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Let G be a discrete group. I am interested in the following: let $\pi$ and $\rho$ be two representations of $G$. Denote by $C^*Ker\pi$ and $C^* Ker \rho$ the kernels of the corresponding representations of $C^*(G)$. Let $\ell^1(G) \cap C^* Ker \pi \subset \ell^1(G) \cap C^* Ker \rho$. Is it true that $KerC^* \pi \subset Ker C^* \rho$? (so the question is the following --- if I need to check that $\rho \prec \pi$, is it enough to check $KerC^* \pi \subset Ker C^* \rho$ only for $\ell^1(G)$?)

Thanks in advance!

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    $\begingroup$ No: the regular representation faithfully represents $\ell^1(G)$, but the regular representation does not weakly contain the universal representation unless $G$ is amenable. In general, one has to be very careful with thinking about the elements of $C^*(G)$ as linear combinations of group elements: this is an appropriate picture for $C^*_r(G)$, but not for $C^*(G)$, which only maps surjectively onto $C^*_r(G)$. $\endgroup$
    – Tobias Fritz
    Commented Dec 20, 2016 at 12:52
  • $\begingroup$ @TobiasFritz Why not make this an answer? $\endgroup$
    – Yemon Choi
    Commented Dec 20, 2016 at 19:26

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For a counterexample, take $\pi$ to be the regular representation and $\rho$ to be a universal representation. Then both $\pi$ and $\rho$ represent $\ell^1(G)$ faithfully; for the regular representation, you can see this by letting some $x\in\ell^1(G)$ act on the neutral element $e\in\ell^2(G)$. However, we have $\rho\prec\pi$ only if $G$ is amenable. Thus you cannot detect $\rho\not\prec\pi$ purely by looking at the kernels at the level of $\ell^1(G)$.

On a related note, it may help to realize that the elements of the full group C*-algebra $C^*(G)$ are not (infinite) linear combinations of group elements, unless $G$ is amenable. The elements of the reduced group C*-algebra $C^*_r(G)$ can indeed be considered as linear combinations of group elements. Moreover, every element of $C^*(G)$ acquires such a linear combination by considering its image in $C^*_r(G)$. But this linear combination determines the original element only modulo the kernel of the quotient map $C^*(G)\to C^*_r(G)$.

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    $\begingroup$ Regarding the second paragraph: Whenever $G$ has an element of infinite order you can not consider elements of $C^*_r(G)$ as infinite linear combinations of group elements. This amounts to the fact that there are continuous functions on the circle that are not uniform limits of their Fourier series $\endgroup$
    – Caleb Eckhardt
    Commented Dec 21, 2016 at 18:47
  • $\begingroup$ @CalebEckhardt: I don't mean to claim that finite linear combinations of group elements are dense. What I'm trying to say is just this: $\delta_e\in\ell^2(G)$ is a separating vector, and thus for every $x\in C^*_r(G)$, the vector $x\delta_e\in\ell^2(G)$ is an infinite linear combination of group elements that uniquely determines $x$. Am I missing something? $\endgroup$
    – Tobias Fritz
    Commented Dec 22, 2016 at 2:44
  • $\begingroup$ No, I don't think you're missing anything. I just misunderstood which convergence ($\ell^2(G)$ not $C*_r(G)$) you were referring to. $\endgroup$
    – Caleb Eckhardt
    Commented Dec 22, 2016 at 13:25

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