How do we explicitly prove that the induced covariant derivative on non-expanding horizons (NEH) reduces to the rotational one-form connection?

Question

I have been studying the formulation of Isolated Horizons recently. As a reference I will follow Isolated horizons in classical and quantum gravity by Engle and Liko. But, I will be using the notation in Matthias Blau lectures (you will see why). In this paper, I do not understand how they get the following relation on the non-expanding horizon $\Delta$, \begin{align} D_a \ell_b = \omega_a \ell_b. \end{align} In this relation, the indices are spacetimeindices which confuses me and there is a hat over the equality stating that the equality only holds on $\Delta$. To avoid confusing myself, I will use $\mu, \nu$ for spacetime indices, k, m for indices over the non-expanding horizon $\Delta$, and a, b for the $S^2$ cross section indices similar to Matthias Blau lectures. I will not use arrows for pull back since it is easy to miss what actually is being pulled back. For pull back, I use the notation in Matthias Blau lectures again. So, we take the spacetime as a Lorentzian manifold, ($\mathcal{M}$,$g_{\mu\nu}$). Define $s_{\mu\nu}=g_{\mu\nu}+(\ell_\mu n_\nu + \ell_\nu n_\mu)$ with the null normal $\ell$ and the auxiliary null vector n. We consider a non-expanding horizon $\Delta$ as defined in the same paper at the end of page 9. Then, I believe the induced covariant derivative in $\Delta$ is defined by the pull back of the covariant derivative in $\mathcal{M}$, $\nabla_\mu$. The induced covariant derivative can be calculated by $D_k=E^\mu_k \nabla_\mu$ corresponding to the induced metric $q_{km}$ on $\Delta$ such that $D_k q_{mn}=0$. Finally, something that I do not understand, we somehow have \begin{align} D_k \ell_\mu=\omega_k \ell_\mu? \end{align} I am pretty sure this is wrong. But, I do not understand how else this induced covariant derivative should be defined.

Answer 1

First, "$D_k = E^\mu_k\nabla_\mu$" is not quite correct. Just pulling back the index of the derivative operator is not enough, you need to pull back the indices of the tensor it operates on. For example, $$D_k t_m = E^\mu_k E^\nu_m \nabla_\mu t_\nu \,.$$ The opertor $D_k$ is completely instrinsic to the 3-manifold and therefore only acts on the tensors intrinsic to it. Therefore the expressions $D_a \ell_b$ and $D_m \ell_\mu$ do not make sense. This is because $\ell_\mu$ is not intrinsic to the 3-manifold. Instead, the result should be written as $$ D_k \ell^m = \omega_k \ell^m\,.$$

You can derive the result using the three properties of the null generator $\ell^\mu$, namely:

1. $\ell^\mu$ is hypersurface orthogonal by definition, and hence twist free ($E_k^{\mu} E_m^{\nu} \nabla_{[\mu} \ell_{\nu]} = 0$)
2. $\ell^\mu$ is expansion-free by definition ($ q^{ab} E_a^{\mu} E_b^{\nu} \nabla_{\mu} \ell_{\nu} = 0$)
3. $\ell^\mu$ is shear-free, which follows from Raychaudhuri equation ( symmetric trace-free part of $E_k^{\mu} E_m^{\nu} \nabla_{\mu} \ell_{\nu}$ vanises)

Together, the three equations imply that $E_k^{\mu} E_m^{\nu} \nabla_\mu \ell_\nu = 0$. This tells us that $E_k^{\mu} \nabla_\mu \ell_\nu$ has vanishing pull-back and therefore must be proportional to $\ell_\nu$. So we get $$E_k^{\mu} \nabla_\mu \ell_\nu = \omega_k \ell_\nu$$ for some one-form $\omega_k$. Raising the index, we get $E_k^{\mu} \nabla_\mu \ell^\nu = \omega_k \ell^\nu$, which can now be pulled back to the 3-manifold because $\ell^\mu$ is tangential, giving us $$D_k \ell^m = \omega_k \ell^m$$