# Characterization of continuous weakly closed 1-forms

Recall that a differential $$k$$-form $$\alpha$$ on a smooth manifold $$M$$ is called weakly closed if

$$\int_M \alpha \wedge d\beta = 0,$$

for all smooth forms $$\beta$$ of degree $$n-k-1$$, where $$n = \dim M$$. My question is:

If $$\alpha$$ is a weakly closed continuous 1-form on a closed manifold $$M$$, can we conclude that $$\alpha$$ is the sum of a smooth closed 1-form and $$d\phi$$, where $$\phi$$ is a $$C^1$$ function on $$M$$?

This appears to follow from the properties of the de Rham regularization operator(s) and associated homotopy operator(s) on forms; see, for instance, Theorem 12.5 in V. Gol'dshtein and M. Troyanov, Sobolev inequalities for differential forms and $$L_{q,p}$$-cohomology, Journal of Geometric Analysis, vol. 6, no. 4, 2006.

Any thoughts on this would be appreciated.

• This is true for $\phi \in L^1_{\rm loc}$ though. – Henri Jun 20 at 7:25
• Thanks! But wouldn't $\phi \in L^1_{loc}$ and $d\phi$ continuous (being equal to a continuous form minus a smooth form) imply that on a compact manifold, $\phi \in W^{1,\infty}$? If so, then $\phi$ is Lipschitz, hence differentiable a.e. in the ordinary sense and therefore (since $d\phi$ is $C^0$) $\phi$ is a.e. equal to a $C^1$ function? – Slobodan Simić Jun 20 at 17:28

The answer is no. If $$\varphi$$ is a continuous function on $$\mathbb{R}^n$$, then the form $$\alpha=\varphi\, dx_1\wedge\ldots\wedge dx_n$$ is weakly closed since it is an $$n$$-form. Then using classical notation solving the equation $$d\omega=\alpha$$ is the same as solving the equation $$\operatorname{div}\Phi=\varphi$$. However it was proved in  that the equation $$\operatorname{div}\Phi=\varphi$$ may have no $$C^1$$ solutions.
• Thanks! It looks like the "no" answer can be obtained for 1-forms (at least in $\mathbb{R}^n$) as follows. Let $\alpha$ be a continuous weakly closed 1-form. Pick any smooth form $\theta$ of degree $n-2$. Then $\alpha \wedge d\theta$ is trivially weakly closed. If $\alpha$ can be written as $\alpha_\ast + d\phi$, for some smooth closed 1-form $\alpha_\ast$ and a $C^1$ function $\phi$, then $\alpha \wedge d\theta = d(\pm \alpha_\ast \wedge \theta +\phi d\theta)$. Since this works for any continuous weakly closed $\alpha$, we get a result which contradicts . Is this correct? – Slobodan Simić Jun 20 at 17:43