Choi matrix with input factor first

Let $H_{\mathrm{i}\mathrm{n}}$ and $H_{\mathrm{o}\mathrm{u}\mathrm{t}}$ be finite-dimensional (have finite bases by Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis) complex inner-product spaces (Real and complex inner-product spaces and their induced length), and fix an ordered orthonormal basis $\mathcal{B}=(e_0,\dots ,e_{n-1})$ of $H_{\mathrm{i}\mathrm{n}}$ (the register basis convention is in Finite registers and their ordered computational bases). For $i,j<n$, let $E_{ij}\in \mathcal{L}(H_{\mathrm{i}\mathrm{n}})$ be the matrix unit, where $\mathcal{L}(H_{\mathrm{i}\mathrm{n}})$ denotes the complex-linear endomorphisms (Linear map between vector spaces over the same field), $E_{ij}x:=⟨x,e_j⟩e_i$, using the root inner product convention that the first argument is linear. Given a complex-linear map $\mathrm{\Phi }:\mathcal{L}(H_{\mathrm{i}\mathrm{n}})\to \mathcal{L}(H_{\mathrm{o}\mathrm{u}\mathrm{t}})$ (Linear map between vector spaces over the same field), its Choi matrix relative to $\mathcal{B}$ is

$J_{\mathcal{B}}(\mathrm{\Phi }):={\sum }_{i,j<n}{E}_{ij}\otimes \mathrm{\Phi }(E_{ij})\in \mathcal{L}(H_{\mathrm{i}\mathrm{n}}\otimes H_{\mathrm{o}\mathrm{u}\mathrm{t}}),$ where the operator tensor is the tensor map of Tensor products of finitely many linear maps and the tensor inner product is the finite product pairing in Ordered tensor registers, their inner product, and wire permutations.

The tensor factors are ordered input first and output second. The coordinate matrix has size $(n\dim H_{\mathrm{o}\mathrm{u}\mathrm{t}})\times (n\dim H_{\mathrm{o}\mathrm{u}\mathrm{t}})$. The chosen input basis is part of this definition. If the input basis is empty, the displayed sum is empty; if $\dim H_{\mathrm{o}\mathrm{u}\mathrm{t}}=0$, each summand is zero and $J_{\mathcal{B}}(\mathrm{\Phi })$ is the unique endomorphism of the zero-dimensional tensor product.