# Is it possible to integrate a function over a null hypersurface?

For $f$ a compactly supported function on a spacetime $(\mathcal{M},g)$, one may define its integral over $\mathcal{M}$ by $$f\longmapsto \int_\mathcal{M}\star f,$$ where $\star$ is the Hodge star of $\mathcal{M}$. Since $\star$ is unique, this map is canonical. Suppose that $\Sigma$ is a null hypersurface of $\mathcal{M}$. (Assume $\Sigma\cap\operatorname{supp}f\ne\emptyset$.) We cannot define the integral of $f$ over $\Sigma$ by $$f\longmapsto \int_\Sigma\star_\Sigma f$$ because the induced metric on $\Sigma$ is degenerate, and hence $\star_\Sigma$ is not defined. Is there a way to "canonically" integrate $f$ over a null hypersurface?

• You cannot without other pieces of information. With a vector field $J$ (typically a timelike vector field) you can, defining the volume form $\omega = \star J$ and next integrating $f\omega$. – Valter Moretti Mar 9 '16 at 12:34
• @ValterMoretti Why timelike? Could one not take the future-directed null normal $l$ to $\Sigma$ and integrate $f\star l$? – Ryan Unger Mar 9 '16 at 16:20
• Indeed it may be spacelike too. I wrote timelike because, n GR, one usually has some notion of time and that notion can be used. – Valter Moretti Mar 9 '16 at 18:26
• However, yes I think you can use $\ell$ – Valter Moretti Mar 9 '16 at 18:47

## 1 Answer

You cannot without other pieces of information. With a vector field J -- typically a timelike vector field-- you can: Defining the volume form $\omega = *J$ and next integrating $f\omega$. I wrote timelike because in GR, one usually has some notion of time and that notion can be used. However one could also take the future-directed null normal $J= \ell$ to $\Sigma$.

ADDENDUM. If $\Sigma$ is an embedded null 3-submanifold in the 4D spacetime, locally it is parametrized by coordinates $x^1,x^2,U$ where $\partial_{x^1},\partial_{x^2}$ are spacelike and $\partial_U$ is light-like. It is possible to complete this set of coordinates with another lightlike coordinate $V$, constant over $\Sigma$, such that $g(\partial_U,\partial_V)=-1$ and $g(\partial_{x^1}\partial_V)= g(\partial_{x^2},\partial_V)= g(\partial_{x^1}\partial_U)= g(\partial_{x^2},\partial_U)=0$.

The normal co-vector to $\Sigma$ is $dV$, however its contravariant form is $-\partial_U$ which is tangent to $\Sigma$.

What I meant above is (a bit improperly perhaps)

$$\ell := \partial_V$$

so that, with may conventions, $*$ transforms $1$-vector ($\ell$) into a $3$-form $\omega$. The latter turns out to be proportional to $\sqrt{|\det g|} \:dx^1\wedge dx^2 \wedge dU$, where $$g = -dU\otimes dV -dV\otimes dU + \sum_{i,j=1}^2h_{ij}dx^i \otimes dx^j$$ so that $\sqrt{|\det g|}= \sqrt{\det h}$.

The form $\omega$ is not canonical as there are many ways to fix coordinates $x^1,x^2, U,V$.

• I think there is a problem using the null normal. Since it is null, $* \ell = \ell \wedge \alpha$ for some $\alpha$, and $\ell$ pulls back to zero on the null surface (by definition of being the normal). You could introduce an auxiliary null vector $k$ such that $k \cdot \ell = -1$, and then define $* k$, but there is considerable ambiguity in the choice of such $k$. – asperanz Mar 9 '16 at 21:16
• By null normal I meant your k! – Valter Moretti Mar 9 '16 at 22:36
• @asperanz Why would $\ell$ pull back to zero on the surface? It's a property of null hypersurfaces that the normal is also a tangent vector. – Ryan Unger Mar 11 '16 at 17:52
• The thing that pulls back to zero is the normal covector $\ell_a$. This is just another manifestation of the induced metric being degenerate on the surface. So we have that $(*\ell)_{abc} = \ell_{[a} \alpha_{bc]}$ for some $\alpha_{bc}$, which pulls back to zero on the null surface. – asperanz Mar 11 '16 at 18:35