If a compact group $G$ acts on vN algebra factor $M\subset B(L^{2}(M))$, what would be the index of subfactor $[M^{G}:M]$? KIndly explain the answer.
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$\begingroup$ Dear Desperate mathematician, it would be useful if you could explain what are the things that you've already studied and understood, that go in the direction of the question asked. For example, do you understand the case when $G$ is $\mathbb Z/2 \mathbb Z$? If yes, do you understand the case when $G$ is finite... etc. $\endgroup$– André HenriquesCommented Jan 30, 2020 at 12:26
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$\begingroup$ No, I do not understand in any case. The definition of index to me is the dimension of $L^{2}(M)$ over $M^{G}$ module. $\endgroup$– sibaniCommented Jan 30, 2020 at 12:40
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1$\begingroup$ Are you asking if it could be infinite? This is indeed the case: consider the very first example (which led to Vaughan Jones' results), the product type action of SU(2) on the $2^{\infty}$ UHF C*-algebra, extended to the action on the hyperfinite factor. If the group is connected and the action is nontrivial, presumably the index is always infinite. $\endgroup$– David HandelmanCommented Jan 30, 2020 at 14:18
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Case $G=\mathbb Z/2$, generated by $g\in G$:
In that case, the action of $G$ on $M$ induces a grading $M=M_0 \oplus M_1$ where $M_0=M^G=\{m\in M: gm=m\}$ and $M_1=\{m\in M: gm=-m\}$. Similarly, the action of $G$ on $L^2M$ induces a grading $L^2M=(L^2M)_0\oplus (L^2M)_1$. The subspace $(L^2M)_0$ is isomorphic to $L^2(M^G)$.
And both $(L^2M)_0$ and $(L^2M)_1$ are isomorphic as $M^G$-modules. So $$ \dim_{M^G}(L^2M) = \dim_{M^G}(L^2M)_0 + \dim_{M^G}(L^2M)_1 = 2\cdot \dim_{M^G}(L^2(M^G))=2. $$
answered Jan 30, 2020 at 12:46