9Dual spaces and tensor products of representations
II Representation Theory
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
1≤i<j≤n
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