Ultralimit of $w^*$-continuous maps

Question

Let $\omega$ be a free ultrafilter on $\mathbb N.$ Let $(\mathcal M_n)$ be a sequence of finite von Neumann algebras. Let $\mathcal N$ be another finite von Neumann algebra and we have maps $\phi_n:\mathcal N\to\mathcal M_n$ which are all $w^*$ continuous. Then can we say that the ultra product of $(\phi_n)$ is again $w^*$-continuous? Or do we have anything like this for under some conditions? Any reference will be welcome.

Answer 1

I'm assuming the ultraproduct is the tracial ultraproduct, in which case $\prod_{\omega} \mathcal{M}_n$ is naturally a finite vNa. And by the ultraproduct of $(\phi_n)$ you mean the map $\mathcal{N} \ni x \mapsto (\phi_n(x)) \in \prod_{\omega} \mathcal{M}_n$.

In that case, no, it is not necessarily $w^*$-continuous. For example, let $\mathcal{N} = L^\infty([0, 1])$ and $\mathcal{M}_n = L^\infty([\frac{1}{2} - \frac{1}{n}, \frac{1}{2} + \frac{1}{n}])$ (for $n \geq 2$). Let $\phi_n: \mathcal{N} \rightarrow \mathcal{M}_n$ be the natural restriction map. Then for any free ultrafilter $\omega$, the ultraproduct $\phi$ of $(\phi_n)$ is not $w^*$-continuous. Indeed, the sequence $f_m = 1_{[\frac{1}{2} - \frac{1}{m}, \frac{1}{2} + \frac{1}{m}]}$ converges weakly to 0, but $\phi(f_m) = 1$ for all $m$ as $\phi_n(f_m) = 1$ whenever $n \geq m$.

The problem here is that the center of $\mathcal{N}$ (which is $\mathcal{N}$ itself in this case) has non-normal traces. Indeed, $\prod_{\omega} \mathcal{M}_n$ is equipped with a natural normal faithful tracial state $\tau_\omega$, so $\tau_\omega \circ \phi$ is a tracial state on $\mathcal{N}$. Hence, it quotients through the center-value trace $E: \mathcal{N} \rightarrow \mathcal{Z}(\mathcal{N})$ by the generalized Dixmier averaging theorem. One can then show that $\phi$ is normal iff $\tau_\omega \circ \phi$ is normal. Since $\tau_\omega \circ \phi$ quotients through $E$, it is normal iff $\tau_\omega \circ \phi$ is normal when restricted to $\mathcal{Z}(\mathcal{N})$. Thus, if $\mathcal{Z}(\mathcal{N})$ has only normal states, equivalently if it is finite-dimensional, equivalently if $\mathcal{N}$ is a finite direct sum of factors, then $\phi$ must be $w^*$-continuous. The converse also holds, at least if $\mathcal{N}$ is separable. Indeed, assume $\mathcal{N}$ is separable and its center has a non-normal state $\psi$. The state space of $\mathcal{Z}(\mathcal{N})$ has normal states as a weak$*$-dense subset. $\mathcal{N}$ being separable means the space of normal states on $\mathcal{Z}(\mathcal{N})$ is separable under the norm-topology, so $S(\mathcal{Z}(\mathcal{N}))$ is separable, whence we may choose a free ultrafilter $\omega$ on $\mathbb{N}$ and a sequence $(\psi_n)$ of normal states on $\mathcal{Z}(\mathcal{N})$ that converges weak$*$ to $\psi$ under $\omega$. Let $\phi_n$ be the GNS contrsuction associated with $\psi_n \circ E$ and $\mathcal{M}_n = \phi_n(\mathcal{N})$. It is then easy to show that $\tau_\omega \circ \phi = \psi$, so $\phi$ is not $w^*$-continuous.