1
$\begingroup$

Let $A,B \subset B(H)$ be two concrete von Neumann algebra. Is $A\cap B$ a von Neumann algebra, too?

What about the intrinsic analogy of this question, as follows:

Let $C$ be a $C^*$ algebra and $A,B \subset C$ be two von Neumann algebras. Is their intersection, a von Neumann algebra, too?

Can one speak of a kind of minimal von Neumann algebra contained in a given $C^*$ algebra?

On the other extreme, can one think of a kind of maximal von Neumann algebra contained in a given $C^*$ algebra?

In particular what are two maximal von neumann algebras in $B(H)$ which are not isomorphic?

$\endgroup$
  • 1
    $\begingroup$ For the first question the answer is yes, because intersection behaves nicely in both algebra and topology. For the rest, I prefer $C^*$-algebras over von-Neumann algebras, so wait for an expert to reply. $\endgroup$ – Aweygan Nov 16 '18 at 8:49
2
$\begingroup$

The intersecion of von Neumann algebras is a von Neumann algebra.

It is possible for a C$^*$-algebra to contain no von Neumann algebra. For example $C_0(\mathbb R)$ has no nonzero projections, so it cannot contain any von Neumann algebra other than $\{0\}$.

If $A$ is unital, then $\mathbb C\subset A$, so there is always a von Neumann algebra. But again many C$^*$-algebras are projectionless, so $\mathbb C$ is the only one.

Even when C$^*$-algebras have many projections, it is very unlikely that they'll contain von Neumann algebras. It is common to find copies of $M_n(\mathbb C)$ (a von Neumann algebra). But any infinite-dimensional von Neumann algebra is non-separable as a C$^*$-algebra, so no separable C$^*$-algebra contains an infinite-dimensional von Neumann algebra.

And many separable C$^*$-algebras contain enough projections that one can find $M_n(\mathbb C)$ for all $n$, so there is certainly no maximal von Neumann subalgebra.

$\endgroup$
  • $\begingroup$ Thank you and +1 for your answer. by Maximal von Neumann algebra i mean a proper subalgebra which is von Neumann and is contained in no proper von Neumann subalgebra. $\endgroup$ – Ali Taghavi Nov 17 '18 at 5:03
  • 1
    $\begingroup$ I see. I don't think that I can say anything meaningful, but I'll think about it. $\endgroup$ – Martin Argerami Nov 17 '18 at 5:29

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.