Topology enabling completion of the unweighted Choi correspondence to all bounded maps on trace-class spaces
Determine a topology on the space of bounded linear maps from the trace-class operators on a Hilbert space to the trace-class operators on another Hilbert space, that is, on B(B^1(H), B^1(Z)), under which the unweighted Choi map that identifies finite-rank maps X ↦ ∑_{j=1}^m tr(A_j X) B_j with the algebraic tensor product B(H) ⊙ B^1(Z) can be completed to a well-defined correspondence on the entire space B(B^1(H), B^1(Z)), overcoming the shortcomings that the operator-norm topology restricts the domain to compact maps and that weak operator–type topologies appear too weak.
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Indeed, the “unweighted” Choi map $\Phi\mapsto\sum_{j,k\in J}|g_j\rangle\langle g_k|\otimes\Phi(|g_j\rangle\langle g_k|)$ establishes a correspondence between maps $:\mathcal B1(\mathcal H)\to\mathcal B1(\mathcal Z)$ of finite rank (i.e.~maps of the form $X\mapsto\sum_{j=1}m{\rm tr}(A_jX)B_j$ with $m<\infty$) and the algebraic tensor product $\mathcal B(\mathcal H)\odot\mathcal B1(\mathcal Z)$. However, it is not clear what topology on the domain would allow for a completion of this correspondence to $\mathcal B(\mathcal B1(\mathcal H),\mathcal B1(\mathcal Z))$: the norm topology leads to a domain $\mathcal K(\mathcal B1(\mathcal H),\mathcal B1(\mathcal Z))$ which is too small, but something like the weak operator topology would likely be too weak and the domain would become too large.