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