THE HEISENBERG

NEIL LYALL

Contents 1. Fourier transform on Rn 1 2. on the Heisenberg group 2 2.1. Representations of the Heisenberg group 2 2.2. Group Fourier transform 3 2.3. Convolution and twisted convolution 5 3. Hermite and Laguerre functions 6 3.1. Hermite polynomials 6 3.2. Laguerre polynomials 9 3.3. Special Hermite functions 9 4. Group Fourier transform of radial functions on the Heisenberg group 12 References 13

1. Fourier transform on Rn

We start by presenting some standard properties of the Euclidean Fourier transform; see for example [6] and [4].

Given f ∈ L1(Rn), we define its Fourier transform by setting Z fb(ξ) = e−ix·ξf(x)dx. Rn

n ih·ξ If for h ∈ R we let (τhf)(x) = f(x + h), then it follows that τdhf(ξ) = e fb(ξ). Now for suitable f the inversion formula Z f(x) = (2π)−n eix·ξfb(ξ)dξ, Rn holds and we see that the Fourier transform decomposes a function into a continuous sum of characters (eigenfunctions for translations).

If A is an orthogonal matrix and ξ is a column vector then f[◦ A(ξ) = fb(Aξ) and from this it follows that the Fourier transform of a radial function is again radial. In particular the Fourier transform −|x|2/2 n of Gaussians take a particularly nice form; if G(x) = e , then Gb(ξ) = (2π) 2 G(ξ). In general the Fourier transform of a radial function can always be explicitly expressed in terms of a Bessel 1 2 NEIL LYALL transform; if g(x) = g0(|x|) for some function g0, then Z ∞ n 2−n n−1 gb(ξ) = (2π) 2 g0(r)(r|ξ|) 2 J n−2 (r|ξ|)r dr, 0 2 where J n−2 is a Bessel function. 2

If f ∈ L2(Rn) then Plancherel’s theorem states that fb ∈ L2(Rn), more precisely 2 n 2 kfbk2 = (2π) kfk2. If for f, g ∈ L1(Rn) we define convolution by Z (f ∗ g)(x) = f(x − y)g(y)dy, then we have f[∗ g(ξ) = fb(ξ)gb(ξ).

2. Fourier analysis on the Heisenberg group

We now turn our attention to the Heisenberg group Hn, see also Stein [5] and the books by Thangavelu [7] and [8].

2.1. Representations of the Heisenberg group. The Heisenberg group Hn is of course Cn ×R endowed with the group law [z, t] · [w, s] = [z + w, t + s + hz, wi], 1 where the symplectic form h·, ·i is defined by hz, wi = 2 Im(z · w¯), with identity the origin and inverses given by [z, t]−1 = [−z, −t].

The following transformations are automorphisms of the group Hn:

• the nonisotropic dilations [z, t] 7→ δ ◦ [z, t] = [δz, δ2t], for all δ > 0; • the rotations [z, t] 7→ [Uz, t], with U a unitary transformation of Cn.

The of the Heisenberg group is well understood, using the Stone-von Neumann theorem we can give a complete classification of all the irreducible unitary representations of Hn; see Folland [1]. Let U(L2(Rn)) denote the group of unitary operators acting on L2(Rn), and for each λ ∈ R define the mapping n 2 n πλ : H → U(L (R )) as follows. For each (z, t) ∈ Hn, z = x + iy, and ϕ ∈ L2(Rn), we let iλ(x·ξ+ 1 x·y+t) πλ(z, t)ϕ(ξ) = e 2 ϕ(ξ + y). n 2 n It is then easy to check that πλ(z, t) is a homomorphism from H to U(L (R )), that is

πλ(z, t)πλ(w, s) = πλ(z + w, t + s + hz, wi), 2 and that πλ(z, t) is unitary. Moreover, it is continuous in the sense that for ϕ ∈ L ,

kπλ(z, t)ϕ − ϕkL2 → 0 as (z, t) → 0. THE GROUP FOURIER TRANSFORM ON Hn 3

n Thus πλ is a of H . Let us now show that they are irreducible when λ 6= 0. iλt Write πλ(z, t) = πλ(z)e , where πλ(z) = πλ(z, 0). So iλ(xξ+ 1 xy) πλ(z)ϕ(ξ) = e 2 ϕ(ξ + y). 2 2 Suppose M ⊂ L (R) is invariant under all πλ(z, t). If M 6= {0} we will show M = L (R) proving 2 the irreducibility of πλ. Suppose M is a proper invariant subspace of L (R), then there exists 2 n functions f, g ∈ L (R ) such that f ∈ M and (πλ(z)f, g) = 0 for all z. Now Z iλx·ξ y y (πλ(z)f, g) = e f(ξ + )¯g(ξ − )dξ. Rn 2 2 Applying the Plancherel theorem for the Fourier transform in the x variable gives Z  n Z 2 2π y 2 y 2 |(πλ(z)f, g)| dz = |f(ξ + )| |g¯(ξ − )| dξdy Cn |λ| R2n 2 2 which after making a change of variables gives 2π n (1) k(π (z)f, g)k2 = kfk2 kgk2 , λ L2(Cn) |λ| L2(Rn) L2(Rn) so we must have kfkL2(R2)kgkL2(R2) = 0, but this is a contradiction since f, g are non-trivial.

The theorem of Stone-von Neumann, states that, up to unitary equivalence these are all the infinite- dimensional irreducible unitary representations of the Heisenberg group.

It follows from (1) by polarization that if ϕ, ψ, f, g ∈ L2(Rn), then 2π n (2) ((π (z)ϕ, ψ), (π (z)f, g)) = (ϕ, f)(g, ψ). λ λ |λ|

Notice that this result has the following immediate consequence. Suppose {ej} is an orthonormal 2 n system in L (R ), then {ejk} defined by n |λ| 2 e = (π (z)e , e ) jk 2π λ j k forms an orthonormal system in L2(Cn). In actual fact, using properties of the group Fourier transform, we will show that {ejk} is an orthonormal basis whenever {ej} is.

2.2. Group Fourier transform. Given f ∈ L1(Hn) the group Fourier transform fb(λ) is defined for each λ 6= 0 as the operator-valued function on the L2(Rn) given by Z fb(λ)ϕ(ξ) = f(z, t)πλ(z, t)ϕ(ξ)dzdt. Hn If ψ is another function in L2(Rn), then Z (fb(λ)ϕ, ψ) = f(z, t)(πλ(z, t)ϕ, ψ)dzdt. Hn Since πλ(z, t) are unitary operators, we have

|(πλ(z, t)ϕ, ψ)| ≤ kϕkL2(Rn)kψkL2(Rn), and so |(fb(λ)ϕ, ψ)| ≤ kϕkL2(Rn)kψkL2(Rn)kfkL1(Hn). 4 NEIL LYALL

2 n Hence fb(λ) is a on L (R ), moreover kfb(λ)kL2(Rn)→L2(Rn) ≤ kfkL1(Hn).

If we define Z f λ(z) = eiλtf(z, t)dt, R then it is clear that λ fb(λ)ϕ(ξ) = Wλ(f )ϕ(ξ), where1 Z λ λ Wλ(f )ϕ(ξ) = f (z)πλ(z)ϕ(ξ)dz. Cn Letting z = x + iy = (x, y) we see that Z 1 iλ(x·ξ+ x·y) fb(λ)ϕ(ξ) = f λ(x, y)e 2 ϕ(ξ + y)dxdy Hn Z = Kλ(ξ, η)ϕ(η)dη, Rn where Z λ iλ( x ·(ξ+η)) Kλ(ξ, η) = f (x, η − ξ)e 2 dx. We can now prove the Plancherel theorem for the group Fourier transform.

Theorem 1. If f ∈ L2(Hn), then fb(λ) is a Hilbert-Schmidt operator and Z Z 2 n n+1 2 kfb(λ)kHS|λ| dλ = (2π) |f(z, t)| dzdt. Hn

Proof. Suppose that f ∈ L1 ∩ L2(Hn), it then follows from Plancherel’s theorem for the Fourier transform that Z  n Z 2 2 2π λ 2 kfb(λ)kHS = |Kλ(ξ, η)| dξdη = |f (x, y)| dxdy. R2n |λ| R2n If we now integrate in λ and using Plancherel’s theorem for the Fourier transform in the t variable we get Z Z Z Z 2 2 n n iλt kfb(λ)kHS|λ| dλ = (2π) e f(z, t)dt dzdλ Cn R Z Z Z 2 n iλt = (2π) e f(z, t)dt dλdz, Cn R Z = (2π)n+1 |f(z, t)|2dzdt. Hn An additional limiting argument then proves the theorem.  2 n Corollary 2. If {ej} is an orthonormal basis for L (R ), then {ejk} defined by n |λ| 2 e = (π (z)e , e ), jk 2π λ j k is an orthonormal basis for L2(Cn).

1 When λ = 1 this is called the Weyl transform of f. THE GROUP FOURIER TRANSFORM ON Hn 5

Proof. Orthonormality follows immediately from (2), to prove completeness suppose g ∈ L2(Cn) is orthogonal to all ejk, that is Z g¯(z)(πλ(z)ej, ek)dz = 0. Cn If we now take h ∈ L2(R) and define f(z, t) =g ¯(z)h(t), it follows that Z Z Z iλt (fb(λ)ej, ek) = f(z, t)(πλ(z, t)ej, ek)dzdt = h(t)e dt g¯(z)(πλ(z)ej, ek)dz = 0. Hn R Cn 2 n The completeness of {ek} in L (R ) then implies fb(λ)ej = 0 which shows fb(λ) = 0. Plancherel’s theorem for the group Fourier transform then implies that f = 0. 

2.3. Convolution and twisted convolution. Consider the operator f 7→ f∗K where convolution is taken with respect to the group structure on Hn, that is Z f(w, s)K([w, s]−1 · [z, t])dwds. Hn n Using the fact that πλ(z, t) is a homomorphism from H to the group of unitary operators on L2(Rn), we see that f\∗ K(λ) = fb(λ)Kb(λ). Now it follows from Plancherel’s theorem for the group Fourier transform that the boundedness of our convolution operators on L2(Hn) is equivalent to the uniform boundedness of the operator norms of Kb(λ) over λ 6= 0. Note that in the previous section we saw Z Kb(λ)ϕ(ξ) = Lλ(ξ, η)ϕ(η)dη, Rn where ZZ iλ( x ·(ξ+η)+t) Lλ(ξ, η) = K(x, η − ξ, t)e 2 dxdt.

Alternatively, in view of the Plancherel theorem for the Fourier transform, the boundedness of our convolution operators on L2(Hn) is also equivalent to the estimate Z Z |(f ∗ K)λ(z)|2dz ≤ C |f λ(z)|2dz, Cn Cn where the constant C is independent of λ. Now as Z (f ∗ K)λ(z) = f λ(w)Kλ(z − w)eiλhz,widw, Cn we are naturally led to consider so called twisted convolutions Z iλhz,wi (g ∗λ h)(z) = g(w)h(z − w)e dw. Cn It then follows that boundedness of our convolution operators on L2(Hn) is equivalent to that of λ 2 n the twisted convolution operator f 7→ f ∗λ K on L (C ). Note that Z λ (f ∗λ K )(z) = Mλ(z, w)f(w)dw, Cn 6 NEIL LYALL where Z iλ(t+hz,wi) Mλ(z, w) = K(z − w, t)e dt. Note also that using the fact that iλhz,wi πλ(z)πλ(w) = πλ(z + w)e , it is easy to see that the operator Wλ bears the same relation to the twisted convolution as the group Fourier transform bears to the convolution, namely

Wλ(g ∗λ h) = Wλ(g)Wλ(h).

n However, if our kernels were chosen radial on H , i.e. K(z, t) = K0(|z|, t) for some function K0, then it is a result of Geller [3] that the operators Kb(λ) are in fact diagonal on the Hermite basis for L2(Rn) and that the diagonal entries can be expressed explicitly in terms a Laguerre functions. Therefore, for radial K, studying the boundedness of convolution operators on L2(Hn) reduces to studying the uniform behavior of these diagonal entries.

The remainder of this appendix shall be devoted to the formulation and proof of this result, we start by introducing the Hermite and Laguerre functions.

3. Hermite and Laguerre functions

Fourier analysis on the Heisenberg group is intimately connected with Hermite expansions in Rn.

3.1. Hermite polynomials. Hermite polynomials are defined, for x ∈ R, by k 2 d 2 H (x) = (−1)kex e−x . k dxk By expanding e−(x−r)2 in a Taylor series, multiply by ex2 and then compare terms one obtains the following generating function identity. If |r| < 1, then ∞ k X r 2 2 H (x) = ex e−(x−r) . k k! k=0 ∞ Proposition 3. The Hermite polynomials {Hk}k=0 form a complete orthogonal set on R with −x2 2 k 1 respect to the weight w(x) = e and kHkkw = 2 k!π 2 .

Proof. If P is any polynomial, then Z Z k Z −x2 k d −x2 (k) −x2 P (x)Hk(x)e dx = (−1) P (x) k e dx = P (x)e dx. R R dx R (k) Now if P is a polynomial of degree less that k, in particular if P = Hj with j < k then P ≡ 0 and the orthogonality of the Hermite polynomials follows. Now if P = Hk, then we have P (x) = (2x)k + ... and hence P (k) ≡ 2kk! and Z 2 k −x2 k 1 kHkkw = 2 k! e dx = 2 k!π 2 . R THE GROUP FOURIER TRANSFORM ON Hn 7

2 2 Finally we point out that our family {Hk} is complete in L (R). That is, if f ∈ L (R) and Z −x2 f(x)Hk(x)e dx = 0, for all k, R then f = 0. This is equivalent to

Z 2 I(r) = f(x)e−x e2rxdx = 0. R Now, I(r) is entire, so this equality holds for purely imaginary r. The inverse Fourier transform −x2 applied to f(x)e then gives that f = 0. 

2 − x 2 It is therefore clear that hk(x) = Hk(x)e 2 forms an orthogonal basis for L (R) and furthermore that for each λ 6= 0 the normalized, rescaled Hermite functions

1 λ k − 1 |λ| 4 1 h (x) = (2 k!) 2 h (|λ| 2 x), k π k n form an orthonormal basis. It follows that if we now have x ∈ R and let α = (α1, . . . , αn) then the functions λ λ λ hα(x) = hα1 (x1) . . . hαn (xn), forms an orthonormal basis for L2(Rn), where α ranges over all multi–indices.

Now given any function f on Rn, the Hermite expansion of f is (formally) given by

∞ ∞ X X λ λ X f(x) = (f, hα)hα(x) = Pkf(x), k=0 |α|=k k=0

λ where Pk is the projection onto the kth eigenspace spanned by {hα : |α| = k}. Of course Pk is an integral operator with kernel λ X λ λ hk(x, y) = hα(x)hα(y). |α|=k Lemma 4. For |r| < 1 we have the following identity known as Mehler’s formula, for x, y ∈ R

∞  1/2 (x−ry)2 X k λ λ |λ| 2 −1/2 |λ| (x2−y2) −|λ| r h (x)h (y) = (1 − r ) e 2 e 1−r2 . k k π k=0

Proof. It clearly suffices to prove the following generating function identity for Hk; for |r| < 1, ∞ k (x−ry)2 X r 2 −1/2 x2 − H (x)H (y) = (1 − r ) e e 1−r2 . k k 2kk! k=0 To see this first recall Z 2 2 e−i2uxe−u du = π1/2e−x . R 8 NEIL LYALL

Now,

 k 2 d 2 H (x) = (−1)kex e−x k dx  k 2 d Z 2 = (−1)kex π−1/2 e−i2uxe−u du dx R 2 Z 2 = ex π−1/2 (i2u)ke−i2uxe−u du. R So,

∞ k ZZ ∞ k X r 1 2 2 X (−2uvr) 2 2 H (x)H (y) = ex +y e−i2uxe−u e−i2vye−v dudv k k 2kk! π k! k=0 k=0   1 2 2 Z 2 Z 2 = ex +y e−i2uxe−u e−i2vye−2uvre−v dv du. π

But,

Z 2 2 2 Z 2 2 2 2 e−i2vye−2uvre−v dv = eu r ei2ury e−i2vye−v dv = π1/2eu r ei2urye−y . R Therefore,

∞ k Z X r 2 2 2 2 H (x)H (y) = π−1/2ex e−i2uxe−u eu r ei2urydu k k 2kk! k=0

2 Z 2 2 = π−1/2ex e−u (1−r )e−i2u(x−ry)du

(x−ry) 2 Z 2 −i2u = π−1/2ex (1 − r2)−1/2 e−u e (1−r2)1/2 du

(x−ry)2 2 −1/2 x2 − = (1 − r ) e e 1−r2 .



The Mehler kernel is defined, for |r| < 1, by

∞ X k λ M(x, y, r) = r hk(x, y). k=0 It therefore follows from Mehler’s formula that,

n/2 |λ| − |λ| ((|x|2+|y|2)(1+r2)−4rx·y) M(x, y, r) = (1 − r2)−n/2e 2(1−r2) . π

λ 2 2 Remark 5. The Hermite functions hα are the normalized eigenfunctions of the operator λ |x| − ∆, corresponding to the eigenvalues |λ|(2|α| + n). THE GROUP FOURIER TRANSFORM ON Hn 9

3.2. Laguerre polynomials. Laguerre polynomials of type δ > −1 are defined, for x ∈ (0, ∞), by 1 dk Lδ (x) = exx−δ (e−xxk+δ). k k! dxk These are clearly polynomials of degree k and from the product rule for derivatives it follows that k X k + δ(−x)j Lδ (x) = . k k − j j! j=0 From this it is easy to see that d Lδ (x) = −Lδ+1 (x). dx k k−1 We also have the following generating function for the Laguerre polynomials; for x > 0 and |r| < 1, ∞ r X δ k −δ−1 − 1−r x (3) Lk(x)r = (1 − r) e . k=0 δ ∞ Proposition 6. The Laguerre polynomials {Lk}k=0 form a complete orthogonal set on (0, ∞) with −x δ δ 2 (k+δ)! respect to the weight w(x) = e x and kLkkw = k! .

Proof. If P is any polynomial, then Z ∞ Z ∞ k k Z ∞ δ −x δ 1 d −x k+δ (−1) (k) −x k+δ P (x)Lk(x)e x dx = P (x) k e x dx = P (x)e x dx. 0 k! 0 dx k! 0 δ (k) Now if P is a polynomial of degree less that k, in particular if P = Lj with j < k then P ≡ 0; this δ (k) k proves the orthogonality of the Laguerre polynomials. Now if P = Lk, then we have P = (−1) and Z ∞ δ 2 1 −x k+δ (k + δ)! kLkkw = e x dx = . k! 0 k! The argument giving completeness is similar to that for Hermite polynomials; see Folland [2].  Definition. Laguerre functions of type δ, δ > −1 are given by 1/2 1 δ δ k!  δ − 2 x 2 Λk(x) = (k+δ)! Lk(x)e x .

δ 2 + It is an immediate consequence of Proposition 6 that Λk(x) form an orthonormal basis for L (R ). In particular, we also have that for each λ 6= 0, the “Laguerre functions” 1−n λ 2 2 n−1 1 2 `k(r) = (|λ|r ) Λk ( 2 |λ|r ), form an orthonormal basis for L2(R+, |λ|nr2n−1dr).

3.3. Special Hermite functions. For each α, β ∈ Nn and z ∈ Cn, we define the special Hermite λ functions hα,β by n |λ| 2 hλ (z) = (π (z)e , e ). α,β 2π λ j k λ 2 n We have therefore already shown that {hα,β} forms an orthonormal basis in L (C ). 10 NEIL LYALL

For a function f ∈ L2(Cn) we have the eigenfunction expansion

X X λ λ f = (f, hα,β)hα,β α β which is called the special Hermite expansion. This series can be put in a compact form once we’ve shown that special Hermite functions can be expressed in terms of Laguerre functions. Proposition 7. For f ∈ L2(Cn) we have ∞ X λ f(z) = f ∗λ ϕk(z), k=0

|λ| n−1 1 2 λ n 1 2 − 4 |λ||z| where ϕk(z) = ( 2π ) Lk ( 2 |λ||z| )e .

This will be an immediate consequence of the following two lemmas. The first will show that our Laguerre functions are in fact special Hermite functions, more precisely Lemma 8.  n/2 n |λ| Y 1 2 hλ (z) = L0 ( 1 |λ||z |2)e− 4 |λ||zj | . α,α 2π αj 2 j j=1

Proof. It suffices to consider the one dimensional case. Mehler’s kernel identity gives,

∞  1/2 |λ| X |λ| − ((x2+y2)(1+r2)−4xyr) hλ(x)hλ(y)rk = (1 − r2)−1/2e 2(1−r2) . k k π k=0 An easy calculation gives that   ∞ 2  1/2 −|λ| ξ2 1−r + y 1+r X y y |λ| 1+r 4 1−r hλ(ξ + )hλ(ξ − )rk = (1 − r2)−1/2e . k 2 k 2 π k=0 Therefore, ∞ 1/2 ∞ X |λ| X hλ (z)rk = (π (z)hλ, hλ))rk kk 2π λ k k k=0 k=0 1/2 ∞ |λ| Z X y y = eiλxξ hλ(ξ + )hλ(ξ − )rkdξ 2π k 2 k 2 R k=0 2 Z |λ| 2 −1/2 −|λ| y 1+r −|λ|ξ2 1−r i|λ|xξ = (2(1 − r )) e 4 1−r e 1+r e dξ π R 1/2   |λ| |λ| −1 − 1+r (x2+y2) = (1 − r) e 4 1−r . 2π But it follows from (3) that ∞ 1 1+r X 0 − x k −1 − 2(1−r) x Lk(x)e 2 r = (1 − r) e . k=0 THE GROUP FOURIER TRANSFORM ON Hn 11

Hence, ∞  1/2 ∞ X λ k |λ| X 0 1 2 2 − 1 |λ|(x2+y2) k h (z)r = L ( |λ|(x + y ))e 4 r . kk 2π k 2 k=0 k=0  Lemma 9. 2π n/2 hλ ∗ hλ = δ hλ , α,β λ µ,ν |λ| β,µ α,ν

Proof. For ϕ, ψ ∈ L2(Rn), Z ¯λ ¯λ (Wλ(hα,β)ϕ, ψ) = hα,β(πλ(z)ϕ, ψ)dz Cn |λ|n/2 = ((π (z)ϕ, ψ), (π (z)hλ, hλ)) 2π λ λ α β 2π n/2 = (ϕ, hλ)(hλ, ψ). |λ| α β Therefore 2π n/2 2π n (W (h¯λ ∗ h¯λ )ϕ, ψ) = (W (h¯λ )ϕ, hλ)(hλ, ψ) = (ϕ, hλ)(hλ, hλ)(hλ, ψ). λ α,β λ µ,ν |λ| λ µ,ν α β |λ| µ ν α β This of course implies 2π n/2 h¯λ ∗ h¯λ = δ h¯λ , α,β λ µ,ν |λ| α,ν µ,β ¯ which gives us our result, since f ∗λ g =g ¯ ∗λ f. 

Proof of Proposition 7. For f ∈ L2(Cn) we have that

X X λ λ f = (f, hα,β)hα,β. α β Lemma 9 therefore implies that n/2 X X 2π  X (4) f ∗ hλ = (f, hλ )hλ ∗ hλ = (f, hλ )hλ , λ µ,µ α,β α,β λ µ,µ |λ| α,µ α,µ α β α and so n/2 ∞ |λ| X X f(z) = f ∗ hλ (z). 2π λ β,β k=0 |β|=k We are therefore required to show that n/2 X 2π  (5) hλ (z) = ϕλ(z). β,β |λ| k |β|=k 12 NEIL LYALL

λ It follows from (3) that ϕk satisfy the generating function ∞  n 1 X λ k |λ| −n − |λ||z|2 1+r (6) ϕ (z)r = (1 − r) e 4 1−r . k 2π k=0 Now it follows from Lemma 8 that  n/2 n |λ| Y 1 2 hλ (z) = L0 ( 1 |λ||z |2)e− 4 |λ||zj | , β,β 2π βj 2 j j=1

1 2 and from (3) we see that each L0 ( 1 |λ||z |2)e− 4 |λ||zj | satisfies βj 2 j ∞ 1 1 X − |λ||z |2 − |λ||z |2 1+r 0 1 2 4 j k −1 4 j 1−r Lk( 2 |λ||zj| )e r = (1 − r) e . k=0 From this it is clear that ∞  n/2 1 X X λ k |λ| −n − |λ||z|2 1+r h (z)r = (1 − r) e 4 1−r , β,β 2π k=0 |β|=k comparing this with (6) we obtained our desired result.  Lemma 10. λ λ λ λ Wλ(ϕk) = Pk and hence ϕk ∗λ ϕj = δkjϕk.

Proof. From the calculation in the proof of Lemma 9 it follows that 2π n/2 W (hλ )ϕ = (ϕ, hλ)hλ, λ α,α |λ| α α in view of identity (5) we therefore have

λ X λ λ Wλ(ϕk)ϕ = (ϕ, hα)hα = Pk, |α|=k and the second part of the claim follows immediately. 

4. Group Fourier transform of radial functions on the Heisenberg group

Recall that the group Fourier transform of an integrable function g on Hn is, for each λ 6= 0, an operator-valued function on the Hilbert space L2(Rn) given by λ gb(λ)ϕ(ξ) = Wλ(g )ϕ(ξ), where Z Z λ λ λ iλt Wλ(f )ϕ(ξ) = g (z)πλ(z)ϕ(ξ)dz and g (z) = g(z, t)e dt. Cn R Now if g is also radial on Hn, which means that it depends only on |z| and t, then it follows that 2 n the operators gb(λ) are diagonal on the Hermite basis for L (R ). THE GROUP FOURIER TRANSFORM ON Hn 13

1 n Theorem 11. If g ∈ L (H ) and g(z, t) = g0(|z|, t), then λ λ gb(λ)hα(x) = Cnµ(|α|, λ)hα(x), where ∞ 1/2 Z 1−n k!  λ 1 2 2 n−1 1 2 2n−1 µ(k, λ) = (k+n−1)! g0 (s)( 2 |λ|s ) Λk ( 2 |λ|s )s ds, 0 and Cn is a constant which depends only on n.

λ λ λ Proof. It is clear that g (z) = g0 (|z|), for some function g0 . We can therefore write ∞ Z ∞ λ X λ λ n 2n−1  λ g0 (r) = g0 (s)`k(s)|λ| s ds `k(r). k=0 0 From this we see that we formally have ∞ λ X λ g (z) = Cn µ(k, λ)ϕk(z), k=0 n 1−n where Cn = (2π) 2 . It now follows from Lemma 10 that λ λ λ g ∗λ ϕk(z) = Cnµ(k, λ)ϕk(z), and hence from Proposition 7 we see that this formal “Laguerre expansion” in fact agrees with the special Hermite expansion, ∞ ∞ λ X λ λ X λ (7) g (z) = g ∗λ ϕk(z) = Cn µ(k, λ)ϕk(z). k=0 k=0 λ Now since gb(λ) = Wλ(g ) we see that Theorem 11 follow immediately from (7) and Lemma 10. 

References

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