Even dimensional real projective spaces cannot be combed

Question

I have to prove that the even-dimensional real projective space cannot be combed, i.e. there isn't any non-vanishing smooth vector field.

(I can't use Hopf theorem since those manifolds are not orientable.)

What I think I need to use is that even-dimensional spheres cannot be combed neither.

Somehow, my idea is the prove the statement by contradiction, using the fact that $\mathbb{R}P^{2n+1}$ is obtained from $S^{2n+1}$ identifying antipodal points.

But, I cannot proceed further.

Any help?

Answer 1

The canonical map $\pi \colon S^k \to \mathbb{R}P^k$ is a local diffeomorphism. Let $v$ be a vector field on the projective space. On every small enough open $U \subset S^k$ (small enough means it doesn't contain any antipodal points) you can define a smooth vector field $\hat{v}_U$ via the push-forward through the inverse of the projection:

$$\hat{v}_U = \bigl((\pi\lvert_U)^{-1})_{\ast}v.$$

The locally defined fields coincide on the intersections of their domains, hence you get a globally defined smooth vector field $\hat{v}$ on the sphere. Since the differential of $\pi$ is an isomorphism of tangent spaces at each point, $\hat{v}$ has zeros if and only if $v$ has zeros.

Answer 2

$\pi:S^{2n}\to \mathbb{R}P^{2n}$ is a local diffeomorphism.

Let $v:\mathbb{R}P^{2n}\to T\mathbb{R}P^{2n}$ be a non-vanishing smooth vector field.

For every $p\in \mathbb{R}P^{2n}$, let's consider $v_p\in T_p\mathbb{R}P^{2n}$ and such that $v_p\ne 0$.

Consider $\pi^{-1}(p)=\{p_1,p_2\}$. Let's choose $p_1$.

Consider $U\subset S^{2n}$ neighborhood of $p_1$ sufficiently small.

We have

$$ \pi_{|U}:S^{2n}\to\mathbb{R}P^{2n} $$ diffeomorphism.

Hence $$ d(\pi_{|U})^{-1}:T\mathbb{R}P^{2n}\to TS^{2n}_{|U} $$ is well-defined.

Define $\tilde{v}:=d(\pi_|U)^{-1}v$. $\tilde{v}$ gives the desired contradiction.