Let $\chi$ be a Killing vector field that is null along a Killing horizon $\Sigma$
Why is $\chi_{[\mu}\nabla_\nu \chi_{\sigma ]} = 0$ at $\Sigma$?
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1 Answer
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This is a partial answer. It assumes that the Killing field $\chi$ is normal to the Killing horizon. This implies that $\chi$ is null along the horizon, so it is consistent with the condition given in the OP. However, a Killing field that is null along the horizon is not necessarily orthogonal to the horizon; that's why this is only a partial answer.
Given a function $\Phi$ on spacetime and a constant $c$, the equation $\Phi=c$ defines a hypersurface. The vector field $$ n_a=\partial_a\Phi=\nabla_a\Phi \tag{1} $$ is orthogonal to that hypersurface. Now, if $\chi_a$ is any vector field orthogonal to this family of hypersurfaces, we must have $$ \chi_a=mn_a \tag{2} $$ for some function $m$. If $\chi$ is normal to the Killing horizon, then it can be written in the form (2) along the horizon.
Equation (2) implies $$ \nabla_b\chi_a=\nabla_b(m\nabla_a\Phi) = (\nabla_b m)(\nabla_a\Phi) + m\nabla_b\nabla_a\Phi. \tag{3} $$ Now consider $\chi_{[c}\nabla_b \chi_{a ]}$, where the square brackets denote complete antisymmetrization (which is what I'm assuming they mean in the OP). According to the preceding equations, $$ \chi_{[c}\nabla_b \chi_{a ]} = m(\nabla_{[c}\Phi)(\nabla_b m)(\nabla_{a]}\Phi) + m^2\nabla_{[c}\Phi\nabla_b\nabla_{a]}\Phi. \tag{4} $$ The first term in (4) is zero because of the identity $$ (\nabla_{[c}\Phi)(\nabla_{a]}\Phi) = 0, \tag{5a} $$ and the second term in (4) is zero because of the identity $$ \nabla_{[b}\nabla_{a]}\Phi = 0 \hskip2cm \text{(zero torsion)}. \tag{5b} $$ Altogether, this shows that $$ \chi_{[c}\nabla_b \chi_{a ]} = 0 \tag{6} $$ whenever the vector field $\chi$ is orthogonal to the given hypersurface. This is (part of) the Frobenius theorem.
This derivation of equation (6) is outlined in section 2.3.3 in Poisson (2002), "An advanced course in general relativity," https://www.physics.uoguelph.ca/poisson/research/agr.pdf.
answered Mar 19, 2019 at 13:17
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$\begingroup$ If we assume that $\chi$ is not only null along $\Sigma$, but is also in the tangent space $T\Sigma$, then it will be normal to $\Sigma$, correct? $\endgroup$– RodrigoCommented Mar 19, 2019 at 16:49
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1$\begingroup$ @Rodrigo Correct. Since $\Sigma$ is a null hypersurface, which by definition has a null normal vector $n$ at each point, the only way another vector $\chi$ at the same point can be both null and orthogonal to $n$ (so that $\chi\in T\Sigma$) is if $\chi$ is proportional to $n$. $\endgroup$ Commented Mar 19, 2019 at 21:06
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1$\begingroup$ and that's because the metric signature is (3,1), right? It wouldn't be true if the signature were (2,2), for example. $\endgroup$– RodrigoCommented Mar 19, 2019 at 22:10
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$\begingroup$ @Rodrigo Yes, that's exactly right. I'm assuming that the signature is Lorentzian, with signature $(n,1)$. $\endgroup$ Commented Mar 19, 2019 at 23:48