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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.

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  • $\begingroup$ In general, $QP, PQ \neq 0$ does not imply $P \wedge Q \neq 0$. Take $\frac{1}{\sqrt{2}}\left( \begin{matrix} 1 \\ 1 \end{matrix} \right) , \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \in \mathbb{C}^2$, and $P$ and $Q$ the projection onto their linear span. $\endgroup$
    – Hetebrij
    May 13, 2016 at 7:41
  • $\begingroup$ @Hetebrij Ok, THX. Then, we back to my first question. What if the measure of $P$ is bigger? $\endgroup$
    – hjinghao
    May 13, 2016 at 7:50

1 Answer 1

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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 $.

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  • $\begingroup$ I see a couple problems here. Let $$ P=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \ \ \ Q=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}. $$ Then $P\wedge (I-Q)=0$, so $R=P$. But $$ P\vee (I-Q)-(I-Q)=\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}1/2&-1/2\\-1/2&1/2\end{bmatrix}=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}=Q $$ (if you look at Theorem 6.1.7 in Kadison-Ringrose, you'll see that one gets equivalence, but not equality; your argument still goes through, though, as equivalence implies equal trace and that's what you need). $\endgroup$ May 16, 2016 at 19:03
  • $\begingroup$ Now, while the above works, I cannot see how from $\tau(R)\leq\tau(Q)$ (they are actually equal, as I mentioned), you conclude that $(Q-R)(H)$ is nonzero (it is always nonempty, as it contains zero). Finally, $$ Q-R=Q-P=\begin{bmatrix}-1/2&1/2\\1/2&1/2\end{bmatrix} $$ is not a projection. And, in the example, $Q^\perp\wedge P=0$. $\endgroup$ May 16, 2016 at 19:04
  • $\begingroup$ @MartinArgerami THX for pointing out 'equivalent'. And the traces of your P and Q are the same. $\tau(P)$ is strictly bigger than $\tau(R)$. $\endgroup$
    – hjinghao
    May 22, 2016 at 10:04

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