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We want to write the intrinsic form of E.L. equations on the tangent bundle. we define the Poincare-Cartan one-form:

$\Theta = \frac{\partial L}{\partial \dot{q}^{i}}d{q}^{i}$ where L in the lagrangian of the system. It can be shown that the E.L. equations can be written in terms of this one-form as:

$\mathcal{L} _{\Gamma} \Theta - dL =0$, where $\mathcal{L}$ is the lie derivative along $\Gamma$, and $\Gamma$ is the vector field on the fiber bundle TQ, obtained by canonical lifting the tangent vector field of the trajectories on the configuration space Q.

My question is: $\Theta$ is a one-form defined on the configuration space Q, does it make sense consider its lie derivative along a field $\Gamma$ defined on the tangent bundle TQ?

amWhy
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Explosiveness
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    But $q^i, \dot{q}^j$ are the coordinates on $TQ$, right? Why do you think that $\Theta$ is not defined on $TQ$? – T.P. Jan 09 '23 at 17:34
  • Beacause $q_{i}$ is only the coordinate on the base space Q (that is the configuration space). The coordinates corresponding to the fiber are missing in the differential form $dq_{i}$ – Explosiveness Jan 09 '23 at 17:48
  • They are missing because their coefficients are zero. It’s the same as the differential form $f(x,y), dx=0 , dy + f(x,y), dx$ in $\mathbb R^2$ with coordinates $(x,y)$. – T.P. Jan 09 '23 at 18:11
  • Note that $\Theta$ is not the pullback of a form from $Q$ (under the map $\pi\colon TQ\to Q$) because the coefficients of $dq^i$ depend on the $\dot q^j$ coordinates (as well as on the $q^j$ coordinates, quite likely). – Ted Shifrin Jan 09 '23 at 22:29

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$\Theta=\Theta_L$ is a 1-form defined on $TQ$, not $Q$. If you want, the more abstract definition of $\Theta_L$ relies on the fiber-derivative: \begin{align} \Theta_L= (\mathbf{F}L)^*(\theta), \end{align} where $\mathbf{F}L:TQ\to T^*Q$ is the Fiber-derivative map, and $\theta$ is the tautological 1-form on $T^*Q$ (in terms of adapted coordinates $(q^i,p_i)$ on $T^*Q$, this is given by $\theta=p_i\,dq^i$); and the tauotological 1-form already has a well-known coordinate-free definition.

I think part of your confusion arises from the overload of notation for $q^i$ as a coordinate; people often use it to mean the base coordinate functions on $Q$, or its pullback to $TQ$ or $T^*Q$. Anyway, this a special 1-form in that the second set of coefficients vanish: $\Theta_L=\frac{\partial L}{\partial\dot{q}^i}\,dq^i+\sum_{i=1}^n0\,d\dot{q}^i$.

Now, because $\Theta_L$ is a $1$-form on $TQ$, its makes sense to consider its Lie-derivative with respect to a vector field on $TQ$.

peek-a-boo
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