Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let S and S' be closed [Edit: orientable] surfaces, then it is well known that for S' to cover S it is necessary and sufficient that chi(S)|chi(S'). (Here 'chi' denotes the Euler characteristic).

However, if S and S' are punctured surfaces then the above condition is necessary but no longer sufficient.

Is the question of determining necessary and sufficient criteria for S' to cover S answered in the research or expository literature?

I think that I know such criteria and want to use them a paper that I'm working on but (a) may be deceiving myself and (b) want to know whether I should write up a proof anew or whether there's a suitable reference. Surely the relevant criteria have been rediscovered many times, but I've never seen them discussed in writing.

Edit: Thanks Pete, I should have demanded that my surfaces be orientable

share|improve this question
Are you sure your criterion is correct in the closed case? Let S be the closed, nonorientable surface of Euler characteristic -2 and let S' be the closed orientable surface of Euler characteristic -6. Then $\chi(S')/\chi(S) = 3$, so your divisibility condition is satisfied, but since $3$ is odd, the covering map does not factor through the orientation cover (a degree 2 cover) and therefore passing to an odd degree cover cannot make a nonorientable surface orientable. –  Pete L. Clark Jan 23 '10 at 11:22
I had a previous answer which Dmitri Panov pointed out was wrong. Take II (as a comment this time): we start with an arbitrary branched covering of compact Riemann surfaces $S' \rightarrow S$. Then we remove a finite set of points on $S$ together with the complete preimage in $S'$. In order to get an unramified covering we must remove at least the branch locus from $S$; we can also remove any finite number of unbranched points if we wish. I think that every topological unramified cover $S' \rightarrow S$ arises in this way. Now we examine cases using Riemann-Hurwitz... –  Pete L. Clark Jan 23 '10 at 12:00
Anyway, I realize I am not answering your actual question: i.e., do I know a reference? Unfortunately no, sorry, although I agree that it must have been worked out dozens of times. –  Pete L. Clark Jan 23 '10 at 12:05
For a reference you might check out the literature on dessin d'enfants; I feel like I've seen such criteria discussed in that more combinatorial context. –  Tom Church Jan 23 '10 at 18:32
add comment

2 Answers

I don't know the reference for this question, but I am pretty sure that it should follow from some known statement. Anyway let me give the answer in the case when S has negative Euler characteristic, orientable and CONNECTED. At least you can compare with your own answer. Denote by p(S) the number of punctures.

1) chi(S')/chi(S)=d with $d$ positive integer

2) $p(S)\le p(S') \le p(S)d$.

3) $p(S)d-p(S')$ should be even.

Moreover, in the case Genus(S)=0 you have an additional condition

$p(S)d-p(S')>d-2$. This condition assures that S' is connected.

I think that these conditions are necessary and sufficient. It is obvious that 1), 2) are necessary. Condition 3) comes from the fact, that the permutation corresponding to going around all punctures on S is a commutator, so it should be even.

In order to show that these conditions are sufficient, you could use the old result of Ore that tells that every even permutation is a commutator of two permutations. Oystein Ore. Some remarks on commutators. Proc. Amer. Math. Soc., 2:307–314, 1951. Let me prove that conditions are sufficient in the case when genus of S is two or more.

Sketch of a proof. We want to show that there exists a collection of permutation in $S_d$, $s_1,...,s_{2g}, t_1,...,t_p$ ($p=p(S)$) that act transitively on ${1,...,d}$ such that $s_1s_2s_1^{-1}s_2^{-1}...=t_1...t_p$, where $t_1,...,t_p$ are given permutations with the product in the alternating group $A_d$. Then we chose $s_1$ and $s_2$ in such a way that $s_1s_2s_1^{-1}s_2^{-1}=t_1...t_p$ (Ore result) and take $s_3=s_4$ - cycles of length $d$, while all other permutations $s_i$ should be trivial. Clearly the action on $\{1,...d\}$ is transitive. Now the existence of a cover follows by standard arguments.

If you manage to make the proof very short it is worth to put it in the article, or at least give a hint. Otherwise, indeed, as Pete said it would be nice to find a reference.

share|improve this answer
@Dmitri: I think what I said above is consistent with this. –  Pete L. Clark Jan 23 '10 at 12:16
Pete, I ageree with you, just the answer is slightly more tricky than I thought in the beguinning. Hope it is correct though –  Dmitri Jan 23 '10 at 12:32
@Dmitri: agreed. The first two conditions are obviously necessary, but I wouldn't have predicted 3). I would say that I changed my mind and that, as a reader, I would prefer to see a reference! –  Pete L. Clark Jan 23 '10 at 12:42
add comment
up vote 2 down vote accepted

In a recent paper by Calegari, Sun, and Wang, the authors cite

W.S. Massey, Finite covering spaces of 2-manifolds with boundary. Duke Math. J. 41 (1974), no. 4, 875-887.

which proves that Dmitri's conditions (1) and (2) are sufficient if the S and S' are compact, orientable with non-empty boundary.

(Dmitri's condition (3) is redundant as Massey points out in the second page of his article: writing chi(S') = 2 - 2g(S') - p(S') and chi(S) = 2 - 2g(S) - p(S) Dimitri's condition (1) gives

2 - 2g(S') - p(S') = d(2 - 2g(S) - p(S))

and reducing (mod 2) gives that p(S') and dp(S) have the same parity. But the proof therein is longer than the one that Dimitri for the case that he treats (if one takes the Ore result as given).

share|improve this answer
That's a neat trick: have people answer your question, wait a while, then write your own equivalent answer and mark it as "correct". How fair is that? –  Victor Protsak Jun 12 '10 at 5:59
I appreciated Dimitri's answer, but my question was asking for a reference that I could cite, and I finally found on my own. –  Jonah Sinick Jun 13 '10 at 20:46
add comment

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.