Timeline for Can the injective envelope ever be injective for $*$-homomorphisms?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2021 at 15:27 | comment | added | Chris Ramsey | Actually, here it is fine. One can prove that there are no proper intermediate operator systems between $A$ and $C(X)$. Thus, $I(A) = C(X)$ | |
| Dec 3, 2021 at 23:30 | comment | added | Chris Ramsey | @NarutakaOZAWA But the injective envelope need not be realized as a C$^*$-subalgebra of $C(X)$ but rather as an intermediate operator system with some other multiplication that turns it into a C$^*$-algebra. Or perhaps things simplify in commutative C$^*$-algebras? | |
| Dec 3, 2021 at 23:12 | comment | added | Narutaka OZAWA | That's because $C(X)$ is injective and there is no proper intermediate subalgebra between $C(Y)\subset C(X)$. | |
| Dec 3, 2021 at 22:57 | comment | added | Chris Ramsey | @NarutakaOZAWA This could indeed work. However, among other questions I have, why is the injective envelope of $C(Y)$ equal to $C(X)$? | |
| Dec 2, 2021 at 23:48 | comment | added | Narutaka OZAWA | Here's a sketch for a (possible) counter-example. For $X$ with $C(X)$ injective and $x,y\in X$ not isolated, $A=\{ f \in C(X) : f(x)=f(y)\} = C(Y)$, $Y=X/x=y$, is (probably) no longer isometrically injective, but any $\ast$-homomorphism from $C(Y)$ into $B(H)$ extends on the C*-algebra of bounded Borel functions on $Y$., which (probably) naturally contains $C(X)$. | |
| Dec 2, 2021 at 22:28 | history | asked | Chris Ramsey | CC BY-SA 4.0 |