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This question already has an answer here:

Does the Lagrangian formalism require a metric on the configuration manifold $Q$ in order to define a Lagrangian $L$ on the tangent bundle $TQ$?

Further, if we specify a metric on the tangent bundle then we can via an isomorphism, move this to the cotangent bundle $T^*Q$. That being said how does this metric structure interplay with the symplectic structure?

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marked as duplicate by ACuriousMind, Kyle Kanos, John Rennie, Manishearth Apr 2 '15 at 7:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Comments to the question (v2):

  1. On one hand, let there be given a configuration space $(Q,g)$ endowed with a metric $g$. (As ACuriousMind points out in a comment, there is a 1-1 correspondence between a metric $g$ and the kinetic term in a Lagrangian.) On the other hand, note that the canonical symplectic 2-form $\omega$ on the cotangent bundle $T^{\ast}Q$ does not depend on the metric $g$ at all, cf. e.g. this Phys.SE post. The lesson is to expect no relation between $g$ and $\omega$ generically.

  2. Of course, Kähler manifolds (which assume a compatibility condition between a symplectic and metric structure) are used in many areas of modern theoretical physics. But that is another story.

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No, but you are most likely to get one from the kinetic term of the Lagrangian itself. In most cases one requires it to be a convex function in the $\dot q$ variables. You then get a metric if such kinetic term is quadratic in $\dot q$ (and of course sensible kinetic energy is positive-definite).

The metric and symplectic structures on a manifold are usually independent and define to preferential ways of realising isomorphisms between tangent and cotangent bundles (since there is no natural choice in the functorial sense).

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