# Covering space of a non-orientable surface

I have the following problem:

Find the 2-sheeted (orientable) cover of the non-orientable surface of genus g.

The cases $g=1,2$ are well-known, we have that the cover of $\mathbb{R}P^2$ is $\mathbb{R}^2$ and the Klein bottle is covered by the torus. My intuition is that the answer is going to be the orientable surface of genus $g-1$ or the $g-1$ torus, and I tried to work with the representation of the non-orientable surface as a $2g$-gon determined by the word $a_1a_1a_2a_2\dots a_g a_g$... where $a_i$ are the vertices. Any hints?

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The orientable double-cover of $\mathbb{R}P^2$ is not $\mathbb{R}^2$ ... –  Neal Jan 15 '13 at 12:59
ok sorry, I meant the sphere. –  Miguel Jan 15 '13 at 13:42

Since Euler characteristic is multiplicative with respect to (finite sheeted) coverings, we see the orientable double cover must have twice the Euler characteristic. This already pins down the homeomorphism type of the cover.

To see it more explicitly, consider the usual embedding of an orientable surface into $\mathbb{R}^3$ (as a long chain-like shape). This embedding can be translated/rotated around so that the reflections in all 3 coordinate planes maps the surface to itself.

With this embedding, there is a $\mathbb{Z}/2$ action on $\mathbb{R}^3$ given by the antipodal map $x \mapsto -x$. This action preserves the embedded surface and acts freely on it so the quotient is a manifold. Since the antipodal map is orientation reversing, the quotient is nonorientable.

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Ok so since the covering has twice the Euler characteristic, I assume that the genus $g$ non-oriented surface has as two-sheeted oriented covering the oriented surface of genus $2g-1$? I find it very interesting that the covering map can be described by means of the antipodal map! Thank you. –  Miguel Jan 15 '13 at 13:40
Well, you're solving $2(2-k) = 2-2g$ where $k$ is the nonorientables surface's genus (the number of $\mathbb{R}P^2$ pieces) and $g$ is the genus of the orientable surface. If I've down my algebra right, this gives $g = k-1$. I also found it interesting that the antipodal map works - I didn't realize it did before composing this answer! –  Jason DeVito Jan 15 '13 at 13:45
Ups, yeah you are right, obviously it finally gives $g=k-1$, thanks again! –  Miguel Jan 15 '13 at 15:47