Globally defined integral curves on the tangent bundle

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?

• Is the word bounded perhaps missing from the first paragraph? There are incomplete vectorfieds on R... also any manifold admits a complete metric. – Thomas Rot Feb 19 '19 at 23:27
• 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? – Piotr Hajlasz Feb 20 '19 at 12:09
• 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. – Foivos Feb 20 '19 at 13:30

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.
• I don't understand how the previous example can be extended for $TM$. Could you write some more details? – Foivos Feb 20 '19 at 16:35