# If $A$ is a $W^*$-algebra, how to construct a nice predual for $M_n(A)$?

Let $$A$$ be a $$W^*$$-algebra. By concretely representing $$A \subseteq B(H)$$ as a von Neumann algebra and using that $$M_n(A) \subseteq B(H^{n})$$, one can check that $$M_n(A)$$ is again a $$W^*$$-algebra such that $$\sigma$$-weak convergence in $$M_n(A)$$ is given by entrywise $$\sigma$$-weak convergence of all entries.

However, I'm wondering if it is also possible to deduce this fact from Sakai's predual theorem. I.e., given a $$W^*$$-algebra $$A$$, it it possible to construct a Banach space $$F$$ together with an isometric isomorphism $$\Phi: M_n(A) \to F^*$$ so that it becomes clear that $$\sigma$$-weak convergence in $$M_n(A)$$ is given entrywise?

Very naively, we can fix a predual $$A_*$$ for $$A$$ and then we want to do something like $$M_n(A) \cong M_n((A_*)^*) \cong M_n(A_*)^*$$ but it is not clear to me how to turn $$M_n(A_*)$$ in a Banach space such that we obtain an isometry.

• It's late here, so just a very short remark: $M_n(A)\cong A\bar\otimes M_n(\mathbb C)$ and the predual of the von Neumann tensor product is described in Takesaki's book in the section/chapter on tensor products. Jan 23, 2022 at 1:54
• @maowao does it have a simple description? If yes would you mind saying a few words about it? Jan 23, 2022 at 9:28
• @JustDroppedIn In this case, it basically boils down to what you wrote. In general, the predual of $M\bar\otimes N$ is the closure of $M_\ast\otimes N_\ast$ inside $(M\otimes_\min N)^\ast$. Jan 25, 2022 at 8:27
• @MaoWao I see, thanks:) Jan 25, 2022 at 11:01

Maybe there is a way to define some norm on $$M_n(X)$$ for a Banach space $$X$$, but personally I don't know how to do this. However, there is one thing that feels natural to do:

Since we know that $$M_n(A)$$ is a von Neumann algebra and thus a Banach space, we can consider its dual space $$M_n(A)^*$$. We will define a linear injective map $$\Phi: M_n(A_*)\to M_n(A)^*$$ which will give rise to a norm by setting $$\|\cdot\|_{M_n(A_*)}:=\|\Phi(\cdot)\|_{M_n(A)^*}$$. This definition automatically makes $$\Phi$$ an isometry.

Let $$x=[x_{i,j}]\in M_n(A_*)$$. Define a map $$\phi_x:M_n(A)\to\mathbb{C}$$ by $$\phi_x([a_{i,j}]):=\sum_{i,j}a_{i,j}(x_{i,j})$$. We first show that $$\phi_x$$ is bounded. Indeed, for $$a=[a_{i,j}]\in M_n(A)$$, we have $$|\phi_x(a)|=|\phi_x([a_{i,j}])|\le\sum_{i,j}|a_{i,j}(x_{i,j})|\le\sum_{i,j}\|a_{i,j}\|\cdot\|x_{i,j}\|\le\sum_{i,j}\|x_{i,j}\|\cdot\|a\|$$ where we have used the classical inequality $$\|a_{i,j}\|_A\le\|a\|_{M_n(A)}$$. So the assignment $$\Phi:M_n(A_*)\to M_n(A)^*$$, $$\Phi(x)=\phi_x$$ is well-defined. Linearity of the map $$\Phi$$ is trivial to verify. Let's check injectivity. Assume that $$\phi_x=0$$ for some $$x=[x_{i,j}]\in M_n(A_*)$$. Fix a pair of indices $$(i,j)$$. By the Hahn-Banach theorem, find a functional $$\alpha\in A$$ such that $$|\alpha(x_{i,j})|=\|x_{i,j}\|$$. Now consider the matrix $$a=[a_{k,l}]\in M_n(A)$$ where each $$a_{k,l}=0$$ for $$(k,l)\ne(i,j)$$ and $$a_{i,j}=\alpha$$. Then $$0=\phi_x(a)=\alpha(x_{i,j})$$, so $$\|x_{i,j}\|=0$$. As $$(i,j)$$ was an arbitrary pair of indices, we conclude that $$x=0$$.

We have thus defined a linear isometry $$\Phi:(M_n(A_*),\|\cdot\|_{M_n(A_*)})\to (M_n(A)^*,\|\cdot\|_{M_n(A)^*})$$ between normed spaces. We now consider the adjoint map $$\Phi^*:M_n(A)^{**}\to M_n(A_*)^*$$. As $$\Phi$$ is an isometry, so is $$\Phi^*$$. Set $$\Psi:=\Phi^*\vert_{M_n(A)}$$ (recall that for any normed space $$X$$ sits inside $$X^{**}$$ as evaluation functionals). If we show that $$\Psi$$ is surjective, we are done, since then we will have identified $$M_n(A_*)$$ as a predual of $$M_n(A)$$.

But indeed, let $$\zeta:M_n(A_*)\to\mathbb{C}$$ be a bounded linear functional. Consider the maps $$h_{i,j}:A_*\to M_n(A_*)$$ that send an element $$z\in A_*$$ to a matrix that is zero everywhere except the $$(i,j)$$ slot, where the entry is $$z$$. I leave it for the end to verify that $$h_{i,j}$$ is bounded. But then $$\zeta\circ h_{i,j}\in (A_*)^*=A$$, so set $$a_{i,j}=\zeta\circ h_{i,j}\in A$$ and set $$a=[a_{i,j}]$$. Then for $$x=[x_{i,j}]\in M_n(A_*)$$ we have $$\Psi(a)(x)=\Psi(a)[x_{i,j}]=\Phi^*(a)[x_{i,j}]=\Phi([x_{i,j}])([a_{i,j}])=\sum_{i,j}a_{i,j}(x_{i,j})=\sum_{i,j}\zeta(h_{i,j}(x_{i,j}))=$$ $$=\zeta(\sum_{i,j}h_{i,j}(x_{i,j}))=\zeta(x)$$ proving the desired surjectivity.

Note that $$M_n(A_*)$$ might not be a priori complete, but we can just consider its completion. It is always the case that a normed space $$X$$ and its completion have isometrically isomorphic dual spaces.

Finally, I guess this is enough to conclude for ultraweak convergence in $$M_n(A)$$: $$[a_{i,j}^\lambda]\to0$$ ultraweakly if and only if for all $$[x_{i,j}]\in M_n(A_*)$$ we have $$\Phi^*([a_{i,j}^\lambda])([x_{i,j}])\to0$$, i.e. if and only if $$\sum_{i,j}a_{i,j}^\lambda(x_{i,j})\to0$$ for all $$[x_{i,j}]\in M_n(A_*)$$. By fixing a pair of indices and taking matrices that are zero everywhere except that pair of indices, this occurs if and only if $$a_{i,j}^\lambda(x)\to0$$ for all $$x\in A_*$$ and all $$i,j$$, i.e. if and only if $$a_{i,j}^\lambda\to0$$ ultraweakly in $$A$$ for all $$i,j$$.

P.S: Verification that $$h_{i,j}$$ are bounded. Let $$z\in A_*$$. By definition, $$\|h_{i,j}(z)\|=\sup_{[a_{k,l}]\in M_n(A)_1}|\phi_{h_{i,j}(z)}([a_{k,l}])|=\sup_{[a_{k,l}]\in M_n(A)_1}|a_{i,j}(z)|$$. But since $$\|a_{i,j}\|\le\|[a_{k,l}]\|\le1$$ we have $$|a_{i,j}(z)|\le\|z\|$$, as we wanted.