Posted on 8th July 2023

Finally, I can combine the last post with a post 3 ago. This argument is from a paper of Elphick and Wocjan, attributed there to Roberson.

Fix a graph \( G \) with adjacency matrix \( A \).

Theorem:Let \( \mathcal C \) be a pinching on \( \mathbb M_n\otimes\mathcal B(H) \) with \( \mathcal C(A\otimes 1_H)=0 \) and \( \mathcal C(D\otimes 1_H)=D\otimes 1_H \) for each diagonal matrix \( D \). Then \( \mathcal C \) arises from a quantum colouring of \( G \), in the manner described before.

Proof:Let the pinching be \( \mathcal C(x) = \sum_k P_k x P_k \) for some projections \( (P_k) \) on \( \mathbb C^n\otimes H \) with \( \sum_k P_k=1 \). As \( P_k=P_k^* \), we see that \( \mathcal C \) is completely positive, and unital as \( \sum_k P_k=1 \). If \( H \) is finite-dimensional, we compute that \[ \newcommand{\tr}{\operatorname{Tr}} \tr(\mathcal C(x)) = \sum_k \tr(P_kx_Pk) = \sum_k \tr(P_k^2x) = \tr(x), \] so \( \mathcal C \) is a quantum channel.(If \( H \) is infinite-dimensional separable, then \( \mathbb C^c\otimes H \cong H \) and we can regard \( \mathcal C \) as acting on \( \mathcal B(H) \). We can find a trace-class operator \( d \) with full support which is diagonal with respect to the orthogonal decomposition given by the \( (P_k) \). Thus \( P_kd=dP_k \) for each \( k \). Then the functional \( \phi(x) = \tr(xd) \) is faithful, and invariant, because \[ \phi(\mathcal C(x)) = \sum_k \tr(P_kxP_kd) = \sum_k \tr(x dP_k) = \tr(dx) = \phi(x). \] If \( H \) is not separable, then we can find a family of invariant functions which together are faithful, and this is sufficient for the proof of our fixed point result to still work.)

As \( \mathcal C \) is written in Kraus form, we conclude that each \( P_k \) commutes with the fixed points, in particular, with \( D\otimes 1_H \) for each diagonal matrix \( D \). This means that each \( P_k \) has the form \[ P_k = \sum_{v\in V} e_v e_v^* \otimes P_{v,k} \] for some projections \( (P_{v,k}) \) on \( H \). (Here recall that \( n=|V| \) so \( \mathbb C^n \) is indexed by \( V \).) Then \[ 1 = \sum_k P_k = \sum_v e_ve_v^* \otimes \sum_k P_{v,k} \] and so \( \sum_k P_{v,k}=1 \) for each \( v \).

Finally, \[ \begin{split} 0 &= \mathcal C(A\otimes 1_H) = \sum_{u,v} e_u e_u^* A e_v e_v^* \otimes \sum_k P_{u,k} P_{v,k} \\ &= \sum_{u\sim v} e_u e_v^* \otimes \sum_k P_{u,k} P_{v,k}, \end{split} \] as \( e_u^*Ae_v = A_{u,v}=1 \) only when \( u\sim v \). Choose \( u\sim v \), so we have that \( \sum_k P_{u,k} P_{v,k} = 0 \). As \( \sum_k P_{u,k} = 1 \), the projections \( (P_{u,k})_k \) are orthogonal, so given \( l \), \[ 0 = \sum_k P_{u,l} P_{u,k} P_{v,k} = P_{u,l} P_{v,l} \] which is the second condition to be a quantum colouring.

I wonder if some similar result holds for quantum graphs.