3
$\begingroup$

Let $X$ be a locally compact Hausdorff space, such that $C_0(X)$ is an injective Banach space, i.e. a $\mathfrak{P}_\lambda$ space for some $\lambda\geq 1$.

  • Is it true that $X$ is compact?

  • If additionally $X$ is a locally compact group, is it true that $X$ is finite?

$\endgroup$
  • 2
    $\begingroup$ NO. $C_0(X\setminus\{x_0\})$ is complemented in $C(X)$. $\endgroup$ – Narutaka OZAWA Jul 3 '14 at 21:51
  • $\begingroup$ @NarutakaOZAWA I agree for the 1st question, but is the 2nd one also obvious? $\endgroup$ – Yemon Choi Jul 3 '14 at 21:55
  • 1
    $\begingroup$ Regarding your second question, one obvious attempt seems to be ruled out by encyclopediaofmath.org/index.php/Extremally-disconnected_space which says, essentially, that the only compact groups $G$ for which $C(G)$ is $1$-injective are the finite ones. $\endgroup$ – Yemon Choi Jul 3 '14 at 22:30
  • $\begingroup$ @Yemon Choi: It's not obvious, but probably not too difficult (modulo a structure theorem of LCA groups) to show it's YES. Every LCA group $G$ has a compact open subgroup $K$ such that $G/K$ is a Lie group. One can probably show $K$ is finite and then $G=K$. $\endgroup$ – Narutaka OZAWA Jul 4 '14 at 0:31
  • $\begingroup$ @NarutakaOZAWA Thanks (by the way, you meant locally compact, not LCA, right?) $\endgroup$ – Yemon Choi Jul 4 '14 at 0:43
3
$\begingroup$

$C_0(G)$ is not injective for infinite $G$, because it is not complemented in $L_\infty(G)$.

See Lau, Anthony To Ming; Losert, Viktor. Complementation of certain subspaces of $L_\infty(G)$ of a locally compact group. Pacific Journal of Mathematics 141 (1990), no. 2, 295--310, corollary 3.

$\endgroup$

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.