0
$\begingroup$

For figure "eight" there is a list of finite sheeted covering discussed in Hatcher's book "Algebraic topology". I was thinking about the following question:

Let $S$ be a finite subset with $|S|>1$ of $\mathbb{C}$ and $n\ge 2$, then what are the $n$-sheeted coverings for $\mathbb{C}-S$?

It is known to me that, for $|S|=1$, finite $n$-sheeted covering for $\mathbb{C}-S$ is itself given by $z\to z^n$.

Thanks in advance!

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ We get one connected n-sheeted covering for each $|S|$-tuple of permutations in $S_n$ that together generate a group that acts transitively on $\{1,\dots, n\}$, up to simultaneous conjugation. So really there is a reasonable classification only for $|S|=1$, where only an $n$-cycle does the trick and all $n$-cycles are conjugate. $\endgroup$
    – Will Sawin
    Commented May 12, 2022 at 21:08
  • 2
    $\begingroup$ Isomorphism classes of finite-to-one covers of $\mathbb{C}$-with-$n$-punctures are in bijection with isomorphism classes of finite $F_n$-sets, where $F_n$ is the free group on $n$ letters. I imagine that, as $n$ grows (perhaps already for $n=2$ or $n=3$), the problem of classifying finite $F_n$-sets becomes intractable, in the sense that the representation theory of $F_n$ becomes "of wild type." Perhaps somebody reading this page who knows more rep. theory than I do can speak up about what the least value of $n$ is, such that classifying the finite $F_n$-sets is a problem "of wild type." $\endgroup$
    – user164898
    Commented May 12, 2022 at 21:50
  • $\begingroup$ @A.S. your $n$ is not the OP's $n$ (it is $|S|$ in the OP's notation). $\endgroup$
    – YCor
    Commented May 13, 2022 at 8:41
  • $\begingroup$ @A.S. I'm not sure this is a representation-theoretic fact, but "classifying" finite index subgroups of $F_2$ is not easier than "classifying" finite index subgroups of $F_d$ for larger $d$, since $F_d$ itself lies as a finite index subgroup of $F_2$. — Note however that the OP also fixes the index ($n$ in OP's notations, assuming we stick to nonempty connected coverings), and hence one could also fix the index and let the number of generators grow. $\endgroup$
    – YCor
    Commented May 13, 2022 at 8:45
  • $\begingroup$ Well can I say atleast that such covering space is $\mathbb{C}$ with finitely many points removed? $\endgroup$
    – piper1967
    Commented May 13, 2022 at 15:37

2 Answers 2

Reset to default
1
$\begingroup$

The fundamental group of the $n$-times punctured plane is a free group of rank $n$. By the so-called Galois correspondence the connected $d$~fold covers of the punctured plane are in bijection with the index $d$ subgroups of the fundamental group. This gives another description of the desired set of covers. Since the set (under either description) is very “wild” I don’t think there is “clean” answer to your question, except on the once-punctured case (where we arrive at the subgroups of $\mathbb{Z}$).

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Note that you need to work in the category of pointed spaces to get a bijection. (In the non-pointed category, if I'm correct, conjugate subgroups of index $d$ yield isomorphic coverings.) $\endgroup$
    – YCor
    Commented Oct 10, 2022 at 11:17
1
$\begingroup$

This is a difficult and very interesting question. As explained in comments and the other answer, the problem is as stated is "wild", but There are many special versions of it where something can be said. For instance, there is the ELSV formula for Hurwitz numbers (https://en.wikipedia.org/wiki/ELSV_formula), relating this question to intersection theory on moduli spaces of curves.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .