Differentials of non-smooth functions, wedge products of currents?

Question

In a paper of McMullen he considers foliations on a manifold determined by a closed 1-form $\rho$. He says an $L^\infty$ function $f$ is constant on the leaves of the foliation if "$df \wedge \rho = 0$ as a current," and I'm having a hard time unwinding exactly what's happening with the currents (which I'm not very familiar with).

My understanding is that currents are functionals on differential forms, and given a function $f$ we can define a current on $n$-forms (for an $n$-dimensional manifold) by integrating $f$ times the $n$-form. Similarly, the 1-form $\rho$ defines a current on the $(n-1)$-forms by integrating the wedge of the form with $\rho$. But what's going on with $df \wedge \rho$? I get that we can define a differential of a current by $[dT](\omega) = T(d\omega)$, so I guess $df$ refers to the current on $n-1$ forms which sends $\omega \mapsto \int f d\omega$, but how is the wedge with $\rho$ defined?

My only guess is that $df \wedge \rho$ is a current acting on $(n-2)$ forms sending $\omega \mapsto \int f \cdot (d\omega \wedge \rho)$. Even if that's correct, why does $f$ being constant on leaves imply this current is zero? It seems strange to me that functions constant on the leaves of the foliation determined by $\rho$, which have codimension $1$, can be expressed in terms of $n-2$ forms.

Answer 1

First of all, when $f$ is smooth, we know that $df$ is a (functional) multiple of $\rho$ (e.g., because the gradient is orthogonal to level hypersurfaces).

If $T$ is a current acting on $k$-forms and $\phi$ is an $\ell$-form (with $\ell\le k$), then $T\wedge\phi$ is defined to be the current acting on $(k-\ell)$-forms given by $(T\wedge\phi)(\omega) = T(\phi\wedge\omega)$ for any $(k-\ell)$-form $\omega$. So you're right that $df\wedge\rho$ is given as a current by $$(df\wedge\rho)(\omega) = df(\rho\wedge\omega) = f\big(d(\rho\wedge\omega)\big) = -f(\rho\wedge d\omega),$$ since $d\rho = 0$.

As a philosophical remark, one often thinks of $(n-k)$-currents (which act on $k$-forms) as $(n-k)$-forms with locally integrable (as opposed to smooth) coefficients.

If $f$ is smooth, then note that $\int_M f\rho\wedge d\omega = \int_M d(f\rho\wedge\omega) = 0$ (either because $\partial M = \emptyset$ or because $\omega$ has compact support, either way by Stokes's Theorem).

To see what's going on in general, by Frobenius, we can choose local coordinates $(x^1,\dots,x^n)$ so that $\rho = dx^n$. It follows from the hypothesis, moreover, that $f=f(x^n)$. By a usual partition of unity argument, we can suppose $\omega$ has compact support in our coordinate chart $U$. Working mod $dx^n$, we may take $\omega = \sum\limits_{i=1}^{n-1} g_i dx^1\wedge\dots\wedge\widehat{dx^i}\wedge\dots\wedge dx^{n-1}$, so that $d\omega = \sum \dfrac{\partial g_i}{\partial x^i} dx^1\wedge\dots\wedge dx^{n-1}$ (again mod $dx^n$). As with the usual proof of Stokes's Theorem, note now that $$\int_U f\rho d\omega = \pm\int_U f(x^n)\sum \frac{\partial g_i}{\partial x^i} dx^1\wedge\dots\wedge dx^n.$$ Now apply Fubini: For the $i$th term, integrate first with respect to $dx^i$, noting that the fact that $f=f(x^n)$ allows us to pull that out of the integral; by compact support in $U$, we get $0$ for every term.