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Let $M$ be a compact manifold and the tangent bundle $TM$ have a Riemannian metric $g$ so that it is isomorphic to the cotangent bundle $T^*M$. Consider the pull-back of the canonical symplectic form $\omega_{std}$ to define a symplectic form $\omega_g$ on $TM. $ Assuming the following theorem:

The geodesics in $M$ are the images by the standard projection $$\pi: TM \to M$$ of the integral curves of the vector field $X_H$ where $H$ is the norm function squared in TM defined by the metric $g$

I want to show that the geodesics exist for all of time.

Please help by giving some ideas!

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The short-time existence & uniqueness theorem for ODE's contains estimates for the length of the interval on which the solution exists in terms of the coefficients and their derivatives and the size of the chart you must remain within.

This implies that an ODE on a compact manifold (not quite your situation) admits a solution for all time: Cover by finitely many half-charts, get the bounds in terms of the coefficients and the sizes of the charts, and get a global $\epsilon > 0$ such that the ODE can always be solved for time $\epsilon$. Then it can be solved for all time just by "moving forward in the interval" and repeatedly solving for $\epsilon$ more time starting from the end of your previous solution.

Your situation is slightly different, because your manifold, $TM$, is not compact. But you can prove that your integral curves always stay the same length because they remain on a level set for $H$. So in fact if you start in the interior of a compact $B_R(TM)$ (consisting of the closed $R$ ball around $0$ in each tangent space) then you will remain there (and also remain at a fixed distance from the boundary of this). So the same proof as in the compact case works.

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  • $\begingroup$ Could you make the argument more precise? Or perhaps you could give me the reference for the proof in the compact case? I really need to understand this! $\endgroup$ – user166467 Dec 14 '14 at 20:13
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    $\begingroup$ The point in the compact case is just that there's a global $\epsilon$ such that you can solve for time $\epsilon$ from any starting point. Then if you solve for time $[0,\epsilon]$, start again at $t = \epsilon$ as your initial value and solve for $\epsilon$ more time. Now you have a solution on $[0,2\epsilon]$. And so on. The theorem sounds fancy but it's all just based on the ODE case in coordinates. I may have time to add more exposition later. I believe Hirsch Differential Topology discusses this sort of stuff. $\endgroup$ – aes Dec 14 '14 at 20:22
  • $\begingroup$ I get the intuition but want to make it more precise. I will look for the reference, but would also appreciate your exposition if you get time. Thanks! $\endgroup$ – user166467 Dec 14 '14 at 20:24
  • $\begingroup$ Btw, I could not find relevant material on the reference. $\endgroup$ – user166467 Dec 14 '14 at 21:09
  • $\begingroup$ @monomorphic Hirsch, Differential Topology, Section 6.2. A more comprehensive, modern, and readable source: Lee, Smooth Manifolds, Chapter 12. $\endgroup$ – aes Dec 14 '14 at 22:07

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