I'm trying to prove that the projective plane $\mathbb{P}^n$ is orientable is and only if $n$ is odd. To do that that, I have a hint,to prove that the antipodal map is orientation preserving if only if $n$ is odd, I've done that, but it don't know how to conclude the result.
[Math] Projective Space orientation
differential-geometry
Related Solutions
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).
This is an elaboration on my comments on the question, and on Jason DeVito's answer.
Firstly, what does it mean to choose an orientation on a manifold $M$?
One way to think of it is that an orientation is a collection of charts covering $M$ such that the Jacobians of the change-of-coordinate map on all overlaps have positive determinants.
Another way to think of an orientation is that we have to choose an orientation for $TM_p$ for each $p \in M$, with the property that given any $p \in M$, there is some chart $U$ containing $p$, with local coordinates $x_1,\ldots,x_n$, such that the orientation on $TM_q$ is the one that contains the basis $\partial_{x_1}, \ldots, \partial_{x_n}$, for each $ q \in U$. (Recall that an orientation on a vector space over $\mathbb R$ is a collection of bases such that all change of basis matrices have positive determinant.)
It's not hard to check that these two notions coincide. Indeed, given a colllection of charts as in the first definition, we can define an orientation on the tangent space $TM_p$ for each $p\in M$ as follows: if $p \in U$ for one chart $U$ in our given collection, and if $x_1,\ldots,x_n$ are the local coordinates on $U$, then we define the orientation on $TM_p$ to be the one containing the basis $\partial_{x_1},\ldots,\partial_{x_n}$. Note that our assumption on the transition maps on the overlaps of the charts means that this really does give a well-defined orientation on the vector space $TM_p$ for each $p$. By construction the resulting set of orientations on the tangents spaces $TM_p$ satisfies the conditions of 2.
Conversely, given a set of orientations on the $TM_p$ as in definition 2, consider the set of charts $U$ whose existence is guaranteed by 2; this collection of charts evidently satisfies the conditions of definition 1.
For the sphere, there is a standard way to choose an orientation: fix a unit normal vector field $\mathbb n$ on $\mathbb S^n$, either the inward pointing normal or the outward pointing normal. Also fix an orientation on $\mathbb R^{n+1}$ as a vector space. If $p \in \mathbb R^{n+1}$, then $T\mathbb R^{n+1}_p \cong \mathbb R^{n+1}$ canonically, and so we get an orientation on $T\mathbb R^{n+1}_p$ for each $p$. (A slightly more long-winded way to describe what I just did, which might nevertheless be helpful, is: I am using $\mathbb R^{n+1}$ as a global chart on itself, and hence defining an orientation on $\mathbb R^{n+1}$ as a manifold as in definition 1. I am then using the procedure described above, of going from 1 to 2, to get an orientation on each $T\mathbb R^{n+1}_p$.)
Now for each $p \in \mathbb S^n$, define an orientation on $T\mathbb S^n_p$ such that the induced orientation on $T\mathbb S^n_p \oplus \mathbb R\mathbb n = T\mathbb R^{n+1}_p$ (induced orientation meaning that we add $\mathbb n$ to any positively oriented basis of $T\mathbb S^n_p$ so as to get a basis for $T\mathbb R^{n+1}_p$) coincides with the given orientation on $T\mathbb R^{n+1}_p$.
Now that we have fixed on orientation on $\mathbb S^n$, we are finally in a position to make a Jacobian computation to compute whether $f$ preserves or reverses orientation.
As noted by the OP, $Dh(p)$ has determinant $(-1)^{n+1}$ for any point $p$. On the other hand, $Dh$ takes the unit normal $\mathbb n(p)$ to the unit normal $\mathbb n(f(p))$. (Draw the picture!)
In other words, when we consider $Dh(p): T\mathbb R^{n+1}_p \to T\mathbb R^{n+1}_{f(p)}$, and we decompose this into the direct sum of $Df(p): T\mathbb S^n_p \to T\mathbb S^n_{f(p)}$ and the map $\mathbb R \mathbb n(p) \to \mathbb R \mathbb n(f(p))$ induced by $Dh(p)$, the latter map has the matrix $1$ with respect to the bases $\mathbb n(p)$ in the source and $\mathbb n(f(p))$ in the target. Thus $Df(p)$ also has determinant $(-1)^{n+1}$ with respect to the positively oriented bases on its source and target.
Thus $f$ is orientation reversing/preserving according to whether $n$ is even/odd.
Best Answer
Let $\{O_p\}_{p \in S^n}$ be an orientation for $S^n$. If $\pi:S^n \rightarrow \mathbb P^{n}$ is the projection (a local diffeomorphism), then the idea is to define a basis $(b_1,\dots,b_n)$ of $T_q\mathbb P^{n}$ to be in $O_q'$ if $\pi_{*,p}^{-1}(b_1,\dots,b_n) \in O_p$ for any one of the two points $p$ in the fibre $\pi^{-1}(q)$.
You have already shown that this is well defined if $n$ is odd!