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 $\xi\in H$.



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


