7
$\begingroup$

It is an old question if every injective Banach space is isomorphic as Banach space to $C(X)$-space.

I would like to know if the weakened module version of this question is answered. More precisely: For which compact Hausdorff spaces $X$ the module $C(X)$ is an isomorphically injective $C(X)$-module. I know that for $X$ Stonean $C(X)$ is even an isometrically injective $C(X)$-module.

$\endgroup$
  • 1
    $\begingroup$ Is this relative injectivity as in Helemskii's theory, or the stricter version? $\endgroup$ – Yemon Choi Dec 6 '14 at 22:47
  • $\begingroup$ In other words: are we testing over what Helemskii calls the admissible embeddings, or just the embeddings with closed range? $\endgroup$ – Yemon Choi Dec 6 '14 at 22:49
  • $\begingroup$ The stricter version, where embeddings have closed range. $\endgroup$ – Norbert Dec 6 '14 at 22:58
  • 1
    $\begingroup$ I think this version is usually called "strict injectivity", at least in Helemskii's book. So just to clarify: you require that whenever $M$ is a closed sub-module of a Banach $C(X)$-module $N$, every bounded $C(X)$-module map $M\to C(X)$ has an extension to a bounded $C(X)$-module map $N\to C(X)$. Is that correct? $\endgroup$ – Yemon Choi Dec 6 '14 at 23:02
  • $\begingroup$ If you use the absolutely strictest version of injectivity (viewing C(X) as a ring, then saying it is an injective module in the ring-theoretic sense---which is how I interpreted the question initially), then the answer when $X$ is compact is simply that $X$ be finite (since a commutative unital ring with no nilpotents, on being injective as a module, must be von Neumann regular (aka absolutely flat) ...) $\endgroup$ – David Handelman Dec 7 '14 at 0:08
7
$\begingroup$

It's equivalent to that $C(X)$ is isometrically ijective (i.e., $X$ is stonean). We take the definition cited by Yemon in the comment and let $\iota\colon C(X)\hookrightarrow C(Y)$ be a faithful $*$-homomorphism from $C(X)$ into an injective abelian $C^*$-algebra $C(Y)$. Then, there is a $C(X)$-module projection $\Phi$ from $C(Y)$ onto $C(X)$. In particular, $C(X)$ is isomorphically injective, but Huruya'e example (Proc AMS 1984) says isomorphic injectivity need not imply isometric injectivity (though his example is isomorphic as a Banach space to an isometrically injective one). We will exploit the fact that the projection $\Phi$ is a $C(X)$-module map. Let $\Phi^*\colon X \to C(Y)^*$ be the continuous map define by $\langle\Phi^*(x),f\rangle = \Phi(f)(x)$. Here $C(Y)^*$ is equipped with the weak$^*$-topology. Note that $\|\Phi(x)\| \geq 1$ for every $x\in X$, and define a continuous map $\Psi^*\colon X\to C(Y)^*$ by $\Psi^*(x)=|\Phi^*(x)|/\|\Phi^*(x)\|$. This in turn gives rise to a unital positive contractive map $\Psi\colon C(Y) \to C(X)$, which is given by $\Psi(f)(x) = \langle\Psi^*(x),f\rangle$. We claim that $\Psi$ is a projection onto $C(X)$. Let $h\in C(X)_+$ be given arbitrary. Since $\Phi(\iota(h)f)(x)=h(x)\Phi(f)(x)$ for all $f\in C(Y)$, one has $\iota(h)\Phi^*(x)=h(x)\Phi^*(x)$ as an element in $C(Y)^*$ (which is also viewed as a complex Radon measure on Y). Since $\iota(h)\geq0$, this implies that $\iota(h)|\Phi^*(x)|=|\iota(h)\Phi^*(x)|=h(x)|\Phi^*(x)|$ and so $\iota(h)\Psi^*(x)=h(x)\Psi^*(x)$ for every $x\in X$. This means that $\Psi(\iota(h))(x)=h(x)$ for every $x \in X$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you prof. Ozawa and excuse me for such late reply. $\endgroup$ – Norbert Jan 25 '15 at 9:33
  • 3
    $\begingroup$ Unfortunately, I do not understand one step of your proof. Why is $\Psi^*$ continuous? The continuity of $\Phi^*$ is clear, but the operation of total variation doesn't preserve weak convergence of measures so I don't see why $\Psi^*(x)=\frac{|\Phi^*(x)|}{\Vert\Phi^*(x)\Vert}$ must be continuous. Could you explain that step? $\endgroup$ – Norbert Feb 2 '15 at 20:12
  • 2
    $\begingroup$ Indeed, you're right. My proof does not stand. $\endgroup$ – Narutaka OZAWA Feb 3 '15 at 6:24
  • 1
    $\begingroup$ Anyway, great thanks for thinking about my problem. $\endgroup$ – Norbert Feb 3 '15 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.