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For every line $l$ in $\mathbb{R}^3$ we write $l^\perp$ for the plane orthogonal to $l$. Let $F$ be : $$F = \{(u,l) | l\in\mathbb{P}^2(\mathbb{R}),u\in l^\perp\}$$

How do you show that this is not isomorphic to the trivial bundle on $\mathbb{P}^2(\mathbb{R})$.

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up vote 3 down vote accepted

Assume there were a non-vanishing section $g: P^2(\mathbb{R}) \rightarrow F$. Precomposing with the quotient map $S^2 \rightarrow P^2(\mathbb{R})$, one gets for each $v \in S^2$ a $w \in \mathbb{R}^3$ - namely the vector appearing in g(v) - which is nonzero and perpendicular to $v$. But this defines a nowhere vanishing vector field on $S^2$.

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Ok thanks a lot, but that uses (I think) the hairy ball theorem which I don't know how to prove (without looking on the internet). I was looking for something a little bit more elementary. If nobody gives a more elementary answer I'll accept yours though because it does indeed answer the question. –  Zorba le Grec Dec 17 '12 at 14:57
    
This would also be a contradiction to the Poincare-Hopf theorem, maybe you know that one? –  Kofi Dec 18 '12 at 8:39
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