Klein-Poincaré uniformization theorem

From Mathematics wiki

Theorem:

Let $(M,g)\,$ be a compact 2-dimensional Riemannian manifold. Then there is a metric $\tilde{g} = e^{2\omega}g\,$ conformal to $g\,$ which has constant Gaussian curvature.

Examples

Torus

Consider a manifold with the topology of a torus $\mathbb{T}^2\,$ . Since the Euler characteristic $\chi = 0\,$ , by the Gauss-Bonnet theorem,

$\int_M\!\sqrt{g} R = 0\,$ .

A natural inner product between functions $h\,$ and $g\,$ is

$\left\langle f, h\right\rangle = \int_M\!\sqrt{g} f h\,$ ,

under which the Laplace-Beltrami operator $\nabla^2\,$ is Hermitian and can be diagonalized. First, $R\,$ is orthogonal to any constant function, thus $R\,$ has no zero mode on the torus. We can therefore formally invert $\nabla^2\,$ on $R\,$ to find some function $f\,$ such that $\nabla^2 f = R\,$ .

Under a Weyl transformation,

$g_{ab} \to e^{2\omega} g_{ab}\,$ ,

the Ricci scalar transforms as

$R \to e^{-2\omega} ( R - 2\nabla^2 \omega)\,$ ,

therefore we can always find some $\omega\,$ such that $2 \nabla^2 \omega = R\,$ , and so it is always possible to set $R = 0\,$ on the torus. In 2 dimensions, this implies that we can set the Riemann tensor to zero also, which means we can always find a metric that is that of Euclidean $\mathbb{R}^2\,$ .