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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.

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How to you define $\mathbb P^n$? – Phira May 20 '12 at 17:35
Identifying antipodal points on the sphere... – Jr. May 20 '12 at 17:38
What do you know about the orientability of the sphere? – Phira May 20 '12 at 17:45
I know that it is a orientable manifold... – Jr. May 20 '12 at 17:50
I think the word "plane" is misleading if we aren't talking about the 2-dimensional object; perhaps projective "space" would be a better term? – Gerry Myerson May 21 '12 at 3:57

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!

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do you have a proof using differential forms aproach? – Jr. May 21 '12 at 15:41
you mean providing a nowhere vanishing $n-$Form? – Blah May 21 '12 at 21:16
yes!! This exercise was in the context of differential forms. – Jr. May 21 '12 at 22:44
You must show a nonwhere vanishing $n$-form: if you have $n$ tangent vectors at $q$, then you could push them up via $\pi^{-1}_{*,p}$ for the two points $p$. On $S^n$ there is a volume form - evaluate. This is well defined if $n$ is odd. (think about the pullback of the volumeform under the antipodal map) – Blah May 22 '12 at 18:11

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