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.

Advertisements