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Let $\Sigma$ be an $n$-dimensional smooth closed manifold ($n\ge 3$) with a non-continuous metric $g\in W^{2,\frac{n}2}\cap L^{\infty}(\Sigma)$. Let $g'$ be a fixed smooth metric on $\Sigma$, there exists $\Lambda>1$ such that $\Lambda^{-1} g'_p(X,X)\le g_p(X,X)\le \Lambda g'_p(X,X)$ for a.e. $p\in \Sigma$ and any $X\in T_p(\Sigma)$. For a differential form $\alpha\in L^2(\Sigma,\Omega^k(\Sigma))$, assume we have $d^{*}\alpha=0$ and $d^*d\alpha=0$, where $d^*$ is the codifferential with respect to $g$. If $\alpha$ is a function, then it's a solution of the divergence-form elliptic PDE $\Delta_g \alpha=0$. A duality argument as in divergence-form regularity implies that $\alpha\in W^{1,p}$ for any $p<\infty$. And we can further prove that $\alpha\in W^{3,q}$ for any $q<\frac n2$.

If $\alpha$ is a $k$-form for $k\ge 1$, I wonder if we can get the same improvement of regularity. We have $\Delta_g \alpha=0$, which in local coordinates is a non-divergence-form elliptic system of equations. But it seems we can only get $\alpha\in W^{1,\frac{2n}{2+n}}$ and $\alpha\in W^{2,p}$ for any $p\ge 1$ satisfying $\frac 1p\ge \frac 12+\frac 2n$. Maybe this is really the best we can do, otherwise there should be other ways to use the condition $d^*\alpha=0$.

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    $\begingroup$ If $\alpha$ is a k-form with $k>0$, then the equation $d^{\ast}d\alpha=0$ is not elliptic. Instead, $(d^{\ast}d+dd^{\ast})\alpha=0$ is. $\endgroup$
    – G. Blaickner
    Commented May 25 at 14:14
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    $\begingroup$ Since $\alpha$ is not a scalar function but, in local coordinates, is specified by a set of scalar functions, it is more comparable to a vecctor-valued function. In any case, have you tried to write the equations in local coordinates and checked whether they form a divergence-free strongly elliptic system of PDEs? What did you get? $\endgroup$
    – Deane Yang
    Commented May 25 at 15:37
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    $\begingroup$ This question seems more suited to math.stackexchange.com $\endgroup$
    – Overflowian
    Commented May 25 at 16:00
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    $\begingroup$ @DeaneYang In the case $n=2$ and $g$ is conformal, I tried to insert $d^*\alpha=0$ in some sense to get a divergence-form system of equations, but it's weird: the system is easy to solve and improve regularity but the coefficients don't satisfy the criterion for strong ellipticity. $\endgroup$
    – Tian LAN
    Commented May 25 at 18:24
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    $\begingroup$ @TianLAN my advice: I think that if you want to get some answers and upvotes you should explain extensively and clearly what are you hypothesis, what you want and what are the common approaches you considered and why they fail. You say that it's less trivial that it what appears to be, this shouldn't be part of the comments but should be explained in the question. That's not what the comment section is for. Furthermore $d^*$ is also the formal adjoint of $d$, so why isn't $d^* d\alpha $ already in divergence form? Making your question appreciated also by non-experts will give you moreupvotes. $\endgroup$
    – Overflowian
    Commented May 26 at 10:21

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