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
products
of
conjugacy classes of finite
groups
4r6radoldbrancf,
of finite group theory. This topic was intensively studied in the 1980's. The book [22] "Products of Conjugacy Classes in Groups," edited by Z. Arad and M.
Herzog, gives period. It
was
comprehensive picture of the results obtained during this
a
realized
by several authors that
this research could be extended to
of irreducible characters. We refer the reader to the papers
products
[1, 2, 11,
13-16,21,23,35,40,51,52,651. In several of these papers the authors found an analogy between prodconjugacy classes and products of irreducible characters which led to
ucts of
the notion of table
algebra,
introduced
H.L.Blau and Z. Arad in
by
[7],
in
uniform way the decomposition of products of conjugacy classes and irreducible characters of finite groups. Since then, the theory order to
study
in
a
algebras was extensively developed in papers of Z. Arad, H. Arisha, H. Blau, F. B-dnger, D. Chillag, M.R. Darafsheh, J. Erez, E. Fisman, V. Miloslavsky, M. Muzychuk, A. Rahnamai, C. Scopolla and B. Xu [3-5,7of table
10,12,17-20,25,29-33,35,41]. algbras, as defined, may be c onsidered a special class of C-algbras by Y. Kawada [49] and G. Hoheisel [48].:More precisely, a table algebra is a C-algebra where the structure constants are nonnegative. Each finite group yields two natural table algebras: the table algebra of conjugacy classes and the table algebra of generalized characters. Table
introduced
Both table
algebras arriving
from group
erty: their structure constants and degrees
theory
have
an
additional prop-
nonnegative integers (we refer the reader to the Introduction where these notions axe defined). Such algebras were defined in [30] as integral table algebras (briefly, ITA). Generalized table algebras (briefly, GT-algebras were introduced in [20]. They generalize properties of such well-known objects, e.g., homogeneous coherent algebras, Iwahori-Hecke algebras, etc. are
integral table algebra may be rescaled to a homogeneous one [32], algebra whose non-trivial degrees are equal. This common degree is a natural parameter which may be used for a classification of integral table algebras. The first result in this direction was obtained by Z. Arad and H. Blau in [7] where homogeneous table algebras of degree 1 were classified. The classification of homogeneous integral table algebras of degree 2 with a faithful element was obtained by H. Blau in [31]. This research was continued in [10] where a complete classification of homogeneous integral table algebras of degree 3 with a faithful element was obtained provided that the algebra Each
i.e.,
an
does not contain linear elements.
VI
important class of ITA is comprised of so-called standard integral algebras (briefly, SITA) which axiomatize tile properties of Bose-Mesner algebras of commutative association schemes. The standard algebras are also involved in the study of homogeneous ITA. Each element of a table algebra is contained in a unique table subalgebra which may be considered as a table subalgebra generated by this element. So it is natural to start the study of integral table algebras from those which are generated by a single element. Table algebras generated by an element of degree 2 were completely classified by H. Blau in [291 under the assumption that either a generating element is real or the algebra does not contain linear elements of degree a power of 2. If a table algebra is generated by an element of degree 3 or greater, then its structure is more complicated. If a generating element is real, then we are'faced with a classifleation. of P-polynomial table algberas which would imply powerful consequences for a classifleation of distance-regular graphs. In contrast, if a generating element is non-real and of small degree, then either a complete classification or important structure information may be obtained. For example, standard integral table algebras generated by a non-real element of degree 3 were classified in (5], [33] under the additional assumption that there is no nontrivial element of degree 1. In this volume we continue the investigation of integral standard table algebras generated by a non-real element of small valency. More precisely, we collect here the recent results about integral standard table algberas generated by a non-real element of degree 4 or 5. In all the examples known to us of SITA generated by a non-real element of degree k, the degrees of all basis elements are bounded by some function f (k). This gives evidence of the following Another
table
I
Conjecture generated by bounded
are
a
There exists
are
function f : N -+ N such that if a SITA is of degree k, then all degrees of the algebra
by f (k).
The results of
and
a
non-real element
[29]
show that this
conjecture
is true if k
=
2. If k
=
3
SITA does not contain nontrivial elements of degree 1, then all degrees bounded by 6 and the conjecture is valid. The paxtial classification of a
standard ITA
generated by
an
element of
degrees 4,5 obtained
in this volume
also supports this conjecture. It is not difficult to show that the conjecture holds for the table algebras of generalized characters of a finite group even without the assumption of
being
non-real.
[22] and the paper [7] countries to work on table algebras, The book
attracted
many researchers from various
products of conjugacy classes
and related
topics. At Bar-Ilan
Z. Arad and his students H.
Arisha, V. Miloslavsky, Fisman, jointly with his colleague M. Muzychuk, performed extensive research on table algebras. In the academic year 1998/99, H. Blau from Northern Illinois University (deKalb) and two postdoctoral University,
and his former student E.
Vii
F. Bfinger from Germany and M. Hirasaka from Japan, joined the University group in order to further advance the theory of table algebras. This volume, together with [5] and [331, collect most of the results obtained in this period at Bar-Ilan University. This volume contains 5 chapters. The first chapter is an Introduction, which contains all necessary definitions and facts about table algebras. The second chapter, Integral Table Algebras with a Faithful Nonreal Element of Degree 4, deals with standard integral table algebras generated by a non-real element of degree 4. The contribution of one of its co-authors, H. Arisha, is a part of his Ph.D thesis. Another co-author, E. Fisman, was supported by the Emmy Noether Research Institute at Bar-Ilan University. The third chapter, Standard Integral Table Algebras with a Faithful Nonreal Element of Degree 5, and the fourth chapter, Standard Integral Table Algebras with a Faithful Real Element of Degree 5 and Width 3, are devoted to standard integral algebras generated by an element of degree 5. F. BiAnger, one of the co-authors of these chapters, was supported by the Minerva Foundation in Germany through the Emmy Noether Research Institute at Bar-Ilan University. The last chapter, The Enumeration of Primitive Commutative Association Schemes with a Non-symmetric Relation of Valency at Most 4, classifies primitive commuta, tive ass'ociation schemes which contain a connected non-symmetric relation of valency 3 or 4. Its author, Mitsugu Hirasaka, was supported by the Japan Society for Promotion of Science, and worked in both the Graduate School of Mathematics at Kyushu University and the Emmy Noether Research Institute at Bar-Ilan University.
students, Bar-Ilan
We also would like to thank Mrs. Miriam Beller who corrected the ous
misprints
in the text and
Ramat-Gan and
July 1999
Netanya,
prepared
the final version of the
numer-
manuscript.
Israel
Z. Arad
M.
Muzychuk
Contents
Introduction
I
Z.
Arad,
..............................................
1. 1 Main Definitions
1.2 Basic 1.3 Basic
examples properties
.............................................
........................................
...........................
1.4 Basic constructions
.........
Arad,
a Faithful Nonreal Element of Degree 4 Muzychuk, H. Arisha, E. Fisman examples .............................................. .........
2.2 Proof of the main results
Z.
SITA with
Arad,
F.
a
..............
E.
Element of Degree 5 Fisman, M. Muzychuk
5
........
.................................................
3.2 General facts and known results
Degree
.........................
Faithful Nonreal
Biinger,
3.1 Introduction
3.3
........
...............
................
....................................................
3.4 Case 3
......................................................
3.5 Case 5
......................................................
4
F.
6
7 13
SITA with
a
Faithful Real Element of
Degree
14 19 43
43 44 61 62 66
5 and Width 3 83'
Bilnger
4.1 Introduction
.................................................
4.2 Case 1
......................................................
4.3 Case 2
......................................................
5
1
4
M.
2.1 Known
3
......
...........................................
2. SITAwith Z.
I
Muzychuk
M.
83
83 87
The Enumeration of Primitive Commutative Association
Schemes
Mitsugu
......................................................
5.1 Introduction 5.2 The
case
5.3 The
case
5.4 The
case
References Index
105
Hirasaka .................................................
of
valency of valency of valency
2
1
or
3
...................
4
.........................................
.....................................
......................
....................................................
105 109 110
117 121
125 ...............
Integral
2
Nonreal
Arad',',
Z.
Algebras of Degree
Table
Element
Muzychukl,2,
M.
Department of Mathematics Bar-Ilan University Ramat-Gan 52900, Israel Department of Mathematics Netanya Academic College Netanya, Israel
2
Arishal,
H.
with
a
Faithful
4
Fisman'
and E.
and
Computer
and
Computer Science
Science
Throughout this chapter we assume that (A, B) is a algebras generated by a non-real element of degree from the class A strongly of the algebras 3. The structure b_ where b is a non-real faithful of the multiset structure
table
degree 4. A direct for [6'b-]: possibilities
shows that
verification
11 4,621, 11 4,43],
It
not
difficult
3).'If
[6-b]
is
tion
=
algebra Bb is exactly 11, cl is uniquely defined we have the following element
basis
Let
1. 2.
If If
[6-b] [6-b]
=
=
b
isomorphic to 1, cl. by bTb
an
(A, B)
of degree
4], [14
4
min(B)
depends
on
element
multisets
! the
of B of
exhibit
all
41],
2
integral
a
[Cb]
an
Assume that
EE
table
standard
be 4.
Z,,, ? 11, cl where [ 14 4', 4', 4'1, [14 81,411,
product
wreath
For
min(B)
algebra
> 3.
with
non-real
a
Then
F,. [14 41,41,41], then Bb F',. [14 41,81], then Bb
(The algebras F, The remaining that
3
4 and
[14 41,41,41], [14 81,41], [14, 121]. to show that [Cb] V j[14 6 2], [14,43], [14 42 41]1 (Proposito this book, [14,34], then by Theorem 5 of the Introduction
the
Theorem 1.
[14
following
the
integral
standard
element
and
F,,n
case
x E
B of
are
[6-b]
=
defined
in the
[14,121]
degree
4
is
if
is'strong
section).
next
the x
most
difficult
is nonreal
one.
and
[xyx]
Let =
us
say
[14, 1211.
c Supp (X2) Of always 2. If x" then by Theorem 5 [Xa-X7a is real, [14 3 41. If Xa is not real, then either [14 3 4] or [Xayal [14, 1211. Thus, in any case, either x" is strong [Xa7Xc, or [Xa-X7a [14 34]. In the second case we have the following:
element For each strong degree 4 (Theorem 3). It
x
there
turns
exists
out
a
that
unique
the
element
width
of x'
x" is
=
=
=
=
Theorem 2. element
Let
b which
(A,B) satisfies
be
the
integral following
an
standard conditions:
Z, Arad, M. Muzychuk (Eds:): LNM 1773, pp. 13 - 41, 2002 © Springer-Verlag Berlin Heidelberg 2002
table
algebra
with
a
strong
Z. Arad et al.
14
min(B)
1.
[6-b]
2. 3
=
b2
.
Then
=d+2f,JfJ=6 sgn
exactly
(see
we
algebra isomorphic
to the
:
in the
1, -11
Z2n
(x)
1 and sgn
-
next
=
is
sgn
sign function
a
(-x).
section).
if and only if n 2. In this An (sgn) is group-like group-like algebra over the group Z4 (1) Z4
case
Z4
(9
2.1.1).
b and b'
the
have
< m< n
is defined
Subsection If both
1, 0
3 4].
[14
where sgn
> 2
n =
The
Remark 1.
[cFdj
An(sgn)
(The algebra is
and
An(sgn), (2m + 1)
Bb _9',;
such that
it
3;
>
[14, 121];
same
then
strong,
are
alternative.
we
Thus there
may consider
b ,2
two
possibilities:
N such that
[F_77
are
(b)'
=
for
which
either
bl'-
is
for
strong
every
natural
(minimal)
m (=-
m
or
there If B is first
case
a
exists
a
group-like If
occurs.
B
algebra over is a group-like
an
abelian
algebra
Z2- 6) Z271 then [F-7177] [14 3 4] for each case Bb is always group-like.
group over
an
[14, 34].
of odd
order,
abelian
then
the
Z271
group
-
=
mG
N. We conjecture
that
in
the first
examples
Known
2.1
algebras
Group-like
2.1.1
Z4
H be an arbitrary subgroup of the symmetZ4 coordinates. The subgroup G by permuting group I (x, y, z, w) I x + y + z + w 01 is an 11-invariant subgroup. So we can define table algebra O(G, H). Let now a standard an S-ring, and, therefore, integral H be an arbitrary subgroup of S4 that contains a. full cycle (1, 2, 3, 4). Then 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
.
Let
on
=
=
=
-
is easy to see that 0 is a non-real faithful element of O(G, H) of denon-trivial elements gree 4. If H G JA4) S41 , then 0 (G, H) does not contain of degree less than 4. The constructed algebra is of dimension RO. In order to on
G. It
build
dimensional
finite
examples one has to take an H-invariant subgroup of example, kG, k E Z) and factor out with respect to the is not difficult to see that such a quotient with respect > 5 gives us a finite example of the above algebra.
(for finite It chosen subgroup. index
G of
to
a
subgroup
2.1.2
Weshall
The build
following multiplication
the
kG,
k
algebras a
series
F.,,, of
6-dimensional table
F'
M
algebras of dimension 4m+ 2, m > algebra (H, E) where E 1, h,
of which looks
as
follows:
I
-
Westart
C1, C2) V) U)
with
1,
the
ITA with
2
h
Faithful
a
C2
Cl
h +
Element
Nonreal
of
Degree 4
U
V
V
3cl + 3C2
h
41+h+c,+C2
h+v
C1.
h +
v
4C2
41 + 4u
3c,
3h + 3v
C21
h +
v
41 + 4u
4c,
3C2
3h + 3v
31 + 2u
3h + 2v
3C2
3h + 3v 3h + 3v 3h + 2v 121 + 9cl
v
+4u +
3c,
+
+4u +
hold
gebras U
=
jh,
v,
By direct
I
I
Define
case.
21 + 16 2
=
I
all
shows that
this
in
s
v
check
A routine
3C2
3c,
V
U V
(CI
C2)7
+
computations
the a
+
h
v
2s + 4t +
S,
3h+v
tj
2v
41+cl
,
3c,
v
A+
+
tables:
2v
v
1 2s + 4tl
4v + 6h
4u +
(2.2)
4s + A
t
S
3h +
3C2+
al-
follows:
4s + A 2s + 4t
5v + 3h
I
+C2 + h
table as
t
S
V -
H
multiplication
following
v
.
10s + 8t + 3h + 2v 5v + 3h 4v + 6h
v
h h
+
V
2s + 4t +
h + 2s
h
U
the
obtain
(2 1)
standard
integral
HQ := Q (9z " + C2)1(Cl 2
U C
subset
t
we
v
+ 9C2 +3h + 8u + 2v
,
of
axioms
15
2v
v
v
3h+v
5v + 3h
2v
4v + 6h
4v + 6h
5v + 3h
3c, + 3C2+ 121 + 9cl + 9C2 +8u + 3h + 2v 4u+v
(2.3) 4u + 3cl
61 + 3c,
+ 3C2
+3C2 + 2u
I
h
let
S
t
2s+4t+v
3h+v
2v
h +
v
3h + 3v
2s + 2t
2s + 2t
C2.
h +
v
3h + 3v
2s + 2t
2s + 2t
v
2v + 3h
V v
+ 4t + 2s
'
13h + 2v
I +8t
2t +
s
I
I
3h + 5v 4v + 6h
+ 10s
(2.4)
t + 2s
j
that QUis an ideal of QE. multiplicatively cyclic group of order m written (a) For each a E C, table algebra. Cm) be the corresponding
from the above formulas
follows Let
61 + 3c, +3C2 + 2u
3C2
Cl
U
It
+
V
h + 2s
h
4u + 3c,
C,,,
(CC,,,,
=
be
and
a
Ua
E,a U,a:7
=
1; 1.
we
set:
Z. Arad et al.
16
Therefore
C,,, a
0 U is
0 h is
A direct of
U
basis
JUj,,cc_
=
4 of the
check shows that fusion
a
The
2.1.3
subalgebra
is
Introduction
of
of
set
a
this
to
C,"-graded
book the
fibres.
F,
set
:=
the
by H(m, n),
denote
following n E N,
H(m, n) C'
e,
(F,,,
\ JC1) C21)
Fn. Its dimension
equal
is
U
fC1
+
C21
is
a
to 4m + 1.
co,
two-parametric
R>O.
mE
I
cn-1
...,
U
of table
series
is defined
It
by
foo, ..-On-11
the
f
U
eo,
algebras following
we
I
en-1
...,
which data:
oa
eg ec,+o
(m (M
is
F' M :=
the set of
,
ec,+,3
where the additions
n)
f
=
cp C,,+O
0,3. Ocl+,8.
an
An(sgn)
algebra
with
Westart
H(m,
that
by Proposition a distinguished
standard table algebra. Since integral non-real a faithful element, the algebra F.. is an example of algebras looking for. The dimension of this algebra is 4m + 2.
we are
basis
(2.2)-(2.4)
from
follows
It
00,+0
1)c,,+j3 1) 06,+0
-
-
(M
is done modulo
algebra
table
2)e,,+,3
-
'
of indepes
standard
(m
+
which
1)oa+,a
-
+ Me",+'3+1
Mcc,+'8+1
A direct
n.
check shows that
if 'M G N. integral Let now C2n be a cyclic by g (E C2n- In a group algebra group generated basis xi (CC2n we choose the following sgn(i)g' where sgn : Z2n -+ 1-1, 11 function which satisfies is an arbitrary We 1, sgn (-x) sgn (0) sgn (x). a
is
=
=
have x,,xp
=
S(a,,3)x,,+,3
S(a, -a)
symmetric,
sgn is
the
Now we consider
and commutative
elements
form
basis
0+
=
4eO, +
Ca
:=
f
a
=
4c,,,
An(sgn).
through the Clearly, lAn(sgn)l
structure
constants
+
+
Oa 00
in the
0-
54 0;
Ce
:34
CO + XO,
a runs
set
+
0).
Since
10, =
0
=
-
5n.
-
'
(CC2,rb
Clearly,
-
check shows that
I
-
this
is
the
following
an
as-
ca+0+1 + (1 + S)e a+o+l 4c+0 + e+, 0 a+ 0 + 1 0 =
2X2ce+l)
-
2ea
0,
f
0, 3co
2X2.,
-
a
=
a
0 0;
0
0; -
xO,
a
=
0
11 C Z2n We denote this show following equalities -
The
above basis
3o,
:=
C
n
-,
=
e,
0; a
(1)
CC2n:
(1)
0
a
=
sgn(a)sgn(,8)/sgn(a
A direct
2X2ce+l,
2e, + 2X2co
ea
where
+
o,,,
H(4, n)
sum
algebra.
H(4, n)
of
=
1.
=
direct
sociative
a
S(a,,3)
where
are
+
nonnegative
s)e,+,,+,,
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
+
3+3so+ 2
Oe0 s
=
S(2a
where
s
=
S(2a
+
+ c,
+,,
10
a
+
0
a
+
)3 +0
+
0
0
0-
a+ 'a,
2
3oa,,O,
+
0
0
1).
+
4. 0
3-3s
+
a+18
3o
1, 2)3
+
s)e-
+
1).
+
3.
where
(3
+
+,
a+
+
3-3so+ 2
a+
e
)3+3+so
-
a+
2
)3,0: O
0
1, 2,3).
5. +
6.
o+c
Oa+)3+Oa+,3,0: O
c+-
0 ,+
+ "3 Oa
,q
=
0.
o-
=
ce
7.
9c,,,+,,+,
0-0
where
s
(9
+
s) ece+,6+1
S(2a
+
1, 2)3
9-3so+ 2 a+
+
Oa e,8 s
S(2a
+
+
0-6a
3ol
0
=
3o,,,
9+sO 2
+
43
)3
-
0,3
3o
a
C+ ,q 6
+
cl
+
Q+0
2
0
a
+
9+3so+ )3
10.
oc
s) e,+,,+,,
a+
0"3
0
0
1, 2,0).
9.
11.
(9
1).
+
8.
where
+
12c++8c-+9e,+,a+0+1=0 0 0 0
=
9-so2
a+,6
10 0
0
0.
3o+
+
3o
2o;;,)3
a+,6 +
3o,+,,,,O =
0
0
0.,
2o,-.
12.
6ct
+
3co+, +0 6c+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 et al.
18
where
s
=
S(2a, 2ß).
13. 0 0"3 6c+ + 4(e+ + 6-), -p ß =
3c+",+ß
where
s
=
0
C,
(2 s)e++ + 4c, +2e.,a+0=0 +
-
ß
a
+
(2
+
s)e-
+
S(2a, 2ß).
14.
e "ß +
ea
=
4(ep
+=
Co
e++", c,+ 2e0
,
0
eß ),
+
a
=:
0
7 01,6:A: 02 C+ 7 0 0. 7 0, ß:7 0, a + ß
a a
=
15.
e
+c-0 ea
ce:7
+ ei, +
3e0
,
a
=
0.
16.
0,a e
==
3e,
e
6c+0 where
s
+ß
0
0
or
+(2+s)e
+ 2c-0 +
a+
2e+, a:7 0
0,
0,
a
+
j3
0
S(2a, 2,6).
17. 0
0,a
+
e-c a
e+"+O,
=:
0
2e0 18.
ei cJ0
=
2e+
+
ej,
a
0
=
:7
* 0, ß * 0, ce 7 01 01 a
ce
,a
+
ß :A 0 0.
0.
19. +
ca
3
=
4c",+" a + ß 7 0; a + ß 4(c+0 + c-), 0
20.
4 cj 21.
c0 ej
=
3co+
+
2co-
3c+,
a
=
Co
,
a
:7 0;
=
0.
=
0.
7
0
ITA with
2
Nonreal
Element
Degree
19
4
Preclassification
2.2.1
Proposition either Aabc
B# be
a, b (=-
Let
1. E
simple
following
from the
Westart
10, 11 for
each
E B
c
elements
two or
the
of
one
of degree 4, a =7 T. cases holds: following
=lhl=4 2f +g+h,lfi=lgl 2f +g,lf l=4,l gl=8 6, I gi4 2f + g, if I
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
=
1 Proof.
Then
(ab, ab)
ab
=
=
1
4
5
Routine.
As
a
consequence
all
b E
be
following
the
obtain
we
(A, B)
Let
Theorem 3.
for
of
main results
of the
Proof
2.2
Faithful
a
exists
JbI
such that
table algebra integral b E B* with b 0 b
standard
a
BO. Assume that there
and
JbI
=
4.
3
Then
either bb
41 +
=
4c, b2= 4d, Icl
=
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 =
product
1: beB#
4,
28
28
28
=
Consider
all
bb:
bL 4 +
(2.6)
V, ESUPP(b-b)#Ab c
/\bbeC.
36 40 52 64
c
G
B#,
there
are
Z. Arad et al.
20
If all
coefficients
nonzero
Therefore,
there
(bb, bb) 28, (b2, b2)
that
=
c
(b2, b2)
E
(b2, b2)
then
ones,
B# with
C-
Abbc :
f 28, 40, 52, 64}.
(2.6)
then
=
axe
exists
holds.
Thus
16
=
(b2, b2)
If
have to
we
contrary
(bb, bb)
to
follows'from
It
2.
64, then (2.5)
=
deny
28. I
holds.
following
the
>
Proposition
If
possibil-
lities.
bb
=
41 +
2 3c, b =4d+e,ldl
bb
=
41 +
2 2c, b =3d+e,lcl
Assume first
equality
(2.7)
of
sides
of this
lcl
4, I\jdc
=
b satisfies
that
db
equality =
Consider be real
that
in bc is not
b
we
0. Therefore
now
in this
by
the
than
greater
It
9,
a
lcl)
=4.
(2-7)'
=4.
(2.8)
from the directly A 36. Multiplying
follows
second
both
=
e)
+
lei
=lei
Idl
Since
121 + 9c.
=
=
3 and
contradiction.
(2.8).
when b satisfies
case
gcd(lbl,
=6,ldl
d(4d
obtain
remaining
=
equivalently,
or,
Aj,
Since
case.
(2-7).
3b,
=
3,lcl
=
Note that
2, the number of
=
c
should
summands
nonzero
Moreover,
2.
(bb, c)
(bc, b)
12 =: -
/\bcb
12 = -
=
=
3.
Therefore, bc The second g
0 b, lgl
=
Af, f =/= b, Alf I
3b +
equality
(2.8)
in
bd
implies
12.
=
3b + g for
==
some
g
E
B#,
4. Wehave
(b2)b
(3d
=
b(bb)
b(4
=
+ +
e)b
2c)
=
=
3(3b
+
g)
2(3b
4b +
eb,
+ +
Af).
Consequently, eb Therefore
f
=
g, A
immediately
b +
=
implies
and for
JR(e some
Therefore
(C2 6b)
to
+
=
Proposition
d)l
sides
both
pf
for
equality,
latter
of the
f
we
satisfies
=
Now Abbe
2.
:5
suitable
a
book
one.
of degree 4 that be 3b + f.
element
21
4
this
to
interesting
12 such that
=
ISupp(be)l
2,
=
3b +
=
Introduction
5 of the
non-real
a
BO, if I
E
Degree
of
Element
is the most
case
BO be
Let b E
Nonreal
Faithful
by Theorem
then
So the second
Then there
Proof. Abeb
a
obtain
1-tif I
12/,t
> 24.
2
=
BO and
G
=
implies
/.t
E N.
12.
The
that
Taking identity
(bb, ei ) implies
=
E
A,e,lcl
12 +
=
r_GSUPP(bb)\j1j The left-hand
Therefore
it
In what rem
3 and
side
Let
if bx
Properties
Proof.
a
2f, R
=
tt
=
b E BO of
element
divisible
by
12.
0
1.
degree
4.
By Theo-
Jbi
3b + g,
=
Idl
=
6, Igi
4, jfj
=
(2.9)
12.
say that
[bb]
that
is
element
an
one
following: -of degree
of the
Supp(bb)
x
G
a
basis
[14,41,41,41], 4 is
a
starting
Ed.
=
of the
starting
If
Supp( b_)
3.
starting
For the
element. contains
of degree 4,
element
then
Supp(66b)
one.
Eb
product
there
possibilites:
two
are
juj=jvj=jwj=4, U:
b
Icl us
30 and is
always'have
Eb=41+u+v+w,
Let
161
-
implies
in turn,
non-real
a
we assume
us
Proposition contains
which,
we
JeJ2
than
of Theorem I
*subsection
[14, 81, 41]. element
d +
=
Proof
In this
greater
24
to
we fix follows, Proposition 2,
b2
2.2.2
is not
equal
is
=
8, jhj
compute the product
(bb)E
=
(d
+
2f)E
=
b _b =
=
4,
V,
V
7-L
W,
-
c
+
h,
41 =
h,7!
=
:7
W
u
(2.11)
c
in two ways:
dE + 2Cb
=
dE + 6b
(2.10)
+
2g.
Z. Arad et al.
22
b(blb-) After
=
reduction
of
d
4b
common
+ 2b +
Assume first
t
+bu
bc + bh if
terms,
bc + bh holds.
Abwg 11 Abug ! 17 Abvg < 1. So, we have the following equalities: '5
=
=
is
w
element
starting holds,
a
(2.11)
If
in this
Abcg 'A
Clearly,
Abcg (2.13)
2.
=
hand, Together
implies
bh
Proposition 1.
Let
4.
x
for
a
=
Further
1.
Supp(bE),
(2.12)
d.
=
6
=
Abcglgl 2A -bc
+ 6b +
(2.13)
2g. is
2 this
=
gives
bc
us
least
at
Abcg
=
=
=
2b +
On
2.
< 3.
Thus
2g. Now El
Supp(bE)
E
=
0(mod 8). Therefore Abcg 32 implies :5 Jbilcl
=-
inequality with Abcb Ed..
=
y + 2f
dT
Abcglgl
But
0.
the
other
=
case.
4b + bc + bh
the
holds
b + g; b + g;
bw Thus
holds
Abub Abvb Abwb renaming of the elements of
up to
bv
holds
(2.10) (2. 11)
if
Then
bu
holds
obtain
we
bu + bv + bw if
2g
(2.10)
that
(2.10) (2.11)
+ bv + bw if
suitable
be
a
element.
starting
B#, jyj
y E
=
Then
4;
2.
Cbd, Ed) Proof.
(f, dY) < Idl
AdTd for
a
2.
(bx, bx)
0, then (d, dY) -
1
suitable
( d, Ed)
Proposition Proof.
=
=
(bx,
if
0
y
d
otherwise
=
Cbd, bx) A
5.
that
[bq
of the elements
(6-b, uUU)
=
(6-b, v-v)
(d T, b2)
clear
in
the
[14 41,41,411, u, =
v
=
element
starting
is =
=
=
(b2, dY)
=
(d
+
2f, dT)
=
B# \ ff I.
y E
d)
the fact that 4AdYd >' 16. But that contradicts 2. Thus dY AdTf 0 0 implying AdTf y + 2f
=
Therefore,
3.
Our claim
assume one
(bx, bx) =24 28
1.
16
-
=
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
(2.14),
with
Together
+
implies
this
uv)
16
=
16
Auvv
+
uv
4w. Romhere it follows
=
Auvw
=
=
4.
that
uw
=
4v.
Thus
(6-b, uw)
(bu, bw) Since
b + g and /\bwb 16 contrary bw)
bu
=
(bw,
and
1. dY
d
2.
to
=
Proposition
If
6.
=
d +
=
x-x
x
1, (bu, bw) Proposition
=
Supp(bE)
E
is
=
12Abwg.
Hence bw
=
b + g
4.
element,
starting
a
16.
4 +
=
then
2f;
=
b.
Proof. 1. Assume the contrary, 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 8] since Abxb 1. Therefore occur
=
to
or
bx are
=
[4 2 4,4].
of type =
b + 2u + v,
distinct.
pairwise
b(bx) (bb)x
=
The first
where u,
Now we
b 2 + 2bu + bv
(d
+
type
2f)x
=
=
cannot
basis
are
v
can
=
elements
of
degree
=
u, v, b
write
d + 2f + 2bu + bv
dx + 2fx
4 such that
d + 2bu + bv
y + 2f + 2fx
=.
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
2bu + bv
b2X On the other b 2X
bEd
=
=
(2.16)
[tGN,zGB of
substitution =
y + 5d +
y + 2f + 6d +
(2.16)
in
(2-15)
we
obtain
2ttz
(2.17)
2/,tz.
(2.18)
hand, 4d + xd +
cd
cESupp( b)\11,xI
=
4d + Y + 2f +
cd.
CESupp(bb)\{I,xj
Z. Arad et al.
24
Comparing this
with
(2.18),
we
(
T
obtain
cESupp(0b)\f1
'.
I
C)
d
=
+ 2d.
2/-tz
in the left-hand side do not exceed 4, /t < 2. By (2.16), /-tlzl Hence, jzj E f6,121. According to (2.17), bv =_ y + d(mo 'd2). Therefore 0. Hence AbvzlZl :5 8. Together with AbvziZl =_ O(mod 12), this implies Abvz d + p z. But now b(u + v + b) 3d + y, bu bv 5d, which is impossible distinct. since u, v, d are pairwise d as required. Thus y
Since coefficients 12.
=
=
=
=
2. Since
y
=
d, (bb,xTY)
(bx, bx)
=
E
Therefore
28.
=
Ax-,,,Icl
12.
=
CESupp(1;b_)\{1j
Together
with
1cl
=
12
ESupp(6_b)\{1j and Theorem 3 this xx
=
that
implies
Ax-x,
=
I for
each
as
before,
c
G
b.
Since
bx
Td, Cbd, Ed)
=
=
Arguing
28.
Supp(bU)
we
obtain
111. that
6-b
Hence
=
du. 0
In
bE
41
-
sum
what -
of two basis
So
we
c
Proof.
for
some v
E
(fd, Td)
=
dh
=
B# \ jhj.
,
set
c
:=
=
=
d,
dh
exists
v
d + 2f.
=
c-
=
Now we
(2.19) =
ff
d +
2f, dU
=
h
2
=
B# such that jvj =
Therefore
Td
(Ff dU)
h. We also
=
There
By (2.19) gives us
as
of
Td
2, this
element
starting
then c is a basis element, otherwise c is a degree 4. Moreover, it follows from the proof of to u + U, u E BO. In both cases x-x o c is proportional 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.
h.
follows, If 6-b
=
12 and
(2.20)
3h +v.
ATdh
3h + pv, can
(2.19)
bb.
3.
Together
with
ged(If 1, Idl)
p
write
jd, Td) (bf,Ef ,6b)
36 + 12 p (2=9) 48
p
=
1, jvj
=
12.
ITA with
2
Proposition
a
Faithful
Nonreal
Element
of
Degree
4
25
8.
hc
2h +
=
2v V
(2.21)
v.
=
Proof. (2.9)
E2 d
=
(2.20)
d3 + 6h ( + 2?)d E2 d (2.19) Ebh=4h+hc =
=
Since
and h
c
Proposition
are
real,
9.
There
is real
v
exists
B* \ jcj
hc
such that
Jul
Cf
61 + 3c +
Arf,.u
C2
81 + 4c +
4Ai!f-uu
Arfu Proof. Weuse the following form: following
U)
=
2h + 2v.
+ h2
well.
as
u E
+ 2v
E
f3,61
and
(2.22)
{1, 21.
(6-b)2
identity:
E
b
=
2(E)2
.
The left-hand
side
has
the
(41+c+h )2 =161+8c+8h+h 2+C2 +2ch The
right-hand
side
is
equal
d3
=
h2,
=
161+c
2
+h2+12h+8c+4v.
to
=O)d3+2(6h+2v)+4Cf.
(2.2
(d+2f)(d+2f)=dd+2(df+f'd_)+4Cf By (2.19)
(2.21)
hence
4Cf
=
161 + 8c +
C2.
Comparing
of c in both sides, the coefficients 8 + Ac,,. we obtain 4Arfc I 7. Thus Acc, E Accc is divisible by 4. On the other hand, Accc < Icl 2 =#0, then Aqc 10741. If Acc, Arflcl 0 0(mod Ifl), a contradiction. Therefore 3. Accc 4, Arfc =
Hence
-
=
=
=
Let
4Arfu 24),
which
On the and
E Supp(c 2) \ 11, cl be an arbitrary basis element. Then Accu Since Arfulul =Ac,ulul 0(mod 6), A,culul =- 0(mod 4Afrfulul. implies 24 < Acculul. other hand, Acculul : IC12 24. Therefore 24 JCJ ACCCICI A,,ulul
now u
=#-
Supp(c 2)
equality
A,,,,
=
-
contains =
4Arf,,
-
=
from cj. exactly one element, say u, distinct the proof. implies that Aff,, E 117 21 finishing
=
The 0
Z. Arad et al.
26
g (xy, xy) (x-x, y-y) 4, Isl unique r, s G B0, Irl
x, y c-
grading G, we
there
exist
The
Proposition
Proof. :
we
x
shall
Thus z-z
41 +
=
Now let
+ h
c
x
x, y,
:7 T, T 0
z,
(y
x *
If
(y
z)
*
If
definition. z
The
h
If
x
Xyzv
=
follows
of
=
zj
that
xy
in
the
gh
Proposition
Proof. other
At the a
a
h otherwise.
=
z
h
*
as
G.
E
x, y
Denote
of
So
*.
z(mod 2), and, consequently, -=
(mod 2).
+ h
c
E G.
G. If
g
gh(mod 2).
=-
unit.
a
from the definition
each g (=-
triple.
x(yz)
=-
h*z
=
*
Denote
h,
=
there
then
\g-g-h
But
1
=
v
*
*
y
:=
from the
y)
*
y,
x *
:=
=
I.
h E
If
z.
following:
z(mod 2).
otherhand, x*(y*z) z by v y 7 v, then y other hand, A-g,z I implies
Onthe
z.
But
(x
=-
On the
h.
u
law follows
(xy)z
1, A-g,z v
Supp(c). (y * z)
=
If
=
Supp(z"T)
Supp(Cc).
h
while
z.
=
*
=
it
=
analogously.
== -
So
Furthermore,
[xyl (x
x
y =7
x *
may
we
assume
now
v
*
y
=
implies
that
* z
[4', 621
=
y)
* z
=
from which
it
h. The
case
of
Y
h,
way.
G,
g
to
=7 h,
that
mention
to
the
equality
x *
=
0
x.
there
exists
an
element
sg
(=-
134 of degree
2sg,
The mapping g
same
contradiction.
h.
=
1. Since
enough
is
g E
g +
x * v =
invertible
7 is
(glh,92h)
10, 41,
g
may be treated
proof
11.
[4,62].
y
similar
a
Assume that
hand,
=
z) \xyr,=-'T(mod 2)
For each element 6 such that
then
=-
xy
g for
=
U
(y
x *
G shows that
E
g
=-
1. Hence
=
To finish
y
-
7 is treated
=
directly
h)2 (mod 2)
+
Ay,v
x * z
U, then
Xyjx-
z)
then
and
(c
h
G whenever
C-
y
case
arbitrary
*
f 1, hj z,
x *
that
Define
+ 28.
r
each
For
implies
I
=
and
with
then the associative
Since v,
=
*
(x*y)*z v.
y
case
'77 % jx,
that
x
*
x(y
=-
Supp(y-y)
c
Therefore
u
'g
=
6-bj.
=
p(mod 2).
an
0 7,
v
z)
*
then
=g,
x
7,
u
*
x
=-
0 h,
-=
G be
z
h
g
h
*
=
xy
x
group
b& Consequently,
=
If
nothing to prove. 1, whence g I\ghg
if
r
follows
77. In this
that
is
Let
that claim
our
x
check
us
y
gT
Proposition
28, 6, such that
abelian
an
x7V-y(mod 2)
=-
zz
is
check
that
assume
may.
h)
*,
F, then
=
4 &
=
and
(2-19).
(G,
we
y. If
x *
I IgI
E B
=
G by setting
on
10.
First x *
=
*
h E G by
Note that
z
G
=
=
operation
binary
Let
group.
have
36
gi
=
time,
-+
92 and sg,.
(h 2, 92711).
[h2l
E
sg, g =
Since
G G
h}
S92. Then
gj,T2
j[j4,4,4,4],[j4,41,81]j.
is
injective.
(g, h,
E G
92
h)
=
24.
On the
and 91 7 92; [92TI] Hence (h2, 9 2 T-)
E
0
ITA with
2
Proposition
Jwl
Faithful
a
arbitrary
Let x, y E G be
12.
hy
If
0 T,
x
then
there
exists
=
q Cz
Y(x
Proof.
Denote
z
x *
z
then
x,
=
Supp(c)
left-hand
the
U jhj.
A,yylyl
8
together
with
(C2, y-y)
A,y,,Iul
==>
Jul
64. Thus
=
+ pq
==>
+ q, cy
=
2y
n
C=
Let
m,
(2.23),
m=
TZ
cy +
2y
=
lcllyl
+
First
and Sh
mn
sh
n
0
there
h
-1c) 2
we
we
=
may
==*
exists
obtain
we
(2.25)-
+ cy.
1, from which elements
follows
it
that
G. Therefore
from
6. We claim E BO, Jul =_ (mod 24). But A,,.,,Iul 24. Therefore 24 which Acyujul (cy, cy) 2y + 4u. But this contradicts ==>. Tz. ==> u (E Supp(hy) hy y + 2u
suitable
a
u
=
then
=
=
=
=
=
1 that
we
obtain
4y
Let
27xw.
=
=
E
now q
2ATwqlql. Together with implies Acyqiql 2,\-Zwllql Supp(cy) f y, qj and Yw
this
=
=
that
(cy, cy)
But
+ cy
Acyqlql
2ATwq =* O(mod 6),
elements
Y
41 + h +
=
=
64
=
==>
1, i.e.,
p
Tw
m
0
Set y
TF.
=
m
n,
x
By
M.
=
2 of the
2
q G
=
as
that
so we
B#, JqJ
=
for
statement, =
2y
(2.26)
to
the
case
3(m
*
n)
each y C G there
Wenote that
2t,.
+
of
th
exists
V.
+ trrt*n-
special case of n Then, by the definition
assume
y,
we'may extend
a
3m + tm,
=4
latter
12 and cy
=
consider
.1c.
So
c.
(2.26)
+ 28m*n-
1c. 2
21 +
=
Vm,nEG Tnsn
13.
21 +
=
+
hy
+
contains
and part
arbitrary
G be
71, then by setting
Proposition Proof.
4y
Supp(cy), Acyylyl
-
2/Lq.
(2.24)
2q.
According to part ty G BO such that Ityl
So
2w,
y +
x *
have
we
m(21
=
27
12,
by Y,
sides
E
u
implies cy 0 Supp(cy)
u
=
mn = m* n
m=
xy
4
(2.23)
does not contain
if
:
6
=
Txy
=
y + 2u, for
=
from (2.25) 2. It follows Supp(Yxw), q: y. Then Aqyq Acyqlql =_ O(mod 8), 1\"Ewqlql (mod 24). Arguing as before,
If
Write
Degree
y),
*
both
side
Supp(c)
But
y=h and hy=Yz. So, z 0 x, whence zY that u Supp(cy). Indeed,
3y 3y
of
Yw=3y+q.
Multiplying
y.
Yz + 2Yw
y E
elements.
BO such that JqJ
cy=2y+2q,
If
Element
Then
6.
2.
Nonreal
h. Wehave
=
of t,,,,
mc
m, * n =
2m +
m*
h
=
m
2t,,,
desired. n
0
may
h. Set
x
=
M,
y
=
apply Proposition
m* n.
12, part
12 such that
Ysn
=
3y
+ q; cy
=
2y
+
2q.
Then xy 2, which
=
n
+
28n,
says that
Z. Arad et al.
28
But
cy
3(m
*
2y
=
n)
+
2ty
==: -
q
Proposition
3Y
=
ty
+
Tnsn
14.
hg
Wehave
ggh
+
13, sgT
(41 Together
g +
=
(g
=
By Proposition
+
+
2-3g)
=
=
3h +
3h + th
h)2
+
c
Aq.u.
61 + 3c +
=
(2.27)
Therefore
2sg.
2sg)(T =
ff
=
g-gh
=
=
g-g +
2(sg
+W-gg)
-9-gg
==*
v
+
4sgg-g.
3h + th
=
+ 12h + 4v +
g g-
A+
v.
Thus
4sg 7gg
with 161 + 8q + 8h + 4h + 4v +
we
Ts,,
0
VgEG sg--g Proof.
Therefore
t..
=
+ tm*n-
c
2+
h
2
(41
=
+
c
+
h)2,
obtain
4sg"-g By (2.22)
right-hand
the
side
of Bh.
The structure
161 + 8c +
=
4fT
exactly
is
The purpose
C2.
of this
section
following
the
is to prove
result:
11, h,
Theorem 4. Our first
v,
step is
to
ul
U Supp(c)
< B.
Jul
show that
=
3, where
u
is the
defined
element
by
(2.22). order
In
show that
to
the
of
case
Arfu
=
1 is
impossible,
we
need the
following Proposition
Let
15.
k, 1,
elements
tains
m,
(A,B)
be
k2
=
n m-
where x
Proof. k
1,
mE
BO,
c
First
ki
==
+
B and either
then 1
we
T1,
ki
m and
=
k (E B
11, 11
that
note
G B.
standard
a
table
algebra
which
81 + 4 k + 41
=
61 + 3 k +
or
k
=
U Supp(k)
AWy-k
is
follows
directly
ISupp(mk)l
2.
It
integral
con-
such that
(2.28)
1,
kj + 71, ki is
a
correctly from
table
E B.
subset
defined
(2.28)
that
for
Iml
If JxI of B.
> 3
each
Ill
y
for
each
E
B if
8. 6, Jkl 4, this implies =
-
Together with Amkm BO\ Iml. Using the identity 4Ta + /-tn, for a suitable Tnk n (mk, mk) 4m + 3n 8, i.e., mk (k 2, rrOM) and formulae (2.28), we obtain M 3, Inj 8. with n C B, Inj If
k c
B, then
=
=
=
=
=
=
ITA with
2
If
2,
k
=
ki
Tnki
:!
(2.28),
mulae
=
n k-j
holds
it
cases
(Trz7k-j,Tr:k-j),
+ tt% obtain
Using
-
M
nk
E B
n
or
n
for
every
According
C-
=
12.
Therefore In2j. (k 2, m7mg) and for
=
-
4 and nj
=
Milnil
n2.
Thus in
-
both
=-
Ini
3n,
(2.29)
8,
=
G B.
0(mod 4). There-n(mod 4). By (2.28), V 0(mod 4), which, in turn, implies that A,,7n--- E
(k2, rrn)
32/\n:n-k
64 +
JnJ2
An-nk
+
Inj
-
7,
=
Jkl
24An7n-1.
10, 41.
A,,7n-1
If
29
E B.
x
(2.28),
to
=-
4m +
=
Since
/\rrn-k
Ind
=
n2 E B and
A2,1nil (mk, mk)
=
identity
+ n2, ni
ni
=
mk
write
can
0(mod 4),Ti-n
=-
10, 4, 81
4
that
where either fore
/-tl
the
3, Inil
=
mk
Now we
Degree
=
where nj,
2m + /-12n2,
=
=
we
of
=
(mki,mki) 4m'+ /inj
=
Element
Nonreal
4 which implies that ki E B, then 1kil jSupp(rakj)j Together with Anikim A,,i-k 'M 2, we obtain
2.
2,rn + pini,
=
Since mk
Tj-,
+
jSupp(mTkj)j
Faithful
a
=
8, then for each
A,,7nxlxl
=-
0(mod 8).
implies
jxj
=
implies
6, this
=
is a unique subset By (2.29), Akkm
k.
=
gives
us
v
cv
hu
==: -
=
h2C yields =
v.
6h + 6v.
Furthermore UV
V2 Now we may
=
complete
41 + h +
c
c
2h+2v
v3c+4u+v
Remark I
If
table
subset
f 1, h,
table
of 11, h,
way.
This
=
3h + 2v,
C
U V
V
3c + 4u +
81+4c+8u
3c
6h + 6v
3c
31 + 2u
3h + 2v
r -6h+6v
13h+2vl
B, then Table 2.27 gives If c % B, then c v, u, cl.
v, u, cl,
=
Tl-}
to.Table
may be. reconstructed
2.3.
v
(2-31)
l2l+9c+3h
G
c
leads
u2h
121 + 9c + 3h + 8u + 2v.
2h + 2v
V
U
=
=
the above table:
h
hl
u2 h2
the cl
multiplication + 71-
from
of the multiplication Table 2.27 in a unique and the
table
ITA with
2
Bb.
of
The structure
gh Therefore
sg
According
(89,g- g,C2)
=
exists 4
=
=
is
sg,
gP(h)
easy to
closed
imply
we
that
to the
a
E
Proof. an
Since
tg
(G,
*,
mapping
+
4rg.
=
b.
gBh
from
polynomial
=
h with
in
(g*f)Bh.
Thus
Moreover,
the
sg 7 rational
tg, rg I
U
UgEGgBh is
Ig,
:=
may be
coefficients.
quotient
algebra
where y
c
U/BK
a
is
h).
W:
written
as
U -+
proof
yP(h)
G & Bh
h & P(h) y &
the
(2.32)
2sg
+
if
hP(h)
as
Y
if
=
11
U G\
jhj.
follows: 1
Y rz G
(2.33)
111.
of Theorem 1.
16.
and injective; well-defined is an homomorphism. algebra W g & v,r o 9 gw g 0 h,s o 9 g & s,ff 9 =
=
=
=
g ot
1, then our claim is evident. (i) If in (2.33) y We have to show, that gP(h) element. arbitrary is positively definite, =
=
gP(h) '
=
W is
gP(h) But g7j
3g + tg
a
=
2sg
g +
U may be
below finishes
Proposition
be
:5 2. Therefore, =
=
contains
(yP(h)) '
(i) (ii) (iii)
(sgc, sgc)
Therefore,
jSupp(sgc)j
4 and
each element
that
group x
may define
The claim
tg
61 + 3c + 2u.
=
=
P(h) is (gBh)(fBh)
of B that
Each element So
By the
=
where
see
subset
isomorphic
+
=
These formulas It
3g
=
h 1.
G
(=-
g
31
4
directly
hsg htg
as
have
Degree
-
hg
written
element we
of
=
other
Ig,
=
2sg; sgh
g +
Element
4sg + Arg, A E N. rg 0 sg and sgc 192: computed from the equality (s g T_g' c 2) 6. Thus Ig,sg,rg,,tg} C gBh, implying IgBhl > 28. Irgi 28. Finally hand, Supp(gg-) C Bh, therefore IgBhl jBhj have the and we equalities: following 1, tg, rg
==>-
On the
gB h
E
A is
The coefficient A
=
to
'
there
an arbitrary Proposition 13,
gBh 14, sg"g-g Proposition 192. Furthermore, \,.,c,_, a unique rg E B# such that
t9
7
Nonreal
Take
of sg, g E G and
definitions
Faithful
a
=
0
(UU) lul
+
=
1-
I)e an
+ 1d.
element
c
C-
B#
SITA with
3
Faithful
a
Nonreal
Element
Degree
of
49
5
Rom
Icl follows
it
or
(ii) lei
=>
(i).
n.
In the first
=
Since
lei
that
assume
(w, w)
6
b272 =.nl
+
b2b
(1
(i) (ii)
b3T
(1
If
B
Cb
If
c
=
!,
Set
n.
+
+
c
=
73 b3
=
the
(1) 72b2 (2) 72b (3) 72b2
+
yields z
EE
n1 +
=
n1 +
(i)
Proof. nb2
n1 +
=
Set y
(n (n l(b6
b2(Cb)
b2C
b4T
-
-
=
(ii)
1
n1 +
=
/F3 b3 holds:
1)b61 1)b7; + b7) -
=
(1
n
odd, 6-b
6-b
then
,
=
T-2b2
=
ln
-
=
E
2n
=
may
nl and
Supp(w)
=
and
n
=
+
(b6772b2)b6
+
moreover,
NB
Cb
Tb,
then
one
(b7,72b2)b7-
Cb and (b6, b2 bb2 b)
=
1) b2.
Then
(I
1)b3T
=
L-fl-y.
(b6 72 b2 6b)
=
+
+
lb4T
=
(1
1
If
gcd(l + 1, 1)'= 1, V Supp(z) implies z C-
c
E
B, then Icl
=
1)2b2
+
From
and I
21n.
If,
-
some c G
then
2
2
(c,T2b2)
we
for
n1 + lc
n(n +1)=(b2b,b2b)=(Cb,72b2)=n
shows that
lei
Hence
Tb-
=
(b2b)T
+ lb2 +
Finally, Izi The equation
either
n
-
n.
-
or
=
NB.
=
-
21 + 1 is
n
cases
b3T
:=
lb2C
=
-
-
=
3
=
following
=
=
12n. By LemmaI there exists d 6 1. Comparing degrees yields 1, idl
76 b2F2 b373 T6 =7 b7 T7,
and
n
Icl
or
=
72 lb5; b5 b6 + 76, b6 0
b6 + b7, b6
moreover
n1 + lc
n2)tt-2 2n, we get b2F2 n1 + le, and we are finished. 2n b272 1-te n1. Then lwl w
T2b2 If of
=
M
lb4-
+
1)b2
((b2T2, b272)
!
1)b3
or
b2b3
is immediate.
-
that
+
1 and
v
(ii)
=
2n,
M,
if
i+j
if
i+j=m,
if
i+j>m,
M,
yield a complex that
(V
that
el)
4(5eo
+
Y)
W, X
E)
-
is
el)
-
Y)
W, X
table
algebra.
(XY)
function
algebra
valued
51 + 4h.
=
integral
an
the constant (D
I. f 1,
=
Wnj
table
Wi+j+l -
'\+I
linear
involutory
standard
WO,
(Wn n
The n-dimensional
2
WO) Wi+j-n-I
and
if if if
integral multiplication i+j < i+j=n, i + i >
defined by T mapping algebra automorphism. -
n,
n
=
1 and U7
Z. Arad et al.
74
Proof.
proof
The
[321 (for
'm
'p(b)
polynomial Wewill
Example
(3)
T-
follows
0')
=
make
Let
3.
lei,
the
:=,8b,,+,
now
:=
the
with
-
construction
,
+
wi+j
-3el
WiWj
if 7if if
wo + w,,,
2wi+j-m-l
-
of
Example
has to
,
in
out
Al
-
of
3.3.
factor
=,3b,,+,
Example as
3
2wi+j+l
-
-wi+j-m is the unit
3T,,,()
X:=
'p(b) in
as
one
one
of
w,,,I,
-
-
that
ab' instead
N and define
mG
wo,
similar
the
as
difference
+ Al +
a
pattern
same
only
-
the
ab'.
0
2.
Example
2 and Y:=
i+j < M, i+j=Tn, i + j > m,
of Y.
where el ated by X and
By V and W, we denote again the algebras generThen the statements Y, respectively. (i) to (iii) of Example 2 remain true. Moreover, T- (3) is not isomorphic to 3T,, (iv) 3Tn( E)3 Tn(3). M 3
(i)
Proof.
(ii)
and
real, z+ m-i
we have 1
X
Y. Choose
c
zj- zj t
=
z-.z-. "
11
ztz,-.
1 E X
=
Y for
X
all
i, j
i
l5eO
+ vo + vn
3zP+ i+j+l
3h +
z-
zi+j-m
+
3zi+j-m-l
vi+j
+
4vi+j+l
zi-+j
+
3zi-+j+l
l5eO +
+
ZP+j-m I
=
(5eo
h2 This
proves
Again, the be linearly This
proves
+
wi
=
wi
=
-
(ii) above
el
)2
are
zm-i, zi elements
-
of
2wi+j+l if
wi+j-m
+
-
i
+j
< M,
)if
i+j=m,
)if
i+j
> M,
if
i+j
< M,
if
i+j=m,
)if
i+j>Ta,
2wi+j-m-l
Zi+j-m-l
wi+j
-
+
-
2wi+j+l
zi+j+l -
+ wo
el
-
wm=
4vi+j-,n-l
+ wi+j-m
3zt+j-m-l
2zP 3zp =
+
3zi-,
+
2zi-
25eo
+
zi+j
+
equations
(V
to
E)
a
2wi+j-m-l
-m -1
1
+ el
=
5(eo
+
el)
+
(V (D W, X Y) yield that the constant complex valued algebra
and shows that
extended that
eo and e]. ::--
the
permutes
zO+ + zz;
51 + 2h +
5vi +
wm-i
wo + wm=
-
-
vo + vm + 3
vi+j-m
=
Hence
-
WjWj
-
hzP
vm-i
+ z,
4vi+j-,,n-l
hzi
+
3e,
-
+
-
V (D W. Since
Z,-.+i+j+l
+
vi+j-m
5vi
for
zP
Furthermore,
+ wi+j
+
=
basis
+ WjWj
zP++j
VjVj
a
10,..., ml. ml. Then
E
4vi+j+l
+
Y is
Y.
10,...
(E
VjVj
=
I
vi+j
X
Clearly,
W, X
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
(iii)
of
is the
Faithful
a
Example
in
same as
Nonreal
Element
Degree
of
5
75
2.
IL, h, zt, zizi- defined that Assume i E T, (3) and ml. 10, si, via 0. Since both table algebras T- (3) are isomorphic are stan3 T,(A) 3 dard and homogeneous, 0 is an exact isomorphism. h2 and Clearly, 0(hi) T- (3)\j h2j with since ro and sm 70 are the only elements z of 3T,, ( )3
(iv)
We change
the
Example above by X2, h2, ri,
defined
by
2
in
Let
notation.
hl,
xi,
pi,
.
.
denote
us
the
elements
,
.
zp,
1, h, 3T,, ( a)3
elements
and the
qj
-
n
=
=
z
E
Supp(z-7)
M
follows
it
from
O(po)O(po)
holds.
0(po)
A:
either
that
O(po75o)
=
=
Let
=sm.
first
us
that
assume
A
case
that
Weclaim
0(pi) induction
For the
0(po)
B:
or
ro
5X2 + 2h2 + O(PO) + O(PO)
=
=
step take i
ri
for
all
E
10,
11
m
-
(3-35)
ml.
10,
i E
and
0 (pi)
that
assume
ri.
=
Then
ri
+
Since
30(pi+l) 0
+
O(qj+j) this
is exact,
the contradiction Next let
us
LpLj+j)
means
sm
=
assume
77o'
for
sm-i
=
E
10,
=
.
.
,
I
I
m
-
.
10,
i e
=
case
.
.
and
.
3rj+j
+
ri
=
(3.35).
o(p,,,)
In this
all
rori
=
proving
rj+j
O(T-o)
=
B holds.
case
step take i
For the induction
=
0(po)
=
that
O(pj)
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
Again
fact
the
But then
ro
A E xy
=
0
that
=
Tm6
R>o
and x, y,
'X+1
-\-'
z
2
Set
Proof.
(A
+
4
+
1)2
*9
z
x-z
A +
-
-
2
O(qj+j)
yields
is exact
be G
=
a
O(To-)
=
1.X 7+
O(popj)
=
sm-i
0(pi+,) O(pm)
such that
V,
A+1F. 2
=
=
2'+1
7 +
2
=
O(po)O(pj)
+ 3sm-i-I
so,
0
A-1 2
xT
x,
7 and z-z
Al + =
holds. 0
contradiction.
Suppose =
smsm-i
(3.36)
Hence
final
a
=
+ rm-i-1.
sm-i-I.
=
=
algebra.
table
standard
y. Then xT
2
u:=
+
O(po)
=
(U, V)
Let
Remark 7.
30(pi+,)
+
ri
that A
1 2
there
(x
+
7)
x-T.
Then
A2
_
4
1
1
(xx)7 (A
+ 4
=
1)
2
x7y A-1
A+ 1 2
u
+
2
xy'
exist
and
Z. Arad et al.
76
i.e.,
u
2
1'gl proving A2 + 1
A
2 so
(x, zT)
that
=
Proof.
If
(h, a7x-) (y, ah)
aTx)
b3 0 b, b4) T Bh 11, hj
b2
(ii)
jdx-j (h, a7a)
=
all
y E
B,
is
a
coset
and
if G
Yy-
=
E also
(bH)+Tb--H)+.
=
=
6b,
x-y consisting
3c, for
=
Now 3b 2+ 2bc
Since
H
set
:=
particular,
=
2c7b supply
b is nonreal
T(bh) (-bb)h
=
=
b(ch)
this
2-bb
+
=
c =
of degree 6-b-
is another
(bH)+G+
of
two
elements
(h, 6-b)
(y, y)
=
2,
B\jbj
,
3-bc
2cb +
c2
b2 and
5h + bh
=
101 +
+ T-h
for
c.
2(b
+
+ 2h
2
+ b +
b41
H
=
ch
(bH)+(bH)+,
(6-b, h2)
5
satisfying
2b + y for
=
=
3c+2b,
from bH consisting =
2G+ + 8E+ for
bh
=
then
some
H-
E+E+
some
y c
NB,
65
i.e.,
c(bh)
true
distinct
of degree
(bh, bh)
3b3
2b+3c,
=
G+G+
that
so
=
5 such that
holds:
H-coset
5 such that
=
and b 2
5 and bh
and
is also
T
of degree
following
2b + 3c and therefore
=
=
=
element
b-c
=
some c c-
bh
36c- +
we
4. In
from
=
basis
a
of degree
20 + y
If
degree
of
a
follows
it
Then the
jx, yj
x,y
(i) Wehave (bh, b) V Supp(y). Now
yields
51 + 4h.
=
51 + 2h + b +
=
subset.
Proof. b
bTb
that
closed
elements
=
1) (X, ZZ)) A,
-
=
that
=
two
xx
(A
4(x, ah) (ah, x) 0. For x =.a, this (x, ah) (h, dx-) 20 (ah, ah) (a7a, 51 + 4h) jahl. Hence a (I + h). 0 (aH)+ then
and
so
1b, cl for some nonreal 3h + c + Z!, c7c C2, b-c
=7 T
c
+
elements
16 that
=
=
=
If of
degree 4,
b4 =7 - b, where h is
=
bH
of
Suppose
Lemma 10.
(i)
(A
h2
h E B satisfies
aH+ for all basis
and
0
=
I for
=
=
obtain
we
0
that =
E B is
5 (h,
(a, ah)
yields
(XY, 27)
=
Romthis
assertion.
24.
x
=
of the
part
1.
-
(aH)+
then
1(aH)+1
A
(XZ, X7)
=
Suppose
Remark 8.
11, hj,
=
the first
=
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,
=
u
(bH)+
:=
y)(Y
+
T)
+
=
c
and
v
u-u
=
10(l
for
and therefore
al
2.
=
Nonreal
G+
:=
h)
+
Ci.
H-cosets
Purthermore,
N<j.
E
v
vU
=
aiCt
uv
i
b +
=
Faithful
a
2
Set
uv
w
2(b +T +
+
Since
(b, vU)
2v
c
+
we
we
may
product
than
greater
degree
5 whose order
Now Remark 8
is
JE+l
yields
%
=
lei
jej,
=
(b
+
c)(x
+
y)
=
4. divisible by 5,'we conclude that IzI 80. Thus all contradicting ajiE+l < Jul of a E cannot consist divisible by 5. Clearly, two possibilities: that we obtain the following not
=
=
uv
=
G+
lG+l
(ii)
4.8
Proposition
Rom (3.42)
=
2v +
8g
2(x
=
+
y)
+
(3.41)
8g.
assume
either
y)
+
=
G+ =yH+ =y+yh
IG+l
(3.42)
2x+3y and yh= 2y+3x.
(h, x-T)
xT
3(x
[201) implies
of
deduce
we
that
=
=
IFy
or
2
=
(h, y`y)
and
Cb.
from
is distinct
(h, x-y)
=
3
=
Rom(3.38)
(h, Zix-). (3.40)
and
obtain:
For d c
1b, cl
and
(dz, dz) Using (3.41),
=
we
z
E
xx
=
yy
=
51 + 2h +
xy
=
yx
=
3h + b +
Ix, yj,
(cFd, z-z)
we
(51
=
=
x
+
2g,
cy
c
=7
+ 2h + b +
b,
have
(a, dz)
shows that
this cx
Next
a
degree
that
note
xh=
us
of
substituent
degree
5 and
10.
x+xh=xH+ (cf.
such that
E+. Then
Set g:=
First
a
lwl
=
=
=
E
contain
11
whose
degree
24
degrees of elements of E are single element of degree 5, so 5 Ifl (a) E fe, f 1, lei
(b)
5 cannot
Then
z
that
assume
..,
=
Ci =A H
have
Eli=2 aiCp.
=
of
and
(3.40)
bH,
640. By Lemma I there exists a j E 12, 80 and (w, w) wlw) element E contains that an := 8. > Suppose Cj aj IWI is not divisible by 5. Since bH and G consist of elements of
of two elements
77
Z).
that =
5
assumption
G =A
means -
The
+ y.
x
=
Degree
of
Element
=
y +
-
as
< 2
for
2g, bx
=
+
c.
b
U,
-
51 + 2h +
all
a
y +
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
78
Z. Arad et al.
i.e., b4
8b + b + 8c + 3b3 +
(cf.
Therefore
VT
-bc
=
b(Cb)
b4
=
7! +
+
T2
+
6b-
follows
Hence it
Let
( t-x, d E
T, cj!J,
now
us
6-b)
55,
=
Supp(xb)
whence 50
have
such that
(xb
=
B/H
quotient
assume
we
-
37b
=
T
+ 3c + b +
+
T+
10b +
=
c
this
yields
b3
'z and
+
a-O 20
-
x)
90,
!
a
(x, xb) 2
>
we
(3.41)
,
3
x
=
.1u, 6
y
Supp(g)
proved
x-x
Bb y-y
=
and
(xb, xb) supplies
I
=
a
e already implies Finally, passing to the Iv and z -1g) yield 6
d
=
=
=
6
uUm
0
Suppose that 6-b'= 51+b+T+2h, b 2 T. Then 2 51 + 4h and (A, B) is one of 3, b3 =/= table algebras of degree 5:
Theorem 5.
=
B
=
11, b, T, hl, 3T
B -2-'x ments
is
a
..
b 2=3h+b+T
( )3
T,,,
(3),
cosetofH:=j1,hj
VjVj
=
1, h,
+
wi+j-m
hvi
=
2vi
+
3wi,
hwi
=
3vi
+
2wi.
3Tm( )3
enumerated
b,
:=
3vi+j-,n-l
wi+j + 3wi+j+i 3h + wo + w,,
Viwj
B --- x
vo
wo,...'
means
that
1, h,
T-
(3),
M
vo
the
vm, wm such
basis
ele-
Ivi, wil
that
and
vi+j-m
(iii)
This
2m + 4.
=
+ wi+j+l vi+j + 3vi+j+l 51 + 2h + vo + vm
wjWj
=
homogeneous
andbh=3-b+2b;
dim A
be enumerated
can
11b, b3, b4jj
3b3+b+b4, following
the
=
(i) (ii)
=
bb.
=
Lemma I
=
that
so
contradiction.
(withA
and Remark 7
that
Thus
3. But then
!
-
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
bTb
(3.37),(3.38),(3-39),(3.42)
from
contradiction
a
-bc
=
5b + 2bh + b 2 +
=
bc
1, h, b,
T, b4,
2g-y. Since b3 0 b,
(3.39)),
:=
b,
+
+ wi+j-m-l
=
+ vi+j-m-l
2m, + 4.
vm, wm such
i
+j < M, i+j=m, i + j > M,
,
The basis that
+j < M, i+j=m, i + j > m, i
)if )if if
+ vi+j+l
3wi+j-m-l
dim A wo,...,
if if if
elements
Ivi, wil
is
a
can
coset
be
of
SITA with
3
11, hj
H :=
ViVj
+ wi+j+l vi+j + 3vi+j+l 3h + wo + vm
=
+
wi+j-m
I
ViWj=
b5T5
and If b5
2b +
=
=
Hence
T,
3wi+j-m-l
3wi+j+l
Let
+
vi+j-m
2vi + 3wi,
hwi
=
3vi + 2wi. we
3vi+j-m-l
+ ch
+
wi+'j-.-l
have h 2= 51 + 4h. Lemma10
3b,5 (b5 -7 b), b5 h Cb.
(i)
and
holds
79
5
=
b52, 6-b-5
2 3b + 2b5, b=
)if 7if if
i+j <M' i+j=m, i + i > m,
if 7if 7if
i+j<m, i+j=m, i +j > M,
provides
following
the
3h + b 5+75
=
(3.44)
assume
multiplication
C
=
cH+
=
1c, dj
=
by (3.44).
C+
1C1
=
(3.45)
b.
and E of
two elements
exactly
determined
completely
is
that
suppose that
containing c
Degree
,
51+2h+vo+wm
=
may
us
+ vi+j-m-l
+ vi+j+l
b5 :
H
of
=
then
we
+
wi+j
hvi
Proof. By Lemma8 equations: bh
Element
and
WiWj
=
Nonreal
Faithful
a
3(c
le, f I
=
+
d)
from
distinct
H-cosets
are
5. Then
degree =
C+
ICI
=
dH+
d + dh
implies
(3.46)
ch= 2c+3d and dh= 2d+3c.
Set eo
:=
-'+h 6
and el
eo
(1
eic
-
=
51-h. 6
eO)C
=
By (3.46)
-(C 2
-
we
have
(3.47)
d).
Therefore, Ce
=
(eO
+
e0ce
((c
+
d)(e
+
f)
+
(c
-
d)(e
-
f))
2
(ce
+
df),
i.e.,
ce=
df,cf
=
de and
(c+d)(e+f)
=
2c(e+f).
(3.48)
Z. Arad et al.
80
Let
that
us
that
now assume
E is the
uniquely
C
J ,75}
=76
determined
Lemma 10).
By (3.48)
c(b From Lemma10
(c, cb) Now it
follows
either
Using
Lemma10 and
Ivil
5
=
=
1wil,
..
vovi
vj
j
(vi
0,
E
:=
viwj
.
,
.
3vi+l
+
vi
we
=
m
-
.
+
1
=
=
+
b5),
so
that
f).
+
F5 (b5) Cb5)
=
(c7c, Cb)
=
define
we can
JB/Hj
m:=
,
=
By (3.48),(3.50),(3-51)
Take i,
4(e
2c(b
=
(3.49)
(b5
7
(d, 65)
C-tC
55, that
=
b,
wo
+
recursively
21
-
in the
(3.50)
H-cosets
following
Ci
=
fvi,
I,
wi
way:
b5) CO := JVO) WOJ) 2& + 8CP+l and
:=
=
3vi+l
0 < i
wnvi+l
wi+1 + 3wi + vi.
=
Now it
=
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
+
=
+ + 3vi+j-,n-l vi+j-,,, for all + vi+j-,n-l + 3wi+j-,n-l wi+j-,n 3 T,,&E) Hence we conclude that B 3
vivj
=
vrnwi+j-,n
i 10'...' Example 2). T,n(3) (cf.
i, j
vnwi+l
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 + 6wi + 3vi+l + wi+j 10vi + (cf (3.52)). Multiplying 3w,vi+l w,,,,wi+l the second yields first vnvi+l wnwi+l equation by 3 and subtracting =
=
=
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. > 3-Tm(sE) G Therefore, 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
Computer
and
Science
University Ramat-Gan 52900, Israel Bar-Ilan
2Vogt-Groth-Weg
44
Hamburg, 22609, Germany
Introduction
4.1 This
(A, B) real
with
deals
chapter
L(B)
with
==
element
basis
Supp(b 2)\111
obtain
two
III
and
degree
b of
JbI
=
Theorem I. sume
Suppose fl,
that
B,,
ad
(A
-
=
3)
such that
=
Az-gxlxl,
x, y,
z
B,
cz
I
[b 2]
(b 2, b2)
[15,
proved by
51]
75
5252]
65
53
Suppose that (A, B) a prime integer
jai
a, d E B with a
=
=
d
=
Al +
Idl
=
U);
is
a
+
a
=
=
fl,
integral
Ivi
A and a7a
Ba
Theorem 2.10.
standard
such that
or
[331,
Blau in
H.I.
: =
a,
\
-
Al +
dj
(A
for -1
and a2
all
1)d. =
As-
algebra.
table 1
v
E
B\f 11.
Then either
Al +
(A
_
I)d,
2)d.
Case 1
4.2 In
was
al (that is, I)a + d, d'
ISuPp(b 2)1
(i.e.,
Table
A > 3 is
that
3
identity
(x, z-y)
=
21 [15 theorem
b G
types for b 2:
possible
1
The next
all
5 and width
(xy, z)
integral GT-algebras B* which contain a faithful
of standard
> 4 for
9 135- Using the basic
A,y-,Izl we
classification
the
case
support
1 of Table
of
b2
or
I it
not.
is useful
to
Wehave the
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
51 + 3b + c, c G B5\f bl, (a) b2 51 + 3c + b, c C- B5\jbJ, (b) b2 (c) b2 =51+3c+d, c,dEB5\jbj, =
=
cod.
of the cases (a) and (b) we only deal with case (c). For a discussion [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 algebras of dimension 4 and 5, respectively, will
Wewill refer
to
was
lead
be closed.
51 If b2 following
Theorem 2.
then
(a) (b)
of
one
B
=
cd
=
B
=
C2
=
de=
11, b,
Proof.
bc
d, d2
for distinct
real
basis
elements
of degree 5,
holds:,
cases
=
3b +
=
51 +
2d, bd c
b, c2
2c + 2d +
=
51 + 2c +
=
2d,
+ d + 2b.
3b + 2d, bd 2e + 2d + b, be 11, b, c, d, ej, bc 2c + 2d + e, 51+2c+2d, cd 2e+2c+d, ce 2b+2d+e, d2 51+c+d+2b, 51+b+c+d+e. 2b+2c+e, e2 =
=
=
=
=
(A, B)
Set
Case 1.
dj,
c,
2b + 2c +
particular,
In
+ 3c + d
=
the
w
bc
w
=
2b2, b2
C2
=
and
homogeneous.
five
3b. Then either
-
=
25 +
and therefore
is real
T2 0
3(c, C2)
w
2b2
=
or
E B
w
or
w
b3 + b4, b3 0 b4
=
b. Then +
(d, C2)
(b 2, C2)
=
(bc, bc)
=
65
=
Now
2c + 2d.
151 + 11c + 6d + dc
5c + 3c
=
2
+ dc
b2C
=
151 + 9c + 3d +
b(bc)
=
=
3b 2+ 2bb2
2bb2
yields 2bb2 In 5 or
particular, -
(d, C2) (d, dc)
=
Case 1. 1.
b3
=
35 +
we =
73 :A
3
have 2 (d,
(d, bb2)
(d, dc)
=
I
=
(d, bb2)
and
d, and bb2 10(d, d2) yields (d, d2) c,
(b2, bd)
2b2 + b
+
=
=
(d, bb2)
6b3
+
(d, dc).
that
Z
2.
=
2c + 2d + =
=
2N, N =.T4- = 6 b,
51 + 9d + 7c +
+
(4.1) (d, dc) : 5 (c, cd) 1 and (d, bb2) (d, de)
Since
either
-
=
=
=
2
-
2c + d By (4. 1) we have dc b3. Hence 45 (dc, dc) (d 2, =
=
1. Romthis
(bd, bd) Since
2c + 3d + de.
3 bb2) (4. 1) shows
3, equation =
=
=
2 and
d. Set
(b 2, d2)
(b, bd)
z :=
follows
it
=
=
5d + 3cd + d 2
=
51 + 7c + 5d +
=
2b3)
C2)
=
45. b 2)
(d,
b2d
2b3
+
that
=
d2 -51 -c-d.
=
=
=
+
b(bd)
2bb4
1, this
means
bd
Then =
b
2
+
2bb2
+
2bb4
=
SITA with
4
yields
z
40
2b5, b5
=
Thus
=
=
5(b2, be)
+
10(b2, b2C)
+
(b2, b2C)
(c2, bb2)
=
we
(bc)b2
=
b(cb2)
=
b2= 51 2 d
b4
=
(cb, cb2)
15(b2, be)
c
+ d +
x
-
so
b(b2C)
=
(bb2)C cb3
that
and 45
=
+ 8b +
Le.,'
bb3
yields
=
b4C
6b4 b2
=
(b4,
+
+ 2cd +
cb3
101 + 8c + 8d + 5b3 +
=
+ 2d +
2b4
11, b,
dj
(b2, bd)
=
=
-
ROM
2b57
c,
=
bd
and
bb2
=
=
2c + 2d +
have obtained
we
2,
=
we
b3
the
have
55,
101 + 8c + 8d +
5b3
+
2b5
101 + 8c + 6d +
4b3
+
cb3
(b, b2d)
(bb2, d)
=
that
we see
+
2b4
+
=
b2b2
=
b(bb2)
b2d
=
=
=
2, (b2, b2d) 2b6, b6
2b + b2 +
=
=
(d, b22)
76 0 b, b2
=
(bd)c
=
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- Wehave b2 + 2b6) -
=
10(b2, cb2)-
2b + 2 b4 + b2
(d2, b2)2 =(62, db2) 0
=
2b2+ 2b4C + b2
2b6
2b + 40
=
(b4, b2C))-
deduce
20 +
implies
=
2d + 2b5 + b3- Sincewe
202
=
30 +
=
+
from
(d2, b2)2 =(b2d, b2d),
=
ROMthis
9b2
=
10((b2, b2C)
+ 6d + 3 b3 +
b3 and b
2(b5,b3)
2C2
10 +
=
=
=
compute
we
b5-
+
(b, db2)
Since
Next
5
3
(a).
in
b5 0 b3- It follows
that
=
b5.
b2 c
b. Hence B
=
45 +
b3
+
2 and
b2b
+
2b3
10(b2, cb2)
+
I? (b4, b2 C)
2b2+ 2b4b
given
0 b2
Case 1. 1. 2. d
Moreover,
=
Degree 5 and Width
2b4, b2C)
3.
6c 3bb2 + 2b2= 2
constants
structure
+
(b4, b2C)
+
and
2b2
+
10(b4, b2C)
b2. Then b5
=
c
(b
=
of
2d +
=
+
(b2, b2 C)
conclude
Case 1.1.1.
implies
(bd, b2C)
=
bb4
and
(dc, bb2)
obtain
we
d,
c,
=
Hence 40
T5 :
=
Real Element
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
get bb6
shows
=
=
b(b2d)
=
that
B
=
Q
=
=
2bb2
+
2bb6
=
3bb4
+
2b2b4
b2 N)
=
2b2
bb2
2c +
+
=
2b3
2b3
+
(b37 b42),
bb4
b(64)
+
=
8b3
=
+ 6d +
2c 2+
=
4b3
+
4b,5
+
2bb6
101 + 8c + 4d +
5b3
+
6b5
b3
+
2b5 and from
=
5c +
=
2dc +
b3C + 2b5c
2b42
2bb4 +
=
C2
+ dc +
d2
+
265
263
=
=
d 2C
+
b4b2
=
3
=
that
=
2b42
c2
=
yield
IwI.
(bc, bc)
c2
51 +
=
51 +
c
5b3 + 8b5
64 + dc
3 Now (b, bd) a (bd,bd) E 1105,85,551, =
0
contradiction.
Case 2
4.3
For Case 2 of Table
b 2= 51 + 2b + b 2= 51 + 2c+
(a) (b) (c)
we
only
treat
of
classification
2c, c E B5\fbl7 2d, c,d E B5\jbj,
Chapter
the
(a)
cases
(b)
case
Finally,
subcases.
elements
(c, cd) then
(c)
(d)
> one
B
=
cd
=
B
=
C2
=
dt
=
B
=
df
d
and
B
=
=
=
(b).
Wewere not able to
Theorem following solve will completely
Theorem 5
3
gives
give
a
answers
complete for
(c) by using
case
many
results
=
c, d
=
11, b,
dj,
bc
11, b, c, dj, 2c + 2d +
b, d2
c,
3d,
2b +
=
bd
2b +
=
3c,
C2
=
b2'
2c +
3b,
51 + 2d +
2b,
cd
=
+ 4d.
11, b,
c,
bc
d,tj,
2b + 2c +
=
d, bd
=
2b + 2d + c,
C2
=
51 + 2c + 2b.
2b + 2t +
bc
51 + 2d +
2c + 2d +
2t, cd 2d + 2h + t, t2 f 1, b, c, d, f 1, bc
=
51 +
c
2b + 2t + c,- bt d2
d, bd b, ct
2c+ 2h + t,
2c + 2d +
51 + 2c +
t, 2t,
+ d + t + h.
2b + 2f + d, bf c + 2d + 2f, f, bd 2f +2d+b, d2 51+b+c+d+f, 2c+d+2f, cf 51+2d+2b, cd 51 + 2c + 2b. 2b + 2c + d, f2 2 51+3c+d, cd 3b+c+d, 2b+3c, c 2b+3d, bd 11, b, c, dj, bc
d2 =
(f)
=A
c
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) true: cases holds of the following
b,
=
B
C2
(e)
=
3.
d2= 51
(b)
Z
but the
Theorem 3.
(a)
main subcases:
following
have the
b2=51+2c+2Z,c(EB5,ZOC-
Wewill
of
I
(c, cd)
=
2b + 2c +
=
=
=
=
51 + 3d + =
1
=
[The tablealgebras
=
=
=
c.
(d, cd),
(c, c2)
defined
in
=
(c)
0
=
and
(d, d2). (d)
are
exactly
isomorphic.]
88
Florian
Proof.
Set
l3iinger be
x :=
20 +
(x, x)
20 +
(y, y) 0
c
b
(bd, bd)
=
(b 2, d2)
=
(x, x), (y, y) Rom (4.2)
(d, cd).
(A) (B) (C) (D)
(c, (c, (c, (c,
cd) cd) cd) cd)
=
=
=
=
one
obtain
(c, cd)
of the
following
(A)
51 +
I
yields
10((C' C2)
+
(d, C2)),
(4.3)
25 +
10((c,
2)
+
(d,d 2)).
(4.4)
G
d
=A (y, y).
35
f 15, 25,
Hence
45 1.
(4.5)
! 2. We can
(c, cd)
that
assume
>
holds:
cases
Case
(B) Applying Set
45 +
(4.3) cd
z :=
(z, z)
=
Since
and
(4.5)
3c.
Then
-
c
(4.4) 9(b2, b3) and
contradiction
the
c.
25 +
(d, cd)
+
51 + 4d.
=
Using (4.3)
(4.2)
51 + 3d +
(4.2)
(d, cd).
=
+ 3d.
c
(d, cd)),
4; 3; 2;
Wehave 0
Case
d2
=
we
Hence
+
=
(x, x) =,4
yields
10((c, cd)
=
,
(b 2, C2)
d
then
(b2 cd)
=
=
0
2b,
-
(be, be)
=
=
bd
:=
(be, bd)
(x) y)
20 +
Note that
2b and y
-
(cd, cd)
(c2,
=
=A d,
=
Theorem I
obtain
we
x
=
yields
cd
3b2 and
y
4c + d, 3b3. Now
==
=
30.
we
get
d 2)
=
[x]
[5 3), (C' C2)
=
25 +
15(d,
d 2) +
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)
and
(c, d2)) yields
(4.5),
f 10, 201
E
we
a
E
J([5 3] 2,2),([5
3], 1,3),([52
51],2,0)1.
contradiction.
(c, c2))
([x],
obtain
E
([5 3],2),([5
2,511,0)1.
5 1 + 2d + 2c. Now (4.2) reads 3 (e, y) 3e, jej 57 C2 (CI) x so that (d,cd) G 10,31. If we assume that (e,y) : - 0 and (equiv10(d,cd), 3e 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 +
2cb, + 3ce
2y,
shows that
c(cb)
=
we see
that
(d, d2)
=
=
=
=
y
C2 b =
4, i.e.,
=
5b + 2db + 2cb
3f for d2
some =
real
51 + 4d.
=
f
13b +'6e + 2y, E
Now
B5\f ej. we
But
compute
SITA with
4
(cd, ed) for-
(c 2, d2)
=
some
real
g E
a
be
=
3c +
gc
us
=
g
derive
=
b,
Case Z :=
so
e
2
3d +
B
(C2) x C2-5172d,
=
2c2
+ 2cd
b 2C
=
5c +
=
101 + 4c + 4d + 3be
3gc
=
2
3gc
c(cd)
=
2c
=
101 + 13d + 4c +
b(bc)
=
2b
=
2
89
3
2c+3g
=
+ 3be
+
=
c2d
=
=
5d + 2cd + 2d
2
6g
2g. Rom
4g
3c 2+
=
11, b,
=
f,
2e +
dj
c,
T
e
=
d
e
(cb)e
=
by
jej
and
=
5
=
as
4g
6c +
implies
multiply
=
f
=
stated
+ c
in
3e2,
and
(a).
0. Set if 1, (c, c2) d (d, 2)d- (c, d 2)C. =
d 2- 51
w:=
2
2be + 3e
elements
=A f =7,
(d, cd)d-2c
=
Theorem 5. This
and the basis
=
cd-
v:=
c(be)
2cg
d2 and therefore
=
that
6g
Hence cd
65.
=
5 and Width
2g. Now
151 + 6c + 12d + we
Degree
of
(51+2c+2d,51+4d) B5\ICI. The equation
101 + 4c + 4d +
gives
Real Element
=
101 + 13c + 4d +
yields
Faithful
-
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 +
+ 2c +
2d)(51
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 4e 2+
(c,
4ef
+
f2)
=
d 2)+
4(c,
2
5
(z,
z
+
(4.8)
V).
(C' f2) 0 4 since 65 > 25 + 10 (c, f2) + (e, f2) is even. Clearly, (f, Cf) Hence 5 (C' f2) E 10,21. We consider the three (fC, fC) : (f, Cf)2. (Z' f2)
Hence
=
=
possible
subcases.,
(C2. 1) (d, cd) (C2.2) (d, cd) (C2.3) (d, cd)
=
=
=
2; 1; 0.
Case
(C2.1)
(4.2) (4.4)
implies (2e + f, y) 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),
Bfinger
Florian
90
we
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,
2ct
=
101 + 5d + 4c + 4t +
=
2d
=
101 + 5c + 4d + 4t +
2
+ 2dc + dh
Next let d 2C2
(51
=
2t) (51
+ 2d +
(2c
the
26 +
2
cd
=
5c +
2c2
Let
us
h2
h)2
+ 2d +
of
occurence
4((t,
h)
+
(C' t2))
4((t,
h)
+
(d, t2))
first
multiply we
that
assume
0
+ 2c +
=,A (h, ct).
(4.9)
2t) 4t2
401 + 24c + 24d + 16t + 8h + 4ch + 4dh + h 2.
=
and d in the expressions
c
(c, a)
=
stated
as
b
have either
(c,)3)
h
a
and 3
8(t, h)
28 +
h 2)
2
=
=
(d,h 2),
yields:
(c, h2)
+
ct
=
[ch]
us
=
2c + 2h +
t,
[5 2 dh
52 =
a
=
=
51]
=
101 + 6d + 9t + 6h +,2c
c
=
=
51 + c i.e., t2 like Bh multiply =
(bh, bh) means
=
[bt]
(b2, c
+ d + t + h. in
h 2)
(c) =
with 65
(d,h 2), i.e.,
=
and the elements
(bh, bh)
of b. Since first
(4.10)
Then
h.
5 1 + 2c + 2 d
(c2,h 2)
2h +
2
=
(4.10)
h 2).
(d,
case
of Bh
(b 2 h2)
=
=
,
done,
we are
65
in the
contradiction.
54
t
dj
c,
In the
+
(c, h2)
implies
11, h,
=
5 2].
[5 3
=
h2
i. e.,
(4.10)
Bh
8(t, h)
28 +
h instead
with
[bhl
or
(ch,ch) shows
(d,,3)
h. Then
=
second case, Lemma2 yields Let us now assume that
(c,
a)
Therefore,
(b)
in =
='(d, t
b 2.
51 +2c+2d=
=
+ 2d
2ct,
and 26 +
2dt
compute
us
=
(cd)2
Counting
(cd)d
=
251 + 18c + 18d + 20t + 4h + 4dt + 4ct +
,8
2+
2dt,
2h + 3c + dh. Hence
=
(h, dt)
a
5d+2d
=
and
101 + 8c + 4t + 4d + 2h + dh
i.e.,
(dc)c=dC2
[dh].
=
45
=
(c2, d2)
=
means
2 f [5 52 51), [102,51]1.
[bh]
Bh
=
we
=
7t + 4d + 4h + Hence
=
[53,
(d, t2)
and
=
ch
2h + d +
=
Finally,
c2t
=
we
2t, compute
5t + 2dt +
2t2
2t2,
f 1, h,
now
=
(cd, ed)
obtain
c(ct)
of b. Let
But
=
2d + 2h + t.
+ 2ch + ct
h instead
(C' t2)
implies b 2. Now
Using (4.9),
+ 2t and dt
2c2
=
5
21
2
c,
us
d, tj
and the
assume
and
yields
that
(bt, bt) b E
=
elements
b
of
Bh. Then
(b 2, t2)
Bh contrary
=
to
45 our
SITA with
4
either we
(d, cd)
(C2.2)
Case
y
=
2f
[y] (d, d2)
have
yield
(C2.2.3)
Hence
2.
(d, d2) (d, d2) (d, d2)
(C2.2.1) (C2.2.2)
(C2.2.1) It B5\jc, d, bl.
Rom (4.2)
1.
=
+ g, g E B5 [5 3], so that
2(h, w)
15 +
20 +
=
bg
b(bd) bg
d 2)
(bf,
45
(be, ec) (be, c2) follows
yield
=
In
10, so that particular,
(4.4)
and
(4.5)
possibilities
+ 2d + 2h for
c
=
=
(bd7 bf)
=
=
10 +
=
20 +
some
real
h E
(cd, cd)
(c25 d2)
(b 2, f2)
25 +
Hence
we
> 4
we see
which
(fC, fC)
=
that
=
(C7 f2)
=
=
(4.12)
=
(4.13)
=
+
2(e, cf),
(z, w),
35 +
10(C' f2)
+
=
=
w
=
0.
b. If 2k
If
(4.15)
(d, f2),
(4.16)
=
z
=
=
so
z,
(C7 f2)
10(d, f2)
51 + 2c + 2k.
(4.17)
2k for that
that
=
+
(Z' f2)
some
real
0, (h, v) (4.15) gives a 0, then (4.8) and
we assume
the contradiction 25 +
(4.14)
-
and therefore
k
have
f2
(g, fd),
2(e, ec) + (f, ec) (z, be) (z, be 2c).
(h, v) 0
supplies
y2' C2)
+
20 + =
(z, z) 0 10, =
(g, f c)
10 +
=
=
that
2(d, f2)
(4.11)
=
=
(4.17)
20 +
=
(be, cf)
=
101+6c+9d+2v+2w, 2w,
2(c5 f2) + (g, fC) 2(e, f d) + (d, f2) 20 + 2(h, v),
(2b + 2e + f, ec) (be, 51 + 2d + z)
=
b 2d=
=
d + 2v +
d}. Our aim is to prove that k and and (4. 11) yield h
85
+4ef
=
Furthermore,
101 + 4c + 4d + 2be +
=
=
X2
two subcases
=
65
the
f2
201 + 8c + 8d + 4bx +
=
z)
+ 2d +
and d's
+
z
we
+ 4bx +
2d)(5
(c, 4e2
4ef
(C2.3. 1) Looking 0 and
X2
+
8be+ 4bf + 4e 2 + 4ef +
2cz + 2dz.
i.e., bf
4b2
(2b
=
+ 2cd
2g
we
+
h,
obtain
Bilnger
Florian
98
(C2.3.2)
2 (mod 4), so that equation of (4.30) supplies (C' f 2) 0 (mod 4). By equation of (4.30) implies (d, f2) this means (d, f 2) 0. Therefore observation we obtain
The first
2. C, f 2) the previous
=
(C' 62
+
ef)
+
20 +
Moreover,
=
The second
=
(g, v)
=
(v, v)
(d, e2
and
(cd, cd)
=
ef)
+
(c2, d2)
=
3 +
=
(g, v)
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 + so -
(d, by (4.32) that
(h, d2) =7
e
2
e
(v,,v)
(e, de)
0 and set
a
(bc, ed)
=
ef)
+
=
(g,
+
=
h be the
Let
15.
this
d
we
obtain
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,d 2)
+
ef)
+
5, (c, e2
'f
+
-
(g,v)
3+
=
(f, de)
+
10(g, d2).
+
Combining
> 1.
(e, de)
5 >
10(d, d2)
5 +
=
then
,
de
+
ae
=
3f
This
.
means
10a +
5,3
Hence
(a,,O)
(4-31))
=
G
we
-
(be, cd)
h
a
1(0, 5), (1, 4), (2, 3) 1.
0. Set
u :=
(be, be)
+ 45
Then
JwJ
and
e2
=
that
Then
w
=
But
v)
(be, v)
20 +
=
(c, be)
as
-
d2
-
(d, u) JwJ
=
=
=
50,
a
=='10 and 2k for
(h
51
+ u, ad +
20 +
:
The
2cl
-
(c, e2)
2 and
=
w)
Jul
and
of
(4.33)
h).
0
=
=
(Cf.
let
2(h, d) i.e., c, gj, 3, (C' f2)
B5\fd, (d, ef)
k E
Now
=
=
2 and
w)-
(a, u,w) < (2, 3). (a, 0) that
assume
(h ', d)
(h
+
that
assume
2e +
(4.33)
,
+ u,
(h+u, w).
de =
=
Then
ad.
-
first
us
us
+
2),
=
Let
Finally, =
e
51
becomes 60
contradiction. reads 40
2-
=
25 +
=
e
10(d,
(d 2, e2) a(h, d) + (h
(de,de)
=
25 +
w:=
(4.33)
some
51 + 2d + 2k.
=
=
h and
-
the choice
by
0
15
(b2, e2)
=
(a2 +'32 )5 25 +
(Recall (1, 4). (d, d)
2c +
have 35
4 clearly degrees of basis elements we deduce in any case, since But this is not possible
x
that
(bx, bx)
the definition
:! 2. Hence table algebras
is
by
the
basic
our
(y, bx)
then
case,
assumption
on
(bx, bx) > (y, bX)21yl > 25.4 e B,\11, bj has degree 5 so nonreal whence (c, xT) as c is
=
b
=
h
)
Let
Set y
or
finally c2 C -
-
that
a
=
(b2, x y)
=
we
=
obtain
25 +
B
yields
that
By (4.38)
we
=
B is
20(c, t-x)
g). that
(x, x)
=
10.
have
< 65
Looking at either Y4 (and
B,.
a:=
(Z,
c
2) 7
0.
ccc-.
45 so
=
asume
us :=
particular
of the
Y5 (and b
is
minimal
2. But
=
(C7,c d-)c
now
(4-38)
c
2, c 2)
yields
+
a2 )5
the contradiction
+
(y, y) 10
=
(2d
+ e,
2-d
+
j). 0
of Primitive
The Enumeration
5
Commutative
with
Association
Non-symmetric
a
Valency,
Schemes of
Relation
Most 4
at
Hirasaka','
Mitsugu
IDepartment
of Mathematics
and
Computer Science
University Ramat-Gan 52900, Israel
Bar-Ilan
2Depattment National
Taipei,
University
Taiwan
Introduction
5.1
(X, G) be (X, g) is
Let
Then to
of Mathematics
Taiwan
all
find
an
hypotheses
about
W.J.
numbers
intersection
the whole structure
and to determine
[62],
scheme in the
association
regular digraph for each regular digraphs which might
g E
a
Wong classified
all
or
sense
of
G. It
is
be
an
an
induced
[641 an
element
where X is finite.
problem
interesting of G under
certain
subgraph of the graph,
(X, G).
of
primitive
permutation groups valency 3, [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, it is known that the of valency 4. As a generalization of a fix-point stabilizer of each primitive transitive cardinality permutation non-trivial valency (see [37]). group is bounded by the minimal In 1981 L.Babai [261 proposed the following In
with
a
2-orbit
Conjecture the maximal
non-trivial
1.
of
(L. Babai) valency
elements
transitive
and in
Let
of G is of G.
(X, G) bounded
be
by
a
primitive function
a
association
of
scheme.
the minimal
Then
valency
of
seems to be quite However, this conjecture open, and we need to make the or comproblem easier, for instance to assume that G is a P-polynomial to impose some restrictions numbers. If G is or on intersection mutative, relation of valency and contains at most 4, commutative a non-symmetric then Conjecture 1 is true (see [46], [47)). In this chapter, commutative association schemes we focus on primitive whose valency is small enough. We observed relation with a non-symmetric series of such schemes with growing I GI are translation that all known infinite 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
moreover
which
the
is
give
be
primitive
IGI
commutative
is
large
asso-
enough,
then
scheme.
Conjecture
that
a
and
non-symmetric
association
prove
we
we
(X, G)
Let
g E G is
the enumeration
scheme of 2-orbits
2 is true
of all
of
a
valency of g is at most schemes, each of group on an elementary
if the
such association
permutation
abelian
group. Another motivation
schemes with all
erties:
a
of this
chapter comes from research of association which satisfy the following points, propall relations of odd valency are connected;
number of
prime
relations
non-trivial
If an association scheme is a translation scheme over a non-symmetric. of association scheme cyclic group of prime order, then the whole structure determined is uniquely elements. Asby the minimal valency of non-trivial schemes of this type are called sociation cyclotomic (see [36, Cor. 2.10.2]). schemes with a prime However, it is still open to determine all association and there are only classifications number of points, if the cardinality of the of the are immediate point set is small enough. The following consequences of this chapter: all association enumeration schemes with a prime number of points relation of valency 3 are cyclotomic; a non-symmetric containing scheme with a prime number of points there is no association a containing relation of valency 4. non-symme ric In contrast to the situation of the previous paragraphs, any primitive scheme with a relation of valency 3 is a P-polynomial, symmetric association which is proved in [63]. of distanceCombining this with the classification regular graphs of valency 3, we see that- there are exactly four isomorphism to the following classes corresponding graphs: a complete graph with 4 points; a Coxeter a Peterson a Biggsgraph with 10 points; graph with 28 points; Smith graph with 126 points (see [36, pp. 1791). In this chapter we use notations of [64] which are slightly from different are
'
used in the Introduction
those
Let X be
a
the empty set. Wedefine 1x Let r C X x X be given. x
E
X,
define
we
Definition
(i) (ii) (iii)
Ix
E
Wedenote
xg
following
E
:
=
called :=
pl99,
a
We set
r*
X1 (x, y)
c
The pair conditions:
of X x X which does not contain
partition
I (x, x) I
E
x
:=
X 1.
J(x,y)
I (y,x)
E
rJ and,
for
each
rJ,
(X, G)
is
called
an
association
scheme
if
it
G;
For each g E G we have g* cFor all d, e, f E G and x,
(X, Y) are
ly
:=
( [64]).
I
the
satisfies
xr
book.
to this
set and G be
finite
E
f
y
E
X,
lxdnye*l
is
whenever
constant
-
the number
the
G;
I xdnye* I
intersection
which
is called
with
numbers the
valency
(x, y)
C
f by
of G. For
of g.
pf, d
and
each g E
Jpf,,dj
G,
we
d,
e,
f
G
GJ
abbreviate
(X, G) (X, G) (X, G)
Wesay that Wesay that We say that
where G'
connected
[64],
1 011.bwing
I
we
=
f P,d for
all
Schemes
d,
f
e,
symmetric primitive G f 1x 1.
is
107
G.
E
if g = g* for each g c G. if, for each g G G' the graph
is
:=
f if Pde
commutative
is
Association
Commutative
of Primitive
The Enumeration
5
(X, g)
is
-
define
product
the
If
de:=
of two elements
Pfe
GI
C
e
E
G to be
7' 01-
d
For each (x, y) E X x X, we (x, y) by r (x., y) Let (X, G) and (Y, H) be two association isom4phic to'(Y, H) if there are two bijections such that (x, V) E g if and only if (O(x), 0(y))
denote
shall
d,
(5.1) element
the
of G
containing
-
each g E G. For each g
G,
E
for
that,
0 E
X
:
p(g)
(X, G)
Wesay that ---+
for
is
Y and p : G --+ H all x, y E X and for
'
define
we
adjacency
the
f Note
schemes.
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 GI is a subalgebra algebra of (X, G),
follows
with
table
Example
We denote
scheme then
Given g
(X, H)
E
be
an
G, a
ation
it
we
shall
g
the Bose-Mesner
orbits
L (X,
write
L(X, g)
is
a
ordering
g)
for
the
set
of association
poset with
respect of partitions
the set
on
(see [28])
scheme
association
schemes with about
on
a
non-symmetric
intersection
numbers
(see [46], re
*lation
of
..There
exists
a
to the
schemes
paxtial
of X
maximal
order
X which
x
element
following
the
[471)
(see [201, [331)
are
valency
as point example frequently
set
commutative primitive of valency at most
relation,
already
the elements appears
in
and the structure
known.
in
These results
associ4.
of
Some a non-
show that
gives a great amount of restricscheme, and makes it Possible of an elementary abelian group. this chapter.
at most 4
of the association
the whole structure
characterize
The
it
-
refinement
symmetric relation a non-symmetric to
by JA_' I
C spanned call
g),
results
tion
we
and A be
set
X and the
which is unique (see [60, pp. 18]). is to determine The goal of this chapter
L (X,
over
scheme.
association
finite
g E H. Then
with
where -< is
might
a
formal
the
[27],
a
an
of
the Bose-Mesner
spanned by
space
a finite transitively group acting of A on X x X is an association if X ts a finite by 2-orb(A; X). Furthermore, group algebra of 2-orb(A; X) is isomorphic to the Schur ring sums of the orbits Of Aid,, on X (see [58]).
Let X be
1.
Then the pair
X.
on
of
the vector
C). Following
by CG. In the sense of [20], CG is a standard basis JAgI g E GJ. The following distinguished
denoted
algebra example
integral is a typical
that
kat(X,
of
Hirasaka
Mitsugu
108
Example 2. Let Fp,,, be a the symmetric group S,,+I, dimensional
product of Zp' for some odd prime p by of which on ZPn is given by its natural n-
the action
Then
representation. scheme with
association
(p, n)
semidirect
2-orb(rp,n;
Zpn)
is
primitive of valency
(3, 2).
=
For each g E G, the ordinary
shall
we
denote
just
write
product
A,
by A
B
A,
of
g instead
of matrices
commutative
a
relation
non-symmetric
a
for
as in (5-1). product of two relations from the theory of standard are results following
and
convenience
B in order
-
unless
+ I
n
distinguish
to
between the The
bras,
which
(H-1.
is
connected
a
isomorphic
rank
at most
The
following
with 3. Let
a
(i) (ii) (iii)
=
an
immediate
(H.I.
Blau)
Then
maxgEGx Pgaa* a
-
relation
3
-
1X,
al
! 2 for
Although
a
properties result
the
Theorem 2.
Let
non-symmetric the following:
e
E
in
-
ai
JgJ
this
-
Xa2
a*
-
fg
E
G,ng
g c
GJr,
relation-
are
(X, G) relation
a
9
of
=
an
determined
chapter
our
mingEGx
3 which is not
-'g
=
necessarily
=
ai-1
a2:7
3 and
a*;
(5.2)
+ ai-2
is
31
=
31
U
13 1xJ; -
(e, f)
then
E
scheme
association
[46],
in
shorter
J(d, d*), (d*, Ix)J.
shall
we
than
one
satisfying
prove
given
in
the
[46]
the same
The
main results:
be
a primitive of valency
x C3; Zp) for some p of Zp and C3 : Aut(Cp);
2-orb(Cp tion
proof
enumerations
a
C
Theorem 1 is
in
scheme
association
3. Assume
Then
a.
non-symmetric
as
since
following
d,
some
-
:=
:=
Jaiji>o
Ifpdf de
commutative
1;
=
=
=
a
of valency of valency
2a* + a2 for some a2 E G with For each i > 2 we set
a
where ao
(i)
be
relation
following:
ai+1
(iv)
(X, G)
Let
have the
we
of Theorem 1.
consequence
non-symmetric G be a non-symmetric
connected.
alge-
[33]).
than 3.
is 1
E
table
Blau).
connected
a
integral
(see [20],
relation
greater
Corollary
schemes
association
M
Introduction odd prime
for
Let (X, G) be a commutative scheme association 3. Then CG of valency 3. Assume MingGGx ',g to one of the following: O(C3, H) for some abelian group H of 2; or 0 (S3, Z,2,,,) for some m 4 where 0 is defined in the to this book. Furthermore, then m is an 'if (X, G) is primitive
Theorem I with
reformulated
are
_=
commutative
3.
Then
(X, G)
1 mod 3 where
association is
Cp
scheme with
isomorphic
is
a
regular
to
one
of
permuta-
2-orb(C'
(ii)
G' Z2) for P
4
V
Of TV,2 of Let -K be
Z3) p
of
index
group
Let
(X, G)
(X, G)
with
a
The
Then
(i)
If
(ii)
E
a
G'
where
Cp
If
G'
c
a
is
(i)
Since Ka
==
X is =
Ka
Cp
a
be
a
valency
in Section
I
Example 2.
in
Cayley graph
the
E
is
1
5.3 and with
a
2-orb(Cp
-
Ma
=
1, and
inductively
minimal we can take finite, are distinct. 1, jxjjo<jO such
that
from
xO
-=
exists
a
in xi-la. xn-
Since that
primitivity
Cay(Zn, 111)
a)
[20,
from that
n
is
desired.
as
Let
valency
a
(=-
G be with
It
follows
exists
a
1.
Ka
Weconclude
(X, a)
=
from
unique
unique element in c Z>O such that
distinct.
p
-
Then there
n
prime
andr
We conclude
(1XiTO1 G is P
and E C K. Wedenote
group
relation
non-symmetric
2-orb(F'p,3;
>
Example
in
as
Schemes
-
(X, G)
Let
Theorem 3. a
finite
a
is
odd prime p'>
some
by Cay (K, E)
K
over
odd primep
-Pp,2
2 where
for Z2) P
2-orb(Fp,2;
(iii)
Some
,
index
Association
Commutative
of Primitive
The Enumeration
5
xi-la xO
from
=
Then, by (i) there
X2 G x1a, -
=
2.
[20, Prop.
xn-
fXi-21Since
-
that
no
=
Wedefine
finite,
X is
Ka
2 and
=
is a
Since
a
non-trivial
a*.
Let
we can =
a*,
1,
a)
-
Cay(Zn, 11,
-
11)
relation
(xO, xj)
inductively
that
primitivity
(1xi 10o suchthat
T(YO, Yj-in)
proof of the first statement. Since jxjjo<j 2.
Thus, Corollary
it
remains
2. If
ai
to
=
ai*
1,
read modulo
are
(i, j, k)
=
have
we
ai* =7
a2i
n.
(i, 0, i) ly (0, 2i),
y
(i, i),
y (0,
0)), r(o, y(O, -i))}
a2i
ajai
=7 a
only.
11x, ail,
=
If
ai
=
0 a
then
and hence a2i
=
a2i
3. 1x
(n, 3)
=
I
Assume the contrary,
have a2i
=
r
(o,
y (2i,
i.e.,
0))
=
r
1 Let (YO Y1 Y2 7 107 O)l fYilo
