# On injectivity of the Banach space $C_0(X)$

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?

• NO. $C_0(X\setminus\{x_0\})$ is complemented in $C(X)$. – Narutaka OZAWA Jul 3 '14 at 21:51
• @NarutakaOZAWA I agree for the 1st question, but is the 2nd one also obvious? – Yemon Choi Jul 3 '14 at 21:55
• 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. – Yemon Choi Jul 3 '14 at 22:30
• @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 $G=K$. – Narutaka OZAWA Jul 4 '14 at 0:31
• @NarutakaOZAWA Thanks (by the way, you meant locally compact, not LCA, right?) – Yemon Choi Jul 4 '14 at 0:43

## 1 Answer

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