Orthogonal bases of certain symmetry classes of tensors associated with Brauer
characters
by
K. A. A. Indika
A dissertation submitted to the Graduate Faculty of
Auburn University
in partial ful llment of the
requirements for the Degree of
Doctor of Philosophy
Auburn, Alabama
August 03, 2013
Keywords: Brauer character, projective indecomposable module, block, dihedral group,
symmetrizer, symmetry class of tensors, o-basis, generalized orthogonality relation
Copyright 2013 by K. A. A. Indika
Approved by
Randall R. Holmes, Professor of Mathematics
Narendra K. Govil, Associate Chair & Alumni Professor of Mathematics
Huajun Huang, Associate Professor of Mathematics
Abstract
The main focus of this dissertation is on the existence of an orthogonal basis consisting
of standard symmetrized tensors (o-basis for short) of a symmetry class of tensors associated
with a Brauer character of a nite group. Most of the work is done for the dihedral group and
some results are given for the symmetric group. The existence of an o-basis of a symmetry
class of tensors associated with an (ordinary) character of a nite group have been studied
by several authors. My study was motivated by the work done on the existence of such a
basis of a symmetry class of tensors associated with an (ordinary) irreducible character of a
dihedral group.
In Chapter 1 we introduce the basic de nitions in character theory. In this a Brauer
characters, character of a projective indecomposable module (PI) and a block of a nite
group will be introduced. Also in this chapter a generalised orthogonality relation of blocks
of a nite group is established. In chapter 2 we introduce the symmetrizer and related
notions. Some general results associated with Brauer characters of a nite group will also
be given in this chapter. Chapter 3 consists of the results associated with Brauer characters,
PIs and blocks of a dihedral group. Finally Chapter 4 lists some result associated with the
Brauer characters of the symmetric group.
ii
Acknowledgments
This dissertation is a partial ful lment of my Ph.D. program. It consists of the work I
did in the last 5 years at Auburn University. It has been a fruitful experience in my academic
life to have worked at Auburn University with a lot of nice people. I would like to take this
opportunity to thank and mention all the people who supported me and encouraged me in
this journey to achieve this special goal of my life.
It is with great honor I mention the support and guidance of Professor Randall R.
Holmes. As my Ph.D. advisor he took all possible e orts to make my work as smooth as
possible. His incomparable intuition of the subject has played a big role in my improvements
as a researcher. The exibility and the nature of his guidance made my research experience
a pleasant one. His advice helped me not only with my research but also with my teaching
and my student life. His teaching inspired me and he always happily shared his teaching
experience with me to help me improve my teaching. He is very understanding and had an
extraordinary ability to know what to say and when to say it. I will ever be so grateful to
him not just for the advisor he is but also for the nice person he is.
I am specially thankful to the Dr. Narendra K. Govil for his continuous support since
the rst day I arrived at the Auburn University. He was always there whenever I needed
his advice. He gladly accepted to be in my Ph.D. advisory committee showing his kindness
to see my success in the program. I will always remember him as a mathematician with a
unique character.
I rst met Dr. Huajun Huang as a teacher. His humble nature and commitment to his
work has helped me motivate myself in my work. He is very approachable because of his
amicable nature. I am so grateful to him for accepting to be in my Ph.D. advisory committee
and for continuing to support me in achieving my goal.
iii
I thank Dr. Rajesh Amin for kindly accepting to be the university reader of my disser-
tation.
I can not nd enough words to express my gratitude to my mother M. Kusumawathi
Wanigarathna and my (late) father K. A. George Martin. Their loving care has made me
the person I am today. I owe them for all the success I have achieved in my life. The trust
and the con dence my mother has in me made me reach this level of education.
My wife Shilpa needs a special mention for being there for me and encouraging me to
achieve these heights. Her support in my decisions gives me con dence to do my work well.
Our loving daughter Savana makes my life the most enjoyable one. She keeps me relaxed
after a tiresome day.
Finally my sincere thank goes to each individual responsible in some way to my success
in the program.
iv
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Character Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 Group representations and Group Algebra . . . . . . . . . . . . . . . . . . . 1
1.2 Character . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Brauer character . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4 PIs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.5 Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.6 Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.7 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2 Symmetrized Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 Symmetrizers associated with ordinary and Brauer characters of G . . . . . . 12
2.2.1 Symmetrizers associated with PIs . . . . . . . . . . . . . . . . . . . . 15
2.2.2 Symmetrizers associated with blocks . . . . . . . . . . . . . . . . . . 16
3 Dihedral group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.1 Brauer characters of Dn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.2 PIs of Dn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.3 Blocks of Dn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.4 Block idempotent symmetrization . . . . . . . . . . . . . . . . . . . . . . . . 26
3.5 Irreducible symmetrization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.6 Projective symmetrization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4 Symmetric group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
v
4.1 Special case S4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
vi
Chapter 1
Character Theory
1.1 Group representations and Group Algebra
Let G be a nite group and V be a nite dimensional vector space over the eld of
complex numbers C. By GL(V) we denote the group of invertible linear transformations
from V to itself. A representation of G is a group homomorphism : G! GL(V). The
degree of the representation is the dimension of V.
Denote by CG the vector space over C with the basis G. CG is a ring with the multi-
plication de ned by,
(
X
a2G
aa)(
X
b2G
bb) =
X
a;b2G
a bab:
A C-algebra is a ring A that is also a vector space over C such that (ab) = ( a)b = a( b)
for all 2C and a;b2A. Note that CG is a C-algebra and is called the group algebra of
G over C. CG has an identity given by 1e6= 0, where e is the identity of G. De ne a map
from C to CG by 7! 1, where 1 6= 0 is the identity of CG. This is a well de ned ring
monomorphism, hence C is viewed as a subring of CG.
Let V be a nite dimensional vector space over C and let : G! GL(V) be a repre-
sentation of G. Then V can be viewed as a (left) CG module by de ning av = (a)(v) for
a2G, v2V and extending linearly to CG.
On the other hand, let V be a CG module. Then V is a vector space over C by viewing
C as a subring of CG. When we say that V is a CG module, we always assume that V is
nite dimensional when viewed as a vector space over C in this way. De ne a map from G
to GL(V) by (a)(v) = av for a2G, v2V. Then is a well de ned group homomorphism
and hence a representation of G called the representation a orded by V.
1
An irreducible representation of G is a representation a orded by a simple CG module.
1.2 Character
Let V be a CG module and let be the representation of G a orded by V. Let 2G,
so that ( )2GL(V).
Let M ( ) be the matrix representation of ( ) corresponding to a xed basis of V. The trace
of ( ) is given by tr( ( )) = tr(M ( )): Note that the value of tr( ( )) does not depend on
the choice of the basis since similar matrices have the same trace.
The (ordinary) character of G a orded by or V is the function : G!C de ned by
( ) = tr( ( )) for 2G. We say (e) the degree of where e is the identity element of G.
Theorem 1.1 ([11, Lemma 2.15, page 20]). Let be a character of G. Let 2G and let
m be the order of . Then
i) ( ) is a sum of mth roots of unity,
ii) ( ) = ( 1).
Two representations and % of the same degree n are said to be similar if there exists
an invertible matrix P of size n n such that M ( ) = P 1M%( )P for all 2G. It is easy
to observe that the following result holds using the property that tr(AB) = tr(BA) for all
square matrices A;B.
Theorem 1.2 ([11, Lemma 2.3, page 14]).
i) Similar representations of G a ord equal characters.
ii) Characters are constant on the conjugacy classes of G.
A class function is a function on G that is constant on conjugacy classes. The theorem states
that the characters of G are class functions.
2
A character of G is called an irreducible character of G if it is a orded by an irreducible
representation of G (or, equivalently, a simple module of CG). Representations of G a orded
by isomorphic CG-modules are similar. There is a one-to-one correspondence between iso-
morphism classes of CG-modules and similarity classes of representations of G (see [11, page
10]). Therefore in light of the theorem above the number of di erent irreducible characters
of a group G is the same as the number of isomorphism classes of simple CG-modules.
Let Irr(G) denote the set of irreducible characters of G. Maschke?s theorem stated below
provides a way to reduce the study of characters of G to the study of irreducible characters
of G.
Theorem 1.3 (Maschke). Let K be a eld. If charK - jGj, then every KG-module is a
direct sum of simple KG-modules.
Since char C = 0, Maschke?s theorem holds for the eld C. Now let V and V0 be
two CG-modules and let and 0 be the representations they a ord respectively. Let
be the representation a orded by the direct sum V V0. Then for any 2G the matrix
representation of ( ) relative to an ordered basis formed by taking an ordered basis for V
and appending an ordered basis for V0 is given by a block diagonal matrix with blocks the
matrix representations of ( ) and 0( ) as given by
M ( ) =
0
B@ M ( ) 0
0 M 0( )
1
CA:
Then the character a orded by the CG-module V V0 is + 0, since trM ( ) = trM ( ) +
trM 0( ).
Because of the above results we see that to study the characters of G it is enough to look
at the irreducible characters of G. Once all the irreducible characters of G are known the
other characters of G are known as well since they are simply sums of irreducible characters.
3
One gets the number of irreducible characters of G from the number of conjugacy classes of
the group G as stated below.
Theorem 1.4 ([11, Corollary 2.7, page 16]). Let G be a group. The number of irreducible
characters of G equals the number of conjugacy classes of G.
1.3 Brauer character
The Brauer characters are the main focus of this entire thesis. These characters are also
known as modular characters. The modular representation theory was founded by Richard
Brauer in the 1930?s. We begin by setting basic de nitions.
Let R be the ring of algebraic integers in C. Fix a prime p, and let M be a maximal
ideal of R such that pR M. Set K = R=M. Then K is a eld. Considering the
natural homomorphism : R!K we have p7!0, so K has characteristic p. The natural
homomorphism is going from characteristic 0 to characteristic p.
Theorem 1.5 ([11, Lemma 15.1, page 263]). Let U =f 2Cj m = 1 for some integer m
with p-mg and let R;K be as above. Then
i) U R,
ii) the natural homomorphism maps U isomorphically onto Knf0g,
iii) K is algebraically closed and algebraic over its prime eld.
An element of G is called a p-regular element if its order is not divisible by p. Denote
by ^G the set of all p-regular elements of G.
Let V be a KG-module of nite dimension n and let be the representation of G
a orded by V. Let 2 ^G and let 1;:::; n 2Knf0g be the eigenvalues of ( ). Then
by the theorem above there exist unique 1;:::; n 2 U such that i 7! i via the natural
4
homomorphism. De ne a function ? : ^G!C by
?( ) =
nX
i=1
i: (1.1)
Then ? is called the Brauer character of G a orded by .
Let 2 ^G and suppose 2K is an eigenvalue of ( ). Then 1 is an eigenvalue of
( 1) since
( 1)v = ( 1) 1 ( )v = 1 (e)v = 1v:
Also if ( ) = ( 2U), then 1 = ( ) = ( ) ( ) = ( ), so ( ) = 1. Therefore
?( ) = ?( 1). If ? is a Brauer character of G, then ?, the complex conjugate of ? is also
a Brauer character [11]. A Brauer character corresponding to a simple KG-module is called
an irreducible Brauer character of the group G. We denote the set of irreducible Brauer
characters of G by IBr(G). The irreducible Brauer characters are linearly independent over
C [11, Theorem 15.5, page 265].
Brauer characters are constant on conjugacy classes. The number of irreducible Brauer
characters is equal to the number of conjugacy classes of G containing p-regular elements of
G as stated by the following theorem.
Theorem 1.6 ([14, Corollary 3, page 150]). The number of classes of simple KG-modules
is equal to the number of p-regular conjugacy classes of G.
Let ^ denote the restriction of an ordinary character of G to the set ^G of p-regular
elements of G. The following result is a well known relationship between the ordinary
characters and the Brauer characters of a group.
Theorem 1.7 ([11, Theorem 15.6, page 265]). Let be an ordinary character of G. Then
^ is a Brauer character of G:
5
The character uniquely determines the Brauer character ^ . We note here that if p
does not divide the order of the group G, then ^G = G and the Brauer characters of G
coincide with the ordinary characters G.
The set of complex valued class functions on ^G form a vector space over C
Theorem 1.8 (R. Brauer). The irreducible Brauer characters of a group G form a basis of
the vector space of complex valued class functions on ^G.
1.4 PIs
We are also interested in the characters associated with projective indecomposable mod-
ules of RG. Note that RG is of nite dimension so satis es the A.C.C. and D.C.C. Then by [2,
Theorem 14.2, page 81] RG can be written as a direct sum of indecomposable RG-modules.
A summand of this direct sum is a principle indecomposable module (a PIM) of RG. Sim-
ilarly KG can be expressed as a direct sum of principle indecomposable KG-modules. As
direct summands of free modules, PIMs of RG and KG are projective. [5, Theorem I.13.7,
page 44] states that there is a one to one correspondence between the isomorphism classes
of PIMs of RG and those of KG.
Theorem 1.9 ([2, Theorem 54.11, page 372]). Let P be a PIM of KG. Then P has a unique
maximal submodule NP. Two PIMs P and Q are isomorphic if and only if the irreducible
modules P=NP and Q=NQ are isomorphic.
Theorem 1.10 ([2, Corollary 54.14, page 374]). There is a one-to-one correspondence be-
tween the isomorphism classes of PIMs and the isomorphism classes of irreducible KG-
modules.
The character a orded by a PIM P of RG (PI for short) is the character a orded by
CNRP . By the theorem above we have that there is a one to one correspondence between
the irreducible Brauer characters of G and the PIs of G. We will denote by ? the PI
6
corresponding to ?2 IBr(G). A PI is a complex valued function de ned on G with the
property that ( ) = 0 for 2Gn ^G [5, Corollary IV.2.5, page 144].
1.5 Relationships
Let 2 Irr(G) and let ^ be the restriction of to ^G. Recall by Theorem 1.7 ^ is a
Brauer character of G, so
^ =
X
?2IBr(G)
d ?? (1.2)
for some uniquely determined nonnegative integers d ?. The integers d ? ( 2 Irr(G);?2
IBr(G)) are called the decomposition numbers of G for the prime p. The matrix of size
jIrr(G)j jIBr(G)j with the d ??s as entries is called the decomposition matrix of G.
Let ?2IBr(G). By [14, page 151] we have the following relationships
? =
X
2Irr(G)
d ? ; (1.3)
? =
X
2IBr(G)
c ? ;
where the coe cients c ? are the entries of the matrix C = DDT, with DT the transpose of
D. The matrix C is called the Cartan matrix.
1.6 Block
Let e be a centrally primitive idempotent of the group algebra CG. The block B = Be
corresponding to e is the category of CG-modules V such that eV = V. A CG-module V is
said to belong to B if it is an object of B, that is, if eV = V. By [5, Theorem 7.8, page 23]
a nitely generated indecomposable CG-module V belongs to a unique block. If V belongs
to B, then every submodule and homomorphic image of V belongs to the same block B.
7
A character or a Brauer character of G is said to belong to a block B if the associated
module belongs to B. If 2Irr(G) belongs to the block B, then ?2IBr(G) belongs to B if
the decomposition number d ? is nonzero. Further each irreducible character and irreducible
Brauer character belongs to a unique block. Two irreducible characters and are in a
same block B of G if there is ?2 IBr(G) such that d ? and d ? are both nonzero. In this
case the block is the unique block that contains the Brauer character ? [11].
If is a character or a Brauer character, we write 2B to mean that belongs to the
block B. More generally, if S is a set of characters or Brauer characters, we write 2B\S
to mean that belongs to the block B and 2S.
Let B be a block of G. The Osima idempotent of CG corresponding to B is given by
sB =
X
2B\Irr(G)
s = 1jGj
X
2B\Irr(G)
X
2G
(e) ( 1) : (1.4)
Theorem 1.11 ([11, Theorem 15.30, page 277]). For blocks B and B0 of G, we have
sBsB0 = BB0sB:
The following theorem holds due to the Equations 1.2 and 1.3.
Theorem 1.12 (Osima).
sB = 1jGj
X
?2B\IBr(G)
X
2G
?(e) ?( 1)
= 1jGj
X
?2B\IBr(G)
X
2^G
?(e)?( 1) :
8
1.7 Orthogonality
Let Fun(G;C) denote the set of all functions from G to C. Fun(G;C) is a vector space
over C. For f;g2Fun(G;C) set,
(f;g) = 1jGj
X
2G
f( )g( ):
( ; ) is an inner product on Fun(G;C). By de nition the characters of G are in Fun(G;C).
Now we will state some known orthogonality relations of characters of G. For a character
of G and 2G, we have ( ) = ( 1) by Theorem 1.1. It is a known fact that Irr(G)
forms a basis for the set of class functions from G to C. It is indeed an orthonormal basis
due to the following result.
Theorem 1.13 ([11, Corollary 2.14, page 20]). Let ; 02Irr(G). Then
( ; 0) = 1jGj
X
2G
( ) 0( 1) = 0:
Theorem 1.14 (Generalized Orthogonality Relation). Let ; 02Irr(G). For any 2G
X
2G
( ) 0( 1) = 0jGj ( ) (e) :
For complex-valued functions f and g on ^G de ne
(f;g)^= 1jGj
X
2^G
f( )g( ):
Theorem 1.15 ([5, Lemma 3.3, page 145]). For ?; 2IBr(G), we have
( ?; )^= 1jGj
X
2^G
?( ) ( 1) = ? :
9
We establish an orthogonality relation associated with Osima idempotents of blocks of
a group G which we call the generalized orthogonality relation of blocks. Below we discuss
the formulation of this new result.
Theorem 1.16. For 2G,
X
2^G
X
?2B\IBr(G)
X
2B0\IBr(G)
?(e)?( ) (e) ( 1 ) = BB0jGj
X
?2B\IBr(G)
?(e) ?( ):
Proof. By using Theorem 1.12 we have
sBsB0 = 1jGj2
X
2^G
X
?2B\IBr(G)
?(e)?( )
X
2G
X
2B0\IBr(G)
(e) ( )
= 1jGj2
X
2^G
X
2G
X
?2B\IBr(G)
X
2B0\IBr(G)
?(e)?( ) (e) ( )
= 1jGj2
X
2G
X
2^G
X
?2B\IBr(G)
X
2B0\IBr(G)
?(e)?( ) (e) ( 1 )
and
BB0sB =
X
2G
BB0 1jGj
X
?2B\IBr(G)
?(e) ?( ) :
Now, by Theorem 1.11, sBsB0 = BB0, so by comparing the coe cients on both sides for a
xed 2G we get,
X
2^G
X
?2B\IBr(G)
X
2B0\IBr(G)
?(e)?( ) (e) ( 1 ) = BB0jGj
X
?2B\IBr(G)
?(e) ?( ):
10
Chapter 2
Symmetrized Tensors
In this chapter we will state some basic de nitions and results of di erent symmetrizers
corresponding to characters discussed in Chapter 1.
2.1 Background
For xed positive integers n;m set
n;m =f = ( 1; 2;:::; n)2Znj1 i mg:
Let G be a subgroup of the symmetric group Sn. De ne a right action on n;m by G as
follows. For 2G and 2 n;m
= ( (e);:::; (n)): (2.1)
Consider the relation for ; 2 n;m given by if there is an element 2G such that
= . This is an equivalence relation on n;m. We x a set of representatives of the
equivalence classes of n;m with respect to .
Let V be a complex inner product space of dimension m with orthonormal basis
fe1;e2;:::;emg. V n = V V V (n factors) is the nth tensor power of V. For
2 n;m let e = e 1 e 2 e n. The inner product induced on V n is given by
he ;e i = Qni=1(e i;e i) where ( ; ) is the inner product of V. Under this inner product
fe j 2 n;mg is an orthonormal basis for V n. V n is a CG-module with the action
e = e 1 for 2G extended linearly to CG.
11
2.2 Symmetrizers associated with ordinary and Brauer characters of G
In the following discussion stands for either an irreducible ordinary character or an
irreducible Brauer character of G. Denote by S a subset of G where S = ^G when 2IBr(G)
and S = G when 2Irr(G).
The symmetrizer corresponding to is de ned by,
s = (e)jSj
X
2S
( ) :
The theorems in this section pertaining to 2 Irr(G) are well known. On the other
hand, we generalize some well-known results for 2Irr(G) to handle the case of 2IBr(G).
Theorem 2.1. The elements s for 2Irr(G) are orthogonal idempotents.
Proof. Let ; 2Irr(G). Then by Theorem 1.14 we get,
s s = (e) (e)jGj2
X
2G
X
2G
( ) ( ) = (e) (e)jGj2
X
2G
X
2G
( ) ( 1 )
= (e) (e)jGj2
X
2G
jGj
(e) ( ) =
(e)
jGj
X
2G
( ) = s :
The symmetry class of tensors V corresponding to is the image of V n under the
symmetrizer s :
V = s V n:
Corollary 2.2. If ; 2Irr(G) and 6= , then the vector spaces V and V are orthogonal.
Proof. If s v = s w for some v;w2V n, then
s v = s (s v) = s (s w) = 0:
12
Let 2 n;m. The standard symmetrized tensor e corresponding to is the image of
e 2V n under s :
e = s e = (e)jSj
X
2S
( ) e = (e)jSj
X
2S
( )e 1: (2.2)
By o-basis of a subspace W of V n we mean an orthogonal basis of W that consists of
standard symmetrized tensors. An interesting question to ask is \For which W does there
exist an o-basis?" In 1991 Wang and Gong gave an example in [16] of such an o-basis for a
symmetry class of tensors V with 2 Irr(G) when G is the dihedral group of order eight.
Ever since there have been papers [1, 3, 4, 7, 8, 10, 15] answering the question when such an
o-basis exists. All these papers however address the problem in the ordinary character case.
This dissertation is devoted to answering the question of when an o-basis exists for a
symmetry class of tensors symmetrized by a Brauer symmetrizer for particular choices of G.
Let V = he j 2Gi and let V = s (V ). Observe that V = he j 2Gi. V is
called the orbital subspace corresponding to . Using the orbital subspaces we can write the
symmetry class of tensors as an orthogonal direct sum.
Theorem 2.3 ([9, Theorem 1.1]). We have
V = _
X
2
V (orthogonal direct sum):
In particular, V has an o-basis if and only if V has an o-basis for each 2 .
Proof. Let 2 n;m. Then = for some 2 and 2G, so that e = e 2V . This
shows that V is contained in (and hence equals) the indicated sum.
The sets E =fe j 2Gg, 2 , are pairwise disjoint subsets of the orthogonal set
fe j 2 n;mg and are therefore pairwise orthogonal. For each 2 the subspace V is
contained in the span of E , so the indicated sum is an orthogonal direct sum.
13
Assume that V has an o-basis B. By the rst paragraph, B is the union of the sets
B = B\V , 2 , and these sets are pairwise disjoint by the second paragraph, so B is
an o-basis for V for each 2 .
Finally, if V has an o-basis for each 2 , then the union of these bases is an o-basis
for V .
For 2 n;m let G =f 2Gj = g the stabilizer subgroup of in G.
Theorem 2.4. For 2 n;m and a xed 2G, we have
(e ;e ) = (e)
2
jSj2
X
2S
X
2 1S\G
( ) ( 1 1):
Proof. Let 2 n;m and 2S be xed. Now by using the Equation 2.2 we get,
(e ;e ) = (e)
2
jSj2
X
2S
X
2S
( ) ( )(e 1;e 1)
= (e)
2
jSj2
X
2S
X
2S
1 2G
( ) ( 1)
= (e)
2
jSj2
X
2S
X
2 1S\G
( ) ( 1 1):
Corollary 2.5. For 2 n;m, 2Irr(G), and a xed 2G
(e ;e ) = (e)jGj
X
2G
( ):
Proof. From the above Theorem 2.4 we get
(e ;e ) = (e)
2
jGj2
X
2G
X
2G
( ) ( 1 1) = (e)
2
jGj2
X
2G
X
2G
( ) ( 1 1):
14
Then by using the generalized orthogonality relation (Theorem 1.14) we get
(e ;e ) = (e)
2
jGj2
X
2G
jGj ( 1 )
(e) =
(e)
jGj
X
2G
( 1 ) = (e)jGj
X
2G
( ):
For 2Irr(G) and for 2 Freese gives the dimension of the orbital subspace V in
[6]:
dimV = (e)jG
j
X
2G
( ): (2.3)
2.2.1 Symmetrizers associated with PIs
Let ?2IBr(G) and put = ?.
The symmetrizer associated with is de ned by
s = ?(e)jGj
X
2G
( ) : (2.4)
Note that s 2 CG. For a PI of G the symmetry class of tensors is de ned by
V = s V n. For 2 n;m the standard symmetrized tensor is de ned by e = s e and the
orbital subspace is de ned by V =he j 2Gi. With a similar argument as in Theorem
2.3 we have
V = _
X
2 V
(orthogonal direct sum): (2.5)
Theorem 2.6. For 2G and 2 n;m
(e ;e ) = ?(e)
2
jGj2
X
2G
X
2G
( 1 ) ( ):
15
Proof.
e
;e
= ?(e)2
jGj2
X
2G
X
2G
( ) ( ) (e ;e )
= ?(e)
2
jGj2
X
2G
X
2G
12G
( ) ( )
= ?(e)
2
jGj2
X
2G
X
2 G 1\G
( 1 ) ( )
= ?(e)
2
jGj2
X
2G
X
2G
( 1 ) ( ):
2.2.2 Symmetrizers associated with blocks
Let B be a block of G. The symmetrizer corresponding to B is the Osima idempotent
sB of B (Equation 1.4). The symmetry class of tensors is de ned by VB = sBV n. For
2 n;m the standard symmetrized tensor is de ned by eB = sBe and the orbital subspace
is de ned by VB =heB j 2Gi.
Lemma 2.7. For 2 n;m
eB =
X
2B\Irr(G)
e :
Proof. By Equation 1.4 we get,
eB = sB(e ) =
X
2B\Irr(G)
s (e ) =
X
2B\Irr(G)
e :
Theorem 2.8.
VB = _
X
2
VB (orthogonal direct sum):
16
Proof. The argument in the proof is similar to that of the proof of Theorem 2.3, and we
omit the details.
Theorem 2.9. For 2
dimVB = 1jG
j
X
2^G
X
?2B\IBr(G)
?(e) ?( 1)
= 1jG
j
X
2^G
X
2B\Irr(G)
(e) ( 1):
Proof. Since sB is an idempotent we have rank (sB) = tr (sB). Then by Equation 1.4 we get,
dimVB = rank (sB) = tr (sB) = tr 1jGj
X
2G
X
2B\Irr(G)
(e) ( 1)
= 1jGj
X
2G
X
2B\Irr(G)
(e) ( 1)tr ( ):
Note here that it makes sense to write tr ( ) by viewing as a linear transformation on V n.
In [6, Equation 13] Freese shows that 1jGjP 2G (e) ( 1)tr ( ) = (e)jG jP 2G ( 1). So we
get,
dimVB =
X
2B\Irr(G)
(e)
jG j
X
2G
( 1) = 1jG
j
X
2G
X
2B\Irr(G)
(e) ( 1):
Now by Equations 1.2 and 1.3 we get, for any 2G,
X
2B\Irr(G)
(e) ( 1) =
X
?2B\IBr(G)
X
2B\Irr(G)
?(e)d ? ( 1) =
X
?2B\IBr(G)
?(e) ?( 1):
17
So it gives
dimVB = 1jG
j
X
2G
X
?2B\IBr(G)
?(e) ?( 1) = 1jG
j
X
2^G
X
?2B\IBr(G)
?(e) ?( 1)
= 1jG
j
X
2^G
X
2B\Irr(G)
(e) ( 1):
Theorem 2.10. For 2G,
(eB ;eB ) = 1jGj
X
2 G
X
?2B\IBr(G)
?(e) ?( ): (2.6)
Proof. Recall that V and V are orthogonal for ; 2Irr(G) with 6= by Corollary 2.2.
Then using Lemma 2.7 and Corollary 2.5 we get
(eB ;eB ) = (
X
2B\Irr(G)
e ;
X
2B\Irr(G)
e ) =
X
; 2B\Irr(G)
(e ;e ) =
X
2B\Irr(G)
(e ;e )
=
X
2B\Irr(G)
(e)
jGj
X
2G
( ) =
X
2B\Irr(G)
(e)
jGj
X
2 G
( )
= 1jGj
X
2 G
X
2B\Irr(G)
(e) ( ) = 1jGj
X
2 G
X
?2B\IBr(G)
?(e) ?( ):
The following is a useful lemma which we call the translation principle of the orthogo-
nality of symmetrized tensors.
Lemma 2.11. If the standard symmetrized tensors eB and eB 0 are orthogonal, then eB
and eB 0 are orthogonal for every 2G.
18
Proof. By Theorem 2.10 we get, for each 2G,
(eB 0 ;eB ) = 1jGj
X
2 0 ( ) 1G
X
?2B\IBr(G)
?(e) ?( )
= 1jGj
X
2 0 1G
X
?2B\IBr(G)
?(e) ?( ) = (eB 0;eB ) = 0:
19
Chapter 3
Dihedral group
In this chapter we focus on the existence of an o-basis of a symmetry class of tensors
associated with a Brauer character, a PI and an Osima idempotent of a block of the dihedral
group.
For an integer n( 3), the dihedral group of degree n is the subgroup Dn of the sym-
metric group Sn, generated by the elements
r =
0
B@ 1 2 ::: n 1 n
2 3 ::: n 1
1
CA s =
0
B@ 1 2 ::: n 1 n
1 n ::: 3 2
1
CA;
That is Dn =frk;srkj0 k n 1g and jDnj= 2n.
Dn with n even has 4 degree one irreducible characters. Let 1; 2; 3; 4 denote these
characters. Dn with n odd has only 2 degree one irreducible characters; we denote them
with 1; 2. For all n the degree two irreducible characters of Dn are given by h where
1 h < n2 . For each integer k we get h(rk) = !hk + ! hk = 2 cos 2 hkn where !n = 1 [14,
page 37]. The character table for Dn is given by
rk srk
1 1 1
2 1 1
3 ( 1)k ( 1)k (n even)
4 ( 1)k ( 1)k+1 (n even)
h 2 cos(2 hkn ) 0 1 h< n2
20
We observe from the table that if a character of degree one or degree two of Dn, then for
2Dn we have
( ) = ( 1): (3.1)
Let G = Dn. For a xed prime p write n = pq? with p - ?. The set ^G of p-regular
elements of G is given by,
^G =
8
><
>:
frapq;srkj0 a>>>
<
>>>>
:
?
2 + 3; if ? even, p6= 2;
? 1
2 + 2; if ? odd, p6= 2;
? 1
2 + 1; if p = 2.
(3.2)
3.1 Brauer characters of Dn
Our e ort in this section is to nd conditions for the existence of an o-basis for the
symmetry class of tensors corresponding to a Brauer character of the dihedral group G = Dn.
We begin by listing the distinct irreducible Brauer characters of G.
Recall for an ordinary character of G the restriction of to ^G is denoted by ^ . By 1.7
^ is a Brauer character of G.
21
For each 1 h n2 , the Brauer character ^ h is of degree 2. The next lemma gives
conditions for when two Brauer characters of degree two are the same.
Lemma 3.1. For 1 i;j < n2 , ^ i = ^ j if and only if either i + j 0 mod ? or i j 0
mod ?.
Proof. Since any degree two character of G is zero on srk for all k it is enough to check
when two characters are the same on the elements rapq 2G. Suppose ^ i = ^ j. Then for any
0 a>>
><
>>>>
:
4; if ? even, p6= 2;
2; if ? odd, p6= 2;
1; if p = 2.
(3.3)
For each 1 j , the Brauer character ?1j = ^ j is of degree 1.
Theorem 3.3. Let G = Dn. The complete list of distinct irreducible Brauer characters of
G is
?1j = ^ j, for 1 j , ?2i = ^ i for 1 i< ?2.
Proof. For each 1 j , the Brauer character ^ j is of degree one and hence is irreducible.
We see the distinctness of these characters by observing the character values for srk in the
character table above. So we have distinct irreducible Brauer character of degree one of G.
To see the distinctness of the degree two Brauer characters in the given list observe that
for any i;j such that 1 i
<
>:
2pt; if G Cn;
pt; otherwise.
29
Proof. First observe that for any 0 >>
><
>>>>
:
0; if a is odd;
2pt; if G Cn, a is even;
pt; if G *Cn, a is even.
Proof. By Theorem 2.9, Theorem 3.5 and the fact that 3( ) + 4( ) = 0 for all 62Cn we
get,
dimVB13 = 1jG
j
X
2^G
X
?2B13\IBr(G)
?(e) ?( ) = 1jG
j
X
2^G
3( ) + 4( ) + 2
pq 1
2X
k=1
?
2 +k?
( )
= 1jG
j
X
2^G \Cn
3( ) + 4( ) + 2
pq 1
2X
k=1
?
2 +k?
( )
:
30
Suppose that a is odd. Let a = a0pt for some odd integer a0. Then ^G \Cn =fr apq t j0
?a0 1g. Note that ?a0 is even.
?
2 +k?
(r apq t) = 2 cos 2 (
?
2 +k?) ap
q t
n = 2 cos 2 (
1
2 +k) a
0 = 2 cos a0;
which is equal to 2 or 2 depending upon whether is even or odd. Also 3(r apq t) =
4(r apq t) equals 1 or 1 according as is even or odd respectively. So the sum 3(r apq t)+
4(r apq t)+2P
pq 1
2
k=1 ?2 +k?(r
apq t) on the right side of the formula for dimVB13 above equals
2pq if is even and 2pq if is odd. Therefore in the case a is odd we get
dimVB13 = 1jG
j
2pq ?2a0 + 2pq ?2a0
= 0:
Now suppose a is even. Then any p-regular element inhraican be expressed as r2 pq for
some integer . We have
?
2 +k?
(r2 pq) = 2 cos 2 (
?
2 +k?)2 p
q
n = 2 cos 2 (
1
2 +k)2 = 2;
in which case the proof is similar to the proof of Lemma 3.8 above. So we get dimVB13 = 2pt
if G is contained in Cn and dimVB13 = pt if G is not contained in Cn.
Theorem 3.10. Let G = Dn. Write B = B11. The symmetry class of tensors VB has an
o-basis.
Proof. By Theorem 2.8 we have
VB = _
X
2
VB ;
so it su ces to show that VB has an o-basis for each 2 . Let 2 . By Lemma 3.8 if
G * Cn, then dimVB = pt. Therefore by part i) of Lemma 3.7 the set feB j 2H0g is an
orthogonal basis. If G Cn, then dimVB = 2pt and in this case feB ;eB s j 2H0g is an
orthogonal basis by part ii) of Lemma 3.7.
31
Theorem 3.11. Let G = Dn with n even. Write B = B13. The symmetry class of tensors
VB has an o-basis.
Proof. Due to Theorem 2.8 it su ces to show that VB has an o-basis for each 2 . Let
2 . We have G \Cn = hrai for some integer a. If a is odd, then by Lemma 3.9
dimVB = 0, so VB has an o-basis. Suppose a is even. Then dimVB = pt if G * Cn, so
by part i) of Lemma 3.7 the set feB j 2H0g is an orthogonal basis and dimVB = 2pt if
G Cn, so feB ;eB s j 2H0g is an orthogonal basis by part ii) of Lemma 3.7.
Now we will bring our attention to the blocks consisting only of degree two characters
of G. For each 1 i< ?2 the block B2i contains ?2i 2IBr(G) and this is the only irreducible
Brauer character of G it contains. Below is a statement for conditions when the dimension
of the orbital subspace VB2i corresponding to a 2 is not zero.
Theorem 3.12. Fix 2 n;m. Then for 1 i < ?2 we have dimVB2i 6= 0 if and only if
^G \Cn hrn0i, where n0 = ngcd(n;i).
Proof. Suppose ^G \Cn6 hrn0i. We have ^G \Cn =hrbiwith bjn, so rb62hrn0i. Fix i with
1 i< ?2. Then for 2 ^G we have by Theorem 3.5 and Lemma 3.4
2i( ) =
X
a2Ai
a( ) =jAij i( ) = pq i( ): (3.4)
Now by Theorem 2.9
dimVB2i = 1jG
j
X
2^G
?2i(e) 2i( ) = ?
2
i(e)
jG j
X
2^G
pq i( ) = p
q?2
i(e)
jG j
X
2^G \Cn
i( )
= p
q?2
i(e)
jG j
n
b 1X
j=0
i(rjb) = p
q?2
i(e)
jG j
n
b 1X
j=0
(!ibj +! ibj);
where !ib is an nbth root of unity. Now Pnb 1j=0 (!ibj + ! ibj) 6= 0 if !ib = 1. But if !ib = 1,
then there is an integer m such that ib = mn, so i00b = mn0 where i00 = igcd(n;i). Now since
32
gcd(n0;i00) = 1 we get n0jb which is a contradiction. Therefore Pnb 1j=0 (!ibj + ! ibj) = 0 and
hence dimVB2i = 0.
Conversely suppose ^G \Cn hrn0i. Note that for any integer c we have
i(rcn0) = 2 cos 2 icn
0
n = 2 cos 2 i
00c = 2;
and in this case dimVB2i = 2i(e)jG j P 2^G \Cnpq i( )6= 0.
Theorem 3.13. Let G = Dn. Fix i with 1 i< ?2. Let G \Cn =hrki with kjn and let t
be the largest integer such that ptjk. If ^G \Cn hrn0i, then
dimVB2i =
8>
<
>:
4pt; if G Cn
2pt; otherwise.
Proof. Let 2 ^G \Cn hrn0i. Then = rcn0 for some integer c and
i( ) = i(rcn0) = 2 cos 2 icn
0
n = 2 cos 2 i
00c = 2:
Therefore by Theorem 2.9 and Equation 3.4
dimVB2i = 1jG
j
X
2^G
?2i(e) 2i( ) = ?
2
i(e)
jG j
X
2^G
pq i( ) = p
q?2
i(e)
jG j
X
2^G \Cn
i( )
= 2p
q
jG j
X
2^G \Cn
2 = 4p
q
jG jj
^G \Cnj= 4pq
jG j
n
apq t =
4
jG j
n
ap
t:
Now if G Cn, then jG j = na, so dimVB2i = 4pt, and if G 6 Cn, then jG j = 2na , so
dimVB2i = 2pt.
Theorem 3.14. Let G = Dn and assume dimV 2. For xed i with 1 i < ?2 write
B = B2i . The space VB has an o-basis if and only if ?0 0 mod 4, where ?0 = ?gcd(?;i).
33
Proof. Suppose VB has an o-basis. Then by Theorem 2.8 it follows that VB has an o-basis
for each 2 . Let = (1;2;2;:::;2). Then G = f1;sg, so G \Cn = hrni. Now since q
is the largest such that pqjn, by Theorem 3.13 dimVB = 2pq. The space VB has an o-basis,
that is, an orthogonal basis of the form E = feB x j x 2G;1 x 2pqg. Consider the
subgroup J =hrpqiG of G. The index of J in G is pq, so by the pigeonhole principle there
is at least one right coset of J containing x and y for some 1 x;y 2pq with x6= y.
Then we have x 1y 2 J, so x 1y = rmpq for some integer m and 2 G . Therefore
x 1y G = rmpq G = rmpqG =frmpq;sr mpqg. Then by Corollary 3.6 and Equation 3.4
0 = (eB x;eB y) = 1jGj
X
2 x 1y G \Cn
?2i(e) 2i( ) = ?
2
i(e)
jGj
2
i(r
mpq) = ?
2
i(e)
jGj p
q i(rmpq):
So we get 0 = i(rmpq) = 2 cos 2 impqn = cos 2 im? which gives 2 im? = (2k + 1) 2 for some
integer k. Let i0 = igcd(?;i). Then
4i0m = 4imgcd(?;i) = (2k + 1)?gcd(?;i) = (2k + 1)?0:
So ?0 is divisible by 4.
Conversely suppose ?0 0 mod 4. Fix 2 such that dim VB 6= 0. Then by Theorem
3.12 we have ^G \Cn hrn0i where n0 = ngcd(n;i). Let ^G \Cn = hrai where ajn. Note
that a = a0n0 for some integer a0. We will rst show that there exists 2 ^Cn such that
(eB ;eB ) = 0. Let = r pq where = ?04 2Z. Note here that the set of p-regular elements of
r pqG \Cn is the same as r pq ^G \Cn since for some 2 ^Cn, 2 ^Cn if and only if 2 ^Cn.
Then by Corollary 3.6
(eB r pq;eB ) = 1jGj
X
2r pqG \Cn
?2i(e) 2i( ) = ?
2
i(e)
jGj
X
2r pq ^G \Cn
2i( )
= ?
2
i(e)
jGj
n
a 1X
=0
2i(r pq+ a) = ?
2
i(e)
jGj
n
a 1X
=0
pq i(r pq+ a):
34
Note that since 4j?0, the number i0 = igcd(?;i) is odd. Therefore using i00 = igcd(n;i) and = ?04
we get
i(r pq+ a) = 2 cos 2 i( p
q + a0n0)
n = 2 cos
2 i0gcd(?;i)
? +
2 i00gcd(n;i) a0n0
n
= 2 cos
i0
2 + 2 i
00 a0
= 2 cos i
0
2 = 0;
so (eB r pq;eB ) = 0. Let t be the largest such that pt ja. Then by Theorem 3.13, we have
dimVB = 2pt if G 6 Cn, in which case feB ;eB j 2H0g is an o-basis by Lemma 3.7 part
iii), and dimVB = 4pt if G Cn, in which case feB ;eB ;eB s ;eB s j 2H0g is an o-basis
by Lemma 3.7 part iv).
3.5 Irreducible symmetrization
In this section we will give necessary and su cient conditions for the existence of an
o-basis of the symmetry class of tensors corresponding to an irreducible Brauer character of
Dn. We will also show the existence of an orthogonal basis for the Brauer symmetry class
of tensors that consists of ordinary standard symmetrized tensors in the case of degree two
irreducible Brauer characters of Dn.
Let G = Dn. Recall that ?2i 2IBr(G) for 1 i< ?2 is of degree two and i; jl i; jl+i2
Irr(G) for 1 j pq 12 are of degree two. Recall also that 2i denotes the PI corresponding
to ?2i and that ^ 2i denotes the restriction of 2i to ^G.
Lemma 3.15. For each 1 i< ?2 we have
^ 2i = pq?2i:
Proof. By using Theorem 3.5 we write
^ 2i = X
a2Ai
^ a =
X
a2Ai
?2i =jAij?2i = pq?2i
35
Lemma 3.16. For each 1 i< ?2,
s?2i = jGjpqj^Gj
X
a2Ai
s a:
Proof. By the de nition of a symmetrizer s?2i = ?2i(e)j^Gj P 2^G?2i( ) . Now by the Lemma 3.15
above
s?2i = 2j^Gj
X
2^G
1
pq
^ 2i( ) = 2
j^Gj
X
2G
1
pq
2
i( ) ;
where the second equality holds because 2i vanishes o of ^G. Now by Theorem 3.5 we get
s?2i = 2pqj^Gj
X
2G
X
a2Ai
a( ) = jGjpqj^Gj
X
a2Ai
2
jGj
X
2G
a( ) = jGjpqj^Gj
X
a2Ai
a(e)
jGj
X
2G
a( )
= jGjpqj^Gj
X
a2Ai
s a;
which completes the proof.
Lemma 3.17. For each 1 i< ?2 we have
V?2i = _
X
a2Ai
V a (orthogonal direct sum):
Proof. Let v2V n. Then s?2i(v)2V?2i. Let g = jGjpqj^Gj. By using Theorem 3.16 we get
s?2i(v) =
X
a2Ai
s a(gv);
so V?2i Pa2AiV a. Now to show the other inclusion, note that for some arbitrary va2V n,
P
a2Ais a(va) is in
P
a2AiV a. Therefore there exists an element
P
a2Ais a(
1
gva) in V
n such
36
that
X
a2Ai
s a(va) = g
X
a2Ai
s a
X
a2Ai
s a(1gva)
= s?2i
X
a2Ai
s a(1gva)
2V?2i;
so Pa2AiV a V?2i as desired. In the above computation we have used the Lemma 3.16
and the fact that the symmetrizers corresponding to ordinary irreducible characters are
orthogonal projections (Theorem 2.1). Since ordinary symmetrized spaces are orthogonal by
Corollary 2.2 we have the result.
Recall that for 2Irr(G) a standard decomposable symmetrized tensor corresponding
to is given by e where 2 n;m.
Theorem 3.18. For 1 i < ?2, V?2i has an orthogonal basis consisting of decomposable
tensors of the form e , 2Irr(G), if and only if ?0 0 mod 4, where ?0 = ?gcd(?;i).
Proof. Suppose V?2i has an orthogonal basis of the stated form. Then in particular V i has an
orthogonal basis due to Lemma 3.17. Therefore by [10, Theorem 3.1] we get n 0 mod 4i2
where i2 is the power of 2 such that ii2 is odd. This means that i2 is a factor of ? and further
gcd(?;i)
i2 = gcd(
?
i2;
i
i2 ) is odd. Then since ?
0 = ?
gcd(?;i), for some integer m we get
4m = ni
2
= p
q?
i2 =
pq?0gcd(?;i)
i2 = p
q?0gcd(?;i)
i2 ;
so 4j?0 as desired.
Conversely suppose ?0 0 mod 4. Let ?2 be the largest factor of ? that is expressed
as a power of 2. Now since 4 j ?0 = ?gcd(?;i) it is the case that 4 j ?2gcd(?2;i2), which gives
gcd(?2;i2) = i2, so 4i2 j?2 and hence 4i2 j?. So we have n 0 mod 4i2 and therefore V i
has an o-basis by [10, Theorem 3.1]. Now, since 4i2 j?, we have ? = a4i2 for some integer a,
so for a xed k where 1 k pq 12 we have
k? i = k(a4i2) i2i20 = i2(ka4 i20);
37
where i20 = ii2 is odd. Then if (k? i)2 is the largest factor of k? i as a power of 2, then
(k? i)2 = i2, so n 0 mod 4(k? i)2. Therefore V k? i has an orthogonal basis by [10,
Theorem 3.1]. This means V a has an o-basis for each a2Ai and hence by Lemma 3.17 we
get the result.
Recall for a character of G, V =he j 2Gi.
Lemma 3.19. For 1 i< ?2 and 2 n;m
V?2i = _
X
a2Ai
V a (orthogonal direct sum):
Proof. Fix 1 i < ?2 and 2 n;m. Take s?2i(w) 2V?2i for some w2V . Let g = jGjpqj^Gj.
Then by Lemma 3.16,
s?2i(w) = g
X
a2Ai
s a(w) =
X
a2Ai
s a(gw);
and this gives the inclusion V?2i Pa2AiV a : For the other inclusion consider arbitrary
wa2V and set
v :=
X
a2Ai
s a(wa)2
X
a2Ai
V a :
Note that v is also in V , so 1gv2V . Then
v = g
X
a2Ai
s a(1gv) = s?2i(1gv)2V?2i ;
so we have Pa2AiV a V?2i .
Since orbital subspaces are orthogonal by Theorem 2.3 we have an orthogonal direct
sum as desired.
38
For 2 n;m recall that G is the stabilizer subgroup of and = f 2 j
P
2G ( )6= 0g. Using Theorem 2.3 and Equation 2.3 we have
V = _
X
2
V = _
X
2
V : (3.5)
For 1 i< ?2 put i = Sa2Ai a.
Theorem 3.20. For 1 i< ?2 we have
V?2i = _
X
2 i
V?2i :
Proof. By using Lemma 3.17 and Equation 3.5 we get,
V?2i = _
X
a2Ai
V a = _
X
a2Ai
_X
2 a
V a = _
X
2 i
_X
a2Ai
V a = _
X
2 i
V?2i ;
where the last equality is due to Lemma 3.19.
Lemma 3.21. For each 1 i< ?2, 2 m;n, and 2G, we have
(e?2i ;e?2i ) = g2
X
a2Ai
(e a ;e a );
where g =jGj=(pqj^Gj).
Proof. Fix 1 i< ?2, 2 m;n, and 2G. By Lemma 3.16 we get,
e?2i = s?2ie = g
X
a2Ai
s ae = g
X
a2Ai
e a ;
so
(e?2i ;e?2i ) = g2(
X
a2Ai
e a ;
X
a2Ai
e a ) = g2
X
a2Ai
(e a ;e a ):
39
Lemma 3.22. For a xed integer a consider the list of numbers of the form k? + a where
k = 0;:::;pq 1. Then for any integer 0 q, there are exactly pq numbers in the list
that are divisible by p .
Proof. Let 0 q. We rst show that for any list of p numbers of the form k?+a where
k = 0;:::;p 1 there is exactly one number in the list divisible by p . We will show that
the remainders when divided by p of the numbers k?+a where k = 0;:::;p 1 are distinct.
Let 0 k1 <
>:
2; if n is odd;
4; if n is even.
Tj =
8>
><
>>:
fk?j1 k pq 12 g; j = 1;2;
f?2 +k?j0 k pq 12 1g; j = 3;4.
(3.6)
Then using Theorem 3.5 we can write
1j = j +
X
t2Tj
t; for 1 j "; (3.7)
2i = i +
pq 1
2X
k=1
( k?+i + k? i); for 1 i< ?2: (3.8)
47
First considering the PIs corresponding to degree one Brauer characters and using the
Equation 2.4 the symmetrizer is given by
s 1i = ?
1
i(e)
jGj
X
2G
1i( ) :
Proposition 3.28. Fix j with 1 j ". Then
s 1j = s j + 12
X
t2Tj
s t:
Proof. By Equations 2.4 and 3.7 we get
s 1j = ?
1
j(e)
jGj
X
2G
1j( ) = ?
1
j(e)
jGj
X
2G
( j( ) +
X
t2Tj
t( ))
= 1jGj
X
2G
j( ) +
X
t2Tj
1
jGj t( )
=
1
j(e)
jGj
X
2G
j( ) +
X
t2Tj
t(e)
2jGj t( )
= s j + 12
X
t2Tj
s t:
Recall that is a set of representatives of the orbits of n;m under the action given in
Equation 2.1. Then for 1 j " by Equation 2.5 we have
V 1j = _
X
2
V 1j :
Theorem 3.29. Fix j with 1 j " and x 2 . Then
V 1j = V j _+ _
X
t2Tj
V t (orthogonal direct sum)
48
Proof. Let s 1j(v)2V 1j . Then it is clear from Proposition 3.28 above that s 1j(v)2V j +
P
t2Tj V
t
. To show the other inclusion consider s j(w) +
P
t2Tj s t(wt)2V
j
+
P
t2Tj V
t
.
Note that s j(w) +Pt2Tj s t(2wt)2V . Then
s j(w) +
X
t2Tj
s t(wt) = s j(w) + 12
X
t2Tj
s t(2wt) = s j(w) + 12
X
t2Tj
X
b
s bs t(2wt)
=
s j + 12
X
b
s b
s j(w) +
X
t2Tj
s t(2wt)
=
s j + 12
X
b
s b
s j(w) +
X
t2Tj
s t(2wt)
= s 1j
s j(w) +
X
t2Tj
s t(2wt)
2V 1j ;
where we have used that s s = s for all ; 2 Irr(G). This shows that V 1j =
V j + Pt2Tj V t . The orthogonality follows from the argument in the proof of Theorem
2.3.
Fix j with 1 j " and 2 n;m. We have by Theorem 2.6
(e 1j ;e 1j ) = ?
1
j(e)
2
jGj2
X
2G
X
2G
1j( 1 ) 1j( ):
Below we state as a corollary a useful form of this inner product.
Corollary 3.30.
(e 1j ;e 1j ) = 12jGj
X
2G
2 j( 1 ) +
X
t2Tj
t( 1 )
= 12jGj
X
2G
j( 1 ) + 1j( 1 )
:
49
Proof. Using Equation 3.7, Theorem 1.13 and Theorem 1.14 we get,
(e 1j ;e 1j ) = ?
1
j(e)
2
jGj2
X
2G
X
2G
1j( 1 ) 1j( )
= ?
1
j(e)
2
jGj2
X
2G
X
2G
j( 1 ) +
X
t2Tj
t( 1 )
j( ) +
X
u2Tj
u( )
= ?
1
j(e)
2
jGj2
X
2G
X
2G
j( 1 ) j( 1) +
X
t2Tj
t( 1 ) t( 1)
= ?
1
j(e)
2
jGj2
X
2G
X
2G
j( 1 ) j( 1) +
X
t2Tj
X
2G
t( 1 ) t( 1)
= ?
1
j(e)
2
jGj2
X
2G
jGj
j(e) j(
1 ) +X
t2Tj
jGj
t(e) t(
1 )
:
Now since ?1j(e) = 1j(e) = 1 and t(e) = 2 for each t2Tj we get,
(e 1j ;e 1j ) = 12jGj
X
2G
2 j( 1 ) +
X
t2Tj
t( 1 )
= 12jGj
X
2G
j( 1 ) + 1j( 1 )
:
Theorem 3.31. For i = 1;2;3;4 the space V 1j has an o-basis if and only if at least one of
the following holds.
i) dimV = 1,
ii) p = 2,
iii) n is not divisible by p.
Proof. If dimV = 1, then V 1j =he 1j i, where = (1;1;:::;1), so V 1j has o-basisfe 1j gor;
according as dimV 1j is 1 or 0.
Assume p = 2. Then 1j is an ordinary irreducible character of the group ^G =hrpqi
Cn. Then using Freese?s result for the dimension of an orbital subspace [6], dimV 1j is at
most one and hence has an o-basis. So V 1j has an o-basis by Equation 2.5.
50
Assume n is not divisible by p. Then ^G = G and hence 1j = j. Since j is of degree
one, each orbital subspace has dimension at most one and hence V 1j has an o-basis by the
same argument as in the above paragraph.
Now assume that none of the three conditions stated in the theorem holds. Let =
(1;2;:::;2). which is in n;m since dimV 2. We show that (e 1j ;e 1j )6= 0 for every 2G.
Note that G =f1;sg.
First let 2Gn ^G. Since p6= 2 we have 2Cn. Using Corollary 3.30 we get,
2jGj(e 1j ;e 1j ) = ( j + 1j)( 1) + ( j + 1j)( 1s) = j( ) + 2 j(s )6= 0:
Now let 2 ^G. Then 2frapq;srb j 0 a < ?;0 b < ng. Assume = rapq for some
0 a 1. So we conclude that V 1j does not have an o-basis and the proof
is complete.
Now we will consider the PIs corresponding to degree two Brauer characters of G = Dn.
Corollary 3.32. For 1 i< ?2
s 2i = p
qj^Gj
jGj s?2i:
Proof. Let 1 i< ?2. By Equation 2.4 and Lemma 3.15,
s 2i = ?
2
i(e)
jGj
X
2G
2i( ) = ?
2
i(e)
jGj
X
2^G
^ 2i( ) = pq?2i(e)
jGj
X
2^G
?2i( ) = p
qj^Gj
jGj s?2i:
Recall that V 2i = s 2i(VNn) and V?2i = s?2i(VNn).
Theorem 3.33. For 1 i< ?2 we have
V 2i = V?2i:
Proof. Take s 2i(w)2V 2i. Then using Corollary 3.32 we get
s 2i(w) = p
qj^Gj
jGj s?2i(w) = s?2i(
pqj^Gj
jGj w)2V?2i;
52
which shows the inclusion V 2i V?2i. Then take s?2i(u) 2V?2i. Again using Corollary 3.32
we see that
s?2i(u) = jGjpqj^Gjs 2i(u) = s 2i( jGjpqj^Gju)2V 2i;
which gives the other inclusion. So we get V 2i = V?2i.
Theorem 3.34. Fix i where 1 i< ?2 and let 2 n;m. Then
e 2i = p
qj^Gj
jGj e
?2i
:
Proof. By Corollary 3.32 we get that,
e 2i = s 2i(e ) = p
qj^Gj
jGj s?2i(e ) =
pqj^Gj
jGj e
?2i
:
Lemma 3.35. Fix i with 1 i < ?2. Then V 2i has an o-basis if and only if V?2i has an
o-basis.
Proof. We observe that,
(e 2i ;e 2i ) = (p
qj^Gj
jGj e
?2i
;
pqj^Gj
jGj e
?2i
) = (
pqj^Gj
jGj )
2(e?2i ;e?2i ):
So (e 2i ;e 2i ) = 0 if and only if (e?2i ;e?2i ) = 0.
Theorem 3.36. Fix i with 1 i < ?2. Then V 2i has an o-basis if and only if either
dimV = 1 or ?0 is divisible by 4, where ?0 = ?=gcd(?;i).
Proof. The result follows from Lemma 3.35 and Theorem 3.26.
Theorem 3.37. Fix i with 1 i < ?2. Then V 2i has an orthogonal basis consisting of
decomposable tensors of the form e if and only if ?0 0 mod 4. Where ?0 = ?gcd(?;i).
Proof. The result follows from Theorem 3.33 and Theorem 3.18.
53
Chapter 4
Symmetric group
In this chapter we will discuss some results associated with Brauer characters of a
symmetric group.
For some positive integer n, the symmetric group Sn of degree n is the group of permu-
tations of the set f1;2;:::;ng with the binary operation de ned by function composition.
The number of elements of Sn is n!. A permutation of Sn is given by
=
0
B@ 1 2 n
(e) (2) (n)
1
CA
or
= (i11;:::;i1r1)(i21;:::;i2r2) (is1;:::;isrs);
where 1 iab n, iab = icd implies a = c and b = d, and (iab) = ia(b+1) (b 1, then V
does not have an o-basis.
Proof. Let = (1;:::;1;2). We can assume to be the representative of the orbit containing
it, so 2 . Observe that G =f 2Snj (n) = ng = Sn 1:
By Theorem 2.4, for any 2G we have
(e ;e ) = (e)
2
j^Gj2
X
2^G
X
2 1 ^G\G
( ) ( 1 1) = (e)
2
j^Gj2
X
2^G
X
2 1G \^G
( ) ( 1)
= 1j^Gj2
X
2^G
j 1G \ ^Gj:
So (e ;e ) = 0 only when 1G \ ^G =;for some 2 ^G. We will show that for all 2G
there is some 2 ^G such that 1G \ ^G6=;, which implies that (e ;e )6= 0.
We claim here that the cyclic group H =h(1;2;:::;n)i is a set of right coset represen-
tatives of G in G. For h1;h2 2H with h1 6= h2 we have h1h 12 (n) 6= n, so G h1 6= G h2.
AlsojG : G j= n =jHj. Now since G H = G we havefe j 2Gg=fe hjh2Hg. So it is
enough to show (e h;e )6= 0 for all h2H. Let h2H. If h = e, then letting = e2 ^G we
get h 1G \ ^G = G \ ^G, and this latter set contains the transposition (1;2), since n 3
and p6= 2, so it is nonempty as desired. Now assume that h6= e. Then h(n)6= n and there is
1 m n 1 such that h(m) = n. Now let = (m;n) and observe that we get h 1(n) = n,
so h 1 2Sn 1 = G . Therefore, 2 ^G, since p6= 2, and h 1G \ ^G = G \ ^G6= ; as
desired.
55
If dimV > 1, then V does not have an o-basis, so by Theorem 2.3 the space V does
not have an o-basis.
The following example is a case, where we do not have an o-basis for V .
Example 4.2. Let G = S3. Assume dimV 2 and p = 3. Let be the principal Brauer
character of G and let = (1;1;2). Then dimV > 1.
Proof. Note since p = 3 we have ^G = f1;(a;b);(a;c);(b;c)g. Write e = e (112). Then for
= (a;b;c)2G we get e = e (211). Now by Equation 2.2 we get
e (112) = 14(2e(112) +e(211) +e(121));
e (211) = 14(2e(211) +e(112) +e(121)):
By inspection we see that e (112) and e (211) are linearly independent, so dimV > 1.
The alternating group G = An is the subgroup of Sn consisting of all the even per-
mutations of Sn. Let be the irreducible Brauer character of G with ( ) = 1 for all
2G.
Theorem 4.3. Let G = An. Assume that dimV 2, n( 3) is odd, and p 6= 2. Then
(e ;e ) 6= 0 for all 2G, where = (1;:::;1;2). In particular, if dimV > 1, then V
does not have an o-basis.
Proof. Let = (1;:::;1;2). We can assume to be the representative of the orbit containing
it, so 2 . Observe that G = f 2 Anj (n) = ng = An 1. Since n is odd H =
h(1;2;:::;n)i G. Following the same argument as in the proof of Theorem 4.1 we can
show that H is a set of right coset representatives of G in G. The argument to show
(e ;e )6= 0 for all 2G is the same as in the proof of Theorem 4.1.
56
4.1 Special case S4
The symmetric group G = S4 of degree 4 is the group of permutations of a setfa;b;c;dg.
Let p = 2. Then there are two, 2-regular conjugacy classes and
^G =f1;(abc);(acb);(abd);(adb);(acd);(adc);(bcd);(bdc)g:
The Brauer character table of G in this case is (see [12, page 431])
( ) ( )
?1 1 1
?2 2 1
Theorem 4.4. Assume that dimV > 1. The space V?i does not have an o-basis for i 2
f1;2g.
Proof. Fix i2f1;2g and put ? = ?i. To show V? does not have an o-basis it is enough by
Theorem 2.3 to show that V? does not have an o-basis for some 2 .
Let = (1;1;1;2). Then G =f1;(ab);(ac);(bc);(abc);(acb)g.
Let H = f1;(ab)(cd);(ac)(bd);(ad)(bc)g and observe that G = G H. So fe? j 2
Gg=fe? j 2Hg. We will compute the value of (e? ;e? ) using the formula
(e? ;e? ) = ?(e)
2
j^Gj2
X
2^G
X
2 1G \^G
?( )?( 1) (4.1)
57
(see proof of Theorem 4.1). The following table lists the products 1, with 2 ^G and
2H.
n 1 (abc) (acb) (abd) (adb) (acd) (adc) (bcd) (bdc)
1 1 (abc) (acb) (abd) (adb) (acd) (adc) (bcd) (bdc)
(ab)(cd) (ab)(cd) (acd) (bcd) (adc) (bdc) (abc) (abd) (acb) (adb)
(ac)(bd) (ac)(bd) (bdc) (abd) (acb) (acd) (adb) (bcd) (adc) (abc)
(ad)(bc) (ad)(bc) (adb) (adc) (bcd) (abc) (bdc) (acb) (abd) (acd)
Now we will look at the partP 2^GP 2 1G \^G?( )?( 1) on the right side of the Equation
4.1. Note from the table above that 1 is even for all and . Now since ^G does not
contain odd cycles we can neglect the products of 1 with the elements of the form ( ) in
G when considering 1G \ ^G.
When = 1, for all cases of 6= 1 we get:
1( ) = ( )( )( ) = ( )( ) and 1( ) = ( )( )( ) = ( ) (two times):
From the table we see if = ( ), then for each 2H we have 1 = ( ). The
table below lists the products 1 (columns) of the form ( ) with the even permutations
in G (rows).
(abc) (acb) (abd) (adb) (acd) (adc) (bcd) (bdc)
1 (abc) (acb) (abd) (adb) (acd) (adc) (bcd) (bdc)
(abc) (acb) 1 (ad)(bc) (bcd) (abd) (ab)(cd) (ac)(bd) (adc)
(acb) 1 (abc) (acd) (ac)(bd) (ad)(bc) (bdc) (adb) (ab)(cd)
58
Assume that ? = ?1. Since ?( ) = 1 for all 2 ^G and 1G \ ^G6=;we get (e? ;e? )6= 0
for all 2H. Now write e? = e?(1112). Then e? (ad) = e?(2111). Now by Equation 2.2 we get
e?(1112) = 19(3e(1112) + 2e(2111) + 2e(1211) + 2e(1121));
e?(2111) = 19(3e(2111) + 2e(1112) + 2e(1211) + 2e(1121)):
By inspection we note that e?(1112) and e?(2111) are linearly independent, so dimV? 2. So
we conclude that V? does not have an o-basis and hence V? does not have an o-basis.
Now assume that ? = ?2. Then to evaluate (e? ;e? ) for each 6= 1 in H we observe
from the computations above in the cases of = 1 = ( ) and of the form ( ) that
X
2^G
X
2 1G \^G
?( )?( 1) = 4?
( )
?
( )
+ 16?
( )
?
( )
:
So by the Brauer character table given above and Equation 4.1 we get
(e? ;e? ) = (e)
2
j^Gj2
4(2)( 1) + 16( 1)( 1)
6= 0:
Now note that by Equation 2.2 we get
e?(1112) = 29( 2e(2111) 2e(1211) 2e(1121));
e?(2111) = 29( 2e(1112) 2e(1211) 2e(1121)):
By inspection e?(1112) and e?(2111) are linearly independent implying dimV? 2. So we
conclude that V? has no o-basis and therefore V? does not have an o-basis.
59
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61