projections in von Neuman algebra

Question

Consider a semifinite von Neumann algebra $\mathcal{M}$ with a semifinite faithful normal trace $\tau$. If $Q, P$ are projections in $\mathcal{M}$ with $\tau(Q)< \tau(P)$, then does $\tau(P\wedge Q^\perp) >0$?

The condition $\tau(Q) <\tau(P)$ is very important. I can prove that for an $n$-dimensional space, it is true, just by some classic differential geometry.

Answer 1

Let $R$ be the projection onto the range of $PQ$. Then, by page 116 of Kadison Ringrose' book. $$R:=P-P\wedge (I-Q). $$ Then, $R \sim P\vee (I-Q) - (I-Q)$. Then,　 $\tau(R) =\tau(P\vee (I-Q) - (I-Q)) =\tau(Q) - \tau(I- P\vee (I-Q) ) \le \tau(Q)<\tau(P)$. Then, $(P-R)(H)$ is not empty and it is easy to check $P-R$ is $Q^\perp \wedge P $.