Rational Sphere Maps [341, 1 ed.] 9783030758080, 9783030758097

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Progress in Mathematics 341

John P. D’Angelo

Rational Sphere Maps

Progress in Mathematics Volume 341

Series Editors Antoine Chambert-Loir

, Université Paris-Diderot, Paris, France

Jiang-Hua Lu, The University of Hong Kong, Hong Kong SAR, China Michael Ruzhansky, Ghent University, Ghent, Belgium, Queen Mary University of London, London, UK Yuri Tschinkel, Courant Institute of Mathematical Sciences, New York, USA

More information about this series at http://www.springer.com/series/4848

John P. D’Angelo

Rational Sphere Maps

John P. D’Angelo Department of Mathematics University of Illinois, Urbana-Champaign Urbana, IL, USA

ISSN 0743-1643 ISSN 2296-505X (electronic) Progress in Mathematics ISBN 978-3-030-75808-0 ISBN 978-3-030-75809-7 (eBook) https://doi.org/10.1007/978-3-030-75809-7 Mathematics Subject Classification: 32M99, 32A10, 15B57, 32V99 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 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

Preface

The unit circle S1 in the complex number system C and its self-mappings have played a major role in the history of mathematics. Below we give many striking examples. The central theme throughout this book will be to understand higher dimensional analogues, where things are more subtle and ideas from many fields of mathematics make their appearance. In one dimension, if f is holomorphic (complex analytic) in a neighborhood of the closure of the unit disk B1 , and f maps the circle to itself, then f is a finite Blaschke product. One can draw the same conclusion assuming only that f is a proper holomorphic mapping from B1 to itself. In particular such functions are rational. Our primary topic will be the study of holomorphic rational maps sending the unit sphere in the source complex Euclidean space Cn to the unit sphere in some target space CN . We call such mappings rational sphere maps. We use the terms monomial sphere map and polynomial sphere map with obvious meaning; even these mappings exhibit remarkably interesting and complicated behavior as the source and target dimensions rise. In this book, a rational sphere map f is complex analytic where it is defined. In other words, f depends on the z variables but not on the
z variables. In Chap. 6 we briefly discuss some differences between holomorphic polynomial sphere maps and real polynomial sphere maps. In particular, in complex dimension n at least 2, the only non-constant holomorphic polynomial maps sending the unit sphere to itself are linear, whereas there are real polynomial sphere maps of every degree. I considered the title Complex Analytic Rational Sphere Maps to prevent possible confusion, but the shorter title seems more appealing. In some sense, this book is a research monograph, as it develops in a systematic fashion most of the research on rational sphere maps done in the last forty years. It differs however from many monographs in several ways, which we now describe. First of all, scattered throughout the book are a large number of computational examples; the author feels that merging the abstract and concrete enhances both. Many times in his work on this subject, a theorem resulted from trying to cast a collection of examples into one framework. Some readers will stare at these

v

vi

Preface

formulas, observe subtle patterns, and pose their own open questions. Other readers may find the formulas distracting. I hope that I have achieved the right balance. Chaps. 3 and 4 include formulas that could not easily be obtained by hand computation. Mathematica was used to help perform some of these calculations. The author acknowledges assistance in coding received from Jiri Lebl, Daniel Lichtblau, Dan Putnam, and Bob Vanderbei. Some results from coding have led to theorems and others have led to unanswered questions. Both types of results appear here. Section 4.9 includes recent code by Lichtblau [1]. Second, I have included more than 100 exercises. Most of these are computational and have a simple purpose: give the reader something to do when things become confusing. These exercises are numbered by Chapter and often appear in the middle of a section. Given the many search tools available, this method seems most appropriate. This book hopes to expose some beautiful mathematics; it is not a calculus text where long lists of exercises appear at the end of each section. The exercises are meant for readers who enjoy them but none are indispensable to the general development. I have posed fifteen open problems here. They belong to many parts of mathematics; the symmetry of the unit sphere is responsible for their variety. These problems appear within the text but are repeated in a short chapter at the end of the book. The author hopes that this book will enhance research by engaging others in both what is known and where this knowledge leads. Section 1.7 provides a kind of global positioning system for the book. It locates where in the book some of the fundamental results are discussed and indicates what happens in each chapter. The author modestly hopes that both experts and novices find this map to be useful both in learning about rational sphere maps and navigating the book. To introduce the subject of rational sphere maps, we provide several examples in one dimension and indicate how to extend the ideas to higher dimensions. Example 1 Many elementary trigonometric identities are easily proved by combining the binomial expansion with de Moivre’s formula ðcosðhÞ þ i sinðhÞÞm ¼ ðeih Þm ¼ eimh ¼ cosðmhÞ þ i sinðmhÞ:

ðÞ

In fact every trig identity follows from the following facts: 1. 2. 3. 4.

The complex numbers are a field. iz iz iz iz cosðzÞ ¼ e þ2e and sinðzÞ ¼ e þ2ie : For complex numbers z; w we have ez þ w ¼ ez ew . Complex conjugation is continuous, and hence e
z ¼ e
z .

It is natural to take (2) as the definition of the trig functions. Combining (2) and (3) yields cos2 ðzÞ þ sin2 ðzÞ ¼ 1. Combining (3) and (4) yields jeih j2 ¼ 1 when h is real. Item (4) is needed because the exponential function is defined by its power series; one needs to know that the conjugate of a convergent infinite sum is the infinite sum of the conjugated terms.

Preface

vii

Formula (*) is closely related to the map z ! zm , which sends the circle to itself, and hence is a monomial sphere map. One higher dimensional analogue of this mapping will be the tensor product z ! zm for z 2 Cn . The tensor product provides a monomial sphere map, but requires a higher dimensional target space. We will encounter restricted tensor products and a kind of tensor division. Example 2 The unit circle can be regarded as the unitary group Uð1Þ. The m-th roots of unity form a finite cyclic subgroup Cm under multiplication. The map z ! zm sends the circle to itself and is invariant under Cm . We will study analogues in higher dimensions in Chap. 5, by associating both invariant and equivariant groups with rational sphere maps. The unitary group UðnÞ and the holomorphic automorphism group of the unit ball arise throughout. In addition, representations of Cm in Uð2Þ lead in Chap. 3 to interesting combinatorial results. Example 3 The theory of Fourier series is based upon the complete orthonormal system feimh g for L2 ðS1 Þ. Closely related is the result that the monomials z ! zm form a complete orthogonal system for L2 ðB1 Þ. The analogous statement for the monomials z ! za holds in any dimension. Example 4 Riemann surfaces arose from trying to visualize the space of solutions to equations such as zm ¼ w. We will study proper mappings from Bn to BN ; the image of the ball is then an n-dimensional complex variety. We also study a subvariety of Bn  CN associated with a rational sphere map. This variety contains the graph of the map, but exceptional fibers often arise. Example 5 Each factor (including the eih term) of the Blaschke product eih

m Y aj  z 1 
aj z j¼1

can be regarded as an automorphism of the unit disk. In n dimensions, the automorphism group of the unit ball Bn is the Lie group SUðn; 1Þ divided by its center. We will see tensor products of automorphisms, but (as in Example 1) new phenomena arise. Not every rational sphere map is a tensor product of automorphisms. Example 6 Example 5 shows that every polynomial q that does not vanish on the closed unit disk is the denominator of a rational sphere map that is reduced to lowest terms. Proving the analogous statement in higher dimensions is much more subtle and seems to require Hermitian analogues of Hilbert’s 17-th problem. Let us elaborate. Suppose z 2 Cn and rðz;
zÞ is a real-valued polynomial. When the values of r are non-negative, we naturally ask whether r is a Hermitian sum of squares; that is, can we write rðz;
zÞ ¼

k X

jf j ðzÞj2

j¼1

for (holomorphic) polynomials f j ? The answer is not necessarily. What can we say? The resulting ideas (see Chap. 2) enable us to prove the following result. Let q

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Preface

be a polynomial that does not vanish on the closed unit ball in Cn . Then there is an integer N and a polynomial mapping p : Cn ! CN such that pq is reduced to lowest terms and defines a rational sphere map. There are no bounds possible on N nor on the degree of p that depend only upon n and the degree of q. We emphasize that the easy proof in one dimension does not require these ideas. This discussion combines with Hermitian linear algebra to give Theorem 2.15, which provides a general description of all rational sphere maps. Example 7 In Chap. 3 we will introduce a class of polynomials in two variables that arise from considering group-invariant monomial sphere maps. These polynomials turn out to be related to Chebyshev polynomials and they exhibit a long list of remarkable properties. One of these properties is that the so-called freshman’s dream: ðx þ yÞd is congruent to xd þ yd modulo d if and only if d ¼ 1 or d is prime, holds for these polynomials f d ðx; yÞ as well. Example 8 In Chap. 7 we establish a sharp bound on the volume of the image of a polynomial sphere map. A one-dimensional version of this result is quite appealing and we discuss it in detail as well. The underlying theme in this book derives from the following simple observations. First, the collection of rational sphere maps with a given source dimension n and target dimension N has little algebraic structure, unless n ¼ N. In Chap. 2, we show that there is considerably more structure to the problem if we regard the target dimension as a variable. Determining the rational sphere maps of degree d in source dimension n and unspecified target dimension leads to a system of linear equations for the inner products of unknown vectors. See Theorem 2.15. If we assume these vectors are orthogonal, then we obtain a linear system for unknown non-negative numbers. This case is equivalent to the study of monomial sphere maps, which we investigate in Chaps. 3 and 4. Even in the monomial case, the dimension of the set of solutions tends to infinity as the degree tends to infinity. As usual in Mathematics, when there are too many solutions to a problem, one can restrict the solutions by optimizing various quantities. For example, in Chap. 7, we discuss the volume of the image of the ball under a polynomial sphere map of degree d. We show that the homogeneous mapping zd provides the maximum volume. Chaps. 3 and 4 consider minimizing two somewhat related quantities for monomial sphere maps of degree d in source dimension n. One of these quantities is the minimum target dimension; the other is the minimum value of the map at the point with coordinates all equal to 1. We obtain some rather difficult combinatorial and asymptotic results about these problems. Let us a say a few words about prerequisites. The author believes that everything in this book should be accessible to most mathematicians, including graduate students. Because of the symmetry of the unit sphere, however, the material interacts with nearly all fields of mathematics. We use basic facts from complex analysis, linear algebra, functional analysis, and algebra. We will sometimes use ideas from elementary differential geometry and we will employ combinatorial reasoning. No deep theorems are required. The only prerequisite is appreciation of the ideas.

Preface

ix

Numbering in this book is done by Chapter. Thus, for example, Proposition 1.5 means the fifth proposition in Chap. 1. It precedes Corollary 1.1, because there are no items called corollary before it. We do not number every displayed equation. This point is worth elaborating. Paul Halmos (I don’t know the precise reference, but he said so!) once suggested that every equation should be numbered, because even if the author never refers to a given equation, someone else might. On the other hand, numbering everything seems to clutter things too much. I hope, unrealistically of course, that I have compromised by numbering an equation if and only if it should be numbered. I wish to acknowledge various mathematicians who have contributed to my understanding of the ideas in this book, or who have coded some computations: Eric Bedford, Dan Burns, Paulo Cordaro, Peter Ebenfelt, Jim Faran, Franc Forstnerič, Dusty Grundmeier, Zhenghui Huo, Bernhard Lamel, Jiri Lebl, Daniel Lichtblau, Han Peters, Dan Putnam, Bob Vanderbei, and Ming Xiao. I also acknowledge Simon Kos, a physicist, who made an important contribution to the ideas in Chap. 3. Quite a few of the results in this book are outgrowths of work I did with Jiri Lebl and other work I did with Ming Xiao. Their contributions have been indispensable. Many other mathematicians have indirectly contributed, primarily via their own inspiring work. I have also benefited from attending meetings and conferences over the years. Let me specifically mention programs at the American Institute of Math, workshops in Serra Negra (Brazil), conferences at the Erwin Schrödinger Institute (Vienna, Austria), and various special sessions at AMS sectional meetings. I thank Chris Tominich of Springer for his role as editor and especially for his solicitation of useful reviews. I thank the anonymous reviewers for their comments; I modestly hope that I have improved the book by dealing with their suggestions and criticisms. During the preparation of this book I have been supported by NSF Grants DMS-1066177 and DMS 13-61001. I also acknowledge support from the Kenneth D. Schmidt Professorial Scholar award from the University of Illinois. I dedicate this book to my wife Annette and our four children John, Lucie, Paul, and Henry. Urbana, USA

Reference 1. D. Lichtblau, personal communication.

John P. D’Angelo

Contents

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1 1 6 10 15 19 22 24

2 Examples and Properties of Rational Sphere Maps . . . . . . 1 Definition and Basic Results about Rational Sphere Maps 2 Sphere-Ranks and Target-Ranks . . . . . . . . . . . . . . . . . . . 3 Ranks of Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Juxtaposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 The Tensor Product Operation . . . . . . . . . . . . . . . . . . . . 6 The Restricted Tensor Product Operation . . . . . . . . . . . . 7 An Abundance of Rational Sphere Maps . . . . . . . . . . . . 8 Some Results in Low Codimension . . . . . . . . . . . . . . . . 9 A Result in Sufficiently High Codimension . . . . . . . . . . . 10 Homotopy and Target-Rank . . . . . . . . . . . . . . . . . . . . . . 11 Remarks on Degree Bounds . . . . . . . . . . . . . . . . . . . . . . 12 Inverse Image of a Point . . . . . . . . . . . . . . . . . . . . . . . . 13 The General Rational Sphere Map . . . . . . . . . . . . . . . . . 14 A Detailed Rational Example . . . . . . . . . . . . . . . . . . . . . 15 An Example in Source Dimension 3 . . . . . . . . . . . . . . . .

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25 25 31 33 37 38 40 44 47 50 53 54 55 57 67 70

3 Monomial Sphere Maps . . . . . . . . . . . . . . . . . 1 Properties of Monomial Sphere Maps . . . . . 2 Some Remarkable Monomial Sphere Maps . 3 More on These Remarkable Polynomials . .

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73 74 77 83

1 Complex Euclidean Space . . . . . . . . . . . . . . . 1 Generalities . . . . . . . . . . . . . . . . . . . . . . . 2 The Groups AutðB1 Þ, SUð2Þ, and SUð1; 1Þ 3 Automorphisms of the Unit Ball . . . . . . . . 4 Hermitian Forms . . . . . . . . . . . . . . . . . . . 5 Proper Mappings . . . . . . . . . . . . . . . . . . . 6 Some Counting . . . . . . . . . . . . . . . . . . . . 7 A GPS for This Book . . . . . . . . . . . . . . .

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Contents

4 5 6 7 8 9 10

Cyclic Groups and Monomial Sphere Maps . . . . . . . . . . . . Circulant Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Pell Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Elaboration of the Method for Producing Sharp Polynomials Additional Tricks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Maps with Source Dimension 2 and Target Dimension 4 . . . Target-Ranks for Monomial Sphere Maps . . . . . . . . . . . . . .

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. 84 . 91 . 94 . 97 . 102 . 103 . 106

4 Monomial Sphere Maps and Linear Programming . . . . . . . . . . 1 Underdetermined Linear Systems . . . . . . . . . . . . . . . . . . . . . 2 An Optimization Problem for Monomial Sphere Maps . . . . . . 3 Two Detailed Examples in Source Dimension 2 . . . . . . . . . . 4 Results of Coding and Consequences in Source Dimension 2 5 Monomial Sphere Maps in Higher Dimension . . . . . . . . . . . . 6 Sparseness in Source Dimension 2 . . . . . . . . . . . . . . . . . . . . 7 Sparseness in Source Dimension at Least Three . . . . . . . . . . 8 The Optimal Polynomials in Degrees 9 and 11 . . . . . . . . . . . 9 Coding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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111 112 113 116 119 132 141 142 145 151

5 Groups Associated with Holomorphic Mappings . . . . . . 1 Five Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Examples of the Five Groups . . . . . . . . . . . . . . . . . . . 3 Hermitian-Invariant Groups for Rational Sphere Maps . 4 Additional Examples . . . . . . . . . . . . . . . . . . . . . . . . . 5 Behavior of Cf Under Various Constructions . . . . . . . 6 Examples Involving the Symmetric Group . . . . . . . . . 7 The Symmetric Group . . . . . . . . . . . . . . . . . . . . . . . . 8 Groups Arising from Rational Sphere Maps . . . . . . . . 9 Different Representations . . . . . . . . . . . . . . . . . . . . . . 10 Additional Results . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 A Criterion for Being a Polynomial . . . . . . . . . . . . . .

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153 153 156 158 162 164 167 170 172 174 176 179

6 Elementary Complex and CR Geometry . . . . . . . . . . . 1 Subvarieties of the Unit Ball . . . . . . . . . . . . . . . . . 2 The Unbounded Realization of the Unit Sphere . . . . 3 Geometry of Real Hypersurfaces . . . . . . . . . . . . . . 4 CR Functions and Mappings . . . . . . . . . . . . . . . . . 5 Strong Pseudoconvexity of the Unit Sphere . . . . . . 6 Comparison with the Real Case . . . . . . . . . . . . . . . 7 Varieties Associated with Rational Sphere Maps . . . 8 Examples of Xf . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 A Return to the Definition of Rational Sphere Map .

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183 183 185 188 193 196 198 200 204 206

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Contents

7 Geometric Properties of Rational Sphere Maps . . . . . . . . . . . . . 1 Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 A Geometric Result in One Dimension . . . . . . . . . . . . . . . . . . 3 An Integral Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Volume Inequalities for Polynomial and Rational Sphere Maps 5 Comparison with a Real Variable Integral Inequality . . . . . . . .

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211 211 214 216 219 223

8 List of Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231

Chapter 1

Complex Euclidean Space

This chapter develops basic properties of complex Euclidean space. Some of the main ideas are unitary transformations, the holomorphic automorphism group of the unit ball, the use of Hermitian forms, and proper holomorphic mappings. We also gather some elementary combinatorial information.

1 Generalities The notation Cn denotes complex Euclidean space of dimension n. As a set, it consists of n-tuples of complex numbers z = (z 1 , ..., z n ). The notation includes the information that Cn is an inner product space. The inner product of vectors z and w is defined by n  z jwj. z, w = j=1

We denote the corresponding squared norm by z2 = z, z =

n 

|z j |2 .

j=1

The set Cn is then a metric space with distance function given by z − w. As a consequence, we have all the usual notions from point-set topology. In particular, a subset  is open if, for each p ∈ , there is a positive  such that z − p <  implies z ∈ . We denote the unit ball, centered at the origin, by Bn . Its boundary is the unit sphere S 2n−1 . Odd dimensional spheres arise throughout this book. We assume the reader knows such terms as connected, compact, limit, sequence, subsequence and © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 J. P. D’Angelo, Rational Sphere Maps, Progress in Mathematics 341, https://doi.org/10.1007/978-3-030-75809-7_1

1

2

1 Complex Euclidean Space

so on. A domain will be an open, connected set. As a metric space Cn is complete, and hence Cn is a Hilbert space of dimension n. Remark 1.1 We use the notations z and z, w for the norm and the inner product in any (unspecified) dimension. The implication  f (z)2 = 1 on z2 = 1 often arises. Here z and f (z) need not live in the same dimensional space. In one dimension, |ζ| denotes the magnitude of a complex number ζ. Several copies of complex Euclidean spaces often arise in the same discussion. Suppose N = K + L. We will write C N = C K ⊕ C L , where the symbol ⊕ denotes orthogonal sum. When A is a subspace of C N , we let A⊥ denote the orthogonal complement of A, and thus C N = A ⊕ A⊥ . In these settings, the Pythagorean theorem holds: z ⊕ w2 = z2 + w2 . When M < N , we often regard C M as a subspace of C N . For w ∈ C M , we write either w ⊕ 0 or (w, 0) for the corresponding element in C N . We say a bit now about tensor products and discuss them in more detail in Chap. 2. If v ∈ C M and w ∈ C N , we can form an element v ⊗ w in C M N whose components are v j wk for 1 ≤ j ≤ M and 1 ≤ k ≤ N in some specified order. It is easy to see that v ⊗ w2 = v2 w2 ; thus, the squared norm of a tensor product is the product of the squared norms. See Lemma 2.1. Let  be an open subset of Cn . Suppose f :  → C N is a function. Then f is holomorphic if, for each p ∈ , f is complex differentiable at p. In other words, there is a (complex) linear map d f ( p) : Cn → C N such that f ( p + h) = f ( p) + d f ( p)h + error( p, h), p,h) = 0. We assume the reader knows possible equivalent definiwhere lim h→0 error( h tions; for example, f is locally given by a convergent (vector-valued) power series, or f satisfies the Cauchy-Riemann equations. On several occasions we will have holomorphic mappings defined on open balls, and we use without comment that the power series expansion converges uniformly on compact subsets of the ball. Assume  is connected. If all the power series coefficients of f vanish at a point p ∈ , then f vanishes identically. The standard proof is to observe that the set of such points is both open and closed in . On occasion, we will need the following form of the maximum principle.

Proposition 1.1 (Maximum principle) Let  be a bounded, open, and connected subset of Cn . Suppose f :  → C N is holomorphic and extends continuously to the boundary ∂. Then: • For all z ∈ ,  f (z) ≤ sup∂  f . • If there exists a z ∈  where  f (z) = sup∂  f , then f is a constant. The following obvious consequence of the maximum principle provides a strong contrast to the real case. For example, the polynomial (x1 )2 + · · · + (xn )2 is constant

1 Generalities

3

on the unit sphere in Rn but it is not constant. A holomorphic function that is constant on too large of a set must itself be constant. We need only the following simple case. Corollary 1.1 Suppose f is holomorphic in a domain containing the closed unit ball, and f is constant on the sphere. Then f is constant. Proof If z → f (z) − c vanishes on the sphere, then (by the first part of the maximum principle)  f (z) − c vanishes on the ball. Hence, f − c vanishes identically on its domain. (Recall that a domain is connected.)  Recall that a linear transformation L : Cn → Cn is unitary if L preserves the inner product. Thus, Lz, Lw = z, w for all z, w. It follows trivially that Lz2 = z2 for all z. The converse also holds. Thus, L is unitary if and only if Lz2 = z2 for all z. In other words, Lz, Lz = z, z for all z implies the apparently stronger statement Lz, Lw = z, w for all z, w. This fact provides an easy example of polarization. Remark 1.2 (Polarization) It is often crucial in complex analysis to regard z and its conjugate z as independent variables. In rather general circumstances, when an identity involving z and z holds for all choices of z, we may replace z by w, and the identity will hold for all z, w. We will often have identities that hold on the unit sphere defined by z2 = 1. We may then replace z in the identity with w and the resulting identity will hold when z, w = 1. Example 1.1 We give a beautiful example of polarization. Consider a harmonic function u defined near the origin in R2 . We wish to find a holomorphic function f whose real part is u. Thus, we must have f (z) + f (z) =u 2



z+z z−z , 2 2i

 = u(x, y).

(1.1)

We polarize (1.1) by assuming that z and z are independent variables. For example, if we suppose f (0) = 0 and set z = 0, then we obtain the formula f (z) = 2u

z . 2 2i

z

,

(1.2)

Formula (1.2) recovers f from u without the usual process involving differentiation, integration, and the Cauchy-Riemann equations. Careful thought shows that we are assuming here that u is real-analytic. Proving that a harmonic function in the plane is real-analytic is usually done by showing that it is the real part of a holomorphic function. The result about real-analyticity holds in all dimensions. It can be proved by estimating the size of the successive derivatives after starting with the mean-value property. The following ideas are equivalent to polarization and arise throughout complex analysis. Consider a polynomial or convergent series

4

1 Complex Euclidean Space

f (z, z) =



cab z a z b

a,b

 in the variables (z, z) ∈ Cn × Cn with  f (0, 0) = 0. The terms in the series ca0 z a are called holomorphic terms, those in c0b z b are called anti-holomorphic terms, and the remaining terms are called mixed terms. One says pure terms to mean any terms that are not mixed. Suppose f (z, z) vanishes identically. Then cab = 0 for all multi-indices a, b. In particular, the series of holomorphic terms, the series of anti-holomorphic terms, and the series of mixed terms all vanish identically. Exercise 1.1 Use the technique of Example 1.1 to find a holomorphic function f whose real part is the given u(x, y). • u(x, y) = x 2 − y 2 . • u(x, y) = e x cos(y). • u(x, y) = ln(x 2 + y 2 ). (Note that we cannot set z equal to 0 here!) Unitary transformations arise throughout this book. The definition implies that the composition of unitary maps is unitary, and the next proposition implies that the collection U(n) of unitary maps on Cn is a group. We can identify U(n), when regarded as unitary matrices, as a subset of complex Euclidean space of dimension n 2 . An n × n matrix is unitary if and only if its column vectors form an orthonormal basis of Cn . Thus, this subset is closed and bounded. The group operations of multiplication and taking inverses are smooth. The next two propositions summarize the basic facts about unitary transformations. Proposition 1.2 Let L : Cn → Cn be linear. The following are equivalent: (1) (2) (3) (4)

L is unitary. For all z, w, we have Lz, Lw = z, w. For all z, we have Lz2 = z2 . L is invertible and L −1 = L ∗ . (Here, L ∗ is the adjoint of L.)

Proposition 1.3 The unitary group U(n) is a compact Lie group. A Lie group G is a smooth manifold endowed with a binary operation (g, h) → gh that makes G into a group, and for which this operation and the operation of taking inverses are smooth maps. We don’t use any major results from the theory of Lie groups or their Lie algebras, but specific examples of Lie groups arise throughout the book. We mention some of the groups that will arise. The special unitary group SU(n) consists of those unitary maps with determinant equal to 1. The n-torus T(n) consists of those diagonal maps z = (z 1 , ..., z n ) → U (z) = (eiθ1 z 1 , ..., eiθn z n ). We often identify the n-torus with U(1) × · · · × U(1). Notice also that operators permuting the variables are unitary. Hence, (by Cayley’s theorem) every finite group is isomorphic to a subgroup of U(n).

2 The Groups Aut(B1 ), SU(2), and SU(1, 1)

5

Let n = p + q. The groups U( p, q) are groups of n-by-n matrices preserving the Hermitian form p n   |z j |2 − |z j |2 . j=1

j= p+1

The subgroup SU( p, q) consists of those matrices in U( p, q) with determinant 1. In this book, we use only SU( p, 1), for the following reason. The quotient of the group SU(n, 1) by its center is isomorphic to the group of holomorphic automorphisms of the unit ball in Cn , but we prefer a more concrete description in terms of linear fractional transformations. See Sects. 2 and 3. An interesting class of examples of finite unitary groups arises in this book. In each case, the group itself is cyclic of order p, but the representations as subgroups of U(2) differ. Example 1.2 Let η be a primitive p-th root of unity and assume q is relatively prime to p. Let ( p, q) denote the cyclic subgroup of U(2) generated by 

 η 0 . 0 ηq

The special cases ( p, 1) and (2r + 1, 2) play surprising major roles in this book. The next proposition has been used innumerable times by the author and will be applied on several occasions in this book. Proposition 1.4 Let B denote a ball in Cn . Assume that f : B → C N and g : B → C M are holomorphic maps. Suppose  f (z)2 = g(z)2 on B. • If M = N , then there is a U ∈ U(N ) such that f (z) = U g(z) for z ∈ B. • If M < N , then there is a U ∈ U(N ) such that f (z) = U (g(z) ⊕ 0). 

Proof See [15] or [19] for a proof.

Exercise 1.2 Determine the real dimensions of U(n) and SU(n). (Proposition 1.7 and Corollary 1.4 give the answers when n = 2.) We conclude this section by defining rational sphere map. The simplest examples are given by unitary maps. In Sect. 3, we find all the equi-dimensional examples (automorphisms of the unit ball when n ≥ 2). Most of our discussion is devoted to rational sphere maps in the positive codimension case; in other words, when the target dimension exceeds the source dimension. Definition 1.1 A rational sphere map is any rational map f = following properties:

p q

satisfying the

• p : Cn → C N is a (holomorphic) polynomial. • q : Cn → C is a (holomorphic) polynomial, with q(z) = 0 when z ≤ 1.

6

1 Complex Euclidean Space

• f = qp is reduced to lowest terms. •  p(z)2 = |q(z)|2 when z2 = 1. Let p : Cn → C N be a polynomial. When p sends the unit sphere in the source to the unit sphere in the target, we call p a polynomial sphere map. When also each component of p is a single monomial, we call p a monomial sphere map. Remark 1.3 By Proposition 1.4, we can draw a very strong conclusion if  p2 = |q|2 holds on the ball. In fact, p = L(q ⊕ 0) for some linear map L. The equality  p2 = |q|2 on the sphere is much weaker; we will spend most of this book studying its solutions!

2 The Groups Aut(B1 ), SU(2), and SU(1, 1) Before discussing the automorphism group of the unit ball in Cn , we recall the situation in one dimension. We also briefly consider the special unitary group SU(2) and the group SU(1, 1). Let Aut(B1 ) denote the set of holomorphic maps f : B1 → B1 such that f is bijective and has a holomorphic inverse. We begin with the famous Schwarz lemma. Proposition 1.5 (Schwarz lemma) Let f : B1 → C be holomorphic. Suppose f (0) = 0 and | f (z)| ≤ 1. Then the stronger inequality | f (z)| ≤ |z| holds on B1 . Proof Since f (0) = 0, the function g defined by g(z) = f (z) is also holomorphic. z For any r < 1, its maximum absolute value on the disk |z| ≤ r is achieved on the circle |z| = r . Therefore, 

| f (z)| |g(z)| ≤ max |z|≤r r

 ≤

1 . r

Letting r tend to 1 in (1.3) shows that |g(z)| ≤ 1 and hence | f (z)| ≤ |z|.

(1.3) 

Corollary 1.2 Suppose f : B1 → B1 is a holomorphic automorphism with f (0) = 0. Then f is a rotation. Thus, f (z) = eiθ z. Proof Applying Proposition 1.5 to both f and f −1 gives | f (z)| = |z|. Either the one-dimensional version of Proposition 1.4 or elementary complex analysis forces f (z) to be a constant map, and the conclusion follows.  z Lemma 1.1 Suppose |w| < 1 , |ζ| < 1, and eiθ is on the unit circle. Then • |eiθ + ζw|2 = |1 + weiθ ζ|2 . • |eiθ w + ζ|2 < |eiθ + ζw|2 .

2 The Groups Aut(B1 ), SU(2), and SU(1, 1)

7

Proof By expanding the squared magnitudes, the equality is equivalent to 1 + 2Re(eiθ ζw) + |ζw|2 = 1 + 2Re(eiθ (wζ)) + |wζ|2 and hence holds. Expanding the squared norms and cancelling the equal real part terms shows that the inequality is equivalent to |w|2 + |ζ|2 < 1 + |wζ|2 , which is equivalent to |w|2 (1 − |ζ|2 ) < 1 − |ζ|2 . This inequality holds because both |w| < 1 and |ζ| < 1.  z−ζ z−w Lemma 1.2 For |w| < 1, |ζ| < 1 put f (z) = eiθ 1−wz and g(z) = eiφ 1−ζz . Then their composition is of the same form:

(g ◦ f )(z) = eiγ e +ζw where eiγ = eiφ 1+we and s = iθ ζ iθ

eiθ w+ζ . eiθ +ζw

z−s 1 − sz

Furthermore, |s| < 1.

Proof Compute g( f (z)). Then clear denominators by multiplying by 1 − wz. Factor (eiθ + ζw) from the numerator and (1 + ζweiθ ) from the denominator. The result is the claimed formula for s. Also, by Lemma 1.1, the factors we extracted have the same magnitude. Hence, eiφ times their quotient is on the unit circle. Finally, note that |s|2 < 1 if and only if |eiθ w + ζ|2 < |eiθ + ζw|2 , which was proved in Lemma 1.1.  Corollary 1.3 The collection of maps in Lemma 1.2 form a group under composition. Proof By Lemma 1.2, the composition of such maps is of the same form. Putting ζ = −eiθ w shows that each such map has an inverse of the same form.  Proposition 1.6 Aut(B1 ) is the set of functions f for which f (z) = eiθ

z−w , 1 − wz

(1.4)

where |w| < 1 and eiθ is on the unit circle. Proof It follows from Corollary 1.3 that each such f is an automorphism. We must show that there are no others. Let h be an automorphism with h(0) = w. For f as in (1.4), f ◦ h is an automorphism sending w to 0. By Corollary 1.2 of Schwarz’s  lemma, f ◦ h is a rotation U , and h = f −1 ◦ U has the desired form. The following proposition will not be explicitly used in the book, but its importance in physics and geometry suggests its inclusion. See [38] and its references for the many uses of SU(2) in physics. It is useful for us because of the major role the unit sphere, especially in two complex dimensions, plays in this book. Furthermore, comparing SU(2) with SU(1, 1) is interesting and SU(1, 1) is closely related to Aut(B1 ).

8

1 Complex Euclidean Space

Recall that sets are diffeomorphic if there is a smooth bijective map with a smooth inverse between them. Proposition 1.7 The sets SU(2) and the unit sphere S 3 are diffeomorphic. Proof Given p = (z, w) ∈ C2 , we define U (z, w) to be the 2-by-2 matrix  z w . −w z

 U ( p) = U (z, w) =

The mapping U is obviously injective on all of C2 . Assume |z|2 + |w|2 = 1. Computing U (z, w)U (z, w)∗ gives 

z w −w z

 

    2  z −w 0 1 0 |z| + |w|2 = = . w z 0 |z|2 + |w|2 0 1

Thus, if (z, w) ∈ S 3 , then U (z, w) is unitary. Furthermore, det(U (z, w)) = 1 as well. Therefore, U : S 3 → SU(2). The map U is obviously smooth; we need to prove that it is bijective with a smooth  inverse.  a b A 2-by-2 matrix M = is unitary if these equations are met: c d |a|2 + |b|2 = |a|2 + |c|2 = |c|2 + |d|2 = 1 ab + cd = 0. It is in SU(2) if also ad − bc = 1. First suppose that b = 0. Then |a| = 1 and hence both c = 0 and |d| = 1. In this case, M = U (a, 0). Suppose that b = 0; then and we get the equation c = ad−1 b ab +

ad − 1 d = 0. b

It follows that 0 = a|b|2 + (ad − 1)d = a(1 − |d|2 ) + a|d|2 − d = a − d and (a, b). Hence, U is surjective. The inverse hence a = d. Thus, c = −b and M= U a b map T sending a unitary matrix to its first row (a, b) is obviously also c d smooth.  Corollary 1.4 The sets S 3 × S 1 and U(2) are diffeomorphic.

2 The Groups Aut(B1 ), SU(2), and SU(1, 1)

9

Proof By the Proposition, it suffices to show that SU(2) ×  S 1 andU(2) are diffeoeiθ 0 M. Then A is a morphic. Given M ∈ SU(2) and eiθ ∈ S 1 , put A(M, eiθ ) = 0 1 diffeomorphism.  Proposition 1.7 implies that the unit sphere S 3 is a Lie Group. The only spheres that are Lie groups are S 1 and S 3 , although this fact is not so easy to prove. Another useful result in physics (which is easy to prove) is that the quotient of SU(2) by the subgroup of two elements ±I is the special orthogonal group SO(3). For us, it will be important (and shown in Chapter 6) that all odd dimensional spheres have an unbounded realization associated with the Heisenberg group. Exercise 1.3 What is the group multiplication on S 3 determined by Proposition 1.7? In other words, let p1 = (z 1 , w1 ) and p2 = (z 2 , w2 ) be points on the sphere. Define p1 ∗ p2 by applying the inverse map T to the product U ( p1 )U ( p2 ). Thus, p1 ∗ p2 = T (U ( p1 )U ( p2 )). Find this formula. Exercise 1.4 In proving Corollary 1.4, what would go wrong if we defined A by A(M, eiθ ) = eiθ M? Let n = p + q. The groups SU( p, q) are groups of n-by-n matrices with determinant 1 and preserving the Hermitian form p  j=1

|z j |2 −

n 

|z j |2 .

j= p+1

In the special case where n = 2, we determinethe relationship between SU(1, 1) and  1 0 the automorphisms of the unit disk. Put J = . A matrix M lies in SU(1, 1) 0 −1   ab if and only if det(M) = 1 and M ∗ J M = J . Assuming that M = , we see, in cd a manner similar to the proof of Proposition 1.7, that M ∈ SU(1, 1) if and only if M has the form   ab (1.5) M= ba where now |a|2 − |b|2 = 1. Note the minus sign and that a = 0. Since |a|2 − |b|2 = 1 allows |a| to be arbitrarily large, formula (1.5) shows that SU(1, 1) is not compact. Notice also that the center of SU(1, 1) consists of the matrices ±I . (The center of a group G is the set of elements g such that gh = hg for all h ∈ G.) Let us identify a fraction wζ with the row vector (ζ w). Thus, the complex number z is identified with the row vector (z 1). Given M in SU(1, 1), we multiply the row vector (z 1) on the right by M to obtain the row vector (az + b bz + a). Our identification ∼ with this row vector as a fraction yields

10

1 Complex Euclidean Space

a (z 1)M ∼ a



z+ 1+

b a b z a

.

(1.6)

Writing aa in the form eiθ , we see that (z 1)M ∼ eiθ ψ(z) where ψ is a holomorphic automorphism of the unit disk. If we replace (a, b) by (−a, −b), then the map on the right-hand side of (1.6) is unchanged. We obtain the following conclusion. Proposition 1.8 The group Aut(B1 ) is isomorphic to the quotient of SU(1, 1) by its center ±I . Proof Let M be as in (1.5). Define a map T : SU(1, 1) → Aut(B1 ) by putting T (M) to be the map on the right-hand side in (1.6). Note that aa is an arbitrary element of the unit circle and ab is an arbitrary element of the unit disk. Thus, by Proposition 1.6, T is surjective. Lemma 1.2 implies that T is a group homomorphism. It is not an isomorphism, because T (−M) = T (M), but it is two-to-one. The conclusion follows.  In dimension n, there is an isomorphism from SU(n, 1) divided by its center to Aut(Bn ). We prefer the concrete expression, from Theorem 1.1 in the next section, in terms of the linear fractional transformations generalizing (1.4).

3 Automorphisms of the Unit Ball Let  be a set and let Aut() denote the group of its automorphisms. Let us clarify what we mean by the term automorphism. An automorphism must be a bijection from  to itself. When  has some given algebraic structure, an automorphism must preserve this structure. An automorphism of a finite set is simply a permutation of that set. An automorphism of a vector space V is an invertible linear map from V to itself. For us,  will be an open subset of complex Euclidean space Cn and a holomorphic automorphism will be a biholomorphic mapping from  to itself. Thus, f :  →  is holomorphic (complex analytic), injective, and surjective. The inverse mapping f −1 is also holomorphic. Since the composition of functions is an associative operation, it follows that Aut() is a group under composition. When  is an open subset of some complex Euclidean space, the notation Aut() denotes the group of holomorphic automorphisms of . Remark 1.4 By definition, a mapping f :  →  is a holomorphic automorphism if f :  →  is holomorphic, injective, surjective, and the inverse mapping is also holomorphic. One can show that the holomorphicity of the inverse mapping follows automatically. By contrast, however, things differ for smooth functions. The function x → x 3 on R is of class C ∞ , injective, and surjective, but the inverse function is not smooth at 0. For holomorphic maps, this situation does not arise.

3 Automorphisms of the Unit Ball

11

Next, we give a concrete description of the automorphism group of the unit ball. The book [71] has a similar treatment and also contains a huge amount of analytic information about holomorphic functions on the unit ball. Let a ∈ Cn satisfy a2 < 1. Write s = 1 − a2 . Define a linear map L a by L a (z) =

z, a a + sz. s+1

(1.7)

a − La z . 1 − z, a

(1.8)

We define a rational map φa by φa (z) =

For each a ∈ Bn , we will show that φa is an automorphism of the unit ball Bn . Furthermore, φa is a rational sphere map. Lemma 1.3 For L a as in (1.7), and φa as in (1.8), the following holds: • • • •

L a (a) = a. L a z, a = z, a. L a (L a (z)) = z, aa + (1 − a2 )z. φa (a) = 0.

Proof The first item: L a (a) =

a2 1 − s2 a, a a + sa = a + sa = a + sa = a. s+1 s+1 1+s

The second item: L a z, a = 

z, a z, aa , a + sz, a = (1 − s 2 ) + sz, a = z, a. s+1 s+1

The third item:  L a (L a (z)) = L a =

z, a a + sz s+1

 =

z, a L a (a) + s s+1



z, a a + sz s+1



z, a (1 + s)a + s 2 z = z, aa + (1 − a2 )z. s+1

The final item follows from the first item.



Lemma 1.4 φa maps the unit sphere to itself. Proof It suffices to show on the unit sphere that (1.9) holds: a − L a z2 = a2 − 2Rea, L a z + L a z2 = |1 − z, a|2 .

(1.9)

12

1 Complex Euclidean Space

In anticipation of a later calculation we will show that a − L a z2 − |1 − z, a|2 = (a2 − 1)(1 − z2 ).

(1.10)

The conclusion follows from (1.10) upon setting z2 = 1. To prove (1.10), we + sz and the second item of Lemma 1.3. We expand everything using L a (z) = z,a s+1 also use 2Rez, a = 2Rea, z. Putting these things together shows that the left-hand side of (1.10) equals  (a − 1) + |z, a| 2

2

 1 − s2 2s − 1 + s 2 z2 = (a2 − 1) + s 2 z2 . + (s + 1)2 s+1

Finally, we use s 2 = 1 − a2 to obtain (1.10).



Lemma 1.5 φa ◦ φa = I . Proof After clearing denominators, we use the items in Lemma 1.3:

φa (φa (z)) =

=

a − La



a−L a (z) 1−z,a



a (z) 1 −  a−L , a 1−z,a

(1 − z, a)a − L a (a) + L a (L a (z)) 1 − z, a − a − L a (z), a =

(1 − a2 )z = z. (1 − a2 ) 

Exercise 1.5 Write down an explicit formula for the automorphism φa when a = (0, ..., 0, α). The automorphism group Aut(Bn ) is the real Lie Group SU(n, 1)/Z . Here Z denotes the center of the group SU(n, 1). See Exercise 1.6. Rather than regarding Aut(Bn ) in this way, we give explicit formulas as rational mappings in Theorem 1.1. To better understand Aut(Bn ), we need Corollary 1.5 below, an analogue of Schwarz’s lemma (Proposition 1.5) in n-dimensions, and also Corollary 1.6. Corollary 1.5 Let f : Bn → B N be holomorphic and f (0) = 0. Then  f (z) ≤ z holds for z ∈ Bn . Proof Choose a non-zero z ∈ Bn . Let l be a linear functional on C N with l( f (z)) = ζz . Then  f (z) ≤ 1 and l ≤ 1. Define the linear map L : C → C N by ζ → z L = 1 as well. Put g = l ◦ f ◦ L. Then g satisfies the hypotheses of Schwarz’s lemma in one dimension and therefore |g(ζ)| ≤ |ζ|. Put ζ = z. We get

3 Automorphisms of the Unit Ball

 f (z) = l( f (z)) = l( f ((Lζ)) = g(ζ) ≤ |ζ| = z.

13



Corollary 1.6 Let f : Bn → Bn be an automorphism with f (0) = 0. Then f is a unitary map. Proof By Corollary 1.5, applied to both f and f −1 , we have z2 ≤  f (z)2 ≤ z2 . Hence,  f (z)2 = z2 and f is unitary.



Theorem 1.1 A holomorphic self-map f of Bn is an automorphism of Bn if and only if there are a unitary U and a φa such that f = U ◦ φa . Proof By Lemma 1.5, φa is its own inverse. By Lemma 1.4, φa maps the unit sphere to itself. Since φa is not a constant, the maximum principle implies that φa maps the ball to itself. Thus, φa is an automorphism. It is obvious that each unitary map is an automorphism of the unit ball. Hence, U ◦ φa is an automorphism. We must show there are no others. Let f be an automorphism. Put a = f −1 (0). Then f ◦ φa = L is an automorphism that preserves the origin. Note that f ◦ φa = L is equivalent to f = L ◦ φa . By Corollary 1.6, which relied on Schwarz’s lemma, L(0) = 0 implies that L is unitary. Thus, f is the composition  of a φa with a unitary map. Corollary 1.7 The sets Aut(Bn ) and U(n) × Bn are diffeomorphic. Hence, the real dimension of Aut(Bn ) is n 2 + 2n. Proof Given (U, a) ∈ U(n) × Bn , the map T defined by T (U, a) = U φa is a diffeomorphism. Thus, each pair (U, a) determines a unique automorphism U φa and each automorphism determines a unitary U and a point a in the ball. By Exercise 1.2, the real dimension of the unitary group U(n) is n 2 . The real dimension of the  open ball is 2n. Hence, the real dimension of Aut(Bn ) is n 2 + 2n.   I 0 The purpose of the next exercise is to investigate SU(n, 1). Put J = , 0 −1 where I is the identity matrix in n space, the 0 in the top right denotes the column vector 0 in n-space, and the 0 in the bottom left denotes the row vector 0 in n-space. Then an (n + 1)-by-(n + 1) matrix is in SU(n, 1) if M ∗ J M = J and det(M) = 1.   Ab Exercise 1.6 Put M = , where A is an n-by-n matrix, b is a column vector, c d c is a row vector and d is a scalar. Determine the conditions of A, b, c, d such that M ∗ J M = J . For M to be in SU(n, 1), we also require det(M) = 1. Show that this condition determines d, and hence c. Now write down the generalization of (1.5). Determine the center of SU(n, 1). Hint: To be in the center, M must be a multiple of the identity. If you are ambitious, use the analogue of (1.6) to discover the automorphisms φa .

14

1 Complex Euclidean Space

Automorphisms of the ball are rational mappings that send the unit sphere to itself. In one dimension, we can multiply automorphisms together and obtain Blaschke products that also map the circle to itself. See Proposition 1.13. The analogous idea for dimensions at least 2 requires allowing higher dimensional target spheres and will be a major theme throughout this book. Remark 1.5 Aut(Bn ) is transitive. Given any pair of points a, b in the ball, there is an automorphism mapping a to b. To see why, first apply φa to move to the origin, and then apply φb to move to b. Remark 1.6 Let  be an open, connected set in Cn for n ≥ 2. When  is bounded, its automorphism group, with the topology of uniform convergence on compact subsets, is a finite-dimensional real Lie group. Exercise 1.7 Assume that the automorphism group G of a domain is transitive. Show that G must be non-compact. Exercise 1.8 Let p be a positive integer. Let  be the set of (z, w) ∈ C2 for which |z|2 + |w|2 p < 1. First show, for |a| < 1, that there is a number c such that 

a−z cw , 1 − az (1 − az) 1p



is a holomorphic automorphism of . Use this formula to show that Aut() is noncompact. Exercise 1.9 Let  be as in Exercise 1.8. Show that there is no linear map mapping  bijectively to the ball. Harder: Show that there is no biholomorphic map from  to the ball. For results on the dimensions of automorphism groups see [37] and [50]. One can also consult [2] for an extensive introductory treatment of real and complex matrix groups. Remark 1.7 The unitary group forms the maximal compact subgroup of Aut(Bn ). The collection of maps φa is identified with the open ball itself and is non-compact. We close this section by mentioning two topics we do not pursue in this book. Wong [80] proved the following result in 1977. If the automorphism group of a smoothly bounded strongly pseudoconvex domain in Cn is non-compact, then the domain is biholomorphic to the unit ball. Since then many additional results have been proved; the main idea is that of a boundary accumulation point. For example, Rosay [70] proved the following statement for an arbitrary bounded domain  ⊆ Cn . Suppose that there is a compact subset K of , a sequence z k ∈ K , and a sequence of automorphisms Tk such that Tk (z k ) converges to a strongly pseudoconvex boundary point. Then  is biholomorphic to the unit ball.

3 Automorphisms of the Unit Ball

15

Among many papers, see [3], [39], [51] for results in the weakly pseudoconvex case, and see also [40] for a broader treatment. The complete characterization of domains  for which Aut() is non-compact remains an open problem. We briefly mention the the so-called Iwasawa decomposition of a semi-simple Lie Group. See [9] for an excellent exposition. The author Bump begins by considering the example of the complex general linear group G L(n, C), whose maximal compact subgroup is the unitary group U(n). Then, using the Gram-Schmidt process, each g ∈ G L(n, C) can be factored as g = α ◦ v ◦ U , where α is diagonal with positive eigenvalues, where v is upper-triangular and with diagonal elements equal to 1, and where U is unitary. The Iwasawa decomposition generalizes this factorization by writing G = A × N × K for appropriate subgroups A, N , K . Here A is Abelian, N unipotent, and K compact.

4 Hermitian Forms Let r (z, z) be a real-valued polynomial on Cn . We can write r (z, z) =



cab z a z b .

(1.11)

a,b

We call (cab ) the underlying matrix of coefficients of r . A polynomial written as in (1.11) is real-valued if and only if the matrix (cab ) is Hermitian; that is cba = cab for all multi-indices a, b. By diagonalizing (cab ), or directly, one sees that there are linearly independent holomorphic polynomials f 1 , · · · , f N+ , g1 , · · · , g N− (where we allow these indices to be 0) such that r (z, z) =  f (z)2 − g(z)2 =

N+  j=1

| f j (z)|2 −

N− 

|gk (z)|2 .

(1.12)

k=1

We call (N+ , N− ) the signature pair of r . We next make the connection with rational sphere maps. Let f = qp be a rational sphere map with source dimension n and target dimension N . Since  p(z)2 − |q(z)|2 vanishes on the unit sphere, as a polynomial in z, z it is divisible by z2 − 1. Write the quotient polynomial as A(z)2 − B(z)2 . Then (A2 − B2 )(z2 − 1) = A2 z2 + B2 − B2 z2 − A2 . We write this equation in cleaner notation: (A2 − B2 )(z2 − 1) = (z ⊗ A) ⊕ B2 − (z ⊗ B) ⊕ A2 .

(1.13)

16

1 Complex Euclidean Space

Thus, when qp is a rational sphere map, there are holomorphic polynomial maps A(z), B(z) such that  p(z)2 − |q(z)|2 = (z ⊗ A) ⊕ B2 − (z ⊗ B) ⊕ A2 .

(1.14)

Conversely, let A and B be vector-valued holomorphic polynomial maps. Denote the right-hand side of (1.14) by ρ(z, z). The following simple result summarizes this discussion. Proposition 1.9 Let A and B be vector-valued holomorphic polynomial maps. The right-hand side of (1.14) determines a rational sphere map if and only if three things are true: • The signature pair of ρ is (N , 1) for some N . • The function q does not vanish on the closed unit ball. • qp is reduced to lowest terms. Example 1.3 Put A(z)2 = 1 + z2 + · · · + z2m−2 and B(z) = 0. Then (A2 − B2 )(z2 − 1) = z2m − 1 = z ⊗m 2 − 1 and we obtain one of the most important examples of a rational sphere map. Let f = qp be a rational sphere map. Assume that the maximum degree of p and q is d. We define the Hermitian norm H( f ) by H( f ) =  p(z)2 − |q(z)|2 =



cab z a z b .

(1.15)

a,b

The underlying matrix C has one negative eigenvalue. We often think of (1.15) as defining a Hermitian form on the vector space of polynomials of degree at most d in n variables. For a unitary map, the underlying Hermitian form is simply z2 − 1. We study what happens when we compose with an automorphism. Proposition 1.10 Let f =

p q

= U φa be an automorphism of Bn . Then

H( f ) = (1 − a2 )(z2 − 1).

(1.16)

More generally, let f = qp be a rational sphere map with f (0) = 0. Let ψa be an automorphism of the source ball with ψa (0) = a. Put F = QP = ψa ◦ f . Then H(F) = P2 − |Q|2 = (1 − a2 )( p2 − |q|2 ).

(1.17)

Proof Formula (1.16) was established in the proof of Lemma 1.4. Formula (1.17) holds via a similar computation, noting that ψa = U ◦ φa where φa is as in (1.8) and U is unitary. 

4 Hermitian Forms

17

Proposition 1.10 and Hermitian norms arise often in this book. Given a rational sphere map f and a source automorphism γ, we will need to know whether there is a target automorphism μ such that f ◦ γ = μ ◦ f . Such a μ exists if and only if H( f ◦ γ) is a constant times H( f ). We can prove the following result now. Proposition 1.11 Let f = qp be a rational sphere map with source dimension n and target dimension N . Suppose γ ∈ Aut(Bn ). Then there is a μ ∈ Aut(B N ) for which f ◦ γ = μ ◦ f if and only if there is a constant cγ such that H( f ◦ γ) = cγ H( f ).

(1.18)

Proof First assume that (1.18) holds. Write f = qp and f ◦ γ = QP . By convention we assume q(0) = Q(0) = 1 and that the fractions are in lowest terms. After composing with automorphisms of the target we may also assume p(0) = 0 and P(0) = 0. Our assumption (1.18) yields P2 − |Q|2 = Cγ ( p2 − |q|2 ),

(1.19)

and thus the constant Cγ must equal 1. Write Q = 1 + H and q = 1 + h and plug in (1.19). Equating pure terms yields 2Re(H ) = 2Re(h). Since H, h are polynomials vanishing at 0 we obtain H = h, and thus Q = q. Equating mixed terms then gives P2 − |H |P2 =  p2 . By Proposition 1.4, P = U p for some unitary U ∈ U(N ). Thus, f ◦ γ = U ◦ f for some unitary automorphism U . The converse is easy: we are given that f ◦ γ = μ ◦ f and we need to prove (1.18). If ϕ is an automorphism of the target ball, then ϕ = U ◦ φa where φa satisfies (1.4). We may assume f (0) = 0. Hence, by (1.17), H(ϕ ◦ f ) = (1 − a2 )H( f ).

(1.20)

The equality f ◦ γ = ψγ ◦ f and (1.17) guarantee that H( f ◦ γ) = H(ψγ ◦ f ) = cγ H( f ) for a non-zero constant cγ . Hence (1.18) holds.

(1.21) 

Corollary 1.8 Let qp be a non-constant rational sphere map with source dimension n and target dimension N . • If N = n, then  p2 − |q|2 is a non-zero constant multiple of (1 − z2 ) if and only if qp is an automorphism.

18

1 Complex Euclidean Space

• If N > n, and p(0) = 0, then  p2 − |q|2 is a non-zero constant multiple of (1 − z2 ) if and only if there is a U ∈ U(N ) such that p = U (z ⊕ 0) and q = 1. Proof First suppose N = n. If qp is an automorphism then the conclusion was established in Proposition 1.10. Conversely, consider a rational sphere map f = qp whose Hermitian norm is a constant multiple of (1 − z2 ). We compose with an automorphism to make ( f ◦ γ)(0) = 0. The Hermitian norm P(z)2 − |Q(z)|2 remains a constant multiple of (1 − z2 ). We may assume Q(0) = 1. Thus, P(z)2 − |Q(z)|2 = z2 − 1. As in the proof of Proposition 1.11, we conclude that Q is a constant (hence Q = 1) and P(z)2 = z2 . Still assuming n = N we conclude that P is unitary and hence qp is an automorphism. Suppose next that N > n. Since p(0) = 0, as before, the condition implies q = q(0) = 1 and  p(z)2 = 1. By Proposition 1.4, p = U (z ⊕ 0) for some U ∈ U(N ). Conversely, if p has this form, and q = 1 is constant, then  p2 − |q|2 = z2 − 1.  In Chapter 6, we will show when N = n ≥ 2 that a non-constant rational sphere map must in fact be an automorphism. Hence, when N = n ≥ 2, the only if part of the Corollary is not particularly interesting. When N = n = 1, it is easy to check that a Blaschke product with more than one factor does not satisfy the condition on its Hermitian norm. Remark 1.8 When γ ∈ U(n), the constant cγ from (1.18) must equal 1. Exercise 1.10 Prove (1.17). Exercise 1.11 Let h and H be holomorphic mappings with the same domain and assume Re(h) = Re(H ). If h(0) = H (0), show that h = H . Exercise 1.12 In each case, verify that the given real-valued polynomial r (z, z) r (z,z) vanishes on the unit sphere. Then compute the quotient z 2 −1 . r (z, z) = |z 1 |2 + |z 1 z 2 |2 + |z 2 |4 − 1. r (z, z) = |z 1 |6 + 3|z 1 |2 |z 2 |2 + |z 2 |6 − 1. r (z, z) = (|z 1 |2 + |z 2 |2 )2 − (|z 1 |2 + |z 2 |2 )3 . Exercise 1.13 In each case, give an example of a real-valued polynomial r (z, z) on C2 satisfying the given property. • The underlying matrix of coefficients of r has eigenvalues of both signs, but r (z, z) ≥ 0 for all z. • r (z, z) > 0 for all z and inf(r (z, z)) = 0. • r (z, z) = 0 for z on the unit sphere, r is of degree 4 but r (z, z) is not (|z 1 |2 + |z 2 |2 − 1)2 .

5 Proper Mappings

19

5 Proper Mappings Let X, Y be locally compact topological spaces. The reader can imagine that they are open subsets of complex Euclidean spaces. A continuous map f : X → Y is called proper if, whenever K is compact in Y , f −1 (K ) is compact in X . There is a simple intuition arising from compactifications. Suppose we take the one point compactifications X ∪ ∞ and Y ∪ ∞, and we extend a continuous map f to a map F between the compactifications by putting F(∞) = ∞ and F(x) = f (x) otherwise. We put the usual topology on the compactifications. Then f is proper if and only if F is continuous on the compactifications. This concept will be particularly useful in this book when X and Y are unit balls. As a consequence, we have the following simple equivalent definition. Proposition 1.12 A holomorphic mapping f : 1 → 2 between domains in complex Euclidean spaces is proper if and only if, whenever {z ν } is a sequence in 1 that tends to the boundary of 1 , the image sequence { f (z ν )} tends to the boundary of 2 . Proof Left to the reader.



Exercise 1.14 Prove Proposition 1.12. Exercise 1.15 In each case decide whether the map f : R2 → R is proper. • • • •

f (x, y) = x 2 + y 2 . f (x, y) = x 2 − y 2 . f (x, y) = |x| + |y|. f (x, y) = (x y − 1)2 + y 2 .

Remark 1.9 Our work emphasizes maps between balls in the positive codimension case. In other words, our main concern is when the source ball lies in a lower dimensional space than the target ball. We pause to mention several standard results for domains in the equi-dimensional case. Let f : 1 → 2 be a proper holomorphic mapping between bounded domains in Cn . • f is an open mapping. If A ⊆ 1 is open, then f (A) is open in 2 . • f is surjective. Thus, f (1 ) = 2 . • Let V denote the zero-set of the Jacobian determinant of f . There is a positive integer m (the multiplicity of f ) such that the following holds. For w ∈ 2 but not in f (V ), the cardinality of f −1 (w) equals m. For w ∈ 2 ∩ f (V ), the cardinality of f −1 (w) is smaller than m. Exercise 1.16 Let 1 be the domain in C2 defined by |z|2a + |w|2b < 1. Let 2 be the unit ball. Put f (z, w) = (z a , w b ). Then f : 1 → 2 is proper. What is the multiplicity of f ? Exercise 1.17 Let 1 be the domain in C2 defined by |z 2 − w 3 |2 + |zw|2 < 1. Let 2 be the unit ball. Put f (z, w) = (z 2 − w 3 , zw). Then f : 1 → 2 is proper. What is the multiplicity of f ?

20

1 Complex Euclidean Space

We return to our discussion of proper mappings between balls. Let f : Bn → B N be a proper holomorphic mapping, and suppose f has a continuous extension to the unit sphere in the source. Then f maps the unit sphere in the source to the unit sphere in the target. When f is a rational function, we get a rational sphere map. Furthermore, Forstneriˇc proved, in source dimension at least 2, that a proper holomorphic mapping between balls with sufficiently many continuous derivatives at the sphere must be rational [36]. When the source and target dimensions both equal 1, every proper holomorphic map between balls is rational. The next result gives a complete description in one dimension. Proposition 1.13 Let f : B1 → B1 be a proper holomorphic mapping. Then f is a finite Blaschke product. Thus, there are finitely many points a j in the unit disk, positive integers m j , and a point eiθ on the circle, such that f (z) = eiθ

 K   aj − z mj . 1 − ajz j=1

(1.22)

Proof Consider f −1 (0). Since f is not identically 0, this set is a discrete collection of points. Since f is proper and a single point is compact, this set is finite. By Exercise 1.21, it is non-empty. To each a j let m j denote the multiplicity of the 0 at a j . Put  K   aj − z mj B(z) = . 1 − ajz j=1 f (z) Note that B is also proper. We will show that B(z) = 1, and hence iθ

f B

is a constant

f B

e . Note that is also holomorphic on the disk, because each zero of B is cancelled by a zero of f . Given  > 0, there is a δ > 0 such that both 1 −  < | f (z)| < 1 and 1 −  < |B(z)| < 1 on 1 − δ ≤ |z| < 1. Thus, f (z) < 1 1−< B(z) 1 −  there. Since Bf is holomorphic, the maximum principle guarantees that the same inequality holds on |z| < 1 − δ and hence on the disk. Letting  tend to 0 shows that | Bf | = 1. Either by standard complex analysis, or by the one-dimensional case  of Proposition 1.4, we conclude that there is an eiθ such that f = eiθ B. Corollary 1.9 Assume f is holomorphic in a neighborhood of the closed unit disk, and suppose that f maps the circle to the circle. Then either f is a constant or f is of the form (1.22). Corollary 1.10 Let q be a polynomial in one variable; assume that q(z) = 0 on the closed unit disk. Then there is a polynomial p such that qp is reduced to lowest terms and maps the circle to itself.

5 Proper Mappings

21

Proof If q = c is constant, put p(z) = cz. If q is not constant, we may assume q(0) = 1. Since q has no zeroes on the closed unit ball, we can factor it, writing q(z) =

m 

(1 − a j z),

j=1

where |a j | < 1 for each j. We allow the a j to be repeated. Define p by p(z) =

m 

(a j − z).

j=1

The fraction qp is reduced to lowest terms. Also | p(z)|2 = |q(z)|2 on the circle, by formula (1.16) in one dimension.  Remark 1.10 The higher dimensional analogues of Proposition 1.11 and its corollaries are much more subtle, and they will occupy most of this book. The heuristic explanation is that multiplication gets replaced by the tensor product and tensor products of automorphisms require larger dimensional target spaces. Furthermore, not all rational sphere maps are constructed via the tensor product operation alone. The analogue of Corollary 1.10 holds, but it is much harder to prove, and there is no analogous formula for p. Both the degree and target dimension of p depend on the coefficients of q. See Corollary 2.6 from the next Chapter. For later purposes we introduce the following standard bit of terminology. Definition 1.2 Assume f, g are proper holomorphic maps from Bn to B N . We say that f is spherically equivalent to g if there are automorphisms γ and ξ such that f ◦ γ = ξ ◦ g. Example 1.4 A proper linear fractional transformation between balls is spherically equivalent to z → (z, 0). Exercise 1.18 Suppose f, g are spherically equivalent rational sphere maps. Show that they are of the same degree. Here the degree means the degree of the numerator. See Definition 2.3. Exercise 1.19 Let f be a Blaschke product with 3 factors. When is f spherically equivalent to z 3 ? Exercise 1.20 Let {an } be a sequence of points in the unit disk. Assume that {an } satisfies the Blaschke condition ∞  (1 − |an |) < ∞. n=1

22

Put einθ = product

1 Complex Euclidean Space |an | an

when an = 0. If an = 0, put einθ = 1. Show that the infinite Blaschke ∞ 

einθ

n=1

an − z 1 − an z

converges to an analytic function B(z) on the unit disk. Explain why z → B(z) is not a proper holomorphic mapping from the disk to itself. Exercise 1.21 Let f be a non-constant holomorphic function in a neighborhood of the closed unit disk. Suppose that | f (z)| = 1 on |z| = 1. Our proof that f is a finite Blaschke product considered f −1 (0), and assumed that it is non-empty. How do we know that f has a zero inside the unit disk? . Express the condition that |z|2 < 1 in terms of w. ComExercise 1.22 Put z = w−i w+i ment: A generalization of this computation will be important for us in Chapter 6, when we discuss an unbounded realization of the unit sphere.

6 Some Counting Throughout this book, we will be studying rational sphere maps. Doing so will lead to various underdetermined linear systems of equations; there will be usually more unknowns than there are equations, and hence there will be many solutions. For that reason, we need to recall some standard combinatorial facts. Lemma 1.6 Let n denote the number of variables. 

independent homogeneous polynomials of degree m. • There are n+m−1 m

n+m • There are m independent polynomials of degree at most m, and  m   n+k−1 k=0

k

=

  n+m . m

Proof The first item has a well-known beautiful proof. Let α be a multi-index of degree m. We count the total of such α by choosing n − 1 dividers among n + m − 1 locations. We put α1 items before the first divider, α2 between the first and second dividers, and so on. There is a one-to-one correspondence between multi-  the

n+m−1 = indices of length m and the choice of dividers. Hence there are n+m−1 n−1 m independent homogeneous polynomials of degree m in n variables. The second item also has an elegant proof. Consider the number

 of homogeneous by the first item. polynomials of degree m in one additional variable, namely n+m m This number equals the number of polynomials of degree at most m in n variables, because we can simply set the additional to 1 to dehomogenize. Alter n+k−1equal   variable . This sum can be simplified to natively, the answer equals the sum m k=0 k

n+m  .  m

6 Some Counting

23

Symmetric polynomials will arise often in this book. Given a monomial f in n variables, we can symmetrize it by considering the average of f ◦ σ over all permutations of the coordinates. Lemma 1.7 Let n be the number of variables. • There are n(n−1) independent quadratic monomials x j xk with j = k. 2 • Put p(x1 , x2 ) = x1a x2b where a = b. The symmetrization of p has n(n − 1) terms. Proof For quadratics, we are simply counting the number of ways to choose 2 indices

 from among n. For the second statement, there are n2 ways to choose the 2 variables, a b b a and each choice

n  contributes the two terms x y + x y , since a, b are distinct. Hence  there are 2 2 = n(n − 1) terms. Lemma 1.8 Assume n = 2. If d = 2r + 1 is odd, then there are (r + 1)(r + 2) independent symmetric monomials of degree at most d. If d = 2r is even, then there are (r + 1)2 independent symmetric monomials of degree at most d. Proof We need to count the number of pairs (a, b) such that a ≥ b and a + b ≤ d. When d = 2r + 1 we get 1 + 1 + 2 + 2 + · · · + r + r + r + 1 + r + 1 = (r + 1)(r + 2). When d = 2r , using the result for d = 2r + 1, we get the sum 1 + 1 + 2 + 2 + · · · + r + r + r + 1 = (r + 1)(r + 2) − (r + 1) = (r + 1)2 .  Exercise 1.23 How many independent symmetric homogeneous polynomials of degree 3 in four variables are there? Exercise 1.24 Show that the generic polynomial of degree at least 2 in two or more complex variables is irreducible. Suggestion: Suppose f is of degree d and f = gh where the degrees of g, h are each at least one and their sum is d. Count the number of constants needed to determine f . Show that this number is smaller than the number

n+d  needed to descrbe polynomials of degree d in n variables. n Exercise 1.25 Suppose f (x) is a polynomial in one real variable of degree at least 2 with non-negative coefficients. For x, y > 0, prove that f (x + y) + f (0) > f (x) + f (y). Can you use this result in the previous exercise?

24

1 Complex Euclidean Space

7 A GPS for This Book We have defined rational sphere map in Definition 1.1. This short section provides the location of where each fundamental result is discussed, thereby giving the reader a kind of global positioning system for the book. We also indicate briefly what happens in each chapter. Let n denote the source dimension and N the target dimension of a rational sphere map f = qp . • • • • • • • • • • • •

When N < n, f is a constant. See Proposition 6.2. When N = n = 1, f is a finite Blaschke product. See Proposition 1.13. When N = n ≥ 2, f is an automorphism of the ball. See Theorem 6.4. If q is a polynomial that does not vanish on the closed unit ball in Cn , then there is a target dimension N and a numerator p such that qp is a rational sphere map (reduced to lowest terms). See Theorem 2.7. If f is a proper rational map between balls, then f is a rational sphere map. See Theorem 6.8. Chapter 2 discusses a wide variety of results about rational sphere maps. In particular, Theorem 2.15 shows how to find a numerator p, given a denominator q, by solving a system of linear equations obtained by equating Hermitian forms. Chapter 3 discusses combinatorial results about monomial sphere maps. In particular, the chapter spends considerable time developing properties of some remarkable polynomials in two variables. Chapter 4 discusses optimization results about monomial sphere maps. This chapter includes output obtained by coding. Some but not all of those results have been proved, and thus this chapter suggests opportunities for research. Chapter 5 discusses groups associated with rational sphere maps. Chapter 6 discusses some of the relevant CR Geometry. Chapter 7 discusses a sharp inequality in the volume of the image of a rational sphere map. Chapter 8 provides a list of 15 open problems with references to where in the text the problem is stated.

Chapter 2

Examples and Properties of Rational Sphere Maps

We recall the definition and then amplify some of the basic issues. Allowing the circle and the torus to act on the sphere yields some fundamental identities for rational sphere maps which get used throughout the book. We discuss a wide variety of results concerning the impact of the target dimension on the possibilities. We introduce juxtaposition and tensor products, both of which get used extensively. We show that the denominator of a rational sphere map can be any polynomial not vanishing on the closed ball, but only if we allow a sufficiently large target dimension. To do so, we prove that the first few components of a rational sphere map can be any rational function that maps the closed ball in the source strictly inside the open ball in the target. This statement shows that the collection of rational sphere maps with a fixed source dimension has an arbitrarily large dimension as the target dimension increases. The chapter also includes a result about the inverse image of a point under a rational sphere map and its relationship with a reflection principle. Secttion 13 provides a universal method to find all rational sphere maps. The idea is to first fix the denominator, and then find a numerator of the lowest degree that works. Then one uses the tensor product operation to put the numerator in a certain form. Doing so reduces the problem of solving a very large system of linear equations for the inner products of unknown vectors. In Sect. 14, we provide a rather detailed complicated example of this method.

1 Definition and Basic Results about Rational Sphere Maps We begin by recalling the definition and conventions of rational sphere maps. We continue by writing down the equations satisfied by the coefficients of the numerator and denominator.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 J. P. D’Angelo, Rational Sphere Maps, Progress in Mathematics 341, https://doi.org/10.1007/978-3-030-75809-7_2

25

26

2 Examples and Properties of Rational Sphere Maps

Definition 2.1 A rational sphere map is any rational map f = following properties: • • • •

p q

satisfying the

p : Cn → C N is a (holomorphic) polynomial. q : Cn → C is a (holomorphic) polynomial, with q(z) = 0 when z ≤ 1. f = qp is reduced to lowest terms.  p(z)2 = |q(z)|2 when z2 = 1.

Definition 2.1 allows constant maps. When n = 1, the second item eliminates maps such as z → 1z (or Blaschke factors with |a| > 1) which map the circle to itself and are rational, but which are not holomorphic on the open unit disk. When n ≥ 2, if f is assumed to be rational and the fourth condition holds, then either f is constant or f is proper. The condition that q(z) = 0 on the closed ball is thus not necessary when n ≥ 2. In Theorem 6.8, we show the following. If f is a proper rational map between balls, reduced to the lowest terms, then q(z) = 0 on the closed ball. If f is proper and q has a zero on the sphere, then p and q must have a common factor. Thus, the assumption that q = 0 on the closed ball is superfluous when n ≥ 2. We will assume that q(0) = 1. When q is a constant, we use the terms polynomial sphere map and monomial sphere map with obvious meaning. We call n the source dimension and N the target dimension. For n ≥ 2, all rational mappings for which the image of the unit sphere in the source lies in the unit sphere in the target are rational sphere maps. Such an f is either constant or proper. As noted above, when f is proper and reduced to the lowest terms, Theorem 6.8 implies the remaining condition. Remark 2.1 The assumption q(0) = 1 has a simple consequence; the numerator of a rational sphere map determines the denominator. Suppose qp and rp are rational sphere maps. Since q, r are not zero on the closed ball, both qr and qr are holomorphic. They have the same absolute value on the sphere. By the maximum principle, r is a constant multiple of q. Since r (0) = q(0), we conclude r = q. Remark 2.2 Let g be a rational sphere map with source dimension k and target dimension n, and let f be a rational sphere map with source dimension n and target dimension N . Then the composition g ∗ ( f ) = f ◦ g defines a rational sphere map with source dimension k and target dimension N . Perhaps one might need to cancel common factors from the numerator and denominator of g ∗ ( f ). The key point of course is that g maps the unit sphere in its source to the unit sphere in its target, and that f maps that unit sphere to the unit sphere in its target. We next turn to a fundamental tool used throughout the book. The unit circle acts on the unit sphere via the map z → eiθ z. After replacing z by eiθ z, we obtain various useful Fourier series. Definition 2.2 Let p : Cn→ C N be a polynomial of degree d. Its homogeneous expansion is written p = dj=0 p j , where p j (ζz) = ζ j p j (z) for ζ ∈ C.

1 Definition and Basic Results about Rational Sphere Maps

27

Exercise 2.1 With p as in Definition 2.2, show that 1 p j (z) = 2π



2π

p(eiθ z)e−i jθ dθ.

0

. Definition 2.3 The degree of a rational sphere map is the degree of the numerator p as a polynomial. We will prove below that the degree of q is always at most the degree of p; in fact, if f (0) = 0, then the degree of q is strictly smaller than the degree of p. When f is a rational sphere map, then either f is a constant or its restriction to the unit ball is a proper holomorphic mapping between unit balls. Conversely, by the theorem of Forstneriˇc, when n ≥ 2, a proper map between balls Bn and B N with sufficiently many continuous derivatives at the boundary sphere is a rational sphere map [36]. We discuss this result in Chap. 6. Proposition 2.1 Let f =

p q

: Bn → B N be a rational sphere map. Then

• The degree of q is less than or equal to the degree of p. • If p(0) = 0, then the degree of q is less than the degree of p.  Proof Let D be the maximum of the two degrees. We let p = Dj=0 p j and q = D j=0 q j be the homogeneous expansions of p and q. For z on the sphere, we have  p(z)2 = |q(z)|2 and hence  p(z)2 =

   p j (z), pk (z) = q j (z)qk (z) = |q(z)|2 . j,k

(2.1)

j,k

Replace z by eiθ z and use homogeneity. Equate Fourier coefficients and replace the index j by k + m. We obtain (for each m) the following identity: D D    pk+m (z), pk (z) = qk+m (z)qk (z). k=0

(2.2)

k=0

When m = D, we must have k = 0 and therefore  p D (z), p0 = q D (z)q0 = q D (z). If q D = 0, then p D = 0 as well, and the first statement holds. If p(0) = 0, then  q D = 0 and the second statement holds.

28

2 Examples and Properties of Rational Sphere Maps

The identities (2.2) and their homogenized form appear throughout this book. They arise from letting the circle act on the sphere. The next two corollaries provide consequences of (2.2) in the polynomial case. In the next section, we give an important consequence in the general rational case and Sect. 13 applies these identities to give a general description of all rational sphere maps. To prepare for later discussion, we will often write D for the degree of p and d for the degree of q. Corollary 2.1 Let p be a polynomial sphere map. Suppose that the order of vanishing of p at 0 is ν and the degree of p is D. Assume that ν < D. (In other words, p is not homogeneous.) Then  p D , pν = 0. Thus, when p is not homogeneous, the lowest and highest order parts of p map into orthogonal subspaces. Corollary 2.1 will have several applications later in the book. The next result, Corollary 2.2, is useful because it provides a necessary and sufficient condition for a polynomial to be a polynomial sphere map. Corollary 2.2 Let p : Cn → C N be a polynomial of degree D. Then p is a polynomial sphere map if and only if the following conditions on its homogeneous expansion hold. First, D   pk 2 z2D−2k = z2D . (2.3) k=0

Also, for each m > 0,

D 

 pk+m , pk z2D−2k = 0.

(2.4)

k=0

Also, if p is not a constant, then the polynomial  p D 2 is divisible by z2 . Proof Put q = 1 in (2.2) and then homogenize. We obtain (2.3) and (2.4). Formula (2.3) implies the last statement, because, if D ≥ 1, then the right-hand side of (2.3) and every term in the sum in (2.3) except for  p D 2 is divisible by z2 . Hence,   p D 2 also is. An analogue of the last statement holds in the rational case; see Theorem 2.3. Exercise 2.2 Let

p q

be a rational sphere map where q is degree 1. Prove that q1 (z) =

 p D (z), p D−1 (z) . z2D−2

One can also let the n-torus T(n) act on the sphere and obtain additional identities. We write them down after we offer a notational warning about multi-index notation and then introduce an efficient way to express the identities. Remark 2.3 Assume z ∈ Cn and that α is an n-tuple of non-negative integers. Then the following holds:

1 Definition and Basic Results about Rational Sphere Maps

zα =

n 

29

α

zj j

j=1

|z|2α = |z α |2 =

n 

|z j |2α j .

j=1

Definition 2.4 Let G(z, z) be a polynomial (possibly vector-valued) that is homogeneous of degree a in the z variables and of degree b in the z variables. In other words, G(sz, sz) = s a s b G(z, z) for all complex numbers s. We say that G is bihomogeneous of bi-degree (a, b). The most important case of bihomogeneity is when a = b. Sometimes the phrase G is bihomogeneous is used to imply that a = b. In order to fully benefit from the various identities satisfied by rational sphere maps, we introduce the following operation L sending (possibly vector-valued) holomorphic polynomials of degree D into Hermitian forms. n N Definition 2.5  Let f : C → C be a holomorphic polynomial of degree at most D. Let f = f j denote its expansion in homogeneous parts. For 0 ≤ k ≤ D, define polynomials in (z, z) by the formulas

Lk ( f ) =

D−k 

 f j+k , f j z2D−2 j−2k

(2.5)

j=0

Then define a real-valued polynomial by D     S( f ) = L0 ( f ) + Lk ( f ) + Lk ( f ) = cαβ z α z β .

(2.6)

k=1

The polynomials Lk ( f ) are of bi-degree (D, D − k). To make the bi-degree homogeneity more transparent, we can also write (2.5) as Lk ( f ) =

D−k 

 f j+k ⊗ z ⊗(D− j−k) , f j ⊗ z ⊗(D− j−k) .

j=0

Notice that L0 ( f ) is real-valued, and hence S( f ) also is. The matrix cαβ in (2.6) is therefore Hermitian symmetric. We call this matrix L( f ). We make an important comment. Definition 2.5 makes sense independent of the target dimension N . When qp is a rational sphere map with target dimension N we will apply the map L in Definition 2.5 both to p (with target dimension N ) and to q (with target dimension 1). The map L depends on the degree D. When we write L(q) the degree D will be the degree of p. Thus, Proposition 2.1 is significant.

30

2 Examples and Properties of Rational Sphere Maps

Theorem 2.1 below will be a crucial tool for us, so we express it in two ways. Theorem 2.1 Let qp be a rational sphere map.   • Put p(z) = Cα z α and q(z) = bα z α . For each multi-index δ, the following holds on the unit sphere:  Cβ+δ , Cβ − bβ+δ bβ z δ |z|2β = 0.

(2.7)

β

• Write p =

D k=0

pk and q =

d k=0

q j . Then L( p) = L(q).

2 2 Proof We write out the condition that  p(z)  iθ − |q(z)|iθ = 0 on the sphere. Let T(n) iθ iθ 1 act by replacing z with e z. Here e z = e z 1 , · · · , e n z n . Expanding the squared norms yields  Cα , Cβ − bα bβ z α z β ei(α−β)θ = 0. (2.8)

This multiple Fourier series vanishes for z on the unit sphere and for all θ. We then replace α by β + δ and equate Fourier coefficients to obtain (2.7). The second item follows by homogenizing (2.2). For 0 ≤ m ≤ D, we obtain D D    p j+m (z), p j (z) z2D−2 j−2m = q j+m (z)q j (z)z2D−2 j−2m j=0

(2.9)

j=0

The equations in (2.9) are precisely Lm ( p) = Lm (q). Hence, L( p) = L(q).  Corollary 2.3 Let p(z) = Cα z α be a polynomial sphere map. Put



m(z) = (. . . , Cα z α , . . . ). Then m is a monomial sphere map. Proof When p is a polynomial sphere map, q(z) = 1. Thus, bβ = 0 for β = 0. Also put δ = 0 in (2.7). We obtain 

Cβ 2 |z|2β = 1

β

on the unit sphere, and the conclusion follows.



Corollary 2.4 Each polynomial sphere map is a composition L ◦ m of a linear map and a monomial sphere map. The linear map L from Corollary 2.4 typically decreases dimensions. Each monomial arising in p contributes to the target dimension of m, but the vector coefficients need not be linearly independent, and hence the target dimension of p is typically smaller.

1 Definition and Basic Results about Rational Sphere Maps

31

Example 2.1 Let u jk be the components of an element of U(2). Define a polynomial sphere map by p(z, w) = (u 11 z + u 12 w, u 21 z 2 + u 22 zw, u 21 zw + u 22 w 2 ). Then p has target dimension 3, but the monomial map m from Corollary 2.3 has target dimension 5. Thus, L : C5 → C3 . The degree of a polynomial is more useful than measurements such as the number of inverse images of points. First, we note the following standard fact. Remark 2.4 A proper holomorphic map is a finite map; the number of inverse images of a point is finite. The inverse image of a single point is a compact complex analytic subvariety. We will see in Chap. 6 that such objects must be finite sets; we allow the empty set as a possibility. Remark 2.5 The degree of a rational sphere map is defined to be the degree of the numerator p. Here, we mean the degree as a polynomial. Because our maps are not in general equi-dimensional, measurements such as the generic or maximum number of inverses images do not in general equal the degree. The next example illustrates why it is difficult to assign multiplicities to points when maps are not equi-dimensional. The reader should contrast this example with Remark 1.9. Example 2.2 Put f (z 1 , z 2 ) = (z 1 , z 1 z 2 , z 1 z 22 , z 23 ). Then f is a monomial sphere map. Consider a point w = (a, b, c, d) ∈ C4 . If a = 0, and w is in the image of f , then f −1 (w) is the single point (a, ab ). What happens when a = 0? If w = (0, 0, 0, 0), then f −1 (w) is the single point (0, 0). If w = (0, 0, 0, d) with d = 0, then f −1 (w) consists of the three points (0, α) where α3 = d. If a = 0 but b = 0 or c = 0, then f −1 (w) is empty. Thus, it is not clear how to assign an integer that measures the singularity at (0, 0).

2 Sphere-Ranks and Target-Ranks Let f be holomorphic. At times it is convenient to consider maps of the form f ⊕ 0 = ( f, 0). Then  f 2 =  f ⊕ 02 . Hence, if f is a rational sphere map, then so is f ⊕ 0. We can think of ( f, 0) as the composition of f with an injection of the target sphere into the equator of a larger dimensional target sphere. Deciding whether to regard f and f ⊕ 0 as essentially the same maps is a persistent nuisance throughout the subject. One way to be precise is to introduce the notion of target-rank. In the next definition, we could replace the target space with any Hilbert space without essential change.

32

2 Examples and Properties of Rational Sphere Maps

Definition 2.6 Let f : Bn → C N be holomorphic. For vectors Ca , we write f (z) =



Ca z a

a

 f (z)2 =

 Ca , Cb z a z b . a,b

• The target-rank of f is the rank of the Hermitian matrix (cab ) defined by cab = Ca , Cb . • The sphere-rank of f is the smallest k for which there is a holomorphic map h with target-rank k and  f 2 = h2 on the unit sphere. If no such k exists, then we say the sphere-rank is infinite. The sphere-rank of a rational sphere map equals 1. Note that the target-rank of f ⊕ 0 equals the target-rank of f . When f (0) = 0, the target-rank of f is the smallest number k for which the image of f lies in a subspace of dimension k. When f is a holomorphic mapping, perhaps taking values in a Hilbert space, its target-rank is the number N+ obtained by putting r (z, z) =  f (z)2 . In this case, the rank of the matrix (cab ) equals the minimum number k of terms for which  f (z)2 =

k 

|h j (z)|2 .

j=1

This rank is sometimes called the Hermitian length or the Pythagoras number of r (z, z). Also, often one says that r is a Hermitian square. The next examples clarify the distinction between target-rank and sphere-rank. Example 2.3 Consider the holomorphic map (z 1 , z 2 , z 3 ) → f (z) = (z 1 , bz 2 , cz 3 ). The following statements are all obvious. • • • • •

If b = c = 0, then the target-rank of f is 1. If b = 0 but c = 0, then the target-rank of f is 2. If b, c = 0, then the target-rank of f is 3. If |b| = |c| = 1, then the sphere-rank of f is 1. If 1 < |b| < |c|, then the sphere-rank of f is 3.

3 3 Example √ 2.4 Put f (z 1 , z 2 ) = (z 1 , cz 1 z 2 , z 2 ). For c = 0 the target-rank of f is 3. If |c| = 3, then the sphere-rank of f is 1, as f is a monomial sphere map in this case. If |c|2 > 3, then the sphere-rank of f is 2. The reason is that, on the sphere, we can write  f (z)2 = 1 + (|c|2 − 3)|z 1 z 2 |2 .

We recall the related idea of signature pairs. Let r (z, z) be real-analytic in a ball centered at 0 in Cn . Assume that r takes real values. The matrix of Taylor coefficients (cab ) of r at 0 is then Hermitian. We can always write

2 Sphere-Ranks and Target-Ranks

r (z, z) = F(z)2 − G(z)2 =

33

 j

|F j (z)|2 −



|G j (z)|2

j

for holomorphic functions F j and G j and where the sums converge in a neighborhood of 0. As in the polynomial case, we may assume that these functions are linearly independent. The signature pair of r is (N+ , N− ), where N+ , N− denote the number (possibly infinite) of positive and negative eigenvalues of (cab ). Thus, N+ + N− is the rank of the matrix (cab ). It is worthwhile to consider both polynomials and real-analytic functions that are not themselves squared norms. Since the focus in this book is rational sphere maps, the most useful signature pairs for us are (N , 0) (corresponding to a squared norm) and (N , 1), corresponding to  p2 − |q|2 . In the next section, we develop the ideas considerably further.

3 Ranks of Products Theorem 2.2 below, from [25], describes ranks of real-analytic functions that are multiples of z2d . Even the special case where d = 1 leads to Theorem 2.3. The interest in the result comes from the comparison with a famous theorem of Pfister [65] concerning sums of squares of real polynomials. We briefly discuss that case. Let r be a polynomial in n real variables, and assume r (x) ≥ 0 for all x ∈ Rn . Then, by the solution to Hilbert’s 17-th problem, r is the sum of squares of rational functions. Taking a common denominator, there is thus a polynomial g such that g 2 r is a sum of squares of polynomials. It is well understood that r itself need not be a sum of squares of polynomials. Pfister proved that one can always choose g such that g 2 r as a sum of at most 2n squares. This result is impressive because the number of terms required is independent of the degree. The analogous statement in the Hermitian case fails. We will prove the following theorem. Theorem 2.2 Let R(z, z) be a real-analytic function defined near the origin in Cn . Assume that R is not identically zero, and that R is a multiple of z2d . Then the rank n+d−1 = N (n, d). Equality is possible, and holds when R = z2d . of R is at least d In particular, when d = 1, the rank of R is at least n. Before giving the proof, we prove the following special case. Even when p is real-valued this result is both appealing and useful. Proposition 2.2 Letp(x) be a homogeneous (possibly complex-valued) polynomial on Rn . Put s(x) = nj=1 x j . Assume that p is a multiple of s d . Then either p is  identically 0 or p has at least n+d−1 = N (n, d) monomials. d Proof When n = 1, the result is trivial. We give a proof by induction on the dimension, after using the method of descent on the degree. The main issue arises already

34

2 Examples and Properties of Rational Sphere Maps

when n = 2. Given p(x, y) we dehomogenize and consider p(x, 1). We want to show that p(x, 1)(1 + x)d has at least d + 1 terms. Suppose that there is an integer d and polynomials q and r such that r (x) = (1 + x)d q(x) and that r has at most d terms. There is then a smallest such d. If the polynomial q is divisible by x, then r also is, and we divide both sides by x. We may therefore assume that either q is identically zero, or that q has a non-zero constant term. In the second situation, differentiate both sides of r (x) = (1 + x)d q(x) to obtain r  (x) = d(1 + x)d−1 q(x) + (1 + x)d q  (x) = (1 + x)d−1 h(x). The constant term goes away upon differentiation, and thus the number of terms drops by 1. Thus, we have an example where we have multiplied by (1 + x)d−1 but with at most d − 1 terms. By the method of descent, d could not have been minimal, unless of course r = 0. Thus the result holds when n = 2. We next extend the result to higher dimensions by induction. Assume that we have proved the result in dimension n. Let y = (y1 , ..., yn ) and put s(y) = nj=1 y j The induction hypothesis guarantees that a non-zero polynomial divisible by s j has at least N (n, j) terms. Let p be homogeneous of degree m + d in the n + 1 variables (y, x). Assume that p is a multiple of (x + s)d . We wish to show that the number of distinct monomials in p is at least N (n + 1, d). In Lemma 1.4, we noted that N (n + 1, d) =

d 

N (n, j).

j=0

We expand p in two ways, writing p(x, y) = (x + s)

d



h j (y)x

j



=

m+d 

A j (y)x j .

(∗)

j=0

As before, after dividing through by a power of x, we may assume without loss of generality that h 0 (y) = 0 and hence that A0 (y) = 0. We claim that at least d + 1 of the A j are not zero. To verify the claim, replace x by ws. Using homogeneity, we obtain a polynomial in the single variable w that is divisible by (1 + w)d . The claim thus follows from the one variable case proved above. Hence, there exist d + 1 integers such that 0 = j0 < j1 < · · · < jd and for which A jk = 0. At each stage, we choose these integers minimally. Next we note that s d−k divides A jk . This result holds by expanding the middle term in (*) by the binomial theorem, which yields an explicit formula for the A j (y), and then proceeding inductively using the minimality. Each expression A jk (y)x jk is divisible by a different power of x and hence all the resulting terms are distinct. Therefore, the number of terms K in p(x, y) satisfies

3 Ranks of Products

35

K ≥

d 

N (n, k) = N (n + 1, d).

k=0



The induction step is complete. Next, we use Proposition 2.2 to prove Theorem 2.2.

Proof Theorem 2.2 is trivial in 1 dimension, as the rank condition simply says that R is not identically 0. Assume n ≥ 2. Write R(z, z) = z2d r (z, z). Consider the lowest order part u(z, z) of the Taylor expansion of r ; the lowest order part of the Taylor expansion of R is given by z2d u(z, z). The rank of the coefficient matrix for R is at least as large as that of z2d u(z, z); it therefore suffices to prove the theorem when u is a homogeneous polynomial. Since n ≥ 2, we may de-homogenize by setting z n = 1. We continue to write z for the variable (z 1 , ..., z n−1 ) ∈ Cn−1 . Put p(z, z) = (1 + z2 )d u(z, z). We need to show that the rank of the matrix of coefficients of p is at least N (n, d). We know that  cαβ z α z β . p(z, z) = (1 + z2 )d |α|+|β|=m

Replacing z by eiθ z = (eiθ1 z 1 , ..., eiθn z n ) gives p(eiθ z, e−iθ z) = (1 + z2 )d



cαβ z α z β ei(α−β)θ = (1 + z2 )d



cβ+γ,β z γ |z|2β eiγθ

Each γ in the sum corresponds to the terms on a diagonal parallel to the main diagonal of a matrix with vector-valued coefficients. In lexicographical ordering, find the largest γ such that  cβ+γ,β |z|2β β

is non-zero. These coefficients are linearly independent because γ was chosen maximally. The rank of the matrix of coefficients of p is thus at least as large as the number of terms in this sum. Put (|z 1 |2 , ..., |z n−1 |2 ) = (x1 , .., xn−1 ) = x and |z|2β is simply x β . We re-homogenize by putting x = (x, xn ). Then 1 + z2 becomes |z n |2 + z2 = s(x). The hypotheses of Proposition 2.2 apply and the conclusion follows.  Our next result follows from this special case of Theorem 2.2 when d = 1. The theorem provides interesting information about the top order terms of the numerator in a rational sphere map. Theorem 2.3 Let qp be a rational sphere map of degree m with p(0) = 0. Let pm denote the terms of degree m in p. Then the target-rank of pm is at least the source dimension n.

36

2 Examples and Properties of Rational Sphere Maps

Proof By Proposition 2.1, the degree of q is less than m. By (2.2) (where the degree is D) we see that m m−1    pk 2 = |qk |2 k=1

k=0

on the sphere. Homogenizing gives m  k=1

 pk 2 z2m−2k =

m−1 

|qk |2 z2m−2k .

k=0

Every term on both sides of this identity except  pm 2 is multiplied by a power of z2 . Therefore,  pm 2 must be divisible by z2 . The result then follows from (the easiest case of) Theorem 2.2 with d = 1.  Remark 2.6 Proposition 2.2 includes the following statement in one dimension. Choose an arbitrary finite collection of numbers c0 , ..., ck such that at least one of them is non-zero. We regard these numbers as the first row, and then generate additional rows by the same method as we generate Pascal’s triangle from an initial row of 1. Then the d-th row has at least d + 1 non-vanishing numbers. We identify the first row with the coefficients of a polynomial, and the analogous triangle is generated by repeatedly multiplying by 1 + x. Exercise 2.3 Let r (z, z) be a polynomial. Show that r is real-valued if and only if its underlying matrix of coefficients is Hermitian symmetric. Exercise 2.4 Let r (z, z) be a real-valued polynomial. • Assume that r (z, z) =  p(z)2 for some vector-valued polynomial map p. Let w1 , ..., wk be points in Cn . Show that the matrix r (w j , w k ) is non-negative definite. • (Harder). Suppose, for every k and every choice of k points, the matrix r (w j , w k ) is non-negative definite. Show that r (z, z) =  p(z)2 for some vector-valued polynomial map p. • Give an example of a polynomial for which the property in the previous item holds for k = 1 and k = 2 but not for k = 3. Exercise 2.5 Show that each non-negative polynomial on R can be written as a sum of two squares of polynomials. Suggestion: First factor p. Then observe that (x − a)2 + b2 = |(x − a) + ib|2 . Finally, for z, w ∈ C, note that |zw| = |z| |w|. Remark 2.7 One consequence of Exercise 2.5 is the following elementary but perhaps counter-intuitive thing. Suppose that p j = u 2j + v 2j for j = 1, 2 for real polynomials. Then the product p1 p2 is also the sum of two squares. The simplest proof is to use complex numbers: note that p j = |u j + iv j |2 and hence p1 p2 = |(u 1 u 2 − v1 v2 ) + i(u 1 v2 + u 2 v1 )|2 = (u 1 u 2 − v1 v2 )2 + (u 1 v2 + u 2 v1 )2 ,

3 Ranks of Products

37

because |zw| = |z| |w| for complex z, w. We see a related idea in Sect. 6 of Chap. 3 when we solve the Diophantine equation A2 − δ B 2 = 1. Exercise 2.6 Let F be a field. Suppose that −1 is a sum of three squares in F. Show that in fact −1 is the sum of two squares in F.

4 Juxtaposition It is natural to ask what sort of algebraic structure the collection of rational sphere maps has. For n > 1, the set of rational sphere maps with source dimension n and target dimension N has little algebraic structure when N > n. We can compose with automorphisms on both sides, but we have no notions of addition, multiplication, or composition. In order for these notions to appear, we regard the target dimension as a variable. Given rational sphere maps f, g with the same source, we will provide several constructions of new rational sphere maps, impacting the target dimension. The target dimensions add when we juxtapose two maps and they multiply when we tensor two maps. We observed in Remark 2.2 that we can compose rational sphere map when the source dimension of one map is the target dimension of the other. These ideas indicate the usefulness of allowing the target dimension to vary. Definition 2.7 Let f, g be rational sphere maps with the same source. Let θ ∈ [0, π2 ]. We define the θ-juxtaposition Jθ ( f, g) by Jθ ( f, g) = cos(θ) f ⊕ sin(θ)g = (cos(θ) f, sin(θ)g). Proposition 2.3 Suppose f, g are rational sphere maps with the same source dimension and target dimensions N f and N g . Then Jθ ( f, g) is a rational sphere map with target dimension N f + N g . Proof For all z, we have Jθ ( f, g)(z)2 = cos2 (θ) f (z)2 + sin2 (θ)g(z)2 . For z2 = 1, the right-hand side becomes cos2 (θ) + sin2 (θ) = 1. The target dimensions add, by definition of orthogonal sum. The conclusion follows.  We can think of Jθ ( f, g) as defining a homotopy between f ⊕ 0 and 0 ⊕ g. Each element of the one-parameter family is itself a rational sphere map. The family depends continuously on the parameter θ. Example 2.5 Put f (z 1 , z 2 ) = (z 1 , z 2 ) and g(z 1 , z 2 ) = (z 1 , z 1 z 2 , z 22 ) Then  Jθ ( f, g) = cos(θ)z 1 , cos(θ)z 2 , sin(θ)z 1 , sin(θ)z 1 z 2 , sin(θ)z 22

38

2 Examples and Properties of Rational Sphere Maps

The target dimension of Jθ is 2 + 3 = 5. Its target-rank is 4. Defined h θ by  h θ (z) = z 1 , cos(θ)z 2 , sin(θ)z 1 z 2 , sin(θ)z 22 . It satisfies h θ 2 = Jθ 2 and has target dimension 4. We obtain a homotopy between (z 1 , z 2 , 0, 0) and (z 1 , 0, z 1 z 2 , z 22 ). It seems to be a difficult problem to decide the following thing. Given rational sphere maps f, g with the same source, what is the smallest dimension in which they are homotopic. See Proposition 2.8 and [26] for more discussion. Remark 2.8 We would like to say rational sphere maps f, g are homotopic in target dimension N if the following holds: there is a one-parameter family of rational sphere maps Ht with target dimension N such that the coefficients of Ht depend continuously on t, and such that H0 = f and H1 = g. For the concept to be useful, however, we need to change the last statement to H0 = f ⊕ 0 and H1 = U (g ⊕ 0) for U unitary. Equivalently, we could say H0 2 =  f 2 and H1 2 = g2 . Exercise 2.7 Suppose f, g are spherically equivalent. Show that they are homotopic. Example 2.5 shows that the converse fails. Exercise 2.8 What information about the unitary group U(n) is required in the previous exercise? Prove it. Open problem. Assume that f, g are rational sphere maps with the same source dimension. What is the smallest target dimension N in which f and g are homotopic? Is there a way to decide using topology?

5 The Tensor Product Operation Let f = ( f 1 , ..., f J ) and g = (g1 , ..., g K ) be holomorphic mappings with values in the indicated complex Euclidean spaces. We define their tensor product f ⊗ g to be the mapping with components f j gk for all possible pairs j, k. It will not matter in what order we list these components, but for the sake of precision, we write f ⊗ g = ( f 1 g1 , f 1 g2 , ..., f 1 g K , f 2 g1 , ..., f 2 g K , ..., f J g1 , ..., f J g K ).

(2.10)

The following trivial lemma provides the connection with rational sphere maps. Lemma 2.1 Let f and g be mappings with the same source but whose targets are complex Euclidean spaces of possibly different dimension. Then  f ⊗ g2 =  f 2 g2 . Hence, if f and g are rational sphere maps, then so is f ⊗ g. Proof  f ⊗ g2 =

 j,k

| f j gk |2 = (

 j

| f j |2 ) (

 k

|gk |2 ) =  f 2 g2 .

5 The Tensor Product Operation

39

In other words, the squared norm of a tensor product is the product of the squared norms. Thus, if f, g take values on the unit sphere, then so does f ⊗ g.  The tensor product of the identity with itself m times, written z ⊗m , plays a major role in this book. It is a monomial sphere map, with components the monomials z α = z 1α1 ...z nαn . It is annoying that these are repeated, and it is therefore natural to consider a symmetric version. First, we recall the multinomial expansion.  Proposition 2.4 (Multinomial theorem) Let mα denote the multinomial coefficient m! . Then α1 !...αn ! ⎞m ⎛ n   m  ⎠ ⎝ x α. xj = α |α|=m j=1 Definition 2.8 Let m be a positive integer. Fix an ordering of the list of monomials of order m in n variables. For z ∈ Cn , define Hm by Hm (z) = (..., cα z α , ...). In (2.11), α has order m and cα is the positive number satisfying |cα |2 =

(2.11) m . α

Note that Hm (z)2 = z ⊗m 2 = (

n  j=1

|z j |2 )m =

 m  |z|2α . α |α|=m

(2.12)

⊗m 2 Thus, Hm is a monomial n+m−1 sphere map with the same squared norm as z  . is the dimension of the space of homogeneous polynomials Recall that m of degree m in n variables. The following result was noted independently by Rudin [73] and the author.

Theorem 2.4 Let f : Cn → C N be a homogeneous polynomial mapping of degree m. Assume f is a polynomial sphere map. • There is a unitary U ∈ U(N ) such that f = U (Hm ⊕ 0).  and f = U Hm . • If the components of f are linearly independent, then N = n+m−1 m n+m−1 ≤ N . • The target-rank of f is m Proof We are given that  f (z)2 = 1 on z2 = 1. The same is true for Hm . Thus, by (2.12)  f (z)2 = Hm (z)2 = z2m , on the sphere, and therefore everywhere by homogeneity. The result then follows by Proposition 1.4. 

40

2 Examples and Properties of Rational Sphere Maps

6 The Restricted Tensor Product Operation Assume that f : Cn → C N and g : Cn → C K are holomorphic mappings. Their tensor product, as defined in (2.7), then maps to C N K . This target dimension is too large for many purposes, and hence we naturally consider the following restricted tensor product operation. This operation will have many uses in this book. Definition 2.9 Let f : Cn → C N and g : Cn → C K be holomorphic mappings. Let A be a subspace of C N . Let π A be the orthogonal projection onto A. We define E A,g ( f ) by (2.13) E A,g ( f ) = (π A f ⊗ g) ⊕ ((1 − π A ) f ). Proposition 2.5 If f, g are rational sphere maps with the same source, then E A ( f, g) is also a rational sphere map. Proof E A,g ( f ) is clearly rational. On the sphere  f 2 = g2 = 1. Using the Pythagorean theorem, on the sphere, we therefore have E A ( f, g)2 = π A f ⊗ g2 + ((1 − π A ) f 2 = π A f 2 g2 + (1 − π A ) f 2 = π A f 2 + (1 − π A ) f 2 =  f 2 = 1.  Example 2.6 (Whitney map) Define W (z) as follows: W (z 1 , ..., z n ) = (z 1 , ..., z n−1 , z 1 z n , z 2 z n , ..., z n2 ).

(2.14)

Then W is a monomial sphere map with target dimension 2n − 1. It is obtained as in Definition 2.7 by tensoring the identity map with the identity map on a onedimensional subspace. Example 2.7 Let us revisit Example 2.1. It is obtained by tensoring a unitary map with the identity map on the second component. Remark 2.9 We write E −1 A,g for the operation of replacing the right-hand side of (2.10) with f . This inverse operation is required in the classification of rational sphere maps, including the monomial case. One cannot obtain all examples via tensor products; one must allow undoing, a kind of √tensor division. The simplest counter-example is given by the map p(z, w) = (z 3 , 3zw, w 3 ). In fact, in degree at least 3, the generic monomial sphere map requires undoing. See also Remark 2.10. The restricted tensor product operations provide a subtle generalization of the notion of Blaschke product in source dimension greater than 1. The following theorem completely describes all polynomial sphere maps. The result, however, does not keep track of the target dimension. In the next result, the subspaces A j are subspaces

6 The Restricted Tensor Product Operation

41

of complex Euclidean spaces of different dimensions. The symbol z stands for the identity map. For clarity, we explain the notation. Given a polynomial sphere map f j , we find a subspace A j of the target space of f j . Let π j denote orthogonal projection onto A j . We define a new polynomial sphere map f j+1 by f j+1 = E(A j , z) f j = (π j f j ⊗ z) ⊕ (1 − π j ) f j . By Proposition 2.5, f j+1 is also a polynomial sphere map. The next result shows that we can apply this process finitely many times to obtain a map we can completely classify, namely the homogeneous map Hm . Theorem 2.5 Let p be a polynomial sphere map of degree m. Let Hm denote the homogeneous sphere map defined by (2.9). Then there is a finite list of operations of the form E j = E A j ,z and a unitary map U such that p = E 1−1 ◦ · · · E k−1 U (Hm ⊕ 0). Equivalently, there is a linear map L such that Hm = L ◦ E k ◦ · · · E 1 ◦ p. Proof Let ν denote the order of vanishing of p at 0. We expand p in terms of homogeneous polynomials: m  pk . p= k=ν

Assume first that ν = m. Then p is homogeneous, and the conclusion follows immediately from Theorem 2.4 without using any of the E j operations. Now suppose that ν < m. By Corollary 2.1, we have  pν , pm = 0. Let A1 denote the span of pν . We consider the map E(A1 , z) p. It is also a polynomial sphere map, and its order of vanishing is now ν + 1 while its degree remains m. As along as the order of vanishing is less than the degree, we continue. After finitely many operations of this type, we obtain a homogeneous polynomial sphere map. Therefore, Hm is obtained from p as claimed.  Example 2.8 Consider the polynomial sphere map f defined by f (z, w) = (z 5 ,

√

5z 3 w,

√

5zw 2 , w 5 ).

We discuss this sort of example in Chap. 3. For now, we show how Theorem 2.5 applies to f . The order of vanishing of f is 3. We tensor on the subspace of terms of order 3, getting the new map E( f )(z, w) = (z 5 ,

√

5z 3 w,

√

5z 2 w 2 ,

√

5zw 3 , w 5 ).

42

2 Examples and Properties of Rational Sphere Maps

Applying E on the subspace of terms of order 4 gives E(E( f ))(z, w) = (z 5 ,

√

5z 4 w,

√

5z 3 w 2 ,

√

5z 3 w 2 ,

√

5z 2 w 3 ,

√

5z 2 w 3 ,

√

5zw 4 , w 5 ).

After an appropriate unitary map U , we see that U E E f (z, w) = (z 5 ,

√ 4 √ 3 2 √ 2 3 √ 5z w, 10z w , 10z w , 5zw4 , w5 , 0, 0) = H5 (z, w) ⊕ 0.

We restate Theorem 2.5 in the following way. In Chap. 7, we use this version to derive a beautiful geometric inequality. An analogous statement holds in the general rational case, but the resulting geometric inequality is less elegant. Theorem 2.6 (Orthogonal homogenization and squared norms) Let p be a polynomial sphere map of degree d and source dimension n. Then there is a polynomial mapping w such that  z2d =  p(z)2 + w(z)2 z2 − 1 .

(2.15)

Proof We againexpress the sphere map condition in terms of the homogeneous expansion p = pk , yielding  p(z)2 = 



pk (z)2 =

  pk (z), p j (z) = 1. k, j

As before, we conclude that p must satisfy certain identities when z = 1: 

 pk 2 = 1,

  pk , pk+l = 0. (l = 0)

(2.16) (2.17)

k

The homogenized forms of these identities hold everywhere. As above, p0 and pd are orthogonal. Let π A denote the projection of C N onto the span A of p0 . Put p = (1 − π A ) p ⊕ π A p.

(2.18)

Consider the map g, defined by g = E A ( p) = (1 − π A ) p ⊕ (π A p ⊗ z). In terms of squared norms, we can describe the process as follows: E A p2 = π A p2 z2 + (1 − π A ) p2 =  p2 + π A p2 (z2 − 1). (2.19) Thus, E A ( p) is also a polynomial sphere map.

6 The Restricted Tensor Product Operation

43

The unit sphere is of course the zero set of z2 − 1; we call this function a defining function. At each stage of the process in the proof of Theorem 2.6, we add a squared norm times the defining function to a squared norm, obtaining a new squared norm. Proceeding in this fashion, we increase the order of vanishing without increasing the degree, stopping when the result is homogeneous. Thus, there is a natural sequence of subspaces A0 , . . . , Ad−1 such that composing these tensor product operations yields something homogeneous of degree d. As the last step, we obtain a homogeneous proper mapping ζ of degree d between balls. It follows that ζ(z)2 = z2d . Hence,  z2d =  p(z)2 + w(z)2 z2 − 1 , where w(z)2 is the sum of the squared norms arising at each step.

(2.20) 

Theorems 2.5 and 2.6 state the same thing in different ways. Each polynomial sphere map of degree d can be turned into Hd by finitely many restricted tensor product operations. In Sect. 12, we will use Eq. (2.17). In Chap. 7, we will see that each of these operations increase the volume of the image of the map. On the other hand, tensoring can impact the target-rank in all three possible ways. Example 2.9 Put n = 2. Put f (z, w) = (z, w). Tensoring f on the subspace generated by (0, 1) gives g(z, w) = (z, zw, w2 ), and hence the target-rank increases. Tensoring g on the subspace generated by (1, 0, 0) gives h(z, w)√= (z 2 , zw, zw, w 2 ), whose target-rank is 3, since h is unitarily equivalent to (z 2 , 2zw, w 2 , 0). Here, the target-rank is unchanged. Finally, in any dimension n, consider the juxtaposi . tion map p(ζ) = λζ ⊕ (1 − λ)ζ ⊗2 for 0 < λ < 1. The target-rank of p is n + n+1 2 Tensoring by the identity map on the image of the linear part yields a map unitar . Thus, tensoring decreased the ily equivalent to ζ ⊗2 , which has target-rank n+1 2 target-rank in this case. Remark 2.10 Theorems 2.5 and 2.6 do not say the following false thing. “Each polynomial proper mapping of degree d can be obtained by finitely many restricted tensor products.” The simplest counterexample is the map defined by p(z, w) = (z 3 ,

√

3zw, w 3 ).

√ √ The theorem tells us that we can tensor it into the map (z 3 , 3z 2 w, 3zw 2 , w 3 ), but we cannot start with (z, w) and perform tensor operations to reach p. Thus, undoing, or partial tensor division, is required in general.

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2 Examples and Properties of Rational Sphere Maps

7 An Abundance of Rational Sphere Maps In this section, we prove a decisive result establishing the existence of many rational sphere maps. This theorem tells us that the first several components of a rational sphere map are essentially arbitrary. We also establish a sufficient (but not necessary) condition for a polynomial map of degree d − 1 to be the Taylor polynomial of a polynomial sphere map of degree d. Both results rely on the identities (2.2) and their corollaries. Our first theorem demonstrates the abundance of rational sphere maps. The flexibility in this result arises because we put no constraint on the target dimension. Theorem 2.7 Let q : Cn → C be a polynomial. Let p : Cn → C N be a polynomial map such that  p(z)2 < |q(z)|2 on the closed unit ball. Then there is an integer M and a polynomial map g : Cn → C M such that •  p ⊕ g2 = |q|2 on the unit sphere. is reduced to lowest terms. • The rational map p⊕g q is a rational sphere map. • p⊕g q Corollary 2.5 Let p : Cn → C N be a polynomial map such that  p(z)2 < 1 on the unit sphere. Then there is an M and a polynomial map g : Cn → C M such that p ⊕ g is a polynomial sphere map. Proof Since  p(z)2 < 1 on the unit sphere, and p is holomorphic, the same inequality holds on the closed unit ball. Hence, we may take q = 1 in the theorem.  Remark 2.11 Suppose f is a holomorphic map from Bn to B N , continuous on the sphere, and  f (z)2 < 1 on the closed ball. It follows from the work in [61] that there is a holomorphic map g such that f ⊕ g is a proper map to some ball. Our result is quite different; we make the stronger assumption that f is rational, and we obtain a stronger conclusion: g is rational and has the same denominator. It does not seem to be possible to derive either result from the other. Example 2.10 Put p(z 1 , z 2 ) = cz 1 z 2 . The condition that | p|2 < 1 on the unit sphere is equivalent to |c|2 < 4. Thus, p is a component of a rational sphere map f = p ⊕ g if |c|2 < 4. We will show that both the degree and minimum target dimension of f tend to infinity as |c|2 approaches 4. For monomial sphere maps, we will give sharp bounds for this constant in terms of the degree. See Proposition 4.2. Corollary 2.6 Let q be an arbitrary polynomial that is not zero on the closed unit ball. Then q is the denominator of a rational proper map between balls. Proof Since q = 0 on the closed ball, |q| achieves a positive minimum. Hence, we can find a non-zero constant c such that |cz 1 | < |q(z)|2 on the closed ball. By the theorem, there is a rational sphere map with the first component czq1 and denominator q. This sphere map is not a constant, and hence it is proper. 

7 An Abundance of Rational Sphere Maps

45

Corollary 2.6 extends Corollary 1.10 to arbitrary source dimension. We discuss the fundamental difference in the results. For n = 1, there is an explicit formula for the numerator p. Its degree is the same as the degree of q, and its target dimension equals 1. For n ≥ 2, both the degree and target dimension of the numerator depend on the values of the denominator. For example, if q(z) = 1 − λz 1 z 2 , then q is the denominator of a rational sphere map whenever |λ|2 < 4. The minimum target dimension of a possible numerator tends to ∞ as |λ|2 tends to 4. We now fill in the necessary steps to prove Theorem 2.7. The first observation is by now well known. Theorem 2.8 is proved for example in [10] and [19]. The special case where r (z, z) is bihomogeneous of bi-degree (d, d) goes back to [67]. Theorem 2.8 Let r (z, z) be a polynomial. Assume that r (z, z) > 0 on the unit sphere. Then there are finitely many holomorphic polynomials g1 , . . . , gk such that r (z, z) =

k 

|g j (z)|2 = g(z)2

j=1

on the unit sphere. Proof (of Theorem 2.7.) Theorem 2.8 implies Theorem 2.7 as follows. Put r (z, z) = |q(z)|2 −  p(z)2 . Theorem 2.8 guarantees that there are polynomials g j in z for which |q(z)|2 =  p(z)2 + g(z)2 on the sphere. When n ≥ 2, the zero-locus of q must intersect the sphere if it intersects the open ball. Since r (z, z) > 0 on the unit sphere, q cannot be 0 there. Hence, q is to be a rational sphere map, we need to ensure not zero on the closed ball. For p⊕g q that the fraction is reduced to the lowest terms. To ensure this conclusion, we include an additional function, say cz 1 , among the p j . Since the hypothesis assumes strict inequality, we can choose c small enough such that  p ⊕ cz 1 2 < |q|2 remains true on the sphere. Since z 1 is not divisible by any polynomial q with q(0) = 0, we obtain the desired conclusion.  n N We next develop these ideas d−1a bit further. Consider a polynomial p : C → C of degree d − 1. Write p = j=0 p j in terms of its homogeneous parts. We naturally ask whether there is a homogeneous polynomial pd such that p + pd is a polynomial sphere map. It is of course a necessary but not sufficient condition that  p(z)2 ≤ 1 on the unit sphere. In Example 2.10, where p(z) = cz 1 z 2 , this necessary condition is that |c|2 ≤ 4. The necessary and sufficient condition such that there is a p3 for which p + p3 is a polynomial sphere map is the inequality |c|2 ≤ 3. A more subtle necessary condition  arises from considering Hermitian forms. Let r (z, z) = a,b cab z a z b be a real-valued bihomogeneous polynomial of degree d in both z and z. Recall from Chap. 1 that (cab ) is called the underlying matrix of coefficients, and we naturally define a Hermitian form by

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2 Examples and Properties of Rational Sphere Maps

Q(ζ, ζ) =



cab ζa ζ b .

a,b

If Q(ζ, ζ) ≥ 0 for all ζ, then r (z, z) ≥ 0 for all z, but the converse fails. Example 2.11 Put n = 2 and write (z, w) for the variables. Put r (z, w, z, w) = |z|4 + |w|4 − λ|zw|2 . The values of r are non-negative for λ ≤ 4. The corresponding Hermitian form is diagonal, with eigenvalues 1, −λ, 1. It is non-negative definite only for λ ≤ 0.  Let us apply these ideas to the polynomial p = d−1 j=0 p j . If there is a pd such that p + pd is a rational sphere map, then  p(z) + pd (z)2 = 1 on the unit sphere. Homogenization yields a large number of relationships among these p j , namely the identities (2.2) or (2.3). The simplest of these identities is that d−1 

 p j (z)2 z2d−2 j +  pd 2 = z2d .

(2.21)

j=0

It follows from (2.21) that z2d −

d−1 

 p j (z)2 z2d−2 j =  pd 2 ,

(2.22)

j=0

and hence the underlying matrix of coefficients on the left-hand side of (2.22) is nonnegative definite. Let us consider Example 2.10 in this light. Put p(z, w) = czw = p2 (z, w). We seek a p3 . After homogenizing, we obtain a diagonal underlying matrix of coefficients whose eigenvalues are 1, 3 − |c|2 , 3 − |c|2 , 1. In this simple case, the other required information from Corollary 2.2 is that p3 be orthogonal to p2 . For |a|2 = 3 − |c|2 , we obtain the following family of maps: f (z, w) = (z 3 , az 2 w, azw 2 , w 3 , czw).

(2.23)

The author finds it interesting that one can choose the target dimension for such a map to be 3 only when |c|2 = 3. For c = 0, there is an example in target dimension 4, and for 0 < |c|2 < 3, the target dimension must be at least 5. We state the main point of this discussion in the following manner. Proposition 2.6 Consider a polynomial p : Cn → C N of degree d − 1. If there exists a homogeneous polynomial pd of degree d such that p + pd is a polynomial sphere map, then the Hermitian form defined by the left-hand side of (2.22) must be non-negative definite. The converse is false.

7 An Abundance of Rational Sphere Maps

47

Proof We have already established the necessity of the condition, but we repeat the idea. We apply identity (2.2) when q = 1 and m = 0. We obtain, for z on the sphere, d−1 

 p j (z)2 +  pd 2 = 1.

(2.24)

j=0

Homogenizing (2.24) gives (2.21) and hence (2.22). Since the left-hand side of (2.22) must be a Hermitian squared norm, the necessity of the condition follows. To show that the condition is not sufficient, consider an arbitrary vector-valued p1 : Cn → C N . For a not identically zero linear complex-valued linear function g, put p2 = gp1 . We seek a p3 such that p1 + p2 + p3 is a polynomial sphere map. By Corollary 2.2, the unknown p3 must satisfy three equations:  p3 , p1 = 0,  p3 , p2 +  p2 , p1 z2 = 0,  p3 2 +  p2 2 z2 +  p1 2 z4 = z6 . Thus, p3 must be orthogonal to p1 , and by our choice of p2 , it must also be orthogonal to p2 . The middle equation therefore yields 0 =  p2 , p1 = g p1 2 , which is possible only if p1 = 0. There exist non-zero p1 for which the Hermitian form defined by the left-hand side of (2.22) is semi-definite. When p2 = gp1 , we have shown that p1 + p2 + p3 cannot be a polynomial sphere map unless p1 = 0. 

8 Some Results in Low Codimension In Chap. 6 we will show, for n ≥ 2, that equi-dimensional rational sphere maps must be automorphisms and hence linear fractional transformations. Webster [77] proved that a three times continuously differentiable map f : S 2n−1 → S 2n+1 , satisfying the tangential Cauchy-Riemann equations, must be a linear fractional transformation when n ≥ 3. In fact, f must be spherically equivalent to the map z → (z, 0). Faran then improved this result in two ways. First, when n = 2, he found that there were precisely four spherical equivalence classes of proper rational maps f : B2 → B3 . [32]. Then he extended Webster’s result to low codimension. [33]. We discuss these results now. Theorem 2.9 (Faran) Suppose f : B2 → B3 is a proper holomorphic map with three continuous derivatives at the boundary sphere. Then f is spherically equivalent to one of the following four maps:

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2 Examples and Properties of Rational Sphere Maps

(z, w) → (z, w, 0) (z, w) → (z, zw, w2 ) (z, w) → (z 2 , (z, w) → (z 3 ,

√ √

2zw, w 2 )

3zw, w 3 ).

Cima-Suffridge [12] reduced the differentiability requirement in Theorem 2.9 to two continuous derivatives. Furthermore, Forstneriˇc showed, for n ≥ 2, that proper holomorphic maps f : Bn → B N with N − n + 1 continuous derivatives at the sphere must be rational [36]. Cima-Suffridge showed, in this setting, that the maps must extend holomorphically past. In our terminology, Theorem 2.9 lists the four spherical equivalence classes of rational sphere maps from B2 to B3 . Lebl [53] and Reiter [68] have each generalized Theorem 2.9 to also classify maps from the sphere S 3 to hyperquadrics in C3 . Lebl follows the same basic ideas as Faran but requires additional Hermitian linear algebra. Reiter’s proof is perhaps conceptually simpler, but requires some computer assistance as the formulas become quite complicated. Rather than listing the seven equivalence class of mappings, we mention two interesting aspects. First of all, there are rational maps not equivalent to monomials. The simplest example is given by 1  f (z, w) = 2 z, w 2 , w = z



1 w2 w , , z z2 z2

 .

(2.25)

Second, there is an obvious collection of examples that cannot arise for the unit sphere. Consider the map f (z, w) = (1, g(z, w), g(z, w)) where g is a completely arbitrary holomorphic function. Then | f1 |2 + | f 2 |2 − | f 3 |2 = 1 on the unit sphere. Exercise 2.9 Show that the map in (2.25) sends the unit sphere to a hyperquadric. Theorem 2.10 below was first proved by Faran, assuming that f is proper and of class C ∞ at the sphere. Huang-Ji [47] then established the result when f is assumed to be proper and has two continuous derivatives at the sphere. Their result establishes rationality, in this special case, with fewer derivatives required than in Forsteneriˇc’s theorem. See also [46]. We return to spheres and the low codimension case when n ≥ 3. The next result holds when N ≤ 2n − 2, but fails in target dimension 2n − 1. Example 2.6 defines the Whitney map W (z) = (z 1 , z 2 , ..., z n−1 , z 1 z n , z 2 z n , ..., z n2 )

8 Some Results in Low Codimension

49

with target dimension 2n − 1. It arises quite often in this book, as it is the simplest example of a restricted tensor product. Theorem 2.10 Assume n ≤ N ≤ 2n − 2. Each rational sphere map with source dimension n and target dimension N is spherically equivalent to the map z → (z, 0). Our next result shows what happens when we increase the target dimension to 2n. In some sense, this simple result shows that irrigidity begins in target dimension 2n. A stronger result about spherical equivalence of homotopic sphere maps was proved in [26]: an arbitrary homotopy of rational sphere maps whose endpoints are spherically inequivalent must contain uncountably many spherically inequivalent maps. Theorem 2.11 For each n ≥ 1, there is an uncountable family of spherical equivalence classes of rational sphere maps with source dimension n and target dimension 2n. Proof It suffices to write down an uncountable family of spherically inequivalent monomial sphere maps. We use the technique of juxtaposition from Definition 2.5. Let f denote the identity map and let g denote the√Whitney map. Consider the juxtaposition h t defined for t ∈ [0, 1] by h t = t f ⊕ 1 − t 2 g. As given, it defines a rational sphere map whose target dimension is too large. We have       h t (z) = t z 1 , ..., t z n , 1 − t 2 z 1 , ...., 1 − t 2 z n−1 , 1 − t 2 z 1 z n , ..., 1 − t 2 z n2 . A glance reveals that h t is unitarily equivalent to the map ζt defined by   ζt (z) = (z 1 , ..., z n−1 , t z n , 1 − t 2 z 1 z n , ..., 1 − t 2 z n2 ) ⊕ 0. The first 2n components of ζt thus define a monomial sphere map with source dimension n and target dimension 2n. We leave the following easy detail to an exercise: monomial sphere maps preserving the origin are spherically equivalent if and only if they are unitarily equivalent. We must show that there are no unitary U ∈ U(n) and V ∈ U(2n) such that V ◦ ζt ◦ U = ζr unless r = t. Doing so is routine and left to the reader.  The reader should observe that the family in the proof of Theorem 2.11 generalizes Example 2.5. We briefly discuss the generalization to arbitrary homotopies of rational sphere maps. Given a family Ht , the first step is to show that the set St of t ∈ [0, 1] for which Ht is spherically equivalent to H0 is closed. The second step is to apply a classical result of Sierpinski: the closed interval [0, 1] is not a countable union of (non-empty) disjoint closed sets. Therefore, either St = [0, 1] and all the maps are spherically equivalent, or there are uncountably many spherical equivalence classes.

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2 Examples and Properties of Rational Sphere Maps

Remark 2.12 Sierpinksi proved the following general result. Let X be a compact, connected Hausdorff space. If {Fn } is a closed cover of pairwise disjoint subsets, then one of these sets is X and the others are empty. Perhaps one could use this result to study higher dimensional families of rational sphere maps. Exercise 2.10 Prove that monomial maps preserving the origin are spherically equivalent only if they are unitary equivalent. Give a counter-example when the origin is not preserved. Exercise 2.11 Complete the last step of the proof of Theorem 2.11. Exercise 2.12 Show that U(n) is path-connected. Thus, show that each unitary map is homotopic to the identity in target dimension n. Show that the analogous statement is false if we consider the group of orthogonal matrices. Exercise 2.13 Show that each Blaschke product of order m is homotopic (in source and target dimensions equal to 1) to the map z → z m . What is the analogue of this result in higher dimensions? To complete our discussion in the low codimension case, we mention the following result. When N > n, that there exist proper holomorphic mappings from Bn to B N that are continuous at the boundary but are not rational. See, for example, [31]. In fact, there is a profusion of examples. It is an open problem to find the minimum regularity that guarantees rationality. In this book, we assume rationality because, even with this assumption, the number of interesting results is astonishing. Furthermore, many connections to other parts of mathematics reveal themselves in the rational case. Although, in codimension 1, there exist many proper maps with very bad boundary behavior, complete flexibility does not happen. For example, in [23] it is proved that there is no one-parameter convex family of proper maps preserving the origin. Open problem. Assume n ≥ 2 and that f : Bn → B N is proper holomorphic. Determine the minimum regularity of f at the sphere needed to prove that f is rational.

9 A Result in Sufficiently High Codimension Theorem 2.10 indicates the appearance of gaps in possible minimal target dimensions. If n ≤ N ≤ 2n − 2, then any non-constant rational sphere map is spherically equivalent to the map z ⊕ 0. Thus, there are no truly new rational sphere maps in the target dimension N when n + 1 ≤ N ≤ 2n − 2. If we allow constant sphere maps, then we see a similar gap. There are no new maps for 2 ≤ N < n. Huang and Ji (see especially [48]) have observed other similar gaps. For example, when 4 ≤ n ≤ N ≤ 3n − 4, the only maps arising are spherically equivalent to those already occurring when the target dimension is 2n. On the other hand, the author and Lebl proved Theorem 2.12 below, which shows the following. In each target dimension N with N ≥ n 2 − 2n + 2, there is a new

9 A Result in Sufficiently High Codimension

51

polynomial sphere map. In other words, for each such N there is a polynomial sphere map whose minimal target dimension is N . See also Theorem 3.16 for a related result. Before proving Theorem 2.12, we make an important observation relating targetrank and the term new map. Let c, s be non-zero numbers with |c|2 + |s|2 = 1. Let f be a rational sphere map with f (0) = 0. Then c ⊕ s f is also a rational sphere map and it is spherically equivalent to f . The target-rank of c ⊕ s f exceeds that of f by 1. Thus, if f has target dimension N , and N is minimal, then c ⊕ s f maps into an affine subspace of dimension N , but its target-rank is N + 1. To give a concrete example, consider f to be the identity map in 3 variables. Then c ⊕ s f has target dimension and target-rank both equal to 4. Since this map is spherically equivalent to (z, 0), it does not count as a new map. In Theorem 3.16, we will focus on target-ranks. We will revisit the effect of allowing f (0) to be non-zero when we prove that result. For now, we repeat this point as a warning. Warning. Let f be a rational sphere map. The target-rank of f is not generally invariant under composition with automorphisms unless f (0) = 0. The target-rank is invariant under unitary automorphisms, but can drop when composed with an automorphism that moves the origin. Exercise 2.14 For z ∈ Cn , prove that z → c ⊕ sz, with c, s as above, is spherically equivalent to the map z → (z, 0). Theorem 2.12 uses the restricted tensor product operation and is completely constructive. We will write down explicit monomial sphere maps that do the job. The motivation comes from the so-called postage stamp problem. Given a pair of postage stamps with integer values a, b, where a, b are relatively prime, there is a largest postage that cannot be made. This postage is ab − a − b, called the Frobenius number; this result is known as Sylvester’s theorem. Given a, b, two things are true: Proposition 2.7 (Sylvester’s theorem) Let a, b be relatively prime positive integers. The following holds: • Whenever N > ab − a − b, there are non-negative integers j, k such that N = ja + kb. • There are no non-negative integers j, k with ab − a − b = ja + kb. Remark 2.13 Results for more than two stamps are not complete to this date! Exercise 2.15 Prove Sylvester’s theorem. Suggestion: Consider the level lines of the function f (x, y) = ax + by. We now use the idea of the postage stamp problem in the trivial situation of consecutive integers. Put a = n and b = n − 1. Then the number from Sylvester’s theorem is n(n − 1) − n − (n − 1) = n 2 − 3n + 1. We will start with the identity map, which has target dimension n, and add n 2 − 3n + 1 terms to it by applying two operations, one n − 1 times and the other n times.

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2 Examples and Properties of Rational Sphere Maps

Theorem 2.12 Put T (n) = n 2 − 2n + 2. For every N at least T (n), there is a polynomial sphere map with source dimension n and minimal target dimension N . Proof For z ∈ Cn , put x = (x1 , ..., xn ) = (|z 1 |2 , ..., |z n |2 )  and put s(x) = nj=1 x j = z2 . As discussed in great detail in the next chapter, and explained below, it suffices to find a real polynomial f (x) such that • f (x) = 1 when s(x) = 1. • The coefficients of f are positive. • f has precisely N terms. Such an f will be the squared norm m(z)2 of a monomial sphere map m(z). N j First we note that  the case n = 1 is easy. The polynomial PN (x) = 1 c j x does the job when c j = 1 and each c j > 0. Hence, we assume that the domain dimension n is at least 2. Consider a polynomial p of degree d satisfying the first two conditions above and containing the monomial cxnd for c > 0. We can perform two distinct restricted tensor product operations on p, as follows: (W p)(x) = p(x) + c(s(x) − 1)xnd c (V p)(x) = p(x) + (s(x) − 1)xnd . 2 By construction, if p satisfies the above conditions, then so do W p and V p. Apply W , the Whitney operation, a total of j times to s(x). We get an example W j s with j (n − 1) + n terms. The idea is to apply W a total of j times and V a total of k times, as in the postage stamp problem, to get postage ja + kb. Each application of V adds n terms, so V k W j s has N = j (n − 1) + n + kn = j (n − 1) + (k + 1)n terms. The total number of terms is thus a positive integer combination of the relatively prime numbers n − 1 and n. (Recall that n ≥ 2.) The beginning polynomial s has n terms. The procedure V k W j s therefore produces an example as long as N > n(n − 1) − n − (n − 1) + n = n 2 − 2n + 1. We conclude, for N ≥ n 2 − 2n + 2, that there is a real polynomial f satisfying the three properties above. As we discuss in detail in the next Chapter, f (x) = m(z)2 . The target-rank of f is N .



10 Homotopy and Target-Rank

53

10 Homotopy and Target-Rank This section has two purposes. First, we give a surprising example from [26]. Given a one-parameter family of rational sphere maps with the same target-rank, depending nicely on a parameter t, we might expect that the degree cannot vary with t. This plausible statement fails! We then prove Proposition 2.8. Example 2.12 We give a pair f, g of monomial sphere maps from B2 to B5 . Each map has target-rank 5. These maps are of different degrees, but they are homotopic in target dimension 5. f (z, w) = (z, zw, zw 2 , zw 3 , w 4 ) g(z, w) = (−w 2 , zw, −zw 2 , z 2 w, z 2 ). With c denoting cosine and s denoting sine, put t = sin(θ). We define a oneparameter family Ht by Ht (z, w) = (cz − sw2 , zw, (cz − sw 2 )(sz + cw 2 ), zw(sz + cw 2 ), (sz + cw 2 )2 ). When t = sin(θ) = 0, we obtain f and, when t = 1, we obtain g. Thus, the degree of the map Ht is 4 for 0 ≤ t < 1, but jumps down to 3 when t = 1. Target-rank plays a key role in the following proposition where the maps need not be rational. Proposition 2.8 Let f, g be proper holomorphic maps between balls, with the same source ball. Write N f for the target dimension of f and N g for that of g. Then there is a continuous one-parameter family of proper holomorphic maps from Bn to B N for N = n + max(N f , N g ) connecting f and g. Rather than writing out a formal proof, we sketch the simple idea. Let z denote the identity map. We consider two families obtained via juxtaposition,  (t z, 1 − t 2 f )  (t z, 1 − t 2 g). Each defines a homotopy between the identity map and either f or g, with target dimensions n + N f and n + N g . Thus, both f and g are homotopic in target dimension N = n + max(N f , N g ) to the identity and hence to each other. If we allow constant sphere maps, then we can replace the identity map in the proof of Proposition 2.8 by the constant map 1. Then we can lower N to 1 + max(N f , N g ). See [26] for additional information about homotopies.

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2 Examples and Properties of Rational Sphere Maps

11 Remarks on Degree Bounds We have seen in source dimension n = 1 that there is a rational sphere map (for example, z d ) of arbitrarily large degree with target dimension 1. For n ≥ 2, there is a bound C(n, N ) on the degree of a rational sphere map. The precise bound is not known, although sharp bounds in the monomial case are known. See Theorem 3.2 for the case n = 2 (the most difficult case) and Theorem 4.6 for n ≥ 3. The author has conjectured that the same bounds hold for general rational sphere maps. In some contexts, it is useful to know that any degree bound exists. Thus, we seek a function C(n, N ) such that the degree of a rational sphere map with source dimension n and target dimension N is at most C(n, N ). Meylan [62] gave such a bound when n = 2 and D’Angelo-Lebl [23] used her bound to obtain a bound in all source dimensions at least 2. That a bound exists also follows from the proof of Forstneriˇc’s theorem in [36]. We say a bit more. For a rational sphere map of degree d, source dimension 2, and target dimension N , Meylan proved that d ≤ N (N2−1) . The author and Lebl showed the following. If ) c(2, N ) is any degree bound for these dimensions, then c(2,N will provide a degree 2n−3 bound in any source dimension n ≥ 2. The idea of the proof uses certain group invariant monomial sphere maps discussed in the next Chapter. As of 2020, sharp bounds are known only in the monomial case. It is plausible that these bounds hold for all rational sphere maps. Open Problem. Find the smallest number c(n, N ) such that d ≤ c(n, N ) holds for all rational sphere maps of degree d, source dimension n, and target dimension −1 for N . The author has long conjectured that c(2, N ) = 2N − 3 and c(n, N ) = Nn−1 n ≥ 3. These bounds are sharp in the monomial case. A degree bound has a consequence for families of rational sphere maps. Theorem 2.13 For each source dimension n ≥ 2 and target dimension N , the number of homotopy classes of rational sphere maps is finite. Proof Let f = qp be a non-constant rational sphere map. Recall, by convention, that q(0) = 1. Since N is assumed fixed here, some degree bound exists. The expression R(z, z) =  p(z)2 − |q(z)|2 is thus also of bounded degree. The vector coefficients of p and the scalar coefficients of q are also bounded. Since R vanishes on the unit sphere, it is divisible as a polynomial by z2 − 1. The identities from Theorem 2.1 show that the coefficients of R are constrained by finitely many polynomial inequalities. Thus, rational sphere maps correspond to a bounded semi-algebraic set in real Euclidean space. Such sets have finitely many components, each of which consists of homotopic rational sphere maps. 

12 Inverse Image of a Point

55

12 Inverse Image of a Point A non-constant rational sphere map is a proper mapping between balls. Hence, the inverse image of a point in the target ball is a compact subset of the ball as well as a complex variety. Such a set must be finite; see [71] or our discussion in Chap. 6. Thus, the inverse image of a point in the target ball is either empty or a finite set in the source ball. We first observe that any finite set is possible. In the next result, each a j can be repeated finitely many times. If some a j is repeated m j times, then our rational sphere map as constructed in the proof will have a zero with multiplicity m j at a j . See, however, Remark 2.14. Proposition 2.9 Let a1 , ..., ak be a finite set of points in Bn with repetitions allowed. Then there is an integer N and a rational sphere map f with target dimension N such that f −1 (0) = {a1 , ..., ak }. Proof Consider the automorphisms ψ j defined by ψ j = φa j . Here φa is the usual ball automorphism satisfying φa (a) = 0. Let f be the full tensor product of the ψ j : f = ψ1 ⊗ · · · ⊗ ψ k . Then f (a j ) = 0 for all j. Note that  f (z)2 =

k 

ψ j (z)2 .

j=1

Thus, f (z) = 0 implies that some ψ j (z) = 0, and therefore z = a j for some j.  Next, we observe what happens in the case of polynomial sphere maps. Proposition 2.10 Let p be a polynomial sphere map. Then p −1 (0) is either empty or the single point 0. Proof The result follows from either Theorem 2.5 or Theorem 2.6. We use Theorem 2.6. Assume p is of degree d. By Theorem 2.6, we can write z2d =  p(z)2 + w(z)2 (z2 − 1)

(2.26)

for some vector-valued polynomial mapping w. Suppose that p(a) = 0 for some nonzero a in the ball. Then the right-hand side of (2.26) is non-positive; if w(a) = 0, then it is negative because a2 < 1, and if w(a) = 0, it is zero. But the left-hand side of (2.26) is positive except at a = 0. Hence, p(a) = 0 implies a = 0. The empty set is also a possibility. For example, the juxtaposition Jθ of the constant map 1 and  the identity map is never zero for 0 < θ < π2 . Remark 2.14 We mentioned in Sect. 2 that it is problematic to define a notion of multiplicity at a point. It is reasonable to say that the multiplicity of the zero at the

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2 Examples and Properties of Rational Sphere Maps

origin for the map z ⊗m equals m, or more generally that the multiplicity at a for φa⊗m is also m. As Example 2.2 shows, however, for a general rational sphere map (and even for a monomial sphere map) it is best to avoid this concept. Exercise 2.16 Derive Proposition 2.10 from Theorem 2.5. Proposition 2.9 is of course obvious for Blaschke products (thus n = N = 1). In this case, the zeroes of the numerator of a Blaschke product are the Schwarz reflections across the circle of the zeroes of the denominator. Proposition 2.10 is also obvious when n = N = 1, as the only polynomial sphere maps are z → z m . When N > n = 1, there are non-constant polynomial maps for which p −1 (0) is empty. Take for example, p(z) = (cos(θ), sin(θ)z). When n = N = 1, of course, f −1 (0) cannot be empty. Recall the proof of Proposition 1.11 and Exercise 1.21. Our next result provides a kind of Schwarz reflection principle; the (finite) zero set of p is contained in the Schwarz reflection across the sphere of the zero set of q. If n ≥ 2, and q is not a constant, then its zero set is of course infinite. The result also gives an alternative way to derive Proposition 2.10. The technique of polarization enables us to generalize the idea of Schwarz reflection to higher dimensions. Theorem 2.14 Let f = qp be a rational sphere map of degree d. Define a polynomial H q(w, z) by homogenizing the denominator q as follows: H q(w, z) =

d 

q j (w)w, z d− j .

j=0

Then the set Sq of z in the ball for which H q(w, z) vanishes identically in w is finite. Furthermore, the zero set of p lies in Sq . Proof We are given  p(z)2 = |q(z)|2 on the sphere. Polarizing yields  p(z), p(ζ) = q(z)q(ζ) on the set z, ζ = 1. We may therefore substitute  p(z), p(

w w,z

for ζ to obtain

w w ) = q(z)q( ). w, z w, z

Writing q in terms of its homogeneous expansion and clearing denominators gives z, w d  p(z), p(

  w = q(z) z, w d + z, w d−1 q1 (w) + · · · + qd (w) . w, z

Suppose for some z with z2 < 1 that p(z) = 0. By definition of rational sphere map, q(z) = 0. Then the left-hand side vanishes and the expression

12 Inverse Image of a Point

57

z, w d + z, w d−1 q1 (w) + · · · + qd (w)

(2.27)

therefore vanishes identically as a homogenous polynomial in w. We call its complex conjugate H q(w, z). Thus, the zero set of p(z) is contained in the set Sq of z for which H q(w, z) vanishes identically. It remains to show that this set is finite. We can expand H q(w, z) as H q(w, z) =



cα (z)w α .

|α|=d

If this polynomial in w vanishes identically, then each cα (z) vanishes. Set z k equal to 0 for k > 1. By (2.27), cα (z 1 , 0, ...0), is a polynomial in z 1 of degree d and hence has at most d roots. The symmetry of (2.27) in its variables gives the same conclusion for each index k with 1 ≤ k ≤ n. Thus, the set of common zeroes of the  (anti-holomorphic) polynomials cα lies in a finite set. If q = 1, note that H q(w, z) = w, z d . This polynomial vanishes identically only if z = 0. Hence, we obtain a second proof of Proposition 2.10. Corollary 2.7 If p is a polynomial sphere map, then p−1 (0) is either 0 or empty. Remark 2.15 We will return to this homogenization technique in Chap. 6.  Exercise 2.17 Suppose n = 1 and q(z) = (1 − a j z). What is H q(w, z)? Exercise 2.18 Suppose q(z) = 1 − z, a . What is H q(w, z)? For what z does it vanish identically? Exercise 2.19 Suppose qp is a rational sphere map and q is a product of irreducible factors. What can you say about H q(w, z)?

13 The General Rational Sphere Map We hope somehow to classify all rational sphere maps. Theorem 2.15 in this section seems to be about the best one can hope for. Let us first recall Corollary 2.6. Let q be a scalar-valued polynomial of degree d on Cn such that q does not vanish on the closed unit ball. Without loss of generality we assume that q(0) = 1. Then there is a non-negative integer m and a positive integer N at least n for which the following holds: there is a polynomial map p : Cn → C N of degree D = m + d such that qp is a rational sphere map. Here, qp is non-constant and reduced to the lowest terms. One can bound neither N nor m in terms of n and d alone. The size of the coefficients of q matter. We mention, however, that when q is a constant or is linear, we may always choose p such that qp is an automorphism and N = n. In all other cases, there is no possible bound on N .

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2 Examples and Properties of Rational Sphere Maps

Corollary 2.6 suggests what to do. First, fix the denominator q of degree d. Then find the smallest D = m + d for which there is a numerator p such that qp is a nonconstant rational sphere map. Then fix D and find all p of degree D such that qp maps the unit sphere in the source to the unit sphere in the target. The target dimension will depend on q. Theorem 2.15 determines a universal procedure for finding all such p, based upon Theorem 2.1. Given an integer D, this procedure assigns a Hermitian matrix to every holomorphic polynomial map of degree at most D. The procedure is independent of the target dimension of the map, and it gets applied both to a given denominator and to all prospective numerators. The resulting linear algebra problem has a huge number of parameters. To some extent, we can decrease the size of the search space by using the restricted tensor product operation. The idea generalizes the role the homogenous map z ⊗m plays in the polynomial case. In the polynomial case, however, there is a unique solution to a certain system of equations. When q is of higher degree, the resulting system will be underdetermined. Even when q is linear, and m = 2, there are more unknowns than equations. See Example 2.16. Let r (z, z) be a complex-valued polynomial of degree at most D in both the z and the z variables. Put  r (z, z) = cαβ z α z β . |α|≤D,|β|≤D

To express the information using matrices requires that we assign an ordering of the multi-indices. Consider all multi-indices α with |α| ≤ D. If |α| > |β|, then α > β. If |α| = |β|, then we order lexicographically. Examples below will make this ordering clear. We label the columns of a matrix by z α where the multi-indices are in decreasing order starting from the left, and we label the rows by z β where the multi-indices are in decreasing order starting from the top. Given r as above, place the complex number cαβ in the column corresponding to α and row corresponding to β. Write (cαβ ) = (cαβ )(r ) for the matrix of coefficients. Let H(n, D) denote the collection of Hermitian matrices indexed by the monomials of degree at most D in n-variables. Given a rational sphere map qp , Theorem 2.2 provides a linear system of equations for the inner products of the vector coefficients of p. The system equates two Hermitian matrices in H(n, D), namely L( p) and L(q). The map L depends on the degree being considered, but it is not sensitive to the target dimension of its input, and hence the same technique determines these matrices for both p and q.  Recall that we have ordered the multi-indices. Given that f (z) = α Cα z α , there is an obvious one-to-one correspondence between the collection of inner products Cα , Cβ with α ≥ β and the entries of the matrix (cαβ ) with α ≥ β. Assume α ≥ β. By formulas (2.5) and (2.6) the map T sending Cα , Cβ to the entry cαβ is linear. (It is conjugate linear when α < β.) By expanding the f j in formula (2.5) in terms of the Cα , we see that the matrix of T has non-negative integer entries. The entries of the matrix L( f ) are thus (very easy) integer linear combinations of the various Cα , Cβ . The next example should help clarify the definition of L and this procedure.

13 The General Rational Sphere Map

59

Example 2.13 Put f (z, w) = Az 2 + Bzw + Cw 2 + Dz + Ew + F, where the coefficients live in the same space of unspecified dimension. Then f 0 = F and f 1 = Dz + Ew and f 2 = Az 2 + Bzw + Cw 2 . The order of the multi-indices is: (2, 0) > (1, 1) > (0, 2) > (1, 0) > (0, 1) > (0, 0). The polynomials Lk ( f ) are given by: L0 ( f ) =  f 2 2 + (|z|2 + |w|2 ) f 1 2 + (|z|2 + |w|2 )2  f 0 2 L1 ( f ) =  f 1 , f 0 (|z|2 + |w|2 ) +  f 2 , f 1 L2 ( f ) =  f 2 , f 0 . Since there are 6 independent polynomials of degree at most 2 in 2 variables, the matrix of the Hermitian form will be six-by-six. The columns and rows will be labeled according to the multi-indices as follows: (2, 0) (1, 1) (0, 2) (1, 0) (0, 1) (0, 0)

(2, 0) (1, 1) (0, 2) (1, 0) (0, 1) (0, 0)

The entry indexed by the pair α, β will be the coefficient of z α1 w α2 z β1 wβ2 in any of the Lk ( f ). As noted above, each such monomial arises in exactly one of the Lk ( f ) or its conjugate. To illustrate the idea, consider the coefficient of wzw. This coefficient gets placed in the column labeled (0, 1) and row labeled (1, 1). The polynomial L1 ( f ) is Dz + Ew, F (|z|2 + |w|2 ) + Az 2 + Bzw + Cw 2 , Dz + Ew and hence the coefficient of wzw (occuring in the congugate) is F, D + E, B . We abbreviate the notation by writing A2 for A2 and AB for A, B and so on. Using this notation we obtain L( f ): ⎛ 2 A + D2 + F 2 BA CA FD + DA EA 2 + D 2 + E 2 + 2F 2 ⎜ AB B C B D B + B E E B + FD ⎜ ⎜ AC BC C2 + E2 + F2 DC F E + EC ⎜ ⎜ D F + AD BD + EF CD 0 0 ⎜ ⎝ AE BE + DF EF + CE 0 0 AF BF CF 0 0

⎞ FA F B⎟ ⎟ FC ⎟ ⎟ 0 ⎟ ⎟ 0 ⎠ 0

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2 Examples and Properties of Rational Sphere Maps

We can think of the various inner products as elements of a vector v with 21 components, ordered according to our ordering above. There is a 21-by-21 matrix T that assigns to v the 21 entries as follows: all six in the top row, the second through sixth in the second row, and so on. (everything to the right of and including the diagonal). The matrix L( f ) is an efficient way of expressing this map T . Here is the crucial idea for rational sphere maps. Given the denominator q of degree d, and an integer D = d + m, we find L(q) as above. We consider an arbitrary vector-valued polynomial p = α Cα z α . Computing L( p) gives a Hermitian form that depends only upon the inner products Cα , Cβ . By Theorem 2.1, L( p) = L(q). These unknown inner-products satisfy the linear equation given by L( p) = L(q). Let q : Cn → C be a holomorphic polynomial of degree d such that q(0) = 1. Consider the possibly larger  degree D = d + m. At the risk of redundancy we write down Lk (q). Let q = dj=0 q j denote its homogeneous expansion. Formula (2.5) determines Lk (q) and makes the bi-degrees evident. L0 (q) = z2m |qd |2 + · · · + z2m+2d−2 |q1 |2 + z2m+2d . .. . L D−2 (q) = z2m qd q 2 + z2m+2 qd−1 q 1 + z2m+4 qd−2 . L D−1 (q) = z2m qd q 1 + z2m+2 qd−1 . L D (q) = z2m qd . Once we have the Lk (q), we expand each polynomial and find the coefficient of z α z β . We put these coefficients into the Hermitian matrix L(q). Typically D > d, and hence many of the entries in the matrix vanish. We can now state and prove the main theorem of this section and one of the main results of the book. Theorem 2.15 Let q : Cn → C be a holomorphic polynomial of degree d such that q(0) = 1 and q(z) = 0 for z2 ≤ 1. The following hold: 1. There is an integer D = d + m for which q is the denominator of a rational sphere map of degree D.  G α z α . Then 2. Let qg be a rational sphere map of degree D = m + d. Put g(z) = L(g) = L(q), and thus the inner products G α , G β are solutions to a universal system of linear equations. 3. There is a rational sphere map qp of degree D such that p vanishes to order at least m − 1 at the origin, p is obtained from g by finitely many partial tensor product operations, and L( p) = L(g) = L(q). Proof By Corollary 2.6, there is a g such that qg is a rational sphere map. Hence (1) holds. Recall that the degree of the numerator is at least as large as the degree of

13 The General Rational Sphere Map

61

the denominator. Choose g of minimum degree D = m + d. For each such  g, either the identities (2.2) or Theorem 2.1 imply that L(g) = L(q). Write g = m+d j=ν g j as its homogeneous expansion. By the identities (2.2), if ν < d, then the term gν maps into a subspace A orthogonal to the subspace where gm+d maps. We may thus take a tensor product on the subspace A and obtain a new numerator E 1 g of the same degree which vanishes to at least one higher order. Proceeding as in the proof of Theorem 2.5, we may successively tensor until we obtain a numerator p = E k ◦ · · · ◦ E 1 (g) whose homogeneous expansion is of the form pm + · · · + pm+d . Also qp is a rational sphere map. Once we know that the system of equations L(g) = L(q) has a solution, Theorem 2.1 implies that the numerator p of each rational sphere map qp of degree D  satisfies the same linear system L( p) = L(q). Put p = Pα z α . The inner products of the coefficient vectors Pα thus satisfy the same linear system and all the items hold.   If p = Pα z α is as in the third item of the theorem, then the inner products Pα , Pβ satisfy the same universal system of linear equations as the G α , G β do. The classification of rational sphere maps becomes the following combinatorial question. Each scalar polynomial q that does not vanish on the closed unit ball determines the right-hand side of a linear system. The unknowns are the inner products Pα , Pβ of vectors in a Hilbert space of some unspecified finite dimension. We solve this system to obtain all the possible numerators given the denominator q. We then determine the smallest N for which there exist vectors satisfying this system of equations for their inner products, and for which the fraction qp is reduced to lowest terms. Without this condition, we could take N = 1 and put p = q. We offer an intuitive summary. Fix a denominator q of degree d with q(0) = 1 and q(z) = 0 on the closed unit ball. We wish to classify all g such that qg maps the unit sphere in the source to the unit sphere in the target. We do not attempt to determine the target dimension. By Corollary 2.6, such a g exists. Choose such a g of lowest degree m + d. We perform finitely many tensor product operations on g until we reach a numerator  to order mα − 1; its homogeneous  p which vanishes p . Put p = expansion has the form p = m+d j |α|≥m Pα z . All the inner products j=m are determined by linear algebra. Tensoring has the following impact. Not only are many of the inner products zero, but many of the vectors themselves are 0. We get a smaller, but still huge, linear algebra problem. Its impact on the target-rank or target dimension is more subtle. By Example 2.9, tensoring can increase the target-rank, decrease it, or keep it the same. In Chap. 7 we show that tensoring always increases the volume of the image of the rational sphere map. Corollary 2.8 There is an explicit map L : V(n, D) → H(n, D) with the following property. If q : Cn → C is a polynomial of degree d that does not vanish on the closed unit ball, then there is an integer D ≥ d for which the following holds. If qg is a rational sphere map of degree D, then L(g) = L(q). Corollary 2.8 suggests an interesting perspective on the basic question of describing rational sphere maps. One should study the problem for each fixed denominator q separately, by regarding L(q) as the right-hand side of the linear system

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2 Examples and Properties of Rational Sphere Maps

L( p) = L(q). The unknowns are the inner products of the coefficients of p. The target dimension of the map will be the smallest dimension into which we can fit vectors whose inner products satisfy these relationships. In this point of view, all polynomial sphere maps are treated in the same fashion. In the polynomial case, the matrix equation determining z ⊗m is easily solved. See Example 2.14. For general denominators, except in trivial cases, the matrix equation is not of full rank, but it has solutions if the degree is sufficiently large. The existence of these solutions is established by the operator theory required to prove Theorem 2.8. Remark 2.16 One can regard Theorem 2.15 as a vector-valued Schwarz reflection principle. Given q, we find all g of a fixed degree such that qg maps the sphere to a sphere. When n = 1, each factor 1 − a j z of q leads to a factor of z − a j in g. We can then multiply by z m to find all examples up to a multiple of eiθ . In higher dimensions we cannot in general factor q. In fact, by Exercise 1.24, the generic polynomial is irreducible. We must choose m large enough to guarantee that some g exists. Once we do so, we perform finitely many tensor products to reach a point where we can describe all possible numerators p. We will use the formula from the next Proposition in Chap. 7. Proposition 2.11 Let qg be a rational sphere map where the degree of g is m + d and the degree of q is d. Then there are polynomial maps p and w such that • The homogenous expansion of p is given by p = pm + · · · + pm+d . •  p2 = g2 + w2 (z2 − 1) As a consequence,  qp 2 =  qg 2 +  wq 2 (z2 − 1). We now provide various examples illustrating Theorem 2.15. In most of them, we assume the source dimension is 2. The ideas are the same in all dimensions, but the number of parameters becomes too large to be useful in examples. Example 2.14 When q is a constant, and p is homogeneous of degree m, the system of linear equations becomes very simple: Aα , Aβ = 0 if α = β   m Aα , Aα = α and we recover the map z ⊗m . Example 2.15 We begin with the easiest case, and it is already quite elaborate. Put n = 2 and put q(z, w) = 1 + q1 (z, w) = 1 + sz + tw. We wish to classify all rational sphere maps of the form qp where p = p1 + p2 . Thus, m = d = 1. We write p2 (z, w) = Az 2 + Bzw + Cw 2 and p1 (z, w) = Dz + Ew. Here A, B, C, D, E are vectors in some complex Euclidean space of unspecified dimension. We have 15 inner

13 The General Rational Sphere Map

63

products, including the squared norms, of these 5 vectors. There are 12 equations. The first 6 equations arise by equating coefficients of the polynomials Az 2 + Bzw + Cw 2 2 + Dz + Ew2 (|z|2 + |w|2 ) = (|z|2 + |w|2 )2 + |sz + tw|2 (|z|2 + |w|2 ). The second set of 6 equations arises from equating coefficients in Az 2 + Bzw + Cw 2 , Dz + Ew = (|z|2 + |w|2 )(sz + tw). A good way to express the equations is to consider the matrices of coefficients. To save space we write AB for A, B and |A|2 for A2 . On the left-hand side of L( p) = L(q) we have ⎛ 2 |A| + |D|2 AB + D E AC ⎜ B A + E D |B|2 + |D|2 + |E|2 D E + BC ⎜ ⎜ CA CB + ED |C|2 + |E|2 ⎜ ⎝ DA DB DC EA DC EC

AD BD CD 0 0

⎞ AE B E⎟ ⎟ C E⎟ ⎟. 0 ⎠ 0

(2.28)

On the right-hand side we have ⎛

1 + |s|2 st 0 2 2 ⎜ st 2 + |s| + |t| st ⎜ ⎜ 0 1 + |t|2 ts ⎜ ⎝ s t 0 s t 0

s t 0 0 0

⎞ 0 s⎟ ⎟ t⎟ ⎟. 0⎠ 0

(2.29)

We equate these matrices, obtaining 12 equations for the unknown 15 inner products. We pause to note why there are only 12 equations. All 5 equations arising from the first row count. Only the last 4 equations arising from the second row count, as the first entry is determined by complex conjugation. Only the last 3 equations arising from the third row count for the same reason. The last two rows do not count for the same reason. An alternative way to see why the number is 12 is to note that the full matrix is Hermitian and hence would typically have 15 independent entries. But 3 of these entries (bottom right corner) yield the equations 0 = 0 and hence don’t provide any information. Our work is not done, because it remains to solve these twelve equations in some reasonable way. We sketch how to solve them. Since there are five unknown vectors, it suffices to assume that the target is fivedimensional. By unitary coordinate changes in the target, since A and C are orthogonal, we may suppose that the vector A has the form (a, 0, 0, 0, 0) and the vector C has the form (0, 0, c, 0, 0). We regard a, c as parameters. Except in trivial cases, the constants a, c are not 0. Furthermore, using A, E = C, D = 0, we may assume that

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2 Examples and Properties of Rational Sphere Maps

D = (d1 , d2 , 0, d4 , 0) and that E = (0, e2 , e3 , 0, e5 ). Using the equations A, D = s and C, E = t, we may eliminate d1 and e3 . The above equations imply that B2 = |a|2 + |c|2 . We have |A|2 + |D|2 = 1 + |s|2 and |C|2 + |E|2 = 1 + |t|2 . Adding these two equations yields |A|2 + |D|2 + |C|2 + |E|2 = 2 + |s|2 + |t|2 = |B|2 + |D|2 + |E|2 . Hence |B|2 = |A|2 + |C|2 = |a|2 + |c|2 . Using A|2 + |D|2 = 1 + |s|2 and |C|2 + |E|2 = 1 + |t|2 , the following two crucial equations hold: |a|2 +

|s|2 = 1 + |s|2 − |d4 |2 − |d2 |2 |a|2

(∗d)

|c|2 +

|t|2 = 1 + |t|2 − |e5 |2 − |e2 |2 |c|2

(∗e)

These equations determine |d4 | and |e5 |, but they do even more. They constrain the parameters |a| and |c|, as we note below. Furthermore there are five equations involving B. The equations ab1 + d2 e2 = st = b3 c + d2 e2 together with B2 = |a|2 + |c|2 simplify things further. At this stage the numerator of our map has the following form (**)   s t az 2 + b1 zw + z, b2 zw + d2 z + e2 w, cw2 + b3 zw + w, b4 zw + d4 z, b5 zw + e5 w . a c

We will regard a, c, d2 , e2 as parameters. In this formula, the b j are determined for j = 2, and |b2 | is determined. The equations (*d) and (*e) determine |d4 | and |e5 |. If we wish, we may apply a unitary matrix in the target and make them real. Let us see how a, c are constrained. For the moment, put |a|2 = x. We are given an 2 equation of the form x + |s|x + |δ|2 = 1 + |s|2 . Here |δ|2 = |d4 |2 + |d2 |2 . We know that 0 ≤ |s| < 1. As s tends to 0, δ tends to 0 and thus 0 < |a|2 < 1. For 0 < x, we have x + x1 ≥ 2. As |s| tends to 1, we therefore see that δ tends to 0 and |a| tends to 1. For each |s| ∈ (0, 1) we obtain an interval in which |a| must lie. By completing the square, one can write down this information exactly. A similar analysis holds for |c| and t. We summarize as follows. Assume n = 2. There are families of rational sphere 1 + p2 with target dimension 5. Put q1 = sz + tw. The numerator p maps of the form p1+q 1 has the form (**), where most (but not all) of the constants are determined. The inner products of the vector coefficients satisfy the equations obtained by setting (2.28) equal to (2.29). When one solves the equations various inequalities arise.

13 The General Rational Sphere Map

65

By setting w = 0 we obtain an interesting family of rational sphere maps with source dimension 1. Corollary 2.9 For |s| < 1 and a = 0, put f (z) =

(az 2 + as z, bz) . (1 + sz)

|s| 2 2 If a and b satisfy the relationship |a|2 + |a| 2 + |b| = 1 + |s| , then f defines a proper map from the unit disk to the unit ball B2 . 2

The relationship in Corollary 2.9 can be rewritten as |b|2 = (1 − |a|2 )(1 −

|s|2 ). |a|2

Exercise 2.20 Consider the map f in Corollary 2.9. Suppose b = 0. Show that f is a rational sphere map. What happens if b = 0? (Recall that a rational sphere map must be reduced to lowest terms.) Exercise 2.21 Consider the map f in Corollary 2.9. For fixed b and s, solve the relationship there for |a|. In Example 2.15, there are infinitely many solutions if we allow the target dimension to be at least five. (Doing so enables us have enough room for the vectors whose inner products we have determined. Increasing the target dimension doesn’t change things significantly.) It is convenient to express the equations using block matrices. When q is linear, and m is arbitrary, each side of the equation equating (2.28) and (2.29) can be expressed in the form 

 L m+1 L m . L ∗m 0

The corresponding linear system is rather simple. If we regard the 15 unknowns as a column vector v, then T (v) = L( f ). The matrix of T consists only of zeroes and ones. In other words, the map that takes the fifteen inner products to the Hermitian matrix in (2.28) consists of only zeroes and ones. Remark 2.17 Let us return to the notation of Theorem 2.15. When q is linear, we may always choose m = 1. Remark 2.18 When q is linear, and p(0) = 0, our sphere map must be of the form p2 + p1 . A theorem of Lebl from [54] states that a map of degree 2 is spherically 1+q1 equivalent to a monomial map. Hence the special case where q is linear could be handled by finding all monomial maps of degree 2 and using spherical equivalence. We emphasize that we spent so much time on the case where q is linear because it illustrates the general case, but with smaller matrices.

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2 Examples and Properties of Rational Sphere Maps

Remark 2.19 We make an observation about the matrices L( p). In Example 2.13, all the coefficients of the matrix were 0, 1 or 2 and in formula (2.28) they were all 0 or 1. In general, all the non-zero coefficients will be binomial or multinomial coefficients. This conclusion follows from formula (2.5). The coefficients arise from the expansion of z2m for various exponents m. It follows that the degree D determines a bound on the entries in these matrices. Example 2.16 To further illustrate Theorem 2.15, consider a given denominator 1 + q1 + q2 when n = 2. To find all such p for which qp maps the unit sphere to some unit sphere, it suffices to assume that p = pm + pm+1 + pm+2 , and then to apply the undoing (tensor division) operation. The polynomials p j depend on j + 1 arbitrary vectors, because the number of homogeneous polynomials of degree j in two variables is j + 1. There are therefore W = (m + 1) + (m + 2) + (m + 3) = 3m + 6  unknown vectors. There are therefore W2 = W (W2 −1) unknown variables. Again we will equate the entries in two large matrices arising from setting L( p) = L(q). The system of equations arising from L( p) = L(q) can be regarded as follows. We are equating two block matrices of the form ⎛

⎞ L m+2 L m+1 L m ⎝ L ∗m+1 0 0 ⎠. 0 0 L ∗m

(2.30)

When q is a constant, and thus p = pm is a polynomial, L(q) is a diagonal square matrix whose entries are the squared absolute values of the multinomial coefficients. We determine p as in Example 2.14. In the last section of this Chapter we consider what happens in the polynomial case if we do not tensor into the homogeneous case. When q is linear, both matrices are of the form 

 L m+1 L m . L ∗m 0

For a general d we get a large block matrix whose top row consists of blocks from L m+d to L m . There will always be a large block matrix of zeroes in the bottom right-hand corner. Exercise 2.22 Put n = 2. Write down the linear system that determines all rational 2 + p3 sphere maps of the form p1+q . If you are ambitious, do the same for maps of the 1 p1 + p2 + p3 form 1+q1 +q2 . pm+d is a rational sphere Exercise 2.23 Consider q of degree d and m such that pd +···+ q map. Find the number of independent parameters in the formula for p.

13 The General Rational Sphere Map

67

Exercise 2.24 Consider the system of (linear) equations A2 = 1, B2 = 1, A, B = α for the inner products of two vectors. For each α, determine the minimum dimension into which we can fit A, B. The answer depends on α. Formulate and solve a similar question for an arbitrary collection of vectors v j . Hint: Consider the matrix v j , vk ) of inner products.

14 A Detailed Rational Example In this section we consider rational sphere maps in source dimension 2 whose denominator is 1 − λzw. This denominator is possible only if |λ|2 < 4, because the maximum of |zw| given that |z|2 + |w|2 = 1 occurs when |z|2 = |w|2 = 21 . As |λ|2 tends to 4, the minimum degree and target dimension of a possible numerator tend to infinity. See the discussions in Sects. 7 and 4 of Chap. 4. In order to keep the degree of the numerator and the target dimension under control, we assume 0 < |λ|2 < 3. In this case we can find examples where the numerator is of degree 3. Notice also that q1 = 0 here. Theorem 2.15 leads us to study all polynomials p = p1 + p2 + p3 such that three equations hold:  p3 2 + (|z|2 + |w|2 ) p2 2 + (|z|2 + |w|2 )2  p1 2 = (|z|2 + |w|2 )3 + |λ|2 |zw|2 (|z|2 + |w|2 )

 p3 , p2 + (|z|2 + |w|2 ) p2 , p1 = 0  p3 , p1 = (|z|2 + |w|2 )λzw. There are solutions with p2 = 0. We make this assumption to simplify things, but we still have a complicated problem. Thus, we set p3 (z, w) = Az 3 + Bz 2 w + C zw 2 + Dw 3 p1 (z, w) = E z + Fw for unknown vectors A, B, ..., F. The system of equations from Theorem 2.15 yields the following information on the inner products of these vectors. As before, we simplify the notation by writing A2 for A2 and AB for A, B and so on. All inner products vanish except for the following relations: A2 + E 2 = 1 D2 + F 2 = 1

68

2 Examples and Properties of Rational Sphere Maps

BE = λ CF = λ B 2 = 3 + |λ|2 C 2 = 3 + |λ|2 . All such polynomial sphere maps have target dimension at most 6. We write the general map as follows, where the small letters are complex numbers and e does not mean the base of the natural logarithm. In fact, e ∈ C and 0 < |e|2 ≤ 1.   λ λ p(z, w) = az 3 , bz 2 w, czw 2 , dw 3 , ez + z 2 w, f w + zw 2 e f

(2.31)

The equation  p2 = |q|2 on the sphere becomes four equations: |a|2 + |e|2 = 1 |d|2 + | f |2 = 1 λ |b|2 + 2|e|2 + | f |2 + | |2 = 3 + |λ|2 e λ |c|2 + |e|2 + 2| f |2 + | |2 = 3 + |λ|2 . f Here none of b, c, e, f can be 0. We may choose e and f otherwise arbitrarily in the closed unit disk, except that not both can be on the unit circle. If both are on the unit circle, then a = b = c = d = 0 and our map will not be reduced to lowest terms. Once we choose e, f , the numbers a, d are determined up to modulus by the first two equations. The numbers b, c are determined up to modulus by the last two equations. The maps for different choices of these moduli will be unitarily equivalent. In order to write down the map in a palatable manner, we put t (x) =

√ 1−x

s(x, y) = 3 + |λ|2 −

|λ|2 − 2x − y. x

The above four equations become |a| = t (|e|2 ) |d| = t (| f |2 )

14 A Detailed Rational Example

69

|b| = s(|e|2 , | f |2 ) |c| = s(| f |2 , |e|2 ). Finally the numerator p(z, w) becomes   λ λ t (|e|2 )z 3 , s(|e|2 , | f |2 )z 2 w, s(| f |2 , |e|2 )zw2 , t (| f |2 )w3 , ez + z 2 w, f w + zw2 . e f

(2.32) We can of course compose on the left with an arbitrary element of U(6). We next make peace with the condition that |λ|2 < 3. Either |e| < 1 or | f | < 1. Without loss of generality, we may assume that | f |2 ≤ |e|2 and | f |2 < 1. Hence 1−|e|2 ≤ 1. We solve the equation 1−| f |2 λ |c|2 + |e|2 + 2| f |2 + | |2 = 3 + |λ|2 f for |λ|2 , obtaining |λ|2 = | f |2
0 cr s x r y s . The polynomials cr s x r y s (x + y)d−r −s are of different degrees in x.

Then f (x, y) is congruent to x d + y d mod (d) if and only if d is prime. Example 3.5 Put f (x, y) = x + x y + x y 2 + y 3 . This polynomial is 1 on the line x + y = 1, and thus the first two items hold. The fourth item holds, as the polynomials x(x + y)2 , x y(x + y), x y 2 , y 3 are of different degrees in x. The third item fails. We have (x + y)3 = x(x + y)2 + x y(x + y) + x y 2 + y 3 The conclusion fails, as f (x, y) is not congruent to x 3 + y 3 mod (3). We needed both x 3 and y 3 to appear. Remark 3.5 For q ≥ 3, the polynomials  will typically have negative coefficients and hence will not be squared norms of monomial sphere maps. We will return to this topic later on. We list the polynomials f p,3 (x, y) for 1 ≤ p ≤ 11 and comment. x+y x 2 + 2x y + y 2 x 3 + 3y − 3y 2 + y 3 x 4 + 4x y − 2x 2 y 2 + y 4 x 5 + 5x 2 y + 5x y 3 + y 5 x 6 + 6x 3 y + 2x 3 y 3 + 3y 2 − 3y 4 + y 6 x 7 + 7x 4 y + 7x y 2 − 7x 2 y 4 + y 7 x 8 + 8x 5 y + 12x 2 y 2 − 2x 4 y 4 + 8x y 5 + y 8 x 9 + 9x 6 y + 18x 3 y 2 + 3y 3 + 9x 3 y 5 − 3y 6 + y 9 x 10 + 10x 7 y + 25x 4 y 2 + 10x y 3 + 2x 5 y 5 − 15x 2 y 6 + y 10 x 11 + 11x 8 y + 33x 5 y 2 + 22x 2 y 3 − 11x 4 y 6 + 11x y 7 + y 11

90

3 Monomial Sphere Maps

The number of terms in f p,3 as a function of p satisfies no obvious pattern; it is 2, 3, 4, 4, 4, 6, 5, 6, 7, 7, 7, . . . . For f p,2 the pattern is simply 2, 3, 3, 4, 4, 5, 5, . . . . A quick glance verifies the conclusion of Theorem 3.6 for the groups (d, 3) for 1 ≤ d ≤ 11. Another quick glance reveals for d = 5 that f 5,2 (y, x) = f 5,3 (x, y). The reason is simple. Let η be a primitive 5-th root of unity. Let G denote the cyclic group generated by A, where

η 0 . A= 0 η3 Then A2 is also a generator for G, but putting ω = η 2 gives 2 ω 0 . A = 0 ω 2

After interchanging the source variables, we obtain the group (5, 2). It follows that f 5,2 (y, x) = f 5,3 (x, y). Because of this phenomenon, we make a convention, which we state in the following definition. Definition 3.2 The notation ( p, q) denotes the cyclic group generated by a diagonal unitary matrix with eigenvalues α and αq where α is a primitive p-th root of unity and q is chosen minimally. We elaborate. Consider the cyclic group generated by the matrix

η 0 , A= 0 ηq where η is a primitive p-th root of unity. When some power of A has the form m ω 0 , 0 ω where ω = 1 and m < q, we regard this group as ( p, m). For the reader’s amusement, we list the polynomials f p,4 for 4 ≤ p ≤ 15 when q = 4. We also comment as before.

4 Cyclic Groups and Monomial Sphere Maps

91

x 4 + 4y − 6y 2 + 4y 3 − y 4 x 5 + 5x y − 5x 2 y 2 + y 5 x 6 + 6x 2 y − 3x 4 y 2 + 2y 3 + 3x 2 y 4 − y 6 x 7 + 7x 3 y + 14x 2 y 3 + 7x y 5 + y 7 x 8 + 8x 4 y + 4y 2 + 8x 4 y 3 − 6y 4 + 4y 6 − y 8 x 9 + 9x 5 y + 9x y 2 + 3x 6 y 3 − 18x 2 y 4 + 3x 3 y 6 + y 9 x 10 + 10x 6 y + 15x 2 y 2 − 25x 4 y 4 + 2y 5 + 10x 2 y 7 − y 10 x 11 + 11x 7 y + 22x 3 y 2 − 11x 6 y 4 + 33x 2 y 5 + 11x y 8 + y 11 x 12 + 12x 8 y + 30x 4 y 2 + 4y 3 − 3x 8 y 4 + 48x 4 y 5 − 6y 6 + 3x 4 y 8 + 4y 9 − y 12 x 13 + 13x 9 y + 39x 5 y 2 + 13x y 3 + 39x 6 y 5 − 39x 2 y 6 + 13x 3 y 9 + y 13 x 14 + 14x 10 y + 49x 6 y 2 + 28x 2 y 3 + 14x 8 y 5 − 98x 4 y 6 + 2y 7 + 21x 2 y 10 − y 14 x 15 + 15x 11 y + 60x 7 y 2 + 50x 3 y 3 + 3x 10 y 5 − 95x 6 y 6 + 60x 2 y 7 + 3x 5 y 10 + 15x y 11 + y 15

Notice that f 7,4 (x, y) = f 7,2 (y, x). The most important thing to notice however is that otherwise, some coefficients are negative. The underlying monomial maps send the unit sphere in C2 to a hyperquadric rather than to a sphere. Many results about this situation appear in the work [41] of Grundmeier. For example, for the groups ( p, q) with q ≥ 3, the fraction of negative coefficients tends to 41 as p tends to infinity. Perhaps the main point is that the problem formulated after Theorem 3.9 below is difficult and the answer is unknown in general.

5 Circulant Matrices One can define the polynomials f p,q using circulant determinants. This approach was used in [60] to give combinatorial interpretations of the coefficients. Let a0 , ..., a p−1 denote complex numbers. Recall that a circulant matrix is a p × p matrix of the form

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3 Monomial Sphere Maps

⎛

a0 ⎜a p−1 A=⎜ ⎝ ... a1

a1 a0 ... a2

a2 a1 ... a3

⎞ . . . a p−1 . . . a p−2 ⎟ ⎟. ... ... ⎠ . . . a0

Note that each row is a permutation of the first row. The connection with cyclic groups of order p should be evident. We want to find the determinant of A. To do so, we first find the eigenvectors (which are independent of the numbers a j ) and then find the corresponding eigenvalues. Let 1 denote the vector in p-dimensional space of all ones. Let ω be a primitive p-th root of unity. Let ω 1 denote the vector (1, ω, ω 2 , ..., ω p−1 ). Let ω k denote the vector (1, ω k , ω 2k , ..., ω k( p−1) ). We are writing these vectors as p-tuples; the reader should perhaps regard them as column vectors. Proposition 3.3 The eigenvectors of the circulant matrix A are the vectors 1 = (1, 1, ..., 1) ω 1 = (1, ω, ω 2 , ..., ω p−1 ) ω 2 = (1, ω 2 , ω 4 , ..., ω 2( p−1) ) ... ω p−1 = (1, ω p−1 , ..., ω ( p−1) ). 2

The corresponding eigenvalues are α(ω j ), where α(t) = A(1) = α(1)1. Furthermore, we have

 p−1 j=0

a j t j . Note that

A(ω 1 ) = α(ω)A(ω 1 ) A(ω 2 ) = α(ω 2 )A(ω 2 ) ... A(ω p−1 ) = α(ω p−1 )A(ω p−1 ). Corollary 3.5 The determinant of the circulant matrix A equals

 p−1 j=0

α(ω j ).

Proof The corollary follows from the proposition because the determinant is the product of the eigenvalues. We prove the proposition. First note that 1 is an eigenvector, because ⎛ ⎞ p−1 p−2 p−1    A(1) = ⎝ a j , a p−1 + a j , ..., a j + a0 ⎠ = α(1) 1. j=0

j=0

j=1

5 Circulant Matrices

93

Each component is simply α(1). In other words, we are simply performing a permutation of the indices of summation. Similarly, we get

 A(ω 1 ) = α(ω), ωα(ω), ..., ω p−1 α(ω) = α(ω) ω 1 and the results for A(ω k ) follow as well. Rewriting the definition of f p,q using Corollary 3.4 gives the following result. Theorem 3.8 Let η be a primitive p-th root of unity and suppose 1 ≤ q < d. Let f p,q (x, y) denote the unique polynomial defined in Theorem 3.6. Then 1 − f p,q (x, y) =

p−1 

(1 − ω j x − ω q j y) = det(A),

j=0

where A is the circulant matrix whose first row is (1, −x, . . . , −y, 0, 0). This theorem provides a simple proof from [60] that the coefficients of f p,q are integers. See [17] for the author’s original more complicated proof. Proposition 3.4 The coefficients of f p,q are integers. Proof Since all the entries in A are integer multiples of x or y, the determinant of A is a polynomial in x, y with integer coefficients. Using circulants and basic properties of determinants, Loehr et al. [60] gave the following beautiful combinatorial interpretation of these integers: Theorem 3.9 The coefficient of x r y s is equal, aside from its sign, to the number of permutations σ of p letters such that for j = 1, . . . , p the differences σ( j) − j mod ( p) take the values 0, 1, q with respective multiplicities p − r − s, r, s. Furthermore, these permutations all have the same sign and the same cycle type. We mention also that [42] extended this result to polynomials of the form p−1 

(1 − xω q1 j − yω q2 j ).

j=0

They also provided detailed information on the signs of the coefficients. Given our emphasis on sphere maps, where all the signs are positive, we do not develop this machinery. We do, however, pose the following open generalization. This problem was solved in [41] for all finite subgroups of SU(2). For the binary icosahedral group , for example, the map  has 40 positive signs and 22 negative signs. Open Problem. Consider the polynomials defined by (3.16). Determine the number of positive and negative signs in the defining hyperquadric Q(A, B) arising in (3.16).

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3 Monomial Sphere Maps

6 The Pell Equation A simple question about the polynomials pd leads to some intriguing number theory. =r +2 By Theorem 3.2, when d = 2r + 1 is odd, the polynomial pd (x, y) has d+3 2 terms, and this number is minimal given the degree. Of course, the polynomial pd (y, x) also has r + 2 terms. One naturally asks whether there are additional polynomials p of the same degree, with non-negative coefficients and r + 2 terms, and satisfying p(x, y) = 1 on x + y = 1. The surprising answer is that it depends in a complicated way on the degree. In this section, we provide a number-theoretic method which discovers infinitely many odd degrees for which such a p exists. There exist odd degrees in which there are no other examples, and it is an open problem whether the number of such degrees is finite or infinite. We state this problem in Sect. 8. Let P(2, d) denote the collection of polynomials p(x, y) with non-negative coefficients and of degree d such that p(x, y) = 1 when x + y = 1. We say that p is a . The sharp polynomial if the number of terms in p is precisely the ceiling of d+3 2 problem is most interesting when d is odd. Can one find all the sharp polynomials of a given degree? For odd degrees d, the polynomials pd studied in detail in Sect. 2 are sharp, by Theorem 3.2. In some degrees, we can construct other sharp polynomials using some elementary number theory. We introduce the idea via an example. The method leads to a quadratic Diophantine equation called Pell’s equation. First consider the polynomial p2 (x, y) = x 2 + 2y − y 2 from Sect. 2. Since p2 (x, y) = 1 when x + y = 1, we have the identity x 2 + 2y = 1 + y 2

(3.21)

on this set. Consider a polynomial p of degree at least 3 in P(2, d). Suppose there are polynomials g, h with non-negative coefficients such that p(x, y) = (x 2 + 2y)g(x, y) + h(x, y).

(3.22)

Then, by (3.21), the polynomial p ∗ defined by p ∗ (x, y) = (1 + y 2 )g(x, y) + h(x, y) is thus also in P(2, d). Furthermore p ∗ has at most the same number of terms as p. Hence, if p had the minimal number of terms for its degree, p ∗ would have the same number of terms, and provide a second example. Said in another way, if p is sharp, then p ∗ is sharp. Example 3.6 We write p7 in the form (3.22): p7 (x, y) = x 7 + 7x 5 y + 14x 3 y 2 + 7x y 3 + y 7 = x 7 + y 7 + 7x y 3 + 7x 3 y(x 2 + 2y).

6 The Pell Equation

95

Using (3.21), we obtain p ∗ (x, y) = x 7 + y 7 + 7x y 3 + 7x 3 y(1 + y 2 ) = x 7 + y 7 + 7x y 3 + 7x 3 y + 7x 3 y 3 . We naturally seek to generalize this trick. We begin with the polynomials pd , where d = 2r + 1 is odd. By Theorem 3.3 and formula (3.8), there is an explicit formula for the coefficients ck . To mimic the idea in Example 3.6, we first ask when there are two consecutive coefficients in ratio 1 : 2. In other words, we need



2r + 1 2r − k − 1 2r + 1 2r − k = . (3.23) 2 k k−1 k+1 k After some simplification, (3.23) gives 2(k + 1)(2r − k) = (2r + 1 − 2k)(2r − 2k),

(3.24)

which further simplifies to 6r k + r = 2r 2 + 3k 2 . We write this last equation in the form 3(r − k)2 =

1 ((2r + 1)2 − 1). 4

Put A = (2r + 1) and B = (r − k). We obtain the Diophantine equation 1 = A2 − 12B 2 .

(3.25)

Equations such as (3.25) are called Pell equations and have a fascinating history. See [52] for a nice account. As a simple check, we note that the case r = 3 and k = 1 corresponds to Example 3.6. Thus, we have a solution pair (A, B) = (7, 2). Since (3.25) has infinitely many integer solutions, we obtain infinitely many degrees in which there are additional polynomials (other than the p2r +1 ) in P(2, d) that minimize the number of terms. In the next section, we discuss the work in [24] which significantly generalizes this method, and we mention work in [55] obtained by coding. For the moment, we stick with the special case of (3.25). How does one find all integer solutions to (3.25)? There are several standard techniques, one of which we discuss next. The equation 1 = A2 − 12B 2 can of course be written √ √ 1 = (A − 12B)(A + 12B). Raising both sides to positive integer powers yields the additional solutions. √ Thus, √ for positive integers k, we need to write (A − 12B)k in the form Ak − 12Bk .

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3 Monomial Sphere Maps

Given a solution pair (A, B), squaring yields a new solution pair (A2 , B2 ) = (A2 + 12B 2 , 2 AB). We had the solution (A, B) = (7, 2). We obtain the new solution (97, 28). Raising to the power three gives the pair (1351, 195). This technique establishes the following result: Theorem 3.10 For a positive integer k, put √ √ (7 + 2 12)k + (7 − 2 12)k d= . 2 Then there are polynomials other than pd (x, y) or pd (y, x) in P(2, d) with terms.

d+3 2

The first few values of d are given by d = 7, 97, 1351, 18817, 262087. Most readers of this book are interested in complex analysis. There are many formal similarities with the number theory associated with Pell’s equation. We pause to briefly discuss these matters. √ One √ standard way to√define C is to adjoin −1 to the field R. Thus, as a field, C = R[ −1]. Put i = −1 as usual. For z = x + i y, we write z for x − i y and we put |z|2 = zz. Note that |zw| = |z| |w|. Consider Pell’s equation A2 − δ B 2 = 1, where δ is a positive integer√but not a √ δ to the ring of integers and work in the ring Z[ δ]. If we square. We can adjoin √ write ζ = A + B δ, then we have √ √ 1 = A2 − δ B 2 = (A + B δ)(A − B δ) = ζζ. √ √ When ζ = A + B δ solves the Pell equation, we put ζ n = An + Bn δ. Then ζ n also does, because A2n − Bn2 δ = 1. We can regard these ideas in terms of matrices. Define L by

A δB L= . (3.26) B A Then the pair (A, B), regarded as a column vector, is the result of applying L to (1, 0), regarded as a column vector. Thus, (An , Bn ) = L n (1, 0). If we think of An as an analogue to the real part of ζ n , then we have An =

1 n n (ζ + ζ ). 2

These remarks essentially establish Theorem 3.10, because the expressions for d are essentially

ζ k +ζ 2

k

, the analogue of the real part of ζ k .

6 The Pell Equation

97

The reader might pause and consider the relationship between√ this discussion √ k k 12) is and the Chebyshev polynomials. For example, the expression (7+2 12) +(7−2 2 simply Tk (7). Exercise 3.4 Show that

√ √ (7+2 12)k +(7−2 12)k 2

is congruent to 1 mod 6.

Exercise 3.5 Define L as in (3.26). Find the eigenvalues of L and the corresponding eigenvectors. Use this information to diagonalize L. Then find a simple formula for the n-th power L n . Exercise 3.6 Show that the number of sharp polynomials in P(2, d) is finite. Exercise 3.7 Find a relationship between pd (1, 1) and the Golden Ratio. In the next section, we turn to rather intricate generalizations of the idea.

7 Elaboration of the Method for Producing Sharp Polynomials Consider the polynomials p2r . By Theorem 3.1, there is a polynomial q with nonnegative coefficients for which q(x, y) = 1 + y 2r on the line x + y = 1. Given any polynomial p ∈ P(2, d), where d > 2r , we attempt, as in (3.22), to write pd (x, y) = q(x, y)g(x, y) + h(x, y) where g and h have non-negative coefficients. If we could do, so we would obtain another example in P(2, d) by replacing q(x, y) by 1 + y 2r . It would seem that we are significantly decreasing the number of terms. To make the idea work, we must allow q and h to include some of the same monomials. We must also carefully analyze the coefficients in p2r . Again we start with an example. We have p4 (x, y) = x 4 + 4x 2 y + 2y 2 − y 4 and therefore, on x + y = 1, we have x 4 + 4x 2 y + 2y 2 = 1 + y 4 . It turns out that one cannot find three consecutive terms in pd , for d odd, whose coefficients are a multiple of (1, 4, 2). BUT, for example, if we found three consecutive

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3 Monomial Sphere Maps

terms whose coefficients are a multiple of (1, 4, 2 + λ), where λ > 0, then we would be in business as in the next example. Example 3.7 Recall that p11 (x, y) = x 11 + 11x 9 y + 44x 7 y 2 + 77x 5 y 3 + 55x 3 y 4 + 11x y 5 + y 11 . Notice that 11x 9 y + 44x 7 y 2 + 77x 5 y 3 = 11x 5 y(x 4 + 4x 2 y + 2y 2 ) + 55x 5 y 3 .

(3.27)

We can therefore replace the three terms in p11 from the left-hand side of (3.27) by the right-hand side of (3.27). We obtain an additional polynomial in P(2, 11) with seven terms, the minimum possible for degree 11. Thus we have found another sharp polynomial in degree 11. The technique in Example 3.7 enables one to prove the following result. Verifying that the coefficients are non-negative is a bit complicated and hence we omit that detail. Example 3.7 is the special case of the theorem when m = 3. Theorem 3.11 Suppose that d is congruent to 3 mod 4 and is at least 7. Then there are at least three sharp polynomials in P(2, d). Proof Write d = 4m − 1. Put p(x, y) = p4m−1 (x, y) − (4m − 1)x 2m−1 y( p2m−2 − 1). Then p(x, y) = 1 on x + y = 1 because both p4m−1 and p2m−2 do. It is clear that p(x, y) is neither p4m−1 (x, y) nor p4m−1 (y, x). To finish the proof one needs to show two things: first, that p has the same number of terms as p4m−1 does, and second that all the coefficients are non-negative. We omit the details and refer to [24]. We can use p4 to prove the following general result. The key idea will be to show for certain d that there are three consecutive terms in pd whose coefficients are multiples of (1 + λ, 4, 2). Theorem 3.12 Suppose that d is congruent to 1 mod 6 and d = 1. Then there are at least 3 sharp polynomials in P(2, d). Proof Put d = 6n + 1. Put r = 3n and s = 2n − 1. We want three consecutive coefficients of pd to satisfy

 K r,s , K r,s+1 , K r,s+2 = c(1 + λ, 4, 2)

for some positive c and λ. Recall (where d = 2r + 1) that K r,s



2r + 1 2r − s = s s−1

(3.28)

7 Elaboration of the Method for Producing Sharp Polynomials

and hence

(2r − s − 2)!s!(2r − s − 1)! s+1 K r,s+2 . = K r,s+1 s + 2 (s + 1)!(2r − 2s − 3)!(2r − s − 1)!

99

(3.29)

After simplifying (3.29), and then using the values r = 3n and s = 2n − 1 we obtain 1 (2r − 2s − 1)(2r − 2s − 2) 1 (2n + 1)(2n) 1 = = . s+2 (2r − s + 1)(s + 1) 2n + 1 4n 2 Now we require that the ratio of the first two coefficients must satisfy K r,s+1 ≤ 4. K r,s Using (3.29) with s replaced by s − 1, we obtain K r,s+1 1 (2r − 2s + 1)(2r − 2s) . = K r,s s+1 (2r − s)(s) Again, substituting r = 3n and s = 2n − 1, we obtain a rational function in n that, for positive integers n, is bounded above by its value at 1 which is 53 ≤ 4. Now that we have established the result about these three coefficients, we can write pd (x, y) = g(x, y)(x 4 + 4x 2 y + 2y 2 ) + h(x, y) for polynomials g, h with non-negative coefficients. Replacing (x 4 + 4x 2 y + 2y 2 ) by (1 + y 4 ) yields a polynomial p ∗ in P(2, d) that has the same number of terms as p. Since p(x, y) is neither pd (x, y) nor pd (y, x), the conclusion follows. Example 3.8 We show how the proof of Theorem 3.11 works when d = 13. p13 (x, y) = x 13 + 13x 11 y + 65x 9 y 2 + 156x 7 y 3 + 182x 5 y 4 + 91x 3 y 5 + 13x y 6 + y 13.

We observe that the terms 156x 7 y 3 + 182x 5 y 4 + 91x 3 y 5 can be written 221 7 3 91 3 3 4 221 7 3 91 7 3 x y + x y + 182x 5 y 4 + 91x 3 y 5 = x y + x y (x + 4x 2 y + 2y 2 ). 2 2 2 2

We replace the terms (x 4 + 4x 2 y + 2y 2 ) with 1 + y 4 and we obtain a polynomial in P(2, 13) also with eight terms. That polynomial is x 13 + 13x 11 y + 65x 9 y 2 +

221 7 3 91 3 3 91 3 7 x y + x y + x y + 13x y 6 + y 13 . 2 2 2

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When d = 13, the numbers r and s in the proof of Theorem 3.11 are 6 and 3. If we didn’t notice by inspection that the coefficients of the terms x 7 y 3 , x 5 y 4 , x 3 y 5 are of the form c(1 + λ, 4, 2), setting s = 3 tells us to examine these three coefficients. We can extend this reasoning by using p2k for larger values of k. For polynomials of large degrees, one can often apply these tricks more than once. The results are extremely complicated. In degree 13, there are several other additional sharp polynomials. In degree 19 there are at least 24 examples. See [53]. Example 3.9 The polynomial p6 (x, y) leads to the identity x 6 + 6x 4 y + 9x 2 y 2 + 2y 3 = 1 + y 6 on the sphere. We can therefore study whether there exists a d such that pd (x, y) = q(x, y)(x 6 + 6x 4 y 2 + 9x 2 y 2 + 2y 3 ) + h(x, y), where the coefficients of q and h are non-negative. There is no d for which there are four consecutive terms in the ratio (1, 6, 9, 2). We can still use the idea as before. Definition 3.3 Consider a degree d. We say that uniqueness fails in degree d if there are at least three distinct sharp polynomials p ∈ P(2, d). Otherwise, we say that uniqueness holds. The reader might wonder why we require three distinct examples rather than just two. Symmetry provides the explanation. Suppose first that d is odd and at least 5. Then there at least two sharp polynomials in P(2, d), namely pd (x, y) and pd (y, x). We want uniqueness to fail only when there is another sharp example. Here is one of the results from [24]. See also [55] and [53]. Lebl posted additional information while introducing sequences A143105 and A143106 into the Online encyclopedia of integer sequences [63]. Theorem 3.13 Uniqueness holds when d = 1, 3, 5, 9, 17. Furthermore, • Uniqueness fails for all even d. • Suppose d is congruent to 3 mod 4. Then uniqueness holds for d = 3 and fails for d ≥ 7. (Theorem 3.12.) • Suppose that d > 1 and that d is congruent to 1 mod 6. Then uniqueness fails. (This result was stated and proved as Theorem 3.12. It includes Theorem 3.10.) We briefly discuss the situation in even degrees, which is less interesting. Example 3.10 We show that uniqueness fails in degrees 2 and 4. In degree 2, we have the following examples with three terms: x 2 + x y + y,

7 Elaboration of the Method for Producing Sharp Polynomials

101

x 2 + 2x y + y 2 , x + x y + y2. In degree 4, we have the following examples with four terms: x(x 3 + 3x y + y 3 ) + y, y(x 3 + 3x y + y 3 ) + x, (x + y)x 3 + 3x y + y 3 , (x + y)y 3 + 3x y + x 3 , Remark 3.6 The statement for even degrees d + 1 (when d + 1 ≥ 6 ) is easy to verify. Start with the polynomial pd (x, y) in degree one less. For d ≥ 5, this polynomial is not symmetric in x and y. It contains the terms x d and y d . If we multiply one of these terms by x + y we obtain a sharp polynomial p in degree d + 1. If we multiply the other term by x + y, we obtain a second sharp polynomial q. Note that p(x, y) = q(y, x). We can do the same with pd (y, x), obtaining at least four sharp examples. We have proved all but the part when d is congruent to 3 mod 4. The ideas are the same, but a bit more complicated, and we refer to [24] for details. The methods of this section have used the coefficients of the polynomials pd , where one studied ratios of successive coefficients. One naturally wonders what happens if one applies the same ideas to the binomial coefficients. The following elementary but beautiful result about Pascal’s triangle holds. We leave the proof to the interested reader. Theorem 3.14 Let (a, b, c) denote three consecutive entries in a row of Pascal’s triangle. Then, for each positive integer k, there is no other row in which the triple (ka, kb, kc) appears as consecutive coefficients. Proof Left as an exercise. Exercise 3.8 Find all solutions to the Pell equation A2 − 2B 2 = 1. Exercise 3.9 Find a sharp polynomial in degree 13 that is neither p13 (x, y) nor p13 (y, x). Comment: there are several examples, all of which have non-integer coefficients. There is one example, however, with half-integer coefficients. Exercise 3.10 Prove Theorem 3.14. Remark 3.7 We discuss the interaction between the pd when d is even and when p is odd. For d odd, the pd give the minimum number of terms. Because of the group invariance, the individual monomials that arise have the form x d−2 j y j . To rewrite some terms it is natural to use other polynomials pk whose monomials have a similar form. The condition on non-negativity suggests why we take k even.

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8 Additional Tricks We have seen the basic idea many times. Suppose f ∈ P(2, d). The coefficients are solutions to an underdetermined linear system, which even in degree 1 has infinitely many solutions. (For 0 ≤ λ < 1, we can use λ + (1 − λ)(x + y).) As the degree increases toward infinity, the dimension of the parameter space of solutions increases as well. Since there are so many solutions, we place additional restrictions on f and study what happens. In each degree, the number of sparse solutions is finite. Only in low degree, there is only one sparse solution and there are infinitely many degrees where there are more than two sparse solutions. For such d, given the sparse solution pd (x, y), we found others by various combinatorial identities. We develop these ideas a bit further. Lemma 3.2 For any non-negative integer k, the following identity holds on the line x + y = 1: x k+4 + y k+4 + (x y)(x k+1 + y k+1 ) = x k+3 + y k+3 + (x y)2 (x k + y k ).

(3.30)

As a consequence, if q(x, y) is any polynomial with non-negative coefficients, we can multiply both sides of (3.30) by q and obtain an additional identity. Proof The identity (3.30) holds if and only if the identity resulting from homogenizing both sides of it holds. Thus, we multiply the term (x y)(x k+1 + y k+1 ) on the left-hand side by (x + y) and we multiply the term x k+3 + y k+3 on the right-hand side by (x + y). In both cases we obtain x k+4 + y k+4 + x k+3 y + x k+2 y 2 + x 2 y k+2 + x y k+3 and (3.30) follows. The second statement is immediate. Putting k = 9 in (3.30) yields an identity we will use below: (x 13 + y 13 ) + (x y)(x 10 + y 10 ) = (x y)2 (x 9 + y 9 ) + (x 12 + y 12 ).

(3.31)

Multiplying both sides of (3.31) by (x y)2 yields (x y)2 (x 13 + y 13 ) + (x y)3 (x 10 + y 10 ) = (x y)4 (x 9 + y 9 ) + (x y)2 (x 12 + y 12 ). (3.32) Given a sharp polynomial p which contains a multiple of the left-hand side of (3.31), we can replace it by the same multiple of the right-hand side of (3.31). It is quite surprising that there exists in degree 19 such a polynomial. Lebl [53] found all the sharp polynomials of degrees 19 and 21; by coding, he noted that the computation takes approximately 5 days for degree 19 on a relatively recent 4-core CPU. For degree 21, the computation took over 8 months, and it verifies that uniqueness holds in this degree. In other words, the only examples of sharp polynomials are p21 (x, y) and p21 (y, x).

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Here is one of the sharp polynomials in degree 19: x 19 + y 19 + +

19 323 (x y)(x 16 + y 16 ) + (x y)2 (x 13 + y 13 ) 2 11

323 323 399 (x y)3 (x 10 + y 10 ) + (x y)(x 5 + y 5 ) + (x y). 11 55 110

are the left-hand side of (3.32). This procedure Notice that the terms multiplied by 323 11 thus leads to another sharp polynomial in degree 19. Symmetric sharp polynomials exist in degree 1, 3, 7, 19 but not in most low degrees. The degrees 1 and 3 are exceptions because they are so small. Degree 7 is the smallest degree in which uniqueness fails, and the smallest degree for which there is a non-trivial symmetric example. The author, based on flimsy evidence, has wondered whether being congruent to 7 mod 12 might be a required condition. The reason is that, if d is congruent to 7 mod 12, then both congruences mod 4 and mod 6 from Theorem 3.13 hold. That suggests that there might be many sharp polynomials, and hence there is a chance symmetric examples might exist. Open problem. Determine whether the number of odd degrees d for which there are at most two sharp polynomials of degree d is finite or infinite. Open problem. Find all degrees d in which there is a sharp symmetric polynomial in P(2, d). Open problem. Find all sharp polynomials with integer coefficients. In the next chapter we will study this sort of procedure with a different goal in mind. Rather than trying to minimize the number of terms in a polynomial in P(2, d), we will seek to minimize its value at the point (1, 1). We give a simple example to anticipate where we are going. By Lemma 3.2 with k = 0, we see that f (x, y) = x 3 + y 3 + 2(x y)2 = x 4 + y 4 + x y(x + y) = x 4 + y 4 + x y = g(x, y) on x + y = 1. Note that f (1, 1) = 4 and g(1, 1) = 3. Assume we are working in degree at least 4, and that p = q f + h for some polynomials with non-negative coefficients. Then p cannot minimize p(1, 1) because we can replace f by g and decrease the value of p(1, 1).

9 Maps with Source Dimension 2 and Target Dimension 4 In the papers [14] and [13] from 1988, the author claimed to list all the spherical equivalence classes of monomial proper maps from B2 to B4 . The author had done the computations by hand and they took many pages. In preparing the list, the author inadvertently omitted one example. He observed so in [15] from 1993 by mentioning the omitted map, but he did not include the complete list. It therefore seems worthwhile to produce the complete list here. The best way to do these computations is

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to solve the linear system from Sect. 1 that defines monomial maps, using symbolic computation. Of course one needs to have a degree bound to do so. In this case, the maximum degree is 5. The list provides opportunities for some additional comments as well. Theorem 3.15 Suppose that f is a monomial sphere map with source dimension 2 and target dimension 4. Assume, without loss of generality, that f (0) = 0. Then f is spherically equivalent to one of the following maps: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

(z, w, 0, 0) zw, w, 0) (z 2 , √ (z 2 , √2zw, w 2 , 0) (z 3 , √3zw, w√3 , 0) (z 3 , 3z 2 w, 3zw 2 , w 3 ) (z 3 , z 2 w, zw, w) z 2 w, zw 2√ , w) (z 2 , √ 2 (z , √2z 2 w, √2zw 2 , w 2 ) 2 w, 2zw 2 , w 2 ) (z 3 , 3z√ 2 (z, z w, √2zw 2 , w 3 ) z 3 w, 3zw, w 3 ) (z 4 , √ 4 zw 3 , w) (z , √3z 2 w, √ 5 3 (z , √5z w, 5zw 2 , w 5 ) (z 2 , 2zw, zw 2 , w 3 ) For each θ ∈ (0, π2 ), the map is (z, cos(θ)w, sin(θ)zw, sin(θ)w 2 )  For each θ ∈ (0, π2 ), the map is (z 2 , 1 + cos2 (θ)zw, cos(θ)w 2 , sin(θ)w).

Remark 3.8 Map number (14) was omitted from the lists in [14] and [13]. We make a large number of comments about this list. The map switching the source variables is of course unitary; hence, we do not list both f (z, w) and f (w, z). Furthermore, there is no loss in generality in assuming that f (0) = 0. The reason is simple. If  f (0) = c, then f would be unitarily equivalent to g ⊕ c where g(0) = 0. Since g2 + |c|2 = 1 on the sphere, we could replace g ⊕ c with the map G = √ g 2 , which is a monomial sphere map spherically equivalent 1−|c|

to g ⊕ c with G(0) = 0. Map (1) is of the form f ⊕ 0 where f has target dimension 2. Maps (2), (3), and (4) are of the form f ⊕ 0 where f has target dimension 3. These four maps all appeared in Theorem 2.9. Map (5) is the map (z, w)⊗3 . Map (6) is a Whitney map and map (7) is the result of tensoring map (2) on its second component. Map (13) is the group invariant map p4 from earlier in this Chapter. It is the only example of degree 5. Furthermore, since there is no example of degree 4 mapping to the three balls, it is not the result of tensoring alone. Map (14) results from tensoring map (3) on its third component. Maps (15) and (16) are juxtapositions of examples with lower target dimension. Map (15) is the juxtaposition of maps (1) and (2) after ignoring the zero components, and map (16) is the juxtaposition of maps (3) and (2) after

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105

ignoring the zero components. In each case, the maps are inequivalent for different values of θ. See Theorem 2.11. Maps (1) and (3) together contain five independent monomials; their juxtaposition, therefore, requires the target dimension to be at least 5, and hence it does not appear on this list. There are polynomial sphere maps with source dimension 2 and target dimension 4 not equivalent to any of these maps. We begin to see the impact of higher codimension. Not all maps are equivalent to monomial maps and there are one-parameter families of inequivalent maps. Wono, in his 1993 thesis [81], gave a similar, but much longer list, in target dimension 5. We briefly summarize that list. There are three examples of degree 7. There are five examples of degree 6. There are 24 examples of degree 5. There are 45 examples of degree 4 plus 2 one-parameter families of degree 4. There are 30 one-parameter families of degree at most 3 and 1 two-parameter family of degree 3. The two-parameter family is easy to explain. For each b, c in the interval [0, 1], consider the real polynomial f (x, y) = bx + cy + (1 − b)x 2 + (2 − b − c)x y + (1 − c)2 . Then f is the squared norm of a monomial sphere map. We can think of it along the lines of Proposition 2.6; to find all maps p of degree 2 with p(0) = 0, we choose the linear part (a1 z, a2 w) and then the quadratic part is essentially determined. Example 3.1 and Exercise 3.1 also show that there is a two-parameter family of monomial sphere maps of degree at most 2 with source dimension 2 and target dimension 5. Example 3.2 lists the maps of highest possible degree, namely 7. It is easy to see that there can be no 1-parameter family of examples of maximum degree. We make one additional remark and pose an open problem below. The squared norms of all 14 of the isolated examples in Theorem 3.15 have integer coefficients. There are 77 isolated examples in Wono’s list, and 68 of their squared norms have integer coefficients. Open problem. Let f (x) be a polynomial of degree d in n variables, with nonnegative coefficients, and with f (x) = 1 on s(x) = 1. If f is isolated (not part of a family) in this set, how likely is f to have integer coefficients? How does this probability depend on n and d? The examples from Theorem 3.15 (and from Wono’s list) always have at least 2 monomials of maximum degree. In fact, for a non-constant rational sphere map with f (0) = 0, the rank of the terms of maximum degree must be at least as large as the source dimension. This result follows either from the so-called Huang lemma in [46] or Theorem 2.3. For a polynomial sphere map, Corollary 2.2 shows that  pd 2 is a multiple of z2 and hence has rank at least n.

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10 Target-Ranks for Monomial Sphere Maps We return to the notion of target-rank. Let f be a rational sphere map with source dimension n and target dimension N . The target-rank of f can of course be smaller than N . The

target-rank of a polynomial sphere map of degree d is bounded by the of holomorphic polynomials of degree at most d in n variables. We number n+d d next consider possible target-ranks for monomial sphere maps of a given degree d and ask what target-ranks are possible. We consider source dimension 2 here, although analogous results hold in dimensions at least 3.

 . Theorem and let b(d) = d+2 Assume n = 2. Let a(d) be the ceiling of d+3 2 2 3.16 states that each integer in the interval [a(d), b(d)] is possible for the target-rank of a monomial sphere map of degree d, and no other values are possible.

 We can , and the even arrange that f (0) = 0, although in that case we cannot achieve d+2 2 right-hand end point becomes b(d) − 1. The discussions just after Definition 2.6 and in Section 9 clarify why demanding that f (0) = 0 is a natural thing to do. We also consider the problem for a natural special class of polynomial sphere maps, namely orthogonal sums of the homogeneous maps Hk . Also in this case we determine the precise possible values of the target-rank of such a sum. In Chapter 5, we will see that this special collection of maps has a nice characterization in terms of Hermitian invariant groups. For d ≥ 2, the Hermitian-invariant group  f of a rational sphere map is the unitary group U(n) if and only if the map is unitarily equivalent to such an orthogonal sum. The proofs for such orthogonal sums rely on elementary combinatorial reasoning. monoWhen n = 2, this reasoning is easy. For k ≥ 0, let Hk denote the homogeneous  mial sphere map of degree k. The target-rank of Hk is k + 1. Assume dk=0 |ck |2 = 1. Each term in the orthogonal sum ⊕

d 

ck Hk

k=0

contributes k + 1 to the target-rank if ck = 0 and contributes nothing if ck = 0. We note and use the following elementary result.  Lemma 3.3 Consider sums dk=0 m k (k + 1) where m k is either 0 or 1.

 ] is such a sum and no others are. • Each integer in the interval [0, d+2 2

 • If we insist that m d = 1, then the set of sums is the integer interval [d + 1, d+2 ] 2 • If we also insist that m = 0, then the set of sums is every integer in the interval 0



 − 1] except for d + 2 and d+2 − 2. [d + 1, d+2 2 2 Proof Consider the first item. When d = 0, the possible sums are 0 and 1; the result thus holds and provides the basic step for an induction. Assume that we have proved the result for some d. Consider the collection of sums with d replaced by d + 1. By setting m d+1 = 0, the induction hypothesis shows that this collection contains the

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107

set Sd of sums for d. By setting m d+1 = 1, we see that it also contains the translate of Sd by d + 2. The smallest element of the translated set is contained in Sd . Also



(d + 2)(d + 3) d +3 d +2 2(d + 2) + (d + 2)(d + 1) = = . d +2+ = 2 2 2 2

 Thus, Sd+1 equals the integer interval [0, d+3 ] and the first item follows for all 2 d by induction. No other numbers are possible because the maximum value of such

 . a sum for a given d is d+2 2 The second item follows because the term d + 1 must be included. To prove the third item, we note that m 0 = 0 precludes 1 as a possible summand. Since m d = 1, we obtain d + 1 as a sum, but we cannot get d + 1 + 1 = d + 2. The largest sum we  get is 2 + 3 + · · · + d + 1 but we cannot get 1 less than this. Therefore,

d+2can − 2 is precluded. When we pass from d to d + 1, the smallest element in the 2 translated set is contained in the previous set. An induction similar to the proof of the first item shows there are no other restrictions. Theorem 3.16 For each degree d ≥ 1 and each integer N in the interval [a(d), b(d)], there is a monomial sphere map f with source dimension 2 such that • f is of degree d. • The target-rank of f is N . Furthermore, no other values are possible for the target-rank of a monomial sphere map of degree d. If we also demand that f (0) = 0, then the set of possible values for the target-rank is the integer interval [a(d), b(d) − 1]. Proof We begin by proving the result in degrees 1 and 2. Suppose first that we allow f (0) = 0. Let c, s be non-zero real numbers with c2 + s 2 = 1. We regard them as cosine and sine, as we juxtapose maps. When d = 1, the interval is [2, 3]. The maps f (z, w) = (z, w) and f (z, w) = (c, sz,  do the job. The second map,

sw) = 3 and hence is disallowed however, has f (0) = 0. When d = 1, however, d+2 2 as a target-rank when f (0) = 0. The result thus holds when d = 1. When d = 2, the interval is [3, 6]. We have seen each of the following maps: • • • •

(z, zw, w2 ) has target-rank 3. 2 4. (z, cw, szw, sw √ ) has target-rank 2 2 sw ) has target-rank 5. (cz, cw, sz , s 2zw, √ (λ, az, aw, bz 2 , b 2zw, bw 2 ) has target-rank 6 if the real parameters are non-zero and λ2 + a 2 + b2 = 1.

 The map of target-rank 6, however, has f (0) = 0. But 6 = 24 and hence is disallowed when f (0) = 0. The conclusion thus holds when d = 2. These examples suggest how to generate maps whose target-rank is at least d + 1. We simply juxtapose homogeneous maps of the appropriate degrees. Namely, for 0 ≤ j ≤ d, let H j denote the usual homogeneous monomial sphere map of degree

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j. (When j = 0, we put H0 = 1.) Choose d + 1 non-negative numbers c j such that  cd = 0 and dj=0 |c j |2 = 1. Put f = c0 H0 ⊕ c1 H1 ⊕ c2 H2 ⊕ · · · ⊕ cd Hd . Then f is a monomial sphere map of degree d. If c j = 0, then the term H j contributes j + 1 to the target-rank. By choosing the non-zero c j appropriately, Lemma 3.3

 allows us to achieve any target-rank in the interval [d + 1, d+2 ]. If we insist that 2

 − 2 as f (0) = 0, however, we cannot achieve the possible values d + 2 and d+2 2 orthogonal sums. We now assume that d ≥ 3. To achieve d + 2, we first consider the homogeneous map Hd−1 . It has target-rank d and contains the monomials z d−1 and w d−1 . We zw d−1 , w d . tensor these to obtain the distinct (since d ≥ 3) monomials z d , z d−1 w, d+2 The resulting map has degree d and target-rank d + 2. To achieve 2 − 2, we first

 find an example of degree d − 1 whose target-rank is d+2 − 3. We then tensor a 2 term of degree d − 1 by the identity map, adding one term to both the degree and the target-rank. We have thus proved the result for target-ranks at least d + 1. It remains to find examples with target-rank close to the left-hand point a(d) of our interval. The polynomials pd from Theorem 3.1 play a key role. First suppose is r + 2. Theorem 3.1 that d = 2r + 1 is odd. Then the ceiling of (the integer) d+3 2 provides a polynomial pd , which we write as pd (x, y) = y 2r +1 + x 2r +1 + x 3 g(x, y) + λx y r , such that pd is the squared norm of a monomial sphere map of degree d. By Theorem 3.1, the target-rank of this monomial map is r + 2. If we multiply the term λx y r by x + y, a simple example of the restricted tensor product operation, we obtain another example of degree d with one additional term. We write this polynomial as y 2r +1 + x 2r +1 + x 3 g(x, y) + λ(x 2 y r + x y r +1 ). This polynomial is the squared norm of a monomial sphere map with target-rank r + 3. We continue in this fashion, by multiplying the term x y j by (x + y). Doing so increases the target-rank by 1 and gives a new map at each stage. As long as this polynomial is of degree at most d we have one of the desired examples. Thus, for r ≤ j ≤ 2r this procedure gives us maps whose target-ranks are r + 2, r + 3, ..., 2r + 2. Thus all target-ranks j are possible for q(d) ≤

j ≤ d + 1. We showed above that all − 1. Thus the desired conclusion target-ranks j are possible for d + 1 ≤ j ≤ d+2 d holds for odd degree d. The proof for even degree is similar. Put d = 2r + 2. Then q(d) = r + 3. We again consider the polynomial p2r +1 . This time we increase the degree by 1 and start with (x + y)x 2r +1 + y 2r +1 + x 3 h(x, y) + λx y r ,

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109

which is of degree 2d + 2 and has r + 3 terms. We follow the above procedure to generate examples with target-rank j for r + 3 ≤ j ≤ d + 1. Each N in the interval [a(n), b(n) − 1] is achieved (by a map with f (0) = 0) for even degree as well. By Theorem 3.2, in source dimension 2, the target dimension for a monomial . The target-rank is sphere map of degree d cannot be smaller than the ceiling of d+3 2 of course at most the target dimension. We have established all the conclusions. When n = 2, we have for each d determined all possible target-ranks for monomial sphere maps of degree d. We have also determined all possible minimal target dimensions, recalling that, when f (0) = 0, the target-rank equals the minimal target dimension. Remark 3.9 When n = 2, the target-rank of the polynomial H j is j + 1. By Lemma  c j H j to obtain any target-rank 3.3, we can choose the c j in an orthogonal sum

 from 1 to d+2 . The analogous statement fails for n ≥ 3. For example, suppose 2 |c0 |2 + |c1 |2 + |c2 |2 = 1. Consider the real polynomial p(x, y, z) = c0 + c1 (x + y + z) + c2 (x + y + z)2 . The number of terms in p can be 1, 3, 4, 6, 7, 9, 10 but cannot be 2, 5, 8.

 Numbers of the form d+1 are known as triangular numbers. They arise for us in 2

 source dimension 3, where there are d+2 independent homogeneous polynomials 2 of degree d in three variables. Considering orthogonal sums of the monomials maps Hk for 0 ≤ k ≤ d leads to a fascinating (but solved) combinatorial problem. What numbers can be written as

d  k+1 , mk 2 k=1 where the m k are either 0 or 1? It is easier to say what numbers cannot be written in this way. The answer is that all non-negative integers can be written in this way except for 2, 5, 8, 12, 23, 33. Each number larger than 33 is such a sum. The sequence 2, 5, 8, 12, 23, 33 is A053614 in the On-Line Encyclopedia of Integer Sequences [63]. We conclude the following curious result. See Exercise 3.12. Proposition 3.5 Let S = {2, 5, 8, 12, 23, 33}. There is a monomial sphere map in source dimension 3 that is an orthogonal sum of the maps Hk (for k ≥ 1) and whose target-rank is N if and only if N is not in S. The collection of orthogonal sums of homogeneous maps will be nicely characterized in Chapter 5, where we assign groups to rational sphere maps. The Hermitianinvariant group of a rational sphere map f is the unitary group U(n) if and only if f is an orthogonal sum of tensor powers and of degree at least 2. We next briefly discuss source dimension 3. There is no rational sphere map in 3 variables with target-rank 2. By the Huang Lemma or by Theorem 2.3, the target-rank must be at least 3 if the map is not a constant. The only maps with target-rank 4 are

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those of the form c ⊕ sU z, where U is unitary, and where |c|2 + |s|2 = 1 and both c, s = 0. Such maps are spherically equivalent to the map z ⊕ 0. Thus, although the target-rank is 4, the map is spherically equivalent to a map whose target-rank is 3. By Theorem 2.12, there are monomial examples with target-ranks equal to each integer at least 5. Example 3.11 We use variables z, w, ζ. 2 • (z, w, zζ, wζ, target-rank √ ζ ) has √ √ 5. 2 • H2 = (z , 2zw, w 2 , 2zζ, 2wζ, ζ 2 ) has target-rank 6. • (z, w, zζ, wζ, zζ 2 , wζ 2 , ζ 3 ) has target-rank 7.

Given a monomial example of degree d and target-rank t, we may (via the partial tensor product operation), replace a monomial m of degree d with the three monomials mz, mw, mζ to create a new example whose target-rank is t + 2. Since there are examples with target-ranks 5, 6, we can achieve examples of any larger target-rank, all of which have f (0) = 0. We recall the discussion from Sect. 9 concerning the impact of allowing f (0) to be non-zero. If we insist that f (0) = 0, then we are not allowed to use the map H0 . The impact is to increase the bad set S. In Remark 3.9, for example, the number of terms in p could only then be 3, 6, 9. Thus there are more possible values of target-rank for maps of a given degree than there are possible values of target dimensions of new maps. The natural combinatorial problem involves target-ranks; it differs slightly from the natural question about minimum target dimension in a given spherical equivalence class. Exercise 3.11 Put n = 3. For each degree d, find the target-rank of Hd . Then prove Proposition 3.5. What happens to the set S if we disallow H0 ? Exercise 3.12 Put n = 3. Write down monomial sphere maps whose target-ranks are 5, 8, 12, 23, 33. Arrange things such that your maps have f (0) = 0. Exercise 3.13 Put n = 3. For which N are there monomial sphere maps of degree 5 with target-rank N ? Exercise 3.14 What is the target-rank of the map (z + w, z + w2 , ..., z + w K )? What is the target-rank of the map (z + w, z + 2w, ..., z + K w)? Remark 3.10 There is an analogue of Proposition

 3.5 when n = 4. In this case, the target-rank of Hk is the tetrahedral number k+3 . Sequence A102806 from [63] 3 lists numbers that are not sums of distinct tetrahedral numbers and includes a proof by Robert Israel that the list is finite. It follows that the orthogonal sums of the Hk achieve all but a finite number of possible target-ranks.

Chapter 4

Monomial Sphere Maps and Linear Programming

This chapter continues our investigation of the linear systems for monomial sphere maps. Since the number of examples tends to infinity as the degree increases, we consider methods for restricting the collection of examples. We do so by considering the problem from the perspective of linear programming. The results in this chapter concern two optimization problems. In Sect. 2, we introduce a natural minimization problem; we wish to minimize the sum of the coefficients of polynomials in a certain collection. We also study sparse solutions, those that minimize the number of terms in such polynomials for a given degree. Theorem 4.9 provides an asymptotic result comparing these problems as the degree tends to infinity. In source dimension two, Theorem 4.3 provides a method for significantly reducing the number of unknown coefficients in the problem of minimizing the sum of the coefficients. This result has several strong consequences, but it is not strong enough to yield a formula for the optimal polynomial in source dimension two. Coding suggests that no simple formula exists. In source dimensions at least three, however, the result from dimension two enables us to prove Theorems 4.4 and 4.6, which provide a simple formula for the polynomial that minimizes the sum of its coefficients. Coding reveals a bevy of incredibly complicated formulas in source dimension n = 2. For example, consider the squared norm of the monomial sphere map with n = 2 and degree d = 121 that minimizes the sum of its coefficients. This real polynomial has rational coefficients for which the numerators and denominators (when reduced to lowest terms) have more than 250 decimal digits! A bounded monotone sequence arises in our discussion, and it would be interesting to establish its limit. The fundamental theorem of linear programming makes an appearance here; in odd degree d at least 3, the real polynomial that minimizes the value at (1, 1) has at most d terms. The author hopes that others get excited about these ideas; they seem to create © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 J. P. D’Angelo, Rational Sphere Maps, Progress in Mathematics 341, https://doi.org/10.1007/978-3-030-75809-7_4

111

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a fertile mix of linear algebra, number theory, and CR geometry. We emphasize that these considerations are easy in all source dimensions except two. It is natural to ask why dimension 2 behaves so differently. One crucial issue arises as follows. When we set a variable equal to 0 to lower the dimension, we continue to have lots of structure when n ≥ 3. When n = 2, however, setting a variable equal to 0 reduces to one dimension and consequently does not help much. We postpone discussion about the corresponding optimization problem for rational sphere maps until Chap. 7. See Corollary 7.4 and Proposition 7.6.

1 Underdetermined Linear Systems Let V, W be vector spaces and assume L : V → W is a linear map. Solutions to the linear equation Lu = w are completely described as follows. Assume w is in the range of L. Then each solution u to Lu = w can be written u = u0 + v, where u0 is a particular solution and v is an arbitrary element of the null space of L. The linear system is called underdetermined when the null space is non-trivial. Depending on the setting, one often determines a unique solution to an underdetermined system based on additional information. Linear ordinary differential equations provide a familiar example. We solve the second-order equation Lu = u  + bu  + cu = f by first solving L y = 0 and then using various tricks to find a particular solution g. The null space of L is two dimensional; each solution to L y = 0 depends on two arbitrary constants. Thus, u = c1 y1 + c2 y2 + g, where the y j are linearly independent, L(y1 ) = L(y2 ) = 0, and L(g) = f . We typically determine these constants by using initial conditions, such as values of u(0) and u  (0). We mention a more sophisticated example from Hodge theory, discussed in detail in [78]. Consider the Laplace equation (u) = α for differential forms on a compact Riemannian manifold. The null space of  consists of the harmonic forms. There is a solution to (u) = α if and only if α is orthogonal to the harmonic forms. One works in L 2 (the forms with square-integrable components) and seeks the solution with minimal L 2 norm. Consider next the linear equation L(u) = v where L : Rn → R N . We assume that the standard bases are used. In compressed sensing, one wishes to choose a solution u for which the fewest number of components are non-zero. This solution is called the sparsest solution. There may be several such solutions, and hence the superlative ending est is slightly misleading. We will instead use the word sparse. We make a surprising connection to monomial sphere maps. Fix a degree d. As in Chap. 3, Sect. 1, we solve the underdetermined linear system

1 Underdetermined Linear Systems d 

113

|Cα |2 xα s(x)d−|α| = s(x)d

(4.1)

|α|=0

for the unknown constants |Cα |2 to determine all monomial sphere maps of degree at most d. Let us assume that Cα = 0 for some α with |α| = d, in order to guarantee that the solution is of degree precisely equal to d. Finding the minimum possible target dimension for the corresponding monomial map is the same as finding a solution for which the number of non-vanishing components is minimal. We discuss the relationship between sparseness and degree estimates in Sect. 7. See especially Theorem 4.9. Remark 4.1 Equation   (4.1) amounts to a system of linear equations. The number , the number of independent polynomials of degree at most of unknowns is n+d d   , the number of independent d in n variables. The number of equations is n+d−1 d homogeneous polynomials of degree d in n variables. For example, when n = 2, unknowns and d + 1 equations. As we there are 1 + 2 + · · · + d + 1 = (d+1)(d+2) 2 proceed, we will see how various natural hypotheses restrict these numbers. In the next several sections, we consider an optimization problem distinct from the sparseness question. When solving a linear system, rather than minimizing the number of non-vanishing components, one could instead minimize other quantities, such as the L 1 -norm. Consider solutions u = (u 1 , ..., u n ) to the system Lu = v. The L 1 -norm is given by u1 = nj=1 |u j |. A famous theorem of Donoho (See [30]) states for most large linear systems that a generic solution with minimum L 1 norm is also a sparse solution. While minimizing the L 1 norm in the monomial sphere map context does not generally give a sparse solution, we are able to relate the two notions asymptotically as the degree tends to infinity. See Theorem 4.9. When n = 2, the polynomials minimizing the L 1 norm exhibit remarkably complicated properties, whereas they are easy to understand in higher dimensions.

2 An Optimization Problem for Monomial Sphere Maps We will consider a natural subset of the monomial sphere maps of degree d in n variables, and determine asymptotic results about an interesting optimization problem. Let m(z) = (. . . , Cα z α , . . . ) denote a monomial sphere map with source dimension n. As usual, we put x = (x1 , . . . , xn ) = (|z 1 |2 , . . . , |z n |2 ) and consider the real polynomial f (x) defined by

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f (x) =



|Cα |2 xα .

As in (4.1), we obtain a linear system for the unknowns |Cα |2 . We want to minimize the L 1 norm. Let 1 denote the point (1, . . . , 1). We have m(1)2 = f (1) =



|Cα |2 .

We naturally seek to optimize the values of f (1) given that f is of degree d. Remark 4.2 Despite the appearance of |Cα |2 here, we are studying the L 1 norm rather than the L 2 norm. The reason is that we are working with the real polynomial f , whose non-negative coefficients happen to be of the form |Cα |2 . We pose the following linear programming problem. It was formulated in [21] and some of the results about it in this chapter come from [20]. Definition 4.1 Let S(n, d) denote the set of polynomials f such that 1. f : Rn → R is of degreed and its coefficients are non-negative. 2. f (x) = 1 when s(x) = nj=1 x j = 1. 3. f contains the monomials x1d , . . . , xnd each with coefficient 1.   denote the dimension of the space of polynomials of degree at Let K = n+d d most d in n real variables. We identify S(n, d) with a subset of real Euclidean space R K . The topology is determined by taking limits of the coefficients. Problem 1. Given the source dimension n and degree d, find the minimum m(n, d) and maximum M(n, d) values of f (1) given that f ∈ S(n, d). The case of source dimension 1 is trivial; the only polynomial in S(1, d) is x d . Thus m(1, d) = M(1, d) = 1 for all d. Whether or not we include item (3), the maximum value M(n, d) of f (1) is n d . This value occurs when f (x) = s(x)d . The same maximum holds for polynomial sphere maps but not for rational sphere maps in general. See the discussion after Corollary 7.4. Lemma 4.1 For all n, d we have M(n, d) = n d .  Proof Assume f ∈ S(n, d). Put f (x) = |cα |2 xα . Then f (x) = 1 when s(x) = 1. Homogenizing gives  (s(x))d = |cα |2 xα (s(x))d−|α| and hence n d = s(1)d =



|cα |2 (n)d−|α| ≥



|cα |2 = f (1).

Since equality holds for f (x) = s(x)d , and s d ∈ S(n, d), the conclusion follows.  Studying the minimum is much more interesting. For n ≥ 3, we prove in Theorem 4.6 that m(n, d) = n + (d − 1)n(n − 1). The same result holds when n = 1,

2 An Optimization Problem for Monomial Sphere Maps

115

but when n = 2, things are extremely complicated and there is no simple formula. After some general discussion, we study dimension 2 in detail. We include item (3) as part of the definition of S(n, d) for the following reason. If we drop item (3), then the minimum value of f (1) does not exist when d ≥ 1. The infimum is 1. For 0 ≤ t < 1, put f t (x) = t + (1 − t)(s(x))d . Then f t satisfies items (1) and (2) of Definition 4.1. We also have f t (1) = t + (1 − t)n d . As t tends to 1, the value tends to 1. The problem is that the family of polynomials here is not closed under taking limits of the coefficients. We naturally restrict to closed sets. We note momentarily that S(n, d) is closed and bounded. Evaluation at a point is a continuous operation, hence the minimum occurs. Lemma 4.2 For each n, d, the set S(n, d) is compact in R K . There exists an f ∈ S(n, d) achieving the minimum value of f (1). Proof The first two properties defining S(n, d) guarantee that this set is bounded. Including the third property guarantees that it is closed. Evaluation at 1 is a linear function of the coefficients and hence is continuous. The second conclusion thus follows by compactness.  It is relatively easy to prove directly that any polynomial satisfying items (1) and (2) from Definition 4.1 must include at least n terms of degree d. This conclusion also follows from Theorem 2.3. In addition, setting all but one variable equal to 0 shows, for each index j, that there must be a monomial of the form x kj for some k. We assume in item (3) the symmetric condition that each x dj appears with coefficient equal to 1. Hence, the monomial x kj does not appear for 0 ≤ k < d. The following lemma suggests why we consider the symmetric condition. Lemma 4.3 There is a symmetric polynomial F ∈ S(n, d) with F(1) = m(n, d). Proof By Lemma 4.2 there is an f in S(n, d) that minimizes f (1). Define F by averaging over the permutations of the coordinates: F=

1  f ◦ σ. n! σ

Then F contains all the monomials x dj because f does. Since s ◦ σ = s for all σ, it follows that ( f ◦ σ)(x) = 1 when s(x) = 1. Since σ(1) = 1 for all σ and F(1) = f (1), it follows that F ∈ S(n, d). 

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Definition 4.2 Let f be a polynomial in n variables. We write Sym( f ) for the symmetric polynomial defined (as in Lemma 4.3) by averaging over the permutations of the coordinates. 1  f ◦ σ. Sym( f ) = n! σ

3 Two Detailed Examples in Source Dimension 2 We discuss degree 4 and degree 7 in source dimension 2. When n = 2 we write (x, y) instead of (x1 , x2 ). We analyze degree 7 twice; first, we avoid using results such as Theorem 4.3 that significantly simplify the solution. Then we provide a shorter solution. In Sect. 8, we analyze degrees 9 and 11. Example 4.1 Degree 4. We show that m(2, 4) = 7. The collection of polynomials satisfying the first two properties of Definition 4.1 involves 15 unknown coefficients and 5 independent equations. The third condition in Definition 4.1 states that the coefficients of x 4 and y 4 both equal 1 and amount to setting 7 coefficients equal to 0. These are the constant term and the coefficients of x, x 2 , x 3 , y, y 2 , y 3 . At this stage, we are considering polynomials of the form p(x, y) = x 4 + y 4 + A(x y) + B(x y)2 + C1 x 2 y + C2 x y 2 + D1 x 3 y + D2 x y 3 . By Lemma 4.3, the minimum of p(1, 1) occurs for a symmetric polynomial. We therefore set C1 = C2 = C and D1 = D2 = D, obtaining four parameters and getting p(x, y) = x 4 + y 4 + A(x y) + B(x y)2 + C(x 2 y + x y 2 ) + D(x 3 y + x y 3 ). We must invoke the condition that p(x, y) = 1 on the line x + y = 1. We homogenize p(x, y) and obtain a linear system for A, B, C, D by equating coefficients in (x + y)4 = x 4 + y 4 + A(x y)(x + y)2 + B(x y)2 + C(x 2 y + x y 2 )(x + y) + D(x 3 y + x y 3 ).

Solving the linear system reduces to two parameters and we obtain p(x, y) = x 4 + y 4 + (4 − C − D)(x y) + (2D − 2)x 2 y 2 + C(x 2 y + x y 2 ) + D(x 3 y + x y 3 ).

Since p(1, 1) = 4 + C + 3D, our objective is to minimize 4 + C + 3D subject to the constraints: (4 − C − D) ≥ 0 2D − 2 ≥ 0 C ≥0

3 Two Detailed Examples in Source Dimension 2

117

D ≥ 0. These constraints determine a feasible region in the (C, D) plane. This region is a triangle with vertices at (0, 1), (3, 1), and (0, 4). Evaluating at these points shows that the minimum occurs at (C, D) = (0, 1). We obtain p(x, y) = x 4 + y 4 + 3x y + x 3 y + x y 3 and p(1, 1) = m(2, 4) = 7. Example 4.2 Brute force solution in degree 7. In degree 7, the number of parameters is much larger. We wish to minimize p(1, 1). The minimum turns out to be 25 . 2 We again use Lemma 4.3 and start with the assumption that f is symmetric. There are 20 independent symmetric polynomials. These are the symmetrizations of x a y b with 0 ≤ a ≤ b and a + b ≤ 7. At this stage, a symmetric polynomial f has 20 coefficients c[a, b]. If f ∈ S(2, 7) then the coefficients are non-negative. We require that x 7 and y 7 appear with coefficient 1; in other words, c[7, 0] = c[0, 7] = 1. As a trivial consequence, the coefficients c[k, 0] all vanish for 0 ≤ k ≤ 6. To enforce the condition that f (x, y) = 1 when x + y = 1, we homogenize as usual. We obtain four independent linear equations on the remaining coefficients c[a, b], by equating the resulting coefficients of x 7 , x 6 y, x 5 y 2 , x 4 y 3 in the homogenized equation with those in (x + y)7 . The other equations are satisfied by symmetry. At this stage, we are considering polynomials with 12 unknown non-negative coefficients. These polynomials have the form p(x, y) =x 7 + y 7 + λ(x y) +

6 

c[ j](x j y + x y j ) + D(x y)3 + D[4](x 4 y 3 + x 3 y 4 )

j=2

+ B(x y)2 + B[3](x 3 y 2 + x 2 y 3 ) + B[4](x 4 y 2 + x 2 y 4 ) + B[5](x 5 y 2 + x 2 y 5 ).

(4.2) These unknown coefficients satisfy a system of three independent equations, arising from homogenization when we equate coefficients of the terms x 6 y, x 5 y 2 , and x 4 y 3 to the binomial coefficients 7, 21, 35. It is not difficult to check that these three equations are independent, and the other equations are determined by symmetry. We thus have a nine-dimensional parameter space. We wish to minimize p(1, 1) over all polynomials in this space. By Theorem 4.3, we could set B, B[3], B[4], B[5], D, D[4] all equal to 0. We do so in the more efficient solution below. Various tricks from [20] show that additional coefficients must vanish for a polynomial minimizing p(1, 1). For example, the coefficient c[2] must be 0 in a minimizer. The reason is that x 2 y + x y 2 = x y on the line x + y = 1. Using this substitution decreases f (1, 1). We therefore may set c[2] = 0. Similar reasoning shows that we may replace (x 3 y + x y 3 ) with x3y + 2 (x 4 y + x y 4 ) and also decrease f (1, 1). Thus, we may set c[3] = 0. 3 Theorem 4.3 and these tricks significantly reduce the source space, but we don’t employ them until we present the second solution. Using Mathematica, we solve the

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linear system and thus get p(x, y) = x 7 + y 7 + λ(x y) + (35 − 10λ − 3B − D)(x 4 y + x y 4 ) +(−49 + 5B − B[5] + 15λ + 2D)(x 5 y + x y 5 ) +(21 − 2B + B[5] − D − 6λ)(x 6 y + x y 6 ) +B(x y)2 + B[5](x 5 y 2 + x 2 y 5 ) + D(x y)3 . The linear program is to minimize the objective function 16 − λ + B + 2B[5] + D given that λ, B, B[5], D ≥ 0 and the next three inequalities also hold 35 − 10λ − 3B − D ≥ 0 49 + 5B − B[5] + 15λ + 2D ≥ 0 21 − 2B + B[5] − D − 6λ ≥ 0. There are four unknown constants and seven linear inequalities. After considerable effort, we discover that the minimum is attained by setting λ = 27 and setting the other three coefficients equal to 0. We plug these values into the long formula for p. The polynomial in S(2, 7) that minimizes p(1, 1) is thus given by 7 x 7 + y 7 + (x y + x 5 y + x y 5 ). 2 Hence m(2, 7) =

25 . 2

By Theorem 4.3, in source dimension 2 and degree d, for a minimizer, the only possible non-zero coefficients are those of x d , y d , x y, and x j y + x y j for 2 ≤ j ≤ d − 1. Theorem 4.6 generalizes this result to source dimensions at least 3, where things are much easier. We now use Theorem 4.3 to simplify the solution in degree 7. Example 4.3 Efficient solution in degree 7. Using Theorem 4.3 we may assume that 6  p(x, y) = x 7 + y 7 + A(x y) + c[k](x k y + x y k ). k=1

After homogenizing, we obtain a system of equations for the unknown coefficients by equating coefficients in (x + y)7 = x 7 + y 7 + A(x y)(x + y)5 +

6  k=2

c[k](x k y + x y k )(x + y)6−k .

3 Two Detailed Examples in Source Dimension 2

119

By linear algebra, we obtain the following equations: c[5] = c[3] = A=

7 5c[6] − 2 2

−3c[4] 5c[6] + 2 2

c[4] 7 − c[2] + − c[6]. 2 2

Since x 2 y + x y 2 = x y on the line x + y = 1, we decrease f (1, 1) by setting c[2] = 0. Our problem becomes minimizing f (1, 1) given the inequalities 0 ≤ c[4] 0 ≤ c[6] 0≤

−3c[4] 5c[6] + 2 2

0≤

7 5c[6] − . 2 2

Put (u, v) = (c[4], c[6]). The feasible region in the (u, v) plane is determined by the lines v = 35 u, v = 75 , and v = 35 u. It suffices to evaluate the objective function at the is achieved at the origin. three vertices: (0, 0), (0, 75 ), and ( 73 , 75 ). The minimum 25 2 Thus c[4] = c[6] = 0; hence c[3] = 0 and c[5] = 27 . We also obtain A = 27 and the optimal polynomial is again 7 x 7 + y 7 + (x y + x 5 y + x y 5 ). 2 Finding the values m(n, d) for n ≥ 2 and d ≥ 3 is similar to these examples, but the number of parameters, even after symmetrizing, is large. Before turning to general theorems, we give some results obtained from coding when n = 2.

4 Results of Coding and Consequences in Source Dimension 2 The author acknowledges Bob Vanderbei, Jiri Lebl, Dan Putnam, and Daniel Lichtblau, who provided independent code for many of the results. In particular, using powerful tools from Mathematica, Lichtblau [59] determined the polynomials that minimize f (1, 1) for degrees up to 201. His results corroborate the independent

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results of the others which were obtained for somewhat smaller degrees. A recent version of Lichtblau’s code appears in Sect. 9. Assume n = 2. Let m(2, d) denote the minimum value of f (1, 1) given that f ∈ S(2, d). For 1 ≤ d ≤ 9, we list m(2, d) and the minimizing polynomials f d whose values at (1, 1) equal m(2, d). m(2, 1) = 2 for x + y m(2, 2) = 4 for x 2 + y 2 + 2x y m(2, 3) = 5 for x 3 + y 3 + 3x y m(2, 4) = 7 for x 4 + y 4 + x y(3 + x 2 + y 2 ) m(2, 5) =

26 for x 5 + y 5 + x y 3

32 for x 6 + y 6 + x y m(2, 6) = 3





29 for x 8 + y 8 + x y 2 

m(2, 9) = 577/35 for x 9 + y 9 + x y



10 5 3 + (x + y 3 ) + x 4 + y 4 3 3

25 for x 7 + y 7 + x y m(2, 7) = 2 m(2, 8) =

10 5 3 + (x + y 3 ) 3 3





7 7 4 + (x + y 4 ) 2 2





7 7 4 + (x + y 4 ) + x 6 + y 6 2 2



 123 6 9 + 3(x 4 + y 4 ) + (x 5 + y 5 ) + (x 7 + y 7 ) . 35 5 7

The reader can surely guess what f 10 must be, but the passage to f 11 really tests one’s pattern-recognition abilities. The polynomial f 11 is given by the following:  x 11 + y 11 + (x y)

 99 33 4 33 55 11 + (x + y 4 ) + (x 5 + y 5 ) + (x 8 + y 8 ) + (x 9 + y 9 ) . 28 14 14 28 14

Let us write out the first few terms of the sequences m(2, d) and 2d − m(2, d). m(2, d) : 2d − m(2, d) :

2, 4, 5, 7,

26 32 25 29 577 , , , , ,... 3 3 2 2 35

4 4 3 3 53 53 43 43 2393 2393 , ... 0, 0, 1, 1, , , , , , , , , 3 3 2 2 35 35 28 28 1540 1540

We continue in dimension 2 and postpone general results to Section 3.

4 Results of Coding and Consequences in Source Dimension 2

121

Lemma 4.4 For each degree d ≥ 1, there is an f ∈ S(2, d) with f (1, 1) = 2d. Hence, for each d we have the inequality 2 ≤ m(2, d) ≤ 2d. Proof For d = 1 the only polynomial in S(2, d) is x + y and the result holds. Define a sequence recursively by f 1 (x, y) = x + y and f k+1 (x, y) = (x + y)(x k + y k ) + f k (x, y) − (x k + y k ). Then f k+1 is of degree k + 1 and f k+1 (1, 1) = 4 + f k (1, 1) − 2 = 2 + f k (1, 1). Since f 1 (1, 1) = 2, we have f k (1, 1) = 2k. Also f k+1 (x, y) = f k (x, y) on the line  x + y = 1. Therefore f k+1 ∈ S(2, k + 1). Hence m(2, k) ≤ f k (1, 1) = 2k. Lemma 4.5 For all d, we have m(2, d + 1)≤m(2, d) + 2. Furthermore, the sequence d → 2d − m(2, d) is monotone. Proof Let f d (x, y) = x d + y d + (x y)gd (x, y) and assume f d (1, 1) = m(2, d). Pass to degree d + 1 by putting Fd+1 (x, y) = (x + y)(x d + y d ) + (x y)gd (x, y)   = x d+1 + y d+1 + (x y) gd (x, y) + x d−1 + y d−1 . The first equality gives Fd+1 = f d on the line x + y = 1. The second equality then shows that Fd+1 ∈ S(2, d + 1). Finally, Fd+1 (1, 1) = 4 + gd (1, 1) = 4 + f d (1, 1) − 2 = f d (1, 1) + 2 = m(2, d) + 2. Thus, the first conclusion holds. Also, 2(d + 1) − m(2, d + 1) ≥ 2d − m(2, d) holds if and only if the first conclusion holds, and hence the second conclusion holds.  The above examples suggest that m(2, d + 1) = 2 + m(2, d) when d is odd, and that m(2, d + 1) < 2 + m(2, d) when d is even. Furthermore, they suggest that the value of m(2, d) is completely determined by the coefficient λ of the x y term. This property will be established in Corollary 4.4. The relationship between λ and f (1, 1) holds even when f is not a minimizer. We note that λ = 2c[1] if we regard c[1] as the coefficient of x j y + x y j for j = 1. Lemma 4.6 Suppose f ∈ S(2, d) and is symmetric. Assume d ≥ 2. Then there are constants λ and c[ j], and a symmetric polynomial h(x, y) such that f (x, y) = x d + y d + λx y +

d−1  j=2

c[ j](x j y + x y j ) + (x y)2 h(x, y).

(4.3)

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 In the notation of (4.3), we must have d = λ + d−1 j=2 c[ j]. As a consequence, f must j j contain a term of the form x y + x y for some j with 1 ≤ j ≤ d − 1. Proof The formula defines h(x, y) as precisely those symmetric terms divisible by (x y)2 . It includes all the symmetric polynomials, without using the information that f (x, y) = 1 on the line x + y = 1. Including this information and homogenizing, we obtain (x + y)d = x d + y d + λ(x y)(x + y)d−2 +

d−1 

c[ j](x j y + x y j )(x + y)d− j−1 + (x y)2 B(x, y)

j=2

for an appropriate homogeneous B. Equating the coefficients of x d−1 y on both sides of this formula yields d−1  c[ j]. d =λ+ j=2

The coefficients cannot all vanish if they sum to d, and the last statement follows.  The next proposition enables us to understand the asymptotic behavior of m(2, d). A later result, Corollary 4.3, yields a stronger conclusion than Corollary 4.2. Namely, with λd chosen maximally, we have equality m(2, d) = 2d + 2 − λd . Proposition 4.1 Assume d ≥ 2. Suppose f ∈ S(2, d) is symmetric, and f (1, 1) is minimal. In the notation of (4.3), f (1, 1) ≥ 2d + 2 − λ. Proof The coefficients of h are non-negative. Lemma 4.6 thus gives f (1, 1) = 2 + λ + 2

d−1 

c[ j] + h(1, 1) = 2 + λ + 2(d − λ) + h(1, 1)

(4.4)

j=2

≥ 2 + λ + 2(d − λ) = 2d + 2 − λ.  Combining Lemma 4.4 and Proposition 4.1 yields an extremely useful result. Recall that λ depends on the degree. Corollary 4.1 Assume d ≥ 2. Let λ = λd be the coefficient of x y in a minimizer. Then 2d + 2 − λd ≤ m(2, d) ≤ 2d. Furthermore, 2 ≤ λd < 4. Since m(2, d) − 2d is a nonincreasing sequence, λd is a nondecreasing sequence. It is natural to give bounds on λd , for general polynomials in S(2, d). Bounds independent of d are required to prove Theorem 4.1. Before stating the general estimate for λ, we note the following: • m(2, 1) = 2 because the only polynomial in S(2, 1) is x + y and λ1 = 0.

4 Results of Coding and Consequences in Source Dimension 2

123

• m(2, 2) = 4 because the only polynomial in S(2, 2) is (x + y)2 and λ2 = 2. • m(2, 3) = 5 because the only polynomials in S(2, 3) are of the form (for 0 ≤ t ≤ 1) x 3 + y 3 + t3x y + (1 − t)(3x 2 y + 3x y 2 ) Hence f (1, 1) = 8 − 3t ≥ 5. The next proposition gives a sharp bound on the value of the coefficient λ of x y for f ∈ S(2, d). Except in low degrees, this sharp bound does not occur for a polynomial minimizing f (1, 1). Nonetheless, any bound on λ is good enough to establish the asymptotic result in Theorem 4.1. Proposition 4.2 Suppose f ∈ S(2, d). Let λ denote the coefficient of x y in f . Then λ≤4−

2 when d = 2r + 1 r +1

λ≤4−

2 when d = 2r. r

Thus, using the floor function, λ ≤ 4 −

2 . [ d+1 2 ]

This result is sharp.

Proof We use homogenization to write x y(x + y)d−2 λ + g(x, y) = (x + y)d − x d − y d ,

(4.5)

where g(x, y) has non-negative coefficients. For each k with 1 ≤ k ≤ d − 1, we require     d −2 d λ ≤ . k−1 k Writing this inequality in terms of factorials and simplifying yields λ≤

d(d − 1) k(d − k)

for each such k. When d = 2r + 1 is odd, the smallest value occurs when k = r +2 2 and we obtain λ ≤ 4rr +1 = 4 − r +1 . When d = 2r is even, the smallest value occurs when k = r and we obtain λ ≤ 2r (2rr 2−1) = 4 − r2 . In both cases, the denominator ]. The next example proves that the inequality is sharp.  equals [ d+1 2 Example 4.4 It is possible for equality to hold in this proposition. Put λ equal to the maximum value, and define g by g(x, y) = (x + y)d − x d − y d − (x + y)d−2 λx y.

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4 Monomial Sphere Maps and Linear Programming

Note that g is homogeneous of degree d. The polynomial given by f (x, y) = λx y + g(x, y) + x d + y d does the job, as f ∈ S(2, d). When d = 9, for example, we put 18 54 42 27 f (x, y) = x 9 + y 9 + x y + (x 8 y + x y 8 ) + (x 7 y 2 + x 2 y 7 ) + (x 6 y 3 + x 3 y 6 ). 5 5 5 5

Note that f (1, 1) =

274 5

> m(2, 9) =

577 35

here.

The following theorem follows from what we have done so far. It is the special case n = 2 of a general result (in all dimensions) that limd→∞ m(n,d) = n(n − 1) that we d will prove in the next section. Theorem 4.1 limd→∞

m(2,d) d

= 2.

Proof By Corollary 4.2, 2d + 2 − λd ≤ m(2, d) ≤ 2d and therefore 2+

2 λd m(2, d) − ≤ ≤ 2. d d d

(4.6)

By Proposition 4.2, λd is bounded independent of d. We may therefore let d tend to ∞ in (4.6) and conclude the result.  Remark 4.3 Assume 0 ≤ λ < 4. Then there is a smallest integer d = dλ such that there is a g ∈ S(2, d) with the coefficient of x y in g equal to λ. As λ increases to 4, dλ tends to infinity. If we could always find such a g which minimizes g(1, 1), then we would see that 2d − m(2, d) would tend to 2. Whether this conclusion holds is an open problem, and it is not clear whether the elaborate coding suggests it. Below we list the values of 2d − m(2, d) for 1 ≤ d ≤ 12. For all d, these values are rational numbers and are known exactly for d ≤ 201. In a subsequent list, we give nine place decimal expansions, including degrees satisfying 171 ≤ d ≤ 202. The author would very much like to know whether the limit of 2d − m(2, d) as d tends to infinity equals 2, and if it does not, what is its value. Remark 4.4 After looking at the values of 2d − m(2, d) up to degree 35, the author wondered whether the limit might be the golden ratio, which is approximately 1.618. Numbers related to the golden ratio arose in Exercise 3.6 when we computed pd (1, 1). This conjecture is false, however, as the value for d = 55 exceeds this number, and the sequence is nondecreasing.

4 Results of Coding and Consequences in Source Dimension 2

125

Open problem. Find the limit of 2d − m(2, d) as d tends to infinity. The first few values of 2d − m(2, d): d = 1. 2d − m(2, d) = 0 d = 2. 2d − m(2, d) = 0 d = 3. 2d − m(2, d) = 1 d = 4. 2d − m(2, d) = 1 d = 5. 2d − m(2, d) =

4 3

d = 6. 2d − m(2, d) =

4 3

d = 7. 2d − m(2, d) =

3 2

d = 8. 2d − m(2, d) =

3 2

d = 9. 2d − m(2, d) =

53 = 1.51428571428571 35

d = 10. 2d − m(2, d) =

53 = 1.51428571428571 35

d = 11. 2d − m(2, d) =

43 = 1.53571428571429 28

d = 12. 2d − m(2, d) =

43 = 1.53571428571429 28

On the next page, we list numerical values of both λd and 2d − m(2, d) for many more values of d. These numbers are rational, but the number of digits is so large that writing them down is not useful. The decimal expansions give a sense of the slow growth of these numbers. Note, however, in every degree that λd = 2d + 2 − m(2, d). We prove this equality in Corollary 4.3.

126

4 Monomial Sphere Maps and Linear Programming

For 3 ≤ d ≤ 46 and 171 ≤ d ≤ 202 here are the decimal values (d, λd , 2d − m(2, d)) (two per line). λd is the coefficient of x y in the polynomial realizing m(2, d). 3, 3.000000000, 1.000000000, 4, 3.000000000, 1.000000000, 5, 3.333333333, 1.333333333, 6, 3.333333333, 1.333333333, 7, 3.500000000, 1.500000000, 8, 3.500000000, 1.500000000, 9, 3.514285714, 1.514285714, 10, 3.514285714, 1.514285714, 11, 3.535714286, 1.535714286, 12, 3.535714286, 1.535714286, 13, 3.553896104, 1.553896104, 14, 3.553896104, 1.553896104, 15, 3.564814815, 1.564814815, 16, 3.564814815, 1.564814815, 17, 3.575396825, 1.575396825, 18, 3.575396825, 1.575396825, 19, 3.584861735, 1.584861735, 20, 3.584861735, 1.584861735, 21, 3.592763625, 1.592763625, 22, 3.592763625, 1.592763625, 23, 3.596362937, 1.596362937, 24, 3.596362937, 1.596362937, 25, 3.599509193, 1.599509193, 26, 3.599509193, 1.599509193, 27, 3.601413742, 1.601413742, 28, 3.601413742, 1.601413742, 29, 3.603922960, 1.603922960, 30, 3.603922960, 1.603922960, 31, 3.605849774, 1.605849774, 32, 3.605849774, 1.605849774, 33, 3.607745297, 1.607745297, 34, 3.607745297, 1.607745297, 35, 3.610180355, 1.610180355, 36, 3.610180355, 1.610180355, 37, 3.611744976, 1.611744976, 38, 3.611744976, 1.611744976, 39, 3.612615931, 1.612615931, 40, 3.612615931, 1.612615931,

4 Results of Coding and Consequences in Source Dimension 2

127

41, 3.613367747, 1.613367747, 42, 3.613367747, 1.613367747, 43, 3.614158840, 1.614158840, 44, 3.614158840, 1.614158840, 45, 3.614794227, 1.614794227, 46, 3.614794227, 1.614794227, ................................................................................................................ 171, 3.625628409, 1.625628409, 172, 3.625628409, 1.625628409, 173, 3.625657771, 1.625657771, 174, 3.625657771, 1.625657771, 175, 3.625689162, 1.625689162, 176, 3.625689162, 1.625689162, 177, 3.625712111, 1.625712111, 178, 3.625712111, 1.625712111, 179, 3.625741003, 1.625741003, 180, 3.625741003, 1.625741003, 181, 3.625769634, 1.625769634, 182, 3.625769634, 1.625769634, 183, 3.625800229, 1.625800229, 184, 3.625800229, 1.625800229, 185, 3.625828976, 1.625828976, 186, 3.625828976, 1.625828976, 187, 3.625851010, 1.625851010, 188, 3.625851010, 1.625851010, 189, 3.625873450, 1.625873450, 190, 3.625873450, 1.625873450, 191, 3.625896653, 1.625896653, 192, 3.625896653, 1.625896653, 193, 3.625919915, 1.625919915, 194, 3.625919915, 1.625919915, 195, 3.625942073, 1.625942073, 196, 3.625942073, 1.625942073, 197, 3.625969137, 1.625969137, 198, 3.625969137, 1.625969137, 199, 3.625995692, 1.625995692, 200, 3.625995692, 1.625995692, 201, 3.626016659, 1.626016659, 202, 3.626016659, 1.626016659

128

4 Monomial Sphere Maps and Linear Programming

Here is another result obtained from coding. For degree 35, the polynomial that minimizes f (1, 1) is given by ⎛ f (x, y) = x 35 + y 35 + (x y)g(x, y) = x 35 + y 35 + (x y) ⎝λ +

33 

⎞ c[ j](x j + y j )⎠ ,

j=4

where g(x, y) has the following ridiculous formula: g(x, y) =

+

6577750203643 33 4444879050537 32 80542310068451 31 (x + y 33 ) + (x + y 32 ) + (x + y 31 ) 6582868396338 4388578930892 88868723350563

123681728937569 30 440190222037753 28 16644755158064 27 (x + y 30 ) + (x + y 28 ) + (x + y 27 ) 88868723350563 177737446701126 88868723350563 +

8589095073713 24 3002335599446 21 6246963922595 25 (x + y 25 ) + (x + y 24 ) + (x + y 21 ) 2278685214117 9114740856468 1001740842921

+

420183145217 20 196888318759 16 2051312808644 15 (x + y 20 ) + (x + y 16 ) + (x + y 15 ) 286211669406 41715061128 9015667586289

+

+

84692790557474 11 269590266034553 10 (x + y 11 ) + (x + y 10 ) 23040039387183 138240236323098

2185836788798179 5 1051926854589677 4 (x + y 5 ) + (x + y 4 ). 414720708969294 2488324253815764 +

17966598676264183 . 4976648507631528

The value of the coefficient of (x y) is the last fraction, whose value to ten places is 3.6101803551. The set of j for which the coefficient of x y(x j + y j ) is not 0 is {0, 4, 5, 10, 11, 15, 16, 20, 21, 24, 25, 27, 28, 30, 31, 32, 33}. At the end of Sect. 8, we pose an open problem about how this set depends on the degree. For now we mention, in degree 121, the non-zero coefficients of the polynomial g are those in the following list. The 0 corresponds to x y, the 4 corresponds to x y(x 4 + y 4 ), etc. 0 4 5 10 11 16 17 22 23 28 29 34 35 39 40 45 46 50 51 56 57 61 62 66 67 70 71 75 76 79 80 83 84 87 88 91 92 94 95 97 98 100 101 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119

4 Results of Coding and Consequences in Source Dimension 2

129

We comment also about the numerical values. After proving Theorem 4.3, we then discuss the number of terms. In degree 121, the coefficient λ of x y is a rational number. When reduced to lowest terms, the numerator of λ is approximately 2.38172 × 10259 and the denominator is approximately 6.57125 × 10258 . The value of λ is approximately 3.6244546525984083525. Remark 4.5 Consider the relationship to sparseness. Let N denote the number of terms. In degrees 3, 5, 9, 11, 13, the maps minimizing f (1, 1) and thus realizing m(2, d) have 3, 5, 9, 11, 13 terms, respectively. Thus, N = d in these cases. In degree 7 equality fails, as N = 5, which is smaller than the degree. There is an extra coefficient equal to zero in degree 7. Corollary 4.5 shows, when the degree is odd and at least 3, that the number of terms in the polynomial realizing m(2, d) has at most d terms. When d is even the number of terms is at most d + 1. In degree 35, the map minimizing f (1, 1) and thus realizing m(2, 35) has 35 terms. In degree 121, the map minimizing f (1, 1) and thus realizing m(2, 121) has 121 terms. The terms x 121 + y 121 contribute two, the x y term contributes one, and the remaining 59 terms in the above list contribute two each. The total is 121. The sparse solution has 62 terms. By Theorem 3.2 of Chap. 3, the minimum number of terms N (d) is the ceiling . By Theorem 4.1, the ratio m(2,d) tends to 2. of d+3 2 d Open problem. Find all degrees for which the optimal polynomial has fewer terms than its degree. Perhaps 7 is the only such odd degree and 8 the only such even degree. We have obtained a decisive relationship between the two minimization problems, which we summarize as a general theorem. Theorem 4.9 gives the corresponding result in higher dimensions. Theorem 4.2 Let m(2, d) denote the minimum of f (1, 1) for f ∈ S(2, d) and let N (2, d) denote the minimum number of terms for a polynomial satisfying the first two properties of Definition 4.1. Then lim

d→∞

m(2, d) 2m(2, d) = lim = 4. N (2, d) d→∞ d + 3

(4.7)

We obtain an asymptotic relationship between solutions to the two minimization problems. An analogous result holds in higher dimensions, see Theorem 4.9. The next result shows how to significantly reduce the search space for finding m(2, d). For a minimizer, the coefficient of any monomial divisible by (x y)2 must vanish. Corollary 4.3 is especially interesting. Theorem 4.3 Let f be a polynomial in S(2, d) for which f (1, 1) is minimal. Then there are no non-zero monomials in f (x, y) that are divisible by (x y)2 . As a consequence, d−1  f (x, y) = x d + y d + λx y + c[ j](x j y + x y j ) (4.8) j=2

for non-negative values of the constants λ and c[ j].

130

4 Monomial Sphere Maps and Linear Programming

Proof We give the proof for d odd; the case for d even is similar. Put d = 2r + 1. We may assume that f is symmetric. Every symmetric polynomial in S(2, d) can be written f (x, y) = x d + y d + Ax y +

d−1 

C[ j](x j y + x y j ) + (x y)2 g(x, y)

(4.9)

j=2

for some symmetric polynomial g with non-negative coefficients. Lemma 4.6 implies  that A + j C[ j] = d. Again using Lemma 4.6, f (1, 1) = 2 + A + 2



C[ j] + g(1, 1) = 2 + 2d − A + g(1, 1).

(4.10)

Assume that g does not vanish identically in (4.9). For 0 ≤ t ≤ 1, consider the family of maps Ft (x, y) = x d + y d + A(t)x y +

d−1 

C[ j, t](x j y + x y j ) + t (x y)2 g(x, y). (4.11)

j=2

We want each Ft ∈ S(2, d). We will solve for those C[ j, t] for r + 1 ≤ j ≤ 2r . We do so by homogenization and then equating coefficients of the terms x 2r +1− j y j for 1 ≤ j ≤ r . Let G(x, y) be the homogenization of (x y)2 g(x, y). We note that A(t) and the other C[ j, t] are not uniquely determined by t. In particular, the notation A(t) allows any value of the coefficient of x y for which (4.11) is in S(2, d). In order for (4.11) to be in S(2, d), we require

A(t)x y(x + y)d−2 +

2r 

C[ j, t](x j y + x y j )(x + y)d− j−1 = (x + y)d − x d − y d − t G(x, y).

j=2

(4.12) We break up the sum on the left-hand side of (4.12) into two parts as r 

C[ j, t](x j y + x y j )(x + y)d− j−1 +

2r 

C[ j, t](x j y + x y j )(x + y)d− j−1 .

j=r +1

j=2

(4.13) For each j with r + 1 ≤ j ≤ 2r , we will equate coefficients of the term x 2r +1− j on the left-hand side of (4.12) with the right-hand side of (4.12). We have already seen, when j = 2r , that C[2r, t] satisfies the equation A[t] +

2r  j=1

C[ j, t] = d.

4 Results of Coding and Consequences in Source Dimension 2

131

The coefficient of C[2r, t] is 1, and we can therefore solve for C[2r, t] and eliminate it from the problem. Equating the coefficients of x 2r −1 y 2 leads to an equation where C[2r, t] does not appear and the coefficient of C[2r − 1, t] again equals 1. Again we solve for C[2r − 1, t]. We continue to do so until we solve for C[r + 1, t]. Plugging in these values yields the most general Ft of the form (4.11) that lies in S(2, d). By (4.10), Ft (1, 1) = 2d + 2 − A(t) + tg(1, 1). Our optimization problem is thus to minimize 2d + 2 − A(t) + tg(1, 1). Because C[ j, t] ≥ 0 for all j, decreasing t provides less of a restriction on the possible values of A(t). Hence, the maximum value of A(0) is at least as large as the maximum value of A(t) for any t ∈ [0, 1]. Since both A(t) and t are non-negative, we minimize by putting t = 0 and choosing the maximum possible value of A(0). The resulting function satisfies (4.8) for some choice of the constants. This proof does not say how to choose them, and the results of coding show how complicated it is to do so. See Section 8 for more information.  Corollary 4.2 For all d, the minimum m(2, d) is 2d + 2 − λ, where λ is the largest number for which there are non-negative c[ j] such that x + y + λx y + d

d

d−1 

c[ j](x j y + x y j ) = 1

j=2

on the line x + y = 1. Proof By the theorem, the minimizer f satisfies (4.8). Hence f (1, 1) = 2 + λ +

d−1 

c[ j].

j=2

 By Lemma 4.6, we can substitute c[ j] = d − λ and obtain f (1, 1) = 2d + 2 − λ. This value is minimized by maximizing λ. The conclusion follows.  This result does not tell us what this maximum λ = λd is. The results of coding do not tell us how to choose the c[ j] so as to maximize λd . Corollary 4.3 If f ∈ S(2, d) has f (1, 1) = m(2, d), then f satisfies (4.8). The fundamental theorem of linear programming (see, for example, [76]) combines nicely with Theorem 4.3 and Corollary 4.3. By Corollary 4.3 our optimization problem is equivalent to maximizing λ, the coefficient of x y. This coefficient satisfies a system of linear equations where the other unknowns are also non-negative. Let us suppose that d = 2r + 1 is odd, as the case for even d is similar. When d = 2r + 1 we have r linear equations in 2r variables (including λ), all assumed to be nonnegative, obtained by equating the coefficients in the usual homogenized equation analogous to (4.5):

132

4 Monomial Sphere Maps and Linear Programming

λx y(x + y)d−2 +

d−1 

c[k](x k y + x y k ) = (x + y)d − x d − y d .

k=2

The fundamental theorem of linear programming (Theorem 3.4 in [76]) guarantees that there is an optimal solution with at least r of the 2r − 1 unknowns c[k] equal to zero. Hence, the number of terms in this optimal polynomial is at most 2 + 1 + 2(r − 1) = 2r + 1, which is the degree. In this expression, the first 2 arises from x d + y d , the 1 arises from λx y, and the 2(r − 1) arises from the terms where c[k] = 0. We illustrated this conclusion in Remark 4.5, especially when we considered d = 121. Corollary 4.4 Assume d is odd and d ≥ 3. Then there is a polynomial p in S(2, d) with p(1, 1) = m(2, d) and with at most d terms. Remark 4.6 It seems to be true that the polynomial realizing m(2, d) is unique, but this conclusion has not been rigorously proved.

5 Monomial Sphere Maps in Higher Dimension In many problems regarding monomial sphere maps, dimension 2 plays a special role. For example, the remarkable polynomials of degree 2r + 1 discussed in detail in Chap. 3 have minimum target dimension r + 2 given the degree, thus providing sparse solutions to the linear system. In some degrees, these are the only sparse solutions; in infinitely many degrees, there are other sparse solutions. We discussed this matter in Sect. 7 of Chap. 3. In dimensions at least 4, the maps with minimum target dimension given the degree have a much simpler structure. Dimension 3 illustrates features both of dimension 2 and of dimensions 4 or more. See Example 4.7. Also, as we have seen in 2 dimensions, the problem of minimizing f (1) given the degree of f is remarkably complicated. In this section, we provide some examples and results in higher dimensions where things are, perhaps surprisingly, much simpler. We will compute m(n, d) for all n at least 3. We recall some notation and introduce certain symmetric polynomials that achieve this value.  Let x denote the real n-tuple (x1 , . . . , xn ). Recall that s(x) = nk=1 xk . We define a sequence of symmetric polynomials as follows: g0 (x) = 0. gd+1 (x) = gd (x) +

n  j=1

⎛

⎞  ⎝x j (xk )d ⎠ . k= j

5 Monomial Sphere Maps in Higher Dimension

133

Recall that 1 denotes the n-tuple of all ones. Thus gd+1 (1) = gd (1) + n(n − 1). When n = 3, put x = (x, y, z) and we have g1 (x, y, z) = x(y + z) + y(x + z) + z(x + y) = 2(x y + x z + yz). g2 (x, y, z) = x(y 2 + z 2 ) + y(x 2 + z 2 ) + z(x 2 + y 2 ) + 2(x y + x z + yz). g3 (x, y, z) = g2 (x, y, z) + x(y 3 + z 3 ) + y(x 3 + z 3 ) + z(x 3 + y 3 ). Next, we define a sequence of polynomials f d by f d (x) =

n  (x j )d + gd−1 (x). j=1

Again, when n = 3, we list the first few of these polynomials: f 1 (x, y, z) = x + y + z. f 2 (x, y, z) = x 2 + y 2 + z 2 + 2(x y + x z + yz). f 3 (x, y, z) = x 3 + y 3 + z 3 + x(y 2 + z 2 ) + y(x 2 + z 2 ) + z(x 2 + y 2 ) + 2(x y + x z + yz).

When n = 3 we have f d (x, y, z) = x d + y d + z d +

d−1  (x j (y + z) + y j (x + z) + z j (x + y)). j=1

Thus f d (1, 1, 1)=3 + 6(d − 1); we will show that these polynomials realize m(3, d). There is an intuition underlying this definition. Recall the restricted tensor product operation discussed in Chap. 2, Sect. 5. In the monomial setting, this operation amounts to the following idea. Suppose a real polynomial is written f = A + B. Then we replace f (x) by s(x)A(x) + B(x). We do not change the value of f on the set where s(x) = 1. Assume f is symmetric and f = A + B. When A is symmetric, the resulting function s A + B is also symmetric. This process is thus a symmetrized version of the method for generating Whitney maps. For n ≥ 3, the polynomials in the subsequent proposition satisfy a remarkable uniqueness property, to be shown for n = 3 in Theorem 4.4. Proposition 4.3 For each source dimension n, there is a polynomial f d (x) satisfying all of the following properties: • f d (x) = 1 on the hyperplane s(x) = 1. • The coefficients of f d are non-negative.

134

• • • •

4 Monomial Sphere Maps and Linear Programming

f d has degree d. ( f d ◦ σ)(x) = f d (x) for all permutations σ of n letters. f d (1) = n + (d − 1)n(n − 1). f d satisfies the explicit formula (4.14).

Proof We could define f d recursively as above and the proofs would follow by induction on the degree. It is a bit cleaner to define f d directly by f d (x) =

n 

x dj +

n  d−1   ( x j )( xkl ). j=1 j=k

j=1

(4.14)

l=1

We write ∼ = to denote equality on the hyperplane given by s(x) = 1. Then we have f d (x) ∼ =

n  j=1

x dj +

n d−1 n n n      (1 − xk )( xkl ) ∼ x dj + (xk − xkd ) = xk ∼ = = 1. k=1

l=1

j=1

k=1

j=1

In this calculation, we have used the finite geometric series (1 − xk )

d−1 

xkl = xk − xkd .

(4.15)

l=1

Formula (4.14) makes it evident that the coefficients of f d are non-negative that f has degree d, and that f is symmetric. Replacing all the variables by 1 shows that f d (1) = n + (d − 1)n(n − 1). The formula for Bd = f d (1) can be found without formula (4.14) as follows: Bd+1 = f d+1 (1) = n + gd (1) + n(n − 1) = n 2 + f d (1) − n = n 2 − n + Bd . (4.16) Equation (4.16) is an easy recurrence relation. The solution must be of the form Bd = αd + β for some numbers α, β. Using B1 = n and B2 = n(n − 1) + n = n 2 , we see that α = n 2 − n and β = 2n − n 2 . The solution to the recurrence is thus given  by Bd = n(n − 1)d + 2n − n 2 = n + n(n − 1)(d − 1). Corollary 4.5 For all n, d, we have m(n, d) ≤ n + n(n − 1)(d − 1). (Later we will show that equality holds when n ≥ 3.) Let Bd = f d (1). When n = 2 we obtain the sequence Bd = 2d, which appeared as an upper bound in Lemma 3.3. When n = 3 we get the sequence 6d − 3 which minimizes the value of f (1), as shown in Theorem 4.4. The first part of this result makes no assumption on the non-negativity of the coefficients. Recall, by definition, for j a positive integer, Sym(x j y) =

1 j (x y + x j z + y j x + y j z + z j x + z j y). 3!

5 Monomial Sphere Maps in Higher Dimension

135

We will multiply by 3! = 6 in our definition of the crucial polynomial f below. When j = 1, this formula gives x y + x z + yx + yz + zx + zy = 2(x y + x z + yz). The proof of minimization follows the same line of reasoning as that of Theorem 4.3. Theorem 4.4 For each integer d ≥ 2, there is a unique polynomial of the form f (x, y, z) = x + y + z + d

d

d

d−1 

6c[ j]Sym(x j y)

(4.17)

j=1

such that f (x, y, z) = 1 when x + y + z = 1. In fact, c[ j] = 1 for all j and thus f (1, 1, 1) = 6d − 3. Furthermore, f minimizes g(1, 1, 1) over all g ∈ S(3, d) and hence m(3, d) = 6d − 3. Proof Suppose that f satisfies (4.17) and equals 1 on x + y + z = 1. We homogenize f to obtain (x + y + z)d = x d + y d + z d +

d−1 

6c[ j]Sym(x j y)(x + y + z)d− j−1 .

(4.18)

j=1

Using the definition of Sym, we get (x + y + z)d = x d + y d + z d +

d−1 

c[ j] (x j (y + z) + y j (x + z) + z j (x + y))(x + y + z)d− j−1 .

j=1

This equation holds for all x, y, z. First put z = −y. After simplifying, we obtain x d = x d + y d (1 + (−1)d ) +

d−1 

  c[ j] y j (x − y) + (−y) j (x + y))x d− j−1 .

j=1

After dropping the term x d from both sides, and writing the sum as two sums, depending on the parity of the index j, we obtain 0 = y d (1 + (−1)d ) + 2

 even

c[ j]y j x d− j − 2



c[ j]y j+1 x d− j−1 .

odd

We group in powers of x. When d = 2r + 1 is odd, we obtain 0=2

r 

(c[2k] − c[2k − 1])x 2r +1−2k y 2k .

k=1

When d = 2r is even we obtain  (c[2k] − c[2k − 1])x d−2k y 2k . 0 = 2y d − 2 k

(4.19)

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4 Monomial Sphere Maps and Linear Programming

These polynomial identities are satisfied for all x, y. In each case, putting y = 1 gives a polynomial in x which must be constant. Equating coefficients yields the equations c[2] = c[1], c[4] = c[3], . . . , c[2r ] = c[2r − 1] when d = 2r + 1. When d = 2r is even, we obtain c[2] = c[1], c[4] = c[3], . . . , c[2r − 2] = c[2r − 3], c[2r − 1] = 1. In both cases, Proposition 4.3 (when n = 3) provides a solution where c[ j] = 1 for all j. To establish uniqueness, consider a solution where the c[ j] are unknown, but satisfy these equations. We homogenize it, and subtract it from the homogenized solution defined in Proposition 4.3. We obtain a polynomial in x, y, z that is identically 0. Put b[ j] = c[ j] − 1. Our goal is to show that b[ j] = 0 for all j. We prove this statement by a kind of backward induction as follows. Put T j (x, y, z) = x j (y + z) + y j (x + z) + z j (x + y) and put s(x, y, z) = (x + y + z). Note that T j (1, 1, −2) = −2 + (−1) j 2 j+1 is not 0 since j = 0. We are given that the following holds for all x, y, z: 0=

d−1 

b[ j]T j (x, y, z)s(x, y, z)d− j−1 .

(4.20)

j=1

The only term not divisible by s(x, y, z) is the term when j = d − 1. Therefore, we put s(x, y, z) = 0 and obtain 0 = b[d − 1]Td−1 (x, y, z). Evaluating at (x, y, z) = (1, 1, −2) gives b[d − 1] = 0. Now we plug this value into (4.20) and obtain a similar identity, except that the sum goes only to d − 2 and every term is divisible by s(x, y, z). Divide by s(x, y, z) and then set s(x, y, z) = 0. We obtain 0 = b[d − 2]Td−2 (x, y, z), and as before b[d − 2] = 0. Proceeding in this fashion, or using the method of descent, we conclude that b[ j] = 0 for all j. Therefore c[ j] = 1 for all j. It remains to prove that f (1, 1, 1) = m(3, d). Choose an arbitrary symmetric g ∈ S(3, d). Write A = 2c[1]. There are then polynomials h and r with non-negative coefficients such that we may write g(x, y, z) = x d + y d + z d + A(x y + x z + yz)+

5 Monomial Sphere Maps in Higher Dimension d−1 

137

  c[ j] (x j (y + z) + y j (x + z) + z j (x + y) + zh(x, y, z) + (x y)2 r (x, y).

j=2

Following the proof of Theorem 4.3, we consider the polynomial gt defined by g(x, y, z) = x d + y d + z d + A(t)(x y + x z + yz)+ d−1 

  c[ j, t] (x j (y + z) + y j (x + z) + z j (x + y) + t (zh(x, y, z) + (x y)2 r (x, y)).

j=2

Setting z = 0 yields a polynomial that is 1 on x + y = 1. By Lemma 4.6, we have d = A + d−1 j=2 c[ j]. Therefore g(1, 1, 1) = 3 + 3A + 6

d−1 

c[ j] + t (h(1, 1, 1) + r (1, 1)) =

j=2

3 + 3A + 6(d − A) + t (h(1, 1, 1) + r (1, 1)) = 6d + 3 − 3A + t (h(1, 1, 1) + r (1, 1)).

As in the proof of Theorem 4.3, we minimize by setting t = 0. This time, however, A is completely determined. It must equal 2, namely, twice c[1]. By the first part of this theorem, when t = 0, the only possible minimizing polynomial is g(x, y, z) = x d + y d + z d + 2(x y + x z + yz) +

d−1  (x j (y + z) + y j (x + z) + z j (x + y)) j=2

and the conclusion follows.



Remark 4.7 We use the symbol A instead of 2c[1] to be consistent with the notation in Theorem 4.3. The factor of 2 arises because 6 times the symmetrization of x y in three variables is 2(x y + x z + yz). Suppose n = 2. The formulas from Section 4 show, even if we consider only polynomials of the form in Theorem 4.4, that the coefficients are not unique. When n = 3, however, we are able to obtain additional equations on the c[ j]. After using them, setting z = 0 reduces to two dimensions. Now the coefficients become unique. Remark 4.8 Let us consider the first part of this result from the point of view of linear algebra. The number of unknown coefficients is d − 1. The number of equations they must satisfy depends on the degree, but this number is larger than d − 1 for d large enough. Assume n = 3. Both existence and uniqueness of the solution seem a bit surprising. Since there is a solution, the equations cannot all be independent. The additional equations c[1] = c[2], c[3] = c[4] and so on manage to create an invertible linear system. When n = 2, however, uniqueness fails.

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The uniqueness result from Theorem 4.4 holds because we assumed that only certain symmetric polynomials appear. Arbitrary symmetric elements of S(3, d) can have other terms, but they will not be minimizers unless they are f d (x, y, z). The next example considers the minimization problem without using Theorem 4.4. Imagine how hard the problem would be if we did not have this theorem! Example 4.5 We provide a two-parameter family Fbc of symmetric polynomials in three variables such that Fbc = 1 on x + y + z = 1. Fbc (x, y, z) = x 4 + y 4 + z 4 + c(x yz) + b(x y + x z + yz)+   c 5b − (x 2 y + x y 2 + x 2 z + y 2 z + x z 2 + yz 2 )+ 6− 2 2    c  3b (x + y)z 3 + y 3 (x + z) + x 3 (y + z) + + −2 + 2 2    3b c  2 (x + y 2 )z 2 + y 2 (x 2 + z 2 ) + x 2 (y 2 + z 2 ) . −3 + + 2 2 Notice that Fbc contains terms not allowed for a minimizer. Let us see what happens. Evaluating at 1 yields the value Fbc (1) = 9 + 6b + 4c. Assuming that all the coefficients are non-negative, we obtain the constraints b ≥ 0, c ≥ 0, 5b + c ≤ 12, 3b + c ≥ 6. The linear programming problem of minimizing 9 + 6b + 4c given these constraints can be done by hand. The constraints define a quadrilateral in the b, c plane. The minimum value is 21, which occurs when b = 2 and c = 0. The corresponding polynomial becomes the polynomial f 4 from Proposition 3.6: x 4 + y 4 + z 4 + x 3 (y + z) + y 3 (x + z) + z 3 (x + y) +x 2 (y + z) + y 2 (x + z) + z 2 (x + y) + 2(x y + x z + yz). In order to establish asymptotic information in all dimensions, we need a lower bound for m(n, d). We will establish the following lower bound. Lemma 4.7 Assume that p ∈ S(n, d) and p is symmetric. Let λ be the coefficient of the quadratic term. Then n + n(n − 1)d − λ

n(n − 1) ≤ p(1). 2

Proof First assume that n ≥ 3. The hypotheses allow us to write

5 Monomial Sphere Maps in Higher Dimension

p(x) =



x dj + λ



x j xk +

j

139 d−1 

j

c[ j]Sym(x1 x2 ) + B(x)

(4.21)

j=2

In (4.21), the term B(x) consists of terms that are either divisible by x j xk xl for some j < k < l or are divisible by some (x j xk )2 . There are n(n−1) quadratic terms and 2 j there are n(n − 1) terms in the symmetrization of x1 x2 . After including the count of these terms and noting that the coefficients of B are non-negative, we have p(1) = n + λ

 n(n − 1) + n(n − 1) c[ j] + B(1) 2

≥n+λ

 n(n − 1) + n(n − 1) c[ j]. 2

(4.22)

Next we evaluate where x3 = x4 = ... = xn = 0. We write (x, y) for (x1 , x2 ). The resulting polynomial p(x, y) lies in S(2, d). By Lemma 4.6 we have λ+

d−1 

c[k] = d.

k=2

We therefore eliminate the constants c[ j] to obtain p(1) ≥ n + λ n+λ

 n(n − 1) + n(n − 1) c[ j] = 2

n(n − 1) n(n − 1) + (d − λ)n(n − 1) = n − λ + dn(n − 1). 2 2

(4.23)

Thus, the right-hand side of (4.23) is the desired lower bound for m(n, d). Finally, we note that this bound holds trivially when n = 1 and agrees with Proposition 4.1 and Corollary 4.2 when n = 2.  We now have both upper and lower bounds for m(n, d) in all dimensions. Theorem 4.5 For all n, d, we have n−λ

n(n − 1) + dn(n − 1) ≤ m(n, d) ≤ n + (d − 1)n(n − 1). 2

(4.24)

As a consequence of (4.24), we obtain lim

d→∞

m(n, d) = n(n − 1). d

Proof Lemma 4.7 provides the left-hand inequality in (4.24) and Corollary 4.5 provides the right-hand inequality. We observe that the case n = 1 is trivial because

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4 Monomial Sphere Maps and Linear Programming

(4.24) then simply states 1 ≤ m(1, d) ≤ 1. Now, the coefficient λ of the quadratic term is bounded above by its value in dimension 2, which by Proposition 4.2 is less than 4. It is, of course, bounded below by 0. Hence, after dividing by d and letting d → ∞ in (4.24), we obtain the value of the limit.  We close this section by establishing the equality m(n, d) = n + (d − 1)n(n − 1) for n ≥ 3. The result is a corollary of Theorem 4.4, the special case when n = 3. Theorem 4.6 For each n ≥ 3 and each degree d, m(n, d) = n + (d − 1)n(n − 1). The value m(n, d) is achieved by the polynomial f (x) =

n 

x dj +

n  d−1   ( x j )( xkl ). j=1 j=k

j=1

l=1

Proof We proved this result when n = 3 in Theorem 4.4. Assume n ≥ 4. We write the variables as x, y, z, w, where w = (w4 , ..., wn ). Suppose that g(x, y, z, w) is a symmetric element of S(n, d). Then we can write, for polynomials h and R with non-negative coefficients, g(x, y, z, w) = x d + y d + z d +

n d−1   (w j )d + c[ j]n!Sym(x j y) + (x y)2 h(x, y) + R, j=4

j=1

where R is in the ideal generated by z, w4 , ...wn . Setting z = 0 and w j = 0 for all j gives a polynomial of the in Lemma 4.3. Hence, we have d = 2c[1] + form d−1 d−1 c[ j]. Thus c[1] = d − c[ j]. Plug this value into g. Reasoning as in j=2 j=1 the proof of Theorem 4.3, we see that h and R must vanish if g is a minimizer. Now set each w j equal to 0 to get a polynomial in three variables. By Theorem 4.4, each c[ j] = 1. Put z and w back into g. Notice that the symmetrization of x j y has n(n − 1) terms for each j. Since c[ j] = 1 for each j, the sum of these terms contributes n(n − 1)(d − 1) to g(1). There are n other terms (the pure monomials of degree d). Thus g(1) = n + n(n − 1)(d − 1). Since h and R vanish, and f must be symmetric, it follows (when n ≥ 3) that f (x) =

n  j=1

x dj +

n  d−1   ( x j )( xkl ). j=1 j=k

l=1

6 Sparseness in Source Dimension 2

141

6 Sparseness in Source Dimension 2 According to item (5) of Theorem 3.1, or equivalently, Corollary 3.3, for each odd d there is a monomial sphere map of degree d with source dimension 2 and target . Our goal in this section is to sketch the complicated proof of dimension N = d+3 2 Theorem 3.2, stating that the bound d ≤ 2N − 3 is best possible. This complicated proof from [22] provides a method for choosing the sparse solution to a linear system. Let p(x, y) denote a polynomial of degree d = 2r + 1 with non-negative coefficients. Assume that p(x, y) = 1 on the line x + y = 1. Then the polynomial p(x, y) − 1 is divisible by x + y − 1. We call the quotient q(x, y). Then q is of degree d. Let G p denote the Newton diagram for q. Thus, G p is the collection of lattice points (a, b) for which the coefficient of x a y b in q is not zero. If the coefficient is positive we label the lattice point with a P, if it is negative we label it with a N, and if zero with a Z. When the point is labeled P we draw directed arrows upward and to the right from (a, b) to (a, b + 1) and to (a + 1, b). When the point is labeled N we draw these arrows from (a, b + 1) and (a + 1, b) to (a, b). We call a point in G p a source if all arrows drawn at that point lead away from it, and we call it a sink if all arrows drawn at that point lead into it. The non-negativity of the coefficients of p implies that the origin is the only source. The first important observation is that the number of terms in p is at least as large as the number of sinks in G p . The reason is that a sink at (a, b) means that the coefficients of x a+1 y b and x a y b+1 in p(x, y) must be positive, as no cancellation can occur. Next one observes that the graph G corresponding to (x + y)d has each lattice point (a, b) labeled P when a + b ≤ d − 1. The reason is simply that the geometric series gives (x + y)d − 1 = 1 + (x + y) + (x + y)2 + · · · + (x + y)d−1 . x +y−1 The partial tensor product operation, which is crucial here, amounts to homogenization. Given any p, multiplying the terms of degree of k by (x + y)d−k for each k homogenizes p. The process of homogenization can be reversed to reach a given polynomial p. We next illustrate the effect of dehomogenization when we pass from (x + y)5 to the sparse polynomial p(x, y) = x 5 + 5x 3 y + 5x y 2 + y 5 . What happens to the number of sinks in the directed graphs when p is a solution of degree d with the minimum number of terms? Assume first that d = 2r + 1 is odd. Then G has 2r + 2 sinks, the points (a, b) where a + b = d. As we dehomogenize, we change the picture. By [22] the following hold. The points (d − 1, 0) and (0, d − 1) must have the labels P and hence there are sinks at (d, 0) and (0, d). These correspond to the terms x d and y d in p. The remaining 2r sinks may coalesce down to r sinks, but no further. We conclude that the total number of sinks is therefore at least 2 + r , and hence the number of terms N in p is a least 2 + r . Thus N ≥ 2 + r

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4 Monomial Sphere Maps and Linear Programming

and hence 2r + 1 = d ≤ 2N − 3. The group-invariant polynomials from Chapter 3 show that this bound is realized. We illustrate with the example given by p(x, y) = x 5 + 5x 3 y + 5x y 2 + y 5 . We begin with (x + y)5 and dehomogenize until we get to p. We use → to denote the effect of either rewriting terms or setting x + y equal to 1. (x + y)5 = x 5 + 5x 4 y + 10x 3 y 2 + 10x 2 y 3 + 5x y 4 + y 5 → x 5 + 5x 3 y(x + y) + 5x 3 y 2 + 5x 2 y 3 + 5x 2 y 3 + 5x y 4 + y 5 → x 5 + 5x 3 y + 5x 2 y 2 (x + y) + 5x y 3 (x + y) + y 5 → x 5 + 5x 3 y + 5x 2 y 2 + 5x y 3 + y 5 → x 5 + 5x 3 y + 5x y 2 (x + y) + y 5 → x 5 + 5x 3 y + 5x y 2 + y 5 . The polynomial q corresponding to p is given by q(x, y) = 1 + x + x 2 + x 3 + x 4 + y + y 2 + y 3 + y 4 + 2x y − 2x y 2 − x y 3 + 3x 2 y + x 2 y 2 − x y 3 .

There are sinks at the points (5, 0), (3, 1), (1, 2), and (0, 5). Two of the original six sinks remain; the other four coalesced into two. The total number of sinks in the graph G p is four, corresponding to the solution of the linear system with the fewest number of terms. When d is even, it is easy to see that the minimum number of terms arises also from these examples. One simply multiplies either x d−1 or y d−1 by x + y to create an example with one more term and of degree d. Notice that this procedure differs from the procedure used in Lemma 4.4. There we multiplied x d + y d by x + y; here, we multiply only one of these terms by x + y. In the language of restricted tensor products, in Lemma 4.4, we are tensoring on a two-dimensional subspace, and here we are tensoring on a one-dimensional subspace. In both situations, however, the process of going from odd degree to the next even degree is much simpler than the process of going from an even degree to the next odd degree.

7 Sparseness in Source Dimension at Least Three Our first example illustrates a property peculiar to dimension 3. We then state two sparseness results from [57]. One result holds for dimensions at least 3. The other requires dimension at least 4. Example 4.6 We write (x, y, z) for the variables in dimension 3. Consider the following maps with source dimension 3:

7 Sparseness in Source Dimension at Least Three

143

f (x, y, z) = (x + z)3 + 3(x + z)y + y 3

(4.25)

g(x, y, z) = x + y + z(x + y) + z 2 (x + y + z).

(4.26)

We easily check that both f and g correspond to monomial sphere maps. Put z = 1 − x − y; both (4.25) and (4.26) evaluate to 1. Each has degree 3 and each has 7 terms. Consider the linear system from (3.3) when n = 3 and d = 7. Assuming that there is no constant term, this system has 19 unknown coefficients which satisfy 10 independent equations. Assuming that at least one coefficient of a term of degree 3 is non-zero, one can check directly (or use Theorem 2.3) that there must be at least three such non-vanishing coefficients. By examining these equations, and doing lots of case analysis, one can see that all solutions of degree 3 must have at least 7 terms. Hence, both f and g are sparse solutions. The solution f arises from the sparse solution in dimension 2 by replacing x by x + z and leaving y alone. Using the notation of Theorem 3.1, f (x, y, z) = p3 (x + z, y). The solution g is an example of a Whitney polynomial, as discussed below; it arises by starting with x + y + z and tensoring as follows: (x + y + z) → x + y + z(x + y + z) = x + y + x z + yz + z 2 → x + y + x z + yz + z 2 (x + y + z) = g(x, y). In dimension 2, the sparse polynomial p3 is not obtained by tensoring. In dimension 4 or more, all sparse examples are obtained by tensoring. We see that dimension 3 exhibits aspects of both dimension 2 and of dimensions at least 4. Remark 4.9 Whitney introduced the map W : Rn → R2n−1 defined by W (x1 , x2 , ..., xn ) = (x1 , x2 , ...xn−1 , x1 xn , ..., xn2 ). This map is proper. For us, it is the simplest example of the restricted tensor product operation for monomial sphere maps. One often calls maps formed by iterating this idea Whitney maps. We iterated the tensor product operation in Theorem 2.12. We have also noted that not every monomial sphere map is obtained in this fashion. We next state two of the results from [57]. Theorem 4.7 (Lebl-Peters) Let p : Rn → R be a polynomial of degree d with nonnegative coefficients such that p(x) = 1 on s(x) = 1. Let N denote the number of distinct monomials in p. Assume n ≥ 3. Then N ≥ 1 + d(n − 1) and this result is sharp.

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Theorem 4.8 (Lebl-Peters) Let p : Rn → R be a polynomial of degree d with nonnegative coefficients such that p(x) = 1 on s(x) = 1. Let N denote the number of distinct monomials in p. Assume n ≥ 4. Then N = 1 + d(n − 1) if and only if p is a generalized Whitney polynomial. For completeness, we remind the reader of Theorem 3.2. In dimension 2, the . This value is realized by the remarkable sphere maps sharp inequality is N ≥ d+3 2 discussed in Chap. 3. On the other hand, it is unknown in which dimensions these maps are the only sharp examples. We also recall that dimension 1 is completely different, as d can be arbitrarily large when n = N = 1. We do not give a formal definition of Whitney polynomial, but we have seen the idea many times thus far. Let p be any polynomial satisfying p(x) = 1 on s(x) = 1. If we write p = f + g, then f + gs = p + g(s − 1) is also 1 there. When p has non-negative coefficients, the trick is to choose g such that p + g(s − 1) also has non-negative coefficients. In this case, we are applying the restricted tensor product operation of Definition 2.6 from Chapter 2 to a monomial sphere map. A generalized Whitney polynomial is any polynomial obtained from s(x) by finitely many operations of the form f + g → f + g(s − 1) where all the polynomials involved have non-negative coefficients, and we always choose g to be a single monomial of highest degree. Suppose p has degree d and N terms. Write p = f + g, where g is a single monomial of degree d in p. If we perform the tensoring operation on g, then the polynomial p + g(s − 1) will have degree d + 1 and precisely N + n − 1 terms. Begin with s. Iterating this operation d − 1 times yields a polynomial with 1 + d(n − 1) terms. We obtain a family of polynomials pn,d satisfying the following: • pn,d is a generalized Whitney polynomial. • pn,d has degree d. • pn,d has N = 1 + d(n − 1) terms. Example 4.7 The polynomial g from Example 4.7 is a Whitney polynomial. Here are two more examples when n = 3: f (x, y, z) = x + y + x z + z 2 + zy 2 + z 2 y + x 2 yz + x y 2 z + x yz 2 . g(x, y, z) = (x + y)

d−1 

z j + zd .

j=0

The reader should note the simple proof that g = 1 on s = 1: replace x + y by (1 − z) and use the finite geometric series. The reader should also note that f has degree 4 and 9 terms; thus N = 9 = 1 + 4(3 − 1). Finally, g has 2d + 1 terms. We refer to the wonderful paper [57] for proofs. We mention that the directed p−1 Newton diagram of the quotient s−1 plays a crucial role, as it does when n = 2. By combining Theorem 4.6 with Theorem 4.7, we obtain a decisive general analogue of Theorem 4.2.

7 Sparseness in Source Dimension at Least Three

145

Theorem 4.9 Let m(n, d) denote the minimum value of f (1) for f ∈ S(n, d). Let N (n, d) denote the minimum number of terms for a polynomial with non-negative coefficients that is 1 on the hyperplane s(x) = 1. If n = 2, then lim

d→∞

m(n, d) = n. N (n, d)

(4.27)

If n = 2, the limit is 4. Proof When n = 1, the only polynomial in question is x d and therefore m(1, d) = N (1, d) = 1 for all d. We established in Theorem 4.2 when n = 2 that the limit is 4. tends to n(n − 1). (By Theorem 4.6, Assume n ≥ 3. Theorem 4.2 tells us that m(n,d) d m(n,d) n d−1 = + n(n − 1) .) By Theorem 4.7, N (n, d) = 1 + (n − 1)d for n ≥ 3. It d d d m(n,d) n(n−1) follows that N (n,d) tends to n−1 = n. Theorem 4.9 provides an asymptotic formula relating the solutions to both optimization problems. The limit of the ratios is the dimension n, except when n = 2.

8 The Optimal Polynomials in Degrees 9 and 11 We return to dimension 2 here in order to provide additional insight into the polynomial of degree d that minimizes f (1, 1). We illustrate in degrees 9 and 11, but the method works in all odd degrees and helps clarify what is going on. We pose an open problem and then prove Proposition 4.4, which provides considerable insight into the general situation. Example 4.8 Degree 9. We study the polynomial p in S(2, 9) whose value at (1, 1) . Recall that this polynomial is is minimal; here m(2, 11) = 577 35  x + y + xy 9

9

 123 6 5 9 7 4 4 5 7 + 3(x + y ) + (x + y ) + (x + y ) . 35 5 7

To find this formula, we use Theorem 4.3 and begin with the assumption that p(x, y) = x 9 + y 9 + A(x y) +

8 

c[k](x k y + x y k ).

k=2

We homogenize and equate coefficients in the identity (x + y)9 = x 9 + y 9 + A(x y)(x + y)9 +

8  k=2

c[k](x k y + x y k )(x + y)8−k .

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4 Monomial Sphere Maps and Linear Programming

We obtain four independent equations in eight unknowns. Using c[2], c[4], c[6], c[8] as parameters yields the four parameter family of solutions (EE9): x 9 + y 9 + (x 2 y + x y 2 )c[2] + (x 4 y + x y 4 )c[4] + (x 6 y + x y 6 )c[6] + (x 8 y + x y 8 )c[8]  +x y

−3 c[4] 17c[8] − c[2] + − c[6] + 4 2 4



9 7c[8] − +(x y + x y ) 2 2 7



7



 + (x 3 y + x y 3 )

21 3c[4] 5c[6] 21c[8] − + − 2 2 2 2



 −21 5c[6] 35c[8] + (x y + x y ) − + . 4 2 4 

5

5

The coefficient of c[2] in (EE9) is x 2 y + x y 2 − x y, which vanishes on the line x + y = 1. Replacing the term x 2 y + x y 2 by x y lowers the value of f (1, 1). Hence, in a minimizer, we may assume c[2] = 0. We plug c[2] = 0 in (EE9) and then look at the coefficient of c[4]. It is xy 3 + (x 4 y + x y 4 ) − (x 3 y + x y 3 ), 2 2 which also vanishes on the line x + y = 1. Replacing x 3 y + x y 3 by 23 ( x2y + x 4 y + x y 4 ) also decreases f (1, 1). Doing so results in the polynomial   5c[6] x 9 + y 9 + (x 6 y + x y 6 )c[6] + (x 8 y + x y 8 )c[8] + (x 4 y + x y 4 ) 7 + − 7c[8] 3



9 7c[8] − +(x y + x y ) 2 2 7

7



 + xy

11 c[6] 3c[8] − + 4 6 4



 −21 5c[6] 35c[8] − + . +(x y + x y ) 4 2 4 

5

5

We write c[6] = u and c[8] = v to work in the (u, v) plane. We have 5 constraints: u, v ≥ 0 v≤

9 7

21 + 5u − 21v ≥ 0 10u + 21 ≤ 35v.

8 The Optimal Polynomials in Degrees 9 and 11

147

The feasible region is a triangle with vertices (0, 97 ), (0, 1), and ( 56 , 97 ). Let A be the coefficient of x y in the above final formula for p(x, y). By Corollary 4.3 we must − u6 + 3v on the feasible have f (1, 1) + A = 2d + 2 = 20. The maximum of 11 4 4 6 9 123 region happens at ( 5 , 7 ). The value is 35 and we obtain the optimal polynomial  x 9 + y9 + x y

 123 6 9 + 3(x 4 + y 4 ) + (x 5 + y 5 ) + (x 7 + y 7 ) . 35 5 7

Example 4.9 Degree 11. We study the polynomial in S(2, 11) whose value at (1, 1) . Recall that this polynomial was determined by is minimal; here m(2, 11) = 573 28 coding to be  x 11 + y 11 + (x y)

 99 33 5 55 8 11 9 33 4 + (x + y 4 ) + (x + y 5 ) + (x + y 8 ) + (x + y 9 ) . 28 14 14 28 14

How could we find it by hand? As above, we use Theorem 4.3 to assume that the polynomial has the form in (4.28), where d = 11: f (x, y) = x d + y d + λx y +

d−1 

c[ j](x j y + x y j ).

(4.28)

j=2

There are 10 unknown coefficients. They satisfy five equations, obtained by equating coefficients in the homogenized equation: (x + y)

11

=x

11

+y

11

+ λx y(x + y) + 9

10 

c[ j](x j y + x y j )(x + y)10− j .

j=2

(4.29) The five equations are obtained by equating the coefficients of x k y 11−k for 1 ≤ k ≤ 5. These equations are independent, and we may parametrize the solutions v by v=u+

5 

tjwj,

j=1

where u is a particular solution and the w j form a basis for the null space. As in the degree 9 case, it becomes natural to let the coefficients c[10], c[8], c[6], c[4], c[2] be the arbitrary parameters in writing the general solution. (A similar result holds in every odd degree d.) The general solution is given by the following formulas:

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4 Monomial Sphere Maps and Linear Programming

121 1 17 − c[2] + c[4] − c[6] + c[8] − 31c[10], 4 2 4 5 21 153 3 c[10], c[3] = −66 − c[4] + c[6] − c[8] + 2 2 2 2 231 5 35 c[5] = − c[6] + c[8] − 63c[10], 4 2 4 −33 7 − c[8] + 21c[10], c[7] = 2 2 11 9 c[9] = − c[10]. 2 2

λ=

We dehomogenize (4.29) by setting the (x + y) terms equal to 1, and substitute these formulas into the result. We get the following elaborate expression (E E):   c[4] 17 121 − c[2] + − c[6] + c[8] − 31c[10] x 11 + y 11 + x y 4 2 4 +(x 2 y + x y 2 )c[2] + (x 4 y + x y 4 )c[4] + (x 6 y + x y 6 )c[6] + (x 8 y + x y 8 )c[8]   231 5 35 +(x 5 y + x y 5 ) − c[6] + c[8] − 63c[10] 4 2 4     11 9 +(x 9 y + x y 9 ) − c[10] + x 10 y + x y 10 c[10] 2 2  −33 7 7 7 − c[8] + 21c[10] +(x y + x y ) 2 2   5 21 153 3 3 3 c[10] . +(x y + x y ) −66 − c[4] + c[6] − c[8] + 2 2 2 2 By the results in this chapter, our goal remains to maximize the coefficient λ of x y given that all the coefficients in EE are non-negative. We have seen that the maximum of λ exists, and for d ≥ 2 it lies in the interval [2, 4). We can write 5  (EE) = H (x, y) + c[2k]h 2k (x, y) k=1

for polynomials H and h 2k . First, we note that the coefficient of c[2], namely, the polynomial h 2 (x, y) in the above, is x 2 y + x y 2 − x y. This term vanishes on the line x + y = 1, but its value at (1, 1) equals the positive number 1. We therefore decrease the value of the entire expression at (1, 1) by setting c[2] = 0. This tiny simplification always arises. We next look at the coefficient of c[4] in the nightmarish formula (EE). It is xy 3 + (x 4 y + x y 4 ) − (x 3 y + x y 3 ). h 4 (x, y) = 2 2

8 The Optimal Polynomials in Degrees 9 and 11

149

Again this expression vanishes on x + y = 1. We get something useful, namely, that x 3 y + x y3 =

1 2 x y + (x 4 y + x y 4 ) 3 3

(∗)

on the line x + y = 1, and the value at (1, 1) decreases from 2 to 53 . It therefore follows, in any degree at least 5, that we may use (*) to eliminate x 3 y + x y 3 in any optimal polynomial. In degree at least 5, it follows that c[3] = 0 for a polynomial minimizing the value at (1, 1). In degree 11, we use (*) to eliminate the terms x 3 y + x y 3 . The resulting polynomial is independent of c[4]. Hence, we may have assumed from the start that c[3] = c[4] = 0 and use the last line of (EE) to express c[4] in terms of c[6], c[8], c[10]. But setting c[3] and c[4] equal to 0 tells us that 21 153 5 c[10] 0 = −66 + c[6] − c[8] + 2 2 2 and determines c[6]. Plugging in its value yields a two-parameter family:    77 132 c[8] 2c[10] 21c[8] 153c[10] + (x 6 y + x y 6 ) + − + − 20 20 5 5 5 5   9c[10] 11 9 9 8 8 10 10 +(x y + x y ) + (x y + x y )c[8] + (x y + x y )c[10] − 2 2 

x 11 + y 11 + x y

 +(x 5 y + x y 5 )

−33 7c[8] 27c[10] − + 4 4 2



 + (x 7 y + x y 7 )

 −33 7c[8] − + 21c[10] . 2 2

Put u = c[8] and v = c[10]. Our problem has now been reduced to maximizing λ=

u 2v 77 + − , 20 20 5

given that each of the following inequalities holds: 0≤u 0≤v −33 7u 27v − + 4 4 2 153v 132 21u + − 0 ≤ c[6] = 5 5 5 −33 7u 0 ≤ c[7] = − + 21v] 2 2 11 9v − . 0 ≤ c[9] = 2 2 0 ≤ c[5] =

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4 Monomial Sphere Maps and Linear Programming

These inequalities determine the feasible region in the (u, v) plane, and the problem can be solved by evaluating the objective function at the vertices of the feasible region. 11 . We The maximum of the objective function occurs when c[8] = 0 and c[10] = 14 obtain the desired polynomial  x 11 + y 11 + (x y)

 99 33 5 55 8 11 9 33 4 + (x + y 4 ) + (x + y 5 ) + (x + y 8 ) + (x + y 9 ) . 28 14 14 28 14

Remark 4.10 Since the rank of the system is 5, and we have seen that c[2] must be 0, we know by Corollary 4.5 that there will be a solution where an additional four of the unknown c[ j] equal 0. We could try c[4] = c[8] = 0. Choosing these to vanish forces c[3] = c[7] = 0. In general, we could consider all the ways for setting half of the c[ j] equal to 0 and solve each of these easier problems. The optimal polynomial in degree 11 has c[ j] = 0 for j = 2, 3, 4, 7, 8. Earlier, for degree 121, we listed the numbers k for which the coefficient of x y(x k + y k ) is not 0. This coefficient is c[k + 1]. The first 12 of these values are precisely the numbers congruent to 4 or 5 modulo 6, but then the pattern breaks down. Thus, for small k, c[k] = 0 if and only if k is congruent to 5 or 6 modulo 6. The same pattern holds in other large degrees, and the elements of these sets are the same for awhile. For example, consider degrees 197 and 201. The first 62 values of k for which c[k] = 0 are the same; the lists differ slightly thereafter. It is natural to seek some sort of number-theoretic interpretation, perhaps asymptotically, so we pose the following problem. Open problem. For each odd degree d, determine the set of j for which c[ j] must be non-zero in the optimal polynomial. If doing so is too difficult, prove a stabilization result and explain it! We conclude this section with an interesting result; the author believes that this proposition should allow a deeper understanding of the problem of finding the polynomial of degree d that minimizes f (1, 1). The subtle idea is to change perspective as follows. Instead of regarding the unknown c[k] as coefficients, we regard as coefficients the polynomials that multiply the c[ j] when we expand. Proposition 4.4 Suppose d = 2r + 1 is odd and p(x, y) is as in Theorem 4.3: p(x, y) = x + y + A(x y) + d

d

d−1 

c[k](x k y + x y k ).

(4.30)

k=2

Assume that p(x, y) = 1 on the line x + y = 1. Then there are polynomials H (x, y) and h 2 j (x, y) for which the following hold: p(x, y) = H (x, y) +

r  j=1

h 2 j (x, y)c[2 j].

(4.31)

8 The Optimal Polynomials in Degrees 9 and 11

151

• For 1 ≤ k ≤ r , we have h 2k = 0 on the line x + y = 1. • Given two degrees d1 = 2r1 + 1 and d2 = 2r2 + 1 with d1 < d2 , the polynomial h 2k arising from degree d1 is the same as the corresponding polynomial arising from degree d2 . Proof Sketch. As usual homogenize (4.30) and set the result equal to (x + y)d . There are r independent equations in 2r unknowns. As before, it is convenient to regard the coefficients c[2k] as parameters and solve for the c[ j] for odd j in terms of them. (See the example in degree 11.) We obtain (4.31) for some polynomials. The key point is to show that the h 2k (x, y) vanish on the line x + y = 1. To see why, note that p(x, y) = 1 on the line, no matter how the c[ j] are chosen. Setting them all equal to 0 shows that H (x, y) = 1 there. Then setting all but one of them equal to 0, say c[2m], and using H (x, y) = 1 there, shows that the resulting polynomial h 2m must vanish on x + y = 1. It is not difficult to see that replacing d by d + 2 leads to the same type of linear system. The polynomial H (x, y) changes and a new polynomial  h 2r +2 arises, but the h 2k for k ≤ r are unchanged. Here are the first few polynomials from Proposition 4.4: p2 (x, y) = x 2 y + x y 2 − x y p4 (x, y) =

−3 3 xy (x y + x y 3 ) + + (x 4 y + x y 4 ) 2 2

5 5 p6 (x, y) = x 6 y + x y 6 + (x 3 y + x y 3 ) − (x y + (x 5 y + x y 5 )) 2 2 p8 (x, y) =

−17x y 21(x 3 y + x y 3 ) 35(x 5 y + x y 5 ) 7(x 7 y + x y 7 ) − + − + (x 8 y + x y 8 ). 4 2 4 2

We have used p2 and p4 , and the others can be used similarly to replace some terms in an example with some other terms and decrease the value of f (1, 1). We conjecture that combining Proposition 4.4 with the process used in degree 11 determines all optimal examples and which c[k] are non-zero for a given degree.

9 Coding We provide Lichtblau’s Mathematica code [59] for finding the polynomial that achieves m(2, d), after making a few comments about it. An earlier version of the code included in [20] includes the coefficients c[i, j] of x i y j . It does not take advantage of Theorem 4.3, which significantly decreases the search space. The improved version is quite similar to our usual (homogenized) Ansatz, as in (4.28) for example. The coefficient we call λ or A is written c[0]. One

152

4 Monomial Sphere Maps and Linear Programming

must decide whether to use x y or x y + yx. We have always used x y and not regarded its coefficient as one of the c[ j]. The code below equates the coefficient of x j y d− j in the homogenized polynod  mial to the binomial coefficient j . We achieve the same thing by equating the homogenized polynomial to (x + y)d . The code seeks to maximize the value of the coefficient λd of x y. Recall from Corollary 4.3 that λd + m(2, d) = 2d + 2 holds, and therefore maximizing λd is equivalent to minimizing f (1, 1). Here is a rather efficient version of the code. The program computes the maximum value of c[0] given that the homogenized version of (4.32) equals (x + y)d . The unknown polynomial is written as ⎛ x d + y d + (x y) ⎝c[0] +

d−2 

⎞ c[ j](x j + y j )⎠ .

(4.32)

j=1

minCoefficientTotalExact[d_Integer /; d > 1 && OddQ[d], {x_, y_}] := Module[ {c, r, vars, sum, poly, coeffs, ca, newvars, max, vals}, vars = Array[c, d - 1, 0]; poly = (x + y)ˆ(d - 2)*c[0] + Rest[vars] . Table[(xˆj + yˆj)*(x + y)ˆ(d - 2 - j), {j, 1, d - 2}]; coeffs = CoefficientList[poly /. y -> 1, x]; {max, vals} = Maximize[{c[0], Reduce[coeffs - Binomial[d, Range[d - 1]] == 0], Thread[vars >= 0]}, vars]; poly = x*y*poly + xˆd + yˆd /. (x + y) -> 1 /. vals; {d, max, poly /. (x + y) -> 1 /. {x -> 1, y -> 1}, poly} ]

Chapter 5

Groups Associated with Holomorphic Mappings

We begin by discussing five groups one can associate with a mapping between domains with interesting automorphism groups. After providing many examples, we study these groups for rational sphere maps. The first few sections closely follow Chap. 5 in [15] but most of the material is based on work the author did with Ming Xiao in [28] and [29]. Several new examples appear as well. We summarize the main results for rational sphere maps in Sect. 8. The most important group associated with a rational sphere map f is the Hermitian-invariant group  f , a subgroup of Aut(Bn ). We characterize when it is non-compact, when it is the unitary group, and when it contains an n-torus. We prove that each finite subgroup of Aut(Bn ) is  f for some rational sphere map f . The proofs rely on much of our earlier work. We also relate the invariant group G f to our work on group-invariant monomials from Chap. 3.

1 Five Groups Let 1 and 2 be open subsets of complex Euclidean spaces. For the sake of intuition, we suppose that both 1 and 2 admit interesting groups of automorphisms. We start with the group Aut(1 ) × Aut(2 ). Let f : 1 → 2 be a function. We will assign various groups to this mapping f . Definition 5.1 (Five groups) Let f : 1 → 2 be a holomorphic map. • • • • •

Let A f = {(γ, ψ) : f ◦ γ = ψ ◦ f }; here γ ∈ Aut(1 ) and ψ ∈ Aut(2 ). Let  f denote the projection of A f onto its first factor. Let T f denote the projection of A f onto its second factor. Let G f = {γ ∈ Aut(1 ) : f ◦ γ = f }. Let H f = {ψ ∈ Aut(2 ) : ψ ◦ f = f }.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 J. P. D’Angelo, Rational Sphere Maps, Progress in Mathematics 341, https://doi.org/10.1007/978-3-030-75809-7_5

153

154

5 Groups Associated with Holomorphic Mappings

An automorphism γ is in  f if and only if there exists an automorphism ψ of the target for which f ◦ γ = ψ ◦ f . An automorphism γ is in G f if f is invariant under γ. Informally, we call G f the invariant group and we call  f the Hermitianinvariant group. For non-constant rational sphere maps, Proposition 5.3 provides a computational way to decide whether a given γ is in  f . The technique involves the Hermitian norm (1.12), hence the name. We continue with the general situation. Lemma 5.1 A f is a subgroup of Aut(1 ) × Aut(2 ). Proof Assume (γ1 , ψ1 ) and (γ2 , ψ2 ) are in A f . Thus f ◦ γ1 = ψ1 ◦ f and f ◦ γ2 = ψ2 ◦ f . Hence f ◦ (γ1 ◦ γ2 ) = ( f ◦ γ1 ) ◦ γ2 = (ψ1 ◦ f ) ◦ γ2 = ψ1 ◦ ( f ◦ γ2 ) = (ψ1 ◦ ψ2 ) ◦ f. Therefore (γ1 ◦ γ2 , ψ1 ◦ ψ2 ) ∈ A f .



Definition 5.2 Let f and g be maps from 1 to 2 . We say that f and g are equivalent if there are automorphisms ψ and ϕ for which ψ ◦ f = g ◦ ϕ. When 1 and 2 are unit balls (in possibly different dimensions), and f and g are equivalent, we say (as in Definition 1.2) that f and g are spherically equivalent. Lemma 5.2 Assume that f and g are equivalent, that is, there are automorphisms for which ψ ◦ f = g ◦ ϕ. Let (γ, ζ) ∈ A f . Then (ϕ ◦ γ ◦ ϕ−1 , ψ ◦ ζ ◦ ψ −1 ) ∈ Ag . In particular,  f and g are conjugate by ϕ: g = ϕ ◦  f ◦ ϕ−1 . Also T f and Tg are conjugate by ψ. Proof The proof is a formal computation. We are given that g = ψ ◦ f ◦ ϕ−1 . Assuming f ◦ γ = ζ ◦ f we must show that g ◦ (ϕ ◦ γ ◦ ϕ−1 ) = (ψ ◦ ζ ◦ ψ −1 ) ◦ g. Starting with f ◦ γ = ζ ◦ f we obtain ψ ◦ f ◦ γ ◦ ϕ−1 = ψ ◦ ζ ◦ f ◦ ϕ−1 . Inserting ϕ−1 ◦ ϕ and ψ −1 ◦ ψ in the right places gives

(1)

1 Five Groups

155

(ψ ◦ f ◦ ϕ−1 ) ◦ (ϕ ◦ γ ◦ ϕ−1 ) = (ψ ◦ ζ ◦ ψ −1 ) ◦ (ψ ◦ f ◦ ϕ−1 ). Since g = ψ ◦ f ◦ ϕ−1 , Eq. (2) implies Eq. (1).

(2) 

Proposition 5.1 Let f : 1 → 2 be holomorphic. The group H f is trivial if and only if, for each γ ∈  f , there is a unique ψ ∈ T f for which f ◦ γ = ψ ◦ f . In this setting, the map γ → ψ is a group homomorphism. Furthermore, the kernel of this map is the group G f . Proof Assume first that H f is trivial. If f ◦ γ = ψ1 ◦ f = ψ2 ◦ f then ψ2−1 ◦ ψ1 ◦ f = f and hence ψ2−1 ◦ ψ1 ∈ H f . Therefore ψ2 = ψ1 . Conversely, suppose that uniqueness holds. Let I2 denote the identity in the target and I1 the identity in the source. Assume ψ ∈ H f . Then ψ ◦ f = f = f ◦ I1 = I2 ◦ f. By uniqueness, ψ = I2 and H f is trivial. Let  denote the map that assigns ψ to γ. Assume f ◦ γ1 = (γ1 ) ◦ f and f ◦ γ2 = (γ2 ) ◦ f . Then (γ2 ◦ γ1 ) ◦ f = f ◦ (γ2 ◦ γ1 ) = (γ2 ) ◦ f ◦ γ1 = (γ2 ) ◦ (γ1 ) ◦ f. Hence (γ2 ◦ γ1 ) = (γ2 ) ◦ (γ1 ), and  is a homomorphism. Finally, note that the image of γ is the identity if and only if ψ is the identity if  and only if f ◦ γ = f if and only if γ ∈ G f . This result is useful for balls, where the following equivalence holds. Proposition 5.2 Let f be a non-constant rational sphere map with target dimension N . Then H f is trivial if and only if N is also the target-rank of f . (In other words, N is minimal.) Proof Suppose first that the target dimension exceeds the target-rank. Then f is spherically equivalent to g ⊕ 0 for some g, and H f is a conjugate of Hg⊕0 . But Hg⊕0 is not trivial, as g ⊕ 0 = (I ⊕ V ) ◦ (g ⊕ 0) . Here I is the identity on the image of g and V is an arbitrary unitary map defined on the orthogonal complement of the image of g. Thus H f also is not trivial. Conversely, suppose that the target dimension is minimal. After composing with an automorphism, we may assume that f (0) = 0. Choose ψ ∈ H f . Thus ψ ◦ f = f and hence ψ(0) = 0. Thus ψ is unitary. After expanding f in a Taylor series at 0, we see that ψ must fix all the (vector) Taylor coefficients. Since the target dimension is minimal, these coefficients span C N . Since ψ fixes every vector in a spanning set, ψ  must be the identity and therefore H f is trivial.

156

5 Groups Associated with Holomorphic Mappings

2 Examples of the Five Groups After a very general but dull example, we return to complex analysis. Example 5.1 Let f :  →  be the identity function. Then A f is isomorphic to Aut() expressed as the diagonal of Aut() × Aut(). We next find the holomorphic automorphism group of C and use it to give some easy examples of the five groups. Lemma 5.3 The holomorphic automorphism group of C consists of the affine transformations z → az + b with a = 0. Proof That these maps are automorphisms follows immediately from the definition. To show that there are no others requires some complex analysis. Let f be an automorphism, and consider the function g defined for z = 0 by g(z) = f ( 1z ). Then g has a singularity at 0. There are three kinds of isolated singularities. If the singularity is removable, then g(z) is bounded for z near 0. Thus, f is bounded near z = ∞, and hence f is bounded on C. By Liouville’s theorem, f is constant and hence not injective. Thus, this type of singularity is impossible. If the singularity is essential, then by the Casorati-Weierstrass theorem (or by the more difficult Picard’s great theorem), the image under g of {|z| < r } is dense in C. Thus, the image under f of {|z| > r } is dense in C. Now consider the image of the disjoint set {|z| < r } under f . It is open and hence intersects the (dense) image of {|z| > r } under f . Thus, f is not injective, and this type of singularity is also impossible. The only remaining possibility is that g has a pole at 0. Since f is entire it has a global power series centered at 0. Thus f (z) =

∞ 

ck z k

k=0 ∞

 1 g(z) = f ( ) = ck z −k . z k=0 Since g has a pole, this sum is finite, and hence the sum for f is finite. Thus f is a polynomial. But the only injective polynomials on C are of first degree.  Remark 5.1 Things are much more complicated when  = Cn for n ≥ 2. The group Aut(Cn ) is very large. We illustrate when n = 2, using z and w for the variables. For any entire function p in one variable, the map (z, w) → (z, w + p(z)) is an automorphism. Such maps are called shears. We can, of course, interchange the roles of z and w. A finite composition of such maps is also an automorphism. Andersen and Lempert introduced a collection of automorphisms called overshears; such maps have the form (z, w) → (z, weh(z) + g(z))

2 Examples of the Five Groups

157

(z, w) → (zeh(w) + g(w), w). There are analogous definitions in higher dimensions. The Andersen-Lempert [1] theorems for n ≥ 2 provide the following information. First, there are automorphisms that are not finite compositions of overshears. Second, the subgroup generated by overshears is dense in Aut(Cn ) in the topology of uniform limits on compact subsets. Finally, an analogue of this second result holds for the group of diffeomorphisms of real Euclidean space. When f : Cn → Cn is a polynomial map, the following result holds. If f is injective, then f is surjective, and f −1 is a polynomial map. The Jacobian conjecture asks the following. Suppose that the Jacobian determinant J ( f ) is a non-zero constant. Must f be injective and therefore have a polynomial inverse? In the complex case, saying that J ( f ) is everywhere non-zero implies that J ( f ) is a constant. In the real case, this conclusion fails. Consider f (x) = x + x 3 . Then f (x) = 1 + 3x 2 = 0 but f −1 is not a polynomial. See [64] for an amazing higher dimensional counterexample. Let us return to C. We find  f and T f for three explicit functions f . Example 5.2 Let 1 = 2 = C. • Put f (z) = z k , for k ≥ 2. Then  f = C∗ , the multiplicative group of non-zero complex numbers. To see this fact, given the automorphism az + b we need to find c, d such that (az + b)k = cz k + d. For k > 1, finding such c, d is possible only when b = 0. Since a is an arbitrary element of C∗ we conclude that  f = C∗ . Also, b = 0 implies d = 0, and hence c = a k is an arbitrary non-zero complex number. Thus T f = C∗ as well. • Put f (z) = e z , then  f consists of translations and is isomorphic to C. The reason is that the equation eaz+b = ce z + d forces a = 1 and d = 0, but allows b to be arbitrary. Here c = eb , and hence T f = C∗ . The automorphism of translation by b gets mapped to the automorphism of multiplication by eb . This map is a group homomorphism whose kernel is isomorphic to Z. The kernel is G f ; it consists of translations by integer multiples of 2πi. See Proposition 5.1. • Put f (z) = z + e z . To determine  f , we consider the equation (az + b) + eaz+b = c(z + e z ) + d.

(3)

Differentiating (3) twice shows that a = 1 and eb = c. Putting a = 1 in (3) gives c = 1 and hence b = 0. We conclude that  f is the trivial group. It follows that T f is trivial and hence all five groups are trivial.

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5 Groups Associated with Holomorphic Mappings

We next provide additional examples of the groups. Later we will show, when f is a non-linear polynomial sphere map preserving the origin, that  f is contained in the unitary group. We assume that result in several of these examples. Example 5.3 Let  be the unit disk. Put f (z) = z m , where m is an integer at least two. Then G f is the cyclic group generated by a primitive m-th root of unity, and  f = U(1). Example 5.4 Let  = C and put f (z) = z m for m ≥ 2. Then  f = C∗ and T f is also C∗ . There is a group homomorphism  :  f → T f given by (a) = a m . The kernel of  is the set of m-th roots of unity, a cyclic group of order m. The kernel is thus the group G f . By Proposition 5.1, this relationship holds rather generally. Example 5.5 Let 1 be the unit disk and let 2 be the unit ball in C2 . Put f (z) = 1 (z + z 2 , z 2 − z 3 ). Then f : B1 → B2 is a proper holomorphic mapping and  f 2 is the trivial group. We will verify this result in Example 5.9 and generalize it in Theorem 5.10. Example 5.6 For m ≥ 2, put f (z) = z ⊗m = Hm (z) as in (2.9). Then f : Bn → B N is a rational sphere map and, by Theorem 5.3,  f = U(n). Example 5.7 Define f : C → C by f (z) = z 2 + z m+2 . Then  f is cyclic of order m. The relevant equation is (az + b)2 + (az + b)m+2 = c(z 2 + z m+2 ) + d. Equating coefficients gives b = d = 0 and a 2 = a m+2 = c = 0. Hence a m = 1. √ Example 5.8 For (z, w) ∈ C2 , put f (z, w) = (z 3 , 3zw, w 3 ). Then f is a monomial sphere map. We will prove in Example 5.10 that  f is the semi-direct product of the torus and the group of two elements permuting the variables. 2

Exercise 5.1 Put f (z) = e z + z 2 as a map from C to itself. Compute T f and G f . Find the group homomorphism  :  f → T f whose kernel is G f .

3 Hermitian-Invariant Groups for Rational Sphere Maps Given our emphasis on rational sphere maps, we will usually restrict our work on these five groups to proper holomorphic mappings between balls. We remind the reader that a non-constant rational sphere map defines a proper holomorphic mapping between balls. Henceforth, we will tacitly assume that a rational sphere map is not constant. This section contains several of the most important results in this book. First, we show that Proposition 1.11 gives a computational way to decide whether a source automorphism lies in  f . We state it using our new terminology.

3 Hermitian-Invariant Groups for Rational Sphere Maps

159

Proposition 5.3 Let f : Bn → B N be a rational proper map. Let  f denote the Hermitian-invariant group from Definition 5.1. Then a source automorphism γ is in  f if and only if there is a constant cγ such that H( f ◦ γ) = cγ H( f ). Remark 5.2 When γ ∈ U(n), the constant cγ necessarily equals 1. Corollary 5.1 When f is a rational sphere map,  f =  f ⊕0 . Proof For g any map, H(g ⊕ 0) = H(g). Therefore H(( f ⊕ 0) ◦ γ) = H(( f ◦ γ) ⊕ 0) = H( f ◦ γ) = cH( f ) = cH( f ⊕ 0).  Remark 5.3 We briefly reconsider the distinction between  f and G f . Let f be a polynomial with f (0) = 0. To be in G f , a unitary map γ must satisfy f ◦ γ = f . To be in  f , it must satisfy the weaker condition  f ◦ γ2 =  f 2 . This last equation is equivalent to the existence of a unitary V such that f ◦ γ = V ◦ f . Thus, equality of squared norms is equivalent to γ being in  f . We now turn to the monomial case. Proposition 5.4 Let f be a monomial sphere map. Then  f contains T(n), namely, the diagonal unitary mappings. Proof Put f (z) = (..., cα z α , ...). Let γ be a diagonal unitary matrix with eigenvalues eiθ j . Then, in multi-index notation, where α · θ = α j θ j , ( f ◦ γ)(z) = (..., cα eiα·θ z α , ...). Therefore,  f ◦ γ2 =  f 2 and thus H( f ◦ γ) = H( f ) =  f (z)2 − 1. By Propo sition 5.3, γ ∈  f . Remark 5.4 For a monomial map f , the group  f can be larger than the torus. For the identity map,  f is the full automorphism group. For the tensor product map z → z ⊗m for m ≥ 2, by Theorem 5.3,  f is the unitary group. Example 5.10 gives a monomial sphere map whose group is generated by the torus and the map that interchanges the variables. For a generic monomial sphere map, however,  f is the n-torus. When f is spherically equivalent to a monomial sphere map, then (by Lemma 5.2)  f contains a conjugate to the torus. A theorem of Lebl [54] states that a rational sphere map of degree 2 is spherically equivalent to a monomial mapping. Combining this result with Proposition 5.4 yields the following corollary.

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5 Groups Associated with Holomorphic Mappings

Corollary 5.2 Let f be a rational sphere map of degree 2. Then  f contains a conjugate of the torus. In particular,  f is infinite. Let f be a proper map between balls. Whether or not f is rational, it can be shown (see [29]) that  f is non-compact if and only if T f is non-compact. We assume here that f is a rational sphere map and ask when is the group  f noncompact. The answer is striking; f must be a linear fractional transformation, and hence spherically equivalent to the injection z → (z, 0). For this map, the group  f is, of course, Aut(Bn ). Therefore, groups intermediate between the unitary group and the full automorphism group do not arise as  f for any rational proper mapping f. We will use Corollary 1.5, the n-dimensional version of Schwarz’s lemma, to establish a strong rigidity result. Theorem 5.1 Let f : Bn → B N be a rational proper map. Then  f is non-compact if and only if f is a linear fractional transformation. If  f is compact, then  f lies in a maximal compact subgroup, that is, a conjugate of U(n). In order to prove Theorem 5.1, we first prove the following auxiliary result. Theorem 5.2 Let f be a rational sphere map of degree d. Put f = qp and assume that f (0) = 0. If  f contains an automorphism γ = U φa that moves the origin, then ( p(a)2 − |q(a)|2 )( p(U a)2 − |q(U a)|2 ) = (1 − a2 )2d .

(4)

Proof Since p(0) = 0, Proposition 2.1 implies that the degree of the denominator q is at most d − 1. We write d  Aα z α p(z) = |α|=1

q(z) =

d−1 

bβ z β .

|β|=0

Without loss of generality we assume b0 = 1. Assuming that γ = U φa exists as hypothesized, we will compute the coefficient cγ in two ways. By definition we have   cγ  p2 − |q|2 = cγ H( f ) = H( f ◦ (U φa )). Putting f ◦ (U φa ) =

P , Q

we have H( f ◦ (U φa )) = P2 − |Q|2 .

We have the following formulas for P and Q:

3 Hermitian-Invariant Groups for Rational Sphere Maps

161

⎛ ⎞ d  1 ⎝ P(z)2 = Aα (U (a − L a (z))α (1 − z, a)d−|α| 2 ⎠ |q(a)|2 |α|=1 ⎛ ⎞ d−1 



1 2 ⎝

|Q(z)|2 = bβ (U (a − L a (z))β (1 − z, a)d−|β| ⎠ . |q(a)|2 |β|=0

The factor of

1 |q(a)|2

arises in order to make Q(0) = 1. We evaluate at 0 to get

P(0)2 − |Q(0)|2 =

1 ( p(U a)2 − |q(U a)|2 ). |q(a)|2

Then we evaluate at a, using L a (a) = a, to get P(a)2 − |Q(a)|2 =

1 (−(1 − a2 )2d ). |q(a)|2

Evaluating H( f ) at 0 and a, and using cγ H( f ) = H( f ◦ γ), yields both formulas cγ ( p(0)2 − 1) = −cγ =

 1   p(U a)2 − |q(U a)|2 2 |q(a)|

cγ ( p(a)2 − |q(a)|2 ) = −

1 (1 − a2 )2d . |q(a)|2 

Formula (4) follows.

Proof We can now prove Theorem 5.1. Let f = qp be of degree d. After composition with an automorphism of the target, we may assume f (0) = 0. Assume U φa ∈  f . In this case, Schwarz’s lemma yields  p(z)2 ≤ |q(z)|2 z2 for z in the ball. Therefore |q(z)|2 −  p(z)2 ≥ |q(z)|2 (1 − z2 ). Since U a2 = a2 , we plug this inequality into (4) to get (1 − a2 )2d ≥ (1 − a2 )2 |q(a)|2 |q(U a)|2 and therefore (1 − a2 )2d−2 ≥ |q(a)|2 |q(U a)|2 .

(5)

If  f is not compact, then we can find a sequence of automorphisms Uk φak ∈  f where ak  tends to 1. Assume d ≥ 2. By (5), a subsequence of q(ak ) or of q(Uk ak ) tends to 0. But the denominator q cannot vanish on the closed ball. This contradiction therefore implies d = 1. The degree of q is smaller than the degree of p. Thus, if the

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5 Groups Associated with Holomorphic Mappings

degree of p is 1, then q is constant. We have shown when f (0) = 0 that f must be linear. Any map spherically equivalent to f must be a linear fractional transformation. Suppose that f is not a linear fractional transformation. Since  f is closed in Aut(Bn ), it is compact. By standard Lie group theory,  f is contained in a maximal compact subgroup which must be a conjugate of U(n).  Corollary 5.3 Let p be a proper polynomial map between balls with p(0) = 0. Unless p is of degree 1, we have  p ⊆ U(n). Proof When q = 1, formula (5) forces a = 0 or d = 1. Thus, unless d = 1, there is no automorphism in  p that moves the origin.  Corollary 5.3 will get used several times. For polynomial maps f of degree at least 2 with f (0) = 0, only subgroups of U(n) are candidates for  f . The next corollary is important enough to be called a theorem. Theorem 5.3 For m ≥ 2, put f (z) = z ⊗m . Then  f = U(n). Proof By the previous corollary,  f is contained in U(n). We show the opposite containment. Let U be unitary. Then ( f ◦ U )(z)2 = (U (z))⊗m 2 = U z2m = z2m =  f (z)2 . By Proposition 1.4, there is a unitary V such that f ◦ U = V ◦ f . Thus U ∈  f . 

4 Additional Examples We compute the group  f in several more cases and make a remark about the polynomial case. Example 5.9 We revisit Example 5.5. Put n = 1, N = 2, and f (z) =

1 (z + z 2 , z 2 − z 3 ). 2

Then f is a polynomial sphere map. We claim  f is the trivial group. To verify the claim, note that f (0) = 0 and hence  f is a subgroup of the circle U(1). The target automorphism must be unitary as well. We study the equation f (eiθ z) = V ◦ f . But V is irrelevant when we compute the Hermitian norm. We must have 1 (|zeiθ + z 2 e2iθ |2 + |z 2 e2iθ − z 3 e3iθ |2 ) = (V f )(z)2 =  f (z)2 . 4 Expanding the left-hand side and equating coefficients show that θ = 0.

4 Additional Examples

163

Example 5.10 Consider the Faran map f given by f (z, w) = (z 3 , Chap. 3, we saw that G f is the cyclic group of order 3 generated by  η 0 , 0 η2

√

3zw, w 3 ). In

(6)

where η is a primitive cube root of unity. We check this fact again below. The group H f is trivial. The target group T f is generated by the three-by-three unitary matrices with eigenvalues e3iθ , ei(θ+φ) , e3iφ and the permutation matrix that interchanges the first and third variables. The Hermitian-invariant group  f is the semi-direct product of the torus and the group of order two that interchanges the source variables. It is non-Abelian and is generated by the matrices (7) and (8):  iθ 0 e 0 eiφ 

0 1 . 1 0

(7)

(8)

Proof The map f is invariant under the transformation in (6), its square, and the identity (its cube), but under no other linear map. Hence, G f is this group of order 3. It is clear that H f is trivial, and hence T f is the image of the homomorphism from Proposition 5.1. Determining  f is a bit subtle. First, the Hermitian form corresponding to the polynomial sphere map f is H( f ) = |z|6 + 3|z|2 |w|2 + |w|6 − 1. We rewrite H( f ) in terms of the defining function ρ = |z|2 + |w|2 − 1 for the ball as (|z|2 + |w|2 )3 + 3|z|2 |w|2 (1 − |z|2 − |w|2 ) − 1 = (1 + ρ)3 − 1 − 3|z|2 |w|2 ρ. (9) The terms in (9) involving ρ are, of course, invariant under U(2). The term |z|2 |w|2 is invariant under both matrices (7) and (8). Thus,  f contains the semi-direct product. To finish, we must show that this term is invariant under no other unitary maps L. Let L(z, w) = (u 11 z + u 12 w, u 21 z + u 22 w). Assume that |z|2 |w|2 is invariant under L. Setting L(z, w) = (ζ, τ ) forces |ζ|2 |τ |2 = |z|2 |w|2 and therefore both u 11 u 21 = u 12 u 22 = 0 |u 11 u 22 + u 12 u 21 |2 = 1.

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5 Groups Associated with Holomorphic Mappings

Because of the first pair of equations and because L is unitary, either u 12 = u 21 = 0 and (7) holds, or u 11 = 0 and u 22 = 0. In this case, |u 12 u 21 | = 1. Since u 11 = 0 and L is unitary, we must have |u 12 | = |u 21 | = 1. Thus, for some θ and φ, L is the matrix product of (7) and (8). The second statement follows.  Remark 5.5 Suppose p is a non-linear polynomial sphere map with p(0) = 0. We know that  p is a subgroup of U(n). The relevant equation for deciding whether γ ∈  p is that p ◦ γ = ψ ◦ p. Using Hermitian forms, we require only that  p ◦ γ2 = ψ ◦ p2 . Since the origin is preserved, ψ must be unitary; we need to study only  p ◦ γ2 =  p2 . Example 5.11 Consider the Whitney map W defined by W (z) = (z 1 , ..., z n−1 , z 1 z n , ..., z n2 ). Then W = U(n − 1) × U(1). By Corollary 5.3, W is a subgroup of U(n). Put z = (z 1 , ..., z n−1 ). Since W (z)2 = z 2 + |z n |2 z2 , W contains U(n − 1) × U(1). Choose L ∈ U(n). Since H(W ◦ L) = H(W ) we have z 2 + |z n |2 z2 = L(z) 2 + |L n (z)|2 Lz2 = L(z) 2 + |L n (z)|2 z2 Setting z = 0 shows that L(z n ) = eiθ z n for some θ, and then setting z n = 0 shows that the restriction of L to the first n − 1 variables is unitary. Exercise 5.2 Put f (ζ) = (ζ13 , ζ1 ζ2 , ζ23 ). For ζ real, show that f is injective. For ζ complex, show that f fails to be injective. Determine the maximum number of inverse images of a point. With ζ real, show that the image of a neighborhood of 0 under f is not a smooth manifold. Exercise 5.3 Define f : C → C by f (z) = z 2 + z 4 . Find  f . Then find a map g for which g is cyclic of order m.

5 Behavior of  f Under Various Constructions The notions of juxtaposition and tensor product have played a major role thus far. It is natural to study how Hermitian-invariant groups behave under them. We begin with a simple observation related to earlier results such as Corollary 2.5 from Chap. 2. There we showed, given a polynomial map p for which  p(z)2 < 1

5 Behavior of  f Under Various Constructions

165

on the closed ball, that we could find a polynomial map g such that p ⊕ g is a polynomial sphere map. We recall from Example 2.10 that one can bound neither the degree nor the target dimension of g in terms of the source dimension and the degree of p. The coefficients of p matter. The subsequent proposition benefits from allowing a small positive . Proposition 5.5 Let p : Cn → C N be a polynomial map of degree K . Then there is an  > 0 and a polynomial map g of degree K such that  p ⊕ g is a polynomial sphere map. We may choose g such that  p ⊕  g is a juxtaposition of tensor powers. Thus, there are positive numbers λk such that |λk |2 = 1 and 2  p(z)2 + g(z)2 =

K 

|λk |2 z2k .

(10)

k=0

Proof Consider  the sum T on the right-hand side of (10), where the λk are all positive and |λ j |2 = 1. As a Hermitian form, T is positive definite on the space of polynomials of degree K . Hence, for sufficiently small , the form determined by T − 2  p2 is also positive definite. Thus, there is a g for which T − 2  p2 = g2 . Thus, (10) holds, and  p ⊕ g is a polynomial sphere map.  The following lemma will be often used. Lemma 5.4 Let f and g be polynomial maps that contain no terms of the same degree. Assume that  f ◦ L2 + g ◦ L2 =  f 2 + g2 for some unitary map L. Then  f ◦ L2 =  f 2 and g ◦ L2 = g2 . Otherwise said, ( f ⊕ g) ◦ L2 =  f ⊕ g2 =⇒  f ◦ L2 =  f 2 and g ◦ L2 = g2 . Proof Since linear maps preserve terms of a given degree, the result follows by equating coefficients.  Theorem 5.4 Let f, g be proper polynomial maps between balls with the same source. Let ν(g) denote the order of vanishing of g and deg( f ) the degree of f . Assume that 2 ≤ deg( f ) < ν(g). For 0 < θ < 2π, put J = Jθ = Jθ ( f, g). Then  J =  f ∩ g . Corollary 5.4 Let f, g be proper polynomial maps between balls with the same source. Let m be larger than the degree of f . For 0 < θ < 2π, put j ( f, g) = Jθ ( f, g ⊗ z ⊗m ). Then  j ( f,g) ∩ U(n) =  f ∩ g ∩ U(n).

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5 Groups Associated with Holomorphic Mappings

Corollary 5.5 Suppose in Theorem 5.4 that f is a polynomial of degree at least 2 with f (0) = 0. With j ( f, g) as in the previous corollary,  j ( f,g) =  f ∩ g . Proof Corollary 5.5 follows from Corollaries 5.3 and 5.4 because the hypothesis on f guarantees that the groups involved are subsets of U(n). Corollary 5.4 follows from Theorem 5.4 because the order of vanishing of g ⊗ z ⊗m is at least m. We now prove the theorem. First let c = cos(θ) and s = sin(θ). Then H(J ) = c2  f 2 + s 2 g2 − 1

(11)

H(J ◦ γ) = c2  f ◦ γ2 + s 2 g ◦ γ2 − 1.

(12)

The hypotheses imply that  f and g are subgroups of the unitary group. If γ lies in both these groups, then the right-hand sides of (11) and (12) are equal, and hence the left-hand sides are also equal. Thus,  f ∩ g ⊆  J . To prove the opposite inclusion, suppose that the left-hand sides are equal. Since γ is unitary (and hence linear), it preserves degrees. The hypotheses prevent any interaction between the terms  f ◦ γ2 and g ◦ γ2 . By Lemma 5.4, we conclude that  f ◦ γ2 =  f 2 and g ◦ γ2 = g2 . If, therefore, γ ∈  J , then γ is in both  f and g .  Next consider the mappings f m defined by z → z ⊗m , for m a positive integer. When m = 1, this map f 1 is the identity map, and therefore its group  f1 is the full automorphism group. For m ≥ 2, by Theorem 5.3,  fm is the unitary group. Corollary 5.5 has, therefore, the following additional corollary.  Corollary 5.6 Assume K ≥ 2, each λ j = 0 and Kj=1 |λ j |2 = 1. For distinct positive integers m 1 , ..., m K , define F, a juxtaposition of tensor powers, by F(z) = λ1 z ⊗m 1 ⊕ ... ⊕ λ K z ⊗m K . Then  F = U(n). Exercise 5.4 Let f be the Faran map as in Example 5.10. Define F by F(z) = 1 + f (z) = (1 + z 13 , 1 +

√ 3z 1 z 2 , 1 + z 23 ).

Expanding the squared norm of F introduces the terms z 13 , z 1 z 2 , z 23 . Verify their invariance under the cyclic subgroup of U(2) generated by the map (z 1 , z 2 ) → (ηz 1 , η 2 z 2 ); here, η is a primitive cube root of 1. Exercise 5.5 Put f (z, w) = (z 2k+1 , zw, w 2k+1 ). Find all U ∈ U(2) for which there is a linear map V with f ◦ U = V ◦ f . Determine the group G f . Exercise 5.6 Show that f ◦ U = f for all unitary U is impossible for a proper holomorphic map f between balls. Give an example of a non-constant holomorphic map for which f ◦ U = f for all U in an infinite subgroup of U(n).

6 Examples Involving the Symmetric Group

167

6 Examples Involving the Symmetric Group Given a subgroup G of Aut(Bn ), we will investigate whether G can be h for some rational sphere map h. One of the key ideas is the use of Corollary 5.3; it enables us to work in U(n) instead of in Aut(Bn ). Another key idea will be to use Corollary 5.5 to cut down the size of groups. Lemma 5.4 and the following lemma provide crucial ideas. When working with the Hermitian-invariant group for a polynomial map f , the definition H =  f 2 − 1 forces us to consider the equality  f ◦ U 2 =  f 2 . Lemma 5.4 tells us what happens when f = g ⊕ h and where g and h have no terms of the same degree. The next lemma enables us to remove squared absolute values. The author feels that this simple lemma is a major part of CR Geometry. One often distinguishes between pure and mixed terms in a Taylor expansion. Lemma 5.5 Let h be a scalar-valued holomorphic function with h(0) = 0 and let U be a unitary map. Assume |1 + h ◦ U |2 = |1 + h|2 . Then h ◦ U = h. Proof The hypothesis gives 1 + 2Re(h(U (z))) + |h(U z)|2 = 1 + 2Re(h(z)) + |h(z)|2 . Looking at pure terms and mixed terms separately shows that |h(U z)|2 = |h(z)|2 and also that 2Re(h(U (z))) = 2Re(h(z)). In particular, h ◦ U = h.  Let us seek a polynomial sphere map h, with source dimension 2, for which h is the symmetric group S2 on two letters. To do so, we will make h(0) = 0 to guarantee that h ⊆ U(2). The definition of h leads us to study h ◦ U 2 = h2 . It is natural to begin with the three polynomials 1, z + w, zw. These generate the algebra of polynomials invariant under S2 . Consider first the map, for small  > 0, defined by f (z, w) = (1 + z + w, 1 + zw) = f 1 (z, w) ⊕ f 2 (z, w). It is invariant under the symmetric group but it is not a sphere map. By Proposition 5.5, we can find a map g such that f ⊕ g is a polynomial sphere map and is also an orthogonal sum of tensor powers. Therefore  λ2j (|z|2 + |w|2 ) j . (13)  f (z, w)2 + g(z, w)2 = j

The right-hand side of (13) is invariant under unitary maps L. Therefore  f ◦ L2 =  f 2 implies g ◦ L2 = g2 .

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5 Groups Associated with Holomorphic Mappings

To invoke Corollary 5.3, we want ( f ⊕ g)(0) = 0. We also require g to have the needed symmetries. We use tensor products. Write Z = (z, w). Thus, we consider h = ( f 1 ⊗ Z ) ⊕ ( f 2 ⊗ Z ⊗2 ) ⊕ (g ⊗ Z ⊗4 ). The polynomial sphere map h now satisfies h(0, 0) = 0 and we have h2 = | f 1 |2 Z 2 + | f 2 |2 Z 4 + g2 Z 8 =

(14)

  2 |1 + z + w|2 (|z|2 + |w|2 ) + |1 + zw|2 (|z|2 + |w|2 )2 + g2 (|z|2 + |w|2 )4 . Next we use Lemmas 5.4 and 5.5. We have constructed things such that the terms |1 + z + w|2 (|z|2 + |w|2 ) and |1 + zw|2 (|z|2 + |w|2 )2 have no terms of the same degree. The other terms in h2 are of even higher degree. Consider a unitary U for which h ◦ U 2 = h2 . Since U is unitary, the terms in (14) involving |z|2 + |w|2 are invariant under U . Therefore Lemma 5.4 guarantees both: |(1 + z + w) ◦ U |2 = |1 + z + w|2 |(1 + zw) ◦ U |2 = |1 + zw|2 . Now Lemma 5.5 implies that (z + w) ◦ U = z + w and also that (zw) ◦ U = zw. Let U have components u jk . Reasoning as in Example 5.10 shows that the second condition forces U to be one of two possibilities (7) or (8). Then, using also the first condition shows that the only for (7) is the identity. Therefore, U is  possibility 01 either the identity or the matrix . The comment following (14) shows that, if 10 f ◦ L = f , then g ◦ L = g also. Hence h = S2 . We state the conclusion of this discussion as a theorem. The functions f 1 , f 2 , g in formula (14) have been defined in the paragraphs after the proof of Lemma 5.5. In the next section, we extend the following result and its proof. Theorem 5.5 Let h be the polynomial sphere map whose squared norm satisfies (14). Then h = S2 . Remark 5.6 One naturally asks what happens in the previous discussion if we began with (1 + zw) but did not include (1 + z + w). We follow essentially the same steps, but at the end we conclude only that the unitary L must satisfy (u 11 z + u 12 w)(u 21 z + u 22 w) = zw. Equating coefficients now gives one of the following two matrices:

6 Examples Involving the Symmetric Group

169

 iθ 0 e 0 e−iθ

 01 10

or

and hence the group h is infinite. In a similar fashion, we could ask what happens if we start with (1 + z + w) and do not include (1 + zw). We conclude that the unitary U must satisfy u 11 z + u 12 w + u 21 z + u 22 w = z + w. In this case, the columns of U must sum to 1. Below we describe this collection of unitary matrices. Example 5.12 Assume U ∈ U(2) and the columns of U sum to 1. Then U is of the following form:  1 1 + eiθ 1 − eiθ . (15) 2 1 − eiθ 1 + eiθ 

z 1−w . Since UU ∗ = I , one 1−z w obtains three equations for the complex numbers z, w. These equations easily imply that z = w. Hence, the rows also sum to 1. Since |z|2 + |1 − z|2 = 1, we see that z lives on a circle of radius 21 centered at 21 . In particular, the dimension of this subgroup of U(2) is 1. To verify (15), assume that U is of the form

We next discuss the generalization of Example 5.12 (and the second part of Remark 5.6) to higher dimension.  Proposition 5.6 Put h(z) = nj=1 z j . Suppose U ∈ U(n) and h ◦ U = h. Then each row and column of U sums to 1. Proof Let the components of U be u jk . We are given that n  n 

u jk z k =

j=1 k=1

n 

zk .

k=1

Equating the coefficients of z k on each side gives, for each k, that n 

u jk = 1.

j=1

Hence, each column sums to 1. Since h ◦ U = h, we also have h = h ◦ U −1 . By the same reasoning, each column of U −1 sums to 1. Since U is unitary, U −1 is the complex conjugate transpose of U , and therefore the complex conjugate of each row of U sums to 1. Since 1 is real, each row sums to 1 as well. 

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5 Groups Associated with Holomorphic Mappings

Reference [75] makes extensive use of the subgroup of U(n) consisting of matrices whose rows and columns all sum to 1. They label this subgroup XU(n). The dimension of this Lie group is (n − 1)2 . This dimension can be computed as follows. Consider elements of U(n) of the form  T

10 T −1 . 0L

Here L is an arbitrary element of U(n − 1) and T is unitary of a special form. All the entries in the first row and first column of T are equal to √1n . Thus, the matrix T is a constant times a dephased complex Hadamard matrix. A computation shows that all members of XU(n) have this form and conversely everything of this form is in XU(n). This formula shows that XU(n) is conjugate (in the group-theoretic sense) to U(n − 1), and hence is of the same dimension as U(n − 1), namely, (n − 1)2 . The results in [75] show that each element of XU(n) can be written as a (particular kind of) weighted sum of permutation matrices and the authors investigate the number of terms in this sum. Exercise 5.7 Fill in the details in Example 5.12. Exercise 5.8 Let L θ denote the matrix in (15). What is L θ L φ ? Exercise 5.9 Fill in details of the above description of XU(n). We state the conclusions of the remark and example in the next proposition. Proposition 5.7 Define polynomial sphere maps f, g as follows, where the functions ζ, ξ are polynomials chosen as in the above discussion and guarantee that f, g are polynomial sphere maps.   f (z, w) =  ((1 + zw)z, (1 + zw)w) ⊕ ζ(z, w) ⊗ (z, w)⊗4 .   g(z, w) =  ((1 + z + w)z, (1 + z + w)w) ⊕ ξ(z, w) ⊗ (z, w)⊗3 .   iθ 01 e 0 and , and g is the group XU(2) Then  f is the group generated by 10 0 eiφ of unitary matrices satisfying (15). Exercise 5.10 Write explicit formulas for the ζ and ξ from Proposition 5.7.

7 The Symmetric Group In this section, we construct a map f whose Hermitian-invariant group is the symmetric group Sn . In the following section, we generalize the proof to construct a map f for which  f is an arbitrary finite subgroup of U(n).

7 The Symmetric Group

171

The proof of Theorem 5.5 used the pair of functions 1 + z + w and 1 + zw. One could have also used the single function h(z, w) = 1 + z + w + zw. The equation h ◦ U = h implies both (z + w) ◦ U = z + w and (zw) ◦ U = zw, by considering degrees. We, therefore, begin with the following result in n dimensions. See also Exercise 5.15.

  2   2 Lemma 5.6 For U ∈ U(n), assume nj=1 1 + (U z) j = nj=1 1 + z j . Then U is a permutation of the coordinates.  



 Proof The hypothesis implies that nj=1 1 + (U z) j = eiθ nj=1 1 + z j for some eiθ . Evaluating at z = 0 forces eiθ = 1. Hence n   j=1

n     1 + (U z) j = 1 + zj . j=1

Expand the products and equate coefficients. It follows for each symmetric polynomial σk that σk (U z) = σk (z). Since these symmetric polynomials precisely generate the algebra of symmetric polynomials, it follows that U itself is a permutation.  Theorem 5.6 There is a polynomial sphere map f with source dimension n whose Hermitian symmetric group  f is the symmetric group Sn .

Proof Put h(z) = nj=1 (1 + z j ) and consider the map p(z) = h(z) ⊗ z. Since  p(z)2 = |h(z)|2 z2 , the following holds: if U ∈ U(n) and h ◦ U = h, then  p ◦ U 2 =  p2 . By Proposition 5.5, for sufficiently small positive , there is a mapping g such that  p ⊕ g is a rational sphere map, and furthermore (10) holds. We put   f (z) =  p(z) ⊕ g(z) ⊗ z ⊗(n+2) . Then f is a rational sphere map. By Corollary 5.3,  f is contained in U(n). Note that  p and g ⊗ z ⊗(n+2) contain no terms of the same degree. Given U ∈ U(n), Lemma 5.4 guarantees both  p ◦ U 2 =  p2 = |h|2 z2 and (g ⊗ z ⊗(n+2) ) ◦ U 2 = g ⊗ z ⊗(n+2) 2 . It follows that g ◦ U 2 = g2 and |h ◦ U |2 = |h|2 . Lemma 5.6 now implies that  U is (an arbitrary) permutation of the coordinates. Thus  f = Sn .

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8 Groups Arising from Rational Sphere Maps Let us summarize the results about the Hermitian-invariant groups for (non-constant) rational sphere maps. We assume the source dimension is n. 1.  f = Aut(Bn ) if and only if f is a linear fractional transformation. 2.  f is non-compact if and only if  f = Aut(Bn ). Otherwise  f is contained in a conjugate of U(n). 3.  f = U(n) if and only if f is a juxtaposition of tensor powers. 4.  f is a conjugate of U(n) if and only if f is spherically equivalent to a juxtaposition of tensor powers. 5.  f contains an n-torus if and only if f is spherically equivalent to a monomial map. 6. Let f be a symmetric monomial map. Then  f is the semi-direct product of the torus and the group of permutations of the source variables. 7. There is a polynomial sphere map f with source dimension n for which  f = Sn . 8. Let G be an arbitrary finite subgroup of Aut(Bn ). Then there is an N and a proper rational map f : Bn → B N such that  f = G. 9. Let G be an arbitrary finite subgroup of U(n). Then there is an N and a proper polynomial map f : Bn → B N such that  f = G. We have proved most of these results already. Items (1) and (2) were proved in Theorem 5.1. Item (3) is Corollary 5.6. Item (4) follows from Item (3) and Lemma 5.2. Item (5) follows from Proposition 5.3 and Lemma 5.2. Item (6) follows the same lines as in Example 5.10. Item (7) is Theorem 5.6. The proof of item (9) follows the same lines as that of Theorem 5.5. We will prove (9) and derive (8) from it using a simple result from Lie group theory. We now prove one of the main result of this chapter. The discussion in the previous section suggests the proof, but it constructs a polynomial whose degree is very large. The open question stated after the proof seems interesting. Theorem 5.7 Let G be a finite subgroup of Aut(Bn ). Then there is a rational proper map f : Bn → B N for which  f = G. When G is a finite subgroup of U(n), we may choose f to be a polynomial. Proof We first assume G is a subset of U(n) and construct the polynomial map f . Let A denote the algebra of G-invariant polynomials. By Noether’s theorem (see [74]), A is finitely generated. Let {1, h 1 , ..., h K } denote a basis for A; assume h i (0) = 0 for each i. Put h = (h 1 , ..., h K ); thus, h : Cn → C K . Then h is precisely G-invariant, that is, for γ ∈ U(n), we have h ◦ γ = h if and only if γ ∈ G. After tensoring each summand with an appropriate tensor power z ⊗m j we consider the map       p = (1 + h 1 ) ⊗ z ⊗m 1 ⊕ (1 + h 2 ) ⊗ z ⊗m 2 ⊕ ... ⊕ (1 + h K ) ⊗ z ⊗m K . The m j are chosen in order to guarantee that all the monomials in each summand are distinct from those in the other summands, in order to invoke Lemma 5.4. By

8 Groups Arising from Rational Sphere Maps

173

also assuming each m j ≥ 1 we obtain p(0) = 0. By Corollary 5.3,  f ⊆ U(n). By Proposition 5.5, for some  > 0, we can find a polynomial q such  p ⊕ q is a proper polynomial map such that (10) holds. Put f =  p ⊕ (q ⊗ z ⊗m ), where m ≥ deg( p) + 1. We claim that  f = G. Let γ ∈ U(n). Then (10) and Lemma 5.4 imply  p ◦ γ2 =  p2 if and only if q ◦ γ2 = q2 . Hence, if γ ∈ G, then γ preserves both  p2 and q2 . Therefore G ⊆ f. To prove the opposite inclusion, let γ ∈  f ⊆ U(n). Then  f ◦ γ2 =  f 2 . Since γ preserves the degrees of polynomials, Lemma 5.4 gives  p ◦ γ2 =  p2 . Therefore, γ preserves (1 + h i ) ⊗ z ⊗m i 2 = |1 + h i |2 z2m i for each i. Since γ is unitary, it preserves z2 and thus also preserves |1 + h i |2 = 1 + h i + h i + |h i |2 . By Lemma 5.5, γ preserves each h i , and therefore γ ∈ G, establishing the claim and proving the theorem when G ⊆ U(n). Next, we assume G is an arbitrary finite subgroup of Aut(Bn ). Hence G is compact. By Lie group theory G is contained in a conjugate of the maximal compact subgroup U(n). For some χ ∈ Aut(Bn ), we thus have G 0 = χ ◦ G ◦ χ−1 ⊆ U(n). By the result proved above for U(n), there is a polynomial proper map f for which  f = G 0 . By  Lemma 5.2, g = G when g = f ◦ χ. Open problem. Let G be a subgroup of Aut(Bn ). Find the smallest degree d(G, n) for which there is a rational sphere map f of degree d(G, n) with  f = G. Here are some examples: For G = Aut(Bn ), we have d(G, n) = 1. For G = U(1) or G = U(1) × · · · × U(1), we have d(G, n) = 2. If G is finite, then d(G, n) ≥ 3. (Recall the result of Lebl mentioned in Corollary 5.2.) Exercise 5.11 Assume f, g are polynomial maps such that f ⊕ g is an orthogonal sum of tensor products. Thus, they satisfy  f 2 + g2 =



|λ j |2 z2 j .

j

Give an example where f ⊕ (g ⊗ z) is not an orthogonal sum of tensor products. Exercise 5.12 Put n = 2 and write (z, w) for the variables. Let η be a primitive odd p-th root of unity with p ≥ 5. Consider the cyclic subgroup G of the unitary group generated by the map (z, w) → (ηz, η 2 w). Find a basis for the algebra of polynomials invariant under G. When G has order 5, construct a proper polynomial map f for which  f = G.

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Exercise 5.13 Consider the dihedral group D4 with eight elements. Represent it as a subgroup of the unitary group U(2). Suggestion: you can do so with matrices whose entries are 0, 1, −1. If you are ambitious, find a proper polynomial map f for which  f = D4 . (The author does not know the smallest possible degree of such a map.)

9 Different Representations Let G be a finite group. By Cayley’s theorem from elementary algebra, G is isomorphic to a subgroup of a permutation group, and we often regard G in this way. A permutation on n letters may be regarded as an element of U(n), as a permutation matrix is unitary. Hence, we may represent G as a subgroup of U(n). Thus, given G, there is an n for which there is an injective group homomorphism π : G → π(G) ⊂ U(n). Such a map is called a unitary representation of G. Many things in this book depend on the representation of the group. For example, each cyclic group of order d can be represented as a subgroup of U(1) generated by a primitive d-th root of unity. On the other hand, in Chap. 3, we considered the groups (d, q). These groups are cyclic subgroups of order d contained in U(2), but the answers to many questions about these groups depend upon q. Recall that the group ( p, q) is generated by the matrix  ω 0 , 0 ωq where ω is a primitive p-th root of unity and chosen such that q is minimal. The next several examples clarify what this minimality convention means.  4πi η 0 5 Example 5.13 Let η = e . Put A = . At first glance, it appears that A gener0 η3 ates the group (5, 3). Consider insteadusing A2 as a generator. The group elements η2 0 , after interchanging the variables, we are now A2 , A4 , A, A3 , I . Since A2 = 0 η obtain  the group (5, 2). It is, of course, obvious that no generator has the form τ 0 . By our convention, we regard this group as (5, 2). 0τ  2πi η 0 . It appears as if A generates the group Example 5.14 Let η = e 7 . Put A = 0 η3 (7, 3) and, in fact, this assertion is correct. Consider instead using Ak as a generator. k 2k 3k 6k The group elements are now A , A , A , . . . , A , I . None of these matrices are  ω 0 , where q = 1 or q = 2. Thus A generates (7, 3). Note that of the form 0 ωq  5 η 0 A = , 0 η 5

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and, after switching variables, we seem to have (7, 5). By our convention, we choose q minimally, and hence equal to 3. Example 5.15 Just after Definition 3.2, we provided a list of the polynomials f p,4 for 4 ≤ p ≤ 15. One of these which has all positive coefficients, and hence corresponds to a monomial sphere map m (thus m2 = f 7,4 ) is x 7 + 7x 3 y + 14x 2 y 3 + 7x y 5 + y 7 . Interchanging the variables yields f 7,4 (x, y) = f 7,2 (y, x). Hence G m = (7, 2). Recall that a source automorphism γ is in G f if and only if f ◦ γ = f . In other words, f is invariant. Recall from Remark 2.4 that a proper holomorphic map between balls is a finite map. We may assume that its target dimension is minimal. By Proposition 5.1 we may identify G f as the kernel of a group homomorphism. By Proposition 3.2 in [29], G f is a finite group. Forstneriˇc [35] found restrictions on the possible invariant groups for (nonconstant) rational sphere maps. Properness forces the group to be conjugate to a fixed-point free finite unitary group. The fixed-point free subgroups of U(n) have been classified in conjunction with determining the spherical space forms. See Wolf’s classic book [79]. Forstneriˇc ruled out certain fixed-point free subgroups as possible for G f using this classification. Lichtblau showed in his thesis that invariant groups for non-constant rational sphere maps must be cyclic. The result appears in [58]. The proof involves showing that a rational proper map cannot be invariant under a certain kind of (non-diagonal) matrix arising in the classification. By a standard result in [79] a fixed-point free finite Abelian subgroup of U(n) is necessarily cyclic; thus, the proof in [58] could have shown instead that the group was Abelian. The author and Lichtblau later gave the complete short list of representations of (necessarily finite) cyclic groups that can arise. See [27] and [15]. The following theorem answers completely the possible groups that can be G f for some rational proper map f between balls. Theorem 5.8 Let f be a proper rational map between balls. Then G f is cyclic. Furthermore, G f must be conjugate to one of the following: • The cyclic subgroup of U(n) generated by ω I , where ω is an m-th root of unity. Here n is arbitrary. • The cyclic subgroup of U(n 1 + n 2 ) generated by ω In 1 ⊕ ω 2 In 2 , where ω is a primitive odd root of unity. Here n 1 and n 2 are arbitrary. • The cyclic subgroup of U(n 1 + n 2 + n 3 ) generated by ω In 1 ⊕ ω 2 In 2 ⊕ ω 4 In 3 , where ω is a primitive seventh root of unity. Here n 1 , n 2 , n 3 are arbitrary. In each case, f may be chosen to be a monomial sphere map.

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Corollary 5.7 The group (d, q) is a subgroup of G f for some rational sphere map f if and only if q = 1 or q = 2. (In case q = 2, we follow the convention of Definition 3.2, discussed in the three previous examples.) Example 5.16 We give a monomial sphere map m for which G m is as in the third item of the theorem. Here n 1 = n 2 = n 3 = 1 and the source dimension is 3. Also f : B3 → B17 . As usual in the monomial case, we write down the formula for r = m2 . Put x = |z 1 |2 , y = |z 2 |2 , and u = |z 3 |2 . r (x, y, u) = x 7 + y 7 + u 7 + 14(x 3 y 2 + x 2 u 3 + y 3 u 2 + x yu) +7(x 5 y + xu 5 + y 5 u + x y 3 + x 3 u + yu 3 ) +7(x y 2 u 4 + x 2 y 4 u + x 4 yu 2 + x 2 y 2 u 2 ). It is worth noticing how closely this mapping is related to the maps invariant under (7, 2). This map was first written down in the thesis [11] of Chiappari, but it is a special case of the general construction (3.16). Remark 5.7 The quotient of the unit sphere S 3 by the group ( p, q) is called the Lens space L( p, q). Topologists long ago determined when Lens spaces are homeomorphic and when they are of the same homotopy type. We do not discuss this point here, but there are homeomorphic Lens spaces S 3 / X 1 and S 3 / X 2 for which X 1 can be the invariant group of a rational sphere map but for which X 2 cannot. The explanation is that the issue involves the C ∞ topology of the space, rather than its usual continuous topology.

10 Additional Results We refer to [28] for the proof of the following theorem. In the proof it is convenient to regard Aut(Bn ) as the Lie group SU(n, 1)/Z . Recall that Z denotes the center of SU(n, 1). When f is rational,  f becomes an algebraic subvariety of SU(n, 1)/Z . Theorem 5.9 Let f be a rational proper map from Bn to B N . Then  f is a Lie subgroup of Aut(Bn ) with finitely many connected components. The next result generalizes Example 5.9. Theorem 5.10 Let m be a positive integer. Let f m : C → C2 be defined by f m (z) = 1 m (z + z 2m , z 2m − z 3m ). Then f m is a proper mapping from the disk to the ball and 2 hence a rational sphere map. With f regarded in this way,  fm is the cyclic subgroup of 2πi the unit circle generated by e m . In particular, when m = 1, the Hermitian-invariant group  f1 is the trivial group.

10 Additional Results

177

Proof A simple computation shows that  f (z)2 < 1 on the disk and that  f (z)2 = 1 on the circle. Therefore, f is a proper mapping from B1 to B2 . Corollary 5.3 implies that  f is a subgroup of the unitary group U(1), which is the unit circle. We apply Proposition 5.3. Suppose 0 ≤ θ < 2π and eiθ ∈  fm . Assume that  f m (eiθ z)2 − 1 = c ( f m (z)2 − 1).

(16)

Putting z = 0 shows that c = 1 and hence  f m (eiθ z)2 =  f m (z)2 . Using the formula for f m yields |z m + z 2m |2 + |z 2m − z 3m |2 = |z m eimθ + z 2m e2imθ |2 + |z 2m e2imθ − z 3m e3imθ |2 . Hence, for every z and each such θ, we have 2Re(z m z 2m − z 2m z 3m ) = 2Re(eimθ (z m z 2m − z 2m z 3m )). Equating coefficients of the z m z 2m term gives eimθ = 1, and the other coefficients are then equal as well. Hence, mθ is a multiple of 2π and the result follows.  Consider the mappings z → z ⊗m , for m a positive integer. When m = 1, the map f is the identity map, and therefore its group  f is the full automorphism group. For m ≥ 2, the group  f is the unitary group. More generally we have seen that  f = U(n) if and only if f is a juxtaposition of tensor powers. To be precise, we allow f to have only one term as long as f is not linear. We naturally ask what happens if we take restricted tensor products as in Chap. 4. Several things can happen. For example, if we begin with the identity map f , then  f is the full automorphism group, and a full or partial tensor product will decrease the group. If n = 2 and f (z, w) = (z 2 , zw, w), then f is a restricted tensor product of the identity map and  f is a torus U(1) × U(1). But, when we take a restricted tensor product on the third component of f , we get a map whose group is U(2). For clarity, we repeat this point in the next remark. Remark 5.8 There exist rational sphere maps f and tensor products E f such that  E f is a proper subset of  f , for which they are equal, and for which  f is a proper subset of  E f . Exercise 5.14 Give examples verifying Remark 5.7. The following lemma allows for a more general unitary U than a permutation. The result is closely related to Example 5.10. Lemma 5.7 For z ∈ Cn , define p by p(z) = (..., z j z k , ....), where 1 ≤ j < k ≤ n. Suppose U is unitary and that  p(U z)2 =  p(z)2 . Then there is a diagonal matrix L and a permutation σ of the coordinates such that U = Lσ.

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Proof First suppose n = 2. Let U be unitary. We are given p(U z) = (u 11 z 1 + u 12 z 2 )(u 21 z 1 + u 22 z 2 ) and  p(U z)2 =  p(z)2 = |z 1 z 2 |2 . The coefficients of |z 1 |4 and |z 2 |4 both vanish, and hence we get two equations: u 11 u 21 = 0 and u 12 u 22 = 0. If u 11 = 0, then U unitary implies u 21 = 0 and u 12 = 0. The second equation yields u 22 = 0. Let L be the diagonal matrix with diagonal entries u 12 and u 21 , and let σ permute the coordinates. The conclusion holds. If u 11 = 0, then u 12 = 0 and hence u 22 = 0. Now the second equation gives u 21 = 0, and U is diagonal. Thus, the conclusion holds with σ the identity. Consider n = 3. We are given p(U z) = (a1 (z), a2 (z), a3 (z)) where a1 (z) = (u 11 z 1 + u 12 z 2 + u 13 z 3 )(u 21 z 1 + u 22 z 2 + u 23 z 3 ) a2 (z) = (u 11 z 1 + u 12 z 2 + u 13 z 3 )(u 31 z 1 + u 32 z 2 + u 33 z 3 ) a3 (z) = (u 21 z 1 + u 22 z 2 + u 23 z 3 )(u 31 z 1 + u 32 z 2 + u 33 z 3 ). The coefficient of each |z j |4 in  p(z)2 is 0. The coefficients cl of |zl |4 in  p(U z)2 are given by c1 = |u 11 u 21 |2 + |u 11 u 31 |2 + |u 21 u 31 |2 c2 = |u 12 u 22 |2 + |u 12 u 32 |2 + |u 22 u 32 |2 c3 = |u 13 u 23 |2 + |u 13 u 33 |2 + |u 23 u 33 |2 . Hence, each of the nine terms separately vanishes. For general n these equations become, for each l with 1 ≤ l ≤ n, 

|u jl u kl |2 = 0.

(17)

1≤ j KKn1 |μ j | for all j. Then (20) is violated, and hence m · (η − μ j ) = 0. But (19) holds, and therefore cη , cμ j  = 0 for all the μ j . Since the cμ j span, we conclude that cη = 0. Hence, there are only finitely many non-vanishing coefficients and f is a polynomial.  The hypothesis of Proposition 5.8 evokes the group  f . Rephrasing things in terms of this group yields the following theorem and several corollaries. Theorem 5.11 Suppose f : Bn → B N is holomorphic and f (0) = 0. Assume that the Lie algebra g of  f ∩ U(n) contains a matrix M such that −i M has all positive eigenvalues. Then f is a polynomial. Proof We may write M = i P D P ∗ where D is diagonal with positive eigenvalues and P is unitary. Let W be the subspace of V spanned by M. Then W is the Lie algebra of a group G ⊆  f ∩ U(n). Its elements are maps of the form Pei Dt P ∗ for t ∈ R. Thus, since f (0) = 0,  f (z)2 =  f (Pei Dt P ∗ z)2 =  f (Pei Dt P −1 z)2 .

(21)

Put z = P(ζ) and put h = f ◦ P. Then h(0) = f (P(0)) = f (0) = 0. We claim that ei Dt lies in h . We must show that h ◦ ei Dt 2 = h2 . But (21) implies h(ei Dt ζ)2 = h(ei Dt P −1 (z))2 =  f ◦ P(ei Dt P −1 (z))2 =  f (z)2 = h(ζ)2 . Note that dtd ei Dt = i Dei Dt and hence the derivative at t = 0 is i D. Hence, the eigenvalues of −i(i D) are positive and Proposition 5.8 implies that h is a polynomial.  Thus, f = h ◦ P −1 is also a polynomial. Corollary 5.8 Suppose f : Bn → B N is holomorphic and  f contains the center of U(n). Then f is spherically equivalent to a polynomial and hence rational. Proof Suppose first that f (0) = 0. The center of U(n) consists of the matrices eiθ I . For θ = 0, consider the curve t → eiθt I = γ(t). Its derivative at t = 0 equals iθ I , and hence the eigenvalues of −iγ (0) are all positive. By Proposition 5.8, f is a polynomial. If f (0) = 0, we may compose with an automorphism and obtain a spherically equivalent map g with g(0) = 0. Since  f is a conjugate of g , the result follows.  Corollary 5.9 Suppose f : Bn → B N is holomorphic and  f contains the center of U(n). If  f is not compact, then f is a linear fractional transformation.

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Proof After composing with an automorphism, we may assume that f (0) = 0 and the condition on  f is unchanged. Theorem 5.11 implies that f is a polynomial. Hence, the result follows from Theorem 5.2.  Remark 5.9 Several of the results summarized in Section 8 assume that the group  f for a rational sphere map f has some property and draws a conclusion about f . Theorem 5.11 sometimes allows one to make no a priori assumption of rationality but to draw the same conclusion nonetheless. Exercise 5.18 Give an example of a holomorphic map f : C2 → C such that f ◦ U = f for some U ∈ T(2) (not the identity) but f is not a polynomial. Give an example for which f ◦ U = f for U in an infinite subgroup of U(2) but f is not a polynomial.

Chapter 6

Elementary Complex and CR Geometry

In this chapter, we will establish that a non-constant equi-dimensional rational sphere map is a ball automorphism when n ≥ 2. The proof of this simple sounding result requires some geometric information which we gather. We will discuss the rudiments of CR geometry, including the Levi form and strong pseudoconvexity. We will encounter an unbounded realization of the unit sphere, the Heisenberg group, and an algebraic variety X f associated with a rational sphere map f . We move even further from the unit sphere by stating and using some results by Baouendi-Rothschild and Baouendi-Huang-Rothschild on complex analogues of the Hopf lemma. We conclude the chapter by proving a basic fact about rational sphere maps. In Definition 2.1, we assumed that the denominator of a rational sphere map does not vanish on the closed ball. If a rational map is proper and reduced to lowest terms, then the denominator cannot vanish on the closed ball. Thus, each rational proper map defines a rational sphere map. The proof uses the unbounded realization of the unit sphere and polarization.

1 Subvarieties of the Unit Ball Zero sets of different classes of functions exhibit rather different behavior. We begin by briefly discussing the set of common zeros of holomorphic functions. Definition 6.1 A subset V of Cn is called a complex analytic subvariety if for each p ∈ V , there is a neighborhood  of p and finitely many complex-valued holomorphic functions f 1 , ..., f k defined in  such that  ∩ V is the set of common zeroes of the f j . When the functions f j can be chosen to be polynomials, V is called a complex algebraic subvariety. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 J. P. D’Angelo, Rational Sphere Maps, Progress in Mathematics 341, https://doi.org/10.1007/978-3-030-75809-7_6

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Definition 6.2 A complex analytic variety V is called a complex submanifold of dimension n − k or codimension k if, at each point p ∈ V we can find functions f j as in Definition 6.1 such that d f 1 ∧ · · · ∧ d f k ( p) = 0. A point p ∈ V where V is a submanifold is called nonsingular; the complement of the nonsingular points consists of singular points. We make a few simple observations. Any open subset of Cn is a complex manifold of dimension n. None of the functions f j are used! We can think of Cn as having n independent variables. Each time we set a function f j equal to 0 we would like to say (locally) that we are eliminating a variable. The condition d f 1 ∧ · · · ∧ d f k = 0 combines with the implicit function theorem to make this intuition precise. One difference between real and complex geometry is that we can define a real submanifold M of Rn of dimension n − k by setting a single smooth function equal to 0; if M is locally given by f 1 = · · · = f k = 0, then M is also locally given by setting F = f 12 + · · · + f k2 equal to 0. Here F is also a smooth function. The condition on the wedge products is violated however. In the complex case, the set of common zeroes of f 1 , ..., f k is, of course, the same as the zero set of the single function  | f j |2 , but this function is not complex analytic unless each f k is constant. Another (huge) difference in the real case is the standard result that every closed subset of Rn is the zero set of a smooth function. The main point in the proof is that there exist smooth functions with constant value 1 on a closed ball but identically zero off a slightly larger closed ball with the same center. Such functions are smooth, but of course not real analytic. In the complex case, varieties are natural objects to study. They are less general than arbitrary closed sets, but general enough to exhibit all sorts of interesting behavior. In Sect. 6, we will associate complex varieties with rational sphere maps. Even these varieties exhibit quite distinct behavior. √ Example 6.1 Let V be the zero set of the function f given by z 23 − 3 3z 1 z 3 . Thus, V is a complex subvariety of C3 . Here √ √ d f = −3 3z 3 dz 1 + 3z 22 dz 2 − 3 3z 1 dz 3 . 2 The origin is the only singular point. √ This3variety is the image of C under the 3 polynomial sphere map ζ → (ζ1 , 3ζ1 ζ2 , ζ2 ).

Exercise 6.1 Define f : C → C2 by f (ζ) = (ζ 2 , ζ 3 ). Show that the image V under f is a complex analytic subvariety of C2 . (In other words, find a function whose zero set is V .) Show that V is not a complex manifold. (There is a singularity at only one point.) Exercise 6.2 Consider the variety in the previous exercise. Show that there exists a rational function f of two complex variables with the following properties. The function f is continuous on V , but has a singularity in C2 at the singular point of V . A complex variety, perhaps with singularities, is called normal if this phenomenon does not happen! This variety V is the simplest example of a non-normal variety.

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185

Exercise 6.3 Consider the Whitney map given by W (z, w) = (z, zw, w2 ). Show that the image of C2 under W is a complex subvariety. What are its singular points? Exercise 6.4 Consider the monomial sphere map (z, w) → (z 2 , z 2 w, zw 2 , w). Show that the image of C2 is a complex analytic subvariety of C4 . What are its singular points? Exercise 6.5 Put f (z, w) = (z, zw). Show that the image of f is not a variety. We need the following result. A positive dimensional complex analytic subvariety of the unit ball cannot be compact. In fact, the only compact complex analytic subvarieties are finite sets. Proposition 6.1 In every dimension n, a compact complex analytic subvariety of an open subset of Cn is a finite set. Hence, when n ≥ 2, a compact complex analytic subvariety of an open subset of Cn cannot be positive dimensional. Proof This result is proved in detail in [71], by induction on the dimension and via a study of projections of varieties. We give a different proof in the special case when V = f −1 (0) for a single holomorphic function f . If V were compact, then 1f would be holomorphic in the complement of a compact subset of a ball. By Hartogs’s extension theorem (see [44], for example), 1f would extend to be holomorphic in the ball, and hence V would be empty.  Proposition 6.2 Suppose f is a rational sphere map whose source dimension exceeds its target dimension. Then f is a constant. Proof We are given that f maps the unit sphere in the source to the unit sphere in the target. By the maximum principle, f maps the closed unit ball in the source to the closed unit ball in the target. If f is not a constant, then f : Bn → B N must be a proper map. The inverse image V of a single point ζ in the image of f is then compact by the definition of proper map. On the other hand, V is the solution set of the N polynomial equations p j (z) − q(z)ζ j = 0. Thus, V is a compact complex algebraic subvariety in the open ball. By Proposition 6.1, V cannot be positive dimensional. Thus, either f is constant or N ≥ n. 

2 The Unbounded Realization of the Unit Sphere In this section, we assume n ≥ 2. The unit sphere is the simplest example of a strongly pseudoconvex real hypersurface in Cn . A real hypersurface M of Cn is a real submanifold of real codimension 1. Typically M is given as the zero set of a smooth real-valued function r with dr = 0 on M. For example, the function r (z, z) = z 2 − 1 is a defining function for the unit sphere. Defining functions are not unique, as one can multiply a given defining function by a non-vanishing function. One can also make biholomorphic changes of coordinates.

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A key point in our discussion will be that an equi-dimensional rational sphere map f provides a defining function ρ defined by ρ = f 2 − 1. To check that ρ is a defining function, we need to know that dρ = 0 on ρ = 0. This result is best understood by considering an unbounded realization of the unit sphere. Remark 6.1 Up to now, we have defined the unit ball defined via the squared norm. Doing so has enabled us to take advantage of homogeneity. The usual conventions in CR geometry suggest, however, that one should study the unbounded realization of the ball described below. The idea extends a venerable method in (both pure and applied) one complex variable: it is often useful to conformally map the unit disk to the upper half-plane. For w ∈ C, note that 4Re(w) = |w + 1|2 − |w − 1|2 . For ζ ∈ Cn , it follows that 4Re(ζn ) +

n−1 

|ζ j |2 = 0

(1)

j=1

if and only if n−1 

|ζ j |2 + |ζn + 1|2 = |ζn − 1|2 .

(2)

j=1

For ζn = 1, we divide (2) by |ζn − 1|2 to see that (1) holds if and only if n−1  j=1

|

ζj 2 ζn + 1 2 | +| | = 1. ζn − 1 ζn − 1

(3)

ζ

j for 1 ≤ j ≤ n − 1 and z n = ζζnn +1 . PlugDefine a change of variables by z j = ζn −1 −1 2 ging into (3) gives the equation of the unit sphere z = 1. Formula (1) defines an unbounded representation of the unit sphere. We have moved the boundary point z = (0, ..., 0, 1) to ζ = (0, ..., 0, ∞) via a linear fractional transformation. This unbounded representation has many applications in CR geometry, a term we will not precisely define. We will use this formula to help us prove that a proper map of a ball in two or more dimensions is actually an automorphism. Note also that the origin lies on the hypersurface M defined by (1). To agree with common usage we make a small notational change in (1). We write wn = −4iζn , and write w j = ζ j for 1 ≤ j ≤ n − 1. Put w = (w1 , ..., wn−1 ). Then (1) becomes n−1  |w j |2 . (4) Im(wn ) = w 2 =

j=1

We let Un = {w ∈ Cn : Im(wn ) > w 2 } and we regard Un as an unbounded realization of the unit ball Bn . Of course Im(wn ) = w 2 defines the boundary ∂Un ,

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187

an unbounded realization of the unit sphere. We next make the standard connection to the Heisenberg group. Definition 6.3 The Heisenberg group Hn−1 is the group Cn−1 × R with multiplication defined by (z, t) ∗ (w, s) = (z + w, t + s + 2Im(z, w)) . The identity element is (0, 0) and the inverse of (z, t) is given by (−z, −t). Suppose (z, t) ∈ Hn−1 and w ∈ Cn . We can move w using (z, t) as follows: Tz,t (w) = (w + z, wn + t + i z 2 + 2iw , z).

(5)

Lemma 6.1 Tz,t maps Un to itself and ∂Un to itself. Proof

Tz,t (w , wn ) = (w + z, wn + t + i z 2 + 2iw , z).

The imaginary part η of the second component is η = Im(wn ) + z 2 + 2Rew , z = Im(wn ) + z + w 2 − w 2 . Then Im(wn ) > w 2 implies η = Im(wn ) + z + w 2 − w 2 > z + w 2 and  Im(wn ) = w 2 implies η = z + w 2 . Theorem 6.1 The map Tz,t is a holomorphic automorphism of Un . The following formula holds: (6) Tz,t ◦ Tζ,τ = T(z,t)∗(ζ,τ ) . Proof Formula (6) shows that Tz,t is bijective, as its inverse is T−z,−t . It is holomorphic because it depends on w but not on w. Checking the formula is routine, but slightly tedious, and left to the reader.  Remark 6.2 For r > 0, define a map on Cn by the anisotropic dilation δr (w) = (r w , r 2 wn ). Then δr is also an automorphism of Un . Exercise 6.6 What is the condition on Heisenberg group elements (z, t) and (w, s) such that they commute? In other words, (z, t) ∗ (w, s) = (w, s) ∗ (z, t). Exercise 6.7 Verify formula (6). Exercise 6.8 Verify the statement in Remark 6.2.

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Remark 6.3 Not all the automorphisms of the ball arise in these two ways, as the unitary maps are excluded. This discussion is closely connected to the Iwasawa decomposition, mentioned in Sect. 2 of Chap. 1. We have made considerable use of Euclidean homogeneity in this book. The author wonders whether it is possible to gain additional insight using the Heisenberg methods in this section. The author himself has been unable to use these ideas to establish many of the results in this book. Open problem. Use the so-called CR vector fields, defined in the next section, on the unbounded realization of the unit sphere to prove Theorems 3.1 and 4.4.

3 Geometry of Real Hypersurfaces We prepare for the required geometry by making some standard remarks that lie at the basis of both complex and CR geometries. We mention a standard amusement; the letters CR stand for both Cauchy-Riemann and complex-real. See [5] and [45] for excellent accounts of various aspects of CR geometry. We start with R2n . Its tangent bundle T (R2n ) is defined as usual. The tangent bundle is simply the product R2n × R2n , and is thus a trivial vector bundle. When we identify R2n with Cn , a process that can be done in multiple ways, interesting and subtle things happen. Let us identify R2n with Cn in the following way. Denote by x1 , ..., x2n the usual coordinates on R2n . Write y j for xn+ j . For 1 ≤ j ≤ n, we define z j = x j + i xn+ j = x j + i y j and the z j becomes holomorphic coordinates on Cn . Of course z j = x j − i y j . We extend the basic notions of real geometry to the complex case in the following obvious way. Assuming we know the meaning of the contraction of real 1-forms and vector fields, we define the contraction of complex-valued forms and fields by α + iβ, v + iw = α, v − β, w + i(α, w + β, v). Remark 6.4 We have used the same notation for contraction and inner product. In the contraction, there is no complex conjugate. Since contraction appears only in this chapter and only a few times, no confusion should result. The exterior derivative d is defined as a map from functions to 1-forms. As usual, d f is defined by its contraction with a vector: d f, v =

∂f = v[ f ], ∂v

the usual directional derivative. When f = g + i h is complex valued, we put d f = dg + idh. Defining the exterior derivative d on complex-valued functions determines the complex-valued differential forms dz j and dz j :

3 Geometry of Real Hypersurfaces

189

dz j = d x j + idy j dz j = d x j − idy j . We obtain a basis of the complex-valued 1-forms and determine a splitting of the complex-valued 1-forms into forms of type (1, 0) (combinations of the dz j ) and those of type (0, 1) (combinations of the dz j ). We define the complex vector fields ∂z∂ j and ∂z∂ j by duality. For example, the

contraction dz j , ∂z∂ k  equals 1 when j = k and is zero otherwise. Also, dz j , ∂z∂ k  equals 0 for all j, k. These ideas determine the following formulas; these formulas are often given as definitions, but they are determined by the above duality:   ∂ 1 ∂ ∂ = −i ∂z j 2 ∂x j ∂yj ∂ 1 = ∂z j 2



∂ ∂ +i ∂x j ∂yj

 .

Exercise 6.9 Derive the formulas for the complex derivatives from the above considerations of duality. Then explain why it is the barred derivative operator with the plus sign in front of i. Once we have the splitting of 1-forms into (1, 0) forms and (0, 1) forms, we obtain the usual notions of ( p, q) forms. As usual, one defines the splitting d = ∂ + ∂ of the exterior derivative. For a function, ∂h is the (0, 1) part of dh, and hence h is holomorphic on an open set if and only if ∂h = 0 there. The operators ∂ and ∂ extend to differential forms of all degrees. We will use differential forms of higher degree in the last chapter. The splitting of differential 1-forms into types (1, 0) and (0, 1) induces a similar splitting of the complex vector fields and thereby defines the vector subbundles T 1,0 (Cn ) and T 0,1 (Cn ) of the complexified bundle T (Cn ) ⊗ C. In coordinates, sections of T 1,0 (Cn ) are functional combinations of the ∂z∂ j . Sections of T 0,1 (Cn ) are functional combinations of the

∂ ∂z j

.

Exercise 6.10 On each tangent space to R2n , there is a (real) linear map J whose square is minus the identity. A choice of such a J at each point is called an almost complex structure. Choose J that is compatible with our discussion above. What does J do to the vector fields ∂x∂ j , ∂∂y j , and ∂z∂ j ? Definition 6.4 A real hypersurface of Cn is a smooth real submanifold of real codimension 1. Let M be a real hypersurface of Cn and suppose p ∈ M. By definition, there is a neighborhood of p on which M is given by the zero set of a smooth real-valued function r such that dr (x) = 0 for x in this neighborhood. Such an r is called a local

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defining function. Usually, we will express what we do using defining functions, but at times it is useful to be more abstract. Definition 6.5 The complexified tangent bundle to a real hypersurface M, written CT M, is defined to be the tensor product T M ⊗ C . Its sections are complex vector fields tangent to M. By definition, T 1,0 (M) = CT M ∩ T 1,0 (Cn ). Also T 0,1 (M) is the complex conjugate bundle, and thus T 0,1 (M) = CT M ∩ T 0,1 (Cn ). Observe that T 1,0 (M) ∩ T 0,1 (M) = {0}. The local sections of T 1,0 (M) are complex vector fields involving only the ∂z∂ j . These local sections must be tangent to  M, and hence they annihilate the defining function. Put L = nj=1 a j ∂z∂ j . The conn dition that L be tangent to the zero set of r becomes j=1 a j r z j = 0 on r = 0. Notice the following implication. Assume r is a defining function for M, and u is a non-vanishing function. Then L(r ) = 0 on M =⇒ L(ur ) = u L(r ) + L(u)r = 0 on M. Therefore, the space T 1,0 (M) does not depend on the choice of defining function. The bundles T 1,0 (M) and T 0,1 (M) are integrable in the sense of Frobenius; thus, they are closed under the Lie bracket operation. In general, and in particular when M is the unit sphere, the bundle T 1,0 (M) + T 0,1 (M) is not integrable. The Levi form measures the extent to which integrability fails. The sum T 1,0 (M) + T 0,1 (M) forms a subbundle of codimension one in CT M. Let us determine local sections in terms of a defining function r for M. Since dr = 0 ∂r = 0 there. on M, we may choose coordinates near a point p ∈ M such that rn = ∂z n We use subscripts for partial derivatives to keep the notation under control. Since r is real valued r j = r j . We then define Lj =

rj ∂ ∂ − , ∂z j rn ∂z n

Lj =

rj ∂ ∂ − , ∂z j rn ∂z n

T =

1 ∂ 1 ∂ − . rn ∂z n rn ∂z n

These vector fields form a local basis for sections of CT M. The L j form a local basis for sections of T 1,0 (M), their complex conjugates L j form a local basis of sections of T 0,1 (M), and the missing direction in CT M is generated by T . Local sections of T 0,1 (M) are called CR vector fields. Exercise 6.11 Verify that the L j , the L j , and T are tangent to M and form a basis for local sections of CT (M).

3 Geometry of Real Hypersurfaces

191

Definition 6.6 The Lie bracket [L , K ], or commutator, of the vector fields L , K is the vector field whose action on functions is given by [L , K ]( f ) = L(K ( f )) − K (L( f )). Exercise 6.12 Why is [L , K ] a vector field? Exercise 6.13 Compute the Lie brackets [L j , L k ] and [L j , L k ] . We now finally define the Levi form on a real hypersurface M. Given that T 1,0 M + T M has codimension one in CT M, this sum is the space of sections annihilated by a non-vanishing 1-form. Following the notation in [15], we call this 1-form η. It is defined only up to a multiple. We may assume that η is purely imaginary. We can fix this multiple by assuming that T has been chosen and that the contraction T, η identically equals 1. We usually choose η = ∂r − ∂r when r is a defining function for M. 0,1

Definition 6.7 The Levi form λ on a real hypersurface M is the Hermitian form defined for L , K ∈ T 1,0 by λ(L , K ) = [L , K ], η. See, among many sources, [15], for more information about the Levi form on a real hypersurface. The Levi form can be regarded (Exercise 6.15) as the restriction of the complex Hessian of a defining function to T 1,0 (M). Definition 6.8 A real hypersurface is called strongly pseudoconvex if λ is definite and pseudoconvex if λ is semi-definite. Remark 6.5 The Levi form is defined only up to a multiple. In case M is a strongly pseudoconvex real hypersurface bounding a bounded domain, one chooses the signs to make λ positive definite. The Hessian of a twice-differentiable function r of n complex variables is the matrix r z j z k for 1 ≤ j, k ≤ n. We sometimes write this matrix as r jk . One often regards the Hessian as the matrix of a Hermitian form. Suppose r is a smooth defining function for a real hypersurface, and u is a non-vanishing smooth function. Then ur is also a defining function. Computing the Hessian of ur gives the matrix (ur ) jk = ur jk + u j rk + u k r j + u jk r. This formula shows that, on the set r = 0, the restriction of the Hessian of ur to vectors in T 1,0 is the same as u times the Hessian of r there. The following exercise provides a useful formula for the Levi form. Exercise 6.14 Use Definition 6.7 to prove that the Levi form on a real hypersurface defined by r = 0, near a point where rn = 0, can be identified with the (n − 1) by (n − 1) matrix whose entries λ jk are (using subscripts to denote partial derivatives):

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|rn |2 r jk − rn rk r jn − r j rn rnk + r j rk rnn . Exercise 6.15 Using either Exercise 6.14 or directly, verify that the Levi form is the restriction of the Hessian of a defining function to the space T 1,0 . Exercise 6.16 Show that λ(L , K ) = λ(K , L) . Thus λ is Hermitian. Exercise 6.17 Suppose λ(L , L)( p) ≥ 0 for all L ∈ T 1,0 and some p ∈ M. Show that |λ(L , K )( p)|2 ≤ λ(L , L)( p) λ(K , K )( p). The author introduced a finite-type condition to describe hypersurfaces for which the Levi form degenerates at a point, but whose geometry there behaves qualitatively like that of a point on a strongly pseudoconvex hypersurface. Perhaps the simplest analogy is to the graphs in the plane defined by y = x 2m , for m an integer at least 1. Each of these curves looks something like a parabola, but is a parabola only when m = 1. The curves become flatter at the origin as m increases. The second derivative is positive there only when m = 1, but for each m, the origin is a strict local minimum for the function x → x 2m . In higher dimensions, finite type is somewhat analogous to considering a critical point p of a smooth function r with a strict local minimum at p. There is no simple test for deciding whether a degenerate critical point is a strict local minimum. The strongly pseudoconvex case is analogous to p being a non-degenerate local minimum. Exercise 6.18 Suppose that M is strongly pseudoconvex and t → z(t) is a holomorphic map whose image lies in M. Show that z is constant. Exercise 6.19 Suppose that  is a bounded domain in Cn with smooth boundary. Show that ∂ contains some strongly pseudoconvex points. The book [15] devotes several lengthy chapters to the concept of D-finite type. We give an informal definition here: M is D-finite type at p if there is a bound on the order of contact of all complex analytic varieties with M at p. The definition allows the varieties to be singular. If M is strongly pseudoconvex at p, then M is D-finite type there. The order of contact is at most 2. We give several examples. Example 6.2 Let  be a neighborhood of 0 ∈ Cn−1 and let f :  → C N be a holomorphic map with f (0) = 0. Define M to be the zero set of r , where r (z) = 2Re(z n ) + f (z) 2 . Then M is D-finite type at 0 if and only if the (germ at 0 of the) variety V ( f ) defined by the f j consists of 0 alone. Commutative algebra provides many tests for deciding whether V ( f ) = {0}. When the number of functions f j is the same as the number of variables, one often says that f has finite multiplicity at the origin. In this situation, the inverse image of each point sufficiently close to 0 is a finite set. The generic number of points in this set can be computed in many ways. See [15] for detailed discussion.

3 Geometry of Real Hypersurfaces

193

Example 6.3 Put r (z) = 2Re(z 2 ) + |z 1 |2m + ..., where the dots denote higher order terms. Then M is D-finite type at 0. Example 6.4 Put n = 3 and let M be the zero set of the function r (z) = 2Re(z 3 ) + |z 12 − z 23 |2 . By Example 6.2, M is not D-finite type at the origin. The complex variety defined by z 12 − z 23 = z 3 = 0 lies in M. The reader should also study the next exercise. Exercise 6.20 For M as in Example 6.4, let p be the origin. Define L j as usual. Show that [[[L 1 , L 1 ], L 1 ], L 1 ], η( p) = 0 [[[[[L 2 , L 2 ], L 2 ], L 2 ], L 2 ], L 2 ], η( p) = 0. Show more generally the following: if L is any type (1, 0) vector field with L( p) = 0, and we take iterated commutators as above, then, after either four or six total brackets, the contraction with η at p is non-zero. Remark 6.6 Example 6.4 and Exercise 6.20 show that one cannot detect whether a hypersurface contains a complex variety by using iterated commutators of vector fields of type (1, 0) at a single point.

4 CR Functions and Mappings CR functions are analogues of the boundary values of holomorphic functions, and CR mappings between hypersurfaces M and M are analogues of holomorphic mappings whose restrictions to M map to M . We assume that these functions are continuously differentiable, although one often defines them (using distribution theory) under weaker regularity assumptions. Definition 6.9 Let M, M be real hypersurfaces in complex Euclidean spaces of possibly different dimension. A continuously differentiable function f : M → C is called a CR function if L( f ) = 0 for all sections L of T 0,1 (M). A continuously differentiable function f : M → M is called a CR mapping if its (real) derivative d f maps T 1,0 (M) → T 1,0 (M ). An equivalent statement is that each component of f is a CR function. Suppose p ∈ M, that  is a neighborhood of p, and h :  → C N is holomorphic. When N = 1, the restriction of h to M is a CR function. When M ⊆ C N is a hypersurface such that the restriction of h to M maps into M , then h is a CR map. Without any geometric conditions on M, one cannot conclude that a CR map is the restriction of a holomorphic map.

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We pause to consider the opposite extreme from strong pseudoconvexity, called Levi flat. Consider the hypersurface M defined by Im(z n ) = 0. It is clear that M is a flat 2n − 1 real dimensional hyperplane in Cn . The CR vector fields on M are given by L j = ∂z∂ j . Hence, any function depending only on Re(z n ) is a CR function. On the other hand, if M is given by Im(z n ) = z 2 , then every CR function on M extends to be holomorphic on at least one side of M . The problem in the Levi flat situation is that M contains a complex analytic hypersurface of complex codimension 1. Such extension results hold much more generally and have been a major part of the literature in CR geometry for years. For example, Trepreau gave a necessary and sufficient condition on the germ at a point of a smooth real hypersurface for every CR function to extend holomorphically to at least one side, namely, there is no complex hypersurface passing through p and lying in M. This condition differs considerably from being D-finite. D-finiteness is much stronger, as it considers complex varieties of all possible dimensions and also demands a bound on the order of contact. Tumanov gave a necessary and sufficient condition for extension of CR functions (to a wedge) in the case of CR manifolds of higher codimension. We refer to [8] and its reference list for more information. Our work on rational sphere maps considers different issues; by assumption such maps are holomorphic in a neighborhood of the closed ball. It is only in the equidimensional case that things are simple. We will use the geometry of hypersurfaces to see why rational maps are automorphisms in the equi-dimensional case when n ≥ 2. By now there are many proofs and many generalizations. We next state without proof several remarkable results of Baouendi-Rothschild ([7] and [8]). They prove several even stronger results. Theorem 6.2 Let H : M → M be a formal CR map, where Mand M are formal hypersurfaces at the origin in Cn+1 . Assume M is of finite n-type. Let w denote a transversal coordinate for M. If H is of finite multiplicity and G is a transversal component of H , then we have ∂G (0) = 0. ∂w While we will not give details here, we explain to some extent what the result says. A formal CR map is simply a collection of formal power series in the z variables, and thus independent of the z variables. A formal hypersurface is defined by giving a formal power series r depending on n + 1 complex variables and their conjugates.  It is assumed that r is formally real valued, that is, the formal series cαβ z α z β satisfies the Hermitian symmetry condition cαβ = cβα . It is assumed that r (0) = 0, that is, r has no constant term. It is also assumed that dr (0) = 0, in other words, some first-order coefficient does not vanish. We make the same assumptions about the formal defining function ρ for M . The assumption that M is of finite n-type is rather weak; it simply means that every complex hyperplane

4 CR Functions and Mappings

195

of dimension n has finite order of contact with M at 0. The notion of transversal coordinate arises throughout CR geometry. It comes about by assuming that (even formal) defining functions are given by equations such as Im(z n+1 ) = φ(z, z, Re(z n+1 )). Here φ contains no pure terms. In this setting, z n+1 is a transversal coordinate. We write the map H as an n + 1-tuple of the form (H1 , ..., Hn , G). The condition that H formally maps M to M means that the power series Im(G(z)) − ψ(H (z), H (z)) vanishes when we plug in the condition Im(z n+1 ) = φ(z, z, Re(z n+1 )). Remark 6.7 The notion of finite multiplicity is mentioned in Example 6.2 for holomorphic germs. Many of the methods for computing the multiplicity extend to the context of formal power series. Perhaps the most natural definition of multiplicity here is that the quotient vector space C[[z]]/(H ) be finite dimensional. Here (H ) denotes the ideal generated by the components of H in the formal power series ring C[[z]]. The simplest case of Theorem 6.2 is when M is the unbounded realization of the unit sphere in the source, M is the unbounded realization of the unit sphere in the target, and f : M → M is holomorphic. Put z = (z , z n+1 ). All the conditions are met when Im(z n+1 ) = z 2 defines M, Im(ζn+1 ) = ζ 2 defines M , and f (z) = ( f (z) , G(z)). Assume Im(G(z)) = f (z) 2 when z ∈ M. The theorem tells us that ∂G (0) = 0 and hence Im(G(z)) − f (z) 2 is a defining equation for the target ∂z n+1 sphere. Once we know that some derivative does not vanish at 0, it will be easy to show that f is an automorphism. Baouendi and Rothschild also established the following result. Either of these theorems imply that an equi-dimensional rational sphere map is an automorphism. Theorem 6.3 Let M be a D-finite real hypersurface in Cn . Then a CR mapping h : M → M is either constant or a local diffeomorphism. Furthermore, if M is real analytic, then h extends to be locally biholomorphic. Corollary 6.1 A non-constant rational sphere map f with the same source and target dimension (at least 2) defines a smooth local diffeomorphism of the unit sphere to itself. Furthermore, for each p in the unit sphere, f is biholomorphic in some neighborhood of p. As a result, d f ( p) is invertible for each p near the sphere. Proof The sphere is strongly pseudoconvex, and hence D-finite. By definition, a rational sphere map f with the same source and target dimension maps the sphere to itself. Since f is holomorphic, its restriction to the sphere is a CR map, and hence is locally biholomorphic. 

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Corollary 6.2 Let f be a proper holomorphic map from Bn to itself that extends to be at least twice differentiable at the boundary. Then d f ( p) has maximal rank at each p on the sphere and det(d f ) does not vanish there. Proof Let F denote the map f expressed in real variables. Then d F( p) has full rank at p because it is a local diffeomorphism. Thus det(d F( p)) = 0. But det(d F( p)) =  | det(d f ( p)|2 and hence det(d f )( p) is also not 0. For n ≥ 2, the next result has a fascinating history. Rather than attempting to discuss it, we refer to several papers among many: [4] and [36]. The paper [4] is one of the wittiest mathematical papers this author has ever read. By now there are innumerable results along the lines that a proper holomorphic self-map of a domain must be an automorphism. We need here only the case of the ball; the result holds for strongly pseudoconvex domains and for many classes of weakly pseudoconvex domains. The most difficult issue is the extension of such maps to the boundary. Once the map extends, the techniques of CR geometry become available. Theorem 6.4 Let f be a rational sphere map with the same source and target dimensions. If n = N = 1, then f is a finite Blaschke product. If N = n ≥ 2, then f is either an automorphism of the ball, and thus a linear fractional transformation, or f is constant. Proof We proved the result when n = 1 in Proposition 1.13. The higher dimensional case relies on a bit of topology. A proper map from a ball to itself is a covering map. Unbranched covering maps of simply connected domains are injective. Hence, it suffices to prove that a non-constant rational sphere map f from Bn to itself is unbranched. To do so, we need to show that det(d f ) does not vanish on the ball. Corollary 6.2 shows that it does not vanish on the sphere. If it vanished somewhere on the ball, its zero set V would be a complex analytic subvariety of positive dimension. By Proposition 6.1, V would be non-compact. Hence, det(d f ) would have to vanish somewhere on the sphere, contradicting Corollary 6.2.  Remark 6.8 This theorem considers the equi-dimensional case. The polynomial sphere map z → z ⊗m is generically m-to-1. When n ≥ 2, of course, its target dimension exceeds the domain dimension. Remark 6.9 The theorem applies to holomorphic rational sphere maps. Later in this chapter we give, for each dimension n and each degree d, an example of a non-holomorphic polynomial sphere map P : S 2n−1 → S 2n−1 of degree d.

5 Strong Pseudoconvexity of the Unit Sphere We wish to understand the previous results without appealing to the theorem of Baouendi-Rothschild. We continue to use the unbounded realization of the unit sphere using the ζ variables. Thus, we define M to be the zero set of r , where

5 Strong Pseudoconvexity of the Unit Sphere

r (ζ, ζ) = 2Re(ζn ) +

197 n−1 

|ζ j |2 .

(7)

j=1

We used twice the real part of ζn instead of its imaginary part as in (4), simply to avoid various factors of 21 and i. The above general formulas simplify when M is defined by (7). The basis of (1, 0) vector fields is given by L j = ∂ζ∂ j − ζ j ∂ζ∂ n for 1 ≤ j ≤ n − 1. Their complex conjugates L j form a basis of sections of T 0,1 M. The missing direction in CT M is generated by T = ∂ζ∂ n − ∂ζ∂ . n The unbounded realization of the sphere is particularly nice because its Levi form is so easy to understand. For r as in (7), the Hessian is the n-by-n matrix 

 I 0 , 0 0

where I denotes the identity in n − 1 dimensions. The subspace of the complexified tangent space to which we must restrict is precisely T 1,0 M, the space spanned by the vectors L j for 1 ≤ j ≤ n − 1. Thus, the Levi form is the identity Hermitian form. We use a standard lemma from potential theory, called the Hopf lemma. Its generalization to the CR setting has led to many variations, often called complex Hopf lemmas. See [34] for an early use of this lemma in a related context, and see [6] and [8] for generalizations and extensions. Our use in the case of the unbounded realization of the sphere is quite easy compared to these results. First consider a harmonic function h on the closed unit disk in C and suppose h ∂h (1) > 0. Note achieves its maximum at the point 1. The Hopf lemma states that ∂x ∂ that ∂x is the derivative in the direction normal to the circle. This result can easily be proved using the mean-value property of harmonic functions. The proof generalizes to subharmonic functions, because the sub-mean-value property is adequate. Now, consider a bounded domain  with smooth boundary in higher dimensional real Euclidean space. We can place an open ball B inside the domain such that the closure of B is tangent to ∂ at a boundary point p. If we consider plurisubharmonic functions on this closed ball that achieve a maximum at p, then we can conclude that the directional derivative normal to ∂ is positive. In the language of the previous section, the transverse derivative does not vanish. We sketch an alternative proof of Corollary 6.2 for the case of a strongly pseudoconvex domain. Our proof for the sphere proves the following more general assertion, although we stick to the sphere. The result also follows from the Baouendi-Rothschild theorem 6.3. Proposition 6.3 Let 1 and 2 be strongly pseudoconvex domains in Cn with smooth boundaries. Suppose that f is a proper holomorphic map between them that extends to be at least twice differentiable at the boundary. Then d f ( p) has maximal rank at each p ∈ ∂1 . As a consequence, det(d f ) does not vanish there.

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Proof We assume that both 1 and 2 are the unbounded realizations of the ball. Put w = ( f (z) , f n (z)). Given a point p in the source sphere we choose coordinates such that p is the origin and f ( p) is the origin in the target. Our assumption amounts to saying that ρ(w) = 2Re( f n (z)) + f (z) 2 = 0 whenever 2Re(z n ) + z 2 = 0. Consider an open ball B contained in the unbounded ball 1 which touches the boundary at the point 0. The restriction of the map z → 2Re( f n (z)) + f (z) 2 is plurisubharmonic and achieves its maximum on this closed ball at 0. By the Hopf lemma, its outer normal derivative is strictly positive. Here this derivative is ∂∂ζfnn . Since this derivative is not zero, dρ is not zero, and we may use ρ as a defining function. If we change coordinates back to the usual bounded version of the sphere, we see that R = f (z) 2 − 1 is a defining function of the sphere. The Hessian R jk is easily seen to be  f z j , f zk . The determinant of the Hessian is | det(d f (z)|2 . The Levi form is positive definite. Computing the Levi form leads to a factor of | det(d f )|2 . Since the Levi form is definite, this factor cannot vanish. 

6 Comparison with the Real Case We briefly discuss some differences between the real and complex cases. Recall that S k−1 denotes the unit sphere in Rk . The unit sphere in Cn is, of course, odd dimensional. Odd and even dimensional spheres exhibit quite different behavior. For example, there is no non-vanishing global tangent vector field to S 2k . On the other hand, for odd dimensional spheres, such vector fields exist. We regard S 2n−1 as sitting in Cn . Let r (z) = z 2 − 1 define the sphere. Put X = L − L, where L=

n  j=1

zj

∂ . ∂z j

 Then, since L(r ) = L(r ) = |z j |2 = 1 on the sphere, we see that X (r ) = 0. The only 0 of X occurs at the origin, which is not on the sphere. Since X is purely imaginary, i X is a global real tangent vector field to S 2n−1 with no zeroes. Exercise 6.21 What is the vector field X (in terms of x, y) when n = 1? Another major difference concerns real polynomials versus holomorphic polynomials. The next example shows, for n ≥ 2, that there are polynomials of degree 2 mapping the unit sphere in Cn to itself. By Theorem 6.4, such polynomials cannot be holomorphic. The subsequent example shows that no degree estimate is possible. Example 6.5 Let us write coordinates in Cn as (z 1 , ..., z n−1 , w) = (z, w). Consider the following polynomial mapping P from Cn to itself: P(z, w) = (2Re(w)z, w 2 − z 2 ).

6 Comparison with the Real Case

199

Of course P is not holomorphic. But,

P(z, w) 2 = |(w + w)|2 z 2 + |w|4 − 2Re(w 2 z 2 ) + z 4 = |w|4 + 2|w|2 z 2 + z 4 = (|w|2 + z 2 )2 , which equals 1 on the unit sphere. Therefore, P is a non-holomorphic polynomial sphere map of degree 2 with the same source and target dimension. Example 6.6 Let p : Rk → Rk be a polynomial that maps the unit sphere to itself. We then have an analogue of the tensor product operation, defined as follows. For a proper subspace A of (the target space) Rk , consider the map E A ( p) = ( x 2 π A p) ⊕ (1 − π A )( p). Then E A ( p) 2 = p 2 = 1 on the unit sphere, and hence E A ( p) is also a polynomial mapping the unit sphere to itself. A big difference from the complex case is that the target dimension is unchanged. Iterating this operation to the function in Example 6.5 leads to polynomial maps of unbounded (even) degree mapping the unit sphere to itself. Hence, there is no possible degree bound. We could also consider g(x) = ( x 2d x1 , x2 , ..., xn ). Then g is a polynomial of degree 2d + 1 mapping the unit sphere to itself. Another issue arises. The function g agrees with the identity map on the sphere. Do we consider g itself to be the identity map? For a second example, put f (x) = (cos θ x 2 , sin(θ)). Then f is a polynomial, and f maps S n−1 to S 1 . But, the restriction of f to S n−1 is a constant map. Do we consider f itself to be a constant map? This problem does not arise for us, because (by Corollary 1.1) a holomorphic polynomial that is constant on the sphere must be constant on Cn . Using the polynomial p and notation from Example 6.5, we put h(x) = (2Re(w)z, ( z 2 + |w|2 )d (w 2 − z 2 ))). Then h is of degree 2d + 2 and maps the unit sphere to itself. We conclude, for each k, that no degree bound exists for real polynomial maps sending the unit sphere S k to itself. One difference between the real and complex cases is that the function z → z 2 is not holomorphic, whereas the map x to x 2 is a polynomial. Hence, the tensor product operation in the complex case must increase the target dimension of a holomorphic rational sphere map.

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6 Elementary Complex and CR Geometry

7 Varieties Associated with Rational Sphere Maps Let f : 1 → 2 be a proper holomorphic mapping. A famous theorem of Remmert states that the image f (V ) of a subvariety V ⊆ 1 is a subvariety of 2 . In particular, in our setting, the proper image of a source ball is a subvariety of the target ball. Our primary interest will be the complex analytic variety from Definition 6.10. √ Example 6.7 Put n = 2 and f (z) = (z 13 , 3z 1 z 2 , z 23 ). Then f is a monomial√sphere map. In this case, V = f (C2 ) is the subvariety of C3 defined by ζ23 = 3 3ζ1 ζ3 . This two-dimensional subvariety has an isolated singular point at the origin. By intersecting V with B3 we find the image f (B2 ). √ √ Example 6.8 Put n = 2 and f (z) = (z 13 , 3z 12 z 2 , 3z 1 z 22 , z 23 ). Again f is a monomial sphere map. Here V = f (C2 ) is the subvariety of C4 defined by the three equations 3ζ1 ζ4 = ζ2 ζ3 ζ22 = ζ32 =

√ √

3ζ1 ζ3

3ζ2 ζ4 .

Although V is two dimensional, and sits in 4-space, it cannot be defined by just two of these equations. For example, points of the form (0, 0, s, t) satisfy the first two equations but not the third when s = 0. Points of the form (s, t, 0, 0) satisfy the first and third equations, but not the second when s = 0, and points of the form (0, s, t.0) satisfy the second and third equations √ but not the first when s = 0. The mapping f (usually defined without the factors of 3) is an example of a Veronese embedding. It is, however, possible to define this variety as a set using only two polynomials; any such description has algebraic drawbacks, as the ideal of functions vanishing on the variety requires at least three generators. See the discussion about the twisted cubic curve in [43]. Remark 6.10 The previous example shows that images of C2 (or of B2 ) under monomial sphere maps can exhibit interesting complicated behavior. Things are, in general, much more complicated, even for monomial sphere maps. In this section, we associate another variety X f with a rational sphere map f . This variety was first defined by Forstneriˇc for a proper map between balls (with sufficiently many continuous derivatives at the sphere) in order to establish rationality [36]. We state his result in Theorem 6.5 below. Assuming the map is rational, the author gave a simplified way to compute this variety by combining polarization and homogenization. Theorem 6.5 (Forstneriˇc) Assume n ≥ 2. Let f : Bn → B N be a proper holomorphic map that extends to the boundary and has N − n + 1 continuous derivatives there. Then f is rational.

7 Varieties Associated with Rational Sphere Maps

201

Furthermore, by a result from [12], such an f extends holomorphically past the sphere, and hence f is a rational sphere map. See also Theorem 6.8 and Corollary 6.6. These results provide some justification for our study of rational sphere maps. A natural setting for the definition of X f is when f is a holomorphic mapping between real-analytic hypersurfaces M and M in different dimensional spaces. Let ρ and ρ denote their defining functions. A holomorphic mapping f such that f (M) ⊆ M then satisfies the condition that ρ ( f (z), f (z)) = 0 when ρ(z, z) = 0. By polarizing these analytic equations, we obtain an often used fact in the theory of holomorphic mappings: Proposition 6.4 (Polarization) Let M ⊆ Cn and M ⊆ C N be real-analytic hypersurfaces defined by equations ρ(z, z) = 0 and ρ (ζ, ζ) = 0 respectively. Let f be a holomorphic mapping with f (M) ⊆ M . Then ρ ( f (z), f (w)) = 0 whenever ρ(z, w) = 0. Corollary 6.3 Let f be a rational sphere map. Then  f (z), f (w) = 1 whenever z, w = 1. Furthermore,  p(z), p(w) − q(z)q(w) = 0 when z, w = 1. Exercise 6.22 Prove Proposition 6.4. Suggestion: First write ρ ( f (z), f (z)) locally as a multiple of ρ(z, z). Then use Remark 1.2 from Chap. 1. Remark 6.11 Polarization is crucial to the proof of Theorem 6.2. We will use it in Theorem 6.8. Given a holomorphic mapping that sends M into M , we consider the complex analytic variety X f in Cn+N associated with the mapping f , defined as follows: X f = {(w, ζ) : ρ ( f (z), ζ) = 0 whenever ρ(z, w) = 0}.

Corollary 6.4 (w, f (w)) ∈ X f , and thus X f contains the graph of f . We study the particularly interesting special case when the hypersurfaces are spheres and the map is rational. In Definition 6.10, we give a slight extension of the above formula allowing w = 0. In most cases, X f is larger than the graph of f ; it is of some interest to know conditions under which X f equals the graph of f . Definition 6.10 Let f be a rational sphere map. Let (w, ζ) ∈ Cn × C N . Then (w, ζ) ∈ X f if either • w = 0 and z, w = 1 implies  f (z), ζ = 1, or • (w, ζ) = (0, f (0)). Remark 6.12 We have defined X f to be a subvariety of Cn × C N . It is also sensible and sometimes preferable to regard X f as a subvariety of Bn × C N .

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We describe X f in several equivalent ways. First we have the following result. Theorem 6.6 Let f be a rational sphere map of degree d. There is a (antiholomorphic) polynomial map w → ψ(w) taking values in the space of homogeneous C N -valued polynomials (in z) of degree d such that (w, ζ) ∈ X f if and only if either • (w, ζ) = (0, f (0)), or • w = 0 and ζ − f (w) is orthogonal to ψ(w). Proof Assume that f has degree d. By Definition 6.10 we have X f = {(w, ζ) :  f (z), ζ = 1 whenever z, w = 1} ∪ {(0, f (0))}. Since (w, f (w)) ∈ X f we have X f = {(w, ζ) :  f (z), ζ − f (w) = 0 whenever z, w = 1} ∪ {(0, f (0))}. (8) Recall that f = qp , where this fraction is reduced to lowest terms, and that p is of degree d. Multiplying through by q(z) gives X f = {(w, ζ) :  p(z), ζ − f (w) = 0 when z, w = 1} ∪ {(0, f (0))}.

(9)

 We expand p = dk=0 pk , where each pk is a vector-valued homogeneous polynomial of degree k. On the set z, w = 1, we have p(z) =

d  k=0

pk (z) =

d 

pk (z)z, wd−k .

(10)

k=0

Substituting (10) in (9) gives a formula for X f : d  {(w, ζ) :  pk (z)z, wd−k , ζ − f (w) = 0 when z, w = 1} ∪ {(0, f (0))}. k=0

(11) Observe that the expression d   pk (z)z, wd−k , ζ − f (w)

(12)

k=0

arising in (11) is homogeneous of degree d in z, and vanishes on the hyperplane z, w = 1. By homogeneity it vanishes on z, w = η for every η ∈ C. Thus, we no longer need to restrict to the hyperplane z, w = 1. We write

7 Varieties Associated with Rational Sphere Maps d 

203



pk (z)z, wd−k =

Cα (w)z α .

(13)

|α|=d

k=0

Note that Cα (w) ∈ C N for each α. Furthermore, it is a polynomial of degree at most d in w. We see for w = 0 that ζ − f (w) ∈ X f if and only if ζ − f (w) is orthogonal  to |α|=d Cα (w)z α , for every z. Using the usual orthonormal basis for C N , we can express this theorem using matrices. Recall that H (n, d) denotes the vector space of complex-valued homogeneous polynomials of degree d in n variables. Theorem 6.7 Let f be a rational sphere map of degree d. There is a linear map C(w) : C N → H (n, d), depending polynomially on w, with the following properties: • For each non-zero w in the domain of f , the fiber X f (w) over w consists of the affine space f (w) + null space(C(w)). • X f equals the graph of f if and only if C(w) has trivial null space at all w = 0 and in the domain of f . Proof Begin with the right-hand side of (13). Then let e1 . . . , e N denote the usual orthonormal basis of C N . Then (13) becomes 

Cα (w)z α =

α

 α

Cαj (w)e j z α .

j

Put μ = ζ − f (w). Using orthonormality, we rewrite the condition that formula (12) equals 0 to get N    0= Cαj (w)e j , μk ek  = Cαj (w) μ j . α

j

k

|α|=d J =1

Thus, μ is in the null space of the matrix C(w). Taking complex conjugates yields the conclusion of the theorem.  Corollary 6.5 X f consists of all pairs (w, ζ) such that w ∈ dom( f ) and • If w = 0, then ζ − f (w) is an arbitrary element of the null space of C(w). • If w = 0, then (w, ζ) = (0, f (0)). We may think of the m vectors Cα (w) in C N as defining a linear transformation from C N → C K for an appropriate K depending on n and d. We call this linear mapping C(w), to indicate its anti-holomorphic dependence on w. Remark 6.13 Our computation of X f presumes that w ∈ dom( f ). Thus, we ignore the zero set of the denominator. Perhaps one could allow some components of f (w) and ζ to take the value ∞, but we don’t attempt to do so. In case n = 1 and f (z) =

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6 Elementary Complex and CR Geometry

a−z φa (z) = 1−az , the matrix C(w) is the one-by-one matrix aw − 1. This matrix fails to be invertible when aw = 1, which is precisely the condition that w not be in the domain of f . Hence X f is the graph of f .

8 Examples of X f We next give some explicit examples. Example 6.9 Suppose that f (z) = z ⊗d . Let N be the number of linearly independent monomials of degree d in n variables. may identify f with the mapping Hd to  We d α N C whose α component is given by α z for each multi-index α of length d. Then f is a rational sphere map. We claim that X f equals the graph of f . For this map (because it is homogeneous), the  matrix C(w) is square. Its only non-zero entries are the diagonal entries, which   are αd . Hence, the null space of C(w) is trivial. Example 6.10 For positive integers j and k with j < k, consider the juxtaposition f defined by f (z) = cos(θ)H j (z) ⊕ sin(θ)Hk (z). Here the variety X f contains points not in the graph of f . The reason is simple. The target dimension N equals the sum of the dimensions of the spaces of homogeneous polynomials of degrees j and k, which exceeds the dimension of the space of such polynomials of degree k. Example 6.11 In order for X f to equal the graph of f it is necessary for f to be target minimal. To illustrate this idea, we consider the map f given by f (z) = (z, 0). Here z ∈ Cn , and N = n + 1. Then, after identifying C N with Cn × C, we obtain X f = {(w, w, η) : η ∈ C}. This formula shows that ζ = (w, η) rather than ζ = (w, 0), which would correspond to the graph of f . More generally, the variety corresponding to the map f ⊕ 0 will always have positive dimensional fibers. Remark 6.14 Proposition 5.2 of Chap. 5 gives a necessary and sufficient condition for a rational sphere map to be target minimal. Example 6.12 Put f (z 1 , z 2 ) = (z 1 , z 1 z 2 , z 22 ). Then X f is larger than the graph of f . The proof of Theorem 3.4 yields ⎛

⎞ w1 0 0 C(w) = ⎝w 2 1 0⎠ . 0 01

8 Examples of X f

205

When w 1 = 0, this matrix has a non-trivial null space. Here X f = {(w, f (w))} ∪ {(0, w2 , t, −tw2 , w22 ) : t ∈ C}. When t = 0, we obtain points not in the graph of f . Example 2.12 from Chap. 2 is somewhat striking. It is a one-parameter family for which the degree is not a homotopy invariant. We compute its X -variety. Put c = cos(θ) and s = sin(θ). Example 6.13 Define a one-parameter family of rational sphere maps by f (z) = (cz 1 − sz 22 , z 1 z 2 , (cz 1 − sz 22 )(sz 1 + cz 22 ), z 1 z 2 (sz 1 + cz 22 ), (sz 1 + cz 22 )2 ). We compute the X -variety for these maps. ⎛ cw13 0 csw12 2 ⎜ 3cw 2 w2 w 2csw 1 w2 1 1 ⎜ ⎜3cw1 w 2 − sw 2 2w1 w2 csw 2 + (c2 − s 2 )w1 1 2 ⎜ 3 2 ⎝cw2 − 2sw1 w2 w22 (c2 − s 2 )w2 2 −sw2 0 −sc

⎞ 0 s 2 w12 sw1 2s 2 w1 w2 ⎟ ⎟ sw2 2scw1 + s 2 w22 ⎟ ⎟. ⎠ c 2scw2 0 c2

The determinant of this matrix is c2 w16 . Thus, unless c = 0, it is generically invertible. When c = 0, there is an exceptional fiber when w1 = 0. When c = 0 we have a map of degree 3; hence, the rank cannot exceed 4.   rows. The number of In general, the matrix C(w) has N columns and n+d−1 d rows is the number of homogeneous monomials of degree d in n variables and f is assumed to be of degree d. Here are some additional remarks about X f . Remark 6.15 C(w) is independent of w if and only if f is homogeneous. Remark 6.16 We can recover f from C. Let e1 , ..., e N be the usual orthonormal basis for C N . Then C(w)(ek ) is the k-th component of the numerator of f , homogenized by writing 1 = z, w. Then we can dehomogenize to determine p. We noted in Remark 2.1 from Chap. 2 that the numerator of a rational sphere map determines the denominator up to a constant factor. Our convention that q(0) = 1 then determines q completely. Thus, we can recover f from p. Example 6.14 Let f : B2 → B4 be the group-invariant map f (z 1 , z 2 ) = (z 15 ,

√

5z 13 z 2 ,

√

5z 1 z 22 , z 25 ).

We compute X f by homogenizing: (z 15 ,

√

5z 13 z 2 (z 1 w 1 + z 2 w 2 ),

√

5z 1 z 22 (z 1 w 1 + z 2 w 2 )2 , z 25 )

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6 Elementary Complex and CR Geometry

We obtain the matrix C(w). ⎛

1 √0 ⎜0 5w1 ⎜ √ ⎜0 5w2 ⎜ ⎜0 0 ⎜ ⎝0 0 0 0

0 √0 2 √ 5w 1 2 √5w 1 w 2 5w 2 2 0

⎞ 0 0⎟ ⎟ 0⎟ ⎟. 0⎟ ⎟ 0⎠ 1

It has maximal rank away from the origin, and hence X f is the graph of f . Exercise 6.23 Compute the X -variety for each of the maps in Theorem 3.15. Open problem. Give a necessary and sufficient condition that X f is the graph of f . (In other words, no exceptional fibers exist.) See [69] for recent related work.

9 A Return to the Definition of Rational Sphere Map We close this chapter by answering a basic question about rational sphere maps. In one dimension, a proper holomorphic map from the disk to itself is automatically rational and the denominator is never 0 on the circle. In Definition 2.1, we defined rational sphere maps to be rational maps qp , sending the source sphere to the target sphere, reduced to lowest terms, and where q does not vanish on the closed ball. It is natural to ask whether a rational proper map between balls must, in fact, be a rational sphere map. The answer is yes, proved in [12] and in [11], although neither reference uses the term rational sphere map. We give a simple proof, along the lines of [11], in Theorem 6.8. The key idea in the proof is polarization, as in Proposition 6.4. We will use the unbounded realization of the unit sphere in the source. Denote the variables in Cn by (z, ζ) where ζ ∈ C. The defining equation of the unbounded realization of the sphere is thus given by r (z, ζ, z, ζ) = ζ + ζ + z 2 = 0.

(14)

We will polarize (14) by replacing ζ with a new variable η. Equation (14) becomes 0 = ζ + η + z 2 = 0, and hence (since z 2 is real) η = −ζ − z 2 . In our proof, p will be the numerator of a rational sphere map. The vector-valued function p(z, η) satisfies (15) p(z, η) = p(z, ζ − r (z, ζ, z, ζ)) = p(z, ζ) + A(z, ζ, z, ζ) where A is divisible by r . Lemma 6.2 For t ∈ C, suppose R(t, t) is smooth, real-valued, and vanishes at the origin. Assume that limt→0 R(t,t) exists for some positive integer k. Then the limit tk

9 A Return to the Definition of Rational Sphere Map

207

must be 0. If A(t, t) is any smooth multiple of R, and limt→0 limit is 0 as well.

A(t,t) tk

exists, then this

Proof If R vanishes to infinite order the result is obvious. Otherwise, consider the lowest order terms B(t, t) in the Taylor expansion of R at 0. Write B(t, t) =

m 

cjt jt

m− j

.

j=0

The hypotheses imply that limt→0 m = k, then

B(t,t) tk

exists. If m exceeds k, then the limit is 0. If

B(t, t)  k− j = c j t j−k t . tk j=0 k

Since limt→0 (t/t) does not exist, the only way the limit can exist is if c j = 0 for j = 0, and then the limit is c0 . But B is real valued and hence c0 = cm . Therefore, the limit is 0. We leave it to the reader to verify the same conclusion when A is a multiple of R.  We are now ready to show for n ≥ 2 that a rational proper map between balls is, in fact, a rational sphere map. Theorem 6.8 Assume n ≥ 2. Let f = qp : Bn → B N be a proper rational map. If there is a boundary point where q(z 0 ) = 0, then p and q have a common factor. Thus, a rational proper map between balls, reduced to lowest terms, is a rational sphere map. Proof We assume a basic fact from algebraic geometry. Let t → z(t) ⊆ Cn be a non-constant holomorphic curve with z(0) = z 0 and whose image lies in the zero set of q. Suppose p(z(t)) = 0 for all t near 0. Then p and q have a common factor. We consider the unbounded realization of the sphere in the source and we assume z 0 = 0. Denote the variables in Cn by (z, ζ) where ζ ∈ C and thus the defining equation of this realization of the sphere satisfies (14). The function qp is well defined on the intersection of an open neighborhood of 0 with the side of the boundary p 2 corresponding to the ball. Since f is proper, |q| 2 is bounded there and (approaching from within)

p(z, ζ) 2 lim = 1. (16) (z,ζ)→0 |q(z, ζ)|2 On the set r = 0 it follows that

p(z, ζ) 2 − |q(z, ζ)|2 = 0.

(17)

The key idea is to polarize (17) by regarding ζ as an independent variable η. Thus

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6 Elementary Complex and CR Geometry

 p(z, ζ), p(z, η) = q(z, ζ)q(z, η)

(18)

on the set where r (z, ζ, z, η) = 0. This set is also defined by ζ + η + z 2 = 0. Hence, we can eliminate η from the problem and obtain (everywhere)  p(z, ζ), p(z, − z 2 − ζ) = q(z, ζ)q(z, − z 2 − ζ).

(19)

Let t → γ(t) be a holomorphic curve defined near 0 ∈ C with γ(0) = 0 and suppose that q(γ(t)) vanishes identically. Then the right-hand side of (19) vanishes identically. We want to show that p(γ(t)) vanishes identically; we know that the lefthand side of (19) vanishes identically. Write − z 2 − ζ = ζ − r (z, ζ, z, ζ)). Using (15) gives  p(z, ζ), p(z, ζ − r (z, z, ζ, ζ)) =  p(z, ζ), p(z, ζ) + A(z, ζ, z, ζ)

(20)

where A is divisible by r . We evaluate along the curve, writing A(t, t) for the pullback of A. If p(γ(t)) does not vanish identically, write p(γ(t)) = Lt k + ... where the dots denote higher order terms. Plug into (20) and divide by |t|2k . We get two terms: 0 = L + ..., L + ... + L + ...,

1 A(t, t). tk

(21)

The limit as t tends to 0 is of course 0, because the expression is zero. The first term in (21) tends to L 2 . Therefore, the second term tends to − L 2 . We are in the situation of Lemma 6.2 and hence the limit must be 0. Hence, L = 0 as well, and p(γ(t)) vanishes identically. Hence, p and q have a common factor.  Corollary 6.6 Assume n ≥ 2. A proper map f : Bn → B N with N − n + 1 continuous derivatives at the sphere extends holomorphically past the sphere. Proof By Forstneriˇc’s theorem (Theorem 6.5), f must be rational. By Theorem 6.8, after reducing to lowest terms, the denominator of f doesn’t vanish on the closed ball. Hence, f is holomorphic in a neighborhood of the closed ball.  Remark 6.17 Chiappari proved a stronger theorem in his thesis [11]. One can replace the source sphere by an arbitrary real-analytic hypersurface, and replace the rational function by a meromorphic function that is holomorphic on one side. The proof is essentially the same, although one must use the implicit function theorem to solve for η in the relation r (z, ζ, z, η) = 0. Similar techniques arise, for example, in [7] and [8]. The next example shows that the target domain matters in this theorem. k

z Example 6.15 Put f (z, w) = ( 1−w , w). Then f : B2 →  is a proper rational map to some domain . The map has a singularity at the boundary point (0, 1). By making k large we obtain a map with as many continuous derivatives (on the closed ball)

9 A Return to the Definition of Rational Sphere Map

209

as we wish at (0, 1). Yet it does not extend past this point. Thus, one needs some assumptions on the target domain to conclude holomorphic extension. Exercise 6.24 Assuming the result for R in Lemma 6.2, verify the result for smooth multiples of R. Give a simple counter-example to Lemma 6.2 when R is not real valued.

Chapter 7

Geometric Properties of Rational Sphere Maps

We wish to discuss areas and volumes. The main result of this chapter, Theorem 7.1, is an integral inequality. Even the one-dimensional result is interesting. We use Theorem 7.1 to show that the homogeneous map z ⊗m solves a natural optimization problem. Namely, of all polynomial sphere maps of degree m, the volume of its image of the ball is maximized by z ⊗m . We compute this volume explicitly. To prove these results, we will require some basic properties of differential forms. This information appears in many sources, such as [19, 49, 78], at varying levels of sophistication.

1 Volumes We first recall the role of the determinant function in computing volumes. Assume that R K is equipped with the standard Euclidean metric. We write the standard basis as e1 , ..., e K , and we assume that these vectors are orthonormal in this metric. Let v1 , ..., v K denote vectors in R K . We express each in terms of the standard basis e1 , ..., e K : K  vj = v kj ek . k=1

The wedge product of these vectors then satisfies v1 ∧ ... ∧ v K = det(v kj )e1 ∧ ... ∧ e K .

(1)

Formula (1) is sometimes taken as the definition of the determinant. Alternatively, the wedge product is alternating and multi-linear, and the scalar on the right-hand side of (1) is therefore alternating and multi-linear in the vectors v j . Thus, it is a © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 J. P. D’Angelo, Rational Sphere Maps, Progress in Mathematics 341, https://doi.org/10.1007/978-3-030-75809-7_7

211

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7 Geometric Properties of Rational Sphere Maps

multiple of the determinant. The definition of the Euclidean metric presumes that e1 ∧ ... ∧ e K equals the volume form, often written d VK . We will assume (1) and define the oriented volume of the K -dimensional box spanned by the v j to be det(v kj ). Assume K = 2n is even, and we are working in complex Euclidean space Cn . Certain simplifications occur, which we next discuss. First, in C, where z = x + i y, we have (2) dz ∧ dz = −2idx ∧ dy. Hence, the area form in C is given by d V2 = 2i dz ∧ dz. Suppose z ∈ Cn . The volume form can now be written i d V2n = ( )n dz 1 ∧ dz 1 ∧ ... ∧ dz n ∧ dz n . 2

(3)

It is quite interesting in complex geometry to regard the volume form as the n-th exterior power of a (1, 1) form. If ω is a (1, 1) form, and 1 ≤ k ≤ n, we let (with k factors) denote its k-th exterior power. Up to a constant k (ω) = ω ∧ ... ∧ ω we may take ω = cn j dz j ∧ dz j and we have n (cn



dz j ∧ dz j ) = d V2n .

j

Most readers will have seen the following standard fact. We prove it to illustrate the power of differential forms. Proposition 7.1 Let  be an open subset of Cn and assume f :  → Cn is holomorphic. Regard  as a subset of R2n and define F :  → R2n by F(x, y) = f (x + iy). Then det(d F) = | det d f |2 . Proof The geometric meaning of det(d F) is the infinitesimal rate of distortion of volume. Put w = f (z). By definition of the derivative, we have dw1 ∧ ... ∧ dwn = det(d f )dz 1 ∧ ... ∧ dz n . Hence, including the conjugated terms, we have dw1 ∧ dw 1 ∧ ... ∧ dwn ∧ dw n = | det(d f (z))|2 dz 1 ∧ dz 1 ∧ ... ∧ dz n ∧ dz n . Thus, | det(d f (z))|2 is the rate of volume distortion at z. But the infinitesimal rate of volume distortion is also det(d F). Hence they are equal.  We will make a change of notation. Let f be a holomorphic mapping with target C N . In the rest of this chapter, F will not mean the F from Proposition 7.1. Sometimes F will denote an arbitrary twice differentiable function, and often it will be the

1 Volumes

213

Euclidean squared norm of a holomorphic mapping f . Thus 

F =  f 2 =

| f j |2 .

Since the target dimension N will typically exceed the source dimension n, we will need several simple facts involving Jacobian determinants. First consider an equi-dimensional holomorphic map f I with components f i1 , ..., f in . We write J ( f I ) for the Jacobian determinant J ( f I ) = det(

∂ fi j ). ∂z k

Suppose f :  ⊆ Cn → C N , and I = (i 1 , ..., i n ) is an n-tuple of indices. Then each J ( f I ) contributes to the volume of the image of  under f . Lemma 7.1 Suppose f : Cn → C N is holomorphic. Put F =  f 2 , Then det(Fz j z k ) =



|J ( f I )|2 .

I



Proof The proof is a simple computation using differential forms. Exercise 7.1 Prove Lemma 7.1.

Consider the usual decomposition of the exterior derivative as d = ∂ + ∂ discussed in Chap. 6. Given a twice differentiable function F, the (1, 1) form ∂∂ F plays a crucial role in what follows. The following proposition provides several ways to compute volumes. When f :  → C N is a holomorphic map, we write V2n ( f, ) for the 2n-dimensional volume of the image of  under f . This number includes multiplicity. See [19] for considerable discussion of this issue. Example 7.1 Let  be the unit disk, and put f (z) = z m . Then 

| f (z)|2 d V2 =



 |mz m−1 |2 d V2 = 0

2π



1

m 2 r 2m−1 dr dθ =

0

2π m 2 = π m. 2m

The image is the disk covered m times. Hence the area of the image must be mπ . The next proposition generalizes Example 7.1 to arbitrary source dimension. Proposition 7.2 Put G = z ⊗m 2 . Then  det(G z j z k ) d V2n = Bn

mn π n . n!

Proof This statement can be proved by direct computation. One can also argue as ( 1 )k follows. The volume of the unit ball in Rk is ( k 2+1) . Replacing k by 2n, and using 2

214

7 Geometric Properties of Rational Sphere Maps

√ n ( 21 ) = π , we see that the volume of the unit ball Bn is πn! . To compute the volume ⊗m of the image of the ball under z → z , one uses the change of variables formula for integrals under a finite-to-one map. The answer is m n times the volume of the ball.  Remark 7.1 The previous proof is interesting even when n = 1. The map z → z m covers the unit disk m times, and hence the value of the integral is mπ . In other words, the result of the computation done in Example 7.1 follows immediately.

2 A Geometric Result in One Dimension We have mentioned that Theorem 2.6 leads to a beautiful, sharp inequality on the volume of the image of a polynomial sphere map. To prepare for this result, we recall some standard notation and then prove a general result about areas in one dimension. We regard the unit disk B as the set where ρ < 0, for ρ(z, z) = zz − 1 = |z|2 − 1. We observe that ρ is a subharmonic function: its Laplacian is the positive constant 4. We also note that the Laplacian in the plane of a twice differentiable function u is given by (u) = 4u zz . This simple fact illustrates the power of working with z and z rather than with x and y. Recall from Chap. 6 the standard definitions of the first-order operators ∂ and ∂: ∂=

∂ 1 ∂ ( −i ) 2 ∂x ∂y

∂=

∂ 1 ∂ ( + i ). 2 ∂x ∂y

We think of ∂ as a mapping from functions to (1, 0)-forms and ∂ as an operator from functions to (0, 1)-forms. We can apply them again, and we obtain d = ∂ + ∂ and 2 0 = ∂ 2 = ∂ . By (2), the area form in C is 2i dz ∧ dz = d x ∧ dy. We then have ∂∂u = 4 u dz ∧ dz. The next proposition holds more generally than we have stated, but it will be adequate for our needs. Proposition 7.3 Let w be a twice differentiable function on the closed unit disk that is non-negative on the unit circle. Then  B

i ∂∂(ρw) ≥ 0. 2

(4)

2 A Geometric Result in One Dimension

215

Proof Note that d∂ = ∂∂. We use Stokes’s theorem in the plane. Since ρ = 0 on the unit circle S 1 , we have      i i i i i ∂(ρw) = ∂∂(ρw) = d∂(ρw) = w∂ρ = wzdz. B 2 B 2 S1 2 S1 2 S1 2 We can evaluate this integral by putting z = eiθ to obtain 

2π

0

1 i w(eiθ ) (−i)dθ = 2 2



2π

w(eiθ )dθ ≥ 0.

0

 Corollary 7.1 Suppose F is twice differentiable on the closed unit disk, and that w is as in Proposition 7.3. Then  B

i ∂∂(F + wρ) ≥ 2

 B

i ∂∂(F). 2

Proof Since the operator ∂∂ is linear, (5) follows from the proposition.

(5) 

Suppose in Corollary 7.1 that F = | f |2 and w = |g|2 are squared absolute values of holomorphic mappings. Note that F is the squared norm of the derivative of f . The left-hand side of (5) measures the area of the image of the disk under a map constructed from f and g. The right-hand side of (5) is the area of the image under the map f . Thus, Proposition 7.3 implies the following proposition, which we will generalize in the next section. Proposition 7.4 Assume that f is holomorphic on the unit disk and twice differentiable on the circle. The following equality holds: (z f ) 2L 2

=

f 2L 2

1 + 2



2π

| f (eiθ )|2 dθ.

(6)

0

Suppose in addition that f is not identically 0. Put E f = z f . Then the area A E f of the image of the ball under E f exceeds the area of the image under f . Proof The left-hand side of (6) is the integral of the Laplacian of |z f |2 over the ball. The first term on the right-hand side of (6) is the integral of the Laplacian of | f |2 over the ball. Their difference is, therefore, the integral (4), where w = | f |2 . The proof of (4) evaluates this difference as the boundary integral of i | f (z)|2 zdz. 2 Putting z = eiθ gives the integral on the right-hand side of (6).

216

7 Geometric Properties of Rational Sphere Maps

Thus, the difference in the areas of the images is the boundary integral in (6), which is strictly positive when f is not identically 0.  Proposition 7.4 has interpretations in terms of two operators on the Hilbert space of square-integrable holomorphic functions on the unit disk. See [19] for considerable discussion. Exercise 7.2 Put f (z) = z m . Verify Proposition 7.4 by direct computation. Exercise 7.3 Fill in the details in the proof of Proposition 7.4. Exercise 7.4 Show that the monomials z m are orthogonal in L 2 (B1 ). Exercise 7.5 Prove (6) by combining Exercises 7.1 and 7.3 with the Taylor expansion of f .

3 An Integral Inequality We now generalize these ideas to arbitrary complex dimension n. Proposition 7.5 provides a way to compute the volume of the image of a domain under a holomorphic mapping f . Our primary interest will be when f is a polynomial sphere map. We will establish a general inequality in Theorem 7.1 and derive from it the results for polynomial and rational sphere maps in the next section. Proposition 7.5 For a general C 2 function F, we have det(Fz j z k ) d V2n

1 = n n!



 i ∂∂(F) . 2

(7)

Suppose  is bounded, and let f :  → C N be holomorphic. Put F =  f 2 . Then  V2n ( f, ) =



 J( f )2 d V2n =



det(Fz j z k ) d V2n .

(8)

Proof Formula (7) follows by combining the definition of the volume form with the definition of the determinant. Formula (8) then restates Lemma 7.1.  Let  be a bounded domain in Cn and assume F is a twice differentiable realvalued function on the closure of . We define a functional  by integrating the Hessian determinant of F as follows.  (F) = det(Fz j z k )d V2n . (9) 

By Proposition 7.5, when F =  f 2 is the squared norm of a holomorphic mapping, (F) equals the volume, with multiplicity counted, of the image of 

3 An Integral Inequality

217

under f . Assume  is an open subset of Cn and f :  → C N is holomorphic. Let I = (i 1 , ..., i n ) be an n-tuple of distinct increasing indices. When N ≥ n, we consider all possible  Jacobians J ( f I ) formed from n of the components of f . We write J( f )2 = I |J ( f I )|2 . Thus V2n ( f, ) = J( f )2L 2 () . This formula is the analogue of the L 2 -norm of the derivative from the one-dimensional case. Definition 7.1 A twice differentiable real-valued function F on a domain  ⊆ Cn is called plurisubharmonic if its Hessian F jk is positive semi-definite. A real-valued function F is plurisubharmonic if and only if 2i ∂∂ F is a positive semi-definite (1, 1) form. The reason is that each statement means n 

F jk a j a k ≥ 0

j,k=1

for all (a1 , ..., an ) ∈ Cn . A squared norm  f 2 of a holomorphic mapping f is plurisubharmonic. If a defining function of a domain is plurisubharmonic, then the domain is pseudoconvex. The converse does not hold, because pseudoconvexity is a positivity condition on only part of the Hessian. Exercise 7.6 Verify that the squared norm of a holomorphic mapping is plurisubharmonic. Exercise 7.7 Show that F is plurisubharmonic if and only if its restriction to each complex line is subharmonic. Exercise 7.8 Give an example of a plurisubharmonic f such that f 2 is not plurisubharmonic. (There are easy examples even in one dimension.) Although the Hessian determinant is non-linear, the analogue of Corollary 7.1 will use the following Lemma about subtracting the n-th exterior powers of the (1, 1)-forms μ and η. Lemma 7.2 Let μ and η be (1, 1)-forms. Then n (μ) − n (η) = (μ − η) ∧

n−1 

k (μ) ∧ n−k−1 (η).

(10)

k=0

Proof Expand the right-hand side of (10). Since μ and η are (1, 1)-forms, μ ∧ η = η ∧ μ. Therefore, all the terms on the expanded right-hand side cancel except the  terms n (μ) and −n (η). Theorem 7.1 Let F, W be twice continuously differentiable non-negative functions on the closed unit ball. Assume F is plurisubharmonic. Let ρ = z2 − 1 be the defining function for the ball. Then (F + Wρ) ≥ (F). On the other words, 

 det((F + Wρ) jk )d V2n ≥

det(F jk )d V2n .

(11)

218

7 Geometric Properties of Rational Sphere Maps

Proof To prove (11), we write the determinants of the Hessians in terms of exterior powers of differential forms. We remark that the pointwise inequality det((F + Wρ)z j z k ) ≥ det(Fz j z k ) does not hold. It is only after integration that we obtain an inequality. The proof follows the same lines as in the one-dimensional case, but is a bit more complicated. First we make a simple remark about surface integrals. Suppose that  is a bounded domain in R2n defined by ρ < 0. Then the differential 1-form dρ is orthogonal to ∂. Therefore, if ω is a 2n − 2 form, then  ∂

ω ∧ dρ = 0.

As a consequence of d = ∂ + ∂, in Cn , we obtain 

 ∂

ω ∧ ∂ρ = −

∂

ω ∧ ∂ρ.

Furthermore, for any 2n − 2 form ω, we have 

 ∂

∂ρ ∧ ω =

∂

∂ρ ∧ ζ,

where ζ is congruent to ω modulo the ideal generated by ρ, ∂ρ, ∂ρ. Write  for the unit ball, ∂ for the unit sphere with its standard orientation, and ρ for the defining function. Let μ = n ( 2i (∂∂ F + Wρ)) and let η = n ( 2i ∂∂ F). Lemma 7.2 yields   i i  (μ) −  (η) = ∂∂(ρW ) ∧ β = d ∂( ρW ) ∧ β , 2 2 n

n

(12)

where β is the (n − 1, n − 1)-form arising from the sum in (10). A (1, 1)-form such as ∂∂α is exact, and a wedge product of exact forms also is exact. Therefore β is exact and hence dβ = 0. Thus, the second equality in (12) holds. We apply Stokes’s theorem to get    i i n n W ∂ρ ∧ β.  (μ) −  (η) = ∂( ρW ) ∧ β = (13) 2 2  b ∂ We have incorporated the factor ( 2i )n−1 with β. Regarding the terms arising from β, we may work modulo the ideal generated by ρ, ∂ρ, ∂ρ. To see why, notice that terms with ρ vanish on ∂, terms with ∂ρ in β get wedged with the other ∂ρ in (13), and terms with ∂ρ are negatives of the corresponding terms with ∂ρ since d = ∂ + ∂ as noted above, and hence also vanish. The only terms remaining in β involve wedge products of exterior powers of 2i ∂∂ F, the non-negative function W , and exterior powers of 2i ∂∂ρ. Recall that F is plurisubharmonic. So is ρ, as

3 An Integral Inequality

219

∂∂ρ =



dz j ∧ dz j .

After incorporating the appropriate powers of 2i with each term in β, we obtain a surface integral where the form 2i ∂ρ is wedged with a positive 2n − 2 form. Evaluate it by iterated integration. The inner 2n − 2-dimensional integral gives a non-negative function. We reduce to the boundary integral from the one-dimensional case. Hence, the right-hand side of (13) is positive and (11) follows by (7). 

4 Volume Inequalities for Polynomial and Rational Sphere Maps We begin with a trivial observation. Among polynomial sphere maps in source dimension n and of degree d > 0, the minimum volume of the image of the ball does not exist. The reason is similar to an observation from Chap. 4. Consider the oneparameter family f λ = (1 − λ) ⊕ λHd , where Hd is the map z ⊗d , and 0 < λ ≤ 1. Then each f λ is a polynomial sphere map of degree d, but the limit as λ tends to 0 is the constant map. It follows that the volume of the image also tends to 0. Hence, the infimum value is 0. For this reason, we care only about the maximum value. Theorem 7.2 Let p be a polynomial sphere map of degree m in source dimension n. Then the volume, with multiplicity counted, of the image of the unit ball under p n n is at most m n!π . Equality holds if and only if p is homogeneous. Proof Theorem 2.6 gives the following equality for a polynomial sphere map of degree m:   z2m =  p(z)2 + w(z)2 z2 − 1 =  p(z)2 + w(z)2 ρ(z, z).

(14)

In (14), w2 is a squared norm of a polynomial mapping arising from restricted tensor product operations. Since  p2 is plurisubharmonic, Theorem 7.1 implies   (z2m ) =   p(z)2 + w(z)2 ρ(z, z) ≥ ( p(z)2 ).

(15)

The far left-hand side of (15) equals m n!π by Proposition 7.2, and the far right-hand side equals the volume of the image of the ball under p. The inequality is strict unless w2 = 0; in other words, the inequality is strict except when p is already homogeneous.  n

n

Corollary 7.2 Let p : Bn → B N be a proper polynomial mapping of degree m. Then ( p2 ) ≤ (z2m ). Corollary 7.3 Let f be a polynomial sphere map with source dimension n and target dimension N . Assume n ≥ 2. Let c(n, N ) denote the sharp degree bound. Then the n n volume V ( f ) of the image of the ball under f is bounded by c(n,Nn!) π .

220

7 Geometric Properties of Rational Sphere Maps

Proof Let m be the degree of f . By the definition of sharp degree bound, m ≤ c(n, N ). Since the function x → x n is increasing, Theorem 7.2 implies that V( f ) ≤

mn π n c(n, N )n π n ≤ . n! n! 

Recall that 1 denotes the n-tuple of all ones. The next result could have been noted just after Theorem 2.6. It shows that the bound M(n, d) from Lemma 4.1 holds also for polynomial sphere maps. We mention it now because of its connection to homogeneous sphere maps. We also briefly discuss the rational case before returning to volumes. Corollary 7.4 Let p be a polynomial sphere map of degree d with source dimension n. Then  p(1)2 ≤ n d . Equality holds when p is homogeneous, but can hold more generally. Proof Formula (14) gives n d = 12d =  p(1)2 + w(1)2 (n − 1) ≥  p(1)2 . Strict inequality holds if w(1) = 0 and n ≥ 2.



Equality can occur in Corollary 7.4 even when p is not homogeneous. We give two examples. When n = 1, put p(z) = (cz m , sz k ), where m > k, where neither c nor s vanishes, and |c|2 + |s|2 = 1. Then p is a sphere map, p is not homogeneous, and  p(1)2 = 1 = n d . Next, we give an example where n = 2 but w(1) = 0. Use (z, w) for the variables and define p by  p(z, w) =

 z−w z+w z+w √ , z( √ ), w( √ ) . 2 2 2

Then p is a sphere map of degree d = 2 and √ √  p(1, 1)2 = ( 2)2 + ( 2)2 = 4 = n d . Finally, we note that  f (1)2 has neither a maximum nor a minimum when f ranges over rational sphere maps of degree d in source dimension at least 2. The failure of the minimum to exist is a consequence of the non-compactness of the automorphism group of the ball. Proposition 7.6 Assume n ≥ 2. • For each integer d at least 1 there is a rational sphere map f of degree d for which  f (1)2 is infinite.

4 Volume Inequalities for Polynomial and Rational Sphere Maps

221

• For the tensor product φa⊗d of the automorphism φa , inf φa⊗d (1)2 = 1. a

Proof Put q(z) = 1 −

n j=1

n

zj

. By the Cauchy-Schwarz inequality, 



z j ≤

√

n z.

√ If n > 1, then n < n, and hence q is never 0 on the closed unit ball. By Theorem 2.7, there is a numerator p such that f = qp is a rational sphere map. In this case, one can choose p to be of first degree (see Exercise 7.11). After taking a tensor product with z ⊗(d−1) , we may assume that the numerator is of degree d. By construction, f has a singularity at 1. To prove the second statement, consider the automorphism φa (z) =

(a − z 1 , −sz 2 , ..., −sz n ) . 1 − z1a

Recall that s 2 = 1 − |a|2 and that |a| < 1. One computes that φa (1)2 = 1 +

(1 − |a|2 )(n − 1) . |1 − a|2

Let a tend to −1 though negative values. Then φa (1)2 tends to 1. Since φa (z)2 = 1 on the sphere, the maximum principle forces φa (1)2 > 1. Therefore, the infimum value as a varies is 1, and this value is not achieved. The d-th tensor power of φa is  a rational sphere map of degree d, and (φa⊗d )(1)2 tends to 1. Proposition 7.6 suggests why we focused on the optimization problems for monomial sphere maps. Let us return to results for the volume of the image. Remark 7.2 Theorem 7.2 is a striking geometric inequality. When the source and target dimensions both equal 1, the only polynomial sphere maps are powers of z, and the result is a simple computation. In higher target dimensions, even when n = 1, there are many more polynomial sphere maps. When n ≥ 2, we must replace the Laplacian with the determinant of the complex Hessian. For clarity, we note the following. When n = 1, the Hessian is simply (four times) the Laplacian. In general, the Laplacian is a constant times the trace of the complex Hessian. The determinant of the Hessian is a non-linear function, and hence things become a bit more complicated. Remark 7.3 The volume of the image of the ball under a rational sphere map is independent of the target dimension. Nonetheless, Corollary 7.3 tells us that the target dimension does provide an upper bound on the volume. The value of c(n, N ) is not yet known.

222

7 Geometric Properties of Rational Sphere Maps

It is easy to generalize (12). For example, we do not require ρ to be z2 − 1. We used only that it is plurisubharmonic. We make a simple observation about the possible values of the volume of the image. Consider the juxtaposition of the constant function 1 with z ⊗m . Thus Jθ = (cos(θ ), sin(θ )z ⊗m ). For each θ ∈ (0, π2 ], the map Jθ is a polynomial sphere map of n n degree m. The volume of the image of the ball under Jθ varies from 0 to m n!π as θ π increases from 0 to 2 . Thus, every such value is achieved. Exercise 7.9 Put n = 2. Carry out the proof of (12) using 2 (μ) − 2 (η) = (μ − η) ∧ (μ + η). Do the analogous thing when n = 3. Exercise 7.10 Put p(z, w) = (z, zw, w2 ). Compute the volume of √ the image of the ball under p by doing the integral. Do the same for p(z, w) = (z 3 , 3zw, w 3 ). Exercise 7.11 Show in the proof of Proposition 7.6 that one can choose p of degree 1. Note that p(0) cannot vanish. We naturally study what happens for general rational sphere maps. An analogous result to Theorem 7.2 holds, but it is less elegant. To see what is going on, we recall a generalization of Theorem 2.5 that was used in Theorem 2.15 and Proposition 2.11. We sketch the proof for convenience. One weakness of this result seems to be in the nature of things. For polynomial sphere maps of degree m, we can estimate the volume using the homogeneous map z ⊗m . The analogue in the rational case requires us to do the same for each possible denominator, and every denominator arises. Theorem 7.3 Let h = qg be a rational sphere map. Assume that q is of degree d and that g is of degree m + d. Then there is a rational sphere map H = Gq , where G is of degree d + m and G vanishes to order m at the origin. Let V f denote the volume of the image of the ball under f . Then VH ≥ Vh . Proof By Theorem 2.1 of Chap. 2, various identities hold. Suppose g vanishes to order ν, and ν < m. Then the homogeneous part of gν of order ν must be orthogonal to gm+d . As in the proof of Theorem 2.5, one tensors on the image of gν . Continuing in this way, one reaches the map Gq , where G G m + ... + G m+d = . q 1 + q1 + ... + qd At this stage no additional orthogonality results hold, and we cannot go on further. For some polynomial map w, we have H 2 = 

G 2 h w  =  2 +  2 (z2 − 1) = h2 + Wρ. q q q

4 Volume Inequalities for Polynomial and Rational Sphere Maps

223

By the proof of Theorem 7.1, (H 2 ) ≥ (h2 ). The proof of (12) applies here because q is never zero on the closed ball. In terms of the holomorphic maps h, H , we have V2n (H, Bn ) ≥ V2n (h, Bn ). Thus, each tensor operation increases the volume (with multiplicity counted) of the image.  Exercise 7.12 For a, b > 0, evaluate the integral  B2

|z|2a |w|2b d V.

Exercise 7.13 For each non-negative multi-index α, evaluate the integral  Bn

|z α |2 d V.

Exercise 7.14 Consider the monomial sphere maps in Theorem 3.15. For as many of them as you can tolerate, compute the volume of the image of the ball.

5 Comparison with a Real Variable Integral Inequality Consider continuously differentiable functions y = y(x) mapping the interval [0, 1] to itself. Suppose that y(0) = 0 and y(1) = 1. Thus, y maps the boundary to the boundary. We regard the graph of y as an arc in the plane R2 . The length of this arc is of course  1 1 + (y (x))2 d x. L(y) = 0

Multiplying y by x preserves the boundary conditions. We call the new function E y in analogy with the tensor operation we have used. The new length is 

1

L(E y) =



1 + (x y (x) + y(x))2 d x.

0

If y is convex, then it is obvious from looking at the graph that L(E y) > L(y). One explanation is that both curves start at (0, 0) and end at (1, 1). But x y(x) < y(x) for 0 < x < 1. Hence, the derivative of E y must grow rapidly as x tends to 1 for the graph to catch up. On average, the arc length form (the integrand) must be larger, although it is smaller for x near 0. A similar explanation applies in the complex case. For example, in Proposition 7.4, f and z f have the same norm at the boundary, but |z f (z)| < | f (z)| on the open ball, except when z = 0 where equality might hold. Hence, the magnitude of the derivative of z f (z) grows more quickly than that of f (z) as we approach the circle. The area of the image under z f is therefore also larger than the area of the image under f .

224

7 Geometric Properties of Rational Sphere Maps

We return to the real case. Consider the convex examples y = x α for α ≥ 1, not necessarily an integer. Then y (x) = αx α−1 and (E y) (x) = (α + 1)x α . The inteα . The integral giving grand for E y exceeds the integrand for y only where x > α+1 the arc length is larger. More generally, y convex implies that L(x y) > L(y), giving an analogue of Theorem 7.1. Convexity is analogous to the plurisubharmonicity of F in Theorem 7.1. Thus, y convex implies L(x y) > L(y), when y is the monomial x k . A big difference from Theorem 7.2 is that multiplying by x does not preserve the degree. Let us look at the role of convexity. When α < 1, convexity fails. The inequality on the arc length need not hold. The larger of the two integrals L(y) and L(x y) is the one whose graph deviates more from the graph of the identity function x. The identity function, whose graph is a line segment, gives the minimum arc length. When we multiply a convex function (satisfying the boundary conditions) by x we make the resulting arc deviate more from the line segment. Thus, the arc length increases. When we multiply a general function by x, the arc length can either increase or decrease. Evaluation of these one-variable integrals is a non-trivial matter as they involve inverse sine functions and hypergeometric functions in general. The integrals would of course be elementary if we considered energy instead of length. One of the advantages of the complex variables problems is that the formulas for volume avoid square roots. We close this section by making a remark along these lines. Let  ⊆ R2 be an open set and assume T :  → R3 is a smooth injective map. Then the area (real two-dimensional volume) of the image set T () is given by  

Tu × Tv dudv =

 Tu 2 Tv 2 − |Tu , Tv |2 dudv.

(16)



Here the subscripts denote partial derivatives. The square root in (16) makes it difficult to compute surface integrals explicitly. In our complex variable setting, the analogous integrals are much easier to compute because formulas such as in Lemma 7.1 and Proposition 7.5 avoid these square roots. Let T :  ⊆ C2 → C N be holomorphic. The (complex two-dimensional) volume of the image is given by  

Tz 2 Tw 2 − |Tz , Tw |2 d V4 .

(17)

Chapter 8

List of Open Problems

We provide a list of open problems; each of these is stated in the text. 1. Chapter 2, Sect. 5. Assume that f, g are rational sphere maps with the same source dimension. What is the smallest target dimension N in which f and g are homotopic? (Each intermediate map is required to be a rational sphere map.) Is there a way to decide using topology? 2. Chapter 2, Sect. 8. Assume n ≥ 2 and that f : Bn → B N is proper holomorphic. Determine the minimum regularity of f at the sphere needed to prove that f is rational. 3. Chapter 2, Sect. 11. Assume n ≥ 2. Find the smallest number c(n, N ) such that d ≤ c(n, N ) holds for all rational sphere maps of degree d, source dimension n, and target dimension N . The author has long conjectured that c(2, N ) = 2N − 3 −1 for n ≥ 3. and c(n, N ) = Nn−1 4. Chapter 2, Sect. 15. Give a complete combinatorial analysis of the map f → L( f ) from polynomials of degree D to the space of Hermitian forms. For example, given the source dimension n, the degree D, and the denominator q of degree d, find the smallest target dimension N for which there are vectors Pα whose inner products satisfy L( p) = L(q), and qp is reduced to lowest terms. Relate this analysis to problem (3) above. 5. Chapter 3, Sect. 5. Consider the polynomials defined by (3.16). Determine the number of positive and negative signs in the defining hyperquadric Q(A, B) arising in (3.16). 6. Chapter 3, Sect. 8. Determine whether the number of odd degrees d for which there are at most two sharp polynomials of degree d is finite. See Sequences A143105 and A143106 in [63] for additional comments. 7. Chapter 3, Sect. 8. Find all degrees d in which there is a sharp symmetric polynomial in P(2, d). © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 J. P. D’Angelo, Rational Sphere Maps, Progress in Mathematics 341, https://doi.org/10.1007/978-3-030-75809-7_8

225

226

8 List of Open Problems

8. Chapter 3, Sect. 8. Find all sharp polynomials with integer coefficients. 9. Chapter 3, Sect. 9. Let f (x) be a polynomial of degree d in n variables, with non-negative coefficients, and with f (x) = 1 on s(x) = 1. If f is isolated (not part of a family) in this set, how likely is f to have integer coefficients? How does this probability depend on n and d? 10. Chapter 4, Sect. 4. Find the limit of 2d − m(2, d) as d tends to infinity. 11. Chapter 4, Sect. 4. Find all degrees for which the polynomial minimizing p(1, 1) has fewer terms than the degree. Perhaps 7 and 8 are the only such degrees. 12. Chapter 4, Sect. 8. For each odd degree d, determine the set of j for which the coefficient c[ j] must be non-zero in the optimal polynomial. If doing so is too difficult, prove a stabilization result about congruences and explain it! 13. Chapter 5, Sect. 5. Let G be a subgroup of Aut(Bn ). Find the smallest degree d(G, n) for which there is a rational sphere map f of degree d(G, n) with  f = G. 14. Chapter 6, Sect. 2. Use the method of CR vector fields on the unbounded realization of the unit sphere to prove Theorems 3.2 and 4.4. 15. Chapter 6, Sects. 6 and 7. Give a necessary and sufficient condition that X f is the graph of f . (In other words, for which rational sphere maps are there no exceptional fibers.) See [69] for related recent information.

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Index

A Andersen, E., 157 Automorphism, 10 Automorphism, holomorphic, 10 B Baouendi, M. S., 183, 194 Bedford, E., 83 Bell, S., 196 Bi-degree of a polynomial, 29 Bihomogeneous, 29 Blaschke condition, 21 Bump, D., 15 C Cauchy-Riemann equations, 2 Cayley’s theorem, 174 Center of a group, 9 Chebyshev polynomials, 78 Chiappari, S., 176, 208 Circulant determinant, 91 Codimension, 5 CR geometry, 186

Equivalent, 154 F Faran, J., 47 Finite type, 192 Fixed-point free, 86 Forstneriˇc, F., 20, 27, 54, 175, 200 Fourier series, 26, 30 Freshman’s dream, 88 G Gaps in target-rank, 51, 106, 107, 109 Golden ratio, 97 Group, Hermitian invariant, 154 Grundmeier, D., 91

D Defining function, 190 Degree bound, 54 Degree of rational sphere map, 27, 31 D finite type, 192 Dilation, 187 Donoho, D., 113

H Hadamard matrix, 170 Heisenberg group, 187 Hermitian form, 15 Hermitian length, 32 Hermitian norm, 16 Hermitian-invariant group, 154 Hessian, 217 Hilbert space, 2 Hilbert’s 17-th problem, 33 Holomorphic, 2 Homogeneous expansion, 26 Huang lemma, 105 Huang, X., 183 Hypersurface, real, 191

E Equivalence, spherical, 154

I Irrigidity, 49

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 J. P. D’ Angelo, Rational Sphere Maps, Progress in Mathematics 341, https://doi.org/10.1007/978-3-030-75809-7

231

232 Israel, R., 110 Iwasawa decomposition, 15, 188

J Jacobian conjecture, 157 Juxtaposition, 37, 166

L Lebl, J., 48, 50, 54, 65, 102, 120, 143, 159, 173 Lempert, L., 157 Lens space, 176 Levi form, 191 Lichtblau, Daniel, 120, 175 Lie bracket, 191 Lie group, 4 Linear programming, 74, 111, 114, 131, 138

M Map, finite, 31 Mathematica, 120 Maximum principle, 2 Meylan, F., 54 Missing direction, 190, 197 Mixed terms, 4 Multiplicity, 55, 192

O Online encyclopedia of integer sequences, 100, 109, 110, 225 Overshear, 157

P Pascal triangle, 36 Pell equation, 94 Peters, H., 143 Pfister’s theorem, 33 Plurisubharmonic, 217 Polarization, 3, 201 Polynomial, bihomogeneous, 45 Postage stamp problem, 51 Proper map, 19 Proper mapping, 19 Pseudoconvex, 217 Pseudoconvexity, strong, 191 Pure terms, 4 Putnam, Dan, 120 Pythagoras number, 32 Pythagorean theorem, 2

Index R Rank, sphere, 32 Rank, target (see also target-rank), 32 Rational sphere map, 5 Rational sphere map, definition, 5, 26 Reiter, M., 48, 206 Remmert proper mapping theorem, 200 Representation, 174 Restricted tensor product, 40 Rosay, J., 14 Rothschild, L. P., 183, 194 Rudin, W., 39

S Schwarz lemma, 6, 7, 12, 160 Schwarz reflection principle, 56, 62 Sharp polynomial, 75, 94 Shear, 156, 157 Sierpinski’s theorem, 50 Sierpinski, W., 50 Signature pair, 15, 32 Sink, 141 Source, 141 Sparse, 75, 102 Sparse solution, 112 Sphere map, monomial, 6 Sphere map, polynomial, 6 Sphere map, rational, 5 Sphere-rank, 32 Spherical equivalence, 154 Spherically equivalent, 21 Sturm-Liouville equation, 78 Submanifold, 184 Subvariety, algebraic, 183 Subvariety, complex analytic, 31, 183 Sylvester’s theorem, 51

T Target-rank, 32, 43, 61, 106 Tensor product, 38 Tensor product, restricted, 40, 43, 61 Tetrahedral numbers, 110 Transitive, 14 Transitive group, 14 Transversal coordinate, 195 Triangular numbers, 109

U Underdetermined, 58, 112 Underlying matrix of coefficients, 15, 45 Undoing, 40

Index

233

Uniqueness, failure of, 100 Unitary, 3

W Whitney map, 49, 143 Wolf, J., 175 Wong, B., 14 Wono, M., 105

V Vanderbei, Bob, 120 Veronese embedding, 200

X X variety, 201 Xiao, M., 153