Timeline for Finite covers of hyperbolic surfaces and the `second systole´
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 10, 2017 at 21:10 | history | edited | HJRW | CC BY-SA 3.0 |
Minor edit.
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| Mar 17, 2017 at 10:06 | history | edited | HJRW | CC BY-SA 3.0 |
Majore edit to correct proof.
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| Mar 16, 2017 at 12:50 | comment | added | Misha | @HJRW: One does not need the orbifold theorem for this, just Maskit combination theorem for the amalgam of two Fuchsian groups, which is quite simple. As for the higher dimensional statement, what I meant is that I think the result is a hyperbolic orbifold group, in which case the group is residually finite because it is linear. | |
| Mar 16, 2017 at 9:22 | comment | added | HJRW | @Misha, I would say that the orbifold theorem is a much bigger hammer than the residual finiteness gluing result I quoted! But I would be very interested in a reference for the fact that killing high powers of nonsimple loops gives orbifolds. (In this case, if they're higher dimensional, is there any reason to think that they're residually finite? In fact, they are residually finite by Agol--Wise, but a proof avoiding cubical machinery would be interesting.) | |
| Mar 16, 2017 at 9:19 | history | edited | HJRW | CC BY-SA 3.0 |
Corrected Lemma 2 and added a detail to the proof.
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| Mar 15, 2017 at 21:16 | comment | added | HJRW | @IanAgol, sorry, yes, it only needs to be true for $r_N(g)$ not a power of $r_N(\alpha)$. I'll edit in the morning. For specific examples, Frobenius complements, for instance, can be used and are overkill. | |
| Mar 15, 2017 at 19:45 | comment | added | Ian Agol | @HJRW: I'm confused about Lemma 2. If $r_N(g)$ centralizes $\alpha$, then this won't be true (assuming $\langle r_N(\alpha)\rangle \neq \langle r_N(\alpha)\rangle^N$, which you seem to be assuming given the later argument). So it seems to me that you need to make some extra assumption here, or maybe I am misunderstanding the quantifiers. Anyway, a reference to this fact about finite groups would be helpful. | |
| Mar 15, 2017 at 16:36 | comment | added | rpotrie | I think I understood. So, the strategy is to first lift to a finite cover that opens every curve but $\alpha$ may have several preimages. It is in this step that the residual finiteness of the quotient is used. Then, one chooses a lift that opens all but one preimage of $\alpha$. Thanks! | |
| Mar 15, 2017 at 16:35 | vote | accept | rpotrie | ||
| Mar 15, 2017 at 15:48 | comment | added | Misha | @HJRW: Very good. The first step can be a bit simplified (depending on one's taste) using the fact that the quotient of $\pi_1(S)$ by $<<\gamma^n>>$, where $\gamma$ is a simple loop, is the fundamental group of a 3-dimensional hyperbolic orbifold and, hence, is residually finite. I do not know if a similar thing is true for nonsimple loops (where instead of 3-d hyperbolic orbifolds one uses higher-dimensional ones). I think, it is true if n is large enough. | |
| Mar 15, 2017 at 14:58 | comment | added | HJRW | The general principle is that if you can kill (a power of) the element that you're interested in and obtain something that's still residually finite, then you are in good shape. In this case, we can do it explicitly because the curve you are interested in is simple and so the quotient is an orbispace. Wise's Malnormal Special Quotient Theorem can be used in much more general situations. | |
| Mar 15, 2017 at 11:25 | comment | added | rpotrie | Thanks. I need to sit on this for a while to see if I understand it, but looks like uses things we had not considered. | |
| Mar 15, 2017 at 9:55 | history | answered | HJRW | CC BY-SA 3.0 |