Let $M$ be a von Neumann algebra in $B(H)$. Let $p$ and $q$ be projections in $M$. Assume that they are equivalent in $B(H)$, i.e there is a partial isometry $u$ in $B(H)$ with $p=uu^*$ and $q=u^*u$.

Question: Are $p$ and $q$ equivalent in $M$?

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    $\begingroup$ No, let $H={\mathbb C}^2$ and $M\subset B(H)\cong M_2({\mathbb C})$ be the diagonal matrices. Then let $p$ and $q$ be the projections to the first and second coordinate. $\endgroup$ – user1688 Jan 14 '16 at 7:46
  • $\begingroup$ Yes, that shows that they are equivalent in $B(H)$. But as $M$ is abelian, they are not equivalent in $M$. $\endgroup$ – user1688 Jan 14 '16 at 10:08
  • $\begingroup$ I'm voting to close this question as off-topic because it should have been thought through more carefully $\endgroup$ – Yemon Choi Jan 16 '16 at 4:12

The answer to your question is negative.
Take two inequivalent projections in a type $II_1$ factor $R\subset B(H)$.
These projections have infinite dimensional range in $H$. They are therefore equivalent in $B(H)$.


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