Is a matrix element of a norm continuous representation always a trigonometric polynomial?

Question

I asked a similar question for the case of compact groups not long ago in math.stackexchange. Now I understand that the answer was "yes", and I want to modify that question. This is also related to my recent question here.

Let us say that a unitary representation $\pi:G\to {\mathcal B}(H)$ of a locally compact group $G$ is norm-continuous, if it is continuous with respect to the usual operator norm in ${\mathcal B}(H)$: $$ t_i\to t\quad\Longrightarrow\quad ||\pi(t_i)-\pi(t)||\to 0. $$

Let $\pi:G\to {\mathcal B}(H)$ be a norm-continuous unitary representation of a locally compact group $G$ in a Hilbert space $H$. Consider a matrix element of $\pi$, i.e. a function of the form $$ f(t)=\langle\pi(t)x,y\rangle,\quad t\in G, $$ where $x,y\in H$.

From the paper by A.I.Shtern it follows that if $G$ is compact, then $f$ is automatically a trigonometric polynomial, i.e. a (finite) linear combination of matrix elements of some unitary irreducible representations $\pi_i:G\to {\mathcal B}(H_i)$: $$ f(t)=\sum_{i=1}^n\lambda_i\cdot\langle\pi_i(t)x_i,y_i\rangle,\quad t\in G, $$ ($x_i,y_i\in H_i$, and $\lambda_i\in{\mathbb C}$).

A question:

Is the same true for wider classes of groups, in particular for Moore groups (i.e. where every unitary irreducible representation is finite-dimensional)?

Answer 1

In general, the answer is "no".

Let $G=\Bbb Z$; then all trigonometric polynomials are linear combinations of $f_\xi\colon t\mapsto\xi^t$, $\xi\in\Bbb C$, $|\xi|=1$. On the other hand, consider the space $L^2(\Bbb Z)$, i.e., the space of sequences $\{a_i\,|\,i\in\Bbb Z\}$ such that $\sum_i|a_i|^2<\infty$. This space has a canonical orthonormal basis $e_i$, $i\in\Bbb Z$, and $\Bbb Z$ acts on it by the shifts $t\colon e_i\mapsto e_{i+t}$. Let, further, $x=y=e_0$; the resulting "matrix entry" $f(t)$ is $0\mapsto1$, $t\mapsto0$ for $t\ne0$. This function is not a trigonometric polynomial, as for any finite collection $f_{\xi_i}$, $i=1,\ldots,N$, the only linear combination taking value $0$ at $t=1,\ldots,N$ is the trivial one (Vandermonde determinant $\ne0$); hence, it cannot take value $1$ at $t=0$.