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The Heisenberg Group Fourier Transform THE HEISENBERG GROUP FOURIER TRANSFORM NEIL LYALL Contents 1. Fourier transform on Rn 1 2. Fourier analysis 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 representation theory 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 unitary representation 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 Hilbert space 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 bounded operator 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).
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