A name for a pseudo-Riemannian manifold that admits no nonzero null vectors Is there a name for a pseudo-Riemannian manifold that admits no nonzero null vectors? More precisely: For a pseudo-Riemannian manifold $(R,g)$, a null vector is a non-zero vector field $X:M \to TM$ such that
$$
g(X,X)(m) = 0,  \forall m \in M.
$$
As this [question][1] shows - vector like this exist. But can there exist manifolds where they do not exist and do such manifolds have a name?
 A: Note first that every pseudo-Riemmanian manifold admits a null vector field which is not identically $0$ (just construct one locally and multiply it by a bump function). So by "non-zero vector field" I assume you mean "nowhere vanishing".
Let $(M,g)$ be a pseudo-Riemannian manifold of signature $(p,q)$. The tangent bundle $TM$ always admits an orthogonal splitting as $E \overset{\perp}{\oplus} F$, where $E$ and $F$ are respectively positive and negative definite (hence of respective rank $p$ and $q$). Moreover this splitting is unique up to homotopy (because, pointwise, the set of such splittings is the symmetric space of the orthogonal group $O(p,q)$, which is contractible).
Proposition: $M$ admits a nowhere vanishing null vector field if and only if $E$ and $F$ both admit nowhere vanishing sections.
Proof: Decompose a nowhere vanishing null vector field $X$ as $X_E + X_F$. Then $g(X_E,X_E) = -g(X_F,X_F)$. If this is $0$ at some point then $X_E$ and $X_F$ vanish at that point (since $g$ is positive definite on $E$ and negative definite on $F$) contradicting the non-vanishing of $X$. Hence $X_E$ and $X_F$ are non-vanishing sections of $E$ and $F$.
Conversely, if $X_E$ and $X_F$ are non-vanishing sections of $E$ and $F$ respectively, then up multiplying $X_F$ them by a positive function, we can assume that $g(X_E,X_E) = -g(X_F,X_F)$. Hence $X_E+X_F$ is a nowhere vanishing null vector field. CQFD
There are thus topological obstructions to the existence of such a vector field (mainly the non-vanishing of the Euler class of $E$ or $F$). For instance, Let $(A,g_A)$ and $(B,g_B)$ be Riemannian manifolds, with $A$ of non-zero Euler characteristic, and consider $(M,g) = (A\times B, g_A \oplus -g_B)$. Then $M$ does not admit a nowhere vanishing null vector field. Indeed, we have the splitting $TM = TA\oplus TB$, and the projection of a null vector field to TA must vanish somewhere since the Euler class of TA is non-zero.
A: Such manifolds are precisely the Riemannian ones and their “opposites”.
Suppose $M$ is a pseudo-Riemannian manifold whose signature is not trivial in one direction or the other. We work in a neighbourhood of a point and will define a vector field with arbitrarily small support, so we may as well consider $M=\mathbb R^d$. Suppose also, after a linear change of coordinates, that the metric at zero is $dx_1^2+\cdots+dx_k^2-dx_{k+1}^2-\cdots-dx_d^2$. By hypothesis $0<k<d$.
For all x sufficiently close to 0, define $X(x)$ as the only vector inbetween $\epsilon_1:=(1,0,\ldots,0)$ and $2\epsilon_d:=(0,\ldots,0,2)$ of norm 0. If you accept for a moment that this is a well-defined smooth vector field, then a suitable multiple of $X$ will have very small support, hence be defined everywhere upon extending by zero, but will be precisely $(2/3,\ldots,2/3)$ at zero.
The fact that it is smooth comes from the inverse function theorem. We are looking for $(1-t,0,\ldots,0,2t)$ where $t$ is solution to
$$ (1-t)^2g_x(\epsilon_1,\epsilon_1) + 4t(1-t)g_x(\epsilon_1,\epsilon_d) + 4t^2g_x(\epsilon_d,\epsilon_d) = 0. $$
Single roots of a polynomial depend smoothly on the coefficients, so if we show that $t$ is a single root, then it will depend smoothly on the metric. But this is an open condition, and at $x=0$ we have
$$ (1-t)^2g_0(\epsilon_1,\epsilon_1) + 4t(1-t)g_0(\epsilon_1,\epsilon_d) + 4t^2g_0(\epsilon_d,\epsilon_d) = (1-t)^2 - 4t^2 = -(3t-1)(t+1) $$
so the root $1/3$ is simple.
