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Let $V$ be an operator System in $B(H)$. By Hamana and Ruan theorems, there is an injective envelope $I(V)$ which is minimal injective subspace of $B(H)$ contains $V$. Thus there is a completely contractive onto projection $\varphi: B(H) \to I(V)$.

By Choi-Effross Theorem, $I(V)$ is a $C^{\star}$-algebra by the new product $T \circ_{\varphi} S=\varphi(TS)$ for each $T, S \in I(V)$.

The enveloping $C^\star$-algebra $C^\star_e(V)$ of $V$ is the $C^*$-subalgebra of $I(V)$ generated by $V$.

$C^\star_e(V)$ has the following property: Suppose that $\psi : V \to A$ is any unital completely isometric map such that $A$ is generated by $\psi(V )$ as a $C^{\star}$-algebra. Then there exists a unique $^\star$-homomorphism $\pi: A → C^*_e(V)$ such that $\pi\circ \psi = id_V$.

Question: Let $V \subseteq B(H)$ be an operator system and $U \in U(H)$ such that $UVU^{\star}=V$. What about $UC^{\star}_e(V)U^{\star}=C^{\star}_e(V)$ or $UC^{\star}_e(V)U^{\star}\subseteq I(V)$ ?

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  • $\begingroup$ Are you asking if the $U$ conjugation of $C^*_e(V)$ is an isomorphism? Doesn't it follow from the functoriality of the construction? $\endgroup$
    – Uri Bader
    Commented Aug 14, 2016 at 11:21
  • $\begingroup$ @UriBader C$^*_e(V)$ here is only an operator system when acted on by $U$ in $B(H)$ since its multiplication is coming from $I(V)$. I suspect that there is a negative answer to the question but nothing comes to mind at the moment. $\endgroup$
    – Chris Ramsey
    Commented Aug 14, 2016 at 23:46

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Since $C^*_e(V) \subseteq V''$ and $U \in V' = (V'')'$, doesn't this follow automatically? The nontrivial question here is whether $U \in I(V)'$, which implies that $C^*_e(V)$ has a weak expectation.

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