Hyperboloids in Minkowski geometry

Question

Let $(\mathbb R^{1+2},\eta)$ be Minkowski with the metric $\eta= -dt^2+(dx^1)^2+(dx^2)^2$. Suppose $\Sigma$ is a smooth timelike hypersurface and denote by $h$ the second fundamental form on $\Sigma$. Assume that $$ h(N,N) \geq 0 \quad \forall N\in L_p\Sigma \quad \text{and}\quad p \in \Sigma,$$ where $L_p\Sigma= \{N \in T_p\Sigma\,:\, \eta(N,N)=0\}$. Assume also that there is no point $p$ on $\Sigma$ with the property that $h(N,N)>0$ for all nontrivial $N \in L_p\Sigma$. Does it follow that $\Sigma$ is part of a hyperboloid?

Answer 1

By hyperboloid, you seem to mean the one-sheeted surface $\{ |x|^2 - t^2 = 1\}$; this surface is umbilical with the induced metric proportional to the mean curvature, and so $h(N,N) = 0$ for all $N\in L_p\Sigma$.

Before answering your question, a couple comments:

1. It is meaningless to specify $h(N,N) \geq 0$ with a particular sign: the choice of the sign of the mean curvature scalar depends on the choice of orientation of the surface.
2. If you have a time-like surface in $\mathbb{R}^{2,1}$, the set $L_p\Sigma$ is the union of two lines. Locally (and in fact globally under mild assumptions) there exists two non-vanishing linearly independent vector fields $\ell, n$ generating $L_p\Sigma$. Your requirements is essentially stating that $h(\ell,\ell)h(n,n) = 0$ and $h(\ell,\ell)+h(n,n)$ is signed.

Certainly it is not necessary for $\Sigma$ to be part of the hyperboloid for this to hold. As Leo Moos pointing out in the comment, just taking $\Sigma$ to be a time-like hyperplane guarantees that $h \equiv 0$ and hence your requirement is satisfied. More generally, if you take any time-like surface $\Sigma$ with the property that it is ruled by null lines, then this guarantees one of $h(\ell,\ell)$ or $h(n,n)$ vanishes identically. It then suffices to bend the surface so that the other family of null curves accelerate only in one way.

For example: you can let $\Sigma$ be the graph $\{x^1 = f(t, x^2)\}$ where $f(t,x^2) = g(t - x^2)$ with $|g'| < 1$ and $g'' \geq 0$.