Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{P}^2(\mathbb{R})$.

share|cite|improve this question
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 $S^2$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.