MMath

Dimensions of Symmetry Classes of Tensors
by
Larry Huang
A thesis
presented to the University of Waterloo
in fulﬁlment of the
thesis requirement for the degree of
Master of Mathematics
in
Pure Mathematics
Waterloo, Ontario, Canada, 1993
c Larry Huang 1993

I hereby declare that I am the sole author of this thesis.
I authorize the University of Waterloo to lend this thesis to other institutions
or individuals for the purpose of scholarly research.
I further authorize the University of Waterloo to reproduce this thesis by pho-
tocopying or by other means, in total or in part, at the request of other institutions
or individuals for the purpose of scholarly research.
ii

The University of Waterloo requires the signatures of all persons using or pho-
tocopying this thesis. Please sign below, and give address and date.
iii

Abstract
This paper will study ways to calculate dimensions of symmetry classes of ﬁnite
dimensional complex tensor products spaces. General results as well as several
methods for explicitly calculating the dimensions of the symmetry classes are pre-
sented.
iv

Acknowledgements
I would like to thank my supervisor Dr. Larry J. Cummings who gave me
excellent guidance and provided me with most of the research material. Also of
invaluable help during preparation of this thesis is fellow graduate student Jun
Wu, who gave me useful pointers.
I would like to thank the Department of Pure Mathematics, University of Wa-
terloo, for their ﬁnancial support which made my graduate study possible.
v

Contents
1 Introduction 1
1.1 Preliminary Deﬁnitions and Notations . . . . . . . . . . . . . . . . 1
1.2 Symmetry Classes of Tensors . . . . . . . . . . . . . . . . . . . . . . 6
2 Dimensions of Symmetry Classes 11
2.1 General Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 Freese’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.3 Using Disjoint Cycle Decomposition of Sm . . . . . . . . . . . . . . 22
2.4 Subspaces of the Orbital Subspaces . . . . . . . . . . . . . . . . . . 24
3 Some Special Results 33
3.1 General Results on Cyclic Groups . . . . . . . . . . . . . . . . . . . 33
3.2 Simpliﬁed Formulas for Cyclic Groups . . . . . . . . . . . . . . . . 36
3.3 Dihedral Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.4 Linear Symmetry Classes . . . . . . . . . . . . . . . . . . . . . . . . 51
3.5 Counting Restricted Orbits . . . . . . . . . . . . . . . . . . . . . . . 53
vi

Chapter 1
Introduction
1.1 Preliminary Definitions and Notations
Let V be a finite dimensional inner product space of dimension n over a field F (for
the purpose of this paper F is the field of complex numbers). The usual Cartesian
Product space of m copies of V is denoted by V m
, and the space of mth
tensor
power over V is denoted by ⊗m
V . The inner product on ⊗m
V is the one induced
from V .
Let Γm,n be the set of all the functions from {1, 2, . . . , m} to {1, 2, . . . , n}. For
α ∈ Γm,n, α is considered as a sequence of positive integers of length m chosen from
{1, 2, . . . , n}.
If V has an ordered basis E = {e1, e2, . . . , en}, then the set of decomposable ten-
sors {eα(1) ⊗ . . . ⊗ eα(m) | α ∈ Γm,n} form a basis of ⊗m
V . If E is orthonormal then
so is {eα(1) ⊗ . . . ⊗ eα(m) | α ∈ Γm,n}. (A common short hand for the decomposable
tensor eα(1) ⊗ . . . ⊗ eα(m) is e⊗
α .)
1

CHAPTER 1. INTRODUCTION 2
Let Sm be the symmetry group on {1, 2, . . . , m}. That is, Sm is the set of all
bijections from {1, 2, . . . , m} to {1, 2, . . . , m}.
For any ﬁxed σ ∈ Sm, deﬁne φ : V m
→ ⊗m
V by φ(v1, . . . , vm) = v⊗
σ−1 . By
straight forward calculations it can be easily shown that φ is multilinear. By the
Universal Factorization Property of ⊗m
V , there is a unique linear operator P(σ) on
⊗m
V such that for all v1, . . . , vm ∈ V , φ(v1, . . . , vm) = P(σ)(v1 ⊗ . . . ⊗ vm). That
is, the following diagram is commutative:
V m -⊗
⊗m
V
⊗m
V
?
P(σ) (unique)
HHH
HHHHj
φ
Obviously for all v1, . . . , vm ∈ V , P(σ)(v1 ⊗ . . . ⊗ vm) = vσ−1(1) ⊗ . . . ⊗ vσ−1(m).
For any α ∈ Γm,n, let µi = vα(i) for i = 1, . . . , m. we have:
P(σ)(v⊗
α ) = P(σ)(vα(1) ⊗ . . . ⊗ vα(m))
= P(σ)(µ1 ⊗ . . . ⊗ µm)
= µσ−1(1) ⊗ . . . ⊗ µσ−1(m). (1.1)
Since µi = vα(i) for i = 1, . . . , m, we have µσ−1(i) = vα(σ−1(i)) = vασ−1(i), for i =
1, . . . , m. Thus (1.1) above becomes
P(σ)(v⊗
α ) = vασ−1(1) ⊗ . . . ⊗ vασ−1(m)
= v⊗
ασ−1 .
Note that for all σ, τ ∈ Sm, P(σ) ◦ P(τ) = P(στ). Further, with respect to the
induced inner product, the adjoint of P(σ) is P(σ)∗
= P(σ−1
) = (P(σ))−1
(where
P(σ)∗
denotes the adjoint of P(σ)).

To show this, let u1 ⊗ . . . ⊗ um and v1 ⊗ . . . ⊗ vm be two decomposable tensors
in ⊗m
V . Observe that
(P(σ)(v1 ⊗ . . . ⊗ vm), u1 ⊗ . . . ⊗ um) = (v1 ⊗ . . . ⊗ vm, P(σ−1
)(u1 ⊗ . . . ⊗ um))
since by expanding the left side, we have
(vσ−1(1) ⊗ . . . ⊗ vσ−1(m), u1 ⊗ . . . ⊗ um) =
m
i=1
(vσ−1(i), ui),
and by expanding the right side we have
(v1 ⊗ . . . ⊗ vm, uσ(1) ⊗ . . . ⊗ uσ(m)) =
m
i=1
(vi, uσ(i)),
and we see that the multipliers and inner products match. Thus P(σ)∗
= P(σ−1
) on
the set of decomposables. But the decomposables span ⊗m
V , so P(σ)∗
= P(σ−1
).
Deﬁnition 1.1.1 Let G be a subgroup of Sm, and λ be an irreducible character of
G. Deﬁne the linear operator T(G, λ) on ⊗m
V by
T(G, λ) =
λ(e)
|G| σ∈G
λ(σ)P(σ)
(where e is the identity element in G). The image of ⊗m
V under T(G, λ) is denoted
by Vλ(G).
In the trivial case where G = {e} (and so the only irreducible character of G is
λ = 1), we have:
T(G, λ) =
λ(e)
|G| σ∈G
λ(σ)P(σ) = I⊗mV .
So in this case Vλ(G) is just ⊗m
V .
By straight-forward calculations, it can be easily shown that for any σ ∈ G,
P(σ) ◦ T(G, λ) = T(G, λ) ◦ P(σ).
We have the following property for T(G, λ).

Proposition 1.1.2 With respect to the induced inner product, T(G, λ) is an or-
thogonal projection.
Proof: We use the following result from group representation theory: Let λ and
χ be irreducible characters of the ﬁnite group G aﬀorded by representations A and
B of G respectively. Then for any π ∈ G,
σ∈G
λ(σ−1
)χ(σπ) =



|G|λ(π)/λ(e) if A ≡ B,
0 otherwise
(1.2)
(see [37], page 11).
We have:
T(G, λ)T(G, λ) =
λ(e)2
|G|2
σ∈G
λ(σ)P(σ)
τ∈G
λ(τ)P(τ)
=
λ(e)2
|G|2
σ,τ∈G
λ(σ)λ(τ)P(στ)
=
λ(e)2
|G|2
π∈G σ∈G
λ(σ)λ(σ−1
π) P(π) by writing π = στ
=
λ(e)
|G| π∈G
λ(π)P(π) by (1.2)
= T(G, λ).
Thus T(G, λ) is a projection.
By straight-forward calculations, it is easily shown that
λ(e)
|G| σ∈G
λ(σ)P(σ)∗
(where λ(σ) denotes the complex conjugate of λ(σ)) has all the properties of
T(G, λ)∗
. By the uniqueness of adjoints and the fact that P(σ)∗
= P(σ−1
), we
have
T(G, λ)∗
=
λ(e)
|G| σ∈G
λ(σ)P(σ)∗
=
λ(e)
|G| σ∈G
λ(σ−1
)P(σ−1
) = T(G, λ). 2

Theorem 1.1.3 Let I(G) denote the set of all irreducible characters of G, where
G is a subgroup of Sm. If λ, χ ∈ I(G) and λ = χ, then T(G, λ) ◦ T(G, χ) = 0.
Moreover,
λ∈I(G)
T(G, λ) is the identity map on ⊗m
V .
Proof: Using (1.2), we have:
T(G, λ)T(G, χ) =
λ(e)χ(e)
|G|2
π∈G σ∈G
λ(σ)χ(σ−1
π) P(π)
= 0.
To show that
λ∈I(G)
T(G, λ) = I⊗mV , we use the following result from group
representation theory:
Let σ ∈ G, where G is any ﬁnite group. Then
χ∈I(G)
χ(e)χ(σ) =



|G| if σ = e,
0 otherwise
(1.3)
(see [37], page 17).
Using (1.3), we get:
λ∈I(G)
T(G, λ) =
λ∈I(G)
(
λ(e)
|G| σ∈G
λ(σ)P(σ))
=
1
|G| λ∈I(G)
λ(e)
σ∈G
λ(σ)P(σ)
=
1
|G| σ∈G
(
λ∈I(G)
λ(e)λ(σ))P(σ)
= P(e)
= I⊗mV . 2
Deﬁnition 1.1.4 Vλ(G) is a subspace of ⊗m
V (since T(G, λ) is linear), and is
called the symmetry class of tensors associated with G and λ.

Here is an easy consequence of the two preceding results:
Corollary 1.1.5 For any subgroup G of Sm, ⊗m
V can be written as
λ∈I(G)
Vλ(G),
an orthogonal direct sum. 2
Thus the study of ⊗m
V can be reduced to the study of the subspaces Vλ(G).
1.2 Symmetry Classes of Tensors
The following result will be used when we attempt to ﬁnd a basis of Vλ(G) in the
next chapter.
Proposition 1.2.1 If λ(e) = 1, that is, λ is one-dimensional, then x ∈ Vλ(G) if
and only if P(σ−1
)(x) = λ(σ)x for all σ ∈ G.
Proof: Let E = {e1, . . . , en} be a basis of V . Then {e⊗
α | α ∈ Γm,n} is a basis of
⊗m
V .
Suppose x ∈ Vλ(G). Write x in general form:
x = T(G, λ)(
α∈Γm,n
cαe⊗
α )
=
λ(e)
|G| τ∈G
λ(τ)P(τ)
α∈Γm,n
cαe⊗
α
=
λ(e)
|G| τ∈G
λ(τ)
α∈Γm,n
cαe⊗
ατ−1 .
Using straight-forward calculations, we have:
P(σ−1
)(x) = P(σ−1
)(
λ(e)
|G| τ∈G
λ(τ)
α∈Γm,n
cαe⊗
ατ−1 )

=
λ(e)
|G| τ∈G
λ(τ)
α∈Γm,n
cαe⊗
ατ−1σ
=
λ(e)
|G| τ∈G
λ(στ)
α∈Γm,n
cαe⊗
ατ−1
=
λ(e)
|G| τ∈G
λ(σ)λ(τ)
α∈Γm,n
cαe⊗
ατ−1 since λ is one-dimensional
= λ(σ)
λ(e)
|G| τ∈G
λ(τ)
α∈Γm,n
cαe⊗
ατ−1
= λ(σ)x.
The converse is easily shown by observing that x = T(G, λ)(x) ∈ Vλ(G):
T(G, λ)(x) =
λ(e)
|G| σ∈G
λ(σ)P(σ)(x)
=
1
|G| σ∈G
λ(σ)λ(σ−1
)(x)
=
1
|G| σ∈G
λ(e)x
=
1
|G| σ∈G
x
= x. 2
Deﬁnition 1.2.2 Denote T(G, λ)(v1 ⊗ . . . ⊗ vm) by v1 ∗ . . . ∗ vm. It is called a
decomposable symmetrized tensor. (For α ∈ Γm,n, a common short hand for
vα(1) ∗ . . . ∗ vα(m) is v∗
α.)
Note that in the above deﬁnition, the ∗ in v1 ∗ . . . ∗ vm depends on G and λ.
We can consider v1 ∗ . . . ∗ vm as the part of v1 ⊗ . . . ⊗ vm that belongs to Vλ(G).
Also note that Vλ(G) is spanned by, but not necessarily equal to, {v1 ∗ . . . ∗ vm |
v1, . . . , vm ∈ V }.

By straight-forward calculations, it is easily shown that P(σ−1
)(v1 ∗ . . . ∗ vm) =
v∗
σ, for all σ ∈ G.
Deﬁnition 1.2.3 Let G be a subgroup of Sm, λ an irreducible character of G, W
a vector space over F. A multilinear map φ : V m
→ W is said to be symmetric
with respect to G and λ if for all v1, . . . , vm ∈ V ,
φ(v1, . . . , vm) =
λ(e)
|G| σ∈G
λ(σ−1
)φ(vσ(1), . . . , vσ(m)).
Proposition 1.2.4 Let G be a subgroup of Sm, λ be a one-dimensional character
of G, and φ : V m
→ Vλ(G) by φ(v1, . . . , vm) = v1 ∗ . . . ∗ vm. Then φ is multilinear
and symmetric with respect to G and λ (This is where the term “symmetry class”
comes from).
Proof: That φ is multilinear comes from straight-forward calculations.
Let v1, . . . , vm ∈ V , we have:
λ(e)
|G| σ∈G
λ(σ−1
)φ(vσ(1), . . . , vσ(m))
=
λ(e)
|G| σ∈G
λ(σ−1
)(vσ(1) ∗ . . . ∗ vσ(m))
=
λ(e)
|G| σ∈G
λ(σ−1
)(
λ(e)
|G| τ∈G
λ(τ)P(τ)(vσ(1) ⊗ . . . ⊗ vσ(m)))
=
λ(e)
|G|2
σ∈G τ∈G
λ(σ−1
)λ(τ)(vστ−1(1) ⊗ . . . ⊗ vστ−1(m))
=
λ(e)
|G|2
σ∈G τ∈G
λ(τσ−1
)P(τσ−1
)(v1 ⊗ . . . ⊗ vm) since λ is one-dimensional
=
λ(e)
|G|2
σ∈G τ∈G
λ(τ)P(τ)(v1 ⊗ . . . ⊗ vm)
=
λ(e)
|G| τ∈G
λ(τ)P(τ)(v1 ⊗ . . . ⊗ vm)

= T(G, λ)(v1 ⊗ . . . ⊗ vm)
= v1 ∗ . . . ∗ vm
= φ(v1, . . . , vm) 2
The following result is known as the Universal Factorization Property for Sym-
metric Multilinear Functions:
Theorem 1.2.5 Let G be a subgroup of Sm, λ be an irreducible character of G, W
a vector space over F, and φ : V m
→ W be multilinear and symmetric with respect
to G and λ. Then there is a unique linear map hλ : Vλ(G) → W such that for all
v1, . . . , vm ∈ V, hλ(v1 ∗ . . . ∗ vm) = φ(v1, . . . , vm). That is, the following diagram is
commutative:
V m -∗ Vλ(G)
W
?
hλ
HHH
HHHHj
φ
Proof: By the universal factorization property of ⊗m
V , there is a unique linear
map h : ⊗m
V → W such that h(v1 ⊗ . . . ⊗ vm) = φ(v1, . . . , vm), for all v1, . . . , vm ∈
V . In particular, for all v1, . . . , vm ∈ V and for all σ ∈ G, we have:
h(vσ(1) ⊗ . . . ⊗ vσ(m)) = φ(vσ(1), . . . , vσ(m)).
Multiply both sides by
λ(e)
|G|
λ(σ−1
), and sum on σ ∈ G, we get:
σ∈G
λ(e)
|G|
λ(σ−1
)h(vσ(1) ⊗ . . . ⊗ vσ(m)) =
σ∈G
λ(e)
|G|
λ(σ−1
)φ(vσ(1), . . . , vσ(m)).
Noting that h is linear and φ is symmetric with respect to G and λ, we re-arrange
the above and get:
h(T(G, λ)(v1 ⊗ . . . ⊗ vm)) = φ(v1, . . . , vm),

or:
h(v1 ∗ . . . ∗ vm) = φ(v1, . . . , vm).
Deﬁne hλ to be the restriction of h to Vλ(G). 2

Chapter 2
Dimensions of Symmetry Classes
In last chapter we showed that ⊗m
V is an orthogonal direct sum of the symmetry
classes Vλ(G) over all irreducible characters λ of G. Indeed, ⊗m
V itself is Vλ(G) in
the case where G = {e} (and hence λ = 1). Thus to study ⊗m
V we need to study
Vλ(G).
One of the most basic concerns about Vλ(G) is how to ﬁnd its dimension. In
the rest of this thesis we will discuss results on this topic.
2.1 General Results
We will ﬁnd a spanning set of Vλ(G) in this section. In addition, a basis of Vλ(G)
is presented in the simpliest case, where the irreducible character λ of G is one-
dimensional.
Let α ∈ Γm,n, and σ ∈ Sm. Note that since α : {1, . . . , m} → {1, . . . , n} and
σ : {1, . . . , m} → {1, . . . , m}, α ◦ σ : {1, . . . , m} → {1, . . . , n}, and so α ◦ σ ∈ Γm,n.
11

CHAPTER 2. DIMENSIONS OF SYMMETRY CLASSES 12
Definition 2.1.1 Given a subgroup G of Sm, define an equivalence relation ≡ on
Γm,n as follows: for α, β ∈ Γm,n, α ≡ β (mod G) if there is σ ∈ G such that
β = α ◦ σ.
It is easy to show that ≡ is indeed an equivalence relation on Γm,n.
Consider each α ∈ Γm,n as a sequence of length m of positive integers chosen
from the set {1, . . . , n}, i.e., α = (α(1), α(2), . . . , α(m)). We can order elements of
Γm,n lexicographically, i.e., α precedes β if the first non-zero difference β(i) − α(i)
is positive.
Let be the set of distinct representatives of the equivalence classes (mod G)
of Γm,n so chosen that α ∈ if and only if α is the first in lexicographical order in
its equivalence class. (Note that depends on m, n, and G.)
From chapter 1 we know that if {e1, . . . , en} is a basis of V , then Vλ(G) has
a spanning set {T(G, λ)(e⊗
α ) | α ∈ Γm,n}. Some elements of this spanning set,
however, are 0, and the rest are not all linearly independent. Thus to find a basis
of Vλ(G) we need to eliminate these elements from our spanning set. We start with
a definition and a very important lemma.
Definition 2.1.2 Let G be a subgroup of Sm, and α ∈ Γm,n. The stabilizer sub-
group of α is Gα = {σ ∈ G | ασ = α}.
It is easy to show that Gα is indeed a subgroup of G.
Lemma 2.1.3 Let G be a subgroup of Sm, λ be an irreducible character of G,
α, β ∈ Γm,n, and {ei | i = 1, . . . , n} be an orthonormal basis of V . Then
(e∗
α, e∗
β) =



λ(e)
|G| σ∈Gα
λ(τ−1
σ) if β = ατ for some τ ∈ G
0 otherwise.

Proof:
(e∗
α, e∗
β) = (T(G, λ)(e⊗
α ), T(G, λ)(e⊗
β ))
= (T(G, λ)∗
◦ T(G, λ)(e⊗
α ), e⊗
β ).
By Proposition (1.1.2), T(G, λ) is idempotent and Hermitian, so by straight-
forward calculations the above reduces to:
(e∗
α, e∗
β) =
λ(e)
|G| σ∈G
λ(σ)(e⊗
ασ−1 , e⊗
β )
=
λ(e)
|G| σ∈G
λ(σ)
m
i=1
(eασ−1(i), eβ(i)) (2.1)
since the inner product on ⊗m
V is the one induced from V . If α ≡ β (mod G),
then (2.1) above is 0 because the ei’s are orthonormal. If β = ατ for some τ ∈ G,
then
m
i=1
(eασ−1(i), eβ(i)) = 1 if and only if ασ−1
= β, which means ασ−1
= ατ, which
is equivalent to τσ ∈ Gα. Therefore (2.1) above becomes
λ(e)
|G| τσ∈Gα
λ(σ),
and can be written as
λ(e)
|G| σ∈Gα
λ(τ−1
σ). 2
By Lemma 2.1.3, e∗
α
2
=
λ(e)
|G| σ∈Gα
λ(σ), and since e∗
α = 0 if and only if
e∗
α = 0, it follows that e∗
α = 0 if and only if
σ∈Gα
λ(σ) = 0.
Deﬁne Ω = {α ∈ Γm,n |
σ∈Gα
λ(σ) = 0}. It follows that {e∗
α | α ∈ Ω} spans
Vλ(G), since {e∗
α | α ∈ Γm,n} does.
Having eliminated the zero elements from our spanning set of Vλ(G), we now
want to ﬁnd a subset of this new spanning set that is a basis of Vλ(G).

Deﬁnition 2.1.4 Let = ∩ Ω = {α ∈ |
σ∈Gα
λ(σ) = 0}.
Note that in the case of λ being identically l (i.e., λ being the principal character
of G), we have
σ∈Gα
λ(σ) =
σ∈Gα
1 = |Gα| = 0
for all α ∈ , which gives = .
The following is an easier and more useful way to look at Ω:
Proposition 2.1.5 Ω = α∈ {ασ | σ ∈ G} = {ασ | α ∈ , σ ∈ G}.
Proof: That α∈ {ασ | σ ∈ G} = {ασ | α ∈ , σ ∈ G} is obvious. We will
show that Ω = {ασ | α ∈ , σ ∈ G}.
Let β ∈ Ω. If β ∈ , then β ∈ Ω ∪ = , and certainly
β = βe ∈ {ασ | α ∈ , σ ∈ G}. (2.2)
If β ∈ , then there is β ∈ and σ ∈ G such that β = β σ. We have:
β ∈ Ω ⇐⇒
τ∈{π∈G|βπ=β}
λ(τ) = 0
⇐⇒
τ∈{π∈G|β σπ=β σ}
λ(τ) = 0
⇐⇒
τ∈{π∈G|β (σπσ−1)=β }
λ(τ) = 0
⇐⇒
τ∈G
β
λ(τ) = 0
⇐⇒ β ∈ Ω.
But β ∈ , so β ∈ Ω ∩ = . Thus
β = β σ ∈ {ασ | α ∈ , σ ∈ G}. (2.3)

Results (2.2) and (2.3) give Ω ⊆ {ασ | α ∈ , σ ∈ G}.
Now let β ∈ {ασ | α ∈ , σ ∈ G}. Then β = ατ, for some α ∈ and some
τ ∈ G. We have:
σ∈Gβ
λ(σ) =
βσ=β
λ(σ)
=
ατσ=ατ
λ(σ)
=
α(τστ−1)=α
λ(σ)
=
α(τστ−1)=α
λ(τστ−1
)
=
ασ=α
λ(σ)
=
σ∈Gα
λ(σ)
= 0 since α ∈ = ∩ Ω ⊆ Ω.
So β ∈ Ω, and hence {ασ | α ∈ , σ ∈ G} ⊆ Ω. 2
The following theorem gives the relationship between Vλ(G) and the induced
basis of ⊗m
V .
Theorem 2.1.6 Let G be a subgroup of Sm, λ be an irreducible character of G,
and E = {e1, ..., en} be a basis of V . Then Vλ(G) is the direct sum of the subspaces
span{e∗
ασ | σ ∈ G}, for α ∈ . That is, Vλ(G) =
α∈
span{e∗
ασ | σ ∈ G}, the sum
being direct.
Proof: From the discussion after lemma 2.1.3, {e∗
α | α ∈ Ω} spans Vλ(G). By
Proposition 2.1.5, Ω = {ασ | α ∈ , σ ∈ G}, so it follows that Vλ(G) is the sum of
the subspaces span{e∗
ασ | σ ∈ G}, for α ∈ .

To see that the sum is direct, take the inner product on V that makes E or-
thonormal. By lemma 2.1.3, the subspaces span{e∗
ασ | σ ∈ G}, α ∈ Γm,n become
orthogonal, and the sum is therefore direct. 2
In next section we will show a formula by R. Freese that calculates the dimension
of span{e∗
ασ | σ ∈ G}, for α ∈ Γm,n.
We can now prove the major result of this section:
Theorem 2.1.7 Let E = {e1, . . . , en} be a basis of V . Then {e∗
α | α ∈ } is a
linearly independent set in Vλ(G). If E is orthogonal then so is {e∗
α | α ∈ }. If
λ(e) = 1 then {e∗
α | α ∈ } is a basis of Vλ(G).
Proof: From theorem 2.1.6 and the discussion after lemma 2.1.3, the e∗
α’s are
non-zero vectors from distinct subspaces that form a direct sum of Vλ(G). So the
e∗
α’s must be linearly independent.
If E is orthogonal, then by lemma 2.1.3, {e∗
α | α ∈ } is an orthogonal set
(since for α, β ∈ , α ≡ β (mod G) if and only if α = β).
If λ(e) = 1, by proposition 1.2.1 e∗
ασ = λ(σ)e∗
α for all σ ∈ G and α ∈ Γm,n. Thus
span{e∗
ασ | σ ∈ G} = span{λ(σ)e∗
α | σ ∈ G} = span{e∗
α},
which means the subspaces span{e∗
ασ | σ ∈ G} making up the direct sum of Vλ(G)
are all one-dimensional. This shows {e∗
α | α ∈ } spans Vλ(G), and so must be a
basis of Vλ(G). 2
Theorem 2.1.6 tell us that Vλ(G) is the direct sum of span{e∗
ασ | σ ∈ G}, for
α ∈ . It is therefore important to ﬁnd the dimension of span{e∗
ασ | σ ∈ G}.
In the remaining sections of this chapter we will discuss methods for calculating
the dimension of span{e∗
ασ | σ ∈ G}.

2.2 Freese’s Theorem
Deﬁnition 2.2.1 Let {e1, . . . , en} be a basis of V , and let α ∈ Γm,n. The subspace
spanned by {e∗
ασ | σ ∈ G} is called an orbital subspace of Vλ(G) corresponding to
α. The dimension of span{e∗
ασ | σ ∈ G} is denoted sα.
Proposition 2.2.2 sα deﬁned above has the following properties:
1. sα = 0 if and only if α ∈ Ω
2. if λ(e) = 1 then sα = 1 for all α ∈ Ω
3. dim Vλ(G) =
α∈
sα
Proof:
1. sα = 0 if and only if {e∗
ασ | σ ∈ G} = {0}, which holds if and only if e∗
ασ = 0
for some σ ∈ G. By the discussion after lemma 2.1.3, e∗
ασ = 0 if and only if
τ∈Gασ
λ(τ) = 0. But we have:
τ∈Gασ
λ(τ) =
αστ=ασ
λ(τ)
=
αστσ−1=α
λ(τ)
=
α(στσ−1)=α
λ(στσ−1
)
=
ατ=α
λ(τ)
=
τ∈Gα
λ(τ),
so
τ∈Gα
λ(τ) = 0, which gives α ∈ Ω.

Conversely, α ∈ Ω means
τ∈Gα
λ(τ) = 0 which, by the discussion after lemma
2.1.3, is equivalent to e∗
α = 0, i.e., e∗
αe = 0. Thus {e∗
ασ | σ ∈ G} = {0}, giving
sα = 0.
2. This is the last part in the proof of theorem 2.1.7.
3. By theorem 2.1.6, Vλ(G) =
α∈
span{e∗
ασ | σ ∈ G}, the sum being direct.
This immediately gives us the result. 2
In [9], R. Freese gave the following general formula for sα:
Theorem 2.2.3 (Freese) For α ∈ Γm,n, sα = λ(e)(λ, 1)Gα , where
(λ, 1)Gα =
1
|Gα| σ∈Gα
λ(σ).
To prove Freese’s theorem, we ﬁrst prove the following well known result in
linear algebra:
Lemma 2.2.4 The rank of an idempotent linear operator on a vector space is its
trace.
Proof: Let L : V → V be an idempotent linear operator on the n-dimensional
space V . Let R and N denote the image and kernel of L respectively.
It is easy to see that v ∈ R if and only if v = L(v), for if v = L(u) for some
u ∈ V , then since L is idempotent, L(v) = L2
(u) = L(u) = v; and if v = L(v), then
(obviously) v ∈ R.
Thus we have R ∩ N = {0}. Further, since every vector v ∈ V can be written
as v = L(v) + (v − L(v)), and v − L(v) ∈ N, we conclude that V is a direct sum of
R and N.

Let {e1, . . . , et} be a basis of R and {et+1, . . . , en} be a basis of N. Then E =
{e1, . . . , en} is a basis of V and the matrix representing L with respect to E is:



It 0
0 0


 ,
where It is the t × t identity matrix. This shows rank(L) = trace(L) = t. 2
Now we can prove Freese’s theorem:
Proof: Since Gα is a subgroup of G, we can decompose G by picking τi ∈ G, i =
1, . . . , r such that G = Gατ1 ∪ . . . ∪ Gατr, where the Gατi’s are all disjoint.
Observe that for all π ∈ G,
e∗
απ = T(G, λ)(e⊗
απ)
=
λ(e)
|G| σ∈G
λ(σ)P(σ)(e⊗
απ)
=
λ(e)
|G| σ∈G
λ(σ)e⊗
απσ−1
=
λ(e)
|G| σ∈G
λ(σ−1
π)e⊗
ασ
=
λ(e)
|G|
r
i=1 τ∈Gα
λ((ττi)−1
π)e⊗
αττi
=
λ(e)
|G|
r
i=1
(
τ∈Gα
λ(τ−1
i τ−1
π))e⊗
ατi
since ατ = α.
In particular,
e∗
ατj
=
λ(e)
|G|
r
i=1
(
τ∈Gα
λ(τ−1
i ττj))e⊗
ατi
. (2.4)
For each σ ∈ G, σ ∈ Gατi for some i ∈ {1, . . . , r}. Thus σ = σ τi for some
σ ∈ Gα. So for σ ∈ G,
e∗
ασ = T(G, λ)(e⊗
ασ)

= T(G, λ)(e⊗
ασ τi
)
= T(G, λ)(e⊗
ατi
)
= e∗
ατi
.
Therefore we have
span{e∗
ασ | σ ∈ G} ⊆ span{e∗
ατi
| i = 1, . . . , r},
hence
span{e∗
ασ | σ ∈ G} = span{e∗
ατi
| i = 1, . . . , r},
and by (2.4),
span{e∗
ασ | σ ∈ G} ⊆ span{e⊗
ατi
| i = 1, . . . , r}.
Since {e⊗
β | β ∈ Γm,n} is a basis of ⊗m
V , {e⊗
ατi
| i = 1, . . . , r} is a linearly indepen-
dent set if the ατi’s are distinct. The ατi’s are indeed distinct because:
If ατi = ατj, then α = ατjτ−1
i , so τjτ−1
i ∈ Gα. This gives τj = (τjτ−1
i )τi ∈ Gατi.
But τj = eτj ∈ Gατj, so by the choice of the τi’s in the decomposition of G we
must have Gατi = Gατj, which means τi = τj. This shows that the ατi’s are indeed
distinct, and hence {e⊗
ατi
| i = 1, . . . , r} is a linearly independent set.
Now let C be the r×r matrix with
λ(e)
|G| σ∈Gα
λ(τ−1
i στj) at the (i, j)th
position (ith
row and jth
column). Consider C as a linear operator on span{e⊗
ατi
| i = 1, . . . , r}.
Clearly for k ∈ {1, . . . , r},
C(e⊗
ατk
) =
r
i=1
(
λ(e)
|G| σ∈Gα
λ(τ−1
i στk))e⊗
ατi
= e∗
ατk
by (2.4).
Thus the image of span{e⊗
ατi
| i = 1, . . . , r} under C is span{e∗
ατi
| i = 1, . . . , r}, so
the dimension of span{e∗
ατi
| i = 1, . . . , r} is just rank(C).

C is idempotent because the (i, j)th
element of C2
is:
(C2
)i,j =
r
k=1
(
λ(e)
|G| σ∈Gα
λ(τ−1
i στk))(
λ(e)
|G| σ∈Gα
λ(τ−1
k στj))
= (
λ(e)
|G|
)2
r
k=1
(
σ,τ∈Gα
λ(τ−1
i στk)λ(τ−1
k ττj))
= (
λ(e)
|G|
)2
r
k=1
(
σ,π∈Gα
λ(τ−1
i στk)λ(τ−1
k σ−1
πτj))
= (
λ(e)
|G|
)2
r
k=1
(
σ,π∈Gα
λ(τ−1
i (στk))λ((στk)−1
πτj))
= (
λ(e)
|G|
)2
π∈Gα σ∈G
λ(τ−1
i σ)λ(σ−1
πτj)
= (
λ(e)
|G|
)2
π∈Gα σ∈G
λ(σ)λ(σ−1
τ−1
i πτj)
= (
λ(e)
|G|
)2
π∈Gα
(
|G|λ(τ−1
i πτj)
λ(e)
) by (1.2)
=
λ(e)
|G| π∈Gα
λ(τ−1
i πτj)
= (C)i,j.
By lemma 2.2.4,
rank(C) = trace(C)
=
r
i=1
(C)i,i
=
r
i=1
λ(e)
|G| σ∈Gα
λ(τ−1
i στi)
=
λ(e)
|G|
r
i=1 σ∈Gα
λ(σ)
=
λ(e)
|G|
r
σ∈Gα
λ(σ)
=
λ(e)
|Gα| σ∈Gα
λ(σ). 2

2.3 Using Disjoint Cycle Decomposition of Sm
In [26], M. Marcus gave a formula for the rank of the linear operator T(G, λ), that
is, the dimension of Vλ(G). The method uses the disjoint cycle decomposition of
elements of Sm. To prove that result we ﬁrst prove the following lemma.
Lemma 2.3.1 Let σ ∈ Sm and let c(σ) denote the number of cycles in the dis-
joint cycle decomposition of σ, including cycles of length one. Then for the linear
operation P(σ) on ⊗m
V , trace(P(σ)) = nc(σ)
.
Proof: Let {e1, . . . , en} be a basis of V , {f1, . . . , fn} be the basis of V ∗
dual to
{e1, . . . , en}, i.e.,
fi(ej) =



1 if i = j
0 otherwise
extended linearly to all of V ∗
.
Let E = {e⊗
α | α ∈ Γm,n} be the corresponding (ordered) basis of ⊗m
V . For
each β ∈ Γm,n, deﬁne a linear functional hβ : ⊗m
V → F by extending
hβ(eα(1) ⊗ . . . ⊗ eα(m)) = fβ(1)(eα(1)) . . . fβ(m)(eα(m)) =
m
i=1
fβ(i)(eα(i))
To all of ⊗m
V . We have
hβ(e⊗
α ) =



1 if α = β
0 otherwise.
This means that {hβ | β ∈ Γm,n} is the basis of (⊗m
V )∗
dual to E.
The elements on the main diagonal of the matrix representing P(σ) with respect
to E are the numbers
hα(P(σ)(e⊗
α )) = hα(e⊗
ασ−1 )

=
m
i=1
fα(i)(eασ−1(i))
=



1 if α = ασ−1
0 otherwise
= δα,ασ−1 .
Thus the trace of P(σ) is
α∈Γm,n
δα,ασ−1 , which is the number of α’s in Γm,n
satisfying α = ασ−1
, or equivalently, α = ασ.
Let (t σ(t) σ2
(t) . . . σl−1
(t)) be a cycle of length l in σ. Then any α ∈ Γm,n
satisfying α = ασ must satisfy
α(t) = ασ(t),
ασ(t) = ασ2
(t),
ασ2
(t) = ασ3
(t),
...
ασl−1
(t) = ασl
(t) = α(t).
This shows α(t) = ασ(t) = ασ2
(t) = . . . = ασl−1
(t) for all α with α = ασ.
Consider each α ∈ Γm,n as a sequence (α(1) . . . α(m)). Any α satisfying
α = ασ must have the same integer k at positions t, σ(t), . . . , σl−1
(t) in the above
sequence. There are n choices for the integer k for each disjoint cycle in σ, and
since there are c(σ) disjoint cycles in σ, there are nc(σ)
choices for α. Hence
trace(P(σ)) =
α∈Γm,n
δα,ασ−1 = nc(σ)
. 2
We can now prove the main result mentioned at the beginning of this section:
Theorem 2.3.2 The dimension of Vλ(G) is
λ(e)
|G| τ∈G
λ(τ)nc(τ)
.

Proof: By proposition 1.1.2 T(G, λ) is idempotent. We have:
dim Vλ(G) = rank(T(G, λ))
= trace(T(G, λ)) by lemma 2.2.4
= trace(
λ(e)
|G| τ∈G
λ(τ)P(τ))
=
λ(e)
|G| τ∈G
λ(τ)trace(P(τ))
=
λ(e)
|G| τ∈G
nc(τ)
by lemma 2.3.1. 2
2.4 Subspaces of the Orbital Subspaces
In [46] B. Y. Wang and M. P. Gong described methods to construct subspaces of
the orbital subspaces span{e∗
ασ | σ ∈ G}, α ∈ , and calculate their dimensions.
Some of their results are summarized in this section.
As before, let G be a subgroup of Sm with identity element e, and let {e1, . . . , en}
be a fixed orthonormal basis of V , so that {e⊗
α | α ∈ Γm,n} is an orthonormal basis
of ⊗m
V . Let A be an irreducible representation of G which affords the irreducible
character λ. For each σ ∈ G and i, j ∈ {1, . . . , λ(e)}, let aij(σ) denote the (i, j)th
element of the matrix A(σ) with respect to the basis {e1, . . . , en}.
Define the operators Tij on ⊗m
V by
Tij =
λ(e)
|G| σ∈G
aij(σ)P(σ)
for each i, j = 1, . . . , λ(e).
By straight-forward calculations we see that
T(G, λ) =
λ(e)
i=1
Tii.

We further have the following identities:
T2
ii = Tii (2.5)
and
TijTkl =



Til if j = k,
0 otherwise
(2.6)
for all i, j, k, l ∈ {1, . . . , λ(e)} (see [37], page V-8).
If A is a unitary representation (that is, A(σ)A(σ)∗
= A(σ)∗
A(σ) = I, or that
A(σ) is a unitary matrix, for every σ ∈ G), then
T∗
ij = Tji (2.7)
for all i, j ∈ {1, . . . , λ(e)}.
Let V ij
A (G) = Tij(⊗m
V ). We have the following theorem:
Theorem 2.4.1 The symmetry class of tensors, Vλ(G), is equal to
λ(e)
i=1
V ii
A (G), a
direct sum, and dim V ii
A (G) =
dim Vλ(G)
λ(e)
. If A is a unitary representation of G,
then the sum is orthogonal. 2
(see [37], page V-8).
Let α ∈ Γm,n. To divide span{e∗
ασ | σ ∈ G} into subspaces, we ﬁrst introduce a
lemma:
Lemma 2.4.2 Let V i
α = span{Ti1(e⊗
α ), . . . , Tiλ(e)(e⊗
α )}, for i = 1, . . . , λ(e). Then
V i
α = span{Tij(e⊗
ασ) | σ ∈ G, j = 1, . . . , λ(e)}, for i = 1, . . . , λ(e) (2.8)
and
V i
ατ = V i
α, for all τ ∈ G and i = 1, . . . , λ(e). (2.9)

Proof: Note that we only need to prove (2.8), since (2.9) follows directly from
(2.8).
Let k ∈ {1, . . . , λ(e)}. By (2.6) we have
Tik(e⊗
α ) = Tij(Tjk(e⊗
α )) for any j ∈ {1, . . . , λ(e)}
= Tij
λ(e)
|G| σ∈G
ajk(σ)e⊗
ασ−1
=
λ(e)
|G| σ∈G
ajk(σ)Tij(e⊗
ασ−1 )
=
λ(e)
|G| σ∈G
ajk(σ−1
)Tij(e⊗
ασ)
∈ span{Tij(e⊗
ασ) | σ ∈ G}.
So we have V i
α ⊆ span{Tij(e⊗
ασ) | σ ∈ G, j = 1, . . . , λ(e)}. To show inclusion in the
other direction, ﬁrst note that for any i, j ∈ {1, . . . , λ(e)} and any π, σ ∈ G,
aij(πσ) =
λ(e)
l=1
ail(π)alj(σ). (2.10)
For i, j ∈ {1, . . . , λ(e)} and σ ∈ G, We have:
Tij(e⊗
ασ) =
λ(e)
|G| τ∈G
aij(τ)P(τ)(e⊗
ασ)
=
λ(e)
|G| τ∈G
aij(τ)P(τ)P(σ−1
)(e⊗
α )
=
λ(e)
|G| τ∈G
aij(τ)P(τσ−1
)(e⊗
α ). (2.11)
By writing π = τσ−1
, (2.11) above becomes:
Tij(e⊗
ασ) =
λ(e)
|G| π∈G
aij(πσ)P(π)(e⊗
α )
=
λ(e)
|G| π∈G


λ(e)
l=1
ail(π)alj(σ)

 P(π)(e⊗
α ) by (2.10)

=
λ(e)
l=1
λ(e)
|G| π∈G
ail(π)alj(σ)P(π)(e⊗
α )
=
λ(e)
l=1
alj(σ)
λ(e)
|G| π∈G
ail(π)P(π)(e⊗
α )
=
λ(e)
l=1
alj(σ)Til(e⊗
α )
∈ V i
α. 2
Now we can divide span{e∗
ασ | σ ∈ G} into subspaces:
Theorem 2.4.3 For any ﬁxed α ∈ Γm,n,
span{e∗
ασ | σ ∈ G} =
λ(e)
i=1
V i
α = span{Tij(e⊗
α ) | i, j = 1, . . . , λ(e)}; (2.12)
and
Vλ(G) =
α∈
λ(e)
i=1
V i
α =
α∈
span{Tij(e⊗
α ) | i, j = 1, . . . , λ(e)}. (2.13)
If A is a unitary representation of G, then the above sums are orthogonal.
Proof: From straight-forward calculations we know that span{e⊗
ασ | σ ∈ G} is
invariant under both T(G, λ) and Tij, for all i, j = 1, . . . , λ(e). That is,
T(G, λ)(span{e⊗
ασ | σ ∈ G}) ⊆ span{e⊗
ασ | σ ∈ G}
and
Tij(span{e⊗
ασ | σ ∈ G}) ⊆ span{e⊗
ασ | σ ∈ G}
for all i, j = 1, . . . , λ(e).
We have:
span{e∗
ασ | σ ∈ G} = span{T(G, λ)(e⊗
ασ) | σ ∈ G}

= span{
λ(e)
i=1
Tii(e⊗
ασ) | σ ∈ G}
= {
σ∈G
(fσ
λ(e)
i=1
Tii(e⊗
ασ)) | fσ ∈ F}
= {
λ(e)
i=1 σ∈G
fσTii(e⊗
ασ) | fσ ∈ F}
=
λ(e)
i=1
span{Tii(e⊗
ασ) | σ ∈ G}
=
λ(e)
i=1
span{Tii(e⊗
ασ) | σ ∈ G}
⊆
λ(e)
i=1
V i
α
= span{Tij(e⊗
α ) | i, j = 1, . . . , λ(e)}. (2.14)
Since span{e⊗
ασ | σ ∈ G} is invariant under Tij for all i, j = 1, . . . , λ(e), and
Tij = TiiTij, we have:
Tij(span{e⊗
ασ | σ ∈ G}) = Tii(Tij(span{e⊗
ασ | σ ∈ G}))
⊆ Tii(span{e⊗
ασ | σ ∈ G}).
This mean the inclusion in (2.14) is also valid in the other direction. This proves
(2.12). Theorem 2.1.6 and (2.12) leads directly to (2.13).
If A is unitary, then by (2.6) and (2.7) the direct sums are orthogonal. 2
Lemma 2.4.4 Let λ ∈ Γm,n. If {T1j1 (e⊗
α ), T1j2 (e⊗
α ), . . . , T1jk
(e⊗
α )} is a basis of V 1
α ,
then
{Tij1 (e⊗
α ), Tij2 (e⊗
α ), . . . , Tijk
(e⊗
α ) | i = 1, . . . , λ(e)}
is a basis of span{e⊗
ασ | σ ∈ G}.

Proof: Denote {Tij1 (e⊗
α ), Tij2 (e⊗
α ), . . . , Tijk
(e⊗
α )} by Ei, for all i = 1, . . . , λ(e).
By theorem 2.4.3, it suﬃces to show that Ei is a basis of V i
α, for all i =
2, . . . , λ(e).
Let i ∈ {2, . . . , λ(e)}. Suppose
k
l=1
ClTijl
(e⊗
α ) = 0, for some Cl ∈ F. Then
k
l=1
ClT1jl
(e⊗
α ) =
k
l=1
ClT1i(Tijl
(e⊗
α ))
= T1i(
k
l=1
ClTijl
(e⊗
α ))
= T1i(0)
= 0.
Since E1 is linearly independent, we must have Cl = 0 for all l = 1, . . . , k. This
shows that Ei is linearly independent for all i = 2, . . . , λ(e).
Now let Tij(e⊗
α ) ∈ V i
α for some j ∈ {1, . . . , λ(e)}. We have:
Tij(e⊗
α ) = Ti1(T1j(e⊗
α )).
But T1j(e⊗
α ) ∈ V 1
α = span E1, so we can write
T1j(e⊗
α ) =
k
l=1
ClT1jl
(e⊗
α )
for some suitable values of Cl ∈ F. We have:
Tij(e⊗
α ) = Ti1(T1j(e⊗
α ))
= Ti1(
k
l=1
ClT1jl
(e⊗
α ))
=
k
l=1
ClTi1(T1jl
(e⊗
α ))
=
k
l=1
ClTijl
(e⊗
α ))
∈ span Ei.

Thus Ei also spans V i
α. Hence Ei is a basis of V i
α. 2
The above lemma shows that if we know how to calculate dim V i
α, we can then
easily ﬁnd the dimension of span{e∗
ασ | σ ∈ G}. The following theorem does exactly
that:
Theorem 2.4.5 dim V i
α = (λ, 1)Gα , where
(λ, 1)Gα =
1
|Gα| σ∈Gα
λ(σ),
for all i = 1, . . . , λ(e).
Proof: By lemma 2.4.4 all subspaces V i
α have the same dimensions. Using this
result and theorem 2.4.3, we have:
dim V i
α =
dim span{e∗
ασ | σ ∈ G}
λ(e)
for each i = 1, . . . , λ(e). Freese’s theorem gives
dim span{e∗
ασ | σ ∈ G} = λ(e)(λ, 1)Gα
which leads directly to our result. 2
If the irreducible representation A of G is unitary, we can then construct or-
thonormal bases of all the orbital subspaces of Vλ(G). Before we do that, ﬁrst we
introduce a couple of notation:
If A is a p × q matrix and i1, . . . , ik, j1, . . . , jl are positive integers with i1 <
. . . < ik ≤ p and j1 < . . . < jl ≤ q, let A[i1, . . . , ik | j1, . . . , jl] denote the k × l
matrix whose element at row s and column t is the element at row is and column
jt of the matrix A.

For α ∈ Γm,n, let
dj =
λ(e)
|G| σ∈Gα
ajj(σ)
for j = 1, . . . , λ(e). We have the following:
Theorem 2.4.6 Let A be an irreducible and unitary representation of G. Let α ∈
Γm,n and k = (λ, 1)Gα . If there exist j1, . . . , jk such that
σ∈Gα
A(σ)[j1, . . . , jk | j1, . . . , jk]
is a diagonal matrix, and djl
= 0 for all l = 1, . . . , k, then
E =



1
djl
Tijl
(e⊗
α ) | i = 1, . . . , λ(e), l = 1, . . . , k



is an orthonormal basis of the orbital subspace span{e∗
ασ | σ ∈ G}.
Proof: By the proof of theorem 2.4.5 above, we know that E contains the
right number of elements. So all we need to do is to show that elements in E are
orthonormal.
Using the identities (2.6) and (2.7), we obtain


1
djt
Tijt (e⊗
α ),
1
djs
Trjs (e⊗
α )

 =
1
djt djs
(Tijt (e⊗
α ), Trjs (e⊗
α ))
=
1
djt djs
(TjsrTijt (e⊗
α ), e⊗
α )
=
1
djt djs
δir(Tjsjt (e⊗
α ), e⊗
α )
=
δir
djt djs
λ(e)
|G| σ∈G
ajsjt (σ)P(σ)(e⊗
α ), e⊗
α
=
δir
djt djs
λ(e)
|G| σ∈G
ajsjt (σ)(e⊗
ασ−1 ), e⊗
α

=
δir
djt djs
λ(e)
|G| σ∈G
ajsjt (σ)(e⊗
ασ−1 , e⊗
α )
=
δir
djt djs
λ(e)
|G| σ∈Gα
ajsjt (σ)(e⊗
ασ−1 , e⊗
α )
=
δir
djt djs
λ(e)
|G| σ∈Gα
ajsjt (σ)
= δirδst. 2

Chapter 3
Some Special Results
3.1 General Results on Cyclic Groups
In [13] R. Holmes and T. Tam discussed some results regarding cyclic groups. Their
results are presented here.
Let r ∈ Sm be an m-cycle (e.g., the cycle (1 2 . . . m)). Then the subgroup
of Sm generated by r is the cyclic group Cm. The elements of Cm are rk
, for
k = 0, . . . , m − 1.
For convenience in notation, in the following discussions we assume that r is
the m-cycle (1 2 . . . m). The result is the same for other m-cycles.
We the have the following result:
Proposition 3.1.1 Using the above notation, for any rk
∈ Cm, c(rk
) = gcd(m, k).
(where c(g) is the number of cycles in the disjoint cycle decomposition of g, as
deﬁned in lemma 2.3.1)
33

CHAPTER 3. SOME SPECIAL RESULTS 34
Proof: Let R(N, D) denote the remainder of N when divided by D, where both
N and D are positive integers. Deﬁne M : {1, 2, . . .} → {1, . . . , m} by
M(i) = 1 + R(i − 1, m).
If k = 0, then rk
= r0
= e = (1)(2) . . . (m). So c(rk
) = m = gcd(m, k).
If gcd(m, k) = 1, then the cycle in rk
containing 1 is
(1 M(1 + k) M(1 + 2k) . . . M(1 + (m − 1)k)),
because if M(1 + tk) = 1 for some t ∈ {1, . . . , m − 1}, then R(tk, m) = 0, which
implies m divides tk. But gcd(m, k) = 1, and therefore m divides t, contradicting
t ∈ {1, . . . , m − 1}. Thus (1 M(1 + k) M(1 + 2k) . . . M(1 + (m − 1)k)) must be
the only cycle in rk
. This gives c(rk
) = 1 = gcd(m, k).
If k = 0 and gcd(m, k) = g = 1, write m = gm1 and k = gk1, where
gcd(m1, k1) = 1. Then the distinct cycle decomposition of rk
is
(1 M(1 + k) . . . M(1 + (
m
g
− 1)k))
(2 M(2 + k) . . . M(2 + (
m
g
− 1)k))
...
(g M(g + k) . . . M(g + (
m
g
− 1)k)).
The cycle containing 1 is as shown above because:
Clearly M(1+(
m
g
)k) = 1. Suppose M(1+tk) = 1 for some t ∈ {1, . . . ,
m
g
−1}.
Then R(tk, m) = 0, i.e., m divides tk, which implies gm1 divides tgk1, which implies
m1 divides tk1. But gcd(k1, m1) = 1, and so m1 must divide t. This contradicts
t ∈ {1, . . . ,
m
g
− 1} = {1, . . . , m1 − 1}. So the cycle containing 1 must be the one
shown above.

By the same reasoning, the cycles containing 2, 3, . . . , g are also as shown above.
Thus c(rk
) = g = gcd(m, k). 2
For π ∈ Sm, deﬁne bi(π) to be the number of cycles of length i in the disjoint
cycle decomposition of π. Clearly for any i > m, bi(π) = 0, and the function c(π)
in the proposition above is precisely
m
i=1
bi(π)
and is called the type of π.
Let r ∈ Sm be an m−cycle. Then the subgroup of Sm generated by r is the
cyclic group Cm. The elements of Cm are rk
, for k = 0, . . . , m − 1.
By proposition 3.1.1, c(rk
) = gcd(m, k) for k = 0, . . . , m − 1, so c(rk
) divides
m. On the other hand, for any k that divides m, k = gcd(m, k) = c(rk
). Thus the
types of elements of G constitute exactly the set of divisors of m, i.e.,
{c(σ) | σ ∈ G} = {d | d divides m}.
Since Cm is Abelian, from group representation theory we know that all of its
irreducible representations are of degree one. Thus for any irreducible character λ
of Cm, λ(Cm) is a group of complex roots of unity. Therefore the m irreducible
characters of Cm are:
λl(rk
) = e2πilk/m
, for l = 0, . . . , m − 1 (3.1)
(see [45], page 35). Since each λl is of degree one, by theorem 2.1.7, {e∗
α | α ∈ }
is a basis of Vλl
(Cm), so dimVλl
(Cm) = | |.
Another way to ﬁnd the dimension of Vλl
(Cm) is to use the following formula
given by R. Holmes and T. Tam in [13]:

Theorem 3.1.2 dim Vλl
(Cm) =
1
m
m−1
k=1
e2πilk/m
ngcd(m,k)
, for l = 0, . . . , m − 1.
Proof: By theorem 2.3.2,
dim Vλl
(Cm) =
λl(e)
|Cm| σ∈Cm
λl(σ)nc(σ)
=
1
m
m−1
k=0
λl(rk
)nc(rk)
=
1
m
m−1
k=0
e2πilk/m
ngcd(m,k)
by (3.1) and proposition 3.1.1. 2
3.2 Simplified Formulas for Cyclic Groups
Although the results presented in the last chapter and the previous section are of
theoretical importance, they are difficult to use to actually calculate the dimen-
sion of some symmetry class. In [5], L. J. Cummings gave three formulas that
greatly simplify the necessary calculations in some special cases. Those results are
presented in this section.
Definition 3.2.1 For each positive integer p, let φ(p) denote the number of integers
in {1, 2, . . . , p} that are coprime with p. The function φ is called the Euler Totient
Function.
Theorem 3.2.2 Let G = Cm be a subgroup of Sm generated by an m−cycle, and
let λ be the character identically 1. Then
dimVλ(G) =
1
m d|m
φ(m/d)nd
,

where φ is the Euler totient function, and d|m means the positive integer d divides
m.
Proof: Let G = Cm be generated by the m−cycle r. By theorem 2.3.2,
dim Vλ(G) =
λ(e)
|G| σ∈G
λ(σ)nc(σ)
=
1
m σ∈G
nc(σ)
. (3.2)
By the discussion after proposition 3.1.1, the set of types of elements of G
constitute exactly the set of divisors of m. So the sum in (3.2) can be rewritten to
sum by the divisors of m:
dim Vλ(G) =
1
m d|m
#(d)nd
where #(d) stands for the number of elements in G having type d, that is, having
d disjoint cycles.
By proposition 3.1.1, for any rk
∈ G = Cm (where k ∈ {0, . . . , m − 1}), c(rk
) =
gcd(m, k). So the number of elements in G having d disjoint cycles is the number
of k’s in {1, . . . , m} satisfying gcd(k, m) = d, or equivalently, the number of k/d’s
satisfying gcd(k/d, m/d) = 1. This number is precisely φ(m/d). So #(d) = φ(m/d)
and the result follows. 2
Definition 3.2.3 Let G be a cyclic subgroup of Sm, and let λ be a one-dimensional
character of Sm. λ is said to be primitive if its value at a generator of G is a
primitive mth
root of unity.
Definition 3.2.4 The Möbius function µ : {1, 2, . . .} → {−1, 0, 1} is defined to be:
µ(n) =



1 if n = 1,
(−1)t
if n is a product of t disctinct primes,
0 if p2
devides n for some prime p.

Theorem 3.2.5 Let G be a cyclic subgroup of Sm generated by an m−cycle and λ
be a primitive one dimensional character of Sm. Then
dim Vλ(G) =
1
m d|m
µ(m/d)nd
,
where µ is the M¨obius function.
Proof: We will use a result by K. Gauss: If r is a primitive mth
root of unity,
then
gcd(k,m)=1
rk
= µ(m). (3.3)
Let G be generated by the m−cycle π. By theorem 2.3.2 and proposition 3.1.1,
dim Vλ(G) =
λ(e)
|G| σ∈G
λ(σ)nc(σ)
=
1
m
m−1
k=0
λ(πk
)nc(πk)
=
1
m
m−1
k=0
λ(πk
)ngcd(k,m)
. (3.4)
We know from the discussion after proposition 3.1.1 that the divisors of m
constitute exactly the set {c(πk
) | k = 0, . . . , m − 1}. So (3.4) above can be re-
written to sum by the divisors of m, as follows:
dim Vλ(G) =
1
m d|m
(
c(πk)=d
λ(πk
))nd
=
1
m d|m
(
gcd(k,m)=d
λ(πk
))nd
. (3.5)
Writing l = k/d for the bracketed sum in (3.5), we get:
dim Vλ(G) =
1
m d|m
(
gcd(k/d,m/d)=1
λ(πld
))nd
=
1
m d|m
(
gcd(l,m/d)=1
λ(πd
)l
)nd
. (3.6)

Since λ(π) is a primitive mth
root of unity, λ(πd
) = λ(π)d
is a primitive (m/d)th
root of unity. By (3.3), the bracketed sum in (3.6) is µ(m/d). The desired result
follows immediately. 2
Theorem 3.2.6 Let G be a subgroup of Sm generated by disjoint cycles π1, . . . , πk
of orders m1, . . . , mk, respectively, and let λ be a character of G. If m =
k
i=1
mi,
then
dim Vλ(G) =
k
i=1
dim Vλ( πi ). (3.7)
Proof: Let
F = {f : {1, . . . , k} →
k
i=1
{1, . . . , mi − 1} | f(i) ∈ {1, . . . , mi − 1}, ∀i = 1, . . . , k},
and for each f ∈ F, deﬁne rf to be
rf =
k
i=1
c(π
f(i)
i ).
By theorem 2.3.2,
dim Vλ( πi ) =
1
mi
mi−1
l=0
λ(πl
i)nc(πl
i)
for i = 1, . . . , k. So the right side of (3.7) becomes:
k
i=1
dim Vλ( πi ) =
k
i=1
(
1
mi
mi−1
l=0
λ(πl
i)nc(πl
i)
)
= (
k
j=1
1
mj
)
k
i=1
(
mi−1
l=0
λ(πl
i)nc(πl
i)
)
= (
1
m
)
f∈F
(
k
i=1
λ(π
f(i)
i ))nrf
=
1
m f∈F
(nrf
λ(π
f(1)
1 . . . π
f(k)
k )). (3.8)

Since the πi’s are disjoint and G = π1, . . . , πk , we must have
{π
f(1)
1 . . . π
f(k)
k | f ∈ F} = G,
and
rf =
k
i=1
c(π
f(i)
i ) = c(π
f(1)
1 . . . π
f(k)
k ).
So equation (3.8) above becomes
k
i=1
dim Vλ( πi ) =
1
m g∈G
nc(g)
λ(g)
= dim Vλ(G). 2
3.3 Dihedral Groups
In [13] R. Holmes and T. Tam analyzed the dihedral subgroup Dm of Sm (m ≥ 3).
Consider the elements r, s ∈ Sm with
r = (1 2 . . . m)
and
s =



(2 m)(3 m − 1) . . . (
m
2
m
2
+ 2) if m is even
(2 m)(3 m − 1) . . . (
m + 1
2
m + 3
2
) if m is odd.
The subgroup of Sm generated by r and s is Dm, the dihedral group of degree m.
Clearly, it contains the cyclic subgroup Cm (since Cm = r ). By straight-forward
calculations we see that r, s satisfy
rm
= s2
= e and srs = r−1
. (3.9)
Further,
Dm = {rk
, srk
| k = 0, . . . , m − 1} (3.10)

(see [14], pages 50-51).
For each integer l with 0 < l < m/2, Dm has an irreducible character χl of
degree 2 which is induced from the irreducible character λl of Cm in (3.1):
χl(rk
) = 2 cos(
2πlk
m
)
χl(srk
) = 0 (3.11)
for k = 0, . . . , m − 1. All other irreducible characters of Dm are of degree 1 (see
[45], page 37).
Also note that for any integer k, by (3.9), we have:
srk
= sr(ss)rk−1
= (srs)srk−1
= r−1
srk−1
...
= r−k
s. (3.12)
Theorem 3.3.1 Let {e1, . . . , en} be an orthonormal basis of V with n ≥ 2, G =
r, s = Dm with m ≥ 3, and χ = χl for some 0 < l < m/2. There is a subset S of
Γm,n for which
{e∗
γ | γ ∈ S}
is an orthogonal basis of Vχ(G), if and only if
m ≡ 0 (mod 4l2)
where l = l2l2 with l2 being the largest power of 2 dividing l, and l2 being odd.
Proof: First suppose Vχ(G) has an orthogonal basis {e∗
γ | γ ∈ S} for some subset
S of Γm,n.

Let γ = (1, 2, . . . , 2) ∈ Γm,n. Clearly, γs = γ, and so s ∈ Gγ.
For any g ∈ G, by (3.10) either g = rk
or g = srk
for some k ∈ {1, . . . , m − 1, }.
If e = g = rk
∈ Cm, we have
γrk
(1) = 2 = 1 = γ(1),
so g = rk
∈ Gγ, and hence Gγ ∩ Cm = {e}.
If g = srk
∈ Gγ then
rk
= erk
= (ss)rk
= s(srk
) ∈ Gγ,
since both s and srk
are in Gγ. But rk
∈ Cm and we already know that Gγ ∩ Cm =
{e}. So we must have rk
= e, and so srk
= s. This gives Gγ = {e, s}.
For all τ, µ ∈ G, we can use straight-forward calculations with the identities
(3.9) and (3.12) to get:
τ−1
Gγµ =



{srj+k
, rk−j
}, if τ = srj
, µ = rk
{rk−j
, srj+k
}, if τ = rj
, µ = rk
{srj+k
, rk−j
}, if τ = rj
, µ = srk
{rk−j
, srj+k
}, if τ = srj
, µ = srk
.
(3.13)
This clearly shows that for all τ, µ ∈ G, |τ−1
Gγµ ∩ Cm| = |{rk−j
}| = 1.
From lemma 2.1.3, we have:
(e∗
γτ , e∗
γµ) =
χ(e)
|G| σ∈Gγµ
χ((µ−1
τ)−1
σ)
=
χ(e)
|G| σ∈Gγµ
χ(τ−1
µσ). (3.14)
For σ ∈ Gγµ, we have:
σ ∈ Gγµ ⇐⇒ γµσ = γµ

⇐⇒ γµσµ−1
= γ
⇐⇒ µσµ−1
∈ Gγ
⇐⇒ σ ∈ µ−1
Gγµ.
So (3.14) above becomes
(e∗
γτ , e∗
γµ) =
χ(e)
|G| σ∈µ−1Gγµ
χ(τ−1
µσ)
=
χ(e)
|G| σ∈τ−1Gγµ
χ(σ)
=
χ(e)
|G|
χ(rj
) (3.15)
for some j ∈ {0, . . . , m − 1} by (3.13) and the fact that χ(srk
) = 0 for all k (by
(3.9)).
Since Vχ(G) is a direct sum of the subspaces span{e∗
ασ | σ ∈ G}, for α ∈
(see theorem 2.1.6), and Vχ(G) has an orthogonal basis consisting of decomposable
symmetrized tensors, each subspace span{e∗
ασ | σ ∈ G}, where α ∈ must also
have an orthogonal basis consisting of decomposable symmetrized tensors.
Clearly γ ∈ . So by Freese’s theorem,
dim span{e∗
γσ | σ ∈ G} =
χ(e)
|Gγ| σ∈Gγ
χ(σ)
=
2
2 σ∈{e,s}
χ(σ)
= χ(e) + χ(s)
= 2 + 0
= 2. (3.16)
Thus span{e∗
γσ | σ ∈ G} must have an orthogonal basis consisting of exactly
two decomposable symmetrized tensors, say e∗
γτ and e∗
γµ, for some τ = µ in G.

If χ = 0 on Cm, then by (3.15) (e∗
γτ , e∗
γµ) = 0 and so e∗
γτ and e∗
γµ can not be
orthogonal to each other. This contradiction shows that χ must vanish for some
element in Cm.
Consequently, we must have
χ(rk
) = 2 cos(
2πlk
m
) = 0
for some k. That is,
2πlk
m
=
(2h + 1)π
2
for some integer h. This means 4lk = (2h + 1)m, or 4l2l2k = (2h + 1)m, where
l2l2 = l and l2 is the largest power of 2 dividing l. Since (2h + 1) is odd, 4l2 must
divide m. Hence m ≡ 0 (mod 4l2).
Conversely, suppose m ≡ 0 (mod 4l2). Let γ ∈ , and let H = Gγ ∩ Cm. We
can easily check that H is a subgroup of Cm.
Recall that χ = χl is the induced character λG
where λ = λl (see (3.11)).
Denote the restrictions to H of χ and λ by χH and λH respectively. Using Mackey’s
subgroup theorem (see [45], page 58) we get
χH = (λG
)H = λH + (λH)s
where (λH)s
is the character of H deﬁned by:
(λH)s
(rk
) = λH(s−1
rk
s)
= λH(s−1
sr−k
) by (3.12)
= λH(r−k
)
= λH(rk
)−1

for any rk
∈ H. This implies λH is identically 1 if and only if (λH)s
is identically
1. Taking the Hermitian inner product on characters, we get:
(λH + (λH)s
, 1)H = (χH, 1)H
=
1
|H| σ∈H
χ(σ)
=
1
|H| σ∈Gγ∩Cm
χ(σ)
=
1
|H| σ∈Gγ
χ(σ) since χ(x) = 0 ∀x ∈ GCm
= 0
since γ ∈ . It follows that λH = 1.
Let rk
∈ H for some k. Then
λ(rk
) = λH(rk
) = 1.
But
λ(rk
) = λl(rk
) = e(2πilk/m)
,
so 2πlk/m = 2πh for some integer h, that is, lk = hm. Setting
l =
l
gcd(m, l)
and m =
m
gcd(m, l)
,
we easily get l k = hm .
If d is a positive integer that divides both m and l , then m = dm and l = dl
for some integers m and l , which means
m
gcd(m, l)
= dm and
l
gcd(m, l)
= dl .
This gives
m = (d gcd(m, l))m and l = (d gcd(m, l))l ,

which implies (d gcd(m, l)) is a common factor of m and l. Thus d must be 1, and
so m and l are coprime.
Since l k = hm , we have k = (hm )/l = m (h/l ). If (h/l ) is not an integer,
then there is a prime factor p of l which divides m but does not divide h. This
cannot happen because m and l are coprime. So (h/l ) must be an integer.
Since k = m (h/l ), and (h/l ) is an integer, we have rk
∈ rm
. But rk
is
arbitrary in H, so we conclude that H is a subgroup of rm
.
Suppose Gγ = H. Then Gγ contains some t ∈ GγCm ⊆ GCm, which means
t must be in the form srk
for some integer k. By (3.12) we have t2
= e, and so
T = {t, e} is a subgroup of Gγ.
For any g ∈ Gγ, if g = rj
for some j, then clearly gH = Hg. If g = srj
for some
j, then for any h = rk
∈ H, by (3.9) and (3.12) we have:
gh = (ghg−1
)g = (srj
rk
(srj
)−1
)g = (srj+k
r−j
s−1
)g = (srk
s−1
)g = r−k
g = h−1
g,
so again we have gH = Hg. This indicates that H is a normal subgroup of Gγ,
and hence HT is a subgroup of Gγ. Since T = {t, e}, we have |HT| = 2|H|.
For any g ∈ G, either g = rk
or g = srk
for some integer k. If g = rk
then
g = rk
e ∈ CmGγ. If g = srk
, let t = srj
∈ Gγ for some integer j (t = rj
for any j
because t ∈ Cm). By (3.9) and (3.12) we have:
g = srk
= srk−j
rj
= srk−j
(ss)rj
= (srk−j
s)(srj
) = (rj−k
ss)t = rj−k
t ∈ CmGγ.
So G ⊆ CmGγ, which gives G = CmGγ.
By the Second Group Isomorphism Theorem (see [14], page 44),
Gγ/H = Gγ/(Cm ∩ Gγ) ∼= CmGγ/Cm = G/Cm.

Since |G/Cm| = 2, |Gγ/H| = 2, hence |Gγ| = 2|H|. But we already showed that
HT is a subgroup of Gγ with |HT| = 2|H|, so we must have Gγ = HT.
Thus we have established that either Gγ = H, or Gγ = HT where T = t with
t2
= e and t ∈ Cm.
By theorem 2.1.6, Vχ(G) is the direct sum of the subspaces span{e∗
γσ | σ ∈ G}
for γ ∈ . Thus to show the existence of a set S ⊆ Γm,n with {e∗
α | α ∈ S} being an
orthogonal basis of Vχ(G), it suﬃces to show that for each γ ∈ , there is a subset
Kγ ⊆ G such that {e∗
γσ | σ ∈ Kγ} is an orthogonal basis of span{e∗
γσ | σ ∈ G}, and
set S =
γ∈
{γσ | σ ∈ Kγ}.
Let γ ∈ . Recall that on H, χH = λH + (λH)s
= 1 + 1 = 2. So if Gγ = H,
then by Freese’s theorem
dim span{e∗
γσ | σ ∈ G} =
χ(e)
|Gγ| σ∈Gγ
χ(σ)
=
2
|H| σ∈H
χH(σ)
=
2
|H|
· 2|H|
= 4.
On the other hand, if Gγ = HT, then we easily see that Gγ = H Ht, and
H Ht = {}. So again by Freese’s theorem
dim span{e∗
γσ | σ ∈ G} =
χ(e)
|Gγ| σ∈Gγ
χ(σ)
=
2
2|H|
(
σ∈H
χ(σ) +
σ∈Ht
χ(σ))
=
1
|H|
(
σ∈H
χH(σ))
=
1
|H|
· 2|H|

= 2.
Let
Kγ =



{e, rm /4
, s, srm /4
} if Gγ = H,
{e, rm /4
} if Gγ = HT.
First we have to make sure the above definition is valid by showing m /4 is an
integer. Since m ≡ 0 (mod 4l2), there is a positive integer p with m = 4pl2. We
have
m =
m
gcd(m, l)
=
4pl2
gcd(4pl2, l2l2)
=
4pl2
l2 gcd(4p, l2)
=
4p
gcd(4p, l2)
.
Since l2 is odd, gcd(4p, l2) must also be odd. Therefore 4 is a factor of m , and m /4
is an integer.
To show that {e∗
γσ | σ ∈ Kγ} is an orthogonal basis of span{e∗
γσ | σ ∈ G}, it
suffices to show that e∗
γτ and e∗
γµ are orthogonal if τ, µ ∈ Kγ and τ = µ.
By (3.15), to show that e∗
γτ and e∗
γµ are orthogonal, it suffices to show that
χ(τ−1
Gγµ) = {0}. We will do this for all combinations of τ and µ in Kγ.
Consider the case Gγ = H first. If τ = rk
∈ Cm and µ = srj
∈ Cm, then
τ−1
Gγµ Cm = {}, which means τ−1
Gγµ ⊆ GCm, and so χ(τ−1
Gγµ) = {0}.
If τ = e and µ = rm /4
, then
τ−1
Gγµ = eHrm /4
= Hrm /4
⊆ rm
= {rm (1+4k)/4
| k = . . . , −1, 0, 1, . . .}.
Thus
χ(rm (1+4k)/4
) = 2 cos(
2πlm (1 + 4k)/4
m
)
= 2 cos(
lm (1 + 4k)π
2m
)
= 2 cos(
l gcd(m, l)m (1 + 4k)π
2m gcd(m, l)
)
= 2 cos(
l (1 + 4k)π
2
). (3.17)

But l is odd since
l =
l
gcd(m, l)
=
l
gcd(4l2k, l2l2)
for some integer k
=
1
l2 gcd(4k, l2)
and l2 is the largest power of 2 that divides l. Hence (3.17) above evaluates to 0.
This again gives χ(τ−1
Gγµ) = {0}.
If τ = s and µ = srm /4
, then
τ−1
Gγµ ⊆ s−1
rm
srm /4
= {s−1
rkm
srm /4
| k = . . . , −1, 0, 1, . . .}
= {r−km +m /4
| k = . . . , −1, 0, 1, . . .}
= {rm (1−4k)/4
| k = . . . , −1, 0, 1, . . .}
and
χ(rm (1−4k)/4
) = 2 cos(
2πlm (1 − 4k)/4
m
)
= 2 cos(
πl gcd(m, l)m (1 − 4k)
2m gcd(m, l)
)
= 2 cos(
l (1 − 4k)π
2
)
= 0,
since we have determined that l is odd. So again χ(τ−1
Gγµ) = {0}.
Finally consider the second case, Gγ = HT = H Ht. We only need to consider
τ = e and µ = rm /4
:
τ−1
Gγµ = τ−1
Hµ τ−1
Htµ = Hrm /4
Htrm /4
.

By (3.17), χ(Hrm /4
) = {0}. Since Htrm /4
Cm = {}, χ(Htrm /4
) = {0}. This
gives χ(τ−1
Gγµ) = {0} and completes the proof. 2
The tensor space ⊗m
V is said to have an orthogonal basis of decomposable sym-
metrized tensors if for each irreducible character λ of G, there is a subset Sχ of Γm,n
for which {e∗
γ | γ ∈ Sχ} is an orthogonal basis of Vλ(G). We have the following
result from theorem 3.3.1:
Corollary 3.3.2 Let G = Dm with m ≥ 3, and suppose dimV = n ≥ 2. Then
⊗m
V has an orthogonal basis of decomposable symmetrized tensors if and only if
m is a power of 2.
Proof: Let m2 denote the largest power of 2 dividing m, and let χ = χm2 . If m
is not a power of 2, then 0 < m2 < m/2, and m ≡ 0 (mod 4m2). Thus by (3.11)
χ is an irreducible character of G. Since m ≡ 0 (mod 4m2), theorem 3.3.1 implies
that there is no subset Sχ of Γm,n for which {e∗
γ | γ ∈ Sχ} is an orthogonal basis of
Vχ(G). This is a contradiction and so m must be a power of 2.
Conversely, suppose m is a power of 2. Then (3.11) indicates that for all l with
0 < l < m/2, χl is an irreducible character of G of degree 2, and all other irreducible
characters of G = Dm are of degree 1.
If λ is a one dimensional character of G, by theorem 2.1.7, {e∗
α | α ∈ } is an
orthogonal basis of Vλ(G). If λ is two dimensional, then λ = χl for some l with
0 < l < m/2. Setting l2 to be the largest power of 2 dividing l, we get l2 ≤ m/4,
and hence m ≡ 0 (mod 4l2). By theorem 3.3.1 there is a subset Sλ of Γm,n for
which {e∗
γ | γ ∈ Sλ} is an orthogonal basis of Vλ(G) = Vχl
(G). 2

3.4 Linear Symmetry Classes
In [7] L. J. Cummings and R. W. Robinson used Pólya’s counting theorem to
calculate the dimension of V1(G), the linear symmetry class associated with the
principal character (i.e., identically 1). The results from their work is summarized
in this section.
We first introduce some notations and Pólya’s counting theorem.
Let G be a subgroup of Sm acting on Γm,n by
g · α = α ◦ g−1
for all g ∈ G and α ∈ Γm,n.
For each g ∈ G, denote the number of cycles of length i in the cyclic decompo-
sition of g by bi(g) (as defined after proposition 3.1.1).
The cycle type of g ∈ G is the monomial
Z(g) = Z(g; a1, . . . , am) =
m
k=1
a
bk(g)
k
in the variables a1, . . . , am. The cycle index of G is the polynomial
Z(G) = Z(G; a1, . . . , am) =
1
|G| g∈G
Z(g).
If A is a commutative ring and w : {1, . . . , n} → A, then the elements w(i) are
called weights. The weight of a function α ∈ Γm,n is
W(α) =
m
i=1
w(α(i)).
It is easy to show that if two functions α, β ∈ Γm,n are in the same orbit of Γm,n
under the action of G then W(α) = W(β).

Note that the orbits of Γm,n under the action of G are precisely the equivalence
classes of Γm,n defined in definition 2.1.1, and is a traversal of these equivalence
classes or orbits.
If W(α) is the weight of a representative of an orbit, then the result of summing
the weights of representatives over all orbits (e.g., summing over ,
α∈
W(α)) is
called the pattern inventory of the equivalence classes. This pattern inventory can
be calculated using Pólya’s Counting Theorem:
Theorem 3.4.1 (Pólya’s Counting Theorem)
α∈
W(α) = Z(G;
n
i=1
w(i), . . . ,
n
i=1
(w(i))m
)
=
1
|G| g∈G
(
m
k=1
(
n
i=1
(w(i))k
)bk(g)
). 2
(For a complete proof of Pólya’s Counting Theorem, see [23], chapter 5.)
Corollary 3.4.2 If G is a subgroup of Sm then the number of G-orbits in Γm,n
(i.e., | |) is Z(G; n, . . . , n).
Proof: Let w : {1, . . . , n} → {1}. Then for all α ∈ Γm,n,
W(α) =
m
i=1
w(α(i)) = 1.
By Pólya’s counting theorem we have
| | =
α∈
1
=
α∈
W(α)
= Z(G;
n
i=1
w(i), . . . ,
n
i=1
(w(i))m
)
= Z(G; n, . . . , n). 2

Theorem 3.4.3 If λ is the principal character of G (i.e., identically 1), then
dim Vλ(G) =
1
|G| g∈G
Z(g; n, . . . , n).
Proof: Since λ is identically 1, by the remark after definition 2.1.4, we have
= . By theorem 2.1.7 and corollary 3.4.2,
dim Vλ(G) = | | = | | = Z(G; n, . . . , n),
which by definition is
1
|G| g∈G
Z(g) =
1
|G| g∈G
Z(g; n, . . . , n). 2
3.5 Counting Restricted Orbits
In [7] section 3 L. J. Cummings and R. W. Robinson studied the action of a subgroup
of Sm acting on a finite set X equipped with a linear (one-dimensional) character,
λ. They developed a formula for counting special orbits of X under the action of
G, and applied that formula to calculate dim Vλ(G) when X = Γm,n. Their results
are presented in this section.
Let G be a subgroup of Sm acting on a finite set X (later the results obtained
here will be applied to X = Γm,n, with g(α) = g · α = α ◦ g−1
for all g ∈ G and
α ∈ Γm,n). For α ∈ X, let Gα = {g ∈ G | g · α = α} be the stabilizer subgroup
of α. For any g ∈ G, let F(g) denote the number of elements in X left fixed by g.
Burnside’s lemma (see [10], pages 134-135) states that the number of orbits of X
under the action of G is
1
|G| g∈G
F(g).

For any positive integer k, denote the group of complex kth
roots of unity by
Ck. If λ is a linear character of G, then λ(G) = Cc for some c. For each α ∈ X,
let σ(α) be determined by λ(Gα) = Cσ(α). If α and β are in the same G-orbit (i.e.,
α = g · β for some g ∈ G) then we can easily show that Gα and Gβ are conjugate
(i.e., Gα = gGβg−1
for some g ∈ G), so λ(Gα) = λ(Gβ), and hence σ(α) = σ(β).
For any g ∈ G, denote the order of λ(g) by λ(g)◦
. A new character λk of G can be
obtained by deﬁning
λk(g) = (λ(g))k
for each g ∈ G. We have the following theorem:
Theorem 3.5.1 With the above set-up, suppose λ(G) = Cc with k|c. Then the
number of G-orbits consisting of α ∈ X with σ(α) = k is
ak =
1
|G| g∈G
F(g)(
v|k
µ(λv(g)◦
)
φ(λv(g)◦
)
)µ(k/v),
where µ is the M¨obius function, and φ is the Euler totient function. 2
(See [7], pages 1315-1317 for a complete proof.)
Now let G act on Γm,n by g(α) = g · α = α ◦ g−1
for all g ∈ G and α ∈ Γm,n.
The basic result of L. J. Cummings and R. W. Robinson (see [7]) implies
dim Vλ(G) = a1. (3.18)
By the second half of the proof of lemma 2.3.1, F(g) = nc(g)
, where c(g) is the
number of cycles in the disjoint cycle decomposition of g. Thus using (3.18) and
straight-forward calculations we obtain the following result:
Theorem 3.5.2 If G is a subgroup of Sm and λ is a one-dimensional character of
G, then
dim Vλ(G) =
1
|G| g∈G
µ(λ(g)◦
)
φ(λ(g)◦
)
nc(g)
. 2

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