Characterizing the absolute continuity of the convolution of orbital measures in a classical Lie algebra
aa r X i v : . [ m a t h . F A ] O c t Characterizing the absolute continuity of the convolution oforbital measures in a classical Lie algebra
Sanjiv Kumar Gupta and Kathryn E. Hare
Abstract.
Let g be a compact, simple Lie algebra of dimension d . It is aclassical result that the convolution of any d non-trivial, G -invariant, orbitalmeasures is absolutely continuous with respect to Lebesgue measure on g andthe sum of any d non-trivial orbits has non-empty interior. The number d was later reduced to the rank of the Lie algebra (or rank +1 in the case oftype A n ). More recently, the minimal integer k = k ( X ) such that the k -foldconvolution of the orbital measure supported on the orbit generated by X isan absolutely continuous measure was calculated for each X ∈ g .In this paper g is any of the classical, compact, simple Lie algebras. Wecharacterize the tuples ( X , . . . , X L ), with X i ∈ g , which have the propertythat the convolution of the L -orbital measures supported on the orbits gen-erated by the X i is absolutely continuous and, equivalently, the sum of theirorbits has non-empty interior. The characterization depends on the Lie type of g and the structure of the annihilating roots of the X i . Such a characterizationwas previously known only for type A n .
1. Introduction
Let G be a compact, connected simple Lie group and g its Lie algebra. Given X ∈ g , we let µ X denote the G -invariant, orbital measure supported on O X , theorbit generated by X under the adjoint action of G . Geometric properties of theLie algebra ensure that if a suitable number of non-trivial orbits are added togetherthe resulting subset of g has non-empty interior and if a suitable number of orbitalmeasures are convolved together, the resulting measure is absolutely continuouswith respect to the Lebesgue measure on g . From the work of Ragozin in [ ] itcan be seen that the dimension of the Lie algebra is a ‘suitable number’.In a series of papers (see [ ] and [ ] and the papers cited therein) the authors,with various coauthors, improved upon Ragozin’s result determining, for each X ∈ g , the integer k ( X ) with the property that µ kX is absolutely continuous for all k ≥ k ( X ) and µ kX is singular to Lebesgue measure otherwise (where µ kX denotes the Mathematics Subject Classification.
Primary 43A80; Secondary 17B45, 58C35.
Key words and phrases. compact Lie algebra, orbital measure, absolutely continuousmeasure.The first author would like to thank the Dept. of Pure Mathematics at the University of Wa-terloo and the second author the School of Mathematics and Statistics at St. Andrews Universityfor their hospitality while some of this research was done. This research was supported in part bythe Edinburgh Math. Society, NSERC and Sultan Qaboos University. k -fold convolution). Furthermore, the k -fold sum of O X has non-empty interior if k ≥ k ( X ) and otherwise has measure zero. A formula was given for k ( X ) dependingon combinatorial properties of the annihilating roots of X . In particular, it wasshown that the convolution of any r orbital measures is absolutely continuous ifand only if r is at least the rank of the Lie algebras when g is of type B n , C n or D n and r is at least rank+1 for the Lie algebras of type A n . The proofs relied heavilyupon representation theory and harmonic analysis.By taking a geometric approach, Wright in [ ] extended these results in thespecial case of the classical Lie algebra g = su ( n ) (type A n − ), proving that µ X ∗· · · ∗ µ X L is absolutely continuous with respect to Lebesgue measure if and only if P Li =1 s i ≥ n ( L −
1) where s i is the dimension of the largest eigenspace of the n × n matrix X i , provided it is not the case that L = 2, n ≥ X , X eachhave two distinct eigenvalues, both of multiplicity n/ ], [ ]), ad-dressed analogous problems in the setting of a non-compact, symmetric space, im-proving upon other work of Ragozin, [ ]. In particular, they characterized whenthe convolution of two (possibly different) bi-invariant measures is absolutely con-tinuous in the symmetric spaces sl ( n, F ) /su ( n, F ) (where the restricted root systemis also type A n − ).Inspired by their methods, in this paper we characterize the L -tuples, ( X , . . . , X L )with X i ∈ g , such that the convolution µ X ∗· · ·∗ µ X L is absolutely continuous whenthe Lie algebra is any one of the classical Lie algebras (those of type A n , B n , C n or D n ) , leaving only one pair in D n where we have been unable to decide the answer.As well, this characterizes the L -tuples such that P Li =1 O X i has non-empty interiorin g as opposed to measure zero. As Wright found with type A n , the characteri-zation can be expressed most simply as a function of the dimensions of the largesteigenspaces of the X i when these are viewed as matrices in the classical matrix Liealgebras (see Section 3 for the precise statement). The characterization can also bedescribed in terms of the root structure of the set of annihilating roots of the X i , as was done in the previous study of convolutions of a single orbital measure. Ourargument is completely different from that used by Wright and from the harmonicanalysis - representation theory approach used by the authors previously. It reliesheavily upon the (algebraic) Lie theory of roots and root vectors.Using these results, we also obtain a similar characterization of the absolutecontinuity of the convolution products of G -invariant measures, µ x i , supported onconjugacy classes C x i in G, for the elements x i ∈ G whose annihilating roots agreewith those of a preimage of x i in g under the exponential map. This extends workof [ ] where the minimal integer k ( x ) with the property that µ k ( x ) x is absolutelycontinuous was determined.In a future paper, we will adapt our general strategy to improve upon Gracyzkand Sawyer’s symmetric space results.Finding the density function, or Radon Nikodym derivative, of the absolutelycontinuous measure µ X ∗ · · · ∗ µ X L is a challenging problem. In the case of theconvolution of two orbital measures in su ( n ) , this has been computed in [ ]. Ageneral formula for the convolution of two orbital measures in terms of the projec-tion of such measures to maximal tori was found in [ ]. The density function forthe analogous problem on non-compact symmetric spaces was studied in [ ] (andsee also the references cited there). In [ ], the sum of two adjoint orbits in su ( n ) BSOLUTE CONTINUITY 3 is explicitly described in terms of a system of linear equations, but for more than2-fold sums this too seems very difficult. Other work investigating the smoothnessproperties of convolutions of measures supported on manifolds whose product hasnon-empty interior was carried out by Ricci and Stein in [ ] and [ ].The paper is organized as follows: In section 2 we review background materialin Lie theory and introduce basic notation. In section 3 we state the main result.The necessity of our characterization is proven in section 4. In section 5 we establishthe general strategy for tackling the absolute continuity problem and then completethe proof of the main theorem in section 6. In section 7 we discuss consequencesof our result and deduce the absolute continuity result for convolutions of orbitalmeasures on Lie groups mentioned above.
2. Notation and Background2.1. Notation.
We begin by establishing notation and reviewing basic factsabout roots and root vectors. Assume G n is a classical, compact, connected simpleLie group of rank n , one of type A n , B n , C n or D n . We denote by g n its (real) Liealgebra, t n a maximal torus of g n and W the Weyl group.We write [ · , · ] for the Lie bracket action. The map ad : g n → g n is given by ad ( X )( Y ) = [ X, Y ]. The exponential function, exp , is a surjection of g n onto G n , and G n acts on g n by the adjoint action, denoted Ad ( · ). Recall that for M ∈ g n , Ad (exp M ) = exp( ad ( M )) = Id + ∞ X k =1 ad k ( M ) k !where ad k ( M ) is the k -fold composition of ad ( M ).By an orbit of an element X ∈ g n , we mean the subset O X := { Ad ( g )( X ) : g ∈ G n } ⊆ g n . There is no loss in assuming X belongs to t n since every orbit contains a toruselement. Orbits are compact manifolds of proper dimension in g n and hence ofLebesgue measure zero. If X = 0, then O X = { } is a singleton, but otherwise O X has positive dimension.By the orbital measure, µ X , we mean the probability measure invariant underthe adjoint action of G n and compactly supported on O X . It integrates bounded,continuous functions f on g n by the rule Z g n f dµ X = Z G n f ( Ad ( g ) X ) dg where dg is the Haar measure on G n . The orbital measures are singular to Lebesguemeasure since their supports have Lebesgue measure zero. Except in the specialcase when X = 0 , µ X is an example of a continuous measure, meaning the µ X -measure of any singleton is zero.The classical Lie groups and algebras are said to be of type A n for n ≥ B n for n ≥ C n for n ≥ D n for n ≥
4. This means that the root system ofthe complexified Lie algebra with respect to the complexified torus, denoted Φ n , isof that Lie type. It is often convenient to refer to type A n as type SU ( n + 1) forreasons that will become clear later.For the convenience of the reader we describe Φ n below for each of the classicaltypes. Note that by e j we mean the j ′ th standard basis vector of R n (or in R n +1 in the case of type A n ). The real span of Φ n , denoted sp Φ n , is equal to R n (or the SANJIV KUMAR GUPTA AND KATHRYN E. HARE subspace of R n +1 spanned by the standard vectors e j − e n +1 for j = 1 , . . . , n in thecase of type A n ). Lie algebra Root system Φ n A n {± ( e i − e j ) : 1 ≤ i < j ≤ n + 1 } B n {± e i , ± e i ± e j : 1 ≤ i = j ≤ n } C n {± e i , ± e i ± e j : 1 ≤ i = j ≤ n } D n {± e i ± e j : 1 ≤ i = j ≤ n } In the case of type A n , the Weyl group is the group of permutations on theletters { , . . . , n + 1 } . For types B n , C n (and D n ), the Weyl groups are the groupof permutations on { , . . . , n } , together with (an even number of) sign changes.These Lie algebras and groups can be identified with the classical matrix al-gebras and groups listed below. All compact, connected simple Lie groups arehomomorphic images by finite subgroups of these classical matrix groups. • su ( n ) – the set of n × n skew-Hermitian, trace zero matrices is the modelwe use for the Lie algebra of type A n − . SU ( n ) - the n × n special unitarymatrices is a compact Lie group of type A n − . • so ( p ) – the set of p × p real, skew-symmetric matrices. When p = 2 n it isthe Lie algebra of type D n and when p = 2 n + 1 it is of type B n . SO ( p ) -the p × p special orthogonal matrices are associated compact Lie groups. • sp ( n ) – the set of 2 n × n matrices of the form (cid:20) A B − B A (cid:21) where
A, B arecomplex n × n matrices with B symmetric and A skew-Hermitian is theLie algebra of type C n . The n ’th order symplectic group, Sp ( n ) , is the setof 2 n × n unitary matrices U satisfying U tr JU = J, where J = (cid:20) − II (cid:21) with I being the n × n identity matrix. Sp ( n ) is a compact Lie group oftype C n . For each root α ∈ Φ n , we let E α denote a corresponding root vector so that if H ∈ t n , then(2.1) [ H, E α ] = iα ( H ) E α . (We make the convention that roots are real valued.) We will choose a collectionof root vectors, { E α } , that form a Weyl basis (see [ , p. 290]). In particular,this ensures that if α , β and α + β are roots, then there are non-zero scalars N α,β satisfying N α,β = N − α, − β and[ E α , E β ] = N α,β E α + β . If α + β is not a root, then [ E α , E β ] = 0.The root vector, E α , can be written in a unique way as E α = RE α + iIE α ,where RE α and IE α both belong to the (real) Lie algebra g n . We refer to theseas the real and imaginary parts of the root vector. We write F E α if we meaneither RE α or IE α . One can easily see that E − α = RE α − iIE α . Furthermore, RE α = ( E α + E − α ) / IE α = ( E α − E − α ) / (2 i ).The vector space spanned by RE α and IE α over various sets of roots α will beimportant to us. In particular, we put(2.2) V n = { RE α , IE α : α ∈ Φ + n } ⊆ g n BSOLUTE CONTINUITY 5 where Φ + n denotes the subset of positive roots. With this notation the Lie algebracan be decomposed as g n = t n M α ∈ Φ + n sp { RE α , IE α } = t n M sp V n where sp denotes the real span. Thus the dimension of g n is equal to n + | Φ n | . From (2.1) it follows that(2.3) [
H, RE α ] = − α ( H ) IE α and [ H, IE α ] = α ( H ) RE α . It is also well known that [ RE α , IE α ] = − i [ E α , E − α ]is a non-zero element of the maximal torus. It should be noted that if { α j : j ∈ J } ⊆ Φ n is a spanning set for sp Φ n , then { RE α j , IE α j ] : j ∈ J } spans t n .Since { E α } is a Weyl basis, we have[ RE α , RE β ] = cRE α + β + dRE β − α , (2.4) [ RE α , IE β ] = cIE α + β + dIE β − α [ IE α , IE β ] = − cRE α + β + dRE β − α , where RE γ and IE γ should be understood to be the zero vector if γ is not a rootand c = N α,β / d = N α, − β / ], [ ] and [ ] for proofs of these well known factsand further details on the representation theory of Lie algebras. We call a root, α, an annihilating root of X ∈ t n if α ( X ) = 0 and call α a non-annihilating root of X otherwise. The set of annihilatingroots of X , Φ X := { α ∈ Φ : α ( X ) = 0 } , is a root subsystem of Φ n . As we will see, these root subsystems are critical forunderstanding properties about orbits and orbital measures, as are the associatedroot vectors. We will denote by(2.5) N X := { RE α , IE α : α / ∈ Φ X } ⊆ V n , the linearly independent subset of V n consisting of the real and imaginary parts ofthe root vectors corresponding to the non-annihilating roots of X . It is known thatdim O X = |N X | [ , VI.4]. Indeed, the tangent space at X to O X is spanned bythe vectors in N X and these are linearly independent (see the proof of Prop. 1). The torus of su ( n ), the classical Lie algebra oftype A n − (or type SU ( n )) consists of the diagonal matrices in su ( n ). After ap-plying a suitable Weyl conjugate, any X in the torus can be identified with the n -vector of the real parts of the diagonal elements, X = ( a , . . . , a | {z } s , . . . , a m , . . . , a m | {z } s m ) , where the a j ∈ R are distinct and P mj =1 s j a j = 0. This means that ia j is aneigenvalue of the n × n matrix X with multiplicity s j . The set of annihilating roots SANJIV KUMAR GUPTA AND KATHRYN E. HARE of X is Φ X = Ψ ∪ · · · ∪ Ψ m whereΨ = { e i − e j : 1 ≤ i = j ≤ s } andΨ l = { e i − e j : s + · · · + s l − < i = j ≤ s + · · · + s l } for l > . Following [ ], we say that X is type SU ( s ) × · · · × SU ( s m ) as this is the Lie typeof its set of annihilating roots.The torus of so (2 n + 1), the classical Lie algebra of type B n , consists of blockdiagonal matrices, with n × (cid:20) b j − b j (cid:21) having b j ≥
0, anda 0 in the final diagonal position. We identify X in the torus with the n -vector( b , . . . , b n ) ∈ R + n . Up to a Weyl conjugate, X can thus be identified with the n -vector(2.6) X = (0 , . . . , | {z } J , a , . . . , a | {z } s , . . . , a m , . . . , a m | {z } s m )where the a j > n +1) × (2 n + 1) matrix X with multiplicity 2 J + 1 and ± ia j are eigenvalues withmultiplicity s j .The set of annihilating roots Φ X = Ψ ∪ Ψ · · · ∪ Ψ m whereΨ = {± e k , ± e i ± e j : 1 ≤ i, j, k ≤ J, i = j } andΨ l = { e i − e j : J + s + · · · + s l − < i = j ≤ J + s + · · · + s l } for l = 1 , . . . , m . We will say that X is type B J × SU ( s ) × · · · × SU ( s m ) , as this is the Lie type of Φ X . Here by B we mean the root subsystem {± e } , while SU (1) , B and SU (0) are empty (and typically omitted in the description).Similarly, if X belongs to the torus of the Lie algebra of type C n or D n then,up to a Weyl conjugate, X can be identified with the n -vector X = (0 , . . . , | {z } J , a , . . . , a | {z } s , . . . , a m , . . . , ( ± ) a m | {z } s m )where the a j > D n and only if J = 0. (This is because the Weyl group in type D n changes onlyan even number of signs.) Viewing X as an 2 n × n matrix in sp ( n ) or so (2 n ), thismeans that 0 is an eigenvalue of X with multiplicity 2 J, and ± ia j are eigenvalueswith multiplicity s j .The set of annihilating roots of X can again be written as Φ X = Ψ ∪ Ψ · · · ∪ Ψ m . In this case Ψ = {± e k , ± e i ± e j : 1 ≤ i, j, k ≤ J, i = j } when the Lie algebra is type C n andΨ = {± e i ± e j : 1 ≤ i, j ≤ J, i = j } when the Lie algebra is type D n . For l ≥
1, the Ψ l are as in type B n , except when X = ( a , . . . , a , . . . , a m , . . . , − a m ) in D n whenΨ m = {± ( e i − e j ) , ± ( e i + e n ) : n − s m < i = j ≤ n − } . We will say X is type C J × SU ( s ) × · · · × SU ( s m ) or D J × SU ( s ) × · · · × SU ( s m ) BSOLUTE CONTINUITY 7 respectively, as these are the Lie types of Φ X . Here C is the subsystem {± e } , C is {± e , ± e , ± e ± e } , D is {± e ± e } (or type A × A ), D is defined inthe obvious way, and D , D , C are empty (and often omitted).Note that there are two distinct subsystems (up to Weyl conjugacy) of annihi-lating roots of elements of type SU ( n ) in D n . Definition . Suppose X is in the torus of the Lie algebra of type B n andis type B J × SU ( s ) × · · · × SU ( s m ). We will say X is dominant B type if2 J ≥ max s j , and is dominant SU type otherwise. We define dominant C and D type similarly for X in C n or D n .It was shown in [ , Thm. 8.2] that for each non-zero X ∈ g n , there is aninteger k ( X ) such that for k ≥ k ( X ), µ kX ∈ L T L ( g n ) (in particular, µ kX isabsolutely continuous with respect to Lebesgue measure) and µ kX is purely singularif k < k ( X ). A formula was given for k ( X ) depending only on the type of X andthe type of the Lie algebra. For example, if X is dominant SU type in the Liealgebra of type B n , C n or D n , and not of type SU ( n ) when the Lie algebra is type D n , then k ( X ) = 2. If X is type B n − , ( C n − , D n − or SU ( n − B n ( C n , D n or SU ( n )), then k ( X ) = n and this is the maximalchoice required for k ( X ).
3. Statement of the Main Result3.1. Eligible and Exceptional Tuples.
We introduce the following termi-nology.
Notation . If X is of type SU ( s ) × · · · × SU ( s m ) in the Lie algebra of type A n , put S X = max s j .If X is type B J × SU ( s ) × · · · × SU ( s m ) in the Lie algebra of type B n , put S X = ( J if X is dominant B typemax s j elseDefine S X similarly when X belongs to the Lie algebras of type C n or D n .If X ∈ so (2 n + 1) is dominant B type, then the dimension of the largesteigenspace of the matrix X is S X + 1, while if X is dominant SU type, then thedimension of the largest eigenspace is S X . In all the other Lie algebras, S X is thedimension of the largest eigenspace when X is viewed as a matrix in the appropriateclassical matrix algebra. Definition . (i) We will say that the L -tuple ( X , X , . . . , X L ) of elementsin the torus of a Lie algebra of type SU ( n + 1) is eligible in g n if L X i =1 S X i ≤ ( L − n + 1) . (ii) We will say that the L -tuple ( X , X , . . . , X L ) of elements in the torus of aLie algebra of type B n , C n or D n is eligible in g n if(3.1) L X i =1 S X i ≤ ( L − n. SANJIV KUMAR GUPTA AND KATHRYN E. HARE
Definition . We will say that ( X , X , . . . , X L ) ∈ t L is an exceptionaltuple if it is any one of the following: • g is type SU (2 n ), L = 2, n ≥ X and X are both of type SU ( n ) × SU ( n ) (i.e., X i = ( a i , . . . , a i | {z } n , − a i , . . . , − a i | {z } n )); • g is type D n , L = 2, X is type SU ( n ) and X is either type SU ( n ) or type SU ( n −
1) (more precisely, type SU ( n − × D or SU ( n − × SU (1)); • g is type D , L = 2, X is type SU (4) and X is either type SU (2) × SU (2)and Φ X is Weyl conjugate to a subset of Φ X , or X is type SU (2) × D ; • g is type D , L = 3 and X , X , X are all of type SU (4) with Weylconjugate sets of annihilators. Definition . We will call ( X , X , . . . , X L ) an absolutely continuous tu-ple if µ X ∗ µ X ∗ · · · ∗ µ X L is an absolutely continuous measure.Our main result is that other than for the exceptional tuples, eligibility char-acterizes absolute continuity of the convolution product. The proof of this theoremwill occupy most of the remainder of the paper. Here is the formal statement ofthe theorem. Theorem . Let g n be one of the classical, compact, connected Lie algebras oftype A n with n ≥ , B n with n ≥ , C n with n ≥ , or D n with n ≥ . Assumenon-zero X i , i = 1 , , . . . , L for L ≥ , belong to the torus of g n .(i) Suppose ( X , X , . . . , X L ) is not an exceptional tuple. The measure, µ X ∗ µ X ∗ · · · ∗ µ X L , is absolutely continuous with respect to Lebesgue measure on g n ifand only if ( X , X , . . . , X L ) is an eligible tuple.(ii) If ( X , X , . . . , X L ) is an exceptional tuple, other than a pair ( X , X ) oftype ( SU ( n ) , SU ( n − in a Lie algebra of type D n with n ≥ , then the measure µ X ∗ µ X ∗ · · · ∗ µ X L is not absolutely continuous. Remark . The characterization of absolute continuity in type A n was pre-viously established by Wright [ ]. We will include a proof in this paper as ourapproach is completely different and requires little additional effort. Remark . (i) We conjecture that a pair of type ( SU ( n ) , SU ( n − D n with n ≥ , Thm. 8.2]), the property of being absolutely continuous does not depend onlyupon the type of the annihilating root systems of the underlying elements, but also,in some cases, upon their Weyl conjugacy class.In proving both absolute continuity and its failure we will rely crucially uponthe following known geometric properties.The notation T Z ( O X ) will denote the tangent space to O X at Z ∈ O X . Proposition . The measure µ X ∗ µ X ∗· · ·∗ µ X L on g n is absolutely continuouswith respect to Lebesgue measure if and only if any of the following hold:(i) P Li =1 O X i ⊆ g n has non-empty interior; When we say a pair (
X, Y ) is of type ( ∗ , ∗∗ ) we mean that X is of type ∗ and Y is of type ∗∗ . BSOLUTE CONTINUITY 9 (ii) P Li =1 O X i ⊆ g n has positive Lebesgue measure;(iii) There exists g i ∈ G n with g = Id , such that (3.2) sp { Ad ( g i )( N X i ) : i = 1 , . . . , L } = g n , (iv) There exists g i ∈ G n with g = Id , such that L X i =1 T Ad ( g i )( X i ) ( O X i ) = g n . Furthermore, if the identity holds in (iii) or (iv) for one choice of ( g , . . . , g L ) ∈ G L − n , then it holds for all ( g , . . . , g L ) in an open dense subset of G L − n of fullmeasure. Remark . We note that (ii) implies that if µ X ∗ µ X ∗ · · · ∗ µ X L is notabsolutely continuous, then µ X ∗ µ X ∗ · · · ∗ µ X L is a purely singular measure. Proof.
This proposition is a compilation of arguments that can be found in[ ], [ ] and [ ]. We include a sketch here for the convenience of the reader. Wewill show that (iii) and (iv) are equivalent and then demonstrate the implications(ii) ⇒ (iv) ⇒ absolute continuity and (iv) ⇒ (i). If µ X ∗ µ X ∗· · ·∗ µ X L is absolutelycontinuous or (i) holds, then (ii) clearly holds so this completes the equivalence.(iii) ⇔ (iv). It is well known (see [ ], [ , VI.4]) that T X ( O X ) = { [ Y, X ] : Y ∈ g n } . Writing Y = P a α RE α + b α IE α + t for some t ∈ t n and a α , b α real, it is easily seenthat T X ( O X ) = sp N X . Further, T Ad ( g ) X ( O x ) = Ad ( g ) ( T X ( O X )) = sp { Ad ( g ) N x } ,proving the equivalence of (iii) and (iv).The final comment is an analyticity argument. Assume (iii) holds, for example,with g = ( Id, g , . . . , g L ). For any h = ( h , h , . . . , h L ) ∈ G Ln , h = Id , considerthe collection Ad ( h j ) Y for Y ∈ N X j and j = 1 , . . . , L , as vectors in R dim g n , andform the associated matrix M ( h ). As (iii) holds with g , there is a suitable squaresubmatrix of M ( g ) with non-zero determinant. By analyticity of the determinantmap, the determinant of the corresponding square submatrix of M ( h ) must be non-zero for an open, dense subset of h ∈ G L − n of full measure. The same argumentapplies to (iv).(ii) ⇒ (iv). Consider the addition map F : O X × · · · × O X L → g n given by F ( Y , . . . , Y L ) = P Lj =1 Y j . The image of F is P Lj =1 O X j . If the rank of F is not fullat any point in its domain, then Sard’s theorem ([ , p. 286]) implies the measureof the image of F is zero. Thus the differential of F at some point Y = ( Y , . . . , Y L ) , where Y j = Ad ( g j ) X j , has full rank. But the range of the differential of F at Y is P Lj =1 T Y j ( O X j ) and hence this sum must be g n .(iv) ⇒ (i). The hypothesis of (iv) guarantees that the map F defined abovehas full rank at some point Y . By the Implicit function theorem, F is an open mapin a neighbourhood of Y and thus Im F has non-empty interior.(iv) ⇒ absolute continuity. This is similar again. To see that the measure µ = µ X ∗ µ X ∗ · · · ∗ µ X L is absolutely continuous with respect to m , we shouldshow that µ ( E ) = 0 whenever m ( E ) = 0. Define f : G Ln → g n by f ( g , . . . , g L ) = F ( Ad ( g ) X , . . . , Ad ( g L ) X L ) . By definition, µ ( E ) = m G Ln ( f − ( E )). By (iv), the differential of f has full rank atsome point. An analyticity argument ensures that this is true on a subset of g ∈ G Ln of full measure. An application of the Implicit function theorem shows f − ( E ) has m G Ln - measure zero. For more details see [ , Thm. 2.2]. (cid:3) An immediate corollary of this proposition and the main theorem is the follow-ing.
Corollary . Suppose ( X , X , . . . , X L ) is eligible and not exceptional. Then P Li =1 O X i has non-empty interior. If ( X , X , . . . , X L ) is either not eligible or isexceptional and not type ( SU ( n ) , SU ( n − in D n , then P Li =1 O X i has measurezero. There is a sufficient condition for absolute continuity, established by Wright in[ ], that we will use in the proof of the main theorem to establish the absolutecontinuity of certain convolution products of orbital measures in small rank Liealgebras. We state this result below. By the rank of a subsystem we mean thedimension of the vector space it spans. Theorem . [ , Thm. 1.3] Let X , . . . , X L belong to the torus of g n . Assume (3.3) ( L −
1) ( | Φ | − | Ψ | ) − ≥ L X i =1 (cid:18) | Φ X i | − min σ ∈ W | Φ X i ∩ σ (Ψ) | (cid:19) for all root subsystems Ψ ⊆ Φ of rank n − and having the property that sp (Ψ) ∩ Φ =Ψ . Then µ X ∗ · · · ∗ µ X L is absolutely continuous.
4. Tuples That Are Not Absolutely Continuous
We begin by establishing the necessity of the conditions which give absolutecontinuity.
Lemma . If ( X , . . . , X L ) is an absolutely continuous L -tuple, then ( X , . . . , X L ) is eligible. Proof.
Suppose the L -tuple, ( X , . . . , X L ) ∈ g Ln , is not eligible, that is, L X i =1 S X i ≥ ( L − n + 1 (or ( L − n + 1) + 1 if g n is type A n . )Let α i be the eigenvalue of X i with greatest multiplicity (where we view each X i asa complex matrix of the appropriate size depending on the Lie type of g n ) and let g i belong to the associated Lie group, G n . Let V i be the eigenspace of Ad ( g i )( X i )corresponding to the eigenvalue α i .If g n is of type C n or D n , then Ad ( g i )( X i ) are 2 n × n matrices and dim V i = S X i , so L X i =1 dim V i = L X i =1 S X i ≥ ( L − n + 1 . BSOLUTE CONTINUITY 11
We deduce thatdim L \ i =1 V i = L X i =1 dim V i − dim( V + V ) + dim(( V ∩ V ) + V ) + · · · + dim( L − \ i =1 V i + V L ) ! ≥ ( L − n + 1 − n ( L − ≥ , and hence the matrices, Ad ( g i )( X i ) , have a common eigenvector, v . As L X i =1 Ad ( g i )( X i )( v ) = L X i =1 α i v, it follows that P i α i is an eigenvalue of P i Ad ( g i )( X i ). Since P i Ad ( g i )( X i ) is anarbitrary element of O X + · · · + O X L , one can see that every element of P i O X i has eigenvalue P i α i . This is impossible if O X + · · · + O X L has non-empty interior,thus an application of Prop. 1(i) allows us to conclude that µ X ∗ · · · ∗ µ X L is notabsolutely continuous.The argument is similar if g n is type A n , viewing X i as matrices in su ( n + 1),acting on R n +1 .In the case when g n is type B n we require a slight variation on the argumentsince every matrix in the Lie algebra so (2 n + 1) (the model for type B n ) has0 as an eigenvalue. We use the same notation as above and first observe thatif all X i are dominant B type, then all α i = 0 and dim V i = S X i + 1. Thus P Li =1 dim V i ≥ ( L − n + L + 1. Since the vector spaces V i are subspaces of R n +1 , it follows thatdim L \ i =1 V i ≥ ( L − n + L + 1 − (2 n + 1)( L − ≥ . Consequently, 0 is an eigenvalue of every element of O X + · · · + O X L of multiplicityat least two. Again, we can conclude that O X + · · · + O X L has empty interior andtherefore ( X , . . . , X L ) is not an absolutely continuous tuple.If, instead, precisely one X i is dominant SU type, with eigenvalue α = 0 ofmaximum multiplicity, then P dim V i ≥ ( L − n + L . This shows that dim L T i =1 V i has dimension at least one and hence every element of O X + · · · + O X L has α as aneigenvalue, again a contradiction if ( X , . . . , X L ) is an absolutely continuous tuple.If two or more X i are dominant SU type, then ( X , . . . , X L ) is automaticallyeligible. (cid:3) Lemma . Suppose ( X , . . . , X L ) is an exceptional tuple and is not a pair ( X , X ) of type ( SU ( n ) , SU ( n − in D n where n ≥ . Then ( X , . . . , X L ) is not an absolutely continuous tuple. Proof.
We will need separate arguments for the various exceptional tuples.(i) Suppose X and X are both type SU ( n ) in the Lie algebra D n . Observethat dim ( sp { Ad ( g i )( N X i ) : i = 1 , } ) ≤ | N X | + |N X | . In this case, |N X i | = | Φ n | /
2. As the dimension of the Lie algebra is | Φ n | + n it isclearly impossible for sp { Ad ( g i )( N X i ) : i = 1 , } to be the full Lie algebra. ThusProp. 1(iii) proves that this pair is not absolutely continuous.(ii) Suppose X and X are of types SU ( n ) and SU ( n − D n with n = 4 or 5. For this problem, we will use the fact that a root system of type SU (4) is isomorphic to one of type D . We will explain the argument for n = 4and leave n = 5 as an exercise.Let π be an automorphism of the root system of type D (an isomorphismthat preserves the Cartan matrix) that maps the annihilating roots of X (thoseof type SU (4)) onto a root subsystem of type D . This automorphism extends toan automorphism on the torus of D which maps X to the element π ( X ) whoseset of annihilating roots is the D root subsystem, and it maps X to the element π ( X ) whose set of annihilating roots is isomorphic to those of X and hence istype SU (3) (as this is unique up to Lie isomorphism). It induces a Lie algebraisomorphism that we also call π . We have π ( O X j ) = O π ( X j ) and π ( T Ad ( g j )( X j ) ( O X j )) = T Ad ( π ( g j ))( π ( X j )) ( O π ( X j ) )where if g j = exp H j , then π ( g j ) = exp π ( H j ).The pair ( π ( X ) , π ( X )) is not eligible in D as S π ( X ) = 6 and S π ( X ) = 3,so by our previous lemma it is not an absolutely continuous pair. Consequently,Prop. 1(iv) implies thatdim X i =1 T Ad ( π ( g i )) π ( X i ) ( O π ( X i ) ) ! < dim D n for any choices of g , g . But then a similar statement holds for P i =1 T Ad ( g j ) X j ( O X j )and thus ( X , X ) is not an absolutely continuous pair.(iii) When ( X , X ) is a pair of type ( SU (4) , SU (2) × D ) in D the argumentsare similar. The Lie isomorphism, π, that maps the subsystem of type SU (4) ontoone of type D must preserve the type of the root subsystem of type SU (2) × D .But the pair ( π ( X ) , π ( X )) is not eligible and hence neither it, nor the originalpair, can be absolutely continuous.Next, suppose X is type SU (4) and X is type SU (2) × SU (2) in D withthe subsystem, Φ X , Weyl conjugate to a subset of the subsystem Φ X . Since anyWeyl conjugate of X generates the same orbit as X there is no loss of generalityin assuming Φ X ⊆ Φ X . Consider the same Lie isomorphism π again. Then π (Φ X ) ⊆ π (Φ X ) has the same Lie type as Φ X . But the only subsystems of type D that are isomorphic to type SU (2) × SU (2) are of the form {± e i ± e j } for some i = j, and hence are type D . Being of type ( D , D ), the pair ( π ( X ) , π ( X )) isnot eligible and therefore ( X , X ) is not absolutely continuous.(iv) Assume X , X , X are each of type SU (4) in D , with Weyl conjugate setsof annihilators. As the annihilators are Weyl conjugate, for each i = 1 , , h i in the Lie group of type D such that Ad ( h i )( N X ) = N X i . Therefore thereexist g i in the group such that sp { Ad ( g i )( N X i ) : i = 1 , , } = g if and only if sp { Ad ( g i h i )( N X ) : i = 1 , , } = g . But the latter was shown to be impossible in the proof of [ , Thm. 8.2]. BSOLUTE CONTINUITY 13 (v) The argument is similar if X and X are both type SU ( n ) × SU ( n ) in the Liealgebra of type SU (2 n ). In this case, N X and N X are Weyl conjugate and it wasshown in [ , Prop. 5.1] that there is no g ∈ SU (2 n ) such that sp { Ad ( g ) N X , N X } = su (2 n ). (cid:3)
5. Proving Absolute Continuity - Main Ideas5.1. General Strategy.
Our proof that the eligible, non-exceptional tuplesare absolutely continuous will proceed by induction on the rank of the Lie algebra.The reduction is based upon the following idea.
Notation . Suppose X in the torus of the Lie algebra of type SU ( n ), B n , C n or D n is identified (after a suitable Weyl conjugate) with the n -vector(0 , . . . , | {z } J , a , . . . , a | {z } s , . . . , a m , . . . , ( ± ) a m | {z } s m )where s = max s j and J = 0 in the case of type SU ( n ). Define the element X ′ ∈ t n − by(5.1) X ′ = (0 , . . . , | {z } J − , a , . . . , a | {z } s , . . . , a m , . . . , ( ± ) a m | {z } s m ) if 2 J ≥ s (0 , . . . , | {z } J , a , . . . , a | {z } s − , . . . , a m , . . . , ( ± ) a m | {z } s m ) if 2 J < s This means, for example, that if X has type B J × SU ( s ) × · · · × SU ( s m ) where s = max s j , then X ′ has type B J − × SU ( s ) × · · · × SU ( s m ) if X is dominant B type and X ′ has type B J × SU ( s − × · · · × SU ( s m ) if X is dominant SU type. If X in SU ( n ) has type SU ( s ) × · · · × SU ( s m ), then S X ′ = S X − s > max j ≥ s j ,and S X ′ = S X otherwise. In the latter case S X ≤ n/ t n − into t n by taking the standard basis vectors e , . . . , e n in R n (or e − e n +1 , . . . , e n − e n +1 in R n +1 in the case of type SU ( n + 1)) as the basis for t n and taking the vectors e , . . . , e n (resp., e − e n +1 , . . . , e n − e n +1 ) as the basis for t n − . This also gives a natural embedding of Φ n − into Φ n and together these givean embedding of g n − into g n , an embedding of V n − into V n and an embeddingof G n − into G n . We will also view X ′ as an element of t n in the natural way.An induction argument will be applicable because of the following lemma. Lemma . If ( X, Y ) is an eligible pair in g n and X, Y are not both of type SU ( m ) × SU ( m ) in the Lie algebra of type SU (2 m ) , then the reduced pair, ( X ′ , Y ′ ) , is eligible in g n − . Proof.
Case 1: g n is type B n , C n or D n .Observe that always S X ′ ≤ S X since the dimensions of the eigenspaces of X ′ can only be at most the dimensions of those of X. If both X and X ′ are dominant B, C or D type, then S X ′ = S X −
2. If X ′ is dominant SU type, then S X ′ ≤ n −
1, regardless of the type of X . Finally, if X is dominant SU type while X ′ is dominant B , C or D type, then S X = s > J = S X ′ ≥ s −
1. Since it is always true that J + s ≤ n , one can check that s ≤ (2 n + 1) / S X ′ ≤ n − . Thus if either X and X ′ or Y and Y ′ are both dominant B, C or D type, then S X ′ + S Y ′ ≤ S X + S Y − ≤ n − . Otherwise, both S X ′ and S Y ′ ≤ n − S X ′ + S Y ′ ≤ n − g n is type SU ( n + 1).If either S X ′ < S X or S Y ′ < S Y , then S X ′ + S Y ′ ≤ S X + S Y − X ′ , Y ′ ) is eligible. Otherwise, S X ′ = S X and S Y ′ = S Y and in that case S X ′ , S Y ′ ≤ ( n + 1) /
2. If n is even, then we must have S X ′ , S Y ′ ≤ n/ S X ′ + S Y ′ ≤ n .If n is odd, it is still true that S X ′ + S Y ′ ≤ n unless S X ′ = S Y ′ = ( n + 1) /
2. Butthat happens only when X and Y are both type SU (( n + 1) / × SU (( n + 1) / (cid:3) Remark . It is easy to see that if X and X ′ are of opposite dominant types,then X is type B J , ( C J or D J ) × SU ( s ) × · · · × SU ( s m ) where 1 ≤ J < P s i . Itfollows from [ , Thm. 8.2] that µ X ∈ L .We record here a well known fact from elementary linear algebra that is aconsequence of the continuity of the determinant function and will be quite usefulfor us. Lemma . If { v , . . . , v n } is a set of linearly independent vectors in vectorspace V and w , . . . , w n ∈ V , then for sufficiently small ε > , the collection { v + εw , . . . , v n + εw n } is also linearly independent. Notation . Given X ′ as defined above, let N X ′ = { RE α , IE α : α / ∈ Φ X ′ } ,(as in 2.5), but viewed as embedded into V n . LetΩ X = N X (cid:31) N X ′ . We will refer to the next result as our general strategy. It will enable us toestablish Prop. 1(iii) holds for a given tuple.
Proposition . (General Strategy) Let X i ∈ t n , i = 1 , . . . , L for L ≥ , andassume ( X ′ , . . . , X ′ L ) is an absolutely continuous tuple in g n − . Suppose Ω is asubset of V n \V n − that contains all Ω X i and has the property that ad ( H )(Ω) ⊆ sp Ω whenever H ∈ g n − . Fix Ω ⊆ Ω X L .Assume there exists g , . . . g L − ∈ G n − and M ∈ g n such that(i) sp { Ad ( g i )(Ω X i ) , Ω X L \ Ω : i = 1 , . . . , L − } = sp Ω; (ii) ad k ( M ) : N X L \ Ω → sp { Ω , g n − } for all positive integers k ; and(iii) The span of the projection of Ad (exp sM )(Ω ) onto the orthogonal com-plement of sp { g n − , Ω } in g n is a surjection for all small s > . Then ( X , . . . , X L ) is an absolute continuous tuple. Proof.
As ( X ′ , . . . , X ′ L ) is an absolutely continuous tuple, Prop. 1(iii) tells usthat sp { Ad ( h i )( N X ′ i ) , N X ′ L : i = 1 , . . . , L − } = g n − for a dense set of ( h , . . . , h L − ) ∈ G L − n − . Given ε >
0, choose such h i = h i ( ε ) ∈ G n − with k Ad ( h i ) − Ad ( g i ) k < ε, where the elements g i ∈ G n − are the ones givenin the hypothesis of the proposition. (The norm can be taken to be the operatornorm.)Lemma 4, together with assumption (i), shows that for sufficiently small ε > sp Ω) = dim ( sp { Ad ( g i )(Ω X i ) , Ω X L \ Ω : i = 1 , . . . , L − } )= dim ( sp { Ad ( h i )(Ω X i ) , Ω X L \ Ω : i = 1 , . . . , L − } ) . BSOLUTE CONTINUITY 15
Since ad ( H )(Ω) ⊆ sp (Ω) for all H ∈ g n − and h i = exp H i for some H i ∈ g n − we have Ad ( h i )(Ω) = Ad (exp H i )(Ω)= exp( ad ( H i )(Ω) ⊆ sp Ωfor all h i ∈ G n − . Thus for sufficiently small ε > sp { Ad ( h i )(Ω X i ) , Ω X L \ Ω : L = 1 , . . . , L − } = sp ΩFor such a choice of ε (hereafter fixed) we have sp { Ad ( h i )( N X i ) , N X L \ Ω : i = 1 , . . . , L − } = sp { Ad ( h i )( N X ′ i ) , Ad ( h i )Ω X i , N X ′ L , Ω X L \ Ω } = sp { Ω , g n − } . Assumption (ii), and the fact that N X L \ Ω ⊆ sp { Ω , g n − } , implies that forany real number s , exp( s · adM ) = Ad (exp sM ) maps N X L \ Ω to sp { Ω , g n − } .Moreover, k Id − Ad (exp sM ) k → s →
0, thus similar reasoning to that aboveshows that for all small enough s > sp { Ω , g n − } = sp { Ad ( h i )( N X i ) , N X L \ Ω : i = 1 , . . . , L } = sp { Ad ( h i )( N X i ) , ( Ad (exp sM ))( N X L \ Ω ) : i = 1 , . . . , L − } . Combined with assumption (iii), this proves that for sufficiently small s > ,sp { Ad ( h i )( N X i ) , Ad (exp sM )( N X L ) : i = 1 , . . . , L − } = g n . Another application of Prop. 1(iii) shows that µ X ∗ · · · ∗ µ X L is absolutely contin-uous. (cid:3) We will occasionally make use of the following specific application of the el-ementary linear algebra property in order to verify the hypothesis of the generalstrategy.
Lemma . Suppose Ω is a subset of V n \V n − that contains both Ω X and Ω Y , and has the property that ad ( H )(Ω) ⊆ sp Ω whenever H ∈ g n − . Fix Ω ⊆ Ω X .Assume Ω ⊆ (Ω Y ∩ Ω X ) (cid:31) Ω and the vectors in { adH (Ω ) , Ω Y (cid:31) Ω , Ω X (cid:31) Ω } span Ω for some H ∈ g n − . Then for sufficiently small t > , sp { Ad (exp tH )(Ω Y ) , Ω X \ Ω } = sp Ω . Proof.
The arguments are similar to that of the general strategy. Since (cid:13)(cid:13) ad ( H ) − t ( Ad (exp tH ) − Id ) (cid:13)(cid:13) and k Id − Ad (exp tH ) k both tend to 0 as t → ad k ( H )(Ω) ⊆ sp Ω for all k , the same argument as used above shows that sp { ( Ad (exp tH ) − Id ) (Ω ) , Ad (exp tH )(Ω Y \ Ω ) , Ω X \ Ω } = sp Ω . But since Ω ⊆ Ω X \ Ω , we can replace ( Ad (exp tH ) − Id ) (Ω ) in the span on theleft hand side by Ad (exp tH ) (Ω ). Hence sp { Ad (exp tH ) (Ω Y ) , Ω X \ Ω } = sp Ω. (cid:3) L = 2 . The following proposi-tion, the ‘induction step’, is the most important ingredient in the proof of the maintheorem.We continue to use the notation Ω X = N X (cid:31) N X ′ , where X ′ was defined in(5.1). Proposition . Suppose ( X, Y ) is an eligible pair in g n other than X, Y bothof type SU ( n ) in D n or type SU ( n/ × SU ( n/ in SU ( n ) . Assume also that thereduced pair, ( X ′ , Y ′ ) , is an absolutely continuous pair in g n − . Then ( X, Y ) is anabsolutely continuous pair in g n . Proof.
The main task of the proof is to show that any eligible pair, otherthan one of the two exceptional pairs mentioned, satisfy properties (i) - (iii) of thegeneral strategy, Prop. 2.
Part I: g n is type B n , C n or D n . The proof is divided into three cases depending on the dominant types of X and Y .Case 1: Neither X nor Y are of dominant SU type.With the notation as before, we have S X = 2 J and S Y = 2 K (meaning X isdominant B J ( C J or D J ) type and Y is dominant B K ( C K or D K ) type). Applyinga Weyl conjugate, if necessary, we can assume without loss of generality thatΩ X = { F Ee ± e j : J < j ≤ n, F = R, I } and similarly Ω Y = { F Ee ± e j : K < j ≤ n, F = R, I } . Case 1(a): g n is type D n .Recall that V n is set of all real and imaginary parts of the chosen Weyl basisof root vectors of g n . PutΩ = V n (cid:31) V n − = { F Ee ± e j : j = 2 , . . . , n, F = R, I } and Ω = { REe + e n , IEe + e n } . If H ∈ g n − , then H is a linear combination of a torus element of g n − and thevectors REe i ± e j , IEe i ± e j with 2 ≤ i < j ≤ n . It follows easily from (2.4) that ad ( H )(Ω) ⊆ sp Ω.Take g ∈ G n − to be the Weyl conjugate that permutes the letters 1 + j and K + j for j = 1 , . . . , J −
1. This is well defined and leaves the letter n unchanged asthe eligibility condition ensures J + K − ≤ n −
1. Consequently, Ad ( g )( F Ee ± e K + j ) = F Ee ± e j for j = 1 , . . . , J − , and all other vectors in Ω are fixed,including F Ee ± e n . Thus { Ad ( g )(Ω Y ) , Ω X (cid:31) Ω } = { F Ee ± e k : k = 2 , . . . , n } proving that (i) of the general strategy, Prop 2 (with L = 2) is satisfied.Let M = REe + e n ∈ g n . Applying (2.4) again, we see that if H = F Ee ± e j for some j < n , then ad ( M )( H ) = cF Ee j ∓ e n ∈ g n − for a non-zero constant c depending on j, n and F . If H = F Ee i ± e n , then ad ( M )( H ) = cF Ee ∓ e i ∈ sp (Ω (cid:31) Ω ). Finally, note that ad ( M )( H ) = 0 if H = F Ee i ± e j for 1 < i, j < n or H = F Ee − e n . This proves ad k ( M ) : N X \ Ω → sp { Ω , g n − } for all positiveintegers k, so that property (ii) of the general strategy is satisfied. BSOLUTE CONTINUITY 17 As sp { Ω , g n − } is of co-dimension one, its orthogonal complement is spannedby the projection onto any element in the complement of sp { Ω , g n − } . The toruselement, ad ( M )( IEe + e n ) = [ REe + e n , IEe + e n ] := t is such an element. Since ad ( M )( t ) = [ REe + e n , t ] = cIEe + e n for some c = 0 (see 2.3), it follows that Ad (exp sM )( IEe + e n ) = a ( s ) IEe + e n + sb ( s ) t where a ( s ) , b ( s ) → s →
0. Therefore hypothesis (iii) of Prop. 2 is also fulfilledwith any s >
0. Applying that proposition, we conclude that µ X ∗ µ Y is absolutelycontinuous.Case 1(b): g n is type B n .Again, we will apply the general strategy, but here withΩ = V n \V n − = { F Ee ± e j , F Ee : j = 2 , . . . , n, F = R, I } and Ω = { REe + e n , IEe + e n } . The fact that ad ( H )(Ω) ⊆ Ω whenever H ∈ g n − follows easily from propertiesof the roots, as with the case D n .For t >
0, let g t = (exp tREe n ) g where g ∈ G n − corresponds to the Weylconjugate that permutes the letters 1 + j and K + j for j = 1 , . . . , J − REe n ∈ g n − , g t ∈ G n − . Observe that[ REe n , F Ee ] = cF Ee + e n + c ′ F Ee − e n and [ REe n , F Ee ± e j ] = ( c ( ± ) F Ee if j = n c, c ′ and c ( ± ) non-zero constants. In particular, this implies Ad (exp tREe n )( F Ee ± e j ) = F Ee ± e j for j = n. Since Ad ( g )( F Ee ± e K + j ) = F Ee ± e j for j = 1 , . . . , J − Ad ( g ) fixes F Ee ± e n , it follows that for j = 1 , . . . , J − Ad ( g t )( F Ee ± e K + j ) = Ad (exp tREe n )( F Ee ± e j )= F Ee ± e j and Ad ( g t )( F Ee ± e n ) = Ad (exp tREe n )( F Ee ± e n )= a ( t ) F Ee ± e n + tb ( t ) F Ee + t c ( t ) F Ee ∓ e n where a ( t ) → t →
0, and b ( t ) and c ( t ) converge to non-zero scalars . All otherchoices of F Ee ± e j are fixed by Ad ( g t ). Hence sp { F Ee − e n , Ad ( g t )( F Ee ± e n ) : F = R, I } = sp { F Ee − e n , F Ee + e n + tb ′ ( t ) F Ee , F Ee + tc ′ ( t ) F Ee + e n : F = R, I } a ( t ) , b ( t ) , c ( t ) depend on F and the choice of ± , as well as t . From here on we will omitnoting this dependence, unless it is important. where b ′ ( t ) and c ′ ( t ) converge to non-zero limits as t →
0. Since { F Ee ± e n , F Ee : F = R, I } is a set of six linearly independent vectors, so too is the collection { F Ee − e n , F Ee + e n + tb ′ ( t ) F Ee , F Ee + tc ′ ( t ) F Ee + e n : F = R, I } for sufficiently small t , and therefore they span the same space. Because Ω X \ Ω contains F Ee − e n , it follows that sp { Ad ( g t )(Ω Y ) , Ω X \ Ω } = sp { Ad ( g t )( F Ee ± e k ) , F Ee ± e j , F Ee − e n : k > K, J < j < n, F = R, I } = sp { F Ee ± e j , F Ee ± e n , F Ee : j ≤ n, F = R, I } = sp Ω . Again, put M = REe + e n ∈ g n . As with type D n , ad k ( M )( F Ee ± e j ) ∈ sp { g n − , Ω } for all k and j < n , and ad ( M )( F Ee − e n ) = 0. Furthermore, ad ( M )( F Ee j ) = 0 if j = 1 , n, ad ( M )( F Ee n ) = cF Ee and ad ( M )( F Ee ) = cF Ee n , so property (ii) of the general strategy holds. As in the first case, sp { g n − , Ω } is of co-dimension one in g n , and just as in type D n property (iii) holds, so we deducethe absolute continuity of µ X ∗ µ Y by appealing to Prop. 2.Case 1(c): g n is type C n .Here we will use a variant on the general strategy. As with type D n we beginwith Ω = { F Ee ± e j : j = 2 , . . . , n, F = R, I } and g the Weyl conjugate permuting the letters 1 + j and K + j for j = 1 , . . . , J − = { F Ee ± e n : F = R, I } . The eligibility condition gives that sp { Ad ( g )(Ω Y ) , Ω X \ Ω } = sp Ω.As with type D n , ad ( F Ee i ± e j )(Ω) ⊆ sp { Ω , g n − } for all 1 < i < j ≤ n andsimilarly, ad ( F E (2 e j ))(Ω) ⊆ Ω for j >
1, so ad ( H )(Ω) ⊆ sp { Ω , g n − } whenever H ∈ g n − . Thus, as in the proof of the general strategy, upon applying the inductionassumption we can deduce there is some h ∈ G n − such that(5.2) sp { Ad ( h )( N Y ) , N X \ Ω } = sp { Ω , g n − } . Once again, we will put M = REe + e n ∈ g n . As with the types B n and D n ,standard facts about roots show that ad ( M )( H ) ∈ sp { Ω , g n − } for all H ∈ N X \ Ω .In fact, for all k ≥ ad k ( M )( H ) ∈ sp { Ω , g n − } for all H ∈ N X \ Ω except for H = F E (2 e n ) as ad k ( M )( F E (2 e n )) has a component in F E (2 e ). (Recall that F E (2 e n ) ∈ N X since the only roots 2 e j ∈ Φ X are those with j ≤ J .) It is becauseof this exception that we cannot appeal directly to the general strategy.Another difference between this set up and the situation for types B n and D n is that here sp { Ω , g n − } has co-dimension three, its orthogonal complementbeing spanned by RE (2 e ) , IE (2 e ) and the projection onto the torus element[ REe + e n , IEe + e n ]. That will also complicate matters.Let Λ be the subspace spanned by the torus of g n − and the vectors RE β and IE β where β ranges over all the positive roots except 2 e , e n ,Λ := sp { Ω , g n − } ⊖ sp { RE (2 e n ) , IE (2 e n ) } . BSOLUTE CONTINUITY 19
Let P be the orthogonal projection onto Λ. Since N X (cid:31) { Ω , F E (2 e n ) } ⊆ Λ, Prop-erty (5.2) implies that sp {P ( Ad ( h )( N Y )) , N X (cid:31) { Ω , RE (2 e n ) , IE (2 e n ) }} = Λ . Choose Y Fβ , Y j ∈ Ad ( h ) N Y and X Fβ , X j ∈ N X (cid:31) { Ω , RE (2 e n ) , IE (2 e n ) } such that(i) Y Fβ + X Fβ = F E β + W Fβ where W Fβ ∈ sp { RE (2 e n ) , IE (2 e n ) } , F = R, I and β ranges over all roots except 2 e , e n , and(ii) Y j + X j = t j + W j where j = 2 , . . . , n , { t , . . . , t n } is a basis for t n − and W j ∈ sp { F E (2 e n ) } .Note that if we put t = [ REe + e n , IEe + e n ], then { t , . . . , t n } is a basis for t n . This collection of vectors { Y Fβ + X Fβ , Y j + X j } is linearly independent and hencefor small enough s >
0, so is also the set { Y Fβ + Ad (exp sM )( X Fβ ) , Y j + Ad (exp sM )( X j ) : β = 2 e , e n , j = 2 , . . . , n, F = R, I } . Observe that Y Fβ + Ad (exp sM )( X Fβ ) = Y Fβ + X Fβ + ( Ad (exp sM ) − Id )( X Fβ )= F E β + W Fβ + sQ Fβ ( s ) , where the vector Q Fβ ( s ) depends on s, but has bounded norm. The projection of Q Fβ ( s ) onto sp { RE e , IE e } is zero since Ad (exp sM ) maps N X (cid:31) { Ω , F E (2 e n ) : F = R, I } into g n ⊖ sp { RE e , IE e } . Also, it is clear from the definitions thatfor β = e − e n , the projection of F E β + W Fβ onto sp { REe − e n , IEe − e n } iszero. Similar statements can be made for Y j + Ad (exp sM )( X j ). Claim:
The collection of vectors, Y Fβ + Ad (exp sM )( X Fβ ), Y j + Ad (exp sM )( X j )over all positive roots β = 2 e , e n , F = R, I , and j = 2 , . . . , n, together with thefour vectors Ad (exp sM )( F E (2 e n )), Ad (exp sM )( F Ee − e n ) for F = R, I , arelinearly independent.To prove this we first observe that[
REe + e n , F E (2 e )] = c F Ee − e n [ REe + e n , F E (2 e n )] = c F Ee n − e , [ REe + e n , F Ee − e n ] = c F E (2 e ) + c F E (2 e n )where c j = 0. Thus(5.3) Ad (exp sM )( F Ee − e n ) = a Fs F Ee − e n + sb Fs F E (2 e ) + sc Fs F E (2 e n )and(5.4) Ad (exp sM )( F E (2 e n )) = sb ′ Fs F Ee − e n + s c ′ Fs F E (2 e ) + a ′ Fs F E (2 e n )where the coefficients, a Fs , a ′ Fs , b Fs , b ′ Fs , c Fs , c ′ Fs , converge to non-zero constants as s → sp { g n − , Ω } , belong to g n ⊖ sp { t } . We view them as vectors in R d with d = dim g n −
1, whose coordinatesare given by the basis for g n ⊖ sp { t } consisting of the torus elements, { t , . . . , t n } ,together with the real and imaginary parts of the Weyl basis { E α } , taking as thefinal six positions the basis vectors F Ee − e n , F E (2 e n ) and F E (2 e ), F = R, I .With this understanding, consider the square matrix whose rows are givenby the vectors Y j + Ad (exp sM )( X j ) for j = 2 , . . . , n ; followed by the vectors Y Fβ + Ad (exp sM )( X Fβ ), β = 2 e , e n , ordered consistently to above so that the finaltwo come from β = e − e n ; and then finally the four vectors Ad (exp sM )( F E (2 e n ))and Ad (exp sM )( F Ee − e n ) (for a small, but fixed, choice of s ).The calculations above show that this matrix, denoted A = ( A ij ) , has the form A = [ I d − + O ( s )] ( d − × ( d − [ O ( s )] ( d − × [ ∗ ] ( d − × [0] ( d − × [ O ( s )] × ( d − [ I + O ( s )] × (cid:20) ∗ ∗∗ ∗ (cid:21) (cid:20) (cid:21) [0] × ( d − (cid:20) sb ′ Rs sb ′ Is (cid:21) (cid:20) a ′ Rs a ′ Is (cid:21) (cid:20) O ( s ) 00 O ( s ) (cid:21) [0] × ( d − (cid:20) a Rs a Is (cid:21) (cid:20) sc Rs sc Is (cid:21) (cid:20) sb Rs sb Is (cid:21) where I m denotes the m × m identity matrix, O ( s k ) means terms dominated by Cs k for some constant C independent of s and ∗ denotes terms that may dependon s , but are bounded independently of s .We estimate the determinant of this matrix using the Leibniz formula: Since | A A · · · A dd | ≥ C s for some C > A σ (1) A σ (2) · · · A dσ ( d ) , where σ is a permutation of { , . . . , d } , are dominated inabsolute value by C s , the determinant is non-zero for sufficiently small s > g n ⊖ sp { t } . Recall that X Fβ , F E (2 e n ) and F Ee − e n all belong to N X (cid:31) { F Ee + e n : F = R, I } , hence sp { Ad ( h ) N Y , Ad (exp sM )( N X (cid:31) { F Ee + e n } ) } = g n ⊖ sp { t } . Finally, our familiar calculation shows Ad (exp sM )( IEe + e n ) = a s IEe + e n + sb s t where b s converges to a non-zero constant. It follows that for small enough s , sp { Ad ( h )( N Y ) , Ad (exp sM )( N X ) } = g n , as we desired to show.Case 2: Both X and Y are dominant SU type.First, assume the Lie algebra is type B n or C n . According to [ , Thm. 8.2] both µ X and µ Y belong to L . Applying Holder’s inequality we see that µ X ∗ µ Y ∈ L .Being compactly supported, it follows that µ X ∗ µ Y is in L , and hence is a measurethat is absolutely continuous with respect to Lebesgue measure. (Note that thesame argument applies to the Lie algebra of type D n unless one of X or Y is oftype SU ( n ).)However, we prefer to give an argument that is independent of [ ] as the tech-niques will then have more general application and such an argument will be neededin the case of type D n , in any case. For this, in the case of type B n , putΩ = { F Ee , F Ee ± e j : j ≥ , F = R, I } andΩ = { REe , IEe } BSOLUTE CONTINUITY 21 while in the case of type C n , putΩ = { F E (2 e ) , F Ee ± e j : j ≥ , F = R, I } andΩ = { RE (2 e ) , IE (2 e ) } . In either case ad ( H )(Ω) ⊆ sp Ω for all H ∈ g n − .As X, Y are dominant SU type, both Ω X and Ω Y contain F E (2) e and all theroots F Ee + e j , j ≥
2. If g ∈ G n − is the Weyl conjugate that changes the signsof the letters 2 , . . . , n , then { Ad ( g )(Ω X ) , Ω Y (cid:31) Ω } = Ω. Now take M = RE (2) e and apply the general strategy.The arguments are similar when the Lie algebra is type D n . LetΩ = { F Ee ± e j : j ≥ , F = R, I } . As we do not permit both X and Y to be of type SU ( n ), without loss of generalityΩ X contains all the roots F Ee + e j for 2 ≤ j ≤ n −
1, as well as both
F Ee ± e n , and Ω Y contains either all F Ee + e j for 2 ≤ j or all F Ee + e j for 2 ≤ j ≤ n − F Ee − e n . Let Ω be the choice of { F Ee + e n } or { F Ee − e n } , dependingon which belongs to Ω Y . Let g ∈ G n − be the Weyl conjugate that changes thesigns 2 , . . . , n − n if needed to be an even sign change). Then Ad ( g )(Ω X ) ⊇{ F Ee − e j , F Ee ± e n : j ≥ } and hence { Ad ( g )(Ω X ) , Ω Y (cid:31) Ω } = Ω. Take M = REe ± e n with the choice of ± depending on which belongs to Ω Y .Case 3: X and Y are of different dominant type.Without loss of generality assume X is dominant SU ( m ) type and Y is dom-inant B J , C J or D J type, depending on the type of the Lie algebra. Eligibilityimplies that 2 J + m ≤ n. Let Ω = { F Ee ± e j , ( F E (2) e ) : j ≥ } . (with the inclusion of F Ee if the Lie algebra is type B n or F E (2 e ) if the Liealgebra is C n ). We haveΩ X = { F Ee + e j , F Ee − e n : j < n, } if X = ( a, . . . , a, − a ) in D n andΩ X = { F Ee + e j , F Ee − e k , ( F E (2) e ) : j ≥ , k > m, F = R, I } otherwise.Put Ω = { F Ee + e n − J +1 } ⊆ Ω X ∩ Ω Y (or Ω = { F Ee − e n } if J = 1 and X = ( a, . . . , a, − a ) in D n ). Applying a Weylconjugate from G n − , we can assumeΩ Y = { F Ee ± e j : 2 ≤ j ≤ n − J + 1 , F = R, I } . If n − J + 1 ≥ m , then we already have { Ω Y , Ω X (cid:31) Ω } = Ω , so property (i) of the general strategy holds with g = Id.
Take M = REe + e n − J +1 (resp., take M = REe − e n ) to complete the argument.Otherwise m + J − n ≥ J ≥ = { F Ee + e k : 2 ≤ k ≤ n − J, F = R, I } ⊆ (Ω Y ∩ Ω X ) (cid:31) Ω
02 SANJIV KUMAR GUPTA AND KATHRYN E. HARE and define H = J − X j =2 REe j + e n − J + j + REe J − e n if X = ( a, . . . , a, − a ) in type D n and H = m + J − n X j =2 REe j + e n − J + j otherwise.As J = n , e j + e n − J + j are roots of the Lie algebra g n − . Let 2 ≤ k ≤ m + J − n .Observe that k = n − J + j for any j ≥
2, for if so, then j = k − n + J ≤ m + 2 J − n and therefore the eligibility condition would imply j ≤
0. Thus, if 2 ≤ k ≤ m + J − n ,then ad ( H )( F Ee + e k ) = c k F Ee − e n − J + k (or ad ( H )( F Ee + e J ) = c J F Ee + e n if X = ( a, . . . , − a )).The eligibility condition also impliesΩ ⊇ { F Ee + e k : 2 ≤ k ≤ m + J − n } , therefore sp { ad ( H )(Ω ) } ⊇ sp { F Ee − e j , F Ee + e n : n − J + 2 ≤ j ≤ n − , F = R, I } if X = ( a, . . . , a, − a ) in type D n and sp { ad ( H )(Ω ) } ⊇ sp { F Ee − e j : n − J + 2 ≤ j ≤ m, F = R, I } otherwise.Since Ω Y (cid:31) Ω = { F Ee − e j , F Ee + e n − J +1 : 2 ≤ j ≤ n − J + 1 } , in eithercase we have sp { ad ( H )(Ω ) , Ω Y (cid:31) Ω , Ω X (cid:31) Ω } = sp Ω . By Lemma 5 there is some g ∈ G n − (namely, g = exp tH for sufficiently small t )such that sp { Ad ( g )(Ω Y ) , Ω X (cid:31) Ω } = sp Ω . Again, take M = REe + e n − J +1 and apply the general strategy to complete theargument. Part II: g n is type SU ( n ) . This is very similar to case 1(a). LetΩ = { F Ee − e j : 2 < j ≤ n, F = R, I } . We have Ω X = { F Ee − e j : S X < j ≤ n, F = R, I } andΩ Y = { F Ee − e j : S Y < j ≤ n, F = R, I } . Put Ω = { F Ee − e n : F = R, I } . Take g ∈ SU ( n −
1) to be the Weyl conjugatethat interchanges the letters S Y + j and 1 + j for j = 1 , . . . , S X −
1. The eligibil-ity condition ensures this is well defined and leaves 1 and n unchanged. Clearly { Ad ( g )(Ω Y ) , Ω X (cid:31) Ω } = Ω. Take M = REe − e n and apply the general strategyin the usual manner. (cid:3) BSOLUTE CONTINUITY 23
6. Proof of the Main Theorem
In this section we will complete the proof of Theorem 1.
Proof of Theorem 1.
Necessary conditions for Absolute continuity:
Lemma 1 shows that absolutely continuous tuples are eligible, while in Lemma 2 wesaw that the exceptional tuples, other than possibly the pairs of type ( SU ( n ) , SU ( n − D n with n ≥
6, are not absolutely continuous.The rest of the proof is devoted to establishing that the eligible, non-exceptionaltuples are absolutely continuous.
Sufficient conditions for Absolute continuity for Lie types A n , B n and C n : Case L = 2. The proof proceeds by induction on the rank n of the Lie algebra.We begin A n with n = 1 (type SU (2)) and B n with n = 2. Although it is customaryto only define C n for n ≥
3, there is no harm in beginning with C , meaning theroot system ±{ e , e , e ± e } , which is Lie isomorphic to B .According to [ , Thm. 8.2], all non-zero pairs ( X, Y ) in the Lie algebras oftype SU (2) and B have the property that both µ X , µ Y ∈ L . Thus µ X ∗ µ Y is acompactly supported measure in L and hence is an absolutely continuous measure.The existence of g , g ∈ G n with X i =1 T Ad ( g i )( X i ) ( O X i ) = g n is a Lie isomorphism invariant, thus from Prop. 1 we can also deduce that µ X ∗ µ Y is an absolutely continuous measure for all non-zero ( X , X ) in the Lie algebra oftype C .Now, inductively assume that all eligible, non-exceptional pairs in SU ( n − ,B n − or C n − , with n ≥ , are absolutely continuous. (Of course, there are noexceptional pairs in B n − or C n − .)Let ( X, Y ) be an eligible, non-exceptional pair in SU ( n ), B n or C n , and formthe reduced pair ( X ′ , Y ′ ). The reduced pair is eligible by Lemma 3. Notice thatonly an element of type SU ( n +12 ) × SU ( n − ) in SU ( n ) will reduce to an elementof type SU ( n − ) × SU ( n − ) in SU ( n − SU ( n +12 ) × SU ( n − ) is not eligible in SU ( n ), thus we can assume ( X ′ , Y ′ )is both eligible and non-exceptional. By the induction assumption, ( X ′ , Y ′ ) is anabsolutely continuous pair. But then the induction step, Prop. 3, implies that( X, Y ) is absolutely continuous.Case L ≥
3. Again, we proceed by induction on n . We remark that as µ ∗ ν isabsolutely continuous if µ is absolutely continuous and ν is an arbitrary measure,the fact that the convolution of any two non-zero orbital measures in type SU (2) , B or C is absolutely continuous, proves that the same is true for the convolution ofany L non-zero orbital measures. This starts the induction.First, suppose ( X , . . . , X L ) is an eligible L -tuple in B n or C n with n ≥
3. Wewill let Ω be as in Prop. 3, depending on whether g is type B n or C n ,Ω = { F Ee ± e j , F E (2) e : j = 2 , . . . , n, F = R, I } . As a pair of elements that is dominant SU type in B n or C n is eligible and notexceptional, the theorem for L = 2 implies the convolution of (even) their two orbital measures is absolutely continuous. Thus we may assume that at most one X i is dominant SU type.Suppose that no X i are dominant SU type and form the corresponding X ′ i . If X ′ i and X ′ j are dominant SU , then the pair ( X i , X j ) is eligible (and not exceptional),thus µ X i ∗ µ X j is absolutely continuous. Hence we can assume that at most one X ′ i is dominant SU type.Since S X ′ = S X − X and X ′ are dominant B (or C ) type it followsthat L X i =1 S X ′ i ≤ L X i =1 S X i − L − ≤ n ( L − − L −
1) = 2( n − L − . This shows that ( X ′ , . . . ., X ′ L ) is eligible in g n − . As it is not exceptional, theinduction assumption implies it is an absolutely continuous tuple.Here Ω X i = { F Ee ± e j : j > J i } where 2 J i = S X i . Taking g i to be the Weylconjugate that switches appropriate letters (and fixes the letters 1 and n ) we canarrange for Ad ( g i )(Ω X i ) = { F Ee ± e j : j = ( i − n − i − X k =1 J k + 2 , . . . , in − i X k =1 J k + 1 } (with suitable modifications if any of the specified choices of j exceed n ).If ( L − n − P L − k =1 J k + 1 ≥ n , then L − [ i =1 Ad ( g i ) (Ω X i ) = { F Ee ± e j : j = 2 , . . . , n } , and this coincides with the set Ω Y for a suitable Y of type B (or C ) (meaningtype B (or C ) × SU (1) × · · · × SU (1)). As always S X ≤ n − Y, X L )is eligible.Otherwise, if we let m = n − ( L − n + P L − i =1 J i and take a suitable choice of Y of type B m (or C m ), then L − [ i =1 Ad ( g i )Ω X i = Ω Y . The eligibility condition ensures that S Y + S X L = 2 n − ( L − n + L − X i =1 J i ! + 2 J L ≤ n − L − n + L X i =1 S X i ≤ n and thus the pair ( Y, X L ) is eligible and clearly not exceptional. The argumentsgiven in the proof of Prop. 3 Case 1 show that then there is some g ∈ G n − , M ∈ g n and Ω ⊆ Ω X L such that(i) sp Ω = sp { Ad ( g )(Ω Y ) , Ω X L \ Ω } ;(ii) ad k ( M ) : N X L \ Ω → sp { Ω , g n − } for all positive integers k ; and(iii) The span of the projection of Ad (exp sM )(Ω ) onto the orthogonal com-plement of sp { g n − , Ω } in g n is a surjection for all small s > . BSOLUTE CONTINUITY 25
Since sp { Ad ( g )(Ω Y ) , Ω X L \ Ω } ⊆ sp { Ad ( gg i )(Ω X i ) , Ω X L \ Ω : i = 1 , . . . ., L − } we can call upon the general strategy, Prop. 2, with g i replaced there by gg i , todeduce that ( X , . . . , X L ) is an absolutely continuous tuple. This completes theargument when no X i are of dominant SU type.Otherwise, there is one X i which is of dominant SU type, say X L . If there isanother index j such that X ′ j is of dominant SU type, then the pair ( X L , X j ) iseligible and not exceptional, hence µ X L ∗ µ X j is absolutely continuous.So we may assume all X ′ j with j = L are of dominant B (or C ) type. Thus X S ′ X i ≤ X S X i − L − ≤ n − L − , so ( X ′ , . . . , X ′ L ) is an eligible L -tuple. Again, taking g i to be suitable Weyl conju-gates we have L − [ i =1 Ad ( g i )(Ω X i ) = { F Ee ± e j : j = 2 , . . . , ( L − n − L − X i =1 J i + 1 } and if we let Y be of type B m where m = n − ( L − n + P L − i =1 J i , then L − [ i =1 Ad ( g i )(Ω X i ) = Ω Y . The eligibility condition ensures that S Y + S X L = 2 n − ( L − n + L − X i =1 J i ! + S X L ≤ n, so the pair ( Y, X L ) is eligible. Complete the proof using the arguments of Prop. 3,but this time using Case 3 as X L and Y are of opposite dominant types.The argument is similar, but easier, if the Lie algebra is type SU ( n ). We firstcheck that ( X ′ , . . . , X ′ L ) is eligible when ( X , . . . , X L ) is eligible. This is clear if atmost one X i has S X i = S X ′ i . If two or more X i have S X i = S X ′ i , then these twosatisfy S X i ≤ n/ S X ′ ≤ n −
2, we have L X i =1 S X ′ i ≤ n/
2) + ( L − n − ≤ ( L − n − , proving eligibility.Set Ω = { F Ee − e j : 2 ≤ j ≤ n } . We have Ω X i = { F Ee − e j : j > S X i } .Upon taking g i suitable Weyl conjugates that permute letters we obtain L − [ i =1 Ad ( g i )Ω X i = Ω Y . where Y is an element of the torus of SU ( n ) of type SU ( m ) with m = n − ( L − n + P L − i =1 J i . The eligibility assumption ensures ( X L , Y ) is an eligible pair andit is clearly not exceptional. Now complete the argument using the L = 2 case inthe same manner as for type B n and C n . (cid:3) The many exceptional pairs in D n , ( n = 4 in particular), cause complicationsin proving the theorem for type D n . We will again prove the main theorem by aninduction argument for L = 2, but it will be convenient to begin the argument withtype D . In the next lemma we will prove that all eligible, non-exceptional pairsin D and D are absolutely continuous. This will start the base case for us.We will actually begin with D . Usually D n is defined for n ≥
4, but that isbecause D is Lie isomorphic to type A . As the problem of characterizing the L -tuples in type A has already been done we can use this characterization, togetherwith the induction step, Prop. 3, to handle most of the eligible, non-exceptionalpairs in D and D . This approach will work whenever the reduced pair is knownto be an absolutely continuous pair (in D or D , respectively). There will still bea few remaining pairs to consider and these will be handled directly by verifyingWright’s criteria for absolute continuity, Thm. 2. Lemma . All the eligible, non-exceptional pairs in D and D are absolutelycontinuous. Proof.
As explained above, we begin the proof by considering D . Underthe Lie isomorphism between D and A , any subsystem of type D in D isisomorphic to one of type A × A , type D is isomorphic to one of type A (or SU (1)) and types SU ( j ) for j = 1 , , A , it is easy to check that all pairs ( X, Y ) in D areabsolutely continuous except those of type ( D , D ), ( D , SU (3)), ( SU (3) , SU (3))and ( SU (3) , SU (2)), the first two of these being not eligible and the latter two,exceptional.Case D : Prop. 3 guarantees that all eligible, non-exceptional pairs, ( X, Y ),in D are absolutely continuous, except when the reduced pair, ( X ′ , Y ′ ) , is one ofthe four pairs listed above. Furthermore, because we have already seen that thepair ( X ′ , Y ′ ) is eligible whenever ( X, Y ) is an eligible, non-exceptional pair, wewill only need to give a special argument for those pairs (
X, Y ) where X ′ is type SU (3) and Y ′ is either type SU (3) or SU (2) (the latter being type SU (2) × D or SU (2) × SU (1)).Thus we are left to study the pairs ( X, Y ) where X is of type SU (4) and Y is one of type SU (4), type SU (3) (to be more precise, either type SU (3) × D or SU (3) × SU (1)), type SU (2) × D or SU (2) × SU (2). However, these are allexceptional pairs except when X is type SU (4), Y is of type SU (2) × SU (2) andΦ Y is not Weyl conjugate to a subset of Φ X .To prove this last pair is absolutely continuous, we verify the criteria of Thm. 2(with X = X and X = Y ) and follow the notation there. Here we have | Φ | = 24and | Φ X | + | Φ X | = 12 + 4 = 16. The rank 3, root subsystems, Ψ of D , are thoseof type D , SU (4) (two non-Weyl conjugate subsystems) and D × SU (2).When Ψ is type D × SU (2), then | Ψ | = 6. Thus we even have | Φ | − | Ψ | − ≥| Φ X | + | Φ X | , so (3.3) clearly holds. When Ψ is type D , then | Ψ | = 12. However, | Φ X ∩ σ (Ψ) | = 6 and | Φ X ∩ σ (Ψ) | ≥ σ ∈ W because σ (Ψ) mustcontain ± e i ± e j , ± e i ± e k , ± e j ± e k for three choices of letters i, j, k . Thus the LHSof (3.3) is 11, while the RHS is at most 8.Now assume Ψ is type SU (4). First, suppose Ψ is Weyl conjugate to the set ofannihilators of X . Since we need to calculate the intersection of Φ X j with all Weylconjugates of Ψ, there is no loss of generality in assuming Φ X = Ψ = { e i − e j : 1 ≤ BSOLUTE CONTINUITY 27 i = j ≤ } . By assumption, Φ X is not Weyl conjugate to a subset of Φ X , thusthere is also no loss of generality in assuming Φ X = {± ( e − e ) , ± ( e + e ) } .The reader can check that | Φ X ∩ σ (Ψ) | is minimal when we take the choice of σ ∈ W that switches two signs and in this case | Φ X ∩ σ (Ψ) | = 4. Similarly, it canbe shown that if σ is any Weyl conjugate, then | Φ X ∩ σ (Ψ) | ≥
2, so that again theLHS of (3.3) is 11 and the RHS is at most 10.Finally, suppose Ψ is not Weyl conjugate to Φ X . Without loss of generalitywe can assume Ψ is as before and Φ X = { e i − e j , ± ( e + e j ) : 1 ≤ i = j ≤ } .Again | Φ X ∩ σ (Ψ) | is minimal when σ is the Weyl element that switches two signs,but in this case | Φ X ∩ σ (Ψ) | = 6. This is already enough to establish (3.3) andcompletes the argument that ( X, Y ) is an absolutely continuous pair.Case D : Again, Prop. 3 implies we only need to study the eligible non-exceptional pairs, ( X, Y ) in D , where the reduced pair has X ′ of type SU (4)and Y ′ one of type SU (4), SU (3), SU (2) × D , or SU (2) × SU (2). Since thepairs ( SU (5) , SU (5) and ( SU (5) , SU (4)) are exceptional and the pair ( SU (5) , D × SU (2)) is not eligible, this reduces the problem to the study of the pairs ( X, Y )where X is of type SU (5) and Y is either of type SU (3) × D or SU (3) × SU (2).Further, since the set of annihilators of an element of type SU (3) × SU (2) is con-tained in the set of annihilators of an element of type SU (3) × D , it will suffice toprove that the pair ( SU (5), SU (3) × D ) is absolutely continuous.For this, we again use Thm. 2. In D , the rank 4 root subsystems Ψ whichwe must study are those of type D , SU (5), D × SU (2) and D × SU (3), withcardinalities 24, 20, 14 and 10, respectively. The cardinality of Φ is 40, while | Φ X | = 20 and | Φ X | = 10.Let Λ be a root subsystem of type D , D or D in D . It is easy to see thatif Λ is a root subsystem of type D j in D , with j = 2 , ,
4, then | Φ X ∩ Λ | = | Λ | .Furthermore, | Φ X ∩ Λ | = 6 whenever Λ is type D . Since the action of a Weylelement preserves the type of a root subsystem, these calculations can be used toshow that (3.3) holds if Ψ is type D , D × SU (2) or D × SU (3).When Ψ is type SU (5), then | Φ X ∩ σ (Ψ) | ≥ σ (with the minimumoccurring when σ is two sign changes). Moreover, | Φ X ∩ σ (Ψ) | ≥ SU (5), SU (3) × D ) is absolutelycontinuous and completes the base case arguments. (cid:3) Further complications arise with type D n because of the fact that when X is of type SU ( n ), then µ X is not absolutely continuous. We have already seenthis complication in the proof of Prop. 3 (when L = 2), but it presents furtherdifficulties when L >
2. To handle this, we introduce the following terminology forthe remainder of the proof.
Definition . We will say that X is almost dominant SU type if X is type D J × SU ( s ) × · · · × SU ( s t ) where J ≤ P s i .Of course, if X is dominant SU type, then it is almost dominant SU type.However, X is also almost dominant SU type if X is dominant D type, but X ′ isdominant SU type, for instance. If X is almost dominant SU type and not type SU ( n ), then [ , Thm. 8.2], implies µ X ∈ L . Here are some additional properties. Lemma . Suppose X , X are almost dominant SU type in D n and X = 0 .(i) If neither X nor X are type SU ( n ) , then µ X ∗ µ X ∈ L . (ii) If X and X are both type SU ( n ) and X is not, then µ X ∗ µ X ∗ µ X ∈ L . (iii) More generally, if X is type SU ( n ) and X is not, then µ X ∗ µ X ∗ µ X ∈ L . (iv) If n ≥ (or n = 4 ) and X (and X ) is almost dominant SU ( n ) type, then µ X ∗ µ X ∗ µ X ( ∗ µ X ) ∈ L . Proof. (i) follows from [ ] as remarked above. The fact that it is absolutelycontinuous, which is actually all we will need for our application, also follows fromthe L = 2 part of the proof of the main theorem since ( X , X ) is an eligible,non-exceptional pair.(iv) holds similarly from [ ] since µ X ∈ L whenever X is almost dominate SU type and n ≥ , and µ X ∈ L when n = 4. (Alternatively, absolute continuity canbe checked from Theorem 2.)For (ii) and (iii) we proceed by induction on n , noting that according to themain theorem, as already established for all L ≥ A n ,all triples in A (equivalently, D ) are absolutely continuous, except when all threeare type SU (3).Now assume n ≥
4. We putΩ = { F Ee ± e j : j = 2 , . . . , n, F = R, I } . (ii): Here X ′ , X ′ will be of type SU ( n −
1) in D n − , while X ′ is not, so theinduction hypothesis applies. Without loss of generality we can assumeΩ X = { F Ee + e j : j ≥ , F = R, I } and Ω X either coincides with Ω X orΩ X = { F Ee + e j , F Ee − e n : j ≤ n − , F = R, I } . As X is not of type SU ( n ), Ω X contains { F Ee ± e n } . Let g be the Weylconjugate changing the signs of 2 , . . . , n − n if needed to be an even signchange). Then Ω X ∪ Ad ( g )(Ω X ) contains all of Ω except for possibly { F Ee − e n : F = R, I } . If Ω = { F Ee + e n } , we haveΩ X ∪ Ad ( g )(Ω X ) ∪ (Ω X (cid:31) Ω ) = Ω . Taking M = REe + e n one can verify that the hypotheses of the general strategy,Prop. 2, are all satisfied. Consequently, ( X , X , X ) is absolutely continuous.(iii): We define X ′ and X ′ as usual, but will re-define X ′ so that it continuesto be of almost dominant SU type and not type SU ( n −
1) (so that we will beable to apply the induction hypothesis). This can be achieved by defining X ′ tobe type D J − × SU ( s ) × · · · × SU ( s t ) if X is type D J × SU ( s ) × · · · × SU ( s t )with J >
1, or defining X ′ to be type D J × SU ( s − ) × · · · × SU ( s t ) if J = 0 or1 and s = max s j . The fact that X is almost dominant SU type ensures that J ≤ n/
2, so whether X is dominant SU type or not, S X ≤ n and thus ( X , X )is an eligible pair. Further, X , X are not both of type SU ( n ).The arguments given in Prop. 3, Case 2 or 3 depending on the situation, can beapplied to prove there is some g ∈ D n − such that sp { Ω X (cid:31) Ω , Ad ( g )(Ω X ) } = sp Ωwhere Ω is taken to be the choice of F Ee + e n or F Ee − e n that belongs to Ω X .Therefore sp { Ω X (cid:31) Ω , Ad ( g ) (Ω X ) , Ω X } = sp Ω . Now take M = REe ± e n (depending on the choice of Ω ) and apply the generalstrategy. (cid:3) BSOLUTE CONTINUITY 29
We are now ready to conclude the proof of Theorem 1 by completing the proofof sufficiency for absolute continuity in type D n . Proof of Theorem 1 continued.
Sufficient conditions for Absolutecontinuity for Lie type D n : Case L = 2. Lemma 6 starts the induction argument for type D n . Now,inductively assume that all eligible, non-exceptional pairs in D n − , with n ≥ X ′ , Y ′ ) is eligible. If it is anexceptional pair, then it must be either of type ( SU ( n − SU ( n − SU ( n − SU ( n − SU ( n −
2) could be type SU ( n − × D or SU ( n − × SU (1)). But then ( X, Y ) must also have been an exceptional pair in D n ,which is a contradiction. By the induction assumption, ( X ′ , Y ′ ) is an absolutelycontinuous pair and hence Prop. 3 implies that ( X, Y ) is absolutely continuous.Case L ≥
3. Again, we give an induction argument. The base case, D , will bediscussed at the conclusion of the proof. So assume n ≥ X , . . . , X L ) is aneligible L -tuple in D n . We note that there are no exceptional L -tuples in D n for n ≥ L ≥ { F Ee ± e j : j = 2 , . . . , n } . More care is needed in this situation then for the Lie algebras of type B n and C n , since the fact that µ X / ∈ L when X is of type SU ( n ) means, for example, that wecannot immediately assume that at most one X i is dominant SU type, as we didin the argument for those Lie types. Here is where Lemma 7 will be useful.If three or more X i are dominant SU type, then the induction argument is noteven necessary as Lemma 7 (iv) implies that their convolution is already in L andhence is absolutely continuous.If two X i (say, X , X ) are both dominant SU type and some X j , say X , isnot, then we call upon one of the first three parts of the lemma.So we can assume there is at most one X i that is dominant SU type, say X . Ifthere is some X j , other than X , with X ′ j of dominant SU type, then X j is almostdominant SU and not type SU ( n ). Apply the appropriate part of Lemma 7 with X , X equal to this X j , and X any other X i .If all X ′ j , other than j = 1 , remain dominant D type, then the calculations usedin the type B n or C n case show that ( X ′ , . . . , X ′ L ) is an eligible, non-exceptionaltuple in D n − . For the induction step we argue in the same fashion as we did forthe Lie algebras of type B n or C n in the same situation.Finally, assume all X j are dominant D type. If two or more X ′ j are dominant SU type, then the corresponding two X j are almost dominant SU type and not oftype SU ( n ). Their convolution is even in L . If at most one X ′ j is dominant SU type, then ( X ′ , . . . , X ′ L ) is an eligible tuple, so the induction hypothesis applies.The induction step is the same as for the corresponding situation with type B n or C n . To conclude, we must establish the base case, n = 4. Since µ X ∈ L forany non-trivial X in the Lie algebra of type D , every L -tuple with L ≥ L = 3. The induction argument above can be applied to( X , X , X ) provided at most one X j is dominant SU type, hence for such triplesit suffices to check that ( X ′ , X ′ , X ′ ) in D is an absolutely continuous triple. But this follows from the main theorem for type A n , since all triples in A , except whenall X i are type SU (3), are absolutely continuous.If two or three X j are dominant SU type, but at least one X i is not of type SU (4), Lemma 7 gives the result.If all three X i are type SU (4) and their annihilating root systems are Weylconjugate, then the triple, ( X , X , X ), is exceptional. Thus we can assume theannihilating root systems are not Weyl conjugate. As the arguments are symmetric,there is no loss of generality in assuming that the set of annihilating roots for X coincides with that of X and is given byΦ X = Φ X = { e i − e j : 1 ≤ i = j ≤ } , while Φ X = { e i − e j , ± ( e + e k ) : 1 ≤ i = j ≤ , k = 1 , , } . We will again call upon Thm. 2 to check the absolute continuity of the triple. Theroot systems Ψ of rank 3 that we must consider are those of type D , SU (4) (twonon-Weyl conjugate root subsystems) and D × SU (2).Of course, P i =1 | Φ X i | = 36 and | Φ | = 24 when Φ is the root system of D .When Ψ is type D × SU (2), then | Ψ | = 6, and as Ψ intersects non-trivially anyroot subsystem of type SU (4), the inequality (3.3) is clear. When Ψ is type D itis easy to see that | Φ X i ∩ σ (Ψ) | ≥ i and any Weyl element σ .When Ψ is type SU (4), then | Φ X i ∩ σ (Ψ) | ≥
4. However, as noted in the L = 2argument, if Ψ and Φ X i are non-Weyl conjugate subsystems of type SU (4), thenthis lower bound can be improved to ˙6. Consequently, 2( | Φ | − | Ψ | ) − − (4 + 4 + 6) = 22, so the inequality holds.This completes the base case argument and hence the proof of Theorem 1. (cid:3)
7. Applications7.1. Consequences of the Main Theorem.
An element X ∈ t n is said tobe regular if its set of annihilating roots is empty. These would be the elementsof type SU (1) × · · · × SU (1) (in any Lie algebra) or D × SU (1) × · · · × SU (1) intype D n , and hence have S X = 1 or 2. In [ ] it was shown that the convolutionof the orbital measures of any two regular elements is absolutely continuous. Themethods used there could be used to prove, more generally, that the convolutionof any orbital measure with the orbital measure of a regular element is absolutelycontinuous. Our theorem shows that more is true. Corollary . Let
X, Y be non-zero elements in the Lie algebra of type B n , C n , or D n . If S Y ≤ , then µ X ∗ µ Y is absolutely continuous and O X + O Y hasnon-empty interior, except if ( X, Y ) is the exceptional pair ( SU (4) , SU (2) × SU (2)) in D where the annihilating roots of Y are a subset of a Weyl conjugate of thoseof X . Proof.
This is immediate from the theorem since any non-zero X has S X ≤ n −
1) (with equality only if X is type B n − ( C n − or D n − )). (cid:3) Corollary . If ( X , . . . , X L ) is an eligible, non-exceptional L -tuple of ma-trices in any of the classical Lie algebras, then there are unitarily similar matrices, g − i X i g i , with the property that P Li =1 g − i X i g i has distinct eigenvalues. BSOLUTE CONTINUITY 31
Proof.
This follows from the main theorem because any subset of these matrixgroups with non-empty interior must contain an element with distinct eigenvalues.Indeed, the elements with distinct eigenvalues are dense. (cid:3)
On the other hand, if X i are matrices in one of the classical Lie algebrasand there are unitarily similar matrices, g − i X i g i ∈ O X i , with the property that P Li =1 g − i X i g i has distinct eigenvalues, then P O X i contains an element Y with S Y ≤
2. (Indeed, Y is either type SU (1) or type B , C or D .) As noted in the firstcorollary, O X + O Y has non-empty interior for any X = 0 and thus µ X ∗· · · µ X L ∗ µ X is absolutely continuous for any X = 0. It would be interesting to characterize the L -tuples for which P O X i contains a matrix with distinct eigenvalues.It is known that any n -fold sum of non-trivial orbits in B n , C n or D n hasnon-empty interior. More can be said. Corollary . Let n ≥ . If X i are non-zero elements in B n ( C n or D n ) for i = 1 , . . . , n − , then O X + · · · + O X n − has empty interior if and only if all X i are type B n − ( C n − or D n − ). Proof.
Suppose some X i , say X n − , is not type B n − . Then S X n − ≤ n − S X i ≤ n − n − X i =1 S X i ≤ n − n −
2) + 2( n − ≤ n ( n − . Thus ( X , . . . , X n − ) is eligible and non-exceptional and hence the sum of the orbitshas non-empty interior. Since all root subsystems of type B n − are Weyl conjugate,the converse follows from the fact that if X is type B n − , then µ n − X is not absolutelycontinuous [ ]. (cid:3) We leave it as an exercise for the reader to determine the choice of n − n ≤ A n .We note that for type A n , B n and C n our proof required the use of [ ] only tostart the induction process. In the proof given in [ ] an induction argument wasalso used and the base cases were simply done directly. That approach could havebeen taken here, as well. For type D n our proof also used [ ] to establish thatwhen X was type SU ( n ), then µ X was absolutely continuous for n ≥ µ X was absolutely continuous when n = 4. In fact, the argument that was given therefor these special types actually showed that Theorem 2 was satisfied. Thus, ourtheorem gives another way to deduce the formulas of [ ]. For example, we have Corollary . Suppose g is type B n and X is type B J × SU ( s ) ×· · ·× SU ( s m ) .(i) If X is dominant B type, then µ LX is absolutely continuous (and ( L ) O X hasnon-empty interior ) if and only if L ≥ n/ ( n − J ) .(ii) If X is dominant SU type, then µ X is absolutely continuous. Similar statements can be made for the other types, taking into account theexceptional cases.
Remark . (i) We have not been able to determine if the pair of type ( SU ( n ), SU ( n − D n for n ≥ n = 6 ,
7. We remark that Prop. 3 shows that if such apair is absolutely continuous for any n , say n = n then, being an eligible pair, itis absolutely continuous for all n > n . (ii) It remains open to solve the analogous problem in the exceptional Lie alge-bras, those of type G , F , E , E , or E . In [ ] the minimal k ( X ) so that µ k ( X ) X is absolutely continuous was determined for each X in the compact exceptionalLie algebras. Although the abstract root theory machinery can be applied in thissetting, there is no underlying classical matrix algebra from which to derive thenecessary conditions. A related, but more challenging problem, is to determine which L -tuples,( x , . . . , x L ) ∈ G L , have the property that µ x ∗ · · · ∗ µ x L is absolutely continu-ous with respect to Haar measure on the group G , when µ x is the probabilitymeasure, invariant under the conjugation action of G on itself, and supported onthe conjugacy class generated by x , C x = { g − xg : g ∈ G } . In [ , Thm. 9.1], theminimum integer k ( x ) for which µ k ( x ) x is absolutely continuous was determined forall the classical Lie groups. The number k ( x ) depended on the type of the set ofannihilating roots of x , where in this setting by the set of annihilating roots wemean Φ x := { α ∈ Φ : α ( x ) ≡ π } . Again, by the type of x , we will mean the type of Φ x .This was extended by Wright [ ] to convolution products of (possibly) different µ x in the case of SU ( n ), obtaining the same characterization as for the Lie algebraproblem. In this subsection, we will obtain a similar result for all the classical Liegroups whenever the group elements x i = exp X i where X i ∈ g and x i ∈ G havethe same type.We need the following preliminary result, analogous to Prop. 1. Given x ∈ G ,we let N x := { RE α , IE α : α ( x ) = 0 mod 2 π } . Lemma . (c.f. [ ] , [ ] ) The measure µ x ∗ · · · ∗ µ x L on G n is absolutelycontinuous with respect to Haar measure on G n if and only if any of the followinghold:(i) The set L Q i =1 C x i ⊆ G n has non-empty interior;(ii) The set L Q i =1 C x i ⊆ G n has positive measure;(iii) There exists g i ∈ G n with g = Id , such that sp { Ad ( g i ) N x i : i = 1 , . . . , L } = g n . Proposition . Let x , . . . , x L ∈ G n and assume x i = exp X i for some X i ∈ g n where x i and X i have the same type. Then µ x ∗· · ·∗ µ x L is absolutely continuouswith respect to Haar measure on G n if and only if µ X ∗ · · · ∗ µ X L is absolutelycontinuous with respect to Lebesgue measure on g n . Moreover, L Q i =1 C x i has non-empty interior in G n if and only if P Li =1 O X i has non-empty interior in g n . Proof. If x i and X i are of the same type, then N x i = N X i . Consequently, sp { Ad ( g i ) N x i : i = 1 , . . . , L } = sp { Ad ( g i ) N X i : i = 1 , . . . , L } BSOLUTE CONTINUITY 33 and thus µ x ∗ · · · ∗ µ x L is absolutely continuous if and only if µ X ∗ · · · ∗ µ X L isabsolutely continuous. The latter statement holds as absolute continuity is equiva-lent to non-empty interior of either the product of conjugacy classes or the sum oforbits, depending on the setting. (cid:3) Remark . If x i = exp X i and µ X ∗ · · · ∗ µ X L is not absolutely continuous,then it still follows µ x ∗ · · · ∗ µ x L is not absolutely continuous. We simply note thatalways Φ X i ⊆ Φ x i .Consider the Lie group SU ( n ). Every conjugacy class contains a diagonalmatrix so in studying the measure µ x there is no loss in assuming x = diag (exp ia , . . . , exp ia n )where a j ∈ [0 , π ) and P a j ≡ π . Notice that x = exp X where X = diag ( ia , . . . ., ia n ) belongs to su ( n ). The root α = e j − e k acts on x (and X ) by α ( x ) = a j − a k . Thus Φ x = Φ X and so the Proposition applies to all L -tuples in SU ( n ).This is not true for the other classical Lie groups. For example, in SO (2 n + 1)(type B n ) there is an element x with Φ x = { e i ± e j : 1 ≤ i = j ≤ n } , i.e., oftype D n . This type does not arise in the Lie algebra. Indeed, the only element X ∈ so (2 n + 1) with Φ X ⊇ Φ x is X = 0. The element x has the property that µ nx ∈ L ( G ), but µ n − x is singular with respect to Haar measure on G . In contrast,any X with exp X = x has µ X ∈ L ( g ).These additional (and often more complicated) types of elements that can arisein the Lie groups make the problem of characterizing absolute continuity of orbitalmeasures on Lie groups more challenging than for Lie algebras. References [1] A. Dooley, J. Repka and N. Wildberger,
Sums of adjoint orbits , Linear and multilinear algebra (1993), 79–101.[2] A. Frumkin and A. Goldberger, On the distribution of the spectrum of the sum of two Her-mitian or real symmetric spaces , Adv. Appl. Math. (2006), 268–286.[3] P. Gratczyk and P. Sawyer, On the kernel of the product formula on symmetric spaces , J.Geom. Anal. (2004), 653–672.[4] P. Gratczyk and P. Sawyer, Absolute continuity of convolutions of orbital measures on Rie-mannian symmetric spaces , J. Func. Anal. (2010), 1759–1770.[5] P. Gratczyk and P. Sawyer,
A sharp criterion for the existence of the density in the productformula on symmetric spaces of Type A n , J. Lie Theory (2010), 751–766.[6] S.K. Gupta and K.E. Hare, Singularity of orbits in classical Lie algebras , Geom. Func. Anal. (2003), 815–844.[7] S.K. Gupta and K.E. Hare, Convolutions of generic orbital measures in compact symmetricspaces , Bull. Aust. Math. Soc. (2009), 513–522.[8] S.K. Gupta and K.E. Hare, L -singular dichotomy for orbital measures of classical compactLie groups , Adv. Math. (2009), 1521–1573.[9] S.K. Gupta, K.E. Hare and S. Seyfaddini, L -singular dichotomy for orbital measures ofclassical simple Lie algebras , Math. Zeit. (2009), 91–124.[10] K.E. Hare, D. Johnstone, F. Shi and M. Yeung, L -singular dichotomy for exceptional Liegroups and algebras , J. Aust. Math. Soc. (2013), 362–382.[11] S. Helgason, Differential geometry, Lie groups and symmetric spaces , Academic Press, NewYork, 1978.[12] J. Humphreys,
Introduction to Lie algebras and representation theory , Springer-Verlag, NewYork, 1972.[13] S. Lang,
Differential and Riemannian manifolds , Springer-Verlag, New York, 1995.[14] A. Knapp,
Lie groups beyond an introduction , Birkhauser, Verlag AG, 2002. [15] A. Knutson and T. Tao,
Honeycombs and sums of Hermitian matrices , Notices Amer. Math.Soc. (2001), 175–186.[16] M. Mimura and H. Toda, Topology of Lie groups , Trans. Math. Monographs 91, Trans. Amer.Math. Soc., Providence, R.I., 1991.[17] D. Ragozin,
Zonal measure algebras on isotropy irreducible homogeneous spaces , J. Func.Anal. (1974), 355–376.[18] D. Ragozin, Central measures on compact simple Lie groups , J. Func. Anal. (1972), 212–229.[19] F. Ricci and E. Stein, Harmonic analysis on nilpotent groups and singular integrals. II.Singular kernels supported on submanifolds , J. Func. Anal. (1988), 56–84.[20] F. Ricci and E. Stein, Harmonic analysis on nilpotent groups and singular integrals. III.Fractional integration along manifolds,
J. Func. Anal. (1989), 360–389.[21] V.S. Varadarajan, Lie groups and Lie algebras and their representations , Springer-Verlag,New York, 1984.[22] N. Wildberger,
On a relationship between adjoint orbits and conjugacy classes of a Lie group ,Can. Math. Bull. (1990), 297–304.[23] A. Wright, Sums of adjoint orbits and L -singular dichotomy for SU ( m ), Adv. Math. (2011) 253–266. Dept. of Mathematics and Statistics, Sultan Qaboos University, P.O. Box 36 AlKhodh 123, Sultanate of Oman
E-mail address : [email protected] Dept. of Pure Mathematics, University of Waterloo, Waterloo, Ont., Canada, N2L3G1
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