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This question is a continuation of what I asked here. Tristan Bice showed the following nice result there:

Let $A$ be a von Neumann algebra and $P$ its projection lattice, ordered by $p\leq q\Leftrightarrow p=pq$. Then $$Q=\{q\in P:pa=qa\}$$ is a complete sublattice of $P$, for any $a\in A$ and $p\in P$.

It is not hard to see that this is also a convex sublattice, that is, if $p \le r \le q$ with $p,q \in Q$, then $r \in Q$.

Now, fix $a \in A$ and consider the following equivalence relation $p \sim q$ iff $p a = qa$. Is this a (lattice) congruence relation on $P$? See for example J.B. Nation's notes on lattice theory (Chapter 5) for the definition of a congruence relation. The reason why I suspect this is that it is known that the congruence classes of a congruence relation on a lattice are convex sublattices of the said lattice. (I don't know if the reverse is true.)

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No. If $p_1\sim p_2$ and similarly for $q$, then $p_1a=p_2a$ and

$$ p_1\vee q_1 \sim p_2\vee q_2 $$ iff $$ (p_1\vee q_1)a=( p_2\vee q_2)a $$ Let $a$ be projection on the $x$-axis, $p_1$ the diagonal, $p_2$ the $y$-axis. Let $q_2=p_2$ and $q_1$ yet another line.

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  • $\begingroup$ I might be misinterpreting your answer, but if by projection onto the diagonal, you mean the 45 degree line in the plane, then it seems that $p_1 a \neq p_2 a$ in this case? We have $p_1 = \frac12 11^T$, $a = e_1 e_1^T$ and $p_2 = e_2 e_2^T$. Then, $p_1 a = \frac12 1 e_1^T$ while $p_2 a = 0$. $\endgroup$
    – passerby51
    Commented Sep 25, 2020 at 1:34
  • $\begingroup$ @passerby51 ah, you're right... maybe some other permutation of these lines will do the trick? $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Sep 25, 2020 at 4:19

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