Let $(M,g)$ be a riemannian manifold and $TM$ its tangent bundle. We know that if, for instance, $M$ is compact, any integral curve of any vector field on $M$ can be defined in the whole $R$ and not just locally. A similar result exists for left invariant vector fields on Lie groups. We know also that if $M$ is complete as a metric space, then it is geodesically complete, i.e. for any $p \in M$ and for any $u \in T_pM$, there exists a geodesic through $p$ in the direction defined by $u$, defined in the whole real axis (geodesics are integral curves of the geodesic spray, which is a vector field of the tangent bundle). I am interested in formulating some suitable assumption for $TM$ that allows to extend each integral curve of $TM$ in the whole real axis ($TM$ is not compact except of some trivial cases). This is because I would like to study global solutions of second order ODEs on riemannian manifolds (which are integral curves of second order vector fields, i.e. vector fields of the tangent bundle) So I need to put a property on the tangent bundle which (almost) directly implies that each integral curve of the tangent bundle can be defined globally. Any idea?
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$\begingroup$ Is the word bounded perhaps missing from the first paragraph? There are incomplete vectorfieds on R... also any manifold admits a complete metric. $\endgroup$– Thomas RotCommented Feb 19, 2019 at 23:27
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$\begingroup$ What do you mean by integral curve of $TM$? We can talk about integral curves of vectorfields and what vectorfield do you have in mind? $\endgroup$– Piotr HajlaszCommented Feb 20, 2019 at 12:09
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$\begingroup$ I mean an integral curve of some vector field in the tangent bundle. I have a second order ODE in the manifold $M$ and I am able to write it as a first order ODE in $TM$, so its solution is an integral curve of the vector field in $TM$, that can be represented locally in a way obtained directly by the second order ODE. The problem is that I have no idea how exactly this vector field looks like, so I need an assumption for $TM$, which provides that we can extend each integral curve of each vector field of $TM$ until infinity. $\endgroup$– FoivosCommented Feb 20, 2019 at 13:30
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There is no assumption on $TM$ that will do the trick for all vector fields, as you can embed the example of Poitr Hajlasz into $TM$. However there is the result that any vectorfield that is bounded w.r.t. a complete Riemannian metric will have complete integral curves (I think this is in Hirsch’ differential topology book). The metric on $M$ defines one on $TM$ So you can check it w.r.t. to that metric.
answered Feb 20, 2019 at 14:13
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$\begingroup$ I don't understand how the previous example can be extended for $TM$. Could you write some more details? $\endgroup$– FoivosCommented Feb 20, 2019 at 16:35
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$\begingroup$ Just writeit out in a chart and extend it by zero using a bumpfunction $\endgroup$ Commented Feb 21, 2019 at 1:39