Can there exist a faithful action of a $C^{*}$-simple group $G$ on a von Neumann algebra $(M,\varphi)$ equipped with faithful normal state $\varphi$ such that action preserves the state $\varphi$ and the identity automorphism can be approximated by non-identity automorphisms i.e, there exists $g_{n}$ $\in$ $G$, such that \begin{align*} \|\alpha_{g_{n}}(x)-x\|_{2}\rightarrow 0, \end{align*} as $n\rightarrow \infty$ $\forall$ $x$ $\in$ $M$ and $\alpha_{g_{n}}\neq I$, $\|\cdot\|_{2}$ is the GNS Hilbert space $L^{2}(M,\varphi)$ norm?
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1$\begingroup$ Some questions which might improve the question: what is a $C^*$-simple group? What is $\|\cdot\|_2$ (asked because you have placed no restriction on your von Neumann algebra $M$). $\endgroup$– Matthew DawsSep 12, 2019 at 9:24
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$\begingroup$ A group is said to be $C^*$-simple if its reduced $C^*$-algebra is simple (every proper quotient is reduced to $\{0\}$). This also means that whenever a nonzero unitary representation is weakly contained in the regular representation, then it weakly contains the regular representation. $\endgroup$– YCorSep 12, 2019 at 9:50
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2$\begingroup$ It's obviously possible for $G=F_\infty$. Also, e.g., if $G$ embeds densely into a compact group $K$, then $G\curvearrowright L^\infty(K)$ is an example. $\endgroup$– Narutaka OZAWASep 13, 2019 at 0:48
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$\begingroup$ @Ozawa Sir, can you please elaborate the reason why this example will work? $\endgroup$– user136400Sep 13, 2019 at 6:54
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