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Let $\Gamma$ be a group, $N(\Gamma)$ its group von Neumann algebra, $\newcommand{\cUG}{{\mathcal U}(\Gamma)}$ and $\cUG$ the ring of all densely-defined, closed operators $\ell^2(\Gamma)\to\ell^2(\Gamma)$ that are affiliated to $N(\Gamma)$.

In some work I've been doing on (algebras of) convolution operators associated to group representations, I'd like to give a clear reference for the fact that $\cUG$ is directly finite, or in alternative terminology Dedekind finite: this just means that whenever $a,b$ belong to $\cUG$ and satisfy $ab=1$, then $ba=1$. (Note: I am only interested in the case of $\cUG$ but the analogous result holds if $N(\Gamma)$ is replaced by any other finite von Neumann algebra.)

The statement of this result is something I learned years ago from reading preprints of Elek, but there he gives as his references an article of Berberian, which in turn relies on various results about Baer $\ast$-rings in Berberian's own book on the same ... so this begins to look less and less like a convenient reference for readers who aren't already happy with von Neumann regular rings, Baer $\ast$-rings, and so on.

In contrast, I've found by doing some foraging online that the fact I need to cite follows straightforwardly from the polar decomposition for unbounded affiliated operators, as explained simply in Lemma 2.2 of this paper by H. Reich:

On the K- and L-theory of the algebra of operators affiliated to a finite von Neumann algebra. K-Theory, 24 (2001), no.4, 303-326 (Copy on author's webpage)

It seems odd to give the preliminary section of a 2001 paper as the citation/reference for something that must surely have been known to Murray and von Neumann. So my question is this:

Can anyone suggest an older reference, accessible for analysts and preferably in a book, for the fact that $ab=1 \implies ba=1$ for all $a,b\in\cUG$? Or is a quick outline, and a citation of Reich's exposition, the best that I can do? I haven't had time to check in Kadison-Ringrose, nor Lück's book on $L^2$-invariants, but I can think of several MO regulars who would be much more familiar with the literature on $\cUG$ than I am.

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I am not sure if Goodearls book on von Neumann regular rings is accessible for analysts, but I would give it a try. It is at least older than from 2001 and it is a book. – Abel Stolz Feb 27 '13 at 10:55
Even if you cannot find a reference, isn't it easy to write your own short proof? If ab=1, then leftsupp(ab)=leftsupp(1)=1, hence leftsupp(a)=1, hence rightsupp(a)=1, because rightsupp(a) is equivalent to leftsupp(a), and the only projection equivalent to 1 in a finite von Neumann algebra M is 1 itself. Similarly, leftsupp(b)=rightsupp(b)=1. Also leftsupp(a)=rightsupp(a)=1 is equivalent to a being invertible in the algebra A(M) of affiliated elements of M. Since a and b are invertible in A(M), we have ab=1 if and only if ba=1. – Dmitri Pavlov Feb 27 '13 at 16:51
Dmitri: this is indeed a nice looking argument, although the polar decomposition is lurking behind everything. I just didn't want to claim any originality on my part, so was planning to cite Reich, but then felt this was a bit silly as explained in my post. – Yemon Choi Feb 27 '13 at 16:58
@Yemon: Again, the relevant argument is short enough to be included in the text: an operator T is affiliated with a von Neumann algebra A if and only if both operators in the polar decomposition of T commute with the commutant of A. Hence the partial isometry part of T belongs to A, which is what we need for the above argument to work. – Dmitri Pavlov Feb 27 '13 at 18:03
@Dmitri: thanks for pointing this out, this is more or less what Reich does (phrased a little differently) – Yemon Choi Feb 27 '13 at 19:21

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