Minimal injective extension is rigid

Question

Let $V$ be an operator system.

Definition 1: A pair $(W, \kappa)$ is called extension of $V$ if $W$ is an operator system and $\kappa: V \to W$ is a unital complete isometry.

Definition 2: An extension $(W,\kappa)$ of $V$ is called injective extension if $W$ is an injective operator system.

Definition 3: An extension $(W, \kappa)$ of $V$ is called rigid if for every unital completely positive map $\varphi: W \to W$ with $\varphi \kappa = \kappa$, we necessarily have $\varphi = \iota_W$.

Definition 4: An injective extension $(W,\kappa)$ of $V$ is called minimal if $\kappa(V)\subseteq W_1 \subseteq W$ with $W_1$ injective implies that $W_1 = W.$

Question: Is a minimal injective extension necessarily rigid? Is there a quick way to see this? I have a rather long argument along the lines of stuff in Paulsen where we use minimal seminorms, but maybe there is an easier way. If it helps, I know that for injective extensions, rigidity = essentiality.

Answer 1

Hamana's proof (Theorem 3.5 in Injective Envelopes of Operator Systems, PubL RIMS, Kyoto Univ. 15 (1979), 773-785) is fairly direct. Consider the partial ordering on the space $\Xi = \{ \phi \in UCP(W, W) \mid \phi \kappa = \kappa \}$ given by $\phi \prec \psi$ when $\| \phi(x) \| \leq \| \psi(x) \|$ for all $x \in W$. First, note that every element in $\Xi$ dominates a minimal element in $\Xi$ since if $\{ \phi_i \}_i$ is a decreasing net in $\Xi$, then taking any concrete realization $W \subset \mathcal B(\mathcal H)$ we have $E \phi \prec \phi_i$ for all $i$, where $E: \mathcal B(\mathcal H) \to W$ is any ucp idempotent and $\phi$ is any limit point of $\{ \phi_i \}_i$ in $UCP(W, \mathcal B(\mathcal H))$, which is compact in the topology of pointwise ultraweak convergence.

Second, note that any minimal element $\phi \in \Xi$ is an idempotent. Indeed, for any $N \geq 1$ and $x \in W$ we have $\| \frac{1}{N} \sum_{n = 1}^N \phi^n( x ) \| \leq \| \phi(x) \|$ and so by minimality it follows that this is equality. Considering $x = y - \phi(y)$ we have $\| \phi(y) - \phi^2(y) \| = \| \frac{1}{N} \sum_{n = 1}^N \phi^n(y - \phi(y) ) \| \leq \frac{2}{N} \| y \| \to 0$.

Thus, if every idempotent in $\Xi$ is the identity, it follows that the identity is the only map in $\Xi$.

Answer 2

Let $\varphi: W \rightarrow W$ be a u.c.p map.

Let Fix$(\varphi)=\{ w \in W: \varphi(w)=w\}$.

Then, Fix$(\varphi)$ is an injective operator system, if $W$ is an injective operator system. Moreover, Fix$(\varphi)$ contains $\kappa(V)$, as $\varphi \circ \kappa= \kappa$.

By minimality of $W$, we must therefore have that Fix$(\varphi)= W$, thereby establishing rigidity.