Chapter 1 Compact Groups

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

1 (x, y) xy, x x− 7→ 7→ 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.

424–II 1–1 424–II 1–2

(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 = U [ i ı I ∈ has a ﬁnite subcovering: X = Ui Ui . 1 ∪ · · · ∪ r (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 VU 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 a subspace 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): T 0T = I . { ∈ }

Here Mat(n, R) denotes the space of all n n real matrices; and T 0 denotes the transpose of T : ×

Tij0 = Tji.

We can identify Mat(n, R) with the Euclidean space En2 , by regarding the 2 n entries tij as the coordinates of T . 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 T 0T = I. It is bounded because each entry

tij 1. | | ≤ 424–II 1–3

In fact, for each i,

2 2 2 t + t + + t = (T 0T )ii = 1. 1i 2i ··· ni 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 2 T 0T = I = det T 0 det T = 1 = (det T ) = 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.

We can identify Mat(n, C) with the Euclidean space E2n2 , by regarding 2 the real and imaginary parts of the n entries tij as the coordinates of T . With this understanding, U(n) is a closed subspace of E2n2 . It is bounded because each entry has absolute value

tij 1. | | ≤ In fact, for each i,

When n = 1, U(1) = x C : x = 1 . { ∈ | | } Thus 1 T1 R Z U(1) = S ∼= = / . 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 2 T ∗T = I = det T ∗ det T = 1 = det T = 1, ⇒ ⇒ | | since det T ∗ = det T . The map U(1) SU(n) U(n):(λ, T ) λT × → 7→ is a surjective homomorphism. It is not bijective, since

λI SU(n) λn = 1. ∈ ⇐⇒ Thus the homomorphism has kernel

Cn = ω , h i 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 en- × tries; and T ∗ denotes the conjugate transpose of T : ¯ Tij∗ = Tji. 424–II 1–5

(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)∗ = 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 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 regarding 2 the coefﬁcients of 1, i, j, k in the n entries tij as the coordinates of T . With this understanding, Sp(n) is a closed subspace of E4n2 . It is bounded because each entry has absolute value

tij 1. | | ≤ In fact, for each i,

2 2 2 t1i + t2i + + tni = (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 + y2 + z2 = 1 . { ∈ | | } { } Thus 3 Sp(1) ∼= S . We leave it to the reader to show that there is in fact an isomorphism Sp(1) = SU(2). 424–II 1–6

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 role—ˆ 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 ∈ ∈

1 1 I = π− U π− 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 En. 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. 424–II 1–7

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 gx0. 7→ Then if x = gx0, 1 Θ− x = gH. { } Lemma 1.1 Each coset gH is connected.

Proof of Lemma B The map

h gh : H gH 7→ → is a continuous bijection. 1 But H is compact, since it is a closed subgroup of G (as H = Θ− 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 1 so φ(U) = Y φ(C) is open in Y . This shows that φ− 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 ). ∩ ∪ ∩ 424–II 1–8

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.

n Proof I Consider the action of SO(n) on R : (T, x) T x. 7→ This action preserves the norm:

T x = x k k k k 2 2 2 (where x = x0x = x + + x ). For k k 1 ··· n 2 T x = (T x)0T x = x0T 0T x = x0x. k k It follows that T sends the sphere

n 1 S − = x R : x = 1 { ∈ k k } n 1 into itself. Thus SO(n) acts on S − . This action is transitive: we can ﬁnd an orthogonal transformation of determi- n 1 nant 1 sending any point of S − into any other. (The proof of this is left to the reader.) n 1 Moreover the space S − is compact, since it is closed and bounded. Thus the conditions of our Proposition hold. Let us take 0   . x =  .  . 0    0     1  Then H(x0) = S(x0) = SO(n 1). ∼ − 424–II 1–9

For 0  .  T . T x = x = T =  1 .  0 0   ⇒  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∈ −

2 2 t + t + + 1 = 1 = tn1 = tn2 = = 0. n1 n2 ··· ⇒ ··· 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.

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

(T, x) T x. 7→ This again preserves the norm

1 2 2 2 x = x1 + xn , k k | | · · · | | since 2 2 T x = (T x)∗T x = x∗T ∗T x = x∗x = x . k k k k 424–II 1–10

Thus SU(n) sends the sphere

2n 1 n S − = x C : x = 1 { ∈ k k } 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 = S1 { ∈ | | } is connected. But this also follows from the connectedness of SU(n) through the homomorphism (λ, T ) λT : U(1) SU(n) U(n) 7→ × → 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, × 2π/n where Cn = ω is the ﬁnite cyclic group generated by ω = e . h i 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) T x 7→ 424–II 1–11 of Sp(n) on Hn. This action sends the sphere

4n 1 n S − = x H : x = 1 { ∈ k k } 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 : q 2 = t2 + x2 + y2 + z2 = 1 = S3. { ∈ k k } ∼ 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 the integral f of a function f. This approach is much simpler, where applicable, R 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.

424–II 2–1 2.1. INTEGRATION ON A COMPACT SPACE 424–II 2–2

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 f in C(X, k). Thus C(X, k) is a complete normed7→ 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(λ1f1 + λ2f2) = λ1µ(f1) + λ2µ(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 Z f(x) dµ(x) 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) S1 = T = R/Z. We parametrise S1 by the angle θ mod 2π. The usual measure dθ is a measure in our sense; in fact 1 2π µ(f) = Z f(θ) dθ 2π 0 is the invariant Haar measure on the group S1 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). 1 Measures can evidently be combined linearly, as for example µ2 = µ + 2 µ1, ie 2π 1 µ2(f) = Z f(θ) dθ + f(π). 0 2 2.2. INTEGRATION ON A COMPACT GROUP 424–II 2–3

2.2 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 ∈

1 (gµ)(f) = µ(g− 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 (gf) dµ = Z f dµ G 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 f dµ Z f ; dµ. | G | ≤ G | |

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) = λ1f(g1x) + + λrf(grx), ··· where g1, . . . , gr G, 0 λ1, . . . , λr 1, λ1 + + λr = 1. ∈ ≤ ≤ ··· These averages have the following properties: 2.2. INTEGRATION ON A COMPACT GROUP 424–II 2–4

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 1 ∈ x (xx0− )U. Since U is open, and X is compact, a ﬁnite number of these transforms∈ cover X, say X g1U grU. ⊂ ∪ · · · ∪ Now consider the average 1 F = (g1f + + grf) , r ··· 2.2. INTEGRATION ON A COMPACT GROUP 424–II 2–5 ie

1 1 1 F (x) = f(g− x) + + f(g− x) . r 1 ··· r

1 1 1 For any x, at least one of g1− x, . . . , gr− x lies in U (since x giU = gi− 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 1 π F (x) = (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 π Z f dg = Z F dg = . 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) f will be sand- R wiched between these 2 bounds,

min F Z f dg max F. ≤ ≤ 2.2. INTEGRATION ON A COMPACT GROUP 424–II 2–6

So we can determine f as a limit in this way. R 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 f. For this, we have to introduce the second action of G on C(G), by right R multiplication, (g, f) f(xg). 7→ This leads to a second way of averaging, using right transforms f(xg). The com- mutation 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 = fx fy < . | − | ⇒ | − | 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 e (the neutral element of G) such that 3 1 x− y U = fx fy < . ∈ ⇒ | − | 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 B ∈ ∈ 1 U(g) = x G : f(x) f(g) < . { ∈ | − | 2 } 2.2. INTEGRATION ON A COMPACT GROUP 424–II 2–7

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) xy. Here W 2 7→ denotes the set w1w2 : 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 = g1W1 g2W2 grWr, ∪ ∪ · · · ∪ where Wi = W (gi). Let W = iWi. ∩ 1 Suppose x− y W , ie ∈ y xW. ∈ Now x lies in some set giWi. Hence

2 x, y giWiW giW ; ∈ ⊂ i 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 = λ1g1f + + λrgrf (0 λ1, . . . , λr 1, λ1 + + λr = 1) ··· ≤ ≤ ··· 2.2. INTEGRATION ON A COMPACT GROUP 424–II 2–8 then

1 1 1 1 F (x) F (y) λ1 f(g− x) f(g− y) + + λr f(g− x) f(g− y) . | − | ≤ | 1 − 1 | ··· | r − r | But 1 1 1 1 1 1 (gi− x)− (gi− y) = x− gigi− y = x− y. Thus

1 1 1 x− y U = f(g− x) f(g− y) < ∈ ⇒ | i − i | = 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 e such that − 3 1 1 x− 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 g1U grU. ⊂ ∪ · · · ∪

One of these transforms, giU say, must contain a point x0 at which F takes its minimal value. But then, within giU, 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 1 F 0 = (g1F + grF )) , r ··· as before, then r 1 1 M m max F 0 − max F + min F + − . ≤ r r 2

Since min F 0 min F , it follows that ≥ 1 1 var F 0 1 var F + var f. ≤ − r 2r 2.2. INTEGRATION ON A COMPACT GROUP 424–II 2–9

A little thought shows that this implies that

var F 0 < var F provided 1 var F > 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

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 2 var F < var f. 3

Repeating the argument, with F in place of f, we ﬁnd a second average F 0 such that 2 2 var F 0 < 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 424–II 2–10 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 F 0, max F 0) (min F 0, max F 0) ⊃ 1 1 ⊃ 2 2 ··· converging on a different real number I0 = I. This will follow at once from the following6 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 × 1 ((g, h)f)(x) = f(g− xh).

Let us temporarily adopt the notation fh for this right action, ie

(fh)(x) = f(xh).

We can use this action to deﬁne right averages

µ (fh ). X j j The point of introducing this complication is that we can use the right averages to reﬁne the left averages, and vice versa. 2.2. INTEGRATION ON A COMPACT GROUP 424–II 2–11

Thus suppose we have a left average

F = λ (g f) X i i and a right average F = µ (fh ). 0 X j j Then we can form the joint average

F = λ µ (g fh ). 00 XX i j i j

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 1 f˜(g) = f(g− ). For it is readily veriﬁed that

˜ ˜ 1 ˜ 1 F = λ1(g1f) + + λr(grf) = F = λ1(fg− ) + + λr(fg− ). ··· ⇒ 1 ··· r Thus if F is a left average then F˜ 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 < . − 2.2. INTEGRATION ON A COMPACT GROUP 424–II 2–12

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

F = λ g f 1 X i i 1 of f1 such that

µ(f1) < min F1 max F1 < µ(f1) + ; − ≤ and similarly we can ﬁnd a right average

F = µ f h 2 X j 2 j of f2 such that

µ(f2) < min F2 max F2 < µ(f2) + . − ≤ Now let F = λ µ (g (f + f )h ) . X X i j i 1 2 j i 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 424–II 2–13 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 = g1U grU. ∪ · · · ∪ Let F be the average

1 1 1 F (x) = f(g− x) + + f(g− x) . r 1 ··· r Then 1 1 x giU = g− x U = f(g− x) δ, ∈ ⇒ i ∈ ⇒ i ≥ and so δ F (x) . ≥ r Hence δ Z f dg = Z F dg > 0. ≥ r Since min f Z f dµ max f, ≤ ≤ 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 < . | | ⇒ | | 2.2. INTEGRATION ON A COMPACT GROUP 424–II 2–14

It follows at once from

min f Z f dg max f ≤ ≤ that 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 integral on G with 1 dg = 1; for any such integral must be sandwiched between R 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:

1 Z f(g− 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 f(gh) dg = Z f(g) dg (h G, f C(G, R)) . G G ∈ ∈

Proof Suppose h G. The map I ∈

µh : f µ(fh) 7→ 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 424–II 2–15

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 (f) C(G) denote the set of averages of f. This set is convex, ie A ⊂

F,F 0 (f) = λF + (1 λ)F 0 (f) (0 λ 1). ∈ A ⇒ − ∈ A ≤ ≤ Let Λ C(G) denote the set of constant functions f(g) = c. Evidently ⊂ R Λ ∼= . We want to show that Λ (f) = , ∩ A 6 ∅ ie the closure of (f) contains a constant function. (In other words, we can find a sequence of averagesA converging on a constant function.) To prove this, we establish that (f) is pre-compact, ie its closure (f) is compact. For then it will follow thatA there is a ‘point’ X (f) (ie a functionA X(g)) which is closest to Λ. But if this point is not in Λ,∈ weA 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 (f) is pre-compact. We note in the first place that the set of transforms of f, A

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). 7→ → Also, (f) is the convex closure of this set Gf, ie the smallest convex set containingAGf (eg the intersection of all convex sets containing G), formed by the points

λ1F1 + + λrFr : 0 λ1, . . . , λr 1; λ1 + + λr = 1 . { ··· ≤ ≤ ··· } Thus (f) is the convex closure of the compact set Gf. But the convex closure of aA compact set in a complete metric space is always pre-compact. That 2.2. INTEGRATION ON A COMPACT GROUP 424–II 2–16 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 (f) is non-empty. We must then show that it consists of a single point. This we∩ A do as in our proof proper, by introducing right averages. Finally, we define f dg to be this point of intersection (or R 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 S1 is 1 dθ. 2π In other words, 1 2 µ(f) = Z πf(θ) dθ. 2π 0 2. Consider the compact group SU(2). We know that

3 SU(2) ∼= S , since the general matrix in SU(2) takes the form x + iy z + it U = ! , x 2 + y 2 + z 2 + t 2 = 1. z + it x iy | | | | | | | | − − The usual volume on S3, 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 x + iy z + it x + iy z + it U ! = 0 0 0 0 ! z + it x iy z0 + it0 x0 iy0 − − − − then 2 2 2 2 2 2 2 2 x0 + y0 + z0 + t0 = x + y + z + t for all (x, y, z, t) R4. ∈ It follows that multiplication by U preserves the volume on S3. In other words, this volume provides an invariant measure on SU(2), which must therefore be—after normalisation—the Haar measure on SU(2). 2.2. INTEGRATION ON A COMPACT GROUP 424–II 2–17

As this example—the simplest non-abelian compact group—demonstrates, 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. 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 n V ∼= k ; and this unique topology on V is just that arising from the product topology on kn.

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 . ⊂

424–II 3–1 3.1. REPRESENTATIONS OF A COMPACT GROUP 424–II 3–2

3.1 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 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 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 424–II 3–3

3.3 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 ﬁnite- dimensional 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 SEMISIMPLE424–II3–4

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 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 u, v = Z p(gu, gv) dg. h i V It is a straightforward matter to verify that this new inner product is invariant:

gu, gv = u, v . h i h i It also follows at once from the positivity of the Haar measure that this inner product is positive, ie v, v 0. h i ≥ It’s a little more difﬁcult to see that the inner product is positive-deﬁnite, ie

v, v = 0 = v = 0. h i ⇒ 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 f(g) dg 0 G ≥ (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 g1U . . . grU. ⊂ ∪ It follows from this that

1 1 f(g− x) + + f(g− x) /2 1 ··· r ≥ 3.6. EVERY REPRESENTATION OF A FINITE GROUP IS SEMISIMPLE424–II3–5 for all x G. For ∈ 1 1 x giU = g− x U = f(g− x) /2. ∈ ⇒ i ∈ ⇒ i ≥ It follows from this, on integrating, that

r Z f(g) dg /2. G ≥ In particular f 0. R Note that our≥ alternative proof of semisimplicity also carries over to the compact case. This proof depended on the fact that if

π : V V → is a projection onto a stable subspace U = π(V ) of V then its average 1 Π = gπg 1 X − 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

1 Π = Z gπg− dg. G For here we are integrating the operator-valued function

1 F (g) = gπg−

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 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 SEMISIMPLE424–II3–6 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 f(g, h) X X 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 ∈ I = Z 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) = Z Z f(g, h) dh dg. G G g G 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:

2 1 1 Π = Z gπg− dg Z hπh− dh g G h G ∈ ∈ 1 1 = Z gπg− hπh− d(g, h) (g,h) G G ∈ × 1 1 = Z gg− hπh− d(g, h) (g,h) G G ∈ × (using the fact that πgπ = gπ, since U = im π is stable under G). Thus

2 1 Π = Z hπh− d(g, h) (g,h) G G ∈ × 1 = Z dg Z hπh− dh g G h G ∈ ∈ = Π. 3.7. UNIQUENESS AND THE INTERTWINING NUMBER 424–II 3–7

3.7 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 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 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 u, v on the representation- space V . We can choose a basis for V so that h i

2 2 v, v = x1 + + xn , h i | | ··· | | 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:

Uv = λv = v∗U ∗ = λv∗ ⇒ = v∗U ∗Uv = λλv∗v ⇒ 2 = v∗v = λ v∗v ⇒ | | = λ = 1. ⇒ | | Hence 1 λ− = λ for each such eigenvalue. J Alternative proof I Recall how we proved this in the ﬁnite case. By Lagrange’s Theorem gm = 1 for some m > 0, for each g G. Hence ∈ α(g)m = I; 3.8. THE CHARACTER OF A REPRESENTATION 424–II 3–8 and so the eigenvalues of α(g) all satisfy

λm = 1.

In particular λ = 1; | | and so 1 λ− = 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 gn of g approach arbitrarily close to the 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 neighbourhood3 of e such that 1 n VV − U. Then the subsets g V are disjoint. For ⊂ m n m n x g V g V = x = g v1 = g v2 ∈ ∩ ⇒ n m 1 = g − = v1v2− ⇒ n m = g − U, ⇒ ∈ contrary to hypothesis. It follows [the details are left to the student] that the subgroup

1 2 g = . . . , g− , e, g, g ,... h i { } 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 e → as i . It→ follows ∞ that α(g)ni I → 3.9. THE REGULAR REPRESENTATION 424–II 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 1 χα(g− ) = χα(g) for all g G ∈

Proof I Suppose the eigenvalues of α(g) are λ1, . . . , λn. Then the eigenvalues of 1 1 1 1 α(g− ) = α(g)− are λ1− , . . . , λn− . Thus

1 1 χα(g− = tr α(g− ) 1 1 = lambda− + + λ− 1 ··· n = λ1 + + λn ··· = λ1 + + λn ··· = tr α(g)

= χα(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. → 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 424–II 3–10 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 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 = g1H grH. ∪ · · · ∪ 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 424–II 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 → H H G hom (X,W ) = hom hom (V,U),W is non-zero. But it is readily veriﬁed that

H G G H hom hom (V,U),W = hom × (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 424–II 3–12

3.12 Real Representations

Everything in this chapter carries over to the compact case, with no especial prob- lems arising. Chapter 4

Representations of U(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, 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(gn) = α(g) n | | | | → ∞ as n , contradicting the boundedness of α(G). On→ the ∞ other hand,

1 1 α(g) < 1 = α(g− = α(g) − > 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). →

424–II 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 an : A A 7→ → is a homomorphism.

Deﬁnition 4.1 For each n Z, we denote by En the representation ∈ eiθ einθ 7→ 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 h∈2 = 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 ∈ ωN = 1 = α(ω)N = 1. ⇒ It follows that 2πni/N n α(ω) = e = ω = En(ω) for some n Z in the range N/2 < n < N/2. We shall deduce from this that ∈ − α = En. Let πi/N ω1 = e . 424–II 4–3

Then

2 2 n ω1 = ω = α(ω1) = α(ω) = ω ⇒ n = α(ω1) = ω , ⇒ 1 since this is the unique square root of ωn in U. Repeating this argument successively with we deduce that if

2π j i ωj = e 2 N then n α(ωj) = ωj = En(ωj) for j = 2, 3, 4,... . But it follows from this that

k k n k α(ωj ) = (ωj ) = En(ωj ) for k = 1, 2, 3,... . In other words

α(eiθ) = E(eiθ) for all θ of the form k θ = 2π 2j But these elements eiθ are dense in U(1). Therefore, by continuity,

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 2π 1 iθ nθ cn = Z α(e )e− 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 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 → iθ i(θ φ) (gf)(e ) = e − .

The Fourier series f(eiθ) = c einθ X n n Z ∈ expresses the splitting of C(U(1)) into 1-dimensional spaces

C (U(1)) = V , M n where inθ inθ Vn = e = ce : c C . h i { ∈ } Notice that with our deﬁnition of group action, the space Vn carries the repre- iφ iθ inθ sentation E n, rather than En. For if g = e , and f(e ) = e , then − iθ inφ iθ iθ (gf)(e ) = e− f(e ) = E n(g)f(e ). − In terms of representations, the splitting of C(U(1)) may be written:

ρ = E . X n 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 1 0 ! ! 0 1 6∼ 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. ≡ 424–II 5–1 5.1. CONJUGACY IN SU(N) 424–II 5–2

Proof We know that the eigenvalues of T U(n) have absolute value 1, since I ∈

T v = λv = v∗T = λv¯ ∗ ⇒ = v∗T ∗T v = λλv¯ ∗v ⇒ = v∗v = λλv¯ ∗v ⇒ = λ 2 = λλ¯ = 1 ⇒ | | = λ = 1 ⇒ | | Thus the eigenvalues of T can be written in the form

iθ1 iθn e , . . . , e (θ1, . . . , θn R). ∈ In particular the diagonal matrices in U(n) are just the matrices

eiθ1   ..  .   iθ   e n  It follows that the homomorphism

eiθ1   U(1)n U(n): eθ1 , . . . , eθn ..  .  → 7→  iθ   e n  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 T denote the closed subgroup generated by T , ie the closure in U(n) of the grouph i 1 2 ...,T − ,I,T,T ,... { } formed by the powers of T . 5.1. CONJUGACY IN SU(N) 424–II 5–3

This group is abelian; and its natural representation in Cn leaves invariant the 2 2 standard positive-deﬁnite hermitian form x1 + + xn , 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 orthonor- mal set of eigenvectors of T . If U is the matrix of change of basis, ie

U = (e1, . . . , en) then eiθ1   U TU = .. ∗  .   iθ   e n  where θi T ei = e ei.

J Lemma 5.3 The diagonal matrices in SU(n) form a subgroup isomorphic to the n 1 n 1 torus group T − U(1) − . ≡

Proof I If eiθ1   T = ..  .   iθ   e n  then i(θ1+ +θn) det T = e ··· . Hence T SU(n) θ1 + + θn = 0 (mod 2π). ∈ ⇐⇒ ··· Thus the homomorphism

eiθ1   .. n 1 θ1 θn 1 . U(1) − SU(n): e , . . . , e −    iθn 1  → 7→  e −   i(θ1+ +θn 1)   e− ··· − 

n 1 maps U(1) − homeomorphically onto the diagonal subgroup of SU(n), allowing us to identify the two: n 1 U(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: ∈ n 1 T D U(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 ∗TU = D (U U(n)). ∈ We know that det U = 1, say | | det U = eiφ.

Let iφ/n V = e− U. Then V SU(n); and ∈ V ∗TV = D.

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

1 D2 = P − D1P.

2 Now P U(n) since permutation of coordinates clearly leaves the form x1 + ∈2 | | + xn 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 iφ/n V = e− U. Then V SU(n); and ∈ T = V ∗SV

5.2 Representations of SU(2)

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

iθ 1. each T SU(2) has eigenvalues e± ∈ 2. with the same notation,

eiθ 0 ! T U(θ) = iθ ∼ 0 e−

3. U( θ) U(θ) − ∼ Thus SU(2) divides into classes C(θ) (for 0 θ π) containing all T with iθ ≤ ≤ eigenvalues e± . 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 z z ! 0 ! = T ! w 7→ w0 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 ∈ a b U = ! ( a 2 + b 2 = 1) ¯b a¯ | | | | − Taking T in this form, its action is given by

(z, w) (az + bw, ¯bz +aw ¯ ) 7→ − By extension, this change of variable deﬁnes an action of SU(2) on polynomials P (z, w) in z and w:

P (z, w) P (az + bw, ¯bz +aw ¯ ). 7→ −

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

2j 2j 1 2j V (j) = z , z − w, . . . , w h i of homogeneous polynomials in z, w of degree 2j.

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

z3, z2w, zw2, w3 form a basis for V(3/2). Consider the action of the matrix 0 i T = ! SU(2). i 0 ∈ We have

T (z3) = (iw)3 = iw3, − T (z2w) = izw2, − T (zw2) = iz2w, − T (w3) = iz3 −

Thus under D 3 , 2 0 0 0 i 0 i  0 0 i −0  !  −  i 0 7→  0 i 0 0   −   i 0 0 0  − 5.2. REPRESENTATIONS OF SU(2) 424–II 5–7

Proposition 5.1 The character χj of Dj is given by the following rule: Suppose iθ T has eigenvalues e± Then

2ijθ 2i(j 1)θ 2ijθ χj(T ) = e + e − + + e− ···

Proof I We know that eiθ 0 ! T U(θ) = iθ . ∼ 0 e−

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

k 2j k k 2j k iθ iθ − U(θ)z w − = e z e− w 2i(k j)θ k 2j k = e − z w − .

Thus under Dj,

e2ijθ  2i(j 2)θ  e − U(θ)    ..  7→  .   2ijθ   e−  whence 2ijθ 2i(j 2)θ 2ijθ χj (U(θ)) = e + e − + + e− . ··· J

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

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

(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) = z2j w2j h i ⊕ · · · ⊕ h i for V (j) as a direct sum of simple U(1)-modules is unique. 5.2. REPRESENTATIONS OF SU(2) 424–II 5–8

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 zawb. In particular z2j must belong either to U or to W . Without loss of generality we may suppose that z2j U ∈ But then T (z2j) U ∈ for all T SU(2). In particular, taking ∈ 1 1 1 T = ! √2 1 1 − (almost any T would do) we see that

2j 2j 2j 1 2j (z + w) = z + 2jz − w + + w 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:

2j k k z − w U for all k. ∈ Hence U = V (j), ie Dj is simple. 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 α = n E , U(1) X j j j where of course all but a ﬁnite number of the nj vanish (and the rest are positive integers). It follows that χ (U(θ)) = n eijθ α X j j 5.2. REPRESENTATIONS OF SU(2) 424–II 5–9

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):

χ (U(θ)) = c χ (U(θ)) . α X j j j

Since each T SU(2) is conjugate to some U(θ) it follows that ∈ χ (T ) = c χ (T ) α X j j 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 DjDk 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

DjDk = Dj+k + Dj+k 1 + + D j k . − ··· | − |

Proof I We may suppose that j k. ≥ iθ Suppose T has eigenvalues e± . 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 χD D (T ) = χj(θ)χk(θ) = L( j, j)L( k, k). j k − − 5.2. REPRESENTATIONS OF SU(2) 424–II 5–10

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 2ikθ 2ikθ L(k) = L(k 1) + (e− + e ). − But

2ikθ L( j, j)e− = L( j k, j k), − − − − L( j, j)e2ikθ = L( j + k, j + k). − − Thus

2ikθ 2ikθ L( j, j)(e− + e ) = 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

2ijθ 2i(j 1)θ 2ijθ χj(θ) = e + e − + + e− ··· is real, since

2ijθ 2i(j 1)θ 2ijθ χj(θ) = e− + e− − + + e = χj(θ). ···

Thus Dj (which we know to be simple) is either real or quaternionic. 5.2. REPRESENTATIONS OF SU(2) 424–II 5–11

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 D 1 is quaternionic. 2

Proof of Lemma Suppose D 1 were real, say B 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 2 2 bring this to x1 + x2, 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 . 2 C

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) gq (g Sp(1), q H). 7→ ∈ ∈ 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 1 It remains to prove that Dj is quaternionic for half-integral j > 2 . Suppose in fact Dj were real; and suppose this were the ﬁrst half-integral j with that property. Then DjD1 = 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 1 if α is real, 2  Z χα(g ) dg =  0 if α is essentially complex, 1 if α is quaternionic.  − iθ 2 Let α = Dj. Suppose g SU(2) has eigenvalues e± . Then g has eigenval- 2iθ ∈ ues e± , and so

2 4ijθ 4i(j 1)θ 4ijθ χj(g ) = e + e − + + e− ··· 2j = χ2j(g) χ2j 1(g) + + ( 1) χ0(g). − − ··· − Thus

2 2j Z χj(g ) dg = Z χ2j(g) dg Z χ2j 1(g) dg + + ( 1) Z χ0(g) dg − − ··· − 2j = I(1,D2j) I(1,D2j 1) + + ( 1) I(1,D0) − − ··· − = ( 1)2jI(1, 1) − +1 if j is integral = ( 1 if j is half-integral − J 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 ⊂ 1 V − V U, ⊂ 1 ie v− v2 U for all v1, v2 V . This follows from the continuity of the map 1 ∈ ∈ 1 (x, y) x− y : G G G. 7→ × → 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 of S. For suppose3 s, t gV , say 3 ∈ s = gv1, t = gv2.

424–II 6–1 424–II 6–2

Then 1 1 s− t = v− v2 U S. 1 ∈ ∩ 1 Thus s− 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 ker Θ = n. k k 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 α0 = α0 α1 = α2. 1 2 ⇐⇒ 3. With the same notation,

(α1 + α2)0 = α10 + α20 , (α1α2)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. 424–II 6–3

Proof All follows from the fact that gv (g G, v V ) is the same whether I ∈ ∈ 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, de- ﬁned by a homomorphism

Θ: 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 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 denote the space of all 2 2 hermitian matrices, ie all matrices of the form H × x y iz A = − ! (x, y, z, t R). y + iz t ∈ Evidently is a 4-dimensional real vector space. (It is not a complex vector space, sinceHA hermitian does not imply that iA is hermitian; in fact

A hermitian = iA skew-hermitian, ⇒ 424–II 6–4

since (iA)∗ = iA∗ for any A.) Now suppose− U SU(2). Then ∈ A = (U ∗AU)∗ = U ∗A∗U ∗∗ = U ∗AU ∈ H ⇒ = U ∗AU . ⇒ ∈ H Thus the action 1 (U, A) U ∗AU = U − AU 7→ of SU(2) on deﬁnes a 4-dimensional real representation of SU(2). This is notH quite what we want; we are looking for a 3-dimensional representation. Let = : tr = T {A ∈ H A 0} denote the subspace of formed by the trace-free hermitian matrices, ie those of the form H x y iz A = − ! (x, y, z, t R). y + iz x ∈ − These constitute a 3-dimensional real vector space; and since

1 tr (U ∗AU) = tr U − AU = tr A this space is stable under the action of SU(2). Thus we have constructed a 3- dimensional 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 , since T x y iz det − ! = x2 y2 z2 y + iz x − − − − Moreover this quadratic form is left invariant by the action of SU(2), since

1 det (U ∗AU) = det U − AU = det A. In other words, SU(2) acts by orthogonal transformations on , so that T 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. 424–II 6–5

Lemma 6.1 ker Θ = I . {± } Proof of Lemma Suppose U ker Θ. In other words, B ∈ 1 U − AU = A for all A . In fact∈ this T will hold for all hermitian matrices A since ∈ H = . H T MhIi 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: ∈ 1 1 A = (A + A∗) + (A A∗) . 2 2 − Since 1 U − AU = A AU = UA, ⇐⇒ 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

eiθ 0 ! U = U(θ) = iθ ; 0 e− and suppose x y iz A = − ! . y + iz x ∈ T − 424–II 6–6

Then e iθ 0 x y iz eiθ 0 − ! ! ! U ∗AU = iθ − iθ 0 e y + iz x 0 e− − x e 2iθ(y iz) = − ! e2iθ(y + iz) x− − XY iZ = ! , Y + iZ −X − where

X = x Y = cos 2θy + sin 2θz Z = sin 2θy cos 2θz. − Thus U(θ) induces a rotation in the space through 2θ about the Ox-axis, say T U(θ) R(2θ, Ox). 7→ As another example, let

1 1 1 V = ! ; √2 1 1 − In this case x y + iz XY + iZ V AV = ! = ! , ∗ y iz x 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 homomor- 1 phism, VU(θ)V − maps onto

1 TR(2θ, Ox)T − = 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 424–II 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

1 TSR(φ, l)(TS)− = R(φ, Ox); and so 1 1 R(φ, l) = S− T − 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 . {± }

3 Recall that SU(2) ∼= S . The result shows that SO(3) is homeomorphic to 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:

3 4 SO(3) = P (R) = (R 0 )/R×. ∼ \{ } 2. We shall see in Part 4 that the space (or more accurately the space i ) is just the Lie algebra of the group SUT(2). Every Lie group acts on itsT 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) ( z, w) 7→ − − and so if P (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 if 2j is even, ie j is integral. J The− following result is almost obvious. 424–II 6–8

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: × 1 ((g, h)f)(x) = f(g− 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

424–II 7–1 7.1. THE FINITE CASE 424–II 7–2

where t0 is the linear map 1 t0(v) = ht(g− v). The expression for τ above can be re-written as

C(G) hom(Vσ ,Vσ ) + + hom(Vσ ,Vσ ), ≡ 1 1 ··· s s where Vσ is the space carrying the simple representation σ. In other words C(G) = C(G)σ C(G)σ , 1 ⊕ · · · ⊕ s 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 × ⊂ × 1 (gf)(x) = f(g− 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)σ C(G)σ , 1 ⊕ · · · ⊕ s 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 1 π = χ(g 1)g. X − G g G | | ∈ It follows that 1 f (x) = χ(g)f(gx). σ X G g G | | ∈