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Do we know examples of separably acting von Neumann algebras $\mathcal{M}$ such that $\mathcal{M}$ is unitarily equivalent to $M_{k}(\mathcal{M})$ for some $k\in\mathbb{N}\,?$

Obviously there are examples, but I mostly care for $II_1$ factors. For example, $\,II_{1}$ factors with trivial fundamental group do not satisfy my question. On the other hand, if a $II_1$ factor $\mathcal{M}$ has fundamental group $\mathcal{F}(\mathcal{M})=\mathbb{R}_{*}^{+}$ and its commutant is a $II_{\infty}$ factor, then $\mathcal{M}$ and $\mathcal{M}_{k}(\mathcal{M})$ are unitarily equivalent for every natural number $k.$

Any other cases?

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    $\begingroup$ To talk about unitary equivalence, you need to specify representations. Do you mean $M$ acting on some separable $H$ and then $M_k(M)$ is understood to act naturally on $\mathbb{C}^k \otimes H$? If so (and ignoring the trivial case $k=1$), in the II$_1$ factor case this happens iff $k$ is in the fundamental group of $M$ and $\dim(_MH) = \infty$, i.e., the commutant of $M$ is II$_\infty$. $\endgroup$
    – David Gao
    Commented Sep 27 at 16:22
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    $\begingroup$ Just for completeness - this is always true when $M$ is a type II$_\infty$ or type III factor. $\endgroup$
    – David Gao
    Commented Sep 27 at 16:29
  • $\begingroup$ @DavidGao, yes I meant that $\mathcal{M}$ acts on a separable Hilbert space $H$. Thank you for your answer. So, if $\mathcal{M}\subseteq \mathcal{B}(H)$ is a $II_1$ factor with fundamental group equals to $\mathbb{R}^+$ and its commutant $\mathcal{M}^\prime$ is also of type $II_1,$ then, for example, $\mathcal{M}$ is isomorphic to $M_2(\mathcal{M})$ but not unitarily equivalent. $\endgroup$
    – E. Papapetros
    Commented Sep 27 at 17:57
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    $\begingroup$ Yes, that is correct. $\endgroup$
    – David Gao
    Commented Sep 27 at 19:01

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