Differential Forms 08: Hodge-Weyl on Riemann Surfaces

Differential forms / 08

Hodge-Weyl on Riemann surfaces

The compact-surface Poisson equation is solved by turning Stokes' theorem into an energy identity and then applying Hilbert-space existence.

Calculus root: solving \(u''=f\)

The Hodge-Weyl theorem has a one-variable shadow. On a circle of length \(L\), one may write the model equation with either sign convention:

\[u''(x)=f(x)\qquad\text{or}\qquad -u''(x)=f(x).\]

Either sign convention can have a periodic solution only if

\[\int_0^L f(x)\,dx=0,\]

because the integral of \(u''\) over one period is zero. Constants are also invisible to the second derivative, so one fixes uniqueness by imposing \(\int_0^L u\,dx=0\).

Example: compatible forcing

For \(f(x)=\sin x\) on \([0,2\pi]\) with periodic boundary conditions, \(\int_0^{2\pi}f\,dx=0\), and \(u(x)=-\sin x\) solves \(u''=f\). With the nonnegative convention \(-u''=f\), the solution is \(u(x)=\sin x\).

Example: incompatible forcing

For \(f(x)=1\), the compatibility condition fails:

\[\int_0^{2\pi}1\,dx=2\pi.\]

No periodic function has second derivative equal to \(1\), and no periodic function satisfies \(-u''=1\) either. On a compact surface, the condition \(\int_M f\,\omega=0\) is the same obstruction expressed through Stokes and the area form.

Chapter 5 of Part III proves a Hodge-Weyl theorem for functions on a compact Riemann surface. Let \(M\) be compact, let \(\omega\) be a smooth Hermitian area form, and consider

\[\Delta_\omega u=f.\]

The obstruction is constants:

\[\int_M \Delta_\omega u\,\omega=0.\]

Thus a solution can exist only if

\[\int_M f\,\omega=0.\]

The normalization

\[\int_M u\,\omega=0\]

removes the freedom to add constants.

Weak formulation

Define

\[\langle u,v\rangle_{L^2}=\int_M uv\,\omega, \qquad \|\nabla u\|_{L^2}^2=\int_M |\nabla u|^2\,\omega.\]

The weak equation is

\[\int_M \langle \nabla u,\nabla\phi\rangle\,\omega =\int_M f\phi\,\omega\]

for all smooth test functions \(\phi\), using the nonnegative Laplacian convention \(\Delta=-\operatorname{div}\nabla\). With the opposite convention the right-hand side changes sign. The identity is integration by parts on a closed surface, hence Stokes without boundary.

Mean-zero Hilbert space

Let

\[W^{1,2}_\perp(M)= \left\{u\in W^{1,2}(M):\int_M u\,\omega=0\right\}.\]

On this space the Poincare inequality says

\(\|u\|_{L^2}\le C_P\|\nabla u\|_{L^2}.\)

Existence by Lax-Milgram

Set

\[A(u,v)=\int_M\langle\nabla u,\nabla v\rangle\,\omega, \qquad \Lambda_f(v)=\int_M fv\,\omega.\]

On \(W^{1,2}_\perp(M)\), the Poincare inequality makes \(A\) coercive:

\[A(u,u)=\|\nabla u\|_{L^2}^2\ge c\|u\|_{W^{1,2}}^2.\]

The functional \(\Lambda_f\) is bounded by Cauchy-Schwarz and Poincare. Lax-Milgram gives a unique weak solution \(u\in W^{1,2}_\perp(M)\) and an energy estimate

\[\|\nabla u\|_{L^2}\le C\|f\|_{L^2}.\]

Weyl’s lemma and elliptic regularity upgrade the weak solution to a smooth solution when \(f\) is smooth.

The theorem

If \(f\in C^\infty(M)\) and \(\int_M f\,\omega=0\), then there is a unique smooth \(u\) satisfying

\[\Delta_\omega u=f,\qquad \int_M u\,\omega=0.\]

The proof is: mean-zero compatibility, Poincare coercivity, Lax-Milgram weak existence, then elliptic regularity.

Model computations

Example 1: circle Fourier mode

On \(S^1=\mathbb R/2\pi\mathbb Z\) with \(\Delta=-d^2/d\theta^2\), solve

\[\Delta u=\sin(3\theta),\qquad \int_0^{2\pi}u\,d\theta=0.\]

Since

\[-{d^2\over d\theta^2}\left({1\over 9}\sin(3\theta)\right)=\sin(3\theta),\]

the normalized solution is

\(u(\theta)={1\over9}\sin(3\theta).\)

Example 2: flat torus mode

On \(T^2=\mathbb R^2/\mathbb Z^2\) with \(\Delta=-(\partial_x^2+\partial_y^2)\), take

\[f(x,y)=\cos(2\pi x)\cos(2\pi y).\]

The mean is zero. Since

\[\Delta\bigl(\cos(2\pi x)\cos(2\pi y)\bigr)=8\pi^2\cos(2\pi x)\cos(2\pi y),\]

the solution is

\(u(x,y)={1\over 8\pi^2}\cos(2\pi x)\cos(2\pi y).\)

Example 3: round sphere height

On the round sphere, the height function \(z\) has mean zero and is a first spherical harmonic. With the nonnegative Laplacian convention,

\[\Delta z=2z.\]

Therefore \(u=z/2\) solves \(\Delta u=z\) with mean zero.

Why this belongs in a forms chapter

The proof is not an isolated PDE result. The weak identity is integration by parts, integration by parts is Stokes, and the mean-zero obstruction is the statement that a divergence integrates to zero on a compact manifold without boundary.