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
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).