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This is a very basic question, but I have trouble finding a concise answert to that without having to go through a 50+ page paper/references. For simplicity, assume we are in 2+1 and consider a simple spherically symmetric metric: $-\alpha^2(t,r) dt^2+a^2(t,r) dr^2 + r^2 d\theta^2$ FInding trapped surfaces in general means that the outgoing expansion $$\theta_- := \nabla_a U_-^a < 0,$$ Now, I could parametrize the codimension 2 surface $S = S_1 \cap S_2$ as $$f_1(t,r,\theta) = c_1, \qquad f_2(t,r,\theta) = c_2,$$ where $c_1 and c_2$ are some constants. Note that in 2+1, this "surface" is really just a curve (dimension 1). I then could define the associated normal vectors $$n_{1,a} := \nabla_a f_1, \qquad n_{2,a} := \nabla_a f_2, \qquad.$$ These covectors are clearly normal to $S_1$ and $S_2$ respectively. However, I don't necessarily have that $n_{1,a}$ is normal to $S_2$, nor $n_{2,a}$ is normal to $S_1$. Not to mention that even if it were so, I would still not have a PDE (or PDI I guess for partial differential inequality?) just in terms of the metric variables?

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  • $\begingroup$ Generically, the $\mathcal{S}$ you construct this way would be co-dimension 2 not 1. $\endgroup$ – mmeent Oct 17 '19 at 11:11
  • $\begingroup$ Yes, sorry for the typo. I corrected it. $\endgroup$ – Patrick.B Oct 17 '19 at 12:54

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