1
$\begingroup$

Suppose I have a bounded, orientable genus 5 surface with 4 boundary circles. Is there a way to determine what surfaces it covers?

First, I know that there is a covering map from the closed orientable surface of genus 5 to the closed non-orientable surface of genus 6. This covering map "corresponds" to the quotient maps that identifies anti-podal points. With this in mind, I think the the bounded, orientable genus 5 surface with 4 boundary circles covers a bounded non-orientable surface of genus 6 with 2 boundary circles.

Am I understanding this correctly? Can I take this further? Thanks for any responses!

$\endgroup$
  • 1
    $\begingroup$ One obvious comment: if $S$ covers $S'$ Then $\chi(S) = d\chi(S')$ where $d>0$ is the degree of the cover. Also if $S$ has non- empty boundary $S'$ must have non-empty boundary. This gives a small number of possibilities to chrck, since for us $\chi(S)=-12$. $\endgroup$ – Nick L Nov 18 '18 at 15:57
3
$\begingroup$

In my answer below, I was implicitly assuming that the covering on the boundary was trivial. As Mike Miller points out below, this is not necessarily the case. See his comments for more details.


Denote the closed orientable surface of genus $g$ with $b$ boundary components by $\Sigma_{g, b}$, and the closed non-orientable surface of genus $g$ with $b$ boundary components by $S_{g,b}$. Recall that $\chi(\Sigma_{g,b}) = 2 - 2g - b$ and $\chi(S_{g,b}) = 2 - g - b$.

If $p : M \to N$ is a covering map between manifolds with boundary, then it restricts to a covering map $p|_{\partial M} : \partial M \to \partial N$ of the same degree. So if $p : \Sigma_{g,b} \to \Sigma_{g',b'}$ is a degree $k$ covering map, then $b = kb'$. Moreover,

\begin{align*} \chi(\Sigma_{g,b}) &= k\chi(\Sigma_{g',b'})\\ 2 - 2g - b &= k(2 - 2g' - b')\\ 2 - 2g - kb' &= k(2 - 2g' - b')\\ 2 - 2g &= k(2 - 2g')\\ \chi(\Sigma_g) &= k\chi(\Sigma_{g'}). \end{align*}

The converse is also true. That is, if $\chi(\Sigma_g) = k\chi(\Sigma_{g'})$ and $b = kb'$, then there is a degree $k$ covering map $\Sigma_{g,b} \to \Sigma_{g',b'}$. To see this, note that if $\chi(\Sigma_g) = k\chi(\Sigma_{g'})$, then there is a degree $k$ covering map $p : \Sigma_g \to \Sigma_{g'}$; see this answer. If $D \subset \Sigma_{g'}$ is the interior of a closed disc in $\Sigma_{g'}$, then $p^{-1}(D)$ is a disjoint union of the interiors of $k$ disjoint closed discs in $\Sigma_g$. So if $D_1, \dots, D_{b'}$ are the interiors of $b'$ disjoint closed discs in $\Sigma_{g'}$, then $p^{-1}(\Sigma_{g'}\setminus(D_1\cup\dots\cup D_{b'})$ is $\Sigma_g$ with $kb' = b$ interiors of disjoint closed discs removed, i.e. $\Sigma_{g,b}$. Therefore the restriction of $p$ to $\Sigma_{g,b}$ is a degree $k$ covering $\Sigma_{g,b} \to \Sigma_{g',b'}$.

Likewise, $\Sigma_{g,b}$ is a $k$-sheeted covering of $S_{g',b'}$ if and only if $\chi(\Sigma_g) = k\chi(S_g)$, $b = kb'$ and $k$ is even. Note that $k$ must be even as any orientable covering of a non-orientable manifold must factor through the orientation double cover.

Therefore, we have the following complete list of coverings:

  • $\Sigma_{5,4} \to \Sigma_{5,4}$ of degree one,
  • $\Sigma_{5,4} \to \Sigma_{3,2}$ of degree two,
  • $\Sigma_{5,4} \to S_{6,2}$ of degree two,
  • $\Sigma_{5,4} \to \Sigma_{2,1}$ of degree four, and
  • $\Sigma_{5,4} \to S_{4,1}$ of degree four.
$\endgroup$
  • $\begingroup$ I believe you've made an assumption here that the covering map is trivial on the boundary, which is not necessary; the condition on boundary components should be $b/k \leq b' \leq b$. I still believe your claim that (in addition to this constraint) the only condition is that $\chi(\Sigma_{g,b}) = k \chi(\Sigma_{g', b'})$, but I do not know a nice reference. I think there is some literature further trying to specify the data of $(g',b')$ and the $k$-partition of $b$ corresponding to the degrees of covers on the boundary, and IIRC all are realizable except possibly some with $g = 0$. $\endgroup$ – user98602 Nov 18 '18 at 20:04
  • $\begingroup$ In particular, I think the following should also be on the list of possible covers: $\Sigma_{2,4}, \Sigma_{1,4}, \Sigma_{1,3}, \Sigma_{1,2}, \Sigma_{0,4}, \Sigma_{1,1}, \Sigma_{0,3},$ and I think all should be realized. I think there is an evenness condition as you say on the number of boundary components and one should have similar results for covering $S_{g',b'}$. If you fill the boundary components in with discs, these covering maps correspond go branched covers with specified branching data, and yours correspond to those with trivial branch data (i.e., are actually covering maps). $\endgroup$ – user98602 Nov 18 '18 at 20:07
  • $\begingroup$ You're right. Can't believe I missed that. $\endgroup$ – Michael Albanese Nov 19 '18 at 0:07
  • 1
    $\begingroup$ That subtlety has bitten me more than once. $\endgroup$ – user98602 Nov 19 '18 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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