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I am having a hard time to understand why the integral of the tautological 1-form is the action of the system.

The tautological one form is defined by :

\begin{align} \theta_{(q,p)} : T_{(q,p)}T^*Q &\to \mathbb{R}\\ X &\mapsto p(\pi_*(X)) \end{align}

where $\pi_*$ is the pushforward/differential of the projection from the bundle to the base space.

Then,

$$\int_{\gamma} \theta = \int \theta_{\gamma(t)}(\dot{\gamma}(t)) \, dt = \int p(\pi_*(\dot{\gamma}(t)))\, dt$$

where $\gamma(t)=(q(t),p(t))$. But how can I show that this is the action, i.e. $\int L$ where $L$ is the Lagrangian?

edit
If $\gamma(t))=(q(t),p(t))$ is an integral curve of $X_H$ it will satisfy the Hamilton equations $\dot{\gamma}(t))=(\dot{q}(t),\dot{p}(t))=(\frac{\partial H}{\partial p}, -\frac{\partial H}{\partial q})$. But then what is $p(\pi_*((\frac{\partial H}{\partial p}, -\frac{\partial H}{\partial q})))$?

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  • $\begingroup$ At the risk of showing my ignorance of both geometry and physics, I'll guess that you're conflating two different situations. The tautological $1$-form is just a matter of geometry, whereas the action in a physical system depends on more than just the geometry of the phase space. For example, if the dynamics of the system is based on conservative forces, then the potential resulting from those forces will contribute to the Lagrangian and therefore to the action. $\endgroup$
    – Andreas Blass
    Commented Jun 4, 2019 at 0:41
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    $\begingroup$ The devil's in the details: if $\gamma$ is an integral curve for $X_H$ the Hamiltonian flow corresponding to the Hamiltonian $H$ (i.e., if $\gamma$ satisfies Hamilton's equations with respect to the $H$), then $\int_\gamma \theta$ recovers the action along $\gamma$ with respect to $H$. $\endgroup$
    – Branimir Ćaćić
    Commented Jun 4, 2019 at 1:27
  • $\begingroup$ will $p(\pi_*((\frac{\partial H}{\partial p}, -\frac{\partial H}{\partial q})))$ be equal to the action? $\endgroup$
    – roi_saumon
    Commented Jun 4, 2019 at 11:51
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    $\begingroup$ Maybe I'm missing something, but isn't the tautological one-form just $\theta = p\cdot dq$ in canonical coordinates? So we have $\int_{\gamma} p(t) \cdot dq(t) = \int_{\gamma} p(t) \cdot \dot{q}(t) dt$, which is the abbreviated action? $\endgroup$
    – pregunton
    Commented Jun 6, 2019 at 10:33

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