Let $X$ be a scheme proper and flat over a complete discrete valuation ring $O$ with finite residue field $k$, and choose a prime $l$ not equal to characteristic of $k$. Consider the Galois representation $H^i_{et}(X_{k^{alg}}, R\Phi \mathbb Q_l)$ where $R\Phi \mathbb Q_l$ denotes the vanishing cycle, are all of its weights less than $i+1$? Assume the generic fiber $X_K$ is proper and smooth, do we know that $H^i_{et}(X_{k^{alg}}, R\Phi \mathbb Q_l)$ is always pure of some weight?

A theorem of Gabber says that $R\Phi \mathbb Q_l$ is perverse (up to shift), maybe this is helpful.

Example: when $X_k$ has ordinary quadratic singularity, then this is true by Picard-Lefschetz formula.

  • $\begingroup$ No, the cohomology of $R \Phi \mathbb{Q}_l$ wouldn't be pure in general. The vanishing cycle sheaf is mixed. It carries a monodromy weight filtration with respect to which the associated graded is pure. Look at Weil II for more details. $\endgroup$ – Donu Arapura Aug 23 '19 at 12:44
  • $\begingroup$ Added: Weil II only talks about the equicharacteristic case. I don't know the state of the art in general. $\endgroup$ – Donu Arapura Aug 23 '19 at 12:50
  • $\begingroup$ No to both questions, but I think it should be true that $H^i$ always has weights $\le 2i$. $\endgroup$ – user45878 Aug 23 '19 at 16:50
  • $\begingroup$ @DonuArapura Thank you! Do you know any example that the vanishing cycle sheaf is not pure? $\endgroup$ – sawdada Aug 23 '19 at 18:35
  • $\begingroup$ @user45878 Can you give a reference? $\endgroup$ – sawdada Aug 23 '19 at 18:37

Here's a typical example of what can go wrong in high dimension:

Let $\mathcal E$ be an elliptic curve degenerating to a nodal cubic. Let $X = \mathcal E^n$.

Then $H^* ( X_{\eta}, \mathbb Q_\ell ) = H^* (E_{\eta}, \mathbb Q_\ell)^{\otimes n}$ where the most interesting piece $H^1( E_{\eta}, \mathbb Q_\ell)$ is a two-dimensional representation with weights $0$ and $2$. Similarly $H^* ( X_{s}, \mathbb Q_\ell ) = H^* (E_{s}, \mathbb Q_\ell)^{\otimes n}$ where $H^1( E_{s}, \mathbb Q_\ell)$ is one-dimensional and has weight zero.

The specialization map $H^*( X_s, \mathbb Q_\ell) \to H^*(X_\eta, \mathbb Q_\ell)$ will be the $n$th tensor power of the specialization map for $\mathcal E$ and in particular will be injective. So the cohomology of vanishing cycles, which is the mapping cone of this map, will just be the cokernel.

Using this, you can see that the weights do all kinds of crazy things and that the weights in degree $n$ go all the way up to $2n$, not just $n+1$.

  • $\begingroup$ 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. $\endgroup$ – sawdada Aug 23 '19 at 20:43
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    $\begingroup$ @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 $\mathbb G_m / q^\ell$. $\endgroup$ – Will Sawin Aug 23 '19 at 20:58
  • $\begingroup$ Is there an easy way to see that the weights must be bounded by 2n (say without the weight spectral sequence)? $\endgroup$ – user45878 Aug 23 '19 at 21:21
  • $\begingroup$ @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. $\endgroup$ – Will Sawin Aug 23 '19 at 21:44

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