Let $M$ be a von Neumann algebra with separable predual. Let us assume that $M$ is of type II$\_1$, meaning that it is finite but has no type I part. Let $\tau$ be a faithful normal tracial state on $M$.


First, I would like to choose some free ultrafilter $\omega$ on $\mathbb N$ and consider the tracial ultrapower
$$
M^\omega = \ell^\infty(\mathbb N, M)/\{ (x\_n)\_n \mid \tau(x\_n^\*x\_n)\to 0 \}.
$$


Next, let $\alpha$ be an automorphism on $M$. It is called properly outer, if for every non-zero projection $p\in M^\alpha$, the induced automorphism on $pMp$ is not inner.


Note that each automorphism gives rise to $\alpha^\omega\in\operatorname{Aut}(M^\omega)$ via pointwise application, and restricts to an automorphism on the relative commutant $M^\omega\cap M'$.


**Theorem:** ([Connes](https://alainconnes.org/docs/automorphismes.pdf)) If $M$ is the hyperfinite II$\_1$-factor, then $\alpha$ is (properly) outer if and only if $\alpha^\omega$ is (properly) outer on $M^\omega\cap M'$.



> 
> **My question:** Is the above also true if $M$ is a hyperfinite type II$\_1$ von Neumann algebra, i.e., not assumed to be a factor? So is an automorphism $\alpha$ properly outer if and only if $\alpha^\omega$ is properly outer on $M^\omega\cap M'$?
> 
> 
> 


I am sure that somebody must have worked this out somewhere, but I cannot seem to find it. I would be grateful either for the argument showing/disproving this statement or a reference with the answer. 


EDIT: I realized that it is not necessary for $\alpha$ to assume that it preserves the trace $\tau$. The question makes perfect without it.



Properly outer automorphisms on type II$_1$ von Neumann algebras
The answer to your question is yes, although I haven't found a precise reference for this fact.


Let $M$ be a hyperfinite, type II$\_1$ von Neumann algebra with separable predual. Later on I will use that it is possible to decompose $M$ as $Z(M) \overline{\otimes} \mathcal{R}$, where $Z(M)$ is the centre of $M$ and $\mathcal{R}$ is the hyperfinite II$\_1$ factor (this is Theorem 1.5 in Chapter XVI of Takesaki's third book on operator algebras)


**Definition 1.** An automorphism $\alpha \in \text{Aut}(M)$ is *centrally (non-)trivial* if $\alpha^\omega$ is (non-)trivial on $M^\omega \cap M'$.
It is *properly centrally non-trivial* if none of its restriction to $pM$, for some $\alpha$-invariant, non-zero, central projection $p \in M$, is centrally trivial.


The proof is composed of two parts:


1. Every properly outer automorphism on $M$ is properly centrally non-trivial.
2. If $\alpha \in \text{Aut}(M)$ is properly centrally non-trivial, then $\alpha^\omega$ is properly outer on $M^\omega \cap M'$.


Part 1 is addressed in the following claim.


**Claim 1.** If $\alpha$ is an outer automorphism of $M$, then $\alpha$ is centrally non-trivial.


*Proof.* If the restriction of $\alpha$ to $Z(M)$ is non-trivial, then the conclusion is obvious, as $Z(M) \subset M^\omega \cap M'$. Suppose that $\alpha$ is trivial on $Z(M) \cong L^\infty(X)$, for some measure space $X$. Then $\alpha$ can be decomposed as
\begin{equation}
\alpha = \int\_X^\oplus \alpha^x,
\end{equation}
where $\alpha^x \in \text{Aut}(\mathcal{R})$ for every $x \in X$ (a good reference for disintegration of von Neumann algebras and of their automorphisms is Chapter IV of Takesaki's first book on operator algerbas; see in particular Theorem 8.23 for the passage above). The claim follows by contradiction using the following three results:


* $\alpha$ is centrally trivial if and only if $\alpha^x$ is centrally trivial for almost every $x \in X$ (I'm not sure who originally proved this, but it is proved as Theorem 9.14 [here](http://www.ams.org/books/memo/1153/)).
* $\beta \in \text{Aut}(\mathcal{R})$ is centrally trivial if and only if it is inner ([Connes](https://alainconnes.org/docs/automorphismes.pdf)).
* $\alpha$ is inner if and only if $\alpha^x$ is inner for almost every $x \in X$ (this is due to [Lance](https://link.springer.com/article/10.1007/BF02547968), Theorem 3.4).


The second part of the proof, namely that if $\alpha$ is properly centrally non-trivial then $\alpha^\omega$ is properly outer on $M^\omega \cap M'$ was proved by [Ocneanu](https://link.springer.com/book/10.1007/BFb0098579) (see Lemma 5.6). One has to backtrack a bit in the paper to make sense of the proof, because of all the notation and definition appearing in the lemma (the setting there is much more general than what we need), but the argument is similar to the one given by Connes to prove the analogous result for $\mathcal{R}$ (Proposition 2.1.2 in [Outer conjugacy classes of automorphisms of factors](https://alainconnes.org/docs/automorphismes.pdf)).


