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?
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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. $\endgroup$– mohanraviCommented Jun 22, 2012 at 17:36
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$\begingroup$ 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 KuznetsovaCommented Jun 22, 2012 at 20:22
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$\begingroup$ 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 DawsCommented Jun 22, 2012 at 20:31
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$\begingroup$ Matthew, mohanravi, I should have written "morphism", so normal and involutive. Then indeed the image is weakly closed since it is the kernel of $F-Id$. $\endgroup$– Yulia KuznetsovaCommented Jun 23, 2012 at 0:57
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1 Answer
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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 (1-z)M. As a 2x2 matrix F has only two nonzero entries, one that corresponds to an idempotent automorphism (hence the identity map) of (1-z)M and another one to an arbitrary morphism from (1-z)M to zM. Thus idempotent morphisms are classified by central projections and morphisms from (1-z)M to zM.
answered Jun 22, 2012 at 18:35
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$\begingroup$ Indeed, $G(X)=F(x)-(1-z)x$ is a morphism but one cannot get anything better... $\endgroup$ Commented Jun 23, 2012 at 0:48