1
$\begingroup$

I want to prove the following statement (which came up in a proof of the operator algebraic Cantor-Bernstein):

Let $H$ be a Hilbert space and consider a von Neumann algebra $\mathcal M \subseteq L(H)$.

Suppose that $p,q$ are Murray-von Neumann equivalent projections, $p \underset{u}{\sim} q$, i.e. there exists a partial isometry $u$ such that $$u^* u = p \quad\text{and}\quad u u^* = q.$$

Furthermore, let $p_\infty \leq p$ and $q_\infty \leq q$ be subprojections. Then, we have $$u\; p_\infty \; u^* = q_\infty \tag{$\star$}$$ if and only if $p_\infty$ and $q_\infty$ are Murray-von Neumann equivalent.

The first implication of the above result is easy, since Murray-von Neumann equivalence already gives us a partial isometry $v$ for which the desired equality $(\star)$ holds, i.e. $$ v \; p_\infty \; v^* = v \; v^* \; v\; v^* = (v \; v^*)^2 = v v^* = q_\infty, $$ since $v v^*$ is an orthogonal projection. But what about the converse? Given a partial isometry $v$ with $v^* v = p$, $v v^* = q$ and $(\star)$, I think we have to restrict $v$ to $\mathrm{range} (p_\infty)$ but I am not sure how to proceed from here.

EDIT

As pointed out in the comments, we cannot prove the following stronger result:

Two projections $x,y \in \mathcal M$ are Murray-von Neumann equivalent iff there exists a partial isometry $v \in \mathcal M$ such that $$v\; x\; v^* = y \tag{$\ast$}.$$

$\endgroup$
6
  • 1
    $\begingroup$ This is not true. Consider $\mathcal M=L(H)$ (with $H$ infinite-dimensional), $x=I$, $y$ is any projection with finite-dimensional range, and $v$ a partial isometry with $vv^*=y$. $\endgroup$
    – Aweygan
    Feb 24, 2017 at 15:46
  • $\begingroup$ @Aweygan: Thank you. So the optimistic statement $(\ast)$ is wrong, but it ignored a central part of the original statement which I need to prove, i.e. if $x,y$ are subprojections of two projections $p,q$ which are Murray-von Neumann equivalent via $v$, does $(\ast)$ imply that $x,y$ are Murray-von Neumann equivalent? $\endgroup$
    – el_tenedor
    Feb 24, 2017 at 15:53
  • $\begingroup$ That is a good question. I'll think about it. $\endgroup$
    – Aweygan
    Feb 24, 2017 at 16:00
  • $\begingroup$ @Aweygan: I have edited the question to reflect your first comment. $\endgroup$
    – el_tenedor
    Feb 24, 2017 at 16:01
  • 1
    $\begingroup$ I don't understand why it is sufficient in the first direction to show that $vp_\infty v^*=q_\infty$ as opposed to $up_\infty u^*=q_\infty$ Could someone elaborate this implication, please ? $\endgroup$
    – user412810
    Mar 28, 2018 at 23:49

1 Answer 1

Reset to default
2
$\begingroup$

You are assuming that $x,y$ are projections, that $x\leq p$, $y\leq q$, $p=v^*v$, $q=vv^*$, and that $vxv^*=y$.

Note that $$ v^*yv=v^*vxv^*v=pxp=x. $$

The partial isometry you need is then $w=yvx$. You have $$ w^*w=xv^*y^*yvx=xv^*yvx=xxx=x. $$ And $$ ww^*=yvxx^*v^*y=yvxv^*y=yyy=y. $$

$\endgroup$
2
  • $\begingroup$ Does this proof even need that $p$ and $q$ are equivalent? Wouldn't it work exactly the same if $p,q$ were sub equivalent to each other? $\endgroup$
    – user412810
    Mar 31, 2018 at 11:02
  • $\begingroup$ The proof does not mention $q$, but one has $y=vxv ^*\leq vv ^*$. And it is required that $pxp=x$, so $x\leq v^*v$. $\endgroup$
    – Martin Argerami
    Mar 31, 2018 at 13:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .