Timeline for Motivic fundamental group of the moduli space of curves?
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8 events
| when toggle format | what | by | license | comment | |
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| Aug 29, 2015 at 15:16 | comment | added | Will Sawin | @BenWieland To show a monodromy group is large, I can work with a particular family of curves and show that has large monodromy. To show the monodromy of a representation is large, I just need to show that the invariants of tensor powers of this representation are small. This is controlled by expected value of powers of the trace of Frobenius acting on the representation of a random point on that family of curves over a finite field. This type of thing can often be controlled with standard methods from probability theory. | |
| Aug 28, 2015 at 19:37 | comment | added | Ben Wieland | I'm not familiar with the moment method, but it sounds pretty difficult to apply. Doesn't it require you to know about the cohomology of all curves? Does it amount to: if most curves have the maximal MT group, then the monodromy is the MT group? But don't other results say that you just need a single curve with the maximal MT group? | |
| Aug 15, 2015 at 12:32 | comment | added | Donu Arapura | Will, no I would agree that these examples aren't exotic. The pure graded part of the moduli of semistable bundles come from $Sp_{2g}$ as you surmised. | |
| Aug 14, 2015 at 22:30 | comment | added | Will Sawin | @DonuArapura Yes, I agree. There should be no problem showing that these Prym-type families have big monodromy either by topology or moment methods. This is an example of what I asked for, but it still doesn't feel "exotic" to me, so probably I should have included this in the "expected" part. For the space of semistable vector bundles, is it the case that the irreducible factors come from SP_{2g} but the extension classes don't? | |
| Aug 14, 2015 at 22:25 | comment | added | Will Sawin | @DanielLitt Good point. I guess I tend to think of the cohomology as a Galois representation rather than a Hodge structure. | |
| Aug 14, 2015 at 17:57 | comment | added | Donu Arapura | Actually, I think that the reductive part of $\pi_1^{mot}(M_g)$ would bigger than $Sp_{2g}$. You can see my answer to mathoverflow.net/questions/186133/… for why I think so. | |
| Aug 14, 2015 at 17:52 | comment | added | Daniel Litt | Nice question! In a weak sense, the answer is "yes" by Torelli, since the (polarized) cohomology knows the curve, hence whatever you construct from it. I don't see how to get at the formalization you ask about (which I would say is more like "can we compute the cohomology of the new variety from the cohomology of the curve, via linear algebra operations?"). | |
| Aug 14, 2015 at 16:58 | history | asked | Will Sawin | CC BY-SA 3.0 |