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I raised a related question but hope to get some answer using the nonvanishing 2 form definition.

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 by showing that any 2 form on P will vanish somewhere?

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

This curve is closed in P and 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. Also V and $\alpha'(t)$ are always linearly independent.

For any 2 form $u$ on P,

$u(a'(0), V(a(0))=-u(a'(\pi), V(a(\pi))$ Therefore $u$ must vanish somewhere on the curve.

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

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If $\omega $ is a nonzero two form on $P$ then define $\pi^\ast \omega$ on $S^2$ where $\pi : S^2\rightarrow P$. Note that $$ \pi^\ast \omega(e_1,e_2)=\omega (d\pi e_1,d\pi e_2)\neq 0$$ where $\{ e_1,e_2\}$ is indepedent.

This implies that $\pi^\ast \theta$ is nonzero form on $S^2$. And note that $$ (\pi^\ast \omega)_p (e_1,e_2) = (\pi^\ast \omega)_{-p}(-e_1,-e_2)\neq 0 $$

Let $N$ is unit normal on $S^2$ and $\theta = (N,\ )$

Then $(\pi^\ast \omega)\wedge\theta$ is a nonzero three form. If $\{ (1,0,0),e,N \}$ is an orthnormal frame along $c(t)=( 0,\sin\ t,\cos\ t),\ 0\leq t\leq \pi$, and if $e(0)=(0,1,0)$, then $$ (\pi^\ast \omega \wedge \theta )_{c(t)} ((1,0,0),e,N)\neq 0 $$

So $(\pi^\ast \omega)_{c(0)} ((1,0,0),(0,1,0))$ and $(\pi^\ast \omega )_{c(\pi)} ((1,0,0),(0,-1,0))$ have same sign. So $$e_1=(1,0,0),\ e_2=(0,1,0),\ p=(0,0,1),\ (\pi^\ast \omega)_p (e_1,e_2) (\pi^\ast \omega)_{-p}(e_1,-e_2)>0 $$ It is a contradiction.

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  • $\begingroup$ I do not understand the last 4 lines $\endgroup$ – noot Oct 15 '13 at 16:38
  • $\begingroup$ Note that $\pi^\ast \omega\wedge \theta$ is nonzero three form since $\pi^\ast \omega$ is a nonzero two from on tangent space of sphere and $\theta$ is dual of $N$. And $\{ (1,0,0),e, N\}$ is linearly independent. $\endgroup$ – HK Lee Oct 15 '13 at 16:43
  • $\begingroup$ can you just use 2 form? i need to prove it with 2 form. $\endgroup$ – noot Oct 15 '13 at 16:47
  • $\begingroup$ how to derive your last equation $\endgroup$ – noot Oct 15 '13 at 16:50

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