Suppose I have a proper, flat family of curves $X \to S$ that has a section. Fix a basepoint $s \in S$ and let $X_s$ denote the corresponding fiber. Let $\mathbb{L}$ be a set of primes which does not contain any of the residue characteristics of $S$. Let $\pi_1(X_s)^{\mathbb{L}}$ be the pro-$\mathbb{L}$ quotient of the etale fundamental group of $X_s$ (here for convenience I'm ignoring basepoints), and define $\pi_1(X)'$ as follows: if $N$ is the largest normal subgroup of the kernel $K$ of $\pi_1(X) \to \pi_1(S, s)$ such that $K/N$ is pro-$\mathbb{L}$, then $\pi_1(X)' = \pi_1(X) / N$. Then clearly we have maps $\pi_1(X_s)^{\mathbb{L}} \to \pi_1(X)' \to \pi_1(S, s)$ whose composition is trivial. Now a result of Grothendieck (SGA1, XIII.4.3) states that, under certain technical conditions on the morphism $X \to S$ (including being 0-acyclic and locally 1-aspheric), these maps form a short exact sequence $$1 \to \pi_1(X_s)^{\mathbb{L}} \to \pi_1(X)' \to \pi_1(S, s) \to 1.$$ (See When can the "homotopy exact sequence" of etale fundamental groups for a smooth curve fail to be exact? for a similar question regarding this result.) It is known that smoothness of $X \to S$ is enough to imply these technical conditions (SGA1, XIII.4.4).

What I want to ask, before trying to work my way through all the technicalities in understanding what kinds of properties imply and are implied by 0-acyclic and 1-aspheric, is this: Does anyone know for what kinds of non-smooth families $X \to S$ does this short exactness still hold? In particular, do we still get short exactness if $X \to S$ has a semistable fiber $X_t$ with no vanishing cycles, i.e. such that the Jacobian of $X_{\eta}$ (where $t \in S$ is a specializataion of $\eta \in S$) has good reduction at $t$? (From what I do understand of the definitions of 0-acyclic and 1-aspheric, these are conditions involving cohomology of sheaves of abelian $\mathbb{L}$-groups, which seems vaguely promising since such mild degeneracy shouldn't be visible in cohomology, yes?)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.