2
$\begingroup$

Let Y be a smooth projective curve over C and prescribe a branch divisor B on Y. I want to know if the number of coverings of Y of fixed degree and branched along B is finite. If so, why? Or, where is the reference for this? I was told it is claimed in an article by Manin ("On branched coverings of algebraic curves", Izv. Akad. Nauk SSSR Ser. Mat., 1961, vol. 25, 789-796), in the last sentence of the paper. My problem is that the article is in Russian (I think it is not translated) and, unfortunately, I can't speak Russian, so I can't check it. Also, I don't know for what characteristic Manin's paper results are. I assume the claim should follow from the theorem stated in the first page (in Section 1), which is about field extensions, but, as I said, I am not sure about the field characteristic. If someone knowing Russian can at least tell me what the statement of the theorem and the last sentence of the article say, I think it would already help me a lot!

$\endgroup$
6
$\begingroup$

This can be understood by looking at fundamental groups.

Covers of $Y$ branched only along $B$ of degree $n$ correspond to homomorphisms $\pi_1(Y-B)\rightarrow S_n$ ($S_n$ is the symmetric group on $n$ letters). (The cover is connected if and only if the image is a transitive subgroup, and the cover is Galois iff the image is transitive of order $n$)

Here, isomorphism classes of covers corresponds to conjugacy classes of the homomorphism by elements of $S_n$.

(This follows from standard results which you can find in Hatcher's "Algebraic Topology" online book)

It's well known that the fundamental groups of curves are finitely generated - if $g$ is the genus of $Y$, then $\pi_1(Y-B)$ is can be generated by $2g + |B|$ elements. (Roughly speaking, the $2g$ generators come from the global topology of $Y$, and the $|B|$ additional generators come from small loops around the points in $B$)

Thus, covers of $Y$ branched only along $B$ of degree $n$ are in bijection with conjugacy classes of homomorphisms from a finitely generated group to a finite group, of which there can be only finitely many.

In the above, technically a "cover" refers to a covering space in the sense of topology, but by the Riemann Existence Theorem, the category of finite etale covers of the algebraic curve is equivalent to the category of finite topological covers of its analytification.

$\endgroup$
  • $\begingroup$ It is not clear that you are answering the question in the same category as the OP is asking it... $\endgroup$ – Igor Rivin Mar 18 '17 at 1:39
  • $\begingroup$ @IgorRivin Sorry, I don't understand what you mean. What sort of answer do you think the OP was looking for? $\endgroup$ – Will Chen Mar 18 '17 at 21:08
  • $\begingroup$ The OP is asking about algebraic curves, so the notion of equivalence might (in principle) be different than equivalence for topological branched covers [it is not, but this needs to be addressed]. $\endgroup$ – Igor Rivin Mar 18 '17 at 21:31
  • 3
    $\begingroup$ It is perhaps worth remarking that this argument nicely points to why the corresponding result fails in characteristic $p$. Namely, $\pi_1(Y - B)$ is not (topologically) finitely generated unless $B$ is empty. For example, $t^p - t = x^m$ ($m$ prime to $p$) is an infinite family of distinct degree $p$ covers of $\mathbb{P}^1$ branched only at infinity. (They can be distinguished by their Swan conductors). Everything should be okay in arbitrary characteristic if we only consider tame covers. $\endgroup$ – Piotr Achinger Mar 18 '17 at 22:05
  • $\begingroup$ Thanks Oxeimon and the others, I think this settles my question. Still, I am interested in the contents of the theorem in the first page (1 Teorema) of Manin's article I refered to previously and in the last sentence of the paper. If someone out there could help in translating those twothings, I will greatly appreciate it! The article is "On branched coverings of algebraic curves", Izv. Akad. Nauk SSSR Ser. Mat., 1961, vol. 25, 789-796 mathnet.ru/php/… $\endgroup$ – Enoch Mar 22 '17 at 20:07

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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