x δ b for some x ∈ A. The relation α ∨ β is formed by repeated products. Theorem 1.3 strengthens this statement.

1.3. Some algebraic concepts

15

The integer n in Theorem 1.3 can be restricted for some congruence joins. We call the congruences α and β permutable if α ∨ β = α ◦ β. A lattice L is congruence permutable if any pair of congruences of L are permutable. The chain Cn is congruence permutable iff n ≤ 2. Con L is called the congruence lattice of L. Observe that Con L is a sublattice of Part L; that is, the join and meet of congruence relations as congruence relations and as equivalence relations (partitions) coincide. If L is nontrivial, then Con L contains the two-element sublattice {0, 1}. If Con L = {0, 1}, we call the lattice L simple. All the nontrivial lattices of Figure 1.5 are simple. Of the many lattices of Figure 4.1, only M3 is simple. Given a, b ∈ L, there is a smallest congruence con(a, b)—called a principal congruence—under which a ≡ b. The formula (3)

α=

_

( con(a, b) | a ≡ b (mod α) )

is trivial but important. For H ⊆ L, W we form the smallest congruence under which H is in one class as con(H) = ( con(a, b) | a, b ∈ H ). Homomorphisms and congruence relations express two sides of the same phenomenon. Let L be a lattice and let α be a congruence relation on L. Let L/α = { a/α | a ∈ L }. Define ∧ and ∨ on L/α by a/α ∧ b/α = (a ∧ b)/α and a/α ∨ b/α = (a ∨ b)/α. The lattice axioms are easily verified. The lattice L/α is the quotient lattice of L modulo α. Lemma 1.4. The map φα : x 7→ x/α

for x ∈ L,

is a homomorphism of L onto L/α. The lattice K is a homomorphic image of the lattice L iff there is a homomorphism of L onto K. Theorem 1.5 (illustrated in Figure 1.7) states that any quotient lattice is a homomorphic image. To state it, we need one more concept: Let φ : L → L1 be a homomorphism of the lattice L into the lattice L1 , and define the binary relation α on L by x α y iff φx = φy; the relation α is a congruence relation of L, called the kernel of φ, in notation, ker(φ) = α. Theorem 1.5 (Homomorphism Theorem). Let L be a lattice. Any homomorphic image of L is isomorphic to a suitable quotient lattice of L. In fact, if φ : L → L1 is a homomorphism of L onto L1 and α is the kernel of φ, then L/α ∼ = L1 ; an isomorphism (see Figure 1.7 ) is given by β : x/α 7→ φx for x ∈ L. We also know the congruence lattice of a homomorphic image.

16

1. Basic Concepts

ϕ L

x

L1

onto

7→ x/α L/α

ψ:

x/

α

7→

ϕx

Figure 1.7: The Homomorphism Theorem Theorem 1.6 (Second Isomorphism Theorem). Let L be a lattice and let α be a congruence relation of L. For any congruence β of L such that β ≥ α, define the relation β/α on L/α by x/α ≡ y/α

(mod β/α)

iff

x≡y

(mod β).

Then β/α is a congruence of L/α. Conversely, every congruence γ of L/α can be (uniquely) represented in the form γ = β/α for some congruence β ≥ α of L. In particular, the congruence lattice of L/α is isomorphic with the interval [α, 1] of the congruence lattice of L. Let L be a bounded lattice. A congruence α of L separates 0 if 0/α = {0}, that is, x ≡ 0 (mod α) implies that x = 0. Similarly, a congruence α of L separates 1 if 1/α = {1}, that is, x ≡ 1 (mod α) implies that x = 1. We call the lattice L nonseparating if there is no congruence α ̸= 0 separating both 0 and 1 (see Figure 1.8).

a Figure 1.8: Illustrating a separating lattice Similarly, a homomorphism φ of the lattices L1 and L2 with zero is 0separating if φ0 = 0, but φx ̸= 0 for x ̸= 0. We also use 1-separating and {0, 1}-separating.

Chapter

2

Special Concepts

In this chapter, we introduce special elements, constructions, and classes of lattices that play an important role in the representation of finite distributive lattices as congruence lattices of finite lattices.

2.1.

Elements and lattices

In a nontrivial finite lattice L, an element a is join-reducible if a = 0 or if a = b ∨ c for some b < a and c < a; otherwise, it is join-irreducible. If a ∈ L is join-irreducible, then it has a unique lower cover, denoted by a∗ . Let J(L) denote the set of all join-irreducible elements of L, regarded as an ordered set under the ordering of L. By definition, 0 ∈ / J(L). For a ∈ L, set J(a) = { x | x ≤ a, x ∈ J(L) } = id(a) ∩ J(L), that is, J(a) is id(a) formed in J(L). In Wa finite lattice, every element is a join of join-irreducible elements (indeed, a = J(a)), and similarly for meets. Dually, we define meet-reducible, meet-irreducible element, its unique upper cover, denoted by a∗ , and M(L) the set of all meet-irreducible elements of L, regarded as an ordered set under the ordering of L. By definition, 1 ∈ / M(L). An element a is an atom if 0 ≺ a and a dual atom if a ≺ 1. Atoms are join-irreducible. A lattice L is atomistic if every element is a finite join of atoms. In a bounded lattice L, the element a is a complement of the element b iff a∧b = 0 and a ∨ b = 1. A complemented lattice is a bounded lattice in which 17 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_2

18

2. Special Concepts

every element has a complement. The lattices of Figure 1.5 are complemented and so are all but one of the lattices of Figure 4.1. The lattice Bn is also complemented. Let a ∈ [b, c]; the element x is a relative complement of a in [b, c] iff a ∧ x = b, and a ∨ x = c. A relatively complemented lattice is a lattice in which every element has a relative complement in any interval containing it. The lattice N5 of the Picture Gallery is complemented but not relatively complemented. In a lattice L with zero, let a ≤ b. A complement of a in [0, b] is called a sectional complement of a in b. A lattice L with zero is called sectionally complemented if a has a sectional complement in b for all a ≤ b in L. The lattice N6 of the Picture Gallery is sectionally complemented but not relatively complemented.

2.2.

Direct and subdirect products

Let L and K be lattices and form the direct product L × K as in Section 1.1.3. Then L × K is a lattice and ∨ and ∧ are computed “componentwise:;” (a, b) ∨ (c, d) = (a ∨ c, b ∨ d), (a, b) ∧ (c, d) = (a ∧ c, b ∧ d). Figure 2.1 shows the example C2 × N5 . Obviously, Bn is isomorphic to a direct product of n copies of B1 . There are two projection maps (homomorphisms) associated with this construction: πL : L × K → L and πK : L × K → K,

defined by πL : (x, y) 7→ x and by πK : (x, y) 7→ y, respectively. Similarly, we can form the direct product L1 × · · · × Ln with elements (x1 , . . . , xn ), where xi ∈ Li for i ≤ n; we denote the projection map (x1 , . . . , xi , . . . , xn ) 7→ xi by πi . If Li = L for all i ≤ n, we get the direct power Ln . By identifying x ∈ Li with (0, . . . , 0, x, 0, . . . , 0) (x is the i-th coordinate), we regard Li as an ideal of L1 × · · · × Ln for i ≤ n. The black-filled elements in Figure 2.1 show how we consider C2 and N5 ideals of C2 × N5 . A very important property of direct products is: Theorem 2.1. Let L and K be lattices, let α be a congruence relation of L, and let β be a congruence relation of K. Define the relation α × β on L × K by (a1 , b1 ) ≡ (a2 , b2 ) (α × β)

iff

a1 ≡ a2 (mod α) and b1 ≡ b2

(mod β).

Then α × β is a congruence relation on L × K. Conversely, every congruence relation of L × K is of this form.

2.2. Direct and subdirect products

19

N5 C2

Figure 2.1: C2 × N5 , a direct product of two lattices A more general construction is subdirect products. If L ≤ K1 × · · · × Kn and the projection maps πi are onto maps for all i ≤ n, then we call L a subdirect product of K1 , . . . , Kn . Trivial examples: L is a subdirect product of L and L if we identify x ∈ L with (x, x) ∈ L2 (diagonal embedding). In this example, the projection map is an isomorphism. To exclude such trivial cases, let us call a representation of L as a subdirect product of K1 , . . . , Kn trivial if one of the projection maps π1 , . . . , πn is an isomorphism. A lattice L is called subdirectly irreducible iff all representations of L as a subdirect product are trivial. For a subdirect product L of K1 and K2 , we have ker(π1 ) ∧ ker(π2 ) = 0. This subdirect product is trivial iff ker(π1 ) = 0 or ker(π2 ) = 0. Conversely, if α1 ∧ α2 = 0 in Con K, then K is a subdirect product of K/α1 and K/α2 , and this representation is trivial iff α1 = 0 or α2 = 0. Every simple lattice is subdirectly irreducible. The lattice N5 is subdirectly irreducible but not simple. There is a natural correspondence between subdirect representations of a lattice L and sets of congruences {γ 1 , . . . , γ n } satisfying γ 1 ∧ · · · ∧ γ n = 0. This representation is nontrivial iff γ i ̸= 0 for all i ≤ n. In this subdirect representation, the factors (the lattices L/γ i ) are subdirectly irreducible iff the congruences γ i are meet-irreducible for all i ≤ n, by the Second Isomorphism Theorem (Theorem 1.6). For a finite lattice L, the lattice Con L is finite, so we can represent 0 as a meet of meet-irreducible congruences, and we obtain Birkhoff Subdirect RT. Theorem 2.2. Every finite lattice L is a subdirect product of subdirectly irreducible lattices. This result is true for any algebra in any variety (a class of algebras defined

20

2. Special Concepts

by identities, such as the class of all lattices or the class of all groups). Finite subdirectly irreducible lattices are easy to recognize. If L is such a lattice, then the meet α of all the > 0 elements is > 0. Obviously, α is an atom, the unique atom of Con L. Conversely, if Con L has a unique atom, then all > 0 congruences are ≥ α, so their meet cannot be 0. We call α the base congruence of L (called monolith in many publications). If u ̸= v and u ≡ v (mod α), then α = con(u, v). So Con L = {0} ∪ [con(u, v), 1], as illustrated in Figure 2.2. 1

1

L

Con L

v u

con(u, v)

0

0

Figure 2.2: A subdirectly irreducible lattice and its congruence lattice Let L be a finite subdirectly irreducible lattice with con(u, v) the base congruence, where u ≺ v ∈ L. By inserting two elements as shown in Figure 2.3, we embed L into a simple lattice. Lemma 2.3. Every finite subdirectly irreducible lattice can be embedded into a simple lattice with at most two extra elements. Note that every finite lattice can be embedded into a finite simple lattice; in general, we need more than two elements. Lemma 14.3 provides a stronger statement.

2.3.

Terms and identities

From the variables x1 , . . . , xn , we can form (n-ary) terms in the usual manner using ∨, ∧, and parentheses. Examples of terms are: x1 , x3 , x1 ∨ x1 , (x1 ∧ x2 ) ∨ (x3 ∧ x1 ), (x3 ∧ x1 ) ∨ ((x3 ∨ x2 ) ∧ (x1 ∨ x2 )). An n-ary term p defines a function in n variables (a term function, or simply, a term) on a lattice L. For example, if p = (x1 ∧ x3 ) ∨ (x3 ∨ x2 )

2.3. Terms and identities

21

1

L a

v u b 0

Figure 2.3: Embedding into a simple lattice and a, b, c ∈ L, then p(a, b, c) = (a ∧ c) ∨ (c ∨ b) = b ∨ c. If p is a unary (n = 1) lattice term, then p(a) = a for any a ∈ L. If p is binary, then p(a, b) = a, or p(a, b) = b, or p(a, b) = a ∧ b, or p(a, b) = a ∨ b for all a, b ∈ L. If p = p(x1 , . . . , xn ) is an n-ary term and L is a lattice, then by substituting some variables by elements of L, we get a function on L of n-variables. Such functions are called term functions. Unary term functions of the form p(x) = p(a1 , . . . , ai−1 , x, ai+1 , . . . , xn ), where a1 , . . . , ai−1 , ai+1 , . . . , an ∈ L, play the most important role (see Section 3.1). A term (function), in fact, any term function, p is isotone; that is, if a1 ≤ b1 , . . . , an ≤ bn , then p(a1 , . . . , an ) ≤ p(b1 , . . . , bn ). Furthermore, a1 ∧ · · · ∧ an ≤ p(a1 , . . . , an ) ≤ a1 ∨ · · · ∨ an . Note that many earlier publications in Lattice Theory and Universal Algebra use polynomials and polynomial functions for terms and term functions. Terms have many uses. We briefly discuss three. (i) The sublattice generated by a set Lemma 2.4. Let L be a lattice and let H be a nonempty subset of L. Then a ∈ sub(H) (the sublattice generated by H) iff a = p(h1 , . . . , hn ) for some integer n ≥ 1, for some n-ary term p, and for some h1 , . . . , hn ∈ H.

22

2. Special Concepts

(ii) Identities A lattice identity (resp., lattice inequality)—also called equation—is an expression of the form p = q (or p ≤ q), where p and q are terms. An identity p = q (or p ≤ q) holds in the lattice L iff p(a1 , . . . , an ) = q(a1 , . . . , an ) (or p(a1 , . . . , an ) ≤ q(a1 , . . . , an )) holds for all a1 , . . . , an ∈ L. The identity p = q is equivalent to the two inequalities p ≤ q and q ≤ p; the inequality p ≤ q is equivalent to the identity p ∨ q = q. The most important properties of identities are given by Lemma 2.5. Identities are preserved under the formation of sublattices, homomorphic images, direct products, and ideal lattices. A lattice L is called distributive if the identities x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z), x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) hold in L. In fact, it is enough to assume one of these identities, because the two identities are equivalent. As we have just noted, the identity x∧(y ∨z) = (x∧y)∨(x∧z) is equivalent to the two inequalities: x ∧ (y ∨ z) ≤ (x ∧ y) ∨ (x ∧ z), (x ∧ y) ∨ (x ∧ z) ≤ x ∧ (y ∨ z). However, the second inequality holds in any lattice. So a lattice is distributive iff the inequality x ∧ (y ∨ z) ≤ (x ∧ y) ∨ (x ∧ z) holds. By duality, we get a similar statement about the second identity defining distributivity. The class of all distributive lattices will be denoted by D. A Boolean lattice is a distributive complemented lattice. A finite Boolean lattice is isomorphic to some Bn . A lattice is called modular if the identity (x ∧ y) ∨ (x ∧ z) = x ∧ (y ∨ (x ∧ z)) holds. Note that this identity is equivalent to the following implication: x ≥ z implies that (x ∧ y) ∨ z = x ∧ (y ∨ z). The class of all modular lattices will be denoted by M. Every distributive lattice is modular. The lattice M3 is modular but not distributive. All the lattices of Figures 1.5 and 4.1 are modular except for Part {1, 2, 3, 4} and N5 . A class of lattices V is called a variety if it is defined by a set of identities. The classes D and M are examples of varieties, and so is the class L, the variety of all lattices, and T, the (trivial) variety of one-element lattices.

2.3. Terms and identities

23

(iii) Free lattices Starting with a set H, we can form the set of all terms over H, collapsing two terms if their equality follows from the lattice axioms. We thus form the “free-est” lattice over H. For instance, if we start with H = {a, b}, then we obtain the four-element lattice, F(2), of Figure 2.4. We obtain more interesting examples if we start with an ordered set P , and require that the ordering in P be preserved. For instance, if we start with P = {a, b, c} with a < b, then we get the corresponding nine-element “free” lattice, F(P ), of Figure 2.4.1

b

c

a

F(2)

F(P )

Figure 2.4: Two free lattices Sometimes, we need free lattices with respect to some special conditions. The following result illustrates this. Lemma 2.6. Let x, y, and z be elements of a lattice L and let x ∨ y, y ∨ z, and z ∨ x be pairwise incomparable. Then sub({x ∨ y, y ∨ z, z ∨ x}) ∼ = B3 . Lemma 2.6 is illustrated by Figure 2.5. We will also need “free distributive lattices,” obtained by collapsing two terms if their equality follows from the lattice axioms and the distributive identities. Starting with a three-element set H = {x, y, z}, we then obtain the lattice, FreeD (3), of Figure 2.6. Similarly, we can define “free modular lattices.” Starting with a threeelement set H = {x, y, z}, we then obtain the lattice, FreeM (3), of Figure 2.7. An equivalent definition of freeness is the following. Let H be a set and let K be a variety of lattices. A lattice FreeK (H) is called a free lattice over K generated by H iff the following three conditions are satisfied: 1What

we call “free,” is called finitely presented in the literature.

24

2. Special Concepts

x∨y∨z

x∨y

(x ∨ y) ∧ (y ∨ z)

y∨z

z∨x (y ∨ z) ∧ (z ∨ x)

(x ∨ y) ∧ (z ∨ x)

(x ∨ y) ∧ (y ∨ z) ∧ (z ∨ x)

Figure 2.5: A free lattice with special relations a = (x ∨ y) ∧ (x ∨ z) ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z) ∨ (y ∧ z) x∨y∨z

b = (x ∨ y) ∧ (y ∨ z) c = (x ∧ y) ∨ (y ∧ z)

x∨y

x∨z b

(x ∨ y) ∧ (x ∨ z)

a

x

(x ∧ y) ∨ (x ∧ z) x∧y

y∨z (x ∨ z) ∧ (y ∨ z)

y

c x∧z

z

(x ∧ z) ∨ (y ∧ z) y∧z

x∧y∧z

Figure 2.6: The free distributive lattice on three generators, Free D (3)

25

2.4. Gluing and generalizations u = (x ∧ y) ∨ (y ∧ z) ∨ (x ∧ z) v = (x ∨ y) ∧ (y ∨ z) ∧ (x ∨ z) x1 = (x ∧ v) ∨ u y1 = (y ∧ v) ∨ u z1 = (z ∧ v) ∨ u

x

x1 x∧v

y∨u

v

x∨u

y1

u

z∨u

y

z1 y∧v

z

z∧v

Figure 2.7: The free modular lattice on three generators, FreeM (3) (i) FreeK (H) ∈ K. (ii) FreeK (H) is generated by H. (iii) Let L ∈ K and let ψ : H → L be a map; then there exists a (lattice) homomorphism φ : FreeK (H) → L extending ψ (that is, satisfying φa = ψa for all a ∈ H).

2.4.

Gluing and generalizations

2.4.1

Gluing

In Section 1.1.3, for an ordered set P with the largest element 1P and an ordered set Q with the smallest element 0Q , we introduced the glued sum

26

2. Special Concepts

P ∔ Q. This applies to any two lattices K with a unit and L with a zero. A natural generalization of this construction is gluing. Let K and L be lattices, let F be a filter of K, and let I be an ideal of L. If F is isomorphic to I (with φ the isomorphism), then we can form the lattice G, the gluing of K and L over F and I (with respect to φ), defined as follows. We form the disjoint union K ∪ L, and identify a ∈ F with φa ∈ I, for all a ∈ F , to obtain the set G. We order G as follows (see Figure 2.8):

a≤b

iff

 a ≤K b if a, b ∈ K;    a ≤ b if a, b ∈ L; L a ≤K x and φx ≤L b if a ∈ K and b ∈ L    for some x ∈ F .

Lemma 2.7. G is an ordered set, in fact, G is a lattice. The join in G is described by   if a, b ∈ K; a ∨K b a ∨G b = a ∨L b if a, b ∈ L;   φ(a ∨K x) ∨L b if a ∈ K and b ∈ L for any b ≥ x ∈ F , and dually for the meet. If L has a zero, 0L , then the last clause for the join may be rephrased: a ∨G b = φ(a ∨K 0L ) ∨L b

if a ∈ K and b ∈ L.

The lattice G contains K and L as sublattices; in fact, K is an ideal and L is a filter of G. An example of gluing is shown in Figure 2.9. There are more sophisticated examples in this book, for instance, in Chapters 12, 17, and Section 2.4 (see also Section 20.3). Lemma 2.8. Let K, L, F , I, and G be given as above. Let A be a lattice containing K and L as sublattices so that K ∩ L = I = F . Then K ∪ L is a sublattice of A and it is isomorphic to G. Now if αK is a binary relation on K and αL is a binary relation on L, we r define the reflexive product αK ◦ αL as αK ∪ αL ∪ (αK ◦ αL ) ∪ (αL ◦ αK ). We can easily describe the congruences of G. Lemma 2.9. A congruence α of G can be uniquely written in the form r

α = αK ◦ αL ,

2.5. Modular and distributive lattices

27

where αK is a congruence of K and αL is a congruence of L satisfying the condition that αK restricted to F equals αL restricted to I (under the identification of elements by φ). Conversely, if αK is a congruence of K and αL is a congruence of L satisfying the condition that αK restricted to F equals αL restricted to I, then r α = αK ◦ αL is a congruence of G. Let A, B, and C be lattices, FA a filter of A, IB an ideal of B, FB a filter of B, and IC an ideal of C. Let us assume that the lattices FA , IB , FB , and IC are isomorphic. We now define what it means to obtain L by double gluing A, B, and C. Let K be the gluing of A and B over FA and IB with the filter FB regarded as a filter FK of K (see Figure 2.11). Then we glue K and C over FK and IC to obtain L, the double gluing of A and B, and C. Let A and B be lattices, FA a filter of A, IB an ideal of B, satisfying that FA and IB are isomorphic as lattices. For k ≥ 1, we now define the k-times gluing of B to A. For k = 1, let L1 be the gluing of A and B over FA and IB with the filter FB regarded as a filter FL1 of L1 . Now if Lk−1 with the filter FLk−1 is the (k − 1)-times gluing of B to A, then we glue Lk−1 and B over FLk−1 and IB , to obtain L, the k-times gluing of B to A with the filter FB regarded as a filter FL of L. Two references on gluings: my joint papers [90] and [91] with E. Fried. 2.4.2

Generalizations

Gluing is one of the most useful constructions for lattices, because it retains so many important properties: distributivity, modularity, semimodularity, planarity. . . So it is not surprising that there are so many generalizations. There are some very easy ones, such as the triple gluing of the section by the same title on page 55. E. Fried, G. Gr¨atzer, and E. T. Schmidt [92] introduces multipasting, a common generalization of S-glued sum (where S is a lattice of finite length) of C. Herrmann [209], and pasting, see V. Slav´ık [248] and my book GLT-[99], Exercise 12 of Section V.4. G. Cz´edli and E. T. Schmidt [74] introduced the patchwork systems, another generalization of gluing.

2.5.

Modular and distributive lattices

2.5.1

The characterization theorems

The two typical examples of nondistributive lattices are N5 and M3 , whose diagrams are given (again) in Figure 2.10. The following characterization theorem follows immediately by inspecting the diagrams of the free lattices in Figures 2.4 and 2.7.

28

2. Special Concepts

b

(a ∨K x) ∨L b b

L

L

y

I

I x

x

a

K

a

K

G

y = a ∨K x

Figure 2.8: Defining gluing

I

L

F

G K

Figure 2.9: An easy gluing example

i i

b c a

a

N5

o

c

b o M3

Figure 2.10: The two characteristic nondistributive lattices

2.5. Modular and distributive lattices

29

Theorem 2.10. (i) A lattice L is modular iff it does not contain N5 as a sublattice. (ii) A modular lattice L is distributive iff it does not contain M3 as a sublattice. (iii) A lattice L is distributive iff L contains neither N5 nor M3 as a sublattice. Theorem 2.11. Let L be a modular lattice, let a ∈ L, and let U and V be sublattices with the property u ∧ v = a for all u ∈ U and v ∈ V . Then sub(U ∪ V ) is isomorphic to U × V under the isomorphism (u ∈ U and v ∈ V ) u ∨ v 7→ (u, v). Conversely, a lattice L satisfying this property is modular. Corollary 2.12. Let L be a modular lattice and let a, b ∈ L. Then sub([a ∧ b, a] ∪ [a ∧ b, b]),

C

B C A K

L

Figure 2.11: Double gluing

30

2. Special Concepts

that is, the sublattice of L generated by [a ∧ b, a] ∪ [a ∧ b, b], is isomorphic to the direct product [a ∧ b, a] × [a ∧ b, b]. In the distributive case, the sublattice generated by [a ∧ b, a] ∪ [a ∧ b, b] is the interval [a ∧ b, a ∨ b]; this does not hold for modular lattices, as exemplified by M3 . Let G be the gluing of the lattices K and L over F and I, as in Section 2.4. Lemma 2.13. If K and L are modular, so is the gluing G of K and L. If K and L are distributive, so is G. The distributive identity easily implies that every n-ary term equals one we get by joining meets of variables (the disjunctive normal form). So we get: Lemma 2.14. A finitely generated distributive lattice is finite. 2.5.2

Finite distributive lattices

For a finite distributive lattice L, set spec(a) = { x ∈ J(L) | x ≤ a } = id(a) ∩ J(L) = ↓ a ∩ J(L), the spectrum of a. The structure of finite distributive lattices is described by the following result. Theorem 2.15. Let L be a finite distributive lattice. Then the map φ : a 7→ spec(a) is an isomorphism between L and Dn(J(L)). Corollary 2.16. The correspondence L 7→ J(L) makes the class of all finite distributive lattices with more than one element correspond to the class of all finite orders; isomorphic lattices correspond to isomorphic orders, and vice versa. ∼ L. Proof. This is obvious from J(Dn P ) ∼ = P and Dn(J(L)) = The RT relates bounded homomorphisms of distributive lattices and isotone maps of ordered sets. Theorem 2.17. Let D and E be finite distributive lattices and set P = J(D) and Q = J(E). Then (i) With every bounded homomorphism φ : D → E, we can associate an isotone map J(φ) : Q → P defined by ^ J(φ)x = ( e ∈ D | x ≤ φe ) for x ∈ Q.

2.5. Modular and distributive lattices

31

(ii) With every isotone map ψ : Q → P , we can associate a bounded homomorphism Dn(ψ) : D → E defined by Dn(ψ)(e) = for e ∈ D.

_

ψ −1 (spec(e))

(iii) The constructions of (i) and (ii) are inverses to one another, and so yield together a bijection between bounded homomorphisms J(φ) : Q → P and isotone maps J(φ) : D → E. (iv) φ is one-to-one iff J(φ) is onto. (v) φ is onto iff J(φ) is an order-embedding. This result is illustrated by Figures 2.12–2.13.

b

c

v

w

a

u

Figure 2.12: An example of the distributive lattices D and E and a bounded homomorphism φ : D → E b

c

w

v

a

u

Figure 2.13: An example of the ordered sets P and Q and an isotone map ψ: Q → P d=b ∨c b

c

b=a∨d a

b∧c=a

d c=a∧d

up

dn

Figure 2.14: [a, b] ∼ [c, d] and [a, b] ∼ [c, d]

32

2. Special Concepts

The two main formulas in this result are easy to visualize. To verbalize the formula in (i), for x ∈ Q, take the set E of all elements of D mapped by φ onto x or above. Since φ is a homomorphism, it follows that the meet of all the elements (let us denote it by x† ) is still mapped by φ onto x or above. The map J(φ) maps x into x† . With every isotone map ψ : Q → P , we can associate a bounded homomorphism Dn(ψ) : D → E defined by _ Dn(ψ)(e) = ψ −1 (spec(e))

for e ∈ D. Now for the formula in (ii), take an element e ∈ D. We form the set of join-irreducible elements of E mapped by ψ to an element ≤ e. The join of this set, denoted by e‡ , is an element of E. The map Dn(ψ) maps e into e‡ . The constructions of (i) and (ii) are inverses to one another, and so yield together a bijection between bounded homomorphisms φ : D → E and isotone maps ψ : Q → P . These results express the duality between finite distributive lattices, with bounded homomorphisms, and finite ordered sets, with isotone maps (“duality” means “equivalence between a category and the opposite — aka dual — of another one”). 2.5.3

Finite modular lattices

Take a look at the two positions of the pair of intervals [a, b] and [c, d] in Figure 2.14. In either case, we will write [a, b] ∼ [c, d], and say that [a, b] is perspective to [c, d]. If we want to show whether the perspectivity is “up” or up

dn

“down,” we will write [a, b] ∼ [c, d] in the first case and [a, b] ∼ [c, d] in the second. If for some natural number n and intervals [ei , fi ], for 0 ≤ i ≤ n, [a, b] = [e0 , f0 ] ∼ [e1 , f1 ] ∼ · · · ∼ [en , fn ] = [c, d], then we say that [a, b] is projective to [c, d] and write [a, b] ≈ [c, d]. One of the most important properties of a modular lattice is stated in the following result (see Figure 2.15). Theorem 2.18 (Isomorphism Theorem for Modular Lattices). Let L be a up modular lattice and let [a, b] ∼ [c, d] in L. Then φc : x 7→ x ∨ c,

x ∈ [a, b],

is an isomorphism of [a, b] and [c, d]. The inverse isomorphism is β b : y 7→ y ∧ b,

y ∈ [c, d].

2.5. Modular and distributive lattices

33

d y x∨c ψb

b

c ϕc

y∧b x a

Figure 2.15: The isomorphisms φc and β b Corollary 2.19. In a modular lattice, projective intervals are isomorphic. Corollary 2.20. In a modular lattice if a prime interval p is projective to an interval q, then q is also prime. A finite lattice L is semimodular (or upper semimodular ), if the covering a ≺ b implies that a ∨ c ≺ b ∨ c or a ∨ c = b ∨ c for a, b, c ∈ L. The dual of an upper semimodular lattice is a lower semimodular lattice. Lemma 2.21. A modular lattice is both upper and lower semimodular. For a finite lattice, the converse also holds: a finite upper and lower semimodular lattice is modular, and conversely. The lattice S8 (see the Picture Gallery) is an example of a semimodular lattice that is not modular. Section 10.2 provides an interesting use of this lattice. The following is even more trivial than Lemma 2.13. Lemma 2.22. If K and L are finite semimodular lattices, so is the gluing G of K and L. A large class of semimodular lattices is provided by Lemma 2.23. Let A be a nonempty set. Then Part A is semimodular; it is not modular unless |A| ≤ 3.

Chapter

3

Congruences

3.1.

Congruence spreading

Let a, b, c, d be elements of a lattice L. If a ≡ b (mod α)

implies that

c ≡ d (mod α),

for any congruence relation α of L, then we say that a ≡ b congruence-forces c ≡ d. In Section 1.3.3 we saw that a ≡ b (mod α) iff a ∧ b ≡ a ∨ b (mod α); therefore, to investigate congruence-forcing, it is enough to deal with comparable pairs, a ≤ b and c ≤ d. Instead of comparable pairs, we will deal with intervals [a, b] and [c, d]. Projectivity (see Section 2.5.3) is sufficient for the study of congruenceforcing (or congruence-spreading) in some classes of lattices (for instance, in the class of modular lattices). In general, however, we have to introduce somewhat more general concepts and notation. As illustrated in Figure 3.1, we say that [a, b] is up congruence-perspective up

up

onto [c, d] and write [a, b] ↠ [c, d] iff there is an a1 ∈ [a, b] with [a1 , b] ∼ [c, d]. Similarly, [a, b] is down congruence-perspective onto [c, d] and we shall write dn

dn

up

[a, b] ↠ [c, d] iff there is a b1 ∈ [a, b] with [a, b1 ] ∼ [c, d]. If [a, b] ↠ [c, d] or dn

[a, b] ↠ [c, d], then [a, b] is congruence-perspective onto [c, d] and we write [a, b] ↠ [c, d]. If for some natural number n and intervals [ei , fi ] for 0 ≤ i ≤ n, [a, b] = [e0 , f0 ] ↠ [e1 , f1 ] ↠ · · · ↠ [en , fn ] = [c, d], 35 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_3

36

3. Congruences

d b

b

c d

a

c

a up

dn

Figure 3.1: [a, b] ↠ [c, d] and [a, b] ↠ [c, d] then we call [a, b] congruence-projective onto [c, d], and we write [a, b] ⇒ [c, d]. If [a, b] ⇒ [c, d] and [c, d] ⇒ [a, b], then we write [a, b] ⇔ [c, d]. up

up

dn

Also if (see Section 2.5.3) [a, b] ∼ [c, d], then [a, b] ↠ [c, d] and [c, d] ↠ [a, b]; dn

up

dn

if [a, b] ∼ [c, d], then [a, b] ↠ [c, d] and [c, d] ↠ [a, b]; if [a, b] ∼ [c, d], then [a, b] ↠ [c, d] and [c, d] ↠ [a, b]; if [a, b] ≈ [c, d], then [a, b] ⇒ [c, d] and [c, d] ⇒ [a, b]. But while ∼, ≈, and ⇔ are symmetric, the relations ↠ and ⇒ are not. In particular, if a ≤ c ≤ d ≤ b, then [a, b] ⇒ [c, d]. If [a, b] ⇒ [c, d], then there is a unary term function p with p(a) = c and p(c) = d. It is easy to see what special kinds of unary term functions are utilized in the relation ⇒. Intuitively, “a ≡ b congruence-forces c ≡ d” iff [c, d] is put together from pieces [c′ , d′ ], each of which satisfies [a, b] ⇒ [c′ , d′ ]. To state this more precisely, we describe the smallest congruence relation under which a ≡ b, denoted by con(a, b), as introduced in Section 1.3.3 (see R. P. Dilworth [79]). We use the Technical Lemma to prove this result. Theorem 3.1. Let L be a lattice, a, b, c, d ∈ L, a ≤ b, c ≤ d. Then c ≡ d (mod con(a, b)) iff, for some sequence c = e0 ≤ e1 ≤ · · · ≤ em = d, we have [a, b] ⇒ [ej , ej+1 ]

for

j = 0, . . . , m − 1.

This result can be usefully augmented by the following lemma (for an application, see Lemma 25.2). Lemma 3.2. Let L be a lattice, a, b, c, d ∈ L with a ≤ b and c ≤ d. Then [a, b] is congruence-projective to [c, d] iff the following condition is satisfied: There is an integer m and there are elements e0 , . . . , em−1 ∈ L such that pm (a, e0 , . . . , em−1 ) = c, pm (b, e0 , . . . , em−1 ) = d,

3.1. Congruence spreading

37

where the term pm is defined by pm (x, y0 , . . . , ym−1 ) = · · · (((x ∨ y0 ) ∧ y1 ) ∨ y2 ) ∧ · · · . Let L be a lattice and H ⊆ L2 . To compute con(H), the smallest congruence relation α under which a ≡ b (mod α) for all (a, b) ∈ H, we use the formula _ con(H) = ( con(a, b) | (a, b) ∈ H ). We also need a formula for joins.

Lemma 3.3. Let L Wbe a lattice and let αi , i ∈ I, be congruence relations of L. Then a ≡ b (mod ( αi | i ∈ I )) iff there is a sequence z0 = a ∧ b ≤ z 1 ≤ · · · ≤ zn = a ∨ b

such that for each j with 0 ≤ j < n, there is an ij ∈ I satisfying zj ≡ zj+1 (mod αij ). This is an easy but profoundly important result. For instance, the result of N. Funayama and T. Nakayama [95]—which provides the foundation for this book—immediately follows. (Another typical application is Lemma 3.12.) Theorem 3.4. The lattice Con L is distributive for any lattice L. By combining Theorem 3.1 and Lemma 3.3, we get: Corollary 3.5. Let L be a lattice, let H ⊆ L2 , and let a, b ∈ L with a ≤ b. Then a ≡ b (mod con(H)) iff, for some integer n, there exists a sequence a = c0 ≤ c1 ≤ · · · ≤ cn = b such that for each i with 0 ≤ i < n, there exists a (di , ei ) ∈ H satisfying [di ∧ ei , di ∨ ei ] ⇒ [ci , ci+1 ]. There is another congruence structure we can associate with a lattice L, the ordered set of principal congruences, Princ L, a subset of Con L. We dedicate Part VI to the study of this structure. Lemma 3.6. For a lattice L, the ordered set Princ L is a directed ordered set with zero. If L is bounded, so is Princ L. Indeed, for a, b, c, d ∈ L, an upper bound of con(a, b) and con(c, d) is con(a ∧ b ∧ c ∧ d, a ∨ b ∨ c ∨ d).

38

3. Congruences

3.2.

Finite lattices and prime intervals

Let Prime(L) denote the set of prime intervals of a finite lattice L. Let p = [a, b] be a prime interval in L (that is, a ≺ b in L) and let α be a congruence relation of L. In a finite lattice L, the formula α=

_

( con(p) | 0p ≡ 1p (mod α) )

immediately yields that the congruences in J(Con L) (which we shall denote by J(Con L)) are the congruences of the form con(p) for some p ∈ Prime(L). Of course, a join-irreducible congruence α can be expressed, as a rule, in many ways in the form con(p). Since a prime interval cannot contain a three-element chain, the following two lemmas easily follow from the Technical Lemma and from Corollary 3.5. Lemma 3.7. Let L be a finite lattice and let p and q be prime intervals in L. Then con(p) ≥ con(q) iff p ⇒ q. This condition is easy to visualize using Figure 3.2; the sequence of congruence-perspectivities, as a rule, has to go through nonprime intervals (intervals of arbitrary size) to get from p to q.

q

p

Figure 3.2: Congruence spreading from prime interval to prime interval Lemma 3.8. Let L be a finite lattice, let p be a prime interval of L, and let [a, b] be an interval of L. If p is collapsed under con(a, b), then there is a prime interval q in [a, b] satisfying q ⇒ p. In view of Theorem 2.15, we get the following.

3.2. Finite lattices and prime intervals

39

Theorem 3.9. Let L be a finite lattice. The relation ⇒ is a quasiordering on Prime(L). The equivalence classes under ⇔ form an ordered set isomorphic to J(Con L). If the finite lattice L is atomistic, then the join-irreducible congruences are even simpler to find. Indeed, if [a, b] is a prime interval, and p is an atom with dn

p ≤ b and p ≰ a, then [a, b] ∼ [0, p], so con(a, b) = con(0, p). Corollary 3.10. Let L be a finite atomistic lattice. Then every join-irreducible congruence can be represented in the form con(0, p), where p is an atom. The relation p ⇒ q, defined as [0, p] ⇒ [0, q], introduces a quasiordering on the set of atoms of L. The equivalence classes under the quasiordering form an ordered set isomorphic to J(Con L). We use Figure 3.3 to illustrate how we compute the congruence lattice of N5 using Theorem 3.9. N5 has five prime intervals: [o, a], [a, b], [b, i], [o, c], [c, i]. The equivalence classes are α = {[a, b]}, β = {[o, c], [b, i]}, and γ = {[o, a], [c, i]}. dn

up

The ordering α < γ holds because [c, i] ↠ [o, b] ↠ [a, b]. Similarly, α < β. It is important to note that the computation of p ⇒ q may involve nonprime intervals. For instance, let p = [o, c] and q = [a, b] in N5 . Then p ⇒ q, up

dn

because p ↠ [a, i] ↠ q, but we cannot get p ⇒ q involving only prime intervals. As another example, we compute the congruence lattice of S8 (see Figure 3.4). By Corollary 2.21, in a modular lattice if p and q are prime intervals, then up

dn

up

dn

p ↠ q implies that p ∼ q and p ↠ q implies that p ∼ q; thus p ⇒ q implies that p ≈ q. Therefore, Theorem 3.9 tells us that J(Con L) is an antichain in a finite modular lattice L, so Con L is Boolean. Corollary 3.11. The congruence lattice of a finite modular lattice is Boolean. By a colored lattice we will mean a finite lattice (some) of whose prime intervals are labeled so that if the prime intervals p and q are of the same color, 1

i β

γ

b α a

β

γ

c γ

β o N5

γ

β α ConJ N5

α 0 Con N5

Figure 3.3: Computing the congruence lattice of N5

40

3. Congruences

β =1 α

β αα α β α α α α β S8

β

α

α

0

ConJ S8

Con S8

Figure 3.4: Computing the congruence lattice of S8 then con(p) = con(q). These labels represent (a subset of) the equivalence classes of prime intervals, as stated in Theorem 3.9. In Figure 3.4, every prime interval of S8 is labeled; in Figure 9.2, only some are labeled. The coloring helps in the intuitive understanding of some constructions.

3.3.

Congruence-preserving extensions and variants

Let L be a lattice and K ≤ L. How do the congruences of L relate to the congruences of K? Every congruence α restricts to K: the relation α ∩ K 2 = α⌉K on K is a congruence of K. So we get the restriction map: re : Con L → Con K, which maps a congruence α of L to α⌉K. Lemma 3.12. Let K ≤ L be lattices. Then re : Con L → Con K is a {∧, 0, 1}homomorphism. For instance, if K = {o, a, i} and L = M3 (see Figure 2.10), then Con K is isomorphic to B2 , but only 0 and 1 are restrictions of congruences in L. As another example, take the lattice L of Figure 3.5 and its sublattice K, the black-filled elements; in this case, Con L ∼ = Con K ∼ = B2 , but again only 0 and 1 are restrictions. There is no natural relationship between the congruences of K and L. If K is an ideal in L (or any convex sublattice), we can say a lot more. Lemma 3.13. Let K ≤ L be lattices. If K is an ideal of L, then re : Con L → Con K is a bounded homomorphism. Let K ≤ L be lattices, and let α be a congruence of K. The congruence conL (α) (the congruence con(α) formed in L) is the smallest congruence γ of L such that α ≤ γ⌉K. Unfortunately, conL (α)⌉K may be different from α, as in the example of Figure 3.5. We say that the congruence α of K extends to L, iff α is the restriction of conL (α). Figure 3.6 illustrates this in part.

3.3. Congruence-preserving extensions and variants

K

41

L

Figure 3.5: A sublattice for the discussion of the map re If a congruence α extends, then the congruence classes of α in K extend to congruence classes in L, but there may be congruence classes in L that are not such extensions. The extension map: ext : Con K → Con L maps a congruence α of K to the congruence conL (α) of L. The map ext is a {∨, 0}-homomorphism of Con K into Con L. In addition, ext preserves the zero, that is, ext is {0}-separating. To summarize, Lemma 3.14. Let K ≤ L be lattices. Then ext : Con K → Con L is a {0}-separating join-homomorphism. The extension L of K is a congruence-reflecting extension (and the sublattice K of L is a congruence-reflecting sublattice) if every congruence of K extends to L. Utilizing the results of Sections 3.1 and 3.2, we can find many equivalent formulations of the congruence-reflecting extension property for finite lattices. Lemma 3.15. Let L be a finite lattice, and K ≤ L. Then the following conditions are equivalent: (i) K is a congruence-reflecting sublattice of L. (ii) L is a congruence-reflecting extension of K. (iii) Let p and q be prime intervals in K; if p ⇒ q in L, then p ⇒ q in K. As an example, the reader may want to verify that any sublattice of a distributive lattice is congruence-reflecting.

42

3. Congruences

L K

Figure 3.6: A congruence extends A much stronger concept—central to this book—is the following. Let K be a lattice. A lattice L is a congruence-preserving extension of K (or K is a congruence-preserving sublattice of L), if L is an extension and every congruence α of K has exactly one extension α to L satisfying α⌉K = α. Of course, α = conL (α). It follows that α 7→ conL (α) is an isomorphism between Con K and Con L. Two congruence-preserving extensions are shown in Figure 3.7, whereas Figure 3.8 shows two other extensions that are not congruence-preserving. We can also obtain congruence-preserving extensions using gluing, based on the following result. Lemma 3.16. Let K and L be lattices, let F be a filter of K, and let I be an ideal of L. Let φ be an isomorphism between F and I. Let G be the gluing of K and L over F and I with respect to φ. If L is a congruence-preserving extension of I, then G is a congruence-preserving extension of K. If I and L are simple, then L is a congruence-preserving extension of I. So we obtain the following result. Corollary 3.17. Let K, L, F , I, and φ be given as above. If I and L are simple lattices, then G is a congruence-preserving extension of K. Lemma 3.18. Let the lattice L be an extension of the lattice K. Then L is a congruence-preserving extension of K iff the following two conditions hold : (i) re(ext α) = α for any congruence α of K; (ii) ext(re α) = α for any congruence α of L.

3.3. Congruence-preserving extensions and variants

K

L

K

43

L

Figure 3.7: Examples of congruence-preserving extensions

K

L

K

L

Figure 3.8: Extensions that are not congruence preserving We can say a lot more for finite lattices. Lemma 3.19. Let L be a finite lattice, and K ≤ L. Then L is a congruencepreserving extension of K iff the following two conditions hold: (a) Let p and q be prime intervals in K; if p ⇒ q in L, then p ⇒ q in K. (b) Let p be a prime interval of L. Then there exists a prime interval q in K such that p ⇔ q in L. Lemma 3.18.(ii) is very interesting by itself. It says that every congruence α of L is determined by its restriction to K. In other words, α = conL (α⌉K). We will call such a sublattice K a congruence-determining sublattice. We can easily modify Lemma 3.19 to characterize congruence-determining sublattices in finite lattices.

44

3. Congruences

Lemma 3.20. Let L be a finite lattice L, and K ≤ L. Then K is a congruence-determining sublattice of L iff for any prime interval p in L, there is a prime interval q in K satisfying p ⇔ q in L. Of course, a congruence-preserving sublattice is always congruence-determining. In fact, a sublattice is congruence-preserving iff it is congruence-reflecting and congruence-determining.

Chapter

4

Planar Semimodular Lattices

4.1.

Planar lattices

A finite lattice L is planar if it is planar as an ordered set (see Section 1.1.2), that is, it has a planar diagram as an ordered set. For planar lattices, the term prime interval is used interchangeably with edge. In Chapters 27 and 29, we use edges. We have quite a bit of flexibility to construct a planar diagram for an ordered set, but we are much more constrained for a lattice because it has a zero, which must be the lowest element and a unit, which must be the highest element—contrast this with Figure 1.1. All lattices with five or fewer elements are planar; all but the five chains are shown in the first two rows of Figure 4.1. The third row of Figure 4.1 provides an example of “good” and “bad” lattice diagrams; the two diagrams represent the same lattice, C23 . Planar diagrams are the best. Diagrams in which meets and joins are hard to figure out are not of much value. For me, even Figure 1.5 is of limited value. In the last row of Figure 4.1 there are two more diagrams. The one on the left is not planar; nevertheless, joins and meets are easy to see (the notation M3 [C3 ] will be explained in Section 6.1). The one on the right is not a lattice: the two black-filled elements have no join. An X-configuration in a planar ordered set P is formed by two edges E and F of P satisfying the following properties: (i) 0E is to the left of 0F ; (ii) 1E is to the right of 1F . 45 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_4

46

4. Planar Semimodular Lattices

B2

B2 + C1

C1 + B2

i

i

a c

a

c

b

b

N5 o

M3 o

M3 [C3 ]

Figure 4.1: More diagrams

4.2. Two acronyms: SPS and SR

47

Lemma 4.1. A diagram of a planar ordered set P is the diagram of a planar lattice iff it does not have an X-configuration. This is almost a tautology. Let us say that a finite ordered set P with (strict) ordering relation < has a complementary (strict) order relation λ, if any two distinct elements of P are comparable by exactly one of < and λ. Theorem 4.2 (Zilber’s Theorem). A finite lattice L is planar iff it has a complementary order relation. This result has immediate interesting ramifications, for instance, planar canonical forms of diagrams and their uniqueness (see my joint paper with K. A. Baker [14]). A planar lattice L has a left boundary chain, Cl (L), and a right boundary chain, Cr (L). A left corner lc(L) (or right corner rc(L)) of L is a doublyirreducible element in L − {0, 1} on the left (or right) boundary of L. We define a covering square S of a lattice L as a four-element Boolean sublattice such that the coverings in S are also coverings in L. If C and D are maximal chains in the interval [a, b], and there is no element of L between C and D, then we call C ∪ D a cell. A four-element cell is a 4-cell. A diagram of M3 has exactly two 4-cells and three covering squares. A planar lattice is called a 4-cell lattice if all of its cells are 4-cells. For example, M3 is a 4-cell lattice but N5 is not.

4.2.

Two acronyms: SPS and SR

There are two classes of planar semimodular lattices we want to designate by acronyms. As defined in my joint paper with E. Knapp [134], a semimodular lattice L is slim, if it contains no cover-preserving M3 sublattice. An SPS lattice is a slim, planar, semimodular lattice. Following my joint paper with E. Knapp [134], a semimodular lattice L is rectangular if its left boundary chain, Cl (L), has exactly one doubly-irreducible element, lc(L), and its right boundary chain, Cr (D), has exactly one doublyirreducible element, rc(D), and these elements are complementary, that is, lc(D) ∨ rc(D) = 1, lc(D) ∧ rc(D) = 0.

An SR lattice is a slim rectangular lattice. By G. Cz´edli and E. T. Schmidt [74], whether a lattice is an SR lattice does not depend on the planar diagram chosen.

48

4.3.

4. Planar Semimodular Lattices

SPS lattices

Let L be a planar semimodular lattice. An internal element of a coverpreserving M3 sublattice is called an eye. Removing all eyes one at a time, we get a planar semimodular lattice, called the slimming of L. We can also add an eye—in the obvious sense—to a 4-cell. In an SPS lattice L, an element u covering n ≥ 3 elements together generate, up to isomorphism, a unique sublattice. For n = 3, it is S7 , for n = 4, we get the lattice of Figure 4.2 (see G. Cz´edli [29]). The following lemma is from my joint paper with E. Knapp [134]. Lemma 4.3. An SPS lattice L is distributive iff S7 (see the Picture Gallery) is not a cover-preserving sublattice of L. Let L be an SPS lattice. Two prime intervals of L are consecutive if they are opposite sides of a 4-cell. As in G. Cz´edli and E. T. Schmidt [71], maximal sequences of consecutive prime intervals form a trajectory. So a trajectory is an equivalence class of the transitive reflexive closure of the consecutive relation (see Figure 4.3 for two examples). We denote by Traj L the set of all trajectories of L. Let C = {o, al , ar , i} be a 4-cell of L and let E = [ar , i] be the prime interval on the upper right of C. We go down by consecutive prime intervals until we reach the left boundary of L. These intervals form the left wing of E and of the 4-cell C. We define the right wing symmetrically. A simple trajectory goes from an upper boundary to a lower boundary; a more interesting trajectory has a top edge, which has a left and a right wing (see the first diagram of Figure 4.3). Lemma 4.4. Let L be an SR lattice. Then for a join-irreducible congruence α exactly one of the following conditions holds. (i) There is a unique edge on the lower left boundary of L generating α, but no such edge on the lower right boundary of L. (ii) There is a unique edge on the lower right boundary of L generating α, but no such edge on the lower left boundary of L. (iii) There is a unique edge on the lower left boundary and a unique edge on the lower right boundary of L both generating α. For a rectangular lattice L, in general, the same statement holds except for the uniqueness. The first and third statement of the next lemma can be found in the literature (see my joint papers [134]–[138] with E. Knapp, and also G. Cz´edli and E. T. Schmidt [73]–[74]). Lemma 4.5. Let L be an SPS lattice.

4.3. SPS lattices

49

(i) An element of L has at most two covers. (ii) If the elements a, u, v, w ∈ L satisfy a = u ∧ v = v ∧ w = w ∧ u, then two of them are comparable. (iii) Let x ∈ L cover three distinct elements u, v, and w. Then the set {u, v, w} generates an S7 sublattice. Lemma 4.5(i) and (ii) state in different ways that there are only two directions “to go up” from an element. The next lemma states this in one more way (see G. Cz´edli and E. T. Schmidt [71]). Lemma 4.6. Let L be an SPS lattice. Let q, q1 , q2 be pairwise distinct prime dn dn intervals of L satisfying q1 ∼ q and q2 ∼ q. Then q1 ∼ q2 .

Figure 4.2: A special semimodular lattice

C

Figure 4.3: A trajectory and a left wing of a 4-cell

50

4. Planar Semimodular Lattices

Call a lattice L meet-semidistributive, if the condition x ∧ y = x ∧ z implies that x ∧ y = x ∧ (y ∨ z)

(SD∧ )

holds. Dually, we define (SD∨ ) and join-semidistributive. SPS lattices are meet-semidistributive (see LTF-[105] for more discussion, especially Section VI.2.7 by K. Adaricheva). For an overview of semidistributivity, see Chapters 3–6 ([8]–[11]) by K. Adaricheva and J. B. Nation in LTS2-[207].

4.4.

Forks

Our goal is to construct all planar semimodular lattices from planar distributive lattices, based on G. Cz´edli and E. T. Schmidt [73]. Let L be an SPS lattice and let C be a 4-cell in L. We construct a lattice extension L[C] of L as follows. We observe that the lower left edge of C is in an interval A = C2 × Cn , stretching from the lower left edge of C to the left boundary of L, see the gray-filled elements in Figure 4.4. We replace the interval A with the interval B = C3 × Cn (as in the third diagram of Figure 4.4, the new elements are black-filled). We do the same on the right side of C and add one more element in C, the join of the two top black-filled elements. Let L[C] denote the lattice we obtain. We say that L[C] is obtained from L by inserting a fork at the 4-cell C. It is easy to see that L[C] is an SPS lattice. For the new elements we shall use the notation of Figure 4.5. The direct product of two nontrivial chains is a grid. Now we can state the main result of G. Cz´edli and E. T. Schmidt [73]. Theorem 4.7. An SPS lattice with at least three elements can be constructed from a grid by the following two steps. (i) Inserting forks. (ii) Followed by removing corners. My joint paper with G. Cz´edli [57] (see also Chapter 3 in LTS1-[206]) presents a twin of this construction, called resection. Sometimes, we can delete a fork (see G. Cz´edli and E. T. Schmidt [73], as illustrated by Figure 4.6). Lemma 4.8. Let L be an SPS lattice and let S be a cover-preserving S7 in L. Let us assume that the top element t of S is minimal, that is, there is no S ′ a covering S7 with top element t′ satisfying that t′ < t. Then L has a sublattice L− with 4-cell C = S − {m, bl , br } such that L = L− [C].

4.4. Forks

51

C

Figure 4.4: Inserting a fork into L at C (the third and fourth diagrams represent the same lattice)

t αr (S)

αl (S) γ(S) al = xl,1 xl,2

bl = zl,1

ar = xr,1 br = zr,1

yl,1 = yr,1

zl,2 xl,nl

m

o yl,2

xr,2

zr,2 xr,nr

y r,2 zr,nr

zl,nl yr,nr yl,nl

Figure 4.5: Notation for the fork construction

52

4. Planar Semimodular Lattices

There are other constructions that yield all planar semimodular lattices. SPS lattices can also be described by Jordan–H¨ older permutations (see the papers G. Cz´edli and E. T. Schmidt [75], G. Cz´edli, L. Ozsv´ art, and B. Udvari [67], and G. Cz´edli [34]). These descriptions are generalized to a larger class of lattices in G. Cz´edli [31] and K. Adaricheva and G. Cz´edli [4] (see also K. Adaricheva, R. Freese, and J. B. Nation [5]). There is a description with matrices in G. Cz´edli [27] and smaller diagrams in G. Cz´edli [37]; they can also be built from smaller building blocks called rectangular lattices, which are discussed in the next section.

4.5.

Rectangular lattices

The following two lemmas are from my joint paper with E. Knapp [138]. Lemma 4.9. Let L be a rectangular lattice. Then the intervals [0, lc(L)], [lc(L), 1], [0, rc(L)], and [rc(L), 1] are chains. So the chains Cl (L) and Cr (L) are split into two, a lower and an upper part: Cll (L) = [0, lc(L)], Cul (L) = [lc(L), 1], Clr (L) = [0, rc(L)], and Cur (L) = [rc(L), 1] (see Figure 4.7). Lemma 4.9 has an interesting form for SR lattices. Lemma 4.10. Let L be an SR lattice. Then for any x ∈ L, the following equation holds: x = (x ∧ lc(L)) ∨ (x ∧ rc(L)). Moreover, the map φ : x → (x ∧ lc(L), x ∧ rc(L)) embeds L into the grid Cl (L) × Cr (L).

The structure of rectangular lattices is easily described utilizing Theorem 4.7 (see G. Cz´edli and E. T. Schmidt [73]). Theorem 4.11 (Structure Theorem for SR Lattices). L is an SR lattice iff it can be obtained from a grid by inserting forks. There is a slightly stronger version of this result, implicit in G. Cz´edli and E. T. Schmidt [69]. Theorem 4.12 (Structure Theorem, Strong Version). For every SR lattice K, there is a grid G and sequences G = K1 , K2 , . . . , Kn−1 , Kn = K of SR lattices and C1 = {o1 , c1 , d1 , i1 }, C2 = {o2 , c2 , d2 , i2 }, . . . , Cn−1 = {on−1 , cn−1 , dn−1 , in−1 } of 4-cells in the appropriate lattices such that G = K1 , K1 [C1 ] = K2 , . . . , Kn−1 [Cn−1 ] = Kn = K. Moreover, the principal ideals id(cn−1 ) and id(dn−1 ) are distributive.

4.5. Rectangular lattices

Figure 4.6: Deleting a fork (a trajectory)

1 Cul (L)

Cur (L) ,y

Ltop

x lc(L),

rc(L)

Lright

Lleft Cll (L)

Lbottom

Clr (L)

0 Figure 4.7: Decomposing an SR lattice

Figure 4.8: Two steps to a congruence-preserving rectangular extension

53

54

4. Planar Semimodular Lattices

Proof. We prove this result by induction on the number n of cover-preserving S7 -s in K. If n = 0, then K is distributive by Lemma 4.3, so the statement is trivial. Now let us assume that the statement holds for n − 1. Let K be an SR lattice with n covering S7 -s. As in Lemma 4.8, we take S, a minimal covering S7 in K. Then we form the sublattice K − by deleting the fork at S. So we get a 4-cell C = Cn−1 = {on−1 , cn−1 , dn−1 , in−1 } of K − such that K = K − [C]. Since K − has n − 1 covering S7 -s, we get the sequence G = K1 , K1 [C1 ] = K2 , . . . , Kn−2 [Cn−2 ] = Kn−1 = K − , which, along with K = K − [C], proves the statement for K. By the minimality of S, it follows from Lemma 4.3 that the principal ideals id(cn−1 ) and id(dn−1 ) are distributive. The following result of G. Cz´edli [29] is a stronger version of the Structure Theorems. Theorem 4.13. Every SR lattice is obtained from a grid by a sequence of fork insertions at distributive 4-cells, and every diagram obtained this way is an SR diagram. A rectangular lattice L is a patch lattice, if the corners are dual atoms. Let L be a nontrivial lattice. If L cannot be obtained as a gluing of two nontrivial lattices, we call L gluing indecomposable. The following is a result of G. Gr¨atzer and E. Knapp [138] (see also G. Cz´edli and E. T. Schmidt [73]); it is an easy consequence of Theorem 4.17. Theorem 4.14. Let L be an SPS lattice with at least four elements. Then L is a patch lattice iff it is gluing indecomposable. The next result of G. Cz´edli and E. T. Schmidt [74] follows easily from Theorem 4.11. Theorem 4.15. A patch lattice L can be obtained from the four-element Boolean lattice by inserting forks. My paper [106] provides an alternative approach to the results of G. Cz´edli and E. T. Schmidt [74]. Finally, start with a planar semimodular lattice L. Can we add a corner? Yes, unless L is rectangular. So we get the following result, The following two lemmas are from my joint paper with E. Knapp [138] (illustrated by Figure 4.8). Three more steps are needed to get a rectangular extension. Lemma 4.16. Let K be a planar semimodular lattice. Then K has a rectangular extension L. In fact, we can construct L as a congruence-preserving extension of K. Note that under reasonable additional assumptions, the lattice K (and even its diagram) is unique, as formulated in G. Cz´edli [40].

4.5. Rectangular lattices

55

Triple gluing For an SR lattice L, let x ∈ Cul (L) − {1, lc(L)}, y ∈ Cur (L) − {1, rc(L)}. We introduce some notation (see Figure 4.7).

(1)

Ltop (x, y) = [x ∧ y, 1], Lleft (x, y) = [lc(L) ∧ y, x], Lright (x, y) = [x ∧ rc(L), y], Lbottom (x, y) = [0, (lc(L) ∧ y) ∨ (x ∧ rc(L))]

The following decomposition theorem is from my joint paper with E. Knapp [138]. Theorem 4.17 (Decomposition Theorem for SR Lattices). Let L be an SR lattice, and let (2) (3)

x ∈ Cul (L)−{1, lc(L)}, y ∈ Cur (L)−{1, rc(L)}.

Then L can be decomposed into four SR lattices Ltop (x, y), Lleft (x, y), Lright (x, y), Lbottom (x, y), and the lattice L can be reconstructed from these by repeated gluing. We now reformulate Theorem 4.17. As in my joint paper with H. Lakser [152], let G, Y, Z, and U be rectangular lattices, arranged in the plane as in Figure 4.9, such that the facing boundary chains have the same number of elements. First, we glue U and Y together over the boundaries, to obtain the rectangular lattice X, then we glue Z and G together over the boundaries, to obtain the rectangular lattice W . Finally, we glue X and W together over the boundaries, to obtain the rectangular lattice V , which we call the triple gluing of G, Y, Z, U . Now we restate Theorem 4.17 using triple gluing, using the notation (1). Theorem 4.18. Let L be an SR lattice and let (4) (5)

x ∈ Cul (L)−{1, lc(L)}, y ∈ Cur (L)−{1, rc(L)}.

Then L is the triple gluing of the four intervals [x ∧ y, 1], [lc(L) ∧ y, x], [x ∧ rc(L), y], [0, (lc(L) ∧ y) ∨ (x ∧ rc(L))].

56

4. Planar Semimodular Lattices

G

G Z

Y

Y

Z

U U Figure 4.9: Triple gluing The basic results on gluing (see Section 2.4) extend to triple gluings. We will use Lemma 2.9 in the following form (see my joint paper with H. Lakser [152]). Lemma 4.19. A congruence α of V is uniquely associated with the four congruences αG of G, αY of Y , αZ of Z, and αU of U , satisfying the condition that αG and αY agree on the facing boundaries, and the same for {G, Z}, {Y, U }, and {Z, U }. Congruences of rectangular lattices There is a “coordinatization” of congruences of rectangular lattices. Theorem 4.20. Let L be a rectangular lattice and let α be a congruence of L. Let αl denote the restriction of α to Cll . Let αr denote the restriction of α to Clr . Then the congruence α is determined by the pair (αl , αr ). In fact, α = con(αl ∪ αr ). Proof. Since α ≥ con(αl ∪ αr ), it is sufficient to prove that (P) if the prime interval p of A is collapsed by the congruence α, then it is collapsed by the congruence con(αl ∪ αr ). First, let L be a slim patch lattice. By Theorem 4.15, we obtain L from the square, C22 , with a sequence of n fork insertions. We induct on n. If n = 0, then L = C22 , and the statement is trivial. Let the statement hold for n − 1 and let K be the patch lattice we obtain by n − 1 fork insertions into C22 , so that we obtain L from K by one fork insertion at the covering square S. We have three cases to consider.

4.6. Rectangular intervals

57

Case 1. p is a prime interval of K. Then the statement holds for p and α⌉K , the restriction of α to K by induction. So p is collapsed by con((α⌉K )l ∪ (α⌉K )r ) in K. Therefore, (P) holds in L. In the next two cases, we assume that p is not in K. Case 2. p is perspective to a prime interval of K. Same proof as in Case 1. This case includes p = [o, a] and any one of the new intervals up-perspective with [o, a]. Case 3. p = [a, c] and any one of the new intervals is up-perspective with [a, c]. Then the fork extension defines the terminating prime interval q = [y, z] on the boundary of L, which is up-perspective with p, verifying (P). Second, let L be a patch lattice, not necessarily slim. This case is obvious because (P) is preserved when inserting an eye. Finally, if L is not a patch lattice, we induct on |L|. By Theorem 4.14, L is the rectangular gluing of the rectangular lattices A and B over the ideal I and filter J. Let p be a prime interval of L. Then p is a prime interval of A or B, say, of A. (If p is a prime interval of B, then the argument is easier.) By induction, p is collapsed by con( α⌉Cll (A) ∪ α⌉Clr (A) ), so it is collapsed by con( α⌉Cll (L) ∪ α⌉Clr (L) ) = con(αl ∪ αr ). Congruences of SR lattices are not that different from congruences of SPS lattices. Finally, start with a planar semimodular lattice L. Can we add a corner? Yes, unless L is rectangular. So we get the following result, illustrated by Figure 4.8. Three more steps are needed there to get a rectangular extension. Lemma 4.21. Let K be a planar semimodular lattice. Then K has a rectangular extension L. In fact, L is a congruence-preserving extension of K.

4.6.

Rectangular intervals

We call the interval I = [o, i] of an SPS lattice L rectangular, if there are complementary a, b ∈ I such that the element a is to the left of the element b. The next result is a new way to construct an SR lattice from an SPS lattice (see my paper [126]). Theorem 4.22. Let L be an SPS lattice and let I be a rectangular interval of L. Then the lattice I is also an SR lattice. Proof. Theorem 20.3 obviously holds for grids. Otherwise, we can assume that the SR lattice K is not a grid, so n = Rank(K) > 1. Let K − be the lattice we obtain by deleting a minimal fork in K − at the covering square Cn−1 = {on−1 , cn−1 , dn−1 , in−1 }.

58

4. Planar Semimodular Lattices

Figure 4.10: Proving Theorem 4.22: Case 1 We obtain K from K − by inserting a fork at Cn−1 . We add the element m in the middle of Cn−1 , and add the sequences of elements x1 , . . . on the left going down and y1 , . . . on the right going down as in Figure 4.4. Let I be a rectangular interval in K with corners a, b, where a is to the left of b. We want to prove that I is an SR lattice. Of course, the lattice I is slim. We induct on n = Rank(K). There are three subcases. Case 1. I is disjoint to id(m), as illustrated in Figure 4.10. Then the interval I is not changed as we add the fork to K − . By induction, I is rectangular in K − , therefore, I is also rectangular in K. Case 2. In Figure 4.11 (and Figure 4.12), the bold lines form the boundary of the rectangular sublattice I in K − , the elements of Cn−1 are gray-filled, and the elements m, x1 , . . . , y1 , . . . are black-filled. The element m is internal in I, so the element a is cn−1 or it is to the left of cn−1 , and symmetrically (see Figure 4.11). Therefore, Cn−1 = [on−1 , in−1 ]K − is a covering square in K − and we obtain the interval [on−1 , in−1 ]K of K by adding a fork to Cn−1 at [on−1 , in−1 ]K − . A fork extension of an SR lattice is also an SR lattice, so we conclude that I is an SR lattice. Case 3. m is not an internal element of I but some xi or yi is (see Figure 4.12), where y2 is an internal element of I. By utilizing that id(dn−1 ) is distributive, we conclude that we obtain I from [o, i]K − by replacing a cover preserving Cm × C2 by Cm × C3 , and so I remains rectangular. We proceed with some applications of this result. The next statement follows directly.

4.6. Rectangular intervals

59

Figure 4.11: Proving Theorem 4.22: Case 2 Corollary 4.23. Let L be an SPS lattice and let I be a rectangular interval of L. Let (P ) be any property of slim rectangular lattices. Then the property (P ) holds for the lattice I. For instance, let (P) be the property: the intervals [o, a] and [o, b] are chains and all elements of the lower boundary of I are meet-reducible, except for a, b. Then we get the main result of G. Cz´edli [52]. Corollary 4.24. Let L be an SPS lattice and let I be a rectangular interval of L with corners a, b. Then [o, a] and [o, b] are chains and all the elements of the lower boundary of I except for a, b are meet-reducible. Another nice application is the following. Corollary 4.25. Let L be an SPS lattice and let I be a rectangular interval of L with corners a, b. Then for any x ∈ I, the following equation holds: x = (x ∧ a) ∨ (x ∧ b). There is a more elegant way to formulate the last result.

60

4. Planar Semimodular Lattices

I

Figure 4.12: Proving Theorem 4.22: Case 3 Corollary 4.26. Let L be an SPS lattice and let a, b, and c be pairwise incomparable elements of L. If a is to the left of b, and b is to the left of c, then b = (b ∧ a) ∨ (b ∧ c).

4.7.

Special diagrams for SR lattices

Two approaches Let L be an SR lattice. When is a diagram L “nice?” Clearly, if the diagram has some special properties that allow us to easily visualize the properties of L and may even help us with some proofs. Our “nice” is not the same as the “nice” of our graphic artist friends. They may draw S7 with the golden ratio, as in the second diagram of Figure 4.13, which may be nicer but not practical. We work with vector graphics apps and grids. The first diagram of Figure 4.13 is easy to draw, but the second diagram is not. Try to draw Figure 29.3 with the golden ratio. The first attempt to define “nice” was in my joint paper with E. Knapp [138], published in 2008; we named it natural. The third diagram of Figure 4.13

4.7. Special diagrams for SR lattices

61

Figure 4.13: 1. S7 ; 2. S7 with golden ratio; 3. a natural diagram of S7 illustrates how we obtain the natural diagram of S7 by deleting two elements from the diagram of C23 . A decade later (in 2017), the second approach to define “nice’ diagrams was presented in G. Cz´edli [40]. A C1 -diagram is defined as a diagram with some very nice properties; the angles of the edges are very restrictive. On the other hand, their existence is not trivial. Natural diagrams In this section, we follow my joint paper with E. Knapp [138]. For an SR lattice L, let Cl (L) be the lower left and Cr (L) the lower right boundary chain of L, respectively, and let lc(L) be the left and rc(L) the right corner of L, respectively. We regard the grid G = Cl (L) × Cr (L) as a planar lattice, with Cl (L) = Cl (G) and Cr (L) = Cr (G). Then the map (6)

ψ : x 7→ (x ∧ lc(L), x ∧ rc(L))

is a meet-embedding of L into G; the map ψ also preserves the bounds. Therefore, the image of L under ψ in G is a diagram of L; we call it the natural diagram representing L. For instance, the third diagram of Figure 4.13 shows the natural diagram representing S7 . Let L be an SR lattice. By the Structure Theorem, Strong Version, we can represent L in the form L = K[C], where K is an SR lattice and C = {o, c, d, i} is a distributive 4-cell of K. Let D be a diagram of K. We form the diagram D[C] by adding the element m in (the center of) C, the elements m, x1 , . . . , and m, y1 , . . . , as in the last diagram of Figure 4.4, so that the lines spanned by the elements m, x1 , . . . and m, y1 , . . . are both normal. Lemma 4.27. Let L, C, K, D, and D[C] be as in the previous paragraph. Then D[C] is a diagram of L. Proof. This is obvious.

62

4. Planar Semimodular Lattices

Lemma 4.28. Let us make the assumptions of Lemma 4.27. If D is a natural diagram of K, then D[C] is a natural diagram of L. Proof. Let D be a natural diagram of K. Let the line m, x1 , . . . terminate with xkl and the line m, y1 , . . . with ykr . We have to show that all the new elements in L can be represented as a join ul ∨ ur , where ul ∈ Cl (L) and ur ∈ Cr (L). Indeed, m = xkl ∨ xkr . The others follow from the distributivity assumptions. C1 -diagrams This research tool, introduced by G. Cz´edli, has been playing an important role in some recent papers (see G. Cz´edli [40]–[52], my joint paper with G. Cz´edli [59], and also G. Gr¨atzer [124]; for the definition, see G. Cz´edli [40] and G. Gr¨atzer [124]). In the diagram of an SR lattice K, a normal edge (line) has a slope of 45◦ or 135◦ . Any edge (line) of slope strictly between 45◦ and 135◦ is steep. Figure 4.14 depicts the lattice S7 . A peak sublattice S7 (peak sublattice, for short) of a lattice L is a sublattice isomorphic to S7 such that the three edges at the top are covers in the lattice L.

b b

a

a

c

c

Figure 4.14: Two diagrams of the lattice S7 as a peak sublattice Definition 4.29. A diagram of an SR lattice L is a C1 -diagram, if the middle edge of a peek sublattice is steep and all other edges are normal. Theorem 4.30. Every slim rectangular lattice L has a C1 -diagram. This was proved in G. Cz´edli [40]. My note [124] presents a short and direct proof. This result also follows from Theorem 4.32.

4.8.

Natural diagrams and C1 -diagrams

We start with a trivial statement. Lemma 4.31. Let us make the assumptions of Lemma 4.27. If D is a C1 diagram of K, then D[C] is a C1 -diagram of L.

4.8. Natural diagrams and C1 -diagrams

63

Now we state our second result on SR lattices. Theorem 4.32. Let L be a SR lattice. Then a natural diagram of L is a C1 -diagram. Conversely, every C1 -diagram is natural. Proof. Let us assume that the SR lattice L can be obtained from a grid G by inserting forks n-times, where n = Rank(L). We induct on n. The case n = 0 is trivial because then L is a grid. So let us assume that the theorem holds for n − 1. By the Structure Theorem, Strong Version (Theorem 4.12), there is an SR lattice K and a distributive 4-cell C = {o, a, b, i} of K such that K can be obtained from the grid G by inserting forks (n − 1)-times and also L = K[C] holds. Now form the natural diagram D of K. By induction, it is a C1 -diagram. By Lemma 4.27, the diagram D[C] is both natural and C1 . We prove the converse the same way. Natural diagrams exist by definition. So as we noted before, Theorem 4.30 also follows from Theorem 4.32. G. Cz´edli [40] also defined C2 -diagrams. A C1 -diagram is C2 , if any two edges on the lower boundary are of the same length. We use Theorem 4.32 to prove the following two results of G. Cz´edli [40]. Theorem 4.33. Let L be a SR lattice. Then L has a C2 -diagram. Proof. Let Cl and Cr be chains of the same length as Cl (L) and Cr (L), respectively. Then Cl (L) × Cr (L) and Cl × Cr are isomorphic, so we can regard the map ψ (see equation (6)) as a map from L into Cl × Cr , a bounded and meet-preserving map. So the natural diagram it defines is the diagram of the lattice L. If we choose Cl and Cr so that the edges are of the same size, we obtain a C2 -diagram of the SR lattice L. Natural diagrams have a left-right symmetry. The symmetric diagram is obtained with the map (7) replacing equation (6).

ψe : x 7→ (x ∧ rc(L), x ∧ lc(L)),

Theorem 4.34 (Uniqueness Theorem). Let L be a slim rectangular lattice. Then the C1 -diagram of L is unique up to left-right symmetry.

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4. Planar Semimodular Lattices

4.9.

Discussion

Semidistributivity is an important lattice property. Can we prove an RT for finite join-semidistributive lattices? The three-element chain cannot be represented as the congruence lattice of a finite semidistributive lattice. In fact, consider the following result. Theorem 4.35. A finite distributive lattice D with a three-element filter cannot be represented as the congruence lattice of a finite semidistributive lattice. We start with an easy statement. Lemma 4.36. Let a be an atom in a finite meet-semidistributive lattice L. Define Pa = id(a∗ ) = { x ∈ L | x ≤ a∗ }. Then Pa is a prime ideal. Proof. Indeed, if Pa is not a prime ideal, then there are elements b, c ∈ L such / Pa , we get that a ∧ b ̸= 0. Since a that b, c ∈ / Pa but b ∧ c ∈ Pa . From b ∈ is an atom, we conclude that a ≤ b. Similarly, a ≤ c. Therefore, a ≤ b ∧ c, contradicting that b ∧ c ∈ Pa . With a prime ideal P , we associate the congruence εP with two congruence classes: P and L − P . By way of contradiction, let the finite meet-semidistributive lattice L represent D, a finite distributive lattice with a there-element filter F . Let α generate F as a filter. Then Con(L/α) represents F . Since the homomorphic image of a finite meet-semidistributive lattice is itself a finite meet-semidistributive lattice (see Corollary 3-1.21 of [8]), we obtain the following. There is a finite meet-semidistributive lattice K representing the threeelement chain. Clearly, K has exactly one nontrivial congruence. If K has exactly one atom a, then K has at least two nontrivial congruences, namely con(0, a) and con(a, 1), a contradiction. If K has at least two distinct atoms a and b, then K has at least two nontrivial congruences, namely εPa and εPb , again a contradiction. This completes the proof of Theorem 4.35. So congruence lattices of finite semidistributive lattices form a proper subclass of the class of finite distributive lattices. Theorem 12 in K. Adaricheva, R. Freese, and J. B. Nation [5] characterize the class of finite semidistributive lattices that can be represented as congruence lattices of finite semidistributive lattices. R. P. Dilworth [78] first studied lattices with unique irreducible decomposition. P. Crawley and R. P. Dilworth [24] present a deeper account. In recent

4.9. Discussion

65

research, finite convex geometries are finite lattices with unique irreducible decomposition. It turns out that a finite lattice is join-distributive iff it has unique irreducible decomposition (see Theorem 3-1.4, K. Adaricheva and J. B. Nation [8]). There is a lot of research on finite semidistributive lattices (see Chapters 3–6 by K. Adaricheva and J. B. Nation in LTS2-[207] for a survey). G. Cz´edli et al. (see [3], [4], [39], [43], [64], and [221]) studied the connection between SPS lattices and finite convex geometries. R. Freese and J. B. Nation [89] prove that the two-element chain can be represented as the congruence lattice of an infinite semidistributive lattice. Problem 1. Can every finite distributive lattice be represented as the congruence lattice of an infinite semidistributive lattice?

Part II

Some Special Techniques

67

Chapter

5

Chopped Lattices

The first basic technique is the use of a chopped lattice, a finite meet-semilattice (M, ∧) regarded as a partial algebra (M, ∧, ∨), where ∨ is a partial operation. It turns out that the ideals of a chopped lattice form a lattice with the same congruence lattice as that of the chopped lattice. So to construct a finite lattice with a given congruence lattice it is enough to construct such a chopped lattice. The problem is how to ensure that the ideal lattice of the chopped lattice has some given properties. As an example, we will look at sectionally complemented lattices. Chopped lattices were introduced by myself and H. Lakser (published in GLT-[99]). They were named in my joint paper with E. T. Schmidt [180] and generalized to the infinite case in another joint paper with E. T. Schmidt [181].

5.1.

Basic definitions

An (n-ary) partial operation on a nonempty set A is a map from a subset of An to A. For n = 2, we call the partial operation binary. A partial algebra is a nonempty set A with partial operations defined on A. A finite meet-semilattice (M, ∧) may be regarded as a partial algebra, (M, ∧, ∨), called a chopped lattice, where ∧ is an operation and ∨ is a partial operation: a ∨ b is the least upper bound of a and b, provided that it exists. We can obtain an example of a chopped lattice by taking a finite lattice with unit, 1, and defining M = L − {1}. The converse also holds: by adding a new unit 1 to a chopped lattice M , we obtain a finite lattice L, and chopping off the unit element, we get M back. 69 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_5

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5. Chopped Lattices

A more useful example is obtained with merging. Let C and D be lattices such that J = C ∩ D is an ideal in both C and D. Then with the natural ordering, Merge(C, D) = C ∪ D, called the merging of C and D, is a chopped lattice. Note that whenever a ∨ b = c in Merge(C, D), then we have a, b, c ∈ C and a ∨ b = c in C or a, b, c ∈ D and a ∨ b = c in D. Among finite ordered sets, chopped lattices are easy to spot. Lemma 5.1. Let M be a finite ordered set and let Max be the set of maximal elements of M . If ↓ m is a lattice, for each m ∈ Max, and if m ∧ n exists, for all m, n ∈ Max, then M is a meet-semilattice. Proof. Indeed, for x, y ∈ M , x ∧ y = (x ∧m a) ∧a (y ∧n a), where m, n ∈ Max, x ≤ m, y ≤ n, a = m ∧ n, and ∧m , ∧n , ∧a denotes the meet in the lattice ↓ m, ↓ n, and ↓ a, respectively. It is easy to see that x ∧ y is the greatest lower bound of x and y in M . A meet-subsemilattice A of a chopped lattice M is a sublattice, if whenever c = a ∨ b in A, then we have c = a ∨ b in M . We define an equivalence relation α to be a congruence of a chopped lattice M as we defined it for lattices in Section 1.3.3: we require that (SP∧ ) and (SP∨ ) hold, the latter with the proviso: whenever a ∨ c and b ∨ d exist. The set Con M of all congruence relations of M ordered by set inclusion is a lattice. Lemma 5.2. Let M be a chopped lattice and let α be an equivalence relation on M satisfying the following two conditions for x, y, z ∈ M : (1) If x ≡ y (mod α); then x ∧ z ≡ y ∧ z (mod α). (2) If x ≡ y (mod α) and x ∨ z and y ∨ z exist, then x ∨ z ≡ y ∨ z (mod α). Then α is a congruence relation on M . Proof. Condition (1) states that α preserves ∧. Now let x, y, u, v ∈ S with x ≡ y (mod α) and u ≡ v (mod α); let x ∨ u and y ∨ v exist. Then x ≡ x ∧ y ≡ y (mod α) and (x ∧ y) ∨ u and (x ∧ y) ∨ v exist. Thus, by condition (2), x ∨ u ≡ (x ∧ y) ∨ u ≡ (x ∧ y) ∨ v ≡ y ∨ v (mod α). A nonempty subset I of the chopped lattice M is an ideal iff it is a down set with the property, (Id) a, b ∈ I implies that a ∨ b ∈ I, provided that a ∨ b exists in M .

5.2. Compatible vectors of elements

71

The set Id M of all ideals of M ordered by set inclusion is a lattice. For I, J ∈ Id M , the meet is I ∩ J, but the join is a bit more complicated to describe. Lemma 5.3. Let I and J be ideals of the chopped lattice M . Define U (I, J)0 = I ∪ J,

U (I, J)i = { x | x ≤ u ∨ v, u, v ∈ U (I, J)i−1 } for 0 < i < ω.

Then S

I ∨J =

[

( U (I, J)i | i < ω ).

Proof. Define U = ( U (I, J)i | i < ω ). If K is an ideal of M , then I ⊆ K and J ⊆ K imply—by induction—that U ⊆ K. So it is sufficient to prove that U is an ideal of M . Obviously, U is a down set. Also, U has property (Id), since if a, b ∈ U and a ∨ b exists in M , then a, b ∈ U (I, J)n , for some 0 < n < ω, and therefor a ∨ b ∈ U (I, J)n+1 ⊆ U . Most lattice concepts and notations for them will be used for chopped lattices without further explanation. Infinite chopped lattices are briefly mentioned on page 113.

5.2.

Compatible vectors of elements

Let M be a chopped lattice, and let Max(M ) (MaxSif M is understood) be the set of maximal elements of M . Then M = ( id(m) | m ∈ Max ) and each id(m) is a (finite) lattice. A vector (associated with M ) is of the form (im | m ∈ Max), where im ≤ m for all m ∈ M . We order the vectors componentwise. With every ideal I of M , we can associate the vector (im | m ∈ Max) S defined by I ∩ id(m) = id(im ). Clearly, I = ( id(im ) | m ∈ M ). Such vectors are easy to characterize. Let us call the vector (jm | m ∈ Max) compatible if jm ∧ n = jn ∧ m for all m, n ∈ Max. Lemma 5.4. Let M be a chopped lattice. (i) There is a one-to-one correspondence between ideals and compatible vectors of M . (ii) Given any vector g = (gm | m ∈ Max), there is a smallest compatible vector g = (im | m ∈ Max) containing g. (iii) Let I and J be ideals of M , with corresponding compatible vectors (im | m ∈ Max) and (jm | m ∈ Max).

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5. Chopped Lattices

Then (a) I ≤ J in Id M iff im ≤ jm for all m ∈ Max.

(b) The compatible vector corresponding to I ∧ J is (im ∧ jm | m ∈ Max). (c) Let a = (im ∨ jm | m ∈ Max). Then the compatible vector corresponding to I ∨ J is a.

Proof. (i) Let I be an ideal of M . Then (im | m ∈ Max) is compatible since im ∧ m ∧ n and in ∧ m ∧ n both generate the principal ideal I ∩ id(m) ∩ id(n) for all m, n ∈ Max. Conversely, let (jm | m ∈ Max) be compatible, and define [ I = ( id(jm ) | m ∈ Max ). Observe that (1)

I ∩ id(m) = id(jm ),

for m ∈ Max. Indeed, if x ∈ I ∩ id(m) and x ∈ id(jn ), for n ∈ Max, then x ≤ m ∧ jn = n ∧ jm (since (jm | m ∈ Max) is compatible), so x ≤ jm , that is, x ∈ id(jm ). The reverse inclusion is obvious. I is obviously a down set. To verify property (Id) for I, let a, b ∈ I and let us assume that a ∨ b exists in M . Then a ∨ b ≤ m, for some m ∈ Max, so a ≤ m and b ≤ m. By (1), we get a ≤ jm and b ≤ jm , therefore, a∨b ≤ jm ∈ I. Since I is a down set, it follows that a ∨ b ∈ I, verifying property (Id). (ii) Obviously, the vector (m | m ∈ Max) contains all other vectors and it is compatible. Since the componentwise meet of compatible vectors is compatible, the statement follows. (iii) This is obvious since, by (ii), we are dealing with a closure system (see Section 1.2.2).

5.3.

Compatible vectors of congruences

Let M be a chopped lattice. With any congruence α of M , we can associate the restriction vector ( α⌉m | m ∈ Max), where α⌉m is the restriction of α to id(m). The restriction α⌉m is a congruence of the lattice id(m). Let γ m be a congruence of id(m) for all m ∈ Max. The congruence vector (γ m | m ∈ Max) is called compatible if γ m restricted to id(m ∧ n) is the same as γ n restricted to id(m ∧ n) for m, n ∈ Max. Obviously, a restriction vector is compatible. The converse also holds. Lemma 5.5. Let (γ m | m ∈ Max) be a compatible congruence vector of a chopped lattice M . Then there is a unique congruence α of M such that the restriction vector of α agrees with (γ m | m ∈ Max).

5.3. Compatible vectors of congruences

73

Proof. Let (γ m | m ∈ Max) be a compatible congruence vector. We define a binary relation α on M as follows. Let m, n ∈ Max. For x ∈ id(m) and y ∈ id(n), let x ≡ y (mod α) iff x ≡ x ∧ y (mod γ m ) and y ≡ x ∧ y (mod γ n ). Obviously, α is reflexive and symmetric. To prove transitivity, let m, n, k ∈ Max, and let x ∈ id(m), y ∈ id(n), z ∈ id(k); let x ≡ y (mod α) and y ≡ z (mod α), that is, x ≡ x ∧ y (mod γ m ),

(2)

y ≡ x ∧ y (mod γ n ),

(3)

y ≡ y ∧ z (mod γ n ),

(4)

z ≡ y ∧ z (mod γ k ).

(5)

Then meeting the congruence (4) with x (in the lattice id(n)), we get x ∧ y ≡ x ∧ y ∧ z (mod γ n ),

(6)

and from (3), by meeting with z, we obtain y∧z ≡x∧y∧z

(7)

(mod γ n ).

Since x ∧ y and x ∧ y ∧ z ∈ id(m), by compatibility, (6) implies that x∧y ≡x∧y∧z

(8)

(mod γ m ).

Now (2) and (8) yield (9)

x≡x∧y∧z

(mod γ m ).

z ≡x∧y∧z

(mod γ k ).

Similarly, (10)

γ m is a lattice congruence on id(m) and x ∧ y ∧ z ≤ x ∧ z ≤ x, so (11)

x≡x∧z

(mod γ m ).

z ≡x∧z

(mod γ k ).

Similarly, (12)

Equations (11) and (12) yield that x ≡ z (mod α), proving transitivity. (SP∧ ) is easy: let x ∈ id(m), y ∈ id(n), z ∈ M ; if x ≡ y (mod α), then x ∧ z ≡ y ∧ z (mod α) because x ∧ z ≡ x ∧ y ∧ z (mod γ m ) and y ∧ z ≡ x ∧ y ∧ z (mod γ n ).

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5. Chopped Lattices

Finally, we verify (SP∨ ). Let x ≡ y (mod α) and z ∈ M , and let us assume that x ∨ z and y ∨ z exist. Then there are p, q ∈ Max such that x ∨ z ∈ id(p) and y ∨ z ∈ id(q). By compatibility, x ≡ x ∧ y (mod γ p ), so x ∨ z ≡ (x ∧ y) ∨ z

(mod γ p ).

Since (x ∧ y) ∨ z ≤ (x ∨ z) ∧ (y ∨ z) ≤ x ∨ z, we also have x ∨ z ≡ (x ∨ z) ∧ (y ∨ z) (mod γ p ). Similarly, y ∨ z ≡ (x ∨ z) ∧ (y ∨ z) (mod γ q ). The last two displayed equations show that x ∨ z ≡ y ∨ z (mod α).

5.4.

From the chopped lattice to the ideal lattice

The map m 7→ id(m) embeds the chopped lattice M with zero into the lattice Id M , so we can regard Id M as an extension. It is, in fact, a congruencepreserving extension (see my joint paper with H. Lakser [139], proof first published in LTFC-[98]). Theorem 5.6. Let M be a chopped lattice. Then Id M is a congruencepreserving extension of M . Proof. Let α be a congruence relation of M . If I, J ∈ Id M , define I ≡ J (mod α)

iff

I/α = J/α,

where I/α denotes I modulo α⌉I and the same for J/α. Obviously, α is an equivalence relation. Let I ≡ J (mod α), N ∈ Id M , and x ∈ I ∩ N . Then x ≡ y (mod α), for some y ∈ J, and so x ≡ x ∧ y (mod α) and x ∧ y ∈ J ∩ N . This shows that (I ∩ N )/α ⊆ (J ∩ N )/α. Similarly, (J ∩ N )/α ⊆ (I ∩ N )/α, so I ∩ N ≡ J ∩ N (mod α). To prove I ∨ N ≡ J ∨ N (mod α), by symmetry, it is sufficient to verify that I ∨ N ⊆ (J ∨ N )/α. By Lemma 5.3, this is equivalent to proving that Un ⊆ (J ∨ N )/α for n < ω. This is obvious for n = 0. Now assume that Un−1 ⊆ (J ∨ N )/α and let x ∈ Un . Then x ≤ t1 ∨ t2 for some t1 , t2 ∈ Un−1 . Thus t1 ≡ u1 (mod α) and t2 ≡ u2 (mod α), for some u1 , u2 ∈ J ∨ N , and so t1 ≡ t1 ∧ u1 (mod α) and t2 ≡ t2 ∧ u2 (mod α). Observe that t1 ∨ t2 is an upper bound for {t1 ∧ u1 , t2 ∧ u2 }; consequently, (t1 ∧ u1 ) ∨ (t2 ∧ u2 ) exists. Therefore, t1 ∨ t2 ≡ (t1 ∧ u1 ) ∨ (t2 ∧ u2 ) (mod α).

5.4. From the chopped lattice to the ideal lattice

75

Finally, x ≡ x ∧ (t1 ∨ t2 ) = x ∧ ((t1 ∧ u1 ) ∨ (t2 ∧ u2 )) (mod α), and x ∧ ((t1 ∧ u1 ) ∨ (t2 ∧ u2 )) ∈ J ∨ N. Thus x ∈ (J ∨N )/α, completing the induction, verifying that α is a congruence relation of Id M . If a ≡ b (mod α) and x ∈ id(a), then x ≡ x ∧ b (mod α). Thus id(a) ⊆ id(b)/α. Similarly, id(b) ⊆ id(a)/α, and so id(a) ≡ id(b) (mod α). Conversely, if id(a) ≡ id(b) (mod α), then a ≡ b1 (mod α) and a1 ≡ b (mod α) for some a1 ≤ a and b1 ≤ b. Forming the join of these two congruences, we get a ≡ b (mod α). Thus α has all the properties required by Lemma 5.3. To show the uniqueness, let γ be a congruence relation of Id M satisfying id(a) ≡ id(b) (mod γ) iff a ≡ b (mod α). Let I, J ∈ Id M , I ≡ J (mod γ), and x ∈ I. Then id(x) ∩ I ≡ id(x) ∩ J

(mod γ),

id(x) ∩ I = id(x),

id(x) ∩ J = id(y)

for some y ∈ J. Thus id(x) ≡ id(y) (mod γ), and so x ≡ y (mod α), proving that I ⊆ J/α. Similarly, J ⊆ I/α, and so I ≡ J (mod α). Conversely, if I ≡ J (mod α), then take all congruences of the form x ≡ y (mod α), x ∈ I, y ∈ J. By our assumption regarding γ, we get the congruence id(x) ≡ id(y) (mod γ), and by our definition of α, the join of all these congruences yields I ≡ J (mod α). Thus γ = α. This result is very useful. It means that in order to construct a finite lattice L to represent a given finite distributive lattice D as a congruence lattice, it is sufficient to construct a chopped lattice M with Con M ∼ = D, since Con M ∼ = Con(Id M ) = Con L, where L = Id M , and L is a finite lattice. This result also allows us to construct congruence-preserving extensions. Corollary 5.7. Let M = Merge(A, B) be a chopped lattice with A = id(a) and B = id(b). If a ∧ b > 0 and B is simple, then Id M is a congruence-preserving extension of A. Proof. Let α be a congruence of A. Then (α, γ) is a compatible congruence vector iff ( 0 if α is discrete on [0, a ∧ b]; γ= 1, otherwise. So γ is determined by α and the statement follows.

76

5.5.

5. Chopped Lattices

Sectional complementation

We introduce sectionally complemented chopped lattices as we did for lattices in Section 2.1. We illustrate the use of compatible vectors with two results on sectionally complemented chopped lattices. The first result is from my joint paper with E. T. Schmidt [184]. Lemma 5.8 (Atom Lemma). Let M be a chopped lattice with two maximal elements m1 and m2 . We assume that id(m1 ) and id(m2 ) are sectionally complemented lattices. If p = m1 ∧ m2 is an atom, then Id M is sectionally complemented. m2 m1 p

M 0 Figure 5.1: The Atom Lemma illustrated

Proof. Figure 5.1 illustrates the setup. To show that Id M is sectionally complemented, let I ⊆ J be two ideals of M , represented by the compatible vectors (i1 , i2 ) and (j1 , j2 ), respectively. Let s1 be the sectional complement of i1 in j1 and let s2 be the sectional complement of i2 in j2 . If p ∧ s1 = p ∧ s2 , then (s1 , s2 ) is a compatible vector, representing an ideal S that is a sectional complement of I in J. Otherwise, without loss of generality, we can assume that p ∧ s1 = 0 and p ∧ s2 = p. Since id(m2 ) is sectionally complemented, there is a sectional complement s′2 of p in [0, s2 ]. Then (s1 , s′2 ) satisfies p ∧ s1 = p ∧ s′2 (= 0), and so it is compatible; therefore, (s1 , s′2 ) represents an ideal S of M . Obviously, I ∧ S = {0}. From p ∧ s2 = p, it follows that p ≤ s2 ≤ j2 . Since J is an ideal and j2 ∧ p = p, it follows that j1 ∧ p = p, that is, p ≤ j1 . Obviously, I ∨ S ⊆ J. So to show that I ∨ S = J, it is sufficient to verify that j1 , j2 ∈ I ∨ S. Evidently, j1 = i1 ∨ s1 ∈ I ∨ S. Note that p ≤ j1 = i1 ∨ s1 ∈ I ∨ S. Thus p, s′2 , i2 ∈ I ∨ S, and therefore p ∨ s′2 ∨ i2 = (p ∨ s′2 ) ∨ i2 = s2 ∨ i2 = j2 ∈ I ∨ S.

5.5. Sectional complementation

77

The second result (G. Gr¨atzer, H. Lakser, and M. Roddy [155]) shows that the ideal lattice of a sectionally complemented chopped lattice is not always sectionally complemented. Theorem 5.9. There is a sectionally complemented chopped lattice M whose ideal lattice Id M is not sectionally complemented. m2

m1

q′

p′ c

d u b

a p

q

L1

L2 0

Figure 5.2: The chopped lattice M Proof. Let M be the chopped lattice of Figure 5.2, where L1 = id(m1 ) and L2 = id(m2 ). Note that p is meet-irreducible in id(m2 ) and q is meet-irreducible in id(m1 ). The unit element of the ideal lattice of M is the compatible vector (m1 , m2 ). We show that the compatible vector (a, b) has no complement in the ideal lattice of M . Assume, to the contrary, that the compatible vector (s, t) is a complement of (a, b). Since (a, b) ≤ (a ∨ u, m2 ), a compatible vector, (s, t) ≰ (a ∨ u, m2 ), that is, (13)

s ≰ a ∨ u.

Similarly, by considering (m1 , b ∨ u), we conclude that (14)

t ≰ b ∨ u.

Now (a, b) ≤ (p′ , q ′ ), which is a compatible vector. Thus (s, t) ≰ (p′ , q ′ ), and so either s ≰ p′ or t ≰ q ′ . Without loss of generality, we may assume that s ≰ p′ . It then follows by (13) that s can be only c or m1 . Then since s ∧ a = 0, we conclude that s = c. Thus s ∧ u = p, and so t ∧ u = p. But p is meet-irreducible in L2 . Thus t = p ≤ b ∨ u, contradicting (14).

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5. Chopped Lattices

This result illustrates that the Atom Lemma (Lemma 5.8) cannot be extended to the case where [0, m1 ∧ m2 ] is a four-element Boolean lattice.

Chapter

6

Boolean Triples

In Part IV, we construct congruence-preserving extensions of finite lattices, extensions with special properties, such as sectionally complemented, semimodular, and so on. It is easy to construct a proper congruence-preserving extension of a nontrivial finite lattice. In the early 1990s, in my joint paper with E. T. Schmidt [180], we raised the question whether every nontrivial lattice has a proper congruence-preserving extension. (See also another joint paper of mine with E. T. Schmidt [181].) It took almost a decade for an answer to appear in my joint paper with F. Wehrung [199]. For infinite lattices, the affirmative answer was provided by the Boolean triples construction, which is described in this chapter. It is interesting that Boolean triples also provide an important tool for finite lattices.

6.1.

The general construction

In this section, I describe a congruence-preserving extension of a (finite) lattice L, introduced in my joint paper with F. Wehrung [199]. We will see that this generalizes a construction of E. T. Schmidt [238] for bounded distributive lattices. For a lattice L, let us call the triple (x, y, z) ∈ L3 Boolean iff (F)

x = (x ∨ y) ∧ (x ∨ z), y = (y ∨ x) ∧ (y ∨ z), z = (z ∨ x) ∧ (z ∨ y), 79

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_6

80

6. Boolean Triples

where F stands for “Fixed point definition.” Note that by Lemma 2.6, if (F) holds, then sub({x, y, z}) is Boolean. We call (F) the “Fixed point definition” because the triple (x, y, z) satisfies condition (F) iff p(x, y, z) = (x, y, z), where p(x, y, z) = ((x ∨ y) ∧ (x ∨ z), (y ∨ x) ∧ (y ∨ z), (z ∨ x) ∧ (z ∨ y)). We denote by M3 [L] ⊆ L3 the ordered set of Boolean triples of L (ordered as an ordered subset of L3 , that is, componentwise). If we apply the construction to an interval [a, b] of L, we write M3 [a, b] for M3 [[a, b]]. Observe that any Boolean triple (x, y, z) ∈ L3 satisfies x ∧ y = y ∧ z = z ∧ x.

(B)

where B stands for “Balanced definition.” Indeed, if (x, y, z) is Boolean, then x ∧ y = y ∧ z = z ∧ x = (x ∨ y) ∧ (y ∨ z) ∧ (z ∨ x). We call such triples balanced. Here are some of the basic properties of Boolean triples. Lemma 6.1. Let L be a lattice. (i) (x, y, z) ∈ L3 is Boolean iff there is a triple (u, v, w) ∈ L3 satisfying (E)

x = u ∧ v,

y = u ∧ w, z = v ∧ w.

(ii) M3 [L] is a closure system in L3 . For (x, y, z) ∈ L3 , the closure is (x, y, z) = ((x ∨ y) ∧ (x ∨ z), (y ∨ x) ∧ (y ∨ z), (z ∨ x) ∧ (z ∨ y)). (iii) If L has 0, then the ordered subset { (x, 0, 0) | x ∈ L } is a sublattice of M3 [L] and γ : x 7→ (x, 0, 0) is an isomorphism between L and this sublattice. (iv) If L is bounded, then M3 [L] has a spanning M3 , that is, a bounded sublattice isomorphic to M3 , namely, {(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)}. Remark. (E) is the “Existential definition” because (x, y, z) is Boolean iff there exists a triple (u, v, w) satisfying (E).

6.1. The general construction

81

Proof. (i) If (x, y, z) is Boolean, then u = x ∨ y, v = x, ∨z, w = y ∨ z satisfy (E). Conversely, if there is a triple (u, v, w) ∈ L3 satisfying (E), then by Lemma 2.6, the sublattice generated by x, y, z is isomorphic to a quotient of B3 and x, y, z are the images of the three atoms of B3 . Thus (x ∨ y) ∧ (x ∨ z) = x, the first equation in (F). The other two equations are proved similarly. (ii) M3 [L] ̸= ∅; for instance, for all x ∈ L, the diagonal element (x, x, x) ∈ M3 [L]. For (x, y, z) ∈ L3 , define u = x ∨ y, v = x ∨ z, w = y ∨ z. Set x1 = u ∧ v, y1 = u ∧ w, z1 = v ∧ w.

Then (x1 , y1 , z1 ) is Boolean by (i) and (x, y, z) ≤ (x1 , y1 , z1 ) in L3 . Now if (x, y, z) ≤ (x2 , y2 , z2 ) in L3 and (x2 , y2 , z2 ) is Boolean, so x2 = (x2 ∨ y2 ) ∧ (x2 ∨ z2 ) ≥ (x ∨ y) ∧ (x ∨ z) = u ∧ v = x1 ,

(by (F)) (by (x, y, z) ≤ (x2 , y2 , z2 ))

and similarly, y2 ≥ y1 , z2 ≥ z1 . Thus (x2 , y2 , z2 ) ≥ (x1 , y1 , z1 ), and so (x1 , y1 , z1 ) is the smallest Boolean triple containing (x, y, z). (iii) and (iv) are obvious. M3 [L] is difficult to draw, in general. Figure 6.1 shows the diagram of M3 [C3 ] with the three-element chain C3 = {0, a, 1}. (1, 1, 1) (1, 1, a) (1, 0, 0)

(a, a, a)

(a, 0, 0)

(a, 1, 1)

h1, a, 1i

h0, a, 0i

(0, 0, 1)

(0, 1, 0)

(0, 0, a) (0, 0, 0)

Figure 6.1: The lattice M3 [C3 ] with a sketch

82

6. Boolean Triples

If C is an arbitrary bounded chain, with bounds 0 and 1, it is easy to picture M3 [C], as sketched in Figure 6.1. The element (x, y, z) ∈ C 3 is Boolean iff it is of the form (x, y, y), or (y, x, y), or (y, y, x), where y ≤ x in C. So the diagram is made up of three isomorphic “flaps” overlapping on the diagonal. Two of the flaps form the “base,” a planar lattice: C23 , the third one (shaded) comes up out of the plane pointing in the direction of the viewer. We get some more examples of M3 [L] from the following observation. Lemma 6.2. Let the lattice L have a direct product decomposition: L = L1 × L2 . Then M3 [L] ∼ = M3 [L1 ] × M3 [L2 ]. Proof. This is obvious, since ((x1 , y1 ), (x2 , y2 ), (x3 , y3 )) is a Boolean triple iff (x1 , x2 , x3 ) and (y1 , y2 , y3 ) are both Boolean triples (xi ∈ L1 and yi ∈ L2 for i = 1, 2, 3).

6.2.

The congruence-preserving extension property

Let L be a nontrivial lattice with zero and let γ : x 7→ (x, 0, 0) ∈ M3 [L] be an embedding of L into M3 [L]. Here is the main result of my joint paper with F. Wehrung [199]. Theorem 6.3. M3 [L] is a congruence-preserving extension of γL. The next two lemmas prove this theorem. For a congruence α of L, let α3 denote the congruence of L3 defined componentwise. Let M3 [α] be the restriction of α3 to M3 [L]. Lemma 6.4. M3 [α] is a congruence relation of M3 [L]. Proof. M3 [α] is obviously an equivalence relation on M3 [L]. Since M3 [L] is a meet-subsemilattice of L3 , it is clear that M3 [α] satisfies (SP∧ ). To verify (SP∨ ) for M3 [α], let (x1 , y1 , z1 ), (x2 , y2 , z2 ) ∈ M3 [L], let (x1 , y1 , z1 ) ≡ (x2 , y2 , z2 ) (mod M3 [α]), and let (u, v, w) ∈ M3 [L]. Set

(x′i , yi′ , zi′ ) = (xi , yi , zi ) ∨ (u, v, w)

(the join formed in M3 [L]) for i = 1, 2. Then using Lemma 6.1.(ii) for x1 ∨ u, y1 ∨ v, and z1 ∨ w, we obtain that x′1 = (x1 ∨ u ∨ y1 ∨ v) ∧ (x1 ∨ u ∨ z1 ∨ w) ≡ (x2 ∨ u ∨ y2 ∨ v) ∧ (x2 ∨ u ∨ z2 ∨ w) = x′2 (mod M3 [α]),

and similarly, y1′ ≡ y2′ (mod M3 [α]), z1′ ≡ z2′ (mod M3 [α]), hence, (x1 , y1 , z1 ) ∨ (u, v, w) ≡ (x2 , y2 , z2 ) ∨ (u, v, w) (mod M3 [α]).

6.2. The congruence-preserving extension property

83

It is obvious that M3 [α] restricted to γL is γα. Lemma 6.5. Every congruence of M3 [L] is of the form M3 [α] for a suitable congruence α of L. Proof. Let γ be a congruence of M3 [L], and let α denote the congruence of L obtained by restricting γ to the sublattice L′ = { (x, 0, 0) | x ∈ L } of M3 [L], that is, for x, y ∈ L, x ≡ y (mod α) iff (x, 0, 0) ≡ (y, 0, 0) (mod γ). We prove that γ = M3 [α]. To show that γ ⊆ M3 [α], let (x1 , y1 , z1 ) ≡ (x2 , y2 , z2 ) (mod γ).

(1)

Meeting the congruence (1) with (1, 0, 0) yields (x1 , 0, 0) ≡ (x2 , 0, 0) (mod γ), and so (x1 , 0, 0) ≡ (x2 , 0, 0) (mod M3 [α]).

(2)

Meeting the congruence (1) with (0, 1, 0) yields (0, y1 , 0) ≡ (0, y2 , 0) (mod γ).

(3) Since

(0, y1 , 0) ∨ (0, 0, 1) = (0, y1 , 1) = (y1 , y1 , 1), and (y1 , y1 , 1) ∧ (1, 0, 0) = (y1 , 0, 0), and similarly for (0, y2 , 0), joining the congruence (3) with (0, 1, 0) and then meeting with (1, 0, 0), yields (y1 , 0, 0) ≡ (y2 , 0, 0) (mod γ), and so (0, y1 , 0) ≡ (0, y2 , 0) (mod M3 [α]).

(4) Similarly, (5)

(0, 0, z1 ) ≡ (0, 0, z2 ) (mod M3 [α]).

Joining the congruences (2), (4), and (5), we obtain (6)

(x1 , y1 , z1 ) ≡ (x2 , y2 , z2 ) (mod M3 [α]),

proving that γ ⊆ M3 [α].

84

6. Boolean Triples

To prove the converse, M3 [α] ⊆ γ, take (x1 , y1 , z1 ) ≡ (x2 , y2 , z2 ) (mod M3 [α])

(7)

in M3 [L]; equivalently, (8) (9) (10)

(x1 , 0, 0) ≡ (x2 , 0, 0) (mod γ), (y1 , 0, 0) ≡ (y2 , 0, 0) (mod γ), (z1 , 0, 0) ≡ (z2 , 0, 0) (mod γ)

in M3 [L]. Joining the congruence (9) with (0, 0, 1) and then meeting the result with (0, 1, 0), we get (as in the computation following (3)): (11)

(0, y1 , 0) ≡ (0, y2 , 0) (mod γ).

Similarly, from (10), we conclude that (12)

(0, 0, z1 ) ≡ (0, 0, z2 ) (mod γ).

Finally, joining the congruences (8), (11), and (12), we get (13)

(x1 , y1 , z1 ) ≡ (x2 , y2 , z2 ) (mod γ),

that is, M3 [α] ⊆ γ. This completes the proof of this lemma.

6.3.

The distributive case

Let D be a bounded distributive lattice. In 1974 (25 years before the publication of my joint paper with F. Wehrung [199]), E. T. Schmidt [238] defined M3 [D] as the set of balanced triples (x, y, z) ∈ D3 (defined in Section 6.1), regarded as an ordered subset of D3 . Schmidt proved the following result. Theorem 6.6. Let D be a bounded distributive lattice. Then M3 [D] (the set of balanced triples (x, y, z) ∈ D3 ) is a modular lattice. The map γ : x 7→ (x, 0, 0) ∈ M3 [D] is an embedding of D into M3 [D], and M3 [D] is a congruence-preserving extension of γD. Proof. Examining Figure 2.6 (page 24), we immediately see that in a distributive lattice D, conditions (B) and (F) are equivalent, so M3 [D] is the Boolean triple construction. Therefore, this result follows from the results in Sections 6.1 and 6.2, except for the modularity. A direct computation of this is not so easy—although entertaining. However, we can do it without computation. Observe that it is enough to prove modularity for a finite D. Now if D is finite, then by Lemma 6.2, M3 [D] can be embedded into M3 [Bn ] ∼ = (M3 [C2 ])n ∼ = Mn3 , a modular lattice; hence, M3 [D] is modular.

6.4. Two interesting intervals

6.4.

85

Two interesting intervals

Let L be a bounded lattice. For an arbitrary a ∈ L, consider the interval M3 [L, a] of M3 [L] (illustrated in Figure 6.2): M3 [L, a] = [(0, a, 0), (1, 1, 1)] ⊆ M3 [L]. Define the map (with image B = γa L in Figure 6.2): γa : x 7→ (x, a, x ∧ a). Lemma 6.7. γa is an embedding of L into M3 [L, a]. Proof. γa is obviously one-to-one and meet-preserving. It also preserves the join because (x ∨ y, a, (x ∧ a) ∨ (y ∧ a)) = (x ∨ y, a, (x ∨ y) ∧ a) for x, y ∈ L. The following result of my joint paper with E. T. Schmidt [186] is a generalization of Theorem 6.3. Theorem 6.8. M3 [L, a] is a congruence-preserving extension of γa L. The proof is presented in the following lemmas. Lemma 6.9. Let (x, y, z) ∈ M3 [L, a]. Then a ≤ y and x ∧ a = z ∧ a.

(14)

(1, 1, 1)

(1, a, a) (1, 0, 0)

M3 [L, a] (0, 1, 0)

B

K

M3 [L]

(0, a, 0)

(0, 0, 1)

(0, 0, 0) Figure 6.2: The shaded area: the lattice M3 [L, a]

86

6. Boolean Triples

Proof. Since (x, y, z) is Boolean, it is balanced, so x ∧ y = z ∧ y. Therefore, x ∧ a = (x ∧ y) ∧ a = (z ∧ y) ∧ a = z ∧ a, as claimed. We need an easy decomposition statement for the elements of M3 [L, a]. Let us use the notation (B and K are marked in Figure 6.2): B = { (x, a, x ∧ a) | x ∈ L } (= γa L), K = { (0, x, 0) | x ∈ L, x ≥ a }, J = { (x ∧ a, a, x) | x ∈ L }. Lemma 6.10. Let v = (x, y, z) ∈ M3 [L, a]. Then v has a decomposition in M3 [L, a]: (15)

v = vB ∨ vK ∨ vJ ,

where (16) (17) (18)

vB = (x, y, z) ∧ (1, a, a) = (x, a, x ∧ a) ∈ B, vK = (x, y, z) ∧ (0, 1, 0) = (0, y, 0) ∈ K, vJ = (x, y, z) ∧ (a, a, 1) = (z ∧ a, a, z) ∈ J.

Proof. (16) follows from (14). By symmetry, (18) follows, and (17) is trivial. Finally, the right side of (15) componentwise joins to the left side—in view of (14). We can now describe the congruences of M3 [L, a]. Lemma 6.11. Let γ be a congruence of M3 [L, a] and let v, w ∈ M3 [L, a]. Then (19)

v≡w

(mod γ),

iff (20) (21) (22)

vB ≡ wB (mod γ), vK ≡ wK (mod γ), vJ ≡ wJ (mod γ).

Proof. (19) implies (20) by (16). Similarly, for (21) and (22). Conversely, (20)–(22) imply (19) by (15). Lemma 6.12. For a congruence α of L, let M3 [α, a] be the restriction of α3 to M3 [L, a]. Then M3 [α, a] is a congruence of M3 [L, a], and every congruence of M3 [L, a] is of the form M3 [α, a], for a unique congruence α of L.

6.4. Two interesting intervals

87

Proof. It follows from Lemma 6.4 that M3 [α, a] is a congruence of M3 [L, a]. Let v = (x, y, z), v′ = (x′ , y ′ , z ′ ) ∈ M3 [L, a]. Let γ be a congruence of M3 [L, a], and let α be the restriction of γ to L with respect to the embedding γa . By Lemma 6.11, v ≡ v′ (mod γ) ′ ′ iff (20)–(22) hold. Note that vB , vB ∈ γa L, so (20) is equivalent to vB ≡ vB (mod α). Now consider p(x) = (x ∨ (0, 1, 0)) ∧ (1, a, a).

Then p((x ∧ a, a, x)) = (x, 1, x) ∧ (1, a, a) = (x, a, x ∧ a). So (22) implies that p(vJ ) ≡ p(vJ′ ) (mod γ), and symmetrically. Thus (22) is equivalent to p(vJ ) ≡ p(vJ′ ) (mod γ), that is, to p(vJ ) ≡ p(vJ′ )

(mod α),

since p(vJ ), p(vJ′ ) ∈ γa L. Now consider q(x) = (x ∨ (a, a, 1)) ∧ (1, a, a). ′ Then q((0, x, 0)) = (x, a, x ∧ a) for x ≥ a. So q(vK ) ≡ q(vK ) (mod γ), that ′ is, q(vK ) ≡ q(vK ) (mod α). Finally, define r(x) = (x ∨ (a, a, 1)) ∧ (0, 1, 0). ′ Then q((x, x ∧ a, a)) = (0, x, 0). So q(vB ) ≡ q(vB ) (mod γ). From these it ′ ′ ) (mod γ) and follows that vK ≡ vK (mod γ) is equivalent to q(vK ) ≡ q(vK ′ ′ ) (mod α). q(vK ), q(vK ) ∈ γa L, so the latter is equivalent to q(vK ) ≡ q(vK ′ We conclude that the congruence v ≡ v (mod γ) in M3 [L, a] is equivalent to the following three congruences in L: ′ vB ≡ vB

p(vJ ) ≡ p(vJ′ )

′ ) q(vK ) ≡ q(vK

concluding the proof of the lemma.

(mod α), (mod α), (mod α),

88

6. Boolean Triples

In Chapter 17, we need a smaller interval introduced in my joint paper with E. T. Schmidt [190]; namely, for a, b ∈ L with a < b, we introduce the interval M3 [L, a, b] of M3 [L]: M3 [L, a, b] = [(0, a, 0), (1, b, b)] ⊆ M3 [L]. Again, γa : x 7→ (x, a, x ∧ a) is a (convex) embedding of L into M3 [L, a, b]. (Note that if L is bounded, then M3 [L, a] = M3 [L, a, 1].) Using this notation (illustrated in Figure 6.4; the black-filled elements form J): B = { (x, a, x ∧ a) | x ∈ L } (= γa L), Ia,b = [(0, a, 0), (0, b, 0)], J = { (x ∧ a, a, x) | x ≤ b }. We can now generalize Lemmas 6.10–6.12. Lemma 6.13. Let v = (x, y, z) ∈ M3 [L, a, b]. Then v has a decomposition in M3 [L, a, b]: v = vB ∨ vIa,b ∨ vJ , where vB = (x, y, z) ∧ (1, a, a) = (x, a, x ∧ a) ∈ B, vIa,b = (x, y, z) ∧ (0, b, 0) = (0, y, 0) ∈ Ia,b , vJ = (x, y, z) ∧ (a, a, b) = (z ∧ a, a, z) ∈ J. (1, 1, 1) (1, b, b) (1, a, a) (1, 0, 0)

Da,b (0, 1, 0) (0, b, 0)

Ia,b

M3 [L]

(0, a, 0)

(0, 0, 0) Figure 6.3: The shaded area: the lattice M3 [L, a, b]

(0, 0, 1)

6.4. Two interesting intervals

89

(1, b, b)

Fa,b

(1, a, a)

(a, a, b)

B

J

(0, b, 0)

Ia,b (0, a, 0) Figure 6.4: The lattice M3 [L, a, b] Lemma 6.14. Let γ be a congruence of M3 [L, a, b] and let v, w ∈ M3 [L, a, b]. Then v ≡ w (mod γ) iff vB ≡ wB (mod γ),

(23)

vIa,b ≡ wIa,b (mod γ),

(24)

v J ≡ wJ

(25)

(mod γ).

Lemma 6.15. For a congruence α of L, let M3 [α, a, b] be the restriction of α3 to M3 [L, a, b]. Then M3 [α, a, b] is a congruence of M3 [L, a, b], and every congruence of M3 [L, a, b] is of the form M3 [α, a, b] for a unique congruence α of L. It follows that γa is a congruence-preserving convex embedding of L into M3 [L, a, b]. We will use the notation Fa,b = [(0, b, 0), (1, b, b)] = { (x, b, x ∧ b) | x ∈ L }. The following two observations are trivial. Lemma 6.16. (i) Ia,b is an ideal of M3 [L, a, b] and Ia,b is isomorphic to the interval [a, b] of L.

90

6. Boolean Triples

(ii) Fa,b is a filter of M3 [L, a, b] and Fa,b is isomorphic to L. We can say a lot more about Fa,b . But first we need another lemma. Lemma 6.17. Let L be a lattice, let [u, v] and [u′ , v ′ ] be intervals of L, and let δ : [u, v] 7→ [u′ , v ′ ] be an isomorphism between these two intervals. Let δ and δ −1 be term functions in L. Then L is a congruence-preserving extension of [u, v] iff it is a congruence-preserving extension of [u′ , v ′ ]. Proof. Let us assume that L is a congruence-preserving extension of [u′ , v ′ ]. Let α be a congruence relation of [u, v] and let δα be the image of α under δ. Since δ is an isomorphism, it follows that δα is a congruence of [u′ , v ′ ], and so δα has a unique extension to a congruence γ of L. We claim that γ extends α to L and extends it uniquely. Step 1. γ extends α. Let x ≡ y (mod α). Then δx ≡ δy (mod δα), since δ is an isomorphism. By definition, δ extends δα, so δx ≡ δy (mod α). Since δ −1 is a term function, the last congruence implies that x ≡ y (mod α). Conversely, let x ≡ y (mod γ) and x, y ∈ [u, v]. Then δx, δy ∈ [u′ , v ′ ]; since γ is a term function, it follows that δx ≡ δy (mod γ). Since γ extends δα, we conclude that δx ≡ δy (mod δα). Using that δ −1 is an isomorphism, we obtain that x ≡ y (mod α), verifying the claim. Step 2. γ extends α uniquely. Let γ extend α to L. As in the previous paragraph—mutatis mutandis—we conclude that γ extends δα to L; hence, α = γ, proving the uniqueness and the claim. By symmetry, the lemma is proved. We have already observed in Lemma 6.12 that M3 [L, a, b] is a congruencepreserving convex extension of [(0, a, 0), (1, a, a)]. Since the isomorphism x 7→ x ∨ (0, b, 0) between [(0, a, 0), (1, a, a)] and [(0, b, 0), (1, b, b)] = Fa,b is a term function, and so is the inverse x 7→ x ∧ (1, a, a), from Lemma 6.17 we conclude the following. Corollary 6.18. The lattice M3 [L, a, b] is a congruence-preserving convex extension of the filter Fa,b . We summarize our results (my joint paper with E. T. Schmidt [190]). Lemma 6.19. Let L be a bounded lattice, and let a, b ∈ L with a < b. Then there exists a bounded lattice La,b (with bounds 0a,b and 1a,b ) and ua,b , va,b ∈ La,b , such that the following conditions are satisfied :

6.4. Two interesting intervals

91

(i) va,b is a complement of ua,b . (ii) Fa,b = [va,b , 1a,b ] ∼ = L. (iii) Ia,b = [0a,b , va,b ] ∼ = [a, b]. (iv) La,b is a congruence-preserving (convex ) extension of [0a,b , ua,b ] and of [va,b , 1a,b ]. (v) The congruences on Ia,b and Fa,b are synchronized, that is, if α is a congruence on L, α is the extension of α to La,b (we map α to Fa,b under the isomorphism, and then by (iv) we uniquely extend it to La,b ), and x, y ∈ [a, b], then we can denote by xFa,b , yFa,b ∈ Fa,b the images of x, y in Fa,b and by xIa,b , yIa,b ∈ Ia,b the images of x, y in Ia,b ; synchronization means that xFa,b ≡ yFa,b (mod α) is equivalent to xIa,b ≡ yIa,b

(mod α).

Proof. Of course, La,b = M3 [L, a, b], 0a,b = (0, a, 0), 1a,b = (1, b, b), ua,b = (1, a, a), and va,b = (0, b, 0). The lattice La,b is illustrated with L = C5 in Figure 6.5, the five-element chain, a is the atom and b is the dual atom of S. This figure is the same as Figure 6.4, only the notation is changed. 1a,b

Fa,b ua,b

va,b

Ia,b

0a,b Figure 6.5: The lattice La,b

92

6.5.

6. Boolean Triples

Discussion

The main result of this chapter is Theorem 6.3: M3 [L] is a congruencepreserving extension of L. We can ask if there is an analogous result for interesting classes of lattices. So let X be one of the following list of classes of lattices (add any other class of interest): (i) the class M of modular lattices, (ii) the class of sectionally complemented lattices, (iii) the class of uniform lattices, (iv) the class of isoform lattices,indexIsoform lattice (v) the class of regular lattices. Problem 2. Let L be a lattice in the class X. Is it true that L has a proper congruence-preserving extension in X?

Chapter

7

Cubic Extensions

In this chapter, for a finite lattice K, we introduce an extension Cube K with the following properties: (i) The lattice K is a congruence-reflecting sublattice of Cube K. (ii) Con(Cube K) is Boolean. (iii) The minimal extension of the meet-irreducible congruences are the dual atoms of Con(Cube K); their ordering is “flattened.”

7.1.

The construction

Let K be a finite lattice. Following my joint paper with E. T. Schmidt [184], for every meet-irreducible congruence γ of K (in symbols, γ ∈ ConM K), we form the quotient lattice K/γ, and extend it to a finite simple lattice Simp Kγ with zero 0γ and unit 1γ , using Lemma 2.3. Let CubeSimp K be the direct product of the lattices Simp Kγ for γ ∈ ConM K: Y CubeSimp K = (Simp Kγ | γ ∈ ConM K). If the function Simp is understood, we write Cube K for CubeSimp K. We regard Simp Kγ , for γ ∈ ConM K, an ideal of Cube K, as in Section 2.2. For a ∈ K, define Diag(a) ∈ Cube K as follows. Diag(a) = (a/γ | γ ∈ ConM K). The lattice K has a natural (diagonal) embedding into Cube K by γ : a 7→ Diag(a)

for a ∈ K.

93 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_7

94

7. Cubic Extensions

1

1

γ

c d

c

a

N6

b

d

a

b

0

f

Simp N6

0

Figure 7.1: The lattices N6 and Simp N6 Let Diag(K) be the sublattice { Diag(a) | a ∈ K } of Cube K, and for a congruence α of K, let Diag(α) denote the corresponding congruence of Diag(K). By identifying a with Diag(a), for a ∈ K, we can view Cube K as an extension of K; we call Cube K a cubic extension of K. Cubic extensions are hard to draw because they are direct products. Here is a small example. We take the lattice N6 (see Figure 7.1). The lattice N6 is subdirectly irreducible; it has two meet-irreducible congruences, 0 and γ. Since N6 /γ = C2 , it is simple, so we can choose Simp N6 /γ = C2 . The other lattice quotient is N6 /0 = N6 and for this we choose a simple extension Simp N6 , by adding an element; (see Figure 7.1). The lattice Cube N6 is shown in Figure 7.2 along with the embedding Diag of N6 into Cube N6 . The images of elements of N6 under Diag are black-filled. As an alternative, you may draw the chopped lattice M whose ideal lattice is Cube N6 (see Figure 7.3). Unfortunately, the embedding Diag is not easy to see with this representation. Diag(1)

F D(d)

B

D(b) = Diag(b) D(c) = Diag(c) D(d) = Diag(d)

D(c) D(a)

D(b)

Diag(0) Figure 7.2: The embedding of N6 into Cube N6

7.2. The basic property

95

Figure 7.3: The chopped lattice M whose ideals represent Cube N6

7.2.

The basic property

The following crucial property of cubic extensions was stated in my joint paper with E. T. Schmidt [184]. Theorem 7.1. Let K be a finite lattice and let Cube K be a cubic extension of K. Then K (= Diag(K)) is a congruence-reflecting extension in Cube K. Proof. For κ ∈ Con K and γ ∈ ConM K, define the congruence Cube(κ, γ) on the lattice Simp Kγ as follows. ( 0 if κ ≤ γ, Cube(κ, γ) = 1 if κ ≰ γ; and define Cube(κ) =

Y

( Cube(κ, γ) | γ ∈ ConM K ),

a congruence of the lattice Cube K. Observe that Cube(0K ) = 0Cube K and Cube(1K ) = 1Cube K , that is, Cube(0) = 0 and Cube(1) = 1. We show that Diag(κ) is the restriction of Cube(κ) to K. First, assume that Diag(u) ≡ Diag(v) (mod Diag(κ)) in Diag(K). Then u ≡ v (mod κ) in K, by the definition of D. The congruence u ≡ v (mod κ) implies that u ≡ v (mod γ), for all γ ∈ ConM K satisfying κ ≤ γ, that is, u/γ = v/γ, for all γ ∈ ConM K satisfying κ ≤ γ. This, in turn, can be written as u/γ ≡ v/γ

(mod Cube(κ, γ)), for all γ ∈ ConM K satisfying κ ≤ γ,

since, by definition, κγ = κ for κ ≤ γ. Again, by definition, κγ = 1 for κ ≰ γ. Therefore, the congruence u/γ ≡ v/γ (mod Cube(κ, γ)) always holds. We conclude that u/γ ≡ v/γ (mod Cube(κ, γ)) for all γ ∈ ConM K. This congruence is equivalent to Diag(u) ≡ Diag(v) (mod Cube(κ)) in Cube K, which was to be proved.

96

7. Cubic Extensions

Second, assume that Diag(u) ≡ Diag(v) (mod Cube(κ)) in Cube K. Then u/γ ≡ v/γ (mod Cube(κ, γ)), for all γ ∈ ConM K; in particular, for all γ ∈ ConM K satisfying κ ≤ γ. Thus u/γ = v/γ, for all γ ∈ ConM K satisfying κ ≤ γ, that is, u ≡ v (mod γ) for all γ ∈ ConM K satisfying κ ≤ γ. Therefore, ^ u ≡ v (mod ( γ ∈ ConM K | κ ≤ γ )). The lattice Con K is finite, so every congruence is a meet of meet-irreducible congruences; therefore, ^ κ = ( γ ∈ ConM K | κ ≤ γ ),

and so u ≡ v (mod κ) in K, that is, Diag(u) ≡ Diag(v) (mod Diag(κ)) in Diag(K). For κ ∈ Con K, the set ∆κ = { γ ∈ ConM K | κ ≰ γ }

is a down set of ConM K, and every down set of ConM K is of the form ∆κ , for a unique κ ∈ Con K. The down set ∆κ of ConM K describes Cube(κ), and conversely. In the example of Section 7.1 (see Figures 7.1–7.3), the congruence lattice Con N6 = {0, γ, 1}. Since Cube(0) = 0 and Cube(1) = 1, we only have to compute Cube(γ). Clearly, Cube(γ) = 0 × 1 (that is, Cube(γ) = 0C2 × 1Simp N6 ); it splits Cube N6 into two parts as shown by the dashed line in Figure 7.2. We now summarize the properties of the lattice Cube K. Theorem 7.2. Let K be a finite lattice with a cubic extension Cube K. Then (i) Cube K is finite. (ii) Choose an atom sγ ∈ Simp Kγ , for each γ ∈ ConM K, and let B be the Boolean ideal of Cube K generated by these atoms. Then B is a congruence-determining ideal of Cube K. (iii) There are one-to-one correspondences among the subsets of ConM K, the sets of atoms of B, and the congruences κ of Cube K; the subset of ConM K corresponding to the congruence κ of Cube K is δ κ = { γ ∈ ConM K | κ ≰ γ }, which, in turn corresponds to the set Sκ = { sγ | sγ ≡ 0 (mod κ) } of atoms of B. Hence, the congruence lattice of Cube K is a finite Boolean lattice.

7.2. The basic property

97

(iv) Every congruence κ of K has an extension Cube(κ) to a congruence of Cube K corresponding to the down set ∆κ of ConM K. In the example of Section 7.1, for B we can choose the shaded ideal of Figure 7.2. Corollary 7.3. Choose a dual atom tγ ∈ Simp Kγ , for each γ ∈ ConM K, and define tγ as the element of Cube K whose γ-component is tγ and all other components are 1. The element tγ is a dual atom of Cube K. Let F be the Boolean filter of Cube K generated by these dual atoms. Then F is a congruence-determining filter of Cube K. There are one-to-one correspondences among the subsets of ConM K, the sets of dual atoms of F , and the congruences κ of Cube K; the subset of ConM K corresponding to the congruence κ of Cube K is Dκ = { γ ∈ ConM K | κ ≰ γ }, which, in turn, corresponds to the set Tκ = { sγ | sγ ≡ 1 (mod 0) } of dual atoms of F . Also, |F | = |B|. In the example of Section 7.1, we can choose for F the shaded filter of Figure 7.2. So a cubic extension Cube K of K (i) has a “cubic” congruence lattice (the Boolean lattice Bn , an “n-dimensional cube”); (ii) K and its cubic extension, Cube K, have the same number of meetirreducible congruences; (iii) the congruences κ of K extend to Cube K. As a rule, the cubic extension has many more congruences than the Cube(κ)-s. There are other small examples in Sections 14.2 and 15.2, in particular, in Figures 14.3 and 15.2.

Part III

RTs

99

Chapter

8

Sectionally Complemented RT

In Parts III–V, we discuss a subfield of Lattice Theory that started with the following results. Theorem 8.1 (Basic RT). Every finite distributive lattice D can be represented as the congruence lattice of a finite lattice L. Theorem 8.2 (Sectionally Complemented RT). Every finite distributive lattice D can be represented as the congruence lattice of a finite sectionally complemented lattice L. In this chapter, we follow my joint paper with H. Lakser [139] (published in my book GLT-[99]), and prove these results based on the discussion of chopped lattices in Chapter 5, a simpler proof than the one in my joint paper with E. T. Schmidt [172].

8.1.

The Basic RT

The Basic RT is an unpublished result of R. P. Dilworth from around 1944. Our presentation is based on in my joint paper with E. T. Schmidt [172], where the first proof appeared. In his book (P. Crawley and R. P. Dilworth [24]), Dilworth reproduces the proof from [172]. It is clear from his recollections in Bogart et al.[22] that our thinking was very close to his. By Theorem 5.6, to prove the Basic RT, it is sufficient to verify the following. Theorem 8.3. Let D be a finite distributive lattice. Then there exists a chopped lattice M such that Con M is isomorphic to D. 101 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_8

102

8. Sectionally Complemented RT

Using the equivalence of nontrivial finite distributive lattices and finite ordered sets (see Section 2.5.2) and using the notation J(Con(M)) (see Section 3.2) for the ordered set of join-irreducible congruences, we can rephrase Theorem 8.3 as follows. Theorem 8.4. Let P be a finite order. Then there exists a chopped lattice M such that J(Con(M)) is isomorphic to P . We are going to prove the Basic RT in this form.

8.2.

Proof-by-Picture

The basic gadget for the construction is the lattice N6 = N (p, q) of Figure 8.1. The lattice N (p, q) has three congruence relations, namely, 0, 1, and α, where α is the congruence relation with congruence classes {0, q1 , q2 , q} and {p1 , p(q)}, indicated by the dashed line. Thus con(p1 , 0) = 1. In other words, p1 ≡ 0 “implies” that q1 ≡ 0, but q1 ≡ 0 “does not imply” that p1 ≡ 0. We will use the “gadget” N6 = N (p, q) to achieve such congruence-forcing. To convey the idea of the proof of Theorem 8.4, we present three small examples in which we construct the chopped lattice M from copies of N (p, q). Example 1. The three-element chain C = C3 . Let P = {a, b, c} with c ≺ b ≺ a. We take two copies of the gadget, N (a, b) and N (b, c); they share the ideal I = {0, b1 } (see Figure 8.2). So we can merge them (in the sense of Section 5.1) and form the chopped lattice M = Merge(N (a, b), N (b, c)) as shown in Figure 8.2. The congruences of M are easy to find. The isomorphism P ∼ = J(Con(M)) is given by x1 7→ con(0, x) for x ∈ P . The definition of x1 is illustrated in Figure 8.1. p(q) q p1

q2

q1 0

Figure 8.1: The lattice N6 = N (p, q) and the congruence α

103

8.2. Proof-by-Picture

The congruences of M can be described by a compatible congruence vector (αa,b , αb,c ) (see Section 5.3), where αa,b is a congruence of the lattice N (a, b) and αb,c is a congruence of the lattice N (b, c), subject to the condition that αa,b and αb,c agree on I. Looking at Figure 8.1, we see that if the shared congruence on I is 0 (= 0I ), then we must have αa,b = 0 (= 0N (a,b) ) and αb,c = 0 (= 0N (b,c) ) or αb,c = α on N (b, c). If the shared congruence on I is 1 (= 1I ), then we must have αa,b = α or αa,b = 1 (= 1N (a,b) ) on N (a, b) and αb,c = 1 (= 1N (b,c) ) on N (b, c). So there are three congruences distinct from 0, namely, (0, α), (α, 1), (1, 1). Therefore, the join-irreducible congruences form the three-element chain. Example 2. The three-element ordered set V of Figure 8.3, the “ordered set V”. We take two copies of the gadget, N (b, a) and N (c, a); they share the ideal J = {0, a1 , a2 , a}. We merge them, to form the chopped lattice MV = Merge(N (b, a), N (c, a)) (see Figure 8.3). Again, the isomorphism V ∼ = J(Con(M)) MV is given by x1 7→ con(0, x) for x ∈ V . Example 3. The three-element ordered set H of Figure 8.4, the “ordered set hat.” We take two copies of the gadget, N (a, b) and N (a, c); they share b(c)

a(b) b

N (a, b)

N (b, c)

b2

a1

c1

b1 b1 I

c2

I

0

0

a(b) a

P

b

N (a, b) a1

b2

c

b(c) b

N (b, c) b1

c

c2

c1

I c

0 Figure 8.2: The chopped lattice M for C = C3

M

104

8. Sectionally Complemented RT

the ideal J = {0, a}; we merge them to form the chopped lattice MV = Merge(N (a, b), N (a, c)) ∼ J(Con(M)) MH is given by (see Figure 8.4). Again, the isomorphism H = x1 7→ con(0, x) for x ∈ V . The reader should now be able to picture the general proof: instead of the few atoms in these examples, we start with enough atoms to reflect the structure of P (see Figure 8.5). Whenever b ≺ a in P , we insert a copy of N (a, b) (see Figure 8.6).

8.3.

Computing

For a finite ordered set P , let Max be the set of maximal elements in P . We form the set [ M0 = {0} ∪ { p1 | p ∈ Max } ∪ ( {a1 , a2 } | a ∈ P − Max ) b(a)

c(a) a

N (b, a) b

c

PV

a1

b1

a

N (c, a) c1

a2

MV

0

Figure 8.3: The chopped lattice for the ordered set V

a(b)

a(c) c

b N (a, b)

PH b

b1

a c

N (a, c)

b2

a

c1

c2

MH 0

Figure 8.4: The chopped lattice for the ordered set hat

8.3. Computing

105

consisting of 0, the maximal elements of P indexed by 1, and two copies of the nonmaximal elements of P , indexed by 1 and 2. We make M0 a meetsemilattice by defining inf{x, y} = 0 if x ̸= y, as illustrated in Figure 8.5. Note that x ≡ y (mod α) and x ̸= y imply that x ≡ 0 (mod α) and y ≡ 0 (mod α) in M0 ; therefore, the congruence relations of M0 are in one-to-one correspondence with subsets of P . Thus Con M0 is a Boolean lattice whose atoms are associated with atoms of M0 ; the congruence γ x associated with the atom x has only one nontrivial block {0, x}. We construct an extension M of M0 as follows. The chopped lattice M consists of four kinds of elements: (i) the zero, 0; (ii) for all maximal elements p of P , the element p1 ; (iii) for any nonmaximal element p of P , three elements: p, p1 , p2 ; (iv) for each pair p, q ∈ P with p ≻ q, a new element, p(q). For p, q ∈ P with p ≻ q, we set N6 = N (p, q) = {0, p1 , q, q1 , q2 , p(q)}. For x, y ∈ M , let us define x ≤ y to mean that either x = y, or for some p, q ∈ P with p ≻ q, we have x, y ∈ N (p, q) and x ≤ y in the lattice N (p, q). It is easily seen that x ≤ y does not depend on the choice of p and q, and that ≤ is an ordering. Under this ordering, all N (p, q) and N (p, q) ∩ N (p′ , q ′ ) ...

q1

p1

a1

a2

b1

b2

...

0 Figure 8.5: The chopped lattice M0

a(b)

c(d) d ...

c1

b

a

d2

d1 a2

a1

b2

0 Figure 8.6: The chopped lattice M

b1

...

106

8. Sectionally Complemented RT

(p ≻ q and p′ ≻ q ′ in P ) are lattices. Also x, y ∈ M , x ∈ N (p, q), and y ≤ x imply that y ∈ N (p, q). So we conclude that M is a chopped lattice. In fact, it is a union of the ideals N (p, q) with p ≻ q in P , and two such distinct ideals intersect in a one-, two-, or four-element ideal. Since the chopped lattice M is atomistic, Corollary 3.10 applies. If pi ⇒ qj in M , for p, q ∈ P and i, j ∈ {1, 2}, then p ≥ q in P , and conversely. So the equivalence classes of the atoms under the quasiordering ⇒ form an ordered set isomorphic to Dn P . This completes the verification that J(Con(M)) M ∼ = P, and therefore, of Theorem 8.4.

8.4.

Sectionally complemented lattices

Let M be the chopped lattice described in Section 8.3. Since all N (p, q)-s are sectionally complemented, so is M . The following result is proved in my joint paper with E. T. Schmidt [172]. Theorem 8.5. L = Id M is sectionally complemented. Combined with Theorem 8.4, this yields the Sectionally Complemented RT Theorem 8.1, the main result of this chapter. To prove Theorem 8.5, we will view L = Id M as a closure system. For an ideal U of M , let Atom(U ) be the set of atoms of M in U . Note that the atoms of M are the {pi }, where p ∈ P and i ∈ {1, 2}. We start with the following two trivial statements (to simplify the notation, we compute with the indices modulo 2). Lemma 8.6. For a set A ⊆ Atom(M ), there is an ideal U with Atom(U ) = A iff A satisfies the condition (Clo)

For p ≻ q in P , if p1 , qi ∈ A, then qi+1 ∈ A.

Let us call a subset A of Atom(M ) closed if it satisfies (Cl). It is obvious that every subset A of Atom(M ) has a closure A. Lemma 8.7. The assignment I 7→ Atom(I) is a bijection between the ideals of M and closed subsets of Atom(M ), and Atom(I ∧ J) = Atom(I) ∩ Atom(J), Atom(I ∨ J) = Atom(I) ∪ Atom(J)

for I, J ∈ Id M . The inverse map assigns to a closed set X of atoms, the ideal id(X) of M generated by X. Lemma 8.7 allows us to regard L as the lattice of closed sets in Atom(M ), so I ∈ L will mean that I is a closed subset of Atom(M ). Thus I ∧ J = I ∩ J and I ∨ J = I ∪ J for I, J ∈ L.

8.4. Sectionally complemented lattices

107

Let I ⊆ J ∈ L. Let us say, that q ∈ P splits over (I, J) if there exists a covering pair p ≻ q in P , with p1 , qi ∈ J − I and qi+1 ∈ I. If there is a q ∈ P that splits over (I, J), then J − I is not closed. Let X = X(I, J) be the set of all elements qi in J − I such that q splits over (I, J). Let S(I, J) = (J − I) − X, that is, S(I, J) is the set of all elements qi in J − I such that q does not split over (I, J). Lemma 8.8. S(I, J) ∈ L. Proof. We have to prove that S = S(I, J) is closed. Let u ≻ v in P, u1 ∈ S, and vi ∈ S. Since, by the definition of S, the / I. element v does not split over (I, J) and u1 , vi ∈ J − I, it follows that vi+1 ∈ Since u1 ∈ J and vi ∈ J and J is closed, we obtain that vi+1 ∈ J. Thus vi+1 ∈ J − I. Since v does not split over (I, J), we get that vi+1 ∈ S by the definition of S. Thus S is closed. We claim that S = S(I, J) is the sectional complement of I in J. By Lemma 8.7, we have to prove that (1) (2)

I ∩ S = ∅, I ∪ S = J.

(1) is obvious from the definition of S. Since I ⊆ J and S ⊆ J, to verify (2), it is sufficient to show that (3)

I ∪ S ⊇ J.

Assume, to the contrary, that there is a q ∈ P and i ∈ {1, 2} such that (4)

qi ∈ J − I ∪ S.

We can choose q so that it is maximal with respect to this property, that is, if p > q and pj ∈ J for some j ∈ {1, 2}, then pj ∈ I ∪ S. It follows from (4) that qi ∈ J − (I ∪ S) = X. So, by the definition of X, there is p ≻ q in P with p1 ∈ J − I and qi+1 ∈ I. Since p1 ∈ J and p ≻ q, by the maximality of q, we have that p1 ∈ I ∪ S. Also, qi+1 ∈ I ⊆ I ∪ S. So qi ∈ I ∪ S by the definition of closure, contradicting (4). This completes the proof of the claim and of Theorem 8.5. The lattice L we construct for the Basic RT (Theorem 8.5) has a very interesting congruence structure: every join-irreducible congruence γ is of the form con(0, pγ ), where pγ is an atom and the pγ -s generate a Boolean ideal. We state this formally. Theorem 8.9 (Full Basic RT). Every finite distributive lattice D can be represented as the congruence lattice of a finite sectionally complemented lattice L with the following property.

108

8. Sectionally Complemented RT

(B) There is a Boolean ideal B in L such that there is a bijection γ 7→ pγ between J(Con L) and the atoms of B satisfying γ = con(0, pγ ). There is a bijection between the congruences of L and “down sets” of the atoms of B, that is, sets H of atoms satisfying the condition: pγ ∈ H and δ ≤ γ imply that pδ ∈ H for γ, δ ∈ J(Con L). Proof. Let P = J(D) and take the sublattice B generated by { p1 | p ∈ P }. It is easy to see that B is an ideal. The statement about the congruences is implicit in the discussion of Section 8.3. The ideal B is, of course, a congruence-determining sublattice of L.

8.5.

The N-relation

The approach to congruences taken in this section evolved from techniques developed over a period of time in the papers R. N. McKenzie [223], R. Freese [83], R. Freese and J. B. Nation [87], and B. J´onsson and J. B. Nation [218]. The closure relation of Section 8.4 has been generalized to arbitrary finite lattices. With the permission of J. B. Nation, this section is based on Chapter 10 of his book [228] see also R. Freese, J. Jeˇzek, and J. B. Nation [86]). We start with some new concepts. Refinements For a finite lattice L and sets X, Y ⊆ J(L), we say that X refines Y , written as X ≪ Y , if for any element x ∈ X, there exists an element y ∈ Y with x ≤ y. The relation ≪ is a quasiorder on the set Pow J(L) of all subsets of J(L). It is not antisymmetric, for instance, for a < b ∈ L, we have {a, b} ≪ {b} ≪ {a, b} but {a, b} = ̸ {b}. We make a few observations. (i) {a} ≪ {b} iff a ≤ b; (ii) X ≪ {a} iff a is an upper bound for X; (iii) X ⊆ Y implies that X ≪ Y ; (iv) X ≪ {a, b} iff X = Xa ∪ Xb and Xa ≪ {a}, Xb ≪ {b}. Join expressions and join covers W A join expression W a = B relates an element a ∈ L and a set B ⊆ L. A join expression a = B is minimal if it is irredundant W and B cannot be properly refined, that is, if b ∈ B and c < b ∈ L, then c ∨ W W (B − {b}) < a. Equivalently, a = B is minimal, if B ⊆ C whenever a = C and C ≪ B. A join cover W W of p ∈ L is a set A such that p ≤ A. A join cover A of p is minimal if A is irredundantWand A cannot be properly refined to another join cover of p, that is, if p ≤ B and B ≪ A, then A ⊆ B.

8.5. The N-relation

109

W Most of the time, we deal with join covers a =W A with A ⊆ J(L). In this case, minimality takes on the simple form b∗ ∨ (A − {b}) < a for all b ∈ A. For an example, take the lattice L = N6 , as in Figure 8.1. Since L is atomistic, J(L) is the set of atoms, so b∗ = 0 for all b ∈ A. It follows that the join expression p(q) = p1 ∨ q1 is minimal. The relation N We define a binary relation N on J(L) as follows:1 p N q if there exists x ∈ L such that p ≤ q ∨ x but p ≰ q∗ ∨ x. In the example of L = N6 , as in Figure 8.1, q1 N q2 and q2 N q1 (using x = p1 for both). We use the following notation for u ∈ L: (5) W

φu = { x ∈ J(L) | x ≤ u } = J(L) ∩ id(u).

Note that u = φu. The following lemma summarizes some properties of finite lattices and the relation N. Lemma 8.10. Let L be a finite lattice. (i) If b ≰ a in L, then there exists an element p ∈ J(L) with p ≤ b and p ≰ a. (ii) Every join expression in L refines to a minimal join expression, and every join cover refines to a minimal join cover. (iii) For p, q ∈ J(L), the relation p N q holds iff q ∈ A for some minimal join cover A of p. W W Proof. (i) Since b = φb, if every p ∈ φb satisfies that p ≤ a, then b = φb ≤ a, a contradiction. (ii) W Let us assume that L contains an element s with a join representation s = F that does not refine to a minimal one. So there is W an element t ≤ s minimal with respect to having a join representation t = A that fails to refine to a minimal one. Clearly, W t is join reducible, and there is a proper, irredundant join expression t = B with B ≪ A. Let B = {b1 , . . . , bk }. We can find c1 ≤ b1 such that t = c1 ∨ b2 ∨ · · · ∨ bk , but c1 cannot be replaced by any lower element: t > u ∨ b2 ∨ · · · ∨ bk whenever u < c1 . Now apply the same argument to b2 and {c1 , b2 , . . . , bk }. After k such steps we obtain a join cover C that refines B and is minimal pointwise: no element can be replaced by a lower element. Every element of C is strictly below t, and hence, has a minimal join expression. Choose a minimal join expression Ec for all c ∈ C. Clearly, 1 In the literature, various notations are used for this relation; we use N for Nation. The relation N we define here is reflexive, whereas the relation D defined in the literature is irreflexive.

110

8. Sectionally Complemented RT

W E = ( Ec | c ∈ C ) is a minimal join expression for t, and E ≪ C ≪ B ≪ A, which contradicts the choice of t and B. W Now let u ∈ L and let A be a join W cover of u, that is, u ≤ A. We can find a subset B ⊆ A such that u ≤ B is irredundant. Again, we refine B to a pointwise minimal join cover C.WWe know that minimal join expressions exist, so we may define again E = ( Ec | c ∈ C ). This is a minimal join cover of u, and E ≪ C ≪ B ≪ A. (iii) Let us assume that p N q, and let x ∈ L be such that p ≤ q ∨ x but p ≰ q∗ ∨ x. By (ii), we can find a minimal join cover A of p with A ≪ {q, x}. So q ∈ A, since p ≰ q∗ ∨ x. Conversely,Wif A is a minimal join cover of p, and q ∈ A, then p N q is verified with x = (A − {q}). A closure operator

Next we define a closure operator on sets of join-irreducible elements of a finite lattice L. For S ⊆ J(L), let _ S = { p ∈ J(L) | p ≤ F for some F ⊆ S }.

The closed sets form a lattice Cl(L). Compare this with the definition of closure in Section 8.4, where we deal with a very special lattice and a particularly nice relation N. To generalize that closure definition, we have to utilize the tools developed in this section. Theorem 8.11. Let L be a finite lattice. Then the map φ (see formula (5)), is an isomorphism of L onto the lattice Cl(L) of closed subsets of J(L). W Proof. If x = A is a minimal join expression, then φx = A, so φx is closed. The map φ is clearly W isotone. It is one-to-one by Lemma 8.10(i). It is also onto, because F = φ ( (F )) for F ⊆ J(L). To use this result, we need a characterization of closed sets.

Theorem 8.12. Let L be a finite lattice. A subset C of J(L) is closed iff the following two properties hold : (i) C is a down set of J(L), (ii) if A is a minimal join cover of p ∈ J(L) and A ⊆ C, then p ∈ C. Proof. It is easy to see that closed sets have both properties. Conversely, let C ⊆ J(L) W satisfy these properties. We want to show that C ⊆ C. If p ∈ C, then p ≤ F for some F ⊆ C. By Lemma 8.10(ii), there is a minimal join cover A of p refining F ; since C is an down set, it follows that A ⊆ C. Then (ii) yields that p ∈ C, as required. So for finite lattices, the closure is determined by the order on J(L) and the minimal join covers of elements of J(L). By Theorem 8.11, so is L.

8.5. The N-relation

111

Theorem 8.13. Let L be a finite lattice. Let σ map Con L to the lattice Pow J(L) of all subsets of J(L) by σα = { p ∈ J(L) | p α p∗ }. Then σ is an embedding. Proof. Clearly, σ is isotone. To see that σ is one-to-one, assume that α ≰ β ∈ Con L. Then there exists a pair of elements a < b ∈ L with a ≡ b (mod α) but a ̸≡ b (mod β). Clearly, x ̸≡ b (mod β) for any x ≤ a. Let p ≤ b be minimal with respect to the property: p ≡ x (mod β) implies that x ≰ a. We claim that p is join-irreducible. Let y1 , . . . , yn ∈ J(P ) and y1 , . . . , yn < p. Then there exist x1 , . . . , xn ∈ J(P ) with x1 , . . . , xn < p such that yi ≡ xi (mod β) for i ≤ n. Hence, _ _ ( yi | i ≤ n ) ≡ ( xi | i ≤ n ) (mod β), _ _ ( yi | i ≤ n ), ( xi | i ≤ n ) ≤ a, W so ( xi | i ≤ n ) < p. Since p = p ∨ b, p ∨ b ≡ p ∨ a (mod β), and p ∨ a ≤ p∗ , it follows that p ≡ p∗ (mod β), that is, p ∈ σα. But p ̸≡ p∗ (mod β), because p∗ ≡ x (mod β) and x ≤ a for some x; thus p ∈ / σβ. Therefore σα ⊈ σβ. It is easy to see that ^ \ σ ( αi | i ≤ n ) = ( σαi | i ≤ n ) for the congruences αi , i ≤ n. Since σ is isotone, we obtain that [ _ ( σαi | i ≤ n ) ⊆ σ ( αi | i ≤ n ).

It remains to show the reverse inclusion. To verify this, let _ p ≡ p∗ (mod ( αi | i ≤ n )).

By Theorem 1.3, there exist

p = x 0 , x 1 , . . . , x k = p∗ ∈ L

αi1 , . . . , αk1 ∈ Con L

such that xi ≡ xi (mod αij ). Let yj = (xj ∨ p∗ ) ∨ p. Then y0 = p, yk = p∗ , and p∗ ≤ yj ≤ p imply that yj ∈ {p∗ , p} for all j ≤ k. Moreover, jj−1 ≡ yj (mod αj ) for j ≤ k. So there must exist a j ≤ k with yj−1W= p and yj = p∗ . We conclude that p ≡ p∗ (mod αij ) and so p ∈ σαij ⊆ ( σαij | j ≤ k ), that is, σ preserves arbitrary joins.

112

8. Sectionally Complemented RT

Next we need to identify the range of σ. Theorem 8.14. Let L be a finite lattice, and let S ⊆ J(L). Then S = σα for some α ∈ Con L iff p N q and q ∈ S imply that p ∈ S. Proof. Let S = σα. If q ∈ S and p N q, then q ≡ q∗ (mod α), and p ≤ q ∨ x, but p ̸= q∗ ∨ x for some x ∈ L. Thus p = p ∨ (q ∨ x), p ∨ (q ∨ x) ≡ p ∨ (q∗ ∨ x)

(mod α),

and p ∨ (q∗ ∨ x) < p, hence, p ≡ p∗ (mod α) and p ∈ σα = S. Conversely, assume we are given S ⊆ J(L) satisfying the condition of the theorem. Then we must produce a congruence relation α such that σα = S. Let T = J(L) − S, and note that T has the property that q ∈ T whenever p N q and p ∈ T . Define x≡y

(mod α)

iff

↓ x ∩ T = ↓ y ∩ T.

Clearly, (a) α is an equivalence relation; (b) x ≡ y (mod α) implies that x ∨ z ≡ y ∨ z (mod α); (c) p ∈ J(L) and p ≡ p∗ (mod α) iff p ∈ / T , that is, p ∈ S. The last statement implies that σα = S. It remains to show that α respects joins. Assume that x ≡ y (mod α), and let z ∈ L. We want to show that (6)

↓ (x ∨ z) ∩ T ⊆ ↓ (y ∨ z) ∩ T.

So let p ∈ T and p ≤ x ∨ z. Then there exists a minimal join cover Q of p with Q ≪ {x, z}. If q ∈ Q and q ≤ z, then q ≤ y ∨ z. Otherwise, q ≤ x. 8.10(iii)), we obtain that qW∈ T . Thus Since p ∈ T and p N q (by Lemma W q ∈ ↓ x ∩ T = ↓ y ∩ T , so q ≤ Q ≤ y ∨ z. It follows that p ≤ Q ≤ y ∨ z. This shows ⊆ in (6); by symmetry, (6) holds. Hence, α respects joins. In order to interpret the consequences of these two theorems, let N denote the transitive closure of N on J(L). Then N is a quasiorder, so it induces an equivalence relation ≡ on J(L) (defined by p ≡ q iff pNq and qNp), modulo which N is a partial order. Let QN denote the ordered set (J(L)/ ≡, N), defined on the set J(L)/ ≡ and ordered by the relation N. Then we get the main result of this section. Corollary 8.15. If L is a finite lattice, then Con L ∼ = Down(QN ).

8.6. Discussion

113

On a finite lattice L, the N relation is easy to determine, so it is not hard to find QN . Hence, this result provides an efficient algorithm for determining the congruence lattice of a finite lattice. J. B. Nation comments: “To me, Theorems 8.11 and 8.12 are fundamental to lattice theory, treating finite lattices as the closed sets of a closure operator. This is analogous to regarding algebraic lattices as ideal lattices of join semilattices. In both cases, the observation changes one’s perspective. “There are some easy consequences of Corollary 8.15. It is straightforward to show that the dependence relation N is symmetric for finite modular or relatively complemented lattices. That makes QN an antichain, and their congruence lattices finite Boolean algebras. Tischendorf’s theorem (see [251] and Section 14.1), is only slightly harder, that every finite lattice has an atomistic extension with the same congruence lattice. You just remove the order from J(L), keep the minimal join covers from L, and show that while N may change (some extra join covers that were not minimal might become so), N does not.”

8.6.

Discussion

The Basic RT is the foundation of all the results in Parts III–V. Moreover, the ETs of Part III derive from the results of Parts IV via the Basic RT. For instance, the ET for semimodular lattices in Chapter 15 combines with the Basic RT to obtain the main result of Chapter 10: the semimodular RT. Of course, this RT is a relatively easy result compared to the ET. Nevertheless, an ET does not directly imply the corresponding RT. Sectionally complemented chopped lattices To verify the sectionally complemented RT, for a finite distributive lattice D, we construct a finite sectionally complemented chopped lattice M and prove that the ideal lattice of M is again sectionally complemented. Theorem 5.9 shows that it is not true in general. Problem 3. Let M be a finite sectionally complemented chopped lattice. Find reasonable sufficient conditions under which Id M is sectionally complemented. There are a number of related problems that arise. As in my joint papers [180] and [181] with E. T. Schmidt, we can generalize chopped lattices from the finite to the general case. Let M be a meet-semilattice satisfying the following condition: (Ch) sup{a, b} exists for any a, b ∈ M having a common upper bound in M .

114

8. Sectionally Complemented RT

We define in M the partial join operation ∨ as follows: a ∨ b = sup{a, b}, whenever sup{a, b} exists in M . This makes M into a partial lattice called a chopped lattice. For a finite M , this is equivalent to the definition in Section 5.1. We define ideals, congruences, and merging in the general case as we did in the finite case. Unfortunately, the fundamental result, Theorem 5.6 of my joint paper with H. Lakser [139], fails for the general case. Problem 4. Investigate generalizations of Theorem 5.6 to the infinite case. See my joint paper with E. T. Schmidt [181] for some related results. As we noteSin Section 5.2, every finite chopped lattice M decomposes into lattices: M = ( id(m) | m ∈ Max ).

Problem 5. Can chopped lattices, in general, be usefully decomposed into lattices? Could this be utilized in Problem 4 by assuming that the chopped lattices decompose into finitely many lattices or into lattices with nice properties? Congruence class sizes Spectra The basic gadget in this chapter is the lattice N6 ; Figure 8.1 shows N6 and its only nontrivial congruence α. We can associate with the congruence α the pair (4, 2) measuring the size of the two congruence classes. Which pairs (t1 , t2 ) can be substituted for (4, 2)? In other words, for which pairs of integers (t1 , t2 ) is there a finite lattice L such that (1) L is sectionally complemented; (2) L has exactly one nontrivial congruence α; (3) α has exactly two congruence classes: a prime ideal P and a prime filter Q satisfying that |P | = t1 and |Q| = t2 ? This question is answered as follows in my joint paper with E. T. Schmidt [172]. Theorem 8.16. Let (t1 , t2 ) be a pair of natural numbers. Then there is a finite lattice L with properties (1)–(3) iff (t1 , t2 ) satisfies the following three conditions: (P1 ) 2 ≤ t1 and t1 ̸= 3. (P2 ) 2 ≤ t2 and t2 ̸= 3. (P3 ) t1 > t2 .

8.6. Discussion

115

What can we say about the cardinalities of the congruence classes of a nontrivial congruence in a finite sectionally complemented lattice? Let L be a finite lattice, and let α be a congruence of L. We denote by Spec α the spectrum of α, that is, the family of cardinalities of the congruence classes of α. In my joint paper with E. T. Schmidt [172], spectra are characterized for finite sectionally complemented lattices. Theorem 8.17. Let S = (mj | j < n) be a family of natural numbers, n ≥ 1. Then there is a nontrivial finite sectionally complemented lattice L and a nontrivial congruence α of L such that S is the spectrum of α iff S satisfies the following conditions: (S1 ) 2 ≤ n and n ̸= 3. (S2 ) 2 ≤ mj and mj ̸= 3 for all j < n. Note that repetition is allowed in S. Corollary 8.18. Let S = (mj | j < n) be a family of natural numbers, n > 1. Then there is a nontrivial finite sectionally complemented lattice L with a unique nontrivial congruence α of L such that S is the spectrum of α iff S satisfies (S1 ) and (S2 ), and additionally: (S3 ) S is not constant, that is, there are j, j ′ < n satisfying that mj ̸= mj ′ . (S4 ) n ̸= 4. We know only that the congruence lattice Con L of the lattice L we construct in Theorem 8.17 has three or more elements. Can we prescribe its structure? Problem 6. Let D be a finite distributive lattice. Can we construct the lattice L of Theorem 8.17 that represents S = (mj | j < n) as the spectrum of a nontrivial congruence α so that L also satisfies D ∼ = Con L? Valuations There is a more sophisticated way of looking at spectra. Let K be a finite sectionally complemented lattice. Let us represent K in the form L/α, where L is a finite sectionally complemented lattice and α is a congruence of L. Then there is a natural map v : K → N (where N is the set of natural numbers) defined as follows. Let a ∈ K; then a is represented by a congruence class A of α, so we can define v(a) = |A|. We call v a valuation on K. The following result is from my joint paper with E. T. Schmidt [174]. Theorem 8.19. Let K be a nontrivial finite sectionally complemented lattice, and let v : K → N. Then there exists a finite sectionally complemented lattice L

116

8. Sectionally Complemented RT

and a nontrivial congruence α of L, such that there is an isomorphism g : K → L/α satisfying v(a) = |g(a)|, for all a ∈ K, iff v satisfies the following two conditions: (V1 ) v is antitone. (V2 ) 2 ≤ v(a) and v(a) ̸= 3 for all a ∈ K. Corollary 8.20. Let K be a nontrivial finite sectionally complemented lattice, and let v : K → N. Then there exists a finite sectionally complemented lattice L and a unique nontrivial congruence α of L, such that there is an isomorphism γ : K → L/α satisfying v(a) = |γa|,

for all a ∈ K,

iff v satisfies the conditions (V1 ) and (V2 ), and additionally, v satisfies the following two conditions: (V3 ) v is not a constant function. (V4 ) K is simple. We can also state Problems 6 for valuations. Problem 7. Let D be a finite distributive lattice. Can we prove Theorem 8.19 ∼ D? with the additional condition: Con K = Here is a congruence-preserving extension variant of the valuation problem. Problem 8. Let L be a finite lattice and let α be a nontrivial congruence of L with spectrum v on K = L/α. Let v ′ : K → N satisfy (V1 ) and (V2 ). If v ≤ v ′ (that is, v(a) ≤ v ′ (a), for all a ∈ K), then does there exist a finite congruence-preserving extension L′ of L such that the spectrum on L′ /α′ is v ′ ?

Chapter

9

Minimal RT

9.1.

The results

In the proof of the Basic RT, we construct, for a distributive lattice D with n ≥ 1 join-irreducible elements, a lattice L satisfying Con L ∼ = D. The size of this lattice is O(22n ). Can we do better? O(22n ) was improved to O(n3 ) in my joint paper with H. Lakser [140]— a substantial improvement from exponential to polynomial size—where it was conjectured that O(n3 ) can be improved to O(n2 ) and that O(n2 ) is best possible. Indeed, O(n2 ) is possible, as proved in G. Gr¨atzer, H. Lakser, and E. T. Schmidt [156]. Theorem 9.1. Let D be a finite distributive lattice with n ≥ 1 join-irreducible elements. Then there exists a lattice L of O(n2 ) elements with Con L ∼ = D. In fact, there is such a planar lattice L. Or, equivalently, Theorem 9.1′ . Let P be a finite ordered set with n elements. Then there exists a lattice L of O(n2 ) elements with J(Con L) ∼ = P . In fact, there is such a planar lattice L. The second part of the conjecture was verified in G. Gr¨ atzer, I. Rival, and N. Zaguia [168]. Theorem 9.2. Let α be a real number satisfying the following condition: Every distributive lattice D with n join-irreducible elements can be represented as the congruence lattice of a lattice L with O(nα ) elements. Then α ≥ 2. 117 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_9

118

9. Minimal RT

p3 p3 p2 p2 p1 p2 p1 p1

p3

P

C

Figure 9.1: P and C

9.2.

Proof-by-Picture for the minimal construction

We present now the Proof-by-Picture of the minimal construction (Theorem 9.1′ ) with the ordered set P (the hat) of Figure 9.1. We form the chain C from P , as shown by the same diagram. The chain C is of length 2|P | = 6, it is colored (in the sense of Section 3.2) with elements of P , as illustrated. We form C 2 , and color the prime intervals in the obvious way: a prime interval of C 2 is of the form [(a, c), (b, c)] or [(c, a), (c, b)], where a ≺ b in C, and we color both intervals the color of [a, b] in C. Figure 9.2 shows the two gadgets (colored lattices) we use: M3 and N5,5 . We proceed as follows: To construct L, we take C 2 with the coloring we have just introduced. If both lower edges of a covering square in C 2 have the same,

N5,5 M3 q p

p

q

p

Figure 9.2: The two gadgets for p < q

9.2. Proof-by-Picture for the minimal construction

119

then we add an element to make it a cover-preserving M3 , the first gadget. If in C 2 we have a cover-preserving C3 × C2 , where the C2 is colored by p, the C3 is colored by q twice, where p < q in P , then we add an element to make it an N5,5 , the second gadget. What do these gadgets do? If p and q are any two prime intervals of M3 , then p ⇔ q (using the notation of Section 3.2). If p and q are any two prime intervals of N5,5 of color p and q with p < q, then q ⇒ p (but p ⇒ q fails). The lattice L we obtain is shown in Figure 9.3. (The copies of N5,5 in the diagram are marked by black-filled elements.) Every prime interval of C 2 is projective to a prime interval in one of the two copies of C in C 2 . The first gadget makes sure that a prime interval in the first copy of C is projective to a prime interval of the same color in the second copy of C. The second gadget orders the prime intervals: if p is of color p and q is of color q, then p > q implies that p ⇒ q. Now Theorem 3.9 shows that the equivalence classes of prime intervals of L form an ordered set isomorphic to P , that is, J(Con L) ∼ = P. In particular, we make both copies of C congruence-determining sublattices of L. What is the size of L? The size of C 2 is (2n + 1)2 . For each color p ∈ P , we add the first gadget four times, adding 4n elements. For all p < q ∈ P , there is only one pair of adjacent prime intervals of color q in the first copy of C, and these can be paired with two prime intervals of color p in the second copy of C, yielding two more elements. Since there are at most n2 /2 such pairs, we get at most n2 new elements. Therefore, |L| ≤ (2n + 1)2 + 4n + n2 , so |L| is obviously O(n2 ). This completes the Proof-by-Picture.

p3

p3 p3

C

p3 p2

p2 p2

p2 p1

p1 p1

p1

Figure 9.3: The minimal lattice L

C

120

9.3.

9. Minimal RT

The formal construction

We now formally prove Theorem 9.1′ . Let P be a finite ordered set of n elements. Let P0 = {a0 , a1 , . . . , ak−1 } be the set of nonminimal elements of P . If k = 0, then P is an antichain, and we can choose L as a chain of length n. Henceforth, we assume that k > 0. Let C0 be a chain of length 2k. We color the prime intervals of C0 as follows: we color the lowermost two prime intervals of C0 with a0 , the next two with a1 , and so on. Thus, for each a ∈ P0 , there are in C0 two prime intervals of color a (and they are successive). So for each a ∈ P0 , there is a unique subchain ab ≺ am ≺ at such that the prime intervals [ab , am ] and [am , at ] have color a, and no other prime interval of C0 has color a. Observe that, for each a distinct from a0 , there is a c ∈ P0 with ab = ct , and, similarly, for each a distinct from ak−1 , there is a c ∈ P0 with at = cb . The elements am , however, are labeled uniquely. Let C1 be a chain of length n = |P |. We color the prime intervals of C1 by an arbitrary bijection with P . Thus, for each a ∈ P , there is in C1 exactly one prime interval of color a; we denote it by [ao , ai ]. We set L0 = C0 × C1 . We will regard C0 and C1 as sublattices of L0 in the usual manner. As discussed in Section 3.2, we can quasiorder the prime intervals of L0 by the quasiorder ⇒ and we can choose the prime intervals in C0 ∪ C1 as representatives of the equivalence classes. Hence, the order of join-irreducible congruences of L0 is an antichain of cardinality 2k + n. Note that |L0 | = (2k + 1)(n + 1). We next extend the lattice L0 to a lattice L1 . For each a ∈ P0 , we adjoin two new elements m0 (a) and m1 (a) to L0 ; we set (ab , ao ) ≺ m0 (a) ≺ (am , ai ),

(am , ao ) ≺ m1 (a) ≺ (at , ai ).

The resulting ordered set L1 is a lattice, and, for each a ∈ P0 , the intervals [(ab , ao ), (am , ai )] = {(ab , ao ), (am , ao ), m0 (a), (ab , ai ), (am , ai )}, [(am , ao ), (at , ai )] = {(am , ao ), (at , ao ), m1 (a), (am , ai ), (at , ai )}

are isomorphic to M3 . By adjoining these elements, we have made congruence equivalent any two prime intervals of L1 in C0 ∪ C1 of the same color. Therefore, the ordered set of join-irreducible congruences of L1 is isomorphic to the antichain P . Note that both C0 and C1 are congruence-preserving sublattices of L1 .

9.4. Proof-by-Picture for minimality

121

Observe that |L1 | = (2k + 1)(n + 1) + 2k. We finally further extend L1 so as to induce the correct order on the join-irreducible congruences. For each pair a ≻ c in P (whereby a ∈ P0 , necessarily), we add a new element n(a, c) to L1 , setting (am , co ) ≺ n(a, c) ≺ (am , ci ). The resulting ordered set L is a lattice. In L, the interval [am , at ] is projective with [n(a, c), (am , ci )], which in turn is projective with [co , ci ]. So the ordered set of join-irreducible congruences of L is isomorphic to P . Note that L is a planar lattice, and that in going from L1 to L, we adjoin no more than kn elements. Thus |L| ≤ (2k + 1)(n + 1) + 2k + kn < 3(n + 1)2 .

9.4.

Proof-by-Picture for minimality

We present a Proof-by-Picture of Theorem 9.2 by contradiction. Let α < 2 be a real number such that every order P with n elements can be represented as J(Con L), for a lattice L with O(nα ) elements, that is, |L| ≤ Cnα for some constant C. Let p = [a, b] and q = [c, d] be prime intervals of L such that con(p) ≻ con(q) in J(Con L). Then by Lemma 3.7, p ⇒ q, for instance, up

up

dn

dn

p = [a, b] ∼ [u1 , v1 ] ↠ [u2 , v2 ] ↠ [u3 , v3 ] ∼ q = [c, d], as illustrated in Figure 9.4. Note the elements hi ∈ [ui , vi ], i = 1, 2, 3, satisfying dn

up

dn

u1 , h1 ] ∼ [u2 , v2 ], [h2 , v2 ] ∼ [u3 , v3 ], h3 , v3 ] ∼ q. Since con(p) > con(q) and con(p) ≥ con(u1 , v1 ) ≥ con(u2 , v2 ) ≥ con(u3 , v3 ) ≥ con(q), we must have a > in the last formula; assume, for instance, that con(p) = con(u1 , v1 ) > con(u2 , v2 ). Then con(u1 , h1 ) = con(u2 , v2 ) < con(p), so con(h1 , v1 ) = con(p). Since q is collapsed by con(u1 , h1 ), by Lemma 3.8, there is a prime interval [e1 , x] in [u1 , h1 ] so that q is collapsed by con(e1 , x). Now con(p) > con(e1 , x) ≥ con(q), so by the assumption con(p) ≻ con(q), it follows that con(e1 , x) = con(q).

122

9. Minimal RT

b p a

v1

v3

e2

h1

h3

con(p)

u1

v2

h

d q c

u3 h2

con(q) e1

u2

e2 con(q) h con(p) e1

Figure 9.4: p ⇒ q illustrated for con(p) ≻ con(q) in J(Con L) Therefore, with h = h1 and e2 = v1 , we found a three-element chain e1 < h < e2 with con(e1 , h) = con(q) and con(h, e2 ) = con(p). Other choices lead to the same result, with maybe con(q) on the top. The rest is easy combinatorics. Let P2n = {k1 , . . . , kn , m1 , . . . , mn } (= Bn ) be the bipartite graph with minimal elements k1 , . . . , kn , maximal elements m1 , . . . , mn , so ki ≺ mj in P , for 1 ≤ i, j ≤ n, and {k1 , . . . , kn }, {m1 , . . . , mn } are antichains. Figure 9.5 shows B3 . Let L2n be a lattice with J(Con L)2n = Bn and satisfying |L2n | ≤ C(2n)α = ′ α C n for some constant C and C ′ = 4C. For 1 ≤ i, j ≤ n, we get in L2n a chain C(i, j): e1 (i, j) < h(i, j) < e2 (i, j) in L. Since there are n2 elements of the form h(i, j) and there are at most C ′ nα elements in L2n , therefore, there is an element h occurring as the middle element in n2 /(C ′ nα ) (that is, C1′ n2−α ) of these chains. For h(i, j), either con(h(i, j), e1 (i, j)) or con(h(i, j), e2 (i, j)) 1 2−α is a minimal element of Bn . So there are 2C of these elements for which ′n these choices are consistent, say, all con(h(i, j), e1 (i, j)) are maximal. Let A be the set of these e1 (i, j)-s. We get the situation depicted in Figure 9.6: for any x ∈ A, the congruence con(h, x) corresponds to a maximal element of Bn . Therefore, A is an antichain. Since a finite join of a subset of an antichain of join-irreducible elements cannot contain an element of the antichain not in the subset, we conclude that A k1

k2

k3

m1

m2

m3

Figure 9.5: The ordered set B3

123

9.5. Computing minimality

...

...

x

A

h Figure 9.6: A configuration that is too large in L under join generates 2|A| − 1 distinct elements. From |A| ≥ follows that 2−α 2|A| − 1 = C1 2n ,

1 2−α , 8C n

it now

2−α

for some constant C1 . So we conclude that L2n has at least C1 2n elements, contradicting that L2n has at most C ′ nα elements, completing the Proof-byPicture.

9.5.

Computing minimality

We start by formalizing the situation depicted in Figure 9.4. Lemma 9.3. Let L be a finite lattice, and let vi , ui ∈ L satisfy vi ≺ ui for i = 1, 2. Let γ i = con(vi , ui ) for i = 1, 2. If γ 1 ≺ γ 2 in J(Con L), then there is a three-element chain {e1 , h, e2 } in L such that γ i = α(h, ei ), for i = 1, 2, and e1 < h < e2 or e2 < h < e1 . Proof. Since v1 ≡ u1 (con(v2 , u2 )) and v1 ≺ u1 , by Theorem 3.1, there is a sequence of congruence-perspectivities up

dn

up

dn

[v2 , u2 ] = [x1 , y1 ] ↠ [x2 , y2 ] ↠ [x3 , y3 ] ↠ . . . ↠ [xn , yn ] = [v1 , u1 ], for some natural number n > 1. Obviously, γ 2 = con(x1 , y1 ) = con(x2 , y2 ) ≥ con(x3 , y3 ) ≥ · · · ≥ con(xn , yn ) = γ 1 . Since γ 2 > γ 1 , there is a smallest i satisfying con(xi , yi ) < γ 2 ; obviously, 3 ≤ i ≤ n. Let i be odd, and let z = xi−1 ∨ yi . Then con(xi−1 , z) = con(xi , yi ) < γ 2 , but con(xi−1 , yi−1 ) = γ 2 , so con(z, yi−1 ) = γ 2 . Since u1 ≡ v 1

(mod con(xi−1 , z))

and u1 ≺ v1 , there are u, v ∈ [xi−1 , z], such that v ≺ u and con(u1 , v1 ) ≤ con(u, v). So γ 1 = con(u1 , v1 ) ≤ con(u, v) < γ 2 ,

124

9. Minimal RT

and con(u, v) is a join-irreducible congruence, hence, by the assumption on γ 1 and γ 2 , it follows that γ 1 = con(u, v). Therefore, we can choose e2 = xi−1 , h = z, and e1 = yi−1 . If i is even, then we proceed dually. The second lemma deals with join-independence. A set A in a finite lattice W L is join-independent if for any a ∈ A and subset A1 ⊆ A, the inequality a ≤ A1 implies that a ∈ A1 . Lemma 9.4. Let L be a finite lattice, let A ⊆ L, and let b ∈ L be a lower bound of A in L. If { con(b, x) | x ∈ A } is join-independent in Con L, then A is join-independent in L. W Proof. Indeed, if a ≤ A1 for some A1 ⊆ A, then _ con(b, a) ≤ { con(b, x) | x ∈ A1 }, a contradiction.

Finally, observe that in a finite distributive lattice, a set A of join-irreducible elements is join-independent iff the elements are pairwise incomparable. So we obtain the following. Corollary 9.5. Let L be a finite lattice, let A ⊆ L, and let b ∈ L be a lower bound of A in L. If { con(b, x) | x ∈ A } is a set of pairwise incomparable join-irreducible congruences, then A is join-independent in L. Now we prove Theorem 9.2 with the combinatorial argument of Section 9.4, assisted by Corollary 9.5.

9.6.

Discussion

History At the start, R. P. Dilworth, G. Gr¨atzer, and E. T. Schmidt used only the N6 gadget for constructing lattices with given congruence lattices. Thus the size O(22n ). About 30 years later, I tackled the old problem of G. Birkhoff [19], characterizing congruence lattices of infinitary algebras (see my paper [101] and GLT2[102]); in fact, in the stronger form conjectured by R. Wille (see K. Reuter and R. Wille [234]): Every complete lattice can be represented as the lattice of complete congruences of a suitable complete lattice. The finite case was settled in S.-K. Teo [249]. With H. Lakser (see our joint paper [142]), we found a substantially simpler proof of my result, using colored chains (see also my joint paper with E. T. Schmidt [177]). Next year, we realized (see our joint paper [143]), the result was announced in my joint paper with H. Lakser [141]) that this technique can

9.6. Discussion

125

be utilized to prove that Every finite distributive lattice D can be represented as the congruence lattice of a finite lattice L of size O(n3 ), where n is the number of join-irreducible elements of D. Soon thereafter, using similar techniques, G. Gr¨atzer, H. Lakser, and E. T. Schmidt [156] found the O(n2 ) result. As we discussed in the Introduction, the papers flowing from the Basic RT split into four topics. The research field of this chapter is part of Topic A. Here atzer, are some additional papers from this field, in chronological order: G. Gr¨ H. Lakser, and B. Wolk [163], R. Freese, G. Gr¨atzer, and E. T. Schmidt [85], my joint paper with H. Lakser [144], and my joint papers with E. T. Schmidt [174], [175], [178], [179], [182], and [188]. The best result in this field is in [178]. Theorem 9.6. Every complete lattice can be represented as the complete congruence lattice of a complete distributive lattice. G. Gr¨atzer [110] and Chapter 10 of LTS1-[206] provide a detailed discussion of the complete case. The class D of distributive lattices is the minimal nontrivial variety, so this result cannot be improved by replacing D by a smaller variety. However, there are special classes of complete distributive lattices. The two best known infinitary identities are the Join-Infinite Distributive Identity: _ _ X = ( a ∧ x | x ∈ X ), a∧ (JID) and its dual, the Meet-Infinite Distributive Identity: ^ ^ X = ( a ∨ x | x ∈ X ). a∨ (MID)

We will denote by (JIDm ) the condition that (JID) holds for sets X satisfying |X| < m, where m is a regular cardinal with m > ℵ0 . We define (MIDm ) dually. The lattice we construct for Theorem 9.6 in my joint paper with E. T. Schmidt [178] fails both (JIDm ) and (MIDm ). In my other joint paper with E. T. Schmidt [182], we prove the following result. Theorem 9.7. Let L be a complete lattice with more than two elements and with a meet-irreducible zero. Then L cannot be represented as the lattice of complete congruence relations of a complete distributive lattice K satisfying (JID) and (MID). So we can raise the following. Problem 9. Characterize the complete congruence lattices of complete distributive lattices satisfying (JID) and/or (MID). Or more generally: Problem 10. Characterize the lattices of m-complete congruences of m-complete distributive lattices satisfying (JIDm ) and/or (MIDm ).

126

9. Minimal RT

Improved bounds For a natural number n and a class V of lattices, define mcr(n, V) (minimal congruence representation) as the smallest integer such that, for any distributive lattice D with n join-irreducible elements, there exists a finite lattice ∼ D and |L| ≤ mcr(n, V). Let mcr(n, L) = mcr(n). L ∈ V satisfying Con L = A somewhat sharper form of Theorem 9.1 is the following. Theorem 9.8. For any integer n ≥ 2, 1 n2 < mcr(n) < 3(n + 1)2 . 16 log2 n The upper bound was proved in G. Gr¨atzer, H. Lakser, and E. T. Schmidt [156] and the lower bound in my joint paper with D. Wang [197]. Different approaches to minimality A different kind of lower bound is obtained in R. Freese [84]; it is shown that if J(Con L) has e edges (e > 2), then e ≤ |L|. 2 log2 e R. Freese also proves that J(Con L) can be computed in time O(|L|2 log2 |L|). Consider the optimal length of L. E. T. Schmidt [239] constructs a finite lattice L of length 5m, where m is the number of dual atoms of D (for finite chains, this was done in J. Berman [18]); S.-K. Teo [249] proves that this result is best possible. Problem 11. In Theorem 9.2, is the construction “best” in some sharper sense? In other words, is there a function f (n) with the properties: (i) Theorem 9.1 holds with O(f (n)) in place of O(n2 ); (ii) f (n) < n2 ; (iii) n2 = O(f (n)) fails? Yet another approach to minimality starts with the elegant inequality for a finite lattice L: |J(L)| ≥ |J(Con L)| of R. Freese, J. Jeˇzek, and J. B. Nation [86]. So if we define je(L) = |J(L)| − |J(Con L)|, then je(L) ≥ 0 and je(L) is one measure of the efficiency of the representation of D = Con L as a congruence lattice. For a finite distributive lattice D, define JE(D) = min(je(L) | Con L ∼ = D),

9.6. Discussion

127

where L ranges over finite lattices. Then from this point of view, the best representation of a finite distributive lattice D as a congruence lattice of a finite lattice L is obtained when je(L) = JE(Con L). This is the approach we take in my joint paper with F. Wehrung [205]. In my joint paper with F. Wehrung [205], we compute the exact value of JE(D), yielding the inequality 0 ≤ JE(D) ≤

2 n, 3

and the constant 2/3 in this estimate is best possible. Problem 12. Determine the least constant k such that for every finite distributive lattice D, there exists a finite sectionally complemented lattice L such that Con L ∼ = D and J(L) ≤ k|J(D)|. By my joint paper with E. T. Schmidt [172], we have k ≤ 2. The value of the constant defined similarly for the class of atomistic lattices (or the class of all lattices as well) equals 5/3, by Corollary 5.4 of my joint paper with F. Wehrung [205]. Problem 13. Let V be a variety of lattices. If D is a finite distributive lattice representable by a finite lattice in V, compute the least possible value of |J(L)|, for a finite lattice L in V such that Con L ∼ = D. Problem 14. Is mcr(n, SecComp) = O(22n ) the best possible result? For any class S of lattices if the Basic RT holds for S, then theoretically, the function mcr(n, S) exists, although it may be difficult to compute. Of course, for any class S of lattices for which the Basic RT holds, we can raise the question what is mcr(n, S), and chances are that we get an interesting problem. In many cases, however, the Basic RT fails for S. Here are some nontrivial examples. (i) The congruence lattice of a finite relatively complemented lattice is Boolean. So SecComp cannot be narrowed to the class of relatively complemented lattices. (ii) The class S = SecComp ∩ DuallySecComp would also be a logical candidate. Note, however, that by Theorem 11 of my joint paper with E. T. Schmidt [172] and Lemma 4.16 of M. F. Janowitz [217], every finite lattice in SecComp ∩ DuallySecComp has a Boolean congruence lattice (see also Theorem II.4.9 of GLT2-[102]). (iii) Similarly, one can ask whether SecComp can be narrowed to the class of semimodular sectionally complemented lattices. The discussion in Section IV.3 of GLT2-[102] (in particular, the top paragraph of p. 240) shows

128

9. Minimal RT

that this cannot be done either. Every finite, semimodular, sectionally complemented lattice has a Boolean congruence lattice.

Chapter

10

Semimodular RT

10.1.

Semimodular lattices

Should we try to get an RT for semimodular lattices? For a class of lattices to be a candidate, it would have to satisfy the following two criteria. First, there ought to be very many finite simple lattices in the class. For semimodular lattices, the finite partition lattices (Section 1.2.1) are all simple semimodular lattices by O. Ore [229], so we have lots of finite simple semimodular lattices. Second, there ought to be finite lattices in the class with a three-element congruence lattice. For semimodular lattices, the lattice S8 of Figure 10.1 is such a lattice. These criteria satisfied, we looked for an RT for semimodular lattices. p p p q

p p q

q

p

p

Figure 10.1: The semimodular gadget S8 , colored with p < q 129 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_10

130

10. Semimodular RT

In fact, in G. Gr¨atzer, H. Lakser, and E. T. Schmidt [159], we found a very good RT for semimodular lattices, also obtaining—to our great surprise— planarity and a small size. Theorem 10.1 (Semimodular RT). Every finite distributive lattice D can be represented as the congruence lattice of a finite semimodular lattice L. In fact, if D has n ≥ 1 join-irreducible elements, then L can be constructed as a planar lattice of size O(n3 ). By Theorem 9.1, the optimal size for any lattice is O(n2 ), so the size we obtain in this result seems very good.

10.2.

Proof-by-Picture

The proof of this result is very similar to the proof of Theorem 9.1. The gadget we use now is the semimodular (colored—in the sense of Section 3.2) lattice S8 of Figure 10.1. To represent the ordered set P of Figure 10.2 as J(Con L), for a finite planar semimodular lattice L, we start out (see Figure 10.3) by constructing the a b

d

e

c Figure 10.2: The ordered set P

a q a

a

d

p

b

a a

b b

d

b

Aa

b

c

Figure 10.3: The lattices Aa and Ab

c

Ab

10.3. Construction and proof

131

lattices Aa and Ab . We obtain Aa by gluing two copies of S8 together, coloring it with {a, b, d} so that con(a) > con(b) is accomplished in the top S8 of Aa and con(a) > con(d) is accomplished in the bottom S8 of Aa . The lattice Ab is S8 colored by {b, c} so that con(b) > con(c) in Ab . Observe that the lattice Aa takes care of all a ≻ x orderings; in the example, there are only two. We could do three coverings by gluing three copies of S8 together, and so on. Form the glued sum S of Aa and Ab ; all the covers of P are taken care of in S. There is only one problem: S is not a colored lattice; in this example, if p is a prime interval of color b in Sb (as in Figure 10.3) and q is a prime interval of color b in Sb (as in Figure 10.3), then in S we have con(p) ∧ con(q) = 0. Of course, we should have con(p) = con(q) = con(b). We accomplish this by extending S to the lattice L of Figure 10.4. In L, the black-filled elements form the sublattice S. As you see, we extend S by adding to it a distributive “grid.” The right corner is C25 colored by {a, b, c, d}; each of the four covering squares colored by the same color twice are made into a cover-preserving M3 . This makes the coloring behave properly in the right corner. In the rest of the lattice we do the same: we look for a covering square colored by the same color twice, and make it into a cover-preserving M3 . This makes L into a colored lattice: any two prime intervals of the same color generate the same congruence. For instance, if p and q are prime intervals as in the previous paragraph (see Figure 10.4), then we find in L the prime intervals r1 , r2 , r3 , r4 (see Figure 10.4), so that up

dn

up

dn

up

p ↠ r1 ↠ r2 ↠ r3 ↠ r4 ↠ q; therefore, conL (p) = conL (q). Finally, we remember the element e ∈ P . We add a “tail” to the lattice and color it e. It is easy to see that the resulting lattice L is planar and semimodular and that J(Con L) is isomorphic to P with the isomorphism x 7→ con(x) for x ∈ {a, b, c, d, e}.

10.3.

Construction and proof

We construct the semimodular lattice L of Theorem 10.1 in several steps. Take the lattice S8 with the notation of Figure 10.5. The lattice S8 has an ideal, I8 , and a filter, F8 , both isomorphic to C2 ; we will utilize these for repeated gluings—defined in Section 2.4. The elements of I8 and F8 are black-filled in Figure 10.5. Let D be a finite distributive lattice, and let P = J(D) be the ordered set of its join-irreducible elements, n = |P |. We enumerate p1 , p 2 , . . . , p m

132

10. Semimodular RT

the non-minimal elements of P . For every pi , for i = 1, 2, . . . , m, let cov(pi ) = {pi1 , p2i , . . . , piki } denote the set of all lower covers of pi in P ; since pi is nonminimal, it follows that ki > 0. Let r1 , r2 , . . . , rt enumerate all elements of P that are incomparable with all other elements.

d

r1 c b

p b a a

r3

c b

r2

a c b b q

c

b d d

r4

d

e

Figure 10.4: The lattice L

133

10.3. Construction and proof

i8 F8

a8 e8

c8 f8 d8

b8 I8 o8

Figure 10.5: The semimodular gadget S8 , with notation In the example of Section 10.2, we have, say, a = p1 , b = p2 , k1 = 2, k2 = 1, b = p11 , d = p12 , c = p21 , cov(a) = {b, d}, and cov(b) = {c}. Step 1: For every i, with 1 ≤ i ≤ m, we construct a lattice Ai with an ideal Ii and a filter Fi , where Ii is a chain of length 2(ki + · · · + km ) and Fi is a chain of length 2(ki+1 + · · · + km ). For the ordered set P of Section 10.2, we construct the lattices A1 and A2 (see Figure 10.6); the elements of the ideals I1 and I2 , and of the filters F1 and F2 are black-filled. Now we will twice use the construction, gluing k times, described in Section 2.4. To form Ai , glue S8 to itself (ki − 1) times with the ideal I8 and

F1

F2 I1

I2

A1

Figure 10.6: The lattices A1 and A2

A2

134

10. Semimodular RT

the filter F8 , to obtain the lattice A1i with a filter FA1i . Now take B2 = {(0, 0), (0, 1), (1, 0), (1, 1)}

with the ideal and the filter

IB2 = {(0, 0), (1, 0)} FB2 = {(0, 1), (1, 1)},

and glue 2(ki+1 + · · · + km ) times B2 to A1i . The ideal Ii is generated by the element (0, 1) of the top B2 , whereas Fi is generated by the unit element of A1i . We define a coloring µi of Ai as follows. On any copy of S8 , µi [o8 , b8 ] = pi and on the j-th copy of S8 , µi [o8 , d8 ] = µi [d8 , c8 ] = pij ; on the first two copies i+1 of B2 , µi [(0, 1), (1, 1)] = pi+1 1 , on the next two copies, µi [(0, 1), (1, 1)] = p2 , i+2 after ki+1 pairs, the next two satisfy µi [(0, 1), (1, 1)] = p1 , and so on. Lemma 10.2. µi is a coloring of Ai . The join-irreducible congruences of Ai are generated by prime intervals of Ii and by [o8 , b8 ] of the bottom S8 in Ai . If p and q are [o8 , b8 ] or a prime interval [o8 , d8 ] or [d8 , c8 ] of a copy of S8 in Ai , then α(p) ≥ α(q) iff µi (p) ≥ µi (q). In particular, con(o8 , b8 ) ≻ con(o8 , d8 ) in J(Con(M)) Ai , where o8 , b8 , d8 are in a copy of S8 in Ai . If p is a prime interval [(0, 1), (1, 1)] in a copy of B2 , then con(p) is incomparable with any con(q), where q is [o8 , b8 ] or a prime interval of Ii different from p. Proof. This is trivial since every prime interval of S8 is projective to one of [o8 , b8 ], [o8 , d8 ], [d8 , c8 ]. Step 2: We define the lattice A by gluing together the (colored) lattices Ai for 1 ≤ i ≤ m. For the ordered set P of Section 10.2, we construct the lattice A (see Figure 10.7). For 1 ≤ i ≤ m, we define, by induction, the lattice Ai , which contains Ai , and, therefore, Fi , as a filter. Let A1 = A1 . Assume that Ai with Fi as a filter has been defined. Observe that both Fi and Ii+1 are chains of length 2(ki+1 + · · · + km ), and so they are isomorphic; in fact, this isomorphism preserves colors. We glue Ai to Ai+1 over Fi and Ii+1 to obtain Ai+1 . Define A = Am and IA = I1 (see Figure 10.7). Observe that µi on Fi agrees with µi+1 on Ii+1 ; therefore, the µi define a coloring µA of A for 1 ≤ i ≤ m. Let FA be the filter of A generated by the element (0, 1) of the top B2 in A1 . FA is a chain of length m. The prime interval [o8 , b8 ] in the bottom S8 in Ai (1 ≤ i ≤ m) is projective to a unique prime interval p of FA ; define µA (p) = µA [o8 , b8 ]. Lemma 10.3. µA is a coloring of A. The join-irreducible congruences of A are generated by prime intervals of IA and FA . Let p and q be prime intervals in IA and FA .

10.3. Construction and proof

135

FA

IA

Figure 10.7: The lattice A (i) If p and q are prime intervals of FA , then con(p) and con(q) are incomparable. (ii) If p is a prime interval of FA and q is a prime interval of IA , then con(p) and con(q) are comparable iff p ⊆ Ai , for some 1 ≤ i ≤ m, q is perspective to some [o8 , d8 ] or [d8 , c8 ] in some S8 in Ai ; in which case, con(p) ≻ con(q) in J(Con(M)) A. (iii) If p and q are prime intervals of IA , then con(p) ≥ con(q) iff p and q are perspective to prime intervals p′ and q′ in some Ai , respectively, for some 1 ≤ i ≤ m, and p′ and q′ are adjacent edges of some S8 in Ai , in which case, con(p) = con(q). Proof. This is obvious from Lemma 2.9. Observe that the congruence lattice of A is still quite different from D in two ways, exactly as we discussed this for S in Section 10.2: the congruences that correspond to the ri -s are still missing; prime intervals in IA ∪ FA of the same color generate incomparable congruences with one exception: they are adjacent intervals in IA , perspective to the two prime intervals of some S8 in some Ai .

136

10. Semimodular RT

Step 3: We extend A to a lattice B with an ideal IB which is a chain and which has the property that every prime interval of B is projective to a prime interval of IB . This step is easy. We form the lattice FA2 with the ideal IFA2 = { (x, 0FA ) | x ∈ FA }, where 0FA is the zero of FA . Let 1FA denote the unit element of FA and, for x ∈ FA , x < 1FA , let x∗ denote the cover of x in FA . For every x ∈ FA , x < 1FA , we add an element xm to FA2 so that the elements (x, x), (x, x∗ ), (x∗ , x), xm , (x∗ , x∗ ) form a sublattice isomorphic to M3 with (x, x) as zero and (x∗ , x∗ ) as unit. Let M be the resulting lattice. Obviously, M is a finite planar modular lattice whose congruence lattice is isomorphic to the congruence lattice of FA . Clearly, IFA2 is also an ideal of M ; we will denote it by IM . We glue A to M over FA and IM to obtain B. Let IB be defined as the ideal generated by (0, 1FA ). We define µB as an extension of µA ; every prime interval p of M is projective to exactly one prime interval p of IM , we define µB (p) = µA (p). Lemma 10.4. µB is a coloring of B. The join-irreducible congruences of B are generated by prime intervals of IB . Let p and q be prime intervals in IB . (i) If p and q are prime intervals of M , then con(p) and con(q) are incomparable. (ii) If p is a prime interval of M and q is a prime interval of IA , then con(p) and con(q) are related as conA (p) and conA (q) are related in A. (iii) If p and q are prime intervals of IA , then con(p) and con(q) are related exactly as conA (p) and conA (q) are related in A. Proof. This is obvious from the congruence structure of M . Step 4: We extend B to the lattice L of the Theorem 10.1, as sketched in Figure 10.8. This is also an easy step. We take a chain C of length n and we color C over P so that the coloring is a bijection. We form the lattice C × IB . For every pair of prime intervals, p = [a, b] of C and q = [c, d] of IB if p and q have the same color, then we add an element m(p, q) to C over P so that the elements (a, c), (b, c), (a, d), m(p, q), (b, d)

10.3. Construction and proof

137

form a sublattice isomorphic to M3 . Let N denote the resulting lattice. N is obviously modular and planar. Set IN = { (x, 0IB ) | x ∈ C },

FN = { (1C , x) | x ∈ IB }, where 0IB is the zero of IB and 1C is the unit of C. Then IN is the ideal of N (isomorphic to C) and FN is a filter of N (isomorphic to IB ). Every prime interval of N is projective to a prime interval of IN , so we have a natural coloring µN on N . Note that this coloring agrees with the coloring µB on FN under the isomorphism with IB . We glue N to B over FN and IB to obtain L with the coloring µL . Set IL = IN . It is clear from the construction and from the lemmas that every prime interval of L is projective to a prime interval of IL and that distinct prime intervals of IL generate distinct join-irreducible congruences of L. It remains to verify that if p and q are distinct prime intervals, then con(p) ≥ con(q) iff µL (p) ≥ µL (q). Since P is finite, it is sufficient to prove that con(p) ≻ con(q) in J(Con L) iff µL (p) ≻ µL (q) in J(D). But this is clear, since if µL (p) ≻ µL (q) in J(D), then µL (p) = pi , for some 1 ≤ i ≤ m, and µL (q) = pij , for some 1 ≤ j ≤ ki , so con(p) ≻ con(q) was guaranteed in Ai . To establish that the size of L is O(n3 ), we give a very crude upper bound for |L|. Since 2n2 + 1 is an upper bound for |Ii |, 1 ≤ i ≤ m, it follows that

M

A N

Figure 10.8: Sketching the lattice L

138

10. Semimodular RT

3(2n2 + 1) is an upper bound for |Ai | and 3(2n2 + 1)n is an upper bound for |A|. Since |FA | ≤ n + 1, we get the upper bound (n + 1)2 + n + 1 for |M |. Finally, |IB | ≤ 2n2 + 1 + n + 1 = 2n2 + n + 2, so |N | ≤ 2(2n2 + n + 2)(n + 1). Therefore, 3(2n2 + 1)n + (n + 1)2 + n + 1 + 2(2n2 + n + 2)(n + 1) is an upper bound for L and it is a cubic polynomial in n. This completes the proof of the Theorem 10.1. It is not difficult to find better upper bounds for |L|; for instance, |L| ≤ 3n3 + 2n2 − 7n + 4.

10.4.

All congruences principal RT for planar semimodular lattices

In this section, we go one step beyond the Semimodular RT, Theorem 10.1. Theorem 10.5. Every finite distributive lattice D can be represented as the congruence lattice of a planar semimodular lattice L in which all congruences are principal. Proof. We have to modify the construction of the planar semimodular lattice L of Section 10.3 to make all congruences principal. In Step 3, we choose a chain C of length four. Observe that the proof of Theorem 10.1 remains valid as long as every color is represented as the coloring of C. Now we change the definition of C. For every x ∈ D, define r(x) = { a ∈ J(D) | x ≤ a }, and let Cx be a chain of |r(x)| + 1 elements, colored by the elements of r(x) (in any order). Let 0x , 1x denote the bounds of Cx . Let C be the glued sum of the chains Cx for x ∈ D (in any order). This chain C obviously satisfies the condition that every color is represented as the color of an edge in C. Therefore, the lattice L constructed in Section 10.3 satisfies the requirements of Theorem 10.5. We only have to observe that all congruences are principal. Let α be a congruence of L. Let x be an element of D that corresponds to the congruence α under an isomorphism between Con L and D. Since Cx is colored by the set r(x), we conclude that in L, we have con(0x , 1x ) = α, completing the proof.

10.5. Discussion

10.5.

139

Discussion

An addendum In Chapter 15, we need a more detailed version of Theorem 10.1. Theorem 10.6. Let P be a finite order. Then there exists a finite semimodular lattice L with the following properties: (i) J(Con L) is isomorphic to P . (ii) |L| is O(|P |3 ). (iii) L has an congruence-determining sublattice C, which is an ideal and a chain. Problems In Chapter 9, the O(n2 ) construction is followed by a proof that this size is optimal. It would be nice to have a similar result for planar semimodular lattices. Problem 15. Is the O(n3 ) result optimal for planar semimodular lattices? Problem 16. Would the O(n3 ) result be optimal for semimodular lattices? These problems for rectangular lattices are solved in the next chapter. Let Planar + SemiMod denote the class planar semimodular lattices. Using the notation of Section 8.6, these problems ask whether mcr(n, SemiMod) = O(n3 ) is optimal, and whether mcr(n, Planar + SemiMod) = O(n3 ) is optimal, where SemiMod is the class of semimodular lattices.

Chapter

11

Rectangular RT

11.1.

Results

In Chapter 10, we proved that every finite distributive lattice D can be represented as the congruence lattice of a finite semimodular lattice L of size O(n3 ). In this chapter we discuss what can we say about rectangular lattices, which form a very small subclass of semimodular lattices. So the first question is, can every finite distributive lattice D be represented as the congruence lattice of a rectangular lattice L? The answer is surprisingly easy in view of Lemma 4.16 and Theorem 10.1. Theorem 11.1 (Rectangular RT). Every finite distributive lattice D can be represented as the congruence lattice of a rectangular lattice L. However, we want more, the analog of Theorem 10.1: a rectangular lattice L of size O(n3 ), where n is the number of join-irreducible elements of D. Unfortunately, applying Theorem 10.1 to a finite distributive lattice D with n join-irreducible elements and then adding corners to obtain a rectangular lattice L, we may get a lattice not of size O(n3 ) but of size O(n4 ) (see my joint paper with E. Knapp [138]). Nevertheless, my joint paper with E. Knapp [137] proved the analog of Theorem 10.1 for rectangular lattices. Theorem 11.2. Every finite distributive lattice D can be represented as the congruence lattice of a rectangular lattice L. In fact, if D has n ≥ 1 joinirreducible elements, then L can be constructed as a planar lattice of size O(n3 ). 141 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_11

142

11. Rectangular RT

To prove Theorem 11.2, we need a new construction. The Proof-by-Picture of this construction is presented in Section 11.2. The main result of this field is in my joint paper with E. Knapp [138]; it proves that O(n3 ) is the best possible in Theorem 11.2. Theorem 11.3. (i) Let P be a finite ordered set with n ≥ 1 elements. Then P can be represented as the ordered set of join-irreducible congruences of a rectangular lattice L satisfying |L| ≤

2 3 4 n + 2n2 + n + 1. 3 3

(ii) Let Ln be a rectangular lattice whose ordered set of join-irreducible congruences is a balanced bipartite ordered set on n elements. Then for some constant k > 0, the inequality |Ln | ≥ kn3 holds. Apart from the Proof-by-Picture section, we do not go into the proofs of these results. We refer the reader to my joint paper with E. Knapp [138] and to my chapter in LTS1-[206].

11.2.

Proof-by-Picture

Let P = {p1 , p2 , p3 } be the ordered set of Figure 11.1, where the enumeration of the elements of P is an extension of the order, that is, if pi < pj , then i < j. We construct a rectangular lattice L = L3 representing P as the ordered set of join-irreducible congruences of L, that is, L satisfies that J(Con L) ∼ = P3 . Let Pi = {p1 , . . . , pi } for i ≤ 3. We inductively construct the lattices Li , for 1 ≤ i ≤ 3, so that the lattice Li is rectangular and it represents Pi as J(Con L)i . We define the rectangular lattice L1 as M3 . Note that J(Con L)1 ∼ = P1 . Let i = 2. We have the ordered set P2 = {p1 , p2 } with p1 ∥ p2 . Let A2 be a 0-stacked S∗7 and B2 = C2 × C2 (as shown in Figure 11.2). We glue A2 and B2 together to form the lattice K2 . We glue the lattice L1 to D2 = C2 × C2 to obtain the lattice K2′ and then glue the K2′ to K2 , to get L2 —the dashed line in the diagram of L2 indicates how L2 was glued together. In the resulting lattice, the top left and top right boundaries are colored with the elements p1 and p2 , so J(Con L)2 ∼ = P2 . Finally, let i = 3. The ordered set P = P3 = {p1 , p2 , p3 } has the two coverings p1 ≺ p3 and p2 ≺ p3 , so we take A3 as a 2-stacked S∗7 . Consider the lattice C3 × C4 . We add an element each to the interiors of the intervals [(0, 1), (1, 2)] and [(1, 2), (2, 3)] to form the lattice B3 and assign a coloring to A3 and B3 , as indicated in Figure 11.3. Glue A3 and B3 to obtain the lattice K3 (the dashed line indicates the gluing). Set D3 = C4 × C3 (not

11.3. All congruences principal RT

143

Figure 11.1: The ordered set P shown in Figure 11.3). Glue L2 to D3 and then the result (also not shown in Figure 11.3) to K3 to form the lattice L = L3 (the dashed line indicates the gluing). Note the coloring of L; it verifies that J(Con L) ∼ = P.

11.3.

All congruences principal RT

Now we want to do the same as in Section 10.4: adding the condition that all congruences are principal to the RT of this chapter. Theorem 11.4. Every finite distributive lattice D can be represented as the congruence lattice of a rectangular lattice K with the property that all congruences are principal. We prove this RT in a much stronger form, as an ET in Section 15.6 (see Theorem 15.3).

11.4.

Discussion

Let L be a rectangular lattice and let α be a join-irreducible congruence of L. We call the congruence α left-sided, if there is a prime interval p ⊆ Cll (L) such that con(p) = α, but there is no such p ⊆ Clr (L). In the symmetric case, we call the congruence α right-sided. The congruence α is one-sided if it is left-sided or right-sided. The congruence α is two-sided if it is not one-sided. Using these concepts, we can further analyze Theorem 4.20 and condition (P). We build a rectangular lattice from a grid (the direct product of two chains) by inserting first forks and then eyes. At the start, all join-irreducible congruences are one-sided. When we insert a fork, we introduce a two-sided congruence. When we insert an eye, we identify two congruences, resulting in a two-sided congruence. What congruence pairs occur in Theorem 4.20? Let β l be a congruence of Cll (L) and let β r be a congruence of Clr (L). Under what conditions is there a congruence α of L such that αl = β l and αr = β r ? Here is the condition. If p is a prime interval of Cll (L) collapsed by β l and there is a prime interval q of Clr (L) with con(p) = con(q), then q is collapsed by β r ; and symmetrically. In Step 3 of the construction, we use the chain Cn+1 . Clearly, Cn would have sufficed. Can we use, in general, shorter chains?

144

11. Rectangular RT

A2

K2

B2

K2′

D2

L2

L1 Figure 11.2: The lattices for i = 2

In a finite sectionally complemented lattice, the congruences are determined around the zero element. So it is clear that for finite sectionally complemented lattices, all congruences are principal. For a finite semimodular lattice, the congruences are scattered all over. So it is somewhat surprising that Theorem 11.4 holds. For modular lattices, the situation is similar to the semimodular case. E. T. Schmidt [238] proved that every finite distributive lattice D can be represented as the congruence lattice of a countable modular lattice K. (See also my joint papers with E. T. Schmidt [185] and [189].) The two major open problems are Problems 15 and 16.

145

11.4. Discussion

A3

B3

K3

L3

L2

Figure 11.3: The lattices for i = 3

Chapter

12

Modular RT

12.1.

Modular lattices

Corollary 3.11 notes that the congruence lattice of a finite modular lattice is Boolean; however, a finite distributive lattice has a representation as the congruence lattice of an infinite modular lattice by E. T. Schmidt [238] and [244]. Theorem 12.1 (Modular RT). Every finite distributive lattice D can be represented as the congruence lattice of a modular lattice L. We are going to prove this result in the following stronger form provided in my joint paper with E. T. Schmidt [189]. Theorem 12.2. Let P be a finite order. Then there exists a lattice L with the following properties: (i) L is a modular lattice. (ii) Con L is finite and J(Con L) is isomorphic to P . (iii) L is countably infinite. (iv) L is weakly atomic. (v) L has a unique complemented pair {a, a′ } = ̸ {0, 1} and id(a), id(a′ ) are chains, in fact, successor ordinals. (vi) L is rigid. 147 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_12

148

12. Modular RT

In this result, a lattice L is weakly atomic if every nontrivial interval contains a prime interval. We are constructing a rigid modular lattice L, laying the foundation for the Independence Theorem for Modular Lattices (see Section 12.4). The lattice L we construct is infinite, but “small.” As an order, it can be embedded in C 3 , where C is a countable ordinal; it is also weakly atomic.

12.2.

Proof-by-Picture

The gadgets in previous constructions were lattices whose congruence lattices are isomorphic to the three-element chain. The gadget in the modular case is the lattice M3 [C] of Chapter 6 for a chain C. We will use the sketch of this lattice as in Figure 6.1. Take the three-element ordered set P = {p1 , p2 , p3 } with the only relation: p1 < p3 (see Figure 12.1). We present a Proof-by-Picture of Theorem 12.2 for the order P . We start by constructing the lattice L1 representing P1 = {p1 } (see Figure 12.2). The lattice L1 , clearly, is simple and has no automorphism. The congruence representing p1 is con(b11 , c11 ). p3 p1

p2

Figure 12.1: The ordered set P From L1 , we obtain the lattice L2 to represent {p1 , p2 }, the two-element antichain (Figure 12.6). In L2 , we designated the elements b21 < c21 < b22 < c22 in the chain fil(a2 ), so that p1 is represented by con(b21 , c21 ) and p2 is represented by con(b22 , c22 ). Finally, we want to add p3 , so that it is over p1 but incomparable with p2 . This is accomplished in L3 (see Figure 12.3), where L2 is only partially drawn. The lattice L3 is glued together from four parts; the boundaries are thick lines. The bottom part is L2 (only partially drawn and shaded), whereas on the upper left edge (the chain E = fil(a2 ) in L2 ) we mark the elements b21 < c21 < b22 < c22 , as described above. We take a chain C of type ω + 2, which we obtain by forming glued sums (introduced in Section 1.1.3) C = [b21 , c21 ] ∔ [b21 , c21 ] ∔ · · · ω-times and adding n0 ≺ n1 to the top.

12.2. Proof-by-Picture

149

11 c11

q1

b11 a′1

a1

01 Figure 12.2: The lattice L1 The left part of L3 , call it Left, and the right part of L3 , call it Right, are both C × E and the top part, call it Top, is M3 [C]. Finally, in Left, each square [b21 , c21 ] × [b21 , c21 ] is made into M3 [b21 , c21 ], and the top square of Left into a cover-preserving M3 . What are the congruences of L3 ? Think of L3 as the base B, a direct product of two chains, B = id(a3 ) × id(a′3 ), with the flaps of the M3 [X]-es sticking out (each X is a chain); then a congruence α of L3 is a congruence of B with the property that it acts the same way on the left edge (a chain) of an M3 [X] as on the right edge (an isomorphic chain). Therefore, every congruence of L2 extends, and extends uniquely to L3 . The extension of con(b22 , c22 ) (which corresponds to p2 ) is easy to see: in Left, for c, d ∈ C, (c, x) ≡ (d, y) is equivalent to c = d and x ≡ y in L2 . Similarly, in Right. No two distinct elements are congruent in Top. The extension of con(b21 , c21 ) (which corresponds to p1 ) is much larger; since it collapses b21 and c21 , it collapses all of C; the quotient lattice is finite, as shown in Figure 12.4. Collapsing two distinct elements of {n0 , n1 } × E is equivalent to collapsing the two corresponding elements of E. If we collapse two elements of Left with the same C-component, this is obviously equivalent to collapsing the two Ecomponents. If we collapse two elements of Left with the same E-component, this is obviously equivalent to collapsing two elements of the left side of an M3 [b1 , c1 ], which makes it equivalent to the collapse of the corresponding two

150

12. Modular RT 13 c33

Top

b33

c32 b32 c31 b31

a3

a′3

c22 b22

C

Left

C

Right

c21

D

b21 a′2

a2

E

E

L2

Figure 12.3: The lattice L3 elements in the lower right edge, which is the last case. So there is exactly one new join-irreducible congruence, con((a2 , 0C ), (n1 , 0C )), which contains con(b21 , c21 ) but not con(b22 , c22 ). In L3 , we marked the elements b31 < c31 < b32 < c32 < b33 < c33 in the chain fil(a3 ), so that p1 is represented by con(b31 , c31 ), p2 is represented by con(b32 , c32 ), and p3 is represented by con(b33 , c33 ) in L3 . An automorphism α either fixes a3 or takes it into a′3 . But if αa3 = a′3 , then α would define an automorphism of L2 that takes a2 into a′2 , a contradiction. So α fixes a2 and a′2 . Therefore, α fixes id(a2 ) and id(a′2 ), and so id(a2 ) × id(a′2 ). Since every element of L3 belongs to this direct product, or is on the third flap of an M3 [X], whose base is in this direct product, we conclude that the base, and therefore, the whole M3 [X] is fixed by α, so α is the identity map, that is, L3 is rigid.

12.3. Construction and proof

151

Figure 12.4: The lattice L3 /con(b21 , c21 )

12.3.

Construction and proof

We begin by constructing the basic building block of the construction (see Figure 12.5). Lemma 12.3. Let U be a chain with zero, 0U , and unit, 1U . Let 0U ≤ u < v ≤ 1U , and let V be the interval [u, v] of U . Construct the lattice L as follows: (i) Form the direct product U × V . (ii) Glue the interval [(0U , u), (u, v)] of U × V with the filter [(u, u), (u, v)] (∼ = V ) to M3 [V ] with the ideal { (0, 0, v) | v ∈ V } ( ∼ = V ). (iii) Glue the lattice we obtained with the filter { (1, x, x) | x ∈ V } ( ∼ = V) to the interval [(v, u), (1U , v)] with the ideal [(v, u), (v, v)] ( ∼ = V ); the resulting lattice is denoted by U ⊛ V . (iv) Consider U as a sublattice of U ⊛ V by identifying x ∈ U with (x, u) if x ≤ u or if v ≤ x; otherwise, x ∈ [u, v], and we identify x with (x, 0, 0) ∈ M3 [V ]. This identifies U with a principal ideal of U ⊛ V .

152

12. Modular RT

(1U , v) (1U , u)

(u, v) M3 [V ] (v, u) U

(v, v) (u, u)

(0U , v) V (0U , u)

Figure 12.5: The lattice U ⊛ V Then every congruence α of U has a unique extension to a congruence α of U ⊛ V ; therefore, Con U ∼ = Con(U ⊛ V ). Proof. By Lemmas 2.13, 2.9, and 6.5. We use induction on n = |P |, the size of P , to construct a rigid modular lattice RP with the following properties: (I1 ) RP is a rigid modular lattice with zero, 0P , and unit, 1P . (I2 ) Every principal congruence of RP is a join of join-irreducible congruences and the join-irreducible congruences of RP form an ordered set isomorphic to P . (I3 ) RP has an element aP with a unique complement a′P such that the ideals id(aP ) and id(a′P ) are isomorphic to a countable successor cardinal αP , and RP has a dual atom qP ≥ aP . (I4 ) Every congruence of RP is determined by its restriction to fil(aP ). Observe that (I4 ) is equivalent to the following condition: (I′4 ) RP contains a chain P P P P P a P ≤ bP 1 < c1 ≤ b2 < c2 ≤ · · · ≤ bn < cn ≤ 1P

such that the join-irreducible congruences of RP are exactly the principal congruences P P P P P con(bP 1 , c1 ), con(b2 , c2 ), . . . , con(bn , cn ).

If |P | = 1, then the lattice of Figure 12.2 satisfies these requirements with αP = 4; this lattice is also simple and rigid.

12.3. Construction and proof

153

12 q2 c22 b22 c21 b21 a2

a′2

11 q1

a′1

a1

02

01

Figure 12.6: The lattice L2 In general, if P is an antichain, we proceed as in the case n = 2 (see Figure 12.6). We form the glued sum K = L1 ∔ L1 ∔ · · · ∔ L1 with n copies of L1 . Then we add elements (completing cover-preserving B2 -s to cover-preserving M3 -s) to the lattice C 2 , where C is a chain of length 3n, so that we can embed K into this lattice. It is clear that the lattice we obtain is a planar lattice satisfying (I1 )–(I4 ). Now let P be a finite order, not an antichain, and let us assume that RS has been constructed for all finite ordered sets S satisfying |S| < |P |. Since P is not an antichain, we can choose a maximal element p of P that is not minimal. Let q1 , q, . . . , qr be the elements of P covered by p, with r ≥ 1.

154

12. Modular RT

For the ordered set Q = P − {p}, there exists a rigid modular lattice RQ such that (I1 )–(I4 ) hold; in particular, there is a chain Q Q Q Q Q a Q ≤ bQ 1 < c1 ≤ b2 < c2 ≤ · · · ≤ bn−1 < cn−1 ≤ 1Q ,

or simply, aQ ≤ b1 < c1 ≤ b2 < c2 ≤ · · · ≤ bn−1 < cn−1 ≤ 1Q , such that the join-irreducible congruences of RQ are the principal congruences con(b1 , c1 ), con(b2 , c2 ), . . . , con(bn−1 , cn−1 ). Let C = fil(a) in Q. By (I3 ), C is a well-ordered chain (observing that, by modularity, id(a′ ) ∼ = fil(a)); it is isomorphic to αQ . Let con(bk1 , ck1 ), con(bk2 , ck2 ), . . . , con(bkr , ckr ) be the join-irreducible congruences corresponding to q1 , q2 , . . . , qr , respectively. Without loss of generality, we can assume that b k 1 < c k 1 ≤ bk 2 < c k 2 ≤ · · · ≤ b k r < c k r . Let F = [bk1 , ck1 ] ∔ [bk2 , ck2 ] ∔ · · · ∔ [bkr , ckr ].

For every natural number i, let Fi be a chain isomorphic to F , and let xi denote the image of x in Fi under this isomorphism. Finally, we consider the glued sum C of these chains, and the chain C that is C with two elements, n0 ≺ n1 , adjoined, C = F1 ∔ F2 ∔ · · · + {n0 , n1 } = C + {n0 , n1 }. We apply Lemma 12.3 first to U1 = E and V1 = [bk1 , ck1 ], to obtain the lattice E ⊛ [bk1 , ck1 ] and filter {ck1 } × E (∼ = E), and second to U2 = E and V2 = [bk2 , ck2 ], in order to obtain the lattice E ⊛ [bk2 , ck2 ] and ideal {bk2 } × E ∼ E); and we glue these two lattices together over the given ideal and filter. (= We proceed similarly and glue E ⊛ [bk3 , ck3 ] to the resulting lattice, and so on. In r steps, we obtain a lattice L with an ideal and a filter both isomorphic to E. Now for each i < ω, we take a copy Li of L, and glue L2 to L1 , L3 to the resulting lattice, and so on; in the last step, we glue Li . Call S the lattice we obtained Li . Obviously, L1 ⊆ L2 ⊆ · · · ⊆ Ln , so we can take ( Li | i < ω ). We adjoin to this lattice {n0 , n1 } × E (where {n0 , n1 } is regarded as the two-element chain with n0 < n1 ) so that (u, c) ∧ (v, d) = (u ∧ v, c ∧ d), (u, c) ∨ (v, d) = (u ∨ v, c ∨ d),

12.3. Construction and proof

155

for u and/or v ∈ {n0 , n1 } and c, d ∈ E. We add one more element. E has a dual atom qC by (I3 ); the lattice {n0 , n1 } × {qC , 1Q } is a cover-preserving four-element Boolean sublattice of the lattice we have constructed. We add an element w so that {n0 , n1 } × {qC , 1Q } with w form a cover-preserving M3 . Let Left denote the lattice we have just obtained. Finally, let us glue RQ and Left together over the filter E and ideal {bk1 }×E (∼ = E) in L0 ; call the resulting lattice Left∗ . To investigate the lattice Left, let us present a more intuitive description. Form the direct product of C and E; we will call this the base of Left. Consider the interval [(bkj )i , (ckj )i ] in C and the corresponding interval [bkj , ckj ] in E. In the direct product C × E, replace [bki j , cki j ] × [bkj , ckj ] with M3 [bkj , ckj ] so that (x, y) ∈ [bikj , cki j ] × [bkj , ckj ] is replaced by (x, x ∧ y, y) and we add the element w in the top prime square so that it forms an M3 . Of course, with this definition it is not clear whether Left is a lattice, whether it is modular, and what are the congruences of Left. It is clear, however, that this definition is the same as the more complicated one given above, so we get all these properties of the construction from the results in Chapter 6 and from Lemma 12.3. Take a congruence γ of Left; then the restriction of γ to the base of Left is of the form γ × α, where γ is a congruence of C and α is a congruence of E. It is clear from Lemma 12.3 that γ and α uniquely determine γ. Moreover, α “almost” determines γ. In fact, a ≡ b (mod γ) is fully determined by α for a, b ∈ C. The same is true for a, b ∈ {n0 , n1 }. Finally, let a ∈ C and b ∈ {n0 , n1 }. Then a ≡ b (mod γ) implies that all Di are collapsed by γ from some i on, and so all of C is collapsed by γ. We conclude that Left has exactly one join-irreducible congruence that is not a minimal extension of a congruence from E, namely, con((0C , 0E ), (n0 , 0E )), and this congruence majorizes all the join-irreducible congruences that are minimal extensions of the congruences con(b1 , c1 ), con(b2 , c2 ), . . . , con(bn−1 , cn−1 ) of E to Left. By the inductive assumption (I3 ), every congruence of RQ is determined by its action on E, and so we obtain that every congruence of RQ can be extended to Left∗ . Therefore, the join-irreducible congruences of Left∗ can be described as follows: they are the minimal extensions of the join-irreducible congruences of RQ to Left∗ and the congruence con((0C , 0E ), (n0 , 0E )). Hence, they form ∼ P , by (I2 ). an ordered set isomorphic to P , and so J(Con(M)) Left∗ =

156

12. Modular RT

The lattice Left∗ does not satisfy all the inductive assumptions, so we will define RP as an appropriate extension of Left∗ . Define the lattice Top as M3 [C] and define the ideal C1 = { (0C , 0C , x) | x ∈ C } of M3 [C]. Define the lattice Right = [a′Q , 1Q ] × C1 and let C (∼ = C) be the filter generated by (1Q , 0C1 ) in Right. Let Right∗ be the gluing of Top and Right over C1 and C. Obviously, the ideal of Right∗ generated by (1C , 0C , 0C ) ∈ Top is isomorphic with the filter fil(a′Q ) of Left∗ . So we can glue Left∗ and Right∗ together to obtain RP . Define aP = (n1 , 0E ) (∈ Left) and let a′P be its unique complement, ′ aP = (a′Q , 0C1 ) (∈ Right). The dual atom (n1 , n0 , n0 ) of Top can serve as the dual atom of RP . Now we verify that RP satisfies the conditions of Theorem 12.2 as well as the inductive assumptions (I1 )–(I4 ). The ideal id(aP ) is a countable well-ordered chain, namely, αQ ∔ C, which ′ is a successor ordinal; the same is true of id(aP ). Let α be an automorphism of RP . Since 1Q is the smallest element x ∈ RP such that there is a sublattice isomorphic to M3 from x to 1P , it follows that α1Q = 1Q . Since RQ is rigid, it follows that αx = x for all x ∈ RQ ⊆ RP . Left is built on C × E, and E is kept fixed by α. Therefore, α maps C into itself. But C is well-ordered, so α is the identity map on C, and so on all of Left. Top = [1Q , 1P ] and both these elements are fixed by α. Next, consider the three atoms in the sublattice isomorphic to M3 , stretching from 1Q to 1P : ′ ; the elements (0C , 1C , 0C ) the element (1C , 0C , 0C ) is fixed because it is in RQ and (0C , 0C , 1C ) cannot be interchanged because 1Q has a complement in the ′ , (0C , 1C , 0C )]. So α fixes the elements interval [a′Q , (0C , 0C , 1C )] but not in [aQ (1C , 0C , 0C ), (0C , 1C , 0C ), (0C , 0C , 1C ) and also the chain C, hence, α fixes all of the lattice Top. Since we already know that in Right, α fixes id(1Q ) and fil(1Q ), it follows that α fixes every element of Right. We conclude that RP is rigid. It remains to verify (I4 ). In RQ , we were given the chain a Q ≤ b 1 < c 1 ≤ b 2 < c 2 ≤ · · · ≤ b n < c n ≤ 1Q such that the join-irreducible congruences of RQ are the principal congruences con(b1 , c1 ), con(b2 , c2 ), . . . , con(bn−1 , cn−1 ). In RP , we define biP = aQ ∨ bi ,

cP i = aQ ∨ c i ,

12.4. Discussion

157

for i = 1, . . . , n − 1, and bP n = (1C , 0C , 0C , ) ∈ Top,

cP n = 1P ∈ Top.

Then condition (I4′ ) is obvious for the chain P P P P P a P ≤ bP 1 < c 1 ≤ b2 < c 2 ≤ · · · ≤ bn < c n = 1 P .

This completes the proof of Theorem 12.2.

12.4.

Discussion

The Independence Theorem for Modular Lattices In GLT-[99], I raised Problem II.18, whether the congruence lattice and the automorphism group of a finite lattice are independent. This problem was solved affirmatively by V. A. Baranski˘ı [15], [16] and A. Urquhart [254]. It is natural to ask whether one could prove the Independence Theorem for modular lattices. This was done in my joint paper with E. T. Schmidt [189]. Theorem 12.4 (Independence Theorem for Modular Lattices). Let D be a nontrivial finite distributive lattice and let G be a finite group. Then there exists a modular lattice M such that the congruence lattice of M is isomorphic to D and the automorphism group of M is isomorphic to G. The proof uses the following statement. Theorem 12.5. Let G be a finite group. Then there exists a simple modular lattice S with an atom p such that the automorphism group of S is isomorphic to G and every automorphism keeps p fixed. A weaker form of this theorem (just constructing a modular lattice) is due to E. Mendelsohn [225]. The present form is in my joint paper with E. T. Schmidt [185] (see also C. Herrmann [210] and E. T. Schmidt [245]). Now it is easy to prove the Independence Theorem for Modular Lattices. Let G and D be given as in the Independence Theorem. Let S be constructed as in Theorem 12.5 for G and let R be the lattice in Theorem 12.2 constructed for D. We take the ideal id(p) in S and the filter fil(q) in R; both are twoelement lattices, so we can glue S and R together over id(p) and fil(q). Let M be the resulting lattice. M is a modular lattice by Lemma 2.22. By Lemma 2.9, Con M ∼ = Con R, so Con M is isomorphic to D. Now let α be an automorphism of S, and define a map α† of M into itself as follows: ( if x ∈ S; αx, † α x= x, if x ∈ R.

158

12. Modular RT

Since p and 0S are kept fixed by α, it is evident that α† is an automorphism of M . Moreover, q and p is the only pair of elements in M satisfying q ≺ p and M = id(p) ∪ fil(q); therefore, every automorphism of M acts as an automorphism on S and R. It is now clear that every automorphism of M is of the form α† for some automorphism α of S, hence, α 7→ α† is an isomorphism between the automorphism group of S and the automorphism group of M . It follows that the automorphism group of M is isomorphic to G. This completes the proof of the Independence Theorem for Modular Lattices. Two stronger results The following two results state Theorem 12.1 in a stronger form. Theorem 12.6 (R. Freese [82]). Every finite distributive lattice D can be represented as the congruence lattice of a finitely generated modular lattice L of breadth 2. Since a modular lattice of breadth 2 is 2-distributive (see Section 18.5), the lattices of Freese are also 2-distributive. Theorem 12.7 (E. T. Schmidt [242]). Every finite distributive lattice D can be represented as the congruence lattice of a complemented modular lattice L. Arguesian lattices In my joint paper with E. T. Schmidt [189], we prove a much stronger form of the Independence Theorem. We construct an arguesian lattice A with a given finite congruence lattice and finite automorphism group. Recall that a triangle a = (a1 , a2 , a3 ) in a lattice L is an ordered triple of element of L. With two triangles a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ), we associate the elements p = (a0 ∨ b0 ) ∧ (a1 ∨ b1 ) ∧ (a2 ∨ b2 ),

m = (c2 ∧ (c0 ∨ c1 ).

A lattice is Arguesian if it satisfies the arguesian identity: p ≤ (a0 ∧ (a1 ∨ m)) ∨ b0 . This is a very strong form of modularity; it is a lattice theoretic formulation of Desargues’ Theorem. I would recommend that the reader consult Section V.5 of LTF-[105] for background. To prove the Independence Theorem for Arguesian Lattices, we need arguesian versions of Theorems 12.2 and 12.5. In fact, in my joint paper with E. T. Schmidt [185], we accomplished this for Theorem 12.5. In my joint paper with E. T. Schmidt [189], we prove that the lattice constructed in Section 12.3 is arguesian, and that the two arguesian theorems can be combined to get the Arguesian Independence Theorem.

12.4. Discussion

159

Problems Part IV of this book deals with ETs. Modular lattices are conspicuous with their absence in that part. Here are some obvious problems to raise. Problem 17. Does every nontrivial modular lattice have a proper, modular, congruence-preserving extension? Problem 18. Let L be a nontrivial modular lattice with a finite congruence lattice and let G be a group. Does L have a modular congruence-preserving extension whose automorphism group is isomorphic to G? In light of Theorem 12.7, we can ask: Problem 19. Does the Independence Theorem hold for Complemented Modular Lattices? Problem 20. Are there any results (paralleling the results of Chapters 13 and 16) about regular, uniform, or isoform modular lattices? Nothing is really known about congruence lattices of infinite modular lattices. It is astonishing that the following problem seems open. Problem 21. Is it true that for every lattice L, there is a modular lattice M ∼ Con M ? with Con L = Can R. Freese’s Theorem 12.6 and E. T. Schmidt’s Theorem 12.7 be combined? Problem 22. Can every finite distributive lattice D be represented as the congruence lattice of a finitely generated, complemented, modular lattice L?

Chapter

13

Uniform RT

13.1.

Uniform lattices

Why couldn’t lattices be more like groups and rings?1 We want RTs by lattices whose congruences behave “as in groups and rings.” We will look at lattices that are regular and uniform. Let L be a lattice. We call a congruence relation α of L regular if any congruence class of α determines the congruence. The lattice L is regular if all congruences of L are regular. Sectionally complemented lattices are regular, so Theorem 8.5 is an RT for finite regular lattices. In this chapter, we consider a concept stronger than regularity. We call a congruence relation α of a lattice L uniform if any two congruence classes of α are of the same size (cardinality). Let us call the lattice L uniform if all congruences of L are uniform. We start with an easy result. Lemma 13.1. A uniform lattice L is always regular. Proof. Let L be a uniform lattice that is not regular. Then L has a congruence relation α and a congruence class A of α, such that the congruence β of L generated by A satisfies that β < α. So there is a congruence class B of α that is not a congruence class of β. Let C be a congruence class of β in B. It follows that |C| < |A|, contradicting that β is uniform. So L is not uniform either. 1“Why

Can’t a Woman Be More Like a Man?” My Fair Lady, lyrics by Alan Jay Lerner.

161 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_13

162

13. Uniform RT

The following result was proved in G. Gr¨atzer, E. T. Schmidt, and K. Thomsen [195]. Theorem 13.2 (Uniform RT). Every finite distributive lattice D can be represented as the congruence lattice of a finite uniform lattice L. So the lattices of Theorem 13.2 are also regular. Figure 13.1 shows the result of the construction for D = C4 .

Figure 13.1: The uniform construction for the four-element chain

13.2.

Proof-by-Picture

Let A be a finite lattice. We introduce a new lattice, A∗ . Form A × C2 . Then for every a ∈ A, we get a prime interval [(a, 0), (a, 1)]. We replace this by a copy of B2 , for all a ∈ A, to obtain A∗ . We identify a ∈ A with (a, 0) and b ∈ B2 with the copy of b in the interval [(0, 0), (0, 1)]; so we can regard A and B2 as sublattices of A∗ . Figure 13.2 shows the diagrams of B∗1 and B∗2 . It is easy to compute that A∗ is a subdirectly irreducible lattice with base congruence con((0, 0), (0, 1)), and A∗ /con((0, 0), (0, 1)) is isomorphic to A;

163

13.2. Proof-by-Picture

the isomorphism provided by a 7→ (a, 0)/con((0, 0), (0, 1)). So Con A∗ is isomorphic to Con A with a new zero added.

B1

B2

B2 B2

Figure 13.2: The lattices B∗1 and B∗2 We present the Proof-by-Picture of Theorem 13.2 in the following form. Theorem 13.3. For any finite ordered set P , there exists a finite uniform lattice L such that J(Con L) is isomorphic to P , and L satisfies the following property: (A) If α1 , α2 , . . . , αn ∈ J(Con L) are pairwise incomparable, then L contains atoms p1 , p2 , . . . , pn that generate an ideal isomorphic to Bn and satisfy αi = con(0, pi ) for all i ≤ n. Let P be a finite ordered set with n elements. If n = 1, then D ∼ = B∗1 (see Figure 13.2). Let us assume that, for all finite distributive lattices with fewer than n join-irreducible elements, there exists a lattice L satisfying Theorem 13.3. Let q be a minimal element of P and let q1 , . . . , qk (k ≥ 0) list all upper covers of q in P . Let P1 = P − {q}. By the inductive assumption, there exists a lattice L1 satisfying J(Con L)1 ∼ = P1 and condition (A). If k = 0, then we define L = B∗1 × L1 . So we assume that 1 ≤ k. The congruences of L1 corresponding to the qi -s are pairwise incomparable and therefore can be written in the form con(0, pi ) so that the pi -s generate an ideal I isomorphic to Bk . Figure 13.3 shows the inductive step. The lattices L1 and I ∗ share the ideal I, so we can form the chopped lattice K = Merge(L1 , I ∗ ), in the notation of Section 5.1. We define L = Id K. The idea is that con((0, 0), (0, 1)) is a join-irreducible congruence of L, below the con(0, pi ), for all i ≤ n, but not below any other minimal join-irreducible congruence of L1 . And there is no other new join-irreducible congruence in L, so J(Con L) ∼ = P. Of course, one has to compute that L is uniform, which is tedious. The induction condition (A) is easy to verify for L.

164

13. Uniform RT

13.3.

The lattice N (A, B)

We introduce the A∗ construction of Section 13.2 in a more general form (see Section 13.5 for an even more general lattice construction). The construction Let A and B be lattices. Let us assume that A is bounded, with bounds 0 and 1, with 0 ̸= 1. Let A− = A − {0, 1}. We introduce a new lattice construction N (A, B)—the A∗ of Section 13.2 is the special case N (A, B2 ). For u ∈ A × B, we use the notation u = (uA , uB ); the binary relation ≤× will denote the ordering on A × B, and ∧× , ∨× the meet and join in A × B, respectively. On the set A × B, we define a binary relation ≤N (A,B) (denoted by ≤N if A and B are understood) as follows: ≤N = ≤× − { (u, v) | u, v ∈ A− × B, uB ̸= vB }. We define N (A, B) = (A × B, ≤N ). Lemma 13.4. The set N (A, B) is ordered by ≤N , in fact, N (A, B) is a lattice. The join and meet in N (A, B) of ≤N -incomparable elements can be computed using the formulas: ( (1, uB ∨ vB ) if u ∨× v ∈ A− × B and uB ̸= vB ; u ∨N v = u ∨× v, otherwise. ( (0, uB ∧ vB ) if u ∧× v ∈ A− × B and uB ̸= vB ; u ∧N v = u ∧× v, otherwise.

L1

I+ I

Figure 13.3: The inductive step: the chopped lattice K

13.3. The lattice N (A, B)

165

̸ uB Proof. Since ≤× is reflexive, it follows that ≤N is reflexive, since uB = fails for all u ∈ A × B. Since ≤× is antisymmetric, so is ≤N . Let u, v, w ∈ A × B; let us assume that u ≤N v and v ≤N w. Since ≤× is transitive, we conclude that u ≤× w. So if u ≤N w fails, then u, w ∈ A− × B and uB ̸= wB . It follows that v ∈ A− × B and either uB ̸= vB or vB ̸= wB , contradicting that u ≤N v or v ≤N w. So ≤N is transitive, proving that N (A, B) is an order. We prove that N (A, B) is a lattice by verifying the join and meet formulas. By duality, we only have to verify one of them; we will prove the meet formula. Let u, v ∈ A × B be ≤N -incomparable elements, and let t be a lower bound of u and v in N (A, B). Case 1. u ∧× v is not a lower bound of both u and v in N (A, B). If u ∧× v is not a lower bound of both u and v in N (A, B), say, u ∧× v ≰N u, then u, u ∧× v ∈ A− × B and uB ̸= (u ∧× v)B (which is the same as uB ̸= vB ). / A− × B Since t ≤× u ∧× v, it follows that tB ≤ (u ∧× v)B < uB , so t ∈ (otherwise, we would have t ≰N u). We conclude that t = (0, tB ), so tB ≤ uB ∧ vB , which yields that t ≤ (0, uB ∧ vB ). So in Case 1, u ∧N v = (0, uB ∧ vB ). Case 2. u ∧× v is a lower bound of both u and v in N (A, B). If t ≰N u ∧× v, then t, u ∧× v ∈ A− × B and tB < (u ∧× v)B , so u ∈ A− × B or v ∈ A− × B, say, u ∈ A− × B. Therefore, the assumption of Case 2, namely, u ∧× v ≤N u, implies that (u ∧× v)B = uB . So t, u ∈ A− × B and tB ̸= uB , contradicting that t ≤N u. Thus t ≰N u ∧× v leads to a contradiction. We conclude that t ≤N u ∧× v. So in Case 2, u ∧N v = u ∧× v. Since the two cases correspond to the two clauses of the meet formula, this verifies the meet formula. We will use the notation: B∗ = {0} × B, B ∗ = {1} × B, and for b ∈ B, Ab = A × {b}. Note that B∗ is an ideal and B ∗ is a filter of N (A, B). Congruences Let K and L be lattices, and let α be an embedding of K into L. Given a congruence α of L, we can define the congruence α1 on K via α, that is for a, b ∈ K, a ≡ b (mod α1 ) iff αa ≡ αb (mod α). We will call α1 the restriction of α transferred via the isomorphism α to K. Let γ be a congruence relation of N = N (A, B). Using the natural isomorphisms of B into N (A, B) with images B∗ and B ∗ , we define γ ∗ and γ ∗ as the restriction of γ to B∗ and B ∗ , respectively, transferred via the natural isomorphisms to B. Let αb be the restriction of γ to Ab , for b ∈ B, transferred via the natural isomorphisms to A.

166

13. Uniform RT

Lemma 13.5. γ ∗ = γ ∗ . Proof. Indeed, if b1 ≡ b2 (mod γ ∗ ), then (0, b1 ) ≡ (0, b2 ) (mod γ). Joining both sides with (1, b1 ∧ b2 ), we obtain that (1, b1 ) ≡ (1, b2 ) (mod γ), that is, b1 ≡ b2 (mod γ ∗ ). This proves that γ ∗ ≤ γ ∗ . We prove the reverse inclusion symmetrically. It is easy to see that γ = γ ∗ = γ ∗ ∈ Con B, and { αb | b ∈ B } ⊆ Con A describe γ, but it is difficult to obtain a description of the congruences of N (A, B), in general. So we impose a very stringent condition on A. Lemma 13.6. Let A and B be lattices with |A| > 2 and |B| > 1; let A be bounded, with bounds 0 and 1. Let us further assume that A is nonseparating. Then the map sending γ ̸= 0N to its restriction to B∗ transferred to B by the natural isomorphism is a bijection between the non-0N congruences of N (A, B) and the congruences of B. Therefore, Con N (A, B) is isomorphic to Con B with a new zero added. Proof. Let γ ̸= 0N be a congruence relation of N (A, B). We start with the following statement. Claim 13.7. There are elements a1 < a2 in A and an element b1 ∈ B such that (a1 , b1 ) ≡ (a2 , b1 ) (mod γ). Proof. Assume that (u1 , v1 ) ≡ (u2 , v2 ) (mod γ) with (u1 , v1 ) 2, we can pick an a ∈ A− . Then (a, v1 ) = (a, v1 ) ∨ (0, v1 ) ≡ (a, v1 ) ∨ (0, v2 ) = (1, v2 ) (mod γ), from which we conclude that (a, v1 ) ≡ (1, v1 ) (mod γ), so the claim is verified with a1 = a, a2 = 1, and b1 = v1 . Second case. u1 < u2 . Since we have assumed that (u1 , v1 ) 1). Meeting both sides with (1, b2 ), we obtain that (1, b2 ) ≡ (0, b2 ) (mod γ), that is, Ab2 is in a single congruence class of γ. Claim 13.9. Ab is in a single congruence class of γ for each b ∈ B. Proof. Let b ∈ B. By Claim 13.8, there is an element b2 ∈ B such that Ab2 is in a single congruence class of γ, that is, (1, b2 ) ≡ (0, b2 ) (mod γ). Therefore, (1, b) = ((1, b2 ) ∨ (0, b ∨ b2 )) ∧ (1, b)

≡ ((0, b2 ) ∨ (0, b ∨ b2 )) ∧ (1, b) = (0, b)

(mod γ),

that is, Ab is in a single congruence class of γ. Now the statement of the lemma is easy to verify. It is clear that the map γ 7→ γ is one-to-one for γ ∈ Con N (A, B) − {0N }. It is also onto: given a congruence γ of B, define γ on N (A, B) by (u1 , v1 ) ≡ (u2 , v2 ) (mod γ) iff

v1 ≡ v2 (mod γ).

Then γ is in Con N (A, B) − {0N } and it maps to γ.

168

13. Uniform RT

Congruence classes Now let U be a finite lattice with an ideal V isomorphic to Bn . We identify V with the ideal (Bn )∗ = id((0, 1)) of N (B2 , Bn ) to obtain the merged chopped lattice K = Merge(U, N (B2 , Bn )) (using the notation of Section 5.1). Let m denote the generator of V = (Bn )∗ . We view Id K as the set of compatible pairs (u, y) with u ∈ U and y ∈ N (B2 , Bn ) (see Section 5.2). Lemma 13.10. Let u ∈ U . Then { y ∈ N (B2 , Bn ) | (u, y) ∈ Id K } is isomorphic to B2 . Proof. This is clear since there are exactly four elements y of N (B2 , Bn ) satisfying that u ∧ m = y ∧ m, namely, the elements of (B2 )u∧m , which form a sublattice isomorphic to B2 . Therefore, { y ∈ N (B2 , Bn ) | (u, y) ∈ Id K } is a four-element set, closed under coordinate-wise meets and joins; the statement follows. Now let us further assume that U is uniform. Lemma 13.11. Id K is uniform. Proof. A congruence Λ of Id K can be described (see Section 5.3) as a congruence vector (α, γ), where α is a congruence of U , γ is a congruence of N (B2 , Bn ), and α and γ restrict to the same congruence of V = (Bn )∗ . The trivial congruences, 0Id K (represented by (0U , 0N (B2 ,Bn ) )) and 1Id K (represented by (1U , 1N (B2 ,Bn ) )), are obviously uniform. Therefore, we only need to look at two cases. First case. Λ is represented by (α, 0). So α⌉V = 0V . Let (x, y) be an element of Id K and note that (x, y)/(α, 0) = { (t, y) ∈ Id K | t ≡ x (α) }. If t ≡ x (mod α), then t ∧ m ≡ x ∧ m (mod α), but α⌉V = 0V , so t ∧ m = x ∧ m. We conclude that (x, y)/(α, 0) = { (t, y) | t ≡ x (α) },

|(x, y)/(α, 0)| = |x/α|.

So Λ is uniform; each congruence class of Λ is of the same size as a congruence class of α. Second case. Λ is represented by (α, γ), where γ ̸= 0.

13.4. Formal proof

169

Let (x, y) be an element of Id K. Then (x, y)/(α, γ) = { (w, z) ∈ Id K | x ≡ w (α) and y ≡ z (γ) }. For a given w if (w, t1 ) and (w, t2 ) ∈ Id K, then t1 ≡ t2 (mod γ) because (B2 )w is in a single congruence class of γ by Lemma 13.6 (in particular, by Claim 13.9) and { t ∈ N (B2 , Bn ) | (w, t) ∈ Id K } = (B2 )w∧m ; thus |{ t ∈ N (B2 , Bn ) | (w, t) ∈ Id K }| = |(B2 )w∧m | = 4. We conclude that (x, y)/(α, γ) = { (w, z) ∈ Id K | x ≡ w (α), z ∈ (B2 )w∧m }, and |(x, y)/(α, γ)| = 4|x/α|. So Λ is uniform; each congruence class of Λ is four times the size of a congruence class of α.

13.4.

Formal proof

Based on the results of Section 13.3, we will prove Theorem 13.3 by induction. Let P be a finite ordered set with n elements. If n = 1, then Dn P ∼ = B1 , so there is a lattice L = B1 that satisfies Theorem 13.3. Let us assume that for all finite ordered sets with fewer than n elements, there exists a lattice L satisfying Theorem 13.3. Let q be a minimal element of P and let q1 , . . . , qk (k ≥ 0) list all upper covers of q in P . Let P1 = P − {q}. By the inductive hypothesis, there exists a lattice L1 satisfying J(Con L)1 ∼ = P1 and also satisfying condition (A). If k = 0, then Dn P ∼ = B1 × Dn P1 , and so L = B1 × L1 obviously satisfies all the requirements of the theorem. So we assume that 1 ≤ k. The congruences of L1 corresponding to the qi -s are pairwise incomparable; therefore, these congruences can be written in the form con(0, pi ) and the pi -s generate an ideal I1 isomorphic to Bk . The lattice N (B2 , Bk ) also contains an ideal (Bk )∗ isomorphic to Bk . Merging these two lattices by identifying I1 and (Bk )∗ , we get the chopped lattice K and the lattice L = Id K. The chopped lattice K is sketched in Figure 13.4. L is uniform by Lemma 13.11. Let α be a join-irreducible congruence of L. Then we can write α as con(a, b), where a is covered by b. We can assume that either a, b ∈ L1 or a, b ∈ N (B2 , Bk ). In either case, we find an atom q in L1 or in N (B2 , Bk ), so that con(a, b) = con(0, q) in L1 or in N (B2 , Bk ). Obviously, q is an atom of L and con(a, b) = con(0, q) in L.

170

13. Uniform RT

L1

I1 = (Bk )∗

N (B2 , Bk )

Figure 13.4: The chopped lattice K for the formal proof Let α1 , α2 , . . . , αt be pairwise incomparable join-irreducible congruences of L. To verify condition (A), we have to find atoms p1 , p2 , . . . , pt of L satisfying αi = con(0, pi ), for all i ≤ t, and such that p1 , p2 , . . . , pt generate an ideal of L isomorphic to Bt . Let p denote an atom in N (B2 , Bk ) − I1 ; there are two, but they generate the same congruence con(0, p). If con(0, p) is not one of α1 , α2 , . . . , αt , then clearly we can find p1 , p2 , . . . , pt in L1 as required, and p1 , p2 , . . . , pt also serves in L. If con(0, p) is one of α1 , α2 , . . . , αt , for instance, con(0, p) = αt , then let {p1 , p2 , . . . , pt−1 } be the set of atoms establishing condition (A) for the congruences α1 , α, . . . , αt−1 in L1 and therefore, in L. Then the elements p1 , p2 , . . . , pt−1 , p represent the congruences α1 , α2 , . . . , αt and generate an ideal isomorphic to Bt . Therefore, L satisfies condition (A). Finally, it is clear from this discussion that J(Con K) has exactly one more element than J(Con L)1 , namely, con(0, p), and this congruence relates to the congruences in J(Con L)1 exactly as q relates to the elements of P . Therefore, P ∼ = J(Con L).

13.5.

Discussion

Isoform lattices The next logical step is to require that any two congruence classes be not only of the same size but isomorphic as lattices. We called such lattices isoform. The following result was proved in my joint paper with E. T. Schmidt [191]. Theorem 13.12 (Isoform RT). Every finite distributive lattice D can be represented as the congruence lattice of a finite isoform lattice L.

13.5. Discussion

171

Since the congruence classes of isomorphic lattices are of the same size, Theorem 13.12 is a much stronger version of Theorem 13.2. Figure 13.1 shows that the lattice we obtained for D = C4 is not isoform. Indeed, the congruence marked by the two dotted ovals has two congruence classes of 16 elements each, but they are not isomorphic; in the top lattice, two dual atoms are join-irreducible, and in the bottom lattice, none are join-irreducible. Note that isoform lattices are not “like groups and rings.” The isoform concept is special to idempotent algebras. The N (A, B, α) construction We could prove Theorem 13.12 based on the following generalization of the N (A, B) construction. Let P = (P, ≤P ) be a finite ordered set. Then the ordering ≤P on P is the reflexive-transitive closure of ≺P , the covering relation in (P, ≤P ), in symbols: rt(≺P ) = ≤P . Now take a subset H of ≺P , and take the reflexive-transitive extension rt(H) of H. Then (P, rt(H)) is also an ordered set; we call it a pruning of P . If you think of P in terms of its diagram, then the terminology is easy to picture: We obtain the diagram of (P, rt(H)) from the diagram of P by cutting out (pruning) some edges (each representing a covering) but not deleting any elements. For instance, the lattice of Figure 13.1 is a pruning of the Boolean lattice B52 . Let L be a finite lattice. We call L discrete-transitive if for any congruence γ of L and for a < b < c in L, whenever γ is discrete on [a, b] and [b, c], then γ is discrete on [a, c]. Let A be a nontrivial finite lattice with bounds 0 and 1; let |A| > 2. Set A− = A − {0, 1}. Let B be a nontrivial finite lattice with a discrete-transitive congruence α. To prune A × B, we define the set: Prune(A, B, α) = { ((a, b1 ), (a, b2 )) | a ∈ A− , b1 ≺ b2 in B, b1 ≡ b2 (α) }. Prune(A, B, α) is a subset of ≺× , so we can define H = ≺× − Prune(A, B, α). Now we take the reflexive-transitive extension rt(H) of H. The set A × B with the ordering rt(H) is N (A, B, α). It is clear that N (A, B, 0) is the direct product A × B and N (A, B, 1) = N (A, B). We can describe the order N (A, B, α): u ≤ v in N (A, B, α) iff uA , vA ∈ A− and [uB , vB ] is α-discrete, or uA or vA ∈ / A− , and we prove that N (A, B, α) is a lattice under this ordering (see my joint paper with E. T. Schmidt [191]).

172

13. Uniform RT

Pruning seldom produces a lattice. Note the very strong condition imposed on α to make the pruned order a lattice. We use the N (A, B, α) construction to obtain the following version of Theorem 13.12 (in my joint paper with E. T. Schmidt [191]). Theorem 13.13. Every finite distributive lattice D can be represented as the congruence lattice of a finite lattice L with the following properties: (i) L is isoform. (ii) For every congruence α of L, the congruence classes of α are projective intervals. (iii) L is a finite pruned Boolean lattice. (iv) L is discrete-transitive. By Properties (i) and (ii), for every congruence relation α of L and for any two congruence classes U and V of α, the congruence classes U and V are required to be isomorphic and projective intervals, but we do not require that there be a projectivity that is also an isomorphism. We refer the reader to my joint paper with E. T. Schmidt [191], for a proof of Theorem 13.13. A stronger form of Theorem 13.13 will be proved in Chapter 16. Problems Let us start with some general problems on uniformity. Problem 23. Develop a theory of uniform lattices. Problem 24. Prove that uniformity is not a first-order property. The answer to the following question is likely to be in the negative. Problem 25. Are infinite relatively complemented lattices uniform? The congruence classes of a congruence of the lattice L in Theorem 13.13 are both isomorphic and projective. Problem 26. Is it possible to sharpen Theorem 13.13 so that the isomorphisms of the congruence classes be projectivities? Can we combine the main result of this chapter with the results of the previous chapters? In other words, Problem 27. Can every finite distributive lattice D be represented as the congruence lattice of a finite uniform (resp., isoform) lattice L with some additional property: L be semimodular, sectionally complemented, or 2-distributive, and so on?

13.5. Discussion

173

Problem 28. Can every finite distributive lattice D be represented as the congruence lattice of a uniform (resp., isoform) modular lattice L? What happens in the infinite case? Problem 29. Is there an analog of Theorem 13.2 (resp., Theorem 13.12) for infinite lattices? Let Uniform and Isoform denote the class of uniform and isoform lattices, respectively. (Recall that the function mcr was introduced in Section 8.6.) Problem 30. What is mcr(n, Uniform)? What is mcr(n, Isoform)? Can one get a good result for O(mcr(n, Uniform)) and O(mcr(n, Isoform))? Let L be a finite uniform lattice and let γ be an isomorphism between a finite distributive lattice D and Con L. Then we can introduce a function s = s(L, γ) from D to the natural numbers, as follows: Let d ∈ D; then γ(d) is a congruence of L. Since L is uniform, all congruence classes of γ(d) are of the same size. Let s(d) be the size of the congruence classes. The function s has the following obvious properties: (s1 ) s(0) = 1. (s2 ) If a < b in D, then s(a) < s(b). Problem 31. Characterize the function s. In other words, let f be a function from a finite distributive lattice D to the natural numbers that satisfies conditions (s1 ) and (s2 ) above. What additional conditions on the function f are required for f = s(L, γ), for some finite uniform lattice L and isomorphism γ : D → Con L? This problem may be too difficult to solve in its full generality. The following lists some special cases that may be easier to attack. Problem 32. Characterize the function s for some special classes of finite distributive lattices: (i) Finite Boolean lattices; (ii) Finite chains; (iii) “Small” distributive lattices.

Part IV

ETs

175

Chapter

14

Sectionally Complemented ET

14.1.

Sectionally complemented lattices

As outlined in the Introduction, Part IV deals with congruence-preserving ETs. We start with sectionally complemented lattices, as in Part III. The reader should compare the relatively small constructs and shorter proofs of Part III with the much more complex ones in this part. For instance, to represent the distributive lattice C1 + B2 as the congruence lattice of a finite sectionally complemented lattice, by Figure 8.3, we take the ideal lattice L of the chopped lattice MV , and we are done. The lattice L has 10 elements. Now start with the lattice K of Figure 14.1, a small seven-element lattice, whose congruence lattice is C1 + B2 . Even though K was chosen to minimize the size of the sectionally complemented congruence-preserving extension L, we find this L much larger than the L of the RT. In general, the extension L we construct is O(|K||K| ) in size. In my joint paper with E. T. Schmidt [184], we proved the congruence-preserving extension theorem for sectionally complemented lattices. Theorem 14.1 (Sectionally complemented ET). Every finite lattice K has a finite, sectionally complemented, congruence-preserving extension L. The proof—as most proofs in Part IV—has two major ingredients: the RT from Part III and the cubic extension from Part II, with a little bit of chopped lattices thrown in. Since every finite sectionally complemented lattice is atomistic, this result contains the first published result of an ET, namely, the theorem of M. Tischendorf [251]. 177 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_14

178

14. Sectionally Complemented ET

Theorem 14.2. Every finite lattice has a finite atomistic congruence-preserving extension.

14.2.

Proof-by-Picture

Let us find a finite, sectionally complemented, congruence-preserving extension for the lattice K of Figure 14.1. The figure also shows Con K, J(Con K), and ConM K; the congruence γ 1 splits K into two classes, marked by the dotted line (γ 2 is symmetric). Note that J(Con K) ∼ = ConM K (see Theorem 2.15). So K has three subdirectly irreducible quotients, K ∼ = K/0, K/γ 1 ∼ = C2 , and K/γ 2 ∼ = C2 . To form a cubic extension (see Section 7), we embed the subdirectly irreducible quotients into simple lattices; C2 being simple, we only have to embed K. Figure 14.2 shows the Simp K we choose (compare it with the construct of Lemma 14.3), adding a single element a to K. So the cubic extension of K is Cube K = Simp K × C2 × C2 . This lattice is not easy to draw, but we easily draw the chopped lattice M of Figure 14.3, whose ideal lattice is Cube K. In this example, the Boolean ideal B—of Theorem 7.2—is generated by the atoms a, sγ 1 , and sγ 2 . 1

γ1

γ2

γ1 γ3

K

Con K

γ2

γ1

γ2

γ1 0

0

Figure 14.1: The example lattice K

a

K

ConJ K

γ3

Simp K

Figure 14.2: A simple extension of K

ConM K

14.2. Proof-by-Picture

a

sγ 1

179

sγ 2

Figure 14.3: The chopped lattice M

Now recall Theorem 8.9: For D = Con K, there is a finite sectionally complemented lattice L0 with Con L0 ∼ = D and with a Boolean ideal B0 , the eight-element Boolean lattice with atoms pγ 1 , pγ 2 , and pγ 3 so that con(0, pγ i ) represents γ i for i = 1, 2, 3 (see Figure 14.4). We merge Simp K and L0 by identifying the zeros and the atom a of Simp K with the atom of pγ 3 of L0 and of B0 (see Figure 14.5). We view M as being a part of the merged chopped lattice by identifying sγ 1 with pγ † = pγ 2 and sγ 2 with pγ † = pγ 1 ; Figure 14.5 shows the merged chopped 1 2 lattice, the black-filled elements form M . Basically, we identify the Boolean ideal B of Cube K with the Boolean ideal B0 of L0 . We define L as the ideal lattice of this chopped lattice. By the Atom Lemma—Lemma 5.8—L is a finite sectionally complemented lattice. The diagonal embedding Diag of K into Id M of Section 7.1 also embeds K into L, so (after identifying a ∈ K with Diag(a)) the lattice L can be regarded as an extension of K.

L0 B0 pγ 1

pγ 2

pγ 3

Figure 14.4: The lattices L0 and B0

180

14. Sectionally Complemented ET

L0 B0

Simp K Figure 14.5: Merging L0 and M What about the congruences of L? Clearly, merging L0 with a simple lattice, we obtain a chopped lattice that is a congruence-preserving extension of L0 , so L is a congruence-preserving extension of L0 and Con L ∼ = Con L0 ∼ = Con K ∼ = D. A congruence of L can be described as a compatible congruence vector (α, γ) of the chopped lattice, where α is a congruence of Cube K and γ is a congruence of L0 agreeing on {o, a} = {0, pγ 3 }. The congruence α is determined on B and the congruence γ is determined on B0 . In L, we dentified B and B0 , so it follows that α = γ. Moreover, (α, α) is a compatible congruence vector, because α acts on B by collapsing pγ and 0 iff γ ≤ α, whereas α acts on B0 by collapsing pγ † and 0 iff γ † ≰ α; also, γ ≤ α is equivalent to γ † ≰ α. Since K has the congruence extension property in Cube K, we conclude that γ : α 7→ con(α) (in L), for α ∈ Con K, is a bijection. On the other hand, Con L ∼ = Con L0 ∼ = Con K, so γ is an isomorphism between Con K and Con L, that is, L is a congruence-preserving extension of K.

14.3.

Easy extensions

To start out in the proof, we construct sectionally complemented cubic extensions. So we need the following statement. Lemma 14.3. For every finite lattice K, there is a finite, simple, sectionally complemented {0, 1}-extension L. Proof. We proceed by induction. Let 0 and 1 be the zero and unit element of L, respectively. If |K| ≤ 2, then we can take L = K. So we can assume that |K| > 2. We distinguish two cases. Case 1. 1 is join-irreducible. Let a be the unique dual atom of K. By the induction hypothesis, id(a) has a finite, simple, sectionally complemented

14.3. Easy extensions

181

{0, 1}-extension L′ . We define L = L′ ∪ {1, u, v}, where u and v are common complements to all elements of L′ − {0} and are complementary, that is, u ∧ x = 0 and u ∨ x = 1 for all x ∈ (L′ ∪ {v}) − {0}; and symmetrically for v. It is easy to see that L satisfies all the requirements. Case 2. 1 is join-reducible. So there are elements x1 , x2 ∈ K − {1} satisfying x1 ∨ x2 = 1. Let N (K) = K − { x | x = 0 or x is an atom }. For every a ∈ N (K), we adjoin an atom pa < a so that if a ̸= b, then pa ̸= pb . We define on Simp K = K ∪ { pa | a ∈ N (K) } (a disjoint union) an ordering by the following rules. (α) K is a suborder of Simp K; (β) if a ∈ N (K) and x ∈ K, then x < pa

iff

x = 0,

pa < x

iff

a ≤ x;

pa ≤ p b

iff

a = b.

(γ) if a and b ∈ N (K), then

It is easy to check that Simp K is a lattice. Obviously, Simp K is a finite lattice; 0 and 1 are the zero and unit elements of Simp K, respectively. Simp K is an extension of K. To show that Simp K is sectionally complemented, take 0 < u < v in Simp K. If u ∈ K, then v ∈ N (K) and pv is a sectional complement of u in [0, v]. / K, then u = pa , for some a ∈ N (K), and v ∈ K satisfies a ≤ v. If u ∈ If a = v, then any x satisfying 0 < x < a is a sectional complement of u in [0, v]; and there is such an x because a ∈ N (K). If a < v, then pv is a sectional complement of u in [0, v]. Finally, L is simple. Indeed, let α > 0 be a congruence of K. We verify that there is an a > 0 in L, such that a ≡ 0 (mod α). Indeed, since α > 0, there are u, v ∈ K such that u < v and u ≡ v (mod α). If u = 0, then a = v satisfies the requirements. If 0 < u, then v ∈ N (K), so there is an element pv ∈ L. If u ∈ K, then u ≡ v (mod α) implies that pv ≡ 0 (mod α), so we can take a = pv . If u ∈ Simp K − K, then u = px for some x ∈ N (K). Obviously, v ∈ K and x ≤ v. Since x ∈ N (K), there is an a ∈ K satisfying 0 < a < x. Now u ≡ v (mod α) implies that px ≡ x (mod α) and so a ≡ 0 (mod α).

182

14. Sectionally Complemented ET

Using the congruence a ≡ 0 (mod α), we conclude that p1 ≡ 1 (mod α). Meeting with x1 , we obtain that x1 ≡ 0 (mod α), and similarly x2 ≡ 0 (mod α). Joining these two congruences, we obtain that 0 ≡ 1 (mod α), that is, α = 1. We should point out that since a finite partition lattice is simple and sectionally complemented by O. Ore [229], Lemma 14.3 follows from the following very deep result of P. Pudl´ ak and J. T˚ uma [231]: Every finite lattice can be embedded into a finite partition lattice. It can also be derived from a much earlier result of R. P. Dilworth (first published in P. Crawley and R. P. Dilworth [24]): Every finite lattice can be embedded into a finite geometric lattice.

14.4.

Formal proof

Let K be a finite lattice. In this section we construct a finite, sectionally complemented, congruence-preserving extension L of K, as required by Theorem 14.1. Using Lemma 14.3, for every γ ∈ ConM K, we select a finite, sectionally complemented, simple extension Kγ = Simp K/γ of K/γ, with an atom, sγ , and zero, 0γ . Let D = Con K, and let L0 be a lattice whose existence was stated in Theorem 8.9 for the finite distributive lattice D. For a congruence α of K, let α0 denote the congruence of L0 determined by the set { qγ | γ ≤ α }, that is, _ α0 = ( con(qγ , 0L0 ) | γ ≤ α );

obviously, α 7→ α0 sets up an isomorphism Con K → Con L0 . Now we inductively construct the lattice L of Theorem 14.1. Let γ 1 , . . . , γ n list the meet-irreducible congruences of K. We apply the Atom Lemma (see Lemma 5.8), with A = L0 , B = Simp K/γ 1 , and the atoms pγ † of L0 1 and sγ 1 of Simp K/γ 1 , to obtain the chopped lattice M1 and L1 = Id M1 . From Corollary 5.7, it follows that L1 is a congruence-preserving extension of L0 with the same Boolean ideal B0 . Moreover, L1 is finite and sectionally complemented by the Atom Lemma. So we can apply again the Atom Lemma with A = L1 , B = Simp K/γ 2 , and the atoms pγ † of L1 and sγ 2 of Simp K/γ 2 , 2 obtaining the chopped lattice M2 and L2 = Id M2 . In n steps, we construct Mn and L = Ln = Id Mn . It is clear that L is a finite sectionally complemented lattice. It is also evident that L is a congruence-preserving extension of L0 ; hence, Con L ∼ = D. So to complete the proof of Theorem 14.1, we have to show that K is congruence reflecting in L. In the next step, we need the following statement.

14.4. Formal proof

183

Lemma 14.4. Let A, B, C be pairwise disjoint finite lattices with zeros 0A , 0B , ′ ′ , sC ∈ C. We construct some 0C , and atoms sA ∈ A, sB , sB ∈ B, sB ̸= sB chopped lattices by merging. (i) Let N1 be the chopped lattice obtained by forming the disjoint union of A and B and identifying 0A with 0B and sA with sB . (ii) Let N be the chopped lattice obtained by forming the disjoint union of ′ with sC . A, B, C and identifying 0A with 0B and 0C , sA with sB , and sB (iii) We consider the ideal lattice Id N1 an extension of N1 . So it has atoms ′ sA = sB and sB . Let N2 be the chopped lattice that is the disjoint union of Id N1 and C with the zeros identified. We also identify the atom ′ ∈ Id N1 with the atom sC ∈ C. sB Then Id N2 is isomorphic to Id N . Proof. The elements of Id N1 are compatible pairs (a, b) ∈ A × B. The elements of Id N2 are compatible pairs ((a, b), c); note that (a, b) ∈ Id N1 iff a ∧ sA = b ∧ sB and note also that (a, b) ∧ s′B = b ∧ s′B . On the other hand, the elements of Id N are compatible triples (a, b, c) ∈ A × B × C. It should now be obvious that ((a, b), c) 7→ (a, b, c) is the required isomorphism Id N2 → Id N . Another view of L is the following. We form the chopped lattice M1 ; instead of proceeding to L1 = Id M1 , let us now form by merging the chopped lattice M2′ from M1 = M1′ and Simp K/γ 2 by identifying the zeros of M1′ and Simp K/γ 2 and the atom pγ † of M1′ with the atom sγ 2 of Simp K/γ 2 . 2 Observe that M2′ is the union of L0 , Simp K/γ 1 , and Simp K/γ 2 . Also, Simp K/γ 1 ∩ Simp K/γ 2 = {0}. By Lemma 14.4, Id M2′ ∼ = L2 . Proceeding thus, we obtain the chopped lattice Mn′ = L0 ∪ Simp K/γ 1 ∪ Simp K/γ 2 ∪ · · · ∪ Simp K/γ n , where, for all 1 ≤ i < j ≤ n, we have Simp K/γ i ∩ Simp K/γ j = {0}, and L∼ = Id Mn′ , again by Lemma 14.4. So the chopped sublattice Simp K/γ 1 ∪ Simp K/γ 2 ∪ · · · ∪ Simp K/γ n of Mn′ is just the lattices Simp K/γ 1 , Simp K/γ 2 , . . . , Simp K/γ n merged by their zeros, and, therefore, Id(Simp K/γ 1 ∪ Simp K/γ 2 ∪ · · · ∪ Simp K/γ n ) ∼ = Cube K. Thus we can view L as the ideal lattice of the chopped lattice obtained by gluing together L0 and Cube K over the Boolean lattices B0 in L0 and B in Cube K. A congruence α of L is then determined by a congruence αL0

184

14. Sectionally Complemented ET

on L0 and a congruence αCube K on Cube K that agree on the Boolean lattice. Since Cube K is a congruence-preserving extension of K ∼ = Diag(K), it follows now that L is a congruence-preserving extension of K, concluding the proof of Theorem 14.1.

14.5.

Discussion

Problem 33. By Lemma 14.3, every finite lattice K has a finite, simple, sectionally complemented extension L. What is the minimum size of such an L? Problem 34. What is the size of the lattice L we construct for Theorem 14.1? What is the minimum size of a lattice L satisfying Theorem 14.1? Can we do better than Theorem 14.1? Problem 35. Is there a natural subclass S of SecComp for which Theorem 14.1 holds (that is, every finite lattice K has a finite, sectionally complemented, congruence-preserving extension L ∈ S)? For infinite lattices, almost nothing is known, as illustrated by the following question. Problem 36. Does every nontrivial sectionally complemented lattice have a proper, sectionally complemented, congruence-preserving extension?

Chapter

15

Semimodular ET

15.1.

Semimodular lattices

In Part III, finite sectionally complemented lattices are followed by finite semimodular lattices. It is the same in this part. In my joint paper with E. T. Schmidt [187], we follow the same path as in Chapter 14. First, we have to find a semimodular cubic extension; luckily, the article P. Crawley and R. P. Dilworth [24] comes to the rescue. Then we utilize the RT for semimodular lattices (Theorem 10.1) to order the congruences. These two lattices have to be connected since there is no obvious way of amalgamating the two; we do this with the “conduit” construction. Thus we obtain: Theorem 15.1 (Semimodular ET). Every finite lattice K has a congruencepreserving embedding into a finite semimodular lattice L.

15.2.

Proof-by-Picture

Let us start with a finite lattice K of Figure 15.1 (the dual of N6 , introduced in Section 7.1 (see Figure 7.1)), with Con K = {0, γ, 1}, the three-element chain. The congruence γ splits K into two classes, marked in Figure 15.1 with the dashed line. First, we define a semimodular extension Cube K. To do this, we have to specify the function Simp. Since ConM K = {γ, 0}, we have to define Simp K/γ and Simp K/0. The subdirectly irreducible quotient, K/γ, is isomorphic to C2 , a simple semimodular lattice, so we can define Simp C2 = C2 . We embed the lattice K/0 ∼ = K into a simple semimodular lattice Simp K 185 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_15

186

15. Semimodular ET

Figure 15.1: The lattice K and a simple semimodular extension Simp K as in Figure 15.1. Therefore, the semimodular cubic extension, Cube K, is defined as C2 × Simp K. We thereby obtain a semimodular congruencereflecting extension Cube K (see Figure 15.2). This figure also shows a “sketch” of the diagram, which we use in Figure 15.5 to visualize the gluings. The black-filled elements of Cube K in Figure 15.2 form the lattice K, identified with Diag(K). To see how the congruences of K extend to Cube K, we only have to describe the extension Cube(γ) of γ, which splits Cube K into two classes as shown by the dashed line in Figure 15.2. The shaded area in Figure 15.2 is (one option for) the filter F (of Corollary 7.3), which is a congruence-preserving sublattice of Cube K. We color the filter F with two colors, p1 and p2 . In Cube K, we have con(p1 ) = Cube(γ). Also, Con(Cube K) = {0, con(p1 ), con(p2 ), 1}.

p2

Cube(γ)

p1

F p1

p2

p1 p2 p2 p1

F

Cube K

Figure 15.2: The semimodular cubic extension Cube K with sketch

15.2. Proof-by-Picture

p1 p2

S

p1 p2

p1 p1

p1 N

187

p2 p1

N

Figure 15.3: The lattice S with sketch We now have to construct a congruence-determining extension of Cube K in which con(p1 ) = 1, so that con(p1 ) > con(p2 ). We borrow from Section 10.2 the semimodular lattice S (see Figure 15.3). We utilize the special circumstances to make S a little smaller. The lattice S contains an ideal N that is a chain of length 2; the two prime intervals colored by p1 and p2 . The ideal N is a congruence-determining sublattice of S. This figure also shows a sketch of the diagram. Note that in S, we have con(p1 ) > con(p2 ). We have to patch Cube K and S together, respecting the colors. We utilize for this the modular lattice M —which we call the ‘conduit’—shown in Figure 15.4 with a sketch. The lattice M has an ideal B, which is a congruencep2

p2

J

p1 J

p1 p1 p2

B

p2

B

p1 p2 p 2 p1

p1 Figure 15.4: The conduit with sketch preserving sublattice isomorphic to F and a filter J, which is a congruencepreserving sublattice isomorphic to N . We color M as shown in Figure 15.4, so that there are color preserving isomorphisms γ : B → F and δ : J → N . We glue Cube K and the conduit over the filter F and ideal B with respect to γ; the resulting lattice T has filter J. We glue T and S over the filter J and ideal N with respect to δ to obtain the lattice L. The two gluings are sketched in Figure 15.5. The semimodularity of L follows by two applications of Lemma 2.22. This semimodular congruence-preserving extension L of the six-element

188

15. Semimodular ET

S S S

p2 p1

N p1

B

p2 p1

p2

N

J

p1 p2 p 2 p1

p1

M

p1

J

p1 p2 p 2 p1

T

p2

p1

p1

p1 p2 p 2 p1

F

p2

p2

p2

T Cube K

Cube K

Cube K

Figure 15.5: Two gluings: before and after; sketches lattice K has 57 elements; compare this with the size of the lattice from Section 10.2: S has 24 elements.

15.3.

The conduit

In this section we give the general construction of the conduit, a modular lattice, which we constructed in a special case in Section 15.2 (see Figure 15.4). Let n ≥ 1 be a natural number, and for every i, with 1Q≤ i ≤ n, we take a copy Ti of M3 , with atoms pi , qi , and ri . We form Mn3 = ( Ti | 1 ≤ i ≤ n ) with zero 0, and regard Ti as an ideal of Mn3 , so pi , qi , and ri are atoms of Mn3 for 1 ≤ i ≤ n. n by { pi | 1 ≤ i ≤ n }; let 1B = W Let B be the sublattice of M3 generated ( pi | 1 ≤ i ≤ n ). Obviously, B is a 2n -element Boolean lattice, an ideal of Mn3 . Define _ q i = ( qj | 1 ≤ j ≤ i ),

for 1 ≤ i ≤ n, and set E = { q i | 0 ≤ i ≤ n }, where q 0 is the zero of Mn3 . Obviously, E is a maximal chain (of length n) in id(q n ) of Mn3 and b ∧ e = 0 for all b ∈ B and e ∈ E.

15.4. The construction

189

By Theorem 2.11, the sublattice A of Mn3 generated by B and E is isomorphic to B × E under the isomorphism b ∨ e 7→ (b, e). Let b ∈ B and let i, with 0 ≤ i < n, satisfy the condition pi+1 ≰ b. Define the element of Mn3 : r(b, i) = b ∨ q i ∨ ri+1 , and the subset M of Mn3 :

M = A ∪ { r(b, i) | b ∈ B, 0 ≤ i < n, pi+1 ≰ b }. The set M is a sublattice of Mn3 ; hence, M is a modular lattice. The lattice M contains B and E as ideals. Let α be a congruence of B. Let αE be the congruence on E satisfying q i ≡ q i+1 (mod αE ) in E iff pi+1 ≡ 0 (mod α) in B for 0 ≤ i < n. Then α × αE is a congruence on B × E. We extend this to a congruence αM of M as follows: let r(b, i) be defined (that is, b ∈ B and pi+1 ≰ b); if b ≡ b ∨ pi+1 (mod α) in B, then r(b, i) ∈ (b ∨ q i )/αM , otherwise, {r(b, i)} is a congruence class. The map α 7→ αM is an isomorphism between Con B and Con M . In fact, M is a congruence-preserving extension of both B and E. Let J = { 1B ∨ e | e ∈ E }. Obviously, E and J are isomorphic chains and J is a filter of M , a congruence-preserving sublattice of M . To summarize. Lemma 15.2. For each n ≥ 1, M3n has sublattices B, J, and M satisfying the following conditions: (i) B is an ideal of M and it is isomorphic to the Boolean lattice Bn . (ii) J is a filter of M and length J = n. (iii) M is a congruence-preserving extension of both B and J. Note that Con M is a Boolean lattice and Con M ∼ = Con B ∼ = Con J.

15.4.

The construction

We are given a finite lattice K. To prove the Semimodular ET (Theorem 15.1), we have to construct a finite, semimodular, congruence-preserving extension L of K. We glue L together from three lattices, sketched in Figure 15.5. The first lattice is S of Section 10.3 (denoted there by L) (see Figure 10.4) with the ideal N , defined at the end of Section 10.3, which is a chain consisting of the gray-filled elements, a congruence-determining sublattice of S.

190

15. Semimodular ET

Let D = Con K; in Section 10.3, we construct a finite semimodular lattice S with D ∼ = Con S. The construction of S in Section 10.3 starts out by representing D as Dn P for P = J(D). By Theorem 2.15, the ordered sets J(D) and M(D) are isomorphic, so we can start the construction of S with P = M(D). Accordingly, we color N with P = M(D); there is a one-to-one correspondence between the prime intervals of N and the elements of P , the meet-irreducible congruences of K. The second lattice is the conduit lattice M of Section 15.3, with the ideal B (which is Boolean) and the filter J (which is a chain), constructed so that J and N are isomorphic. This (unique) isomorphism allows us to color J so that the isomorphism preserves colors. The coloring of J extends to a coloring of all of M , in particular, to a coloring of B. Both B and J are congruence-preserving sublattices of M . The third lattice is a cubic extension, so we have to define the function Simp. For a finite lattice A, let Simp A be a finite partition lattice extending A, ak a semimodular lattice by Lemma 2.23. Such an extension exists by P. Pudl´ and J. T˚ uma [231]—alternatively, use the more accessible result of P. Crawley and R. P. Dilworth [24] (Theorem 14.1) to obtain a simple semimodular extension. In the cubic extension Cube K, we have the filter F generated by dual atom tγ ∈ Simp Kγ , for γ ∈ ConM K = P , by Corollary 7.3. We color [tγ , 1F ] by γ, for γ ∈ ConM K = P , and thereby color all of F . Observe that B and F are isomorphic finite Boolean lattices, both are colored by P , so there is an isomorphism (unique!) α between B and F that preserves colors. First gluing. We glue M and S together over J and N to obtain the lattice T . The lattice T is also colored by P and T has an ideal B, a congruence-determining sublattice. The lattice T is obtained by gluing together two semimodular lattices, hence, it is semimodular by Lemma 2.22. Second gluing. We have the color preserving isomorphism between the Boolean ideal B of the lattice T of the first step and the filter F in Cube K; we do the second gluing of T and B with Cube K and F with respect to this isomorphism α, to obtain the lattice L. The lattice L is obtained by gluing together two semimodular lattices, hence, it is semimodular by Lemma 2.22.

15.5.

Formal proof

To prove the Semimodular ET (Theorem 15.1), we have to verify that L is a finite, semimodular, congruence-preserving extension of K. We already know that L is finite and semimodular. We view K as a sublattice of Cube K (as in Section 7.1), so K is a sublattice of L. Let κ be a congruence of K. We have to show that κ has one and only one extension to L.

15.6. Rectangular ET

191

Let γ ∈ ConM K be the color of the prime interval p of L. Let qγ be the unique prime interval in N of color γ. We claim that con(p) = con(qγ ). To prove this claim, let p be a prime interval in L. Then p is a prime interval in S, or M , or Cube K. So we have three cases to consider. First, if p is a prime interval in S, this follows from the fact that N is a congruence-determining sublattice of S (see Lemma 3.20 and the last paragraph of Section 10.3). Second, if p is a prime interval in M , this follows from the fact that J is a congruence-preserving sublattice of M , by Lemma 15.2. Third, if p is a prime interval in Cube K, then there is a prime interval p1 in F with con(p) = con(p1 ), since F is a congruence-determining sublattice of Cube K. Because of the gluing, the prime interval p1 can be regarded as a prime interval of M , and now the claim follows from the second case. This completes the proof of the claim. It follows from this claim that a congruence α of L can be described by the set Σ of prime intervals of N collapsed by α, or equivalently, by the set of colors of the prime intervals: C = { γ | pγ ∈ Σ }. Now by the construction of S (see Section 10.3), the set C is a down set of P = ConM K. And conversely, if C is a down set of P , then it defines a congruence αC on S, as computed in Section 10.3. This congruence αC spreads uniquely to M because J is a congruence-preserving sublattice of M . Finally, from the description of Cube 0 in Section 7.2, we see that the congruence uniquely spreads to Cube K. So there is a one-to-one correspondence between congruences of L and down sets of P . For κ ∈ Con K, its extension Cube(κ) to Cube K corresponds to the down set { γ | κ ≰ γ }. So every congruence of K extends. But since there are no congruences other than those that correspond to down sets of P , the extension must be unique. This completes the proof of the Semimodular ET (Theorem 15.1).

15.6.

Rectangular ET

In Section 11.3, Theorem 11.4 states that Every finite distributive lattice D can be represented as the congruence lattice of a rectangular lattice K with the property that all congruences are principal. This is a trivial consequence of the following ET. Theorem 15.3. Let L be a planar semimodular lattice. Then L has an extension K satisfying the following conditions: (i) K is a rectangular lattice; (ii) K is a congruence-preserving extension of L; (iii) K is a cover-preserving extension of L;

192

15. Semimodular ET

(iv) Every congruence relation of K is principal. Observe that we only have to prove Theorem 15.3. Indeed, let Theorem 15.3 hold and let D be a finite distributive lattice. By the Rectangular RT (Theorem 11.1), there is a planar semimodular lattice K1 whose congruence lattice is isomorphic to D. By Theorem 15.3, the lattice K1 has a congruence-preserving extension K in which every congruence relation is principal. This lattice K satisfies the conditions of Theorem 11.4. Now we proceed with the construction for the planar semimodular lattice L for Theorem 15.3. Step 1. We apply Theorem 28.1 to get a rectangular, cover-preserving, and congruence-preserving extension K1 of K. Step 2. Let D = Clr (K1 ). We form the lattice D2 , and insert eyes into the b (see Figure 15.6). covering squares of the main diagonal, obtaining the lattice D b Now we do a rectangular gluing of K1 and D, obtaining the lattice K2 . Here is the crucial statement.

Lemma 15.4. K2 is an extension of L; it is rectangular, cover-preserving, and congruence-preserving. For every join-irreducible congruence α of L, there is a prime interval pα of C = Cll (K2 ) such that con(pα ) in K2 is the unique extension of α to K2 .

Proof. Indeed, by Theorem 4.20, there is a prime interval qlα of Cll (K1 ) or a prime interval qrα of Clr (K1 ) such that con(qlα ) or con(qrα ) in K1 is the unique extension of α to K1 . If we have qlα ⊆ Cll (K1 ) ⊆ C, set qlα = pα and we are done. If we have qrα ⊆ Clr (K1 ) with con(qrα ) the unique extension of α to K1 , b ⊆ Cll (K2 ) such that in D, b the then in K2 there is a unique q ⊆ Cll (D) r prime intervals qα and q are connected by an M3 on the main diagonal (see Figure 15.7 for an illustration). Now clearly, we can set pα = q. Note: Lemma 15.4 is a variant of several published results. Maybe G. Cz´edli [27, Lemma 7.2] is its closest predecessor. Step 3. For the final step of the construction, take the chain C = Cll (K2 ) and a congruence α of L. We can view α as a congruence of K2 and let α = γ 1 ∨· · ·∨γ n be a join-decomposition of α into join-irreducible congruences. By Theorem 4.20 and (P), we can associate with each γ i , for i = 1, . . . , n, a prime interval pi of C so that con(pi ) = γ i . b as follows. We construct a rectangular lattice C[α] (a cousin of D) Let Cn+1 = {0 < 1 < · · · < n}. Take the direct product C × Cn+1 . We think of this direct product as consisting of n columns, column 1 (the bottom one), . . . , column n (the top one). In column i, for 1 ≤ i ≤ n, we take the covering square whose upper right edge is perspective to pi and insert an eye. In the cover-preserving M3 sublattice

15.6. Rectangular ET

K1

b D

Figure 15.6: Step 2 of construction b D

qrα

q

b Figure 15.7: Step 2 of construction: a detail of the lattice D γ1

p1 γ3

γ1 γ2

b γ3

p3

γ1 a

γ2

γ2

p2

Figure 15.8: Step 3 of construction: the lattice C[α]

193

194

15. Semimodular ET

we obtain, every prime interval p satisfies con(p) = γ i . Figure 15.8 has an illustration with n = 3; a prime interval p is labelled with γ i if con(p) = γ i . Let b denote the top element of the M3 we constructed for pn , clearly, we have b ∈ Cur (C[α]). Take the element a ∈ Cll (C[α]) so that the interval [a, b] is a chain of length n. Then the n prime intervals q1 , . . . , qn of [a, b] satisfy con(q1 ) = γ 1 , . . . , con(qn ) = γ n , so con([a, b]) = α, finding that in the lattice C[α], the congruence α is principal. We identify C with Cur (C[α]); note that this is a “congruence-preserving” isomorphism: for a prime interval p of C, the image p′ of p in Cur (C[α]) satisfies con(p) = con(p′ ). Now we form the rectangular gluing of C[α] with filter C and K2 with the ideal C to obtain the lattice K2 [α]. Obviously, K2 [α] is a rectangular lattice, it is a cover-preserving, congruence-preserving extension of K2 and, therefore, of L. It is easy to see that Cll (K2 [α]) is still (congruence) isomorphic to C; for a rigorous treatment see the Corner Lemma and the Eye Lemma in G. Cz´edli [27] as they are used in the proof of Lemma 7.2 in [27]. We can continue this expansion with all the congruences of L. In the last step, we get the lattice K3 = K, satisfying all the conditions of Theorem 15.3.

15.7.

Discussion

Using P. Crawley and R. P. Dilworth [24], we get a very large simple semimodular lattice extending the lattice K. Of course, from P. Pudl´ ak and J. T˚ uma [231], we get an even larger lattice. So it is natural to ask: Problem 37. Let f (n) be the smallest integer with the property that every lattice K of n elements can be embedded into a simple semimodular lattice of size f (n). Give an asymptotic estimate of f (n). Problem 38. What is the size of the lattice L we construct for Theorem 15.1 expressed in terms of f (n) of Problem 37? What is the minimum size of a lattice L satisfying Theorem 15.1? Can we do better than Theorem 15.1? Problem 39. Is there a natural subclass S of the class of semimodular lattices for which Theorem 15.1 holds (that is, every finite lattice K has a finite, semimodular, congruence-preserving extension L ∈ S)? How about planar lattices? Problem 40. Does every planar lattice K have a congruence-preserving embedding into a finite, planar, semimodular lattice L?

Chapter

16

Isoform ET

16.1.

Isoform lattices

In Chapter 13, we proved the RT for finite uniform lattices. In Section 13.5, we discussed the RT for finite isoform lattices, but we did not prove it, for a reason I am about to explain. In G. Gr¨atzer, R. W. Quackenbush, and E. T. Schmidt [167], we proved the following. Theorem 16.1 (Isoform ET). Every finite lattice K has a congruencepreserving extension to a finite isoform lattice L. This problem was raised for uniform lattices as Problem 1 in G. Gr¨atzer, E. T. Schmidt, and K. Thomsen [195] and for isoform lattices as Problem 2 in my joint paper with E. T. Schmidt [191]. See also Problem 9 in my joint paper with E. T. Schmidt [192]. The proof of this theorem breaks the earlier pattern: we start constructing a congruence-preserving extension by a cubic extension and then we utilize the RT. For isoform lattices, we again start with a cubic extension, but we do not know how /to proceed utilizing the RT—this is why it was not important to have the RT proved in Chapter 13. Instead, we use “pruning.” Special cases of pruning are the A∗ construction of Section 13.2, the N (A, B) construction of Section 13.3, and the N (A, B, α) construction of Section 13.5.

16.2.

Proof-by-Picture

In this section we construct the lattice L of the Isoform ET (Theorem 16.1) for the lattice K = N6 of Figure 16.1. 195 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_16

196

16. Isoform ET

1 γ d v

a b

c

0 Figure 16.1: The lattices K and S0 As always, when we construct a congruence-preserving extension, we start with a cubic extension. To construct the cubic extension of K, we have to define the function Simp on the subdirectly irreducible quotients. The lattice K has only one nontrivial congruence γ, splitting K into two classes, indicated by the dashed line; it has two meet-irreducible congruences, γ and 0. We use the notation: Simp K/0 = S0 and Simp K/γ = Sγ . The choice for S0 is easy; add a single element v as in Figure 16.1 (the black-filled element). The lattice K/γ is simple, so we can choose Sγ = C2 . 1 10

1γ 0 Figure 16.2: The cubic extension S of K Figure 16.2 shows the cubic extension S = S0 × Sγ of K; the black-filled elements represent K. On the diagram, we labeled four elements: the 0 and 1 of S, and the unit elements of the two direct factors: 10 and 1γ . The congruences of S are in one-to-one correspondence with subsets of {10 , 1γ }, and Cube γ (the extension of γ to S) corresponds to {10 }, that is, Cube γ = con(0, 10 ). So we have to “kill” the congruence con(0, 1γ ); in other words, we have to make sure that 1γ ≡ 0 implies that 10 ≡ 0.

16.2. Proof-by-Picture

197

In Chapters 14 and 15, such a congruence-forcing was accomplished by a suitable extension of S utilizing the appropriate representation theorems. The inspiration for the isoform result comes from the RT: the use of pruning, although the RT itself is not needed. We construct L from S by pruning (deleting) a single edge, the dashed edge of Figure 16.2; the resulting ordered set L is depicted in Figure 16.3.

Figure 16.3: The lattice L It has to be checked that L is a lattice (pruning an edge may not define a lattice). Then we have to see that 1γ ≡ 0 implies that 10 ≡ 0 in L, so L has only one nontrivial congruence and it splits L into two classes as indicated by the dashed line. The two classes are both isomorphic to S0 , so L is isoform. Computing this example in detail, one comes to the conclusion that everything works out because the element v in S0 is so special. We axiomatize the important property of v as follows. For a finite lattice A, let us call an element v a separator if 0 ≺ v ≺ 1. A lattice A with a separator is called separable; we also allow C2 as a separable lattice. The general construction starts with a finite ordered set P (which plays the role of Con Q M K), and a family Sp , for p ∈ P , of separable lattices. We form the set S = ( Sp | p ∈ P ). In each Sp ̸= C2 , with p ∈ P , we fix a separator vp . Let us describe the edges we delete from S. Let a = (ap )p∈P ≺ b = (bp )p∈P in S. Then there is exactly one q ∈ P with aq ≺ bq in Sq , and ap = bp for all p ̸= q. We delete the edge a, b if there is a p < q such that Sp has a separating element vp and vp = ap = bp . The pruned diagram defines the ordered set L. Here is a small example. The ordered set is P = {p, q} with p < q and the two lattices are Sp = C3 and Sq = C2 . Figure 16.4 illustrates the construction; the edge of S = C3 × C2 to be pruned is dashed.

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(1, 1)

(v, 1) (v, 0)

S

(1, 1)

p 0 for some p < p′ with p′ ∈ D. But this cannot happen either since D is a down set.

16.4. The congruences

16.4.

205

The congruences

Now let each Sp be simple and let each Sp ̸= C2 ; then S has a congruence lattice isomorphic to B such that uQ ∈ B is associated with the smallest congruence on S collapsing 0 and uQ , denoted αQ . Clearly, S is isoform. We are going to prove that the congruences of L are just the congruences αD of S for D a down set of P . As an illustration, in Figure 16.6, the lattice has exactly one nontrivial congruence, “projecting” L onto Mq . The five congruence classes have five elements each, labeled x, with x ∈ {0, v, a, b, 1}. For p ∈ P , the lattice Sp is assumed to be simple, so Sp − {0, 1, v} = ̸ ∅. Therefore, we can select wp ∈ Sp − {0, 1, v}; we will write w for wp if the index is understood. We define wq by (wq )q = w and otherwise, (wq )p = 0. Lemma 16.7. Let α be a congruence of L, and let 0 ≡ up (mod α) for some p ∈ P . If q < p, then 0 ≡ uq (mod α). Proof. From 0 ≡ up (mod α) and q < p, the join formula yields the congruence vq = 0 ∨ vq ≡ up ∨ vq = u{p,q}

(mod α).

Similarly, we get the congruence 0 = wq ∧ vq ≡ wq ∧ u{p,q} = wq (mod α). Lemma 16.5 implies that [0, uq ] is a simple sublattice of L, so we obtain that 0 ≡ uq (mod α). Lemma 16.8. Let α be a congruence of L and D a down set of P . Let a, b ∈ L with ap = 0 and bp = 1, for p ∈ D, and ap = bp for p ∈ / D. Then a ≤ b and 0 ≡ uD (mod α) iff a ≡ b (mod α). Proof. Clearly, b = a ∨S uD = a ∨ uD , by Lemma 16.5. Thus a ≤ b. Let 0 ≡ uD (mod α). Then a = 0 ∨ a ≡ uD ∨ a = b

(mod α).

Conversely, if a ≡ b (mod α), then 0 = a ∧ uD ≡ b ∧ uD = uD

(mod α),

proving the lemma. Lemma 16.9. Let α be a congruence of L and let a ≺ b in L with a ≡ b (mod α). Then there is a unique p ∈ P with ap ≺ bp and aq = bq for p = ̸ q; moreover, 0 ≡ up (mod α). Proof. If a ≺ b, then from Lemma 16.3, there is a unique p ∈ P with ap ≺ bp and aq = bq otherwise. Moreover, if aq = bq = v, then q ≰ p. If bp = v so that ap = 0, then we get (a ∨ wp )p = w < 1 = (b ∨ wp )p , and otherwise (a ∨ wp )q = aq = bq = (b ∨ wp )q , since aq = bq = v implies that q ≰ p, so that (wp )r = 0, for all q < r. This yields the covering a ∨ wp ≺ b ∨ wp and the

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congruence a ∨ wp ≡ b ∨ wp (mod α). Thus we may assume that bp ̸= v. Then 0 ≤ a ∧ up < b ∧ up ≤ up and a ∧ up ≡ b ∧ up (mod α). By Lemma 16.5, the interval [0, up ] is a simple sublattice of L. Hence, 0 ≡ up (mod α). Theorem 16.10. Let α be a congruence of L; then there is a down set D of P such that α = αD . Conversely, let D be a down set of P ; then αD is a congruence of L. Proof. Let α be a congruence of L. Define Prec(α) to be the set of all (a, b) ∈ L2 such that a ≡ b (mod α) and a ≺ b. Define D to be the set of all p ∈ P such that ap ≺ bp for some (a, b) ∈ Prec(α). By Lemma 16.9, 0 ≡ up (mod By Lemma 16.7, D is a down set of P . By Lemma 16.5, W p α) for all p ∈ D. ( u | p ∈ D ) = uD ; hence, 0 ≡ uD (mod α). By Lemma 16.8, αD ⊆ α. On the other hand, since L is finite, α is the smallest equivalence relation containing Prec(α), so we must have α ⊆ αD . Thus α = αD . Conversely, let D be a down set of P . A typical αD -class is of the form [a, b], where for p ∈ D, ap = 0 and bp = 1, and otherwise aq = bq . By Lemma 16.6, a ≤ a∨uD = a∨S uD = b. Let c ∈ L; it suffices to show that a∨c ≡ b∨c (mod α) and a ∧ c ≡ b ∧ c (mod α). For the join, we may take a ≤ c, so that a ∨ c = c ≤ b ∨ c. Let us assume that cp < (b∨c)p ; we must show that p ∈ D. If cp < bp ∨cp , then from ap ≤ cp , we conclude that ap ̸= bp , and so p ∈ D. Otherwise, cp = bp ∨ cp < (b ∨ c)p so that cp = v = bp ∨ cp < 1 = (b ∨ c)p . Again, if ap ̸= bp , then p ∈ D; so assume that ap = bp ≤ cp = v. As D is a down set, we must have aq = bq for all q ≥ p. Now, bp ∨ cp = v < 1 = (b ∨ c)p implies that one of Cases (1∨ )–(3∨ ) holds for {b, c} for some p′ ≥ p. Note that bp ≤ cp implies that Case (3∨ ) cannot occur. If Case (1∨ ) occurs, then ap′ = bp′ = cp′ = v; since p′ is a {b, c}-fork, this forces p′ to be an {a, c}-fork, contradicting that a ≤ c. If Case (2∨ ) occurs, then ap′ = bp′ ≰ cp′ , again contradicting a ≤ c. This proves that a ∨ c ≡ b ∨ c (mod α). For the meet, we proceed similarly.

16.5.

The isoform property

As in Section 16.4, we assume that each Sp is simple and Sp ̸= C2 . Theorem 16.11. L is isoform. Proof. Let D be a down set of P ; in S every block of αD can be written in the form [a, a ∨S uD ] for some a ≤S (uD )′ . Hence, if b, c ∈ [a, a ∨S uD ] and bp ̸= cp , then p ∈ D. The map φa : [0, uD ] → [a, a ∨S uD ] defined by φa (x) = x ∨S uD is an isomorphism of sublattices of S. By Lemma 16.6, a ≤ a ∨ uD = a ∨S uD ; thus the blocks of αD in L are also of the form [a, a ∨S uD ] for some a ≤ (uD )′ (which is the same as a ≤S (uD )′ ). We will show that φa is also an isomorphism of L-sublattices; this will prove that L

16.6. Discussion

207

is isoform. We will make use without further mention that for b, c ∈ L, (b ∨S c)p = bp ∨ cp for all p ∈ P . Obviously, φa is a bijection. Let us assume that 0 ≤ b ≺ c ≤ uD . Then by Lemma 16.6 and the fact that φa is an S-isomorphism, it follows that b ∨S a ≺S c ∨S a. If we do not have b ∨S a ≺ c ∨S a, then by Lemma 16.3, there are p < p′ with bp ∨ ap = v = cp ∨ ap and bp′ ∨ ap′ ≺ cp′ ∨ ap′ . From the latter, we get p′ ∈ D; since D is a down set and p < p′ , we have p ∈ D. But then a ≤ (uD )′ implies that ap = ap′ = 0; therefore, v = bp ∨ ap = bp and v = cp ∨ ap = cp , so that bp = cp = v. Since b ≺ c, we have that bp′′ = cp′′ for all p < p′′ . For p′′ = p′ , this yields bp′ ∨ ap′ = bp′ = cp′ = cp′ ∨ ap′ , contradicting that bp′ ∨ ap′ ≺ cp ∨ ap . Thus b ∨S a ≺ c ∨S a. Now assume that a ≤ b′ ≺ c′ ≤ a ∨S uD ; hence, a ≤S b′ ≺S c′ ≤S a ∨S uD . Since γa is an S-isomorphism, there are unique 0 ≤S b ≺S c ≤S uD such that b′ = b ∨S a and c′ = c ∨S a. Hence, there is a unique p ∈ D such that bp ≺ cp , bp ∨ ap ≺ cp ∨ ap , and bq ∨ aq = cq ∨ aq for p ̸= q. We need to show that b ≤ c. So let br = v = cr ; then r ̸= p. We have to verify that bq = cq , for r < q. This can fail only if r < p. But then r < p ∈ D, a down set; hence, r ∈ D. Since a ≤ (uD )′ , it follows that ar = ap = 0. But then br ∨ ar = br = v = cr = cr ∨ ar , whereas bp ∨ ap ≺ cp ∨ ap , contradicting that b′ ≤ c′ . Thus b ≺ c. We conclude that L is isoform. This completes the proof of the Isoform ET (Theorem 16.1).

16.6.

Discussion

16.6.1

Variants

Regular lattices Sectionally complemented lattices are regular, so we already have a congruence-preserving ET for regular lattices in Chapter 14. We would like to point out that Isoform ET (Theorem 16.1) also contains this result. Lemma 16.12. Every finite isoform lattice is regular. Proof. Let L be an isoform lattice. Let α and β be congruences sharing the block A. Then A is also a block of α ∧ β; in other words, we can assume that α ≤ β. Let B be an arbitrary α block. Since α ≤ β, there is a (unique) β block B containing B. Since L is isoform, A ∼ = B and A ∼ = B; by finiteness, B = B, that is, α = β. Corollary 16.13. The lattice L of Theorem 16.1 is regular.

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Permutable congruences Since sectionally complemented lattices have permutable congruences, the ET for congruences permutable lattices follows from the result in Chapter 14. We would like to point out that Theorem 16.1 contains this result. This statement follows from Lemma 16.14. The lattice L of Isoform ET is congruence permutable. Proof. This is obvious since the congruences of L are congruences of S, which is a direct product of simple lattices. Problem 2 of G. Gr¨atzer, R. W. Quackenbush, and E. T. Schmidt [167] raised the question of whether this is typical. K. Kaarli [219] provided an affirmative solution. Theorem 16.15. Every finite uniform lattice is congruence permutable. Kaarli also described a countably infinite isoform lattice that is not congruence permutable. Similarly, if a finite algebra has a majority term, then any two congruences whose blocks are two-element sets permute (see G. Cz´edli [25], [26], and [49]). Deterministic isoform lattices Let L be an isoform lattice, let γ be a congruence of L, and let Aγ be a congruence class of γ. Then any congruence class of γ is isomorphic to Aγ . However, if γ and δ are congruences of L, it may happen that Aγ and Aδ as lattices are isomorphic. Let us call an isoform lattice L deterministic if this cannot happen; in other words if γ ̸= δ are congruences of L, then Aγ and Aδ are not isomorphic. Lemma 16.16. The lattice L of Theorem 16.1 can be constructed to be deterministic. Proof. The size |Aγ | is the product of the |Sγ | for γ ∈ ConM K. Since we can easily construct the Sγ -s so that all |Sγ |-s are distinct primes; the statement follows. Naturally isoform lattices Let L be a finite lattice. Let us call a congruence relation α of L naturally isoform if any two congruence classes of α are naturally isomorphic (as lattices) in the following sense: if a ∈ L is the smallest element of the class a/α, then x 7→ x ∨ a is an isomorphism between 0/α and a/α. Let us call the lattice L naturally isoform if all congruences of L are naturally isoform. The lattice L of Theorem 16.1 is not naturally isoform. There is a good reason for this.

16.6. Discussion

209

Theorem 16.17. Let L be a finite lattice. If L is naturally isoform, then Con L is Boolean. Proof. If L is simple, the statement is trivial. Let α be a nontrivial congruence relation of L. Let a be the largest element of the α-class 0/α and let b be the smallest element of the α-class 1/α. Then obviously, a ∨ b = 1. If a ∧ b > 0, then b ∨ 0 = b ∨ (a ∧ b) (= b); therefore, x 7→ x ∨ b is not an isomorphism between 0/α and b/α. Thus a ∧ b = 0. We prove that L ∼ = id(a) × id(b). For c ∈ L, we get c ∧ b ≡ c ∧ 1 = c (mod α), and so c ∧ b ∈ c/α. If d < c ∧ b is the smallest element of c/α, then the natural isomorphism between 0/α and d/α would force that c ∧ b = d ∨ x, for some 0 < x ≤ a, contradicting that (c∧b)∧a = 0. The natural isomorphism between 0/α and c/α yields that c = (c ∧ b) ∨ x for some unique x ≤ a. Since x ≤ c, clearly, x ≤ c ∧ a. Therefore, (c ∧ b) ∨ (c ∧ a) ≤ c = (c ∧ b) ∨ x ≤ (c ∧ b) ∨ (c ∧ a), that is, c = (c ∧ b) ∨ (c ∧ a). This proves that L ∼ = id(a) × id(b). Thus α is the kernel of the projection of L onto id(a) and so has a complement, the kernel of the projection of L onto id(b). We conclude that Con L is Boolean. A generalized construction We defined an element v of a finite lattice A to be a separator if 0 ≺ v ≺ 1. Instead, we can have the more general definition: a separator v is a doubly irreducible element. Then there is a unique v∗ ≺ v and a unique v ∗ ≻ v, and these take over the role of 0 and 1 in the proofs. Let us restate Theorem 16.4. Theorem 16.18. L is a lattice. Let a = (ap )p∈P , b = (bp )p∈P ∈ L. Then  ∗  if ap ∨ bp = v and, for some p′ ≥ p, v     (1∨ ) p′ is an {a, b}-fork, or  (a ∨ b)p = (2∨ ) bp ≤ ap and bp′ ≰ ap′ , or    (3∨ ) ap ≤ bp and ap′ ≰ bp′ ;    a ∨ b , otherwise; p

and

(a ∧ b)p =

16.6.2

Problems

  v∗      

p

      a ∧ b , p p

if ap ∧ bp = v and for some p′ ≥ p, (1∧ ) p′ is an {a, b}-fork, or (2∧ ) bp ≥ ap and bp′ ≱ ap′ , or (3∧ ) ap ≥ bp and ap′ ≱ bp′ ; otherwise.

Problem 41. Develop a theory of (deterministic) isoform lattices.

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16. Isoform ET

Problem 42. Let f (n) be the smallest integer with the property that every lattice K of n elements has a congruence-preserving extension into an isoform lattice of size f (n). Estimate f . Let L be a finite lattice. A congruence α of L is term isoform if, for every a ∈ L, there is a unary term function p(x) that is an isomorphism between 0/α and a/α. The lattice L is term isoform if all congruences are term isoform. Problem 43. Does every finite lattice have a congruence-preserving extension to a term isoform finite lattice? Problem 44. Can we carry out the construction for Theorem 16.1 in case Con K is finite (rather than K is finite)? Problem 45. Is there an analog of Theorem 16.1 for infinite lattices? By Lemma 16.16, every finite lattice has a congruence-preserving extension to a deterministic lattice. Can this result be extended to infinite lattices? Problem 46. Does every lattice have a congruence-preserving extension to a deterministic lattice? 16.6.3

The Congruence Lattice and the Automorphism Group

As discussed in Section 12.4, Problem II.18 of GLT-[99] was solved for finite lattices by V. A. Baranski˘ı [15], [16] and A. Urquhart [254]. Theorem 16.19 (Baranski˘ı–Urquhart Theorem). Let D be a nontrivial finite distributive lattice and let G be a finite group. Then there exists a finite lattice L such that the congruence lattice of L is isomorphic to D and the automorphism group of L is isomorphic to G. This is an RT; the ET variant was published in my joint paper with E. T. Schmidt [180]. Theorem 16.20 (Strong Independence Theorem). Let K be a nontrivial finite lattice and let G be a finite group. Then K has a congruence-preserving extension L whose automorphism group is isomorphic to G. There is a stronger form of this theorem; to state it, we need the analog of the congruence-preserving extension concept for automorphisms. Let K be a lattice. The lattice L is an automorphism-preserving extension of K if L is an extension and every automorphism of K has exactly one extension to L; moreover, every automorphism of L is the extension of an automorphism of K. Of course, then the automorphism group of K is isomorphic to the automorphism group of L. Now we state the stronger version of Theorem 16.20.

16.6. Discussion

211

Theorem 16.21 (Strong Independence Theorem, Full Version). Let KC and KA be nontrivial finite lattices. Let us assume that KC ∩ KA = {0}. Then there exists a lattice L such that the following conditions hold. (1) L is a finite, atomistic, {0}-extension of both KA and KC . (2) L is a congruence-preserving extension of KC . (3) L is an automorphism-preserving extension of KA . Of course, then the congruence lattice of L is isomorphic to the congruence lattice of KC , and the automorphism group of L is isomorphic to the automorphism group of KA . 16.6.4

More problems

We do not have an independence theorem for lattices with any special properties. Let us mention two possibilities. Problem 47. Is there an independence theorem for sectionally complemented lattices? Problem 48. Is there an independence theorem for semimodular lattices? The combinatorial questions are all open. Problem 49. What is the size of the lattice L we construct for Theorem 16.20? What is the minimum size of a lattice L (as a function of |K| and |G|) satisfying Theorem 16.20? Problem 50. What is the size of the lattice L we construct for Theorem 16.21? What is the minimum size of a lattice L—as a function of |KC | and |KA |— satisfying Theorem 16.21? Theorem 16.20 can also be proved for general lattices (see my joint paper with F. Wehrung [203]). Unfortunately, this proof is about 30 pages long and complex. Problem 51. Is there a different proof for the Independence Theorem for general lattices? Notice how the topic is shifting. In Chapters 14 and 15, we seek congruence-preserving extensions into first-order definable classes: sectionally complemented and semimodular lattices. In Chapter 16, we deal with the properties of congruences: uniform and isoform congruences. In this chapter, for the first time, we go “outside” the lattice, looking at automorphism groups. Naturally, the techniques that worked so well no longer apply, in particular, we do not use cubic extensions. Chopped lattices, however, continue to play a crucial role.

Chapter

17

Magic Wands

17.1.

Constructing congruence lattices

Bijective maps A typical way of constructing an algebra B with a given congruence lattice C is to construct an algebra A with a much larger congruence lattice and then “collapsing” sufficiently many pairs of congruences of the form con(a, b) and con(c, d) in B, so that the congruence lattice “shrinks” to C. To do this we need a “magic wand” that will make a ≡ b equivalent to c ≡ d. Such a magic wand may be a pair of partial operations f and g such that f (a) = c, f (b) = d, and g(c) = a, g(d) = b. This is the start of the Congruence Lattice Characterization Theorem of Universal Algebras in my joint paper with E. T. Schmidt (see [173], and also my books UA-[97], UA2-[100]). If we want to construct a lattice L with a given congruence lattice C, how do we turn the action of the “magic wand” into lattice operations? To construct a simple modular lattice, E. T. Schmidt [238] started with the rational interval K = [0, 1] and by a “magic wand” he required that all [a, b] (0 ≤ a < b ≤ 1) satisfy that a ≡ b be equivalent to 0 ≡ 1. The action of the magic wand was realized with the gadget M3 [K] (see Section 6.3). In Sections 17.2 and 17.3 we prove that one can apply the magic wand to arbitrary lattices with zero (see my joint paper with E. T. Schmidt [190]). If we are considering a “magic wand” (an extension L of the lattice K) that will realize that a ≡ b is equivalent to c ≡ d in the lattice L, we immediately notice that we have to say something about the intervals [a, b] and [c, d]. For instance, if a ≡ b (mod α ∨ β) in K, for congruences α and β of L, then c ≡ d (mod α ∨ β), for congruences α = conL (α) and β = conL (β) of L, 213 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_17

214

17. Magic Wands

therefore, the sequence in [a, b] that forces a ≡ b (mod α ∨ β) in K (see Theorem 1.3) must somehow be mapped in L to a sequence in [c, d]L to force c ≡ d (mod α ∨ β). So for lattices, “magic wands” must act on intervals, not on pairs of elements. To set up “magic wands”—as (convex) extensions—for lattices formally, let K be a bounded lattice, let [a, b] and [c, d] be intervals of K, and let δ : [a, b] → [c, d] be an isomorphism between these two intervals. We can consider δ and δ −1 as partial unary operations. Let us call a congruence α of K a δ-congruence iff α satisfies the Substitution Property with respect to the partial unary operations δ and δ −1 . Let Kδ denote the partial algebra obtained from K by adding the partial operations δ and δ −1 . Thus a congruence relation of the partial algebra Kδ is the same as a δ-congruence of the lattice K. We call an extension L of the lattice K a δ-congruence-preserving extension of K if a congruence of K extends to L iff it is a δ-congruence and every δ-congruence of K has exactly one extension to L. As a special case, we get the well-known concept of a congruence-preserving extension (in case, δ is trivial). Let us call δ (resp., δ −1 ) a term function iff there is a unary term function p(x) such that δx = p(x) for all x ∈ [a, b] (resp., δ −1 x = p(x) for all x ∈ [c, d]). In my joint paper with E. T. Schmidt [190], we proved the following result. Theorem 17.1. Let K be a bounded lattice, let [a, b] and [c, d] be intervals of K, and let φ : [a, b] → [c, d] be an isomorphism between these two intervals. Then K has a φ-congruence-preserving convex extension into a bounded lattice L such that both φ and φ−1 are term functions in L. In particular, the congruence lattice of the partial algebra Kφ is isomorphic to the congruence lattice of the bounded lattice L. So the lattice L constructed in this result is the magic wand for φ. Surjective maps G. Gr¨atzer, M. Greenberg, and E. T. Schmidt [132] generalized to surjective maps the approach of Section 17.1. Let K be a bounded lattice, let [a, b] and [c, d] be intervals of K, let φ be a homomorphism of [a, b] onto [c, d]. We can consider φ as a partial unary operation. Let us call a congruence α of K a φ-congruence iff α satisfies the Substitution Property with respect to the partial unary operation φ, that is, x ≡ y (mod α) implies that φx ≡ φy (mod α) for all x, y ∈ [a, b]. Let Kφ denote the partial algebra obtained from K by adding the partial operation φ. Thus a congruence relation of the partial algebra Kφ is the same as a φ-congruence of the lattice K. We call an extension L of the lattice K a φ-congruence-preserving extension of K if a congruence of K extends to L iff it is a φ-congruence and every

17.2. Proof-by-Picture for bijective maps

215

φ-congruence of K has exactly one extension to L. If L is a φ-congruencepreserving extension of K, then the congruence lattice of the partial algebra Kφ is isomorphic to the congruence lattice of the lattice L. Theorem 17.1 generalizes as follows. Theorem 17.2. Let K be a bounded lattice, let [a, b] and [c, d] be intervals of K, and let φ be a homomorphism of [a, b] onto [c, d]. Then K has a φ-congruence-preserving convex extension into a bounded lattice L such that φ is a term function in L. In particular, the congruence lattice of the partial algebra Kφ is isomorphic to the congruence lattice of the bounded lattice L. If K is finite, then L can be constructed as a finite lattice. So L realizes the “magic wand:” a ≡ b implies that c ≡ d in L. To illustrate the use of such a result, we reprove the Basic RT. Let D be a finite distributive lattice and P = J(D). Let K be the Boolean lattice with atoms ai , i ∈ P . For i > j in P , define φi,j as the only homomorphism of [0, ai ] onto [0, aj ]. Apply Theorem 17.2 to these homomorphisms one at a time to construct a lattice L whose congruence lattice is isomorphic to D.

17.2.

Proof-by-Picture for bijective maps

In this section we present the Proof-by-Picture of Theorem 17.1; we build the lattice L. We use two gadgets: the Boolean triple construction of Chapter 6 and its special intervals of Section 6.4. Let the bounded lattice K, the intervals [a, b] and [c, d], and let the isomorphism φ : [a, b] → [c, d] be given as in Theorem 17.1. We start the construction with four gadgets, which are assumed to be pairwise disjoint (see the top left of Figure 17.1). (i) A = M3 [K]. Let {0A , p1 , p2 , p3 , 1A } be the spanning M3 in A—as in Lemma 6.1.(iv). (ii) B = M3 [a, b], with the spanning M3 , {0B , q1 , q2 , q3 , 1B }. (iii) The lattice Ka,b —as in Lemma 6.19. (iv) The lattice Kc,d —as in Lemma 6.19. Some notation: We describe an element of one of the four gadgets as a triple (x, y, z) ∈ K 3 . Note that a triple (x, y, z) may belong to two or more gadgets. For a triple (x, y, z), we use subscripts to indicate which gadget it belongs to: (x, y, z)A (x, y, z)a,b (x, y, z)c,d (x, y, z)B

for for for for

(x, y, z) (x, y, z) (x, y, z) (x, y, z)

as as as as

an an an an

element element element element

of of of of

A; Ka,b ; Kc,d ; B.

216

17. Magic Wands A = M3 [K]

A = M3 [K]

V p2

p1

p1

p3 ϕ2

Ka,b

Fa,b

Fa,b Ic,d Ia,b

Kc, d

q2

D

I ψ

Ic,d

Kc, d

Ia,b

ϕ1 q1

p3 Fc,d

Ka,b

Fc,d

p2

q1

U

q2

q3

q3

B = M3 [a, b]

B = M3 [a, b] The four gadgets.

Two gluings.

A = M3 [K]

L p2

p1 Fa,b

Ka,b

p3 Fc,d

Ia,b q1

Ic,d q2

Kc, d q3

B = M3 [a, b] The final gluing.

Figure 17.1: Constructing the lattice L We start the construction by gluing together B and Ka,b to obtain the lattice U (see the top right of Figure 17.1). In B, we use the filter fil(q1 ) = { (1, x, x) | a ≤ x ≤ b }, whereas, in Ka,b we utilize the ideal Ia,b = { (0, x, 0) | a ≤ x ≤ b }, and we consider the isomorphism φ1 : (1, x, x)B 7→ (0, x, 0)a,b ,

x ∈ [a, b],

between the filter fil(q1 ) of B and the ideal Ia,b of Ka,b to glue B and Ka,b together to obtain the lattice U . Similarly, we glue Kc,d and A over the filter Fc,d = { (x, d, x ∧ d) | x ∈ K }

17.2. Proof-by-Picture for bijective maps

217

of Kc,d and the ideal id(p3 ) = { (0, 0, x) | x ∈ K } of A, with respect to the isomorphism φ2 : (x, d, x ∧ d)c,d 7→ (0, 0, x)A ,

x ∈ K,

to obtain the lattice V . In U , we define the filter F = [q3 , 1B ] ∪ Fa,b , which is the union of [q3 , 1B ] and Fa,b , with the unit of [q3 , 1B ] identified with the zero of Fa,b . In V , we define the ideal I = Ic,d ∪ [0A , p1 ], which is the union of Ic,d and [0A , p1 ], with the unit of Ic,d identified with the zero of [0A , p1 ]. Next we set up an isomorphism φ : F → I. Since [q3 , 1B ] = { (x, x, b) | a ≤ x ≤ b } and Ic,d = { (0, x, 0) | c ≤ x ≤ d }, we define φ on [q3 , 1B ] by φ : (x, x, b)B 7→ (0, φx, 0)c,d ,

x ∈ [a, b],

where φ : [a, b] → [c, d] is the isomorphism given in Theorem 17.1. This “twist” by φ—hidden in the last displayed formula—is the pivotal idea of the proof. We define φ on Fa,b by φ : (x, b, x ∧ b)a,b 7→ (x, 0, 0)A ,

x ∈ K.

It is clear that φ : F → I is well defined and that it is an isomorphism. Finally, we construct the lattice L of Theorem 17.1 by gluing U over I with V over F with respect to the isomorphism φ : F → I (see the bottom half of Figure 17.1). By Lemma 6.1.(iii), the map x 7→ (x, 0, 0)A is an isomorphism between the lattice K and the ideal id(p1 ) of the lattice A; this gives us a convex embedding of K into A. We identify K with its image, and regard the lattice K as a convex sublattice of the lattice A and therefore of the lattice L. So the lattice L is a convex extension of the lattice K. We have completed the construction of the bounded lattice L of Theorem 17.1.

218

17. Magic Wands

x6

L p1

p2

d b

p3

x7 x

c

x4

x2

(b, a, a) (a, a, b)

a x1

x5

(d, c, c) (c, c, d)

(c, c, c)

(a, a, a) q2

q1

q3

x3

Figure 17.2: The seven steps Now we have to show that φ and φ−1 are term functions in the lattice L. We define p(x) = ((((((x∧(a, a, b)a,b )∨(b, a, a)a,b )∧q2 )∨(d, c, c)c,d )∧(c, c, d)c,d )∨p2 )∧p1 . Figure 17.2 shows that φx = p(x) for all x ∈ [a, b]. The proof for φ−1 is similar.

17.3.

Verification for bijective maps

We now verify that the lattice extension L of K satisfies the conditions stated in Theorem 17.1. First, we will describe the congruences of L. We need some notation.

17.3. Verification for bijective maps

219

For a congruence α of K, αA denotes the congruence α3 restricted to A; αa,b denotes the congruence α3 restricted to Ka,b ; αc,d denotes the congruence α3 restricted to Kc,d ; αB denotes the congruence α3 restricted to B. Let us start with U . The lattices B and Ka,b are glued together over fil(q1 ) and Ia,b with the isomorphism φ1 : (1, x, x)B 7→ (0, x, 0)a,b , and obviously (1, x, x) ≡ (1, y, y) (mod α3 ) iff (0, x, 0) ≡ (0, y, 0) (mod α3 ). Hence, by r r Lemma 2.9, the congruences of U are of the form αB ◦ αa,b (the symbol ◦ is defined in Section 2.4). The lattice B is a congruence-preserving extension of fil(q1 ) (formed in B) by Theorem 6.3; therefore, by Lemma 3.16 and Corollary 6.18, the lattice U is a congruence-preserving extension of Ka,b , which, in turn, is a congruence-preserving extension of Fa,b (∼ = K). So U is a congruence-preserving extension ∼ K). Similarly, V is a congruence-preserving extension of A, which, of Fa,b (= in turn, is a congruence-preserving extension of [0A , p1 ] = K, so V is a congruence-preserving extension of [0A , p1 ] = K. We glue U and V together over F and I over φ; equivalently, we identify the filter Fa,b = { (x, b, x ∧ b) | x ∈ K } ⊆ U of U with [0A , p1 ] = { (x, 0, 0) | x ∈ K } ⊆ A, and note that for any congruence α of K, (x, b, x∧b) ≡ (y, b, y ∧b)

(mod α3 )

iff

(x, 0, 0) ≡ (y, 0, 0) (mod α3 ),

so α3 restricted to Fa,b is mapped by φ to α3 restricted to [0A , p1 ]—and we identify the filter [q3 , 1B ] = { (x, x, b) | a ≤ x ≤ b } of B with the ideal Ic,d = { (0, x, 0) | c ≤ x ≤ d } of Kc,d by identifying (x, x, b) with (0, φx, 0) for x ∈ [a, b]. So in the lattice L, (x, x, 1) ≡ (y, y, 1)

iff

(0, φx, 0) ≡ (0, φy, 0);

translating this back to K, we obtain that x ≡ y (mod α) iff φx ≡ φy (mod α). This condition is equivalent to the statement that α has the Substitution Property with respect to the partial unary operations φ and φ−1 . This proves, on the one hand, that if α extends to the lattice L, then α is a φ-congruence, and, on the other hand, that a φ-congruence α extends

220

17. Magic Wands

uniquely to L, that is, the lattice L is a φ-congruence-preserving extension of the lattice K. Second, we have to show that φ and φ−1 are term functions in the lattice L. We define p(x) = ((((((x∧(a, a, b)a,b )∨(b, a, a)a,b )∧q2 )∨(d, c, c)c,d )∧(c, c, d)c,d )∨p2 )∧p1 . For x ∈ [a, b], we want to compute p(x). There are seven steps in the computation of p(x) (see Figure 17.2): x = (x, 0, 0)A = (x, b, x)a,b x1 = x ∧ (a, a, b) x2 = x1 ∨ (b, a, a) x 3 = x 2 ∧ q2 x4 = x3 ∨ (d, c, c) x5 = x4 ∧ (c, c, d) x 6 = x 5 ∨ p2 x 7 = x 6 ∧ p1

(in A and in Ka,b ), (computed in Ka,b ), (computed in Ka,b ), (computed in U ), (computed in L), (computed in Kc,d ), (computed in V ), (computed in A).

Our goal is to prove that x7 = φx. By the definition of φ, when gluing U and V together, we identify x = (x, 0, 0) ∈ A with (x, b, x ∧ b) = (x, b, x) ∈ Ka,b , so x = (x, b, x) ∈ Ka,b . Therefore, x1 = (x, b, x) ∧ (a, a, b) = (a, a, x), computed in Ka,b . We compute x2 completely within Ka,b , utilizing Lemma 6.1: x2 = x1 ∨ (b, a, a) = (a, a, x) ∨ (b, a, a) = (b, a, x) = (b, x, x). The element x3 = x2 ∧ q2 is computed in U , which we obtained by gluing Ka,b and B together with respect to the isomorphism φ1 : (1B , x, x)B 7→ (0, x, 0)a,b ,

x ∈ [a, b],

between the filter fil(q1 ) of B and the ideal Ia,b of Ka,b . So x3 is computed in two steps. First, in Ka,b : x2 ∧ va,b = (b, x, x) ∧ (0, b, 0) = (0, x, 0). The image of (0, x, 0) under φ−1 1 is (b, x, x), so in B: x3 = (b, x, x) ∧ q2 = (b, x, x) ∧ (a, b, a) = (a, x, a). Now comes the crucial step. To compute x4 = x3 ∨(d, c, c), we first compute in B: x3 ∨ q3 = (a, x, a) ∨ (a, a, b) = (a, x, b) = (x, x, b).

17.4. 2/3-Boolean triples

221

Take the image of (x, x, b) under φ and join it with (d, c, c) in Kc,d : x4 = φ(x, x, b) ∨ (d, c, c) = (0, φx, 0) ∨ (d, c, c) = (d, φx, c) = (d, φx, φx). So x5 = x4 ∧ (c, c, d) = (d, φx, φx) ∧ (c, c, d) = (c, c, φx),

computed in Kc,d . To compute x6 = x5 ∨ p2 , we note that

x5 ∨ vc,d = (c, c, φx) ∨ (0, d, 0) = (c, d, φx) = (φx, d, φx). Then we take the image of (φx, d, φx)c,d under φ2 and join it with p2 in A: x6 = φ2 (φx, d, φx) ∨ p2 = (0, 0, φx) ∨ (0, 1, 0) = (0, 1, φx) = (φx, 1, φx). Finally, in A, x7 = x6 ∧ p1 = (φx, 1, φx) ∧ (1, 0, 0) = (φx, 0, 0), and φx is identified with (φx, 0, 0), so x7 = φx, as claimed. The proof for φ−1 is similar, using the term function q(y) = ((((((y ∨p2 )∧(c, c, d)c,d )∨(d, c, c)c,d )∧q2 )∨(b, a, a)a,b )∧(a, a, b)a,b )∨a. This completes the proof of Theorem 17.1. As you can see, the four sublattices of L isomorphic to M3 play a crucial role in the proof; the elements forming these M3 -s are gray-filled in Figure 17.2.

17.4.

2/3-Boolean triples

To prove the surjective result, we need all the gadgets from the bijective case, and a new one. The new gadget will be described in Lemma 17.14. This new gadget is based on the 2/3-Boolean triple construction, which we proceed to describe. Let N6 = {o, p, q1 , q2 , r, i} denote the six-element lattice depicted in Figure 17.3, with o the zero, i the unit element, p, q1 , q2 the atoms, satisfying the relations q1 ∨ q2 = r, p ∧ q1 = p ∧ q2 = p ∧ r = o, and p ∨ q1 = p ∨ q2 = p ∨ r = i. Following G. Gr¨ atzer, M. Greenberg, and E. T. Schmidt [132], for a bounded lattice P , we introduce 2/3-Boolean triples: the element (x, y, z) ∈ P 3 is called a 2/3-Boolean triple iff y = (y ∨ x) ∧ (y ∨ z), z = (z ∨ x) ∧ (z ∨ y). We denote by N6 [P ] the set of all 2/3-Boolean triples ordered componentwise. We prove that N6 [P ] is a lattice and describe the congruences of this lattice.

222

17. Magic Wands

i

i

xr

xp p

r q1

p

q2

r

xp

q1 xq1

q2 xq2

o

o

Figure 17.3: Illustrating N6 and N6 [P ] Lemma 17.3. The subset N6 [P ] of P 3 is a closure system; let (x, y, z) denote the closure of (x, y, z) ∈ P 3 and call it the 2/3-Boolean closure of (x, y, z). Then (x, y, z) = (x, (y ∨ x) ∧ (y ∨ z), (z ∨ x) ∧ (z ∨ y)). Proof. To simplify the notation, let y = (y ∨x)∧(y ∨z) and z = (z ∨x)∧(z ∨y). We have to verify that (x, y, z) = (x, y, z) is the closure of (x, y, z). The triple (x, y, z) is 2/3-Boolean closed. Indeed, y ≤ y, so x ∨ y = x ∨ y. Also, z ≤ z, so y ∨ z = y ∨ z. Therefore, (y ∨ x) ∧ (y ∨ z) = y, verifying the first half of the definition of 2/3-Boolean triples; the second half is proved similarly. So (x, y, z) ≤ (x, y, z) ∈ N6 [P ]. To prove that N6 [P ] is a closure system in P 3 and that (x, y, z) is the closure of (x, y, z), it suffices to verify that if (x1 , y1 , z1 ) ∈ N6 [P ] and (x, y, z) ≤ (x1 , y1 , z1 ), then (x, y, z) ≤ (x1 , y1 , z1 ), which is obvious. Corollary 17.4. N6 [P ] is a lattice. Meet is componentwise and join is the closure of the componentwise join. Moreover, N6 [P ] has a spanning N6 (see Figure 17.3), with the elements o = (0, 0, 0), p = (1, 0, 0), q1 = (0, 1, 0), q2 = (0, 0, 1), r = (0, 1, 1), and i = (1, 1, 1). Lemma 17.5. (i) The interval [o, p] of N6 [P ] is isomorphic to P under the isomorphism xp = (x, 0, 0) 7→ x,

x ∈ P.

17.4. 2/3-Boolean triples

223

(ii) The interval [o, q1 ] of N6 [P ] is isomorphic to P under the isomorphism xq1 = (0, x, 0) 7→ x,

x ∈ P.

(iii) The interval [o, q2 ] of N6 [P ] is isomorphic to P under the isomorphism x ∈ P.

xq2 = (0, 0, x) 7→ x,

(iv) The interval [p, i] of N6 [P ] is isomorphic to P under the isomorphism xp = (1, x, x) 7→ x,

x ∈ P.

(v) The interval [o, r] of N6 [P ] is isomorphic to P 2 under the isomorphism (0, x, y) 7→ (x, y),

x, y ∈ P.

(vi) The interval [r, i] of N6 [P ] is isomorphic to P under the isomorphism xr = (x, 1, 1) 7→ x,

x ∈ P.

Proof. By trivial computation. For instance, to prove (iv), observe that (1, x, y) is closed iff x = y. For the five isomorphic copies of P in N6 [P ], we use the notation: Pp = [o, p],

Pq1 = [o, q1 ],

Pq2 = [o, q2 ],

with zero o and unit elements, p, q1 , q2 , respectively, and P p = [p, i],

P r = [r, i],

with unit i and zero elements, p, r, respectively. We describe the congruence structure of N6 [P ] based on the following decomposition of elements. Lemma 17.6. Every α ∈ N6 [P ] has a decomposition α = (α ∧ p) ∨ (α ∧ q1 ) ∨ (α ∧ q2 ), where α ∧ p ∈ Pp , α ∧ q1 ∈ Pq1 , and α ∧ q2 ∈ Pq2 . Proof. Indeed, the componentwise join of the right side equals α. For a congruence γ of N6 [P ], let γ p denote the restriction of γ to Pp , ˆ p denote γ p regarded as a congruence of P ; the same for γ q1 and γ q2 . Let γ ˆ q1 and γ ˆ q2 . Similarly, let γ p and γ r denote the restriction the same for γ ˆ p and γ ˆ r denote the corresponding of γ to P p and P r , respectively, and let γ congruences of P . Then we obtain:

224

17. Magic Wands

Lemma 17.7. Let α, α′ ∈ N6 [P ] and γ ∈ Con N6 [P ]. Then α ≡ α′ (mod γ) iff α ∧ p ≡ α′ ∧ p (mod γ p ),

α ∧ q1 ≡ α′ ∧ q1 (mod γ q1 ),

α ∧ q2 ≡ α′ ∧ q2 (mod γ q2 ). Proof. This is clear from Lemma 17.6. Lemma 17.8. Let α = (x, y, z), α′ = (x′ , y ′ , z ′ ) ∈ N6 [P ], and γ ∈ Con N6 [P ]. Then α ≡ α′ (mod γ) iff ˆ p ), x ≡ x′ (mod γ

ˆ q1 ), y ≡ y ′ (mod γ ˆ q2 ). z ≡ z ′ (mod γ

Proof. This is clear from Lemma 17.7. Now we have the tools to describe the congruences. The next four lemmas provide the description. ˆ q1 = γ ˆ q2 in P . Lemma 17.9. Let γ ∈ Con N6 [P ]. Then γ

ˆ q1 ), then (0, u, 0) ≡ (0, u′ , 0) (mod γ q1 ), so Proof. Indeed, if u ≡ u′ (mod γ ′ (0, u, 0) ≡ (0, u , 0) (mod γ). Therefore, (0, 0, u) = ((0, u, 0) ∨ (1, 0, 0)) ∧ (0, 0, 1)

≡ ((0, u′ , 0) ∨ (1, 0, 0)) ∧ (0, 0, 1) = (0, 0, u′ )

(mod γ).

ˆ q2 ), We conclude that (0, 0, u) ≡ (0, 0, u′ ) (mod γ q2 ), that is, u ≡ u′ (mod γ ˆ q1 ≤ γ ˆ q2 . By symmetry, γ ˆ q1 ≥ γ ˆ q2 , so γ ˆ q1 = γ ˆ q2 . proving that γ ˆp ≤ γ ˆ q2 holds. Lemma 17.10. For γ ∈ Con N6 [P ], the congruence inequality γ

ˆ p ), then (u, 0, 0) ≡ (u′ , 0, 0) (mod γ p ). ThereProof. Indeed, if u ≡ u′ (mod γ fore, (0, 0, u) = (u, 1, u) ∧ (0, 0, 1) = ((u, 0, 0) ∨ (0, 1, 0)) ∧ (0, 0, 1)

≡ ((u′ , 0, 0) ∨ (0, 1, 0)) ∧ (0, 0, 1)

= (u′ , 1, u′ ) ∧ (0, 0, 1) = (0, 0, u′ ) ˆ q2 ), proving that γ ˆp ≤ γ ˆ q2 . that is, u ≡ u′ (mod γ

(mod γ),

17.4. 2/3-Boolean triples

225

Lemma 17.11. Let α ≤ β ∈ Con P . Then there is a unique γ ∈ Con N6 [P ], ˆ q2 = γ. ˆ p = α and γ ˆ q1 = γ such that γ Proof. The uniqueness follows from the previous lemmas. To prove the existence, for α ≤ β ∈ Con P , define a congruence γ on N6 [P ] by (x, y, z) ≡ (x′ , y ′ , z ′ ) (mod γ)

iff x ≡ x (mod α),

y ≡ y ′ (mod β),

z ≡ z ′ (mod β).

It is obvious that γ is an equivalence relation and it satisfies the Substitution Property for meet. To verify the Substitution Property for join, let (x, y, z) ≡ (x′ , y ′ , z ′ ) (mod γ) and let (u, v, w) ∈ N6 [P ]. Then (x, y, z) ∨ (u, v, w) = (x ∨ u, y ∨ v, z ∨ w)

= (x ∨ u, (x ∨ y ∨ u ∨ v) ∧ (y ∨ z ∨ v ∨ w), (x ∨ z ∨ u ∨ w) ∧ (y ∨ z ∨ v ∨ w)).

Similarly, (x′ , y ′ , z ′ ) ∨ (u, v, w)

= (x′ ∨ u, (x′ ∨ y ′ ∨ u ∨ v) ∧ (y ′ ∨ z ′ ∨ v ∨ w), (x′ ∨ z ′ ∨ u ∨ w) ∧ (y ′ ∨ z ′ ∨ v ∨ w)). Since x ≡ x′ (mod α), we also have

x ∨ u ≡ x′ ∨ u (mod α).

From x ≡ x′ (mod α) and α ≤ β, it follows that x ≡ x′ (mod β). Also, y ≡ y ′ (mod β), so x∨y ≡ x′ ∨y ′ (mod β). Therefore, x∨y∨u∨v ≡ x′ ∨y ′ ∨u∨v (mod β). Similarly (or even simpler), y ∨ z ∨ v ∨ w ≡ y ′ ∨ z ′ ∨ v ∨ w (mod β). Meeting the last two congruences, we obtain that (x ∨ y ∨ u ∨ v) ∧ (y ∨ z ∨ v ∨ w) ≡ (x′ ∨ y ′ ∨ u ∨ v) ∧ (y ′ ∨ z ′ ∨ v ∨ w) (mod β). Similarly, (x ∨ z ∨ u ∨ w) ∧ (y ∨ z ∨ v ∨ w) = (x′ ∨ z ′ ∨ u ∨ w) ∧ (y ′ ∨ z ′ ∨ v ∨ w) (mod β). The last three displayed equations verify that (x, y, z) ∨ (u, v, w) ≡ (x′ , y ′ , z ′ ) ∨ (u, v, w) (mod β). Now note that for x, y ∈ P and congruence γ of P , xr ≡ y r

xp ≡ yp (mod γ)

iff

xp ≡ y p (mod γ)

iff

(mod γ)

and

p

r

x q1 ≡ yq1

(mod γ).

ˆ =γ ˆ q1 = γ ˆ q2 and γ ˆ =γ ˆ p , so Lemmas 17.9–17.11 can be It follows that γ restated as follows.

226

17. Magic Wands

Corollary 17.12. There is a one-to-one correspondence between the congruences of N6 [P ] and pairs of congruences α ≤ β of Con P , defined by ˆ p ). γ 7→ (ˆ γr, γ ˆr ≤ γ ˆ p can be established in a stronger form by exhibiting The inequality γ a term function t(x) on N6 [P ] such that t(ur ) = up for u ∈ P . Lemma 17.13. There is a term function t(x) on N6 [P ] such that t(ur ) = up for u ∈ P . Proof. Define t(x) = (((x ∧ p) ∨ q1 ) ∧ q2 ) ∨ p.

Indeed, if u ∈ P , then ur = (u, 1, 1) and so t(ur ) = t((u, 1, 1))

= ((((u, 1, 1) ∧ (1, 0, 0)) ∨ (0, 1, 0)) ∧ (0, 0, 1)) ∨ (1, 0, 0) = (((u, 0, 0) ∨ (0, 1, 0)) ∧ (0, 0, 1)) ∨ (1, 0, 0) = ((u, 1, 0) ∧ (0, 0, 1)) ∨ (1, 0, 0)

= ((u, 1, u) ∧ (0, 0, 1)) ∨ (1, 0, 0) = (0, 0, u) ∨ (1, 0, 0) = (1, 0, u)

= (1, u, u) = up , as required. Actually, to prove Theorem 17.2, we need not the lattice N6 [P ] but a quotient thereof, which we now proceed to construct. Let P and Q be nontrivial bounded lattices and let φ : P → Q be a homomorphism of P onto Q. By Corollary 17.12, there is a unique congruence γ of N6 [P ] corresponding to the congruence pair 0 ≤ ker(φ) of P . Define B = N6 [P ]/γ. This is our new gadget. It is useful to note that B can be represented as { (x, y, z) ∈ P × Q × Q | y = (y ∨ φx) ∧ (y ∨ z) and z = (z ∨ φx) ∧ (z ∨ y) }. By the Second Isomorphism Theorem (Theorem 1.6), there is a one-to-one correspondence between the congruences of B and congruence pairs α ≤ β of P satisfying ker(φ) ≤ β. For x ∈ N6 [P ], let x denote the congruence class x/β. Using this notation, utilizing the results of this section, we state some important properties of the lattice B.

17.5. Proof-by-Picture for surjective maps

227

Lemma 17.14. Let P and Q be bounded lattices and let φ : P → Q be a homomorphism of P onto Q. Then there is a lattice B, with the following properties: (i) B has a spanning sublattice {o, p, q1 , q2 , s, i} isomorphic to N6 . (ii) The interval [s, i] is isomorphic to P under the map x 7→ xr , x ∈ P . (iii) The interval [p, i] is isomorphic to Q under the map y 7→ y p , y ∈ Q, where x ∈ P with φx = y. (iv) The congruences Σ of B are in one-to-one correspondence with pairs of congruences (α, β), where α is a congruence of P and β is a congruence of Q satisfying α ≤ φ−1 β, where up to isomorphism, Σ restricted to [r, i] is α and Σ restricted to [p, i] is β. (v) There is a term function t(x) such that t(ur ) = xp for u ∈ P , where x ∈ P with φx = u.

17.5.

Proof-by-Picture for surjective maps

We proceed with the Proof-by-Picture of Theorem 17.2 as in Section 17.2. We can assume, without loss of generality, that [a, b] and [c, d] are nontrivial intervals, that is, a < b and c < d.

A = M3 [K] p2

p1

p3

Fc,d

Fa,b Ka,b

Ic,d

Ia,b

Kc,d

1

r

p 0

B

Figure 17.4: The four gadgets to construct L

228

17. Magic Wands

We take the four gadgets: A = M3 [K], Ka,b , B, and Kc,d , except that now B is not M3 [a, b] but the lattice B = N6 [a, b]/γ (see Lemma 17.14), constructed from P = [a, b], Q = [c, d], and the homomorphism φ from [a, b] onto [c, d] given in Theorem 17.2. We do three gluings. First gluing. In B, we use the filter fil(r) = { (x, 1, 1) | a ≤ x ≤ b } (which is isomorphic to [a, b] since γ on [r, 1] is 0), whereas in Ka,b we utilize the ideal Ia,b = { (0, x, 0) | a ≤ x ≤ b } (which is obviously isomorphic to [a, b]), and we consider the isomorphism φ1 : (x, 1, 1)B 7→ (0, x, 0)Ka,b ,

x ∈ [a, b],

between the filter fil(r) of B and the ideal Ia,b of Ka,b to glue B and Ka,b together to obtain the lattice U . (As in Section 17.2, we use the following notation: to indicate whether a triple (x, y, z) belongs to A = M3 [K], Ka,b , B, or Kc,d , we subscript it: with (x, y, z)A , (x, y, z)a,b , (x, y, z)B , or (x, y, z)c,d , respectively.) Second gluing. We glue Kc,d and A over the filter Fc,d = { (x, c, x ∧ c) | x ∈ K } of Kc,d and the ideal id(p3 ) = { (0, 0, x) | x ∈ K } of A, with respect to the isomorphism φ2 : (x, c, x ∧ c)Kc,d 7→ (0, 0, x)A ,

x ∈ K,

to obtain the lattice V . Third gluing. In U , we define the filter F = [p, 1] ∪ Fa,b , which is the union of [p, 1] and Fa,b , with the unit of [p, 1] identified with the zero of Fa,b . In V , we define the ideal I = Ic,d ∪ [0A , p1 ], which is the union of Ic,d and [0A , p1 ], with the unit of Ic,d identified with the zero of [0A , p1 ].

17.6. Verification for surjective maps

229

Next we set up an isomorphism φ : F → I. Since [p, 1] = { (1, x, x)B | a ≤ x ≤ b } and Ic,d = { (0, x, 0)c,d | c ≤ x ≤ d }, we define φ on [p, 1] by φ : (1, x, x)B 7→ (0, γx, 0)c,d , where φ : [a, b] → [c, d] is the homomorphism given in Theorem 17.2. We define φ on Fa,b by φ : (x, b, x ∧ b)a,b 7→ (x, 0, 0)A ,

x ∈ K.

It is clear that φ : F → I is well defined and that it is an isomorphism. Finally, we construct the lattice L of Theorem 17.2 by gluing U over I with V over F with respect to the isomorphism φ : F → I. The map x 7→ (x, 0, 0)A is an isomorphism between K and the principal ideal id(p1 ) of A; this gives us a convex embedding of K into A. We identify K with its image, and regard K as a convex sublattice of A and therefore of L. So L is a convex extension of K. We have completed the construction of the bounded lattice L of Theorem 17.2.

17.6.

Verification for surjective maps

The proof in Section 17.3 heavily depended on the fact that we glued over ideals and filters of which the building components were congruence-preserving extensions. This is no longer the case; however, a modification of Lemma 2.9 comes to the rescue. A congruence κ of L can be described by four congruences, κA , the restriction of κ to A, κa,b , the restriction of κ to Ka,b , κc,d , the restriction of κ to Kc,d , κB , the restriction of κ to B. These congruences satisfy a number of conditions: (i) (0, x, 0)a,b ≡ (0, y, 0)a,b (mod κa,b ) iff (x, 1, 1)B ≡ (y, 1, 1)B for x, y ∈ [a, b].

(mod κB )

230

17. Magic Wands

(ii) (0, 0, x)A ≡ (0, 0, y)A (mod κA ) iff (x, c, x ∧ c)c,d ≡ (y, c, y ∧ c)c,d (mod κc,d ) for x, y ∈ K. (iii) (x, 0, 0)A ≡ (y, 0, 0)A (mod κA ) iff (x, a, x ∧ a)a,b ≡ (y, a, y ∧ a)a,b (mod κa,b ) for x, y ∈ K. (iv) (0, γx, 0)c,d ≡ (0, γy, 0)c,d (mod κc,d ) iff (1, x, x)B ≡ (1, y, y)B (mod κB ) for x, y ∈ [a, b]. Conversely, if we are given congruences κA on A, κa,b on Ka,b , κc,d on Kc,d , κB on B, then by (i), we can define a congruence κU on U . By (ii), we can define a congruence κV on V . By (iii) and (iv), we can define a congruence κL on L. Now it is clear that if we start with a congruence σ of K, then we can define the congruences σ A on A, σ a,b on Ka,b , σ c,d on Kc,d componentwise, and σ B on B as in Section 17.3. Conditions (i)–(iii) trivially hold (since σ A , σ a,b , and σ c,d are defined componentwise). Finally, (iv) holds if σ is a γ-congruence. So every γ-congruence of K has an extension to L. Let σ be a congruence of K that extends to L. Since A is a congruencepreserving convex extension of K = [0A , p1 ], further, Ka,b is a congruencepreserving extension of Fa,b , and Kc,d is a congruence-preserving extension of Fc,d , the congruence σ uniquely extends to A as σ A , to Ka,b as σ a,b , and to Kc,d as σ c,d . Therefore, σ uniquely extends to the intervals [r, 1] and [p, 1] of B, and so by Lemma 17.14 to B. We conclude that if a congruence σ of K extends to L, then it extends uniquely. To complete the proof, we prove that γ is a term function. Define p(x) = ((((((((x ∧ (a, a, b)a,b ) ∨ (b, a, a)a,b ) ∧ p) ∨ q1 ) ∧ q2 )

∨ (d, c, c)c,d ) ∧ (c, c, d)c,d ) ∨ p2 ) ∧ p1 .

By Lemma 17.13, p(x) behaves properly in B, whereas outside of B, p(x) is the same term function as in Section 17.3. This completes the proof of Theorem 17.2.

17.7.

Discussion

Note that Theorem 17.2 implies Theorem 17.1. However, the two theorems have different generalizations. In my joint paper with E. T. Schmidt [190], we proved the generalizations stated in the three following subsections.

17.7. Discussion

231

First generalization of Theorem 17.1 Let K be a bounded lattice, let [ai , bi ], i < α, be intervals of K (α is an initial ordinal ≥ 2), and let γi,j : [ai , bi ] → [aj , bj ],

for i, j < α,

be an isomorphism between the intervals [ai , bi ] and [aj , bj ]. For notational convenience, we write [a, b] for [a0 , b0 ]. Let γ = { γi,j | i, j < α } be subject to the following conditions for i, j < α: (1) γi,i is the identity map on [ai , bi ]. −1 = γj,i . (2) γi,j

(3) γj,k ◦ γi,j = γi,k . Let Kγ denote the partial algebra obtained from K by adding the partial operations γi,j for i, j < α. Let us call a congruence α of K a γcongruence iff α satisfies the Substitution Property with respect to the partial unary operations γi,j ∈ γ. Thus a congruence relation of the partial algebra Kγ is the same as a γ-congruence of the lattice K. We call the lattice L a γ-congruence-preserving extension of K if a congruence of K extends to L iff it is a γ-congruence of Kγ and every γ-congruence of K has exactly one extension to L. Theorem 17.15. Let K be a bounded lattice, and let α ≥ 2 be an ordinal. Let [ai , bi ], for i < α, be intervals of K, and let γi,j : [ai , bi ] → [aj , bj ] be an isomorphism between the intervals [ai , bi ] and [aj , bj ], for i, j < α, subject to the conditions (1)–(3), where γ = { γi,j | i, j < α }. Then the partial algebra Kγ has a γ-congruence-preserving convex extension into a bounded lattice L such that all γi,j , for i, j < α, are term functions in L. In particular, the congruence lattice of the partial algebra Kγ is isomorphic to the congruence lattice of the lattice L. It would be nice to be able to claim that this theorem can be proved by applying Theorem 17.1 to the isomorphisms γi,j one at a time, and then forming a direct limit. Unfortunately, the direct limit at ω produces a lattice with no zero or unit, so we cannot continue with the construction. Second generalization of Theorem 17.1 Let K be a bounded lattice, let α be an ordinal, and for i < α, let γi be an isomorphism between the interval [ai , bi ] and the interval [ci , di ]: γi : [ai , bi ] → [ci , di ].

232

17. Magic Wands

Let γ = { γi | i < α }, and let Kγ denote the partial algebra obtained from K by adding the partial operations γi for i < α. Let us call a congruence α of the lattice K a γ-congruence iff α satisfies the Substitution Property with respect to the partial unary operations γi , for i < α, that is, x ≡ y (mod α) implies that γi x ≡ γi y (mod α) for all x, y ∈ [ai , bi ] and i < α. Thus a congruence relation of the partial algebra Kγ is the same as a γ-congruence of the lattice K. We call the lattice L a γ-congruence-preserving extension of the lattice K if a congruence of K extends to L iff it is a γ-congruence of K and every γ-congruence of K has exactly one extension to L. Theorem 17.16. Let K be a bounded lattice, let γ be given as above. Then the partial algebra Kγ has a γ-congruence-preserving convex extension into a lattice L such that all γi , for i ∈ I, are term functions in L. In particular, the congruence lattice of the partial algebra Kγ is isomorphic to the congruence lattice of the lattice L. A generalization of Theorem 17.2 Let K be a bounded lattice, and for i ∈ I, let γi be a homomorphism of the interval [ai , bi ] onto the interval [ci , di ]. Let γ = { γi | i ∈ I }, and let Kγ denote the partial algebra obtained from K by adding the partial operations γi for i ∈ I. Let us call a congruence α of the lattice K a γ-congruence iff α satisfies the Substitution Property with respect to the partial unary operations γi , for i ∈ I, that is, x ≡ y (mod α) implies that γi x ≡ γi y (mod α) for all x, y ∈ [ai , bi ] and i ∈ I. Thus a congruence relation of the partial algebra Kγ is the same as a γ-congruence of the lattice K. We call the lattice L a γ-congruence-preserving extension of the lattice K if a congruence of K extends to L iff it is a γ-congruence of K and every γ-congruence of K has exactly one extension to L. Theorem 17.17. Let K be a bounded lattice, let γ be given as above. Then the partial algebra Kγ has a γ-congruence-preserving convex extension into a lattice L such that all γi , for i ∈ I, are term functions in L. In particular, the congruence lattice of the partial algebra Kγ is isomorphic to the congruence lattice of the lattice L. Theorem 17.17 easily implies Theorems 17.15 and 17.16, with one important difference: In Theorem 17.15, we obtain a bounded lattice L. Problem 52. Can we ensure that the lattice L is bounded in Theorem 17.17?

17.7. Discussion

233

Magic wands with special properties In Theorems 17.1 and 17.2, can we construct a lattice L with special properties, such as being semimodular? Obviously not if we insist on convex embeddings, since a convex sublattice of a semimodular lattice is semimodular again. Let Theorem 17.1∗ denote Theorem 17.1 with “convex” deleted, and the same for Theorem 17.2∗ . Let us call a class C of lattices a CPE-class if every finite lattice A has a finite congruence-preserving extension B ∈ C. Theorem 17.18. Let K be a finite lattice, and let C be a CPE-class of lattices. Then in Theorems 17.1∗ and 17.2∗ , for a finite K, we can assume that L ∈ C. Problem 53. What can we say about CPE-classes? By Theorem 14.1, sectionally complemented lattices; by Theorem 15.1, semimodular lattices; by Theorem 16.1, isoform lattices; by Theorem 16.19, lattices with a given automorphism group form such classes. So there is a sectionally complemented magic wand, a semimodular magic wand, and so on. Problem 54. Is it possible to get a magic wand combining any two of these properties? Fully invariant congruences As usual, let us call a congruence α fully invariant iff a ≡ b (mod α) implies that δa ≡ δb (mod α) for any automorphism δ. For a lattice K, let Coninv K denote the lattice of fully invariant congruences of K, and let Aut K denote the set (group) of automorphisms of K. For a bounded lattice K, we can apply Theorem 17.17 to I = Aut K; for δ ∈ Aut K, let [aδ , bδ ] = [cδ , dδ ] = [0, 1], and let γδ = δ. Theorem 17.19. Let K be a bounded lattice. Then K has a convex extension into a lattice L such that a congruence of K extends to L iff it is fully invariant and a fully invariant congruence of K extends uniquely to L. In particular, Coninv K is isomorphic to Con L. The 1/3-Boolean triple construction The reader may ask, what is a 1/3-Boolean triple construction? For a lattice P , let us call the element (x, y, z) ∈ P 3 a 1/3-Boolean triple iff z = (z ∨ x) ∧ (z ∨ y).

234

17. Magic Wands

Then instead of N6 , we now get the lattice Nδ7 (see Figure 17.5), and the 1/3-Boolean triples form a lattice N7 [P ]. Problem 55. Is there some use for the 1/3-Boolean triple construction?

Figure 17.5: The lattice Nδ7

Part V

Congruence Lattices of Two Related Lattices

235

Chapter

18

Sublattices

18.1.

The results

The simplest connection between two lattices K and L is the sublattice relation: K ≤ L. How then does Con K relate to Con L? As we discussed in Section 3.3, the relation K ≤ L induces a map ext of Con K into Con L: For a congruence relation α of K, the image ext α is the congruence relation of L generated by α, that is, ext α = conL (α). The map ext is a {0}-separating join-homomorphism. (If we want to emphasize the embedding idK : K → L, we write ext idK for ext.) In 1974, A. P. Huhn [213] stated the converse. Theorem 18.1. Let D and E be finite distributive lattices, and let γ : D → E be a {0}-separating join-homomorphism. Then there are finite lattices K ≤ L, and isomorphisms α : D → Con K and β : E → Con L satisfying βγ = (ext idK )α, where idK is the embedding of K into L; that is, such that the diagram

is commutative.

D   ∼ = yα

γ

−−−−→ ext id

E   ∼ = yβ

K Con K −−−−−→ Con L

In this chapter we prove a much stronger version of this theorem from G. Gr¨atzer, H. Lakser, and E. T. Schmidt [157]. 237 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_18

238

18. Sublattices

Theorem 18.2. Let K be a finite lattice, let E be a finite distributive lattice, and let γ : Con K → E be a {0}-separating join-homomorphism. Then there is a finite extension L of K and an isomorphism β : E → Con L satisfying ext idK = βγ, that is, such that the diagram γ

Con K −−−−→   =y ext id

is commutative.

E   ∼ = yβ

K Con L Con K −−−−−→

Although Theorem 18.1 claims the existence of finite lattices K ≤ L, Theorem 18.2 claims that for every K, there is such an extension L. This stronger form allows us to claim the existence of K and L in any CPEclass C (introduced in the subsection: Magic wands with special properties of Section 17.7). Theorem 18.3. Let D and E be finite distributive lattices, and let γ : D → E be a {0}-separating join-homomorphism. Let C be a CPE-class of lattices. Then there are finite lattices K, L ∈ C with K ≤ L, and isomorphisms α : D → Con K and β : E → Con L satisfying γβ = (ext idK )α, where idK is the embedding of K into L. So there are such lattices K ≤ L that are sectionally complemented, by Theorem 14.1; semimodular, by Theorem 15.1; isoform, by Theorem 16.1; with a given automorphism group, by Theorem 16.19. Theorem 18.3 trivially follows from Theorem 18.2. Indeed, let D, E, and γ be given as in Theorem 18.3. By the Basic RT, there is a finite lattice K1 ∼ D with the isomorphism φ1 . By the definition of a CPEsatisfying Con K1 = class, the lattice K1 has a congruence-preserving extension K ∈ C; let φ2 be an isomorphism Con K → Con K1 . Then γ ′ = φ2 φ1 γ is a {0}-separating join-homomorphism of Con K into E. Applying Theorem 18.2 to K, E, and γ ′ , there is a finite extension L1 of K, an embedding idK : K → L1 , and an isomorphism β : E → Con L1 with ext idK = γβ. We use once more that C is a CPE-class: there is a finite congruence-preserving extension L ∈ C of L1 . Obviously, K and L satisfy the requirements of Theorem 18.3. So we only have to prove Theorem 18.2.

18.2. Proof-by-Picture

18.2.

239

Proof-by-Picture

This Proof-by-Picture is based on the ideas in G. Gr¨atzer, H. Lakser, and F. Wehrung [162] (see Section 18.5 for the history of the proofs). Let K, E, and γ : Con K → E be given as in Theorem 18.2. We want to construct a finite extension L of K, an embedding idK : K → L, and an isomorphism β : E → Con L with ext idK = βγ. First, let E = C2 = {0, 1}. Since γ is a {0}-separating join-homomorphism, it follows that γ0 = 0, and γα = 1, for all α > Con K −{κK }. By Lemma 14.3, the lattice K has a finite simple extension L, with the embedding idK : K → L. Observe that ext 0K = κL and ext 1K = 1L , since L is simple. Define β : E → Con L by β0 = κL and β1 = 1L . So ext idK = βγ is obvious; it is illustrated in Figure 18.1. Second, let E = C2 × C2 . We get two projection maps πi : E → C2 for i = 1, 2. So we can apply the E = C2 case—the last paragraph—to K, C2 , and γπi : Con K → C2 , and we obtain the simple extensions Li of K. Obviously, L = L1 × L2 will do the job. In general, if E = Bn , we can take L = L1 × L2 × · · · × Ln , where the finite lattices L1 , L2 , . . . , Ln are simple extensions of K. Now let us take the smallest non-Boolean case: E = C3 = {0, a, 1}. 1K

Con K

Con K − {0K }

ψ

1 E β

ψ

0

0K ext id K β 1L ext idK

Con L

0L Figure 18.1: The case E = C2

240

18. Sublattices

We embed E into B = B2 = {0, a, b, 1}, and take the lattice LB we have just constructed for B. Let Con LB = {0, αB , γ B , 1} and choose in LB a Boolean filter F = {0F , uF , vF , 1B } so that con(uF , 1B ) = αB and con(vF , 1B ) = γ B . Obviously, F is a congruence-determining sublattice of LB . By Theorem 8.9, we can represent E as the congruence lattice of a finite lattice LE with Con LE = {0, αE , 1} so that LE has a Boolean ideal I = {0, uE , vE , 1I } and con(0, uE ) = αLE and con(0, vE ) = 1LE . Obviously, I is a congruence-determining sublattice of LE .

LE 1I αE

vE

1

I 1 αE

LE

uE

0

α

v

1B

vF

αB β B

F

β B αB

α

0F uF

LB

0F LB

Figure 18.2: Constructing L

u

L

18.3. Multi-coloring

241

Now we construct the lattice L by gluing LB (with the filter F ) to the lattice LE (with the ideal I) with respect to the obvious isomorphism between F and I that maps uF to uI and vF to vI . In L, we use the notation u = uF = uI and v = vF = vI . Since αB restricted to F agrees with αI restricted to I, it is clear that L has only one nontrivial congruence α, which agrees with αB on LB and with αE on LE . The maps work out with no difficulty (almost the same way as in Figure 18.1), verifying that L satisfies the requirements of Theorem 18.2. Since a formal proof of Theorem 18.2 is almost the same as that of this special case, we will not present a formal version of this Proof-by-Picture.

18.3.

Multi-coloring

We introduced colored lattices in Section 3.2. Now we need a generalization from G. Gr¨atzer, H. Lakser, and E. T. Schmidt [157]. Let M be a finite lattice and let C be a finite set; the elements of C will be called colors. Recall that Prime(M ) denotes the set of prime intervals of M . A multi-coloring of M over C is a map µ from Prime(M ) into Pow∗ C (all nonempty subsets of C ordered by set-inclusion) satisfying the condition that if p, q ∈ Prime(M ) and con(p) ≤ con(q), then µp ⊆ µq. Coloring is the special case when all the µp are singletons. Equivalently, a multi-coloring is an isotone map of the ordered set J(Con M ) into the ordered set Pow∗ C. We will now show that a multi-colored lattice has a natural extension to a colored lattice. Lemma 18.4. Let M be a finite lattice with a multi-coloring µ over the set C. Then there exists a lattice M ∗ with a coloring µ∗ over C such that the following conditions holds: (i) M ∗ is the direct product of the lattices Mc , c ∈ C, where Mc is a homomorphic image of M colored by {c}. (ii) There is a lattice embedding a 7→ a∗ of M into M ∗ . (iii) For every prime interval p = [a, b] of M , µp = { µ∗ (q) | q ∈ Prime(M ∗ ) and q ⊆ [a∗ , b∗ ] } and the minimal extension of con(p) under this embedding into M ∗ is of the form Y (con(pc ) | c ∈ C),

where pc is a prime interval of Mc iff c ∈ µp and pc is a trivial interval otherwise (in which case, con(pc ) = 0Mc ).

242

18. Sublattices

Proof. For c ∈ C, define the binary relation γ c on M as follows: u ≡ v (mod γ c ) iff c ∈ / µp, for every prime interval p ⊆ [u ∧ v, u ∨ v]. This relation is obviously reflexive and symmetric. To show transitivity, assume that u ≡ v (mod γ c ) and v ≡ w (mod γ c ), and let q be a prime interval in [u ∧ w, u ∨ w]. Then q is collapsed by con(u, v) ∨ con(v, w), hence, there is a prime interval p in [u ∧ v, u ∨ v] or in [v ∧ w, v ∨ w] satisfying con(q) ≤ con(p). It follows from the definition of multi-coloring that µq ⊆ µp; since c ∈ / µp, we conclude that c ∈ / µq, hence, u ≡ w (mod γ c ). The proof of the Substitution Property is similar. ForQc ∈ C, we define the lattice Mc as M/γ c . A prime interval p of M ∗ = (Mc | c ∈ C) is uniquely associated with a c ∈ C and a prime interval of Mc . We define µ∗ (p) = c. It is easy to see that µ∗ is a coloring of M ∗ over C, establishing the first condition. To establish the second condition for a ∈ M , define a∗ so that its Mc component be a/γ c . The mapping a 7→ a∗ is obviously a lattice homomorphism. We have to prove that it is one-to-one. Let a, b ∈ M and a ̸= b; we have to prove that a∗ ̸= b∗ . Let p be a prime interval in [a ∧ b, a ∨ b]. Since µ∗ is a multi-coloring, there is a c ∈ µ∗ (p). Obviously, then a ̸≡ b (mod γ c ), from which the statement follows. Finally, the third condition is trivial from the definition of M ∗ and µ∗ .

18.4.

Formal proof

Now we are ready for the second proof of Theorem 18.2 as presented in G. Gr¨ atzer, H. Lakser, and E. T. Schmidt [157]. Let K, E, and γ be given as in Theorem 18.2. Step 1. We define a map µ of Prime(K) to subsets of J(E): µp = J(E) ∩ id(γ(con(p))). The map µ is obviously isotone. The join-homomorphism γ separates 0, therefore, µp ̸= ∅. Therefore, µ is a multi-coloring of K over J(E). We apply Lemma 18.4 to obtain the lattice Y K∗ = (Kc | c ∈ J(E)). Step 2. Any finite lattice M can be embedded in a finite simple lattice, Simp M , with the same zero and unit (see Lemma 14.3). Use such an extension for each Kc to obtain a simple lattice Simp Kc , then define: Y L0 = (Simp Kc | c ∈ J(E)),

18.4. Formal proof

243

and extend the coloring so that Simp Kc is also colored by {c}. Since L0 is a direct product of simple lattices, it follows that J(Con(M)) L0 is an antichain; the congruence lattice of L0 is a Boolean lattice with |J(E)| atoms. K is a sublattice of K ∗ and K ∗ is a sublattice of L0 , so we obtain an embedding γ : K → L0 . Finally, we construct a special ideal of L0 . Let pc be an arbitrary atom of the direct component Simp Kc ; then the prime interval [0, pc ] of L0 has color c. The atoms pc , for c ∈ J(E), generate an ideal B0 of L0 , which is a Boolean lattice satisfying the following properties: (1) any two distinct atoms have different colors; (2) every color c ∈ J(E) occurs in B0 . ∼ Step 3. We continue by forming a finite atomistic lattice L1 with E = Con L1 under the isomorphism β1 . For L1 , we construct a chopped lattice P1 as in Section 8.2, except that we use a uniform “tripling;;”—as opposed to “doubling”—of non-maximals (for every join-irreducible element p of E, we take three atoms p1 , p2 , and p3 , so that in P1 they generate a sublattice isomorphic ∼ J(Con L)1 to M3 ). Let L1 be the ideal lattice of P1 . The isomorphism J(E) = is given as follows: for p ∈ J(E), the congruence con(0, p) of L1 corresponds to p. Let β1 denote the corresponding isomorphism β1 : E → Con L1 . We consider on L1 the natural coloring over J(E) (a prime interval p is colored by β1−1 (con(p)) ∈ J(E)). Note that L0 and L1 are colored over the same set, J(E). Let B1 be the ideal of L1 generated by the atoms p2 for p ∈ J(E). Then the ideal B1 is a Boolean lattice satisfying the properties (1) and (2) stated in Step 2. Step 4. We have the lattice L0 with the ideal B0 and L1 with an ideal B1 . Note that B0 and B1 are isomorphic finite Boolean lattices with the same coloring. Take the dual L2 of L1 ; in this lattice B1 corresponds to a filter B2 . Again, note that B0 and B2 are isomorphic finite Boolean lattices with the same coloring. Glue together L0 and L2 by a color-preserving identification of B0 and B2 . The resulting lattice is L. The prime intervals of L are colored by J(E), and we have the isomorphism β : E → Con L. Since L0 is a sublattice of L, we may view γ as an embedding of K into L. Step 5. Finally, we have to verify that ext γ = βγ. It is enough to prove that (ext γ)(α) = βγ(α) for join-irreducible congruences α in K. So let α = con(p), where p = [a, b] ∈ Prime(K). By Lemma 18.4, ext γ(con(p)) = con(a∗ , b∗ ) collapses in K ∗ the prime intervals of color ≤ γα; the same holds in L0 and in L. Computing βγ(α) we get the same result, hence, (ext γ)(α) = βγ(α), completing the proof of Theorem 18.2.

244

18. Sublattices

18.5.

Discussion

History A. P. Huhn’s 1983 paper [213]—and his later papers [214] and [215] (published posthumously)—attacked the Congruence Lattice Characterization Problem (CLP) from below (see the discussion in the Introduction on page xxx). He observed that a distributive algebraic lattice D with countably many compact elements is the direct limit (union) of an increasing countable family ( Di | i < ω ) of finite distributive {∨, 0}-subsemilattices of D. The Di -s are, of course, finite distributive lattices. For each i < ω, let us denote by γi : Di → Di+1 the {∨, 0}-embedding. Huhn constructs a sequence ( Li | i < ω ) of finite lattices with lattice embeddings γi : Li → Li+1 such that ext γi : Con Li → Con Li+1 represents γi . Then denoting by L the direct limit of the sequence ( Li | i < ω ), ∼ D. he finds the representative D of L, that is, a lattice L satisfying Con L = The construction of the Li and γi is the most complicated part of his papers. However, using our Theorem 18.2, we can proceed in a straightforward manner. We first represent D0 by a finite lattice L0 , and, inductively, given Li , we immediately get a finite lattice Li+1 and an embedding γi : Li → Li+1 with ext γi representing γi . P. Pudl´ak [230] showed that every finite subset of a distributive algebraic lattice D is contained in a finite distributive {∨, 0}-subsemilattice S of D. Of course, S is a finite distributive lattice. P. Pudl´ak used this to find a new approach to E. T. Schmidt’s result discussed in the Introduction. Huhn’s result also follows from Theorems 5.5 and 5.6 of M. Tischendorf’s 1992 thesis [251]. In 1996, G. Gr¨atzer, H. Lakser, and E. T. Schmidt set out to give a short proof of Huhn’s result in [157]. The result published was not only a short proof but also a stronger form, with a number of applications in subsequent papers. The Proof-by-Picture we present in this chapter originated in the paper [252], in which J. T˚ uma proved the following result. Theorem 18.5. Let L0 , L1 , L2 be finite atomistic lattices and let η1 : L0 → L1 , η2 : L0 → L2

be {0}-embeddings such that ext η1 and ext η2 are injective. Let D be a finite distributive lattice, and, for i ∈ {1, 2}, let γi : Con Li → D be {∨, 0}-embeddings such that γ1 (ext η1 ) = γ2 (ext η2 ). Then there is a finite atomistic lattice L, there are {0}-embeddings γi : Li → L, for i ∈ {1, 2}, satisfying γ1 η1 = γ2 η2 , and there is an isomorphism α : Con L → D such that α ext γi = γi for i ∈ {1, 2}. In G. Gr¨atzer, H. Lakser, and F. Wehrung [162], T˚ uma’s result was extended as follows.

18.5. Discussion

245

Theorem 18.6. Let L0 , L1 , L2 be lattices and let η1 : L0 → L1 , η2 : L0 → L2

distributive lattice and, for be lattice homomorphisms. Let D be a finite W i ∈ {1, 2}, let γi : Con Li → D be complete -homomorphisms such that γ1 (ext η1 ) = γ2 (ext η2 ).

There is then a lattice L, there are lattice homomorphisms γi : Li → L, for i ∈ {1, 2}, with γ1 η1 = γ2 η2 , and there is an isomorphism α : Con L → D such that α ext γi = γi , for i ∈ {1, 2}. If L0 , L1 , L2 have zero and both η1 and η2 preserve the zero, then L can be chosen to have a zero and γ1 and γ2 can be chosen to preserve the zero. If L1 and L2 are finite, then L can be chosen to be finite and atomistic. uma’s theorem—we need only This theorem is indeed an extension of T˚ observe that if the γi are injective, then the γi must be lattice embeddings. This fact follows from the elementary fact that a lattice homomorphism γ : K → L is an embedding iff ext γ separates 0. Our Proof-by-Picture is the special case: L0 = L1 = L2 is finite and γ1 = γ2 . Applications As outlined in the history subsection, for every distributive algebraic lattice D with countably many compact elements, we can find a lattice L representing D, where L is a ω-union of finite lattices from a CPE-class C. Sometimes, we atzer, H. Lakser, and get even more as in the following two results from G. Gr¨ F. Wehrung [162]. Theorem 18.7. Let D be a distributive algebraic lattice. If D has at most ℵ1 compact elements, then there exists a locally finite, relatively complemented lattice L with zero such that Con L ∼ = D. A lattice L is congruence-finite if Con L is finite; it is 0-congruence-finite if L can be written as a union, [ L = ( Ln | n < ω ),

where ( Ln | n < ω ) is an increasing sequence of congruence-finite sublattices of L. We also apply Theorem 18.6 to prove the following.

Theorem 18.8. Every 0-congruence-finite lattice K has a 0-congruence-finite, relatively complemented congruence-preserving extension L. Furthermore, if K has a zero, then L can be taken to have the same zero.

246

18. Sublattices

Isotone maps With restrictions, we get the “dual” of Lemma 3.14. Lemma 18.9. Let K ≤ L be finite lattices. Then re : Con L → Con K is a {∧, 0, 1}-homomorphism. And we can obviously “dualize” Theorems 18.1 and 18.2. So what happens if we compose an extension and a restriction? Between the first and last congruence lattices, we surely get {0}-isotone map. The converse of this was proved in G. Gr¨ atzer, H. Lakser, and E. T. Schmidt [158]. Theorem 18.10. Let D1 and D2 be finite distributive lattices, and let the map γ : D1 → D2 be {0}-isotone. Then there are finite lattices L1 , L2 , L, lattice embeddings η1 : L0 → L1 , η2 : L0 → L2

γ1 : L1 → L and γ2 : L2 → L, and isomorphisms αi : Di → Con Li , for i = 1, 2, such that α2 γ = (re γ2 )(ext γ1 )α1 , that is, such that the diagram D1   ∼ = y α1

γ

−−−−→ ext γ1

Con L1 −−−−→

Con L

re γ2

D2   ∼ = y α2

−−−−→ Con L2

is commutative. In G. Gr¨atzer, H. Lakser, and E. T. Schmidt [161], there is a “concrete” version of this result. Theorem 18.11. Let L1 and L2 be arbitrary lattices with finite congruence lattices Con L1 and Con L2 , respectively, and let γ : Con L1 → Con L2 be an isotone map. Then there is a lattice L with a finite congruence lattice, a lattice embedding γ1 : L1 → L, and a homomorphism γ2 : L → L2 such that γ = (re γ2 )(ext γ1 ). Furthermore, γ2 is also an embedding iff γ preserves 0. If L1 and L2 are finite, then L can be chosen to be finite and atomistic. Theorem 18.6 enabled us to prove Theorem 18.11 rather easily.

18.5. Discussion

247

Size and breadth We now have several constructions to verify Huhn’s Theorem 18.1; in all of them, the lattices K and L are very large and very “wide.” We can measure the “width” of a finite lattice L as follows. A lattice L is of breadth p if p is the smallest integer with the property that W forWevery finite X ⊆ L, there exists a Y ⊆ X such that |Y | ≤ p and X = Y . Note that this concept is self-dual. If a finite lattice L is of breadth p, then there is an element a ∈ L with at least p covers. The breadth of the Boolean lattice Bn is n. A “small” version of Theorem 18.1 was proved in [160] by G. Gr¨atzer, H. Lakser, and E. T. Schmidt. Theorem 18.12. Let D be a finite distributive lattice with n join-irreducible elements, let E be a finite distributive lattice with m join-irreducible elements, let k = max(m, n), and let γ : D → E be a {0}-separating join-homomorphism. Then there is a planar lattice K with O(n2 ) elements, a finite extension L ≥ K (with embedding idK : K → L) of breadth 3 with O(k 5 ) elements, and isomorphisms α : E → Con L, β : D → Con K with αγ = (ext idK )β, that is, such that the diagram D   ∼ = yβ

γ

−−−−→ ext id

E   ∼ = yα

K Con L Con K −−−−−→

is commutative. The proof uses the planar construction of Chapter 9 and multi-coloring. Problem 56. Is O(k 5 ) optimal for the lattice L in Theorem 18.12? In other words, can one prove (analogously to G. Gr¨atzer, I. Rival, and N. Zaguia [168] discussed in Chapter 9) that the size O(k 5 ) cannot be replaced by the size O(k α ) for any α < 5? Problem 57. Is breadth 3 optimal for L? This is almost certainly so since a breadth 2 lattice cannot contain B3 as a sublattice, making it very difficult to manipulate the congruences. However, in view of the result of the next chapter (Theorem 19.3), one can ask: Problem 58. Is there a planar version of Theorem 18.1?

248

18. Sublattices

There is an interesting way of measuring the complexity of an order. The order dimension of a finite ordered set (P, ≤) is the smallest integer n ≥ 1 such that ≤ can be represented as the intersection of the orderings of n chains defined on the set P . The order dimension of a finite lattice L is 1 iff L is a chain. The order dimension of L is 2 iff L is planar. Problem 59. Prove that the lattice L we construct for Theorem 18.12 is of order dimension 3. The construction of the lattice L for Theorem 18.12 starts with the planar lattice of Chapter 9. Problem 60. Can one construct L for Theorem 18.12 starting from a different lattice K? 2-distributive lattices A. P. Huhn introduced n-distributivity in [211] and [212]. Let n ≥ 1 be an integer. A lattice L is n-distributive if for all x, y1 , . . . , yn+1 ∈ L, x∧(

n+1 _ i=1

yi ) =

n+1 _ i=1

(x ∧ (

n+1 _

yj )).

j=1 j̸=i

In particular, a lattice L is 1-distributive iff it is distributive; it is 2-distributive iff it satisfies the identity x ∧ (y1 ∨ y2 ∨ y3 ) = (x ∧ (y1 ∨ y2 )) ∨ (x ∧ (y1 ∨ y3 )) ∨ (x ∧ (y2 ∨ y3 )). We will call a lattice L doubly 2-distributive if it satisfies the 2-distributive identity and its dual. For instance, N5 and M3 are doubly 2-distributive lattices. In my joint paper with [183] E. T. Schmidt proved the following. Theorem 18.13. The lattices K and L in Theorem 18.1 can be constructed as finite doubly 2-distributive lattices. The proof is quite complex and uses multi-coloring. Problem 61. Can the construction for Theorem 18.13 be continued with the lattice L serving as the starting lattice? If this could be done, and repeated ω times, then we could represent distributive algebraic lattice with countably many compact elements as congruence lattices of doubly 2-distributive lattices.

Chapter

19

Ideals

19.1.

The results

There is another natural connection between two lattices K and L. If the lattice K is an ideal of the lattice L, how does Con K relate to Con L? In Section 3.3, we discussed the connection between Con K and Con L, for the lattice L and its sublattice K. In Chapter 18, we proved the corresponding RT for finite lattices. Lemma 3.13 states that if I is an ideal of the lattice L, then the restriction map re : Con L → Con I is a bounded homomorphism. The corresponding RT for finite lattices was proved in my joint paper with H. Lakser [140]. Theorem 19.1. Let D and E be finite distributive lattices. Let φ be a bounded homomorphism of D into E. Then there exists a finite lattice L and an ideal I of L such that D ∼ = Con L, E ∼ = Con I, and φ is represented by the restriction map, re. E. T. Schmidt [243] provides an alternative proof of this result. In my joint survey paper with E. T. Schmidt [192], Problem 15 asks (in part), whether this result can be proved for sectionally complemented lattices. At first glance, we may think that we can solve this problem by combining Theorem 19.1 with Theorem 14.1: Every finite lattice has a finite, sectionally complemented, congruence-preserving extension. We obtain a stronger form of Theorem 19.1: the lattice L can be assumed to be sectionally complemented. Unfortunately, in the congruence-preserving extension of L constructed in Theorem 14.1, I is no longer an ideal, so this method fails to solve this problem. In my joint paper with H. Lakser [147], using quasiorders, we answer the question in the affirmative. 249 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_19

250

19. Ideals

Theorem 19.2. Let D and E be finite distributive lattices. Let φ be a bounded homomorphism of D into E. Then there exists a sectionally complemented finite lattice L and an ideal I of L such that D ∼ = Con L, E ∼ = Con I, and φ is represented by re, the restriction map. As an ideal of a sectionally complemented lattice, of course, the ideal I is also a sectionally complemented lattice. In my joint paper with H. Lakser [145], we took a different approach to the congruence restriction problem—the strange story of how this came about is told in Section 19.5. Theorem 19.3. Let D and E be finite distributive lattices. Let φ be a bounded homomorphism of D into E. Then there exists a finite planar lattice L and an ideal I of L such that D ∼ = Con L, E ∼ = Con I, and φ is represented by re, the restriction map. This is, of course, much stronger than Theorem 19.1: we obtain planar lattices (and very small ones). However, Theorem 19.1 is the foundation on which Theorem 19.2 is built. One cannot obtain Theorem 19.2 based on Theorem 19.3.

19.2.

Proof-by-Picture for the main result

For the Proof-by-Picture, take the five-element distributive lattices D and E and the bounded homomorphism φ at the top of Figure 19.1. We choose a bounded homomorphism φ, which is not an onto map. We associate with φ an “inverse map” ϱ : J(E) → J(D) (also shown in Figure 19.1), where we define ϱ on an x ∈ J(E) as the smallest element of D that is mapped to an element ≥ x by φ (see Section 2.5.2). Our plan is the following: We construct a chopped lattice M as in Section 8.2—using the gadget N6 —except that we triple, not double, every join-irreducible element, starting with the order J(D) ∪ J(E) (disjoint union). Every join-irreducible element x has three copies, xL (left), xM (middle), and xR (right). We further add two more copies of all these elements as in Section 8.2. So the new gadget (for a ∈ J(D)) is the chopped lattice G(a) in Figure 19.1, which already guarantees that aL ≡ 0 is equivalent to aM ≡ 0 is equivalent to aR ≡ 0. We take six gadgets: G(x), for each x ∈ J(D) ∪ J(E)—they are pairwise disjoint except they share the 0. We now have three chopped lattices: IC = G(d) ∪ G(e) ∪ G(i), L′C = G(a) ∪ G(b) ∪ G(c), and L′′C = IC ∪ L′C . We order the congruences in L′C to agree with the ordering in J(D) as in Section 8.2, using the middle elements. For instance, to achieve that b ≡ 0 implies that a ≡ 0, we insert the element bM (aM ) so that N (bM , aM ) becomes a sublattice, as illustrated in Figure 19.1. So Con L′C ∼ = D.

19.2. Proof-by-Picture for the main result

i

i E

D b

c

f

a

d

e

o

o

the map ϕ J(D) b

J(E) c

i

̺ a

d

e

the map ̺ aL (aM )

aR (aL ) aL

aR

aM aL2 aM

aL1

aM (aR )

aR1

aM2

1

aR2

0 the new gadget G(a) aM (bM ) aM

aM2

bM

bM2

aM1

bM1

0

inserting N (aM , bM ) Figure 19.1: Proof-by-Picture for the main result

251

252

19. Ideals

Next, we order the congruences in IC to agree with the ordering in J(D). ∼ E. At this stage, there is no congruence connection between L′ Thus Con IC = C and IC , so Con L′′C is isomorphic to the direct product Con L′C × Con IC ∼ = D × E. Finally, we want to achieve that x ≡ 0 is equivalent to ϱx ≡ 0, for x ∈ J(D), for instance, that e ≡ 0 is equivalent to c ≡ 0. To accomplish this, we add the element eL (cL ) to L′′C so that N (eL , cL ) becomes a sublattice; therefore, in the new chopped lattice, eL ≡ 0 implies that cL ≡ 0. For the reverse implication, we add the element cR (eR ) so that N (cR , eR ) becomes a sublattice; now cR ≡ 0 implies that eR ≡ 0. Since the gadget G(e) ensures that eL ≡ 0 is equivalent to eR ≡ 0, we conclude that eL ≡ 0 is equivalent to cL ≡ 0. We proceed similarly with all x ∈ J(D). Let LC denote the chopped lattice we obtain after adding all these elements. ∼ E, By construction, Con LC ∼ = D and LC contains an ideal IC with Con IC = and the congruences of LC restrict to IC as determined by ϱ. Now we invoke Theorem 5.6, and conclude that L = Id LC and I = Id IC satisfy the requirements of Theorem 19.1.

19.3.

Formal proof

Categoric preliminaries In our proof, we will make use of several classes of algebraic structures. It will be useful to use the language of (concrete) category theory. A concrete category is a class K of algebras (partial algebras, relational systems) called objects, along with a designated class of morphisms between any two objects. It is assumed that the identity map on an object is a designated morphism, and if K, L, M are objects, φ : K → L and ψ : L → M are designated morphisms, then so is ψφ : K → M . For categories K and N, the map F : K → N is a functor if F maps objects to objects, morphisms to morphisms, and whenever φ : K → L and ψ : L → M are morphisms in K, then F (ψ)F (φ) = F (ψφ). A contravariant functor F reverses this, that is, it satisfies F (φ)F (ψ) = F (ψφ). Also, if φ : K → L is a morphism in K, then F (φ) : F (K) → F (L) for a functor and F (φ) : F (L) → F (K) for a contravariant functor. Moreover, F (idK ) = idF (K) for both functors and contravariant functors. Finally, we need the concept of natural equivalence. Let K and N be categories, and let F : K → N and G : K → N be functors. Then η : F → G is a natural transformation if for every object A of K, there is a morphism ηA : F (A) → G(A) such that for every morphism α : A → B in K, the diagram

19.3. Formal proof

253

F (α)

F (A) −−−−→ F (B)     ηA y y ηB G(α)

G(A) −−−−→ G(B)

is commutative, that is, ηB F (α) = G(α)ηA . A natural transformation η : F → G is a natural equivalence if ηA is an isomorphism for each object A of K. We now introduce the various categories and some of the associated functors we will use. Note that all the objects are finite structures. 1. Let DI (for DIstributive) denote the category of finite distributive lattices. The morphisms are bounded homomorphisms. 2. Let LA (for LAttice) denote the category whose objects are the finite lattices. A morphism is an ideal-embedding (see Section 1.3.2). There is a contravariant functor Con : LA → DI that associates with each lattice L its congruence lattice Con L, that is, Con(L) = Con L. If K and L are finite lattices and ε : K → L is an ideal-embedding, then Con(ε) denotes the restriction map (denoted by re in Section 3.3 and elsewhere), that is, Con(ε)(α) = α⌉K , for a congruence α of L. By Lemma 3.13, Con(ε) is a morphism in DI . 3. Let OR (for ORder) denote the category whose objects are finite orders and whose morphisms are isotone maps. We have a contravariant functor Dn : OR → DI; if φ : P → Q is an isotone map of orders, then the map Dn φ : Dn Q → Dn P is defined by (Dn φ)(H) = φ−1 H, for each down-set H ⊆ Q. 4. Let CH (for CHopped) denote the category whose objects are chopped lattices. An embedding ε in CH is a one-to-one meet-homomorphism that preserves ∨ in the strong sense: εx ∨ εy exists iff x ∨ y exists and then ε(x ∨ y) = εx ∨ εy. The morphisms of CH are ideal-embeddings ε : S → T . We will show that the functor Con : CH → DI that associates with each object S of CH its congruence lattice Con S is a contravariant functor. As in the case of LA, if ε : S → T is an ideal-embedding, then Con(ε) : Con T → Con S denotes the restriction map. 5. The last category we consider is HE (for HEmi order); it is of a more technical nature. An object of HE is a finite nonempty set Q with a binary relation ρ satisfying the following three conditions: (a) ρ is irreflexive, that is, x ρ x fails, for all x ∈ Q; (b) ρ is antisymmetric, that is, x ρ y implies that y ρ x fails, for all x, y ∈ Q; (c) ρ is cycle-rich, that is, for each x ∈ Q, there is a y ∈ Q with y ρ x.

254

19. Ideals

If Q and R are objects of HE, a morphism φ : Q → R in HE is a one-to-one map that preserves ρ in the following strong sense: φx ρ φy

iff

x ρ y.

Let Q be an object of HE; a subset H of Q is a down-set if x ∈ H and x ρ y imply that y ∈ H. The lattice Dn Q of down-sets in Q is distributive with ∅ as 0 and Q as 1. If φ : Q → R is a morphism in HE, then define (Dn φ)(H) = φ−1 (H) for H ∈ Dn R. It is easy to see that Dn φ : Dn R → Dn Q. So we have a contravariant functor Dn : HE → DI. Note that we use the same symbol Dn for the functors Dn : OR → DI and Dn : HE → DI. From DI to OR Let us recall and rephrase the results from Section 2.5.2. Given an object D of DI, that is, a finite distributive lattice, we consider the order J(D).VIf φ : D → E is a morphism in DI, then J(φ) : J(E) → J(D), with J(φ)(x) = φ−1 (x), is an isotone map. Thus J : DI → OR is a contravariant functor. Let idDI be the identity map on DI, regarded as a functor DI → DI. Note that Dn J is also a functor DI → DI. Lemma 19.4. There is a natural equivalence ψ : idDI → Dn J that associates with each object D of DI the isomorphism ψD : D → Dn J(D) defined by ψD (x) =↓ J(D) for x ∈ D, that is, if D and E are finite distributive lattices and φ : D → E is a bounded homomorphism, then the diagram ψD

D −−−−→ Dn J(D)   Dn J(φ)  φy y ψE

E −−−−→ Dn J(E)

commutes, in other words, Dn J(φ)ψD = ψE φ, and ψD , ψE are lattice isomorphisms. From OR to HE Let P and Q be finite orders and let φ : Q → P be an isotone map. For P , Q, and φ, we construct two HE-objects, B(Q) and A(φ), and define the HE-morphism εφ : B(Q) → A(φ); the notation reflects the fact that B(Q) depends only on Q, whereas A(φ) and εφ depend on φ. Set B(Q) = Q × {L, M, R}; for a ∈ Q, denote the ordered pairs (a, L), (a, M ), (a, R) by aL , aM , aR , respectively. Define ρ on B(Q) by setting:

19.3. Formal proof

255

(α) aL ρ aM ρ aR ρ aL , for a ∈ Q; (β) aM ρ bM , whenever a ≻ b in Q, where ≻ denotes the cover relation in Q. Set A(φ) = (P × {L, M, R}) ∪ (Q × {L, M, R}), where we assume that P and Q are disjoint. Define ρ on A(φ) by setting: (1) aL ρ aM ρ aR ρ aL if a ∈ P ∪ Q; (2) aM ρ bM if a ≻ b, and a, b ∈ P or a, b ∈ Q; (3) aL ρ φaL , for all a ∈ Q; (4) φaR ρ aR , for all a ∈ Q. Define εφ : B(A) → A(φ) by setting εφ aK = aK for a ∈ Q, K ∈ {L, M, R}. Then εφ is a HE-morphism. We define the maps υ : B(Q) → Q and ϑ : A(φ) → P by setting υaK = a ϑaK = φa ϑaK = a

for a ∈ Q, K ∈ {L, M, R};

for a ∈ Q, K ∈ {L, M, R}; for a ∈ P , K ∈ {L, M, R}.

These determine the maps υ ′ : Dn Q → Dn B(Q) with υ ′ H = υ −1 H, for H ∈ Dn Q, and ϑ′ : Dn P → Dn A(φ) with ϑ′ H = ϑ−1 H for H ∈ Dn P . The following statement is easy to verify. Lemma 19.5. The maps υ ′ : Dn Q → Dn B(Q) and ϑ′ : Dn P → Dn A(φ) are lattice isomorphisms and the diagram ϑ′

Dn P −−−−→ Dn A(φ)     Dn φy yDn εφ υ′

Dn Q −−−−→ Dn B(Q)

commutes, that is, (Dn εφ )ϑ′ = υ ′ Dn φ. From CH to DI We need only one observation. Lemma 19.6. Con : CH → DI is a contravariant functor.

256

19. Ideals

Proof. Since the objects S of CH are meet-semilattices, congruence relations are determined by pairs x, y with x ≤ y. Since id(y) is a sublattice of S, we get, exactly as in the case of lattices, that for x ≤ y, the congruence x ≡ y (mod α ∨ β) holds iff there is a sequence x = z0 ≤ z1 ≤ · · · ≤ zn = y, with zi ≡ zi+1 (α) or zi ≡ zi+1 (β) for each 0 ≤ i < n. Then we can establish that Con S is a distributive lattice and that Con(ε) is a DI-morphism for any CH-morphism ε, exactly as for lattices (see Theorem 3.1). From HE to CH We describe a functor S : HE → CH. Let Q be an object in HE and set S(Q) = {0} ∪ {a1 , a2 , a | a ∈ Q} ∪ { b(a) | a, b ∈ Q, with b ρ a }, where 0 is distinct from the other elements. Define an ordering ≤ on S(Q): 0 < ai ai < a b1 < b(a) a < b(a)

for i = 1, 2; for i = 1, 2; if b ρ a; if b ρ a.

The maximal elements of S(Q) are then of the form b(a) with b ρ a, and b, a ∈ Q, since for each a ∈ Q there is a b ∈ Q with b ρ a. Each id(b(a)) is isomorphic to the lattice N6 = N (b, a) depicted in Figure 8.1. If φ : Q → R is a HE-morphism, then define S(φ) : S(Q) → S(R) by setting S(φ)(0) = 0; S(φ)(ai ) = φai for i = 1, 2; S(φ)(a) = φa; S(φ)(b(a)) = φb(φa). Note that b ρ a implies that φb ρ φa, justifying the last equation. Lemma 19.7. S : HE → CH is a functor. Proof. In S(Q), d(c) ∧ b(a) = 0 c(b) ∧ b(a) = b1 c(a) ∧ b(a) = a c(a) ∧ c(b) = c1

if if if if

b ρ a, d ρ c and a, b, c, d are all distinct; c ρ b ρ a, and a, b, c are all distinct; c ρ a, b ρ a, and a, b, c are all distinct; c ρ a, c ρ b, and a, b, c are all distinct.

There appears to be one more case to check: a(b) ∧ b(a); however, this cannot occur by property (b) in the definition of HE.

19.3. Formal proof

257

Thus S(Q) is a meet-semilattice and it is easy to see that S(Q) is indeed a chopped lattice. Clearly, S(φ) is an ideal-embedding of S(Q) into S(R), whenever φ : Q → R is a HE-morphism. Given an object Q of HE, we define a map ψQ : Dn Q → Con S(Q) by setting ψQ (H) = con(H). Note that each subset of Q is a subset of S(Q). We now prepare the proof of Theorem 19.10 with two lemmas. Given a, b ∈ Q with b ρ a, the principal ideal N (b, a) generated by b(a) has exactly three congruence relations: 1b,a , collapsing all of N (b, a), 0b,a , the identity relation, αb,a , depicted in Figure 8.1 (where it is denoted by α), with congruence classes {b1 , b(a)} and {0, a1 , a2 , a}. We first show that ψQ is surjective. Lemma 19.8. Let Q be an object of HE and let α be a congruence relation on S(Q). Then H = { a ∈ Q | a ≡ 0 (α) } is a down-set in Q and α = con(H). Proof. Let b ∈ H and let b ρ a. Then b ≡ 0 (mod α), so b1 ≡ 0 (mod α). Thus α⌉N (b,a) = 1b,a , and we conclude that a ∈ H. Consequently, H is a down-set in Q. Since S(Q) is sectionally complemented, α is determined by its congruence class containing 0. However, for any congruence, a1 ≡ 0 iff a2 ≡ 0 iff a ≡ 0, and b(a) ≡ 0 iff b ≡ 0. Thus α is determined by H = { a ∈ Q | a ≡ 0 (α) }, that is, α = con(H), concluding the proof. We now characterize con(H).

258

19. Ideals

Lemma 19.9. Let Q be an object of HE and let H be a down-set in Q. Let a, b ∈ Q with b ρ a. Then  1b,a if b ∈ H;  con(H)⌉N (b,a) = αb,a if b ∈ / H and a ∈ H;   0b,a if a ∈ / H.

Proof. The set Max = Max(S(Q)) of maximal elements of S(Q) consists of all elements of the form b(a), with b ρ a. For each b(a) ∈ Max, define the congruence relation αb(a) on the ideal N (b, a) by setting  1b,a , if b ∈ H;  αb(a) = αb,a , if b ∈ / H and a ∈ H;   0b,a , if a ∈ / H.

Then

αb(a) ⌉N (b,a)∩N (d,c) = αd(c) ⌉N (b,a)∩N (d,c) ,

for any b(a), d(c) ∈ Max. Therefore, (αb(a) | b ρ a) is a congruence-vector, introduced in Section 5.3, so by Lemma 5.5, there is a unique congruence relation α on S(Q) with α⌉N (b,a) = αb(a)

for all b(a) ∈ M .

By Lemma 19.9, α = con(H1 ),

with H1 = {a ∈ Q | a ≡ 0(α)}.

For each a ∈ Q, there is a b ∈ Q with b ρ a. Then a ∈ H1 iff a ≡ 0 (mod α) iff a ≡ 0 (mod αb(a) ) iff a ∈ H; the last equivalence follows by the definition of αb(a) . Thus H = H1 . We are now ready for the main result of this section. Theorem 19.10. The map ψ establishes the natural equivalence of the functors Dn : HE → DI and Con S : HE → DI. That is, for every object Q in HE, ψQ : Dn Q → Con S(Q) is an isomorphism and, given a HE-morphism φ : Q → R, the diagram in DI: ψR

Dn R −−−−→ Con S(R)    Con S(φ) Dn φy y ψQ

Dn Q −−−−→ Con S(Q)

commutes, that is, Con S(φ)ψR = ψQ Dn φ.

19.3. Formal proof

259

Proof. We first show that, for each object Q of HE, ψQ : Dn Q → Con S(Q) is an isomorphism. Clearly, H1 ⊆ H2 implies that con(H1 ) ⊆ con(H2 ) and, by Lemma 19.8, ψQ is surjective. We need to show that ψQ is an embedding, that is, that con(H1 ) ⊆ con(H2 ) implies that H1 ⊆ H2 . Let con(H1 ) ⊆ con(H2 ) for H1 , H2 ∈ Dn Q. Take a ∈ H1 , and let b ∈ Q with b ρ a; such a b exists by property (HE.c). Then a ≡ 0 (mod con(H2 )) and so con(H2 )⌉N (b,a) ̸= 0b,a . By Lemma 19.9, a ∈ H2 . Thus H1 ⊆ H2 , concluding the proof that ψQ is an isomorphism. Next we verify that the diagram is commutative. Let H ∈ Dn R. Then by Lemma 19.9, (Con S(f )ψR )(H) = con(H1 ), for some H1 ∈ Dn Q. But a ∈ H1 iff a ≡ 0 (mod (Con S(φ)ψR )(H)) iff S(φ)(a) ≡ 0 (mod ψR (H)), that is, φa ≡ 0 (mod con(H)) iff φa ∈ H. Thus H1 = Dn φH. But then (Con S(φ)ψR )(H) = (ψQ Dn φ)(H), and so Con S(φ)ψR = ψQ Dn φ. From CH to LA The functor Id : CH → LA associates with each object S in CH the lattice Id(S). If φ : S → T is a CH-morphism, then Id(φ) : Id S → Id T is the ideal-embedding Id(I) = φI. Consider the functors Con : CH → DI and Con Id : CH → DI. For any chopped lattice M , Theorem 5.6 establishes that Con M and Con Id M are isomorphic; we obtain an isomorphism by assigning to the congruence α of M , the congruence σS (α) = con(α), the congruence generated by α in Id M . (We regard M as a sublattice of Id M by identifying m with id(m), for m ∈ M .)

Lemma 19.11. σ is a natural equivalence Con → Con Id. That is, for each object S of CH, the map σS : Con S → Con(Id S) is an isomorphism and, given a CH-morphism φ : S → T , the diagram σ

Con T −−−T−→ Con(Id T )    Con(Id(φ)) Con(φ)y y σ

Con S −−−S−→ Con(Id S)

260

19. Ideals

in DI commutes, that is, Con(Id(φ))σT = σS Con(φ). The final step By combining the natural equivalences of Lemmas 19.4, 19.11, and Theorem 19.10, and the commutative diagram of Lemma 19.5, we get our main result (for Con(α), reference Lemma 19.6). Theorem 19.12. Let D and E be finite distributive lattices and let φ : D → E be a bounded homomorphism. Then there exist finite lattices I, L, an ideal-embedding α of I into L, and isomorphisms β : D → Con L, γ : E → Con I such that the diagram β D −−−−→ Con L   Con(α)  φy y γ

is commutative, that is,

E −−−−→ Con I Con(α)β = γφ.

Proof. Consider the following diagram: ψD

ϑ′

ψA(J(φ))

ψE

v′

ψB(J(E))

D −−−−→ Dn J(D) −−−−→ Dn A(J(φ)) −−−−−→ . . .       φy Dn εJ(φ) y Dn J(φ)y

E −−−−→ Dn J(E) −−−−→ Dn B(J(E)) −−−−−→ . . . σS(A(J(φ)))

. . . −−−−→ Con S(A(J(φ))) −−−−−−−→ Con Id S(A(J(φ)))     Con SεJ(φ) y Con(Id(S(EJ(φ) ))y σS(B(J(E)))

. . . −−−−→ Con S(B(J(E))) −−−−−−−→ Con Id S(B(J(E)))) By Lemma 19.4, ψD and ψE are isomorphisms and the left-most square commutes. By Lemma 19.5, the maps v ′ and ϑ′ are isomorphisms and the next square commutes. By Theorem 19.10, ψA(J(φ)) and ψB(J(E)) are isomorphisms and the corresponding square commutes. Finally, by Lemma 19.11, σS(A(J(φ))) and σS(B(J(E))) are isomorphisms and the right-most square commutes. Set L = Id S(A(J(φ)), I = Id S(B(J(E))), α = Id S(εJ(φ) )). If we set ′ β = σS(A(J(φ))) ψA(J(φ)) ϑψ D

and

γ = σS(B(J(E)) ψB(J(E)) v ′ ψE ,

the theorem follows. Observe that we can prove that the lattice L is sectionally complemented by adjusting the proof of Theorem 8.5 to the present more general setup.

19.4. Proof-by-Picture for planar lattices

19.4.

261

Proof-by-Picture for planar lattices

The Proof-by-Picture will use the example distributive lattices D and E and the bounded homomorphism φ of Figure 19.2. The figure also shows the “inverse map” ϱ : J(E) → J(D), as it was done in Section 19.2 (see also Section 2.5.2). i i E

D c d

b

a

o J(D) c a

o ϕ: D → E

J(E) i d

b ̺ : J(E) → J(D)

Figure 19.2: Distributive lattices for the planar example As in Section 9.2, we base the construction on the direct product of two chains C and C ′ . Again we use two gadgets: cover-preserving M3 -s to make a prime interval p of the first chain congruence-equivalent to a prime interval q of the second chain; and N5 (rather than the N5,5 of Section 19.2) to force that con(p) < con(q), for the prime intervals p and q of the first chain. Let C = C7 and C ′ = C6 . For every cover x ≺ y in J(D) ∪ J(E), we color three adjacent prime intervals of C with y, x, and y. We start with J(E); it has one cover: d ≺ i, so we color the first three prime intervals with i, d, i. Then we take the only cover of J(E): b ≺ c, and we color the next three prime intervals of C with b, b, c. We color C ′ by setting up a bijection between the prime intervals of C ′ and J(D) ∪ J(E). We start with J(E), then with J(D). We start by adding the M3 gadgets. First, the M3 gadget is used to “fill in” every covering square whose two sides have the same color; there are six. Second, we “identify” by ϱ, that is, we fill in every covering square whose two sides have color x and xϱ, for some x ∈ J(E); there are two.

262

19. Ideals

c b

c c

b i

a d

i i

d

Figure 19.3: The planar lattice L and ideal I

Next comes the N5 gadget: for every cover x ≺ y in J(D) ∪ J(E), take the three adjacent prime intervals of C colored with y, x, and y; add an element so that they will form an N5 . Note that the prime interval colored by a is not related to any other prime interval by the N5 or M3 gadget because a is not comparable to any other element of J(D) and because a is not in the image of ϱ. This completes the construction of the lattice L. The principal ideal generated by the black-filled element is I (see Figure 19.3). It is really easy to compute that I represents E, L represents D, and the restriction map represents φ. A formal proof Theorem 19.3 is very similar to proofs we have already presented in detail, so we leave it to the reader.

19.5. Discussion

19.5.

263

Discussion

In my joint paper with H. Lakser [148], we prove a very hard variant of Theorem 19.1, constructing the lattices K and L as isoform lattices (see Section 13.5 and Chapter 16). The proof is very technical and there is no Proof-by-Picture. In general, automorphisms of a lattice do not restrict to automorphisms of an ideal. In my joint paper with H. Lakser [145], we construct lattices where this does happen. Theorem 19.13. Let D and E be finite distributive lattices with more than one element, and let ψ : D → E be a bounded homomorphism. Let G and H be groups, and let η : G → H be a group homomorphism. Then there exist a lattice L, an ideal I in L, lattice isomorphisms ϱD : D → Con L,

ϱE : E → Con I,

and group isomorphisms τG : G → Aut L,

τH : H → Aut I

such that, for each x ∈ D, the congruence relation ϱE (ψx) on I is the restriction to I of the congruence relation ϱD (x) on L, and, for each g ∈ G, the automorphism τH (ηg) of I is the restriction of the automorphism τG (g) of L. If G and H are finite, then the lattice L can be chosen to be finite. By identifying D with Con L, E with Con I, G with Aut L, and H with Aut I, Theorem 19.13 can be paraphrased as follows: any pair ψ, a bounded homomorphism of finite distributive lattices, and ϱ, a homomorphism of groups, can be simultaneously realized as the respective restrictions Con L → Con I and Aut L → Aut I for some lattice L and some ideal I in L. Note that this result is a far reaching generalization of the Baranski˘ı-Urquhart Theorem. In Section 18.5 we discussed the (doubly) 2-distributive lattices of A. P. Huhn. Problem 62. Can Theorem 19.1 be proved for (doubly) 2-distributive lattices? Problem 63. What can we say about semimodular (doubly) 2-distributive lattices? Problem 64. Can Theorem 19.1 be proved for (semi) modular lattices?

Chapter

20

Two Convex Sublattices

20.1.

Introduction

Let L be a finite lattice and let G be a convex sublattice of L. Then the restriction map re, defined as re : α → α⌉G (the congruence α of L is mapped to its restriction to G), is a bounded (that is, {0, 1}-) lattice homomorphism of Con L into Con G. In my joint paper with H. Lakser [140], we proved the converse. Theorem 20.1. Let D and E be finite distributive lattices and let φ : D → E be a bounded lattice homomorphism. Then there exist a finite lattice L, a convex sublattice G of L that can be chosen to be either an ideal or a filter of L, and isomorphisms η : D ∼ = Con L and ξ : E ∼ = Con G, making the following diagram commutative: η D −−− −→ Con L ∼ =     φy rey ξ

E −−− −→ Con G ∼ =

where re is the restriction map: for a congruence α of L, re(α) is α restricted to G. E. T. Schmidt [243] provides an alternative proof. Theorem 20.1 is an abstract/abstract result. The congruence lattices are given as abstract finite distributive lattices, whereas the finite lattices L and G are constructed. We can improve on Theorem 20.1 in two ways. 265 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_20

266

20. Two Convex Sublattices

First, we can construct the finite lattices L and G in smaller classes of lattices. In my joint paper with H. Lakser [145], we constructed them as planar lattices. 16 years later, my joint paper with E. E. Knapp [137] represented finite distributive lattices as congruence lattices of rectangular lattices. Theorem 20.1 for rectangular lattices was proved in G. Cz´edli [27]. Second, we can try to loosen the abstract/abstract representation. This was also done in G. Cz´edli [27]. Theorem 20.2. Let G be a rectangular lattice, and let φ be a {0, 1}-lattice homomorphism of a finite distributive lattice D to Con G. Then there is a rectangular lattice L containing G as a filter and a lattice isomorphism η : D → Con L such that φ = re ◦ η, where re is the restriction map of congruences from L to G. That is, the diagram D   φy

η

−−− −→ Con L ∼ =   rey =

Con G −−−−→ Con G is commutative. Thus, G is made into a filter and φ is represented as restriction of congruences to G. This is a concrete/abstract result: instead of the finite distributive lattice E, we are given the rectangular lattice G and Con G plays the role of E. E. T. Schmidt [246] proved this result for the special case: φ is injective. In this chapter, we present the following result from my joint paper with H. Lakser [145]. Theorem 20.3. Let F and G be rectangular lattices and let φ : Con F → Con G be a bounded lattice homomorphism. Then there is a rectangular lattice L with G as a filter and F as a convex sublattice satisfying the following two conditions: (i) the lattice L is a congruence-preserving extension of F ; (ii) for α ∈ Con L, the map φ is α⌉F → α⌉G. This is a concrete/concrete result: instead of the finite distributive lattice D, we are given the rectangular lattice F and Con F plays the role of D; instead of the finite distributive lattice E, we are given the rectangular lattice G and Con G plays the role of E. The lattices constructed are rectangular and G is constructed as a filter, so Theorem 20.3 is a stronger form of Cz´edli’s result. This approach also provides a short, elementary, and direct proof based on the construction in my joint paper [193] with E. T. Schmidt.

20.2. Proof-by-Picture

a

F

b

u

c

v

w

G

E

D b

c

v

w

a

b

P

267

u

c

w

v

a

u

Q

Figure 20.1: The top diagrams: The rectangular lattices F and G The middle diagrams: The distributive lattices D = Con F , E = Con G and the bounded homomorphism φ : D → E (iii) The bottom diagrams: The ordered sets P = J(D), Q = J(E), and the isotone map J(φ) : Q → P

20.2.

Proof-by-Picture

For the Proof-by-Picture, we choose F = G = S7 . Let (1)

D = Con F, E = Con G, P = J(D), Q = J(E),

and let φ be the bounded homomorphism D → E, as in Figure 20.1(ii). Note that S7 is “square,” as a result all the rectangular lattices look “square” in this and the subsequent diagrams of this chapter.

268

20. Two Convex Sublattices

a

a c

b

A a

F

b

b

c

c a

a

a c

b

b a

a c

b

a c

b

c

a

c

b

F

b

B a a

b

c

c a

a

w

ub

a b

c

b

G

c

c

a b

v

u

a c

b c

b

C

c

u

a

b

a

v

b

a

c c

v

u

b

F

w

b c

b

a

u

u v

b

a a

w a

c c

b

b a

w

v

u

a b c

v

b

b

c c

u

a

a

b c

u

L

c

G

b

b

v

u

a c

F

b b

w

c

a

u

u v a

w

b

a

a

c c

b

b a

a b

c

b c

c

Figure 20.2: The rectangular lattices A, B, C, and L

20.3. Proof

269

x

Figure 20.3: An M3 , all the edges are colored by x We construct the rectangular lattice L of Theorem 20.3 in a few steps, as illustrated in Figure 20.2. Step 1. We construct a congruence-preserving extension B of F in which all join-irreducible congruences of F appear on both upper boundaries of B. The distributive lattice D = Con F has 3 join-irreducible congruences, so we start with the glued sum of 3 copies of M3 and F and extend this to a rectangular lattice A (see Figure 20.2). We add three more eyes to A to form M3 -s, to make sure that any two edges of the same color generate the same congruence; thus we obtain the rectangular lattice B of Figure 20.2. In Figures 20.2–20.5, we use the “shorthand” of Figure 20.3. Step 2. We form a rectangular lattice C containing all the congruences of F and G. To do this, we form the glued sum of B and G and extend this to a rectangular lattice C as in Figure 20.2. Step 3. We go back to Figure 20.1, to the element v of Q. The element b of D is the smallest element mapped by φ to an element e of E with e ≥ v; the element b is join-irreducible, so it is in P . We identify b and v (as per the RT), by finding a cover preserving C22 colored by b and v and adding an eye (colored gray in Figure 20.2). We proceed the same way with the element w ∈ Q, again finding b ∈ E, and identify b and w. Finally, we do the same for the element u ∈ Q and identify the congruences represented by u and a. Thereby, L is a congruence-preserving extension of F , since each join-irreducible congruence of L is one of a, b, c. This finishes the construction of the rectangular lattice L, as in Figure 20.2. Now we infer from the Birkhoff RT, that the map φ is represented by α⌉F → α⌉G, for α ∈ Con F , as required in Theorem 20.3.

20.3.

Proof

We use the notation (1). The first triple gluing For the rectangular lattice F of Theorem 20.3, let blF , brF , tlF , and trF denote the number of elements of the bottom left and right, and top left and right boundaries, respectively, and let j denote the number of join-irreducible congruences. If the lattice F is understood, the subscripts may be omitted. We prove the following statement.

270

20. Two Convex Sublattices

F b a

Z

Y

b c

a b

ac c

c

F

aa ab

b a

c cb

c a

c

a

a

b b

b

a c

a c

a a

c c b

U a a

b c

ac c c

a

c b a a

c

b

b

c

b

a b a

c

a b b a c c a b c b a

a

c

F

ab

V

c

a

a b

a

c

b a b a c b c a c a b c a b

R

Figure 20.4: The first triple gluing Lemma 20.4. The rectangular lattice F has a congruence-preserving rectangular extension R such that all join-irreducible congruences of R appear on both upper boundaries of R. Proof. Let us define the rectangular lattices, Y and Z as follows. Y = Cbl × Cj+1 ,

Z = Cj+1 × Cbr .

We also need the rectangular lattice U we obtain from C2j+1 by adding eyes j-times to the covering C22 -s on the main diagonal. We label F ; there are j labels. We use these to label the M3 -s on the main diagonal of U , bijectively. Now we form the triple gluing of F, Y, Z, U to form the rectangular lattice V , as illustrated in Figure 20.4. The labeling of V is not a coloring, because two edges of the same label do not necessarily generate the same congruence. Let x be the label of a join-irreducible congruence of F . There is an edge Ax of label x in the lower boundary of F , say, in the lower left boundary. For two such distinct edges Ax and Bx labeled by x let A′x and Bx′ be the corresponding edges of Y that are identified with Ax and Bx , respectfully, in the triple gluing. Let

20.3. Proof

271

αx and β x denote the congruences of Y generated by A′x and Bx′ , respectfully. Note that, after the gluing, A′x and Bx′ will generate the same congruence in the resulting lattice, since Ax and Bx generate the same congruence in F . We claim that for the color x, there is a covering square Sx = C22 in Y , that is colored by both αx and β x . Indeed, take the trajectory r of Y containing A′x ; it is a normal-up trajectory. Take the trajectory t of Y containing Bx′ ; it is a normal-down trajectory. Therefore, r and t intersect in a covering square in Y colored by both αx and β x , as claimed. We add an eye to Sx , as in Figure 20.4. We do it for all colors x in F , to obtain the rectangular lattice R, illustrated in Figure 20.4. Now the labeling is a coloring, the color x of F is the same as the color x of U . Let γ l be the restriction of αx to CblY , the lower left boundary of Y . Similarly, for each join-irreducible congruence x of U , there is a unique edge Cx in the upper left boundary of U . Take the congruence of Y generated by the corresponding edge Cx′ in the lower right boundary of Y and denote its restriction to that lower right boundary by γ r . Finally, R is a congruence-preserving extension of F , as claimed in this lemma. To accomplish this, we define the congruences αF of F , αY of Y , αZ of Z, and αU of U , as follows: (i) The congruence αF of F is x; equivalently, αF is generated in F by the edge Ax . (ii) We define αY = γ l × γ r . (iii) The congruence αZ of Z is defined symmetrically. (iv) The congruence αU of U is generated by the edge Bx ⊆ U . By construction, these congruences satisfy the conditions of Lemma 4.19, so there is a congruence αR extending all four, and this is the extension of the color x to R. This gives us that there is at least one such extension. The uniqueness follows from the fact that every edge of R is perspective to an edge of F or U . We conclude by observing that the congruence extension property for join-irreducible congruences is equivalent to the congruence extension property (for all congruences). The second triple gluing We prove the following statement in this section. Lemma 20.5. The rectangular lattice G has a congruence-preserving extension L such that id(0G ) = R (the rectangular lattice of Lemma 20.4 ).

272

20. Two Convex Sublattices

Proof. The technical aspects of the proof are very similar to the proof of Lemma 20.4, mutatis mutandis. We use three auxiliary rectangular lattices, defined as follows: the rectangular lattice R constructed in the previous section, and Y = CblG × CtlR ,

Z = CtrR × CbrG .

Then we form the triple glued sum of G, Y, Z, R to form the rectangular lattice U , as illustrated in Figure 20.5.(ii). Recall that we use the notation (1). Let x be the color of a join-irreducible congruence of G, that is, x ∈ Q. As in Section 20.2, there is a smallest element y ∈ D for which φy ≥ x holds, namely, J(φ)x ∈ P . There is an edge Ax of color x in the lower boundary of G, say, in the lower left boundary. There is also an edge By of color y in the upper left boundary of R. Let A′x and By′ be the edges of Y that are identified with Ax and By , respectfully, in the second triple gluing. Let αx and β y denote the congruences of Y generated by A′x and By′ , respectfully. As the proof of Lemma 20.4, we identify x and y by finding a cover preserving C22 in Y colored by x and y and adding an eye (colored gray in Figure 20.2). We do it for all colors x in G, to obtain the rectangular lattice L, illustrated in Figure 20.5. We verify the properties of L as we verified the properties of R in the previous section. Completing the proof In the previous section, we constructed a rectangular lattice L containing the filter G and the convex sublattice F such that 20.3(i) holds. To complete the proof of Theorem 20.3, we have to verify that 20.3(ii) also holds for L. By the Birkhoff RT, it is equivalent to verify that for α ∈ Q, the map J(φ) is represented by the construction, which is evident.

20.4.

Discussion

G. Cz´edli [27] had a good reason to change “ideal” in Theorem 20.1 to “filter” for rectangular lattices in Theorem 20.2. Indeed, Theorems 20.2 and 20.3 fail in general with ideals. First some terminology. Let G be a rectangular lattice. We say G is simple-embeddable, if it is embeddable as an ideal in a simple rectangular lattice. The rectangular lattice G is abstractly ideal-representable if, for any finite distributive lattice D and any bounded lattice homomorphism φ : D → Con G, there is a rectangular lattice L containing G as an ideal, and an isomorphism η : D → Con L such that φ = re ◦ η.

273

20.4. Discussion

G

Y

Z b a b c

G w

U b

u w b

v b

v

u

w

a

a b b a c b c a b c a b c a

b a

c a

cb a c

b

b

c

v c

a

c

F

ab

a

u a u a v b a c b a u b v c F c v b aa u b c ab c a v a a c cb c a c c b a c b c b a

a b aa

a c c c

c

a

c

b

R

u

c

c

u a au w b a w b a b a c b c a b a c b c a

c

w

b

b c

u c

u

w

G w

R a

b u

u w b

v b

v

u v w

a v u a c ub v a a b c F aa u cv b b ab c a v a a c cb c a c c b a c b c b b

L c

u a au b w bu c a c a b w b a c w b c a b a c b c a

c

w

b

R

Figure 20.5: The second triple gluing Top diagram: Setting up the second triple gluing Middle diagram: The result of the second triple gluing Bottom diagram: The rectangular lattice L

274

20. Two Convex Sublattices

Finally, G is concretely ideal-representable if, for any rectangular lattice F and any bounded lattice homomorphism φ : Con F → Con G, there is a rectangular congruence-preserving extension L of F that contains G as an ideal and such that for any α of Con L, the map φ is α⌉F 7→ α⌉G. The following theorem characterizes those rectangular lattices that can be represented as ideals (see my joint paper [152] with H. Lakser). Theorem 20.6. Let G be a rectangular lattice. The following conditions are equivalent: (i) each non-trivial congruence of G collapses some edge in one of its upper chains; (ii) G is concretely ideal-representable; (iii) G is abstractly ideal-representable; (iv) G is simple-embeddable.

Chapter

21

Tensor Extensions

21.1.

The problem

Let A and B be nontrivial finite lattices. Then Con A × Con B is a finite distributive lattice, so by the Basic RT, the lattice Con A × Con B can be represented as Con L, for some finite lattice L. How can we construct the lattice L from the lattices A and B? By Theorem 2.1, (1)

Con(A × B) ∼ = Con A × Con B,

so we can take L = A × B. For a similar isomorphism, we now introduce tensor products. For a nontrivial lattice K with zero, we denote by K − the ∨-subsemilattice of K defined on K − {0}. Let A and B be nontrivial lattices with zero. We denote by A ⊗ B the tensor product of A and B, defined as the “free” {∨, 0}-semilattice generated by the set A− × B − (see the footnote on page 23) and subject to the relations (2)

(a, b1 ) ∨ (a, b2 ) = (a, b1 ∨ b2 ),

(a1 , b) ∨ (a2 , b) = (a1 ∨ a2 , b),

for a ∈ A− , b1 , b2 ∈ B − ;

for a1 , a2 ∈ A− , b ∈ B − .

If A and B are finite nontrivial lattices, then A ⊗ B is a finite lattice since it is a finite ∨-semilattice with zero. Since A ⊗ B just “happens to be a lattice,” it is quite surprising that the isomorphism (3)

Con A ⊗ Con B ∼ = Con(A ⊗ B), 275

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_21

276

21. Tensor Extensions

holds (see G. Gr¨atzer, H. Lakser, and R. W. Quackenbush [153]). The proof of this result is very complicated and not appropriate for presentation in this book. There is, however, a closely related result, with an approachable proof. If L is a finite lattice and D is a finite distributive lattice, then we can define L[D] as the lattice of all isotone maps from J(D) to L. By E. T. Schmidt [240], we obtain that ∼ (Con L)[Con D]. Con L[D] = (See also Section 6.3 and my joint paper with E. T. Schmidt [176].) So here is the question: Let A and B be nontrivial finite lattices. Then (Con A)[Con B] is a finite distributive lattice. Can we represent it as Con L, for a finite lattice L constructed from A and B? In my joint paper with M. Greenberg [128], we introduced a construction: the tensor extension A[B] for nontrivial finite lattices A and B. In this chapter we will prove that (4)

Con(A[B]) ∼ = (Con A)[Con B],

providing an answer to the above question. In Section 21.7 we discuss in detail the history of this isomorphism and how it relates to tensor products. In [128], we prove that for finite lattices, a tensor extension is isomorphic to a lattice tensor product, introduced in my joint paper with F. Wehrung [200], so our result is the finite case of the main result of [200]. The reader should view this chapter as an introduction to a very technical chapter of congruence lattice theory, and should consult Section 21.7 for further readings. This elementary introduction is based on my joint papers with M. Greenberg [128] and [131].

21.2.

Three unary functions

To prepare for the introduction of tensor extensions, we define some useful functions. Let A and B be nontrivial finite lattices. We are going to define three unary functions, m, j, and p, on B A , the set of all maps from B into A. First, for a ∈ A and α ∈ B A , we define—computed

21.2. Three unary functions

277

B A

α a (A − fil(a))α

A − fil(a)

V

ma (α) =

(A − fil(a))α

B A ja (α) = A − id(a)

(A − id(a))α

(A − id(a))α

α

a

W

Figure 21.1: Illustrating ma (α) and ja (α) in B (see Figure 21.1): ^ ^ (5) ma (α) = α(A − fil(a)) = αx, (6) (7)

ja (α) = pa (α) =

_

x≱a

α(A − id(a)) =

^

x≱a

jx (α) =

^_

x≱a

_

αy,

y≰a

α(A − id(x)) =

^ _

x≱a y≰x

αy.

278

21. Tensor Extensions

The following statements about these functions are easy to verify. Lemma 21.1. For a, b ∈ A and α ∈ B A , ma (α) ∧ mb (α) = ma∨b (α),

(8)

ja (α) ∨ jb (α) = ja∧b (α).

(9)

Moreover, m0 (α) = 1B , j1 (α) = 0B , and p0 (α) = 1B . For A = M3 = {0, a, b, c, 1}, we get ma (α) = αb ∧ αc ∧ α0, and symmetrically for mb (α) and mc (α). These are the expressions that occur in the existential definition of Boolean triples. For A = M3 , pa (α) = jb (α) ∧ jc (α) ∧ j0 (α)

= (αa ∨ αc ∨ α1) ∧ (αb ∨ αc ∨ α1) ∧ (αa ∨ αb ∨ αc ∨ α1)

= (αa ∨ αc ∨ α1) ∧ (αb ∨ αc ∨ α1),

and symmetrically for pb (α) and pc (α). These are the same expressions that occur in the fixed-point definition of Boolean triples. Now we define the function m : B A → B A by (10)

m(α)(a) = ma (α),

for a ∈ A and α ∈ B A . We similarly define the functions j and p: (11)

j(α)(a) = ja (α),

(12)

p(α)(a) = pa (α).

21.3.

Defining tensor extensions

In Section 6.1 we had three definitions of M3 [B]: (F), the fixed point definition; (E), the existential definition; the closure definition in Lemma 6.1.(ii). The elements of M3 [B] are 3-tuples of elements of B. Similarly, the elements of the tensor extension A[B] are defined as A-tuples of elements of B, that is, elements of B A .

21.3. Defining tensor extensions

279

Let A and B be finite nontrivial lattices. The existential definition presents the tensor extension of A by B: A[B] = { m(α) | α ∈ B A }.

(13)

In other words, for γ ∈ B A ,

γ ∈ A[B] iff there exists α ∈ B A satisfying γ = m(α).

To obtain a “fixed-point description,” we introduce the notation p(α) = α (for α ∈ B A ) and call α the closure of α (in B A ). We now verify that this is indeed a closure operation on B A . Lemma 21.2. For α, β ∈ B A , (i) α ≤ α.

(ii) If α ≤ β and β ∈ A[B], then α ≤ β. (iii) α ∈ A[B] iff α = α. Proof. (i) WLet a ∈ A. Then for any x ≱ a, we have a ∈ A − id(x), and so αa ≤ α(A − id(x)). Therefore, by (7), ^_ αa ≤ α(A − id(x)) = pa (α) = α(a). x≱a

(ii) By the definition of A[B], there is a γ ∈ B A such that β = m(γ). Choose a ∈ A and x ≱ a. Thus for each y ∈ A − id(x), we have ^ αy ≤ βy = my (γ) = γ(A − fil(y)) ≤ γx, since x ∈ A − fil(y). Joining these inequalities for y ∈ A − id(x), we obtain that _ α(A − id(x)) ≤ γx. Meeting these inequalities for x ≱ a, we get ^_ ^ pa (α) = α(A − id(x)) ≤ γx = ma (γ) = βa, x≱a

x≱a

or α(a) ≤ βa, proving that α ≤ β. A (iii) W Let α = α, that is, αa = pa (α) for all a ∈ A. Define β ∈ B by βb = α(A − id(b)) for b ∈ A. Then ^ ^ ^_ ma (β) = β(A − fil(a)) = βx = α(A − id(x)) = pa (α) = αa, x≱a

x≱a

for any a ∈ A; therefore, α ∈ A[B]. Conversely, assume that α ∈ A[B]. By (i), α ≤ α holds. The reverse inequality holds by (ii) with β = α, so α = α.

280

21. Tensor Extensions

We have established the following result. Theorem 21.3. Let A and B be nontrivial finite lattices. The tensor extension A[B] is a lattice with meets computed pointwise and joins computed as the closures of the pointwise joins, that is, according to the formula α ∨A[B] β = α ∨B A β for α, β ∈ A[B]. The main result on finite tensor extensions is the following. Theorem 21.4. Let A and B be nontrivial finite lattices. Then the following isomorphism holds: Con(A[B]) ∼ = (Con A)[Con B]. If A is simple, then Con A ∼ = C2 , so (Con A)[Con B] ∼ = Con B. We have obtained the following generalization of Theorem 6.3. Corollary 21.5. Let A and B be nontrivial finite lattices. If A is simple, then A[B] is a congruence-preserving extension of B.

21.4.

Computing

Some special elements Let A and B be nontrivial finite lattices. For any q ∈ B, we define the (almost) constant function, κq , in B A with value q: ( q for x ∈ A − {0A }; κq (x) = 1B for x = 0A . Clearly, κq ∈ A[B], for any q ∈ B, since p(κq ) = κq . We will call { κq | q ∈ B } the diagonal of A[B]. For a ∈ A, we define the element χa ∈ B A —the characteristic function of id(a)—by (see Figure 21.2) ( 1B , if x ∈ id(a); χa (x) = 0B , otherwise. Lemma 21.6. χa is an element of A[B] for any a ∈ A. Proof. Define γ ∈ B A by γfil(a) = 0B and γ(A − fil(a)) = 1B . Clearly, mb (γ) = χa (b), for any b ∈ A, that is, χa = m(γ), so χa ∈ A[B]. If p ∈ B and a, b ∈ A with a < b, we define p⃗ab = χb ∧ (χa ∨B A κp ) ∈ B A (see Figure 21.2).

281

21.4. Computing

0B

A

A

b

0B

χa

p~ab a

p a

1B

1B

Figure 21.2: Illustrating χa and p⃗ab Lemma 21.7. p⃗ab ∈ A[B]. Proof. Under these hypotheses, χa ∨B A κp = χa ∨A[B] κp ∈ A[B], therefore, p⃗ab ∈ A[B]. The proof of the following statement is trivial since the joins and the relevant meets in A[B] are computed pointwise. Lemma 21.8. Let a, b ∈ A with a < b. Then the map γab : p 7→ p⃗ab is an embedding of B into the sublattice [χa , χb ] of A[B]; in fact, γab is a bounded embedding. Proof. This is obvious. The last statement adds the facts that (⃗0B )ab = χa and (⃗1B )ab = χb . We will denote by Bab the image of B under the map γab . Note that the diagonal, { κq | q ∈ B }, can also be defined as B0A 1A . Moreover, if a is covered by b and b is join-irreducible, then id(b) − id(a) = {b}, so Bab = [χa , χb ]. The next lemma will be useful in computing some joins in A[B]. Lemma 21.9. Let a, b, c ∈ Define α ∈ B A by  1B     q tα =  p    0B

A with c = a ∨ b and let p, q ∈ B with p ≤ q. for for for for

Then α = ⃗qac . (See Figure 21.3.)

t ∈ id(a); t ∈ id(b) − id(a); t ∈ id(c) − (id(a) ∪ id(b)); t ∈ A − id(c).

282

21. Tensor Extensions

0B

A

0B

c

A

c p

a

q

a

b q

b

1B

1B α

α = ~qac

Figure 21.3: Illustrating α and α Proof. Since α ≤ ⃗qac , it follows that α ≤ ⃗qac . So it is sufficient to prove that α(t) = q for t ∈ id(c) − (id(a) ∪ id(b)). To see this, let e ∈ id(c) − (id(a) ∪ id(b)), that is, e ≤ c, but e ≰ a and e ≰ b. Let f ∈ A − fil(e), that is, e ≰ f . Then either a ≰ f or b ≰ f ; otherwise, we would get a ∨ b = c ≤ f , implying that e ≤ f , a contradiction. Therefore, _ α(A − id(f )) ≥ αa ∧ αb = 1B ∧ q = q,

from which it follows that

α(e) =

^

f ∈A−fil(e)

_

α(A − id(f )) ≥ q,

for any e ∈ id(c) − (id(a) ∪ id(b)). Since αe ≤ ⃗qac , it follows that eα ≤ ⃗qac , which implies that eα ≤ q. We conclude that eα = q, as claimed. Lemma 21.10. The map a 7→ χa is an embedding of A into A[B]. Proof. This map is clearly one-to-one and meet-preserving. To compute χa ∨ χb , let α be the pointwise join. Applying Lemma 21.9, we obtain that α = χa∨b , proving that χa ∨ χb = χa∨b . Note that while we get many copies of B in A[B], we obtain only one copy of A in A[B]—this is not surprising since there is, in general, only one copy of M3 in M3 [B] and, further, A[C2 ] is isomorphic to A. An embedding Let A and B be finite nontrivial lattices. Then for an ideal I of A, we have a natural embedding of I[B] into A[B].

21.4. Computing

283

Lemma 21.11. Let I be an ideal of A. Then there exists a unique embedding f : I[B] → A[B], such that f (⃗ pab ) = p⃗ab , for p ∈ B and a < b in I, where in the formula f (⃗ pab ) = p⃗ab , the first p⃗ab is in I[B], whereas, the second is in A[B]. Proof. For α ∈ I[B], we define αz (∈ B A ), the “zero-padded” version of α, by ( αx, if x ∈ I; z α (x) = 0B , otherwise. It is obvious that αz ∈ A[B]. Define f : I[B] → A[B], pab ) = p⃗ab . By Lemma 21.22, by α 7→ αz . Clearly, f is an embedding and f (⃗ we have the uniqueness. Distributive lattices For the lattices X and Y , let hom{∨,0} (X, Y ) denote the set of {∨, 0}-homomorphism of X into Y , regarded as a {∨, 0}-semilattice. The following statement is a generalization of the fact that balanced triples are Boolean for a distributive lattice B (see Section 6.3). Lemma 21.12. Let B be a distributive lattice. Then A[B] = hom{∨,0} (A, B δ ).

(14)

Proof. By Lemma 21.1, for any lattice B, we always have A[B] ⊆ hom{∨,0} (A, B δ ). To prove the reverse containment in (14), let us assume that B is distributive. Let α ∈ hom{∨,0} (A, B δ ), and let p be the term introduced in (7). By Lemma 21.2, it is sufficient to show that p(α) ≤ α. For a ∈ A, compute: ^ _ p(α)(a) = αy x≱a y≰x

=

_ ^ Y ( αν(x) | ν ∈ (A − id(x)) ) x≱a

=

_ _ Y ( αν(x) | ν ∈ α(A − id(x) ) x≱a

(B is distributive)

x≱a

x≱a

(α ∈ hom{∨,0} (A, B δ )).

284

21. Tensor Extensions

To show that p(α) ≤ α, that is, that p(α)a ≤ αa, for all a ∈ A, it suffices to show that _ α( (15) ν(x)) ≤ αa, x≱a

for any ν∈

(16)

Y

x≱a

(A − id(x)).

W So let (16) hold, and let b = x≱a νx. We claim that b ≥ a. Indeed, otherwise, from the definition of b, we see that νb ∈ id(b), contradicting (16). Since α is antitone, we obtain that αb ≤ αa, completing the proof of (15).

21.5.

Congruences

In this section we compute many properties of the congruences of A[B], to lay the groundwork for the proof of the isomorphism (4). Congruence spreading We show that congruence-perspectivities and congruence-projectivities in A naturally translate into a family of congruence-perspectivities and congruenceprojectivities in A[B]. Lemma 21.13. Let a, b, c, d ∈ A with a < b and c < d. Let p, q ∈ B with p ≤ q. If [a, b] is congruence-perspective onto [c, d] in A, then [⃗ pab , ⃗qab ] is congruence-perspective onto [⃗ pcd , ⃗qcd ] in A[B]. Proof. For a down congruence-perspectivity of [a, b] onto [c, d], we only have to compute four inequalities and a componentwise meet; these are trivial. For the up congruence-perspectivity, we have to verify that p⃗ab ∨ ⃗qcd = ⃗qab ; we perform the componentwise join and then apply Lemma 21.9. Since congruence-projectivity is the transitive extension of congruence-perspectivity, we obtain: Corollary 21.14. If the interval [a, b] is congruence-projective onto the interval [c, d] in A, then the interval [⃗ pab , ⃗qab ] is congruence-projective onto the interval [⃗ pcd , ⃗qcd ] in A[B]. If a, b ∈ A with a < b and γ is a congruence of A[B], we define γab as the congruence of B that corresponds to the restriction of γ to Bab (defined in Section 21.4) under the isomorphism γab : p 7→ p⃗ab of B and Bab . The following statement easily follows from Lemma 21.13.

21.5. Congruences

285

Lemma 21.15. Let a, b, c ∈ A with a ≤ b ≤ c. Let γ be a congruence of A[B]. Then γac = γab ∧ γbc in B. Lemma 21.16. Let a, b, c, d ∈ A with a ≤ b and c ≤ d. If con(c, d) ≤ con(a, b), then γab ≤ γcd in B. Proof. First, let us assume that c ≺ d. Then it follows from con(c, d) ≤ con(a, b) that [a, b] is congruence-projective onto [c, d]. Let p ≤ q in B and p ≡ q (mod γab ). Then p⃗ab ≡ ⃗qab (mod γ). By Corollary 21.14, [⃗ pab , ⃗qab ] pcd , ⃗qcd ], so p⃗cd ≡ ⃗qcd (mod γ), that is, p ≡ q is congruence-projective onto [⃗ (mod γcd ), verifying that γab ≤ γcd . Second, in the general case, let c = c0 < c 1 < · · · < c n = d be a maximal chain in the interval [c, d] in B. Then con(cj−1 , cj ) ≤ con(c, d) ≤ con(a, b) for every 0 < j ≤ n. Hence, by the previous paragraph, γab ≤ γcj−1 cj . Applying Lemma 21.15 (n − 1) times, we obtain that ^ γcj−1 cj = γcd . γab ≤ 0 0 of a finite lattice L is bound isolating (BI, for short), if {0} and {1} are congruence blocks of α. The lattice S(p, q) has two BI congruences: con(ap , bp ) < con(aq , bq ) (see Figure 22.3). The lattice S(p, q) has two BI congruences: con(ap , bp ) < con(aq , bq ) (see Figure 22.3). In addition, it has the two trivial congruences 0 and 1, so Con S(p, q) is the four-element chain. The lattice K We are going to construct the lattice K (for the Representation Theorem) as an extension of Frame P . (The principal congruence of K representing p ∈ P − will be con(ap , bp ).) We add the set {cp,q , dp,q , ep,q , fp,q , gp,q } to the sublattice {o, ap , bp , aq , bq , i} −

of F for p < q ∈ P , as illustrated in Figure 22.2, to form the sublattice S(p, q). ∥ For p ∈ P , let Cp = {o < ap < bp < i} be a four-element chain. We define the set [ [ ∥ K = ( S(p, q) | p < q ∈ P − ) ∪ ( Cp | p ∈ P ) ∪ {a0 , a1 }. Now we are ready to define the lattice K. We make the set K into a lattice by the following nine rules:

22.2. Proving the RT

299

i

bp

a0 = b0

a1 = b1

ap

bq

aq

o

Figure 22.1: The lattice Frame P i gp,q bp

bq fp,q ep,q dp,q

ap

aq cp,q o

Figure 22.2: The lattice S(p, q) for p < q ∈ P i

i

bq

bp

aq

ap

o

con(ap , bp )

bq

bp

aq

ap

o

Figure 22.3: The BI congruences of S(p, q)

con(aq , bq )

300

22. The RT for Principal Congruences i

bq

bp

bq ′

aq

ap

aq′

o

Figure 22.4: The lattice SC = S(p < q, q < q ′ )

i

bq

bp

bq ′

aq

ap

aq′

o

Figure 22.5: The lattice SV = S(p < q, p < q ′ ) with q ̸= q ′

i

bp

bq

ap

aq

bp′

ap′

o

Figure 22.6: The lattice SH = S(p < q, p′ < q) with p ̸= p′

22.2. Proving the RT

301

(i) The operations ∨ and ∧ are idempotent and commutative and o is the zero and i is the unit of K. ∥

(ii) For p ∈ P and x, y ∈ Cp ⊆ K, we define x ∨ y, x ∧ y in K as in the chain Cp . (So Cp is a sublattice of K.) (iii) For p < q ∈ P − and x, y ∈ S(p, q) ⊆ K, we define x ∨ y, x ∧ y in K as in the lattice S(p, q). (So S(p, q) is a sublattice of K.) ∥

(iv) For p ∈ P , x ∈ Cp− , and y ∈ K − Cp , the elements x and y are complementary in K, that is, x ∨ y = i and x ∧ y = o. (v) For x = a0 and for x = a1 , the element x is complementary to any element y ̸= x ∈ K − . In the following four rules, let p < q, p′ < q ′ ∈ P − , x ∈ S(p, q)− , and y ∈ S(p′ , q ′ )− . By rule (iii), we can assume that {p, q} ̸= {p′ , q ′ }.

(vi) If {p, q} ∩ {p′ , q ′ } = ∅, then the elements x and y are complementary in K. (vii) If q = p′ , we form x ∨ y and x ∧ y in K in the lattice SC = S(p < q, q < q ′ ), illustrated in Figure 22.4. (viii) If p = p′ and q ̸= q ′ , we form x ∨ y and x ∧ y in K in the lattice SV = S(p < q, p < q ′ ), illustrated in Figure 22.5. (ix) If q = q ′ and p ̸= p′ , we form x ∨ y and x ∧ y in K in the lattice SH = S(p < q, p′ < q), as illustrated in Figure 22.6. In the last three rules, C for chain, V for V-shaped, H for Hat-shaped refer to the shape of the three-element order {p, q} ∪ {p′ , q ′ } in P − , respectively. Observe that Rules (vi)–(ix) exhaust all possibilities under the assumption that {p, q} = ̸ {p′ , q ′ }. Note that S SC SV SH

= S(p, q), = S(p < q, q < q ′ ), = S(p < q, p < q ′ ), = S(p < q, p′ < q)

302

22. The RT for Principal Congruences

are sublattices of K. Informally, these rules state that to form K, we add elements to Frame P so that we get the sublattices listed in the previous paragraph. Let us summarize some of our observations. Lemma 22.2. For p ∈ P − , the principal congruence con(ap , bp ) is a BI congruence and every principal BI congruence is of this form. There are two principal congruences that are not BI, namely, 0 and 1. ∥

∥

For p ∈ P , let εp = con(ap , bp ), be viewed as a partition. Let H ⊆ P , and let εH denote the equivalence relation on L: _ εH = ( εp | p ∈ H ). Let β be a BI congruence of the lattice L. We associate with β a subset of the ordered set P − : Base(β) = { p ∈ P − | ap ≡ bp (mod β) }. Then Base(β) is a down set of P − . The following lemma is easy to compute. Lemma 22.3. Let H be a down set of P − . Then the binary relation: [ [ β H = εH ∪ ( conS(p,q) (aq , bq ) | q ∈ H ) ∪ ( conS(p,q) (ap , bp ) | p ∈ H )

is a BI congruence on L and the correspondence γ : β 7→ Base(β)

is an isotone bijection between the ordered set of BI congruences of L and the ordered set of down sets of P − . We extend γ by 0 7→ {0} and 1 7→ P . Then γ is an isomorphism between Con(L) and Dn− P , the ordered set of nonempty down sets of P verifying Theorem 22.1, the Representation Theorem for Principal Congruences. Let P be an ordered set and let D be a finite distributive lattice with J(D) = P . Then P is bounded iff D has a unique dual atom. So we get an important consequence of the Representation Theorem (see my joint paper with H. Lakser [149]). Corollary 22.4. Let P be a bounded ordered set and let D be a finite distributive lattice with J(D) = P . Then the lattice K of the Representation Theorem satisfies that Con K = D and Princ K = P . Proof. By Lemma 22.2.

22.3. An independence theorem

22.3.

An independence theorem

22.3.1

Frucht lattices

303

For a finite lattice L, G. Cz´edli [35] proved that the two related structures Princ L and Aut L are independent. To discuss his result, we’ll need the following. By a graph C = (V, E), we mean a nonempty set V of vertices and a set E of edges; every e ∈ E is a two-element subset of V . Figure 22.7 illustrates how we draw graphs: the vertices are represented by small circles ; the circles representing two distinct vertices x and y are connected by a line segment iff {x, y} is an edge. An automorphism α of C is a bijection of V preserving the edges, that is, {x, y} is an edge iff {αx, αy} is an edge for x ̸= y ∈ V . The automorphisms of C form a group, denoted by Aut C. Now let C be a graph (V, E), and we order the set Frucht C = {0, 1}∪V ∪E by 0 < v, e < 1, for all v ∈ V and e ∈ E, and v < e for v ∈ V and e ∈ E iff v ∈ e (as in R. Frucht’s [93]). The ordered set Frucht C is a lattice; it is in fact atomistic and of length 3. Figure 22.7 shows a small example. It is clear from the picture that C is coded into Frucht C, so Aut(Frucht C) = Aut C. The lattice Frucht C is almost always simple. So we obtain the following result (R. Frucht [94]). 1 c w y

d

C

x

u

v

a

b

x

y

w

v b

c

d

u a

Frucht C 0

Figure 22.7: Constructing the Frucht lattice Theorem 22.5. For every graph C, there is an atomistic lattice F of length 3 with Aut C ∼ = Aut F . 22.3.2

An independence result

Theorem 22.6. Let P be a finite ordered set with at least two elements and let G be a group. Then there exists a finite lattice L such that Princ L is order isomorphic to P and Aut L is group isomorphic to G. I will sketch a new approach to this result, based on my paper [119].

304

22. The RT for Principal Congruences

Note that Cz´edli proved more than what is stated in Theorem 22.6 by constructing a lattice L that is self-dual and of length 16. To prove this, he had to work much harder than I. Let us call a family { Ki | i ∈ I } of finite rigid lattices mutually rigid if Ki has no embedding into Kj for i ̸= j ∈ I. The following is a special case of some results from the 1970s (see also G. Cz´edli [38]). Theorem 22.7. For every cardinal m, there exists a mutually rigid and simple family { Ki | i ∈ I } of finite lattices satisfying |I| = m. In the early 1970s, much more sophisticated results were published dealing with endomorphism semigroups and full embeddings of categories (see for example, my joint paper with J. Sichler [196] and also the paper Z. Hedrlin and J. Sichler [208]). Let P and G be given as in Theorem 22.6. The lattice K we have just constructed satisfies that Princ K ∼ = P . However, Aut K ∼ = Aut P . − Let I = P . For every [o, ap ], we insert the lattice Kp into K provided by Theorem 22.7, identifying the bounds, obtaining the lattice K. We then have Princ K ∼ = P and Aut K is rigid. Now take the Frucht lattice, Frucht G (see Section 22.3.1); it is a lattice of length 3, satisfying Aut(Frucht G) ∼ = G. We obtain the lattice L by forming the disjoint union of K and Frucht G and identifying the bounds (see Figure 22.8). The diagram does not show the lattices Kp . i

Frucht G

a0 = b0

bq

bp a1 = b1 ap

aq Kq o

Figure 22.8: The final step

22.4.

Discussion

We can easily represent the finite ordered set C22 as Princ K of a lattice K (see the first diagram of Figure 22.9). Indeed, take the lattice K = C22 and then Princ K = C22 , in fact, Princ K = Con K. In general, however, Princ K = C22 does not imply that Princ K = Con K. For an example, take

22.4. Discussion

305

the lattice K, the third diagram in Figure 22.9. Then we have Princ K = C22 but Princ K ̸= Con K, the second diagram in Figure 22.9. So we can ask for the characterization of Princ K as a finite ordered subset of Con K for a finite lattice K. Clearly, J(Con K) ⊆ Princ K. For a finite lattice L, define J+ (L) = {0, 1} ∪ J(L). Problem 71. Let D be a finite distributive lattice. Let Q be a subset of D satisfying J+ (D) ⊆ Q ⊆ D. When is there a finite lattice K such that Con K is isomorphic to D and under this isomorphism Princ K corresponds to Q? An affirmative solution for the special case J+ (D) = Q is presented in the proof of Theorem 22.1. The most natural problem concerning Theorem 22.1 is the following. Problem 72. Can we characterize the ordered set Princ L for a lattice L as a directed ordered set with zero? G. Cz´edli [36] solved this problem in the affirmative for countable lattices. Even more interesting would be to characterize the pair P = Princ L in S = Conc L (the join-semilattice of compact congruences of L) by the properties that P is a directed ordered set with zero that join-generates S. We rephrase this to make this independent of the congruence lattice characterization problem (see LTF-[105] and LTS1-[206]). Problem 73. Let S be a representable join-semilattice. Let P ⊆ S be a directed ordered set with zero and let P join-generate S. Under what conditions is there a lattice K such that Conc K is isomorphic to S and under this isomorphism Princ K corresponds to P ? For a lattice L, let us define a valuation v on Conc L as follows.

Figure 22.9: A Princ K with a Con K and a lattice K

306

22. The RT for Principal Congruences

For a compact congruence α of L, let v(α) be the smallest integer n such that the congruence α is the join of n principal congruences. A valuation v has some obvious properties, for instance, v(0) = 0 and v(α ∨ β) ≤ v(α) + v(β). Note the connection with Princ L: Princ L = { α ∈ Conc L | v(α) ≤ 1 }.

Problem 74. Let S be a representable join-semilattice. Let v map S to the natural numbers. Under what conditions is there an isomorphism γ of S with Conc K for some lattice K so that under γ the map v corresponds to the valuation on Conc K? Let D be a finite distributive lattice. In Chapter 8, we represent D as the congruence lattice of a finite (sectionally complemented) lattice K in which all congruences are principal (that is, Con K = Princ K). In Chapter 10, we construct a planar semimodular lattice K with Con K = Princ K ∼ = D. We discuss this and some related questions in Part VI. In the finite variant of Problem 74, we need an additional property. Problem 75. Let S be a finite distributive lattice. Let v be a map of D to the natural numbers satisfying v(0) = 0, v(1) = 1, and v(a ∨ b) ≤ v(a) + v(b) for a, b ∈ D. Is there an isomorphism γ of D with Con K for some finite lattice K such that under γ the map v corresponds to the valuation on Con K? Problem 76. Let K be a bounded lattice. Does there exist a complete lattice L such that Con K ∼ = Con L? What can we say about the (planar) semimodular case? Problem 77. Characterize Princ L for (i) a finite semimodular lattice L; (ii) a planar semimodular lattice L; (iii) an SPS lattice L. Problem 71 becomes much harder for planar semimodular lattices. Problem 78. Let D be a finite distributive lattice. Let Q be a subset of D satisfying J+ (D) ⊆ Q ⊆ D. When is there a finite semimodular lattice L (or a finite planar semimodular lattice or, the best, an SPS lattice) such that Con L is isomorphic to D and under this isomorphism Princ L corresponds to Q? We state an interesting special case. Problem 79. Let D be a finite distributive lattice. When is there a finite planar semimodular lattice L such that Con L is isomorphic to D and under this isomorphism Princ L corresponds to J+ (D)? Contrast this with the note following Problem 71.

Chapter

23

Minimal RTs

23.1.

The Minimal RT

Recall that for a finite distributive lattice D, J+ (D) = J(D) ∪ {0, 1} (see Section 22.4). For a finite lattice L, (1)

J+ (Con L) ⊆ Princ L ⊆ Con L,

since every join-irreducible congruence is generated by a prime interval; furthermore, 0 = con(x, x) for any x ∈ L and 1 = con(0, 1). Let us call the subset Q of D principal congruence representable, if there is a finite lattice L such that Con L is isomorphic to D and Princ L corresponds to Q under this isomorphism. Of course, such a subset Q satisfies (2)

J+ (D) ⊆ Q ⊆ D.

We call such a subset Q ⊆ D a candidate for representability. In this chapter, we deal with an extreme case of (2): all proper principal congruences are join-irreducible, that is, (3)

Princ L = J+ (Con L).

In this case, we say that a finite lattice L has a minimal set of principal congruences. Similarly, for a finite distributive lattice D, we call the finite 307 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 G. Grätzer, The Congruences of a Finite Lattice, https://doi.org/10.1007/978-3-031-29063-3_23

308

23. Minimal RTs

lattice L a minimal representation of D, if D and Con L are isomorphic and (3) holds for L. In my paper [107], I raised the following question. Let D be a finite distributive lattice. Under what conditions does D have a minimal representation? We solve this problem as follows. Theorem 23.1 (Minimal Representation Theorem, MRT). Let D be a finite distributive lattice. Then D has a minimal representation L iff D has at most two dual atoms. MRT naturally splits up into three statements. Theorem 23.2 (MRT1, my joint paper with H. Lakser [150]). Let D be a finite distributive lattice with three or more dual atoms. Then D does not have a minimal representation L. Theorem 23.3 (MRT2, my paper [107]). Let D be a finite distributive lattice with exactly one dual atom. Then D has a minimal representation L. Theorem 23.4 (MRT3, my joint paper with H. Lakser [150]). Let D be a finite distributive lattice with exactly two dual atoms. Then D has a minimal representation L. Note that Theorem 23.3 is the same as Corollary 22.4. To illustrate the concept of minimal representation, let us look at some examples. Example 23.5. C22 = B2 is minimally represented by itself. Example 23.6. C3 × C2 has a minimal representation as illustrated by Figure 23.1. Example 23.7. The finite distributive lattice B3 has no minimal representation. Let us assume that L is a finite lattice minimally representing B3 , that is, with Con L = B3 and Princ L = J+ (B3 ). Let α1 , α2 , α3 be the three join-irreducible congruences of L, note that α1 ∨ α2 ∨ α3 = 1 and so 0≡1

(mod α1 ∨ α2 ∨ α3 ).

By Theorem 3.1, there is a finite chain 0 = x0 < x1 < · · · < xn = 1 such that for every 0 ≤ i < n, there is 1 ≤ ji ≤ 3 satisfying xi ≡ xi+1 (mod αji ) and j0 ̸= j1 ̸= . . . ̸= jn−1 . Without loss of generality, let j0 = 1 and j1 = 2. Observe that 0 < con(x0 , x1 ) ≤ α1

23.2. Three or more dual atoms

309

b a a b

a

b

a a

b b a

b

b

c c

Figure 23.1: A minimal representation and so con(x0 , x1 ) = α1 because α1 is an atom. Since n > 1, it follows that α1 ∨ α2 = con(x0 , x2 ) ∈ Princ L, contrary to the assumption that J+ (B3 ) = Princ L. Example 23.8. The nine-element distributive lattice D = C23 has a minimal representation. We take N6 (as in the first diagram of Figure 23.2) as a minimal representa. tion of the chain C3 . Then the glued sum N6 +N6 is a congruence representation of D = C23 but it is not minimal (see the second diagram of Figure 23.2). Indeed, con(a, b), con(b, c) < con(a, c) so con(a, b) ∨ con(b, c) = con(a, c) is principal and join-irrreducible. The third diagram of Figure 23.2 provides a minimal representation of D = C23 .

23.2.

Three or more dual atoms

We begin with the following result. Lemma 23.9. For any finite distributive lattice D, there is a one-to-one correspondence between the set of dual atoms of D and the set of maximal elements of the ordered set J(D). Proof. By the Basic RT, to each dual atom a of D, there corresponds a unique p ∈ J(D) with p ≰ a, which is perforce maximal. The inverse correspondence assigns to each maximal element of J(D) the join of all the other elements of J(D), which is a dual atom of D.

310

23. Minimal RTs

c

b

a

Figure 23.2: Illustrating Example 23.8 In the proof of MRT1 (Theorem 23.2), we use the following result. Theorem 23.10. Let L be a finite lattice and x < y be elements of L. Let A be W an antichain of size at least 2 of join-irreducible congruences of L with A = con(x, y). Then for each α ∈ A, there is a join-irreducible congruence β on L such that the congruence α ∨ β is principal and join-reducible. Proof. Let x = c0 ≺ c1 ≺ · · · ≺ cn = y be a (maximal) chain C in the interval [x, y] and let β i = con(ci , ci+1 ) for 0 ≤ i < n. Then β i is a join-irreducible congruence of L and _ _ ( β i | 0 ≤ i < n ) = con(x, y) = A. Let J ⊆ {0, 1, . . . n − 1} so that { con(ci , ci+1 ) | i ∈ J } are the maximal elements in the ordered set { con(ci , ci+1 ) | 0 ≤ i < n }. Then A = { con(ci , ci+1 ) | i ∈ J } (see, for instance, Corollary 111 in LTF-[105]). So let α = con(cj , cj+1 ). Let [ck , cl ], with k ≤ j and j + 1 ≤ l, be a maximal subinterval of C with con(ck , cl ) = α. We cannot have both k = 0

23.3. Exactly two dual atoms

311

W and l = n, otherwise, α = A, contradicting the assumptions on A. Without loss of generality, let l < n and define β = con(cl , cl+1 ). By the definition of l, it follows that β ≰ α. So β is join-irreducible and α ∨ β is principal (indeed, α ∨ β = [ck , cl+1 ]). Corollary 23.11. Let D be a finite distributive lattice W with an antichain A of join-irreducible elements with at least 3 elements. If A = 1, then D does not have a minimal representation. Proof. Assume that the finite lattice L with bounds o and i provides a minimal representation of D, that is, Princ L = {0, 1} ∪ J(Con L) and there is an isomorphism between D and Con L. Let Q ⊆ D correspond to Princ L under this isomorphism. By Theorem 23.10 applied to the interval [o, i], for each a ∈ A there is a b ∈ J(D), such that a ∨ b is a join-reducible element of Q. Since A has at least 3 elements, a ∨ b ̸= 1, a contradiction. From Corollary 23.11, we prove MRT1 as follows. Let M be W the set of maximal elements of J(D). Then M is an antichain in D and M = 1. By Lemma 23.9, M has at least three elements. Thus by Corollary 23.11, D does not have a minimal representation.

23.3.

Exactly two dual atoms

23.3.1

Constructing the lattice L

Let D be a finite distributive lattice with exactly two dual atoms and P = J(D). By Lemma 23.9, P has exactly two maximal elements, p0 , p1 . Let P0 =↓ p0 and P1 =↓ p1 . Our construction is based on the frame lattice, Frame P , introduced in Section 22.2, illustrated in Figure 22.1 with the chains Cp = {0, ap , bp , 1} for p ∈ P and the lattice S(p, q) (see Figure 22.2). For all p < q ∈ P , we insert S(p, q) into Frame P , to form Frame S P , the frame lattice with S(p, q) (see Figure 23.3). Let L0 be the lattice Frame S P0 , with zero o and unit i, and let L1 , with zero i′ and unit o′ , be the dual of the lattice Frame S P1 , where we denote by x′ the element of L1 corresponding to x ∈ Frame S P1 under the duality. . Now we are ready to construct, Base P , the base lattice for P , as L0 + L1 (see Figure 23.4). We then show that Base P is a minimal representation of the ordered set . P0 ∪ P1 , the free union of P0 and P1 with p ∥ q for p ∈ P0 and q ∈ P1 . This is easy since con(x, y) = 1Base P if x < i < y. Now each element r ∈ P0 ∩ P1 determines two distinct congruences of Base P , the first, con(ar , br ), in the sublattice L0 , and the second, con(a′r , b′r ),

312

23. Minimal RTs

br

a0 = b0

a1 = b1 ar

Figure 23.3: Frame S P : adding S(p, q) to Frame P for p < q ∈ P in the sublattice L1 . Our task is to identify these two congruences, which we do with the bridge construction. A bridge lattice, Bridge, is C23 with an additional element m, turning the covering square of the right corner into an M3 (see the first diagram in Figure 23.5). An r-bridge lattice, Bridge(r), for r ∈ P0 ∩ P1 , is a bridge lattice with the elements subscripted with r (see Figure 23.5). We then obtain the desired lattice L for MRT2 (Theorem 23.3), by adding a bridge for each r ∈ P0 ∩ P1 to the base lattice Base P by forming the disjoint union of Base P and Bridge(r), and then identifying the five elements ar , br , i = i′ , b′r , a′r (see Figure 23.6). We first must show that adding a bridge results in a lattice. Then it is clear that adding a bridge will identify con(ar , br ) with con(a′r , b′r ) for each r ∈ P0 ∩ P1 . Our task will then be to show that no other congruences collapse and that all principal congruences distinct from 1L remain join-irreducible. We do this in Sections 23.3.2–23.3.6. We need an easy statement from the folklore. Lemma 23.12. Let K be a lattice and let a ≺ c ≺ b in K. Let K ∗ = K ∪ {u} and define u ∧ b = a and u ∨ b = c. Then K ∗ is a lattice extension of K and, for x ∈ K, ( u for x ≤ a; u∨x= b ∨ x, otherwise, and dually. We apply Lemma 23.12 five times for each r ∈ R, that is, we add tr , ur , u′r , sr , and mr successively, to conclude that L is a lattice.

23.3. Exactly two dual atoms

313

o′

a′0 = b′0

a′q

a′p

b′q

b′p

a′r

a′1 = b′1

b′r

i = i′ br

a1 = b1 a0 = b0 ar

Figure 23.4: The base lattice of P , Base P 23.3.2

Fusion of ordered sets

We present a construction for ordered sets to lay the foundation for the Bridge Theorem. Let P be an ordered set and let A be a nonempty convex subset of P . We define an ordered set Fuse(P, A) that is obtained in a natural manner by fusing the subset A to a single element ιA ; if there is no danger of confusion, we write ι for ιA . That is, we let F = Fuse(P, A) = (P − A) ∪ {ι} and define an order on F . We work with the strict order