Compact Groups

8/3/2019 Compact Groups

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Chapter 1

Compact Groups

Most infinite groups, in practice, come dressed in a natural topology, with re-

spect to which the group operations are continuous. All the familiar groups

in particular, all matrix groupsare locally compact; and this marks the

natural boundary of representation theory.

A topological group G is a topological space with a group structure definedon it, such that the group operations

(x, y) xy, x x1

of multiplication and inversion are both continuous.

Examples:

1. The real numbersR 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 multipli-cation, under the same metric.

3. The strictly-positive reals R+ = {x R : x > 0} form a closed subgroupofR, 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 closedsubgroups. 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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424II 12

(b) Note too that if a subgroup H

G is open then it is also closed. Forthe cosets gH are all open; and so H, as the complement of the unionof 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 =I

Ui

has a finite subcovering:

X = Ui1

Uir .

(The space X is hausdorff if given any 2 points x, y X there exist open setsU, 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 En. In such a case it is usually easy to determine compactness, since asubspace X En 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) : TT = I}.

HereMat(n,R) denotes the space of all nn real matrices; and T denotesthe transpose ofT:

Tij = Tji.

We can identifyMat(n,R) with the Euclidean space En2

, by regarding the

n2 entries tij as the coordinates ofT.

With this understanding, O(n) is a closed subspace of En2

, since it is the

set of points satisfying the simultaneous polynomial equations making up

the matrix identity TT = I. It is boundedbecause each entry

|tij| 1.


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424II 13

In fact, for each i,

t21i + t22i + + t2ni = (TT)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

TT = I = det T det T = 1 = (det T)2 = 1,since det T = det T. Thus O(n) splits into 2 parts: SO(n) where det T =1; and a second part where det T = 1. Ifdet T = 1 then it is easy to seethat this second part is just the coset TSO(n) ofSO(n) in O(n).

We shall find that the groups SO(n) play a more important part in represen-tation theory than the full orthogonal groups O(n).

3. The unitary group

U(n) = {T Mat(n,C) : TT = I}.Here Mat(n,C) denotes the space of n n complex matrices; and Tdenotes the conjugate transpose ofT:

Tij = Tji.

We can identify Mat(n,C) with the Euclidean space E2n2

, by regarding

the real and imaginary parts of the n2 entries tij as the coordinates ofT.

With this understanding, U(n) is a closedsubspace ofE2n2 . It is boundedbecause each entry has absolute value

|tij| 1.In fact, for each i,

|t1i|2 + |t2i|2 + + |tni|2 = (TT)ii = 1.Thus the unitary group U(n) is compact.


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424II 14

When n = 1,U(1) = {x C : |x| = 1}.

Thus

U(1) = S1 = T1 = R/Z.Note that this group (which we can denote equally well by U(1) or T1) isabelian (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

TT = I = det T det T = 1 = | det T|2 = 1,since det T = det T.

The map

U(1) SU(n) U(n) : (, T) Tis a surjective homomorphism. It is not bijective, since

I SU(n) n = 1.Thus the homomorphism has kernel

Cn = ,where = e2/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 repre-sentation theory than the full unitary groups U(n).

5. The symplectic group

Sp(n) = {T Mat(n,H) : TT = I}.Here Mat(n,H) denotes the space of n n matrices with quaternion en-tries; and T denotes the conjugate transpose ofT:

Tij = Tji.


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(Recall that the conjugate of the quaternion

q = t + xi + yj + zk

is the quaternion

q = t xi yj zk.Note that conjugacy is an anti-automorphism, ie

q1q2 = q2q1.

It follows from this that

(AB) = BA

for any 2 matrices A, B whose product is defined. This in turn justifies ourimplicit assertion that Sp(n) is a group:

S, T Sp(n) = (ST)(ST) = TSST = TT = I = ST Sp(n).

Note too that while multiplication of quaternions is not in general commu-

tative, q and q do commute:

qq = qq = t2 + x2 + y2 + z2 = |q|2,defining the norm, or absolute value,

|q

|of a quaternion q.)

We can identify Mat(n,H) with the Euclidean space E4n2 , by regardingthe coefficients of1, i , j , k in the n2 entries tij as the coordinates ofT.

With this understanding, Sp(n) is a closed subspace ofE4n2

. It is bounded

because each entry has absolute value

|tij| 1.In fact, for each i,

|t1i|2 + |t2i|2 + + |tni|2 = (TT)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 + y2 + z2 = 1}.Thus

Sp(1) = S3.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 roleconnectivity.

Recall that the space X is said to be disconnected if it can be partitioned into2 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 joining x to y, ie a continuous map : [0, 1] X

with

(0) = x, (1) = y.

It is easy to see that

pathwise-connected = connected.For ifX = U V is a disconnection of X, and we choose points u U, v V,then there cannot be a path joining u to v. If there were, then

I = 1U 1Vwould be a disconnection of the interval [0, 1]. But it follows from the basic prop-erties of real numbers that the interval is connected. (Suppose I = U V. Wemay suppose that 0 U. Let

l = infx 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-connectedwill 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 En. This will become apparent muchlater 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 compactspace X. Letx0 X; and let

H = S(x0) = {g G : gx0 = gx0}be the corresponding stabiliser subgroup. Then

X connected &H connected = G connected.

Proof By a familiar argument, the action of G on X sets up a 1-1 correspon-dence between the cosets gH ofH in G and the elements ofX. In fact, let

: G Xbe the map under which

g gx0.Then ifx = gx0,

1{x} = gH.Lemma 1.1 Each cosetgH is connected.

Proof of Lemma The map

h gh : H gHis a continuous bijection.

But H is compact, since it is a closed subgroup of G (as H = 1{x0}).Now a continuous bijection of a compact space K onto a hausdorff space Yis necessarily a homeomorphism. For if U K is open, then C = K \ U isclosed and therefore compact. Hence (C) is compact, and therefore closed; andso (U) = Y \ (C) is open in Y. This shows that 1 is continuous, ie is ahomeomorphism.

Thus H = gH; and soH connected = gH connected.

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 : G X they define asplitting ofX:

X = U V, U V = .Since U and V are closed (as the complements of each other) and therefore com-pact, it follows that U and V are compact and therefore closed. Hence each isalso open; so X is disconnected.

This is contrary to hypothesis. We conclude that G is connected.

Corollary 1.1 The special orthogonal group SO(n) is connected for each n.

Proof Consider the action ofSO(n) on Rn:

(T, x) T x.This action preserves the norm:

T x = x(where

x2 = xx = x21 +

+ x2n). For

T x2 = (T x)T x = xTT x = xx.It follows that T sends the sphere

Sn1 = {x R : x = 1}into itself. Thus SO(n) acts on Sn1.

This action is transitive: we can find an orthogonal transformation of determi-

nant 1 sending any point of Sn1 into any other. (The proof of this is left to thereader.)

Moreover the space Sn1

is compact, since it is closed and bounded.Thus the conditions of our Proposition hold. Let us take

x0 =

0...

01

.

Then

H(x0) = S(x0) = SO(n 1).


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For

T x0 = x0 = T =

0

T1...

00 0 1

where T1 SO(n 1). (Since T x0 = x0 the last column ofT consists of 0s anda 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 alln.

Remark: Although we wont make use of this, our Proposition could be slightly

extended, to state that if X is connected, then the number of components of Hand 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 thisfollows from the connectedness ofSO(n), since we know that O(n) splits into 2parts, SO(n) and a coset ofSO(n) (formed by the orthogonal matrices T withdet T = 1) homeomorphic to SO(n).

Corollary 1.2 The special unitary group SU(n) is connected for each n.

Proof This follows in exactly the same way. SU(n) acts on Cn by

(T, x) T x.

This again preserves the norm

x =|x1|2 + |xn|2

12 ,

since

T x2 = (T x)T x = xTT x = xx = x2.


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Thus SU(n) sends the sphere

S2n1 = {x Cn : x = 1}

into itself. As before, the stabiliser subgroup

S

0...

01

= SU(n 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.

Remark: The same argument shows that the full unitary group U(n) is connectedfor all n, since

U(1) = {x C : |x| = 1} = S1

is connected.

But this also follows from the connectedness ofSU(n) through the homomor-phism

(, T) T : 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

I SU(n) n = 1.

It follows thatU(n) = (U(1) SU(n)) /Cn,

where Cn = is the finite cyclic group generated by = e2/n.

Corollary 1.3 The symplectic group Sp(n) is connected for each n.

Proof The result follows in the same way from the action

(T, x) T x


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ofSp(n) on Hn. This action sends the sphere

S4n1 = {x Hn : x = 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 : q2 = t2 + x2 + y2 + z2 = 1} = S3.

So again, the induction starts; and we conclude that Sp(n) is connected for all n.


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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 (S) of a subset S.Secondly, there is the Bourbaki approach, in which the fundamental notion is

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 thevector 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| = maxxX

|f(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 424II 22

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 sequencein k for each x X. Since R and C are complete metric spaces, this sequenceconverges, to f(x), say; and it is a simple technical exercise to show that the limitfunction f(x) is continuous, and that fi f in C(X, k).

Thus C(X, k) is a complete normed vector spacea Banach space, in short.A measure on X is defined to be a continuous linear functional

: C(X, k) k (k = R or C).More fully,

1. is linear, ie

(1f1 + 2f2) = 1(f1) + 2(f2);

2. is continuous, ie given > 0 there exists > 0 such that

|f| < = |(f)| < .We often write

Xf d or

X

f(x) d(x)

in place of(f).Since a complex measure splits into real and imaginary parts,

= R + iI,where the measures R and I are real, we can safely restrict the discussion toreal measures.

Example: Consider the circle (or torus)

S1 = T = R/Z.

We parametrise S1 by the angle mod 2. The usual measure d is a measure inour sense; in fact

(f) =1

2

20

f() d

is the invariant Haar measure on the group S1

whose existence and uniqueness onevery compact group we shall shortly demonstrate.

Another measurea point measureis defined by taking the value of f at agiven point, say

1(f) = f().

Measures can evidently be combined linearly, as for example 2 = +12

1,ie

2(f) =20

f() d +1

2f().


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2.2. INTEGRATION ON A COMPACT GROUP 424II 23

2.2 Integration on a compact group

Suppose now G is a compact group. If is a measure on G, and g G, then wecan define a new measure g by

(g)(f) = (g1f) =G

f(gx) dg.

(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 realmeasure on G such that

1. is invariant on G, ie G

(gf) d =G

f d

for all g G, f C(G,R).2. is normalised so thatG has volume 1, ie

G1 d = 1.

Moreover,

1. this measure is strictly positive, ie

f(x) 0 for all x =

f d 0,

with equality only if f = 0, ie f(g) = 0 for all g.

2.

|G

f d| G

|f|; d.

Proof

The intuitive idea. As the proof is long, and rather technical, it may help tosketch 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 averageof transforms off, ie a function of the form

F(x) = 1f(g1x) + + rf(grx),where

g1, . . . , gr G, 0 1, . . . , r 1, 1 + + r = 1.These averages have the following properties:


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An average of an average is an average, ie if F is an average of f, then anaverage ofF is also an average off.

If there is an invariant measure on F, then averaging leaves the integralunchanged, ie ifF is an average off then

F dg =

f dg.

Averaging smoothes, in the sense that if F is an average off then

min f

min F

max F

max f.

In particular, if we define the variation off by

var f = max f min f

then

var F var f.Now suppose a positive invariant measure exists. Then

min f

f dg

max f,

ie the integral off is sandwiched between its bounds.Iff 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

U = {x G : f(x) < 12

(m + M)}.

The transforms of U (as of any non-empty set) cover X; for if x0 U thenx (xx10 )U. Since U is open, and X is compact, a finite number of thesetransforms cover X, say

X g1U grU.Now consider the average

F =1

r(g1f + + grf) ,


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ie

F(x) =1

r

f(g11 x) + + f(g1r x)

.

For any x, at least one ofg11 x , . . . , g1r x lies in U (since x giU = g1i x U).

Hence

F(x) 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 mustuse the fact that f is uniformly continuous.

Recall that a function f : R R is said to be uniformly continuous on theinterval I R if given > 0 we can always find > 0 such that

|x y| < = |f x f y| < .

We can extend this concept to a function f : G R on a compact group G asfollows: f is said to be uniformly continuous on G if given > 0 we can find anopen set U e (the neutral element ofG) such that

x1y U = |f x f y| < .

Lemma 2.1 A continuous function on a compact group is necessarily uniformly

continuous.

Proof of Lemma Suppose f C(G, k). For each point g G, let

U(g) = {x G : |f(x) f(g)| < 12

}.


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By the triangle inequality,

x, y U(g) = |f(x) f(y)| < .Now each neighbourhood U ofg in G is expressible in the form

U = gV

where V is a neighbourhood ofe in G.Furthermore, for each neighbourhood V ofe, we can find a smaller neighbour-

hood W ofe such thatW2

V.

(This follows from the continuity of the multiplication (x, y) xy. Here W2denotes the set {w1w2 : w1, w2 W}.)

So for each g G we can find an open neighbourhood W(g) ofe such thatgW(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 com-

pact, we can find a finite subcover, say

G = g1W1 g2W2 grWr,where Wi = W(gi).

Let

W = iWi.Suppose x1y W, ie

y xW.Now x lies in some set giWi. Hence

x, y giWiW giW2i ;and so

|f(x) f(y)| < .

Now we observe that this open set U will serve not only for f but also forevery average F off. For if

F = 1g1f + + rgrf (0 1, . . . , r 1, 1 + + r = 1)


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then

|F(x) F(y)| 1|f(g11 x) f(g11 y)| + + r|f(g1r x) f(g1r y)|.But

(g1i x)1(g1i y) = x

1gig1i y = x

1y.

Thus

x1y U = |f(g1i x) f(g1i y)| < = |F(x) F(y)| < (1 + r) = .

Returning to our construction of an improving average F, let us take =(M m)/2; then we can find an open set U e such that

x1y U = |F(x) F(y)| < 12

(M m)

for every average F off. In other words, the variation of F on any transform gUis less than half the variation off on G.

As before, we can find a finite number of transforms of U covering G, say

G g1U grU.

One of these transforms, giU say, must contain a point x0 at which F takes itsminimal value. But then, within giU,

|F(x) F(x0)| < 12

(M m);

and so

F(x) < min F +1

2(M m).

If now we form the new average

F = 1r

(g1F + grF)) ,

as before, then

max F r 1r

max F +1

r

min F +

M m2

.

Since min F min F, it follows that

var F

1 1r

var F +

1

2rvar f.


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A little thought shows that this implies that

var F < var F

provided

var F >1

2var f.

At first sight, this seems a weaker result than our earlier one, which showed that

var F < var F in all cases! The difference is, that r now is independent of F.Thus we can find a sequence of averages

F0 = f, F1, F2, . . .

(each an average of its predecessor) such that var Fi is decreasing to a limit satisfying

1 1r

+

1

2rvar(f),

ie

12

var f.

In particular, we can find an average F with

var F 0. Then we can find an open set U containing g such that

f(x) > 0

for x U. Now we can find g1, . . . , gr such that

G = g1U grU.

Let F be the average

F(x) =1

rf(g

11 x) +

+ f(g1r x) .

Then

x giU = g1i x U = f(g1i x) ,and so

F(x) r

.

Hence f dg =

F dg

r> 0.

Since

min f f d max f,it follows at once that

|

f d| |f|.It is now easy to show that is continuous. For a linear function is continuous

if it is continuous at 0; and we have just seen that

|f| < = |

f d| < .


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It follows at once from

min f

f dg max f

that

|

f dg| |f|.Finally, since f and gf (for f C(G), g G) have the same transforms,

they have the same (left) averages. Hencegf dg =

f dg,

ie the integral is left-invariant.

Moreover, it follows from our construction that this is the only left-invariant

integral on G with

1 dg = 1; for any such integral must be sandwiched betweenmin F and max F for all averages F of f, and we have seen that these intervalsconverge on a single real number.

The Haar measure, by definition, is leftinvariant:f(g1x) d(x) =

f(x) d(x).

It followed from our construction that it is also rightinvariant:f(xh) d(x) =

f(x) d(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, ieG

f(gh) dg =G

f(g) dg (h G, f C(G,R)) .

Proof Suppose h G. The maph : f (f h)

defines a left invariant measure on G. By the uniqueness of the Haar measure, andthe fact that

h(1) = 1

(since the constant function 1 is right as well as left invariant),

h = ,


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ie is right invariant. 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 A(f) = F + (1 )F A(f) (0 1).Let C(G) denote the set of constant functions f(g) = c. Evidently

= R.

We want to show that A(f) = ,

ie the closure ofA(f) contains a constant function. (In other words, we can finda sequence of averages converging on a constant function.)

To prove this, we establish that A(f) is pre-compact, ie its closure A(f) iscompact. For then it will follow that there is a point X A(f) (ie a functionX(g)) which is closest to . But if this point is not in , we will reach a con-tradiction; 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 . (We actu-ally 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 av-erages. 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 anotheraverage closer to than X(g).)

It remains to show that A(f) is pre-compact. We note in the first place thatthe 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 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 thepoints

{1F1 + + rFr : 0 1, . . . , r 1; 1 + + r = 1}.Thus A(f) is the convex closure of the compact set Gf. But the convex clo-

sure of a compact set in a complete metric space is always pre-compact. That


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follows (not immediately, but by a straightforward argument) from the following

lemma in the theory of metric spaces: A subsetS X of a complete metric spaceis 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 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 introduc-

ing right averages. Finally, we define

f dg to be this point of intersection (orrather, the corresponding real number); and we show as before that this defines an

invariant integral (f) with the required properties.

Examples:

1. As we have already noted, the Haar measure on S1 is

1

2d.

In other words,

(f) =1

2

20

f() d.

2. Consider the compact group SU(2). We know that

SU(2) = S3,since the general matrix in SU(2) takes the form

U =

x + iy z + itz + it x iy

, |x|2 + |y|2 + |z|2 + |t|2 = 1.

The usual volume on S3, when normalised, gives the Haar measure onSU(2). To see that, observe that multiplication by U SU(2) defines adistance preserving linear transformationan isometryofR4, ie if

U

x + iy z + itz + it x iy

=

x + iy z + it

z + it x iy

then

x2

+ y2

+ z2

+ t2

= x2 + y2 + z2 + t2

for all (x , y, z, t) R4.It follows that multiplication by U preserves the volume on S3. In otherwords, this volume provides an invariant measure on SU(2), which musttherefore beafter normalisationthe Haar measure on SU(2).


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As this examplethe simplest non-abelian compact groupdemonstrates,

concrete computation of the Haar measure is likely to be complicated. For-

tunately, the mere existence of the Haar measure is usually sufficient for our

purpose.


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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 ourmain results wereor can beestablished 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 restrictiongreatly simplifies the story, for three reasons:

1. Each finite-dimensional vector space over k carries a unique hausdorff topol-ogy under which addition and scalar multiplication are continuous. If V isn-dimensional then

V = kn;and this unique topology on V is just that arising from the product topologyon kn.

2. If U and V are finite-dimensional vector spaces over k, then every linearmap

t : U Vis continuous. Continuity is automatic in finite dimensions.

3. IfV is a finite-dimensional vector space over k, then every subspace U Vis closed in V.

424II 31


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3.1. REPRESENTATIONS OF A COMPACT GROUP 424II 32

3.1 Representations of a Compact Group

We have agreed that a representation of a topological group G in a finite-dimensionalvector 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 equiv-alent ways:

1. by a linear action

G

V

V;

2. by a homomorphism

G GL(V),whereGL(V) denotes the group of invertible linear maps t : V V.

We again have the same choice. We have chosen (1) as our fundamental defi-

nition 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 ofall linear maps t : V

V; ifdim V = n then

GL(V) Mat(n, k) = kn2.

This n2-dimensional vector space has a unique hausdorff topology, as we haveseen; and this induces a topology on GL(V).

We know that there is a one-one correspondence between linear actions of Gon V and homomorphisms G GL(V). It is a straightforward matter to verifythat 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 , of a group G in thefinite-dimensional vector spaces U, V over k holds for all groups, and so extendswithout question to compact groups.

We note that the map : U V defining such an equivalence is necessarilycontinuous, 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

addthe requirement that should be continuous.


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3.3. SIMPLE REPRESENTATIONS 424II 33

3.3 Simple Representations

Recall that the representation of a group G in the finite-dimensional vector spaceV over k is said to be simple if no proper subspace U V is stable under G. Thisdefinition 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 ofV. 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 , are representations of the group G in the finite-dimensional vectorspaces U, V over k. We have defined the representations + , , in thevector spaces U V, U V, U, respectively. These definitions hold for allgroups G.

However, there is something to verify in the topological case, even if it is

entirely straightforward. We must show that if and are continuous then so are + , , and . (This is left as an exercise to the student.)

3.5 Semisimple Representations

The definition of the semisimplicity of a representation of a group G in a finite-dimensional vector space V over k makes no restriction on G, and so extends tocompact groups (and indeed to all topological groups); is semisimple if andonly if it is expressible as a sum of simple representations:

= 1 + + m.

Recall that a finite-dimensional representation ofG in V is semisimple if andonly if each stable subspace U V has at least one stable complementary sub-space 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,


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3.6. EVERY REPRESENTATION OF A FINITE GROUP IS SEMISIMPLE424II 34

3.6 Every Representation of a Finite Group is Semisim-

ple

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 is arepresentation 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 anew inner product

u, v =V

p(gu,gv) dg.

It is a straightforward matter to verify that this new inner product is invariant:

gu,gv = u, v.It also follows at once from the positivity of the Haar measure that this inner

product is positive, ie

v, v 0.Its a little more difficult to see that the inner product is positive-definite, ie

v, v = 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 functionon G such that f(g) 0 for all g G then not only is

Gf(g) dg 0

(this is the positivity of the measure) but also

G f(g) dg = 0 = f(g) = 0

for allg.

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 iscompact) G can be covered by a finite number of transforms ofU:

G g1U . . . grU.It follows from this that

f(g11 x) + + f(g1r x) /2


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for all x

G. For

x giU = g1i x U = f(g1i x) /2.It follows from this, on integrating, that

rG

f(g) dg /2.

In particular

f 0.Note that our alternative proof of semisimplicity also carries over to the com-

pact case. This proof depended on the fact that if

: V Vis a projection onto a stable subspace U = (V) ofV then its average

=1

|G|gG

gg1

is also a projection onto U; and

W = ker

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

=G

gg1 dg.

For here we are integrating the operator-valued function

F(g) = gg1

However, there is little difficulty in extending the concept of measure to vector-

valued functions F on G, ie maps

F : G V,where V is a finite-dimensional vector space over k. This we can do, for exam-ple, 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


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straightforward, The case of a function with values in hom(U, V), where U, V arefinite-dimensional vector spaces over k, may be regarded as a particular case ofthis, 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 gG

hG

f(g, h)

over G. The easiest way to extend such an argument to compact groups is toconsider the corresponding integral

GG f(g, h) d(g, h)of the continuous function f(g, h) over the product group G G.

In such a case, let us set

F(g) =hG

f(g, h) dh

for each g G. Then it is readily shown that F(g) is continuous, so that we cancompute

I =gG

F(g) dg

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 thatGG

f(g, h) d(g, h) =gG

hG

f(g, h) dh

dg.

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 is itself a projection, we argue as follows:

2 =gG

gg1 dghG

hh1 dh

= (g,h)GG

gg1hh1 d(g, h)

=(g,h)GG

gg1hh1 d(g, h)

(using the fact that g = g, since U = im is stable under G). Thus

2 =(g,h)GG

hh1 d(g, h)

=gG

dghG

hh1 dh

= .


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3.7. UNIQUENESS AND THE INTERTWINING NUMBER 424II 37

3.7 Uniqueness and the Intertwining Number

The definition of the intertwining number I(, ) does not presuppose that G isfinite, 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 de-

pend 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 proofrequires a little more thought.

Proposition 3.1 Suppose is an n-dimensional representation of a compact groupG overR orC; and suppose g G. Let the eigenvalues of (g) be 1, . . . , n.Then

|i| = 1 (i = 1, . . . , n).

ProofWe know that there exists an invariant inner product u, v on the representation-space V. We can choose a basis for V so that

v, v = |x1|2

+ + |xn|2

,

where v = (x1, . . . , xn). Since (g) leaves this form invariant for each g G,

it follows that the matrix A(g) of (g) with respect to this basis is orthogonal ifk = R, or unitary ifk = C.

The result now follows from the fact that the eigenvalues of an orthogonal or

unitary matrix all have absolute value 1:

Uv = v = vU = v= vUU v = vv=

vv =

|

|2vv

= || = 1.Hence

1 =

for each such eigenvalue.

Alternative proof Recall how we proved this in the finite case. By Lagranges

Theorem gm = 1 for some m > 0, for each g G. Hence(g)m = I;


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3.8. THE CHARACTER OF A REPRESENTATION 424II 38

and so the eigenvalues of(g) all satisfy

m = 1.

In particular

|| = 1;and so

1 = .

We cannot say that an element g in a compact group G is necessarily of finiteorder. However, we can show that the powers gn ofg approach arbitrarily close tothe identity e

G. (In other words, some subsequence of

{g, g2, g3, . . .

}tends to

e.)For suppose not. Then we can find an open set U e such that no power

of g except g0 = e lies in U. Let V be an open neighbourhood of e such thatV V1 U. Then the subsets gnV are disjoint. For

x gmV gnV = x = gmv1 = gnv2= gnm = v1v12= gnm U,

contrary to hypothesis.

It follows [the details are left to the student] that the subgroup

g = {. . . , g1, e , g , g2, . . . }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 < n2 < . . .such that

gni eas i .

It follows that

(g)ni I


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3.9. THE REGULAR REPRESENTATION 424II 39

as i

. Hence if is any eigenvector of(g) then

ni 1.This implies in particular that

|| = 1.

Corollary 3.1 If is a finite-dimensional representation of a compact group overR orC then

(g1) = (g)

for all g G

Proof Suppose the eigenvalues of(g) are 1, . . . , n. Then the eigenvalues of(g1) = (g)1 are 11 , . . . ,

1n . Thus

(g1 = tr (g1)

= lambda11 + + 1n= 1 + + n= 1 + + n= tr (g)= (g).

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.IfG is discrete (in particular ifG is finite) then every map f : G k is continu-ous; 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. [Weleave 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


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3.10. INDUCED REPRESENTATIONS 424II 310

do this, rather briefly, in Chapter 7 below, when we consider the Peter-Weyl Theo-

rem. 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 notextend to the compact case.

As we shall see in Chapter 7, a compact but non-finite group G has an infinitenumber of distinct simple finite-dimensional representations 1, 2, . . . . So anyargument 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 RepresentationsThe results of this chapter have only a limited application in the topological case,

since they apply only where we have a subgroup H G offinite index in G; thatis, G is expressible as the union of a finite number of H-cosets:

G = g1H grH.

In this limited case each finite-dimensional representation of H induces asimilar representation G ofG.

For example, SO(n) is of index 2 in O(n); so each representation ofSO(n)

defines a representation ofO(n).

3.11 Representations of Product Groups

If , are finite-dimensional representations of the groups G, H in the vectorspaces U, W over k then we have defined the representation of G Hin U W. This extends without difficulty to the topological case; and it is astraightforward matter to verify that is continuous, in the finite-dimensionalcase.

Recall our main result in this context; ifk =

C then

is simple if and

only if and are both simple; and furthermore, every simple representation ofG H over C arises in this way.

The proof that is simple if and only if if and only if and are bothsimple 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 1, . . . , m. Similarly if H has n classes, then it has nsimple representations 1, . . . , m.


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3.11. REPRESENTATIONS OF PRODUCT GROUPS 424II 311

But now G

H has mn classes; and so the mn simple representations i

jprovide all the representations ofG H.

This argument fails in the compact case, since m and n are infinite (unless Gor H is finite).

We must turn therefore to our second proof that a simple representation ofG H over C is necessarily of the form . Recall that this alternative proofwas based on the natural equivalence

hom(hom(V, U), W) = hom(V, U W).This proofdoes carry over to the compact case.

Suppose the representation-space of is the G

H-space V. Consider V as aG-space (ie forget for the moment the action of H on V). Let U V be a simpleG-subspace of V. Then there exists a non-zero G-map t : V U (since theG-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 ofX. Then there exists anon-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

= homGH(V, U W).Thus there exists a non-zero G H-map T : V U W. Since V and U Ware both simple G H-spaces, T must be an isomorphism:

V = U W.In particular

= ,where is the representation ofG in U, and is the representation ofH in W.

Thus ifG and Hare compact groups then every simple representation ofGHover C is of the form .

[Can you see where we have used the fact that G and H are compact in ourargument above?]


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3.12. REAL REPRESENTATIONS 424II 312

3.12 Real Representations

Everything in this chapter carries over to the compact case, with no especial prob-

lems arising.


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Chapter 4

Representations ofU(1)

The group U(1) goes under many names:

U(1) = SO(2) = S1 = T1 = R/Z.

Whatever it is called, U(1) is abelian, connectedandabove allcompact.As an abelian group, every simple representation ofU(1) (overC) is 1-dimensional.

Proposition 4.1 Suppose : G C is a 1-dimensional representation of thecompact group G. Then

|(g)| = 1 for all g G.

Proof Since G is compact, so is its continuous image (G). In particular (G)is bounded.

Suppose |(g)| > 1. Then

|(gn)| = |(g)|n

as n , contradicting the boundedness of(G).On the other hand,

|(g)| < 1 = |(g1| = |(g)|1 > 1.

Hence |(g)| = 1.

Corollary 4.1 Every 1-dimensional representation of a compact group G is ahomomorphism of the form

: G U(1).

424II 41


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424II 42

In particular, the simple representations ofU(1) are just the homomorphisms

U(1) U(1).

But ifA is an abelian group then for each n Z the map

a an : A A

is a homomorphism.

Definition 4.1 For each n Z, we denote by En the representation

ei ein

ofU(1).

Proposition 4.2 The representations En are the only simple representations ofU(1).

Proof Suppose is a 1-dimensional representation ofU(1), ie a homomor-phism

: U(1) U(1).Let U U(1) be the open set

U = {ei : /2 < < /2}.

Note that each g U has a unique square root in U, ie there is one and only oneh U such that h2 = g.

Since is continuous at 1, we can find > 0 such that

< < = (ei) U.

Choose N so large that 1/N < . Let = e2i/N. Then ()

U; while

N = 1 = ()N = 1.

It follows that

() = e2ni/N = n = En()

for some n Z in the range N/2 < n < N/2. We shall deduce from this that = En.

Let

1 = ei/N.


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424II 43

Then

21 = = (1)2 = () = n= (1) = n1 ,

since this is the unique square root of n in U.Repeating this argument successively with we deduce that if

j = e2

2jNi

then

(j) =

n

j = En(j)for j = 2, 3, 4, . . . .

But it follows from this that

(kj ) = (kj )

n = En(kj )

for k = 1, 2, 3, . . . . In other words

(ei) = E(ei)

for all of the form

= 2k

2j

But these elements ei are dense in U(1). Therefore, by continuity,

(g) = En(g)

for all g U(1), ie = En. Alternative proof Suppose

: U(1) U(1)

is a representation ofU(1) distinct from all the En. Then

I(En, ) = 0

for all n, ie

cn =1

2

20

(ei)en d = 0.

In other words, all the Fourier coefficients of (ei) vanish.But this implies (from Fourier theory) that the function itself must vanish,

which is impossible since (1) = 1.


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424II 44

Remark: As this proof suggests, the representation theory ofU(1) is just theFourier theory of periodic functions in disguise. (In fact, the whole of group rep-

resentation theory might be described as a kind of generalised Fourier analysis.)

Let denote the representation ofU(1) in the space C(U(1)) of continuousfunctions f : U(1) C, with the usual action: ifg = ei then

(gf)(ei) = ei().

The Fourier series

f(ei) =nZ

cnein

expresses the splitting ofC(U(1)) into 1-dimensional spaces

C(U(1)) =

Vn,

where

Vn = ein = {cein : c C}.Notice that with our definition of group action, the space Vn carries the repre-

sentation En, rather than En. For ifg = ei, and f(ei) = ein, then

(gf)(ei) = einf(ei) = En(g)f(ei).

In terms of representations, the splitting ofC(U(1)) may be written:

=nZ

En.

We must confess at this point that we have gone out of bounds in these re-

marks, 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 represen-

tations. We shall see in Chapter 7 how we can justify this extension.


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Chapter 5

Representations ofSU(2)

5.1 Conjugacy in SU(n)

Since characters are class functions, our first step in studying the representations

of a compact group Gas of a finite groupis to determine how G divides intoconjugacy classes.

We know that if 2 matrices S, T GL(n, k) are similar, ie conjugate inGL(n, k), then they will have the same eigenvalues 1, . . . , n. So this gives anecessary 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 10 1

1 00 1

in GL(2, C), although both matrices have eigenvalues 1, 1. However we shallsee 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 inwhat group we are taking conjugates. Two matrices S, T G GL(n, k) maywell be conjugate in GL(n, k) without being conjugate in G.

Secondly, the concepts of eigenvalue and eigenvector really belong to a repre-

sentation of a group rather than the group itself. So for example, when we speak

of an eigenvalue ofT U(n) we really shouldthough we rarely shallsay aneigenvalue ofT in the natural representation ofU(n) in Cn.

Lemma 5.1 The diagonal matrices in U(n) form a subgroup isomorphic to thetorus group Tn U(1)n.

424II 51


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5.1. CONJUGACY INSU(N) 424II 52

Proof We know that the eigenvalues ofTU(n) have absolute value 1, since

T v = v = vT = v= vTT v = vv= vv = vv= ||2 = = 1= || = 1

Thus the eigenvalues ofT can be written in the form

ei1, . . . , ein (1, . . . , nR).

In particular the diagonal matrices in U(n) are just the matrices

ei1

. . .

ein

It follows that the homomorphism

U(1)n

U(n) : e1 , . . . , en

ei1

. .. ein

maps U(1)n homeomorphically onto the diagonal subgroup ofU(n), allowing usto identify the two:

U(1)n U(n).

Lemma 5.2 Every unitary matrix T U(n) is conjugate (inU(n)) to a diagonalmatrix:

T

DU(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 Let T denote the closed subgroup generated by T, ie the closure inU(n) of the group

{. . . , T 1, I , T , T 2, . . . }formed by the powers of T.


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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 ofunitary matrices.

It follows that this representation splits into a sum of 1-dimensional represen-

tations, 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 orthonor-mal set of eigenvectors ofT. IfU is the matrix of change of basis, ie

U = (e1, . . . , en)

then

UT U =

ei1

. . .

ein

where

T ei = eiei.

Lemma 5.3 The diagonal matrices in SU(n) form a subgroup isomorphic to thetorus group Tn1 U(1)n1.

Proof If

T =

ei1

. . .

ein

then

det T = ei(1++n).

Hence

T SU(n) 1 + + n = 0 (mod 2).Thus the homomorphism

U(1)n1 SU(n) :

e1, . . . , en1

ei1

. . .

ein1

ei(1++n1)

mapsU(1)n1 homeomorphically onto the diagonal subgroup ofSU(n), allowingus to identify the two:

U(1)n1 SU(n).


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Lemma 5.4 Every matrix TSU(n) is conjugate (in SU(n)) to a diagonal

matrix:

T D U(1)n1.

Proof From the corresponding lemma for U(n) above, T is conjugate in the fullgroup U(n) to a diagonal matrix:

UT U = D (U U(n)).

We know that | det U| = 1, say

det U = ei

.

Let

V = ei/nU.

Then V SU(n); andVT V = D.

Lemma 5.5 Let G = U(n) orSU(n). Two matrices U, V G are conjugate ifand only if they have the same eigenvalues

{ei1 , ei2 , . . . , ein}.

Proof Suppose U, V G. If U V then certainly they must have the sameeigenvalues.

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 D2contain the same entries, perhaps permuted. So we can find a permutation matrix

P (with just one 1 in each row and column, and 0s elsewhere) such that

D2 = P1D1P.

Now P U(n) since permutation of coordinates clearly leaves the form |x1|2 + + |xn|2 unchanged. Thus ifG = U(n) we are done:

S D1 D2 T.


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5.2. REPRESENTATIONS OFSU(2) 424II 55

Finally, suppose G = SU(n). Then

T = USU

for some U U(n). Suppose

det U = ei.

Let

V = ei/nU.

Then V SU(n); andT = V

SV

5.2 Representations ofSU(2)

Summarising the results above, as they apply to SU(2):

1. each T SU(2) has eigenvalues ei

2. with the same notation,

T U() =

ei 00 ei

3. U() U()Thus SU(2) divides into classes C() (for 0 ) containing all T witheigenvalues ei.

The classes

C(0) = {I}, C(1) = {I},

constituting the centre of SU(2), each contain a single element; all other classesare infinite, and intersect the diagonal subgroup in 2 elements:

C() U(1) = {U()}.

Now let denote the natural representation ofSU(2) in C2, defined by theaction

zw

z

w

= T

zw


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Explicitly, recall that the matrices TSU(2) are just those of the form

U =

a bb a

(|a|2 + |b|2 = 1)

Taking T in this form, its action is given by

(z, w) (az + bw, bz + aw)By extension, this change of variable defines an action of SU(2) on polyno-

mials P(z, w) in z and w:

P(z, w) P(az + bw, bz + aw).Definition 5.1 For each half-integerj = 0, 1/2, 1/, 3/2, . . . we denote by Dj therepresentation ofSU(2) in the space

V(j) = z2j, z2j1w , . . . , w2jof homogeneous polynomials in z, w of degree 2j.

Example: Let j = 3/2. The 4 polynomials

z3

, z2

w,zw2

, w3

form a basis for V(3/2).

Consider the action of the matrix

T =

0 ii 0

SU(2).

We have

T(z3) = (iw)3 = iw3,T(z2w) =

izw2,

T(zw2) = iz2w,T(w3) = iz3

Thus under D32

,

0 ii 0

0 0 0 i0 0 i 00 i 0 0i 0 0 0


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Proposition 5.1 The characterj of Dj is given by the following rule: SupposeT has eigenvalues ei Then

j(T) = e2ij + e2i(j1) + + e2ij

Proof We know that

T U() =

ei 00 ei

.

Hence

j(T) = j (U()) .The result follows on considering the action of U() on the basis {z2j , . . . , w2j}ofV(j). For

U()zkw2jk =

eizk

eiw2jk

= e2i(kj)zkw2jk.

Thus under Dj,

U()

e2ij

e2i(j2). . .

e2ij

whence

j (U()) = e2ij + e2i(j2) + + e2ij .

Proposition 5.2 For each half-integerj, Dj is a simple representation ofSU(2),of dimension 2j + 1.

Proof On restricting to the diagonal subgroup U(1) SU(2),

(Dj)U(1) = E2j + E2j+2 + + E2j.

Since the simple parts on the right are distinct, it follows that the corresponding

expression

V(j) = z2j w2jfor V(j) as a direct sum of simple U(1)-modules is unique.


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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 obtainan expression for V(j) as a direct sum of simple U(1)-spaces. It follows from theuniqueness of this expression that each of U and W must be the spaces spannedby some of the monomials zawb. In particular z2j must belong either to U or toW. Without loss of generality we may suppose that

z2j U

But thenT(z2j) U

for all T SU(2). In particular, taking

T =1

2

1 1

1 1

(almost any T would do) we see that

(z + w)2j = z2j + 2jz2j1w + + w2j U.Each of the monomials of degree 2j occurs here with non-zero coefficient. Itfollows that each of these monomials must be in U:

z2jkwk U for all k.Hence U = V(j), ie Dj is simple.

Proposition 5.3 The Dj are the only simple representations ofSU(2).

Proof Suppose is a simple representation of SU(2) distinct from the Dj.Then in particular

I(, Dj) = 0.

In other words, is orthogonal to each j,Consider the restriction of to the diagonal subgroup U(1). Suppose

U(1) =j

njEj,

where of course all but a finite number of the nj vanish (and the rest are positiveintegers). It follows that

(U()) =j

njeij


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Lemma 5.6 For any representation ofSU(2),

nj = nj,

ie Ej andEj occur with the same multiplicity in U(1).

Proof This follows at once from the fact that

U() U()in SU(2).

Since nj = nj, we see that (U()) is expressible as a linear combination of

the j(U()) (in fact with integraland not necessarily positivecoefficients):

(U()) =j

cjj (U()) .

Since each T SU(2) is conjugate to some U() it follows that(T) =

j

cjj(T)

for all T SU(2). But this contradicts the proposition that the simple charactersare linearly independent (since they are orthogonal).

We know that every finite-dimensional representation ofSU(2) is semi-simple.In particular, each product DjDk is expressible as a sum of simple representations,ie as a sum ofDns.

Theorem 5.1 (The Clebsch-Gordan formula) For any pair of half-integers j, k

DjDk = Dj+k + Dj+k1 + + D|jk|.

Proof We may suppose that j k.Suppose T has eigenvalues ei. For any 2 half-integers a, b such that a

b, a b N

, let

L(a, b) = e2ia + e2i(a+1) + + e2ib.(We may think ofL(a, b) as a ladder linking a to b on the axis, with rungs everystep, at a + 1, a + 2, . . . .) Thus

j() = L(j,j);and so

DjDk(T) = j()k() = L(j,j)L(k, k).


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We have to show that

L(j,j)L(k, k) = L(jk, j+k)+L(jk+1, j+k1)+ +L(j+k, jk).

We argue by induction on k. The result holds trivially for k = 0.By our inductive hypothesis,

L(j,j)L(, k+1, k1) = L(jk+1, j+k1)+ +L(j+k1, jk+1).

Now

L(k) = L(k 1) + (e2ik + e2ik).

But

L(j,j)e2ik = L(j k, j k),L(j,j)e2ik = L(j + k, j + k).

Thus

L(j,j)(e2ik + e2ik) = 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.

Proposition 5.4 The representation Dj ofSU(2) is real for integralj and quater-nionic for half-integral j.

Proof The character

j() = e2ij + e2i(j1) + + e2ij

is real, since

j() = e2ij + e2i(j1) + + e2ij = j().

Thus Dj (which we know to be simple) is either real or quaternionic.


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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

x1y2 x2y1 + x3y4 x4y3 +

But

dim Dj = 2j + 1

is odd for integral j. Hence Dj must be real in this case.

Lemma 5.7 The representation D12

is quaternionic.

Proof of Lemma Suppose D 12

were real, say

D 12

= C,

where

: SU(2) GL(2,R)is a 2-dimensional representation ofSU(2) over R. We know that this representa-tion carries an invariant positive-definite form. By change of coordinates we can

bring this to x2

1 + x2

2, so that im O(2).Moreover, since SU(2) is connected, so is its image. Hence

im SO(2).

Thus defines a homomorphism

SU(2) SO(2) = U(1),

ie a 1-dimensional representation ofSU(2), which must in fact be D0 = 1. It

follows that = 1 + 1, contradicting the simplicity ofD 12

.

Remark: It is worth noting that the representation D 12

is quaternionic in its original

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) gq (g Sp(1), q H).


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(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 12

.

It remains to prove that Dj is quaternionic for half-integral j >12

. Suppose in

fact Dj were real; and suppose this were the first half-integralj with that property.Then

DjD1 = Dj+1 + Dj + Dj1

would also be real (since the product of 2 real representations is real). But Dj1is 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. Alternative Proof Recall that if is a simple representation then

(g

2) dg =

1 if is real,0 if is essentially complex,

1 if is quaternionic.

Let = Dj. Suppose g SU(2) has eigenvalues ei. Then g2 has eigenval-ues e2i, and so

j(g2) = e4ij + e4i(j1) +

+ e4ij

= 2j(g) 2j1(g) + + (1)2j0(g).

Thusj(g

2) dg =

2j(g) dg

2j1(g) dg + + (1)2j

0(g) dg

= I(1, D2j) I(1, D2j1) + + (1)2jI(1, D0)= (1)2jI(1, 1)=

+1 ifj is integral1 ifj is half-integral


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Chapter 6

Representations ofSO(3)

Definition 6.1 A covering of one topological group G by anotherC is a continu-ous homomorphism

: C Gsuch that

1. ker is discrete;

2. is surjective, ie im = G.

Proposition 6.1 A discrete subgroup is necessarily closed.

Proof Suppose S G is a discrete subgroup. Then by definition we can findan 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 theintersection of an open set in G with S.)

We can find an open set V G containing 1 such that

V1V U,ie v11 v2 U for all v1, v2 V. This follows from the continuity of the map

(x, y) x1y : G G G.Now suppose g G \ S. We must show that there is an open set O g not

intersecting S. The open set gV g contains at most 1 element ofS. For supposes, t gV, say

s = gv1, t = gv2.

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424II 62

Then

s1t = v11 v2 U S.Thus s1t = 1, ie s = t.

IfgV 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 takeO = gV W.

Corollary 6.1 A discrete subgroup of a compact group is necessarily finite.

Remark: We say that

: C Gis an n-foldcovering if ker = n.

Proposition 6.2 Suppose : C G is a surjective (and continuous) homomor-phism of topological groups. Then

1. Each representation of G in V defines a representation of C in V bythe composition

: C G GL(V).

2. If the representations 1, 2 of G define the representations

1

, 2

of C inthis way then

1 = 2 1 = 2.

3. With the same notation,

(1 + 2) = 1 +

2, (12)

= 12, (

) = ().

4. A representation ofC arises in this way from a representation of G if andonly if it is trivial on ker, ie

g ker = (g) = 1.5. The representation ofC is simple if and only if is simple. Moreover, if

that is so then is real, quaternionic or essentially complex if and only ifthe same is true of .

6. The representation ofC is semisimple if and only if is semisimple.


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424II 63

Proof All follows from the fact that gv (g

G, v

V) is the same whetherdefined through or .

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 of G with the corresponding representation of C; so that the representation theory of G is included, in this sense, in therepresentation theory ofC.

The following result allows us, by applying these ideas, to determine the rep-resentations ofSO(3) from those ofSU(2).

Proposition 6.3 There exists a two-fold covering

: SU(2) SO(3).

Remark: We know that SU(2) has the real 2-dimensional representation D1, de-fined by a homomorphism

: SU(2)

GL(3,R).

Since SU(2) is compact, the representation-space carries an invariant positive-definite quadratic form. Taking this in the form x2 + y2 + z2, we see that

im O(3).

Moreover, since SU(2) is connected, so is its image. Thus

im SO(3).

This is indeed the covering we seek; but we prefer to give a more constructive

definition.

Proof Let H denote the space of all 2 2 hermitian matrices, ie all matrices ofthe form

A =

x y iz

y + iz t

(x , y, z, t R).

Evidently H is a 4-dimensional real vector space. (It is not a complex vectorspace, since A hermitian does not imply that iA is hermitian; in fact

A hermitian = iA skew-hermitian,


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424II 64

since (iA) =

iA for any A.)Now suppose U SU(2). Then

A H = (UAU) = UAU = UAU= UAU H.

Thus the action

(U, A) UAU = U1AUofSU(2) on H defines a 4-dimensional real representation ofSU(2).

This is not quite what we want; we are looking for a 3-dimensional represen-

tation. Let

T = {A H : tr A = }denote the subspace ofH formed by the trace-free hermitian matrices, ie those ofthe form

A =

x y iz

y + iz x

(x , y, z, t R).These constitute a 3-dimensional real vector space; and since

tr (UAU) = tr

U1AU

= tr A

this space is stable under the action of SU(2). Thus we have constructed a 3-dimensional representation ofSU(2) over R, defined by a homomorphism

: SU(2) GL(3,R).The determinant defines a negative-definite quadratic form on T, since

det

x y iz

y + iz x

= x2 y2 z2

Moreover this quadratic form is left invariant by the action ofSU(2), since

det(UAU) = det

U1AU

= det A.

In other words, SU(2) acts by orthogonal transformations on

T, so that

im O(3).Moreover, since SU(2) is connected, its image must also be connected, and so

im SO(3).We use the same symbol to denote the resulting homomorphism

: SU(2) SO(3).We have to show that this homomorphism is a covering.


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424II 65

Lemma 6.1 ker ={

I}

.

Proof of Lemma Suppose U ker. In other words,

U1AU = A

for all A T.In fact this will hold for all hermitian matrices A H since

H = TI.But now the result holds also for all skew-hermitian matrices, since they areof 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

2(A + A) +

1

2(A A) .

Since

U1AU = A AU = U A,we are looking for matrices U which commute with all 2 2-matrices A. It isreadily verified that the only such matrices are the scalar multiples of the identity,

ieU = I.

But now,

U SU(2) = det U = 1= 2 = 1= = 1.

Lemma 6.2 The homomorphism is surjective.

Proof of Lemma Let us begin by looking at a couple of examples. Suppose first

U = U() =

ei 00 ei

;

and suppose

A =

x y iz

y + iz x

T.


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424II 66

Then

UAU =

ei 0

0 ei

x y iz

y + iz x

ei 00 ei

=

x e2i(y iz)

e2i(y + iz) x

=

X Y iZ

Y + iZ X

,

where

X = xY = cos 2y + sin 2z

Z = sin 2y cos2z.Thus U() induces a rotation in the space T through 2 about the Ox-axis, say

U() R(2,Ox).As another example, let

V =1

2 1 1

1 1 ;

In this case

VAV =

x y + iz

y iz x

=

X Y + iZ

Y iZ X

,

where

X = yY = x

Z = z.

Thus (V) is a rotation through /2 about Oz.It is sufficient now to show that the rotations R(,Ox) about the x-axis, to-

gether with T = R(/2, Oz), generate the group SO(3). Since is a homomor-phism, V U()V1 maps onto

T R(2,Ox)T1 = R(2, T(Ox)) = R(2,Oy).

Thus im contains all rotations about Ox and about Oy. It is easy to see thatthese generate all rotations. For consider the rotation R(, l) about the axis l. We


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424II 67

can find a rotation S about Ox bringing the axis l into the plane Oxz; and then arotation T about Oy bringing l into the coordinate axis Ox. Thus

T SR(, l)(T S)1 = R(,Ox);

and so

R(, l) = S1T1R(,Ox)T S.

These 2 lemmas show that defines a covering ofSO(3) by SU(2).

Remarks:

1. We may express this result in the succinct form:

SO(3) = SO(2)/{I}.

Recall that SU(2) = S3. The result shows that SO(3) is homeomorphicto the space resulting from identifying antipodal points on the sphere S3.Another way of putting this is to say that SO(3) is homeomorphic to 3-dimensional real projective space:

SO(3) = P3(R) = (R4 \ {0})/R.

2. We shall see in Part 4 that the space T (or more accurately the space iT) isjust the Lie algebra of the group SU(2). Every Lie group acts on its ownLie algebra. This is the genesis of the homomorphism .

Proposition 6.4 The simple representations ofSO(3) are the representations Dj for integral j:

D0 = 1, D1, D2, . . .

Proof We have established that the simple representations ofSO(3) are just

those Dj which are trivial on {I}. But under I,(z, w) (z, w)

and so ifP(z, w) is a homogeneous polynomial of degree 2j,

P(z, w) = (1)2jP(z, w).

Thus I acts trivially on Vj if and only if2j is even, ie j is integral. The following result is almost obvious.


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Proposition 6.5 Let be the natural representation ofSO(3) is R3. Then

C = D1.

Proof To start with, is simple. For if it were not, it would have a 1-dimensionalsub-representation. In other words, we could find a direction in R3 sent into itself

be every rotation, which is absurd.

It follows thatC 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.


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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

gG |f(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 ofG G:((g, h)f) (x) = f(g1xh).

Recall that the corresponding representation ofG G splits into simple parts = 1 1 + + s s

where 1, . . . , s are the simple representations ofG (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 secondon V. A little thought shows that this action can be defined as follows. Supposet : V V is a linear map, ie an element ofhom(V, V). Then

(g, h)t = t

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7.1. THE FINITE CASE 424II 72

where t is the linear mapt(v) = ht(g1v).

The expression for above can be re-written as

C(G) hom(V1 , V1) + + hom(Vs, Vs),

where V is the space carrying the simple representation .In other words

C(G) = C(G)1 C(G)s,where

C(G) hom(V, V).Since the representations ofG G are simple and distinct, it follows thatthe subspaces C(G) C(G) are the isotypic components ofC(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(g1x),

then each subspace V V is isomorphic (as a G-space) to dim V. Thus itremains isotypic, while ceasing (unless dim = 1) to be simple. It follows thatthe expression

C(G) = C(G)1 C(G)s,can equally well be regarded as the splitting of the G-space C(G) into its isotypicparts.

Whichever way we look at it, we see that each function f(x) on G splits intocomponents f(x) corresponding to the simple representations ofG.

What exactly is this component f(x) off(x)? Well, recall that the projection of the G-space V onto its -component V is given by

=1

|G

| gG(g1)g.

It follows that

f(x) =1

|G|gG

(g)f(gx).

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