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The Gibbons-Hawking-York term which supplements the Einstein-Hilbert action is,

$$S_{GH} = \frac{1}{8\pi G} \int_{\partial M} d^3 x\sqrt{-h} \, K$$

where $\partial M$ is the boundary of the manifold $M$, $K$ is the trace of the extrinsic curvature, and $h_{\mu \nu}$ is the induced metric on the boundary of the manifold. My questions are:

  1. Is there a general formula for the metric $h_{\mu \nu}$ in terms of the metric $g_{\mu \nu}$ of the manifold M?

  2. Is there a general formula for the inward/outward normal?

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    $\begingroup$ As usual, thank you Qmechanic for adding the right tags :) $\endgroup$ – user32361 Mar 10 '14 at 15:48
  • $\begingroup$ For 1, have you taken a look at en.wikipedia.org/wiki/Induced_metric ? The first equation is the standard expression provided you have embedding functions in hand. $\endgroup$ – joshphysics Mar 10 '14 at 18:28
  • $\begingroup$ By the way, the induced metric and the metric $h_{\mu \nu}$ are not equivalent. The former has indices running from 0, to D-1, whilst the latter (i.e. first fundamental form) has indices going over the 'intrinsic coordinates' of the submanifold. $\endgroup$ – user32361 Mar 10 '14 at 21:26
  • $\begingroup$ @user32361 The induced metric can be determined for an embedded submanifold of any dimension. I'm not sure what $D$ is in your notation, but the expression for the induced metric in the linked wiki is not restricted to manifolds of a particular codimension. By the way, if you want to make sure another user in comments is pinged, make sure to use an @<user>. $\endgroup$ – joshphysics Mar 10 '14 at 23:49
  • $\begingroup$ @user1997744 Ok but does the question not ask about computing the induced metric? Are you saying there is some issue in computing the GBH boundary term by integrating over $\partial M$ using the induced metric? $\endgroup$ – joshphysics Mar 11 '14 at 20:55
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You want to take a look at Eric Poisson's Advanced General Relativity: https://www.physics.uoguelph.ca/poisson/research/agr.pdf

The induced metric can be obtained by looking at the invariant interval $ds^2$, and then setting $dx_i=0$ for one of your coordinates. Note that you also forgot an $\varepsilon$ in your equation which is -1 for spacelike boundaries and +1 for timelike boundaries. Typically your unit normal will point "outward" if you're talking about a convex spacelike surface, and for timelike surfaces it always points forward in time (so not really "outward"). To see why, insert an arbitrary spacelike boundary into some region and show the contributions cancel out.

In general for the unit normal, write its equation in the embedding space as a vector, and then make it covariant by multiplying by $g^{\mu\nu}$.

The simplest example to play around with is the $(3+1)$-dimensional de Sitter manifold (spherical foliation) bounded by two constant-time spacelike surfaces.

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If you can define a unit vector $n^{\mu}$ normal to the boundary, then you can use it to relate the metric on $M$ and the induced metric on $\partial M$.

Let's assume that $n^{\mu}$ is a spacelike unit vector, so $n^{\mu} n_{\mu} = 1$. Then the induced metric on $\partial M$ is $$h_{\mu\nu} = g_{\mu\nu} - n_{\mu} n_{\mu}$$ If the unit normal is timelike, we flip the sign in the second term.

But how do we obtain $n^{\mu}$ in the first place? Locally, we can always express $\partial M$ as an isosurface of some scalar function -- let's call it $\rho$. So our description of $\partial M$ would take the form $$\rho - \rho_{0} = 0$$ where $\rho_{0}$ is a constant. The direction normal to this surface is given by the gradient $\partial_{\mu} \rho$. We construct $n_{\mu}$ by normalizing this gradient. Let's assume $\partial_{\mu}\rho$ is spacelike, as we did earlier. If we define $$\alpha^2 = g^{\mu\nu} \partial_{\mu} \rho \partial_{\nu} \rho$$ then the unit normal is $$n_{\mu} = \frac{1}{\alpha} \partial_{\mu} \rho ~.$$ By convention, signs are usually chosen so that the normal $n^{\mu}$ is 'outward pointing'. In other words, if $M$ corresponds to $\rho < \rho_0$, then $n^{\mu}$ will point towards $\rho > \rho_0$. And if $M$ corresponds to $\rho > \rho_0$ (as is often the case when, say, we are working in coordinates where $\partial M$ is at $\rho=0$) then $n^{\mu}$ will point towards $\rho < \rho_0$.

(Edit: There are a few obvious sign changes in the construction above if $n^{\mu}$ is timelike. A compact review, along with a discussion of adapted coordinates and other useful stuff, can be found here.)

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