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)



Can a finite von Neumann algebra be strongly morita equivalent to a properly infinite von neumann algebra?
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.


