# Are there connected closed 4-manifolds admitting a regular Almost Lagrangian distribution, and which are not Lorentzian?

In the category of real differential manifolds, connected (of $$C ^ {\infty}$$ class in the sequel), closed of dimension 4, is there any manifold admitting a regular Almost Lagrangian distribution and which are not Lorentzian?

Let $$M$$ be a real differential manifold closed (compact without boundary) of dimension 4.

By Almost Lagrangian distribution on $$M$$, we mean a differentiable field of 2-plane $$L$$ tangent to $$M$$ defined everywhere over $$M$$. $$L$$ is regular if the rank of $$L$$ is 2 at any point of $$M$$.

A Lorentzian metric $$g$$ on $$M$$ is a differentiable field of non-degenerated quadratic form of signature (+++—) defined on the tangent bundle $$TM$$ of $$M$$. Two Lorentzian metrics $$g\text { and } g'$$ on $$M$$ are conformal if there is a smooth positive function $$f$$ defined all over on $$M$$ such that $$g'= f^ 2g.$$ This defines an equivalence relationship on the space of the Lorentzian metrics on $$M$$. A conformal Lorentzian structure on $$M$$ is a class of equivalence for this relationship (when the space of the lorentzian metrics on $$M$$ is non void; that is, when the euler characteristic of $$M$$ is zero (for $$M$$ orientable).) Let $$(M, g)$$ be a Lorentzian manifold of real dimension 4. A null frame on an open set $$U$$ of $$M$$ is a frame $$(e_l) _l$$ of $$M$$ such that: $$g \text{ is positive definite on the span } \langle e_1,e_2\rangle,\; g(e_3, e_3) = g (e_4; e_4) = 0; g (e_3; e_4) = -2\text { on } U.$$ Considering a cover of $$(M, g)$$ by $$g$$-convex open sets, I think we can get a canonical Lagrangian distribution by taking at each point of $$M$$, the 2-plane $$\langle e_1, e_2\rangle$$ spanned by $$e_1 \text { and } e_2$$.

The question I'm asking is whether the reciprocal is true; namely: Is the existence of a regular Almost Lagrangian distribution on the closed differential manifold $$M$$ leads to the existence of a conformal Lorentzian structure on $$M$$?

This question arises as I'm studying the existence and the stability of a Lorentzian metric $$g$$ in its conformal class, or in general, when one want (I think), to study the existence of a conformal Lorentzian structure on a given manifold of real dimension 4.
Indeed, it seems to me that the solution to the problem of the formation, development and stability of the singularities in general relativity, as formulating in the epilogue of D. Christodoulou; "The Formation of Black Hole in General Relativity", and in general, the problem of the geometrization of a 4-differential manifold necessarily passes through:

• the study of the existence of a regular Almost Lagrangian distribution on any differential manifold of real dimension 4;
• when the manifold has a regular Almost Lagrangian distribution, study the link between this Lagrangian distribution and the existence of a conformal Lorentzian structure on the said manifold.
• More generally, study the singularities of an Almost Lagrangian distribution (eventually non regular) on a given class of manifold and the relationship between the singularities of the distribution and the geometry of a representative of the class of manifolds.
The geometrization method is provided, for conformal Lorentzian manifolds, by the Einstein-Yang-Mills equations in a well-adapted gauge (space-temporal-lorenz; double null foliation) coupled with good conditions on the space of twistors of the manifolds (which we suppose orientable).

I think that by this procedure, one can attack with success, the stability of the family of Kerr solutions of vacuum Einstein field equations, in the lines given by S. Klainerman and J. Szeftel, "Kerr stability for small angular momentum" arxiv:2104.11857v1 (if I understood well what I learned at the introduction of that monograph).

All assistance, especially an example, even abstract, of a connected, closed manifold of real dimension 4, admitting a Lagrangian distribution on which there is no class of conformal Lorentzian metrics will be of the utmost importance (any reference is welcome, as is any indication of the demonstration method (cohomological, analytical, geometric...) with some details.) Thank you heartfully for the time you will take to answer me and the answer you will give me.

• Thanks Daniele Tampieri for the editing. I still have to master the task of well formatting. Nov 20, 2021 at 14:27
• I don't see why a 2-plane field is called Lagrangian. Nov 20, 2021 at 14:47
• You can't get a canonical Lagrangian distribution out of a conformal Lorentz metric. The conformal compactification of Minkowski space has no Lagrangian distribution invariant under its conformal isomorphisms. Nov 20, 2021 at 14:48
• Indeed, as @BenMcKay points out, the adjective 'Lagrangian' is usually only used with respect to a symplectic $2$-form on the manifold. In any case, the manifold $S^2\times S^2$ clearly has a non-singular 2-plane field, but doesn't have a Lorentzian structure because its Euler characteristic is not zero. Nov 20, 2021 at 14:49
• You distinguish $L$ being a 2-plane and being of rank 2 (which you call regular); what do you mean by a 2-plane which is not of rank 2? How do you measure rank? Mar 10, 2022 at 10:08