For $p, q$ projections in a factor $M$, there is a $x \in M$ such that $qxp \neq 0$

Question

Going through my lecture notes I found the following claim, that he didn't prove: If $(M, \tau)$ is a II$_1$-factor and $p, q \in M$ are projections but neither $0$ nor $1$ such that they have equal trace $\tau(q) = \tau(p)$. Then, using that $M$ is a factor, we can find an $x \in M$ such that $qxp \neq 0$.

I don't really see how this should work. How does one show that?

Thanks!

Answer 1

We don't need $p$ and $q$ of equal trace: only that they are nonzero.

Assume that $M\subset B(H)$. The key observation is that the orthogonal projection onto $K=\overline{MpH}$ is in the center of $M$: indeed, for any $x\in M$, $xMpH\subset MpH$, so $xp_K=p_Kxp_K$ for all $x\in M$. Using $x$ selfadjoint, we get that $xp_K=p_Kx$, and it follows that $p_K\in M'$. If $y\in M'$, then $yMpH=MpyH\subset MpH$; so $yp_K=p_Kyp_K$, and as before we get that $p_K\in M''=M$. So $p_K$ is a central projection in $M$.

Suppose that $qxp=0$ for all $x\in M$ (so $qMp=0$) and that $M$ is a factor. By the above, $p_K$ is central, so $p_K=I$. Thus $\overline{MpH }=H$. As $qMpH=0$, we get $qH=0$, a contradiction. It follows that there exists $x\in M$ with $qxp\ne0$.

Answer 2

A projection $p$ is said to be Murray--von Neumann equivalent to $q$, in symbols $p \sim q$, if there is a $u$ with $u^*u=p$ and $uu^* = q$. E.g. $|0\rangle\langle0| \sim |1\rangle\langle1|$ via $|1\rangle\langle0|$. We write $p \lesssim q$, when $p \sim q'$ for some $q' \leq q$. This is a partial order and the study of it is called the comparison theory of projections. For a good introduction see for instance chapter 6 of Kadison and Ringrose.

One of the early results proven is that in a factor each pair of non-zero projections have equivalent non-zero subprojections, see e.g. Corollary 6.1.9 of Kadison and Ringrose. (This is a stepping stone to show that projections in a factor are up-to-equivalence totally ordered.) So for your $p$ and $q$ there is a $u$ with $uu^* \leq p$ and $u^*u \leq q$. Note $uu^*p = uu^*$ and $qu^*u=u^*u$. With the C*-identity one can show $u^*uu^* = u^*$. Thus $qu^*p = qu^*uu^*p = u^*uu^* = u^* \neq 0$.

Note that we did not use that $p$ and $q$ have equal trace.

In fact the following are equivalent for projections $p$ and $q$ in any von Neumann algebra

1. $pxq \neq 0$ for some $x$
2. $p$ and $q$ have non-zero equivalent subprojections
3. $C_p C_q \neq 0$, where $C_p$ denotes the least central projection above $p$

Answer 3

Let $M \subset B(H)$ and suppose that $pxq = 0$ for all $x \in M$. Then, in particular, $$ u^*puq = 0, \quad \text{ for all } u \in M \text{ unitary}. $$ But since $M$ is a factor, we have $$q = I q = \left(\bigvee_u u^* p u \right) q = 0,$$ since the central support of $p$ is $I$. But $q$ was assumed to be nonzero, thus we have arrived at a contradiction.