1 Non separated points in the dual spaces of semi simple Lie groups Let G be a connected semi simple Lie group with finite center, g its Lie algebra, g c = g R C, G the universal enveloping algebra of g c and a maximal compact subgroup of G. Let π be a continuous representation of G on a Banach space H. For ˆ let H denote the isotypic submodule of H; H is the range of the continuous projection = dim() ξ (k)π(k)dk P π where ξ is the character of and dk is the normalized Haar measure on. We call π admissible if the subspace H is finitedimensional for every ˆ. In this case set (π : ) = dim H dim and H = ˆ H (algebraic direct sum). Then H is a G module anbd we denote the corresponding representation of G again by π. For any finite set S ˆ we set H S = S H. Let π and π be admissible representations of G on H and H, respectively. They are called infinitesimally equivalent if the corresponding G modules H and H are isomorphic. If π and π are unitary and irreducible, this is equivalent to the unitary equivalence. Let g = k p be the Cartan decomposition of g coirresponding to. Choose a maximal Abelian subspace a in p and set M = Z (a) and m = Z k (a) (Z denotes centralizers). Let d be a Cartan subalgebra of m. Set R = R(g c, h c ) h 0, h = d a, h 0 = id a, R 0 = R(m c, d c ) d, Σ=R(g, a) a. Let W be the Weyl group of R acting on h c and h c. Choose compatible orders on h 0,id and a and let R +,R 0 + and Σ + denote the corresponding positive roots. Set n = g α, N = exp n, A = exp a; α Σ + m α = dim g α, α Σ; ρ c = 1 α, ρ 0 = 1 α, ρ = 1 m α α. α R + α Σ + α R + 0
Let H : G a, ν: G N, κ : G be the analytic maps corresponding to the Iwasawa decomposition G = AN : x = exp(h(x))ν(x)κ(x), x G. Let σ ˆM be realized on a finitedimensional Hilbert space V σ and set H σ = {f L (, V σ ); f(mk) =σ(m)f(k), m M, k } (a closed subspace of the Hilbert space L (, V σ )). For any λ a c define the so called elementary representation π σ,λ of G on H σ by [π σ,λ (x)f](k) =e (λ+ρ)(h(kx)) f(κ(kx)), k, x G, f H σ. Then π σ,λ is an admissible representation of G. The restriction π σ,λ is independent of λ and it is the representation of induced by σ. Furthermore, by Frobenius reciprocity (π σ,λ : ) =( : σ), ˆ, σ ˆM, λ a c. Let Z be the center of G and let ν χ ν be the mapping from h c to Hom(Z, C) which induces the Harish Chandra s bijection from the set h c/w of W orbits in h c onto Hom(Z, C). The representation π σ,λ admits infinitesimal character and it is equal to χ ν(σ,λ), where ν(σ, λ) d c =Λ σ + ρ 0, ν(σ, λ) a c = λ; Λ σ is the highest weight (with respect to R + 0 ) of the representation of m c derived from σ. We shall need the following result by W. Casselman: Theorem 1. Every irreducible admissible representation π of G admits an infinitesimal embedding into some π σ,λ (i.e. π is infinitesimally equivalent to a subrepresentation of some π σ,λ ). As far as I know the proof of this theorem has not been published yet, but my colleague D. Miličić has written down the complete proof, and it turns out to be considerably simpler then the proof of Harish Chandra s subqotient theorem. It is based on the following: 1. (Osborne s theorem) H is finitely generated as U(n c ) module, where U(n c ) is the universal enveloping algebra of n c.. The study of asymptotic behaviour of finite matrix coefficients of π shows that the subspace H (n) =span {π(x)v; v H, X n} is different from H. Let V be the quotient space H /H (n). By 1. V if finite dimensional and it is naturally a MAN module on which N acts trivially. 3. (Casselman s reciprocity theorem) Let σ ˆM and λ a c and let V σ be endowed with MAN module structure by man e (λ+ρ)(log a) σ(m), m M, a A, n N. For T Hom G (H, H σ,λ ) let T : V V σ be defined by T (v + H (n)) = (Tv)(e), v H. Then T T is an isomorphism from Hom G (H, H σ,λ ) onto Hom MAN(V,V σ ).
3 4. By. we can find σ and λ such that Hom MAN (V,V σ ) {0}. By 3. it follows that Hom G (H, H σ,λ ) {0}. πbeing irreducible, there exists an infinitesimal embeding of π into π σ,λ. Let D(G) be the space of compactly supported C functions on G and let D (G) denote the space of distributions on G endowed with the weak topology. Let π be an irreducible admissible representation of G on a Banach space H. By theorem 1. up to infinitesimal equivalence we can suppose that H is a Hilbert space. By Harish Chandra s results for any f D(G) π(f) is a trace class operator and if we set Θ π (f) =Trπ(f), f D(G), then Θ π D (G). Furthermore, if π and π are infinitesimally equivalent, then Θ π =Θ π, and if π 1,π,...,π n are pairwise infinitesimally inequivalent, then Θ π1, Θ π,...,θ πn are linearly independent. be the corre- For an admissible representation π on a Banach space H and for ˆ let Φ π sponding trace spherical function For f D(G) set Φ π (x) =Tr(P π π(x)p π ), x G. Φ π (f) =Tr(P π π(f)p π )= Of course, dx denotes a fixed Haar measure on G. G f(x)φ π (x)dx. If π is of finite length, Θ π is also well defined and it equals the sum of characters of the successive subquotients in any Jordan Hölder series of π. Obvously we have Θ π (f) = Φ π (f), f D(G). ˆ We shall need the following result from PhD thesis of D. Miličić: Theorem. (i) Let ξ : Ĝ D (G) be the canonical injection, i.e. ξ(π) =Θ π. The closure ξ(ĝ) of ξ(ĝ) in D (G) consists of distributions of the form m(ω)θ ω, m(ω) N, Ω Ĝ finite. ω Ω (ii) If Θ= ω Ω m(ω)θ ω is in the closure ξ(ĝ) of ξ(ĝ) set T (Θ) = Ω. For any subset S of Ĝ its closure in Ĝ is S = { } T (Θ); Θ ξ(s). (iii) If (π n ) n N is a sequence in Ĝ and if Ω denotes the set of all its limits, there exists a subsequence (π nk ) k N such that the sequence ( Θ πnk )k N converges in D (G). For any such subsequence ( ) Ω=T lim Θ π nk. k
4 (iv) Let (π n ) n N be a sequence in Suppose that Ĝ, let Ω be a finite subset of Ĝ and m(ω) N for any ω Ω. lim Θ pi n = m(ω)θ ω. o Ω Then lim Φπn (f) = o Ω m(ω)φ ω (f), ˆ, f D(G). This theorem is deduced from Theorem 7. in [1] by showing that C (G) isac algebra with bounded trace and that D(G) is an admissible subalgebra of C (G). The proof of these facts is an adaptation of Harish Chandra s proof that Θ π are distributions. Let α denote the representation k (Ad k) p c of on p c. For ˆ set S() ={γ ˆ; (α : γ) > 0}. Lemma 1. Let π be an admissible representation of G on H, ˆ and X g c. Then π(x)h H S(). Proof: This follows from g c = k c p c and from the fact that X i v i π(x i )v i, i i X i p c, v i H, is a morphism from p c H into H. Lemma. Let σ ˆM, ˆ and X g c. Then (λ, f) π σ,λ (X)f is a continuous mapping from a c H σ into Hσ S(). Proof: For k, f H σ and X g c we have [ π σ,λ (X)f ] (k) = d dt e(λ+ρ)(h(k exp tx)) f(k(k exp tx)) = t=0 where =(λ + ρ)(h(k, X))f(k)+(X(f k))(k), H(k, X) = d dt H(k exp tx)) t=0, k, X g c. Now, H σ consists of C (even analytic) functions V σ. Furthermore, the space H σ S is finite dimensional for any finite S ˆ, hence the usual topology on this finite dimensional space is the one induced from C (, V σ ). The assertion follows from this and from Lemma 1. Lemma 3. For ϕ D(G), σ ˆM and ˆ the mapping (λ, f, g) ( π σ,λ (ϕ)f g ) from a c Hσ Hσ into C is continuous. Proof: We have ( π σ,λ (ϕ)f g ) = The assertion follows by easy estimation. G ϕ(x)e (λ+ρ)(h(kx)) (f(k(kx)) g(k)) V σdk dx. We are now able to prove a result on non separated points in Ĝ.
5 Theorem 3. Let Ω be a finite subset of Ĝ such that there exists a sequence (π n) n N in Ĝ having Ω as the set of all limits. Then there exist σ ˆM and λ 0 a c and a subrepresentation π 0 of π σ,λ such that Ω is the set of all infinitesimal equivalence classes of irreducible subquotients of π 0. Proof: Let (π n ) n N be a sequence in Theorem. we can suppose that Ĝ such that Ω is the set of all its limits. By (iii) of and by (iv) of the same theorem we have then lim Θ π n = m(ω)θ ω, m(ω) N, o Ω lim Φπn (ϕ) = o Ω m(ω)φ ω (ϕ), ˆ, ϕ D(G). (1) By Theorem 1. for every n N there exist σ n ˆM and λ n a c such that π n is infinitesimally equivalent to a subrepresentation of π σn,λn. Hence, we can and shall identify H (π n ) with a G submodule (with respect to π σn,λ:n )ofh σn. Thus, χ ν(σ n,λ n) is the uinfinitesimal character of the representation π n. Let χ 0 be the infinitesimal character of the members of Ω. Due to a theorem by P. Bernat and J. Dixmier χ 0 (z) = lim χ ν(σn,λ n)(z) z Z. Now, ν χ ν induces a homeomorphism of h c/w (with the quotient topology) onto Hom(Z, C) (with the pointwise convergence topology). Therefore, the sequence (Wν(σ n,λ n )) n N in h c /W converges to the W orbit in h c associated to χ 0.W being finite, we can suppose (by passing to a subsequence of (π n ) if necessary) that the sequence (ν(σ n,λ n )) n N converges in h c. Set ν 0 = lim ν(σ n,λ n ). Then ν(σ n,λ n ) d c =Λ σn + ρ 0 converges to ν 0 d c in d c. But the set {Λ σ; σ ˆM} is discrete in d c, hence we can assume (by passing to a subsequence again) that Λ σ n + ρ 0 = ν o d c for every n N. The group M has finitely many connected components, hence for a given σ ˆM the set N(σ) ={τ ˆM; Λ τ =Λ σ } is finite and sup {#N(σ); σ ˆM} < +. Therefore, by taking a subsequence again we can suppose that σ n = σ m for all n, m N. Set σ = σ n,n N. Furthermore, set π λ = π σ,λ,λ a c, and H = H σ. Now, H πn H for every n N and H πn = H πn H, ˆ. Let λ 0 = lim λ n = ν 0 a c. Fix ˆ. The space Grass(H ) of all subspaces of H with the usual topology is compact. Hence, a subsequence of (H πn ) n N convereges in Grass(H ). ˆ being countabel, using a diagonal procedure we can pass to a subsequence of (π n ) n N (denoted again by (π n ) n N ) such that the sequence (H πn ) n N converges in Grass(H ). Denote by H 0 its limit and set V = ˆ H 0.
6 Let ˆ and f 0 H 0 be arbitrary. Choose f n H πn,n N, so that f 0 = lim f n. For any X g c we have by Lemma 1. π λ 0 (X)f 0 = lim π λn (X)f n lim H πn S() = Hγ 0 V. γ S() It follows that V is a G submodule of H eith respect to π λ 0, hence its closure H 0 in H is a G submodule for π λ 0. Let π 0 be the corresponding subrepresentation of π λ 0. Let Ω 1 be the set of all infinitesimal equivalence classes of irreducible subquotients of π 0 and for ω Ω 1 let n(ω) denote the multiplicity of ω in a Jordan Hölder series of π 0. Fix ˆ. Let n 0 N be such that dim H πn = dim H 0 = m n n 0. For any n n 0 choose an orthonormal basis (f1 n,fn,...,fn m )ofhπn in such a way that this sequence of orthornormal bases converges to an orthonormal basis (f 1,f,...,f n )ofh 0 : f j = lim fj n, By Lemma 3. for any ϕ D(G) we have lim Φπn (ϕ) = lim = m j=1 m j=1 j =1,,...,m. (π n (ϕ)ϕ n j f n j ) = lim m (π λn (ϕ)fj n ϕn j )= j=1 (π λ 0 (ϕ)f j f j )=Φ π 0 (ϕ) = ω Ω 1 n(ω)φ ω (ϕ). Therefore, for arbitrary ˆ and ϕ D(G) it follows from (1) m(ω)φ ω (ϕ) = n(ω)φ ω (ϕ). ω Ω 1 Summing over all ˆ we get Thus, Ω 1 =Ω. REFERENCES: ω Ω m(ω)θ ω = n(ω)θ ω. o Ω 1 ω Ω [1] D. Miličić, On C algebras with bounded trace, Glasnik Mat. 8(1973),7 1 Comment: In my first letter I have mentioned that π, π Ĝ are non separated if and only if they are infinitesimally equivalent to subquotients of the same elementary representation. The only if part follows from Theorem 3. But unfortunately the if part is not true. The simplest counter example is found in the case G = Sl(, R). Namely, in this case dte discrete series representations occur in elementary representations in pairs which are separated in Ĝ. Similar thing happens in the cases G = SU(n, 1) and G = SO(n, 1), n. But in all these cases if and only if holds true in Ĝ for representations with trivial infinitesimal character, even in Ĝ \ Ĝd, where Ĝd denotes the set of infinitesimal equivalence classes of discrete series representations. If we consider the non unitary dual space of G introduced by Fell instead of Ĝ, the if part is easily proven, because for any σ ˆM the set {λ a c ; πσ,λ is irreducible} is dense in a c. Probably the only if part holds true also in the non unitary dual space, but to prove this it is necessary to generalize the Miličić s theorem to include the case of non unitary representations.