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Prove that an even dimensional Real Projective space is not orientable.

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What have you tried? What tools do you have at your disposal? –  Jason DeVito May 1 '12 at 19:19
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When you post some question, please do not do this in an imperative fashion. Show what was your effort. –  matgaio May 1 '12 at 19:25
    
Here's the answer in 2 dimensions. See if you can generalize to higher dimensions on your own. (The basic arguments are the same). math.stackexchange.com/q/133274/22405 –  Brett Frankel May 1 '12 at 20:23
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You can follow Lee's scketch (Lee's "Introduction to Smooth Manifolds", p. 346-7 - exercises 13-4 and 13-5): first, if $M$ is a connected smoot manifold and $\Gamma$ is a group acting smoothly, freely, and properly on $M$, then $M/\Gamma$ is orientable if and only if every element on $\Gamma$ preserves orientation on $M$. Then, apply this to the case of projevtive spaces, where $M=\mathbb{S}^n$ $\Gamma$ will be $I,-I$ and the quotient will be the projective space itself. –  matgaio May 1 '12 at 20:27

1 Answer 1

Following the comment by matgaio: Suppose $\mathbb RP^{2n}$ admits a nonvanishing $2n$-form $\omega$. Consider the quotient map $\pi:S^{2n}\to \mathbb RP^{2n}$ associated to the action of antipodal map $f(x)=-x$. The pullback $\zeta=\pi^* \omega$ is a nonvanishing $2n$-form on $S^{2n}$. Since $\pi\circ f=\pi$, it follows that $f^*\zeta=\zeta$: in other words, $f$ is orientation-preserving map on $S^{2n}$. But the latter statement is false: $f$ reverses the sign of the standard volume form on $S^{2n}$.

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