# Metric and the Lagrangian [duplicate]

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?

## marked as duplicate by ACuriousMind♦, Kyle Kanos, John Rennie, ManishearthApr 2 '15 at 7:55

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## 2 Answers

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.

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).