I know there already exist some questions about this and some very good answers. However, I am still having trouble understanding one part of the calculation. The GHY term is given by
$S_{GHY}=-\frac{1}{8 \pi G} \int_{\partial M} d^dx \sqrt{-h} K$
I'm really confused about how to compute $\delta K$:
1, I would have thought $K$ as the trace of the extrinsic curvature $K^{\mu \nu}=-\frac{1}{2} (\nabla^\mu n^\nu + \nabla^\nu n^\mu)$ would be given by $K=K^{\mu \nu} g_{\mu \nu}$. I believe I'm supposed to contract with $g_{\mu \nu}$ rather than $h_{\mu \nu}$ since the normal vector is not confined to $\partial M$ - can someone confirm this please?
2, If I then vary $K$ I find $\delta K = K^{\mu \nu} \delta g_{\mu \nu} + \delta K^{\mu \nu} g_{\mu \nu}$ but according to Explicit Variation of Gibbons-Hawking-York Boundary Term, I should have $-\frac{1}{2} K^{\mu \nu} \delta g_{\mu \nu}$ on the first term. It appears I'm out by a minus sign and a factor of $\frac{1}{2}$ but I don't know why though?
3, We have $\delta K = \dots + g_{\mu \nu} \delta K^{\mu \nu}$ and I don't know how he gets this expression for the variation. Could somebody please explain how he does the variation of the normal vector and picks up this conrtibution $c_\mu$. N.B. I tried to follow the advice in one of the comments to consider a specific example of $n_\mu$ but I couldn't find the correct variation of $\alpha_\mu$?
Thanks.
