I realize I’ve been slow as of late.
Though the decomposition of the square-integrable 1-forms was a Hilbert space, -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
term. The ultimate goal is the existence theorems for harmonic functions on Riemann surfaces.
Smooth differentials in
I now claim that any smooth differential is exact and any smooth differential in
is co-exact. This is nontrivial because of the way we took completions. I will only prove the claim for
.
We already have closedness of by the previous post. To show exactness, it will be sufficient to show that for any closed smooth curve
,
because we could use a path integral to define the antiderivative. We can approximate by a simple closed curve homotopic to
, so we can assume at the outset that
is a homeomorphism
.
Proposition 1 Given
as above, there is a closed differential
such that
![]()
for any closed
.
To prove this, find an annular neighborhood of the compact set , which is divided into
as in the figure.
Choose smooth on
taking the value
in a neighborhood of the outside part of
,
on a neighborhood of the inside part of
.Define the differential
Now
by Stokes theorem and since is closed. But this is
proving the proposition.
Now fix a smooth ; from its orthogonality to
and smoothness is seen to be exact.
Smoothness of components
Fix a differential , say
where . We want conditions that will give local smoothness of
in case
is in some neighborhood;
is always smooth.
Proposition 2 If
is smooth and closed in a neighborhood
, so is
.
This is local, so we can assume is a coordinate neighborhood with coordinate
. Write
We are given that are smooth; we need to prove that
are. Take any smooth
supported in
and consider
Similarly
The first equals . The second is zero because
is closed in
. So
But we also have
if we apply the same reasoning to itself! So as a result
is harmonic in
, hence smooth, and so is
. The other part of the proof is similar (replace
with
, etc.).
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