Segal-Bargmann transformation

Universiteit GentFaculteit Wetenschappen

Vakgroep Wiskundige Analyse

The quantum mechanicalSegal-Bargmann transform using

Jordan algebrasSigiswald Barbier

Academiejaar 2013-2014

Promotoren:Prof. Dr. H. De BieDr. K. Coulembier

Masterproef ingediend tot het behalenvan de academische graad van master inde wiskunde, afstudeerrichting wiskundigenatuurkunde en sterrenkunde.

Dankwoord

In de eerste plaats wil ik mijn beide promotoren bedanken. Zowel voor het aanreiken vanhet onderwerp van mijn thesis, dat op een natuurlijke manier analyse, algebra en wiskundigenatuurkunde verbindt, de drie takken van de wiskunde die me het meest boeien, als voor debegeleiding tijdens het voorbije jaar. Dankzij hun verbeteringen en aanwijzingen op wiskundigen taalkundig vlak, is het werk dat nu voor u ligt veel beter dan dat ik het ooit alleen zoukunnen maken hebben.Ook wil ik mijn ouders bedanken. Ze hebben van jongs af aan mijn wetenschappelijk denkengestimuleerd en hebben me ook alle kansen gegeven om te studeren. Verder bedank ik hen ookvoor hun oprechte interesse in mijn thesis, hoewel ik weet dat die voor niet-wiskundigen zeerlastig te begrijpen is.Als laatste wil ik mijn medestudenten bedanken, niet alleen voor hun hulp bij wiskundige enLATEX-problemen, maar vooral voor hun vriendschap de voorbije vijf jaar. In het bijzonder wilik mijn vriendin Melissa bedanken, die een veel grotere toeverlaat was en is dan ze zelf ooit zalbeseffen.

Toelating tot bruikleen

De auteur geeft de toelating deze masterproef voor consultatie beschikbaar te stellen en delenvan de masterproef te kopieren voor persoonlijk gebruik. Elk ander gebruik valt onder debeperkingen van het auteursrecht, in het bijzonder met betrekking tot de verplichting de bronuitdrukkelijk te vermelden bij het aanhalen van resultaten uit deze masterproef.

The author gives his permission to make this work available for consultation and to copy partsof the work for personal use. Any other use is bound by the restrictions of copyright legislation,in particular regarding the obligation to specify the source when using the results of this work.

30 mei 2014Sigiswald Barbier

iii

Contents

Preface vii

1 The classical Segal-Bargmann transform 1

1.1 The quantum harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 The Segal-Bargmann transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 The weight function ρn . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.2 The integral kernel An . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 The Fock space F(Cn) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3.1 Reproducing kernel Hilbert spaces . . . . . . . . . . . . . . . . . . . . . 4

1.3.2 The inner product on the Fock space . . . . . . . . . . . . . . . . . . . . 6

1.3.3 An orthonormal basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4 Unitarity of the Segal-Bargmann transform . . . . . . . . . . . . . . . . . . . . 8

1.4.1 Holomorphic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.4.2 Injectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.4.3 Unitarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.5 Orthonormal functions on L2(Rn) . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.6 The harmonic oscillator on the Fock space . . . . . . . . . . . . . . . . . . . . . 12

2 Jordan algebras 13

2.1 The symmetric matrices Sym(n,R) . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2 The operator L(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.3 The trace form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.4 The Peirce decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.4.1 Idempotents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.4.2 Jordan frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.5 The structure algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5.1 Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.5.2 The structure algebra is equal to gl(n,R). . . . . . . . . . . . . . . . . . 26

3 The Bessel operators 29

3.1 Orbits of Sym(n,R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.1.1 The open cone Ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.1.2 The minimal orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.1.3 Integration on the minimal orbit . . . . . . . . . . . . . . . . . . . . . . 34

3.2 Operators on Sym(n,R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.2.1 The gradient ∇ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.2.2 The operator P . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.2.3 The Bessel operators Bλ . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.3 The Bessel operator is tangential . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.3.1 The determinant det(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.3.2 The zeta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

v

vi

3.3.3 Tangential operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.4 The Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4 The Schrodinger model 514.1 The conformal algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.2 A representation of the conformal algebra on C∞(O) . . . . . . . . . . . . . . . 52

4.2.1 The representation dπ is well defined . . . . . . . . . . . . . . . . . . . . 544.2.2 The representation dπ is a Lie algebra representation . . . . . . . . . . . 56

4.3 A representation on L2(O,dµ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.3.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.3.2 A subrepresentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.3.3 Extension to L2(O, dµ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5 The Fock model 735.1 The complex symmetric matrices . . . . . . . . . . . . . . . . . . . . . . . . . . 735.2 The complex minimal orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745.3 The Fock space on the complex minimal orbit . . . . . . . . . . . . . . . . . . . 765.4 The Lie algebra automorphism c of gC . . . . . . . . . . . . . . . . . . . . . . . 785.5 The Lie algebra representation dπC . . . . . . . . . . . . . . . . . . . . . . . . . 81

6 An intertwining operator 856.1 The new Segal-Bargmann transform . . . . . . . . . . . . . . . . . . . . . . . . 856.2 Connection with the classical Segal-Bargmann transform . . . . . . . . . . . . . 876.3 The Segal-Bargmann transform as an intertwining operator . . . . . . . . . . . 886.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

A Nederlandstalige samenvatting 95

Bibliography 97

Preface

The classical Segal-Bargmann is an integral transform between the Schrodinger space of square-integrable functions and the Fock space of holomorphic functions. It was introduced byBargmann [1] in 1961 and gives an isomorphism between two formulations of quantum mechan-ics. Recently, in Mollers [6] and Hilgert et al. [4], the Segal-Bargmann transform is reinterpretedas an intertwining operator between different realizations of the minimal representation of aLie algebra. This Lie algebra is constructed using Jordan algebras.

In [6] Mollers associates with a simple Euclidean Jordan algebra certain algebraic structureslike the structure group, the conformal group and their corresponding Lie algebras. He alsostudies the minimal orbit of the structure group, which is an important subset of the Jordanalgebra. This minimal orbit is used to define a minimal representation of (a finite cover of)the conformal group on the space of square-integrable functions defined on the minimal orbit.This gives us the so-called Schrodinger model.

In [4], the authors construct another realization of this minimal representation, this time onthe complexification of the minimal orbit. This gives rise to the so-called Fock model. Theythen use the differential representations of these two models to define a generalization of theSegal-Bargmann transform as an integral transform that intertwines the Schrodinger and Fockmodel.

The aim of this thesis is to gain a better understanding of this construction by studying indetail the specific case where the Jordan algebra consists of the real, symmetric n×n-matricesSym(n,R). In our case, it will be sufficient to consider only Lie algebras and Lie algebrarepresentations and not their corresponding Lie groups and integrated representations. Theadvantage of this approach is that we will only be needing basic analysis and algebraic tools.Moreover, a lot of proofs are considerably shorter and more elementary than in the general case.By looking at a concrete example, it is also possible to prove some theorems by straightforwardcalculations, because we have a good understanding of the objects we are working with.

The contents of this thesis are structured as follows. In the first chapter we will introduce theclassical Segal-Bargmann in the same way as it was first studied in the article of Bargmann [1].In particular we will prove that it connects the ladder operators on the Schrodinger space,which are well known from quantum mechanics, with the operators z and ∂z on the Fockspace. We will also show that the Segal-Bargmann transform is a unitary mapping betweenthese two spaces.

In the second chapter we will introduce Jordan algebras and look in detail at Sym(n,R). Wewill define an inner product and give a decomposition using idempotent elements. We will alsodefine the structure algebra.

In the third chapter we will continue our study of Jordan algebras. We will consider certainorbits which are subsets of the Jordan algebra Sym(n,R). Of these orbits, we will be mostlyinterested in the minimal orbit. In particular we will find a relation between functions on theminimal orbit and even functions on Rn. We will also study the Bessel operators, which willplay an important role in the definition of the representations later on. We will be spendingquite some time to prove that these Bessel operators are tangential to the orbits of the Jordan

vii

viii

algebra.In the fourth chapter we will be able to construct the Schrodinger model. First we will definethe conformal algebra associated with the Jordan algebra Sym(n,R). Then we will constructthe Schrodinger representation as a Lie algebra representation of the conformal algebra onthe space of smooth functions on the minimal orbit. Last we will show that we can redefinethe Schrodinger representation as a representation on the square-integrable functions on theminimal orbit. The material of chapter two to four is mostly based on Mollers [6] and Farautand Koranyi [3], where it is treated in a more general setting.We will start the fifth chapter with a study of the Jordan algebra of the complex symmet-ric matrices and the complexification of the minimal orbit. We will define a Fock space onthis complex minimal orbit and find an isomorphism between this Fock space and the evenpart of the Fock space we studied in the first chapter. We will then consider a Lie algebraautomorphism of the complex conformal algebra and combine this automorphism with thecomplexification of the representation found in the fourth chapter to obtain a representationwhich leads to the Fock model.In the final chapter we will construct an integral transform between the square-integrablefunctions on the minimal orbit and the Fock space on the complex minimal orbit. We willshow that there is a precise connection with the Segal-Bargmann transform we studied in thefirst chapter. We will then prove that this integral transform gives an isomorphism betweenthe two representations we studied in chapter four and five. These last two chapters are basedon Hilgert et al. [4].

Chapter 1

The classical Segal-Bargmanntransform

The goal of this chapter is to find a unitary integral transformation from the Hilbert spaceof square-integrable functions to a space of holomorphic functions, called the Fock space. Wewant this transformation to be such that the ladder operator a† corresponds to the operator z,the ladder operator a corresponds to the operator ∂z and that the operators z and ∂z areadjoint operators with respect to the inner product on the Fock space.

We will first discuss the quantum harmonic oscillator and define the associated ladder operatorsa† and a. Then we will use the conditions that the Segal-Bargmann transform has to satisfyto establish an expression for the integral kernel of the Segal-Bargmann transform and forthe weight function of the Fock space. We will examine this Fock space in detail and provethat it is a reproducing kernel Hilbert space. We will then prove the main theorem of thischapter, namely that the Segal-Bargmann transform is unitary. Finally, we will see that theHermite functions arise naturally using the Segal-Bargmann transform and we will find thatthe quantum harmonic oscillator has an easy expression on the Fock space.

In this chapter we follow mainly the article of Bargmann [1], although the proof of unitarity isbased on the book of Olafsson [7].

1.1 The quantum harmonic oscillator

The Hamiltonian of the n-dimensional harmonic oscillator with unit mass, unit angular fre-quency and the reduced Planck constant ~ equal to one is given by

H =n∑i=1

p2i

2+q2i

2,

where qi is the position operator and pi = −i∂qi the momentum operator. These operators areself-adjoint and satisfy the following commutation relations:

[qi, pj ] = iδij , [qi, qj ] = 0, [pi, pj ] = 0.

We define the ladder operators by

ai =1√2

(qi + ipi) =1√2

(qi + ∂qi) (1.1a)

a†i =1√2

(qi − ipi) =1√2

(qi − ∂qi). (1.1b)

1

2 Chapter 1. The classical Segal-Bargmann transform

The ladder operators are adjoint with respect to the standard inner product on L2(Rn). Itfollows from the commutation relations of qi and pi that

[ai, a†j ] = δij , [ai, aj ] = 0, [a†i , a

†j ] = 0.

We can express the Hamiltonian with these ladder operators. This yields

H =n

2+

n∑i=1

a†iai.

We also have the following commutation relations between H and the ladder operators:

[H, a†i ] = a†i , [H, ai] = −ai.

If |ψ〉 is an eigenstate of H with energy E, then ai|ψ〉 and a†i |ψ〉 are also eigenstates with energyE − 1 and E + 1, because

Hai|ψ〉 = aiH|ψ〉+ [H, ai]|ψ〉= (E − 1)ai|ψ〉,

and

Ha†i |ψ〉 = a†iH|ψ〉+ [H, a†i ]|ψ〉

= (E + 1)a†i |ψ〉.

Hence we observe that ai destroys a unit of energy, while a†i creates a unit of energy. Therefore

ai is called an annihilation operator and a†i a creation operator.

1.2 The Segal-Bargmann transform

In this section we will determine how the Segal-Bargmann transform SB and the Fock spacelook like. In particular we want to find a weight function ρn that gives the inner product onthe Fock space and an integral kernel An such that SB is given by

SB(ψ)(z) =

∫RnAn(z, q)ψ(q)dnq,

where z = (z1, . . . , zn)t with zk = xk + iyk in C. We will further use the notations ∂zk =12(∂xk − i∂yk) and zz =

∑nk=1 zkzk =

∑nk=1(x2

k + y2k).

1.2.1 The weight function ρn

The operators zk and ∂zk have to be adjoint on the Fock space F(Cn). This leads to thefollowing equation, which ρn has to satisfy,

(zkf, g)F(Cn) =

(f,

∂g

∂zk

)F(Cn)

(1.2)

or

∫Cnzkfgρndnz =

∫Cnf∂g

∂zkρndnz.

Using integration by parts, we get∫Cnf∂g

∂zkρndnz =

∫Cn

∂

∂zk(fgρn)dnz −

∫Cn

∂f

∂zkgρndnz −

∫Cnfg∂ρn∂zk

dnz.

1.2. The Segal-Bargmann transform 3

If we only consider rapidly decreasing functions for ρn, i.e. functions that go to zero whenz goes to infinity, then the first integral on the right-hand side vanishes. For a holomorphicfunction f , we also have ∂f/∂zk = 0. In that case the equation for ρn reduces to∫

Cnzkfgρndnz = −

∫Cnfg∂ρn∂zk

dnz. (1.3)

This suggestsρn = ce−zz, (1.4)

with c a constant that we choose equal to π−3n2 .

Indeed, a calculation gives us

∂ρn∂zk

=1

2

(∂ρn∂xk− i∂ρn

∂yk

)=

1

2(−2xkρn + i2ykρn)

= −zkρn,

and we conclude that our choice for ρn satisfies (1.3).

1.2.2 The integral kernel An

In this section we will obtain an expression for the integral kernel An(z, q). The Segal-Bargmanntransform has to satisfy the following two diagrams:

L2(Rn) F(Cn)

L2(Rn) F(Cn)

ak

SB

∂zk

SB

and

L2(Rn) F(Cn)

L2(Rn) F(Cn)

a†k

SB

zk

SB

.

Here L2(Rn) is the Hilbert space of square-integrable functions on Rn and F(Cn) is the Fockspace we will introduce later in this chapter. We can express the two conditions that theSegal-Bargmann transform has to satisfy as∫

RnAn(z, q)

(akψ(q)

)dnq =

∂

∂zk

∫RnAn(z, q)ψ(q)dnq

and ∫RnAn(z, q)

(a†kψ(q)

)dnq = zk

∫RnAn(z, q)ψ(q)dnq.

Using these two equations and the fact that ak and a†k are adjoint with respect to the L2-innerproduct, we get ∫

Rn

(akAn(z, q)

)ψ(q)dnq =

∫RnAn(z, q)

(a†kψ(q)

)dnq

=

∫RnzkAn(z, q)ψ(q)dnq

and ∫Rn

(a†kAn(z, q)

)ψ(q)dnq =

∫RnAn(z, q)

(akψ(q)

)dnq

=

∫Rn

∂An(z, q)

∂zkψ(q)dnq.


We conclude

zkAn(z, q) = akAn(z, q) =1√2

(qk + ∂qk)An(z, q)

and

∂An(z, q)

∂zk= a†kAn(z, q) =

1√2

(qk − ∂qk)An(z, q),

where we have used the definitions of the ladder operators (1.1). We can rewrite this as

∂An(z, q)

∂qk= (√

2zk − qk)An(z, q) (1.5a)

∂An(z, q)

∂zk= (√

2qk − zk)An(z, q). (1.5b)

An integral kernel of the form

An(z, q) = c exp

(−1

2(z2 + q2) +

√2z · q

)satisfies (1.5), where c is a constant which we choose equal to one and where we use the notationq2 =

∑nk=1 q

2k, z

2 =∑n

k=1 z2k and q · z =

∑nk=1 qkzk. We remark that we have chosen different

constants for the integral kernel An(z, q) and the weight function ρn than the ones in the articleof Bargmann [1], but that this makes no significant difference.

We can calculate the following integral:∫RnAn(z, q)An(w, q)dnq =

∫Rne−

12

(z2+q2)+√

2z · qe−12

(w2+q2)+√

2w · qdnq

= e−12

(z2+w2)e(z+w)2

2

∫Rne−(q− z+w√

2)2

dnq

= ezw∫Rne−t

2dnt

= πn2 ezw, (1.6)

which we will use later on.

1.3 The Fock space F(Cn)

1.3.1 Reproducing kernel Hilbert spaces

We will begin our study of the Fock space by giving some definitions and stating some basicfacts concerning reproducing kernel Hilbert spaces. These definitions and results can be foundin section 2.2 of [7].

Consider a locally compact Hausdorff topological space X and a Hilbert space H consisting offunctions on X. We will assume that for this Hilbert space the maps

evx : H → C, f → f(x),

which are called point evaluation maps, are well defined. For such a Hilbert space we have thefollowing definition.

1.3. The Fock space F(Cn) 5

Definition 1.1 (Reproducing kernel Hilbert space). A Hilbert space is a reproducing kernelHilbert space if all the point evaluation maps are continuous. Hence for all x ∈ X and for allg ∈ H, it holds that

∀f ∈ H, ∀ε > 0, ∃δ > 0 such that (‖f − g‖H < δ ⇒ |f(x)− g(x)| < ε).

By the Riesz-representation Theorem (see [2], theorem 6.1), this implies that for each evx thereexists a Kx in H such that, for all f in H,

f(x) = evx(f) = (Kx, f)H. (1.7)

For a reproducing kernel Hilbert space we define the reproducing kernel K as follows:

K : X ×X → C, (x, y)→ Ky(x).

For the reproducing kernel K the following holds (see [7], Lemma 2.24).

1. K(x, y) = (Kx,Ky)H and K(y, x) = K(x, y).

2. K(x, x) = (Kx,Kx)H ≥ 0 and K(x, x) = 0 if and only if f(x) = 0 for all f ∈ H.

3. Let H0 denote the space of finite linear combinations∑j

αjKxj , αj ∈ C and xj ∈ X.

Then H0 is dense in H.

Because H0 is dense in H, it follows that H is determined by the reproducing kernel. We statethe following theorem. For a proof of this theorem see [7], Theorem 2.38.

Theorem 1.2. For H, a Hilbert space of holomorphic functions on M , with M a complexconnected manifold, the following holds:

1. H is a reproducing kernel Hilbert space, with kernel given by K(z, w).

2. H is separable.

3. If (ϕn)n∈I is an orthonormal basis for H, then (ϕn(w))n∈I ∈ `2(I) for all w ∈ M .Moreover, it holds that

∑n∈I |ϕn(w)|2 = K(w,w) and the convergence is uniform on

compact subsets of M .

4. Let F ∈ H. Then

F (z) =∑n∈I

(ϕn, F )ϕn(z)

uniformly on compact subsets of M .

5. Let z, w ∈M . Then

K(z, w) =∑n∈I

ϕn(z)ϕn(w)

and the sum converges uniformly on compact subsets of M ×M .


1.3.2 The inner product on the Fock space

Consider a space of holomorphic functions f defined on Cn. We will use the weight function ρn,which we found earlier, to define an inner product as follows:

(f, g)F(Cn) :=

∫Cnf(z)g(z)ρndz = π

−3n2

∫Cnf(z)g(z)e−zzdz. (1.8)

Here z is the complex vector (z1, z2, · · · , zn)t and dz =∏ni=1 dxi

∏ni=1 dyi is a volume element.

The Fock space F(Cn) consists of the holomorphic functions f : Cn → C for which

(f, f)F(Cn) <∞.

We will use the multi-index notation [m] = (m1,m2, · · · ,mn), where the mi are non-negativeintegers. We shall also write [m!] for m1!m2! · · ·mn!, |m| for m1 +m2 + · · ·+mn and z[m] forzm1

1 zm22 · · · zmnn . A holomorphic function f can be expressed as a power series:

f(z) =∑[m]

α[m]z[m]

where the sum goes over all possible [m]’s. We will show that it is possible to rewrite the innerproduct on F(Cn) in terms of the expansion coefficients α[m].

To do this we will calculate (1.8) for g = f using polar coordinates zk = rkeiϕk and the power

series expression of f . We then get

(f, f)F(Cn) =

∫Cn

∑m,m′

α[m]α[m′]z[m]z[m′]ρndz

=∑m,m′

α[m]α[m′]θm,m′ ,

where

θm,m′ = π−3n2

n∏k=1

∫ 2π

0ei(m

′k−mk)ϕkdϕk

∫ +∞

0e−r

2krmk+m′k+1

k drk.

To calculate θm,m′ we will use ∫ 2π

0ei(m

′k−mk)ϕkdϕk = 2πδmkm′k

and ∫ +∞

0e−r

2rmk+m′k+1dr =

1

2

∫ +∞

0e−tt

mk+m′k

2 dt

=1

2Γ

(mk +m′k

2+ 1

),

where Γ is the gamma function defined by Γ(z) =∫ +∞

0 e−ttz−1dt. It has the property thatΓ(z) = (z − 1)! for z ∈ N. So we get

θm,m′ = δmm′π−n2 [m!]

yielding

(f, f)F(Cn) = π−n2

∑m

[m!]α[m]α[m],

1.3. The Fock space F(Cn) 7

from which we conclude that the condition for a function f to be an element of the Fock spaceis given by ∑

m

[m!]∣∣α[m]

∣∣2 <∞.Using the linearity of the inner product

(f + g, f + g)F(Cn) = (f, f)F(Cn) + (f, g)F(Cn) + (g, f)F(Cn) + (g, g)F(Cn),

we also find

(f, g)F(Cn) = π−n2

∑m

[m!]α[m]β[m]. (1.9)

It is also possible to obtain (1.9) without explicitly calculating the integral. Using (1.2), onecan show that (

z[n], z[m])F(Cn)

= π−n2 δ[n][m][m!],

from which (1.9) immediately follows.

Now we want to prove that F(Cn) is a reproducing kernel Hilbert space.

Theorem 1.3. Let F(Cn) be the space of holomorphic functions for which (f, f)F(Cn) < ∞.Then F(Cn) is a reproducing kernel Hilbert space.

Proof. We find the following bound for f(z):

|f(z)|2 ≤(∑

[m]

∣∣∣α[m]z[m]∣∣∣)2

≤(∑

[m]

[m!]∣∣a[m]

∣∣2)∑[m]

∣∣z[m]∣∣2

[m!]

,

where we have used Schwarz’ inequality in the second step. Substituting

(∑[m]|z[m]|2

[m!]

)= ezz,

leads to

|f(z)| ≤ πn4 e

zz2 ‖f‖F(Cn) ,

which implies

|f(z)− g(z)| ≤ πn4 e

zz2 ‖f − g‖F(Cn) .

Thus the evaluation functions evz are continuous for every z in Cn and F(Cn) is a reproducingkernel Hilbert space.

1.3.3 An orthonormal basis

We will now construct an orthonormal sequence in F(Cn), which we will use to compute thereproducing kernel of F(Cn). Define the following functions in F(Cn):

u[m](z) = πn4z[m]√[m!]

. (1.10)

From (1.9) it is clear that (u[m], u[n]

)F(Cn)

= δ[m][n].


So we found an orthonormal sequence in F(Cn). Since every function in F(Cn) is holomorphicand thus can be expressed as a power series, this sequence is also a basis of F(Cn). Theorem 1.2implies that the kernel of F(Cn) is given by

K(z, w) =∑[m]

u[m](z)u[m](w)

= πn2

∑[m]

1

[m!]

n∏i=1

zmii wimi

= πn2

∞∑k=0

1

k!

∑|m|=k

k!

[m!]

n∏i=1

(ziwi)mi

= πn2

∞∑k=0

1

k!

(n∑i=1

ziwi

)k= π

n2 ezw, (1.11)

which is the same expression as the one we found in (1.6).

1.4 Unitarity of the Segal-Bargmann transform

Consider the Hilbert space L2(Rn) of functions ψ : Rn → C for which (ψ,ψ)L2(Rn) <∞, with

(ψ,ϕ)L2(Rn) =

∫Rnψ(q)ϕ(q)dnq.

We will now define the classical Segal-Bargmann transform.

Definition 1.4. The Segal-Bargmann transform is an integral transformation from L2(R) toF(Cn) defined as

SB(ψ)(z) =

∫RnAn(z, q)ψ(q)dnq for z ∈ Cn and ψ ∈ L2(Rn), (1.12)

with An(z, q) = exp(−12(z2 + q2) +

√2z · q) the integral kernel found in section 1.2.2.

We will use the following theorem ([2], Theorem 10.2) to prove that An(z, q)ψ(q) is indeedintegrable for ψ in L2(Rn).

Theorem 1.5 (Schwarz’ inequality). Let ψ, ϕ ∈ L2(Rn), then we have ψϕ ∈ L1(Rn) and∣∣∣∣∫Rnψϕ

∣∣∣∣ ≤ ‖ψ‖L2(Rn) ‖ϕ‖L2(Rn) .

From (1.6) we get that, for every z in Cn,(An(z, q), An(z, q)

)L2 = π

n2 ezz.

Therefore An(z, q) is in L2(Rn) for every z in Cn. Combining this with Schwarz’ inequalityand the fact that ψ ∈ L2(Rn) implies ψ ∈ L2(Rn), we find that SB(ψ)(z) is well defined for allψ in L2(Rn).From Schwarz’ inequality it also follows that

|SB(ϕ)(z)| ≤ πn4 e

zz2 ‖ψ‖L2(Rn) , (1.13)

since ‖An(z, q)‖2L2(Rn) = πn2 ezz.

1.4. Unitarity of the Segal-Bargmann transform 9

We will now first show that the image SB(ψ) is indeed a holomorphic function. Then ourmain goal is to prove that the Segal-Bargmann transform is a unitary transformation, i.e. anisomorphism between L2(Rn) and F(Cn) that preserves the inner product.For this we will first show that the Segal-Bargmann transform is injective. We will use thisfact to define an inner product on the image SB(L2(Rn)) of the Segal-Bargmann transformso that the Segal-Bargmann transform is unitary between L2(Rn) and SB(L2(Rn)). We thenprove that this inner product makes SB(L2(Rn)) into a reproducing kernel Hilbert space withthe same kernel as F(Cn). Because two reproducing kernel Hilbert spaces with the same kernelare equal, this implies that SB(L2(Rn)) is equal to F(Cn). Therefore the Segal-Bargmanntransform is a surjective map between L2(Rn) and F(Cn) that preserves the inner product.

1.4.1 Holomorphic

To show that the image of the Segal-Bargmann transform gives a holomorphic function we willuse the following theorem ([7], Lemma 1.47).

Theorem 1.6. Let (X,µ) be an open and non-empty measure space and let µ be a Radonmeasure on X. Assume that Ω is a non-empty open subset of VC, where VC is a complex vectorspace, and f : X × Ω→ C is such that

1. For all z ∈ Ω the function x→ f(x, z) is integrable.

2. For all x ∈ X the function z → f(x, z) is holomorphic.

3. For each zo ∈ Ω there exists an open neighbourhood W ⊂ Ω of zo and a non-negativefunction g ∈ L1(X,dµ) such that for all z ∈W

|f(x, z)| ≤ g(x).

Let F (z) :=∫X f(x, z)dµ(x). Then F is holomorphic on Ω, and for all v ∈ VC and zo ∈ Ω we

have

∂vF (zo) =

∫X∂2,vf(x, zo)dµ(x), (1.14)

where ∂2,v refers to the directional derivative in the second variable.

Let X be Rn, Ω be Cn and f(q, z) be

An(z, q)ψ(q) = exp

(−1

2(z2 + q2) +

√2z · q

)ψ(q).

We have already proven that An(z, q)ψ(q) is integrable for all z in Cn. The second conditionfollows from the fact that the exponential function is holomorphic. Because∣∣∣√2z · q

∣∣∣ ≤ 2z · z +1

4q2,

we have

|An(z, q)ψ(q)| ≤ exp

(5

2α2 − 1

4q2

)|ψ(q)| for zz ≤ α2.

The function exp(52α

2− 14q

2) |ψ(q)| is non-negative and integrable. Therefore the third conditionis also satisfied and we conclude that

SB(ψ)(z) =


is a holomorphic function.


1.4.2 Injectivity

In this section we will show that the Segal-Bargmann transform is injective. Because the Segal-Bargmann transform is a linear transformation, it is sufficient to prove that SB(ψ) = 0 impliesψ = 0. For z = i y√

2we have

SB(ψ)

(iy√2

)= e

y2

4

∫Rne−

q2

2 ψ(q)eiqydnq

= ey2

4 F(e−

q2

2 ψ(q)

)(y),

where F is the Fourier transform on L1(Rn).

The function exp(− q2

2 )ψ(q) is an element of L1(Rn) ∩ L2(Rn) and the Fourier transform is a

bijective operator on L2(Rn). Hence SB(ψ)(z) = 0 implies exp(− q2

2 )ψ(q) = 0. Consequently,we also have ψ(q) = 0, which proves injectivity.

1.4.3 Unitarity

Because of linearity, the image SB(L2(Rn)) of the Segal-Bargmann transform is a vector space.We define an inner product on it as follows:

(f, g)SB = (ψ,ϕ)L2(Rn),

where ψ and ϕ are chosen such that SB(ψ) = f and SB(ϕ) = g. Because the Segal-Bargmanntransform is injective, this is well defined and it is clear that the Segal-Bargmann transformis an unitary mapping from L2(Rn) to SB(L2(Rn)). From (1.13) we deduce that, for f inSB(L2(Rn)) and z in Cn,

|f(z)| ≤ πn4 e

zz2 ‖f‖SB .

This implies that all evaluation maps are continuous. Hence SB(L2(Rn)) is a reproducingkernel Hilbert space. Now we want to show that SB(L2(Rn)) = F(Cn). Because reproducingkernel Hilbert spaces are determined by their kernel, it suffices to show that their kernels areequal.

We have, see (1.7),

f(z) = (Kz, f)SB,

and also by definition of the Segal-Bargmann transform,

f(z) =


=(An(z), ψ

)L2(Rn)

=(SB(An(z)), SB(ψ)

)SB

=(SB(An(z)), f

)SB,

for all f in SB(L2(Rn)). Thus we find that

K(w, z) = Kz(w) = SB(An(z))(w)

=

∫RnAn(z, q)An(w, q)dnq

= πn2 ewz,

1.5. Orthonormal functions on L2(Rn) 11

where we have used (1.6). This kernel is equal to the kernel we found for F(Cn) in (1.11).Therefore, we have proven that SB(L2(Rn)) is equal to F(Cn) and

(f, g)F(Cn) = (f, g)SB.

From the definition of the inner product on SB(L2(Rn)) we also have

(f, g)SB = (ψ,ϕ)L2 ,

and hence we obtained the following theorem.

Theorem 1.7 (Unitarity of the Segal-Bargmann transform). The Segal-Bargmann transformdefined by

f(z) =

∫RnAn(z, q)ψ(q)dnq,

is a unitary mapping between the Hilbert spaces L2(Rn) and F(Cn).

1.5 Orthonormal functions on L2(Rn)

The functions u[m] defined in (1.10) form an orthonormal basis of F(Cn). In this section welook for functions ϕ[m] such that SB(ϕ[m]) = u[m]. Because the Segal-Bargmann transform isunitary, we then immediately have that the ϕ[m] form an orthonormal basis of L2(Rn).For the sake of simplicity take n = 1. The functions um are defined by

um = π14zm√m!.

We can rewrite this cleverly using

zm =∂m

∂bmebz∣∣∣b=0

.

Because of (1.6), we get

ebz = π−12

∫RA1(z, q)A1(b, q)dq,

which yields

um = π−14

1√m!

∂m

∂bm

(∫RA1(z, q)A1(b, q)dq

) ∣∣∣∣b=0

= π−14

1√m!

∫RA1(z, q)

∂m

∂bmA1(b, q)

∣∣b=0

dq

=

∫RA1(z, q)ϕm(q)dq.

Switching the integral and derivative is allowed because of (1.14). Hence the functions

ϕm(q) = π−14

1√m!

∂m

∂bmA1(b, q)

∣∣b=0

are a set of orthonormal functions in L2(Rn). Using the definition of A1 and setting b =√

2γ,we can rewrite these ϕm as

ϕm(q) =1√

2mm!√πe−q22

∂m

∂γme2γq−γ2∣∣

γ=0.

The generating function of the Hermite polynomials is given by Hm(q) = ∂m

∂γm e2γq−γ2 |γ=0. We

conclude that the ϕm(q) we found are the normalized Hermite functions

ϕm(q) =1√

2mm!√πe−q22 Hm(q),

which are indeed known to form an orthonormal basis of L2(R).


1.6 The harmonic oscillator on the Fock space

In this last section we will obtain a harmonic oscillator H on F(Cn) such that the followingdiagram commutes.

L2(Rn) F(Cn)

L2(Rn) F(Cn)

H

SB

H

SB

Here H is the harmonic oscillator on L2(Rn) given by

H =n

2+

n∑i=1

a†iai.

The correspondence between ai and ∂zi and between a†i and zi yields

H =n

2+

n∑i=1

zi∂zi

for the harmonic oscillator on F(Cn).

Consider f(z) =∑

m α[m]z[m] in F(Cn). Applying the harmonic oscillator H on this f , we get

Hf(z) =∑m

(n2

+ |m|)α[m]z

[m].

So the eigenvalues are given by n/2 + l, l ∈ N, and the eigenfunctions are homogeneouspolynomials of degree l. Hence the eigenfunctions are linear combinations of u[m] with |m| = l.The corresponding eigenfunctions of H on L2(Rn) will be linear combinations of the ϕ[m] wefound in the previous section. This agrees with the well-known fact that the Hermite functionsare eigenfunctions of the harmonic oscillator on L2(Rn).

Chapter 2

Jordan algebras

In this chapter we will introduce Jordan algebras. We will mainly focus on one specific Jordanalgebra, which has as underlying vector space the symmetric matrices Sym(n,R). A moregeneral treatment can be found in the first chapter of Mollers [6] or in the book of Faraut andKoranyi [3].We will begin this chapter by stating the definition and some basic properties of Jordan algebrasand defining a linear operator L(x) for each element in Sym(n,R). Then we will introduce aninner product and give a decomposition of Sym(n,R) based on idempotent elements. Finally,we will define the structure algebra and prove that for Sym(n,R) it is equal to gl(n,R).

Let us start with recalling the definition of an algebra.

Definition 2.1. An algebra V is a vector space over a field K equipped with a bilinear product ∗;in other words the product satisfies:

left distributivity : x ∗ (y + z) = x ∗ y + x ∗ z, ∀x, y, z ∈ V ,

right distributivity: (x+ y) ∗ z = x ∗ z + y ∗ z ∀x, y, z ∈ V ,

compatibility with scalars: λ(x ∗ y) = (λx) ∗ y = x ∗ (λy) ∀x, y ∈ V, λ ∈ K.

If there exists an element e such that e ∗x = x = x ∗ e for all x in V , we call V a unital algebraand the element e the unit of the algebra.We will now give the definition of a Jordan algebra.

Definition 2.2. A Jordan algebra V is an algebra for which the multiplication satisfies:

commutativity: x ∗ y = y ∗ x ∀x, y ∈ V ,

Jordan identity: x ∗ (x2 ∗ y) = x2 ∗ (x ∗ y) ∀x, y ∈ V ,

where x2 = x ∗ x.

Note that a Jordan algebra is in general not associative. When we will consider a general Jordanalgebra, we will always assume that the underlying vector space is real and finite-dimensionaland that the algebra contains a unit.

2.1 The symmetric matrices Sym(n,R)

We will now look at an important example that will be our main object of study.

Theorem 2.3. The space of symmetric n × n-matrices Sym(n,R) is a Jordan algebra withproduct given by

x ∗ y =xy + yx

2,

where xy denotes standard matrix multiplication.

13

14 Chapter 2. Jordan algebras

Proof. It is clear that

(x ∗ y)t =(xy)t + (yx)t

2=yx+ xy

2= x ∗ y,

thus (x ∗ y) is an element of Sym(n,R). The product is easily seen to be commutative. Bydefinition we have

x ∗ (λy + µz) =x(λy + µz) + (λy + µz)x

2

= λ(xy + yx)

2+ µ

(xz + zx)

2= λ(x ∗ y) + µ(x ∗ z),

so ∗ is a bilinear multiplication. It is also clear that x ∗ x = xx and hence the notation x2 isunambiguous. We then obtain

x ∗ (x2 ∗ y) =1

4x(x2y + yx2) +

1

4(x2y + yx2)x

=1

4(x3y + xyx2 + x2yx+ yx3)

=1

4(x2(xy + yx) + (xy + yx)x2)

= x2 ∗ (x ∗ y).

Hence the Jordan identity is satisfied and we conclude that Sym(n,R) is indeed a Jordanalgebra.

In the proof we observed that the notation x2 is unambiguous, because x ∗ x = xx. We cangeneralize this to all powers of x.

Lemma 2.4. The Jordan algebra Sym(n,R) is power associative, i.e. the subalgebra generatedby one element is associative. Furthermore, for n in N and x in Sym(n,R), it holds that

x ∗ x ∗ · · · ∗ x︸︷︷︸n−times

= xn, (2.1)

where xn is the nth power of standard matrix multiplication.

Proof. We will first prove (2.1) using induction on n. We have already shown that, for n = 2,x ∗ x = x2. Assume that (2.1) holds for n. Then we get

x ∗ (x ∗ x ∗ · · · ∗ x︸︷︷︸n−times

) =1

2(xxn + xnx) = xn+1.

Thus (2.1) also holds for n+ 1. Consequently, we also have that the algebra generated by x isassociative because the standard matrix multiplication is associative.

Let W be a vector space over a field K. Then we denote by End(W ) the space of all linearoperators on W , i.e. operators T : W →W such that

T (λx+ µy) = λT (x) + µT (y),

for x, y ∈W and λ, µ ∈ K. We can equip End(W ) with the structure of an algebra by definingaddition as pointwise addition in the vector space W and multiplication as composition ofoperators, i.e.

T + T ′(x) := T (x) + T ′(x)

2.1. The symmetric matrices Sym(n,R) 15

andT T ′(x) := T (T ′(x)),

for T, T ′ ∈ End(W ) and x ∈W .If W is finite-dimensional, we know that End(W ) is isomorphic to the space of k × k-matricesover K, where k is the dimension of W . We define the trace and determinant of operators inEnd(W ) as the trace and determinant of the corresponding matrix. Throughout this thesis wewill use the notation Tr for the trace of a linear operator and the notation tr for the trace of amatrix.We can see Sym(n,R) as a subspace of End(Rn). With each orthogonal change of basis in Rncorresponds a linear transformation of End(Rn) and thus also of Sym(n,R). More concretely,if P is the matrix of the orthogonal transformation on Rn:

q → Pq for all q ∈ Rn,

then the corresponding transformation on Sym(n,R) is given by

x→ PxP−1 for all x ∈ Sym(n,R).

This is indeed an endomorphism of the vector space Sym(n,R), since

(PxP−1)t = P−1txP t

= PxP−1,

where we used that P is an orthogonal matrix: P−1 = P t. It also holds that

2P (x ∗ y)P−1 = PxP−1PyP−1 + PyP−1PxP−1

= 2(PxP−1) ∗ (PyP−1).

Therefore, the transformation is even an endomorphism of the Jordan algebra Sym(n,R).

Definition 2.5. For x, y in Sym(n,R) we say that x is conjugate to y if there exists anorthogonal matrix P such that x = PyP−1.

From linear algebra we know that every symmetric matrix is conjugate to a diagonal matrix.Thus for a particular element x of Sym(n,R), we can always choose a basis of Rn such that xis a diagonal matrix.

In the following lemma we establish the unit of Sym(n,R).

Lemma 2.6. The identity matrix In is the unique unit of Sym(n,R).

Proof. It is clear that In is a unit since x ∗ In = x+x2 = x = In ∗ x. Let e be an arbitrary unit,

then we have e ∗ In = e because In is a unit, and e ∗ In = In because e is a unit. Hence weconclude that e = In.

We will now define what an invertible element is.

Definition 2.7. An element x of a Jordan algebra V is invertible if there exists a y in V suchthat

y ∗ x = e = x ∗ y,

where e is the unit of the Jordan algebra. We call y an inverse of x.

For Sym(n,R) we have a simple criterion to decide whether an element is invertible.

Theorem 2.8. An element x of Sym(n,R) is invertible if and only if det(x) 6= 0. In that casean inverse is given by x−1, the inverse of x for the standard matrix multiplication.


Proof. If det(x) 6= 0, then we know that the standard inverse x−1 exists. We immediately getthat x ∗ x−1 = In.Conversely, if det(x) = 0, then we can choose a basis of Rn such that x is a diagonal matrix.Because det(x) = 0, at least one of the diagonal elements will be zero. Fix i such that thediagonal element λi = 0. Then for all y in Sym(n,R) it holds that

(x ∗ y)ii =1

2

n∑r=1

(xiryri + yirxri) = 0,

because xir = xri = 0 for all r = 1, . . . , n. Hence x ∗ y 6= In and we conclude that x is notinvertible.

Note that we did not prove that x−1 is the only inverse. This is also not the case as thefollowing example shows. Define

x =

(1 00 −1

)and

y =

(1 aa −1

),

then we have for each a in R that

x ∗ y =xy + yx

2= I2.

When we talk about the inverse of x we will always mean the standard matrix inverse x−1.In general it is also not true that x ∗ (x−1 ∗ y) = y, but it is possible to switch the positions ofx and x−1.

Lemma 2.9. For x, y in Sym(n,R) the following holds:

x ∗ (x−1 ∗ y) = x−1 ∗ (x ∗ y).

Proof. We have

x ∗ (x−1 ∗ y) =1

4(xx−1y + xyx−1 + x−1yx+ yx−1x),

and

x−1 ∗ (x ∗ y) =1

4(x−1xy + x−1yx+ xyx−1 + yxx−1).

The two right-hand sides are clearly equal to each other.

2.2 The operator L(x)

We define for each x of a Jordan algebra V an operator by (left) multiplication:

L(x) : V → V, y → x ∗ y.

This is a linear operator on V , which follows immediately from the bilinearity of the Jordanmultiplication. In the following lemma we give a relation between the matrix trace of x andthe trace of the linear operator L(x).

Lemma 2.10. For each x in Sym(n,R) it holds that

Tr(L(x)

)=n+ 1

2tr(x),

where Tr denotes the trace of the linear operator and tr is the standard matrix trace.

2.2. The operator L(x) 17

Proof. We define a basis Eij , 1 ≤ i ≤ j ≤ n of Sym(n,R), where the Eij are matrices that areonly different from zero on the (i, j)th and (j, i)th entry:

(Eij)kl =1

2(δikδjl + δilδjk). (2.2)

The expression for x in Sym(n,R) in this basis is as follows:

x =∑

1≤i≤j≤n(2xij − δijxij)Eij .

The trace of a linear operator T is defined as the trace of the associated matrix. The matrixof size n(n+1)

2 × n(n+1)2 associated with L(x) has the form

A(kl)(ij) = (2− δkl)(L(x)Eij

)kl, (2.3)

where (L(x)Eij)kl is the (k, l)th entry of the matrix L(x)Eij . This entry is equal to(L(x)Eij

)kl

= (x ∗ Eij)kl

=1

2(xEij + Eijx)kl

=1

4(xkiδjl + xkjδil + xjlδik + xilδjk). (2.4)

Hence we obtain

Tr(L(x)

)=

∑1≤i≤j≤n

A(ij)(ij)

=∑

1≤i≤j≤n(2− δij)

(L(x)Eij

)ij

=∑

1≤i<j≤n

1

2(xiiδjj + xijδij + xjjδii + xijδji) +

n∑i=1

xii

=n∑i=1

n∑j=i+1

xjj2

+n∑j=1

j−1∑i=1

xii2

+n∑i=1

xii

=n∑i=1

n∑j=1

1

2xii +

n∑i=1

1

2xii

=n+ 1

2tr(x),

thus completing the proof.

We also have the following identity for L(x) in Sym(n,R), which we will use later on.

Lemma 2.11. For the operator L and for x, y, z in Sym(n,R) the following holds:

L((x ∗ y) ∗ z

)−L(x ∗ y)L(z) =

(L(x ∗ z)−L(x)L(z)

)L(y) +

(L(y ∗ z)−L(y)L(z)

)L(x). (2.5)

Proof. One can check that((x∗y)∗z

)∗a−(x∗y)∗(z ∗a) = (x∗z)∗(y ∗a)−x∗

(z ∗(y ∗a)

)+(y ∗z)∗(x∗a)−y ∗

(z ∗(x∗a)

)holds for all x, y, z, a in Sym(n,R) by a long but straightforward calculation.


2.3 The trace form

We define the trace form τ on a Jordan algebra V by

τ : V × V → R, (x, y)→ tr(x ∗ y).

Note that we did not define tr for an arbitrary Jordan algebra. Such a definition is given in[3], section II.2. For Sym(n,R) this definition coincides with the standard matrix trace.We also remark that tr(x ∗ y) = tr(xy) for Sym(n,R), since tr(xy) is equal to tr(yx). We willnow show that τ is an inner product on Sym(n,R).

Theorem 2.12. The trace form τ is

symmetric: τ(x, y) = τ(y, x),

linear: τ(λx+ µy, z) = λτ(x, z) + µτ(y, z),

positive definite: τ(x, x) ≥ 0 and τ(x, x) = 0 if and only if x = 0,

associative: τ(x ∗ y, z) = τ(x, y ∗ z).

The first three properties imply that the trace form defines an inner product on Sym(n,R).

Proof. We immediately have symmetry because tr(xy) = tr(yx). Linearity follows from linear-ity of the matrix trace. Because x is a symmetric matrix, we have

τ(x, x) = tr(x2)

=

n∑i,j=1

xijxji =

n∑i,j=1

x2ij ≥ 0,

and τ(x, x) = 0 if and only if xij = 0 for all i, j ∈ 1 . . . n. Hence the trace form is positivedefinite.Because the trace is invariant under cyclic permutations, we obtain

tr((x ∗ y) ∗ z

)=

1

4tr((xy)z + (yx)z + z(xy) + z(yx)

)=

1

4tr(xyz + xzy + yzx+ zyx)

= tr(x ∗ (y ∗ z)

).

Thus the trace form τ is associative.

We call a Jordan algebra semisimple if the trace form is non-degenerate, and Euclidean if thetrace form is positive definite and the field K = R. Hence, from Theorem 2.12 it follows thatSym(n,R) is a Euclidean Jordan algebra. A Jordan algebra is called simple if it is semisimpleand does not contain non-trivial ideals for the Jordan product.From now on we will assume that we are working with simple Euclidean Jordan algebras. Thefollowing theorem asserts that Sym(n,R) is indeed a simple Jordan algebra.

Theorem 2.13. The Jordan algebra Sym(n,R) is a simple Jordan algebra.

Proof. We want to prove that every ideal of Sym(n,R) is either equal to zero or to Sym(n,R).Consider an arbitrary non-zero ideal I. We will show that I contains the identity matrix Inwhich implies that I is equal to Sym(n,R), since x ∗ In = x. Let x be an arbitrary element ofI. By choosing an appropriate basis of Rn, we can turn x into a diagonal matrix. Thus x is ofthe form

x =

λ1

. . .

λkOn−k

,

2.3. The trace form 19

where On−k is an (n− k)× (n− k) zero matrix. Define y as the matrix with y11 = 2λ−11 and

zeros on the other entries. We then get that

y ∗ x =

2

0. . .

0

is an element of I. Let k be in 1, . . . , n and multiply y ∗x twice with the matrix z which hasone on the entries (z)1k and (z)k1 and zeros on the other entries. Then we also have that

z ∗(z ∗ (y ∗ x)

)=

1

Ok−2

1On−k

is an element of I. Because the ideal I is a linear subspace, we have shown that for each k thematrix Ekk, as defined in (2.2), is in I. Then the sum

∑nk=1Ekk = In is also an element of I,

which completes the proof.

Set k equal to n(n + 1)/2, the dimension of the vector space Sym(n,R). We will show thatthe trace form τ gives actually the same inner product as the standard inner product on Rk.To obtain this connection between Rk and Sym(n,R) we need an orthonormal basis (eα)α ofSym(n,R) with respect to τ , i.e. a basis such that

τ(eα, eβ) = δαβ.

In an orthonormal basis (eα)α we can write every x in Sym(n,R) as

x =∑α

xαeα.

The coefficients xα are equal to τ(x, eα). This follows from the orthonormality of the basis andthe linearity of τ since

τ(x, eα) = τ(∑

β

xβeβ, eα

)=∑β

xβτ(eβ, eα) =∑β

xβδαβ = xα.

Using an orthonormal basis, we can construct an isomorphism (of vector spaces) betweenSym(n,R) and Rk by mapping the matrix x =

∑α xαeα to its coordinates xα = τ(x, eα) in this

basis. It follows immediately from the fact that the set (eα)α forms a basis of Sym(n,R) andthat the dimension of Sym(n,R) is equal to k that this is indeed a vector space isomorphism.This isomorphism gives us another inner product on Sym(n,R), namely the one correspondingto the standard inner product of Rk defined by

〈x, y〉 =k∑i=1

xiyi,

for x = (x1, x2, . . . , xk)t and y = (y1, y2, . . . , yk)

t in Rk. The corresponding inner product onSym(n,R) is then given by

〈x, y〉 =∑α

xαyα,

for x and y in Sym(n,R) and an orthonormal basis (eα)α.The following lemma states that this inner product is equal to the trace form τ .


Lemma 2.14. Let (eα)α be an orthonormal basis of Sym(n,R) and let k be the dimension ofSym(n,R). Then the inner product defined by the trace form τ and the standard inner producton Rk are equal, i.e. for all x, y in Sym(n,R) the following holds:

τ(x, y) = 〈x, y〉.

Proof. We have

τ(x, y) = τ(∑

α

xαeα,∑β

yβeβ

)=∑α,β

xαyβδαβ =∑α

xαyα = 〈x, y〉.

Hence the two inner products are equal.

This lemma has the immediate consequence that the topology of Sym(n,R) induced by thetrace form τ is equivalent to the standard topology on Rk. Using this equivalence, we canapply our knowledge of the standard Euclidean space Rk to the Jordan algebra Sym(n,R).For example, the partial derivative ∂/∂xα is just the standard partial derivative to the αth

component of Rk and integration over a measurable subset of Sym(n,R) reduces to integrationover the corresponding measurable subset of Rk.We also give the following lemma, which we will use later.

Lemma 2.15. Let (eα)α be an orthonormal basis with respect to the trace form τ of the Jordanalgebra V . Then we have the following property of τ :∑

α

τ(x, eα)τ(eα, y) = τ(x, y),

for all x, y in V .

Proof. We have that y =∑

α yαeα, where yα = τ(y, eα). Using this we get

τ(x, y) =∑α

τ(x, yαeα) =∑α

τ(x, eα)yα =∑α

τ(x, eα)τ(eα, y).

This proves the statement.

For further reference we will now explicitly construct an orthonormal basis with respect to thetrace form τ . In (2.2) we have defined a basis (Eij)i≤j of Sym(n,R). One can calculate that

tr(Eij ∗ Ekl) =n∑

r,s=1

(Eij)rs (Ekl)sr =1

2(δikδjl + δilδjk). (2.6)

Hence this basis is orthogonal. We also have

Eij ∗ Eij =

14(Eii + Ejj) if i 6= j

Eij if i = j.

Thus τ(Eii, Eii) = 1 and τ(Eij , Eij) = 1/2. So we can define an orthonormal basis (Eij)i≤j ofSym(n,R) by setting

Eij :=

√2Eij for i 6= j

Eij for i = j.(2.7)

2.4. The Peirce decomposition 21

2.4 The Peirce decomposition

We will now look at decompositions of Sym(n,R) based on idempotent elements. First we willconstruct a decomposition based on one idempotent. Then we will look at a decompositionbased on a so-called Jordan frame, a certain collection of idempotents.

2.4.1 Idempotents

Let c be an idempotent element of Sym(n,R), i.e. c2 = c. If we choose a basis of Rn such thatc is diagonal, then the condition that c is idempotent translates into the following equationsfor the diagonal elements of c:

λ2i = λi for i = 1, . . . , n.

So we have that λi = 0 or λi = 1. By reordering of the basis of Rn we obtain

c =

(Ik

On−k

),

where Ik is the k × k identity matrix, and On−k the (n − k) × (n − k) zero matrix. For anidempotent c in this form it is easy to determine the eigenvalues of L(c).

Lemma 2.16. The only possible eigenvalues of L(c), where c is an idempotent, are 0, 1/2 and1.

Proof. Choose a basis of Rn such that c is of the form

c =

(Ik

On−k

),

and let x be a symmetric matrix that we write as(A BBt C

),

where A is a symmetric k×k-matrix, C a symmetric (n−k)×(n−k)-matrix and B an arbitraryk × (n− k)-matrix. We then get

L(c)x =cx+ xc

2

=1

2

(A B0 0

)+

1

2

(A 0Bt 0

)=

(A B

2Bt

2 0

).

Hence we may observe that the possible eigenvalues of L(c) are 0, 1/2 and 1 with correspondingeigenspaces:

V (c, 1) =

(A 00 0

)|A ∈ Sym(k,R)

V

(c,

1

2

)=

(0 BBt 0

)|B ∈ Mat(k × n− k,R)

V (c, 0) =

(0 00 C

)|C ∈ Sym(n− k,R)

.

In this lemma we have established the Peirce decomposition corresponding to an idempotent.


Theorem 2.17. With each idempotent c corresponds a decomposition of the Jordan algebraSym(n,R) given by

V = V (c, 1)⊕ V(c,

1

2

)⊕ V (c, 0),

where V (c, λi) is the eigenspace belonging to the eigenvalue λi. This decomposition is orthogonalwith respect to the trace form τ and is called the Peirce decomposition corresponding to c.

Proof. The existence of this decomposition follows from the proof of the above lemma. Sinceτ is associative, we have

τ(L(c)x, y) = τ(x, L(c)y).

It also holds that

τ(L(c)x, y) = λ1τ(x, y)

τ(x, L(c)y) = λ2τ(x, y),

where λ1 is the eigenvalue of x and λ2 is the eigenvalue of y. For x and y in different eigenspacesof L(c) we then get

(λ1 − λ2)τ(x, y) = 0.

Thus τ(x, y) = 0 and the decomposition is orthogonal.

2.4.2 Jordan frame

We will now construct a decomposition of Sym(n,R) based on a Jordan frame, a collection oforthogonal primitive idempotents. First we will define what orthogonal primitive idempotentsare.

Definition 2.18. An idempotent element of a Jordan algebra V is called primitive if it isnon-zero and cannot be written as the sum of two non-zero idempotents. Two idempotents c1

and c2 are called orthogonal if c1 ∗ c2 = 0.

The definition of a Jordan frame is then as follows.

Definition 2.19. A collection of orthogonal primitive idempotents c1, . . . , ck in V is called aJordan frame if c1 + . . .+ ck = e, where e is the unit of V .

The following lemma characterizes the primitive idempotents of Sym(n,R).

Lemma 2.20. Every primitive idempotent of Sym(n,R) is conjugate to a matrix of the form

c1 =

(1

On−1

). (2.8)

Proof. We already saw that every idempotent is conjugate to a matrix of the form

ck =

(Ik

On−k

).

If k 6= 1, we can write ck = c1 + c′k−1 where

c′k−1 =

0Ik−1

On−k

.

Hence in that case ck is not primitive. Now we will prove that c1 is indeed a primitive idem-potent. Suppose c1 = c+ c′, with c2 = c and c′2 = c′. We have

c21 = (c+ c′) ∗ (c+ c′) = c+ 2c ∗ c′ + c′ = c1.

2.5. The structure algebra 23

Thus we find that c ∗ c′ = 0. This is equivalent to c ∗ c1 = c. Using the matrix form (2.8) of c1,we see that this implies that

c = c1, c′ = 0 or c = 0, c′ = c1,

and we conclude that c1 is primitive.

Consider the set (Eii)ni=1, with the Eii defined as in (2.2). Every element of this set is conjugate

to(

1On−1

), hence they are primitive idempotents. We also have that Eii ∗ Ejj = 0 for i 6= j

and∑n

i=1Eii = In. Thus the set (Eii)ni=1 is a Jordan frame.

The Jordan frame (Eii)ni=1 leads to a decomposition of Sym(n,R) called the Peirce decomposi-

tion:V =

⊕1≤i≤j≤n

Vij ,

where

Vii = V (Eii, 1) = REii

Vij = V

(Eii,

1

2

)∩ V

(Ejj ,

1

2

)= REij .

We have already proven in (2.6) that the basis (Eij)i≤j is orthogonal with respect to the traceform τ . Therefore the Peirce decomposition is also orthogonal with respect to τ .

Every Jordan frame in Sym(n,R) is actually related to this specific Jordan frame (Eii)ni=1.

Theorem 2.21. Every Jordan frame c1, . . . , ck is conjugate to the Jordan frame (Eii)ni=1.

This means that there exists an orthogonal matrix P such that PciP−1 = Eii for all i in

1, . . . , n.

Proof. Consider a Jordan frame c1, c2, . . . , cn. By changing the basis of Rn, we can makec1 = E11. Because c1 ∗ ci = 0, for i = 2, . . . , n, we have that in this basis the ci are of the form(

0c′i

),

where c′i ∈ Sym(n − 1,R). By changing the basis of Rn−1 we can bring c2 into the form E22

and because c2 ∗ ci = 0, we find for the ci, i = 3, . . . , n,00

c′′i

,

where c′′i ∈ Sym(n − 2,R). By repeating this procedure, we observe that we can bring theJordan frame c1, . . . , cn into the form E11, . . . , Enn by transformations of the basis of Rn.

Note that this theorem implies that every Jordan frame consists of exactly n elements.

2.5 The structure algebra

In this section we will define the structure algebra and show that for Sym(n,R) this algebra isequal to the Lie algebra gl(n,R). This is the Lie algebra of the real n × n-matrices with Liebracket given by

[x, y] = xy − yx, (2.9)

for x, y in gl(n,R). Although the structure algebra can be defined for any simple real Jordanalgebra V , in the following we will always assume that V = Sym(n,R).


2.5.1 Derivations

The structure algebra will consist of two parts, where one part will be the space of derivations.Therefore we start by defining what a derivation is.

Definition 2.22 (Derivation). A linear operator on a Jordan algebra V is called a derivationif D(x ∗ y) = D(x) ∗ y + x ∗D(y) for all x, y in V .

We write Der(V ) for the space of all derivations on V . We can reformulate the definition of aderivation in the following way:

D ∈ Der(V )⇐⇒ L(D(x)) = [D,L(x)] for all x ∈ V . (2.10)

The commutator of the operator L gives us an important class of derivations. In fact, we willshow later that every derivation is a finite sum of such commutators.

Lemma 2.23. For x, y in Sym(n,R) the commutator [L(x), L(y)] is a derivation.

Proof. Using (2.5) we get, for x, y, z in Sym(n,R),

L((x ∗ z) ∗ y

)− L(x ∗ z)L(y) =

(L(x ∗ y)− L(x)L(y)

)L(z) +

(L(z ∗ y)− L(z)L(y)

)L(x)

L((y ∗ z) ∗ x

)− L(y ∗ z)L(x) =

(L(x ∗ y)− L(y)L(x)

)L(z) +

(L(z ∗ x)− L(z)L(x)

)L(y).

Subtracting these two equations from each other results in

L((x∗z)∗y

)−L((y∗z)∗x

)= −L(x)L(y)L(z)−L(z)L(y)L(x)+L(y)L(x)L(z)+L(z)L(x)L(y).

This we can rewrite asL([L(x), L(y)]z

)=[[L(x), L(y)], L(z)

],

which (see 2.10) is equivalent to [L(x), L(y)] being a derivation.

We call a derivation an inner derivation if it is a finite sum of derivations of type [L(x), L(y)].To prove that every derivation is inner we will need the following lemma.

Lemma 2.24. One has

Tr(L(x)L(y)

)=n+ 2

4tr(xy) +

1

4tr(x) tr(y),

for x, y in Sym(n,R). Here Tr is the trace of a linear operator and tr is the matrix trace.

Proof. We choose a basis of Rn such that x is a diagonal matrix with diagonal elements λi,i = 1, . . . , n. We then find(

L(x)L(y)Eij)ij

=

λi+λj

8 (yii + yjj) for i < j

λiyii for i = j.

Substituting this in (2.3), we get for the trace

Tr(L(x)L(y)) =∑

1≤i<j≤n

λi + λj4

(yii + yjj) +n∑i=1

λiyii

=

n∑i=1

n∑j=i+1

λjyjj4

+

n∑j=1

j−1∑i=1

λiyii4

+

n∑i=1

λi4

n∑j=i+1

yjj +

n∑j=1

λj4

j−1∑i=1

yii +

n∑i=1

λiyii

=n∑

i,j=1

λiyii4

+n∑i=1

λi4

n∑j=1

yjj +1

2

n∑i=1

λiyii

=n+ 2

4tr(xy) +

1

4tr(x) tr(y),

which proves the lemma.


This lemma allows us to prove that every derivation is inner.

Theorem 2.25. Every derivation on Sym(n,R) is an inner derivation.

Proof. Let D be a derivation and (eα)α an orthonormal basis with respect to the trace form τ .

We can express D(eα) in this basis as D(eα) =∑

β Dβαeβ, with Dβ

α in R. Since D is a linearoperator, we have for an arbitrary element x of Sym(n,R)

D(x) =∑α

xαD(eα) =∑α,β

xαDβαeβ.

We define symmetric matrices Uβ by Uβ =∑

αDβαeα. Using these Uβ we can rewrite D(x) as

D(x) =∑β

τ(x, Uβ)eβ, (2.11)

because from the linearity of τ it follows that

τ(x, Uβ) =∑α,γ

xαUβγ τ(eα, eγ) =

∑α

xαDβα.

Combining (2.10) and Lemma 2.10, we get

n+ 1

2tr(D(x)) = Tr

(L(D(x))

)= Tr

([D,L(x)]

)= 0,

since the trace of a commutator vanishes. Applying this to Lemma 2.24, we obtain for arbitraryx, y in Sym(n,R)

n+ 2

4τ(Dx, y) = Tr

(L(D(x))L(y)

)−

=0︷︸︸︷1

4tr(D(x)) tr(y)

= Tr([D,L(x)]L(y)

)= Tr

(D[L(x), L(y)]

).

In the second step we used (2.10) and in the last step we used the fact that the trace is invariantunder cyclic permutations. The trace Tr

(D[L(x), L(y)]

)is given by∑

α

τ(D([L(x), L(y)]eα

), eα

).

Substituting in this the expression we found in (2.11) for D , we get

n+ 2

4τ(Dx, y) =

∑α,β

τ(τ(

[L(x), L(y)]eα, Uβ)eβ, eα

)=∑α,β

τ(

[L(x), L(y)]eα, Uβ)δαβ

=∑α

(τ (y ∗ eα, x ∗ Uα)− τ (x ∗ eα, y ∗ Uα)

)= τ

(∑α

[L(eα), L(Uα)]x, y),

where we used the associativity of τ . We conclude that

D =4

n+ 2

∑α

[L(eα), L(Uα)],

which proves the theorem.


We note that from the fact that every derivation is a finite sum of commutators, it follows that

Tr(D) = 0, (2.12)

for all D in Der(Sym(n,R)). In the proof of the theorem we have also obtained that

tr(D(x)) = 0 (2.13)

for all x in Sym(n,R) and D in Der(Sym(n,R)).

2.5.2 The structure algebra is equal to gl(n,R).

We will now define the structure algebra.

Definition 2.26 (Structure algebra). The structure algebra of V is a Lie subalgebra of gl(V )defined by

str(V ) := h + q,

where

h := Der(V )

q := L(V ) = L(x) | x ∈ V .

The Lie bracket is given by

[L(x) +D,L(x′) +D′] = L(D(x′)−D′(x)

)+ [L(x), L(x′)] + [D,D′]. (2.14)

For Sym(n,R) this structure algebra is isomorphic to the Lie algebra gl(n,R).

Theorem 2.27. The structure algebra str(V ) of Sym(n,R) is given by gl(n,R) which acts onV as follows:

A ·x = Ax+ xAt for all A ∈ gl(n,R), x ∈ V . (2.15)

Proof. We will first show that every element of str(V ) corresponds to an element of gl(n,R).Consider an element A of str(V ). It is of the form A = L(x) + D, for x in Sym(n,R) and Din Der(V ). Since every derivation is inner, we can also write A as L(x) +

∑i[L(yi), L(zi)] for

x, yi, zi in Sym(n,R). The element A acts on V by

A · v = L(x)v +∑i

[L(yi), L(zi)]v

=xv + vx

2+

1

4

∑i

yi(ziv + vzi) + (ziv + vzi)yi − zi(yiv + vyi)− (yiv + vyi)zi

=

(x

2+

1

4

∑i

yizi − ziyi)v + v

(x

2+

1

4

∑i

yizi − ziyi)t.

Hence, with each element A of str(V ) corresponds a matrix(x2 + 1

4

∑i yizi − ziyi

)in gl(n,R)

and we have proven that str(V ) ⊂ gl(n,R).

Now we will show that we can write every matrix in gl(n,R) as(x2 + 1

4

∑i yizi − ziyi

), with

x, yi, zi in Sym(n,R). This will imply that every element of gl(n,R) is of the form L(x) + Dand gl(n,R) ⊂ str(V ). Let a be an arbitrary matrix in gl(n,R), then

a =a+ at

2+a− at

2,


where (a+at)/2 is a symmetric matrix and (a−at)/2 is antisymmetric. So it will suffice to provethat we can write every antisymmetric matrix as a finite sum of commutators of symmetricmatrices. Consider an arbitrary antisymmetric matrix b. We define the symmetric matrices bl

as

blij =

bij if l ≤ i ≤ j−bij if l ≤ j < i

0 else.

These bl are symmetric and we have that b =∑n

i=1Eiibi − biEii, with the Eii as defined in

(2.2).A calculation shows that the Lie bracket for the structure algebra as defined in (2.14) cor-responds to the Lie bracket of gl(n,R) as defined in (2.9). Hence we have proven thatstr(V ) = gl(n,R) as Lie algebras.

We remark that from this proof it follows that the linear operator L(x) in str(V ) correspondsto the matrix x/2 in gl(n,R) and the operator [L(x), L(y)] in str(V ) corresponds to the matrix[x, y]/4 in gl(n,R). Hence, for V = Sym(n,R), the components h = Der(V ) and q = L(V ) of thestructure algebra str(V ) are given by the antisymmetric and symmetric matrices respectively.This implies that the sum

str(V ) = h + q

is a direct sum of vector spaces.For the action of the structure algebra on Sym(n,R) we have the following two lemmas.

Lemma 2.28. Let A,B in str(Sym(n,R)) and x in Sym(n,R), then

A · (B ·x)−B · (A ·x) = [A,B] ·x.

Proof. We haveA · (B ·x) = A(Bx+ xBt) + (Bx+ xBt)At,

andB · (A ·x) = B(Ax+ xAt) + (Ax+ xAt)Bt.

Subtracting these two equations, we get

A · (B ·x)−B · (A ·x) = ABx+ xBtAt −BAx− xAtBt

= [A,B]x+ x[A,B]t

= [A,B] ·x,

thus proving the lemma.

The following lemma gives the adjoint of an element of the structure algebra.

Lemma 2.29. For the action of the structure algebra on Sym(n,R) the adjoint of A instr(Sym(n,R)) with respect to the trace form τ is the transpose At, i.e.

τ(A ·x, y) = τ(x,At · y),

for all x, y ∈ Sym(n,R).

Proof. Using A ·x = Ax+ xAt, we get

τ(A ·x, y) = tr(Axy) + tr(xAty) = tr(yAx) + tr(Atyx) = τ(At · y, x) = τ(x,At · y),

where we used the invariance of the trace under cyclic permutations and the symmetry of thetrace form.

Chapter 3

The Bessel operators

We will start this chapter by introducing certain orbits of Sym(n,R) under the action of thegroup R+SL(n,R). We will be mainly interested in two special orbits: the open cone Ω andthe minimal orbit O.

We will then define the gradient ∇ and the operator P on a Jordan algebra V . Using these,we can define the Bessel operators Bλ for λ in C. For the value λ = 1/2 we will examine theaction of the Bessel operator on radial functions that are defined on the minimal orbit.

Our main goal is then to prove that for certain values of λ the Bessel operator on Sym(n,R) istangential to the orbit corresponding with this λ.

Finally, we will also define a family of Bessel functions for further reference.

3.1 Orbits of Sym(n,R)

Let L = R+SL(n,R) be the space of real n× n matrices with positive determinant:

L = g ∈ M(n,R)|det(g) > 0,

where M(n,R) are the real n× n-matrices. The action of L on Sym(n,R) is given by

g ·x = gxgt,

for x in Sym(n,R) and g in L.

Let c1, . . . cn be a Jordan frame of Sym(n,R). We define ek for k = 1, . . . , n as

ei = c1 + · · ·+ ck.

We then define for λ ∈ 1/2, 1, . . . , n/2 the orbit Oλ as the orbit of L containing the elemente2λ, thus

Oλ := L · e2λ.

For the case λ = n/2 we call the orbit On2

the open cone, which we denote by Ω, and for thecase λ = 1/2 we call the orbit O 1

2the minimal orbit, which we denote by O. We will now take

a closer look at these two special orbits.

3.1.1 The open cone Ω

Since en = c1 + · · · + cn = e, we have that the open cone Ω is the orbit of L containing theidentity element e, i.e.

Ω = L · e.

We have the following characterization for the open cone.

29

30 Chapter 3. The Bessel operators

Theorem 3.1. The open cone Ω consists of the symmetric matrices for which all eigenvaluesare positive.

Proof. By definition Ω consists of elements g · e, where g is a matrix with positive determinant.We have

g · e = gegt = ggt.

ThusΩ = ggt | g ∈ R+SL(n,R).

Consider a diagonal matrix x with eigenvalues λi, i = 1, . . . , n, which are all positive. Define gas the diagonal matrix (g)ii =

√λi. We then have that x = ggt and g is an element of L, since

det(g) =

n∏i=1

√λi > 0.

Let y be a symmetric matrix with positive eigenvalues. We know that there exists an (orthog-onal) matrix P such that

y = P txP,

where x is a diagonal matrix with positive eigenvalues. We have already shown that for this xthere exists a g in L, such that x = ggt. Define g′ as P tgP . We then have that g′ in L because

det(g′) = det(P t) det(g) det(P ) = det(g) > 0,

and y = g′g′t. Hence the symmetric matrices with positive eigenvalues are a subset of Ω.

We will now show that every element of Ω is a symmetric matrix with positive eigenvalues.Consider an element x of Ω. It is of the form ggt with g in L. Because

xt = (ggt)t = ggt = x,

we see that x is symmetric. Let P be an orthogonal matrix such that x′ = P txP is diagonal.Then x has the same eigenvalues as x′ and x′ = g′g′t, where g′ = P tgP . We then have

λi = x′ii =

n∑k=1

g′ikg′tki =

n∑k=1

(g′ik)2 ≥ 0.

If λi = 0, then g′ki = 0 for all k = 1, . . . , n. This implies that det(g′) = det(g) = 0, whichcontradicts g ∈ L. Therefore, we can conclude that all eigenvalues of x are positive.

We called Ω an open cone. We will now show that Ω is indeed a cone and open in Sym(n,R).First we will give the definition of a cone.

Definition 3.2. Let W be a finite-dimensional real vector space. A subset of W is called acone if for all elements x in the subset and λ in R, it holds that λ > 0 implies that λx is in thesubset.

With this definition we can easily prove that Ω is indeed a cone.

Lemma 3.3. The open cone Ω is a cone in Sym(n,R).

Proof. Let x be an arbitrary element of Ω and λ ∈ R+. Because of the definition of Ω, we canwrite x as x = ggt with g an element in L. Since λ is positive,

√λg is also in L. Hence we

have thatλx =

√λg(√λg)t

=(√λg)· e.

We conclude that λx is an element of Ω and thus that Ω is a cone.

3.1. Orbits of Sym(n,R) 31

We will now prove that Ω is open.

Lemma 3.4. It holds that Ω is an open subset of Sym(n,R).

Proof. Consider the function f from Sym(n,R) to R that maps a matrix x to its smallesteigenvalue. Because x is symmetric, we know that its eigenvalues are real and thus the smallesteigenvalue is well defined. The eigenvalues are given by a polynomial equation in the matrixentries and f is the minimum of the finite set of continuous functions that associate which eachmatrix x an eigenvalue. Hence we know that f is a continuous function.If we consider the open interval ]0,∞[ of R, we then have that f−1(]0,∞[) is an open subset ofSym(n,R), since f is continuous. But from Theorem 3.1 we know that f−1(]0,∞[) is exactlyΩ. Therefore we conclude that Ω is an open subset of Sym(n,R).

Because Ω is an open subset of Sym(n,R), we have that Ω is Lebesgue measurable. So we candefine integration on Ω as the integration of the corresponding measurable subset of Rk, wherek is the dimension of Sym(n,R), using the isomorphism defined by an orthonormal basis, i.e.∫

Ω⊂Sym(n,R)f(x)dx :=

∫Ω⊂R

n(n+1)2

f(x)dx,

where f is a Lebesgue integrable function from Ω to C.

One can show ([6], remark under Proposition 1.5.3) that the the measure det(x)−n+12 dx is

invariant under the action of L, i.e.∫Ωf(g ·x) det(x)−

n+12 dx =

∫Ωf(x) det(x)−

n+12 dx for all g in L.

From this and det(g ·x) = det(g) det(x) det(gt) = det(g)2 det(x) it then follows that∫Ωf(g ·x)dx = det(g)−(n+1)

∫Ωf(g ·x) det(g ·x)

n+12 det(x)−

n+12 dx

= det(g)−(n+1)

∫Ωf(x)dx,

for all g in L. Hence the integration over Ω is L−equivariant.

3.1.2 The minimal orbit

We called the orbit Oλ for the value λ = 1/2 the minimal orbit O. We have the followingcharacterization for O.

Theorem 3.5. The minimal orbit is given by

O =qqt|q ∈ Rn\0

.

Proof. We can choose a basis of Rn such that e1 = c1 has the following matrix form:

c1 =

(1

On−1

).

For an element g in L we then get

g · c1 = gc1gt

= g1gt1,

where g1 is the first column of g. Because det(g) > 0, we know that g1 is different from zero.


We remark that the norm of an element η = qqt in the minimal orbit is given by

|η| = qtq,

since|η| =

√tr(η2) =

√tr(qqtqqt) =

√(qtq)(qtq) = qtq.

Note that this implies that the trace is also equal to the norm, because

tr(η) =n∑i=1

q2i = qtq. (3.1)

The following theorem gives an alternative characterization of the minimal orbit in terms ofthe eigenvalues.

Theorem 3.6. The minimal orbit O of Sym(n,R) is given by the symmetrical matrices withexactly one positive eigenvalue different from zero. This positive eigenvalue is given by thenorm

|η| = q21 + · · ·+ q2

n = qtq,

for η in O.

Proof. We will first show that every η = qqt in the minimal orbit has indeed exactly oneeigenvalue λ = |q|2. We will prove this by induction on the dimension n of Sym(n,R). Theminimal orbit was given by (see Theorem 3.5)

O =qqt|q ∈ Rn\0

.

For n = 1, the lemma is true because every element in the minimal orbit is of the form q2, q inR\0. For n = 2 we can easily calculate the eigenvalues. Let q = (q1, q2)t ∈ R2, q 6= 0, then

det(qqt − λI2) =

∣∣∣∣q21 − λ q1q2

q1q2 q22 − λ

∣∣∣∣ = λ2 − λ(q21 + q2

2).

Hence the eigenvalues are λ = q21 + q2

2 = |q|2 and λ = 0.Assume the lemma is true for n = k. Consider the vectors q = (q1, . . . , qk+1)t in Rk+1\0 andq′ = (q2, . . . , qk+1)t in Rk. If q′ = 0 we immediately see that

qqt =

(q2

1

Ok

).

Thus in that case qqt has exactly one eigenvalue different from zero and it is equal to |q|2.Consider the case q′ 6= 0. Then we know by the induction hypothesis that q′q′t has oneeigenvalue |q′|2 and that the other eigenvalues are zero. Because q′q′t is a symmetrical matrix,there exists an orthogonal matrix P such that

Pq′q′tP t =

(|q′|2

Ok−1

). (3.2)

So we findk∑j=1

Pijq′j = (Pq′)i = δi1

∣∣q′∣∣ . (3.3)

Define the matrix Q as

Q =

(1

P

).


Note that Q is orthogonal since P is orthogonal. We know that the eigenvalues remain invariantunder orthogonal transformations. Hence it is sufficient to calculate the eigenvalues of QqqtQt.We have

QqqtQt =

(1

P

)(q2

1 q1q′t

q1q′ q′q′t

)(1

P t

)=

(q2

1 q1q′t

q1Pq′ Pq′q′t

)(1

P t

)=

(q2

1 q1(Pq′)t

q1Pq′ Pq′q′tP t

)

=

q21 q1 |q′|

q1 |q′| |q′|2

Ok−1

,

where we used (3.2) and (3.3). So the characteristic equation is given by

det(QqqtQt − λIk+1) =

∣∣∣∣∣∣q2

1 − λ q1 |q′|q1 |q′| |q′|2 − λ

−λIk−1

∣∣∣∣∣∣=(

(q21 − λ)

(∣∣q′∣∣2 − λ)− q21

∣∣q′∣∣2) (−λ)k−1

= (−λ)k(q21 +

∣∣q′∣∣2 − λ).

We conclude that qqt has one eigenvalue equal to q21 + |q′|2 = |q|2. The other eigenvalues are

zero.To complete the proof we have to show that every matrix η with one positive eigenvalue belongsto the minimal orbit O. Let λ be the eigenvalue of η and P the orthogonal matrix such that

η = P

(λ

On−1

)P t.

Define q as q = (√λ, 0, . . . , 0)t. Then we have that η = Pq(Pq)t and Pq 6= 0. Thus η is an

element in O.

This characterization allows us to prove that the minimal orbit is not invariant under the actionof the structure algebra on Sym(n,R).

Lemma 3.7. For n > 2, there exists an element A in the structure algebra str(Sym(n,R)) andan element η in the minimal orbit O, such that A · η is not in the minimal orbit O, where

A · η = Aη + ηAt

is the action of the structure algebra on Sym(n,R).

Proof. If η is of the form

η =

(|η|

On−1

),

then A · η is given by

Aη + ηAt = |η|(

2a11 a1

a1 On−1

),

where a1 = (a21, a31, . . . , an1)t. The eigenvalues are solutions of∣∣∣∣2a11 − λ a1

a1 −λIn−1

∣∣∣∣ = (2a11 − λ)(−λ)n−1 −n∑i=2

a2i1(−λ)n−2 = 0.


This implies that λ = 0 is a solution with algebraic multiplicity n− 2 and that the other twoeigenvalues are solutions of

λ2 − 2a11λ−n∑i=2

a2i1 = 0.

Hence, if∑n

i=2 a2i1 6= 0, then A · η has two eigenvalues different from zero. From Theorem 3.6

it follows that A · η is not in the minimal orbit.

3.1.3 Integration on the minimal orbit

We get as a consequence of Theorem 3.6 and the fact that every symmetric matrix is diago-nalizable the following polar decomposition of the minimal orbit O:

η = tζ, t ∈ R+, ζ ∈ S,

where t is the norm of η and where we set

S = ζ ∈ O | |ζ| = 1 = P · c1 | P ∈ SO(n,R) .

Here P · c1 is defined as the matrix product Pc1Pt and

c1 =

(1

On−1

).

Therefore we haveO ∼= R+ × S.

Using this polar decomposition, we get the following definition for integration on the minimalorbit (Theorem 1.5.10 in [6]).

Definition 3.8. The integration on the minimal orbit is given by∫Oψ(η)dµ(η) :=

Vol(Sn−1)

2

∫SO(n,R)

∫ ∞0

ψ(tk · c1)tn2−1dtdk,

where SO(n,R) is the special linear group of n× n orthogonal matrices with determinant equalto one and dk is the Haar measure on SO(n,R) normalized to one, i.e.∫

SO(n,R)dk = 1.

We define L2(O,dµ) as the space of square-integrable functions on the minimal orbit, i.e.

ψ(η) ∈ L2(O, dµ)⇐⇒∫O|ψ(η)|2 dµ(η) <∞.

We will now prove that there is a tight relation between L2(O,dµ) and L2(Rn).Define the following map between Rn\0 and the minimal orbit:

p : q → qqt. (3.4)

From Theorem 3.5 it follows that p is surjective. For q, q ∈ Rn\0, we find that p(q) = p(q)implies

qiqj = qiqj for i, j ∈ 1, . . . , n.

From q2i = q2

i we conclude that qi = ±qi and from q1qj = q1qj for j ∈ 1, . . . , n, we concludethat

q = q or q = −q. (3.5)

The map p allows us to establish a connection between integration on the minimal orbit andintegration on Rn.


Lemma 3.9. Let ψ : O → C be an integrable function on O. Then f = ψ p : Rn → C isintegrable and ∫

Rnf(q)dnq =

∫Oψ(η)dµ(η).

Proof. We know that there is an isomorphism between the (n − 1)-dimensional sphere andSO(n,R)/SO(n− 1,R). Set e1 = (1, 0, . . . , 0)t, then we have for k in SO(n− 1) that ke1 = e1.Using this and using spherical coordinates for the integration over Rn, we get∫

Rnψ p(q)dnq = Vol(Sn−1)

∫SO(n)

∫ ∞0

ψ p(tke1)tn−1dtdk

= Vol(Sn−1)

∫SO(n)

∫ ∞0

ψ(tke1et1ktt)tn−1dtdk

= Vol(Sn−1)

∫SO(n)

∫ ∞0

ψ(t2k · c1)tn−1dtdk

=Vol(Sn−1)

2

∫SO(n)

∫ ∞0

ψ(t′k · c1)t′n2−1dt′dk

=

∫Oψ(η)dµ(η),


We can now show that there is a unitary isomorphism between L2(O, dµ) and L2even(Rn).

Theorem 3.10. The map

p∗ : L2(O,dµ)→ L2even(Rn), ψ → ψ p, (3.6)

is unitary.

Proof. From Lemma 3.9 we know∫Rn|p∗ψ(q)|2 dnq =

∫O|ψ(η)|2 dµ(η),

thus p∗ preserves the norm.Because p is surjective, it follows that p∗ is injective. Indeed, if for ψ,ϕ in L2(O, dµ) one hasp∗(ψ) = p∗(ϕ), then ψ(qqt) = ϕ(qqt) for all q in Rn\0 and thus also ψ(η) = ϕ(η) for all η inO. Therefore ψ must be equal to ϕ.To prove that p∗ is surjective we associate with each f in L2

even(Rn) a function ψf in L2(O,dµ)in the following way:

ψf (η) := f(q) for all η ∈ O,

where we choose q such that qqt = η and q1 > 0. Because of (3.5), this is well defined. Wethen have

p∗(ψf )q =

f(q) for q1 > 0

f(−q) for q1 < 0.

The function f is even, so p∗(ψf )q = f and p∗ is surjective. We conclude that

p∗ : L2(O, dµ)→ L2even(Rn)

is unitary.


3.2 Operators on Sym(n,R)

The Bessel operator will be a composition of the gradient and the operator P . Therefore, wewill first introduce these two operators.

3.2.1 The gradient ∇

For a Jordan algebra V we define the gradient ∇ as a map from C∞(V ) to C∞(V ) ⊗ V suchthat the following holds:

τ(y,∇f(x)) =d

dt

∣∣∣t=0

f(x+ ty) for all y in V ,

where f ∈ C∞(V ) is a smooth complex-valued function, and τ the trace form, which for theJordan algebra Sym(n,R) is defined as τ(x, y) = tr(x ∗ y).If (eα)α is an orthonormal basis of V with respect to τ , we can decompose ∇f(x) as follows:

∇f(x) =∑α

(∇f(x)

)αeα, (3.7)

where the (∇f(x))α are in C∞(V ). These coefficients are the partial derivatives, since

(∇f(x)

)α

= τ(eα,∇f(x)

)=

d

dt

∣∣∣t=0

f(x+ teα) =∂f

∂xα.

To be able to calculate the action of the Bessel operator on radial functions, i.e. functions thatonly depend on the norm |x| =

√τ(x, x), we will need the gradient of radial functions. To

obtain this gradient we first establish the following lemma, which gives us the gradient of thenorm.

Lemma 3.11. The gradient of the norm is given by

∇ |x| = x

|x|.

Proof. Consider an orthonormal basis (eα)α with respect to the trace form τ . We have

∇ |x| =∑α

∂|x|∂xα

eα =∑α

∂√τ(x, x)

∂xαeα =

∑α

1

2√τ(x, x)

∂τ(x, x)

∂xαeα.

Using the linearity of the trace, we get

∂τ(x, x)

∂xα= lim

t→0

τ(x+ teα, x+ teα)− τ(x, x)

t

= limt→0

2t τ(x, eα) + t2τ(eα, eα)

t

= 2τ(x, eα),

which yields

∇ |x| =∑α

2τ(x, eα)

2√τ(x, x)

eα =x

|x|.

Hence the claim is proved.

Using the chain rule, this immediately leads to the following.

3.2. Operators on Sym(n,R) 37

Lemma 3.12. For a radial function ψ(x) = f(|x|), f ∈ C∞(R+), the gradient is given by

∇ψ(x) = f ′(|x|) x|x|.

We will also calculate the gradient of the determinant in Sym(n,R). We will need this resultlater to prove that the Bessel operator Bλ is tangential to the orbit Oλ.

Lemma 3.13. The gradient of the determinant is given by

∇ det(x) = det(x)x−1,

for all x in Sym(n,R) with det(x) 6= 0.

Proof. Consider an arbitrary element x of Sym(n,R) for which det(x) 6= 0. We choose a basisof Rn such that x is a diagonal matrix. Then we have that

det(x) =n∏i=1

λi,

where λi are the eigenvalues of x. Since det(x) 6= 0, these eigenvalues are all different fromzero.

Let (Eij)i≤j be the basis of Sym(n,R) defined in (2.7). To calculate the gradient we need an

expression for det(x+ tEij). For i = j we can expand the determinant along the ith row. Wethen get

det(x+ tEij) = (λi + t)n∏

k=1,k 6=iλk = det(x) + t

n∏k=1,k 6=i

λk.

For i 6= j we can expand the determinant along all rows except the ith and jth. This gives

det(x+ tEij) =

n∏k=1,k 6=i,j

λk

∣∣∣∣∣ λit√2

t√2

λj

∣∣∣∣∣ = det(x)− t2

2

n∏k=1,k 6=i,j

λk.

Therefore

∂

∂xijdet(x) = lim

t→0

det(x+ tEij)− det(x)

t

= δij

n∏k=1,k 6=i

λk

= det(x) δijλ−1i .

Because x−1 is the diagonal matrix with eigenvalues λ−1i , we obtain

∇ det(x) = det(x)x−1,

which had to be proven.

3.2.2 The operator P

In this section we will define the operator P which is a composition of the operator L.


Definition 3.14. For each element x of a Jordan algebra V the linear operator P (x) : V → Vis defined as

P (x) = 2L(x)2 − L(x2).

Thus, for all y in V ,P (x)y = 2

(x ∗ (x ∗ y)

)− (x ∗ x) ∗ y,

where the brackets cannot be interchanged.

We also define for each x, y in V the operator P (x, y) in the following manner.

Definition 3.15. For each x, y of a Jordan algebra V the linear operator P (x, y) : V → V isdefined as

P (x, y) := L(x)L(y) + L(y)L(x)− L(x ∗ y).

Thus, for all z in V ,

P (x, y)z = x ∗ (y ∗ z) + y ∗ (x ∗ z)− (x ∗ y) ∗ z.

It is clear that P (x, y) = P (y, x) and P (x, x) = P (x). We also have the following bilinearity.

Lemma 3.16. For x, y, z in V and λ, µ in R we have

P (λx+ µy, z) = λP (x, z) + µP (y, z).

Proof. This follows from

P (λx+ µy, z)v = (λx+ µy) ∗ (z ∗ v) + z ∗((λx+ µy) ∗ v

)−((λx+ µy) ∗ z

)∗ v,

for all v in V , and the bilinearity of the Jordan product.

If V = Sym(n,R) we can express the operator P using standard matrix multiplication.

Lemma 3.17. Let x, y, z be elements of Sym(n,R). Then we have

P (x, y)z =1

2(xzy + yzx),

where xy denotes standard matrix multiplication. In particular we have

P (x)z = xzx.

Proof. Using the definition of the Jordan product we find

x ∗ (y ∗ z) =1

4(xyz + xzy + yzx+ zyx)

y ∗ (x ∗ z) =1

4(yxz + yzx+ xzy + zxy)

(x ∗ y) ∗ z) =1

4(xyz + yxz + zxy + zyx).

Combining these expressions, we get

P (x, y)z = x ∗ (y ∗ z) + y ∗ (x ∗ z)− (x ∗ y) ∗ z)

=1

4(xzy + yzx+ yzx+ xzy)

=1

2(xzy + yzx),



We immediately have the following corollary.

Lemma 3.18. For an invertible x in Sym(n,R) and an arbitrary y in Sym(n,R) the followingholds for the operator P (x, y):

P (x−1, y)x = y.

Proof. Computing this with the previous lemma, we get

P (x−1, y)x =1

2(x−1xy + yxx−1) = y,

which proves the statement.

Consider a function f in C∞(V ) and an orthonormal basis (eα)α of V with respect to the traceform τ . We will use the operator P and the gradient to define

P (∇)x : C∞(V )→ C∞(V )⊗ V,

which maps complex-valued functions to vector-valued functions, as follows:(P (∇)x

)(f) = 2

(∇ ∗ (∇ ∗ x)

)(f)−

((∇ ∗∇) ∗ x

)(f)

=∑α,β

(2eα

∂

∂xα∗(eβ

∂

∂xβ∗ x)−(eα

∂

∂xα∗ eβ

∂

∂xβ

)∗ x)

(f)

=∑α,β

∂2f(x)

∂xα∂xβ

(2eα ∗ (eβ ∗ x)− (eα ∗ eβ) ∗ x

)=∑α,β

(∂2f(x)

∂xα∂xβeα ∗ (eβ ∗ x) +

∂2f(x)

∂xβ∂xαeβ ∗ (eα ∗ x)− ∂2f(x)

∂xα∂xβ(eα ∗ eβ) ∗ x

)

=∑α,β

∂2f(x)

∂xα∂xβP (eα, eβ)x. (3.8)

Note that the derivatives in the gradient only work on f and not on the variable x, consequentwith the fact that ∇ ∗ x should be the same as x ∗ ∇.

3.2.3 The Bessel operators BλWe are now ready to define for each λ in C a second order differential operator

Bλ : C∞(V )→ C∞(V )⊗ V

byBλ := P (∇)x+ λ∇.

This operator is called the Bessel operator. If (eα)α is an orthonormal basis of the Jordanalgebra V with respect to the trace τ , then we find from (3.7) and (3.8)

Bλf(x) =∑α,β

∂2f(x)

∂xα∂xβP (eα, eβ)x+ λ

∑α

∂f(x)

∂xαeα. (3.9)

Since the partial derivative is linear, the Bessel operator is also linear. For the Bessel operatorwe have the following product rule.

Theorem 3.19. For f, g in C∞(V ) it holds that

Bλ(fg) = Bλ(f)g + 2P (∇f,∇g)x+ fBλ(g).


Proof. We know from the standard product rule that

∂fg

∂xα=

∂f

∂xαg + f

∂g

∂xα

and∂2fg

∂xα∂xβ=

∂2f

∂xα∂xβg +

∂f

∂xα

∂g

∂xβ+

∂f

∂xβ

∂g

∂xα+ f

∂2g

∂xα∂xβ.

Substituting this in (3.9), we subsequently get

Bλ(fg) =∑α,β

∂2fg

∂xα∂xβP (eα, eβ)x+ λ

∑α

∂fg

∂xαeα

=∑α,β

(∂2f

∂xα∂xβg +

∂f

∂xα

∂g

∂xβ+

∂f

∂xβ

∂g

∂xα+ f

∂2g

∂xα∂xβ

)P (eα, eβ)x

+ λ∑α

(∂f

∂xαg + f

∂g

∂xα

)eα

= Bλ(f)g + 2∑α,β

∂f

∂xα

∂g

∂xβP (eα, eβ)x+ fBλ(g)

= Bλ(f)g + 2P (∇f,∇g)x+ fBλ(g),

thus completing the proof.

For Sym(n,R) we want to calculate the action of the Bessel operator Bλ for λ = 1/2 on radialfunctions defined on the minimal orbit O. To do this we first need the following lemma.

Lemma 3.20. For the operator P and the orthonormal basis (Eij)i≤j, defined in (2.7), thefollowing holds for every x in Sym(n,R):

∑1≤i≤j≤n

P (Eij)x =1

2(x+ tr(x)e), (3.10)

where e is the unit of Sym(n,R).

Proof. First we set x(ij) = τ(x, Eij), such that x =∑

i≤j x(ij)Eij . Note that x(ii) = xii but,

for i 6= j, it does not hold that x(ij) = xij , where xij the (i, j)th matrix entry is. Instead

x(ij) =√

2xij , since Eij got a factor 1/√

2.

From Lemma 3.17 we get

P (Eij)x = EijxEij .

For i 6= j, multiplying on the left with√

2Eij switches the ith and jth row and gives zero for

the other rows, while multiplying on the right with√

2Eij switches the ith and jth column andgives zero for the other columns.

For i = j, multiplying on the left with Eii just gives the ith row and multiplying with Eij onthe right gives the ith column. Hence,

EijxEij =

12

(x(ij)Eij + x(ii)Ejj + x(jj)Eii

)for i 6= j

x(ii)Eii for i = j.


Therefore, we get∑1≤i≤j≤n

P (Eij)x =∑

1≤i<j≤n

1

2

(x(ij)Eij + x(ii)Ejj + x(jj)Eii

)+

n∑i=1

x(ii)Eii

=x

2+

1

2

n∑i=1

x(ii)

n∑j=i+1

Ejj +1

2

n∑j=1

x(jj)

j−1∑i=1

Eii +1

2

n∑i=1

x(ii)Eii

=x

2+

1

2

n∑i=1

x(ii)

n∑j=1

Ejj

=1

2(x+ tr(x)e),

which completes the proof.

We will use this lemma to prove the following theorem.

Theorem 3.21. Consider a radial function ψ : O → C, i.e. ψ(η) = f(|η|), where f is afunction in C∞(R+). The action of the Bessel operator for the value 1/2 on ψ is given by

B 12ψ(η) = f ′′(|η|)η +

1

2f ′(|η|)e.

Proof. Using the gradient of radial functions (Lemma 3.12) and the definition of the Besseloperator, we get

B 12ψ(η) =

(P (∇)η

)ψ +

1

2∇ψ

=∑α,β

∂

∂xα

(f ′(|η|)|η|

ηβ

)P (eα, eβ)η +

1

2f ′(|η|) η

|η|

=∑α,β

((f ′′(|η|)|η|

− f ′(|η|)|η|2

)ηα|η|ηβ +

f ′(|η|)|η|

δαβ

)P (eα, eβ)η +

1

2f ′(|η|) η

|η|

=

(f ′′(|η|)|η|2

− f ′(|η|)|η|3

)P (η)η +

∑α

(f ′(|η|)|η|

)P (eα)η +

1

2f ′(|η|) η

|η|, (3.11)

where we also used

∂ηα∂xβ

= limt→0

τ(η + teβ, eα)− τ(η, eα)

t= τ(eα, eβ) = δαβ. (3.12)

Since Sym(n,R) is power associative (Lemma 2.4), we can easily calculate P (η)η as

P (η)η = 2(η ∗ (η ∗ η)

)− (η ∗ η) ∗ η = η3.

Further from (3.1), it follows that |η| = tr(η) and

η ∗ η = η2 = qqtqqt = |η| η.

If we consider the orthonormal basis defined in (2.7), we can apply Lemma 3.20 to the secondterm in the right-hand side of (3.11). This leads to

B 12ψ(η) = f ′′(|η|)η − f ′(|η|) η

|η|+

1

2f ′(|η|)

(η

|η|+ e

)+

1

2f ′(|η|) η

|η|

= f ′′(|η|)η +1

2f ′(|η|)e,

which proves the theorem.


3.3 The Bessel operator is tangential

3.3.1 The determinant det(x)

Before we go on to prove that the Bessel operator Bλ is tangential to Oλ, we will first need anexpression for the action of the Bessel operator on powers of the determinant. To be able tocalculate this action, we will need the following lemmas.

Lemma 3.22. Let (eα)α be an orthonormal basis of Sym(n,R). The gradient of the inversex−1β = τ(x−1, eβ) is given by

∇(x−1β ) = −P (x−1)eβ.

For the coefficients of the gradient we also have

∂x−1β

∂xα= −τ(P (x−1)eα, eβ).

Proof. Consider an invertible element x in Sym(n,R). Denote by x−1 its inverse for matrixmultiplication. Taking the partial derivative ∂/∂xα of xx−1 = In, we get

∂x

∂xαx−1 + x

∂x−1

∂xα= 0. (3.13)

We have

∂x

∂xα=∑β

∂xβ∂xα

eβ =∑β

δαβeβ = eα.

Consequently, multiplying (3.13) with x−1 leads to

∂x−1

∂xα= −x−1eαx

−1 = −P (x−1)eα,

where we used Lemma 3.17. Taking the inner product of this expression with eβ, we get

∂x−1β

∂xα= −τ

(P (x−1)eα, eβ

),

which proves the second part of the lemma. Multiplying this with eα and summing over α, weobtain

∇x−1β =

∑α

∂x−1β

∂xαeα = −

∑α

τ(P (x−1)eα, eβ

)eα = −P (x−1)eβ,

sinceτ(P (x−1)eα, eβ

)= tr(x−1eαx

−1eβ) = tr(x−1eβx−1eα) = τ

(P (x−1)eβ, eα

),

which holds because of Lemma 3.17 and the invariance of the trace under cyclic permutations.

Using this lemma we can calculate the derivatives of det(x)µ.

Lemma 3.23. Let (eα)α be an orthonormal basis. Then we have for µ in C that

∂

∂xαdet(x)µ = µ det(x)µx−1

α

∂2

∂xα∂xβdet(x)µ = µ det(x)µ

(µx−1

α x−1β −

(P (x−1)eα

)β

),

where x is an element in Sym(n,R) for which det(x) 6= 0.

3.3. The Bessel operator is tangential 43

Proof. We have already proven in Lemma 3.13 that ∇ det(x) = det(x)x−1. By the chain rule,we then immediately find

∂

∂xαdet(x)µ = µdet(x)µ−1 det(x)x−1

α = µ det(x)µx−1α .

Using the previous lemma, we then find for the second order derivative

∂2

∂xα∂xβdet(x)µ = µ2 det(x)µx−1

α x−1β − µdet(x)µτ

(P (x−1)eα, eβ

),

from which the result follows.

We will also need the following lemma.

Lemma 3.24. For an orthonormal basis (eα)α in Sym(n,R) the following holds:∑α

e2α =

n+ 1

2e, (3.14)

where e is the unit of Sym(n,R).

Proof. Consider the orthonormal basis (Eij)i≤j defined in (2.7) . We have already calculatedthat

Eij ∗ Eij =

12(Eii + Ejj) for i 6= j

Eij for i = j.

Substituting this in the left-hand side of (3.14), gives us

∑1≤i≤j≤n

E2ij =

1

2

n∑j=1

j−1∑i=1

Eii +1

2

n∑i=1

n∑j=i+1

Ejj +n∑i=1

Eii

=

(1

2n+

1

2

) n∑i=1

Eii

=n+ 1

2e.

Consider another orthonormal basis (eα)α in Sym(n,R). We can express Eij in this basis as∑α

τ(Eij , eα

)eα.

This leads to ∑1≤i≤j≤n

E2ij =

∑1≤i≤j≤n

∑α,β

τ(Eij , eα

)eατ

(Eij , eβ

)eβ.

Using Lemma 2.15, this reduces to∑1≤i≤j≤n

E2ij =

∑α,β

τ(eα, eβ)eαeβ =∑α

e2α.

Hence, (3.14) holds for all orthonormal basis in Sym(n,R).

We can finally calculate the action of the Bessel operator Bλ on det(x)µ, where we will restrictthe determinant to the open cone Ω.


Theorem 3.25. The action of the Bessel operator Bλ : C∞(Ω) → C∞(Ω) ⊗ Ω on det(x)µ isgiven by

Bλ (det(x)µ) = µ

(µ+ λ− n+ 1

2

)det(x)µx−1, (3.15)

where λ and µ are complex constants.

Proof. From Theorem 3.1 we know that det(x) > 0 for all x in Ω. Thus the derivatives ofdet(x)µ are well defined. From Lemma 3.18 we know that

P (x−1, x−1)x = x−1. (3.16)

We also have, see Lemma 3.17,

P (eα, x−1eαx

−1)x =1

2

(eαxx

−1eαx−1 + x−1eαx

−1xeα)

=1

2

(e2αx−1 + x−1e2

α

)= e2

α ∗ x−1. (3.17)

We can calculate Bλ(det(x)µ) by using (3.16), (3.17), Lemma 3.23 and Lemma 3.24:

Bλ(det(x)µ) =∑α,β

∂2

∂xα∂xβ(det(x)µ)P (eα, eβ)x+ λ

∑α

∂

∂xα(det(x)µ) eα

= µ2 det(x)µP(x−1, x−1

)x− µdet(x)µ

∑α

P(eα, x

−1eαx−1)x+ λµ det(x)µx−1

= µ det(x)µ(µx−1 − n+ 1

2x−1 + λx−1

)= µ

(µ+ λ− n+ 1

2

)det(x)µx−1,

which proves the statement.

3.3.2 The zeta function

The zeta function is defined as the following integration over the open cone Ω:

Z(ψ, λ) :=

∫Ωψ(x) det(x)λ−

n+12 dx, (3.18)

with λ ∈ C and <(λ) > (n− 1)/2. Here <(λ) stands for the real part of λ. We will show thatthe Bessel operator is symmetric with respect to the zeta function Z. This will be the crucialpart in the proof of the theorem that the Bessel operator is tangential.

Let S(V ) be the Schwartz space of rapidly decreasing functions on Sym(n,R). Then we havethe following lemma, which states that the zeta function is well defined (Proposition VII.2.1in [3]).

Lemma 3.26. For a function ψ in S(V ), the integral∫Ωψ(x) det(x)λ−

n+12 dx

converges absolutely for <(λ) > (n− 1)/2.

We can use this result to prove that the zeta function is holomorphic as a function of λ.


Lemma 3.27. The complex function

Z(ψ, · ) : λ→∫

Ωψ(x) det(x)λ−

n+12 dx,

is holomorphic for <(λ) > (n − 1)/2 and ψ in S(V ), where S(V ) is the Schwartz space ofrapidly decreasing functions on Sym(n,R).

Proof. For α, β in C we haveαβ := eβ ln(α),

where ln is the complex logarithm defined using the negative numbers ]−∞, 0] as the branchcut. From [8], Theorem 2.3.5 we know that

z → az

is holomorphic for all a in R+ and(az)′ = az ln(a).

Set µ = λ− (n+ 1)/2. To show that∫

Ω ψ(x) det(x)µdx is holomorphic for <(µ) > −1 we willuse the Theorem of Morera ([8], Theorem 1.5.17). This theorem states that∮

γf(z)dz = 0,

for all closed piecewise continuous functions γ in a connected open set D, implies that f isholomorphic on D.Consider a disc D in the complex plane with an arbitrary origin and a finite radius such that,for all z in this disc, we have <(z) > −1. Then we need to prove that, for every closed piecewisecontinuous function γ in this disc,∮

γ

∫Ωψ(x) det(x)zdxdz = 0.

If we can change the order of the integration, this follows from the fact that on Ω we havedet(x) > 0 and thus det(x)z is holomorphic which implies∮

γdet(x)zdz = 0.

To see that we can indeed change the order of integration, we need to check that the conditionto apply the Theorem of Fubini is satisfied, i.e.∫

γ×Ω|ψ(x) det(x)z|dz × dx <∞.

We know that ∣∣det(x)R∣∣ =

∣∣∣eR ln(det(x))∣∣∣

is an increasing function of R for det(x) > 1 and a non-increasing function of R for det(x) ≤ 1.Take R ∈ R+ and ε ∈ R+ such that

−1 + ε < <(z) < R

for all z ∈ D. Then we get

|ψ(x) det(x)z| ≤

∣∣ψ(x) det(x)R∣∣ for x such that det(x) > 1∣∣ψ(x) det(x)−1+ε∣∣ for x such that det(x) ≤ 1.


The piecewise-defined function on the right-hand side is an integrable function on γ×Ω becauseof Lemma 3.26. Therefore Fubini is satisfied and we can conclude that the function

µ→∫

Ωψ(x) det(x)µdx

is holomorphic for <(µ) > −1.

We want to extend the zeta function to the whole complex plane. To do this, we first mentionthe following result (Corollary VII.1.3 in [3]).

Theorem 3.28. The gamma function ΓΩ of the symmetric cone, defined as

ΓΩ(λ) :=

∫Ωe− tr(x) det(x)λ−

n+12 dx,

converges absolutely for <(λ) > (n− 1)/2. It has the value

ΓΩ(λ) = (2π)n2

2 Γ (λ) Γ

(λ− 1

2

)· · ·Γ

(λ− n− 1

2

). (3.19)

Here Γ(λ) is the classical gamma function defined as∫ +∞

0 e−ttλ−1dt.

Since the classical gamma function admits a meromorphic extension to the whole complexplane with poles at the negative integers, it is also possible to extend the gamma function ΓΩ

of the symmetric cone to a meromorphic function on the whole complex plane using (3.19).This extension will have poles at the negative integers and negative half-integers and also for

λ ∈

0,1

2, . . . ,

n− 1

2

.

We will also use the following lemma (Theorem VII.2.2 in [3]).

Lemma 3.29. For a function ψ in the Schwartz space S(V ), the integral

1

ΓΩ(λ)


n+12 dx (3.20)

admits an analytic continuation as a holomorphic function of λ to the whole complex plane.

We can define the extension of the zeta function as the product of (3.20) and the gammafunction ΓΩ. Because this is the product of a holomorphic function and a meromorphic function,we get the following lemma.

Lemma 3.30. Let the function ψ be an element of S(V ), where S(V ) is the Schwartz space ofrapidly decreasing functions on Sym(n,R). Then the complex function

Z(ψ, · ) : λ→∫

Ωψ(x) det(x)λ−

n+12 dx,

is holomorphic for <(λ) > (n − 1)/2 and has a unique meromorphic extension to the wholecomplex plane. It has poles for the negative integers and negative half-integers and for

λ ∈

0,1

2, . . . ,

n− 1

2

.

Proof. We can define the extension as the product of the holomorphic function

1

ΓΩ(λ)


n+12 dx,

and the meromorphic function ΓΩ(λ). The product of a holomorphic function and a meromor-phic function is again meromorphic and it has the same poles as ΓΩ.


We can now prove that the Bessel operator is symmetric with respect to the zeta function. Weuse the notation C∞c (Sym(n,R)) for the space of smooth functions on Sym(n,R) with compactsupport.

Theorem 3.31. Let ψ be in C∞(Sym(n,R)) and ϕ in C∞c (Sym(n,R)), or let ψ and ϕ be bothin the Schwartz space of Sym(n,R). For λ in C it holds that

Z(Bλ(ψ)ϕ, λ

)= Z

(ψBλ(ϕ), λ

),

i.e. ∫ΩBλ(ψ(x)

)ϕ(x) det(x)λ−

n+12 dx =

∫Ωψ(x)Bλ

(ϕ(x)

)det(x)λ−

n+12 dx, (3.21)

where for λ ≤ (n− 1)/2 the integral denotes the meromorphic extension.

Proof. Set µ = λ− n+12 and assume first µ > 1.

Using integration by parts we have, for ψ and ϕ in C∞(Sym(n,R)),∫Ω

∂ψ

∂xαϕdx = −

∫Ωψ∂ϕ

∂xαdx+

∫∂Ωψϕdx.

The open cone Ω consists of matrices for which all eigenvalues are positive (Lemma 3.1). If x isan element of the boundary of Ω, then it will have a zero eigenvalue and hence its determinantwill be zero. For µ > 1 this implies that the integrals∫

∂Ωψ(x)ϕ(x) det(x)µdx

and ∫∂Ωψ(x)ϕ(x)

∂det(x)µ

∂xadx

are zero. Hence ∫Ω

∂ψ(x)

∂xαϕ(x) det(x)µdx = −

∫Ωψ∂ϕ(x) det(x)µ

∂xαdx,

and ∫Ω

∂2ψ(x)

∂xα∂xβϕ(x) det(x)µdx =

∫Ωψ∂2ϕ(x) det(x)µ

∂xα∂xβdx.

We have∫ΩBλ(ψ(x)

)ϕ(x) det(x)µdx

=∑α,β

∫Ω

∂2ψ(x)

∂xα∂xβP (eα, eβ)x ϕ(x) det(x)µdx

︸︷︷︸I

+λ∑α

∫Ω

∂ψ(x)

∂xαeαϕ(x) det(x)µdx︸︷︷︸II

.

By applying integration by parts twice, we find for the first term on the right-hand side

I =∑α,β

∫Ωψ(x)

∂2

∂xα∂xβ

(P (eα, eβ)x ϕ(x) det(x)µ

)dx

=∑α,β,γ

∫Ωψ(x)P (eα, eβ)eγ

(2∂xγ∂xα

∂

∂xβ

(ϕ(x) det(x)µ

)+ xγ

∂2

∂xα∂xβ

(ϕ(x) det(x)µ

))dx

=∑α,β

∫Ωψ(x)2P (eα, eβ)eα

∂

∂xβ

(ϕ(x) det(x)µ

)dx+

∫Ωψ(x)B0(ϕ(x) det(x)µ)dx,


where we used ∂xγ/∂xα = δγα (see 3.12) and the definition of the Bessel operator Bλ for thevalue λ = 0. From Lemma 3.17 and Lemma 3.24 we deduce∑

α

P (eα, eβ)eα =1

2

∑α

eαeαeβ + eβeαeα =∑α

e2α ∗ eβ =

n+ 1

2eβ.

Substituting this in the expression for I, we get

I =

∫Ωψ(x)

(B0

(ϕ(x) det(x)µ

)+ (n+ 1)∇

(ϕ(x) det(x)µ

))dx.

For the second term on the right-hand side we find

II = −λ∑α

∫Ωψ(x)

∂

∂xα

(eαϕ(x) det(x)µ

)dx

= −λ∫

Ωψ(x)∇

(ϕ(x) det(x)µ

)dx.

Combining the results for I and II we get∫ΩBλ(ψ(x)

)ϕ(x) det(x)µdx =

∫Ωψ(x)Bn+1−λ

(ϕ(x) det(x)µ

)dx.

Applying the product rule for the Bessel operator (Theorem 3.19), we obtain

Bn+1−λ(ϕ(x) det(x)µ

)= Bn+1−λ

(ϕ(x)

)det(x)µ + 2P

(∇ϕ(x),∇ det(x)µ

)x+ ϕ(x)Bn+1−λ

(det(x)µ

).

The last term of the right-hand side is zero because µ + n + 1 − λ − (n + 1)/2 = 0 (seeTheorem 3.25). We also have

2P (∇ϕ,∇ det(x)µ)x = 2µdet(x)µP(∇ϕ, x−1

)x

= 2µdet(x)µ∇ϕ,

because of Lemma 3.18 and Lemma 3.23. So we find

Bn+1−λ(ϕ(x) det(x)µ

)= det(x)µBn+1−λ+2µ

(ϕ(x)

).

Since n+ 1− λ+ 2µ = λ, we finally obtain, for µ > 1,∫ΩBλ(ψ(x))ϕ(x) det(x)µdx =

∫Ωψ(x)Bλ(ϕ(x)) det(x)µdx.

Since ψ is in C∞(Sym(n,R)) and ϕ is in C∞c (Sym(n,R)), or since ψ and ϕ are both inS(Sym(n,R)), we have that Bλ(ψ)ϕ is in S(Sym(n,R)). From Lemma 3.30, it follows thenthat ∫

ΩBλ(ψ(x)

)ϕ(x) det(x)µdx−

∫Ωψ(x)Bλ

(ϕ(x)

)det(x)µdx

has a unique meromorphic extension to the whole complex plane. Because it is zero for µ > 1and the meromorphic extension is unique, we conclude that∫

ΩBλ(ψ(x)

)ϕ(x) det(x)µdx−

∫Ωψ(x)Bλ

(ϕ(x)

)det(x)µdx = 0

for all µ in C, thus completing the proof.

3.4. The Bessel functions 49

3.3.3 Tangential operators

We will use the zeta function defined in the previous section to prove that the Bessel operatorBλ is tangential to the orbit Oλ. We first state the following theorem for measures dµλ on theorbits Oλ (Proposition 1.5.12 in [6]).

Theorem 3.32. Let V be the Euclidean Jordan algebra Sym(n,R). Then the measure dµλ,for λ in 1

2 , 1, . . .n−1

2 , is a constant multiple of the residue of the zeta function

Z(f, ν) :=

∫Ωf(x) det(x)ν−

n+12 dx (3.22)

at the value ν = λ.

So for integration on the orbit Oλ we have∫Oλf dµλ := const · resλ=νZ(f, ν).

The definition of a tangential operator is as follows.

Definition 3.33. An operator B : C∞(V ) → C∞(V ) is tangential to the subspace W ⊂ V iffor all f in C∞(V ) we have

f |W = 0 =⇒ B(f)|W = 0.

We will now prove that the Bessel operator Bλ is tangential to the orbit Oλ.

Theorem 3.34. For every λ in 1/2, 1, . . . , n/2 the Bessel operator Bλ is tangential to theorbit Oλ.

Proof. First, for λ = n/2, we have that Oλ = Ω. Since Ω is open (Lemma 3.4), the differentialoperator Bλ will be tangential.For λ 6= n/2, consider a function ϕ in C∞(V ) such that ϕ|Oλ = 0 and let ψ in C∞c (V ) bearbitrary. We have, using Theorem 3.31 and Theorem 3.32,∫

OλBλ(ϕ)ψdµλ = const · resλ=νZ(Bλ(ϕ)ψ, ν)

= const · resλ=νZ(ϕBλ(ψ), ν)

=

∫OλϕBλ(ψ)dµλ

= 0,

because ϕ = 0 on Oλ. As ψ is arbitrary we conclude that

Bλ(ϕ) = 0

on Oλ. Hence the Bessel operator is tangential to Oλ.

3.4 The Bessel functions

In this section we will define a family of Bessel functions which we will use later on. Theseresults can be found in [6], appendix D.The modified Bessel function of the first kind or I-Bessel function is defined as

Iα(z) =(z

2

)α ∞∑n=0

1

n!Γ(n+ α+ 1)

(z2

)2n,


for z in C. It is a meromorphic function in z and α and solves the following second orderdifferential equation:

z2 du

dz2+ z

du

dz− (z2 + α2)u = 0.

The modified Bessel function of the third kind or K-Bessel function defined as

Kα(z) =π

2 sin(πα)

(I−α(z)− Iα(z)

)also solves this differential equation. We will be using the following renormalizations:

Iα(z) :=(z

2

)−αIα(z), (3.23)

and

Kα(z) :=(z

2

)−αKα(z).

We will be mostly interested in the case α = −1/2. Using the following property of the gammafunction for non-negative integer values of n:

Γ

(n+

1

2

)=

(2n)!

4nn!

√π,

we get

I− 12(z) =

∞∑n=0

1

n!Γ(n+ 12)

(z2

)2n

=∞∑n=0

1

n!(2n)!√π

4nn!(z

2

)2n

=1√π

∞∑n=0

z2n

2n!

=cosh(z)√

π. (3.24)

In the same way we also get, using Γ(n+ 32) = (n+ 1

2)Γ(n+ 12),(z

2

) 12I 1

2(z) =

z

2

∞∑n=0

1

n!(n+ 12)Γ(n+ 1

2)

(z2

)2n

=∞∑n=0

1

n!(n+ 12)(2n)!

√π

4nn!(z

2

)2n+1

=1√π

∞∑n=0

z2n+1

(2n+ 1)2n!

=sinh(z)√

π.

Thus for the K-Bessel function we find

K− 12(z) =

π

2 sin(−π2 )

((z2

) 12I 1

2− I− 1

2(z)

)= −√π

2(sinh(z)− cosh(z))

=

√π

2e−z. (3.25)

Chapter 4

The Schrodinger model

In this chapter we will define the conformal algebra, also called the Tits-Kantor-Koecher alge-bra, for the Jordan algebra Sym(n,R). We will show that it is a Lie algebra and we will definean action of this algebra on Sym(n,R).

We will use the conformal algebra to construct the Schrodinger model. First, we define a Liealgebra representation of the conformal algebra on the smooth functions defined on the minimalorbit. We will then look at a subrepresentation and show that this subrepresentation can beextended to a representation of the conformal algebra on L2(O, dµ).

4.1 The conformal algebra

For the Jordan algebra Sym(n,R) the conformal algebra will be isomorphic to the Lie algebraof symplectic matrices. Therefore, we first recall the definition of this Lie algebra.

The Lie algebra of symplectic matrices sp(2n,R) is defined by

sp(2n,R) =

(A bc −At

)| A ∈ gl(n,R) and b, c ∈ Sym(n,R)

.

The Lie bracket for X and Y in sp(2n,R) is given by

[X,Y ] = XY − Y X,

where XY is standard matrix multiplication. For

X =

(A bc −At

)and Y =

(T uv −T t

),

where A, T ∈ gl(n,R) and b, c, u, v ∈ Sym(n,R), we get

[X,Y ] =

(A bc −At

)(T uv −T t

)−(T uv −T t

)(A bc −At

)=

(AT + bv − TA− uc Au− bT t − Tb+ uAt

cT −Atv − vA+ T tc cu+AtT t − vb− T tAt)

=

([A, T ] + bv − uc A ·u− T · b−At · v + T t · c −[A, T ]t − (bv − uc)t

). (4.1)

We will now define the conformal algebra and establish an isomorphism between the conformalalgebra and sp(2n,R).

51

52 Chapter 4. The Schrodinger model

Definition 4.1. The conformal algebra co(V ), also called the Tits-Kantor-Koecher algebra, ofthe Jordan algebra V = Sym(n,R) is given by

co(V ) = n + l + n, (4.2)

where

n := (u, 0, 0) | u ∈ V l := (0, T, 0) | T ∈ str(V )n := (0, 0, v) | v ∈ V .

The Lie bracket on co(Sym(n,R)) is defined as follows:

[(b, A, c), (u, T, v)] = (A ·u− T · b, [A, T ] + bv − uc, T t · c−At · v).

Here A ·u denotes the action of the structure algebra (2.15) and bv is standard matrix multi-plication. Hence bv is interpreted as an element of gl(n,R) ∼= str(V ) (Theorem 2.27).

We have the following isomorphism (of vector spaces) between co(Sym(n,R)) and sp(2n,R):

co(Sym(n,R))→ sp(2n,R), (u, T, v)→(T uv −T t

).

From (4.1) we see that this isomorphism is a Lie algebra isomorphism, which implies thatco(Sym(n,R)) is a Lie algebra.The action of an element X = (u, T, v) in co(Sym(n,R)) on Sym(n,R) is given as follows:

X(z) = u+ T · z − P (z)v for all z in Sym(n,R),

where P is the operator defined in the previous chapter (see 3.14) as

P (z)v = 2z ∗ (z ∗ v)− (z ∗ z) ∗ v,

andT · z = Tz + zT t

is the action of the structure algebra on Sym(n,R) (see 2.15).Further, we will denote the conformal algebra co(Sym(n,R)) by g. We will also denote theBessel operator B 1

2by B.

4.2 A representation of the conformal algebra on C∞(O)

We will now construct the Schrodinger representation. It is a Lie algebra representation of theconformal algebra g on the space of smooth functions which are defined on the minimal orbitO. We remark that in this section we will work with the real symmetric matrices and the realLie algebra g, but that all the proofs also hold, mutatis mutandis, if we would work over thecomplex field. We will need these results over the complex field in chapter 5.

First, we will recall the definition of a Lie algebra representation.

Definition 4.2. A Lie algebra representation ρ of a Lie algebra g on a vector space V is a Liealgebra homomorphism

ρ : g→ End(V ),

from the Lie algebra g to the space of endomorphisms of V . This means that ρ is a linear mapand that

ρ([x, y]g

)= [ρ(x), ρ(y)]End(V ) = ρ(x)ρ(y)− ρ(y)ρ(x),

for all x and y in g.

4.2. A representation of the conformal algebra on C∞(O) 53

We define the map dπ : g→ End(C∞(O)) as follows:

dπ(b, 0, 0)ψ(η) = iτ(ηψ(η), b

)for (b, 0, 0) ∈ n

dπ(0, A, 0)ψ(η) = DAt · ηψ(η) +1

2tr(A)ψ(η) for (0, A, 0) ∈ l

dπ(0, 0, c)ψ(η) = iτ(Bψ(η), c

)for (0, 0, c) ∈ n,

with ψ(η) in C∞(O). Here τ is the trace form on Sym(n,R), B is the Bessel operator (for thevalue λ = 1/2), tr is the matrix trace of A, where we interpret A as an element of gl(n,R) and

At ·x = Ax+ xAt

is the action of the structure algebra on Sym(n,R). For y in Sym(n,R), we define Dyψ(x) as

Dyψ(x) = τ(y,∇ψ(x)

).

Note that this definition coincides with the standard definition of a directional derivative if thefunction ψ is defined on an open subset of Sym(n,R). Indeed, for an orthonormal basis (eα)αwith respect to the trace form τ and y =

∑α yαeα, we have


)=∑α,β

yα∂ψ(x)

∂xβτ(eα, eβ)

=∑α

yα∂ψ(x)

∂xα.

For the directional derivative, on the other hand, it holds that

limt→0

ψ(x+ ty)− ψ(x)

t=∑α

limt→0

ψ(x+ tyαeα)− ψ(x)

t

=∑α

limyαt→0

yαψ(x+ tyαeα)− ψ(x)

yαt

=∑α

yα∂ψ(x)

∂xα.

So we conclude that they are equal.For a function ψ that is only defined on O we need to be careful, since O is not an open subsetof Sym(n,R) and thus η + teα is not necessarily an element of the orbit, even for t very small.(For an open subset W of Sym(n,R), x + teα would be an element of W if t was sufficientlysmall.) In that case we will consider an extension ψ of ψ for which the derivative is well definedand then restrict

DAt ·xψ(x) +1

2tr(A)ψ(x)

to the minimal orbit. We will show later that, for functions ψ in C∞(O), the definition

DAt · ηψ(η) +1

2tr(A)ψ(η)

is independent of the choice of the extension, by proving that it is a tangential operator.

We have the following rules for Dy.

Lemma 4.3. Let V be a Jordan algebra. For x, y, z ∈ V and ψ,ϕ ∈ C∞(V ) and λ ∈ C thefollowing holds:


1. Dy

(ψ(x) + ϕ(x)

)= Dyψ(x) +Dyϕ(x)

2. Dy

(λψ(x)

)= λDyψ(x)

3. Dy+zψ(x) = Dyψ(x) +Dzψ(x)

4. Dλyψ(x) = λDyψ(x).

Proof. This follows easily from the definition


),

and linearity of the trace form τ and the gradient ∇.

With these rules for Dy we can verify that the map dπ is indeed a well-defined map from g toEnd(C∞(O)).

4.2.1 The representation dπ is well defined

To show that dπ is well defined, we will prove that dπ(X) is an element of End(C∞(O)) forall X in g, i.e.

dπ(X)(ψ) ∈ C∞(O) for all ψ ∈ C∞(O),

and

dπ(X)(λψ + µϕ) = λdπ(X)(ψ) + µdπ(X)(ϕ) for all ψ,ϕ ∈ C∞(O) and λ, µ ∈ C.

Since DAt · η, the trace form and the Bessel operator are all linear, it follows that dπ(X) actslinearly on C∞(O).

Using the fact that τ(η, b) is in C∞(O) for all b in Sym(n,R) and the product of two smoothfunctions is again smooth, we get that

τ(ηψ(η), b) = ψ(η)τ(η, b)

is a smooth function.

To prove that DAt · ηψ(η)+ 12 tr(A)ψ(η) is a well-defined function in C∞(O) that is independent

of the chosen extension, we first have to show that this operator is symmetric with respect tothe zeta function defined in (3.18).

Theorem 4.4. Let ψ be in C∞(Sym(n,R)) and ϕ in C∞c (Sym(n,R)), or let ψ and ϕ be bothelements of the Schwartz space S(Sym(n,R)). For λ in C it holds that

Z((DAt ·xψ(x) + λ tr(A)ψ(x)

)ϕ(x), λ

)= −Z

(ψ(x)

(DAt ·xϕ(x) + λ tr(A)ϕ(x)

), λ),

where Z is the zeta function:

Z(f, λ) =

∫Ωf(x) det(x)λ−

n+12 dx.


Proof. Set µ = λ − n+12 and assume µ > 0. Then we find, using integration by parts and the

fact that det(x)µ is zero on the boundary of Ω,∫ΩDAt ·xψ(x)ϕ(x) det(x)µdx

=∑α

∫Ωτ(At ·x, eα

) ∂ψ(x)

∂xαϕ(x) det(x)µdx

= −∑α

∫Ωψ(x)

∂

∂xα

(τ(At ·x, eα

)ϕ(x) det(x)µ

)dx

= −∑α

∫Ωψ(x)

(τ(At · eα, eα

)ϕ(x) det(x)µ + τ

(At ·x, eα

) ∂ϕ(x)

∂xαdet(x)µ

+τ(At ·x, eα

)ϕ(x)µ det(x)µτ

(x−1, eα

))dx, (4.3)

where we used∂

∂xατ(At ·x, eα

)= τ

(At · eα, eα

),

and Lemma 3.23:∂

∂xαdet(x)µ = µ det(x)µτ

(x−1, eα

).

From Lemma 2.15, the definition of the trace form τ(x, y) = tr(xy) and Lemma 3.24, we getfor the first and last term on the right-hand side of (4.3)∑

α

(τ(At · eα, eα

)+ µτ

(At ·x, eα

)τ(eα, x

−1))

= tr

(At(∑

α

eαeα

)+(∑

α

eαeα

)A

)+ µτ(At ·x, x−1)

=n+ 1

2tr(At +A

)+ µ tr

(Atxx−1 + x−1xA

)=(n+ 1 + 2λ− (n+ 1)

)tr(A)

= 2λ tr(A). (4.4)

Substituting (4.4) in (4.3) and adding Z(λ tr(A)ψ(x)ϕ(x), λ), we obtain


)ϕ(x), λ

)=

∫Ω

(DAt ·xψ(x) + λ tr(A)ψ(x)

)ϕ(x) det(x)µdx

=

∫Ω−ψ(x)τ

(At ·x,∇ϕ(x)

)det(x)µ − ψ(x)2λ tr(A)ϕ(x) det(x)µ + λ tr(A)ψ(x)ϕ(x) det(x)µdx

= − Z(ψ(x)


), λ).

Since (DAt ·xψ(x) + λ tr(A)ψ(x)

)ϕ(x) + ψ(x)


)is a function in the Schwartz space, it follows from Lemma 3.30 that


)ϕ(x), λ

)+ Z

(ψ(x)


), λ)

has a meromorphic extension for λ ≤ (n + 1)/2. By the uniqueness of the meromorphicextension we find


)ϕ(x), λ

)+ Z

(ψ(x)


), λ)

= 0,

for all λ in C, which proves the lemma.


We will now show that dπ(0, A, 0)ψ(x) is tangential.

Theorem 4.5. For every A in str(Sym(n,R)) it holds that

DAt ·xψ(x) +1

2tr(A)ψ(x)

is tangential to the minimal orbit O, i.e. let ψ be in C∞(Sym(n,R)), then ψ|O = 0 implies(DAt ·xψ(x) +

1

2tr(A)ψ(x)

) ∣∣∣∣O

= 0.

Proof. Consider a smooth function ψ in C∞(Sym(n,R)) such that ψ|O = 0. Let ϕ be anarbitrary smooth function on Sym(n,R) with compact support, i.e. ϕ ∈ C∞c (Sym(n,R)).Then we have, using Theorem 3.32,∫

O

(DAt · ηψ(η) +

1

2tr(A)ψ(η)

)ϕ(η)dµ(η)

= const · resν= 12Z

((DAt ·xψ(x) +

1

2tr(A)ψ(x)

)ϕ(x), ν

)= − const · resν= 1

2Z

(ψ(x)

(DAt ·xϕ(x) +

1

2tr(A)ϕ(x)

), ν

)= −

∫Oψ(η)

(DAt · ηϕ(η) +

1

2tr(A)ϕ(η)

)dµ(η)

= 0,

because ψ = 0 on O. As ϕ is arbitrary, we conclude that(DAt ·xψ(x) +

1

2tr(A)ψ(x)

)= 0

on O.

From this theorem and linearity, it follows that(DAt · ηψ(η) +

1

2tr(A)ψ(η)

)= 0

is independent of the chosen extension.Since we have already proven that the Bessel operator is a tangential operator (Theorem 3.34),we also have that

τ(Bψ(η), c) ∈ C∞(O),

is well defined. So we conclude that dπ(X)(ψ) is a smooth function on the minimal orbit forall X in g and thus

dπ(X) ∈ End(C∞(O)) for all X in g.

4.2.2 The representation dπ is a Lie algebra representation

In this section we will prove that dπ is a Lie algebra representation. To do this, we will firstestablish certain lemmas about how the trace form τ , the Bessel operator and the operator Dy

interact with each other.We begin by a lemma that states how the operator Dy acts on the trace form τ .

Lemma 4.6. For x, y in Sym(n,R), A in str(Sym(n,R)) and ψ(x) in C∞(Sym(n,R)) we have

DA ·x

(τ(xψ(x), y)

)= τ(x, y)DA ·xψ(x) + ψ(x)τ(A ·x, y).


Proof. Using the linearity of τ , we get

∂

∂xα

(τ(xψ(x), y)

)= τ

(∂

∂xα

(xψ(x)

), y

)= τ

(eαψ(x), y

)+ τ

(x∂ψ(x)

∂xα, y

).

Hence we find

DA ·x

(τ(xψ(x), y)

)=∑α

τ

(A ·x, eα

∂

∂xατ(xψ(x), y

))=∑α

τ(A ·x, eα)

(τ(eαψ(x), y

)+ τ

(x∂ψ(x)

∂xα, y

))= ψ(x)τ(A ·x, y) +

∑α

τ(A ·x, eα)∂ψ(x)

∂xατ (x, y)

= ψ(x)τ(A ·x, y) + τ (x, y)DA ·xψ(x),

where we used Lemma 2.15.

The following lemma expresses a commutation rule for the operator Dy.

Lemma 4.7. Let A, T in str(Sym(n,R)), x in Sym(n,R) and ψ in C∞(Sym(n,R)). We thenhave the following expression for the commutator of DA ·x and DT ·x:

[DA ·x, DT ·x]ψ(x) = D[T,A] ·xψ(x). (4.5)

Proof. We will first calculate the gradient of DT ·xψ(x):

∇(DT ·xψ(x)

)= ∇τ

(T ·x,∇ψ(x)

)=∑α

eα

(τ

(∂T ·x

∂xα,∇ψ(x)

)+ τ

(T ·x,∇∂ψ(x)

∂xα

))=∑α

eατ(T · eα,∇ψ(x)

)+∑α,β

eα∂2ψ(x)

∂xα∂xβτ (T ·x, eβ) .

Taking the inner product with A ·x, we get

DA ·x

(DT ·xψ(x)

)= τ

(A ·x,∇(DT ·xψ(x))

)=∑α

τ(A ·x, eα

)τ(T · eα,∇ψ(x)

)+∑α,β

τ(A ·x, eα

) ∂2ψ(x)

∂xα∂xβτ(T ·x, eβ

).

In the same manner we also have

DT ·x

(DA ·xψ(x)

)=∑α

τ(T ·x, eα

)τ(A · eα,∇ψ(x)

)+∑α,β

τ(T ·x, eα

) ∂2ψ(x)

∂xα∂xβτ(A ·x, eβ

).

Subtracting this two equations, we find

[DA ·x, DT ·x]ψ(x) = DA ·x

(DT ·xψ(x)

)−DT ·x

(DA ·xψ(x)

)=∑α

(τ(A ·x, eα


)− τ(T ·x, eα

)τ(A · eα,∇ψ(x)

))= τ

(T · (A ·x),∇ψ(x)

)− τ(A · (T ·x),∇ψ(x)

)= τ

([T,A] ·x,∇ψ(x)

)= D[T,A] ·xψ(x),


where we used Lemma 2.28 and∑α

τ(A ·x, eα


)=∑α

τ(A ·x, eα

)τ(eα, T

t·∇ψ(x)

)= τ

(A ·x, T t ·∇ψ(x)

)= τ

(T · (A ·x),∇ψ(x)

),

which follows from Lemma 2.15 and Lemma 2.29. This completes the proof.

We can also calculate a composition of the Bessel operator and the operator Dy.

Lemma 4.8. The action of the Bessel operator on DAt ·xψ(x) is given by

Bλ(DAt ·xψ(x)

)= A ·Bλψ(x) +

∑α

τ(At ·x, eα)∂Bλψ(x)

∂xα, (4.6)

where A ∈ str(Sym(n,R)), x ∈ Sym(n,R), λ ∈ C and ψ ∈ C∞(Sym(n,R)).

Proof. We have

DAt ·xψ(x) =∑α

τ(At ·x, eα)∂ψ(x)

∂xα.

Using the product rule (Lemma 3.19) and the linearity of the Bessel operator, we obtain

Bλ(DAt ·xψ(x)

)=∑α

(Bλ(τ(At ·x, eα)

)∂ψ(x)

∂xα

+ 2P

(∇τ(At ·x, eα),∇∂ψ(x)

∂xα

)x +τ(At ·x, eα)Bλ

(∂ψ(x)

∂xα

)). (4.7)

We can easily calculate that

∂τ(At ·x, eα)

∂xβ= τ(At · eβ, eα),

∂2τ(At ·x, eα)

∂xβ∂xγ= 0.

Using this and the expression for the Bessel operator (3.9) and Lemma 2.29, we find for thefirst term on the right-hand side of (4.7)

Bλ(τ(At ·x, eα)

)= λ

∑β

τ(At · eβ, eα)eβ = λ∑β

τ(eβ, A · eα)eβ = λA · eα. (4.8)

From the bilinearity of P and Lemma 2.29, we get for the second term of the right-hand sideof (4.7)

2P

(∇τ(At ·x, eα),∇∂ψ(x)

∂xα

)x = 2

∑β,γ

∂2ψ(x)

∂xα∂xγP(τ(At · eβ, eα)eβ, eγ

)x

= 2∑γ

∂2ψ(x)

∂xα∂xγP (A · eα, eγ)x. (4.9)


For the last term of the right-hand side of (4.7), we can calculate the action of the Besseloperator on ∂ψ(x)/∂xα. We then find

Bλ(∂ψ(x)

∂xα

)=∑β,γ

∂3ψ(x)

∂xα∂xβ∂xγP (eβ, eγ)x+ λ

∑β

∂2ψ(x)

∂xα∂xβeβ

=∂

∂xα

∑β,γ

∂2ψ(x)

∂xβ∂xγP (eβ, eγ)x+ λ

∑β

∂ψ(x)

∂xβeβ

−∑β,γ

∂2ψ(x)

∂xβ∂xγP (eβ, eγ)eα

=∂Bλψ(x)

∂xα−∑β,γ

∂2ψ(x)

∂xβ∂xγP (eβ, eγ)eα, (4.10)

where we used∂

∂xαP (eβ, eγ)x = P (eβ, eγ)eα.

Combining (4.9) and the second term on the right-hand side of (4.10), we obtain

∑α

(2∑γ

∂2ψ(x)

∂xα∂xγP (A · eα, eγ)x− τ(At ·x, eα)

∑β,γ

∂2ψ(x)


)

=∑α,β

(2∂2ψ(x)

∂xα∂xβP (A · eα, eβ)x− ∂2ψ(x)

∂xα∂xβP (eα, eβ)(At ·x)

).

We can simplify this further using Lemma 3.17:

2P (A · eα, eβ)x− P (eα, eβ)(At ·x)

= Aeαxeβ + eαAtxeβ + eβxAeα + eβxeαA

t − 1

2

(eαA

txeβ + eαxAeβ + eβAtxeα + eβxAeα

).

Multiplying with ∂2ψ(x)∂xα∂xβ

, summing over α and β and changing some indices, yields

∑α,β

∂2ψ(x)

∂xα∂xβ

(2P (A · eα, eβ)x− P (eα, eβ)(At ·x)

)=∑α,β

∂2ψ(x)

∂xα∂xβ

(Aeαxeβ + eαxeβA

t)

=∑α,β

∂2ψ(x)

∂xα∂xβA ·P (eα, eβ)x. (4.11)

Substituting (4.8),(4.9),(4.10) and (4.11) in (4.7), we conclude

Bλ(DAt ·xψ(x)

)=∑α

(λA · eα

∂ψ(x)

∂xα+ 2

∑γ

∂2ψ(x)

∂xα∂xγP (A · eα, eγ)x

+τ(At ·x, eα)∂Bλψ(x)

∂xα− τ(At ·x, eα)

∑β,γ

∂2ψ(x)


= λA ·∇ψ(x) +

∑α,β

∂2ψ(x)

∂xα∂xβA ·P (eα, eβ)x+

∑α


∂xα

= A ·Bλψ(x) +∑α


∂xα.

This proves the lemma.

The Bessel operator and the trace form interact in the following way.


Lemma 4.9. Let x, a, b ∈ Sym(n,R), λ ∈ C and ψ ∈ C∞(Sym(n,R)). Then one has

τ(Bλτ

(xψ(x), a

), b)

= τ(x, a)τ(Bλψ(x), b

)+ τ(∇ψ(x), (ab)t ·x

)+ λψ(x)τ(a, b).

Proof. We will first calculate Bλτ(xψ(x), a). Because of the linearity of the trace form, we have

∂

∂xατ(xψ(x), a

)= τ

(eαψ(x), a

)+ τ

(x∂ψ(x)

∂xα, a

)∂2

∂xα∂xβτ(xψ(x), a

)= τ

(eα∂ψ(x)

∂xβ, a

)+ τ

(eβ∂ψ(x)

∂xα, a

)+ τ

(x∂2ψ(x)

∂xα∂xβ, a

).

Substituting this in expression (3.9) for the Bessel operator, we get

Bλτ(xψ(x), a) =∑α,β

∂2τ(xψ(x), a

)∂xα∂xβ

P (eα, eβ)x+ λ∑α

∂

∂xατ(xψ(x), b

)eα

=∑α,β

(∂ψ(x)

∂xβτ(eα, a) +

∂ψ(x)

∂xατ(eβ, a) +

∂2ψ(x)

∂xα∂xβτ(x, a)

)P (eα, eβ)x

+ λ∑α

(ψ(x)τ(eα, a) +

∂ψ(x)

∂xατ(x, a)

)eα

= 2P(a,∇ψ(x)

)x+ λψ(x)a+ τ(x, a)Bλ

(ψ(x)

).

Taking the inner product with b we find

τ(Bλτ

(xψ(x), a

), b)

= τ(

2P(a,∇ψ(x)

)x, b)

+ λψ(x)τ(a, b) + τ(x, a)τ(Bλ(ψ(x)

), b). (4.12)

Expressing the operator P with matrix products (Lemma 3.17) and using the invariance of thetrace under cyclic permutations, we obtain

τ(

2P(a,∇ψ(x)

)x, b)

= tr(ax∇ψ(x)b

)+ tr

(∇ψ(x)xab

)= tr

(∇ψ(x)(bax+ xab)

)= τ

(∇ψ(x), (ab)t ·x

),

where we used (ab)t ·x = bax+ xab since a and b are symmetric matrices. Substituting this in(4.12), we find

τ(Bλτ

(xψ(x), a

), b)

= τ(x, a)τ(Bλψ(x), b

)+ τ(∇ψ(x), (ab)t ·x

)+ λψ(x)τ(a, b),


All these lemmas allow us to prove that the map dπ is indeed a Lie algebra representation.

Theorem 4.10. The map dπ : g→ End(C∞(O)) defined as

dπ(b, 0, 0)ψ(η) = iτ(ηψ(η), b

)for (b, 0, 0) ∈ n (4.13a)

dπ(0, A, 0)ψ(η) = DAt · ηψ(η) +1

2tr(A)ψ(η) for (0, A, 0) ∈ l (4.13b)

dπ(0, 0, c)ψ(η) = iτ(Bψ(η), c

)for (0, 0, c) ∈ n, (4.13c)

with ψ(η) in C∞(O), is a Lie algebra representation of g on C∞(O). We call this map theSchrodinger representation.


Proof. We have already proven that dπ is well defined. By definition we know that

dπ(b, A, c) = dπ(0, A, 0) + dπ(b, 0, 0) + dπ(0, 0, c).

So to show that dπ is a linear map, it suffices to show it for these components. For X in l thisfollows from the linearity of the trace form τ and the operator Dy. For X in n or X in n thisfollows from the linearity of τ .To prove that dπ is a Lie algebra representation, we thus only have to show that

dπ([X,Y ]) = dπ(X)dπ(Y )− dπ(Y )dπ(X)

for all X,Y in g. Since the Lie bracket is linear, it will again be sufficient to prove this for itscomponents.

First, consider X = (0, A, 0) and Y = (0, T, 0). Then the Lie bracket is given by

[X,Y ] = (0, [A, T ], 0).

For ψ in C∞(O) we have

dπ(X)dπ(Y )(ψ(η)

)= DAt · η

(DT t · ηψ(η) +

1

2tr(T )ψ(η)

)+

1

2tr(A)

(DT t · ηψ(η) +

1

2tr(T )ψ(η)

)= DAt · η

(DT t · ηψ(η)

)+

1

2tr(T )DAt · ηψ(η) +

1

2tr(A)DT t · ηψ(η) +

1

4tr(A) tr(T )ψ(η).

We also have

dπ(Y )dπ(X)(ψ(η)

)= DT t · η

(DAt · ηψ(η)

)+

1

2tr(A)DT t · ηψ(η) +

1

2tr(T )DAt · ηψ(η) +

1

4tr(T ) tr(A)ψ(η).

Subtracting these two equations and using Lemma 4.7, we get

[dπ(X),dπ(Y )](ψ(η)

)= DAt · η

(DT t · ηψ(η)

)−DT t · η

(DAt · ηψ(η)

)= D[T t,At] · ηψ(η)

= D[A,T ]t · ηψ(η) +1

2tr([A, T ])ψ(η)

= dπ(0, [A, T ], 0)ψ(η),

because tr([A, T ]) = 0.

Let X = (0, A, 0) and Y = (u, 0, 0), then we get for the Lie bracket

[X,Y ] = (A ·u, 0, 0).

For ψ in C∞(O) we have


)= DAt · η

(iτ(ηψ(η), u

))+

1

2tr(A)iτ

(ηψ(η), u

),

and


)= iτ

(ηDAt · ηψ(η), u

)+ iτ

(η

1

2tr(A)ψ(η), u

)= iDAt · ηψ(η)τ(η, u) + i

1

2tr(A)τ

(ηψ(η), u

).


Subtracting these two equations and using Lemma 4.6:

DAt · η

(τ(ηψ(η), u

))= τ(η, u)DAt · ηψ(η) + ψ(η)τ

(At · η, u

),

we obtain

[dπ(X), dπ(Y )](ψ(η)

)= iψ(η)τ

(At · η, u

)= iψ(η)τ(η,A ·u) = dπ(A ·u, 0, 0)

(ψ(η)

),

where we used lemma 2.29.

Let X = (0, A, 0) and Y = (0, 0, v), then we have for the Lie bracket

[X,Y ] = (0, 0,−At · v).

For ψ in C∞(O) we find


)= DAt · η

(iτ(Bψ(η), v)

)+

1

2tr(A)

(iτ(Bψ(η), v)

),

and


)= iτ

(B(DAt · ηψ(η)

), v)

+1

2tr(A)

(iτ(Bψ(η), v)

).

Subtracting these two equations and using Lemma 4.8:

Bλ(DAt · ηψ(η)

)= A ·Bλψ(η) +

∑α

τ(At · η, eα

)∂Bλψ(η)

∂xα,

we get


)= DAt · η

(iτ(Bψ(η), v

))− iτ

(A ·Bψ(η), v

)− i∑α

τ(At · η, eα

)τ

(∂Bψ(η)

∂xα, v

)

= DAt · η

(iτ(Bψ(η), v

))− iτ

(Bψ(η), At · v

)− iτ

(At · η,

∑α

eα∂τ(Bψ(η), v)

∂xα

)= τ

(At · η,∇

(iτ(Bψ(η), v)

))− iτ

(Bψ(η), At · v

)− iτ

(At · η,∇

(τ(Bψ(η), v)

))= iτ

(Bψ(η),−At · v

)= dπ(0, 0,−At · v)

(ψ(η)

).

Let X = (b, 0, 0) and Y = (u, 0, 0). Then the Lie bracket is given by

[X,Y ] = (0, 0, 0).

We have for ψ in C∞(O)


)= iτ

(ηiτ(ηψ(η), u

), b)

= −ψ(η)τ(η, b)τ(η, u)

and


)= iτ

(ηiτ(ηψ(η), b

), u)

= −ψ(η)τ(η, u)τ(η, b).

Subtracting these two equations we get zero, which corresponds to the Lie bracket whichis also zero.

4.3. A representation on L2(O, dµ) 63

Let X = (b, 0, 0) and Y = (0, 0, v). Then we have

[X,Y ] = (0, bv, 0).

We find for ψ in C∞(O)


)= iτ

(ηiτ(Bψ(η), v

), b)

= −τ(η, b)τ(Bψ(η), v

)and


)= iτ

(iBτ(ηψ(η), b

), v).

Using Lemma 4.9:

τ(Bλτ

(ηψ(η), a

), b)

= τ(η, a)τ(Bλψ(η), b

)+ τ(∇ψ(η), (ab)t · η

)+ λψ(η)τ(a, b),

we obtain


)= τ

(∇ψ(η), (bv)t · η

)+

1

2ψ(η)τ(b, v)

= D(bv)t · ηψ(η) +1

2tr(bv)ψ(η)

= dπ(0, bv, 0)(ψ(η)

).

Finally for X = (0, 0, c) and Y = (0, 0, v) we have

[X,Y ] = (0, 0, 0).

We find for ψ in C∞(O)


)= iτ

(B(iτ(Bψ(η), v)

), c)

and


)= iτ

(B(iτ(Bψ(η), c)

), v).

From the linearity of τ and the Bessel operator B, it follows that both equations are equalto each other.

Since the Lie bracket is linear and [X,Y ] = −[Y,X], we can conclude that for all X in g

[dπ(X), dπ(Y )](ψ(η)

)= dπ([X,Y ])

(ψ(η)

).

So we have proven that dπ is indeed a Lie algebra representation.

4.3 A representation on L2(O, dµ)

4.3.1 Polynomials

Before we go on to refine the Schrodinger representation into a representation on L2(O,dµ), wewill say something about polynomials on the Jordan algebra Sym(n,R) and its complexification.Consider Sym(n,K), the symmetric n× n-matrices over the field K where K = R or K = C.

We define k as the dimension of the vector space Sym(n,K), i.e. k = n(n+1)2 . Let (ei)

ki=1 be a

basis for Sym(n,K), then we can write every x ∈ Sym(n,K) as

x =k∑i=1

xiei,


where xi ∈ K. We say that the map P : Sym(n,K)→ C, is a polynomial if it is of the form

P (x) =∑m

λ[m]x[m],

where the sum is a finite sum over multi-indices [m] ∈ Nk, the λ[m] are complex numbers and

where we define the monomial x[m] for m = (m1,m2, . . . ,mk) as

x[m] = xm11 xm2

2 · · ·xmkk .

We define the degree of the monomial x[m] as |m| = m1 + · · · + mk. We say that the map Pis a homogeneous polynomial of degree l if it as finite sum of monomials of degree l. It is clearthat every polynomial is a linear combination of homogeneous polynomials. The definition of ahomogeneous polynomial (and thus also of a polynomial) is independent of the choice of basis.To see this, let (fj)

kj=1 be another basis of Sym(n,K). We can express the basis elements ei in

this new basis as ei =∑n

j=1 Λjifj , with Λji ∈ K. Then we get

x =n∑i=1

xiei =n∑i=1

n∑j=1

xiΛjifj =

n∑j=1

xjfj ,

and hencen∑i=1

xiΛji = xj .

So let x[m] be a monomial of degree l in the new basis, then

x[m] = xm11 xm2

2 · · · xmkk ,

with |m| = l. In the old basis this monomial has the expression

x[m] =

(n∑i=1

xiΛ1i

)m1(

n∑i=1

xiΛ2i

)m2

· · ·

(n∑i=1

xiΛki

)mk.

If we work this out, then we get a finite linear combination of monomials of degree l andthus a homogeneous polynomial of degree l. Therefore, we can conclude that a homogeneouspolynomial in one basis is also a homogeneous polynomial in another basis.We will use the notation P(sym(n,K)) for the space of all polynomials and the notationP l(sym(n,K)) for the space of all homogeneous polynomials of degree l. We have

P(sym(n,K)) =

∞⊕l=0

P l(sym(n,K)).

We remark that for K = R polynomials are smooth functions and for K = C polynomials areholomorphic functions.

4.3.2 A subrepresentation

We will now look at the subrepresentation of the Schrodinger representation dπ that we becomeby letting the representation act on the function ψ0(η) := K−1

2(|η|). From (3.25) we know that

ψ0(η) =

√π

2e−|η|.

Let U(g) be the universal enveloping algebra of the conformal algebra g. (See [5], Chapter III,for more information on universal enveloping algebras.) Then dπ is also a representation of


this universal enveloping algebra. We define L2(O, dµ)k as the space that arises by letting thisrepresentation act on ψ0, i.e.

L2(O,dµ)k = dπ(U(g))ψ0.

We will later show that L2(O,dµ)k is dense in L2(O,dµ).The conformal algebra is composed as g = p + k, where

p = (b, L(a), 0) for b, a in Sym(n,R)

k = (b,D,−b) for b in Sym(n,R) and D in Der(Sym(n,R)).

From the Poincare-Birkhoff-Witt Theorem (Theorem 3.8 in [5]) it then follows that

dπ(U(g))ψ0 = dπ(U(p))dπ(U(k))ψ0.

We will first calculate dπ(U(k))ψ0.

Lemma 4.11. One has

dπ(U(k))ψ0 = Cψ0.

Proof. Since ψ0 is a radial function, its gradient is given by (see Lemma 3.12)

∇ψ0(η) = −ψ0(η)η

|η|.

For the action of the Bessel operator we find (see Theorem 3.21)

Bψ0(η) =

√π

2

(e−|η|

)′′η +

√π

2

1

2

(e−|η|

)′In

=

√π

2e−|η|η −

√π

4e−|η|In.

Therefore, we find for b in Sym(n,R)

dπ(b, 0,−b)ψ0(η) = iτ(η, b)ψ0(η) + iτ(Bψ0(η),−b)

= iτ(η, b)

√π

2e−|η|η − iτ(η, b)

√π

2e−|η|η + i

√π

4e−|η|τ(In, b)

= i tr(b)

√π

4e−|η|

= itr(b)

2ψ0(η).

From (2.12) we know that tr(D) = 0 and from (2.13) we deduce

τ(D(η), η

)=

1

2tr(D(η ∗ η)

)= 0.

For D in Der(Sym(n,R)) we thus get

dπ(0, D, 0)ψ0(η) = τ(D(η),∇ψ0

)+

1

2tr(D)ψ0(η)

= −τ(D(η),

η

|η|

)ψ0(η)

= 0.

Hence we conclude dπ(U(k))ψ0 = Cψ0.


Denote by P(O) the space of polynomials on Sym(n,R) restricted to the minimal orbit. Notethat this restriction is not a bijection. For example the polynomial η2

ij and the polynomialηiiηjj restrict to the same function in P(O), since

η2ij = (qiqj)

2 = q2i q

2j = ηiiηjj ,

for η in the minimal orbit. We will use the notation ψ0 ⊗ P for the space of polynomialsmultiplied with ψ0. A function in this space is thus of the form

ψ0 ⊗ ρ : O → C, η → ψ0(η)ρ(η),

for a polynomial ρ in P(O). We will now prove that L2(O, dµ)k is given by ψ0 ⊗ P.

Theorem 4.12. The space L2(O, dµ)k is equal to ψ0 ⊗ P.

Proof. From the Poincare-Birkhoff-Witt Theorem and the previous lemma, we get

L2(O,dµ)k = dπ(U(p))dπ(U(k))ψ0 = dπ(U(p))Cψ0.

We haveτ(η, b) =

∑i,j

ηijbij .

Hence τ( · , b) is a polynomial of degree one. From this, we conclude that applying dπ(b, 0, 0) ona function ϕ amounts to multiplying ϕ with a polynomial of degree one. Because dπ(U(k))ψ0 =Cψ0, we conclude that we can create every element of ψ0⊗P by applying several times dπ(b, 0, 0)on Cψ0 with appropriate choices for b and then taking linear combinations. So we have

ψ0 ⊗ P ⊂ dπ(U(g))ψ0.

To obtain the other inclusion, we only need to prove that dπ(0, L(a), 0)ϕ is in ψ0⊗P for all ϕin ψ0 ⊗ P and a in Sym(n,R).Let ϕ = ψ0 ⊗ ρ, with ρ in P(O). Then we have to show that

dπ(0, L(a), 0

)ϕ(η) = DL(a)ηϕ(η) +

1

2tr(a

2

)ϕ(η) =

∑α

τ(L(a)η, eα

)∂ϕ(η)

∂xα+

1

2tr(a

2

)ϕ(η)

is an element of ψ0 ⊗ P. The last term on the right-hand side is clearly in ψ0 ⊗ P. We have

∂ϕ(η)

∂xα= −ψ0(η)

ηα|η|ρ(η) + ψ0(η)

∂ρ(η)

∂xα. (4.14)

Combining the first term on the right-hand side of (4.14) with τ(L(a)η, eα), we get

−ψ0(η)ρ(η)τ

(L(a)η,

η

|η|

)= −ψ0(η)ρ(η)τ

(L(a),

η ∗ η|η|

)= −ψ0(η)ρ(η)τ

(L(a), η

),

since (η ∗ η)/ |η| = η for an element in the minimal orbit. The function ρ(η)τ (L(a), η) is apolynomial and thus

ψ0(η)ρ(η)τ

(L(a)η,

η

|η|

)is again an element of ψ0 ⊗ P.Since the trace from is associative, we have

τ(L(a)η, eα) = τ(η, L(a)eα),

which is a polynomial of degree one. Combining this with the second term on the right-handside of (4.14), we also get an element of ψ0 ⊗P. Hence we conclude that dπ(0, L(a), 0)ϕ(η) isin ψ0 ⊗ P for all a in Sym(n,R), which proves the theorem.


To show that L2(O,dµ)k is dense in L2(O,dµ), we will first prove that it is a subset.

Lemma 4.13. One hasL2(O,dµ)k ⊂ L2(O, dµ).

Proof. We have to prove that for ψ0 ⊗ ρ in ψ0 ⊗ P it holds that∫O|ψ0(η)ρ(η)|2 dµ(η) <∞.

Because of Lemma 3.9 about the connection between integration on O and Rn, it suffices toshow that the function q → ψ0(qqt)ρ(qqt) is in L2(Rn). Since ρ(η) is a polynomial on O it hasan expression of the form

ρ(η) =∑[m]

α[m]η[m].

We get

ρ(qqt) =∑[m]

α[m]q[m](qt)[m],

which is a polynomial on Rn. We also find ψ0(qqt) =√π

2 e−q2 , because

∣∣qqt∣∣ = qtq = q2. Henceψ0(qqt)ρ(qqt) is the product of a polynomial and the Gaussian function exp(−q2). This is arapidly decreasing function, and thus certainly in L2(Rn). Therefore ψ0(η)ρ(η) is in L2(O,dµ),which proves the lemma.

We will be needing the following lemma about dense subspaces.

Lemma 4.14. Consider a dense subspace X of a Hilbert space Y . Let A,B be two orthogonalsubspaces which span Y , i.e A+B = Y and 〈a, b〉 = 0 for all a in A and b in B. Then X ∩Ais dense in A and X ∩B is dense in B.

Proof. We will prove that X ∩A is dense in A. Let a be an arbitrary element in A, we have toshow that there exists a sequence in X ∩ A that converges to a. Because X is dense in Y weknow that there exists a sequence (xn)n in X that converges to a. Note that X = X∩A+X∩B,thus we have xn = an + bn with an ∈ A ∩ X and bn ∈ B ∩ X. Let ε > 0 be arbitrary, for nlarge enough we have

ε ≥ ‖xn − a‖2 = 〈an + bn − a, an + bn − a〉 = 〈an − a, an − a〉+ 〈bn, bn〉 ≥ ‖an − a, an − a‖2 ,

where we used 〈a, b〉 = 0 for all a in A and b in B. Hence the sequence (an)n converges to aand it is in X ∩A. Therefore we conclude that X ∩A is dense in A.

We know that the space of even functions L2even(Rn) and the space of odd functions L2

odd(Rn)span L2(Rn), because we can write every ψ(q) as

ψ(q) + ψ(−q)2

+ψ(q)− ψ(−q)

2.

We also know that the inner product of an even ψ and odd function ϕ is zero, since∫Rnψ(q)ϕ(q)dnq =

∫Rnψ(−q)ϕ(−q)dnq = −

∫Rnψ(q)ϕ(q)dnq.

HenceL2(Rn) = L2

even(Rn)⊕ L2odd(Rn), (4.15)

and we can apply Lemma 4.14 to L2(Rn). This allows us to prove the following theorem.

Theorem 4.15. It holds that L2(O,dµ)k is dense in L2(O,dµ).


Proof. Consider p∗ as defined in Theorem (3.6). Since p∗ is a unitary isomorphism betweenL2(O,dµ) and L2

even(Rn), it suffices to show that p∗(L2(O,dµ)k

)is dense in L2

even(Rn).

We know that the space e−q2 ⊗ P(Rn) of polynomials multiplied with the Gaussian exp(−q2)

is dense in L2(Rn), because the Hermite functions are already dense in L2(Rn). Hence byapplying Lemma 4.14 to (4.15), we get that e−q

2 ⊗ Peven(Rn) is dense in L2even(Rn) and we

only have to prove that p∗(L2(O,dµ)k

)is equal to the space e−q

2 ⊗ Peven(Rn).

We will first show that every homogeneous polynomials of degree l in P(O) is mapped by p∗ toa homogeneous polynomial of degree 2l in P(Rn). Consider a monomial η[m] of degree |m| = l,then we find

p∗(η[m]

)= p∗

(∏i≤j

ηmijij

)=∏i≤j

(qiqj)mij .

Because every exponent mij appears twice, the degree of p∗η[m] is

2∑i≤j

mij = 2 |m| .

We also have that p∗(ψ0) = (√π/2) exp(−q2). Hence we conclude that a function in L2(O, dµ)k

is mapped to a function in e−q2 ⊗ Peven(Rn).

We will now prove that p∗ restricted to L2(O, dµ)k is also surjective on e−q2 ⊗ Peven(Rn). Let

q[m] =n∏i=1

qmii

be a polynomial of even degree in P(Rn), then we can split every qi in an even and odd part,i.e.

q[m] =

n∏i=1

q2mii

n∏i=1

qmii ,

where mi ∈ 0, 1 and 2mi + mi = mi. Because q[m] is even, the mi, i = 1, . . . , n come in pairsof two and we can associate with each pair a ηij . Then we can define the following polynomialin P(O):

η[m′] =∏i

ηmiii∏pairs

ηij ,

where the ηij are associated with the pair qiqj . It has the property that p∗(η[m′]

)= q[m].

Hence

p∗(ψ0η

[m′])

=

√π

2e−q

2q[m],

and we conclude that the map p∗ is surjective. This proves the theorem.

4.3.3 Extension to L2(O, dµ)

We would like to extend the Schrodinger representation from L2(O,dµ)k to L2(O,dµ). Since weknow that L2(O,dµ)k is dense in L2(O,dµ), this is possible if dπ(X) : L2(O,dµ)k → L2(O, dµ)kis a continuous map for all X in the conformal algebra g. We will not directly show that dπ(X)is continuous but we will use the isomorphism between L2(O, dµ) and L2

even(Rn) to induce aSchrodinger representation on e−q

2 ⊗ Peven(Rn) ⊂ L2even(Rn). We will then show that this is

a continuous representation, which implies that the original Schrodinger representation is alsocontinuous.


We define the Schrodinger representation d$ on L2even(Rn) such that the following diagram

commutes for all X in g, where p∗ is the isomorphism from Theorem 3.6.

L2(O,dµ)k L2even(Rn)

L2(O,dµ)k L2even(Rn)

p∗

dπ(X) d$(X)

p∗

(4.16)

We find the following expression for d$.

Theorem 4.16. The Lie algebra representation d$ of the conformal algebra g on the spacee−q

2 ⊗ Peven(Rn) is given by

d$(b, 0, 0)f(q) = in∑

i,j=1

bijqiqjf(q) for (b, 0, 0) ∈ n

d$(0, A, 0)f(q) =n∑

i,j=1

Aijqi∂f(q)

∂qj+

1

2tr(A)f(q) for (0, A, 0) ∈ l

d$(0, 0, c)f(q) =i

4

n∑i,j=1

cij∂2f(q)

∂qi∂qjfor (0, 0, c) ∈ n,

for f(q) in e−q2 ⊗ Peven(Rn).

Proof. We will check (4.16) for the three different components of g.

First, consider (b, 0, 0) in n and ψ in L2(O, dµ)k. Then we get

p∗(dπ(b, 0, 0)ψ(η)

)= p∗

(iτ(η, b)ψ(η)

)= iτ(qqt, b)ψ(qqt)

= iqtbqψ(qqt)

= d$(b, 0, 0)p∗(ψ(η)

).

So (4.16) commutes.

Secondly, we will show that (4.16) holds for (0, A, 0) in l. Let ψ be an element ofL2(O, dµ)k. We find

p∗(dπ(0, A, 0)ψ(η)

)= p∗

(DAt · ηψ(η) +

1

2tr(A)ψ(η)

). (4.17)

Applying the chain rule to ∂ψ(qqt)/∂qi, we get

∂ψ(qqt)

∂qi=∑α

∂ψ(qqt)

∂xα

∂xα∂qi

=∑α

∂ψ(qqt)

∂xα

∂τ(qqt, eα)

∂qi

=∑α

∂ψ(qqt)

∂xα

∂qteαq

∂qi

=∑α

∂ψ(qqt)

∂xα2(eαq)i. (4.18)


Here (eαq)i is the ith component of the vector eαq. Consequently, we find

p∗(DAt · ηψ(η)

)= τ

(At · qqt,∇xψ(qqt)

)=∑α

tr((Atqqt + qqtA)eα

)∂ψ(qqt)

∂xα

=∑α

n∑i=1

2(Atq)i(eαq)i∂ψ(qqt)

∂xα

=n∑i=1

(Atq)i∂ψ(qqt)

∂qi

=n∑

i,j=1

Ajiqj∂ψ(qqt)

∂qi.

Substituting this in (4.17), we conclude

p∗(dπ(0, A, 0)ψ(η)

)=

n∑i,j=1

Ajiqj∂ψ(qqt)

∂qi+

1

2tr(A)ψ(qqt)

= d$(0, A, 0)p∗ψ(η).

Finally, let us prove (4.16) for (0, 0, c) in n. For an element ψ in L2(O, dµ)k we have

p∗(dπ(0, 0, c)ψ(η)

)= p∗

(iτ(Bψ(η), c)

)= i∑α,β

∂2ψ(qqt)

∂xα∂xβτ(P (eα, eβ)qqt, c

)+i

2

∑α

∂ψ(qqt)

∂xατ(eα, c).

From Lemma 3.16 we deduce

τ(P (eα, eβ)qqt, c

)=

1

2tr(eαqq

teβc+ eβqqteαc) =

n∑i,j=1

(eαq)i(eαq)jcij .

So we find

p∗(dπ(0, 0, c)ψ(η) = i∑α,β

n∑i,j=1

(eαq)i(eαq)jcij∂2ψ(qqt)

∂xα∂xβ+i

2

∑α

∂ψ(qqt)

∂xατ(eα, c).

We also have, using (4.18),

d$(0, 0, c)p∗ψ(η) =i

4

n∑i,j=1

cij∂2ψ(qqt)

∂qi∂qj

=i

4

n∑i,j=1

cij∂

∂qi

(∑α

∂ψ(qqt)

∂xα2(eαq)j

)

=i

4

n∑i,j=1

cij

∑α,β

∂2ψ(qqt)

∂xα∂xβ4(eαq)j(eβq)i +

∑α

∂ψ(qqt)

∂xα2(eα)ij

.

We see that d$(0, 0, c)p∗ψ(η) = p∗(dπ(0, 0, c)ψ(η). Thus (4.16) commutes, which provesthe theorem.

Before we prove that d$ is continuous, we will show that derivation and multiplying with qiare continuous operators on e−q

2 ⊗ Peven(Rn).


Lemma 4.17. The linear map

qi : e−q2 ⊗ Peven(Rn)→ e−q

2 ⊗ Peven(Rn), f(q)→ qif(q)

is continuous.

Proof. For the linear map qi, the condition to be continuous can be reformulated as: thereexists a positive constant C such that

‖qif(q)‖L2 ≤ C ‖f(q)‖L2 ,

for all f in e−q2 ⊗ Peven(Rn). We will first calculate an expression for the the norm on

e−q2 ⊗ Peven(Rn). To do this, we will use∫

Re−2r2rmdr = 2

−m−12

∫Re−r

2rmdr =

2−m−1

2 Γ(m+12 ) if m is even

0 if m is odd.

Consider a function f in e−q2 ⊗Peven(Rn). It is of the form e−q

2∑[m] α[m]q

[m], where the sumis finite. We obtain for the norm of f

‖f(q)‖2L2 =

∫Rne−2q2

∣∣∣∑[m]

α[m]q[m]∣∣∣2dnq

=∑[m]

∑[k],[l]

[k]+[l]=[m]

α[k]α[l]

∫Rne−2q2q[m]dnq

=∑[m]

[m] is even

∑[k],[l]

[k]+[l]=[m]

α[k]α[l]2−|m|−n

2

[Γ

([m] + 1

2

)],

where we introduced the notation[Γ(

[m]+12

)]for the product

Γ

(m1 + 1

2

)· · ·Γ

(mn + 1

2

).

We will now examine the norm of qif(q). Set [1i] = (0, . . . , 1, . . . , 0), where the one is on theith position. We then get

‖qif(q)‖2L2 =

∫Rnq2i e−2q2

∣∣∣∑[m]

α[m]q[m]∣∣∣2dnq

=

∫Rne−2q2

∣∣∣∑[m]

α[m]q[m]+[1i]

∣∣∣2dnq

=∑[m]

[m] is even

∑[k],[l]

[k]+[l]=[m]

α[k]α[l]2−|m|−2−n

2

[Γ

([m] + 2[1i] + 1

2

)]

≤ max[m]|m(m+ 1)| ‖f(q)‖2L2 ,

because Γ(z + 1) = zΓ(z). Thus we have shown that qi is a continuous linear map.

We also have that the derivative is continuous on e−q2 ⊗ Peven(Rn).



∂

∂qi: e−q

2 ⊗ Peven(Rn)→ e−q2 ⊗ Peven(Rn)

is continuous.

Proof. Let f be an arbitrary function in e−q2 ⊗ Peven(Rn). So f = e−q

2ρ(q), with ρ ∈ Peven.

Then we get∂f

∂qi= −2qie

−q2ρ(q) + e−q2 ∂ρ(q)

∂qi.

The triangle inequality implies∥∥∥∥ ∂f∂qi∥∥∥∥L2

≤∥∥∥−2qie

−q2ρ(q)∥∥∥L2

+

∥∥∥∥e−q2 ∂ρ(q)

∂qi

∥∥∥∥L2

.

The proof that the right-hand side is bounded by the norm of f can be done in the same wayas in the proof of Lemma 4.17.

These two lemmas imply that the representation d$ is continuous.

Theorem 4.19. For every element X in g, it holds that the map

d$(X) : e−q2 ⊗ Peven(Rn)→ e−q

2 ⊗ Peven(Rn),

is continuous.

Proof. The map d$(X) is a finite linear combination of compositions of the operators qi and∂∂qi. In Lemma 4.17 and Lemma 4.18, we have shown that these operators are continuous.

Consequently, d$(X) is also continuous.

We can use this to extend the representation d$ to the space L2even(Rn) as follows. Consider a

function f in L2even(Rn). Since e−q

2 ⊗Peven(Rn) is dense in L2even(Rn), there exists a sequence

(fn)n in e−q2 ⊗ Peven(Rn) such that f = limn→∞ fn. For X in g we define d$(X)(f) as the

limit of d$(X)(fn).We can see as follows that this limit exists. We have shown in Theorem 4.19 that d$(X) iscontinuous. This implies that d$(X)(fn) is a Cauchy sequence, since for every ε > 0 we have,for n,m large enough and a positive constant C,

‖d$(X)(fn)− d$(X)(fm)‖L2 ≤ C ‖fn − fm‖L2 ≤ ε.

The last inequality holds because (fn)n is a convergent sequence and hence a Cauchy sequence.We conclude that the limit of d$(X)(fn) indeed exists.We will now show that this definition is also independent of the sequence fn. Consider anothersequence gn in e−q

2 ⊗Peven(Rn) that converges to f . Then we can also construct the sequencehn, with h2n = fn and h2n+1 = gn. This sequence will converge to f . Furthermore d$(X)(fn)and d$(X)(gn) are converging subsequences of the converging sequence d$(X)(hn). Thisimplies that they will all have the same limit.Hence the extension of the representation d$ is well defined. We can then use the isomorphismp∗ between L2(O, dµ) and L2

even(Rn) to extend the representation dπ to the whole L2(O, dµ)space.

To conclude this chapter we remark that it is possible to integrate dπ to a unique unitaryirreducible representation of a finite cover of the conformal group (Section 2.1.3 in [6]). Us-ing the isomorphism between the conformal algebra and the symplectic Lie algebra and theisomorphism between L2(O,dµ) and L2

even(Rn), this will lead to a representation of the meta-plectic group Mp(n,R) on L2

even(Rn). This representation is known as (the even part of) theSegal-Shale-Weil representation.

Chapter 5

The Fock model

In this chapter we will construct the Fock representation of the conformal algebra. To achievethis goal, we will first study the Jordan algebra of complex symmetric matrices and the complexminimal orbit. We will define a Fock space of holomorphic functions on this complex minimalorbit. This Fock space is similar to the Fock space encountered in the first chapter and we willgive a connection between these two Fock spaces.We will then construct the Fock model starting from a complexification of the Schrodingerrepresentation. We will combine this with a Lie algebra automorphism of the conformal alge-bra to obtain the Fock representation. We then prove that the Fock representation is also aLie algebra representation. In the last chapter we will prove that the Schrodinger and Fockrepresentation are actually two realizations of the same representation.

5.1 The complex symmetric matrices

Consider the complexification of Sym(n,R) as a vector space. This complexification is given bythe space of complex symmetric n×n−matrices, for which we will use the notation Sym(n,C).We can make Sym(n,C) into a Jordan algebra by defining the Jordan product in the followingway:

y ∗ z =yz + zy

2,

for y, z ∈ Sym(n,C). This is indeed a (complex) Jordan algebra, as can be proven in exactlythe same manner as for the real symmetric matrices (Theorem 2.3). The fact that it is a powerassociative algebra can also be proven in the same way as for Lemma 2.4.We denote by τC the C-linear extension of the trace form τ . So τC(y, z) is given by tr(y ∗ z) =tr(yz) and it is still symmetric and bilinear which follows immediately from the linearity ofthe trace form and the fact that tr(yz) = tr(zy). It is, however, no longer an inner product onSym(n,C), since it is not conjugate symmetric, i.e.

τC(y, z) 6= τC(z, y),

and it is not positive definite, since τC(z, z) < 0 for a matrix z that has only an imaginary part.We can define an inner product on Sym(n,C) by

〈y, z〉 := tr(yz),

for y, z ∈ Sym(n,C). This inner product corresponds to the standard inner product on Cn(n+1)

2 ,as can be proven in the same manner as for Lemma 2.14. We will define the norm on Sym(n,C)by

|z| =√〈z, z〉,

for z in Sym(n,C).

73

74 Chapter 5. The Fock model

Let (eα)α be an orthonormal basis of Sym(n,R) with respect to the trace form τ . Then it isalso an orthonormal basis of Sym(n,C) with respect to the inner product 〈 · , · 〉. We have, forevery z in Sym(n,C),

z =∑α

zαeα =∑α

(xα + iyα)eα = x+ iy,

where x, y are elements of Sym(n,R). We define ∂/∂zα as

∂

∂zα=

1

2

(∂

∂xα− i ∂

∂yα

),

and the gradient ∇z as

∇z =1

2(∇x − i∇y).

Using this gradient, we also define the (complex) Bessel operator Bλ as

Bλ = P (∇z) + λ∇z. (5.1)

In coordinates this becomes

Bλf(z) =∑α,β

∂2f(z)

∂zα∂zβP (eα, eβ)z + λ

∑α

∂f

∂zαeα.

5.2 The complex minimal orbit

We will now define the complexification of the minimal orbit.

Definition 5.1. The complex minimal orbit O is given by

X :=uut | u ∈ Cn\0

.

Let pC be the map from Cn\0 to X defined by

pC : u→ uut for u in Cn\0. (5.2)

It is the complexification of the map p defined in (3.4). From the definition of X it follows thatthis map is surjective.Consider

u = (r1eiθ1 , . . . , rne

iθn)

andv = (s1e

iη1 , . . . , sneiηn),

with ri, si in R+ and θi, ηi in ]− π, π]. Then uut = vvt implies

rjrkei(θj+θk) = sjske

i(ηj+ηk+2πl),

for j, k ∈ 1, . . . , n and l ∈ Z. From the case j = k we conclude rj = sj and θj = ηj orθj = ηj ±π for all j ∈ 1, . . . , n. From the case j 6= k we conclude that if θ1 = η1 then θj = ηjfor all j ∈ 1, . . . , n and if θ1 = η1 ± π then θj = ηj ± π for all j ∈ 1, . . . , n. Hence, for u, vin Cn\0, we obtain that

pC(u) = pC(v) implies u = v or u = eiπv = −v. (5.3)

Using the map pC we can prove an analogue of Theorem 3.10, namely that there is an isomor-phism (as vector spaces) between the space F(X) of complex-valued functions defined on the

5.2. The complex minimal orbit 75

complex minimal orbit and the space Feven(Cn) of even complex-valued functions defined onCn. We can equip these spaces with a vector space structure in the following manner:

(λψ + µϕ)(ζ) := λψ(ζ) + µϕ(ζ),

for ψ,ϕ ∈ F(X), ζ ∈ X and λ, µ ∈ C. In the same way we set

(λf + µg)(u) := λf(u) + µg(u),

for f, g ∈ Feven(Cn), u ∈ Cn and λ, µ ∈ C.

Theorem 5.2. Let F(X) be the space of functions ψ : X→ C and Feven(Cn) the space of evenfunctions f : Cn\0 → C. Let pC : Cn\0 → X be the map defined in (5.2). Then the map

p∗C : F(X)→ Feven(Cn)

ψ → ψ pC (5.4)

is an isomorphism of vector spaces.

Proof. We first observe that the map p∗C is linear, because

p∗C(ψ + ϕ)(u) = (ψ + ϕ)(uut) = ψ(uut) + ϕ(uut) = p∗C(ψ)(u) + p∗C(ϕ)(u)

and

p∗C(λψ)(u) = (λψ)(uut) = λψ(uut) = λp∗C(ψ)(u),

for all ψ,ϕ in F(X), u in Cn\0 and λ in C. From the surjectivity of pC it follows that p∗C isinjective. Indeed, if p∗Cψ(u) = 0 for all u in Cn\0, then we also have ψ(ζ) = 0 for all ζ in Xand hence ψ = 0.

Consider a function f in Feven(Cn). We will construct a function ψ in F (X) such thatp∗C(ψ) = f . Define ψ in F(X) as follows:

ψ(ζ) := f(u),

where we choose u = (r1eiθ1 , . . . , rne

iθn) such that uut = ζ and θ1 ∈]0, π]. Because of (5.3), ψis a well-defined function on X. We find, for u = (r1e

iθ1 , . . . , rneiθn),

p∗C(ψ)(u) = ψ(uut) =

f(u) if θ1 ∈]0, π]

f(−u) if θ1 ∈]− π, 0].

Since f is even, we conclude p∗C(ψ) = f . Hence the map p∗C is surjective, which proves thetheorem.

We can use this correspondence between F (X) and Feven(Cn) to define an integral on thecomplex minimal orbit

Definition 5.3. Let ψ be a function on X for which the corresponding p∗Cψ is integrable. Thenthe integral of ψ is defined by ∫

Xψ(ζ)dν(ζ) =

∫Cnψ(uut)du. (5.5)

In the next section we will use this integral to define a Fock space which is similar to the Fockspace of chapter 1.


5.3 The Fock space on the complex minimal orbit

We define the following space of polynomials on the complex minimal orbit X.

Definition 5.4. The space P(X) is the space of (holomorphic) polynomials on Sym(n,C) re-stricted to X, i.e.

P(X) :=P |X

∣∣ P ∈ P(Sym(n,C)).

Note that different polynomials on Sym(n,C) can give rise to the same function in P(X). Forexample, for i and j in 1, . . . , n, the polynomial z2

ij is equal to the polynomial ziizjj , becausefor all ζ = uut in X we have

ζ2ij = uiujuiuj and ζiiζjj = uiuiujuj .

The space P(X) is related to the space of polynomials on Cn in the following way.

Theorem 5.5. The map p∗C, defined in (5.4), is an isomorphism of vector spaces between P(X)and the space of polynomials of even degree on Cn.

Proof. We remark that this proof is very similar to the proof of Theorem 4.15.From Theorem 5.2, we already know that p∗C restricted to P(X) is a well-defined, injective,linear map to the space of even functions on Cn.We will first show that every homogeneous polynomial of degree l in P(X) is mapped to ahomogeneous polynomial on Cn of degree 2l. Consider a monomial ζ [m] of degree |m| = l, thenwe find

p∗C

(ζ [m]

)= p∗C

(∏i≤j

ζmijij

)=∏i≤j

(uiuj)mij .

Because every exponent mij appears twice, the degree of p∗Cζ[m] is

2∑i≤j

mij = 2 |m| .

Hence, we conclude that a polynomial in P(X) is mapped to a polynomial of even degree inF(Cn).We will now prove that p∗C is surjective on the even polynomials defined on Cn. Let

u[m] =

n∏i=1

umii

be an even polynomial, then we can split every ui in an even and an odd part, i.e.

u[m] =

n∏i=1

u2mii

n∏i=1

umii ,

where mi ∈ 0, 1 and 2mi + mi = mi. Because u[m] is even, the mi, i = 1, . . . , n come in pairsof two and we can associate with each pair a ζij . Then we can define the following polynomialin P(X):

ζ [m′] =∏i

ζmiii

∏pairs

ζij ,

where the ζij are associated with the pair uiuj . Since

p∗C

(ζ [m′]

)= u[m],

we conclude that the map p∗C is surjective on the even polynomials defined on Cn. This provesthe theorem.

5.3. The Fock space on the complex minimal orbit 77

We will use the integral of (5.5) to define the following inner product on P(X):

(ψ,ϕ)P(X) := π−3n2

∫Xψ(ζ)ϕ(ζ)e−|ζ|dν(ζ).

The inner product on F(Cn) is given by (see 1.8)

π−3n2

∫Cnf(u)g(u)e−uudu.

From the definition of the integral on X, we get∫Xψ(ζ)ϕ(ζ)e−|ζ|dν(ζ) =

∫Cnψ(uut)ϕ(uut)e−|uut|du.

Therefore, we conclude

(ψ,ϕ)P(X) = (p∗Cψ, p∗Cϕ)F(Cn). (5.6)

Because p∗C induces an isomorphism of vector spaces between P(X) and a subset of F(Cn), andF(Cn) is a Hilbert space, we see that ( · , · )P(X) is indeed an inner product.

We define F(X) as the completion of P(X). Then F(X) is a Hilbert space, which we callthe Fock space on the complex minimal orbit or simply the Fock space. This Fock space isisomorphic to the even part of the Fock space we saw in 1.3.

Theorem 5.6. Let Feven(Cn) be the subspace of even functions of F(Cn). Then the map p∗Cfrom F(X) to Feven(Cn) defined by

p∗Cψ(u) = ψ(uut), (5.7)

is a unitary isomorphism.

Proof. The Stone-Weierstrass Theorem implies that the polynomials are a dense subset ofF(Cn). It then follows from Lemma 4.14 that the even polynomials are a dense subset ofFeven(Cn). Hence Feven(Cn) is the completion of the even polynomials.From (5.6) and Theorem 5.5, it follows that p∗C is a unitary isomorphism between P(X) and thespaces of even polynomials defined on Cn. As F(X) is defined as the completion of P(X) andFeven(Cn) is the completion of the even polynomials, it follows that p∗C is a unitary isomorphismbetween F(X) and Feven(Cn).

We have the following characterization of the Fock space.

Theorem 5.7. The Fock space F(X) is given by

F(X) =

ψ ∈ O(X)

∣∣∣∣ ∫X|ψ(ζ)|2 e−|ζ|dν(ζ) <∞

,

where O(X) is the space of holomorphic functions on X.

Proof. Because p∗C is a unitary isomorphism between F(X) and Feven(Cn), and a function f inFeven(Cn) satisfies ∫

Cn|f(u)|2 e−|u|

2

du <∞,

it follows that a function ψ in F(X) has to satisfy∫X|ψ(ζ)|2 e−|ζ|dν(ζ) =

∫Cn|p∗Cψ(u)|2 e−|u|

2

du <∞.


It also follows that, for a holomorphic function in Feven(Cn), the corresponding function inF(X) will also be holomorphic. So we have

F(X) ⊂ψ ∈ O(X)

∣∣∣∣ ∫X|ψ(ζ)|2 e−|ζ|dν(ζ) <∞

.

To show the other inclusion, we remark that every holomorphic function ψ defined on X, forwhich (ψ,ψ)F(X) < ∞, corresponds to (an even) holomorphic function f on Cn, for which(f, f)F(Cn) <∞. Hence the theorem is proved.

5.4 The Lie algebra automorphism c of gC

Before we can construct the Fock representation, we will need an automorphism of the confor-mal algebra. First we revise some properties of the structure algebra.The structure algebra str = h + q of Sym(n,R) is defined in 2.26. We have proven inTheorem 2.27 that the structure algebra is equal to gl(n,R). From the proof of this theo-rem it follows that the isomorphism between str and gl(n,R) is given by

str −→ gl(n,R)

L(x) −→ x

2for x ∈ Sym(n,R)

[L(x), L(y)] −→ 1

4(xy − yx) for x, y ∈ Sym(n,R).

Since every derivation is inner, i.e. a finite sum of elements of the type [L(x), L(y)] (Theorem2.25), this completely determines the isomorphism.The complexification of gl(n,R) is given by gl(n,C) and we have the following similar isomor-phism between gl(n,C) and strC:

strC −→ gl(n,C)

L(x) −→ x

2for x ∈ Sym(n,C) (5.8a)

[L(x), L(y)] −→ 1

4(xy − yx) for x, y ∈ Sym(n,C). (5.8b)

Consider the conformal algebra g defined in 4.2. Then the complexification gC is given bynC + lC + nC, where

nC := (u, 0, 0) | u ∈ Sym(n,C)lC := (0, T, 0) | T ∈ strCnC := (0, 0, v) | v ∈ Sym(n,C).

The Lie bracket on gC is given by

[(b, A, c), (u, T, v)] = (A ·u− T · b, [A, T ] + bv − uc, T t · c−At · v),

where bv and uc are interpreted as elements of gl(n,C).

We will now give a Lie algebra automorphism of gC.

Theorem 5.8. The map c : gC → gC defined by

c(b, 0, 0) =

(b

4, iL(b), b

)b ∈ Sym(n,C) (5.9a)

c(0, L(a) +D, 0) =(ia

4, D,−ia

)L(a) ∈ qC, D ∈ hC (5.9b)

c(0, 0, b) =

(b

4,−iL(b), b

)b ∈ Sym(n,C), (5.9c)

is a Lie algebra automorphism.

5.4. The Lie algebra automorphism c of gC 79

Proof. We will not prove this theorem directly but instead use the Lie algebra isomorphismbetween gC and sp(2n,C) given by

gC → sp(2n,C)

(u, T, v)→(T uv −T t

),

for u, v in Sym(n,C) and T in gl(n,C). We can split every matrix X in sp(2n,C) in a symmetricand an antisymmetric part as

X =X +Xt

2+X −Xt

2.

The operator L(a) corresponds to the symmetric matrix a/2 and the operator D correspondsto an antisymmetric matrix d. So we can write (b, L(a) +D, c) ∈ gC as(

b+ c

2,a

2,b+ c

2

)+

(b− c

2, d,

c− b2

).

The first term corresponds to the symmetric part in sp(2n,C) and the second term corre-sponds to the antisymmetric part in sp(2n,C). The expression for the map c on sp(2n,C)corresponding to c is then given by

c : sp(2n,C)→ sp(2n,C)(A BB −A

)→(

0 12(B + iA)

2(B − iA) 0

)for the symmetric part(

D C−C D

)→(D + iC 0

0 D − iC

)for the antisymmetric part,

where A,B,C are symmetric n× n-matrices and D is an antisymmetric n× n-matrix.It is clear that c is linear. Assume c(X) = 0 for an arbitrary X in sp(2n,C). Then we getD + iC = 0 and D − iC = 0, so we conclude C = D = 0. We also get B + iA = 0 andB − iA = 0. Hence, A = B = 0 and thus X = 0. This implies that c is injective. Sincesp(2n,C) is finite-dimensional, we also have that c is surjective and thus it is a vector spaceisomorphism.To show that it also is a Lie algebra homomorphism, i.e.

c([X,Y ]) = [c(X), c(Y )],

for all X,Y in sp(2n,C), we will consider the cases where X and Y are both symmetric or bothantisymmetric and the case where one is symmetric and the other one is antisymmetric.

First, assume that X =(A BB −A

)and Y =

(T UU −T

)are symmetric. Then A,B, T, U are

symmetric n× n−matrices. For [X,Y ] we get

[X,Y ] =

[(A BB −A

),

(T UU −T

)]=

([A, T ] + [B,U ] AU + UA− TB −BT

−AU − UA+ TB +BT [A, T ] + [B,U ]

).

This is an antisymmetric matrix, hence

c([X,Y ]) =

[A, T ] + [B,U ]

+i(AU + UA− TB −BT )0

0[A, T ] + [B,U ]−i(AU + UA− TB −BT )

.


We also have

[c(X), c(Y )] =

[(0 1

2(B + iA)2(B − iA) 0

),

(0 1

2(U + iT )2(U − iT ) 0

)]

=

(B + iA)(U − iT )−(U + iT )(B − iA)

0

0(B − iA)(U + iT )−(U − iT )(B + iA)

.

Comparing the two matrix expressions, we conclude that [c(X), c(Y )] = c([X,Y ]).

Secondly, assume that X =(A BB −A

)is symmetric and Y =

(D C−C D

)is antisymmetric.

Then A,B,C are symmetric n × n−matrices and D is an antisymmetric n × n-matrix.For [X,Y ] we have

[X,Y ] =

[(A BB −A

),

(D C−C D

)]=

([A,D]−BC − CB AC + [B,D] + CA[B,D] +AC + CA −[A,D] +BC + CB

).

Since this is a symmetric matrix, we get

c([X,Y ]) =

012(AC + [B,D] + CA)

+ i2([A,D]−BC − CB)

2(AC + [B,D] + CA)−2i([A,D]−BC − CB)

0

.

We also have

[c(X), c(Y )] =

[(0 1

2(B + iA)2(B − iA) 0

),

(D + iC 0

0 D − iC

)]

=

012(B + iA)(D − iC)−1

2(D + iC)(B + iA)2(B − iA)(D + iC)−2(D − iC)(B − iA)

0

.

Therefore, we can conclude that also in this case [c(X), c(Y )] = c([X,Y ]).

Last, assume that X =(D C−C D

)and Y =

(T U−U T

)are both antisymmetric. Then C,U

are symmetric n× n−matrices and D,T are antisymmetric n× n−matrices. For [X,Y ]we get

[X,Y ] =

[(D C−C D

),

(T U−U T

)]=

([D,T ]− [C,U ] [D,U ]− [T,C]−[D,U ] + [T,C] [D,T ]− [C,U ]

).

This is an antisymmetric matrix, hence

c([X,Y ]) =

[D,T ]− [C,U ]

+i([D,U ]− [T,C])0

0[D,T ]− [C,U ]−i([D,U ]− [T,C])

.

5.5. The Lie algebra representation dπC 81

We also have

[c(X), c(Y )] =

[(D + iC 0

0 D − iC

),

(T + iU 0

0 T − iU

)]=

([D + iC, T + iU ] 0

0 [D − iC, T − iU ]

).

So in this case we also get [c(X), c(Y )] = c([X,Y ]).

We have proven that c is a Lie algebra automorphism of sp(2n,C). Consequently, we also havethat c is a Lie algebra automorphism of gC.

5.5 The Lie algebra representation dπC

We will now consider a complexification of the Schrodinger representation dπ. The mapdπC : gC → End(P(X)) is defined as

dπC(b, 0, 0)ψ(ζ) = iτC(ζ, b)ψ(ζ) (5.10a)

dπC(0, A, 0)ψ(ζ) = DAt · ζψ(ζ) +1

2tr(A)ψ(ζ) (5.10b)

dπC(0, 0, c)ψ(ζ) = iτC(Bψ(ζ), c

), (5.10c)

for all ψ in P(X). Here B is the (complex) Bessel operator for the value λ = 1/2 (see 5.1)and DAt · ζψ(ζ) := τC(At · ζ,∇zψ(ζ)), where At · ζ = Atζ + ζA is the action of the (complex)structure algebra on Sym(n,C). We will first show that this is a well-defined linear map.

Linearity follows from the linearity of the gradient, the Bessel operator and the trace form τC.We have

τC(ζ, b) = tr(ζb) =n∑

i,j=1

ζijbij ,

which is a homogeneous polynomial of degree one. So iτC(ζ, b)ψ(ζ) is an element of P(X) forψ in P(X). We also know that for k ∈ N

∂ζkβ∂zα

= δαβkζk−1β ,

from which it follows that the partial derivative of a polynomial is again a polynomial. So

DAt · ζψ(ζ) =∑α

τC(At · ζ, eα)∂ψ(ζ)

∂zα=∑α

τC(ζ,A · eα)∂ψ(ζ)

∂zα

is the sum over α of the product of two polynomials, which is again a polynomial. Finally wehave

τC(Bψ(ζ), c

)=∑α,β

∂ψ(ζ)

∂zα∂zβτC(P (eα, eβ)ζ, c

)+

1

2

∑α

∂ψ(ζ)

∂zατC(eα, c).

Using Lemma 3.17, we find

τC(P (eα, eβ)ζ, c

)=

1

2

n∑i,j,k,l=1

(eαijζjkeβklcli + eβijζjkeαklcli),


which is a homogeneous polynomial of degree one. Hence τC(Bψ(ζ), c) is a polynomial and thusdπC(X)ψ in P(X) for all X in gC. Therefore, dπC is a well-defined map from the (complex)conformal algebra gC to End(P(X)). The proof that dπC is a Lie algebra homomorphism, i.e

dπC([X,Y ]) = dπC(X)dπC(Y )− dπC(Y )dπC(X),

for all X,Y in gC, is completely similar to the proof of theorem 4.10. So we conclude that dπCis a Lie algebra representation of gC on P(X).

We can use the isomorphism p∗C of (5.7) to get a representation d$C on Feven(Cn).

Theorem 5.9. The Lie algebra representation d$C of the conformal algebra g on the subspaceof even polynomials of Feven(Cn) is given by

d$C(b, 0, 0)f(u) = in∑

i,j=1

uibijujf(u) for (b, 0, 0) ∈ nC

d$C(0, A, 0)f(u) =

n∑i,j=1

Aijui∂f(u)

∂uj+

1

2tr(A)f(u) for (0, A, 0) ∈ lC

d$C(0, 0, c)f(u) =i

4

n∑i,j=1

cij∂2f(u)

∂ui∂ujfor (0, 0, c) ∈ nC,

for f a polynomial in Feven(Cn).

Proof. This proof is essentially the same as the one of Theorem 4.16 by changing the realderivative ∂/∂qi to the complex derivative ∂/∂ui.

Just as in section 4.3, it is possible to extend dπC to the whole space F(X). To do this, wefirst prove that the complex derivative is bounded on the polynomials in Feven(Cn).


∂

∂ui: Feven(Cn) ∩ P(Cn)→ Feven(Cn) ∩ P(Cn),

is continuous.

Proof. Consider a polynomial ρ in Feven(Cn) ∩ P(Cn). It is of the form ρ(u) =∑

[m] α[m]u[m]

and we have∂ρ(u)

∂ui=∑[m]

miα[m]u[m]−[1i],

where [1i] = (0, . . . , 1, . . . , 0) with the one on the ith place. From expression (1.9) for the innerproduct on F(Cn) we get∥∥∥∥∂ρ(u)

∂ui

∥∥∥∥2

F(Cn)

= π−n2

∑[m]

[m!]

mim2i

∣∣α[m]

∣∣2≤ π

−n2 max

[m]|m|

∑[m]

[m!]∣∣α[m]

∣∣2≤ max

[m]|m| ‖ρ(u)‖2F(Cn) .

This proves the lemma.

Similar we can also prove that the operator ui is continuous.

5.5. The Lie algebra representation dπC 83


ui : Feven(Cn) ∩ P(Cn)→ Feven(Cn) ∩ P(Cn),

is continuous.

Proof. Consider a polynomial ρ(u) =∑

[m] α[m]u[m] in Feven(Cn) ∩ P(Cn). Then uiρ(u) =∑

[m] α[m]u[m]+[1i]. Using (1.9), we get

‖uiρ(u)‖2F = π−n2

∑[m]

[m!](mi + 1)∣∣α[m]

∣∣2≤ π

−n2 (max

[m]|m|+ 1)

∑[m]

[m!]∣∣α[m]

∣∣2≤ (max

[m]|m|+ 1) ‖ρ(u)‖2F ,


These two lemmas can be used to prove the following theorem.

Theorem 5.12. For every element X in gC, it holds that the map

d$C(X) : Feven(Cn) ∩ P(Cn)→ Feven(Cn) ∩ P(Cn),

is continuous.

Proof. The map d$C(X) is a finite linear combination of compositions of the operators ui and∂/∂ui. These operators are continuous, which implies that d$(X)C is also continuous.

For a function f in Feven(Cn) we can define d$C(X)f(u) as the limit of d$Cfn(u), where fnis a sequence in Feven(Cn)∩P(Cn) that converges to f . It is possible to show in the same wayas in section 4.3 that this is well defined. Hence we have extended the representation d$C tothe whole space Feven(Cn).We can then use the isomorphism between F(X) and Feven(Cn) (Theorem 5.6) to extend therepresentation dπ to the whole space F(X).

We define the representation dρ asdρ = dπC c. (5.11)

Because c is a Lie algebra automorphism of gC and dπC is a Lie algebra representation of gCon F(X), it is clear that dρ is also a Lie algebra representation of gC on F(X). We will call dρthe Fock representation.

We can combine Theorem 5.9 with the automorphism c to obtain the following expression forthe Fock representation d% on Feven(Cn).

Theorem 5.13. The Lie algebra representation d% of the conformal algebra g on the subspaceof even polynomials of Feven(Cn) is given by

d%(b, a+ d, c)f(u) =i

4

n∑i,j=1

(i2aij + bij + cij)uiujf(u)

+n∑

i,j=1

(i

2(bij − cij) + dij

)ui∂f(u)

∂uj+i

4tr(b− c)f(u)

+i

4

n∑i,j=1

(−i2aij + bij + cij)∂2f(u)

∂ui∂uj,

for a, b, c in Sym(n,R), d an antisymmetric n× n-matrix and f a polynomial in Feven(Cn).


Proof. The representation d% is equal to d$C c, with d$C as in Theorem 5.9. From thedefinition of c we obtain

c(b, a+ d, c) =

(i2a+ b+ c

4, ib− c

2+ d,−i2a+ b+ c

).

Substituting this in Theorem 5.9 and using tr(d) = 0 gives the desired result.

Chapter 6

The Segal-Bargmann transform asan intertwining operator

We are now ready to reinterpret the Segal-Bargmann transform as an intertwining operatorbetween the Schrodinger model and the Fock model.

First we will define an integral transform from the square-integrable functions on the minimalorbit to the Fock space on the complex minimal orbit. Then we will see how this integraltransform is related to the classical Segal-Bargmann transform. Finally we will prove that thisnew integral transform is indeed an isomorphism between the Schrodinger representation andthe Fock representation.

6.1 The new Segal-Bargmann transform

We define the following complex function B from C to C:

B(t) = Γ(λ)Iλ−1(2√t),

where Iα is the renormalized I-Bessel function (3.23) and λ is a complex constant. We willonly be interested in the case λ = 1/2. Then B reduces to

B(t) = Γ(

12

)I− 1

2(2√t) = cosh(2

√t),

where we used (3.24). The square root is not holomorphic on the whole complex plane. How-ever, because of

cosh(2√t) =

∞∑n=0

(2√t)2n

2n!=

∞∑n=0

(4t)n

2n!,

we see that B is holomorphic on the whole complex plane. For y, z in Sym(n,C) we will writeB(y|z) for B(τC(y, z)).

Using this notation we can define the integral transform SB for functions ψ in L2(O,dµ) asfollows:

SBψ(z) := 2n4

∫Oe−

12

tr(z)B(η|z)e− tr(η)ψ(η)dµ(η) for z ∈ Sym(n,C). (6.1)

We call this transform the Segal-Bargmann transform. We can immediately prove the followinglemma, which states that this transform is well defined.

Lemma 6.1. The Segal-Bargmann transform SB is a well-defined linear map from L2(O,dµ)to the space of holomorphic functions on Sym(n,C).

85

86 Chapter 6. An intertwining operator

Proof. The linearity of this map follows from the linearity of the integral. To show that SB(ψ)is a well-defined holomorphic function on Sym(n,C), for every ψ in L2(O,dµ), we will apply

Theorem 1.6 to the function f : O×Sym(n,C)→ C where f(η, z) = e−12

tr(z)B(η|z)e− tr(η)ψ(η).So we have to prove

η → f(η, z) is integrable for all z in Sym(n,C).

z → f(η, z) is holomorphic for all η in O.

For each z in Sym(n,C) there exists an open neighbourhood W ⊂ Sym(n,C) of z and anon-negative function g in L1(O,dµ) such that for all z′ in W∣∣f(η, z′)

∣∣ ≤ g(η).

From the Schwarz’ inequality∣∣∣∣∫OB(η|z)e− tr(η)ψ(η)dµ(η)

∣∣∣∣ ≤ ∥∥∥B(η|z)e− tr(η)∥∥∥L2(O)

‖ψ(η)‖L2(O) ,

it follows that f(η, z) is integrable for all z in Sym(n,C). From the fact that the exponentialfunction, the function B and the trace form τC are holomorphic, it follows that f(η, z) is alsoholomorphic for all η in O. To find a non-negative function g we will first establish certaininequalities.

In section 5.1 we defined the inner product and the norm on Sym(n,C) as 〈x, z〉 = tr(xz) and|z| =

√〈z, z〉 =

√tr(zz). Then we find, for x in Sym(n,R) and z in Sym(n,C),

|tr(z)|2 =∣∣∣ n∑i=1

zii

n∑j=1

zjj

∣∣∣ ≤ n∑i,j=1

|zii| |zjj | ≤n∑

i,j=1

(maxi|zii|

)2= n2

(maxi|zii|

)2 ≤ n2 |z|2 ,

since |zii|2 ≤ |z|2. We also have

|tr(xz)| = |tr(xz)| = |〈x, z〉| ≤ |x| |z| ,

where we used the fact that x is real in the first step and the Schwarz’ inequality in the laststep. Using these inequalities, we get∣∣∣e− 1

2tr(z)B(x|z)

∣∣∣ ≤ e−n2 |z|B(|x| |z|).Let x = η be an element of the minimal orbit O, then tr(η) = |η| and we find for |z| < R,where R is a positive real number,∣∣∣e− 1

2tr(z)B(η|z)e− tr(η)ψ(η)

∣∣∣ ≤ e−n2RB(R |η|)e−|η| |ψ(η)| .

Because B(R |η|)e−|η| and ψ(η) are in L2(O, dµ), this function is a non-negative function inL1(O,dµ) and we conclude that SB(ψ) is a holomorphic function on Sym(n,C).

We remark that from this proof and theorem 1.6 it also follows that we can bring the derivativewith respect to the complex variable z under the integration, i.e.

∂

∂zαSBψ(z) = 2

n4

∫O

∂

∂zα

(e−

12

tr(z)B(η|z)e− tr(η)ψ(η))

dµ(η). (6.2)

6.2. Connection with the classical Segal-Bargmann transform 87

6.2 Connection with the classical Segal-Bargmann transform

We will now establish a connection between the Segal-Bargmann transform defined in theprevious section and the classical Segal-Bargmann transform, which we studied in chapter 1.We remind that the classical Segal-Bargmann transform is defined as

SB : L2(Rn)→ F(Cn)

ψ(q)→∫Rne−

12u2+√

2u · q− 12q2ψ(q)dnq.

Define the map r as

r : L2(Rn)→ L2(Rn)

ψ(q)→ 2−n4 ψ

(q√2

).

Then r is a vector space isomorphism and we have

‖r(ψ)‖2L2(Rn) = 2−n2

∫Rn

∣∣∣∣ψ( q√2

)∣∣∣∣2 dnq

= 2−n2 2

n2

∫Rn

∣∣ψ (q′)∣∣2 dnq′

= ‖ψ‖2L2(Rn) .

We define SB as SB r. Because SB is unitary and r is unitary, SB is also unitary. We have,for u in Cn,

SBψ(u) = 2−n4

∫Rne−

12u2+√

2u · q− 12q2ψ

(q√2

)dnq

= 2−n4 2

n2

∫Rne−

12u2+2u · q′−q′2ψ

(q′)

dnq′. (6.3)

We remark that sometimes SB is used as the definition of the classical Segal-Bargmann trans-form. We also note that if we restrict the Segal-Bargmann transform to the even functions onL2(Rn), we get a unitary isomorphism between L2

even(Rn) and Feven(Cn). This follows from:

SBψ(−u) =

∫Rne−

12u2−√

2u · q− 12q2ψ(q)dnq

=

∫Rne−

12u2+√

2u · q− 12q2ψ(−q) |−1|n dnq.

Hence SBψ(−u) = SBψ(u) for ψ in L2even(Rn) and SBψ(−u) = −SBψ(u) for ψ in L2

odd(Rn).So the Segal-Bargmann transform preserves the composition of the space in an even and oddpart.We find the following relation between SB and SB.

Theorem 6.2. Let p∗ the unitary isomorphism between L2(O, dµ) and L2even(Rn) defined in

3.6, and p∗C the unitary isomorphism between F(X) and Feven(Cn) defined in 5.7. Then wehave

p∗C SB = SB p∗.So the following diagram commutes.

L2(O) L2even(Rn)

F(X) Feven(Cn)

SB

p∗

SB

p∗C


Proof. Consider an element ψ of L2(O,dµ). On the one hand we have, for u in Cn,

SB(p∗(ψ)

)u = 2

n4

∫Rne−

12u2+2u · q−q2ψ

(qqt)

dnq, (6.4)

where we used (6.3) and the definition of p∗. On the other hand we get, using (6.1) and thedefinition of p∗C,

p∗C(SB(ψ)( · )

)u = p∗C

(2n4

∫Oe−

12

tr( · )B( · |η)e− tr(η)ψ(η)dµ(η)

)(u)

= 2n4

∫Oe−

12

tr(uut)B(uut|η)e− tr(η)ψ(η)dµ(η)

= 2n4

∫Rnp∗(e−

12

tr(uut)B(uut| · )e− tr( · )ψ( · ))

(q)dnq

= 2n4

∫Rne−

12

tr(uut)B(uut|qqt)e− tr(qqt)ψ(qqt)dnq,

where we used Lemma 3.9 about the correspondence between integration over Rn and over theminimal orbit O. Because B(uut|qqt) = cosh(2

√(tr(uutqqt)) and

tr(uutqqt) = tr(qtuutq) =

n∑i=1

qiui

n∑j=1

ujqj = (u · q)2,

and

tr(qqt) =

n∑i=1

qiqi = q2, tr(uut) =

n∑i=1

uiui = u2,

we can simplify this to

p∗C (SB(ψ)( · ))u = 2n4

∫Rne−

12u2 cosh(2u · q)e−q

2ψ(qqt)dnq

= 2n4

∫Rne−

12u2 e

2u · q + e−2u · q

2e−q

2ψ(qqt)dnq

= 2n4

∫Rne−

12u2e2u · qe−q

2ψ(qqt)dnq.

Comparing this with (6.4), we conclude that SB p∗ is equal to p∗C SB.

We remark that from the fact that SB, p∗ and p∗C are all unitary, it follows immediately thatSB is a unitary isomorphism between L2(O,dµ) and F(X). If we combine Lemma 6.1 withTheorem 5.7, this also implies that every function in F(X), which is holomorphic on O, can beextended to a function that is holomorphic on the whole space Sym(n,C).

6.3 The Segal-Bargmann transform as an intertwining operator

In this section we will show that SB is an intertwining operator between the Schrodinger andFock representation. To be able to prove this, we will first calculate the action of the gradientand the Bessel operator on the kernel of the integral transform SB. Let us start by proving thefollowing lemma.

Lemma 6.3 (Complex version). Let z be an arbitrary element of Sym(n,C). Then

∇zeλ tr(z) = λeλ tr(z)In,

where In is the unity matrix and λ is an arbitrary complex constant.

6.3. The Segal-Bargmann transform as an intertwining operator 89

Lemma 6.4 (Real version). Let x be an arbitrary element of Sym(n,R). Then

∇xeλ tr(x) = λeλ tr(x)In,

where In is the unity matrix and λ is an arbitrary real constant.

Proof. The two lemmas can be proved in the same manner. We will only give a proof for thecomplex version. Let (Eij)ij be the orthonormal basis defined in (2.7). Then we have

z =∑

1≤i≤j≤nz(ij)Eij

and

tr(z) =n∑i=1

zii =n∑i=1

z(ii).

Note that for i 6= j, it does not hold that zij = z(ij). We get

∇zeλ tr(z) =∑

1≤i≤j≤nEij

∂

∂z(ij)exp(λ

n∑k=1

z(kk)

)=

∑1≤i≤j≤n

Eijλδij exp(λ

n∑k=1

z(kk)

)= λIne

λ tr(z),


With these lemmas we can calculate the action of the Bessel operator on the exponential.

Lemma 6.5. Let y be an element of Sym(n,R) or Sym(n,C) and B the real or complex Besseloperator (for the value 1/2). Then it holds

Beλ tr(y) = λeλ tr(y)(λy + 12In),

where In is the unity matrix and λ is a complex constant.

Proof. For y in Sym(n,R) and B the real Bessel operator we have, using expression 3.9 andthe orthonormal basis defined in (2.7),

Beλ tr(y) =∑

(i,j),(k,l)

∂2eλ tr(y)

∂x(ij)∂x(kl)P (Eij , Ekl)y + 1

2∇xeλ tr(y)

=∑

(i,j),(k,l)

λ2δijδkleλ tr(y)P (Eij , Ekl)y + 1

2λeλ tr(y)In

=∑i,k

λ2eλ tr(y)P (Eii, Ekk)y + 12λe

λ tr(y)In

= λ2eλ tr(y)y + 12λe

λ tr(y)In,

which proves the lemma. The complex version for y in Sym(n,C) can be proven similar.

To calculate the action of the Bessel operator on B(x|z), we will first calculate the gradient ofB(x|z).


Lemma 6.6. Let x in Sym(n,R) and z in Sym(n,C). Then the following holds

∇xB(x|z) =z√

tr(xz)sinh(2

√tr(xz)),

and∇zB(x|z) =

x√tr(xz)

sinh(2√

tr(xz)).

Proof. For an orthonormal basis (eα)α with respect to the trace form τ we have

∇z tr(xz) =∑α

eα∂

∂zα

(∑β

xβzβ

)=∑

eαxα = x,

for x in Sym(n,R) and z in Sym(n,C). Using the chain rule and cosh′ = sinh, we obtain

∇zB(x|z) = ∇z cosh(2√

tr(xz))

=sinh

(2√

tr(xz))√

tr(xz)∇z tr(xz) =

x√tr(xz)

sinh(2√

tr(xz)).

The proof of

∇xB(x|z) =z√

tr(xz)sinh

(2√

tr(xz))

is similar.

This can be used to compute the action of the Bessel operator on B(η|ζ).

Lemma 6.7. Let η in O and ζ in X. For the real Bessel operator at the value 1/2 one has

BB(η|ζ) = ζB(η|ζ).

For the complex Bessel operator at the value 1/2 one has

BB(η|ζ) = ηB(η|ζ).

Proof. From the previous lemma and the product rule, it follows that

∂2B(η|ζ)

∂xα∂xβ=

∂

∂xα

(sinh

(2√

tr(ηζ))√

tr(ηζ)ζβ

)=

cosh(2√

tr(ηζ))

tr(ηζ)ζβζα −

1

2

sinh(2√

tr(ηζ))

tr(ηζ)32

ζβζα.

Using this, we obtain

BB(η|ζ) =∑α,β

∂2B(η|ζ)

∂xα∂xβP (eα, eβ)η +

1

2∇xB(η|ζ)

=

(cosh

(2√

tr(ηζ))

tr(ηζ)− 1

2

sinh(2√

tr(ηζ))

tr(ηζ)32

)P (ζ, ζ)η +

1

2

ζ√tr(ηζ)

sinh(2√

tr(ηζ)).

An element ζ of X is of the form uut, with u in Cn. Hence

P (ζ, ζ)η

tr(ηζ)=uutηuut

utηu= uut = ζ,

where we used lemma 3.17 for the expression of P . This leads to

BB(η|ζ) = cosh(2√

tr(ηζ))ζ − 1

2

sinh(2√

tr(ηζ))

tr(ηζ)12

ζ +1

2

sinh(2√

tr(ηζ))√

tr(ηζ)ζ = ζB(η|ζ).

One can show the version for the complex Bessel operator in a similar way.


Combining the previous lemmas, we can calculate the action of the Bessel operator on theintegral kernel of SB.

Lemma 6.8. For η in O and ζ in X, one has that the action of the complex Bessel operator Bon the integral kernel e−

12

tr(ζ)B(η|ζ)e− tr(η) is equal to

e−12

tr(ζ)B(η|ζ)e− tr(η)

(ζ − In

4+ η

)− e−

12

tr(ζ)−tr(η) sinh(2√

tr(ηζ))√

tr(ηζ)η ∗ ζ.

Proof. We will use the product rule for the Bessel operator (Theorem 3.19). Then we get

B(e−

12

tr(ζ)B(η|ζ)e− tr(η))

=

B(e−

12

tr(ζ))B(η|ζ)e− tr(η) + 2P

(∇ze−

12

tr(ζ),∇zB(η|ζ))ζe− tr(η) + e−

12

tr(ζ)B(B(η|ζ)

)e− tr(η).

Using the Lemmas 6.3, 6.5, 6.6 and 6.7, we find

B(e−

12

tr(ζ)B(η|ζ)e− tr(η))

=

14e− 1

2tr(ζ)(ζ − In)B(η|ζ)e− tr(η) + 2P

(−1

2e− 1

2tr(ζ)In,

sinh(2√

tr(ηζ))√

tr(ηζ)η

)ζe− tr(η)

+ e−12

tr(ζ)ηB(η|ζ)e− tr(η).

Because P (In, η)ζ = η ∗ ζ, this proves the lemma.

For the real Bessel operator we obtain a similar lemma.

Lemma 6.9. For η in O and ζ in X, one has that the action of the real Bessel operator B onthe integral kernel e−

12

tr(ζ)B(η|ζ)e− tr(η) is equal to

e−12

tr(ζ)B(η|ζ)e− tr(η)(η − 1

2In + ζ)− 2e−

12


tr(ηζ))√

tr(ηζ)η ∗ ζ.

Proof. The proof goes the same as for the complex Bessel operator.

Using these lemmas we can prove the main theorem of this chapter.

Theorem 6.10. Let g be the conformal algebra, dπ the Schrodinger representation of g onL2(O,dµ), dρ the Fock representation of g on F(X) and SB the Segal-Bargmann transformfrom L2(O, dµ) to F(X). Then we have for all X in g

SB dπ(X) = dρ(X) SB, (6.5)

i.e. the following diagram commutes.

L2(O) L2(O)

F(X) F(X)

dπ(X)

SB SBdρ(X)

Proof. We know that g = n + l + n. So we will split the proof of (6.5) in three parts. Usingthe definition of the Fock representation dρ = dπC c with dπC the complexification of theSchrodinger representation defined in (5.10) and c the Lie algebra automorphism defined in(5.9), we have to prove, for all a in Sym(n,R) and D in Der(Sym(n,R)),

Part A: SB dπ(a, 0, 0)(ψ) = dπC(a4 , iL(a), a

) SB(ψ)

Part B: SB dπ(0, L(a) +D, 0)(ψ) = dπC(ia4 , D,−ia

) SB(ψ)

Part C: SB dπ(0, 0, a)(ψ) = dπC(a4 ,−iL(a), a

) SB(ψ),

for all ψ in L2(O, dµ). Since L2(O,dµ)k is dense in L2(O,dµ) and SB is a unitary isomorphism,it suffices to prove it for all ψ in L2(O,dµ)k.


Part A

Consider a function ψ(η) in L2(O,dµ)k, then we get

dπ(a, 0, 0)ψ(η) = iτ(η, a)ψ(η),

from which follows:

SB dπ(a, 0, 0)ψ(ζ) = 2n4

∫Oe−

12

tr(ζ)B(η|ζ)e− tr(η)iτ(η, a)ψ(η)dµ(η). (6.6)

On the other hand we have

SB(ψ)(ζ) = 2n4

∫Oe−

12

tr(ζ)B(η|ζ)e− tr(η)ψ(η)dµ(η),

from which we obtain

dπC

(a4, iL(a), a

) SBψ(ζ)

= iτ(ζ,a

4

)SBψ(ζ) +DiL(a) · ζSBψ(ζ) + 1

2 tr(iL(a)

)SBψ(ζ) + iτ

(B(SBψ(ζ)

), a). (6.7)

HereDiL(a) · ζSBψ(ζ) = iτ

(a ∗ ζ,∇zSBψ(ζ)

),

B is the complex Bessel operator for the value 1/2 (5.1) and tr(L(a)) = tr(a/2) (see 5.8a).Using (6.2), Lemma 6.3 and Lemma 6.6, we get

∇zSBψ(ζ) = 2n4

∫O∇z(e−

12

tr(ζ)B(η|ζ)e− tr(η))ψ(η)dµ(η)

= 2n4

∫O−1

2Ine− 1

2tr(ζ)B(η|ζ)e− tr(η)ψ(η)dµ(η)

+ 2n4

∫Oe−

12

tr(ζ)−tr(η) η√tr(ηζ)

sinh(2√

tr(ηζ))ψ(η)dµ(η). (6.8)

From Lemma 6.8 we deduce

B(SBψ(ζ)

)= 2

n4

∫OB(e−

12

tr(ζ)B(η|ζ)e− tr(η))ψ(η)dµ(η)

= 2n4

∫Oe−

12

tr(ζ)B(η|ζ)e− tr(η)

(ζ − In

4+ η

)ψ(η)dµ(η)

− 2n4

∫Oe−

12


tr(ηζ))√

tr(ηζ)η ∗ ζψ(η)dµ(η). (6.9)

Inserting (6.8) and (6.9) in (6.7), we get

dπC

(a4, iL(a), a

) SBψ(ζ)

= 2n4

∫O

(iτ(ζ,a

4

)B(η|ζ)− i

2τ(a, ζ)B(η|ζ) + iτ(a ∗ ζ, η)

sinh(2√

tr(ηζ))√

tr(ηζ)+i

4tr(a)B(η|ζ)

+ iB(η|ζ)τ

(ζ − In

4+ η, a

)−i

sinh(2√

tr(ηζ))√

tr(ηζ)τ(η ∗ ζ, a)

)e−

12

tr(ζ)−tr(η)ψ(η)dµ(η)

= 2n4

∫Oe−

12

tr(ζ)B(η|ζ)e− tr(η)iτ(η, a)ψ(η)dµ(η), (6.10)

where we used τ(a ∗ ζ, In) = τ(a, ζ) = 2τ (ζ, a/4) + 2τ (ζ/4, a) and τ(η ∗ ζ, a) = τ(a ∗ ζ, η).Comparing (6.6) and (6.10) we conclude that

SB dπ(a, 0, 0) = dπC(a4 , iL(a), a

) SB,

which proves part A.


Part B

We have

dπ(0, L(a) +D, 0

)ψ(η) = DL(a)η−D(η)ψ(η) + 1

2 tr(a2

)ψ(η),

where we used (5.8a) and tr(D) = 0 (2.12). Since ψ ∈ L2(O,dµ)k is a combination of apolynomial with the Gaussian exp(− tr(η)), it is a rapidly decreasing function in the Schwartz

space. The integral kernel e−12

tr(z)B(η|z)e− tr(η) is also a rapidly decreasing function in theSchwartz space. Therefore, we can apply Theorem 4.4. Combining this with Lemma 6.4 andLemma 6.6, we get

SB dπ(0, L(a) +D, 0)ψ(ζ)

= 2n4

∫Oe−

12

tr(ζ)B(η|ζ)e− tr(η)(DL(a)η−D(η)ψ(η) + 1

4 tr(a)ψ(η))

dµ(η)

= − 2n4

∫ODL(a)η−D(η)

(e−

12

tr(ζ)B(η|ζ)e− tr(η))ψ(η) + e−

12

tr(ζ)B(η|ζ)e− tr(η) 14 tr(a)ψ(η)dµ(η)

= − 2n4

∫O

(τ(L(a)η −D(η), ζ

)sinh(2√

tr(ηζ))√

tr(ηζ)

− τ(L(a)η −D(η), In

)B(η|ζ) + 1

4 tr(a)B(η|ζ)

)e−

12


= − 2n4

∫O

(τ(L(a)η −D(η), ζ

)sinh(2√

tr(ηζ))√

tr(ηζ)

− τ(a, η)B(η|ζ) + 14 tr(a)B(η|ζ)

)e−

12

tr(ζ)−tr(η)ψ(η)dµ(η), (6.11)

where we used τ(D(η), In) = −τ(η,D(In)) = 0, since D(In) = 0. We also have, using (6.9),

dπC

(ia

4, D,−ia

) SBψ(ζ)

= − τ(ζ,a

4

)SBψ(ζ)− τ

(D(ζ),∇zSBψ(ζ)

)+ τ(B(SBψ(ζ)

), a)

= 2n4

∫O

(−τ(ζ,a

4

)B(η|ζ) + 1

2τ(D(ζ), In

)B(η|ζ)− τ

(D(ζ), η

)sinh(2√

tr(ηζ))√

tr(ηζ)

+B(η|ζ)τ

(ζ − In

4+ η, a

)−

sinh(2√

tr(ηζ))√


)e−

12


= 2n4

∫O

(τ(−a ∗ η +D(η), ζ

)sinh(2√

tr(ηζ))√

tr(ηζ)

+τ(a, η)B(η|ζ)− 14 tr(a)B(η|ζ)

)e−

12

tr(ζ)−tr(η)ψ(η)dµ(η), (6.12)

where we used τ(D(ζ), In) = −τ(ζ,D(In)) = 0 and τ(D(ζ), η) = −τ(ζ,D(η)). Because (6.11)is equal to (6.12), we have proven part B.

Part C

We have

dπ(0, 0, a)ψ(η) = iτ(Bψ(η), a

),


where B is the real Bessel operator. Since ψ and e−12

tr(z)B(η|z)e− tr(η) are elements of theSchwartz space, we can apply Theorem 3.31. Then we get, using Lemma 6.9,

SB dπ(0, 0, a)ψ(ζ) = 2n4 i

∫Oe−

12

tr(ζ)B(η|ζ)e− tr(η)τ(Bψ(η), a

)dµ(η)

= τ

(2n4 i

∫Oe−

12

tr(ζ)B(η|ζ)e− tr(η)Bψ(η)dµ(η), a

)= τ

(2n4 i

∫OB(e−

12

tr(ζ)B(η|ζ)e− tr(η))ψ(η)dµ(η), a

)= 2

n4 i

∫O

(e−

12

tr(ζ)B(η|ζ)e− tr(η)τ(η − 1

2In + ζ, a)

− 2e−12


tr(ηζ))√


)ψ(η)dµ(η). (6.13)

On the other hand we get in the same way as in part A

dπC

(a4,−iL(a), a

) SBψ(ζ)

= 2n4

∫O

(iτ(ζ, 1

4a)B(η|ζ) +

i

2τ(a, ζ)B(η|ζ) − iτ(a ∗ ζ, η)

sinh(2√

tr(ηζ))√

tr(ηζ)− i

4tr(a)B(η|ζ)

+ iB(η|ζ)τ

(ζ − In

4+ η, a

)−i

sinh(2√

tr(ηζ))√


)e−

12


= 2n4 i

∫O

(e−

12

tr(ζ)B(η|ζ)e− tr(η)τ(ζ − 1

2In + η, a)

− 2e−12


tr(ηζ))√tr(ηζ)

τ(η ∗ ζ, a)

)ψ(η)dµ(η), (6.14)

where we used tr(a) = τ(a, In) and τ(a ∗ ζ, In) = τ(ζ, a). We see that (6.14) is equal to (6.13).Hence part C is proven.

6.4 Conclusion

Using the Jordan algebra Sym(n,R), we defined the structure algebra and the conformal alge-bra. We then constructed two representations of the conformal algebra: the Schrodinger modeland the Fock model. In the last chapter we proved that this two representations are actuallytwo different realizations of the same representation and that the isomorphism is given by theSegal-Bargmann transform. In this way we have reinterpreted the classical Segal-Bargmanntransform, which we studied in the first chapter, as an intertwining operator between theSchrodinger model and the Fock model.Of course, this thesis has not exhausted the subject at all. It is for example possible to integratethe representations we constructed to Lie group representations. One can then show that theseintegrated representations are unitary, irreducible and minimal. These representations willturn out to be realizations of the well-known Segal-Shale-Weil representation of the metaplecticgroup.As remarked in the preface, it is also possible to do the whole construction that we have donefor Sym(n,R) for an arbitrary simple Euclidean Jordan algebra. This gives us generalizationsof the Segal-Bargmann transform.

Appendix A

Nederlandstalige samenvatting

De klassieke Segal-Bargmanntransformatie is een integraaltransformatie van de Schrodinger-ruimte van kwadratisch integreerbare functies naar de Fockruimte van holomorfe functies.De Segal-Bargmanntransformatie, voor het eerst geıntroduceerd door Bargmann [1] in 1961,geeft een isomorfisme tussen twee equivalente formuleringen van de kwantummechanica. Inde doctoraatsscriptie van Mollers [6] en het artikel van Hilgert et al. [4] wordt de Segal-Bargmanntransformatie geherinterpreteerd als een intertwining operator tussen twee verschil-lende realisaties van de minimale representatie van een Lie-algebra. Deze Lie-algebra wordtgeconstrueerd door gebruik te maken van Jordan-algebra’s.

In [6] beschouwt de auteur simpele Euclidische Jordan-algebra’s en associeert hiermee bepaaldealgebraısche structuren zoals de conformaalgroep, de structuurgroep en de corresponderendeLie-algebra’s. De minimale baan van de actie van de structuurgroep op de Jordan-algebraspeelt ook een belangrijke rol. De ruimte van kwadratisch integreerbare functies gedefinieerdop deze minimale baan wordt namelijk gebruikt om een minimale representatie te construeren.Deze minimale representatie is een Liegroep representatie van (een eindige bedekking van) deconformaalgroep op deze ruimte. Dit geeft ons het Schrodingermodel.

In [4] construeren de auteurs een andere realisatie van deze minimale representatie. Ze makengebruik van de complexificatie van de minimale baan om een Fockruimte te creeren. Deze Fock-ruimte is dan de vectorruimte waarop de representatie van de conformaalgroep gedefinieerd is.Dit geeft ons het Fockmodel. Het is mogelijk om het Fockmodel met het Schrodingermodelte verstrengelen door gebruik te maken van een integraaltransformatie. Deze integraaltrans-formatie, die een veralgemening geeft van de Segal-Bargmanntransformatie, is een isomorfismetussen de afgeleide representaties van de conformaalalgebra.

Het doel van deze thesis is om meer inzicht te krijgen door de constructie volledig uit tewerken voor een concreet geval waarbij de Jordan-algebra gelijk is aan de reele symmetrischen × n-matrices Sym(n,R). In dat geval zal het voldoende blijken te zijn om enkel de Lie-algebra’s en de Lie-algebrarepresentaties te bestuderen en niet de overeenkomstige Liegroepenen Liegroeprepresentaties. Dit heeft als voordeel dat een elementaire kennis van analyse enalgebra voldoende is om deze thesis te kunnen volgen. Door gebruik te maken van een concreetvoorbeeld zullen voor ons geval ook veel bewijzen korter en eenvoudiger zijn dan voor hetalgemeen geval.

De inhoud van deze thesis is als volgt gestructureerd. In het eerste hoofdstuk introducerenwe de klassieke Segal-Bargmanntransformatie zoals ze ook bestudeerd werd in het artikel vanBargmann [1]. In het bijzonder zullen we aantonen dat de ladderoperatoren op de Schrodinger-ruimte omgezet worden in de operatoren z en ∂z op de Fockruimte. We zullen ook bewijzendat de Segal-Bargmanntransformatie een unitaire afbeelding is van de Schrodingerruimte naarde Fockruimte.

In het tweede hoofdstuk beginnen we met de studie van Jordan-algebra’s, waarbij we vooral

95

96 Appendix A. Nederlandstalige samenvatting

aandacht zullen schenken aan Sym(n,R). We zullen een inproduct definieren en een decom-positie van de ruimte door middel van idempotente elementen. In dit hoofdstuk zullen we ookde structuuralgebra definieren.Daarna gaan we verder met het bestuderen van Jordan-algebra’s. We bekijken de banen onderde actie van een bepaalde groep op Sym(n,R). Daarbij zijn we vooral geınteresseerd in deminimale baan en we zullen een isomorfisme vinden tussen kwadratisch integreerbare functiesgedefinieerd op deze minimale baan en kwadratisch integreerbare, even functies gedefinieerd opRn. Het hoofddoel van het derde hoofdstuk is het definieren van de Besseloperator en aantonendat deze operator tangentieel is aan de minimale baan.In het vierde hoofdstuk kunnen we dan het materiaal van de vorige hoofdstukken gebruiken omhet Schrodingermodel te construeren. We zullen eerst de conformaalalgebra geassocieerd aanSym(n,R) definieren. Dan geven we een Lie-algebrarepresentatie van de conformaalalgebraop de ruimte van gladde functies gedefinieerd op de minimale baan. Daarna tonen we aandat deze constructie ook aanleiding geeft tot een representatie op de ruimte van kwadratischeintegreerbare functies gedefinieerd op de minimale baan. De inhoud van hoofdstuk twee totvier is hoofdzakelijk gebaseerd op Mollers [6] en Faraut and Koranyi [3], waar het behandeldwordt in een meer algemene context.We beginnen het vijfde hoofdstuk met het bestuderen van de complexe symmetrische matricesSym(n,C) en de overeenkomstige complexe minimale baan. We gebruiken deze complexe mini-male baan om een Fockruimte te definieren en zullen aantonen dat deze Fockruimte isomorf ismet het even deel van de Fockruimte die we in het eerste hoofdstuk zagen. Daarna bekijken weeen Lie-algebra automorfisme van de complexe conformaalalgebra. We zullen dit combinerenmet een complexificatie van de representatie uit hoofdstuk vier om tot het Fockmodel te komen.In het laatste hoofdstuk bestuderen we een integraaltransformatie die gaat van de kwadratischintegreerbare functies gedefinieerd op de minimale baan naar de Fockruimte uit hoofdstukvijf. We zullen eerst het verband tussen deze transformatie en de klassieke Segal-Bargmann-transformatie geven. Daarna zullen we aantonen dat deze transformatie een isomorfisme geefttussen de representatie van het Schrodingermodel en de representatie van het Fockmodel. Dezelaatste twee hoofdstukken zijn gebaseerd op Hilgert et al. [4] waar de constructie, die wij hebbenuitgewerkt voor het geval Sym(n,R), voor een algemene simpele Euclidische Jordan-algebrawordt gedaan.

Bibliography

[1] V. Bargmann. On a Hilbert space of analytic functions and an associated integral transformpart I. Communications on pure and applied mathematics, 14:187–214, 1961.

[2] N. De Schepper and H. De Bie. Wiskundige analyse V. 2011. Cursus 3de bachelor Wiskunde.

[3] J. Faraut and A. Koranyi. Analysis on symmetric cones. Oxford University Press, 1994.

[4] J. Hilgert, T. Kobayashi, J. Mollers, and B. Ørsted. Fock model and Segal-Bargmann trans-form for minimal representations of Hermitian Lie groups. Journal of functional analysis,263:3492–3563, 2012.

[5] A. Knapp. Lie groups beyond an introduction, second edition. Birkhauser, 2002.

[6] J. Mollers. Minimal representations of conformal groups and generalized Laguerre functions.PhD thesis, Universitat Paderborn, 2010.

[7] G. Olafsson. The Segal-Bargmann transform on Euclidean space and generalizations. 2013.Preprint of book.

[8] H. Vernaeve. Wiskundige analyse II Deel b. 2010. Cursus 1ste bachelor Wiskunde.

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