TITLE: Automorphism of $W^*$ algebra
QUESTION [5 upvotes]: Let $\mathfrak{A}$ be von Neumann algebra. It is in particular $C^*$ algebra.
Is it true that every $*$-isomorphism of $\mathfrak{A}$ is also $W^*-$isomorphism?
(Note that every $*$-isomorphism of $C^*$ algebra is $C^*-$isomorphism).
I'm especially interested in application to concrete von Neumann algebra (weakly closed subalgebra of $B(\mathcal{H})$ - the algebra of bounded operators on Hilbert space $\mathcal{H}$). Is it true that any isomorphism of this algebra is automatically continuous in weak, strong and ultraweak operator topologies on $\mathcal{H}$.

REPLY [2 votes]: Your question raises the point of what is meant by an "isomorphism of von Neumann algebras". 
Any C$^*$-isomorphism will of course preserve order, and then it will be normal: if $\varphi:A\to B$ is a $*$-isomorphism and $\{a_j\}\subset A^{\rm sa}$ is a bounded increasing net with least upper bound $a$, then $\varphi(a)$ is the least upper bound for $\{\varphi(a_j)\}\subset B$. 
And a normal map is strong and weak-operator continuous on bounded sets. I would say that this is the natural notion of isomorphism of von Neumann algebras. But I stand to be corrected.