Let $(M,g)$ be a $d$-dimensional Lorentzian manifold and let $\Sigma \subset M$ be a null hypersurface, which therefore has dimension $(d-1)$. We know that its normal vector $k^\mu$ is null and since it is null, this normal vector is also tangent to the hypersurface. Its integral lines are null geodesics which are the generators of $\Sigma$.

My question here is essentially whether or not each connected component of $\Sigma$ can be foliated by spacelike sections indexed by some parameter along the generator. I've tried formalizing this as follows.

At each point $\sigma \in \Sigma$ we can pick some $(d-2)$-dimensional spacelike subspace $\Delta_\sigma\subset T_\sigma \Sigma$ which is a complement to the space $L_\sigma$ spanned by $k_\sigma\in T_\sigma \Sigma$, meaning that $T_\sigma \Sigma$ decomposes as a direct sum $$T_\sigma\Sigma\simeq \Delta_\sigma \oplus L_\sigma,\quad L_\sigma = \{\alpha k_\sigma:\alpha \in \mathbb{R}\}.$$

This gives rise to a $(d-2)$-dimensional spacelike distribution $\sigma\mapsto \Delta_\sigma$ over $\Sigma$.

Question: Is it always possible to pick $\Delta_\sigma$ so that the resulting distribution is integrable in each connected component of $\Sigma$? If in general $\Delta$ is not integrable globally inside each connected component of $\Sigma$, around each $\sigma\in \Sigma$ can we find one neighborhood of it $U\subset \Sigma$ so that $\Delta$ restricted to $U$ is integrable?

As an example this is trivially true for the double lightcone of the origin ${\cal C}$ in Minkowski spacetime. It has two connected components ${\cal C}^\pm$ and in each of them we can pick the spacelike complement at each $\sigma\in {\cal C}^\pm$ to be spanned by the angular vectors $\partial_\theta,\partial_\phi$ in spherical coordinates. Since $[\partial_\theta,\partial_\phi]=0$ the resulting distribution is integrable. In the end each component can indeed be foliated by spacelike sections which are diffeomorphic to $S^2$ and where the indexing is by the parameter along the generators. This renders the components with topology $\mathbb{R}\times S^2$. The question is essentially if this admits some generalization to arbitrary null hypersurfaces.

  • $\begingroup$ In the spirit of “Taub–NUT space as a counterexample to almost anything” can you check if it is possible to foliate Cauchy horizon of Taub–NUT? $\endgroup$
    – A.V.S.
    May 26, 2020 at 8:35

1 Answer 1


The result is true at least locally. I do not think that it is valid globally.

I assume that $\Sigma$ is an immersed (at least) submanifold.

Take $p\in \Sigma$, then there is a local coordinate system $(u,x,y,z)$ in $M$ with domain an open neighborhood of $p$ such that a neighborhood $S\subset \Sigma$ of $p$ is represented by $u=0$. Since $\Sigma$ is lightlike, $g(du^\sharp,du^\sharp) =0$.

The vectors $\partial_x,\partial_y,\partial_z$ are therefore tangent to $\Sigma$ in $S$ and $x,y,z$ are coordinates in $S$ (viewed as an embedded submanifold).

Now observe that $$0= g(du^\sharp, du^\sharp) = \langle du^\sharp, du \rangle\:,$$ so that $du^\sharp \in TS$ as well. This smooth vector field can be integrated in $S$ since the conditions of Frobenius theorem are trivially satisfied. This means that we can change coordinates $x,y,z$ in $S$, passing to a new local coordinate system $v,r,s$ around $p$ such that $\partial_v = du^\sharp$.

Let us study the nature of the remaining coordinates $r,s$.

By construction $\partial_v$ is lightlike. Therefore for every $q\in S$ we can arrange an orthonormal basis of $T_qM$ where, for some constant $k\neq 0$, $$\partial_v \equiv k(1,0,0,1)^t\:.$$ Just in view of the definition of dual basis, we have that $$\langle \partial_r, du\rangle =0 \:, $$ which means $$g(\partial_r, \partial_v)=0\:.$$ Using the said basis and assuming $$\partial_r \equiv (a,b,c,d)^t$$ the orthogonality condition implies $$\partial_r \equiv (a,b,c,a)^t\:.$$ Hence $$g(\partial_r,\partial_r) = b^2+c^2 \geq 0$$ However, if $b=c=0$, we would have that $\partial_r$ is linearly dependent from $\partial_v$ which is not possible by construction. We conclude that $$g(\partial_r,\partial_r) = b^2+c^2 > 0$$ Therefore $\partial_r$ is spacelike. The same argument proves that $\partial_s$ is spacelike as well. Obviously these two vectors are also linearly independent as they arise from a coordinate system.

In summary, the surfaces in $S$ at $v=const$ are spacelike and $S$ is therefore foliated by spacelike surfaces (embedded submanifolds of $S$).

The procedure generalises to every dimension.


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