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.).

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