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)?
2 Answers
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.
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$\begingroup$ Anton, What do you mean by "espace to infinity"? Also, where did you use the hypothesis of completeness of the manifold? $\endgroup$– GigouApr 13, 2011 at 23:14
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$\begingroup$ I use that any curve of finite length has the end point inside the manifold. $\endgroup$ Apr 13, 2011 at 23:22
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$\begingroup$ @Ken, I assume we have completeness. You might use Hopf–Rinow to show that "geodesic completeness" $\Leftrightarrow$ "completeness". $\endgroup$ Apr 14, 2011 at 0:50
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$\begingroup$ @Ken, I'm confused about your comment. @Anton, Sorry, but I still don't understand your argument. What do you mean by "escape to infinity"? Also, have you considered that failure to completeness might occur because the curve is defined only on an interval of the form (a,+\infty) and thus have infinite lenght, and not necessarily on an interval of the form (a,b). $\endgroup$– GigouApr 14, 2011 at 2:10
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$\begingroup$ I add few words, now it should be totally clear. $\endgroup$ Apr 14, 2011 at 13:05
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 $a=-\infty$, $b=\infty$.