When can a cone be a lightcone?

Question

Suppose I am given a cone of vectors in $\mathbb{R}^n$ with $n\geq3$, which we can take to be an $(n-2)$-parameter family of vectors $V^a(\theta_1,\ldots,\theta_{n-2}$) along with all scalar multiples of these vectors. Is there a way to determine whether there exists a Lorentzian metric $\eta_{ab}$ (signature $(-,+,+,\ldots)$) such that this cone of vectors is the lightcone for the metric, i.e. the set of null vectors satisfying $\eta_{ab} V^a V^b = 0$?

There is obviously a brute force method where you pick $n$ vectors and solve for the metric components (up to an overall conformal factor) required to make these $n$ vectors null, and then check whether the rest of the vectors in the cone also are null, but this is rather cumbersome, and I was wondering if there is a more direct way to test whether the cone can be a lightcone of some Lorentzian metric.

Answer 1

The key necessary and sufficient condition is that there exist an invertible linear transformation $T$ such that the image of the cone under $T$ is equal to the "standard" light cone $-x_1^2+x_2^2+...+x_n^2=1$. The reason this holds is that every Lorentz metric is equivalent to the standard one under some linear transformation.

You could now try to translate this necessary and sufficient condition into other languages. For example, one such condition is that in projective space $\mathbb{R}P^{n-1}$ the projectivization of your cone is equivalent, under a projective transformation, to the standard embedding of $\mathbb{R}P^{n-2}$.

You could also translate this into the language of "conic sections". That is, the intersection of your cone with any codimension-1 plane not passing through the origin is a conic section (as appropriately defined in dimension $n$).