4Schur's lemma

II Representation Theory



4 Schur’s lemma
The topic of this chapter is Schur’s lemma, an easy yet extremely useful lemma
in representation theory.
Theorem (Schur’s lemma).
(i)
Assume
V
and
W
are irreducible
G
-spaces over a field
F
. Then any
G-homomorphism θ : V W is either zero or an isomorphism.
(ii)
If
F
is algebraically closed, and
V
is an irreducible
G
-space, then any
G-endomorphism V V is a scalar multiple of the identity map ι
V
.
Proof.
(i)
Let
θ
:
V W
be a
G
-homomorphism between irreducibles. Then
ker θ
is
a
G
-subspace of
V
, and since
V
is irreducible, either
ker θ
= 0 or
ker θ
=
V
.
Similarly,
im θ
is a
G
-subspace of
W
, and as
W
is irreducible, we must
have
im θ
= 0 or
im θ
=
W
. Hence either
ker θ
=
V
, in which case
θ
= 0,
or ker θ = 0 and im θ = W , i.e. θ is a bijection.
(ii)
Since
F
is algebraically closed,
θ
has an eigenvalue
λ
. Then
θ λι
V
is a
singular
G
-endomorphism of
V
. So by (i), it must be the zero map. So
θ = λι
V
.
Recall that the
F
-space
Hom
G
(
V, W
) is the space of all
G
-homomorphisms
V W . If V = W , we write End
G
(V ) for the G-endomorphisms of V .
Corollary. If V, W are irreducible complex G-spaces, then
dim
C
Hom
G
(V, W ) =
(
1 V, W are G-isomorphic
0 otherwise
Proof.
If
V
and
W
are not isomorphic, then the only possible map between
them is the zero map by Schur’s lemma.
Otherwise, suppose
V
=
W
and let
θ
1
, θ
2
Hom
G
(
V, W
) be both non-
zero. By Schur’s lemma, they are isomorphisms, and hence invertible. So
θ
1
2
θ
1
End
G
(V ). Thus θ
1
2
θ
1
= λι
V
for some λ C. Thus θ
1
= λθ
2
.
We have another less obvious corollary.
Corollary.
If
G
is a finite group and has a faithful complex irreducible repre-
sentation, then its center Z(G) is cyclic.
This is a useful result it allows us transfer our representation-theoretic
knowledge (the existence of a faithful complex irreducible representation) to
group theoretic knowledge (center of group being cyclic). This will become
increasingly common in the future, and is a good thing since representations are
easy and groups are hard.
The converse, however, is not true. For this, see example sheet 1, question
10.
Proof.
Let
ρ
:
G GL
(
V
) be a faithful irreducible complex representation. Let
z Z
(
G
). So
zg
=
gz
for all
g G
. Hence
φ
z
:
v 7→ zv
is a
G
-endomorphism
on
V
. Hence by Schur’s lemma, it is multiplication by a scalar
µ
z
, say. Thus
zv = µ
z
v for all v V .
Then the map
σ : Z(G) C
×
z 7→ µ
g
is a representation of
Z
(
G
). Since
ρ
is faithful, so is
σ
. So
Z
(
G
) =
{µ
z
:
z
Z(G)} is isomorphic to a finite subgroup of C
×
, hence cyclic.
Corollary.
The irreducible complex representations of a finite abelian group
G
are all 1-dimensional.
Proof.
We can use the fact that commuting diagonalizable matrices are simulta-
neously diagonalizable. Thus for every irreducible
V
, we can pick some
v V
that is an eigenvector for each
g G
. Thus
hvi
is a
G
-subspace. As
V
is
irreducible, we must have V = hvi.
Alternatively, we can prove this in a representation-theoretic way. Let
V
be
an irreducible complex representation. For each g G, the map
θ
g
: V V
v 7→ gv
is a
G
-endomorphism of
V
, since it commutes with the other group elements.
Since V is irreducible, θ
g
= λ
g
ι
V
for some λ
g
C. Thus
gv = λ
g
v
for any g. As V is irreducible, we must have V = hvi.
Note that this result fails over
R
. For example,
C
3
has a two irreducible real
representations, one of dimension 1 and one of dimension 2.
We can do something else. Recall that every finite abelian group
G
isomorphic
to a product of abelian groups. In fact, it can be written as a product of
C
p
α
for various primes
p
and
α
1, and the factors are uniquely determined up to
order.
This you already know from IB Groups Rings and Modules. You might be
born knowing it it’s such a fundamental fact of nature.
Proposition.
The finite abelian group
G
=
C
n
1
× ··· × C
n
r
has precisely
|G|
irreducible representations over C.
This is not a coincidence. We will later show that the number of irreducible
representations is the number of conjugacy classes of the group. In abelian
groups, each conjugacy class is just a singleton, and hence this result.
Proof. Write
G = hx
1
i × ··· × hx
r
i,
where
|x
j
|
=
n
j
. Any irreducible representation
ρ
must be one-dimensional. So
we have
ρ : G C
×
.
Let
ρ
(1
, ··· , x
j
, ··· ,
1) =
λ
j
. Then since
ρ
is a homomorphism, we must have
λ
n
j
j
= 1. Therefore λ
j
is an n
j
th root of unity.
Now the values (λ
1
, ··· , λ
r
) determine ρ completely, namely
ρ(x
j
1
1
, ··· , x
j
r
r
) = λ
j
1
1
···λ
j
r
r
.
Also, whenever
λ
i
is an
n
i
th root of unity for each
i
, then the above formula
gives a well-defined representation. So there is a one-to-one correspondence
ρ (λ
1
, ··· , λ
r
), with λ
n
j
j
= 1.
Since for each
j
, there are
n
j
many
n
j
th roots of unity, it follows that there
are |G| = n
1
···n
r
many choices of the λ
i
. Thus the proposition.
Example.
(i)
Consider
G
=
C
4
=
hxi
. The four 1-dimensional irreducible representations
are given by
1 x x
2
x
3
ρ
1
1 1 1 1
ρ
2
1 i 1 i
ρ
2
1 1 1 1
ρ
2
1 i 1 i
(ii)
Consider the Klein four group
G
=
V
R
=
hx
1
i × hx
2
i
. The irreducible
representations are
1 x
1
x
2
x
1
x
2
ρ
1
1 1 1 1
ρ
2
1 1 1 1
ρ
2
1 1 1 1
ρ
2
1 1 1 1
These are also known as character tables, and we will spend quite a lot of
time computing these for non-abelian groups.
Note that there is no “natural” one-to-one correspondence between the
elements of
G
and the representations of
G
(for
G
finite-abelian). If we choose
an isomorphism
G
=
C
n
1
× ···C
n
r
, then we can identify the two sets, but it
depends on the choice of the isomorphism (while the decomposition is unique,
we can pick a different generator of, say,
C
n
1
and get a different isomorphism to
the same decomposition).
Isotypical decompositions
Recall that we proved we can decompose any
G
-representation into a sum of
irreducible representations. Is this decomposition unique? If it isn’t, can we say
anything about, say, the size of the irreducible representations, or the number of
factors in the decomposition?
We know any diagonalizable endomorphism
α
:
V V
for a vector space
V
gives us a vector space decomposition
V =
M
λ
V (λ),
where
V (λ) = {v V : α(v) = λv}.
This is canonical in that it depends on α alone, and nothing else.
If
V
is moreover a
G
-representation, how does this tie in to the decomposition
of V into the irreducible representations?
Let’s do an example.
Example.
Consider
G
=
D
6
=
S
3
=
hr, s
:
r
3
=
s
2
= 1
, rs
=
sr
1
i
. We have
previously seen that each irreducible representation has dimension at most 2.
We spot at least three irreducible representations:
1 triad r 7→ 1 s 7→ 1
S sign r 7→ 1 s 7→ 1
W 2-dimensional
The last representation is the action of
D
6
on
R
2
in the natural way, i.e. the
rotations and reflections of the plane that corresponds to the symmetries of the
triangle. It is helpful to view this as a complex representation in order to make
the matrix look nice. The 2-dimensional representation (
ρ, W
) is defined by
W = C
2
, where r and s act on W as
ρ(r) =
ω 0
0 ω
2
, ρ(s) =
0 1
1 0
,
and
ω
=
e
2πi/3
is a third root of unity. We will now show that these are indeed
all the irreducible representations, by decomposing any representation into sum
of these.
So let’s decompose an arbitrary representation. Let (
ρ
0
, V
) be any complex
representation of
G
. Since
ρ
0
(
r
) has order 3, it is diagonalizable has eigenvalues
1
, ω, ω
2
. We diagonalize
ρ
0
(
r
) and then
V
splits as a vector space into the
eigenspaces
V = V (1) V (ω) V (ω
2
).
Since
srs
1
=
r
1
, we know
ρ
0
(
s
) preserves
V
(1) and interchanges
V
(
ω
) and
V (ω
2
).
Now we decompose
V
(1) into
ρ
0
(
s
) eigenspaces, with eigenvalues
±
1. Since
r
has to act trivially on these eigenspaces,
V
(1) splits into sums of copies of the
irreducible representations 1 and S.
For the remaining mess, choose a basis
v
1
, ··· , v
n
of
V
(
ω
), and let
v
0
j
=
ρ
0
(
s
)
v
j
. Then
ρ
0
(
s
) acts on the two-dimensional space
hv
j
, v
0
j
i
as
0 1
1 0
, while
ρ
0
(
r
) acts as
ω 0
0 ω
2
. This means
V
(
ω
)
V
(
ω
2
) decomposes into many copies
of W.
We did this for
D
6
by brute force. How do we generalize this? We first have
the following lemma:
Lemma. Let V, V
1
, V
2
be G-vector spaces over F. Then
(i) Hom
G
(V, V
1
V
2
)
=
Hom
G
(V, V
1
) Hom
G
(V, V
2
)
(ii) Hom
G
(V
1
V
2
, V )
=
Hom
G
(V
1
, V ) Hom
G
(V
2
, V ).
Proof. The proof is to write down the obvious homomorphisms and inverses.
Define the projection map
π
i
: V
1
V
2
V
i
,
which is the G-linear projection onto V
i
.
Then we can define the G-homomorphism
Hom
G
(V, V
1
V
2
) 7→ Hom
G
(V, V
1
) Hom
G
(V, V
2
)
ϕ 7→ (π
1
ϕ, π
2
ϕ).
Then the map (ψ
1
, ψ
2
) 7→ ψ
1
+ ψ
2
is an inverse.
For the second part, we have the homomorphism
ϕ 7→
(
ϕ|
V
1
, ϕ|
V
2
) with
inverse (ψ
1
, ψ
2
) 7→ ψ
1
π
1
+ ψ
2
π
2
.
Lemma.
Let
F
be an algebraically closed field, and
V
be a representation of
G
.
Suppose
V
=
L
n
i=1
V
i
is its decomposition into irreducible components. Then
for each irreducible representation S of G,
|{j : V
j
=
S}| = dim Hom
G
(S, V ).
This tells us we can count the multiplicity of
S
in
V
by looking at the
homomorphisms.
Proof.
We induct on
n
. If
n
= 0, then this is a trivial space. If
n
= 1, then
V
itself is irreducible, and by Schur’s lemma,
dim Hom
G
(
S, V
) = 1 if
V
=
S
, 0
otherwise. Otherwise, for n > 1, we have
V =
n1
M
i=1
V
i
!
V
n
.
By the previous lemma, we know
dim hom
G
S,
n1
M
i=1
V
i
!
V
n
!
= dim Hom
G
S,
n1
M
i=1
V
i
!
+ dim hom
G
(S, V
n
).
The result then follows by induction.
Definition
(Canonical decomposition/decomposition into isotypical compo-
nents)
.
A decomposition of
V
as
L
W
j
, where each
W
j
is (isomorphic to)
n
j
copies of the irreducible
S
j
(with
S
j
6
=
S
i
for
i 6
=
j
) is the canonical decomposition
or decomposition into isotypical components.
For an algebraically closed field F, we know we must have
n
j
= dim Hom
G
(S
j
, V ),
and hence this decomposition is well-defined.
We’ve finally all the introductory stuff. The course now begins.