0
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ dear martin thank you very much I now understand it. $\endgroup$ – user197041 Dec 3 '14 at 12:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.