Is there any description of unital idempotent ($F^2(x)=F(x)$) morphisms of a von Neumann algebra into itself? Or, equivalently, of weakly closed subalgebras which are retracts as von Neumann algebras?

2$\begingroup$ Naive question perhaps  Why are the two questions equivalent? I don't see why the range of an unital idempotent is weakly closed. $\endgroup$ – mohanravi Jun 22 '12 at 17:36

$\begingroup$ I had a doubt about that when asking. Indeed, these might be wo different questions. Just in my case I know that the image is weakly closed. $\endgroup$ – Yulia Kuznetsova Jun 22 '12 at 20:22

$\begingroup$ So can I ask a naive question: is it correct that $F$ is just an algebra homomorphism (not assumed normal, or a $*$map, etc.?) $\endgroup$ – Matthew Daws Jun 22 '12 at 20:31

$\begingroup$ Matthew, mohanravi, I should have written "morphism", so normal and involutive. Then indeed the image is weakly closed since it is the kernel of $FId$. $\endgroup$ – Yulia Kuznetsova Jun 23 '12 at 0:57
Yes. The kernel of F is an ultraweakly closed *ideal of M generated by some central projection z. M splits as a direct sum of zM and (1z)M. As a 2x2 matrix F has only two nonzero entries, one that corresponds to an idempotent automorphism (hence the identity map) of (1z)M and another one to an arbitrary morphism from (1z)M to zM. Thus idempotent morphisms are classified by central projections and morphisms from (1z)M to zM.

$\begingroup$ Indeed, $G(X)=F(x)(1z)x$ is a morphism but one cannot get anything better... $\endgroup$ – Yulia Kuznetsova Jun 23 '12 at 0:48