*This question is cross-posted at <https://math.stackexchange.com/questions/3234895/existence-of-transverse-null-vector-bundle>*


Let $(M,g)$ be a Lorentzian manifold. Let $\dim(M)=d$. Given a null hypersurface, i.e., a smooth embedding $(H, \phi )$ of $M$ s.t. $\dim(H)=d-1$, $h:= \phi ^\* g$ over $H$ is degenerate with $rank(h\_p)=d-2$ at each $p \in H$. Let $\phi^\*(TM)$ be the induced bundle of $M$ over $H$.


Define the degenerate subspace as $C\_p H=\{X\_p\in T\_p H : h(X\_p,Y\_p) =0, \forall Y\_p\in T\_pH\}$, and the corresponding $CH= \sqcup\_{p\in H} C\_p H$. It can be shown that this defines a vertical subbundle of $TH$. Consider the complementary bundle of $\phi'(TH)$ in the induced bundle $\phi ^\*(TM)$, as say $RH$. There is an arbitrary choice of such an $RH$; they are all isomorphic to the normal bundle say $NH$, but the choice is by no means unique.


Now the question is about the existence of some specific kind of $RH$. **Does there exist a choice of $RH$ s.t. $CH\oplus RH$ is non-degenerate with respect to the metric $g$ induced over it? Further, can you make such a choice with the additional constraint that $RH$ is itself degenerate w.r.t $g$ induced over it? ([1] for details).** I need the existence; the uniqueness I'm sure does not hold.


[1] By $g$ induced over a vertical subbundle $E$ of $\phi^\*(TM)$, I mean the following: let $\iota:E \rightarrow \phi^\*(TM) $ denote the embedding map. Now, it is known that the map $\phi' :TH \rightarrow TM$ splits into a vertical and horizontal part over $\phi^\*(TM)$. Use the horizontal part $\phi'\_h:\phi^\*(TM) \rightarrow TM$, to induce a metric on $E$ as $\iota^\*(\phi\_v^\*g)$. It is with respect to such induced metric that I say that $RH\oplus CH$ should be non-degenerate, and $RH$ should be degenerate.


**Addendum**. The corresponding statement on the level of linear subspaces is true, and I know how to do that. It's at the level of bundles that it doesn't seem to work.



Existence of transverse null vector bundle in a degenerate Lorentzian hypersurface
Note that $CH$ is [complemented](https://ncatlab.org/nlab/show/direct+sum+of+vector+bundles#TopologicalSubBundlesOverParacompactHausdorffSpacesAreDirectSummands) in $TH = CH \oplus SH$, with $SH$ some non-unique choice of complementary sub-bundle. Then the restricted metric on $SH$ is non-degenerate (of Riemannian signature). The orthogonal complement $S^\perp H$ in $\phi^\*(TM)$ is a rank-2 bundle and the restricted metric on it is Lorentzian. The null cone of a rank-2 Lorentzian metric is the union of two null lines. Obviously, globally the null cone bundle of $S^\perp H$ then decomposes into the union of two null vector sub-bundles of $S^\perp H$, with $CH$ being one of them (by construction), while you can take the other one to be the desired $RH$.


