Extension of Choi-Jamiolkowski isomorphism to not completely positive maps In quantum physics, the concept of Channel-State duality is of prime importance in understanding the final state of the system after passing through a channel or performing quantum operations. The result uses Choi-Jamiolkowski isomorphism since completely positive (CP) maps are involved.

Given a map $T:\mathcal{M}_d\to\mathcal{M}_d$ (matrices), then $T$ is completely positive iff
  $$ \tau:=(T\otimes \mathrm{id}_d)(|\Omega\rangle\langle \Omega|)$$
  is a positive semidefinite matrix, where $|\Omega\rangle=\sum_{i=1}^d |ii\rangle$ the maximally entangled state.

I know that this theorem works fine for CP maps. Is there a generalization under the weaker assumption that one of the eigenvalues is negative? Essentially, does this theorem have a counterpart for not completely positive maps?
 A: The following extension is the only extension that I know in the direction you want.
Given a $k-$positive map $T:\mathcal{M}_d\to\mathcal{M}_d$, then 
$$ \tau:=(\mathrm{id}_m \otimes T^*)(\rho)$$
has at most $(m-k)(d-k)$ negative eigenvalues, where $\rho$ is any state of $\mathcal{M}_m\otimes \mathcal{M}_d$.
Check the third paragraph of page 5 and theorem 1 of page 3 of this this paper.
Note that a $d-$positive map $T:\mathcal{M}_d\to\mathcal{M}_d$ is simply a completely positive map.
A: Your question is rather vague to me but I'll give it a shot, hoping I understood your intents correctly. In the CJ-isomorphism one considers the map $\mathfrak C:\mathcal L(\mathbb C^{n\times n})\to \mathbb C^{n^2\times n^2}$ which takes any linear operator $T:\mathbb C^{n\times n}\to \mathbb C^{n\times n}$ and maps it to the $n^2\times n^2$-matrix
$$
\mathfrak C(T):=\sum\nolimits_{j,k=1}^n E_{jk}\otimes T(E_{jk})=\begin{pmatrix} T(E_{11})&T(E_{12})&\ldots&T(E_{1n})\\ T(E_{21})&T(E_{22})&\ddots&\vdots\\ \vdots&\ddots&\ddots&T(E_{(n-1)n})\\ T(E_{n1})&\cdots&T(E_{n(n-1)})&T(E_{nn}) \end{pmatrix}
$$
with $(E_{jk})_{j,k=1}^n$ being the standard basis of $\mathbb C^{n\times n}$ and $\otimes$ the usual Kronecker product. Bilinearity of the tensor product shows that $\mathfrak C$ is linear itself (i.e. $\mathfrak C(\lambda S+T)=\lambda \mathfrak C(S)+\mathfrak C(T)$ for all $\lambda\in\mathbb C$, $S,T\in\mathcal L(\mathbb C^{n\times n})$). Now one readily verifies that the map
$$
\big(\mathfrak C^{-1}(A)\big)(X)=\operatorname{tr}_1\big((X^T\otimes\operatorname{id}_n)A\big)\quad \text{ for any }A\in\mathbb C^{n^2\times n^2}\text{ and all }X\in\mathbb C^{n\times n}
$$
is the inverse of $\mathfrak C$, where $\operatorname{tr}_1$ is the partial trace on the first tensor component. Altogether, this shows the following:

$\mathfrak C$ is an isomorphism between $\mathcal L(\mathbb C^{n\times n})$ and $\mathbb C^{n^2\times n^2}$ (and not "only" between the convex cones of completely positive maps and the positive semi-definite $n^2\times n^2$-matrices)

This should not be surprising, given that $\mathfrak C(T)$ contains the action of $T$ on a basis of $\mathbb C^{n\times n}$ which of course ($\to$ linearity!) uniquely determines every such $T$ so one gets a one-to-one correspondence between said spaces.
