Is $x \mapsto x \otimes 1$ $\sigma$-weakly continuous?

Question

Let $M\subseteq B(H)$ be a von Neumann algebra. Is it true that the mapping $$\psi: M \to B(H \otimes H): m \mapsto m \otimes \text{id}_H$$ is $\sigma$-weakly continuous? Here the $\sigma$-weak topology can be described in two ways:

(1) Let $M_*$ be any predual of $M$. Then the $\sigma$-weak topology is the weak$^ *$-topology on $M$ coming from the isomorphism $M \cong (M_*)^*$.

(2) It is the topology generated by the normal functionals. I.e. if $x \in M$ and $(x_i)$ is a net in $M$, then $x_i \to x$ if and only if $\text{Tr}((x_i-x)y) \to 0$ for all trace class operators $y$ on $H$.

We have to show that if $m_i \to m$ in the $\sigma$-weak topology, then also $m_i \otimes \text{id}_H \to m \otimes \text{id}_H$. I guess we will need some result that connects trace class operators on $H \otimes H$ and trace class operators on $H$?

Answer 1

There are many ways to do this. Maybe the quickest is to notice that $\psi$ is a $*$-isomorphism between $M$ and $\psi(M)$, hence an order isomorphism, hence normal. (For this reason, $*$-isomorphisms between von Neumann algebras are always weak* continuous.)

Answer 2

I want to note that the final question in the OP has a simple answer. For a Hilbert space $H$ and $\xi,\eta\in H$ let $\omega_{\xi,\eta}$ be the trace-class operator, which gives the normal function $\newcommand{\mc}{\mathcal}\mc B(H) \rightarrow\mathbb C; T\mapsto (T\xi|\eta)$. Any trace-class operator $\omega\in\mc T(H)$ can be written as a normal convergent sum $$ \omega = \sum_{n=1}^\infty \omega_{\xi_n, \eta_n} \qquad\text{with}\qquad \sum_n \|\xi_n\| \|\eta_n\| < \infty. $$ So for example Takesaki, Chapter II, Theorem 1.6 (which gives a stronger statement).

Now consider a Hilbert space $K$ and the Hilbert space tensor product $H\otimes K$. If $(e_i)$ is any orthonormal basis of $K$ then any $\xi\in H\otimes K$ can be (uniquely) written as $$ \xi = \sum_i \xi_i \otimes e_i \qquad\text{with}\qquad \|\xi\| = \sum_i \|\xi_i\|^2. $$ Thus, given $T\in\mc B(H)$ and $\xi,\eta\in H\otimes K$, then $$ \langle T\otimes 1 , \omega_{\xi,\eta} \rangle = \sum_{i,j} \big( (T\otimes 1)(\xi_i\otimes e_i) \big| \eta_j\otimes e_j \big) = \sum_i (T(\xi_i)|\eta_i). $$ Notice that this sum converges absolutely, as $$ \sum_i \big| (T(\xi_i)|\eta_i) \big| \leq \|T\| \sum_i \|\xi_i\| \|\eta_i\| \leq \|T\| \Big(\sum_i \|\xi_i\|^2\Big)^{1/2} \Big(\sum_i \|\eta_i\|^2\Big)^{1/2}, $$ by Cauchy-Schwarz. Notice that we need something like this argument to show that $T\otimes 1$ is even defined! Thus $\omega_{\xi,\eta}$ induces the functional $$ \sum_i \omega_{\xi_i, \eta_i} \in \mc T(H). $$ We extend to $\mc T(H\otimes K)$ by absolute convergence.