CHAPTER 6. Representations of compact groups

CHAPTER 6 Representations of compact groups Throughout this chapter, denotes a compact group. 6.1. Examples of compact groups A standard theorem in elementary analysis says that a subset of C m (m a positive integer) is compact if and only if it is closed and bounded. The group U(n) := { g L n (C) t ḡg = 1 } is a closed and bounded subset of M n n (C) C n2. Hence U(n) is a compact group. The real orthogonal group O n (R) := U(n) L n (R) is another example of a compact group. Because of the definition of group of t.d. type, there are many such groups which are compact. Some examples of compact groups of t.d. type are L n (Z p ) := { g L n (Q p ) g M n n (Z p ), det g Z p } Sp 2n (Z p ) = Sp 2n (Q p ) M 2n 2n (Z p ) { g Sp 2n (Q p ) g 1 M 2n 2n (p j Z p ) }, j Z, j > 0. 6.2. Finite-dimensional representations of compact groups If happens to be finite, then normalized Haar measure on assigns measure 1 X to each nonempty subset X of. When discussing representations of finite groups, we often used left and right invariance of Haar measure on finite groups. Many results concerning finite-dimensional continuous representations of compact groups are proved the same way as the analogous results for finite groups, simply by replacing the normalized Haar measure on the finite group by normalized Haar measure on the compact group. Note that the continuity of a finite-dimensional representation of implies continuity of its matrix coefficients and character, so these functions are integrable. A finite-dimensional representation of is said to be continous if all of its matrix coefficients are continuous functions. Let A() be the space spanned by all matrix coefficients of continuous finite-dimensional representations of. This is a subspace of the space C() of continuous complex-valued functions on, and we will see in 6.3 that A() is dense in C(). If is finite, the results of Chapter 2 tell us that A() is the set of complex-valued functions on, so this notation is consistent with the notation used for finite groups. Theorem. Let (π, V ) be a continuous finite-dimensional representation of. Then π is unitary. Proof. Let, 1 be any inner product on V. By continuity of π, given v, w V, the function g π(g)v, π(g)w 1 is a continuous function. Hence it is integrable. Set 1

v, w = π(g)v, π(g)w 1 dg, where dg is Haar measure on. This defines an inner product on V. It follows from -invariance of dg that π(g)v, π(g)w = v, w. qed As shown in Chapter 1, a finite-dimensional unitary representation is completely reducible. Corollary. π is completely reducible. Let L 2 () be the space of functions on that are square-integrable relative to Haar measure. Then L 2 () is a Hilbert space relative to the inner product f 1, f 2 L 2 = f 1(g) f 2 (g) dg. Of course this inner product restricts to an inner product on the subspace A(). Theorem (Orthogonality relations for matrix coefficients). Let (π 1, V 1 ) and (π 2, V 2 ) be irreducible continuous finite-dimensional representations of. Let a i jk (g) be the matrix entries of the matrix of π i (g) relative to an orthonormal basis of V i (relative to an inner product on V i which makes π i unitary). Then (1) If π 1 π 2, then a 1 jk, a2 lm L2 = 0 for all j, k, l and m. (2) a 1 jk, a1 lm L 2 = δ jlδ km /n 1, where n 1 = dim V 1. Proof. Mimic the proof of Theorem 3 of Chapter 2, except where B is a linear transformation from V 2 to V 1, set A = π 1(g)Bπ 2 (g) 1 dg. Throughout the rest of the proof, replace the expression 1 g by integration over (relative to normalized Haar measure on ). qed Corollary. Let (π 1, V 1 ) and (π 2, V 2 ) be as in the theorem. If π 1 π 2, then the subspace of A() spanned by all matrix coefficients of π 1 is orthogonal to the subspace spanned by all matrix coefficients of π 2. The following is equivalent to part 2 of the above theorem. Corollary. Let (π, V ) be an irreducible continuous finite-dimensional representation of. Let v, w, v, w V, and let, be a -invariant inner product on V. Then π(g)w, v π(g)w, v dg = (dim V ) 1 w, w v, v. Theorem. Let (π 1, V 1 ) and (π 2, V 2 ) be (finite-dimensional continuous) representations of. Then χ π1 (g)χ π2 (g) dg = dim Hom(π 1, π 2 ). If π 1 and π 2 are irreducible, then χ π1 (g)χ π2 (g) dg = 2 { 1, if π1 π 2 0, otherwise..

6.3. The Peter-Weyl Theorem As above, let L 2 () be the space of measurable functions on for which f(g) 2 dg <. (Here, the integral is relative to Haar measure on ). If f L 2 (), set f 2 = ( f(g) 2 dg ) 1/2. Let L 1 () be the space of measurable functions on for which f(g) dg <. For f L 1 (), set f 1 = f(g) dg. If f 1, f 2 L 2 (), then f 1 f 2 L 1 () and f 1 f 2 1 f 1 2 f 2 2. iven f 1, f 2 L 2 (), the inner product f 1, f 2 L 2 is defined by f 1, f 2 L 2 = f 1 (g) f 2 (g) dg. With this inner product, L 2 () is a Hilbert space. The inequality f 1, f 2 L 2 f 1 2 f 2 2 is known as the Schwarz inequality. If f : C and g, set (L(g)f)(g 0 ) = f(g 1 g 0 ), g 0. Lemma 1. Let f L 2 (). Then the map g L(g)f is a continuous map from to L 2 (). Proof. It is clear that the function L(g)f belongs to L 2 (). Let ε > 0. Using the fact that C() is dense in L 2 (), choose a continuous function ϕ C() such that f ϕ 2 < ε/3. Because is compact and ϕ is continuous, ϕ is uniformly continuous. Hence there exists an open neighbourhood U of 1 in such that if g1 1 g 2 U, then ϕ(g1 1 g) ϕ(g 1 2 g) < ε/3 for all g. Now L(g 1 )f L(g 2 )f 2 L(g 1 )f L(g 1 )ϕ 2 + L(g 1 )ϕ L(g 2 )ϕ 2 + L(g 2 )ϕ L(g 2 )f 2 2 ϕ f 2 + L(g 1 )ϕ L(g 2 )ϕ 2 < 2ε/3 + sup g L(g 1 )ϕ L(g 2 )ϕ < ε Lemma 2. Let f L 2 (). For every ε > 0, there exist finitely many g i and Borel sets S i such that is the disjoint union of the S i s and L(g)f L(g i )f 2 < ε for all i and for all g S i. Proof. By Lemma 1, there exists an open neighbourhood U of 1 such that L(g)f f 2 < ε for all g U. Let g U and g 0. Then L(gg 0 )f L(g 0 )f 2 < ε. Now Ug 0 is an open neighbourhood of g 0. Hence { Ug 0 g 0 } is an open cover of. By compactness of ( 1., there are finitely many g 1,..., g n such that = n i=1 Ug i. Let S j = U j \ j 1 i=1 i) U qed Lemma 3. Let f L 2 () and f L 1 (). Set F (g 0 ) = f(g)f(g 1 g 0 ) dg. Then F is a limit in L 2 () of a sequence of functions, each of which is a finite linear combination of left translates of f. 3

Proof. Let ε > 0. Choose g i and S i as in Lemma 2. Set c i = S i f(g) dg. Then n F c i L(g i )f 2 i=1 n f(g) L(g)f L(g i )f 2 dg S i n f(g) ε dg = ε f 1. S i i=1 i=1 qed As in 6.2, the space spanned by all matrix coefficients of continuous finite-dimensional representations of is denoted by A(). Theorem (Peter-Weyl Theorem). A() is dense in L 2 (). Proof. Let (π, V ) be a continuous finite-dimensional representation that is unitary relative to an inner product, on V. Then, given v, w V, the function g π(g)v, w is a matrix coefficient of π. Let f(g) = π(g)v, w, g. The functions g f(g 1 ), g f(gg 0 ) and g f(g 0 g) are also matrix coefficients of π. Hence the space A() is closed under the map taking f to the function g f(g 1 ) and also under left and right translation. The closure A() of A() in L 2 () is also invariant under these operations. Suppose that A() L 2 (). Then A() {0}, and A() is closed under the above operations. Suppose that f 0 A() is nonzero. If U is an open neighbourhood of 1, let 1 U be the characteristic function of U, and let U be the Haar measure of U. Define f U (g) = U 1 1 U (g 0 )f 0 (g0 1 g) dg 0. Since 1 U, f 0 L 2 (), the Schwarz inequality can be used to prove that f U is a continuous function. As U shrinks to {1}, f U f 0 in L 2 (). Because f 0 0, we know that some f U is nonzero. Since A() is -invariant (that is, invariant under left translation by elements of ) and the left regular representation is unitary, it follows that A() is also -invariant. Hence linear combinations of translates of f 0 belong to A(). Applying Lemma 3, we see that f U A(). Therefore A() contains a nonzero continuous function. After translating and multiplying by a scalar, if necessary, we can produce a continuous function f 1 A() such that f 1 (1) is a nonzero real number. Set f 2 (g) = f 1(g 0 gg0 1 ) dg 0. Then f 2 is continuous, f 2 is a class function, and f 2 (1) is a nonzero real number. To see that f 2 A(), note that if h A() then f 2, h L 2 = f 2 (g) h(g) dg = f 1 (g 0 gg0 1 ) h(g) dg 0 dg = f 1 (g)h(g0 1 gg 0) dg dg 0 = 0 because the function g h(g 1 0 gg 0) belongs to A(). 4

Next, set f 3 (g) = f 2 (g) + f 2 (g 1 ). Then f 3 (g) = f 3 (g 1 ). Let K(g, g 0 ) = f 3 (g 1 g 0 ), g,, g 0. Set (T f)(g) = K(g, g 0 )f(g 0 ) dg 0, f L 2 (). Then T is an operator on L 2 (). Note that K(g, g 0 ) = f 3 (g 1 g 0 ) = f 3 (g0 1 g) = K(g 0, g). Thus T is a Hilbert-Schmidt operator(see comments following the proof). Now T is nonzero, so, according to the Hilbert-Schmidt Theorem, T has a nonzero real eigenvalue λ. and the eigenspace V λ L 2 () is finite-dimensional. If f V λ, then (T L(g 0 )f)(g) = K(g, g 1 )(L(g 0 )f)(g 1 ) dg 1 = f 3 (g 1 g 1 )f(g0 1 g 1) dg 1 = f 3 (g 1 g 0 g 1 )f(g 1 ) dg 1 = K(g0 1 g, g 1)f(g 1 ) dg 1 = (T f)(g 1 0 g) = λ f(g 1 0 g) = λ (L(g 0)f)(g). Hence T L(g 0 )f = λl(g 0 )f for all f V λ. Therefore V λ is invariant under left translation. According to Lemma 1, g L(g)f is continuous. Because the restriction of left translation to V λ is a finite-dimensional representation of, there exists a finitedimensional irreducible -invariant (nonzero) subspace W of V λ. Choose an orthonormal basis { e 1,..., e n } of W (with respect to which left translation is unitary). Then g L(g)e j, e i L 2 = e j(g 1 g 0 )e i (g 0 ) dg 0 is a matrix coefficient of this representation, hence belongs to A(). Since f 3 A(), we have ( ) ( ) 0 = f 3 (g) e j (g 1 g 0 )e j (g 0 ) dg 0 dg = f 3 (g)e j (g 1 g 0 ) dg e j (g 0 ) dg 0 ( ) ( ) = f 3 (g 0 g 1 )e j (g) dg e j (g 0 ) dg 0 = f 3 (g 1 g 0 )e j (g 0 )dg 0 e j (g) dg = (T e j )(g)e j (g) dg = λ e j, e j L 2 (For the fourth equality above, we used the fact that f 3 is a class function: f 3 (g 0 g 1 ) = f 3 (g 1 (g 0 g 1 )g) = f 3 (g 1 g 0 )). Hence λ = 0. Contradiction. Hence the assumption that A() is nonzero is false. qed If K L 2 ( ), then the operator on L 2 () defined by (T f)(g) = K(g, g 0)f(g 0 ) dg 0 is called a Hilbert-Schmidt operator. Such an operator is self-adjoint when K(g, g 0 ) = K(g 0, g) for all g, g 0. The Hilbert-Schmidt Theorem is a theorem from functional analysis which says that a nonzero self-adjoint Hilbert-Schmidt operator has some nonzero eigenvalues, and the eigenspace corresponding to each nonzero eigenvalue is finitedimensional. For a proof of the Hilbert-Schmidt Theorem, see Appendix B of the book by Simon (listed in the references). 5

As we will soon see, the Peter-Weyl Theorem can be applied to prove that an infinitedimensional unitary representation of a compact group is a direct sum of finite-dimensional irreducible representations. Let Ĝ be a collection of finite-dimensional irreducible unitary representations of having the property that each finite-dimensional irreducible representation of is equivalent to exactly one of the representations in Ĝ. If ρ Ĝ, let d ρ be the degree of ρ, and let a ρ (g) ij be the matrix coefficients of ρ relative to an orthonormal basis of the space of ρ, such that the corresponding inner product on the space of ρ is -invariant. Let H be a separable Hilbert space. Let π be a unitary representation of in the space H, having the property that the map (g, v) π(g)v is a continuous map from H to H. Define a linear operator P ρ ij on H by P ρ ij (v) = d 1 α aρ (g) ij π(g)v dg, v H. This is the analogue of the definition used in Chapter 2 when studying the decompositions of representations of finite groups. Arguing as in Chapter 2, we can easily prove: Lemma 4. Let ρ Ĝ. Let π be as above. (1) If g, then π(g)p ρ ij = k aρ (g) ki P ρ kj (2) Pmn ρ P ρ ij = δρ ni P ρ mj. If ρ Ĝ and ρ ρ, then Pmn ρ P ρ ij = 0 for all m, n, i, and j. (3) (P ρ ij ) = P ρ ji Let P ρ = i P ρ ii = d 1 ρ χ ρ(g)π(g) dg. (Here, χ ρ denotes the character of ρ.) Lemma 5. (1) P ρ Hom (π, π), (2) If ρ Ĝ and ρ ρ, then P ρ P ρ = 0. (3) (P ρ ) 2 = P ρ and (P ρ ) = P ρ. It follows from Lemma 5 that { P ρ ρ Ĝ } is a set of mutually orthogonal projections. Theorem. Let H be a separable Hilbert space. Let π be a unitary representation of in the space H, having the property that the map (g, v) π(g)v is a continuous map from H to H. Then H = n=1h n, where each H n is a finite-dimensional -invariant irreducible subspace of H. Proof. Let, be a -invariant inner product on H. Suppose that v H belongs to P ρ (H) for all ρ Ĝ. Because P ρ (v), v = 0 and I P ρ is the orthogonal projection of H onto P ρ (H), we have (I P ρ )(v) = v. From Lemma 4(2), we can see that Pmn(I ρ P ρ ) = 0. It follows that Pmn(v) ρ = 0 for all ρ Ĝ and all m and n. Let { e k } be an orthonormal basis of H. Set f k (g) = e k, π(g)v. Our hypotheses guarantee that f k is a continuous function, hence lies in L 2 (). By the Peter-Weyl theorem, we have an L 2 -expansion f k (g) = c ρ;ij (k)a ρ (g) ij, ρ;ij 6

with c ρ;ij (k) = e k, P ρ ij (v). Because P ρ ij (v) = 0 for all ρ, i and j, it follows that c ρ;ij(k) = 0 for all k. Therefore f k (g) = 0 for almost all g. Continuity of f k then guarantees that f k is identically zero for all k. In particular, f k (1) = e k, v = 0 for all k. Hence v = 0. We have shown that the only vector with is orthogonal to every P ρ (H) is the zero vector. Hence H is the direct sum of the subspaces P ρ (H). According to Lemma 5(1), P ρ (H) is a -invariant subspace of H. Hence to complete the proof of the theorem it suffices to prove that the restriction π ρ of π to P ρ (H) is a direct sum of irreducible representations of, for every ρ Ĝ. Let m ρ = dim P ρ 11 (H). Arguing as in Chapter 2, we can show that dim P ρ ii (H) = m ρ, for all i, and π ρ is equivalent to the representation m ρ ρ (the m ρ -fold direct sum ρ ρ -m ρ times). qed Corollary. A continuous irreducible unitary representation of is finite-dimensional. 6.4. Weyl s character formula Let be a compact connected Lie group. A Cartan subgroup of such a group is a closed subgroup of that is isomorphic to a direct product of the form R/Z R/Z and is maximal with respect to this property. (That is, it cannot be a proper subgroup of another closed subgroup that is also isomorphic to a direct product of the same form). A Cartan subgroup is a maximal abelian subgroup and consists of semisimple elements. (Recall that an element of a matrix Lie group is semisimple if it is semisimple as a matrix, that is, it is diagonalizable over the complex numbers.) A Cartan subgroup is equal to its own centralizer in. The Cartan subgroups of are all conjugate in. Examples: (1) = U(n). The subgroup T of diagonal matrices in consists of diagonal matrices in L n (C) whose diagonal entries are e iθ 1,..., e iθ n for some θ 1,..., θ n R. The group T is a Cartan subgroup of U(n). (2) Let = SO 4 (R) = O 4 (R) SL 4 (R). The subgroup cos θ 1 sin θ 1 0 0 sin θ T = 1 cos θ 1 0 0 θ 0 0 cos θ 2 sin θ 1, θ 2 R 2 0 0 sin θ 2 cos θ 2 is a Cartan subgroup of Fix a Cartan subgroup T of. The set ˆT of unitary characters of T is a locally compact group with the discrete topology. Let N (T ) be the normalizer of T in, and W (, T ) = N (T )/T. The group W (, T ) is a finite group, called the Weyl group of T. Each w W (, T ) acts on T and hence on ˆT by (w χ)(t) = χ(w 1 t), χ ˆT, t T. Let ˆT be the set of χ ˆT such that w χ χ for any nontrivial w W. 7

The set of irreducible unitary representations of can be parametrized by the set of W (, T )-orbits in ˆT. iven χ ˆT, there is an irreducible unitary representation π χ of. Every irreducible unitary representation of is equivalent to π χ for some χ ˆT. Furthermore, given χ, χ ˆT, π χ and π χ are equivalent if and only if χ = w χ for some w W (, T ). Weyl s character formula. Let χ ˆT. Let Θ χ be the character of π χ. The function Θ χ is determined by its values on T. There is a sign ɛ(χ) = ±1 such that Θ χ (t) = ɛ(χ) w W (,T ) = ɛ(χ) (t) 1 (w t) 1 χ(w t), w W (,T ) det(w)(w 1 χ)(t), t T. Here, T = { t T w W (, T ), w 1 w t t, } and is the so-called Weyl denominator. It has the property that (w t) = det(w) (t), t T, with det(w) = ±1. 8