Let $M\subset B(H)$ be a von Neumann algebra. Also $p$ and $q$ are projections in $M$ such that $pMp$ is star isomorphic to $qMq$. Does this imply that $p$ and $q$ are equivalent in the Murray-von Neumann sense?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Do you require $M$ to be a factor? In any case there are counterexamples. For instance take $M = R$ the hyperfinite factor. Any $p R p$ gives another hyperfinite factor. $\endgroup$– Adrián González PérezCommented Jul 19, 2018 at 14:44
-
$\begingroup$ The notion you are after is probably that of "fundamental group" $\endgroup$– Adrián González PérezCommented Jul 19, 2018 at 14:44
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
No. Take $M=\mathbb C^2\subset M_2(\mathbb C)$, $p=E_{11}$, $q=E_{22}$. Then $$ pMp=\mathbb C\oplus 0\simeq 0\oplus\mathbb C=qMq, $$ but $p$ and $q$ are not equivalent (equivalence in a commutative algebra is equality).
Even with factors, the assertion is not true. If $M$ is the hyperfinite II$_1$-factor and $p\in M$ is any projection (alternatively, $M$ is any II$_1$-factor and $p$ is in the fundamental group) then $M\simeq pMp$ while $p$ is not equivalent to the identity.
answered Jul 18, 2018 at 15:23