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Q: Show that $\mathbb {RP}^n$ is not orientable for $n$ even.

First I looked at the definition for orientability for manifolds of higher degree than 2, because for surfaces I know the definition with the Möbius strip.

A n-dimensional manifold is non-orientable if it contains a homeomorphic image of the space formed by taking the direct product of a (n-1)-dimensional ball B and the unit interval [0,1] and gluing the ball B×{0} at one end to the ball B×{1} at other end with a single reflection.

The projective space can be seen as $\mathbb S^n/\mathbb Z_2$ or $\mathbb D^n$ with antipodal points identified on the boundary.

I have looked at other similar questions like Why isn't $\mathbb{RP}^2$ orientable?

In that thread someone states that the antipodal map is orientation preserving if $n$ is odd and reversing when $n$ is even. Why is this? Intuitively the antipodal map can be constructed by combining $n+1$ reflections in the hyperspaces $x_i=0$ and if $n$ is odd $n+1$ is even and combining an even amount of reflections is orientation preserving and the other way around for $n$ even and $n+1$ odd. I need a hint for a rigorous proof.

Also after I establish this for myself, I need to show that in $\mathbb D^n$ with n even and with antipodal points identified on the boundary the higher dimensional analogue of Möbius strip, which has just one reflection, can be embedded.

The subset to which it would be homeomorphic is obvious, but I would like a hint in showing that the the different reflections yield the same space.

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  • $\begingroup$ Hmm, I tend to make long questions. Should I be more concise? $\endgroup$
    – bbnkttp
    Apr 10 '13 at 10:11
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According to Poincare Duality, a compact $n$-manifold $M$ is orientable iff $H_n(M,\mathbb{Z}) \neq 0$. The homology groups of real projective space can be computed using a cell decomposition with only one cell in each dimension. This wikipedia article has a nice explanation of why this leads to a vanishing top dimensional homology group or $\mathbb{R} \mathbb{P}^n$ iff $n$ is even.

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  • $\begingroup$ This seems a very clean way of proving it, but I am yet to study homology groups. $\endgroup$
    – bbnkttp
    Apr 17 '13 at 9:19
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An orthonomal frame at the north pole of the 2 sphere can be brought to the south pole either by parallel translation along a great circle or by the antipodal map. If the projective plane were orientable then these two resulting frames would be the same.

But the translated frame and the identified frame have opposite orientation.

For odd spheres they have the same orientation so the odd projective spaces are orientable.

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As you mentioned in the question, $\pi:\mathbb{S}^n\rightarrow\mathbb{R}\mathbb{P}^n$ is a $\mathbb{Z}_2$-principal covering, with deck group $\{1,\alpha\}$, where $\alpha$ is the antipode map of $\mathbb{S}^n$. Since $\pi\circ\alpha=\pi$ we get pullbacks satisfying $\alpha^*\circ\pi^*=\pi^*$, which shows if $\mathbb{R}\mathbb{P}^n$ is orientable then $\alpha$ must be orientation preserving by pullbacking the same orientation on $\mathbb{R}\mathbb{P}^n$ to $\mathbb{S}^n$. But the antipode map $\alpha$ is orientation preserving iff $n$ is odd.

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