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CHAPTER 1. GROUP REPRESENTATIONS Proof. We prove only the first assertion, leaving the second one for the reader, It is known from the theory of vector spaces that ker 6 is a subspace of V since. e is linear. So we only need to show closure under the action of G. But if ve kere, then for any ge G, 22 0(gv) = g(v) (8 is a G-homomorphism) (v € ker) = 90 =0. and so gv E kere, as desired. It is now an easy matter to prove Schur's lemma, which characterizes G homomorphisms of irreducible modules. This result plays a crucial role when we discuss the commutant algebra in the next section. Theorem 1.6.5 (Schur's Lemma) Let V and W be two irreducible G. modules. If : VW is a G-homomorphism, then either. 1.0 is a G-isomorphism, or 2.0 is the zero map. Proof. Since V is irreducible and kere is a submodule (by the previous. proposition), we must have either ker 0 (0) or ker 0 = V. Similarly, the irreducibility of W implies that im 0 = (0) or W. If ker 0= V or im 0= {0}, then must be the zero map. On the other hand, if ker = (0) and im 0 W, then we have an isomorphism. It is interesting to note that Schur's lemma continues to be valid over arbitrary fields and for infinite groups. In fact, the proof we just gave still works. The matrix version is also true in this more general setting. Corollary 1.6.6 Let X and Y be two irreducible matrix representations of G. IfT is any matris such that TX (g) = Y(g)T for all ge G, then either 1. T is invertible, or 2. T is the zero matriz. We also have an analogue of Schur's lemma in the case where the range module is not irreducible. This result s conveniently expressed in terms of the vector space Hom(V, W) of all G-homomorphisms from V to W Corollary 1.6.7 Let V and W be two G-modules with V being irreducible. Then dim Hom(V, W)=0 if and only if W contains no submodule isomorphic to V. ■ When the field is C, however, we can say more. Suppose that 7 is a matrix such that TX(g) = X(g)T (1.10) 1.7. COMMUTANT AND ENDOMORPHISM ALGEBRAS for all g E G. It follows that t 23 (T-cl)X = X(T-cl), where I is the appropriate Identity matrix and c E C is any scalar. Now C is algebraically closed, so we can take e to be an eigenvalue of T. Thus T-cl satisfies the hypothesis of Corollary 1.6.6 (with XY) and is not invertible. by the choice of c. Our only alternative is that T-cl=0. We have proved the following result: Corollary 1.6.8 Let X be an irreducible matriz representation of G over the complex numbers. Then the only matrices T that commute with X (g) for all gG are those of the form T-cl-ie., scalar multiples of the identity matriz, a (14) Prove the following converse of Schur's lemma. Let X be a represen- tation of G over C with the property that only scalar multiples cl commute with X (g) for all g E G. Prove that X is irreducible. 15. Let X and Y be representations of G. The inner tensor product, XOY, assigns to each g E G the matrix (XôY) (g) = X(g) Y (g). (a) Verify that X&Y is a representation of G. (b) Show that if X, Y, and X&Y have characters denoted by x, , and xô, respectively, then (x) (g) = x(g)(g). (c) Find a group with irreducible representations X and Y such that is not irreducible.