The Heat equation, the Segal-Bargmann transform and generalizations. B. Krötz B. Ørsted H. Schlichtkrull R. Stanton

Transcription

1 The Heat equation, the Segal-Bargmann transform and generalizations Based on joint work with B. Krötz B. Ørsted H. Schlichtkrull R. Stanton - p. 1/54

2 Organizations 1. The heat equation on R n. 2. The Fock space and the Segal-Bargmann Transform. 3. Remarks and Comments. 4. Generalizations and the Restriction Principle. 5. Structure Theory. 6. Spherical Functions and the Fourier Transform. 7. The Crown and the Heat Kernel. 8. The Abel Transform and the Heat Kernel. 9. The Faraut-Gutzmer Formula and the Orbital Integral. 10. The Image of the Segal-Bargmann transform on G/K. 11. The K-invariant case (more than one section) - p. 2/54

3 1. The heat equation on R n Consider the Laplace operator = on R n. n j=1 2 x 2 j - p. 3/54

4 1. The heat equation on R n Consider the Laplace operator = n j=1 2 x 2 j on R n. The heat equation is the Cauchy problem u(x, t) = t u(x, t) lim u(x, t) = t 0 + f(x) where we can take f L 2 (R n ), a distribution, a hyperfunction, or from another class of analytic objects. - p. 3/54

5 Formally we write u(x, t) = e t f(x) - p. 4/54

6 Formally we write u(x, t) = e t f(x) = H t f(x) - p. 4/54

7 Formally we write u(x, t) = e t f(x) = H t f(x) where {e t = H t } t 0 is the heat semigroup. It is a linear map H t : L 2 (R n ) L 2 (R n ), but also a smoothing operator as we will see. - p. 4/54

8 Formally we write u(x, t) = e t f(x) = H t f(x) where {e t = H t } t 0 is the heat semigroup. It is a linear map H t : L 2 (R n ) L 2 (R n ), but also a smoothing operator as we will see. To actually solve the equation, we proceed by applying the Fourier transform f F(f) = ˆf, λ (2π) n/2 f(x)e ix λ dx using that F( f)(λ) = λ 2 ˆf(λ) and get the simple differential equation for t û(λ, t): - p. 4/54

9 Formally we write u(x, t) = e t f(x) = H t f(x) where {e t = H t } t 0 is the heat semigroup. It is a linear map H t : L 2 (R n ) L 2 (R n ), but also a smoothing operator as we will see. To actually solve the equation, we proceed by applying the Fourier transform f F(f) = ˆf, λ (2π) n/2 f(x)e ix λ dx using that F( f)(λ) = λ 2 ˆf(λ) and get the simple differential equation for t û(λ, t): t û(λ, t) = λ 2 û(λ, t), û(λ, 0) = ˆf(λ). - p. 4/54

10 The solution to this differential equation is û(λ, t) = ˆf(λ)e t λ 2 - p. 5/54

11 The solution to this differential equation is û(λ, t) = ˆf(λ)e t λ 2 and we get the Fourier Transform Formula for the solution: H t f(x) = (2π) n/2 ˆf(λ)e λ 2t e iλ x dλ (0.1) - p. 5/54

12 The solution to this differential equation is û(λ, t) = ˆf(λ)e t λ 2 and we get the Fourier Transform Formula for the solution: H t f(x) = (2π) n/2 ˆf(λ)e λ 2t e iλ x dλ (0.1) The heat kernel h t is the solution to the heat equation with f = δ 0. Using that the δ-distribution has Fourier transform ˆδ 0 (λ) = (2π) n/2 we get H t δ 0 (x) = h t (x) = (2π) n e λ 2t e ix λ dλ = - p. 5/54

13 The solution to this differential equation is û(λ, t) = ˆf(λ)e t λ 2 and we get the Fourier Transform Formula for the solution: H t f(x) = (2π) n/2 ˆf(λ)e λ 2t e iλ x dλ (0.1) The heat kernel h t is the solution to the heat equation with f = δ 0. Using that the δ-distribution has Fourier transform ˆδ 0 (λ) = (2π) n/2 we get H t δ 0 (x) = h t (x) = (2π) n e λ 2t e ix λ dλ = (4πt) n/2 e x 2 /4t - p. 5/54

14 The solution to this differential equation is û(λ, t) = ˆf(λ)e t λ 2 and we get the Fourier Transform Formula for the solution: H t f(x) = (2π) n/2 ˆf(λ)e λ 2t e iλ x dλ (0.1) The heat kernel h t is the solution to the heat equation with f = δ 0. Using that the δ-distribution has Fourier transform ˆδ 0 (λ) = (2π) n/2 we get H t δ 0 (x) = h t (x) = (2π) n e λ 2t e ix λ dλ = (4πt) n/2 e x 2 /4t It is clear from this formula, that R n x h t (x) R + has a holomorphic extension to C n given by h t (z) = (4πt) n/2 e z2 /4t, z 2 = z z 2 n. - p. 5/54

15 Note and t (f h t ) = f ( t h t ) = f ( h t ) = (f h t ) lim f h t = f δ 0 = f t p. 6/54

16 Note and t (f h t ) = f ( t h t ) = f ( h t ) = (f h t ) lim f h t = f δ 0 = f t 0 + and we get the heat kernel formula for the solution: H t f(x) = f h t (x) = (4πt) n/2 f(y)e (x y)2 /(4t) dy (0.2) - p. 6/54

17 Note and t (f h t ) = f ( t h t ) = f ( h t ) = (f h t ) lim f h t = f δ 0 = f t 0 + and we get the heat kernel formula for the solution: H t f(x) = f h t (x) = (4πt) n/2 f(y)e (x y)2 /(4t) dy (0.2) We read of from (0.1) or (0.2) that the function x H t f(x) has a holomorphic extension to C n : H t f(z) = (4πt) n/2 f(y)e (z y)2 /4t dy = - p. 6/54

18 Note and t (f h t ) = f ( t h t ) = f ( h t ) = (f h t ) lim f h t = f δ 0 = f t 0 + and we get the heat kernel formula for the solution: H t f(x) = f h t (x) = (4πt) n/2 f(y)e (x y)2 /(4t) dy (0.2) We read of from (0.1) or (0.2) that the function x H t f(x) has a holomorphic extension to C n : H t f(z) = (4πt) n/2 f(y)e (z y)2 /4t dy = (2π) n/2 ˆf(λ)e λ2t e iz λ dλ - p. 6/54

19 Note and t (f h t ) = f ( t h t ) = f ( h t ) = (f h t ) lim f h t = f δ 0 = f t 0 + and we get the heat kernel formula for the solution: H t f(x) = f h t (x) = (4πt) n/2 f(y)e (x y)2 /(4t) dy (0.2) We read of from (0.1) or (0.2) that the function x H t f(x) has a holomorphic extension to C n : H t f(z) = (4πt) n/2 f(y)e (z y)2 /4t dy = (2π) n/2 ˆf(λ)e λ2t e iz λ dλ where z λ = n j=1 z jλ j. - p. 6/54

20 The map H t : L 2 (R n ) O(C n ) is the Segal-Bargmann transform. - p. 7/54

21 The map H t : L 2 (R n ) O(C n ) is the Segal-Bargmann transform. Note, that we have really only used the following: - p. 7/54

22 The map H t : L 2 (R n ) O(C n ) is the Segal-Bargmann transform. Note, that we have really only used the following: 1. We have a Fourier transform that such that F( g)(λ) = λ 2 F(g)(λ) to derive the Fourier transform form for the solution and to find an explicit expression for the heat kernel. - p. 7/54

23 The map H t : L 2 (R n ) O(C n ) is the Segal-Bargmann transform. Note, that we have really only used the following: 1. We have a Fourier transform that such that F( g)(λ) = λ 2 F(g)(λ) to derive the Fourier transform form for the solution and to find an explicit expression for the heat kernel. 2. In using (0.1) that the exponential function λ e λ (z) = e iz λ grows much slower than λ ˆf(λ)e λ 2 t to show that the solution extends to a holomorphic function on C n. - p. 7/54

24 The map H t : L 2 (R n ) O(C n ) is the Segal-Bargmann transform. Note, that we have really only used the following: 1. We have a Fourier transform that such that F( g)(λ) = λ 2 F(g)(λ) to derive the Fourier transform form for the solution and to find an explicit expression for the heat kernel. 2. In using (0.1) that the exponential function λ e λ (z) = e iz λ grows much slower than λ ˆf(λ)e λ 2 t to show that the solution extends to a holomorphic function on C n. 3. Or in using (0.2) that heat kernel h t has a holomorphic extension to C n and y h t (z y) grows much slower than y f(y)e y2 /(4t). - p. 7/54

25 2. The Fock space and the Segal-Bargmann Transform - p. 8/54

26 2. The Fock space and the Segal-Bargmann Transform We will now describe the image of the Segal-Bargmann transform. For that we define a positive weight function by ωt Rn (y) = ω t (x) = (2πt) n/2 e y2 /2t = h t/2 (y) - p. 8/54

27 2. The Fock space and the Segal-Bargmann Transform We will now describe the image of the Segal-Bargmann transform. For that we define a positive weight function by and a measure on C n by ωt Rn (y) = ω t (x) = (2πt) n/2 e y2 /2t = h t/2 (y) dµ t (x + iy) = ω t (y) dxdy. - p. 8/54

28 2. The Fock space and the Segal-Bargmann Transform We will now describe the image of the Segal-Bargmann transform. For that we define a positive weight function by and a measure on C n by ωt Rn (y) = ω t (x) = (2πt) n/2 e y2 /2t = h t/2 (y) dµ t (x + iy) = ω t (y) dxdy. Set H t (C n ) = {F O(C n ) F 2 t := C n F(x + iy) 2 dµ t < }. - p. 8/54

29 Theorem 0.1 (Segal-Bargmann, /1961,... ). The following holds: 1. H t (C n ) is a Hilbert space with continuous point evaluation, i.e., the maps H t (C n ) F ev z (F) = F(z) C, z C n are continuous. In particular, with L y F(x) = F(x y) and K w (z) = K(z,w) := H t (L w h t )(z) = (8πt) n/2 e (z w)2 /8t, we have K w H t (C n ) and F(w) = (F,K w ) for all F H t (C n ), i.e., K(z,w) is the reproducing kernel for H t (C n ) 2. H t : L 2 (R n ) H t (C n ) is an unitary isomorphism. 3. If f S(R n ), then f(x) = H t f(x + iy)h t (y)dy. R n - p. 9/54

30 Few words on the proof, but note that I will not prove the surjectivity or that H t is an isometry. = = = - p. 10/54

31 Few words on the proof, but note that I will not prove the surjectivity or that H t is an isometry. For (2) we simply calculate the norm on the right hand side. Note that all functions involved are positive, so that there is no problem using Fubini s Theorem: = = = - p. 10/54

32 Few words on the proof, but note that I will not prove the surjectivity or that H t is an isometry. For (2) we simply calculate the norm on the right hand side. Note that all functions involved are positive, so that there is no problem using Fubini s Theorem: Let c = (2πt) n/2 = ( e y2 /2t dy) 1 : c H t f(x + iy) 2 dxe y2 /2t dy = c = Ĥtf(λ) 2 e 2y λ e y2 /2t dλdy = = = - p. 10/54

33 Few words on the proof, but note that I will not prove the surjectivity or that H t is an isometry. For (2) we simply calculate the norm on the right hand side. Note that all functions involved are positive, so that there is no problem using Fubini s Theorem: Let c = (2πt) n/2 = ( e y2 /2t dy) 1 : c H t f(x + iy) 2 dxe y2 /2t dy = c = c = Ĥtf(λ) 2 e 2y λ e y2 /2t dλdy ˆf(λ) 2 e (2tλ2 +2y λ+ y2 2t ) dλdy = = - p. 10/54

34 Few words on the proof, but note that I will not prove the surjectivity or that H t is an isometry. For (2) we simply calculate the norm on the right hand side. Note that all functions involved are positive, so that there is no problem using Fubini s Theorem: Let c = (2πt) n/2 = ( e y2 /2t dy) 1 : c H t f(x + iy) 2 dxe y2 /2t dy = c Ĥtf(λ) 2 e 2y λ e y2 /2t dλdy = c ˆf(λ) 2 e (2tλ2 +2y λ+ y2 2t ) dλdy ) = (c ˆf(λ) 2 e (y+2tλ)2 /2t dy dλ = = - p. 10/54

35 Few words on the proof, but note that I will not prove the surjectivity or that H t is an isometry. For (2) we simply calculate the norm on the right hand side. Note that all functions involved are positive, so that there is no problem using Fubini s Theorem: Let c = (2πt) n/2 = ( e y2 /2t dy) 1 : c H t f(x + iy) 2 dxe y2 /2t dy = c Ĥtf(λ) 2 e 2y λ e y2 /2t dλdy = c ˆf(λ) 2 e (2tλ2 +2y λ+ y2 2t ) dλdy ) = (c ˆf(λ) 2 e (y+2tλ)2 /2t dy dλ = ˆf(λ) 2 dλ = - p. 10/54

36 Few words on the proof, but note that I will not prove the surjectivity or that H t is an isometry. For (2) we simply calculate the norm on the right hand side. Note that all functions involved are positive, so that there is no problem using Fubini s Theorem: Let c = (2πt) n/2 = ( e y2 /2t dy) 1 : c H t f(x + iy) 2 dxe y2 /2t dy = c Ĥtf(λ) 2 e 2y λ e y2 /2t dλdy = c ˆf(λ) 2 e (2tλ2 +2y λ+ y2 2t ) dλdy ) = (c ˆf(λ) 2 e (y+2tλ)2 /2t dy dλ = ˆf(λ) 2 dλ = f p. 10/54

37 The proof of the inversion formula is similar. Let c = (4πt) n/2 : ( ) (0.1) H t f(x + iy)h t (y) dy = e tλ2 ˆf(λ)e i(x+iy) λ dλ h t (y) dy = = = = - p. 11/54

38 The proof of the inversion formula is similar. Let c = (4πt) n/2 : ( ) (0.1) H t f(x + iy)h t (y) dy = e tλ2 ˆf(λ)e i(x+iy) λ dλ h t (y) dy = c ( e tλ2 ix λ ˆf(λ)e ) e y λ e y2 /4t dy dλ = = = - p. 11/54

39 The proof of the inversion formula is similar. Let c = (4πt) n/2 : ( ) (0.1) H t f(x + iy)h t (y) dy = e tλ2 ˆf(λ)e i(x+iy) λ dλ h t (y) dy = c = c ( e tλ2 ix λ ˆf(λ)e ˆf(λ)e ix λ ( e (2tλ+y)2 4t ) e y λ e y2 /4t dy ) dy dλ dλ = = - p. 11/54

40 The proof of the inversion formula is similar. Let c = (4πt) n/2 : ( ) (0.1) H t f(x + iy)h t (y) dy = e tλ2 ˆf(λ)e i(x+iy) λ dλ h t (y) dy = c ( e tλ2 ix λ ˆf(λ)e ( = c ˆf(λ)e ix λ = ˆf(λ)e ix λ dλ e (2tλ+y)2 4t ) e y λ e y2 /4t dy ) dy dλ dλ = - p. 11/54

41 The proof of the inversion formula is similar. Let c = (4πt) n/2 : ( ) (0.1) H t f(x + iy)h t (y) dy = e tλ2 ˆf(λ)e i(x+iy) λ dλ h t (y) dy = c ( e tλ2 ix λ ˆf(λ)e ( = c ˆf(λ)e ix λ = ˆf(λ)e ix λ dλ e (2tλ+y)2 4t ) e y λ e y2 /4t dy ) dy dλ dλ = f(x). - p. 11/54

42 For the formula for the reproducing kernel we again assume that H t is an unitary isomorphism. - p. 12/54

43 For the formula for the reproducing kernel we again assume that H t is an unitary isomorphism. So let F H t (C n ) and f L 2 (R n ) such that F = H t f. Then F(w) = = H t f(w) f(x)h t (x w) dx h t even = (f, L w h t ) L 2 = (H t f, H t (L w h t )) Ht H t unitary = (F, H t (L w h t )) Ht. - p. 12/54

44 For the formula for the reproducing kernel we again assume that H t is an unitary isomorphism. So let F H t (C n ) and f L 2 (R n ) such that F = H t f. Then F(w) = = H t f(w) f(x)h t (x w) dx h t even = (f, L w h t ) L 2 = (H t f, H t (L w h t )) Ht H t unitary = (F, H t (L w h t )) Ht. Thus K(z, w) = H t (λ( w)h t )(z) = (λ( w)h t ) h t (z) = h t h t (z w) = h 2t (z w) the semigroup property. - p. 12/54

45 3. Remarks and Comments Note first of all, that we can interpret C n as the cotangent bundle T (R n ), where the y-variable in z = x + iy is an element of T xr n. Hence the Segal-Bargmann transform is some kind of quantization. - p. 13/54

46 3. Remarks and Comments Note first of all, that we can interpret C n as the cotangent bundle T (R n ), where the y-variable in z = x + iy is an element of T xr n. Hence the Segal-Bargmann transform is some kind of quantization. Note also, that in the definition of the norm and in the inversion formula we only weight the cotangent variable y T x(r n ) and the weights are given by the heat kernel. - p. 13/54

47 3. Remarks and Comments Note first of all, that we can interpret C n as the cotangent bundle T (R n ), where the y-variable in z = x + iy is an element of T xr n. Hence the Segal-Bargmann transform is some kind of quantization. Note also, that in the definition of the norm and in the inversion formula we only weight the cotangent variable y T x(r n ) and the weights are given by the heat kernel. There are other versions of the Segal-Bargmann transform in the literature. In particular, for the physics and infinite dimensional analysis, as well as in the original works, the space L 2 (R n ) was replaced by the weighted L 2 -space L 2 (R n, dν n ), where the heat kernel measure on R n. dν n (x) = h t (x)dx - p. 13/54

48 3. Remarks and Comments Note first of all, that we can interpret C n as the cotangent bundle T (R n ), where the y-variable in z = x + iy is an element of T xr n. Hence the Segal-Bargmann transform is some kind of quantization. Note also, that in the definition of the norm and in the inversion formula we only weight the cotangent variable y T x(r n ) and the weights are given by the heat kernel. There are other versions of the Segal-Bargmann transform in the literature. In particular, for the physics and infinite dimensional analysis, as well as in the original works, the space L 2 (R n ) was replaced by the weighted L 2 -space L 2 (R n, dν n ), where dν n (x) = h t (x)dx the heat kernel measure on R n. On the image side the measure is then dσ n t (z) = (2πt) n e z 2 /2t dxdy - p. 13/54

49 Denote the corresponding space of L 2 -holomorphic functions by F t (C n ). It is still holds, that the Segal-Bargmann transform L 2 (R n, dν) f f h t F t (C n ) is an unitary isomorphism. This normalization has several advantages. - p. 14/54

50 Denote the corresponding space of L 2 -holomorphic functions by F t (C n ). It is still holds, that the Segal-Bargmann transform L 2 (R n, dν) f f h t F t (C n ) is an unitary isomorphism. This normalization has several advantages. It is very easy to find the reproducing kernel for the space F t (C n ). It is K t (z, w) = (2πt) n e z w/2t. - p. 14/54

51 Denote the corresponding space of L 2 -holomorphic functions by F t (C n ). It is still holds, that the Segal-Bargmann transform L 2 (R n, dν) f f h t F t (C n ) is an unitary isomorphism. This normalization has several advantages. It is very easy to find the reproducing kernel for the space F t (C n ). It is K t (z, w) = (2πt) n e z w/2t. For the proof, one notice that the polynomials ζ α (z) = z α are dense in F(C n ) and therefore one only has to prove z α = (ζ α, K z ) for all α, and this reduces to a simple one-dimensional integral. - p. 14/54

52 Denote the corresponding space of L 2 -holomorphic functions by F t (C n ). It is still holds, that the Segal-Bargmann transform L 2 (R n, dν) f f h t F t (C n ) is an unitary isomorphism. This normalization has several advantages. It is very easy to find the reproducing kernel for the space F t (C n ). It is K t (z, w) = (2πt) n e z w/2t. For the proof, one notice that the polynomials ζ α (z) = z α are dense in F(C n ) and therefore one only has to prove z α = (ζ α, K z ) for all α, and this reduces to a simple one-dimensional integral. Connection to the theory of orthogonal polynomials: There are constants c α (easy to calculate) such that {c α ζ α } α N0 is an orthogonal basis for H t (C n, dσ t ) and there are constants (again easy to calculate) such that H t (ζ α ) = d α h α, where h α is the Hermite polynomial. - p. 14/54

53 Furthermore, the measures on both sides are probability measures, and n-fold product of the one-dimensional measures in the coordinates. - p. 15/54

54 Furthermore, the measures on both sides are probability measures, and n-fold product of the one-dimensional measures in the coordinates. We can take the limit as n. Consider the projections pr n : R n R n 1. This gives us isometric maps maps pr n : L 2 (R n 1, dν n 1 ) L 2 (R n, dν n ), f f pr n and we have a sequence of commutative diagrams L 2 (R n 1, dν n 1 ) pr n L 2 (R n, dν n ) pr n+1 L 2 (R, dν ) H n 1 t H t (C n 1, dσt n 1 ) prn H t (C n, dσt n ) H n t pr n+1 H t H t (C, dσ t ) - p. 15/54

55 Sometimes, in particular studying the Schrödinger representation of the Heisenberg group, one uses the Segal-Bargmann transform S t : L 2 (R n, dx) F t (C n ). One of the idea is, that F t (C n ) is a much simpler space than L 2 (R n ) to work with. Also, the canonical commutation rules, the creation operator and the annulation operator have simpler form in F t (C n ). - p. 16/54

56 Sometimes, in particular studying the Schrödinger representation of the Heisenberg group, one uses the Segal-Bargmann transform S t : L 2 (R n, dx) F t (C n ). One of the idea is, that F t (C n ) is a much simpler space than L 2 (R n ) to work with. Also, the canonical commutation rules, the creation operator and the annulation operator have simpler form in F t (C n ). In this case - as I will prove later - the Segal-Bargmann transform is given by: S t (f)(z) = (πt) n/4 f(y)e 1 2t (y2 2xy+ x2 2 ) dy. - p. 16/54

57 Sometimes, in particular studying the Schrödinger representation of the Heisenberg group, one uses the Segal-Bargmann transform S t : L 2 (R n, dx) F t (C n ). One of the idea is, that F t (C n ) is a much simpler space than L 2 (R n ) to work with. Also, the canonical commutation rules, the creation operator and the annulation operator have simpler form in F t (C n ). In this case - as I will prove later - the Segal-Bargmann transform is given by: S t (f)(z) = (πt) n/4 f(y)e 1 2t (y2 2xy+ x2 2 ) dy. The connection to the theory of special functions is, this case, a multiple of the Hermite functions are mapped into a multiple of the polynomials ζ α. - p. 16/54

58 4. Generalizations and the Restriction Principle. - p. 17/54

59 4. Generalizations and the Restriction Principle. Before I prove the unitarity of the Segal-Bargmann transform S t let me make some general remarks about the underlying idea and general principle behind a transform like S t. - p. 17/54

60 4. Generalizations and the Restriction Principle. Before I prove the unitarity of the Segal-Bargmann transform S t let me make some general remarks about the underlying idea and general principle behind a transform like S t. Let M C be a complex analytic manifold (i.e., M C = C n ) and M M C a totally real analytic submanifold. Thus the restriction map is injective. O(M C ) F F M A(M) - p. 17/54

61 4. Generalizations and the Restriction Principle. Before I prove the unitarity of the Segal-Bargmann transform S t let me make some general remarks about the underlying idea and general principle behind a transform like S t. Let M C be a complex analytic manifold (i.e., M C = C n ) and M M C a totally real analytic submanifold. Thus the restriction map is injective. O(M C ) F F M A(M) Let F(M C ) be a Hilbert space of holomorphic function on M C such that the point-evaluation maps F F(w) are continuous and hence given by the inner product with an element K w F(M C ): F F(M C ) : F(w) = (F, K w ) - p. 17/54

62 The function K : M C M C C, K(z, w) = K w (z) is reproducing kernel of F(M C ). It satisfies: 1. K is holomorphic in the first variable and anti-holomorphic in the second variable. 2. K(z, w) = K(w, z) because 3. K w 2 = K(w, w) 0. K w (z) = (K w, K z ) = (K z, K w ) = K z (w) = K(w, z). - p. 18/54

63 The function K : M C M C C, K(z, w) = K w (z) is reproducing kernel of F(M C ). It satisfies: 1. K is holomorphic in the first variable and anti-holomorphic in the second variable. 2. K(z, w) = K(w, z) because 3. K w 2 = K(w, w) 0. K w (z) = (K w, K z ) = (K z, K w ) = K z (w) = K(w, z). Furthermore, the linear hull of {K x } x M is dense in F(M C ) and hence: the reproducing kernel determines F(M C ), knowing K(z, w) is the same as knowing F(M C ). - p. 18/54

64 The function K : M C M C C, K(z, w) = K w (z) is reproducing kernel of F(M C ). It satisfies: 1. K is holomorphic in the first variable and anti-holomorphic in the second variable. 2. K(z, w) = K(w, z) because 3. K w 2 = K(w, w) 0. K w (z) = (K w, K z ) = (K z, K w ) = K z (w) = K(w, z). Furthermore, the linear hull of {K x } x M is dense in F(M C ) and hence: the reproducing kernel determines F(M C ), knowing K(z, w) is the same as knowing F(M C ). Assume that F is orthogonal to the linear span of {K x } x M. Then F(x) = (F, K x ) = 0 for all x M and hence F M = 0. As M is a totally real submanifold, it follows that F = 0. - p. 18/54

65 We now make the following assumption: There exists a measure µ on M and a holomorphic function D : M C C, D M > 0, D(z) 0, such that - p. 19/54

66 We now make the following assumption: There exists a measure µ on M and a holomorphic function D : M C C, D M > 0, D(z) 0, such that 1. For all F F(M C ) we have R(F) := (DF) M L 2 (M, µ). 2. R(F(M C )) is dense in L 2 (M) (can be dropped, but then we have only a partial isometry later). 3. R : F(M C ) L 2 (M) is a closed operator. - p. 19/54

67 We now make the following assumption: There exists a measure µ on M and a holomorphic function D : M C C, D M > 0, D(z) 0, such that 1. For all F F(M C ) we have R(F) := (DF) M L 2 (M, µ). 2. R(F(M C )) is dense in L 2 (M) (can be dropped, but then we have only a partial isometry later). 3. R : F(M C ) L 2 (M) is a closed operator. Then R : L 2 (M, dµ) F(M C ) is densely defined and R = U RR where U : L 2 (M, dµ) F(M C ) is an unitary isomorphism by definition. - p. 19/54

68 We now make the following assumption: There exists a measure µ on M and a holomorphic function D : M C C, D M > 0, D(z) 0, such that 1. For all F F(M C ) we have R(F) := (DF) M L 2 (M, µ). 2. R(F(M C )) is dense in L 2 (M) (can be dropped, but then we have only a partial isometry later). 3. R : F(M C ) L 2 (M) is a closed operator. Then R : L 2 (M, dµ) F(M C ) is densely defined and R = U RR where U : L 2 (M, dµ) F(M C ) is an unitary isomorphism by definition. We call U the generalized Segal-Bargmann transform. - p. 19/54

69 Note the following: R f(w) = (R f, K w ) F = (f, RK w ) = and hence RR f(x) = Hence RR is always an integral operator. M M f(y)d(y)k(w, y) dx f(y)d(x)d(y)k(y, x) dµ(y). - p. 20/54

70 Note the following: and hence R f(w) = (R f, K w ) F = (f, RK w ) = RR f(x) = Hence RR is always an integral operator. M M f(y)d(y)k(w, y) dx f(y)d(x)d(y)k(y, x) dµ(y). Furthermore, by multiplying by U, and then using that RR is self-adjoint, we get the following formula for RU and then U: U R = RU = RR or Uf(x) = D(x) 1 RR (f)(x). }{{} holomorphic - p. 20/54

71 Note the following: and hence R f(w) = (R f, K w ) F = (f, RK w ) = RR f(x) = Hence RR is always an integral operator. M M f(y)d(y)k(w, y) dx f(y)d(x)d(y)k(y, x) dµ(y). Furthermore, by multiplying by U, and then using that RR is self-adjoint, we get the following formula for RU and then U: U R = RU = RR or Uf(x) = D(x) 1 RR (f)(x). }{{} holomorphic But what is RR? - p. 20/54

72 We apply this now to M = R n C n, F = F t and D(z) = h t (z). Then: RR f(x) = f(y)h t (x)h t (y)k(x, y) dy = 2 n (πt) 3n/2 f h t (x). - p. 21/54

73 We apply this now to M = R n C n, F = F t and D(z) = h t (z). Then: RR f(x) = f(y)h t (x)h t (y)k(x, y) dy = 2 n (πt) 3n/2 f h t (x). As {H t } t 0 is a semi-group it follows that RR f(x) = 2 n/2 (πt) 3n/4 f h t/2 (x) and hence Uf(x) = h t (x) 1 2 n/2 (πt) 3n/4 f h t2 (x) = (πt) n/4 f(y)e 1 2t (y2 2xy+ x2 2 ) dy. - p. 21/54

74 We apply this now to M = R n C n, F = F t and D(z) = h t (z). Then: RR f(x) = f(y)h t (x)h t (y)k(x, y) dy = 2 n (πt) 3n/2 f h t (x). As {H t } t 0 is a semi-group it follows that RR f(x) = 2 n/2 (πt) 3n/4 f h t/2 (x) and hence Uf(x) = h t (x) 1 2 n/2 (πt) 3n/4 f h t2 (x) = (πt) n/4 f(y)e 1 2t (y2 2xy+ x2 2 ) dy. Thus U = S t and S t is an unitary isomorphism. - p. 21/54

75 Let me know recollect the problems/phylosophy for the general case: - p. 22/54

76 Let me know recollect the problems/phylosophy for the general case: If M is a Riemannian manifold, then the elliptic differential operator is well defined and invariant under isometries of M. Let dσ be the volume form on M. Then the heat equation is given as before: u(x, t) = t u(x, t), and the solution is (by definition) given by lim u(x, t) = f(x) t 0 L2 (M, dσ). + H t f(x) = e t f(x). - p. 22/54

77 Let me know recollect the problems/phylosophy for the general case: If M is a Riemannian manifold, then the elliptic differential operator is well defined and invariant under isometries of M. Let dσ be the volume form on M. Then the heat equation is given as before: u(x, t) = t u(x, t), and the solution is (by definition) given by lim u(x, t) = f(x) t 0 L2 (M, dσ). + H t f(x) = e t f(x). But more importantly, there exists a function h t (x, y), the heat kernel, such that: h t (x, y) = h t (y, x) 0; dµ t (y) = h t (x, y)dσ(y) is a probability measure on M; If g : M M is an isometry, then h t (gx, gy) = h(x, y). H t f(x) = f(y)h t (x, y) dσ(y); M - p. 22/54

78 But to generalize the previous results one need to find a natural complexification M C for M such that h t, and H t f extend to holomorphic functions on M C. - p. 23/54

79 But to generalize the previous results one need to find a natural complexification M C for M such that h t, and H t f extend to holomorphic functions on M C. Then we have to find a Hilbert space H t (M C ) O(M C ) such that the transform L 2 (M) f H t f H t (M C ) becomes an unitary isomorphism. In particular, what is the natural generalization of the measure µ t? - p. 23/54

80 But to generalize the previous results one need to find a natural complexification M C for M such that h t, and H t f extend to holomorphic functions on M C. Then we have to find a Hilbert space H t (M C ) O(M C ) such that the transform L 2 (M) f H t f H t (M C ) becomes an unitary isomorphism. In particular, what is the natural generalization of the measure µ t? There is one class of Riemannian manifolds where a natural complexification exists. Those are the Riemannian symmetric spaces G/K, where G is a connected and semisimple Lie group. - p. 23/54

81 But to generalize the previous results one need to find a natural complexification M C for M such that h t, and H t f extend to holomorphic functions on M C. Then we have to find a Hilbert space H t (M C ) O(M C ) such that the transform L 2 (M) f H t f H t (M C ) becomes an unitary isomorphism. In particular, what is the natural generalization of the measure µ t? There is one class of Riemannian manifolds where a natural complexification exists. Those are the Riemannian symmetric spaces G/K, where G is a connected and semisimple Lie group. B. Hall in 1997 for compact connected Lie groups. Here G = M G C = M C T G. Here G C is a complex Lie group with Lie algebra g C = g R C, i.e., G = SO(n) SO(n, C) SO(n) exp{x im(n, R) X = X}. - p. 23/54

82 M.B. Stenzel in 1999 for symmetric spaces M = G/K, where G is compact. Here M C = G C /K C T(G/K). Here G is a compact connected Lie group, τ : G G is a non-trivial involution and i.e, S n = SO(n + 1)/SO(n). K = G τ = {g G τ(g) = g} Note, that Hall s result is a special case as G G G/G with τ(a, b) = (b, a). - p. 24/54

83 M.B. Stenzel in 1999 for symmetric spaces M = G/K, where G is compact. Here M C = G C /K C T(G/K). Here G is a compact connected Lie group, τ : G G is a non-trivial involution and i.e, S n = SO(n + 1)/SO(n). K = G τ = {g G τ(g) = g} Note, that Hall s result is a special case as G G G/G with τ(a, b) = (b, a). The restriction principle was formulated in general by G. Ólafsson and B. Ørsted in 1996, but it had been applied earlier by G. Zhang and B. Ørsted in special cases. - p. 24/54

84 M.B. Stenzel in 1999 for symmetric spaces M = G/K, where G is compact. Here M C = G C /K C T(G/K). Here G is a compact connected Lie group, τ : G G is a non-trivial involution and i.e, S n = SO(n + 1)/SO(n). K = G τ = {g G τ(g) = g} Note, that Hall s result is a special case as G G G/G with τ(a, b) = (b, a). The restriction principle was formulated in general by G. Ólafsson and B. Ørsted in 1996, but it had been applied earlier by G. Zhang and B. Ørsted in special cases. B. Hall and J.J. Mitchell did the case M = G/K where G is complex or of rank one in p. 24/54

85 B. Krötz, G. Ólafsson, and R. Stanton: 2005 the general case G/K where G is non-compact and semisimple and K is a maximal compact subgroup, i.e., SL(n, R)/SO(n). - p. 25/54

86 B. Krötz, G. Ólafsson, and R. Stanton: 2005 the general case G/K where G is non-compact and semisimple and K is a maximal compact subgroup, i.e., SL(n, R)/SO(n). A different formula for the K-invariant function on G/K and generalization to arbitrary non-negative multiplicity functions by G. Ólafsson and H. Schlichtkrull (Copenhagen) in 2005, to appear in Adv. Math. - p. 25/54

87 B. Krötz, G. Ólafsson, and R. Stanton: 2005 the general case G/K where G is non-compact and semisimple and K is a maximal compact subgroup, i.e., SL(n, R)/SO(n). A different formula for the K-invariant function on G/K and generalization to arbitrary non-negative multiplicity functions by G. Ólafsson and H. Schlichtkrull (Copenhagen) in 2005, to appear in Adv. Math. One of the reasons, that it took so long to get from the compact case to the non-compact case is, that it was not so clear, what the right complexification of G/K is. It is the Akhiezer-Gindikin domain also called the complex crown which I will define in a moment. But first we will need some basic structure theory for semisimple symmetric space of the non-compact type. - p. 25/54

88 5. Structure Theory - p. 26/54

89 5. Structure Theory G a connected, non-compact semisimple Lie group with finite center - p. 26/54

90 5. Structure Theory G a connected, non-compact semisimple Lie group with finite center K G a maximal compact subgroup, and θ : G G the corresponding Cartan involution: K = G θ = {g G θ(g) = g}. - p. 26/54

91 5. Structure Theory G a connected, non-compact semisimple Lie group with finite center K G a maximal compact subgroup, and θ : G G the corresponding Cartan involution: K = G θ = {g G θ(g) = g}. Denote the corresponding involution on the Lie algebra g by the same letter θ and let k = {X g θ(x) = X} p = {X g θ(x) = X} - p. 26/54

92 5. Structure Theory G a connected, non-compact semisimple Lie group with finite center K G a maximal compact subgroup, and θ : G G the corresponding Cartan involution: K = G θ = {g G θ(g) = g}. Denote the corresponding involution on the Lie algebra g by the same letter θ and let k = {X g θ(x) = X} p = {X g θ(x) = X} We have the Cartan decomposition g = k p.. - p. 26/54

93 Our standard example is G = SL(n, R), K = SO(n) and θ(g) = (g 1 ) T. The corresponding involution on the Lie algebra sl(n, R) = {X M n (R) Tr(X) = 0} is θ(x) = X T. The decomposition g = k p corresponds to the decomposition of sl(n, R) into skew-symmetric (= k) and symmetric (= p) matrices. - p. 27/54

94 Our standard example is G = SL(n, R), K = SO(n) and θ(g) = (g 1 ) T. The corresponding involution on the Lie algebra sl(n, R) = {X M n (R) Tr(X) = 0} is θ(x) = X T. The decomposition g = k p corresponds to the decomposition of sl(n, R) into skew-symmetric (= k) and symmetric (= p) matrices. Recall the linear map ad(x) : g g, Y [X, Y ] and define an inner product on g by (X, Y ) = Tr(ad(X), ad(θ(y ))) - p. 27/54

95 Our standard example is G = SL(n, R), K = SO(n) and θ(g) = (g 1 ) T. The corresponding involution on the Lie algebra sl(n, R) = {X M n (R) Tr(X) = 0} is θ(x) = X T. The decomposition g = k p corresponds to the decomposition of sl(n, R) into skew-symmetric (= k) and symmetric (= p) matrices. Recall the linear map ad(x) : g g, Y [X, Y ] and define an inner product on g by (X, Y ) = Tr(ad(X), ad(θ(y ))) On sl(n, R) this is (X, Y ) = 2nTr(XY T ). - p. 27/54

96 Our standard example is G = SL(n, R), K = SO(n) and θ(g) = (g 1 ) T. The corresponding involution on the Lie algebra sl(n, R) = {X M n (R) Tr(X) = 0} is θ(x) = X T. The decomposition g = k p corresponds to the decomposition of sl(n, R) into skew-symmetric (= k) and symmetric (= p) matrices. Recall the linear map ad(x) : g g, Y [X, Y ] and define an inner product on g by (X, Y ) = Tr(ad(X), ad(θ(y ))) On sl(n, R) this is (X, Y ) = 2nTr(XY T ). If X p then ad(x) = ad(x), i.e., ad(x) is symmetric. - p. 27/54

97 Let a R n (for some n) be a maximal abelian subspace of p, i.e., all diagonal matrices with trace zero. m = = - p. 28/54

98 Let a R n (for some n) be a maximal abelian subspace of p, i.e., all diagonal matrices with trace zero. Then {ad(x) X a} is a commuting family of symmetric operators and has therefore a joint basis for g of eigenvectors. Thus we set: g α = m = = - p. 28/54

99 Let a R n (for some n) be a maximal abelian subspace of p, i.e., all diagonal matrices with trace zero. Then {ad(x) X a} is a commuting family of symmetric operators and has therefore a joint basis for g of eigenvectors. Thus we set: g α = {X g ( H a) [H, X] = α(h)x}, α a \ {0} m = = - p. 28/54

100 Let a R n (for some n) be a maximal abelian subspace of p, i.e., all diagonal matrices with trace zero. Then {ad(x) X a} is a commuting family of symmetric operators and has therefore a joint basis for g of eigenvectors. Thus we set: g α = {X g ( H a) [H, X] = α(h)x}, α a \ {0} m = z k (a) = {X g ( H a) [H, X] = 0} z g (a) = m a = - p. 28/54

101 Let a R n (for some n) be a maximal abelian subspace of p, i.e., all diagonal matrices with trace zero. Then {ad(x) X a} is a commuting family of symmetric operators and has therefore a joint basis for g of eigenvectors. Thus we set: g α = {X g ( H a) [H, X] = α(h)x}, α a \ {0} m = z k (a) = {X g ( H a) [H, X] = 0} z g (a) = m a = {α a \ {0} g α = {0}}. - p. 28/54

102 Let a R n (for some n) be a maximal abelian subspace of p, i.e., all diagonal matrices with trace zero. Then {ad(x) X a} is a commuting family of symmetric operators and has therefore a joint basis for g of eigenvectors. Thus we set: g α = {X g ( H a) [H, X] = α(h)x}, α a \ {0} m = z k (a) = {X g ( H a) [H, X] = 0} z g (a) = m a = {α a \ {0} g α = {0}}. Let X a be such that α(x) 0 for all α. This is possible as set {H a ( α ) α(h) = 0} is a finite union of hyperplanes and hence nowhere dense. - p. 28/54

103 Let a R n (for some n) be a maximal abelian subspace of p, i.e., all diagonal matrices with trace zero. Then {ad(x) X a} is a commuting family of symmetric operators and has therefore a joint basis for g of eigenvectors. Thus we set: g α = {X g ( H a) [H, X] = α(h)x}, α a \ {0} m = z k (a) = {X g ( H a) [H, X] = 0} z g (a) = m a = {α a \ {0} g α = {0}}. Let X a be such that α(x) 0 for all α. This is possible as set {H a ( α ) α(h) = 0} is a finite union of hyperplanes and hence nowhere dense. Let + := {α α(x) > 0}. Then as α θ = α we have = ( + ) and ( ) +. - p. 28/54

104 As [g λ, g µ ] g µ+λ it follows that n := α + g α is a nilpotent Lie algebra such that [m a, n] n = - p. 29/54

105 As [g λ, g µ ] g µ+λ it follows that is a nilpotent Lie algebra such that and with n := θ(n) = α g α n := α + g α [m a, n] n g = α g α m a }{{} the zero eigenspace = = n m a n = - p. 29/54

106 As [g λ, g µ ] g µ+λ it follows that is a nilpotent Lie algebra such that and with n := θ(n) = α g α n := α + g α [m a, n] n g = α g α } m {{ a } = n m a n the zero eigenspace = ( (id + θ)(g α ) m) a n α } + {{} =k = - p. 29/54

107 As [g λ, g µ ] g µ+λ it follows that is a nilpotent Lie algebra such that and with n := θ(n) = α g α n := α + g α [m a, n] n g = α g α } m {{ a } = n m a n the zero eigenspace = ( (id + θ)(g α ) m) a n α } + {{} =k = k a n the Iwasawa decomposition - p. 29/54

108 On the group level this corresponds to Theorem 0.2 (Iwasawa Decomposition). The map N A K (n,a,k) nak G is an analytic isomorphism. We write x = N n(x) A a(x) K k(x) for the unique decomposition of x G. In particular G/K N A. - p. 30/54

109 On the group level this corresponds to Theorem 0.3 (Iwasawa Decomposition). The map N A K (n,a,k) nak G is an analytic isomorphism. We write x = N n(x) A a(x) K k(x) for the unique decomposition of x G. In particular G/K N A. We assume that G G C, where Lie(G C ) = g C. Then we can complexify all the groups under consideration and obtain N C, A C and K C. Then N C A C K C G C is open and dense but not equal to G C. Furthermore, the decomposition x = n(x)a(x)k(x) N C A C K C is not unique in general. - p. 30/54

110 For our standard example this corresponds to: a = {diag(x i ) x i = 0} = {x R n x 1 + x x n = 0} R n 1 A = = + = N = - p. 31/54

111 For our standard example this corresponds to: a = {diag(x i ) x i = 0} = {x R n x 1 + x x n = 0} R n 1 A = {diag(a i ) ( i)a i > 0, a 1 a n = 1} = + = N = - p. 31/54

112 For our standard example this corresponds to: a = {diag(x i ) x i = 0} = {x R n x 1 + x x n = 0} R n 1 A = {diag(a i ) ( i)a i > 0, a 1 a n = 1} = set of α ij : diag(x k ) x i x j, i j + = N = - p. 31/54

113 For our standard example this corresponds to: a = {diag(x i ) x i = 0} = {x R n x 1 + x x n = 0} R n 1 A = {diag(a i ) ( i)a i > 0, a 1 a n = 1} = set of α ij : diag(x k ) x i x j, i j + = {α ij i < j} N = - p. 31/54

114 For our standard example this corresponds to: a = {diag(x i ) x i = 0} = {x R n x 1 + x x n = 0} R n 1 A = {diag(a i ) ( i)a i > 0, a 1 a n = 1} = set of α ij : diag(x k ) x i x j, i j + = {α ij i < j} N = {(x ij ) i > j : x ij = 0, x ii = 1} - p. 31/54

115 For our standard example this corresponds to: a = {diag(x i ) x i = 0} = {x R n x 1 + x x n = 0} R n 1 A = {diag(a i ) ( i)a i > 0, a 1 a n = 1} = set of α ij : diag(x k ) x i x j, i j + = {α ij i < j} N = {(x ij ) i > j : x ij = 0, x ii = 1} The Iwasawa decomposition follows directly from the Gram-Schmidt orthogonalization. - p. 31/54

116 For our standard example this corresponds to: a = {diag(x i ) x i = 0} = {x R n x 1 + x x n = 0} R n 1 A = {diag(a i ) ( i)a i > 0, a 1 a n = 1} = set of α ij : diag(x k ) x i x j, i j + = {α ij i < j} N = {(x ij ) i > j : x ij = 0, x ii = 1} The Iwasawa decomposition follows directly from the Gram-Schmidt orthogonalization. Note, for n = 2 this is ( ) ( ) ( ac+bd a b 1 1 = c 2 +d 2 c d and this breaks down as c 2 + d 2 = 0. c 2 +d 2 0 c2 + d 2 ) ( d c2 +d 2 c c 2 +d 2 c c2 +d 2 d c 2 +d 2 ) - p. 31/54

117 6. Spherical Functions and the Fourier Transform - p. 32/54

118 6. Spherical Functions and the Fourier Transform We will also need the Weyl group. It is the finite reflection group in O(a) generated by the reflections r α in the hyperplanes α = 0. It is denoted by W. We have W N K (a)/m M = Z K (a). Permutation of the coordinates for our standard case. - p. 32/54

119 6. Spherical Functions and the Fourier Transform We will also need the Weyl group. It is the finite reflection group in O(a) generated by the reflections r α in the hyperplanes α = 0. It is denoted by W. We have W N K (a)/m M = Z K (a). Permutation of the coordinates for our standard case. For a differential operator D : C c (G/K) C c (G/K) and g G, let (g D)(f) = D(f L g 1f) L g. Then D is G-invariant if g D = D for all g G. Thus D is G-invariant if and only if D commutes with translation D(f L g ) = [D(f)] L g. Denote by D(G/K) the commutative algebra of all invariant differential operators on G/K. On R n this is just the algebra of constant coefficient differential operators D(R n ) = C[ 1,..., n ]. - p. 32/54

120 For λ a C let ϕλ(x) := K a(kx) λ+ρ dk. The functions ϕ λ are the spherical functionson G/K. We have ϕ λ = ϕ µ w W : λ = wµ. - p. 33/54

121 For λ a C let ϕλ(x) := K a(kx) λ+ρ dk. The functions ϕ λ are the spherical functionson G/K. We have ϕ λ = ϕ µ w W : λ = wµ. The spherical functions are K-invariant eigenfunctions of D(G/K). In particular for the Laplace operator G/K D(G/K): G/K ϕ λ = (λ 2 ρ 2 )ϕ λ where m α = dimg α and ρ = 1 m α α. 2 α + - p. 33/54

122 For λ a C let ϕλ(x) := K a(kx) λ+ρ dk. The functions ϕ λ are the spherical functionson G/K. We have ϕ λ = ϕ µ w W : λ = wµ. The spherical functions are K-invariant eigenfunctions of D(G/K). In particular for the Laplace operator G/K D(G/K): G/K ϕ λ = (λ 2 ρ 2 )ϕ λ where m α = dimg α and ρ = 1 2 α + m α α. In the harmonic analysis of K-invariant functions on G/K they play the same role as the exponential functions e λ (x) = e λ x on R n. We will discuss that in more details later on. - p. 33/54

123 Let - p. 34/54

124 Let B = M\K where M = Z K (A), - p. 34/54

125 Let B = M\K where M = Z K (A), a + = {λ a ( α + )(λ, α) > 0} it is a fundamental domain for the W -action on a. - p. 34/54

126 Let B = M\K where M = Z K (A), a + = {λ a ( α + )(λ, α) > 0} it is a fundamental domain for the W -action on a. dσ(b, λ) = db dλ c(λ) 2 where c(λ) is the Harish-Chandra c-function. - p. 34/54

127 Let B = M\K where M = Z K (A), a + = {λ a ( α + )(λ, α) > 0} it is a fundamental domain for the W -action on a. dσ(b, λ) = db dλ c(λ) where c(λ) is the Harish-Chandra c-function. 2 For f C c (G/K) define the Fourier transform ˆf : B a C, of f by ˆf(b, λ) = f(x)a(bx) ρ iλ dx. G/K - p. 34/54

128 Let B = M\K where M = Z K (A), a + = {λ a ( α + )(λ, α) > 0} it is a fundamental domain for the W -action on a. dσ(b, λ) = db dλ c(λ) where c(λ) is the Harish-Chandra c-function. 2 For f C c (G/K) define the Fourier transform ˆf : B a C, of f by ˆf(b, λ) = f(x)a(bx) ρ iλ dx. G/K Theorem 0.3 (Helgason). 1. The Fourier transform extends to an unitary isomorphism F : L 2 (G/K) L 2 (B a,dσ)+some W -invariance. 2. If f C c (G/K) then f(x) = c ˆf(b,λ)a(bx) iλ+ρ dσ. B a 3. We have F( G/K f)(b,λ) = (λ 2 ρ 2 ) ˆf(b,λ). - p. 34/54

129 For K-invariant functions, this reduces to the Harish-Chandra spherical Fourier transform ˆf(λ) = f(x)ϕ iλ (x) dx. and the spherical Fourier transform extends to an unitary isomorphism L 2 (G/K) K f ˆf L 2 (a, dλ c(λ) 2)W L 2 (a dλ +, W c(λ) 2) - p. 35/54

130 For K-invariant functions, this reduces to the Harish-Chandra spherical Fourier transform ˆf(λ) = f(x)ϕ iλ (x) dx. and the spherical Fourier transform extends to an unitary isomorphism L 2 (G/K) K f ˆf L 2 (a, dλ c(λ) 2)W L 2 (a dλ +, W c(λ) 2) with inversion formula f(x) = 1 W a + ˆf(λ)ϕ iλ (x) dλ c(λ) 2. - p. 35/54

131 Using the Fourier transform and part (3) of Helgason s Theorem we get the following form for the solution of the heat equation: H t f(x) = e (λ2 +ρ 2 )t ˆf(b, λ) a(bx) iλ+ρ dσ(b, λ) = f h t (x). Note the ρ 2 -shift! - p. 36/54

132 Using the Fourier transform and part (3) of Helgason s Theorem we get the following form for the solution of the heat equation: H t f(x) = e (λ2 +ρ 2 )t ˆf(b, λ) a(bx) iλ+ρ dσ(b, λ) Note the ρ 2 -shift! = f h t (x). For the heat kernel we get the expression: 1 h t (x) = e ( λ 2 + ρ 2 )t ϕ iλ (x) W a + 1 = 2 + ρ 2 )t W 2 ϕ iλ (x) a e ( λ dλ c(λ) 2 dλ c(λ) 2. - p. 36/54

133 Using the Fourier transform and part (3) of Helgason s Theorem we get the following form for the solution of the heat equation: H t f(x) = e (λ2 +ρ 2 )t ˆf(b, λ) a(bx) iλ+ρ dσ(b, λ) Note the ρ 2 -shift! = f h t (x). For the heat kernel we get the expression: 1 h t (x) = e ( λ 2 + ρ 2 )t ϕ iλ (x) W a + 1 = 2 + ρ 2 )t W 2 ϕ iλ (x) a e ( λ dλ c(λ) 2 dλ c(λ) 2. So, how far does x h t (x) extend? Or, how far does x ϕ λ (x) extend, and what is the growth of the extension? - p. 36/54

134 7. The Crown and the Heat Kernel - p. 37/54

135 7. The Crown and the Heat Kernel We define Ω = {X a ( α ) α(x) < π/2} W invariant polytope Ξ = Gexp(iΩ) x o G C /K C where x o is the base point ek C G C /K C. Then Ξ is an open G-invariant subset of G C /K C, the Akhiezer-Gindikin domain or complex crown. It has been studied by several group of people: Barchini, Burns + Halverscheid + Hind, Huckleberry, Krötz + Stanton, Wolf and others. - p. 37/54

136 7. The Crown and the Heat Kernel We define Ω = {X a ( α ) α(x) < π/2} W invariant polytope Ξ = Gexp(iΩ) x o G C /K C where x o is the base point ek C G C /K C. Then Ξ is an open G-invariant subset of G C /K C, the Akhiezer-Gindikin domain or complex crown. It has been studied by several group of people: Barchini, Burns + Halverscheid + Hind, Huckleberry, Krötz + Stanton, Wolf and others. Its importance in harmonic analysis on G/K comes from the following. - p. 37/54

137 7. The Crown and the Heat Kernel We define Ω = {X a ( α ) α(x) < π/2} W invariant polytope Ξ = Gexp(iΩ) x o G C /K C where x o is the base point ek C G C /K C. Then Ξ is an open G-invariant subset of G C /K C, the Akhiezer-Gindikin domain or complex crown. It has been studied by several group of people: Barchini, Burns + Halverscheid + Hind, Huckleberry, Krötz + Stanton, Wolf and others. Its importance in harmonic analysis on G/K comes from the following. Theorem 0.4 (Krötz+Stanton,...). 1. We have Ξ N C A C x o and the Iwasawa projection Ξ ξ a(ξ) A C is well defined and holomorphic. 2. Ξ is a maximal G-invariant domain in G C /K C such that all the joint eigenfunctions for D(G/K) extends to holomorphic functions on Ξ. - p. 37/54

138 It follows that the spherical functions extends to Ξ. With some extra work, involving the the growth of the spherical functions we have: Theorem 0.5 (Krötz+Stanton). The heat kernel extends to a holomorphic function on Ξ given by the same formula h t (ξ) = 1 W a + e ( λ 2 + ρ 2 )t ϕ iλ (ξ)dσ(λ). - p. 38/54

139 It follows that the spherical functions extends to Ξ. With some extra work, involving the the growth of the spherical functions we have: Theorem 0.5 (Krötz+Stanton). The heat kernel extends to a holomorphic function on Ξ given by the same formula h t (ξ) = 1 W a + e ( λ 2 + ρ 2 )t ϕ iλ (ξ)dσ(λ). As a consequence we have that each solution to the heat equation f h t, f L 2 (G/K) extends to a holomorphic function on Ξ: H t f(ξ) = f(gx o )h t (g 1 ξ)dg. G As before, the problem is then to determine the image of H t : L 2 (G/K) O(Ξ). - p. 38/54

140 8. The Abel Transform and the Heat Kernel - p. 39/54

141 8. The Abel Transform and the Heat Kernel Recall that f(x) dx = G/K A N f(na x o )a 2ρ dnda = A N f(an x o )a 2ρ dnda. For a K-invariant f function on G/K, say of compact support, define the Abel transform of f by A(f)(a) = a ρ f(an) dn = a ρ f(na) dn (0.3) N N }{{} The Radon Transform using the notation (exp(x)) λ = e λ(x). Then A(f) is a W -invariant function on A. - p. 39/54

142 8. The Abel Transform and the Heat Kernel Recall that f(x) dx = G/K A N f(na x o )a 2ρ dnda = A N f(an x o )a 2ρ dnda. For a K-invariant f function on G/K, say of compact support, define the Abel transform of f by A(f)(a) = a ρ f(an) dn = a ρ f(na) dn (0.3) N N }{{} The Radon Transform using the notation (exp(x)) λ = e λ(x). Then A(f) is a W -invariant function on A. and we have A( G/K f) = ( A ρ 2 )A(f). (0.4) - p. 39/54

143 8. The Abel Transform and the Heat Kernel Recall that f(x) dx = G/K A N f(na x o )a 2ρ dnda = A N f(an x o )a 2ρ dnda. For a K-invariant f function on G/K, say of compact support, define the Abel transform of f by A(f)(a) = a ρ f(an) dn = a ρ f(na) dn (0.3) N N }{{} The Radon Transform using the notation (exp(x)) λ = e λ(x). Then A(f) is a W -invariant function on A. and we have A( G/K f) = ( A ρ 2 )A(f). (0.4) - p. 39/54

144 We have the following Fourier slice theorem for K-invariant functions: ˆf(λ) = f(x)ϕ iλ (x) dx = = G/K G/K A f(x)a(k 1 x) iλ+ρ dx (a ρ N = F A (A(f))(λ) ) f(na x o ) dn a iλ da - p. 40/54

145 We have the following Fourier slice theorem for K-invariant functions: ˆf(λ) = f(x)ϕ iλ (x) dx = = G/K G/K A f(x)a(k 1 x) iλ+ρ dx (a ρ N = F A (A(f))(λ) ) f(na x o ) dn a iλ da Or F G/K = F A A. - p. 40/54

146 We have the following Fourier slice theorem for K-invariant functions: ˆf(λ) = f(x)ϕ iλ (x) dx = = G/K G/K A f(x)a(k 1 x) iλ+ρ dx (a ρ N = F A (A(f))(λ) ) f(na x o ) dn a iλ da Or F G/K = F A A. Equation (0.4) implies also that h t (expx) = e ρ 2 t }{{} the ρ shift A 1 }{{} a shift operator ((4πt) n/2 e X 2 /4t ) } {{ } the heat kernel on A. - p. 40/54

147 Define a holomorphic function on a C. ψ λ (expx) = 1 W w W e wλ(x) - p. 41/54

148 Define a holomorphic function on a C. ψ λ (expx) = 1 W w W e wλ(x) Then the equation F G/K = F A A implies that A (ψ λ ) = ϕ λ. The Abel/Radon transform shifts the flat case to the curved case! - p. 41/54

149 Define a holomorphic function on a C. ψ λ (expx) = 1 W w W e wλ(x) Then the equation F G/K = F A A implies that A (ψ λ ) = ϕ λ. The Abel/Radon transform shifts the flat case to the curved case! Define now the pseudo-differential operator D on A by: D = F 1 A 1 c(λ) 2 F G/K - p. 41/54

150 Define a holomorphic function on a C. ψ λ (expx) = 1 W w W e wλ(x) Then the equation F G/K = F A A implies that A (ψ λ ) = ϕ λ. The Abel/Radon transform shifts the flat case to the curved case! Define now the pseudo-differential operator D on A by: D = F 1 A 1 c(λ) 2 F G/K or for good W -invariant functions: Dh(a) = a + then the multiplier {}}{ F G/K (h)(λ) }{{} Firt the FT on G/K 1 c(λ) 2 ψ iλ(a) dλ } {{ } back using F 1 A. - p. 41/54

151 9. The Faraut-Gutzmer Formula and the Orbital Integral - p. 42/54

152 9. The Faraut-Gutzmer Formula and the Orbital Integral For sufficiently decreasing functions h : Ξ C we define the G-orbital integral O h : 2iΩ C by ( ) i O h (Y ) = h(g exp 2 Y x o ) dg. G - p. 42/54