I have been trying to understand the Legendre transformation (in mechanics, in the hyperregular case: when the Legendre transformation is one-to-one) and the correspondence between symmetry $\to$ conserved current on the Lagrangian side and conserved charge $\to$ symmetry on the Hamiltonian side. I found a lot of interesting elements here on Physics Stack Exchange:

among others. Let's say we have a configuraiton space $E$. So here is the picture I get:

  • From a (quasi-)symmetry $Y$ of the Lagrangian $L:TE\to\mathbb{R}$, one gets the conserved Noether charge $Q_Y$. As a conserved charge, from the Hamiltonian formulation we obtain a corresponding (infinitesimal) symmetry $X_{Q_Y}$ acting on the phase space $T^*E$.
  • From the Hamiltonian $H:T^*E\to\mathbb{R}$ we construct the Hamiltonian Lagrangian $L_H(q,\dot q,p) = p\dot q - H(q,p)$. Its quasi-symmetries are in one-to-one correspondance with the conserved charge of the Hamiltonian.
  • A quasi-symmetry $X$ of the Hamiltonian Lagrangian provides a quasi-symmetry of the initial Lagrangian directly (by pullback) through the Legendre transformation $(q,\dot q)\mapsto (\mathcal{L}(q,\dot q),\dot q)=(q,\partial_{\dot q}L(q,\dot q),\dot q)$.

But I found myself unable to show that the quasi-symmetry of the initial Lagrangian then obtained matches with the initial one. From the second point I know they have the same Noether charge. I'd be happy with a more direct argument but I would be already satisfied if one can show that it does imply that the (Lagrangian) quasi-symmetry are the same. To show that $X_{Q_Y} = \mathcal{L}_*Y$ or that $\delta_{\mathcal{L}_*Y} L_H = f$ would be enough for what I am looking for.

Another point that is not clear to me are the so called generalized variational symmetries $Y$ defined here. They are symmetries depending on the first order derivative, e.g. the symmetry associated with the Laplace-Runge-Lenz vector. Geometrically, should they be viewed as a vector field of $E$ defined on $TE$ (section of the pullback $TE\times_E TE\to TE$)? Or a vector field of $TE$?

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/426160/2451 and links therein. $\endgroup$ – Qmechanic May 27 '19 at 13:14
  • $\begingroup$ I am probably interested in details about "similar bijective correspondence [...] for the corresponding quasi-symmetries" between the Lagrangian and Hamiltonian formulations you mention there. I essentially reached the same results as OP, but I can't really make sense of requiring the equations of motion to get the two vector fields matching $\endgroup$ – jpdm May 27 '19 at 15:12
  • $\begingroup$ Actually if the variation of $q(t)$ depends on $\dot q(t)$, the variation of $\dot q(t)$ will depend on $\ddot q(t)$, which is a second order dependence that I can't find in the Hamiltonian symmetry generated by the charge. From this remark it makes sense to require a condition constraining the value of $\ddot q(t)$ to get the Hamiltonian vector ffield and the original quasi-symmetry matching. But then we wouldn't have a one-to-one correspondence between Lagrangian quasi-symmetries and we're not assured that a given Noether charge is generated by a unique Lagrangian quasi-symmetry. $\endgroup$ – jpdm May 27 '19 at 15:27

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