Timeline for The length is bounded
Current License: CC BY-SA 4.0
15 events
when toggle format | what | by | license | comment | |
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Oct 29 at 18:43 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 4.0 |
Typo in title corrected
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Oct 29 at 14:16 | history | edited | Adam | CC BY-SA 4.0 |
edited title
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Oct 28 at 14:55 | vote | accept | Adam | ||
Oct 28 at 14:55 | history | undeleted | Adam | ||
Oct 21 at 19:42 | history | deleted | Adam | via Vote | |
Oct 21 at 18:23 | answer | added | Moishe Kohan | timeline score: 0 | |
Oct 21 at 15:17 | comment | added | Moishe Kohan | First, ask yourself how a homeomorphism of $\Sigma$ acts on the Teichmuller space. | |
Oct 21 at 15:16 | comment | added | Adam | @MoisheKohan : could you please let me know why it is true trivially ? | |
Oct 21 at 15:15 | comment | added | Moishe Kohan | Are you sure you are using the correct codomain for your $f$? Because, as written, the answer is trivially positive. (Your map hits only finitely many mapping classes.) | |
Oct 21 at 15:09 | history | edited | Adam | CC BY-SA 4.0 |
added 57 characters in body
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Oct 21 at 14:24 | comment | added | Adam | @AidanBackus : Thanks for your comment. Yes! You are right! | |
Oct 21 at 14:17 | comment | added | Aidan Backus | The Thurston metric generates the same topology as the usual topology on Teichmueller space, so while I might be misunderstanding the question, I think it is equivalent to: given $x_0$, under what minimal/generic conditions must a compact family of self-homeomorphisms of $\Sigma$ not map $x_0$ to itself? | |
Oct 21 at 13:50 | history | edited | Adam | CC BY-SA 4.0 |
added 202 characters in body
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Oct 21 at 13:42 | history | edited | Adam | CC BY-SA 4.0 |
added 102 characters in body
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Oct 21 at 13:35 | history | asked | Adam | CC BY-SA 4.0 |