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Jun 25, 2020 at 17:26 answer added Will Sawin timeline score: 3
Jun 25, 2020 at 17:01 comment added Daniil Rudenko I see, thank you.
Jun 25, 2020 at 17:00 comment added Will Sawin You want the Hodge structure to be mixed Tate type then? It is for $n=4$, because $\mathcal M_{0,4}^s $ is $ \mathbb P^1$ minus six points. Probably is for $n=5$ as well. I guess this should stop at some point though, but I don't know.
Jun 25, 2020 at 16:41 comment added Daniil Rudenko Thank you! I guess that I was implicitly working over C.
Jun 25, 2020 at 16:39 comment added Will Sawin This definition doesn't define a finite cover of $\mathcal M_{0,n}$ over $\mathbb Q$ - you need to pick a base point. However, it's possible to check, regardless of the base point, that the cohomology of $\mathcal M_{0,4}^s$ is Artin-Tate, by just bashing it out.
Jun 25, 2020 at 15:28 history asked Daniil Rudenko CC BY-SA 4.0