5Character theory
II Representation Theory
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
g∈G
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
g∈G
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
2mπ
3
−sin
2mπ
3
sin
2mπ
3
cos
2mπ
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.