# Trigonometric polynomials on non-compact and non-abelian groups

I asked this initially in math.stackexchange, but it disappeared almost immediately, so I hope it will be proper to aks this here.

Hewitt and Ross define trigonometric polynomial on a locally compact group $G$ as a (finite) linear combination of matrix elements of continuous unitary irreducible representations $\pi_i:G\to B(H_i)$ (not necessarily finite dimensional) of $G$: $$f(t)=\sum_{i=1}^n \lambda_i\cdot\langle\pi_i(t)x_i,y_i\rangle,\qquad t\in G, \quad \lambda_i\in{\mathbb C},\quad x_i,y_i\in H_i.$$ If $G$ is abelian or compact then the space ${\tt Trig}(G)$ of trigonometric polynomials on $G$ is an algebra with respect to the pointwise multiplication.

This is strange, I can't find mentionings of the same proposition in the non-abelian and non-compact case. Is it possible that in general case (for arbitrary locally compact group $G$) the space ${\tt Trig}(G)$ is not an algebra?

I would be grateful, if somebody could advice reading on this theme.

• Although I cannot think of a counterexample right out of the box, but I think the problem is that, in general, there's no "good" representation theory and the tensor product of two representations (where products of coefficients would appear) does not need to decompose into irreducibles. – Alex Degtyarev Jan 25 '15 at 11:20
• Are there at least some natural classes of groups, where this is true? – Sergei Akbarov Jan 25 '15 at 11:22
• Abelian (all irreducibles are of dimension $1$) or compact (there are transfers and all classical representation theory works). Otherwise, I have no idea :) – Alex Degtyarev Jan 25 '15 at 11:28
• Хм... Большое спасибо! :) – Sergei Akbarov Jan 25 '15 at 11:30
• Is this true for Moore groups (where every irreducible unitary representation is finite dimensional)? – Sergei Akbarov Jan 25 '15 at 11:49

It isn't an algebra: the tensor product of irreducibles does not always decompose as a finite sum of irreducibles, although in some of those cases it may decompose as a direct integral of irreducibles.

For instance, take $G=H_3({\bf R})$, the group of upper-triangular matrices with real entries and with $1$ on the diagonal. This has very nice representation theory from the point of view of harmonic analysis (it is not only Type I, but it is what Kaplansky called CCR, that is $\pi(L^1(G))\subseteq {\mathcal K}(H_\pi)$ for every continuous irreducible unitary representation $\pi: G \to {\mathcal U}(H_\pi)$). There is a continuous quotient homomorphism $q:G \to G/Z(G) \cong {\bf R}^2$.

Now for $t\neq 0$ there is an infinite irreducible unitary representation $\pi_t: G \to {\mathcal U}(L^2({\bf R}))$ (in some realizations called the Schrödinger representation). It is true that if $s$, $t$ and $s+t$ are all non-zero, and $f$ is a coefficient function of $\pi_s$ and $g$ is a coefficient function of $\pi_t$, then $fg$ is the norm-limit inside $B(G)$ of sums of coefficient functions of $\pi_{s+t}$. I don't know if $fg$ is actually a coefficient function itself.

However, the representation $\pi_t\otimes \pi_{-t}$ is equivalent to the representation $\lambda_{\bf R{^2}} \circ q: G \to {\mathcal U}(L^2({\bf R}^2))$ which is not (quasi-)equivalent to a direct sum of irreducibles. Consequently there should exist $f$ a coefficient function of $\pi_t$ and $g$ a coefficient function of $\pi_{-t}$ such that $fg$ does not belong to the closure of ${\rm Trig}(G)$ inside $B(G)$.

Similar example should exist for e.g. ${\rm SL}(2,{\bf R})$ but I don't remember the fusion rules'' for its irreducible representations off the top of my head.

On the positive side: if $G$ is a Moore group then the completion of ${\rm Trig}(G)$ inside $B(G)$ is indeed a subalgebra of $B(G)$. This may be folklore: I learned of it from the paper http://arxiv.org/abs/1208.1519 which has more information on "negative examples".

• Yemon, for a long time I am dreaming of getting acquaintance with you. :) – Sergei Akbarov Jan 25 '15 at 16:17
• Thank you :) Most of these things I learned from discussions with Nico Spronk and Brian Forrest and their (former) students – Yemon Choi Jan 25 '15 at 16:26
• OK. They will be my heroes as well. :) – Sergei Akbarov Jan 25 '15 at 16:27
• Yemon, could you, please, make a little paper from this (for example, in arxiv), so that I could cite it? A separate request: if ${\tt Trig}(G)$ is not an algebra even for Moore groups, then this should be mentioned there, because this is an important proposition for me. – Sergei Akbarov Jan 26 '15 at 7:11
• @LSpice I don't know if Nico has done anything. I have been busy not getting anything written for the last 3 years (O joys of life in UK academia) – Yemon Choi Nov 18 at 16:44