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Can a finite (by finite I mean when the projection $1$ is finite) von Neumann algebra be strongly morita equivalent to a properly infinite von Neumann algebra?

(Strong morita equivalence is the same as Morita equivalence as a ring for $C^*$-algebras according to the theorem on page 253 of http://www.sciencedirect.com/science/article/pii/0022404982901098, see remark 1.5 for definition of strong Morita equivalence)

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  • $\begingroup$ For those of us without access to that paper: are you talking about Rieffel's notion of Morita equivalence for vN algebras, or his notion of Morita equivalence for Cstar algebras? $\endgroup$
    – Yemon Choi
    Mar 13, 2015 at 1:35
  • $\begingroup$ I'm talking about his notion of strong Morita equivalence for $C^*$-algebras, given by the existence of an imprimitivity bimodule. $\endgroup$
    – Louis A
    Mar 13, 2015 at 5:14
  • $\begingroup$ Here's a definition of imprimitivity bimodule that maybe you can access tinyurl.com/nolv2f2, it's on page 287, the third page. $\endgroup$
    – Louis A
    Mar 13, 2015 at 5:16
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    $\begingroup$ @LouisA: Please do not use URL shorteners like tinyurl. I edited your post to replace the shortened URL with a proper link. $\endgroup$ Mar 15, 2015 at 14:26

1 Answer 1

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Assume $A$ and $B$ are two unital $C^*$-algebras which are strongly Morita equivalent.

This means that there exists a Hilbert $A$-module $H$ such that $B$ is the algebra of compact operators on $H$. But as $B$ is unital it implies that the unit of $B$ has to act as the identity of $H$ (because $B$ contains all "rank one operators") hence the identity of $H$ is a compact operator.

One now has the following lemma:

Lemma : For any $C^*$-algebra $A$, If the identity of $H$, a Hilbert $A$-module, is compact then $H$ is a direct (orthogonal) summand of $A^n$ for some integer $n$

This implies that $B$ is a corner of $M_n(A)$. Hence if $A$ is a finite von Neuman algebra, $B$ also is: $M_n(A)$ is finite* and any projection of a finite von Neuman algebra is finite hence $B \simeq p.M_n(A).p$ is also finite.

Proof of the lemma: The identity of $H$ is compact, hence it can be approximated by a "finite rank operator" and if this approximation is close enough it will be invertible. Hence there exists an operator $k:A^n \rightarrow H$ such that $kk^*$ is invertible. Let $v = (kk^*)^{-1/2}$, then $p=vk$ is an operator $A^n \rightarrow H$ such that $pp^* = Id$ and hence $p^* p $ is a symetric projection and $H \simeq_p (p^*p).A^n$ is a direct summand.

*: Note that the fact that $M_n(A)$ is finite is well known but non trivial. It fails for general $C^*$-algebras, and corresponds to the fact (proved in Murray and Von neuman original paper) that in von Neuman algebras the sum of two orthogonal finite projection is again a finite projection.

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  • $\begingroup$ Thank you so much for taking the time to answer, one thing I was wondering: for $C^*$-algebras, does "algebraic" Morita equivalence imply strong Morita equivalence? That is, if two $C^*$-algebras $A$ and $B$ are Morita equivalent as rings are they also strongly Morita equivalent? $\endgroup$
    – Louis A
    Aug 28, 2015 at 21:50
  • $\begingroup$ Yes they are (at least for unital algebra) but unfortunately I can't remember why exactly. $\endgroup$ Aug 31, 2015 at 7:42
  • $\begingroup$ By the way, when you say "Note that the fact that $M_n(A)$ is finite is well known but non trivial. It fails for general $C^*$-algebras..." you mean that there are unital $C^*$-algebras $X$ which are finite for which $M_n(X)$ is not a finite $C^*$-algebra? $\endgroup$
    – Louis A
    Sep 1, 2015 at 0:21
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    $\begingroup$ Indeed, see for example math.ku.dk/~rordam/manus/shanghai.pdf $\endgroup$ Sep 1, 2015 at 14:54
  • $\begingroup$ Whoops, I hadn't seen this, thanks, let me give that a read. $\endgroup$
    – Louis A
    Sep 8, 2015 at 4:57

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