Let $M \subseteq B(H)$ be a von Neumann algebra. Let $U: H \to K$ be a unitary where $K$ is another Hilbert space and consider a faithful $*$-representation $$\pi: M \hookrightarrow B(K): m \mapsto UmU^*.$$ Is $\pi$ normal (= $\sigma$-weakly continuous = weak$^*$-continuous = ultra-weakly continuous)?
Attempt: It suffices to show that $\pi(M)= UMU^*$ is a von Neumann algebra, since a $*$-isomorphism between von Neumann algebras is automatically normal. There are now several ways to proceed:
- Show that $\pi(M)$ is strongly/weakly closed.
- Show that $\pi(M)'' = \pi(M)$.
However, none of these seem easy to figure out.
I think things are simpler: First observe that $\mathbb{B}(H)\to\mathbb{B}(K)$, $T\mapsto UTU^*$ is a $*$-isomorphism between von Neumann algebras, thus, as you said, normal. Hence $\pi$, which is merely a restriction of this map is normal.
If you want to verify that $\pi(M)$ is a von Neumann algebra, simply show that $\pi(M)$ is storngly closed: Let $\{x_\lambda\}\subset M$ be a net such that $\pi(x_\lambda)\to y\in\mathbb{B}(K)$ strongly, so we have that $ux_\lambda u^*\to y$. We want to show that $y\in\pi(M)$. Multiplicaton is separately continuous, so $$u^*\cdot(ux_\lambda u^*)\cdot u\to u^*yu$$ strongly. But $u^*ux_\lambda u^*u=x_\lambda$, so $x_\lambda\to u^*yu$ strongly hence $u^*yu\in M$, since $M$ is strongly closed. But since it is obvious that $y=u(u^*yu)u^*=\pi(u^*yu)$, we have that $y\in\pi(M)$.