4
$\begingroup$

$\newcommand{\dz}{{\rm d}z}$ $\newcommand{\dbz}{{\rm d}\bar z}$ I am reading this version of the Gauss-Bonnet theorem (in the book Compact Riemann Surfaces by Jurgen Jost), which is regarding compact Riemann surfaces without boundary equipped with a conformal metric:

Theorem (Gauss-Bonnet) Let $\Sigma$ be a compact Riemann surface without boundary, of genus $p$, with a conformal Riemannian metric given in local coordinates by $\rho^2(z)\dz\dbz$. Let the curvature be $K_\rho$. Then $$\int_\Sigma K_\rho\rho^2(z)\frac{i}{2}\dz\wedge\dbz=2\pi(2-2p)$$

Before giving the proof I should note that the cases in which the curvature is constant have been proved, so we can make use of them, namely, these cases:

  • The unit sphere $\Sigma=S^2,\ K\equiv 1$ (Induced Eulidean metric)
  • The torus $\Sigma=\mathbb C/M,\ K\equiv 0$ (Induced Eucliean metric, and $M$ is a rank-2 lattice)
  • $\Sigma=H/\Gamma,\ K\equiv -1$ (Hyperbolic metric given by $\frac{1}{y^2}\dz\dbz$, and $\Gamma$ is a discrete group of isometries, acting without fixed points)

The proof in the book goes like this:

proof. By the uniformization theorem we first assume $\Sigma$ is diffeomorphic to $S^2$ or $\mathbb C/M$ or $H/\Gamma$. We put another metric $\lambda^2(z)\dz\dbz$ of constant curvature $K$. Now the quotient $\rho^2(z)/\lambda^2(z)$ is invariant under coordinate transformations, i.e. behaves like a function. We compute now $$\int K\lambda^2\frac{i}{2}\dz\wedge\dbz-\int K_\rho\rho^2\frac{i}{2}\dz\wedge\dbz\\ =-4\int\frac{\partial^2}{\partial z\partial\bar z}\log\lambda\frac{i}{2}\dz\wedge\dbz+4\int\frac{\partial^2}{\partial z\partial\bar z}\log\rho\frac{i}{2}\dz\wedge\dbz\\ =4\int\frac{\partial^2}{\partial z\partial\bar z}\log\frac{\rho}{\lambda}\frac{i}{2}\dz\wedge\dbz$$ which vanishes by Gauss' Divergence Theorem. The theorem then follows from the cases with constant curvatures.

Thoughts and Questions:

(1) How can we ensure the existence of a metric with constant curvature? Can I say that the diffeomorphism $f:\Sigma\to S^2$ (or $\mathbb C/M$ or $H/\Gamma$, by the uniformization theorem) equips $S^2$ with a metric induced by that on $\Sigma$, and the curvature integrals on them are the same, so as to focus on $S^2$ instead of $\Sigma$?

(2) I understand why $\rho^2(z)/\lambda^2(z)$ is invariant under a coordinate transformation. But it doesn't seem to be useful in the following proof. What does it do?

(3) Why does the last integral vanish? The Gauss divergence theorem turns an integral over a domain to its boundary. However, $\Sigma$ is a surface without a boundary, so I am not sure how the divergence theorem applies here. Does this have something to do with $\rho^2/\lambda^2$ being invariant under coordinate changes?

$\endgroup$
3
+50
$\begingroup$

(1) The existence of metric with constant curvature comes from the pull-back metric by the diffeomorphism $f:\Sigma\to (S^2,g_{std})$, where $g_{std}$ is the standard metric with constant curvature in $S^2$.

(2), (3) In the boundaryless case, the divergence theorem says the $\int \nabla\cdot F$ is zero for vector fields $F$. To apply the theorem, we must show that the integrand is of this form. Notice that we have $\dfrac{\partial}{\partial z}\dfrac{\partial}{\partial \bar{z}}u=4\nabla\cdot\nabla u$, for any smooth function $u$.

As in the last line of the proof, it suffice to show that $\dfrac{\rho}{\lambda}$ is a function, so that $\log(\dfrac{\rho}{\lambda})$ is also a function. That's why $\dfrac{\rho^2}{\lambda^2}$ behaves like a function is useful. (Notice that $\rho,\lambda>0$, we are safe to take square root.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.