Hendrik De Bie. Hong Kong, March 2011

Transcription

1 A Ghent University (joint work with B. Ørsted, P. Somberg and V. Soucek) Hong Kong, March 2011 A

2 Classical FT New realizations of sl 2 in harmonic analysis A

3 Outline Classical FT New realizations of sl 2 in harmonic analysis A

4 Problem: Find complex-valued functions f (z, w), g(z, w) with (z, w) R + [ 1, 1], satisfying (m 2 + c)g + (1 + c)z z g + 1 z w f + icf iczwg = 0 cz z f w w f czwg cz 2 w z g +z(w 2 1) w g + icz 2 g = 0 Here: m N, c > 0 A

5 Problem: Find complex-valued functions f (z, w), g(z, w) with (z, w) R + [ 1, 1], satisfying (m 2 + c)g + (1 + c)z z g + 1 z w f + icf iczwg = 0 cz z f w w f czwg cz 2 w z g +z(w 2 1) w g + icz 2 g = 0 Here: m N, c > 0 Observe: c = 1, g = 0 then f = e izw Fourier kernel! A

6 Problem: Find complex-valued functions f (z, w), g(z, w) with (z, w) R + [ 1, 1], satisfying (m 2 + c)g + (1 + c)z z g + 1 z w f + icf iczwg = 0 cz z f w w f czwg cz 2 w z g +z(w 2 1) w g + icz 2 g = 0 Here: m N, c > 0 Observe: c = 1, g = 0 then f = e izw Fourier kernel! Why is this system interesting? Can we find nice other solutions? Are there symmetries present? A

7 Outline Classical FT New realizations of sl 2 in harmonic analysis Classical FT New realizations of sl 2 in harmonic analysis A

8 Classical FT New realizations of sl 2 in harmonic analysis 4 definitions of classical FT in R m : F1 F(f )(y) = e i x,y f (x) dx R m F2 F(f )(y) = R K(x, y) f (x) dx with K(x, y) unique m solution of F3 F4 yj K(x, y) = ix j K(x, y), j = 1,..., m. K(x, y) = F = e iπ 4 ( x 2 ) (k + λ)( i) k (z) λ J k+λ (z) Ck λ (w) k=0 with z = x y, w = x, y and λ = (m 2)/2. Each with its specific uses A

9 Classical FT New realizations of sl 2 in harmonic analysis F3 F = e iπ 4 ( x 2 ) easiest to generalize connects FT with representation theory of sl 2 : r 2 E = m i=1 2 x i, Laplace operator = x 2 = m i=1 x 2 i = m i=1 x i xi, Euler operator, r 2 and E + m/2 generate the Lie algebra sl 2 : [, r 2 ] = 4(E + m 2 ) [, E + m ] [ 2 r 2, E + m ] 2 = 2 = 2r 2 A

10 Classical FT New realizations of sl 2 in harmonic analysis Overview of possible deformations: κ x 2 Dunkl deformation x 2 Clifford deformation a - deformation D 2 + (1 + c) 2 x 2 x 2 a x a A

11 Classical FT New realizations of sl 2 in harmonic analysis Dunkl operators: reduce O(m) symmetry to finite reflection group symmetry change the structure of functions on the sphere (e.g. spherical harmonics Dunkl harmonics)? Natural question can we preserve the spherical symmetry? change the radial structure A

12 Classical FT New realizations of sl 2 in harmonic analysis Deforming the operators in R m Introduce a parameter a > 0 and substitute r 2 r a r 2 a E + m E + a + m A

13 Classical FT New realizations of sl 2 in harmonic analysis Deforming the operators in R m Introduce a parameter a > 0 and substitute r 2 r a r 2 a E + m E + a + m The sl 2 relations also hold for r a, r 2 a and E + a+m 2 2 : [ r 2 a, r a] = 2a (E + a + m 2 ) [ 2 r 2 a, E + a + m 2 ] = a r 2 a 2 [ r a, E + a + m 2 ] = a r a 2 Ben Saïd S., Kobayashi, T. and Ørsted B., Laguerre semigroup and Dunkl operators. Preprint. arxiv: , 74 pages. A

14 Classical FT New realizations of sl 2 in harmonic analysis The associated Fourier transform F a = e iπ 2a (r 2 a k r a ) Main question: write F a as F a = K(x, y)f (x)h a (x)dx R m Here, h a (x) = r a 2 is the measure naturally associated with the deformation Resulting kernel: explicitly known if m = 1 or a = 1, 2 other values: infinite series of Bessel functions times Gegenbauer polynomials Boundedness? A

15 Classical FT New realizations of sl 2 in harmonic analysis New approach: try to find (and solve) system of PDEs for kernel K(x, y) (i.e. formulation F2) A

16 Classical FT New realizations of sl 2 in harmonic analysis New approach: try to find (and solve) system of PDEs for kernel K(x, y) (i.e. formulation F2) equivalent with factorizing r 2 a as r 2 a = m i=0 D 2 i with D i family of commuting differential operators (D i D j = D j D i ) Example (classical Fourier transform, formulation F2) In the case a = 2, D i = xi and one has xj K(x, y) = iy j K(x, y), j = 1,..., m leading to K(x, y) = e i x,y A

17 Outline Classical FT New realizations of sl 2 in harmonic analysis A

18 Factorization of and r 2 Introduce orthogonal Clifford algebra Cl m with generators e i, i = 1,..., m: e i e j + e j e i = 0 ei 2 = 1 We have Cl m = m k=0 Cl k m with Cl k m := span{e i1 e i2... e ik, i 1 <... < i k }. Then put x = x = m e i xi i=1 m e i x i i=1 Dirac operator vector variable A

19 We have x 2 = x 2 = r 2 and x, x generate the Lie superalgebra osp(1 2) A

20 We have x 2 = x 2 = r 2 and x, x generate the Lie superalgebra osp(1 2) Questions: Can we factorize r 2 a and r a? Do CA methods provide insight in radial deformation? A

21 Definition of the operators Factorization of r a : x a = r a 2 1 x A

22 Definition of the operators Factorization of r a : x a = r a 2 1 x Factorization of r 2 a : Ordinary Dirac operator given by x = 1 [x, ] 2 Hence first Ansatz: D a = 1 2 [x a, r 2 a ] = r 1 a 2 x ( a 2 1 ) ( a 2 + m 1 ) r a 2 1 x + ( a 2 1 ) r a 2 1 xe A

23 We can work even slightly more general, by replacing r 1 a 2 x ( a ) ( a ) ( m 1 r a a ) 2 1 x r a 2 1 xe D = r 1 a 2 x + br a 2 1 x + cr a 2 1 xe with a > 0 and b, c C a-deformed Dirac operator (depending on 3 parameters) A

24 We can work even slightly more general, by replacing r 1 a 2 x ( a ) ( a ) ( m 1 r a a ) 2 1 x r a 2 1 xe D = r 1 a 2 x + br a 2 1 x + cr a 2 1 xe with a > 0 and b, c C a-deformed Dirac operator (depending on 3 parameters) If a = 2, b = c = 0 then D = x A

25 Theorem The operators D and x a D = r 1 a 2 x + br a 2 1 x + cr a 2 1 xe, x a = r a 2 1 x generate for each value of a, b and c a copy of osp(1 2): with {x a, D} = 2(1 + c) ( E + 2) δ [ E + δ 2, D] = a 2 [ D x 2 a, D ] [ = a(1 + c)x a E + δ 2, x ] a = a 2 x a [ D 2 ] [, x a = a(1 + c)d E + δ 2, D2] = ad 2 [ D 2, x 2 ] a = 2a(1 + c) 2 ( E + δ ) [ 2 E + δ 2, x 2 a] = ax 2 a, δ = a 2 + 2b + m c δ = dimension of theory! A

26 However, a computation shows D 2 r 2 a Hence, we have NOT found a factorization of r 2 a Nevertheless, new operators interesting in their own right instead of radial deformation of Laplace, radial deformation of underlying Dirac we continue by developing related function theory A

27 Intertwining operators We want to reduce D to its simplest form A

28 Intertwining operators We want to reduce D to its simplest form Let P and Q be two operators defined by ( (a Pf (x) = r b f 2 Qf (x) = r ab 2 f ( (2 a ) 1 a xr 2 a 1 ) ) 1 2 xr a 2 1 ). These two operators act as generalized Kelvin transformations: QP = PQ = ( ) b 2 2. a A

29 Proposition One has the following intertwining relations Q ( x + br 2 x + cr 2 xe ) P = r 1 a 2 x + βr a 2 1 x + γr a 2 1 xe with Q x P = x a β = 2b + 2c γ = 2 (1 + c) 1. a we can get rid of r 1 a 2 similarly technique: we can make b = 0 A

30 So we are reduced to D = x + cr 2 xe Theorem The operators D and x generate osp(1 2), with δ = 1 + m 1 1+c : {x, D} = 2(1 + c) ( E + 2) δ [ E + δ 2, D] = D [ x 2, D ] [ = 2(1 + c)x E + δ 2, x] = x [ D 2, x ] [ = 2(1 + c)d E + δ 2, D2] = 2D 2 [ D 2, x 2] = 4(1 + c) 2 ( E + δ ) [ 2 E + δ 2, x 2] = 2x 2. c is deformation parameter A

31 The operator D = x + cr 2 xe, also appears in a different context in Cação, I., Constales, D., and Krausshar, R. S. On the role of arbitrary order Bessel functions in higher dimensional Dirac type equations. Arch. Math. (Basel) 87 (2006), Its square is complicated and NOT scalar: D 2 = (cm c) r 1 r ( c 2 + 2c ) r 2 cr 2 e i e j (x i xj x j xi ). i<j Recall Γ = i<j e i e j (x i xj x j xi ) is Gamma operator A

32 The measure associated to D: Proposition If c > 1, then for suitable differentiable functions f and g one has (Df ) g h(r)dx = f (Dg) h(r)dx R m R m 1+mc 1 with h(r) = r 1+c, provided the integrals exist.. is the main anti-involution on Cl m Measure looks complicated. However, radial part: h(r)dx r δ 1 dr A

33 Function space We use L 2 c(r m ) = L 2 (R m, h(r)dx) Cl m with inner product [ ] f, g = f c g h(r)dx R m satisfying Df, g = f, Dg xf, g = f, xg. The related norm is defined by f 2 = f, f. 0 A

34 We need a basis for L 2 c(r m ): A

35 We need a basis for L 2 c(r m ): functions φ t,l,m (t, l N and m = 1,... dim M l ) defined as γ l φ 2t,l,m = L 2 1 t (r 2 )r β l M (m) l e r 2 /2 γ l φ 2t+1,l,m = L 2 t (r 2 )xr β l M (m) l e r 2 /2 with L β α the Laguerre polynomials and β l = c 1 + c l γ l = c and with {M (m) l } a basis of M l ( l + µ 2 2 ) + c c M l = ker x P l : spherical monogenics of degree l A

36 Theorem (Orthogonality of basis) After suitable normalization, one has φ t1,l 1,m 1, φ t2,l 2,m 2 = δ t1 t 2 δ l1 l 2 δ m1 m 2. Every f L 2 c(r m ) can hence be decomposed as f = t,l,m a t,l,m φ t,l,m, a t,l,m R with a t,l,m 2 < t,l,m A

37 Creation and annihilation operators in this context: A + = D (1 + c)x A = D + (1 + c)x satisfying A + φ t,l,m = φ t+1,l,m and A φ t,l,m = φ t 1,l,m. Lead to generalized harmonic oscillator: Theorem The functions φ t,l,m satisfy the following second-order PDE ( D 2 (1 + c) 2 x 2) φ t,l,m = (1 + c) 2 (γ l + 2t)φ t,l,m. Recall γ l = 2 ( l + m 2 ) + c c c A

38 We want to study the holomorphic semigroup attached to this harmonic oscillator ω FD ω = e 2(1+c) 2 (D 2 (1+c) 2 x 2 ). ω complex number, Rω 0 ω = iπ/2 generalized Fourier transform e ω( x 2 ) is so-called Hermite semigroup A

39 Theorem Suppose c > 1. Then 1. {φ t,l,m } is an eigenbasis of F ω D : F ω D (φ t,l,m) = e ωt e ωl (1+c) φ t,l,m. 2. F ω D is a continuous operator on L2 c(r m ) for all ω with Rω 0, in particular FD ω (f ) f 3. If Rω > 0, then F ω D is a Hilbert-Schmidt operator on L2 c(r m ). 4. If Rω = 0, then F ω D is a unitary operator on L2 c(r m ). A

40 Explicit formula for semigroup We want to find K(x, y; ω) such that F ω D (f ) = ω e 2(1+c) 2 (D 2 (1+c) 2 x 2 ) f (x) = K(x, y; ω)f (x) h(r)dx R m Note that in general K(x, y; ω) takes values in Cl m Moreover, K(x, y; ω) K(y, x; ω) We use techniques developed in H. De Bie and Y. Xu, On the Clifford-Fourier transform IMRN, arxiv: , 30 pages. A

41 Theorem (Rω > 0 and c > 1) Put K(x, y; ω) = e coth ω 2 ( x 2 + y 2 ) ( A(z, w) + x yb(z, w) ) with A(z, w) = + k=0 ( + α k 1 4 sinh ω k + 2λ α k 2λ z k 1+c J γ k2 1 k λ z k+c 1+c J γ k 1 2 ( iz sinh ω ( iz sinh ω ) )) Ck λ (w) and z = x y, w = x, y /z and α k = 2e ωδ 2 (2 sinh ω) γk/2. These series are convergent and the transform defined on L 2 c(r m ) by Fc ω (f ) = K(x, y; ω)f (x)h(r x )dx R m coincides with the operator FD ω = e 2(1+c) 2 (D 2 (1+c) 2 x 2 ) on {φt,l,m }. ω A

42 If Rω = 0 the kernel follows by taking a limit We are mostly interested in 1 specific value: ω = i π 2 the Clifford deformed Fourier transform We denote it by F c A

43 Theorem (Fourier transform; ω = iπ/2) Put K(x, y) = A(z, w) + x yb(z, w) with A(z, w) = + k=0 z δ 2 2 ( α k k + 2λ 2λ J γ k2 1 (z) iα k k 1 2λ J γ k 1 2 ) (z) Ck λ (w) and z = x y, w = x, y /z and α k = e iπk 2(1+c). These series are convergent and the transform defined on L 2 c(r m ) by F c (f ) = K(x, y)f (x)h(r x )dx R m coincides with the operator F D = e iπ 4(1+c) 2 (D 2 (1+c) 2 x 2 ) on {φt,l,m }. A

44 Theorem (Properties of Fourier transform) The operator F c defines a unitary operator on L 2 c(r m ) and satisfies the following intertwining relations on a dense subset: F c D = i(1 + c)x F c F c x = i 1 + c D F c F c E = (E + δ) F c. Moreover, F c is of finite order if and only if 1 + c is rational. A

45 Proposition (Bochner formulae) Let M l M l be a spherical monogenic of degree l. Let f (x) = f (r) be a radial function. Then: iπl F c (f (r)m l ) = e 2(1+c) M l (y ) iπl + F c (f (r)xm l ) = ie 2(1+c) y M l (y ) 0 r l f (r)z δ 2 2 J γ k2 1 (z)h(r)r m 1 dr + with y = sy, y S m 1 and z = rs. h(r) is the measure associated with D. 0 r l+1 f (r)z δ 2 2 J γ k2 (z)h(r)r m 1 d connects deformed Fourier transform with Hankel transform A

46 Proposition (Heisenberg inequality) For all f L 2 c(r m ), the deformed Fourier transform satisfies x f (x). x (F c f ) (x) δ 2 f (x) 2. The equality holds if and only if f is of the form f (x) = λe r 2 /α. Heisenberg inequality = statement about lowest eigenvalue of D 2 (1 + c) 2 x 2 A

47 Theorem (Master formula) Let s > 0. Then one has with 2s = sinh ω. R m K(y, x; i π 2 )K(z, y; i π 2 )e sr 2 y h(r y )dy = cst K(z, x; ω)e x 2 + z cosh ω sinh ω A

48 Theorem (Master formula) Let s > 0. Then one has with 2s = sinh ω. R m K(y, x; i π 2 )K(z, y; i π 2 )e sr 2 y h(r y )dy = cst K(z, x; ω)e x 2 + z cosh ω sinh ω connects our semigroup with the fundamental solution of the heat equation defines a generalized translation order of variables is important A

49 Series representation of kernel is nice for L 2 -theory If we want to go beyond that, we need more explicit expressions Kernel satisfies system of PDEs: D y K(x, y) = i(1 + c)k(x, y)x (K(x, y)d x ) = i(1 + c)yk(x, y) As we have series representation, we have Ansatz for solution: K(x, y) = f (z, w) + x y g(z, w) with z = x y and w = x, y /z A

50 This leads us back to the beginning of the talk! Problem: Find complex-valued functions f (z, w), g(z, w) with (z, w) R + [ 1, 1], satisfying (m 2 + c)g + (1 + c)z z g + 1 z w f + icf iczwg = 0 Here: m N, c > 0 cz z f w w f czwg cz 2 w z g +z(w 2 1) w g + icz 2 g = 0 Moreover, series representation learns that solution in R m is determined by solution in R m 2 sufficient to solve PDEs for m = 2, 3! A

51 and outlook Clifford analysis study of new Dirac operator determination of the Fourier kernel for special values (easier than in Dunkl case?) heat equation and translation operator connection with Hecke algebras and DAHA investigation of more general deformations H. De Bie, B. Ørsted, P. Somberg and V. Souček, Dunkl operators and a family of realizations of osp(1 2). Preprint: arxiv: , 25 pages. H. De Bie, B. Orsted, P. Somberg and V. Soucek, The. Preprint, 27 pages, arxiv: A