A von Neumann algebra M is said to be finite, infinite, properly infinite, or purely infinite according to the property of the identity projection 1. I think that this classification of types are independent from choosing algebra, since the identity projection is unique for all von Neumann algebras. is it true?
1 Answer
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No. It wouldn't make sense if it were independent of the algebra.
The properties you mention depend on the relation of the identity with the rest of the projections.
For example, $M_2(\mathbb C)$ is finite because $I$ is not equivalent to any proper projection (because equivalence of projections is given by rank).
On the other hand, on $B(\ell^2(\mathbb N))$, the identity is properly infinite because it is equivalent to proper subprojections: $I=S^*S$, $SS^*=I-E_{11}$ where $S$ is the unilateral shift. It is not purely infinite because there are subprojections not equivalent to $I$: for instance, $I$ is not equivalent to any finite-rank projection.
answered Dec 2, 2014 at 22:25
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$\begingroup$ dear martin thank you very much I now understand it. $\endgroup$ Dec 3, 2014 at 12:53