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How general is the following statement about Murray-von Neumann equivalence of projections in a von Neumann algebra?

Let $M$ be a von Neumann algebra and let $p,q\in M$ be projections. If there exists a sequence $\{x_n\}_{n\in\mathbb N}\subset M$ such that $$p=\operatorname*{{\scriptsize SOT}\ lim}_{n\to\infty}x_n^*x_n\qquad\text{and}\qquad q=\operatorname*{{\scriptsize SOT}\ lim}_{n\to\infty}x_nx_n^*$$ then $p$ and $q$ are Murray-von Neumann equivalent projections.

This seems to be true at least when $M$ is a factor and $\tau$ is a trace on $M$, for in this case $p\sim q$ iff $\tau(p)=\tau(q)$.

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It fails badly on $B(H)$: let $p=I$, $q=0$, and $x_n=\sum_{k=1}^\infty E_{n+k,k}$, where $\{E_{k,j}\}$ is a set of matrix units. Then $$ x_n^*x_n=\sum_{k,j=1}^\infty E_{k,n+k}E_{n+j,j}=\sum_{k=1}^\infty E_{kk}=I, $$ $$ x_nx_n^*=\sum_{k,j=1}^\infty E_{n+k,k}E_{j,n+j}=\sum_{k=1}^\infty E_{n+k,n+k}=\sum_{k=n+1}^\infty E_{kk}\xrightarrow[n\to\infty]{sot}0. $$ This makes it fail in any infinite von Neumann algebra.

As you mention, it works on finite factors. I'm not sure about finite von Neumann algebras, because there equal trace is not enough for equivalence.

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  • $\begingroup$ Many thanks for your example in answer to my original question. I was just wondering what would happen if the von Neumann algebra is the bidual of some separable C*-algebra $A$ and the sequence $x_n$ is now taken in $A$. What I'm actually interested into is the situation where the projection $p$ is open, i.e. it is the support projection of some positive element from $A$. $\endgroup$
    – Phoenix87
    Commented Jun 9, 2015 at 15:07
  • $\begingroup$ Same thing. If you replace all the $\infty $ above with $n $, you can take $x_n\in A=K (H) $; the example works the same, and the bidual is $B (H ) $. $\endgroup$
    – Martin Argerami
    Commented Jun 9, 2015 at 15:22
  • $\begingroup$ Right. In this case every projection is also open, as $B(H) = M(K)$. Thanks again. $\endgroup$
    – Phoenix87
    Commented Jun 9, 2015 at 16:33
  • $\begingroup$ My pleasure. ${ }$ $\endgroup$
    – Martin Argerami
    Commented Jun 9, 2015 at 16:37

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