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This is a follow-up question on this. Let $A$ be a von Neumann algebra and $P$ be its projection lattice.

For $p,s,q \in P$, let us define $ p \perp q \mid s \iff ps^\perp q = 0$ where $s^\perp = 1-s$.

Let $P_0 \subset P$ be a complete sublattice of $P$. Is the following true over $P_0$ (i.e., for any $p,q,r,s \in P_0$): \begin{align*} r \perp p \mid (q \vee s), \quad r \perp q \mid (p \vee s) \implies r \perp (p\vee q) \mid s\;? \end{align*}

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    $\begingroup$ What conditions are you having in mind? You must formulate your question more clearly, because, being a mathematician, I would say the conditions are precisely those that you gave. $\endgroup$
    – user1688
    Commented Feb 3, 2016 at 7:32
  • $\begingroup$ @Anton, not sure what you mean. Let me ask this: Is the statement true for every complete sublattice $P_0$, or one needs to assume something more about $P_0$? $\endgroup$
    – passerby51
    Commented Feb 3, 2016 at 14:35
  • $\begingroup$ @Anton, OK, I understand what you mean. I just reworded it as a yes/no question. But my guess is the answer is no. I was hoping that there is a simpler condition that one can verify for $P_0$, from which mine follows, but perhaps not. In other words, I was hoping that someone recognize a connection with some other more well-understood condition. Otherwise, I will delete the question. $\endgroup$
    – passerby51
    Commented Feb 3, 2016 at 14:44

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I think it is false - let $r=1$ and take any distinct rank $1$ projections $p,q,s\in M_2$ so $p\vee q=p\vee s=q\vee s=1$.

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    $\begingroup$ Or just $p = q = r = 1$ and $s = 0$. $\endgroup$
    – Nik Weaver
    Commented Feb 3, 2016 at 16:36
  • $\begingroup$ @Tristan, how can $r=1$ and a rank 1 projection at the same time? $\endgroup$
    – passerby51
    Commented Feb 3, 2016 at 16:38
  • $\begingroup$ Yep, that'll do it! $\endgroup$
    – Tristan Bice
    Commented Feb 3, 2016 at 16:39
  • $\begingroup$ Oops, I meant $s$ for the second $r$. In any case, Nik's answer is even simpler. $\endgroup$
    – Tristan Bice
    Commented Feb 3, 2016 at 16:42
  • $\begingroup$ @Nik and Tristan, OK... fair enough. Maybe I should have mentioned distinct elements $p,q,r,s$, but Tristan's example seem to cover that. Though not sure if I understand that example correctly. Isn't $M_2$ 4-dimensional? Then, why $p \vee q = 1$ for example? $\endgroup$
    – passerby51
    Commented Feb 3, 2016 at 16:48

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