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Suppose M is a von Neumann algebra. Denote by L its maximal noncommutative localization, i.e., the Ore localization with respect to the set of all left and right regular elements, i.e., elements whose left and right support equals 1.

Denote by A the set of all closed unbounded operators with dense domain affiliated with the standard representation of M on a Hilbert space, i.e., L^2(M), also known as the standard form of M.

Von Neumann proved that if M is finite, then L and A are canonically isomorphic.

What can we say about the relationship of L and A when M has type III?

I am also interested in the properly infinite semifinite case.

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up vote 3 down vote accepted

I think the question is not well-posed or has a negative answer.

One first has to deal with the question whether the left-right-regular elements satisfy the Ore condition, or equivalently, we have to ask: Can we find common denominators? If one is not in the finite case, this is not possible.

For $B(H)$, let us take injective bounded operators $T$ and $S$ such that the images are dense but intersect only in the zero vector. In order to find a common denominator, we need to find an operator $R$ (bounded, injective, dense image) and bounded operators $X$ and $Y$ such that $R = TX$ and $R= SY$. This cannot possibly work since $T$ and $S$ have disjoint image.

Since $B(H)$ sits inside any type $III$-factor, no Ore localization in the above sense exists.

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Can we say then that M satisfies the Ore condition if and only if M is finite? –  Dmitri Pavlov Aug 5 '10 at 17:37
    
I think that is true. –  Andreas Thom Aug 5 '10 at 20:16
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