Equivalent projections and 2-sided ideals in von Neumann algebras

Question

Let $M$ be a von Neumann algebra. For given $x$ in $M$, we put $I_x=MxM$ (the algebraic ideal generated by $x$).

We know that if two projections $p$ and $q$ are equivalent (that is $\exists u\in M$ with $p=u^*u,q=uu^*$) then $I_p=I_q$. What about the converse?

Martin Argerami · Accepted Answer · 2017-02-26T16:50:57.663

1

The converse does not hold. Take $M=B(H)$. For any two nonzero finite-rank projections $p$ and $q$ (in particular, with different trace, so non-equivalent), you have $$ MpM=K_0(H)=MqM, $$ the (algebraic) ideal of finite-rank operators.

edited Feb 26 '17 at 16:50

answered Jan 18 '16 at 20:25

Martin Argerami

205,756

Thanks, You are right. Just, it seems that $MpM=MqM$ is the space of all finite rank operators (not compact operators). – ABB Jan 19 '16 at 09:00
Yes, of course. I was thinking of the closure. – Martin Argerami Jan 19 '16 at 12:35

Equivalent projections and 2-sided ideals in von Neumann algebras

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