Symmetry and uniformization

by Willie Wong

A bit of trivia (can’t think of any use of it now)

Let (M,g) be a two-dimensional compact Riemannian manifold with negative Euler characteristic. Let V be a Killing vector field on it. By the uniformisation theorem, there exists a conformal factor u: M\to \mathbb{R} such that e^{2u}g has constant negative scalar curvature. Then V(u) = 0.

The statement follows from considering the conformal change of curvature formula in two dimensions. By a rescaling, we can assume e^{2u}g has Gauss curvature -1. Then u must solve

\displaystyle \triangle_g u - e^{2u} = K_g

where \triangle_g, K_g are the Laplace-Beltrami operator and Gauss curvature associated with g. Now hit the equation with the Lie derivative \mathcal{L}_V. Since V is Killing relative to g, its Lie derivative commutes with the Laplace-Beltrami operator, and kills K_g. So we have

\displaystyle \triangle_g V(u) - 2 e^{2u}V(u) = 0

Now we multiply by V(u) throughout, and integrate over M, we have that

\displaystyle \int_M |\nabla V(u)|^2 + 2 e^{2u} V(u)^2 dvol_M = 0

which leads to our conclusion. Q.E.D.

Observe that the same argument works for conformally flat Riemann surfaces. That is: let (M,g) be a 2-dimensional compact Riemannian manifold with non-positive Euler characteristic, and let g_0 be the conformal metric with constant Gauss curvature. Then any infinitesimal symmetry of g is a symmetry of g_0.

For surfaces of positive Euler characteristic (well, the topological 2-sphere), this argument doesn’t work. And in fact, the statement is false. This is related to the fact that the 2-sphere has a really huge conformal group (any fractional-linear/Moebius transformation of the complex plane gives rise to one such) (which also causes a bit of technical difficulty when considering the uniformisation problem for the sphere). For example, under the stereographic projection of the 2-sphere to the complex plane \mathbb{C}, the canonical rotation vector field is given by \frac{\partial}{\partial\theta} in polar coordinates. The metric can be written as 16dz d\bar{z}/(4 + |z|^2)^2 . Now take the conformal transformation w = (z+1)/(z+2), which implies that dw = dz/(z+2)^2. The pull-back metric 16dwd\bar{w} / (4 + |w|^2)^2 differs from the standard metric by a factor of [(4 + |z|^2) / (4|z+2|^2 + |z+1|^2)]^2, which is emphatically not constant along integral curves of \frac{\partial}{\partial\theta}.