Timeline for Surface automorphisms and conformal automorphisms
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Feb 25, 2011 at 20:38 | comment | added | Andy Putman | @Sam : I just left a comment explaining where Marius's criticism of my answer went wrong. Now that Marius has modified his question, my answer no longer answers it, but given that it contains correct mathematics and answers (one interpretation of) a previous version of the question, I think it's worth leaving up. | |
| Feb 25, 2011 at 20:37 | comment | added | Andy Putman | (continued) The following might make this less confusing. In my answer, $f : S \rightarrow S$ is a biholomorphic map between two different Riemann surfaces on the same topological surface $S$ (they are isomorphic Riemann surfaces, but the surfaces are "wearing" their Riemann surfaces structures on different ways). This is exactly what happens when the whole mapping class group acts on Teichmuller space. Two points in the same mapping class group orbit on Teichmuller space are isomorphic Riemann surfaces, but they are not the same Riemann surface! | |
| Feb 25, 2011 at 20:34 | comment | added | Andy Putman | @Marius : It doesn't matter whether $f$ is of finite order or not. If $S$ is a Riemann surface and $\phi : S' \rightarrow S$ is a diffeomorphism, then we can pull back the Riemann surface structure on $S$ to get a Riemann surface structure on $S'$. With this Riemann surface structure, $\phi : S' \rightarrow S$ will be a biholomorphism. Moreover, if $\tau : S \rightarrow S$ is a biholomorphic map, then $\tau \circ \phi \circ \tau^{-1} : S' \rightarrow S'$ will be a biholomorphic map. I'm just applying this fact when $S' = S$. | |
| Feb 25, 2011 at 20:20 | comment | added | Sam Nead | @Andy - I've removed the downvote and I apologize for doing so. A vote, up or down, is a single bit and thus is a limited form of communication. Let me instead ask you directly: why is your answer on-point, given Marius' edit of his post and then his comment just above? | |
| Feb 24, 2011 at 17:52 | comment | added | Andy Putman | @Sam -- Scott and I gave answers to a reasonable interpretation of the question. It isn't fair to downvote us for not reading the OP's mind correctly. | |
| Feb 24, 2011 at 17:46 | comment | added | Marius Overholt | @Andy - I think you misunderstood my question. Unfortunately I did not formulate the question precisely enough, so that there are now several different answers. It looks like the problem with your answer, from my point of view, is that the conjugating map $f$ may not be of finite order. If it were, your answer probably would give $m(\sigma) \leq 3$, though I am not quite sure of this. | |
| Feb 24, 2011 at 17:14 | history | answered | Andy Putman | CC BY-SA 2.5 |