Sources on Screens and the Gibbons-Hawking term In the theory of junction conditions, the Riemann tensor in terms of the Heaviside theta function is expressed as follows
$$R^{\alpha}_{\beta\lambda\mu} = \Theta(l)R^{\alpha^{+}}_{\beta\lambda\mu} + \Theta(-l)R^{\alpha^{-}}_{\beta\lambda\mu} + \delta(l)A^{\alpha}_{\beta\lambda\mu}.$$
Now, the trace of tensor $A^{\alpha}_{\beta\lambda\mu}$ reads 
$$A = \epsilon\left(\kappa_{\rho\nu}n^{\rho}n^{\nu} - \epsilon\kappa \right),$$
where $\epsilon = n^{\lambda} n_{\lambda}$ and hence, the Ricci scalar has the form, $R = \Theta(l)R^{+} + \Theta(-l)R^{-} + \delta(l)A$. Would the Action of the Ricci scalar yield the Gibbons-Hawking term?
My attempt: 
I am able to obtain the term but seem to be missing a factor of $2$.
$$S = \frac{1}{16\pi G}\int_{M}d^{4}x\ \sqrt{g}R = \frac{1}{16\pi G}\left[\underbrace{\int_{M}d^{4}x\sqrt{g} \left(\Theta(l)R^{+} +\Theta(-l)R^{-} \right)}_{=0} + \int_{M}d^{4}x\sqrt{g}\delta(l)A\right]$$
$$ = \frac{1}{16\pi G}\int_{M}d^{4}x\ \sqrt{g}\frac{d}{dl}\left(\Theta(l)A \right) \underbrace{=}_{Gauss'\ th.} \frac{1}{16\pi G}\int_{\partial M}d^{3}x\ \sqrt{h}n_{\mu}A^{\mu} \Theta(l),$$
since $\delta(l) = \frac{d\Theta(l)}{dl}$. Expanding $A$, we end up with the Gibbons-hawking term and the counter term, but I am ending up with a $1/16\pi G$ instead of a $1/8\pi G$,
$$S = \frac{1}{16\pi G}\int_{\partial M}d^{3}x\ \sqrt{h}\ \epsilon\  \kappa_{\alpha \beta}n^{\alpha}n^{\beta} \Theta(l) - \frac{1}{16\pi G}\int_{\partial M}d^{3}x\ \sqrt{h}\ \epsilon^{2}\kappa \Theta(l). $$
 A: I don't think the GHY term can be obtained this way, at least not directly.


*

*The expression $A$ does not contain the trace of the second fundamental form, but rather the jump of the trace of the second fundamental form. If $\kappa_{\mu\nu}$ is defined as $[\partial_\kappa g_{\mu\nu}]=\epsilon n_\kappa g_{\mu\nu}$, then $$ [K_{\mu\nu}]=h^\kappa_\mu h^\lambda_\nu [\nabla_\kappa n_\lambda]=-h^\kappa_\mu h^\lambda_\nu [\Gamma^\rho_{\kappa\lambda}]n_\rho=\frac{1}{2}h^\kappa_\mu h^\lambda_\nu \kappa_{\kappa\lambda}=\frac{1}{2}\kappa^\parallel_{\mu\nu}. $$ Now $$ \kappa=\kappa_{\mu\nu}g^{\mu\nu}=\kappa_{\mu\nu}(h^{\mu\nu}+\epsilon n^\mu n^\nu)=\kappa^\parallel+\epsilon\kappa_{\mu\nu}n^\mu n^\nu, $$ thus $$ A=\epsilon(\kappa_{\mu\nu}n^\mu n^\nu-\epsilon(\kappa^\parallel+\epsilon \kappa_{\mu\nu}n^\mu n^\nu)) = -\kappa^\parallel =-2[K]. $$

*With this, the action is (without the factor of $(16\pi G)^{-1}$) $$ S=\int\mathrm d^4x\sqrt{-g}(\bar R-2[K]\delta(l))=\int\mathrm d^4x\sqrt{-g} \bar R-2\int_\Sigma\mathrm d^3\xi\int_\mathbb R\mathrm dl\sqrt{|h|}[K]\delta(l) \\ =\int\mathrm d^4x\sqrt{-g} \bar R-2\int_\Sigma\mathrm d^3\xi\sqrt{|h|}[K], $$ where $\bar R=R^+\Theta(l)+R^-\Theta(-l)$ is the regular part of the scalar curvature $R$.

*From this one may infer that the GHY term is $\int_{\partial M}\mathrm d^3\xi\sqrt{|h|}(-2)K$, but what have been obtained is a difference of two such terms, and if all quantities involved are well-behaved, then on the boundary of $M$ such term would not appear, unless $g_{\mu\nu}$ happens to be nondifferentiable at the boundary. After all, if $g_{\mu\nu}$ is $C^1$, then we have $K^+=K^-$ and as such $[K]=0$.

*This approach can, however, be used to derive junction conditions if we suppose that along a hypersurface $\Sigma$ that cuts $M$ into two halves, $g_{\mu\nu}$ is continuous but not necessarily differentiable.

*The junction conditions can also be derived if one ignores the distributional contribution of $R$ to the action by integrating separately on $M^+$ and $M^-$, however then the action needs to be appended with the GHY term from the beginning. The procedure here allows one to derive the junction conditions while being able to get away with not using the boundary term and not splitting the integral from the get-go.
