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.