Can we find an invertible projection in an arbitrary von Neumann algebra?

Question

I am looking for an answer for this question:

Let $\mathcal{A}$ be an arbitrary von Neumann algebra, can we say there is an invertible projection ($P\neq I$) in $\mathcal{A}$?

I think, if there is an invertible pojection $P\in \mathcal{A}$ then $P$ is injective so $ker P=0$ therefore we can write $$H=kerP\oplus \overline{ran P}$$ But is $ran P$ closed!? in this case we can say $P=I$

Thanks.

Answer 1

This has nothing to do with topology nor von Neumann algebras. Any invertible idempotent in a ring is equal to the identity:

Let $P$ be an invertible idempotent. Then there exists $X$ with $XP=PX=I$. So $I=XP=XPP=IP=P$.