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I'm trying to see if there is a "nice-enough" way of describing/constructing the universal cover for a compact surface with $n$ boundary components. Clearly, if $n=0 $, the classification theorem for surfaces gives us a nice recipe; using either $S^1 $, or for genus $g$, using $\mathbb R^g$. I guess we could just cap a disk on each boundary component, end up with a genus$-g$ surface, find its cover, and then consider the restriction of the cover to the capped disks, but this does not seem to help. Any ideas? Thanks.

InsideOut
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User Some Number
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  • Well, up to homotopy it will be contractible for $n\geq 1$ because a punctured surface is homotopy equivalent to a wedge of circles and graphs have contractible universal covers. Homeomorphism type is a more interesting question. The restricted covers you mention (which for $g\geq 1$ looks like $\mathbb{R}^2$ with a countable collection of removed open disks) is a cover still, but not a universal cover because it has non-trivial fundamental group. – Dan Rust Feb 12 '14 at 02:33
  • @Daniel Rust: I'm sorry, I don't get why you're referring to punctured surfaces. – User Some Number Feb 12 '14 at 02:45

1 Answers1

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The universal covering space for all compact surfaces with nonempty boundary (except for the disk, annulus and Moebius band) is the following. Take the closed 2-disk and then remove a Cantor set from its boundary. Here Cantor set is any set homeomorphic to the standard Cantor set. Describing the covering transformation group is much more difficult, although possible.

Edit. Here are some steps towards the proof assuming you do not know any hyperbolic geometry. First of all, the universal cover is contractible and has nonempty boundary. One then sees that it's interior is homeomorphic to an open disk; call the boundary circle of the disk C. Then by working a bit more you see that it (let's call the universal cover U) is an open disk plus some disjoint union (call it A) of (open) boundary arcs.

Using the fact that fundamental group is free nonabelian you see that A consists of countably infinitely many boundary arcs. Then you check that topology of U does not change if you collapse each component of C minus A to a point. Now the set C minus A is compact, metrizable, totally disconnected and infinite, call this set K. The next and the most difficult step is to show that K is perfect. Lastly, you quote a theorem from general topology that such K is homeomorphic to the standard Cantor set.

Edit: As an alternative to the topological argument, you can use hyperbolic geometry as follows: First, you realize that your compact surface with boundary admits a complete hyperbolic metric with geodesic boundary. Then, represent the surface equipped with this metric as the quotient of a closed convex domain $D$ in the hyperbolic plane ${\mathbb H}^2$ by a "Fuchsian group of the 2nd kind" $\Gamma$. I will think of the hyperbolic plane as the open unit disk equipped with the Poincare metric.Consider the closure $\bar{D}$ of $D$ in the complex plane. It is homeomorphic to the closed unit disk. The frontier of $D$ is the disjoint union of two sets: A countably infinite disjoint union of open circular arcs and a certain set $\Lambda(\Gamma)$, the intersection of $\bar{D}$ with the unit circle. The set $\Lambda(\Gamma)$ is compact, perfect, totally disconnected and, of course, metrizable. Therefore, it is homeomorphic to the standard Cantor set. See for instance the book

S. Katok, "Fuchsian Groups",

especially Theorem 3.4.5 and section 3.6, for (many) more details.

For the annulus and Moebius band the universal covering space is the product of a closed interval and the real line. The proof is elementary, by describing the covering map. In the case of annulus you can use polar coordinates to define this map.

Moishe Kohan
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  • Thank you; would you give me an argument; at least a sketch for why this is the case? – User Some Number Feb 12 '14 at 03:42
  • All proofs I know are complicated. If you were to know some hyperbolic geometry, it would help. ( The keyword then is "Fuchsian group of second kind".) Or, do you know about "ends of groups"? I can quote some theorems but then you would not know what am I talking about. – Moishe Kohan Feb 12 '14 at 03:51
  • @MoisheKohan I have a question. Is it true that the universal cover of a non-compact surface with non-empty boundary is of the form: delete a closed subset from the boundary of the closed unit disc. Any reference will be helpful. Thanks. (+1) – Sumanta Nov 28 '20 at 05:33
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    @0-thUserSumanta: Yes, this is true (for connected surfaces). It follows from classification theorems for noncompact surfaces with boundary. See here, Jack Lee's answer and my comments. – Moishe Kohan Nov 28 '20 at 08:26
  • Thanks for your comment. So, universal cover of any connected surface, possibly non-compact with boundary is either $\Bbb S^2$ or a convex subset of $\Bbb R^2$. Am I right? – Sumanta Nov 28 '20 at 08:29
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    @0-thUserSumanta: Yes, more precisely, is homeomorphic to $S^2$ or to a convex subset of the plane. – Moishe Kohan Nov 28 '20 at 08:32