Let M be some manifold, and TM the tangent bundle. Let $\gamma : [a,b] \to M$ be a smooth curve on M defined on an interval on $\mathbb{R}$. Let $J$ be another interval in $\mathbb{R}$ containing 0. A 'deformation of $\gamma(t)$ with fixed endpoints' is a curve $\overline{\gamma}:[a,b]\times J \to M : (t,\epsilon) \mapsto \overline{\gamma}_{\epsilon}(t)$ such that

  • $\overline{\gamma}_{0}(t)=\gamma(t), \forall t \in [a,b]$
  • $\overline{\gamma}_{\epsilon}(a) = \gamma(a)$ and $\overline{\gamma}_{\epsilon}(b) = \gamma(b)$ for all $\epsilon \in J$

Let L be a lagrangian, i.e. a smooth map $L : TM \to \mathbb{R} : (p,\dot{p}) \mapsto L(p,\dot{p})$.

For $M = \mathbb{R}^n$ it is simple to prove that $\gamma$ fulfills the variational principle

$$\left. \frac{d}{d\epsilon} \right |_{\epsilon=0} \int_a^b L(\overline{\gamma}_{\epsilon}(t),\dot{\overline{\gamma}}_{\epsilon}(t)) dt = 0$$

for every deformation of $\gamma$, if and only if $\gamma$ satisfies the Euler-Lagrange equations

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{p}}(\gamma(t),\dot{\gamma}(t)) - \frac{\partial L}{\partial {p}}(\gamma(t),\dot{\gamma}(t)).$$

My question

In many references (any book on geometric mechanics), it is stated that this equivalence hold on any manifold - not just the euclidean case. And in several places (for examples Marsden and Ratiu's book on geometric mechanics) I have seen it stated that this can be prooved in coordinates. However, this is only done for the case where $\gamma$ is contained in a single chart. I am trying to prove, or looking for a reference that proves, the general case. Preferably in coordinates, or in a relatively 'simple' intrinsic way. Can anyone help with this?

My attempt

Say we want to prove the following direction; let $\gamma : [a,b]\to M$ fulfill the Euler-Lagrange equation. I.e. it fulfills the equation in every chart. We want to show that the variation of the integral is 0.

Choose a cover of M, and let $\gamma$ be covered by 3 charts, as in the figure below (copied from the book 'Geometric mechanics and symmetry' by Holm et al). Then its deformations (for small enough $\epsilon$) is also covered by these charts.

Depiction of a curve and deformations covered by 3 charts

Choose one such deformation. Then we can split it into three subcurves defined on the intervals $[a,t_1],[t_1,t_2],[t_2,b]$, respectively, such that each is contained in a single chart. Likewise, we can split up the integral into three integrals

\begin{align} \left. \frac{d}{d\epsilon} \right |_{\epsilon=0} \int_a^b L(\overline{\gamma}_{\epsilon}(t),\dot{\overline{\gamma}}_{\epsilon}(t)) dt =& \left. \frac{d}{d\epsilon} \right |_{\epsilon=0} \int_a^{t_{1}} L(\overline{\gamma}_{\epsilon}(t),\dot{\overline{\gamma}}_{\epsilon}(t)) dt \\ &+ \left. \frac{d}{d\epsilon} \right |_{\epsilon=0} \int_{t_{1}}^{t_{2}} L(\overline{\gamma}_{\epsilon}\nonumber(t),\dot{\overline{\gamma}}_{\epsilon}(t)) dt \\ &+ \left. \frac{d}{d\epsilon} \right |_{\epsilon=0} \int_{t_{2}}^b L(\overline{\gamma}_{\epsilon}(t),\dot{\overline{\gamma}}_{\epsilon}(t)) dt. \end{align}

In each integral, we can use the coordinates of the suitable chart. However, for each such curve/deformation in $\mathbb{R}^n$, the endpoints will not be fixed, except at a and b. From the proof of the equivalence on $M = \mathbb{R}^n$, one can deduce that if an arbitrary deformation $\overline{g} : [T_1,T_2]\times J \to \mathbb{R}^n$ (not necessarily with fixed endpoints) fulfills the E-L equations, then

\begin{align*} \left. \frac{d}{d\epsilon} \right |_{\epsilon=0} \int_{T_1}^{T_2} L(\overline{g}_{\epsilon}(t),\dot{\overline{g}}_{\epsilon}(t)) dt = \left[ \frac{\partial L}{\partial \dot{p}}(g(t),\dot{{g}}(t)) \cdot \left.\frac{d}{d\epsilon}\right|_{\epsilon=0} \overline{g}_{\epsilon}(t)\right]_{T_1}^{T_2} \end{align*}

This can be used on the previous equation to get

\begin{align} \left. \frac{d}{d\epsilon} \right |_{\epsilon=0} \int_a^b L(\overline{\gamma}_{\epsilon}(t),\dot{\overline{\gamma}}_{\epsilon}(t)) dt =& \left[ \frac{\partial L'}{\partial \dot{p}}(\gamma(t)',\dot{{\gamma}}'(t)) \cdot \left.\frac{d}{d\epsilon}\right|_{\epsilon=0} \overline{\gamma}'_{\epsilon}(t)\right]_{a}^{t_1} \\ &+ \left[ \frac{\partial L''}{\partial \dot{p}}(\gamma(t)'',\dot{{\gamma}}(t)'') \cdot \left.\frac{d}{d\epsilon}\right|_{\epsilon=0} \overline{\gamma}_{\epsilon}''(t)\right]_{t_1}^{t_2} \\ &+ \left[ \frac{\partial L'''}{\partial \dot{p}}(\gamma(t)''',\dot{{\gamma}}(t)''') \cdot \left.\frac{d}{d\epsilon}\right|_{\epsilon=0} \overline{\gamma}_{\epsilon}'''(t)\right]_{t_2}^{b} \end{align}

where the clumsy '-notation denotes that in each term of the sum we use a different coordinate representation of L, $\gamma$ and $\overline{\gamma}$, since they belong to different charts.

In the case where $M = \mathbb{R}^n$ we can use a single chart, so the sum telescopes. But on a general manifold, the sum does not necessarily telescope due to the different coordinate maps.

Is there a way to fix this?


Just take advantage of a partition of unit for the family of the three open sets you are using: the intermediate boundary terms vanish because the maps of the partition of the unit smoothly vanish approaching the frontier of the corresponding open sets.


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