Timeline for On injectivity of the Banach space $C_0(X)$
Current License: CC BY-SA 3.0
14 events
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| Jan 6, 2016 at 17:28 | history | edited | Tomasz Kania | CC BY-SA 3.0 |
correcting grammar as my colleague would like to refer to this post in his paper
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| Jul 7, 2014 at 15:05 | vote | accept | Norbert | ||
| Jul 7, 2014 at 15:04 | answer | added | Norbert | timeline score: 3 | |
| Jul 5, 2014 at 13:26 | history | undeleted |
Gerry Myerson Emil Jeřábek Bill Johnson |
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| Jul 4, 2014 at 15:05 | history | deleted | user53043 | via Vote | |
| Jul 4, 2014 at 7:06 | history | edited | Norbert |
edited tags
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| Jul 4, 2014 at 1:01 | comment | added | Narutaka OZAWA | @Yemon Choi: Oh. Somehow I assumed $G$ was abelian. | |
| Jul 4, 2014 at 0:43 | comment | added | Yemon Choi | @NarutakaOZAWA Thanks (by the way, you meant locally compact, not LCA, right?) | |
| Jul 4, 2014 at 0:31 | comment | added | Narutaka OZAWA | @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 $G=K$. | |
| Jul 3, 2014 at 22:30 | comment | added | Yemon Choi | 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. | |
| Jul 3, 2014 at 21:55 | comment | added | Yemon Choi | @NarutakaOZAWA I agree for the 1st question, but is the 2nd one also obvious? | |
| Jul 3, 2014 at 21:51 | comment | added | Narutaka OZAWA | NO. $C_0(X\setminus\{x_0\})$ is complemented in $C(X)$. | |
| Jul 3, 2014 at 21:47 | review | First posts | |||
| Jul 3, 2014 at 22:11 | |||||
| Jul 3, 2014 at 21:29 | history | asked | Norbert | CC BY-SA 3.0 |