Gelfand Pairs, Representation Theory of the SymmetricGroup, and the Theory of Spherical Functions

John RyanStanford University

June 3, 2014

Abstract

This thesis gives an introduction to the study of Gelfand pairsand their applications. We begin with a brief introduction to thenotion of a Gelfand pair and then move to some of the foundationalresults concerning Gelfand pairs. Next, we explore specific examples ofGelfand pairs, developing tools of independent interest as we progress.We find that consideration of a specific example of a Gelfand pair andof the tools used in our study naturally leads us to a discussion ofthe representation theory of the symmetric group. We then concludeour study by developing the theory of spherical functions on groups,which gives us a glimpse of the relevance of Gelfand pairs to areas ofmathematics 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

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1 Introduction

The aim of this thesis is to give an introduction to the theory of Gelfandpairs and to explore related topics that emerge naturally from a discussion ofGelfand pairs. Attempts have been made to make this work as self-containedand accessible as possible. Very little background is necessary to understandthe results and arguments here beside command of basic linear algebra, grouptheory, and elementary notions from representation theory.

There are two natural and equivalent definitions of a Gelfand pair thatwe give here (equivalence of the definitions will be proved later).

The first definition is natural and easy to define. Let G be a finite groupand H ≤ G a subgroup. Then we say that (G,H) is a Gelfand pair if forevery 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∣Hcontains the trivial representation 1H with multiplicity at most 1.

The second (equivalent) definition of a Gelfand pair requires the notionof an induced representation, which we recall here:

Let G be a finite group, H ≤ G a subgroup of G and π ∶ H → GL(V ) arepresentation of H. We define the induced representation IndGHπ to be thevector space of all functions f ∶ G→ V such that f(hx) = π(h)f(x) for h ∈Hand x ∈ G. Now we define, for g ∈ G:

(πG(g)f)(x) = f(xg).

Under this definition, πG is a representation defined by g acting on IndGHπby right translation. We will usually abuse notation and write πG simply asπ 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 ele-mentary fact that π can be written uniquely as a direct sum of irreduciblerepresentations

π =n

⊕i=1diπi

where π1, . . . , πn are the distinct irreducible representations of G and thedi 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 IndGH1H constructed by in-ducing the trivial representation of H, is decomposed as a direct sum of ir-reducible representations, each irreducible occurs with multiplicity less than

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or equal to one, i.e., if we write

IndGH1H =n

⊕i=1diπi

then di ≤ 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 ringstructure is endowed by the multiplication rule

(ϕ1 ∗ ϕ2)(g) =1

∣H ∣ ∑s∈Gϕ1(s)ϕ2(s−1g).

This algebra is known as the Hecke algebra of the pair (G,H). The first majorresult of this paper will be to show that it is equivalent to say that (G,H) isa Gelfand pair and that the Hecke algebra of (G,H) is a commutative ring.

We will then explore some interesting examples of multiplicity free in-duced representations including the so-called Gelfand-Graev Representationof the general linear group over a finite field GLn(Fq) and the representationof the symmetric group induced by the trivial representation of the subgroupSm × Sn ≤ Sm+n where m and n are positive integers. In exploring theseexamples, we develop general tools which are useful for detecting Gelfandpairs.

This second example, in which we prove that (Sm+n, Sm×Sn) is a Gelfandpair, leads us naturally to generalize the subgroup Sm × Sn ≤ Sm+n. That is,we consider the subgroup Sλ1×⋯×Sλk ≤ Sn where (λ1, . . . , λk) is a partition ofa positive integer n. Although (Sn, Sλ1 ×⋯×Sλk) is not in general a Gelfandpair, we may utilize some of the same tools we used to study Gelfand pairsto deduce an interesting connection between the study of partitions of n andthe irreducible representations of Sn, demonstrating the power of some of thetools used to study Gelfand pairs. What we will show is that there exists anatural bijection between the partitions of n and irreducible representationsof Sn.

The final major result of this paper comes from a discussion of sphericalfunctions, a special class of functions from a group G into the complex num-bers. We will see that the discussion of these spherical functions leads to an

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interesting theory that can be viewed as a generalization of Fourier analysis.It is a classical result from Fourier analysis that if we let

en(x) =1√2πeinx, −π < x < π

then en∞−∞ forms an orthonormal basis for L2([−π,π]). As a result, if welet

fm =m

∑n=−m

⟨f, en⟩en

then fm → f as m → ∞ for a reasonably nice function f . This fact has thefollowing generalization, namely the Schur Orthogonality Relation: If G isa compact group, then the irreducible characters of G form an orthonormalbasis of the space of class functions on G.

Thus, Schur Orthogonality amounts to Fourier Analysis on finite or com-pact groups. This fact has a further significant generalization to Gelfandpairs, which is as follows. Given a Gelfand pair (G,H), if (π,V ) is a repre-sentation of G with a unique H-fixed vector v ∈ V , the associated sphericalfunction will be given by σ(g) = ⟨π(g)v, v⟩. We begin our consideration ofspherical functions by studying the example of diagonally embedding a groupG 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 torecover classical character theory, i.e., we find that the spherical functionsare the irreducible characters of G, which form an orthonormal basis of thespace of class functions on G.

Generalizing this example to the more abstract setting of spherical func-tions on arbitrary Gelfand pairs (G,H) will allow us to construct an orthonor-mal basis of the space C(H/G/H) consisting of H-bi-invariant functions fromG into the complex numbers. This result, a generalization of the classicalresults from Fourier analysis and character theory, cuts to the core of whyGelfand pairs are important not just to representation theory but to otherareas of mathematics.

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2 Foundations

We lay some of the foundations here that are necessary for the study ofGelfand pairs. We begin our theory with a proposition:

Proposition 1. Let H ≤ G be a subgroup of a finite group G. Let (ψ1, V1)and (ψ2, V2) be representations of H. LetM =HomG(IndGHψ1, IndGHψ2) and

H = ϕ ∶ G→ EndC(V1, V2) ∶ ϕ(hgh′) = ψ2(h)ϕ(g)ψ1(h′) for h,h′ ∈H,g ∈ G.

ThenM and H are isomorphic C-algebras, where the multiplicative structurein M is composition of homomorphisms and that in H is convolution:

(ϕ1 ∗ ϕ2)(g) =1

∣H ∣ ∑s∈Gϕ1(s) ϕ2(s−1g).

Proof. For M ∈M, define ϕM ∶ G → EndC(V1, V2) by ϕM(g)v = (Mfv)(g)for v ∈ V , where fv(h) = ψ1(h)v for h ∈H and fv(g) = 0 for g ∉H. It is clearthat under this definition fv ∈ IndGHψ1.

That ϕM ∈ H is a computation. Let h,h′ ∈H and g ∈ G. Then:

ϕM(hgh′)v = (Mfv)(hgh′)= ψ2(h)(Mfv)(gh′)= ψ2(h)(Mψ1(h′)fv)(g)= ψ2(h)(Mfψ1(h′)v)(g)= ψ2(h)ϕM(g)ψ1(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:

ϕM+cM ′(g)v = ((M + cM ′)fv)(g)= (Mfv + (cM ′)fv)(g)= (Mfv)(g) + c(M ′fv)(g)= ϕM(g)v + cϕM ′(g)v= (ϕM(g) + cϕM ′(g))v.

So indeed the map M ↦ ϕM is a vector space homomorphism.

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To show it is isomorphism, we construct its two sided inverse. Let ϕ ∈ H.Then define Mϕ by (Mϕf)(g) = ∑r ϕ(gr−1)f(r) where r runs over a set ofcoset representatives of G/H and f ∈ IndGHψ1. Now let h ∈ H and g ∈ G.Then

(Mϕf)(hg) =∑r

ϕ(hgr−1)f(r)

= ψ1(h)∑r

ϕ(gr−1)f(r)

= ψ1(h)(Mϕf)(g)

so that indeed Mϕf ∈ IndGHψ2

Now for g, g′ ∈ G it holds:

(MϕψG1 (g′)f)(g) =∑

r

ϕ(gr−1)(ψG1 (g′)f)(r)

=∑r

ϕ(gr−1)f(rg′) since f ∈ IndGHψ1

=∑r

ϕ(gg′r−1)f(r)

= (Mϕf)(gg′)= (ψG2 (g′)Mϕf)(g)

proving that Mϕ ∈M. Now we check that the maps are indeed inverses, i.e.,that MϕM =M for M ∈M and ϕMϕ = ϕ for ϕ ∈ H. Now for v ∈ V and g ∈ G,we have:

ϕMϕ(g)v = (Mϕfv)(g)=∑

r

ϕ(gr−1)fv(r)

= ϕ(g)v

where the last equality holds because fv(r) = 0 for all r ∉ H and for the onecoset representative h ∈H, we have

ϕ(gh−1)fv(h) = ϕ(g)ψ1(h−1)ψ1(h)v = ϕ(g)v.

That MϕM =M for M ∈M is a similarly straightforward calculation and isthus omitted.

To complete our proof we only need to check that this map respects themultiplicative structure, i.e., ϕMM ′ = ϕM∗ϕM ′ . This is another computation.For any g ∈ G and any v ∈ V we have:

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ϕM ∗ ϕM ′(g)v = 1

∣H ∣ ∑s∈GϕM(s) ϕM ′(s−1g)v

= 1

∣H ∣ ∑s∈GϕM(s) (M ′fv)(s−1g)

= 1

∣H ∣ ∑s∈Gg∈sH

ϕM(s) (M ′ψ1(s−1g))v

= 1

∣H ∣ ∑s∈Hg∈sH

Mψ1(s)(M ′ψ1(s−1g))v

= (M M ′)fv(g) if g ∈H0 otherwise

= ϕMM ′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 = EndG(IndGHψ) and

H = ϕ ∶ G→ EndC(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 ofthe pair (G,H).

As noted in the introduction, the Hecke algebra of a pair (G,H) will beuseful in detecting whether or not the (G,H) is a Gelfand pair. In order touncover this connection, we must first prove some foundational facts:

Proposition 3. Suppose G is a finite group and H ≤ G a subgroup. Supposefurther that (π1, V1) and (π2, V2) are representation of H and H is the associ-ated Hecke algebra. Now let Hg =H ∩ gHg−1 for g ∈ G. Then the subspace ofH consisting of all functions supported on the coset HgH is isomorphic as avector space to HomHg(πg1 , π2∣Hg) where πg1 , π2∣Hg are the two representationsof Hg defined by πg1(h) = π1(g−1hg) and π2∣Hg is just the restriction of π2 toHg.

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Proof. Let g ∈ G be fixed and ϕ ∈ H supported on HgH be given. Define amap Mϕ ∶ V1 → V2 by Mϕ(v) = ϕ(g)v for v ∈ V1. Then for h ∈ Hg it is truethat:

Mϕ(πg1(h)v) = ϕ(g)(π1(g−1hg)v) = ϕ(hg)v = π2(h)ϕ(g)v = π2(h)∣HgMϕ(v).

So Mϕ is a Hg-module homomorphism, i.e., Mϕ ∈HomHg(πg1 , π2∣Hg).Now given M ∈ HomHg(πg1 , π2∣Hg), define ϕM ∶ G → EndC(V1, V2) by

ϕM(hgh′)v = π2(h)ϕ(g)π1(h′)v if h,h′ ∈H and ϕM(g)v = 0 if g ∉HgH. It iseasy to see that under this definition, ϕM is an element of H supported onHgH.

It is also easy to see that the maps ϕ ↦ Mϕ and M ↦ ϕM are mutualinverses. This is sufficient to prove the vector space isomorphism.

Proposition 4 (Mackey’s Theorem). Suppose G is a finite group and H ≤ Ga subgroup. Suppose further that (π1, V1) and (π2, V2) are representations ofH. For g ∈ G, let Hg be defined as in the previous proposition. Similarly, letπg1 , π2∣Hg be defined as in the previous proposition. Then

HomG(IndGHπ1, IndGHπ2) ≅ ⊕g∈HgH

HomHg(πg1 , π2∣Hg)

and as a consequence

dim HomG(IndGHπ1, IndGHπ2) = ∑g∈HgH

dim HomHg(πg1 , π2∣Hg).

Proof. This is clear from what we have shown so far.

The following result, known as Schur’s Lemma, is important in grouprepresentation theory and will be used here throughout.

Proposition 5 (Schur’s Lemma). Let (ρ1, V1) and (ρ2, V2) be irreduciblerepresentations of a group G. Then any element of HomG(V1, V2) is eitherthe zero map or an isomorphism. As a corollary, HomG(π,π) ≅ C for anyirreducible representation π of G.

Proof. Suppose A ∈ HomG(V1, V2) is nonzero. Since A is a G-module ho-momorphism, image(A) is a G-invariant subspace of V2. But that V2 isirreducible forces that either image(A) = 0 or image(A) = V2. The first caseis false since A is nonzero. Thus, the second case holds, i.e., A is surjective.

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Similarly, ker(A) is a G-invariant subspace of V1, and, thus, either ker(A) = 0or ker(A) = V1. The second case is false since A is nonzero; thus the fist caseholds, 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 ∈ HomG(π,π) benonzero. Let λ be an eigenvalue of ψ (we are assured that such an eigenvalueexists since π is finite dimensional and since C is an algebraically closed field).Then ψ − λI ∈ HomG(π,π) is not invertible so by what we have just provenit 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 HomG(ρ, ρ) is commutative. In thiscase, dim HomG(ρ, ρ) is equal to the number of irreducible constituents inthe direct sum decomposition of ρ.

Proof. Let ρ = ⊕mi=1 ρi be the direct sum decomposition of ρ into irreducible

representations ρi with the ρi’s ordered so that equivalent representationsoccur consecutively. Now any σ ∈ HomG(ρ, ρ) can be written as a blockdiagonal matrix A = (aij), where aij ∈ HomG(ρj, ρi). But Schur’s lemmagives us that

HomG(ρj, ρi) ≅ C if ρi ≅ ρj0 otherwise.

Thus, we can see that HomG(ρ, ρ) is isomorphic to an algebra of block diag-onal matrices over the complex numbers where the size of the blocks corre-sponds to the multiplicity of each irreducible representation ρi. Compositionof homomorphisms corresponds to matrix multiplication so that HomG(ρ, ρ)will indeed be commutative if and only if each block is of size one, i.e., if andonly if ρ is multiplicity free as claimed.

If this is the case, then HomG(ρ, ρ) is isomorphic to the algebra of m×mdiagonal matrices over the complex numbers so that indeed dim HomG(ρ, ρ) =m, which completes the proof.

Proposition 7. Let H ≤ G be a subgroup of the finite group G and letψ ∶H → V be a finite dimensional representation. Then IndGHψ is multiplicityfree if and only if the associated Hecke algebra H is commutative.

Proof. This follows from Propositions 2 and 6.

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Definition 2. Suppose G is a group, H ≤ G a subgroup and (ψ,V ) a rep-resentation of H. If IndGHψ is multiplicity free, we say that (G,H,ψ) is aGelfand triple. In the case that ψ = 1H we say that (G,H) is a Gelfand pairor that H is a Gelfand subgroup of G.

From these results, we can see that the study of Gelfand pairs is intimatelylinked with the study of Hecke algebras. In the following section, we willexplore specific examples of Gelfand pairs and will produce general toolsthat allow us to exploit this fundamental connection.

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3 Examples

We now consider some specific examples of Gelfand triples/pairs.Let Fq be any finite field, n ∈ N and let G = GLn(Fq). Now let N ≤

G consist of the subgroup of all upper triangular matrices with 1’s on thediagonal and let ψ ∶ Fq → C∗ be a nontrivial linear character. Define ψ0 ∶N → C by

ψ0(a) = ψ(a12 + a23 +⋯ + an−1,n) for a = (aij) ∈ N.

Our goal will be to show that the induced representation IndGNψ0 is mul-tiplicity free. To this end, we will show that the associated Hecke algebra His commutative. First we need a proposition:

Proposition 8 (Bruhat Decomposition). G = GLn(Fq) can be decomposedas a disjoint union of double cosets with representatives of the following form:

GLn(Fq) = ∐w∈WBwB

where B ≤ G is the Borel subgroup of upper triangular matrices and W is thegroup 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 onn. For n = 1, B = G so the result holds trivially. Now let n > 1 and let g ∈ Gbe arbitrary. I will prove that there exists a permutation matrix w that isan element of BgB, which suffices to prove our assertion. We consider twocases:

Case 1: gn,1 ≠ 0. In this case, we may multiply g by appropriate elementsb0, b1 ∈ B on the left and right so that b0gb1 is 0 in every entry in the firstcolumn and last row besides the n,1th entry which is equal to gn,1. We maythen normalize this to 1 by dividing by gn,1. Now by disregarding the firstcolumn and last row, we have an n − 1 × n − 1 matrix g∗ to which we mayapply our induction hypothesis to find 2 n−1×n−1 upper triangular matricesb′, b′′ such that b′g∗b′′ is an n − 1 × n − 1 permutation matrix. By expandingthese matrices b′b0gb1b′′ will give an n × n permutation matrix.

Case 2: gn,1 = 0. Then pick i as great as possible and j as least as possiblesuch that gi,1 ≠ 0 and gn,j ≠ 0. By multiplying by elements of B on the rightand left, we can clear the first and jth columns and the ith and last rows,except for the entries gi,1 and gn,j, which we can normalize to 1. We can

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then apply the induction hypothesis to the matrix obtained by removingthese rows and columns, to create a permutation matrix. This completes theinduction.

Now I demonstrate that the union is indeed disjoint. Let w1,w2 ∈ Wbe representatives of the same double coset. Then there exists b ∈ B suchthat w1bw−1

2 ∈ B. Now since w1bw−12 is just b with some rows and columns

permuted, if we changed some of the nonzero entries of b to any arbitraryfield element, the result would still be upper triangular, i.e., still in B. Solet us then replace b with the identity matrix. Then: w1w−1

2 ∈ B. But sincew1,w2 ∈ W , it follows that w1w−1

2 ∈ B ∩W = In, where In is the n x nidentity matrix. Thus w1 = w2 and the result is proven.

Proposition 9 (Modified Bruhat Decomposition). GLn(Fq) may be decom-posed as a disjoint union of double cosets in the following fashion:

GLn(Fq) = ∐m∈MNmN

where N is the subgroup of upper triangular matrices with 1’s on the diago-nal and M is the subgroup of monomial matrices (matrices with exactly onenonzero entry in every row and column).

Proof. Let D ≤ G be the subgroup of diagonal matrices. The result, ignoringdisjointness of the union, follows from the Bruhat decomposition becauseB = DN = ND and M = DW = WD. That the union is disjoint can bededuced as before.

Now we introduce a powerful tool for proving that a Hecke algebra H iscommutative. We define an involution of a group G to be a map ι ∶ G → Gthat satisfies ι2 = id and ι(g1g2) = ι(g2)ι(g1) for g1, g2 ∈ G. Similarly, a mapι ∶ H → H is called an involution of H if ι2 = id and ι(ϕ1ϕ2) = ι(ϕ2)ι(ϕ1) forany ϕ1, ϕ2 ∈ H.

Proposition 10. Let H be a subgroup of a finite group G and (ψ,V ) aone-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 allh ∈ H. Then the map ι ∶ H → H defined by ι(ϕ(g)) = ϕ(ι(g)) where g ∈ G isan involution of H.

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Proof. Observe that if ϕ ∈ H, h1, h2 ∈ H, and g ∈ G, then

ι(ϕ(h1gh2) = ϕ(ι(h1gh2))= ϕ(ι(h2)ι(g)ι(h1)) since ι reverses multiplication in G

= ψ(h2)ϕ(ι(g))ψ(h1) since ϕ ∈ H and ψ(ι(h)) = ψ(h) for all h ∈H= ψ(h1)ι(ϕ(g))ψ(h2)

where the last equality holds since V is 1-dimensional implies that EndCVis commutative. This proves that ι(ϕ) ∈ H.

Further, note that:

ι2(ϕ(g)) = ι(ϕ(ι(g)))= ϕ(ι2(g))= ϕ(g) since ι is an involution.

Thus, ι2 = id. Finally, that ι reverses multiplication follows easily from thatfact that EndCV is commutative, which completes the proof.

Proposition 11. Suppose that there is an involution ι ∶ G → G such thatHgH =Hι(g)H for all g ∈ G. Then (G,H) is a Gelfand pair.

Proof. We wish to show that IndGNψ is multiplicity-free, where ψ = 1H . Weconstruct ι according to Proposition 10 and note that since ψ is the trivialrepresentation that the requirement that ψ(ι(h)) = ψ(h) for all h ∈ H holdstrivially.

Now H consists of all function ϕ ∶ G → C such that ϕ is constant onH−H double cosets (this is because ψ is trivial). Thus, ι acts by the identityon H. Finally, observe that ι = id being an involution implies that H iscommutative.

We now return to our study of the example introduced. We will applythe involution method to attain our desired result in this example, but firstwe will need this basic fact:

Proposition 12. Let F be any field and suppose m = (mij) ∈ Fn×n is amonomial matrix and that mij and mi+1,k are nonzero implies that k ≤ j + 1.Then m is of the form

m =⎛⎜⎜⎜⎝

D1

D2

⋰Dp

⎞⎟⎟⎟⎠

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where p is a positive integer and the Di are diagonal matrices for i = 1, . . . , p.

Proof. The proof is by induction on n. For n = 1, the conclusion followstrivially, so suppose n > 1 and that our conclusion holds for all matricesthat satisfy the above conditions and of are size less than n. Now since m ismonomial, there exists a unique i ∈ 1, . . . , n such that m1i ≠ 0. If i = n, thenwe note that the n − 1 × n − 1 matrix constructed by removing the first rowand last column of m is of the desired form by the induction hypothesis. Sosuppose that i < n. We claim then that m2,k ≠ 0 forces k = i + 1. If not, thenk < i. But then for any j ≥ 2, we have that mjl ≠ 0 implies that l < i by theassumption that mij and mi+1,k both nonzero implies that k ≤ j +1. But nowthis implies that column i + 1 is composed purely of zeros, a contradiction.This proves now that m2,i+1 is nonzero. Repeating this argument n− i timesproves that m1+k,i+k is nonzero for k = 1, . . . , n − i so that indeed m is of theform:

m = (0 D∗ 0

)

where D is an n− i+1×n− i+1 diagonal matrix. Now applying the inductionhypothesis to ∗, our conclusion follows.

Proposition 13. Let G = GLn(Fq), and let N and ψ0 be as above. ThenIndGNψ0 is multiplicity free.

Proof. Let g ∈ G and consider the double cosetNgN . By the modified Bruhatdecomposition proposition, we may find a monomial matrix m = (mij) suchthat NgN = NmN . Assume that there is some ϕ ∈ H and some g ∈ NmNwith ϕ(g) ≠ 0. Now I claim that if mij and mi+1,k are both nonzero entriesof m, then k ≤ j + 1 so that m is of the form

m =⎛⎜⎜⎜⎝

D1

D2

⋰Dp

⎞⎟⎟⎟⎠

where the Di are diagonal matrices for i = 1, . . . , p.Now suppose for a contradiction that mij ≠ 0 and mi+1,k ≠ 0 with k > j+1.

Since mij ≠ 0 and since ψ is non-trivial, we may select t ∈ Fq such thattmij ∉ kerψ, i.e., ψ(tmij) ≠ 1. (This is because Fq is a field). Now let

x = In + tmijei,i+1,

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y = In + tmi+1,kejk

where ejk is an n × n matrix that is 1 at position j, k and zero elsewhere.It is easy to see that xm = (In + tmijei,i+1)m =m + tmijmi,i+1eik =my.Now, it is a simple computation that ψ0(x) = ψ(tmij) ≠ 1 and ψ0(y) =

ψ(0) = 1.Now since x, y ∈ N and ϕ ∈ H, it holds that

ψ0(x)ϕ(m) = ϕ(xm) = ϕ(my) = ϕ(m)ψ0(y).

But then(ψ0(x) − ψ0(y))ϕ(m) = 0

which forces ϕ(m) = 0 since our above computation demonstrates that ψ0(x)−ψ0(y) is nonzero. But then ϕ(NmN) = 0, a contradiction. This shows thatm does indeed adopt the given form.

Next, I claim that each Di is a homothety, i.e., a scalar matrix, for eachi = 1, . . . , p. This is equivalent to the statement that mij ≠ 0 and mi+1,j+1 ≠ 0implies mij =mi+1,j+1. So suppose that mij and mi+1,j+1 are nonzero. Further,suppose for a contradiction that mij ≠ mi+1,j+1. Then mij −mi+1,j+1 ≠ 0 sothat we may select s ∈ Fq such that s(mij −mi+1,j+1) ∉ kerψ. So if we alter xand y from above to be

x = In + smijei,i+1,

y = In + smi+1,j+1ej,j+1

then as above a simple calculation gives

(ψ0(x) − ψ0(y))ϕ(m) = 0.

Now it is easy to see from the formulation of x and y that ψ0(x) = ψ(smij)and ψ0(y) = ψ(smi+1,j+1). But now since

ψ0(x) − ψ0(y) = ψ(smij) − ψ(smi+1,j+1) = ψ(s(mij −mi+1,j+1)) ≠ 0

it follows that ϕ(m) = 0, a contradiction. Thus mij = mi+1,j+1 and indeedeach Di is a homothety.

Finally, we consider the involution ι ∶ G → G defined by ι(g) = hgThwhere

h =⎛⎜⎝

1⋰

1

⎞⎟⎠.

15

It is a simple computation to see that ι thus defined fixes matrices m of theform described above (for this, note that right multiplication by h reversesthe order of columns, while left multiplication by h reverses the order ofrows) and that ψ0(ι(n)) = ψ0(n) for all n ∈ N . Thus, we find that ι asdefined earlier is an involution of H. Now we claim that ι is the identitymap on H. Once we have substantiated this claim, we will be done provingthat H is commutative because indeed this implies that for any ϕ1, ϕ2 ∈ H,it holds:

ϕ1 ∗ ϕ2 = ι(ϕ1 ∗ ϕ2) = ι(ϕ2) ∗ ι(ϕ1) = ϕ2 ∗ ϕ1

where the second equality holds since ι 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 fixesthese matrices, i.e., for m a matrix of the above form it holds:

ι(ϕ(m)) = ϕ(ι(m)) = ϕ(m)

and our proof is complete.

Thus, we have seen in action the power of the involution method as atool to study Gelfand pairs. We will again apply this method to anotherinteresting example: that of the symmetric group Sn. There is a naturalembedding Sn × Sm Sn+m in which Sn acts on the first n elements of1, . . . , n +m and Sm acts on the last m. We will again use the involutionmethod to prove that Sn × Sm Sn+m is a Gelfand subgroup.

Proposition 14. Sn × Sm is a Gelfand subgroup of Sn+m.

Proof. Take H = Sn × Sm and G = Sn+m. Define the map ι ∶ G→ G to be theinvolution g ↦ g−1. We must check that each double coset HgH is fixed byι.

We may identify the elements of Sn+m with permutation matrices. Weclaim that each double coset HgH has a representative of the form

⎛⎜⎜⎜⎝

Il 0 0 00 0n−l 0 In−l0 0 Im−n+l 00 In−l 0 0n−l

⎞⎟⎟⎟⎠.

We may represent g in block form

g = (A BC D

)

16

where A, B, C, and D are matrices with only 1’s and 0’s and at most one 1 inany row or column, i.e., as permutation matrices. A is an n×n matrix and Dis a m×m matrix. Let l denote the rank of A. Then B and C have rank n− lsince g is a permutation matrix. Thus D has rank m − n + l. Multiplying onthe left by elements of Sn corresponds to permuting rows, while multiplyingon the right by elements of Sn corresponds to permuting columns. Thus,multiplying on the left and right by elements of Sn allows us to rearrange thenonzero elements of A so that they are in the top left hand corner. Similarly,multiplying by elements of Sm will rearrange D so that its nonzero elementsare in the top left hand corner. This yields a matrix of the form

⎛⎜⎜⎜⎝

Ul 0 0 00 0n−l 0 Wn−l0 0 Vm−n+l 00 Xn−l 0 0n−l

⎞⎟⎟⎟⎠

where Ul, Vm−n+l,Wn−l,Xn−l are all permutation matrices. Note that we maynaturally identify Sl×Sn−l×Sm−n+l×Sn−l as a subgroup of Sn×Sm. Multiplyingour matrix on the left and right by elements of Sl × Sn−l × Sm−n+l × Sn−l willput it in the form

⎛⎜⎜⎜⎝

Il 0 0 00 0n−l 0 In−l0 0 Im−n+l 00 In−l 0 0n−l

⎞⎟⎟⎟⎠

which verifies the claim. It is easy to see that squaring such a matrix givesan n + m × n + m identity matrix. That is, ι fixes such matrices. Thus,Sn × Sm ≤ Sn+m is a Gelfand subgroup.

We have found that if λ ∈ Z+ and m,n ∈ Z+ satisfy n +m = λ, then theinduced representation, IndSλSm×Sn1Sm×Sn is multiplicity free. This motivatesus to wonder about the behavior of a more general case. That is, if λ is apositive integer and λi positive integers satisfying ∑i λi = λ, then what doesthe representation of Sλ induced by the trivial character on ∏i Sλi look like?We explore the answer to this question in the following section, finding thatsuch a question leads to an interesting story that classifies the irreduciblerepresentations of the symmetric group Sλ.

17

4 Representation Theory of the Symmetric Group

To discuss the representation theory of the symmetric group, we introducethe notion of a partition. Suppose n is a positive integer. Then we say thatλ = (λ1, . . . , λk) forms a partition of n if n = ∑k

i=1 λi and if λ1 ≥ ⋯ ≥ λk > 0. Forexample, if n = 3, then λ = (3), µ = (2,1), ν = (1,1,1) are all of the partitionsof 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 Snare determined by cycle type so that indeed they are in bijective correspon-dence to the distinct partitions of n. Thus, the number of distinct irreduciblerepresentations will be equal to the number of partitions of n.

Now suppose λ = (λ1, . . . , λk) is a partition of n. Then define the sub-group Sλ = Sλ1 × ⋯ × Sλk ≤ Sn. Given such a partition λ, we will study therepresentations of Sn that are induced by certain one-dimensional represen-tations of Sλ. Now consider the following one-dimensional representations ofSλ.

The trivial representation:

ρλ ∶ Sλ → GL1(C) given by σ ↦ 1 for all σ ∈ Sλand the sign representation:

πλ ∶ Sλ → GL1(C) given by σ ↦ sgn(σ) for all σ ∈ Sλ.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 IndSnSλρλ, IndSnSλπλ. Fur-

ther, we let ψ∣H denote the restriction of a representation ψ ofG to a subgroupH ≤ G.

So let λ ⊢ n,µ ⊢ n be partitions of n. Now by Mackey’s theorem, it holds:

dim HomSn(IndSnSλρλ, IndSnSµρµ) = ∑

SλσSµ

dim Hom(ρλ∣Sλ∩σSµσ−1 , ρµ∣Sλ∩σSµσ−1)

= ∑SλσSµ

dim Hom(ρSλ∩σSµσ−1 , ρSλ∩σSµσ−1)

= ∑SλσSµ

1.

18

Similarly, we can compute

dim Hom(IndSnSλρλ, IndSnSµπµ) = ∑

SλσSµ

dim Hom(ρSλ∩σSµσ−1 , πSλ∩σSµσ−1).

Now both ρSλ∩σSµσ−1 and πSλ∩σSµσ−1 are irreducible. Further, it is clearthat these two representations are equal if and only if the intersection Sλ ∩σSµσ−1 is contained in the alternating group, i.e., Sλ ∩ σSµσ−1 ≤ An. Butnow we note that since Sλ ∩σSµσ−1 is isomorphic to a product of symmetricgroups ∏m

i=1 Sxi , we will have Sλ ∩ σSµσ−1 ≤ An if and only if xi = 1 fori = 1, . . . ,m, i.e., the intersection is trivial Sλ∩σSµσ−1 = 1. Thus, with thisextra condition we deduce:

dim Hom(IndSnSλρλ, IndSnSµπµ) = ∑

SλσSµSλ∩σSµσ−1=1

1.

We see then that our investigation about the symmetric group reduces tothe combinatorial problem of counting double cosets of the form SλσSµ.

To this end, we first introduce some terminology and prove a useful result.Given a positive integer n and λ = (λ1, . . . , λk) a partition of n, we let nλi for1 ≤ i ≤ k denote subsets of 1, . . . , n that satisfy ∣nλi ∣ = λi for all 1 ≤ i ≤ k,ni ∩ nj = ∅ for i ≠ j, and ⋃ki=1 nλi = 1, . . . , n. We call such a collection of nλifor i = 1, . . . , k a dissection of 1, . . . , n. Now, we prove the following usefulfact:

Proposition 15. Let n be a positive integer. Let λ = (λ1, . . . , λk) and µ =(µ1 . . . , µl) be partitions of n. Let Sλ, Sµ be defined as above. Then τ ∈ SλσSµif and only if for every 1 ≤ i ≤ k and 1 ≤ j ≤ l it holds: ∣nλi ∩ σ(n

µj )∣ =

∣nλi ∩ τ(nµj )∣.

Proof. Assume τ ∈ SλσSµ. Say τ = ψσφ where ψ ∈ Sλ and φ ∈ Sµ. Then foreach j ∈ 1, . . . , l it will hold:

τ(nµj ) = ψσφ(nµj ) = ψσ(n

µj ).

Thus, if i ∈ 1, . . . , k and j ∈ 1, . . . , l, it holds:

nλi ∩ τ(nµj ) = nλi ∩ ψσ(n

µj ) = ψ(nλi ∩ σ(n

µj ))

since ψ ∈ Sλ. Thus, since σ is a bijection, this yields ∣nλi ∩σ(nµj )∣ = ∣nλi ∩τ(n

µj )∣,

as desired.

19

Now conversely, suppose that ∣nλi ∩ σ(nµj )∣ = ∣nλi ∩ τ(n

µj )∣. Then for each

fixed i, the subsets nλi ∩σ(nµj ) and the subsets nλi ∩ τ(n

µj ) form dissections of

nλi which can be put into ordered pairs of the form (nλi ∩ σ(nµj ), nλi ∩ τ(n

µj ))

of subsets of equal order. To see that these are indeed dissections, we noteit is clear that they are pairwise disjoint for distinct j1, j2 and that indeed:

l

⋃j=1

(nλi ∩ τ(nµj )) = nλi ∩

l

⋃j=1τ(nµj )

= nλi ∩ 1, . . . , n= nλi

and similarly for σ. Now for each i then we pick ψi ∈ Sλ such that ψi fixes1, . . . , n − nλi and such that

ψi(nλi ∩ σ(nµj )) = nλi ∩ τ(n

µj )

for each j = 1, . . . , l. Now define ψ = ψ1⋯ψk. Then we see that for everyj = 1, . . . , k:

ψσ(nµj ) = τ(nµj )

so that indeed there exists φ ∈ Sµ that satisfies τ = ψσφ, i.e., τ ∈ SλσSµ andthe proof is complete.

This shows that for two fixed partitions λ ⊢ n and µ ⊢ n the double cosetsSλσSµ are characterized by the numbers

xij = ∣nλi ∩ σ(nµj )∣,1 ≤ i ≤ k,1 ≤ j ≤ l.

Now let m =maxk, l. Then by letting

xij = ∣nλi ∩ σ(nµj )∣ if 1 ≤ i ≤ k,1 ≤ j ≤ l

0 if i > k or j > l

we attain an injective map g ∶ Sλ/Sn/Sµ →Mm(N) defined by SλσSµ ↦ (xij)where Mm(N) is the set of m ×m matrices over the natural numbers. It isclear that the image of g is

im(g) = (xij) ∈Mm(N)∣m

∑i=1xij = µj and

m

∑j=1xij = λi and xij = 0 if i > k or j > l.

This proves the following result:

20

Proposition 16. Suppose λ = (λ1, . . . , λk) and µ = (µ1, . . . , µl) are two par-titions of a positive integer n. Then there exists a bijection:

g ∶ Sλ/Sn/Sµ →X

where

X = (xij) ∈Mm(N)∣m

∑i=1xij = µj and

m

∑j=1xij = λi and xij = 0 if i > k or j > l

and g is given by SλσSµ ↦ (xij) where

xij = ∣nλi ∩ σ(nµj )∣ 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 byMackey’s intertwining number formula.

Now if we restrict our view to double cosets SλσSµ with the trivial inter-section property:

Sλ ∩ σSµσ−1 = 1and restrict g to this subset, then we obtain the following corollary:

Proposition 17. The number of double cosets SλσSµ with the property Sλ∩σSµσ−1 = 1 is equal to the number of n×n 0− 1 matrices with row sums λiand column sums µj.

Thus, we have that dim Hom(IndSnSλρλ, IndSnSµπµ) is equal to the number

of n × n 0 − 1 matrices with row sums λi and column sums µj.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 arepartitions of n. Then we say that λ ≥ µ if ∑k

i=1 λi ≥ ∑ki=1 µi for all k.

Proposition 18. Let n be a positive integer. Then ≥ defines a partial orderon the set of partitions of n.

Proof. Reflexivity is obvious. To prove antisymmetry, suppose λ ≥ µ andµ ≥ λ. Then for any j, we have ∑j

i=1 λi ≥ ∑ji=1 µi and ∑j−1

i=1 λi ≤ ∑j−1i=1 µi so

that λj = ∑ji=1 λi −∑

j−1i=1 λi ≥ ∑

ji=1 µi −∑

j−1i=1 µi = µj. Similarly µj ≥ λj and thus

λj = µj for each j, i.e., λ = µ. Transitivity also follows directly from thedefinition of ≥.

21

Now for a partition λ = (λ1, . . . , λk) of a positive integer n, define D(λ)to be the diagram consisting of rows of blocks where the ith row consists ofλi blocks. For example, if n = 7 and λ = (3,2,1,1) then D(λ) is given by

Now noting that λi ≥ λi+1 for each i, we see that the lengths of the columnsof D(λ) form a partition λ′ of n. This partition is called the conjugatepartition of λ. D(λ′) can be clearly obtained by simply rotating D(λ) aroundit’s main diagonal or by interchanging rows and columns. For example withλ = (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 writer(A)

Proposition 19. Let λ = (λ1, . . . , λk), µ = (µ1, . . . , µl) be partitions of apositive integer n. Then there exists a 0,1 matrix with c(A) = λ and r(A) = µif and only if λ′ ≥ µ.

Proof. To prove one direction, assume that A = (aij) is a k × l matrix withc(A) = λ and row sums r(A) = µ. Now if there is no i ≤ k, j < h ≤ l such thataij = 0 and aih = 1 then it is easy to see that λ′ = µ.

Now if, on the contrary, there exists (i, j) satisfying the conditions aboveso that aij = 0 is a gap in the matrix, then let h > j be maximal such thataih = 1. Then swapping aij and aih we attain a matrix A′ with c(A′) = λ andthat satisfies r(A′) ≥ r(A). Thus, it follows by induction on the number ofsuch gaps that λ′ ≥ µ.

Now to prove the converse, suppose that λ′ ≥ µ. Then we note that thereexists a 0-1 k × l matrix A with c(A) = λ and r(A) ≥ µ namely the matrixwhose ith row is 1 for 1 ≤ j ≤ λi and 0 for j ≥ λi, i.e., this is the matrixconstructed by putting 1’s on the diagram D(λ) and 0 elsewhere.

22

Now I claim that given a k × l matrix A such that c(A) = λ and r(A) ≥ µand such that r(A) ≠ µ, we can find a k× l 0-1 matrix A′ such that c(A′) = λand r(A′) ≥ µ that satisfies ∣∣r(A′) − µ∣∣ ≥ ∣∣r(A) − µ∣∣ where ∣∣ ⋅ ∣∣ is the usualEuclidean norm. Note that since ∣∣r(A)−µ∣∣2 is an integer, this process musteventually terminate after a finite number of steps, i.e., we will be able tofind a matrix A such that c(A) = λ and r(A) = µ. Thus, to complete ourproof all we need to do is verify this claim.

So let r(A) = (r1, . . . , rl). Now let i be minimal such that ri > µi, andlet j be minimal such that rj < µj. Then i < j since r(A) ≥ µ. Now letr = (r′1, . . . , r′n) = (r1, . . . , ri−1, ri − 1, ri+1, . . . , rj−1, rj + 1, rj+1, . . . , rl then it isclear that ∣∣r′ − µ∣∣ < ∣∣r − µ∣∣. Further, it is clear that r′ ≥ µ.

Now since ri > qi ≥ qj > rj, we can find a number h ∈ 1, . . . , k such thatahi = 1 and ahj = 0. For such an h, define the matrix A′ = (a′st) by

a′st =⎧⎪⎪⎪⎨⎪⎪⎪⎩

1 if (s, t) = (h, j)0 if (s, t) = (h, i)

ast otherwise

By briefly inspecting this definition of A′, it is easy to see that indeedr(A′) = r′ and c(A) = λ which verifies our claim and thus completes theproof.

We note that it is an immediate consequence of this result that λ′ ≥ µ ifand only if µ′ ≥ λ, so that indeed λ = µ′ if and only if λ′ = µ . We will use thisresult and the following to gain some interesting insight into the irreduciblerepresentations of the symmetric group.

Proposition 20. Let λ = (λ1, . . . , λk) 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 is1 where there is a node in D(λ) and 0 elsewhere. More formally, A willbe constructed by letting the first λi elements of the ith row be 1 and theremaining elements of the row be 0 for i = 1, . . . , k. It is clear that this matrixwill satisfy c(A) = λ and r(A) = λ′.

To see that this is the only such matrix, suppose that there is a 0-1 matrixB with c(B) = λ and r(B) = λ′ such that there exists i ∈ 1, . . . , k with bij = 0for some j < λi. Then it is easy to see that ∑k

l=1 alj < λ′j, a contradiction. Soindeed any matrix satisfying our conditions must be of the form of A and weare done.

23

Thus, we know that if λ ⊢ n then dim Hom(IndSnSλρ, iSnSλ′π) = 1 so that

IndSnSλρ and iSnSλ′π contain precisely one equivalent irreducible representationof multiplicity one in their direct sum decompositions. We will let Πλ denotethis irreducible representation.

Proposition 21. Let λ ⊢ n and µ ⊢ n be two partitions of a positive integern. Then Πλ = Πµ implies λ = µ.

Proof. If Πλ = Πµ then we may deduce that IndSnSλρ, iSnSλ′π, IndSnSµρ, i

SnSµ′π share

precisely one irreducible representation in common. Thus, dim Hom(IndSnSλρ, iSnSµ′π) >

0 so that λ′ ≥ µ′ or equivalently µ = µ′′ ≥ λ′′ = λ. Similarly, dim Hom(IndSnSµρ, iSnSλ′π) >

0 so that it holds that µ′ ≥ λ′ or equivalently that λ ≥ µ. Thus, since ≥ is apartial order, this forces λ = µ.

Now we have proven that for any positive integer n, there exists an in-jective map from the set of partitions λ ⊢ n of n into the set of irreduciblerepresentations of Sn defined by λ↦ Πλ. To see that this is a surjection, werecall the basic group theoretic fact that the number of conjugacy classes ofSn is equal to the number of partitions of n and the basic fact from the repre-sentation theory of finite groups that the number of inequivalent irreduciblerepresentations of a finite group G is equal to the number of conjugacy classesof G, so that indeed the set of partitions of n and the set of inequivalent ir-reducible representations of Sn are two finite sets of equal cardinality. Wesummarize what we have proved in the following proposition:

Proposition 22. For any positive integer n, there exists a natural bijectionbetween the partitions of n and the set of inequivalent irreducible representa-tions of the symmetric group Sn, given by λ↦ Πλ.

24

5 Spherical Functions

We begin this section with an important example. Our first goal will be toprove 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 diagonalembedding, G is a Gelfand subgroup of G ×G.

We provide multiple different proofs of this result that are enlighteningin different ways.

Proof 1 (Involution Method). Consider the map ι ∶ G ×G → G ×G given by(g1, g2)↦ (g−12 , g−11 ). Observe that given (g1, g2) ∈ G ×G it holds:

ι2((g1, g2)) = ι((g−12 , g−11 )) = (g1, g2)

so that indeed ι2 = id. Further, if (g1, g2), (g3, g4) ∈ G ×G, then

ι((g1, g2)(g3, g4)) = ι((g1g3, g2g4))= (g−14 g−12 , g−13 g−11 )= (g−14 , g−13 )(g−12 , g−11 )= ι((g3, g4))ι((g1, g2))

so that indeed ι reverses multiplication and is thus an involution. To see thatι fixes each double coset G(g1, g2)G, let (g1, g2) ∈ G ×G be arbitrary. Nowlet a = g−11 and b = g−12 and consider the elements (a, a), (b, b) ∈ G. It is easyto see that

ι((g1, g2)) = (g−12 , g−11 ) = (g−12 g1g−11 , g−12 g2g−11 ) = (b, b)(g1, g2)(a, a) ∈ G(g1, g2)G

so that ι acts trivially on double cosets. The result follows.

Our other proofs require some extra machinery. Let (πi, Vi) be a repre-sentation of the group Gi for i = 1,2. We may define a representation π1⊗π2of the direct product G1 ×G2 by

(π1 ⊗ π2)(g1, g2)(v1 ⊗ v2) = π1(g1)v1 ⊗ π2(g2)v2where g1 ∈ G1, g2 ∈ G2 and where v1⊗v2 ∈ V1⊗V2 is a simple tensor. Since anyelement of V1 ⊗ V2 is a linear combination of simple tensors extending thisrepresentation by linearity does indeed give a representation of G1 ×G2. Wegive conditions on π1 and π2 that are necessary and sufficient so that π1⊗π2is an irreducible representation of G ×G.

25

Proposition 24. Let (πi, Vi) be a finite dimensional linear representation ofthe group Gi for i = 1,2. Then π1⊗π2 is an irreducible representation of G1×G2 if and only if π1, π2 are irreducible representations of G1,G2 respectively.

Proof. Suppose without loss of generality that (π1, V1) is reducible, say V1 ≅U⊕W where U,W ⊆ V1 are nontrivial G1 invariant subspaces. Then it is easyto see that V1⊗V2 ≅ (U⊕W )⊗V2 ≅ (U⊗V2)⊕(W⊗V2) so that (π1⊗π2, V1⊗V2)is reducible.

Now suppose that π1, π2 are irreducible. Let n = dimV2. ThenHomG1(π1, π1)n ≅HomG1(π1, πn1 ) via the isomorphism A1⊕⋯⊕An ↦ B, where B is defined byB(v) = A1(v)⊕ . . .⊕An(v).

Because V2 ≅ Cn and C ≅HomG1(π1, π1), it holds that V2 ≅HomG1(π1, π1⊗1n), where π1 ⊗ 1n is the representation of G1 on V1 ⊗ V2 defined by (π1 ⊗1n)(g1)(v1⊗v2) = π1(g1)v1⊗v2 for v1 ∈ V1, v2 ∈ V2. (Note that this representa-tion can be identified with the restriction of π1⊗π2 to the subgroup G1×1 ofG1×G2. If m is a positive integer, then the map T ∶ V1⊗HomG1(π1, πm1 )→ V m

1

defined by v ⊗A↦ A(v) is an isomorphism.These facts show that there is a bijection

G1 − invariant subspaces of V1 ⊗ V2↔ C − vector subspaces of V2

given by V1 ⊗W ←W and X →HomG1(π1,X) ⊆HomG1(π1, π1 ⊗ 1n) = V2.Suppose that X ⊆ V1 ⊗ V2 is a nonzero G1 ×G2-invariant subspace. Then

X is also a G1-invariant subspace, so that by what we have shown it holdsthat X = V1 ⊗W for some complex subspace W ⊆ V2. That π2 is irreducibleguarantees

span(π1 ⊗ π2)(1, g2)x ∶ x ∈X,g2 ∈ G2 = V1 ⊗ spanπ2(g2)w ∶ w ∈W,g2 ∈ G2= V1 ⊗ V2

We see then that G1 ×G2-invariance of X forces that X = V1 ⊗ V2, i.e., thetensor product π1 ⊗ π2 is irreducible.

Proposition 25. Let (π,V ) be an irreducible representation of the directproduct group G1 ×G2. Then there exist irreducible representations π1 andπ2 of G1 and G2 respectively such that π ≅ π1 ⊗ π2.

Proof. For g1 ∈ G1, g2 ∈ G2, and v ∈ V , define π1(g1)v = π(g1,1)v andπ2(g2)v = π(1, g2)v. It is easy to see that these define representations of

26

G1,G2 respectively. It is not hard to see that we can pick a nonzero G1 in-variant subspace V1 of V such that π1 = π1∣V1 is an irreducible representationof G1. Now let v ∈ V1 be nonzero and define V2 = spanπ2(g)v ∶ g ∈ G2. V2is clearly G2 invariant and π2 = π2∣V2 is a representation of G2.

Now for any v1 ∈ V1 and v2 ∈ V2, there exist numbers ai, bi ∈ C and elementsg(i)1 , g

(i)2 ∈ G1 ×G2 such that v1 = ∑i aiπ1(g

(i)1 )v and v2 = ∑i biπ2(g

(i)2 )v. Given

such elements v1, v2, we define a map T ∶ V1 ⊗ V2 → V by the rule

T (v1 ⊗ v2) =∑i

∑j

aibjπ(g(i)1 , g(j)2 )v

on simple tensors and then extend by linearity to define a map on all elementsof V1 ⊗ V2.

It is not difficult to verify that T is a well-defined G1 × G2-module ho-momorphism, i.e., T ∈ HomG1×G2(V1 ⊗ V2, V ). Further T is nonzero. Forexample, T (v ⊗ v) = v and thus image(T ) ⊆ V is a nontrivial G1 × G2-invariant subspace. By irreducibility of V , this forces image(T ) = V , i.e., Tis surjective. It is similarly straightforward to check T is injective.

Now we have shown that the irreducible representations of the group G×Gare precisely those representations πi ⊗ πj where πi and πj are irreduciblerepresentations of G. In order to utilize this information to prove that Gembedded diagonally in G ×G is a Gelfand subgroup, we must utilize sometools from character theory. We introduce some basic tools from charactertheory below (such as Frobenius reciprocity and Schur orthogonality) in orderto 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 avector space isomorphism

HomG(W,IndGHπ) ≅HomH(W,V )

where the vector space isomorphism and its inverse are given thus: for σ ∈HomG(W,IndGHπ) define φ ∈ HomH(W,V ) by φ(w) = σ(w)(1). For φ ∈HomH(W,V ) define σ ∈HomG(W,IndGHπ) by σ(w)(g) = φ(ψ(g)w).

Proof. Suppose that σ ∶ W → IndGHπ is a G-module homomorphism. Thenfor h ∈H, we have

φ(ψ(h)w) = σ(ψ(h)w)(1) = (πG(h)σ(w))(1) = σ(w)(1 ⋅ h) = π(h)σ(w)(1).

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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 isan H-module homomorphism then σ(w)(g) = φ(ψ(g)w) gives a G-modulehomomorphism σ ∶W → IndGHπ and that indeed the maps σ ↦ φ and φ ↦ σare mutual inverses.

The following corollary, also referred to as Frobenius reciprocity, is theform of the result as we will use it.

Proposition 27 (Frobenius reciprocity). Let G be a finite group and H ≤ Ga subgroup. Let (π,V ) be a representation of H and (ψ,W ) a representationof G, and let χπ and χψ be the associated characters. Then

⟨χIndGHπ, χψ⟩G = ⟨χπ, χψ∣H ⟩H .

Proof. This is immediate from the previous proposition.

The following result is a consequence of Frobenius reciprocity and is ac-tually a statement of the equivalence of the two definitions of Gelfand pairswhich were provided in the introduction.

Proposition 28. Let G be a finite group and H ≤ G a subgroup. Then H isa 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, orequivalently, the restriction π∣H contains the trivial representation 1H withmultiplicity at most 1.

Proof. Note that V H ≅HomH(1H , V ). By Frobenius reciprocity, it holds:

dim HomH(1H , V ) = ⟨χ1H , χπ ∣H⟩H = ⟨χIndGH1H, χπ⟩G.

The expression on the left is the multiplicity of π in IndGH1H , so the resultfollows from the definition of a Gelfand subgroup (Definition 2).

The following special type of representation will be useful in our exampleof interest:

Definition 4. Let G be a a group and (π,V ) a linear representation of G.The representation π of G on V ∗ the dual space of V defined by

π(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.

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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∣ ∑g∈G⟨⟨π(g)v, π(g)v′⟩⟩

for v, v′ ∈ V defines a Hermitian inner product on V . It is G-invariant byconstruction.

Proposition 30. Suppose that (π,V ) is a representation of a finite group G

and χπ is its character. Then χ(g−1) = χ(g) for all g ∈ G and the character ofthe contragradient representation π is the complex conjugate of the characterof π, i.e., χπ = χπ.

Proof. Let ⟨, ⟩ be a G-invariant inner product and let v ∈ V be an eigenvectorof π(g) with eigenvalue λ (existence of such a v and λ is guaranteed by basiclinear 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 λ1, . . . , λn denote theeigenvalues of π(g). Then

tr(π(g−1)) =n

∑i=1λ−1i =

n

∑i=1λi = χ(g).

Now π(g) is the adjoint of π(g−1), so its trace is equal to tr(π(g−1)) whichcompletes the proof.

Proposition 31. Let (π,V ) and (π′, V ′) be irreducible representations of afinite 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 ′

byµ(v) = ∑

g∈G⟨π(g)v, y⟩π′(g−1)y′.

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We claim that µ(π(h)v) = π′(h)µ(v) to see this, we make the variable changeg ↦ gh−1 and see that

µ(π(h)v) = ∑gh−1∈G

⟨π(gh−1)v, y⟩π′(hg−1)y′

= π′(h)∑g∈G

⟨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 eitheridentically 0 or an isomorphism, so if π /≅ π′ then it holds:

0 = ⟨µ(x), x′⟩

= 1

∣G∣ ∑g∈G⟨π(g)x, y⟩⟨π′(g−1)y′, x′⟩

= 1

∣G∣ ∑g∈G⟨π(g)x, y⟩⟨y′, π′(g)x′⟩

= 1

∣G∣ ∑g∈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 toprove.

Proposition 32. Let (π,V ) be an irreducible representation of the finitegroup G, with G-invariant inner product ⟨, ⟩. Then there exists a positivereal constant d ∈ R+ such that

∑g∈G

⟨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-modulehomomorphism, 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′)xfor all x ∈X. Thus, it holds

c(y, y′)⟨x,x′⟩ = ⟨µ(x), x′⟩ = ∑g∈G

⟨π(g)x, y⟩⟨π(g−1)y′, x′⟩

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making the variable change g ↦ g−1, conjugating the inner product and takingadjoints gives:

c(y, y′)⟨x,x′⟩ = ∑g∈G

⟨π(g)x, y⟩⟨π(g)x′, y′⟩

Applying the same argument allows us to see that there exists a constantc(x,x′) such that

c(x,x′)⟨y′, y⟩ = ∑g∈G

⟨π(g)x, y⟩⟨π(g)x′, y′⟩

Together, these identities give the formula in the proposition. Setting x =x′ = y = y′, we see that the constant d must be real and positive.

Proposition 33. Suppose G is a finite group and (π,V ) is an irreduciblerepresentation of G of dimension n. Let χπ denote the character of π. Then

1

∣G∣ ∑g∈Gχπ(g) =

⎧⎪⎪⎨⎪⎪⎩

1 if π is the trivial representation;

0 otherwise.

More generally, if (π,V ) is any representation of G and χπ its character then

1

∣G∣ ∑g∈Gχπ(g) = dim(V G)

where V G denotes the subspace of V consisting of G-fixed vectors.

Proof. In the case that π is irreducible, we note that if π is trivial, thenχπ(g) = 1 for all g ∈ G so that the conclusion is obvious. If π is not trivial,then we note that χπ(g) is a sum of functions of the form L(π(g)x) wherex ∈ V . Indeed if we write pick a basis v1, . . . , vn of V and write

π(g) =⎛⎜⎝

π11(g) ⋯ π1n(g)⋮ ⋮

πn1(g) ⋯ πnn(g)

⎞⎟⎠

then πij(g) = Li(π(g)vj) where Li(∑j cjvj) = ci. Thus, since 1∣G∣ ∑g∈G χπ(g) is

the inner product of χπ with the trivial character (which is a function of theform L(1Gv) = 1) we may extend this inner product by linearity to see that1∣G∣ ∑g∈G χπ(g) is a sum of inner products of the form ⟨L(π(g)vj),1G⟩ which

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are all zero since π and the trivial representation are not isomorphic. Thisproves the first statement.

More generally, if (π,V ) is any representation, we write π = ⊕idiπi wheredi are nonnegative integers as the decomposition of π into irreducibles. Thenχπ = ∑i diχπi so that

∑g∈G

χπ(g) = ∑g∈G∑i

diχπi(g) =∑i

∑g∈G

diχπi(g) = ∑g∈G

d1Gχ1G(g)

where d1G is the multiplicity of the trivial representation. The last equalityholds by what we proved in the irreducible case. Recognizing that the mul-tiplicity of the trivial representation in (π,V ) is the dimension of V G yieldsthe result.

We now have the tools to prove the following important result known asSchur orthogonality:

Proposition 34 (Schur Orthogonality). Suppose G is a finite group withrepresentations (π1, V ) and (π2, V2) and associated characters χ1, χ2. Then⟨χ1, χ2⟩ = dim HomG(V1, V2). By extension, if π1 and π2 are irreducible, then

⟨χ1, χ2⟩ =⎧⎪⎪⎨⎪⎪⎩

1 if π1 ≅ π2;0 otherwise.

Proof. Consider the vector space of all linear transformationsA =HomC(V1, V2)from V1 to V2. We may define a representation Ω of G on A by the ruleΩ(g)T = π2(g) T π1(g−1). It is not hard to prove that χ2(g)χ1(g) is thecharacter of Ω(g). The space of G-fixed vectors ΩG consists precisely of thoselinear transformations that commute with the action of G, i.e., the G-modulehomomorphisms. By proposition two,

⟨χ1, χ2⟩ = dim HomG(V1, V2).

The second statement about irreducible representations is a direct conse-quence of Schur’s lemma.

Proposition 35. Let (πi, Vi), (πj, Vj) be irreducible representations of a fi-nite group G. Then the representation (πi ⊗ πj)∣G of G ×G restricted to thediagonal subgroup G contains exactly one copy of the trivial representation 1Gif and only if πi = πj. Otherwise, the multiplicity of the trivial representationin (πi ⊗ πj)∣G is 0. Thus, G G ×G is a Gelfand subgroup.

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We already gave one proof of this fact. Here are two more which areenlightening in different ways.

Proof 2. Let χi, χj denote the characters of πi, πj respectively. Then χiχj isthe character of (πi ⊗ πj)∣G. Now let

χiχj = d ⋅ 1G + ∑χl≠1G

dlχl

be the decomposition of χiχj into irreducible characters. Now note that

d = ⟨χiχj,1G⟩G = 1

∣G∣ ∑g∈Gχiχj(g) = ⟨χi, χj⟩G

Observe that the expression on the left is the multiplicity of 1G in (πi⊗πj)∣Gand that Schur orthogonality implies that the expression on the right is 0 ifχi ≠ χj and 1 if χi = χj. We already saw that χj is the character of πj. Theconclusion follows.

Proof 3. Note that Vi ⊗ V ∗j ≅ HomC(Vj, Vi) = V where (gi, gj) ∈ G ×G acts

on T ∈HomC(Vj, Vi) by the rule

(gi, gj)T = πi(gi)Tπj(g−1j ).

Now observe that saying that T is a G-fixed vector is equivalent to sayingthat T is a G module homomorphism. That is, (g, g)T = T if and only ifT (π2(g−1)v) = π1(g−1)T (v). This shows that V G = HomG(Vj, Vi) where V G

denotes the subspace of V consisting of G-fixed vectors. But Schur’s lemmasays that

dimHomG(Vj, Vi) =⎧⎪⎪⎨⎪⎪⎩

1 if Vj ≅ Vi.0 otherwise.

Thus, for any irreducible representation V of G × G, V G is at most onedimensional, i.e., G G ×G is indeed a Gelfand subgroup.

We now introduce the notion that is the purpose of this section:

Definition 5. Let G be a finite group and H ≤ G a Gelfand subgroup. Givenan irreducible representation (π,V ) of G with a unique H-fixed vector v ∈ V ,a spherical function σ associated with this representation, is a function σ ∶G→ C which satisfies:

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1. σ is H bi-invariant, i.e., σ(hgh′) = σ(g) for all g ∈ G, h,h′ ∈H,

2. σ is of the form σ(g) = f(π(g) ⋅ v) for g ∈ G) where f ∶ V → C is alinear functional.

Proposition 36. The G be a group and (π,V ) a finite dimensional irre-ducible representation of G with a unique H-fixed vector v ∈ V . Let ⟨, ⟩ bea G-invariant inner product on V . Then the spherical function σ as definedabove is uniquely determined by conditions 1) and 2) and is given by theformula σ(g) = ⟨π(g)v, v⟩.

Proof. It is easy to see that σ(g) = ⟨π(g)v, v⟩ satisfies condition 2) above.That it satisfies condition 1) follows by taking adjoints on the inner productand noting that v is H-fixed, i.e.,

⟨π(hgh′)v, v⟩ = ⟨π(g)v, π(h)∗v⟩ = ⟨π(g)v, v⟩

To see that 1) and 2) force σ to take this form observe that 2) implies thatwe may write σ(g) = ⟨π(g)v, y⟩ for some y ∈ V . But 1) forces that for anyh ∈H and any g ∈ G ⟨π(g)v, π(h)∗y⟩ = ⟨π(g)v, y⟩. But since π is irreducible,this implies that ⟨z, π(h)∗y−y⟩ = 0 for all z ∈ V so that indeed π(h)∗y = y forall h ∈ H. Thus, π(h)y = y for all h ∈ H, i.e., y is H-fiixed. By assumption,there is only one H-fixed vector v, so y = v which completes the proof.

We note then, that it makes sense to talk about the spherical functionσπ associated with an irreducible representation (π,V ) of G with a uniqueH-fixed vector without any ambiguity.

Now we have already shown that G embedded diagonally in H = G×G isa Gelfand subgroup. So suppose (π,V ) is an irreducible representation of Gso that (π⊗ π, V ⊗V ∗) has a unique G-fixed vector. Identifying V ⊗V ∗ withHomC(V,V ) and letting Π denote the representation isomorphic to π ⊗ πgiven by Π((g1, g2)) ⋅T = π(g1)Tπ(g−12 ), we see that idV ∈HomC(V,V ) is theunique G-fixed vector, since for (g, g) ∈ G it holds that

Π((g, g)) ⋅ idV = π(g)idV π(g−1) = π(gg−1) = idV .

Now let χπ denote the character of π and define σ((g1, g2)) = χπ(g1g−12 ).It is easy to see that σ is G bi-invariant since for (g, g), (g′, g′) ∈ G and

34

(g1, g2) ∈H, it holds

σ((g, g)(g1, g2)(g′, g′)) = χπ(gg1g′g′−1g−12 g−1)= χπ(g1g−12 )= σ((g1, g2))

where the second to last equality holds since χπ is a class function. Further,for h = (g1, g2) ∈H we see that

σ(h) = χπ(g1g−12 ) = tr(π(g1)idV π(g−12 )) = tr(Π(h) ⋅ idV )

so that since we saw that idV is G-fixed and since the trace operator is alinear functional, we see that σ satisfies condition 2) of being a sphericalfunction. Thus, we have established that for an irreducible representationπ of G, σ((g1, g2)) = χπ(g1g−12 ) defines the spherical function on H = G ×Gassociated with π ⊗ π.

As it turns out, as πi ranges over the irreducible representations of G,the spherical functions associated with the πi’s form an orthonormal basisfor the space of functions C(G/H/G) of all G bi-invariant functions from Hinto the complex numbers.

To prove this, we require the following proposition:

Proposition 37. Let G be a group and let H = G ×G. Identify G with thediagonal subgroup of H given by g ↦ (g, g). Then the set of double cosetsG/H/G is in bijection with the conjugacy classes of G.

Proof. First, observe that the left cosets H/G are in bijection with G itself,since for any left coset representative (g, h), multiplying on the right by(h−1, h−1) ∈ H gives (gh−1,1). Now suppose that (g,1) ∈ G/(g′,1)/G whereg, g′ ∈ G, i.e., there exist h, l ∈ G with (g,1) = (h,h)(g′,1)(l, l). Multiplyingon the right by (l−1h−1, l−1h−1) keeps us within the same double coset andgives (g,1) = (hg′h−1,1), i.e., g and g′ are in the same conjugacy class.

It follows from this result that counting dimensions yields a vector spaceisomorphism L2

class(G) ≅ C(G/H/G) where L2

class(G) denotes the set of

class functions on G. Now it is a well known fact that the irreducible charac-ters χπi of G form an orthonormal basis of L2

class(G) where the inner product

35

in L2

class(G) is given by

⟨φ,ψ⟩ = 1

∣G∣ ∑g∈Gφ(g)ψ(g)

for φ,ψ ∈ L2

class(G).

Motivated by this, we note that

⟨F,F ′⟩ = 1

∣H ∣ ∑h∈HF (h)F ′(h)

for F,F ′ ∈ C(G/G/H) defines a Hermitian inner product on C(G/H/G).Now given a class function, f ∶ G → C, define a function F ∶ H → C by

F (g1, g2) = f(g1g−12 ). To see that F (g1, g2) ∈ C(G/H/G) observe that for anyh,h′ ∈ G it holds:

F ((h,h)(g1, g2)(h′, h′)) = f(hg1g−12 h−1) = f(g1g−12 ) = F (g1, g2)

since f is a class function. Thus, f ↦ F gives a well-defined map fromL2

class(G) to C(G/H/G).

This map preserves inner products since given f, f ′ ∈ L2

class(G) and the

associated functions F,F ′ ∈ C(G/H/G), we observe that

⟨F,F ′⟩ = 1

∣H ∣ ∑h∈HF (h)F ′(h)

= 1

∣G∣2 ∑g1,g2∈G

F (g1, g2)F ′(g1, g2)

= 1

∣G∣2 ∑g1,g2∈G

f(g1g−12 )f ′(g1g−12 )

= 1

∣G∣ ∑g∈Gf(g)f ′(g)

= ⟨f, f ′⟩.

Further, we see that given the character χπi of an irreducible character πi,this map will give us the spherical function σi(g1, g2) = χπi(g1g−12 ). Since theinner product is preserved under this mapping, the spherical functions willthus be be orthonormal in C(G/H/G) and by the vector space isomorphismestablished, they compose an orthonormal basis for C(G/H/G).

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This interesting example motivates us to ask the question, of whethergiven a Gefland pair (G,H) it holds in general that the associated spheri-cal functions provide an orthonormal basis for C(H/G/H), where the innerproduct on C(H/G/H) is given by

⟨f, f ′⟩ = 1

∣G∣ ∑g∈Gf(g)f ′(g).

As it turns out, the answer is yes, which gives an interesting characterizationof this function space, analogous to the theory of irreducible characters withinthe space of class functions. The proof of this fact is analogous to that ofthe example. We begin by showing the following fact.

Proposition 38. Let H ≤ G be a Gelfand subgroup and let IndGH1H = ⊕ni=1πibe the decomposition of IndGH1H into irreducible representations, which weknow to be multiplicity free since H is a Gelfand subgroup. Then the cardi-nality of the set of double cosets H/G/H is given by ∣H/G/H ∣ = n.

Proof. This is a simple consequence of Mackey’ theorem. That is,

dim HomG(IndGH1H , IndGH1H) = ∑

HgH

dim Hom(1H ∣H∩g−1Hg,1H ∣H∩g−1Hg) = ∑HgH

1.

Recognizing that the left hand side is equal to n, the result follows.

Now from Proposition 31 and 38, we can see that for a Gelfand pair (G,H)the spherical functions form an orthogonal basis for the space of functionsC(H/G/H). Indeed, we can scale the spherical functions by a constant, sothat they form an orthonormal basis.

We verify here that indeed the constant in Proposition 32 is d = dim(V ),which is the last tool we need in constructing an orthonormal basis of C(H/G/H).

Proposition 39. The constant in Proposition 32 is dim(V ).

Proof. Let v1, . . . , vn be an orthonormal basis of V , where n = dim(V ). Thenχπ(g) = ∑n

i=1⟨π(g)vi, vi⟩. Combining this with Schur orthogonality gives:

1 = 1

∣G∣ ∑g∈Gχπ(g)χπ(g) = ∑

i,j=1,...,n

1

∣G∣ ∑g∈G⟨π(g)vi, vi⟩⟨π(g)vj, vj⟩

Of the n2 terms on the right hand side, only n are nonzero (those with i = j)and each of those is equal to the constant d−1 of Proposition 32. Thus,nd−1 = 1. That is, d = dim(V ) which is what we wanted.

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Using this, we attain the following, our desired result.

Proposition 40. Let G be a finite group and H ≤ G a Gelfand subgroup. Let(π1, V1), . . . , (πn, Vn) denote the irreducible representations of G with uniqueH-fixed vectors and σ1, . . . , σn denote the associated spherical functions. Then√d1σ1, . . . ,

√dnσn (where di is the dimension of πi for each i = 1, . . . , n)

forms an orthonormal basis for the vector space C(H/G/H) of all functionsf ∶H/G/H → C.

Proof. We have already seen that d1σ1, . . . , dnσn form an orthogonal basis.Using Proposition 39, we have that for any σi, it holds

⟨√diσi,

√diσi⟩ = di⟨σi, σi⟩ = di(

1

di⟨v, v⟩⟨v, v⟩) = 1

where v denotes the unique H-fixed vector of unit length. Thus, the basis isorthonormal as desired.

Although we have developed our theory of spherical functions in the set-ting of finite groups, it should be noted that the finiteness of the groupswas not used in any of the proofs. Indeed in the setting that G is a locallycompact abelian group, replacing the finite sums in many of these proofswith integrals will give the same results (left to the reader to verify). In thespecial case that G = R and H = 2πk ∶ k ∈ Z ≤ R, it is not hard to see thatthe spherical functions are of the form σn(x) = einx for n ∈ Z. Identifying thespaces L2(H/G/H) and L2([−π,π]) (these are different ways of describingthe space of 2π periodic functions), we see that indeed the classical theoryof Fourier series is recovered as a special case of our more general theory, aswas claimed in the introduction.

6 Acknowledgements

I would like to thank my thesis advisor, Professor Daniel Bump, for suggest-ing this project, generously offering his time and guidance as I developed thisthesis, and consistently supporting my mathematical study throughout theyear. Many thanks are also owed to my parents, whose support made myundergraduate career and this thesis possible.

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References

[1] D. Bump. Lie Groups, volume 225 of Graduate Texts in Mathematics.Springer-Verlag, New York, 2013.

[2] D. Dummit and R. Foote. Abstract Algebra. John Wiley and Sons, NewYork, 2004.

[3] S. Lang. Algebra, volume 211 of Graduate Texts in Mathematics.Springer-Verlag, New York, 2002.

[4] M. Takeuchi. Modern Spherical Functions. Iwanami-Shoten Publishers,Tokyo, 1975.

[5] N. Young. An Introduction to Hilbert Space. Cambridge University Press,1988.

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