Gelfand pairs and spherical means on H-type groups

Abstract

We study the injectivity of the spherical mean operator associated to the Gelfand pairs (U, N), where N is a Heisenberg type group and U the subgroup of the group of orthogonal transformations of N that act trivially on its centre. We prove that when the dimension of the centre of N is 3, these spherical mean operator is injective on $L^p(N)$ for the optimal range $1 \le p \le 3$.

1 Introduction and preliminaries

Finding out if a function can be reconstructed from its averages on spheres with a definite radius $r > 0$ is one of the challenges in integral geometry. By the average of the function f over the sphere of radius r centered at the point x, we mean the integral of f with respect to the normalised surface measure on the sphere $\{y \in {\mathbb {R}}^n:\vert y-x\vert =r \}$ in ${\mathbb {R}}^n$. This can be written as the convolution with the normalized surface measure $\nu _r$ on the sphere $\{x\in {\mathbb {R}}^n:\vert x\vert =r \}$ as

$$\begin{aligned} f*\nu _r(x)=\int \limits _{\vert x\vert =r}f(x-y)d\nu _r(y). \end{aligned}$$

The function can be recovered from its spherical averages only if this spherical mean operator is injective. That is,

$$\begin{aligned} f*\nu _r(x)=0,\, \forall x\implies f\equiv 0. \end{aligned}$$

But this operator is not injective in general. For $\lambda >0$, consider the function

$$\begin{aligned} \varphi _\lambda (x)= c~\frac{J_{\frac{n}{2}-1}(\lambda \vert x\vert )}{(\lambda \vert x\vert )^{\frac{n}{2}-1}},\,~x \in {\mathbb {R}}^n, \end{aligned}$$

where $J_\alpha$ denotes the Bessel function of order $\alpha$ and c is a constant to normalize $\varphi _\lambda$ in such a way that $\varphi _\lambda (0) =1$. Then it is well known that

$$\begin{aligned} \varphi _\lambda *\nu _r(x)=\varphi _\lambda (r)\varphi _\lambda (x),~x \in {\mathbb {R}}^n. \end{aligned}$$

Hence, if we choose $r> 0$ to be a zero of the function $s \rightarrow J_{\frac{n}{2}-1}(\lambda s)$, then $\varphi _\lambda *\nu _r$ vanishes identically(see [1]). Since $\varphi _\lambda \in L^p({\mathbb {R}}^n)$ when $p>\frac{2n}{n-1}$, the operator $f\mapsto f*\nu _r$ is fails to be injective in $L^p({\mathbb {R}}^n)$ for $\frac{2n}{n-1}<p\le \infty$. When $1 \le p \le \frac{2n}{n-1}$, these operators are injective. That is, for a continuous function $f \in L^p(\mathbb R^n)$ with $1 \le p \le \frac{2n}{n-1}$, if $f *\nu _r$ is identically zero for a fixed radius $r > 0,$ then f vanishes identically (see [2]).

Let $H^n$ denote the Heisenberg group ${\mathbb {C}}^n\times {\mathbb {R}}$ with the group law

$$\begin{aligned} (z,t)(w,s)=(z+w,t+s+\frac{1}{2}{\text {Im}}(z\cdot {\overline{w}})). \end{aligned}$$

The Lie algebra of this Lie group is ${\mathfrak {h}}^n = {\mathbb {C}}^n\times {\mathbb {R}}$ with the Lie bracket

$$\begin{aligned}{}[(z,t),(w,s)] =(0,{\text {Im}}(z\cdot {\overline{w}})). \end{aligned}$$

This makes ${\mathfrak {h}}^n$ a step two nilpotent Lie algebra and therefore $H^n$ a two-step nilpotent Lie group. The spherical means of a function f on $H^n$ can also be written in terms of the convolution as

$$\begin{aligned} f*\mu _r(z,t)=\int \limits _{\vert w\vert =r}f(z-w,t-\frac{1}{2}{\text {Im}}(z\cdot {\overline{w}}))\, d\mu _r(w). \end{aligned}$$

(1)

where $\mu _r$ is the normalized surface measure on the sphere $\{(z,0)\in H^n: \vert z\vert =~r\}$ of radius r in $H^n$. Unlike the Euclidean case, the spherical mean operators are injective on $L^p(H^n)$ for all p such that $1\le p<\infty$. This was proved by Thangavelu [2] using the spectral decomposition of the sublaplacian on the Heisenberg group provided by Strichartz in [3].

Notice that the unitary group U(n) acts on the Heisenberg $H^n$ by $\sigma (z,t) = (\sigma z, t).$ Hence the spherical means in (1) can be seen as the averages of the function f over U(n)-orbits. Since the sub-algebra of the U(n) invariant functions in $L^1(H^n)$ is commutative, the pair $(H^n, U(n))$ forms a Gelfand pair. Also, the spectral decomposition studied by Strichartz [3], coincides with the expansion in terms of the spherical functions associated with this Gelfand pair. This point of view led to a general result in [4].

The Heisenberg type groups, or H-type groups, were introduced by A. Kaplan in 1980 [5] as a class of two step nilpotent groups that includes the Heisenberg groups. Let ${\mathfrak {n}}$ be a real two-step nilpotent Lie algebra endowed with an inner product $\langle ,\rangle$. Let ${\mathfrak {z}}$ be the center of ${\mathfrak {n}}$ and ${\mathfrak {v}}$ be its orthogonal complement. For a unit vector $v\in {\mathfrak {v}}$, let ${\mathfrak {f}}_v$ be the kernel of the adjoint map $ad_v:{\mathfrak {v}}\rightarrow {\mathfrak {z}}$ defined by

$$\begin{aligned} ad_v(v')=[v,v']\quad v'\in {\mathfrak {v}}. \end{aligned}$$

Then the Lie algebra ${\mathfrak {n}}$ is said to be Heisenberg type or H-type if the adjoint map restricted to the orthogonal complement of its kernel is a surjective isometry. That is, if $ad_v:{\mathfrak {v}}_v\rightarrow {\mathfrak {z}}$ is a surjective isometry where

$$\begin{aligned} {\mathfrak {v}}={\mathfrak {v}}_v\oplus {\mathfrak {f}}_v. \end{aligned}$$

A connected and simply connected Lie group N is said to be a Heisenberg type group or H-type group if its Lie algebra is of H-type.

If ${\mathfrak {n}}$ is an H-type Lie algebra, for non-zero $z\in {\mathfrak {z}}$ we can define a skew-symmetric linear operator $J_z:{\mathfrak {v}}\rightarrow {\mathfrak {v}}$ by

$$\begin{aligned} \langle J_z(v),v'\rangle =\langle z,[v,v']\rangle \qquad \text{ for } \text{ all } v,v'\in {\mathfrak {v}}. \end{aligned}$$

It can be proved that ${\mathfrak {n}}$ is a H-type algebra if and only if $J_z^2=-\vert z\vert ^2I$, for every nonzero $z\in {\mathfrak {n}}$ [5]. For $\vert z\vert =1$, $J_z$ defines a complex structure on ${\mathfrak {v}}$ and hence $\dim {\mathfrak {v}}$ has to be even, say $\dim {\mathfrak {v}}=2n$. We will identify ${\mathfrak {v}}$ with ${\mathbb {C}}^n$ and ${\mathfrak {z}}$ with $\mathbb R^m$. Since we can identify the connected and simply connected Lie group N with its nilpotent Lie algebra ${\mathfrak {n}}$ via the exponential map, we will write (z, t) for points in N, where $z\in {\mathbb {C}}^n$ (identified with ${\mathfrak {v}}$) and $t\in {\mathbb {R}}^m$ (identified with ${\mathfrak {z}}$). The Haar measure on N is given by the Lebesgue measure on ${\mathfrak {n}}$ and will be denoted by dzdt. The group law is then given by

$$\begin{aligned} (z,t)(w,s)=(z+w,t+s+\frac{1}{2}[z,w]), \end{aligned}$$

where $[\;,\;]$ denotes the Lie bracket. For any fixed $a\in {\mathfrak {z}}\setminus \{0\}$ we can choose a basis with some properties ( see [6, p. 294]) so that

$$\begin{aligned} \langle a,[z,w]\rangle = \vert a\vert {\text {Im}}(z \cdot {\bar{w}}). \end{aligned}$$

Let f, g be functions on N with $g(z,t)=\exp (-i\langle a,t\rangle )\varphi (z)$, then,

$$\begin{aligned} \begin{aligned} f*g(z,t)&= \int \limits _N f((z,t)(w,s)^{-1})g(w,s)\, dwds\\&=\int \limits _{{\mathbb {C}}^n}\int \limits _{{\mathbb {R}}^m}f(z-w,t-s-\frac{1}{2}[z,w])g(w,s)\, dwds\\&=\int \limits _{{\mathbb {C}}^n}\int \limits _{{\mathbb {R}}^m}f(z-w,t-s-\frac{1}{2}[z,w])\exp (-i\langle a, s\rangle )\varphi (w)\, dwds\\&=\int \limits _{{\mathbb {C}}^n}\int \limits _{{\mathbb {R}}^m}f(z-w,s)\exp (-i\langle a, t-s-\frac{1}{2}[z,w]\rangle )\varphi (w)\, dwds\\&=\exp (-i\langle a,t\rangle )\int \limits _{{\mathbb {C}}^n}f^{a}(z-w)\varphi (w)\exp \left( \frac{i\vert a\vert }{2}{\text {Im}}(z \cdot {\bar{w}})\right) \, dw\\&=\exp \left( -i\langle a,t\rangle \right) f^{a}\times _{\vert a\vert }\varphi (z). \end{aligned} \end{aligned}$$

where $f^{a}$ is the Fourier transform of f in the central variable t and $\times _{\lambda }$ denote the twisted convolution on ${\mathbb {C}}^n$ of order $\lambda$, defined by,

$$\begin{aligned} F\times _\lambda G(z)=\int \limits _{{\mathbb {C}}^n}F(z-w)G(w)\exp \left( \frac{i\lambda }{2}Im(z\cdot {\overline{w}})\right) \, dw. \end{aligned}$$

The irreducible unitary representations of N that are not one dimensional are parameterized by $a\in {\mathfrak {z}} \setminus \{0\}$. For each $a\in {\mathfrak {z}}\backslash \{0\}$, we can define the Hilbert space

$$\begin{aligned} {\mathcal {F}}_a({\mathfrak {v}})=\left\{ F:{\mathfrak {v}} \equiv {\mathbb {C}}^n \rightarrow {\mathbb {C}}:F\text { is holomorphic},\int _{\mathfrak {v}}\vert F(w)\vert ^2e^{\frac{-\vert a\vert \vert w\vert ^2}{2}}\, d{\mathfrak {v}}(w)<\infty \right\} . \end{aligned}$$

These Hilbert spaces support the irreducible representation $\pi _a$ of N, known as the Bargmann representation, defined by,

$$\begin{aligned} \pi _a(v,t)F(w)=\exp (i\langle a,t\rangle -\frac{1}{4}\vert a\vert (\vert v\vert ^2+2\langle w,v \rangle -i\langle b,[w,v] \rangle ) )F(w+v) \end{aligned}$$

for $v\in {\mathfrak {v}}, t\in {\mathfrak {z}}$, where $b=\frac{a}{\vert a\vert }$. Moreover, any infinite dimensional unitary representation $\pi$ of N such that $\pi \vert _{\mathfrak {z}}=e^{i<a,t>}{\text {Id}}$ is equivalent to $\pi _a$ [7, p. 420].

Let A(N) be the group of orthogonal transformations of $N={\mathfrak {v}}\oplus {\mathfrak {z}}$, which are automorphisms of N. Let U be the subgroup of A(N) that act trivially on ${\mathfrak {z}}$. That is,

$$\begin{aligned} U=\{k\in A(N): k (z) = z, \text { for all } z \in {\mathfrak {z}} \}. \end{aligned}$$

For $z\in {\mathfrak {z}}$, the map $J_z:{\mathfrak {v}}\rightarrow {\mathfrak {v}}$ extends to N as an automorphism by defining

$$\begin{aligned} J_z(z)=z \text { and } J_z(z')=-z' \text { if } z'\perp z. \end{aligned}$$

We denote by ${\text {Pin}}(m)$ the subgroup of A(N) generated by $\{J_z:z\in {\mathfrak {z}} \}$. Then U and ${\text {Pin}}(m)$ commute and their intersection contains at most four elements [8]. Also $A(N) = U \cdot {\text {Pin}}(m)$ unless $m \equiv 1 \pmod 4$ and in that case $A(N)/\left( U\cdot {\text {Pin}}(m)\right)$ has two elements [7].

We recall that for any Lie group N and any compact subgroup K of its automorphism group, the pair (K, N) is said to be a Gelfand pair if the set $L_K^1(N)$ of integrable K-invariant functions on N forms a commutative algebra under convolution. For the particular case of Heisenberg type groups we have the following classification theorem[9, p. 266].

Theorem 1.1

The groups N of H-type for which $L_{A(N)}^1(N)$ is commutative, that is (A(N),N) is a Gelfand pair, are those for which

$$\begin{aligned} \dim ({\mathfrak {z}})=m=\left\{ \begin{aligned}&1,2\text { or }3\\ {}&5,6\text { or }7\text { and }{\mathfrak {v}}\text { is irreducible}\\ {}&7, {\mathfrak {v}} \text { is isotypic and } \dim ({\mathfrak {v}})=16. \end{aligned} \right. \end{aligned}$$

Here irreducibility of ${\mathfrak {v}}$ means irreducible under the action of the associated Clifford algebra. See [9] for more details. From the proof of the above theorem we obtain the following corollary[9, p. 268],

Corollary 1.1

Let U be the subgroup of A(N) that act trivially on the center ${\mathfrak {z}}$. Then (U, N) is a Gelfand pair if and only if $\dim {\mathfrak {z}}=1,2,\text { or }3$.

We consider the above cases in some detail. First we notice that when $m=1$, N is the Heisenberg group ${\mathbb {C}}^k \oplus {\mathbb {R}}$ and $U=U(k)$ is the unitary group. The U-averages give rise to the spherical means on N and the injectivity result follow from [2, Theorem 5.1]. See also [4, Theorem 5.2].

When $m=2$, $N\cong {\mathbb {H}}^k \oplus {\mathbb {R}}^2$ (See [9, p. 268] ) where ${\mathbb {H}}$ is the space of quaternions and $U=Sp(k)$, the compact symplectic group. In this case U acts transitively on the spheres centered at origin in ${\mathbb {H}}^k(\cong {\mathbb {C}}^{2k})$. Therefore the averages over U-orbits coincide with the following spherical means

$$\begin{aligned} f*\mu _r(z,t)=\int \limits _{\vert w\vert =r}f(z-w,t-\frac{1}{2}[z,w])\, d\mu _r(w) \end{aligned}$$

defined in [10] in terms of the normalised surface measure $\mu _r$ on the sphere $\{(z,0)\in N : \vert z\vert =r \}$ for a continuous function f on N. This is one of the three spherical mean operators which were shown to be injective on $L^p(N)$ for $1 \le p \le 2m/(m-1)$, where $m = \dim {\mathfrak {z}}$ (See [10, Theorem 1.1]).

When $m=3$, ${\mathfrak {v}}\cong {\mathbb {H}}^k\oplus {\mathbb {H}}^l$ and $U=Sp(k)\times Sp(l)$ (see [9, p. 268]). The orbit of U in ${\mathfrak {v}}$ is the product of spheres in ${\mathbb {H}}^k$ and ${\mathbb {H}}^l$. So the U- averages give rise to new type of spherical means, not considered in [10].

We consider the above case $m=3$ and prove injectivity result for averages over U-orbit. Fix $r_2,r_2>0$. Let $S_{r_1}^k$ and $S_{r_2}^l$ be the spheres of radii $r_1$ and $r_2$ centered at the origin in ${\mathbb {H}}^k$ and ${\mathbb {H}}^l$, respectively. Let $\mu _{r_1}^k$ and $\mu _{r_2}^l$ be the normalized surface measures on $S_{r_1}^k$ and $S_{r_2}^l$ respectively and let $\nu _{r_1,r_2}=\mu _{r_1}^k~\times ~\mu _{r_2}^l$ realised as a measure on $N={\mathbb {H}}^k\oplus {\mathbb {H}}^l\oplus {\mathbb {R}}^3$. Then the U-spherical means can be defined as

$$\begin{aligned} f*\nu _{r_1,r_2}(z,w,t)=\int \limits _{S_{r_1}^k\times S_{r_2}^l}f(z-u,w-v,t-\frac{1}{2}[(z,w),(u,v)])\, d\mu _{r_1}^k(u)d\mu _{r_2}^l(v) \end{aligned}$$

Our result is the following:

Theorem 1.2

If $f\in L^p(N)$ for $1\le p\le 3$ and $f*\nu _{r_1,r_2}\equiv 0$ then $f\equiv 0$.

For the proof of the above, we closely follow the arguments in [2] and [10]. First we compute the spherical functions for the Gelfand pair $(Sp(k)\times Sp(l),N)$ (see 2). Then we obtain an expansion of $L^2$-functions in term of the spherical functions and establish the Abel summability of this expansion in $L^p$ (see Theorem 3.4). Then the proof of the injectivity will follow as in [10].

2 Spherical functions for the case $m=3$

Let (K, N) be a Gelfand pair and $\pi$ be an irreducible unitary representation of N on a Hilbert space ${\mathcal {H}}_\pi$. Define,

$$\begin{aligned} K_\pi =\{k\in K: \pi \circ k \text { unitarily equivalent to }\pi . \} \end{aligned}$$

Let ${\mathcal {H}}=\oplus _\alpha P_\alpha$ be the decomposition into the $K_\pi$-irreducible subspaces. The following theorem was proved in [11, p. 415].

Theorem 2.1

If $\phi$ is a bounded K-spherical function on N, then there exist a unique (up to unitary equivalence) irreducible representation $\pi$ and a subspace ${\mathcal {P}}_\alpha$ in the decomposition of the representation space ${\mathcal {H}}=\oplus _\alpha {\mathcal {P}}_\alpha$ into $K_\pi$-irreducible subspaces, such that,

$$\begin{aligned} \phi (x)=\phi _{\pi ,v}(x)=\int _K\langle \pi (k\cdot x)v,v\rangle \, dk, \end{aligned}$$

for any unit vector $v\in {\mathcal {P}}_\alpha$ and $x\in N$. In particular, if $K=K_\pi$ and $\{v_1,v_2,\ldots , v_l \}$ is any orthonormal basis for ${\mathcal {P}}_\alpha$, then

$$\begin{aligned} \phi _{\pi ,\alpha }(x)=\frac{1}{l}\sum \limits _{j=1}^l\langle \pi (x)v_j,v_j\rangle . \end{aligned}$$

When ${\mathfrak {v}} = {\mathbb {H}}^k\oplus {\mathbb {H}}^l$, the action of $U=Sp(k)\times Sp(l)$ on the space ${\mathcal {P}}({\mathfrak {v}})$ of holomorphic polynomials on ${\mathfrak {v}}$, decomposes as

$$\begin{aligned} {\mathcal {P}}({\mathfrak {v}})=\bigoplus \limits _{p=0, q=0}^\infty {\mathcal {P}}^p({\mathbb {H}}^k)\otimes {\mathcal {P}}^q({\mathbb {H}}^l) \end{aligned}$$

where ${\mathcal {P}}^p({\mathbb {H}}^k)={\mathcal {P}}^p({\mathbb {C}}^{2k})$ is the space of homogeneous polynomials of degree p and ${\mathcal {P}}^q({\mathbb {H}}^l)={\mathcal {P}}^q({\mathbb {C}}^{2l})$ is the space of homogeneous polynomial of degree l [9, p. 268].

The U action on ${\mathfrak {n}}$ is via the Sp(k) action on ${\mathcal {P}}^p({\mathbb {H}}^k)$ and the Sp(l) action on ${\mathcal {P}}^q({\mathbb {H}}^l)$ and so is trivial on the centre ${\mathfrak {z}}$. Hence for every $k \in U$, $\left. ( \pi _a \circ k)\right| _{{\mathfrak {z}}} =~\left. \pi _a\right| _{{\mathfrak {z}}}$. That is,

$$\begin{aligned} U_{\pi _a } = \{k \in U: \pi _a \circ k \equiv \pi _a\} = U. \end{aligned}$$

Hence, by Theorem 2.1 every spherical function is of the form $\phi _{\pi _a,v}$, for some $a\in {\mathfrak {z}}\setminus \{0\}$ and a unit vector $v = (p,q) \in {\mathcal {P}}^p({\mathbb {H}}^k)\otimes {\mathcal {P}}^q({\mathbb {H}}^l)$. Hence the spherical functions are parameterised by (p, q) as

$$\begin{aligned} e_{p,q}^a(z,w,t)=\phi _{\pi _a,v}(z,w,t) \end{aligned}$$

where $(z,w)\in {\mathbb {H}}^k\times {\mathbb {H}}^l={\mathbb {C}}^{2k}\times {\mathbb {C}}^{2l}$.

To obtain the explicit expression for $e_{p,q}^a$, consider an orthonormal basis $\{u_\alpha (\xi )=\xi ^\alpha , \xi \in {\mathbb {C}}^{2k} :\alpha \in {\mathbb {N}}^{2k}, \vert \alpha \vert =p \}$ for ${\mathcal {P}}^p({\mathbb {C}}^{2k})$ and an orthonormal basis $\{v_\beta (\eta )=\eta ^\beta , \eta \in {\mathbb {C}}^{2l} :\eta \in {\mathbb {N}}^{2l}, \vert \beta \vert =q \}$ for ${\mathcal {P}}^q({\mathbb {C}}^{2l})$. Then $\{u_\alpha \otimes v_\beta :\alpha ~\in ~{\mathbb {N}}^{2k},\beta \in {\mathbb {N}}^{2l},\vert \alpha \vert =p,\vert \beta \vert =q\}$ is an orthonormal basis for ${\mathcal {P}}^p({\mathbb {C}}^k)~\otimes ~{\mathcal {P}}^q({\mathbb {C}}^l)$. Let $d_p=\dim {\mathcal {P}}^p({\mathbb {C}}^{2k})$ and $d_p=\dim {\mathcal {P}}^q({\mathbb {C}}^{2l})$, then

$$\begin{aligned} e_{p,q}^a(z,w,t)&=\frac{1}{d_pd_q}\,\,\sum _{{\begin{array}{c} \vert \alpha \vert =p\\ \vert \beta \vert =q \end{array}}}<\pi _a(z,w,t)u_\alpha \otimes v_\beta ,u_\alpha \otimes v_\beta>\\&=e^{i<a,t>}\left( \frac{1}{d_p} \sum _{\vert \alpha \vert =p}\left\langle \pi _{a}(z, 0,0) u_{\alpha }, u_{\alpha }\right\rangle \right) \\&\qquad \times \left( \frac{1}{d_{q}} \sum _{\vert \beta \vert =q}\left\langle \pi _{a}(0, w,0) v_{\beta }, v_{\beta }\right\rangle \right) \\ \end{aligned}$$

Since the action of $\pi _a(z,0,0)$ on ${\mathcal {P}}({\mathbb {C}}^{2k})$ is same as the action of the representation $\pi _{\vert a\vert }(z,0)$ of the Heisenberg group $H^{2k}={\mathbb {C}}^{2k}\times {\mathbb {R}}$ on ${\mathcal {P}}({\mathbb {C}}^{2k})$, we have

$$\begin{aligned} \frac{1}{d_p}\sum \limits _{\vert \alpha \vert =p}\langle \pi _a(z,0,0)u_\alpha ,u_\alpha \rangle =\frac{1}{d_p}\sum \limits _{\vert \alpha \vert =p}\langle \pi _{\vert a\vert }(z,0)u_\alpha ,u_\alpha \rangle . \end{aligned}$$

Then by [11, Proposition 6.2],

$$\begin{aligned} \frac{1}{d_p}\sum \limits _{\vert \alpha \vert =p}\langle \pi _a(z,0,0)u_\alpha ,u_\alpha \rangle =L_p^{2k-1}\left( \frac{\vert a\vert }{2}\vert z\vert ^2\right) e^{-\frac{\vert a\vert }{4}\vert z\vert ^2}, \end{aligned}$$

where $L_r^\delta$ is the r-th Laguerre polynomial of type $\delta >-1$. Similarly,

$$\begin{aligned} \frac{1}{d_q}\sum \limits _{\vert \beta \vert =q}\langle \pi _a(0,w,0)v_\beta ,v_\beta \rangle =L_q^{2l-1}\left( \frac{\vert a\vert }{2}\vert w\vert ^2\right) e^{-\frac{\vert a\vert }{4}\vert w\vert ^2}. \end{aligned}$$

Therefore, we have,

$$\begin{aligned} e_{p,q}^a(z,w,t)&=e^{i\langle a, t\rangle } L_{p}^{2 k-1}\left( \frac{1}{2}\vert a\vert \vert z\vert ^{2}\right) L_{q}^{2 l-1}\left( \frac{1}{2}\vert a\vert \vert w\vert ^{2}\right) e^{-\frac{1}{4}\vert a\vert \left( \vert z\vert ^{2}+\vert w\vert ^2\right) } \\&=e^{i\langle a, t\rangle }\varphi _{p,q}^a(z,w) \end{aligned}$$

where,

$$\begin{aligned} \varphi _{p,q}^a(z,w) = L_{p}^{2 k-1}\left( \frac{1}{2}\vert a\vert \vert z\vert ^{2}\right) L_{q}^{2 l-1}\left( \frac{1}{2}\vert a\vert \vert w\vert ^{2}\right) e^{-\frac{1}{4}\vert a\vert \left( \vert z\vert ^{2}+\vert w\vert ^2\right) } \end{aligned}$$

Therefore,

$$\begin{aligned} e_{p,q}^a(z,w,t) = e^{i\langle a, t\rangle } \varphi _p^{\vert a\vert , 2k}(z) \varphi _q^{\vert a\vert , 2l}(w) \end{aligned}$$

(2)

where $\varphi _j^{\lambda , n}(z)= L_j^{n-1}\left( \frac{1}{2}\lambda \vert z\vert ^2\right) e^{-\frac{1}{4}\lambda \vert z\vert ^2}, \lambda > 0$ is the scaled Laguerre function on ${\mathbb {C}}^n$.

3 Spectral decomposition and Abel summability

An important step in obtaining the injectivity results for spherical means in [10] is the spectral decomposition. For $f \in L^2(N)$ in the H-type group $N~\cong ~{\mathbb {C}}^n \times {\mathbb {R}}^m$,

$$\begin{aligned} f(z,t) = \frac{1}{(2\pi )^{n+m}} \sum _{r=0}^\infty ~\int \limits _{{\mathbb {R}}^n}~f *e_r^a(z, t)~\vert a\vert ^n~da \end{aligned}$$

and the fact that the eigenfunctions $e_r^a$ satisfy

$$\begin{aligned} e_r^a *\mu _k(z,t) = c_r e_r^a(k,0) e_r^a(z,t)\qquad \text {for }(z,t)\in N \end{aligned}$$

where $\mu _k$ is the surface measure on the sphere of radius k and $c_r$ an appropriate constant. When $m = 3$, elements in N can be written as (z, w, t) with $z~\in ~{\mathbb {H}}^k \equiv {\mathbb {C}}^{2k}, w\in {\mathbb {H}}^l \equiv {\mathbb {C}}^{2l}, 2k+2l = n$ and the above decomposition becomes

$$\begin{aligned} f(z,w,t)=\frac{1}{(2\pi )^{2k+2l+3}}\int \limits _{{\mathbb {R}}^3\setminus \{0\}}\sum _{r=0}^\infty f*e_r^a(z,w,t)\vert a\vert ^{2k+2l}\, da . \end{aligned}$$

(3)

In order to obtain a similar expansion for $f \in L^2(N)$ in terms of the U-spherical functions, we first prove the following Lemma.

Lemma 3.1

$$\begin{aligned} \begin{aligned}&\sum _{p+q=k} L_{p}^{2k-1}\left( \frac{1}{2}\vert a\vert \vert z\vert ^{2}\right) L_{q}^{2l -1}\left( \frac{1}{2}\vert a\vert \vert w\vert ^{2}\right) e^{-\frac{1}{4}\vert a\vert (\vert z\vert ^{2}+\vert w\vert ^2)} \\&\qquad =L_{j}^{n-1}\left( \frac{1}{2}\vert a\vert \left( \vert z\vert ^{2}+\vert w\vert ^{2}\right) \right) e^{-\frac{1}{4}\vert a\vert \left( \vert z\vert ^{2}+\vert w\vert ^{2}\right) }\qquad j=0,1,2,\ldots \end{aligned} \end{aligned}$$

where $n=2k+2l$.

Proof

We use the generating function of the Laguerre polynomials of type $\alpha >-1$,

$$\begin{aligned} \sum _{k=0}^\infty L_k^\alpha (x) r^k=(1-r)^{-\alpha -1}e^{-\frac{r}{1-r}x}\qquad \vert r\vert <1. \end{aligned}$$

Hence for $x\ge 0$,

$$\begin{aligned} \sum _{k=0}^\infty L_k^\alpha (x) e^{-\frac{x}{2}}r^k=(1-r)^{-\alpha -1}e^{-\frac{1}{2}\left( \frac{1+r}{1-r}\right) x}\qquad \vert r\vert <1, \end{aligned}$$

since $2k+2l=2n$, for $x,y\ge 0$,

$$\begin{aligned} \left( \sum _{p=0}^\infty L_p^{2k-1}(x) e^{-\frac{x}{2}}r^p\right) \left( \sum _{q=0}^\infty L_q^{2l-1}(y) e^{-\frac{y}{2}}r^q\right)&=(1-r)^{-2k-2l}e^{-\frac{1}{2}\left( \frac{1+r}{1-r}\right) (x+y)}\\&= (1-r)^{-n}e^{-\frac{1}{2}\left( \frac{1+r}{1-r}\right) (x+y)}\\&= \sum _{j=0}^\infty L_j^{n-1}(x+y) e^{-\frac{x+y}{2}}r^j \end{aligned}$$

Since the power series expansion is unique, by comparing the coefficients we get,

$$\begin{aligned} \sum _{p+q=j}L_p^{2k-1}(x)L_q^{2l-1}(y)e^{-\frac{x+y}{2}}=L_j^{n-1}(x+y)e^{-\frac{x+y}{2}}. \end{aligned}$$

The lemma will follow by taking $x=\frac{1}{2}\vert a\vert \vert z\vert ^2$ and $y=\frac{1}{2}\vert a\vert \vert w\vert ^2$. $\square$

Since

$$\begin{aligned} e_r^a(z,w,t)=e^{i\langle a,t\rangle } L_r^{n-1}\left( \frac{1}{2}\vert a\vert (\vert z\vert ^2+\vert w\vert ^2)\right) e^{-\frac{1}{4}\vert a\vert (\vert z\vert ^2+\vert w\vert ^2)} \end{aligned}$$

using the Lemma 3.1 we can write,

$$\begin{aligned} e_r^a(z,w,t)=\sum _{p+q=r}e_{p,q}^a(z,w,t)\qquad \text{ for } r=0,1,2,\ldots \end{aligned}$$

(4)

Hence from (3) we get the following

Proposition 3.1

If $f\in L^2(N)$ we have

$$\begin{aligned} f(z,w,t)=\int \limits _{{\mathbb {R}}^3\setminus \{0\}}\sum _{p,q} f*e_{p,q}^a(z,w,t)\vert a\vert ^n\, da \end{aligned}$$

where the above expansion converges in $L^2(N)$.

When f is a Schwartz class function on N,

$$\begin{aligned} f*e_{p,q}^a(z,w,t)=e^{i\langle a,t\rangle }f^a\times _{\vert a\vert }\varphi _{p,q}^{\vert a\vert }(z,w). \end{aligned}$$

The functions $\varphi _{p,q}^{\vert a\vert }$ satisfy the orthogonality relation

$$\begin{aligned} \varphi _{p,q}^{\vert a\vert } \times _{\vert a\vert }\varphi _{r,s}^{\vert a\vert }(z,w)&=\int \limits _{{\mathbb {C}}^{2k}\times {\mathbb {C}}^{2l}}\varphi _{p,q}^{\vert a\vert }(z-u,w-v)\varphi _{r,s}^{\vert a\vert }(u,v)e^{\frac{i\vert a\vert }{2}Im(z\cdot {\overline{u}}+w\cdot {\overline{v}})}\, du\, dv\\&=\int \limits _{{\mathbb {C}}^{2k}\times {\mathbb {C}}^{2l}}\left( \varphi _{p}^{\vert a\vert ,2k}(z-u)\varphi _{q}^{\vert a\vert ,2l}(w-v)\varphi _{r}^{\vert a\vert ,2k}(u)\varphi _{s}^{\vert a\vert ,2l}(v)\right. \\&\qquad \qquad \left. e^{\frac{i\vert a\vert }{2}Im(z\cdot {\overline{u}}+w\cdot {\overline{v}})}\right) \, du\, dv\\&=\varphi _{p}^{\vert a\vert ,2k}\times _{\vert a\vert }\varphi _{r}^{\vert a\vert ,2k}(z)\, \varphi _{q}^{\vert a\vert ,2l}\times _{\vert a\vert }\varphi _{s}^{\vert a\vert ,2l}(w)\\&=\frac{(2\pi )^{n}}{\vert a\vert ^{n}}\,\delta _{p,r}\,\delta _{q,s}\,\varphi _{p,q}^{\vert a\vert }(z,w) \end{aligned}$$

which follows from the the orthogonality property of the Lagurre functions

$$\begin{aligned} \varphi _i^{\lambda , d} \times _{\lambda } \varphi _j^{\lambda , d} = \frac{(2\pi )^{d}}{\lambda ^{d}} \delta _{ij} \varphi _i^{\lambda , d} \end{aligned}$$

As a consequence of the above orthogonality property of $\varphi _{p,q}^{\vert a\vert }$, we see that the operator

$$\begin{aligned} {\mathcal {P}}_{p,q}:f\mapsto \int \limits _{{\mathbb {R}}^3\setminus \{0\}}f*e_{p,q}^a(z,w,t)\vert a\vert ^n\, da \end{aligned}$$

are projection operators.

Our aim is to write the spectral projection operator ${\mathcal {P}}_{p,q}$ as a convolution operator and prove its $L^p$ boundedness. To write this operator as a convolution operator, we define the kernel,

$$\begin{aligned} P_{p,q}(z,w,t)&=\int \limits _{{\mathbb {R}}^3\setminus \{0\}} ~e_{p,q}^a(z,w,t)~\vert a\vert ^n~da \\&= \int \limits _{{\mathbb {R}}^3\setminus \{0\}}~ e^{-i \langle a, t \rangle }~ \varphi _{p,q}^{\vert a\vert }(z,w)\vert a\vert ^n~ da. \end{aligned}$$

Since $\varphi _{p,q}^{\vert a\vert }(z, w) = L_p^{2k-1}\left( \frac{\vert a\vert \vert z\vert ^2}{2}\right) L_q^{2l-1}\left( \frac{\vert a\vert \vert w\vert ^2}{2}\right) e^{-\frac{\vert a\vert }{4}\left( \vert z\vert ^2+\vert w\vert ^2\right) }$, the kernel $P_{p,q}(z,w,t)$ is a linear combination of functions of the form

$$\begin{aligned} P_{p,q}^{i,j}(z,w,t) = \vert z\vert ^{2i}\vert w\vert ^{2j} \int \limits _{{\mathbb {R}}^3\setminus \{0\}}~ e^{-i \langle a, t \rangle }~ e^{-\frac{\vert a\vert }{4}(\vert z\vert ^2+\vert w\vert ^2)} ~\vert a\vert ^{n+i+j}\, da, \end{aligned}$$

$i=1,2,\ldots ,p$ and $j=1,2,\ldots ,q$. A simple change of variables shows that

$$\begin{aligned} P_{p,q}^{i.j}(sz,sw,,s^2t) = s^{-(2n+6)} P_{p,q}^j(z,w,t), \end{aligned}$$

which is the required homogeneity for singular integral operators on $N~=~{\mathbb {C}}^n~\oplus ~{\mathbb {R}}^3$

Since $P_{p,q}^{i,j}(z,w,t)$ is radial in z and w, we can write

$$\begin{aligned} P_{p,q}^{i,j}(z,w,t) = c \vert z\vert ^{2i}\vert w\vert ^{2j}\int _0^{\infty }\frac{J_{\frac{1}{2}}(\lambda \vert t\vert )}{(\lambda \vert t\vert )^{\frac{1}{2}}}e^{-\frac{\lambda }{4}\left( \vert z\vert ^2+\vert w\vert ^2\right) } \lambda ^{n+i+j+2}\, d\lambda , \end{aligned}$$

where c is a constant. We prove that $P_{p,q}(z,w,t)$ is a Calderón-Zygmund kernel by showing that each $P_{p,q}^j(z,w,t)$ is. Since $P_{p,q}^j(z,w,t)$ is homogeneous of degree $-Q=-2n-6$ and belongs to $C^{\infty }(N \setminus \{0\})$, by the Lemma 2.2 in [10], the required cancellation condition will be obtained from the following lemma.

Lemma 3.2

For $i=1,2,\ldots ,p$ and $j=1,2,\ldots ,q$,

$$\begin{aligned} \int \limits _{{\mathbb {C}}^{2k}}\int \limits _{{\mathbb {C}}^{2l}} P_{p,q}^{i,j}(z,w,1)\, dzdw = 0. \end{aligned}$$

Proof

We start with the integral

$$\begin{aligned} I(\tau ) = \int _0^{\infty }~ \frac{J_{\frac{1}{2}}(\lambda )}{\lambda ^{\frac{1}{2}}} ~e^{-\tau \lambda } ~\lambda ^{2}\; d\lambda , ~\tau > 0. \end{aligned}$$

(5)

Then for any $t \in {\mathbb {R}}^3$ such that $\vert t\vert = 1$, it is easy to see that (up to a constant)

$$\begin{aligned} I(\tau ) = \int \limits _{{\mathbb {R}}^3}~e^{-i \langle x, t \rangle } ~e^{-\tau \vert x\vert }~dx. \end{aligned}$$

The above equals the Poisson kernel,

$$\begin{aligned} c \frac{\tau }{(1+\tau ^2)^2} \end{aligned}$$

for some constant c.

Now,

$$\begin{aligned} \int _0^{\infty }~ \frac{J_{\frac{1}{2}}(\lambda )}{\lambda ^{\frac{1}{2}}} ~e^{-\tau \lambda }~ \lambda ^{n+i+j+2}~ d\lambda&= \frac{d^{n+i+j}}{d\tau ^{n+i+j}}\left( I (\tau )\right) \\&= I^{(n+i+j)}(\tau ). \end{aligned}$$

Hence, to prove the lemma, we need to show that

$$\begin{aligned} \int \limits _{{\mathbb {C}}^{2k}}\int \limits _{{\mathbb {C}}^{2l}}~\vert z\vert ^{2i}\vert w\vert ^{2j}~ I^{(n+i+j)}\left( \frac{\vert z\vert ^2+\vert w\vert ^2}{2}\right) \; dz\, dw = 0, ~j = 0, 1, 2, \dots , k. \end{aligned}$$

Since the integrand is radial in z and w, this reduces to showing that

$$\begin{aligned} \int \limits _0^{\infty }\int \limits _0^{\infty }~ I^{(n+i+j)}\left( \frac{r^2+s^2}{4}\right) ~r^{4k+2i-1}s^{4l+2j-1}~ dr\, ds = 0. \end{aligned}$$

By taking $r=\rho \cos {\theta },\, s=\rho \sin {\theta }$, $\rho >0$ and $0\le \theta \le \frac{\pi }{2}$, we obtain,

$$\begin{aligned} \int \limits _0^{\infty }\int \limits _0^{\infty }~ I^{(n+i+j)}\left( \frac{r^2+s^2}{4}\right)&~r^{4k+2i-1}s^{4l+2j-1}~ dr\, ds\\&= \left( \int \limits _{0}^{\frac{\pi }{2}}\cos ^{4k+2i-1}(\theta )\sin ^{4l+2j-1}(\theta )\, d\theta \right) \\&\qquad \times \left( \int \limits _{0}^\infty I^{(n+i+j)}\left( \frac{\rho ^2}{4}\right) \,\rho ^{4k+4l+2i+2j-2} d\rho \right) \\&= 2^{4k+4l+2i+2j-2}\left( \int \limits _{0}^{\frac{\pi }{2}}\cos ^{4k+2i-1}(\theta )\sin ^{4l+2j-1}(\theta )\, d\theta \right) \\&\qquad \times \left( \int \limits _{0}^\infty I^{(n+i+j)}\left( \rho \right) \,\rho ^{n+i+j-1} d\rho \right) \\ \end{aligned}$$

Now, writing

$$\begin{aligned} \Psi (\rho ) = \frac{1}{(1+\rho ^2)^{2}}, \end{aligned}$$

we get,

$$\begin{aligned} I^{(n+i+j)}(\rho ) = \rho \Psi ^{(n+i+j)}(\rho ) + (n+i+j)\Psi ^{(n+i+j-1)}(\rho ). \end{aligned}$$

Hence

$$\begin{aligned} \int \limits _0^{\infty }~ I^{(n+i+j)}(\rho ) ~\rho ^{n+i+j-1}\; d\rho&= \int _0^{\infty }~ \Psi ^{(n+i+j)}(\rho )~ \rho ^{n+i+j}\; d\rho \\&\quad + (n+i+j) \int _0^{\infty }~ \Psi ^{(n+i+j-1)}(\rho )~\rho ^{n+i+j-1}\; d\rho \\&= \lim \limits _{\rho \rightarrow \infty } \rho ^{n+i+j} \Psi ^{(n+j-1)}(\rho ) \end{aligned}$$

which is easily verified to be zero. This proves the lemma. $\square$

Since the kernel is radial in t, it follows from the Lemma 3.2, that

$$\begin{aligned} \int \limits _{{\mathbb {C}}^k}\int \limits _{{\mathbb {C}}^l}\int \limits _{S^2} P_{p,q}^{i,j}(z,w,t)\, dz\,dw\,d\sigma (t) = 0,~ j = 0, 1, 2, \dots , p+q. \end{aligned}$$

where $\sigma$ is the normalised surface measure on the unit sphere in ${\mathbb {R}}^3$. We need the following well-known theorem.

Theorem 3.1

Let N be a connected, simply connected H-type group. Let $K~\in ~ C^{\infty }(G \setminus \{0\})$ be a kernel which is homogeneous of degree $-Q$. Assume that K satisfies the cancellation condition

$$\begin{aligned} \int \limits _{a< \vert (z,t)\vert<b} K(z,t)\; dzdt = 0, \forall \; 0< a< b < \infty . \end{aligned}$$

Then the singular integral operator

$$\begin{aligned} f \mapsto f*K \end{aligned}$$

is bounded on $L^2(N)$.

Proof

This is a special case of Theorem 1 in [12, p. 494]. $\square$

The next theorem says that for the above operators, the $L^2$-boundedness imply the $L^p$-boundedness.

Theorem 3.2

Let N be an H-type group and $K \in C^{\infty }(N \setminus \{0\})$ be a kernel that satisfy the cancellation condition and is homogeneous of degree $-Q$. If the operator $f \mapsto f*K$ is bounded on $L^2(N)$, then it is bounded on $L^p(N)$ for $1< p< \infty$.

Proof

Follows from Theorem 5.1 of [13]. $\square$

From Theorem 3.1 and Theorem 3.2, we obtain the following result.

Theorem 3.3

For each (p.q) the spectral projection operator

$$\begin{aligned} {\mathcal {P}}_{p,q}:f\mapsto \int \limits _{{\mathbb {R}}^3\setminus \{0\}}f*e_{p,q}^a(z,w,t)\vert a\vert ^n\, da \end{aligned}$$

is a bounded operator on $L^r(N)$ for $1<r<\infty$.

Next we show the Abel summability of the spectral decomposition for $f~\in ~L^p(N)$

Theorem 3.4

For $2 \le p < \infty$ we have the Abel summability

$$\begin{aligned} \lim _{s\rightarrow 1}\sum _{d=0}^\infty s^d\sum _{p+q=d}\, \int \limits _{{\mathbb {R}}^3\setminus \{0\}} f*e_{p,q}^a(z,w,t)\vert a\vert ^n\, da=f(z,w,t) \end{aligned}$$

Proof

From Theorem 3.2 in [10] we have, for $2 \le p < \infty$ and $f \in L^p(N)$,

$$\begin{aligned} \lim \limits _{s \rightarrow 1} \sum \limits _{d=0}^{\infty } s^d \int \limits _{{\mathbb {R}}^m}f *e_k^a(z,t)~ \vert a\vert ^n~ da = f(z, t) \end{aligned}$$

in the $L^p$ norm. Then the result follows from (4). $\square$

4 Spherical means and injectivity

Recall that $U=Sp(k)\times Sp(l)$. An orbit of U is of the form $S_r\times S_s$, where $S_r$ is the sphere of radius r in ${\mathbb {C}}^{2k}$ and $S_s$ is the sphere of radius s in ${\mathbb {C}}^{2l}$. Let $\mu _{r,s}$ be the normalized surface measure on the product of $S_r$ and $S_s$.

If f is of the form $f(z,w,t)=e^{i<a,t>}g(z)h(w)$, for $(z,w,t)\in N$ then

$$\begin{aligned} f*\mu _{r,s}(z,w,t)&= \int \limits _{{\mathbb {C}}^k}\int \limits _{{\mathbb {C}}^l}f\left( (z,w,t)(\xi ,\eta ,0)^{-1}\right) \, d\mu _{r,s}(\xi ,\eta )\\&= \int \limits _{{\mathbb {C}}^k}\int \limits _{{\mathbb {C}}^l}f\left( z-\xi ,w-\eta ,t-\frac{1}{2}[z,\xi ]-\frac{1}{2}[w,\eta ]\right) \, d\mu _{r,s}(\xi ,\eta )\\&= c\int \limits _{{\mathbb {C}}^k}\int \limits _{{\mathbb {C}}^l}g(z-\xi )h(w-\eta )e^{i\langle a,t-\frac{1}{2}[z,\xi ]-\frac{1}{2}[w,\eta ]\rangle }\, d\mu _r(\xi )\,d\mu _s(\eta )\\&= c\,e^{i\langle a,t\rangle }\int \limits _{{\mathbb {C}}^k}g(z-\xi )e^{-i\langle a,\frac{1}{2}[z,\xi ]\rangle }\, d\mu _r(\xi )\\&\qquad \times \int \limits _{{\mathbb {C}}^l}h(w-\eta )e^{-i\langle a,\frac{1}{2}[w,\eta ]\rangle }\,d\mu _s(\eta )\\&=c\, e^{i\langle a,t\rangle } \left( g\times _{\vert a\vert }\mu _r\right) (z) \left( h\times _{\vert a\vert }\mu _s\right) (w) \end{aligned}$$

Since (see [2, Proposition 5.1])

$$\begin{aligned} \varphi _k^{\vert a\vert } \times _{\vert a\vert } \mu _r(z) = \frac{k! (n-1)!}{(k+n-1)!} \varphi _k^{\vert a\vert }(r) \varphi _k^{\vert a\vert } (z) \end{aligned}$$

we obtain,

$$\begin{aligned} e_{p,q}^a*\mu _{r,s}(z,w,t)=e_{p,q}^a(r,s,0)e_{p,q}^a(z,w,t) \end{aligned}$$

(6)

We need the following result.

Theorem 4.1

Let $f \in L^p({\mathbb {R}}^m)$ and support of ${\widehat{f}}$ (distributional Fourier transform of f) is contained in a $C^1$-manifold of dimension d, $0< d < m$. Then f vanishes identically provided $1 \le p \le \frac{2m}{d}.$ If $d = 0,$ f vanishes identically provided $1 \le p < \infty .$

Proof

When the support is a sphere, this follows from [2] (see Lemma 2.2 and Theorem 2.2 there). For the general case see [14] (Theorem 1). $\square$

Theorem 4.2

Let $f\in L^p(N), 1\le p\le 3$. If $f*\mu _{r,s}=0$ then $f\equiv 0$.

Proof

Let $f\in L^p(N), 1\le p \le 3$ and assume that $f *\mu _{r,s}$ vanishes identically. Convolving f with a smooth approximate identity, we may assume that $f\in L^p$ for $2 \le p \le 3$. From (6), the spectral decomposition of $f*\mu _{r,s}$ is given by

$$\begin{aligned} f*\mu _{r,s}(z,w,t)=\sum \limits _{p,q=0}^{\infty } \int \limits _{{\mathbb {R}}^3\setminus \{0\}}c\, e_{p,q}^a(r,s,0)~f*e_{p,q}^a(z,w,t)~\vert a\vert ^n\, da . \end{aligned}$$

If $f*\mu _{r,s}(z,w,t)=0$ for all (z, w, t), by Theorem 3.4,

$$\begin{aligned} \lim \limits _{u\rightarrow 1^{-}}\sum \limits _{d=0}^\infty u^d\sum \limits _{p+q=d} \int \limits _{{\mathbb {R}}^3-\{0\}}~e_{p,q}^a(r,s,0)~f*e_{p,q}^a(z,w,t)~\vert a\vert ^n\, da=0 \end{aligned}$$

where the convergence is in $L^p(N)$. Applying the (p, q)-th spectral projection operator ${\mathcal {P}}_{p,q}$ and using Theorem 3.3 we obtain that, for all $(z,w,t)\in N$ and for all $p,q=1,2,\ldots$,

$$\begin{aligned} \int \limits _{{\mathbb {R}}^3\setminus \{0\}}\varphi _p^{\vert a\vert }(r)\varphi _q^{\vert a\vert }(s) f^a \times _{\vert a\vert } \varphi _{p,q}^{\vert a\vert }(z,w)~ e^{-i\langle a, t \rangle } \vert a\vert ^n\, da=0. \end{aligned}$$

Arguing as in [2, p.276] (also see [4, pp.257-258]), we obtain that, for almost all $(z,w)\in {\mathbb {C}}^k\times {\mathbb {C}}^l,$ the support of $f^a \times _{\vert a\vert }\varphi _{p,q}^{\vert a\vert }(z,w)\vert a\vert ^n$, the distributional Fourier transform of ${\mathcal {P}}_{p,q}f(z,w,\cdot )$, is contained in the zero set of $a \mapsto L_p^{2k-1}(\frac{1}{2}\vert a\vert r^2)L_p^{2l-1}(\frac{1}{2}\vert a\vert s^2)$, which is a finite union of spheres in ${\mathbb {R}}^m.$ But this implies, by Theorem 4.1, that ${\mathcal {P}}_{p,q}f(z,w,t)$ is zero as ${\mathcal {P}}_{p,q}f\in L^p$ for $1\le p \le 3$. This completes the proof. $\square$