Automorphism of $W^*$ algebra 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 $\mathcal{H}$. 
 A: 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 $\{\varphi(a_j)\}\subset B$.
The issue is how this is linked to the algebra's environment. The above paragraph is how one would think intrinsically to the algebra. But it does not necessarily link to the relation of the algebra with its environment. What I mean is this: it is possible to take a von Neumann algebra $M$, say a II$_1$-factor, and represent it via an irreducible representation $\pi$. Because  $M$ is simple, the representation is faithful. But the image, being irreducible and not type I, is a wot dense subalgebra of $B(H_\pi)$, so it is not a von Neumann algebra.
When both domain and codomain are von Neumann algebras,  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.
