Analysis on Finite Gel′fand Spaces

by Michael S. Venn

Bachelor of Arts University of South Carolina, 1991

Submitted in Partial Fulfillment of the Requirements for the Degree of Master of Science in the Department of Mathematics University of South Carolina 2001

Department of Mathematics Director of Thesis Department of Mathematics Second Reader

Dean of The Graduate School

Acknowledgements

I would like to thank, first, Professor Ralph Howard for introducing finite Gel′fand spaces to me. Further, I am thankful for his mathematical guidance and patience, although I suspect that he would rather be remembered as a difficult and impatient thesis director. So I will thank him then, as well, for being such a peevish and merciless director—although, of course, he was verily rather the opposite of that. Professor Howard is a fine neighbor, an excellent mathematician, and, in the words of Ernest Hemingway, a very good drunk. Next, I appreciate the helpful comments of Professor Konstantin Oskolkov, perhaps the greatest story-teller I have ever met, who gave me valuable advice on the particulars of my defense. Also, Professor George McNulty proved a great help in working out the technical difficulties of typesetting my thesis. Finally, to my wife, Mary Elizabeth Brodie Venn, I give thanks for enduring so many sleepless nights with me and for brooking what must have seemed interminable, senseless half-babble about mathematical places that may, or may not, it turns out, even exist.

Abstract

We consider the discrete analogue, called the finite Radon transform, of the classical Radon transform from functional analysis. While others have approached the invertibility question of the finite Radon transform using techniques from graph theory and lattice theory, we approach the problem by considering spaces, which we call Gel′fand spaces in which we can show that finite Radon transforms are invertible. We start with a vector space and add some group structure, as well as some representation theory, to build, largely axiomatically, what we call a convolution algebra of a finite set. The convolution algebra will serve as a foundation for the Gel′fand space. We also include a few examples of finite Radon transforms and their invertibility formulæ.

Contents

Acknowledgements 2

Abstract 3

Chapter 1 Introduction 1 1.1. The Radon Transform 1 1.2. The Finite Radon Transform 2

Chapter 2 The Convolution Algebra of a Finite Set X 6 2.1. Preliminary Definitions 6 2.2. Representation of a Group on ℓ^{2}(X) 7 2.3. The Subgroup G_{o} and G-space Isomorphisms 10 2.4. The G-Space ℓ^{2}(X) and its Isotropic Functions 11 2.5. G-invariant Subspaces of ℓ^{2}(X) 13 2.6. Schur’s Lemma 15 2.7. An Injectivity Theorem 17 2.8. Radon Transforms Between Finite Grassmanians 18 2.9. Radon Injectivity Results for Grassmanians 19 2.10. Convolution in ℓ^{2}(X × X) 23 2.11. The Convolution Ring of X 25 2.12. The Convolution Algebra ℓ^{2}(X × X)^{G} 27 2.13. The Relationship Between ℓ^{2}(X)^{G_{o}} and ℓ^{2}(X × X)^{G} 29

Chapter 3 Finite Gel′fand Spaces 31 3.1. Collections of Linear Operators 31 3.2. Symmetric G-Spaces 34 3.3. The Cartan-Gel′fand Theorem 36 3.4. G-Invariant Linear Operators and Inversion Formulae 41

Chapter 4 Doubly Transitive Group Actions 45 4.1. Introduction and Definitions 45 4.2. The Radon Transform on X = {1, ⋯, n} 46 4.3. The Radon Transform for Doubly Transitive Actions 50

Bibliography 54

Chapter 1

Introduction

1.1. The Radon Transform

Recall from elementary linear algebra and analysis that we give the name Hilbert space to any complete inner product space (See the beginning of Chapter 2 for the definition of an inner product space). Furthermore, recall that a hyperplane of an n-dimensional vector space is a subspace of dimension n − 1. Finally, we will call a subset, A, of a vector space, V, affine if there is a vector subspace, W, of V and an element, v, in V such that A = v + W. The affine subsets, then, are the translates of the subspaces, and we call the translate of an (n − 1)-dimensional subspace an affine hyperplane. Now let ϕ : ℝ^{n} ↦ ℂ be a complex-valued function on an n-dimensional Hilbert space, where ℝ and ℂ denote the real and complex numbers, respectively. Then we will call T ϕ the Radon transform of ϕ, where T ϕ is the complex-valued transformation, defined on the set of affine hyperplanes in ℝ^{n} by,

(T ϕ)(H) = ∫_{H} ϕ.

We integrate, above, with respect to Lebesgue measure, over the affine hyperplane H.

The Radon transform has enjoyed a position of fundamental importance to many applied problems in mathematics and physics and questions in functional analysis. The problems and applications usually appear in some manifestation of the following question. When can the function, ϕ, be reconstructed from its Radon transform, T ϕ? The Radon transform gets its name from Johann Radon, who first, in 1917, derived an explicit formula for the function on the plane, if integrals over all lines through the plane are given [18]. Since then applications of the Radon formula have appeared in radio astronomy, electron micrography, and many other fields of science and mathematics.

Undoubtedly, however, the most famous application of the Radon transform’s invertibility has been in x-ray tomography. In 1970 G. N. Hounsfield and A. M. Cormack introduced a computed tomograph, which physicians could use in a clinical setting, for which, in 1979, they were awarded the Nobel Prize in medicine. For an excellent introduction to the history of the Radon transform in computed tomography, see A. M. Cormack’s treatment in [5].

1.2. The Finite Radon Transform

But what if we consider the analogues of the Radon transform in the discrete setting? To wit, then, can we define the Radon transform on a function space in which the functions are all defined on a finite set? As the finite analogue of integration is summation, we can rewrite the above definition of the Radon transform.

Let 𝒞 𝒞 𝒞 be a collection of subsets of a finite set, X, and let ℓ^{2}(X) denote the set of all complex functions defined on X. (See Section 2.1 for a precise definition of ℓ^{2}(X).) Now if ϕ is in ℓ^{2}(X), then we will call the linear transformation T : ℓ^{2}(X) ↦ ℓ^{2}(𝒞 𝒞 𝒞) the finite Radon transform defined, for C in 𝒞 𝒞 𝒞, by

(Tϕ)(C) = ∑_{x∈C}ϕ(x).

Henceforth, when we refer to the Radon transform, we mean the finite Radon transform as defined immediately above, even if we omit the word finite, inasmuch as in the following, we work solely with the finite Radon transform.

According to Joseph Kung in [15], the first to consider the finite Radon transform was Ethan Bolker circa 1976. Bolker writes in [1] that, asked by a mathematician working in classical Radon transforms, he began thinking about the Radon transform on finite sets and looking for structure in order “to motivate theorems about the classical Radon transform and its relatives.” Bolker’s beautifully written mathematics in [1] is the most important survey on the topic of the finite Radon transform, and, as Kung states in [15],

Bolker’s work is focused on finite analogues of the central ideas in the theory of Radon transforms in analysis: inversion formulas, relation to the Laplacian and other differential operators, ranges of Radon transforms, group actions and homogeneous spaces, and relation to group representation theory.

Kung adds that the importance of Bolker’s work in the field “cannot be overemphasized.” In [1], Bolker explores the relationship between finite Radon transforms and areas of interest to combinatorists, including geometry over finite fields and the Kirkman schoolgirl problem. Bolker continues his work in this area, first, with Eric Grinberg and Kung in [2] and then with Patrick O’Neil and Jan Bowman in [3].

Kung’s own work [15] in which he praises Bolker is itself an exceptional overview of the finite Radon transform. Like Bolker, Kung surveys the work in the finite Radon transform in combinatorics, and much of the paper he devotes to applications to lattice theory. Kung ends the paper with an impressive list of unsolved problems, most—if not all—of which, to our knowledge, remain unsolved.

One approach to solving the injectivity question of the finite Radon transform is to convert the question “When can a function, ϕ, be reconstructed from its Radon transform, T ϕ? ” into a question that might be answered through techniques in graph theory. If we look at the matrix of the linear transformation T, relative to the standard basis,

δ_{x}(y) = \{ \begin{matrix}1 & x = y \\ 0 & x ≠ y. \end{matrix} ,

then we have the incidence matrix with rows indexed by 𝒞 𝒞 𝒞 and columns indexed by C, whose C, x-entry is 1 if x is in C and 0 if x is not in C. Therefore, the study of finite Radon transforms can be reduced to the study of incidence matrices.

Still, as Bolker writes in [1], the question “When is the Radon transform injective?” is too hard in its general form. Therefore, we find in the literature answers, rather, to narrower questions of the following sort: “When is a particular Radon transform injective?” or “Under what conditions can a function in the image of a Radon transform be recovered?” We take a different approach here. Our question is “Can we build a space in which finite Radon transforms are invertible?” The affirmative answer to that question we present in the following pages, in a more analytic approach than others have employed as regards the finite Radon transform problem. We will consider the case in which we can simplify calculations using symmetry in the form of group actions.

The first step in the construction of such a space, which we will call a Gel′fand space, is to find a suitably structured environment in which to start building. We find that a vector space is the environment most suitable and flexible to our needs. Next we will add some group structure, followed by some representation theory to build what we will call a convolution algebra of a finite set X. We dedicate Chapter 2 to a largely axiomatic construction of a convolution algebra, which will serve as the keystone for the Gel′fand space, developed in Chapter 3, built on our work with the convolution algebra. We claim now, and will prove presently, that a Gel′fand space is a mathematical structure in which the finite Radon transform is, indeed, invertible. Finally, in Chapter 4, we present our injectivity results in the special case of doubly transitive group actions. We also present two examples, one a special case of the other.

The impetus behind the following pages, as well as, indeed, much of their content, comes from a set of unpublished notes [13] written by Ralph Howard and a class, entitled Groups and Graphs, he taught at the University of South Carolina in the fall of 2000. We try to provide here a largely axiomatic construction of finite Gel′fand spaces, without assuming that the reader has any prior knowledge of or initiation in their structure or application. Although all of the representation theory and most of the group theory and linear algebra used we try to present from the fundamentals of the definitions that we provide; nevertheless, we do assume that the reader has experience with a few definitions and results from elementary group theory and linear algebra. We find the structure behind the analysis of finite Gel′fand spaces and the proofs supporting that structure, by the nature of the order seemingly inherent in them, often elegant and surprisingly intuitive, with a touch of what seems, at times, an eldritch simplicity. We present then, in the words of the great Twentieth-Century novelist, Salman Rushdie, what we regard as an eff of the ineffable.

The Convolution Algebra of a Finite Set X

2.1. Preliminary Definitions

Let V be a vector space over the field of complex scalars, ℂ. Suppose ⟨·, ·⟩ : V ↦ ℂ is a function that assigns to each ordered pair of vectors v_{1} and v_{2} in V a scalar ⟨v_{1}, v_{2}⟩ in ℂ. Now suppose that our function has the following properties for each v_{1}, v_{2}, and v_{3} in V and for every c_{1}, c_{2}, and c_{3} in ℂ : ⟨c_{1}v_{1} + c_{2}v_{2}, v_{3}⟩ = c_{1}⟨v_{1}, v_{3}⟩ + c_{2}⟨v_{2}, v_{3}⟩; ⟨v_{1}, v_{2}⟩ = \overline{⟨v_{2}, v_{1}⟩}; and ⟨v_{1}, v_{1}⟩ ≥ 0 with equality precisely when v_{1} = 0. Then we call the function, ⟨·, ·⟩, an inner product. Further, we shall call any vector space equipped with an inner product an inner product space.

Recall that a basis of a vector space, V, is a linearly independent spanning set of vectors in V. Also, if the dimension of V is n, then we say that a basis {ϕ_{1}, ϕ_{2}, ⋯, ϕ_{n}} is unitary or orthogonal if

⟨ϕ_{i}, ϕ_{j}⟩ = δ_{ij} = \{ \begin{matrix}1 & i = j \\ 0 & i ≠ j. \end{matrix}

We know from elementary linear algebra that any finite-dimensional vector space has a basis, which we can make unitary by the famous Gram-Schmidt process or some similar algorithm.

Let X be a finite set. Then if ℓ^{2}(X) is the set of all complex-valued functions on X and if we equip ℓ^{2}(X) with the standard inner product, ⟨ϕ, ψ⟩ = ∑_{x} ϕ(x)\overline{ψ(x)}, then ℓ^{2}(X) is an inner product space. As usual, we denote the general linear group, the group of invertible linear transformations from V to V, by GL(V), where V is a finite-dimensional vector space. We will call a linear operator, T, unitary if T^{∗}T = T T^{∗} = I. The unitary group, denoted 𝒰(V), is the subgroup of GL(V) whose elements are, in addition to being invertible, unitary. Let G be a group and X a set. If we let g be in G and x in X, then we call a map (g, x) ↦ gx an action of G on X if, when e is the identity of G, ex = x for all x in X and g_{1}(g_{2}x) = (g_{1}g_{2})x for every g_{1} and g_{2} in G and x in X. Furthermore, if we have an action of G on X, we often write G acts on X and call X a G-space. If H is a subgroup of G and G/H = {ξH : ξ ∈ G} is the set of left cosets, then G/H is a G-space via the action g(ξH) = (gξ)H.

Let X be a G-space. Then if x is an element of X we call Gx = {gx : g ∈ G} the orbit of x under G. Further, we say G acts transitively on X or that there is a transitive group action of G on X, when all elements of X are in the same orbit under G, or, what is the same, when for all x_{1} and x_{2} in X, there exists a group element g such that gx_{1} = x_{2}. Notice that a set X being a transitive G-space makes no guarantee that the action of a subgroup of G on X will also be transitive. Let us call the number of orbits of a group or a subgroup, H, on X the rank of H on X, which we will denote, Rank_{H}(X). We next give a name to the set of G-space members that a group element g leaves fixed, or, in symbols, {x ∈ X : gx = x}, which we shall denote X^{g}; these fixed point sets of g we extend in a natural way to subsets of G; hence, when we write X^{H} and H is a subset of G, we mean the set of members of X that the action of every member of H keeps fixed; in symbols, we have, then, X^{H} = {x ∈ X : gx = x for all g ∈ H}.

2.2. Representation of a Group on ℓ^{2}(X)

Let G be a finite group and V a finite-dimensional vector space. If ρ : G ↦ GL(V) is a group homomorphism—that is, ρ(g_{1}g_{2}) = ρ(g_{1})ρ(g_{2}) for every g_{1} and g_{2} in G— then we call ρ a representation of G or, when there is no confusion about the group to which we refer, a group representation. Often we will abuse notation slightly and write g for ρ(g). Our standard example, which will prove useful, of a group representation is the following. Let us assume that X is a G-space; then for x in X and g in G, define τ : G ↦ GL(ℓ^{2}(X)) by (τ_{g}ϕ)(x) = ϕ(g^{−1}x). Surely, then, the map τ_{g} : ℓ^{2}(X) ↦ ℓ^{2}(X) is linear, as the space ℓ^{2}(X) itself is linear. In addition, τ is a group homomorphism, inasmuch as (τ_{g_{1}})(τ_{g_{2}}ϕ)(x) = (τ_{g_{2}}ϕ)(g_{1}^{−1}x) = ϕ(g_{2}^{−1}g_{1}^{−1}x) = ϕ((g_{1}g_{2})^{−1}x) = (τ_{g_{1}g_{2}}ϕ)(x), and thence we have shown that τ is a representation of G. Finally, since τ is a group representation, and therefore a group homomorphism, we have the usual added structural advantage τ_{g}^{−1}= τ_{g^{−1}} for all group elements g, because for any representation ρ of a group G and for any g in G, we have ρ(g^{−1})ρ(g) = ρ(gg^{−1}) = ρ(e), which is the identity element in GL(V). Accordingly, in our example, τ, of a group representation, we know that every τ_{g} has an inverse, namely, τ_{g^{−1}}, in GL(ℓ^{2}(X)).

We will say that a representation ρ : G ↦ GL(V) is unitary if there exists an inner product on V such that ⟨ρ(g)v, ρ(g)w⟩ = ⟨v, w⟩ for all v and w in V and g in G.

Proposition 2.2.1. If ρ : G ↦ GL(V) is a representation, then there exists an inner product that makes ρ unitary.

Proof. Let ⟨·, ·⟩_{0} be any inner product on V, and define a new inner product on V by

⟨v_{1}, v_{2}⟩_{ν} = \frac{1}{|G|} ∑_{ξ∈G}⟨ρ(ξ)v_{1}, ρ(ξ)v_{2}⟩_{0}

for v_{1} and v_{2} in V. We will show, first, that ⟨·, ·⟩_{ν} is an inner product. Let v_{1}, v_{2} and v_{3} be in V and let c_{1} and c_{2} be in ℂ. Then

⟨c_{1}v_{1} + c_{2}v_{2}, v_{3}⟩_{ν} = \frac{1}{|G|} ∑_{ξ∈G}⟨ρ(ξ)(c_{1}v_{1} + c_{2}v_{2}), ρ(ξ)v_{3}⟩_{0}

= \frac{1}{|G|} ∑_{ξ∈G}⟨ρ(ξ)c_{1}v_{1} + ρ(ξ)c_{2}v_{2}, ρ(ξ)v_{3}⟩_{0}

= \frac{1}{|G|} ∑_{ξ∈G} (⟨ρ(ξ)c_{1}v_{1}, ρ(ξ)v_{3}⟩_{0} + ⟨ρ(ξ)c_{2}v_{2}, ρ(ξ)v_{3}⟩_{0})

= \frac{1}{|G|} ∑_{ξ∈G}⟨ρ(ξ)c_{1}v_{1}, ρ(ξ)v_{3}⟩_{0} + \frac{1}{|G|} ∑_{ξ∈G}⟨ρ(ξ)c_{2}v_{2}, ρ(ξ)v_{3}⟩_{0}

= c_{1} \frac{1}{|G|} ∑_{ξ∈G}⟨ρ(ξ)v_{1}, ρ(ξ)v_{3}⟩_{0} + c_{2} \frac{1}{|G|} ∑_{ξ∈G}⟨ρ(ξ)v_{2}, ρ(ξ)v_{3}⟩_{0}

= c_{1}⟨v_{1}, v_{3}⟩_{ν} + c_{2}⟨v_{2}, v_{3}⟩_{ν}

and, thus, ⟨·, ·⟩_{ν} satisfies the first property of inner products. For the second, note that

\overline{⟨v_{2}, v_{1}⟩}_{ν} = \frac{1}{|G|} ∑_{ξ∈G} \overline{⟨ρ(ξ)v_{2}, ρ(ξ)v_{1}⟩}_{0}

= \frac{1}{|G|} ∑_{ξ∈G}⟨ρ(ξ)v_{1}, ρ(ξ)v_{2}⟩_{0} = ⟨v_{1}, v_{2}⟩_{ν}

Finally, we find ⟨v_{1}, v_{1}⟩_{ν} ≥ 0 with equality holding precisely when v_{1} is the zero vector, because the inner product ⟨·, ·⟩_{0} with which it is defined has the desired property and because \frac{1}{|G|} is never zero if G is nonempty.

Now we will show that ⟨·, ·⟩_{ν} makes ρ unitary. For g in G we have

⟨ρ(g)v_{1}, ρ(g)v_{2}⟩_{ν} = \frac{1}{|G|} ∑_{ξ∈G}⟨ρ(ξ)ρ(g)v_{1}, ρ(ξ)ρ(g)v_{2}⟩_{0}

= \frac{1}{|G|} ∑_{ξ∈G}⟨ρ(ξg)v_{1}, ρ(ξg)v_{2}⟩_{0}

= \frac{1}{|G|} ∑_{η∈G}⟨ρ(η)v_{1}, ρ(η)v_{2}⟩_{0} if we substitute ηg^{−1} for ξ

= ⟨v_{1}, v_{2}⟩_{ν}

Therefore, our candidate, ⟨·, ·⟩_{ν}, makes ρ unitary, and we have proven the proposition. Q.E.D.

By Proposition 2.2.1, then, every representation of a finite group is unitary. Further, we can represent any finite group, with a group representation, not only as a subset of the general linear group but also as a subset of the unitary group. Now let us reconsider our example, (τ_{g}ϕ)(x) = ϕ(g^{−1}x), of a group representation using the claim that every representation of a finite group is unitary. Hence, we can choose to define our representation, τ, from G to 𝒰(ℓ^{2}(X)) instead of to GL(ℓ^{2}(X)).

2.3. The Subgroup G_{o} and G-space Isomorphisms

Now we are ready to provide the basic environment for defining a Gel′fand space, and we establish the convention that any set, X, and any group, G, mentioned henceforth we will assume are finite, unless otherwise specified. Let X be a set on which a group G has a transitive group action. In other words, our set X is a transitive G-space. Now pick any element of X to serve as the origin, and call it o. If gx = x for a given x in X and g in G, we shall call g a stabilizer or, in this case, an x-stabilizer, and the set of x-stabilizers in G we shall denote G_{x}. Let o be any fixed member of X. Now consider the set of o-stabilizers, namely, G_{o}, which comprises the set of all elements of G that leave the origin, o, fixed, or, in symbols, G_{o} = {g ∈ G : go = o}. What should be unsurprising is that the stabilizers in G are subgroups. If g_{1} and g_{2} are in G_{x}, we have g_{1}x = g_{2}x = x, and since (g_{1}g_{2})x = g_{2}(g_{1}x) = g_{2}x = x, we surely have that G_{x} is closed. Also, the identity, e, of G is in G_{x}, as ex = x. Now if we demonstrate G_{x} is closed under taking inverses, as well, we have shown that G_{x} is a subgroup. To that end, let g ∈ G_{x}. Then gx = x, which implies g^{−1}gx = g^{−1}x, and, thence, x = g^{−1}x, which, of course, gives us g^{−1} ∈ G_{x}. Therefore, for any x ∈ X we know that G_{x} is a subgroup.

Let X and Y be G-spaces. Then a map ϕ : X ↦ Y is a G-morphism if ϕ(gx) = gϕ(x) for all x in X and g in G. Furthermore, if there exists a bijective G-morphism ϕ from X into Y, then we call ϕ an isomorphism and say X and Y are isomorphic as Gspaces. We find, when G/G_{o} is the quotient of G modulo the subgroup of o-stabilizers, that X and G/G_{o} are isomorphic as G-spaces. To see this, define Φ : G/G_{o} ↦ X by Φ(gG_{o}) = go, where gG_{o} is a member of the left cosets of the subgroup G_{o} and, therefore, a member of the quotient G/G_{o}. Surely, Φ is well-defined, for if g_{1}G_{o}= g_{2}G_{o} for some g_{1} and g_{2} in G, then g_{1} = g_{2}g_{0} for some g_{0} in G_{o}. But as g_{0} is in G_{o}, we know g_{1}o = g_{2}g_{0}o = g_{2}o, which confirms, indeed, that Φ is well-defined. Now Φ is a G-morphism, inasmuch as

Φ(g_{1}g_{2}G_{o}) = (g_{1}g_{2})o = g_{1}(g_{2}o) = g_{1}Φ(g_{2}G_{o}).

Next we must show that Φ is injective. To that end, let Φ(g_{1}G_{o}) = Φ(g_{2}G_{o}), which means g_{1}o = g_{2}o and, hence, g_{2}^{−1}g_{1}o = o, which, in turn, evinces that g_{2}^{−1}g_{1} is in G_{o} or, equivalently, that g_{1}G_{o} = g_{2}G_{o}. Thence we have shown that Φ(g_{1}G_{o}) = Φ(g_{2}G_{o}) implies g_{1}G_{o} = g_{2}G_{o}, which reveals that Φ is injective. To show surjectivity, let x be an element of X. Then we know there exists a g in G such that go = x, for the action of G on X is transitive. But then Φ(gG_{o}) = go = x, which guarantees the surjectivity of Φ and which finishes our justification of the claim that the transitive G-space, X, and the quotient, G/G_{o}, are isomorphic as G-spaces.

2.4. The G-Space ℓ^{2}(X) and its Isotropic Functions

Now we will reconsider ℓ^{2}(X), the set of all complex-valued functions on a set X and claim that if G acts on X, then G acts on ℓ^{2}(X) by (τ_{g}ϕ)(x) = (gϕ)(x) = ϕ(g^{−1}x), where g is in G and ϕ(x) is in ℓ^{2}(X). For let ϕ(x) be in ℓ^{2}(X). Then eϕ(x) = ϕ(e^{−1}x) = ϕ(ex) = ϕ(x), the last equality holding because X is a G-space. Now let g_{1} and g_{2} be members of G. We will show g_{1}(g_{2}ϕ) = (g_{1}g_{2})ϕ. By the definition of our proposed action of G on ℓ^{2}(X), we have g_{1}(g_{2}ϕ)(x) = g_{2}ϕ(g_{1}^{−1}x) = ϕ(g_{2}^{−1}g_{1}^{−1}x), which equals ϕ(g_{1}g_{2})^{−1}x = (g_{1}g_{2})ϕ(x), and gives us (g_{1}g_{2})ϕ. Hence, if X is, then ℓ^{2}(X) is also, a G-space.

Next we turn our attention to a subspace of the inner product space, ℓ^{2}(X), namely, the set of all functions on X that G_{o} keeps fixed—that is, {ϕ ∈ ℓ^{2}(X) : gϕ = ϕ for all g ∈ G_{o}}. We call these special fixed points in ℓ^{2}(X) isotropic functions and write ℓ^{2}(X)^{G_{o}} for the collection of isotropic functions. We note that although X may be a transitive G-space, and, thus, the action of G on X, by definition, only produces one orbit; nevertheless, the action of G_{o}, a subgroup of G, on X often produces more than one orbit. In fact, we will find, in all cases we consider and, furthermore, in all cases in which |X| ≥ 2, that Rank_{G_{o}}(X) > 1, although the number of orbits of X under the transitive action of G is, by definition, 1.

Proposition 2.4.1. The dimension of the subspace of isotropic functions on X is equal to the number of orbits that result when G_{o} acts on X, or, what is the same, in our notation above, dim ℓ^{2}(X)^{G_{o}} = Rank_{G_{o}}(X).

Proof. To begin our proof, we claim that the action of a group on a set, X, is an equivalence relation and, thus, partitions the set. To see this, let ∼_{𝒪} be the relation such that if X is a G-space, then x_{1} ∼_{𝒪} x_{2} exactly when x_{1} and x_{2} are in the same orbit under G, or, equivalently, when gx_{1} = x_{2} for some g. We will show that ∼_{𝒪} is an equivalence relation.

By the definition of group action we have ex = x for all x; thence x ∼_{𝒪} x and, accordingly, ∼_{𝒪} is reflexive. Next, suppose x_{1} ∼_{𝒪} x_{2}. Then gx_{1} = x_{2} for some g, which means, after multiplication on both sides by the group element g^{−1}, that g^{−1}gx_{1} = g^{−1}x_{2}, which reduces to x_{1} = g^{−1}x_{2}, which, in turn, tells us g^{−1}x_{2} = x_{1}; hence x_{2} ∼_{𝒪} x_{1}. We have shown, then, that ∼_{𝒪} is symmetric. Now we check the transitivity of ∼_{𝒪}. Let x_{1} ∼_{𝒪} x_{2} and x_{2} ∼_{𝒪} x_{3}. We will show x_{1} ∼_{𝒪} x_{3}. Since x_{1} ∼_{𝒪} x_{2}, we know g_{1}x_{1} = x_{2}, and since x_{2} ∼_{𝒪} x_{3}, we have g_{2}x_{2} = x_{3}, whence x_{2} = g_{2}^{−1}x_{3}. Now substituting this value for x_{2} into the equation g_{1}x_{1} = x_{2}, we get g_{1}x_{1} = g_{2}^{−1}x_{3}, which is the same as (g_{2}g_{1})x_{1} = x_{3} and, since G is a group and is therefore closed, setting g_{3} = g_{2}g_{1} evinces g_{3}x_{1} = x_{3}. Therefore, x_{1} ∼_{𝒪} x_{3}. Reflexivity, symmetry and transitivity hold, and thence, ∼_{𝒪} is, indeed, an equivalence relation.

Therefore, we know that if X_{1} = {o}, X_{2}, ⋯, X_{r} form a partition of X, then X_{1} = {o}, X_{2}, ⋯, X_{r} are disjoint. Now let

ϕ_{i}(x) = \{ \begin{matrix}1 & x ∈ X_{i} \\ 0 & x ∉ X_{i}, \end{matrix}

with 1 ≤ i ≤ r. Inasmuch as the sets X_{1} = {o}, X_{2}, ⋯, X_{r} are disjoint, then, surely, {ϕ_{1}, ϕ_{2}, ⋯, ϕ_{r}} are linearly independent, since we can think of each ϕ_{i_{0}}—if |X| = n, where |X| denotes the number of elements in X—as an n-dimensional column vector in ℓ^{2}(X)^{G_{o}} with entries of ones and zeros. The set of all such vectors, that is, {ϕ_{1}, ϕ_{2}, ⋯, ϕ_{r}}, is linearly independent, inasmuch as X_{1}, X_{2}, ⋯, X_{r} are disjoint sets. Our column vector interpretation also reveals that the vectors {ϕ_{i} : 1 ≤ i ≤ r} span ℓ^{2}(X)^{G_{o}}, and because there are r vectors in the linearly independent spanning set, {ϕ_{i}}, we have shown that {ϕ_{i}} forms an r-dimensional basis for ℓ^{2}(X)^{G_{o}} and, in so doing, have shown that dim ℓ^{2}(X)^{G_{o}} = r = Rank_{G_{o}}(X). Q.E.D.

2.5. G-invariant Subspaces of ℓ^{2}(X)

We call a linear transformation from V into V a linear operator. Let 𝒞 𝒞 𝒞 be a collection of linear operators on a finite-dimensional vector space. Then we say that a subspace, W, of V is invariant under 𝒞 𝒞 𝒞 if C[W] ⊆ W for all C in 𝒞 𝒞 𝒞. Furthermore, we say, in a natural extension of the definition of invariant subspaces above, that, if G acts on V, then a subspace W for which gw ∈ W, for all w in W and g in G, is G-invariant. Next we prove a proposition that guarantees every G-invariant subspace of ℓ^{2}(X) whose members G_{o} leaves fixed is not the trivial subspace, {0}.

Proposition 2.5.1. Let W ≠ {0} be a G-invariant subspace of ℓ^{2}(X). Then W^{G_{o}} = {ϕ ∈ W : gϕ = ϕ for all g ∈ G_{o}} ≠ {0}. In fact, there exists a ϕ in W^{G_{o}} with ϕ(o) = 1.

Proof. If we let ϕ_{0} be in W and stipulate that ϕ_{0} ≠ 0, then because W is G-invariant and the action of G on X is transitive, we can assume, with no loss of generality, that ϕ_{0}(o) ≠ 0. For if ϕ_{0}(ξ) ≠ 0 for some ξ in G_{o}, then ξ^{−1}ϕ(o) ≠ 0, and, therefore, we could simply replace ϕ_{0} with ξ^{−1}ϕ_{0}. Now let

ϕ(x) = \frac{1}{|G_{o}|} ∑_{g∈G_{o}}ϕ_{0}(g^{−1}x).

If ξ is in G_{o}, then

(ξϕ)(x) = \frac{1}{|G_{o}|} ∑_{g∈G_{o}}ϕ_{0}(g^{−1}ξ^{−1}x)
= \frac{1}{|G_{o}|} ∑_{g∈G_{o}}ϕ_{0} ((ξg)^{−1}x)
= \frac{1}{|G_{o}|} ∑_{(ξ^{−1}g)∈G_{o}}ϕ_{0} (ξξ^{−1}g)^{−1}x),

by substitution of (ξ^{−1}g) for g,

= \frac{1}{|G_{o}|} ∑_{g∈G_{o}}ϕ_{0}(g^{−1}x),

inasmuch as summing over (ξ^{−1}g) is the same as summing over g, since G_{0} is a subgroup and is, accordingly, closed. But as

\frac{1}{|G_{o}|} ∑_{g∈G_{o}}ϕ_{0}(g^{−1}x) = ϕ(x),

we have shown that ϕ(x) is in W^{G_{0}}. Furthermore,

ϕ(o) = \frac{1}{|G_{o}|} ∑_{g∈G_{o}}ϕ_{0}(g^{−1}o)
= \frac{1}{|G_{o}|} ∑_{g∈G_{o}}ϕ_{0}(o)

because, as G_{o} is a subgroup of G, g ∈ G_{o} implies g^{−1} is in G_{o} and thence we know g^{−1}o = o. But

\frac{1}{|G_{o}|} ∑_{g∈G_{o}}ϕ_{0}(o) = ϕ_{0}(o),

which, by our construction, is not equal to 0, nor is, in that case, ϕ, and we have shown that W^{G_{0}} ≠ \{0\}. Finally, the element \frac{1}{ϕ(o)}ϕ has the value 1 at o, as required. Q.E.D.

2.6. Schur’s Lemma

From elementary linear algebra we recall that if T is a linear transformation, then Im(T) and Ker(T) are subspaces, where Im(T) and Ker(T) represent the image of T and the kernel of T, respectively. Now we prove that under certain conditions Im(T) and Ker(T) are G-invariant.

Lemma 2.6.1. If T : V_{1} ↦ V_{2} is a G-morphism, then Im(T) and Ker(T) are G-invariant subspaces.

Proof. First, we will show that Ker(T) is G-invariant. Let v_{1} be in Ker(T), which is a subspace of V_{1}. Then T v_{1} = 0, as v_{1} is in Ker(T). Now we know that, since T is a G-morphism,

T (gv_{1}) = gT v_{1} = g0 = 0,

if g is in G. Thence, gv_{1} is in Ker(T) and so Ker(T) is G-invariant. Now to show that Im(T) is G-invariant, let v_{2} be in Im(T), which is a subspace of V_{2}. Consequently, we have that, for some v in V_{1},

gv_{2} = gT v = T (gv),

which is, indeed, a member of the image of T; therefore, we have shown that Im(T) is G-invariant. Q.E.D.

We will call a representation ρ : G ↦ GL(V) irreducible if the only subspaces of V invariant under G are {0} and V. If ρ_{1} : G ↦ GL(V_{1}) and ρ_{2} : G ↦ GL(V_{2}) are two representations of G, then we say ρ_{1} and ρ_{2} are equivalent if there exists an invertible linear transformation L : V_{1} ↦ V_{2} so that Lρ_{1}(g) = ρ_{2}(g)L for all g in G, or, what is the same, if ρ_{1}(g) and ρ_{2}(g) are similar for all g in G. In this case, we call L an equivalence between ρ_{1} and ρ_{2}.

Schur’s Lemma. Let ρ_{1} : G ↦ GL(V_{1}) and ρ_{2} : G ↦ GL(V_{2}) be two irreducible representations of G. Then any G-invariant linear transformation L : V_{1} ↦ V_{2} is either the zero transformation, L = 0, or is an equivalence between ρ_{1} and ρ_{2}. To wit, L is either an isomorphism or the zero transformation.

Proof. If L = 0, then there is nothing to prove. Assume, then, L ≠ 0. Now by Lemma 2.6.1, we know Ker(L) is a G-invariant subspace of V_{1}, and, as L ≠ 0, we know Ker(L) ≠ V_{1}. Because V_{1} is irreducible, Ker(L) = {0}, and, thence, L is injective. Similarly, Im(L) is a G-invariant subspace of V_{2} and Im(L) ≠ {0}, since L ≠ 0 and, thus, L is surjective. Therefore, L is bijective and is an equivalence between ρ_{1} and ρ_{2}. Q.E.D.

2.7. An Injectivity Theorem

We use Lemma 2.6.1 to prove the following theorem, which will prove useful when we try to determine whether or not a linear transformation in general, and, specifically, when a Radon Transform, from ℓ^{2}(X) to a vector space is injective and, therefore, invertible.

Theorem 2.7.1. Let ρ : G ↦ GL(V) be a representation and T : ℓ^{2}(X) ↦ V be a G-morphism. Then T is injective if and only if T |_{ℓ^{2}(X)^{Go}} is injective.

Proof. That T is injective evidently guarantees that T |_{ℓ^{2}(X)^{Go}} is injective. To prove the converse, assume to the contrary that T |_{ℓ^{2}(X)^{Go}} is, but T is not, injective. Then Ker(T) ≠ {0}. From Lemma 2.6.1 we know that the subspace, Ker(T), is Ginvariant, and, consequently, by Proposition 2.5.1, Ker(T)^{G_{0}} ≠ {0}, a result that produces the following contradiction:

{0} ≠ Ker(T)^{G_{0}} ⊆ Ker (T |_{ℓ^{2}(X)^{Go}}) = {0},

which proves the assertion that T |_{ℓ^{2}(X)^{Go}} is injective implies T is injective. Q.E.D.

2.8. Radon Transforms Between Finite Grassmanians

As an application, we consider, from [13], Radon transforms between finite Grassmanians. Let 𝔽 be a finite field and 𝔽^{n} the vector space of dimension n over 𝔽. Then GL(𝔽^{n}) is the group of all invertible linear transformations of 𝔽^{n} and Aff(𝔽^{n}) is the group of all invertible affine transformations of 𝔽^{n}. The set of all k-dimensional linear subspaces of 𝔽^{n} we denote G_{k}(𝔽^{n}) and call the Grassmanian of k-dimensional subspaces. With this notation the n-dimensional projective space over 𝔽 is G_{1}(𝔽^{n+1}). Also, AG_{k}(𝔽^{n}) is the set all k-dimensional affine subspaces of 𝔽^{n} and is called the Grassmanian of affine k-planes. Recall from elementary linear algebra that if V and W are two inner product spaces and T : V ↦ W is a linear transformation, then the transformation T^{∗} : W ↦ V for which ⟨T v, w⟩ = ⟨v, T^{∗}w⟩ for all v in V and w in W is the adjoint of T.

For 0 ≤ k < l ≤ n − 1 define the Radon transform R_{k,l} : ℓ^{2}(AG_{k}(𝔽^{n})) ↦ ℓ^{2}(AG_{l}(𝔽^{n})) and its dual, R_{k,l}^{∗} : ℓ^{2}(AG_{l}(𝔽^{n})) ↦ ℓ^{2}(AG_{k}(𝔽^{n})), by

(R_{k,l}ϕ)(P) = ∑_{x⊂P} ϕ(x)

and

(R_{k,l}^{∗}F)(x) = ∑_{P ⊃x} F (P),

respectively. Likewise, for 1 ≤ k < l ≤ n − 1 the projective versions of these transforms, P_{k,l} : ℓ^{2}(G_{k}(𝔽^{n})) ↦ ℓ^{2}(G_{l}(𝔽^{n})) and P_{k,l}^{∗} : ℓ^{2}(G_{l}(𝔽^{n})) ↦ ℓ^{2}(G_{k}(𝔽^{n})), we define by

(P_{k,l}ϕ)(L) = ∑_{x⊂L} ϕ(x)

and

(P_{k,l}^{∗}F)(x) = ∑_{L⊂x} F (L),

respectively.

Define an inner product ℓ^{2}(X) in the usual manner:

⟨ϕ_{1}, ϕ_{2}⟩ := ∑_{x∈X} ϕ_{1}(x)ϕ_{2}(x).

Then the linear transformations R_{k,l} and R_{k,l}^{∗} are adjoint in the sense that

⟨R_{k,l}ϕ, F ⟩ = ∑_{P ⊂Q} ϕ(P)F (Q) = ⟨ϕ, R_{k,l}^{∗}F ⟩.

Therefore, R_{k,l} is injective if and only if R_{k,l}^{∗} is surjective and R_{k,l} is surjective if and only if R_{k,l}^{∗} is injective. Likewise, the maps P_{k,l} and P_{k,l}^{∗} are adjoint.

Theorem 2.8.1. Let 0 ≤ k < l ≤ n − 1.

(a) If k + l ≤ n, then R_{k,l} : ℓ^{2}(AG_{k}(𝔽^{n})) ↦ ℓ^{2}(AG_{l}(𝔽^{n})) is injective, and the

dual map R_{k,l}^{∗} : ℓ^{2}(AG_{l}(𝔽^{n})) ↦ ℓ^{2}(AG_{k}(𝔽^{n})) is surjective.

(b) If k + l ≥ n then R_{k,l} : ℓ^{2}(AG_{k}(𝔽^{n})) ↦ ℓ^{2}(AG_{l}(𝔽^{n})) is surjective, and the

dual map R_{k,l}^{∗} : ℓ^{2}(AG_{l}(𝔽^{n})) ↦ ℓ^{2}(AG_{k}(𝔽^{n})) is injective.

Theorem 2.8.2. Let 1 ≤ k < l ≤ n − 1.

(a) If k + l ≤ n, then P_{k,l} : ℓ^{2}(G_{k}(𝔽^{n})) ↦ ℓ^{2}(G_{l}(𝔽^{n})) is injective, and the dual

map P_{k,l}^{∗} : ℓ^{2}(G_{l}(𝔽^{n})) ↦ ℓ^{2}(G_{k}(𝔽^{n})) is surjective.

(b) If k + l ≥ n then P_{k,l} : ℓ^{2}(G_{k}(𝔽^{n})) ↦ ℓ^{2}(G_{l}(𝔽^{n})) is surjective, and the dual

map P_{k,l}^{∗} : ℓ^{2}(G_{l}(𝔽^{n})) ↦ ℓ^{2}(G_{k}(𝔽^{n})) is injective.

We defer the proof of Theorems 2.8.1 and 2.8.2 until the next section.

2.9. Radon Injectivity Results for Grassmanians

Continuing our example from [13], we claim that the group GL(𝔽^{n}) has a transitive action on G_{k}(𝔽^{n}). Fix L_{0} in G_{k}(𝔽^{n}), and let K = {a ∈ GL(𝔽^{n}) : aL_{0} = L_{0}} be the stabilizer of L_{0}.

Proposition 2.9.1. The orbits of G_{k}(𝔽^{n}) under the action of K are

X_{i} = {L : dim(L ∩ L_{0}) = i} for max(0, 2k − n) ≤ i ≤ k.

Accordingly, the number of orbits of G_{k}(𝔽^{n}) under K is k + 1 for 1 ≤ k ≤ n/2 and 2k − n + 1 for n/2 < k ≤ n − 1. In other notation, Rank_{K}(G_{k}(𝔽^{n})) = k + 1 if 1 ≤ k ≤ n/2 and Rank_{K}(G_{k}(𝔽^{n})) = 2k − n + 1 if n/2 < k ≤ n − 1.

Proof. We will merely sketch the straightforward proof of the proposition. For some L_{1} and L_{2}, say, we can extend a basis of L_{0} ∩ L_{1} to a basis, call it {v_{i}}, for L_{0}. Similarly, we extend a basis of L_{0} ∩ L_{2} to a basis, call it {u_{i}}, for L_{0}, keeping, all the while, the spans of {u_{i}} and {v_{i}} equal. Then we define a unique linear transformation, ℋ, so that ℋv_{i} = u_{i}, whence we determine that ℋ is in K. Q.E.D.

The affine Grassmanians, AG_{k}(𝔽^{n}), are somewhat more complicated. Every P in AG_{k}(𝔽^{n}) is the translation of some k-dimensional linear subspace of 𝔽^{n}. Let ℒ ℒ ℒ(P) in G_{k}(𝔽^{n}) be the translate of P that contains the origin and is, therefore, a linear subspace of 𝔽^{n}. Choose P_{0} in AG_{k}(𝔽^{n}) with 0 in P_{0} so that ℒ ℒ ℒ(P_{0}) = P_{0}, and let K = {a ∈ Aff(𝔽^{n}) : aP_{0} = P_{0}} be the stabilizer of P_{0}.

Proposition 2.9.2. The orbits of AG_{k}(𝔽^{n}) under the action of K are

\begin{matrix}X_{0,i} = \{P : P ∩ P_{0} = ∅, dim(ℒ ℒ ℒ(P) ∩ P_{0}) = i\} \\ X_{1,i} = \{P : P ∩ P_{0} ≠ ∅, dim(ℒ ℒ ℒ(P) ∩ P_{0}) = i\} \end{matrix}

where max(0, 2k − n) ≤ i ≤ k. Hence, Rank_{K}(AG_{k}(𝔽^{n})) = 2(k + 1) if 0 ≤ k ≤ n/2 and Rank_{K}(AG_{k}(𝔽^{n})) = 2(2k − n + 1) if n/2 < k ≤ n − 1.

Proof. This follows from the last proposition by considering the two cases where P ∩ P_{0} = ∅ and P ∩ P_{0} ≠ ∅. Q.E.D.

Now we return to the proof of the theorems we deferred from the previous section.

Proof of Theorem 2.8.1. We first prove (a). Let k+l ≤ n and 0 ≤ k < l ≤ n−1. Choose P_{0} in AG_{k}(𝔽) to use as an origin. We assume that 0 in P_{0} so that ℒ ℒ ℒ(P_{0}) = P_{0}, and let K be the stabilizer of P_{0}. Let X_{0,i} and X_{1,i} be as in (2.9.2). Define functions ϕ_{i} for 0 ≤ i ≤ 2k + 1 by

ϕ_{i}(P) := \{ \begin{matrix}1 & 0 ≤ i ≤ k and P ∈ X_{0,i} \\ 1 & k + 1 ≤ i ≤ 2k + 1 and P ∈ X_{1,i−(k+1)} \\ 0 & otherwise. \end{matrix}

These are the functions that are 1 on precisely one orbit of K and 0 on all other orbits. Because of the condition k +l ≤ n we can choose Q_{j} ∈ AG_{l}(𝔽^{n}) such that Q_{j} ∩P_{0} = ∅ and dim(ℒ ℒ ℒ(Q_{j}) ∩ P_{0}) = j for 0 ≤ j ≤ k and so that if k + 1 ≤ j ≤ 2k + 1, then Q_{j} contains 0 and, hence, ℒ ℒ ℒ(Q_{j}) = Q_{j} and dim(P_{0} ∩ Q_{j}) = j − (k + 1). If P ∈ AG_{k}(𝔽^{n}), P ∈ Q_{j}, and i > j, then ϕ_{i}(P) = 0. For example, if k ≥ i > j then P ⊂ Q_{j} implies P ∩P_{0} = ∅ and ℒ ℒ ℒ(P)∩P_{0} ⊆ ℒ ℒ ℒ(Q)∩P_{0}; thence, dim(ℒ ℒ ℒ(P)∩P_{0}) ≤ dim(ℒ ℒ ℒ(Q_{j})∩P_{0}) = j < i. Thus, P is not in X_{0,i}, so that ϕ_{i}(P) = 0. Similar considerations work in the cases j ≤ k < i and k ≤ j < i. Therefore, R_{k,l}ϕ_{i}(Q_{j}) = 0 whenever j < i. On the other hand, when 0 ≤ i ≤ k, we have c_{i} = |{P ⊂ Q_{j} : P ∈ X_{0,i}}| > 0 and when k + 1 ≤ i ≤ 2k + 1, we also have c_{i} = |{P ⊂ Q_{j} : P ∈ X_{1,i−(k+1)}}| > 0. Therefore, the matrix [R_{k,l}ϕ_{i}(Q_{j})] is triangular, namely,

[R_{k,l}ϕ_{i}(Q_{j})] = [ \begin{matrix}c_{0} & 0 & 0 & ⋯ & 0 \\ ∗ & c_{1} & 0 & ⋯ & 0 \\ ∗ & ∗ & c_{2} & ⋯ & 0 \\ ⋮ & ⋮ & ⋮ & ... & 0 \\ ∗ & ∗ & ∗ & ⋯ & c_{2k+1} \end{matrix} ]

and, as the c_{i} are nonzero, this matrix is nonsingular. But then the functions R_{k,l}ϕ_{i}, when i = 0, ⋯, 2k−1, are linearly independent (If ∑_{i=0}^{2k+1} a_{i}ϕ_{i} = 0, then by evaluating at the Q_{j}s we get a nonsingular system for the a_{i}s). As the functions ϕ_{0}, ⋯, ϕ_{2k+1} are a basis of ℓ^{2}(𝔽^{n})^{K}, the restriction of R_{k,l} to ℓ^{2}(𝔽^{n})^{K}, or, in other notation, R_{k,l}|_{ℓ^{2}(𝔽^{n})^{K}}, is injective. Therefore, by Theorem 2.7.1, R_{k,l} is injective, and R_{k,l}^{∗} surjective, when k + l ≤ n.

Now assume 0 ≤ k < l ≤ n − 1 and k + l ≥ n. We will show R_{k,l}^{∗} is injective. These conditions imply l ≥ n/2. Let Q_{0} in AG_{l}(𝔽^{n}) be so that 0 is in Q_{0} and, thus, ℒ ℒ ℒ(Q_{0}) = Q_{0}. Also, let K = {a ∈ Aff(𝔽^{n}) : aQ_{0} = Q_{0}} be the stabilizer of Q_{0}. Then l ≤ n/2 implies K has (2l − n + 1) orbits on AG_{l}(𝔽^{n}). To simplify notation let r = 2n − l be the codimension of Q_{0}. Then proposition 2.9.2 implies that the orbits of K are

Y_{0,i} = {Q : Q ∩ Q_{0} ≠ ∅, dim(ℒ ℒ ℒ(Q) + Q_{0}) = l + i}
Y_{1,i} = {Q : Q ∩ Q_{0} = ∅, dim(ℒ ℒ ℒ(Q) + Q_{0}) = l + (i − r − 1)}

for 0 ≤ i ≤ r. Define functions F_{i} on AG_{l}(𝔽^{n}) by

F_{i}(Q) := \{ \begin{matrix}1 & 0 ≤ i ≤ r and Q ∈ Y_{0,i} \\ 1 & r + 1 ≤ i ≤ 2r + 1 and P ∈ Y_{1,i−(r+1)} \\ 0 & otherwise. \end{matrix}

Then F_{0}, ⋯, F_{2k+1} is a basis of the isotropic functions ℓ^{2}(AG_{l}(𝔽^{n}))^{K}. Because of the dimension restriction, k + l ≥ n, we can choose elements P_{j} in AG_{k}(𝔽^{n}) so that P_{j} ∩ Q_{0} ≠ ∅, dim(ℒ ℒ ℒ(P_{j}) + Q_{0}) = l + j for 0 ≤ j ≤ r and P_{j} ∩ Q_{0} = ∅, dim(ℒ ℒ ℒ(P_{j}) + Q_{0}) = l + (j − r − 1) for r + 1 ≤ j ≤ 2r + 1. But then by considering the cases 0 ≤ i < j ≤ r, 0 ≤ i ≤ r < j ≤ 2r + 1 and r + 1 ≤ i < j ≤ 2r + 1, we determine that if i < j and Q ⊃ P_{j}, then F_{i}(Q) = 0. Thence, i < j implies R_{k,l}^{∗}F_{i}(P_{j}) = 0. But clearly R_{k,l}^{∗}F_{i}(P_{i}) ≠ 0; whence, [R_{k,l}^{∗}F_{i}(Q_{j})] is a triangular matrix with non-zero elements along the diagonal and is, accordingly, nonsingular, which implies, just as in the previous case, that R_{k,l}^{∗}F_{0}, ⋯, R_{k,l}^{∗}F_{2r+1} are independent which, in turn, implies the restriction of R_{k,l}^{∗} to the isotropic functions ℓ^{2}(𝔽^{n})^{K}, or, equivalently, R_{k,l}^{∗}|_{ℓ^{2}(𝔽^{n})^{K}}, is injective. Therefore, Theorem 2.7.1 implies R_{k,l}^{∗} is injective, and R_{k,l} is surjective by duality. Q.E.D.

Proof of Theorem 2.8.2. An easy variant on the last proof. Q.E.D.

2.10. Convolution in ℓ^{2}(X × X)

Let k be a a function from X × X into ℂ, the complex scalars, or, equivalently, let k be in ℓ^{2}(X × X). Now define 𝒯 𝒯 𝒯_{k} : ℓ^{2}(X) ↦ ℓ^{2}(X) by

(𝒯 𝒯 𝒯_{k}ϕ)(x) = ∑_{y}k(x, y)ϕ(y).

If we view ϕ in ℓ^{2}(X) as a column vector and k as a matrix with entries indexed by X × X, then the linear operator 𝒯 𝒯 𝒯_{k} is matrix multiplication by k.

The following observation will help us prove the subsequent proposition.

Lemma 2.10.1. If 𝒯 𝒯 𝒯_{k}ϕ = 0 for all ϕ in ℓ^{2}(X), then k = 0.

Proof. To see this, let 𝒯 𝒯 𝒯_{k}ϕ = 0. Then 𝒯 𝒯 𝒯_{k}ϕ = ∑_{y}k(x, y)ϕ(y) = 0. If we let z be in X and set

ϕ(y) = \{ \begin{matrix}1 & y = z \\ 0 & y ≠ z, \end{matrix}

then

0 = (𝒯 𝒯 𝒯_{k})(x)

= ∑_{y} k(x, y)ϕ(y) = k(x, z).

Hence, k(x, z) = 0 for all x and z in X, which guarantees k = 0. Q.E.D.

Now define Ψ : ℓ^{2}(X × X) ↦ Hom (ℓ^{2}(X), ℓ^{2}(X)) by Ψ(k) = 𝒯 𝒯 𝒯_{k}. Note that Hom (ℓ^{2}(X), ℓ^{2}(X)) represents the set of all linear operators on ℓ^{2}(X). If Ψ(k) = 0, then 𝒯 𝒯 𝒯_{k} = 0, which means 𝒯 𝒯 𝒯_{k}ϕ = 0 for all ϕ in ℓ^{2}(X). Then, by Lemma 2.10.1, we know that k = 0. Furthermore, without difficulty we see not only that Ψ(k) = 0 implies k = 0, but also that the converse holds: namely, k = 0 implies Ψ(k) = 0. That Ψ(k) = 0 and k = 0 are equivalent reveals Ker(Ψ) = {0}. Now we make a claim about 𝒯 𝒯 𝒯_{k}. If ℒ ℒ ℒ is a linear operator on ℓ^{2}(X), then there exists a unique k such that ℒ ℒ ℒ = 𝒯 𝒯 𝒯_{k}. The truth of the claim we show by noting that dim ℓ^{2}(X × X) = dim Hom (ℓ^{2}(X), ℓ^{2}(X)) = |X|^{2}.

Further, from our remarks above, we know that Ker(Ψ) = 0 and, therefore, since ℓ^{2}(X) is finite-dimensional, the dimension of Im(ℒ ℒ ℒ) is |X|^{2}, by a famous theorem from elementary linear algebra, and, thus, we have shown that ℒ ℒ ℒ is surjective. Thence come the existence and uniqueness of k.

Next, define

k_{1} ∗ k_{2}(x, z) = ∑_{y}k_{1}(x, y)k_{2}(y, z).

We will call the operation, ∗, convolution because of analogues from functional analysis, a good treatment of which can be found in [17] and [4]. If we compose 𝒯 𝒯 𝒯_{k_{1}} with 𝒯 𝒯 𝒯_{k_{2}}, then we have exactly the operator 𝒯 𝒯 𝒯_{k_{1}∗k_{2}} on ℓ^{2}(X), a claim which we prove easily. Let ϕ be in ℓ^{2}(X). Then

(𝒯 𝒯 𝒯_{k_{1}} ∘ 𝒯 𝒯 𝒯_{k_{2}}ϕ)(y) = 𝒯 𝒯 𝒯_{k_{1}}(𝒯 𝒯 𝒯_{k_{2}}ϕ)(y)
= ∑_{y}k_{1}(z, y)(𝒯 𝒯 𝒯_{k_{2}}ϕ)(y)
= ∑_{y}k_{1}(z, y)∑_{x}k_{2}(y, x)ϕ(x)
= ∑_{y}∑_{x}k_{1}(z, y)k_{2}(y, x)ϕ(x)

= ∑_{x} (∑_{y}k_{1}(z, y)k_{2}(y, x)) ϕ(x)
= (𝒯 𝒯 𝒯_{k_{1}∗k_{2}}ϕ)(y).

2.11. The Convolution Ring of X

Now we confirm a few properties of convolution, which afford to the space ℓ^{2}(X × X) much algebraic structure. The first property is that for all k_{1}, k_{2}, and k_{3} in ℓ^{2}(X ×X) we have (k_{1} + k_{2}) ∗ k_{3} = k_{1} ∗ k_{3} + k_{2} ∗ k_{3} and k_{1} ∗ (k_{2} + k_{3}) = k_{1} ∗ k_{2} + k_{1} ∗ k_{3}. In other words, we must show that convolution distributes over addition. If k_{1}, k_{2}, and k_{3} are in ℓ^{2}(X × X), then

k_{1} ∗ (k_{2} + k_{3})(x, y) = ∑_{z}k_{1}(x, z)(k_{2} + k_{3})(z, y)
= ∑_{z}k_{1}(x, z) (k_{2}(z, y) + k_{3}(z, y))
= ∑_{z}k_{1}(x, z)k_{2}(z, y) + ∑_{z}k_{1}(x, z)k_{3}(z, y)
= k_{1} ∗ k_{2}(x, y) + k_{1} ∗ k_{3}(x, y)

Now for the second distributive property:

(k_{1} + k_{2}) ∗ k_{3}(x, y) = ∑_{z}(k_{1} + k_{2})(x, z)k_{3}(z, y)
= ∑_{z}(k_{1}(x, z) + k_{2}(x, z)) k_{3}(z, y)
= ∑_{z}k_{1}(x, z)k_{3}(z, y) + ∑_{z}k_{2}(x, z)k_{3}(z, y)
= k_{1} ∗ k_{3}(x, y) + k_{2} ∗ k_{3}(x, y)

For justification of associativity, recall from above that 𝒯 𝒯 𝒯_{k_{1}} ∘ 𝒯 𝒯 𝒯_{k_{2}} = 𝒯 𝒯 𝒯_{k_{1}∗k_{2}}, which we use to show that

𝒯 𝒯 𝒯_{(k_{1}∗k_{2})∗k_{3}} = 𝒯 𝒯 𝒯_{k_{1}∗k_{2}} ∘ 𝒯 𝒯 𝒯_{k_{3}} = (𝒯 𝒯 𝒯_{k_{1}} ∘ 𝒯 𝒯 𝒯_{k_{2}}) ∘ 𝒯 𝒯 𝒯_{k_{3}} = 𝒯 𝒯 𝒯_{k_{1}} ∘ (𝒯 𝒯 𝒯_{k_{2}} ∘ 𝒯 𝒯 𝒯_{k_{3}}) = 𝒯 𝒯 𝒯_{k_{1}∗(k_{2}∗k_{3})},

and 𝒯 𝒯 𝒯_{(k_{1}∗k_{2})∗k_{3}}= 𝒯 𝒯 𝒯_{k_{1}∗(k_{2}∗k_{3})} reveals that, indeed, (k_{1}∗k_{2})∗k_{3} = k_{1}∗(k_{2}∗k_{3}). In addition to the structure that convolution adds to ℓ^{2}(X × X), we also know that ℓ^{2}(X × X) is a group under function addition, which we will not prove here. The properties and structure we have detailed above evince that (ℓ^{2}(X × X), +, ∗) form a ring, which we will call the convolution ring of X. Of course, we have merely disguised matrix multiplication in a form that will prove useful for later computations.

If V and W are two inner product spaces and T : V ↦ W is a linear transformation, then the transformation T^{∗} : W ↦ V for which ⟨T v, w⟩ = ⟨v, T^{∗}w⟩ for all v in V and w in W is the adjoint of T. Furthermore, if T^{∗} = T, then we say that T is self-adjoint. When a linear transformation is defined on finite inner product spaces, its adjoint always exists and is always unique, although we will not provide a proof here. We can also extend the definition of self-adjoint to make sense in terms of a collection of linear operators. Thus, if 𝒜 𝒜 𝒜 is a collection of linear operators on a finite-dimensional inner product space, then 𝒜 𝒜 𝒜 is self-adjoint if A is a member of 𝒜 𝒜 𝒜 if and only if A^{∗} is in 𝒜 𝒜 𝒜. Finally, we will call a linear operator, T, unitary if T^{∗}T = T T^{∗} = I.

Next we define, if k is in ℓ^{2}(X × X), a sort of transpose of k, denoted k^{∗}, such that k^{∗}(x, y) = \overline{k(y, x)}. Now since 𝒯 𝒯 𝒯_{k} is a linear operator on ℓ^{2}(X), we know that 𝒯 𝒯 𝒯_{k}, because dim ℓ^{2}(X) < ∞, must have an adjoint. We will prove that 𝒯 𝒯 𝒯_{k}^{∗} = 𝒯 𝒯 𝒯_{k^{∗}}. For the proof, let ϕ_{1} and ϕ_{2} be in ℓ^{2}(X). Then, using our standard inner product,

⟨𝒯 𝒯 𝒯_{k}ϕ_{1}, ϕ_{2}⟩ = ∑_{x}(𝒯 𝒯 𝒯_{k}ϕ_{1}) (x)\overline{ϕ_{2}(x)}

= ∑_{x}(∑_{y}k(x, y)ϕ_{1}(y))\overline{ϕ_{2}(x)}
= ∑_{y}ϕ_{1}(y)∑_{x}k(x, y)\overline{ϕ_{2}(x)}
= ∑_{y}ϕ_{1}(y)∑_{x}\overline{k^{∗}(y, x)} \overline{ϕ_{2}(x)}
= ∑_{y}ϕ_{1}(y)(\overline{𝒯 𝒯 𝒯_{k^{∗}}ϕ_{2})(y)} = ⟨ϕ_{1},𝒯 𝒯 𝒯_{k^{∗}}ϕ_{2}⟩

We have already shown that if X is a G-space, then ℓ^{2}(X) is also a G-space. We extend the action of G, presently, a step further by asserting that if ℓ^{2}(X) is a G-space, then ℓ^{2}(X × X) is likewise. For if X is a G-space, then so is X × X via g(x, y) = (gx, gy). Thus, ℓ^{2}(X × X) is a G-space by (gk)(x, y) = k(g^{−1}x, g^{−1}y).

2.12. The Convolution Algebra ℓ^{2}(X × X)^{G}

Finally, we are in a position to consider the set ℓ^{2}(X × X)^{G}, which is

{k ∈ ℓ^{2}(X × X) : k(gx, gy) = k(x, y) for all x, y ∈ X and g ∈ G}.

In words, then, ℓ^{2}(X × X)^{G} is the subspace of ℓ^{2}(X × X) of functions invariant under the group action of G. We conclude easily that ℓ^{2}(X × X)^{G} is closed under convolution. For if k_{1} and k_{2} are in ℓ^{2}(X × X)^{G}, then

(k_{1} ∗ k_{2})(gx, gy) = ∑_{z}k_{1}(gx, z)k_{2}(z, gy)
= ∑_{gz}k_{1}(gx, gz)k_{2}(gz, gy)
= ∑_{z}k_{1}(gx, gz)k_{2}(gz, gy)
= ∑_{z}k_{1}(x, z)k_{2}(z, y) = (k_{1} ∗ k_{2})(x, y).

Therefore, the convolution of two functions in ℓ^{2}(X × X)^{G} is again in ℓ^{2}(X × X)^{G}. We present, next, a proposition that guarantees that the action of G commutes with 𝒯 𝒯 𝒯_{k} if k is in ℓ^{2}(X × X)^{G}, or, symbolically, 𝒯 𝒯 𝒯_{k}gϕ = g𝒯 𝒯 𝒯_{k}ϕ for all g in G and for every ϕ in ℓ^{2}(X).

Proposition 2.12.1. Let k be in ℓ^{2}(X ×X), g in G, and x and y be in X. Suppose X be a G-space. Then the following are equivalent.

(1) 𝒯 𝒯 𝒯_{k} ∘ τ_{g} = τ_{g} ∘ 𝒯 𝒯 𝒯_{k}

(2) k(gx, gy) = k(x, y)

(3) k(gx, y) = k(x, g^{−1}y)

Proof. Suppose, first, that (3) holds. Then

k(x, y) = k(g^{−1}gx, y) = k(gx, (g^{−1})^{−1}y) = k(gx, gy);

thence, we have shown that (3) implies (2). Next, we suppose that (2) holds. Then k(gx, y)=k(g^{−1}gx, g^{−1}y)=k(x, g^{−1}y), and we see that, indeed, (3) holds. Finally, we will show that (1) and (3) are equivalent. Let ϕ be in ℓ^{2}(X) and g in G. Note, first, that

(𝒯 𝒯 𝒯_{k} ∘ τ_{g})ϕ(x) = (𝒯 𝒯 𝒯_{k}(τ_{g}ϕ))(x)
= ∑_{y}k(x, y)τ_{g}ϕ(y)
= ∑_{y}k(x, y)ϕ(g^{−1}y)
= ∑_{y}k(x, gy)ϕ(g^{−1}gy)
= ∑_{y}k(x, gy)ϕ(y)
= 𝒯 𝒯 𝒯_{k_{1}}ϕ(x)

where k_{1}(x, y) = k(x, gy). Likewise,

(τ_{g} ∘ 𝒯 𝒯 𝒯_{k})ϕ(x) = (𝒯 𝒯 𝒯_{k}ϕ)(g^{−1}x)
= ∑_{y}k(g^{−1}x, y)ϕ(y)
= 𝒯 𝒯 𝒯_{k_{2}}ϕ(y)

where k_{2}(x, y) = k(g^{−1}x, y). Therefore, 𝒯 𝒯 𝒯_{k} ∘ τ_{g} = τ_{g} ∘𝒯 𝒯 𝒯_{k} if and only if 𝒯 𝒯 𝒯_{k_{1}} = 𝒯 𝒯 𝒯_{k_{2}}. That is, (1) holds if and only if k_{1} = k_{2}, which implies (1) and (3) are equivalent, and we have completed our proof. Q.E.D.

Proposition 2.12.1, in addition to proving that the action, τ, of G commutes with 𝒯 𝒯 𝒯_{k} for k in ℓ^{2}(X × X)^{G}, gives us different ways to characterize that the action of G commutes with 𝒯 𝒯 𝒯_{k}. Considering our work above, we note that the subspace, ℓ^{2}(X × X)^{G}, of ℓ^{2}(X × X), inasmuch as it is closed under convolution, is an algebra in which the function

δ(x, y) = \{ \begin{matrix}1 & x = y \\ 0 & x ≠ y. \end{matrix}

serves as the identity. We call the algebra ℓ^{2}(X × X)^{G} the convolution algebra of X.

2.13. The Relationship Between ℓ^{2}(X)^{G_{o}} and ℓ^{2}(X × X)^{G}

Let us review our basic structural setup so far. We have a finite set, X, which is a transitive G-space. Further, we have fixed a point, call it o, in X as our origin. Then ℓ^{2}(X)^{G_{o}} is the subspace of ℓ^{2}(X) consisting of functions that are fixed by the elements of G_{o}, the o-stabilizers in G. Recall from 2.4.1 that dim ℓ^{2}(X)^{G_{o}} = Rank_{G_{o}}(X), which is the number of orbits that result when G_{o} acts on X.

We will proceed to show that the Rank_{G_{o}}(X) = dim ℓ^{2}(X)^{G_{o}} = r is also equal to the dimension of our convolution algebra, ℓ^{2}(X ×X)^{G}. To demonstrate this, we define a linear transformation E : ℓ^{2}(X)^{G_{o}} ↦ ℓ^{2}(X × X)^{G} by (Eϕ)(x, y) = ϕ(ξ^{−1}y), where ξ is in G and ξo = x. We see straightaway that E is well-defined, for if ξ′o = x = ξo, then surely ξ^{−1}ξ′ = g_{0} for some g_{0} in G_{o}, which gives us ξ′ = ξg_{0}. To finish our demonstration that E is well-defined, we remark that ϕ(ξ′^{−1}y) = ϕ((ξg_{0})^{−1}y) from our work above. But ϕ((ξg_{0})^{−1}y) = ϕ(g_{0}^{−1}ξ^{−1}y) = ϕ(ξ^{−1}y), the last equality holding because g_{0}, and therefore g_{0}^{−1}, is in G_{o}. Next, consider the linear transformation F : ℓ^{2}(X × X)^{G} ↦ ℓ^{2}(X)^{G_{o}} defined by (F k)(y) = k(o, y), for k in ℓ^{2}(X × X)^{G}. We will show that F does, in fact, send k to an isotropic function. To that end, let g_{0} be in G_{o} and consider (F k)(g_{0}y)=k(o, g_{0}y)=k(g_{0}^{−1}o, y)= k(o, y) = (F k)(y), and so F behaves correctly. We are in a position, now, to prove a proposition about the linear transformations F and E.

Proposition 2.13.1. F and E, as defined above, are inverses to each other. Therefore, dim ℓ^{2}(X)^{G_{o}} = dim ℓ^{2}(X × X)^{G} = Rank_{G_{o}}(X).

Proof. Let ϕ be in ℓ^{2}(X)^{G_{o}}, and consider (F Eϕ)(y) = Eϕ(o, y) = ϕ(ξ^{−1}y), where ξ is in G_{o}. Next, if we set ξ equal to e, the identity in G, which, as G_{o} is a subgroup of G, is also in G_{o}, we get ϕ(ξ^{−1}y) = ϕ((e)^{−1}y) = ϕ(ey) = ϕ(y). Accordingly, (F Eϕ)(y) = ϕ(y). What remains to be shown is that (EF k)(x, y) = k(x, y). Let k be in ℓ^{2}(X × X), our convolution algebra, and consider (EF k)(x, y) = (F k)(ξ^{−1}y) with ξo = x. Then (F k)(ξ^{−1}y) = k(o, ξ^{−1}y) = k(ξo, y) = k(x, y). Thence, we have shown that F and E are inverses to each other, and, as dim ℓ^{2}(X)^{G_{o}} = Rank_{G_{o}}(X), by Proposition 2.4.1, the present proposition holds. Q.E.D.

Finite Gel′fand Spaces

3.1. Collections of Linear Operators

Let T be a linear operator on a vector space V and λ a complex scalar. If there exists a vector v in V for which the equation T v = λv holds, then we will call λ an eigenvalue and v and eigenvector of T. We will always associate eigenvectors to corresponding eigenvalues, and accordingly, we will call the set of all eigenvectors of an eigenvalue, λ, the eigenspace of λ. Without proof, we comment that eigenspaces are subspaces. The preceding definitions direct us to the following observation. Let T_{1} and T_{2} be linear operators on a vector space V. If T_{1} and T_{2} commute, then any eigenspace of T_{1} is invariant under T_{2}. The claim is easily proven, for if E_{λ} is the eigenspace corresponding to λ, an eigenvector of T_{1}—that is, E_{λ}= {v ∈ V : T_{1}v = λv}, then T_{1}T_{2}v = T_{2}T_{1}v = T_{2}λv, which evinces that T_{2}v is, indeed, in E_{λ}.

Recall from Chapter 2 that if 𝒜 𝒜 𝒜 is a collection of linear operators on a finitedimensional inner product space, then 𝒜 𝒜 𝒜 is self-adjoint if A is a member of 𝒜 𝒜 𝒜 if and only if A^{∗} is in 𝒜 𝒜 𝒜.

Proposition 3.1.1. Let 𝒜 𝒜 𝒜 be a set of commuting linear operators on a vector space, V. Then 𝒜 𝒜 𝒜 has a common eigenspace, or, in symbols, there exists an eigenspace E corresponding some λ in V, with E ≠ {0}, such that every v in E is an eigenvector for all A in 𝒜 𝒜 𝒜.

Proof. We proceed with induction on dim V. If dim V = 1, the result is clear. Let us assume, then, that the proposition is true for all spaces of dimension less than dim V. If every A is of the form A = αI, where I is the identity in V, then every v in V is an eigenvector of every A in 𝒜 𝒜 𝒜. In that case, we use E = V. Else, there exists an A with an eigenspace, E_{A} = {v ∈ V : Av = λv}, such that dim E_{A} < dim V. Now let 𝒜 𝒜 𝒜|_{E_{A}} = {B|_{E_{A}} : B ∈ 𝒜 𝒜 𝒜}. By our claim in the preceding paragraph, E_{A} is invariant under each B in A, and, therefore, 𝒜 𝒜 𝒜|_{E_{A}} is a commuting set of linear operators on E_{A}. By the induction hypothesis, 𝒜 𝒜 𝒜|_{E_{A}} has a common eigenspace, which will also be a common eigenspace for 𝒜 𝒜 𝒜. Q.E.D.

Let S be a subset of a vector space V. Then we call {v ∈ V : ⟨v, s⟩ = 0 for all s in S} the orthogonal complement of S, which we denote S^{⊥}.

Proposition 3.1.2. Let 𝒜 𝒜 𝒜 be a self-adjoint collection of linear operators on a finite-dimensional inner product space, V. If a subspace, W, of V is invariant under 𝒜 𝒜 𝒜, then W^{⊥} is also invariant under 𝒜 𝒜 𝒜.

Proof. Let w_{0} be in W^{⊥}, w in W, and let A be any member of 𝒜 𝒜 𝒜. Then ⟨Aw_{0}, w⟩ = ⟨w_{0}, A^{∗}w⟩ = 0, as A^{∗} is in 𝒜 𝒜 𝒜, and, thus, A^{∗}w is in W. Accordingly, ⟨Aw_{0}, w⟩ = 0 for all w in W. Thence, we know that Aw_{0} is in W^{⊥}, and, because we chose A arbitrarily from 𝒜 𝒜 𝒜, we have shown that W^{⊥} is invariant under 𝒜 𝒜 𝒜. Q.E.D.

Proposition 3.1.3. If ρ : G ↦ GL(V) is a unitary representation, then the set of all ρ(g) such that g is in G is self-adjoint collection of linear operators on V.

Proof. Let 𝒜 𝒜 𝒜 = {ρ(g) : g ∈ G}. We will show that for all g in G, we have ρ(g)^{∗} = ρ(g^{−1}) = ρ(g)^{−1}. Note that ρ(g^{−1}) = ρ(g)^{−1}, as ρ is a representation and, therefore, a homomorphism. Thence,

⟨ρ(g)w_{1}, w_{2}⟩ = ⟨ρ(g^{−1})ρ(g)w_{1}, ρ(g^{−1})w_{2}⟩
= ⟨ρ(g^{−1}g)w_{1}, ρ(g^{−1})w_{2}⟩
= ⟨ρ(e)w_{1}, ρ(g^{−1})w_{2}⟩
= ⟨ew_{1}, ρ(g^{−1})w_{2}⟩
= ⟨w_{1}, ρ(g^{−1})w_{2}⟩

which implies ρ(g)^{∗} = ρ(g^{−1}). Now ρ(g) is a member of 𝒜 𝒜 𝒜 if and only if ρ(g^{−1}) = ρ(g)^{∗} is in 𝒜 𝒜 𝒜. Therefore, 𝒜 𝒜 𝒜 is self-adjoint. Q.E.D.

As a corollary to the above proposition, we note that if ρ : G ↦ GL(V) is a unitary representation and W is a subspace of V invariant under ρ, then W^{⊥} is invariant under ρ.

Let V be an inner product space, and suppose V can be written as a direct sum, V = V_{1} ⊕ V_{2} ⊕ ⋯ ⊕ V_{r} with the added property that if v_{i} is in V_{i} and v_{j} in V_{j}, then ⟨v_{i}, v_{j}⟩ = 0 when i ≠ j. Then we will call

V_{1} ⊕ V_{2} ⊕ ⋯ ⊕ V_{r} = ⨁_{n=1}^{r} V_{n}

the orthogonal decomposition of V. We now present the following formulation of the Spectral Theorem from elementary linear algebra.

Spectral Theorem for Commuting, Self-Adjoint Linear Operators. Let V be a finite-dimensional inner product space, and let 𝒜 𝒜 𝒜 be a self-adjoint and commutative collection of linear operators on V. Then there exist nonzero functionals, α_{1}, ⋯, α_{r} : 𝒜 𝒜 𝒜 ↦ ℂ, so that if

V_{α_{i}} = {v ∈ V : Av = α_{i}(A)v for all A ∈ 𝒜 𝒜 𝒜},

then V_{α_{i}} ≠ {0} and there exists the following orthogonal decomposition:

V = V_{0} ⊕ ⨁_{i=1}^{r} V_{α_{i}}

where V_{0} = {v ∈ V : Av = 0 for all A ∈ 𝒜 𝒜 𝒜} and V_{0} is, possibly, {0}.

Proof. We prove our Spectral Theorem based on induction on dim V. Surely, the base case of dim V = 1 is true. Assume that dim V = n and that the result holds on all inner product spaces of dimension less than n. If 𝒜 𝒜 𝒜 is only scalar multiples of the identity, I_{V} , then the result is trivial. Let us assume, then, that there exists a A_{0} in 𝒜 𝒜 𝒜 that is not a scalar multiple of I_{V} and, therefore, has at least one eigenvalue, call it λ. Let E be the eigenspace corresponding to λ. Then {0} ≠ E ≠ V. Now for any B in 𝒜 𝒜 𝒜 and v in E we have A_{0}Bv = BA_{0}v = λBv, as all members of 𝒜 𝒜 𝒜 commute. Thence, E is invariant under all elements of 𝒜 𝒜 𝒜. Furthermore, by Proposition 3.1.2, E^{⊥} is also invariant under 𝒜 𝒜 𝒜. Now if we apply the induction hypothesis to 𝒜 𝒜 𝒜|_{E} = {B|_{E^{⊥}} : B ∈ 𝒜 𝒜 𝒜} and 𝒜 𝒜 𝒜|_{E^{⊥}}, we have completed the proof. Q.E.D.

We will say that a linear operator, call it A, is normal if AA^{∗} = A^{∗}A. Easily, we see that if we let 𝒜 𝒜 𝒜 = {A, A^{∗}}, then 𝒜 𝒜 𝒜 is self-adjoint and we obtain a special case of our Spectral Theorem for normal operators.

3.2. Symmetric G-Spaces

We reconsider, next, our G-space, X, and say that it is symmetric if for every x and y in X, there exists an element of G, call it g, such that gx = y and gy = x. Restating the definition in plainer terms, we consider a special type of G-space with the property that, given any two elements of X, there exists a group element whose action interchanges them. We call such an X a symmetric G-space. We see easily that if X is symmetric, then X is also a transitive G-space. Now we are ready to prove a theorem that relates the symmetry of X with the commutativity of its convolution algebra. The following theorem is attributable to I. M. Gel′fand .

Theorem 3.2.1. If X is a symmetric G-space, then the convolution algebra of X, ℓ^{2}(X × X)^{G}, is commutative.

Proof. Let k be a member of ℓ^{2}(X ×X)^{G} and X be a symmetric G-space. To begin our proof, we will remark that k(x, y) = k(y, x). For k(x, y) = k(gx, gy) = k(y, x), since we have chosen the g in G that interchanges x and y. Now let k_{1} and k_{2} be in ℓ^{2}(X × X)^{G}, and consider

(k_{1} ∗ k_{2})(x, y) = ∑_{z}k_{1}(x, z)k_{2}(z, y)
= ∑_{z}k_{2}(z, y)k_{1}(x, z)
= ∑_{z}k_{2}(y, z)k_{1}(z, x) , by our remark above,
= (k_{2} ∗ k_{1})(y, x),

which is again in ℓ^{2}(X × X)^{G}, as ℓ^{2}(X × X)^{G} is closed under convolution. Again, then, by our remark, (k_{2}∗k_{1})(y, x) = (k_{2}∗k_{1})(x, y), which proves that (k_{1}∗k_{2})(x, y) = (k_{2} ∗ k_{1})(x, y) and, therefore, the theorem. Q.E.D.

We mention, now, a few examples of symmetric G-spaces. The first is the size k subsets of the set of n letters, {1, ⋯, n}, under the natural action of the group of permutations, S_{n}. (For the definition of S_{n} see Section 4.1.) Next, let G_{k}(𝔽^{n}) and AG_{k}(𝔽^{n}) be as in Section 2.8. Then the actions of GL(𝔽^{n}) on G_{k}(𝔽^{n}) and Aff(𝔽^{n}) on AG_{k}(𝔽^{n}) are symmetric.

In Chapter 2 we defined, for k ∈ ℓ^{2}(X × X), a linear operator, 𝒯 𝒯 𝒯_{k}, on ℓ^{2}(X) by

(𝒯 𝒯 𝒯_{k}ϕ)(x) = ∑_{y}k(x, y)ϕ(y).

Theorem 3.2.1 above shows that the set {𝒯 𝒯 𝒯_{k} : k ∈ ℓ^{2}(X × X)^{G}} is a commuting, self-adjoint set of linear operators.

Inspired by Theorem 3.2.1, we formulate the following definition. Let X be a transitive G-space. Then we will call X a Gel′fand space if ℓ^{2}(X × X)^{G} is commutative. If X is a symmetric G-space, then we know from Theorem 3.2.1 that X is a Gel′fand space.

3.3. The Cartan-Gel′fand Theorem

We now fix and review some notation. As above, let us choose an origin, call it o, in X, and let G_{o} be the set of o-stabilizers in G. Next, if E is a G-invariant subspace of ℓ^{2}(X), then let

E^{G_{o}} = {ϕ ∈ E : τ_{g}ϕ = ϕ for all g ∈ G_{o}}

be the isotropic functions of E. Because 𝕋 = {𝒯 𝒯 𝒯_{k} : k ∈ ℓ^{2}(X × X)^{G}} is a commuting, self-adjoint set of linear operators, ℓ^{2}(X) can be diagonalized simultaneously or, equivalently, can be decomposed into an orthogonal direct sum by our Spectral Theorem on page 33. We will call a nonzero linear functional α : ℓ^{2}(X × X)^{G} ↦ ℂ a weight if

E_{α} = {ϕ : 𝒯 𝒯 𝒯_{k}ϕ = α(k)ϕ for all k ∈ ℓ^{2}(X × X)^{G}} ≠ {0}.

Therefore, by our Spectral Theorem, we know that if X is a Gel′fand space, then 𝕋 is commutative and self-adjoint, and there exist weights, call them α_{1}, ⋯, α_{r}, such that

ℓ^{2}(X) = E_{0} ⊕ E_{α_{1}} ⊕ ⋯ ⊕ E_{α_{r}}

where E_{0} = {ϕ ∈ ℓ^{2}(X) : 𝒯 𝒯 𝒯_{k}ϕ = 0 for all k ∈ ℓ^{2}(X × X)^{G}} ≠ {0}.

Furthermore, we will call E_{α_{i}}, if 0 ≤ i ≤ r, the weight space corresponding to α_{i}.

We provide, now, two lemmata, which we will find useful in our proof of the Cartan-Gel′fand Theorem to follow.

Lemma 3.3.1. If ϕ is in E_{α}^{G_{o}} and ϕ(o) = 0, then ϕ ≡ 0.

Proof. Let ϕ be in ℓ^{2}(X)^{G_{o}}. Then we know from remarks leading up to Proposition 2.13.1 that there exists a k_{0} in ℓ^{2}(X × X)^{G} so that k_{0}(o, y) = \overline{ϕ(y)}. Now

0 = α(k_{0})ϕ(o) = (𝒯 𝒯 𝒯_{k_{0}}ϕ)(o) = ∑_{y}k_{0}(o, y)ϕ(y) = ∑_{y}|ϕ(y)|^{2},

which shows that ϕ ≡ 0. Q.E.D.

Lemma 3.3.2. Let ϕ_{1} and ϕ_{2} be in ℓ^{2}(X). Then there exists a constant c_{α}(ϕ_{1}, ϕ_{2}) so that for all ϕ in E_{α},

∑_{g∈G} ∑_{y∈X}ϕ_{1}(g^{−1}x)ϕ_{2}(g^{−1}y)ϕ(y) = c_{α}(ϕ_{1}, ϕ_{2})ϕ(x).

Proof. Suppose h(x, y) = ∑_{g∈G}ϕ_{1}(g^{−1}x)ϕ_{2}(g^{−1}y). We will show that h is in ℓ^{2}(X × X)^{G}. To that end, let ξ be in G and consider

h(ξx, ξy) = ∑_{g∈G}ϕ_{1}(g^{−1}ξx)ϕ_{2}(g^{−1}ξy)
= ∑_{g∈G}ϕ_{1}((ξ^{−1}g)^{−1}x)ϕ_{2}((ξ^{−1}g)^{−1}y)
= ∑_{ξg∈G}ϕ_{1}((ξ^{−1}ξg)^{−1}x)ϕ_{2}((ξ^{−1}ξg)^{−1}y), if we set g = ξg
= ∑_{g∈G}ϕ_{1}(g^{−1}x)ϕ_{2}(g^{−1}y) = h(x, y).

Thence, h is in ℓ^{2}(X × X)^{G} as claimed. Therefore, for any ϕ in E_{α}, we have that 𝒯 𝒯 𝒯_{h}ϕ = α(h)ϕ, which is equivalent to the statement of the lemma, with c_{α}(ϕ_{1}, ϕ_{2}) = α(h). Q.E.D.

The following theorem is a discrete analogue of the results due to E. J. Cartan and I. M. Gel′fand .

Theorem 3.3.1. (Cartan-Gel′fand) If X is a finite Gel′fand space, then there exist weights, α_{1}, ⋯, α_{r}, so that

(1) ℓ^{2}(X) = E_{α_{1}} ⊕ ⋯ ⊕ E_{α_{r}} (orthogonal direct sum)

(2) Each E_{α_{i}}^{G_{o}} is one-dimensional and is spanned by a unique element, p_{α_{i}}, with p_{α_{i}}(o) = 1, called the spherical function in E_{α_{i}}^{G_{o}}.

(3) Each weight space, E_{α_{i}}, with 1 ≤ i ≤ r, is irreducible.

(4) If i ≠ j, then E_{α_{i}} and E_{α_{j}} are not equivalent as representations.

(5) If E ≠ {0} is any irreducible, G-invariant subspace of ℓ^{2}(X), then E = E_{α_{i_{0}}} for some i_{0}.

(6) r = Rank_{G_{o}}(X).

Proof. We already know that ℓ^{2}(X) = E_{0} ⊕ E_{α_{1}} ⊕ ⋯ E_{α_{r}}. Therefore to prove (1), we need merely to show that E_{0} = {0}. Let ϕ be in E_{0}. Then 𝒯 𝒯 𝒯_{k}ϕ = 0 for all k in ℓ^{2}(X × X)^{G}. Now set

k(x, y) = δ(x, y) = \{ \begin{matrix}1 & x = y \\ 0 & x ≠ y. \end{matrix} ,

the identity in the convolution algebra, ℓ^{2}(X × X)^{G}.

Hence,

0 = 𝒯 𝒯 𝒯_{k}ϕ(x) = ∑_{y}k(x, y)ϕ(y)

= ∑_{y}δ(x, x)ϕ(y)
= ϕ(y),

whereby we have shown ϕ = 0 and (1) holds.

To prove that each E_{α_{i}}^{G_{o}} is one-dimensional and is spanned by a unique element we will use Lemma 3.3.1. We know that we can always choose a p_{α} in E_{α_{i}}^{G_{o}} such that p_{α}(o) = 1, by Proposition 2.5.1. Now let ϕ be in E_{α_{i}}^{G_{o}}. Note that ϕ_{1}(o) = ϕ(o) − ϕ(o)p_{α}(o) = 0, as p_{α}(o) = 1. Further, by Lemma 3.3.1, we have shown that ϕ_{1} = 0 and so ϕ = ϕ(o)p_{α} and, hence, that (2) holds.

Next, we show that each E_{α_{i}}^{G_{o}} is irreducible. To that end, assume to the contrary that a weight space, E_{α_{i_{0}}}, is reducible. Then we can decompose E_{α_{i}}^{G_{o}} orthogonally as follows, E_{α_{i}}^{G_{o}} = W_{1} ⊕ W_{2}, where W_{1} and W_{2} are G-invariant subspaces of ℓ^{2}(X). But we know that W_{1}^{G_{o}} ≠ {0} ≠ W_{2}^{G_{o}} by Proposition 2.5.1. Therefore, E_{α_{i}}^{G_{o}} ≥ 2, which is a contradiction, inasmuch as, by (2) above, each E_{α_{i}} is one-dimensional. Accordingly, we have shown that (3) holds.

Now we show that if α and β are weights, then E_{α} and E_{β} are not isomorphic and, therefore, are not equivalent representations. Let ρ be a representation and define χ_{ρ} : G ↦ ℂ by χ_{ρ}(g) = Trace(ρ(g)). Then we call χ_{ρ} the character of ρ. If two linear transformations are isomorphic, then they have the same character, because isomorphic transformations are similar and similar transformations have the same trace.

Let χ_{α}(g) = Trace(τ_{g}|_{E_{α}}) and χ_{β}(g) = Trace(τ_{g}|_{E_{β}}) be the characters of the representation τ restricted to E_{α} and E_{β}. Let {ϕ_{1 α}, ϕ_{2 α}, ⋯, ϕ_{l α}} and {ϕ_{1 β}, ϕ_{2 β}, ⋯, ϕ_{m β}} be unitary bases of E_{α} and E_{β}, respectively. Then, since the trace is the sum of the diagonal elements of the matrix of τ|_{E_{α}}, we have that

χ_{α}(g) = ∑_{i}⟨τ_{g}ϕ_{i α}, ϕ_{i α}⟩.

Similarly, we recognize that

χ_{β}(g) = ∑_{j}⟨τ_{g}ϕ_{j β}, ϕ_{j β}⟩.

Consider, next,

⟨χ_{α}, χ_{β}⟩ = ∑_{g∈G} χ_{α}(g)\overline{χ_{β}(g)}

= ∑_{g∈G} ∑_{i} ∑_{j}⟨τ_{g}ϕ_{i α}, ϕ_{i α}⟩\overline{⟨τ_{g}ϕ_{j β}, ϕ_{j β}⟩}

= ∑_{g∈G} ∑_{i} ∑_{j} ∑_{x} ∑_{y} ϕ_{i α}(g^{−1}x)\overline{ϕ_{i α}(x)} \overline{ϕ_{j β}(g^{−1}y)}ϕ_{j β}(y)

= ∑_{i} ∑_{j} ∑_{x} (∑_{g∈G} ∑_{y} ϕ_{i α}(g^{−1}x)\overline{ϕ_{j β}(g^{−1}y)}ϕ_{j β}(y)) \overline{ϕ_{i α}(x)}

= ∑_{i} ∑_{j} ∑_{x} c_{α}(ϕ_{i α}, \overline{ϕ}_{j β})ϕ_{j β}(x)\overline{ϕ_{i α}(x)}, by Lemma 3.3.2,

= ∑_{i} ∑_{j} c_{α}(ϕ_{i α}, \overline{ϕ}_{j β}) ∑_{x} ϕ_{j β}(x)\overline{ϕ_{i α}(x)}

= ∑_{i} ∑_{j} c_{α}(ϕ_{i α}, \overline{ϕ}_{j β})⟨ϕ_{j β}, ϕ_{i α}⟩

= 0,

since ϕ_{j β} is in E_{β} and ϕ_{i α} is in E_{α} and E_{α} is orthogonal to E_{β}. Now suppose to the contrary that E_{α} and E_{β} are isomorphic. Then χ_{α} = χ_{β}. But then

0 = ∑_{g∈G}|χ_{α}(g)|^{2} > 0,

a contradiction, which evinces, then, that E_{α} and E_{β} are not isomorphic, or, what is the same, E_{α} and E_{β} are not equivalent as representations; thus, (4) holds.

Let us suppose that E ≠ {0} is an irreducible, G-invariant subspace of ℓ^{2}(X), and let π_{i} : ℓ^{2}(X) ↦ E_{α_{i}} be the orthogonal projection. Then π_{i} is G-invariant. As E is a G-invariant subspace of ℓ^{2}(X), we know by Schur’s Lemma that for each i, the projection restricted to E, π|_{E} : E ↦ E_{α_{i}}, is either an isomorphism or the zero transformation. Because E ≠ {0} and, by (4), no two weight spaces are isomorphic, we conclude that for every i but one, call it i_{0}, π_{i} = 0 and π_{i_{0}} is an isomorphism. Therefore, E = E_{α_{i_{0}}}, and we have proven (5).

To prove (6), we recall from Proposition 2.4.1 that Rank_{G_{o}}(X) = dim ℓ^{2}(X)^{G_{o}}. Now dim ℓ^{2}(X)^{G_{o}} = dim(E_{α_{1}} ⊕ ⋯ ⊕ E_{α_{r}})^{G_{o}} by (1) above. Further,

dim(E_{α_{1}} ⊕ ⋯ ⊕ E_{α_{r}})^{G_{o}} = dim(E_{α_{1}}^{G_{o}} ⊕ ⋯ ⊕ E_{α_{r}}^{G_{o}}) = r,

inasmuch as, by (3), we know that each of the weight spaces, E_{α_{i}}, where 1 ≤ i ≤ r, is one-dimensional, and we have, indeed, shown that (6) holds. Accordingly, we have proven the Cartan-Gel′fand Theorem.

3.4. G-Invariant Linear Operators and Inversion Formulae

We next turn our attention to results obtained when we consider the G-invariant operators in the mathematical setting described in Theorem 3.3.1, the Cartan-Gel′fand Theorem, and we will use the notation of that theorem throughout the following. First, we see that if L is a G-invariant linear operator on ℓ^{2}(X) and X is a Gel′fand space, then we can realize the restriction of L on E_{α_{i}} by scalar multiplication and find an explicit inverse for L.

Theorem 3.4.1. Let X be a Gel′fand space and let L : ℓ^{2}(X) ↦ ℓ^{2}(X) be a Ginvariant linear operator—that is, Lτ_{g} = τ_{g}L for all g in G. Then for all i, LE_{α_{i}} ⊆ E_{α_{i}} holds and L|_{E_{α_{i}}} = c_{i}I_{E_{α_{i}}}, where c_{i} = (Lp_{α_{i}})(o). In particular, L is invertible if and only if (Lp_{α_{i}})(o) ≠ 0 for all i. In this case, the inverse of L is given by

L^{−1} = ∑_{i=1}^{r} \frac{1}{Lp_{α_{i}}(o)}π_{i},

where π_{i} : ℓ^{2}(X) ↦ E_{α_{i}} is an orthogonal projection.

Proof. Because LE_{α_{i}} is a G-invariant subspace of ℓ^{2}(X) and, as E_{α_{i}} is irreducible, LE_{α_{i}} is either {0} or is an isomorphic to E_{α_{i}}. Suppose, first, that LE_{α_{i}} = {0}. Then LE_{α} ⊆ E_{α_{i}} and, therefore, L|_{E_{α_{i}}} = 0I_{E_{α_{i}}}. The first part of the theorem, then, holds for the case LE_{α_{i}} = {0}.

Now let us consider the case when LE_{α_{i}} is isomorphic to E_{α_{i}}. Then LE_{α_{i}} = E_{α_{i}} by parts (4) and (5) of Theorem 3.3.1. Note that τ_{g}Lp_{α_{i}} = Lτ_{g}p_{α_{i}} = Lp_{α_{i}}, for g in G_{o}, the last equality holding because p_{α_{i}} is in E_{α_{i}}^{G_{o}}. Thence, Lp_{α_{i}} is in E_{α_{i}}^{G_{o}}, and because E_{α_{i}}^{G_{o}} is one-dimensional, Lp_{α_{i}} = c_{i}p_{α_{i}} for some scalar c_{i}. Then ker(L|_{E_{α_{i}}} − c_{i}I_{E_{α_{i}}}) is a G-invariant subspace of E_{α_{i}}, and since E_{α_{i}} is irreducible, we know that ker(L|_{E_{α_{i}}} − c_{i}I_{E_{α_{i}}}) = E_{α_{i}}; hence, ker(L|_{E_{α_{i}}} − c_{i}I_{E_{α_{i}}}) must be the zero transformation, and we L|_{E_{α_{i}}} = c_{i}I_{E_{α_{i}}}. From above we know that Lp_{α_{i}}(o) = c_{i}p_{α_{i}}(o), and so c_{i} = (Lp_{α_{i}})(o) follows, as p_{α_{i}}(o) = 1.

We will show, next, that the inverse of L is as claimed in the statement of the theorem. Let

S = ∑_{i=1}^{r} \frac{1}{Lp_{α_{i}}(o)}π_{i}.

Then

LS = ∑_{i=1}^{r} c_{i}π_{i} ∑_{j=1}^{r} \frac{1}{Lp_{α_{j}}(o)}π_{j}

= ∑_{i=1}^{r} c_{i}π_{i} ∑_{j=1}^{r} \frac{1}{c_{j}} π_{j}, since Lp_{α_{j}}(o) = c_{j},

= ∑_{i=1}^{r} \frac{c_{i}}{c_{i}} π_{i}^{2}, because π_{i}π_{j} = 0 for i ≠ j

= ∑_{i=1}^{r} π_{i}, as π_{i} is a projection,

= I_{ℓ^{2}(X)}.

Similarly, SL = I_{ℓ^{2}(X)} and, therefore, the inverse of L is given by

L^{−1} = ∑_{i=1}^{r} \frac{1}{Lp_{α_{i}}(o)}π_{i},

as claimed. Q.E.D.

Let V and W be inner product spaces and L : V ↦ W be a linear transformation. We claim that L is injective if and only if L^{∗}L, a linear operator on V, is injective; for Lv = 0 is equivalent to ∥Lv∥^{2} = 0, where ∥v_{0}∥, called the norm of v_{0}, we define, as usual, to be \sqrt{⟨v_{0}, v_{0}⟩}. Now ∥Lv∥^{2} = 0 is the same as ⟨Lv, Lv⟩ = 0. Finally, we have 0 = ⟨Lv, Lv⟩ = ⟨L^{∗}Lv, v⟩, which proves the claim, which, in turn, we use in the proof of the theorem that follows.

Theorem 3.4.2. Let L : ℓ^{2}(X) ↦ V be a G-invariant linear transformation. Then the following are equivalent.

(1) L is injective

(2) Lp_{α_{i}}(o) ≠ 0, if 1 ≤ i ≤ r

(3) The restriction, L|_{ℓ^{2}(X)^{Go}}, of L to the isotropic functions, ℓ^{2}(X)^{G_{o}}, is injective.

Furthermore, if L is injective, any ϕ in ℓ^{2}(X) is recovered from Lϕ by

ϕ = (∑_{i=1}^{r} \frac{1}{L^{∗}Lp_{α_{i}}(o)}π_{i}L^{∗}) Lϕ.

Proof. By Theorem 2.7.1 , we have already shown that (1) is equivalent to (3). Now by the claim above, we know that L is injective if and only if L^{∗}L is injective, which is equivalent to the statement

(L^{∗}Lp_{α_{i}})(o) ≠ 0, 1 ≤ i ≤ r.

To prove that the inversion formula holds, let L be injective. Now if we apply the inversion results in Theorem 3.4.1 to the linear operator, L^{∗}L, on ℓ^{2}(X), then

(L^{∗}L)^{−1} = (∑_{i=1}^{r} \frac{1}{L^{∗}Lp_{α_{i}}(o)}π_{i}L^{∗}) L.

Therefore,

ϕ = (L^{∗}L)^{−1}ϕ = (∑_{i=1}^{r} \frac{1}{L^{∗}Lp_{α_{i}}(o)}π_{i}L^{∗}) Lϕ.

and we have proven the theorem. Q.E.D.

Doubly Transitive Group Actions

4.1. Introduction and Definitions

Heretofore we have been interested in transitive group actions on sets and have constructed our injectivity theorems under the assumption that the set X with which we have been working is a transitive G-space. Now let X be a G-space, and suppose that, for all (x_{1}, x_{2}) and (y_{1}, y_{2}) in X × X such that x_{1} ≠ y_{1} and x_{2} ≠ y_{2}, there exists a g in G such that gx_{1} = x_{2} and gy_{1} = y_{2}. Then we will say that the action of G on X is doubly transitive and will call X a doubly transitive G-space.

Evidently, if X is a doubly transitive G-space, then X is a symmetric G-space. Consequently, by Theorem 3.2.1, if the action of G on X is doubly transitive, then X is a Gel′fand space. Therefore, the injectivity and orthogonal decomposition results from Chapter 3 hold. As in the previous chapters, fix some o in X to serve as the origin, and let G_{o} = {g ∈ G : go = o} be the o-stabilizers in G. Then we claim that X is a doubly transitive G-space if and only if there are exactly two orbits of X under the action of the o-stabilizers, G_{o} : namely, the singleton orbit X_{1} = {o} and the orbit X_{2} = X − {o} or, what is the same, all of X except the origin.

Recall from elementary group theory that the group of permutations on a set, X = {1, 2, ⋯, n}, of n letters is the collection, S_{n}, of bijections from X to X. Furthermore, the natural action of S_{n} on X we will describe as follows. If σ is a member of S_{n} such that

σ = ( \begin{matrix}1 & 2 & 3 & ⋯ & n \\ σ(1) & σ(2) & σ(3) & ⋯ & σ(n) \end{matrix} ) ,

then define Ψ : S_{n} × X ↦ X by Ψ(σ, x) = σ(x), for all x in X and σ in S_{n}, or, as we will write the action Ψ henceforth, σx = σ(x).

4.2. The Radon Transform on X = {1, ⋯, n}

Now set G = S_{n} and X = {1, 2, ⋯, n}; then X is a doubly transitive G-space under the natural action of G = S_{n} on X described in the previous paragraph. Next, let Y = {S ⊆ X : |S| = k}. Then G acts on Y in the obvious way, a natural extension of the action of G on X—that is, if g is in G, then gS = {gs : s ∈ S}. Set, finally, o = 1 as our origin in X. Thus,

G_{o} = {g ∈ G : g1 = 1}
= {σ ∈ S_{n} : σ(1) = 1}
= {σ ∈ S_{n} : σ({2, ⋯, n}) = {2, ⋯, n}},

The last equality, with some renaming, evinces that G_{o} = S_{n−1}.

Next, define R : ℓ^{2}(X) ↦ ℓ^{2}(Y) by the natural Radon transform,

(Rϕ)(S) = ∑_{x∈S} ϕ(x).

Now let

ϕ_{1}(x) = \{ \begin{matrix}1 & x = 1 \\ 0 & x ≠ 1 \end{matrix}

and

ϕ_{2}(x) = \{ \begin{matrix}0 & x = 1 \\ 1 & x ≠ 1 \end{matrix}

Then {ϕ_{1}, ϕ_{2}} is a basis of ℓ^{2}(X)^{G_{o}}. The image of the basis vectors, ϕ_{1} and ϕ_{2}, is

(Rϕ_{1})(S) = ∑_{x∈S} ϕ_{1}(x) = \{ \begin{matrix}1 & 1 ∈ S \\ 0 & 1 ∉ S \end{matrix}

and

(Rϕ_{2})(S) = ∑_{x∈S} ϕ_{2}(x) = \{ \begin{matrix}k − 1 & 1 ∈ S \\ k & 1 ∉ S \end{matrix} .

If we suppose that ϕ_{0} = c_{1}ϕ_{1} + c_{2}ϕ_{2} and that ϕ_{0} ∈ Ker (R|_{ℓ^{2}(X)^{Go}}) , then

Rϕ_{0} = c_{1}Rϕ_{1} + c_{2}Rϕ_{2} = 0.

Let S_{1} = {1, ⋯, k} and S_{2} = {2, ⋯, k + 1}. Now if we evaluate Rϕ_{0} at S_{1} and S_{2}, we get the following pair of homogeneous equations.

Rϕ_{0}(S_{1}) = c_{1}Rϕ_{1}(S_{1}) + c_{2}Rϕ_{2}(S_{1}) = 0
Rϕ_{0}(S_{2}) = c_{1}Rϕ_{1}(S_{2}) + c_{2}Rϕ_{2}(S_{2}) = 0,

which, by our results above, show that c_{1} = c_{2} = 0, which, in turn, reveals that {Rϕ_{1}, Rϕ_{2}} is linearly independent. Therefore,

Ker (R|_{ℓ^{2}(X)^{Go}}) = {0}

and, because dim ℓ^{2}(X) < ∞, we know, by a famous theorem from elementary linear algebra, that R|_{ℓ^{2}(X)^{Go}} is injective.

By Theorem 3.4.2, then, R is injective.

Because X is a doubly transitive G-space, we know from our remarks above that the orbits of X under G_{o} must be X_{1} = {o} = {1} and X_{2} = X −{o} = {2, 3, ⋯, n}. By the results, then, of Theorem 3.3.1, ℓ^{2}(X) = E_{1} ⊕ E_{2}, as r = 2, in the notation of that theorem. Because these two subspaces of ℓ^{2}(X) are both G-invariant, the weight space E_{1} comprises the set of all constant functions and E_{2}, the set of all functions that sum to zero, or, in symbols,

E_{2} = {ϕ ∈ ℓ^{2}(X) : ∑_{x∈X} ϕ(x) = 0}.

Because ℓ^{2}(X) is the orthogonal sum E_{1} ⊕ E_{2}, to see that E_{2} is as we have described, we must simply show that E_{2} is orthogonal to E_{1}. To that end, let ϕ_{0} be in E_{2} and consider

0 = ⟨ϕ_{0}, 1⟩ = ∑_{x∈X} ϕ_{0}(x),

which implies that E_{2} is, indeed, the set of functions that sum to zero.

Certainly, as E_{1} is the set of constant functions on X, we have that p_{1}(x) ≡ 1 is the spherical function corresponding to E_{1}. Furthermore, the spherical function of E_{2} will be of the form

p_{2}(x) = \{ \begin{matrix}1 & x = 1 \\ C_{2} & x ≠ 1 \end{matrix} , where C_{2} = \frac{−1}{|X| − 1}.

That p_{2}(x) is the spherical function of E_{2} follows easily from

0 = p(1) + p(2) + ⋯ + p(|X|)
= 1 + (|X| − 1)C_{2},

which gives us the value for C_{2}. The orthogonal projections of ℓ^{2}(X) onto E_{1} and E_{2} we claim, respectively, are

π_{1}ϕ(x) = \frac{1}{|X|} ∑_{y∈X} ϕ(y)

and

π_{2}ϕ(x) = ϕ(x) − \frac{1}{|X|} ∑_{y∈X} ϕ(y),

the latter projection following because, as π_{1} and π_{2} are orthogonal projections, we can decompose ϕ into π_{1}ϕ + π_{2}ϕ; thence, ϕ = π_{1}ϕ + π_{2}ϕ. Solving for π_{2}ϕ gives us

π_{2}ϕ(x) = ϕ(x) − π_{1}ϕ(x) = ϕ(x) − \frac{1}{|X|} ∑_{y∈X} ϕ(y),

with substitution of π_{1} from above justifying the latter equality. We next give the image of the spherical functions, p_{1}(x) and p_{2}(x), under the Radon transform R : namely,

(Rp_{1})(S) = ∑_{x∈S} p_{1}(x) = |S| = k

and

(Rp_{2})(S) = ∑_{x∈S} p_{2}(x) = \{ \begin{matrix}1 + (k − 1)C_{2} & 1 ∈ S \\ kC_{2} & 1 ∉ S \end{matrix} .

There exists, as well, a natural adjoint of the Radon transform R. Define R^{∗} : ℓ^{2}(Y) ↦ ℓ^{2}(X) by

(R^{∗}F)(x) = ∑_{S∋x} F (S),

if ϕ is in ℓ^{2}(X) and F is in ℓ^{2}(Y), for

⟨ϕ, R^{∗}F ⟩ = ∑_{x∈X} ϕ(x)\overline{(R^{∗}F)(x)}

= ∑_{x∈X} ϕ(x)\overline{∑_{S∋x} F (S)}

= ∑_{S∋x} ∑_{x∈X} ϕ(x)\overline{F (S)}

= ∑_{S∋x}(Rϕ)\overline{F (S)} = ⟨Rϕ, F ⟩.

Finally, we claim that the image of p_{1} and p_{2} under the adjoint, R^{∗}, evaluated at 1 are

(R^{∗}Rp_{1})(1) = ∑_{S∋1}(Rp_{1})(S)

= ∑_{S∋1} k, from our work above,

= \binom{|X| − 1}{k − 1}k

and

(R^{∗}Rp_{2})(1) = ∑_{S∋1}(Rp_{2})(S)

= \binom{|X| − 1}{k − 1} [1 + (k − 1)C_{2}] .

4.3. The Radon Transform for Doubly Transitive Actions

Notice that in the example above of the Radon transform on the space X = {1, ⋯, n}, we used no special characteristics of S_{n} or X other than that X is doubly transitive under the action of S_{n}. Therefore, the resulting decomposition of ℓ^{2}(X) into weight spaces, the spherical functions, and the adjoint of the Radon transform all generalize nicely as follows. We will omit some of the details, inasmuch as many of these specifics are directly analogous to those in the special case outlined above.

Let X be any doubly transitive G-space and let G_{o} be as before for a fixed origin, o. Then, again, G_{o} has two orbits, X_{1} = {o} and X_{2} = X − {o}, and from the Cartan-Gel′fand Theorem, that is Theorem 3.3.1, ℓ^{2}(X) = E_{1} ⊕ E_{2}, and, once more,

E_{1} = {ϕ ∈ ℓ^{2}(X) : ϕ(x) = c_{0} for all x ∈ X},

where c_{0} is a constant—that is, E_{1} is the set of all constant functions—and

E_{2} = {ϕ ∈ ℓ^{2}(X) : ∑_{x∈X} ϕ(x) = 0}.

Then the spherical functions are, as above,

p_{1}(x) ≡ 1

and

p_{2}(x) = \{ \begin{matrix}1 & x = o \\ C_{2} & x ≠ o \end{matrix} , where C_{2} = \frac{−1}{|X| − 1}.

Furthermore, the orthogonal projections of ℓ^{2}(X) onto E_{1} and E_{2} are given exactly as above by, respectively,

π_{1}ϕ(x) = \frac{1}{|X|} ∑_{y∈X} ϕ(y)

and

π_{2}ϕ(x) = ϕ(x) − \frac{1}{|X|} ∑_{y∈X} ϕ(y).

Let L_{0} be a nonempty subset of X other than X itself, and let \overline{X} = \{gL_{0} : g ∈ G\} be the set of G-translates of L_{0}. If \overline{K} = \{g ∈ G : gL_{0} = L_{0}\}, then |\overline{X}| = |G||\overline{K}|, as G/\overline{K} ≅ \overline{X}. There exists, then, a natural Radon transform R R R : ℓ^{2}(X) ↦ ℓ^{2}(\overline{X}) given by

(R R Rϕ)(L) = ∑_{x∈L} ϕ(x).

Furthermore, R R R has a dual transformation R R R^{∗} : ℓ^{2}(\overline{X}) ↦ ℓ^{2}(X) defined by

(R R R^{∗}F)(x) = ∑_{L∋x} F (L).

In this case, we remark that R R R^{∗} is the adjoint of R R R in the following sense:

⟨R R Rϕ, F ⟩_{ℓ^{2}(\overline{X})} = ∑_{x∈L} ϕ(x)\overline{F (L)} = ⟨ϕ,R R R^{∗}F ⟩_{ℓ^{2}(X)}.

Therefore, R R R is injective if and only if R R R^{∗} is surjective.

The image of the spherical functions, p_{1} and p_{2}, under R R R is

(R R Rp_{1})(L) = ∑_{x∈L} p_{1}(x) = |L| = |L_{0}|

and

(R R Rp_{2})(L) = ∑_{x∈L} p_{2}(x) = \{ \begin{matrix}\frac{|X| − |L_{0}|}{|X| − 1} & o ∈ L \\ \frac{−|L_{0}|}{|X| − 1} & o ∉ L \end{matrix}

If x is in X, let m = |\{L ∈ \overline{X} : x ∈ L\}| be the number of elements of \overline{X} that contain x. Then by counting the pairs (x, L) with x in L in two ways—that is, first summing on x and then L, or vice versa, we have

m = \frac{|L_{0}||\overline{X}|}{|X|}.

Then the images of R R Rp_{1} and R R Rp_{2} under R R R^{∗} are

R R R^{∗}R R Rp_{1}(o) = m|L_{0}|

and

R R R^{∗}R R Rp_{2}(o) = m\frac{|X| − |L_{0}|}{|X|}.

As the operator R R R^{∗}R R R is G-invariant, if we apply the results of Theorem 3.4.2, we establish

Theorem 4.3.1. let X be a doubly transitive G-space. Then the Radon transform R R R : ℓ^{2}(X) ↦ ℓ^{2}(\overline{X}) is injective and any ϕ in ℓ^{2}(X) we can recover from R R Rϕ by

ϕ = \frac{1}{m} (\frac{1}{|L_{0}|}π_{1}R R R^{∗} + \frac{|X|}{|X| − |L_{0}|}π_{2}R R R^{∗})R R Rϕ,

where

m = \frac{|L_{0}||\overline{X}|}{|X|}

and π_{1} : ℓ^{2}(X) ↦ E_{1} and π_{2} ↦ E_{2} are orthogonal projections. By duality, the transform R R R^{∗} : ℓ^{2}(\overline{X}) ↦ ℓ^{2}(X) is surjective.

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