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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – MaoWao
    Jan 23, 2022 at 1:54
  • $\begingroup$ @maowao does it have a simple description? If yes would you mind saying a few words about it? $\endgroup$ Jan 23, 2022 at 9:28
  • $\begingroup$ @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 $(M\otimes_\min N)^\ast$. $\endgroup$
    – MaoWao
    Jan 25, 2022 at 8:27
  • $\begingroup$ @MaoWao I see, thanks:) $\endgroup$ Jan 25, 2022 at 11:01

1 Answer 1


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.


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