# Legendre transformation and correspondance between Noether charges and quasi-symmetries

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 $$TE$$?

• Possible duplicates: physics.stackexchange.com/q/426160/2451 and links therein. – Qmechanic May 27 '19 at 13:14
• 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 – jpdm May 27 '19 at 15:12
• 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. – jpdm May 27 '19 at 15:27