# Can differential forms be exact and positive on a distribution?

Let $$M$$ be a manifold of dimension $$d$$, and let $$\mathscr D$$ be a distribution of rank $$d - 1$$ on $$M$$ (I would also be interested in lower rank distributions, but mainly I am interested in codimension $$1$$). We say that $$F$$ is positive on $$\mathscr D$$, if for some (and hence every) frame $$X_1, \dots, X_{d - 1}$$ of $$\mathscr D$$, $$F(X_1, \dots, X_{d - 1}) > 0.$$

My question is:

Does there exist a closed manifold $$M$$, a distribution $$\mathscr D$$ of rank $$d - 1$$, and a $$d - 1$$-form $$F$$, such that $$F$$ is exact and $$F$$ is positive on $$\mathscr D$$?

I think that such a distribution is necessarily nonintegrable (at least if $$F, \mathscr D$$ are smooth -- but if something weird happens in lower regularity I would be interested to know about it). Indeed, by work of Harvey and Lawson we can find a Riemannian metric $$g$$ such that $$F$$ calibrates $$\mathscr D$$ with respect to $$g$$. Therefore if $$\mathscr F$$ is a foliation of integral hypersurfaces of $$\mathscr D$$, in some Riemannian metric every leaf is locally $$F$$-calibrated. I think that we should be able to find a transverse measure $$\mu$$ on some sublamination $$\lambda \subseteq \mathscr F$$. Then $$(\lambda, \mu)$$ defines a Ruelle-Sullivan current $$T$$ which is $$F$$-calibrated. But $$F$$ is exact and nonzero so this is a contradiction.

So we are looking for nonintegrable distributions, and it seems natural to me to look for a contact $$1$$-form $$\alpha$$ on a closed manifold $$M$$ such that $$F := \star \alpha$$ with respect to some Riemannian metric is exact. Thus a positive answer to the following question implies a positive answer to my main question:

Does there exist a closed manifold $$M$$, a contact $$1$$-form $$\alpha$$, and a Riemannian metric $$h$$ such that $$H^{d - 1}(M, \mathbb R) = 0$$ and $$\alpha$$ is $$h$$-coclosed?

My expertise is pretty far from contact geometry, so my intuition could be faulty, but I suspect that this second question should have a positive answer even when $$d = 3$$. Indeed, there are lots of closed $$3$$-manifolds with vanishing Betti number, every such manifold has a contact structure, and then we have a lot of freedom to choose the contact $$1$$-form and the Riemannian metric. On the other hand, maybe the Reeb vector field is somehow "topologically nontrivial" in such a way as to rule this phenomenon out. In any case, I was not able to find an explicit example when $$M = \mathbf S^3$$.

Finally, let me remark why I am interested in a positive solution to the first question. For notational simplicity I take $$d = 3$$. Again, using the theorem of Harvey and Lawson, we can find a Riemannian metric so that $$F$$ calibrates $$\mathscr D$$. Then $$F$$ solves the PDE $$dF = 0, \quad (\nabla_i F_{jk}) F^{jk} {F^i}_{\ell} = 0.$$

This is a generalization of the PDE $$dF = 0, \quad (\nabla_i F_j) F^j F^i = 0$$ solved by $$F = du$$ where $$u$$ is $$\infty$$-harmonic and $$d = 2$$. This equation seems to be a nice model system in the $$L^\infty$$ calculus of variations. If the kernel bundle $$\mathscr D$$ of $$\star F$$ is integrable, then I can show that solutions of this PDE are characterized by absolute minimality of $$\|F\|_{C^0}$$ on small balls, not unlike the $$\infty$$-Laplacian. But if $$\mathscr D$$ is nonintegrable (so, in particular, $$d \geq 3$$), I highly suspect that this is not true. A positive solution to my first question would yield a solution which has zero data but which is itself nonzero, something that doesn't happen for the scalar $$\infty$$-Laplacian.

It turns out your question (in dmension 3) is asking about the existence of a Beltrami field on say $$\mathbb{S}^3$$. An example is given in this paper: Contact structures and Beltrami fields on the torus and the sphere, by Daniel Peralta-Salas, Radu Slobodeanu https://arxiv.org/pdf/2004.10185.pdf