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Feb 21, 2013 at 14:51 comment added Joe Thanks again for the clarification. I should have written "when the curvature of metric is degenerate appropriately at higher order"
Feb 21, 2013 at 12:35 comment added Robert Bryant @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).
Feb 21, 2013 at 10:33 comment added Joe 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.
Feb 21, 2013 at 10:01 vote accept Joe
Feb 21, 2013 at 0:44 history answered Robert Bryant CC BY-SA 3.0