Part II Representation Theory
Based on lectures by S. Martin
Notes taken by Dexter Chua
Lent 2016
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
Linear Algebra and Groups, Rings and Modules are essential
Representations of finite groups
Representations of groups on vector spaces, matrix representations. Equivalence of
representations. Invariant subspaces and submodules. Irreducibility and Schur’s
Lemma. Complete reducibility for finite groups. Irreducible representations of Abelian
groups.
Character theory
Determination of a representation by its character. The group algebra, conjugacy classes,
and orthogonality relations. Regular representation. Permutation representations and
their characters. Induced representations and the Frobenius reciprocity theorem.
Mackey’s theorem. Frobenius’s Theorem. [12]
Arithmetic properties of characters
Divisibility of the order of the group by the degrees of its irreducible characters.
Burnside’s p
a
q
b
theorem. [2]
Tensor products
Tensor products of representations and products of characters. The character ring.
Tensor, symmetric and exterior algebras. [3]
Representations of S
1
and SU
2
The groups
S
1
,
SU
2
and
SO
(3), their irreducible representations, complete reducibility.
The Clebsch-Gordan formula. *Compact groups.* [4]
Further worked examples
The characters of one of GL
2
(F
q
), S
n
or the Heisenberg group. [3]
Contents
0 Introduction
1 Group actions
2 Basic definitions
3 Complete reducibility and Maschke’s theorem
4 Schur’s lemma
5 Character theory
6 Proof of orthogonality
7 Permutation representations
8 Normal subgroups and lifting
9 Dual spaces and tensor products of representations
9.1 Dual spaces
9.2 Tensor products
9.3 Powers of characters
9.4 Characters of G × H
9.5 Symmetric and exterior powers
9.6 Tensor algebra
9.7 Character ring
10 Induction and restriction
11 Frobenius groups
12 Mackey theory
13 Integrality in the group algebra
14 Burnside’s theorem
15 Representations of compact groups
15.1 Representations of SU(2)
15.2 Representations of SO(3), SU(2) and U(2)
0 Introduction
The course studies how groups act as groups of linear transformations on vector
spaces. Hopefully, you understand all the words in this sentence. If so, this is a
good start.
In our case, groups are usually either finite groups or topological compact
groups (to be defined later). Topological compact groups are typically subgroups
of the general linear group over some infinite fields. It turns out the tools we
have for finite groups often work well for these particular kinds of infinite groups.
The vector spaces are always finite-dimensional, and usually over C.
Prerequisites of this course include knowledge of group theory (as much as the
IB Groups, Rings and Modules course), linear algebra, and, optionally, geometry,
Galois theory and metric and topological spaces. There is one lemma where
we must use something from Galois theory, but if you don’t know about Galois
theory, you can just assume the lemma to be true without bothering yourself
too much about the proofs.
1 Group actions
We start by reviewing some basic group theory and linear algebra.
Basic linear algebra
Notation. F always represents a field.
Usually, we take
F
=
C
, but sometimes it can also be
R
or
Q
. These fields
all have characteristic zero, and in this case, we call what we’re doing ordinary
representation theory. Sometimes, we will take
F
=
F
p
or
¯
F
p
, the algebraic
closure of F
p
. This is called modular representation theory.
Notation.
We write
V
for a vector space over
F
this will always be finite
dimensional over
F
. We write
GL
(
V
) for the group of invertible linear maps
θ
:
V V
. This is a group with the operation given by composition of maps,
with the identity as the identity map (and inverse by inverse).
Notation.
Let
V
be a finite-dimensional vector space over
F
. We write
End
(
V
)
for the endomorphism algebra, the set of all linear maps V V .
We recall a couple of facts from linear algebra:
If
dim
F
V
=
n <
, we can choose a basis
e
1
, ··· , e
n
of
V
over
F
. So we
can identify
V
with
F
n
. Then every endomorphism
θ GL
(
V
) corresponds to a
matrix A
θ
= (a
ij
) M
n
(F ) given by
θ(e
j
) =
X
i
a
ij
e
i
.
In fact, we have
A
θ
GL
n
(
F
), the general linear group. It is easy to see the
following:
Proposition.
As groups,
GL
(
V
)
=
GL
n
(
F
), with the isomorphism given by
θ 7→ A
θ
.
Of course, picking a different basis of
V
gives a different isomorphism to
GL
n
(F), but we have the following fact:
Proposition.
Matrices
A
1
, A
2
represent the same element of
GL
(
V
) with respect
to different bases if and only if they are conjugate, namely there is some
X
GL
n
(F) such that
A
2
= XA
1
X
1
.
Recall that
tr
(
A
) =
P
i
a
ii
, where
A
= (
a
ij
)
M
n
(
F
), is the trace of
A
. A
nice property of the trace is that it doesn’t notice conjugacy:
Proposition.
tr(XAX
1
) = tr(A).
Hence we can define the trace of an operator
tr
(
θ
) =
tr
(
A
θ
), which is
independent of our choice of basis. This is an important result. When we study
representations, we will have matrices flying all over the place, which are scary.
Instead, we often just look at the traces of these matrices. This reduces our
problem of studying matrices to plain arithmetic.
When we have too many matrices, we get confused. So we want to put a
matrix into a form as simple as possible. One of the simplest form a matrix
can take is being diagonal. So we want to know something about diagonalizing
matrices.
Proposition.
Let
α GL
(
V
), where
V
is a finite-dimensional vector space over
C and α
m
= id for some positive integer m. Then α is diagonalizable.
This follows from the following more general fact:
Proposition.
Let
V
be a finite-dimensional vector space over
C
, and
α
End
(
V
), not necessarily invertible. Then
α
is diagonalizable if and only if there
is a polynomial f with distinct linear factors such that f (α) = 0.
Indeed, we have x
m
1 =
Q
(x ω
j
), where ω = e
2πi/m
.
Instead of just one endomorphism, we can look at many endomorphisms.
Proposition.
A finite family of individually diagonalizable endomorphisms of
a vector space over
C
can be simultaneously diagonalized if and only if they
commute.
Basic group theory
We will not review the definition of a group. Instead, we look at some of our
favorite groups, since they will be handy examples later on.
Definition
(Symmetric group
S
n
)
.
The symmetric group
S
n
is the set of all
permutations of
X
=
{
1
, ··· , n}
, i.e. the set of all bijections
X X
. We have
|S
n
| = n!.
Definition
(Alternating group
A
n
)
.
The alternating group
A
n
is the set of
products of an even number of transpositions (
i j
) in
S
n
. We know
|A
n
|
=
n!
2
.
So this is a subgroup of index 2 and hence normal.
Definition (Cyclic group C
m
). The cyclic group of order m, written C
m
is
C
m
= hx : x
m
= 1i.
This also occurs naturally, as
Z/mZ
over addition, and also the group of
n
th roots
of unity in
C
. We can view this as a subgroup of
GL
1
(
C
)
=
C
×
. Alternatively,
this is the group of rotation symmetries of a regular
m
-gon in
R
2
, and can be
viewed as a subgroup of GL
2
(R).
Definition (Dihedral group D
2m
). The dihedral group D
2m
of order 2m is
D
2m
= hx, y : x
m
= y
2
= 1, yxy
1
= x
1
i.
This is the symmetry group of a regular
m
-gon. The
x
i
are the rotations and
x
i
y
are the reflections. For example, in
D
8
,
x
is rotation by
π
2
and
y
is any
reflection.
This group can be viewed as a subgroup of
GL
2
(
R
), but since it also acts on
the vertices, it can be viewed as a subgroup of S
m
.
Definition (Quaternion group). The quaternion group is given by
Q
8
= hx, y : x
4
= 1, y
2
= x
2
, yxy
1
= x
1
i.
This has order 8, and we write i = x, j = y, k = ij, 1 = i
2
, with
Q
8
= 1, ±i, ±j, ±k}.
We can view this as a subgroup of GL
2
(C) via
1 =
1 0
0 1
, i =
i 0
0 i
, j =
0 1
1 0
, k =
0 i
i 0
,
1 =
1 0
0 1
, i =
i 0
0 i
, j =
0 1
1 0
, k =
0 i
i 0
.
Definition (Conjugacy class). The conjugacy class of g G is
C
G
(g) = {xgx
1
: x G}.
Definition (Centralizer). The centralizer of g G is
C
G
(g) = {x G : xg = gx}.
Then by the orbit-stabilizer theorem, we have |C
G
(g)| = |G : C
G
(G)|.
Definition
(Group action)
.
Let
G
be a group and
X
a set. We say
G
acts on
X if there is a map : G × X X, written (g, x) 7→ g x = gx such that
(i) 1x = x
(ii) g(hx) = (gh)x
The group action can also be characterised in terms of a homomorphism.
Lemma.
Given an action of
G
on
X
, we obtain a homomorphism
θ
:
G
Sym(X), where Sym(X) is the set of all permutations of X.
Proof.
For
g G
, define
θ
(
g
) =
θ
g
Sym
(
X
) as the function
X X
by
x 7→ gx
.
This is indeed a permutation of X because θ
g
1
is an inverse.
Moreover, for any
g
1
, g
2
G
, we get
θ
g
1
g
2
=
θ
g
1
θ
g
2
, since (
g
1
g
2
)
x
=
g
1
(
g
2
x
).
Definition
(Permutation representation)
.
The permutation representation of a
group action G on X is the homomorphism θ : G Sym(X) obtained above.
In this course,
X
is often a finite-dimensional vector space over
F
(and we
write it as
V
), and we want the action to satisfy some more properties. We will
require the action to be linear, i.e. for all g G, v
1
, v
2
V , and λ F.
g(v
1
+ v
2
) = gv
1
+ gv
2
, g(λv
1
) = λ(gv
1
).
Alternatively, instead of asking for a map
G Sym
(
X
), we would require a
map G GL(V ) instead.
2 Basic definitions
We now start doing representation theory. We boringly start by defining a
representation. In fact, we will come up with several equivalent definitions of a
representation. As always,
G
will be a finite group and
F
will be a field, usually
C.
Definition
(Representation)
.
Let
V
be a finite-dimensional vector space over
F. A (linear) representation of G on V is a group homomorphism
ρ = ρ
V
: G GL(V ).
We sometimes write
ρ
g
for
ρ
V
(
g
), so for each
g G
,
ρ
g
GL
(
V
), and
ρ
g
ρ
h
=
ρ
gh
and ρ
g
1
= (ρ
g
)
1
for all g, h G.
Definition
(Dimension or degree of representation)
.
The dimension (or degree)
of a representation ρ : G GL(V ) is dim
F
(V ).
Recall that
ker ρ C G
and
G/ ker ρ
=
ρ
(
G
)
GL
(
V
). In the very special case
where ker ρ is trivial, we give it a name:
Definition
(Faithful representation)
.
A faithful representation is a representa-
tion ρ such that ker ρ = 1.
These are the representations where the identity is the only element that
does nothing.
An alternative (and of course equivalent) definition of a representation is to
observe that a linear representation is “the same” as a linear action of G.
Definition
(Linear action)
.
A group
G
acts linearly on a vector space
V
if it
acts on V such that
g(v
1
+ v
2
) = gv
1
+ gv
2
, g(λv
1
) = λ(gv
1
)
for all g G, v
1
, v
2
V and λ F. We call this a linear action.
Now if
g
acts linearly on
V
, the map
G GL
(
V
) defined by
g 7→ ρ
g
,
with
ρ
g
:
v 7→ gv
, is a representation in the previous sense. Conversely, given a
representation
ρ
:
G GL
(
V
), we have a linear action of
G
on
V
via
gv
=
ρ
(
g
)
v
.
In other words, a representation is just a linear action.
Definition
(
G
-space/
G
-module)
.
If there is a linear action
G
on
V
, we say
V
is a G-space or G-module.
Alternatively, we can define a
G
-space as a module over a (not so) cleverly
picked ring.
Definition
(Group algebra)
.
The group algebra
FG
is defined to be the algebra
(i.e. a vector space with a bilinear multiplication operation) of formal sums
FG =
X
gG
α
g
g : α
g
F
with the obvious addition and multiplication.
Then we can regard
FG
as a ring, and a
G
-space is just an
FG
-module in
the sense of IB Groups, Rings and Modules.
Definition
(Matrix representation)
. R
is a matrix representation of
G
of degree
n if R Is a homomorphism G GL
n
(F).
We can view this as a representation that acts on
F
n
. Since all finite-
dimensional vector spaces are isomorphic to
F
n
for some
n
, every representation
is equivalent to some matrix representation. In particular, given a linear repre-
sentation
ρ
:
G GL
(
V
) with
dim V
=
n
, we can get a matrix representation
by fixing a basis
B
, and then define the matrix representation
G GL
n
(
F
) by
g 7→ [ρ(g)]
B
.
Conversely, given a matrix representation
R
, we get a linear representation
ρ
in the obvious way ρ : G GL(F
n
) by g 7→ ρ
g
via ρ
g
(v) = R
g
v.
We have defined representations in four ways as a homomorphism to
GL
(
V
), as linear actions, as
FG
-modules and as matrix representations. Now
let’s look at some examples.
Example
(Trivial representation)
.
Given any group
G
, take
V
=
F
(the one-
dimensional space), and
ρ
:
G GL
(
V
) by
g 7→
(
id
:
F F
) for all
g
. This is
the trivial representation of G, and has degree 1.
Despite being called trivial, trivial representations are highly non-trivial
in representation theory. The way they interact with other representations
geometrically, topologically etc, and cannot be disregarded. This is a very
important representation, despite looking silly.
Example.
Let
G
=
C
4
=
hx
:
x
4
= 1
i
. Let
n
= 2, and work over
F
=
C
. Then
we can define a representation by picking a matrix
A
, and then define
R
:
x 7→ A
.
Then the action of other elements follows directly by
x
j
7→ A
j
. Of course, we
cannot choose
A
arbitrarily. We need to have
A
4
=
I
2
, and this is easily seen to
be the only restriction. So we have the following possibilities:
(i) A
is diagonal: the diagonal entries can be chosen freely from
1
, ±i}
.
Since there are two diagonal entries, we have 16 choices.
(ii) A
is not diagonal: then it will be equivalent to a diagonal matrix since
A
4
= I
2
. So we don’t really get anything new.
What we would like to say above is that any matrix representation in which
X
is not diagonal is “equivalent” to one in which
X
is. To make this notion
precise, we need to define what it means for representations to be equivalent, or
“isomorphic”.
As usual, we will define the notion of a homomorphism of representations,
and then an isomorphism is just an invertible homomorphism.
Definition
(
G
-homomorphism/intertwine)
.
Fix a group
G
and a field
F
. Let
V, V
0
be finite-dimensional vector spaces over
F
and
ρ
:
G GL
(
V
) and
ρ
0
:
G GL
(
V
0
) be representations of
G
. The linear map
ϕ
:
V V
0
is a
G-homomorphism if
ϕ ρ(g) = ρ
0
(g) ϕ ()
for all g G. In other words, the following diagram commutes:
V V
V
0
V
0
ρ
g
ϕ ϕ
ρ
0
g
i.e. no matter which way we go form
V
(top left) to
V
0
(bottom right), we still
get the same map.
We say
ϕ
intertwines
ρ
and
ρ
0
. We write
Hom
G
(
V, V
0
) for the
F
-space of all
these maps.
Definition
(
G
-isomorphism)
.
A
G
-homomorphism is a
G
-isomorphism if
ϕ
is
bijective.
Definition
(Equivalent/isomorphic representations)
.
Two representations
ρ, ρ
0
are equivalent or isomorphic if there is a G-isomorphism between them.
If ϕ is a G-isomorphism, then we can write () as
ρ
0
= ϕρϕ
1
. ()
Lemma.
The relation of “being isomorphic” is an equivalence relation on the
set of all linear representations of G over F.
This is an easy exercise left for the reader.
Lemma.
If
ρ, ρ
0
are isomorphic representations, then they have the same di-
mension.
Proof.
Trivial since isomorphisms between vector spaces preserve dimension.
The converse is false.
Example. C
4
has four non-isomorphic one-dimensional representations: if
ω = e
2πi/4
, then we have the representations
ρ
j
(x
i
) = ω
ij
,
for 0 i 3, which are not equivalent for different j = 0, 1, 2, 3.
Our other formulations of representations give us other formulations of
isomorphisms.
Given a group
G
, field
F
, a vector space
V
of dimension
n
, and a representation
ρ
:
G GL
(
V
), we fix a basis
B
of
V
. Then we get a linear
G
-isomorphism
ϕ
:
V F
n
by
v 7→
[
v
]
B
, i.e. by writing
v
as a column vector with respect to
B
.
Then we get a representation
ρ
0
:
G GL
(
F
n
) isomorphic to
ρ
. In other words,
every representation is isomorphic to a matrix representation:
V V
F
n
F
n
ρ
ϕ ϕ
ρ
0
Thus, in terms of matrix representations, the representations
R
:
G GL
n
(
F
)
and
R
0
:
G GL
n
(
F
) are
G
-isomorphic if there exists some non-singular matrix
X GL
n
(F) such that
R
0
(g) = XR(g)X
1
for all g.
Alternatively, in terms of linear
G
-actions, the actions of
G
on
V
and
V
0
are
G-isomorphic if there is some isomorphism ϕ : V V
0
such that
gϕ(v) = ϕ(gv).
for all
g G, v V
. It is an easy check that this is just a reformulation of our
previous definition.
Just as we have subgroups and subspaces, we have the notion of sub-
representation.
Definition
(
G
-subspace)
.
Let
ρ
:
G GL
(
V
) be a representation of
G
. We
say W V is a G-subspace if it is a subspace that is ρ(G)-invariant, i.e.
ρ
g
(W ) W
for all g G.
Obviously, {0} and V are G-subspaces. These are the trivial G-subspaces.
Definition
(Irreducible/simple representation)
.
A representation
ρ
is irreducible
or simple if there are no proper non-zero G-subspaces.
Example.
Any 1-dimensional representation of
G
is necessarily irreducible, but
the converse does not hold, or else life would be very boring. We will later see
that D
8
has a two-dimensional irreducible complex representation.
Definition
(Subrepresentation)
.
If
W
is a
G
-subspace, then the corresponding
map
G GL
(
W
) given by
g 7→ ρ
(
g
)
|
W
gives us a new representation of
W
.
This is a subrepresentation of ρ.
There is a nice way to characterize this in terms of matrices.
Lemma.
Let
ρ
:
G GL
(
V
) be a representation, and
W
be a
G
-subspace of
V
. If
B
=
{v
1
, ··· , v
n
}
is a basis containing a basis
B
1
=
{v
1
, ··· , v
m
}
of
W
(with 0
< m < n
), then the matrix of
ρ
(
g
) with respect to
B
has the block upper
triangular form
∗ ∗
0
for each g G.
This follows directly from definition.
However, we do not like block triangular matrices. What we really like is
block diagonal matrices, i.e. we want the top-right block to vanish. There is
no a priori reason why this has to be true it is possible that we cannot find
another G-invariant complement to W .
Definition
((In)decomposable representation)
.
A representation
ρ
:
G GL
(
V
)
is decomposable if there are proper G-invariant subspaces U, W V with
V = U W.
We say ρ is a direct sum ρ
u
ρ
w
.
If no such decomposition exists, we say that ρ is indecomposable.
It is clear that irreducibility implies indecomposability. The converse is
not necessarily true. However, over a field of characteristic zero, it turns out
irreducibility is the same as indecomposability for finite groups, as we will see in
the next chapter.
Again, we can formulate this in terms of matrices.
Lemma.
Let
ρ
:
G GL
(
V
) be a decomposable representation with
G
-invariant
decomposition
V
=
U W
. Let
B
1
=
{u
1
, ··· , u
k
}
and
B
2
=
{w
1
, ··· , w
`
}
be
bases for
U
and
W
, and
B
=
B
1
B
2
be the corresponding basis for
V
. Then
with respect to B, we have
[ρ(g)]
B
=
[ρ
u
(g)]
B
1
0
0 [ρ
u
(g)]
B
2
Example.
Let
G
=
D
6
. Then every irreducible complex representation has
dimension at most 2.
To show this, let
ρ
:
G GL
(
V
) be an irreducible
G
-representation. Let
r G
be a (non-identity) rotation and
s G
be a reflection. These generate
D
6
.
Take an eigenvector
v
of
ρ
(
r
). So
ρ
(
r
)
v
=
λv
for some
λ 6
= 0 (since
ρ
(
r
) is
invertible, it cannot have zero eigenvalues). Let
W = hv, ρ(s)vi V
be the space spanned by the two vectors. We now check this is fixed by
ρ
. Firstly,
we have
ρ(s)ρ(s)v = ρ(e)v = v W,
and
ρ(r)ρ(s)v = ρ(s)ρ(r
1
)v = λ
1
ρ(s)v W.
Also,
ρ
(
r
)
v
=
λv W
and
ρ
(
s
)
v W
. So
W
is
G
-invariant. Since
V
is
irreducible, we must have W = V . So V has dimension at most 2.
The reverse operation of decomposition is taking direct sums.
Definition
(Direct sum)
.
Let
ρ
:
G GL
(
V
) and
ρ
0
:
G GL
(
V
0
) be
representations of G. Then the direct sum of ρ, ρ
0
is the representation
ρ ρ
0
: G GL(V V
0
)
given by
(ρ ρ
0
)(g)(v + v
0
) = ρ(g)v + ρ
0
(g)v
0
.
In terms of matrices, for matrix representations
R
:
G GL
n
(
F
) and
R
0
: G GL
n
0
(F), define R R
0
: G GL
n+n
0
(F) by
(R R
0
)(g) =
R(g) 0
0 R
0
(g)
.
The direct sum was easy to define. It turns out we can also multiply two
representations, known as the tensor products. However, to do this, we need to
know what the tensor product of two vector spaces is. We will not do this yet.
3 Complete reducibility and Maschke’s theorem
In representation theory, we would like to decompose a representation into sums
of irreducible representations. Unfortunately, this is not always possible. When
we can, we say the representation is completely reducible.
Definition
(Completely reducible/semisimple representation)
.
A representation
ρ
:
G GL
(
V
) is completely reducible or semisimple if it is the direct sum of
irreducible representations.
Clearly, irreducible implies completely reducible.
Not all representations are completely reducible. An example is to be found
on example sheet 1. These are in fact not too hard to find. For example, there
are representations of
Z
over
C
that are not completely reducible, and also a
non-completely reducible representation of C
p
over F
p
.
However, it turns out we have the following theorem:
Theorem.
Every finite-dimensional representation
V
of a finite group over a
field of characteristic 0 is completely reducible, namely,
V
=
V
1
··· V
r
is a
direct sum of irreducible representations.
By induction, it suffices to prove the following:
Theorem
(Maschke’s theorem)
.
Let
G
be a finite group, and
ρ
:
G GL
(
V
)
a representation over a finite-dimensional vector space
V
over a field
F
with
char F
= 0. If
W
is a
G
-subspace of
V
, then there exists a
G
-subspace
U
of
V
such that V = W U.
We will prove this many times.
Proof.
From linear algebra, we know
W
has a complementary subspace. Let
W
0
be any vector subspace complement of W in V , i.e. V = W W
0
as vector
spaces.
Let
q
:
V W
be the projection of
V
onto
W
along
W
0
, i.e. if
v
=
w
+
w
0
with w W, w
0
W
0
, then q(v) = w.
The clever bit is to take this q and tweak it a little bit. Define
¯q : v 7→
1
|G|
X
gG
ρ(g)q(ρ(g
1
)v).
This is in some sense an averaging operator, averaging over what
ρ
(
g
) does. Here
we need the field to have characteristic zero such that
1
|G|
is well-defined. In
fact, this theorem holds as long as char F - |G|.
For simplicity of expression, we drop the ρ’s, and simply write
¯q : v 7→
1
|G|
X
gG
gq(g
1
v).
We first claim that
¯q
has image in
W
. This is true since for
v V
,
q
(
g
1
v
)
W
,
and gW W . So this is a little bit like a projection.
Next, we claim that for
w W
, we have
¯q
(
w
) =
w
. This follows from the
fact that
q
itself fixes
W
. Since
W
is
G
-invariant, we have
g
1
w W
for all
w W . So we get
¯q(w) =
1
|G|
X
gG
gq(g
1
w) =
1
|G|
X
gG
gg
1
w =
1
|G|
X
gG
w = w.
Putting these together, this tells us ¯q is a projection onto W .
Finally, we claim that for
h G
, we have
h¯q
(
v
) =
¯q
(
hv
), i.e. it is invariant
under the G-action. This follows easily from definition:
h¯q(v) = h
1
|G|
X
gG
gq(g
1
v)
=
1
|G|
X
gG
hgq(g
1
v)
=
1
|G|
X
gG
(hg)q((hg)
1
hv)
We now put
g
0
=
hg
. Since
h
is invertible, summing over all
g
is the same as
summing over all g
0
. So we get
=
1
|G|
X
g
0
G
g
0
q(g
0−1
(hv))
= ¯q(hv).
We are pretty much done. We finally show that
ker ¯q
is
G
-invariant. If
v ker ¯q
and h G, then ¯q(hv) = h¯q(v) = 0. So hv ker ¯q.
Thus
V = im ¯q ker ¯q = W ker ¯q
is a G-subspace decomposition.
The crux of the whole proof is the definition of
¯q
. Once we have that,
everything else follows easily.
Yet, for the whole proof to work, we need
1
|G|
to exist, which in particular
means
G
must be a finite group. There is no obvious way to generalize this to
infinite groups. So let’s try a different proof.
The second proof uses inner products, and hence we must take
F
=
C
. This
can be generalized to infinite compact groups, as we will later see.
Recall the definition of an inner product:
Definition
(Hermitian inner product)
.
For
V
a complex space,
h·, ·i
is a
Hermitian inner product if
(i) hv, wi = hw, vi (Hermitian)
(ii) hv, λ
1
w
1
+ λ
2
w
2
i = λ
1
hv, w
1
i + λ
2
hv, w
2
i (sesquilinear)
(iii) hv, vi > 0 if v 6= 0 (positive definite)
Definition
(
G
-invariant inner product)
.
An inner product
h·, ·i
is in addition
G-invariant if
hgv, gwi = hv, wi.
Proposition.
Let
W
be
G
-invariant subspace of
V
, and
V
have a
G
-invariant
inner product. Then W
is also G-invariant.
Proof.
To prove this, we have to show that for all
v W
,
g G
, we have
gv W
.
This is not hard. We know
v W
if and only if
hv, wi
= 0 for all
w W
.
Thus, using the definition of G-invariance, for v W
, we know
hgv, gwi = 0
for all g G, w W .
Thus for all
w
0
W
, pick
w
=
g
1
w
0
W
, and this shows
hgv, w
0
i
= 0.
Hence gv W
.
Hence if there is a
G
-invariant inner product on any complex
G
-space
V
,
then we get another proof of Maschke’s theorem.
Theorem
(Weyl’s unitary trick)
.
Let
ρ
be a complex representation of a finite
group
G
on the complex vector space
V
. Then there is a
G
-invariant Hermitian
inner product on V .
Recall that the unitary group is defined by
U(V ) = {f GL(V ) : hf(u), f(v)i = hu, vi for all u, v V }
= {A GL
n
(C) : AA
= I}
= U(n).
Then we have an easy corollary:
Corollary.
Every finite subgroup of
GL
n
(
C
) is conjugate to a subgroup of U(
n
).
Proof.
We start by defining an arbitrary inner product on
V
: take a basis
e
1
, ··· , e
n
. Define (
e
i
, e
j
) =
δ
ij
, and extend it sesquilinearly. Define a new inner
product
hv, wi =
1
|G|
X
gG
(gv, gw).
We now check this is sesquilinear, positive-definite and
G
-invariant. Sesquilin-
earity and positive-definiteness are easy. So we just check
G
-invariance: we
have
hhv, hwi =
1
|G|
X
gG
((gh)v, (gh)w)
=
1
|G|
X
g
0
G
(g
0
v, g
0
w)
= hv, wi.
Note that this trick also works for real representations.
Again, we had to take the inverse
1
|G|
. To generalize this to compact groups,
we will later replace the sum by an integral, and
1
|G|
by a volume element. This is
fine since (
g
0
v, g
0
w
) is a complex number and we know how to integrate complex
numbers. This cannot be easily done in the case of ¯q.
Recall we defined the group algebra of G to be the F-vector space
FG = he
g
: g Gi,
i.e. its basis is in one-to-one correspondence with the elements of
G
. There is a
linear G-action defined in the obvious way: for h G, we define
h
X
g
a
g
e
g
=
X
g
a
g
e
hg
=
X
g
0
a
h
1
g
0
e
g
0
.
This gives a representation of G.
Definition
(Regular representation and regular module)
.
The regular represen-
tation of a group
G
, written
ρ
reg
, is the natural action of
G
on
FG
.
FG
is called
the regular module.
It is a nice faithful representation of dimension
|G|
. Far more importantly, it
turns out that every irreducible representation of
G
is a subrepresentation of
the regular representation.
Proposition.
Let
ρ
be an irreducible representation of the finite group
G
over
a field of characteristic 0. Then ρ is isomorphic to a subrepresentation of ρ
reg
.
This will also follow from a more general result using character theory later.
Proof.
Take
ρ
:
G GL
(
V
) be irreducible, and pick our favorite 0
6
=
v V
.
Now define θ : FG V by
X
g
a
g
e
g
7→
X
a
g
(gv).
It is not hard to see this is a
G
-homomorphism. We are now going to exploit the
fact that
V
is irreducible. Thus, since
im θ
is a
G
-subspace of
V
and non-zero,
we must have
im θ
=
V
. Also,
ker θ
is a
G
-subspace of
FG
. Now let
W
be
the
G
-complement of
ker θ
in
FG
, which exists by Maschke’s theorem. Then
W FG is a G-subspace and
FG = ker θ W.
Then the isomorphism theorem gives
W
=
FG/ ker θ
=
im θ = V.
More generally,
G
doesn’t have to just act on the vector space generated by
itself. If
G
acts on any set, we can take that space and create a space acted on
by G.
Definition
(Permutation representation)
.
Let
F
be a field, and let
G
act on a
set X. Let FX = he
x
: x Xi with a G-action given by
g
X
x
a
x
e
x
=
X
x
a
x
e
gx
.
So we have a
G
-space on
FX
. The representation
G GL
(
FX
) is the corre-
sponding permutation representation.
4 Schur’s lemma
The topic of this chapter is Schur’s lemma, an easy yet extremely useful lemma
in representation theory.
Theorem (Schur’s lemma).
(i)
Assume
V
and
W
are irreducible
G
-spaces over a field
F
. Then any
G-homomorphism θ : V W is either zero or an isomorphism.
(ii)
If
F
is algebraically closed, and
V
is an irreducible
G
-space, then any
G-endomorphism V V is a scalar multiple of the identity map ι
V
.
Proof.
(i)
Let
θ
:
V W
be a
G
-homomorphism between irreducibles. Then
ker θ
is
a
G
-subspace of
V
, and since
V
is irreducible, either
ker θ
= 0 or
ker θ
=
V
.
Similarly,
im θ
is a
G
-subspace of
W
, and as
W
is irreducible, we must
have
im θ
= 0 or
im θ
=
W
. Hence either
ker θ
=
V
, in which case
θ
= 0,
or ker θ = 0 and im θ = W , i.e. θ is a bijection.
(ii)
Since
F
is algebraically closed,
θ
has an eigenvalue
λ
. Then
θ λι
V
is a
singular
G
-endomorphism of
V
. So by (i), it must be the zero map. So
θ = λι
V
.
Recall that the
F
-space
Hom
G
(
V, W
) is the space of all
G
-homomorphisms
V W . If V = W , we write End
G
(V ) for the G-endomorphisms of V .
Corollary. If V, W are irreducible complex G-spaces, then
dim
C
Hom
G
(V, W ) =
(
1 V, W are G-isomorphic
0 otherwise
Proof.
If
V
and
W
are not isomorphic, then the only possible map between
them is the zero map by Schur’s lemma.
Otherwise, suppose
V
=
W
and let
θ
1
, θ
2
Hom
G
(
V, W
) be both non-
zero. By Schur’s lemma, they are isomorphisms, and hence invertible. So
θ
1
2
θ
1
End
G
(V ). Thus θ
1
2
θ
1
= λι
V
for some λ C. Thus θ
1
= λθ
2
.
We have another less obvious corollary.
Corollary.
If
G
is a finite group and has a faithful complex irreducible repre-
sentation, then its center Z(G) is cyclic.
This is a useful result it allows us transfer our representation-theoretic
knowledge (the existence of a faithful complex irreducible representation) to
group theoretic knowledge (center of group being cyclic). This will become
increasingly common in the future, and is a good thing since representations are
easy and groups are hard.
The converse, however, is not true. For this, see example sheet 1, question
10.
Proof.
Let
ρ
:
G GL
(
V
) be a faithful irreducible complex representation. Let
z Z
(
G
). So
zg
=
gz
for all
g G
. Hence
φ
z
:
v 7→ zv
is a
G
-endomorphism
on
V
. Hence by Schur’s lemma, it is multiplication by a scalar
µ
z
, say. Thus
zv = µ
z
v for all v V .
Then the map
σ : Z(G) C
×
z 7→ µ
g
is a representation of
Z
(
G
). Since
ρ
is faithful, so is
σ
. So
Z
(
G
) =
{µ
z
:
z
Z(G)} is isomorphic to a finite subgroup of C
×
, hence cyclic.
Corollary.
The irreducible complex representations of a finite abelian group
G
are all 1-dimensional.
Proof.
We can use the fact that commuting diagonalizable matrices are simulta-
neously diagonalizable. Thus for every irreducible
V
, we can pick some
v V
that is an eigenvector for each
g G
. Thus
hvi
is a
G
-subspace. As
V
is
irreducible, we must have V = hvi.
Alternatively, we can prove this in a representation-theoretic way. Let
V
be
an irreducible complex representation. For each g G, the map
θ
g
: V V
v 7→ gv
is a
G
-endomorphism of
V
, since it commutes with the other group elements.
Since V is irreducible, θ
g
= λ
g
ι
V
for some λ
g
C. Thus
gv = λ
g
v
for any g. As V is irreducible, we must have V = hvi.
Note that this result fails over
R
. For example,
C
3
has a two irreducible real
representations, one of dimension 1 and one of dimension 2.
We can do something else. Recall that every finite abelian group
G
isomorphic
to a product of abelian groups. In fact, it can be written as a product of
C
p
α
for various primes
p
and
α
1, and the factors are uniquely determined up to
order.
This you already know from IB Groups Rings and Modules. You might be
born knowing it it’s such a fundamental fact of nature.
Proposition.
The finite abelian group
G
=
C
n
1
× ··· × C
n
r
has precisely
|G|
irreducible representations over C.
This is not a coincidence. We will later show that the number of irreducible
representations is the number of conjugacy classes of the group. In abelian
groups, each conjugacy class is just a singleton, and hence this result.
Proof. Write
G = hx
1
i × ··· × hx
r
i,
where
|x
j
|
=
n
j
. Any irreducible representation
ρ
must be one-dimensional. So
we have
ρ : G C
×
.
Let
ρ
(1
, ··· , x
j
, ··· ,
1) =
λ
j
. Then since
ρ
is a homomorphism, we must have
λ
n
j
j
= 1. Therefore λ
j
is an n
j
th root of unity.
Now the values (λ
1
, ··· , λ
r
) determine ρ completely, namely
ρ(x
j
1
1
, ··· , x
j
r
r
) = λ
j
1
1
···λ
j
r
r
.
Also, whenever
λ
i
is an
n
i
th root of unity for each
i
, then the above formula
gives a well-defined representation. So there is a one-to-one correspondence
ρ (λ
1
, ··· , λ
r
), with λ
n
j
j
= 1.
Since for each
j
, there are
n
j
many
n
j
th roots of unity, it follows that there
are |G| = n
1
···n
r
many choices of the λ
i
. Thus the proposition.
Example.
(i)
Consider
G
=
C
4
=
hxi
. The four 1-dimensional irreducible representations
are given by
1 x x
2
x
3
ρ
1
1 1 1 1
ρ
2
1 i 1 i
ρ
2
1 1 1 1
ρ
2
1 i 1 i
(ii)
Consider the Klein four group
G
=
V
R
=
hx
1
i × hx
2
i
. The irreducible
representations are
1 x
1
x
2
x
1
x
2
ρ
1
1 1 1 1
ρ
2
1 1 1 1
ρ
2
1 1 1 1
ρ
2
1 1 1 1
These are also known as character tables, and we will spend quite a lot of
time computing these for non-abelian groups.
Note that there is no “natural” one-to-one correspondence between the
elements of
G
and the representations of
G
(for
G
finite-abelian). If we choose
an isomorphism
G
=
C
n
1
× ···C
n
r
, then we can identify the two sets, but it
depends on the choice of the isomorphism (while the decomposition is unique,
we can pick a different generator of, say,
C
n
1
and get a different isomorphism to
the same decomposition).
Isotypical decompositions
Recall that we proved we can decompose any
G
-representation into a sum of
irreducible representations. Is this decomposition unique? If it isn’t, can we say
anything about, say, the size of the irreducible representations, or the number of
factors in the decomposition?
We know any diagonalizable endomorphism
α
:
V V
for a vector space
V
gives us a vector space decomposition
V =
M
λ
V (λ),
where
V (λ) = {v V : α(v) = λv}.
This is canonical in that it depends on α alone, and nothing else.
If
V
is moreover a
G
-representation, how does this tie in to the decomposition
of V into the irreducible representations?
Let’s do an example.
Example.
Consider
G
=
D
6
=
S
3
=
hr, s
:
r
3
=
s
2
= 1
, rs
=
sr
1
i
. We have
previously seen that each irreducible representation has dimension at most 2.
We spot at least three irreducible representations:
1 triad r 7→ 1 s 7→ 1
S sign r 7→ 1 s 7→ 1
W 2-dimensional
The last representation is the action of
D
6
on
R
2
in the natural way, i.e. the
rotations and reflections of the plane that corresponds to the symmetries of the
triangle. It is helpful to view this as a complex representation in order to make
the matrix look nice. The 2-dimensional representation (
ρ, W
) is defined by
W = C
2
, where r and s act on W as
ρ(r) =
ω 0
0 ω
2
, ρ(s) =
0 1
1 0
,
and
ω
=
e
2πi/3
is a third root of unity. We will now show that these are indeed
all the irreducible representations, by decomposing any representation into sum
of these.
So let’s decompose an arbitrary representation. Let (
ρ
0
, V
) be any complex
representation of
G
. Since
ρ
0
(
r
) has order 3, it is diagonalizable has eigenvalues
1
, ω, ω
2
. We diagonalize
ρ
0
(
r
) and then
V
splits as a vector space into the
eigenspaces
V = V (1) V (ω) V (ω
2
).
Since
srs
1
=
r
1
, we know
ρ
0
(
s
) preserves
V
(1) and interchanges
V
(
ω
) and
V (ω
2
).
Now we decompose
V
(1) into
ρ
0
(
s
) eigenspaces, with eigenvalues
±
1. Since
r
has to act trivially on these eigenspaces,
V
(1) splits into sums of copies of the
irreducible representations 1 and S.
For the remaining mess, choose a basis
v
1
, ··· , v
n
of
V
(
ω
), and let
v
0
j
=
ρ
0
(
s
)
v
j
. Then
ρ
0
(
s
) acts on the two-dimensional space
hv
j
, v
0
j
i
as
0 1
1 0
, while
ρ
0
(
r
) acts as
ω 0
0 ω
2
. This means
V
(
ω
)
V
(
ω
2
) decomposes into many copies
of W.
We did this for
D
6
by brute force. How do we generalize this? We first have
the following lemma:
Lemma. Let V, V
1
, V
2
be G-vector spaces over F. Then
(i) Hom
G
(V, V
1
V
2
)
=
Hom
G
(V, V
1
) Hom
G
(V, V
2
)
(ii) Hom
G
(V
1
V
2
, V )
=
Hom
G
(V
1
, V ) Hom
G
(V
2
, V ).
Proof. The proof is to write down the obvious homomorphisms and inverses.
Define the projection map
π
i
: V
1
V
2
V
i
,
which is the G-linear projection onto V
i
.
Then we can define the G-homomorphism
Hom
G
(V, V
1
V
2
) 7→ Hom
G
(V, V
1
) Hom
G
(V, V
2
)
ϕ 7→ (π
1
ϕ, π
2
ϕ).
Then the map (ψ
1
, ψ
2
) 7→ ψ
1
+ ψ
2
is an inverse.
For the second part, we have the homomorphism
ϕ 7→
(
ϕ|
V
1
, ϕ|
V
2
) with
inverse (ψ
1
, ψ
2
) 7→ ψ
1
π
1
+ ψ
2
π
2
.
Lemma.
Let
F
be an algebraically closed field, and
V
be a representation of
G
.
Suppose
V
=
L
n
i=1
V
i
is its decomposition into irreducible components. Then
for each irreducible representation S of G,
|{j : V
j
=
S}| = dim Hom
G
(S, V ).
This tells us we can count the multiplicity of
S
in
V
by looking at the
homomorphisms.
Proof.
We induct on
n
. If
n
= 0, then this is a trivial space. If
n
= 1, then
V
itself is irreducible, and by Schur’s lemma,
dim Hom
G
(
S, V
) = 1 if
V
=
S
, 0
otherwise. Otherwise, for n > 1, we have
V =
n1
M
i=1
V
i
!
V
n
.
By the previous lemma, we know
dim hom
G
S,
n1
M
i=1
V
i
!
V
n
!
= dim Hom
G
S,
n1
M
i=1
V
i
!
+ dim hom
G
(S, V
n
).
The result then follows by induction.
Definition
(Canonical decomposition/decomposition into isotypical compo-
nents)
.
A decomposition of
V
as
L
W
j
, where each
W
j
is (isomorphic to)
n
j
copies of the irreducible
S
j
(with
S
j
6
=
S
i
for
i 6
=
j
) is the canonical decomposition
or decomposition into isotypical components.
For an algebraically closed field F, we know we must have
n
j
= dim Hom
G
(S
j
, V ),
and hence this decomposition is well-defined.
We’ve finally all the introductory stuff. The course now begins.
5 Character theory
In topology, we want to classify spaces. To do so, we come up with invariants of
spaces, like the number of holes. Then we know that a torus is not the same
as a sphere. Here, we want to attach invariants to a representation
ρ
of a finite
group G on V .
One thing we might want to do is to look at the matrix coefficients of
ρ
(
g
).
However, this is bad, since this is highly highly highly basis dependent. It is not
a true invariant. We need to do something better than that.
Let
F
=
C
, and
G
be a finite group. Let
ρ
=
ρ
V
:
G GL
(
V
) be a
representation of
G
. The clue is to look at the characteristic polynomial of the
matrix. The coefficients are functions of the eigenvalues on one extreme, the
determinant is the product of all eigenvalues; on the other extreme, and the
trace is the sum of all of them. Surprisingly, it is the trace that works. We don’t
have to bother ourselves with the other coefficients.
Definition
(Character)
.
The character of a representation
ρ
:
G GL
(
V
),
written χ
ρ
= χ
v
= χ, is defined as
χ(g) = tr ρ(g).
We say ρ affords the character χ.
Alternatively, the character is
tr R
(
g
), where
R
(
g
) is any matrix representing
ρ(g) with respect to any basis.
Definition (Degree of character). The degree of χ
v
is dim V .
Thus, χ is a function G C.
Definition
(Linear character)
.
We say
χ
is linear if
dim V
= 1, in which case
χ is a homomorphism G C
×
= GL
1
(C).
Various properties of representations are inherited by characters.
Definition
(Irreducible character)
.
A character
χ
is irreducible if
ρ
is irreducible.
Definition (Faithful character). A character χ is faithful if ρ is faithful.
Definition
(Trivial/principal character)
.
A character
χ
is trivial or principal if
ρ is the trivial representation. We write χ = 1
G
.
χ
is a complete invariant in the sense that it determines
ρ
up to isomorphism.
This is staggering. We reduce the whole matrix into a single number the
trace, and yet we have not lost any information at all! We will prove this later,
after we have developed some useful properties of characters.
Theorem.
(i) χ
V
(1) = dim V .
(ii) χ
V
is a class function, namely it is conjugation invariant, i.e.
χ
V
(hgh
1
) = χ
V
(g)
for all g, h G. Thus χ
V
is constant on conjugacy classes.
(iii) χ
V
(g
1
) = χ
V
(g).
(iv) For two representations V, W , we have
χ
V W
= χ
V
+ χ
W
.
These results, despite being rather easy to prove, are very useful, since they
save us a lot of work when computing the characters of representations.
Proof.
(i) Obvious since ρ
V
(1) = id
V
.
(ii) Let R
g
be the matrix representing g. Then
χ(hgh
1
) = tr(R
h
R
g
R
1
h
) = tr(R
g
) = χ(g),
as we know from linear algebra.
(iii)
Since
g G
has finite order, we know
ρ
(
g
) is represented by a diagonal
matrix
R
g
=
λ
1
.
.
.
λ
n
,
and χ(g) =
P
λ
i
. Now g
1
is represented by
R
g
1
=
λ
1
1
.
.
.
λ
1
n
,
Noting that each λ
i
is an nth root of unity, hence |λ
i
| = 1, we know
χ(g
1
) =
X
λ
1
i
=
X
λ
i
=
X
λ
i
= χ(g).
(iv)
Suppose
V
=
V
1
V
2
, with
ρ
:
G GL
(
V
) splitting into
ρ
i
:
G GL
(
V
i
).
Pick a basis
B
i
for
V
i
, and let
B
=
B
1
B
2
. Then with respect to
B
, we
have
[ρ(g)]
B
=
[ρ
1
(g)]
B
1
0
0 [ρ
2
(g)]
B
2
.
So χ(g) = tr(ρ(g)) = tr(ρ
1
(g)) + tr(ρ
2
(g)) = χ
1
(g) + χ
2
(g).
We will see later that if we take characters
χ
1
, χ
2
of
G
, then
χ
1
χ
2
is also a
character of
G
. This uses the notion of tensor products, which we will do later.
Lemma.
Let
ρ
:
G GL
(
V
) be a complex representation affording the character
χ. Then
|χ(g)| χ(1),
with equality if and only if
ρ
(
g
) =
λI
for some
λ C
, a root of unity. Moreover,
χ(g) = χ(1) if and only if g ker ρ.
Proof.
Fix
g
, and pick a basis of eigenvectors of
ρ
(
g
). Then the matrix of
ρ
(
g
)
is diagonal, say
ρ(g) =
λ
1
.
.
.
λ
n
,
Hence
|χ(g)| =
X
λ
i
X
|λ
i
| =
X
1 = dim V = χ(1).
In the triangle inequality, we have equality if and only if all the
λ
i
’s are equal,
to λ, say. So ρ(g) = λI. Since all the λ
i
’s are roots of unity, so is λ.
And, if
χ
(
g
) =
χ
(1), then since
ρ
(
g
) =
λI
, taking the trace gives
χ
(
g
) =
λχ
(1).
So λ = 1, i.e. ρ(g) = I. So g ker ρ.
The following lemma allows us to generate new characters from old.
Lemma.
(i) If χ is a complex (irreducible) character of G, then so is ¯χ.
(ii)
If
χ
is a complex (irreducible) character of
G
, then so is
εχ
for any linear
(1-dimensional) character ε.
Proof.
(i)
If
R
:
G GL
n
(
C
) is a complex matrix representation, then so is
¯
R
:
G
GL
n
(C), where g 7→ R(g). Then the character of
¯
R is ¯χ
(ii)
Similarly,
R
0
:
g 7→ ε
(
g
)
R
(
g
) for
g G
is a representation with character
εχ.
It is left as an exercise for the reader to check the details.
We now have to justify why we care about characters. We said it was a
complete invariant, as in two representations are isomorphic if and only if they
have the same character. Before we prove this, we first need some definitions
Definition
(Space of class functions)
.
Define the complex space of class functions
of G to be
C(G) = {f : G C : f(hgh
1
) = f(g) for all h, g G}.
This is a vector space by f
1
+ f
2
: g 7→ f
1
(g) + f
2
(g) and λf : g 7→ λf(g).
Definition
(Class number)
.
The class number
k
=
k
(
G
) is the number of
conjugacy classes of G.
We can list the conjugacy classes as
C
1
, ··· , C
k
. wlog, we let
C
1
=
{
1
}
. We
choose
g
1
, g
2
, ··· , g
k
to be representatives of the conjugacy classes. Note also
that
dim C
(
G
) =
k
, since the characteristic function
δ
j
of the classes form a basis
where
δ
j
(g) =
(
1 g C
j
0 g 6∈ C
j
.
We define a Hermitian inner product on C(G) by
hf, f
0
i =
1
|G|
X
gG
f(g)f
0
(g)
=
1
|G|
k
X
j=1
|C
j
|f(g
j
)f
0
(g
j
)
By the orbit-stabilizer theorem, we can write this as
=
k
X
j=1
1
|C
G
(g
j
)|
f(g
j
)f
0
(g
j
).
In particular, for characters,
hχ, χ
0
i =
k
X
j=1
1
|C
G
(g
j
)|
χ(g
1
j
)χ
0
(g
j
).
It should be clear (especially using the original formula) that
hχ, χ
0
i
=
hχ
0
, χi
.
So when restricted to characters, this is a real symmetric form.
The main result is the following theorem:
Theorem
(Completeness of characters)
.
The complex irreducible characters of
G form an orthonormal basis of C(G), namely
(i)
If
ρ
:
G GL
(
V
) and
ρ
0
:
G GL
(
V
0
) are two complex irreducible
representations affording characters χ, χ
0
respectively, then
hχ, χ
0
i =
(
1 if ρ and ρ
0
are isomorphic representations
0 otherwise
This is the (row) orthogonality of characters.
(ii)
Each class function of
G
can be expressed as a linear combination of
irreducible characters of G.
Proof is deferred. We first look at some corollaries.
Corollary.
Complex representations of finite groups are characterised by their
characters.
Proof.
Let
ρ
:
G GL
(
V
) afford the character
χ
. We know we can write
ρ
=
m
1
ρ
1
··· m
k
ρ
k
, where
ρ
1
, ··· , ρ
k
are (distinct) irreducible and
m
j
0
are the multiplicities. Then we have
χ = m
1
χ
1
+ ··· + m
k
χ
k
,
where χ
j
is afforded by ρ
j
. Then by orthogonality, we know
m
j
= hχ, χ
j
i.
So we can obtain the multiplicity of each
ρ
j
in
ρ
just by looking at the inner
products of the characters.
This is not true for infinite groups. For example, if
G
=
Z
, then the
representations
1 7→
1 0
0 1
, 1 7→
1 1
0 1
are non-isomorphic, but have the same character 2.
Corollary
(Irreducibility criterion)
.
If
ρ
:
G GL
(
V
) is a complex repre-
sentation of
G
affording the character
χ
, then
ρ
is irreducible if and only if
hχ, χi = 1.
Proof.
If
ρ
is irreducible, then orthogonality says
hχ, χi
= 1. For the other
direction, suppose hχ, χi = 1. We use complete reducibility to get
χ =
X
m
j
χ
j
,
with
χ
j
irreducible, and
m
j
0 the multiplicities. Then by orthogonality, we
get
hχ, χi =
X
m
2
j
.
But
hχ, χi
= 1. So exactly one of
m
j
is 1, while the others are all zero, and
χ = χ
j
. So χ is irreducible.
Theorem.
Let
ρ
1
, ··· , ρ
k
be the irreducible complex representations of
G
, and
let their dimensions be n
1
, ··· , n
k
. Then
|G| =
X
n
2
i
.
Recall that for abelian groups, each irreducible character has dimension 1,
and there are |G| representations. So this is trivially satisfied.
Proof.
Recall that
ρ
reg
:
G GL
(
CG
), given by
G
acting on itself by multipli-
cation, is the regular representation of
G
of dimension
|G|
. Let its character be
π
reg
, the regular character of G.
First note that we have
π
reg
(1) =
|G|
, and
π
reg
(
h
) = 0 if
h 6
= 1. The first
part is obvious, and the second is easy to show, since we have only 0s along the
diagonal.
Next, we decompose π
reg
as
π
reg
=
X
a
j
χ
j
,
We now want to find a
j
. We have
a
j
= hπ
reg
, χ
j
i =
1
|G|
X
gG
π
reg
(g)χ
j
(g) =
1
|G|
· |G|χ
j
(1) = χ
j
(1).
Then we get
|G| = π
reg
(1) =
X
a
j
χ
j
(1) =
X
χ
j
(1)
2
=
X
n
2
j
.
From this proof, we also see that each irreducible representation is a subrep-
resentation of the regular representation.
Corollary.
The number of irreducible characters of
G
(up to equivalence) is
k
,
the number of conjugacy classes.
Proof.
The irreducible characters and the characteristic functions of the conju-
gacy classes are both bases of C(G).
Corollary.
Two elements
g
1
, g
2
are conjugate if and only if
χ
(
g
1
) =
χ
(
g
2
) for
all irreducible characters χ of G.
Proof.
If
g
1
, g
2
are conjugate, since characters are class functions, we must have
χ(g
1
) = χ(g
2
).
For the other direction, let
δ
be the characteristic function of the class of
g
1
.
Then since δ is a class function, we can write
δ =
X
m
j
χ
j
,
where χ
j
are the irreducible characters of G. Then
δ(g
2
) =
X
m
j
χ
j
(g
2
) =
X
m
j
χ
j
(g
1
) = δ(g
1
) = 1.
So g
2
is in the same conjugacy class as g
1
.
Before we move on to the proof of the completeness of characters, we first
make a few more definitions:
Definition
(Character table)
.
The character table of
G
is the
k × k
matrix
X
= (
χ
i
(
g
j
)), where 1 =
χ
1
, χ
2
, ··· , χ
k
are the irreducible characters of
G
, and
C
1
= {1}, C
2
, ··· , C
k
are the conjugacy classes with g
j
C
j
.
We seldom think of this as a matrix, but as an actual table, as we will see in
the table below.
We will spend the next few lectures coming up with techniques to compute
these character tables.
Example.
Consider
G
=
D
6
=
S
3
=
hr, s | r
3
=
s
2
= 1
, srs
1
=
r
1
i
. As in all
computations, the first thing to do is to compute the conjugacy classes. This is
not too hard. We have
C
1
= {1}, C
2
= {s, sr, sr
2
}, C
2
= {r, r
1
}.
Alternatively, we can view this as
S
3
and use the fact that two elements are
conjugate if and only if they have the same cycle types. We have found the
three representations: 1, the trivial representation; S, the sign; and W, the
two-dimensional representation. In
W
, the reflections
sr
j
acts by matrix with
eigenvalues
±
1. So the sum of eigenvalues is 0, hence
χ
w
(
sr
2
) = 0. It also also
not hard to see
r
m
7→
cos
2
3
sin
2
3
sin
2
3
cos
2
3
.
So χ
w
(r
m
) = 1.
Fortunately, after developing some theory, we will not need to find all the
irreducible representations in order to compute the character table.
1 C
2
C
3
1 1 1 1
S 1 1 1
χ
w
2 0 1
We see that the sum of the squares of the first column is 1
2
+ 1
2
+ 2
2
= 6 =
|D
6
|
,
as expected. We can also check that W is genuinely an irreducible representation.
Noting that the centralizers of
C
1
, C
2
and
C
3
are of sizes 6
,
2
,
3. So the inner
product is
hχ
W
, χ
W
i =
2
2
6
+
0
2
2
+
(1)
2
3
= 1,
as expected.
So we now need to prove orthogonality.
6 Proof of orthogonality
We will do the proof in parts.
Theorem
(Row orthogonality relations)
.
If
ρ
:
G GL
(
V
) and
ρ
0
:
G
GL
(
V
0
) are two complex irreducible representations affording characters
χ, χ
0
respectively, then
hχ, χ
0
i =
(
1 if ρ and ρ
0
are isomorphic representations
0 otherwise
.
Proof.
We fix a basis of
V
and of
V
0
. Write
R
(
g
)
, R
0
(
g
) for the matrices of
ρ
(
g
)
and ρ
0
(g) with respect to these bases respectively. Then by definition, we have
hχ
0
, χi =
1
|G|
X
gG
χ
0
(g
1
)χ(g)
=
1
|G|
X
gG
X
1in
0
1jn
R
0
(g
1
)
ii
R(g)
jj
.
For any linear map
ϕ
:
V V
0
, we define a new map by averaging by
ρ
0
and
ρ
.
˜ϕ : V V
0
v 7→
1
|G|
X
ρ
0
(g
1
)ϕρ(g)v
We first check ˜ϕ is a G-homomorphism if h G, we need to show
ρ
0
(h
1
) ˜ϕρ(h)(v) = ˜ϕ(v).
We have
ρ
0
(h
1
) ˜ϕρ(h)(v) =
1
|G|
X
gG
ρ
0
((gh)
1
)ϕρ(gh)v
=
1
|G|
X
g
0
G
ρ
0
(g
0−1
)ϕρ(g
0
)v
= ˜ϕ(v).
(i)
Now we first consider the case where
ρ, ρ
0
is not isomorphic. Then by
Schur’s lemma, we must have ˜ϕ = 0 for any linear ϕ : V V
0
.
We now pick a very nice
ϕ
, where everything disappears. We let
ϕ
=
ε
αβ
,
the operator having matrix
E
αβ
with entries 0 everywhere except 1 in the
(α, β) position.
Then ˜ε
αβ
= 0. So for each i, j, we have
1
|G|
X
gG
(R
0
(g
1
)E
αβ
R(g))
ij
= 0.
Using our choice of ε
αβ
, we get
1
|G|
X
gG
R
0
(g
1
)
R(g)
βj
= 0
for all i, j. We now pick α = i and β = j. Then
1
|G|
X
gG
R
0
(g
1
)
ii
R(g)
jj
= 0.
We can sum this thing over all i and j to get that hχ
0
, χi = 0.
(ii)
Now suppose
ρ, ρ
0
are isomorphic. So we might as well take
χ
=
χ
0
,
V
=
V
0
and ρ = ρ
0
. If ϕ : V V is linear, then ˜ϕ End
G
(V ).
We first claim that tr ˜ϕ = tr ϕ. To see this, we have
tr ˜ϕ =
1
|G|
X
gG
tr(ρ(g
1
)ϕρ(g)) =
1
|G|
X
gG
tr ϕ = tr ϕ,
using the fact that traces don’t see conjugacy (and
ρ
(
g
1
) =
ρ
(
g
)
1
since
ρ is a group homomorphism).
By Schur’s lemma, we know
˜ϕ
=
λι
v
for some
λ C
(which depends on
ϕ). Then if n = dim V , then
λ =
1
n
tr ϕ.
Let ϕ = ε
αβ
. Then tr ϕ = δ
αβ
. Hence
˜ε
αβ
=
1
|G|
X
g
ρ(g
1
)ε
αβ
ρ(g) =
1
n
δ
αβ
ι.
In terms of matrices, we take the (i, j)th entry to get
1
|G|
X
R(g
1
)
R(g)
βj
=
1
n
δ
αβ
δ
ij
.
We now put α = i and β = j. Then we are left with
1
|G|
X
g
R(g
1
)
ii
R(g)
jj
=
1
n
δ
ij
.
Summing over all i and j, we get hχ, χi = 1.
After learning about tensor products and duals later on, one can provide a
shorter proof of this result as follows:
Alternative proof. Consider two representation spaces V and W . Then
hχ
W
, χ
V
i =
1
|G|
X
χ
W
(g)χ
V
(g) =
1
|G|
X
χ
V W
(g).
We notice that there is a natural isomorphism
V W
=
Hom
(
W, V
), and
the action of
g
on this space is by conjugation. Thus, a
G
-invariant element
is just a
G
-homomorphism
W V
. Thus, we can decompose
Hom
(
V, W
) =
Hom
G
(
V, W
)
U
for some
G
-space
U
, and
U
has no
G
-invariant element. Hence
in the decomposition of
Hom
(
V, W
) into irreducibles, we know there are exactly
dim Hom
G
(
V, W
) copies of the trivial representation. By Schur’s lemma, this
number is 1 if V
=
W , and 0 if V 6
=
W .
So it suffices to show that if χ is a non-trivial irreducible character, then
X
gG
χ(g) = 0.
But if
ρ
affords
χ
, then any element in the image of
P
gG
ρ
(
g
) is fixed by
G
.
By irreducibility, the image must be trivial. So
P
gG
ρ(g) = 0.
What we have done is show the orthogonality of the rows. There is a similar
one for the columns:
Theorem (Column orthogonality relations). We have
k
X
i=1
χ
i
(g
j
)χ
i
(g
`
) = δ
j`
|C
G
(g
`
)|.
This is analogous to the fact that if a matrix has orthonormal rows, then it
also has orthonormal columns. We get the extra factor of
|C
G
(
g
`
)
|
because of
the way we count.
This has an easy corollary, which we have already shown previously using
the regular representation:
Corollary.
|G| =
k
X
i=1
χ
2
i
(1).
Proof of column orthogonality.
Consider the character table
X
= (
χ
i
(
g
j
)). We
know
δ
ij
= hχ
i
, χ
j
i =
X
`
1
|C
G
(g
`
)|
χ
i
(g
`
)χ
k
(g
`
).
Then
¯
XD
1
X
T
= I
k×k
,
where
D =
|C
G
(g
1
)| ··· 0
.
.
.
.
.
.
.
.
.
0 ··· |C
G
(g
k
)|
.
Since
X
is square, it follows that
D
1
¯
X
T
is the inverse of
X
. So
¯
X
T
X
=
D
,
which is exactly the theorem.
The proof requires that
X
is square, i.e. there are
k
many irreducible repre-
sentations. So we need to prove the last part of the completeness of characters.
Theorem.
Each class function of
G
can be expressed as a linear combination
of irreducible characters of G.
Proof.
We list all the irreducible characters
χ
1
, ··· , χ
`
of
G
. Note that we don’t
know the number of irreducibles is
k
. This is essentially what we have to prove
here. We now claim these generate C(G), the ring of class functions.
Now recall that
C
(
G
) has an inner product. So it suffices to show that the
orthogonal complement to the span
hχ
1
, ··· , χ
`
i
in
C
(
G
) is trivial. To see this,
assume f C(G) satisfies
hf, χ
j
i = 0
for all
χ
j
irreducible. We let
ρ
:
G GL
(
V
) be an irreducible representation
affording χ {χ
1
, ··· , χ
`
}. Then hf, χi = 0.
Consider the function
ϕ =
1
|G|
X
g
f(g)ρ(g) : V V.
For any h G, we can compute
ρ(h)
1
ϕρ(h) =
1
|G|
X
g
¯
f(g)ρ(h
1
gh) =
1
|G|
X
g
¯
f(h
1
gh)ρ(h
1
gh) = ϕ,
using the fact that
¯
f
is a class function. So this is a
G
-homomorphism. So as
ρ
is irreducible, Schur’s lemma says it must be of the form λι
V
for some λ C.
Now we take the trace of this thing. So we have
= tr
1
|G|
X
g
f(g)ρ(g)
!
=
1
|G|
X
g
f(g)χ(g) = hf, χi = 0.
So
λ
= 0, i.e.
P
g
f(g)ρ
(
g
) = 0, the zero endomorphism on
V
. This is valid for
any irreducible representation, and hence for every representation, by complete
reducibility.
In particular, take ρ = ρ
reg
, where ρ
reg
(g) : e
1
7→ e
g
for each g G. Hence
X
f(g)ρ
reg
(g) : e
1
7→
X
g
f(g)e
g
.
Since this is zero, it follows that we must have
P
f(g)e
g
= 0. Since the
e
g
’s are
linearly independent, we must have f(g) = 0 for all g G, i.e. f = 0.
7 Permutation representations
In this chapter, we are going to study permutation representations in a bit more
detail, since we can find some explicit formulas for the character of them. This
is particularly useful if we are dealing with the symmetric group.
Let
G
be a finite group acting on a finite set
X
=
{x
1
, ··· , x
n
}
(sometimes
known as a
G
-set). We define
CX
to be the complex vector space with basis
{e
x
1
, ··· , e
x
n
}. More explicitly,
CX =
X
j
a
j
e
x
j
: a
j
C
.
There is a corresponding representation permutation
ρ
X
: G GL(CX)
g 7→ ρ(g)
where
ρ
(
g
) :
e
x
j
7→ e
gx
j
, extended linearly. We say
ρ
X
is the representation
corresponding to the action of G on X.
This is a generalization of the group algebra if we let
G
act on itself by
left multiplication, then this is in fact the group algebra.
The corresponding matrix representations
ρ
X
(
g
) with respect to the basis
{e
x
}
xX
are permutation matrices it is 0 everywhere except for one 1 in each
row and column. In particular, (
ρ
(
g
))
ij
= 1 precisely when
gx
j
=
x
i
. So the
only non-zero diagonal entries in
ρ
(
g
) are those
i
such that
x
i
is fixed by
g
. Thus
we have
Definition (Permutation character). The permutation character π
X
is
π
X
(g) = |fix(g)| = |{x X : gx = x}|.
Lemma. π
X
always contains the trivial character 1
G
(when decomposed in the
basis of irreducible characters). In particular,
span{e
x
1
+
···
+
e
x
n
}
is a trivial
G-subspace of CX, with G-invariant complement {
P
x
a
x
e
x
:
P
a
x
= 0}.
Lemma. hπ
X
,
1
i
, which is the multiplicity of 1 in
π
X
, is the number of orbits
of G on X.
Proof.
We write
X
as the disjoint union of orbits,
X
=
X
1
··· X
`
. Then
it is clear that the permutation representation on
X
is just the sum of the
permutation representations on the X
i
, i.e.
π
X
= π
X
1
+ ··· + π
x
`
,
where
π
X
j
is the permutation character of
G
on
X
j
. So to prove the lemma, it
is enough to consider the case where the action is transitive, i.e. there is just
one orbit.
So suppose
G
acts transitively on
X
. We want to show
hπ
X
,
1
i
= 1. By
definition, we have
hπ
X
, 1i =
1
|G|
X
g
π
X
(g)
=
1
|G|
|{(g, x) G × X : gx = x}|
=
1
|G|
X
xX
|G
x
|,
where
G
x
is the stabilizer of
x
. By the orbit-stabilizer theorem, we have
|G
x
||X|
=
|G|. So we can write this as
=
1
|G|
X
xX
|G|
|X|
=
1
|G|
· |X| ·
|G|
|X|
= 1.
So done.
Lemma. Let G act on the sets X
1
, X
2
. Then G acts on X
1
× X
2
by
g(x
1
, x
2
) = (gx
1
, gx
2
).
Then the character
π
X
1
×X
2
= π
X
1
π
X
2
,
and so
hπ
X
1
, π
X
2
i = number of orbits of G on X
1
× X
2
.
Proof.
We know
π
X
1
×X
2
(
g
) is the number of pairs (
x
1
, x
2
)
X
1
× X
2
fixed by
g
. This is exactly the number of things in
X
1
fixed by
g
times the number of
things in X
2
fixed by g. So we have
π
X
1
×X
2
(g) = π
X
1
(g)π
X
2
(g).
Then using the fact that π
1
, π
2
are real, we get
hπ
X
1
, π
X
2
i =
1
|G|
X
g
π
X
1
(g)π
X
2
(g)
=
1
|G|
X
g
π
X
1
(g)π
X
2
(g)1
G
(g)
= hπ
X
1
π
X
2
, 1i
= hπ
X
1
×X
2
, 1i.
So the previous lemma gives the desired result.
Recall that a 2-transitive action is an action that is transitive on pairs, i.e.
it can send any pair to any other pair, as long as the elements in the pair are
distinct (it can never send (
x, x
) to (
x
0
, x
00
) if
x
0
6
=
x
00
, since
g
(
x, x
) = (
gx, gx
)).
More precisely,
Definition
(2-transitive)
.
Let
G
act on
X
,
|X| >
2. Then
G
is 2-transitive on
X
if
G
has two orbits on
X × X
, namely
{
(
x, x
) :
x X}
and
{
(
x
1
, x
2
) :
x
i
X, x
1
6= x
2
}.
Then we have the following lemma:
Lemma. Let G act on X, with |X| > 2. Then
π
X
= 1
G
+ χ,
with χ irreducible if and only if G is 2-transitive on X.
Proof. We know
π
X
= m
1
1
G
+ m
2
χ
2
+ ··· + m
`
χ
`
,
with 1
G
, χ
2
, ··· , χ
`
distinct irreducible characters and
m
i
N
are non-zero.
Then by orthogonality,
hπ
X
, π
X
i =
j
X
i=1
m
2
i
.
Since
hπ
X
, π
X
i
is the number of orbits of
X × X
, we know
G
is 2-transitive on
X if and only if ` = 2 and m
1
= m
2
= 1.
This is useful if we want to understand the symmetric group, since it is
always 2-transitive (modulo the silly case of S
1
).
Example. S
n
acting on
X
=
{
1
, ··· , n}
is 2-transitive for
n
2. So
π
X
= 1+
χ
,
with
χ
irreducible of degree
n
1 (since
π
X
has degree
n
). This works similarly
for A
n
, whenever n 3.
This tells us we can find an irreducible character of
S
n
by finding
π
X
, and
then subtracting 1 off.
Example.
Consider
G
=
S
4
. The conjugacy classes correspond to different
cycle types. We already know three characters the trivial one, the sign, and
π
X
1
G
.
1 3 8 6 6
1 (1 2)(3 4) (1 2 3) (1 2 3 4) (1 2)
trivial χ
1
1 1 1 1 1
sign χ
2
1 1 1 1 1
π
X
1
G
χ
3
3 1 0 1 1
χ
4
χ
5
We know the product of a one-dimensional irreducible character and another
character is another irreducible character, as shown on the first example sheet.
So we get
1 3 8 6 6
1 (1 2)(3 4) (1 2 3) (1 2 3 4) (1 2)
trivial χ
1
1 1 1 1 1
sign χ
2
1 1 1 1 1
π
X
1
G
χ
3
3 1 0 1 1
χ
3
χ
2
χ
4
3 1 0 1 1
χ
5
For the last representation we can find the dimension by computing 24
(1
2
+
1
2
+ 3
2
+ 3
2
) = 2
2
. So it has dimension 2. To obtain the whole of
χ
5
, we can
use column orthogonality for example, we let the entry in the second column
be
x
. Then column orthogonality says 1 + 1
3
3 + 2
x
= 0 . So
x
= 2. In the
end, we find
1 3 8 6 6
1 (1 2)(3 4) (1 2 3) (1 2 3 4) (1 2)
trivial χ
1
1 1 1 1 1
sign χ
2
1 1 1 1 1
π
X
1
G
χ
3
3 1 0 1 1
χ
3
χ
2
χ
4
3 1 0 1 1
χ
5
2 2 1 0 0
Alternatively, we have previously shown that
χ
reg
=
χ
1
+
χ
2
+ 3
χ
3
+ 3
χ
4
+ 2
χ
5
.
By computing χ
reg
directly, we can find χ
5
.
Finally, we can obtain
χ
5
by observing
S
4
/V
4
=
S
3
, and “lifting” characters.
We will do this in the next chapter. In fact, this
χ
5
is the degree-2 irreducible
representation of S
3
=
D
6
lifted up.
We can also do similar things with the alternating group. The issue of course
is that the conjugacy classes of the symmetric group may split when we move to
the alternating group.
Suppose g A
n
. Then
|C
S
n
(g)| = |S
n
: C
S
n
(g)|
|C
A
n
(g)| = |A
n
: C
A
n
(g)|.
We know
C
S
n
(
g
)
C
A
n
(
g
), but we need not have equality, since elements needed
to conjugate
g
to
h C
S
n
(
g
) might not be in
A
n
. For example, consider
σ = (1 2 3) A
3
. We have C
A
3
(σ) = {σ} and C
S
3
(σ) = {σ, σ
1
}.
We know
A
n
is an index 2 subgroup of
S
n
. So
C
A
n
(
g
)
C
S
n
(
g
) either has
index 2 or 1. If the index is 1, then
|C
A
n
|
=
1
2
|C
S
n
|
. Otherwise, the sizes are
equal.
A useful criterion for determining which case happens is the following:
Lemma.
Let
g A
n
,
n >
1. If
g
commutes with some odd permutation in
S
n
,
then
C
S
n
(
g
) =
C
A
n
(
g
). Otherwise,
C
S
n
splits into two conjugacy classes in
A
n
of
equal size.
This is quite useful for doing the examples on the example sheet, but we will
not go through any more examples here.
8 Normal subgroups and lifting
The idea of this section is that there is some correspondence between the
irreducible characters of
G/N
(with
N C G
) and those of
G
itself. In most
cases,
G/N
is a simpler group, and hence we can use this correspondence to find
irreducible characters of G.
The main result is the following lemma:
Lemma.
Let
N C G
. Let
˜ρ
:
G/N GL
(
V
) be a representation of
G/N
. Then
the composition
ρ : G G/N GL(V )
natural
˜ρ
is a representation of G, where ρ(g) = ˜ρ(gN). Moreover,
(i) ρ is irreducible if and only if ˜ρ is irreducible.
(ii) The corresponding characters satisfy χ(g) = ˜χ(gN).
(iii) deg χ = deg ˜χ.
(iv) The lifting operation ˜χ 7→ χ is a bijection
{irreducibles of G/N} {irreducibles of G with N in their kernel}.
We say ˜χ lifts to χ.
Proof.
Since a representation of
G
is just a homomorphism
G GL
(
V
), and
the composition of homomorphisms is a homomorphisms, it follows immediately
that ρ as defined in the lemma is a representation.
(i) We can compute
hχ, χi =
1
|G|
X
gG
χ(g)χ(g)
=
1
|G|
X
gNG/N
X
kN
χ(gk)χ(gk)
=
1
|G|
X
gNG/N
X
kN
˜χ(gN) ˜χ(gN)
=
1
|G|
X
gNG/N
|N|
˜χ(gN) ˜χ(gN)
=
1
|G/N|
X
gNG/N
˜χ(gN) ˜χ(gN)
= h˜χ, ˜χi.
So
hχ, χi
= 1 if and only if
h˜χ, ˜χi
= 1. So
ρ
is irreducible if and only if
˜ρ
is irreducible.
(ii) We can directly compute
χ(g) = tr ρ(g) = tr(˜ρ(gN)) = ˜χ(gN)
for all g G.
(iii) To see that χ and ˜χ have the same degree, we just notice that
deg χ = χ(1) = ˜χ(N) = deg ˜χ.
Alternatively, to show they have the same dimension, just note that
ρ
and
˜ρ map to the general linear group of the same vector space.
(iv)
To show this is a bijection, suppose
˜χ
is a character of
G/N
and
χ
is its
lift to
G
. We need to show the kernel contains
N
. By definition, we know
˜χ
(
N
) =
χ
(1). Also, if
k N
, then
χ
(
k
) =
˜χ
(
kN
) =
˜χ
(
N
) =
χ
(1). So
N ker χ.
Now let
χ
be a character of
G
with
N ker χ
. Suppose
ρ
:
G GL
(
V
)
affords χ. Define
˜ρ : G/N GL(V )
gN 7→ ρ(g)
Of course, we need to check this is well-defined. If
gN
=
g
0
N
, then
g
1
g
0
N
. So
ρ
(
g
) =
ρ
(
g
0
) since
N ker ρ
. So this is indeed well-defined.
It is also easy to see that
˜ρ
is a homomorphism, hence a representation of
G/N .
Finally, if
˜χ
is a character of
˜ρ
, then
˜χ
(
gN
) =
χ
(
g
) for all
g G
by
definition. So
˜χ
lifts to
χ
. It is clear that these two operations are inverses
to each other.
There is one particular normal subgroup lying around that proves to be
useful.
Lemma. Given a group G, the derived subgroup or commutator subgroup
G
0
= h[a, b] : a, b Gi,
where [
a, b
] =
aba
1
b
1
, is the unique minimal normal subgroup of
G
such that
G/G
0
is abelian. So if G/N is abelian, then G
0
N.
Moreover,
G
has precisely
`
=
|G
:
G
0
|
representations of dimension 1, all
with kernel containing G
0
, and are obtained by lifting from G/G
0
.
In particular, by Lagrange’s theorem, ` | G.
Proof. Consider [a, b] = aba
1
b
1
G
0
. Then for any h G, we have
h(aba
1
b
1
)h
1
=
(ha)b(ha)
1
b
1

bhb
1
h
1
= [ha, b][b, h] G
0
So in general, let [a
1
, b
1
][a
2
, b
2
] ···[a
n
, b
n
] G
0
. Then
h[a
1
, b
1
][a
2
, b
2
] ···[a
n
, b
n
]h
1
= (h[a
1
, b
1
]h
1
)(h[a
2
, b
2
]h
1
) ···(h[a
n
, b
n
]h
1
),
which is in G
0
. So G
0
is a normal subgroup.
Let
N C G
. Let
g, h G
. Then [
g, h
]
N
if and only if
ghg
1
h
1
N
if
and only if ghN = hgN , if and only if (gN)(hN) = (hN)(gN) by normality.
Since
G
0
is generated by all [
g, h
], we know
G
0
N
if and only if
G/N
is
abelian.
Since
G/G
0
, is abelian, we know it has exactly
`
irreducible characters,
˜χ
1
, ··· , ˜χ
`
, all of degree 1. The lifts of these to
G
also have degree 1, and by the
previous lemma, these are precisely the irreducible characters
χ
i
of
G
such that
G
0
ker χ
i
.
But any degree 1 character of
G
is a homomorphism
χ
:
G C
×
, hence
χ
(
ghg
1
h
1
) = 1. So for any 1-dimensional character,
χ
, we must have
G
0
ker χ. So the lifts χ
1
, ··· , χ
`
are all 1-dimensional characters of G.
Example.
Let
G
be
S
n
, with
n
3. We claim that
S
0
n
=
A
n
. Since
sgn
(
a
) =
sgn
(
a
1
), we know that [
a, b
]
A
n
for any
a, b S
n
. So
S
0
n
A
n
(alternatively,
since S
n
/A
n
=
C
2
is abelian, we must have S
0
n
A
n
). We now notice that
(1 2 3) = (1 3 2)(1 2)(1 3 2)
1
(1 2)
1
S
0
n
.
So by symmetry, all 3-cycles are in
S
0
n
. Since the 3-cycles generate
A
n
, we know
we must have S
0
n
= A
n
.
Since
S
0
n
/A
n
=
C
2
, we get
`
= 2. So
S
n
must have exactly two linear
characters, namely the trivial character and the sign.
Lemma. G
is not simple if and only if
χ
(
g
) =
χ
(1) for some irreducible character
χ 6
= 1
G
and some 1
6
=
g G
. Any normal subgroup of
G
is the intersection of
the kernels of some of the irreducible characters of G, i.e. N =
T
ker χ
i
.
In other words,
G
is simple if and only if all non-trivial irreducible characters
have trivial kernel.
Proof.
Suppose
χ
(
g
) =
χ
(1) for some non-trivial irreducible character
χ
, and
χ
is afforded by
ρ
. Then
g ker ρ
. So if
g 6
= 1, then 1
6
=
ker ρ C G
, and
ker ρ 6
=
G
.
So G cannot be simple.
If 1
6
=
N C G
is a non-trivial proper subgroup, take an irreducible character
˜χ
of
G/N
, and suppose
˜χ 6
= 1
G/N
. Lift this to get an irreducible character
χ
,
afforded by the representation
ρ
of
G
. Then
N ker ρ C G
. So
χ
(
g
) =
χ
(1) for
g N.
Finally, let 1
6
=
N C G
. We claim that
N
is the intersection of the kernels of
the lifts
χ
1
, ··· , χ
`
of all the irreducibles of
G/N
. Clearly, we have
N
T
i
ker χ
i
.
If
g G \ N
, then
gN 6
=
N
. So
˜χ
(
gN
)
6
=
˜χ
(
N
) for some irreducible
˜χ
of
G/N
.
Lifting
˜χ
to
χ
, we have
χ
(
g
)
6
=
χ
(1). So
g
is not in the intersection of the
kernels.
9 Dual spaces and tensor products of represen-
tations
In this chapter, we will consider several linear algebra constructions, and see
how we can use them to construct new representations and characters.
9.1 Dual spaces
Our objective of this section is to try to come up with “dual representations”.
Given a representation
ρ
:
G GL
(
V
), we would like to turn
V
=
Hom
(
V, F
)
into a representation. So for each
g
, we want to produce a
ρ
(
g
) :
V
V
. So
given
ϕ V
, we shall get a
ρ
(
g
)
ϕ
:
V F
. Given
v V
, a natural guess for
the value of (ρ
(g)ϕ)(v) might be
(ρ
(g)ϕ)(v) = ϕ(ρ(g)v),
since this is how we can patch together
ϕ, ρ, g
and
v
. However, if we do this, we
will not get
ρ
(
g
)
ρ
(
h
) =
ρ
(
gh
). Instead, we will obtain
ρ
(
g
)
ρ
(
h
) =
ρ
(
hg
),
which is the wrong way round. The right definition to use is actually the
following:
Lemma.
Let
ρ
:
G GL
(
V
) be a representation over
F
, and let
V
=
Hom
F
(V, F) be the dual space of V . Then V
is a G-space under
(ρ
(g)ϕ)(v) = ϕ(ρ(g
1
)v).
This is the dual representation to ρ. Its character is χ(ρ
)(g) = χ
ρ
(g
1
).
Proof. We have to check ρ
is a homomorphism. We check
ρ
(g
1
)(ρ
(g
2
)ϕ)(v) = (ρ
(g
2
)ϕ)(ρ(g
1
1
)(v))
= ϕ(ρ(g
1
2
)ρ(g
2
1
)v)
= ϕ(ρ((g
1
g
2
)
1
)(v))
= (ρ
(g
1
g
2
)ϕ)(v).
To compute the character, fix a
g G
, and let
e
1
, ··· , e
n
be a basis of eigenvectors
of V of ρ(g), say
ρ(g)e
j
= λ
j
e
j
.
If we have a dual space of
V
, then we have a dual basis. We let
ε
1
, ··· , ε
n
be
the dual basis. Then
(ρ
(g)ε
j
)(e
i
) = ε
j
(ρ(g
1
)e
i
) = ε
j
(λ
1
i
e
i
) = λ
1
i
δ
ij
= λ
1
j
δ
ij
= λ
1
j
ε
j
(e
i
).
Thus we get
ρ
(g)ε
j
= λ
1
j
ε
j
.
So
χ
ρ
(g) =
X
λ
1
j
= χ
ρ
(g
1
).
Definition
(Self-dual representation)
.
A representation
ρ
:
G GL
(
V
) is
self-dual if there is isomorphism of G-spaces V
=
V
.
Note that as (finite-dimensional) vector spaces, we always have
V
=
V
.
However, what we require here is an isomorphism that preserves the
G
-action,
which does not necessarily exist.
Over F = C, this holds if and only if
χ
ρ
(g) = χ
ρ
(g) = χ
ρ
(g
1
) = χ
ρ
(g).
So a character is self-dual if and only if it is real.
Example.
All irreducible representations of
S
n
are self-dual, since every element
is conjugate to its own inverse. This is not always true for A
n
.
9.2 Tensor products
The next idea we will tackle is the concept of tensor products. We first introduce
it as a linear algebra concept, and then later see how representations fit into this
framework.
There are many ways we can define the tensor product. The definition we
will take here is a rather hands-on construction of the space, which involves
picking a basis. We will later describe some other ways to define the tensor
product.
Definition
(Tensor product)
.
Let
V, W
be vector spaces over
F
. Suppose
dim V
=
m
and
dim W
=
n
. We fix a basis
v
1
, ··· , v
m
and
w
1
, ··· , w
n
of
V
and W respectively.
The tensor product space
V W
=
V
F
W
is an
nm
-dimensional vector
space (over F) with basis given by formal symbols
{v
i
w
j
: 1 i m, 1 j n}.
Thus
V W =
n
X
λ
ij
v
i
w
j
: λ
ij
F
o
,
with the “obvious” addition and scalar multiplication.
If
v =
X
α
i
v
i
V, w =
X
β
j
w
j
,
we define
v w =
X
i,j
α
i
β
j
(v
i
w
j
).
Note that note all elements of
V W
are of this form. Some are genuine
combinations. For example,
v
1
w
1
+
v
2
w
2
cannot be written as a tensor
product of an element in V and another in W .
We can imagine our formula of the tensor of two elements as writing
X
α
i
v
i
X
β
j
w
j
,
and then expand this by letting distribute over +.
Lemma.
(i) For v V , w W and λ F, we have
(λv) w = λ(v w) = v (λw).
(ii) If x, x
1
, x
2
V and y, y
1
, y
2
W , then
(x
1
+ x
2
) y = (x
1
y) + (x
2
y)
x (y
1
+ y
2
) = (x y
1
) + (x y
2
).
Proof.
(i) Let v =
P
α
i
v
i
and w =
P
β
j
w
j
. Then
(λv) w =
X
ij
(λα
i
)β
j
v
i
w
j
λ(v w) = λ
X
ij
α
i
β
j
v
i
w
j
v (λw) =
X
α
i
(λβ
j
)v
i
w
j
,
and these three things are obviously all equal.
(ii) Similar nonsense.
We can define the tensor product using these properties. We consider the
space of formal linear combinations
v w
for all
v V, w W
, and then
quotient out by the relations re had above. It can be shown that this produces
the same space as the one constructed above.
Alternatively, we can define the tensor product using its universal property,
which says for any
U
, a bilinear map from
V ×W
to
U
is “naturally” equivalent
to a linear map
V W U
. This intuitively makes sense, since the distributivity
and linearity properties of
we showed above are exactly the properties required
for a bilinear map if we replace the with a ,”.
It turns out this uniquely defines
V W
, as long as we provide a sufficiently
precise definition of “naturally”. We can do it concretely as follows in our
explicit construction, we have a canonical bilinear map given by
φ : V × W V W
(v, w) 7→ v w.
Given a linear map
V W U
, we can compose it with
φ
:
V × W V W
in order to get a bilinear map
V ×W U
. Then universality says every bilinear
map V × W U arises this way, uniquely.
In general, we say an equivalence between bilinear maps
V × W U
and
linear maps
V W U
is “natural” if it is be mediated by such a universal
map. Then we say a vector space
X
is the tensor product of
V
and
W
if there
is a natural equivalence between bilinear maps
V × W U
and linear maps
X U.
We will stick with our definition for concreteness, but we prove basis-
independence here so that we feel happier:
Lemma.
Let
{e
1
, ··· , e
m
}
be any other basis of
V
, and
{f
1
, ··· , f
m
}
be another
basis of W . Then
{e
i
f
j
: 1 i m, 1 j n}
is a basis of V W .
Proof. Writing
v
k
=
X
α
ik
e
i
, w
`
=
X
β
j`
f
`
,
we have
v
k
w
`
=
X
α
ik
β
jl
e
i
f
j
.
Therefore
{e
i
f
j
}
spans
V W
. Moreover, there are
nm
of these. Therefore
they form a basis of V W .
That’s enough of linear algebra. We shall start doing some representation
theory.
Proposition. Let ρ : G GL(V ) and ρ
0
: G GL(V
0
). We define
ρ ρ
0
: G GL(V V
0
)
by
(ρ ρ
0
)(g) :
X
λ
ij
v
i
w
j
7→
X
λ
ij
(ρ(g)v
i
) (ρ
0
(g)w
j
).
Then ρ ρ
0
is a representation of g, with character
χ
ρρ
0
(g) = χ
ρ
(g)χ
ρ
0
(g)
for all g G.
As promised, the product of two characters of G is also a character of G.
Just before we prove this, recall we showed in sheet 1 that if
ρ
is irreducible
and
ρ
0
is irreducible of degree 1, then
ρ
0
ρ
=
ρ ρ
0
is irreducible. However, if
ρ
0
is not of degree 1, then this is almost always false (or else we can produce
arbitrarily many irreducible representations).
Proof.
It is clear that (
ρ ρ
0
)(
g
)
GL
(
V V
0
) for all
g G
. So
ρ ρ
0
is a
homomorphism G GL(V V
0
).
To check the character is indeed as stated, let
g G
. Let
v
1
, ··· , v
m
be
a basis of
V
of eigenvectors of
ρ
(
g
), and let
w
1
, ··· , w
n
be a basis of
V
0
of
eigenvectors of ρ
0
(g), say
ρ(g)v
i
= λ
i
v
i
, ρ
0
(g)w
j
= µ
j
w
j
.
Then
(ρ ρ
0
)(g)(v
i
w
j
) = ρ(g)v
i
ρ
0
(g)w
j
= λ
i
v
i
µ
j
w
j
= (λ
i
µ
j
)(v
i
w
j
).
So
χ
ρρ
0
(g) =
X
i,j
λ
i
µ
j
=
X
λ
i
X
µ
j
= χ
ρ
(g)χ
ρ
0
(g).
9.3 Powers of characters
We work over C. Take V = V
0
to be G-spaces, and define
Notation.
V
2
= V V.
We define τ : V
2
V
2
by
τ :
X
λ
ij
v
i
v
j
7→
X
λ
ij
v
j
v
i
.
This is a linear endomorphism of
V
2
such that
τ
2
= 1. So the eigenvalues are
±1.
Definition
(Symmetric and exterior square)
.
We define the symmetric square
and exterior square of V to be, respectively,
S
2
V = {x V
2
: τ(x) = x}
Λ
2
V = {x V
2
: τ(x) = x}.
The exterior square is also known as the anti-symmetric square and wedge power.
Lemma. For any G-space V , S
2
V and Λ
2
V are G-subspaces of V
2
, and
V
2
= S
2
V Λ
2
V.
The space S
2
V has basis
{v
i
v
j
= v
i
v
j
+ v
j
v
i
: 1 i j n},
while Λ
2
V has basis
{v
i
v
j
= v
i
v
j
v
j
v
i
: 1 i < j n}.
Note that we have a strict inequality for
i < j
, since
v
i
v
j
v
j
v
i
= 0 if
i = j. Hence
dim S
2
V =
1
2
n(n + 1), dim Λ
2
V =
1
2
n(n 1).
Sometimes, a factor of
1
2
appears in front of the definition of the basis elements
v
i
v
j
and v
i
v
j
, i.e.
v
i
v
j
=
1
2
(v
i
v
j
+ v
j
v
i
), v
i
v
j
=
1
2
(v
i
v
j
v
j
v
i
).
However, this is not important.
Proof.
This is elementary linear algebra. For the decomposition
V
2
, given
x V
2
, we can write it as
x =
1
2
(x + τ(x))
| {z }
S
2
V
+
1
2
(x τ(x))
| {z }
Λ
2
V
.
How is this useful for character calculations?
Lemma.
Let
ρ
:
G GL
(
V
) be a representation affording the character
χ
.
Then
χ
2
=
χ
S
+
χ
Λ
where
χ
S
=
S
2
χ
is the character of
G
in the subrepresentation
on
S
2
V
, and
χ
Λ
= Λ
2
χ
the character of
G
in the subrepresentation on Λ
2
V
.
Moreover, for g G,
χ
S
(g) =
1
2
(χ
2
(g) + χ(g
2
)), χ
Λ
(g) =
1
2
(χ
2
(g) χ(g
2
)).
The real content of the lemma is, of course, the formula for χ
S
and χ
Λ
.
Proof.
The fact that
χ
2
=
χ
S
+
χ
Λ
is immediate from the decomposition of
G-spaces.
We now compute the characters
χ
S
and
χ
Λ
. For
g G
, we let
v
1
, ··· , v
n
be
a basis of V of eigenvectors of ρ(g), say
ρ(g)v
i
= λ
i
v
i
.
We’ll be lazy and just write
gv
i
instead of
ρ
(
g
)
v
i
. Then, acting on Λ
2
V
, we get
g(v
i
v
j
) = λ
i
λ
j
v
i
v
j
.
Thus
χ
Λ
(g) =
X
1i<jn
λ
i
λ
j
.
Since the answer involves the square of the character, let’s write that down:
(χ(g))
2
=
X
λ
i
2
=
X
λ
2
i
+ 2
X
i<j
λ
i
λ
j
= χ(g
2
) + 2
X
i<j
λ
i
λ
j
= χ(g
2
) + 2χ
Λ
(g).
Then we can solve to obtain
χ
Λ
(g) =
1
2
(χ
2
(g) χ(g
2
)).
Then we can get
χ
S
= χ
2
χ
Λ
=
1
2
(χ
2
(g) + χ(g
2
)).
Example. Consider again G = S
4
. As before, we have the following table:
1 3 8 6 6
1 (1 2)(3 4) (1 2 3) (1 2 3 4) (1 2)
trivial χ
1
1 1 1 1 1
sign χ
2
1 1 1 1 1
π
X
1
G
χ
3
3 1 0 1 1
χ
3
χ
2
χ
4
3 1 0 1 1
χ
5
We now compute χ
5
in a different way, by decomposing χ
2
3
. We have
1 3 8 6 6
1 (1 2)(3 4) (1 2 3) (1 2 3 4) (1 2)
χ
2
3
9 1 0 1 1
χ
3
(g
2
) 3 3 0 1 3
S
2
χ
3
6 2 0 0 2
Λ
2
χ
3
3 1 0 1 1
We notice Λ
2
χ
3
is just
χ
4
. Now
S
2
χ
3
is not irreducible, which we can easily
show by computing the inner product. By taking the inner product with the
other irreducible characters, we find
S
2
χ
3
= 1 +
χ
3
+
χ
5
, where
χ
5
is our new
irreducible character. So we obtain
1 3 8 6 6
1 (1 2)(3 4) (1 2 3) (1 2 3 4) (1 2)
trivial χ
1
1 1 1 1 1
sign χ
2
1 1 1 1 1
π
X
1
G
χ
3
3 1 0 1 1
χ
3
χ
2
χ
4
3 1 0 1 1
S
3
χ
3
1 χ
3
χ
5
2 2 1 0 0
9.4 Characters of G × H
We have looked at characters of direct products a bit before, when we decomposed
an abelian group into a product of cyclic groups. We will now consider this in
the general case.
Proposition.
Let
G
and
H
be two finite groups with irreducible characters
χ
1
, ··· , χ
k
and
ψ
1
, ··· , ψ
r
respectively. Then the irreducible characters of the
direct product G × H are precisely
{χ
i
ψ
j
: 1 i k, 1 j r},
where
(χ
i
ψ
j
)(g, h) = χ
i
(g)ψ
j
(h).
Proof.
Take
ρ
:
G GL
(
V
) affording
χ
, and
ρ
0
:
H GL
(
W
) affording
ψ
.
Then define
ρ ρ
0
: G × H GL(V W )
(g, h) 7→ ρ(g) ρ
0
(h),
where
(ρ(g) ρ
0
(h))(v
i
w
j
) 7→ ρ(g)v
i
ρ
0
(h)w
j
.
This is a representation of
G × H
on
V W
, and
χ
ρρ
0
=
χψ
. The proof is
similar to the case where
ρ, ρ
0
are both representations of
G
, and we will not
repeat it here.
Now we need to show
χ
i
ψ
j
are distinct and irreducible. It suffices to show
they are orthonormal. We have
hχ
i
ψ
j
, χ
r
ψ
s
i
G×H
=
1
|G × H|
X
(g,h)G×H
χ
i
ψ
j
(g, h)χ
r
ψ
s
(g, h)
=
1
|G|
X
gG
χ
i
(g)χ
r
(g)
1
|H|
X
hH
ψ
j
(h)ψ
s
(h)
!
= δ
ir
δ
js
.
So it follows that
{χ
i
ψ
j
}
are distinct and irreducible. We need to show this is
complete. We can consider
X
i,j
χ
i
ψ
j
(1)
2
=
X
χ
2
i
(1)ψ
2
j
(1) =
X
χ
2
i
(1)
X
ψ
2
j
(1)
= |G||H| = |G × H|.
So done.
9.5 Symmetric and exterior powers
We have introduced the objects
S
2
V
and Λ
2
V
. We can do this in a more general
setting, for higher powers. Consider a vector space
V
over
F
, and
dim
F
V
=
d
.
We let {v
1
, ··· , v
d
} be a basis, and let
T
n
V = V
n
= V ··· V
| {z }
n times
.
The basis for this space is just
{v
i
1
··· v
i
n
: i
1
, ··· , i
n
{1, ··· , d}}.
First of all, there is an obvious
S
n
-action on this space, permuting the multipli-
cands of
V
n
. For any
σ S
n
, we can define an action
σ
:
V
n
V
n
given
by
v
1
···v
n
7→ v
σ(1)
··· v
σ(n)
.
Note that this is a left action only if we compose permutations right-to-left. If
we compose left-to-right instead, we have to use
σ
1
instead of
σ
(or use
σ
and
obtain a right action). In fact, this defines a representation of
S
n
on
V
n
by
linear representation.
If
V
itself is a
G
-space, then we also get a
G
-action on
V
n
. Let
ρ
:
G
GL(V ) be a representation. Then we obtain the action of G on V
n
by
ρ
n
: v
1
··· v
n
7→ ρ(g)v
1
··· ρ(g)v
n
.
Staring at this carefully, it is clear that these two actions commute with each other.
This rather simple innocent-looking observation is the basis of an important
theorem by Schur, and has many many applications. However, that would be
for another course.
Getting back on track, since the two actions commute, we can decompose
V
n
as an
S
n
-module, and each isotypical component is a
G
-invariant subspace
of
V
n
. We don’t really need this, but it is a good generalization of what we’ve
had before.
Definition (Symmetric and exterior power). For a G-space V , define
(i) The nth symmetric power of V is
S
n
V = {x V
n
: σ(x) = x for all σ S
n
}.
(ii) The nth exterior power of V is
Λ
n
V = {x V
n
: σ(x) = (sgn σ)x for all σ S
n
}.
Both of these are G-subspaces of V
n
.
Recall that in the
n
= 2 case, we have
V
2
=
S
2
V
Λ
2
V
. However, for
n >
2, we get
S
n
V
Λ
n
V V
n
. In general, lots of others are obtained from
the S
n
-action.
Example.
The bases for
S
n
V
and Λ
n
V
are obtained as follows: take
e
1
, ··· , e
d
a basis of V . Then
(
1
n!
X
σS
n
e
i
σ(1)
··· e
i
σ(n)
: 1 i
1
··· i
n
d
)
is a basis for S
n
V . So the dimension is
dim S
n
V =
d + n 1
n
.
For Λ
n
V , we get a basis of
(
1
n!
X
σS
n
(sgn σ)e
i
σ(1)
··· e
i
σ(n)
: 1 i
1
< ··· < i
n
d
)
.
The dimension is
dim Λ
n
V =
(
0 n > d
d
n
n d
Details are left for the third example sheet.
9.6 Tensor algebra
This part is for some unknown reason included in the schedules. We do not need
it here, but it has many applications elsewhere.
Consider a field F with characteristic 0.
Definition
(Tensor algebra)
.
Let
T
n
V
=
V
n
. Then the tensor algebra of
V
is
T
·
(V ) = T (V ) =
M
n0
T
n
V,
with T
0
V = F by convention.
This is a vector space over
F
with the obvious addition and multiplication by
scalars.
T
(
V
) is also a (non-commutative) (graded) ring with product
x·y
=
xy
.
This is graded in the sense that if
x T
n
V
and
y T
m
V
,
x·y
=
xy T
n+m
V
.
Definition
(Symmetric and exterior algebra)
.
We define the symmetric algebra
of V to be
S(V ) = T (V )/(ideal of T (V ) generated by all u v v u).
The exterior algebra of V is
Λ(V ) = T (V )/(ideal of T (V ) generated by all v v).
Note that
v
and
u
are not elements of
V
, but arbitrary elements of
T
n
V
for
some n.
The symmetric algebra is commutative, while the exterior algebra is graded-
commutative, i.e. if x Λ
n
(V ) and y Λ
m
(V ), then x · y = (1)
n+m+1
y · x.
9.7 Character ring
Recall that
C
(
G
) is a commutative ring, and contains all the characters. Also,
the sum and products of characters are characters. However, they don’t quite
form a subring, since we are not allowed to subtract.
Definition
(Character ring)
.
The character ring of
G
is the
Z
-submodule of
C(G) spanned by the irreducible characters and is denoted R(G).
Recall we can obtain all characters by adding irreducible characters. However,
subtraction is not allowed. So the charactering ring includes something that is
not quite a character.
Definition
(Generalized/virtual characters)
.
The elements of
R
(
G
) are called
generalized or virtual characters. These are class functions of the form
ψ =
X
χ
n
χ
χ,
summing over all irreducibles χ, and n
χ
Z.
We know
R
(
G
) is a ring, and any generalized character is a difference of two
characters (we let
α =
X
n
χ
0
n
χ
χ, β =
X
n
χ
<0
(n
χ
)χ.
Then
ψ
=
α β
, and
α
and
β
are characters). Then the irreducible characters
{χ
i
}
forms a
Z
-basis for
R
(
G
) as a free
Z
-module, since we have shown that
they are independent, and it generates R(G) by definition.
Lemma.
Suppose
α
is a generalized character and
hα, αi
= 1 and
α
(1)
>
0.
Then α is actually a character of an irreducible representation of G.
Proof. We list the irreducible characters as χ
1
, ··· , χ
k
. We then write
α =
X
n
i
χ
i
.
Since the χ
i
’s are orthonormal, we get
hα, αi =
X
n
2
i
= 1.
So exactly one of
n
i
is
±
1, while the others are all zero. So
α
=
±χ
i
for some
i
.
Finally, since
α
(1)
>
0 and also
χ
(1)
>
0, we must have
n
i
= +1. So
α
=
χ
i
.
Henceforth we don’t distinguish between a character and its negative, and
often study generalized characters of inner product 1 rather than irreducible
characters.
10 Induction and restriction
This is the last chapter on methods of calculating characters. Afterwards, we
shall start applying these to do something useful.
Throughout the chapter, we will take
F
=
C
. Suppose
H G
. How do
characters of
G
and
H
relate? It is easy to see that every representation of
G
can be restricted to give a representation of
H
. More surprisingly, given a
representation of H, we can induce a representation of G.
We first deal with the easy case of restriction.
Definition
(Restriction)
.
Let
ρ
:
G GL
(
V
) be a representation affording
χ
.
We can think of
V
as an
H
-space by restricting
ρ
’s attention to
h H
. We get
a representation
Res
G
H
ρ
=
ρ
H
=
ρ
H
, the restriction of
ρ
to
H
. It affords the
character Res
G
H
χ = χ
H
= χ
H
.
It is a fact of life that the restriction of an irreducible character is in general
not irreducible. For example, restricting any non-linear irreducible character to
the trivial subgroup will clearly produce a reducible character.
Lemma.
Let
H G
. If
ψ
is any non-zero irreducible character of
H
, then there
exists an irreducible character
χ
of
G
such that
ψ
is a constituent of
Res
G
H
χ
, i.e.
hRes
G
H
χ, ψi 6= 0.
Proof.
We list the irreducible characters of
G
as
χ
1
, ··· , χ
k
. Recall the regular
character π
reg
. Consider
hRes
G
H
π
reg
, ψi =
|G|
|H|
ψ(1) 6= 0.
On the other hand, we also have
hRes
G
H
π
reg
, ψi
H
=
k
X
1
deg χ
i
hRes
G
H
χ
i
, ψi.
If this sum has to be non-zero, then there must be some
i
such that
hRes
G
H
χ
i
, ψi 6
=
0.
Lemma. Let χ be an irreducible character of G, and let
Res
G
H
χ =
X
i
c
i
χ
i
,
with χ
i
irreducible characters of H, and c
i
non-negative integers. Then
X
c
2
i
|G : H|,
with equality iff χ(g) = 0 for all g G \ H.
This is useful only if the index is small, e.g. 2 or 3.
Proof. We have
hRes
G
H
χ, Res
G
H
χi
H
=
X
c
2
i
.
However, by definition, we also have
hRes
G
H
χ, Res
G
H
χi
H
=
1
|H|
X
hH
|χ(h)|
2
.
On the other hand, since χ is irreducible, we have
1 = hχ, χi
G
=
1
|G|
X
gG
|χ(g)|
2
=
1
|G|
X
hH
|χ(h)|
2
+
X
gG\H
|χ(g)|
2
=
|H|
|G|
X
c
2
i
+
1
|G|
X
gG\H
|χ(g)|
2
|H|
|G|
X
c
2
i
.
So the result follows.
Now we shall go in the other direction given a character of
H
, we want to
induce a character of the larger group
G
. Perhaps rather counter-intuitively, it is
easier for us to first construct the character, and then later find a representation
that affords this character.
We start with a slightly more general notion of an induced class function.
Definition
(Induced class function)
.
Let
ψ C
(
H
). We define the induced
class function Ind
G
H
ψ = ψ
G
= ψ
G
by
Ind
G
H
ψ(g) =
1
|H|
X
xG
˚
ψ(x
1
gx),
where
˚
ψ(y) =
(
ψ(y) y H
0 y 6∈ H
.
The first thing to check is, of course, that Ind
G
H
is indeed a class function.
Lemma. Let ψ C
H
. Then Ind
G
H
ψ C(G), and Ind
G
H
ψ(1) = |G : H|ψ(1).
Proof.
The fact that
Ind
G
H
ψ
is a class function follows from direct inspection of
the formula. Then we have
Ind
G
H
ψ(1) =
1
|H|
X
xG
˚
ψ(1) =
|G|
|H|
ψ(1) = |G : H|ψ(1).
As it stands, the formula is not terribly useful. So we need to find an
alternative formula for it.
If we have a subset, a sensible thing to do is to look at the cosets of
H
.
We let
n
=
|G
:
H|
, and let 1 =
t
1
, ··· , t
n
be a left transversal of
H
in
G
, i.e.
t
1
H = H, ··· , t
n
H are precisely the n left-cosets of H in G.
Lemma. Given a (left) transversal t
1
, ··· , t
n
of H, we have
Ind
G
H
ψ(g) =
n
X
i=1
˚
ψ(t
1
i
gt
i
).
Proof.
We can express every
x G
as
x
=
t
i
h
for some
h H
and
i
. We then
have
˚
ψ((t
i
h)
1
g(t
i
h)) =
˚
ψ(h
1
(t
1
i
gt
i
)h) =
˚
ψ(t
1
i
gt
i
),
since
ψ
is a class function of
H
, and
h
1
(
t
1
i
gt
i
)
h H
if and only if
t
1
i
gt
i
H
,
as h H. So the result follows.
This is not too useful as well. But we are building up to something useful.
Theorem (Frobenius reciprocity). Let ψ C(H) and ϕ C(G). Then
hRes
G
H
ϕ, ψi
H
= hϕ, Ind
G
H
ψi
G
.
Alternatively, we can write this as
hϕ
H
, ψi = hϕ, ψ
G
i.
If you are a category theorist, and view restriction and induction as functors,
then this says restriction is a right adjoint to induction, and thsi is just a special
case of the tensor-Hom adjunction. If you neither understand nor care about
this, then this is a good sign [sic].
Proof. We have
hϕ, ψ
G
i =
1
|G|
X
gG
ϕ(g)ψ
G
(g)
=
1
|G||H|
X
x,gG
ϕ(g)
˚
ψ(x
1
gx)
We now write
y
=
x
1
gx
. Then summing over
g
is the same as summing over
y
.
Since ϕ is a G-class function, this becomes
=
1
|G||H|
X
x,yG
ϕ(y)
˚
ψ(y)
Now note that the sum is independent of x. So this becomes
=
1
|H|
X
yG
ϕ(y)
˚
ψ(y)
Now this only has contributions when y H, by definition of
˚
ψ. So
=
1
|H|
X
yH
ϕ(y)ψ(y)
= hϕ
H
, ψi
H
.
Corollary. Let ψ be a character of H. Then Ind
G
H
ψ is a character of G.
Proof. Let χ be an irreducible character of G. Then
hInd
G
H
ψ, χi = hψ, Res
G
H
χi.
Since
ψ
and
Res
G
H
χ
are characters, the thing on the right is in
Z
0
. Hence
Ind
G
H
is a linear combination of irreducible characters with non-negative coefficients,
and is hence a character.
Recall we denote the conjugacy class of
g G
as
C
G
(
g
), while the centralizer
is
C
G
(
g
). If we take a conjugacy class
C
G
(
g
) in
G
and restrict it to
H
, then
the result
C
G
(
g
)
H
need not be a conjugacy class, since the element needed
to conjugate
x, y C
G
(
g
)
H
need not be present in
H
, as is familiar from the
case of
A
n
S
n
. However, we know that
C
G
(
g
)
H
is a union of conjugacy
classes in H, since elements conjugate in H are also conjugate in G.
Proposition. Let ψ be a character of H G, and let g G. Let
C
G
(g) H =
m
[
i=1
C
H
(x
i
),
where the
x
i
are the representatives of the
H
conjugacy classes of elements of
H
conjugate to g. If m = 0, then Ind
G
H
ψ(g) = 0. Otherwise,
Ind
G
H
ψ(g) = |C
G
(g)|
m
X
i=1
ψ(x
i
)
|C
H
(x
i
)|
.
This is all just group theory. Note that some people will think this proof
is excessive everything shown is “obvious”. In some sense it is. Some steps
seem very obvious to certain people, but we are spelling out all the details so
that everyone is happy.
Proof.
If
m
= 0, then
{x G
:
x
1
gx H}
=
. So
˚
ψ
(
x
1
gx
) = 0 for all
x
. So
Ind
G
H
ψ(g) = 0 by definition.
Now assume m > 0. We let
X
i
= {x G : x
1
gx H and is conjugate in H to x
i
}.
By definition of
x
i
, we know the
X
i
’s are pairwise disjoint, and their union is
{x G : x
1
gx H}. Hence by definition,
Ind
G
H
ψ(g) =
1
|H|
X
xG
˚
ψ(x
1
gx)
=
1
|H|
m
X
i=1
X
xX
i
ψ(x
1
gx)
=
1
|H|
m
X
i=1
X
xX
i
ψ(x
i
)
=
m
X
i=1
|X
i
|
|H|
ψ(x
i
).
So we now have to show that in fact
|X
i
|
|H|
=
|C
G
(g)|
|C
H
(x
i
)|
.
We fix some 1
i m
. Choose some
g
i
G
such that
g
1
i
gg
i
=
x
i
. This exists
by definition of x
i
. So for every c C
G
(g) and h H, we have
(cg
i
h)
1
g(cg
i
h) = h
1
g
1
i
c
1
gcg
i
h
We now use the fact that c commutes with g, since c C
G
(g), to get
= h
1
g
1
i
c
1
cgg
i
h
= h
1
g
1
i
gg
i
h
= h
1
x
i
h.
Hence by definition of X
i
, we know cg
i
h X
i
. Hence
C
G
(g)g
i
H X
i
.
Conversely, if
x X
i
, then
x
1
gx
=
h
1
x
i
h
=
h
1
(
g
1
i
gg
i
)
h
for some
h
. Thus
xh
1
g
1
i
C
G
(g), and so x C
G
(g)g
i
h. So
x C
G
(g)g
i
H.
So we conclude
X
i
= C
G
(g)g
i
H.
Thus, using some group theory magic, which we shall not prove, we get
|X
i
| = |C
G
(g)g
i
H| =
|C
G
(g)||H|
|H g
1
i
C
G
(g)g
i
|
Finally, we note
g
1
i
C
G
(g)g
i
= C
G
(g
1
i
gg
i
) = C
G
(x
i
).
Thus
|X
i
| =
|H||C
G
(g)|
|H C
G
(x
i
)|
=
|H||C
G
(g)|
|C
H
(x
i
)|
.
Dividing, we get
|X
i
|
|H|
=
|C
G
(g)|
|C
H
(x
i
)|
.
So done.
To clarify matters, if H, K G, then a double coset of H, K in G is a set
HxK = {hxk : h H, k K}
for some x G. Facts about them include
(i) Two double cosets are either disjoint or equal.
(ii) The sizes are given by
|H||K|
|H xKx
1
|
=
|H||K|
|x
1
Hx K|
.
Lemma.
Let
ψ
= 1
H
, the trivial character of
H
. Then
Ind
G
H
1
H
=
π
X
, the
permutation character of
G
on the set
X
, where
X
=
G/H
is the set of left
cosets of H.
Proof.
We let
n
=
|G
:
H|
, and
t
1
, ··· , t
n
be representatives of the cosets. By
definition, we know
Ind
G
H
1
H
(g) =
n
X
i=1
˚
1
H
(t
1
i
gt
i
)
= |{i : t
1
i
gt
i
H}|
= |{i : g t
i
Ht
1
i
}|
But t
i
Ht
1
i
is the stabilizer in G of the coset t
i
H X. So this is equal to
= |fix
X
(g)|
= π
X
(g).
By Frobenius reciprocity, we know
hπ
X
, 1
G
i
G
= hInd
G
H
1
H
, 1
G
i
G
= h1
H
, 1
H
i
H
= 1.
So the multiplicity of 1
G
in π
X
is indeed 1, as we have shown before.
Example.
Let
H
=
C
4
=
h
(1 2 3 4)
i G
=
S
4
. The index is
|S
4
:
C
4
|
= 6. We
consider the character of the induced representation
Ind
G
H
(
α
), where
α
is the
faithful 1-dimensional representation of C
4
given by
α((1 2 3 4)) = i.
Then the character of α is
1 (1 2 3 4) (1 3)(2 4) (1 4 3 2)
χ
α
1 i 1 i
What is the induced character in S
4
? We know that
Ind
G
H
(α)(1) = |G : H|α(1) = |G : H| = 6.
So the degree of
Ind
G
H
(
α
) is 6. Also, the elements (1 2) and (3 4) are not
conjugate to anything in C
4
. So the character vanishes.
For (1 2)(3 4), only one of the three conjugates in
S
4
lie in
H
(namely
(1 3)(2 4)). So, using the sizes of the centralizers, we obtain
Ind
G
H
((1 3)(2 4)) = 8
1
4
= 2.
For (1 2 3 4), it is conjugate to 6 elements of
S
4
, two of which are in
C
4
, namely
(1 2 3 4) and (1 4 3 2). So
Ind
G
H
(1 2 3 4) = 4
i
4
+
i
4
= 0.
So we get the following character table:
1 6 8 3 6
1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)
Ind
G
H
(α) 6 0 0 2 0
Induced representations
We have constructed a class function, and showed it is indeed the character of
some representation. However, we currently have no idea what this corresponding
representation is. Here we will try to construct such a representation explicitly.
This is not too enlightening, but its nice to know it can be done. We will also
need this explicit construction when proving Mackey’s theorem later.
Let
H G
have index
n
, and 1 =
t
1
, ··· , t
n
be a left transversal. Let
W
be
a H-space. Define the vector space
V = Ind
G
H
W = W (t
2
W ) ··· (t
n
W ),
where
t
i
W = {t
i
w : w W }.
So dim V = n dim W .
To define the
G
-action on
V
, let
g G
. Then for every
i
, there is a unique
j
such that t
1
j
gt
i
H, i.e. gt
i
H = t
j
H. We define
g(t
i
w) = t
j
((t
1
j
gt
i
)w).
We tend to drop the tensor signs, and just write
g(t
i
w) = t
j
(t
1
j
gt
i
w).
We claim this is a G-action. We have
g
1
(g
2
t
i
w) = g
1
(t
j
(t
1
j
g
2
t
i
)w),
where j is the unique index such that g
2
t
i
H = t
j
H. This is then equal to
t
`
((t
1
`
gt
j
)(t
1
j
g
2
t
i
)w) = t
`
((t
1
`
(g
1
g
2
)t
i
)w) = (g
1
g
2
)(t
i
w),
where
`
is the unique index such that
g
1
t
j
H
=
t
`
H
(and hence
g
1
g
2
t
i
H
=
g
1
t
j
H = t
`
H).
Definition
(Induced representation)
.
Let
H G
have index
n
, and 1 =
t
1
, ··· , t
n
be a left transversal. Let
W
be a
H
-space. Define the induced
representation to be the vector space
Ind
G
H
W = W t
2
W ··· t
n
W,
with the G-action
g : t
i
w 7→ t
j
(t
1
j
gt
i
)W,
where t
j
is the unique element (among t
1
, ··· , t
n
) such that t
1
j
gt
i
H.
This has the “right” character. Suppose
W
has character
ψ
. Since
g
:
t
i
w 7→
t
j
(
t
1
j
gt
i
), the contribution to the character is 0 unless
j
=
i
, i.e. if
t
1
i
gt
i
H
.
Then it contributes
ψ(t
1
i
gt
i
).
Thus we get
Ind
G
H
ψ(g) =
n
X
i=1
˚
ψ(t
1
i
gt
i
).
Note that this construction is rather ugly. It could be made much nicer if we
knew a bit more algebra we can write the induced module simply as
Ind
G
H
W = FG
FH
W,
where we view
W
as a left-
FH
module, and
FG
as a (
FG, FH
)-bimodule. Alter-
natively, this is the extension of scalars from FH to FG.
11 Frobenius groups
We will now use character theory to prove some major results in finite group
theory. We first do Frobenius’ theorem here. Later, will prove Burnside’s
p
a
q
b
theorem.
This is a theorem with lots and lots of proofs, but all of these proofs involves
some sort of representation theory it seems like representation is an unavoidable
ingredient of it.
Theorem
(Frobenius’ theorem (1891))
.
Let
G
be a transitive permutation group
on a finite set
X
, with
|X|
=
n
. Assume that each non-identity element of
G
fixes at most one element of
X
. Then the set of fixed point-free elements
(“derangements”)
K = {1} {g G : gα 6= α for all α X}
is a normal subgroup of G with order n.
On the face of it, it is not clear
K
is even a subgroup at all. It turns out
normality isn’t really hard to prove the hard part is indeed showing it is a
subgroup.
Note that we did not explicitly say
G
is finite. But these conditions imply
G S
n
, which forces G to be finite.
Proof.
The idea of the proof is to construct a character Θ whose kernel is
K
.
First note that by definition of K, we have
G = K
[
αX
G
α
,
where
G
α
is, as usual, the stabilizer of
α
. Also, we know that
G
α
G
β
=
{
1
}
if
α 6= β by assumption, and by definition of K, we have K G
α
= {1} as well.
Next note that all the
G
α
are conjugate. Indeed, we know
G
is transitive,
and
gG
α
g
1
=
G
gα
. We set
H
=
G
α
for some arbitrary choice of
α
. Then the
above tells us that
|G| = |K| |X|(|H| 1).
On the other hand, by the orbit-stabilizer theorem, we know
|G|
=
|X||H|
. So it
follows that we have
|K| = |X| = n.
We first compute what induced characters look like.
Claim. Let ψ be a character of H. Then
Ind
G
H
ψ(g) =
(1) g = 1
ψ(g) g H \ {1}
0 g K \ {1}
.
Since every element in
G
is either in
K
or conjugate to an element in
H
, this
uniquely specifies what the induced character is.
This is a matter of computation. Since [
G
:
H
] =
n
, the case
g
= 1
immediately follows. Using the definition of the induced character, since any
non-identity in
K
is not conjugate to any element in
H
, we know the induced
character vanishes on K \ {1}.
Finally, suppose
g H \{
1
}
. Note that if
x G
, then
xgx
1
G
. So this
lies in H if and only if x H. So we can write the induced character as
Ind
G
H
ψ(g) =
1
|H|
X
gG
˚
ψ(xgx
1
) =
1
|H|
X
hH
ψ(hgh
1
) = ψ(g).
Claim. Let ψ be an irreducible character of H, and define
θ = ψ
G
ψ(1)(1
H
)
G
+ ψ(1)1
G
.
Then θ is a character, and
θ(g) =
(
ψ(h) h H
ψ(1) k K
.
Note that we chose the coefficients exactly so that the final property of
θ
holds. This is a matter of computation:
1 h H \ {1} K \ {1}
ψ
G
(1) ψ(h) 0
ψ(1)(1
H
)
G
(1) ψ(1) 0
ψ(1)1
G
ψ(1) ψ(1) ψ(1)
θ
i
ψ(1) ψ(h) ψ(1)
The less obvious part is that
θ
is a character. From the way we wrote it, we
already know it is a virtual character. We then compute the inner product
hθ, θi
G
=
1
|G|
X
gG
|θ(g)|
2
=
1
|G|
X
gK
|θ(g)|
2
+
X
gG\K
|θ(g)|
2
=
1
|G|
n|ψ(1)|
2
+ n
X
h6=1H
|ψ(h)|
2
=
1
|G|
n
X
hH
|ψ(h)|
2
!
=
1
|G|
(n|H|hψ, ψi
H
)
= 1.
So either θ or θ is a character. But θ(1) = ψ(1) > 0. So θ is a character.
Finally, we have
Claim.
Let
ψ
1
, ··· , ψ
t
be the irreducible representations of
H
, and
θ
i
be the
corresponding representations of G constructed above. Set
Θ =
t
X
i=1
θ
i
(1)θ
i
.
Then we have
θ(g) =
(
|H| g K
0 g 6∈ K
.
From this, it follows that the kernel of the representation affording
θ
is
K
, and
in particular K is a normal subgroup of G.
This is again a computation using column orthogonality. For 1
6
=
h H
, we
have
Θ(h) =
t
X
i=1
ψ
i
(1)ψ
i
(h) = 0,
and for any y K, we have
Θ(y) =
t
X
i=1
ψ
i
(1)
2
= |H|.
Definition
(Frobenius group and Frobenius complement)
.
A Frobenius group
is a group
G
having a subgroup
H
such that
H gHg
1
= 1 for all
g 6∈ H
. We
say H is a Frobenius complement of G.
How does this relate to the previous theorem?
Proposition.
The left action of any finite Frobenius group on the cosets of the
Frobenius complement satisfies the hypothesis of Frobenius’ theorem.
Definition
(Frobenius kernel)
.
The Frobenius kernel of a Frobenius group
G
is
the K obtained from Frobenius’ theorem.
Proof.
Let
G
be a Frobenius group, having a complement
H
. Then the action of
G
on the cosets
G/H
is transitive. Furthermore, if 1
6
=
g G
fixes
xH
and
yH
,
then we have
g xHx
1
yHy
1
. This implies
H
(
y
1
x
)
H
(
y
1
x
)
1
6
= 1.
Hence xH = yH.
Note that J. Thompson (in his 1959 thesis) proved any finite group having
a fixed-point free automorphism of prime order is nilpotent. This implies
K
is
nilpotent, which means K is a direct product of its Sylow subgroups.
12 Mackey theory
We work over
C
. We wish to describe the restriction to a subgroup
K G
of an
induced representation
Ind
G
H
W
, i.e. the character
Res
G
K
Ind
G
H
W
. In general,
K
and
H
can be unrelated, but in many applications, we have
K
=
H
, in which
case we can characterize when Ind
G
H
W is irreducible.
It is quite helpful to first look at a special case, where
W
= 1
H
is the trivial
representation. Thus
Ind
G
H
1
H
is the permutation representation of
G
on
G/H
(coset action on the left cosets of H in G).
Recall that by the orbit-stabilizer theorem, if
G
is transitive on the set
X
,
and
H
=
G
α
for some
α X
, then the action of
G
on
X
is isomorphic the action
on G/H, namely the correspondence
gα gH
is a well-defined bijection, and commutes with with the
g
-action (i.e.
x
(
gα
) =
(xg)α x(gH) = (xg)H).
We now consider the action of
G
on
G/H
and let
K G
. Then
K
also acts
on
G/H
, and
G/H
splits into
K
-orbits. The
K
-orbit of
gH
contains precisely
kgH for k K. So it is the double coset
KgH = {kgh : k K, h H}.
The set of double cosets
K\G/H
partition
G
, and the number of double cosets
is
|K\G/H| = hπ
G/K
, π
G/H
i.
We don’t need this, but this is true.
If it happens that
H
=
K
, and
H
is normal, then we just have
K\G/H
=
G/H.
What about stabilizers? Clearly,
G
gH
=
gHg
1
. Thus, restricting to the
action of K, we have K
gH
= gHg
1
K. We call H
g
= K
gH
.
So by our correspondence above, the action of
K
on the orbit containing
gH
is isomorphic to the action of
K
on
K/
(
gHg
1
K
) =
K/H
g
. From this,
and using the fact that
Ind
G
H
1
H
=
C
(
G/H
), we get the special case of Mackey’s
theorem:
Proposition.
Let
G
be a finite group and
H, K G
. Let
g
1
, ··· , g
k
be the
representatives of the double cosets K\G/H. Then
Res
G
K
Ind
G
H
1
H
=
k
M
i=1
Ind
K
g
i
Hg
1
i
K
1.
The general form of Mackey’s theorem is the following:
Theorem
(Mackey’s restriction formula)
.
In general, for
K, H G
, we let
S = {1, g
1
, ··· , g
r
} be a set of double coset representatives, so that
G =
[
Kg
i
H.
We write
H
g
=
gHg
1
K G
. We let (
ρ, W
) be a representation of
H
. For
each
g G
, we define (
ρ
g
, W
g
) to be a representation of
H
g
, with the same
underlying vector space W , but now the action of H
g
is
ρ
g
(x) = ρ(g
1
xg),
where h = g
1
xg H by construction.
This is clearly well-defined. Since
H
g
K
, we obtain an induced representa-
tion Ind
K
H
g
W
g
.
Let G be finite, H, K G, and W be a H-space. Then
Res
G
K
Ind
G
H
W =
M
g∈S
Ind
K
H
g
W
g
.
We will defer the proof of this for a minute or two. We will first derive some
corollaries of this, starting with the character version of the theorem.
Corollary. Let ψ be a character of a representation of H. Then
Res
G
K
Ind
G
H
ψ =
X
g∈S
Ind
K
H
g
ψ
g
,
where ψ
g
is the class function (and a character) on H
g
given by
ψ
g
(x) = ψ(g
1
xg).
These characters
ψ
g
are sometimes known as the conjugate characters. This
obviously follows from Mackey’s restriction formula.
Corollary
(Mackey’s irreducibility criterion)
.
Let
H G
and
W
be a
H
-space.
Then V = Ind
G
H
W is irreducible if and only if
(i) W is irreducible; and
(ii)
For each
g S\H
, the two
H
g
spaces
W
g
and
Res
H
H
g
W
have no irreducible
constituents in common, where H
g
= gHg
1
H.
Note that the set
S
of representatives was chosen arbitrarily. So we may
require (ii) to hold for all
g G \ H
, instead of
g S \ H
. While there are
more things to check in
G
, it is sometimes convenient not have to pick out an
S
explicitly.
Note that the only
g S H
is the identity, since for any
h H
,
KhH
=
KH = K1H.
Proof.
We use characters, and let
W
afford the character
ψ
. We take
K
=
H
in
Mackey’s restriction formula. Then we have H
g
= gHg
1
H.
Using Frobenius reciprocity, we can compute the inner product as
hInd
G
H
ψ, Ind
G
H
ψi
G
= hψ, Res
G
H
Ind
G
H
ψi
H
=
X
g∈S
hψ, Ind
H
H
g
ψ
g
i
H
=
X
g∈S
hRes
H
H
g
ψ, ψ
g
i
H
g
= hψ, ψi +
X
g∈S\H
hRes
H
H
g
ψ, ψ
g
i
H
g
We can write this because if g = 1, then H
g
= H, and ψ
g
= ψ.
This is a sum of non-negative integers, since the inner products of characters
always are. So
Ind
G
H
ψ
is irreducible if and only if
hψ, ψi
= 1, and all the other
terms in the sum are 0. In other words,
W
is an irreducible representation of
H
,
and for all g 6∈ H, W and W
g
are disjoint representations of H
g
.
Corollary.
Let
H C G
, and suppose
ψ
is an irreducible character of
H
. Then
Ind
G
H
ψ
is irreducible if and only if
ψ
is distinct from all its conjugates
ψ
g
for
g G \ H (where ψ
g
(h) = ψ(g
1
hg) as before).
Proof.
We take
K
=
H C G
. So the double cosets are just left cosets. Also,
H
g
= H for all g. Moreover, W
g
is irreducible since W is irreducible.
So, by Mackey’s irreducible criterion,
Ind
G
H
W
is irreducible precisely if
W 6
=
W
g
for all g G \ H. This is equivalent to ψ 6= ψ
g
.
Note that again we could check the conditions on a set of representatives of
(double) cosets. In fact, the isomorphism class of
W
g
(for
g G
) depends only
on the coset gH.
We now prove Mackey’s theorem.
Theorem
(Mackey’s restriction formula)
.
In general, for
K, H G
, we let
S = {1, g
1
, ··· , g
r
} be a set of double coset representatives, so that
G =
[
Kg
i
H.
We write
H
g
=
gHg
1
K G
. We let (
ρ, W
) be a representation of
H
. For
each
g G
, we define (
ρ
g
, W
g
) to be a representation of
H
g
, with the same
underlying vector space W , but now the action of H
g
is
ρ
g
(x) = ρ(g
1
xg),
where h = g
1
xg H by construction.
This is clearly well-defined. Since
H
g
K
, we obtain an induced representa-
tion Ind
K
H
g
W
g
.
Let G be finite, H, K G, and W be a H-space. Then
Res
G
K
Ind
G
H
W =
M
g∈S
Ind
K
H
g
W
g
.
This is possibly the hardest and most sophisticated proof in the course.
Proof.
Write
V
=
Ind
G
H
W
. Pick
g G
, so that
KgH K\G/H
. Given a left
transversal T of H in G, we can obtain V explicitly as a direct sum
V =
M
t∈T
t W.
The idea is to “coarsen” this direct sum decomposition using double coset
representatives, by collecting together the t W ’s with t KgH. We define
V (g) =
M
tKgH∩T
t W.
Now each
V
(
g
) is a
K
-space given
k K
and
t w t W
, since
t KgH
,
we have
kt KgH
. So there is some
t
0
T
such that
ktH
=
t
0
H
. Then
t
0
ktH KgH. So we can define
k · (t w) = t
0
(ρ(t
0−1
kt)w),
where t
0
kt H.
Viewing V as a K-space (forgetting its whole G-structure), we have
Res
G
K
V =
M
g∈S
V (g).
The left hand side is what we want, but the right hand side looks absolutely
nothing like Ind
K
H
g
W
g
. So we need to show
V (g) =
M
tKgH∩T
t W
=
Ind
K
H
g
W
g
,
as K representations, for each g S.
Now for
g
fixed, each
t KgH
can be represented by some
kgh
, and by
restricting to elements in the traversal
T
of
H
, we are really summing over cosets
kgH
. Now cosets
kgH
are in bijection with cosets
k
(
gHg
1
) in the obvious way.
So we are actually summing over elements in
K/
(
gHg
1
K
) =
K/H
g
. So we
write
V (g) =
M
kK/H
g
(kg) W.
We claim that there is a isomorphism that sends
k W
g
=
(
kg
)
W
. We define
k W
g
(
kg
)
W
by
k w 7→ kg w
. This is an isomorphism of vector spaces
almost by definition, so we only check that it is compatible with the action. The
action of x K on the left is given by
ρ
g
(x)(k w) = k
0
(ρ
g
(k
0−1
xk)w) = k
0
(ρ(g
1
k
0−1
xkg)w),
where
k
0
K
is such that
k
0−1
xk H
g
, i.e.
g
1
k
0−1
xkg H
. On the other
hand,
ρ(x)(kg w) = k
00
(ρ(k
00
x
1
(kg))w),
where
k
00
K
is such that
k
00−1
xkg H
. Since there is a unique choice of
k
00
(after picking a particular transversal), and
k
0
g
works, we know this is equal to
k
0
g (ρ(g
1
k
0−1
xkg)w).
So the actions are the same. So we have an isomorphism.
Then
V (g) =
M
kK/H
g
k W
g
= Ind
K
H
g
W
g
,
as required.
Example.
Let
g
=
S
n
and
H
=
A
n
. Consider a
σ S
n
. Then its conjugacy
class in S
n
is determined by its cycle type.
If the class of
σ
splits into two classes in
A
n
, and
χ
is an irreducible character
of
A
n
that takes different values on the two classes, then by the irreducibility
criterion, Ind
S
n
A
n
χ is irreducible.
13 Integrality in the group algebra
The next big result we are going to have is that the degree of an irreducible
character always divides the group order. This is not at all an obvious fact. To
prove this, we make the cunning observation that character degrees and friends
are not just regular numbers, but algebraic integers.
Definition
(Algebraic integers)
.
A complex number
a C
is an algebraic integer
if
a
is a root of a monic polynomial with integer coefficients. Equivalently,
a
is
such that the subring of C given by
Z[a] = {f(a) : f Z[X]}
is finitely generated. Equivalently,
a
is the eigenvalue of a matrix, all of whose
entries lie in Z.
We will quote some results about algebraic integers which we will not prove.
Proposition.
(i) The algebraic integers form a subring of C.
(ii)
If
a C
is both an algebraic integer and rational, then
a
is in fact an
integer.
(iii)
Any subring of
C
which is a finitely generated
Z
-module consists of algebraic
integers.
The strategy is to show that given a group
G
and an irreducible character
χ
,
the value
|G|
χ(1)
is an algebraic integer. Since it is also a rational number, it must
be an integer. So |G| is a multiple of χ(1).
We make the first easy step.
Proposition.
If
χ
is a character of
G
and
g G
, then
χ
(
g
) is an algebraic
integer.
Proof.
We know
χ
(
g
) is the sum of roots
n
th roots of unity (where
n
is the order
of
g
). Each root of unity is an algebraic integer, since it is by definition a root of
x
n
1. Since algebraic integers are closed under addition, the result follows.
We now have a look at the center of the group algebra
CG
. Recall that the
group algebra
CG
of a finite group
G
is a complex vector space generated by
elements of G. More precisely,
CG =
X
gG
α
g
g : α
g
C
.
Borrowing the multiplication of
G
, this forms a ring, hence an algebra. We now
list the conjugacy classes of G as
{1} = C
1
, ··· , C
k
.
Definition
(Class sum)
.
The class sum of a conjugacy class
C
j
of a group
G
is
C
j
=
X
g∈C
j
g CG.
The claim is now
C
j
lives in the center of
CG
(note that the center of the
group algebra is different from the group algebra of the center). Moreover, they
form a basis:
Proposition.
The class sums
C
1
, ··· , C
k
form a basis of
Z
(
CG
). There exists
non-negative integers a
ij`
(with 1 i, j, ` k) with
C
i
C
j
=
k
X
`=1
a
ij`
C
`
.
The content of this proposition is not that
C
i
C
j
can be written as a sum of
C
`
’s. This is true because the
C
`
’s form a basis. The content is that these are
actually integers, not arbitrary complex numbers.
Definition
(Class algebra/structure constants)
.
The constants
a
ij`
as defined
above are the class algebra constants or structure constants.
Proof.
It is clear from definition that
gC
j
g
1
=
C
j
. So we have
C
j
Z
(
CG
).
Also, since the
C
j
’s are produced from disjoint conjugacy classes, they are linearly
independent.
Now suppose z Z(CG). So we can write
z =
X
gG
α
g
g.
By definition, this commutes with all elements of
CG
. So for all
h G
, we must
have
α
h
1
gh
= α
g
.
So the function
g 7→ α
g
is constant on conjugacy classes of
G
. So we can write
α
j
= α
g
for g C
j
. Then
g =
k
X
j=1
α
j
C
j
.
Finally, the center Z(CG) is an algebra. So
C
i
C
j
=
k
X
`=1
a
ij`
C
`
for some complex numbers
a
ij`
, since the
C
j
span. The claim is that
a
ij`
Z
0
for all
i, j`
. To see this, we fix
g
`
C
`
. Then by definition of multiplication, we
know
a
ij`
= |{(x, y) C
i
× C
j
: xy = g
`
}|,
which is clearly a non-negative integer.
Let
ρ
:
G GL
(
V
) be an irreducible representation of
G
over
C
affording
the character
χ
. Extending linearly, we get a map
ρ
:
CG End V
, an algebra
homomorphism. In general, we have the following definition:
Definition
(Representation of algebra)
.
Let
A
be an algebra. A representation
of A is a homomorphism of algebras ρ : A End V .
Let
z Z
(
CG
). Then
ρ
(
z
) commutes with all
ρ
(
g
) for
g G
. Hence, by
Schur’s lemma, we can write
ρ
(
z
) =
λ
z
I
for some
λ
z
C
. We then obtain the
algebra homomorphism
ω
ρ
= ω
χ
= ω : Z(CG) C
z 7→ λ
z
.
By definition, we have ρ(C
i
) = ω(C
i
)I. Taking traces of both sides, we know
X
g∈C
i
χ(g) = χ(1)ω(C
i
).
We also know the character is a class function. So we in fact get
χ(1)ω(C
i
) = |C
i
|χ(g
i
),
where g
i
C
i
is a representative of C
i
. So we get
ω(C
i
) =
χ(g
i
)
χ(1)
|C
i
|.
Why should we care about this? The thing we are looking at is in fact an
algebraic integer.
Lemma. The values of
ω
χ
(C
i
) =
χ(g)
χ(1)
|C
i
|
are algebraic integers.
Note that all the time, we are requiring χ to be an irreducible character.
Proof.
Using the definition of
a
ij`
Z
0
, and the fact that
ω
χ
is an algebra
homomorphism, we get
ω
χ
(C
i
)ω
χ
(C
j
) =
k
X
`=1
a
ij`
ω
χ
(C
`
).
Thus the span of
{ω
(
C
j
) : 1
j k}
is a subring of
C
and is finitely generated
as a
Z
-module (by definition). So we know this consists of algebraic integers.
That’s magic.
In fact, we can compute the structure constants
a
ij`
from the character table,
namely for all i, j, `, we have
a
ij`
=
|G|
|C
G
(g
i
)||C
G
(g
j
)|
k
X
s=1
χ
s
(g
i
)χ
s
(g
j
)χ
s
(g
1
`
)
χ
s
(1)
,
where we sum over the irreducible characters of
G
. We will neither prove it nor
use it. But if you want to try, the key idea is to use column orthogonality.
Finally, we get to the main result:
Theorem. The degree of any irreducible character of G divides |G|, i.e.
χ
j
(1) | |G|
for each irreducible χ
j
.
It is highly non-obvious why this should be true.
Proof. Let χ be an irreducible character. By orthogonality, we have
|G|
χ(1)
=
1
χ(1)
X
gG
χ(g)χ(g
1
)
=
1
χ(1)
k
X
i=1
|C
i
|χ(g
i
)χ(g
1
i
)
=
k
X
i=1
|C
i
|χ(g
i
)
χ(1)
χ(g
i
)
1
.
Now we notice
|C
i
|χ(g
i
)
χ(1)
is an algebraic integer, by the previous lemma. Also,
χ
(
g
1
i
) is an algebraic
integer. So the whole mess is an algebraic integer since algebraic integers are
closed under addition and multiplication.
But we also know
|G|
χ(1)
is rational. So it must be an integer!
Example.
Let
G
be a
p
-group. Then
χ
(1) is a
p
-power for each irreducible
χ
.
If in particular
|G|
=
p
2
, then
χ
(1) = 1
, p
or
p
2
. But the sum of squares of the
degrees is
|G|
=
p
2
. So it cannot be that
χ
(1) =
p
or
p
2
. So
χ
(1) = 1 for all
irreducible characters. So G is abelian.
Example.
No simple group has an irreducible character of degree 2. Proof is
left as an exercise for the reader.
14 Burnside’s theorem
Finally, we get to Burnside’s theorem.
Theorem
(Burside’s
p
a
q
b
theorem)
.
Let
p, q
be primes, and let
|G|
=
p
a
q
b
,
where a, b Z
0
, with a + b 2. Then G is not simple.
Note that if a + b = 1 or 0, then the group is trivially simple.
In fact even more is true, namely that
G
is soluble, but this follows easily
from above by induction.
This result is the best possible in the sense that
|A
5
|
= 60 = 2
2
·
3
·
5, and
A
5
is simple. So we cannot allow for three different primes factors (in fact, there
are exactly 8 simple groups with three prime factors). Also, if
a
= 0 or
b
= 0,
then
G
is a
p
-group, which have non-trivial center. So these cases trivially work.
Later, in 1963, Feit and Thompson proved that every group of odd order is
soluble. The proof was 255 pages long. We will not prove this.
In 1972, H. Bender found the first proof of Burnside’s theorem without the
use of representation theory, but the proof is much more complicated.
This theorem follows from two lemmas, and one of them involves some Galois
theory, and is hence non-examinable.
Lemma. Suppose
α =
1
m
m
X
j=1
λ
j
,
is an algebraic integer, where
λ
n
j
= 1 for all
j
and some
n
. Then either
α
= 0 or
|α| = 1.
Proof (non-examinable).
Observe
α F
=
Q
(
ε
), where
ε
=
e
2πi/n
(since
λ
j
F
for all j). We let G = Gal(F/Q). Then
{β F : σ(β) = β for all σ G} = Q.
We define the “norm”
N(α) =
Y
σ∈G
σ(α).
Then N (α) is fixed by every element σ G. So N(α) is rational.
Now
N
(
α
) is an algebraic integer, since Galois conjugates
σ
(
α
) of algebraic
integers are algebraic integers. So in fact
N
(
α
) is an integer. But for
α G
, we
know
|σ(α)| =
1
m
X
σ(λ
j
)
1.
So if α 6= 0, then N (α) = ±1. So |α| = 1.
Lemma.
Suppose
χ
is an irreducible character of
G
, and
C
is a conjugacy class
in G such that χ(1) and |C| are coprime. Then for g C, we have
|χ(g)| = χ(1) or 0.
Proof. Of course, we want to consider the quantity
α =
χ(g)
χ(1)
.
Since
χ
(
g
) is the sum of
deg χ
=
χ
(1) many roots of unity, it suffices to show
that α is an algebraic integer.
By ezout’s theorem, there exists a, b Z such that
(1) + b|C| = 1.
So we can write
α =
χ(g)
χ(1)
= (g) + b
χ(g)
χ(1)
|C|.
Since χ(g) and
χ(g)
χ(1)
|C| are both algebraic integers, we know α is.
Proposition.
If in a finite group, the number of elements in a conjugacy class
C is of (non-trivial) prime power order, then G is not non-abelian simple.
Proof.
Suppose
G
is a non-abelian simple group, and let 1
6
=
g G
be living in
the conjugacy class
C
of order
p
r
. If
χ 6
= 1
G
is a non-trivial irreducible character
of
G
, then either
χ
(1) and
|C|
=
p
r
are not coprime, in which case
p | χ
(1), or
they are coprime, in which case |χ(g)| = χ(1) or χ(g) = 0.
However, it cannot be that
|χ
(
g
)
|
=
χ
(1). If so, then we must have
ρ
(
g
) =
λI
for some λ. So it commutes with everything, i.e. for all h, we have
ρ(gh) = ρ(g)ρ(h) = ρ(h)ρ(g) = ρ(hg).
Moreover, since
G
is simple,
ρ
must be faithful. So we must have
gh
=
hg
for all
h
. So
Z
(
G
) is non-trivial. This is a contradiction. So either
p | χ
(1) or
χ
(
g
) = 0.
By column orthogonality applied to C and 1, we get
0 = 1 +
X
16=χ irreducible, p|χ(1)
χ(1)χ(g),
where we have deleted the 0 terms. So we get
1
p
=
X
χ6=1
χ(1)
p
χ(g).
But this is both an algebraic integer and a rational number, but not integer.
This is a contradiction.
We can now prove Burnside’s theorem.
Theorem
(Burside’s
p
a
q
b
theorem)
.
Let
p, q
be primes, and let
|G|
=
p
a
q
b
,
where a, b Z
0
, with a + b 2. Then G is not simple.
Proof.
Let
|G|
=
p
a
q
b
. If
a
= 0 or
b
= 0, then the result is trivial. Suppose
a, b >
0. We let
Q Syl
q
(
G
). Since
Q
is a
p
-group, we know
Z
(
Q
) is non-trivial.
Hence there is some 1
6
=
g Z
(
Q
). By definition of center, we know
Q C
G
(
g
).
Also, C
G
(g) is not the whole of G, since the center of G is trivial. So
|C
G
(g)| = |G : C
G
(g)| = p
r
for some 0 < r a. So done.
This is all we’ve got to say about finite groups.
15 Representations of compact groups
We now consider the case of infinite groups. It turns out infinite groups don’t
behave well if we don’t impose additional structures. When talking about
representations of infinite groups, things work well only if we impose a topological
structure on the group, and require the representations to be continuous. Then
we get nice results if we only concentrate on the groups that are compact. This
is a generalization of our previous results, since we can give any finite group the
discrete topology, and then all representations are continuous, and the group is
compact due to finiteness.
We start with the definition of a topological group.
Definition
(Topological group)
.
A topological group is a group
G
which is also
a topological space, and for which multiplication
G × G G
((
h, g
)
7→ hg
) and
inverse G G (g 7→ g
1
) are continuous maps.
Example.
Any finite group
G
with the discrete topology is a topological group.
Example.
The matrix groups
GL
n
(
R
) and
GL
n
(
C
) are topological groups
(inheriting the topology of R
n
2
and C
n
2
).
However, topological groups are often too huge. We would like to concentrate
on “small” topological groups.
Definition
(Compact group)
.
A topological group is a compact group if it is
compact as a topological space.
There are some fairly nice examples of these.
Example. All finite groups with discrete topology are compact.
Example.
The group
S
1
=
{z C
:
|z|
= 1
}
under multiplication is a compact
group. This is known as the circle group, for mysterious reasons. Thus the torus
S
1
× S
1
× ··· × S
1
is also a compact group.
Example.
The orthogonal group O(
n
)
GL
n
(
R
) is compact, and so is
SO
(
n
) =
{A
O(
n
) :
det A
= 1
}
. These are compact since they are closed and bounded
as a subspace of
R
n
2
, as the entries can be at most 1 in magnitude. Note that
SO(2)
=
S
1
, mapping a rotation by θ to e
.
Example.
The unitary group U(
n
) =
{A GL
n
(
C
) :
AA
=
I}
is compact,
and so is SU(n) = {A U (n) : det A = 1}.
Note that we also have
U
(1)
=
SO
(2)
=
S
1
. These are not only isomorphisms
as group, but these isomorphisms are also homeomorphisms of topological spaces.
The one we are going to spend most of our time on is
SU(2) = {(z
1
, z
2
) C
2
: |z
1
|
2
+ |z
2
|
2
= 1} R
4
= C
2
.
We can see this is homeomorphic to S
3
.
It turns out that the only spheres on which we can define group operations
on are
S
1
and
S
3
, but we will not prove this, since this is a course on algebra,
not topology.
We now want to develop a representation theory of these compact groups,
similar to what we’ve done for finite groups.
Definition
(Representation of topological group)
.
A representation of a topolog-
ical group on a finite-dimensional space
V
is a continuous group homomorphism
G GL(V ).
It is important that the group homomorphism is continuous. Note that for a
general topological space
X
, a map
ρ
:
X GL
(
V
)
=
GL
n
(
C
) is continuous if
and only if the component maps x 7→ ρ(x)
ij
are continuous for all i, j.
The key that makes this work is the compactness of the group and the
continuity of the representations. If we throw them away, then things will go
very bad. To see this, we consider the simple example
S
1
= U(1) = {g C
×
: |g| = 1}
=
R/Z,
where the last isomorphism is an abelian group isomorphism via the map
x 7→ e
2πix
: R/Z S
1
.
What happens if we just view
S
1
as an abstract group, and not a topological
group? We can view
R
as a vector space over
Q
, and this has a basis (by Hamel’s
basis theorem), say,
A R
. Moreover, we can assume the basis is ordered, and
we can assume 1 A.
As abelian groups, we then have
R
=
Q
M
α∈A\{1}
Qα,
Then this induces an isomorphism
R/Z
=
Q/Z
M
α∈A\{1}
Qα.
Thus, as abstract groups,
S
1
has uncountably many irreducible representations,
namely for each
λ A \{
1
}
, there exists a one-dimensional representation given
by
ρ
λ
(e
2π
) =
(
1 µ 6∈ Qλ
e
2π
µ Qλ
We see
ρ
λ
=
ρ
λ
0
if and only if
Qλ
=
Qλ
0
. Hence there are indeed uncountably
many of these. There are in fact even more irreducible representation we haven’t
listed. This is bad.
The idea is then to topologize
S
1
as a subset of
C
, and study how it acts
naturally on complex spaces in a continuous way. Then we can ensure that we
have at most countably many representations. In fact, we can characterize all
irreducible representations in a rather nice way:
Theorem.
Every one-dimensional (continuous) representation
S
1
is of the form
ρ : z 7→ z
n
for some n Z.
This is a nice countable family of representations. To prove this, we need
two lemmas from real analysis.
Lemma.
If
ψ
: (
R,
+)
(
R,
+) is a continuous group homomorphism, then
there exists a c R such that
ψ(x) = cx
for all x R.
Proof.
Given
ψ
: (
R,
+)
(
R,
+) continuous, we let
c
=
ψ
(1). We now claim
that ψ(x) = cx.
Since ψ is a homomorphism, for every n Z
0
and x R, we know
ψ(nx) = ψ(x + ··· + x) = ψ(x) + ··· + ψ(x) = (x).
In particular, when x = 1, we know ψ(n) = cn. Also, we have
ψ(n) = ψ(n) = cn.
Thus ψ(n) = cn for all n Z.
We now put x =
m
n
Q. Then we have
(x) = ψ(nx) = ψ(m) = cm.
So we must have
ψ
m
n
= c
m
n
.
So we get
ψ
(
q
) =
cq
for all
q Q
. But
Q
is dense in
R
, and
ψ
is continuous. So
we must have ψ(x) = cx for all x R.
Lemma. Continuous homomorphisms ϕ : (R, +) S
1
are of the form
ϕ(x) = e
icx
for some c R.
Proof.
Let
ε
: (
R,
+)
S
1
be defined by
x 7→ e
ix
. This homomorphism wraps
the real line around S
1
with period 2π.
We now claim that given any continuous function
ϕ
:
R S
1
such that
ϕ
(0) = 1, there exists a unique continuous lifting homomorphism
ψ
:
R R
such that
ε ψ = ϕ, ψ(0) = 0.
(R, +)
0
(R, +)
S
1
ε
ϕ
!ψ
The lifting is constructed by starting with
ψ
(0) = 0, and then extending a small
interval at a time to get a continuous map
R R
. We will not go into the
details. Alternatively, this follows from the lifting criterion from IID Algebraic
Topology.
We now claim that if in addition
ϕ
is a homomorphism, then so is its
continuous lifting
ψ
. If this is true, then we can conclude that
ψ
(
x
) =
cx
for
some c R. Hence
ϕ(x) = e
icx
.
To show that
ψ
is indeed a homomorphism, we have to show that
ψ
(
x
+
y
) =
ψ(x) + ψ(y).
By definition, we know
ϕ(x + y) = ϕ(x)ϕ(y).
By definition of ψ, this means
ε(ψ(x + y) ψ(x) ψ(y)) = 1.
We now look at our definition of ε to get
ψ(x + y) ψ(x) ψ(y) = 2kπ
for some integer
k Z
, depending continuously on
x
and
y
. But
k
can only be
an integer. So it must be constant. Now we pick our favorite
x
and
y
, namely
x = y = 0. Then we find k = 0. So we get
ψ(x + y) = ψ(x) + ψ(y).
So ψ is a group homomorphism.
With these two lemmas, we can prove our characterization of the (one-
dimensional) representations of S
1
.
Theorem.
Every one-dimensional (continuous) representation
S
1
is of the form
ρ : z 7→ z
n
for some n Z.
Proof.
Let
ρ
:
S
1
C
×
be a continuous representation. We now claim that
ρ
actually maps
S
1
to
S
1
. Since
S
1
is compact, we know
ρ
(
S
1
) has closed and
bounded image. Also,
ρ(z
n
) = (ρ(z))
n
for all
n Z
. Thus for each
z S
1
, if
|ρ
(
z
)
| >
1, then the image of
ρ
(
z
n
) is
unbounded. Similarly, if it is less than 1, then
ρ
(
z
n
) is unbounded. So we must
have ρ(S
1
) S
1
. So we get a continuous homomorphism
R S
1
x 7→ ρ(e
ix
).
So we know there is some c R such that
ρ(e
ix
) = e
icx
,
Now in particular,
1 = ρ(e
2πi
) = e
2πic
.
This forces c Z. Putting n = c, we get
ρ(z) = z
n
.
That has taken nearly an hour, and we’ve used two not-so-trivial facts from
analysis. So we actually had to work quite hard to get this. But the result is
good. We have a complete list of representations, and we don’t have to fiddle
with things like characters.
Our next objective is to repeat this for
SU
(2), and this time we cannot just
get away with doing some analysis. We have to do representation theory properly.
So we study the general theory of representations of complex spaces.
In studying finite groups, we often took the “average” over the group via
the operation
1
|G|
P
gG
something
. Of course, we can’t do this now, since the
group is infinite. As we all know, the continuous analogue of summing is called
integration. Informally, we want to be able to write something like
R
G
dg.
This is actually a measure, called the “Haar measure”, and always exists as
long as you are compact and Hausdorff. However, we will not need or use this
result, since we can construct it by hand for S
1
and SU(2).
Definition (Haar measure). Let G be a topological group, and let
C(G) = {f : G C : f is continuous, f(gxg
1
) = f(x) for all g, x G}.
A non-trivial linear functional
R
G
: C(G) C, written as
Z
G
f =
Z
G
f(g) dg
is called a Haar measure if
(i) It satisfies the normalization condition
Z
G
1 dg = 1
so that the “total volume” is 1.
(ii) It is left and right translation invariant, i.e.
Z
G
f(xg) dg =
Z
G
f(g) dg =
Z
G
f(gx) dg
for all x G.
Example. For a finite group G, we can define
Z
G
f(g) dg =
1
|G|
X
gG
f(g).
The division by the group order is the thing we need to make the normalization
hold.
Example. Let G = S
1
. Then we can have
Z
G
f(g) dg =
1
2π
Z
2π
0
f(e
) dθ.
Again, division by 2π is the normalization required.
We have explicitly constructed them here, but there is a theorem that
guarantees the existence of such a measure for sufficiently nice topological group.
Theorem.
Let
G
be a compact Hausdorff topological group. Then there exists
a unique Haar measure on G.
From now on, the word “compact” will mean “compact and Hausdorff”, so
that the conditions of the above theorem holds.
We will not prove theorem, and just assume it to be true. If you don’t like
this, whenever you see “compact group”, replace it with “compact group with
Haar measure”. We will explicitly construct it for
SU
(2) later, which is all we
really care about.
Once we know the existence of such a thing, most of the things we know for
finite groups hold for compact groups, as long as we replace the sums with the
appropriate integrals.
Corollary
(Weyl’s unitary trick)
.
Let
G
be a compact group. Then every
representation (ρ, V ) has a G-invariant Hermitian inner product.
Proof.
As for the finite case, take any inner product (
·, ·
) on
V
, then define a
new inner product by
hv, wi =
Z
G
(ρ(g)v, ρ(g)w) dg.
Then this is a G-invariant inner product.
Theorem
(Maschke’s theoerm)
.
Let
G
be compact group. Then every repre-
sentation of G is completely reducible.
Proof.
Given a representation (
ρ, V
). Choose a
G
-invariant inner product. If
V
is not irreducible, let
W V
be a subrepresentation. Then
W
is also
G-invariant, and
V = W W
.
Then the result follows by induction.
We can use the Haar measure to endow
C
(
G
) with an inner product given by
hf, f
0
i =
Z
G
f(g)f
0
(g) dg.
Definition
(Character)
.
If
ρ
:
G GL
(
V
) is a representation, then the char-
acter
χ
ρ
=
tr ρ
is a continuous class function, since each component
ρ
(
g
)
ij
is
continuous.
So characters make sense, and we can ask if all the things about finite groups
hold here. For example, we can ask about orthogonality.
Schur’s lemma also hold, with the same proof, since we didn’t really need
finiteness there. Using that, we can prove the following:
Theorem
(Orthogonality)
.
Let
G
be a compact group, and
V
and
W
be
irreducible representations of G. Then
hχ
V
, χ
W
i =
(
1 V
=
W
0 V 6
=
W
.
Now do irreducible characters form a basis for C(G)?
Example.
We take the only (infinite) compact group we know about —
G
=
S
1
.
We have found that the one-dimensional representations are
ρ
n
: z 7→ z
n
for
n Z
. As
S
1
is abelian, these are all the (continuous) irreducible representa-
tions given any representation
ρ
, we can find a simultaneous eigenvector for
each ρ(g).
The “character table of
S
1
has rows
χ
n
indexed by
Z
, with
χ
n
(
e
) =
e
inθ
.
Now given any representation of
S
1
, say
V
, we can break
V
as a direct sum
of 1-dimensional subrepresentations. So the character χ
V
of V is of the form
χ
V
(z) =
X
nZ
a
n
z
n
,
where
a
n
Z
0
, and only finitely many
a
n
are non-zero (since we are assuming
finite-dimensionality).
Actually,
a
n
is the number of copies of
ρ
n
in the decomposition of
V
. We
can find out the value of a
n
by computing
a
n
= hχ
n
, χ
V
i =
1
2π
Z
2π
0
e
inθ
χ
V
(e
) dθ.
Hence we know
χ
V
(e
) =
X
nZ
1
2π
Z
2π
0
χ
V
(e
0
)e
inθ
0
dθ
0
e
inθ
.
This is really just a Fourier decomposition of
χ
V
. This gives a decomposition
of
χ
V
into irreducible characters, and the “Fourier mode”
a
n
is the number of
each irreducible character occurring in this decomposition.
It is possible to show that
χ
n
form a complete orthonormal set in the Hilbert
space
L
2
(
S
1
), i.e. square-integrable functions on
S
1
. This is the Peter-Weyl
theorem, and is highly non-trivial.
15.1 Representations of SU(2)
In the rest of the time, we would like to talk about G = SU(2).
Recall that
SU(2) = {A GL
2
(C) : A
A = I, det A = I}.
Note that if
A =
a b
c d
SU(2),
then since det A = 1, we have
A
1
=
d b
c a
.
So we need d = ¯a and c =
¯
b. Moreover, we have
a¯a + b
¯
b = 1.
Hence we can write SU(2) explicitly as
G =

a b
¯
b ¯a
: a, b C, |a|
2
+ |b|
2
= 1
.
Topologically, we know G
=
S
3
C
2
=
R
4
.
Instead of thinking about
C
2
in the usual vector space way, we can think of
it as a subgroup of M
2
(C) via
H =

z w
¯w ¯z
: w, z C
M
2
(C).
This is known as Hamilton’s quaternion algebra. Then
H
is a 4-dimensional
Euclidean space (two components from z and two components from w), with a
norm on H given by
kAk
2
= det A.
We now see that
SU
(2)
H
is exactly the unit sphere
kAk
2
= 1 in
H
. If
A SU
(2) and
x H
, then
kAxk
=
kxk
since
kAk
= 1. So elements of
G
acts
as isometries on the space.
After normalization (by
1
2π
2
), the usual integration of functions on
S
3
defines
a Haar measure on
G
. It is an exercise on the last example sheet to write this
out explicitly.
We now look at conjugacy in G. We let
T =

a 0
0 ¯a
: a C, |a|
2
= 1
=
S
1
.
This is a maximal torus in
G
, and it plays a fundamental role, since we happen
to know about S
1
. We also have a favorite element
s =
0 1
1 0
SU(2).
We now prove some easy linear algebra results about SU(2).
Lemma (SU(2)-conjugacy classes).
(i) Let t T . Then sts
1
= t
1
.
(ii) s
2
= I Z(SU(2)).
(iii) The normalizer
N
G
(T ) = T sT =

a 0
0 ¯a
,
0 a
¯a 0
: a C, |a| = 1
.
(iv)
Every conjugacy class
C
of
SU
(2) contains an element of
T
, i.e.
C T 6
=
.
(v) In fact,
C T = {t, t
1
}
for some t T , and t = t
1
if and only if t = ±I, in which case C = {t}.
(vi) There is a bijection
{conjugacy classes in SU(2)} [1, 1],
given by
A 7→
1
2
tr A.
We can see that if
A =
λ 0
0
¯
λ
,
then
1
2
tr A =
1
2
(λ +
¯
λ) = Re(λ).
Note that what (iv) and (v) really say is that matrices in
SU
(2) are diago-
nalizable (over SU(2)).
Proof.
(i) Write it out.
(ii) Write it out.
(iii) Direct verification.
(iv)
It is well-known from linear algebra that every unitary matrix
X
has an
orthonormal basis of eigenvectors, and hence is conjugate in
U
(2) to one
in T , say
QXQ
T.
We now want to force Q into SU(2), i.e. make Q have determinant 1.
We put
δ
=
det Q
. Since
Q
is unitary, i.e.
QQ
=
I
, we know
|δ|
= 1. So
we let ε be a square root of δ, and define
Q
1
= ε
1
Q.
Then we have
Q
1
XQ
1
T.
(v)
We let
g G
, and suppose
g C
. If
g
=
±I
, then
C T
=
{g}
. Otherwise,
g
has two distinct eigenvalues
λ, λ
1
. Note that the two eigenvlaues must
be inverses of each other, since it is in SU(2). Then we know
C =
h
λ 0
0 λ
1
h
1
: h G
.
Thus we find
C T =

λ 0
0 λ
1
,
λ
1
0
0 λ

.
This is true since eigenvalues are preserved by conjugation, so if any
µ 0
0 µ
1
,
then
{µ, µ
1
}
=
{λ, λ
1
}
. Also, we can get the second matrix from the
first by conjugating with s.
(vi) Consider the map
1
2
tr : {conjugacy classes} [1, 1].
By (v), matrices are conjugate in
G
iff they have the same set of eigenvalues.
Now
1
2
tr
λ 0
0 λ
1
=
1
2
(λ +
¯
λ) = Re(λ) = cos θ,
where λ = e
. Hence the map is a surjection onto [1, 1].
Now we have to show it is injective. This is also easy. If
g
and
g
0
have the
same image, i.e.
1
2
tr g =
1
2
tr g
0
,
then g and g
0
have the same characteristic polynomial, namely
x
2
(tr g)x + 1.
Hence they have the same eigenvalues, and hence they are similar.
We now write, for t (1, 1),
C
t
=
g G :
1
2
tr g = t
.
In particular, we have
C
1
= {I}, C
1
= {−I}.
Proposition. For t (1, 1), the class C
t
=
S
2
as topological spaces.
This is not really needed, but is a nice thing to know.
Proof. Exercise!
We now move on to classify the irreducible representations of SU(2).
We let
V
n
be the complex space of all homogeneous polynomials of degree
n
in variables x, y, i.e.
V
n
= {r
0
x
n
+ r
1
x
n1
y + ··· + r
n
y
n
: r
0
, ··· , r
n
C}.
This is an
n
+ 1-dimensional complex vector space with basis
x
n
, x
n1
y, ··· , y
n
.
We want to get an action of
SU
(2) on
V
n
. It is easier to think about the
action of
GL
2
(
C
) in general, and then restrict to
SU
(2). We define the action of
GL
2
(C) on V
n
by
ρ
n
: GL
2
(C) GL(V
n
)
given by the rule
ρ
n

a b
c d

f(x, y) = f(ax + cy, bx + dy).
In other words, we have
ρ
n

a b
c d

f

x y

= f
x y
a b
c d

,
where the multiplication in f is matrix multiplication.
Example. When n = 0, then V
0
=
C, and ρ
0
is the trivial representation.
When n = 1, then this is the natural two-dimensional representation, and
ρ
1

a b
c d

has matrix
a b
c d
with respect to the standard basis {x, y} of V
1
= C
2
.
More interestingly, when n = 2, we know
ρ
2

a b
c d

has matrix
a
2
ab b
2
2ac ad + bc 2bd
c
2
cd d
2
,
with respect to the basis
x
2
, xy, y
2
of
V
2
=
C
3
. We obtain the matrix by
computing, e.g.
ρ
2
(g)(x
2
) = (ax + cy)
2
= a
2
x
2
+ 2acxy + c
2
y
2
.
Now we know
SU
(2)
GL
2
(
C
). So we can view
V
n
as a representation of
SU(2) by restriction.
Now we’ve got some representations. The claim is that these are all the
irreducibles. Before we prove that, we look at the character of these things.
Lemma.
A continuous class function
f
:
G C
is determined by its restriction
to T , and F |
T
is even, i.e.
f

λ 0
0 λ
1

= f

λ
1
0
0 λ

.
Proof.
Each conjugacy class in
SU
(2) meets
T
. So a class function is determined
by its restriction to
T
. Evenness follows from the fact that the two elements are
conjugate.
In particular, a character of a representation (ρ, V ) is also an even function
χ
ρ
: S
1
C.
Lemma.
If
χ
is a character of a representation of
SU
(2), then its restriction
χ|
T
is a Laurent polynomial, i.e. a finite N-linear combination of functions
λ 0
0 λ
1
7→ λ
n
for n Z.
Proof.
If
V
is a representation of
SU
(2), then
Res
SU(2)
T
V
is a representation
of
T
, and its character
Res
SU(2)
T
χ
is the restriction of
χ
V
to
T
. But every
representation of T has its character of the given form. So done.
Notation. We write N[z, z
1
] for the set of all Laurent polynomials, i.e.
N[z, z
1
] =
n
X
a
n
z
n
: a
n
N : only finitely many a
n
non-zero
o
.
We further write
N[z, z
1
]
ev
= {f N[z, z
1
] : f(z) = f(z
1
)}.
Then by our lemma, for every continuous representation
V
of
SU
(2), the
character χ
V
N[z, z
1
]
ev
(by identifying it with its restriction to T ).
We now actually calculate the character
χ
n
of (
ρ
n
, V
n
) as a representation of
SU(2). We have
χ
V
n
(g) = tr ρ
n
(g),
where
g
z 0
0 z
1
T.
Then we have
ρ
n

z 0
0 z
1

(x
i
y
j
) = (zx
i
)(z
1
y
j
) = z
ij
x
i
y
j
.
So each
x
i
y
j
is an eigenvector of
ρ
n
(
g
), with eigenvalue
z
ij
. So we know
ρ
n
(
g
)
has a matrix
z
n
z
n2
.
.
.
z
2n
z
n
,
with respect to the standard basis. Hence the character is just
χ
n

z 0
0 z
1

= z
n
+ z
n2
+ ··· + z
n
=
z
n+1
z
(n+1)
z z
1
,
where the last expression is valid unless z = ±1.
We can now state the result we are aiming for:
Theorem.
The representations
ρ
n
:
SU
(2)
GL
(
V
n
) of dimension
n
+ 1 are
irreducible for n Z
0
.
Again, we get a complete set (completeness proven later). A complete list of
all irreducible representations, given in a really nice form. This is spectacular.
Proof.
Let 0
6
=
W V
n
be a
G
-invariant subspace, i.e. a subrepresentation of
V
n
. We will show that W = V
n
.
All we know about
W
is that it is non-zero. So we take some non-zero vector
of W .
Claim. Let
0 6= w =
n
X
j=0
r
j
x
nj
y
j
W.
Since this is non-zero, there is some
i
such that
r
i
6
= 0. The claim is that
x
ni
y
i
W .
We argue by induction on the number of non-zero coefficients
r
j
. If there
is only one non-zero coefficient, then we are already done, as
w
is a non-zero
scalar multiple of x
ni
y
i
.
So assume there is more than one, and choose one
i
such that
r
i
6
= 0. We
pick z S
1
with z
n
, z
n2
, ··· , z
2n
, z
n
all distinct in C. Now
ρ
n

z
z
1

w =
X
r
j
z
n2j
x
nj
y
j
W.
Subtracting a copy of w, we find
ρ
n

z
z
1

w z
n2i
w =
X
r
j
(z
n2j
z
n2i
)x
nj
y
j
W.
We now look at the coefficient
r
j
(z
n2j
z
n2i
).
This is non-zero if and only if
r
j
is non-zero and
j 6
=
i
. So we can use this to
remove any non-zero coefficient. Thus by induction, we get
x
nj
y
j
W
for all j such that r
j
6= 0.
This gives us one basis vector inside W , and we need to get the rest.
Claim. W = V
n
.
We now know that x
ni
y
i
W for some i. We consider
ρ
n
1
2
1 1
1 1

x
ni
y
i
=
1
2
(x + y)
ni
(x + y)
i
W.
It is clear that the coefficient of
x
n
is non-zero. So we can use the claim to
deduce x
n
W .
Finally, for general a, b 6= 0, we apply
ρ
n

a
¯
b
b ¯a

x
n
= (ax + by)
n
W,
and the coefficient of everything is non-zero. So basis vectors are in
W
. So
W = V
n
.
This proof is surprisingly elementary. It does not use any technology at all.
Alternatively, to prove this, we can identify
C
cos θ
=
A G :
1
2
tr A = cos θ
with the 2-sphere
{|Im a|
2
+ |b|
2
= sin
2
θ}
of radius
sin θ
. So if
f
is a class function on
G
,
f
is constant on each class
C
cos θ
.
It turns out we get
Z
G
f(g) dg =
1
2π
2
Z
2π
0
1
2
f

e
e

4π sin
2
θ dθ.
This is the Weyl integration formula, which is the Haar measure for
SU
(2). Then
if
χ
n
is the character of
V
n
, we can use this to show that
hχ
n
, χ
n
i
= 1. Hence
χ
n
is irreducible. We will not go into the details of this construction.
Theorem.
Every finite-dimensional continuous irreducible representation of
G
is one of the ρ
n
: G GL(V
n
) as defined above.
Proof.
Assume
ρ
V
:
G GL
(
V
) is an irreducible representation affording a
character
χ
V
N
[
z, z
1
]
ev
. We will show that
χ
V
=
χ
n
for some
n
. Now we see
χ
0
= 1
χ
1
= z + z
1
χ
2
= z
2
+ 1 + z
2
.
.
.
form a basis of
Q
[
z, z
1
]
ev
, which is a non-finite dimensional vector space over
Q. Hence we can write
χ
V
=
X
n
a
n
χ
n
,
a finite sum with finitely many
a
n
6
= 0. Note that it is possible that
a
n
Q
. So
we clear denominators, and move the summands with negative coefficients to
the left hand side. So we get
V
+
X
iI
m
i
χ
i
=
X
jJ
n
j
χ
j
,
with I, J disjoint finite subsets of N, and m, m
i
, n
j
N.
We know the left and right-hand side are characters of representations of
G
.
So we get
mV
M
I
m
i
V
i
=
M
J
n
j
V
j
.
Since
V
is irreducible and factorization is unique, we must have
V
=
V
n
for some
n J.
We’ve got a complete list of irreducible representations of
SU
(2). So we can
look at what happens when we take products.
We know that for V, W representations of SU(2), if
Res
SU(2)
T
V
=
Res
SU(2)
T
W,
then in fact
V
=
W.
This gives us the following result:
Proposition.
Let
G
=
SU
(2) or
G
=
S
1
, and
V, W
are representations of
G
.
Then
χ
V W
= χ
V
χ
W
.
Proof.
By the previous remark, it is enough to consider the case
G
=
S
1
. Suppose
V and W have eigenbases e
1
, ··· , e
n
and f
1
, ··· , f
m
respectively such that
ρ(z)e
i
= z
n
i
e
i
, ρ(z)f
j
= z
m
j
f
j
for each i, j. Then
ρ(z)(e
i
f
j
) = z
n
i
+m
j
e
i
f
j
.
Thus the character is
χ
V W
(z) =
X
i,j
z
n
i
+m
j
=
X
i
z
n
i
!
X
j
z
m
j
= χ
V
(z)χ
W
(z).
Example. We have
χ
V
1
V
1
(z) = (z + z
1
)
2
= z
2
+ 2 + z
2
= χ
V
2
+ χ
V
0
.
So we have
V
1
V
1
=
V
2
V
0
.
Similarly, we can compute
χ
V
1
V
2
(z) = (z
2
+ 1 + z
2
)(z + z
1
) = z
3
+ 2z + 2z
1
+ z
3
= χ
V
3
+ χ
V
1
.
So we get
V
1
V
2
=
V
3
V
1
.
Proposition (Clebsch-Gordon rule). For n, m N, we have
V
n
V
m
=
V
n+m
V
n+m2
··· V
|nm|+2
V
|nm|
.
Proof.
We just check this works for characters. Without loss of generality, we
assume n m. We can compute
(χ
n
χ
m
)(z) =
z
n+1
z
n1
z z
1
(z
m
+ z
m2
+ ··· + z
m
)
=
m
X
j=0
z
n+m+12j
z
2jnm1
z z
1
=
m
X
j=0
χ
n+m2j
(z).
Note that the condition
n m
ensures there are no cancellations in the sum.
15.2 Representations of SO(3), SU(2) and U (2)
We’ve now got a complete list of representations of
S
1
and
SU
(2). We see if
we can make some deductions about some related groups. We will not give too
many details about these groups, and the proofs are just sketches. Moreover, we
will rely on the following facts which we shall not prove:
Proposition. There are isomorphisms of topological groups:
(i) SO(3)
=
SU(2)/I} = PSU(2)
(ii) SO(4)
=
SU(2) × SU(2)/(I, I)}
(iii) U(2)
=
U(1) × SU(2)/(I, I)}
All maps are group isomorphisms, but in fact also homeomorphisms. To show
this, we can use the fact that a continuous bijection from a Hausdorff space to a
compact space is automatically a homeomorphism.
Assuming this is true, we obtain the following corollary;
Corollary. Every irreducible representation of SO(3) has the following form:
ρ
2m
: SO(3) GL(V
2m
),
for some m 0, where V
n
are the irreducible representations of SU(2).
Proof.
Irreducible representations of
SO
(3) correspond to irreducible representa-
tions of
SU
(2) such that
I
acts trivially by lifting. But
I
acts on
V
n
as
1
when n is odd, and as 1 when n is even, since
ρ(I) =
(1)
n
(1)
n2
.
.
.
(1)
n
= (1)
n
I.
For the sake of completeness, we provide a (sketch) proof of the isomorphism
SO(3)
=
SU(2)/I}.
Proposition. SO(3)
=
SU(2)/I}.
Proof sketch.
Recall that
SU
(2) can be viewed as the sphere of unit norm
quaternions H
=
R
4
.
Let
H
0
= {A H : tr A = 0}.
These are the “pure” quaternions. This is a three-dimensional subspace of
H
. It
is not hard to see this is
H
0
= R

i 0
0 i
,
0 1
1 0
,
0 i
i 0

= R hi, j, ki,
where R··i is the R-span of the things.
This is equipped with the norm
kAk
2
= det A.
This gives a nice 3-dimensional Euclidean space, and
SU
(2) acts as isometries
on H
0
by conjugation, i.e.
X · A = XAX
1
,
giving a group homomorphism
ϕ : SU(2) O(3),
and the kernel of this map is
Z
(
SU
(2)) =
I}
. We also know that
SU
(2) is
compact, and O(3) is Hausdorff. Hence the continuous group isomorphism
¯ϕ : SU(2)/I} im ϕ
is a homeomorphism. It remains to show that im ϕ = SO(3).
But we know
SU
(2) is connected, and
det
(
ϕ
(
X
)) is a continuous function
that can only take values 1 or
1. So
det
(
ϕ
(
X
)) is either always 1 or always
1.
But det(ϕ(I)) = 1. So we know det(ϕ(X)) = 1 for all X. Hence im ϕ SO(3).
To show that equality indeed holds, we have to show that all possible rotations
in
H
0
are possible. We first show all rotations in the
i, j
-plane are implemented
by elements of the form
a
+
bk
, and similarly for any permutation of
i, j, k
. Since
all such rotations generate SO(3), we are then done. Now consider
e
0
0 e
ai b
¯
b ai
e
0
0 e
=
ai e
2
b
¯
be
2
ai
.
So
e
0
0 e
acts on
Rhi, j, ki
by a rotation in the (
j, k
)-plane through an angle 2
θ
. We can
check that
cos θ sin θ
sin θ cos θ
,
cos θ i sin θ
i sin θ cos θ
act by rotation of 2
θ
in the (
i, k
)-plane and (
i, j
)-plane respectively. So done.
We can adapt this proof to prove the other isomorphisms. However, it is
slightly more difficult to get the irreducible representations, since it involves
taking some direct products. We need a result about products
G × H
of two
compact groups. Similar to the finite case, we get the complete list of irreducible
representations by taking the tensor products
V W
, where
V
and
W
range
over the irreducibles of G and H independently.
We will just assert the results.
Proposition.
The complete list of irreducible representations of
SO
(4) is
ρ
m
×ρ
n
,
where m, n > 0 and m n (mod 2).
Proposition. The complete list of irreducible representations of U(2) is
det
m
ρ
n
,
where
m, n Z
and
n
0, and
det
is the obvious one-dimensional representation.