15Representations of compact groups
II Representation Theory
15 Representations of compact groups
We now consider the case of infinite groups. It turns out infinite groups don’t
behave well if we don’t impose additional structures. When talking about
representations of infinite groups, things work well only if we impose a topological
structure on the group, and require the representations to be continuous. Then
we get nice results if we only concentrate on the groups that are compact. This
is a generalization of our previous results, since we can give any finite group the
discrete topology, and then all representations are continuous, and the group is
compact due to finiteness.
We start with the definition of a topological group.
Definition
(Topological group)
.
A topological group is a group
G
which is also
a topological space, and for which multiplication
G ×G → G
((
h, g
)
7→ hg
) and
inverse G → G (g 7→ g
−1
) are continuous maps.
Example.
Any finite group
G
with the discrete topology is a topological group.
Example.
The matrix groups
GL
n
(
R
) and
GL
n
(
C
) are topological groups
(inheriting the topology of R
n
2
and C
n
2
).
However, topological groups are often too huge. We would like to concentrate
on “small” topological groups.
Definition
(Compact group)
.
A topological group is a compact group if it is
compact as a topological space.
There are some fairly nice examples of these.
Example. All finite groups with discrete topology are compact.
Example.
The group
S
1
=
{z ∈ C
:
|z|
= 1
}
under multiplication is a compact
group. This is known as the circle group, for mysterious reasons. Thus the torus
S
1
× S
1
× ··· × S
1
is also a compact group.
Example.
The orthogonal group O(
n
)
≤ GL
n
(
R
) is compact, and so is
SO
(
n
) =
{A ∈
O(
n
) :
det A
= 1
}
. These are compact since they are closed and bounded
as a subspace of
R
n
2
, as the entries can be at most 1 in magnitude. Note that
SO(2)
∼
=
S
1
, mapping a rotation by θ to e
iθ
.
Example.
The unitary group U(
n
) =
{A ∈ GL
n
(
C
) :
AA
†
=
I}
is compact,
and so is SU(n) = {A ∈ U(n) : det A = 1}.
Note that we also have
U
(1)
∼
=
SO
(2)
∼
=
S
1
. These are not only isomorphisms
as group, but these isomorphisms are also homeomorphisms of topological spaces.
The one we are going to spend most of our time on is
SU(2) = {(z
1
, z
2
) ∈ C
2
: |z
1
|
2
+ |z
2
|
2
= 1} ≤ R
4
= C
2
.
We can see this is homeomorphic to S
3
.
It turns out that the only spheres on which we can define group operations
on are
S
1
and
S
3
, but we will not prove this, since this is a course on algebra,
not topology.
We now want to develop a representation theory of these compact groups,
similar to what we’ve done for finite groups.
Definition
(Representation of topological group)
.
A representation of a topolog-
ical group on a finite-dimensional space
V
is a continuous group homomorphism
G → GL(V ).
It is important that the group homomorphism is continuous. Note that for a
general topological space
X
, a map
ρ
:
X → GL
(
V
)
∼
=
GL
n
(
C
) is continuous if
and only if the component maps x 7→ ρ(x)
ij
are continuous for all i, j.
The key that makes this work is the compactness of the group and the
continuity of the representations. If we throw them away, then things will go
very bad. To see this, we consider the simple example
S
1
= U(1) = {g ∈ C
×
: |g| = 1}
∼
=
R/Z,
where the last isomorphism is an abelian group isomorphism via the map
x 7→ e
2πix
: R/Z → S
1
.
What happens if we just view
S
1
as an abstract group, and not a topological
group? We can view
R
as a vector space over
Q
, and this has a basis (by Hamel’s
basis theorem), say,
A ⊆ R
. Moreover, we can assume the basis is ordered, and
we can assume 1 ∈ A.
As abelian groups, we then have
R
∼
=
Q ⊕
M
α∈A\{1}
Qα,
Then this induces an isomorphism
R/Z
∼
=
Q/Z ⊕
M
α∈A\{1}
Qα.
Thus, as abstract groups,
S
1
has uncountably many irreducible representations,
namely for each
λ ∈ A\{
1
}
, there exists a one-dimensional representation given
by
ρ
λ
(e
2πiµ
) =
(
1 µ 6∈ Qλ
e
2πiµ
µ ∈ Qλ
We see
ρ
λ
=
ρ
λ
0
if and only if
Qλ
=
Qλ
0
. Hence there are indeed uncountably
many of these. There are in fact even more irreducible representation we haven’t
listed. This is bad.
The idea is then to topologize
S
1
as a subset of
C
, and study how it acts
naturally on complex spaces in a continuous way. Then we can ensure that we
have at most countably many representations. In fact, we can characterize all
irreducible representations in a rather nice way:
Theorem.
Every one-dimensional (continuous) representation
S
1
is of the form
ρ : z 7→ z
n
for some n ∈ Z.
This is a nice countable family of representations. To prove this, we need
two lemmas from real analysis.
Lemma.
If
ψ
: (
R,
+)
→
(
R,
+) is a continuous group homomorphism, then
there exists a c ∈ R such that
ψ(x) = cx
for all x ∈ R.
Proof.
Given
ψ
: (
R,
+)
→
(
R,
+) continuous, we let
c
=
ψ
(1). We now claim
that ψ(x) = cx.
Since ψ is a homomorphism, for every n ∈ Z
≥0
and x ∈ R, we know
ψ(nx) = ψ(x + ··· + x) = ψ(x) + ··· + ψ(x) = nψ(x).
In particular, when x = 1, we know ψ(n) = cn. Also, we have
ψ(−n) = −ψ(n) = −cn.
Thus ψ(n) = cn for all n ∈ Z.
We now put x =
m
n
∈ Q. Then we have
mψ(x) = ψ(nx) = ψ(m) = cm.
So we must have
ψ
m
n
= c
m
n
.
So we get
ψ
(
q
) =
cq
for all
q ∈ Q
. But
Q
is dense in
R
, and
ψ
is continuous. So
we must have ψ(x) = cx for all x ∈ R.
Lemma. Continuous homomorphisms ϕ : (R, +) → S
1
are of the form
ϕ(x) = e
icx
for some c ∈ R.
Proof.
Let
ε
: (
R,
+)
→ S
1
be defined by
x 7→ e
ix
. This homomorphism wraps
the real line around S
1
with period 2π.
We now claim that given any continuous function
ϕ
:
R → S
1
such that
ϕ
(0) = 1, there exists a unique continuous lifting homomorphism
ψ
:
R → R
such that
ε ◦ ψ = ϕ, ψ(0) = 0.
(R, +)
0
(R, +)
S
1
ε
ϕ
∃!ψ
The lifting is constructed by starting with
ψ
(0) = 0, and then extending a small
interval at a time to get a continuous map
R → R
. We will not go into the
details. Alternatively, this follows from the lifting criterion from IID Algebraic
Topology.
We now claim that if in addition
ϕ
is a homomorphism, then so is its
continuous lifting
ψ
. If this is true, then we can conclude that
ψ
(
x
) =
cx
for
some c ∈ R. Hence
ϕ(x) = e
icx
.
To show that
ψ
is indeed a homomorphism, we have to show that
ψ
(
x
+
y
) =
ψ(x) + ψ(y).
By definition, we know
ϕ(x + y) = ϕ(x)ϕ(y).
By definition of ψ, this means
ε(ψ(x + y) − ψ(x) − ψ(y)) = 1.
We now look at our definition of ε to get
ψ(x + y) − ψ(x) − ψ(y) = 2kπ
for some integer
k ∈ Z
, depending continuously on
x
and
y
. But
k
can only be
an integer. So it must be constant. Now we pick our favorite
x
and
y
, namely
x = y = 0. Then we find k = 0. So we get
ψ(x + y) = ψ(x) + ψ(y).
So ψ is a group homomorphism.
With these two lemmas, we can prove our characterization of the (one-
dimensional) representations of S
1
.
Theorem.
Every one-dimensional (continuous) representation
S
1
is of the form
ρ : z 7→ z
n
for some n ∈ Z.
Proof.
Let
ρ
:
S
1
→ C
×
be a continuous representation. We now claim that
ρ
actually maps
S
1
to
S
1
. Since
S
1
is compact, we know
ρ
(
S
1
) has closed and
bounded image. Also,
ρ(z
n
) = (ρ(z))
n
for all
n ∈ Z
. Thus for each
z ∈ S
1
, if
|ρ
(
z
)
| >
1, then the image of
ρ
(
z
n
) is
unbounded. Similarly, if it is less than 1, then
ρ
(
z
−n
) is unbounded. So we must
have ρ(S
1
) ⊆ S
1
. So we get a continuous homomorphism
R → S
1
x 7→ ρ(e
ix
).
So we know there is some c ∈ R such that
ρ(e
ix
) = e
icx
,
Now in particular,
1 = ρ(e
2πi
) = e
2πic
.
This forces c ∈ Z. Putting n = c, we get
ρ(z) = z
n
.
That has taken nearly an hour, and we’ve used two not-so-trivial facts from
analysis. So we actually had to work quite hard to get this. But the result is
good. We have a complete list of representations, and we don’t have to fiddle
with things like characters.
Our next objective is to repeat this for
SU
(2), and this time we cannot just
get away with doing some analysis. We have to do representation theory properly.
So we study the general theory of representations of complex spaces.
In studying finite groups, we often took the “average” over the group via
the operation
1
|G|
P
g∈G
something
. Of course, we can’t do this now, since the
group is infinite. As we all know, the continuous analogue of summing is called
integration. Informally, we want to be able to write something like
R
G
dg.
This is actually a measure, called the “Haar measure”, and always exists as
long as you are compact and Hausdorff. However, we will not need or use this
result, since we can construct it by hand for S
1
and SU(2).
Definition (Haar measure). Let G be a topological group, and let
C(G) = {f : G → C : f is continuous, f (gxg
−1
) = f(x) for all g, x ∈ G}.
A non-trivial linear functional
R
G
: C(G) → C, written as
Z
G
f =
Z
G
f(g) dg
is called a Haar measure if
(i) It satisfies the normalization condition
Z
G
1 dg = 1
so that the “total volume” is 1.
(ii) It is left and right translation invariant, i.e.
Z
G
f(xg) dg =
Z
G
f(g) dg =
Z
G
f(gx) dg
for all x ∈ G.
Example. For a finite group G, we can define
Z
G
f(g) dg =
1
|G|
X
g∈G
f(g).
The division by the group order is the thing we need to make the normalization
hold.
Example. Let G = S
1
. Then we can have
Z
G
f(g) dg =
1
2π
Z
2π
0
f(e
iθ
) dθ.
Again, division by 2π is the normalization required.
We have explicitly constructed them here, but there is a theorem that
guarantees the existence of such a measure for sufficiently nice topological group.
Theorem.
Let
G
be a compact Hausdorff topological group. Then there exists
a unique Haar measure on G.
From now on, the word “compact” will mean “compact and Hausdorff”, so
that the conditions of the above theorem holds.
We will not prove theorem, and just assume it to be true. If you don’t like
this, whenever you see “compact group”, replace it with “compact group with
Haar measure”. We will explicitly construct it for
SU
(2) later, which is all we
really care about.
Once we know the existence of such a thing, most of the things we know for
finite groups hold for compact groups, as long as we replace the sums with the
appropriate integrals.
Corollary
(Weyl’s unitary trick)
.
Let
G
be a compact group. Then every
representation (ρ, V ) has a G-invariant Hermitian inner product.
Proof.
As for the finite case, take any inner product (
·, ·
) on
V
, then define a
new inner product by
hv, wi =
Z
G
(ρ(g)v, ρ(g)w) dg.
Then this is a G-invariant inner product.
Theorem
(Maschke’s theoerm)
.
Let
G
be compact group. Then every repre-
sentation of G is completely reducible.
Proof.
Given a representation (
ρ, V
). Choose a
G
-invariant inner product. If
V
is not irreducible, let
W ≤ V
be a subrepresentation. Then
W
⊥
is also
G-invariant, and
V = W ⊕ W
⊥
.
Then the result follows by induction.
We can use the Haar measure to endow
C
(
G
) with an inner product given by
hf, f
0
i =
Z
G
f(g)f
0
(g) dg.
Definition
(Character)
.
If
ρ
:
G → GL
(
V
) is a representation, then the char-
acter
χ
ρ
=
tr ρ
is a continuous class function, since each component
ρ
(
g
)
ij
is
continuous.
So characters make sense, and we can ask if all the things about finite groups
hold here. For example, we can ask about orthogonality.
Schur’s lemma also hold, with the same proof, since we didn’t really need
finiteness there. Using that, we can prove the following:
Theorem
(Orthogonality)
.
Let
G
be a compact group, and
V
and
W
be
irreducible representations of G. Then
hχ
V
, χ
W
i =
(
1 V
∼
=
W
0 V 6
∼
=
W
.
Now do irreducible characters form a basis for C(G)?
Example.
We take the only (infinite) compact group we know about —
G
=
S
1
.
We have found that the one-dimensional representations are
ρ
n
: z 7→ z
n
for
n ∈ Z
. As
S
1
is abelian, these are all the (continuous) irreducible representa-
tions — given any representation
ρ
, we can find a simultaneous eigenvector for
each ρ(g).
The “character table of
S
1
” has rows
χ
n
indexed by
Z
, with
χ
n
(
e
iθ
) =
e
inθ
.
Now given any representation of
S
1
, say
V
, we can break
V
as a direct sum
of 1-dimensional subrepresentations. So the character χ
V
of V is of the form
χ
V
(z) =
X
n∈Z
a
n
z
n
,
where
a
n
∈ Z
≥0
, and only finitely many
a
n
are non-zero (since we are assuming
finite-dimensionality).
Actually,
a
n
is the number of copies of
ρ
n
in the decomposition of
V
. We
can find out the value of a
n
by computing
a
n
= hχ
n
, χ
V
i =
1
2π
Z
2π
0
e
−inθ
χ
V
(e
iθ
) dθ.
Hence we know
χ
V
(e
iθ
) =
X
n∈Z
1
2π
Z
2π
0
χ
V
(e
iθ
0
)e
−inθ
0
dθ
0
e
inθ
.
This is really just a Fourier decomposition of
χ
V
. This gives a decomposition
of
χ
V
into irreducible characters, and the “Fourier mode”
a
n
is the number of
each irreducible character occurring in this decomposition.
It is possible to show that
χ
n
form a complete orthonormal set in the Hilbert
space
L
2
(
S
1
), i.e. square-integrable functions on
S
1
. This is the Peter-Weyl
theorem, and is highly non-trivial.