3Complete reducibility and Maschke's theorem
II Representation Theory
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
g∈G
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
g∈G
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
g∈G
gq(g
−1
w) =
1
|G|
X
g∈G
gg
−1
w =
1
|G|
X
g∈G
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
g∈G
gq(g
−1
v)
=
1
|G|
X
g∈G
hgq(g
−1
v)
=
1
|G|
X
g∈G
(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
g∈G
(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
g∈G
((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.