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Let $G$ be a locally compact group and $I$ the set of matrix coefficient of irreducible unitary matrix coefficients of $G$. By Gelfand-Raikov's theorem and Stone-Weirestrass's theorem, for a compact $K$ of $G$, the algebra generated by $I$ is dense in $C(K)$ the space of continuous functions on $K$, so in particular the space of all matrix coefficient is dense.

I was wondering if we can get a stronger result : is the vector space $V$ generated by $I$ dense ?

For compact groups this is true by Peter-Weyl, since the vector space and the algebra coïncide.

The idea I had was a proof by contradiction. If we assume that $V$ is not dense, we can consider a nonzero continuous linear form $f$ on $C(K)$ which is $0$ on $V$. By density of all matrix coefficients, there is $\varphi$ a matrix coefficient such that $f(\varphi)\neq 0$.

Then, if we assume that $G$ is second countable, there is a standard Borel space $X$ and a $\sigma$-finite measure $\mu$ on $X$ such that for any $g\in G$, $$\varphi(g)=\int_X \varphi_x(g)d\mu(x)$$where $\varphi_x\in S$. Since $\varphi_x\in S$, we have $f(\varphi_x)=0$ for all $x\in X$.

But then is there a way to exchange the integral and $f$ ?

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  • $\begingroup$ Integrals and continuous linear forms can be switched because integration is weighted addition followed by a limiting process. (But I haven't checked the rest of your argument.) $\endgroup$
    – LSpice
    Commented Apr 26, 2023 at 19:40
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    $\begingroup$ Do you allow infinite dimensional unitary reps? If you only allow finite dimensional you run into groups with no nontrivial finite dimensional reps $\endgroup$
    – Benjamin Steinberg
    Commented Apr 26, 2023 at 20:54
  • $\begingroup$ @BenjaminSteinberg In this context (abstract harmonic analysis) one usually allows unitary reps to be infinite-dimensional unless specified. Indeed, for the reasons you mention, the Gelfand-Raikov theorem would be false if one only allowed finite-dimensional unitary irreps $\endgroup$
    – Yemon Choi
    Commented Apr 27, 2023 at 2:09
  • $\begingroup$ @BenjaminSteinberg Yes, I want infinite dimensional representation $\endgroup$
    – Pople
    Commented Apr 27, 2023 at 17:23
  • $\begingroup$ @LSpice It felt like since $f$ applies to functions, but I wrote the integral pointwise, there might be a problem (if I approximate the at each $g$, can I apply $f$ then ?). But on the other hand, $f$ should be the integral against a finite complex Radon measure on $K$ so with this point of view it should just amount to Fubini $\endgroup$
    – Pople
    Commented Apr 27, 2023 at 17:26

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