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 1


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.

  • $\begingroup$ dear martin thank you very much I now understand it. $\endgroup$
    – user197041
    Dec 3, 2014 at 12:53

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