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Let P be the real projective plane obtained by identifying antipodal points on the unit sphere of $R^3$.

How to prove that P is nonorientable in a rigorous and elementary way? I do not want mere intuition.

My idea is to consider the closed curve $a(t)=(\cos t,\sin t, 0)$ , $0 \leq t \leq \pi$

Then the tangent vectors $a'(0)$ and $a'(\pi)$ are identical.

However, for a vector field V on $a(t)$ defined by $V(a(t))=(0,0,1)$, the tangent vectors at $a(0)$ and $a(\pi)$ differ by a sign. Therefore the normal vector cannot be continuously defined.

Is the constant vector field V continuous on the curve? Are my arguments right?

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Preliminary Result
If $M$ is an orientable manifold and if $(U,\phi),(V,\psi)$ are two charts with $U,V$ connected, then the jacobian determinant of the change of coordinates $\psi\circ \phi^{-1}$ has constant sign (even if $U\cap V$ is not connected)

Application
Let us apply this to projective space and to its two charts $$\phi : U=\{x\neq 0\}\to \mathbb R^2:[x:y:z]\mapsto (y/x,z/x)$$

$$\psi : V=\{y\neq 0\}\to \mathbb R^2:[x:y:z]\mapsto (x/y,z/y)$$
The change of coordinates $\psi\circ \phi^{-1}$ is the morphism $$\mathbb R^2\setminus \{u=0\}\to \mathbb R^2\setminus \{u=0\}:(u,v)\mapsto (\frac 1u,\frac vu)$$
whose jacobian determinant is $\frac {-1}{u^3}$.
Since this jacobian determinant changes sign on $\mathbb R^2\setminus \{x=0\}$, $\mathbb P^2$ is non-orientable as a consequence of the Preliminary Result.

Remark
The Preliminary Result seems not to be often mentioned in books on differential geometry.
It is a pity since as a consequence non orientability of manifolds is often treated in a hand-waving fashion.

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  • $\begingroup$ Do you think that my arguments are correct? $\endgroup$ – noot Oct 14 '13 at 8:18
  • $\begingroup$ Dear noot, I'm not sure I understand it. $\endgroup$ – Georges Elencwajg Oct 14 '13 at 8:24
  • $\begingroup$ I wanted to show that there cannot be 2 form on the projective plane that is nonzero on any two linearly independent tangent vectors. I wanted to show such 2 form must be zero on two linearly indepent tangent vectors somewhere. $\endgroup$ – noot Oct 14 '13 at 8:34
  • $\begingroup$ But you didn't prove that every two-form vanishes on $\mathbb P^2$. Anyway, I wrote a proof answering the question in your second sentence and that's final as far as I'm concerned. I'll let other interested users discuss your arguments. $\endgroup$ – Georges Elencwajg Oct 14 '13 at 8:46

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