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Aug 27, 2019 at 20:21 history bounty ended Zhiyu
Aug 23, 2019 at 21:44 comment added Will Sawin @user45878 Possibly one can deduce it from Poincare duality (reducing it to giving a lower bound on the weights) and the fact that the Frobenius eigenvalues on the compactly supported cohomology of a variety must be integral (hence have weight at least zero). In the geometric setting, you can spread out to a punctured curve and compute the compactly supported cohomology of the total space, and you'll see the local monodromy invariant parts at the puncture show up.
Aug 23, 2019 at 21:21 comment added Pol van Hoften Is there an easy way to see that the weights must be bounded by 2n (say without the weight spectral sequence)?
Aug 23, 2019 at 20:58 comment added Will Sawin @sawdada Another method is to use purity considerations - we can take $E$ to be regular, meaning the pushforward is a pure complex, which forces skyscraper components to have the expected weight. But if you like the Tate curve, one could write down a $\mathbb Z/\ell$-torsor over the special fiber, extend it to the generic point, and note that it remains a nontrivial torsor. In fact you should obtain the covering of $\mathbb G_m/q$ by $\mathbb G_m / q^\ell$.
Aug 23, 2019 at 20:43 comment added Zhiyu Thanks for a good example by taking product, is there an easy way to see $H^1(E_s) \rightarrow H^1(E_{\eta})$ is injective (without applying Picard-Lefschetz formula)? The Galois action on $H^1(E_s)$ and $H^1(E_{\eta})$ is easy to compute using Tate curve and restriction exact sequence.
Aug 23, 2019 at 19:42 history answered Will Sawin CC BY-SA 4.0