# Gauss' theorem for null boundaries

Prenote: I have asked this question first on math stackexhange, but a user suggested that mathoverflow might be a better place for this question. Upon thinking about it I have agreed with him and copy-pasted the question here.

Note: I have solved this problem on my own, mostly while actually typing it in here, as I was stuck with this problem previously. This is however quite important for my research, so I nontheless would like to verify the correctness of my solution, hence I went forth with posting this "question". If this goes against site principles then please vote close on the question.

Introductions: Let $(M,g)$ be a Riemannian manifold with boundary (or a Lorentzian/pseudo-Riemannian manifold whose boundary is timelike or spacelike everywhere). Gauss' theorem then states that $$\int_M\text{div}(X)\mu_g=\int_{\partial M}\langle X,n\rangle\mu_h,$$ where $\mu_g$ is the volume form associated with $g$, $\mu_h$ is the volume form associated with the induced metric $h$, and $n$ is the unit normal vector field along the boundary.

This theorem is also available in an intermediate step between Stokes' theorem and the form I have given above. Since $\mathcal{L}_X\mu_g=\text{div}(X)\mu_g$, but $\mathcal{L}_X\mu_g=\mathrm{d}i_X\mu_g+i_X\mathrm{d}\mu_g=\mathrm{d}i_X\mu_g$, we have $$\int_M\text{div}(X)\mu_g=\int_M\mathrm{d}i_X\mu_g=\int_{\partial M}\phi^*(i_X\mu_g),$$ where $\phi$ is the inclusion $\phi:\partial M\rightarrow M$.

The problem: I wish to use Gauss' theorem on a Lorentzian manifold, whose boundary is a null surface. Then the normal vector is also tangent, the induced metric is degenerate, and the induced volume is zero. Durr. The "intermediate" form of the Gauss' theorem still applies though, so I seek to use it to define a version of Gauss' theorem which, instead of using the normal vector $n$, it will use an arbitrary transverse vector field $N$, which is surely not tangent to $\partial M$.

Proposed solution: If $e_1,...,e_{n-1}$ is a positively oriented frame on $\partial M$, then $X$ is locally decomposible as $X=\langle X,N\rangle N+Y$, where $Y=Y^ie_i$, furthermore, I assume if $N$ is globally defined then this decomposition also applies globally in the general form $X=\langle X,N\rangle N+Y$, where $Y$ is tangent to $\partial M$. (Is this assumption correct?).

Then plugging $X$ into $\mu_g$ gives $$i_X\mu_g(e_1,...,e_{n-1})=\mu_g(\langle X,N\rangle N+Y,e_1,...,e_{n-1})=\langle X,N\rangle\mu_g(N,e_1,...,e_{n-1})= \\=\langle X,N\rangle i_N\mu_g(e_1,...,e_{n-1})$$ where the term involving $Y$ is annihilated, because if $Y$ is tangent to $\partial M$, then the arguments of $\mu_g$ are linearly dependent.

So from this we have $$\int_M\text{div}(X)\mu_g=\int_{\partial M}\langle X,N\rangle \phi^*(i_N\mu_g),$$ so apparantly $i_N\mu_g$ is such a volume form for $\partial M$, that Gauss' theorem with $N$ instead of $n$ applies.

I also want to obtain a local coordinate formula for this, so if $(U,x^\mu)$ is a local chart for $M$, and $(V,y^i)$ is a local chart for $\partial M$, then in the relevant chart domains we have $$i_N\mu_g=i_N(\sqrt{-g}\mathrm{d}x^1\wedge...\wedge\mathrm{d}x^n)=i_N(n!^{-1}\sqrt{-g}\epsilon_{\mu_1...\mu_n}\mathrm{d}x^{\mu_1}\wedge...\wedge\mathrm{d}x^{\mu_n})=\\=i_N(\sqrt{-g}\epsilon_{\mu_1...\mu_n}\mathrm{d}x^{\mu_1}\otimes...\otimes\mathrm{d}x^{\mu_n})=\sqrt{-g}N^{\mu_1}\epsilon_{\mu_1...\mu_n}\mathrm{d}x^{\mu_2}\otimes...\otimes\mathrm{d}x^{\mu_n}=\\=\frac{1}{(n-1)!}\sqrt{-g}N^\mu\epsilon_{\mu\mu_2...\mu_n}\mathrm{d}x^{\mu_2}\wedge...\wedge\mathrm{d}x^{\mu_n}.$$ Now pulling back gives $$\phi^*(i_N\mu_g)=\frac{1}{(n-1)!}\sqrt{-g}\epsilon_{\mu\mu_2...\mu_n}N^\mu\frac{\partial x^{\mu_2}}{\partial y^{i_1}}...\frac{\partial x^{\mu_n}}{\partial y^{i_{n-1}}}\mathrm{d}y^{i_1}\wedge...\wedge\mathrm{d}y^{i_{n-1}} .$$ Now, the matrix $e^\mu_{(i)}=\partial x^\mu/\partial y^i$ maybe identified (with a bit of abuse of notation regarding tangent maps) as the $\mu$th component of the $i$-th coordinate basis vector field on the boundary $\partial M$, with the components being taken with respect to the coordinate system $\{x^\mu\}$.

Then, by the definition of determinants we have $\epsilon_{\mu\mu_2...\mu_n}N^\mu e^{\mu_2}_{i_1}...e^{\mu_n}_{i_{n-1}}=\det(N,e_{(i_1)},...,e_{(i_{n-1})})$, so $$\phi^*(i_N\mu_g)=\frac{1}{(N-1)!}\sqrt{-g}\det(N,e_{(i_1)},...,e_{(i_{n-1})})\mathrm{d}y^{i_1}\wedge...\wedge\mathrm{d}y^{i_{n-1}}=\\=\sqrt{-g}\det(N,e_1,...,e_{n-1})d^{n-1}y,$$ so the modified Gauss' theorem in coordinates is (assuming $M$ can be covered by a single chart) $$\int_M \text{div}(X)\sqrt{-g}d^nx=\int_{\partial M}X^\mu N_\mu\sqrt{-g}\det(N,e_1,...,e_{n-1})d^{n-1}y$$.

Questions:

• Is this derivation correct?

• I can work with the form I have gotten, but I don't like the determinant factor in it. Is there any way I can express that in a more pleasant form?

• In case the derivation is incorrect, how can I get a formula I can work with in coordinate notation that does what I want?

• I am not convinced about the vector field decomposition $X = \langle X, N \rangle N + Y$ because, if I understand correctly, $N$ is not Lorentz orthogonal to your frame of vectors $e_1,..,e_{n-1}$. I would say $X = \lambda N + Y$ where $Y$ is light-like. Then $\langle X, Y \rangle = \lambda \langle N, Y \rangle + \langle Y, Y \rangle = \lambda \langle N, Y \rangle$ and solve for $\lambda$. – Futurologist Aug 4 '16 at 22:52
• @Futurologist You are right, however I do not think it is because $N$ is not orthogonal, but because $N$ might not be normed to unity. In fact, it should be possible to take $N$ to be also a null vector, just not the one that is tangent/normal to $\partial M$, which would further complicate things. I am working on this right now, but I am very tired so probably will postpone till tomorrow. – Bence Racskó Aug 4 '16 at 23:10

Gauss' Theorem has nothing to do with the (pseudo-)metric. Is just a consequence of Stokes' theorem.

Stokes's theorem says that, for any $n-1$ form $\omega$,

$$\int_M d\omega = \int_{\partial M} \omega.$$

Now fix any smooth measure $\mu$ (i.e. given by a smooth non-vanishing top dimensional form, or a density if $M$ is not orientable). It might be the Riemannian measure of a Riemannian structure, but it's not relevant.

Let $X$ be a vector field. If you apply the above formula to the $n-1$ form $\omega:=\iota_X \mu$, you get, using one of the many definitions of divergence (associated with the measure $\mu$)

$$\int_M \mathrm{div}(X) \mu = \int_{\partial M} \iota_X \mu, \qquad (\star),$$

where $\iota_X$ is the contraction. This holds whatever is $X$, on any smooth manifold with boundary.

Now it all boils down to how you want to define a "reference" measure on $\partial M$. As you propose, you can pick a transverse vector $N$ to $\partial M$, and define a reference measure on $\partial M$ as $\eta:=\iota_T \mu$. Then

$$\int_M \mathrm{div}(X) \mu = \int_{\partial M} f\, \iota_T \mu,$$

where

$$f := \frac{\mu(X,Y_1,\ldots,Y_{n-1})}{\mu(T,Y_1,\ldots,Y_{n-1})}.$$

for any arbitrary local frame $Y_1,\ldots,Y_{n-1}$ tangent to $\partial M$ (oriented, otherwise take densities and absolute values). In particular, you can easily compute $f$ by looking at the "$T$" component of "$X$":

$$X = f\, T \mod \mathrm{span}\{Y_1,\ldots,Y_{n-1}\}.$$

There is no need to use orthonormality, or a (pseudo)-metric.

However, if you are on a Riemannian manifold, $\mu = \mu_g$ is the Riemannian measure, and $N$ is the normal vector to $\partial M$, then $f= g(N,X)$, and $\iota_N \mu$ is Riemannian measure of the induced Riemannian metric on $\partial M$, recovering the classical statement.

• Thank you, you have put it way more succintly than I did. I guess my execuse is I discovered many of this as I wrote up my posts. Just a quick question, your $T$ is the same as my "transversal" $N$ right? And well, the way you put it is obvious, but I have never much heard any reference to Gauss' theorem aside from Riemannian geometry. The form $\int_M\text{div}(X)\mu=\int_{\partial M}i_X\mu$ aside, of course. Such manipulations were quite new to me :) . – Bence Racskó Aug 5 '16 at 20:26
• Yes, $T$ is any transverse vector to $\partial M$. That's true, Gauss' theorem is usually presented in relation with a Riemannian metric. – Raziel Aug 5 '16 at 20:31

Okay, as Futurologist pointed out, the decomposition $X=\langle X,N\rangle N+Y$ is incorrect. In fact, this was quite a rookie mistake.

Instead, let $\{e_1,...,e_{n-1}\}=\{\partial/\partial y^1,...,\partial/\partial y^{n-1}\}$ be a positively oriented holonomic frame on $\partial M$, and let $N$ be a transversal along $\partial M$. Let $N$ be extended in such way, that $\phi^N_\lambda(p)=\exp_p(\lambda N)$, where $\phi^N_\lambda(p)$ is the flow of $N$. Then the coordinate system $\{\lambda,y^1,...,y^{n-1}\}$ is some sort of "oblique gaussian normal coordinate system" near $\partial M$. The vector field $X$ then may be decomposed as $$X=\mathrm{d}\lambda(X)N+Y.$$ It is clear that the vector $N^*=\sharp\mathrm{d}\lambda$ is a null vector, because this is normal to all vectors tangential to $\partial M$.

We have then $i_X\mu_g=\langle X,N^*\rangle i_N\mu_g$, so Gauss' theorem is $$\int_M\text{div}(X)\mu_g=\int_{\partial M}\langle X,N^*\rangle\phi^*(i_N\mu_g),$$ or, in local coordinates $$\int_M\nabla_\mu X^\mu\sqrt{-g}d^nx=\int_{\partial M}X^\mu N^*_\mu\sqrt{-g}\det(N,e_1,...,e_{n-1})d^{n-1}y,$$ where clearly $N^*_\mu=(\mathrm{d}\lambda)_\mu=\partial_\mu\lambda$.

The coordinate $\lambda$ is I guess might be difficult to calculate, so alternatively the normal form $\mathrm{d}\lambda$ can be given as $$\mathrm{d}\lambda=\frac{\mu_g(.,e_1,...,e_{n-1})}{\mu_g(N,e_1,...,e_{n-1})}=\frac{\det(.,e_1,...,e_{n-1})}{\det(N,e_1,...,e_{n-1})},$$ which I guess is hardly surprising, since plugging this back into the equation gives just $i_X\mu_g$ on the right-hand side... (I love it when I seem to discover something deep, only for it to end up being a triviality or a tautology...)

On the other hand, this modified version of Gauss' theorem does not require the metric $g$ actually, any preferred volume form will do instead of $\mu_g$, which is, I guess, is pretty interesting.

Meh, I am not sure this is of much use to me in this form, but I answered my own question because it seems this is the answer to my question, however, if anyone has any comment/observation, or any way to express this in a different form, they are welcome to do so.