# Minimal or Maximal Von Neumann algebra contained in a given $C^*$ algebra

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?

• 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. – Aweygan Nov 16 '18 at 8:49

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.