Computing Normals from Metric Tensor I asked a similar question on the Physics Stack Exchange, but unfortunately I have had no reply. It may be more suited for the Math section, as it focuses on the mathematical interpretation of GR. Just to motivate the question, recall the boundary term in the Einstein-Hilbert action:
$$S = \frac{1}{8\pi G} \int_{\partial M} \mathrm{d}^3 x \sqrt{|h|}\, K$$
where the integration is over the boundary of a manifold, $K$ is the trace of the extrinsic curvature, and $h$ is the first fundamental form, 'physically' equivalent to the induced metric on the boundary of the manifold.
To compute the extrinsic curvature, one requires the inward/outward normals. Is there a straightforward way to compute the normals directly from the metric tensor?
In the end, I just want to plug the normals $n_{\alpha}$ into the expression $K_{\mu \nu} = (1/2)n_{\alpha}g^{\alpha \beta}\partial_{\beta}g_{\mu \nu}.$
 A: The manifold should have associated with it a 3-vector quantity, a rank-3 antisymmetric tensor $M^{abc}$, that describes the orientation of the manifold.  If this is not given to you, you can construct it using a parameterization:  let $u, v, w$ be coordinates parameterizing this manifold, so that $s(u, v, w)$ is some point on the manifold.  Let $\mu^a = \partial s^a/\partial u$, and let $\nu^b$, $\omega^c$ be similarly defined as partial derivatives with respect to $v, w$.  Then one choice of orientation is
$$M^{abc} = \mu^{[a} \nu^{b} \omega^{c]}$$
You can then use the four-dimensional Levi-Civita tensor to get the normal covector $\chi_d$ associated with this orientation:
$$\chi_d = \epsilon_{abcd} M^{abc}$$
You can then raise the index to get the vector components using the metric.
Geometrically, the tensor $M^{abc}$ describes the tangent space at any point on the manifold.  It's sometimes called the pseudoscalar that describes the manifold.  The Levi-Civita tensor performs a duality operation:  converting 3-vectors like $M$ to their dual covectors.  The metric is then needed to convert back to a vector, but usually, normal vectors are actually covectors.
A: If the submanifold in the spacetime is defined by some condition on the coordinates,
$$\Phi(x^\mu) = 0$$
then the unit normal is given by,
$$n_\mu = \pm \frac{(\partial_\mu \Phi)}{\lvert g^{\lambda\sigma}\partial_\lambda \Phi \partial_\sigma \Phi \rvert^{1/2}}$$
If the submanifold is null, $(\partial \Phi)^2$ is zero, and instead a normal $n_\mu = - \partial_\mu \Phi$ may be defined. The sign is chosen so that the normal is future-directed. Alternatively, if the submanifold is specified by a set of embedding functions, $X^\mu(\sigma^A)$ where $\{\sigma^A\}$ are the intrinsic coordinates, then the unit normal satisfies the algebraic equation,
$$\frac{\partial X^\mu}{\partial \sigma^A}n_\mu = 0$$
which can be solved in some cases with additional information for the normal, $n_\mu$. Additionally, for a manifold $M$ with boundary $\Sigma$, if the metric on the boundary $h_{\mu\nu}$ is obvious from the metric on the manifold, that is, $g_{\mu\nu}$, one can find certain components of the normal through the relation,
$$h_{\mu\nu} = g_{\mu\nu}-n_\mu n_\nu.$$
For example, it immediately follows $n_0 = \pm \sqrt{g_{00}-h_{00}}$.
