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)?



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


