Vector fields on pseudo-Riemmannian manifolds For a pseudo-Riemannian manifold $(R,g)$, is it possible to have a non-zero vector field $X:M \to TM$ such that
$$
g(X,X)(m) = 0,  \forall m \in M?
$$
Edit: As the comment says, I am asking if a pseudo-Riemannian manifold admits a global non-zero lightlike vector.
 A: For arbitrary pseudo-Riemannian manifolds: the answer is no in general.
Take the standard Moebius strip, parametrized as $(t,x) \in \mathbb{R}\times [0.1]$ with $(t,0)$ identified with $(-t,1)$.
One can check that the flat metric $m = -dt^2 + dx^2$ is Lorentzian and smooth on the strip.
A global non-vanishing light-like vector will have a global non-vanishing projection on the $t$ component, but as is well-known for the Moebius strip this is impossible.

This construction, however, is an issue specific to Lorentzian manifolds.
Let $N_p M$ denote the set of all nonzero null vectors at $p\in M$.

*

*When $M$ is Riemannian, $N_pM = \emptyset$

*When $M$ is Lorentzian, $N_pM$ has multiple connected components (when dimension is 2 there are 4 components, and when the dimension is >2 there are 2 components).

*When $M$ is neither Riemannian nor Lorentzian, $N_pM$ is connected.

Our topological construction relies entirely on the non-time-orientability of the given Lorentzian manifold, which implies that lack of a continuous (over $p$) selection of a component of $N_pM$. For non-Lorentzian pseudo-Riemannian manifolds, this is obviously not a problem.

There can be, however, further topological constraints: for example, on $\mathbb{R}\times\mathbb{S}^2$ with $g = -dt^2 + g_{\mathbb{S}^2}$, if you take a global null vector field, its spatial projection will be a non-vanishing vector field on $\mathbb{S}^2$, which can be ruled out by the hairy ball theorem.
