3
$\begingroup$

Let $A$ and $B$ be two purely infinite von Neumann algebras. Under which conditions $A$ can be embedded in $B$? For example is it true that every infinite factor can be embedded into the another infinite factor?

$\endgroup$
4
$\begingroup$

The only real restriction here is cardinality.

Any two properly infinite von Neumann algebras (i.e. no finite summand) with separable predual are biembeddable. The reason is that a vNa with separable predual is properly infinite (has no finite summand) if and only if it absorbs B(H) tensorially (a relatively standard exercise which can be deduced from Proposition V.1.22 of Takesaki's Theory of operator algebras). One can then do the embedding you ask for into the B(H) tensor factor.

(Note that purely infinite (i.e. type III vNas) are properly infinite (no finite summand)).

In generality the answer is no: let M be a type III factor with separable predual. The tensor product of M and the bounded operators on a Hilbert space with uncountable dimension will not embed in M.

$\endgroup$
  • $\begingroup$ Thank you for the answer. By the phrase "absorbs tensorially" you mean that it is of the form $M \otimes B(H)$ for some M? $\endgroup$ – truebaran Nov 1 '16 at 21:38
  • $\begingroup$ Yes (and in fact $M $is the original vna, so M is properly infinite if and only if $ $\endgroup$ – Stuart White Nov 1 '16 at 22:54
  • $\begingroup$ Yes (and in fact $M$ is the original vna, so $M$ is properly infinite if and only if $M\cong M\,\overline{\otimes}\,B(H)$. $\endgroup$ – Stuart White Nov 1 '16 at 22:55

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.