Standard Integral Table Algebras Generated by a Non-real Element of Small Degree (Lecture Notes in Mathematics, 1773) 3540428518, 9783540428510

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Lecture Notes in Mathematics Editors: J.–M. Morel, Cachan F. Takens, Groningen B. Teissier, Paris

1773

3 Berlin Heidelberg New York Barcelona Hong Kong London Milan Paris Tokyo

Zvi Arad Mikhail Muzychuk (Eds.)

Standard Integral Table Algebras Generated by a Non-real Element of Small Degree

13

Editors Zvi Arad Netanya Academic College Kibbutz Galuyot, St. 16 Netanya 42365, Israel and Dept. of Mathematics and Computer Science Bar-Ilan University Ramat-Gan 52900, Israel e-mail: aradtzvi.biu.ac Mikhail Muzychuk Netanya Academic College Kibbutz Galuyot, St. 16 Netanya 42365, Israel e-mail: [email protected]

Cataloging-in-Publication Data applied for Die Deutsche Bibliothek - CIP-Einheitsaufnahme Standard integral table algebras generated by a non-real element of small degree / Zvi Arad ; Mikhail Muzychuk (ed.). - Berlin ; Heidelberg ; New York ; Barcelona ; Hong Kong ; London ; Milan ; Paris ; Tokyo : Springer, 2002 (Lecture notes in mathematics ; 1773) ISBN 3-540-42851-8 Mathematics Subject Classification (2000): 13A99, 20C05, 20C99, 05E30, 16P10 ISSN 0075-8434 ISBN 3-540-42851-8 Springer-Verlag Berlin Heidelberg New York This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specif ically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microf ilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer-Verlag. Violations are liable for prosecution under the German Copyright Law. Springer-Verlag Berlin Heidelberg New York a member of BertelsmannSpringer Science + Business Media GmbH © Springer-Verlag Berlin Heidelberg 2002 Printed in Germany The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specif ic statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typesetting: Camera-ready TEX output by the editors SPIN: 10856649

41/3143/du - 543210 - Printed on acid-free paper

Preface

of

Properties of finite

group

[22]

book

"Products

Herzog, period.

gives

M.

It

of finite classes 4r6radoldbrancf, conjugacy groups in the 1980©s. studied The topic was intensively of Conjugacy edited in Groups," Classes by Z. Arad and of the results obtained comprehensive picture during this

products theory.

by several

realized

was

products

a

of

This

of irreducible

authors

that

this

We refer

characters.

could

research

reader

the

to

the

be extended

to

[1, 2, 11,

papers

13-16,21,23,35,40,51,52,651. several

In

of

ucts

of these

conjugacy

the

papers

classes

authors

products

and

found

analogy

an

prod-

between

characters

of irreducible

which

led

to

and Z. Arad in [7], in by H.L.Blau of products of conjugacy order to study in a uniform way the decomposition Since characters of finite the theory and irreducible classes then, groups. of Z. Arad, in papers H. Arof table was developed algebras extensively F. D. M.R. E. J. H. Darafsheh, Chillag, Blau, B-dnger, Erez, Fisman, isha, A. Rahnamai, C. Scopolla M. Muzychuk, and B. Xu [3-5,7V. Miloslavsky,

the

algebra,

of table

notion

introduced

10,12,17-20,25,29-33,35,41]. class of C-algbras as defined, a special algbras, may be c onsidered a table by Y. Kawada [49] and G. Hoheisel precisely, [48].:More where the structure constants Each is a C-algebra are nonnegative. algebra the table table two natural finite algebras: algebra of conjugacy group yields characters. and the table classes algebra of generalized Table

introduced

their

the

reader

bras

constants

the

Introduction

to

algebras of properties Iwahori-Hecke Each is

an a

table

where

theory

group

degrees

are

these

have

in

table

parameter

algebras.

The first

[7]

algebra non-trivial which

result

may be rescaled

degrees may be used in this direction

to

for

refer

axe

equal.

are

prop-

(we

integers

Such algedefined). (briefly, ITA). Generalized in [20]. They generalize

notions

[30]

whose

additional

an

nonnegative

(briefly,

integral algebra

natural

from and

table algebras as integral introduced were GT-algebras such well-known homogeneous objects, e.g., etc. algebras,

defined

were

arriving

structure

table

i.e.,

algebras

table

Both

erty:

a

was

a

algebras,

coherent

homogeneous This

classification

obtained

one

common

of

by

[32],

degree integral

Z. Arad

and

of degree 1 were classified. homogeneous table algebras of homogeneous table of degree 2 with The classification algebras integral element faithful was obtained was continued by H. Blau in [31]. This research classification table of homogeneous in [10] where a complete integral algebras element that the algebra 3 with a faithful of degree was obtained provided H. Blau

does

not

in

contain

where

linear

elements.

a

VI

class of ITA is comprised of so-called standard important integral tile properties of Bose-Mesner algebras SITA) which axiomatize (briefly, of commutative association schemes. The standard also are algebras algebras in the study of homogeneous ITA. involved of a table Each element in a unique table algebra is contained subalgebra which may be considered So as a table subalgebra generated by this element. of integral the study it is natural to start table from those which algebras Table algebras are by a single element. generated generated by an element of classified by H. Blau in [291 under the assumption degree 2 were completely is real or the algebra element does not contain that either linear a generating of degree a power of 2. If a table elements algebra is generated by an element then its structure is more complicated. of degree 3 or greater, If a generating then we are©faced with a classifleation. is real, element of P-polynomial tawhich would imply powerful ble algberas of for a classifleation consequences In if element non-real and is a distance-regular graphs. generating contrast, then either classification of small degree, structure a complete or important For example, standard information table algebras integral may be obtained. element of degree 3 were classified in (5], [33] under by a non-real generated that there the additional is no nontrivial element of degree 1. In assumption the investigation of integral this volume we continue table standard algebras of small valency. element More precisely, we collect by a non-real generated results about integral standard table here the recent algberas by a generated 4 or 5. element of degree non-real known to us of SITA generated In all the examples element by a non-real bounded by some function of degree are k, the degrees of all basis elements of the following f (k). This gives evidence

Another

table

Conjecture by generated

There

a

non-real

exists

a

element

function f : N -+ of degree k, then

N such

all

if of

that

degrees

a

SITA is

the

algebra

by f (k).

bounded

are

I

of

The results

SITA does

[29]

show that

this

conjecture

is

true

if

k

of

If

2.

=

k

=

3

degree 1, degrees The paxtial is valid. classification of by 6 and the conjecture ITA generated standard in this volume by an element of degrees 4,5 obtained It is not difficult to show that this conjecture. also supports the conjecture of characters holds for the table of a finite algebras generalized group even of being non-real. the assumption without and

a

contain

not

nontrivial

elements

then

all

bounded

are

The book

countries

[22]

to work

and the on

[7] attracted

paper

algebras,

table

products

many researchers

of

conjugacy

from

classes

various

and related

topics. At Bar-Ilan and

his

performed H.

Blau

former

student

extensive

from

Z. Arad and his

University, E.

research

Northern

on

Illinois

students with

Arisha, colleague

H.

jointly table algebras. In the University (deKalb)

Fisman,

his

V. M.

academic and

two

Miloslavsky, Muzychuk,

year

1998/99,

postdoctoral

Vii

F. Bfinger from Germany and M. Hirasaka from Japan, joined the in order to further advance the theory of table University group This volume, with [5] and [331, collect most of the results algebras. together in this obtained period at Bar-Ilan University. This volume contains 5 chapters. The first is an Introduction, chapter all necessary which contains definitions and facts about table The algebras. Table Algebras second with a Faithful Nonreal Element chapter, Integral of table Degree 4, deals with standard integral algebras generated by a non-real of degree 4. The contribution element of one of its co-authors, H. Arisha, is a of his Ph.D Another thesis. E. Fisman, was supported part co-author, by the at Bar-Ilan The third EmmyNoether Research Institute University. chapter, Table Algebras Standard with a Faithful Nonreal Element Integral of Degree 5, and the fourth Standard Table Algebras with a Faithful Real chapter, Integral Element of Degree 5 and Width 3, are devoted to standard integral algebras of these by an element of degree 5. F. BiAnger, one of the co-authors generated was in Germany through supported chapters, by the Minerva Foundation the EmmyNoether Research Institute at Bar-Ilan The last chapUniversity. Commutative Association Schemes with a of Primitive ter, The Enumeration Relation commuta, Non-symmetric of Valency at Most 4, classifies primitive schemes which contain tive a connected relation ass©ociation non-symmetric of valency 3 or 4. Its author, was supported Mitsugu Hirasaka, by the Japan of Science, and worked in both the Graduate School Society for Promotion of Mathematics at Kyushu University and the Emmy Noether Research Inat Bar-Ilan stitute University.

students,

Bar-Ilan

We also ous

misprints

Ramat-Gan

July

1999

would

like

in the

and

to thank text

Netanya,

and

Mrs.

prepared

Miriam the

Beller final

who corrected version

of the

the

numer-

manuscript.

Israel

Z.

M.

Arad

Muzychuk

Contents

Introduction

I

Z.

Arad,

1. 1 Main Definitions

1.3 Basic 1.4 Basic

constructions

2. SITAwith Z.

Arad,

SITA with

Arad,

...........................

.........

6

........

7

...........................................

Faithful

Element

Nonreal

3.2 General

of

Degree

4

13

.........

a

E.

Nonreal M.

Fisman,

19

.........................

..............

Faithful

14

Element Muzychuk

of

Degree

5

43

........

43

.................................................

facts 5

Degree

results

main

Biinger,

F.

3.1 Introduction 3.3

4

......

E. Fisman Muzychuk, H. Arisha, examples ..............................................

of the

2.2 Proof

Z.

........................................

M.

2.1 Known

3

a

1

.............................................

examples properties

1.2 Basic

I

..............................................

Muzychuk

M.

and known results

44

................

...............

61

....................................................

3.4 Case

3

......................................................

62

3.5 Case

5

......................................................

66

4

F.

SITA with

a

Faithful

Real

Element

of

Degree

5 and

Width

3

83©

Bilnger

4.1 Introduction

83

.................................................

4.2 Case

1

......................................................

83

Case

2

......................................................

87

4.3 5

Mitsugu

Primitive

Commutative

Association 105

......................................................

Hirasaka

5.1 Introduction

valency valency of valency

5.2 The

case

5.3 The

case

of

5.4 The

case

References

105

.................................................

of

Index

of

Enumeration

The

Schemes

2

1

or

3

...................

4

.........................................

.....................................

....................................................

...............

......................

109 110

117 121

125

Introduction

1

Arad©2

Z.

MUZyChUkl,2

and M.

of Maihematics Department Bar-Ilan University Ramat-Gan 52900, Israel

2Department

of Mathematics

R be

Let

called

a

Science

free

a

all

for

R-module

F

distinguished

a

GT-algebra)

with

a

B is

basis

distinguished

axioms:

with

basis

a

B.

1 G B.

1, and

unit

antiautomorphism

an

A and

E R be the

Aabc

with

A with

(briefly,

left

exists E

a

R-algebra

algebra following

R-algebra

an

GT2. There

An

the

table

satisfies

A is

GTI.

Let

Computer

domain.

integral generalized

GTO. A is

holds

and

Definitions

an

B if it

basis

Science

Israel

Main

1.1

Computer

College

Academic

Netanya Netanya,

and

-+

a

-d,

E

a

A,

(a)

that

such

a

B.

=

of A in the

constants

structure

E /\abec,

ab

B, i.e.,

basis

a, b c B.

cEB

follows,

In what

distinguished the

(A, 13)

notation

We also

set

Xb and (b, x) width of b c B as

Supp(x)

=

If

:=

E, D

C

B,

F_, ©Ec

E

c

A real

In what

N,,,o

Ix

NI

E

every

be called

will

B.

:=

ED

will

B#

lb

I

E B

=

mean

B

:=

Xb

a

\ 11}. =A 01. Following For

UcEE,dEDSupp(cd). C C

B,

a

we

x

=

A with

EbEIB [32],

Xbb

A,

define

we

We shall

C+ for

write

the e

aC

write

the

following

>

0, for each

A.

b, c E follows,

a,

set

For

we

C C B.

A GT-algebra

triple

jSupp(6-b)j.

then

JaJC,

of

instead sum

the

=7 b. GT-algebra

O,if

Aba:L, and Aabl

=

B.

basis

set

we

B1. Aabl

a, b E

GT3. For each

we

use

x a

01,

real

if R

the

following

where

a

=

R, /\abc

of the

one

is

called

Aa3. a

algebra.

table

R,,o :=F Ix (E R I binary following

notation:

is

0 and

>

GT-algebra

commutative

x a

01

and

relations:

1 >. Let

direct

t

:

A

-4

R be the of

consequence

We say that each b E B. In

non-degenerate

a

this

linear

GT3,

basis case

associative

we

B

function

form

on

t(xy)

A)

(an algebra_

A becomes

by

defined

that

obtain

a

is

=

t(EbEB Xbb)

t(yx),

non-singular algebra,

Robenius

A.

Z, Arad, M. Muzychuk (Eds:): LNM 1773, pp. 1 - 11, 2002 © Springer-Verlag Berlin Heidelberg 2002

=

x:t.

As

a

x, y E

if

\6-bl

since

0 for

t(xy)

is

a

Z. Arad

We define

and

M.

Muzychuk

bilinear

a

form

IX) Y1 According computed

to GT3, is by the following

a

[E D c B is

=

t

setting

(X-Y)

-

bilinear

symmetric

form,

values

of which

may be

formula:

Xb b,

bEB

A subset

A by

on

said

E Yb bEB

be

to

a

b]

bEB

table

subset

XbYbX b-l if the

(OIED

R-submodule

with distinguished D. basis GT-algebra In what follows, the notation D < B will mean that D is a table subset of B. A routine check show ©s that the intersection of two table subsets is a table subset This justifies the following as well. definition. Given b E B, we define b. We say that Bb as the minimal table subset containing Bb, 6 E B is subset a table generated by b. An element b G B will be called faithful (see [7]) if Bb B. We say that if all non-identity elements of B are faithful. An element (A, B) is primitive b E B is called real (or symmetric), if T b, [7]. If (A; B) is a real GT-algebra, then, by [20, Theorem 3.14], there exists a unique algebra homomorphism I I : A -4 R such that lbl ITI > 0, b c B. We call it the degree homomorphism. The positive real numbers f lblJbE13 are called of (A, B). the degrees In what follows, the notation we use 1XI, X EbEB Xb b for the sum EbEB Xblbl. If C c B, then ICI will stand for the sum Eccc lei. We also write [x] for the following multiset: [IblXb ]bESIJPP(X)Following [30] we call a basis B standard if lbl A6-b:L for each b E B. We is

a

=

=

=

=

=

also

If

R C

C,

then

definite

positive An

(A, B)

that

say

we

set

Hermitian

if B is standard.

(x,y) form

GT-algebra

integral

is

a

A commutative

(briefly,

ITA)

There

are

(A, B): min(B) integral

as

real

defined

:=

in

GT-algebra

EbEBXbYb*A6-b1

denotes

are

complex

such

that

all

which

x

the

(A, B) [32]).

[29].

in

is

algebra

as defined integers an integral table exactly

of

\ f 111, max(B) is called

structure

rational

parameters E B

the

is

conjugation).

[30].

numerical

minf lxl I

=

*

GT-algebra

lbl

GT-algebra

integral two

[X)Y*l (here

=

on A

Aab, and all the degrees

constants

An

is standard

GT-algebra

integral

an :=

maxf lxl I

homogeneous

of

x

degree

c

B1.

R A, the concept of Thus, we have defined and. homogeneous table homogeneous integral algebra. Given A standard real GT-algebra (A, B), one can define the SchurHadamard product o by setting aob to A. Jaba and then extending linearly An easy check shows that A is an associative and commutative algebra with B+ is the unit of the Schur-Hadamard The element to this product. respect if

min(B)

=

max(B) GT-algebra

=

Introduction

A considered

An algebra product. in the nius algebra with

respect

sense

with

[501.

multiplications

two

of B

The elements

is

a

double

Robe-

idempotents

primitive

are

product.

Schur-Hadamard

the

to

of

basis element each table by (replacing Any table algebra may be rescaled and any ITA can be to one which is homogeneous, scalar multiple) positive ITA to a homogeneous rescaled [32, Theorem 1]). The number a

JbJ2

o(B)

A6-bl

bE33

does

Let It

depend

not

order

the is

(A; B). (A, B) be a check

to

easy

is of the

C-coset

A(b, C)

exists

the

into

Cbi,Cbj double

left

disjoint

Three

b C B be

pression:

U

U

...

element.

Cb,,,,

is

an

b c B there

partitioned

where

cosets

two

right

and

following

ex-

define

can

0

Each left

Thus B is

one

subset.

:

if A is commutative.

coincide

of cosets

arbitrary

Cbi

=

each

for

[201.

0 j. Analogously,

if i

types

an

B

C-cosets:

n C

C-cosets.

Moreover,

A(b, C)(Cb)+

=

left

called

is

JBI.

table

a

Supp(ab)

0

rb

each

b E B.

If

pjs

a

(or (A, 13©) all

factors

then bijection, is a rescaling b E B

©rb,

algebras (A, B) B©. B) if W(B)

The B if

=

only

and

if

W is

elements

(A©, 13)

and

In other

isomorphism

an

GT-isomorphic

are

be mentioned

that

in this

case

of R.

called

are

words,

(A©, 13)

and

should

It

isomorphic (denoted exact isomorphism associated algebras

exactly

W : A --+ A© is

of double

an

ftobenius

A©, i.e.,

A and

with

(A, B)).

invertible

are

(A, B)

say that

we

of

O(x Y) W(X) -©P(Y); W(X Y) W(X) W(Y); ©P(X) W(X)=

-

-

-

=

=

The most

famous

is

complex an integral

of

conjugacy

the

examples

Basic

1.2

group

examples algebra

of ITA of

come

finite

a

from finite H. It

group

theory.

group is easy

Let

check

to

t

CHbe

©hat

CH

where all basic elements GT-algebra linear. are The center of CH consisting of complex-valued Z(CH) is the subalgebra class functions. It contains two distinguished bases which define the structure of ITA on Z(CH). The first is formed one functions by the characteristic standard

classes.

the

first

the

corresponding

The second

degree

the

case

a

element

coincides the

with

first

the

basis,

©Association

one

basis

In the

degree

usual

algebra

the

schemes

of irreducible

consists

element

class.

conjugacy

respect-to 1.2.1

of

and

is

equal

second

of

characters.

cardinality degree of a

In

the

the

irreducible

an

Z(CH)

case

to

character.

of basis With

is standard.

algebras

centralizer

R {Ro,..., on X is called Rdj of binary relations a homogeneous (equivalently, coherent conEguration [44]) with d classes if the following conditions satisfied are ( [27]): AS 1. RO I (x, x) I x (=- Xj and X x X Ro U U Rd is a partition of

Let X be an

a

finite

A set

set.

=

scheme

association

=

=

...

X

X

X;

AS2. For each

f (y, x) I (x, V)

E

i G

Rij

10, 1,

=

Rj,;

...©

dj

there

exists

V

(E

f 0, 1,

...©

dj

such

that

RZ

Introduction

I

AS3. the

triple

each

For

i, j, k

E

10, 1,

dl

...,

arbitrary

and

(x, y)

pair

E

Rk

number

Aijk does

depend

not

a

on

Ri,

The relations

I Jz

--z

(X, R).

of X and the

(x, y)-entry

d

...,

(x, y)

called

equal

is

E

to

Rj

E

-

relations

are

(x, y)

of the

coherent

Ri, i.e., Ai is the labelled by the elements of

matrix

of which 1 if

Rk

basic

the

adjacency

and columns

rows

Ri and (z, y)

pair

are

Ai be the

Let-

E

E

Ri,

and zero,

by

the

matrices

otherwise.

It

AS3 that

from

follows

0,

=

I (x, z)

of the

choice

i

configuration matrix, square

the

G X

d

1:

AiAj

AijkAk-

k=O

Therefore is

complex vector of Mx(C). It

the

subalgebra

a

A spanned

space

is called©the

Ai,

i

of

algebra

Bose-Mesner

d

0,

=

associ-

an

scheme.

ation

It

is

to

easy

GT-algebra,

the

the

(A,

pair

then

SITA). algebra (briefly, of homogeneous Many examples from transitive permutation groups. G

group

on

finite

a

set

these

orbits

diagonal

(X;

that

fact

well-known

always

We shall

relation.

-

2

-

the

(G; X) JRo, Ri,..., X) a

of G

on

is

an

table

may be

built

RdJ one

RO is such

a

be the

[59],

relation. on

a

com-

we

of 2-orbits

scheme

association

of

action

X2. Following

G is transitive, that

integral

transitive

=

integral

intersection

standard

a

be

Since

X))

standard

cohfigurations

coherent

assume

orb(G;

a

with

is

Z=

coincide

it becomes

Let

action

(G; X)

©of

2-orbits

the

fAilid-0)

=

2-orb(G;

induced

of the

of orbits

plete

X and

set

B

of which

constants

structure

If A is commutative,

Aiik.

numbers

that

see

It

call is

a

is

a

X. The

with the centralizer algebra of algebra of this scheme coincides of G on X. to the action of G corresponding the permutation representation then the Bose-Mesner is multiplicity-free, If the corresponding representation table standard it is an algeintegral and, therefore, algebra is commutative Let the in of this situation example. following case bra. A particular appears the of H action x H on H deAned the Consider by H be a finite group. Bose-Mesner

It

is

2-orbit

[43]:

rule

following

easy

of

to

check

H

x

X(hi,h2) that

H if

h, lxh2,

X

E H. 2

pairs (X1, X2)) (Y1) Y2) G H belong to the same X1X2-11Y1Y2 1 belong to the only if the_ elements

two

and

=

in one-to-one of this action are class of H. Thus the 2-orbits conjugacy of classes of conjugacy of the set Cla(H) with the elements correspondence has C E.Cla(H) class to a conjugacy RC which corresponds H; the .2-orbit 2 form Rc the following f (x, y) E H 1 Xy-1 E C1. Moreover, this correbetween the Bose-Mesner exact algebra of defines an isomorphism spondence 2 orb(H x H; H) and the center of the group algebra CH.

same

=

-

Z. Arad

H be

write

A

subalgeb a

for

each

X

exists

basis

a

EhEH Xhh

=

h E H. It

is

subsets

B

G

to

easy

a

Schur

=

ITO

which satisfies the following quantities S1. H is a partition of H; To U....Td

simple

a

there

simple

(briefly

111)

simple

that

exists

a

We

quantities.

t-

ring =

CH is called

see

of H and

EtET

sum

A C CH is called

there

H if

group

An element

group.

fO, 11

Xh G

between correspondence T, T C H for the formal

one-to-one

shall

finite

a

if

quantity

Muzychuk

rings

Schur

1.2.2

Let

and M.

)

...

S-ring)

an

Id I

of A

the

over

of

consisting

[56]:

conditions

=

S2.

ft-1

I

For

each

TJ

t E

Ti,

The sets It

i

0,

=

evident

is

10,

i E

(A, B)

becomes

d

...,

that

there

V

exists©

called

are

algebra

the

GT-algebra.

standard

dl

fO,

c

dJ

...,

such

T(-©)

that

Ti,.

=

the

basic

A with

of A.

sets

B

as

table

a

basis

(for example,

If A is commutative

is

if H is

integral

an

abelain),

then

table standard We say that integral algebra. an integral standard is group-like if it is exactly GT-algebra to a Schur ring. isomorphic Let A < Aut(H) and To,-, set of A-orbits Td be the complete of the Aaction Then the vector on H. spanned by the simple quantities space Ti i is a Schur 0,...,d ring over H. We shall write O(A,H) for the set of this case construction 71dl. A particular f1:0) of gives us the center the group algebra H if we take A Inn (H). an

z )

=

...

=

G be

Let

subgroups,

a

in

disjoint

Theorem a

a

FH, form Fh, =

defined

[56].

The vector

S-ring

1.

G=Hx A,A

H,

To,...,

Td

Each Schur basic

are

Basic

1.3 In

2.

this

algebras.

section

the.

a

a

product

a

Since

I.

double

left

Then H =

=

FTi.

intersecting a unique

has

may be written

U F9dF be

FgiF

trivially F-coset

FgF

coset

F U FgIF U...

F-cosets.

of two

each

(To

a =

One

111)

U

...

easy

can

0


jai),

p:!

pjjdjj

=

the

main

operations easily

in

the

of inte-

class

generalized

may be

operations

for

special

(A+ 1)/2.

of

class

table

integral

0 < n, I < Let A be

m.

a

Fix

real

C-vector

algberas

introduced

in

the

jAuj 10 set


3.

with

a

non-real

Let

us

Then

exists out

integral following

most

difficult

is nonreal

x

one.

and

[xyx]

=

say

[14, 1211.

Of c Supp (X2) always 2. If x" [14 3 41. If Xa is not real, then either Thus, in any case, either x" is strong the following: we have

that

case

section).

next

the

if

is©strong

=

an

! the

F,.

[Xa-X7a [14, 1211.

second

the

[Cb]

For

[14

product

min(B)

[14,121]

turns

5

element

Introduction

wreath

standard

defined

there

x

It

[14 [Xayal [Xa7Xc, [14 34]. In the [Xa-X7a

or

or

=

degree

Theorem

3 4]

=

are

[6-b]

on

F©,.

Bb

F,,n

depends

multisets

[14,43],

6 2],

a

min(B)

41],

2

5 of the

that

Bb

then

element

strong

(Theorem

4

an

Assume

then

case

x E

integral

be 4.

4

4 and

[14, 121].

[14 81,41],

then

the

Theorem

4], [14

3

[Cb] V j[14

show that

to =

following

the

[14

[14 41,41,41], It

that

shows

integral

standard

a

unique

the

element

width

standard conditions:

Z, Arad, M. Muzychuk (Eds:): LNM 1773, pp. 13 - 41, 2002 © Springer-Verlag Berlin Heidelberg 2002

of x©

table

x" is

algebra

with

a

strong

Z. Arad

14

min(B)

1.

[6-b]

2. 3

b2

.

=

=d+2f,JfJ=6

that

Remark

(see

in the

1, -11

Z2n

(x)

=

is

function

sign

a

(-x).

sgn

section).

next

2. In this

to the

Z4

case

Z4

(1)

Z4

(9

2.1.1).

Subsection b and

have the

:

1 and sgn

-

if and only if n An (sgn) is group-like group-like algebra over the group

algebra isomorphic

If both we

sgn

< m< n

defined

is

The

1.

1, 0

3 4].

[14

where

> 2

n =

An(sgn)

exactly

is

[cFdj

and

An(sgn), (2m + 1)

sgn

(The algebra it

3;

>

[14, 121];

Bb _9©,;

Then such

al.

et

b©

then

strong,

are

alternative.

same

we

Thus there

b ,2

may consider are

two

possibilities:

that

[F_77

(b)©

=

for

which

either

bl©-

is

for

strong

every

natural

(minimal)

m (=-

m

or

there If B is first

exists

algebra over is a group-like

If B

occurs.

case

a

group-like

a

Z2- 6) Z271 then [F-7177] the first case Bb is always -

group

algebra

3 4] for

each

over m

an

G N.

[14, 34].

of odd

order,

abelian

then

the

Z271

group

We conjecture

that

in

group-like.

algebras

Group-like

2.1.1

[14

abelian

examples

Known

2.1

=

N such

an

Z4

H be an arbitrary of the symmetsubgroup Z4 by permuting coordinates. The subgroup G So we can define I (x, y, z, w) I x + y + z + w 01 is an 11-invariant subgroup. table a standard an S-ring, and, therefore, integral algebra O(G, H). Let now H be an arbitrary subgroup of S4 that contains cycle (1, 2, 3, 4). Then a. full the set 0 J(1, 1, 1 3)) (1,1, -3,1), (1, -3,1, 1), (-3, 1, 1, 1)1 is an H-orbit

Consider

the

group

ric

S4.

H acts

group

Let

.

on

=

=

=

-

G. It is easy gree 4. If H G

to

on

of

degree

see

that

JA4) S41

than

0 is

then

,

0

a

non-real

(G, H)

faithful

does

not

element contain

O(G, H)

of

non-trivial

of de-

elements

RO. In order to algebra is of dimension examples one has to take an H-invariant subgroup of index (for example, G of finite to the kG, k E Z) and factor out with respect It is not difficult to see that chosen subgroup. such a quotient with respect to a subgroup kG, k > 5 gives us a finite example of the above algebra. build

less

2.1.2

We shall

The build

following multiplication

the

4. The constructed

dimensional

finite

algebras a

series

F.,,, of

6-dimensional table

F©

M

of dimension 4m + 2, m > algebras algebra (H, E) where E 1, h,

of which

looks

as

follows:

I

-

We start

C1, C2) V) U)

with

1,

the

ITA with

2

h

h +

v

of

Element

Nonreal

C2

Cl

h+v

41+h+c,+C2

h

Faithful

a

Degree

U

V

V

3cl + 3C2 +4u +

h +

v

4C2

41 + 4u

3c,

3h + 3v

C21

h +

v

41 + 4u

4c,

3C2

3h + 3v

3c,

3C2

31 + 2u

3h + 2v

3C2

3h + 3v 3h + 3v 3h + 2v 121 + 9cl

V

U

3c,

+

+4u + check

A routine hold

gebras U

=

jh,

v,

I

Define

case.

21 + 16 2

=

I

(CI

C2)7

+

computations

I

all

that

shows

this

in

s

By direct

v

the a

+

t

h

S,

3h+v

tj

2v

41+cl

I

3c,

3c,

+

H

multiplication

table as

+

tables:

2v

v

1 2s

4tl

+

4s + A

t

S

3h +

3C2+

(2.2)

4s + A 2s + 4t

4v + 6h

4u +

al-

follows:

t

S

5v + 3h

,

+C2 + h v

HQ + C2)1-

A+

v

V -

(Cl

Q (9z

10s + 8t + 3h + 2v 5v + 3h 4v + 6h

v

h h

+ "2

.

standard

:=

V

2s + 4t +

v

integral

following

2s + 4t +

h + 2s

h

U

the

obtain

we

U C

subset

(2 1)

+ 9C2 +3h + 8u + 2v

,

of

axioms

15

v

C1.

V

4

2v

v

v

4u+v

+8u + 3h + 2v

3h+v

5v + 3h

2v

4v + 6h

4v + 6h

5v + 3h

3C2+ 121 + 9cl + 9C2

(2.3) 4u + 3cl

61 + 3c,

+ 3C2

+3C2 + 2u

I

h h +

C2.

h +

v

v

It

Let

let

from

follows

(CC,,,,

C,,,

the

(a)

2s + 2t

2s + 2t

v

3h + 3v

2s + 2t

2s + 2t

2v + 3h

2t +

+ 4t + 2s

above be

2v

3h + 3v

©

formulas

a

Ua

QU is

of order table

E,a U,a:7

I

an

ideal

m written

algebra. =

1; 1.

(2.4)

t + 2s

I

+ 10s

that

s

3h + 5v 4v + 6h

13h + 2v

I +8t

cyclic group Cm) be the corresponding =

3h+v

v

V

U

t

S

V

Cl

61 + 3c, +3C2 + 2u

3C2

+

2s+4t+v

h + 2s

h

4u + 3c,

For

j

QE. multiplicatively each a E C,

of

and we

set:

Z. Arad

16

a

0 U is

0 h is

faithful

a

A direct basis

of

for.

with

We start

shows

the

by H(m, n),

denote

C©

e,

n)

Let

additions

is

arbitrary

an

sgn

x,,xp we

f co,

a

cyclic following

the

and commutative

elements

form

basis

ea

:=

=

2e,

+

4eO,

a

f

4c,,,

by

the

U

f eo,

fC1

U

+

C21

is

a

4m + 1.

to

algebras following

we

I

en-1

...,

which data:

+

e+,0

(1

A direct

n.

where

shows

check

if ©M G integral by g (E C2n- In

that

N. a

sgn

algebra

group

Z2n

:

1, sgn (-x) sgn(a)sgn(,8)/sgn(a

H(4, n)

:34

=

=

1-1, 11 We sgn (x). + 0). Since -+

0-

-

-

I

n

Clearly,

-

shows

that

this

is

the

following

an

as-

=

+ 0

2X2ce+l)

-

2ea

0,

f

0, 3co

-

a

=

2X2.,

0 0;

a

0

0; -

xO,

a

=

0

11 C Z2n We denote following equalities

this

-

are

a+o+l

+ 1

-

The

basis

3o,

:=

C

©

-,

5n.

S)e 0

0

=

=

e,

0;

10,

(CC2,rb

check

CC2n:

54 0;

a

(1)

A direct (1)

Ce

=

+

=

+ Me",+©3+1

Mcc,+©8+1

(0)

sgn

2X2ce+l,

set

a+

is

1)oa+,a

-

sgn(i)g©

=

S(a,,3) sum

above

in the

ca+0+1 4c+0 +

xi

0

a

constants

+

:=

1.

CO + XO,

structure

modulo

satisfies

2X2co =

(M

generated

basis

+

2)e,,+,3

-

which

group

=

o,,,

runs

00

equal

of table

series

foo, ..-On-11

done

is

H(4, n)

through the Clearly, lAn(sgn)l

+

U

algebra.

An(sgn).

Oa

\ JC1) C21) is

is defined

©

direct

of

0+

a

(F,,,

dimension

It

(m

where

sociative

where

I

+

which

S(a, -a)

+

R>O.

algebra

table

consider

Ca

Its

cn-1

...,

S(a,,3)x,,+,3

a

F© M :=

set

two-parametric mE

1)c,,+j3 1) 06,+0

function

=

symmetric,

is

Now

-

the

choose

fibres.

F,

set

standard table Since integral algebra. the algebra F.. is an example of algebras of this algebra is 4m + 2.

Fn.

of indepes

C2n be

now

we

the

00,+0 -

standard

a

C,"-graded

book

oa

(m (M

is

this

an

,

eg ec,+o

(CC2n have

=

ec,+,3

the

H(m,

the

of

set

a

to

An(sgn)

cp C,,+O

0,3. Ocl+,8.

of

of

following n E N,

H(m, n)

where

that

is

Introduction

element,

subalgebra

algebra

The

2.1.3

basis

JUj,,cc_

=

The dimension

check fusion

a

U

4 of the

non-real

looking

we are

that

by Proposition a distinguished

Therefore

C,,,

(2.2)-(2.4)

from

follows

It

al.

et

nonnegative

s)e,+,,+,,

show

basis

that

integers.

a

+

0

+

10 0;

by the

ITA with

2

where

s

=

S(2a

1, 2,3

+

a

Faithful

Nonreal

Element

of

Degree

417

1).

+

2.

where

(3 s)e+ 4co3c +,+,+3eo,a+,3+1=0 +

0+0-

=

S(2a

s

-

1, 2,3

+

where

s

=

S(2a

3+3so+ 2

where

s

=

S(2a

+ c,

3-3so+ 2

e

=

ce

10

a

+

0

a

+

)3 +0

+

0

0-

0

©a,

a+

0

0

)3+3+so

a+

-

a+

2

)3,0: O

0

1, 2,3).

+

+

Oa+)3+Oa+,3,0: O

c+-

0 ,+

o+c

2

3oa,,O,

+

5.

6.

+,,

1).

+

4. 0

3-3s

+

a+18

3o

0

1, 2)3

+

s)e-

+

1).

+

3.

Oe

(3

+

+,

a+

+

+ Oa

,q

"3

0.

=

o-

7.

9c,,,+,,+,

0-0

where

s

(9

+

s) ece+,6+1

S(2a

1, 2)3

+

+

Oa s

S(2a

+

9-3so+ 2 a+

+

e,8

3ol

0-6a

0

a

0

9+sO 2

+

43

3o,,,

+

)3

-

9+3so+

0,3

=

3o

a

C+,q6

a+

0"3

0

0

=

9-so2

a+,6

10 0

0

0.

3o+

+

3o

2o;;,)3

a+,6 +

3o,+,,,,O =

0

0

0.,

2o,-.

+

cl

+

Q+0

2

)3

10.

oc

s) e,+,,+,,

1, 2,0).

9.

11.

+

1).

8.

where

(9

+

12c++8c-+9e,+,a+0+1=0 0 0 0

12.

6ct

3co+, 6c+0

+ +0

4(et +

(2

+ +

+ 2c-0 +

cc-,),,3

=

s)e+a+O

2e+,0

a

+

+0

0

(2 =

-

0

s)e- a+ ©e,

a

= A 0,)3 =A 0,

a

+0

Z. Arad

18

where

s

=

al.

et

S(2a,

2û).

13. 0 0"3 6c+ + 4(e+ + 6-), -p û =

3c+",+û 4c, where

s

=

S(2a,

0

C,

(2 s)e++ + +2e.,a+0=0 +

-

+

û

a

(2

+

s)e-

+

2û).

14.

e "û +

ea

+=

Co

0

=

4(ep

eû ),

+

e++", c,+ 2e0

,

a

=:

0

7 01,6:A: 02 C+ 7 0, û:7 0, a + û

7

a a

=

0

0.

15.

e

ea+c-0

ce:7

+ ei,

+

3e0

,

a

=

0.

16.

0,a e

6c+0 where

s

S(2a,

0

==

3e,

e

+û

0

or

+(2+s)e

+ 2c-0 +

a+

2e+,0 a:7

0,

0,

a

+

j3

0

2,6).

17. 0 e-c a

0,a

+

e+"+O,

=:

0

2e0 18.

ei cJ0

=

2e+

+

ej,

a

0

=

* 0, û * 0, ce 7 01 01 a

ce

,a

:7

+

û :A 0 0.

0.

19. +

ca

3

=

4c",+" a + û 7 0; 4(c+0 + c-),0 a + û

20.

4 cj 21.

c0 ej

=

3co+

+

2co-

=

3c+, Co

,

a a

:7 0;

=

0.

=

0.

7

0

ITA with

2

Nonreal

Element

of

19

4

Degree

results

main

of the

Proof

2.2

Faithful

a

Preclassification

2.2.1

Proposition either Aabc

a, b

Let

1. E

simple

following

the

from

We start

10, 11 for

(=-

each

B# be E B

c

elements

two or

the

of

one

of degree 4, cases following

24

8 2f,I fI 2f + 2g, i f I= IgI 3f + g, if I= IgI4 4 f,IfI + 3, IgI 4f g, if I

32

24 28

=

=

32

4

40

64

=

Proof.

1

4

=

=

5

Routine.

As

a

consequence

Theorem

for

Then

holds:

(ab, ab)

ab

=lhl=4 2f +g+h,lfi=lgl 2f +g,lf l=4,l gl=8 6, I gi4 2f + g, if I

1

=7 T.

a

all

b E

(A, B)

Let

3.

be

integral

standard

a

BO. Assume that

following

the

obtain

we

there

b E

exists

table algebra such B* with b 0 b and JbI

JbI

that =

4.

3

Then

either bb

4c, b2= 4d, Icl

41 +

=

=

3, Idl

=

(2.5)

4,

or

b2= Proof. only

Since 7

IxI

d + >

2e, Idi

3,

possibilities

G

x

for

=

4, jej

=

6 and

B# and 4 divides

Ab;,ICI

for

f

+ g +

4 +

f

h, If I

(bb, bb) =

IgI

=

8, IgI if I 4 + f, If I 12, IgI 2f + g, if I 6 4 + 2f, I f 1 4 4 + 3f, i f 1 3 4 + 4f, i f 1 + g,

=

IhI

=

=

4,

=

4

4 +

now

the

of the

decomposition b

2

=

1: beB#

4,

28

28

28

=

Consider

all

bb:

bL 4 +

(2.6)

V, ESUPP(b-b)#Ab c

product /\bbeC.

36 40 52 64

c

G

B#,

there

are

Z. Arad

20

If all

coefficients

nonzero

Therefore,

al.

et

there

(bb, bb) 28, (b2, b2)

that

=

(b2, b2)

(b2, b2)

Abbc :

f 28, 40, 52, 64}.

E

(2.6)

then

=

B# with

C-

c

then

ones,

axe

exists

holds.

Thus

have

contrary

deny

to

bb Assume

equality sides

lcl

of

2 3c, b =4d+e,ldl

=

41 +

2 2c, b =3d+e,lcl

first

that

b satisfies

that

db

equality

4, I\jdc

=

Consider be real

41 +

(2.7)

of this

=

=

in bc is not

b

we

0. Therefore

now

in this

by

the

than

greater

9,

e)

(2-7)©

=4.

(2.8) second

both

Idl

Since

3 and

=

contradiction.

a

lcl)

=4.

121 + 9c.

=

I

If

possibil-

=

+

28.

holds.

from the directly A 36. Multiplying

(2.8).

when b satisfies

case

gcd(lbl,

=lei

follows

d(4d

obtain

remaining

=6,ldl It

lei

=

equivalently,

or,

Aj,

Since

case.

(2-7).

3b,

=

3,lcl

=

(2.5)

following

the

>

Proposition

64, then

=

lities.

bb

(bb, bb)

to

follows©from

It

(b2, b2)

If

we

16

=

2.

2, the number

=

Note that

of

should

c

summands

nonzero

Moreover,

2.

(bb, c)

(bc, b)

12 =: -

/\bcb

12 = -

=

=

3.

Therefore, bc The g

0 b, lgl

(2.8)

in

bd

implies

12.

=

3b + g for

==

some

g

E

B#,

We have

4.

=

Af, f =/= b, Alf I

3b +

equality

second

(b2)b

(3d

=

b(bb)

b(4

=

e)b

+ +

2c)

3(3b

+

4b +

2(3b

=

=

g)

eb,

+ +

Af).

Consequently, eb

f

Therefore

=

g, A

immediately

b +

=

implies

and for

JR(e some




(bb, ei )

ISupp(be)l

2,

=

pf

for

equality,

latter

of the

f

we

this

book

one.

satisfies

=

Now Abbe

2.

:5

to

21

4

of degree 4 that be 3b + f.

that

suitable

a

Introduction

element

12 such

=

Degree

of

interesting

most

non-real

a

BO, if I

E

3b +

=

sides

both =

f

exists

is the

case

BO be

b E

Let

2.

Then there

second

Element

5 of the

by Theorem

then

So the

.

Nonreal

Faithful

a

obtain

1-tif I

12/,t

> 24.

=

implies

2

=

BO and

G

/.t

E N.

12.

The

that

Taking identity

implies

E

A,e,lcl

=

than

JeJ2

12 +

r_GSUPP(bb)\j1j The left-hand

Therefore

it

In what

3 and

rem

side

follows, Proposition

b2

In this

Let

if bx

element

Properties

Proof.

a

we

fix

2,

2f, R

=

say that

If

Supp( b_)

3.

starting

For the

Jbi

3b + g,

Idl

=

[bb]

that

element

is

divisible

by

12.

0

degree

6, Igi

4, jfj

=

=

following: -of degree

of the

one

Supp(bb)

x

G

4.

By Theo-

(2.9)

12.

a

basis

[14,41,41,41], 4 is

a

starting

element. contains

of degree

element

4, then

Supp(66b)

one.

Eb

product

there

possibilites:

two

are

juj=jvj=jwj=4, U:

b

Icl us

is

1.

I

Eb=41+u+v+w,

Let

=

b E BO of

element

Ed.

starting

30 and tt

always©have

an

of the

161

-

implies

in turn,

non-real

a we

we assume us

=

Proposition contains

which,

24

of Theorem

*subsection

[14, 81, 41].

greater

not to

d +

=

Proof

2.2.2

is

equal

is

compute

the

(bb)E

=

8, jhj

=

product

(d

+

2f)E

=

b _b =

=

V,

V

7-L

W,

-

c

+

h,

41

4,

=

h,7!

=

:7

W

u

(2.11)

c

in two ways:

dE

+

2Cb

=

dE

+ 6b +

(2.10)

2g.

Z. Arad

22

al.

et

b(blb-) After

reduction

of

d Assume

bc + bh

holds.

bu bv

=

=

bw is

w

a

(2.11)

If

element

starting holds,

in this

Abcg ©A

Clearly, other

Abcg (2.13)

2.

=

implies

bh

Proposition

Let

4.

x

for

a

Abvb

=

=

of the

Abwb

Supp(bE),

of

b + g; b + g;

(2.12)

d.

=

6

=

+ 6b +

(2.13)

2g. is

=

least

at

Abcg

=

=

Further

1.

=

elements

0(mod 8). Therefore Abcg 32 implies Abcglgl :5 Jbilcl 2 this gives us bc 2A -bc

2b +

=

On

2.

< 3.

2g.

Thus Now El

Supp(bE)

E

holds

=-

inequality with Abcb Ed..

=

y + 2f

dT

1.

Abcglgl

But

0.

hand, the Together

Abub renaming

holds

case.

4b + bc + bh

the

holds

(2.10) (2. 11)

if

Then

up to

holds

obtain

we

bc + bh

©5

Thus

if

bu + bv + bw if

2g

(2.10)

that

(2.10) (2.11)

+ bv + bw if

terms,

common

+ 2b +

first

t

+bu

4b

11 Abug ! 17 Abvg < 1. So, have the following equalities:

Abwg we

=

suitable

be

element.

starting

a

B#, jyj

y E

=

Then

4;

2.

Cbd, Ed) Proof.

(f, dY) AdTd < Idl a

-

16
© 16. But that contradicts 2. Thus dY AdTf 0 0 implying AdTf y + 2f

=

Therefore,

3.

y E

Our

assume one

(bx, bx) =24 28

1.

If

for

=

should

(bv, bv),

is case

i.e., be real.

(d, y)

+ 24.

always of

=

[bE]

real. =

[14 81,41].

(2.12).

b satisfies

W.l.o.g.

0

u

=

If

w

U. Since

Thus is

16

we

nonreal, =

may

then

(bu, bu)

ITA with

2

A,-,-, A,,7u-, A,uu Av-,, Av-vw

Nonreal

Faithful

a

=

0

=

0

A... AvUU Awuu Auvv Avv, AWvv

==>-

0

=

0

=

0

=

0

=

== ©

Element

Degree

of

4

23

0

=

0

07

(2.14)

0

0 0.

=

Further (2.12)

(bu, bv)

Auvu

Together

(2.14),

with

+

implies

this

uv)

16

=

Auvv

+

uv

4w. Rom here

=

Auvw

16

=

4.

=

it follows

that

uw

=

4v.

Thus

(6-b, uw)

(bu, bw) Since

bu

b + g and /\bwb 16 contrary

=

(bw, bw)

and

Proposition 1.

dY

2.

d

If

6. =

d +

=

x-x

2f;

to

=

Supp(bE)

E

x

1, (bu, bw) Proposition

=

=

is

=

4 +

=

12Abwg.

Hence

bw

=

b + g

4.

element,

starting

a

16.

then

b.

1. Assume the contrary, Proof. i.e., dY y + 24. By Proposition 4, (bx, bx) Proposition

2f and y 0 d. Then, according 1, bx is either of type [4 2 since 1. Therefore occur Abxb

=

to

or

bx are

=

[4 2 4,4].

of type

b + 2u + v,

=

=

where

distinct.

pairwise

b(bx) (bb)x

The first u,

+

2f)x

=

=

cannot

basis

are

v

Now we

b 2 + 2bu + bv

(d

type

elements

of

degree

d + 2f + 2bu + bv

that

d + 2bu + bv

y + 2f + 2fx

=

4 such

u, v, b

write

can

dx + 2fx

8]

=

=.

y + 2fx.

(2.15) It

follows

from

Afff

=

Afxd

2 that

=

3.

Together

with

gcd(If 1, IxI)

=

2, this

implies

fx=3d+[tz, for

a

suitable

z

E

BO \ Idl.

After

On the b 2X

=

y + 5d +

y + 2f + 6d +

(2.16)

in

(2-15)

we

obtain

2ttz

(2.17)

2/,tz.

(2.18)

hand,

other

bEd

=

of

substitution

2bu + bv

b2X

(2.16)

[tGN,zGB

=

4d + xd +

cd

cESupp( b)\11,xI

=

4d + Y + 2f +

cd.

CESupp(bb)\{I,xj

Z. Arad

24

this

Comparing

Together

8.

since

Thus

y

d

=

Since

2.

we

(

T

y

=

obtain

cESupp(0b)\f1

with

d

+ 2d.

2/-tz

=

4, /t < 2. By (2.16), /-tlzl bv =_ y + d(mo ©d2). Therefore (2.17), 0. Hence AbvziZl =_ O(mod 12), this implies Abvz But now b(u + v + b) 5d, which is impossible side

According

d + p

C)

I

©.

left-hand

f6,121.

E

3d + y, bu u, v, d are

=

(2.18),

in the

jzj

Hence,

AbvzlZl :5 bv

with

coefficients

Since 12.

al.

et

do not exceed

to

=

z.

distinct. pairwise required.

as

=

d, (bb,xTY)

(bx, bx)

=

E

Therefore

28.

=

Ax-,,,Icl

12.

=

CESupp(1;b_)\{1j

Together

with

1cl

=

12

ESupp(6_b)\{1j 3 this

and Theorem xx

=

that

implies

Ax-x,

I for

=

each

c

G

b.

Since

bx

Td, Cbd, Ed)

=

=

Arguing

28.

before,

as

Supp(bU)

we

obtain

111. that

Hence

6-b

=

du. 0

In

bE

what 41

-

h.

-

of two basis

sum

So

we

c

then

of

Proof.

for

some

v

(fd, Td)

E

=

dh

=

B# \ jhj.

,

We also

set

c

:=

c

is

a

basis

=

d,

dh

exists

v

=

d + 2f.

Td

(Ff dU)

h.

=

There

By (2.19) gives us

as

4.

=

c-

Td

2, this

element

starting

otherwise c is a element, it follows from the proof of Moreover, degree x-x o c is proportional to u + U, u E BO. In both cases x-T o c Ad-x, by the equality A, yxcc. bh

7.

the

(2.11),

satisfies

elements

may define

Proposition

denote

we

5 that

Proposition c.

follows, If 6-b

=

=

2f, dU

=

h

B# such that =

Therefore

Now we

(2.19)

d +

=

jvj

(2.19)

bb.

=

12 and

(2.20)

3h +v.

ATdh

3h + pv, can

2

3.

Together

with

ged(If 1, Idl)

p

write

jd, Td) f f ,6b) (bf,Ef

36 + 12 p (2=9) 48

p

=

1, jvj

=

12.

ITA with

2

Proposition

Faithful

a

Nonreal

Element

of

Degree

4

25

8.

hc

2h +

=

2v V

(2.21)

v.

=

Proof.

E2 d

(2.9) =

(2.20)

d3 + 6h ( + 2?)d (2.19) E2 d Ebh=4h+hc

+ 2v

=

Since

and h

c

Proposition

are

real,

9.

There

v

is real

exists

hc

that

Jul

=

2h + 2v.

well.

as

B* \ jcj

u E

such

Cf

61 + 3c +

Arf,.u

C2

81 + 4c +

4Ai!f-uu

Arfu We use the following Proof. form: following

U)

+ h2

=

E

f3,61

and

(2.22)

{1, 21.

(6-b)2

identity:

E

b

=

2(E)2

.

The left-hand

has

side

the

(41+c+h )2 =161+8c+8h+h right-hand

The

side

is

equal

2+C2

+2ch

(2.21)

d3

=

h2,

coefficients

the

+h2+12h+8c+4v.

=O)d3+2(6h+2v)+4Cf.

(2.2

hence

4Cf

Comparing

2

to

(d+2f)(d+2f)=dd+2(df+f©d_)+4Cf By (2.19)

161+c

=

of

=

c

161 + 8c + in

both

Accc is divisible by 4. On the other 2 =#0, then Aqc 10741. If Acc, Therefore 3. Accc 4, Arfc

Hence

=

sides, hand,

Arflcl

C2. we

obtain

Icl I 0 0(mod Ifl), Accc




(UU) lul

+

supplies

=

1-

I)e an

+ 1d.

element

c

C-

B#

SITA with

3

Faithful

a

Nonreal

Element

of

Degree

49

5

Rom

Icl follows

it

or

(ii) lei

=>

(i).

n.

In the

=

Since

lei

that

assume

(w, w)

6

b272 =.nl

+

(i) (ii)

b3T If

(1 G

c

(iii)

If

c

=

+

b2F2 b272 w

that

(1) 72b2 (2) 72b (3) 72b2

c

or

+ =

73 b3

=

nb2

+

yields z

EE

n

n1 +

n1 +

=

3

=

n1 +

Set

y

b2C

b4T

-

=

2

n(n

that

lwl

=

lei

Hence

we

n

-

ln

-

=

d E

exists

1, idl

6

2n

=

may

nl and

Supp(w) =

and

n

M

for

n1 + lc

=

some

c

G

NB

T-2b2

=

=

If,

-

moreover,

Cb

Tb,

then

one

then +

(b7,72b2)b7-

Cb and (b6, b2 bb2 b)

=

(b6 72 b2 6b)

=

1

Tb-

=

(1

+

(b2b)T n

6-b

then

,

1) b2.

Then

(I

1)b3T

=

L-fl-y.

+

gcd(l V Supp(z)

From

and

I

+

lb4T

=

+ 1, 1)'= implies

(1

21n.

If

c

E

B, then

Icl

we

get

+

(1

y

+

=

lz

1)y for

+

lb4T

some

C- B.

2

=

1)2b2

+

1, z

+1)=(b2b,b2b)=(Cb,72b2)=n

(c,T2b2)

2-

n

=

either

holds:

-

+ lb2 +

Finally, Izi The equation 1

/F3 b3

or

1)b61 1)b7; + b7)

b3T

:=

b2(Cb)

-

6-b

odd,

(b6772b2)b6

n1 +

=

cases

NB.

shows

get

finished.

-

=

=

(n (n l(b6

lb2C

(ii)

we

n.

=

72 b6 0 76

b2F2 b373 T6 =7 b7 T7,

and

following

=

21 + 1 is

n

lb5; b5 b6 + 76,

b6 + b7, b6

=

(i)

Proof.

2n,

=

we are

Then

n1.

Icl

lb4-

+

1)b2

+

moreover

the

-

or

n1 + lc

n2)tt-2

le, and

1-te

-

=

=

T2b2 If of

-

n1 +

-

-

1)b3

+

B

Cb

then

b2b3

is immediate.

((b2T2, b272)

!,

Set

n.

1 and

v

(ii)

=

2n,





I. f 1,

-

'\+I

linear

involutory

Wi+j+l

(Wn n

Wnj

WO,

=

2

WO) Wi+j-n-I

and

if i+j < if i+j=n, if i + i >

defined mapping algebra automorphism. -

integral multiplication

by T

standard table

n,

n

=

1 and

U7

Z. Arad

74

Proof.

proof

The

[321 (for

'm

'p(b)

We will

Example

(3)

follows

0')

=

polynomial

T-

al.

et

make

Let

3.

lei,

m

wo,

-

ab'

similar

,

(iv)

(i) we

(ii)

and

have 1

=

z-.z-. 11

ztz,-.

1 E X

all

i, j

i

+

l5eO

+ vo + vn

3zP+i+j+l

3h +

z-

zi+j-m

+

3zi+j-m-l

2 and

Y:=

,

vi+j

+

4vi+j+l

zi-+j

+

3zi-+j+l

+

ZP+j-m I

hzP

=

5vi +

=

(5eo

proves

Again, the be linearly proves

+

wi

=

wi

=

-

zP

vm-i

Hence

-

e].

are

zm-i, zi elements

::--

the

permutes

and

eo

wm-i

-

of

2wi+j+l if

wi+j-m

+

)2

el

wi+j

-

+

-

i

+j

< M,

)if

i+j=m,

)if

i+j

> M,

if i+j

< M,

2wi+j-m-l

Zi+j-m-l

zO+

+

=

3zi-,

+

2zi-

25eo

and shows

equations

(V

to

E)

a

2wi+j+l

+ wo

el

-

wm =

if i+j=m,

zz;

+

(ii)

extended

-

+ wi+j-m

3zt+j-m-l

2zP 3zp

-

zi+j+l

4vi+j-,n-l

above that

V (D W. Since

wo + wm =

-

-

vo + vm + 3

51 + 2h +

hzi

for

2

WjWj

-

vi+j-m

This

Example

+ z,

4vi+j-,,n-l

-

+

3e,

-

+

l5eO +

This

2.

Z,-.+i+j+l

+

vi+j-m

5vi

basis

Furthermore,

+ wi+j

zP++j

=

a

10,..., ml. ml. Then

E

10,...

(E

Y is

Y.

4vi+j+l

+

vi+j

VjVj

h2

0

VjVj + WjWj

=

I

"

X

Clearly,

=

Y for

X

c

zj- zj t

ab'.

-

of Y.

unit

Y. Choose

X

in

Al

-

the

out

X and

remain

real, z+m-i

=,3b,,+,

of

3.3.

factor

By V and W, we denote again the algebras generThen the statements Y, respectively. (i) to (iii) of Example true. Moreover, T- (3) is not isomorphic to 3T,, 3Tn( E) 3 Tn(3). M 3

el

by

Proof.

2wi+j-m-l

-

Example to

if i+j < M, 7if i+j=Tn, if i + j > m,

wo + w,,,

-

has

Example as

3

2wi+j+l

+

-3el

'p(b) in

as

of

one

one

of

3T,,,()

X:=

-wi+j-m

ated

instead

w,,,I,

-

WiWj

is the

that

construction

wi+j

where

the

as

difference

N and define

G -

only

+ Al +

a

pattern

same

the

:=,8b,,+,

now

:=

the

with

+

zi+j

-m

+

2wi+j-m-l )if

-1

1

+ el

=

5(eo

+

el)

+

(V (D W, X Y) yield that the constant complex valued algebra that

W, X

i+j>Ta,

Y)

is standard

4(5eo is

an

-

el)

=

integral

function

51 + 4h.

algebra.

table

(XY)

homomorphism. and homogeneous

x

f 51

can

of V G W. of

degree

5.

SITA with

3

proof

The

(iv)

in

2

T,(A)3

dard since z

We

the

Let

notation.

by

hl,

xi,

pi,

Element

Example

in

as

same

Nonreal

us

qj

the

T-n

i

si,

(3)

.

.

Supp(z-7)

0(po)

A:

either

We claim

holds.

with

O(po75o)

=

0(po)

B:

or

ro

=

5X2 + 2h2 + O(PO) + O(PO)

=

Let

=sm.

first

us

that

assume

A

case

that

induction

the

h2j

from

0(pi) For

h2 and

=

M

follows

it

stan-

are

are

O(po)O(po) that

zp,

1, h,

,

.

75

IL, h, zt, zizi- defined T, (3) and

elements

elements

the

=

E

5

2.

denote

and

Degree

of

E 10, ml. Assume that 3T,, ( a)3 via 0. Since both table algebras isomorphic and homogeneous, Clearly, 0(hi) 0 is an exact isomorphism. T- (3)\j z of 3T,, ( )3 70 are the only elements ro and sm

above -

is the

change Example by X2, h2, ri,

defined 3

(iii)

of

Faithful

a

=

take

step

for

ri

all

10,

i E

m

11

-

(3-35)

ml.

10,

i E

and

0 (pi)

that

assume

ri.

=

Then

ri

30(pi+l)

+

0

Since the

+

O(qj+j) this

is exact,

contradiction let

Next

=

assume

us

77o' that

induction

for

sm-i

=

i E

10,

.

.

,

I

I

m

case

10,

i e

-

.

=

.

.

and

.

3rj+j

+

ri

=

(3.35).

o(p,,,)

=

In this

all

rori

=

proving

rj+j

O(T-o)

=

B holds.

case

take

step

=

0(po)

=

O(pj) For the

LpLj+j)

means

sm

0(po)O(pi)

0(popi)

=

+ si+,.

(3.35)

But

yields

rm. we

claim

that

(3.36)

MI.

,

assume

0(pi)

that

sm-i.

=

Then +

ri

Again

ro

Remark A E xy

fact

the

then

But

=

R>o and 'X+1

z

2

Set

Proof.

(A

+

4

+

Tm6

be G

z

=

-

A +

-

2

yields

a

O(To-)

=

A+1F. 2

1.X 7+

=

=

O(popj)

=

sm-i

that 2'+1

7 +

2

=

O(po)O(pj)

+ 3sm-i-I

0(pi+,) O(pm) 0

A-1 2

7 and

=

z-z

that

Al + =

holds. 0

contradiction.

Suppose

xT

x,

(3.36)

Hence

final

a

so,

smsm-i

=

+ rm-i-1.

sm-i-I.

=

=

algebra.

table

standard

V, such

y. Then xT x-z

O(qj+j)

is exact

(U, V) 2

*9

+

O(po)

=

x, y, -\-'

u:=

1)2

0

that

=

Let

7.

30(pi+,)

A

1 2

there

(x

+

exist

7)

x-T.

Then

A2

_

4

1

1

(xx)7 (A

+ 4

=

1)

2

x7y A-1

A+ 1 2

u

+

2

xy'

and

Z. Arad

76

i.e.,

u

al.

et

1'gl proving

2

A2 + 1

A

2 so

(x, zT)

that

8.

Suppose

then

(aH)+

Remark

11, hj,

1(aH)+1

=

Proof.

A

=

If

yields

that

of

degree

jdx-j (h, a7a)

=

all

B,

y E

Suppose b3 0 b, b4) T b4 =7 - b,

11, hj

=

(i)

bH

b2

(ii)

is

coset

and

Yy-

=

E also

=

x,y

6b,

x-y consisting

=

=

3c, for

Now 3b 2+ 2bc

Since

we

4. In

set

H

:=

particular,

=

2c7b supply

=

=

and

(bH)+G+

of

two

elements

=

2,

B\jbj

this

2-bb

+

2cb +

c2

b2 and

3-bc

=

for

c.

2(b

+

true

101 +

T-h

+ 2h

2

+ b +

that

b41

H

=

ch

=

from bH consisting

(bH)+(bH)+,

=

2G+ + 8E+ for

bh

(6-b, h2)

5

=

then H-

some

E+E+

satisfying

2b + y for

=

3c+2b,

some

y c

NB,

65

i.e.,

c(bh)

is also

5h + bh +

,

distinct

G+G+

that

(bh, bh)

2b+3c,

=

of degree

so

3b3

=

5 such

holds:

5 and bh

b-c

=

b2

and

of degree

that

2b + 3c and therefore =

T

H-coset

5 such

bh

b is nonreal

T(bh) (-bb)h

is another

c

b(ch)

=

following

some

=

=

=

element

of degree 6-b-

c =

(y, y) c-

from 4(x, ah) (ah, x) 0. For x =.a, this (h, dx-) 51 + 4h) 20 jahl. Hence + h). 0

=

Then the

(h, 6-b)

=

follows

it

basis

a

of degree

20 + y

If

degree

of

a

51 + 2h + b +

=

h is

jx, yj

(i) We have (bh, b) Now V Supp(y).

36c- +

bTb

subset.

Proof.

yields

that

where

if G

(bH)+Tb--H)+. b

and

so

that

closed

a

elements

=

1) (X, ZZ)) A,

-

51 + 4h.

=

(x, ah) (ah, ah) (a7a, a (I (aH)+

=

two

xx

h

elements

4, then

1b, cl for some nonreal 3h + c + Z!, c7c C2, b-c

=7 T

c

(A

+

2

16 that

=

=

=

If of

h E B satisfies

aH+ for all basis

Lemma 10.

Bh

(A

0

and

=

(y, ah)

=

obtain

we

1.

-

E B is

5

I for

=

(XY, 27)

=

Rom this

assertion.

24.

x

=

of the

part

(XZ, X7)

=

=

(h, aTx) 0 (a, ah)

(h, a7x-)

first

the

=

ch

=

3C2 and 3bTb

Cc

=

(3-37)

3c + 2b. +

2c7b

=

T(ch)

=

6-b.

c(Th) (3-38)

Now

T)

+ 4h +

101 +

2(b

+

3-bc

b)

+

3(c

+

Z)

+ 13h

implies Te

=

M+

c

+'E.

(3.39)

SITA with

3

(ii) (i) yield

Set

(x Clearly, for

all

C,

=

(bH)+

:=

y)(Y

+

v

and therefore

al

(w, w)

By

640.

=

(a) (b)

E

=

fe, f 1, lei

E

=

jej,

Set

g:=

uv

lei

5

=

=

w

Cj

:=

is

2(b +T +

+

Since

that

2v

c

contains

E

12,

of elements

5 cannot

contain

a

are so

lwl

Then

11

..,

such

whose

that

degree 5 and

degree of

substituent

4. that divisible IzI by 5,'we conclude 80. Thus all contradicting ajiE+l < Jul of a E cannot consist divisible by 5. Clearly, two possibilities: the following that we obtain =

=

Ifl

=

10.

c)(x

+

y)

+

uv

=

2v +

=

8g

2(x

=

+

y)

+

(3.41)

8g.

that

note

x+xh=xH+

=

G+

lG+l

(ii)

4.8

Proposition

(3.42)

assume

either

that

xT

y)

+

=

G+

IG+l

=yH+ =y+yh

=

(h, y`y)

=

and

Cb.

from

is distinct

IFy

or

2

(3.42)

yh= 2y+3x.

and

2x+3y

(h, x-T)

deduce

we

3(x

=

[201) implies

of xh=

(h, x-y) Rom

=

3

=

(3.38)

(h, Zix-). (3.40)

and

obtain:

For

1b, cl

d c

and

(dz, dz) Using

(3.41),

=

we

z

E

xx

=

yy

=

51 + 2h +

xy

=

yx

=

3h + b +

Ix, yj,

(cFd, z-z)

=

x

+

we

(51

=

2g,

cy

c

=7

+ 2h + b +

b,

have

(a, dz)

that

shows

this cx

Next

a

degree

not

24

5,

that

assume

z

of

Ci =A H

have

we

element

an

and

(3.40)

%

j

a

77

Z).

Eli=2 aiCp.

=

exists

G consist

+

may

we

5

assumption

=A bH,

G

means -

The

+ y.

x

=

Degree

of

Element

E+. Then

(b

us

Set

degree

of

degree

of

element

Rom

2

bH and

of E

of elements

degrees single

Ci.

E

JE+l

yields

8

h)

+

(b, vU) 2.

order

5 whose

Now Remark

10(l

=

G+

:=

H-cosets

that

Since

5.

v

Nonreal

Lemma I there

Suppose

8.

by

than

greater

u-u

=

=

and

c

for

of two elements

product

(cf.

vU

=

aiCt

=

is not

b +

=

Purthermore,

N

First

T)

+

uv

i

80 and

aj

u

Faithful

a

=

y +

as

-

+

c.

b

U,

-

51 + 2h +

< 2

for

all

2g, bx

=

y +

a

c

c

+

Supp(dz)

2g, by

=

Z)

2g.

3b3

+

compute

b2 +6-b

b(:i-y)

=

3bh +

(bx)-g

=

y-y + 2g-y

=

=

51 + 2h + 8b +

51 + 2h +

c

+ V+

T+

2jy-,

9c +

45.

and therefore

+

x

=

b4

Let we

Z. Arad

78

i.e., b4

al.

et

8b + b + 8c + 3b3 +

(cf.

Therefore

VT

-bc

=

b(Cb)

=

b4

T2

+

+

6b-

Hence

Let

55,

=

Supp(xb)

whence

50

have

such

(xb

=

B/H

quotient

assume

we

that

T

+ 3c + b +

bTb

37b

=

+

T+

10b +

=

c

this

yields

b3

'z and

+

a-O 20

!

-

-

and Remark

x)

a

(x, xb) 2

>

=

that

so

(3.41)

,

x

3

.1u, 6

=

y

Supp(g)

proved

we

Bb

x-x

=

=

bb.

(xb, xb)

and

I

y-y

=

Lemma I

=

supplies

a

e already implies to the Finally, passing Iv and z -1g) yield 6

d

contradiction.

(withA

7

that

Thus

3. But then

90,

!

Cb,

(3.43)

Since

holds

6b-

6c + 37! +

+ Z.

b is faithful.

as

that (b) (xb-x,xb-x 1XI (d, xb x)

x, A

-

T, b4,

b3 0 b,

(3.37),(3.38),(3-39),(3.42)

from

contradiction

a

now

us

( t-x, 6-b) d E

follows

it

T, cj!J,

-bc

=

5b + 2bh + b 2 +

bc

1, h, b,

Since

2g-y.

7! +

=

(3.39)),

=

=

=

6

uUm

0

that 6-b'= 51+b+T+2h, 5. Suppose b2 2 T. Then 51 + 4h and (A, B) is one of 3, b3 =/= table of degree 5: algebras

Theorem

=

=

(i) (ii)

B

11, b, T, hl,

=

3T

B -2-'x ments

is

a

..

can

b 2=3h+b+T

( )3

T,,,

(3),

1, h,

cosetofH:=j1,hj

VjVj

=

+

wi+j-m

hvi

=

2vi

+

3wi,

hwi

=

3vi

+

2wi.

3Tm( ) 3

enumerated

3vi+j-,n-l

wi+j + 3wi+j+i 3h + wo + w,,

Viwj

B --- x

b,

:=

This

wo,...'

that

means

vm, wm such

1, h,

T-

(3),

M

vo

the

basis

ele-

Ivi, wil

that

and

vi+j-m

(iii)

vo

+ wi+j+l vi+j + 3vi+j+l 51 + 2h + vo + vm

wjWj

=

homogeneous

andbh=3-b+2b;

dim A = 2m + 4.

be enumerated

11b, b3, b4jj

3b3+b+b4, following

the

:=

b,

+

+ wi+j-m-l

)if i +j < M, )if i+j=m, if i + j > M,

+ vi+j+l

3wi+j-m-l

dim A wo,...,

if i +j < M, if i+j=m, if i + j > m,

=

+ vi+j-m-l

2m, + 4.

vm, wm such

The that

,

basis

elements

Ivi, wil

is

can a

coset

be

of

SITA with

3

11, hj

H :=

ViVj

I

ViWj=

bh

b5T5

and If b5

T,

=

Hence

+

3wi+j-m-l

3wi+j+l

Let

=

2vi + 3wi,

=

3vi + 2wi.

(i)

holds

and

assume

that

=

C

that

=

,

if i+j M,

wi+'j-.-l

Lemma 10

b52, 6-b-5

2 3b + 2b5, b=

multiplication

cH+

=

1c, dj

provides

following

the

3h + b 5+75

=

(3.44)

of

=

C+

1C1

=

3(c

by (3.44).

determined

completely

is

(3.45)

b.

and E

two elements

exactly

+ ch

+

h 2= 51 + 4h.

have

we

3vi+j-m-l

+

vi+j-m

hvi

suppose

containing c

79

5

=

may

us

+ vi+j-m-l

+ vi+j+l

b5 :

H

Degree

)if i+j

=

v,vi+j-,

=

=

m.

-

we

obtain

vown

=

vivn-i

=

wiwn-i

51 + 2h + b +

=

T for all

vovn

i E

=

f 0,...,

b_5 MI.

Then

wi+j)

=

vn(vi

+

=

vi(3h

+ wo +

vm)

=

w,(vi

3vi+l

+

=

vi(51

+

wnvi+l

Now it

wi+1 + 3wi + vi.

=

+ + 3vi+j-,n-l vi+j-,,, for all + vi+j-,n-l + 3wi+j-,n-l wi+j-,n B that 3 T,,&E) Hence we conclude 3

vivj

vrnwi+j-,n

i 10'...' T,n(3) (cf. Example 2).

i, j

vnwi+l

wiwj

=

3vi+l

+

=

=

+

3vnvi+l

10wi + 6vi + 3wi_ j

wi+j)

+ 2h + vo +

vrnvi

wn)

=

=

wnvi

+

+ vrnwi+l + vi+j

3wnvi+l

10vi + 6wi + 3vi+l

+ vivm,

+ Wnwi+i + wi+1 + viwn,

10wi + 6vi + 3wi+l + vi+j and 3vnwi+l + vnvi+l i.e., 3vnvi+l + vnwi+l the Multiplying 10vi + 6wi + 3vi+l + wi+j (cf (3.52)). + w,,,,wi+l 3w,vi+l the second and 3 first yields subtracting vnvi+l wnwi+l by equation =

=

=

wi+j-m

wi+j-

Tm-(3)

=

Applying also .(3.54) shows that wiwj vmvi+j-m vivj and therefore + vi+j-m+3vi+j-m-l wivj +vi+j-,n-l +3wi+j-m-l B > m. Therefore, 3-Tm(sE) G 0,---'mwithi+j --,for allij 3 0 (cf. Example 3).

wi+1 + 3wi + vi.

M

=

=

=

=

Integral

Standard

4

Faithful

a

Width

Element

of

Algebras Degree 5

with and

3

1,2

Bfinger

F.

Real

Table

1Department

of Mathematics

University Ramat-Gan 52900,

Computer

and

Science

Bar-Ilan

2Vogt-Groth-Weg Hamburg,

Germany

22609,

Introduction

4.1 This

L(B)

with

Supp(b 2)\111

we

obtain

two

JbI

and

degree

b of

9 135- Using

A,y-,Izl

classification

the

III

==

element

basis

with

deals

chapter

(A, B) real

Israel 44

=

possible

(xy, z)

(x, z-y)

=

Theorem sume

I.

was

Suppose

A > 3 is

that

a

proved that

prime

ISuPp(b 2)1

(i.e.,

Az-gxlxl,

=

Table

I

[b 2]

(b 2, b2)

[15,

21 [15 theorem

3

=

3)

such

x, y,

z

B,

cz

51]

53

75

5252]

65

Blau

by H.I.

(A, B)

is

integer

such

a

in

[331,

Theorem

standard that

integral

Ivi

:

\

-

2.10.

for

=

=

=

all

v

E

B\f 11.

In

Case case

support

-1

=

either _

I)d,

1

1 of Table

of

=

=

=

-

4.2

=

As-

algebra.

table 1

Al + (A A and a7a 1)d. Then Suppose that a, d E B with jai Idl Al + (A d fl, a, dj and a2 fl, al (that is, a U); or Ba ad Al + a + 2)d. (A I)a + d, d'

B,,

that

b 2:

for

types

b G

identity

basic

1

The next

all

5 and width

the

integral GT-algebras B* which contain a faithful

of standard

> 4 for

b2

or

I it not.

useful

to

We have

the

is

whether destinguish possibilities: following

Z, Arad, M. Muzychuk (Eds:): LNM 1773, pp. 83 - 103, 2002 © Springer-Verlag Berlin Heidelberg 2002

b is contained

in the

84

Bfinger

Florian

b2 51 + 3b + c, c G B5\f bl, b2 51 + 3c + b, c C- B5\jbJ, b2 =51+3c+d, c,dEB5\jbj,

(a) (b) (c)

=

=

cod.

of the cases only deal with case (c). For a discussion (a) and (b) we [41) The proof of the next theorem that completely solves this case 2.3 of [41]. Some gaps in this proof that inspired by that of Proposition to two exceptional table of dimension 4 and 5, respectively, algebras will

We will refer

to

was

lead

be closed.

Theorem

then

(a) (b)

2.

of

one

B

=

cd

=

B

=

C2

=

de=

11, b,

Proof. Case

+ 3c + d

bc

d, d2

=

3b +

=

51 +

bc

w

w

=

-

2b2, b2 25 +

C2

and therefore

=

basis

elements

of degree

5,

b, c2

2c + 2d +

=

=

51 + 2c +

2d,

bd

2e + 2d +

b, be d2

2b+2d+e,

ce

=

2c + 2d + e,

51+c+d+2b,

=

51+b+c+d+e. and

homogeneous.

five

3b. Then either

=

real

+ d + 2b.

3b + 2d, 2e+2c+d,

real

is

bd

2d, c

=

=

=

(A, B)

for distinct

holds:,

cases

11, b, c, d, ej, bc cd 51+2c+2d, e2 2b+2c+e,

Set 1.

dj,

c,

2b + 2c +

particular,

In

51 If b2 following =

the

T2 0

w

2b2

=

or

E B

w

or

w

b3

=

+ b4,

b3 0 b4

b. Then

3(c, C2)

+

(d, C2)

(b 2, C2)

=

(bc, bc)

=

65

=

Now

2c + 2d.

151 + 11c + 6d + dc

2

5c + 3c

=

+ dc

b2C

=

151 + 9c + 3d +

b(bc)

=

=

3b 2+ 2bb2

2bb2

yields 2bb2 In 5 or

particular, -

(d, C2) (d, dc)

we =

=

1. 1.

73 :A

35 +

10(d,

=

(d, bb2) =

=

(d, bb2)

I and

d, and bb2 d2) yields (d, d2) c,

(b2, bd)

2b2 + b

+

(d, bb2)

=

=

2N, N =.T4- = 6 b,

51 + 9d + 7c +

6b3

+

(d, dc).

that

Z

2.

=

2c + 2d +

=

=

+

(4.1) (d, dc) : 5 (c, cd) 1 and (d, bb2) (d, de)

Since

either

=

-

=

=

2

-

1.

d.

z

(b, bd) :=

=

5d + 3cd + d 2

=

51 + 7c + 5d +

=

have

45

=

2c + d + =

2b3)

(d 2, C2)

=

that

b 2)

(d,

=

-c-d.

b2d

2b3

dc

(dc, dc)

=

45.

=

d2 -51

=

we

follows

it

(b 2, d2)

=

2 and

Set

By (4. 1) b3. Hence

Rom this

(bd, bd) Since

2c + 3d + de.

3 bb2) (4. 1) shows

have

3, equation

(d, dc)

Case

b3

=

3

2 (d,

=

=

+

b(bd)

2bb4

1, this

means

bd

Then =

b

2

+

2bb2

+

2bb4

=

SITA with

4

yields

z

40

Thus

we

=

5(b2, be)

+

10(b2, b2C)

+

(b2, b2C)

(c2, bb2)

=

(bc)b2

=

b(cb2)

=

d

(cb, cb2)

15(b2, be)

+

x

-

cb3

that

and 45

=

bb3

yields

(bb2)C

=

=

b4C

6b4 b2

=

(b4,

+

+

2b6

2b4

B

=

+ 2d +

2b4

11, b,

dj

c,

2,

=

=

202

=

b2C)

30 +

=

b2 c

=

(b4, b2C))-

+

10(b2, cb2)-

2b + 2 b4 + b2

bd

and

we

(d2, b2)2 =(62,

=

b2

+

+ 2cd +

cb3

=

-

ROM

2b57

bb2

=

=

2c + 2d +

obtained

have

we

b3

the

have

db2) 0 55,

=

101 + 8c + 8d +

5b3

+

2b5

101 + 8c + 6d +

4b3

+

cb3

(b, b2d)

Since-

(bb2, d)

=

b2d

that

we see

+

=

b2b2

=

b(bb2)

=

=

=

2, (b2, b2d) 2b6, b6

2b + b2 +

=

=

(d, b22)

76 0 b, b2

=

=

be +

b(dc)

=

2bc + bd + 2bb3

2b2 + b4

2bc + 2bd +

bb3

=

8b + 8b2 + 4b4 +

2b6. Therefore,

+

4(b6, b4)

2b2C

+

3b4C

=

=

7b + 4b2 + 4b4 + 2b4C

7b + 8b2 + 6b4 + 4b6

2b6. The equality =

(2b

=

(b2b,d 2)

+

b2

this b4 0 b6 In particular, 0- We have b2 + 2b6) -

=

2 and

=

(b2, bd)

=

(bd)c

2b + 40

10((b2,

from

2b2+ 2b4C 2C2

10 +

=

deduce

20 +

implies

2(b5,b3)

2d + 2b5 + b3we

+ 8b +

Le.,'

=

compute

we

(a).

(d2, b2)2 =(b2d, b2d),

=

ROMthis

9b2

b(b2C)

Next

101 + 8c + 8d + 5b3 +

b3 and b

=

(b, db2)

b5 0 b3- It follows

that

=

b5.

5

3

b5-

+

Hence in

Since

45 + so

b.

=

+

+ 6d + 3 b3 +

b2b

+

2b3

10(b2, cb2)

+

I? (b4, b2 C)

b3

+ d +

c

given

0 b2

d

1. 1. 2.

Moreover,

=

5 and Width

Degree

2b4, b2C)

3.

2b2+ 2b4b

constants

structure

+

(b4, b2C)

6c 3bb2 + 2b2= 2

and

2b2

+

10(b4, b2C)

b2. Then b5

=

c

=

(b

=

of

2d +

=

+

(b2, b2 C)

b2= 51 2

b4

implies

=

conclude

1.1.1.

(bd, b2C)

Element

bb4

and

=

obtain

Case

d,

c,

(dc, bb2)

we

Case

T5 :

=

=

Hence 40

b5

2b5,

=

Real

Faithful

a

+ =

2b6, b 20 +

means

+

2b2 + 2b4)

2(b57 b3) (d, b2 b4)

=

(b2d, bd)

20

(b4, b2 d)

=

(b4c, bb3)

=

(be, b4b3)

=

3(bb4, b3)

+

2(b2, b4b3)

=

30 +

2(b2, b4b3)

7

bb37

Bilnger

Florian

86

i.e., (b2, b4b3) b5 =A c, b3 and

(bb4, b2 b4)

we

b2b4

deduce

4c + 6d +

we

get bb6

shows

=

=

b(b2d)

=

that

us

2b2

bb2

2c +

B

=

Q

+

2b3

2b3

=

=

+

+

2bb6

=

3bb4

+

2b2b4

+

bb4

8b3

=

=

2bb4 +

4b3

+

4b,5

101 + 8c + 4d +

5b3

+

6b5

b3

+

2b5 and from

=

5c +

=

2dc +

C2

+ dc +

d2

263

+

265 =

=

d 2C

3

2b42

2b5c

+

+

=

b4b2

101 + 4c + 6d +

2b42

=

that

=

=

(d, bb2).

=

yield

c2

=

IwI.

(bc, bc)

c2

51 +

=

51 +

c

(w, bd)


3. If

=

3 and

=

Real

(d, C2)

2 and

=

(51

=

Faithful

a

Case

2

2 of Table

For

Case

(a) (b) (c)

b 2= 51 + 2b + b 2= 51 + 2c+

only

of

Chapter

Theorem

(c, cd) then

> one

B

=

B

=

cd

=

B

=

C2

=

dt

=

B

=

C2

df

(e)

B

=

=

=

d2 =

(f)

(c, cd)

(a)

cases

(b)

but

=A

c

=

d

and

(b).

Theorem

5

able

We were not

Theorem following solve will completely

the

3

to

gives

give

a

complete for

answers

(c) by using

case

many

results

distinct real basis 51 + 2c + 2d for pairwise Suppose that b2 of degree 5. Then (c, cd) + (d, cd) > 2. If we assume that (d, cd), (d, cd) if (c, cd) :A (d, cd) and that (c, c2) -. (d, d 2) if (c, cd) holds true: cases of the following 3.

=

c, d

=

11, b,

dj,

bc

11, b, c, dj, 2c + 2d +

b, d2

d2= 51

(d)

Z

subcases:

main

3.

b,

elements

(c)

following

the

2c, c E B5\fbl7 2d, c,d E B5\jbj,

case

Finally,

subcases.

(b)

have

we

the

treat

classification

(a)

I

b2=51+2c+2Z,c(EB5,ZOC-

We will

of

a

0

contradiction.

4.3

I =

c,

3d,

2b +

=

bd

2b +

=

C2

3c,

=

b2'

2c +

3b,

51 + 2d +

2b,

cd

=

+ 4d.

11, b,

c,

bc

d,tj,

2b + 2c +

=

2b + 2t +

bc

51 + 2d +

2t, cd 2d + 2h + t, t2 f 1, b, c, d, f 1, bc

2c + 2d +

=

51+2d+2b,

cd

2b + 2c +

d, f2

11, b,

c,

dj,

51 + 3d + =

1

[The tablealgebras

=

bc

d, bd

2b + 2d + c,

=

C2

=

51 + 2c + 2b.

=

51 +

=

c

2b + 2t + c,- bt d2

d, bd b, ct

2c + 2d +

2c+ 2h + t,

51 + 2c +

+ d + t + h.

2b + 2c +

2c+d+2f,

f, cf

bd

2b + 2f 2f +2d+b,

=

=

c + 2d + 2f, d, bf d2 51+b+c+d+f,

+

=

51 + 2c + 2b.

2b+3d,

bd

=

2b+3c,

0

=

(d, d2).

c

2

=

51+3c+d,

cd

c.

(d, cd), defined

t, 2t,

(c, c2) in

=

(c)

and

(d)

are

exactly

isomorphic.]

=

3b+c+d,

88

Florian

Proof.

Set

l3iinger be

x :=

20 +

(x, x)

20 +

(y, y)

that

0

c

(bd, bd)

=

(b 2, d2)

=

(x, x), (y, y) (4.2)

Rom

(d, cd).

(A) (B) (C) (D)

(c, (c, (c, (c,

cd) cd) cd) cd)

=

=

=

=

Case

(A)

d2

51 +

=

we

Hence

obtain

(c, cd)

of the

following

one

I

0

+ 3d.

(4.2)

25 +

10((C'

C2)

+

(d, C2)),

(4.3)

25 +

10((c,

d

2)

+

(d,d 2)).

(4.4)

=A (y, y).

35

G

f 15, 25,

(d, cd)

+

45

Hence

1.

(4.5)

! 2. We can

(c, cd)

that

assume

>

holds:

cases

Case

(B) Applying Set

(4.3)

z

cd

:=

(z, z)

45 +

(4.3)

=

and

(4.5)

3c.

Then

-

(cd, cd)

we =

we

(c2,

=

=A d,

c

(4.4) 9(b2, b3) and

contradiction

the

Since

51 + 4d.

=

Using

yields

c.

(d, cd)),

(d, cd).

=

(4.2)

51 + 3d +

+

4; 3; 2;

We have c

cd)

=

(x, x) =,4

yields

10((c,

=

,

(b 2, C2)

d

then

(b2 cd)

=

=

0

b

2b,

-

(be, be)

=

=

bd

:=

(be, bd)

(x) y)

20 +

Note

2b and y

-

x

=

I

yields

3b2 and

cd y

=

==

4c +

3b3.

d,

Now

30.

[x]

get

d 2)

Theorem

obtain

25 +

=

[5 3), (C' C2)

=

d 2) +

15(d,

1 and

=

5(c,

d

0

2

Rom

(z, z) and

(4.4),

In each

Case

we

case

(C)

deduce

(4-2)

([y],

d 2),

(d,

immediately

Rom (4.3)

(4.5),

and

f 10, 201

E

(c, d2)) yields

we

E a

J([5 3] 2,2),([5

3], 1,3),([52

51],2,0)1.

contradiction.

(c, c2))

([x],

obtain

E

2,511,0)1.

([5 3],2),([5

5 1 + 2d + 2c. Now (4.2) reads 3 (e, y) 3e, jej 57 C2 (CI) x that If so G that we assume (d,cd) 10(d,cd), 10,31. (e,y) : - 0 and (equiv3e and (d, d2) x 3, then (4.4) and (4.5) yield y alently) (d, cd) 0 contradicting (4.2). Hence (e, y) (d, cd). Since

Case

=

=

=

=

=

=

4b + 6e + 3ce

i.e., then

3ce

(4.4)

=:

=

9b + shows

2cb, + 3ce

2y, that

we

=

see

(d, d2)

c(cb) that =

=

=

=

y

C2 b =

4, i.e.,

=

5b + 2db + 2cb

3f for d2

some =

real

51 + 4d.

=

f

13b +'6e E

Now

+

B5\f ej. we

2y, But

compute

SITA with

4

(cd, ed) for-

(c 2, d2)

=

real

some

g E

Faithful

a

be

=

3c +

gc

us

=

we

g

derive

=

b,

Case Z :=

e

so

3d +

B

2c2

+ 2cd

b 2C

5c +

=

101 + 4c + 4d + 3be

=

b(bc)

=

cd

Hence

65.

=

=

3gc

2

2b

=

2

89

3

2c+3g

=

+ 3be

4g

=

=

101 + 13d + 4c +

+

2+ 2cg

3c

11, b,

=

T

c2d

=

=

5d + 2cd + 2d

2

6g

=

(cb)e

by

Theorem elements

e

d

=

=A f =7,

(d, cd)d-2c

jej

and

2be + 3e

=

and the basis

=

cd-

v:=

c(be) e

dj

c,

f,

2e +

=

c(cd)

2c

2g. Rom

=

(C2) x C2-5172d,

3gc

=

d2 and therefore

=

that

6g

5 and Width

Degree

2g. Now

151 + 6c + 12d + 2

of

(51+2c+2d,51+4d) B5\ICI. The equation

101 + 4c + 4d +

gives

Element

=

101 + 13c + 4d +

yields

Real

2 =

implies

5. This

multiply 5

=

as

f

4g =

stated

+ c

in

3e2,

and

(a).

0. Set if 1, (c, c2) d (d, 2)d- (c, d 2)C. =

=

d 2- 51

w:=

6c +

-

We compute 9c + 6d + 2v + 2z

-

b 2C

=

b(bc)

(4.6)

101+4d+4c+2be+bf 2be+bf =5c+2d+2v+2z. and

(be)2

=

(2b

+ 2e +

f)2

b2C2

(51

=

=

4b2

+ 8be +

2d)(51

+ 2c +

4bf

+

4e2

+

z)

+ 2d +

4ef

+

f2 (4.7)

=

2 251 + 20d + 10c + 5z + 4d + 4cd + 2cz + 2dz.

Counting

c's

in

(4-7)

10 +

yields

(c,

(e, f2) is even. (fC, fC) : (Z' f2)

2+ 4ef

Cf)2.

5 (f,

=

+

f2)

(f, Cf)

Clearly,

Hence

possible

4e

=

d 2)+

4(c,

2

5

(z,

z

+

(4.8)

V).

(C' f2) 0 4 since 65 > 25 + 10 (c, f2) + (C' f2) E 10,21. We consider the three

=

Hence

subcases.,

(C2. 1) (d, cd) (C2.2) (d, cd) (C2.3) (d, cd)

=

=

=

2; 1; 0.

Case

(C2.1)

(4.2) (4.4)

(2e + f, y) implies 0. Set yield (d, d 2)

Here

we

have

cd

=

=

45

=

=

2c + 2d +

w :=

(cd, cd)

d 2=

51

h,

h E

[y]

20 and therefore -

(c2, d2)

2c. =

=

h. Equation B5\fc, dj, 25 and [5 2 51]. But (y, y) =

Rom 25 +

(z, w),

we

Bfinger

Florian

90

obtain

z

w

=

2t,

=

B5\jc,

t G

dj,

Using

t.

the

law,

associativity

we

derive

2c2+2dc+ch=

101+8d+4t+4c+2h+ch=

2dt

i.e.,

2h + 3d +

=

ch,

=

101 + 5d + 4c + 4t +

=

2d

=

101 + 5c + 4d + 4t +

2ct

2h + 3c + dh.

=

2

+ 2dc + dh

let

d 2C2

a

(cd)2

(51

=

Counting

2t) (51

+ 2d +

(2c

the

26 +

2

cd

=

5c +

2c2

(4.9)

+ 2c +

2t) 4t2

h)2

+ 2d +

of

occurence

4((t,

h)

+

(C' t2))

4((t,

h)

+

(d, t2))

401 + 24c + 24d + 16t + 8h + 4ch + 4dh + h 2.

=

and d in the

c

(c, a)

=

expressions

(c,)3)

a

and

8(t, h)

28 +

3

yields:

(c, h2)

+

and 26 +

Let h2

first

us

multiply we

Let h 2)

stated

as

have

second

(c,

that

assume

b

either

2

=

=

(d,h 2),

h2

i. e.,

ct

=

[ch]

us

=

2c + 2h +

t,

[5 2 dh

52 =

51]

=

2h +

c

=

=

51 + c i.e., t2 like Bh multiply =

(bh, bh) means

=

[bt]

(b2, c

+ d + t + h. in

h 2)

(c) =

with 65

first

case

(d,h 2), i.e.,

=

elements

(bh, bh)

of b. Since

(4.10)

Then

h.

5 1 + 2c + 2 d

(c2,h 2)

=

101 + 6d + 9t + 6h +,2c

54

=

2

=

and the

(4.10)

h 2).

(d,

of Bh

(b 2 h2)

=

=

,

done,

we are

65

in the

contradiction.

a

t

dj

c,

In the

+

(c, h2)

implies

11, h,

=

5 2].

[5 3

8(t, h)

28 +

(4.10)

Bh

h instead =

that

(ch,ch) shows

(d,,3)

Then

with

yields

assume

now

h.

=

[bhl

or

Lemma 2

case, us

h

a)

Therefore,

(b)

in =

='(d, t

b 2.

51 +2c+2d=

=

+ 2d

2ct,

=,A (h, ct).

0

251 + 18c + 18d + 20t + 4h + 4dt + 4ct +

,8

(cd)d

=

compute

us

=

2dt

2dt,

Hence

(h, dt) Next

2+

5d+2d

=

and

101 + 8c + 4t + 4d + 2h + dh

i.e.,

(dc)c=dC2

45

=

[dh].

Using

+ 2t and

2c2

dt

(c2, d2)

=

(4.9), =

Bh

means

2 f [5 52 51), [102,51]1.

=

we

=

7t + 4d + 4h + Hence

=

But

[53,

(d, t2)

and

=

ch

2h + d +

=

Finally,

c2t

=

we

5t + 2dt

2t, compute +

2t2

2t2,

f 1, h,

now

=

(cd, ed)

obtain

c(ct)

of b. Let

[bh]

=

2d + 2h + t.

+ 2ch + ct

h instead

(C' t2)

implies b 2. Now

=

5

21

2

d, tj

c,

us

assume

and

yields

and

(bt, bt) b E

the

that

elements

b =

of

Bh. Then

(b 2, t2)

Bh contrary

=

to

45 our

SITA with

4

Hence

assumption.

either we

y

=

2f

+ g,

(C2.2.3)

It (C2.2.1) B5\jc, d, bl.

=

=

=

y

or

e

obtain

we

1, y 2f 1, u 2g, 0, (U7 U)

2(h, w)

15 +

(4.6)

=

bg

b(bd) bg

d 2)

(bf,

=

20 +

45

(be, ec) (be, c2)

yield

for

+ 2d + 2h

c

some

=

10 +

=

20 +

=

(2b + 2e (be, 51

(4.17)

(c25 d2)

(b 2, f2)

25 +

+

f, ec)

+ 2d +

=

that

=

z)

(g, f c)

we

> 4

see

real

h

E

which

(fC, fC)

=

supplies

y2' C2)

=

=

=

(z, w),

35 +

+

w

=

0.

2k

If

(4.15)

(d, f2),

z

=

(4.16)

=

(4.17)

=

so

z,

(C7 f2)

contradiction

25 +

10(d, f2)

+

2k for

some

real

0, (h, v) that (4.15) gives a 0, then (4.8) and

we assume

the

51 + 2c + 2k.

(4.14)

-

b. If

have

f2

2(e, cf),

and therefore

k

(4.12)

(4.13)

=

+

10(C' f2)

(4.11)

=

=

(h, v) 0

that

=

(g, fd),

2(e, ec) + (f, ec) (z, be) (z, be 2c).

(z, z) 0 10, =

we

+

20 + =

Our aim is to prove that k and and (4. 11) yield h Hence

(C7 f2)

10 +

=

(cd, cd)

=

2(d, f2)

2(c5 f2) + (g, fC) 2(e, f d) + (d, f2) 20 + 2(h, v),

(be, cf)

=

=

=

101+6c+9d+2v+2w, 2w,

20 +

=

d}.

85
2.

On the

(bf,bf) shows 7

0 and




(cf.

2(d, ef)

(d 27f2)

=

2d,

+ 2c +

and be= 2c+d+z.

(d, ef )

from

even,

(51

=

+

=

(df,df)

55 < Hence

2(d, z)

2(c, ef)

=

Let us now (cf. (4.7)). jwj ! (z, w) (v, v) G 10 (4.8) supplies (z, v) reads Now (4.6) v z. z

4z + 5c + 2d and since

bf =c+2z

have

we

=

=

45 +

=

=

p

SITA with

4

k 2) ::-

(d,

Hence

(C2.3)

Case

Faithful

a

From this

2.

Real

obtain

we

(b2, k2)

=

(51

+ 2c +

(kb,kb)

=

(2f

+ e+pl

In this

(4.2)

case

Element

the

2d, k 2)

Degree

of

5 and Width

3

97

contradiction

> 45

+P2,2f

+

+P2)

+pl

e

35.

=

means

Y)

(4.29)

0

=

We compute

(bC)2

i.e.,

(51

+ 2c +

251 + 20d + 18c + 5z + 4v + 4d 2

10 +

4e

2

h,

z

=

2g,

g E

may

the

2be + =

c

+

4g

+

2+

2cz +

2dz,

4bx + X2 . 51 + 12d+ 10c+ 5z + 4v + 4d 2+ both sides yields

on

4ef

+

f2)

12 +

2(z,z +v) 5 + (z, d 2))

=

2

and

v) 5 ((z,

+

(4.29)

(4.30)

d 2).

4(d,

B5\fc,

B5\fc, dj, dj, 71

first

equation

g, h C-

at the

w.l.o.g.

assume

=

(ed, cd)

51 + 4d.

=

bf

be

=

(c

2 ,

=

h

(4.30)

of

we

(c,

obtain

4e

also

bf

d2)

=

+2(d,

25

2)

d

2

bf

(g

+

b(bc)

=

2b

=

101+4d+8g+2h+9c,

+ 2be +

=

we

have

+ 2h +

=

(e, bg)

the

second =

(e, bd)

+4ef

+ f2)

3g. Hence

8g + 2h. Therefore, c cd, bf 3g

2c +

5 >

65

2

h,

+

d2),

Furthermore,

5c +

=

or

=

v

54

g g.

=

=

b

2

C=

either

be

5c +

=

2g. In the first

contradiction

Using

X2

two subcases

101 + 4c + 4d + 2be +

i.e., bf

=

d's

f2)

+

g +

d2

f2

+

201 + 8c + 8d + 4bx +

=

z)

+ 2d +

and

(c, 4e2

=

65 whence

+ 4bx +

2d)(5

4ef

c's

z

(C2.3. 1) Looking we

=

+

4ef

+

We consider

0 and

X2

=

Counting

(d,

4b2

+

8be+ 4bf + 4e

2cz + 2dz.

(C2.3.1) (C2.3.2)

3 )2

(2b

=

=

b2C2

+

(cd, be) +

(f, bg)

leads =

2(e, de)

to

(g7 be) a

(g, bf

6.

contradiction:

(cb, de) +

+

(f, de)

=

=

(2b

+ 2e, +

10(e, de)

f, de)

+

5(f, de).

2c2

2c +

case

+ 2cd

2g

we

+

obtain

h,

Bilnger

Florian

98

(C2.3.2)

of (4.30) 2 (mod 4), so that supplies equation (C' f 2) 0 (mod 4). By of (4.30) implies equation (d, f2) this means (d, f 2) 0. Therefore observation we obtain

The first

2. f 2) previous

C,

the

=

(C' 62

+

ef)

+

(g, v)

=

(v, v)

20 +

Moreover,

=

The second

=

(d, e2

and

(cd, cd)

=

ef)

+

(c2, d2)

=

(g, v)

3 +

=

10(d, d2)

25 +

=

(g, d2)

+

(d, d2). (4.31)

+

10(g, d2) gives

+

us

(v, v) (d, d2)

Therefore,

+

(g, d2)

(c, e2

4 +

-

(d, by (4.32) that

(h, d2) =7

2

e

e

(v,,v)

0 and

set

(e, de)

a

ef)

+

=

h be the

Let

15.

this

d

obtain

we

2+ ef)

(d,d 2)

+

2)

(d,d 2)

+

0, (g, v) (unique)

1

=

5,

>

(g, d2) + (d, d 2) and of Ig, dj satisfying

=

element

(f de),

0

and

e

(4.32) (4.31)

with

(d,

=

(g,

+

10(g, d2).

+

(g,d 2)

+

ef)

+

5, (c, e2

'f

+

-

(f, de)

+

(g,v)

3+

=

10(d, d2)

Combining

> 1.

(e, de)

5 >

so

5 +

=

then

,

de

3f

+

ae

=

This

.

means

10a +

5,3

Hence

(a,,O)

(4-31))

(bc, ed)

=

h

a

Set

0.

u :=

(be, be)

+ 45

Then

JwJ

and

e2 the

w

=

(d, u) JwJ

that

Then

=

But

d2

-

=

=

=

50,

=='10

a

and

2k for

(h

51

=

v)

20 +

=

(be, v)

(c, be)

as

-

reads

B5\fd, (d, ef)

of

20 +

:

The

2cl

-

(c, e2)

2 and

=

=

10(d,

2-

e

e

51

0

=

=

(Cf.

Let

let

2(h, d) i.e., c, gj, 3, (C' f2) =

60

us

+

=

=

2 and

w)-

(h ', d)

(a, u,w) < (2, 3). (a, 0) that

assume

3f

,

d2 =

=

0

=

(ef, ef)

=

(e 2, f2)

=

25 +

2(k, f2

-

51

-

2c) : -

d,

u

=

=

45.

3k

3k, (c, k) yield

51 + d +

contradiction 55


3wi+j-m-l

+

+

)if i+j 7if i +j if i + j

+ wi+j+l

+ vi+j-m-l

3vi+j-m-l

,

M,

> m,

,

if +j )if i+j=m, if i + j > i

+ wi+j-m-l

< M, =

< M,

,

3wi

51 + 4h.

Suppose and

+

,

< M,

=Tn'

wml,

wi+j + 3wi+j+l 51 + 2h + vo + wm vi+j-m

Theorem

7if i+j if i+j if i + j

+ vi+j-m-l

M,

> M,

7

+ vi+j+l

viwj

< Tn, =

2wi

11, h,

:=

3wi+j-m-l

+

hwi

h2

=

3T

2h

+ vi+j+l wi+j + 3wi+j+l 3h + wo + wm

wi+j-,,,

3Tm( )3

=

+ b +

T+

+ wi+j+l vi+j + 3vi+j+l 5. 1 + 2h + vo + vm

viwj

(vi)

Z)

+ h.

bb

b2= 3h

(v)

2(c

g)

+

hl,

11, b,

Z4

=

+

--

that 4

for

(A, B)

is

all

B\111.

x

G

a

standard

If there

integral is

a

table nonreal

algebra c

with

E B and

a

SiTA

4

real

h G

one

of

in

We are

algebras

example

(1).

able

now

Theorem

b2

solve

to

T2m(5)j

Set

Proof.

b,

then

B,

exactly

is

introduction

satisfy lbl isomorphic Example (1)).

B

exactly

5 and Width

isomorphic

3T,,&2)3

of this

Icl

=

to

T-M(3)

section.

b, Z 54 c and 5, T of the table algebras

=

to

103

3

=

one

We have

2b.

-

Degree

3T,(L ) 3 T,,,,(3),

of the

G

c

101y Y42 Y5 (cf.

bc

u

(c)

case

that

Then B is

E NU

Ta

(c +7!),

51 + 2h +

=

of

Element

Y4, Y52 Y62 Z4,

Suppose

5.

51 + 2c + 2-c.

=

c7c

table

Real

Faithful

a

B5 such that

the

given

with

(b2, C2) (bc, Fc) (u, U) 10(C' C2) + 10(-E, C2)

20 +

=

=

(4.38)

and

(u, u)

Since

2, then

c7c

!

[bxl that

JxJ

us

=

x

defined'in Our aim is this

For

[x]

reason

degree this

is

we

by

our

basic

b2

=

even

so

d,

Chapter

3

(b2' XX)

=

yields

definition h

=

)

have

to

=

(4.39)

1, then Set

e

on

all

in

prove

y of the

(y, bx)

then

assumption

Let

Set

y

:=

of the

Y5 (and

us

finally

c2

-

C

that

a

=

we

first

us

(u, u)

implies

:=

c7c

look

at

x

51

-

the

c

-

[x)

case

25,

=

Z. Then

-

=

[52],

for

that

cases

each

of bx is

support

x

Bc\fbl

E

divisible

divisible by 5 and clearly of basis elements minimal degrees in any 100. But this is not possible is

=

(bx, bx)

that

:! 2. Hence table algebras

(b2, x y)

=

we

obtain

yields

that

By (4.38)

we

=

25 +

B

=

B is

as

5.

If

jyj

>

4

deduce

we

t-x)

(x, x)

=

10.

have

< 65

Looking at either Y4 (and

B,.

g). that

since

case,

20(c,

the

by

a:=

(Z,

c

2) 7

0.

ccc-.

45 so

particular b

asume -

and

h E

element

case,

or

and

65

=

B,'

b E=-

=

b

=

that B, is one B5. Then Theorem 4 yields T-M(3) 3T,,, Y4, Y5) Y6) Z4, 3Tm 3 T,,(3), 3 that we see B" is five homogenous. (1). In particular, b E Be. Lemma (2) of Chapter 3 in order to prove that real

(bx, bx) > (y, bX)21yl > 25.4 x e B,\11, bj has degree 5 so nonreal whence (c, xT) as c is the

(c, c7c)

If

=

is

(bx, bx)

that

11, 21.

G

Chapter 3 yields B, five homogeneous. For

2 of

B,

(4-39)

b is real

as

(c, c7c)

that

distinct

some

the

not

dc-

=

(c, c7c)

that

particular,

In

20(c, c7c).

25 +

=

(4.39)

E,B5\fbl. [517 511, [101] 1. Let

for

element

of each

C7 C)

,

Now Lemma 2 of

1[52],

E

101.

necessarily

algebras example to apply

table

of the

x-T

assume

2h for

=

have

e

2

P. Now Theorem

G NU

2d +

(b

=

from

obtain

we

m

m is

now

u

(bc, bc)

=

[53,52].

=

10 and

=

i.e.,

15,

=

we

conclude Let

Jul

some

therefore

whence

(u, u)

51 + 2c + 27!

=

Tm(5) for x E Bc\fbl

20 +

2. But

=

(C7,c d-)c

now

(4-38)

c

2, c 2)

yields

the

+

a2 )5

contradiction

+

(y, y) 10

=

(2d

+ e,

2-d

+

j). 0

of Primitive

The Enumeration

5

Association

Commutative

with

Schemes

Non-symmetric

a

Valency,

Most

at

of

Relation

4

Hirasaka','

Mitsugu

IDepartment

of Mathematics

University Ramat-Gan 52900,

and

Computer

Science

Bar-Ilan

2Depattment National

Taiwan

Taipei,

University

Taiwan

Introduction

5.1

(X, G) be (X, g) is

Let

Then

all

find

to

Israel

of Mathematics

hypotheses

an

about

and to determine

[62],

scheme

association

the

in

numbers

intersection

the

whole

structure

or

of

of

sense

g E G. It

regular digraph for each which might regular digraphs a

is

be

an

an

induced

[641 an

element

where

X is

of G under

subgraph

finite.

problem

interesting

of the

certain

graph,

(X, G).

all primitive transitive permutation Wong classified groups valency 3, and in [541, [551 W.L. Quirin and C.C. Sims of primitive transitive permutation groups with a 2-orbit gave a classification of the above results, 4. As a generalization it is known that of valency the of a fix-point stabilizer of each primitive transitive cardinality permutation by the minimal non-trivial valency (see [37]). group is bounded In 1981 L.Babai [261 proposed the following In

with

a

Conjecture the

maximal

non-trivial

W.J.

2-orbit

1.

of

(L. Babai) valency

elements

Let

of of G.

(X, G)

G is bounded

be

primitive by function a

a

association

of

the

scheme.

minimal

Then

valency

of

and we need to make the to be quite seems However, this conjecture open, G is a P-polynomial that for instance to assume or comproblem easier, to impose numbers. restrictions If G is or on intersection some mutative, relation of valency and contains at most 4, commutative a non-symmetric then Conjecture 1 is true (see [46], [47)). In this commutative association schemes we focus on primitive chapter, whose valency relation is small enough. We observed with a non-symmetric all known infinite of such schemes with growing series that I GI are translation the following schemes in a sense of [36, pp. 661 (see Example 2). This validates

Z, Arad, M. Muzychuk (Eds:): LNM 1773, pp. 105 - 119, 2002 © Springer-Verlag Berlin Heidelberg 2002

Mitsugu

106

Conjecture

(M. Muzychuk)

2.

If

scheme.

ciation

(X, G)

is

In this

chapter

4,

Hirasaka

translation

a

the

is

abelian

scheme

primitive

IGI

and

commutative

large

is

asso-

enough,

then

of 2-orbits

2 is true

of all

enumeration

of

such

valency of g is at most schemes, each of on an elementary group

if the

association

permutation

a

group.

Another schemes

of this

motivation

with all

erties:

If

cyclic group of prime determined is uniquely schemes

of this

still

scheme

association

an

order, by

of association

of

relations

non-trivial

sociation

from research comes chapter which satisfy the points, all relations are connected;

number

prime

a

non-symmetric.

are

a

scheme.

Conjecture

that

the

give

we

be

non-symmetric

association

prove

we

moreover

which

g

(X, G)

Let

G is

E

then the

whole

the

minimal

type

translation

a

prop-

of odd

valency

scheme

over

of association

structure

valency of cyclotomic

called

are

is

following

elements.

non-trivial

[36,

(see

a

scheme As-

2.10.2]).

Cor.

all association schemes with a prime if the cardinality of the only classifications immediate of the are point set is small enough. The following consequences of this all association enumeration schemes with a prime number chapter: of points relation of valency a non-symmetric 3 are cyclotomic; containing is no association scheme with a prime number of points there a containing relation of valency 4. non-symme ric In contrast to the situation of the previous paragraphs, any primitive scheme with a relation association of valency 3 is a P-polynomial, symmetric

However,

it

number

of

is

determine

to

open

and there

points,

are

is proved

in [63]. of distanceCombining this with the classification four isomorphism graphs of valency 3, we see that- there are exactly to the following classes corresponding graphs: a complete graph with 4 points; with 10 points; a Coxeter a Peterson a Biggsgraph graph with 28 points; Smith graph with 126 points (see [36, pp. 1791). In this of [64] which are slightly notations we use from different chapter

which

regular

'

used

those

in the

Let X be the

E

we

Definition

Ix

each

For

all

We denote are

xg

called :=

pl99,

:

ly

E

I (x, x) I

=

We set

r* c

E

x

:=

does

not

contain

X1.

J(x,y)

I (y,x)

E

rJ and,

for

each

rJ,

(X, G)

The pair

of X x X which

partition

X1 (x, y)

given.

:=

book.

this a

is

called

association

an

scheme

if

it

conditions:

G;

For

(X, Y)

1x

( [64]).

I

E

xr

following

the

satisfies

(i) (ii) (iii)

We define

define

to

and G be

set

C X x X be

r

X,

Introduction

finite

set.

empty Let

x

a

E

g E G

d, f

the

the

e,

f

we

E

have

g*

G and

c-

G;

x,

y

E

X,

lxdnye*l

is

whenever

constant

-

number

I xdnye* I

intersection

which

is called

with

numbers the

valency

(x, y) of G.

of g.

C

f by

For

pf,

each

and

d

g

E

Jpf,,dj G,

we

d,

e,

f

G

abbreviate

GJ

(X, G) (X, G) (X, G)

We say that We say that We say

that

connected

[64],

1 011.bwing

I

G'

where

f

for

P,d

all

d,

f

e,

E

107

G.

(X, g)

is

-

define

we

=

Schemes

if g = g* for each g c G. if, for each g G G' the graph

symmetric is primitive G f 1x 1.

is

:=

f if Pde

commutative

is

Association

Commutative

of Primitive

The Enumeration

5

the

product

If

de:=

Pfe

GI

C

d,

of two elements

e

G to be

E

7' 01-

d

(5.1)

(x, y) E X x X, we shall denote the element of G containing (x, y) by (x., y) schemes. We say that (X, G) is Let (X, G) and (Y, H) be two association isom4phic to'(Y, H) if there are two bijections 0 : X ---+ Y and p : G --+ H such that (x, V) E g if and only if (O(x), 0(y)) E p(g) for all x, y E X and for each

For

r

each

-

g E G. For each

'

G,

g E

define

we

adjacency

the

f Note

for

that,

d,

all

E

e

matrix

(X, y)

if

1, 0,

as

follows:

EE g

otherwise.

G we have

PfdeAf

AdAe fEG

(see [27, pp. 53]). It follows that the vector of kat(X, C). Following GI is a subalgebra algebra of (X, G), denoted by CG. In the integral is a typical

table

Example

1.

the

X be

Let the

it

by the formal

Given

(X, H)

g E

where

-
3. It follows

moreover

are

+ cy-x,z

there

=

E

is

is

a

I

Pc*" axer,,,,JKe-

12.

12 in G. This

Hence

d E H

completes

-

the

G is

proof.

a

disjoint

union

0

Mitsugu

118

It

easily

is

checked

of Theorem

Proof

If p

3 then

=

CUD, dC-D andaE

allcE

For

Lemma 9.

Proof.

Hirasaka

(X, H)

since

< 1.

Z3). P

3.

0 and G

D=

and

We may

nf

=

12.

all,

we

shall

8(i), as desired. (X, H). By G such that d.,yz

A U B U C by Lemma with (X, G) --

=

(X, G) E L(X, a) d.,.,,, E H G there

=

havePa dc

we

2-orb(l'p,3;

--

Assume p > 3. Let for each Lemma 8 (iii), n,

A,

exist

-

assume

u

f

e,

(x,

:=

E

z)

y,

E

without

oe

eU f

=

loss

of

generality. of

First a

show that

uniquely

is

oe

determined.

Now we consider

triple

(w,

v)

a,

U oc)

E

(od

Ax

x

ul)

-

cEC

such that

u,

r(o, u)

i.e.,

E odnwa

v

f since,

=

If there

-

2

pc-.

=

p'e-a.

Lemma 9. contradicting The following is a set of

((0,

y,

z),

a,,,

(-x,

((x,

y,

0),

a-,,

(x

e

such

y

=

PCW.

v)

a,

Aacw

=

then

r

(o, u) = k

e,

7

triples:

x,

-

z, y

-

=

PC-ar.-Ir"

=

ea

a triple (w, 2, and hence

such

exists

otherwise

z

z,

-

((x,y7x)7az-x,(z,y,x));

((x, 0, z),

a,,

((x,

ay,

x));

-

-z));

z),

x,

(x

-

(y,

y, -y,

z

x,

z));

Z'

-Z),

-

y));

((x,y,y),a,-y,(x,z,y)).

Thus,

f(-X)

Y

X,

X), (X

z

(Y'

(x,

Replacing

y,

z)

E oe

f(-Y'

(5.7)

(-Z'

X

and

(5.8)

Secondly,

Y,

z

-

Z' Y

prove we

shall

e =.

Since W,

:=

p >

3, there

(wi, 0, 0).

z,

Z), (z the

-

z), (z,

x,

x)

or

exists

Y

E

y)

-

of

uniqueness

wi

x,

(X, e) r(z,

u

+

Zx such

or

x),

(x, (z,

z, x, we

X), (z

Y,

y) y)

E

of

E oe.

applying the Again, applying

and

obtain

X, X

(5.7)

of.

C

Z), (Y -

Z' Y

Y) I

E oe.

X

X, -X,

=

z,

(z,

Z'

x)

y,

(y,

Z'

show that

r(o, u)

X), (X,

Y,

obtain

we

Y), (Y

X

-

I

by (y,

for them, argument for (y, same argument

same

the

Z) (Z'

X,

Y), (X

Y, -Y'

z

X,

Y,

-

-Y)l

-X), C

of

(5.8) -

oe.

is

isomorphic

z) that

for r

each

(o,

u

to

z

+

E

wi)

Cay(Z3 P) oe), i.e.,

ZP3.

(5.9) A U B U C where

that if r(x, y) (x + wi, y + wi) (y + wi) e*. This implies

We claim

and y Since

e

=

Otherwise

f

E

xaw., n Lemma 9.

Similarly,

there

(X + W3) Y + is jWi}i=1,2,3 1, 2,3).

W3)

of Z3

basis

a

claim,

By the above

r(o, U))

e

r(o,

=

W2

such

(07

Ei=

1,2,3

r(o

u

e.

E

contradicting

W2) E e and (07 07 W3)- Since SiWi (8i (E ZP I

U

+

u

+

7

r(E

SiWi)

U+

8jWj)) 3

3

siwi,

SjWj)

i=1

i=1

r(z,

E v

:=

+ 81W,

2

siwi,

exists

+ W2i Y + W3

Wj))

+

U

2

r(j:

wl)d*,

and

as

there

119

WI)

+ W1, y + e,

-

+

(X

0)

P

(y

n

that

W2)

aw,

=

Schemes

have

we

+ W1,

'=

Z3 is written

E

z

P7

aw,.

-

(X

then

u,

=

e

+ wi G xaw,.

x

v,

where

e

x

-

.

" W2, W3 E ZP

exists E

Association

Commutative

of Primitive

The Enumeration

5

z),

+

desired.

as

Finally,

we

shall

for

each

Zpx

G

w

0. Assume the contrary, i.e., h from Lemma 1. It follows Pa,,,g Lemma 9. contradicting h

=

Applying

the

above

jr(o,

(u,

claim

v,

w))l 11u,

jr ((), (U, for

the

I ju,

If

jv, wl

=

w)

v,

wj

v,

Zp'.

E

n

f

E

G

wl

n

Ix,

e, v,

8(iii)

and

H,

-

we

that

Pga,*, h

we

obtain

zjj

y,

have awg n E) Lemma 9 we have

H, by

-

Then,

n D.

awe,

1(ii)

r(o,

(x,

by (5.10)

y,

zj I

loss

of

jx,

0 without

that

claim

E

w))

y,

W)) I (U,

V,

i.e.,

contrary,

(u,

some

for

g E G

and each

2}

',: :f

=

n E)

pha,,,gKhlr-g

=

=

0.

2,

(5-10)

show that

We shall

Assume

0.

Gn D

show that

that,

We claim

V,

duvw

1, then

=

G

-

E E

Z3 1, P

H since

x

=

=

11u,

u

(5-11)

0. v,

assume

follows

It

n -D

G and

may

we

generality.

E

W)

wj

n

=

u

x

from

(5.10)

0

Again,

w.

jx,

zJ1

y,

and

and

ly, zj

n

the

above

by the

above

claim, du,w a

=

dxvw

(x,

r(o,

w))a,-,y

y,

n D

=

0,

contradiction.

I Ju,

if

G

-

r(o,

v,

H. Since

(u,

v,

z))

wj n Ix, y, zj I r(o, (u, v, z)) C-

G

-

H. It

du,w a

E

Therefore,

contradiction.

steps,

together

association

we

scheme

conclude in

a,-yr(o,

follows E

aw-,r(o,

G n E) that

(u,

from

y,

the

(u, =

the

by

0, then

=

G

v,

above

z)) by

above

z))

claim

the

above

(u,

claim,

y,

z))

we

E

have

that

claim

n D=

r(o,

0,

second and final 0. Combining the first, commutative is a unique primitive the proof. is not (X, H). This completes

(X, G)

L(X, a) which.

0

References

1.

Characters Conjugacy Classes and Irreducible with the Summer Institute Notes, prepared of Finite on Groups, S6-ZA-1-12 Characters and Irreducible Classes of Conjugacy On Products Arad Z. (1987): in Finite Groups. Proc. of Simposia in Pure Math. 47, 3-9 to Finite and Applications Arad Z. (1989): Group Survey on Table Algebras vol. Conference Mathematics In: Israel 1, Weizmann Proceedings, Theory. 96-110 Science Press, of Degrees Table Algebras Arad Z. (1999): Two, Three Homogeneous Intgeral Element. and Four with a Faithful Groups St. Andrews 1997 in Bath, 1, Lonvol. 260, 20-29 Notes Series, don Math. Soc. Lecture Table Standard Arad Z., Arisha H., Fisman E., Muzychuk M. (2000): Integral 231 (2), of Degree 3. J. of Algebra Element Nonreal Algebras with a Faithful

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Bose-Mesner with a Algebras Algebra 231 (2), 484-545 Lecture Notes of Theory Algebras. University 16, 119-134 of Characters and Powers of Conjuand

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a

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19 Vertices.

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Rahnamai

M.R.

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M.

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M.

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Heft

4

Algebra

Linear

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Representations,

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with

Number

Prime

a

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and its

of Points.

Preprint 46.

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Polynomial

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C.C.

J.

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(1986):

1.

Other

Ones.

a

Subgroup

Self-centralizing

of Order

370-376

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Yamazaki

Zisser

Soc.

of

Class

a

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235-246

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(1964): (1969):

no.

An

Association

Schemes

with

k,

=

3.

J.

of

8, 73-100

Algebraic

Approach

to

Association

Schemes.

1628

Numbers of the Sporadic Covering Tel-Aviv Univ. (Hebrew) Thesis,

The

M.Sc.

Symmetric

Groups

and

Simple

Index

a-chain,

fusion

110

P-polynomial, B-graded set 2-orbits

of

fusion

106

of

fibres,

(G; X),

schemes, subalgebra,

8

8

10

table generalized algebra, 6 group-like, 1 GT-algebra, V GT-algebras, 4 GT-homomorphism,

5

1

-

matrix Ag, 107 adjacency 16 algebra A,,,(sgn), 14, F.,,, F' algebras Arad, Z., V Arisha, H., V association scheme, 4 association schemes, 4, 106 M,

Babai, basic basic

L., 105 relations,

5

6

sets,

graph,

Biggs-Smith Blau, H.I.,

106

integral integral

108

algebra, algebras,

Bose-Mesner Bose-M6sner

F.,

Biinger,

Herzog, M., V M., VII Hirasaka, G., V Hoheisel, V coherent algebras, homogeneous configuration, homogeneous coherent 2 GT-algebra, homogeneous table algebra, homogeneous integral table V algebras, homogeneous

5, 107

intersection

VI

isomorphic, ITA, 2, V

V

graph of E centralizer algebras, D., V Chillag, Cayley

over

K,

Kawada,

scheme,

3, 6

VI

106

Miloslavsky, Muzychuk,

4

2

E., V 9

of

Frobenius

algebra,

V., V M., 106, 1

(A; B),

3

105

8-

I

V

VI table algberas, P-polynomial Peterson graph, 106 2, 107 primitive, association scheme, 105 primitive transitive primitive permutation groups,

schemes,

3

non-singular, order

fission

Y., V

C-cosets, line, 112 3 linear,

'

106 enumeration, Erez, J., V isomorphic, exactly

faithful, Fisman, fission,

107

V

algebras,

left

association

coset,

2

5, 106,

4

M.R., V Darafsheh, degrees of (A, B), 2 graphs, distance-regular I basis, distinguished double

2

107

Iwahori-Heeke

109

107 commutative, 106 Coxeter graph,

cyclotomic

GT-algebra, algebra, numbers,

table

Quirin,

W.L.,

105

4 2

Index

126

algebra

quotient

of

(A, B),

3

13 strong, 13 strong element, symmetric, 2, 107

A., V Rahnamai, real, 1, 2 105 digraphs, regular 3 rescaled, 4 rescaling, 3 right coset,

table table tensor

the S -r i ng,

Schur

6

ring,

-

width

6

simple quantity, Sims, C.C., 105 SITA, , VI 2 standard, standard integral

Verarbeitung:

generated I algebra, subset, 2, 3 9 product,

trivial

Schur-H 'adamard, C., V Scopolla,

Druck:

subset

table

2

valency, valency

of b E

fusions,

B, I

9

105 of g,

106

6

Wong. W.J., 105 wreath product, table

Strauss

Schdffer,

algebras,

Offsetdruck, Grtinstadt

VI

M8rlenbach

Xu, B.,

V

10

by b,

2