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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.

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  • $\begingroup$ This is true for $\phi \in L^1_{\rm loc}$ though. $\endgroup$ – Henri Jun 20 at 7:25
  • $\begingroup$ 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? $\endgroup$ – Slobodan Simić Jun 20 at 17:28
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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 [1] that the equation $\operatorname{div}\Phi=\varphi$ may have no $C^1$ solutions.

[1] D. Preiss, Additional regularity for Lipschitz solutions of PDE. J. Reine Angew. Math. 485 (1997), 197-207.

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  • $\begingroup$ 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 [1]. Is this correct? $\endgroup$ – Slobodan Simić Jun 20 at 17:43

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