### 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 .

Oh god I am an idiot. I completely forgot about Bochner’s Lemma when I wrote the above: on a compact Riemannian manifold there cannot exist a K.v.f. such that throughout unless is parallel. Furthermore the exceptional case cannot exists unless vanishes if the Ricci curvature is strictly negative on the manifold. See, for example, http://www.jstor.org/pss/1969418

So my trivial result just says that if is a 2-d compact Riemannian manifold with negative Euler characteristic, it cannot have a non-trivial Killing vector field.

Should be useful somewhere else.