# MAT 445/ INTRODUCTION TO REPRESENTATION THEORY

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1 MAT 445/ INTRODUCTION TO REPRESENTATION THEORY CHAPTER 1 Representation Theory of Groups - Algebraic Foundations 1.1 Basic definitions, Schur s Lemma 1.2 Tensor products 1.3 Unitary representations 1.4 Characters of finite-dimensional representations CHAPTER 2 Representations of Finite Groups 2.1 Unitarity, complete reducibility, orthogonality relations 2.2 Character values as algebraic integers, degree of an irreducible representation divides the order of the group 2.3 Decomposition of finite-dimensional representations 2.4 Induced Representations, Frobenius reciprocity, and Frobenius character formula CHAPTER 3 Representations of SL 2 (F q ) CHAPTER 4 Representations of Finite Groups of Lie Type CHAPTER 5 Topological Groups, Representations, and Haar Measure 5.1 Topological spaces 5.2 Topological groups 5.3 General linear groups and matrix groups 5.4 Matrix Lie groups 5.5 Finite-dimensional representations of topological groups and matrix Lie groups 5.6 Groups of t.d. type 5.7 Haar measure on locally compact groups 5.8 Discrete series representations 5.9 Parabolic subgroups and representations of reductive groups CHAPTER 6 Representations of Compact Groups 6.1 Examples of compact groups 6.2 Finite-dimensional representations of compact groups 6.3 The Peter-Weyl Theorem 6.4 Weyl s character formula 1

2 CHAPTER 1 Representation Theory of Groups - Algebraic Foundations 1.1. Basic definitions, Schur s Lemma We assume that the reader is familiar with the fundamental concepts of abstract group theory and linear algebra. A representation of a group G is a homomorphism from G to the group GL(V ) of invertible linear operators on V, where V is a nonzero complex vector space. We refer to V as the representation space of π. If V is finite-dimensional, we say that π is finite-dimensional, and the degree of π is the dimension of V. Otherwise, we say that π is infinite-dimensional. If π is one-dimensional, then V C and we view π as a homomorphism from G to the multiplicative group of nonzero complex numbers. In the above definition, G is not necessarily finite. The notation (π, V ) will often be used when referring to a representation. Examples: (1) If G is a group, we can define a one-dimensional representation of G by π(g) = 1, g G. This representation is called the trivial representation of G. (2) Let G = R and z C. The function t e zt defines a one-dimensional representation of G. If n is a positive integer and C is the field of complex numbers, let GL n (C) denote the group of invertible n n matrices with entries in C. If (π, V ) is a finite-dimensional representation of G, then, via a choice of ordered basis β for V, the operator π(g) GL(V ) is identified with the element [π(g)] β of GL n (C), where n is the degree of π. Hence we may view a finite-dimensional representation of G as a homomorphism from G to the group GL n (C). Examples: (1) The self-representation of GL n (C) is the n-dimensional representation defined by π(g) = g. (2) The function g det g is a one-dimensional representation of GL n (C). (3) Let V be a space of functions from G to some complex vector space. Suppose that V has the property that whenever f V, the function g 0 f(g 0 g) also belongs to V for all g G. Then we may define a representation (π, V ) by (π(g)f)(g 0 ) = f(gg 0 ), f V, g, g 0 G. For example, if G is a finite group, we may take V to be the space of all complex-valued functions on G. In this case, the resulting representation is called the right regular representation of G. Let (π, V ) be a representation of G. A subspace W of V is stable under the action of G, or G-invariant, if π(g)w W for all g G and w W. In this case, denoting the 2

3 restriction of π(g) to W by π W (g), (π W, W ) is a representation of G, and we call it a subrepresentation of π (or a subrepresentation of V ). If W W are subrepresentations of π, then each π W (g), g G, induces an invertible linear operator π W/W (g) on the quotient space W/W, and (π W/W, W/W ) is a representation of G, called a subquotient of π. In the special case W = V, it is called a quotient of π. A representation (π, V ) of G is finitely-generated if there exist finitely many vectors v 1,..., v m V such that V = Span{ π(g)v j 1 j m, g G }. A representation (π, V ) of G is irreducible if {0} and V are the only G-invariant subspaces of V. If π is not irreducible, we say that π is reducible. Suppose that (π j, V j ), 1 j l, are representations of a group G. Recall that an element of the direct sum V = V 1 V l can be represented uniquely in the form v 1 + v v l, where v j V j. Set π(g)(v v l ) = π 1 (g)v π l (g)v l, g G, v j V j, 1 j l. This defines a representation of G, called the direct sum of the representations π 1,..., π l, sometimes denoted by π 1 π l. We may define infinite direct sums similarly. We say that a representation π is completely reducible (or semisimple) if π is (equivalent to) a direct sum of irreducible representations. Lemma. Suppose that (π, V ) is a representation of G. (1) If π is finitely-generated, then π has an irreducible quotient. (2) π has an irreducible subquotient. Proof. For (1), consider all proper G-invariant subspaces W of V. This set is nonempty and closed under unions of chains (uses finitely-generated). By Zorn s Lemma, there is a maximal such W. By maximality of W, π V/W is irreducible. Part (2) follows from part (1) since of v is a nonzero vector in V, part (1) says that if W = Span{ π(g)v g G }, then π W has an irreducible quotient. qed Lemma. Let (π, V ) be a finite-dimensional representation of G. irreducible subrepresentation of π. Then there exists an Proof. If V is reducible, there exists a nonzero G-invariant proper subspace W 1 of V. If π W1 is irreducible, the proof is complete. Otherwise, there exists a nonzero G-invariant subspace W 2 of W 1. Note that dim(w 2 ) < dim(w 1 ) < dim(v ). Since dim(v ) <, this process must eventually stop, that is there exist nonzero subspaces W k W k 1 W 1 V, where π Wk is irreducible. qed Lemma. Let (π, V ) be a representation of G. Assume that there exists an irreducible subrepresentation of π. The following are equivalent: 3

4 (1) (π, V ) is completely reducible. (2) For every G-invariant subspace W V, there exists a G-invariant subspace W such that W W = V. Proof. Assume that π is completely reducible. Without loss of generality, π is reducible. Let W be a proper nonzero G-invariant subspace of V. Consider the set of G-invariant subspaces U of V such that U W = {0}. This set is nonempty and closed under unions of chains, so Zorn s Lemma implies existence of a maximal such U. Suppose that W U V. Since π is completely reducible, there exists some irreducible subrepresentation U such that U W U. By irreduciblity of U, U (W U) = {0}. This contradicts maximality of U. Suppose that (2) holds. Consider the partially ordered set of direct sums of families of irreducible subrepresentations: α W α = α W α. Zorn s Lemma applies. Let W = α W α be the direct sum for a maximal family. By (2), there exists a subrepresentation U such that V = W U. If U {0}, according to a lemma above, there exists an irreducible subquotient: U U 1 U 2 such that π U1 /U 2 is is irreducible. By (2), W U 2 has a G-invariant complement U 3 : V = W U 2 U 3. Now U 3 V/(W U 2 ) = (W U)/(W U 2 ) U/U 2 U 1 /U 2. Identifying π U1 /U 2 with an irreducible subrepresentation π U4 of π U3, we have W U 4 contradicting maximality of the family W α. qed Lemma. Subrepresentations and quotient representations of completely reducible representations are completely reducible. Proof. Let (π, V ) be a completely reducible representation of G. Suppose that W is a proper nonzero G-invariant subspace of W. Then, according to the above lemma, there exists a G-invariant subspace U of V such that V = W U. It follows that the subrepresentation π W is equivalent to the quotient representation π V/U. Therefore it suffices to prove that any quotient representation of π is completely reducible. Let π V/U be an arbitrary quotient representation of π. We know that π = α I π α, where I is some indexing set, and each π α is irreducible. Let pr : V V/U be the canonical map. Then V/U = pr(v ) = α I pr(v α ). Because pr(v α ) is isomorphic to a quotient of V α (pr(v α ) V α /ker (pr V α )) and π α is irreducible, we have that pr(v α ) is either 0 or irreducible. Hence π V/U is completely reducible. qed Exercises: (1) Show that the self-representation of GL n (C) is irreducible. ( ) 1 t (2) Verify that π : t defines a representation of R, with space C 0 1 2, that is a two-dimensional representation of R. Show that there is exactly one one-dimensional 4

5 subrepresentation, hence π is not completely reducible. Prove that the restriction of π to the unique one-dimensional invariant subspace W is the trivial representation, and the quotient representation π V/W is the trivial representation. If (π 1, V 1 ) and (π 2, V 2 ) are representations of a group G, a linear transformation A : V 1 V 2 intertwines π 1 and π 2 if Aπ 1 (g)v = π 2 (g)av for all v V 1 and g G. The notation Hom G (π 1, π 2 ) or Hom G (V 1, V 2 ) will be used to denote the set of linear transformations from V 1 to V 2 that intertwine π 1 and π 2. Two representations (π 1, V 1 ) and (π 2, V 2 ) of a group G are said to be equivalent (or isomorphic) whenever Hom G (π 1, π 2 ) contains an isomorphism, that is, whenever there exists an invertible linear tranformation A : V 1 V 2 that intertwines π 1 and π 2. In this case, we write π 1 π 2. It is easy to check that the notion of equivalence of representations defines an equivalence relation on the set of representations of G. It follows from the definitions that if π 1 and π 2 are equivalent representations, then π 1 is irreducible if and only if π 2 is irreducible. More generally, π 1 is completely reducible if and only if π 2 is completely reducible. Lemma. Suppose that (π 1, V 1 ) and (π 2, V 2 ) are finite-dimensional representations of G. Then the following are equivalent: (1) π 1 and π 2 are equivalent. (2) dim V 1 = dim V 2 and there exist ordered bases β 1 and β 2 of V 1 and V 2, respectively, such that [π 1 (g)] β1 = [π 2 (g)] β2 for all g G. Proof. Assume (1). Fix ordered bases γ 1 for V 1 and γ 2 for V 2. Via these bases, identifying any invertible operator in Hom G (π 1, π 2 ) as a matrix A in GL n (C), we have [π 1 (g)] γ1 = A 1 [π 2 (g)] γ2 A, g G. Let β 1 = γ 1. Because A GL n (C), there exists an ordered basis β 2 of V 2 such that A is the change of basis matrix from β 2 to γ 2. With these choices of β 1 and β 2, (2) holds. Now assume that (2) holds. Let A be the unique linear transformation from V 1 to V 2 which maps the jth vector in β 1 to the jth vector in β 2. qed A representation (π, V ) of G has a (finite) composition series if there exist G-invariant subspaces V j of V such that {0} V 1 V r = V each subquotient π Vj+1 /V j, 1 j r 1, is irreducible. The subquotients π Vj+1 /V j called the composition factors of π. are Lemma. Let (π, V ) be a finite-dimensional represntation of G. Then π has a composition series. Up to reordering and equivalence, the composition factors of π are unique. Proof left as an exercise. 5

6 Schur s Lemma. Let (π 1, V 1 ) and (π 2, V 2 ) be irreducible representations of G. Then any nonzero operator in Hom G (π 1, π 2 ) is an isomorphism. Proof. If Hom G (π 1, π 2 ) = {0} there is nothing to prove, so assume that it is nonzero. Suppose that A Hom G (π 1, π 2 ) is nonzero. Let g G and v 2 A(V 1 ). Writing v 2 = A(v 1 ), for some v 1 V 1, we have π 2 (g)v 2 = Aπ(g)v 1 A(V 1 ). Hence A(V 1 ) is a nonzero G-invariant subspace of V 2. By irreducibility of π 2, we have A(V 1 ) = V 2. Next, let W be the kernel of A. Let v 1 W. Then A(π 1 (g)v 1 ) = π 2 (g)(a(v 1 )) = π 1 (g)0 = 0 for all g G. Hence W is a G-invariant proper subspace of V 1. By irreducibility of π 1, W = {0}. qed Corollary. Let (π, V ) be a finite-dimensional irreducible representation of G. Then Hom G (π, π) consists of scalar multiples of the identity operator, that is, Hom G (π, π) C. Proof. Let A Hom G (π, π). Let λ C be an eigenvalue of A (such an eigenvalue exists, since V is finite-dimensional and C is algebraically closed). It is easy to see that A λi Hom G (π, π). But A λi is not invertible. By the previous lemma, A = λi. qed Corollary. If G is an abelian group, then every irreducible finite-dimensional representation of G is one-dimensional. Proof left as an exercise. Exercise: Prove that an irreducible representation of the cyclic group of order n > 1, with generator g 0, has the form g k 0 e 2πimk/n for some m { 0, 1,..., n 1 }. (Here i is a complex number such that i 2 = 1 and π denotes the area of a circle of radius one). Let (π, V ) be a representation of G. A matrix coefficient of π is a function from G to C of the form g λ(π(g)v), for some fixed v V and λ in the dual space V of linear functionals on V. Suppose that π is finite-dimensional. Choose an ordered basis β = { v 1,..., v n } of V. Let β = { λ 1,..., λ n } be the basis of V which is dual to β: λ j (v i ) = δ ij, 1 i, j n. Define a function a ij : G C by [π(g)] β = (a ij (g)) 1 i,j n. Then it follows from π(g)v i = n l=1 a il(g)v l that a ij (g) = λ j (π(g)v i ), so a ij is a matrix coefficient of π. If g G and λ V, define π (g)λ V by (π (g)λ)(v) = λ(π(g 1 )v), v V. Then (π, V ) is a representation of G, called the dual (or contragredient) of π. Exercises: (1) Let (π, V ) be a finite-dimensional representation of G. Choose β and β as above. Show that [π (g)] β = [π(g 1 )] t β, for all g G. Here the superscript t denotes transpose. (2) Prove that if (π, V ) is finite-dimensional then π is irreducible if and only if π is irreducible. 6

7 (3) Determine whether the self-representation of GL n (R) (restrict the self-representation of GL n (C) to the subgroup GL n (R)) is equivalent to its dual. (4) Prove that a finite-dimensional representation of a finite abelian group is the direct sum of one-dimensional representations Tensor products Let (π j, V j ) be a representation of a group G j, j = 1, 2. Recall that V 1 V 2 is spanned by elementary tensors, elements of the form v 1 v 2, v 1 V 1, v 2 V 2. We can define a representation π 1 π 2 of the direct product G 1 G 2 by setting (π 1 π 2 )(g 1, g 2 )(v 1 v 2 ) = π 1 (g 1 )v 1 π 2 (g 2 )v 2, g j G j, v j V j, j = 1, 2, and extending by linearity to all of V 1 V 2. The representation π 1 π 2 of G 1 G 2 is called the (external or outer) tensor product of π 1 and π 2. Of course, when π 1 and π 2 are finite-dimensional, the degree of π 1 π 2 is equal to the product of the degrees of π 1 and π 2. Lemma. Let (π j, V j ) and G j, j = 1, 2 be as above. Assume that each π j is finitedimensional. Then π 1 π 2 is an irreducible representation of G 1 G 2 if and only if π 1 and π 2 are both irreducible. Proof. If π 1 or π 2 is reducible, it is easy to see that π 1 π 2 is also reducible. Assume that π 1 is irreducible. Let n = dim V 2. Let Hom G1 (π 1, π 1 ) n = Hom G1 (π 1, π 1 ) Hom G1 (π 1, π 1 ), and π n 1 = π 1 π 1, where each direct sum has n summands. Then Hom G1 (π 1, π 1 ) n Hom G1 (π 1, π n 1 ), where the isomorphism is given by A 1 A n B, with B(v) = A 1 (v) A n (v). By (the corollary to) Schur s Lemma, Hom G1 (π 1, π 1 ) C. Irreducibility of π 1 guarantees that given any nonzero v V 1, V 1 = Span{ π 1 (g 1 )v g 1 G 1 }, and this implies surjectivity. Because V 2 C n and C Hom G1 (π 1, π 1 ), we have (i) V 2 Hom G1 (π 1, π 1 1 n ), where π 1 1 n is the representation of G 1 on V 1 V 2 defined by (π 1 1 n )(g 1 )(v 1 v 2 ) = π 1 (g 1 )v 1 v 2, v 1 V 1, v 2 V 2. (Note that this representation can be identified with the restriction of π 1 π 2 to the subgroup G 1 {1} of G 1 G 2 ). If m is a positive integer, then (ii) V 1 Hom G1 (π 1, π m 1 ) V m 1 v A A(v) 7

8 is an isomorphism. Next, we can use (i) and (ii) to show that { G 1 invariant subspaces of V 1 V 2 } { C subspaces of V 2 } V 1 W W X Hom G1 (π 1, X) Hom G1 (π 1, π 1 1 n ) = V 2 As any (G 1 G 2 )-invariant subspace X of V 1 V 2 is also a G 1 -invariant subspace, we have X = V 1 W for some complex subspace W of V 2. If X {0} and π 2 is irreducible, then Span{ (π 1 π 2 )(1, g 2 )X g 2 G 2 } = V 1 Span{ π 2 (g 2 )W g 2 G 2 } = V 1 V 2. But G 1 G 2 -invariance of X then forces X = V 1 V 2. It follows that if π 1 and π 2 are irreducible, then π 1 π 2 is irreducible (as a representation of G 1 G 2 ). qed Proposition. Let (π, V ) be an irreducible finite-dimensional representation of G 1 G 2. Then there exist irreducible representations π 1 and π 2 of G 1 and G 2, respectively, such that π π 1 π 2. Proof. Note that π 1(g 1 )v = π((g 1, 1))v, g 1 G 1, v V, and π 2(g 2 )v = π((1, g 2 ))v, g 2 G 2, v V, define representations of G 1 and G 2, respectively. Choose a nonzero G 1 -invariant subspace V 1 such that π 1 V1 is an irreducible representation of G 1. Let v 0 be a nonzero vector in V 1. Let V 2 = Span{ π 2(g 2 )v 0 g 2 G 2 }. Then V 2 is G 2 -invariant and π 2 := π 2 V2 is a representation of G 2, which might be reducible. Define A : V 1 V 2 V as follows. Let v 1 V 1 and v 2 V 2. Then there exist complex numbers c j and elements g (j) 1 G 1 such that v 1 = m j=1 c jπ 1 (g (j) 1 )v 0, as well as complex numbers b l and elements g (l) 2 G 2 such that v 2 = n l=1 b lπ 2 (g (l) 2 ). Set A(v 1 v 2 ) = m n j=1 l=1 c j b l π(g (j) 1, g(l) 2 )v 0. Now π(g (j) 1, g(l) 2 )v 0 = π 1 (g (j) 1 )π 2(g (l) 2 )v 0 = π 2 (g (l) 2 )π 1(g (j) 1 )v 0. Check that the map A is well-defined, extending to a linear transformation from V 1 V 2 to V. Also check that A Hom G1 G 2 (V 1 V 2, V ). Because A(v 0 v 0 ) = v 0, we know that A is nonzero. Combining G 1 G 2 -invariance of A(V 1 V 2 ) with irreducibility of π, we have A(V 1 V 2 ) = V. If A also happens to be one-to-one, then we have π 1 π 2 π. 8

9 Suppose that A is not one-to-one. Then Ker A is a G 1 G 2 -invariant subspace of V 1 V 2. In particular, Ker A is a G 1 -invariant subspace of V 1 V 2. Using irreducibility of π 1 and arguing as in the previous proof, we can conclude that Ker A = V 1 W for some complex subspace W of V 2. We have an equivalence of the representations (π 1 π 2 ) (V1 V 2 )/Ker A and π of G 1 G 2. To finish the proof, we must show that the quotient representation (π 1 π 2 ) V/(V1 W ) is a tensor product. If v 1 V 1 and v 2 V 2, define B(v 1 (v 1 + W )) = v 1 v 2 + V 1 W. This extends by linearity to a map from V 1 (V 2 /W ) to the quotient space (V 1 V 2 )/(V 1 W ) and it is a simple matter to check that B is an isomorphism and B Hom G1 G 1 (π 1 (π 2 ) V2 /W, (π 1 π 2 ) (V1 V 2 )/(V 1 W ). The details are left as an exercise. qed If (π 1, V 1 ) and (π 2, V 2 ) are representations of a group G, then we may form the tensor product representation π 1 π 2 of G G and restrict to the subgroup δg = { (g, g) g G } of G G. This restriction is then a representation of G, also written π 1 π 2. It is called the (inner) tensor product of π 1 and π 2. Using inner tensor products gives ways to generate new representations of a group G. However, it is important to note that even if π 1 and π 2 are both irreducible, the inner tensor product representation π 1 π 2 of G can be reducible. Exercise: Let π 1 and π 2 be finite-dimensional irreducible representations of a group G. Prove that the trivial representation of G occurs as a subrepresentation of the (inner) tensor product representation π 1 π 2 of G if and only if π 2 is equivalent to the dual π 1 of π Unitary representations Suppose that (π, V ) is a representation of G. If V is a finite-dimensional inner product space and there exists an inner product, on V such that π(g)v 1, π(g)v 2 = v 1, v 2, v 1, v 2 V, g G. then we say that π is a unitary representation. If V is infinite-dimensional, we say that π is pre-unitary if such an inner product exists, and if V is complete with respect to the norm induced by the inner product (that is, V is a Hilbert space), then we say that π is unitary. Now assume that π is finite-dimensional. Recall that if T is a linear operator on V, the adjoint T of T is defined by < T (v), w >=< v, T (w) > for all v, w V. Note that π is unitary if and only if each operator π(g) satisfies π(g) = π(g) 1, g G. Let n be a positive integer. Recall that if A is an n n matrix with entries in C, the adjoint A of A is just A = t Ā. 9

10 Lemma. If (π, V ) is a finite-dimensional unitary representation of G and β is an orthonormal basis of V, then [π(g)] β = [π(g)] 1 β. Proof. Results from linear algebra show that if T is a linear operator on V and β is an orthonormal basis of V, then [T ] β = [T ] β. Combining this with π(g) = π(g) 1, g G, proves the lemma. qed Exercises: (1) If (π, V ) is a representation, form a new vector space V as follows. As a set, V = V, and V has the same vector addition as V. If c C and v V, set c v = cv, where c is the complex conjugate of c and cv is the scalar multiplication in V. If g G, and v V, π(g)v = π(g)v. Show that ( π, V ) is a representation of V. (2) Assume that (π, V ) is a finite-dimensional unitary representation. Prove that π π. Lemma. Let W be a subspace of V, where (π, V ) is a unitary representation of G. Then W is G-invariant if and only if W is G-invariant. Proof. W is G-invariant if and only if π(g)w W for all g G and w W if and only if π(g)w, w = 0 for all w W, w W and g G if and only if w, π(g 1 )w = 0 for all w W, w W and g G, if and only if W is G-invariant. qed Corollary. A finite-dimensional unitary representation is completely reducible. Lemma. Suppose that (π, V ) is a finite-dimensional unitary represetantion of G. Let W be a proper nonzero G-invariant subspace of V, and let P W be the orthogonal projection of V onto W. Then P W commutes with π(g) for all g G. Proof. Let w W and w W. Then qed P W π(g)(w + w ) = P W π(g)w + P W π(g)w = π(g)w + 0 = π(g)p W (w + w ). Lemma. Let (π, V ) be a finite-dimensional unitary representation of G. Then π is irreducible if and only if Hom G (π, π) C (every operator which commutes with all π(g) s is a scalar multiple of the identity operator). Proof. One direction is simply the corollary to Schur s Lemma (using irreducibility of π). For the other, if π is reducible, and W is a proper nonzero G-invariant subspace of V, Then P W Hom G (π, π) and P W is not a scalar multiple of the identity operator. qed Suppose that (π 1, V 1 ) and (π 2, V 2 ) are representations of G and V 1 and V 2 are complex inner product spaces, with inner products, 1 and, 2, respectively. Then π 1 and π 2 are unitarily equivalent if there exists an invertible linear operator A : V 1 V 2 such that Av, Aw 2 = v, w 1 for all v and w V 1 and A Hom G (π 1, π 2 ). 10

11 Lemma. Let (π 1, V 1 ) and (π 2, V 2 ) be finite-dimensional unitary representations of G. Then π 1 π 2 if and only if π 1 and π 2 are unitarily equivalent. Proof. Assume that π 1 π 2. Let A : V 1 V 2 be an isomorphism such that A Hom G (π 1, π 2 ). Recall that the adjoint A : V 2 V 1 is defined by the condition A v 2, v 1 1 = v 2, Av 1 2 for all v 1 V 1 and v 2 V 2. By assumption, we have (i) π 1 (g) = A 1 π 2 (g)a, g G. Taking adjoints, we have π 1 (g) = A π 2 (g) (A ) 1 for all g G. Since π j is unitary, we have π j (g) = π j (g 1 ). Replacing g 1 by g, we have (ii) π 1 (g) = A π 2 (g)(a ) 1, g G. Expressing π 2 (g) in terms of π 1 (g) using (i), we can rewrite (ii) as π 1 (g) = A Aπ 1 (g)a 1 (A ) 1, g G, or π 1 (g) 1 A Aπ 1 (g) = A A, g G. Now A A is positive definite (that is, self-adjoint and having positive (real) eigenvalues), and so has a unique positive definite square root, say B. Note that π 1 (g) 1 Bπ 1 (g) is also a square root of A A and it is positive definite, using π 1 (g) = π 1 (g) 1. Hence π 1 (g) 1 Bπ 1 (g) = B for all g G. Writing A in terms of the polar decomposition, we have A = UB, with B as above, and with U an isomorphism from V 1 V 2 such that Uv, Uw 2 = v, w 1 for all v and w V 1. Next, note that π 2 (g) = UBπ 1 (g)b 1 U 1 = Uπ 1 (g)u 1, g G. Hence U Hom G (π 1, π 2 ), and π 1 and π 2 are unitarily equivalent. qed 1.4. Characters of finite-dimensional representations Let (π, V ) be a finite-dimensional representation of a group G. The function g tr π(g) from G to C is called the character of π. We use the notation χ π (g) = tr π(g). Note that we can use any ordered basis of V to compute χ π (g), since the trace of an operator depends only on the operator itself. Note that if π were infinite-dimensional, the operator π(g) would not have a trace. 11

12 Lemma. Let (π, V ) be a finite-dimensional representation of G. (1) If π π, then χ π = χ π. (2) The function χ π is constant on conjugacy classes in G. (3) Let π be the representation dual to π. Then χ π (g) = χ π (g 1 ), g G. (4) If π is unitary, then χ π (g 1 ) = χ π (g), g G. (5) Suppose that (π, V ) has a composition series {0} V 1 V r = V, with composition factors π V1,π V2 /V 1,..., π Vr /V r 1 (see page 5). Then χ π = χ πv1 + χ πv2 /V χ πv r /V r 1. (6) The character χ π1 π r of a tensor product of finite-dimensional representations π 1,..., π r of G 1,..., G r, respectively, is given by χ π1 π r (g 1,..., g r ) = χ π1 (g 1 )χ π2 (g 2 ) χ πr (g r ), g 1 G 1,..., g r G r. Proof. By an earlier result, if π π, then π and π have the same matrix realization (for some choice of bases). Part (1) follows immediately. Note that χ π (g 1 gg 1 1 ) = tr (π(g 1)π(g)π(g 1 ) 1 ) = tr π(g) = χ π (g), g, g 1 G. Recall that if β is an ordered basis of V and β is the basis of V dual to β, then [π(g)] β = t [π (g 1 )] β. This implies (3). Suppose that π is unitary. Let β be an orthonormal basis of V. Then [π(g 1 )] β = [π(g)] β = t [π(g)] implies part (4). For (5), it is enough to do the case r = 2. Let β be an ordered basis for V 1. Extend β to an ordered basis γ for V 2 = V. Let γ be the ordered basis for V 2 /V 1 which is the image of γ under the canonical map V V 2 /V 1. Then it is easy to check that [π(g)] γ is equal to ( [π V1 (g)] β 0 [π V2 /V 1 (g)] γ For (6), it is enough to do the case r = 2. Let β = { v 1,..., v n } and γ = { w 1,..., w m } be ordered bases of V 1 and V 2, respectively. Then ). { v j w l 1 j n, 1 l m } is an ordered basis of V 1 V 2. Let a ij (g 1 ) be the ijth entry of [π 1 (g 1 )] β, g 1 G 1, and let b ij (g 2 ) be the ijth entry of [π 2 (g 2 )] γ, g 2 G 2. We have π 1 (g 1 )v j = a 1j (g 1 )v 1 + a 2j (g 1 )v a nj (g 1 )v n, g 1 G r π 2 (g 2 )w l = b 1l (g 2 )w 1 + b 2l (g 2 )w b ml (g 2 )w m, g 2 G 2. 12

13 Hence π 1 (g 1 )v j π 2 (g 2 )w l = n t=1 s=1 m a tj (g 1 )b sl (g 2 )(v t w s ), and, as the coefficient of v j w l on the right side equals a jj (g 1 )b ll (g 2 ), we have χ π1 π 2 (g 1, g 2 ) = n j=1 l=1 m a jj (g 1 )b ll (g 2 ) = χ π1 (g 1 )χ π2 (g 2 ), g 1 G 1, g 2 G 2. Example: The converse to part (1) is false. Consider the Example (2) on page 4. We have χ π (t) = 2 for all t R. Now take π 0 π 0, where π 0 is the trivial representation of R. This clearly has the same character as π, though π 0 π 0 is not equivalent to π. In many cases, for example, if G is finite, or compact, two irreducible finite-dimensional representations having the same character must be equivalent. 13

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