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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)?

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No. since genus is at least 3, it has to have zero. Once you have zero, the universal cover of your surface has to be sphere. –  Anton Petrunin Feb 20 '13 at 19:55
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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.

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Thanks for the answer! The coefficients of a Killing field $v$ and its first derivative satisfy a closed system of DE's of the form $dv=\omega v$, and since $\omega$, which determines the rate of change of $v$, is bounded in a neighborhood of the puncture, $v$ must be bounded. I mistakenly thought that there might exist a Killing field with a pole-type singularity at a point, when the metric is degenerate appropriately at higher order. –  Joe Feb 21 '13 at 10:33
    
@Joe: Well, if the metric actually did degenerate, you certainly could have poles for solutions to the Killing equation. For example, if $M$ is a Riemann surface of genus $g>1$ and $\zeta$ is a nonzero holomorphic $1$-form on $M$, then the pseudometric $ds^2 = \zeta\circ\bar\zeta$ has a meromorphic Killing field $X=\mathrm{Re}(Z)$ where $Z$ is the meromorphic vector field that satisfies $\zeta(Z)=1$; $X$ will have 'poles' where $\zeta$ has zeros. As long as the metric is nondegenerate though, you're OK (as my parenthetical remark and your expansion of it in your comment both indicate). –  Robert Bryant Feb 21 '13 at 12:35
    
Thanks again for the clarification. I should have written "when the curvature of metric is degenerate appropriately at higher order" –  Joe Feb 21 '13 at 14:51
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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 $M$.

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You just repeated my comment :) –  Anton Petrunin Feb 21 '13 at 0:44
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