Decomposition of the space of square-integrable 1-forms

Yesterday I defined the Hilbert space of square-integrable 1-forms ${L^2(X)}$ on a Riemann surface ${X}$ . Today I will discuss the decomposition of it. Here are the three components:

1) ${E}$ is the closure of 1-forms ${df}$ where ${f}$ is a smooth function with compact support.

2) ${E^*}$ is the closure of 1-forms ${{}^* df}$ where ${f}$ is a smooth function with compact support.

3) ${H}$ is the space of square-integrable harmonic forms.

Today’s goal is:

Theorem 1 As Hilbert spaces,

$\displaystyle L^2(X) = E \oplus E^* \oplus H.$

The proof will be divided into several steps.

Computation of ${E^{\perp}}$

I claim that the smooth, compactly supported forms orthogonal to ${E}$ are precisely the co-closed ones. Indeed, if ${f}$ is smooth and compactly supported, and ${\omega}$ is arbitrary:

$\displaystyle ( df, {}^* \omega) = - \int_M df \wedge \overline{ {}^* {}^* d\omega } = \int_M df \wedge \bar{\omega}$

By the identities proved yesterday (essentially Stokes formula) this is

$\displaystyle \int_M f d \bar{\omega} - \int_{\partial M} f \omega = \int_M f d \overline{\omega} \ \mathrm{since} \ \partial M = \emptyset.$

It is now clear that this vanishes for all ${f}$ smooth and compactly supported if f ${d \bar{w} = \overline{ d \omega} = 0}$ .

Similarly the smooth 1-forms orthogonal to ${E^*}$ are precisely the closed ones.

Corollary 2 ${E,E^*}$ are orthogonal.

Computation of ${E^{\perp} \cap E^{* \perp}}$

This is the key place where we will use Weyl’s lemma. Let ${U}$ be an arbitrary coordinate neighborhood with coordinate ${z = x+iy}$ . Let ${\omega \in E^{\perp} \cap E^{* \perp}}$ , and suppose ${\omega = p dx + qdy }$ on ${U}$ .

Lemma 3 ${\omega}$ is smooth in ${U}$ .

As a result, we see by the previous section that ${\omega}$ is closed and co-closed, so harmonic.

Now we prove the lemma.

If ${f}$ is an arbitrary smooth real function supported in ${U}$ , then

$\displaystyle (\omega, -{}^* df_x) = \int_U (p dx + qdy) \wedge (f_{xx}dx + f_{xy} dy) = \int_U (p f_{xy} - q f_{xx}) dx dy.$

Also

$\displaystyle (\omega, df_y) = \int_U (p dx + qdy) \wedge (- f_{yy}d x + f_{xy} dy) = \int_U (p f_{xy} + q f_{yy}) dx dy.$

Both these equal zero though by assumption. Subtracting yields

$\displaystyle \int_U q \Delta f dx dy = 0$

which since ${f}$ was arbitrary implies harmonicity of ${q}$ by Weyl’s lemma. A similar argument with ${{}^* df_x, {}^* df_y}$ shows the same thing for ${p}$ . As a result, we get smoothness.

Climbing Mount Bourbaki