Gelfand Pairs, Representation Theory of the Symmetric Group, and the Theory of Spherical Functions. John Ryan Stanford University.

Transcription

1 Gelfand Pairs, Representation Theory of the Symmetric Group, and the Theory of Spherical Functions John Ryan Stanford University June 3, 2014 Abstract This thesis gives an introduction to the study of Gelfand pairs and their applications. We begin with a brief introduction to the notion of a Gelfand pair and then move to some of the foundational results concerning Gelfand pairs. Next, we explore specific examples of Gelfand pairs, developing tools of independent interest as we progress. We find that consideration of a specific example of a Gelfand pair and of the tools used in our study naturally leads us to a discussion of the representation theory of the symmetric group. We then conclude our study by developing the theory of spherical functions on groups, which gives us a glimpse of the relevance of Gelfand pairs to areas of mathematics outside of representation theory. Contents 1 Introduction 2 2 Foundations 5 3 Examples 11 4 Representation Theory of the Symmetric Group 18 5 Spherical Functions 25 6 Acknowledgements 38 1

2 1 Introduction The aim of this thesis is to give an introduction to the theory of Gelfand pairs and to explore related topics that emerge naturally from a discussion of Gelfand pairs. Attempts have been made to make this work as self-contained and accessible as possible. Very little background is necessary to understand the results and arguments here beside command of basic linear algebra, group theory, and elementary notions from representation theory. There are two natural and equivalent definitions of a Gelfand pair that we give here (equivalence of the definitions will be proved later). The first definition is natural and easy to define. Let G be a finite group and H G a subgroup. Then we say that (G, H) is a Gelfand pair if for every irreducible representation (π, V ) of G, the subspace V H V of H- fixed vectors is at most one dimensional, or equivalently, the restriction π H contains the trivial representation 1 H with multiplicity at most 1. The second (equivalent) definition of a Gelfand pair requires the notion of an induced representation, which we recall here: Let G be a finite group, H G a subgroup of G and π H GL(V ) a representation of H. We define the induced representation Ind G Hπ to be the vector space of all functions f G V such that f(hx) = π(h)f(x) for h H and x G. Now we define, for g G: (π G (g)f)(x) = f(xg). Under this definition, π G is a representation defined by g acting on Ind G H π by right translation. We will usually abuse notation and write π G simply as π when it is clear that we are talking about the induced representation. Now if G is any group with a representation (π, V ), then it is an elementary fact that π can be written uniquely as a direct sum of irreducible representations π = n d i π i i=1 where {π 1,..., π n } are the distinct irreducible representations of G and the d i are non-negative integers. The second and equivalent definition is: a Gelfand pair is a pair (G, H) where G is a group with subgroup H G that satisfies a certain property. This property is that when the representation Ind G H 1 H constructed by inducing the trivial representation of H, is decomposed as a direct sum of irreducible representations, each irreducible occurs with multiplicity less than 2

3 or equal to one, i.e., if we write Ind G H1 H = n d i π i i=1 then d i 1 for i = 1,..., n. It turns out that this multiplicity-free property can be detected by a certain ring of functions on G. Consider the set of H-bi-invariant functions, H = {ϕ G C ϕ(hgh ) = ϕ(g) for h, h H, g G}. We can consider H as an algebra over the complex numbers in which the ring structure is endowed by the multiplication rule (ϕ 1 ϕ 2 )(g) = 1 H ϕ 1 (s)ϕ 2 (s 1 g). s G This algebra is known as the Hecke algebra of the pair (G, H). The first major result of this paper will be to show that it is equivalent to say that (G, H) is a Gelfand pair and that the Hecke algebra of (G, H) is a commutative ring. We will then explore some interesting examples of multiplicity free induced representations including the so-called Gelfand-Graev Representation of the general linear group over a finite field GL n (F q ) and the representation of the symmetric group induced by the trivial representation of the subgroup S m S n S m+n where m and n are positive integers. In exploring these examples, we develop general tools which are useful for detecting Gelfand pairs. This second example, in which we prove that (S m+n, S m S n ) is a Gelfand pair, leads us naturally to generalize the subgroup S m S n S m+n. That is, we consider the subgroup S λ1 S λk S n where (λ 1,..., λ k ) is a partition of a positive integer n. Although (S n, S λ1 S λk ) is not in general a Gelfand pair, we may utilize some of the same tools we used to study Gelfand pairs to deduce an interesting connection between the study of partitions of n and the irreducible representations of S n, demonstrating the power of some of the tools used to study Gelfand pairs. What we will show is that there exists a natural bijection between the partitions of n and irreducible representations of S n. The final major result of this paper comes from a discussion of spherical functions, a special class of functions from a group G into the complex numbers. We will see that the discussion of these spherical functions leads to an 3

4 interesting theory that can be viewed as a generalization of Fourier analysis. It is a classical result from Fourier analysis that if we let e n (x) = 1 2π e inx, π < x < π then {e n } forms an orthonormal basis for L 2 ([ π, π]). As a result, if we let m f m = f, e n e n n= m then f m f as m for a reasonably nice function f. This fact has the following generalization, namely the Schur Orthogonality Relation: If G is a compact group, then the irreducible characters of G form an orthonormal basis of the space of class functions on G. Thus, Schur Orthogonality amounts to Fourier Analysis on finite or compact groups. This fact has a further significant generalization to Gelfand pairs, which is as follows. Given a Gelfand pair (G, H), if (π, V ) is a representation of G with a unique H-fixed vector v V, the associated spherical function will be given by σ(g) = π(g)v, v. We begin our consideration of spherical functions by studying the example of diagonally embedding a group G into the direct product G G. As it turns out, (G G, G) is a Gelfand pair, and the study of spherical functions in this important example allows us to recover classical character theory, i.e., we find that the spherical functions are the irreducible characters of G, which form an orthonormal basis of the space of class functions on G. Generalizing this example to the more abstract setting of spherical functions on arbitrary Gelfand pairs (G, H) will allow us to construct an orthonormal basis of the space C(H/G/H) consisting of H-bi-invariant functions from G into the complex numbers. This result, a generalization of the classical results from Fourier analysis and character theory, cuts to the core of why Gelfand pairs are important not just to representation theory but to other areas of mathematics. 4

5 2 Foundations We lay some of the foundations here that are necessary for the study of Gelfand pairs. We begin our theory with a proposition: Proposition 1. Let H G be a subgroup of a finite group G. Let (ψ 1, V 1 ) and (ψ 2, V 2 ) be representations of H. Let M = Hom G (Ind G H ψ 1, Ind G H ψ 2) and H = {ϕ G End C (V 1, V 2 ) ϕ(hgh ) = ψ 2 (h)ϕ(g)ψ 1 (h ) for h, h H, g G}. Then M and H are isomorphic C-algebras, where the multiplicative structure in M is composition of homomorphisms and that in H is convolution: (ϕ 1 ϕ 2 )(g) = 1 H ϕ 1 (s) ϕ 2 (s 1 g). s G Proof. For M M, define ϕ M G End C (V 1, V 2 ) by ϕ M (g)v = (Mf v )(g) for v V, where f v (h) = ψ 1 (h)v for h H and f v (g) = 0 for g H. It is clear that under this definition f v Ind G H ψ 1. That ϕ M H is a computation. Let h, h H and g G. Then: ϕ M (hgh )v = (Mf v )(hgh ) = ψ 2 (h)(mf v )(gh ) = ψ 2 (h)(mψ 1 (h )f v )(g) = ψ 2 (h)(mf ψ1 (h )v)(g) = ψ 2 (h)ϕ M (g)ψ 1 (h )v so that indeed this map is well-defined. To see that it is a vector space homomorphism, let c C and M, M M. Then for v V, we simply use the definition of addition of functions to attain: ϕ M+cM (g)v = ((M + cm )f v )(g) = (Mf v + (cm )f v )(g) = (Mf v )(g) + c(m f v )(g) = ϕ M (g)v + cϕ M (g)v = (ϕ M (g) + cϕ M (g))v. So indeed the map M ϕ M is a vector space homomorphism. 5

6 To show it is isomorphism, we construct its two sided inverse. Let ϕ H. Then define M ϕ by (M ϕ f)(g) = r ϕ(gr 1 )f(r) where r runs over a set of coset representatives of G/H and f Ind G H ψ 1. Now let h H and g G. Then so that indeed M ϕ f Ind G H ψ 2 Now for g, g G it holds: (M ϕ f)(hg) = ϕ(hgr 1 )f(r) r = ψ 1 (h) ϕ(gr 1 )f(r) r = ψ 1 (h)(m ϕ f)(g) (M ϕ ψ1 G (g )f)(g) = ϕ(gr 1 )(ψ1 G (g )f)(r) r = ϕ(gr 1 )f(rg ) since f Ind G Hψ 1 r = ϕ(gg r 1 )f(r) r = (M ϕ f)(gg ) = (ψ G 2 (g )M ϕ f)(g) proving that M ϕ M. Now we check that the maps are indeed inverses, i.e., that M ϕm = M for M M and ϕ Mϕ = ϕ for ϕ H. Now for v V and g G, we have: ϕ Mϕ (g)v = (M ϕ f v )(g) = ϕ(gr 1 )f v (r) r = ϕ(g)v where the last equality holds because f v (r) = 0 for all r H and for the one coset representative h H, we have ϕ(gh 1 )f v (h) = ϕ(g)ψ 1 (h 1 )ψ 1 (h)v = ϕ(g)v. That M ϕm = M for M M is a similarly straightforward calculation and is thus omitted. To complete our proof we only need to check that this map respects the multiplicative structure, i.e., ϕ M M = ϕ M ϕ M. This is another computation. For any g G and any v V we have: 6

7 ϕ M ϕ M (g)v = 1 H ϕ M (s) ϕ M (s 1 g)v s G = 1 H ϕ M (s) (M f v )(s 1 g) s G = 1 H = 1 H ϕ M (s) (M ψ 1 (s 1 g))v s G g sh Mψ 1 (s)(m ψ 1 (s 1 g))v s H g sh = { (M M )f v (g) if g H 0 otherwise = ϕ M M v and the C-algebra isomorphism is established. Proposition 2. Let H G be a subgroup of a finite group G and ψ H GL(V ) a representation of H. Let M = End G (Ind G Hψ) and H = {ϕ G End C (V ) ϕ(hgh ) = ψ(h)ϕ(g)ψ(h ) for h, h H, g G}. Then M H as C-algebras. Proof. This is an immediate consequence of proposition 1. Definition 1. H as defined in the proposition is called the Hecke algebra of the pair (G, H). As noted in the introduction, the Hecke algebra of a pair (G, H) will be useful in detecting whether or not the (G, H) is a Gelfand pair. In order to uncover this connection, we must first prove some foundational facts: Proposition 3. Suppose G is a finite group and H G a subgroup. Suppose further that (π 1, V 1 ) and (π 2, V 2 ) are representation of H and H is the associated Hecke algebra. Now let H g = H ghg 1 for g G. Then the subspace of H consisting of all functions supported on the coset HgH is isomorphic as a vector space to Hom H g(π g 1, π 2 H g) where π g 1, π 2 H g are the two representations of H g defined by π g 1 (h) = π 1(g 1 hg) and π 2 H g is just the restriction of π 2 to H g. 7

8 Proof. Let g G be fixed and ϕ H supported on HgH be given. Define a map M ϕ V 1 V 2 by M ϕ (v) = ϕ(g)v for v V 1. Then for h H g it is true that: M ϕ (π g 1 (h)v) = ϕ(g)(π 1(g 1 hg)v) = ϕ(hg)v = π 2 (h)ϕ(g)v = π 2 (h) H gm ϕ (v). So M ϕ is a H g -module homomorphism, i.e., M ϕ Hom H g(π g 1, π 2 H g). Now given M Hom H g(π g 1, π 2 H g), define ϕ M G End C (V 1, V 2 ) by ϕ M (hgh )v = π 2 (h)ϕ(g)π 1 (h )v if h, h H and ϕ M (ĝ)v = 0 if ĝ HgH. It is easy to see that under this definition, ϕ M is an element of H supported on HgH. It is also easy to see that the maps ϕ M ϕ and M ϕ M are mutual inverses. This is sufficient to prove the vector space isomorphism. Proposition 4 (Mackey s Theorem). Suppose G is a finite group and H G a subgroup. Suppose further that (π 1, V 1 ) and (π 2, V 2 ) are representations of H. For g G, let H g be defined as in the previous proposition. Similarly, let π g 1, π 2 H g be defined as in the previous proposition. Then and as a consequence Hom G (Ind G Hπ 1, Ind G Hπ 2 ) dim Hom G (Ind G Hπ 1, Ind G Hπ 2 ) = Hom H g(π g 1, π 2 H g) g HgH Proof. This is clear from what we have shown so far. dim Hom H g(π g 1, π 2 H g). g HgH The following result, known as Schur s Lemma, is important in group representation theory and will be used here throughout. Proposition 5 (Schur s Lemma). Let (ρ 1, V 1 ) and (ρ 2, V 2 ) be irreducible representations of a group G. Then any element of Hom G (V 1, V 2 ) is either the zero map or an isomorphism. As a corollary, Hom G (π, π) C for any irreducible representation π of G. Proof. Suppose A Hom G (V 1, V 2 ) is nonzero. Since A is a G-module homomorphism, image(a) is a G-invariant subspace of V 2. But that V 2 is irreducible forces that either image(a) = 0 or image(a) = V 2. The first case is false since A is nonzero. Thus, the second case holds, i.e., A is surjective. 8

9 Similarly, ker(a) is a G-invariant subspace of V 1, and, thus, either ker(a) = 0 or ker(a) = V 1. The second case is false since A is nonzero; thus the fist case holds, i.e., A is injective. This proves that A is a bijective homomorphism, i.e., an isomorphism, which proves the fist statement. To see that the second statement holds, we let ψ Hom G (π, π) be nonzero. Let λ be an eigenvalue of ψ (we are assured that such an eigenvalue exists since π is finite dimensional and since C is an algebraically closed field). Then ψ λi Hom G (π, π) is not invertible so by what we have just proven it is zero, i.e. ψ = λi. Proposition 6. Let ρ be a finite dimensional representation of a group G. Then ρ is multiplicity free if and only if Hom G (ρ, ρ) is commutative. In this case, dim Hom G (ρ, ρ) is equal to the number of irreducible constituents in the direct sum decomposition of ρ. Proof. Let ρ = m i=1 ρ i be the direct sum decomposition of ρ into irreducible representations ρ i with the ρ i s ordered so that equivalent representations occur consecutively. Now any σ Hom G (ρ, ρ) can be written as a block diagonal matrix A = (a ij ), where a ij Hom G (ρ j, ρ i ). But Schur s lemma gives us that Hom G (ρ j, ρ i ) { C if ρ i ρ j {0} otherwise. Thus, we can see that Hom G (ρ, ρ) is isomorphic to an algebra of block diagonal matrices over the complex numbers where the size of the blocks corresponds to the multiplicity of each irreducible representation ρ i. Composition of homomorphisms corresponds to matrix multiplication so that Hom G (ρ, ρ) will indeed be commutative if and only if each block is of size one, i.e., if and only if ρ is multiplicity free as claimed. If this is the case, then Hom G (ρ, ρ) is isomorphic to the algebra of m m diagonal matrices over the complex numbers so that indeed dim Hom G (ρ, ρ) = m, which completes the proof. Proposition 7. Let H G be a subgroup of the finite group G and let ψ H V be a finite dimensional representation. Then Ind G Hψ is multiplicity free if and only if the associated Hecke algebra H is commutative. Proof. This follows from Propositions 2 and 6. 9

10 Definition 2. Suppose G is a group, H G a subgroup and (ψ, V ) a representation of H. If Ind G Hψ is multiplicity free, we say that (G, H, ψ) is a Gelfand triple. In the case that ψ = 1 H we say that (G, H) is a Gelfand pair or that H is a Gelfand subgroup of G. From these results, we can see that the study of Gelfand pairs is intimately linked with the study of Hecke algebras. In the following section, we will explore specific examples of Gelfand pairs and will produce general tools that allow us to exploit this fundamental connection. 10

11 3 Examples We now consider some specific examples of Gelfand triples/pairs. Let F q be any finite field, n N and let G = GL n (F q ). Now let N G consist of the subgroup of all upper triangular matrices with 1 s on the diagonal and let ψ F q C be a nontrivial linear character. Define ψ 0 N C by ψ 0 (a) = ψ(a 12 + a a n 1,n ) for a = (a ij ) N. Our goal will be to show that the induced representation Ind G N ψ 0 is multiplicity free. To this end, we will show that the associated Hecke algebra H is commutative. First we need a proposition: Proposition 8 (Bruhat Decomposition). G = GL n (F q ) can be decomposed as a disjoint union of double cosets with representatives of the following form: GL n (F q ) = w W BwB where B G is the Borel subgroup of upper triangular matrices and W is the group of n n permutation matrices. Proof. I will first prove that G can indeed be decomposed as such a union. Then I will prove that the union is disjoint. I will proceed by induction on n. For n = 1, B = G so the result holds trivially. Now let n > 1 and let g G be arbitrary. I will prove that there exists a permutation matrix w that is an element of BgB, which suffices to prove our assertion. We consider two cases: Case 1: g n,1 0. In this case, we may multiply g by appropriate elements b 0, b 1 B on the left and right so that b 0 gb 1 is 0 in every entry in the first column and last row besides the n, 1th entry which is equal to g n,1. We may then normalize this to 1 by dividing by g n,1. Now by disregarding the first column and last row, we have an n 1 n 1 matrix g to which we may apply our induction hypothesis to find 2 n 1 n 1 upper triangular matrices b, b such that b g b is an n 1 n 1 permutation matrix. By expanding these matrices b b 0 gb 1 b will give an n n permutation matrix. Case 2: g n,1 = 0. Then pick i as great as possible and j as least as possible such that g i,1 0 and g n,j 0. By multiplying by elements of B on the right and left, we can clear the first and jth columns and the ith and last rows, except for the entries g i,1 and g n,j, which we can normalize to 1. We can 11

12 then apply the induction hypothesis to the matrix obtained by removing these rows and columns, to create a permutation matrix. This completes the induction. Now I demonstrate that the union is indeed disjoint. Let w 1, w 2 W be representatives of the same double coset. Then there exists b B such that w 1 bw2 1 B. Now since w 1 bw2 1 is just b with some rows and columns permuted, if we changed some of the nonzero entries of b to any arbitrary field element, the result would still be upper triangular, i.e., still in B. So let us then replace b with the identity matrix. Then: w 1 w2 1 B. But since w 1, w 2 W, it follows that w 1 w2 1 B W = {I n }, where I n is the n x n identity matrix. Thus w 1 = w 2 and the result is proven. Proposition 9 (Modified Bruhat Decomposition). GL n (F q ) may be decomposed as a disjoint union of double cosets in the following fashion: GL n (F q ) = m M NmN where N is the subgroup of upper triangular matrices with 1 s on the diagonal and M is the subgroup of monomial matrices (matrices with exactly one nonzero entry in every row and column). Proof. Let D G be the subgroup of diagonal matrices. The result, ignoring disjointness of the union, follows from the Bruhat decomposition because B = DN = ND and M = DW = W D. That the union is disjoint can be deduced as before. Now we introduce a powerful tool for proving that a Hecke algebra H is commutative. We define an involution of a group G to be a map ι G G that satisfies ι 2 = id and ι(g 1 g 2 ) = ι(g 2 )ι(g 1 ) for g 1, g 2 G. Similarly, a map ῑ H H is called an involution of H if ῑ 2 = id and ῑ(ϕ 1 ϕ 2 ) = ῑ(ϕ 2 )ῑ(ϕ 1 ) for any ϕ 1, ϕ 2 H. Proposition 10. Let H be a subgroup of a finite group G and (ψ, V ) a one-dimensional representation of H. Let H be the associated Hecke algebra. Now suppose that ι G G is an involution satisfying ψ(ι(h)) = ψ(h) for all h H. Then the map ῑ H H defined by ῑ(ϕ(g)) = ϕ(ι(g)) where g G is an involution of H. 12

13 Proof. Observe that if ϕ H, h 1, h 2 H, and g G, then ῑ(ϕ(h 1 gh 2 ) = ϕ(ι(h 1 gh 2 )) = ϕ(ι(h 2 )ι(g)ι(h 1 )) since ι reverses multiplication in G = ψ(h 2 )ϕ(ι(g))ψ(h 1 ) since ϕ H and ψ(ι(h)) = ψ(h) for all h H = ψ(h 1 )ῑ(ϕ(g))ψ(h 2 ) where the last equality holds since V is 1-dimensional implies that End C V is commutative. This proves that ῑ(ϕ) H. Further, note that: ῑ 2 (ϕ(g)) = ῑ(ϕ(ι(g))) = ϕ(ι 2 (g)) = ϕ(g) since ι is an involution. Thus, ῑ 2 = id. Finally, that ῑ reverses multiplication follows easily from that fact that End C V is commutative, which completes the proof. Proposition 11. Suppose that there is an involution ι G G such that HgH = Hι(g)H for all g G. Then (G, H) is a Gelfand pair. Proof. We wish to show that Ind G N ψ is multiplicity-free, where ψ = 1 H. We construct ῑ according to Proposition 10 and note that since ψ is the trivial representation that the requirement that ψ(ι(h)) = ψ(h) for all h H holds trivially. Now H consists of all function ϕ G C such that ϕ is constant on H H double cosets (this is because ψ is trivial). Thus, ῑ acts by the identity on H. Finally, observe that ῑ = id being an involution implies that H is commutative. We now return to our study of the example introduced. We will apply the involution method to attain our desired result in this example, but first we will need this basic fact: Proposition 12. Let F be any field and suppose m = (m ij ) F n n is a monomial matrix and that m ij and m i+1,k are nonzero implies that k j + 1. Then m is of the form m = D p D 2 D 1 13

14 where p is a positive integer and the D i are diagonal matrices for i = 1,..., p. Proof. The proof is by induction on n. For n = 1, the conclusion follows trivially, so suppose n > 1 and that our conclusion holds for all matrices that satisfy the above conditions and of are size less than n. Now since m is monomial, there exists a unique i {1,..., n} such that m 1i 0. If i = n, then we note that the n 1 n 1 matrix constructed by removing the first row and last column of m is of the desired form by the induction hypothesis. So suppose that i < n. We claim then that m 2,k 0 forces k = i + 1. If not, then k < i. But then for any j 2, we have that m jl 0 implies that l < i by the assumption that m ij and m i+1,k both nonzero implies that k j +1. But now this implies that column i + 1 is composed purely of zeros, a contradiction. This proves now that m 2,i+1 is nonzero. Repeating this argument n i times proves that m 1+k,i+k is nonzero for k = 1,..., n i so that indeed m is of the form: m = ( 0 D 0 ) where D is an n i+1 n i+1 diagonal matrix. Now applying the induction hypothesis to, our conclusion follows. Proposition 13. Let G = GL n (F q ), and let N and ψ 0 be as above. Then Ind G N ψ 0 is multiplicity free. Proof. Let g G and consider the double coset NgN. By the modified Bruhat decomposition proposition, we may find a monomial matrix m = (m ij ) such that NgN = NmN. Assume that there is some ϕ H and some g NmN with ϕ(g) 0. Now I claim that if m ij and m i+1,k are both nonzero entries of m, then k j + 1 so that m is of the form m = D p D 2 D 1 where the D i are diagonal matrices for i = 1,..., p. Now suppose for a contradiction that m ij 0 and m i+1,k 0 with k > j +1. Since m ij 0 and since ψ is non-trivial, we may select t F q such that tm ij ker ψ, i.e., ψ(tm ij ) 1. (This is because F q is a field). Now let x = I n + tm ij e i,i+1, 14

15 y = I n + tm i+1,k e jk where e jk is an n n matrix that is 1 at position j, k and zero elsewhere. It is easy to see that xm = (I n + tm ij e i,i+1 )m = m + tm ij m i,i+1 e ik = my. Now, it is a simple computation that ψ 0 (x) = ψ(tm ij ) 1 and ψ 0 (y) = ψ(0) = 1. Now since x, y N and ϕ H, it holds that But then ψ 0 (x)ϕ(m) = ϕ(xm) = ϕ(my) = ϕ(m)ψ 0 (y). (ψ 0 (x) ψ 0 (y))ϕ(m) = 0 which forces ϕ(m) = 0 since our above computation demonstrates that ψ 0 (x) ψ 0 (y) is nonzero. But then ϕ(nmn) = 0, a contradiction. This shows that m does indeed adopt the given form. Next, I claim that each D i is a homothety, i.e., a scalar matrix, for each i = 1,..., p. This is equivalent to the statement that m ij 0 and m i+1,j+1 0 implies m ij = m i+1,j+1. So suppose that m ij and m i+1,j+1 are nonzero. Further, suppose for a contradiction that m ij m i+1,j+1. Then m ij m i+1,j+1 0 so that we may select s F q such that s(m ij m i+1,j+1 ) ker ψ. So if we alter x and y from above to be x = I n + sm ij e i,i+1, y = I n + sm i+1,j+1 e j,j+1 then as above a simple calculation gives (ψ 0 (x) ψ 0 (y))ϕ(m) = 0. Now it is easy to see from the formulation of x and y that ψ 0 (x) = ψ(sm ij ) and ψ 0 (y) = ψ(sm i+1,j+1 ). But now since ψ 0 (x) ψ 0 (y) = ψ(sm ij ) ψ(sm i+1,j+1 ) = ψ(s(m ij m i+1,j+1 )) 0 it follows that ϕ(m) = 0, a contradiction. Thus m ij = m i+1,j+1 and indeed each D i is a homothety. Finally, we consider the involution ι G G defined by ι(g) = hg T h where h =

16 It is a simple computation to see that ι thus defined fixes matrices m of the form described above (for this, note that right multiplication by h reverses the order of columns, while left multiplication by h reverses the order of rows) and that ψ 0 (ι(n)) = ψ 0 (n) for all n N. Thus, we find that ῑ as defined earlier is an involution of H. Now we claim that ῑ is the identity map on H. Once we have substantiated this claim, we will be done proving that H is commutative because indeed this implies that for any ϕ 1, ϕ 2 H, it holds: ϕ 1 ϕ 2 = ῑ(ϕ 1 ϕ 2 ) = ῑ(ϕ 2 ) ῑ(ϕ 1 ) = ϕ 2 ϕ 1 where the second equality holds since ῑ is an antihomomorphism. But the claim follows trivially from what we have already proven: that ϕ is determined by its values on the monomial matrices m and that ι fixes these matrices, i.e., for m a matrix of the above form it holds: and our proof is complete. ῑ(ϕ(m)) = ϕ(ι(m)) = ϕ(m) Thus, we have seen in action the power of the involution method as a tool to study Gelfand pairs. We will again apply this method to another interesting example: that of the symmetric group S n. There is a natural embedding S n S m S n+m in which S n acts on the first n elements of {1,..., n + m} and S m acts on the last m. We will again use the involution method to prove that S n S m S n+m is a Gelfand subgroup. Proposition 14. S n S m is a Gelfand subgroup of S n+m. Proof. Take H = S n S m and G = S n+m. Define the map ι G G to be the involution g g 1. We must check that each double coset HgH is fixed by ι. We may identify the elements of S n+m with permutation matrices. We claim that each double coset HgH has a representative of the form We may represent g in block form I l n l 0 I n l. 0 0 I m n+l 0 0 I n l 0 0 n l g = ( A B C D ) 16

17 where A, B, C, and D are matrices with only 1 s and 0 s and at most one 1 in any row or column, i.e., as permutation matrices. A is an n n matrix and D is a m m matrix. Let l denote the rank of A. Then B and C have rank n l since g is a permutation matrix. Thus D has rank m n + l. Multiplying on the left by elements of S n corresponds to permuting rows, while multiplying on the right by elements of S n corresponds to permuting columns. Thus, multiplying on the left and right by elements of S n allows us to rearrange the nonzero elements of A so that they are in the top left hand corner. Similarly, multiplying by elements of S m will rearrange D so that its nonzero elements are in the top left hand corner. This yields a matrix of the form U l n l 0 W n l 0 0 V m n+l 0 0 X n l 0 0 n l where U l, V m n+l, W n l, X n l are all permutation matrices. Note that we may naturally identify S l S n l S m n+l S n l as a subgroup of S n S m. Multiplying our matrix on the left and right by elements of S l S n l S m n+l S n l will put it in the form I l n l 0 I n l 0 0 I m n+l 0 0 I n l 0 0 n l which verifies the claim. It is easy to see that squaring such a matrix gives an n + m n + m identity matrix. That is, ι fixes such matrices. Thus, S n S m S n+m is a Gelfand subgroup. We have found that if λ Z + and m, n Z + satisfy n + m = λ, then the induced representation, Ind S λ S m S n 1 Sm Sn is multiplicity free. This motivates us to wonder about the behavior of a more general case. That is, if λ is a positive integer and {λ i } positive integers satisfying i λ i = λ, then what does the representation of S λ induced by the trivial character on i S λi look like? We explore the answer to this question in the following section, finding that such a question leads to an interesting story that classifies the irreducible representations of the symmetric group S λ. 17

18 4 Representation Theory of the Symmetric Group To discuss the representation theory of the symmetric group, we introduce the notion of a partition. Suppose n is a positive integer. Then we say that λ = (λ 1,..., λ k ) forms a partition of n if n = k i=1 λ i and if λ 1 λ k > 0. For example, if n = 3, then λ = (3), µ = (2, 1), ν = (1, 1, 1) are all of the partitions of n. We write λ n to mean λ is a partition of n. Now it is a basic fact from group theory that the conjugacy classes of S n are determined by cycle type so that indeed they are in bijective correspondence to the distinct partitions of n. Thus, the number of distinct irreducible representations will be equal to the number of partitions of n. Now suppose λ = (λ 1,..., λ k ) is a partition of n. Then define the subgroup S λ = S λ1 S λk S n. Given such a partition λ, we will study the representations of S n that are induced by certain one-dimensional representations of S λ. Now consider the following one-dimensional representations of S λ. The trivial representation: ρ λ S λ GL 1 (C) given by σ 1 for all σ S λ and the sign representation: π λ S λ GL 1 (C) given by σ sgn(σ) for all σ S λ. Where the sign of σ is defined by sgn(σ) = { 1 if σ is even 1 if σ is odd. Now we can consider the induced representations Ind Sn S λ ρ λ, Ind Sn S λ π λ. Further, we let ψ H denote the restriction of a representation ψ of G to a subgroup H G. So let λ n, µ n be partitions of n. Now by Mackey s theorem, it holds: dim Hom Sn (Ind Sn S λ ρ λ, Ind Sn S µ ρ µ ) = dim Hom(ρ λ Sλ σs 1, ρ µσ µ Sλ σs 1) µσ S λ σs µ = S λ σs µ dim Hom(ρ Sλ σs µσ 1, ρ S λ σs µσ 1) = S λ σs µ 1. 18

19 Similarly, we can compute dim Hom(Ind Sn S λ ρ λ, Ind Sn S µ π µ ) = dim Hom(ρ Sλ σs 1, π µσ S λ σs 1). µσ S λ σs µ Now both ρ Sλ σs µσ 1 and π S λ σs µσ 1 are irreducible. Further, it is clear that these two representations are equal if and only if the intersection S λ σs µ σ 1 is contained in the alternating group, i.e., S λ σs µ σ 1 A n. But now we note that since S λ σs µ σ 1 is isomorphic to a product of symmetric groups m i=1 S xi, we will have S λ σs µ σ 1 A n if and only if x i = 1 for i = 1,..., m, i.e., the intersection is trivial S λ σs µ σ 1 = {1}. Thus, with this extra condition we deduce: dim Hom(Ind Sn S λ ρ λ, Ind Sn S µ π µ ) = 1. S λ σs µ S λ σs µσ 1 ={1} We see then that our investigation about the symmetric group reduces to the combinatorial problem of counting double cosets of the form S λ σs µ. To this end, we first introduce some terminology and prove a useful result. Given a positive integer n and λ = (λ 1,..., λ k ) a partition of n, we let n λ i for 1 i k denote subsets of {1,..., n} that satisfy n λ i = λ i for all 1 i k, n i n j = for i j, and k i=1 n λ i = {1,..., n}. We call such a collection of nλ i for i = 1,..., k a dissection of {1,..., n}. Now, we prove the following useful fact: Proposition 15. Let n be a positive integer. Let λ = (λ 1,..., λ k ) and µ = (µ 1..., µ l ) be partitions of n. Let S λ, S µ be defined as above. Then τ S λ σs µ if and only if for every 1 i k and 1 j l it holds: n λ i σ(n µ j ) = n λ i τ(nµ j ). Proof. Assume τ S λ σs µ. Say τ = ψσφ where ψ S λ and φ S µ. Then for each j {1,..., l} it will hold: τ(n µ j ) = ψσφ(nµ j ) = ψσ(nµ j ). Thus, if i {1,..., k} and j {1,..., l}, it holds: n λ i τ(n µ j ) = nλ i ψσ(n µ j ) = ψ(nλ i σ(n µ j )) since ψ S λ. Thus, since σ is a bijection, this yields n λ i σ(nµ j ) = nλ i τ(nµ j ), as desired. 19

20 Now conversely, suppose that n λ i σ(nµ j ) = nλ i τ(nµ j ). Then for each fixed i, the subsets n λ i σ(nµ j ) and the subsets nλ i τ(nµ j ) form dissections of n λ i which can be put into ordered pairs of the form (n λ i σ(nµ j ), nλ i τ(nµ j )) of subsets of equal order. To see that these are indeed dissections, we note it is clear that they are pairwise disjoint for distinct j 1, j 2 and that indeed: l j=1 (n λ i τ(n µ j )) = nλ i l τ(n µ j ) j=1 = n λ i {1,..., n} = n λ i and similarly for σ. Now for each i then we pick ψ i S λ such that ψ i fixes {1,..., n} n λ i and such that ψ i (n λ i σ(n µ j )) = nλ i τ(n µ j ) for each j = 1,..., l. Now define ψ = ψ 1 ψ k. Then we see that for every j = 1,..., k: ψσ(n µ j ) = τ(nµ j ) so that indeed there exists φ S µ that satisfies τ = ψσφ, i.e., τ S λ σs µ and the proof is complete. This shows that for two fixed partitions λ n and µ n the double cosets S λ σs µ are characterized by the numbers x ij = n λ i σ(n µ j ), 1 i k, 1 j l. Now let m = max{k, l}. Then by letting x ij = { nλ i σ(nµ j ) if 1 i k, 1 j l 0 if i > k or j > l we attain an injective map g S λ /S n /S µ M m (N) defined by S λ σs µ (x ij ) where M m (N) is the set of m m matrices over the natural numbers. It is clear that the image of g is im(g) = {(x ij ) M m (N) x ij = µ j and m i=1 This proves the following result: 20 m j=1 x ij = λ i and x ij = 0 if i > k or j > l}.

21 Proposition 16. Suppose λ = (λ 1,..., λ k ) and µ = (µ 1,..., µ l ) are two partitions of a positive integer n. Then there exists a bijection: where X = {(x ij ) M m (N) x ij = µ j and m i=1 g S λ /S n /S µ X and g is given by S λ σs µ (x ij ) where m j=1 x ij = λ i and x ij = 0 if i > k or j > l} x ij = { nλ i σ(nµ j ) if 1 i k, 1 j l 0 if i > k or j > l. With this we now have the information to compute the formulas given by Mackey s intertwining number formula. Now if we restrict our view to double cosets S λ σs µ with the trivial intersection property: S λ σs µ σ 1 = {1} and restrict g to this subset, then we obtain the following corollary: Proposition 17. The number of double cosets S λ σs µ with the property S λ σs µ σ 1 = {1} is equal to the number of n n 0 1 matrices with row sums λ i and column sums µ j. Thus, we have that dim Hom(Ind Sn S λ ρ λ, Ind Sn S µ π µ ) is equal to the number of n n 0 1 matrices with row sums λ i and column sums µ j. To deal with the combinatorial problem of counting such matrices, we define the following relation on partitions of n: Definition 3. Let n be a positive integer and suppose that λ, µ n are partitions of n. Then we say that λ µ if k i=1 λ i k i=1 µ i for all k. Proposition 18. Let n be a positive integer. Then defines a partial order on the set of partitions of n. Proof. Reflexivity is obvious. To prove antisymmetry, suppose λ µ and µ λ. Then for any j, we have j i=1 λ i j i=1 µ i and j 1 i=1 λ i j 1 i=1 µ i so that λ j = j i=1 λ i j 1 i=1 λ i j i=1 µ i j 1 i=1 µ i = µ j. Similarly µ j λ j and thus λ j = µ j for each j, i.e., λ = µ. Transitivity also follows directly from the definition of. 21

22 Now for a partition λ = (λ 1,..., λ k ) of a positive integer n, define D(λ) to be the diagram consisting of rows of blocks where the ith row consists of λ i blocks. For example, if n = 7 and λ = (3, 2, 1, 1) then D(λ) is given by Now noting that λ i λ i+1 for each i, we see that the lengths of the columns of D(λ) form a partition λ of n. This partition is called the conjugate partition of λ. D(λ ) can be clearly obtained by simply rotating D(λ) around it s main diagonal or by interchanging rows and columns. For example with λ = (3, 2, 1, 1) as above, we have λ = (4, 2, 1) and D(λ ) is given by. Now to denote the sum of the column vectors of a matrix A we write c(a) and to denote the sum of the row vectors of the same matrix we write r(a) Proposition 19. Let λ = (λ 1,..., λ k ), µ = (µ 1,..., µ l ) be partitions of a positive integer n. Then there exists a 0,1 matrix with c(a) = λ and r(a) = µ if and only if λ µ. Proof. To prove one direction, assume that A = (a ij ) is a k l matrix with c(a) = λ and row sums r(a) = µ. Now if there is no i k, j < h l such that a ij = 0 and a ih = 1 then it is easy to see that λ = µ. Now if, on the contrary, there exists (i, j) satisfying the conditions above so that a ij = 0 is a gap in the matrix, then let h > j be maximal such that a ih = 1. Then swapping a ij and a ih we attain a matrix A with c(a ) = λ and that satisfies r(a ) r(a). Thus, it follows by induction on the number of such gaps that λ µ. Now to prove the converse, suppose that λ µ. Then we note that there exists a 0-1 k l matrix A with c(a) = λ and r(a) µ namely the matrix whose ith row is 1 for 1 j λ i and 0 for j λ i, i.e., this is the matrix constructed by putting 1 s on the diagram D(λ) and 0 elsewhere. 22

23 Now I claim that given a k l matrix A such that c(a) = λ and r(a) µ and such that r(a) µ, we can find a k l 0-1 matrix A such that c(a ) = λ and r(a ) µ that satisfies r(a ) µ r(a) µ where is the usual Euclidean norm. Note that since r(a) µ 2 is an integer, this process must eventually terminate after a finite number of steps, i.e., we will be able to find a matrix Ā such that c(ā) = λ and r(ā) = µ. Thus, to complete our proof all we need to do is verify this claim. So let r(a) = (r 1,..., r l ). Now let i be minimal such that r i > µ i, and let j be minimal such that r j < µ j. Then i < j since r(a) µ. Now let r = (r 1,..., r n) = (r 1,..., r i 1, r i 1, r i+1,..., r j 1, r j + 1, r j+1,..., r l then it is clear that r µ < r µ. Further, it is clear that r µ. Now since r i > q i q j > r j, we can find a number h {1,..., k} such that a hi = 1 and a hj = 0. For such an h, define the matrix A = (a st) by 1 if (s, t) = (h, j) a st = 0 if (s, t) = (h, i) a st otherwise By briefly inspecting this definition of A, it is easy to see that indeed r(a ) = r and c(a) = λ which verifies our claim and thus completes the proof. We note that it is an immediate consequence of this result that λ µ if and only if µ λ, so that indeed λ = µ if and only if λ = µ. We will use this result and the following to gain some interesting insight into the irreducible representations of the symmetric group. Proposition 20. Let λ = (λ 1,..., λ k ) be a partition of a positive integer n. Then there is precisely one 0-1 matrix A with c(a) = λ and r(a) = λ. Proof. One such matrix A is the matrix constructed using D(λ), i.e., it is 1 where there is a node in D(λ) and 0 elsewhere. More formally, A will be constructed by letting the first λ i elements of the ith row be 1 and the remaining elements of the row be 0 for i = 1,..., k. It is clear that this matrix will satisfy c(a) = λ and r(a) = λ. To see that this is the only such matrix, suppose that there is a 0-1 matrix B with c(b) = λ and r(b) = λ such that there exists i {1,..., k} with b ij = 0 for some j < λ i. Then it is easy to see that k l=1 a lj < λ j, a contradiction. So indeed any matrix satisfying our conditions must be of the form of A and we are done. 23

24 Thus, we know that if λ n then dim Hom(Ind Sn S λ ρ, i Sn S λ π) = 1 so that π contain precisely one equivalent irreducible representation Ind Sn S λ ρ and i Sn S λ of multiplicity one in their direct sum decompositions. We will let Π λ denote this irreducible representation. Proposition 21. Let λ n and µ n be two partitions of a positive integer n. Then Π λ = Π µ implies λ = µ. Proof. If Π λ = Π µ then we may deduce that Ind Sn S λ ρ, i Sn S λ π, IndSn S µ ρ, i Sn S π share µ precisely one irreducible representation in common. Thus, dim Hom(Ind Sn S λ ρ, i Sn S π) > µ 0 so that λ µ or equivalently µ = µ λ = λ. Similarly, dim Hom(Ind Sn S µ ρ, i Sn S π) > λ 0 so that it holds that µ λ or equivalently that λ µ. Thus, since is a partial order, this forces λ = µ. Now we have proven that for any positive integer n, there exists an injective map from the set of partitions λ n of n into the set of irreducible representations of S n defined by λ Π λ. To see that this is a surjection, we recall the basic group theoretic fact that the number of conjugacy classes of S n is equal to the number of partitions of n and the basic fact from the representation theory of finite groups that the number of inequivalent irreducible representations of a finite group G is equal to the number of conjugacy classes of G, so that indeed the set of partitions of n and the set of inequivalent irreducible representations of S n are two finite sets of equal cardinality. We summarize what we have proved in the following proposition: Proposition 22. For any positive integer n, there exists a natural bijection between the partitions of n and the set of inequivalent irreducible representations of the symmetric group S n, given by λ Π λ. 24

25 5 Spherical Functions We begin this section with an important example. Our first goal will be to prove the following proposition. Proposition 23. Let G be a finite group, and embed G diagonally into G G, that is by the map g (g, g). Identifying G with the image of this diagonal embedding, G is a Gelfand subgroup of G G. We provide multiple different proofs of this result that are enlightening in different ways. Proof 1 (Involution Method). Consider the map ι G G G G given by (g 1, g 2 ) (g2 1, g 1 1 ). Observe that given (g 1, g 2 ) G G it holds: ι 2 ((g 1, g 2 )) = ι((g 1 2, g 1 1 )) = (g 1, g 2 ) so that indeed ι 2 = id. Further, if (g 1, g 2 ), (g 3, g 4 ) G G, then ι((g 1, g 2 )(g 3, g 4 )) = ι((g 1 g 3, g 2 g 4 )) = (g4 1 g2 1, g3 1 g1 1 ) = (g4 1, g3 1 )(g2 1, g1 1 ) = ι((g 3, g 4 ))ι((g 1, g 2 )) so that indeed ι reverses multiplication and is thus an involution. To see that ι fixes each double coset G(g 1, g 2 )G, let (g 1, g 2 ) G G be arbitrary. Now let a = g1 1 and b = g2 1 and consider the elements (a, a), (b, b) G. It is easy to see that ι((g 1, g 2 )) = (g 1 2, g 1 1 ) = (g 1 2 g 1 g 1 1, g 1 2 g 2 g 1 1 ) = (b, b)(g 1, g 2 )(a, a) G(g 1, g 2 )G so that ι acts trivially on double cosets. The result follows. Our other proofs require some extra machinery. Let (π i, V i ) be a representation of the group G i for i = 1, 2. We may define a representation π 1 π 2 of the direct product G 1 G 2 by (π 1 π 2 )(g 1, g 2 )(v 1 v 2 ) = π 1 (g 1 )v 1 π 2 (g 2 )v 2 where g 1 G 1, g 2 G 2 and where v 1 v 2 V 1 V 2 is a simple tensor. Since any element of V 1 V 2 is a linear combination of simple tensors extending this representation by linearity does indeed give a representation of G 1 G 2. We give conditions on π 1 and π 2 that are necessary and sufficient so that π 1 π 2 is an irreducible representation of G G. 25

26 Proposition 24. Let (π i, V i ) be a finite dimensional linear representation of the group G i for i = 1, 2. Then π 1 π 2 is an irreducible representation of G 1 G 2 if and only if π 1, π 2 are irreducible representations of G 1, G 2 respectively. Proof. Suppose without loss of generality that (π 1, V 1 ) is reducible, say V 1 U W where U, W V 1 are nontrivial G 1 invariant subspaces. Then it is easy to see that V 1 V 2 (U W ) V 2 (U V 2 ) (W V 2 ) so that (π 1 π 2, V 1 V 2 ) is reducible. Now suppose that π 1, π 2 are irreducible. Let n = dimv 2. Then Hom G1 (π 1, π 1 ) n Hom G1 (π 1, π1 n) via the isomorphism A 1 A n B, where B is defined by B(v) = A 1 (v)... A n (v). Because V 2 C n and C Hom G1 (π 1, π 1 ), it holds that 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 for 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 the map T V 1 Hom G1 (π 1, π1 m) V 1 m defined by v A A(v) is an isomorphism. These facts show that there is a bijection {G 1 invariant subspaces of V 1 V 2 } {C vector subspaces of V 2 } given by V 1 W W and X Hom G1 (π 1, X) Hom G1 (π 1, π 1 1 n ) = V 2. Suppose that X V 1 V 2 is a nonzero G 1 G 2 -invariant subspace. Then X is also a G 1 -invariant subspace, so that by what we have shown it holds that X = V 1 W for some complex subspace W V 2. That π 2 is irreducible guarantees span{(π 1 π 2 )(1, g 2 )x x X, g 2 G 2 } = V 1 span{π 2 (g 2 )w w W, g 2 G 2 } = V 1 V 2 We see then that G 1 G 2 -invariance of X forces that X = V 1 V 2, i.e., the tensor product π 1 π 2 is irreducible. Proposition 25. Let (π, V ) be an irreducible representation of the direct product group 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. For g 1 G 1, g 2 G 2, and v V, define π 1 (g 1 )v = π(g 1, 1)v and π 2 (g 2 )v = π(1, g 2 )v. It is easy to see that these define representations of 26

27 G 1, G 2 respectively. It is not hard to see that we can pick a nonzero G 1 invariant subspace V 1 of V such that π 1 = π 1 V1 is an irreducible representation of G 1. Now let v V 1 be nonzero and define V 2 = span{ π 2 (g)v g G 2 }. V 2 is clearly G 2 invariant and π 2 = π 2 V2 is a representation of G 2. Now for any v 1 V 1 and v 2 V 2, there exist numbers a i, b i C and elements g (i) 1, g(i) 2 G 1 G 2 such that v 1 = i a i π 1 (g (i) 1 )v and v 2 = i b i π 2 (g (i) 2 )v. Given such elements v 1, v 2, we define a map T V 1 V 2 V by the rule T (v 1 v 2 ) = i a i b j π(g (i) 1, g(j) 2 )v j on simple tensors and then extend by linearity to define a map on all elements of V 1 V 2. It is not difficult to verify that T is a well-defined G 1 G 2 -module homomorphism, i.e., T Hom G1 G 2 (V 1 V 2, V ). Further T is nonzero. For example, T (v v) = v and thus image(t ) V is a nontrivial G 1 G 2 - invariant subspace. By irreducibility of V, this forces image(t ) = V, i.e., T is surjective. It is similarly straightforward to check T is injective. Now we have shown that the irreducible representations of the group G G are precisely those representations π i π j where π i and π j are irreducible representations of G. In order to utilize this information to prove that G embedded diagonally in G G is a Gelfand subgroup, we must utilize some tools from character theory. We introduce some basic tools from character theory below (such as Frobenius reciprocity and Schur orthogonality) in order to prove our desired result. Proposition 26. Let G be a finite group and H G a subgroup. Let (π, V ) be a representation of H and (ψ, W ) a representation of G. Then there is a vector space isomorphism Hom G (W, Ind G Hπ) Hom H (W, V ) where the vector space isomorphism and its inverse are given thus: for σ Hom G (W, Ind G H π) define φ Hom H(W, V ) by φ(w) = σ(w)(1). For φ Hom H (W, V ) define σ Hom G (W, Ind G Hπ) by σ(w)(g) = φ(ψ(g)w). Proof. Suppose that σ W Ind G Hπ is a G-module homomorphism. Then for h H, we have φ(ψ(h)w) = σ(ψ(h)w)(1) = (π G (h)σ(w))(1) = σ(w)(1 h) = π(h)σ(w)(1). 27

28 The right hand side is π(h)φ(w) so that φ is an H-module homomorphism. It is a similarly straightforward computation to show that if φ W V is an H-module homomorphism then σ(w)(g) = φ(ψ(g)w) gives a G-module homomorphism σ W Ind G Hπ and that indeed the maps σ φ and φ σ are mutual inverses. The following corollary, also referred to as Frobenius reciprocity, is the form of the result as we will use it. Proposition 27 (Frobenius reciprocity). Let G be a finite group and H G a subgroup. Let (π, V ) be a representation of H and (ψ, W ) a representation of G, and let χ π and χ ψ be the associated characters. Then χ Ind G H π, χ ψ G = χ π, χ ψ H H. Proof. This is immediate from the previous proposition. The following result is a consequence of Frobenius reciprocity and is actually a statement of the equivalence of the two definitions of Gelfand pairs which were provided in the introduction. Proposition 28. Let G be a finite group and H G a subgroup. Then H is a Gelfand subgroup if and only if for every irreducible representation (π, V ) of G, the subspace V H V of H-fixed vectors is at most one dimensional, or equivalently, the restriction π H contains the trivial representation 1 H with multiplicity at most 1. Proof. Note that V H Hom H (1 H, V ). By Frobenius reciprocity, it holds: dim Hom H (1 H, V ) = χ 1H, χ π H H = χ Ind G H 1 H, χ π G. The expression on the left is the multiplicity of π in Ind G H 1 H, so the result follows from the definition of a Gelfand subgroup (Definition 2). The following special type of representation will be useful in our example of interest: Definition 4. Let G be a a group and (π, V ) a linear representation of G. The representation ˆπ of G on V the dual space of V defined by ˆπ(g) = π(g 1 ) for all g G (where denotes taking the adjoint of a linear transformation) is called the representation of G contragradient to π. 28

29 Proposition 29. Let G be a finite group and (π, V ) a representation of G. Then there exists a G-invariant inner product on V. Proof. Let, be any inner product on V. Then it is easy to see that v, v = 1 G π(g)v, π(g)v for v, v V defines a Hermitian inner product on V. It is G-invariant by construction. Proposition 30. Suppose that (π, V ) is a representation of a finite group G and χ π is its character. Then χ(g 1 ) = χ(g) for all g G and the character of the contragradient representation ˆπ is the complex conjugate of the character of π, i.e., χˆπ = χ π. Proof. Let, be a G-invariant inner product and let v V be an eigenvector of π(g) with eigenvalue λ (existence of such a v and λ is guaranteed by basic linear algebra). Then λ 2 v, v = λv, λv = π(g)v, π(g)v = v, v. Thus, every eigenvalue is of norm 1. So, if we let λ 1,..., λ n denote the eigenvalues of π(g). Then tr(π(g 1 )) = n i=1 λ 1 i = n i=1 λ i = χ(g). Now ˆπ(g) is the adjoint of π(g 1 ), so its trace is equal to tr(π(g 1 )) which completes the proof. Proposition 31. Let (π, V ) and (π, V ) be irreducible representations of a finite group G. Suppose that L V C and L V C are linear functionals. Then either π π or L(π(g)x) and L (π (g)x) are orthogonal. Proof. Since L and L are linear functionals, they are of the form L(x) = x, y, L (x ) = x, y for some y V and y V. Define a map µ V V by µ(v) = π(g)v, y π (g 1 )y. 29

30 We claim that µ(π(h)v) = π (h)µ(v) to see this, we make the variable change g gh 1 and see that µ(π(h)v) = π(gh 1 )v, y π (hg 1 )y gh 1 G = π (h) π(g)v, y π(g 1 )y = π (h)µ(v) so that µ is a G-module homomorphism. By Schur s Lemma, µ is either identically 0 or an isomorphism, so if π / π then it holds: 0 = µ(x), x = 1 G π(g)x, y π (g 1 )y, x = 1 G π(g)x, y y, π (g)x = 1 G π(g)x, y π (g)x, y = L(π(g)x), L (π (g)x ) where the third equality comes from taking adjoints. Thus, we have shown, π π or L(π(g)x) and L (π (g)x ) are orthogonal, which is what we had to prove. Proposition 32. Let (π, V ) be an irreducible representation of the finite group G, with G-invariant inner product,. Then there exists a positive real constant d R + such that π(g)x, y π(g)x, y = 1 d x, x y, y. Indeed we will later show that this formula holds with d = dim(v ). Proof. µ as defined in the proof of the previous proposition is a G-module homomorphism, so by Schur s lemma it is a scalar multiple of the identity. That is, for fixed y, y there is a constant c = c(y, y ) such that µ(x) = c(y, y )x for all x X. Thus, it holds c(y, y ) x, x = µ(x), x = π(g)x, y π(g 1 )y, x 30