Projectivization of the tangent bundle to $S^2$ is nontrivial I want to prove that the projectivization of the tangent bundle to $S^2$ is nontrivial. It looks very similar to the hairy ball theorem.
Also, I would like to understand if the projectivization of the tangent bundle to $X$ is nontrivial for arbitrary closed orientable surface $X$ of genus $g>1$.
These facts seem to be well-known and elementary, so I would be grateful for references.

Lemma. Let $M$ be a smooth compact connected manifold. Then $PT(M)$ admits a section if and only if $M$ has a nonzero vector field. 
Proof. Suppose that $PT(M)$ admits a section $s$. This section yields a real line  bundle $L\to M$ over $M$. This bundle may or may not be orientable. If it is orientable then we obtain an orientation on the  line field on $M$ corresponding. Equip $M$ with a Riemannian metric. For each $p\in M$ take the unit tangent vector $X_p$ whose span is the section $s(p)\subset TM$ and which is consistent with the orientation of $s(p)$. Thus, $M$ admits a nonvanishing vector field. 
Suppose next that $L\to M$ is nonorientable. The obstruction to orientability is the orientation homomorphism $\rho: \pi_1(M)\to {\mathbb Z}_2$. Let $\tilde M\to M$ be the 2-fold covering corresponding to the kernel of $\rho$. Then the pull-back of $L$ to $\tilde M$ is an orientable line bundle. Thus, by the 1st part of the proof, $\tilde M$ admits a nonvanishing vector field, i.e. $\chi(\tilde M)=0$ (P-H theorem).  Since $\chi$ is multiplicative under covering maps, $\chi(M)= \frac{1}{2}\chi(\tilde M)=0$, i.e. $M$ also admits a nonvanishing vector field. 
The opposite implication (the existence of a nonvanishing vector field implies a section of $PT(M)$) is obvious. qed 
In particular, if $S$ is a connected closed surface with $\chi(S)\ne 0$ then $S$ does not admit a line field; in particular, $PT(S)$ is nontrivial. One can do a bit better: $e(PT(S))=2 \chi(S)$ where $e$ is the Euler number (assuming that $S$ is orientable). 
This argument is frequently used to prove that closed surfaces with $\chi\ne 0$ do not admit smooth foliations. 
