Let $(A,H)$ be a von Neumann algebra in standard form (which means that $H=L^2A$), and
recall that the automorphism group $Aut(A)$ acts on both $A$ and $H$.
Let
$$N:=\{u\in U(H): uAu^\*=A\}$$
be the normalizer of $A$, equipped with the strong topoloy (the subspace topology from $U(H)$).
The group $N$ is canonically isomorphic to the semidirect product $$N\cong Aut(A)\ltimes U(A').$$


Equip $Aut(A)$ with Haagerup's *u*-topology, i.e. the topology of pointwise convergence on $A\_\*$, equivlalently, it is the topology of pointwise convergence on $H$.


***Question:*** is the projection map $$N\to Aut(A)$$ continuous?



Normalizer of a von Neumann algebra
Julien's comment to my previous answer leads to an even easier solution.


Consider the isomorphism map $\psi:Aut(A)\ltimes U(A^\prime)\rightarrow N$. The $u$-topology on $Aut(A)$ is the topology of pointwise convergence on $H$, which is the same as strong convergence when we consider the elements of $Aut(A)$ as unitaries on $H$. Since $N$ is a topological group, it follows that $\psi$ is continuous. Julien's comment shows that it is in fact a homeomorphism. So certainly the quotient $N\rightarrow Aut(A)$ is continuous.


