In extension to this question Positive but not completely positive? I'd like to know, for $k>1$, examples of $k$-positive linear maps of a matrix algebra into itself that are not $k+1$-positive. (I don't know a single one.) By M.D. Choi's theorem the size of the matrices involved must grow with $k$, but how fast?

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    $\begingroup$ I took the liberty of fixing the link and retagging ("complete positivity" is a bit over-specialised as a tag, in my view.) $\endgroup$
    – Yemon Choi
    Jul 14 '15 at 15:20

The "canonical" example of a map that is $k$-positive but not $(k+1)$-positive is the map defined by $$ \Phi_k(X) = k\cdot\mathrm{Tr}(X)I_n - X. $$

Above, $n$ denotes the size of $X$ (i.e., $X \in M_n$) and $I_n$ is the $n \times n$ identity matrix. This map was introduced in "J. Tomiyama. On the geometry of positive maps in matrix algebras II. Linear Algebra Appl., 69:169–177, 1985." (though it had been considered by Choi earlier in the special case when $k = n-1$).

It's not difficult to prove that this map is $k$-positive just by elementary linear algebra. I won't post details here since it's a bit long and messy, but it's almost exactly the same as the proof that starts at the bottom of page 4 of these notes.

To show that this map is not $(k+1)$-positive (when $k < n$), simply let $\mathbf{v} = \sum_{i=1}^{k+1} \mathbf{e}_i \otimes \mathbf{e}_i \in \mathbb{C}^{k+1} \otimes \mathbb{C}^n$ (here $\{\mathbf{e}_i\}$ is the standard basis of $\mathbb{C}^{k+1}$ on the first subsystem, and it's just embedded into $\mathbb{C}^n$ in the natural way on the second subsystem). Then compute that $$ (id_{k+1} \otimes \Phi_k)(\mathbf{v}\mathbf{v}^*) = k(I_{k+1} \otimes I_n) - \mathbf{v}\mathbf{v}^*, $$ which has $-1$ as an eigenvalue and is thus not positive.

A couple of side notes:

  1. This map shows that Choi's theorem on complete positivity is optimal in some sense: if $k \geq n$ then $k$-positivity implies complete positivity, but if $k < n$ then $k$-positivity and $(k+1)$-positivity are indeed different sets.

  2. In the $k = 1$ case this map comes up frequently in quantum information theory under the name of the "reduction map".

  • $\begingroup$ Hi Nathaniel, I just want to ask what does K positive map or in general K map means and I do not understand what SN means in your notes. $\endgroup$
    – Ka-Wa Yip
    Feb 1 '17 at 22:12
  • $\begingroup$ @walker - $k$-positive map means that $(id_k \otimes \Phi)(X)$ is a positive semidefinite matrix whenever $X$ is a positive semidefinite matrix. In my notes, SN means "Schmidt number". $\endgroup$ Feb 2 '17 at 16:57
  • $\begingroup$ does k here means the dimension is k? $\endgroup$
    – Ka-Wa Yip
    Feb 4 '17 at 11:29
  • $\begingroup$ @walker - Yes, $k$ refers to the dimension of the space on which the identity transformation acts. So if $\Phi$ acts on n-by-n matrices, then $id_k \otimes \Phi$ acts on kn-by-kn matrices. $\endgroup$ Feb 6 '17 at 0:13

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