If $M$ is in standard form, consider the action of a finite group on $M$, does the fixed point subalgebra under the action is in standard form? What we can say if $M$ is hyperfinite $\mathrm{II}_{1}$ factor?
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3$\begingroup$ No; if the action of a finite group $G$ is outer then you get a subfactor of index equal to the order of the group; denote the inclusion by $N \subset M$. It means that the dimension of $L^{2}(M)$ as an $N$-module is equal to $|G|$ and the dimension of the standard form is equal to $1$. $\endgroup$– Mateusz WasilewskiCommented Apr 9, 2019 at 14:26
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