Denition 1: A (unitary) representation of G is a continuous homomorphism from G

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1 The Peter-Weyl Theorem for Compact roups The following notes are from a series of lectures I gave at Dartmouth College in the summer of 989. The general outline is provided by an introductory section in [] but with considerable detail added by myself. The mistakes of course are mine. x Preliminaries. Dana P. Williams Hanover, June 99 We begin with some warm-up exercises on locally compact groups a.k.a., a long series of denitions! At this point, is meant to be an arbitrary locally compact group. Denition : A (unitary) representation of is a continuous homomorphism from to the unitary group U(H ) on a (complex) Hilbert space H equipted with the strong operator topology. Remark 2: The condition that be continuous merely means that g 7! (g) is continuous from to H for each 2 H. There are many equivalent conditions: the weakest I'm aware of is to insist that g 7! h(g) i be Borel for each 2 H. Example 3: Let H L 2 (). The left regular representation :!B(H) isgiven by (g)f(t) f(g ; t). Denition 4: Two representations and 2 are said to be (unitarily) equivalent ifthere is a unitary operator U : H! H 2 such that (g) U 2 (g)u for all g 2. In this event, we write 2 and let [] denote the (unitary) equivalence class of. If :! U(H ) is a representation of, then we'll write d for the dimension (in f 0 2 ::: g)ofh. Fortunately, whenever 2, then it is clear that d d 2. Thus we will often write d [] to denote the dimension of each representation in the same equivalence class as. Denition 5: A non-zero representation is called irreducible if H has no no-trivial closed invariant subspaces Remark 6:Ifd <, then the word \closed" in redundent in the above denition. Denition 7: The symbol b is used to denote the collection of equivalence classes of irreducible representations of. Example 8: If is abelian, then every irreducible representation is one-dimensional. In particular, b coincides with the character group of.

2 Denition 9: If and are representations of, then denotes the representation on H H dened by (g)( ) ((g) (g)): If n 2 + [fg, then n n i. Denition 0: If is a representation of, then () 0 f A 2 B(H ):A(g) (g)a for all g 2 g: Remark: Since is unitary, () 0 is a self-adjoint. In fact it is not hard to check that () 0 is a -subalgebra of B(H )which is closed in the weak operator topology. Theorem 2: Suppose that is a representation of. Then the following are equivalent. () is irreducible. (2) () 0 C I. (3) Every non-zero 2 H is cyclic for (i.e., () spanf (g) : g 2 g H ). Proof: Suppose that is irreducible. Let A 2 () 0 and suppose for the moment that A is normal: A A AA. Then the norm closed unital *-subalgebra generated by A that is the C -algebra C (I A) is contained in () 0. Since A is normal, C (I A) is commutative and is isomorphic to C ; (A) by the spectral theorem. If (A) 6 f pt g, then we can nd nonzero self-adjoint operators B and B 2 in C (I A) so that B B 2 B 2 B 0. Thus hb B 2 i 0 for all 2 H. In particular, the closures V and V 2 of the ranges of B and B 2, respectively, are closed, non-zero, orthogonal, invariant subspaces for. This contradics the irreducibility of therefore (A) must be a single point and therefore A I for some 2 C.For a general A 2 () 0,we apply the above reasoning to AA and A A. Thus, we have AA I and A A I with > 0 (since A 6 0). Since A A(A A)A, wehave and A is normal. This shows that () implies (2). Since V () is a closed, non-zero, invarinat subspace, in order to show that (2) implies (3) it will suce to show that the orthogonal projection P onto any closed, non-zero, invarinat subspace V is in () 0. But since is unitary, V? is also invariant. Thus if 2 H, then hp(g) i h(g) Pi which, since (g)(i ; P ) 2 V? and P 2 V, h(g)p + (g)(i ; P ) Pi 2

3 h(g)p Pi hp(g)p i h(g)p i because (g)p 2 V. That (3) implies () is clear. 3

4 x2 The Peter-Weyl Theorem. Now we'll specialize to compact groups. Compact groups are characterized by the fact that any Haar measure on satises () <. It is customary to normalize Haar measure on a compact group by choosing the unique measure such that (). Since there is now no possibility of confusion, I'll simply write for the integral of f 2 L (). f(g) dg Now suppose that is a nite dimensional representation of. If b f e ::: e d g is an orthonormal basis for H, then for each g 2 the matrix of (g) with respect to b has ij th coordinate h(g)e j e i i. The function ij (g) he i (g)e j i is called a coordinate function for. (It will be convenient to use this convention even though ij is the complex conjugate of what you might expect. This usage and terminolgy will be justied, somewhat, by Remark 4 below.) Notice that ij 2 C(). We'll write E, or just E when no confusion is likely to arise, for the linear span of the all the functions (g) h (g)i, where ranges over all irreducible representations of and and range over H. Since every nite dimensional representation is the direct sum of irreducibles, notice that (g) h (g)i denes an element ofe for every nite dimensional representation irreducible or not. Remark 3: When is abelian, and sometimes in general, the functions in E are called the trigonometric polynomials. The motivation for this probably comes from the case where T R2. Then each f 2 E T f() nk n;k c n exp(in) has the form k n0 d n cos(n) +b n sin(n): Remark 4: It is clear that E is self-adjoint that is, if f 2 E, then so is f 2 E, where f (g) f(g ; ). This is because matrix coecients are themselves self-adjoint: ij ( ij ). It is also true that if f 2 E, then so is f, where f(g) f(g ; ). To see this we need to introduce the conjugate Hilbert space e H to a given Hilbert space H. The space e H coincides with H as an additive group. If j : H! e H denotes the identity map, then the Hilbert space structure on e H is given by the formulas j() j(), and 4

5 hj() j()i e H h i H. If is a given representation of on H, then we can dene a representation e on e H in the obvious way: e(g)j() j ; (g). The assertion follows from the fact that e ij (g) ij (g; ). g 7! h (g)i a coordinate function. Since ij (g; ) ji (g), it is reasonable to call Denition 5: Let M n be the n n complex matrices. If A (a ij ) 2 M n, then the Hilbert-Schmidt norm of A is k(a ij )k H:S: ij ja ij j 2 : Remark 6:IfA2M n, then kak H:S: tr(a A). In particular, if B U AU for some unitary matrix U, then kak H:S: kbk H:S:. It follows that if is nite dimensional, then k(g)k H:S: is well dened and depends only on []. Our object here is to prove the following theorem known as the Peter-Weyl Theorem. Theorem 7: Let be a compact group. () Every irreducible representation of is nite dimensional. (2) If is the left-regular representation of, then M d []2 b (3) iven g 2, there is a [] 2 b such that (g) 6 I. (4) E is dense in C() (and hence in L p () for p<). (5) If f 2 L 2 (), then kfk 2 2 d tr (f)(f) ) d k(f)k H:S:: []2 b []2 b We'll need the following preliminary results, some of which may be of interest by themselves. 5

6 Lemma 8: Let and 2 be representations of a locally compact group. If A : H! H 2 is a bounded linear operator satisfying A (g) 2 (g)a for all g 2, then A A (g) (g)a A for all g 2. Proof: One computes as follows: ha A (g) i ha (g) Ai h 2 (g)a Ai ha 2 (g ; )Ai ha A (g ; )i h (g)a A i Lemma 9: Suppose that and 2 are representations of a locally compact group, that is irreducible, and that A : H! H 2 is any non-zero bounded linear operator such that A (g) 2 (g)a for all g 2. Then AH is a closed invariant subspace for 2, and 2 j AH, where 2 j AH is the subrepresentation of 2 corresponding to AH. Proof: By Lemma 8, A A 2 () 0. By Theorem 2, A A I. Thus B ; 2 A is an isometery, and hence a unitary from H onto AH. (Note that AH is actually closed in H 2 since A isamultiple of an isometery.) Again Lemma 8 shows that BB 2 2 () 0. Since BB is the orthogonal projection onto BH AH, it follows that AH is invariant for 2 and the assertion follows. The next result is crucial, and depends heavily on the fact that is compact. Proposition 20: Suppose that and 2 are irreducible representations of a compact group. Fix orthonormal bases f e i k gd i for H i, and put i kl (g) hei k i(g)e i l i. () If 6 2, then ij (g)2 jl (g) dg 0 for all i j d,and k l d 2. (2) If is nite dimensional, then ij (g) kl (g) dg ik jl d 6

7 for all i j k l d. Proof: Let B be any bounded linear operator from H to H 2.Thenwe can dene Then A (r) A 2 (g)b (g ; ) ds: 2 (g)b (g ; r) dg 2 (rg)b (g ; ) dg 2 (r)a: If 6 2,thenA 0by Lemma 9. Now suppose that B B ij is the rank-one operator dened by B ij () h e j ie2 l. Then, 0hAe i e2 k i This proves (). hb ij (g ; )e i 2(g ; )e 2 k i dg h (g ; )e i e j ihe2 l 2(g ; )e 2 k i dg ij (g)2 kl (g) ds: Now assume that d <. By the above and Lemma 9, for any B. Taking traces, A tr(a) d (g)b (g ; ) dg I hae k e k i d hb (g ; )e k (g ; )e ki dg tr(b) dg tr(b): Since tr(a) tr(i) d and tr(b jl ) jl,wehave jl d when B B jl.onthe other hand, ij(g) kl (g) dg hae i e ki he i e ki ik jl : d 7

8 Proof of Theorem 7: Let E f be the subset of E consisting of the collection of matrix coecients of the form (g) h (g)i for irreducible and d <. The rst part of the proof will consist of showing that it suces to show that E f is dense in C(). Since C() is dense in L 2 (), it follows that L 2 () H has an orthonormal basis consisting of normalized matrix coecients d 2 ij (g) d 2 he i (g)e j i (where, of course, f e ::: e d g denotes an orthonormal basis for H ). In fact, if is any irreducible representation of and if f e g 2A is an orthonormal basis for H, then it follows from Proposition 20 that? ij, where (g) he (g)e i and d <. Thus we must have 0, and () follows. If d <, then Proposition 20 shows that the d 2 functions f p d ij g areanorthonormal basis for a subspace H [] of L 2 (). Now observe that ij(g ; t)h(g)e i (t)e j i d d h(g)e i e kihe k (t)e j i ik(g ; ) kj(t): Therefore if, for each j d,we dene A j : H! H [] by A j e i ij,then (g)(ae i )(t) ij (g; t) d d ik (g; ) kj (t) he i (g ; )e kia j e k(t) d A h(g)e i e k ie k (t) ; A (g)e i (t): That is, A intertwines the irreducible representation and the subrepresentation jh [] j, where H [] j spanf j 2j ::: d j g. Therefore Lemma 9 implies that jh [] j, and thus, jh [] d. Since we're assuming that E f is dense, we have H M []2 b H [] 8

9 and we have proved (2). Wehavenowshown that f d 2 [] ij g []2 b forms an orthonormal basis for L 2 (). (Under the assumption that E f is dense in C().) Thus, if f 2 L 2 (), we can write f c ; [] i j d 2 [] ij: Furthermore, kfk 2 2 d [] []2 b i j jc ; [] i j j 2 : Now (5) follows from the fact that ; c [] i j d 2 [] f(g)d 2 [] ij (g) dg f(g)h(g)e j e i i dg d 2 [] h(f)e j e i i dg d 2 [] (f) Of course, (3) follows from (4) (otherwise, E wouldn't separate e and g), so it only remains to prove that E f is dense in C(). Towards this end, we need to recall some basic facts about so-called Hilbert-Schmidt operators. If ( M ) is a measure space and if K 2 L 2 ( ), then we can dene a bounded linear operator T : L 2 ()! L 2 () by Tf(x) ij : K(x y)f(y) dy: It is not hard to see that T is self-adjoint ifk(x y) K(y x). An operator of this form is called a Hilbert-Schmidt operator and all such operators are self-adjoint compact operators (). In particular, each eigenspace H f f 2 L 2 () :Tf f g () In our case, we'll only be interested in the case where, is normalized Haar measure, and K is continuous. Then the Stone-Weierstrass Theorem implies that there are functions i 2 C() such that K(x y) i i(x) i(y) () i2i 9

10 is nite dimensional and there is an orthonormal sequence f i g of eigenvectors with eigenvalues i so that every f 2 L 2 () can be written uniquely as where T 0 0 and c i hf i i. f c i i + 0 Our interest in such operators is as follows. Let k be any element of C() which satises k(g) k(g ; ). Therefore f k(g) f(r)k(r ; g) dr K(g r)f(r) dr where K(g r) k(r ; g), is a self-adjoint Hilbert-Schmidt operator. Let C kkk max x y2 jk(y ; x)j. Notice that ktfk Ckfk Ckfk 2. A moments reections allows one to see that this implies that Tf 2 C(). (Of course, Tf is only dened almost everywhere, but I mean it agrees almost everywhere with a continuous function on. Since this function is uniquely determined, it makes sense to treat Tf as a continuous function. This is standard practice.) It follows that each eigenfunction of T is continuous. Lemma2: Let k be as above and let T be the Hilbert-Schmidt operator on L 2 () dened by Tf f k. Then for each 2 Crf 0 g H f f 2 L 2 () :Tf f ge f : Proof: By the above remarks, H is nite dimensional and consists of continuous functions. Suppose that f 2 H.ThenT ; (g)f ; (g)f k (g) ; f k (g)(tf) ; (g)f. That is, H is invariant for. Let f f ::: f r g be an orthonormal basis for H. Dene continuous functions ki by Since (g)f i 2 H, ki(g) h(g)f i f k i: () f i (g ; t) r ki(g)f k (t): (2) uniformly. Notice that for each nite subset F I the operator T F corresponding to K F (x y) i i(x) i(y) i2f is a nite rank operator. Since the convergence in () is uniform, it follows that T F! T in the operator norm hence T is compact. (The remaining assertions in the paragraph follow from the Spectral Theorem.) 0

11 (A priori, Equation (2) is an equality inl 2 (), and so would yeild pointwise equality only almost everywhere. But since both sides are continuous, the equality must hold everywhere.) Now dene (g) ; to be the operator on H whose r r matrix with respect to the basis f f ::: f r g is ij. Since we have ij (g) ji (g ; ), it follows from Equation () that (g) (g ; ). Similarly, ij(gt) h(t)f j (g ; )f i i r r hf k (g ; )f i ih(t)f j f k i ik(g) kj (t): Therefore (gt) (g)(t). It follows that is a nite dimensional (unitary) represention of. Using Equation (2), we see that f i (g) r ki(g ; )f k (e) r f k (e)(g) where ki (g) ki (g ; )hf i (g)f k i. Wehave shown that each f i, and hence H,is in the span of the matrix coecients of nite dimensional representations of. Since every nite dimensional representation is the direct sum of irreducible (nite dimensional) representaions, we have H E f as desired. Lemma 22: Let k and T be as above. If f 2 L 2 (), then Tf 2 E f. Proof: Let f i g be a complete orthonormal set of eigenvectors for the eigenspaces with non-zero eigenvalues of T. By the spectral theorem, we can write f k c k k + 0 where T 0 0 and kfk 2 2 P k jc kj 2.LetT k k k.iven >0, there is an N so that k P k>n c k k k 2 <. In particular, kt N T c k k P k>n c k k k 2 C. But N k c k k 2 E f

12 by Lemma 2. This suces as ktf ; N k c k k k kt c k k k C: k>n The Peter-Weyl theorem now follows as C() always contains a self-adjoint approximate identity. Specically, we have the following. Proposition 23: If is a compact group, then there is a net f k g in C() which satises k k, and such that both f k f g and f f k g converge uniformly to f for each f 2 C(). Proof: For each neighborhood R U of e 2, letk U be a continuous non-negative function which satises k U (e), k U(g) dg,k U k U, and supp k U. Then f k U g U is a net in C() directed by reverse inclusion: i.e., U V if and only if U V. Fix f 2 C(). Since is compact, given >0, there is a neighborhood W of e 2 such that jf(t ; g) ; f(g)j <for all g 2 and t 2 W. Therefore if U W, then jk U f(g) ; f(g)j k U (t)f(t ; g) dt ; f(g) k U (t) f(t ; g) ; f(g) dt k U (t) dt : It follows that k U f! f uniformly, as claimed. On the other hand, (k U f) f k U implies that f k U! f uniformly as well. This proves the lemma. Let T U be the Hilbert-Schmidt operator corresponding to k U. Thus, T U f converges to f uniformly for each f 2 C(). Thus, f 2 E f by Lemma 22, and Theorem 7 is proved. 2

13 References [] Elias Stein, Topics in harmonic analysis, related to the Littlewood-Paley theory, Annals of mathematics studies, no. 63, Princeton University Press, New Jersey, 970 3