# Hamilton's equations arising from a variational principle.

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?

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 $$d\alpha =\omega$$.)
• I see. So just to confirm, the problem is that my initial guess using $\mathbb{F}H$ was simply wrong? Jul 1 '19 at 5:23
• @IvoTerek I think so. The first term should not involve $H$. Jul 1 '19 at 5:25
• 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. Jul 1 '19 at 6:18