# Integration on Hypersurfaces

Consider a pseudo-Riemannian manifold $$(M,g)$$ of dimension $$d$$. The natural volume form induced by the metric is $$\boldsymbol{\omega} = \sqrt{(-1)^s \mathrm{det}(g_{ij})} dx^1 \wedge ...\wedge dx^d$$ where $$s$$ is the number of negatives in the signature. Such a $$d$$-form enables integration of functions $$f : M \to \mathbb{R}$$ over a chart domain $$\mathcal{U}$$ as we define $$\int_{\mathcal{U}} f := \int_{x(\mathcal{U})} d\alpha_1... d\alpha_d \omega(x^{-1}(\alpha))f_{(x)}(\alpha)$$ which is invariant under the choice of chart $$x$$ (non-boldface $$\omega$$ being the components of the volume form). Now, given an embedded submanifold of dimension $$(d-1)$$, can someone show explicitly how to go from the data above to a similar notion of integration on hypersurfaces? For example, in the sort of flux integrals given by the generalized divergence theorem $$\int_{M} \mathrm{div}(X) = \int_{\partial M} \langle X,\vec{n} \rangle$$ what is the volume form on the rhs? I know both the interior product and hodge star operations can be used to get a $$(d-1)$$ form from a $$d$$-form and and a $$(d-1)$$ form from a vector field respectively, and that somehow the pullback of $$g$$ to the hypersuface must show up, but not well enough to implement them myself. i.e. the goal is to understand what really is the volume form in flux-integral resembling expressions like $$E = \int T_{\mu \nu} \xi^\mu n^\nu \sqrt{h} d^3 x$$ and how to derive it from the definition of integration given above. Thank you.

• The metric is there when you do the Hodge star operator. The natural way to get a volume form on a hypersurface (with orientation) is to contract the volume form of the ambient manifold with the unit normal vector of the hypersurface. Of course, the best thing is to integrate a $(d-1)$-form in the first place. If you're interested in flux, you take $\star\omega$, where $\omega$ is the $1$-form corresponding to your vector field. With indefinite metrics, you have to be careful with the correspondence, of course. (I guess the same is true with the contraction above.) Aug 26, 2022 at 18:31
• So for the first example, after forming $\imath_{n}(\omega)$ one would have a $(d-1)$-form over the ambient manifold. We then just pull this form back along the embedding $\phi$ to get the job done, right? Under a chart then, I should be able to show it has components $n^{\mu} \sqrt{\phi^* g} \epsilon_{\mu \nu \sigma \rho}$, or have I misunderstood? Aug 26, 2022 at 19:37
• What are you using as local coordinates on your hypersurface? Aug 26, 2022 at 21:11
• The ones inherited from the ambient manifold by giving the hypersurface as $x^{n}$ = const. for some $n \in \{1,...,d \}$ Aug 26, 2022 at 21:23

Let $$X$$ be a a smooth vector field on a pseudo-Riemannian manifold $$(M,g)$$, and let $$\omega_g$$ denote the volume form on $$M$$ induced by the metric $$g$$ and the orientation on $$M$$. Then, we have \begin{align} \text{div}(X)\,\omega_g&=\mathscr{L}_X(\omega_g)\\ &=d(\iota_X(\omega_g)) + \iota_X(d\omega_g)\tag{Cartan's formula}\\ &=d(\iota_X(\omega_g))+0, \end{align} where the first equal sign is one of the ways of defining the divergence of a vector field (or you can adopt the definition $$\text{div}(X)=\nabla_aX^a=\text{trace}(\nabla X)$$, where $$\nabla$$ is the Levi-Civita connection for $$g$$, and then try to prove the equivalence of definitions). The last equality used that $$\omega_g$$ is a top-degree form, so its exterior derivative vanishes. Now, by Stokes' theorem, for any open set $$\Omega\subset M$$ with smooth boundary and compact closure, we have (keeping in mind the Stokes orientation) \begin{align} \int_{\Omega}\text{div}(X)\,\omega_g&=\int_{\Omega}d(\iota_X(\omega_g))=\int_{\partial\Omega}\iota_X\omega_g.\tag{*} \end{align} If you're happy to deal with differential forms alone, then we can stop here. On the left, you can think of it as either the integral of the $$n$$-form as I've written, or you can think of it as the Lebesgue integral $$\int_{\Omega}\text{div}(X)\,dV_g$$ of the smooth function $$\text{div}(X)$$ with respect to the Riemannian-Lebesgue measure $$dV_g$$. So far, the only thing we've done is apply Stokes' theorem and Cartan's magic formula (and utilized the equivalence in definitions for the divergence of vector field), so this goes through completely for non-degenerate $$g$$ of arbitrary signature.

Non-degenerate hypersurfaces.

The signature of $$g$$ and the nature of $$\partial\Omega$$ become troublesome issue if we want to interpret the RHS of $$(*)$$ as the integral of $$\langle X,n\rangle\cdot \text{some volume form}$$, because the question, as you've noted, becomes: what is the volume form on $$\partial\Omega$$? Suppose $$\partial\Omega$$ is a non-degenerate hypersurface, meaning the pullback $$i^*g$$ is non-degenerate, where $$i:\partial\Omega\to \Omega$$ is the inclusion. Then coupled with the Stokes-induced orientation on $$\partial\Omega$$, the unique $$(n-1)$$-volume form $$\omega_{i^*g}$$ on $$\partial\Omega$$ is available to us. Hence, we should be able to write \begin{align} \iota_X(\omega_g)&=f\cdot\omega_{i^*g}, \end{align} for some smooth function $$f:\partial\Omega\to\Bbb{R}$$. Let us try to figure out what this smooth function is. First, we write \begin{align} X&=X_{\parallel}+\langle X,\nu\rangle_g\nu, \end{align} where $$X_{\parallel}$$ is the component of $$X$$ tangential to $$\partial\Omega$$. Note that this requires non-degeneracy of $$\partial\Omega$$ because in this case the "outward normal" $$\nu$$ is actually transversal to $$\partial\Omega$$ (otherwise, it might happen that the 'normal' is also tangential, in which case there's also no notion of 'outwardness'; this is what happens for null hypersurfaces in GR). Next, we note that the outward normal $$\nu$$ is such that if $$\{\nu(p),\xi_1,\dots,\xi_{n-1}\}$$ is a positively oriented $$g$$-orthonormal basis for $$T_pM$$ for $$p\in \partial\Omega$$, then $$\{\xi_1,\dots,\xi_{n-1}\}$$ is positively oriented for $$\partial\Omega$$ (that's how 'outward' and the Stokes orientation are defined). Thus, \begin{align} \iota_X(\omega_g)(\xi_1,\dots, \xi_{n-1})&=\omega_g(X,\xi_1\cdots,\xi_{n-1})\\ &=\omega_g(X_{\parallel}(p)+ \langle X(p),\nu(p)\rangle_g\nu(p),\xi_1,\dots,\xi_{n-1})\\ &=0+\langle X(p),\nu(p)\rangle_g\cdot\omega_g(\nu(p),\xi_1,\dots,\xi_{n-1})\\ &=\langle X(p),\nu(p)\rangle_g\cdot 1.\tag{**}\\ &=\langle X(p),\nu(p)\rangle_g\cdot \omega_{i^*g}(\xi_1,\dots,\xi_{n-1}) \end{align} Note that the the tangential component of $$X$$ doesn't contribute because we can write it as a linear combination of the $$\xi$$'s, and hence by the alternating nature of $$\omega_g$$, we'd get $$0$$. Also, the $$\cdot 1$$ in $$(**)$$ is because $$\{\nu(p),\xi_1,\dots,\xi_{n-1}\}$$ is a positively oriented $$g$$-orthonormal basis of $$T_pM$$, so by definition $$\omega_g$$ evaluates to $$1$$ on it (see the first link). The last step is also by definition of $$\omega_{i^*g}$$ and my previous remarks about orientation (so this is a proof that $$\omega_{i^*g}=\omega_g(\nu,\cdots)$$). Since $$\{\xi_1,\dots,\xi_{n-1}\}$$ is a basis of $$T_p(\partial\Omega)$$, and $$p\in\partial\Omega$$ was arbitrary, it follows that $$\iota_X(\omega_g)=\langle X,\nu\rangle_g\omega_{i^*g}$$. Thus, Stokes' theorem $$(*)$$ becomes now the divergence theorem \begin{align} \int_{\Omega}\text{div}(X)\,\omega_g&=\int_{\partial\Omega}\langle X,\nu\rangle_g\,\omega_{i^*g}. \end{align} As it stands, these are integrals of differential forms; we can also write it as Lebesgue integrals of smooth functions with respect to the corresponding measures \begin{align} \int_{\Omega}\text{div}(X)\,dV_g&=\int_{\partial\Omega}\langle X,\nu\rangle_g\,dV_{i^*g}, \end{align} (where of course the usual notation for the measures is just $$dV$$ and $$dA$$ respectively).

Null Hypersurfaces.

I shall only make some brief comments here. Let us stick to $$g$$ of Lorentz signature $$(-,+,\dots ,+)$$ (since I once worked out details in this case). In this case, we can no longer write the interior product $$\iota_X\omega_g$$ as above, because $$i^*g$$ becomes a degenerate tensor field so it no longer induces a volume form on $$\partial\Omega$$. Suppose $$\nu$$ is null on portion of the boundary $$\partial\Omega$$ (so $$g(\nu,\nu)=0$$ there, so $$\nu$$ is both normal and tangential to that potion of the boundary). In this case, a common way of inducing a volume form is to fix a transversal field $$\overline{\nu}$$ such that $$g(\nu,\overline{\nu})=-1$$ and $$g(\overline{\nu},\overline{\nu})=0$$ (so $$\overline{\nu}$$ is also a null vector field). Then, consider the $$(n-1)$$-form defined by interior product $$\mu=\pm\iota_{\overline{\nu}}\omega_g$$ (I'm pretty sure $$-$$ sign is correct, but I've lost the computations I've done a while back and am too lazy to check). This will be no-where vanishing on that null piece of the boundary, so it serves as a volume form in that region. Thus (after some computation), we have \begin{align} \iota_X\omega_g&=\langle X,\nu\rangle\mu. \end{align} So, the point is that we use a null frame (a null vector field, together with a transversal null vector field, satisfying some joint-normalization) in order to define the volume form. In GR, this is described as: fixing a null generator on a null hypersurface defines a volume form on it.

• I tried (as much as my sanity would permit) to keep minus signs and orientations straight, and midway I changed notation for the normal from $n$ to $\nu$ to avoid confusion with the dimension. So if you think there are some errors (these or something else I might have overlooked), do let me know. Aug 27, 2022 at 0:35
• You exactly fill in the details of what I was looking for; thanks you! Aug 27, 2022 at 4:32
• I just remembered btw that in the non-Riemannian case sometimes we may have to choose the inward normal because that's the one compatible with the Stokes orientation... but I don't want to edit that now because I think you get the gist and if you really want to work out the signs, you're better off verifying them yourself. Aug 27, 2022 at 14:13