Is the bitranspose continuous for the $\sigma$-strong topology?

Question

Let $\varphi\colon A\to B$ be a bounded, linear map between C*-algebras. Is the bitranspose $\varphi^{**}\colon A^{**}\to B^{**}$ continuous when the von Neumann algebras $A^{**}$ and $B^{**}$ are equipped with their $\sigma$-strong topologies?

Motivation/Background: Note that $\varphi^{**}$ is clearly continuous when $A^{**}$ and $B^{**}$ are equipped with their $\sigma$-weak topologies, since these agree with the weak${}^*$-topology from the preduals, and $\varphi^{**}$ is weak${}^*$-continuous (that is, $\sigma(A^{**},A^*)-\sigma(B^{**},B^*)$-continuous).

If $\varphi$ if completely positive, then it follows that $\varphi^{**}$ is a completely positive, normal map, and therefore is continuous for the $\sigma$-strong topologies. Thus, the question is only interesting if $\varphi$ is not completely positive.

On bounded sets of a von Neumann algebra, the $\sigma$-strong topology agrees with the strong (operator) topology (SOT). If $M\subseteq B(H)$ is a von Neumann algebra, then a net $(a_j)_j$ in $M$ SOT-converges to $a\in M$ if $\|a_j\xi-a\xi\|\to 0$ for every $\xi\in H$.

Answer 1

I think the transpose map on the compacts gives a counterexample.

Let $K(H)$ be the compacts on a separable infinite dimensional Hilbert space with orthonormal basis $\{ e_n \}.$ Let $T:K(H)\rightarrow K(H)$ be the transpose map (i.e. $T(e_{n,m})=e_{m,n}$ on matrix units). Then $T$ is $\sigma$-weakly continuous so $T^{**}:B(H)\rightarrow B(H)$ will also be the transpose map. Let $S$ be the shift $Se_n=e_{n+1}.$ Then $T(S^{*n})=S^n$ for all $n.$ Finally $S^{*n}\rightarrow 0$ strongly while $T(S^{*n})=S^n$ does not converge strongly to anything.

Since all of these maps $S^n, S^{*n}$ are in the unit ball, as you mention the strong and $\sigma$-strong topologies coincide.