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How to prove that $\mathbb{RP}^2$ isn't orientable? My book (do Carmo "Riemannian Geometry") gives a hint: "Show that it has a open subset diffeomorphic to the Möbius band", but I don't know even who is the "open subset".

azimut
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Lonely Penguin
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  • Try following the hint: look for an open subset of the 2-sphere which becomes a Mobius band after you identify antipodal points. – Matt E Apr 18 '12 at 16:46

3 Answers3

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For geometric understanding of the real projective plane, I prefer to think of it as gotten from a closed circular disk by identifying opposite points on the boundary. If you accept this, then a Möbius band subset is easily found: take any diameter of your original circle, and widen it to a strip.

Lubin
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$\mathbb{RP}^2$ is the 2-sphere, modulo the relation identifying opposite points. The open subset referred to is a neighborhood of the equator on the sphere. Consider the tangent vector at the equator pointing upwards, and rotate the sphere horizontally by $\pi$. Using the quotient map from the sphere to $\mathbb{RP}^2$, you will see that your tangent vector is in the same fiber as your original tangent vector, but which way will the tangent vector point now?

Brett Frankel
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  • thanks for the comment, your answer is intuitive but I'd like a analytical proof. – Lonely Penguin Apr 18 '12 at 02:59
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    If you've learned about tangent spaces, then everything I have said can be made precise. If not, try translating my explanation into local coordinates (one coordinate pointing west, one pointing north to start). – Brett Frankel Apr 18 '12 at 03:01
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Here are some general facts which will help us to solve the problem on a more general level.

  1. The Antipodal map $a:S^n\longrightarrow S^n$ on the n-sphere is defined by $$a(x^1,\dots,x^{n+1})=(-x^1,\dots,-x^{n+1})$$ is orientation preserving map iff n is odd.

  2. Suppose $f(x^1,\dots,x^{n+1})$ is a $\mathbb{C}^{\infty}$ function on $\mathbb{R}^{n+1}$ with 0 as a regular value. Then the zero set of $f$ is an orientable submanifold of $\mathbb{R}^{n+1}$, In particular $S^n$ is orientable in $\mathbb{R}^{n+1}$ So use these facts and you will get any odd dimensional projective space is orientable.

Myshkin
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