Compact surface with genus$\geq 2$ with Killing field

Question

Let M be a compact Riemannian surface of genus$\geq 2$.

Can M have a globally defined Killing field ? Can M have a Killing field defined on M-(finite set of points)?

Answer 1

Of course, there's always the Killing field $X\equiv0$. :)

Seriously, here's a different proof and an argument that addresses the 'suppose one leaves out a finite number of points' question:

Taking the orientation double cover if necessary, we can assume that $M$ is connected and orientable with $g\ge 2$. The hypotheses imply that the Euler characteristic of $M$ is negative. However, if $X$ were a nonzero Killing field on $M$, then it would have isolated zeros of index $+1$, and, since the sum of the indices of the zeros of $X$ equals the Euler characteristic, we have a contradiction.

As for omitting points, any Killing vector field $X$ defined on a punctured disk must extend smoothly across the puncture (and remain Killing). To see this, remember that $X$ is also a conformal vector field and hence is the real part of a holomorphic vector field $Z$ on the punctured disk (once one fixes the orientation and hence the underlying complex structure). However, $X = \mathrm{Re}(Z)$ must remain bounded in size near the puncture (because the Killing equation is an overdetermined linear system of finite type with coefficients that are smooth across the puncture), so the usual removable singularities theorem in complex analysis tells us that $Z$, and hence $X$, extends smoothly across the puncture. Since $X$ is Killing everywhere except at the puncture, it must be Killing there, too.

Answer 2

By passing to the double cover of $M$, we may assume that $M$ is orientable (compact of genus $n\geq2$); note that a Killing field on $M$ would lift to a Killing field on the double covering. A Riemannian metric together with an orientation defines an almost complex structure on $M$, simply by declaring that the complex structure on each tangent plane is rotation by ninety degrees, say, in the counter-clockwise sense. Since the dimension of $M$ is two, this structure is integrable, i.e. there is an atlas of complex charts on $M$ inducing it so that $M$ becomes a compact Riemann surface.

A Killing field on $M$ would generate a continuous one-parameter group of isometries of $M$, and any isometry of $M$ (or, for that matter, a conformal transformation) is a holomorphic automorphism. However, a theorem of H. A. Schwarz states that the group of holomorphic automorphisms of a compact Riemann surface of genus $\geq2$ is finite. Hence there are no Killing fields on $M$.