Let $M$ be a type $II_1$ factor von Neumann algebra, and let $G$ be a discrete group acting on $M$ which is free and ergodic. Is the crossed product von Neumann algebra $M \rtimes G$ type $II_1$ factor?


One assumption that would guarantee that the crossed product is a factor is that the action be properly outer. Incidentally, I am not sure what you mean by ``ergodic'' in this setting.

The crossed product is automatically a finite von Neumann algebra (if $M$ is any von Neumann algebra and $G$ is a discrete group, then the projection from $L^2(M\rtimes G)$ onto $L^2(M)$ gives rise to a conditional expectation $E:M\rtimes G\to M$; if $M$ has a $G$-invariant trace it is immediate to verify that $\tau\circ E$ is a trace). Jon Bannon's comment does not apply in this case since you assume $M$ to be type II$_1$ factor so that the trace on $M$ is unique and thus preserved by any automorphism of $M$. There are no trace-scaling automorphisms of II$_1$ factors; they only exist for factors of type II$_\infty$. Of course if $M$ is not a factor it is possible for $G$ to act non-trivially on the center $Z(M)$ and this can lead to crossed products which are not finite (e.g. take $M$ abelian).

  • $\begingroup$ @JonBannon: If $M$ is type II$_1$ (not II$_\infty$) the trace is normalized by $\tau(1)=1$, and it is the unique linear map (no continuity assumptions needed!) satisfying $\tau(xy)=\tau(yx)$ for all $x,y\in M$ and such that $\tau(1)=1$ (in other words, $M/[M,M] = \mathbb{C}$). In particular, any automorphism of $M$ must leave its (unique normalized trace) invariant. Of course you can have an endomorphism, e.g. an isomorphism of $M$ with $pMp$ for some $p<1$ ... and you can have trace-scaling automorphisms of $M\otimes B(H)$, a II$_\infty$ factor. $\endgroup$
    – user75274
    Nov 1 '15 at 5:13
  • $\begingroup$ That is right! I'm losing it. Thanks! $\endgroup$
    – Jon Bannon
    Nov 1 '15 at 11:30

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