6
$\begingroup$

Let $f: Y\to X$ be a finite separable morphism of curves over the finite field $\mathbb{F}_q$. Is there a simple condition under which the arithmetic and geometric monodromy of the covering are equal (i.e. $\mathbb{F}_q$ is algebraically closed in the Galois closure of the extension $\mathbb{F}_q(Y)/\mathbb{F}_q(X)$)?

It is relatively easy to construct examples for which that condition is not satisfied: for example, a Kummer covering of $\mathbb{P}^1$ of degree $n$ coprime to $q-1$. However, I have the vague feeling that for many "natural" coverings I come across, arithmetic and geometric monodromy are indeed equal (although I'm not sure how I'd define "natural" here). Is it possible to give a simple criterion? I'm especially interested in the case when $X$ is an elliptic curve (and if that helps, the characteristic of $\mathbb{F}_q$ is much larger than the degree/genus of everything involved).

$\endgroup$
5
  • 2
    $\begingroup$ "... the function field $\mathbb{F}_q(X)$ is algebraically closed in the Galois closure of the extension $\mathbb{F}_q(Y)/\mathbb{F}_q(X)$" I do not know what this is supposed to say, but this cannot be right: the algebraic closure of $\mathbb{F}_q(X)$ will certainly contain the algebraic extension $\mathbb{F}_q(Y)$. Do you want to say that there is a finite Galois extension $K/\mathbb{F}_q(X)$ generated by $\mathbb{F}_q(Y)/\mathbb{F}_q(X)$ such that $K$ is linearly disjoint from $\overline{\mathbb{F}}_q$ in $K^{\text{sep}}$? $\endgroup$ – Jason Starr Feb 20 '14 at 18:14
  • 2
    $\begingroup$ @JasonStarr — A field $K$ is algebraically closed in an overfield $M$ if every algebraic subextension $K \subset L \subset M$ is equal to $K$ (i.e., $K = L$). $\endgroup$ – jmc Feb 20 '14 at 19:49
  • $\begingroup$ Do you have a reference why this (the function field $\mathbb{F}_{q}(X)$ is algebraically closed in the Galois closure of the extension $\mathbb{F}_{q}(Y)/\mathbb{F}_{q}(X)$) implies that arithmetic and geometric monodromy are equal? If so, maybe you can add it to the question. (I understand that this depends on definitions...) $\endgroup$ – jmc Feb 20 '14 at 19:55
  • 1
    $\begingroup$ @jmc: I know the meaning of "$K$ is algebraically closed in $M$". However, what the OP wrote in the post does not make sense. $\endgroup$ – Jason Starr Feb 20 '14 at 23:50
  • $\begingroup$ Indeed, I obviously wasn't thinking clearly when I wrote that. Edited. $\endgroup$ – Mehdi Tibouchi Feb 20 '14 at 23:54
6
$\begingroup$

I have worked on reasonably similar problems to this, so hopefully what I have to say will be of some use.

In general there is no easy way to do this, as far as I know. There are techniques for proving this kind of statement, and you could phrase an individual technique as a criterion, but it would be unhelpful, because most covers won't satisfy the criterion. This is the sort of statement that is usually true, but can be false for arbitrarily subtle reasons.

That said, here is how a typical proof of this might look:

First, lower bound the geometric monodromy group. The most powerful technique to do this is by evaluating the local monodromy at each ramification point. This is usually much easier to understand then the full monodromy group, and gives you conjugacy classes in the group. You can also use more global facts about the cover - if you can prove it is geometrically irreducible, for instance, then the monodromy must be transitive.

Second, upper bound the arithmetic monodromy group. If there are any automorphisms of your cover defined over $\mathbb F_q$, the arithmetic monodromy must commute with them. If your cover is defined by adding some extra structure to a moduli space, then the arithmetic monodromy group must act on that structure.

If there are any intermediate curves you know exist between $X$ and $Y$, you can use those to both lower bound and upper bound the groups. This might follow from the way they are constructed. Can you compute the discriminant?

Third, use group theory to close the gap between the two groups. The key fact here is that the geometric monodromy group is a normal subgroup of the arithmetic monodromy group, with cyclic quotient. You hope to prove that no pair of groups satisfying all the conditions you set up beforehand can exist. Alternately, prove that one can exist, and try to come up with clever new arguments to rule them out!

This advice might not be too helpful, because it basically says "solve this math problem like you would any other math problem, using math," but I hope that something I said here clicks with something about the "natural" coverings you are defining.

$\endgroup$
3
  • 2
    $\begingroup$ Another way to get bounds is to determine the irreducible components of the fibered product $Y\times_X Y$. If the diagonal and its complement are both geometrically irreducible, then the geometric monodromy group is doubly transitive (and then often the local monodromies will force the group to be the full symmetric group). If the fibered product has more than two irreducible components over $\mathbf{F}_q$ then one gets an upper bound on the arithmetic monodromy group, namely the group of permutations in $S_n$ which preserve the partition of the $n$ letters which corresponds to...[cont.] $\endgroup$ – Michael Zieve Feb 21 '14 at 3:57
  • 2
    $\begingroup$ [cont.] the decomposition of the fibered product into components. This is most useful if there are no intermediate curves between $X$ and $Y$, since then the arithmetic monodromy group is primitive but not doubly transitive, which is a big constraint on the group. For instance, if the fibered product has just three components then the arithmetic group is primitive of rank three, and such groups have been classified. $\endgroup$ – Michael Zieve Feb 21 '14 at 4:00
  • $\begingroup$ Oops -- at the end of my first comment, I meant the group of permutations in $S_n$ which preserve the partition of $\{1,2,\dots,n\}\times\{1,2,\dots,n\}$ which corresponds to the decomposition of the fibered product into components. $\endgroup$ – Michael Zieve Feb 21 '14 at 5:14
2
$\begingroup$

I think generically the geometric monodromy group is the symmetric group $S_n$, acting on the $n=[\mathbb F_q(Y):\mathbb F_q(X)]$ conjugates of $\mathbb F_q(Y)$. But the arithmetic monodromy group acts faithfully on the same set, so it can't be bigger.

@JasonStarr: I believe the OP wanted to say that $\mathbb F_q$ is algebraically closed in the Galois closure of $\mathbb F_q(Y)/\mathbb F_q(X)$ if and only if the two groups are equal.

$\endgroup$
2
  • $\begingroup$ "I believe the OP wanted to say that $\mathbb{F}_q$ is algebraically closed ..." That's what I think as well. I believe that is equivalent to the statement that the Galois closure is linearly disjoint from $\overline{\mathbb{F}}_q$ inside $\mathbb{F}_q(Y)^{\text{sep}}$. $\endgroup$ – Jason Starr Feb 20 '14 at 23:49
  • $\begingroup$ @Peter Mueller: That's a good point, thanks. However, the collection of those "natural" coverings I come across does contain examples with non-generic monodromy. Let me try to think some more about what those "natural" examples look like. $\endgroup$ – Mehdi Tibouchi Feb 21 '14 at 0:02

Your Answer

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

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