# The Hairy Ball theorem and (non-orientable) real projective plane

Is it possible to prove the Hairy Ball theorem via non-orientability of $P^2(\mathbb{R})$?

That is, the non-vanishing section $s \colon S^2 \to TS^2$ would induce (via “2-to-1” bundle $p \colon S^2 \to P^2(\mathbb{R})$) the non-vanishing “two-dimensional” field on $P^2(\mathbb{R})$, which could not exist? Drawing the simple picture hints that it could be possible, but I can't make the argument rigorous. Maybe the geometric intuition is misleading in this case? If yes, where is the obstruction?

(I hope the question, however amateur, is valuable enough.)

• What do you mean by "two-dimensional field"? And I know what is a pull-back of section, but you seem to be using push-forward... – Grigory M Jun 8 '11 at 10:53

Nonorientability of $\mathbb{RP}^2$ means that there is no nonvanishing section of $\bigwedge^2 T (\mathbb{RP}^2)$.
Suppose that $\sigma$ were a nonzero section of $T(S^2)$. For every point $x$ in $\mathbb{RP}^2$, there are two points $x_1$ and $x_2$ of $S^2$ lying above $x$. We can take the two vectors $\sigma(x_1)$ and $\sigma(x_2)$ and push them forward to tangent vectors at $x$; call the results $v_1$ and $v_2$. Note that there is no canonical way to number these; we just get an unordered pair of tangent vectors at $x$. We would like to build a section of $\bigwedge^2 T (\mathbb{RP}^2)$.
The only thing I see to do is look at $v_1 \wedge v_2$. But this doesn't work for two reasons (1) Since we don't have a well defined ordering of $v_1$ and $v_2$, the result is only defined up to sign. (2) Even if you could resolve the sign ambiguity, there is no particular reason that $v_1 \wedge v_2$ should be nonvanishing. Why couldn't you have a vector field $\sigma$ on $S^2$ that is everywhere nonvanishing but such that, at some pair of antipodal points $(x_1, x_2)$, the vectors $\sigma(x_1)$ and $\sigma(x_2)$ are parallel?