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.

Background

In finite dimensions, the Choi formalism establishes a one-to-one correspondence between completely positive maps and positive semidefinite operators. Extending such correspondences to infinite-dimensional settings is delicate. The paper proves that certain weighted Choi constructions lose surjectivity in infinite dimensions and examines alternative approaches.

The authors note that the unweighted Choi map naturally associates finite-rank maps from trace-class to trace-class operators with elements of the algebraic tensor product B(H) ⊙ B1(Z). However, they highlight that a suitable topology on the space of superoperators is needed to complete this correspondence to all bounded maps between trace-class spaces; standard choices either restrict the domain too much (operator norm yielding only compact maps) or are too weak (e.g., weak operator topology).

References

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.