For any non-orientable surface (compact,connected) $X$ with genus $h$, we have a $2n$-sheeted cover of $X$ by an orientable surface $Y$ first by covering $X$ by $\Sigma_{h-1}$ (a double cover) and then taking an $n$-sheeted orientable cover of $\Sigma_{h-1}$ and composing. Also there is an $n$-sheeted cover of $X$ by a non-orientable manifold, and a double cover of that by an orientable manifold. These give isomorphic coverings.

My question is that do all coverings of non-orientable surfaces by orientable surfaces arise in this fashion? Specifically, I want to know if there are odd-sheeted coverings of non-orientable surfaces by orientable surfaces, and if not why there are none. (It seems highly unlikely that there are any to me, and it seems a clever algebraic trick should show this, but I do not know how to do it.)

If this can be done without cohomology, I would appreciate it.


I don't think Daniel's link is really a propos -- Mednykh is solving a much harder problem. It is fairly easy to show that every covering map from an orientable surface [this holds in much greater generality] to an orientable surface factors through the orientatation cover. Since the orientation cover is a double cover you are done.

  • $\begingroup$ Definitely the better answer. $\endgroup$ – Dan Rust Dec 27 '13 at 16:52

The answer to your question is 'No. A non-orientable surface does not admit an odd-sheeted covering by an orientable surface', and the proof involves some rather in-depth study of the fundamental groups of the involved spaces. See for instance here.

Given the reference above, I thought I would instead just prove some rather simple conditions that would need to be satisfied by such an odd-sheeted covering, which only uses the Euler characteristic.

Let $N$ be a non-orientable surfaces and let $M$ be an orientable surface which admits a $2k+1$ cover of $N$. Suppose $M$ has genus $g$ and $N$ has (non-orientable) genus $l$.

From standard Euler characteristic formulae, we get that $$\chi (N)=2-l\\ \chi (M)=2-2g\\ \chi (M)=(2k+1)\chi(N)$$

which gives $$2g=2k(l-2)+l$$

Immediately then, we see that $l$ must be even so write $l=2j$ giving $$g=2k(j-1)+j$$ and so $g\equiv j\mod 2$.

If $g$ is odd then writing $g=2g'+1$ and $j=2j'+1$ we get $$g'=2j'(k+1)$$ and so $g'$ is even. So $g$ can not be congruent to $3\mod 4$.


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