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A nonorientable surface $S$ is homeomorphic to the $k$-th connected sum $\mathbb{R}P^2 \sharp \ldots \sharp \mathbb{R}P^2$.

For each nonorientable surface $S$ there exists an oriented $2$-fold covering $\tilde S$, the covering bundle. This is isomorphic to the determinant bundle.

Now the number of $2$-fold coverings on $S$ can be computed with the help of $H^1(S;\mathbb{Z}/2)$, that is $2^k$.

The question is: is there a possibility to identify the orientation bundle $\tilde S$ within the set of all $2$-fold coverings on $S$ uniquely.

It is clear that it has to be oriented and connected.

Or another question is, if there is a surface $X$ given. Is it possible to find out if it is the orientation bundle $\tilde S$.

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This is not a question: you have a construction of a particular 2-fold covering space as the orientation cover (= sphere bundle of determinant bundle), and want to know whether it identifies a unique double cover. Of course it does, it is a construction. –  Oscar Randal-Williams Feb 4 '11 at 14:08
    
I think my question was not precise. In fact, I was asking for the complex structure. The topological classification is clear as you all pointed out. –  berl1313 Feb 5 '11 at 16:43
    
It's not at all clear to me that "the complex structure" makes sense. There are many complex structures on a compact, orientable surface. How would you associate one such structure to a surface, just based on the fact that it double covers something non-orientable? What would you like this structure to satisfy? –  Dan Ramras Feb 6 '11 at 18:19
    
You are right, so perhaps a better question is the following. Let there be a given surface $X$. Is there some method to find out if it is the orientation covering of some nonorientable surface? –  berl1313 Feb 6 '11 at 21:13
    
Yes, if it admits an orientation-reversing involution. –  Oscar Randal-Williams Feb 7 '11 at 8:00

3 Answers 3

If I interpret your question correctly, you are asking which class in $H^1(S; \mathbb{Z}/2 \mathbb{Z})$ corresponds to the orientation covering. This is the first Stiefel-Whitney class of $TS$, and there are many constructions for it.

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Maybe the simplest way to think about this class is in terms of the universal coefficient theorem, which tells you that this class is really just a map from $H_1(S; Z)$ to $Z/2Z$. As $H_1$ is the abelianization of the fundamental group, this function can be thought of as the answer to a yes/no question, asked about loops $\gamma$ in $S$. That question is: is the tangent bundle orientable when restricted to $\gamma$? –  Dan Ramras Feb 4 '11 at 20:10

Here's a slightly different perspective. Double coverings are classified by subgroups of index two in the fundamental group. So it makes sense to ask for a description of the index two subgroup corresponding to the orientation double cover. This subgroup consists of loops is $S$ that lift to loops in the orientation double cover, and hence these are exactly those loops along which the tangent bundle of $S$ is orientable. This is of course closely related to the comment I made on Andrea's answer.

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Any connected orientable surface $X$ which doubly covers $S$ must be homeomorphic to the orientation cover $\tilde{S}$.

Proof: Connected orientable surfaces are classified up to homeomorphism by their Euler characteristic. Since the Euler characteristic is multiplicative,

$$\chi(X)=\chi(\tilde{S})=2\cdot\chi(S).$$

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I guess it is implied, but it may be worth to note that yet, the covering may well NOT be the orientation covering. –  Andrea Ferretti Feb 4 '11 at 14:14
    
Hmm yes. Can you think of an example? –  Mark Grant Feb 4 '11 at 15:14
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An orientable double cover $X \to S$ lifts to a map $X \to \tilde{S}$ by the universal property of the orientation cover. This map is a continuous bijection, and so a homeomorphism as both source and target and compact Hausdorff. –  Oscar Randal-Williams Feb 4 '11 at 15:25

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