Timeline for The congruence subgroup property for mapping class groups and a conjecture of Grothendieck
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 28 at 16:50 | comment | added | HJRW | @BenWieland: Sorry for the slow response. This sounds interesting, but it's quite quick in a comment! I would be happy to hear more details, explained at greater length. | |
| Oct 15 at 15:13 | comment | added | Ben Wieland | There is a sequence of maps $\widehat{M_g}\rtimes G\to Out(\hat\pi_g)\rtimes G\to G$. The section conjecture and the Grothendieck conjecture are about sections of these two maps. If they are both true, then sections to Out lift in at most one way to sections to M. What I meant is that this is surprising and the obvious way for it to be true is if the groups are the same. But there is a similar situation with $M_{1,1}$: $\widehat{GL_2(Z)}\rtimes G\to GL_2\hat Z\rtimes G\to G$ where sections lift in at most one way to (although usually don't lift at all). | |
| Oct 1 at 14:55 | history | edited | HJRW | CC BY-SA 4.0 |
Added sharpened questions
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| Sep 20 at 19:51 | answer | added | coLaideronnette | timeline score: 7 | |
| Sep 20 at 9:54 | comment | added | HJRW | @BenWieland: Thanks, this is also interesting. Is the connection between the Grothendieck conjecture, the section conjecture and CSP written down anywhere? | |
| Sep 19 at 19:42 | comment | added | Ben Wieland | CSP is exactly what you would want to use to prove the Grothendieck conjecture from the section conjecture for $M_g$, to lift an action on the fundamental group to an action through the MCG. This seems silly because the section conjecture is generally harder, but maybe you can use the section conjecture for affine curves to prove it for $M_g$ to prove the Grothendieck conjecture for closed curves? Also, can you run this backwards to prove CSP? | |
| Sep 19 at 12:49 | comment | added | Will Sawin | Yeah, the coverings I'm considering are stacks over $\mathbb Q$ whose $\mathbb C$-points are orbifold covers of $\mathcal M_g$. Depending on the covering the $\mathbb Q$-structure may or may not be unique. | |
| Sep 19 at 7:56 | comment | added | HJRW | @WillSawin: thanks, this sounds like a start. Excuse my ignorance: when you say one rational point lifts to a cover but another one doesn't, what does this mean? Should I picture an orbifold cover of $\mathcal{M}_g$ (over $\mathbb{C}$), for which one point has rational preimages and the other doesn't? I share your puzzlement: there are very few methods of generating covers of $\mathcal{M}_g$ that aren't a priori congruence (which is why the CSP is hard). | |
| Sep 19 at 0:36 | comment | added | Will Sawin | This seems to rely on the claim that for every two different $\mathbb Q$-points of $\mathcal M_g$, there is a finite \'{e}tale cover of $\mathcal M_g$ that separates them (for example, to which one of them lifts but the other does not). What confuses me here is that I can't think of a method to prove this without using congruence subgroups (i.e. without proving the Grothendieck conjecture directly). | |
| Sep 18 at 18:50 | history | asked | HJRW | CC BY-SA 4.0 |