A von Neumann algebra, or $W^*$-algebra, is a $*$-algebra of bounded operators on a Hilbert space that is closed in the weak operator topology and contains the identity operator. It is a special type of $C^*$-algebra.
I want to ask some things
- Why a $W^*$-algebra is closed in the weak operator topology?
- Why a $W^*$-algebra contains the identity operator?
- If a $W^*$-algebra were open in the weak operator topology, what would happen? There would always be the identity operator?
- If a $W^*$-algebra did not contain identity operator, what happens?
- If a $W^*$-algebra were closed in the weak operator topology but it didn't contain identity operator, what happens ?
- If a $W^*$-algebra were open in the weak operator topology but it didn't contain identity operator, what happens ?
- If a $W^*$-algebra were open in the weak operator topology and contain identity operator, what happens ?
That's the nature of von Neumann algebras. They are the ones equal to they double commutants, which makes them closed in all the usual topologies (via von Neumann's Double Commutant Theorem).
There are several ways to see this. The identity is in the commutant of anything, so it is in particular in the double commutant. Or one can see that von Neumann algebras are generated by projections, that they have approximate units (all C$^*$-algebras do) and that that any approximate unit converges in the sot or wot topologies to the identity.
If any subspace $S$ of $B(H)$ contains a wot neighbourhood $N$ of $0$, then for any $T\in B(H)$ there is $c>0$ with $cT\in N\subset S$. Then $T=\tfrac1c\,(cT)\in S$, so $S=B(H)$.
A von Neumann algebra $M$ always contains its identity. If that's not the identity of $B(H)$, it means that $M$ is degenerate (that is $\overline{MH}\subsetneq H$).
No answer. This never happens.
This never happens.
It equals $B(H)$. You don't need to assume that it contains the identity, that happens on its own.