Rigging technique introduced in [20] is a convenient way to address the study of null hypersurfaces. It offers in addition the extra benefit of inducing a Riemannian structure on the null hypersurface which is used to study geometric and topological properties on it. In this paper we develop this technique showing new properties and applications. We first discuss the very existence of the rigging fields under prescribed geometric and topological constraints. We consider the completeness of the induced rigged Riemannian structure. This is potentially important because it allows to use most of the usual Riemannian techniques.Keywords. Lorentzian manifolds, null hypersurface, normalization, rigging vector field, rigged vector field, completeness. MSC(2010). 53C50, 53B30, 53B50.Our aim is to show that in a 4-dimensional Lorentzian manifold, compact null hypersurfaces with finite fundamental group can not admit such normalizations. We prove first the following:Proposition 3.7. Let ζ be a rigging for a compact null hypersurface M in a Lorentzian manifold (M , g) of constant curvature. If the screen S (ζ) is conformal and the first De Rham cohomology group H 1 (M, R) is trivial, then M is totally geodesic.Proof. Since M is screen conformal there exists a non vanishing function ρBut the left hand side of (3.2) vanishes since M has constant curvature, and ξ is orthogonal to both X and Y . Moreover,MANUEL GUTIÉRREZ, AND RAYMOND HOUNNONKPE since A N = ρ ⋆ A ξ . Using the fact that H 1 (M, R) is trivial, there exists a function (say) φ defined on M such that τ = dφ. Define a new rigging vector field by ζ = exp(−φ)ζ, so N = exp(−φ)N . Moreover, it follows (from [2], Lemma 2.1) that τ = τ + d(ln(exp(−φ)) = 0 as τ = dφ. Denote respectively by ξ, H, ⋆ A ξ the rigged vector field, the mean curvature function and the screen shape operator form of ζ, we have ([5], Remark 3)But Ric( ξ) = 0 since M has constant curvature and τ ( ξ) = 0, it fol-We conclude that M is totally geodesic.We can get now the following result. Proposition 3.8. Let (M 4 , g) be a 4-dimensional Lorentzian manifold of constant curvature and M a compact null hypersurface. If M has finite fundamental group then there is no normalization such that M is screen conformal.Proof. Let M be as above. Suppose there is a normalization such that M is screen conformal. Since M has finite fundamental group, the first De Rham cohomology group H 1 (M, R) is trivial. It follows from Proposition 3.7 that M is totally geodesic. Elsewhere, M being screen conformal, S (ζ) is integrable and induces a foliation on M . We show that the leaves of the screen distribution S (ζ) are totally geodesic in (M, g). For this, recall (from [20], Proposition 3.7)that for X and Y in S (ζ),In other words, the second fundamental form of each leaf of S (ζ) in (M, g) is B and then each of them is totally geodesic in (M, g) as M is totally geodesic in (M 4 , g). It follows that there exits a totally geodesic codimension one foliation on the compact 3-manifold M , hence M must have infinite fundamental g...