Slice maps on a von Neumann algebra completely positive?

Question

Let $M\subseteq B(H)$ and $N\subseteq B(K)$ be von Neumann algebra and let $\omega$ be a positive (normal) functional on $M$. Consider the von Neumann algebraic tensor product $M \boxtimes N \subseteq B(H \otimes K)$. We have a slice map $$\omega\boxtimes \iota_N: M \boxtimes N \to N: m \otimes n \mapsto \omega(m)n$$

Is this slice map completely positive? I know that the slice map $$\omega \otimes \iota_N: M \otimes N \to N$$ on the minimal tensor product is completely positive, maybe that's useful here.

Answer 1

Let's prove something more general. First, we need to easy lemmas:

Lemma: Let $\sigma:M\to N$ be a normal map between von Neumann algebras. Then $\sigma^{(n)}:M_n(M)\to M_n(N)$ is also normal.

Proof: This is an immediate corollary of the fact that a net of matrices $\{[x_{i,j}^\lambda]_{i,j=1}^n\}_{\lambda\in\Lambda}\subset M_n(N)$ converges to $0$ in the ultraweak topology of $M_n(N)$ if and only if each entry net $\{x_{i,j}^\lambda\}_{\lambda\in\Lambda}$ converges to $0$ in the ultraweak topology of $N$.

Lemma: Let $N$ be a von Neumann algebra. Then the set of positive elements in $N$ is ultraweakly closed.

Proof: Let $(x_i)\subset N_+$ be a net of positive elements with $x_i\to x\in N$ ultraweakly. We take a concrete representation $N\subset B(H)$. If $\xi\in H$, then since $x_i\geq0$ we have that $\langle x_i\xi,\xi\rangle\geq0$ for all $i\in I$. Since the ultraweak topology is stronger than the weak topology, we also have that $x_i\to x$ weakly, so $\langle x_i\xi,\xi\rangle\to\langle x\xi,\xi\rangle$, thus $\langle x\xi,\xi\rangle\geq0$. As $\xi$ was an arbitrary vector, this shows that $x\geq0$ in $B(H)$, thus $x\in N_+$.

Now we will prove the following, which will take care of your problem:

Proposition: Let $A$ be a $C^*$-algebra and take a non-degenerate representation, say $A\subset B(H)$ and let $M=\overline{A}^{wot}$, so $M$ is a von Neumann algebra. Let $N$ be another von Neumann algebra and let $\phi:A\to N$ be a c.p. map that is normal. Then $\phi$ extends uniquely to a normal map $\psi:M\to N$ and that $\psi$ is c.p.

Proof: Let $x\in M$. By Kaplansky's theorem we can find a norm-bounded net $(a_i)\subset A$ so that $a_i\to x$ in the weak topology, which is the same with the weak-star topology on norm-bounded sets. We define $$\psi(x):=\text{weak*-}\lim_{i}\phi(a_i)$$ First one needs to check that the limit exists: to do that, one must show that the net $\{\phi(a_i)\}$ is weak-star Cauchy; I think this is routine, so I will skip it. Then, $\psi$ is a well-defined normal map and all we have to do is show that $\psi$ is c.p.

So let $n\geq1$ and let $x\in M_n(M)_+$ be a positive element. It is not hard to see that $M_n(M)=\overline{M_n(A)}^{wot}=\overline{M_n(A)}^{sot}\subset B(H^n)$, so by Kaplansky's density theorem we can approximate $x^{1/2}$ in the strong operator topology by a norm bounded net of self-adjoint elemenets $(a_i)\subset M_n(A)$. Since multiplication is jointly continuous on bounded sets for the strong operator topology, we have that $a_i^2\to x$ strongly, hence weakly and since everything is bounded the weak topology is the same as the ultraweak topology. In other words we can approximate $x$ in the ultraweak topology by a norm bounded net of positive elements of $M_n(A)$. Now $$\psi^{(n)}(x)=\text{weak*-}\lim_{i}\psi^{(n)}(a_i^2)\geq0 $$ where we used the two lemmas in the first equality and the second inequality respectively.

Now we apply the above proposition in our case: By definition, $M\boxtimes N$ is the weak closure of the minimal tensor product $M\otimes N$. The slice map $\omega\boxtimes\iota_N$ is the unique normal extension of the normal and c.p. map $\omega\otimes\iota_N:M\otimes N\to N$. So $\omega\boxtimes\iota_N$ is c.p.