*Prenote: I have asked this question first on [math stackexhange](https://math.stackexchange.com/questions/1882475/gauss-theorem-for-null-boundaries), 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?



Gauss' theorem for null boundaries
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.


