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This is an improved and hopefully a more precise version of the question Covering spaces of surfaces.

Question: Given a regular covering map $\pi:\Sigma_g\to\Sigma_h$, where $\Sigma_n$ denotes a surface of genus $n$, is it possible to describe the covering map?

One example of such a description is the following. In the decomposition of $\Sigma_h$ into the connected sum of tori $T_1$#...#$T_h$, one torus, say $T_1$, is covered $k$-times by another torus $T_1'$, which appears in a similar decomposition of $\Sigma_g$; every other torus $T_i$ in the decomposition of $\Sigma_h$ is covered by $k$ different tori $T_{i_1}'$,...,$T_{i_k}'$ (each covering $T_i$ identically) in the decomposition of $\Sigma_g$.

Any other explicit description would likely also be useful.

A weaker version of the question would be the following: given a regular covering $\rho:\Sigma_l\to \Sigma_h$, is there another regular covering $\rho':\Sigma_g\to \Sigma_l$, such that the composition $\pi=\rho\rho'$ has a description as above (for example)?

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  • $\begingroup$ 1. What do you mean by "describe"? If you mean "realize surface in 3-space so that the covering group action extends", then the answer is "of course not". I do not see a real question here. 2. Your weaker question has obvious negative answer: just take any $\rho$ with nonabelian finite covering group. $\endgroup$
    – Misha
    Oct 7, 2012 at 12:17
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    $\begingroup$ I am interested in any kind of description, really, the simpler the better. In particular, is the description stated in the example above valid in general? $\endgroup$
    – George
    Oct 7, 2012 at 12:54
  • $\begingroup$ Cut your surface $\Sigma_h$ into a $2h$-sided polygon $P$ glued in the traditional pattern $a_1b_1a_1^{-1}b_1^{-1}\ldots$. Letting the degree of the covering by $D$ (which could be infinity), take $D$ copies of $P$. Glue them up appropriately to form $\Sigma_g$. Map each copy to $P$. The covering map has been described. $\endgroup$
    – Lee Mosher
    Oct 7, 2012 at 12:58
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    $\begingroup$ Not every torus is necessarily covered by another torus. If the circles that separate the toruses in the connect-sum decomposition pull back to longer circles in the cover, rather than just disjoint unions of a bunch of copies of themselves, then the covering of the torus, which you get by gluing discs onto those circles, will be a ramified covering (ramified at the center of the disc), which necessarily increases the genus of the torus. $\endgroup$
    – Will Sawin
    Oct 7, 2012 at 16:08
  • $\begingroup$ George: It is a good exercise to see that in the example of a covering (call it $q$) that you gave, the covering group is abelian, see Will's comments. Therefore, no covering $\rho$, with nonabelian deck-group, appears in $q=\rho\circ \rho'$. $\endgroup$
    – Misha
    Oct 7, 2012 at 21:37

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I agree with the commenters that the question is a bit vague, but the most concrete description I know of is the permutation representation of the fundamental group (permuting the "sheets" of the covering. This tends to give a rather concrete "cut and paste" description of the cover (and in addition, I believe that this description, going back to Hurrwitz, might have been historically the first way to look at covering maps). For more along these lines, check out "On extension of coverings" by M. Droste and I. Rivin, and references therein.

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  • $\begingroup$ Igor, I think you just mean that the deck-group $G$ of a regular covering $M\to N$ admits a fundamental domain $F$ in $M$ (translates $g(F), g\in G$, are the "sheets" of the covering). This is valid for arbitrary (not necessarily finite) regular coverings of arbitrary smooth/PL/topological manifolds. I guess, in the surface case you can also say that permutation representation is transitive and is given by a collection of permutations s.t. product of their commutators is $1$. Is there more to it? $\endgroup$
    – Misha
    Oct 7, 2012 at 21:33
  • $\begingroup$ @Misha: there is a lot more to it: in particular, the inverse problem (trying to reconstruct the covering from the monodromy at a collection of points) is the "Riemann-Hilbert problem" in one of its guises, and is still open in important special cases. $\endgroup$
    – Igor Rivin
    Oct 7, 2012 at 23:07

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