weyl group representations and signatures of intertwining operators

Weyl group representations and

signatures of intertwining operators

Alessandra Pantano

A Dissertation

Presented to the Faculty

of Princeton University

in Candidacy for the Degree

of Doctor of Philosophy

Recommended for Acceptance

by the Department of

Mathematics

November 2004

c© Copyright by Alessandra Pantano, 2004.

All Rights Reserved

Abstract

We discuss the non-unitarity of a spherical principal series of a real split group by

means of Weyl group calculations.

For a good choice of the parameters, a spherical principal series admits a non-

degenerate invariant Hermitian form. To discuss unitarity, one needs to compute

the signature of such a form. Computations can be done separately on the isotypic

of each K-type and when the K-type is “petite”, they can be reduced to Weyl group

calculations. It follows that, to compute the signature on the isotypic of a petite

K-type µ, one only needs to understand the representation ψµ of the Weyl group on

the space of M -invariants in µ.

For type A, the set of Weyl group representations ψµ that arise from petite K-types

can be identified with the set of partitions of n in at most two parts. We give an

inductive algorithm to extend each Weyl group representation in this class to a pe-

tite K-type. This construction produces all the K-types used by Barbasch to detect

unitarity in SL(n, R).

Our inductive algorithm appears to be generalizable to other split groups with one

root-length. The final result will be a list of petite K-types on which the signature can

be computed with Weyl group calculations; hence, a non-unitarity test for a spherical

principal series for split groups of types D, E6, E7 and E8.

iii

Acknowledgements

After spending five years in graduate school, there are many people I would like to

take the opportunity to thank. I want to start with my advisor, David Vogan, for

his constant and invaluable support in every aspect of my academic career: from

identifying my thesis problem and helping me tackle and finally solve it, to providing

financial support during my last semester at MIT; but more than everything else, for

trusting me and constantly encouraging me: if I am so excited now about math, I

owe it to him.

My home institution, Princeton University, deserves a special acknowledgment for

making it possible for me to pursue my research at MIT while still providing financial

support. A special thanks goes to Jill LeClair, for helping with every bureaucratic

aspect: without her, I would still be struggling to fill out the absentia forms.

If I enjoyed my stay at Princeton, it is thanks to my precious friends; in particular

Jose Luis Rodrigo and Liz Hough, Maria del Mar Gonzales, Carolina Araujo and Leo

Carvalho, Antonio Rojas-Leon, Yumna Masarwa and Eduardo Duenez. Almost all of

us started grad school together, and we have shared so much in these years. A special

thanks to those of you that helped in that crazy night before my general exams.

Leaving my family behind and moving to the United States was a tough choice for

me. If I managed to feel at home in this country, it is thanks to Julie and Charlie

Fefferman, who so tenderly “adopted” me. I will never forget you.

Finally, at the origin of everything is my family. I want to thank my Dad, my

wonderful brother and sister, and my Mum, who was my first mathematical muse:

without her, I would have never discovered the beauty of mathematics. The fact that

she is not here with me in this moment veils this nice moment with sadness. I so

much wish I could thank you.

This “American adventure” would not have been the same without Lorenzo. We

shared every moment of our life in the past three years. Thanks for being with me.

iv

To my Mom

v

Contents

Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

1 Introduction 1

2 Irreducible Admissible Representations 10

2.1 Principal Series Representations . . . . . . . . . . . . . . . . . . . . . 10

2.2 The Standard Intertwining Operator . . . . . . . . . . . . . . . . . . 14

2.3 A Langlands Quotient . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.4 Generalization to non-minimal parabolic . . . . . . . . . . . . . . . . 18

2.5 Langlands Classification of Irreducible Admissible Representations . . 22

3 Unitary Irreducible Representations 23

3.1 The formal symmetry condition . . . . . . . . . . . . . . . . . . . . . 24

3.2 The operator B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3 Unitary Langlands Quotients . . . . . . . . . . . . . . . . . . . . . . 28

4 The example of SL(2, R) 30

4.1 The data for SL(2, R) . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.2 Principal Series induced from the Minimal Parabolic . . . . . . . . . 32

4.2.1 The Principal Series Representation πP (δ ⊗ λ) . . . . . . . . . 32

4.2.2 The restriction of πP (δ ⊗ λ) to K . . . . . . . . . . . . . . . . 34

vi

4.3 Study of the irreducibility of πP (δ ⊗ λ) . . . . . . . . . . . . . . . . . 35

4.3.1 Irreducibility of πP (δ+ ⊗ λ) . . . . . . . . . . . . . . . . . . . 38

4.3.2 Irreducibility of πP (δ+ ⊗ λ) . . . . . . . . . . . . . . . . . . . 39

4.4 Study of the Unitarity of πP (δ ⊗ ν) . . . . . . . . . . . . . . . . . . . 40

4.4.1 The Spherical Case . . . . . . . . . . . . . . . . . . . . . . . . 42

4.4.2 The Non-Spherical Case . . . . . . . . . . . . . . . . . . . . . 47

5 Invariant Hermitian Forms 51

5.1 The finite-dimensional case . . . . . . . . . . . . . . . . . . . . . . . . 51

5.1.1 Signature of a Hermitian operator on a f.d. vector space . . . 51

5.1.2 Signature of a Hermitian form on a f.d. vector space . . . . . 52

5.2 Signature of an invariant Hermitian form on an admissible (gC0 , K)

module w.r.t. a K-type . . . . . . . . . . . . . . . . . . . . . . . . . . 54

6 The Split Case 63

6.1 Spherical Unitary Dual for Real Split Semisimple Lie Groups . . . . . 64

6.1.1 The rank-one subgroup Gβ attached to a (simple) root . . . . 68

6.1.2 The operator Rµ(ω0, ν) . . . . . . . . . . . . . . . . . . . . . . 69

6.1.3 The operator Rµ(sβ, γ) . . . . . . . . . . . . . . . . . . . . . . 73

6.1.4 Final considerations . . . . . . . . . . . . . . . . . . . . . . . 82

6.2 Spherical representations and the Weyl group . . . . . . . . . . . . . 84

7 The example of SL(3, R) 90

7.1 The data for SL(3, R) . . . . . . . . . . . . . . . . . . . . . . . . . . 90

7.2 The irreducible representations of SO(3) . . . . . . . . . . . . . . . . 92

7.2.1 A closer look at HN . . . . . . . . . . . . . . . . . . . . . . . 92

7.2.2 Spherical K-types . . . . . . . . . . . . . . . . . . . . . . . . . 94

7.2.3 Petite K-types . . . . . . . . . . . . . . . . . . . . . . . . . . 98

7.3 The representations of the Weyl group . . . . . . . . . . . . . . . . . 101

vii

7.3.1 ψH2 : the representation of S3 on (H2)M . . . . . . . . . . . . . 103

7.4 Signature on the K-type H2 . . . . . . . . . . . . . . . . . . . . . . . 105

7.4.1 The “analytic” operator RH2(ω0, ν) . . . . . . . . . . . . . . 105

7.4.2 The “algebraic” operator ψH2(A(ν)) . . . . . . . . . . . . . . . 112

7.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

8 The petite spherical representations of SO(2n, R) 117

8.1 The petite representations of SO(2n, R) . . . . . . . . . . . . . . . . 118

8.1.1 The restricted root system of sl(2n,R) . . . . . . . . . . . . . 120

8.1.2 The SO(2) attached to a restricted root . . . . . . . . . . . . 121

8.1.3 Petite representations . . . . . . . . . . . . . . . . . . . . . . . 123

8.2 The petite spherical representations of SO(2n, R) . . . . . . . . . . . 127

8.3 The irreducible representations of the symmetric group . . . . . . . . 144

8.4 The “petite” Weyl group representations . . . . . . . . . . . . . . . . 152

8.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

9 Constructing petite K-types 156

9.1 Assumptions on ρ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

9.2 Sketch of the construction . . . . . . . . . . . . . . . . . . . . . . . . 158

9.3 Step 1: the representation Fρ′ of M ′ . . . . . . . . . . . . . . . . . . . 161

9.4 Step 2: A quotient of Fρ′ . . . . . . . . . . . . . . . . . . . . . . . . . 167

9.4.1 The restriction to M ′ ∩SO(3) of an arbitrary representation of

M ′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

9.4.2 The restriction of Fρ′ to M ′α, β, γ . . . . . . . . . . . . . . . . . 172

9.4.3 Eliminating the sign from Fρ′ . . . . . . . . . . . . . . . . . . 177

9.5 Step 3: The action of Lie(K) on Fρ′ . . . . . . . . . . . . . . . . . . . 181

9.5.1 The (?)-relations are preserved . . . . . . . . . . . . . . . . . 189

9.5.2 Checking bracket relations-PART 1 . . . . . . . . . . . . . . . 191

viii

9.5.3 Checking bracket relations-PART 2 . . . . . . . . . . . . . . . 201

9.5.4 Checking the eigenvalue condition . . . . . . . . . . . . . . . . 204

9.5.5 Lifting ρ′ to K . . . . . . . . . . . . . . . . . . . . . . . . . . 210

9.5.6 It is an extension of ρ . . . . . . . . . . . . . . . . . . . . . . . 211

9.6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

10 Further results and examples 215

10.1 The highest weight decomposition of Fρ′ . . . . . . . . . . . . . . . . 216

10.2 The construction of ρ′: some examples . . . . . . . . . . . . . . . . . 231

10.2.1 Construction for the trivial representation of S3 . . . . . . . . 231

10.2.2 Construction for the standard representation of S3 . . . . . . . 231


10.2.4 Construction for the representation S2,2 of S4 . . . . . . . . . 241


10.2.6 Construction for the representation S(3,3) of S6 . . . . . . . . . 253

10.2.7 Construction for the representation S(3,2) of S5 . . . . . . . . . 266

10.3 Extension of non-spherical representations . . . . . . . . . . . . . . . 266

10.3.1 Construction for ν2 . . . . . . . . . . . . . . . . . . . . . . . . 267

10.3.2 Construction for ν1 . . . . . . . . . . . . . . . . . . . . . . . . 270

11 A generalization to the non-spherical case 272

11.1 An inductive argument for the construction . . . . . . . . . . . . . . 276

11.2 Suggestions for further generalizations . . . . . . . . . . . . . . . . . 279

ix

Chapter 1

Introduction

One of the most interesting problems in representation theory is the study of the

unitary dual of a real reductive Lie group G. It is a theorem by Harish-Chandra that

every irreducible unitary representation of G is admissible, and that two irreducible

unitary representations are unitarily equivalent if and only if they are infinitesimally

equivalent. Therefore, the problem of classifying the unitary dual is equivalent to:

1. describing the set of irreducible admissible representations of G, up to infinites-

imal equivalence;

2. understanding which irreducible admissible representations of G admit an in-

variant Hermitian form;

3. deciding whether the invariant Hermitian form on an admissible irreducible

representation of G is definite.

The first problem was solved in the early 1970s by Langlands, who showed that every

irreducible admissible representation of G is infinitesimally equivalent to a Langlands

quotient JP (δ ⊗ ν).

Let P = MAN be a parabolic subgroup of G, let δ be an irreducible tempered unitary

representation of M and let ν be a complex linear functional of A, with real part in

1

the open dominant Weyl chamber corresponding to N . Let IP (δ⊗ν) be the principal

series obtained by inducing the representation δ ⊗ ν of P to G. The assumptions on

the parameters δ and ν guarantee that IP (δ ⊗ ν) has a unique irreducible quotient,

JP (δ ⊗ ν), which we call a Langlands quotient with data (P, δ, ν).

We can describe JP (δ ⊗ ν) as the quotient of IP (δ ⊗ ν) modulo the kernel of the

intertwining operator

A(P : P : δ : ν) : IP (δ ⊗ ν) → IP (δ ⊗ ν), F 7→∫

Θ(N)

F (xn) dn.

Please, refer to chapter 2 for more details.

A few years later, Knapp and Zuckerman understood that a Langlands quotient

JP (δ ⊗ ν) is Hermitian if and only of there exists an element ω in K satisfying the

formal symmetry condition

ω · δ ' δ, ωPω−1 = P , ω · ν = −ν. (1.1)

The existence of a non-composition on the rangedegenerate invariant Hermitian form

on JP (δ⊗ν) is equivalent to the existence of a Hermitian (g, K)-map that intertwines

JP (δ⊗ν) and its Hermitian dual JP (δ⊗ (−ν)). The symmetry condition (1.1) follows

from the fact that two Langlands quotients JP (δ ⊗ ν) and JP ′(δ′ ⊗ ν ′) can only be

equivalent if there exists an element ω of K such that ωPω−1 = P ′, ω · δ = δ′ and

ω · ν = ν ′.

For each Hermitian Langlands quotient, Knapp and Zuckerman constructed a

Hermitian intertwining operator B on the principal series IP (δ⊗ ν) whose kernel co-

incides with the kernel of the operator A(P : P : δ : ν). Such an operator induces a

non-degenerate Hermitian form on the Langlands quotient JP (δ⊗ν) = IP (δ⊗ν)

Ker(A(P :P :δ:ν)).

Please, refer to chapter 3 for details.

2

We are left with the problem of deciding whether a Hermitian Langlands quotient

JP (δ ⊗ ν) is unitary. This is equivalent to understanding whether the Hermitian in-

tertwining operator B (on IP (δ ⊗ ν)) is semi-definite.

Because the representation is admissible, the signature of B can be computed sep-

arately on the isotypic component of each K-type appearing in the principal series,

which is a finite-dimensional vector space. Of course there are infinitely many K-

types, so the problem remains complicated.

Please, refer to chapter 5 for a short description of how to study the signature of an

Hermitian intertwining operator defined on an admissible representation.

For each K-type (µ, Eµ) appearing in the principal series, we obtain an intertwin-

ing operator

Iµ(ω, ν) : HomK(Eµ, IP (δ ⊗ ν)) → HomK(Eµ, IP (δ ⊗ ω · ν))

(induced from B by composition on the range) whose signature equals the signature

of B w.r.t. the K-type µ. Using Frobenius reciprocity, we can interpret Iµ(ω, ν) as

an endomorphism of HomM(ResKM Eµ, V δ).

Problem Compute the signature of Iµ(ω, ν), for each K-type (µ, Eµ) appearing

in the principal series IP (δ ⊗ ν).

The example of SL(2, R) (chapter 4) shows that it is fairly easy to solve this problem

when G is a split group of real rank one.

For G = SL(2, R), we let P = MAN be the minimal parabolic consisting of upper

triangular matrices. We assume δ = δ+ to be the trivial representation of M and ν to

3

be a real dominant character of A satisfying the symmetry condition ω · ν = −ν. We

identify the irreducible representations of K = SO(2) with integers, and the charac-

ters of A with complex numbers. In particular, since the long Weyl group element ω

is the negative of the identity, we identify ν with a real positive number number.

For all λ > 0, we construct the spherical principal series IP (δ+ ⊗ λ), and we observe

that IP (δ+ ⊗ λ) contains every even K-type ξ2m with multiplicity one, and no odd

K-type at all.

For all integers m, the operator B acts on the the isotypic of ξ2m by a scalar (be-

cause this isotypic is one-dimensional), and we denote this scalar by c2m. A long but

classical computation shows that

c2m = c−2m =(1− λ)(3− λ) · · · (2m− 1− λ)

(1 + λ)(3 + λ) · · · (2m− 1 + λ)∀ m > 0.

By definition, B is positive semi-definite if and only if c2m > 0, for all integers m, and

this happens if and only of 0 < λ 6 1. We notice that this condition is equivalent to

the one we obtain by requiring the signature to be positive only on the K-types ξ0,

ξ+2 and ξ−2 (which are petite).

When the group G is split1, and P is a minimal parabolic we can always reduce the

computations for the construction of the operator Iµ(ω, ν) to rank-one calculations.

For simplicity, we assume that δ is the trivial representation of M and that ν is a

real character of A. The element ω that appears in the symmetry condition 1.1 is a

Weyl group element, hence it admits a minimal decomposition as a product of simple

reflections. We decompose Iµ(ω, ν) accordingly, and notice that the intertwining op-

erator corresponding to a simple root β can be interpreted an operator for a principal

series of a rank-one subgroup L(β) (which is essentially a product of M and the SL(2)

1Write g0 = k0 ⊕ p0 for the Cartan decomposition of Lie(G), and choose a maximal abeliansubspace a0 of p0. The group G is called “split” if the centralizer of a0 inside k0 is zero. As aconsequence, the group M is finite (and abelian).

4

attached to the root β), and can therefore be computed using the results already

known for SL(2, R). Here are more details.

If ω = sαr · · · sα2sα1 is a minimal decomposition of ω0 as a product of simple reflec-

tions, we can write:

Iµ(ω, ν) = Iµ(sαrsαr−1 · · · sα2sα1 , ν) =

= Iµ(sαr , (sαr−1 · · · sα2sα1) · ν) · · · Iµ(sα2 , sα1 · ν)Iµ(sα1 , ν). (1.2)

Each factor of this decomposition is an operator of the form Iµ(sβ, γ), with sβ a

simple reflection and γ a real character of A satisfying 〈γ, β〉 > 0.

For β and γ as above, and for each integer l, we let ϕl be the isotypic component of the

character ξl of K(β) ' SO(2) inside µ, so that Eµ =⊕

l∈Z ϕl is the decomposition

of µ in K(β)-types, and

HomM(ResKM Eµ, V δ) =

⊕

m∈NHomM(ResMK(β)

M (ϕ2m + ϕ−2m), C)

︸︷︷︸Um

is a decomposition of HomM(ResKM Eµ, V δ) in MK(β)-stable subspaces. The operator

Iµ(sβ, γ) : HomM(ResKM Eµ, V δ) → HomM(ResK

M Eµ, V δ)

preserves the decomposition of HomM(ResKM Eµ, V δ) in MK(β)-stable subspaces. Pre-

cisely, for all m > 0, Iµ(sβ, γ) acts on Um as scalar multiplication by

c2m,γ =(1− 〈γ, β〉)(3− 〈γ, β〉) · · · (2m− 1− 〈γ, β〉)(1 + 〈γ, β〉)(3 + 〈γ, β〉) · · · (2m− 1 + 〈γ, β〉) . (1.3)

Please, refer to chapter 6 for details.

5

We obtain a decomposition of Iµ(ω, ν) as a product of operators corresponding

to simple reflections, and for each of these operators an explicit formula exists. The

difficulty is that this formula depends on the decomposition of Eµ in irreducible K(β)-

types (with K(β) the SO(2) attached to β) and this decomposition changes when

β varies. It is very hard to keep track of these different decompositions when you

multiply the various rank-one operators to obtain Iµ(ω, ν). Therefore, we look for an

alternative approach to construct the operator Iµ(ω, ν).

Strategy For some K-types µ, construct the operator Iµ(ω, ν) by means of Weyl

group calculations.

Definition 1. Let G be a real reductive group, and let K be a maximal compact

subgroup of G. A K-type (µ, Eµ) is called petite if the SO(2) subgroup attached to

every restricted root (of G) only acts with characters 0, ±1, ±2.

When µ is a petite K-type, the operator Iµ(sβ, γ) only depends on the represen-

tation ψµ of the Weyl group on the space of M -fixed vectors in Eµ.

We can construct Iµ(ω, ν) in a purely algebraic fashion, as follows:

1. We factorize Iµ(ω, ν) as a product of rank-one operators (as in 1.2)

2. For each simple root β, and for each real character γ of A satisfying 〈γ, β〉 > 0,

we write:2

Iµ(sβ, γ) =

+1 on the (+1)-eigenspace of ψµ(sβ)

1−〈γ, β〉1+〈γ, β〉 on the (−1)-eigenspace of ψµ(sβ).

2Recall that Iµ(sβ , γ) is defined on the space of M -fixed vectors in Eµ. Because µ is petite

HomM (ResKM Eµ, V δ) = HomM (ResMK(β)

M ϕ0, C)︸︷︷︸U0

⊕HomM (ResMK(β)

M (ϕ2 ⊕ ϕ−2), C)︸︷︷︸U2

.

We can identify U0 with the (+1)-eigenspace of ψµ(sβ), and U2 with the (−1)-eigenspace of ψµ(sβ).It follows from 1.3, that Iµ(sβ , γ) acts by c0,γ = 1 on U0, and by c2m,γ = 1−〈γ, β〉

1+〈γ, β〉 on U2.

6

3. We multiply the various rank-one operators Iµ(sβ, γ) to obtain Iµ(ω, ν).

Please, refer to chapter 6 for details, and to chapter 7 for an application of these

considerations to SL(3, R).

Final Results: When G is a real split group, and X(ν) is a unitary spherical

Langlands quotient, the Hermitian operator Iµ(ω, ν) must be semi-definte for all K-

types µ. When µ is petite, we can compute Iµ(ω, ν) by Weyl group calculations, so

it should not be too hard to understand its signature.

The result is a non-unitarity test for a spherical principal series of G, in the sense

that this test can be used to show that some representations are not unitary: if the

“algebraic” operator Iµ(ω, ν) fails to be semi-definite for some petite K-type µ, then

we can conclude that X(ν) is not unitary.

It is quite an amazing fact that this test also detects unitarity when the group G

is a classical group. This beautiful result is due to Barbasch and can can be stated

as follows:

Theorem. If G is a classical (real or p-adic) split group, then the spherical Lang-

lands quotient X(ν) is unitary if and only if the invariant Hermitian form is positive

semi-definite on the petite K-types.

Inspired by this result, we decide to explore the relation between petite K-types

and Weyl group representations. The main result is an inductive algorithm that

extends a certain class of representations of the Weyl group to petite K-types. We

have formulated this result for the group SL(n), but it appears to be generalizable to

other split groups whose root-system admits only one root-length (this includes the

non-classical groups E6, E7, E8).

7

In chapter 8 we give an explicit description of the petite spherical K-types of

SL(2n), and we compute the corresponding representations of the Weyl group on the

space of M -fixed vectors. The following table summarizes the results of the chapter

petite K-type Weyl group repr. dim.

0 = 0ψ1 + · · ·+ 0ψn S(2m) = trivial 1

2ψ1 + · · ·+ 2ψk S(2n−k,k)(2nk

)− (2n

k−1

)

0 < k < n

2ψ1 + · · ·+ 2ψn−1 ± 2ψn S(n,n)(2nn

)− (2n

n−1

)

With standard notations, we have identified the Weyl group of SO(2n) with the

symmetric group in 2n letters, and the irreducible representations of S2n with the

partitions of 2n. We notice that the Weyl group representations ρ arising from the

petite spherical representations are exactly the Specht modules corresponding to the

partitions in at most two parts. This condition is equivalent to requiring that ρ do

not contain any copy of the sign representation of S3.

In chapter 9 we run this process backwards, and for each Weyl group representation

ρ not containing the sign of S3, we construct a spherical petite K-type µρ such that

ρ is a submodule of the representation of the Weyl group on the space of M -fixed

vectors of µρ. This construction produces all the petite spherical K-types.

In chapter 10 we determine the highest weight decomposition of µρ and provide a

number of examples.

In chapter 11 we generalize this construction to non-spherical representations of M ′,

8

obtaining an inductive algorithm to construct petite K-types.

For type A, our algorithm provides a list of petite K-types on which the inter-

twining operator for a spherical principal series can be tested by means of Weyl group

computations; in other words, a list of K-types on which we should be able to explic-

itly calculate the signature.

Generalizing the algorithm, we will obtain a similar output for other groups with

one root-length. The final result will be a non-unitarity test for a spherical principal

series for split groups of type A, D, E6, E7 and E8.

9

Chapter 2

Classifications of Irreducible

Admissible Representations

2.1 Principal Series Representations

In this section we introduce the notion of a Principal Series Representation induced

from a minimal parabolic subgroup1, and we list its main properties. A detailed

description of the material can be found in Knapp[21].

Let P = MAN be a minimal parabolic subgroup of G. Fix an irreducible repre-

sentation (δ, Vδ) of M and a character ν of A. We consider δ ⊗ ν as a representation

of P , with N acting trivially.

The Principal Series Representation

πP (δ ⊗ ν) = IndGP (δ ⊗ ν)

is obtained by inducing δ ⊗ ν from P to G. We give two different realizations of this

1There is a more general notion of Principal Series Representation, in which the parabolic sub-group is not required to be minimal. We will discuss it later.

10

representation:

Induced Picture. The representation space is the completion of the vector

space

F : G → Vδ continuous | F (gman) = e−(ν+ρ) log(a)δ(m)−1F (g),

∀man ∈ P, ∀ g ∈ G

with respect to the L2 norm

‖f‖2 =∫

K‖f(k)‖2

Vδdk,

where dk is the normalized Haar measure on K. It can be identified with the Hilbert

space

HPδ⊗ν = F : G → Vg : F |K ∈ L2(K, V δ) and F (gman) = e−(ν+ρ) log(a)δ(m)−1F (g),

∀m ∈ M, a ∈ A, n ∈ N, g ∈ G.

This space is invariant by left translation in G, so we can define a representation of

G on HPδ⊗ν by

[πP (δ ⊗ ν)(g) · F ](x) = F (g−1x) ∀ x, g ∈ G.

We call πP (δ⊗ ν) the principal series representation with parameters P, δ and ν.

We also denote it by IndGP (δ ⊗ ν), to emphasize the fact that it is induced from the

representation δ ⊗ ν of P .

Notice that the restriction of πP (δ ⊗ ν) to the maximal compact subgroup K is

independent of the character ν, and it is given by:

ResGK(IndG

P (δ ⊗ ν)) = IndKM δ.

11

Compact Picture. The representation space is obtained from HPδ⊗ν by re-

striction to K:

HPδ = f ∈ L2(K, Vδ) | f(km) = δ(m)−1f(k), ∀m ∈ M, ∀ k ∈ K.

The action of G on Hδ is defined by

[πP (δ ⊗ ν)(g) · f ](k) = e−(ν+ρ)H(g−1k)f(κ(g−1k)) ∀ x ∈ G, ∀ k ∈ K,

where x = κ(x)eH(x)n(x) denotes the Iwasawa decomposition of an element x of

G = KAN .

These two pictures of the principal series representation are isomorphic. Indeed,

restriction to K defines a Hilbert space isomorphism from HPδ⊗ν onto Hδ, which

intertwines the action of G.

The inverse mapping associates to an element f : K → Vδ of Hδ the unique ex-

tension F : G → Vδ endowed with the desired transformation property under right

multiplication by elements of P :

F (g) = F (kan) = e−(ν+ρ) log(a)f(k)

(the unicity of the extension follows from the unicity of the Iwasawa decomposition

of G = KAN).

We will continuously switch between the induced picture and the compact one.

The compact picture has the advantage that the representation space is independent

of ν, but the action of G is much simpler in the induced picture.

Now that the definition of a principal series representation is understood, we

describe its main properties. For clarity, we subdivide the results in three categories:

Admissibility Any principal series representation πP (δ ⊗ ν) is admissible.

Indeed, by Frobenius reciprocity, the multiplicity of a K-type µ in

12

ResGK(IndG

P (δ ⊗ ν)) = IndKM δ

equals the multiplicity of δ in ResKM(µ), which is finite.

We denote by X(δ ⊗ ν) the corresponding Harish-Chandra module.

It is a theorem by Harish-Chandra that every irreducible admissible represen-

tation of G is infinitesimally equivalent to a composition factor of a principal

series representation.

This is the famous “Harish-Chandra subquotient theorem”. It states that an

irreducible representation of G is admissible if and only if the corresponding

Harish-Chandra module is equivalent to an irreducible quotient M1/M2, where

M1 k M2 are two submodules of a principal series induced from a minimal

parabolic.

Equivalence Two principal series representations πP (δ ⊗ ν) and πP ′(δ′ ⊗ ν ′) have

equivalent composition series if and only if there exists a Weyl group element

ω such that

(δ′, ν ′) = (ω · δ, ω · ν).

We make the Weyl group act on equivalence classes of irreducible representations

of M and A as follows:

(ω · δ)(m) = δ(x−1mx) ∀m ∈ M ,

with x a representative for ω in the normalizer of a0 in K, and

(ω · ν)(eH) = ν(eAd(ω−1)H) ∀H ∈ a0.

13

Unitarity If ν is imaginary, the principal series representation πP (δ ⊗ ν) is unitary.

Notice this condition on ν makes δ⊗ν unitary, for each irreducible representation

δ of M . Indeed, when P = MAN is minimal parabolic, M is included in K

and therefore is compact.

2.2 The Standard Intertwining Operator

In this section we discuss the existence of an intertwining operator between two prin-

cipal series representations πMAN(δ ⊗ ν) and πMAN(δ ⊗ ν), that are obtained by

inducing the same representation δ ⊗ ν of MA from two different minimal parabolic

subgroups P = MAN and P = MAN . Results are stated without a proof. A good

reference is Knapp[21].

Let P = MAN be a minimal parabolic subgroup of G, with

Lie(N) =⊕

α∈∆+ g0α.

The choice of a new positive system ∆+ ⊂ ∆(g0, a0) determines a new minimal

parabolic subgroup P = MAN , with

Lie(N) =⊕

α∈∆+ g0α.

Fix an irreducible representation (δ, Vδ) of M and a character ν of A. The rep-

resentation δ ⊗ ν of MA can be regarded both as a representation of P and as a

representation of P (with N and N acting trivially).

Induction to G gives rise to two different principal series representations πMAN(δ⊗ν) and πMAN(δ ⊗ ν), defined on Hilbert spaces HP

δ⊗ν and HPδ⊗ν .

Consider the operator

A(P : P : δ : ν) : HPδ⊗ν −→ HP

δ⊗ν

14

defined formally2 by

[A(P : P : δ : ν)F ](x) =∫

N∩NF (xn) dn,

where N = Θ(N) denotes the analytic subgroup of G with Lie algebra

Lie(N) = ϑ(Lie(N)) =⊕

α∈∆− g0α =⊕

α∈∆+ g0−α.

It is not too hard to check that

• for each F in HPδ⊗ν , the function

A(P : P : δ : ν)F : G → Vδ

has the proper behavior under right multiplication by elements of P = MAN :

A(P : P : δ : ν)F (xm) = δ(m)−1(A(P : P : δ : ν)F (x)) ∀m ∈ M

A(P : P : δ : ν)F (xa) = e−(ν+ρ)A(P : P : δ : ν)F (x) ∀ a ∈ A

A(P : P : δ : ν)F (xn) = A(P : P : δ : ν)F (x) ∀ n ∈ N

(here ρ denotes the semi-sum of the roots in ∆+, counting multiplicity);

• for each F in HPδ⊗ν and each g in G,

A(P : P : δ : ν)[πMAN(δ ⊗ ν)(g) · F ] = πMAN(δ ⊗ ν)(g) · [A(P : P : δ : ν)F ].

So for each choice of the parameters δ and ν, A(P : P : δ : ν) is a formal inter-

twining operator from πMAN(δ ⊗ ν) to πMAN(δ ⊗ ν).

It is an actual intertwining operator only if the integral

A(P : P : δ : ν)F (x) =∫

N∩NF (xn) dn

2At this stage, we only consider A(P : P : δ : ν) as a formal operator; the attribute “formal”means that we do not worry about convergence.

15

converges, for each F in HPδ⊗ν .

Unfortunately, convergence happens only for certain values of ν.

If P = MAN and P = MAN are minimal parabolic, a sufficient condition for the

convergence is that 〈Re(ν), β〉 > 0, for each β which is positive for N and negative

for N :

Theorem 1. Let P = MAN and P = MAN be minimal parabolic subgroups, and

suppose that ν satisfies the condition

〈Re(ν), β〉 > 0 ∀ β ∈ ∆+ ∩ ∆−.

Then, the integral defining A(P : P : δ : ν)F (x) is convergent for every continuous

function F in HPδ⊗ν and for all x in G.

In this case, we refer to A(P : P : δ : ν) as the standard intertwining operator

from πMAN(δ ⊗ ν) to πMAN(δ ⊗ ν).

We also mention a result about the factorization of standard intertwining opera-

tors:

Theorem 2. Let P = MAN , P ′ = MAN ′ and P ′′ = MAN ′′ be minimal parabolic

subgroups of G. Suppose that

Lie(N ′′) ∩ Lie(N) j Lie(N ′) ∩ Lie(N)

and that 〈Re(ν), β〉 > 0, for each β which is positive for N and negative for N ′′.

Then

A(P ′′ : P : δ : ν) = A(P ′′ : P ′ : δ : ν) A(P ′ : P : δ : ν).

16

2.3 A Langlands Quotient

We use the standard intertwining operator A(P : P : δ : ν), with P equals the mini-

mal parabolic opposite to P , to construct an irreducible representation of G. Such

representation is called a Langlands quotient, because it was obtained by Langlands

as a quotient of the principal series representation πP (δ ⊗ ν) module the kernel of

A(P : P : δ : ν). The main reference for this section is [9].

With the same notations used the previous section, we consider the case in which

P is the minimal parabolic opposite to P :

P = Θ(P ) = MAN.

Since every root positive for N is automatically negative for N , the convergence

condition for A(P : P : δ : ν) is

〈Re(ν), β〉 > 0 ∀ β ∈ ∆+.

As a result, we obtain the following theorem:

Theorem 3. Let P = MAN be a minimal parabolic subgroup.

For each irreducible representation δ of M , and for those characters ν of A such that

Re(ν) is strictly dominant, there exists an intertwining operator

A(P : P : δ : ν)

from πP (δ ⊗ ν) to πP (δ ⊗ ν), with P the opposite parabolic.

The intertwining operator A(P : P : δ : ν) was the object of many studies in the

’70s and ’80’s, because of its intimate connection with the irreducibility of the prin-

cipal series representation πP (δ ⊗ ν).

The following results are mainly due to Langlands (and Milicic), and can be found

in [9]:

17

• The operator A(P : P : δ : ν) is not identically zero.

• If F ∈ HPδ⊗ν but F 6∈ Kernel(A(P : P : δ : ν)), then F is a cyclic vector for

πP (δ ⊗ ν).

• The image of the operator A(P : P : δ : ν) is an irreducible submodule of πP (δ⊗ν).

• πP (δ ⊗ ν) has a unique non-zero irreducible quotient:

JP (δ ⊗ ν) =HP

δ⊗ν

Kernel(A(P : P : δ : ν)),

which is isomorphic to A(P : P : δ : ν)(HPδ⊗ν).

The irreducible admissible representation JP (δ⊗ν) is called a Langlands Quotient

with data (P, δ, ν).

If we allow the parabolic subgroup P to be non-minimal, we basically obtain all

the irreducible admissible representations of G in the form of Langlands quotients.

The next step is therefore a generalization of the principal series to the case “P

non-minimal parabolic”.

2.4 Generalization to non-minimal parabolic

Under some additional conditions on δ, all the constructions carried out in the pre-

vious sections (principal series representation, standard intertwining operator, Lang-

lands quotient) can be generalized to the case in which the parabolic subgroup P is

not a minimal parabolic. A good reference is again Knapp[21].

Let P 1 = M1A1N1 be a parabolic subgroup of G, containing a minimal parabolic

P = MAN (so A1 j A, N1 j N and M1 k M).

18

Let ∆1 = ∆(g0, a10) be the set of roots:

∆1 = β ∈ (a10)′: (g0)

1β 66= ∅,

where we have denoted by (g0)1β the root space corresponding to β:

(g0)1β = X ∈ g0 : [H, X] = β(H)X, ∀H ∈ a1

0.

The roots in ∆1 = ∆(g0, a10) are basically 3 obtained from ∆ = ∆(g0, a0) by restriction

to a10. When P 1 is not a minimal parabolic, they do not necessarily form a root system.

We introduce a notion of positivity in ∆1 by calling a root β positive if the cor-

responding root space (g0)1β is included in n0. We denote by ρ1 the semi-sum (with

multiplicities) of the positive roots in ∆1.

The construction of the principal series mimics the one done in the case of a

minimal parabolic:

• We fix a character ν1 of a10 and a unitary4 representation δ1 of M1, and we

regard ν1 ⊗ δ1 as a representation of P 1 = M1A1N1 (with N1 acting trivially);

• We describe the induced picture of the principal series representation πP 1(δ1⊗ν1) as the action of G by left translation on the Hilbert space:

HP 1

δ1⊗ν1 = F : G → V δ1: F |K∈ L2(K, V δ1

) and

F (xman) = e−(ν1+ρ1) log(a)δ1(m)−1F (x), ∀m ∈ M1 ∩K, ∀ an ∈ A1N1,∀x ∈ G.

• In the compact picture of πP 1(δ1 ⊗ ν1), the Hilbert space is simply

HP 1

δ1|M1∩K= f ∈ L2(K, V δ1

) : f(xm) = δ1(m)−1f(x), ∀m ∈ M1 ∩K3Just make sure to disregard the roots of ∆ that are identically zero on a1

0.4In the minimal parabolic case, we did not require δ be unitary. Indeed, this was automatically

true, given the compactness of M .

19

and we define the action of G by:

[πP 1(δ1 ⊗ ν1)(g) · f ](k) = e−(ν1+ρ1)H′(g−1k)δ1(µ′(g−1k))−1f(κ′(g−1k))

(where κ′(x)ν ′(x)eH′(x)n′(x) denotes the decomposition of an element x of G

under G = KM1A1N1).

Restriction to K is an isomorphism from the induced picture to the compact

one. The inverse mapping sends f : K → V δ1to the unique extension F : G → V δ1

satisfying the desired transformation properties under right multiplication by elements

of P 1:

F (g) = F (κ′(g)ν ′(g)eH′(g)n′(g)) = e−(ν1+ρ1)H′(g)δ1(µ′(g))−1f(κ′(g)).

Just like in the minimal parabolic case, the principal series representation πP 1(δ1⊗ν1) is admissible for any value of the parameters δ1 (unitary) and ν1, and it is unitary

if ν1 is purely imaginary.

Encouraged by these strong similarities, we introduce the formal intertwining

operator

A(P 1 : P 1 : δ1 : ν1) : HP 1

δ1⊗ν1 −→ HP 1

δ1⊗ν1 (2.1)

defined by:

A(P 1 : P 1 : δ1 : ν1)F (x) =∫

N1 F (xn) dn.

For convergence (on K-finite vectors) we need Re(ν1) to be in the open positive

Weyl chamber relative to N1, and also an additional condition: we need δ1 to be

tempered.

20

Theorem 4. Let P 1 = M1A1N1 be a (not necessarily minimal) parabolic subgroup

of G. Let δ1 be an irreducible tempered unitary representation of M1, and let ν1 be

a characters of A1 such that Re(ν1) is in the open positive Weyl chamber determined

by N1. Define the parameters (P 2, δ2, ν2) is a similar fashion. Then

1. there exists an intertwining operator

A(P 1 : P 1 : δ1 : ν1)

from πP 1(δ1 ⊗ ν1) to πP 1(δ1 ⊗ ν1), with P the opposite parabolic.

A(P 1 : P 1 : δ1 : ν1) is defined on the space of K-finite vectors, and since πP 1(δ1⊗ν1) and πP 1(δ1⊗ν1) are both admissible, it gives rise to an intertwining operator

for the corresponding actions of g0.

2. πP 1(δ1 ⊗ ν1), as a representation of g0 on the space of K- finite vectors, has a

unique irreducible quotient which we denote by J(P 1, δ1, ν1).

3. J(P 1, δ1, ν1) is infinitesimally equivalent with the image of the intertwining

operator A(P 1 : P 1 : δ1 : ν1) defined above.

4. J(P 1, δ1, ν1) is infinitesimally equivalent to J(P 2, δ2, ν2) if and only if there is

an element of K carrying P 1 to P 2, δ1 to δ2 (up to equivalence), and ν1 to ν2.

Definition 2. Let P 1 = M1A1N1 be a parabolic subgroup of G. Let δ1 be an irre-

ducible tempered unitary representation of M1, and let ν1 be a characters of A1 with

real part in the open positive Weyl chamber determined by N . We call J(P 1, δ1, ν1)

the Langlands quotient with parameters (P 1, δ1, ν1).

21

2.5 Langlands Classification of Irreducible Admis-

sible Representations

We just report the main result, as stated in Theorem 8.54, Knapp[21].

Theorem 5. Fix a minimal parabolic subgroup P = MAN of G.

The infinitesimal equivalence classes of irreducible admissible representations of

G stand in one-one correspondence with all triples (P 1, [δ1], ν1) such that

• P 1 = M1A1N1 is a parabolic subgroup of G containing P ;

• δ1 is an irreducible tempered unitary representation of M1, and [δ1] is its equiv-

alence class;

• ν1 is a complex linear functional on a10, with Re(ν1) in the open positive Weyl

chamber corresponding to N1.

22

Chapter 3

Classification of Unitary

Irreducible Representations

It is a theorem by Harish-Chandra that each irreducible unitary representation is

admissible and two irreducible unitary representations are unitarily equivalent if and

only if they are infinitesimally equivalent. Therefore, classifying the irreducible uni-

tary representations of a semi-simple group amounts to deciding which Langlands

quotients are infinitesimally unitary1.

This problem was solved by Knapp and Zuckerman in 1976. They proved that:

Theorem 6. JP=MAN(δ ⊗ ν) is infinitesimally unitary if and only if

(i) the formal symmetry conditions hold: there exists ω in K normalizing a0 with

ω P ω−1 = P , ω · δ ' δ and ω · ν = −ν

(ii) the Hermitian intertwining operator

B = δ(ω) R(ω) A(P : P : δ : ν)

is positive or negative semi-definite.

1An admissible representation π of G on a Hilbert space V is said to be infinitesimally unitary ifits space V0 of K-finite vectors admits an Hermitian inner product with respect to which π(gC0 ) actsby skew Hermitian operators.

23

We direct the reader to [25] for a detailed proof of this result. We will only add a

few comments.

3.1 The formal symmetry condition

The aim of this section is to explain why the formal symmetry condition

ω P ω−1 = P ω · δ ' δ ω · ν = −ν (3.1)

(for some ω in K) is necessary for the unitarity of the Langlands quotient JP (δ⊗ ν).

Let JP (δ⊗ν) be infinitesimally unitary, and let 〈 , 〉 be a non degenerate (positive

definite) invariant form on JP (δ ⊗ ν). The mapping

T : JP (δ ⊗ ν) → (JP (δ ⊗ ν))h, x 7→ 〈x, ·〉

defines an invertible (g, K)-module map from JP (δ ⊗ ν) to its Hermitian dual.2

Because δ is unitary, the Hermitian dual of JP (δ ⊗ ν) is the Langlands quotient

JP (δ⊗−ν).3 We conclude that JP (δ⊗ν) and JP (δ⊗−ν) are infinitesimally equivalent.

It then follows from 4 that the formal symmetry condition 3.1 holds.

Finally, we notice that the element ω in (3.1) has the property that ω2 lies in M1∩K.

2As a vector space, the Hermitian dual of an admissible (g,K)-module X is the set of K-finiteconjugate linear maps X → C. The actions of K and g are defined by:

(k · f)(x) = f(k−1 · x) (Z · f)(x) = −f(Z · x)

for all k in K, Z in g and x in X. Notice that if X is the Harish-Chandra module of an admissiblerepresentation π of G on a Hilbert space, then Xh is the Harish-Chandra module of πh. We havedenoted by πh the representation of G defined by: πh(g) = (π(g−1))∗. Please, refer to [24] for moreinformation on the Hermitian dual.

3See [24], Proposition 11.26, or [26], lemma 20.

24

3.2 The operator B

Let P be a (not necessarily minimal) parabolic subgroup of G, with Langlands de-

composition P = MAN . Fix a unitary irreducible tempered representation δ of M ,

and a character ν of A in the open positive Weyl chamber determined by N . These

assumptions on δ and ν guarantee the convergence of the standard intertwining op-

erator (2.1)

A(P : P : δ : ν) : HP (δ ⊗ ν) −→ HP (δ ⊗ ν).

Suppose that there exists an element ω in K satisfying the formal parity condition

(3.1). By construction, ω · δ is unitarily equivalent to δ, and ω2 belongs to M ∩K.

Hence, it is possible to define δ(ω) in exactly two ways (differing by a minus sign)

so that δ extends to a unitary representation of the subgroup of K generated by M

and ω.4 Because ω normalizes a0, the operator δ(ω) is a well defined intertwining

operator from πP (y · δ ⊗ ν) to πP (δ ⊗ ν).

We also consider the operator R(ω) of right multiplication by ω, which is an inter-

twining operator from πP (δ ⊗ ν) to πωPω−1(ω · δ ⊗ ω · ν) = πP (ω · δ ⊗−ν).

Composing the standard intertwining operator A(P : P : δ : ν) with the operator

R(ω), we obtain an operator

R(ω) A(P : P : δ : ν) : HP (δ ⊗ ν) −→ HωPω−1=P (ω · δ ⊗ ω · ν)

which intertwines the representations πP (δ ⊗ ν) and πP (ω · δ ⊗−ν).

We then compose [R(ω) A(P : P : δ : ν)] with δ(ω) to obtain an intertwining

operator

B = δ(ω) R(ω) A(P : P : δ : ν) : HP (δ ⊗ ν) −→ HP (δ ⊗−ν) (3.2)

from πP (δ ⊗ ν) to πP (δ ⊗−ν).

4See [21], lemma 14.22 or [22], lemma 18.

25

It is important to notice that the operator B = δ(ω) R(ω) A(P : P : δ : ν) only

depends on the equivalence class of ω modulo M . Indeed, for each m0 in M , the

element ωm0 also satisfies the formal symmetry condition (3.1) and

δ(ωm0) R(ωm0) A(P : P : δ : ν) = δ(ω) R(ω) A(P : P : δ : ν).

Since ω2 lies in M ∩K, this implies that

δ(ω) R(ω) A(P : P : δ : ν) = δ(ω−1) R(ω−1) A(P : P : δ : ν).

The other fundamental remark is that the operator B is Hermitian (we use the com-

pact picture for the principal series representations πP (δ ⊗ ν) and πP (δ ⊗ −ν), so

that domain and codomain of B are identifies with HPδ ). This result follows from the

following lemma:5

Lemma 1. Let G be a semi-simple Lie group, and let P1 = MAN1 and P2 = MAN2

be associated minimal parabolic subgroups of G.

Fix an irreducible (tempered) unitary representation δ of M , and a character ν of A

satisfying

〈Re(ν), α〉 > 0

for each α positive for P1 and negative for P2. Let A(P2 : P1 : δ : ν) be the standard

intertwining operator from πP1(δ ⊗ ν) to πP2(δ ⊗ ν). Then

A(P2 : P1 : δ : ν)∗ = A(P1 : P2 : δ : − ν)

with the adjoint computed K-type by K-type in the compact picture.

We now prove that

5Please, refer to [23] (proposition 7.1 (iv)) for a proof.

26

Proposition 1. The operator B : HPδ 7→ HP

δ is Hermitian.

Proof. This is not hard to prove:

B∗ = [δ(ω) R(ω) A(P : P : δ : ν)]∗

= A(P : P : δ : ν)∗ R(ω)∗ δ(ω)∗

= A(P : P : δ : − ν) R(ω−1) δ(ω−1) (?)

= R(ω−1) [R(ω) A(P : P : δ : − ν) R(ω−1)] δ(ω−1)

= R(ω−1) A(P : P : ω · δ : ν) δ(ω−1) (??)

= R(ω−1) δ(ω−1) [δ(ω) A(P : P : ω · δ : ν) δ(ω−1)]

= R(ω−1) δ(ω−1) A(P : P : δ : ν) (? ? ?)

= δ(ω) R(ω) A(P : P : δ : ν) = B.

Just a few remarks:

• (?) follows from the fact that the operators R(ω) and δ(ω) are unitary, and from

the the previous lemma;

• (??) follows from the identity6

A(S2 : S1 : σ : λ) = R(y−1)A(yS2y−1 : yS1y

−1 : y · σ : y · λ)R(y)

6See [21], p.182.

27

for y = ω−1, S2 = P , S1 = P , σ = ω · δ, and λ = ν;

• (? ? ?) is the easy identity

δ(ω) A(P : P : ω · δ : ν) δ(ω−1) = A(P : P : δ : ν).

3.3 Unitary Langlands Quotients

In this section we sketch the proof of theorem 6. A detailed proof can be found in

[25].

Theorem (6). JP=MAN(δ ⊗ ν) is infinitesimally unitary if and only if


ω P ω−1 = P , ω · δ ' δ and ω · ν = −ν


B = δ(ω) R(ω) A(P : P : δ : ν)


Proof. Let JP (δ ⊗ ν) be a Langlands quotient and suppose that there exists an ele-

ment ω in K satisfying the formal symmetry condition 3.1. Construct the Hermitian


B = δ(ω) R(ω) A(P : P : δ : ν) : HPδ −→ HP

δ

as instructed in the previous section. Because δ(ω) and R(ω) are invertible, we

can identify the kernel of B with the kernel of the standard intertwining operator

A(P : P : δ : ν). By construction, B gives rise to an invariant Hermitian form on

HP (δ ⊗ ν), and to a non degenerate Hermitian form 〈 , 〉 on the quotient space:

28

JP (δ ⊗ ν) = HP (δ⊗ν)

ker(A(P : P : δ : ν)).

Because JP (δ⊗ ν) is irreducible, the form 〈 , 〉 is unique up to a non-zero real factor.

We conclude that the Langlands quotient JP (δ ⊗ ν) is infinitesimally unitary if and

only if the invariant Hermitian form 〈 , 〉 is definite, i.e. if and only if the operator B

is semi-definite.

29

Chapter 4

The example of SL(2, R)

4.1 The data for SL(2, R)

In this section we briefly recall the data for the group SL(2, R), and we fix the

notations that will be used throughout the chapter.

• G = SL(2, R)

• K = SO(2) =

kϑ =

cos(ϑ) sin(ϑ)

− sin(ϑ) cos(ϑ)

: ϑ ∈ R

• g0 = sl(2, R) = a0 ⊕ (g0)α ⊕ (g0)−α, with

a0 = RHα, Hα =

1 0

0 −1

α = 2ε, ρ = ε, ε : a0 → R, tHα 7→ t

30

(g0)α = RXα, Xα =

0 1

0 0

(g0)−α = RYα, Yα =

0 0

1 0

• M =

1 0

0 1

,

−1 0

0 −1

• A =

et 0

0 e−t

: t ∈ R

=

ay =

y 0

0 y−1

: y ∈ R, y > 0

• N =

nx =

1 x

0 1

: x ∈ R

• P =

b r

0 b−1

: b, r ∈ R, b 66= 0

•

b r

0 b−1

=

sgn(b) 0

0 sgn(b)

| b | 0

0 | b |−1

1 r/b

0 1

the Iwasawa decomposition of an element of P (where sgn(b) = b/ | b |)

31

•

a b

c d

= kϑaynx =

cos(ϑ) sin(ϑ)

− sin(ϑ) cos(ϑ)

y 0

0 y−1

1 x

0 1

with eiϑ =a− ic√a2 + c2

and y =√

a2 + c2 (4.1)

the Iwasawa decomposition of an arbitrary element of G

• M = δ±, with δ+ and δ− respectively the trivial and the sign representation

of M

• (a′0)C = Cε, with ε : a0 → R, tHα 7→ t

• K ' Z. For each integer n we denote by ξn the character:

ξn : K → C, kϑ =

cos(ϑ) sin(ϑ)

− sin(ϑ) cos(ϑ)

7→ einϑ.

4.2 Principal Series induced from the Minimal Parabolic

4.2.1 The Principal Series Representation πP (δ ⊗ λ)

Fix an irreducible representation δ of M (δ = δ±), and a complex linear functional λε

on a0. We denote by πP (δ ⊗ λ) the principal series induced from the representation

(δ ⊗ λε) of the minimal parabolic subgroup P = MAN (as usual, N acts trivially).

The Hilbert space HP(δ⊗λ) of this representation is the completion with respect to the

norm

‖F‖2 =∫

K| f(k) |2 dk

32

of the vector space of the continuous functions F : G → C satisfying:

F (gman) = δ(m)−1e−(λε+ρ) log(a)F (g) ∀ g ∈ G, ∀man ∈ P.

In the notations introduced above, we can describe this transformation property by:

F

g

b r

0 b−1

= δ

sgn(b) 0

0 sgn(b)

| b |−λ−1 F (g)

for all b and r in R, with b non-zero. G acts on HP(δ⊗λ) by left translation:

[g · F ](x) = F (g−1x).

It is convenient to introduce also the compact picture for the principal series. In the

compact picture, the Hilbert space is

HPδ = f ∈ L2(K,C) : f(−g) = δ

−1 0

0 −1

f(g).

We notice that HPδ is the space of all even functions in L2(K,C) when δ = δ+, and

it is the space of all odd functions in L2(K,C), when δ = δ−.

Restriction to K is an isomorphism from HP(δ⊗λ) to HP

δ . The inverse mapping

carries an element f of HPδ to the function

F : G → C, g = kϑaynx 7→ F (kϑaynx) = y−λ−1f(kϑ).

33

4.2.2 The restriction of πP (δ ⊗ λ) to K

By Frobenius reciprocity, the multiplicity of a K-type ξn in the restriction of πP (δ⊗λ)

to K equals the multiplicity of δ in the restriction of ξn to M . Since

ξn

−1 0

0 −1

= ei n π =

+1 if n is even

−1 if n is odd,

it follows immediately that

• if δ is the trivial representation of M , then the restriction of πP (δ ⊗ λ) to K

contains only even characters, each occurring with multiplicity one:

ResK(πP (δ+ ⊗ λ)) =⊕

n∈Z ξ2n

• if δ is the sign representation of M , then the restriction of πP (δ ⊗ λ) to K

contains only odd characters, each occurring with multiplicity one:

ResK(πP (δ− ⊗ λ)) =⊕

n∈Z ξ2n+1.

We now describe representatives for the various K-types in both the compact and

the induced picture.

Assume that n has the same parity as δ. The function

fn : K → C, kϕ 7→ fn(kϕ) = e−i n ϕ

belongs to HPδ and it transforms under K according to the character ξn:

[kϑ · fn](kϕ) = fn(k−1ϑ kϕ) = e+i n ϑfn(kϕ) = [ξn(kϑ) · fn](kϕ).

In the induced picture, a generator for the K-type ξn is given by the function:

Fn : G → C, g = kϕaynx 7→ y−λ−1e−i n ϕ.

34

4.3 Study of the irreducibility of πP (δ ⊗ λ)

The trick is to look at the action of gC0 in the compact picture of πP (δ ⊗ λ).

In gC0 we pick the basis

H =

0 −i

i 0

X =

1

2

1 i

i −1

Y =

1

2

1 −i

−i −1

which satisfies the bracket relations:

[H, X] = 2X [H, Y ] = −2Y [X, Y ] = H.

We let H, X, Y act on the orthonormal basis fn of HPδ (n runs over 2Z when δ

is the trivial representation of M , and over 2Z+1 when δ is the sign representation).

We find that

1. H fixes each K-type: H · fn = n fn

ªª ªª ªª ªª ªª. . . Cfn−4

FF

Cfn−2

FF

Cfn

FF

Cfn+2

FF

Cfn+4

FF

. . .

2. X pushes each K-type forward: X · fn = an fn+2 an = (λ + n + 1)/2

. . .--Cfn−4

-- Cfn−2++ Cfn

--Cfn+2

-- Cfn+4** . . .

3. Y pushes each K-type backwards: Y · fn = bn fn−2 bn = (λ− n + 1)/2

. . . Cfn−4jj Cfn−2mm Cfnmm Cfn+2kk Cfn+4mm . . .mm

Proof. 1. H · fn = n fn

35

The element iH =

0 1

−1 0

belongs to the Lie algebra g0 and the vector fn

is K-finite (hence smooth). So we can write:

(iH) · fn = ddt|t=0 eitH · fn = d

dt|t=0

cos(t) sin(t)

− sin(t) cos(t)

· fn =

= ddt|t=0 [eintfn] = infn.

The element H of gC0 acts on each K-type ξn as scalar multiplication by n.

2. X · fn = λ+n+12

fn+2

From the commutator relation [H, X] = 2X it follows immediately that X · fn

is an eigenvector of H of eigenvalue n + 2. So X · fn must belong to the

isotypic component of the K-type ξn+2. Since this isotypic is one-dimensional

and generated by fn+2, there exists a constant an such that

X · fn = anfn+2.

In particular, X · fn(1) = anfn+2(1) = an.

To calculate an = X ·fn(1), we decompose X as a linear combination of elements

of the Lie algebra:

X =1

2

1 i

i −1

=

1

2

1 0

0 −1

− i

2

0 1

−1 0

+ i

0 1

0 0

and we compute the action of each summand on the smooth vector fn:

1 0

0 −1

· fn(1) = d

dt|t=0

et 0

0 e−t

· fn(1) = d

dt|t=0 at · fn(1)

36

= ddt|t=0 fn(a−t1) = d

dt|t=0 fn(1a−t) = d

dt|t=0 e−(λ+1)(−t) fn(1)

= +(λ + 1)fn(1) = (λ + 1).

0 1

−1 0

· fn(1) = iH · fn(1) = in fn(1) = in.

0 1

0 0

· fn(1) = d

dt|t=0

1 t

0 1

· fn(1) = d

dt|t=0 fn(1) = 0.

This gives:

an = X · fn(1) = [12(λ + 1)− i

2in] = λ+n+1

2

so X · fn = λ+n+12

fn+2. Finally, we prove that:

3. Y · fn = λ−n+12

fn−2

Computations are similar to the previous case: since [H, Y ] = −2Y , there exists

a constant bn = Y · fn(1) such that Y · fn = fn+2. Since

Y =1

2

1 −i

−i −1

=

1

2

1 0

0 −1

+

i

2

0 1

−1 0

− i

0 1

0 0

we obtain:

bn = Y · fn(1) = [12(λ + 1) + i

2in] = λ+1−n

2.

We now go back to the problem of discussing the irreducibility of the principal series

representation πP (δ ⊗ λ). Of course, we need to distinguish between the cases“ λ =

trivial” and“ λ = sign.”

37

4.3.1 Irreducibility of πP (δ+ ⊗ λ)

The action of gC0 on the Hilbert space HPδ+ =

⊕m∈ZCf2m is given by:

H · f2m = 2mf2m

X · f2m = (λ+2m+1)2

f2m+2

Y · f2m = (λ−2m+1)2

f2m−2.

If λ is not an odd integer, the representation πP (δ+ ⊗ λ) is clearly irreducible.

If λ is a positive odd integer, say λ = 2m0 − 1 with m0 > 1, the representation

πP (δ+ ⊗ λ) is reducible. There are three proper invariant subspaces:

• ⊕m>m0

Cf2m

(invariance follows from Y · f2m0 = 0)

• ⊕m>m0

Cf−2m

(invariance follows from X · f−2m0 = 0)

• (⊕m>m0

Cf−2m

)⊕ (⊕m>m0

Cf2m

).

If λ is a negative odd integer, say λ = −2m0−1 with m0 > 0, then the representation

πP (δ+ ⊗ λ) is reducible. There are three proper invariant subspaces:

• ⊕−m06m6m0

Cf2m

(invariance follows from X · f+2m0 = Y · f−2m0 = 0)

• ⊕−m06mCf2m

• ⊕m6m0

Cf2m.

38

4.3.2 Irreducibility of πP (δ+ ⊗ λ)

The action of gC0 on the Hilbert space HPδ+ =

⊕m∈ZCf2m−1 is given by:

H · f2m−1 = (2m− 1) f2m−1

X · f2m−1 = λ+2m2

f2m+1

Y · f2m−1 = λ−2(m−1)2

f2m−3.

If λ is not an even integer, the representation πP (δ+ ⊗ λ) is clearly irreducible.

If λ is a positive even integer, say λ = 2m0 with m0 > 1, then the representation

πP (δ− ⊗ λ) is reducible. There are three proper invariant subspaces:

• ⊕m>m0

Cf2m+1

(invariance follows from Y · f2m0+1 = 0)

• ⊕m>m0

Cf−2m−1

(invariance follows from X · f−2m0−1 = 0)

• (⊕m>m0

Cf−2m−1

)⊕ (⊕m>m0

Cf2m+1

).

If λ is zero, then the representation πP (δ− ⊗ 0) is reducible. There are two proper

invariant subspaces:

• ⊕m>0Cf2m+1

(invariance follows from Y · f+1 = 0)

• ⊕m>0Cf−2m−1

(invariance follows from X · f−1 = 0).

If λ is a negative even integer, say λ = −2m0 with m0 > 0, then the representation

πP (δ− ⊗ λ) is reducible. There are three proper invariant subspaces:

39

• ⊕−m0+16m6m0

Cf2m−1

(invariance follows from X · f2m0−1 = Y · f−2m0+1 = 0)

• ⊕−m0+16mCf2m−1

• ⊕m6m0

Cf2m−1.

4.4 Study of the Unitarity of πP (δ ⊗ ν)

Let P = MAN be a minimal parabolic subgroup of G, let δ be an irreducible rep-

resentation of M and let ν be a strictly dominant character of A. A principal series

representation πP (δ ⊗ ν) gives rise to a unitary Langlands quotient if and only if


ω P ω−1 = P , ω · δ ' δ and ω · ν = −ν and


B = δ(ω) R(ω) A(P : P : δ : ν)


When P is the minimal parabolic, the condition ω P ω−1 = P implies that ω is a

representative in K for the long Weyl group element (the reflection sα). So we can

take

ω =

0 1

−1 0

= kπ/2.

It is easy to check that ω stabilizes both the trivial and the sign representation of

M , so there are no conditions on δ. Let us now look at the conditions imposed on ν.

40

Write ν = λ ε, so that

ν : a0 → C, Ht =

t 0

0 −t

7→ λ t.

Since

ω · ν(Ht) = ν(k−π/2Htk+π/2) = ν(H−t) = −λ t

it immediately follows that ν = λ ε satisfies the condition ω · ν = −ν if and only if λ

is real. And of course λ must be strictly positive, because we want the character λ ε

of A to be in the open positive Weyl chamber determined by N .

For δ = δ+ or δ−, ν = λ ε (with λ > 0) and for ω = kπ/2, we have to study the

signature of the Hermitian intertwining operator

B = δ(ω) R(ω) A(P : P : δ : λ).

Notice that ω2 =

−1 0

0 −1

, so δ(ω2) = +1 if δ is trivial, and δ(ω2) = −1 if δ is

the sign representation. We take the operator δ(ω) to be the identity if δ is trivial,

and scalar multiplication by i if δ is sign. In any case, B is a multiple of the operator:

Aω : HPδ⊗λ → HP

δ⊗−λ, F 7→ Aω · F

where (Aω · F )(x) =∫

NF (xω n) dn.

The operator Aω can be computed K-type by K-type, and since HPδ+⊗λ and HP

δ−⊗λ

have different K-types, it is appropriate to to distinguish between the spherical and

the non-spherical case.

41

4.4.1 The Spherical Case

We use for both πP (δ+ ⊗ λ) and πP (δ+ ⊗ −λ) the compact picture, and we identify

domain and codomain of Aω with the Hilbert space:

HPδ+ =

⊕

m∈Zξ2m =

⊕

m∈ZCF2m.

The intertwining operator Aω preserves the decomposition of HPδ+ in isotypical com-

ponents of K-types, and since each isotypical is one-dimensional, Aω acts as scalar

multiplication on each K-type.

For each integer m, and for each kϕ in K, we want to compute the integral

Aω · F−2m(kϕ) =

∫

N

F−2m(kϕ ω n) dn

To proceed, we need to understand the Iwasawa decomposition of (kϕ ω n). Let

n =

1 0

t 1

, and let n = kθ(t)ay(t)n be its Iwasawa decomposition. Then

y(t) =√

1 + t2 and ei θ(t) = 1−i t√1+t2

(tan θ(t) = −t).

Since ω = kπ/2, we obtain for (kϕ ω n) the Iwasawa decomposition

kϕ ω n = kϕ+θ(t)+π/2 ay(t) n.

Therefore

Aω · F−2m(kϕ) =∫R y(t)−1−λf−2m(kϕ+θ(t)+π/2) dt =

=∫R

(√1 + t2

)−1−λe2m i(ϕ+θ(t)+π/2) dt

42

= (−1)mf−2m(kϕ)∫R(√

1 + t2)−1−λ e2m i θ(t) dt

= (−1)mF−2m(kϕ)

∫

R(√

1 + t2)−1−λ

(1− i t√1 + t2

)2m

dt. (4.2)

We perform the change of variable t = − tan θ, so that

θ = − arctan(t) (with − π/2 6 θ 6 +π/2).

This gives:

√1 + t2 = 1

cos θ

1−i t√1+t2

= cos θ + i tan θ cos θ = ei θ

d t = − 1(cos θ)2

d θ.

Finally, we get:

Aω · F−2m(kϕ) = (−1)mF−2m(kϕ)∫ π/2

−π/2(cos θ)1+λ e2m i θ 1

(cos θ)2d θ

= (−1)mF−2m(kϕ)∫ π/2

−π/2(cos θ)λ−1 e2m i θ d θ.

With the change of variable θ → x = θ + π/2, we get

Aω · F−2m(kϕ) = F−2m(kϕ)

∫ π

0

(sin x)λ−1 e2m i x d x.

To conclude, we make use of the following result (see e.g. [10]):

∫ π

0

(sin t)a ei b t d t =π Γ(1 + a) ei π b/2

2a Γ(1 + a+b2

) Γ(1 + a−b2

)

for each b in R, and for each a in C such that Re(a) > −1. We obtain:

43

Aω · F−2m = F−2mπ Γ(λ) eiπm

2λ−1 Γ(1+λ+2m−12 )Γ(1+λ−2m−1

2 )

= F−2mπ Γ(λ) eiπm

2λ−1 Γ(λ+12

+m)Γ(λ+12−m)

.

It is a convention to normalize the operator so that it acts trivially on the fine K-type

Cf0. Dividing by the constant

C = Aω · F0(1) =π Γ(λ)

2λ−1 Γ(

λ+12

)Γ

(λ+1

2

) ,

we get:

1CAω · F−2m = F−2m

eiπm Γ(λ+12 )Γ(λ+1

2 )Γ(λ+1

2+m)Γ(λ+1

2−m)

= (−1)m F−2mΓ(λ+1

2 )Γ(λ+1

2+m)

Γ(λ+12 )

Γ(λ+12−m)

.

So the operator 1CAω acts on the K-type ξ−2m as scalar multiplication by

d−2m = (−1)m Γ(

λ+12

)

Γ(

λ+12

+ m) Γ

(λ+1

2

)

Γ(

λ+12−m

) .

To simplify this expression we recall the factorization property of the Γ function:

Γ(z + 1) = z Γ(z)

and we introduce the notation

(z)n = z(z + 1)(z + 2) · · · (z + n− 1)

44

for each z in C, and for every positive integer n. Then

Γ(z)Γ(z)

Γ(z + n) Γ(z − n)=

Γ(z) (z − n)n Γ(z − n)

(z)n Γ(z) Γ(z − n)=

(z − n)n

(z)n

=(z − 1)(z − 2) · · · (z − n)

z(z + 1) · · · (z + n− 1).

Setting z = λ+12

and n =| m |, we find:

d2m = d−2m = (−1)m (λ− 1)(λ− 3) · · · (λ− 2m + 1)

(λ + 1)(λ + 3) · · · (λ + 2m− 1)=

(1− λ)(3− λ) · · · (2m− 1− λ)

(1 + λ)(3 + λ) · · · (2m− 1 + λ).

Conclusions. Fix λ in C with positive real part. If we use the compact picture

for both πP (δ+⊗ λ) and πP (δ+⊗ω · λ), we can identify domain and codomain of the


Aω : HPδ+⊗λ −→ HP

δ+⊗ω·λ

with the Hilbert space HPδ+ =

⊕m∈Z ξ2m.

The operator Aω preserves this decomposition and acts on the K-types ξ2m and ξ−2m

as multiplication by the scalar

C(1− λ)(3− λ) · · · (2m− 1− λ)

(1 + λ)(3 + λ) · · · (2m− 1 + λ)

where C is the value of Aω on the fine K-type ξ0 = CF0:

C = Aω · F0(1) =π Γ(λ)

2λ−1 Γ(

λ+12

)Γ

(λ+1

2

) .

Since C is real and positive when λ is real and positive, Aω has the same signature

as 1CAω (with respect to all K-types). The signature of the operator B = Aω on a

K-type ξ2m is simply given by the sign of the scalar

d2m =(1− λ)(3− λ) · · · (2 | m | −1− λ)

(1 + λ)(3 + λ) · · · (2 | m | −1 + λ).

45

We notice that

• if 0 < λ < 1, then d2m > 0 for all integers m:

because d2m = d−2m, we just look at the case m > 0. Then it is clear that

d2m = (1−λ)(3−λ)···(2m−1−λ)(1+λ)(3+λ)···(2m−1+λ)

> 0 for 0 < λ < 1.

• if λ = 1, then d2m > 0 for all integers m:

precisely, d0 = 1 and d2m = 0 otherwise.

• if λ > 1, then the integers d2m are neither all non-negative, nor all non-

positive:

for instance, d0 = 1 > 0, while d2 = (1−λ)(1+λ)

< 0.

So the Hermitian operator B on HP (δ+ ⊗ λ) is positive definite when 0 < λ < 1,

positive semi-definite when λ = 1 and indefinite when λ > 1. As a result we obtain

the following theorem:

Theorem 7. Let P be a minimal parabolic subgroup of SL(2, R), let δ+ be the trivial

representation of M and let λ > 0 be a strictly dominant character of A. The Lang-

lands quotient JP (δ+ ⊗ λ) is infinitesimally unitary if and only if 0 < λ 6 1. When

λ = 1, JP (δ+ ⊗ 1) is the trivial representation of SL(2, R).

We conclude the analysis of the spherical case with a remark:

Remark 1. The Hermitian operator B on HP (δ+⊗λ) is positive semi-definite if and

only if it has non negative signature with respect to the K-types ξ0, ξ2 and ξ−2.

Indeed, the previous analysis of the sign of the integers d2m shows that

d2m > 0 ∀m ∈ Z ⇔ d0, d2, d−2 > 0.

46

4.4.2 The Non-Spherical Case

In the non-spherical case, B = iAω with

Aω : HPδ−⊗λ → HP

δ−⊗−λ

Using the compact picture for each principal series representation, we regard Aω as

an endomorphism on

HPδ− =

⊕

m∈Zξ2m+1 =

⊕

m∈ZCF2m+1.

Being an intertwining operator, Aω acts on (the isotypical component of) each K-type

as multiplication by a scalar. We want to compute this scalar explicitly.

Aω · F−2m−1(1) =∫

NF−2m−1(ω n) dn

=∫R y(t)−1−λ f−2m−1(kθ(t)+π/2) dt

=∫R(√

1 + t2)−1−λ e(2m+1) i(θ(t)+π/2) dt

= e(2m+1)iπ/2 f−2m−1(1)∫R(√

1 + t2)−1−λ e(2m+1) i θ(t) dt

= e(2m+1)iπ/2 F−2m−1(1)∫R(√

1 + t2)−1−λ(

1−i t√1+t2

)2m+1

dt

= e(2m+1)iπ/2 F−2m−1(1)∫ π/2

−π/2(cos θ)1+λ e(2m+1) i θ 1

(cos θ)2d θ

= e(2m+1)iπ/2 F−2m−1(1)∫ π/2

−π/2(cos θ)λ−1 e(2m+1) i θ d θ

= F−2m−1(1)∫ π

0(sin x)λ−1 e(2m+1) i x d x

47

= F−2m−1(1) π Γ(λ) ei(2m+1)π/2

2λ−1 Γ(1+λ+2m2 )Γ(1+λ−2m−2

2 )

= F−2m−1(1) π Γ(λ) ei(2m+1)π/2

2λ−1 Γ(λ2+m+1)Γ(λ

2−m)

.

In particular,

D ≡ Aω · F1(1) = F1(1)π Γ(λ) ei(−1)π/2

2λ−1 Γ(

λ2

)Γ

(λ2

+ 1) = e−iπ/2 π Γ(λ)

2λ−1 Γ(

λ2

)Γ

(λ2

+ 1) .

Then the operator 1D

Aω acts on the K-type ξ−2m−1 as scalar multiplication by

d−2m−1 = (−1)m+1 Γ(

λ2

+ 1)

Γ(

λ2

+ 1 + m) Γ

(λ2

)

Γ(

λ2−m

) .

This expression is not symmetric with respect to m. We distinguish various cases:

• d1 = 1

• d−1 = −1

• if m > 0, then d−2m−1 = (−1)m+1 (λ2−m)

m

(λ2+1)

m

= (−1)m (λ−2)(λ−4)···(λ−2m)(λ+2)(λ+4)···(λ+2m)

• if m < 0, then d−2m−1 = (−1)m+1 (λ2+1−(−m))−m

(λ2 )−m

= (−1)m (λ−2)(λ−4)···(λ−2(−m)+2)(λ+2)(λ+4)···(λ+2(−m)−2)

.

Shifting indices, we find:

d2m+1 = d−2(−m−1)−1 = (−1)m (λ− 2)(λ− 4) · · · (λ− 2m)

(λ + 2)(λ + 4) · · · (λ + 2m)= −d−(2m+1)

for each positive integer m.

Conclusions. Fix λ in C with positive real part. If we use the compact picture

for both πP (δ−⊗ λ) and πP (δ−⊗ω · λ), we can identify domain and codomain of the

48


Aω : HPδ−⊗λ −→ HP

δ−⊗ω·λ

with the Hilbert space HPδ− =

⊕m∈Z ξ2m+1.

The operator Aω preserves this decomposition, and it acts on each K-type ξj as

multiplication by a scalar (that we call bj).

In the previous notations, bj = D · dj, so we have:

b1 = −b−1 = D

and, for each positive integer m,

b2m+1 = −b−(2m+1) = (−1)mD(λ− 2)(λ− 4) · · · (λ− 2m)

(λ + 2)(λ + 4) · · · (λ + 2m).

We have denoted by D the value of Aω on the fine K-type ξ1 = CF1:

D = e−iπ/2 π Γ(λ)

2λ−1 Γ(

λ2

)Γ

(λ2

+ 1) .

The operator B = iAω acts on the K-type ξ2m+1 as scalar multiplication by

π Γ(λ)

2λ−1 Γ(

λ2

)Γ

(λ2

+ 1) d2m+1

which is actually a real number (in agreement to the fact that B is Hermitian).

Since π Γ(λ)

2λ−1 Γ(λ2 )Γ(λ

2+1)

> 0, the signature of B is determined by the sign of the integers

d2m+1. Because

d1 = 1 and d−1 = −1

for all values of λ, B is always indefinite.

Theorem 8. Let P be a minimal parabolic subgroup of SL(2, R), let δ− be the sign

49

representation of M and let λ > 0 be a strictly dominant character of A. The Lang-

lands quotient JP (δ− ⊗ λ) is never infinitesimally unitary.

50

Chapter 5

Signatures of Invariant Hermitian

Forms on an Admissible (gC0 , K)

module

The aim of this chapter is to discuss the signature of an invariant Hermitian form on

an admissible (gC0 , K) module with respect to a given K-type.

We start by recalling some general results about Hermitian forms on finite-dimensional

vector spaces.

5.1 The finite-dimensional case

5.1.1 Signature of a Hermitian operator on a f.d. vector

space

Let V be a finite-dimensional vector space with a Hermitian inner product 〈 , 〉PD

(we use the notation “PD” to distinguish an inner product from an arbitrary, not

necessarily positive definite, Hermitian form).

51

Let T be a Hermitian linear operator on V , so that

〈Tx, y〉PD = 〈x, Ty〉PD

for all x and y in V . By the spectral theorem, T is diagonalizable. It has real

eigenvalues and orthogonal (distinct) eigenspaces.

We define the signature of T to be the triple (n+, n−, n0), where

n+ = the number of positive eigenvalues of T , counting multiplicities;

n− = the number of negative eigenvalues of T , counting multiplicities;

n0 = dimension of the kernel of T = multiplicity of the eigenvalue zero.

Since T is diagonalizable, we obtain a decomposition of V in direct sum of orthog-

onal eigenspaces. Collecting all the eigenspaces with positive eigenvalues (V +), the

eigenspaces with negative eigenvalues (V −) and then adding the zero eigenspace (V 0),

we can write:

V = V + ⊕ V − ⊕ V 0

with dim(V +) = n+, dim(V −) = n−, dim(V 0) = n0, and

T positive definite on V +

T negative definite on V −

T identically zero on V 0.

5.1.2 Signature of a Hermitian form on a f.d. vector space

Let V be a finite-dimensional vector space with a Hermitian inner product 〈 , 〉PD.

There exists a one-one correspondence between Hermitian forms on V ×V and linear

operators V → V which are Hermitian with respect to 〈 , 〉PD. The correspondence

being:

52

T 7−→ 〈 , 〉T

with 〈x, y〉T = 〈Tx, y〉PD for all x and y in V . The inverse mapping is

〈 , 〉 7−→ T〈 , 〉

with T〈 , 〉 the Hermitian operator constructed as follows: fix an orthonormal basis vjwith respect to 〈 , 〉PD; the matrix of T〈 , 〉 in this basis is given by (T〈 , 〉)i, j = 〈vj, vi〉.

If 〈 , 〉 is a Hermitian form on V , we define the signature of 〈 , 〉 to be the signature

of the corresponding Hermitian operator T . This definition makes sense only if we

prove that it is independent of the choice of the Hermitian inner product 〈 , 〉PD. So,

we fix another Hermitian inner product ( , )PD on V , and we let S be the Hermitian

operator corresponding to 〈 , 〉 with respect to ( , )PD. There exists an automorphism

C of V relating the two Hermitian inner products:

〈x, y〉PD = (Cx, Cy)PD

for all x and y in V . Since

〈Tx, y〉PD = 〈x, y〉T = (Sx, y)PD = 〈CSx, Cy〉PD

for all x and y in V , we deduce that T = C∗CS. We need to prove that T and S

have the same signature. Notice that

- since T and S are both Hermitian, the operator CSC−1 = (C−1)∗TC−1 is

Hermitian

- the Hermitian operators S and CSC−1 have the same signatures, because they

have the same eigenvalues

- the Hermitian operators T and CSC−1 also have the same signature, because

they give rise to equivalent quadratic forms. Indeed, since CSC−1 = (C−1)∗TC−1,

the change of variable x 7→ y = Cx gives:

53

〈y, y〉CSC−1 = 〈CSC−1y, y〉PD = 〈(C−1)∗TC−1y, y〉PD = 〈TC−1y, C−1y〉PD =

= 〈Tx, x〉PD = 〈x, x〉T .

Therefore, the operators T and S have the same signature, and our definition of sig-

nature of an Hermitian form on a finite dimensional vector space makes sense.

The next step is to discuss the signature of an invariant Hermitian form on an

admissible (gC0 , K) module with respect to a K-type.

5.2 Signature of an invariant Hermitian form on

an admissible (gC0 , K) module w.r.t. a K-type

Let V be a (gC0 , K) module. An invariant Hermitian form on V is a sesquilinear

pairing V ⊗ V → C satisfying:

(i) 〈v, w〉 = 〈w, v〉

(ii) 〈X · v, w〉 = −〈v, X · w〉

(iii) 〈k · v, k · w〉 = 〈v, w〉

for all v, w in V , for all X in g0 and for all k in K. In other words, it is a Hermitian

form on V with respect to which K acts by orthogonal operators and g0 acts by

skew-symmetric operators.1

The definition of positive/negative definite (or semi-definite) invariant Hermitian form

is the standard one.1Equivalently, you can ask that K acts by orthogonal operators and that gC0 acts by skew-

Hermitian operators. Indeed, if you write gC0 = g0 ⊕ ig0, and define conjugation accordingly, thencondition (ii) is equivalent to:

〈X · v, w〉 = −〈v, X · w〉for all X in gC0 .

54

Remark 2. The Harish-Chandra module of a unitary irreducible representation of G

has a positive definite invariant Hermitian form. The converse is less trivial, but also

true: for G reductive, any irreducible (gC0 , K) module admitting a positive definite

invariant Hermitian form is the Harish-Chandra module of a (unique) irreducible

unitary representation of G.

We now discuss how to determine whether an invariant Hermitian form on an

admissible (gC0 , K) module is positive semi-definite.2

Suppose that (π, V ) is an admissible (gC0 , K) module with an invariant Hermitian

form 〈 , 〉, and that V has a Hilbert space structure. Since K is compact, we can

always assume that K acts unitarily with respect to the Hermitian inner product

〈 , 〉PD on V . Then, by the Peter Weyl theorem, V decomposes the orthogonal direct

sum of the K isotypical components:

V =⊕µ∈K∗

V (µ).

Since the representation π is assumed to be admissible, each K-type µ has finite

multiplicity and each isotypic V (µ) has finite dimension (equal to dim(µ) ·mV (µ)).

Moreover, the isotypics of two distinct K-types are orthogonal with respect to 〈 , 〉PD.

We show that they are also orthogonal with respect to the invariant Hermitian form

〈 , 〉:let (µ1, Eµ1) and (µ2, Eµ2) be non equivalent irreducible representations of K that

appear in V , and let W = Eµ1 ⊕ Eµ2 be their direct sum. The finite dimensional

vector space W inherits by restriction the two Hermitian forms 〈 , 〉 and 〈 , 〉PD (the

latter is positive definite).

2The results we find have an obvious analogous for negative semi-definite forms and for positive ornegative definite forms. We focus on positive definite forms, because in order to detect the unitarityof a Langlands quotient JP (δ⊗ν) one needs to verify that an invariant Hermitian form on HP (δ⊗ν)is positive definite.

55

There exists a linear operator L on W , Hermitian with respect to 〈 , 〉PD, such that

〈x, y〉 = 〈Lx, y〉PD

for all x and y in W . Since both 〈 , 〉 and 〈 , 〉PD are invariant under K, L is a

self-intertwining operator for the action of K on W :

〈Lx, y〉PD = 〈x, y〉 = 〈k · x, k · y〉

= 〈Lk · x, k · y〉PD = 〈k−1Lk · x, y〉PD

for all x and y in W . It easily follows that L = k−1Lk (for any k in K), because

〈 , 〉PD is non degenerate.

Since µ1 and µ2 are irreducible and inequivalent, any self-intertwining operator on

W = Eµ1 ⊕ Eµ2 must be of the form c1IEµ1⊕ c2IEµ2

. So for all vectors v = v1 + v2

and u = u1 + u2 in W , we have:

〈u, v〉 = 〈L(u1 + u2), (v1 + v2)〉PD = 〈c1u1 + c2u2, (v1 + v2)〉PD

= c1〈u1, v1〉PD + c2〈u2, v2〉PD.

We have used the fact that Eµ1 and Eµ2 are orthogonal with respect to 〈 , 〉PD.

In particular, 〈u, v〉 = 0 if u is in Eµ1 and v is in Eµ2 . This shows that any two

distinct K-isotypics are also orthogonal with respect to 〈 , 〉. As a consequence:

〈 , 〉 is positive semi-definite on V ⇔ its restriction to each K-isotypic is such .

56

This reduction is very advantageous: since any K-isotypic V (µ) is finite-dimensional,

we can use the signature to discriminate the sign of the form on V (µ). So if

n+V (µ) n−V (µ) n0

V (µ)

denotes the signature of the restriction of 〈 , 〉 to an K-isotypic V (µ), we can say that

〈 , 〉 is positive semi-definite on V ⇔ n0V (µ) = 0 for all µ.

The following lemma shows that the integers n+V (µ), n−V (µ), n0

V (µ) are all di-

visible by the dimension of µ, so one only needs to know the value of the triple:

pV (µ) =n+

V (µ)

dim(µ)qV (µ) =

n−V (µ)

dim(µ)zV (µ) =

n0V (µ)

dim(µ)

which we call the signature of V with respect to the K-type µ.

Lemma 2. Let (π, W ) be a finite-dimensional representation of K, which is isotypic

of a single K-type µ. Suppose that W has an invariant Hermitian form 〈 , 〉. Then,

there exists a decomposition

W = W+ ⊕W− ⊕W 0

of W in K-invariant subspaces such that

〈 , 〉 is positive definite on W+

〈 , 〉 is negative definite on W−

〈 , 〉 is identically zero on W 0.

Moreover, if W = W+1 ⊕W−

1 ⊕W 01 is a similar decomposition, then

mW+(µ) = mW+1(µ)

57

mW−(µ) = mW−1

(µ)

W 0 = W 01 .

Proof. Since W is finite-dimensional and K is compact, we can find an invariant inner

product 〈 , 〉PD on W and a Hermitian endomorphism T on W , such that

〈x, y〉 = 〈Tx, y〉PD

for all x and y in W . The Hermitian forms 〈 , 〉 and 〈 , 〉PD are both invariant under

K, so T must be a self-intertwining operator on W . It follows in particular that each

eigenspace of T is stable under the action of K. Write

W = W+ ⊕W− ⊕W 0

for the orthogonal decomposition of W induced by the Hermitian operator T , so that

W+ is the direct sum of the positive eigenspaces of T , W− is the direct sum of the

negative eigenspaces of T , and W 0 is the kernel of T . Each of these subspaces is

stable under K, hence it decomposes in a direct sum of copies of µ (which is the

unique K-type contained in W ).

The decomposition W = W+ ⊕ W− ⊕ W 0 clearly has the property that the

form 〈 , 〉 is positive definite on W+, negative definite on W− and zero on W 0. We

must show that if W = W+1 ⊕W−

1 ⊕W 01 has the same property, then the “unicity

conditions” listed above hold.

This is very easy, indeed if n+, n−, n0 is the signature of 〈 , 〉 on W , then:

• dim(W+) = dim(W+1 ) = n+, so

mW+(µ) =dim(W+)

dim(µ)=

dim(W+1 )

dim(µ)= mW+

1(µ)

58

• dim(W−) = dim(W−1 ) = n−, so

mW−(µ) = mW+1(µ) =

n−

dim(µ)

• W 0 = W 01 = Ker(T ).

The proof the lemma is now complete.

Definition 3. Let (π, V ) be an admissible (gC0 , K) module. Suppose that 〈 , 〉 is an

invariant Hermitian form on X, and that µ is a K-type appearing in V . Define the

signature of 〈 , 〉 with respect to µ to be the triple:

pV (µ) =n+

V (µ)

dim(µ)qV (µ) =

n−V (µ)

dim(µ)zV (µ) =

n0V (µ)

dim(µ)

where n+V (µ), n−V (µ), n0

V (µ) is the signature of 〈 , 〉 on the µ- isotypic part of V .

We notice that the triple pV (µ), qV (µ), zV (µ) consists of non negative integers,

and satisfies:

• pV (µ) + qV (µ) + zV (µ) = mV (µ) = with mV (µ) the multiplicity of the K-type

µ in V

• 〈 , 〉 is positive semi-definite if and only if qV (µ) = 0, for all µ.

Next, we give another interpretation for pV (µ), qV (µ), zV (µ). There is a natural

way to define a Hermitian form 〈 , 〉µ on the vector space HomK(Eµ, V ), which is

finite-dimensional, of dimension mV (µ). The signature of 〈 , 〉 with respect to µ

equals the signature of 〈 , 〉µ. Therefore:

pV (µ) = dim. of a maximal positive definite subspace of HomK(Eµ, V )

qV (µ) = dim. of a maximal negative definite subspace of HomK(Eµ, V )

zV (µ) = dim. of the radical of the induced form 〈 , 〉µ on HomK(Eµ, V ).

59

The key fact here is that HomK(Eµ, V ) ' E∗µ ⊗ V . We provide more details:

since µ is an irreducible representation of the compact group K, the finite-dimensional

vector space Eµ admits a positive definite Hermitian form 〈 , 〉Eµ which is invariant

under K. The dual space E∗µ inherits a Hermitian form (also positive definite and

K-invariant) in a natural way: for each F in E∗µ, there exists a unique element vF of

Eµ such that F (u) = 〈u, vF 〉Eµ , for all u in Eµ. Define

〈F1, F2〉E∗µ = 〈vF2 , vF2〉Eµ

for any F1, F2 in E∗µ. The tensor product of 〈 , 〉E∗µ and 〈 , 〉 gives a Hermitian form

( , ) on E∗µ ⊗ V . The form ( , ) is defined on tensors by the formula:

(F1 ⊗ x1, F2 ⊗ x2) = 〈F1, F2〉E∗µ 〈x1, x2〉

and it is extended by linearity on the first component and conjugate linearity on the

second one to obtain a Hermitian form on all E∗µ ⊗ V .

Since 〈 , 〉 and 〈 , 〉Eµ are both invariant under K, so are the forms 〈 , 〉E∗µ on E∗µ,

and ( , ) on E∗µ⊗ V . Then ( , ) descends to a Hermitian form on the space of K-fixed

vectors in E∗µ ⊗ V , which is isomorphic to HomK(Eµ, V ).

We obtain a Hermitian form on HomK(Eµ, V ), which we denote by 〈 , 〉µ. The

next step is to show that 〈 , 〉µ has signature equal to the signature of 〈 , 〉 with

respect to the K-type µ. Write V (µ) for the µ-isotypic part of V , and write V (µ) =

V (µ)+⊕V (µ)−⊕V (µ)0 for a decomposition of V (µ) into the direct sum of a positive

definite subspace, a negative definite subspace and the radical of the restriction of

〈 , 〉 to V (µ), so that

n+V (µ) = dim(V (µ)+) n−V (µ) = dim(V (µ)−) n0

V (µ) = dim(V (µ)0)

60

is the signature of 〈 , 〉 on V (µ). Write

s+(µ) s−(µ) s0(µ)

for the signature of 〈 , 〉µ on HomK(Eµ, V ). We must show that:

s+(µ) =n+

V (µ)

dim(µ)= pV (µ)

s−(µ) =n+

V (µ)

dim(µ)= qV (µ)

s0(µ) =n+

V (µ)

dim(µ)= zV (µ).

Because the decomposition V (µ) = V (µ)+⊕V (µ)−⊕V (µ)0 is stable under the action

of K, we obtain a similar decomposition for HomK(Eµ, V ):

HomK(Eµ, V ) = HomK(Eµ, V (µ)) =

= HomK(Eµ, V (µ)+)⊕ HomK(Eµ, V (µ)−)⊕ HomK(Eµ, V (µ)0).

We notice that

• on HomK(Eµ, V (µ)+), the form 〈 , 〉µ is the tensor product of two positive

definite Hermitian forms, hence it is positive definite;

• on HomK(Eµ, V (µ)−), the form 〈 , 〉µ is the tensor product of a positive definite

Hermitian form and a negative definite Hermitian form, hence it is negative

definite;

• on HomK(Eµ, V (µ)0), the form 〈 , 〉µ is the tensor product of a positive definite

Hermitian form and an identically zero Hermitian form, hence it is identically

zero.

The result follows:

61

s+(µ) = dim(HomK(Eµ, V (µ)+)) = mV (µ)+(µ) = pV (µ)

s−(µ) = dim(HomK(Eµ, V (µ)−)) = mV (µ)−(µ) = qV (µ)

s0(µ) = dim(HomK(Eµ, V (µ)0) = mV (µ)0(µ) = zV (µ).

62

Chapter 6

Weyl Group Representations and

Signatures of Intertwining

Operators.

In this chapter we discuss the unitarity of the Langlands quotient with parameters

(P, δ, ν) for a real semi-simple Lie group G, under the assumptions that:

(i) G is split and P = MAN is a minimal parabolic subgroup of G

(ii) δ is the trivial representation of M and ν is a real character of A.

As shown in chapters 3 and 5, the unitarity of JP (δ ⊗ ν) depends on the signature

of a Hermitian intertwining operator for the principal series, which can be computed

separately on the various K-types. We now show how to determine some of this

signature by means of Weyl group computations.

63

6.1 Spherical Unitary Dual for Real Split Semisim-

ple Lie Groups

Let G be a real split group. Let P = MAN be a minimal parabolic subgroup, so

that M is a finite abelian subgroup of K. Let (δ, V δ) be the trivial representation of

M , and let IndGP (δ⊗ ν) be a spherical principal series representation, with Langlands

quotient X(ν). For simplicity, we assume that ν is real, and of course dominant:

〈ν, α〉 > 0 ∀α ∈ ∆+.

As seen in chapter 3, the irreducible representation X(ν) of G is unitary if and only

if the following two conditions are satisfied:

1. ω0 · ν = −ν, with ω0 the long element of the Weyl group;1

2. the Hermitian intertwining operator

B = δ(ω0)R(ω0)A(P : P : δ : ν) : HP (δ ⊗ ν) → HP (δ ⊗ ω0 · ν)

is positive semi-definite or negative semi-definite.

Since δ is the trivial representation of M , we let δ(ω0) be the identity, and we write

B = R(ω0)A(P : P : δ : ν) = R(ω0)A(ω0Pω−10 : P : δ : ν)

not.= AP (ω0, ν).

It follows from the considerations in chapter 5 that the Hermitian operator B is

positive (or negative) semi-definite if and only if for each K-type (µ, Eµ) that appears

1If gC0 is not of type Dn with n odd, then ω0 is minus the identity, and the condition ω0 · ν = −νbecomes trivial.

64

in the principal series, the induced intertwining operator

Iµ(ω0, ν) : HomK(Eµ, HP (δ ⊗ ν)) → HomK(Eµ, HP (δ ⊗ ω0 · ν))

is positive (or negative) semi-definite. Indeed, the signature of Iµ(ω0, ν) equals (by

definition) the signature of B with respect to the K-type µ, and B is positive (or

negative) semi-definite if and only if it is such on each K-type.

Using Frobenius reciprocity, we can interpret Iµ(ω0, ν) as an endomorphism of

HomM(ResKM Eµ, V δ), which we denote by Rµ(ω0, ν). The operator Rµ(ω0, ν) is

again Hermitian, and its signature determines the unitarity of the principal series.2

The example of SL(2, R) (chapter 4) shows that it is fairly easy to determine the

signature when the real split semi-simple Lie group has rank one. Therefore, we try

to reduce computations for the general case to rank-one calculations. The method

of rank-one reduction has proven to be very effective for the study of intertwining

operators3 and it can be briefly described as follows:

step 1 Find a minimal decomposition of ω0 as a product of simple reflections

ω0 = sαr · · · sα2sα1

(such a decomposition is called minimal if ω0 has length r).

step 2 Find the corresponding decomposition of B=AP (ω0, ν):

AP (ω0, ν) = AP (sαrsαr−1 · · · sα2sα1 , ν) =

2Notice that since δ is the trivial representation of M , the space HomM (ResKM Eµ, V δ) is simply

the space of M -invariants vectors in Eµ.3For instance, it was used by Stein and Knapp in [22, 23] to gain information on the irreducibility

of the principal series.

65

= AP (sαr , (sαr−1 · · · sα2sα1) · ν)AP (sαr−1 , (sαr−2 · · · sα2sα1) · ν) · · ·

· · ·AP (sα2 , sα1 · ν)AP (sα1 , ν).

step 3 When sβ is a simple reflection and γ is a real character of A satisfying 〈γ, β〉 > 0,

interpret the Hermitian operator AP (sβ, γ) as an intertwining operator for a

rank-one subgroup, and compute the corresponding operator Rµ(sβ, γ).

step 4 Find Rµ(ω0, ν) = Rµ(sαrsαr−1 · · · sα2sα1 , ν) =

= Rµ(sαr , (sαr−1 · · · sα2sα1) · ν) Rµ(sαr−1 , (sαr−2 · · · sα2sα1) · ν) · · ·

· · ·Rµ(sα2 , sα1 · ν) Rµ(sα1 , ν).

The key-point in this method is that to each simple root β we can associate a subgroup

Lβ of G with real rank one, so that the the operator Rµ(sβ, ν) (for the spherical

principal series G with parameter ν) can actually be regarded as an operator for a

principal series of Lβ and can therefore be computed using the results already known

for SL(2, R).

We define Lβ to be the group MGβ, with Gβ the analytic subgroup of G whose Lie

algebra is the subalgebra of g0 generated by the weight vectors of weight ±β. The

subgroup

Kβ = exp(RZβ) = exp(R(Eβ + ϑEβ))

is a maximal compact subgroup of Gβ isomorphic to SO(2,R), and it is included in

K. Let

Eµ =⊕

m∈Zϕm

66

be the decomposition of µ in isotypic components of irreducible representations of Kβ:

for each integer m, we have denoted by ϕm the isotypic component of the character

ξm : exp(tZβ) 7→ ei m t inside µ. The decomposition

Eµ =⊕

m∈N(ϕm + ϕ−m)

is invariant under MKβ, which is a maximal compact subgroup of Lβ = MKβ, and

the corresponding decomposition of HomM(ResKM Eµ, V δ) is

⊕

m∈NHomM

(ResMKβ

M (ϕ2m + ϕ−2m), C)

.

The operator Rµ(sβ, γ) preserves this decomposition, and acts on each of these sub-

spaces as scalar multiplication. We normalize Rµ(sβ, γ) so that it acts trivially

on HomM(ResMKβ

M ϕ0, C). The action on HomM

(ResMKβ

M (ϕ2m + ϕ−2m), C)

is then

given by the scalar

(1− 〈γ, β〉)(3− 〈γ, β〉) · · · (2m− 1− 〈γ, β〉)(1 + 〈γ, β〉)(3 + 〈γ, β〉) · · · (2m− 1 + 〈γ, β〉) .

for all non negative integers m.

Remark 3. Rµ(ω0, ν) can be decomposed as a product of operators corresponding to

simple reflections sβ, and for each of these operators an explicit formula exists. This

formula depends on the decomposition of Eµ in irreducible Kβ-types.

Although each Kβ is isomorphic to SO(2), the decomposition changes when β varies.

It is very hard to keep track of these different decompositions when you multiply the

various rank-one operators to obtain Rµ(ω0, ν).

In this section we have intentionally omitted the details, so to convey the general

idea in a more effective fashion. Please refer to the following subsections for all the

67

details.

6.1.1 The rank-one subgroup Gβ attached to a (simple) root

In this subsection we construct a subgroup Gβ of G of real rank one, for each root β.

This construction is general, and does not actually require the semi-simple group G

be split nor β be simple.

Without loss of generality we can assume that the restricted root β is reduced,

i.e. that 12β is not a root. We define Gβ to be the analytic subgroup of G whose Lie

algebra is the the subalgebra of g0 generated by the root vectors of weights ±β and

±2β. Being ϑ-stable, the Lie algebra gβ0 is reductive (actually semi-simple), and the

restriction of ϑ to gβ0 gives a Cartan involution on gβ

0 .

If g0 = k0⊕p0 is the Cartan decomposition of g0, we set kβ0 = k0∩gβ

0 and pβ0 = p0∩gβ

0 , so

that gβ0 = kβ

0⊕pβ0 is the Cartan decomposition of gβ

0 . The subspace aβ0 = a0∩gβ

0 = RHβ

is a maximal abelian subspace of pβ0 . The restricted root space decomposition of gβ

0

with respect to aβ0 is given by:

gβ0 = aβ

0 ⊕mβ0 ⊕ [(g0)β ⊕ (g0)−β ⊕ (g0)2β ⊕ (g0)−2β]

with mβ0 = m0 ∩ gβ

0 = Zkβ0(aβ

0 ). If we choose β and 2β to be positive, then nβ0 =

(g0)β ⊕ (g0)2β and

nβ0 = ϑ(nβ

0 ) = (g0)−β⊕ (g0)−2β. The Iwasawa decomposition of Gβ is compatible with

the one of G:

Gβ = KβAβNβ = (K ∩Gβ)(A ∩Gβ)(N ∩Gβ).

It follows that the Iwasawa decomposition of an element x of Gβ = KβAβNβ is the

same as the Iwasawa decomposition of x regarded as an element of G = KAN . In

particular, the function Hβ (which maps x into the log of the Aβ-part of x) is just

the restriction to Gβ of the H function on G.

68

We now compare the “ρ” functions:

ρ = 12

∑α∈∆+ dim((g0)α) α : a0 → R

ρβ = 12[dim((g0)β) β + 2 dim((g0)2β) β] : aβ

0 = RHβ → R.

In particular, we ask whether ρβ equals the restriction of ρ to aβ0 :

ρ(Hβ)?= ρβ(Hβ) =

1

2[ dim((g0)β) + 2 dim((g0)2β) ] · ‖β‖2.

The answer to this question is in general negative. It is instead affirmative when the

root β is simple. Indeed, if β is simple, the reflection sβ carries each multiple of β

into its opposite and permutes the other positive roots. The equation

sβ(ρ) = ρ− [ dim((g0)β) + 2 dim((g0)2β) ] β

implies that [ dim((g0)β) + 2 dim((g0)2β) ] = 2 〈β, ρ〉‖β‖2 . Hence

ρ(Hβ) = 〈β, ρ〉 =1

2[ dim((g0)β) + 2 dim((g0)2β) ] · ‖β‖2.

This shows that ρ |aβ0= ρβ, for each simple root β.

6.1.2 The operator Rµ(ω0, ν)

In this subsection, we give an explicit construction for the operator Rµ(ω0, ν).

The restriction of a principal series representation IndGP=MAN(δ ⊗ ν) of G to the

maximal compact subgroup K is isomorphic to the induced representation IndKM(δ).

The linear isomorphism

T1 : HP (δ ⊗ ν) → Hδ, F 7→ f = F |K

69

intertwines the action of K on the two spaces and it is actually an isometry. The

inverse mapping T2 : Hδ → HP (δ ⊗ ν) is defined by

F = T2(f) : G → V δ, g = (kan) 7→ e−(ρ+ν) log(a)f(k)

for each f : K → V δ in Hδ.4 Consider the composition

B : HδT2→ HP (δ ⊗ ν)

B→ HP (δ ⊗ ω0 · ν)T1→ Hδ

where B = δ(ω0)R(ω0)A(P = ω0Pω−10 : P : δ : ν) is the intertwining operator intro-

duced in chapter 3. When δ is the trivial representation of M , we can take δ(ω0) to

be the identity, and for each F in HP (δ ⊗ ν) we have:

B(F ) : G → V δ, g 7→∫

N

F (xω0n) dn.

The operator B = T1BT2 : Hδ → Hδ is a self-intertwining operator for the represen-

tation IndKM(δ) of K. It is convenient to write down B more explicitly. If f belongs

to Hδ, then

Bf : K → V δ, k 7→∫

N

T2(f)(kω0n) dn =

∫

N

e−(ρ+ν) log(a(n))f(kω0κ(n)) dn.5

Let (µ, Eµ) be a K type appearing in the principal series, and consider the homo-

morphism

Hom(Eµ,Hδ) → Hom(Eµ,Hδ)

induced from B by composition on the range. Since B is an intertwining operator,

4T2(f) is the unique extension of f to G that has the suitable transformation properties underright multiplication by elements of P .

5We have denoted by κ(n)a(n)n(n) the Iwasawa deconposition of an element n of N . Then(kω0κ(n))a(n)n(n) is the corresponding Iwasawa decomposition of kω0n.

70

the subspace of K-fixed vectors is preserved6, so we obtain a homomorphism

Iµ(ω0, ν) : HomK(Eµ,Hδ) → HomK(Eµ,Hδ).

By Frobenius reciprocity, we can interpret Iµ(ω0, ν) as an endomorphism Rµ(ω0, ν)

of HomM(ResKM Eµ, V δ) = (EM

µ )∗. Indeed Hδ = IndKM(Vδ) is the representation space

of the induced representation IndKM δ and Frobenius reciprocity states that the space

HomK(Eµ, IndKM(V δ)) is isomorphic to HomM(ResK

M Eµ, V δ). Let us explain how one

gets this isomorphism.

Let T belong to HomK(Eµ, IndKM(V δ)). For each v in Eµ, T (x) is a function from K

to V δ; in particular, T (v)(1) lives in V δ. Define L2(T ) to be the linear mapping

L2(T ) : Eµ → V δ, v 7→ T (v)(1).

Claim 1. L2(T ) is a fixed point for the action of M on Hom(ResKM Eµ, V δ).

Proof. For all m in M

(m · L(T ))(v) = δ(m)(L2(T )(µ(m−1) · v)) =

= δ(m)(T (µ(m−1) · v)(1)) = T (v)(1) = L2(T )(v)

because T belongs to HomK(Eµ, IndKM V δ).

We obtain a linear mapping

L2 : HomK(Eµ, IndKM(V δ)) → HomM(ResK

M Eµ, V δ).

6The action of K on Hom(Eµ,Hδ) is given by

(k · T )(v) = IndKM (δ)(k) · (T (µ(k−1 · v))

for all v in Eµ, T in Hom(Eµ,Hδ) and k in K.

71

Its inverse L1 is defined as follows: let S : Eµ → V δ be an element of HomM(ResKM Eµ, V δ).

For each v in Eµ consider the function

L1(S)(v) : K → V δ, k 7→ L1(S)(v)(k) = S(µ(k−1) · v).

Claim 2. L1(S)(v) belongs to Hδ = IndKM V δ.

Proof. For all m in M

L1(S)(v)(km) = S(µ(m−1)µ(k−1) · v) = δ(m−1)[δ(m)S(µ(m−1)µ(k−1) · v)] =

= δ(m−1)S(µ(k−1) · v) = δ(m−1)L1(S)(v)(k).

because S is a fixed point for the action of M on Hom(ResKM Emu, V δ).

The result is a well defined linear mapping

L1(S) : Eµ → IndKM V δ, v 7→ L1(S)(v).

Claim 3. L1(S) belongs to HomK(Eµ, IndKM V δ), i.e.

(IndKM δ)(k)(L1(S)(µ(k−1) · v)) = L1(S)(v)

for each k in K and v in Eµ.

Proof. This is easy, because for any x in K

(IndKM δ)(k)(L1(S)(µ(k−1) · v))(x) = L1(S)(µ(k−1) · v)(k−1x) =

= S(µ(k−1x)−1µ(k−1) · v) = S(µ(x−1) · v) = L1(S)(v)(x).

72

We obtain a linear mapping

L1 : HomM(ResKM Eµ, V δ) → HomK(Eµ, IndK

M V δ)

which can be easily checked to be the inverse of L2.

Now that the Frobenius reciprocity is understood, we go back to the construction

of the operator Rµ(ω0, ν). We define Rµ(ω0, ν) to be the composition

HomM(ResKM Eµ, V δ)

L2−→ HomK(Eµ, IndKM V δ)

Iµ(ω0, δ)−→

Iµ(ω0, δ)−→ HomK(Eµ, IndKM V δ)

L1−→ HomM(ResKM Eµ, V δ).

More explicitly, the image of an element T of HomM(ResKM Eµ, V δ) under Rµ(ω0, ν)

is the function

Rµ(ω0, ν)T : Eµ → V δ, v 7→∫

N=N∩(ω0Nω−10 )

e−(ρ+ν) log(a(n))T ((ω0κ(n))−1 · v) dn.

6.1.3 The operator Rµ(sβ, γ)

In this subsection, we compute the operator Rµ(sβ, γ) under the assumptions that

the semi-simple Lie group is split and the root β is simple. These assumptions are

crucial. We assume γ to be a real character of A satisfying 〈γ, β〉 > 0.

An argument similar to the one used in the previous subsection shows that the op-

erator Rµ(sβ, γ) is the endomorphism of HomM(ResKM Eµ, V δ) that carries an element

T into the mapping

Rµ(sβ, γ)T : Eµ → V δ, v 7→∫

N∩(σβNσ−1β )

e−(ρ+γ) log(a(n))T ((σβκ(n))−1 · v) dn,

73

with σβ a representative in M ′ for the root reflection sβ.

Since β is simple, using results described in subsection 6.1.1 we can write:

(Rµ(sβ, γ)T )(v) =∫

Nβ e−(ρ+γ)(H(n))T ((σβκ(n))−1 · v) dn =

=∫

Nβ e−(ρβ+γ|

aβ0

)(Hβ(n))T ((σβκβ(n))−1 · v) dn

and this is exactly the value at T of the “R-operator” (with parameters µ, sβ and

γ |aβ0) for the rank-one subgroup Lβ = MGβ of G.

Notice that, as a representation of MKβ (which is a maximal compact subgroup of

MGβ), the representation µ is reducible. Let us discuss its decomposition in irre-

ducible summands.

When G is split, each restricted root is reduced and each root space is one-

dimensional. In particular, the Lie algebra of Gβ

gβ0 = aβ

0 ⊕ (g0)β ⊕ (g0)−β = RHβ + REβ + Rϑ(Eβ)

is isomorphic to sl(2, R). We have denoted by Eβ a non-zero element of (g0)β that

satisfies the normalizing condition B(Eβ, ϑ(Eβ)) = − 2‖β‖2 , with B the Killing form.

Eβ is a generator for the root space of weight β and is uniquely determined up to a

sign. The element Zβ = Eβ + ϑ(Eβ) is a generator for kβ0 , so

Kβ = exp(RZβ) ' SO(2, R)

is a maximal compact subgroup of Gβ. The group MKβ is a maximal compact

subgroup of Lβ = MGβ, and its structure is described by the following lemma:

Lemma 3. Let β(m) = ±1 denote the scalar by which an element m of M acts on

the root vector Eβ. Then

74

(a) β(m) = (−β)(m)

(b) Ad(m)Zβ = β(m)Zβ

(c) mσβm−1 = σβ(m)β .

Proof. We first show that M acts on the vector Eβ by a scalar. Since the root space

(g0)β is one-dimensional, it is enough to show that Ad(m)(Eβ) belongs to (g0)β, for

each m in M . This is easy, because

[H, Ad(m)(Eβ)] = Ad(m)[Ad(m−1)H, Eβ] = Ad(m)[H, Eβ] = β(H) Ad(m)(Eβ)

for all H in a0. For each m in M , define β(m) by the equation Ad(m)(Eβ) = β(m)Eβ.

Since Ad(m) commutes with the Cartan involution7, we easily deduce that (−β)(m) =

β(m), for all m:

Ad(m)(E−β) = Ad(m)(ϑ(Eβ)) = ϑ(Ad(m)(Eβ)) = β(m)E−β.

It follows that

B(Eβ, ϑ(Eβ)) = B(Ad(m)(Eβ), Ad(m)(ϑ(Eβ))) = (β(m))2B(Eβ, ϑ(Eβ))

and, being B(Eβ, ϑ(Eβ)) = − 2‖β‖2 6= 0 (by definition of Eβ), we conclude that

(β(m))2 = 1, so the function m 7→ β(m) only takes the values +1 or −1 on M .8

Proving that Ad(m)Zβ = β(m)Zβ is trivial, because Zβ = Eβ + E−β and (−β)(m) =

7Let Θ be the global Cartan involution. Being Θ an involutive automorphism of G which fixesK (hence M), we have

mxm−1 = Θ(mΘ(x)m−1) ∀x ∈ G∀m ∈ M.

Differentiating at x = 1 we find that Ad(m) = ϑ Ad(m)ϑ, for all m in M . The results follows fromthe fact that also ϑ is an involution.

8Using the fact that being G split each restricted root is the restriction to a0 of one (and onlyone) root in ∆(gC0 , aC0 ), and some standard results from the representation theory of SL(2,C), youcan prove that if m = mα = exp(πZα) then β(mα) = (−1)

2‖α‖2 〈α, β〉 = (−1)〈α, β〉.

75

β(m). Finally, setting σβ = exp(π2Zβ) and exponentiating, we find:

mσβm−1 = σβ(m)β = σ±1

β .

and this concludes the proof.

More generally, m exp(tZβ) m−1 = exp(tZβ)β(m), so the structure of the group

MKβ is understood.

We now go back to the problem of discussing the decomposition of the (irreducible)

representation µ of K into isotypic components of irreducible representations of the

subgroup MKβ. First, we decompose µ under the action of Kβ ' SO(2, R), obtain-

ing:

Eµ =⊕

l∈Zϕl

where, for each integer l, we have denoted by ϕl the isotypic component of the char-

acter ξl of SO(2). Then we notice that for all t in R and all v in ϕl

exp(tZβ) · (m · v) = m · [(m−1 exp(tZβ)m) · v] = m · [exp(tβ(m−1)Zβ) · v] =

= m · [exp(tβ(m−1)Zβ) · v] = ξl(exp(tβ(m)Zβ)) (m · v) = ξβ(m)l(exp(tZβ)) (m · v)

hence m · ϕl = ϕβ(m)l = ϕ± l. The decomposition

Eµ =⊕

l∈N(ϕl + ϕ−l)

is clearly invariant under MKβ, and it coincides with the decomposition of µ in MKβ-

types.9 Because each subspace (ϕl+ϕ−l) is M-stable and δ is the trivial representation,

9Unless, of course, M stabilizes Zβ , so each ϕl is already stable.

76

we can write:


⊕

l∈NHomM

(ResMKβ

M (ϕl + ϕ−l), C)

When l is odd, the space HomM(ResMKβ

M (ϕl + ϕ−l), C) is zero. Indeed, for each

M -fixed linear map T : ϕl ⊕ ϕ−l → C, and for each v in ϕl ⊕ ϕ−l, we must have:

T (v) = (mβ · T )(v) = T (m−1β · v) = T (−v) = −T (v).

So we can write: HomM(ResKM Eµ, V δ) =

⊕l∈2NHomM

(ResMKβ

M (ϕl + ϕ−l), C)

=

= HomM(ResMKβ

M ϕ0, C)⊕

l∈2N∗HomM

(ResMKβ

M (ϕl ⊕ ϕ−l), C)

.

We will compute the operator Rµ(sβ, γ) separately on each component.

If T belongs to HomM

(ResMKβ

M ϕ0, C)

and v is in ϕ0, then

(Rµ(sβ, γ)T )(v) =∫

Nβ e−(ρβ+γ|

aβ0

)(Hβ(n))T ((σβκβ(n))−1 · v) dn =

=∫

Nβ e−(ρβ+γ|

aβ0

)(Hβ(n))T (v) dn =

=

[∫Nβ e

−(ρβ+γ|aβ0

)(Hβ(n))dn

]T (v).

If T belongs to HomM

(ResMKβ

M (ϕl ⊕ ϕ−l), C)

and v = (v++v−) is in (ϕl⊕ϕ−l), then

(Rµ(sβ, γ)T )(v) =∫

Nβ e−(ρβ+γ|

aβ0

)(Hβ(n))T ((σβκβ(n))−1 · v) dn =

=∫

Nβ e−(ρβ+γ|

aβ0

)(Hβ(n))T

(ξ+l(σβκβ(n))−1v+ + ξ−l(σβκβ(n)

)−1v−) dn =

77

=

[∫Nβ e

−(ρβ+γ|aβ0

)(Hβ(n))ξ+l(σβκβ(n))−1dn

]T (v+) +

+

[∫Nβ e

−(ρβ+γ|aβ0

)(Hβ(n))ξ−l(σβκβ(n))−1dn

]T (v−).

To conclude the computations we need one final observation. Consider the Lie algebra

isomorphism

ψβ : sl(2, R) → gβ0 = RHβ + REβ + Rϑ(Eβ)

defined by:

e =

0 1

0 0

7→ Eβ, f =

0 0

1 0

7→ −ϑ(Eβ), h =

1 0

0 −1

7→ +2

‖β‖2Hβ.

When G has a complexification10, ψβ lifts to a group homomorphism

Ψβ : SL(2, R) → Gβ.

It follows from the formula 4.1 for the Iwasawa decomposition in SL(2, R) that

10This is always the case if G is semi-simple. Indeed every adjoint group has a complexification:if G = Ad(g0), you can take GC to be Ad(gC0 ).It also true, more generally, if the group G is real reductive and satisfies the condition

Z(G) ∩K = 1.

Indeed, if G = K exp(p0) is the Cartan decomposition of G, and ζ is the center of the Lie algebraof G (so that g0 = [g0, g0]⊕ ζ), then we can write

G = K exp(p0 ∩ [g0, g0])︸︷︷︸G1

exp(p0 ∩ ζ)︸︷︷︸Z1

. (>)

with G1 real reductive (of the same rank as G) and Z1 a vector group included in the center.Because Z(G1) = Z(G)∩K = 1, the group G1 is actually semi-simple. So (>) is a decomposition ofG as a direct product of an adjoint group and a vector group, both of which have a complexification.As a result, we obtain a complexification for G.Finally, we notice that Z(G) acts by scalars on any irreducible representation of G (this is Schur’slemma), and that Z(G)∩K acts trivially on the trivial K-type included in any irreducible sphericalrepresentation ρ of G (hence on the whole representation space Eρ). So, when dealing with sphericalrepresentations, we can assume w.l.o.g. that the condition Z(G) ∩K = 1 is satisfied.

78

n = exp(tϑ(Eβ) = exp(−tψβ(f)) = Ψβ(exp(−t f)) = Ψβ

1 0

−t 1

=

= Ψβ

cos(arctan(t)) sin(arctan(t))

− sin(arctan(t)) cos(arctan(t))

√

1 + t2 0

0 1/√

1 + t2

1 x

0 1

=

= Ψβ

(exp(arctan(t) (e− f)

)Ψβ

(exp(ln(

√1 + t2) h)

)Ψβ (exp(x e)) =

= exp(ψβ(arctan(t) (e− f))

)exp

(ψβ(ln(

√1 + t2) h)

)exp (ψβ(x e)) =

= exp (arctan(t) Zβ) exp(ln(√

1 + t2) 2‖β‖2 Hβ

)exp (xEβ) .

This is the Iwasawa decomposition of an arbitrary element n = exp(tϑ(Eα)) of

Nβ. We notice, in particular, that Hβ(n) = ln(√

1 + t2)

2‖β‖2 Hβ and that κβ(n) =

exp (arctan(t) Zβ). Therefore

• ρβ(Hβ(n)) = 12β

(ln(√

1 + t2) 2‖β‖2 Hβ

)= ln(

√1 + t2)

• γ |aβ0

(Hβ(n)) = γ(ln(√

1 + t2) 2‖β‖2 Hβ

)= ln(

√1 + t2) 〈γ, 2

‖β‖2 β〉 =

= ln(√

1 + t2) 〈γ, β〉

• ξ2l(σβκβ(n))−1 = ξ2l

(exp

(−(π2

+ arctan(t))Zβ

))= (−1)le−2l i [arctan(t)] =

= (−1)l(

1+i t√1+t2

)−2l

, for all l in Z.

79

We can now compute the action of Rµ(sβ, γ) on the generic element T of

HomM(ResMKβ

M ϕ0, C):

Rµ(sβ, γ)T =

[∫

Nβ

e−(ρβ+γ|

aβ0

)(Hβ(n))dn

]T =

[∫

R(√

1 + t2)−(1+〈γ, β〉)dt

]T.

We have already computed this integral in chapter 4, its value being the scalar by

which the standard intertwining operator Aω of SL(2, R) acts on the trivial K-type

of the (minimal) spherical principal series with parameter λ = 〈γ, β〉:11

∫

R(√

1 + t2)−(1+〈γ, β〉)dt =π Γ(〈γ, β〉)

2〈γ, β〉−1 Γ(〈γ, β〉+1

2

)Γ

(〈γ, β〉+1

2

) .

We will denote this constant by C〈γ, β〉. For each positive integer l, we now compute

(Rµ(sβ, γ)T )(v+), for v+ in ϕ2l and T in HomM

(ResMKβ

M (ϕ2l ⊕ ϕ−2l), C):

(Rµ(sβ, γ)T )(v+) =

[∫Nβ e

−(ρβ+γ|aβ0

)(Hβ(n))ξ+2l(σβκβ(n))−1dn

]T (v+) =

=

[(−1)l

∫R(√

1 + t2)−(1+〈γ, β〉)(

1+i t√1+t2

)−2l

dt

]T (v+).

After a simple change of variable (t 7→ −t), we can identify the integral

[(−1)l

∫

R(√

1 + t2)−(1+〈γ, β〉)(

1 + i t√1 + t2

)−2l

dt

]

with the one (already computed in chapter 4) that defines the scalar by which the

standard intertwining operator Aω of SL(2, R) acts on the “+2l” K-type of the

(minimal) spherical principal series with parameter λ = 〈γ, β〉.12 The value of this

11Compare this integral with the one appearing in formula 4.2 of chapter 4, for m = 0.12Compare this integral with the one appearing in formula 4.2 of chapter 4, for m = −l.

80

integral is

C〈γ, β〉 ·(1− 〈γ, β〉)(3− 〈γ, β〉) · · · (2l − 1− 〈γ, β〉)(1 + 〈γ, β〉)(3 + 〈γ, β〉) · · · (2l − 1 + 〈γ, β〉) .

For brevity of notation, denote this quantity by C〈γ, β〉d2l(〈γ, β〉).Similarly, for v− in ϕ−2l, we find:

(Rµ(sβ, γ)T )(v−) =

[∫Nβ e

−(ρβ+γ|aβ0

)(Hβ(n))ξ−2l(σβκβ(n))−1dn

]T (v−)=

=

[(−1)l

∫R(√

1 + t2)−(1+〈γ, β〉)(

1+i t√1+t2

)+2l

dt

]T (v−) =

=[C〈γ, β〉d2l(〈γ, β〉)

]T (v−).

It follows that (Rµ(sβ, γ)T ) =[C〈γ, β〉d2l(〈γ, β〉)

]T (v), for any positive integer l,

for all v = v+ + v− in (ϕl ⊕ ϕ−l) and all T in HomM

(ResMKβ

M (ϕl ⊕ ϕ−l), C).

Conclusions. For each simple root β, and for each real character γ of A such that

〈γ, β〉 > 0, the operator

Rµ(sβ, γ) : HomM(ResKM Eµ, V δ) → HomM(ResK

M Eµ, V δ)

preserves the decomposition of HomM(ResKM Eµ, V δ) in MKβ-stable subspaces:

HomM(ResMKβ

M ϕ0, C)⊕

l∈N∗HomM

(ResMKβ

M (ϕ2l ⊕ ϕ−2l), C)

.

Precisely, Rµ(sβ, γ) acts on HomM(ResMKβ

M ϕ0, C) as scalar multiplication by

C〈γ, β〉 =π Γ(〈γ, β〉)

2〈γ, β〉−1 Γ(〈γ, β〉+1

2

)Γ

(〈γ, β〉+1

2

)

81

and on each subspace HomM

(ResMKβ

M (ϕ2l ⊕ ϕ−2l), C), with l in N∗, as scalar mul-

tiplication by

C〈γ, β〉d2l(〈γ, β〉) = C〈γ, β〉 ·(1− 〈γ, β〉)(3− 〈γ, β〉) · · · (2l − 1− 〈γ, β〉)(1 + 〈γ, β〉)(3 + 〈γ, β〉) · · · (2l − 1 + 〈γ, β〉) .

6.1.4 Final considerations

By assumption γ is a real dominant character of A satisfying 〈γ, β〉 > 0. Be-

ing 〈γ, β〉 real and positive, also the constant C〈γ, β〉 is real and positive. Write

Rµ(sβ, γ) = 1C〈γ, β〉

Rµ(sβ, γ) for the normalized operator. Rµ(sβ, γ) acts trivially on

HomM(ResMKβ

M ϕ0, C), and acts on each subspace HomM

(ResMKβ

M (ϕ2m ⊕ ϕ−2m), C)

-with m ∈ N∗- as scalar multiplication by

∏mj=1((2j − 1)− 〈γ, β〉)∏mj=1((2j − 1) + 〈γ, β〉) .

Collecting all the normalizing factors, we can write Rµ(ω0, ν) as a real positive mul-

tiple of a product of Rµ-operators corresponding to simple reflections, and because

multiplication by a real positive constant does not affect the signature, we can as well

assume that this constant is one.

The following theorem summarizes all the information we gathered so far about

the spherical unitary dual of a split semi-simple Lie group.

Theorem 9. Let G be a real split semi-simple Lie group, let P = MAN be a minimal

parabolic subgroup of G and let ω0 be the long element of the Weyl group. For each

real dominant character ν of A satisfying the parity condition ω0 · ν = −ν, denote

by X(ν) the Langlands quotient of G with parameters (P, δ, ν), with δ the trivial

representation of M . Then, X(ν) is unitary if and only if for each K-type µ a certain

Hermitian operator Rµ(ω0, ν) of HomM(ResKM Eµ, V δ) is positive semi-definite. The

operator Rµ(ω0, ν) can be constructed as follows:

82

• Find a minimal decomposition of ω0 as a product of simple reflections:

ω0 = sαr · · · sα2sα1

• Write Rµ(ω0, ν) = Rµ(sαrsαr−1 · · · sα2sα1 , ν) =

= Rµ(sαr , (sαr−1 · · · sα2sα1) · ν)Rµ(sαr−1 , (sαr−2 · · · sα2sα1) · ν) · · ·

· · · Rµ(sα2 , sα1 · ν)Rµ(sα1 , ν)

and observe that each factor is an operator of the form Rµ(sβ, γ), with sβ a

simple reflection and γ a real character of A satisfying 〈γ, β〉 > 0.

• For β and γ as above, define Rµ(sβ, γ) as follows: for each integer l, let ϕl be

the isotypic component of the character ξl : exp(tZβ) 7→ ei l t of Kβ ' SO(2)

inside µ, so that

Eµ =⊕

l∈Zϕl

is the decomposition of µ in Kβ-types, and


⊕

m∈NHomM

(ResMKβ

M (ϕ2m + ϕ−2m), C)

is a decomposition of HomM(ResKM Eµ, V δ) in MKβ-stable subspaces. Define

the operator Rµ(sβ, γ) to to act on HomM

(ResMKβ

M (ϕ2m + ϕ−2m), C)

as scalar

multiplication by ∏mj=1((2j − 1)− 〈γ, β〉)∏mj=1((2j − 1) + 〈γ, β〉) .

for all non-negative integers m.

We emphasize one more time that this construction is less trivial than it looks like,

because the definition of Rµ(sβ, γ) depends on the decomposition of µ in Kβ-types,

and this decomposition depends on the choice of the simple root β. So to compute

Rµ(ω0, ν) one needs to keep track of the many different decompositions. Another

difficulty consists of the fact that we are required to compute the signature of the

83

operator Rµ(ω0, ν) on all the K-types µ that appear in the spherical principal series,

whose number is in general infinite.

In the next section we give an alternative method to compute the operator Rµ(ω0, ν),

which uses only Weyl group calculations.

6.2 Spherical representations and the Weyl group

We start by recalling the formal construction of the Weyl group.

Let V be a finite-dimensional real vector space, with inner product 〈 , 〉 and norm

squared ‖ ‖2. A root system R is a finite set of non-zero elements of V satisfying the

following conditions:

(i) R spans V ,

(ii) for each α in R, the reflection sα : V → V, γ 7→ γ − 2〈γ, α〉‖α‖2 α is an orthogonal

transformation of V that carries R into itself,

(iii) 2〈α, β〉‖α‖2 is an integer, for all α and β in R.

We say that an element α of R is a root, and that sα is a root reflection. We define

the Weyl group W of the root system (R, V ) to be the subgroup of O(V ) generated

by the the root reflections.

By construction, the (real) group algebra R[W ] of W lies in the algebra of all endo-

morphisms of V . We introduce a notion of positivity in V , and we call a root simple

if it is positive and cannot be expressed as a sum of two positive roots. The simple

roots form a basis for the root system, and the corresponding reflections (that we call

“simple reflections”) generate the Weyl group.

W contains a unique element ω0 (of order two) that carries R+ into R−. We call

ω0 the long Weyl group element, because a minimal decomposition of ω0 as a product

84

of simple reflections has length N = #R+.13

An element ν of V is dominant if, for any root β in R, 〈ν, β〉 ≡ 〈ν, 2β‖β‖2 〉 > 0.

Definition 4. For any simple root α in R, and for any element γ in V satisfying

〈γ, β〉 > 0, define

Asα(γ) =1

1 + 〈γ, α〉 e +〈γ, α〉

1 + 〈γ, α〉 sα ∈ R[W ]

where e is the identity of W .

Definition 5. Let ν be a dominant element of V , and let

ω0 = sαNsαN−1

· · · sα1

be a minimal decomposition of the long Weyl group element ω0 as a product of simple

reflections. Define

A(ν) = AsαN((sαN−1

sαN−2· · · sα1)(ν)) · AsαN−1

((sαN−2· · · sα1)(ν)) · · ·

· · ·Asα2(sα1(ν)) · Asα1

(ν).

A(ν) is also an element of the group algebra R[W ]. The following lemma describes

its properties:

Lemma 4. For all dominant ν in V

1. A(ν) is a rational function of ν

2. A(0) = e

13Each element x of W can be written as a product of simple reflections. This decompositionis called minimal if it involves the least possible number of reflections; the length of a minimaldecomposition of x is defined to be the length of x. It can be shown that the length of x equals thenumber of positive roots that are mapped by x into negative roots. Therefore the length of x is lessor equal than the cardinality of R+, for all x in W .

85

3. limν→∞ A(ν) = ω0

4. A(ν) is invertible in R[W ] provided that 〈ν, α〉 6= 1, for all α in R+

(the inverse of Asα(ξ) is Asα(−ξ) = 11−〈ν, α〉e− 〈ν, α〉

1−〈ν, α〉sα)

5. A(ν) is self-adjoint if ω0(ν) = −ν

6. A(ν) is independent of the choice of the reduced decomposition for ω0.

If (ψ, Eψ) is any representation of the Weyl group W , ψ(x) ∈ Aut(Eψ) for all x

in W and ψ(y) ∈ End(Eψ) for all y in R[W ]. In particular,

ψ(Asα(γ)) =Id + 〈γ, α〉ψ(sα)

1 + 〈γ, α〉

is a well defined endomorphism of Eψ, for all α and γ in V such that 〈γ, α〉 > 0, with

α simple. We notice that ψ(Asα(γ)) is the identity on the (+1)-eigenspace of ψ(sα),

and acts as scalar multiplication by 1−〈γ, α〉1+〈γ, α〉 on the (−1)-eigenspace of ψ(sα).

Because the operator ψ(sα) is unitary of order two, Eψ is the orthogonal direct sum

of the (+1) and (−1) eigenspaces of ψ(sα), so we can equivalently define ψ(Asα(γ))

by:

ψ(Asα(γ))v =

v for v in the (+1)-eigenspace of ψ(sα)

1−〈ξ, α〉1+〈ξ, α〉v for v in the (−1)-eigenspace of ψ(sα).

Similarly we can consider the endomorphism ψ(A(ν)), for ν real and dominant, which

-by definition of A(ν)- admits a decomposition

ψ(A(ν)) = ψ(AsαN((sαN−1

sαN−2· · · sα1)(ν))) · ψ(AsαN−1

((sαN−2· · · sα1)(ν))) · · ·

· · ·ψ(Asα2(sα1(ν))) · ψ(Asα1

(ν)).

ψ(A(ν)) is well defined for each ν dominant, it is independent of the choice of the

reduced decomposition for ω0 and it is Hermitian whenever ω0 · ν = −ν.

86

Theorem 10 (Barbasch-Moy [7]). If G is a split p-adic group, the spherical irre-

ducible representation X(ν) is unitary if and only if the following two conditions are

satisfied:

(i) ω0 · ν = −ν;

(ii) for each irreducible Weyl group representation ψ, the Hermitian operator ψ(A(ν))

is positive-semidefinite.

In the real case, the Hermitian operator ψ(A(ν)) does not generally detect uni-

tarity.14 Still, it can be used to create a “non-unitarity test”. To formulate this idea

we need a preliminary definition.

Definition 6. An irreducible representation (µ, Eµ) of K is called petite if for ev-

ery simple root β, the representation µ |Kβ of Kβ ' SO(2) does only contain the

characters 0, ±1, ±2.

Let us look at the operator Rµ(sβ, ξ) when (µ, Eµ) is a petite representation of

K, β is a simple root, and γ is a real character of A satisfying 〈γ, β〉 > 0.

Decomposing the domain of Rµ(sβ, γ) in MKβ-stable subspaces, we obtain:


= HomM(ResMKβ

M ϕ0, C)⊕ HomM

(ResMKβ

M (ϕ2 ⊕ ϕ−2), C)

(there are no other components because µ is petite). The operator Rµ(ω0, γ) acts

trivially on HomM(ResMKβ

M ϕ0, C) and it acts as scalar multiplication by 1−〈γ, β〉1+〈γ, β〉 on

HomM

(ResMKβ

M (ϕ2 ⊕ ϕ−2), C).

The element σβ = exp(

π2Zβ

)is a representative in M

′for the root reflection sβ.

Clearly, σβ acts by (+1) on ϕ0, and by (−1) on (ϕ2⊕ϕ−2). It easily follows that it acts

14Unless G is a classical group. See [5]

87

by (+1) on HomM(ResMKβ

M ϕ0, C), and by (−1) on HomM

(ResMKβ

M (ϕ2 ⊕ ϕ−2), C).

Hence, if we regard HomM(ResKM Eµ, V δ) as a representation ψµ of the Weyl group

(which might be reducible), then we can identify HomM(ResMKβ

M ϕ0, C) with the (+1)-

eigenspace of ψµ(sβ), and HomM

(ResMKβ

M (ϕ2 ⊕ ϕ−2), C)

with the (−1)-eigenspace

of ψµ(sβ). We obtain that:

Rµ(sβ, γ) =

+1 on the (+1)-eigenspace of ψµ(sβ)

1−〈γ, β〉1+〈γ, β〉 on the (−1)-eigenspace of ψµ(sβ).

Therefore, Rµ(sβ, γ) = ψµ(Asβ(γ)) , for each simple root β, and for each real char-

acter γ of A satisfying 〈γ, β〉 > 0.

It follows that for every dominant real character ν, the full intertwining operator

Rµ(ω0, ν) coincides with the p-adic operator ψµ(A(ν)):

Rµ(ω0, ν) = ψµ(A(ν))

(because of the common pattern in the factorization as a product of operators relative

to simple reflections). We have proved the following theorem:

Theorem 11. Suppose G is a real split group, and (µ, Eµ) is a petite irreducible

representation of K. Let ψµ be the representation of the Weyl group W on the space

of M-fixed vectors of µ. Then Rµ(ω0, ν) = ψµ(A(ν)), for all real dominant characters

ν of A.

Corollary 1. Let G be a real split group, let ν be a real dominant characters of A

and let (µ, Eµ) be a petite K-type of the spherical Langlands quotient X(ν). The

Hermitian intertwining operator on the K-type µ can be computed by means of Weyl

group calculations.

Corollary 2. Let G be a real split group, let ν be a real dominant characters of A and

88

let X(ν) be the corresponding spherical Langlands quotient. For each petite K-type µ

in X(ν), denote by ψµ the representation of W on the space of M-fixed vectors µM .

If X(ν) is unitary, then the algebraic operator ψµ(A(ν)) is positive-semidefinte.

This is a non-unitarity test , in the sense that it can be used to show that some

representations are not unitary: if the algebraic operator ψ(A(ν)) fails to be positive

semi-definite for some representation ψ that arises from a petite K-type µ in X(ν),

then we can conclude that X(ν) is not unitary.

It is quite an amazing fact that this test also detects unitarity when the group G

is a classical group. This beautiful result is due to Barbasch and can be found in [5].

It can be stated as follows:

Theorem 12. If G is a classical (real or p-adic) split group, then the spherical Lang-

lands quotient X(ν) is unitary if and only if the invariant Hermitian form is positive

semi-definite on the petite K-types.

Actually, Barbasch uses a subset of the set of petite K-types, that he calls “rel-

evant” K-types. For each relevant K-type µ, the representation ψµ of W on µM

is irreducible, and computing the signature of the algebraic operators ψµ(A(ν)) (for

all relevant µ) is enough to detect unitarity. Therefore, we can study the spherical

unitary dual of a split classical group by means of Weyl group computations. Even if

Barbasch’s result has a very simple form, it is really hard to explain. So we will not

do any attempt in this direction.

Instead, in the next chapter, we will give an application of the result and show how

to use the Weyl group to study the spherical unitary dual of SL(3, R).

89

Chapter 7

The example of SL(3, R)

7.1 The data for SL(3, R)

In this section we briefly recall the data for the group SL(3, R) and fix the notations

that we will be using throughout the chapter.

• G = SL(3, R) = 3× 3 real matrices with determinant 1

• g0 = sl(3, R) = 3× 3 real matrices with trace 0

• ϑ : g0 → g0, X 7→ −XT

• k0 = 3× 3 real skew-symmetric matrices with trace 0

• p0 = 3× 3 real symmetric matrices with trace 0

• a0 = 3× 3 real diagonal matrices with trace 0

• m0 = 0

• εj : a0 → R,

a1 0 0

0 a2 0

0 0 a3

7→ aj (j = 1, 2, 3)

90

• ∆ = ∆(g0, a0) = ±(ε1 − ε2), ±(ε2 − ε3), ±(ε1 − ε3)

• ∆+ = ∆+(g0, a0) = ε1 − ε2, ε2 − ε3, ε1 − ε3

• Π = ε1 − ε2, ε2 − ε3 (simple roots)

• (g0)εi−εj= REi,j (i, j = 1, 2, 3; i 6= j)

• K = SO(3, R) = 3× 3 real orthogonal matrices with determinant 1

• A = 3× 3 diagonal matrices with determinant 1 and non-negative entries

• M = 3× 3 diagonal matrices with determinant 1 and diag. entries = ±1

=

m1 0 0

0 m2 0

0 0 m3

: m1m2m3 = 1, mj = ±1 ∀ j = 1, 2, 3

• N = 3× 3 upper triangular real matrices with diagonal entries = 1

• P = MAN = 3× 3 upper triangular real matrices

• M′=

ε1 eσ(1)

ε2 eσ(2)

ε3 eσ(3)

: σ ∈ S3, εj = ±1 and (ε1ε2ε3)sgn(σ) = 1

• W ' S3 (symmetric group on 3 letters)

• ω0 = sε1−ε3 (the long element of W )

• M = δ0, δ1, δ2, δ3 with δ0 the trivial representation of M , and

δj : M → R,

m1 0 0

0 m2 0

0 0 m3

7→ mj, ∀ j = 1, 2, 3.

91

7.2 The irreducible representations of SO(3)

The group K = SO(3) has no irreducible representations of even dimension and, up

to equivalence, a unique irreducible representation of each odd dimension.

For each integer N > 0, consider the space HN of harmonic homogeneous poly-

nomials of degree N in three variables, with complex coefficients. HN is a complex

vector space of dimension 2N + 1, and SO(3) acts on it by:

(g · F )(x, y, z) = F ((x, y, z)g) ∀ g ∈ SO(3), ∀F ∈ HN .

This representation is irreducible, and any irreducible representation of SO(3) is

isomorphic to some HN .

7.2.1 A closer look at HN

The following remark is helpful to understand the structure of HN .

Remark 4. Every harmonic homogeneous polynomial of degree N in three variables

can be written uniquely in the form:

F (x, y, z) =N∑

k=1

xk

k!Fk(y, z)

with F0 . . . FN homogeneous polynomials in two variables y and z (of degree deg(Fk) =

N − k) satisfying:

Fk(y, z) = −∆y,zFk−2(y, z) ∀ k > 2.

We have denoted by ∆y,z the Laplacian d2

dy2 + d2

dz2 .

Proof. Let F be a harmonic homogeneous polynomial of degree N in x, y, z. Because

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F is homogeneous, we can write

F (x, y, z) =N∑

k=1

xk

k!Fk(y, z)

with each Fk homogeneous of degree N−k. We now show that because F is harmonic,

the condition

Fk(y, z) = −∆y,zFk−2(y, z) ∀ k > 2

must hold. We do it by writing down an explicit expression for the Laplacian of F :

∆x,y,zF =(

d2

dx2 + d2

dy2 + d2

dz2

)(∑N

k=1xk

k!Fk(y, z)) =

=∑

k>01k!

(d2xk

dx2

)Fk(y, z) +

∑k>0

xk

k!

(d2

dy2 + d2

dz2

)(Fk(y, z)) =

=∑

k>2k(k−1)

k!xk−2Fk(y, z) +

∑k>0

xk

k!∆y,zFk(y, z) =

=∑

k>2xk−2

(k−2)![Fk(y, z) + ∆y,zFk−2(y, z)] + xN−1

(N−1)!∆y,zFN−1(y, z)+

+xN

N !∆y,zFN(y, z) = (>)

=∑

k>2xk−2

(k−2)![Fk(y, z) + ∆y,zFk−2(y, z)]

In (>) we have used the fact that FN and FN−1 have degree less than two, so

their Laplacian is zero. It is then clear that ∆x,y,zF = 0 if and only if [Fk(y, z) +

∆y,zFk−2(y, z)] = 0, for all k > 2.

Corollary 3. HN has dimension 2N + 1.

Proof. Let F (x, y, z) =∑N

k=1xk

k!Fk(y, z) be a harmonic homogeneous polynomial of

degree N in x, y, z. F is completely determined by F0 =∑N

j=0 cj yjzN−j (which is a

93

homogeneous polynomial of degree N in y, z) and by F1 =∑N−1

j=0 dj yjzN−j (which

is a homogeneous polynomial of degree N − 1 in y, z). So F depends on a total of

(N + 1) + N = 2N + 1 complex parameters.

7.2.2 Spherical K-types

In this subsection we study restriction of (µn, Hn) to M , and its decomposition in

isotypic components of irreducible representations of M . In particular, we want to

determine whether Hn contains the trivial M -type.

It is easy to see the action of M on Hn is given by

µn

m1 0 0

0 m2 0

0 0 m3

· F (x, y, z) = F (m1x, m2y, m3z)

for all m = diag(m1, m2, m3) in M , and all F in Hn. We write F =∑N

k=0xk

k!Fk(y, z)

and observe that, when N is even

• µn

1 0 0

0 −1 0

0 0 −1

· F =

+F if F1 = 0

−F if F0 = 0

• µn

−1 0 0

0 1 0

0 0 −1

· F =

+F if F0 only contains even powers of z,

and F1 only contains odd powers of z

−F if F0 only contains odd powers of z,

and F1 only contains even powers of z

94

• µn

−1 0 0

0 −1 0

0 0 1

· F =

+F if F0 only contains even powers of y,

and F1 only contains odd powers of y

−F if F0 only contains odd powers of z,

and F1 only contains even powers of y.

Comparing these results with the character table of M

1 0 0

0 1 0

0 0 1

1 0 0

0 −1 0

0 0 −1

−1 0 0

0 1 0

0 0 −1

−1 0 0

0 −1 0

0 0 1

δ0 1 1 1 1

δ1 1 1 −1 −1

δ2 1 −1 1 −1

δ3 1 −1 −1 1

we conclude that

• F spans a copy of δ0 if and only if F1 = 0 and F0 only involves even powers of

95

both y and z. The mapping F 7→ F0 defines an isomorphism from the isotypic

component of δ0 in Hn to the vector space (of dimension N2

+ 1) of all the

homogeneous polynomials of degree N in y, z that only contain even powers of

both y and z:

V (δ0) ' a0yN + a2y

N−2z2 + . . . aN−2y2zN−2 + aNzN.

• F spans a copy of δ1 if and only if F1 = 0 and F0 only involves odd powers

of both y and z. The mapping F 7→ F0 defines an isomorphism from the

isotypic component of δ1 in Hn to the vector space (of dimension N2) of all the

homogeneous polynomials of degree N in y, z that only contain odd powers of

both y and z.

V (δ1) ' a1yN−1z + a3y

N−3z3 + . . . aN−3y3zN−3 + aN−1yzN−1.

• F spans a copy of δ2 if and only if F0 = 0 and F1 only involves even powers

of y and odd powers of z. The mapping F 7→ F1 defines an isomorphism from

the isotypic component of δ2 in Hn to the vector space (of dimension N2) of all

the homogeneous polynomials of degree N − 1 in y, z that only contain even

powers of y and odd powers of z.

V (δ2) ' a1yN−2z + a3y

N−4z3 + . . . aN−3y2zN−3 + aN−1z

N−1.

• F spans a copy of δ3 if and only if F0 = 0 and F1 only involves odd powers of y

and even powers of z. The mapping F 7→ F1 defines an isomorphism from the

isotypic component of δ2 in Hn to the vector space (of dimension N2) of all the

homogeneous polynomials of degree N − 1 in y, z that only contain odd powers

96

of y and even powers of z.

V (δ3) ' a0yN−1 + a2y

N−3z2 + . . . aN−3y3zN−4 + aN−2yzN−2.

The situation for N odd is slightly different, but can be analyzed in a similar

way. The results are as follows:

• The isotypic component of δ0 in Hn has dimension N−12

, and the mapping

V (δ0) → a1yN−2z + a3y

N−4z3 + . . . aN−4y3zN−4 + aN−2yzN−2, F 7→ F1

is an isomorphism.

• The isotypic component of δ1 in Hn has dimension N+12

, and the mapping

V (δ1) → a0yN−1 + a2y

N−3z2 + . . . aN−3y2zN−3 + aN−1z

N−1, F 7→ F1

is an isomorphism.


, and the mapping

V (δ2) → a0yN + a2y

N−2z2 + . . . aN−2y2zN−2 + aN−1yzN−1, F 7→ F0

is an isomorphism.


, and the mapping

V (δ3) → a1yN−1z + a3y

N−3z3 + . . . aN−2y2zN−2 + aNyzN, F 7→ F0

is an isomorphism.

We have proved the following claim:

97

Claim 4. The irreducible representation HN is spherical for each N 6= 1. Indeed, for

all m > 0, we have

H2m = δ m+10 ⊕ δ m

1 ⊕ δ m2 ⊕ δ m

3

H2m+1 = δ m0 ⊕ δ m+1

1 ⊕ δ m+12 ⊕ δ m+1

3 .

7.2.3 Petite K-types

In this subsection we determine the values of N for which the irreducible representa-

tion (µN , HN) of K = SO(3) is petite.

For each simple root α, we must look at the SO(2) subgroup attached to α

Kα = exp(RZα) = exp(R(Eα + ϑEα))

and check whether the restriction of µN to Kα only contains the characters 0, ±1 and

±2 of SO(2). This is equivalent to verify that for each simple root α, dµN(Zα) only

acts with eigenvalues 0, ±i and ±2i.

We start with an explicit description of the subgroups Kα.

• For α = ε1 − ε2, we take Zα =

0 1 0

−1 0 0

0 0 0

. Therefore

Kα =

cos(t) sin(t) 0

− sin(t) cos(t) 0

0 0 1

: t ∈ R

.

98

• For α′ = ε2 − ε3, we take Zα′ =

0 0 0

0 0 1

0 −1 0

. Therefore

Kα′ =

1 0 0

0 cos(t) sin(t)

0 − sin(t) cos(t)

: t ∈ R

.

We notice that Zα and Zα′ are conjugate (via an element of M′):

0 0 1

0 −1 0

1 0 0

0 1 0

−1 0 0

0 0 0

0 0 1

0 −1 0

1 0 0

=

0 0 0

0 0 1

0 −1 0

hence dµN(Zα) and dµN(Zα′) have the same eigenvalues. We just need to consider

one of the two simple roots. We will focus on α′ = ε2 − ε3, and we will assume that

N is even (the odd case is similar).

An easy computation shows that

dµN(Zα′) : HN → HN , F (x, y, z) 7→ y

(dF

dz

)− z

(dF

dy

).

Since dµN(Zα′) does not touch the variable x, it preserves the subspaces

H(1)N = V (δ0)⊕ V (δ1) H(2)

N = V (δ2)⊕ V (δ3)

99

of HN .1 Identifying each element F of H(1)N with its F0-part

a0zN + a1yzN−1 + · · ·+ aN−1y

n−1z + aNyN

and writing down the explicit matrix for the action of dµN(Zα′) on H(1)N , we find that

dµN(Zα′) has even eigenvalues

−Ni, −(N − 2)i, . . . , −2i, 0, 2i, . . . , (N − 2)i, Ni.

This is compatible with the fact that

mα′ = exp(πZα′) = diag(1, −1, −1)

acts on V (δ0)⊕V (δ1) as scalar multiplication by +1. Similarly2, we find that dµN(Zα′)

acts on H(2)N with odd eigenvalues

−(N − 1)i, −(N − 3)i, . . . , −i, i, . . . , (N − 3)i, (N − 1)i,

compatibly with the fact that mα′ acts by −1 on V (δ2)⊕ V (δ3). We conclude that

Claim 5. For each simple root α, the restriction of HN to Kα ' SO(2) is given by

HN |Kα=N⊕

l=−N

ξl .

Therefore, HN is petite if and only if N = 0, 1 or 2.

Corollary 4. The only irreducible representations of SO(3) that are both petite and

1Because N is even, we can identify H(1)N with the subspace of HN that consists of polynomials

with even powers of x (and arbitrary powers of y or z) and H(2)N with the subspace of HN that

consists of polynomials with odd powers of x (and arbitrary powers of y or z).2After identifying each element F of H(2)

N with its F1-part.

100

spherical are H0 and H2.

7.3 The representations of the Weyl group

We identify the Weyl group of SO(3) with the symmetric group in three letters

W = S3. There are three conjugancy classes, corresponding to the three cycle struc-

tures (1), (23) and (123), hence three irreducible inequivalent representations that we

denote by U , U ′ and V . More precisely:

• U is the trivial representation of S3. Every permutation acts by +1.

• U ′ is the sign representation of S3. A permutation σ acts by sgn(σ).

• W is “the other” representation of S3, necessarily3 of dimension 2.

The character table of S3 can easily be computed using the equation

∑

ν irred.

dim(ν)Xν(σ) = 0 ∀σ ∈ S3 − 1.

We obtain:

X 1 (23) (123)

U 1 1 1

U ′ 1 −1 1

V 2 0 −1

.

We now give a more explicit realization of the two-dimensional representation V of

S3. Let EV be the hyperplane

−→v ∈ C3 : v1 + v2 + v3 = 03The sum of the square of the dimensions of the irreducible representations equals the size of the

group.

101

and let a permutation π of S3 act on EV by

π · (v1, v2, v3) =(vπ−1(1), vπ−1(2), vπ−1(3)

)

for all −→v in EV . We notice that if ω is a primitive cubic root of 1, then 1+ω+ω2 = 0.

So the vectors

−→x = (1, ω2, ω) −→y = (1, ω, ω2)

belong to EV , and actually form a basis of EV . It is easy to check that −→x and −→yare eigenvectors of τ=(123) of eigenvalue ω and ω2 respectively, and that σ=(23)

permutes −→x and −→y (hence the vectors −→x +−→y and −→x −−→y are eigenvectors of σ=(23)

of eigenvalue +1 and −1 respectively).

It follows that σ=(23) acts on EV with trace zero, and τ=(123) acts on EV with

trace ω + ω2 = −1. Therefore EV is a realization of the irreducible two-dimensional

representation V of S3.

We conclude this analysis of the representations of the symmetric group S3 by

giving an explicit formula for the decomposition of a representation (ψ, Eψ) of S3

into isotypic components of irreducible representations:

ψ = Ua ⊕ (U ′)b ⊕ V c

with

a = m1(ψ(σ))−mω(ψ(τ))

b = m−1(ψ(σ))−mω(ψ(τ))

c = mω(ψ(τ)).

We have denoted by mz(ψ(π)) the algebraic multiplicity of the eigenvalue z for ψ(π),

for each permutation π = τ or σ, and each complex number z = 1, −1 or ω. This

formula easily follows form the facts that

102

• σ has eigenvalues

+1 on each copy of U

−1 on each copy of U ′

+1, −1 on each copy of V

• τ has eigenvalues

+1 on each copy of U

+1 on each copy of U ′

ω, ω2 on each copy of V .

7.3.1 ψH2: the representation of S3 on (H2)

M

In this subsection we study the representation ψν of the Weyl group on the space of

M invariants of a petite spherical representation µ of SO(3).

If µ = H0, then µ is the trivial representation of K. Clearly µM = C and W acts

trivially on it. The case µ = H2 is more interesting. Recall that

EH2 = ay2 + bz2 − (a + b)x2 : a, b ∈ C︸︷︷︸V (δ0)

⊕ C yz︸︷︷︸V (δ1)

⊕ C xz︸︷︷︸V (δ2)

⊕ Cxy︸︷︷︸V (δ3)

with K acting by:

(g · F )(x, y, z) = F ((x, y, z)g) ∀ g ∈ K, ∀F ∈ H2.

We want to determine the representation of W = S3 on the two-dimensional subspace

(EH2)M = ay2 + bz2 − (a + b)x2 : a, b ∈ C.

103

It is enough to understand the action of σ=(23) and τ=(123). The first step is to

pick representatives in M′for the two permutations. We choose

σ =

1 0 0

0 0 1

0 −1 0

τ =

0 0 1

−1 0 0

0 −1 0

.

For all F = ay2 + bz2 − (a + b)x2 in (EH2)M , we have

ψH2(σ) · F = F

(x, y z)

1 0 0

0 0 1

0 −1 0

= F (x, −z, y) = by2 + az2 − (a + b)x2.

Therefore, the matrix associated to ψH2(σ) with respect to the basis y2−x2, z2−x2

is

0 1

1 0

. It has eigenvalues +1 and −1. Similarly, for τ , we have:

ψH2(τ)·F = F

(x, y z)

0 0 1

−1 0 0

0 −1 0

= F (−y, −z, x) = −(a+b)y2+az2+bx2.

We associate to ψH2(τ) the matrix

−1 −1

1 0

, with eigenvalues ω and ω2.

We conclude that

a = m1(ψH2(σ))−mω(ψH2(τ)) = 0

b = m−1(ψH2(σ))−mω(ψH2(τ)) = 0

c = mω(ψH2(τ)) = 1.

104

Hence, ψH2 = V .

Conclusions. The representation of the Weyl group on the space of M-invariants in

H2 is irreducible, and isomorphic to V .

7.4 Signature on the K-type H2

In this section we compute the signature w.r.t. the K-type H2 of the standard inter-

twining operator for a principal series IP (δ ⊗ ν) of SL(3, R), under the assumptions

that P = MAN is a minimal parabolic subgroup, δ is the trivial representation of M

and ν is a real dominant character of A.

Because H2 is a petite representation, there are two ways to do this:4

(1) construct the “analytic” operator RH2(ω0, ν) and compute its signature

(2) construct the “algebraic” operator ψH2(ν) and compute its signature.

In both cases ν is intended to be a real dominant character of A

ν = λ1ε1 + λ2ε2 + λ3ε3

(3∑

j=1

λj = 0 and λ1 > λ2 > λ3

)

satisfying the “symmetry” condition ω0 · ν = −ν. As usual, we have denoted by ω0

the long element of the Weyl group: ω0 = sε1−ε3 . It is easy to check that ν must be

of the form

ν = λε1 − λε3

for some λ real and positive.

7.4.1 The “analytic” operator RH2(ω0, ν)

Recall the main steps for the construction of RH2(ω0, ν):

4Please, refer to chapter 6 for notations.

105

(i) Find a minimal decomposition of ω0 as a product of simple reflections:

ω0 = sαr · · · sα2sα1

(ii) Decompose RH2(ω0, ν) accordingly:

RH2(ω0, ν) = RH2(sαrsαr−1 · · · sα2sα1 , ν) =

= RH2(sαr , (sαr−1 · · · sα2sα1) · ν) · · · RH2(sα2 , sα1 · ν)Rµ(sα1 , ν)

(iii) For each simple root β and for each real character γ of A (satisfying 〈γ, β〉 > 0),

define RH2(sβ, γ) as follows: for all integers l, let ϕl be the isotypic component

of the character ξl : exp(tZβ) 7→ ei l t of Kβ ' SO(2) inside H2, so that

EH2 =⊕

l∈Zϕl

is the decomposition of H2 in Kβ-types, and

HomM(ResKM EH2 , V δ) =

⊕

m∈NHomM

(ResMKβ

M (ϕ2m + ϕ−2m), C)

︸︷︷︸Um

is a decomposition of HomM(ResKM EH2 , V δ) in MKβ-stable subspaces.

Define the operator RH2(sβ, γ) to act on Um as scalar multiplication by

∏mj=1((2j − 1)− 〈γ, β〉)∏mj=1((2j − 1) + 〈γ, β〉)

for all non-negative integers m.

Steps (i) and (ii) are really easy:

w0 = sε1−ε3 = sε1−ε2 sε2−ε3 sε1−ε2

106

and the corresponding decomposition of the operator RH2(sε1−ε3 , ν) is

RH2(sε1−ε2 , (sε2−ε3sε1−ε2) · ν) RH2(sε2−ε3 , sε1−ε2 · ν) RH2(sε1−ε2 , ν).

We observe that

ν = λε1 − λε3

sε1−ε2 · ν = λε2 − λε3

(sε2−ε3 sε1−ε2) · ν = −λε2 + λε3

so can write:

RH2(ω0, ν) = RH2(sε1−ε3 , λε1 − λε3) =

= RH2(sε1−ε2 , −λε2 + λε3) RH2(sε2−ε3 , λε2 − λε3) RH2(sε1−ε2 , λε1 − λε3).

To continue, we need to understand the decomposition of (H2, EH2) in irreducible

representations of the SO(2) subgroup associated to each simple root.

We start with α = ε1 − ε2. Then

Kα =

kθ =

cos(θ) sin(θ) 0

− sin(θ) cos(θ) 0

0 0 1

: θ ∈ R

.

An easy computations shows that

kθ · F (x, y, z) = F (x cos(θ)− y sin(θ), x sin(θ) + y cos(θ), z)

for all F in EH2 . In particular

107

kθ · (yz − i xz) = ei θ(yz − i xz)

kθ · (yz + i xz) = e−i θ(yz + i xz)

kθ · (2z2 − x2 − y2) = +(2z2 − x2 − y2)

kθ · (y2 − x2 − 2i xy) = e2i θ(y2 − x2 − 2i xy)

kθ · (y2 − x2 + 2i xy) = e−2i θ(y2 − x2 + 2i xy).

The decomposition of EH2 in Kα-types is therefore given by:

EH2 = C(2z2 − x2 − y2)︸︷︷︸ϕ0

⊕ C(y2 − x2 − 2i xy)︸︷︷︸ϕ2

⊕ C(y2 − x2 + 2i xy)︸︷︷︸ϕ−2

⊕

⊕ C(yz − i xz)︸︷︷︸ϕ+1

⊕ C(yz + i xz)︸︷︷︸ϕ−1

.

The corresponding decomposition of HomM(ResKM EH2 , V δ) in MKα-stable subspaces

is

HomM(ResMKα

M ϕ0, C) ⊕ HomM

(ResMKα

M (ϕ2 ⊕ ϕ−2), C)

with

• HomM(ResMKα

M ϕ0, C) ' C(2z2 − x2 − y2)

• HomM

(ResMKα

M (ϕ2 ⊕ ϕ−2)) ' C(y2 − x2)

• HomM(ResKM EH2 , V δ) ' C(y2 − x2) ⊕ C(z2 − x2) ' (EH2)

M .

Once this decomposition is understood, it is really easy to compute the operator

RH2(sε1−ε2 , γ), for each real character γ satisfying 〈γ, ε1−ε2〉 > 0. Indeed, RH2(sε1−ε2 , γ)

acts trivially HomM(ResMKα

M ϕ0, C), and it acts on HomM

(ResMKα

M (ϕ2 ⊕ ϕ−2))

as

108

scalar multiplication by

1− 〈γ, 2(ε1−ε2)‖ε1−ε2‖2 〉

1 + 〈γ, 2(ε1−ε2)‖ε1−ε2‖2 〉

=1− 〈γ, ε1 − ε2〉1 + 〈γ, ε1 − ε2〉 .

The matrix associated to RH2(sε1−ε2 , γ) with respect to the basis

B = (y2 − x2), (z2 − x2)

of HomM(ResKM EH2 , V δ) ' (EH2)

M is therefore given by

RH2(sε1−ε2 , γ) Ã

1−〈γ, ε1−ε2〉1+〈γ, ε1−ε2〉

−〈γ, ε1−ε2〉1+〈γ, ε1−ε2〉

0 1

.

When γ = −λε2 + λε3 or γ = +λε1 − λε3, we have 〈γ, ε1 − ε2〉 = +λ. Hence:

RH2(sε1−ε2 , −λε2 + λε3) = RH2(sε1−ε2 , λε1 − λε3) Ã

1−λ1+λ

−λ1+λ

0 1

.

The construction for α′ = ε2 − ε3 is very similar. The subgroup

Kα′ =

1 0 0

0 cos(θ) sin(θ)

0 − sin(θ) cos(θ)

: θ ∈ R

acts on EH2 by

kθ · F (x, y, z) = F (x, y cos(θ)− z sin(θ), y sin(θ) + z cos(θ))

for all F in EH2 . In particular

kθ · (xy + i xz) = ei θ(xy + i xz)

109

kθ · (xy − i xz) = e−i θ(xy − i xz)

kθ · (y2 + z2 − 2x2) = +(y2 + z2 − 2x2)

kθ · (y2 − z2 + 2i yz) = e2i θ(y2 − z2 + 2i yz)

kθ · (y2 − z2 − 2i yz) = e−2i θ(y2 − z2 − 2i yz).

The decomposition of EH2 in Kα′-types is therefore given by:

EH2 = C(y2 + z2 − 2x2)︸︷︷︸ϕ0

⊕ C(y2 − z2 + 2i yz)︸︷︷︸ϕ2

⊕ C(y2 − z2 − 2i yz)︸︷︷︸ϕ−2

⊕

⊕ C(xy + i xz)︸︷︷︸ϕ+1

⊕ C(xy − i xz)︸︷︷︸ϕ−1

.

The corresponding decomposition of HomM(ResKM EH2 , V δ) in MKα′-stable subspaces

is

HomM(ResMKα′

M ϕ0, C) ⊕ HomM

(ResMKα′

M (ϕ2 ⊕ ϕ−2), C)

with

• HomM(ResMKα′

M ϕ0, C) ' C(y2 + z2 − 2x2)

• HomM

(ResMKα′

M (ϕ2 ⊕ ϕ−2), C)' C(y2 − z2)

• HomM(ResKM EH2 , V δ) ' C(y2 − x2) ⊕ C(z2 − x2) ' (EH2)

M .

For each real character γ satisfying 〈γ, ε2− ε3〉 > 0, the operator RH2(sε2−ε3 , γ) acts

trivially HomM(ResMKα′

M ϕ0, C), and it acts on HomM

(ResMKα′

M (ϕ2 ⊕ ϕ−2), C)

as

scalar multiplication by

1− 〈γ, 2(ε2−ε3)‖ε2−ε3‖2 〉

1 + 〈γ, 2(ε2−ε3)‖ε2−ε3‖2 〉

=1− 〈γ, ε2 − ε3〉1 + 〈γ, ε2 − ε3〉 .

110

The matrix associated to RH2(sε2−ε3 , γ) with respect to the basis

B = (y2 − x2), (z2 − x2)

of HomM(ResKM EH2 , V δ) ' (EH2)

M is therefore given by

RH2(sε1−ε2 , γ) Ã

11+〈γ, ε2−ε3〉

〈γ, ε2−ε3〉1+〈γ, ε2−ε3〉

〈γ, ε2−ε3〉1+〈γ, ε2−ε3〉

11+〈γ, ε2−ε3〉

.

When γ = λε2 − λε3 , we have 〈γ, ε2 − ε3〉 = 2λ. Hence, we obtain:

RH2(sε2−ε3 , λε2 − λε3) Ã

11+2λ

2λ1+2λ

2λ1+2λ

11+2λ

.

Let us now consider the the full intertwining operator:

RH2(ω0, ν) = RH2(sε1−ε3 , λε1 − λε3) =

= RH2(sε1−ε2 , −λε2 + λε3) RH2(sε2−ε3 , λε2 − λε3) RH2(sε1−ε2 , λε1 − λε3).

If we denote by R be the matrix associated to RH2(ω0, ν) with respect to the ba-

sis B of (EH2)M , then

R = 1(1+λ)2

1(1+2λ)

1− λ −λ

0 1 + λ

1 2λ

2λ 1

1− λ −λ

0 1 + λ

111

=(

1−λ1+λ

)1

1+2λ

1− 2λ 0

2λ 1 + 2λ

.

An easy study of the eigenvalues of R

η1 =(

1−λ1+λ

) (1−2λ1+2λ

)

η2 = 1−λ1+λ

shows that (for λ real and positive)

• R is positive definite if 0 < λ < 12

• R is positive semi-definite if λ = 12

• R is indefinite if 12

< λ < 1 or λ > 1

• R is zero if λ = 1.

7.4.2 The “algebraic” operator ψH2(A(ν))

These are the main steps for the construction of ψH2(A(ν)) :

(i) Find a minimal decomposition of ω0 as a product of simple reflections:

w0 = sε1−ε3 = sε1−ε2 sε2−ε3 sε1−ε2

(ii) Decompose A(ν) accordingly:

A(ν) = Asε1−ε2((sε2−ε3sε1−ε2) · ν) Asε2−ε3

(sε1−ε2 · ν) Asε1−ε2(ν).

(iii) For each simple root β and for each real character γ of A (satisfying 〈γ, β〉 > 0)

112

that appear in the above decomposition, construct the element

Asβ(γ) =

1

1 + 〈γ, β〉 e +〈γ, β〉

1 + 〈γ, β〉 sβ

of C[W ]. More explicitly, set

•Asε1−ε2((sε2−ε3sε1−ε2) · ν) = Asε1−ε2

(−λε2 + λε3) =e+λ sε1−ε2

1+λ

•Asε2−ε3(sε1−ε2 · ν) = Asε2−ε3

(λε2 − λε3) =e+2λ sε2−ε3

1+2λ

•Asε1−ε2(ν) = Asε1−ε2

(λε1 − λε3) =e+λ sε1−ε2

1+λ.

(iv) With calculations in C[W ], compute A(ν):

A(ν) = Asε1−ε2((sε2−ε3sε1−ε2) · ν) Asε2−ε3

(sε1−ε2 · ν) Asε1−ε2(ν)

=(

e+λ sε1−ε2

1+λ

)(e+2λ sε2−ε3

1+2λ

)(e+λ sε1−ε2

1+λ

)

= (1+λ2) e+2λ2 (s1,2 s2,3+s2,3 s1,2)+2λ (s1,2+s2,3)+2λ3 s1,3

(1+λ)2(1+2λ)

(for brevity of notations we have set sεi−εj= si,j, for all i, j = 1 . . . 3).

(v) Evaluate the representation ψH2 of W at the element A(ν) of C[W ].

Recall that ψH2 = V (the two-dimensional irreducible representation of the Weyl

group S3), and that EV has a basis −→x , −→y with the following properties:

113

• σ = (23) permutes −→x and −→y

• −→x and −→y are eigenvectors of τ = (123), of eigenvalues ω and ω2 respec-

tively. Here ω a primitive cubic root of 1.

With respect to the basis −→x , −→y we can write:

ψH2((12)) Ã

0 ω

ω2 0

ψH2((23)) Ã

0 1

1 0

ψH2((13)) Ã

0 ω2

ω 0

ψH2(e) Ã

1 0

0 1

obtaining the following matrix for ψH2(A(ν)):

ψH2(A(ν)) Ã A =1

(1 + λ)2(1 + 2λ)

(1− λ2) −2λ(1− λ2)ω2

−2λ(1− λ2)ω (1− λ2)

=

=

(1− λ

1 + λ

)1

1 + 2λ

1 −2λω2

−2λω 1

.

The eigenvalues of A are

ζ1 =(

1−λ1+λ

) (1−2λ1+2λ

)

ζ2 = 1−λ1+λ

.

Therefore (for λ real and positive)

• A is positive definite if 0 < λ < 12

• A is positive semi-definite if λ = 12

• A is indefinite if 12

< λ < 1 or λ > 1

114

• A is zero if λ = 1.

7.5 Conclusions

In this section, we summarize the information gathered in the chapter about the uni-

tarity of a spherical principal series of SL(3, R).

Let P = MAN be the minimal parabolic subgroup of SL(3) consisting of real

3× 3 upper triangular matrices. Let ω0 be the long Weyl group element, and let ν be

a real dominant character of A satisfying the formal symmetry condition ω0 ·ν = −ν.

More explicitly, set ω0 = sε1−ε3 and set ν = λε1 − λε3, for λ > 0.

Denote by X(ν) the Langlands quotient of SL(3) with parameters (P, δ0, ν),

with δ0 the trivial representation of M . X(ν) is unitary if and only if the Hermitian

operator

Rµ(ω0, ν) : HomM(ResKM Eν , V δ0) → HomM(ResK

M Eν , V δ0)

is positive semi-definite for all spherical K-types (µ, Eµ).

By a theorem of Barbasch, it is enough to check the signature of Rµ(ω0, ν) for the

petite K-types. Apart from the trivial one, K = SO(3) has only one spherical petite

irreducible representation: the K-type H2.5

So it all comes down to computing the signature of the operator

RH2(ω0, µ) = RH2(sε1−ε3 , λε1 − λε3).

5(H2, EH2) is the 5-dimensional representation of SO(3) on the space of homogeneous harmonicpolynomials of degree 2 in three variables.

115

We find that

• RH2(ω0, ν) is positive definite if 0 < λ < 12

• RH2(ω0, ν) is positive semi-definite if λ = 12

• RH2(ω0, ν) is indefinite if 12

< λ < 1 or λ > 1

• RH2(ω0, ν) is zero if λ = 1.

We therefore conclude that

the Langlands quotient X(ν) is unitary for 0 < λ ≤ 12

or λ = 1 .

There is an alternative method to construct the operator RH2(ω0, ν), which is

purely algebraic in nature. We identify HomM(ResKM EH2 , V δ0) with the space of M

invariants in EH2 , which carries a representation ψH2 of the Weyl group. Once this

representation is understood, we replace the analytic operator RH2(ω0, ν) with an

algebraic operator ψH2(A(ν)), that is also Hermitian and has the same signature as

RH2(ω0, ν).6 The remarkable fact is that ψH2(A(ν)) can be computed by means only

of Weyl group calculations.

6The fact that H2 is petite is crucial for this construction.

116

Chapter 8

The petite spherical

representations of SO(2n, R)

In this chapter we study the irreducible representations µ of SO(2n, R) that satisfy

the following requirements:

• for each restricted root α of SL(2n, R), the restriction of µ to the SO(2) sub-

group associated to α only contains the characters zero, plus or minus one, plus

or minus two (i.e. µ is petite)

• µ contains an M -fixed vector, where M is the subgroup of diagonal matrices in

SO(2n,R) (i.e. µ is spherical).

These are exactly the K-types that appear in a spherical principal series of SL(2n, R)

on which the intertwining operator can be computed by means of Weyl group com-

putations.1

1Please refer to chapter 6 for details.

117

8.1 The petite representations of SO(2n, R)

First, we fix some notations. Let K be the group SO(2n, R), and let H be the

maximal torus

H =

cos(θ1) sin(θ1) . . . 0 0

− sin(θ1) cos(θ1) . . . 0 0

......

. . ....

...

0 0 . . . cos(θn) sin(θn)

0 0 . . . − sin(θn) cos(θn)

: θ1, . . . , θn ∈ R

.

We denote by k0 and h0 the Lie algebras of K and H, and by k and h their complex-

ification (so k = kC0 and h = hC0 ).

h is a Cartan subalgebra of the complex semi-simple Lie algebra k = so(2n, C), and

it is explicitly given by

h =

0 θ1 . . . 0 0

−θ1 0 . . . 0 0

......

. . ....

...

0 0 . . . 0 θn

0 0 . . . −θn 0

: θ1, θ2, . . . , θn ∈ C

.

For brevity of notations, it is convenient to introduce a basis of h. We pick the

elements

Hj = E2j−1,2j − E2j,2j−1 (j = 1 . . . n)

where as usual Ea,b stands for an elementary matrix2.

Since Hjj=1...n is a basis of h, we write h =⊕n

j=1CHj.

2For each a, b = 1, . . . , 2n, Ea,b is a square matrix of size 2n with entry 1 where the ath row andthe bth column meet, all other entries being 0.

118

For each j = 1 . . . n, let ψj be the complex linear functional on h defined by

ψj(∑n

k=1 θkHk) 7→ −iθj (j = 1 . . . n).

More explicitly,

ψj

0 θ1 . . . 0 0

−θ1 0 . . . 0 0

......

. . ....

...

0 0 . . . 0 θn

0 0 . . . −θn 0

= −iθj

for all θ1, . . . , θn in C.

There exists a one-one correspondence between the set of dominant analytically inte-

gral forms λ on h and the set of irreducible representations µλ of K, the correspon-

dence being that λ is the highest weight of µλ.

The dominant analytically integral forms on h are all the expressions

λ = a1ψ1 + · · ·+ an−1ψn−1 + anψn

with integer coefficients a1, . . . , an−1, an satisfying

a1 > · · · > an−1 >| an |.

We ask which of these forms give rise to petite representations of SO(2n,R).

The answer to this question is easy to state:

µa1ψ1+···+anψn is petite ⇔ aj ∈ 0, ±1, ±2 ∀ j = 1, . . . , n

but the proof requires some preparation. Once we have a deep understanding of

the restricted root space decomposition of sl(2n,R), and an explicit construction for

the SO(2) subgroup Kα associated to each restricted root α, we can look at the

119

restriction of µ to each Kα and impose that this restriction only contains the 0, ±1

or ±2 characters of SO(2).

8.1.1 The restricted root system of sl(2n,R)

In this section we describe the various constituents of the restricted root space de-

composition of g0 = sl(2n,R):

g0 = k0 ⊕ p0 = a0 ⊕m0 ⊕α∈∆(g0, a0) (g0)α.

Fix the Cartan involution ϑ : g0 → g0, X 7→ −XT . Then:

• k0 = so(2n,R) is the (+1)-eigenspace of ϑ

• p0 = g0 ∩ real symmetric matrices is the (-1)-eigenspace of ϑ

• a0 = g0 ∩ real diagonal matrices is a maximal abelian subspace of p0

a0 =

b1 0 . . . 0 0

0 b2 . . . 0 0

......

. . ....

...

0 0 . . . b2n−1 0

0 0 . . . 0 b2n

: b1, . . . , b2n ∈ R s.t.2n∑

j=1

bj = 0

We pick the basis Hj = E2j,2j − E2j+1,2j+1 : j = 1, . . . , 2n− 1 in a0 .

• m0 = 0 is the centralizer of a0 in k0

• ∆ = ∆(g0, a0) = ±(εi − εj)i, j=1,..., 2n, i<j is the set of restricted roots,

where for each j we have denoted by εj the real linear functional on a0 defined

by

εj : a0 → R, diag(b1, . . . , b2n) 7→ bj

120

• ∆+ = εi − εji, j=1,..., 2n, i<j is the set of positive restricted roots

• Π = εi − εi+1i, j=1,..., 2n−1 is the set of simple restricted roots

• (g0)εi−εj= REi,j is the restricted root space associated to a root εi − εj.

8.1.2 The SO(2) attached to a restricted root

To each restricted root α in ∆, we attach a subgroup Kα isomorphic to SO(2,R) in

the following way:

1 We pick an element Eα in (g0)α that satisfies the normalizing condition

B(Eα, ϑ(Eα)) = − 2‖α‖2 ,

where B is the Killing form on g0 and ‖α‖2 = B(Hα, Hα), with Hα the unique

element of a0 such that

B(Hα, H) = α(H) ∀H ∈ a0.

2 We construct the skew-symmetric matrix

Zα = Eα + ϑ(Eα) = Eα − (Eα)T ,

which generates a one-dimensional Lie algebra kα isomorphic to so(2, R).

3 We define Kα to be the analytic subgroup of SO(2n,R) with Lie algebra kα, so

Kα = expRZα.

The first step is to understand the Killing form on g0 = sl(2n, R). For each X and Y

in g0, we have3:

B(X, Y ) = Tr(ad(X) ad(Y )) = (2n)Tr(XY ).

3See [16], Ch.III

121

In particular, if H =∑2n

j=1 xjEj,j and H ′ =∑2n

j=1 x′jEj,j, then

B(H, H ′) = (2n)∑2n

j=1 xj · x′j.

It easily follows that

Hεk−εl= 1

2n(Ek,k − El,l)

for all integers l, k (l, k = 1 . . . 2n, l 6= k). Therefore

‖α‖2 = B(Hα, Hα) = 22n

for any root α = εk − εl.

Since any root space is one-dimensional, the vector Eεk−εlmust be a multiple of the

generator Ek,l, say Eεk−εl= cEk,l, and because

B(Ek,l, ϑ(Ek,l)) = −(2n)Tr(Ek,l(Ek,l)T ) = −(2n)Tr(Ek,lEl,k) = −(2n)

the complex number c must satisfy the equation

−(2n)‖c‖2 = − 22/(2n)

(whose solutions are ±1).

For each root α = εk − εl, we set Eα = Ek,l and Zα = Ek,l + ϑ(Ek,l) = Ek,l − El,k.

Zα is a generator for the so(2) associated to the root α, the corresponding SO(2)

subgroup is therefore Kα = expRZα. Here is an explicit description of Kα: if

122

α = ±(εk − εl), with 1 6 k < l 6 2n, then

1 . . . 0 0 0 . . . 0 0 0 . . . 0

.... . .

......

.... . .

......

.... . .

...

0 . . . 1 0 0 . . . 0 0 0 . . . 0

0 . . . 0 cos(θ) 0 . . . 0 sin(θ) 0 . . . 0

0 . . . 0 0 1 . . . 0 0 0 . . . 0

.... . .

......

.... . .

......

.... . .

...

0 . . . 0 0 0 . . . 1 0 0 . . . 0

0 . . . 0 − sin(θ) 0 . . . 0 cos(θ) 0 . . . 0

0 . . . 0 0 0 . . . 0 0 1 . . . 0

.... . .

......

.... . .

......

.... . .

...

0 . . . 0 0 0 . . . 0 0 0 . . . 1

← 1

...

← k

...

← l

...

← 2n

with θ in R.

8.1.3 Petite representations

Let λ = a1ψ1 + · · ·+ an−1ψn−1 + anψn be an analytically integral dominant form on

h, so that

a1, . . . , an−1, an ∈ Z

a1 > · · · > an−1 > | an | (>),

and let µλ be the irreducible representation of SO(2n,R) with highest weight λ. Let α

be a restricted root for SL(2n,R) and let Kα be the corresponding SO(2) subgroup.

In this section we study the restriction of µλ to Kα. In particular, we ask whether

this restriction consists only of the characters 0, ±1, ±2 of SO(2).

123

Notice that, for each α in ∆, the group Kα is isomorphic to SO(2) via

exp(θZα) ↔

cos(θ) sin(θ)

− sin(θ) cos(θ)

so, if we want Kα to act by the characters

ξ0 :

cos(θ) sin(θ)

− sin(θ) cos(θ)

→ 1

ξ±1 :

cos(θ) sin(θ)

− sin(θ) cos(θ)

→ e±iθ

ξ±2 :

cos(θ) sin(θ)

− sin(θ) cos(θ)

→ e±2iθ

then Zα must act with eigenvalues 0, ±i, ±2i.For each j = 1, . . . , n, consider the element Hj = E2j−1,2j − E2j,2j−1 of the Cartan

subalgebra h of so(2n,C). Because Hj = Zε2j−1−ε2j, the eigenvalues of Hj must also

lie in the set 0, ±i, ±2i. The list of eingenvalues of Hj in dµλ can be found by

looking at the coefficient of ψj in all the possible weights of µλ, and multiplying this

coefficient by −i.4

We conclude that µλ is petite if and only if all the coefficients of all the weights of

dµλ lie in the set 0, ±1, ±2.In particular, the coefficients ajj=1,...,n of the highest weight of µλ must belong to

this set. Taking also the condition (>) into account, we see that µλ is petite only if

its highest weight λ falls in this list:

4If y is a weight vector of weight b1ψ1 + · · ·+ bnψn, then

dµλ(Hj) · y = (b1ψ1 + · · ·+ bnψn)(Hj)y = bjεj(Hj)y = (−ibj)y

so (−ibj) is an eigenvalue of dµλ(Hj). Since the weight vectors from a basis of the representationspace, it is clear that all the eigenvalues of (Hj) are obtained this way.

124

• 0ψ1 + · · ·+ 0ψn

• 1ψ1 + · · ·+ 1ψk 0 < k < n

• (1ψ1 + · · ·+ 1ψn−1)± 1ψn

• 2ψ1 + · · ·+ 2ψk 0 < k < n

• (2ψ1 + · · ·+ 2ψa) + (1ψa+1 + · · ·+ 1ψb) 0 < a, b < n

• (2ψ1 + · · ·+ 2ψk) + (1ψk+1 + · · ·+ 1ψn−1)± 1ψn 0 < k < n

• (2ψ1 + · · ·+ 2ψn−1)± 2ψn.

The next step is to show that each representation in this list is actually petite. We

need a case by case analysis.

If λ = 0ψ1 + · · ·+ 0ψn, then µλ is the trivial representation of SO(2n,R), and

it is of course petite.

If λ = 1ψ1 + · · · + 1ψk, with 0 < k < n, then µλ = ΛkC2n is the kth wedge

product of the standard representation of SO(2n,R), and it is also petite.

We start by determining the weights of the standard representation of SO(2n,R).

Let −→ekk=1,...,2n be the canonical basis of C2n; we notice that the vector (−−→e2j−1+

−→e2j) is a (+1)-eigenvector of Hj and a 0-eigenvector of Hl for l 6= j. Hence it is

a weight vector for h of weight

0ψ1 + · · ·+ 0ψj−1 + 1ψj + 0ψj+1 · · ·+ 0ψn = +ψj.

Similarly, (−−→e2j−1−−→e2j) is a (−1)-eigenvector of Hj and a 0-eigenvector of Hl for

l 6= j, so it is a weight vector (for h) of weight

0ψ1 + · · ·+ 0ψj−1 − 1ψj + 0ψj+1 · · ·+ 0ψn = −ψj.

125

Because the vectors −−→e2j−1±−→e2jj=1,...,n form a basis of C2n, we deduce that the

weights of the standard representation of SO(2n, R) are ±ψjj=1,...,n.

Let us now look at the weights of ΛkC2n. It follows from the product rule for

differentiation that the wedge product of k weight vectors of C2n is a weight

vector for ΛkC2n (relative to the sum of the weights of the k vectors). For

dimensional reasons, all the weight vectors of ΛkC2n are obtained this way. We

therefore conclude that each weight of ΛkC2n is of the form

±ψi1 ± ψi2 · · · ± ψir

for some r 6 k and some (distinct) indices i1, i2, . . . , ir in 1, 2, . . . , n.Every ψj appears with coefficient 0 or ±1, hence the representation ΛkC2n is

petite.

If λ = (1ψ1+· · ·+1ψn−1)±1ψn, then µλ is one of the two irreducible summands

of ΛnC2n. In particular every weight of µλ is also a weight of ΛnC2n, and

therefore the same argument used above shows that the coefficient of ψj in each

weight lies in the set 0, +1, −1. We conclude that µλ is petite.

If λ = 2ψ1 + · · · + 2ψk, with 0 < k < n, then µλ is an irreducible summand of

(ΛkC2n) ⊗ (ΛkC2n). It is easy to check that the weights of (ΛkC2n) ⊗ (ΛkC2n)

are of the form ν +ν ′, with ν and ν ′ (not necessarily distinct) weights of ΛkC2n.

So they are of the form

(±2ψi1 · · · ± 2ψir) + (±ψj1 · · · ± ψjs)

for some (distinct) indices i1, . . . , ir, j1, . . . , js in 1, 2, . . . , n, with r + s 6 k.

Since every weight of µλ is also a weight of (ΛkC2n)⊗ (ΛkC2n), the coefficient of

ψj in a weight of µλ lies in the set 0, ±1, ±2. We conclude that µλ is petite.

If λ = (2ψ1+· · ·+2ψn−1)±2ψn, then µλ is an irreducible summand of (ΛnC2n)⊗(ΛnC2n), and the same argument used above shows that it is petite.

126

If λ = (2ψ1 + · · ·+ 2ψa) + (1ψa+1 + · · ·+ 1ψb), with 0 < a, b < n, then µλ is an

irreducible summand of (ΛaC2n)⊗ (ΛbC2n) and it is again petite.

If λ = (2ψ1 + · · · + 2ψk) + (1ψk+1 + · · · + 1ψn−1)± 1ψn, with 0 < k < n, then

µλ is an irreducible summand of (ΛkC2n)⊗ (ΛnC2n) and it is also petite.

This concludes the proof. We obtain:

Proposition 2. The petite representations of SO(2n, R) are exactly the ones with

highest weight

• 0ψ1 + · · ·+ 0ψn

• 1ψ1 + · · ·+ 1ψk 0 < k < n

• (1ψ1 + · · ·+ 1ψn−1)± 1ψn

• 2ψ1 + · · ·+ 2ψk 0 < k < n

• (2ψ1 + · · ·+ 2ψa) + (1ψa+1 + · · ·+ 1ψb) 0 < a < b < n

• (2ψ1 + · · ·+ 2ψk) + (1ψk+1 + · · ·+ 1ψn−1)± 1ψn 0 < k < n

• (2ψ1 + · · ·+ 2ψn−1)± 2ψn.

8.2 The petite spherical representations of SO(2n, R)

Finding the petite spherical representations of SO(2n, R) accounts to determining

which petite representations contain an M -fixed vector.

Here M is the group of diagonal matrices of size 2n with entries ±1 and determinant

one. It is a finite abelian group of order 22n−1 and its irreducible representations are

the form

δS : M → C, m = diag(m1, m2, . . . , m2n) 7→ δS(m) =∏j∈S

mj1 (>)

127

with S a subsets of 1, 2, . . . , 2n of cardinality less or equal to n.5 In particular, δ∅

is the trivial representation of M .

A petite irreducible representation of SO(2n) is spherical if and only if it contains

δ∅. The previous proposition contains a list of the highest weights of the petite rep-

resentations; for each λ in this list we study the decomposition of µλ in M -types.

If λ = 0ψ1 + · · · + 0ψn, then µλ is the trivial representation of SO(2n, R),

µλ |M= δ∅ and of course µλ is spherical.

If λ = 1ψ1 + · · · + 1ψk, with 0 < k < n, then µλ = ΛkC2n. We will show that

restriction of µλ to M does not include δ∅, hence µλ is not spherical.

Let m = diag(m1, m2, . . . , m2n) be an arbitrary element of M . We notice that m

acts on a basis vector −→ej = −→ej1 ∧ −→ej2 ∧ · · · ∧ −→ejkof ΛkC2n as scalar multiplication

by (mj1 ·mj2 · · ·mjk). It follows that, for every multi-index j = (j1, j2, . . . , jk), the

one-dimensional subspace C−→ej is a copy of the representation δj1, j2,..., jk of M .

Therefore:

µ1ψ1+···+1ψk|M=

⊕S⊆1...2n : |S|=k δS.

Being 0 < k < n, any of these summands is equivalent to δ∅. So µλ is not spherical.

If λ = (1ψ1 + · · · + 1ψn−1) ± 1ψn, then µλ is an irreducible summand of ΛnC2n

and the same argument used above shows that µλ is not spherical.

If λ = (2ψ1 + · · · + 2ψa) + (1ψa+1 + · · · + 1ψb), with 0 < a < b < n, then µλ

is an irreducible summand of ΛaC2n ⊗ ΛbC2n. We will show that the restriction of

5It is clear that, for each S ⊆ 1, 2, . . . , 2n, (>) defines a character of M . It is also easyto check that δS w δS′ if and only if S

′is the complement of S (because any element of M has

determinant one). This gives a list of 1222n = 22n−1 inequivalent representations of M , basically all

the representations of M because M has cardinality 22n−1.

128

ΛaC2n ⊗ ΛbC2n to M does not contain the trivial representation. Hence µλ is not

spherical.

To prove that ΛaC2n ⊗ ΛbC2n does not include the M -type δ∅, we notice that every

basis vector

(−→ei1 ∧ · · · ∧ −→eia)⊗ (−→ej1 ∧ · · · ∧ −→ej b)

spans a copy of an irreducible representation δS not isomorphic to the trivial M -type.

Indeed, the cardinality of S satisfies:

0 < b− a < #S = # (i1, i2, . . . , ia4j1, j2, . . . , jb) < b < n

(4 being the symmetric difference).

If λ = 2ψ1 + · · · + 2ψk, with 0 < k < n, then µλ is an irreducible summand of

ΛkC2n⊗ΛkC2n. Showing that the representation ΛkC2n⊗ΛkC2n is spherical is easy,

but proving that also µλ is spherical is quite harder.

We start with ΛkC2n⊗ΛkC2n. The isotypic of the trivial M -type in ΛkC2n⊗ΛkC2n

has dimension(2nk

)and is spanned by the vectors

(−→ei1 ∧ · · · ∧ −→eik)⊗ (−→ei1 ∧ · · · ∧ −→eik)

with 1 6 i1 < i2 < · · · < ik 6 2n. The next step is to show that some of these copies

of δ∅ actually lie in µλ. This requires a long argument, that we explain the following

lemmas.

Lemma 5. As a representation of SO(2n,R),

ΛkC2n⊗ΛkC2n = µ∑kj=1 2ψj

⊕ µ∑k−1j=1 2ψj

⊕· · ·⊕ µ2ψ1 ⊕ µ0 ⊕ some non spherical types.

Corollary 5. The trivial M-type appears in the irreducible representation µ∑kj=1 2ψj

of SO(2n,R) with multiplicity(2nk

)-

(2n

k−1

).

We start with the proof of lemma 5.

129

Proof. For brevity of notations, we set ΛkC2n ⊗ ΛkC2n = ρ. Recall that the weights

of ρ are of the form

(±2ψi1 · · · ± 2ψir) + (±ψj1 · · · ± ψjt)

for some (distinct) indices i1, . . . , ir, j1, . . . , jt in 1, 2, . . . , n, with r + t 6 k. It

follows that ΛkC2n ⊗ΛkC2n can only include K-types whose highest weight is of the

form

υ0 = 0

υs = 2ψ1 + · · ·+ 2ψs

γc = ψ1 + · · ·+ ψc

γa+b + γb = (+2ψ1 + · · ·+ 2ψa) + (ψa+1 + · · ·+ ψa+b).

Since the K-types µγc and µγa+γa+bare non-spherical (for all possible a, b and c), ρ

can only include the spherical K-types µυ0 , µυ1 , . . . , µυk.

We want to prove that, for every s = 0, 1, . . . , k, the K-type µυs appears in the

decomposition of ρ in irreducible representations of SO(2n, R) with multiplicity one.

With this in mind, we study the decomposition of ρ in irreducible so(2n, C)-modules.6

When s = 0, i.e. υs is the 0 weight, the result follows from the fact that ΛkC2n is

self dual:7

(ΛkC2n)∗ = Λk(C2n)∗ = ΛkC2n

6Look at ρ = ΛkC2n ⊗ ΛkC2n as a so(2n, C)-representation; all the weights have integers coeffi-cients. In particular, the highest weight of each irreducible summand is analytically integral. Hence,any decomposition of ρ in so(2n, C)-modules lifts to a decomposition in SO(2n, R)-representations.

7This, in turn, follows from the fact that the standard representation of SO(2n, R) is self dual:let η = C2n, and let η∗ be its dual. Then, for each x of SO(2n, R), we have

η∗(x) = η(x−1)T = (x−1)T = x = η(x)

hence η is self dual.

130

and from this general formula for the decomposition of a tensor product in highest

weight representations:8

multV µ⊗V ν (V λ) =∑τ∈W

sgn(τ) mµ(λ + ρ− τ · (ν + ρ))

(where of course W is the Weyl group). When V ν = V µ = ΛkC2n and λ = 0, we find:

mµ(λ + ρ− τ · (ν + ρ)) = mµ(ρ− τ · (ν + ρ)) = mµ(τ−1 · ρ− (ν + ρ)) >

= mµ(−ν − ρ + τ−1 · ρ) = mµ(ν + (ρ− τ−1 · ρ)) >>

=

1 if τ = Id

0 if τ 6= Id.>>>, where

(>) follows from the fact that weights that are conjugate under W have the same

multiplicity

(>>) follows from the fact that the weights of a self-dual representation are symmet-

ric with respect to zero9, and finally

(> > >) follows from the fact that if τ is any non-trivial Weyl group element, then

(ρ−τ−1 ·ρ) is a sum of positive roots so µ+(ρ−τ−1 ·ρ) cannot be a weight. Instead,

if τ is the identity of W , then µ+(ρ−τ−1 ·ρ) = µ, and it has multiplicity one because

it is the highest weight.

Therefore:

multΛk(C2n)⊗Λk(C2n)(V0) = 1.

We now discuss the case s > 1. We intend to show that for each s = 1 . . . k, µυs =

µ∑sj=1 2ψj

is an irreducible summand of ΛkC2n ⊗ ΛkC2n.

8See [15], corollary 7.1.6.9For any representation µ, the weights of µ∗ are the negative of the weights of µ.

131

When k = 1 the result is obvious, because υ1 = 2ψ1 is the highest weight of Λ1(C2n)⊗Λ1(C2n) = C2n ⊗ C2n. Similarly, if s = k then υk =

∑kj=1 2ψj is the highest weight

of Λk(C2n)⊗ Λk(C2n).

So we assume 1 6 s 6 k − 1. The trick in this case is to pass from a calculation on

the Lie algebra so(2n, C) to a computation on gl(s, C)⊕ so(2n− 2s, C), which is the

Levi factor of a suitable parabolic subalgebra of so(2n, C). Here are all the details...

Let h be the Cartan subalgebra of k = so(2n, C) already introduced at the begin-

ning of the chapter:

h =

0 θ1 . . . 0 0

−θ1 0 . . . 0 0

......

. . ....

...

0 0 . . . 0 θn

0 0 . . . −θn 0

: θ1, θ2, . . . , θn ∈ C

and let Ψ = Ψ(k, h) be the corresponding root system:

Ψ = ±(ψi ± ψj)i, j=1,..., n, i<j.

We denote by Ψ+ the set of positive roots:

Ψ+ = ψi ± ψji, j=1,..., n, i<j

and by b the corresponding Borel subalgebra: b = h ⊕α∈Ψ+ kα. The parabolic subal-

gebras of k containing b are parameterized by the class of subsets of Ψ including Ψ+;

the one corresponding to a subset Υ is of the form q = h⊕α∈Υ kα.

The subalgebras

l = h⊕( ⊕

α∈Υ∩−Υ

kα

)u =

( ⊕

α∈Υ: −α 6∈Υ

kα

)

are respectively called the Levi factor of q and the nilpotent radical of q, and they

132

are a key tool for the study of the finite-dimensional irreducible representations of k.

Indeed each representation of k on V gives rise to a representation of l on the space

of u invariants

V u = v ∈ V : X · v = 0 ∀X ∈ u

and the structure of the l -module V u completely determines the structure of the

k -module V . In more details:10

• l is reductive. It has center h′′ =⋂

α∈Υ∩−Υ ker(α) ⊆ h and semi-simple part

lss = h′ ⊕

( ⊕α∈Υ∩−Υ

kα

)

with h′=

⊕α∈Υ∩−ΥCHα is a Cartan subalgebra of lss

• for each irreducible finite-dimensional k -module V , the corresponding l -module

V u is irreducible and has the same highest weight as V

• two irreducible finite-dimensional k -modules V1 and V2 are equivalent if and

only if the corresponding l -modules V u1 and V u

2 are equivalent

• for every irreducible finite-dimensional l -module E whose highest weight λ is al-

gebraically integral and dominant for Ψ+(k, h), there exists a finite dimensional

k -module V , with highest weight λ, such that E = V u.

We want to apply these remarks to the parabolic subalgebra corresponding to the set

of roots

Υ = Ψ+ ⊕( ⊕

16i<j6s

−(ψi − ψj)

)⊕

(n⊕

k, l=s+1, k<l

−(ψk ± ψl)

)

so that

10This material can be found in [20].

133

Υ ∩ −Υ = ±(ψi − ψj))16i<j6s ∪ ±(ψk ± ψl)k, l=s+1,...,n, k<l .

We notice that the Levi factor

l = h⊕

α∈Υ∩−Υ

kα =

(( ∑16i<j6s

H±(ψi−ψj)

)⊕

( ⊕16i<j6s

k±(ψi−ψj)

))⊕

⊕(( ∑

s+16k<l6n

H±(ψk±ψl)

)⊕

( ⊕

s+16k<l6n

k±(ψk±ψl)

))

is isomorphic to the reductive Lie algebra gl(s, C)⊕ so(2n− 2s, C). 11

Each dominant weight a1ψ1 + · · · + asψs + as+1ψs+1 + · · · + anψn of k = so(2n, C)

can be regarded as a weight of l = gl(s, C) ⊕ so(2n − 2s, C) by grouping the first s

variables and the last n− s variables together. We write

k1ξ1 + · · ·+ ksξs (k1 > k2 > · · · > ks; ki − ki+1 ∈ Z ∀ i)

for the dominant weights of gl(s, C), and

k1ζ1 + · · ·+ kn−sζn−s (k1 > k2 > · · · > kn−s; 2ki ∈ Z & ki − kj ∈ Z ∀ i, j).

the ones of so(2n− 2s, C).12 If

ν = a1ψ1 + · · ·+ asψs + as+1ψs+1 + · · ·+ anψn

is a weight of so(2n, C), we denote by

ν = (a1ξ1 + · · ·+ asξs) + (as+1ζ1 + · · ·+ anζn−s)

the corresponding weight of l = gl(s, C)⊕ so(2n− 2s, C).

It follows from the previous remarks that the so(2n, C)-representation with highest

weight υs is an irreducible summand of Λk(C2n) ⊗ Λk(C2n), with some multiplicity

ms, if and only if the gl(s, C)⊕ so(2n− 2s, C)-representation with highest weight υs

is an irreducible summand of (Λk(C2n)⊗ Λk(C2n))u = (Λk(C2n))u ⊗ (Λk(C2n))u, with

11It is gl(s, C) instead of sl(s, C) because the ψj ’s are linear independent (here there is no relation∑sj=1 ψj = 0).12Here n− s > 2, because n > k > s.

134

the same multiplicity ms.

So it all comes to understanding the decomposition of (Λk(C2n))u ⊗ (Λk(C2n))u in

irreducible gl(s, C)⊕ so(2n− 2s, C)- modules.

Since Λk(C2n) is the irreducible representation of so(2n, C) with highest weight

ψ1 + · · ·+ψs +ψs+1 + · · ·+ψk, the space of u -invariants (Λk(C2n))u is the irreducible

representation of gl(s, C)⊕ so(2n− 2s, C) with highest weight

(ξ1 + · · ·+ ξs) + (ζ1 + · · ·+ ζk−s).

It is a standard fact that the irreducible representation of so(2n−2s, C) with highest

weight (ζ1 + · · ·+ ζk−s) is Λk−sC2n−2s. The following claim identifies the irreducible

representation of gl(s, C) with highest weight (ξ1 + · · ·+ ξs).

Claim 6. The irreducible representation of gl(s, C) with highest weight (ξ1+· · ·+ξs)

is the the differential of the determinant representation of GL(s, C).

Proof. Let ξ = c1 ξ1 + · · · + cs−1 ξs−1 + cs ξs be an analytically integral form for

gl(s, C), and let Φξ be the irreducible representation of GL(s, C) with highest weight

ξ. We can describe Φξ as follows:

• The restriction of Φξ to SL(s, C) is the unique irreducible representation Φξ of

SL(s, C) with highest weight ξ = c1 ξ1 + · · ·+ cs−1 ξs−1 + cs(−ξ1 − · · · − ξs−1).

• The representation Φξ of GL(s, C) is the unique extension of Φξ to GL(s, C)

that satisfies the condition:

Φξ(zIs×s) = zc1+c2+···+csId ∀ z ∈ C∗

with Is×s the identity matrix of size s.

If A is an element of GL(s, C) and z is an sth-root of the determinant of A, so that

the matrix 1zA belongs to SL(s, C), then

135

Φξ(A) = zc1+c2+···+csΦξ

(1zA

).

Let ξ = ξ1 + · · · + ξs−1 + ξs, then ξ = 0 and Φξ is the trivial representation of

SL(s, C). For A in GL(s, C), we choose z s.t. det(A) = zs, and we obtain

Φξ(A) = z

s︷︸︸︷1 + 1 + · · ·+ 1 Φξ

(1

zA

)= zs Id = det(A)Id.

This shows that Φξ1+···+ξs−1+ξs is the differential of determinant representation of

GL(s, C), as claimed.

As a consequence, we obtain that the representation of GL(s, C)×SO(2n−2s, C)

on the space of u -invariants in Λk(C2n) is equal to (det)⊗ Λk−sC2n−2s . Hence

(Λk(C2n))u ⊗ (Λk(C2n))u = [(det)⊗ Λk−sC2n−2s]⊗ [(det)⊗ Λk−sC2n−2s] =

= [(det)2 ⊗ (Λk−sC2n−2s)⊗2].

We notice that (Λk−sC2n−2s)⊗2 contains the trivial representation of SO(2n−2s, C)

with multiplicity one.13 Therefore, the representation (Λk(C2n))u ⊗ (Λk(C2n))u of

GL(s, C)×SO(2n−2s, C) contains the irreducible representation (det)2⊗ (trivial),

also with multiplicity one.

Because (det)2 ⊗ (trivial) has highest weight

2ξ1 + · · ·+ 2ξs−1 + 2ξs + 0ζ1 + 0ζ2 + · · ·+ 0ζn−s = υs

it follows that the representation Λk(C2n) ⊗ Λk(C2n) of SO(n, C) contains the ir-

reducible representation with highest weight υs (with multiplicity one). This finally

ends the proof of the lemma.

We now give the proof of corollary 5.

13Apply the same argument used in the discussion of the case s = 0.

136

Proof. We must show that, for each k = 1 . . . n, the irreducible representation of

SO(2n,R) with highest weight∑k

j=1 2ψj contains the trivial M -type δ∅ with mul-

tiplicity(

2nk

)− (2n

k−1

).

Recall that, for each k = 1 . . . n, the multiplicity of δ∅ in ΛkC2n⊗ΛkC2n is equal to(

2nk

), and that:

ΛkC2n ⊗ ΛkC2n = µ∑kj=1 2ψj

⊕ µ∑k−1j=1 2ψj

⊕ · · · ⊕ µ 2ψ1 ⊕ µ0 ⊕ Ω

with Ω is a sum of non spherical K-types. It follows that if we denote by

cs: the multiplicity of δ∅ in µ∑sj=1 2ψj

, for all s = 1 . . . n− 1

c0: the multiplicity of δ∅ in µ0 (c0 = 1)

ck: the multiplicity of δ∅ in ΛkC2n ⊗ ΛkC2n ( ck =(2nk

))

then (2n

k

)= ck = c0 + c1 + · · ·+ ck = 1 +

k∑s=1

cs

for all k = 1 . . . n. Solving this equations inductively we find that

cs =

(2n

s

)−

(2n

s− 1

)

for all s = 1 . . . n, and this ends the proof of the corollary, and shows that µ∑sj=1 2ψj

is spherical.

To conclude our discussion of the petite spherical representations of SO(2n, R)

we must still consider the petite representations with highest weight

(2ψ1 + · · ·+ 2ψn−1)± 2ψn.

137

An analysis similar to the one carried out for the K-type 2ψ1 + · · · + 2ψk will show

that they are both spherical.

For brevity reasons, let us introduce some notations:

υs = 2ε1 + 2ε2 + · · ·+ 2εs (s = 1 . . . n− 1)

υ+n = 2ε1 + 2ε2 + · · ·+ 2εn−1 + 2εn

υ−n = 2ε1 + 2ε2 + · · ·+ 2εn−1 − 2εn

γs = 12υs = ε1 + ε2 + · · ·+ εs (s = 1 . . . n− 1)

γ+n = 1

2υ+

n = ε1 + ε2 + · · ·+ εn−1 + εn

γ−n = 12υ−n = ε1 + ε2 + · · ·+ εn−1 − εn.

We notice that, for each s = 1 . . . n− 1, µγs is the irreducible representation Λs(C2n)

of SO(2n, C), while µγ+n

and µγ−n are the two irreducible summands of Λn(C2n).

We also recall that µγs is self-dual, for each s = 1 . . . n−1. It is interesting to observe

that µγ+n

and µγ−n are not necessarily such. Indeed:

Lemma 6. If n is odd, the dual of µγ+n

is equal to µγ−n . If n is even, µγ+n

and µγ−n

are both self-dual.

Proof. It is a standard fact14 that, if V is an irreducible representation of highest

weight λ and lowest weight λ′, then V ∗ has highest weight −λ

′and lowest weight −λ.

Also, λ′= ω0 · λ, where ω0 is the long Weyl group element. It follows that Vλ is self-

dual if and only if its lowest and highest weights satisfy the formal symmetry condition

λ = −λ′= −ω0 · λ. Therefore, computing the dual of an irreducible representation

comes down to determining its lowest weight.

When n is even, the long element of the Weyl group W (so(2n, C)) is the negative of

the identity, hence λ′= ω0 ·λ = −λ, and every irreducible representation is self-dual.

14Please, refer to [15], chapter 5 for details.

138

When n is odd, we notice that the negative of a weight ω of µγ+n

is a weight of µγ−n .

Indeed, since the Weyl group of so(2n, C) only contains the sign changes that involve

an even number of signs, −ω is not conjugate to ω under W (so(2n, C)). It follows

that −ω is a weight of ΛnC that does not lie in the same irreducible representation

as ω.

Setting ω = γ+n , we deduce that −γ+

n is a weight of µγ−n . It is the lowest weight

because, for every weight ν of µγ−n , the negative −ν is a weight of µγ+n, and hence we

have:

(−ν) 4 γ+n ⇔ (−γ+

n ) 4 ν.

Since the lowest weight of µγ−n is equal to −γ+n , the dual representation has highest

weight −(−γ+n ) = + γ+

n . This concludes the proof.

Corollary 6. If n is even, both µγ+n⊗µγ+

nand µγ−n ⊗µγ−n contain the trivial K-type.

If n is odd, none of them does.

Proof. When n is even, the result is trivial: since µγ+n

and µγ−n are both self-dual,

the representations µγ+n⊗µγ+

nand µγ−n ⊗µγ−n (of K) contain the trivial K-type with

multiplicity one.15 When n is odd, we compute the multiplicity of the trivial K-type

in µγ+n⊗ µγ+

nand show that it is zero:

multµγ+n⊗µ

γ+n(µ0) =

∑τ∈W (K) sgn(τ) mµ

γ+n(0 + ρ− τ · (γ+

n + ρ)) =

=∑

τ∈W (K) sgn(τ)mµγ+n(τ−1 · ρ− (γ+

n + ρ)) =

=∑

τ∈W (K) sgn(τ) mµγ+n(− γ+

n − (ρ− τ−1 · ρ)) = >

=∑

τ∈W (K) sgn(τ) mµγ+n(+ γ−n + ρ− (ω0τ

−1ω−10 ) · ρ) =

15Apply the same argument used in the proof of lemma 5 for s = 0.

139

=∑

σ∈W (K) sgn(σ) mµγ+n(+ γ−n + (ρ− σ · ρ)) = 0 >>

Just a few comments are necessary:

in (>), we apply the long Weyl group element ω0 to the weight − γ+n − (ρ− τ−1 · ρ),

obtaining a weight with the same multiplicity. Since

ω0 · γ+n = − γ−n = the lowest weight of µγ+

n

ω0 · ρ = −ρ (because ω0 carries ∆+ in ∆−)

ω0 · (τ−1 · ρ) = −(ω0τ−1ω−1

0 ) · ρ

we find that ω0 · (− γ+n − (ρ− τ−1 · ρ)) = + γ−n + ρ− (ω0τ

−1ω−10 ) · ρ.

In (>>) we use the fact that (+ γ−n + (ρ− σ · ρ)) is never a weight of µγ+n. Indeed, if

σ is the identity element, then (+ γ−n + (ρ− σ · ρ)) = + γ−n and if γ−n were a weight

of µγ+n, then the condition γ−n ¹ γ+

n would make γ+n − γ−n = +2εn a sum of positive

roots. Similarly, if σ is a non trivial element of the Weyl group, then

γ+n − ( γ−n + (ρ− σ · ρ)) = +2εn − (ρ− σ · ρ)︸︷︷︸

a sum of pos. roots

cannot be a sum of positive roots, or 2εn would be.

We have proved that µγ+n⊗ µγ+

ndoes not contain the trivial K-type. The argument

for µγ−n ⊗ µγ−n is similar, so we omit the details.

Corollary 7. If n is odd, both µγ+n⊗µγ−n and µγ−n ⊗µγ+

ncontain the trivial K-type.

If n is even, none of them does.

Proof. Since µγ+n⊗µγ−n and µγ−n ⊗µγ+

nare isomorphic, the multiplicity of the trivial

K-type is the same in the two representations. Let us show that this multiplicity is

equal to one if n is odd, and to zero if n is even.

140

multµγ+n⊗µ

γ−n(µ0) =

∑τ∈W sgn(τ) mµ

γ+n(0 + ρ− τ · (γ−n + ρ)) =

=∑

τ∈W sgn(τ) mµγ+n(τ−1 · ρ− (γ−n + ρ)) =

=∑

τ∈W sgn(τ) mµγ+n(− γ−n − (ρ− τ−1 · ρ)) =

=∑

τ∈W sgn(τ) mµγ+n(−ω0 · γ−n + ρ− (ω0τ

−1ω−10 ) · ρ) =

=∑

τ∈W sgn(τ) mµγ+n(−ω0 · γ−n + (ρ− σ · ρ)) =

=

∑σ∈W sgn(σ) mµ

γ+n(+ γ−n + (ρ− σ · ρ)) = 0 if n is even

∑σ∈W sgn(σ) mµ

γ+n(+ γ+

n + (ρ− σ · ρ)) = 1 if n is odd.

Once these results have been established, it is pretty easy to prove that:

Lemma 7. As a representation of SO(2n, R),

µγ+n⊗ µγ+

n= (

⊕

s = 0 . . . n− 1

n− s even

µυs)⊕ µυ+n⊕ some non spherical types

µγ−n ⊗ µγ−n = (⊕

s = 0 . . . n− 1

n− s even

µυs)⊕ µυ−n ⊕ some non spherical types

µγ+n⊗ µγ−n = (

⊕

s = 0 . . . n− 1

n− s odd

µυs)⊕ some non spherical types.

141

We omit the proof of this lemma, because it is similar16 to the one already given

for the decomposition in irreducible summands of the tensor product µγk⊗ µγk

=

ΛkC2n ⊗ ΛkC2n with k < n. Once again, we need to use the “gl(s) ⊕ so(2n − 2s)

trick”.

Corollary 8. The irreducible representation µγ±n of SO(2n,R) contains the trivial

M-type with multiplicity(

2nn

)− (2n

n−1

).

Proof. Let us introduce some additional notations:

c±n : the multiplicity of δ∅ in µυ±n

cn: the multiplicity of δ∅ in ΛnC2n ⊗ ΛnC2n.

Recall that cs denotes the multiplicity of δ∅ in µ∑sj=1 2ψj

, for all s = 1 . . . n− 1, and

that cs =(

2ns

)− (2n

s−1

). Therefore,

∑n−1s=1 cs =

(2n

n−1

).

The aim of this corollary is to compute c±n . In order to find an equation for c±n , we

look at the decomposition

ΛnC2n ⊗ ΛnC2n = (µγ+n⊗ µγ+

n)⊕ (µγ−n ⊗ µγ−n )⊕ (µγ+

n⊗ µγ−n )⊕ (µγ−n ⊗ µγ+

n)

= µυ+n⊕ µυ−n ⊕ 2

( ⊕s=0...n−1

µυs

)⊕ some non spherical types

which gives:

cn = c+n + c−n + 2

n−1∑s=0

cs = c+n + c−n + 2

(2n

n− 1

). (8.1)

We start by computing cn. Consider all the vectors of the form

(−→ei1 ∧ −→ei2 ∧ · · · ∧ −→ein)⊗ (−→ei1 ∧ −→ei2 ∧ · · · ∧ −→ein)

or

(−→ei1 ∧ −→ei2 ∧ · · · ∧−→ein) ⊗ (−→ej1 ∧ −→ej2 ∧ · · · ∧ −→ejn)

16The few elements of difference have been fully outlined.

142

with i1 < i2 < · · · < in and j1 < j2 < · · · < jn complementary sets in 1 . . . 2n.These vectors form a basis for the isotypic component of the trivial M -type in

ΛnC2n ⊗ ΛnC2n. Hence, cn = 2(2nn

).

The next step is to show that c+n = c−n .

Let C be the orthogonal matrix C = diag(1, 1, . . . , 1,−1) and let Φ be the (irre-

ducible) representation of O(2n, C) on ΛnC2n. Conjugation by C preserves the sub-

group SO(2n, C) of O(2n, C), therefore Φ(C) carries a SO(2n, C)-stable subspace

of ΛnC2n into another SO(2n, C)-stable subspace. Conjugation by C also preserves

the Cartan subalgebra h, sending an element of coordinates θ1, . . . , θn−1, θn into the

element of coordinates θ1, . . . , θn−1, −θn. So Φ(C) carries a weight vector of weight

a1ψ1+· · ·+an−1ψn−1+anψn into a weight vector of weight a1ψ1+· · ·+an−1ψn−1−anψn.

It follows that Φ(C) defines an isomorphism from µυ+n

to µυ−n and, given that all

diagonal matrices commute, Φ(C) intertwines the actions of M on the two representa-

tions. We conclude that the restriction to M of µυ+n

and µυ−n are isomorphic. Hence

the trivial M -type appears in the two representations with the same multiplicity.

To conclude the proof, we use equation 8.1 to deduce that the multiplicity of the

trivial M -type in µγ±n is equal to(

2nn

)− (2n

n−1

).

It is a consequence of this corollary that µγ+n

and µγ−n are both spherical. We

obtain a complete classification of the petite spherical representations of SO(2n, R):

Proposition 3. The petite spherical representations of SO(2n, R) are exactly the

ones with highest weight

• 0ψ1 + · · ·+ 0ψn

• 2ψ1 + · · ·+ 2ψk 0 < k < n

• 2ψ1 + · · ·+ 2ψn−1 ± 2ψn.

143

Our next goal is to describe the representation of the Weyl group W = W (SL(2n, R))

on the space of M -fixed vectors of each spherical petite representation. Being W iso-

morphic to S2n, we start by recalling the representation theory of the symmetric

group.

8.3 The irreducible representations of the sym-

metric group

Definition 7. A Young diagram is a collection of boxes, arranged in left -justified

rows, with a (weakly) decreasing number of boxes in each row.

The correspondence

Young diagrams with m boxes ↔ partitions of m

is clearly bijective.

Definition 8. Let λ be a partition of m. A Young tableau is an array obtained by

filling the m boxes of the Young diagram corresponding to λ with the integers 1 . . . m

(no repetitions allowed).

For instance:

1 3 2 6 8 is a Young tableau of shape (5, 3).4 7 5

Definition 9. A standard Young tableau is a Young tableau with a filling that is

increasing across each row and also down each column.

For instance:

1 3 4 5 8 is a standard Young tableau of shape (5, 3).2 6 7

144

For any given partition λ of m, there is a natural action of the symmetric group

Sm on the set of Young tableaux of shape λ: we define σ · T to be the filling that

puts σ(i) in the box where T puts i.

For example, applying the permutation (1, 7) to the Young tableau

1 3 4 5 8 gives: 7 3 4 5 82 6 7 2 6 1 .

Fix a Young tableau T of shape λ = (λ1 > . . . λk > 0). The row group R(T ) of

T is the subgroup of Sm that consists of those permutations that permute the en-

tries of each row among themselves. It is clearly the product of symmetric groups

Sλ1 ×Sλ2 × . . .Sλk. Similarly, we define the column group C(T ) of T as the subgroup

of Sm that stabilizes each column of T .

Definition 10. Two tableaux T and T′

(of shape λ) are called row equivalent if

corresponding rows contain the same entries, i.e. if T = σ · T ′for some σ in R(T ).

The row equivalence class of T is called a tabloid (of shape λ) and is denoted by T.

A tabloid T is displayed by omitting the vertical lines between boxes, empha-

sizing that only the content of each row matters:

1 3 4 5 8 = 1 8 5 3 42 6 7 2 7 6 .

The action of Sm on the set of all tableaux with m boxes of shape λ descends to

an action on tabloids:

σ · T = σ · T

and gives rise to a representation of Sm of dimension m!(λ1!)(λ1!)···(λk!)

, that we denote by

Mλ.17

Definition 11. We call Mλ the permutation module corresponding to λ.

17Mλ is the complex vector space with basis the tabloids of shape λ.

145

An important remark:

Mλ = IndSmSλ1

×Sλ2×···×Sλk

(trivial)

where λ1 > · · · > λk > 0 are the parts of λ.

The permutation representation Mλ is (in general) reducible. We now discuss its

decomposition in irreducible summands.

For each tableau T , let κT be the sum with sign of all the permutations that stabilize

the columns of T :

κT =∑

q∈C(T )

sgn(q)q.

κT is an element of the group algebra C[Sm] and is called a Young symmetrizer.

Definition 12. If T is a tableau, we call eT = κT · T the polytabloid associated

to T .

Let us illustrate this definition by an example:

if T= 4 1 2 then κT = 1− (4, 3)− (1, 5) + (4, 3)(1, 5) and eT is given by:3 5

4 1 2 - 3 1 2 - 3 1 2 - 3 1 2

3 5 4 5 4 5 4 5 .

We notice that σ · eT = eσ·T , for each permutation σ in Sm, and for each tableau

T . Therefore, the subspace of Mλ spanned by the polytabloids is stable under the

action of Sm and defines a submodule of Mλ.

Definition 13. The Specht module Sλ corresponding to a partition λ is the sub-

module of Mλ spanned by the polytabloids of shape λ.

Theorem 13. For each partition λ of m, the Specht module Sλ is an irreducible

representation of the symmetric group Sm.

146

Theorem 14. Every irreducible representation of Sm is equivalent to a Specht module

Sλ, for some partition λ of m.

In particular, every irreducible summand of a permutation representation Mµ is

a Specht module. If we write

Mµ =⊕

λ`m

mλµ Sλ

for the decomposition of Mµ in isotypic components, then the multiplicity mλµ of Sλ

in Mµ can be explicitly described as follows:

mλµ = #semi-standard tableaux of shape λ and content ν .

More explicitly, mλµ is the number of fillings of a Young diagram of shape λ with

integers 1 . . . r, r being the number of parts of ν, so that

• the rows weakly increase;

• the columns strictly increase; and

• if µ = (µ1 > µ2 > · · · > µr > 0), then µj equals the numbers of j’s in the filling.

Suppose for example that µ = (m− k, k). For each partition λ of m, mλµ counts the

numbers of fillings of the Young diagram of λ with (m − k) integers equal to 1 and

k integers equal to 2, in such a way that the entries increase weakly across each row

and strictly across each column. Let us count these fillings.

If λ has only one row, then there exists one (and only one) filling with these properties,

namely:m−k︷︸︸︷

1 1 · · · 1 1

k︷︸︸︷2 2 · · · 2 2

If λ has three or more rows, then mλµ = 0, because to make the first column strictly

increasing, its third box should be filled up with a 3 (which is not allowed).

147

Let us now look at partitions λ with two rows, say λ = (m − s, s). Since all the

(m− k) entries equal to 1 must sit in the first row, s must be smaller or equal to k.

If this condition is met, the partition (m − s, s) admits one (and only one) filling of

the prescribed form, namely

m−k︷︸︸︷1 1 1 1 1 1 · · · 1 1

k−1︷︸︸︷2 2 · · · 2 2 if s = 1

2

m−k︷︸︸︷1 1 1 1 1 1 · · · 1 1

k−2︷︸︸︷2 2 · · · 2 if s = 2

2 2

. . .

. . .

m−k︷︸︸︷1 1 1 1 1 1 · · · 1 1

1︷︸︸︷2 if s = k − 1

2 · · 2 2︸︷︷︸k−1

m−k︷︸︸︷1 1 1 1 1 1 · · · 1 1 if s = k2 · · 2 2 2︸︷︷︸

k

We conclude that for each s = 0 . . . k, the Specht module S(m−s,s) appears in

M (m−k,k) with multiplicity one. There are no other irreducible summands. Therefore:

M (m−k,k) =k⊕

s=0

S(m−s,s).

For future reference we recall a few more facts about Specht modules.

Each box of a Young diagram determines a hook, which consists of that box and all

the boxes in its row or in its columns below the box. The hook-length of a box is the

number of boxes in its hook. We denote by hi,j the hook-length of the box in the ith

row and the jth column. For instance, in

148

· · • • • • • the box (1, 3) has hook-length 7.· · • · ·· · • ·

Theorem 15 (Hook Formula). Let λ be a partition of m. The Specht module Sλ

has dimension equal to m! divided by the product of the hook-lengths of the boxes:

dim(Sλ

)=

m!∏i,j hi,j

.

As an example, we compute the dimension of the Specht modules corresponding

to the partitions of m in at most two rows.

Assume k ≥ 1. Filling up each box of the Young diagram of λ = (m− k, k) with the

corresponding hook-length, we find:

m-k+1 m-k · · · m-2k+3 m-2k+2 m-2k · · · 2 1 .k k-1 · · · 2 1

Therefore, the product of the hook-lengths equals k!(m−k+1)!m−2k+1

, and the dimension

of the Specht module is:

dim(S(m−k,k)

)=

m!(m− 2k + 1)

k!(m− k + 1)!=

(m

k

)−

(m

k − 1

).

When λ = (m) has only one row, the computation is even easier. Indeed the hook-

length of the box in position (1, j) is equal to m + 1 − j. Hence, the product of the

hook-lengths is equal to∏m

j=1(m + 1 − j) = m! and the Specht module S(m) has

dimension 1. This is not surprising, because S(m) is the trivial representation of the

symmetric group Sm.

Theorem 16 (A basis for Sλ ). Let Sλ be the Specht module corresponding to a

149

partition λ of m. The set

eT : T is a standard tableau of shape λ

is a basis for Sλ. It consists of the polytabloids associated to all the standard tableaux

of shape λ.

Let us describe the action of a transposition (k, k + 1) on the Specht module Sλ

with respect to this basis.18 For any standard tableau T , three possibilities can occur:

(1) The indices k and k + 1 lie in the same column of T , then

(k, k + 1) · eT = − eT

(2) The indices k and k + 1 lie in the same row of T , then

(k, k + 1) · eT = eT + (a sum of other standard polytabloids eT′ , with T

′. T )

(3) The indices k and k + 1 are nor in the same row neither in the same column of

T , then

(k, k + 1) · eT = eT ′

for some standard tableau T′ 6= T .

For instance, the polytabloid eT ∈ S(m−k,k) associated to the standard tableau

T= 1 3 5 · · · 2k-3 2k-1 2k+1 2k+2 2k+3 · · · m-1 m-02 4 6 · · · 2k-2 2k .

is a simultaneous (−1)-eigenvector for the transpositions (1, 2), (3, 4) . . . (2k−3, 2k−2)

18Please, see [27] for details.

150

and (2k − 1, 2k). It is also a simultaneous (+1)-eigenvector for the transpositions

(2k + 1, 2k + 2), (2k + 3, 2k + 4) . . . .

It is clear from this example and from the considerations above that a polytabloid

of shape (m,m − k) can be a simultaneous (−1)-eigenvector of at most k disjoint

transpositions.

We conclude this section on the representations of the symmetric group Sm by

mentioning the branching rule, which describes what happens when a representation

Sλ of Sm is restricted to a subgroup isomorphic to Sm−1.

Definition 14. If λ is a Young diagram with m boxes, we define an inner corner

of λ to be a box (i, j) of λ whose removal makes λ into a Young diagram of a partition

of m− 1. Any partition obtained by such a removal will be denoted by λ−.

The inner corners of λ are exactly those boxes that sit simultaneously at the end

of a row and at the end of a column. For instance, if we mark with a bullet the inner

corners of the partition (4, 4, 1) of m = 9 we obtain:

· · · · so λ− can be (4, 3, 1) or (4, 4).· · · ••

Theorem 17 (Branching Rule). For each partition λ of m, the decomposition of

the Specht module Sλ in irreducible Sm−1 representations is given by:

Sλ ↓Sm−1=⊕

λ−Sλ− .

For instance, S(4,4,1) ↓S8= S(4,3,1) ⊕ S(4,4).

As an application of the branching rule we notice that, for every m ≥ 3, the represen-

tations of Sm whose restriction to S3 does not contain the sign representation (1, 1, 1)

correspond to partitions of m with at most two rows.

151

8.4 The “petite” Weyl group representations

Let (Eµ, µ) be a spherical representation of K, and let c be the dimension of the

isotypical component of the trivial M -type in µ. The Weyl group W of SL(2n, R)

acts on the space of M -fixed vectors EMµ , giving rise to a c-dimensional representation

that we denote by ψµ. In this section we describe the Weyl group representation ψµ

corresponding to each petite spherical K-type (Eµ, µ).

We identify the Weyl group of SL(2n, R) with the symmetric group S2n and the set

of inequivalent irreducible representations of W with the set of partitions of 2n, as

illustrated in the previous section.

If µ has highest weight 0ψ1 + · · ·+ 0ψn, then µ = µ0 is the trivial representation

of SO(2n, R). It is clear that the corresponding Weyl group representation ψµ is the

trivial representation of S2n, that we identify with the partition (2n).

If µ has highest weight 2ψ1 + · · · + 2ψk , with 0 < k < n, then µ = µυkis

an irreducible summand of ΛkC2n ⊗ ΛkC2n. The isotypic of the trivial M -type in

ΛkC2n ⊗ ΛkC2n has dimension(

2nk

)and is spanned by the vectors

(−→ei1 ∧ · · · ∧ −→eik)⊗ (−→ei1 ∧ · · · ∧ −→eik)

with 1 6 i1 < i2 < · · · < ik 6 2n. Denote by ψk the representation of Weyl group on

the space of M -fixed vectors of ΛkC2n ⊗ ΛkC2n.

Let Hk be the subgroup of S2n consisting of all the permutations of the indices

1, . . . , k, and let L2n−k be the subgroup consisting of permutations of k+1, . . . , 2n.Clearly Hk w Sk and L2n−k w S2n−k, so we write S2n−k × Sk for the direct product

L2n−k ×Hk.19 The vector

(−→e1 ∧ −→e2 ∧ −→e3 ∧ · · · ∧ −→ek )⊗ (−→e1 ∧ −→e2 ∧ −→e3 ∧ · · · ∧ −→ek )

19S2n−k × Sk consists of permutations of 1, . . . , 2n whose cycle decomposition only involvescycles that lie either in S2n−k or in Sk.

152

spans a copy of the trivial representation of the subgroup S2n−k × Sk of S2n, so ψk

contains the trivial representation of S2n−k × Sk.

It follows by Frobenius reciprocity that ψk is included in the permutation represen-

tation

M (2n−k,k) = IndS2nS2n−k×Sk

(trivial).

By dimensional reasons, ψk ≡ M (2n−k,k).

Putting together all the information we have gathered so far, we obtain:

• M (2n−k,k) =⊕

s=0...k S(2n−s,s) = S(2n) ⊕ (⊕s=1...k S(2n−s,s)

)

• ψk = ψµ0 ⊕(⊕

s=1...k ψµυs

)

(because ΛkC2n ⊗ ΛkC2n = µ0 ⊕ (⊕

s=1...k µυs)⊕ non spherical K-types)

• S(2n) = ψµ0 = the trivial repr. of W

• dim(S(2n−s,s)

)=

(2ns

)− (2n

s−1

)= dim(ψµs)

for any 0 < K < n.

It easily follows that for each s = 1 . . . k, the Weyl group representation ψµυs(on the

space of M fixed vectors of µυs) is equal to the Specht module S(2n−s,s) corresponding

to the partition (2n− s, s) of 2n. This result holds for every k, hence:

ψµυs= S(2n−s,s) ∀ s = 1, . . . , n− 1.

Finally we discuss the case in which µ = µυ+n

or µ = µυ−n , so µ has highest weight

υ±n = 2ψ1 + · · ·+ 2ψn−1 ± 2ψn and sits inside ΛnC2n ⊗ ΛnC2n.

We have already observed that the space of M -fixed vectors in ΛnC2n ⊗ ΛnC2n has

153

dimension 2 · ( 2nk

), and it is spanned by all the vectors of the form

(−→ei1 ∧ −→ei2 ∧ · · · ∧ −→ein)⊗ (−→ei1 ∧ −→ei2 ∧ · · · ∧ −→ein)

or

(−→ei1 ∧ −→ei2 ∧ · · · ∧−→ein) ⊗ (−→ej1 ∧ −→ej2 ∧ · · · ∧ −→ejn)

with i1 < i2 < · · · < in and j1 < j2 < · · · < jn complementary sets in 1 . . . 2n.Denote by ψk the corresponding Weyl group representation.

We define the subgroups Ln w Sn and Hn w Sn as above, and we write Sn×Sn for

their direct product. The vector

(−→e1 ∧ −→e2 ∧ −→e3 ∧ · · · ∧ −→en)⊗ (−→e1 ∧ −→e2 ∧ −→e3 ∧ · · · ∧ −→en)

spans a copy of the trivial representation of Sn × Sn, and so does the vector

(−−→en+1 ∧ −−→en+2 ∧ −−→en+3 ∧ · · · ∧ −→e2n)⊗ (−−→en+1 ∧ −−→en+2 ∧ −−→en+3 ∧ · · · ∧ −→e2n).

So ψn contains two copies of the trivial representation of Sn × Sn. It follows20 that

ψn = M (n,n) ⊕M (n,n).

We notice that

• M (n,n) =⊕

s=0...n S(2n−s,s) =(⊕

s=0...n−1 S(2n−s,s))⊕ S(n,n)

• ψn = 2(⊕

s=1...n−1 ψµυs

)⊕ ψµυ+

n⊕ ψµ

υ−n, because

ΛnC2n ⊗ ΛnC2n = µυ+n⊕ µυ−n ⊕ 2

( ∑s=0...n−1

µυs

)⊕ some non spherical types

20Use Frobenius reciprocity and the fact that dim(ψn) = 2 dim(IndS2n

Sn×Sn(trivial)).

154

• ψµυs= S(2n−s,s) for all s = 0 . . . n− 1, and

• dim(ψµυ±n

) =(

2nn

)− (2n

n−1

)= dim

(S(n,n).

)

Therefore, ψµυ+

n= ψµ

υ−n= S(n,n). In other words, the Weyl group representation ψµ

υ±n

(on the space of M fixed vectors of µυ±n ) is equal to the Specht module S(n,n).

8.5 Conclusions

The following chart describes the set of petite spherical representations of SL(2n, R)

and summarizes the results of the chapter.

highest weight Weyl group repr. dim.

0 = 0ψ1 + · · ·+ 0ψn S(2m) = trivial 1

2ψ1 + · · ·+ 2ψk S(2n−k,k)(2nk

)− (2n

k−1

)

0 < k < n

2ψ1 + · · ·+ 2ψn−1 ± 2ψn S(n,n)(

2nn

)− (2n

n−1

)

Conclusions. For each petite spherical representation µ of SL(2n, R), denote by ψµ

the representation of the Weyl group W ' S2n on the space of M-fixed vectors in µ.

If we identify the irreducible representations of W with partitions of 2n, the set

ψµ : µ petite and spherical

consists of the partitions of 2n in at most two parts.

155

Chapter 9

Constructing petite K-types

For each representation ( ρ, Fρ) of the Weyl group of SL(n, R) whose restriction to any

W (SL(3)) does not contain the sign representation, we construct a finite-dimensional

representation (µρ, Eµρ) of SO(n, R) with the following properties:

• for each root α of SL(n), the subgroup Kα ' SO(2) only acts with characters

zero, plus or minus one, plus or minus two (i.e. µρ is petite)

• µρ contains a M -fixed vector (i.e. µρ is spherical)

• ρ is a submodule of the representation of W (SL(n)) on the space of M -fixed

vectors of µρ .

When ρ varies in the set of Weyl group representations that do not contain any copy

of the sign representation of W (SL(3)), this construction produces all the petite

spherical representations of SO(n).

9.1 Assumptions on ρ

In this section we motivate the assumptions on the Weyl group representation ρ.

Let µ be a petite representation of K = SO(n, R), and let H be a subgroup of

K isomorphic to SO(3). The restriction of µ to L must be petite, hence it can only

156

contain the irreducible representations H0, H1 and H2 of SO(3).1

We look at the restriction of µ to M ′ ∩ SO(3), and ask whether it contains the sign

representation of S3.2 The answer is negative, because the restriction of H0, H1 and

H2 to M ′ ∩ SO(3) does not contain the sign representation.3

Hence, if we want a Weyl group representation ρ to extend to a petite (and of

course spherical) representation of K we must assume that the restriction of ρ to

any W (SO(3)) does not contain any copy of the sign representation.

Next, we interpret these assumptions on ρ in terms of partitions. We identify the

Weyl group of SL(n, R) with the symmetric group of n letters, and each irreducible

representation of W with a Specht module Sλ, for some partition λ of n.4 It follows

from the branching rule that the representation Sλ of Sn contains the sign represen-

tation (1, 1, 1) of S3 if and only if the partition λ has at least three rows.5 Hence, our

assumptions on ρ can be rephrased by saying that

ρ ' Sλ, for some λ ` n with at most two rows .

1For all N ≥ 0, the representation HN of SO(3) contains the character N of SO(2). So H0,H1 and H2 are the only petite representations of SO(3). Please, refer to chapter 7 for details andnotations.

2We can of course regard any representation of W (SO(3)) = S3 as a representation of M ′∩SO(3).3M ′ ∩ SO(3) has a total of 5 inequivalent irreducible representations: U = trivial, U ′ = sign

and V , which are the three irreducible representations of the Weyl group S3, and two additionalrepresentations, both 3-dimensional, that we will denote by ν1 and ν2. They can be realized asthe two irreducible summands of the (six-dimensional) reducible representation ν = Ind

M ′∩SO(3)M∩SO(3) δ1,

withδ1 : M → ±1, m = diag(m1, m2, m3) 7→ m1.

In particular, we denote by ν1 the irreducible summand of ν on which the element σ1,2 =exp(π

2 Zε1−ε2) acts with eigenvalues 1, ±i and by ν2 the irreducible summand of ν on whichσ1,2 acts with eigenvalues −1, ±i. Then

• H0 ↓M ′∩SO(3)= U

• H1 ↓M ′∩SO(3)= ν1

• H2 ↓M ′∩SO(3)= V ⊕ ν2.

The sign representation U1 makes its first appearance in H3.4Please, refer to chapter 8 for a quick description of Sλ, and to [13] or [27] for more detailed

analysis of the representation theory of the symmetric group.5Again, refer to chapter 8 for details.

157

Equivalently, ρ ' S(n−k,k), for some k = 0, . . . ,[

n2

].

Finally, we give some notations.

Notations. We denote by ∆ the set of restricted roots of SL(n R), and by ∆+ a

choice of positive roots. For each α ∈ ∆+, we let Eα be an element of (g0)α satisfying

the normalizing condition

B(Eα, ϑ(Eα)) =−2

‖α‖2.

We also take

Zα = Eα + ϑ(Eα)

σα = exp(

π2Zα

)

mα = exp (πZα) = σ2α.

We recall that Zα is an element of k0 = Lie(K), mα is an element of M and σα is a

representative in M ′ for the root reflection sα.

9.2 Sketch of the construction

As anticipated in the introduction, the main result of this chapter is an algorithm

that extends a certain class of representations of the Weyl group W (SL(n)) to petite

spherical representations of K = SO(n). This algorithm actually produces all the

petite spherical representations of K.

Because the construction is somehow complicated (although very natural), we provide

a sketch that the reader can keep in mind while reading through the details.

• Input: A representation ρ of W = Sn that satisfies a very natural assumption,

i.e. partition of n in at most two parts.

158

• Output: A “not too large” representation ρ′ of K that extends ρ.

More precisely, a petite spherical representation ( ρ′, Fρ′) of K such that the

representation of W on (Fρ′)M contains ρ.

• Value: On the K-type ρ′, the intertwining operator for a spherical principal

series can be computed using only the Weyl group representation ρ.

• Basic idea: As a Lie algebra representation, the differential of a petite spherical

representation of K is generated by the M -fixed vectors (through an iterated

application of the Zαs).

• Main Steps. They can be summarized as follows:

1. We regard ( ρ, Fρ) as a representation of M ′, and enlarge it by adding to

Fρ the linear span of all the formal strings

S = Zα1 . . . Zαrv

where α1, . . . , αr ∈ ∆+ are mutually orthogonal positive roots and v ∈ Fρ

is a simultaneous (−1)-eigenvector for the root reflections σα1 , . . . , σαr . We

define an action of M ′ on the space

Fρ′ ≡ Fρ ⊕ Span( strings )

by:

σ · (Zα1 . . . Zαrv) = (Ad(σ)(Zα1)) · · · (Ad(σ)(Zαr)) (σ · v).

The result of this first extension is a much larger6 representation of M ′,

that we denote by ρ′.

6Still finite-dimensional.

159

Remark: ρ′ might contain the sign representation of M ′ ∩ SO(3), so it

cannot yet be extended to a petite representation of K.

2. We take a quotient of Fρ′ modulo some suitable equivalence relations, so

to remove any copy of the sign representation of M ′ ∩ SO(3) from Fρ′ .

The result is a representation(

ρ′, F ρ′ =Fρ′R

)of M ′, that can be extended

to a petite K-type.

3. We define a representation of Lie(K) on space Fρ′ such that:

¦ as a Lie algebra representation, ρ′ is generated by Fρ. In particular,

we require that

[ ZβZα1 . . . Zαrv] = Zβ · [Zα1 . . . Zαrv] = · · · = (ZβZα1 . . . Zαr) · [v ]

¦ the representation of Lie(K) on ρ′ lifts to a petite representation of K.

In particular, we require that for each positive root β, the eigenvalues

of ρ′(Zβ) lie in the set 0, ±i, ±2i. To be more precise, we want Zβ

to act by:

0 on the (+1)-eigenspace of σβ

+i on the (+i)-eigenspace of σβ

−i on the (−i)-eigenspace of σβ

±2i on the (−1)-eigenspace of σβ.

Remark These assumptions basically determine the entire action of Zβ on

the quotient space F′( ρ) = Fρ ⊕ Spanstrings

R.

160

4. We lift the representation ρ′ of Lie(K) to a representation of K, and con-

clude the construction.

9.3 Step 1: the representation Fρ′ of M ′

Let ( ρ, Fρ) be an irreducible representation of W (SL(n)) whose restriction to any

W (SL(3)) does not contain any copy of the sign representation of W (SL(3)).7 We

can of course regard ρ as a representation of M ′.

The first step is to extend Fρ to a much larger representation of M ′, that we will

denote by Fρ′ . This extension is obtained by first adding to Fρ the formal linear span

of all the strings of the form

(a1Zα1)(a2Zα2) . . . (akZαk)v

with k > 1 and

a1, a2, . . . , ak : complex numbers

α1, α2, . . . , αk : mutually orthogonal positive roots

v ∈ Fρ : a simultaneous (−1)-eigenvector for the root reflections σα1 , σα2 , . . . , σαk,

and then imposing on the vector space

Fρ ⊕ Span(a1Zα1)(a2Zα2) . . . (akZαk)v

the following “linearity and commutativity equivalence relations”:

• (a1Zα1)(a2Zα2) . . . (akZαk)v = (a1a2 . . . ak)Zα1Zα2 . . . Zαk

v

• Zα1Zα2 . . . Zαk(av + a′v′) = a (Zα1Zα2 . . . Zαk

v) + a′ (Zα1Zα2 . . . Zαkv′)

• Zα1Zα2 . . . Zαkv = Zατ(1)

Zατ(2). . . Zατ(k)

v

7At this stage, our assumption on ρ is not critical.

161

for each choice of k > 1, τ ∈ Sk, and a, a′, a1, a2, . . . , ak ∈ C,

for all α1, α2, . . . , αk ∈ ∆+, mutually orthogonal,

and for every v, v′ ∈ Fρ, simultaneous (−1)-eigenvectors of σα1 , σα2 , . . . , σαk.

We denote the quotient space by Fρ′ :

Fρ′ =Fρ ⊕ Span(a1Zα1)(a2Zα2) . . . (akZαk

)vlinearity relations & commutativity relations

.

Because the relations are only imposed on strings of length k ≥ 1, they do not involve

elements of the original space. Therefore, we can regard the vector space Fρ′ as an

extension of Fρ. The next step is to define a representation ρ′ of M ′ on Fρ′ , which

extends the original representation ρ (of M ′ on Fρ).

For each σ in M ′, we let σ act on elements of Fρ via ρ, and on strings (of orthogonal

roots) by:

ρ′(σ) · (Zα1Zα2 · · ·Zαkv) ≡ (Ad(σ)Zα1)(Ad(σ)Zα2) · · · (Ad(σ)Zαk

)(σ · v) .

To show that this action is well defined we must check that

(i) for all strings, and all σ in M ′, ρ′(σ) · (Zα1Zα2 · · ·Zαkv) is again a well defined

string;

(ii) the linearity and commutativity equivalence relations are preserved

(iii) for all strings, and all σ1, σ2 in M ′

( ρ′(σ1 σ2)) · (Zα1Zα2 · · ·Zαkv) = ρ′(σ1) · ( ρ′(σ2) · (Zα1Zα2 · · ·Zαk

v)) .

Proof. (i) Fix an element σ of M ′. For all j = 1 . . . k, there exists a scalar cj = ±1

such that Ad(σ)(Zαj) = cjZβj

, and since

162

(Ad(σ)Zα1)(Ad(σ)Zα2) · · · (Ad(σ)Zαk)(σ · v) = (c1Zβ1)(c2Zβ2) · · · (ckZβk

)(σ · v) =

= (c1 c2 . . . ck)Zβ1Zβ2 · · ·Zβk(σ · v)

we just need to check that the roots βj s are mutually orthogonal and that the vector

(σ · v) is a simultaneous (−1)-eigenvector for σβ1 , σβ2 , . . . , σβk. We denote by w the

projection of σ in the Weyl group, then

〈βi, βj〉 = 〈w(αi), w(αj)〉 = 〈αi, αj〉 = 0.

and,

σβj· (σ · v) = σ · (σ−1σβj

σ) · v = σ · (σcjαj· v )︸︷︷︸

−v

= −σ · v.

Hence both conditions are satisfied.

(ii) It is easy to show that

• σ · ((a1Zα1)(a2Zα2) · · · (akZαk)v) = (a1 a2 . . . ak) σ · (Zα1Zα2 · · ·Zαk

v)

• σ·(Zα1Zα2 · · ·Zαk(cv + dw)) = c (σ · (Zα1Zα2 · · ·Zαk

v))+d (σ · (Zα1Zα2 · · ·Zαkw))

• σ ·(Zατ(1)

Zατ(2)· · ·Zατ(k)

v)

= σ · (Zα1Zα2 · · ·Zαkv)

for every string, for every permutation τ and for all σ in M ′. We omit the details.

(iii) Because Ad is a representation of M ′, we can write:

( ρ′(σ1 σ2)) · (Zα1Zα2 · · ·Zαkv) =

= (Ad(σ1 σ2)(Zα1)) (Ad(σ1 σ2)(Zα2)) · · · (Ad(σ1 σ2)(Zαk)) (σ1 σ2 · v) =

163

= (Ad(σ1) (Ad(σ2)(Zα1))) (Ad(σ1) (Ad(σ2)(Zα2))) · · ·· · · (Ad(σ1) (Ad(σ2)(Zα1))) (σ1 σ2 · v) =

= ρ′(σ1) · ((Ad(σ2)(Zα1)) (Ad(σ2)(Zα2)) · · · (Ad(σ2)(Zαk)) (σ1 · (σ2 · v))) =

= ρ′(σ1) · ( ρ′(σ2) · (Zα1Zα2 · · ·Zαkv))

for all σ1, σ2 in M ′, and all strings. The proof is now complete.

We conclude that:

Proposition 4. For every irreducible Weyl group representation ( ρ, Fρ), the repre-

sentation ( ρ′, Fρ′) of M ′ is well defined.

Remark 5. Fρ′ is finite-dimensional.

Proof. Because Fρ′ is a quotient of Fρ ⊕ Span(a1Zα1)(a2Zα2) . . . (akZαk)v, it is

enough to prove that there are only finitely many strings. This is obvious: there

are finitely many strings of each given length and, since the roots in each string are

required to be mutually orthogonal, the length of a string cannot exceed the maximum

number of orthogonal roots in sl(n), which is equal to[

n2

].

When ρ is a partition of n in at most two parts, it is possible to get a better

estimate of the maximum length of a string appearing in Fρ′ .

Remark 6. If ρ = (n), then Fρ′ = Fρ and there are no strings at all.

If ρ = (n − s, s), for some 1 ≤ s ≤ [n2

], then Fρ′ only contains strings of length less

than or equal to s.

Proof. The strings allowed in Fρ′ are of the form Zα1Zα2 · · ·Zαkv, with α1, α2, . . . , αk

mutually orthogonal positive roots, and v ∈ Fρ a simultaneous (−1)-eigenvector for

the root reflections σα1 , σα2 , . . . , σαk.

164

Let us look at the possible (−1)-eigenvectors of a transposition (j, j + 1) in Fρ. Pick

a basis eT of Fρ that consists of standard polytabloids.8 Then

• If j and j + 1 lie in the same column of T , then (j, j + 1) · eT = −eT .

• If j and j + 1 do not lie in the same column nor in the same row of T , then

(j, j + 1) · eT is still a standard polytabloid. We notice that j and j + 1 do

not lie in the same column nor in the same row of (j, j + 1) · eT , and that the

difference eT − (j, j + 1) · eT is a (−1)-eigenvector of (j, j + 1).

• If j and j + 1 lie in the same row of T , then (j, j + 1) · eT is not a multiple

of a standard polytabloid. The difference eT − (j, j + 1) · eT is a standard

polytabloid eS and a (−1)-eigenvector of (j, j + 1). We notice that j and j + 1

lie in the same column of S.

Therefore, if we order the basis vectors by putting first the polytabloids in which j

and j + 1 lie in the same column, then the ones in which j and j + 1 are not in the

same row, nor in the same column, and finally the ones in which j and j + 1 belong

to the same row, we obtain for ρ(j, j + 1) a matrix of the form:

ρ(j, j + 1) Ã

−I O A

O B O

O O I

in which of course I and O are respectively the identity and the zero matrix, and B

8Please, refer to section 8.3 for terminology.

165

is given by

B =

0 1 0 0 . . . 0 0

1 0 0 0 . . . 0 0

0 0 0 1 . . . 0 0

0 0 1 0 . . . 0 0

......

......

. . ....

...

0 0 0 0 . . . 0 1

0 0 0 0 . . . 1 0

.

It follows that the (−1)-eigenvectors for ρ(j, j + 1) are either of the form eT with j

and j + 1 in the same column of T , or of the form eT − eT ′ , with eT ′ = (j, j + 1) · eT

and j, j + 1 not in the same column nor in the same row of either T or T ′.9 This

argument shows that any (−1)-eigenvector of (j, j + 1) can be expressed as linear

combination of standard polytabloids in which j and j + 1 sit in different rows.

We now go back to the problem of determining the simultaneous (−1)-eigenvectors

in Fρ for k orthogonal root reflections σα1 , σα2 , . . . , σαk. Without loss of generality,

we can assume that σαt is the transposition (2t− 1, 2t), for all t = 1 . . . k.10

Any simultaneous (−1)-eigenvector for σα1 , σα2 , . . . , σαklives in the subspace of Fρ

generated by the polytabloids in which, for all t = 1 . . . k, 2t−1 and 2t sit in different

rows. We notice that

¦ because ρ = (2n− s, s), every polytabloid has exactly two rows

¦ if a polytabloid with two rows satisfies the above condition, then each of the two

rows must contain exactly one of the two elements 2t−1, 2t (for all t = 1 . . . k).

9This can be easily shown by computing the kernel of

ρ(j, j + 1) + Id Ã

O O AO B + I OO O 2I

.

10Indeed, the k orthogonal root-reflections σα1 , σα2 , . . . , σαkare conjugate in the Weyl group to

the k disjoint transpositions (1, 2), (3, 4), . . . , (2k − 1, 2k).

166

Hence the polytabloid must have at least k columns (of length two)

¦ because ρ = (2n−s, s), every polytabloid has exactly s columns (of length two).

We deduce that if k > s then there is no set of k orthogonal roots admitting a

simultaneous (−1)-eigenvector. As a consequence, F′(2n−s,s) does not contain any

string of length strictly bigger than s.

9.4 Step 2: A quotient of Fρ′

So far we have constructed an extension of the Weyl group representation ρ to a

representation ρ′ of M ′. The aim of the this section is to get rid of the copies of the

sign representation in ρ′ |M ′∩SO(3). This step is necessary, if we want to be able to

extend ρ′ to a petite representation of K.

For clarity, we divide the argument in several parts:

1. We study the restriction to M ′ ∩ SO(3) of an arbitrary representation of M ′

2. We study the restriction to M ′ ∩ SO(3) of Fρ′

3. We get rid of all the copies of the sign representation in ρ′ |M ′∩SO(3), by taking

the quotient of Fρ′ with some suitable equivalence relations.

9.4.1 The restriction to M ′ ∩ SO(3) of an arbitrary represen-

tation of M ′

Let α, β, γ be three positive restricted roots of SL(n) such that

±α, ± β, ± γ

167

is a root system of type A2.11 By construction, the elements Zα, Zβ and Zγ of so(n)

generate an so(3). We denote the subgroup M ′ ∩ SO(3) of M ′ by M ′α, β, γ.

In this subsection we study the restriction to M ′α, β, γ of an arbitrary representation

of M ′. In particular, for each (possibly reducible) representation ( η, Fη) of M ′ and

every element x of Fη, we describe the smallest M ′α, β, γ-invariant subspace Vx of Fη

containing x, and the action of M ′α, β, γ on this subspace.

Without loss of generality, we can assume that one of the following possibilities

holds:

1. σα · x = σβ · x = + x

2. σα · x = σβ · x = − x

3. σα · x = +x, σβ · x 6= +x, σ2β · x = + x

4. σα · x = − x, σβ · x 6= − x, σ2β · x = + x

5. σ2α · x = σ2

β · x = + x, x not a (±1)-eigenvector of σα, σβ, or σγ

6. σα · x = +x, σ2β · x = − x

7. σα · x = − x, σ2β · x = − x

8. σ2α · x = + x, σ2

β · x = − x, x not a (±1)-eigenvector of σα

9. none of the above, i.e. x not stable under the action of Mα, β, γ.

We discuss each case separately.

11This condition basically means that α, β and γ are mutually not orthogonal, and that γ is equalto ±(α + β) or ±(α− β). In other words, if we write

α = εi1 − εj1 β = εi2 − εj2 γ = εi3 − εj3

then the set i1, i2, i3, j1, j2, j3 has cardinality 3.

168

The first two cases are trivial: Vx = Cx and M ′α, β, γ acts on it as the trivial

representation in case (1) and the sign representation in case (2).

In case (3), Vx = < x, σβ · x, σγ · x >C , and it is either two or three-dimensional.

We need to distinguish between these two possibilities.

If dim(Vx) = 2, then x and σβ · x form a basis of Vx.12 Because σ2

β · x = + x, σβ acts

on Vx with eigenvalues ±1. We write σγ · x = a x + b σβ · x, and observe that w.r.t.

the basis x, σβ · x of Vx, the root reflections σα and σγ act by:

σα Ã

1 a

0 b

σγ Ã

a 0

b 1

.

Since σα, σβ and σγ are all conjugate, they must act with the same eigenvalues. We

deduce that a = b = −1, and the relation

σα · x + σβ · x + σγ · x = 0

holds. Therefore, Vx is a copy of the standard representation of M ′α, β, γ.

If dim(Vx) = 3, then x, σβ ·x and σγ ·x are linearly independent. We can decompose

Vx as the sum of two irreducible M ′α, β, γ-invariant subspaces:

Vx = C(x + σβ · x + σγ · x) ⊕ < x− σβ · x, x− σγ · x >C

that are respectively a copy of the trivial representation and a copy of the standard

representation of M ′α, β,γ.

Case (4) is similar: Vx is again equal to < x, σβ · x, σγ · x >C , and it is either two

12Notice that σβ · x 6= x by assumption, and that σβ · x 6= −x or Cx would be a one-dimensionalrepresentation of M ′

α, β, γ different from both the trivial and the sign representation. Because σ2β ·x =

+ x, it cannot be any other multiple of x.

169

or three-dimensional.

If dim(Vx) = 2, we choose the basis x, σβ ·x of Vx, and we write σγ ·x = a x+b σβ ·x.

It is easy to see that the root reflections σα, σβ and σγ act by:

σα Ã

−1 −a

0 −b

σβ Ã

0 1

1 0

σγ Ã

a 0

b −1

.

Since σα, σβ and σγ are all conjugate, they must act with the same eigenvalues. We

deduce that a = 1 and b = −1. Again, the relation

σα · x + σβ · x + σγ · x = 0

holds, and Vx is a copy of the standard representation of M ′α, β, γ.

If dim(Vx) = 3, then x, σβ ·x and σγ ·x are linearly independent. We can decompose

Vx as the sum of two irreducible M ′α, β, γ-invariant subspaces:

Vx = C(x− σβ · x− σγ · x) ⊕ < x + σβ · x, x + σγ · x >C

that are respectively a copy of the sign representation and a copy of the standard

representation of M ′α, β,γ.

In case (5), Vx = x, σα ·x, σβ ·x, σγ ·x, (σασβ)·x, (σασγ)·x. We set y = x+σα ·xand z = x−σα ·x, and we notice that Vx decomposes as the direct sum of two M ′

α, β, γ-


Vx = < y, σβ · y, σγ · y >C︸︷︷︸V +

x

⊕ < z, σβ · z, σγ · z >C︸︷︷︸V −x

.

By the previous arguments, V +x is either a copy of the trivial representation, or a copy

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of the standard representation or a sum of the two. Similarly, V −x is either a copy of

the sign representation, or a copy of the standard representation or a sum of the two.

Cases (6) and (7) are similar. In both cases Vx = < x, σβ · x, σγ · x >C. This

space is always three-dimensional, and it is an irreducible representation of M ′α, β, γ

isomorphic to ν1 in case (6), and to ν2 in case (7).13

In case (8), Vx = x, σα ·x, σβ ·x, σγ ·x, (σασβ)·x, (σασγ)·x. We set y = x+σα ·xand z = x−σα ·x, and we notice that Vx decomposes as the direct sum of two M ′

α, β, γ-


Vx = < y, σβ · y, σγ · y >C︸︷︷︸V +

x

⊕ < z, σβ · z, σγ · z >C︸︷︷︸V −x

.

By the previous arguments, V +x is always a copy of the three-dimensional represen-

tation ν1, and V −x is always a copy of the three-dimensional representation ν2.

Case (9) can be reduced to the previous cases on the basis of the following obser-

vations: the elements σ2α, σ2

β, σ2γ of Mα, β, γ commute, each of them has order two and

their product equals the identity. Therefore, for any representation η of M ′ we have

η(σ2α) = η(σ2

β) η(σ2γ) = η(σ2

γ) η(σ2β) = η(σ2

α)−1

η(σ2β) = η(σ2

α) η(σ2γ) = η(σ2

γ) η(σ2α) = η(σ2

β)−1

η(σ2γ) = η(σ2

β) η(σ2α) = η(σ2

α) η(σ2β) = η(σ2

γ)−1.

It follows that each of the vectors

13Recall that ν1 and ν2 are the two irreducible summands of IndM ′M δ1. A root reflection acts on

ν1 with eigenvalues 1, ±i and on ν1 with eigenvalues −1, ±i.

171

x0 = x + σ2α · x + σ2

β · x + σ2γ · x

x1 = x + σ2α · x − σ2

β · x − σ2γ · x

x2 = x − σ2α · x + σ2

β · x − σ2γ · x

x3 = x − σ2α · x − σ2

β · x + σ2γ · x

spans a one-dimensional representation of Mα, β, γ.14 Hence, we can study the space

Vxjby means of the same techniques used in (1)-(8).

This concludes our analysis of the restriction of an arbitrary representation (η, Fη)

of M ′ to M ′α, β, γ. In the next subsection, we set Fη = Fρ ⊕ Spanstrings.

9.4.2 The restriction of Fρ′ to M ′α, β, γ

With abuse of notations we identify a string with its equivalence class modulo the

commutativity and linearity relations. In particular, we think of Fρ′ as the vector

space generated by Fρ and by all the strings of the form

Zα1 . . . Zαrv

with α1 . . . αr mutually orthogonal positive roots , and v an element of Fρ satisfying

σα1 · v = · · · = σαr · v = −v.

The purpose of this section is the study of the action of M ′α, β, γ on Fρ′ . Precisely, we

ask the following questions:

1. When is C(Zα1 . . . Zαrv) stable under the action of Mα, β, γ?

2. When is C(Zα1 . . . Zαrv) a copy of the trivial representation of M ′α, β, γ?

14The representations Cxjj=0...3 are isomorphic to δ0, δ1, δ2, δ3.

172

3. When is C(Zα1 . . . Zαrv) a copy of the sign representation of M ′α, β, γ?

The answer to this question is the same for all the split groups with one root lenght,

so we formulate it in general terms:

Proposition 5. Let G be a semi-simple split group whose root system admits only

one root-length. Fix a Weyl group representation ρ that does not contain any copy

of the sign representation of W (SL(3)), and form strings of orthogonal roots S =

Zα1 . . . Zαrv, just like in the case of SL(n). Define an action of M ′ on the linear span

of the strings by

σ · (Zα1 . . . Zαrv) = (Ad(σ)Zα1) · · · (Ad(σ)Zαr)(σ · v).

For each triple of positive roots α, β, γ forming an A2, consider the subgroup M ′α, β, γ

of M ′ generated by σα, σβ and σγ. Then

1. C(Zα1 . . . Zαrv) is always stable under the action of Mα, β, γ, hence it defines a

one-dimensional irreducible representation of Mα, β, γ.

2. C(Zα1 . . . Zαrv) can never be a copy of the sign representation of M ′α, β, γ

3. C(Zα1 . . . Zαrv) is a copy of the trivial representation of M ′α, β, γ if and only if the

roots α, β, γ are all orthogonal to α1 . . . αr, and the following condition holds:

σα1 · v = · · · = σαr · v = + v.

These results follow easily from some properties of the split groups whose root

systems have one root-length. We describe such properties in the next few lemmas.

Lemma 8. Let G be a split group whose root system ∆ admits only one root-length.

173

Then for all α, β in ∆, we have:

Ad(σ2β)(Zα) =

+Zα if α = β or α ⊥ β

−Zα otherwise.

Proof. For any split group, and all roots α, β in ∆, the element mα = σ2α acts on Zβ

by the scalar ±1. Precisely15

Ad(σ2β)(Zα) = (−1)〈α, β〉Zα.

Let us focus on the case in which the root system has one root-length. The Cartan

integer 〈α, β〉 takes the value:

±2 if α = ± β

0 if α ⊥ β

±1 otherwise.

This makes Ad(σ2β)(Zα) =

+Zα if α = β or α ⊥ β

−Zα otherwise.

Corollary 9. For every string S = Zα1 . . . Zαrv and for every positive root β

σ2β · (Zα1Zα2 · · ·Zαk

v) = (−1)#j : [ Zβ , Zαj ]6=0 (Zα1Zα2 · · ·Zαkv) .

Proof. It follows from the previous lemma and from the fact that σ2β · v = v, because

ρ is a Weyl group representation.

Remark 7. This result does not extend to the case in which ρ is a representation of

M ′, but not a Weyl group representation.

Remark 8. Corollary 9 shows part (1) of proposition 5.

15Please, refer to [30], lemma 4.3.19 for a proof. In this lemma the result is stated for G quasi-split(i.e. with m0 abelian) and β real, but of course both assumption are satisfied when G is split.

174

Lemma 9. Let Φ be a root system with one root-length.16 For all α, β in Φ, sβ(α)

cannot be orthogonal to α.

Proof. Set γ = sβ(α) = α− 2 〈β, α〉〈β, β〉 β, and suppose by contradiction that γ is orthog-

onal to α. Then

‖γ‖2 = 〈 γ, α− 2 〈β, α〉〈β, β 〉 β〉 = −2 〈β, α〉

〈β, β〉〈γ, β〉 ⇔

⇔ 2 =(−2 〈β, α〉

‖β‖2)(

2 〈γ, β〉‖γ‖2

)⇔

⇔ 2 =(−2 〈β, α〉

‖β‖2)(−2 〈β, α〉

‖β‖2)⇔ (>)

⇔ 2 =(2 〈β, α〉‖β‖2

)2

.

We reach a contradiction, because the quantity(2 〈β, α〉‖β‖2

)is an integer.

In (>) we used the two equalities

‖β‖2 = ‖γ‖2(= ‖α‖2) 〈γ, β〉 = −〈sβ(α), sβ(β)〉 = −〈α, β〉.

Remark 9. This result does not extend to the case in which Φ has two root-lengths.

For instance, in type B2

sε1(ε1 − ε2) = −(ε1 + ε2) ⊥ (ε1 − ε2)

and in type C2

s2ε1(ε1 − ε2) = −(ε1 + ε2) ⊥ (ε1 − ε2).

16Φ can be of type An, Dn, E6, E7, E8.

175

If Φ has two root-lengths, then we can conclude that sβ(α) is not orthogonal to α only

when α and β have the same length.

Corollary 10. Let S = Zα1 . . . Zαrv be a well defined string, i.e. let α1, . . . , αr be

mutually orthogonal positive roots and let v be an element of Fρ satisfying

σα1 · v = · · · = σαr · v = − v.

Let ν be any positive root. Then

(i) S = Zα1 . . . Zαrv is a (+1)-eigenvector of σν if and only if the following condi-

tions are satisfied:

σν · v = + v

ν ⊥ α1, . . . , αr.

(ii) S = Zα1 . . . Zαrv is a (−1)-eigenvector of σν if and only of the following condi-

tions are satisfied:

σν · v = − v

ν belongs to the set α1, . . . , αr, or it is orthogonal to it.

Proof. (i). Suppose that σν · (Zα1 . . . Zαrv) = +(Zα1 . . . Zαrv), i.e.

(Ad(σν)(Zα1))(Ad(σν)(Zα2)) · · · (Ad(σν)(Zαr))(σν · v) = + Zα1 . . . Zαrv.

We notice that for all j = 1 . . . r, Ad(σν)(Zαj) = ±Zsν(αj). So if sν(αj) 6= ±αj, then

sν(αj) must belong to the set α1, . . . , αr − αj and hence must be orthogonal to

αj. We reach a contradiction. Hence, either ν belongs to the set α1, . . . , αr, or it

is orthogonal to it.

We exclude the first possibility, because it would make Zα1 . . . Zαrv a (−1)-eigenvector

176

of σν . Then ν must be orthogonal to α1, . . . , αr (and v must necessarily be a (+1)-

eigenvector of σν). This concludes the proof of (i). The proof of (ii) is similar.

As a consequence, we immediately obtain that

Corollary 11. For any string S = Zα1 . . . Zαrv

(a) C(Zα1 . . . Zαrv) can never be a copy of the sign representation of M ′α, β, γ.

17

(b) C(Zα1 . . . Zαrv) is a copy of the trivial representation of M ′α, β, γ if and only if the

roots α, β, γ are all orthogonal to α1 . . . αr, and the following condition holds:

σα1 · v = · · · = σαr · v = + v.

Remark 10. This corollary concludes the proof of proposition 5.

9.4.3 Eliminating the sign from Fρ′

A combined application of the results of the past subsections shows that the sign

representation of M ′α, β, γ appears in Fρ′ in one of the following forms:

(a) C(S − σβ · S − σγ · S), for a string S satisfying σα · S = −S and σ2β · S = S

(b) C(S−σα ·S), for a string S satisfying σ2α ·S = σ2

β ·S = S and σβ · (S−σα ·S) =

−(S − σα · S)

(c) C((S − σα · S) − σβ · (S − σα · S) − σγ · (S − σα · S)), for a string S satisfying

σ2α · S = σ2

β · S = S and σβ · (S − σα · S) 6= −(S − σα · S).

17Or Cv would be a copy of sign in Fρ.

177

We look for equivalence relations to be imposed on Fρ′ in order to annihilate each of

these possible copies of the sign representation. The first step is to set

S = σβ · S + σγ · S (?)

for all strings S satisfying σα · S = −S and σ2β · S = S.

It is a good choice, because the (?)-relations annihilate all the sign representations of

type (a), and they span a subspace of Fρ′ which is invariant under M ′.18

A natural question arises:

Question. What happens to the sign representations of type (b) and (c) when we

impose the (?)-relations?

Surprisingly enough, they are also annihilated.

Proof. Suppose that the string S satisfies σ2α ·S = σ2

β ·S = S and σβ · (S− σα ·S) =

−(S − σα · S), so that (S − σα · S) is a sign representation of type (b). Let us show

that the equivalence class of (S − σα · S) modulo the (?)-relations is zero.

We start by observing that, if we write S = Zα1 . . . Zαrv, then α cannot be orthogonal

to α1 . . . αr. Indeed, if α were orthogonal to α1 . . . αr, then the difference S−σα ·S =

Zα1 . . . Zαr(v − σα · v) would be a string19 on which both σα and σβ act by (−1).

18For all positive roots ν, α, β, γ such that ±α, ±β, ± γ = A2 and for all strings S such thatσα · S = −S and σ2

β · S = S, we have:

σν · S = σν · (σβ · S) + σν · (σγ · S) .

This relation is equivalent to

(σν · S) = σsν(β) · (σν · S) + σsν(γ) · (σν · S) ,

which is again a relation of type (?), for the roots sν(α), sν(β), sν(γ) (that form an A2) and thestring (σν · S) (that satisfies sν(α) · (σν · S) = −(σν · S) and sν(β)2 · (σν · S) = +(σν · S)).

19If α is orthogonal to α1 . . . αr, then σα · v is a simultaneous (−1)-eigenvector of σα1 . . . σαr .Therefore, it follows from the linearity conditions that

S − σα · S = Zα1 . . . Zαrv − Zα1 . . . Zαr (σα · v) = Zα1 . . . Zαr (v − σα · v).

178

As a consequence, the element v − σα · v of Fρ would generate a copy of the sign

representation of M ′α,β,γ. We reach a contradiction.

Therefore, we can assume that α is not orthogonal to α1. Notice that α1 cannot be

equal to α, or σα would act by (−1)-on the string S = Zα1 . . . Zαrv. So we can find a

positive root α′1 such that the set ±α, ±α1,±α′1 forms a root system of type A2.

Because σα1 · S = −S and σ2α · S = S, also σ2

α′1· S = S and we have a relation:

S = σα · S + σα′1 · S.

Let us look at T = σα′1 · S. By construction, the string T includes the root α, so

σα ·T = −T , and σβ ·T 6= ±T . Since σ2β ·T = σ2

β · (σα′1 ·S) = +T , we have a relation:

T = σβ · T + σγ · T.

In the quotient space of Fρ′ modulo the (?)-relations, we obtain:

[S − σα · S] = [T ] = σβ · [T ] + σγ · [T ] = σβ · [S − σα · S] + σγ · [S − σα · S] =

= [σβ · (S − σα · S)] + [σγ · (S − σα · S)] = −2 [S − σα · S]

hence [S − σα · S] = 0.

This shows that passing to the quotient modulo the (?)-relations eliminates all the

sign representations of type (b). We must prove that also the sign representations of

type (c) are annihilated.

Suppose that the string S satisfies σ2α · S = σ2

β · S = S and

σβ · (S − σα · S) 6= −(S − σα · S), so that

C((S − σα · S)− σβ · (S − σα · S)− σγ · (S − σα · S))

179

is a sign representation of type (c). The same argument used above shows that

S − σα · S is again a string.20 We call such a string T . Because σα · T = −T and

σ2β · T = T , we have a relation T = σβ · T + σγ · T . In the quotient space:

[(S − σα · S)− σβ · (S − σα · S)− σγ · (S − σα · S)] =

= [S − σα · S]− σβ · [S − σα · S]− σγ · [S − σα · S] =

= [T ]− σβ · [T ]− σγ · [T ] = [T − σβ ·T − σγ ·T ] = [ 0 ]. The answer to our question

is now complete.

Corollary 12. The (?)-relations are the correct set of relations that one needs to

impose in order to remove all the copies of the sign representation from Fρ′.

Definition 15. Let R be the subspace of Fρ′ generated by all the relations:

σα · S + σβ · S + σγ · S = 0 ⇔ S = σβ · S + σγ · S (?)

where S is a string satisfying the following two conditions:

1. S is a (−1)-eigenvector of the root reflections σα

2. σ2β · S = +S.

If we write S = Zα1 . . . Zαrv, the first condition means that σα · v = −v and that

the root α either belongs to the set α1, . . . , αr, or it is orthogonal to it. The second

condition means that β fails to be orthogonal to an even number of αis. Notice

that γ automatically satisfies the condition σ2γ · S = +S, hence CS is the trivial

representation of Mα,β,γ.

20This time α is allowed to be orthogonal to α1 . . . αr. If this is the case, then S − σα · S =Zα1 . . . Zαr (v − σα · v). Therefore S − σα · S is a string containing only roots orthogonal to α.

180

Definition 16. Let Fρ′ be the quotient space: Fρ′ =Fρ′

R.

Because R is stable under M ′, the action of M ′ on Fρ′ descends to an action on

the quotient space. The result is a well defined representation of M ′ on Fρ′, that we

denote by ρ′.

Remark 11. ρ′ does not contain any copy of the sign representation of M ′∩SO(3).

Also notice that, for any positive root ν, all the (−1)-eigenvectors of σν on Fρ′ are

of the form

[Zα1 . . . Zαrv]

where v ∈ Fρ is a simultaneous (−1)-eigenvector of sν , sα1 , . . . , sαr , and I = α1, . . . , αris a set of mutually orthogonal positive roots such that ν either belongs to I or it

orthogonal to it.

9.5 Step 3: The action of Lie(K) on Fρ′

The purpose of this section is to define an action of Lie(K) on Fρ′ that satisfies the

following properties:21

(1) as a Lie algebra representation, it is generated by Fρ

(2) for each positive root α, the eigenvalues of Zα ∈ Lie(K) on Fρ′ belong to the

set 0, ±i, ±2i

(3) the representation of Lie(K) on Fρ′ lifts to a representation µ of K such that

¦ µ is petite

¦ µ is spherical

¦ the restriction of µ to M ′ coincides with ρ′.

21These properties are necessary for the existence of a lifting of the action of Lie(K) to a petitespherical representation of K that extends ρ.

181

In particular, the action of Lie(K) must satisfy the condition:

exp((π

2Zβ

)· [ x ]

)=

(exp

(π

2Zβ

))· [ x ] = ρ′(σβ) · [ x ]

for all [ x ] in Fρ′ (i.e. for all x in Fρ′ ), and for all positive roots β. Adding the

“petite-ness” requirement, we obtain the the following set of necessary equivalences:

E1 : (0)-eigenspace of Zβ ≡ (+1)-eigenspace of σβ

E2 : (i)-eigenspace of Zβ ≡ (i)-eigenspace of σβ

E3 : (−i)-eigenspace of Zβ ≡ (−i)-eigenspace of σβ

E4 : (2i)-eigenspace of Zβ ⊕ (−2i)-eigenspace of Zβ ≡ (−1)-eigenspace of σβ .

Let us analyze each of these equivalences in more details.

• E1 implies that for any x ∈ Fρ′ fixed by σβ, we must define Zβ · [ x ] = 0.22

• E2 and E3 imply that each (±i)-eigenvector of σβ in Fρ′ is also an eigenvector

of Zβ (relative the same eigenvalue).

Hence, for all vectors x in Fρ′ satisfying σ2β · x = − x , we must have:

Zβ · [ σβ · x− i x ] = −i[ σβ · x− i x ]

Zβ · [ σβ · x + i x ] = +i[ σβ · x + i x ]

As a result, we need to set: Zβ · [ x ] = σβ · [ x ], for all x ∈ Fρ′ s.t. σ2β ·x = − x .

22Recall that any string fixed by σβ is of the form

Zα1Zα2 · · ·Zαrv

with β is orthogonal to α1 . . . αr and σβ(v) = +v.

182

• E4 is slightly more subtle. Because we want the Lie algebra representation to

be generated by Fρ, we must define Zβ · [ v ] = [ Zβv ] for all v in Fρ for which

the string Zβv makes sense, i.e. for all (−1)-eigenvectors of σβ in Fρ.

Similarly, if Zα1Zα2 · · ·Zαrv is a string such that β is orthogonal to α1 . . . αr

and σβ(v) = −v (so that ZβZα1Zα2 · · ·Zαrv is a well defined string), we expect

Zβ · [Zα1Zα2 · · ·Zαrv] = [ ZβZα1Zα2 · · ·Zαrv ].

The two dimensional subspace

Span ( [ Zα1Zα2 · · ·Zαrv ], [ ZβZα1Zα2 · · ·Zαrv] )

is included in the (−1)-eigenspace of σβ, so we want Zβ to act on it with

eigenvalues ±2 i. This can be achieved by defining23

Zβ · [ ZβZα1Zα2 · · ·Zαrv ] = −4[ Zα1Zα2 · · ·Zαrv ].

These remarks basically determine the entire action of Zβ on the space

F ρ′ = Fρ ⊕ SpanstringsR

.

The action of Zβ on Fρ

The root reflection σβ acts semi-simply on Fρ, with eigenvalues +1 and −1. Therefore

we can decompose Fρ as the direct sum of the (+1)− and the (−1)-eigenspace of σβ.

We define the action of Zβ separately on the two eigenspaces.

23With respect to the basis [ Zα1Zα2 · · ·Zαrv ], [ ZβZα1Zα2 · · ·Zαrv], we will have:

Zβ Ã(

0 −41 0

).

183

• If v ∈ Fρ is a (+1)-eigenvector of σβ, we define Zβ · [ v ] = 0.

• If v ∈ Fρ is a (−1)-eigenvector of σβ, we define Zβ · [ v ] = [ Zβv ].

The action of Zβ on a string [ S ] = [ Zα1Zα2 · · ·Zαrv ]

We distinguish various cases:

• If σ2β · [ S ] = − [ S ] , then we set Zβ · [ S ] = [ σβ · S ].

• If σ2β · [ S ] = +[ S ], then we distinguish three cases:

1. σβ · [ S ] = +[ S ] , then we define Zβ · [ S ] = [ 0 ].

2. σβ · [ S ] = − [ S ] . Then there are two possibilities:

(i) [ S ] = [ Z α1Zα2 · · ·Zαrv ], with β orthogonal to α1 . . . αr and σβ(v) =

−v. In this case, we define

Zβ · [ S ] = Zβ · [ Zα1Zα2 · · ·Zαrv ] = [ ZβZα1Zα2 · · ·Zαrv ]

(ii) = [ Zα1Zα2 · · ·Zαrv ], with β = α1 and σβ(v) = −v. We define

Zβ · [ S ] = Zβ · [ ZβZα2 · · ·Zαrv ] = −4[Zα2Zα3 · · ·Zαrv ].

3. If σ2β · [ S ] = +[ S ] but σβ · x 6= ±x , then the vector [S + (σβ · S) ] is a

(+1)-eigenvector for σβ. Therefore, Zβ · [ S ] = −Zβ · [ σβ · S ], and we can

write:

Zβ · [ S ] =1

2(Zβ · [ S ]− Zβ · [ σβ · S ]) =

1

2Zβ · [S − (σβ · S) ].

Let us look at [S − (σβ · S) ] , which is a (−1)-eigenvector for σβ in F ρ′ .

There are two possibilities:

184

(i) β is orthogonal to α1 . . . αr and of course σβ(v−σβ ·v) = −(v−σβ ·v).

Then

[S − (σβ · S) ] = [ Zα1Zα2 · · ·Zαr(v − σβ · v) ]

and, by 2(i), we must define:

Zβ · [ S ] =1

2Zβ · [S − (σβ · S) ] =

1

2[ ZβZα1Zα2 · · ·Zαr(v − σβ · v) ].

(ii) There exists at least one root αi1 to which β is not orthogonal. Because

β 6= αi124, and (−1)#j : [ Zβ , Zαj ]6=0 = +1 25, there exists one (and only

one) other root αi2 that is not orthogonal to β, and different from β.

Without loss of generality we can assume that αi1 = 1 and αi2 = 2, so

that:

β 6⊥ α1, β 6⊥ α2 β 6= α1, β 6= α2 β ⊥ α3, α4 . . . αr.

For i = 1, 2, let βi be the positive root such that αi, β, βi is a root

system of type A2, and let εi = ±1 be such that [ Zβi, Zαi

] = εiZβ. We

notice that for both i = 1, 2 we have

σαi· S = −S

σ2β · S = + S

so, when we pass to the quotient, we must impose the relations

[ S ] = σβ · [ S ] + σβ1 · [ S ]

24Or Zβ · [ Zα1Zα2 · · ·Zαrv ] = −[Zα1Zα2 · · ·Zαrv ].25Indeed

σ2β · (Zα1Zα2 · · ·Zαk

v) = (−1)#j : [ Zβ , Zαj]6=0 (Zα1Zα2 · · ·Zαk

v) = + (Zα1Zα2 · · ·Zαkv)

by assumption.

185

[ S ] = σβ · [ S ] + σβ2 · [ S ].

Therefore, we should define:

Zβ · [ S ] ≡ 1

2Zβ · [ S− (σβ · S) ] =

1

2Zβ · [σβ1 ·S] =

1

2Zβ · [σβ2 ·S]. (>)

We notice that:

[ σβ1 · S ] = σβ1 · [ Zα1Zα2 · · ·Zαrv ] =

= [ (Ad(σβ1)Zα1) (Ad(σβ1)Zα2) · · · (Ad(σβ1)Zαr) (σβ1 · v) ] =

= [ (ε1Zβ) (Ad(σβ1)Zα2) · · · (Ad(σβ1)Zαr) (σβ1 · v) ].

Therefore:

1

2Zβ · [ σβ1 · S ] = −2 ε1 σβ1 · [ Zα1︸︷︷︸

omitted

Zα2Zα3 · · ·Zαrv ].

Similarly, 12Zβ · [ σβ2 · S ] = −2 ε2 σβ2 · [ Zα1 Zα2︸︷︷︸

omitted

Zα3 · · ·Zαrv ].

So the definition in (>) actually makes sense only if we show that

ε1σβ1 · [ Zα2Zα3 · · ·Zαrv ] = ε2σβ2 · [ Zα1Zα3 · · ·Zαrv ].

Since both β1 and β2 are orthogonal to α3 . . . αr, this is equivalent to

proving that

[ε1σβ1 · (Zα2v)] = [ε2σβ2 · (Zα1v)].

186

This is not hard to check. Comparing the equations26

[ Zα1Zα2v ] = σβ · [ Zα1Zα2v ] + σβ1 · [ Zα1Zα2v ] (I)

[ Zα1Zα2v ] = σβ · [ Zα1Zα2v ] + σβ2 · [ Zα1Zα2v ] (II)

we find:

σβ1 · [ Zα1Zα2v ] = σβ2 · [ Zα1Zα2v ] ⇔

⇔ [ε1Zβ(σβ1(·Zα2v))] = [ε2Zβ(σβ2 · (Zα1v))].

Now we just have to apply Zβ to obtain the result:

−2ε1[σβ1 · (Zα2v)] = −2ε2[σβ2 · (Zα1v)].X

This concludes the case-by-case analysis of the definition of the action of Z(β) on Fρ′ .

Because the elements Zββ∈∆+ generate the Lie algebra of K, the entire action of

Lie(K) is determined.

A picture for action of Zβ on Fρ′

The following diagram shows how to define the action of Zβ on a string

[ S ] = [ Zα1 . . . Zαrv ].

26The first equation is the equivalence relation ? for the triple α1, β, β1 (which of course gen-erates an so(3)), and the string S = Zα1Zα2v. It holds because:

σα · S = −S σ2β · S = σ2

γ · S = S σβ · S 6= ±S σγ · S 6= ±S.

The second equation is the corresponding equation for the triple α2, β, β2.

187

N

jjjjjjjjjjjjjjjj

UUUUUUUUUUUUUUUUUUUU

σ2β · [ S ] = [ S ] σ

2β · [ S ] = −[ S ]

def.

®¶• Zβ · [ S ] = σβ · [ S ]

•

kkkkkkkkkkkkkkkk

TTTTTTTTTTTTTTTTTTTTTT

σβ · [ S ] = +[ S ]

def.

®¶

σβ · [ S ] = −[ S ] • σβ · [ S ] 6= ±[ S ]

def.

®¶Zβ · [ S ] = 0 •

llllllllllllllllll

SSSSSSSSSSSSSSSSSS

em S S S S S S S S S

S S S S S S S S SZβ · [ S ] = 1

2 Zβ · [ S − σβ · S ]_ _ _ _ _ _ _ __ _ _ _ _ _ _ _

ÂÂÂÂ

β = α1

def.

®¶

β 6∈ α1, . . . , αr

def.

®¶ ÂÂÂÂ

Zβ · [ Zα1 · · ·Zα1v ] =

= −4 [ Zα2Zα3 · · ·Zαr v ]

Zβ · [ Zα1 · · ·Zα1v ] =

[ ZβZα1 · · ·Zαr v ]

ÂÂÂÂ

h h h h h h h h h h

ÂÂÂ

β ⊥ α1, . . . , αr

®¶ÂÂÂ

ÂÂÂ

β 6⊥ α1

®¶ÂÂÂ

ÂÂÂ

Zβ · [ Zα1 · · ·Zα1v ] =

12 [ ZβZα1 · · ·Zαr (v − σβ · v) ]

Zβ · [ Zα1 · · ·Zα1v ] =

−2 ε1σβ1 · [ Zα2 · · ·Zαr v ]

β1, β, α1 Ã A2

OO

Some remarks

The action of Lie(K) on Fρ′ is now understood. Of course, we still have to show

that this definition gives rise to as well defined Lie algebra representation, which is

endowed of all the desired properties.

188

In particular, we must verify that:

¦ The action of Lie(K) preserves the (?)-relations.

¦ The action of Lie(K) satisfies the bracket relations.

¦ The following “eigenvalue condition” holds: for all positive roots β, the element

Zβ of Lie(K) acts on Fρ′ with eigenvalues in the set 0,±i,±2i.

¦ The action of Lie(K) lifts to a representation of K.

¦ The lifting is petite.

¦ The restriction of the lifting to M ′ coincides with ρ′.

We will prove these properties in the following subsections.

9.5.1 The (?)-relations are preserved

For any triple of positive roots α, β, γ forming an A2 and for every string S such that

σα · S = −S and σ2β · S = +S, we have a (?)-relation

[ S ] = [ σβ · S ] + [ σγ · S ].

In this subsection, we show that this relation is preserved by the action of Zν on Fρ′ ,

for every positive root ν. Equivalently

Zν · [ S ] = Zν · [ σβ · S ] + Zν · [ σγ · S ]. (>)

We distinguish a few cases:

1. σ2ν · S = −S and ν ⊥ α, β, γ

2. σ2ν · S = −S and ν 6⊥ α, β, γ

189

3. σ2ν · S = +S and ν ⊥ α, β, γ

4. σ2ν · S = +S and ν 6⊥ α, β, γ

and we discuss each such case separately.

1. Because the roots β and γ are both orthogonal to ν, we have:

σ2ν · (σβ · S) = σβ · (σ2

ν · S) = −(σβ · S)

σ2ν · (σγ · S) = σγ · (σ2

ν · S) = −(σγ · S).

The condition (>) is therefore equivalent to

[ σν ·S ] = [ σν · (σβ ·S) ]+ [ σν · (σγ ·S) ] ⇔ σν · [ S ] = σν · [ σβ ·S ] +σν · [ σγ ·S ],

and it is certainly satisfied because σν preserves the (?)-relations. X

2. We apply the same argument used in (1): we suppose that β is not orthogonal

to ν, and show that σ2ν still acts by (−1) on (σβ · S).

σ2ν · (σβ · S) = σβ · ((σ−1

β σ2νσβ) · S) = σβ · (σ±2

sν(β) · S︸︷︷︸−S

) = −(σβ · S)

because β, ν, σsν(β) give rise to anA2, hence the commuting product σ2sν(β)σ

2βσ2

ν

must act by +1 on each representation.

Similarly, σ2ν · (σγ · S) = −(σγ · S).27 X

3. It follows easily from the definition of the action of Lie(K) on Fρ′ that

σ · (Zν · S) = (Ad(σ)(Zν)) · (σ · S)

27γ is not necessarily orthogonal to α, but we have shown the result in both cases.

190

for all σ in M ′, all positive roots ν and every string S. Therefore, when β and

γ are both orthogonal to ν we can re-write the condition (>) as

[ Zν · S ] = [ (Ad(σβ)(Zν)) · (σβ · S) ] + [ ((Ad(σγ)(Zν)) · (σγ · S) ] ⇔

⇔ [ Zν · S ] = [ σβ · (Zν · S) ] + [ σγ · (Zν · S) ].

⇔ [ Zν · S ] = σβ · [ Zν · S ] + σγ · [ Zν · S ].

We deduce that (>) is simply the (?)-relation for the triple α, β, γ and the

string Zν · S. This relation holds because, since α, β, γ are orthogonal to ν, we

have

σα · (Zν · S) = Zν · (σα · S) = −(Zν · S)

σ2β · (Zν · S) = Zν · (σ2

β · S) = +(Zν · S). X

4. We notice that the roots ν, α, β, γ give rise to an A3. Because σ2α, σ2

β, σ2γ and

σ2ν all act by +1 on S, the string S belongs to a Weyl group representation of

S4. So it is enough to show that the action of Lie(K) on Fρ′ is well defined for

every representation ρ of S4. We will do it in the next chapter, in the section

dedicated to the examples.

9.5.2 Checking bracket relations-PART 1

Let α and β be not orthogonal positive roots, and let γ be a positive root in δ such

that ±α, ± β, ± γ is a root system of type A2. Without loss of generality we can

191

assume that the following bracket relations hold:28

[Zβ, Zα] = Zγ [Zα, Zγ] = Zβ [Zβ, Zγ] = −Zα .

We want to verify that

Zβ · (Zα · x)− Zα · (Zβ · x) = Zγ · x

for all x in Fρ′ . It is of course enough to consider the case in which x is either an

element of the original space Fρ or the equivalence class of a string:

x = [ S ] = [ Zα1 . . . Zαrv ]

for some mutually orthogonal positive roots α1, . . . , αr and a simultaneous (−1)-

eigenvector of σα1 , . . . , σαr .

When x belongs to Fρ, we decompose Fρ in isotypic components of irreducible repre-

sentations of the S(3) corresponding to α, β, γ:

ρ = (trivial)mt ⊕ (standard)ms = Umt ⊕ V ms

and we write x = xt + xs for the corresponding decomposition of x. We only need

to check that the bracket relations hold for xt and xs, and this problem is equivalent

to showing that our construction produces a well defined Lie algebra representation

when ρ is either the trivial or the standard representation of S3. We will do this in

28As a consequence, we have:

σβσασ−1β = σ−1

α σβσα = σγ

σασγσ−1α = σ−1

γ σασγ = σβ

σγσβσ−1γ = σ−1

β σγσβ = σα.

192

the next chapter, in the section dedicated to the examples.

Now we discuss the case in which x = [ S ] = [ Zα1 . . . Zαrv ] is the equivalence

class of a string. Define εα, εβ, εγ ∈ ±1 by

σ2α · S = εα S

σ2β · S = εβ S

σ2γ · S = εγ S

so that

εα = #αj : [Zαj, Zα] 6= 0

εβ = #αj : [Zαj, Zβ] 6= 0

εγ = #αj : [Zαj, Zγ] 6= 0.

Because εα εβ εγ = 1,29 we can assume w.l.o.g that either σ2α ·S = σ2

β ·S = σ2γ ·S = + S

or σ2α · S = −σ2

β · S = −σ2γ · S = + S.

We discuss the two cases separately. The trick will be once again to perform a

reduction to small rank cases.

The case εα = εβ = εγ = +1

Without loss of generality, we can assume that one of the following subcases holds:

1. σα · S = σβ · S = +S

2. σα · S = +S, σβ ·+S 6= S, σ2β · S = +S

3. σα · S = −S, σ2β · S = +S, α ∈ α1, . . . , αr

4. σα · S = −S, σ2β · S = +S, α ⊥ α1, . . . , αr

29Indeed the elements mα = σ2α, mβ = σ2

β , mγ = σ2γ of SO(3) satisfy the relations mαmβmγ = I.

193

5. σα · S 6= ±S, σβ · S 6= ±S, σγ · S 6= ±S, σ2α · S = σ2

β · S = +S.

We analyze each of these subcases separately. In all of them we reduce computations

to a construction for SL(k) for some k 6 6 which is explicitly carried out in the next

chapter.

? ? ? ?

Subcase #1

Because σα, σβ and σγ all act by (+1) on the string S = Zα1 . . . Zαrv, each root αj is

orthogonal to the triple α, β, γ. So computations are reduced to the case in which

S is an element of F(3), with (3) the trivial representation of S3.

? ? ? ?

Subcase #2

We assume that the string S satisfies

σα · S = +S

σβ · S 6= +S

σ2β · S = +S.

Because S = Zα1 . . . Zαrv is a (+1)-eigenvector for σα, the root α is orthogonal to the

set α1, . . . , αr, and since we are dealing with a root system of type A, the roots β

and γ must have the same property.30

Therefore, computations are reduced to the case in which S is an element of a Weyl

group representation of S3 (not containing the sign).

? ? ? ?30Suppose that β is not orthogonal to the entire set α1, . . . , αr. Then, since εβ = 1 and ∆ is of

type A, there are exactly two roots αi1 and αi2 to which β is not orthogonal. So β is of the formεa1 − εa2 , with aj an index appearing in αij but not in α (because α is orthogonal to αij ) for allj = 1, 2. It follows that β is orthogonal to α and we reach a contradiction.

194

Subcase #3

We assume that the string S = Zα1 . . . Zαrv satisfies:

σα · S = −S

σ2β · S = +S

and that α ∈ α1, . . . , αr .

Without loss of generality we can assume that α = α1. Because σ2β ·S = +S, there ex-

ists also another root, say α2, which is not orthogonal to β. We can write S = ZαZα2T ,

and ignore the string T which only involves roots orthogonal to α, β, γ. We must

show that:

Zβ · (Zα · x)− Zα · (Zβ · x) = Zγ · x

for x = [ ZαZα2v ]. By construction, the roots α, β, γ, α2 give rise to a root system

of type A3, so we basically need to do a calculation for SL(4). We notice that

the vector v is a simultaneous (−1)-eigenvector for the two orthogonal roots α, α2,

therefore it can only live in (a direct sum of copies of) F(2,2).31

It all comes down to showing that bracket relations hold when ρ = (2, 2) is the

two-dimensional irreducible representation of S4.

? ? ? ?

Subcase #4


σα · S = −S

σ2β · S = +S

and that α ⊥ α1, . . . , αr .

By assumption, the root α is orthogonal to the set α1, . . . , αr. Since we are dealing

31Recall that an element of the Specth module corresponding to a partition (n − k, k) can be asimultaneous (−1)-eigenvector of at most k disjoint transpositions.

195

with a root system of type A, the roots β and γ must have the same property. It

follows that we can reduce computations to the case in which S is an element of F(2,1),

with (2, 1) the standard representation of S3.

? ? ? ?

Subcase #5

We assume that S satisfies

σα · S 6= ±S, σβ · S 6= ±S, σγ · S 6= ±S

σ2α · S = σ2

β · S = +S.

We notice that α, β, γ ∩ α1, . . . , αr = ∅ (because the string S is not a (−1)-

eigenvector of σζ for ζ = α, β, γ), and that the number of αj s not orthogonal to α

(or β, or γ) is even, and hence equal to either zero or two. There are basically two

possibilities:

(i) α, β, γ ⊥ α1, . . . , αr

(ii) α, β, γ ⊥ α4, . . . , αr and

α 6⊥ α1, α 6⊥ α2, α ⊥ α3

β ⊥ α1, β 6⊥ α2, β 6⊥ α3

γ 6⊥ α1, γ 6⊥ α2, γ ⊥ α3.

We discuss these two possibilities separately.

The first one is easy, because we reduce computations to the case in which S is

an element of a Weyl group representation of S3 (not containing the sign).

In the second case, we write S = Zα1Zα2Zα3T , and ignore the string T which only

involves roots orthogonal to α, β, γ. We must show that:

Zβ · (Zα · x)− Zα · (Zβ · x) = Zγ · x

196

for x = [ Zα1Zα2Zα3v ]. By construction, the roots

α, β, γ, α1, α2, α3

give rise to a root system of type A5, so we basically need to do a calculation for

SL(6). We notice that the vector v is a simultaneous (−1)-eigenvector for the three

(orthogonal) root reflections σα1 , σα2 , , σα3 , therefore it can only live in (a direct sum

of copies of) F(3,3). It all comes down to showing that bracket relations hold when ρ

is the Specth module S(3,3) of S6.

The case εα = − εβ = − εγ = +1

We now discuss the case σ2α · S = − σ2

β · S = − σ2γ · S = +S.

Without loss of generality, we can assume that one of the following subcases holds:

1. σα · S = +S, σ2β · S = −S

2. σα · S = −S, σ2β · S = −S, α ∈ α1, . . . , αr .

3. σα · S = −S, σ2β · S = −S, α ⊥ α1, . . . , αr .

4. σα · S 6= ±S, σ2α · S = +S = −σ2

β · S.

We analyze each of these subcases separately.

? ? ? ?

Subcase #1


σα · S = +S

σ2β · S = −S.

W.l.o.g. we can also assume that

197

α ⊥ αj for all j = 1 . . . r

β 6⊥ α1 β ⊥ αj for all j 6= 1

γ 6⊥ α1 γ ⊥ αj for all j 6= 1.

If we write S = Zα1T , we can ignore the string T because it only involves roots

that are orthogonal to α, β, γ. Then, we must show that

Zβ · (Zα · x)− Zα · (Zβ · x) = Zγ · x

for x = [ Zα1v ]. By construction, the roots

α, β, γ, α1


SL(4). We notice that the two (orthogonal) root reflections σα, σα1 act on v by (+1)

and (−1) respectively. Therefore,32 v can only live in (a direct sum of copies of) F(3,1).

It all comes down to showing that bracket relations hold when ρ is the Specth module

S(3,1) of S4.

? ? ? ?

Subcase #2


σα · S = −S

σ2β · S = −S

and that α ∈ α1, . . . , αr .

W.l.o.g. we can suppose that α = α1, so that α, β, γ are orthogonal to αj for all

j > 1. If we write S = ZαT , we can ignore the string T because it only involves roots

32In F(2,2), disjoint transpositions have the same (−1) eigenspace.

198

that are orthogonal to α, β, γ, and we must show that

Zβ · (Zα · x)− Zα · (Zβ · x) = Zγ · x

for x = [ Zαv ]. This is a calculation for SL(3). Because v is a (−1)-eigenvector for

σα, v can only live in (a direct sum of copies of) F(2,1).


S(2,1) of S3.

? ? ? ?

Subcase #3


σα · S = −S

σ2β · S = −S

and that α ⊥ α1, . . . , αr . W.l.o.g. we can also assume that

β 6⊥ α1 β ⊥ αj for all j 6= 1

γ 6⊥ α1 γ ⊥ αj for all j 6= 1.

If we write S = Zα1T , we can ignore the string T because it only involves roots

that are orthogonal to α, β, γ. Then, we must show that

Zβ · (Zα · x)− Zα · (Zβ · x) = Zγ · x


α, β, γ, α1


199

SL(4). We notice that v is a simultaneous (−1)-eigenvector for the two orthogonal

root reflections σα and σα1 . Therefore, v can only live in (a direct sum of copies of)

F(2,2). It all comes down to showing that bracket relations hold when ρ is the Specth

module S(2,2) of S4.

? ? ? ?

Subcase #4


σα · S 6= ±S

σ2α · S = −σ2

β · S = +S.

Because σ2α · S = +S, but σα · S 6= −S the root α is either orthogonal to the set

α1, . . . , αr or not orthogonal to exactly two roots in this set. We need to distin-

guish between these two cases.

(i) α ⊥ α1, . . . , αr.Let α1 be the (only) root to which β and γ are not orthogonal. As usual, we write S =

Zα1T , and we ignore the string T because it only involves roots that are orthogonal

to α, β, γ. We must show that

Zβ · (Zα · x)− Zα · (Zβ · x) = Zγ · x


α, β, γ, α1

give rise to a root system of typeA3, so we basically need to do a calculation for SL(4).

The vector v is an element of a Weyl group representation of S4 (not containing the

sign, nor the trivial representation).33

It all comes down to showing that bracket relations hold when ρ is the Specth modules

33Because σα1 · v = −v

200

S(2,2) or S(3,1) of S4.

(ii) α 6⊥ α1, α2.We can assume that β is orthogonal to α2, but not to α1, and that γ is orthogonal to

α1, but not to α2. In any case, the roots α, β, γ are orthogonal to αj for all j > 2.

We write S = Zα1Zα2T , and we ignore the string T because it only involves roots

that are orthogonal to α, β, γ. We must show that

Zβ · (Zα · x)− Zα · (Zβ · x) = Zγ · x

for x = [ Zα1Zα2v ]. By construction, the roots

α, β, γ, α1, α2

give rise to a root system of typeA4, so we basically need to do a calculation for SL(5).

We notice that the v is a simultaneous (−1)-eigenvector for the two (orthogonal) root

reflections σα1 and σα2 . Therefore, v can only live in (a direct sum of copies of) F(3,2).


S(3,2) of S5.

9.5.3 Checking bracket relations-PART 2

Let α and β be mutually orthogonal positive roots. We want to verify that

Zβ · (Zα · x)− Zα · (Zβ · x) = Zγ · x

for all x in Fρ′ . It is of course enough to consider the case in which x is either an

element of the original space Fρ or the equivalence class of a string:

x = [ S ] = [ Zα1 . . . Zαrv ]

201

for some mutually orthogonal positive roots α1, . . . , αr and a simultaneous (−1)-

eigenvector of σα1 , . . . , σαr .

When x belongs to Fρ, we decompose Fρ in isotypic components of irreducible

representation of the S4 determined by α and β

ρ =(S( 4)

)m0 ⊕ (S( 3,1)

)m1 ⊕ (S( 2,2)

)m2

and we write x = x0+x1+x2 for the corresponding decomposition of x. We just need

to check that the bracket relations hold for the xjs, and this problem is equivalent

to showing that our construction produces a well defined Lie algebra representation

when ρ is a Weyl group representation of S4 not containing the sign. We will do this

in the next chapter, in the section dedicated to the examples.

We now discuss the case in which x = [ S ] = [ z ] is the equivalence class of a

string. The trick is again to perform a reduction to a calculation for a small Weyl

group, namely some symmetric group Sk, with k 6 8 depending on the number of

αj s not orthogonal to α and β. We distinguish several subcases:

1. When α and β are both orthogonal to α1, . . . , αr, we reduce computations to

the case in which ρ is a representation of S4 not containing the sign.

2. When α is orthogonal to α1, . . . , αr and β belongs to this set, we reduce

computations to the case in which ρ is a sum of copies of the representations

S (3,1) and S (2,2) of S4.

3. When α is orthogonal to α1, . . . , αr and there is exactly one root in this set

which is not orthogonal to β (and different from β), we reduce computations to

the case in which ρ is a a sum of copies of the representations S (3,2) and S (4,1)

of S5.

4. When α is orthogonal to α1, . . . , αr and there are two roots in this set that

202

are not orthogonal to β, we reduce computations to the case in which ρ is a

sum of copies of the representations S (4,2) and S (3,3) of S6.

5. When both α and β belong to the set α1, . . . , αr, we reduce computations to

the case in which ρ is the representation S (2,2) of S4.

6. When only α belongs to the set α1, . . . , αr, and there is one root in this

set not orthogonal to β, we reduce computations to the case in which ρ is the

representation S (3,2) of S5.

7. When only α belongs to the set α1, . . . , αr, and there are two roots in this

set not orthogonal to β, we reduce computations to the case in which ρ is the

representation S (3,3) of S6.

8. When neither α nor β belong to the set α1, . . . , αr, and they are both not

orthogonal a specific root, we reduce computations to the case in which ρ is a

sum of copies of the representations S 3,1) and S (2,2) of S4.

9. When neither α nor β belong to the set α1, . . . , αr, and there are two roots

αj1 , αj2 such that α is not orthogonal to a root αj1 and β is not orthogonal to

αj2 , we reduce computations to the case in which ρ is a sum of copies of the

representations S (4,2) and S (3,3) of S6.

10. When neither α nor β belong to the set α1, . . . , αr and there are two roots

αj1 , αj2 such that α is not orthogonal to αj1 , and β is not orthogonal to αj1 and

αj2 , is not orthogonal to one of the roots, we reduce computations to the case

in which ρ is a sum of copies of the representations S (4,2) and S (3,3) of S6.

11. When neither α nor β belong to the set α1, . . . , αr, and there are three roots

αj1 , αj2 , αj3 such that α is not orthogonal to αj1 , and β is not orthogonal to

αj2 , αj3 , we reduce computations to the case in which ρ is a sum of copies of

the representation S (4,3) of S7.

203

12. When there are two roots to which neither α nor β are orthogonal, we reduce

computations to the case in which ρ is the representation S(2,2) of S4.

13. When there are three roots αj1 , αj2 , αj3 such that α is not orthogonal to αj1 ,

αj2 , and β is not orthogonal to αj2 , αj3 , we reduce computations to the case in

which ρ is the representation S(3,3) of S6.

14. When there are two distinct pairs of roots to which α and β are not orthogonal,

we reduce computations to the case in which ρ is the representation S(4,4) of S8.

Most of the examples we refer to are explicitly worked out in the next chapter. The

only exceptions are S(4,2), S(4,3), and S(4,4), but the computations in these cases are

very similar to the ones done for S(3,3).

Claim 7. Alternatively, we can check the bracket relations using the definition of the

the action of Lie(K) on Fρ′, with a case-by-case analysis.

9.5.4 Checking the eigenvalue condition

The purpose of this subsection is to show that, for all positive roots β, the eigenvalues

of ρ′(Zβ) lie in the set 0,±i,±2i.Because the action of Zβ on F

ρ′strongly depends on how the corresponding element

σβ of M ′ acts on the same space, it is convenient to decompose Fρ′

in σβ-stable

subspaces and to study the eigenvalues of Zβ separately on each of these subspaces.

We start by decomposing Fρ′ = Fρ ⊕ Span( strings ) in σβ-stable subspaces.34For

brevity of notations, let S = Zα1 . . . Zαrv be an arbitrary string of orthogonal roots.

Then:

34The existence of this decomposition follows from the fact that for any string S, σ2β · S = ±S.

Once again we identify a string with its equivalence class modulo the commutativity and linearityrelations, so thatFρ′ is identified with Fρ ⊕ Span( strings ).

204

Fρ′ = Fρ ⊕ Span( strings )

= Fρ ⊕ SpanS : σβ · S = +S)︸︷︷︸UI

⊕ SpanS : σβ · S = −S︸︷︷︸UII

⊕

⊕ SpanS : σβ · S 6= ±S, σ2β · S = +S, β 6⊥ α1, . . . , αr︸︷︷︸

UIII

⊕

⊕ SpanS : σ2β · S = −S︸︷︷︸

UIV

. (♣)

Let us explain why we have imposed the extra condition “β 6⊥ α1, . . . , αr” on

the strings generating UIII. If β is orthogonal to α1, . . . , αr and v is an element of Fρ

such that σβ · v = ±v, then using the linearity conditions35 we can write:

S = Zα1 . . . Zαrv =1

2Zα1 . . . Zαr(v + σβ · v)

︸︷︷︸in UI

+1

2Zα1 . . . Zαr(v − σβ · v)

︸︷︷︸in UII

so the string S satisfies σβ · S 6= ±S and σ2β · S = S, but it already lies in UI ⊕ UII.

We now give a detailed description of each summand of the decomposition (♣):

• UI is spanned by those strings on which σβ acts as scalar multiplication by +1.36

It is stable under the action of σβ, and it is included in the (+1)-eigenspace of

σβ.

• UII is spanned by those strings on which σβ acts as scalar multiplication by −1.

Recall that a string S = Zα1 . . . Zαrv is a (−1)-eigenvectors of σβ if and only if

35If v is a (−1)-eigenvector of α1, . . . , αr in Fρ and β is orthogonal to α1, . . . , αr, then the vectorsw± = 1

2 (v ± σβ · v) are also simultaneous (−1)-eigenvector of α1, . . . , αr. Therefore, the following“linearity condition” holds:

Zα1 . . . Zαrv = (Zα1 . . . Zαrw+) + (Zα1 . . . Zαrw

−).

36A string S = Zα1 . . . Zαrv is a (+1)-eigenvector of σβ if and only if β ⊥ α1, . . . , αn andσβ · v = +v.

205

one of the following two conditions occur:

1. β ⊥ α1, . . . , αn and σβ · v = −v

2. β ∈ α1, . . . , αn and σβ · v = −v.

So we can write: UII = SpanS = Zα1 . . . Zαrv : σβ · S = −S =

= SpanS : σβ · S = −S; β 6∈ α1, . . . , αn︸︷︷︸UII1

⊕

⊕ SpanS : σβ · S = −S; β ∈ α1, . . . , αn︸︷︷︸UII2

.

Both subspaces are included in the (−1)-eigenspace of σβ, hence they are stable

under σβ.

• UIII is also stable under σβ. Indeed for each string S = Zα1 . . . Zαrv in UIII we

have

σβ · (σβ · S) = S 6= ± (σβ · S) σ2β · (σβ · S) = σβ · (σ2

β(S)) = + (σβ · S).

Therefore, we can decompose UIII as the direct sum of (two-dimensional) sub-

spaces of the form

< S, (σβ · S) >C

with S = Zα1 . . . Zαrv a string of orthogonal roots such that (σβ ·S) 6= ±S and

σ2β ·S = +S. Because σβ acts on each of these subspaces by

0 1

1 0

, the vector

(S + σβ · S) is a (+1)-eigenvector of σβ, and (S − σβ · S) is a (−1)-eigenvector.

We can write:

206

UIII = SpanS = Zα1 . . . Zαrv : σβ ·S 6= ±S, σ2β ·S = +S, β 6⊥ α1, . . . , αr =

=

⊕

S : σβS 6=±S

σ2βS=+S

β 6⊥α1,...,αr

C(S + σβ · S)

︸︷︷︸UIII1

⊕

⊕

S : σβS 6=±S

σ2βS=+S

β 6⊥α1,...,αr

C(S − σβ · S)

︸︷︷︸UIII2

.

Both subspaces are stable under σβ, and they are included in the (+1)-eigenspace

and (−1)-eigenspace of σβ respectively.

• UIV admits a similar decomposition:

UIV = SpanS = Zα1 . . . Zαrv : σβ · S 6= ±S, σ2β · S = −S =

=

⊕

S : σ2βS=−S

C(S − i σβ · S)

︸︷︷︸UIV1

⊕ ⊕

S : σ2βS=−S

C(S + i σβ · S)

︸︷︷︸UIV2

.

Both subspaces are stable under σβ. Indeed they are the (+i)-eigenspace and

(−i)-eigenspace of σβ respectively.

The description of the decomposition of Fρ′ in σβ-stable subspaces is now complete.

Question. What happens when we impose the (?)-relations (i.e. we pass to the

quotient Fρ′ = Fρ′/R)?

Claim 8. The subspaces UII2 and UIII2 have the same image under the projection

π : Fρ′ → Fρ′ = Fρ′/R.

Proof. Suppose that S = Zα1 . . . Zαrv is a string of orthogonal roots that lies in UIII.

207

By definition, we have:

σβ · S 6= ±S σ2β · S = S β 6⊥ α1, . . . , αr.

We notice that β does not belong to the set α1, . . . , αr37, and that there are exactly

two roots in this set, say αi1 and αi2 , to which β is not orthogonal. W.l.o.g. we can

assume that i1 = 1 and i2 = 2. Hence β 6⊥ α1, β 6⊥ α2, but β ⊥ α3, . . . , αr.

Let β1 be a positive root such that the set ±α1, ± β, ± β1 is a root system of type

A2. Because

σα1 · S = −S σ2β · S = +S

the following relation holds in the quotient space: [ S ] = [ σβ · S, ] + [ σβ1 · S ].

Therefore we can write:

[ S − σβ · S ] = [ σβ1 · S ] = [ σβ1 · (Zα1 . . . Zαrv) ] =

= [± Zβ(Ad(σβ1) · Zα2)Zα3 · · ·Zαr (σβ1 · v)︸︷︷︸in UII2

].

This proves the claim.

As a consequence, we obtain that:

Fρ′

= Fρ

⊕ Span( strings )R

= Fρ ⊕ Span[ S ] : σβ · S = +S︸︷︷︸[ UI ]⊆ (+1)-eigenspace of σβ

⊕

⊕Span [ S ] : σβ · S = −S; β 6∈ α1, . . . , αn︸︷︷︸

[ UII1]⊆ (−1)-eigenspace of σβ

⊕

37Or σβ · (Zα1 . . . Zαrv) = −Zα1 . . . Zαrv.

208

⊕Span [ S ] : σβ · S = −S; β ∈ α1, . . . , αn︸︷︷︸

[ UII2]⊆ (−1)-eigenspace of σβ

⊕

⊕ ⊕

S : σ2βS=−S

C(S − i σβ · S)

︸︷︷︸[ UIV1

]

⊕ ⊕

S : σ2βS=−S

C(S + i σβ · S)

︸︷︷︸[ UIV2

]

⊕

⊕

⊕

S : σβS 6=±S

σ2βS=+S

β 6⊥α1,...,αr

C(S + σβ · S)

︸︷︷︸[ UIII1

]

.

Recall that, by definition, Zβ acts by 0, +i and −i on the (+1), (+i) and (−i)-

eigenspace of σβ. Therefore:

• Zβ only admits the eigenvalue (+1) on v ∈ Fρ : σβ · v = +v ∪ [ UI ] ∪ [ UIII1 ]

• Zβ only admits the eigenvalue (+i) on the subspace [ UIV1 ] of Fρ′

• Zβ only admits the eigenvalue (−i) on the subspace [ UIV2 ] of Fρ′

.

We still have to compute the eigenvalues of Zβ on the set

v ∈ Fρ : σβ · v = −v ∪ [ UII ].

Identifying the elements of Fρ with strings of length zero, we can decompose this set

as a direct sum of two-dimensional subspaces of the form

< [ Zα1 . . . Zαrv ] , [ ZβZα1 . . . Zαrv] >C

209

where α1, . . . , αr are mutually orthogonal roots (all orthogonal to β), and v ∈ Fρ is

a simultaneous (−1)-eigenvector of σα1 , . . . , σαr , σβ.

It is easy to check that Zβ acts on each of these subspace by

0 −4

1 0

. Hence it has

eigenvalues ±2i.

We conclude that Zβ on Fρ′ with eigenvalues 0, ±i, ±2i.

9.5.5 Lifting ρ′ to K

The purpose of this subsection is to show that the representation ( ρ′, Fρ′

) of Lie(K) =

so(n, R) can be lifted to a representation of K = SO(n).

We notice that ρ′ might be reducible, and that a lifting exists if and only if the highest

weight of each irreducible summand of ρ′ is analytically integral.

Of course, the odd and the even cases should be treated separately. We cover only

the even case, the odd one being similar.

Following the notations of chapter 8, we set k = so(2n, C) and we let h be the

Cartan subalgebra

h = Hθ1...θn

0 θ1 . . . 0 0

−θ1 0 . . . 0 0

......

. . ....

...

0 0 . . . 0 θn

0 0 . . . −θn 0

: θ1, θ2, . . . , θn ∈ C

.

For all j = 1 . . . n, we denote by ψj the linear functional h : Hθ1...θn 7→ −i θj, so

that

Ψ = ±(ψi ± ψnj)i, j=1,..., n, i<j

is the root system of g w.r.t. h, and

210

∆+ = ψi ± ψji, j=1,..., n, i<j

is a choice of positive roots.

We say that an element λ = a1ψ1 + · · ·+ an−1ψn−1 + anψn of h∗ is

• algebraically integral if aj ∈ Z, ∀ j = 1 . . . n, or aj ∈(Z+ 1

2

), ∀ j

• analytically integral if a1, . . . , an−1, an ∈ Z

• dominant if a1 > · · · > an−1 > | an | .

Let υ = b1ψ1 + · · · + bnψn be the highest weight of an irreducible summand µυ of

ρ′, and let yυ be a weight vector of weight υ. We notice that, for all j = 1 . . . n, the

element

Zj ≡ Zψ2j−1−ψ2j= E2j−1,2j − E2j,2j−1

belongs to the Cartan subalgebra h, and

µυ(Zj) · yυ = (

∑nk=1 bkψk(Z

j)) yυ = (bjψj(Zj))yυ = (−ibj)yυ.

So there exists a positive root βj = ψ2j−1 − ψ2j such that −ibj is an eigenvalue

of µυ(Zβj). Because the eigenvalues of Zβ on Fρ′ lie in the set 0, ±i, ±2i, we

conclude that bj belongs to 0, ±1, ±2. Therefore, υ has integer coefficients, and it

is analytically integral.

9.5.6 It is an extension of ρ

Let (Θ, FΘ) be the representation of K whose differential is ρ′. Then FΘ = Fρ′

contains the space Fρ on which the original Weyl group representation ρ is defined.38

Let us verify that

ResKM ′Θ |Fρ= ρ.

38Because all the relations are imposed at the level of strings of positive length.

211

Because M ′ is generated by the elements σαα∈∆+ , it is enough to prove that

Θ(σα) = exp(π

2ρ′(Zα)

)(v) = ρ(σα)(v) (>)

for all positive roots α and all v in Fρ.

For any fixed α, we just need to check that (>) holds for v in a basis of Fρ and

because σα acts semi-simply on Fρ, we can pick a basis consisting of (+1)- and(−1)-

eigenvectors of ρ(σα):

B = u1, u2, . . . , ut︸︷︷︸+1 eig.s

, v1, v2, . . . , vs︸︷︷︸−1 eig.s

.

Then, we are reduced to show that Θ(σα)(ui) = + ui and Θ(σα)(vj) = −vj, for all

i = 1 . . . t and all j = 1 . . . r.

Let V be the subspace of Fρ′ defined by:

V = Span(u1, u2, . . . , ut, v1, v2, . . . , vs, Zαv1, Zαv2, . . . , Zαvs).

The strings Zαv1 . . . Zαvs are well defined and they are linearly independent39, so V

has dimension 2s + t. Because, by definition,

• Zα · ui = 0 ∀ i = 1 . . . t

• Zα · vj = Zαvj ∀ j = 1 . . . s

• Zα · (Zαvj) = −4vj ∀ j = 1 . . . s

39There is no relation among them.

212

the action of Zα on V is described by the matrix:

ρ′(Zα) |V Ã

Or×r Os×s Os×s

Os×r Os×s −4Is×s

Os×r Is×s Os×s

with the first two blocks corresponding to Fρ. We obtain:

Θ(σα) = exp(π

2ρ′(Zα)

)|V Ã

Ir×r Os×s Os×s

Os×r −Is×s Os×s

Os×r Os×s −Is×s

.

In particular, Θ(σα) acts on Fρ via:

Θ(σα) |FρÃ

Ir×r Os×s

Os×r −Is×s

and this shows that Θ(σα) |Fρ= ρ.

9.6 Conclusions

For every representation (ρ, Fρ) of the Weyl group of SL(n, R) whose restriction

to any W (SL(3)) does not contain the sign representation, we have constructed a

finite-dimensional representation ( ρ′, Fρ′) of SO(n) with the following properties:

• for each root α of SL(n), the subgroup Kα ' SO(2) only acts with characters

213

zero, plus or minus one, plus or minus two (i.e. ρ′ is petite)

• the space of M -fixed vectors in ρ′ is non-zero (i.e. ρ′ is spherical)

• ρ is a submodule of the representation of W (SL(n)) on the space of M -fixed

vectors of ρ′ .

Some natural questions arise:

¦ Is ρ′ irreducible?

¦ Is (Fρ′)M = ρ?

¦ What is the highest weight decomposition of ρ′?

In particular, how does ρ′ fit in the classification of petite spherical K-types

described in chapter 8 (for n even)?

We will answer these questions in the next chapter.

214

Chapter 10

Further results and examples

In the first section, we descibe the petite representation (ρ, Fρ′) of SO(2n,R) that

arises from an irreducible representation (ρ, Fρ) of the Weyl group through the con-

struction described in the previous chapter.

As usual, we identify the Weyl group of SL(2n) with the symmetric group in 2n let-

ters, and the irreducible representations of W with partitions of 2n. The main results

are the following:

1. For every representation ρ of S2n, we compute the highest weight decomposition

of the petite K-type Fρ′ :

¦ if ρ = (2n), then Fρ′ = L(0ψ1 + 0ψ2 + · · ·+ 0ψn)

¦ if ρ = (2n− k, k), then Fρ′ = L(2ψ1 + 2ψ2 + · · ·+ 2ψk)

for all 0 < k < n

¦ if ρ = (n, n), then Fρ′ = L(2ψ1+ · · ·+2ψn−1−2ψn)⊕L(2ψ1+ · · ·+2ψn−1+

2ψn).

2. For every representation ρ of S2n, we compute the Weyl group representation

on the the space of M -invariants of Fρ′ :

¦ if ρ = (2n), then(Fρ′

)M= (2n)

215

¦ if ρ = (n− k, k), then(Fρ′

)M= (2n− k, k)

for all 0 < k < n

¦ if ρ = (n, n), then(Fρ′

)M= (n, n)⊕ (n, n).

The result for SO(2n + 1) is even simpler, because the representation Fρ′ is always

irreducible:

¦ if ρ = (2n + 1), then Fρ′ = L(0ψ1 + 0ψ2 + · · ·+ 0ψn)

¦ if ρ = (2n + 1− k, k), then Fρ′ = L(2ψ1 + 2ψ2 + · · ·+ 2ψk)

for all 0 < k ≤ n

and

¦ if ρ = (2n + 1), then(Fρ′

)M= (2n + 1)

¦ if ρ = (2n + 1− k, k), then(Fρ′

)M= (2n + 1− k, k)

for all 0 < k ≤ n.

In the second section we provide a number of examples. The construction is explicitly

carried out for partitions (in at most two parts) of n = 3, n = 4, n = 5 and n = 6.

These examples give a useful insight on the general case. In the last section we

construct the extension for the non-spherical representations of M ′ ⊆ SL(3).

10.1 The highest weight decomposition of Fρ′

Throughout the section W = S2n is the Weyl group of SL(2n, R), ρ is an irreducible

representation of W (i.e. a partition of 2n) and Fρ′ is its extension to a petite K-type.

We want to determine the highest weight decomposition of Fρ′ .

216

The case ρ = (2n)

When ρ is the trivial representation of W , i.e. ρ = (2n), a root-reflection does not

have any (−1)-eigenvector. So the extension Fρ Ã Fρ′ is trivial: Fρ′ = Fρ. It is

easy to see that ρ is the trivial representation of SO(2n), and it has highest weight

0ψ1 + 0ψ2 + · · ·+ 0ψn:

ρ = (2n) ⇒ Fρ′ = L(0ψ1 + 0ψ2 + · · ·+ 0ψn) .

The case ρ = (2n− k, k), for 1 ≤ k < n

We will show that Fρ′ is irreducible, and equal to the irreducible representation of

SO(2n) of highest weight 2ψ1 + 2ψ + · · ·+ 2ψk:

ρ = (2n− k, k) ⇒ Fρ′ = L(2ψ1 + 2ψ + · · ·+ 2ψk) .

For clarity, we split the argument it in several steps:

Lemma 10. For ρ = (2n − k, k), with k = 1, . . . , n − 1, the representation of the

Weyl group on the space of M-invariants in Fρ′ is irreducible and equal to (2n−k, k):

ρ = (2n− k) ⇒ (Fρ′

)M= F(2n−k,k).

Lemma 11. If ρ = (2n−k, k), with k = 1, . . . , n−1, then Fρ′ = L(2ψ1 +2ψ2 + · · ·+2ψk).

Corollary 13. If ρ = (2n− k, k), with k = 1, . . . , n− 1, then Fρ′ contains a weight

vector of weight 2ψ1 + 2ψ2 + · · ·+ 2ψk.

? ? ?

Proof of lemma 10. We set ρ = (2n − k, k) and we look at the the space of M -

invariants in Fρ′ , which is the vector space generated by Fρ and all the strings of or-

217

thogonal roots.1 The first step is to observe that every string defines a one-dimensional

representation of M . Because M is generated by the the elements mβs, with β vary-

ing in the set of positive roots, it is sufficient to prove that each mβ acts on a string

by a scalar. This is easy to do.

For all roots α and β (of SL(2n)), we have:

Ad(mβ)(Zα) = εα,βZα

with εα,β = +1 if α = β or α ⊥ β, and εα,β = −1 otherwise.2 Hence

mβ · (Zα1Zα2 . . . Zαrv) = (εα1,βZα1)(εα2,βZα2) . . . (εαr,βZαr)(mβ · v) =

= εα1,βεα2,β · · · εαr,β(Zα1Zα2 . . . Zαrv) = ±(Zα1Zα2 . . . Zαrv).

The second step is to understand whether the one-dimensional representation of M

spanned by a root is the trivial M -type.

We write αl = εil − εjl, for all l = 1 . . . r and notice that being r ≤ k < n, the set

I = il, . . . , ir, jl, . . . , jr is properly contained in 1, 2, . . . , 2n. Hence we can pick

an index s in [1, 2n]−I. By construction, β = ±(εi1−εs) is orthogonal to α2, . . . , αr;

it is different from α1 and not orthogonal to it. Therefore

mβ · (Zα1Zα2 . . . Zαrv) = −(Zα1Zα2 . . . Zαrv).

This shows that for all r = 1 . . . k, C(Zα1Zα2 . . . Zαrv) is not the trivial M -type.

Therefore

ρ = (2n− k) ⇒ (Fρ′)M = F(2n−k,k).

1To simplify notations, we are identifying a string with its equivalence class modulo the linearityand orthogonality relations. At the stage, there are no (?)-relations.

2mβ = σ2β is the square a representative in M

′for the root reflection β.

218

Passing to the quotient modulo the (?)-relations does not introduce additional trivial

M -types, so (Fρ′)M also coincides with Fρ. We conclude that the representation of

the Weyl group on (Fρ′)M is irreducible, and equal to (2n− k, k).

? ? ?

Proof of lemma 11. We set ρ = (2n − k, k) and we look at the decomposition of

Fρ′ in irreducible summands. Because Fρ′ is a petite representation of SO(2n), this

decomposition can only include the K-types with highest weight3

• 0ψ1 + · · ·+ 0ψn

• 1ψ1 + · · ·+ 1ψs 0 < s < n

• 1ψ1 + · · ·+ 1ψn−1 ± 1ψn

• 2ψ1 + · · ·+ 2ψs 0 < s < n

• 2ψ1 + · · ·+ 2ψa + 1ψa+1 + · · ·+ 1ψb 0 < a, b < n

• 2ψ1 + · · ·+ 2ψk + 1ψk+1 + · · ·+ 1ψn−1 ± 1ψn 0 < k < n

• 2ψ1 + · · ·+ 2ψn−1 ± 2ψn.

Therefore, we can write

Fρ′ = L(0)⊕m0 ⊕(

n−1⊕s=0

L(2ψ1 + · · ·+ 2ψs)⊕ms

)⊕ L(2ψ1 + · · ·+ 2ψn−1 + 2ψn)⊕m+

n⊕

⊕L(2ψ1 + · · ·+ 2ψn−1 − 2ψn)⊕m−n ⊕ some non spherical K-types.

We decompose (Fρ′)M accordingly:

(Fρ′)M = (2n)⊕m0 ⊕

(n−1⊕s=1

(2n− s, s)⊕ms

)⊕ (n, n)⊕(m+

n +m−n ).

3Please, refer to chapter 8 for a classification of the petite spherical representations of SO(2n),and an analysis of the corresponding representation of the Weyl group on the space of M -fixedvectors.

219

By the previous lemma (Fρ′)M = (2n− k, k), so

m+n = m−

n = mj = 0 ∀j ∈ 0, 1, . . . n− 1 − k

and mk = 1. We obtain:

Fρ′ = L(2ψ1 + · · ·+ 2ψk)⊕ some non spherical K-types.

The next step is to show that Fρ′ does not contain any non-spherical K-type, and to

do this we need a better understanding of the structure of Fρ′ .

Fρ′ is spanned by Fρ and all the strings of the form

[Zα1Zα2 . . . Zαrv]

where α1, . . . , αr are mutually orthogonal positive roots and v is a simultaneous

(−1)-eigenvector of σα1 , . . . , σαr in Fρ. Because

[Zα1Zα2 . . . Zαrv] = Zα1 · (Zα2 · (. . . (Zαr · [v]) · · · ))

we conclude that, as a representation of Lie(K), Fρ′ is generated by the elements

of Fρ. We write Fρ′ = V1 ⊕ V2 ⊕ · · · ⊕ Vt for the decomposition of Fρ′ in isotypic

components of K-types (with V1 = L(2ψ1 + · · ·+ 2ψk)), and we identify each Vj with

the image of the projection

πj : Fρ′ →Fρ′⊕

i∈1,..., t−j Vi

.

220

By construction, Vj is generated by

πj(Fρ) =Fρ

Fρ ∩(⊕

i∈1,..., t−j Vi

)

and because Vi is non-spherical for all i 6= 1, we get:

• Fρ ∩(⊕

i∈1,..., t−j Vi

)= 0 if j = 1

• Fρ ∩(⊕

i∈1,..., t−j Vi

)= Fρ if j 6= 1.

Therefore πj(Fρ) = Fρ if j = 1, and πj(Fρ) = 0 if j 6= 1.

It follows that Vj = 0 for all j = 2, . . . , t, hence Fρ′ = L(2ψ1 + · · ·+ 2ψk).

? ? ?

Proof of corollary 13. Set ρ = (2n − k, k). By the previous lemma, Fρ′ contains a

weight vector z of weight 2ψ1 + 2ψ2 + · · ·+ 2ψk. The following argument is meant to

provide an explicit construction for z.

Let v = eT be the standard polytabloid associated to the standard tableau

T = 1 3 5 · · · 2k-3 2k-1 2k+12k+22k+3 · · · 2n-1 2n .2 4 6 · · · 2k-2 2k

We notice that v is a simultaneous (−1)-eigenvector in Fρ for the transpositions

(1, 2) (3, 4) (5, 6) . . . (2k − 3, 2k − 2) (2k − 1, 2k)

and a simultaneous (+1)-eigenvector for the transpositions

(2k + 1, 2k + 2) (2k + 3, 2k + 4) . . . (2n− 1, 2n).

221

The string Z(ε1−ε2)Z(ε3−ε4) · · ·Z(ε2k−1−ε2k)v is therefore well defined. So is any string

of the form

Z(ε2j1−1−ε2j1)Z(ε2j2−1−ε2j2

) · · ·Z(ε2js−1−ε2js )v

for 1 ≤ j1 < j2 < · · · < js ≤ k. For brevity of notations, we set Z(ε2m−1−ε2m) = Zm,

for all integers m = 1 . . . n. We also set4

z =k∑

s=0

(−2i)k−s

( ∑

1≤j1<j2<···<js≤k

Zj1Zj2 · · ·Zjsv

).

With abuse of notations, we denote the equivalence class of z in the quotient space

Fρ′ with the same symbol. We claim that z ∈ Fρ′ is a weight vector for the Cartan

subalgebra

h =

0 θ1 . . . 0 0

−θ1 0 . . . 0 0

......

. . ....

...

0 0 . . . 0 θn

0 0 . . . −θn 0

: θ1, θ2, . . . , θn ∈ C

of = so(2n, C), of weight 2ψ1 + 2ψ2 + · · · + 2ψk. Indeed, the element Hm =

E2m−1,2m − E2m,2m−1 of h acts on z by −2i if m ≤ k, and by 0 if k < m ≤ n.

Because Hm = Zε2m−1−ε2m , to compute the action of Hm on a string Zj1Zj2 · · ·Zjsv

we must first understand how the reflection σ(ε2m−1−ε2m) acts on it. This is easy5:

σε2m−1−ε2m · (Zj1Zj2 · · ·Zjsv) = Zj1Zj2 · · ·Zjs

(σ(ε2m−1−ε2m) · v

)=

4A string Zj1Zj2 · · ·Zjsv is intended to be equal to v when s = 0.5Because Ad(σε2m−1−ε2m)(Zj) = Ad(σε2m−1−ε2m)(Z(ε2j−1−ε2j)) = Z(ε2j−1−ε2j) = Zj for all j.

222

=

−Zj1Zj2 · · ·Zjsv if m ≤ k

+Zj1Zj2 · · ·Zjsv if k < m ≤ n.

It follows that:

Zε2m−1−ε2m·(Zj1 · · ·Zjr · · ·Zjsv) =

ZmZj1 · · ·Zjr · · ·Zjsv if m ≤ k and m 6∈ j1 . . . js

−4Zj1 · · · Zjr · · ·Zjsv if m ≤ k and m = jr

0 if m > k .

As a result, we obtain that Zε2m−1−ε2m · z = 0 for all m > k, and that the vectors

um,0 ≡ (−2i)v + Zmv

um,s,j ≡ (−2i)Zj1Zj2 · · ·Zjs−1v + ZmZj1Zj2 · · ·Zjs−1v

are (−2i)-eigenvectors of Zε2m−1−ε2m , for all m ≤ k, for all 1 ≤ s < k and for every

multi-index j of length s− 1 with entries in 1, . . . , k −m.

Because

z = (−2i)k−1um,0 +k−1∑s=1

(−2i)k−s

∑

j

um,s,j

also z is a (−2i)-eigenvector of Zε2m−1−ε2m , for all m ≤ k. This ends the proof.

The case ρ = (n, n)

When ρ = (n, n), the representation Fρ′ fails to be irreducible:

ρ = (n, n) ⇒ Fρ′ = L(2ψ1 + · · ·+ 2ψn−1 − 2ψn)⊕ L(2ψ1 + · · ·+ 2ψn−1 + 2ψn) .

223

As a representation of the Weyl group, (Fρ′)M consists of two copies of (n, n).

As usual, it is convenient to split the proof in several parts.

Lemma 12. Fρ′ contains a weight vector of weight 2ψ1 + · · ·+2ψn−1 +2ψn, and also

a weight vector of weight 2ψ1 + · · ·+ 2ψn−1 − 2ψn.

Lemma 13. Fρ′ ⊇ L(2ψ1 + · · ·+ 2ψn−1 − 2ψn)⊕ L(2ψ1 + · · ·+ 2ψn−1 + 2ψn).

Lemma 14. As a representation of the Weyl group,(Fρ′

)M= (n, n)⊕ (n, n).

Corollary 14. Fρ′ ≡ L(2ψ1 + · · ·+ 2ψn−1 − 2ψn)⊕ L(2ψ1 + · · ·+ 2ψn−1 + 2ψn).

Proof of lemma 12. The existence of a weight vector z of weight 2ψ1 + 2ψ2 + · · · +2ψn−1 + 2ψn is proven just like in the case ρ = (2n− k, k) for k < n. The vector z is

explicitly given by:

z =n∑

s=0

(−2i)n−s

( ∑1≤j1<j2<···<js≤n

Zj1Zj2 · · ·Zjsv

)

with v = eT the standard polytabloid associated to the standard tableau

T = 1 3 5 · · · 2n-3 2n-12 4 6 · · · 2n-2 2n

and with Zm = Z(ε2m−1−ε2m), for all integers m = 1 . . . n.6

We now construct a weight vector of weight 2ψ1 + 2ψ2 + · · · + 2ψn−1 − 2ψn. By

the same argument used before, the vector

x =n−1∑s=0

(−2i)n−1−s

( ∑1≤j1<j2<···<js≤n−1

Zj1Zj2 · · ·Zjsv

)

6As usual, a string Zj1Zj2 · · ·Zjsv is intended to be equal to v when s = 0.

224

is a simultaneous (−2i)-eigenvector of

Hm = E2m−1,2m − E2m,2m−1 = Zε2m−1−ε2m

for all m = 1, . . . , n− 1. We also introduce the element

y = Zε2n−1−ε2n · x + (2i)x

which is an eigenvector of Hn = Zε2n−1−ε2n of eigenvalue (+2i):7

Zε2n−1−ε2n · y = −4x + (2i)Zε2n−1−ε2n · x = (2i)(Zε2n−1−ε2n · x + (2i)x

).

We notice that y is also a (−2i)-eigenvector of H1, H2, . . . , Hn−1:

Zε2m−1−ε2m · y = Zε2n−1−ε2n ·(Zε2m−1−ε2m · x

)+ (2i)

(Zε2m−1−ε2m · x

)=

= (−2i)[ Zε2n−1−ε2n · x + 2i x] = (−2i)y ∀m = 1, . . . , n− 1.

Therefore, y is a weight vector for the Cartan subalgebra h of so(2n, C), of weight

2ψ1 + 2ε2 + · · ·+ 2ψn−1 − 2ψn.

? ? ?

Proof of lemma 13. Let ρ = (n, n). As seen in the previous lemma, ρ′ contains both

7Notice that x only involves strings in the roots α1 = ψ1−ε2, α2 = ε3−ε4 . . . αn−1 = ε2n−3−ε2n−2.Being all these roots orthogonal to ε2n−1 − ε2n, we have

Zε2n−1−ε2n · x =n−1∑s=0

(−2i)n−1−s

∑

1≤j1<j2<···<js≤n−1

ZmZj1Zj2 · · ·Zjsv

.

225

weights

υ−n = 2ψ1 + 2ε2 + · · ·+ 2ψn−1 − 2ψn

υ+n = 2ψ1 + 2ε2 + · · ·+ 2ψn−1 + 2ψn.

The purpose of this lemma is to show that Fρ′ actually contains the irreducible rep-

resentation with highest weight υ−n and υ+n . More precisely, we will prove that if an

irreducible submodule φ of Fρ′ contains the weight υ±n , then φ ≡ L(υ±n ).

We start with υ+n . We assume that φ ⊆ Fρ′ is irreducible and contains the weight

υ+n , and we write

λ = b1ψ1 + b2ε2 + · · ·+ bnψn

for the highest weight of φ.8 Then

• b1 = 2, because the coefficient of ψ1 in

λ− υ+n = (b1 − 2)ψ1 + (b2 − 2)ψ2 + · · ·+ (bn − 2)ψn

must be non negative.9

• b1 ≤ 2, because φ ⊆ Fρ′ is petite, hence the coefficients of any weight of φ

cannot exceed 2.10

We deduce that b1 = 2, and that

λ− υ+n = (b2 − 2)ψ2 + (b3 − 2)ψ3 + · · ·+ (bn − 2)ψn

is a sum of positive roots. Iterate this argument to show that bj = 2 for all j =

1, . . . , n − 1 . Finally, we obtain that (bn − 2)ψn is a sum of positive roots, and of

8Of course b1 ≥ b2 · · · ≥ bn−1 ≥| bn |.9Since υk ¹ λ, the difference λ− υ+

n is a sum of positive roots. As usual, we choose the positivesystem: ψi±ψji < j, so the first non-zero coefficient in a sum of positive roots must be positive.

10By construction, −ib1 is an eigenvalue of H1 = Zψ1−ψ2 .

226

course this can only happen if bn = 2, so λ = υ+n .

The argument for υ−n is similar. Indeed,

λ− υ−n = (b1 − 2)ψ1 + (b2 − 2)ψ2 + · · ·+ (bn−1 − 2)ψn−1 + (bn + 2)ψn

can only be a sum of positive roots if bj = 2 for all j = 1, . . . , n− 1 and bn = −2.

? ? ?

Proof of lemma 14. We fix ρ = (n, n) and show that the representation of the Weyl

group on (Fρ′)M is equal to the direct sum of two copies of (n, n).

We apply the same argument used in the proof of lemma 10, and show that the

equivalence class of a string of orthogonal roots of length 1 < r < n − 1 spans

a one-dimensional irreducible representation of M which is not equal to the trivial

representation. Hence, we just need to consider strings of length n.

If β is an arbitrary root and Zα1Zα2 . . . Zαnv is a string of orthogonal roots (of length

n), two possibilities can occur:

1. β = αj for some j = 1, . . . , n, and β ⊥ αi for all i 6= j

2. there are exactly two roots, say α1 and α2, to which β is not orthogonal (so

β ⊥ αi for all i > 2.

In the first case, mβ acts trivially on all the Zαis, hence

mβ · (Zα1Zα2 . . . Zαnv) = +Zα1Zα2 . . . Zαnv.

In the second case we have

Ad(mβ) (Zα1) = − (Zα1) Ad(mβ) (Zα2) = − (Zα2)

227

and Ad(mβ) (Zαi) = + (Zαi

) for all i = 3, . . . , n.

Once again

mβ · (Zα1Zα2 . . . Zαnv) = +Zα1Zα2 . . . Zαnv.

This proves that M acts trivially on every string of length n. Therefore

ρ = (n, n) ⇒ (Fρ′

)M= Fρ ⊕ Spann-strings.

Passing to the quotient does not introduce additional trivial M -types, so

(Fρ′

)M= Fρ ⊕ Span[ n-strings ].

To conclude the proof we must show that the representation of the Weyl group on the

set of linear combinations of equivalence classes of n-strings is isomorphic to (n, n).

We will make use of the following theorem:11

Theorem 18. Let Sλ be the Specht module associated to a partition λ. Then Sλ is

the vector space with generators the polytabloids eT , as T varies over all the tableaux

of shape λ, and relations of the form eT−∑

S eS, where the sum is over all S obtained

from T by exchanging the top k elements of one column with any k elements of the

preceding column, maintaining the vertical order of each set exchanged.

There is one such relation for each tableau T , each choice of adjacent columns, and

each k at least equal to the length of the shorter column.

We intend to apply these remarks to the Specht module S(n,n), which is equal to

Fρ. A set of generators for S(n,n) consists of the polytabloids eT associated to all the

tableaux of the form

T = j 1 j 3 · · · j 2n-3 j 2n-1 with j1 . . . j2n = 1 . . . 2n.j 2 j 4 · · · j 2n-2 j 2n

11See [13], §7.4.

228

Let us explore the relations among these polytabloids. When k = 1 and the adja-

cent columns under consideration are the first and second one, we can only take

• S = j 3 j 1 · · · j 2n-3 j 2n-1 = σεj1−εj3

· T, or

j 2 j 4 · · · j 2n-2 j 2n

• S = j 1 j 2 · · · j 2n-3 j 2n-1 = σεj2−εj3

· T .

j 3 j 4 · · · j 2n-2 j 2n

We obtain the relation:

e T = e [σ(εj1−εj3

)·T ] + e [σ(εj2−εj3

)·T ]

which we can rewrite as:12

(σ(εj1

−εj2) · eT

)+

(σ(εj1

−εj3) · eT

)+

(σ(εj2

−εj3) · eT

)= 0.

Please notice the analogy with the (?)-relation

(σ(εj1

−εj2) · S

)+

(σ(εj1

−εj3) · S

)+

(σ(εj2

−εj3) · S

)= 0

where S is the string Zεj1−εj2

Zεj3−εj4

· · ·Zεj2n−1−εj2n

v.13

Similarly, when k = 1 and we pick the ith and (i+1)th column, we find the relation:

(σ(εj2i−1

−εj2i)

)· eT ) +

(σ(εj2i−1

−εj2i+1) · eT

)+

(σ(εj2i

−εj2i+1) · eT

)= 0.

When k = 2 and the adjacent columns under considerations are the first and second

12Because eT = −σ(εj1−εj2 ) · eT .13This relation holds because the string S satisfies: σ(εj1−εj2) · S = −S and σ2

(εj1−εj3)· S = +S.

229

one, we can only take

• S = j 3 j 1 · · · j 2n-3 j 2n-1 (switch 1st and 2nd columns of T ).

j 4 j 2 · · · j 2n-2 j 2n

Therefore, we obtain the relation eT = eS, which is analogous to the commutativity

relation:

Zεj1−εj2

Zεj3−εj4︸︷︷︸

switch

Zεj5−εj6


v =

= Zεj3−εj4

Zεj1−εj2

Zεj5−εj6


v.

Finally, when k = 2 and we pick the ith and (i + 1)th column, we find the relation

eT = eS, where S is the tableau obtained from T by switching the ith and the (i+1)th

columns.

Consider that the mapping

< polytabloids >C → < n-strings >C

that carries the polytabloid eT associated to a tableau

T = j 1 j 3 · · · j 2n-3 j 2n-1

j 2 j 4 · · · j 2n-2 j 2n

into the n-string of orthogonal roots

Zεj1−εj2

Zεj3−εj4


eT .

It follows from the previous remarks that this mapping descends to an isomorphism

between Fρ = S(n,n) and the set of linear combinations of equivalence classes of n-

strings. Hence(Fρ′

)M= (n, n)⊕ (n, n) .

230

10.2 The construction of ρ′: some examples

10.2.1 Construction for the trivial representation of S3

When (ρ, Fρ) is the trivial representation of S3, each root reflection σα acts trivially

on Fρ ' C. In particular, there are no (−1)-eigenvectors, so Fρ′ ≡ F′ρ ≡ Fρ.

By definition, Zα · v = 0, for all v ∈ Fρ and all α ∈ ∆. Hence, ρ′ is the trivial repre-

sentation of so(3), and it lifts to the trivial representation of SO(3).

In general, if (ρ, Fρ) is the trivial representation of W (SL(n)), then (ρ′, Fρ′) is the

trivial representations of SO(n).

10.2.2 Construction for the standard representation of S3

Let (ρ, Fρ) be the two-dimensional irreducible representations of S3.14 We fix a basis

v, u of Fρ = C2 that consists of a (+1) and a (−1) eigenvector of σ12. With respect

to this basis v, u we can write:

σ12 Ã

1 0

0 −1

u : (−1) eigenvector of σ12

σ13 Ã

−1/2 −1/2

−3/2 1/2

u + v : (−1) eigenvector of σ13

σ23 Ã

−1/2 1/2

3/2 1/2

v − u : (−1) eigenvector of σ23

14It can be realized as the action of S3 on the hyperplane x1 + x2 + x3 = 0 of C3. Thenv = (1, 1, −2) and u = (1, −1, 0).

231

Since there is no pair of distinct orthogonal roots, we only have strings of length at

most one. Therefore Fρ′ = Cv ⊕ Cu⊕ CZ12(u)⊕ CZ13(u + v)⊕ CZ23(v − u).

We notice that

σ12(Z12u) = −Z12u

σ213(Z12u) = −Z12u

σ223(Z12u) = −Z12u

, so there are no relations. We get:

Fρ′ = Fρ′ = Cv ⊕ Cu⊕ CZ12(u)⊕ CZ13(u + v)⊕ CZ23(v − u).

Let us compute the action of the element Z12 of so(3) on Fρ′ :

¦ Z12 · v = 0, because σ12v = v

¦ Z12 · u = Z12u, because σ12u = −u

¦ Z12 · (Z12u) = −4u

¦ Z12 · (Z13(u + v)) = σ12 · (Z13(u + v)) = −Z23(v − u)

because σ212(Z13(u + v)) = −Z13(u + v)

¦ Z12 · (Z23(v − u)) = σ12 · (Z23(v − u)) = Z13(u + v)

because σ212(Z23(v − u)) = −Z23(v − u).

232

Hence,

Z12 Ã

0 0 0 0 0

0 0 −4 0 0

0 1 0 0 0

0 0 0 0 1

0 0 0 −1 0

with respect to the basis v, u, Z12(u), Z13(u + v), Z23(v − u) of Fρ′ .

The action of Z13 is given by:

¦ Z13 · v = 12Z13(v − σ13v) = 3

4Z13(v + u)

because σ13v 6= ±v and σ213v = v

¦ Z13 · u = 12Z13(u− σ13u) = 1

4Z13(u + v)

because σ13u 6= ±u and σ213u = u

¦ Z13 · (Z12u) = σ13 · (Z12u) = −12Z23(v − u)

because σ213(Z12u) = −Z12u

¦ Z13 · (Z13(u + v)) = −4(u + v)

¦ Z13 · (Z23(v − u)) = σ13 · (Z23(v − u)) = 2Z12u

because σ213(Z23(v − u)) = −Z23(v − u).

233

Hence,

Z13 Ã

0 0 0 −4 0

0 0 0 −4 0

0 0 0 0 2

3/4 1/4 0 0 0

0 0 −1/2 0 0

w.r.t. v, u, Z12(u), Z13(u + v), Z23(v − u).

Finally, we compute the action of Z23:

¦ Z23 · v = 12Z23 · (v − σ23v) = 3

4Z23(v − u)

¦ Z23 · u = 12Z23 · (u− σ23u) = −1

4Z23(v − u)

¦ Z23 · (Z12u) = σ23(Z12u) = −12Z13(u + v)

¦ Z23 · (Z13(u + v)) = σ23(Z13(u + v)) = 2Z12u

¦ Z23 · (Z13(v − u)) = −4(v − u).

234

And hence we have

Z23 =

0 0 0 0 −4

0 0 0 0 4

0 0 0 2 0

0 0 −1/2 0 0

3/4 −1/4 0 0 0

.

We notice that CZ12 is a Cartan subalgebra of so(3). Therefore, to identify the

representation, we just have to look at the eigenvalues of Z12. An easy computation

shows that they are given by 0, ±i, ±2i.

We conclude that ρ′ is an irreducible representation of SO(3), isomorphic to H2.


The standard representation of S4 can be realized as the action of the symmetric

group in four letters on the hyperplane

Fρ = z = (z1, z2, z3, z4) ∈ C4 : z1 + z2 + z3 + z4 = 0

of C4. The action is explicitly given by:

σ · (z1, z2, z3, z4) = (zσ−1(1), zσ−1(2), zσ−1(3), zσ−1(4))

for all σ ∈ S4 and all z ∈ Fρ. We choose the basis:

v1 = (1, −1, 0, 0) v2 = (0, 1, −1, 0) v3 = (0, 0, 1, −1)

235

of Fρ, and write down the matrix of each root reflection w.r.t. this basis.

σ12 Ã

−1 1 0

0 1 0

0 0 1

with eigenvalues

1 ⇒ v3

1 ⇒ v1 + 2v2

−1 ⇒ v1

σ13 Ã

0 −1 1

−1 0 1

0 0 1

with eigenvalues

1 ⇒ v1 + v3

1 ⇒ v2 + v3

−1 ⇒ v1 + v2

σ23 Ã

1 0 0

1 −1 1

0 0 1

with eigenvalues

1 ⇒ 2v1 + v2

1 ⇒ v3 − v1

−1 ⇒ v2

σ14 Ã

0 0 −1

−1 1 −1

−1 0 0

with eigenvalues

1 ⇒ v2

1 ⇒ v1 − v3

−1 ⇒ v1 + v2 + v3

σ24 Ã

1 0 0

1 0 −1

1 −1 0

with eigenvalues

1 ⇒ v1 + v2

1 ⇒ v1 + v3

−1 ⇒ v2 + v3

σ34 Ã

1 0 0

0 1 0

0 1 −1

with eigenvalues

1 ⇒ 2v2 + v3

1 ⇒ v1

−1 ⇒ v3

236

We notice that there is no pair of mutually orthogonal roots with a simultaneous

(−1)-eigenvector, so there are no strings of length bigger than one. We can write:

Fρ′ = C v1 ⊕ C v2 ⊕ C v3 ⊕ CZ12(v1) ⊕ CZ13(v1 + v2) ⊕ CZ23(v2)⊕

⊕CZ14(v1 + v2 + v3) ⊕ CZ24(v2 + v3) ⊕ CZ34(v3).

Because

σ12(Z12v1) = −Z12v1

σ13(Z12v1) 6= ±Z12v1

σ213(Z12v1) = −Z12v1

, no relations are imposed on the strings

Z12v1. The situation for the other strings is similar, so we conclude that there are no

relations and Fρ′ = Fρ′ .

We now compute the action of so(4) on Fρ′ w.r.t. to the basis

v1, , v2, , v3, , Z12(v1), , Z13(v1 +v2), Z23(v2), Z14(v1 +v2 +v3), Z24(v2 +v3), Z34(v3),

starting with the element Z12.

¦ Z12 · v1 = Z12v1

¦ Z12 · v2 = 12Z12 · (v2 − σ12v2) = −1

2Z12v1

¦ Z12 · v3 = 0

¦ Z12 · (Z12v1) = −4v1

237

¦ Z12 · (Z13(v1 + v2)) = σ12 · (Z13(v1 + v2)) = −Z23v2

¦ Z12 · (Z23v2) = σ12 · (Z23v2) = Z13(v1 + v2)

¦ Z12 · (Z14(v1 + v2 + v3)) = σ12 · (Z14(v1 + v2 + v3)) = −Z24(v2 + v3)

¦ Z12 · (Z24(v2 + v3)) = σ12 · (Z24(v2 + v3)) = Z14(v1 + v2 + v3)

¦ Z12 · (Z34v3) = 0.

Hence we have

Z12 Ã

0 0 0 −4 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

1 −1/2 0 0 0 0 0 0 0

0 0 0 0 0 1 0 0 0

0 0 0 0 −1 0 0 0 0

0 0 0 0 0 0 0 1 0

0 0 0 0 0 0 −1 0 0

0 0 0 0 0 0 0 0 0

In particular, we notice that Z12 acts on Fρ′ with eigenvalues

0, 0, 0, −i, −i, +i, +i, −2i, +2i.

The action of Z34 is given by:

238

¦ Z34 · v1 = 0

¦ Z34 · v2 = 12Z34 · (v2 − σ34v2) = −1

2Z34v3

¦ Z34 · v3 = Z34v3

¦ Z34 · (Z12v1) = 0

¦ Z34 · (Z13(v1 + v2)) = −Z14(v1 + v2 + v3)

¦ Z34 · (Z23v2) = σ34 · (Z23v2) = −Z24(v2 + v3)

¦ Z34 · (Z14(v1 + v2 + v3)) = σ34 · (Z14(v1 + v2 + v3)) = Z13(v1 + v2)

¦ Z34 · (Z24(v2 + v3)) = Z23(v2)

¦ Z34 · (Z34v3) = −4v3

239

Hence

Z34 Ã

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 −4

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 1 0 0

0 0 0 0 0 0 0 1 0

0 0 0 0 −1 0 0 0 0

0 0 0 0 0 −1 0 0 0

0 −1/2 1 0 0 0 0 0 0

and again the eigenvalues are

0, 0, 0, −i, −i, +i, +i, −2i, +2i.

If we denote by ξ1, ξ2 the linear functionals on h

ξ1 : aZ12 + b Z34 7→ −i a

ξ2 : aZ12 + b Z34 7→ −i b,

then we can write

0 ± 2ξ1 ± 2ξ2 ± (ξ1 + ξ2) ± (ξ1 − ξ2)

for the list of weights of ρ′.

We notice that ρ′ has highest weight 2ξ1, so we compare it with the irreducible

representation ρ2ξ1 of SO(4) of highest weight 2ξ1. Because also ρ2ξ1 has dimension

240

equal to 9, there is no doubt that ρ ≡ ρ2ξ1 .

10.2.4 Construction for the representation S2,2 of S4

We can choose a basis v, u of Fρ = C2 that consists of a (+1) and a (−1) eigenvector

of σ12. With respect to this basis, we have:

σ12, σ34 Ã

1 0

0 −1

u : (−1) eigenvector of σ12, σ34

σ13, σ24 Ã

−1/2 −1/2

−3/2 1/2

u + v : (−1) eigenvector of σ13, σ24

σ23, σ14 Ã

−1/2 1/2

3/2 1/2

v − u : (−1) eigenvector of σ23, σ14.

Therefore, we can write:

Fρ′ = Cu ⊕ Cv ⊕ CZ12u ⊕ CZ34u ⊕ CZ13(u + v) ⊕ CZ24(u + v)⊕

⊕CZ23(v − u) ⊕ CZ14(v − u) ⊕ CZ12Z34u ⊕ CZ13Z24(v + u) ⊕ CZ23Z14(v − u).

Claim 9. There are no relations among strings of length one.

Proof. We prove that there are no relations involving the string Z12u. The other cases

are similar.

The string Z12u is a (−1)-eigenvector for the root reflections σ12 and σ34. We notice

that there are two so(3)s containing Z12, and they are generated by Z12, Z13, Z23 and

241

Z12, Z14, Z24 respectively. Because

σ213(Z12u) = σ2

14(Z12u) = −Z12u

none of them gives rise to a relation. Similarly, the two so(3)s containing Z34 are

generated by Z34, Z13, Z14 and Z34, Z23, Z24, and because

σ213(Z12u) = σ2

23(Z12u) = −Z12u

none of them gives rise to a relation.

Claim 10. There is only one relation among strings of length two, namely:

Z12Z34u = −1

2Z23Z14(v − u)− 1

2Z13Z24(u + v) (∗) .

Proof. We look for possible relations involving the string Z12Z34u. Since Z12Z34u is

a (−1) eigenvector for σ12 and σ34, we must consider the so(3)s containing either Z12

or Z34. There are 4 of them, and they are generated by

1. Z12, Z13, Z23

2. Z12, Z14, Z24

3. Z34, Z13, Z14

4. Z34, Z23, Z24

We start with (1). Because

σ13(Z12Z34u) 6= ±(Z12Z34u) σ23(Z12Z34u) 6= ±(Z12Z34u)

but

σ213(Z12Z34u) = σ2

23(Z12Z34u) = +Z12Z34u

242

the following relation holds:

Z12Z34u = σ13(Z12Z34u) + σ23(Z12Z34u).

Being

σ13 · (Z12Z34u) = −12Z23Z14(v − u), and

σ23 · (Z12Z34u) = −12Z13Z24(u + v)

our relation becomes:

Z12Z34u = −1

2Z23Z14(v − u)− 1

2Z13Z24(u + v). (∗)

Let us now look at (2), which is the so(3) generated by Z12, Z14, Z24 . Because

σ14(Z12Z34u) 6= ±Z12Z34u σ24(Z12Z34u) 6= ±Z12Z34u

but

σ214(Z12Z34u) = σ2

24(Z12Z34u) = Z12Z34u

we have the relation:

Z12Z34u = σ14(Z12Z34u) + σ24(Z12Z34u).

Because σ14 acts like σ23, and σ24 acts like σ13, this relation is the same as (*).

243

Similarly, (3) gives rise to the relation

Z12Z34u = σ14 · (Z12Z34u) + σ13 · (Z12Z34u)

which is equivalent to

Z12Z34u = σ14(Z12Z34u) + σ24(Z12Z34u)

because σ13 acts as σ24, and (4) gives

Z12Z34u = σ23(Z12Z34u) + σ24(Z12Z34u)

which is equivalent to

Z12Z34u = σ14 · (Z12Z34u) + σ13 · (Z12Z34u).

In both cases we obtain the relation (*).

This proves that there is a unique relation among the 11 generators of F ′ρ. In

other words, the quotient space Fρ′ is the vector space with generators

v, u, Z12u, Z34u, Z23(v − u), Z14(v − u), Z13(v + u) . . .

. . . Z24(v + u), Z12Z34u, Z23Z14(v − u), Z13Z24(v + u)

and one relation:

Z12Z34u = −1

2Z23Z14(v − u)− 1

2Z13Z24(u + v) (∗) .

It is clear that Fρ′ has dimension 10. We pick the first ten elements in the above list as

244

a basis, and we use (*) to express Z13Z24(v+u) as a linear combination of basis vectors.

We now compute the action of so(4) on Fρ′ , starting with Z12.

¦ Z12 · v = 0

¦ Z12 · u = Z12u

¦ Z12 · (Z12u) = −4u

¦ Z12 · (Z34u) = Z12Z34u

¦ Z12 · (Z13(u + v)) = σ12 · (Z13(u + v)) = −Z23(v − u)

¦ Z12 · (Z24(u + v)) = σ12 · (Z24(u + v)) = Z14(v − u)

¦ Z12 · (Z23(v − u)) = σ12 · (Z23(v − u)) = Z13(v + u)

¦ Z12 · (Z14(v − u)) = σ12 · (Z14(v − u)) = −Z24(v + u)

¦ Z12 · (Z12Z34u) = −4Z34u

¦ Z12 · (Z23Z14(v − u)) = 4Z34u.

Only the last equation requires a comment. Using the relation (*) we obtain:

Z12 · (Z23Z14(v − u)) = 12Z12 · (Z23Z14(v − u)− σ12 · (Z23Z14(v − u))) =

245

= 12Z12 · (Z23Z14(v − u) + Z13Z24(v + u)) =

= 12Z12 · (−2 Z12Z34u) = 4Z34u X.

Hence:

Z12 Ã

0 0 0 0 0 0 0 0 0 0

0 0 −4 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 −4 4

0 0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 −1 0 0

0 0 0 0 −1 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0

0 0 0 1 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

We notice, in particular that Z12 acts with eigenvalues

0, 0, i, i, −i, −i, 2i, 2i, −2i, −2i.

Let us compute the action of Z34.

¦ Z34 · v = 0

¦ Z34 · u = Z34u

¦ Z34 · (Z12u) = Z12Z34u

¦ Z34 · (Z34u) = −4u

246

¦ Z34 · (Z13(u + v)) = σ34 · (Z13(u + v)) = −Z14(v − u)

¦ Z34 · (Z24(u + v)) = σ34 · (Z24(u + v)) = Z23(v − u)

¦ Z34 · (Z23(v − u)) = σ34 · (Z23(v − u)) = −Z24(v + u)

¦ Z34 · (Z14(v − u)) = σ34 · (Z14(v − u)) = Z13(v + u)

¦ Z34 · (Z12Z34u) = −4Z12u

¦ Z34 · (Z23Z14(v − u)) = 4Z12u.

A comment on the last equation:

Z34 · (Z23Z14(v − u)) = 12Z34 · (Z23Z14(v − u)− σ34 · (Z23Z14(v − u)) =

= 12Z34 · (Z23Z14(v − u) + Z24Z13(v + u)) =

= Z34 · (−Z12Z34u) = +4Z12u. X

247

Therefore:

Z34 Ã

0 0 0 0 0 0 0 0 0 0

0 0 0 −4 0 0 0 0 0 0

0 0 0 0 0 0 0 0 −4 4

0 1 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 1 0 0

0 0 0 0 0 0 −1 0 0 0

0 0 0 0 0 1 0 0 0 0

0 0 0 0 −1 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

and it acts with eigenvalues

0, 0, i, i, −i, −i, 2i, 2i, −2i, −2i.

To identify the representation, we compute the list of eigenvalues of

ρ′(b Z12 + aZ34):

0 0 ± 2i(a + b) ± 2i(a− b) i(a + b) ± i(a− b).

We conclude that ρ′ has weights:

0 0 ± (2ξ1 + 2ξ2) (2ξ1 − 2ξ2) ± (ξ1 + ξ2) ± (ξ1 − ξ2),

therefore

ρ′ = ρ2ε1+2ε2 ⊕ ρ2ε1−2ε2 .

248


The standard representation of S5 can be realized as the action of the symmetric

group in four letters on the hyperplane

Fρ = z = (z1, z2, z3, z4, z5) ∈ C5 : z1 + z2 + z3 + z4 + z5 = 0

of C5. The action is explicitly given by:

σ · (z1, z2, z3, z4, z5) = (zσ−1(1), zσ−1(2), zσ−1(3), zσ−1(4), zσ−1(5))

for all σ ∈ S5 and all z ∈ Fρ. We choose the basis:

v1 = (1, −1, 0, 0, 0) v2 = (0, 1, −1, 0, 0)

v3 = (0, 0, 1, −1, 0) v4 = (0, 0, 0, 1, −1)

of Fρ, and notice that each transpositions acts on Fρ with eigenvalues +1, +1, +1, −1.

For each transposition, we list the (−1)-eigenvector, and the corresponding string:

σ12 Ã v1 Ã Z12(v1)

σ13 Ã v1 + v2 Ã Z13(v1 + v2)

σ14 Ã v1 + v2 + v3 Ã Z14(v1 + v2 + v3)

σ15 Ã v1 + v2 + v3 + v4 Ã Z15(v1 + v2 + v3 + v4)

σ23 Ã v2 Ã Z23(v2)

249

σ24 Ã v2 + v3 Ã Z24(v2 + v3)

σ25 Ã v2 + v3 + v4 Ã Z25(v2 + v3 + v4)

σ34 Ã v3 Ã Z34(v3)

σ35 Ã v3 + v4 Ã Z35(v3 + v4)

σ45 Ã v4 Ã Z45(v4).

We notice that there is no pair of mutually orthogonal roots with a simultaneous

(−1)-eigenvector, so there are no strings of length bigger than one and the extension

Fρ′ has dimension equal to 14. Since there are no relations among strings, Fρ′ = Fρ′ .

If we choose in so(4) the Cartan subalgebra h = CZ12 ⊕ CZ34, we can obtain

the the weights of ρ′ simply by computing the eigenvalues of the generic element

(aZ12 + bZ34) of h on Fρ′ .

We start by computing the action of Z12 on Fρ′

¦ Z12 · v1 = Z12v1

¦ Z12 · v2 = 12Z12 · (v2 − σ12v2) = −1

2Z12v1

¦ Z12 · v3 = 0

250

¦ Z12 · v4 = 0

¦ Z12 · (Z12v1) = −4v1

¦ Z12 · (Z13(v1 + v2)) = σ12 · (Z13(v1 + v2)) = −Z23v2

¦ Z12 · (Z14(v1 + v2 + v3)) = σ12 · (Z14(v1 + v2 + v3)) = −Z24(v2 + v3)

¦ Z12 · (Z15(v1 + v2 + v3 + v4)) = σ12 · (Z15(v1 + v2 + v3)) = −Z25(v2 + v3 + v4)

¦ Z12 · (Z23v2) = σ12 · (Z23v2) = Z13(v1 + v2)

¦ Z12 · (Z24(v2 + v3)) = σ12 · (Z24(v2 + v3)) = Z14(v1 + v2 + v3)

¦ Z12 · (Z25(v2 + v3 + v4)) = σ12 · (Z25(v2 + v3 + v4)) = Z15(v1 + v2 + v3 + v4)

¦ Z12 · (Z34v3) = 0

¦ Z12 · (Z35(v3 + v4)) = 0

¦ Z12 · (Z45v4) = 0.

251

Hence we have

Z12 Ã

0 0 0 0 −4 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 −1/2 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 0 0 0 0 1 0 0 0

0 0 0 0 0 −1 0 0 0 0 0 0 0 0

0 0 0 0 0 0 −1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 −1 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

In particular, we notice that Z12 acts on Fρ′ with eigenvalues

0, 0, 0, 0, 0, 0, −i, −i, −i, +i, +i, +i, −2i, +2i.

The action of the other Zαs is computed similarly.

We find that ρ′ has highest weight 2ξ1, so we compare it with the irreducible repre-

sentation ρ2ξ1 of SO(5) of highest weight 2ξ1. Because ρ2ξ1 has also dimension 14,

there is no doubt that ρ ≡ ρ2ξ1 .

252

10.2.6 Construction for the representation S(3,3) of S6

We start by giving an explicit description of the representation of S6 associated to

the partition (3, 3).

Let F be the Specht module S(3,3). It follows from the Hook formula that F has

dimension five: filling up each box of the Young diagram T3,3 with the corresponding

hook length we obtain:

| 4 | 3 | 2 || 3 | 2 | 1 | hence dim(F ) = 6!

4·3·2·1·3·2·1 = 5.

We pick a basis of F consisting of standard polytabloids:15

v1 = P

(1 3 52 4 6

)v2 = P

(1 2 53 4 6

)

v3 = P

(1 2 43 5 6

)v4 = P

(1 3 42 5 6

)

v5 = P

(1 2 34 5 6

)

To describe the representation of S6 on F , it is sufficient to explain how the transposi-

15Recall that for any tableau T =a1 a2 a3a4 a5 a6

of shape (3, 3), the corresponding

polytabloid P (T ) is a linear combination of tabloids. More precisely:

P (T ) = a1 a2 a3a4 a5 a6 − a4 a2 a3

a1 a5 a6 − a1 a5 a3a4 a2 a6 − a1 a2 a6

a4 a5 a3

+ a4 a5 a3a1 a2 a6 + a1 a5 a6

a4 a2 a3 + a4 a2 a6a1 a5 a3 − a4 a5 a6

a1 a2 a3 .

We draw a tabloid using bars instead of boxes because the element of each row can be per-muted without changing the tabloid. A polytabloid P (T ) is called standard if the tableau T isstandard, i.e. if the rows and the columns of T are both increasing. The standard polytabloids ofshape (3, 3) form a basis of the Specht module S(3,3).Also recall that the for any σ in S(3,3), and all tableaux T , we define σ · P (T ) = P (σ · T ), and thetableau σ · T is obtained from T by applying the permutation σ to its entries.

253

tions (1 2), (1 3), (1 4), (1 5), (1 6) act on F . However, for the purpose of constructing

the extension Fρ′ , it is convenient to have an explicit expression for the action of each

transposition σ of S6.

For all σ, we give the matrix (w.r.t. the basis v1, v2, v3, v4, v5) of the action of

σ on F , and we list the corresponding (−1)-eigenvectors:

(1 2) ⇒

−1 −1 0 0 1

0 1 0 0 0

0 0 1 0 0

0 0 −1 −1 −1

0 0 0 0 1

(−1)-eigenvectors:

v1, v4

(1 3) ⇒

1 0 0 0 0

−1 −1 0 0 0

0 0 −1 −1 −1

0 0 0 1 0

0 0 0 0 1


v2, v3

(1 4) ⇒

0 1 0 −1 0

1 0 0 1 0

0 0 1 0 0

0 0 0 1 0

0 0 −1 −1 −1


v5, v1 − v2

254

(1 5) ⇒

0 −1 0 0 −1

0 1 0 0 0

1 0 0 1 0

0 1 1 0 1

−1 −1 0 0 0


v1 − v3 + v5

v3 − v4

(1 6) ⇒

1 1 0 1 0

0 0 0 −1 0

−1 0 0 0 1

0 −1 0 0 0

1 1 1 1 0


v1 − v2 + v3 − v4

v3 − v5

(2 3) ⇒

0 1 0 0 1

1 0 0 0 −1

0 0 0 1 0

0 0 1 0 0

0 0 0 0 1


v3 − v4, v1 − v2

(2 4) ⇒

1 0 0 1 −1

−1 −1 −1 −1 0

0 0 1 0 0

0 0 0 0 1

0 0 0 1 0


v2, v1 − v4 + v5

255

(2 5) ⇒

1 0 0 0 0

0 1 0 0 0

−1 −1 −1 −1 0

0 0 0 1 0

1 0 0 0 −1


v3, v5

(2 6) ⇒

0 0 0 −1 −1

0 0 1 1 1

1 1 0 0 0

0 0 0 1 0

−1 0 0 −1 0


v1 − v2 + v5, v2 − v3

(3 4) ⇒

−1 −1 0 −1 0

0 1 0 0 0

0 0 0 0 1

0 0 0 1 0

0 0 1 0 0


v3 − v5, v1

(3 5) ⇒

1 1 0 0 −1

0 0 0 0 1

0 0 1 0 0

−1 −1 −1 −1 0

0 1 0 0 0


v1 − v2 + v5, v4

256

(3 6) ⇒

0 −1 0 1 0

0 1 0 0 0

0 0 1 0 0

1 1 0 0 0

0 −1 −1 0 −1


v1 − v4, v5

(4 5) ⇒

0 0 0 1 1

0 0 1 0 0

0 1 0 0 0

1 0 0 0 −1

0 0 0 0 1


v1 − v4, v2 − v3

(4 6) ⇒

1 0 0 0 0

0 1 0 0 0

0 −1 −1 0 −1

−1 0 0 −1 0

0 0 0 0 1


v3, v4

(5 6) ⇒

−1 0 0 −1 1

0 −1 −1 0 −1

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1


v1, v2

Now that the structure of F = S(3,3) is understood, we proceed to the construction

of the extension Fρ′ . The first step is to add all the strings of length one. Since we

257

have 15 transpositions and each of them has two (−1)-eigenvectors, there is a total

of 30 strings of length one. They are:

Z1 2v1 Z1 3v2 Z1 4v5

Z1 2v4 Z1 3v3 Z1 4(v1 − v2)

Z2 3(v1 − v2) Z1 6(v3 − v5) Z1 5(v1 − v3 + v5)

Z1 5(v3 − v4) Z2 3(v3 − v4) Z1 6(v1 − v2 + v3 − v4)

Z2 5v3 Z2 6(v2 − v3) Z2 4(v1 − v4 + v5)

Z2 4v2 Z2 5v5 Z2 6(v1 − v2 + v5)

Z3 4v1 Z3 6(v1 − v4) Z3 5(v1 − v2 + v5)

Z3 5v4 Z3 4(v3 − v5) Z3 6v5

Z4 6v3 Z4 6v4 Z4 5(v1 − v4)

Z5 6v2 Z5 6v1 Z4 5(v1 − v4).

Claim 11. The 30 strings of length one are linearly independent..

Proof. This is very easy, because a string S = Zµv (of length one) is included in a

258

relation only if there exists a triple of roots α, β, γ, forming an A2, such that

σα · Zµv = −Zµv σ2β · Zµv = +Zµv σ2

γ · Zµv = +Zµv

and σβ · Zµv 6= ±(Zµv), σγ · Zµv 6= ±(Zµv). Suppose that all these assumption are

satisfied by a triple α, β, γ. Then, because Zµv is a (−1)-eigenvector of σα, the root

α is either equal to µ or orthogonal to it. If α = µ, then β does not commute with µ,

and we reach a contradiction:

σ2β · (Zµv) = (−Zµ)(+v) = −Zµv.

If α is orthogonal to µ and β is not orthogonal to µ, then again we reach a contradic-

tion. Finally we discuss the case in which α, β (and of course also γ) are orthogonal

to µ. Because

σβ · (Zµv) = Zµ(σβ · v) σγ · (Zµv) = Zµ(σγ · v)

the relation “Zµv + σβ · (Zµv) + σγ · (Zµv) = 0” is just linearity. Again, we do not

obtain a linear dependence condition on the string Zµv.

The first step of the construction of Fρ′ is now complete. So far we have obtained

an extension of dimension 35. The second step consists of adding strings of pairs of

orthogonal roots “ZαZβv” (with α ⊥ β and σα · v = σβ · v = −v). Writing α = εi− εj

and β = εk− εl, the problem is reduced to finding simultaneous (−1)-eigenvectors for

the disjoint transpositions (i j), (k l).

These are all the possible simultaneous (−1)-eigenvectors for disjoint transpositions:16

• v1

−1 eigenvector- - - - - - - > (1 2), (3 4), (5 6)

16The arrow joins a vector v to a group of disjoint transpositions that acts by (−1) on v.

259

• v2 - - - - - - - > (1 3), (2 4), (5 6)

• v3 - - - - - - - > (1 3), (2 5), (4 6)

• v4 - - - - - - - > (1 2), (4 6), (3 5)

• v5 - - - - - - - > (1 4), (2 5), (3 6)

• (v1 − v2)−1 eigenvector- - - - - - - > (2 3), (1 4), (5 6)

• (v3 − v4) - - - - - - - > (2 3), (1 5), (4 6)

• (v3 − v5) - - - - - - - > (1 6), (3 4), (2 5)

• (v1 − v4) - - - - - - - > (1 2), (3 6), (4 5)

• (v2 − v3) - - - - - - - > (2 6), (1 3), (4 5)

• (v1 − v3 + v5)−1 eigenvector- - - - - - - > (1 5), (3 4), (2 6)

• (v1 − v4 + v5) - - - - - - - > (2 4), (3 6), (1 5)

• (v1 − v2 + v5) - - - - - - - > (1 4), (2 6), (3 5)

• (v1 − v2 + v3 − v4)−1 eigenvector- - - - - - - > (1 6), (2 3), (4 5)

• (v1 − v2 − v4 + v5) - - - - - - - > (1 6), (2 4), (3 5)

We obtain a total of 45 strings of length two:

Z1 2Z3 4v1 Z1 2Z5 6v1 Z3 4Z5 6v1

Z1 3Z2 4v2 Z1 3Z5 6v2 Z2 4Z5 6v2

Z1 3Z2 5v3 Z1 3Z4 6v3 Z2 5Z4 6v3

260

Z1 2Z4 6v4 Z1 2Z3 5v4 Z3 5Z4 6v4

Z1 4Z2 5v5 Z1 4Z3 6v5 Z2 5Z3 6v5

Z1 4Z2 3(v1 − v2) Z1 4Z5 6(v1 − v2) Z2 3Z5 6(v1 − v2)

Z1 5Z2 3(v3 − v4) Z1 5Z4 6(v3 − v4) Z2 3Z4 6(v3 − v4)

Z1 6Z3 4(v3 − v5) Z1 6Z2 5(v3 − v5) Z3 4Z2 5(v3 − v5)

Z1 2Z3 6(v1 − v4) Z1 2Z4 5(v1 − v4) Z3 6Z4 5(v1 − v4)

Z1 3Z2 6(v2 − v3) Z1 3Z4 5(v2 − v3) Z2 6Z4 5(v2 − v3)

Z1 5Z2 6(v1 − v3 + v5) Z1 5Z3 4(v1 − v3 + v5) Z2 6Z3 4(v1 − v3 + v5)



Z1 6Z2 3(v1 − v2 + v3 − v4) Z1 6Z4 5(v1 − v2 + v3 − v4)

Z2 3Z4 5(v1 − v2 + v3 − v4) Z1 6Z2 4(v1 − v2 − v4 + v5)

Z1 6Z3 5(v1 − v2 − v4 + v5) Z2 4Z3 5(v1 − v2 − v4 + v5).

Claim 12. There are 15 linearly independent relations among strings of length two.

261

Hence, there are 30 linearly independent strings of length two.

Proof. A string S = Zµ1Zµ2v (of length two) is included in a relation

S = σβ · S + σγ · S (∗)

only if there exists a triple of roots α, β, γ, forming an A2, such that

σα · S = −S σ2β · S = +S σ2

γ · S = +S

and σβ · S 6= ±S, σγ · S 6= ±S. Suppose that all these assumption are satisfied by

a triple α, β, γ. Then, because S = Zµ1Zµ1v is a (−1)-eigenvector of σα, the root α

either belongs to the set µ1, µ2 or it is orthogonal to it. We discuss the two cases

separately.

If α = µ1, then β cannot be orthogonal to µ2 (or we would have

σ2β · S = σ2

β · (ZαZµ2v) = (−Zα)(+Zµ2v)(+v) = −S,

which is a contradiction). Therefore there exists a subset I = i1, i2, i3, i4 of

1, . . . , 6 such that the roots α = µ1, β, γ, µ2 can all be written in the form εk − εl,

with k, l in I. It follows that the relation (∗) is of the type

aZεi1−εi2

Zεi3−εi4

+ b Zεi1−εi3

Zεi2−εi4

+ c Zεi1−εi4

Zεi2−εi3

.

We have one such relation for each subset I = i1, i2, i3, i4 of 1, . . . , 6. It is

clear that different Is give rise to linearly independent relations, so we get a total of

15 =(64

)relations of this type.

We now discuss the case in which α is orthogonal to both µ1 and µ2, and show that

there are no other relations. Because σ2β · S = +S, the root β is either orthogonal to

262

both µ1 and µ2, or is not orthogonal to any of them. The latter case gives rise to a

contradiction, because it requires β to be of the form ±(εk− εl) with k an index of µ1

and l an index of µ2, making β orthogonal to α. Therefore we must assume that α,

β and γ are all orthogonal to the set µ1, µ2. In this case, the relation (∗) becomes

Zµ1Zµ2(v) = Zµ1Zµ2(σβ · v) + Zµ1Zµ2(σγ · v)

and it is just linearity. This concludes the proof.

The second step of the construction is now complete, and we have introduce 30

additional generators, increasing the dimension of the extension to 65. The next step

is to look at strings of length three (this will be of course the last step, because three

is the maximum number of disjoint transpositions in S6).

We have already listed the possible simultaneous eigenvectors of disjoint transposi-

tions, so writing down the strings of length three is easy:

Z1 2Z3 4Z5 6v1 Z1 3Z2 4Z5 6v2

Z1 3Z2 5Z4 6v3 Z1 2Z3 5Z4 6v4

Z1 4Z2 5Z3 6v5 Z1 4Z2 3Z5 6(v1 − v2)

Z1 5Z2 3Z4 6(v3 − v4) Z1 6Z3 4Z2 5(v3 − v5)

Z1 3Z2 6Z4 5(v2 − v3) Z1 5Z2 6Z3 4(v1 − v3 + v5)

Z1 5Z2 4Z3 6(v1 − v4 + v5) Z1 4Z2 6Z3 5(v1 − v2 + v5)

263

Z1 6Z2 3Z4 5(v1 − v2 + v3 − v4) Z1 6Z2 4Z3 5(v1 − v2 − v4 + v5)

Z1 2Z3 6Z4 5(v1 − v4).

Claim 13. There are 15 relations among the strings of length three, but only 10 of

them are linearly independent. Hence there is a total of 5 linearly independent strings

of length three.

Proof. We list the 15 relations, to prove that only 10 of them are linearly independent

is just a linear algebra exercise.

• Z1 2Z3 4Z5 6v1 = +Z2 3Z1 4Z5 6(v1 − v2)− Z1 3Z2 4Z5 6v2

• Z1 2Z3 4Z5 6v1 = +Z1 2Z4 5Z3 6(v1 − v4)− Z1 2Z3 5Z4 6v4

• Z1 2Z3 4Z5 6v1 = +Z2 5Z3 4Z1 6(v3 − v5)− Z1 5Z3 4Z2 6(v1 − v3 + v5)

• Z1 2Z3 5Z4 6v4 = +Z1 5Z2 3Z4 6(v3 − v4)− Z1 3Z2 5Z4 6v3

• Z1 2Z3 5Z4 6v4 = +Z1 6Z3 5Z2 4(v1 − v2 − v4 + v5)− Z1 4Z2 6Z3 5(v1 − v2 + v5)

• Z1 3Z2 4Z5 6v2 = +Z1 3Z2 6Z4 5(v2 − v3)− Z1 3Z2 5Z4 6v3

• Z1 3Z2 4Z5 6v2 = +Z1 6Z2 4Z3 5(v1 − v2 − v4 + v5)− Z1 5Z2 4Z3 6(v1 − v3 + v5)

• Z1 3Z2 5Z4 6v3 = +Z1 6Z2 5Z3 4(v3 − v5)− Z1 4Z2 5Z3 6v5

264

• Z1 4Z2 3Z5 6(v1 − v2) = +Z2 3Z4 5Z1 6(v1 − v2 + v3 − v4)− Z1 5Z2 3Z4 6(v3 − v4)

• Z1 4Z2 3Z5 6(v1 − v2) = +Z1 4Z3 5Z2 6(v1 − v2 + v5)− Z1 4Z2 5Z3 6v5

• Z1 5Z2 3Z4 6(v3 − v4) = +Z1 5Z3 4Z2 6(v1 − v3 + v5)− Z1 5Z2 4Z3 6(v1 − v4 + v5)

• Z1 4Z2 5Z3 6v5 = +Z1 5Z2 4Z3 6(v1 − v4 + v5) + Z1 2Z4 5Z3 6(v1 − v4)

• Z1 5Z3 4Z2 6(v1 − v3 + v5) = −Z1 4Z3 5Z2 6(v1 − v2 + v5) + Z1 3Z4 5Z2 6(v2 − v3)

• Z1 6Z2 5Z3 4(v3 − v5) = +Z1 6Z2 4Z3 5(v1 − v2 − v4 + v5)+

−Z1 6Z2 3Z4 5(v1 − v2 + v3 − v4)

• Z1 6Z2 3Z4 6(v1 − v2 + v3 − v4) = +Z1 3Z2 6Z4 5(v2 − v3)− Z1 2Z3 6Z4 5(v1 − v4).

Notice that each relation is characterized by the presence of one repeated root.

We conclude that the extension Fρ′ of F has dimension equal to 70.

This is exactly what we expected. Indeed we should obtain

Fρ′ = L(2ε1 + 2ψ2 + 2ε3)⊕ L(2ε1 + 2ψ2 − 2ε3)

and the irreducible representations of SO(6) of highest weight 2ε1 + 2ψ2 ± 2ε3 have

both dimension equal to 35.

265

10.2.7 Construction for the representation S(3,2) of S5

The Specht module S(3,2) of S5 is simply the restriction of the Specht module S(3,3)

of S6 to S5. So the construction of the extension of S(3,2) follows directly from the

computations done for S(3,3). We find that

S(3,2) has dimension 5.

There are 20 strings of length one, and no relations among them.

There are 15 strings of length two, and 5 relations among them.

There are no strings of length three or more.

The extension Fρ′ has dimension 35, and highest weight 2ξ1 + 2ξ2.

For dimensional reasons, ρ′ coincides with the irreducible representation of SO(5) of

highest weight 2ξ1 + 2ξ2.

? ? ??

We conclude our list of examples of construction of Fρ′ with a remark.

Remark. The algorithm described in chapter 9 extends every Weyl group represen-

tations ρ that does not include the sign of S3, to a well defined representation of

Lie(K).

If we carry out the same construction for the sign representation of S3, we obtain

a 4-dimensional extension that fails to be a Lie algebra representation, because the

bracket relations do not hold.

10.3 Extension of non-spherical representations

It is remarkable that our algorithm to construct petite K-types also works when

the representation ρ of M ′ is not spherical (but still satisfies the requirement of not

266

including the sign).

We will discuss this generalization in the next chapter. As an example, we construct

the extension for the two three dimensional irreducible representations ν1 and ν2 of

M ′ ⊆ SO(3).17

10.3.1 Construction for ν2

We can find a basis of Fν2 ' C3 such that the action of M ′ is given by:

σ12 Ã

0 0 −1

0 −1 0

1 0 0

⇒ (−1)-eigenvector: e2

σ13 Ã

0 −1 0

1 0 0

0 0 −1

⇒ (−1)-eigenvector: e3

σ23 Ã

−1 0 0

0 0 1

0 −1 0

⇒ (−1)-eigenvector: e1.

Hence we obtain three strings of length one: Z12e2, Z13e3 and Z23e1. There is

17Recall that ν1 and ν2 are the two irreducible summands of IndM ′M δ1, where δ1 the character of M

that maps a diagonal matrix in SO(3) into its first diagonal entry. Alternatively, we can constructν1 and ν2 by inducing to M ′ respectively the trivial and the sign representation of the subgroup H1

of M ′ defined as follows:let α1 = ψ2− ε3 be the only good root for δ1 (the attribute “good” refers to the fact that δ1(mα1) =Identity), and let W1 be the subgroup of the Weyl group generated by the root reflection sα1 . Letπ : M ′ → W = M ′/M be the projection. We define H1 to be the inverse image of W1 via π.

267

one relation among the one-strings, given by:

Z12e2 = σ13 · (Z12e2) + σ23 · (Z12e2) ⇔ Z12e2 = −Z23e1 + Z13e3.

We also notice that there are no strings of length two, because there is no pair of

orthogonal roots in A2. The extension Fρ′ is therefore five-dimensional. With respect

to the basis e1, e2, e3, Z12e2, Z23e1, we can write

• Z12 · e1 = σ12 · e1 = e3

• Z12 · e2 = Z12e2

• Z12 · e3 = σ12 · e3 = −e1

• Z12 · (Z12e2) = −4e2

• Z12 · (Z23e1) = 12Z12 · (Z23e1 − σ12 · (Z23e1)) =

= 12Z12 · (Z23e1 − Z13e3) = 1

2Z12 · (−Z12e2) = +2e2

• Z13 · e1 = σ13 · e1 = e2

• Z13 · e2 = σ13 · e2 = −e1

• Z13 · e3 = Z13e3 = Z12e2 + Z23e1

• Z13 · (Z12e2) = 12Z13 · (Z12e2 − σ13 · (Z12e2)) =

= 12Z13 · (Z12e2 + Z23e1) = 1

2Z13 · (Z13e3) = −2e3

• Z13 · (Z23e1) = 12Z13 · (Z23e1 − σ13 · (Z23e1)) =

= 12Z13 · (Z23e1 + Z12e2) = 1

2Z13 · (Z13e3) = −2e3

268

• Z23 · e1 = Z23e1

• Z23 · e2 = σ23 · e2 = −e3

• Z23 · e3 = σ23 · e3 = +e2

• Z23 · (Z12e2) = 12Z23 · (Z12e2 − σ23 · (Z12e2)) =

= 12Z23 · (Z12e2 − Z13e3) = 1

2Z13 · (−Z23e1) = +2e1

• Z23 · (Z23e1) = −4e1.

We obtain the matrices:

Z12 Ã

0 0 −1 0 0

0 0 0 −4 2

1 0 0 0 0

0 1 0 0 0

0 0 0 0 0

Z13 Ã

0 −1 0 0 0

1 0 0 0 0

0 0 0 −2 −2

0 0 1 0 0

0 0 1 0 0

269

Z23 Ã

0 0 0 2 −4

0 0 1 0 0

0 −1 0 0 0

0 0 0 0 0

1 0 0 0 0

It is easy to check that:

the bracket relations hold

each Zα acts with eigenvalues 0, ±i, ±2i

Fρ′ has highest weight 2ξ1, and for dimensional reasons, it coincides with the

irreducible representation of SO(3) with highest weight 2ξ1, which is H2.18

10.3.2 Construction for ν1

The construction for ν1 is trivial. We can find a basis of Fν2 ' C3 such that the

action of M ′ is given by:

σ12 Ã

0 0 −1

0 1 0

1 0 0

σ13 Ã

0 −1 0

1 0 0

0 0 1

18We have denoted by H2 the representation of SO(3) on the space of homogeneous harmonicpolynomials of degree two in three variables.

270

σ23 Ã

1 0 0

0 0 −1

0 1 0

.

For each positive root α, σα acts with eigenvalues 1, +i, −i. So there are no strings

at all, and Fρ′ = Fρ. By definition, each Zα acts by 0, +i, −i on the eigenspace of σα

of eigenvalue 1, +i, −1, so we find:

Z12 Ã

0 0 −1

0 0 0

1 0 0

Z13 Ã

0 −1 0

1 0 0

0 0 0

Z23 Ã

0 0 0

0 0 −1

0 1 0

.

It is easy to check that Fρ′ has highest weight ξ1, and for dimensional reasons, it

coincides with the irreducible representation of SO(3) of highest weight ξ1, which is

H1.

271

Chapter 11

A generalization to the

non-spherical case

Let G be the group SL(n, R), and let (ρ, Fρ) be any representation of M ′ that does

not contain the sign of W (SL(3)).1 It is possible to extended ρ to a petite K-type

using “almost” the same construction outlined in chapter 9.

Delicate Point. If ρ is reducible, or ρ is not a Weyl group representation, M does

not necessarily act by ±1 on strings.

Let S = Zα1 . . . Zαrv be a string, i.e. let α1, . . . , αr be mutually orthogonal

positive roots and let v ∈ Fρ be a simultaneous (−1)-eigenvector of σα1 , . . . , σαr . For

every positive root β, we have

σ2β · (Zα1 . . . Zαrv) = Zα1 . . . Zαr︸︷︷︸

=±Zα1 ...Zαr

(σ2β · v︸︷︷︸=?

).

Because (σ2β · v) might not be a multiple of v, σ2

β · S might be different from ±S.

How do we overcome this difficulty?

1We do not require ρ to be irreducible, nor to be a Weyl group representation.

272

Consider the elements v+ = (v + σ2β · v) and v− = (v − σ2

β · v). They belong to

Fρ, and they clearly satisfy the condition σ2β · (v±) = ±(v±). We claim that v+ and

v− are again simultaneous (−1)-eigenvectors of σα1 , . . . , σαr .

If β is orthogonal to the αjs, this claim is obvious, because σαj·(σ2

β ·v) = σ2β ·(σαj

·v) =

−σ2β · v. Let us show that the same results holds when β is not orthogonal to one of

the roots, say to αj. Let βj be a positive root such that the set

±β, ±αj, ±βj

is a root system of type A2. The elements σ2β, σ2

αj, σ2

βjof M are all of order two, they

commute and their product equals the identity. Therefore, we can write:

σαj· (σ2

β · v) = (σαjσ2

βσ−1αj

) · (σαj· v) = −σ±2

βj· v =

= −σ2βj· v = −(σ2

βσ2αj· v) = −σ2

β · v.

It follows that σαj· (v±) = −(v±), for all j = 1 . . . r. Hence, the strings Zα1 . . . Zαrv

+

and Zα1 . . . Zαrv− are well defined. By linearity:

Zα1 . . . Zαrv =1

2(Zα1 . . . Zαrv

+) +1

2(Zα1 . . . Zαrv

−).

This shows that any string S that does not satisfy the condition σ2β · S = ±S, can be

decomposed as a linear combination of strings on which σ2β acts by ±1. Therefore, a

generalization of the construction of ρ′ is possible, and it is fairly easy.

We follows exactly the same steps. The only elements of difference are the following:

1. When looking for copies of the sign representation of S3 inside Fρ′ , we also

need to check the case in which a string S does not generate a one-dimensional

irreducible representation of Mα,β,γ.

273

2. When defining the action of Zα on Fρ, we also need to consider the (± i)-

eigenspace of σα.

3. When defining the action of Zα on a string S, we also need to contemplate the

option σ2α · S 6= ±S.

4. When checking the bracket relations, there are more cases to be considered.

We discuss each of these points separately.

(1) If S = Zα1 . . . Zαrv does not generate a one-dimensional irreducible represen-

tation of Mα,β,γ (i.e. if the subspace Cv of Fρ is not stable under Mα,β,γ), then we

consider the vectors:

v0 = v + σ2α · v + σ2

β · v + σ2γ · v

v1 = v + σ2α · v − σ2

β · v − σ2γ · v

v2 = x− σ2α · v + σ2

β · v − σ2γ · v

v3 = x− σ2α · v − σ2

β · v + σ2γ · v,

which do have this property.2 Because vi is a (−1)-eigenvectors of σαj, for all

j = 1, . . . , r and all i = 1, . . . , 4, we obtain a decomposition of S = Zα1 . . . Zαrv as a

linear combination of strings that generate one-dimensional representations of Mα,β,γ:

Zα1 . . . Zαrv = 14(Zα1 . . . Zαrv0) + 1

4(Zα1 . . . Zαrv1)+

+14(Zα1 . . . Zαrv2) + 1

4(Zα1 . . . Zαrv3).

2For instance, we have:

σα · (v1) = σα · ((v − σ2β · v) + σ2

α · (v − σ2β · v)) = +v1

σβ · (v1) = σα · ((v + σ2α · v)− σ2

β · (v + σ2α · v)) = −v1

σγ · (v1) = σα · ((v + σ2α · v)− σ2

γ · (v + σ2α · v)) = −v1.

274

For each of these strings, the same arguments used in chapter 10 apply, and show

that the extension ρ′ only contains sign representations of type (a), (b) or (c).3

Hence we can remove every copy of the sign representation from ρ′ by imposing

relations of the form

S = σβ · S + σγ · S (?)

for all the strings S such that σα · S = −S and σ2β · S = +S.

(2) We let Zα act on the (± i)-eigenspace of σα by Zα · v = σα · v.

(3) If σ2α does not act by ±1 on a string S = Zα1 . . . Zαrv, we write

Zα1 . . . Zαrv =1

2(Zα1 . . . Zαrv

+) +1

2(Zα1 . . . Zαrv

−)

with v± = v ± σ2α · v, and we observe that since

σ2α · (Zα1 . . . Zαrv

+) = +(Zα1 . . . Zαrv+)

σ2α · (Zα1 . . . Zαrv

−) = −(Zα1 . . . Zαrv−)

the action of Zα on these two strings is well defined. Then we set

Zα · (Zα1 . . . Zαrv) = Zα · (Zα1 . . . Zαrv+) + Zα · (Zα1 . . . Zαrv

−).

(4) Basically, we need to check that the bracket relations hold for the extensions

of the two irreducible non-spherical representations of M ′ ⊆ SL(3). Please, refer to

chapter 10 for an explicit construction of these extensions.

3Please, refer to 9.4.3 for this terminology.

275

11.1 An inductive argument for the construction

The generalization of the algorithm to non-spherical representations of M ′ leads to

an inductive argument for the construction of petite K-types.

Let ρ be any representation of M ′ that does not contain the sign of S3. We can

construct the extension ρ′ by induction. The first step consists of adding to Fρ the

linear span of all the strings of the form Zνv, where ν is a positive root and v is a

(−1)-eigenvector of σν .

We define an action of M ′ on strings by

σ · (Zνv) = (Ad(σ)(Zν))(σ · v)

and we impose two kinds of relations:

• Zν(a1v1 + a2v2) = a1Zνv1 + a2Zνv2 “ linearity relations ′′

for all a1, a2 in C, and all (−1)-eigenvectors v1, v2 of σν in Fρ

• Zνv = σβ · (Zνv) + σγ · (Zνv) “(?)-relations ′′

for all triples of positive roots α, β, γ forming an A2, such that σα · (Zνv) =

−(Zνv) and σ2β · (Zνv) = +(Zνv).

When α, β and γ are all orthogonal to ν, a (?)-relation is just a linearity relation.4

So we can as well assume that ν 6⊥ α, β, γ. Then the condition

σα · (Zνv) = −(Zνv) = −σ2β · (Zνv)

can only occur if

α = ν or α ⊥ ν ; β 6⊥ ν (so automatically γ 6⊥ ν)

and

σα · v = −v; σ2β · v = −v (so automatically σ2

γ · v = −v).

The representation of M ′ on Fρ ⊕ Span(strings) descends to a representation ρ1 on

4Because v, σβ · v and σγ · v are all (−1)-eigenvectors of σν .

276

the quotient space

Fρ1 =Fρ ⊕ Span(strings)

∼.

We notice that, for any positive root ν, the (−1)-eigenvectors of σν in Fρ1 are either

(−1)-eigenvectors of σν in Fρ or equivalence classes of strings of the form Zαv with

v ∈ Fρ, σν · v = σα · v = −v and α either equal to ν or orthogonal to it.

By construction, ρ1 does not contain any sign representation of S3, so we can iterate

this construction. It is at this stage that the real induction begins, because all the

next steps resemble the second one.

We add to Fρ1 the linear span of all the strings of the form Zνv, with v a (−1)-

eigenvector of σν in Fρ1 ,5 and we extend the action of M ′ in the obvious way.

At this stage (and in all the following ones), we impose four kinds of relations:

• Zν(a1v1 + a2v2) = a1Zνv1 + a2Zνv2 “ linearity relations ′′

for all a1, a2 in C, and all (−1)-eigenvectors v1, v2 of σν in Fρ1 ;

• Zν1Zν2v = Zν2Zν1v “ commutativity relations ′′

for all mutually orthogonal positive roots ν1, ν2 and all simultaneous (−1)-

eigenvectors of σν1 , σν2 in Fρ;

• ZνZνv = −4v “ no-repetitions relations ′′

for all positive roots ν and all (−1)-eigenvectors of σν in Fρ;

• Zνv = σβ · (Zνv) + σγ · (Zνv) “ (?)-relations ′′

for all positive roots ν, all (−1)-eigenvectors v of σν in Fρ1 , and all triples of

positive roots α, β, γ forming an A2, such that

α = ν or α ⊥ ν ; β 6⊥ ν (so automatically γ 6⊥ ν)

and

5Equivalently, we add to Fρ1 strings of the form (Zν1Zν2u), with u in Fρ, σν1 · u = σν2 · u = −uand either ν1 = ν2 or ν1 ⊥ ν2.

277

σα · v = −v; σ2β · v = −v (so automatically σ2

γ · v = −v).

The representation of M ′ on Fρ1 ⊕ Span(strings) descends to a representation ρ2 on

the quotient space

Fρ2 =Fρ1 ⊕ Span(strings)

∼

We notice that Fρ2 the (−1)-eigenvectors of σν in Fρ2 are either (−1)-eigenvectors of σν

in Fρ1 or equivalence classes of strings of the form Zαv with v ∈ Fρ1 , σν ·v = σα·v = −v

and α either equal to ν or orthogonal to it.

By construction, ρ2 does not contain any sign representation of S3, so we can iterate

this construction.

At the rth step, we obtain equivalence classes of strings of the form Zα1 . . . Zαrv

with v ∈ Fρ, σα1 · v = · · · = σαr · v = −v and [Zαi, Zαj

] = 0 for all i, j = 1 . . . r.6

Using the “no-repetition” relations, we get rid of all the repetitions, so we are left

with strings that only involve mutually orthogonal roots. Hence, the construction

can be completed in finitely many steps.

This inductive construction of ρ′ has many advantages, for instance we only need

to define the action of Lie(K) on strings of length one

Zα · (Zνv) = 0 if σα · (Zνv) = +(Zνv)

Zα · (Zνv) = ZαZνv if σα · (Zνv) = −(Zνv)

Zα · (Zνv) = σα · (Zνv) if σ2α · (Zνv) = −(Zνv).

Notice that the definition of the action of Zα on the (−1)-eigenspace of σα is simplified

by the fact that strings of repeated roots are allowed.7

6This condition means that either α1 = α2 or α1 ⊥ α2.7This definition is consistent with the one given for Weyl group representations, because by the

“no-repetitions” relations Zα · (Zα) = σαZα = −4v.

278

11.2 Suggestions for further generalizations

Our method for constructing petite K-types appears to be generalizable to other split

semi-simple Lie groups, other than SL(n), whose root system admits one root length.

As seen in chapter 9, this assumption guarantees that we can annihilate every copy

of the sign representation of Fρ′ by imposing the (?)-equivalence relations. Extending

this construction to other groups accounts to verifying that the bracket relations hold,

and this problem can be solved by working out some small rank examples.

As ρ varies in the set of Weyl group representations that do not contain the sign of

S3, the output will be a list of petite K-types on which the intertwining operator for

a spherical principal series can be tested by means of Weyl group computations. In

other words, it is a list of K-types on which we should be able to explicitly calculate

the signature.

The final result will be a non-unitarity test for a spherical principal series for split

groups of type A, D, E6, E7 and E8.

279

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weyl group representations and signatures of intertwining operators

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