# Geodesic completeness and complete Killing fields

Hi,

I would like to know why the Killings fields on on a complete riemannian manifold are themselves complete (that is, the integral curves of the Killing fields are defined for all time).

Thanks

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I don't understand this question. What does it mean for a Killing field to exist for all time? A Killing field is a vector field on the manifold and has no dependence on time. –  Deane Yang Apr 13 '11 at 22:08

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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Anton, What do you mean by "espace to infinity"? Also, where did you use the hypothesis of completeness of the manifold? –  Gigou Apr 13 '11 at 23:14
I use that any curve of finite length has the end point inside the manifold. –  Anton Petrunin Apr 13 '11 at 23:22
@Ken, I assume we have completeness. You might use Hopf–Rinow to show that "geodesic completeness" $\Leftrightarrow$ "completeness". –  Anton Petrunin Apr 14 '11 at 0:50
@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). –  Gigou Apr 14 '11 at 2:10
I add few words, now it should be totally clear. –  Anton Petrunin Apr 14 '11 at 13:05