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Let $A$ be a von Neumann algebra. Let $p$ be a projection in $A$. Suppose that $e$ is a finite projection. Can we determine all types of vn-algebras in which $p-p\wedge(1-e)$ is a finite projection?

Rem. It seems that when $A=B(H)$, the range of the projection $p-p\wedge(1-e)$ is just $\overline{peH}$ which is clearly finite dimentional subspace, since $e$ is a finite projection.

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  • $\begingroup$ Does finiteness of projection $e\in B(H)$ imply that $e(H)$ is finite dimensional? $\endgroup$
    – MSMalekan
    Commented Aug 15, 2018 at 8:34
  • $\begingroup$ I have also another one which probably you have an idea about. Let's consider projections $p,q$ with $p\leq q$. For a given a projection $e$, can we say that $q\wedge e-p\wedge e\leq q-p$? $\endgroup$
    – ABB
    Commented Aug 15, 2018 at 11:52

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By parallelogram rule in "Murray-von Numann equivalency" we have $p-p\wedge(1-e)\sim e-e\wedge(1-p)$. Hence, $p-p\wedge(1-e)$ is always finite, if $e$ is finite.

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