Timeline for Are all endomorphisms of C^* just power maps?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 10, 2016 at 14:30 | comment | added | Andreas Thom | @DenisNardin: You are right, my mistake. | |
| Mar 10, 2016 at 13:35 | comment | added | Denis Nardin | I know this question is old, but the comment above by @AndreasThom is wrong: all the continous endomorphisms are of the form $|z|^\alpha z^n$ for $\alpha\in \mathbb{C}$ and $n\in\mathbb{Z}$. This is not hard to prove using the isomorphism $\mathbb{C}^\times\cong \mathbb{R}\times S^1$ | |
| Feb 11, 2013 at 21:39 | comment | added | HNuer | @Andreas, why is that? I've heard this before and I can easily prove the holomorphic case, but I can't think off the top of my head why the continuous result is true. | |
| Feb 9, 2013 at 10:44 | comment | added | Andreas Thom | All continuous endomorphisms are of the form $z\mapsto z^n\bar z^m$, for some $n,m \in \mathbb Z$. | |
| Feb 5, 2013 at 22:35 | comment | added | Jim Humphreys | @HNuer: This very old question has come up in other forums like math.stackexchange.com, and I'm not convinced it's "research-level". Anyway, the elementary expository article linked by Chris is a good standard source (though I'm prejudiced by the fact that Paul Yale was an undergraduate teacher of mine). Like other MAA publications, it usually requires JSTOR access. | |
| Feb 5, 2013 at 11:15 | history | edited | Andrej Bauer | CC BY-SA 3.0 |
added 11 characters in body
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| Feb 5, 2013 at 10:17 | answer | added | Chris Heunen | timeline score: 5 | |
| Feb 4, 2013 at 22:45 | comment | added | ACL | @HNuer: Well, you did not specify yourselves the kind of structure on $\mathbf C^*$ for which you wanted to determine the endomorphisms. | |
| Feb 4, 2013 at 21:57 | comment | added | HNuer | Fair enough. I didn't know what group structure you were using. | |
| Feb 4, 2013 at 21:27 | comment | added | Mariano Suárez-Álvarez | No, I mean $\mathbb R$, just because the multiplicative group $\mathbb R_{>0}$ and the additive group $\mathbb R$ are isomorphic. | |
| Feb 4, 2013 at 21:27 | comment | added | HNuer | Although I think you mean $R_{>0}$, I see what you're saying, so thanks for the help! | |
| Feb 4, 2013 at 21:16 | comment | added | Mariano Suárez-Álvarez | $C^\times$ is isomorphic as a group to $S^1\times\mathbb R$, and $\mathbb R$, as any infinite dimensional vector space over $\mathbb Q$, has lots of endomorphisms as an abelian group! | |
| Feb 4, 2013 at 21:09 | history | asked | HNuer | CC BY-SA 3.0 |