Timeline for Do mapping classes have gonality?
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14 events
| when toggle format | what | by | license | comment | |
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| Apr 27, 2014 at 21:10 | answer | added | IMeasy | timeline score: 6 | |
| Jul 30, 2011 at 1:45 | comment | added | JSE | Yes, that's right -- Joe Harris explained this to me at PCMI -- you can see that pi_1(U_g) can't actually be the SAME as pi_1(M_g) because it doesn't have enough finite subgroups (there are lots of finite groups that can be Aut(G) for a genus-g curve but not for a trigonal genus-g curve.) So injective implies not surjective. | |
| Jul 29, 2011 at 23:16 | comment | added | Ian Agol | Ok, that helps clarify things. One issue that makes this trickier than the hyperelliptic case is that the hyperelliptic maps of moduli spaces are $\pi_1$-injective (in fact, give isometric embeddings), whereas the $U_g$ map is not an isometric embedding (although I'm not sure if anything is in the kernel of $\pi_1$). If it were $\pi_1$-injective, then I would suspect it couldn't be surjective. | |
| Jul 29, 2011 at 1:39 | comment | added | JSE | Agol -- yeah, that's why I hedged on "equivalent." Two comments. 1. If the genus is at all large, the trigonal map is unique; 2. The trigonal genus-g curves which are simply branched (all monodromy of order 2) are parametrized an open subvariety U_g of the whole trigonal locus T_g; the map from pi_1(U_g) to pi_1(M_g) factors through pi_1(T_g) -> pi_(M_g) so if the former is surjective, the latter is too. | |
| Jul 29, 2011 at 0:18 | comment | added | Ian Agol | Also, to clarify, the trigonal locus is a map of a finite cover of moduli space of an n-punctured sphere (decorated by a homomorphism to S_3) to the moduli space of a genus g surface. This map might not be an embedding, there may be multiple ways that certain surfaces may branch. Also, according to my other comment, there may be multiple components of the moduli space of branchings. I was wondering if you wanted to consider the image in the mapping class group of the corresponding cover of moduli space, or do you want to consider the fundamental group of the total trigonal locus? | |
| Jul 29, 2011 at 0:14 | comment | added | Ian Agol | I'm not sure that your characterization is quite right. A branched cover is a cover in all but a finite number of points. Removing these points, you get a cover of a punctured sphere. This is determined by a homomorphism to $S_3$, however I don't see why some punctures can't map to an element of order 3 (clearly they can't be trivial, or the cover wouldn't branch over those points)? | |
| Jul 28, 2011 at 13:58 | comment | added | Jason Starr | I am not quite sure what the caution is about. Of course the MHS on the fundamental group of a variety depends on the variety. Even if two different varieties have the same fundamental group, the associated MHS's can be different. I guess the real question is whether the MHS on the fundamental group of a variety is covariant for morphisms of varieties. If so, then since the trigonal locus is unirational (rational?), its MHS should be "special". And you can use that to try to prove it cannot surject onto the MHS of $M_g$. | |
| Jul 28, 2011 at 3:49 | comment | added | algori | Jason -- some caution is needed when talking about MHS's on fundamental groups. For example, $\mathbb{Z}\times\mathbb{Z}$ is the fundamental group of $\mathbb{C}^*\times \mathbb{C}^*$ and also of an elliptic curve. | |
| Jul 26, 2011 at 12:46 | comment | added | Jason Starr | Don't fundamental groups of algebraic varieties carry mixed Hodge structures? If so, perhaps you can use the fact that the hyperelliptic and trigonal loci are Zariski open in projective rational varieties to show that these Hodge structures have "smaller coniveau" than the Hodge structure for all of $M_g$. | |
| Jul 26, 2011 at 10:47 | comment | added | algori | JSE -- thanks. The questions are indeed similar. Hopefully, yours will be luckier than the one I asked. | |
| Jul 26, 2011 at 3:40 | comment | added | JSE | algori -- you are very kind not to mention that in addition to my failure to TeX, I failed to notice that you already asked a morally equivalent question a year and a half ago! mathoverflow.net/questions/14179/… | |
| Jul 26, 2011 at 3:36 | comment | added | algori | JSE -- I've taken the liberty of latex'ing the unlatexed formulas. This is an interesting question, but it was a bit difficult to read. | |
| Jul 26, 2011 at 3:35 | history | edited | algori | CC BY-SA 3.0 |
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| Jul 26, 2011 at 2:50 | history | asked | JSE | CC BY-SA 3.0 |