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Let $S$ be a genus 3 closed orientable surface. Let $R\rightarrow S$ be a degree 2 covering map. What is the genus of $S$ ? Do I have to use the Euler characteristic of a surface presentation which is a topological invariant? I guess the question should be what is the question of R?

Thanks

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  • $\begingroup$ I'm no expert in this but doesn't it depend on the covering map? Is it ramified anywhere? $\endgroup$
    – fretty
    Dec 8, 2013 at 10:36
  • $\begingroup$ What do you want to know? Genus of $S$ or $R$? $\endgroup$
    – user99914
    Dec 8, 2013 at 10:44
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    $\begingroup$ In any case, you might need en.wikipedia.org/wiki/Riemann%E2%80%93Hurwitz_formula and the ramification, as fretty suggested. $\endgroup$
    – user99914
    Dec 8, 2013 at 10:45
  • $\begingroup$ So we would have $\chi (R) = N \chi(S)$ where N is the degree of the cover. In this case $N = 2$, and $g_{S} = 3$, then $\chi (R) = 2(2-2g_{S})2(2-2(3)) = -8$ then $\chi(R) = 2-2g_{R} = -8$ therefore $g_{R} = 5$. Am I right? $\endgroup$
    – Gottfried
    Dec 8, 2013 at 18:17

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Let us assume that $R$ and $S$ have triangulations, and further that the covering map $\pi\colon R\to S$ of degree two respects the triangulation on $S$ (that is, $\pi$ is a cellular-map). The Betti numbers of $R$ (resp. $S$) are $1,2g_R,1,0,\ldots$ (resp. $1,2g_S,1,0,\ldots$) which can be found using standard techniques. So, the Eular-characteristics are given by $$\chi(R)=1+2g_R-1=2g_R\\ \chi(S)=1+2g_S-1=2g_S.$$

Now, we can also give the Euler characteristic of a surface in terms of the vertices, edges, and faces of a triangulation, and so we also have $$2-2g_R=\chi(R)=V_R+E_R-F_R\\ 2-2g_S=\chi(S)=V_S+E_S-F_S.$$

Given that $\pi$ is assumed to take vertices to vertices, etc, and is of degree two, so the preimage of every $k$-cell is a disjoint union of two $k$-cells, we should have $$V_R=2V_S,\: E_R=2E_S,\:F_R=2F_S$$ and so $2-2g_R=V_R+E_R-F_R=2(V_S+E_S-F_S)=2(2-2g_S)=2(-4)=-8\implies g_R=5$.

It wasn't clear from your question if you wanted to find $g_R$ from $g_S$ or vice-versa, but from the above you can see that $g_R=2g_S-1$ in all cases.

There is a small concern in the above that the covering map $\pi$ given may not in fact be a cellular map. This concern is justified, and I should have really shown that we can take a suitably fine triangulation of $S$ which properly lifts up to a triangulation on $R$ via $\pi$. The easiest way to do this is to put a triangulation on $R$, consider the induced partition by polygons of $S$ by $\pi$ (which may not be a triangulation) that we get by mapping the $1$-skeleton of $R$ to $S$. We can then refine this partition so that it really is a triangulation, and then lift this new triangulation back up to $S$ via $\pi$. This new triangulation on $R$ will then, by construction, respect the cellular structure of $S$ with respect to $\pi$.

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  • $\begingroup$ Thank you, but what I do not understand is why I did not get the same answer from the Hurwitz formula, do you have any idea? I think you made a mistake in calculating the Euler characteristic, since it is : $\chi(R) = 1 -2g_{R} +1 = 2-2g_{R}$, so $g_{R} = 5$ and same for S. $\endgroup$
    – Gottfried
    Dec 8, 2013 at 22:31
  • $\begingroup$ @Gottfried Sorry, you're of course right - I used the wrong formula (for some reason which I don't know) for the Euler characteristic in terms of the genus. I've fixed this now hopefully. $\endgroup$
    – Dan Rust
    Dec 8, 2013 at 22:41
  • $\begingroup$ Do not worry, your mistake make me understand this topic even more, and again thanks for everything. $\endgroup$
    – Gottfried
    Dec 8, 2013 at 22:58

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