Intertwining integrals on completely solvable Lie groups

Transcription

1 s on s on completely solvable Lie June 16, 2011

2 In this talk I shall explain a topic which interests me in my collaboration with Ali Baklouti and Jean Ludwig. s on

3 In this talk I shall explain a topic which interests me in my collaboration with Ali Baklouti and Jean Ludwig. Let G be a locally compact topological group and H a closed subgroup of G. We note G, H the modular functions of G, H and put H,G (h) = H(h) G (h) s on for h H. We denote by K(G, H) the space of all C-valued continuous functions φ on G with compact support modulo H from right and satisfying the covariance relation φ(gh) = H,G (h)φ(g) (g G, h H).

4 Then, G acts on K(G, H) by left translation and there exists on K(G, H) an G-invariant positive linear form ν G,H, unique up to a scalar multiplication. s on

5 Then, G acts on K(G, H) by left translation and there exists on K(G, H) an G-invariant positive linear form ν G,H, unique up to a scalar multiplication. We write ν G,H (φ) = G/H φ(g)dν G,H (g) s on for φ K(G, H).

6 Then, G acts on K(G, H) by left translation and there exists on K(G, H) an G-invariant positive linear form ν G,H, unique up to a scalar multiplication. We write ν G,H (φ) = G/H φ(g)dν G,H (g) s on for φ K(G, H). Now, let G = exp g be an exponential solvable Lie group with Lie algebra g. We denote by g the dual vector space of g. G acts on g by the coadjoint action and it is well known that the unitary dual Ĝ is realized by the orbit space g /G. This orbit method due to Kirillov, Bernat and Pukanszky provides us a powerful tool to study various questions concerning unitary representations of G.

7 s on Let H = exp h be an analytic subgroup of G and χ a unitary character of H. We consider the monomial representation τ = ind G Hχ of G induced up from χ. In order to study τ, we employ the orbit method. There exists a linear form f in g such that χ(exp X ) = e if (X ) (X h, i = 1). Then we write χ as χ f.

8 s on Here we have some questions on τ: irreducible decomposition, Penney s Plancherel formula, explicit intertwining operator between τ and its irreducible decomposition and the algebra of invariant differential operators associated with τ. Our field of research on these questions is the affine space Γ = f + h of g.

9 s on Here we have some questions on τ: irreducible decomposition, Penney s Plancherel formula, explicit intertwining operator between τ and its irreducible decomposition and the algebra of invariant differential operators associated with τ. Our field of research on these questions is the affine space Γ = f + h of g. At l g we take a (real) polarization b satisfying the Pukanszky condition and construct the monomial representation π l = ind G B χ l, where B = exp b. The Pukanszky condition means B l = l + b or equivalently that π l is the irreducible unitary representation of G corresponding to the coadjoint orbit G l. We denote by H l the Hilbert space of π l.

10 Now we are interested in a formal intertwining integral from π l to τ given by R(τ, π l )(ϕ)(g) = ϕ(gh)χ f (h) 1/2 H,G (h)dν H,H B(h) H/(H B) for ϕ H l. This intertwining integral or its associated generalized vector s on a l (ϕ) = R(τ, π l )(ϕ)(e) = ϕ(h)χ f (h) 1/2 H,G (h)dν H,H B(h), H/(H B) e being the unit element of G, comes in various aspects of our questions.

11 It was first suggested by Michèle Vergne and proved by Gérard Lion when G was nilpotent, l = f and h was a polarization. In that case, the space Hl of C -vectors of π l coincides with the Schwartz space of the rapidly decreasing functions and we have no problem to see the convergence of the integral on this space. s on

12 It was first suggested by Michèle Vergne and proved by Gérard Lion when G was nilpotent, l = f and h was a polarization. In that case, the space Hl of C -vectors of π l coincides with the Schwartz space of the rapidly decreasing functions and we have no problem to see the convergence of the integral on this space. Once we leave the nilpotent case, it becomes difficult to analyze this object. Is it possible to apply the linear form, namely the function in question does it satisfy the required covariance relation? If this is the case, what can we say about the convergence of the integral? Today from now on we assume that h is a polarization at f satisfying the Pukanszky condition. Hence, τ π l and translating the structure by an element of H, we also assume that f = l. s on

13 Let us make our situation more precise. Let G = exp g be an exponential solvable Lie group with Lie algebra g, f g and h 1, h 2 two polarizations at f satisfying the Pukanszky condition. Put H j = exp h j (j = 1, 2) and induce two monomial representations π 1 = ind G H 1 χ f and π 2 = ind G H 2 χ f of G. Thus, π 1 π 2. We denote by H j the Hilbert space of π j for j = 1, 2. s on

14 Let us make our situation more precise. Let G = exp g be an exponential solvable Lie group with Lie algebra g, f g and h 1, h 2 two polarizations at f satisfying the Pukanszky condition. Put H j = exp h j (j = 1, 2) and induce two monomial representations π 1 = ind G H 1 χ f and π 2 = ind G H 2 χ f of G. Thus, π 1 π 2. We denote by H j the Hilbert space of π j for j = 1, 2. First we have: s on Lemma For any X h 1 h 2, Tr ad h1 /(h 1 h 2 )X + Tr ad h2 /(h 1 h 2 )X = 0.

15 Let ϕ H 1 and g G. Using the above trace relation, we easily see that the function Φ g : H 2 h ϕ(gh)χ f (h) 1/2 H 2,G (h) satisfies the required covariance relation Φ g (hx) = (H1 H 2 ),H 2 (x)φ(h) (h H 2, x H 1 H 2 ). s on

16 Let ϕ H 1 and g G. Using the above trace relation, we easily see that the function Φ g : H 2 h ϕ(gh)χ f (h) 1/2 H 2,G (h) satisfies the required covariance relation Φ g (hx) = (H1 H 2 ),H 2 (x)φ(h) (h H 2, x H 1 H 2 ). s on Therefore, we can consider the formal intertwining operator T 21 (ϕ)(g) = H 2 /(H 1 H 2 ) ϕ(gh)χ f (h) 1/2 H 2,G (h)dν H 2,H 1 H 2 (h) for ϕ H 1. The main problem is the convergence of this integral.

17 s on It is easy to see that H 2 H 1 is locally closed in G so that the homogeneous space H 2 /(H 1 H 2 ) is homeomorphic to the quotient space H 2 H 1 /H 1. So, if the simple product H 2 H 1 is closed in G, then our integral would be convergent for ϕ H 1 having compact support modulo H 1. Once we believed to have proved this property, but unfortunately the proof was wrong.

18 s on It is easy to see that H 2 H 1 is locally closed in G so that the homogeneous space H 2 /(H 1 H 2 ) is homeomorphic to the quotient space H 2 H 1 /H 1. So, if the simple product H 2 H 1 is closed in G, then our integral would be convergent for ϕ H 1 having compact support modulo H 1. Once we believed to have proved this property, but unfortunately the proof was wrong. The Pukanszky condition gives us H 2 H 1 = {g G; g (f + h 1 ) (f + h 2 ) }, but the behavior of the affine space g (f + h 1 ) is difficult to see.

19 s on Here we list up the cases where H 2 H 1 turns out to be closed:

20 s on Here we list up the cases where H 2 H 1 turns out to be closed: (1) G is nilpotent (the product set of two analytic sub is always closed),

21 s on Here we list up the cases where H 2 H 1 turns out to be closed: (1) G is nilpotent (the product set of two analytic sub is always closed), (2) h 1, h 2 are admissible for a same nilpotent ideal n containing [g, g], it means that h j n (j = 1, 2) are polarizations at f n.

22 s on Here we list up the cases where H 2 H 1 turns out to be closed: (1) G is nilpotent (the product set of two analytic sub is always closed), (2) h 1, h 2 are admissible for a same nilpotent ideal n containing [g, g], it means that h j n (j = 1, 2) are polarizations at f n. (3) G is algebraic,

23 s on Here we list up the cases where H 2 H 1 turns out to be closed: (1) G is nilpotent (the product set of two analytic sub is always closed), (2) h 1, h 2 are admissible for a same nilpotent ideal n containing [g, g], it means that h j n (j = 1, 2) are polarizations at f n. (3) G is algebraic, (4) h 1 + h 2 is a subalgebra of g,

24 s on Here we list up the cases where H 2 H 1 turns out to be closed: (1) G is nilpotent (the product set of two analytic sub is always closed), (2) h 1, h 2 are admissible for a same nilpotent ideal n containing [g, g], it means that h j n (j = 1, 2) are polarizations at f n. (3) G is algebraic, (4) h 1 + h 2 is a subalgebra of g, (5) h 1 or h 2 is abelian,

25 s on (6) h 1 or h 2 is a Vergne polarization. Namely we take a sequence of subalgebras {0} =g 0 g 1 g n = g, dim(g j ) = j (0 j n)

26 s on (6) h 1 or h 2 is a Vergne polarization. Namely we take a sequence of subalgebras {0} =g 0 g 1 g n = g, dim(g j ) = j (0 j n) such that if g j is not an ideal then g j±1 are ideals and the adjoint action of g on g j+1 /g j 1 is irreducible. Put l j = l gj. Then n j=1 g j(l j ) gives a polarization at l satisfying the Pukanszky condition. Polarizations constructed in such a fashion are called Vergne polarizations.

27 When we apply the orbit method to exponential solvable Lie, we usually employ the induction on the dimension of G. If there exists a non-trivial ideal a of g on which f vanishes, we go down to the quotient algebra g/a. s on

28 When we apply the orbit method to exponential solvable Lie, we usually employ the induction on the dimension of G. If there exists a non-trivial ideal a of g on which f vanishes, we go down to the quotient algebra g/a. If there is no such ideal, then the center of g is at most one-dimensional and we take a minimal non-central ideal a of g. Then, s on a f = {X g; f ([X, a]) = {0}} is a proper subalgebra of g and we try to go down to G 0 = exp a f by using the modification h j h j = (h j a f ) + a (1 j 2) if h j is not contained in a f.

29 If we proceed by this process to prove the closeness of H 2 H 1, there is no problem when at least one of h j (1 j 2) is contained in a f. Suppose h j a f for 1 j 2. s on

30 If we proceed by this process to prove the closeness of H 2 H 1, there is no problem when at least one of h j (1 j 2) is contained in a f. Suppose h j a f for 1 j 2. Three cases are left to be examined. We can treat two cases: either h 1 h 2 a f, or and where h 0 j = h j a f. h 1 h 2 a f a h h 0 2, s on

31 s on We finally come to the last case where we are blocked: h 1 h 2 a f and a h h 0 2.

32 s on We finally come to the last case where we are blocked: h 1 h 2 a f and a h h 0 2. In many cases listed above where H 2 H 1 turns out to be closed, the Pukanszky condition allowed us to choose convenient coexponential basis. But actually I have no idea to treat the last case in this line.

33 Looking for some ideas on what happens, we calculate un example. we assume that g is completely solvable. Let n be a nilpotent ideal of g containing [g, g]. s on

34 Looking for some ideas on what happens, we calculate un example. we assume that g is completely solvable. Let n be a nilpotent ideal of g containing [g, g]. We do not enter into the details, but first we construct a certain sequence S of ideals of g: S : {0} =n 0 n 1 n 2 n m 1 n m = n, dim(n j /n j 1 ) = 1 (1 j m). s on

35 Looking for some ideas on what happens, we calculate un example. we assume that g is completely solvable. Let n be a nilpotent ideal of g containing [g, g]. We do not enter into the details, but first we construct a certain sequence S of ideals of g: S : {0} =n 0 n 1 n 2 n m 1 n m = n, dim(n j /n j 1 ) = 1 (1 j m). s on For 1 j m, we set m j = n j + h 1, M j = exp m j and I h 2 = {1 j m; h 2 m j h 2 m j 1 }.

36 s on Now we propose to establish the following assertion: Assertion Let j I h 2. There exists X j in h 2 m j outside of h 2 m j 1 such that [X j, n j ] n j 1.

37 s on Now we propose to establish the following assertion: Assertion Let j I h 2. There exists X j in h 2 m j outside of h 2 m j 1 such that [X j, n j ] n j 1. If we get this assertion, we can prove inductively on j that H 2 H 1 M j is closed. However, we guess that this assertion is much stronger than the closeness of H 2 H 1. In any event, we get by this way the:

38 Theorem Let G = exp g be a group with Lie algebra g, f g and h j (j = 1, 2) two polarizations of g at f satisfying the Pukanszky condition. Put H j = exp h j for j = 1, 2. Let n be a nilpotent ideal of g containing [g, g]. If one of the following three conditions is satisfied, then the simple product set H 2 H 1 is closed in G: s on

39 Theorem Let G = exp g be a group with Lie algebra g, f g and h j (j = 1, 2) two polarizations of g at f satisfying the Pukanszky condition. Put H j = exp h j for j = 1, 2. Let n be a nilpotent ideal of g containing [g, g]. If one of the following three conditions is satisfied, then the simple product set H 2 H 1 is closed in G: (1) [h 1, h 1 ] h 2 or [h 2, h 2 ] h 1 ; s on

40 Theorem Let G = exp g be a group with Lie algebra g, f g and h j (j = 1, 2) two polarizations of g at f satisfying the Pukanszky condition. Put H j = exp h j for j = 1, 2. Let n be a nilpotent ideal of g containing [g, g]. If one of the following three conditions is satisfied, then the simple product set H 2 H 1 is closed in G: (1) [h 1, h 1 ] h 2 or [h 2, h 2 ] h 1 ; (2) [h i, h i ] n ( f n ) for i = 1 or i = 2, more precisely s on [(h i + n) h j, h j ] n f + (h 1 h 2 ) for i = 1, j = 2 or i = 2, j = 1;

41 Theorem Let G = exp g be a group with Lie algebra g, f g and h j (j = 1, 2) two polarizations of g at f satisfying the Pukanszky condition. Put H j = exp h j for j = 1, 2. Let n be a nilpotent ideal of g containing [g, g]. If one of the following three conditions is satisfied, then the simple product set H 2 H 1 is closed in G: (1) [h 1, h 1 ] h 2 or [h 2, h 2 ] h 1 ; (2) [h i, h i ] n ( f n ) for i = 1 or i = 2, more precisely s on [(h i + n) h j, h j ] n f + (h 1 h 2 ) for i = 1, j = 2 or i = 2, j = 1; (3) n is abelian or two-step nilpotent.

42 We also have a small remark. s on

43 We also have a small remark. Proposition Let G = exp g be a group with Lie algebra g, f g and h j (j = 1, 2) two polarizations of g at f satisfying the Pukanszky condition. Assume that the coadjoint orbit G f of G has the dimension strictly smaller than 10. Then the product H 2 H 1 is closed in G. s on

44 We also have a small remark. Proposition Let G = exp g be a group with Lie algebra g, f g and h j (j = 1, 2) two polarizations of g at f satisfying the Pukanszky condition. Assume that the coadjoint orbit G f of G has the dimension strictly smaller than 10. Then the product H 2 H 1 is closed in G. s on Finally, without the closeness of H 2 H 1, what can we say on the convergence of our integral? For instance, we recently got almost everywhere on Γ the convergence of our integral R(τ, π l )(ϕ) for exponential where ϕ H l has compact support modulo B, b is a Vergne polarization and τ = ind G H χ f has multiplicities of discrete type.

45 In the preceding situation, we put as before. a l (ϕ) = R(τ, π l )(ϕ)(e) s on

46 In the preceding situation, we put as before. a l (ϕ) = R(τ, π l )(ϕ)(e) Does a l extend to a generalized vector belonging to (H l ) H,χ f 1/2 H,G? s on

47 In the preceding situation, we put as before. a l (ϕ) = R(τ, π l )(ϕ)(e) Does a l extend to a generalized vector belonging to (H l ) H,χ f 1/2 H,G? This is an interesting problem on the way to a distribution version of the Frobenius reciprocity. s on

48 In the preceding situation, we put as before. a l (ϕ) = R(τ, π l )(ϕ)(e) Does a l extend to a generalized vector belonging to (H l ) H,χ f 1/2 H,G? This is an interesting problem on the way to a distribution version of the Frobenius reciprocity. Let τ Ĝ m(π)πdµ(π) be the canonical central decomposition of τ. s on

49 In the preceding situation, we put as before. a l (ϕ) = R(τ, π l )(ϕ)(e) Does a l extend to a generalized vector belonging to (H l ) H,χ f 1/2 H,G? This is an interesting problem on the way to a distribution version of the Frobenius reciprocity. Let τ Ĝ m(π)πdµ(π) be the canonical central decomposition of τ. Then, concerning the multiplicity m(π), do we have µ-almost everywhere the equality s on m(π) = dim((h l ) H,χ f 1/2 H,G )?

50 When G is nilpotent, we know that a l belongs to (H l ) H,χ f 1/2 H,G in this sense. and that the Frobenius reciprocity holds s on

51 When G is nilpotent, we know that a l belongs to (H l ) H,χ f 1/2 H,G in this sense. and that the Frobenius reciprocity holds Moreover, we denote by D τ (G/H) the algebra of G-invariant differential operators on the line bundle over G/H associated with the data (H, χ f ). When τ has finite multiplicities, a l turns out to be an eigen-distribution for all the elements of D τ (G/H). I believe that this fact is useful in order to attack the s on

52 When G is nilpotent, we know that a l belongs to (H l ) H,χ f 1/2 H,G in this sense. and that the Frobenius reciprocity holds Moreover, we denote by D τ (G/H) the algebra of G-invariant differential operators on the line bundle over G/H associated with the data (H, χ f ). When τ has finite multiplicities, a l turns out to be an eigen-distribution for all the elements of D τ (G/H). I believe that this fact is useful in order to attack the Polynomial conjecture Assume that G is nilpotent and τ has finite multiplicities. The algebra D τ (G/H) is isomorphic to the algebra C[Γ] H of H-invariant polynomial functions on Γ. s on

53 s on We can also consider a polynomial conjecture about the restrictions π K of irreducible unitary representation π of G on an analytic subgroup K = exp k of G. We denote by Ω the coadjoint G-orbit corresponding to π and by U(g) the enveloping algebra of g C.

54 s on We can also consider a polynomial conjecture about the restrictions π K of irreducible unitary representation π of G on an analytic subgroup K = exp k of G. We denote by Ω the coadjoint G-orbit corresponding to π and by U(g) the enveloping algebra of g C. Let U π (g) k = {X U(g); [k, X ] ker(π)} and D π (G) K = π(u π (g) k ).

55 s on We can also consider a polynomial conjecture about the restrictions π K of irreducible unitary representation π of G on an analytic subgroup K = exp k of G. We denote by Ω the coadjoint G-orbit corresponding to π and by U(g) the enveloping algebra of g C. Let U π (g) k = {X U(g); [k, X ] ker(π)} and D π (G) K = π(u π (g) k ). Polynomial conjecture Assume that π K has finite multiplicities in its canonical central decomposition. The algebra D π (G) K is isomorphic to the algebra C[Ω] K of K-invariant polynomial functions on Ω.

56 s on In conclusion, there are still many problems to be studied.

57 s on THANK YOU!