Why is $\chi_{[\mu}\nabla_\nu \chi_{\sigma ]} = 0$ at the Killing horizon?

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 $$\Sigma$$?

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.
• 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? Mar 19 '19 at 16:49
• @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 $n$. Mar 19 '19 at 21:06
• @Rodrigo Yes, that's exactly right. I'm assuming that the signature is Lorentzian, with signature $(n,1)$. Mar 19 '19 at 23:48