No Lorentzian metric on $S^{2}$

Question

In the book Riemannian Geometry by Gallot et al there is a remark at the beginning that there is no Lorentzian metric on $S^{2}$. Is it a difficult theorem? Or there is an easy solution? Any hint/idea how to prove this?

Answer 1

See this MathOverflow answer by Andrei Moroianu. The intuition here is that on simply connected manifolds the Euler characteristic is the obstruction because, roughly speaking, a Lorentz metric must be able to assign a consistent timelike direction. Moroianu's answer makes this intuition precise.

Answer 2

A Lorentzian structure would produce in every tangent space a light cone. In a $2$-dim case like $S^2$ that would be a pair of $1$-dimensional subspaces. Since $S^2$ is simply connected you can smoothly choose them apart. So now you have two $1$-dimensional linear distributions (subspace in the tangent space at each point). From this you can get a nonzero vector field on the sphere, by taking an orientation on the line ( again possible by simple-connectedness) and then taking a unit vector ( for say the standard Riemann metric). But this is not possible.