Ultraweak convergence on matrix over a von Neumann algebra Let $N$ be a von Neumann algebra and $n\geq1$be an integer. Let $x_\lambda=[x_{i, j}^\lambda]_{i,j} $ be a net in $M_n(N)$. Is it true that $(x_\lambda)_\lambda$ converges ultraweakly to zero in $M_n(N)$ if and only if each entry-net $(x_{i,j}^\lambda)_\lambda$ converges to 0 in the ultraweak topology of $N$?
I was trying to prove that the maps $N\to M_n(N)$ sending an element of $N$ to the matrix with 0 everywhere except the $i, j$ position and the element in the $i, j$ position are ultraweakly continuous and the same for the maps $M_n(N) \to N$ sending a matrix to one of its entries, but I could not do it. Any hint would be appreciated.
 A: I believe the simplest answer goes as follows.
First of all, note that this is easily deduced when $N=\mathcal{B(H)}$ because we have a nice description of the predual of $\mathcal{B(H)}$: the trace-class operators, or, if you prefer,
$$\mathcal{B(H)}_*=\overline{\{\phi\in(\mathcal{B(H)},\|\cdot\|)^*:\phi\text{ is WOT continuous}\}}^{\|\cdot\|_{\mathcal{B(H)}^*}}$$
I actually think that this description is more practical here than the trace class operators. Now the $*$-isomorphism $M_n(\mathcal{B(H)})\cong\mathcal{B(H^{(n)})})$ can be used to obtain the result.
Now let $N$ be an arbitrary von Neumann algebra. Then choose a concrete representation of $N$, i.e. find a faithful representation $\pi:N\to\mathcal{B(H)}$ so that $\pi(N)$ is SOT-closed in $\mathcal{B(H)}$. Then $\pi(N)$ is ultraweakly closed: Indeed, if $(\pi(x_\lambda))\subset\pi(N)$ is a net such that $\pi(x_\lambda)\to y$ ultraweakly, then $\pi(x_\lambda)\to y$ in (WOT) because the ultraweak topology is stronger than the weak topology. But $\pi(N)$ is a convex set and (SOT) closed, thus it is also (WOT) closed, so $y\in\pi(N)$. My point is, $\pi(N)\subset\mathcal{B(H)}$ is an inclusion of von Neumann algebras. Since $N\cong\pi(N)$ and $*$-isomorphisms are topological homeomorphisms with respect to the ultraweak topologies, we may simplify this to $N\subset\mathcal{B(H)}$ as von Neumann algebras. But now $M_n(N)\subset M_n(\mathcal{B(H)})$ as von Neumann algebras and the result follows from the one we proved in the special case of $\mathcal{B(H)}$.
This was hinted to me on MO by Nik Weaver, I deleted my post there because on second thought it was way too elementary for math overflow.
