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Is the following true? What's a nice proof?

Let $M$ and $N$ be von Neumann algebras, and let $\phi:M\rightarrow N$ be a normal, surjective, *-homomorphism. Is there a normal *-homomorphism $\theta:N\rightarrow M$ with $\phi\circ\theta$ being the identity? If I cannot choose $\theta$ as a *-homomorphism, can I at least get a normal complete contraction?

This is, well, hinted at in the proof of Lemma 3.2 of http://pjm.math.berkeley.edu/pjm/2002/205-1/p09.xhtml but I don't follow the details (and they are proving it for weak* TROs: I think surely the von Neumann algebra case should be easier). Normal *-homomorphisms between von Neumann algebras have a very nice structure theorem, and maybe if I stared at that long enough I'd see an answer, but I thought I'd ask on MO...

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2 Answers 2

up vote 4 down vote accepted

Yes you can get a $\phi$ that is a homomorphism. Here is a quick sketch.

First let $p=sup$ {$p_\alpha,$ projections in $Ker \theta$}. So $p\in Ker \theta$. Furthermore $p\in Z(M)$, the center of M.

To see this note that if this were not true then we could find a unitary $u\in M$ with $p\neq upu^\star$. So then $p\wedge upu^\star$ would be a projection in $Ker \theta$ bigger than $p$.

From here you can get that $Ker \theta=pMp$, and so we can decompose $M=Ker \theta \oplus M_1$ and $\theta|_{M_1}$ is injective and thus an isomorphism, thus $\phi$ can be chosen to just be the inverse of $\theta$ on $M_1$.

Note that if we demand that $\phi$ be unital, this doesn't work and I don't think it can be done in general. I will have to give it more thought.

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Yep, that works! (Modulo swapping $\theta$ and $\phi$ ;-> ) Thanks. –  Matthew Daws May 6 '10 at 18:37
    
Haha, yeah i totally messed that up –  Owen Sizemore May 6 '10 at 18:47
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Morphisms of von Neumann algebra have very nice properties. More precisely, the kernel of a morphism f: M→N of von Neumann algebras is a σ-weakly closed two-sided ideal. Such ideals are in bijective correspondence with central projections of M. Moreover, you have a direct sum decomposition M=N⊕ker f. If you allow non-unital morphisms then the map n→(n,0) solves the problem. Otherwise choose an arbitrary morphism f: N→ker f and the map n→(n,f(n)) solves the problem. However, it might happen that there are no morphisms from N to ker f, and hence you have an obstruction. For example, there are no morphisms from a non-trivial factor to C.

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