Chapter 1
Compact Groups
Most infinite groups, in practice, come dressed in a natural topology, with respect to which the group operations are continuous. All the familiar groups—
in particular, all matrix groups—are locally compact; and this marks the
natural boundary of representation theory.
A topological group G is a topological space with a group structure defined
on it, such that the group operations
(x, y) 7→ xy,
x 7→ x−1
of multiplication and inversion are both continuous.
Examples:
1. The real numbers R form a topological group under addition, with the usual
topology defined by the metric
d(x, y) = |x − y|.
2. The non-zero reals R× = R \ {0} form a topological group under multiplication, under the same metric.
3. The strictly-positive reals R+ = {x ∈ R : x > 0} form a closed subgroup
of R× , and so constitute a topological group in their own right.
Remarks:
(a) Note that in the theory of topological groups, we are only concerned
with closed subgroups. When we speak of a subgroup of a topological
group, it is understood that we mean a closed subgroup, unless the
contrary is explicitly stated.
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(b) Note too that if a subgroup H ⊂ G is open then it is also closed. For
the cosets gH are all open; and so H, as the complement of the union
of all other cosets, is closed.
So for example, the subgroup R+ ⊂ R× is both open and closed.
Recall that a space X is said to be compact if it is hausdorff and every open
covering
[
X = Ui
ı∈I
has a finite subcovering:
X = Ui1 ∪ · · · ∪ Uir .
(The space X is hausdorff if given any 2 points x, y ∈ X there exist open sets
U, V ⊂ X such that
x ∈ U, y ∈ V U ∩ V = ∅.
All the spaces we meet will be hausdorff; and we will use the term ‘space’ or
‘topological space’ henceforth to mean hausdorff space.)
In fact all the groups and other spaces we meet will be subspaces of euclidean
space E n . In such a case it is usually easy to determine compactness, since a
subspace X ⊂ E n is compact if and only if
1. X is closed; and
2. X is bounded
Examples:
1. The orthogonal group
O(n) = {T ∈ Mat(n, R) : T 0 T = I}.
Here Mat(n, R) denotes the space of all n×n real matrices; and T 0 denotes
the transpose of T :
Tij0 = Tji .
2
We can identify Mat(n, R) with the Euclidean space E n , by regarding the
n2 entries tij as the coordinates of T .
2
With this understanding, O(n) is a closed subspace of E n , since it is the
set of ‘points’ satisfying the simultaneous polynomial equations making up
the matrix identity T 0 T = I. It is bounded because each entry
|tij | ≤ 1.
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In fact, for each i,
t21i + t22i + · · · + t2ni = (T 0 T )ii = 1.
Thus the orthogonal group O(n) is compact.
2. The special orthogonal group
SO(n) = {T ∈ O(n) : det T = 1}
is a closed subgroup of the compact group O(n), and so is itself compact.
Note that
T ∈ O(n) =⇒ det T = ±1,
since
T 0 T = I =⇒ det T 0 det T = 1 =⇒ (det T )2 = 1,
since det T 0 = det T . Thus O(n) splits into 2 parts: SO(n) where det T =
1; and a second part where det T = −1. If det T = −1 then it is easy to see
that this second part is just the coset T SO(n) of SO(n) in O(n).
We shall find that the groups SO(n) play a more important part in representation theory than the full orthogonal groups O(n).
3. The unitary group
U(n) = {T ∈ Mat(n, C) : T ∗ T = I}.
Here Mat(n, C) denotes the space of n × n complex matrices; and T ∗
denotes the conjugate transpose of T :
Tij∗ = Tji .
2
We can identify Mat(n, C) with the Euclidean space E 2n , by regarding
the real and imaginary parts of the n2 entries tij as the coordinates of T .
2
With this understanding, U(n) is a closed subspace of E 2n . It is bounded
because each entry has absolute value
|tij | ≤ 1.
In fact, for each i,
|t1i |2 + |t2i |2 + · · · + |tni |2 = (T ∗ T )ii = 1.
Thus the unitary group U(n) is compact.
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When n = 1,
U(1) = {x ∈ C : |x| = 1}.
Thus
U(1) = S 1 ∼
= T1 = R/Z.
Note that this group (which we can denote equally well by U(1) or T1 ) is
abelian (or commutative).
4. The special unitary group
SU(n) = {T ∈ U(n) : det T = 1}
is a closed subgroup of the compact group U(n), and so is itself compact.
Note that
T ∈ U(n) =⇒ | det T | = 1.
since
T ∗ T = I =⇒ det T ∗ det T = 1 =⇒ | det T |2 = 1,
since det T ∗ = det T .
The map
U (1) × SU(n) → U(n) : (λ, T ) 7→ λT
is a surjective homomorphism. It is not bijective, since
λI ∈ SU(n) ⇐⇒ λn = 1.
Thus the homomorphism has kernel
Cn = hωi,
where ω = e2π/n . It follows that
U(n) = (U(1) × SU(n)) /Cn .
We shall find that the groups SU(n) play a more important part in representation theory than the full unitary groups U(n).
5. The symplectic group
Sp(n) = {T ∈ Mat(n, H) : T ∗ T = I}.
Here Mat(n, H) denotes the space of n × n matrices with quaternion entries; and T ∗ denotes the conjugate transpose of T :
Tij∗ = T̄ji .
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(Recall that the conjugate of the quaternion
q = t + xi + yj + zk
is the quaternion
q̄ = t − xi − yj − zk.
Note that conjugacy is an anti-automorphism, ie
q1 q2 = q2 q1 .
It follows from this that
(AB)∗ = B ∗ A∗
for any 2 matrices A, B whose product is defined. This in turn justifies our
implicit assertion that Sp(n) is a group:
S, T ∈ Sp(n) =⇒ (ST )∗ (ST ) = T ∗ S ∗ ST = T ∗ T = I =⇒ ST ∈ Sp(n).
Note too that while multiplication of quaternions is not in general commutative, q and q̄ do commute:
q̄q = q q̄ = t2 + x2 + y 2 + z 2 = |q|2 ,
defining the norm, or absolute value, |q| of a quaternion q.)
2
We can identify Mat(n, H) with the Euclidean space E 4n , by regarding
the coefficients of 1, i, j, k in the n2 entries tij as the coordinates of T .
2
With this understanding, Sp(n) is a closed subspace of E 4n . It is bounded
because each entry has absolute value
|tij | ≤ 1.
In fact, for each i,
|t1i |2 + |t2i |2 + · · · + |tni |2 = (T ∗ T )ii = 1.
Thus the symplectic group Sp(n) is compact.
When n = 1,
Sp(1) = {q ∈ H : |q| = 1} = {t + xi + yj + zk : t2 + x2 + y 2 + z 2 = 1}.
Thus
Sp(1) ∼
= S 3.
We leave it to the reader to show that there is in fact an isomorphism
Sp(1) = SU(2).
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Although compactness is by far the most important topological property that
a group can possess, a second topological property plays a subsidiary but still
important rôle—connectivity.
Recall that the space X is said to be disconnected if it can be partitioned into
2 non-empty open sets:
X = U ∪ V,
U ∩ V = ∅.
We say that X is connected if it is not disconnected.
There is a closely related concept which is more intuitively appealing, but is
usually more difficult to work with. We say that X is pathwise-connected if given
any 2 points x, y ∈ X we can find a path π joining x to y, ie a continuous map
π : [0, 1] → X
with
π(0) = x, π(1) = y.
It is easy to see that
pathwise-connected =⇒ connected.
For if X = U ∪ V is a disconnection of X, and we choose points u ∈ U, v ∈ V ,
then there cannot be a path π joining u to v. If there were, then
I = π −1 U ∪ π −1 V
would be a disconnection of the interval [0, 1]. But it follows from the basic properties of real numbers that the interval is connected. (Suppose I = U ∪ V . We
may suppose that 0 ∈ U . Let
l = inf x ∈ V .
Then we get a contradiction whether we assume that x ∈ V or x ∈
/ V .)
Actually, for all the groups we deal with the 2 concepts of connected and
pathwise-connected will coincide. The reason for this is that all our groups will
turn out to be locally euclidean, ie each point has a neighbourhood homeomorphic
to the open ball in some euclidean space E n . This will become apparent much
later when we consider the Lie algebra of a matrix group.
We certainly will not assume this result. We mention it merely to point out
that you will not go far wrong if you think of a connected space as one in which
you can travel from any point to any other, without ‘taking off’.
The following result provides a useful tool for showing that a compact group
is connected.
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Proposition 1.1 Suppose the compact group G acts transitively on the compact
space X. Let x0 ∈ X; and let
H = S(x0 ) = {g ∈ G : gx0 = gx0 }
be the corresponding stabiliser subgroup. Then
X connected &H connected =⇒ G connected.
Proof I By a familiar argument, the action of G on X sets up a 1-1 correspondence between the cosets gH of H in G and the elements of X. In fact, let
Θ:G→X
be the map under which
g 7→ gx0 .
Then if x = gx0 ,
Θ−1 {x} = gH.
Lemma 1.1 Each coset gH is connected.
Proof of Lemma B The map
h 7→ gh : H → gH
is a continuous bijection.
But H is compact, since it is a closed subgroup of G (as H = Θ−1 {x0 }).
Now a continuous bijection φ of a compact space K onto a hausdorff space Y
is necessarily a homeomorphism. For if U ⊂ K is open, then C = K \ U is
closed and therefore compact. Hence φ(C) is compact, and therefore closed; and
so φ(U ) = Y \ φ(C) is open in Y . This shows that φ−1 is continuous, ie φ is a
homeomorphism.
Thus H ∼
= gH; and so
H connected =⇒ gH connected.
C
Now suppose (contrary to what we have to prove) that G is disconnected, say
G = U ∪ V,
U ∩ V = ∅.
This split in G will split each coset:
gH = (gH ∩ U ) ∪ (gH ∩ V ).
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But hG is connected. Hence
gH ⊂ U or gH ⊂ V.
Thus U and V are both unions of cosets; and so under Θ : G → X they define a
splitting of X:
X = ΘU ∪ ΘV, ΘU ∩ ΘV = ∅.
Since U and V are closed (as the complements of each other) and therefore compact, it follows that ΘU and ΘV are compact and therefore closed. Hence each is
also open; so X is disconnected.
This is contrary to hypothesis. We conclude that G is connected. J
Corollary 1.1 The special orthogonal group SO(n) is connected for each n.
Proof I Consider the action of SO(n) on Rn :
(T, x) 7→ T x.
This action preserves the norm:
kT xk = kxk
(where kxk2 = x0 x = x21 + · · · + x2n ). For
kT xk2 = (T x)0 T x = x0 T 0 T x = x0 x.
It follows that T sends the sphere
S n−1 = {x ∈ R : kxk = 1}
into itself. Thus SO(n) acts on S n−1 .
This action is transitive: we can find an orthogonal transformation of determinant 1 sending any point of S n−1 into any other. (The proof of this is left to the
reader.)
Moreover the space S n−1 is compact, since it is closed and bounded.
Thus the conditions of our Proposition hold. Let us take


0
 . 
 .. 


.
 0 
x0 = 
Then
1
H(x0 ) = S(x0 ) ∼
= SO(n − 1).
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For



T x0 = x0 =⇒ T = 



0
.. 
. 

T1
0 
0 ··· 0 1

where T1 ∈ SO(n − 1). (Since T x0 = x0 the last column of T consists of 0’s and
a 1. But then
t2n1 + t2n2 + · · · + 1 = 1 =⇒ tn1 = tn2 = · · · = 0.
since each row of an orthogonal matrix has norm 1.)
Our proposition shows therefore that
SO(n − 1) connected =⇒ SO(n) connected.
But
SO(1) = {I}
is certainly connected. We conclude by induction that SO(n) is connected for all
n. J
Remark: Although we won’t make use of this, our Proposition could be slightly
extended, to state that if X is connected, then the number of components of H
and G are equal.
Applying this to the full orthogonal groups O(n), we deduce that for each n
O(n) has the same number of components as O(1), namely 2. But of course this
follows from the connectedness of SO(n), since we know that O(n) splits into 2
parts, SO(n) and a coset of SO(n) (formed by the orthogonal matrices T with
det T = −1) homeomorphic to SO(n).
Corollary 1.2 The special unitary group SU(n) is connected for each n.
Proof I This follows in exactly the same way. SU(n) acts on Cn by
(T, x) 7→ T x.
This again preserves the norm
kxk = |x1 |2 + · · · |xn |2
1
2
,
since
kT xk2 = (T x)∗ T x = x∗ T ∗ T x = x∗ x = kxk2 .
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Thus SU (n) sends the sphere
S 2n−1 = {x ∈ Cn : kxk = 1}
into itself. As before, the stabiliser subgroup


0
 . 
 .. 
∼
S

 = SU(n − 1);
 0 
1
and so, again as before,
SU(n − 1) connected =⇒ SU(n) connected.
Since
SU(1) = {I}
is connected, we conclude by induction that SU(n) is connected for all n.
J
Remark: The same argument shows that the full unitary group U(n) is connected
for all n, since
U (1) = {x ∈ C : |x| = 1} = S 1
is connected.
But this also follows from the connectedness of SU(n) through the homomorphism
(λ, T ) 7→ λT : U(1) × SU(n) → U(n)
since the image of a connected set is connected (as is the product of 2 connected
sets).
Note that this homomorphism is not quite an isomorphism, since
λI ∈ SU(n) ⇐⇒ λn = 1.
It follows that
U(n) = (U(1) × SU(n)) /Cn ,
where Cn = hωi is the finite cyclic group generated by ω = e2π/n .
Corollary 1.3 The symplectic group Sp(n) is connected for each n.
Proof I The result follows in the same way from the action
(T, x) 7→ T x
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of Sp(n) on Hn . This action sends the sphere
S 4n−1 = {x ∈ Hn : kxk = 1}
into itself; and so, as before,
Sp(n − 1) connected =⇒ Sp(n) connected.
In this case we have
Sp(1) = {q = t + xi + yj + zk ∈ H : kqk2 = t2 + x2 + y 2 + z 2 = 1} ∼
= S 3.
So again, the induction starts; and we conclude that Sp(n) is connected for all n.
J
Chapter 2
Invariant integration on a compact
group
Every compact group carries a unique invariant measure. This remarkable
and beautiful result allows us to extend representation theory painlessly from
the finite to the compact case.
2.1
Integration on a compact space
There are 2 rival approaches to integration theory.
Firstly, there is what may be called the ‘traditional’ approach, in which the
fundamental notion is the measure µ(S) of a subset S.
Secondly,
there is the ‘Bourbaki’ approach, in which the fundamental notion is
R
the integral f of a function f . This approach is much simpler, where applicable,
and is the one that we shall follow.
Suppose X is a compact space. Let C(X, k) (where k = R or C) denote the
vector space of continuous functions
f : X → k.
Recall that a continuous function on a compact space is bounded and always
attains its bounds. We set
|f | = max |f (x)|
x∈X
for each function f ∈ C(X, k).
This norm defines a metric
d(f1 , f2 ) = |f1 − f2 |
on C(X, k), which in turn defines a topology on the space.
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2.1. INTEGRATION ON A COMPACT SPACE
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The metric is complete, ie every Cauchy sequence converges. This is easy to
see. If {fi } is a Cauchy sequence in C(X, k) then {fi (x)} is a Cauchy sequence
in k for each x ∈ X. Since R and C are complete metric spaces, this sequence
converges, to f (x), say; and it is a simple technical exercise to show that the limit
function f (x) is continuous, and that fi 7→ f in C(X, k).
Thus C(X, k) is a complete normed vector space—a Banach space, in short.
A measure µ on X is defined to be a continuous linear functional
µ : C(X, k) → k
(k = R or C).
More fully,
1. µ is linear, ie
µ(λ1 f1 + λ2 f2 ) = λ1 µ(f1 ) + λ2 µ(f2 );
2. µ is continuous, ie given > 0 there exists δ > 0 such that
|f | < δ =⇒ |µ(f )| < .
We often write
Z
f dµ or
X
Z
f (x) dµ(x)
X
in place of µ(f ).
Since a complex measure µ splits into real and imaginary parts,
µ = µR + iµI ,
where the measures µR and µI are real, we can safely restrict the discussion to
real measures.
Example: Consider the circle (or torus)
S 1 = T = R/Z.
We parametrise S 1 by the angle θ mod 2π. The usual measure dθ is a measure in
our sense; in fact
1 Z 2π
µ(f ) =
f (θ) dθ
2π 0
is the invariant Haar measure on the group S 1 whose existence and uniqueness on
every compact group we shall shortly demonstrate.
Another measure—a point measure—is defined by taking the value of f at a
given point, say
µ1 (f ) = f (π).
Measures can evidently be combined linearly, as for example µ2 = µ + 12 µ1 ,
ie
µ2 (f ) =
Z
0
2π
1
f (θ) dθ + f (π).
2
2.2. INTEGRATION ON A COMPACT GROUP
2.2
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Integration on a compact group
Suppose now G is a compact group. If µ is a measure on G, and g ∈ G, then we
can define a new measure gµ by
(gµ)(f ) = µ(g −1 f ) =
Z
f (gx) dg.
G
(Since we are dealing with functions on a space of functions, g is inverted twice.)
Theorem 2.1 Suppose G is a compact group. Then there exists a unique real
measure µ on G such that
1. µ is invariant on G, ie
Z
Z
(gf ) dµ =
G
f dµ
G
for all g ∈ G, f ∈ C(G, R).
2. µ is normalised so that G has volume 1, ie
Z
1 dµ = 1.
G
Moreover,
1. this measure is strictly positive, ie
f (x) ≥ 0 for all x =⇒
Z
f dµ ≥ 0,
with equality only if f = 0, ie f (g) = 0 for all g.
2.
|
Z
G
f dµ| ≤
Z
G
|f |; dµ.
Proof I
The intuitive idea. As the proof is long, and rather technical, it may help to
sketch the argument first. The basic idea is that averaging smoothes.
By an average F (x) of a function f (x) ∈ C(G) we mean a weighted average
of transforms of f , ie a function of the form
F (x) = λ1 f (g1 x) + · · · + λr f (gr x),
where
g1 , . . . , gr ∈ G, 0 ≤ λ1 , . . . , λr ≤ 1, λ1 + · · · + λr = 1.
These averages have the following properties:
2.2. INTEGRATION ON A COMPACT GROUP
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• An average of an average is an average, ie if F is an average of f , then an
average of F is also an average of f .
• If there is an invariant measure on F , then averaging leaves the integral
unchanged, ie if F is an average of f then
Z
F dg =
Z
f dg.
• Averaging smoothes, in the sense that if F is an average of f then
min f ≤ min F ≤ max F ≤ max f.
In particular, if we define the variation of f by
var f = max f − min f
then
var F ≤ var f.
Now suppose a positive invariant measure exists. Then
min f ≤
Z
f dg ≤ max f,
ie the integral of f is sandwiched between its bounds.
If f is not completely smooth, ie not constant, we can always make it smoother,
ie reduce its variation, by ‘spreading out its valleys’, as follows. Let
m = min f,
M = max f ;
and let U be the set of points where f is ‘below average’, ie
1
U = {x ∈ G : f (x) < (m + M )}.
2
The transforms of U (as of any non-empty set) cover X; for if x0 ∈ U then
x ∈ (xx−1
0 )U . Since U is open, and X is compact, a finite number of these
transforms cover X, say
X ⊂ g1 U ∪ · · · ∪ gr U.
Now consider the average
F =
1
(g1 f + · · · + gr f ) ,
r
2.2. INTEGRATION ON A COMPACT GROUP
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ie
F (x) =
1 −1
f (g1 x) + · · · + f (gr−1 x) .
r
For any x, at least one of g1−1 x, . . . , gr−1 x lies in U (since x ∈ gi U =⇒ gi−1 x ∈ U ).
Hence
1
1
F (x) <
(r − 1)M + (m + M )
r
2
1
1
= 1 − M + m.
2r
2r
Thus
1
var F < 1 −
(M − m) < var f.
2r
If we could find an average that was constant, say F (x) = c, then (always
assuming the existence of an invariant measure) we would have
Z
f dg =
Z
F dg = c
Z
1 dg = c.
An example: Let G = U(1); and let f be the saw-tooth function
f (eiθ ) = |θ| (−π < θ ≤ π).
Let g = eπi , ie rotation through half a revolution. Then
F (x) =
1
π
(f (x) + f (gx)) =
2
2
for all x ∈ U(1). So in this case, we have found a constant function; and we
deduce that if an invariant integral exists, then
Z
f dg =
Z
F dg =
π
.
2
But it is too much in general to hope that we can completely smooth a function
by averaging. However, we can expect to make the variation as small as we wish,
so that
var F = max F − min F < ,
say. But then (always assuming there is an invariant measure)
wiched between these 2 bounds,
min F ≤
Z
f dg ≤ max F.
R
f will be sand-
2.2. INTEGRATION ON A COMPACT GROUP
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R
So we can determine f as a limit in this way.
That’s the idea of the proof. Surprisingly, the most troublesome detail to fill
in
is to show that 2 different averaging limits cannot lead to different values for
R
f . For this, we have to introduce the second action of G on C(G), by right
multiplication,
(g, f ) 7→ f (xg).
This leads to a second way of averaging, using right transforms f (xg). The commutation of multiplication on the left and right allows us to play off these 2 kinds
of average against one another.
Proof proper: Suppose f ∈ C(G, R). By the argument above, we can find a
sequence of averages
F0 = f, F1 , F2 , . . .
(each an average of its predecessor) such that
var F0 > var F1 > var F2 > · · ·
(or else we reach a constant function Fr = c).
However, this does not establish that
var Fi → 0
as i → ∞. We need a slightly sharper argument to prove this. In effect we must
use the fact that f is uniformly continuous.
Recall that a function f : R → R is said to be uniformly continuous on the
interval I ⊂ R if given > 0 we can always find δ > 0 such that
|x − y| < δ =⇒ |f x − f y| < .
We can extend this concept to a function f : G → R on a compact group G as
follows: f is said to be uniformly continuous on G if given > 0 we can find an
open set U 3 e (the neutral element of G) such that
x−1 y ∈ U =⇒ |f x − f y| < .
Lemma 2.1 A continuous function on a compact group is necessarily uniformly
continuous.
Proof of Lemma B Suppose f ∈ C(G, k). For each point g ∈ G, let
1
U (g) = {x ∈ G : |f (x) − f (g)| < }.
2
2.2. INTEGRATION ON A COMPACT GROUP
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By the triangle inequality,
x, y ∈ U (g) =⇒ |f (x) − f (y)| < .
Now each neighbourhood U of g in G is expressible in the form
U = gV
where V is a neighbourhood of e in G.
Furthermore, for each neighbourhood V of e, we can find a smaller neighbourhood W of e such that
W 2 ⊂ V.
(This follows from the continuity of the multiplication (x, y) 7→ xy. Here W 2
denotes the set {w1 w2 : w1 , w2 ∈ W }.)
So for each g ∈ G we can find an open neighbourhood W (g) of e such that
gW (g)2 ⊂ U (g);
and in particular
x, y ∈ gW (g)2 =⇒ |f (x) − f (y)| < .
The open sets gW (g) cover G (since g ∈ W (g)). Therefore, since G is compact, we can find a finite subcover, say
G = g1 W1 ∪ g2 W2 ∪ · · · ∪ gr Wr ,
where Wi = W (gi ).
Let
W = ∩i Wi .
Suppose x−1 y ∈ W , ie
y ∈ xW.
Now x lies in some set gi Wi . Hence
x, y ∈ gi Wi W ⊂ gi Wi2 ;
and so
|f (x) − f (y)| < .
C
Now we observe that this open set U will serve not only for f but also for
every average F of f . For if
F = λ1 g1 f + · · · + λr gr f
(0 ≤ λ1 , . . . , λr ≤ 1, λ1 + · · · + λr = 1)
2.2. INTEGRATION ON A COMPACT GROUP
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then
|F (x) − F (y)| ≤ λ1 |f (g1−1 x) − f (g1−1 y)| + · · · + λr |f (gr−1 x) − f (gr−1 y)|.
But
(gi−1 x)−1 (gi−1 y) = x−1 gi gi−1 y = x−1 y.
Thus
x−1 y ∈ U =⇒ |f (gi−1 x) − f (gi−1 y)| < =⇒ |F (x) − F (y)| < (λ1 + · · · λr ) = .
Returning to our construction of an ‘improving average’ F , let us take =
(M − m)/2; then we can find an open set U 3 e such that
1
x−1 y ∈ U =⇒ |F (x) − F (y)| < (M − m)
2
for every average F of f . In other words, the variation of F on any transform gU
is less than half the variation of f on G.
As before, we can find a finite number of transforms of U covering G, say
G ⊂ g1 U ∪ · · · ∪ gr U.
One of these transforms, gi U say, must contain a point x0 at which F takes its
minimal value. But then, within gi U ,
1
|F (x) − F (x0 )| < (M − m);
2
and so
1
F (x) < min F + (M − m).
2
If now we form the new average
F0 =
1
(g1 F + · · · gr F )) ,
r
as before, then
r−1
1
M −m
max F +
min F +
.
max F ≤
r
r
2
0
Since min F 0 ≥ min F , it follows that
var F 0 ≤ 1 −
1
1
var F +
var f.
r
2r
2.2. INTEGRATION ON A COMPACT GROUP
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A little thought shows that this implies that
var F 0 < var F
provided
1
var f.
2
At first sight, this seems a weaker result than our earlier one, which showed that
var F 0 < var F in all cases! The difference is, that r now is independent of F .
Thus we can find a sequence of averages
var F >
F0 = f, F1 , F2 , . . .
(each an average of its predecessor) such that var Fi is decreasing to a limit `
satisfying
`≤ 1−
1
1
`+
var(f ),
r
2r
ie
`≤
1
var f.
2
In particular, we can find an average F with
var F <
2
var f.
3
Repeating the argument, with F in place of f , we find a second average F 0
such that
2
2
0
var F <
var f ;
3
and further repetition gives a new sequence of averages
F0 = f, F1 , F2 , . . . ,
with
var Fi → 0,
as required.
This sequence gives us a nest of intervals
(min f, max f ) ⊃ (min F1 , max F1 ) ⊃ (min F2 , max F2 ) · · ·
2.2. INTEGRATION ON A COMPACT GROUP
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whose lengths are tending to 0. Thus the intervals converge on a unique real
number I.
We want to set
Z
f dg = I.
But before we can do this, we must ensure that no other sequence of averages can
lead to a nest of intervals
(min f, max f ) ⊃ (min F10 , max F10 ) ⊃ (min F20 , max F20 ) · · ·
converging on a different real number I 0 6= I.
This will follow at once from the following Lemma.
Lemma 2.2 Suppose F, F 0 are two averages of f . Then
min F ≤ max F 0 .
In other words, the minimum of any average is ≤ the maximum of any other
average.
Proof of Lemma B The result would certainly hold if we could find a function F 00
which was an average both of F and of F 0 ; for then
min F ≤ min F 00 ≤ max F 00 ≤ maxF 0 .
However, it is not at all clear that such a ‘common average’ always exists. We
need a new idea.
So far we have only been considering the action of G on C(G) on the left. But
G also acts on the right, the 2 actions being independent and combining in the
action of G × G given by
((g, h)f ) (x) = f (g −1 xh).
Let us temporarily adopt the notation f h for this right action, ie
(f h)(x) = f (xh).
We can use this action to define right averages
X
µj (f hj ).
The point of introducing this complication is that we can use the right averages to
refine the left averages, and vice versa.
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Thus suppose we have a left average
F =
X
λi (gi f )
F0 =
X
µj (f hj ).
and a right average
Then we can form the joint average
F 00 =
XX
λi µj (gi f hj ).
We can regard F 00 as arising either from F by right-averaging, or from F 0 by
left averaging. In either case we conclude that F 00 is ‘smoother’ (ie has smaller
variation) than either F of F 0 ; and
min F ≤ min F 00 ≤ max F 00 ≤ max F 0 .
Thus the minimum of any left average is ≤ the maximum of any right average.
Similarly
min F 0 ≤ max F ;
the minimum of any right average is ≤ the maximum of any left average.
In fact, the second result follows from the first; since we can pass from left
averages to right averages, and vice versa, through the involution
f → f˜ : C(G) → C(G),
where
f˜(g) = f (g −1 ).
For it is readily verified that
F = λ1 (g1 f ) + · · · + λr (gr f ) =⇒ F̃ = λ1 (f˜g1−1 ) + · · · + λr (f˜gr−1 ).
Thus if F is a left average then F̃ is a right average, and vice versa.
Now suppose we have 2 left averages F1 , F2 such that
max F1 < min F2 .
Let
min F2 − max F1 = .
Let F 0 be a right average with
var F 0 = max F 0 − min F 0 < .
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Then we have a contradiction; for
min F2 ≤ max F 0 < min F 0 + ≤ max F1 + .
C
We have shown therefore that there is no ambiguity in setting
µ(f ) = I,
where I is the limit of a sequence of averages F0 = f, F1 , . . . with var Fi → 0;
for any two such sequences must converge to the same value.
It remains to show that this defines a continuous and linear function
µ : C(G, R) → R.
Let us consider linearity first. It is evident that
µ(λf ) = λµ(f ),
since multiplying f by a scalar will multiply all averages by the same number.
Suppose f1 , f2 ∈ C(G, R). Our argument above showed that the right averages of f converge on the same constant value µ(f ) = I. So now we can take a
left average of f1 and a right average of f2 , and add them to give an average of
f1 + f2 . More precisely, given > 0 we can find a left average
F1 =
X
λi gi f1
of f1 such that
µ(f1 ) − < min F1 ≤ max F1 < µ(f1 ) + ;
and similarly we can find a right average
F2 =
X
µj f2 hj
of f2 such that
µ(f2 ) − < min F2 ≤ max F2 < µ(f2 ) + .
Now let
F =
XX
i
λi µj (gi (f1 + f2 )hj ) .
j
Then we have
min F1 + min F2 ≤ min F ≤ µ(f + g) ≤ max F ≤ max F1 + max F2 ;
2.2. INTEGRATION ON A COMPACT GROUP
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from which we deduce that
µ(f + g) = µ(f ) + µ(g).
Let’s postpone for a moment the proof that µ is continuous.
It is evident that a non-negative function will have non-negative integral, since
all its averages will be non-negative:
f ≥ 0 =⇒
Z
f dµ ≥ 0.
It’s perhaps not obvious that the integral is strictly positive. Suppose f ≥ 0, and
f (g) > 0. Then we can find an open set U containing g such that
f (x) ≥ δ > 0
for x ∈ U . Now we can find g1 , . . . , gr such that
G = g1 U ∪ · · · ∪ gr U.
Let F be the average
F (x) =
1 −1
f (g1 x) + · · · + f (gr−1 x) .
r
Then
x ∈ gi U =⇒ gi−1 x ∈ U =⇒ f (gi−1 x) ≥ δ,
and so
δ
F (x) ≥ .
r
Hence
Z
δ
> 0.
r
f dg =
Z
F dg ≥
min f ≤
Z
f dµ ≤ max f,
Since
it follows at once that
|
Z
f dµ| ≤ |f |.
It is now easy to show that µ is continuous. For a linear function is continuous
if it is continuous at 0; and we have just seen that
|f | < =⇒ |
Z
f dµ| < .
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It follows at once from
min f ≤
that
Z
f dg ≤ max f
Z
| f dg| ≤ |f |.
Finally, since f and gf (for f ∈ C(G), g ∈ G) have the same transforms,
they have the same (left) averages. Hence
Z
gf dg =
Z
f dg,
ie the integral is left-invariant.
Moreover, it follows
from our construction that this is the only left-invariant
R
integral on G with 1 dg = 1; for any such integral must be sandwiched between
min F and max F for all averages F of f , and we have seen that these intervals
converge on a single real number. J
The Haar measure, by definition, is left invariant:
Z
f (g −1 x) dµ(x) =
Z
f (x) dµ(x).
It followed from our construction that it is also right invariant:
Z
f (xh) dµ(x) =
Z
f (x) dµ(x).
It is worth noting that this can be deduced directly from the existence of the Haar
measure.
Proposition 2.1 The Haar measure on a compact group G is right invariant, ie
Z
G
f (gh) dg =
Z
f (g) dg
G
(h ∈ G, f ∈ C(G, R)) .
Proof I Suppose h ∈ G. The map
µh : f 7→ µ(f h)
defines a left invariant measure on G. By the uniqueness of the Haar measure, and
the fact that
µh (1) = 1
(since the constant function 1 is right as well as left invariant),
µh = µ,
2.2. INTEGRATION ON A COMPACT GROUP
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ie µ is right invariant. J
Outline of an alternative proof Those who are fond of abstraction might prefer
the following formulation of the first part of our proof, set in the real Banach space
C(G) = C(G, R).
Let A(f ) ⊂ C(G) denote the set of averages of f . This set is convex, ie
F, F 0 ∈ A(f ) =⇒ λF + (1 − λ)F 0 ∈ A(f )
(0 ≤ λ ≤ 1).
Let Λ ⊂ C(G) denote the set of constant functions f (g) = c. Evidently
Λ∼
= R.
We want to show that
Λ ∩ A(f ) 6= ∅,
ie the closure of A(f ) contains a constant function. (In other words, we can find
a sequence of averages converging on a constant function.)
To prove this, we establish that A(f ) is pre-compact, ie its closure A(f ) is
compact. For then it will follow that there is a ‘point’ X ∈ A(f ) (ie a function
X(g)) which is closest to Λ. But if this point is not in Λ, we will reach a contradiction; for by the same argument that we used in our proof, we can always
improve on a non-constant average, ie find another average closer to Λ. (We actually need the stronger version of this using uniform continuity, since the ‘closest
point’ X(g) is not necessarily an average, but only the limit of a sequence of averages. Uniform continuity shows that we can improve all averages by a fixed
amount; so if we take an average sufficiently close to X(g) we can find another
average closer to Λ than X(g).)
It remains to show that A(f ) is pre-compact. We note in the first place that
the set of transforms of f ,
Gf = {gf : g ∈ G}
is a compact subset of C(G), since it is the image of the compact set G under the
continuous map
g 7→ gf : G → C(G).
Also, A(f ) is the convex closure of this set Gf , ie the smallest convex set
containing Gf (eg the intersection of all convex sets containing G), formed by the
points
{λ1 F1 + · · · + λr Fr : 0 ≤ λ1 , . . . , λr ≤ 1; λ1 + · · · + λr = 1}.
Thus A(f ) is the convex closure of the compact set Gf . But the convex closure of a compact set in a complete metric space is always pre-compact. That
2.2. INTEGRATION ON A COMPACT GROUP
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follows (not immediately, but by a straightforward argument) from the following
lemma in the theory of metric spaces: A subset S ⊂ X of a complete metric space
is pre-compact if and only if it can be convered by a finite number of balls of
radius ,
S ⊂ B(x1 , ) ∪ · · · ∪ B(xr , ),
for every > 0.
Accordingly, we have shown that Λ ∩ A(f ) is non-empty. We must then show
that it consists of a single point. This we
do as in our proof proper, by introducR
ing right averages. Finally, we define f dg to be this point of intersection (or
rather, the corresponding real number); and we show as before that this defines an
invariant integral µ(f ) with the required properties.
Examples:
1. As we have already noted, the Haar measure on S 1 is
1
dθ.
2π
In other words,
1 Z2
µ(f ) =
πf (θ) dθ.
2π 0
2. Consider the compact group SU(2). We know that
SU(2) ∼
= S 3,
since the general matrix in SU(2) takes the form
U=
x + iy z + it
−z + it x − iy
!
|x|2 + |y|2 + |z|2 + |t|2 = 1.
,
The usual volume on S 3 , when normalised, gives the Haar measure on
SU (2). To see that, observe that multiplication by U ∈ SU (2) defines a
distance preserving linear transformation—an isometry—of R4 , ie if
U
x + iy z + it
−z + it x − iy
!
=
x0 + iy 0 z 0 + it0
−z 0 + it0 x0 − iy 0
!
then
2
2
2
2
x0 + y 0 + z 0 + t0 = x2 + y 2 + z 2 + t2
for all (x, y, z, t) ∈ R4 .
It follows that multiplication by U preserves the volume on S 3 . In other
words, this volume provides an invariant measure on SU(2), which must
therefore be—after normalisation—the Haar measure on SU(2).
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As this example—the simplest non-abelian compact group—demonstrates,
concrete computation of the Haar measure is likely to be complicated. Fortunately, the mere existence of the Haar measure is usually sufficient for our
purpose.
Chapter 3
From finite to compact groups
Almost all the results established in Part I for finite-dimensional representations
of finite groups extend to finite-dimensional representations of compact groups.
For the Haar measure on a compact group G allows us to average over G; and our
main results were—or can be—established by averaging.
In this chapter we run very rapidly over these results, and their extension to
the compact case. This may serve (if nothing else) as a review of the main results
of finite-dimensional representation theory.
The chapter is divided into sections corresponding to the chapters of Part I, eg
section 3.5 covers the results established in chapter 5 of Part I.
We assume, unless the contrary is explicitly stated, that we are dealing with
finite-dimensional representations over k (where k = R or C). This restriction
greatly simplifies the story, for three reasons:
1. Each finite-dimensional vector space over k carries a unique hausdorff topology under which addition and scalar multiplication are continuous. If V is
n-dimensional then
V ∼
= kn;
and this unique topology on V is just that arising from the product topology
on k n .
2. If U and V are finite-dimensional vector spaces over k, then every linear
map
t:U →V
is continuous. Continuity is automatic in finite dimensions.
3. If V is a finite-dimensional vector space over k, then every subspace U ⊂ V
is closed in V .
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3.1. REPRESENTATIONS OF A COMPACT GROUP
3.1
424–II
3–2
Representations of a Compact Group
We have agreed that a representation of a topological group G in a finite-dimensional
vector space V over k (where k = R or C) is defined by a continuous linear action
G × V → V.
Recall that a representation of a finite group G in V can be defined in 2 equivalent ways:
1. by a linear action
G×V →V;
2. by a homomorphism
G → GL(V ),
where GL(V ) denotes the group of invertible linear maps t : V → V .
We again have the same choice. We have chosen (1) as our fundamental definition in the compact case, where we chose (2) in the finite case, simply because
it is a little easier to discuss the continuity of a linear action.
However, there is a natural topology on GL(V ). For we can identify GL(V )
with a subspace of the space of all linear maps t : V → V ; if dim V = n then
2
GL(V ) ⊂ Mat(n, k) ∼
= kn .
This n2 -dimensional vector space has a unique hausdorff topology, as we have
seen; and this induces a topology on GL(V ).
We know that there is a one-one correspondence between linear actions of G
on V and homomorphisms G → GL(V ). It is a straightforward matter to verify
that under this correspondence, a linear action is continuous if and only if the
corresponding homomorphism is continuous.
3.2
Equivalent Representations
The definition of the equivalence of 2 representations α, β of a group G in the
finite-dimensional vector spaces U, V over k holds for all groups, and so extends
without question to compact groups.
We note that the map θ : U → V defining such an equivalence is necessarily
continuous, since U and V are finite-dimensional. In the infinite-dimensional case
(which, we emphasise, we are not considering at the moment) we would have to
add the requirement that θ should be continuous.
3.3. SIMPLE REPRESENTATIONS
3.3
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Simple Representations
Recall that the representation α of a group G in the finite-dimensional vector space
V over k is said to be simple if no proper subspace U ⊂ V is stable under G. This
definition extends to all groups G, and in particular to compact groups.
In the infinite-dimensional case we would restrict the requirement to proper
closed subspaces of V . This is no restriction in our case, since as we have noted,
all subspaces of a finite-dimensional vector space over k are closed.
3.4
The Arithmetic of Representations
Suppose α, β are representations of the group G in the finite-dimensional vector
spaces U, V over k. We have defined the representations α + β, αβ, α∗ in the
vector spaces U ⊕ V , U ⊗ V , U ∗ , respectively. These definitions hold for all
groups G.
However, there is something to verify in the topological case, even if it is
entirely straightforward. We must show that if α and β are continuous then so are
α + β, αβ, and α∗ . (This is left as an exercise to the student.)
3.5
Semisimple Representations
The definition of the semisimplicity of a representation α of a group G in a finitedimensional vector space V over k makes no restriction on G, and so extends to
compact groups (and indeed to all topological groups); α is semisimple if and
only if it is expressible as a sum of simple representations:
α = σ1 + · · · + σm .
Recall that a finite-dimensional representation of G in V is semisimple if and
only if each stable subspace U ⊂ V has at least one stable complementary subspace W ⊂ V :
V = U ⊕ W.
We shall see later that this provides us with a definition of semisimplicity which
extends easily to infinite-dimensional representations,
3.6. EVERY REPRESENTATION OF A FINITE GROUP IS SEMISIMPLE
424–II 3–4
3.6
Every Representation of a Finite Group is Semisimple
This result is the foundation-stone of our theory; and its extension from finite to
compact groups is a triumph for Haar measure.
Let us imitate our first proof of the result in the finite case. Suppose α is a
representation of G in the finite-dimensional vector space V over k (where k =
R or C).
Recall that we start by taking any positive-definite inner product (quadratic if
k = R, hermitian if k = C) P (u, v) on V . Next we average P over G, to give a
new inner product
Z
hu, vi =
p(gu, gv) dg.
V
It is a straightforward matter to verify that this new inner product is invariant:
hgu, gvi = hu, vi.
It also follows at once from the positivity of the Haar measure that this inner
product is positive, ie
hv, vi ≥ 0.
It’s a little more difficult to see that the inner product is positive-definite, ie
hv, vi = 0 =⇒ v = 0.
However, this follows at once from the fact that the Haar measure on a compact
group is itself positive-definite, in the sense that if f (g) is a continuous function
on G such that f (g) ≥ 0 for all g ∈ G then not only is
Z
G
f (g) dg ≥ 0
(this is the positivity of the measure) but also
Z
f (g) dg = 0 =⇒ f (g) = 0 for all g.
G
This follows easily enough from the fact that if f (g0 ) = > 0, then f (g) ≥
/2 for all g ∈ U where U is an open neighbourhood of g0 . But then (since G is
compact) G can be covered by a finite number of transforms of U :
G ⊂ g1 U ∪ . . . gr U.
It follows from this that
f (g1−1 x) + · · · + f (gr−1 x) ≥ /2
3.6. EVERY REPRESENTATION OF A FINITE GROUP IS SEMISIMPLE
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for all x ∈ G. For
x ∈ gi U =⇒ gi−1 x ∈ U =⇒ f (gi−1 x) ≥ /2.
It follows from this, on integrating, that
r
Z
f (g) dg ≥ /2.
G
In particular f ≥ 0.
Note that our alternative proof of semisimplicity also carries over to the compact case. This proof depended on the fact that if
R
π:V →V
is a projection onto a stable subspace U = π(V ) of V then its average
Π=
1 X
gπg −1
|G| g∈G
is also a projection onto U ; and
W = ker Π
is a stable complementary subspace:
V = U ⊕ W.
This carries over without difficulty, although a little care is required. First we
must explain how we define the average
Π=
Z
gπg −1 dg.
G
For here we are integrating the operator-valued function
F (g) = gπg −1
However, there is little difficulty in extending the concept of measure to vectorvalued functions F on G, ie maps
F : G → V,
where V is a finite-dimensional vector space over k. This we can do, for example, by choosing a basis for V , and integrating each component of F separately.
We must show that the result is independent of the choice of basis; but that is
3.6. EVERY REPRESENTATION OF A FINITE GROUP IS SEMISIMPLE
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straightforward, The case of a function with values in hom(U, V ), where U, V are
finite-dimensional vector spaces over k, may be regarded as a particular case of
this, since we can regard hom(U, V ) as itself a vector space over k.
There is one other point that arises: in this proof (and elsewhere) we often
encounter double sums
X X
f (g, h)
g∈G h∈G
over G. The easiest way to extend such an argument to compact groups is to
consider the corresponding integral
Z
f (g, h) d(g, h)
G×G
of the continuous function f (g, h) over the product group G × G.
In such a case, let us set
F (g) =
Z
f (g, h) dh
h∈G
for each g ∈ G. Then it is readily shown that F (g) is continuous, so that we can
compute
Z
I=
F (g) dg
g∈G
But then it is not hard to see that I = I(f ) defines a second Haar measure on
G × G; so we deduce from the uniqueness of this measure that
Z
f (g, h) d(g, h) =
G×G
Z
Z
g∈G
f (g, h) dh
dg.
h∈G
This result allows us to deal with all the manipulations that arise (such as
reversal of the order of integration). For example, in our proof of the result above
that the averaged projection Π is itself a projection, we argue as follows:
Π2 =
Z
=
Z
gπg hπh−1 d(g, h)
=
Z
gg −1 hπh−1 d(g, h)
gπg −1 dg
g∈G
Z
hπh−1 dh
h∈G
−1
(g,h)∈G×G
(g,h)∈G×G
(using the fact that πgπ = gπ, since U = im π is stable under G). Thus
Π2 =
Z
hπh−1 d(g, h)
(g,h)∈G×G
=
Z
g∈G
= Π.
dg
Z
h∈G
hπh−1 dh
3.7. UNIQUENESS AND THE INTERTWINING NUMBER
3.7
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Uniqueness and the Intertwining Number
The definition of the intertwining number I(α, β) does not presuppose that G is
finite, and so extends to the compact case, as do all the results of this chapter.
3.8
The Character of a Representation
The definition of the character of a finite-dimensional representation does not depend in any way on the finiteness of the group, and so extends to the compact
case.
There is one result, however, which extends to this case, but whose proof
requires a little more thought.
Proposition 3.1 Suppose α is an n-dimensional representation of a compact group
G over R or C; and suppose g ∈ G. Let the eigenvalues of α(g) be λ1 , . . . , λn .
Then
|λi | = 1
(i = 1, . . . , n).
Proof I We know that there exists an invariant inner product hu, vi on the representationspace V . We can choose a basis for V so that
hv, vi = |x1 |2 + · · · + |xn |2 ,
where v = (x1 , . . . , xn )0 . Since α(g) leaves this form invariant for each g ∈ G,
it follows that the matrix A(g) of α(g) with respect to this basis is orthogonal if
k = R, or unitary if k = C.
The result now follows from the fact that the eigenvalues of an orthogonal or
unitary matrix all have absolute value 1:
U v = λv =⇒
=⇒
=⇒
=⇒
v ∗ U ∗ = λv ∗
v ∗ U ∗ U v = λλv ∗ v
v ∗ v = |λ|2 v ∗ v
|λ| = 1.
Hence
λ−1 = λ
for each such eigenvalue.
J
Alternative proof I Recall how we proved this in the finite case. By Lagrange’s
Theorem g m = 1 for some m > 0, for each g ∈ G. Hence
α(g)m = I;
3.8. THE CHARACTER OF A REPRESENTATION
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and so the eigenvalues of α(g) all satisfy
λm = 1.
In particular
|λ| = 1;
and so
λ−1 = λ.
We cannot say that an element g in a compact group G is necessarily of finite
order. However, we can show that the powers g n of g approach arbitrarily close to
the identity e ∈ G. (In other words, some subsequence of {g, g 2 , g 3 , . . . } tends to
e.)
For suppose not. Then we can find an open set U 3 e such that no power
of g except g 0 = e lies in U . Let V be an open neighbourhood of e such that
V V −1 ⊂ U . Then the subsets g n V are disjoint. For
x ∈ gmV ∩ gnV
=⇒ x = g m v1 = g n v2
=⇒ g n−m = v1 v2−1
=⇒ g n−m ∈ U,
contrary to hypothesis.
It follows [the details are left to the student] that the subgroup
hgi = {. . . , g −1 , e, g, g 2 , . . . }
is
1. discrete,
2. infinite, and
3. closed in G.
But this implies that G has a non-compact closed subgroup, which is impossible.
Thus we can find a subsequence
1 ≤ n1 < n 2 < . . .
such that
g ni → e
as i → ∞.
It follows that
α(g)ni → I
3.9. THE REGULAR REPRESENTATION
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3–9
as i → ∞. Hence if λ is any eigenvector of α(g) then
λni → 1.
This implies in particular that
|λ| = 1.
J
Corollary 3.1 If α is a finite-dimensional representation of a compact group over
R or C then
χα (g −1 ) = χα (g)
for all g ∈ G
Proof I Suppose the eigenvalues of α(g) are λ1 , . . . , λn . Then the eigenvalues of
−1
α(g −1 ) = α(g)−1 are λ−1
1 , . . . , λn . Thus
χα (g −1 =
=
=
=
=
=
tr α(g −1 )
−1
lambda−1
1 + · · · + λn
λ 1 + · · · + λn
λ 1 + · · · + λn
tr α(g)
χα (g).
J
3.9
The Regular Representation
Suppose G is a compact group. We denote by
C(G) = C(G, k)
(where k = R or C) the space of all continuous maps
f : G → k.
If G is discrete (in particular if G is finite) then every map f : G → k is continuous; so our definition in this case coincides with the earlier one.
If G is not finite then the vector space C(G, k) is infinite-dimensional. [We
leave the proof of this to the student.] So if we wish to extend our results from the
finite case we are forced to consider infinite-dimensional representations. We shall
3.10. INDUCED REPRESENTATIONS
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3–10
do this, rather briefly, in Chapter 7 below, when we consider the Peter-Weyl Theorem. For the moment, however, we are restricting ourselves to finite-dimensional
representations, as we have said; so in this context our results on the regular (and
adjoint) representations do not extend to the compact case.
As we shall see in Chapter 7, a compact but non-finite group G has an infinite
number of distinct simple finite-dimensional representations σ1 , σ2 , . . . . So any
argument relying on this number being finite (as for example the proof of the
fundamental result on the representations of product-groups, discussed below)
cannot be relied on in the compact case.
3.10
Induced Representations
The results of this chapter have only a limited application in the topological case,
since they apply only where we have a subgroup H ⊂ G of finite index in G; that
is, G is expressible as the union of a finite number of H-cosets:
G = g1 H ∪ · · · ∪ gr H.
In this limited case each finite-dimensional representation α of H induces a
similar representation αG of G.
For example, SO(n) is of index 2 in O(n); so each representation of SO(n)
defines a representation of O(n).
3.11
Representations of Product Groups
If α, β are finite-dimensional representations of the groups G, H in the vector
spaces U, W over k then we have defined the representation α × β of G × H
in U ⊗ W . This extends without difficulty to the topological case; and it is a
straightforward matter to verify that α × β is continuous, in the finite-dimensional
case.
Recall our main result in this context; if k = C then α × β is simple if and
only if α and β are both simple; and furthermore, every simple representation of
G × H over C arises in this way.
The proof that α × β is simple if and only if if and only if α and β are both
simple remains valid. However, our first proof that every simple representation of
G × H is of this form fails, although the result is still true.
Let us recall that proof. We argued that if G has m classes, then it has m
simple representations σ1 , . . . , σm . Similarly if H has n classes, then it has n
simple representations τ1 , . . . , τm .
3.11. REPRESENTATIONS OF PRODUCT GROUPS
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3–11
But now G × H has mn classes; and so the mn simple representations σi × τj
provide all the representations of G × H.
This argument fails in the compact case, since m and n are infinite (unless G
or H is finite).
We must turn therefore to our second proof that a simple representation γ of
G × H over C is necessarily of the form α × β. Recall that this alternative proof
was based on the natural equivalence
hom(hom(V, U ), W ) = hom(V, U ⊗ W ).
This proof does carry over to the compact case.
Suppose the representation-space of γ is the G × H-space V . Consider V as a
G-space (ie forget for the moment the action of H on V ). Let U ⊂ V be a simple
G-subspace of V . Then there exists a non-zero G-map t : V → U (since the
G-space V is semisimple). Thus the vector space
X = homG (V, U )
formed by all such G-maps is non-zero.
Now H acts naturally on X:
(ht)(v) = t(hv).
Thus X is an H-space. Let W be a simple H-subspace of X. Then there exists a
non-zero H-map u : X → W (since the H-space X is semisimple). Thus
homH (X, W ) = homH homG (V, U ), W
is non-zero. But it is readily verified that
homH homG (V, U ), W = homG×H (V, U ⊗ W ).
Thus there exists a non-zero G × H-map T : V → U × W . Since V and U ⊗ W
are both simple G × H-spaces, T must be an isomorphism:
V = U ⊗ W.
In particular
γ = α × β,
where α is the representation of G in U , and β is the representation of H in W .
Thus if G and H are compact groups then every simple representation of G×H
over C is of the form α × β.
[Can you see where we have used the fact that G and H are compact in our
argument above?]
3.12. REAL REPRESENTATIONS
3.12
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3–12
Real Representations
Everything in this chapter carries over to the compact case, with no especial problems arising.
Chapter 4
Representations of U(1)
The group U(1) goes under many names:
U(1) = SO(2) = S 1 = T1 = R/Z.
Whatever it is called, U(1) is abelian, connected and—above all—compact.
As an abelian group, every simple representation of U(1) (over C) is 1-dimensional.
Proposition 4.1 Suppose α : G → C× is a 1-dimensional representation of the
compact group G. Then
|α(g)| = 1 for all g ∈ G.
Proof I Since G is compact, so is its continuous image α(G). In particular α(G)
is bounded.
Suppose |α(g)| > 1. Then
|α(g n )| = |α(g)|n → ∞
as n → ∞, contradicting the boundedness of α(G).
On the other hand,
|α(g)| < 1 =⇒ |α(g −1 | = |α(g)|−1 > 1.
Hence |α(g)| = 1.
J
Corollary 4.1 Every 1-dimensional representation α of a compact group G is a
homomorphism of the form
α : G → U(1).
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4–1
424–II
4–2
In particular, the simple representations of U(1) are just the homomorphisms
U(1) → U(1).
But if A is an abelian group then for each n ∈ Z the map
a 7→ an : A → A
is a homomorphism.
Definition 4.1 For each n ∈ Z, we denote by En the representation
eiθ 7→ einθ
of U(1).
Proposition 4.2 The representations En are the only simple representations of
U(1).
Proof I Suppose α is a 1-dimensional representation of U(1), ie a homomorphism
α : U(1) → U(1).
Let U ⊂ U(1) be the open set
U = {eiθ : −π/2 < θ < π/2}.
Note that each g ∈ U has a unique square root in U , ie there is one and only one
h ∈ U such that h2 = g.
Since α is continuous at 1, we can find δ > 0 such that
−δ < θ < δ =⇒ α(eiθ ) ∈ U.
Choose N so large that 1/N < δ. Let ω = e2πi/N . Then α(ω) ∈ U ; while
ω N = 1 =⇒ α(ω)N = 1.
It follows that
α(ω) = e2πni/N = ω n = En (ω)
for some n ∈ Z in the range −N/2 < n < N/2. We shall deduce from this that
α = En .
Let
ω1 = eπi/N .
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4–3
Then
ω12 = ω =⇒ α(ω1 )2 = α(ω) = ω n
=⇒ α(ω1 ) = ω1n ,
since this is the unique square root of ω n in U .
Repeating this argument successively with we deduce that if
2π
ωj = e 2j N i
then
α(ωj ) = ωjn = En (ωj )
for j = 2, 3, 4, . . . .
But it follows from this that
α(ωjk ) = (ωjk )n = En (ωjk )
for k = 1, 2, 3, . . . . In other words
α(eiθ ) = E(eiθ )
for all θ of the form
k
2j
But these elements eiθ are dense in U(1). Therefore, by continuity,
θ = 2π
α(g) = En (g)
for all g ∈ U(1), ie α = En .
J
Alternative proof I Suppose
α : U(1) → U(1)
is a representation of U(1) distinct from all the En . Then
I(En , α) = 0
for all n, ie
1 Z 2π
α(eiθ )e−nθ dθ = 0.
2π 0
In other words, all the Fourier coefficients of α(eiθ ) vanish.
But this implies (from Fourier theory) that the function itself must vanish,
which is impossible since α(1) = 1. J
cn =
424–II
4–4
Remark: As this proof suggests, the representation theory of U(1) is just the
Fourier theory of periodic functions in disguise. (In fact, the whole of group representation theory might be described as a kind of generalised Fourier analysis.)
Let ρ denote the representation of U(1) in the space C(U(1)) of continuous
functions f : U(1) → C, with the usual action: if g = eiφ then
(gf )(eiθ ) = ei(θ−φ) .
The Fourier series
f (eiθ ) =
X
cn einθ
n∈Z
expresses the splitting of C(U(1)) into 1-dimensional spaces
C (U(1)) =
M
Vn ,
where
Vn = heinθ i = {ceinθ : c ∈ C}.
Notice that with our definition of group action, the space Vn carries the representation E−n , rather than En . For if g = eiφ , and f (eiθ ) = einθ , then
(gf )(eiθ ) = e−inφ f (eiθ ) = E−n (g)f (eiθ ).
In terms of representations, the splitting of C(U(1)) may be written:
ρ=
X
En .
n∈Z
We must confess at this point that we have gone ‘out of bounds’ in these remarks, since the vector space C(G) is infinite-dimensional (unless G is finite),
whereas all our results to date have been restricted to finite-dimensional representations. We shall see in Chapter 7 how we can justify this extension.
Chapter 5
Representations of SU(2)
5.1
Conjugacy in SU(n)
Since characters are class functions, our first step in studying the representations
of a compact group G—as of a finite group—is to determine how G divides into
conjugacy classes.
We know that if 2 matrices S, T ∈ GL(n, k) are similar, ie conjugate in
GL(n, k), then they will have the same eigenvalues λ1 , . . . , λn . So this gives a
necessary condition for conjugacy in any matrix group G ⊂ GL(n, k):
S ∼ T (in G) =⇒ S, T have same eigenvalues.
In general this condition is not sufficient, eg
1 1
0 1
!
6∼
1 0
0 1
!
in GL(2, C), although both matrices have eigenvalues 1, 1. However we shall
see that the condition is sufficient in each of the classical compact matrix groups
O(n), SO(n), U(n), SU(n), Sp(n).
Two remarks: Firstly, when speaking of conjugacy we must always be clear in
what group we are taking conjugates. Two matrices S, T ∈ G ⊂ GL(n, k) may
well be conjugate in GL(n, k) without being conjugate in G.
Secondly, the concepts of eigenvalue and eigenvector really belong to a representation of a group rather than the group itself. So for example, when we speak
of an eigenvalue of T ∈ U(n) we really should—though we rarely shall—say an
eigenvalue of T in the natural representation of U(n) in Cn .
Lemma 5.1 The diagonal matrices in U(n) form a subgroup isomorphic to the
torus group Tn ≡ U(1)n .
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5–1
5.1. CONJUGACY IN SU(N )
424–II
5–2
Proof I We know that the eigenvalues of T ∈ U(n) have absolute value 1, since
T v = λv =⇒
=⇒
=⇒
=⇒
=⇒
v ∗ T = λ̄v ∗
v ∗ T ∗ T v = λ̄λv ∗ v
v ∗ v = λ̄λv ∗ v
|λ|2 = λ̄λ = 1
|λ| = 1
Thus the eigenvalues of T can be written in the form
eiθ1 , . . . , eiθn
(θ1 , . . . , θn ∈ R).
In particular the diagonal matrices in U(n) are just the matrices

eiθ1




...
eiθn



It follows that the homomorphism

U(1)n → U(n) : eθ1 , . . . , eθn 7→ 


eiθ1
..

.
eiθn



maps U(1)n homeomorphically onto the diagonal subgroup of U(n), allowing us
to identify the two:
U(1)n ⊂ U(n).
J
Lemma 5.2 Every unitary matrix T ∈ U(n) is conjugate (in U(n)) to a diagonal
matrix:
T ∼ D ∈ U(1)n .
Remark: You are probably familiar with this result: Every unitary matrix can be
diagonalised by a unitary transformation. But it is instructive to give a proof in
the spirit of representation theory.
Proof I Let hT i denote the closed subgroup generated by T , ie the closure in
U (n) of the group
{. . . , T −1 , I, T, T 2 , . . . }
formed by the powers of T .
5.1. CONJUGACY IN SU(N )
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5–3
This group is abelian; and its natural representation in Cn leaves invariant the
standard positive-definite hermitian form |x1 |2 + · · · + |xn |2 , since it consists of
unitary matrices.
It follows that this representation splits into a sum of 1-dimensional representations, mutually orthogonal with respect to the standard form. If we choose a
vector ei of norm 1 in each of these 1-dimensional spaces we obtain an orthonormal set of eigenvectors of T . If U is the matrix of change of basis, ie
U = (e1 , . . . , en )
then

eiθ1

...
U ∗T U = 





eiθn
where
T ei = eθi ei .
J
Lemma 5.3 The diagonal matrices in SU(n) form a subgroup isomorphic to the
torus group Tn−1 ≡ U(1)n−1 .
Proof I If

eiθ1
..

T =


.
eiθn



then
det T = ei(θ1 +···+θn ) .
Hence
T ∈ SU(n) ⇐⇒ θ1 + · · · + θn = 0
(mod 2π).
Thus the homomorphism

n−1
U(1)
θ1
→ SU(n) : e , . . . , e
θn−1


7 
→


eiθ1
..

.
eiθn−1
e−i(θ1 +···+θn−1 )





maps U(1)n−1 homeomorphically onto the diagonal subgroup of SU(n), allowing
us to identify the two:
U(1)n−1 ⊂ SU(n).
J
5.1. CONJUGACY IN SU(N )
424–II
5–4
Lemma 5.4 Every matrix T ∈ SU(n) is conjugate (in SU(n)) to a diagonal
matrix:
T ∼ D ∈ U(1)n−1 .
Proof I From the corresponding lemma for U(n) above, T is conjugate in the full
group U(n) to a diagonal matrix:
U ∗T U = D
(U ∈ U(n)).
We know that | det U | = 1, say
det U = eiφ .
Let
V = e−iφ/n U.
Then V ∈ SU(n); and
V ∗ T V = D.
J
Lemma 5.5 Let G = U(n) or SU(n). Two matrices U, V ∈ G are conjugate if
and only if they have the same eigenvalues
{eiθ1 , eiθ2 , . . . , eiθn }.
Proof I Suppose U, V ∈ G. If U ∼ V then certainly they must have the same
eigenvalues.
Conversely, suppose U, V ∈ G have the same eigenvalues. As we have seen,
U and V are each conjugate in G to diagonal matrices:
U ∼ D1 , V ∼ D2 .
The entries in the diagonal matrices are just the eigenvalues. Thus D1 and D2
contain the same entries, perhaps permuted. So we can find a permutation matrix
P (with just one 1 in each row and column, and 0’s elsewhere) such that
D2 = P −1 D1 P.
Now P ∈ U(n) since permutation of coordinates clearly leaves the form |x1 |2 +
· · · + |xn |2 unchanged. Thus if G = U(n) we are done:
S ∼ D1 ∼ D2 ∼ T.
5.2. REPRESENTATIONS OF SU(2)
424–II
5–5
Finally, suppose G = SU(n). Then
T = U ∗ SU
for some U ∈ U(n). Suppose
det U = eiφ .
Let
V = e−iφ/n U.
Then V ∈ SU(n); and
T = V ∗ SV
J
5.2
Representations of SU(2)
Summarising the results above, as they apply to SU(2):
1. each T ∈ SU(2) has eigenvalues e±iθ
2. with the same notation,
T ∼ U (θ) =
eiθ 0
0 e−iθ
!
3. U (−θ) ∼ U (θ)
Thus SU(2) divides into classes C(θ) (for 0 ≤ θ ≤ π) containing all T with
eigenvalues e±iθ .
The classes
C(0) = {I}, C(1) = {−I},
constituting the centre of SU (2), each contain a single element; all other classes
are infinite, and intersect the diagonal subgroup in 2 elements:
C(θ) ∩ U(1) = {U (±θ)}.
Now let ρ denote the natural representation of SU(2) in C2 , defined by the
action
!
!
!
z
z0
z
7→
=T
w0
w
w
5.2. REPRESENTATIONS OF SU(2)
424–II
5–6
Explicitly, recall that the matrices T ∈ SU(2) are just those of the form
U=
a b
−b̄ ā
!
(|a|2 + |b|2 = 1)
Taking T in this form, its action is given by
(z, w) 7→ (az + bw, −b̄z + āw)
By extension, this change of variable defines an action of SU(2) on polynomials P (z, w) in z and w:
P (z, w) 7→ P (az + bw, −b̄z + āw).
Definition 5.1 For each half-integer j = 0, 1/2, 1/, 3/2, . . . we denote by Dj the
representation of SU(2) in the space
V (j) = hz 2j , z 2j−1 w, . . . , w2j i
of homogeneous polynomials in z, w of degree 2j.
Example: Let j = 3/2. The 4 polynomials
z 3 , z 2 w, zw2 , w3
form a basis for V(3/2).
Consider the action of the matrix
0 i
i 0
T =
!
∈ SU(2).
We have
T (z 3 )
T (z 2 w)
T (zw2 )
T (w3 )
=
=
=
=
(iw)3 = −iw3 ,
−izw2 ,
−iz 2 w,
−iz 3
Thus under D 3 ,
2
0 i
i 0
!




7→ 
0 0 0 −i
0 0 −i 0 


0 −i 0 0 
−i 0 0 0

5.2. REPRESENTATIONS OF SU(2)
424–II
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Proposition 5.1 The character χj of Dj is given by the following rule: Suppose
T has eigenvalues e±iθ Then
χj (T ) = e2ijθ + e2i(j−1)θ + · · · + e−2ijθ
Proof I We know that
eiθ 0
0 e−iθ
T ∼ U (θ) =
!
.
Hence
χj (T ) = χj (U (θ)) .
The result follows on considering the action of U (θ) on the basis {z 2j , . . . , w2j }
of V (j). For
U (θ)z k w2j−k =
eiθ z
k e−iθ w
2j−k
= e2i(k−j)θ z k w2j−k .
Thus under Dj ,



U (θ) →
7 


e2ijθ

e2i(j−2)θ
..
.
e−2ijθ





whence
χj (U (θ)) = e2ijθ + e2i(j−2)θ + · · · + e−2ijθ .
J
Proposition 5.2 For each half-integer j, Dj is a simple representation of SU(2),
of dimension 2j + 1.
Proof I On restricting to the diagonal subgroup U(1) ⊂ SU(2),
(Dj )U(1) = E−2j + E−2j+2 + · · · + E2j .
Since the simple parts on the right are distinct, it follows that the corresponding
expression
V (j) = hz 2j i ⊕ · · · ⊕ hw2j i
for V (j) as a direct sum of simple U(1)-modules is unique.
5.2. REPRESENTATIONS OF SU(2)
424–II
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Now suppose that V (j) splits as an SU(2)-module, say
V (j) = U ⊕ W.
If we expressed U and W as direct sums of simple U(1)-spaces, we would obtain
an expression for V (j) as a direct sum of simple U(1)-spaces. It follows from the
uniqueness of this expression that each of U and W must be the spaces spanned
by some of the monomials z a wb . In particular z 2j must belong either to U or to
W . Without loss of generality we may suppose that
z 2j ∈ U
But then
T (z 2j ) ∈ U
for all T ∈ SU(2). In particular, taking
1
T =√
2
1 1
−1 1
!
(almost any T would do) we see that
(z + w)2j = z 2j + 2jz 2j−1 w + · · · + w2j ∈ U.
Each of the monomials of degree 2j occurs here with non-zero coefficient. It
follows that each of these monomials must be in U :
z 2j−k wk ∈ U
Hence U = V (j), ie Dj is simple.
for all k.
J
Proposition 5.3 The Dj are the only simple representations of SU(2).
Proof I Suppose α is a simple representation of SU(2) distinct from the Dj .
Then in particular
I(α, Dj ) = 0.
In other words, χα is orthogonal to each χj ,
Consider the restriction of α to the diagonal subgroup U(1). Suppose
αU(1) =
X
nj Ej ,
j
where of course all but a finite number of the nj vanish (and the rest are positive
integers). It follows that
X
χα (U (θ)) =
nj eijθ
j
5.2. REPRESENTATIONS OF SU(2)
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Lemma 5.6 For any representation α of SU(2),
n−j = nj ,
ie Ej and E−j occur with the same multiplicity in αU(1) .
Proof I This follows at once from the fact that
U (−θ) ∼ U (θ)
in SU(2). J
Since n−j = nj , we see that χα (U (θ)) is expressible as a linear combination of
the χj (U (θ)) (in fact with integral—and not necessarily positive—coefficients):
χα (U (θ)) =
X
cj χj (U (θ)) .
j
Since each T ∈ SU(2) is conjugate to some U (θ) it follows that
χα (T ) =
X
cj χj (T )
j
for all T ∈ SU(2). But this contradicts the proposition that the simple characters
are linearly independent (since they are orthogonal). J
We know that every finite-dimensional representation of SU(2) is semi-simple.
In particular, each product Dj Dk is expressible as a sum of simple representations,
ie as a sum of Dn ’s.
Theorem 5.1 (The Clebsch-Gordan formula) For any pair of half-integers j, k
Dj Dk = Dj+k + Dj+k−1 + · · · + D|j−k| .
Proof I We may suppose that j ≥ k.
Suppose T has eigenvalues e±iθ . For any 2 half-integers a, b such that a ≤
b, a − b ∈ N, let
L(a, b) = e2iaθ + e2i(a+1)θ + · · · + e2ibθ .
(We may think of L(a, b) as a ‘ladder’ linking a to b on the axis, with ‘rungs’ every
step, at a + 1, a + 2, . . . .) Thus
χj (θ) = L(−j, j);
and so
χDj Dk (T ) = χj (θ)χk (θ) = L(−j, j)L(−k, k).
5.2. REPRESENTATIONS OF SU(2)
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We have to show that
L(−j, j)L(−k, k) = L(−j−k, j+k)+L(−j−k+1, j+k−1)+· · ·+L(−j+k, j−k).
We argue by induction on k. The result holds trivially for k = 0.
By our inductive hypothesis,
L(−j, j)L(, −k+1, k−1) = L(−j−k+1, j+k−1)+· · ·+L(−j+k−1, j−k+1).
Now
L(k) = L(k − 1) + (e−2ikθ + e2ikθ ).
But
L(−j, j)e−2ikθ = L(−j − k, j − k),
L(−j, j)e2ikθ = L(−j + k, j + k).
Thus
L(−j, j)(e−2ikθ + e2ikθ ) = L(−j − k, j − k) + L(−j + k, j + k)
= L(−j − k, j + k) + L(−j + k, j − k).
Gathering our ladders together,
L(−j, j)L(−k, k) = L(−j − k + 1, j + k − 1) + · · · + L(−j + k − 1, j − k + 1)
+L(−j − k, j + k) + L(−j + k, j − k)
= L(−j − k, j + k) + · · · + L(−j + k, j − k),
as required.
J
Proposition 5.4 The representation Dj of SU(2) is real for integral j and quaternionic for half-integral j.
Proof I The character
χj (θ) = e2ijθ + e2i(j−1)θ + · · · + e−2ijθ
is real, since
χj (θ) = e−2ijθ + e−2i(j−1)θ + · · · + e2ijθ = χj (θ).
Thus Dj (which we know to be simple) is either real or quaternionic.
5.2. REPRESENTATIONS OF SU(2)
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A quaternionic representation always has even dimension; for it carries an
invariant non-singular skew-symmetric form, and such a form can only exist in
even dimension, since it can be reduced to the form
x1 y2 − x2 y1 + x3 y4 − x4 y3 + · · ·
But
dim Dj = 2j + 1
is odd for integral j. Hence Dj must be real in this case.
Lemma 5.7 The representation D 1 is quaternionic.
2
Proof of Lemma B Suppose D 1 were real, say
2
D 1 = Cβ,
2
where
β : SU(2) → GL(2, R)
is a 2-dimensional representation of SU(2) over R. We know that this representation carries an invariant positive-definite form. By change of coordinates we can
bring this to x21 + x22 , so that
im β ⊂ O(2).
Moreover, since SU(2) is connected, so is its image. Hence
im β ⊂ SO(2).
Thus β defines a homomorphism
SU(2) → SO(2) = U(1),
ie a 1-dimensional representation γ of SU(2), which must in fact be D0 = 1. It
follows that β = 1 + 1, contradicting the simplicity of D 1 . C
2
Remark: It is worth noting that the representation D 1 is quaternionic in its original
2
sense, in that it arises from a representation in a quaternionic vector space. To see
this, recall that
SU(2) = Sp(1) = {q ∈ H : |q| = 1}.
The symplectic group Sp(1) acts naturally on H, by left multiplication:
(g, q) 7→ gq
(g ∈ Sp(1), q ∈ H).
5.2. REPRESENTATIONS OF SU(2)
424–II
5–12
(We take scalar multiplication in quaternionic vector spaces on the right.) It is
easy to see that this 1-dimensional representation over H gives rise, on restriction
of scalars, to a simple 2-dimensional representation over C, which must be D 1 .
2
It remains to prove that Dj is quaternionic for half-integral j > 21 . Suppose in
fact Dj were real; and suppose this were the first half-integral j with that property.
Then
Dj D1 = Dj+1 + Dj + Dj−1
would also be real (since the product of 2 real representations is real). But Dj−1
is quaternionic, by assumption, and so must appear with even multiplicity in any
real representation. This is a contradiction; so Dj must be quaternionic for all
half-integral j. J
Alternative Proof I Recall that if α is a simple representation then
Z


 1
if α is real,
0 if α is essentially complex,
χα (g ) dg =


−1 if α is quaternionic.
2
Let α = Dj . Suppose g ∈ SU(2) has eigenvalues e±iθ . Then g 2 has eigenvalues e±2iθ , and so
χj (g 2 ) = e4ijθ + e4i(j−1)θ + · · · + e−4ijθ
= χ2j (g) − χ2j−1 (g) + · · · + (−1)2j χ0 (g).
Thus
Z
χj (g 2 ) dg =
Z
χ2j (g) dg −
Z
χ2j−1 (g) dg + · · · + (−1)2j
Z
= I(1, D2j ) − I(1, D2j−1 ) + · · · + (−1)2j I(1, D0 )
= (−1)2j I(1, 1)
(
+1 if j is integral
=
−1 if j is half-integral
J
χ0 (g) dg
Chapter 6
Representations of SO(3)
Definition 6.1 A covering of one topological group G by another C is a continuous homomorphism
Θ:C→G
such that
1. ker Θ is discrete;
2. Θ is surjective, ie im Θ = G.
Proposition 6.1 A discrete subgroup is necessarily closed.
Proof I Suppose S ⊂ G is a discrete subgroup. Then by definition we can find
an open subset U ⊂ G such that
U ∩ S = {1}.
(For if S is discrete then {1} is open in the induced topology on S, ie it is the
intersection of an open set in G with S.)
We can find an open set V ⊂ G containing 1 such that
V −1 V ⊂ U,
ie v1−1 v2 ∈ U for all v1 , v2 ∈ V . This follows from the continuity of the map
(x, y) 7→ x−1 y : G × G → G.
Now suppose g ∈ G \ S. We must show that there is an open set O 3 g not
intersecting S. The open set gV 3 g contains at most 1 element of S. For suppose
s, t ∈ gV , say
s = gv1 , t = gv2 .
424–II
6–1
424–II
6–2
Then
s−1 t = v1−1 v2 ∈ U ∩ S.
Thus s−1 t = 1, ie s = t.
If gV ∩ S = ∅ then we can take O = gV . Otherwise, suppose gV ∩ S = {s}.
We can find an open set W ⊂ G such that g ∈ W, s ∈
/ W ; and then we can take
O = gV ∩ W . J
Corollary 6.1 A discrete subgroup of a compact group is necessarily finite.
Remark: We say that
Θ:C→G
is an n-fold covering if k ker Θk = n.
Proposition 6.2 Suppose Θ : C → G is a surjective (and continuous) homomorphism of topological groups. Then
1. Each representation α of G in V defines a representation α0 of C in V by
the composition
Θ
α
α0 : C → G → GL(V ).
2. If the representations α1 , α2 of G define the representations α10 , α20 of C in
this way then
α10 = α20 ⇐⇒ α1 = α2 .
3. With the same notation,
(α1 + α2 )0 = α10 + α20 , (α1 α2 )0 = α10 α20 , (α∗ )0 = (α0 )∗ .
4. A representation β of C arises in this way from a representation of G if and
only if it is trivial on ker Θ, ie
g ∈ ker Θ =⇒ β(g) = 1.
5. The representation α0 of C is simple if and only if α is simple. Moreover, if
that is so then α0 is real, quaternionic or essentially complex if and only if
the same is true of α.
6. The representation α0 of C is semisimple if and only if α is semisimple.
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Proof I All follows from the fact that gv (g ∈ G, v ∈ V ) is the same whether
defined through α or α0 . J
Remark: We can express this succinctly by saying that the representation-ring of
G is a sub-ring of the representation-ring of C:
R(G, k) ⊂ R(C, k).
We can identify a representation α of G with the corresponding representation
α0 of C; so that the representation theory of G is included, in this sense, in the
representation theory of C.
The following result allows us, by applying these ideas, to determine the representations of SO(3) from those of SU (2).
Proposition 6.3 There exists a two-fold covering
Θ : SU(2) → SO(3).
Remark: We know that SU(2) has the real 2-dimensional representation D1 , defined by a homomorphism
Θ : SU(2) → GL(3, R).
Since SU(2) is compact, the representation-space carries an invariant positivedefinite quadratic form. Taking this in the form x2 + y 2 + z 2 , we see that
im Θ ⊂ O(3).
Moreover, since SU(2) is connected, so is its image. Thus
im Θ ⊂ SO(3).
This is indeed the covering we seek; but we prefer to give a more constructive
definition.
Proof I Let H denote the space of all 2 × 2 hermitian matrices, ie all matrices of
the form
!
x
y − iz
A=
(x, y, z, t ∈ R).
y + iz
t
Evidently H is a 4-dimensional real vector space. (It is not a complex vector
space, since A hermitian does not imply that iA is hermitian; in fact
A hermitian =⇒ iA skew-hermitian,
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since (iA)∗ = −iA∗ for any A.)
Now suppose U ∈ SU(2). Then
A ∈ H =⇒ (U ∗ AU )∗ = U ∗ A∗ U ∗∗ = U ∗ AU
=⇒ U ∗ AU ∈ H.
Thus the action
(U, A) 7→ U ∗ AU = U −1 AU
of SU(2) on H defines a 4-dimensional real representation of SU(2).
This is not quite what we want; we are looking for a 3-dimensional representation. Let
T = {A ∈ H : tr A = 0}
denote the subspace of H formed by the trace-free hermitian matrices, ie those of
the form
!
x
y − iz
(x, y, z, t ∈ R).
A=
y + iz −x
These constitute a 3-dimensional real vector space; and since
tr (U ∗ AU ) = tr U −1 AU = tr A
this space is stable under the action of SU(2). Thus we have constructed a 3dimensional representation of SU(2) over R, defined by a homomorphism
Θ : SU(2) → GL(3, R).
The determinant defines a negative-definite quadratic form on T , since
det
x
y − iz
y + iz −x
!
= −x2 − y 2 − z 2
Moreover this quadratic form is left invariant by the action of SU(2), since
det (U ∗ AU ) = det U −1 AU = det A.
In other words, SU(2) acts by orthogonal transformations on T , so that
im Θ ⊂ O(3).
Moreover, since SU(2) is connected, its image must also be connected, and so
im Θ ⊂ SO(3).
We use the same symbol to denote the resulting homomorphism
Θ : SU(2) → SO(3).
We have to show that this homomorphism is a covering.
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Lemma 6.1 ker Θ = {±I}.
Proof of Lemma B Suppose U ∈ ker Θ. In other words,
U −1 AU = A
for all A ∈ T .
In fact this will hold for all hermitian matrices A ∈ H since
H=T
M
hIi.
But now the result holds also for all skew-hermitian matrices, since they are
of the form iA, with A hermitian. Finally the result holds for all matrices A ∈
GL(2, C), since every matrix is a sum of hermitian and skew-hermitian parts:
A=
1
1
(A + A∗ ) + (A − A∗ ) .
2
2
Since
U −1 AU = A ⇐⇒ AU = U A,
we are looking for matrices U which commute with all 2 × 2-matrices A. It is
readily verified that the only such matrices are the scalar multiples of the identity,
ie
U = ρI.
But now,
U ∈ SU(2) =⇒ det U = 1
=⇒ ρ2 = 1
=⇒ ρ = ±1.
C
Lemma 6.2 The homomorphism Θ is surjective.
Proof of Lemma B Let us begin by looking at a couple of examples. Suppose first
U = U (θ) =
eiθ 0
0 e−iθ
!
;
and suppose
A=
x
y − iz
y + iz −x
!
∈T.
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6–6
Then
U ∗ AU =
e−iθ 0
0 eiθ
!
x
y − iz
y + iz −x
=
x
e−2iθ (y − iz)
e2iθ (y + iz)
−x
=
X
Y − iZ
Y + iZ
−X
!
!
eiθ 0
0 e−iθ
!
!
,
where
X = x
Y = cos 2θy + sin 2θz
Z = sin 2θy − cos 2θz.
Thus U (θ) induces a rotation in the space T through 2θ about the Ox-axis, say
U (θ) 7→ R(2θ, Ox).
As another example, let
1
V =√
2
1 1
−1 1
!
;
In this case
∗
V AV =
x
y + iz
y − iz −x
!
=
X
Y + iZ
Y − iZ
−X
!
,
where
X = −y
Y = x
Z = z.
Thus Θ(V ) is a rotation through π/2 about Oz.
It is sufficient now to show that the rotations R(φ, Ox) about the x-axis, together with T = R(π/2, Oz), generate the group SO(3). Since Θ is a homomorphism, V U (θ)V −1 maps onto
T R(2θ, Ox)T −1 = R(2θ, T (Ox)) = R(2θ, Oy).
Thus im Θ contains all rotations about Ox and about Oy. It is easy to see that
these generate all rotations. For consider the rotation R(φ, l) about the axis l. We
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6–7
can find a rotation S about Ox bringing the axis l into the plane Oxz; and then a
rotation T about Oy bringing l into the coordinate axis Ox. Thus
T SR(φ, l)(T S)−1 = R(φ, Ox);
and so
R(φ, l) = S −1 T −1 R(φ, Ox)T S.
C
These 2 lemmas show that Θ defines a covering of SO(3) by SU(2).
J
Remarks:
1. We may express this result in the succinct form:
SO(3) = SO(2)/{±I}.
Recall that SU(2) ∼
= S 3 . The result shows that SO(3) is homeomorphic
to the space resulting from identifying antipodal points on the sphere S 3 .
Another way of putting this is to say that SO(3) is homeomorphic to 3dimensional real projective space:
SO(3) ∼
= P 3 (R) = (R4 \ {0})/R× .
2. We shall see in Part 4 that the space T (or more accurately the space iT ) is
just the Lie algebra of the group SU(2). Every Lie group acts on its own
Lie algebra. This is the genesis of the homomorphism Θ.
Proposition 6.4 The simple representations of SO(3) are the representations Dj
for integral j:
D0 = 1, D1 , D2 , . . .
Proof I We have established that the simple representations of SO(3) are just
those Dj which are trivial on {±I}. But under −I,
(z, w) 7→ (−z, −w)
and so if P (z, w) is a homogeneous polynomial of degree 2j,
P (−z, −w) = (−1)2j P (z, w).
Thus −I acts trivially on Vj if and only if 2j is even, ie j is integral.
The following result is almost obvious.
J
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Proposition 6.5 Let ρ be the natural representation of SO(3) is R3 . Then
Cρ = D1 .
Proof I To start with, ρ is simple. For if it were not, it would have a 1-dimensional
sub-representation. In other words, we could find a direction in R3 sent into itself
be every rotation, which is absurd.
It follows that Cρ is simple. For otherwise it would split into 2 conjugate parts,
which is impossible since its dimension is odd.
The result follows since D1 is the only simple representation of dimension 3.
J
Chapter 7
The Peter-Weyl Theorem
7.1
The finite case
Suppose G is a finite group. Recall that
C(G) = C(G, C)
denotes the banach space of maps f : G → C, with the norm
|f | = sup |f (g)|.
g∈G
(For simplicity we restrict ourselves to the case of complex scalars: k = C.)
The group G acts on C(G) on both the left and the right. These actions can be
combined to give an action of G × G:
((g, h)f ) (x) = f (g −1 xh).
Recall that the corresponding representation τ of G × G splits into simple parts
τ = σ1 ∗ ×σ1 + · · · + σs ∗ ×σs
where σ1 , . . . , σs are the simple representations of G (over C).
Suppose V is a G-space. We have a canonical isomorphism
hom(V, V ) = V ∗ ⊗ V.
Thus G × G acts on hom(V, V ), with the first factor acting on V ∗ and the second
on V . A little thought shows that this action can be defined as follows. Suppose
t : V → V is a linear map, ie an element of hom(V, V ). Then
(g, h)t = t0
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7.1. THE FINITE CASE
424–II
7–2
where t0 is the linear map
t0 (v) = ht(g −1 v).
The expression for τ above can be re-written as
C(G) ≡ hom(Vσ1 , Vσ1 ) + · · · + hom(Vσs , Vσs ),
where Vσ is the space carrying the simple representation σ.
In other words
C(G) = C(G)σ1 ⊕ · · · ⊕ C(G)σs ,
where
C(G)σ ≡ hom(Vσ , Vσ ).
Since the representations σ ∗ × σ of G × G are simple and distinct, it follows that
the subspaces C(G)σ ⊂ C(G) are the isotypic components of C(G).
If we pass to the (perhaps more familiar) regular representation of G in C(G)
by restricting to the subgroup e × G ⊂ G × G, so that G acts on C(G) by
(gf )(x) = f (g −1 x),
then each subspace V ∗ ⊗ V is isomorphic (as a G-space) to dim σV . Thus it
remains isotypic, while ceasing (unless dim σ = 1) to be simple. It follows that
the expression
C(G) = C(G)σ1 ⊕ · · · ⊕ C(G)σs ,
can equally well be regarded as the splitting of the G-space C(G) into its isotypic
parts.
Whichever way we look at it, we see that each function f (x) on G splits into
components fσ (x) corresponding to the simple representations σ of G.
What exactly is this component fσ (x) of f (x)? Well, recall that the projection
π of the G-space V onto its σ-component Vσ is given by
π=
1 X
χ(g −1 )g.
|G| g∈G
It follows that
fσ (x) =
1 X
χ(g)f (gx).
|G| g∈G