Proof of Choi-Effros Theorem for Operator Systems in Paulsen's Book

Question

I am trying to understand the proof of Theorem 13.1 in Paulsen's Book ``Completely Bounded Maps and Operator Algebras''.

Theorem If $S$ is a matrix-ordered $*$-vector space with an Archimedean matrix order unit $e$, then there exists a Hilbert space $\mathcal{H}$, an operator system $S_1\subseteq\mathcal{B}(\mathcal{H})$, and a complete order isomorphism $\varphi\colon S\rightarrow S_1$ with $\varphi(e)=I_\mathcal{H}$.

The proof starts as follows: Let $$P_n=\{\varphi\colon S\rightarrow M_n\,|\,\varphi\text{ completely positive,} \varphi(e)=I_n\}$$ where $I_n$ is the identity matrix. Define $$ J:=\sum_{n=1}^\infty\sum_{\varphi\in P_n}\oplus\,\varphi\colon S\rightarrow\sum_{n=1}^\infty\sum_{\varphi\in P_n}\oplus\, M_n^\varphi $$ where the latter direct sum is in the $l^\infty$ sense. Clearly, by the choice of $\varphi$'s, $(x_{ij})\in\mathcal{C}_n$ implies $(J(x_{ij}))\geq 0$ where $\mathcal{C}_n$ denotes a cone from the matrix ordering. And the proof goes on...

(1) First of all I am a bit confused by the notation. Is $\sum_{n=1}^\infty\sum_{\varphi\in P_n}\oplus M_n^\varphi$ simply a notation for the direct sum , i.e. could we also write $$ \bigoplus_{n=1}^\infty\bigoplus_{\varphi\in P_n} M_n^\varphi $$ for this space?

(2) Secondly, how do we know that $J$ is well defined, i.e. why is $\Vert J(s)\Vert_\infty$ for $s\in S$ bounded?

(3) And finally, why is it clear that $(J(x_{ij}))\geq 0$ holds? Since multiplication in $\sum_{n=1}^\infty\sum_{\varphi\in P_n}\oplus M_n^\varphi$ is defined component wise, is it, because we know $\varphi$ is completely positive, each component of $(J(x_{ij}))$ can be decomposed into an Hermitian square, i.e. $aa^*$ for some $a$ and thus $(J(x_{ij}))$ can be decomposed into the component wise product of the the tuples consisting of those roots?

Answer 1

1. Yes.

2. When $s$ is selfadjoint the order unit property gives you $c>0$ such that $-c\,e≤s≤c\,e$ and so $-c\,I_n≤\varphi(s)≤c\,I_n$, so $\|\varphi(s)\|≤c$. For arbitrary $s$ you have $s=s_1+i s_2$ with $s_1,s_2$ selfadjoint, and so $$ \|\varphi(s)\|≤\|\varphi(s_1)\|+\|\varphi(s_2)\|≤c_1+c_2. $$

3. Yes, positivity in an $\ell^\infty$-direct sum is defined component-wise. If $s≥0$ then all components will be of the form $\varphi(s)$ with $\varphi$ ucp and hence positive. For the complete positivity you need to notice that $$ M_k\big(\bigoplus_j M_{n(j)}(\mathbb C)\big)\simeq \bigoplus_j M_k(M_{n(j)}(\mathbb C)\big). $$

Answer 2

(1) Yes, you are right here. The point is that the direct sum runs over all 'unital' completely positive maps $S\to M_n$, where $n$ can vary. I admit that the notation is a bit awkward. The following would perhaps be a bit cleaner:

Let $P= \bigcup_{n=1}^\infty P_n$ and then consider the space $\bigoplus_{\varphi \in P} M_{n(\varphi)}$.

I will stick with this proposed notation.

(2) Your map is $J(x) = (\varphi(x))_{\varphi \in P}, x\in S$. Note then that $\|\varphi\|= 1$ for all $\varphi \in S$ (this is a good exercise) Hence, $\|J(x)\|= \sup_{\varphi\in P}\|\varphi(x)\| \le \|x\|$ for $x\in S$ and thus $J$ is a contraction.

(3) The way I see this, the following happens:

We want to show that the inflated map $$M_n(J): M_n(S)\to M_n\left(\bigoplus_{\varphi\in P} M_{n(\varphi)}\right)$$ is positive. However, note that $$M_n\left(\bigoplus_{\varphi\in P} M_{n(\varphi)}\right) \cong \bigoplus_{\varphi \in P} M_n(M_{n(\varphi)})$$ so it suffices to show that the composition $$M_n(S)\to \bigoplus_{\varphi \in P} M_n(M_{n(\varphi)})$$ is positive. This composition is precisely the map $$z \mapsto (M_n(\varphi)(z))_{\varphi \in P}$$ which is positive because each $M_n(\varphi)$ is positive.