Convolution of Orbital Measures on Symmetric Spaces of type Cp and Dp

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OF TYPE Cp AND Dp

P. GRACZYK AND P. SAWYER

Abstract. We study the absolute continuity of the convolutionδ^\

e^X? δ^\

e^Y of two orbital measures on the symmetric spacesSO0(p, p)/SO(p)×SO(p),SU(p, p)/S(U(p)×U(p)) andSp(p, p)/Sp(p)×Sp(p). We prove sharp conditions onX,Y ∈afor the existence of the density of the convolution measure. This measure intervenes in the product formula for the spherical functions.

1. Introduction

The spacesG/K =SO0(p, p)/SO(p)×SO(p) are Riemannian symmetric spaces of non-compact type corresponding to root systems of typeDp. The spacesSU(p, p)/S(U(p)×U(p)) andSp(p, p)/Sp(p)×Sp(p) correspond to root systems of typeCp.

ConsiderX,Y ∈a and letm_K denote the Haar measure of the group K. We defineδ_e^\_X =m_K? δ_eX ? m_K. It is the uniform measure on the orbitKe^XK. The problem of the absolute continuity of the convolution

mX,Y =δ_e^\X ? δ_e^\Y

of two orbital measures that we address in our paper has important applications in harmonic analysis of spherical functions on G/K and in probability theory. Let λ be a complex-valued linear form on a and φ_λ(e^X) be the spherical function, which is the spherical Fourier transform of the measure δ_e^\X. The product formula for the spherical functions states that

φλ(e^X)φλ(e^Y) = Z

a

φλ(e^H)dµX,Y(H)

where µX,Y is the projection of the measure mX,Y on a via the Cartan decomposition G = KAK. Then the existence of a density ofµX,Y, equivalent to the absolute continuity ofmX,Y, is of great importance.

It was proven in [3] that as soon as the spaceG/K is irreducible and one of the elements X, Y is regular and the other nonzero, then the convolutionδ_e^\_X ? δ_e^\_Y has a density. The density can however still exist when bothX andY are singular. It is a challenging problem to characterize all such pairsX,Y.

This problem was solved for symmetric spaces of typeA_n and for the exceptional spaceSL(3,O)/SU(3,O) of typeE6 in [4], and for symmetric spaces of typeBp andBCp in [6]. In the present paper, we present the solution of the problem for Riemannian symmetric spaces of typeCp andDp.

For the good comprehension of the methods of this paper, it is useful to know the paper [6]. However, the cases Cp andDprequire many original ideas that did not appear in the caseBp. We refer to S. Helgason’s books [7] and [8] for the standard notation and results.

In Section 2, we are reminding the reader of the basic information about the Lie group SO(p, p) and its Lie algebraso(p, p). We also provide the necessary notation to describe the configuration of an element of the Cartan subalgebraa ofso(p, p). This configuration notion allows us to “measure” how singular an element ofa is and to describe in a precise manner which pairsX,Y ∈aare “eligible”, the sharp criterion that we establish in the paper for the absolute continuity ofµX,Y. The following theorem is the main result of the paper:

Theorem A.LetG/K=SO0(p, p)/SO(p)×SO(p)andX,Y ∈a. The density of the convolutionδ^\_e_X? δ_e^\_Y exists if and only ifX andY are eligible (see Definition 2.4).

In the following Section 3, a series of definitions and accessory results are given to set the stage for the proof of Theorem A. In Section 4, we show that (X, Y) has to be an eligible pair for the measureµX,Y to be absolutely continuous. In Section 5, we then show that the eligibility condition is also sufficient.

1991Mathematics Subject Classification. Primary 43A90, Secondary 53C35.

Key words and phrases. root systems, spehrical functions, symmetric spaces, non-compact type.

Supported by l’Agence Nationale de la Recherche ANR-09-BLAN-0084-01.

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In the last section, as in [4] and [6], we extend our results to the complex and quaternionic cases. Again, the richness of the root structure comes into play: in the table in Remark 6.1, we find that the complex and quaternionic cases have much more in common with the casesq > pthan with the real caseSO(p, p).

We conclude the paper with a discussion of the absolute continuity of powers ofδ^\_e_X for a nonzeroX∈a.

2. Preliminaries and definitions

We start by reviewing some useful information on the Lie group SO0(p, p), its Lie algebra so(p, p) and the corresponding root system. Most of this material was given in [9]. For the convenience of the reader, we gather below the properties we will need in the sequel. In this paper, Eij is a rectangular matrix with 0’s everywhere except at the position (i, j) where it is 1. Recall that SO(p, p) is the group of matrices g ∈ SL(2p,R) such that g^TIp,pg =Ip,p where Ip,p =

−Ip 0_p×p 0_p×p I_p

. Unless otherwise specified, all 2×2 block decompositions in this paper follow the same pattern. The groupSO₀(p, p) is the connected component of SO(p, p) containing the identity. The Lie algebraso(p, p) ofSO₀(p, p) consists of the matrices

A B B^T D

whereAandD are skew-symmetric. A very important element in our investigations is the Cartan decomposition ofso(p, p) andSO(p, p). The maximal compact subgroupK is the subgroup ofSO(p, p) consisting of the matrices

A 0

0 D

of size (2p)×(2p) such thatA,D∈SO(p) (henceK'SO(p)×SO(p)). Ifkis the Lie algebra ofK andpis the set of matrices

0 B B^T 0

(2.1)

then the Cartan decomposition is given by so(p, p) =k⊕p with corresponding Cartan involution θ(X) =−X^T. To shorten the notation, forX∈p as in (2.1), we will writeX^s=B.

The Cartan spacea⊂p is the set of matrices H =

0_p×p D_H D_H 0_p×p

whereDH = diag[H1, . . . , Hp]. Its canonical basis is given by the matrices Ai:=Ei,p+i+Ep+i,i,1≤i≤p.

The restricted roots and associated root vectors for the Lie algebraso(p, p) with respect toa are given in Table 1. The positive roots can be chosen as α(H) = Hi ±Hj, 1 ≤ i < j ≤ p. The simple roots are given by

rootα multiplicity root vectorsXα

α(H) =±(Hi−Hj) 1 Y_i,j^±=±(Ei,j−Ej,i+Ep+i,p+j−Ep+j,p+i) +Ei,p+j+Ep+j,i

1≤i, j≤p,i < j +Ej,p+i+Ep+i,j

α(H) =±(Hi+H_j) 1 Z_i,j^± =±(Ei,j−E_j,i−E_p+i,p+j+E_p+j,p+i)−(E_i,p+j+E_p+j,i)

1≤i, j≤p,i < j +E_j,p+i+E_p+i,j

Table 1. Restricted roots and associated root vectors

α_i(H) =H_i−H_i+1,i= 1, . . . , p−1 andα_p(H) =H_p−1+H_p. We therefore have the positive Weyl chamber a⁺={H ∈a:H₁> H₂>· · ·> H_p−1>|H_p|}.

The elements of the Weyl groupW act as permutations of the diagonal entries ofDX with eventual sign changes of any even number of these entries. The Lie algebra kis generated by the vectors X_α+θX_α. We will use the notation

k^t_X_α=e^t(X^α^+θX^α⁾.

The linear spacep has a basis formed by Ai ∈a, 1≤i≤pand by the symmetric matrices X_α^s := ¹₂(Xα−θXα) which have the following form

Y_i,j: =E_i,p+j+E_j,p+i+E_p+j,i+E_p+i,j, 1≤i < j≤p;

Zi,j: =−Ei,p+j+Ej,p+i−Ep+j,i+Ep+i,j, 1≤i < j≤p.

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Thus, ifX ∈p is as in (2.1), then the vectors Ai generate the diagonal entries of B =X^s and Yi,j and Zi,j the non-diagonal entries.

We now recall the useful matrixS ∈SO(p+q) which allows us to diagonalize simultaneously all the elements ofa. Let

S=

" ^√

2 2 I_p

√ 2 2 J_p

√ 2 2 I_p −

√ 2 2 J_p

#

whereJ_p= (δ_i,p+1−i) is a matrix of sizep×p. IfH =

0 DH

D_H 0

withDH = diag[H₁, . . . , H_p] thenS^TH S= diag[H₁, . . . , H_p,−Hp, . . . ,−H1]. The “group” version of this result is as follows:

S^Te^HS= diag[e^H¹, . . . , e^H^p, e^−H^p, . . . , e^−H¹].

Remark 2.1. The Cartan projection a(g) on the groupSO0(p, p), defined as usual by g=k1e^a(g)k2, a(g)∈a⁺, k1, k2∈K

is related to the singular values ofg∈SO(p, p) in the following way. Recall that the singular values ofgare defined as the non-negative square roots of the eigenvalues ofg^Tg. Let us writeH=a(g). We have

g^Tg=k^T₂ e²^Hk2= (k₂^TS) (S^Te²^HS) (S^Tk2)

where S^Te²^HS is a diagonal matrix with nonzero entries e²^H¹, . . . , e²^H^p, e⁻²^H^p, . . . , e⁻²^H¹, satisfying H1 ≥ . . .≥H_p−1≥ |Hp|. Let us writeaj =e^H^j forj= 1, . . . , p−1 andap=e^|H^p^|. Thus the set of 2psingular values of gcontains the valuesa1≥. . .≥ap≥a⁻¹_p ≥. . .≥a⁻¹₁ witha1≥. . .≥ap≥1.

Then

a(g) =

0 D_a(g) D_a(g) 0

with D_a(g)= diag[loga1, . . . ,loga_p−1,sgn(Hp) logap].

Note that this method does not allow us to distinguish between the situations whereH_p is positive or negative.

Singular elements ofa. In what follows, we will consider singular elementsX,Y ∈∂a⁺. We need to control the irregularity ofX andY, i.e. consider the simple positive roots annihilatingX andY. A special role is played by the last simple rootα_p=H_p−1+H_p, different from the simple rootsα_i(H) =H_i−H_i+1,i= 1, . . . , p−1. Note thatα_p(X) = 0 implies that the last diagonal entry ofD_X is negative or 0.

We introduce the following definition of the configuration ofX ∈a⁺.

Definition 2.2. Let X ∈a⁺. In what follows,x_i>0,x_i> x_j fori < j,s_i ≥1,u≥0 andPr

i=1 s_i+u=p. Let s= (s₁, . . . , s_r). We define the configuration ofX by:

[s;u] if DX=diag[

s1

z }| { x1, . . . , x1,

s2

z }| { x2, . . . , x2, . . . ,

sr

z }| { xr, . . . , xr,

u

z }| { 0, . . . ,0 ] (2.2)

[s] if DX=diag[

s1

z }| { x1, . . . , x1,

s2

z }| { x2, . . . , x2, . . . ,

s_r−1

z }| { x_r−1, . . . , x_r−1,

s_r=1

z}|{−xr ] (2.3)

[s]⁻ if DX=diag[

s₁

z }| { x1, . . . , x1,

s₂

z }| { x2, . . . , x2, . . . ,

s_r

z }| { xr, . . . , xr,−xr].

(2.4)

We extend naturally the definition of configuration to any X ∈ a, whose configuration is defined as that of the projectionπ(X) ofX ona⁺.

Remark 2.3. We will often writeX =X[. . .] when [. . .] is a configuration ofX. For the configuration (2.4), we write the−superscript inX[s]⁻ to indicate thatX contains nonzero opposite entries. We omit the−superscript in (2.3) and writeX[s] because we are essentially in the same case as in (2.2) withu= 0 andsr= 1.

If the number of zero entriesu= 0, it may be omitted in (2.2). In particular, in the configurations (2.3) and (2.4),u= 0. Note thatX = 0 has configuration [0;p]. A regularX ∈a⁺ has the configuration [1^p; 0] or [1^p−1; 1], where 1^k = (1, . . . ,1) with 1 repeatedktimes.

In what follows, we will write maxs= maxisi and max(s, u) = max(maxs, u). We will show that in the case of the symmetric spacesSO0(p, p)/SO(p)×SO(p), the criterion for the existence of the density of the convolution δ_e^\X ? δ_e^\Y is given by the following definition of an eligible pairX andY:

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Definition 2.4. LetX andY be two elements of awith configurations[s;u]or[s]⁻ and[t;v]or[t]⁻ respectively.

We say thatX andY are eligible if one of the two following cases holds

u≤1,v≤1 and max(s) + max(t)≤2p−2, (2.5)

u≥2 orv≥2 and max(s,2u) + max(t,2v)≤2p (2.6)

and, forp= 4,

{DX,DY} 6={diag[a, a, a, a],diag[b, b, c, c]} nor{diag[a, a, a,−a],diag[b, b, c,−c]}, (2.7)

for anya6= 0, b6= 0,|b| 6=|c|.

Observe that if X and Y are eligible, then X 6= 0 and Y 6= 0. The non-eligible pairs given by (2.7) are {[4],[(2,2)]},{[4]⁻,[(2,2)]⁻} withu=v= 0 or{[4],[2; 2]}and{[4]⁻,[2; 2]}withu= 0 andv= 2.

Remark 2.5. It is interesting to note that the definition of eligible pairs is more complicated for the spaceSO(p, p) than for the spacesSO(p, q) withp < q(recall that the latter spaces have a much richer root structure). As for the spacesSO(p, q) withp < q, the number of zeroes on the diagonal ofDX is important. Unlike in the caseSO(p, q) withp < q, this only becomes a factor when the number of zeroes is greater than 1 (as opposed to greater than 0).

In [6], we showed that ifp < qandX andY ∈a were such that theDX andDY have no zero diagonal elements on the diagonal thenµX,Y was absolutely continuous. This is no longer the case whenp=qand this is one of main differences between theSO(p, p) andSO(p, q) cases. Another difference is the anomalous case whenp= 4 seen in (2.7) and the fact that lower dimensional cases require different proofs. On the other hand, when the number of zeroes on the diagonal of eitherDX or DY is at least 2, then the proof of Theorem A is similar to the one found in [6] but requires considering separately a low dimension caseX[5], Y[3; 2].

3. Basic tools and reductions Definition 3.1. ForZ ∈a, letVZ be the subspace ofp defined by

V_Z=span{X_α−θ(X_α)| α(Z)6= 0} ⊂p.

We denote by|V_Z|the dimension ofV_Z. It equals the number of positive rootsαsuch that α(Z)6= 0.

The following definition and lemmas will help reduce the number of cases of configurations of (X, Y) to verify.

Definition 3.2. We will say thatX andX⁰∈aarerelativesif exactly one of the diagonal entries ofDX andDX⁰

differs by sign. IfX is a relative to X⁰ andY is a relative to Y⁰ then we will say that(X, Y)is a relative pair of (X⁰, Y⁰).

The properties listed in the following lemma are straightforward.

Lemma 3.3.

(1) X is a relative of X if and only if DX has at least one diagonal entry equal to 0, i.e. u≥1 in X[s;u].

Thus X∈a⁺ has no other relatives in a⁺. (2) IfX andX⁰ are relatives then |VX|=|VX⁰|.

(3) If(X, Y)and(X⁰, Y⁰)are relative pairs then(X, Y)is an eligible pair if and only if(X⁰, Y⁰)is an eligible pair.

(4) IfX =X[s;u]∈a⁺ withu >0then all Xi’s are non-negative.

(5) IfX =X[s;u],Y[t;v]∈a⁺ are such thatu >0 orv >0 then eitherDX,DY have no negative entries or we can choose a relative pairX⁰,Y⁰∈a⁺ in such a way thatDX⁰,DY⁰ have no negative entries.

(6) IfX is a relative ofX⁰ and X⁰ is a relative ofX⁰⁰ thenX andX⁰⁰ are in the same Weyl-group orbit.

Lemma 3.4. If (X, Y)and (X⁰, Y⁰) are relative pairs then either both setsKe^XK e^YK andKe^X⁰K e^Y⁰K have nonempty interiors or neither has.

Proof. LetJ₀ = diag{

2p−1

z }| {

1, . . . ,1,−1} and note that J₀ is orthogonal and that J₀² ∈SO(p)×SO(p). Suppose that X,Y, X⁰ andY⁰ are as in the statement of the lemma. Suppose that w₁,w₂∈W ⊂K are such that the element ofDw1·X⁰ (resp. Dw2·Y⁰) that differs by a sign from the corresponding element ofDX (resp. DY) is placed at the end. Then

K e^XK e^Y K=J0K J0w1e^Xw⁻¹₁ J0K J0w2e^Y w₂⁻¹J0K J0

=J₀K(J₀w₁e^Xw⁻¹₁ J₀)K(J₀w₂e^Y w⁻¹₂ J₀)K J₀=J₀K e^X⁰K e^Y⁰K J₀

which allows us to conclude.

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In the sequel we use some ideas, results and notations of [4, Section 3], that we strengthen and complete.

Proposition 3.5.

(i) The density of the measurem_X,Y exists if and only if its supportKe^XKe^YK has nonempty interior.

(ii) Consider the analytic mapT:K×K×K→Gdefined by T(k1, k2, k3) =k1e^Xk2e^Y k3.

The setT(K×K×K) =Ke^XKe^YK contains an open set if and only if the derivative ofT is surjective for some choice ofk= (k1, k2, k3).

Proof. Part (i) follows from arguments explained in [3] in the case of the support of the measureµX,Y, equal to

a(e^XKe^Y).Part (ii) is justified for example in [8, p. 479].

Corollary 3.6. Let (X, Y)and(X⁰, Y⁰)be relative pairs. The measure mX,Y is absolutely continuous if and only if the measuremX⁰,Y⁰ is absolutely continuous.

Proof. We use Proposition 3.5(i) and Lemma 3.4.

Proposition 3.7. Let U_Z =k+Ad(e^Z)k. The measure m_X,Y is absolutely continuous if and only if there exists k∈K such that

U_−X+Ad(k)UY =g.

(3.1)

Proof. We want to show that this condition is equivalent to the derivative ofT atkbeing surjective. We have dTk(A, B, C) = d

dt

_t=0 e^tAk1e^Xe^tBk2e^Ye^tCk3

=A k1e^Xk2e^Y k3+k1e^XB k2e^Y k3+k1e^Xk2e^Y C k3

(3.2)

We now transform the space of all matrices of the form (3.2) without modifying its dimension:

dim{A k1e^Xk2e^Y k3+k1e^XB k2e^Y k3+k1e^Xk2e^Y C k3: A, B, C∈k}

= dim{k⁻¹₁ A k1e^Xk2e^Y +e^XB k2e^Y +e^Xk2e^YC:A, B, C∈k}

= dim{A e^Xk₂e^Y +e^XB k₂e^Y +e^Xk₂e^Y C:A, B, C ∈k}

= dim{e^−XA e^X+B+k2e^Y C e^−Yk₂⁻¹: A, B, C∈k}

The space in the last line equalsk+ Ad(e^−X)(k) + Ad(k2) (Ad(e^Y)(k)) =U_−X+ Ad(k2)UY. In order to apply the condition (3.1), we will consider convenient symmetrized root vectors and the spacesV_Z generated by them.

Lemma 3.8. Let Z∈a. ThenU_Z =k+V_Z=U_−Z.

Proof. Clearly VZ =V_−Z. We show that VZ ⊂ UZ and therefore that k+VZ ⊂ UZ: let α be a root such that α(Z)6= 0. Note that [Z, Xα] =α(Z)Xαand [Z, θ(Xα)] =−α(Z)θ(Xα). LetU =Xα+θ(Xα)∈k. Now,

Ad(e^Z)U =eadZ(X_α+θ(X_α)) =e^α(Z)X_α+e^−α(Z)θ(X_α).

Therefore Xα = (e^α(Z)−e^−α(Z))⁻¹ −e^−α(Z)U+ Ad(e^Z)U

∈ k+ Ad(e^Z)(k) = UZ. The vector θXα is a root vector for the root−α, so we also haveθXα∈UZ.

It remains to show thatUZ ⊂k+VZ. It suffices to show that Ad(e^Z)k⊂k+VZ: for everyα, Ad(e^Z) (Xα+ θ(Xα)) =e^α(Z)Xα+e^−α(Z)θ(Xα) = ^e^α(Z)^+e₂^−α(Z)(Xα+θ(Xα)) +^e^α(Z)^−e₂^−α(Z) (Xα−θ(Xα))∈k+VZ.

The following corollary is then straightforward:

Corollary 3.9. The measuremX,Y is absolutely continuous if and only if there exists k∈K such that V_X+Ad(k)V_Y =p.

(3.3)

Corollary 3.10. The measuremX,Y is absolutely continuous if and only if there exists a dense open subsetU ⊂K such that for everyk∈U

(1) Vw₁·X+Ad(k)Vw₂·Y =p for every w1,w2∈W,

(2) For every r <2p, the matrix obtained by removing the first rrows and r columns ofkis non-singular.

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Proof. First we note that condition (3.3) for somekis actually equivalent to the existence of a dense open subset U ⊂K such that (3.3) holds for everyk∈U. Indeed, since the equality (3.3) can be expressed in terms of nonzero determinants, if it is satisfied for one value ofk, it will be satisfied for every kin a dense open subset ofK.

In addition, (3.3) is equivalent to the fact thata(e^XK e^Y) has non-empty interior which, in turn, implies that a(e^w¹^·XK e^w²^·Y) has non-empty interior for any givenw1,w2∈W and for everyk∈Uw₁,w₂ whereUw₁,w₂ is open and dense. Hence, for any givenw1,w2∈W, there is a dense open setUw₁,w₂ withVw₁·X+ Ad(k)Vw₂·Y =p. For similar reasons, there exists a dense open subset of K such that the second condition is satisfied (the condition being satisfied by the identity matrix). Given that a finite intersection of dense open sets is a dense open set, the

statement follows.

Remark 3.11. Corollary 3.9 and the fact thatV_w·X = Ad(w)V_X forw∈W andX ∈a( [4, Lemma 3.3]) imply that in the proof of Theorem A one can assume that X has a configuration [s;u] with s₁ ≥s₂ ≥. . . ≥ s_r or a configuration [s]⁻ withs₁≥s₂≥. . .≥s_r−1. The same remark applies to the configuration ofY.

The following necessary criterion for the existence of the density will be very useful:

Corollary 3.12. If m_X,Y is absolutely continuous then|V_X|+|V_Y| ≥dimp=p².

The following definition and results will be helpful in resolving the exceptional case indicated in (2.7).

Definition 3.13. Forn≥1, letZ(n)be the group formed by the matrices of the form







cosθ₁ −sinθ₁ sinθ₁ cosθ₁

. ..

cosθr −sinθr

sinθr cosθr







where the last block is replaced by 1isn is odd.

Remark 3.14. Note that dimZ(n)≤n/2 and that each element inZ(n) has a square root inZ(n).

Lemma 3.15. Consider n≥2 andk∈SO(n). Then there exists A∈SO(n)such thatA⁻¹k A∈ Z(n).

Proof. Consult for example [2]. Recall that the eigenvalues ofkaree^±iθ^j, θj ∈R.

Corollary 3.16. Every matrix

A₁ 0 0 A₂

∈SO(p)×SO(p) can be written in the following format A1 0

0 A2

=

A 0

0 A

B 0

0 B⁻¹

C 0

0 C

withA,C∈SO(p)andB ∈ Z(p).

Proof. According to Lemma 3.15, there exists a matrixA ∈ SO(p) such that B⁰ =A⁻¹A₁A⁻¹₂ A∈ Z(p). Pick B∈ Z(p) such thatB²=B⁰ and letC=B⁻¹A⁻¹A₁. ThenA B C=A₁ and

A B⁻¹C=A B⁻¹(B⁻¹A⁻¹A1) =A(B²)⁻¹A⁻¹A1

=A(B⁰)⁻¹A⁻¹A1=A(A⁻¹A1A⁻¹₂ A)⁻¹A⁻¹A1=A2

which proves the lemma.

Remark 3.17. The matrices

B 0 0 B⁻¹

in the last corollary can be written asQ[p/2]

i=1 k^tⁱ

Z₂⁺_i−1,2_ifor an appropriate choice oft_i’s.

In the proof of the necessity of the eligibility condition, we will use the following result stated in [5, Step 1, page 1767]. Let the Cartan decomposition ofSL(N,F) be written asg=k₁e^˜^a(g)k₂.

Lemma 3.18. LetU =diag([

r

z }| {

u0, . . . , u0, u1, . . . , uN−r]andV =diag([

N−s

z }| {

v0, . . . , v0, v1, . . . , vs]. where s+ 1≤r < N, s≥1, and the ui’s and vj’s are arbitrary. Then each element of the diagonal of ˜a(e^USU(N,F)e^V)has at least r−s entries equal tou0+v0.

We will use Lemma 3.18 withN =p+qin the proofs of Proposition 4.5 and Theorem 6.7.

In the proof of Theorem A we will need the following technical lemma from [6]:

Lemma 3.19.

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(1) For the root vectors Z_i,j⁺ andY_i,j⁺, we have

Ad(e^t^(Yî,j⁺^+θ(Yî,j⁺⁾⁾)(Yi,j) = cos(4t)Yi,j+ 2 sin(4t) (Ai−Aj), Ad(e^t(Zî,j⁺^+θ(Zî,j⁺⁾⁾)(Z_i,j) = cos(4t)Z_i,j+ 2 sin(4t) (A_i+A_j).

(2) The operators Ad(e^t(Yî,j⁺^+θ(Yî,j⁺⁾⁾) and Ad(e^t^(Zî,j⁺^+θ(Zî,j⁺⁾⁾) applied to the other symmetrized root vectors do not produce any components ina.

In the proof of the existence of the density for the pairsX[4], Y[2,2]⁻ and X[5], Y[3; 2] without predecessors, we will need the following elementary lemma. Recall thatZk,l=¹₂(Z_k,l⁺ −θ(Z_k,l⁺))∈p.

Lemma 3.20. Let 1≤i < j≤pand1≤k < l≤p. We have

[Z_i,j⁺ +θ(Z_i,j⁺), Z_k,l] =











0 if {i, j} ∩ {k, l}=∅, 4 (A_i+A_j) if {i, j}={k, l}, 2Ymin(j,l),max(j,l) if i=k, j6=l, 2Ymin(i,k),max(i,k) if i6=k, j=l,

−2Y_i,l if j=k,

−2Yk,j if i=l.

Proof. We apply the well known fact that [g_α,g_β]⊂g_α+β whenα+β is a root and [g_α,g_β] = 0 otherwise. For the

computation of exact coefficients in the formulas, we use Table 1.

4. Necessity of the eligibility condition

Proposition 4.1. If X =X[s;u] andY =Y[t;v]∈a withu≤1 andv≤1are such that maxs+ maxt>2p−2 then|VX|+|VY|< p².

Proof. Assume that X =X[s;u] and Y =Y[t;v] ∈a⁺. Without loss of generality, assume that maxs≥maxt.

We then have maxs=pand maxt≥p−1. The only possible pairs are X[p], Y[p] and the relative pairX⁰[p]⁻, Y⁰[p]⁻

X[p], Y[p−1,1] and the relative pairX⁰[p]⁻, Y⁰[1, p−1]⁻ (4.1)

X[p], Y[p]⁻ and the relative pair (X⁰, Y⁰) = (Y, X) X[p], Y[1, p−1]⁻ and the relative pairX⁰[p]⁻, Y⁰[p−1,1].

By Remark 3.11 we do not need to consider the configuration [1, p−1]. We have|VX|= ^p(p−1)₂ and |V_Y_[p−1,1]|=

p(p−1)

2 +p−1. We apply Lemma 3.3 and we find by examination that|VX|+|VY| ≤p²−1 in all cases.

Corollary 4.2. Let p ≥ 2. Consider a pair X = X[s;u] and Y = Y[t;v] with u ≤ 1 and v ≤ 1. Then

|VX|+|VY| ≥p² if and only if

max(s) + max(t)≤2p−2.

(4.2)

Proof. By Proposition 4 only the sufficiency of condition (4.2) needs to be proven. Suppose max(s)+max(t)≤2p−2 and that maxs≥maxt. If maxs=pthen|VX|=p(p−1)/2 and maxt≤p−2 implies|VY| ≥p(p−1)/2+2(p−2).

If both maxs≤p−1 and maxt≤p−1 then|VX| ≥p(p−1)/2 +p−1 and|VY| ≥p(p−1)/2 +p−1. In both

cases, the result follows.

Definition 4.3. We will callexceptionalthe set of configurations listed in (4.1) and denote it by E.

Proposition 4.4. Let X,Y ∈ a be such that DX =diag[b, b, c, c] with b > c > 0 and DY =diag[a, a, a, a] with a >0. Thenδ_e^\_X? δ^\_e_Y has no density.

Proof. According to Corollary 3.16, we can write K e^XK e^YK=K e^X

A 0

0 A

k^t¹

Z_1,2⁺ k^t²

Z⁺_3,4

C 0

0 C

e^Y K

=K e^X exp(







0 R 0 0

−R 0 0 0

0 0 0 R

0 0 −R 0





 )k^t¹

Z_1,2⁺ k^t²

Z⁺_3,4 e^Y K=K e^Xk^r¹

Y_1,3⁺ k^r²

Y_2,4⁺ k^t¹

Z⁺_1,2k^t²

Z_3,4⁺ e^YK

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(hereZ_1,2⁺ ,Z_3,4⁺ ,Y_1,3⁺,Y_2,4⁺ are exactly as in Table 1). We used the fact thate^Y and

C 0

0 C

commute, the Cartan decomposition A=

A1 0 0 A2

exp(

0 R

−R 0

)

C1 0 0 C2

(R= diag[r1, r2], Ai, Ci∈SO(2)), and the facts

thate^X,







A₁ 0 0 0

0 A₂ 0 0

0 0 A1 0

0 0 0 A2







commute and







C₁ 0 0 0

0 C₂ 0 0

0 0 C1 0

0 0 0 C2







commutes withk^t¹

Z⁺_1,2k^t²

Z_3,4⁺ and withe^Y. Now it is easy to see by considering the proof of Proposition 3.5(ii) that for these particular X and Y, the conditionV_X+ Ad(k)V_Y =p must be satisfied by k of the form k₀ =k^r¹

Y_1,3⁺ k^r²

Y_2,4⁺ k^t¹

Z_1,2⁺ k^t²

Z_3,4⁺ . On the other hand, VY =hZi,j, i < jiand Ad(k0)(Zi,j),i < j, can only produce diagonal elements which satisfyH1+H3=H2+H4

as can be checked by Lemma 3.19 and using the fact that Ad(k0) = Ad(k^r¹

Y_1,3⁺) Ad(k^r²

Y_2,4⁺) Ad(k^t¹

Z⁺_1,2) Ad(k^t²

Z_3,4⁺ ).

Consequently,a6⊂V_X+ Ad(k₀)V_Y and the density cannot exist.

Proposition 4.5. If X andY are not eligible then the density does not exists.

Proof. Let the configuration of X be [s;u] or [s]⁻ and the configuration of Y be [t;v] or [t]⁻. Proposition 4.1, Proposition 4.4 and Corollary 3.6 imply the non-existence of density whenX andY are not eligible andu≤1 and v≤1.

Suppose then that u≥ 2 or v ≥ 2 and max(s,2u) + max(t,2v) > 2pand consider the matrices a(e^Xk e^Y), k∈SO(p)×SO(p). Using (5) of Lemma 3.3 and Lemma 3.4, we may assume that the diagonal entries ofDX and DY are non-negative. Applying Remark 2.1, we have

˜

a(e^Xk e^Y) = ˜a(

e^{ST X S}

z }| { (S^Te^XS)

∈SO(p+q)

z }| { (S^Tk S)

e^{ST Y S}

z }| { (S^Te^Y S))

where ˜a(g) is the diagonal matrix with the singular values of g on the diagonal, ordered decreasingly (see the explanation before Lemma 3.18).

Ifu+v > pthen there arer−s=r+ (N−s)−N = 2u+ 2v−2p= 2 (u+v−p) repetitions of 0 + 0 = 0 in coefficients of ˜a(e^Xke^Y). Therefore 0 occurs at least u+v−p >0 times as a diagonal entry ofDH for every H ∈a(e^XK e^Y) which implies that a(e^XK e^Y) has empty interior. If 2u+ max(t)>2pdenote t= max(t). Let Yi 6= 0 be repeated t times in DY (or, if t = tr and Y = Y[t]⁻, we have t−1 times Yr and once −Yr in DY).

Then there arer−s=r+ (N−s)−N= 2u+t−2prepetitions ofYi+ 0 in coefficients of ˜a(e^Xke^Y). Therefore Yi occurs at least 2u+t−2p >0 times as a diagonal entry of DH for everyH ∈a(e^XK e^Y) which implies that

a(e^XK e^Y) has empty interior.

5. Sufficiency of the eligibility condition 5.1. Case u≤1 and v≤1.

Remark 5.1. In our proof, the caseu≤1 andv ≤1 is equivalent to the caseu=v= 0. Indeed, for H ∈a⁺, if the sole diagonal entry 0 inDH is replaced by a positive entry different from the existing diagonal entries ofDH, thenVH is unchanged. We will therefore assume in this section thatu= 0 andv= 0.

Definition 5.2. Let s= (s₁, s₂. . . , s_m) andt= (t₁, t₂, . . . , t_n) be two partitions ofp (P

i s_i =p=P

i t_j). We will say that sis finer than tif the t_i’s are sums of disjoint subsets of the s_j’s (for example, s= [3,2,2,1,1,1]is finer thant= [5,3,2] = [3 + 2,2 + 1,1 + 1]).

Remark 5.3. IfX =X[s] andY =Y[t] andsis finer thantthenVY ⊂VX.

In the following lemma, we reduce in a significant way the number of elements for which we must prove the existence of the density.

Lemma 5.4. Forp≥5, it is sufficient to prove the existence of the density in the following cases:

S1: X[p], Y[p−k, k]forp−k≥k≥2 S2: X[p]⁻, Y[p−k, k]forp−k≥k≥2 S3: X[1, p−1]⁻, Y[p−1,1]

S4: X[p−1,1], Y[p−1,1].

Forp= 4 the same is true provided the caseS₁ is replaced by the casesX[4], Y[2,1,1]andX[3,1], Y[2,2].

Proof. Supposep≥5. Let us callA0the configurations of the form [s] andA1all the others, i.e. the configurations of the form [s]⁻.

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(a) We first observe that if the density exists for S1 then it follows that it exists for all pairs {X, Y} such thatX,Y ∈A0, except whenX or Y have configurations [p] or [p−1,1]. This comes from the fact that all theseX, Y have structures that are finer and, consequently, the corresponding VX and VY are larger.

Thus, existence of the density in the casesS1 together withS4 will imply the existence of the density in all the cases whenX,Y ∈A0, except when{X, Y} ∈ E.

(b) By switching to relatives and changing the order of X and Y, we see that it implies the existence of the density in all the cases whenX,Y ∈A1, except when{X, Y} ∈ E.

(c) It remains to show that the cases S₂ and S₃ imply the existence of the density in all the cases when X ∈ A₁ and Y ∈ A₀, except when {X, Y} ∈ E. Note first that if X = [s]⁻ then either [s]⁻ = [p]⁻ or V_X⁰ ⊂V_X withX⁰ =X⁰[1, p−1]⁻. In the first case, we observe that the caseS₂ implies the casesX[p]⁻ andY ∈A₀\ {[p],[p−1,1]}for the same reason as in (a). The only cases that remain withY[p−1,1] are covered byS₃. Finally, switching to relatives, we get the pairsX ∈A₁, Y[p] which are not inE.

We illustrate the proof of the Lemma in the casep= 5.

[2,1³] [2,2,1] [3,1,1] [3,2] [4,1] [5] [1³,2]⁻ [2,1,2]⁻ [1,1,3]⁻ [3,2]⁻ [2,3]⁻ [1,4]⁻ [5]⁻

√ √ √ √ √ √ √ √ √ √ √ √ √

[2,1³]

√ √ √ √ √ √ √ √ √ √ √ √

[2,2,1]

√ √ √ √ √ √ √ √ √ √ √

[3,1,1]

√ √

S₁ √ √ √ √ √ √

S₂ [3,2]

S₄ X √ √ √ √ √

S₃ X [4,1]

X √ √ √ √ √

X X [5]

√ √ √ √ √ √ √

[1³,2]⁻

√ √ √ √ √ √

[2,1,2]⁻

√ √ √ √ √

[1,1,3]⁻

√ √ √ √

[3,2]⁻

√ √ √

[2,3]⁻

√ X [1,4]⁻ X [5]⁻ In the above table, √

indicates that the pair is eligible, X indicates that the pair is not eligible and therefore that the density does not exist (the cases identified in (4.1)) and the Si’s correspond to the notation above (the pair is eligible where theSi’s appear). We use the reduction from Remark 3.11.

Theorem 5.5. Let p≥2 and suppose thatX =X[s;u]or [s]⁻ andY =Y[t;v] or[t]⁻ are such that u≤1 and v≤1. If the pair {X, Y} does not belong to the set E ∪ {[4],[2,2]} ∪ {[4]⁻,[2,2]⁻} then the density exists.

Proof. The proof proceeds by induction similarly as in [6], but with a different “asymmetric” technique of executing Steps 2 and 3. Also, for small values of p the proofs must be led separately, due the lack of available good predecessors. With some exceptions in the starting phase of the induction, and in the caseS₃ of the Lemma 5.4, the elementsX andY will be ina⁺ and their “usual” predecessors will be obtained by skipping the first diagonal terms ofDX andDY.

The only case of existence of the density forp= 2 is for regularX[1,1] andY[1,1] (according to (4.1), only the pairX[1,1], Y[1,1] of two regular elements verifies|VX|+|VY| ≥p²).

Forp= 3, we have four possible configurations of nonzero singular elements: [2,1], [3], [1,2]⁻, [3]⁻.

All exceptional cases listed in (4.1) appear and only three pairs of singular configurations, namely (X[2,1], Y[2,1]), (X[2,1],Y[1,2]⁻) and (X[1,2]⁻,Y[1,2]⁻), verify |V_X|+|V_Y| ≥p². Given that the pairs (X[2,1], Y[2,1]) and (X[1,2]⁻, Y[1,2]⁻) are relatives, we only have to check the cases (X[2,1], Y[2,1]) and (X[2,1], Y[1,2]⁻).

In the caseX[2,1],Y[2,1], we writeDX= diag[a, a, b],DY = diag[c, c, d] and the predecessorsDX⁰ = diag[a, b], DY⁰ = diag[c, d], obtained by skipping the first coordinates, are regular. In the case X[2,1],Y[1,2]⁻ we consider DX = diag[a, a, b], Dw·y = diag[−d, c, d] and only now go to regular predecessors DX⁰ = diag[a, b], D_(w·Y₎⁰ = diag[c, d]. The general proof given below applies in these cases.

Whenp= 4, by Lemma 5.4, we must show the existence of the density for:

(1) X[4], Y[2,1,1] andX[2,2], Y[3,1],

(2) X[4]⁻, Y[2,2] or equivalently the relative pairX[4], Y[2,2]⁻, (3) X[1,3]⁻, Y[3,1],

(4) X[3,1], Y[3,1].

In the cases (1), (3) and (4), the usual predecessors have density whenp= 3. The general proof given below applies in these cases.

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For the case (2), observe that whenp= 3, the configuration X⁰[3] never gives the existence of density, when Y⁰ is singular. That is why the second case X[4], Y[2,2]⁻ has no good predecessors and this case must be proved separately. We will do it after the general proof.

Starting fromp = 5, the general proofby induction applies, the exceptions due to small values of p being taken care of. We present this proof now.

Step 1. LetY ∈ a⁺ be such thatDY = diag[

p−k

z }| { a, . . . , a,

k

z }| {

b, . . . , b] and let its predecessor Y⁰ be such thatDY⁰ = diag[

p−k−1

z }| { a, . . . , a,

k

z }| {

b, . . . , b]. The spaceVY is generated by completing a basis ofVY⁰ with NY ={Y_1,p−k+1, . . . , Y1,p, Z1,2, . . . , Z1,p}.

We choose the predecessor X⁰ of X in the same manner, except in the case S3, where we first write DX = diag[b, b, . . . , b, a,−b] wherea > b >0 and skip the first termb inDX.

It is easy to see that the predecessors ofX andY are in the corresponding classes S_i forp−1, so with density, except forX[5],Y[3,2], due to the non-eligible caseX[4], Y[2,2]. In this last case we arrangeDX= diag[a, a, a, a, a]

andD_wY = diag[−b, b, b, c,−c] and go down to good predecessorsX⁰[4], (wY)⁰[2,2]⁻. The proof described below leads to the existence of the density.

By the induction hypothesis and considering Corollary 3.10, there exists an open dense subset D⁰ of SO(p− 1)×SO(p−1) such that for allw⁰ ∈W⁰ andk0∈D⁰,

Vw⁰·X⁰+ Ad(k0)VY⁰ =p⁰ (5.1)

andk0verifies condition (2) of Corollary 3.10.

We embedK⁰ =SO(p−1)×SO(p−1) inSO(p)×SO(p) in the following manner:

K⁰3k⁰ =





 1

k0,1

1 k0,2







∈

SO(p)

, k_0,1, k_0,2∈SO(p−1).

Hence, we have (identifyingp⁰ with its natural embedding into p) V1:=Vw⁰·X⁰+ Ad(k0)VY⁰ =p⁰=

0 B⁰ B^0T 0

(5.2)

for any w⁰ ∈ W⁰, where B⁰ =

0_1×1 0_1×(p−1) 0_(p−1)×1 B(p−1)×(p−1)⁰⁰

, and the matrix B⁰⁰ is arbitrary (note that p⁰ is of dimension (p−1)²). We must show that for somek∈K, the spaceVX+ Ad(k)VY =p.

Step 2. The elementY is always of the same form, so the next step of the proof is common for all the 4 cases.

Similarly as in [6], Step 2 of the proof of Theorem 4.8, case (i), we prove that fork₀∈D⁰ ⊂SO(p−1)×SO(p−1) the following property holds:

The space Ad(k₀)span(N_Y) is of dimensionp+k−1 and its elements can be written in the form







0 a1 . . . ap−1

τ₁ ...

τ_p−1−k 0

a_p+k−1 ... ap







s

withai ∈R arbitrary andτj =τj(a1, . . . , a_p+k−1). We will not need to write explicitly the functionsτj. For the sake of completeness, we give a proof of Step 2.

Step 2 comes from the fact that the action of Ad(k0) on the elements of NY gives the linearly independent matrices

Ad(k0)Y1,i=

0 β_i−1^T α_i−1 0

s

, i=p−k+ 1, . . . , p, Ad(k0)Z1,i=

0 −β_i−1^T α_i−1 0

s

, i= 2, . . . , p (5.3)

where theαi’s are the columns ofk0,1 and the βi’s are the columns ofk0,2. Let us write α⁰_i for a columnαi with the firstp−1−k entries omitted. In order to prove the statement of Step 2, we must show that the matrices

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obtained by replacingαi byα⁰_iin (5.3) are still linearly independent. This is equivalent to the linear independence of the matrices

0 −β^T_i α⁰_i 0

^s

, i= 1, . . . , p−k−1,

0 β_i^T

0 0

^s

, i=p−k, . . . , p−1,

0 0 α⁰_i 0

^s

i=p−k, . . . , p−1.

(5.4)

The matrices in (5.4) are linearly independent given that the matrix k0 was assumed to satisfy condition (2) of Corollary 3.10.

Observe that contrary to [6], we have filled the zero margins of the matrixB⁰ asymetrically, which is why we call this method “asymmetric”. The reason for doing this will be clear from the structure of the setNX that we study now.

Step 3. Let us write the setNX in the four casesSi: (1) S₁: N_X ={Z_1,2, . . . , Z_1,p}

(2) S2: NX ={Z1,2, . . . , Z_1,p−1, Y1,p} (3) S3: NX ={Z1,2, . . . , Z_1,p−1, Y_1,p−1, Y1,p} (4) S4: NX ={Z1,2, . . . , Z1,p, Y1,p}.

We will now use the elements ofN_X in order to generate thw missing p−k−1 dimensionsτ_j in the margins of B⁰. We use for this the vectorsZ_1,2, . . . ,Z_1,p−k available in all four cases fork≥1. We proceed as follows:

Ifτ₁(1,0, . . . ,0) =−1, the vectorZ_1,2 ∈N_X is unhelpful. We changeX⁰ into X⁰⁰ by putting the sign−before the second and the last term of X⁰. We obtain X⁰⁰ = w⁰⁰·X⁰ such that N_X contains Y_1,2 instead of Z_1,2 and w⁰⁰∈W⁰ changes two signs ofX⁰. This manipulation is justified by the fact that (5.1) holds for anyw⁰⁰∈W⁰. We repeat this procedure, if needed, wheneverτj(ej) =−1 and obtain, from elements ofN_w·X and Ad(k0) (NY),







0 a1 . . . a_p−1 a2p−2

...

a_p+k−1 0

... a_p







s

(5.5)

forw·X withw∈W and where thea_i’s are arbitrary.

Step 4. Noting that we have at least one element ofN_X that has not been used, eitherY_1,porZ_1,p, combining (5.1) and (5.5), we have

V₀:=V]_w·X+ Ad(k₀)(V_Y) =







0 ∗ . . . ∗

∗ ∗ . . . ∗ ... ... . .. ...

∗ ∗ . . . ∗





 (5.6)

where V]w·X corresponds to the all of VX without using the remaining Y1,p or Z1,p. To fix things, let us assume that the unused element isZ1,p, the reasoning being similar if it isY1,p instead.

The end of the proof is similar to the final step of the proof in [6]. Refer to Lemma 3.19 and note that fortsmall, Ad(e^t^(Zî,j⁺^{+θ Z}⁺î,j⁾)(V]_w·X) + Ad(k0)(VY) = V0. Indeed, the lemma shows that no new element is introduced and, fort small, the dimension is unchanged. On the other hand, V0 is strictly included in Ad(e^t^(Z⁺î,j^{+θ Z}î,j⁺⁾)(hZ1,pi ∪ V]_w·X) + Ad(k0)(VY) since, still by Lemma 3.19, a new diagonal element is introduced. We conclude that fortsmall enough, Ad(e^t(Zî,j⁺^{+θ Z}î,j⁺⁾)(hZ1,pi ∪V]_w·X) + Ad(k₀)(V_Y) =p. Finally,

V_w·X+ Ad(e^−t^(Z^i,j⁺^{+θ Z}^i,j⁺⁾k₀)(V_Y) =p.

The caseX[2,2]⁻,Y[4]: This case is awkward since (X, Y) do not have eligible predecessors. We selectX and Y such thatDX= diag[a, a, b,−b] andDY = diag[c, c, c, c] (assuminga,b,c6= 0 anda6=b). ThenVY is generated by the basisBY composed of all the 6 vectorsZi,j,i < j, while the basisBX ofVX contains the vectorsY1,3,Y1,4, Y2,3,Y2,4,Y3,4 and all the vectorsZi,j exceptZ3,4. Note that|VX|= 10.

For a root vectorZ_i,j⁺ denoteZk

i,j =Z_i,j⁺ +θ(Z_i,j⁺)∈k. DefineZ0=Zk

1,2+Zk

2,3+Zk

1,4+Zk

2,4∈k. We denote F_t=V_X+ Ad(e^tZ⁰)V_Y =V_X+e^tadZ₀(V_Y),

Et=VX+h{v+t[Z0, v] : v∈VY}i.