# von Neumann automorphisms: does convergence on a dense algebra imply $u$-convergence?

Let $M$ be a separable von Neumann algebra and let $A$ be a (von Neumann-)dense *-subalgebra. Suppose that $\alpha,\alpha_1,\alpha_2,\dots$ are automorphisms of $M$, such that for every $a \in A$, $$\alpha_n(a) = \alpha(a)$$ for all $n$ sufficiently large. Does it follow that $\alpha_n$ converges to $\alpha$ in the $u$-topology?

(Also: what about if we weaken the hypothesis to just assuming that $\alpha_n(a)$ converges in norm to $\alpha(a)$, for all $a \in A$?)

This question is inspired by a related MO question, where an explicit example was requested of a sequence of inner automorphisms on the hyperfinite $III_1$- (or $II_1$-) factor which converge to an outer automorphism. An answer involved a sequence of inner automorphisms satisfying, in particular, the hypotheses of my question, although the proof of convergence uses more information. Perhaps other examples answering that question would be available if the answer to my question is yes.

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What do you mean by "von Neumann dense"? Dense in the norm topology, the strong, or the weak topology? –  doug Jul 11 '12 at 17:26
Dense in the strong topology, or equivalently, the weak one or any of those other von Neumann algebra topologies except the norm topology. Also equivalently, $A''=M$. –  Aaron Tikuisis Jul 11 '12 at 18:03

No. For an example, consider $M=L^\infty([0,1])$ with the Lebesgue measure, take $A$ to be the functions that are piecewise constant on dyadic intervals and $\alpha_n(f)=f\circ \phi_n^{-1}$ where $\phi_n(t)=k/2^n + (2^n t-k)^2/2^n$ if $t \in [k/2^n,(k+1)/2^n[$. In words, $\phi_n$ acts as some fixed transformation (here $t \mapsto t^2$, but it could be anything but the identity) but at a smaler and smaler scale.
Since $\phi_n$ preserves the intervals $[k/2^n,(k+1)/2^n[$, $\alpha_n(a)=a$ for all $a$ $\mathcal F_n$-measurable, where $\mathcal F_n = \sigma([k/2^n,(k+1)/2^n[, 0\leq k<2^n)$. So $\alpha_n$ satisfies your assumption with $\alpha=id$.
However, consider $1 \in L^1([0,1])=L^\infty([0,1])_*$. Then $(\alpha_n)_* 1= \phi_n'$, and $\|1 - \phi_n'\|_{L^1} = \|1 - \phi_0'\|_{L^1} = \int_0^1 |2u-1| dt=1/2$ does not converge to $0$.
But the answers to both your questions become yes if you assume that $M$ is finite and $\alpha_n$ preserves a trace (for example if $M$ is a $II_1$ factor). Or more generally if the $\alpha_n$'s preserve a normal faithful state $\phi$. Indeed, then on the one hand by Hahn-Banach $\{ \phi(a \cdot), a \in A\}$ forms a norm dense subspace of $M_*$, and on the other hand $(\alpha_n)_*(\phi(a \cdot)) = \phi(\alpha_n^{-1}(a) \cdot)$. Hence $(\alpha_n)_*$ converges pointwise in norm on a dense subspace of $M_*$, so it converges pointwise in norm on $M_*$.
$L^\infty([0,1])$ is not separable... –  Jochen Wengenroth Jul 12 '12 at 8:02