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?

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    $\begingroup$ 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

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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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