Timeline for Which mapping class group representations come from algebraic geometry?
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5 events
| when toggle format | what | by | license | comment | |
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| Jul 27, 2022 at 18:25 | comment | added | Ben Wieland | To put it another way, if you have a representation of $M_{g,n}$, you can take the cohomology of the kernel (equivalently, cohomology of the configuration space fibers) to get a representation of $M_g$. If you do this with the trivial representation, you get something close to the standard representation. Kontsevich proposes doing this with the Prym representation corresponding to a double cover with $n=4g-4$ ramification points. There isn't a canonical Prym representation, but it's only defined on sufficiently small finite index subgroups. But take any of those and induce it up to $M_{g,n}$. | |
| Nov 5, 2014 at 4:43 | comment | added | Ian Agol | It would be interesting if you could flesh out the details of this approach. | |
| Nov 5, 2014 at 2:57 | comment | added | Ben Wieland | Yes, it works. There are two difficulties. The first is that there are lots of choices; you just have to make all of them. That is: take the universal curve over the moduli stack of double covers of a genus $g$ curve ramified in $n=4g-4$ points. This maps to $M_g$, so its cohomology is a local system there. The second problem is the stackiness: when you take cohomology, it's like taking $\mathbb Z/2$ invariants, but the interesting cohomology all has action by $-1$. But you can just take the tensor square, or some other ad hoc option. | |
| Nov 5, 2014 at 2:37 | comment | added | Ryan Budney | Is it clear they're actually representations of the mapping class group? I think I might have asked Kontsevich back in 2006 and I recall he shrugged. | |
| Nov 5, 2014 at 0:33 | history | answered | Ben Wieland | CC BY-SA 3.0 |