# Why are trapped surfaces required to be closed and orientable?

On pgs. 172 and 203 of A Relativist's Toolkit, Poisson defines a trapped surface to be closed. The Wikipedia article additionally requires trapped surfaces to be compact and orientable. What is the motivation for these definitions? It seems to me that, say, a hemispherical half of a trapped surface surrounding a black hole singularity is just as useful of a concept as the full spherical surface. In particular, you could still fill up the interior of a trapping horizon with open as well as closed trapped surfaces.

• It's a shame this hasn't had an answer because it's a really good question – tfb Apr 25 '17 at 21:51
• @tfb Feel free to offer a bounty! – tparker Apr 26 '17 at 1:09
• I'm looking for my copy of Hawking & Ellis to see if I can remember how it works. – tfb Apr 26 '17 at 22:26

"Open trapped surfaces are not a useful concept because any point in any spacetime (including Minkowski) lies on an open trapped surface. This is quite easy to see in Minkowski space: consider two spacelike separated points and let $M$ be the intersection of their past lightcones (note that $M$ is by construction everywhere spacelike). By construction, the two orthogonal congruences to $M$ are these past lightcones, which have negative future-directed expansions, so $M$ is an (open) trapped surface. But any point in Minkowski space lies on such an M, so open trapped surfaces fill all of Minkowski space. ... The global properties of trapped surfaces (specifically their compactness) are crucial for obtaining the singularity theorems."
Just to give some intuition for what $M$ looks like in $D$-dimensional Minkowski spacetime: if we work in a Lorentz frame in which the two spacelike-separated points lie at $x = \pm a$, with all other coordinates zero, then $M$ is given by $x = 0,\ t = -\sqrt{a^2 + {\bf y}^2}$, where ${\bf y}$ is the codimension-2 spatial vector of directions perpendicular to the $x$-direction. $M$ is therefore one sheet of a codimension-2 hyperboloid of two sheets embedded within the timelike hypersurface $x = 0$. $M$ can be equivalently reparameterized as $x = 0,\ t \leq -a,\ |{\bf y}| = \sqrt{t^2 - a^2}$. For an observer confined to the timelike hypersurface $x = 0$ (the plane bisecting the two points, in 4D), $M$ looks like a spatial codimension-3 sphere (a circle, in 4D) centered at the origin, which starts at $t = -\infty$ with infinite radius and contracting at the speed of light, then contracts faster and faster and shrinks down to a point and vanishes at $(t = -a, {\bf y} = {\bf 0})$, when its radius is contracting infinitely quickly. If this shrinking ring were shining, then all of the light rays it emitted would reach the two points $(\pm a, {\bf 0})$ at the exact same time $t = 0$, so the null congruences have negative expansion. $M$ can be trapped even in the absence of a black hole because it is not compact: it extends infinitely far into the past, and becomes infinitely large there.