Symmetry and uniformization
by Willie Wong
A bit of trivia (can’t think of any use of it now)
Let be a two-dimensional compact Riemannian manifold with negative Euler characteristic. Let be a Killing vector field on it. By the uniformisation theorem, there exists a conformal factor such that has constant negative scalar curvature. Then .
The statement follows from considering the conformal change of curvature formula in two dimensions. By a rescaling, we can assume has Gauss curvature -1. Then must solve
where are the Laplace-Beltrami operator and Gauss curvature associated with . Now hit the equation with the Lie derivative . Since is Killing relative to , its Lie derivative commutes with the Laplace-Beltrami operator, and kills . So we have
Now we multiply by throughout, and integrate over , we have that
which leads to our conclusion. Q.E.D.
Observe that the same argument works for conformally flat Riemann surfaces. That is: let be a 2-dimensional compact Riemannian manifold with non-positive Euler characteristic, and let be the conformal metric with constant Gauss curvature. Then any infinitesimal symmetry of is a symmetry of .
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 , the canonical rotation vector field is given by in polar coordinates. The metric can be written as . Now take the conformal transformation , which implies that . The pull-back metric differs from the standard metric by a factor of , which is emphatically not constant along integral curves of .