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Let $X_k$ be the connected sum of $k$ projective planes. I am interested in necessary and sufficient conditions for the existence of a covering $\pi: X_{k'} \to X_k$, where $k$ and $k'$ are integers.

A necessary condition is that the Euler characteristic of $X_{k'}$ is a multiple of the Euler characteristic of $X_k$. Though obtaining a sufficient condition seems more difficult.

The case of orientable surfaces is easy, but this is more difficult than I think, and I can't find it anywhere.

I would appreciate any hint about this.

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1 Answer 1

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Your necessary condition, rephrased slightly, is sufficient. It should say that the Euler characteristic of $X_{k'}$ is a positive multiple of $\chi(X_k)$. This modification is to take care of the possibility that $X_k$ is $RP^2$, which is not covered by any other non-orientable surface. If $\chi(X_k)= \chi(X_{k'}) = 0$, then both are Klein bottles, and certainly there is a covering. Finally if $\chi(X_k)<0$, then it is a connected sum of a torus with some number of projective planes. So it has non-orientable covers of all degrees; these are determined by their Euler characteristics. So, given $X_{k'}$ with $\chi(X_{k'}) = n \chi(X_k)<0$, choose a degree n cover of $X_{k}$; it will be homeomorphic to $X_{k'}$.

The case of orientable surfaces follows by the same argument.

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  • $\begingroup$ I don't see this implication: "Finally if χ(Xk)<0, then it is a connected sum of a torus with some number of projective planes. So it has non-orientable covers of all degrees; these are determined by their Euler characteristics." How can I construct that coverings? Is that a known result that I can find in some book or something? In the case of orientable surfaces it is easy to construct the coverings because it is a kind of glue-rotation, but in this case I can't visualize anything. Thank you! $\endgroup$
    – juan rojo
    Nov 4, 2013 at 9:28
  • $\begingroup$ If $\chi(X_k) < 0$ then the abelianisation of its fundamental group is infinite, and in particular there exists a surjective homomorphism $\pi_1(X_k) \to \mathbb{Z}$. Composing this with $\mathbb{Z} \to \mathbb{Z}/d$ for any $d$ gives a homomorphism $\phi_d$ which classifies a $d$-fold covering space, which will again be nonorientable as $\phi_d$ factored through a torsionfree group. $\endgroup$ Nov 4, 2013 at 11:58
  • $\begingroup$ Thanks for clarifying, Oscar. To say this more geometrically, the torus has a cyclic cover of any order $d$. These extend to the connected sum $T^2\#RP^2$, which is then covered by $T^2\# (\#_d RP^2)$. $\endgroup$ Nov 4, 2013 at 12:29

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