# Normalizer of a von Neumann algebra

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?

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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.

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Wonderful. This is very elegant! –  André Henriques Feb 7 '12 at 17:21
Let $(u_n)_n$ be a sequence in $N$ that converges to $1$ in strong operator topology. Then we have to show that $\varphi(u_n^\ast\cdot u_n)$ converges to $\varphi$ for all $\varphi\in A_\ast$.
We can assume that $\varphi(x)=\langle\xi,x\eta\rangle$ for some $\xi,\eta\in L^2(A)$. But we know that $u_n\xi\rightarrow\xi$ and $u_n\eta\rightarrow\eta$. So $\langle u_n\xi, xu_n\eta\rangle$ converges to $\langle\xi,x\eta\rangle$ uniformly over $x\in (A)_1$. In other words, the functionals $\varphi(u_n^\ast\cdot u_n)=\langle u_n\xi, \cdot u_n\eta\rangle$ converge to $\langle\xi,\cdot\eta\rangle=\varphi$ in $A_\ast$.