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Let $X\to S$ be a morphism of smooth connected varieties over an algebraically closed field $k$; let $j:\eta\to S$ be the inclusion of the generic point into $S$ (not a geometric generic point) and let $\mathscr F$ be a constructible étale sheaf of $\Lambda$-modules on $X_\eta$. If $\Lambda=\mathbf{Z}/\ell^n\mathbf{Z}$, for example, then (letting $j$ also denote its base extension $X_\eta\to X$), $j_*\mathscr F$ is a constructible sheaf. But if $\Lambda=\mathbf Z_\ell$ or $\mathbf Q_\ell$, is $j_*\mathscr F$ still constructible? When I think about applying $j_*$ to a projective system (of constructible sheaves of modules for finite coefficients), the result is a projective system of constructible sheaves. When I think about it when $X=S$ is a smooth curve and think about $\mathscr F$ as a $\mathbf Q_\ell$-representation $V$ of $\operatorname{Gal}(\overline\eta/\eta)$, however, then the condition is equivalent to the condition that the inertia of all but finitely many points of $S$ act trivially on $V$—this seems dodgy. Thanks for your help.

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A necessary condition is that for each stratum $X_{\eta}^i$ of some stratification of $X_{\eta}$ on which $\mathcal F$ is lisse, the associated representation of $\pi_1(X_{\eta}^i)$ is unramified away from a closed subset $Z^i$ of the closure $\overline{X_\eta^i}$ of $X_{\eta}^i$ in $X$, such that the projection of $Z_i$ to $S$ is not dense.

The fact that this condition is necessary is clear - if the pushforward is constructible, it's lisse on a stratification, and then intersecting each stratum with the generic fiber, you get a stratification with this property.

This can indeed fail, so the very general statement you wrote is not true. Your idea is precisely the right one - it's easy to construct a Galois representation that is not finitely ramified, for instance by constructing a suitable extension of $\mathbb Z_\ell$ by $\mathbb Z_\ell(1)$ using Kummer theory.

What will go wrong in the limiting proof you want to write down is that the pushforward of the associated $\mathbb Z/\ell^n$ pushforward, modulo $\ell$, will not be the pushforward of the associated $\mathbb Z_\ell$-sheaf, and will not even stabilize - it will be a sheaf with zero stalk at more and more ramification points as $n$ goes to $\infty$.

I think this condition is also suffiicent, but I didn't check it outside the case where $X =S$, where it can be proven by first pushing forward to the open set where the Galois representation is unramified, and then pushing tot he whole space.

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  • $\begingroup$ Kudos, Will; thanks! $\endgroup$ – delgato Apr 6 at 20:24

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