MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I make the following sort of argument work in characteristic p?

Let $f:X \to Y$ be a proper equidimensional map of smooth algebraic varieties, assume all fibres are reduced. Say at some point $y \in Y$, I have computed the differential and know that $df(T_x X) \supset V$ for all $x \in X_y$ and some vector subspace $V \subset T_y Y$. Then over a sufficiently small polydisc $D$ such that $T_y D + V = T_y Y$, the total space of $X_D$ is smooth. Thus the same is true of a (say) one-dimensional thickening $\tilde{D}$ of $D$, and so moving $D$ a little bit in $\tilde{D}$ produces many deformation equivalent smooth manifolds (with boundary) $X_{D'}$.

In particular if I want to know something about $H^*(X_y)$, I can first thicken to the smooth $X_D$ which is homotopy equivalent to it, then move $X_D$ to the diffeomorphic $X_{D'}$, which I prefer because say $D'$ avoids some bad points in $Y$.

share|cite|improve this question
Are you looking for a nearby cycles functor in étale cohomology? – S. Carnahan Jun 23 '11 at 9:31
I don't know, you tell me... – Vivek Shende Jun 23 '11 at 12:03
Vivek -- if you're still in London then you should find Wansu Kim and ask him. He organised a study group on nearby cycles last term. – Kevin Buzzard Jun 24 '11 at 18:04
up vote 10 down vote accepted

I think it would be difficult to give a general result that covers everything you want and can do but there is a collection of techniques (maybe better described as a dictionary) that works in many cases.

  • A polydisc should be replaced by the strict Henselisation of some (smooth) subvariety $T$ passing through $y$. The fact that $X_y$ is homotopy equivalent to $X_D$ is then replaced by the proper base change theorem.
  • The transversality property has a direct analogue in that one may arrange that $T$ is transversal to $f$ and hence that $X_T$ is smooth.
  • Deformation of $D$ is replaced by choosing, locally on $Y$, a smooth map $Y\to Z$ such that $T$ is one of the fibres. To get closer to the topological situation one should probably also assume that a section has been chosen. Deformations correspond to fibres over other other points of $Z$ (or rather the Henselisation at the corresponding point of the section).
  • To compare (the analogues of) $X_D$ and $X_{D'}$ one probably also would need the smooth base change theorem together with the proper base change theorem.

As I said to see how this works in any particular situation one would need to have more details on it, no a priori success is guaranteed.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.