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In page 16 of McDuff & Salamon's Introduction to Symplectic Topology they prove that the critical points $z = (x,y)\colon [a,b] \to \Bbb R^{2n}$ of the action integral $$\Phi_H(z) = \int_a^b \langle y, \dot{x}\rangle - H(t,x,y)\,{\rm d}t$$satisfy Hamilton's equations $\dot{x} = \partial_yH$ and $\dot{y} = -\partial_xH$. Namely, they compute the first variation of $\Phi_H(z)$ as $$\widehat{\Phi_H}(z) = \int_a^b \langle \eta,\dot{x}-\partial_yH\rangle - \langle \xi, \dot{y}+\partial_xH\rangle\,{\rm d}t,$$where $(\xi,\eta)$ is the variational vector field of a variation of $z$ with fixed endpoints. I have no problems whatsoever with the proof they present, but I am having trouble formulating a generalization of this statement in the setting of differentiable manifolds. What I have in mind is:

Let $Q$ be a differentiable manifold and $H\colon T^*Q \to \Bbb R$ a smooth Hamiltonian. Consider the action integral $$\mathscr{A}^H(x,{\sf p}) = \int_a^b \mathbb{F}H(x(t),{\sf p}(t)){\sf p}(t) - H(x(t),{\sf p}(t))\,{\rm d}t,$$where $(x,{\sf p})\colon [a,b] \to T^*Q$ is a smooth curve and $\mathbb{F}H$ is the fiber derivative of $H$, given in coordinates by $$\mathbb{F}H(x,{\sf p}) = \sum_{k=1}^n \frac{\partial H}{\partial p_k}(x,{\sf p})\frac{\partial}{\partial q^k}\bigg|_x.$$Then if $(x,{\sf p})$ is a critical point of $\mathscr{A}^H$ and we write $$(x(t),{\sf p}(t)) = (q^1(t),\ldots, q^n(t),p_1(t),\ldots, p_n(t)),$$we have Hamilton's equations $$\frac{{\rm d}q^k}{{\rm d}t}(t) = \frac{\partial H}{\partial p_k}(x(t),{\sf p}(t)) \qquad\mbox{and}\qquad \frac{{\rm d}p_k}{{\rm d}t}(t) = -\frac{\partial H}{\partial q^k}(x(t),{\sf p}(t)).$$

However, I'm not entirely sure of this, as a priori there's no relation between $x(t)$ and ${\sf p}(t)$ (except for ${\sf p}(t) \in T_{x(t)}^*Q$), in contrast with curves $(x(t),\dot{x}(t))$ in $TQ$, when analyzing the Lagrangian case. In particular, I cannot reproduce the proof given in McDuff & Salamon, since there's no apparent way to use integration by parts, as we have no $t$-derivatives in the integrand.

I don't really expect anyone to fix the statement and provide the computation (although it would be nice), but I need help figuring out what sort of relation I am missing here (with that I should be able to run the computation of the first variation myself). In particular, I am not assuming that $H$ is hyperregular, and we do not have Legendre transformations at our disposal. What to do?

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Perhaps it is best to compute things intrinsically.

The action functional in the cotangent bundle is given by $$A_H(x)=\int_{[0,1]} x^* \alpha-\int_{[0,1]} H \circ x,$$ where $\alpha$ is the canonical $1$-form in the cotangent bundle. You can check that this coincides with your formula in the case of $\mathbb{R}^{2n}$.

Differentiating and using Cartan's magic formula yields that the stationary points of the action functional satisfy Hamilton's equations. (Recall that $d\alpha =\omega$.)

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  • $\begingroup$ I see. So just to confirm, the problem is that my initial guess using $\mathbb{F}H$ was simply wrong? $\endgroup$ – Ivo Terek Jul 1 at 5:23
  • $\begingroup$ @IvoTerek I think so. The first term should not involve $H$. $\endgroup$ – Aloizio Macedo Jul 1 at 5:25
  • $\begingroup$ Alright, the math checks out. Indeed, to produce the first term of the integrand without using $H$, there's only one sensible choice: since we start with $(x(t),{\sf p}(t))$ and ${\sf p}(t) \in T_{x(t)}^*Q$, we need to produce an element of $T_{x(t)}Q$ for ${\sf p}(t)$ to act on. Well, we have one: $\dot{x}(t)$. And indeed the pull-back of the tautological form ends up being ${\sf p}(t)\big(\dot{x}(t)\big)$. So we're good. Thanks. $\endgroup$ – Ivo Terek Jul 1 at 6:18

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