In this section we describe all symmetric monomials on \(N^d\). We shall use the following notation: for \(d\ge 2\) and \(1\le k\le d\) we write \([x]_k\) for the element of \(N^k\) whose components are all equal to x. Hence \([x]_1=x\), \([x]_2=(x,x)\), etc. If \(A:N^d\rightarrow \mathbb {C}\) is a d-additive function and it is symmetric in its variables, then the function \(x\mapsto A([x]_d)\) is called its diagonalization.
Theorem 11
[1, 4] Let N be a commutative locally compact topological group and d a positive integer. Every symmetric spherical function s on \(N^d\) has the form \(s=m^{\#}\) with some exponential m on \(N^d\). In other words,
$$\begin{aligned} s(x)=\frac{1}{d!}\sum _{\sigma \in {\mathscr {P}}_d} m(\sigma x) \end{aligned}$$
for each x in \(N^d\).
Lemma 1
[4] A function \(\varphi {:}\,N^{d}\rightarrow \mathbb {C}\) fulfills
$$\begin{aligned} \mathrm {D}_{1; y}*\varphi (x)=0 \end{aligned}$$
that is,
$$\begin{aligned} \int _{\mathscr {P}_{d}}\varphi (x+\sigma y)d\omega _d(\sigma )= \varphi (x) \end{aligned}$$
for each x, y in \(N^d\) if and only if for each \(k=1,2, \ldots , d!-1\) there exist symmetric, k-additive functions \(A_{k}:(N^{d})^{k}\rightarrow \mathbb {C}\) and a complex constant \(A_{0}\) such that, for each x, y in \(N^d\) we have
$$\begin{aligned} \varphi (x)= A_{0}+\sum _{k=1}^{d!-1} A_{k}([x]_k) \end{aligned}$$
and
$$\begin{aligned} \int _{\mathscr {P}_{d}}A_{k}\left( [x]_{k-i}, \left[ \sigma y\right] _{i}\right) d\omega _{d}(\sigma )=0. \end{aligned}$$
Remark
We remark that using some results of [7] we immediately get that symmetric 1-monomials are symmetric generalized polynomials. Indeed, if \(\varphi {:}\,N^{d}\rightarrow \mathbb {C}\) is a symmetric 1-monomial, then there exists a nonnegative integer n such that
$$\begin{aligned} D_{1; y_{1}, y_2,\ldots , y_{n+1}}*\varphi (x)=0 \end{aligned}$$
holds for any \(x, y_{1}, \ldots , y_{n+1}\) in \(N^{d}\). Clearly, in this case the less restrictive equation,
$$\begin{aligned} D_{1; y}^{n+1}*\varphi (x)=0 \end{aligned}$$
is also satisfied for each x, y in \(N^d\). After computing the symmetric modified differences we obtain that a functional equation of the form
$$\begin{aligned} \varphi (x)+\sum _{i=1}^{k}\varphi _{i}\left( x+\psi _{i}(y)\right) =0 \end{aligned}$$
holds for each x, y in \(N^d\) yielding that \(\varphi \) is of degree n, which, in view of Theorem 3.6 of [7], immediately implies that \(\varphi \) is a generalized polynomial. Here we have for each x in \(N^d\) and for \(i=1,2, \ldots , k\):
$$\begin{aligned} \varphi _{i}(x)= \alpha _{i}\varphi (x) \quad \text {and} \quad \psi _{i}(x)=\beta _{i}\sigma _i x \qquad \left( x\in G^{d}\right) \end{aligned}$$
with certain complex constants \(\alpha _{i}, \beta _{i}\).
In [4] Theorem 2.10 describes the solutions of the functional equation
$$\begin{aligned} D_{s; y}*\varphi (x)=0, \end{aligned}$$
where x, y are in \(N^d\). More exactly we have the following result:
Theorem 12
The non-zero functions \(\varphi {:}\,N^{d}\rightarrow \mathbb {C}\) and \(s{:}\,N^{d}\rightarrow \mathbb {C}\) satisfy the functional equation
$$\begin{aligned} \int _{\mathscr {P}_{d}}\varphi (x+\sigma y)d\omega _d(\sigma )= s(y)\varphi (x) \end{aligned}$$
for each x, y in \(N^d\) if and only if there exists an exponential \(m{:}\,N^{d}\rightarrow \mathbb {C}\) such that \(s=m^{\#}\) and there are complex constants \(A_{0}^{\sigma }\) and symmetric, k-additive functions \(A^{\sigma }_{k}\) such that for each x, y in \(N^d\) we have
$$\begin{aligned} \varphi (x)= \sum _{\sigma \in \mathscr {P}_{d}^{*}}m(\sigma x)\cdot \left( A_{0}^{\sigma } +\sum _{i=1}^{\left| \mathscr {P}_{d}^{0}\right| -1}A_{i}^{\sigma }\left( [x]_{i}\right) \right) , \end{aligned}$$
and
$$\begin{aligned} \sum _{\sigma \in \mathscr {P}_{d}^{0}}A_{k}^{\sigma }\left( [x]_{k-i}, \left[ \xi y\right] _{i}\right) =0 \qquad \left( \xi \in \mathscr {P}_{d}^{*}, 1 \le i\le k\le \left| \mathscr {P}_{d}^{0}\right| -1\right) . \end{aligned}$$
Here \(\mathscr {P}_{d}^{0}= \left\{ \sigma \in \mathscr {P}_{d}\, \vert \, m\circ \sigma =m\right\} \), and \(\mathscr {P}_{d}^{*}\) stands for the quotient group \({\mathscr {P}_{d}}\big /{\mathscr {P}_{d}^{0}}\).
Theorem 13
Let n be a positive integer and \(s{:}\,N^{d}\rightarrow \mathbb {C}\) be a symmetric spherical function. If a symmetric function \(\varphi :N^{d}\rightarrow \mathbb {C}\) satisfies
$$\begin{aligned} D_{s; y_{1},y_2, \ldots , y_{n+1}}*\varphi (x)=0 \end{aligned}$$
(7)
for each \(x, y_{1},y_2 \ldots , y_{n+1}\) in \(N^d\), then \(\varphi \) is a generalized exponential polynomial.
Proof
Let n be a positive integer, \(s{:}\,N^{d}\rightarrow \mathbb {C}\) a symmetric spherical function, and assume that for the mapping \(\varphi {:}\,N^{d}\rightarrow \mathbb {C}\) equation (7) holds for each \(x, y_{1},y_2, \ldots , y_{n+1}\) in N. In view of Theorem 11, there exists an exponential m on \(G^{d}\) such that \(s=m^{\#}\).
On computing the symmetric modified differences, Eq. (7) implies that the function \(\varphi \) has to fulfill
$$\begin{aligned}&\sum _{i=0}^{n+1}\sum _{1\le j_{1}< \cdots < j_{l}\le n+1}(-1)^{j_{1}+\cdots +j_{l}} s\left( y_{j_{1}}\right) \cdot s\left( y_{j_2}\right) \cdots s\left( y_{j_{l}}\right) \nonumber \\&\quad \cdot \,\sum _{\sigma _{1}\in \mathscr {P}_{d}}\cdots \sum _{\sigma _{j+1}\in \mathscr {P}_{d}} \varphi \left( x+\varepsilon _{1}\sigma _{1}\left( y_{1}\right) +\varepsilon _{2}\sigma _{2}\left( y_{2}\right) +\cdots +\varepsilon _{j+1}\sigma _{j+1}(y_{j+1})\right) =0\nonumber \\ \end{aligned}$$
(8)
for all \(x, y_{1}, y_2,\ldots , y_{n+1}\) in \(N^{d}\), where in each summand \(\varepsilon _{i}\) in \(\{0, 1\}\) depends on whether i does or does not belong to the set \(\left\{ j_{1}, j_2,\ldots , j_{l}\right\} \). Using this equation, our point is to apply different substitutions for the variables \(x, y_{1}, y_2,\ldots , y_{n+1}\) in \(N^d\) to arrive at the conclusion that the mapping
$$\begin{aligned} \left( x_{1},x_2, \ldots , x_{n+2}\right) \longmapsto \varphi \circ \pi _{1}\left( x_{1}+x_2+\cdots +x_{n+2}\right) \end{aligned}$$
is decomposable, and then we apply Theorem 10.
We introduce the following technical notation: if k and \(1\le n_{1}< n_2<\cdots < n_{k}\le d\) are positive integers, then \(\pi _{n_{1}, n_2,\ldots , n_{k}} y\) denotes that element in \(N^d\) which may differ from y in only those coordinates which do not belong to the set \(\{n_{1},n_2,\dots , n_{k}\}\), and those coordinates are zero. For instance \(\pi _1y=(y^{(1)},0,0,\dots ,0)\), or \(\pi _{1,d}y=(y^{(1)},0,\dots ,0,y^{(d)})\), etc., for each \(y=(y^{(1)}, y_2,\ldots , y^{(d)})\) in \(N^{d}\). In addition, let \(\pi _0 y=0\) for each y.
We have to consider the cases \(n+2\ge d\) and \(n+2<d\) separately. If \(n+2\ge d\), then the substitutions
$$\begin{aligned} \pi _{1}x\quad \text {for}\quad x, \quad \pi _{2,3, \ldots , d}y_{1}\quad \text {for}\quad y_1, \quad \text { and } \quad \pi _0y_{j}\quad \text {for}\quad y_j \quad \text {if} \quad 2\le j\le n+1 \end{aligned}$$
into (8) yield that the mapping \(\varphi \) can be expressed with the one-variable function \(\varphi \circ \pi _{1}\) and the \((d-1)\)-variable function \(\varphi \circ \pi _{2, \ldots , d}\). The substitutions
$$\begin{aligned} \pi _{1}x\quad \text {for}\quad x, \quad \pi _{3,4, \ldots , d}y_{1}\quad \text {for}\quad y_1, \quad \text { and } \quad \pi _0y_{j}\quad \text {for}\quad y_j \quad \text {if} \quad 2\le j\le n+1 \end{aligned}$$
into (8) show that \(\varphi \circ \pi _{2,3, \ldots , d}\) can be expressed with the one-variable function \(\varphi \circ \pi _{1}\), the \((d-2)\)-variable function \(\varphi \circ \pi _{3,4, \ldots , d}\), and the exponential m. Continuing this descending argument, we finally get that the mapping \(\varphi \) can be written with the aid of the one-variable function \(\varphi \circ \pi _{1}\) and the exponential m.
On the other hand, if \(n+2< d\), then the substitutions
$$\begin{aligned} \pi _{1}x\quad \text {for}\quad x, \quad \pi _{2}y_1\quad \text {for}\quad y_1, \quad \ldots \quad , \pi _{n+1}y_n\quad \text {for}\quad y_n \end{aligned}$$
and
$$\begin{aligned} \pi _{n+2, \ldots , d}y_{n+1}\quad \text {for}\quad y_{n+1} \end{aligned}$$
into (8) yield that the mapping \(\varphi \) can be expressed with the one-variable function \(\varphi \circ \pi _{1}\) and the \((d-n-2)\)-variable function \(\varphi \circ \pi _{n+2,n+3, \ldots , d}\). Further, the substitutions
$$\begin{aligned} \pi _{1}x\quad \text {for}\quad x, \quad \pi _{2}y_1\quad \text {for}\quad y_1, \quad \ldots \quad , \pi _{n+1}y_n\quad \text {for}\quad y_n \end{aligned}$$
and
$$\begin{aligned} \pi _{n+2,n+3, \ldots , d-1}y_{n+1}\quad \text {for}\quad y_{n+1} \end{aligned}$$
into (8) show that \(\varphi \circ \pi _{n+2,n+3, \ldots , d}\) can be expressed with the one-variable function \(\varphi \circ \pi _{1}\) and the \((d-n-3)\)-variable function \(\varphi \circ \pi _{n+2,n+3, \ldots , d-1}\) and the exponential m. Continuing this descending argument, we finally get that the mapping \(\varphi \) can be expressed in terms of the one-variable function \(\varphi \circ \pi _{1}\) and the exponential m, in this case, too.
Finally we conclude that the mapping \(\varphi \) can be expressed in terms of the one-variable function \(\varphi \circ \pi _{1}\) and the exponential m, in this case, too. Writing this form back into Eq. (7) we obtain that
$$\begin{aligned} \left( \varphi \circ \pi _{1}\right) \left( x_{1}+x_2+\cdots + x_{n+2}\right)&= \sum _{i=1}^{n+1}\Phi _{i, 1}\left( x_{i}\right) \Psi _{i, n+1}\left( \sum _{j\ne i}x_{j}\right) \\&\quad +\,\sum _{i\ne j}\Phi _{i, j, 2}\left( x_{i}+x_{j}\right) \cdot \Psi _{i, j, n}\left( \sum _{k\ne i, j}x_{k}\right) + \cdots \end{aligned}$$
for each \(x_{1},x_2, \ldots , x_{n+2}\) in \(N^{d}\), showing that the mapping
$$\begin{aligned} \left( x_{1}, x_2,\ldots , x_{n+2}\right) \longmapsto \varphi \circ \pi _{1}\left( x_{1}+x_2+\cdots +x_{n+2}\right) \end{aligned}$$
is decomposable. Using Theorem 10, we obtain that \(\varphi \circ \pi _{1}\) is a generalized exponential polynomial on \(N^{d}\). Since \(\varphi \) is a (continuous) polynomial of \(\varphi \circ \pi _{1}\), we get that \(\varphi \) is a generalized exponential polynomial, as well. The proof is complete. \(\square \)
We shall use the following symmetrization operator: for each positive integer k and continuous function \(F{:}\,(N^d)^k\rightarrow \mathbb {C}\) we write
$$\begin{aligned}&\mathrm {sym\,}_{u_1,u_2,\dots ,u_k} F\left( u_1,u_2,\dots ,u_k\right) \\&\quad =\int _{{\mathscr {P}}_n} \dots \int _{{\mathscr {P}}_n} F\left( \sigma _1 u_1,\sigma _2 u_2,\dots ,\sigma _k u_k\right) \,d\omega \left( \sigma _1\right) \,d\omega \left( \sigma _2\right) \dots \,d\omega \left( \sigma _k\right) . \end{aligned}$$
The subscripts of \(\mathrm {sym\,}\) indicate which variables are subjected to symmetrization. For instance, we have for each f in \({\mathscr {C}}(N^d)\) and exponential m on \(N^d\):
$$\begin{aligned}&\mathrm {sym\,}_{x,y} \left[ f(x+y)-m(y)f(x)\right] \\&\quad = \int _{{\mathscr {P}}_d} \int _{{\mathscr {P}}_d} \left[ f\left( \sigma _1x+\sigma _2y\right) -m\left( \sigma _2y\right) f\left( \sigma _1x\right) \right] \,d\omega \left( \sigma _1\right) \,d\omega \left( \sigma _2\right) \\&\quad = \int _{{\mathscr {P}}_d} \left( \int _{{\mathscr {P}}_d}\left[ f\left( \sigma _1\left( x+\sigma _1^{-1}\sigma _2y\right) -f\left( \sigma _1x\right) m\left( \sigma _2y\right) \right] \,d\omega \left( \sigma _1\right) \right) \,d\omega \left( \sigma _2\right) \right. \\&\quad = \int _{{\mathscr {P}}_d} \left[ f^{\#}\left( x+\sigma y\right) -f^{\#}(x)m(\sigma y)\right] \,d\omega (\sigma ) \\&\quad =\int _{{\mathscr {P}}_d} f^{\#}\left( x+\sigma y\right) \,d\omega (\sigma )-f^{\#}(x)m^{\#}(y)=\mathrm {D}_{m^{\#};y}*f^{\#}(x), \end{aligned}$$
or
$$\begin{aligned} \mathrm {sym\,}_{x,y}\, \left[ \Delta _{m;y}*f(x)\right] =\mathrm {D}_{m^{\#};y}*f^{\#}(x) \end{aligned}$$
for each x, y in \(N^d\). More generally, the modified differences have the following remarkable property.
Proposition 1
Let n be a natural number, \(m{:}\,N^d\rightarrow \mathbb {C}\) an exponential and f in \({\mathscr {C}}(N^d)\). Then we have for each \(x,y_1,y_2,\dots ,y_{n+1}\) in N
$$\begin{aligned} \mathrm {D}_{m^{\#};y_1,y_2,\dots ,y_{n+1}}*f^{\#}(x)=\mathrm {sym\,}_{x, y_1,y_2,\dots ,y_{n+1}}\left[ \Delta _{m;y_1,y_2,\dots ,y_{n+1}}*f(x)\right] . \end{aligned}$$
(9)
Proof
Let \(m{:}\,N^d\rightarrow \mathbb {C}\) be an exponential and f be in \({\mathscr {C}}(N^d)\). We prove the identity by induction on n. Due to the above remark, the statement holds true for \(n=1\).
Suppose now that there exists a natural number n such that the above identity holds for n, that is, we have
$$\begin{aligned} \mathrm {D}_{m^{\#};y_1,y_2,\dots ,y_{n}}*f^{\#}(x)=\mathrm {sym\,}_{x, y_1,y_2,\dots ,y_{n}}\left[ \Delta _{m;y_1,y_2,\dots ,y_{n}}*f(x)\right] . \end{aligned}$$
for each \(x, y_{1},y_2, \ldots , y_{n}\) in \(N^{d}\).
Let \(x, y_{1},y_2, \ldots , y_{n}, y_{n+1}\) be in \(N^{d}\), then due to the induction hypothesis, we have
$$\begin{aligned}&\mathrm {D}_{m^{\#};y_1, y_2,\dots , y_{n}, y_{n+1}}*f^{\#}(x)\\&\quad = \left( \mathrm {D}_{m^{\#}; y_{n+1}}*\mathrm {D}_{m^{\#};y_1,y_2,\dots ,y_{n}}\right) *f^{\#}(x) \\&\quad = \mathrm {D}_{m^{\#}; y_{n+1}}*\left( \mathrm {D}_{m^{\#};y_1,y_2,\dots ,y_{n}}\right) *f^{\#}(x) \\&\quad = \mathrm {D}_{m^{\#}; y_{n+1}}*\left( \mathrm {sym}_{x, y_{1},y_2, \ldots , y_{n}}\Delta _{m; y_{1}, y_2,\ldots , y_{n}}*f(x)\right) \\&\quad = \left( \delta ^{\#}_{-y_{n+1}}-m^{*}(y_{n+1})\delta _{0}\right) *\left( \mathrm {sym}_{x, y_{1}, y_2,\ldots , y_{n}}\Delta _{m; y_{1},y_2, \ldots , y_{n}}*f(x)\right) \\&\quad = \delta ^{\#}_{-y_{n+1}}*\left( \int _{\mathscr {P}_{n}}\cdots \int _{\mathscr {P}_{n}}\Delta _{m; \sigma _1 y_{1},\ldots , \sigma _{n}y_{n}}*f(\sigma x)\,d\omega (\sigma )\,d\omega (\sigma _1)\ldots \omega (\sigma _{n})\right) \\&\qquad -\int _{\mathscr {P}_{d}}m\left( \sigma _{n+1}y_{n+1}\right) d\omega \left( \sigma _{n+1}\right) \\&\qquad \times \,\left( \int _{\mathscr {P}_{n}}\cdots \int _{\mathscr {P}_{n}} \Delta _{m; \sigma _{1}y_{1}, \ldots , \sigma _{n}y_{n}}*f(\sigma x)d\omega (\sigma )d\omega \left( \sigma _1\right) \ldots \omega \left( \sigma _{n}\right) \right) \\&\quad = \int _{\mathscr {P}_{n}}\cdots \int _{\mathscr {P}_{n}}\Delta _{m; \sigma _1 y_{1}, \ldots , \sigma _{n}y_{n}}*f\left( \sigma x+\sigma _{n+1}y_{n+1}\right) d\omega (\sigma )d\omega \left( \sigma _1\right) \ldots \omega \left( \sigma _{n+1}\right) \\&\qquad - \int _{\mathscr {P}_{n}}\cdots \int _{\mathscr {P}_{n}} m\left( \sigma _{n+1}y_{n+1}\right) \cdot \Delta _{m; \sigma _{1}y_{1},\ldots , \sigma _{n}y_{n}}*f(\sigma x)d\omega (\sigma )d\omega \left( \sigma _1\right) \\&\qquad \quad \ldots \omega \left( \sigma _{n+1}\right) \\&\qquad = \int _{\mathscr {P}_{n}}\cdots \int _{\mathscr {P}_{n}}\Delta _{m; \sigma _1 y_{1},\sigma _2y_2, \ldots , \sigma _{n+1}y_{n+1}}*f\left( \sigma x\right) d\omega (\sigma )d\omega \left( \sigma _1\right) \,d\omega \left( \sigma _2\right) \\&\quad \qquad \ldots \omega \left( \sigma _{n+1}\right) = \mathrm {sym}_{x, y_{1}, y_2,\ldots , y_{n+1}}\left[ \Delta _{m;y_1,y_2, \dots ,y_{n+1}}*f(x)\right] , \end{aligned}$$
proving that the statement holds also for \(n+1\). \(\square \)
In view of Theorem 13, every symmetric monomial on \(N^{d}\) is a generalized exponential polynomial. Knowing this, we are able to describe symmetric monomials on \(N^{d}\).
Assume therefore that a symmetric function \(\varphi {:}\,N^{d}\rightarrow \mathbb {C}\) is a spherical monomial, that is, there exists a symmetric spherical function s on \(N^{d}\) and a positive integer n such that
$$\begin{aligned} \mathrm {D}_{s; y_{1},y_2, \ldots , y_{n+1}}*\varphi (x)=0 \end{aligned}$$
hold for each \(x, y_{1}, y_2,\ldots , y_{n+1}\). By Theorem 13, the function \(\varphi \) in \(\mathscr {C}(N^{d})\) is a generalized exponential polynomial, so there is a positive integer k, there exist different exponentials \(m_{1}, m_2,\ldots , m_{k}\) on \(N^{d}\) and there are non-zero generalized polynomials \(P_{1}, P_2,\ldots , P_{k}\) on \(N^{d}\) such that for each x in \(N^d\)
$$\begin{aligned} \varphi (x)= \sum _{i=1}^{k}P_{i}(x)m_{i}(x). \end{aligned}$$
Since \(\varphi \) is symmetric, the representation
$$\begin{aligned} \varphi (x)= \sum _{i=1}^{k}\left( P_{i}m_{i}\right) ^{\#}(x) \end{aligned}$$
also holds for each x in \(N^{d}\). First we show that \(k\le d!\) and for each \(i=1,2, \ldots , k\) there exists a \(\sigma _{i}\) in \(\mathscr {P}_{d}\) such that
$$\begin{aligned} m_{i}(x)=m(\sigma x) \end{aligned}$$
holds for each x in \(N^d\), where the exponential m is determined by \(s=m^{\#}\). To do so, the following lemmata are needed.
Hereinafter Lemma 4.3 from [8], that is, the statement below will be used several times.
Proposition 2
Let G be an Abelian group and n be a positive integer. Suppose that \(\sum _{k=1}^n P_k m_k=0\), where \(m_{1}, m_2,\ldots , m_{n}:G\rightarrow \mathbb {C}\) are different exponentials, and \(P_{1}, P_2,\ldots , P_{n}{:}\,G\rightarrow \mathbb {C}\) are (generalized) polynomials. Then for each \(i=1,2, \ldots , n\) the polynomial \(P_{i}\) is identically zero.
It follows that the representation \(f=\sum _{k=1}^n P_k m_k\) of the generalized exponential polynomial f with different exponentials is unique up to the order of the terms, and it is called the canonical representation.
Lemma 2
Let \(m_{1}, m_{2}\) be exponentials on \(N^{d}\). If \(m^{\#}_1=m^{\#}_{2}\), then there is a \(\sigma \) in \(\mathscr {P}_{d}\) such that \(m_2=\sigma m_1\).
Proof
It is known (see e.g. [8, Theorem 6.11]) that there are exponentials \(m_{1,j}, m_{2,j}:N\rightarrow \mathbb {C}\) for \(j=1,2,\dots ,d\) on N such that
$$\begin{aligned} m_1\left( x_1,x_2,\dots ,x_d\right) =m_{1,1}\left( x_1\right) \cdot m_{1,2}\left( x_2\right) \cdots m_{1,d}\left( x_d\right) , \end{aligned}$$
and
$$\begin{aligned} m_2\left( x_1,x_2,\dots ,x_d\right) =m_{2,1}\left( x_1\right) \cdot m_{2,2}\left( x_2\right) \cdots m_{2,d}\left( x_d\right) , \end{aligned}$$
holds for each \(x=(x_1,x_2,\dots ,x_d)\) in \(N^d\). As \(m_{1}^{\#}=m_{2}^{\#}\), substituting \(\pi _{1}x\) for x we have for each x in \(N^d\)
$$\begin{aligned} \sum _{i=1}^{d}m_{1,i}\left( x_{1}\right) =\sum _{i=1}^{d}m_{2,i}\left( x_{1}\right) . \end{aligned}$$
Using Proposition 2 and the uniqueness of the canonical representation we infer that the exponentials \(m_{2,i}\) are the same as the exponentials \(m_{1,i}\), possibly listed in a different order. But this is exactly the statement of the lemma. \(\square \)
Lemma 3
Let n be a positive integer, \(m_k\) different exponentials on \(N^{d}\), and \(\alpha _k\) nonzero complex numbers \((k=1,2,\dots ,n)\). If \(\sum _{k=1}^n \alpha _k m_k^{\#}=0\), then there is an exponential m on \(N^d\) such that, for each \(k=1,2,\dots ,n\) there exists \(\sigma _k\) in \({\mathscr {P}}_d\) for which \(m_k=\sigma _k m\). In particular, \(n\le d!\).
Proof
We prove by induction on n and the case \(n=1\) is clear, by the previous lemma. Suppose that we have proved the statement for some \(n\ge 1\), and assume that \(\sum _{k=1}^{n+1} \alpha _k m_k^{\#}=0\). We define \(f=\sum _{k=1}^{n+1} \alpha _k m_k\), where the \(m_k\)’s are different exponentials on \(N^d\) and the \(\alpha _k\)’s are nonzero complex numbers. Then, for each \(j=1,2, \ldots , n+1\) and x, y in \(N^d\), we have
$$\begin{aligned} 0=\mathrm {D}_{m_{j}^{\#}; y}*f^{\#}(x)= \mathrm {sym}_{x, y}\left[ \Delta _{m_{j}; y}*f(x)\right] . \end{aligned}$$
With the notation \(\widetilde{\alpha _{i}}(y)=\alpha _{i} \bigl (m_i^{\#}(y)-m_j^{\#}(y)\bigr )\) \((i=1,2,\dots ,n+1,i\ne j)\) this can be written as
$$\begin{aligned} g_j(x)= & {} \sum _{\begin{array}{c} i=1 \end{array}, i\ne j}^{n+1}\widetilde{\alpha _{i}}(y)m_{i}^{\#}(x)\\= & {} \sum _{\begin{array}{c} i=1 \\ i\ne j \end{array}}^{n}\alpha _{i}\left( \int _{\mathscr {P}_{d}}\left[ m_{i}\left( \sigma _{2}y\right) -m_{j}\left( \sigma _{2}y\right) \right] d\omega _d(\sigma _{2})\right) \cdot \left( \int _{\mathscr {P}_{d}}m_{i}(\sigma _{1} x)d\omega _d\left( \sigma _{1}\right) \right) =0, \end{aligned}$$
whenever x, y is in \(N^d\). First we suppose that for any choice of j, there is a y such that \(\widetilde{\alpha _{i}}(y)\ne 0\) for each \(i\ne j\). Then, by induction, we conclude that
$$\begin{aligned} m_i^{\#}=m_j^{\#}\quad \text {whenever}\quad i=1,2,\dots ,n+1, i\ne j. \end{aligned}$$
Interchanging j with an \(i\ne j\) and applying the same argument, by Lemma 2, we obtain the statement of the theorem.
Otherwise we have that, for each \(j=1,2,\dots ,n+1\), there is \(i\ne j\) such that \(\widetilde{\alpha _{i}}=0\). Then, for that pair i, j we have \(m_i^{\#}=m_j^{\#}\), and, by Lemma 2, we infer that \(m_i=\sigma m_j\). Clearly, this will lead to the same conclusion stated by the theorem.
Finally, if we assume indirectly that \(n>d!\), then, by the previous step, there exists an exponential m on \(N^{d}\) so that for each \(i=1, 2,\ldots , n\), there is a certain \(\sigma \in \mathscr {P}_{d}\) such that \(m_i=\sigma m\). Since \(\left| \mathscr {P}_{d}\right| =d!<n\), this would mean that the exponentials \(m_{1}, m_2,\ldots , m_{n}\) cannot be different, contrary to our assumption. \(\square \)
Proposition 3
Let n be a positive integer, let \(m_{1}, m_2,\ldots , m_{n}\) be different exponentials on \(N^{d}\) further let \(P_{1}, P_2,\dots ,P_n{:}\, N^{d}\rightarrow \mathbb {C}\) be nonzero generalized polynomials. If \(f=\sum _{k=1}^n P_k m_k\), and \(f^{\#}=0\), then there is an exponential m on \(N^d\), and for each \(i=1,2,\dots ,n\) there exists a \(\sigma _i\) in \({\mathscr {P}}_d\) such that \(m_i=\sigma _im\).
Proof
The statement is obvious for \(n=1\), hence we assume that \(n\ge 2\). As the projection of the function f is identically zero, we have that for each x in \(N^{d}\)
$$\begin{aligned} \sum _{i=1}^{n}\sum _{j=1}^{d!}P_{i}\left( \sigma _{j} x\right) m_{i}\left( \sigma _{j} x\right) =0 \end{aligned}$$
holds. Since the functions \(m_{1},m_2, \ldots , m_{n}\) are exponentials for all possible values of \(i=1,2, \ldots , n\) and \(j=1,2 \ldots , d!\), it follows that the mappings \(m_{i}\circ \sigma _{j}\) are exponentials on \(N^{d}\), as well. In view of Proposition 2, the set
$$\begin{aligned} \left\{ \sigma _jm_i{:}\,\,i=1,2, \ldots , n; j=1,2 \ldots , d!\right\} \end{aligned}$$
cannot consist of different exponentials. Similarly as in the proof of Lemma 2, we conclude that this can happen in two different ways: either
$$\begin{aligned} \sigma _j m_i= \sigma _l m_k \end{aligned}$$
holds for some i, k in \(\left\{ 1,2, \ldots , n\right\} \) with \(i\ne k\) and j, l in \(\left\{ 1,2, \ldots , d!\right\} \) with \(j\ne l\), or at least one of the coefficient polynomials is identically zero. In any of these two cases the above sum reduces to a sum with fewer terms but the same form, and, by induction, our statement follows.
The statement \(n\le d!\) can be proved as in the previous lemma. \(\square \)
Theorem 14
Let n be a positive integer, let \(m_{1}, m_2,\ldots , m_{n}\) be different exponentials on \(N^{d}\), further let \(P_{1},P_2, \ldots , P_{n}{:}\,N^{d}\rightarrow \mathbb {C}\) be nonzero generalized polynomials. If \(s{:}\,N^d\rightarrow \mathbb {C}\) is a symmetric spherical function and \(f=\sum _{k=1}^n P_k m_k\) is a symmetric monomial corresponding to s, then \(n\le d!\), and there exists an exponential m on \(N^{d}\) such that for each \(i=1,2, \ldots , n\) there is a \(\sigma _{i}\) in \(\mathscr {P}_{d}\) for which \(m_i=\sigma m\) and \(m^{\#}=s\).
Proof
For \(n=1\) the statement follows from the previous theorem. Let \(n\ge 2\). By Theorem 11, \(s=m^{\#}\) for some exponential m on \(N^d\), and by assumption, there is a natural number k such that
$$\begin{aligned} 0= & {} \mathrm {D}_{s; y_{1},y_2, \ldots , y_{k+1}}*f(x) = \mathrm {D}_{s; y_{1},y_2, \ldots , y_{k+1}}*f^{\#}(x) \\= & {} \sum _{i=1}^{n}\mathrm {D}_{s; y_{1}, y_2,\ldots , y_{n+1}}*\left( P_{i}\cdot m_{i}\right) ^{\#}(x)\\= & {} m(x+y_{1}+y_2+\cdots +y_{k+1})\\&\times \sum _{i=1}^{n}\mathrm {sym}_{x, y_{1},y_2, \ldots , y_{k+1}}\left[ \Delta _{y_{1}, y_2,\ldots , y_{k+1}}*\left( P_{i}\cdot m_{i}\cdot \check{m}\right) (x)\right] \end{aligned}$$
for each \(x, y_{1}, y_2,\ldots , y_{k+1}\) in \(N^{d}\). This means that
$$\begin{aligned} \sum _{i=1}^{n}\mathrm {sym}_{x, y_{1},y_2, \ldots , y_{k+1}}\left[ \Delta _{y_{1},y_2, \ldots , y_{k+1}}*\left( P_{i}\cdot m_{i}\cdot \check{m}\right) (x)\right] =0 \end{aligned}$$
holds whenever \(x, y_{1},y_2, \ldots , y_{k+1}\) is in \(N^{d}\). Since \(m_{1},m_2, \ldots , m_{n}\) and also \(\check{m}\) are exponentials, after computing the left hand side of this identity, we can conclude that Lemma 2 can be applied to deduce that for each \(i=1,2, \ldots , n\) there exists \(\sigma _{i}\in \mathscr {P}_{d}\) with \(m_i=\sigma m\). \(\square \)