9Dual spaces and tensor products of representations
II Representation Theory
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
.