Timeline for Growth of smallest closed geodesic in congruence subgroups?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jan 16, 2012 at 4:19 | comment | added | Vitali Kapovitch | @Agol Thanks a lot for the clarification. | |
| Jan 15, 2012 at 16:45 | history | edited | Ian Agol | CC BY-SA 3.0 |
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| Jan 15, 2012 at 16:04 | comment | added | Ian Agol | @Vitali: Ok, I think I see how to show it. In fact, one can show e.g. that the systole of $\Gamma_0(N!)$ goes to $\infty$. | |
| Jan 15, 2012 at 15:54 | comment | added | Vitali Kapovitch | @Agol Thanks, yes, this seems unlikely but it would be nice to have a proof. I can't come up with one myself. But number theory is not my field and I can't even judge if this question is hard or not. | |
| Jan 15, 2012 at 7:25 | comment | added | Ian Agol | @Vitali: I think not. If the systole were uniformly bounded, then one can show that there is a finite list of numbers, such that a least one number is among the list is a quadratic residue (mod N) for all N. But I think this is impossible. | |
| Jan 14, 2012 at 21:58 | comment | added | Vitali Kapovitch | @Agol: Is the limsup of the length of the shortest hyperbolic element of $\Gamma_0(N)$ finite? your example shows that the liminf is. | |
| Jan 14, 2012 at 21:08 | history | edited | Ian Agol | CC BY-SA 3.0 |
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| Jan 14, 2012 at 18:06 | comment | added | Ian Agol | I made a mistake, the systoles of $\Gamma_0(N)$ can remain bounded. | |
| Jan 14, 2012 at 18:06 | history | edited | Ian Agol | CC BY-SA 3.0 |
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| Jan 14, 2012 at 17:11 | comment | added | Marc Palm | This is perfect, so like kassabov points out $\approx \log N$ | |
| Jan 14, 2012 at 17:10 | vote | accept | Marc Palm | ||
| Jan 14, 2012 at 16:37 | history | edited | Ian Agol | CC BY-SA 3.0 |
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| Jan 14, 2012 at 16:23 | history | answered | Ian Agol | CC BY-SA 3.0 |