# How to solve Poisson's equation on compact Riemann surfaces of genus greater than one?

$$M$$ 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?

• In what sense do you want to solve the equation? Do you just want existence of a solution? Or do you also want to say something about uniqueness and/or regularity properties in terms of $f$? – Strants May 2 at 16:08
• @Strants I just want existence of a smooth solution. I know this is related to Hodge decomposition, but I hope the tools used are not too deep, the best thing is to be able to calculate it. – Xingying Li May 2 at 16:56
• I'm not confident enough in my understand of PDE on manifolds to post this as an answer. It seems to me, however, that we could get existence of a solution by variational techniques. Once we have that, by going to a local chart we see should have that $\phi$ satisfies an elliptic equation (with coefficients depending on the metric data for $M$), so we can use the 'flat' elliptic regularity theory to get $\phi \in C^\infty$ on this chart. Since this is true on every chart, it seems that $\phi$ should be smooth. – Strants May 2 at 18:28

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 $$\varphi\in\bigcap_n H^{2n}(M)=C^\infty(M)$$.
• Section 1.3 of Steve Rosenberg's book (Laplacian on a Riemannian manifold) is a good summary of the analytic results needed here (except he used the heat operator $e^{-t\Delta}$ instead). – user10354138 May 4 at 13:57
• Why is $I+\Delta$ invertible? – Xingying Li May 4 at 17:54
• If you know some unbounded operator theory, it follows from $d\colon L^2(M)\to L^2(M;T^*M)$ is a densely-defined closed operator (so $(I+d^*d)^{-1}$ is bounded everywhere defined). If not, just using one-sided inverse (from $I+\Delta$ injective) is sufficient for the purpose here. – user10354138 May 7 at 16:23