# Can a null hypersurface be foliated by spacelike sections?

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.

• 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? – A.V.S. May 26 at 8:35

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.