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Among the best-known examples of nonorientable, compact manifolds are projective spaces. However for these one has the fact that $\mathbb RP^n$ is orientable iff $n$ is odd, so that only "half" of them are nonorientable. This statement relies on the fact that the antipodal map $\alpha$ on $S^n$ is orientation-preserving iff $n$ is odd, and $\mathbb RP^n = S^n/\{id_{S^n}, \alpha\}$ is a quotient manifold.

Now, since $S^n$ is precisely the orientation covering of $\mathbb R P^n$, and $\alpha$ is the nontrivial covering transformation on $S^n$, I wondered, whether one could generalise this to arbitrary compact, connected manifolds $M^n$ of dimension $n$. I.e. whether it is true that if $n$ is odd, the nontrivial covering transformation cannot be orientation reversing, which would mean that every such $M$ has to be orientable.

I guess this cannot be true so generally, so I went looking for some counterexamples, but neither I myself nor google seems to find any. Would anyone here be able to help me out with an example, maybe?

Thanks in advance,

Sam

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Your question isn't precise. Are you asking if non-orientable manifolds exist in odd dimensions? Yes, they do, and examples are easy to construct and find. Perhaps you're asking for something else? –  Ryan Budney Apr 23 '11 at 4:43
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Hint: The product of a nonorientable manifold and an orientable manifold is nonorientable. –  Jason DeVito Apr 23 '11 at 4:44
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@Ryan: Strange, to me, "I wondered, whether one could generalise this to arbitrary compact, connected manifolds Mn of dimension n. I.e. whether it is true that if n is odd, the nontrivial covering transformation cannot be orientation reversing" seems unambiguous. –  Sam Apr 23 '11 at 7:05
    
You haven't specified which covering transformation you're talking about. There is no "the covering transformation" without further initial data. –  Ryan Budney Apr 23 '11 at 18:21
    
@Ryan: Ah, indeed. It did not come out as clearly as I thought that I was only talking about orientation coverings. On these, one of course only has one nontrivial transformation... I apologize for that. –  Sam Apr 24 '11 at 3:24

2 Answers 2

up vote 6 down vote accepted

As I mention in the comment above, the product of any number of manifolds is orientable iff each one of them is. In particular, considering $\mathbb{R}P^2\times S^n$ for $n\geq 0$ gives an example of a nonorientable manifold in every dimension.

Here's a potential reason for what you've noticed. There are two theorems due to Synge which deal with compact Riemannian manifolds with positive sectional curvature (both spheres and projective spaces, with their usual metrics, are compact Riemannian manifolds with positive sectional curvature).

The first says the following: Suppose $M$ is a compact Riemannian manifold of positive sectional curvature and the dimension of $M$ is even. Then $M$ is either orientable and simply connected or nonorientable with $\pi_1(M) =\mathbb{Z}/2\mathbb{Z}$.

The second says the following: Suppose $M$ is compact Riemannian manifold of positive sectional curvature and the dimension of $M$ is odd. Then $M$ is orientable.

Since spheres of odd dimension have positive curvature, this indicates that at least the "usual" things they cover must be orientable. (It's a priori possible for $S^n$ to cover a nonorientable $X$ for which the deck group of $X$ doesn't act by isometries on $S^n$ for any positively curved metric on $S^n$. I don't know if this happens or not, though).

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Thanks for the response. I've seen the second theorem you cite in doCarmo's book I think (But I unfortunately never had the time to properly work through it...). I hadn't really thought about what effect taking products has on orientability, but after you pointed it out, it indeed seems fairly obvious. ;-) Thanks. –  Sam Apr 23 '11 at 7:03
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Looking back at this a year later, it's clear that for $n$ odd, $S^n$ can't cover anything nonorientable. For if $S^n$ covers $M$, consider an element $f$ of the Deck group. If $f$ is orientation reversing, then it has Lefschetz number $2$, so has a fixed point, contradicting the fact that it's in the deck group. Thus the Deck group consists of orientation preserving diffeos, so $M$ inherits an orientation from $S^n$. –  Jason DeVito Dec 4 '12 at 18:12

Here are some examples of odd dimensional non-orientable flat manifolds. Each of these is covered by a flat torus by an orientation reversing isometry of order 2.

Start with an odd dimensional flat torus (the quotient of R^n by the standard lattice). Choose two axis directions, and take the quotient of the flat torus by the orientation reversing isometry that rotates by 180 degrees in the first axis direction and reflects in the second. The quotient is the Klein bottle Cartesian product a flat torus, a non-orientable manifold.

There are infinitely many other examples of flat Riemannian manifolds that are non-orientatble. They exist in all dimensions and are all covered by flat tori. There are many possible groups of isometries that act on the flat torus, not only Z2.

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