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Schur's Lemma LECTURE 21: SCHUR ORTHONONALITY 1. Schur's Lemma Lemma 1.1 (Schur's Lemma). Let V , W be irreducible representations of G. (1) If f : V ! W is a G-morphism, then either f ≡ 0, or f is invertible. (2) If f1; f2 : V ! W are two G-morphisms and f2 6= 0 , then there exists λ 2 C such that f1 = λf2. Proof. (1) Suppose f is not identically zero. Since ker(f) is a G-invariant subset in V , it must be f0g. So f is injective. In particular, f(V ) is a nonzero subspace of W . On the other hand, it is easy to check that f(V ) is a G-invariant subspace of W . It follows that f(V ) = W , and thus f is invertible. −1 (2) Since f2 6= 0, it is invertible. So f = f2 ◦f1 is a G-morpism from V to V itself. Let λ be one of the eigenvalues of the linear map f. Then f − λ·Id is a G-morphism from V to V which is not invertible. It follows that f −λ·Id ≡ 0, and thus f1 = λf2. Note that in the proof we showed in particular Corollary 1.2. Let V be an irreducible representation of G, then HomG(V; V ) = C·Id. Conversely, we have Lemma 1.3. If (π; V ) is a unitary representation of G, and HomG(V; V ) = C·Id, then (π; V ) is an irreducible representation of G. Proof. Let 0 6= W ⊂ V be a G-invariant subspace. We need to show that W = V . Let P : V ! W be the orthogonal projection (with respect to the given G-invariant inner product). Since both W and W ? are G-invariant, we have for any g 2 G and any v = w + w? 2 V , P (g · v) = P (g · w + g · w?) = g · w = g · P (v); i.e. P : V ! W ⊂ V is a G-morphism. It follows that P = λ·Id for some λ 2 C. Now 2 P = P implies λ = 1, and thus W = V . Recall that the center Z(G) of a Lie group G is Z(G) = fh 2 G : gh = hg; 8g 2 Gg: Corollary 1.4. If (π; V ) is an irreducible representation of G, then for any h 2 Z(G), π(h) = λ·Id for some λ 2 C. 1 2 LECTURE 21: SCHUR ORTHONONALITY Proof. Suppose h 2 Z(G), then for any g 2 G, π(h)π(g) = π(hg) = π(gh) = π(g)π(h): In other words, π(h): V ! V is a G-morphism, and the conclusion follows. Corollary 1.5. Any irreducible representation of an abelian Lie group is one dimen- sional. Proof. Since G is abelian, Z(G) = G. By the previous corollary, for any g 2 G, π(g) is a multiple of the identity map on V . It follows that any subspace of V is G-invariant. So V has no nontrivial subspace, which is equivalent to dim V = 1. 2. Schur Orthogonality for Matrix Coefficients Let (V; π) be a representation of a Lie group G. If we choose a basis e1; ··· ; en of V , we can identify V with Cn, and represent any g 2 G by a matrix: 0π11(g) ··· π1n(g)1 0v11 . π(g)v = @ . A @ . A πn1(g) ··· πnn(g) vn P for v = viei. So if we take Lj : V ! C be the function X Lj( viei) = vj; i then πij is the function on G given by πij(g) = Li(π(g)ej): Definition 2.1. For any v 2 V; L 2 V ∗, the map φ : G ! C; φ(g) = L(π(g)v) is called a matrix coefficient of G. Obviously any matrix coefficients of G is a continuous function on G. In fact, they form a subring of C(G): Proposition 2.2. If φ1; φ2 are matrix coefficients for G, so are φ1 + φ2 and φ1 · φ2. ∗ Proof. Let (πi;Vi) be representations of G, vi 2 Vi, Li 2 Vi such that φi(g) = ∗ ∗ Li(πi(g)vi). Then (π1 ⊕ π2;V1 ⊕ V2) is a representation of G, L1 ⊕ L2 2 V1 ⊕ V2 = ∗ (V1 ⊕ V2) and (L1 ⊕ L2)((π1 ⊕ π2)(g)(v1; v2)) = φ1(g) + φ2(g): Similarly we have a linear functional L1 ⊗ L2 on V1 ⊗ V2 satisfying (L1 ⊗ L2)(v1 ⊗ v2) = L1(v1)L2(v2), and thus (L1 ⊗ L2)((π1 ⊗ π2)(g)(v1 ⊗ v2)) = φ1(g)φ2(g): LECTURE 21: SCHUR ORTHONONALITY 3 Now suppose G is a compact Lie group, and dg the normalized Haar measure on G. Recall that L2(G), the space of square-integrable functions with respect to this Haar measure, is the completion of the space of continuous functions on G with respect to the inner product Z hf1; f2iL2 = f1(g)f2(g)dg: G Theorem 2.3 (Schur's Orthogonality I). Let (π1;V1) and (π2;V2) are two non-isomorphic irreducible representations of a compact Lie group G. Then every matrix coefficient of 2 π1 is orthogonal in L (G) to every matrix coefficient of π2. Proof. Fix G-invariant inner products on V1 and V2 respectively. Suppose φi(g) = hπi(g)vi; wii; i = 1; 2 are matrix coefficients for πi, where vi; wi 2 Vi. Fix a basis of V1 such that e1 = v1. Define a linear map f : V1 ! V2 by f(e1) = v2 and f(ek) = 0 for all k ≥ 2. Consider the map Z −1 F : V1 ! V2; v 7! F (v) = π2(g)f(π1(g )v)dg: G F is linear since f is. It is also G-equivariant, since Z Z −1 −1 −1 −1 −1 F (π1(h )v) = π2(g)f(π1(hg) v)dg = π2(h ) π2(hg)f(π1(hg) v)dg = π2(h )F (v): G G By Schur's lemma, F (v) = 0 for any v, and in particular, hF (v); w2i = 0. On the other hand, for any j, −1 X −1 −1 π2(g)f(π1(g )ej) = π2(g)f( π1(g )kjek) = π1(g )1jπ2(g)(v2); k −1 −1 where π1(g )kj = hπ1(g )ej; eki is the matrix coefficients of π1 with respect to the basis fe1; ··· ; eng. It follows that Z −1 hπ1(g )ej; e1ihπ2(g)v2; w2idg = 0 G for any j. So by linearity, Z −1 hπ1(g )w1; v1ihπ2(g)v2; w2idg = 0: G Note that the inner product is G-invariant, −1 hπ1(g )w1; v1i = hw1; π1(g)v1i = hπ1(g)v1; w1i = φ1(g): So the theorem follows. Theorem 2.4 (Schur's Orthogonality II). Let (π; V ) be an irreducible representation of a compact Lie group G, with G-invariant inner product h·; ·i. Then Z 1 hπ(g)w1; v1ihπ(g)w2; v2idg = hw1; w2ihv1; v2i: G dim V 4 LECTURE 21: SCHUR ORTHONONALITY Proof. Define the linear maps f; F : V ! V as above. Then F is G-equivariant, and thus F = λ·Id for some λ = λ(v1; v2) 2 C. On the other hand, when we take π1 = π2 = π, the computation in the previous proof shows Z −1 λ(v1; v2)hw1; w2i = hF (w1); w2i = hπ(g )w1; v1ihπ(g)v2; w2idg: G Since the Haar measure is invariant under the inversion map g ! g−1, the right hand side is invariant if we exchange w1 with v2, and exchange w2 with v1. It follows that λ(v1; v2) = Chv2; v1i = Chv1; v2i for some constant C. Finally if we take v1 = v2 = e1 a unit vector, then Z Z Z Z Tr(F ) = Tr π(g)◦f◦π(g−1)dg = Tr(π(g)◦f◦π(g−1))dg = Tr(f)dg = 1dg = 1: G G G G 1 On the other hand, in this case F = C·Id. It follows from Tr(F ) = 1 that C = dim V . .
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