$M$ is a compact Riemann surface, $f\in C^{\infty}(M)$. I want to find the solution of $\Delta \varphi=f$.
When $M=T^2=\mathbb{R}^2/2\pi \mathbb{Z}^2$, I can use Fourier series on $\mathbb{R}^2$ to solve this equation.
When $M$ are compact Riemann surfaces of genus greater than one, how to solve it?
By the uniformization theorem, compact Riemann surfaces of genus greater than one have simply connected universal covering surface given by the unit disk $D$(poincare metric).
And $M=D/\Gamma$, $\Gamma$ is a Fuchsian group of Mobius transformations. So maybe I can solve possion's equation on the Poincare disk?
Best Answer
Unfortunately You can't just use $D/\Gamma$ because then the boundary $S^1$ will be very badly behaved (e.g., nonexistence of nontangential limit by choosing the appropriate point on each fundamental $4g$-gon).
The usual introductory way is to just use Hilbert space theory with minimal amount of Sobolev spaces thrown in. $(I+\Delta)^{-1}\colon L^2(M)\to H^2(M)$ is continuous, so $(I+\Delta)^{-1}\colon L^2(M)\to H^2(M)\hookrightarrow L^2(M)$ is compact (Rellich-Kondrakov) self-adjoint (easy) and positive (also easy), hence you have an orthonormal basis of eigenfunctions $h_i$, and $\Delta h_i=\lambda_i h_i$ with $0=\lambda_0\leq\lambda_1\leq\dots\leq\lambda_i\uparrow +\infty$. Expressing $f=\sum a_ih_i$, you get a solution $\varphi=\sum(a_i/\lambda_i)h_i$. Note that $\lambda_i=0$ iff $h_i$ is constant (by Green's first identity), so a necessarily and sufficient condition is $\int_M f=0$, and if you want uniqueness you can impose $\int_M\varphi=0$. Since $\Delta^n\varphi=\Delta^{n-1}f\in C^\infty(M)\subset L^2(M)$ for every $n$, we have $\varphi\in\bigcap_n H^{2n}(M)=C^\infty(M)$.