Geodesic completeness and complete Killing fields Why are the Killing fields on a complete Riemannian manifold themselves complete (that is, the integral curves of the Killing fields are defined for all time)?
 A: The corresponding flow, say $\Phi^t: M\to M$ preserves the metric and the field.
Thus, for any $x\in M$, the curve $\alpha_x\colon t\mapsto \Phi^t(x)$ has constant speed.
Therefore it can not escape to infinity in finite time.
More precisely: 
if $\alpha_x$ is defined on a bounded interval $(a,b)$ 
then the restriction $\alpha_x|(a,b)$ has finite length,
and from completeness it can be extended to a neighborhood of $[a,b]$.
This implies that $\alpha_x$ is defined on whole $\mathbb R$;
i.e., the vector field is complete.
A: This is quite the answer as Anton gave years ago, but I just wanted to be a bit more detailed.
Let $M$ be a complete Riemannian manifold and $X$ a Killingfield, that is, $\nabla_{\_} X$ is skew-symmetric.
Now let $\gamma: (a,b) \to M$ be an integral curve of $X$. We got to show $a = -\infty$, $b=\infty$. First of all, notice
$\frac{d}{dt} \Vert \gamma'(t) \Vert^2 = \frac{d}{dt}\Vert X(\gamma(t)) \Vert^2 = \frac{d}{dt}\langle X(\gamma(t)),X(\gamma(t))\rangle = 2\langle\frac{D}{dt}X(\gamma(t)),X(\gamma(t))\rangle 
 = 2\langle\nabla_{\gamma'(t)}X, X(\gamma(t))\rangle = \langle\nabla_{\gamma'(t)}X, X(\gamma(t))\rangle - \langle\gamma'(t),\nabla_{X(\gamma(t))}X\rangle = 0,$
so $\gamma$ has constant speed; assume $\Vert \gamma'(t) \Vert \equiv 1$. Then the length of $\gamma$ is $L(\gamma) = b-a$, so 
$\text{Im}(\gamma) \subset B_{b-a}\left(\gamma\left(\frac{b-a}{2}\right)\right)$
by the definition of the metric on $M$ induced by the Riemannian metric. Completeness implies that the closed ball
$\overline{B_{b-a}\left(\gamma\left(\frac{b-a}{2}\right)\right)}$
is compact. But $\gamma$ has to leave every compact set. Therefore $a=-\infty$, $b=\infty$.
