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But I can't follow the last statement. First, I have no idea what is wrong with having two linearly independent null tangent vectors in a null surface and, second, I have no idea how it is related to the fact that ##\chi## is normal to ##\Sigma##.

One example related to this issue is the case of a Kerr BH. For a Kerr BH, we can think of a Killing vector $$\chi=\partial_t+\Omega_H\partial_\phi,$$ and it is straightforward to show ##\chi^\mu\chi_\mu=0## on $$\Sigma:~r=r_H=M+\sqrt{M^2-a^2},$$ so ##\Sigma## is a Killing horizon of ##\chi##. But this ##\chi## doesn't look like normal to ##\Sigma##, since the tangent vectors on ##\Sigma##, such as ##\partial_t,~\partial_\theta,~\partial_\phi## which are obviously orthogonal to the normal vector of ##\Sigma##, $$n_\mu=\nabla_\mu f(r)=(0,1,0,0)~~(f(r)=r-r_H)$$, are not orthogonal to ##\chi##. The same issue can be rephrased as "##\chi_\mu## doesn't look parallel to ##n_\mu##."

I would really appreciate if you help me out with this trouble...! Even if you are not sure about the general argument in Carroll's, answer for the Kerr BH example would be just enough!