# Derivatives of distributions in general relativity

I am having some trouble when trying to reproduce some calculations involving the description of distributions (mostly used in spacetime junction conditions).

I am trying to reproduce the calculations in this paper. In the appendix, I can't derive this equality:

$$\nabla_\nu([R]n_\mu \, \underline{\delta}^\Sigma)= \underline{\delta}^\Sigma \left( [R]K^\Sigma_{\mu\nu} - [R]K^\rho_{\,\,\rho}n_\mu n_\nu +n_\mu \overline{\nabla}_\nu [R] \right) + \underline{\Delta}_{\mu\nu},$$ where an underline designates a distribution, $\overline{\nabla}_\nu$ is the covariant derivative on the hypersurface, $[R]$ is the scalar curvature jump across the hypersurface $[R]\equiv R^+|_\Sigma- R^-|_\Sigma$, the extrinsic curvature is $$K_{\mu\nu}=h^\rho_{\,\,\mu}h^\sigma_{\,\,\nu}\nabla_\rho n_\sigma,$$ where $n$ is the normal to the hypersurface and $h$ is the projector to the hypersurface $\Sigma$, $h_{\mu\nu}=g_{\mu\nu}-n_\mu n_\nu$. The distribution $\underline{\Delta}$ includes the $\delta'$ term. This is all in Eq.(37) of the paper, and the (unnumbered) equation right before it.

My problem is, instead of getting the above equality, I keep getting only $$\nabla_\nu([R]n_\mu \, \underline{\delta}^\Sigma) = \underline{\delta}^\Sigma \left( [R]K^\Sigma_{\mu\nu}+n_\mu \overline{\nabla}_\nu [R] \right) + \underline{\Delta}_{\mu\nu}.$$

If someone could point me out to what I might be doing wrong, I'd appreciate it.

The problem actually extends to eqn (38), mainly the $-2\alpha \delta$ term, which corresponds to the $f''(R)$ term in the eqn below eqn (31). I can't find where the $-K^\rho_{\,\,\rho} n_\mu n_\nu$ comes from, nor why there is no $-K^\rho_{\,\,\rho}g_{\mu \nu}$ term arising from $-g_{\alpha \beta}\nabla^\rho \nabla_\rho R$ of eqn (31).