I realize I’ve been slow as of late.

Though the decomposition of the square-integrable 1-forms was a Hilbert space, $L^2$-decomposition, it reflects many smoothness properties in which we are interested.  The goal of this post is twofold: first, show that the smooth differentials in the first two components are respectively the exact and coexact ones; and second, show that the closedness of a square-integrable 1-form in some neighborhood  implies the smoothness of the corresponding $E$ term.  The ultimate goal is the existence theorems for harmonic functions on Riemann surfaces.

Smooth differentials in ${E, E^*}$

I now claim that any smooth differential ${\omega \in E}$ is exact and any smooth differential in ${E^*}$ is co-exact. This is nontrivial because of the way we took completions. I will only prove the claim for ${E}$.

We already have closedness of ${\omega}$ by the previous post. To show exactness, it will be sufficient to show that for any closed smooth curve ${c}$,

$\displaystyle \int_c \omega = 0$

because we could use a path integral to define the antiderivative. We can approximate ${c}$ by a simple closed curve homotopic to ${c}$, so we can assume at the outset that ${c}$ is a homeomorphism ${S^1 \rightarrow M}$

Proposition 1 Given ${c}$ as above, there is a closed differential ${\eta_c}$ such that$\displaystyle ( \omega, {}^* \eta_c) = \int_c \omega$

for any closed ${\omega}$

To prove this, find an annular neighborhood of the compact set ${c(S^1) \subset M}$, which is divided into ${R_1, R_2}$ as in the figure.

Choose ${\varphi}$ smooth on ${R_1}$ taking the value ${0}$ in a neighborhood of the outside part of ${\partial R_1}$, ${1}$ on a neighborhood of the inside part of ${R_1}$.Define the differential ${\eta_c}$$\displaystyle \eta_c = d \varphi \ \mathrm{inside} \ R_1 , \ \ 0 \ \text{elsewhere} .$

Now

$\displaystyle ( \omega, {}^* \eta_c) = - \int_{\overline{R_1}} \omega \wedge d \varphi = - \int_{\partial \overline{R_1}} \varphi \omega$

by Stokes theorem and since ${\omega}$ is closed. But this is

$\displaystyle \int_c \omega,$

proving the proposition.

Now fix a smooth ${\omega \in E}$; from its orthogonality to ${E^*}$ and smoothness is seen to be exact.

Smoothness of components

Fix a differential ${\theta \in L^2(M)}$, say$\displaystyle \theta = \alpha + \beta + \gamma$

where ${\alpha \in E, \beta \in E^*, \gamma \in H}$. We want conditions that will give local smoothness of ${\alpha, \beta}$ in case ${\theta}$ is in some neighborhood; ${\gamma}$ is always smooth.

Proposition 2 If ${\theta}$ is smooth and closed in a neighborhood ${U}$, so is ${\alpha}$.

This is local, so we can assume ${U}$ is a coordinate neighborhood with coordinate ${z = x+iy}$. Write$\displaystyle \theta = t_1 dx + t_2 dy, \ \alpha = a_1 dx + a_2 dy.$

We are given that ${t_1, t_2}$ are smooth; we need to prove that ${a_1, a_2}$ are. Take any smooth ${\phi}$ supported in ${U}$ and consider$\displaystyle (\theta, d \phi_x) = \int_U (t_1 dx +t_2 dy) \wedge ( -\phi_{xy} dx + \phi_{xx} dy) = \int_U t_1 \phi_{xx} + t_2 \phi_{xy} dx dy$

Similarly

$\displaystyle (\theta, {}^* d \phi_y) = - \int_U t_1 \phi_{yy} + t_2 \phi_{yx} dx dy$

The first equals ${(\alpha, d \phi_x)}$. The second is zero because ${\theta}$ is closed in ${U}$. So

$\displaystyle (\alpha, d \phi_x) = \int_U t_1 \Delta \phi .$

But we also have

$\displaystyle (\alpha, d \phi_x) = \int_U a_1 \Delta \phi$

if we apply the same reasoning to ${\alpha}$ itself! So as a result ${a_1 - t_1}$ is harmonic in ${U}$, hence smooth, and so is ${p}$. The other part of the proof is similar (replace ${d \phi_x}$ with ${{}^* d \phi_x}$, etc.).