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Let $\Gamma$ be a discrete group and let $\Lambda$ be its subgroup. The function $\mathbf{1}_{\Lambda}$ is positive definite and therefore gives rise to a unitary representation $\pi: \Gamma \to \mathcal{U}(\ell^2(\Gamma / \Lambda))$, which is the quasiregular representation. The function $\mathbf{1}_{\Lambda}$ can be viewed as a state on the von Neumann algebra generated by $\pi(\Gamma)$. My question is: is this state faithful?

The answer is obviously yes when $\Lambda$ is a normal subgroup, since in this case the state is tracial, but I have no idea what is the general behaviour.

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    $\begingroup$ Assume $\Lambda<\Gamma$ is not normal. Find $\lambda \in \Lambda$ and $\gamma\in \Gamma$ such that $\lambda^\gamma\notin\Lambda$. Then $\pi(\lambda)\neq 1$. Consider the positive element $a=(1-\lambda)^*(1-\lambda)$. Denote $v=1_\Lambda$. Note that $(1-\lambda)v=0$ thus $\langle av,v \rangle =0$. $\endgroup$
    – Uri Bader
    Commented Mar 20, 2017 at 10:06
  • $\begingroup$ @UriBader: thanks a lot. This completely settles the problem. $\endgroup$
    – Mateusz Wasilewski
    Commented Mar 20, 2017 at 11:47
  • $\begingroup$ I actually realised now that quasiregular representation can even be irreducible. An example is $\mathbb{F}_2 = \langle a, b\rangle$ and $\mathbb{Z} = \langle a \rangle$. $\endgroup$
    – Mateusz Wasilewski
    Commented Mar 20, 2017 at 18:19
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    $\begingroup$ Yes, $\ell^2(\Gamma/\Lambda)$ will be irreducible iff for every $\gamma\in \Gamma-\Lambda$, $\Lambda\cap\Lambda^\gamma$ is not of finite index in both $\Lambda$ and $\Lambda^\gamma$. This is usually called "Mackey criterion for irreducibility". $\endgroup$
    – Uri Bader
    Commented Mar 20, 2017 at 19:11

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