# Why is any von Neumann algebra SOT-closed?

I am trying to figure out why any von Neumann is SOT closed, which is stated in Conway. Do we need to use the Double Commutant Theorem on this implication aswell?

Is there a simpler explanation?

A von Neumann algebra is often defined as a follows:

Let $A$ be a $*$-subalgebra of $\mathcal{B(H)}$ acting non-degenerately on $\mathcal{H}$. Then $A$ is a von Neumann algebra if $A=A''$.

The bicommutant theorem says that

Let $A$ be a $*$-subalgebra of $\mathcal{B(H)}$ acting non-degenerately on $\mathcal{H}$. Then the following are equivalent:

1. $A=A''$,
1. $A$ is weakly closed,
2. $A$ is strongly closed.

The implication from $2$ to $3$ is trivial. Showing $1$ to $2$ is an easy exercise.

Could you specify the definition you use and which step is causing troubles?

• I use the same definitons, Conway states that any von Neumann is SOT closed. And after that he also states the converse that any SOT closed is von Neumann and refers to double commutnat motivating this. He does not refer to this theorem in the first case and I wounder why this is "trival" w.o useing the double commutnat theorem – Vlad May 10 '17 at 11:25
• His version of double commutnat is "If A is unital C*-subalg. then SOT closure = double commutant". And I can see why any von Neumann is SOT-closed by using this theorem. But not w.o using it. What confuses me is that he states this fact w.o refering to the theorem. – Vlad May 10 '17 at 11:28
• So to get the answer you should prove the implications $1\Rightarrow 2\Rightarrow 3$. The first implication is quite fun to do. – Mathematician 42 May 10 '17 at 11:31
• I will! But it would also follow directly fromt he double commutnat theorem right? – Vlad May 10 '17 at 11:33
• Ah yes definitely! I misread that. – Mathematician 42 May 10 '17 at 11:34