2
$\begingroup$

Noether‘s Theorem says that every continuous symmetry of a physical system (i.e., a Lie group action on phase space ${\bf R}^{2n}$ preserving a Hamiltonian $H$) leads to a conservation law (i.e., a function $I$ satisfying $\left\{I,H\right\}=0$).

In the literature one finds this as an application of the method of moment maps. The general result is that for a group action of a Lie group $G$ on a symplectic manifold $(M,\omega)$ one has the associated moment map $\mu\colon M\to {\mathfrak g}^*$, in particular for fixed $X\in {\mathfrak g}$ a map $\mu^X\colon M\to{\bf R}$, and then $\left\{\mu^X,H\right\}=\omega(X_H,X)$ for the Hamiltonian vector field $X_H$. The right hand side equals $\iota_X dH$, which vanishes because the Lie group action is assumed to preserve $H$.

I do not understand, why the literature makes the assumption that the Lie group preserves the symplectic form and, more seriously, I do not understand why this assumption should be true in the setting of Noether‘s Theorem. That theorem is just about continuous symmetries preserving the given Hamiltonian on ${\bf R}^{2n}$, so why should they preserve the symplectic form $\sum_k dx_kdy_k$?

$\endgroup$
2
  • $\begingroup$ I’m trying to reconcile several expositions on this topic and it’s confusing the heck out of me. Don’t be surprised if I ask you about it because it looks like you might be able to set me straight. $\endgroup$
    – rschwieb
    Dec 19, 2020 at 3:28
  • 2
    $\begingroup$ If the action admits a moment map, then it necessarily preserves the symplectic form (at least for connected G). This is the case because the infinitesimal generators of the action are Hamiltonian, hence symplectic vector fields. $\endgroup$
    – studiosus
    Dec 19, 2020 at 19:01

2 Answers 2

1
$\begingroup$

That theorem is just about continuous symmetries preserving the given Hamiltonian on ${\bf R}^{2n}$, so why should they preserve the symplectic form $\sum_k dx_kdy_k$?

No it is not. In classical mechanics appropriate coordinate changes are the canonical transformations or, in mathematical language, symplectomorphisms, of the phase space.

This is maybe an important point: we want transformations that preserve the nature of the evolution equations (aka Hamilton equations); these turn out to be symplectomorphisms of the phase space.

Such a transformation is a symmetry if it preserves the Hamiltonian.

Now, any diffeomorphism on the configuration space lifts to a canonical transformation.

If your Hamiltonian is of the form "potential + kinetic" and the potential energy $U$ depends only on the position variables, then any diffeo of the configuration space will in fact lift to a symmetry (in the above sense) of the mechanical system. More generally, in the Lagrangian formulation here, perturbing the $q$ coordinates implicitly perturbs "their derivative" $p$ coordinates. If the Lagrangian is "invariant under small perturbations" you will get that the corresponding "perturbation" on the phase space is a symmetry -- i.e. is given by a symplectic vector field (one generating symplectomorphisms) and preserving the Hamiltonian; you will have to go through the Legendre transform from the Lagrangian to the Hamiltonian picture first, though.

On the other hand, there are symplectomorphisms that are not lifts of diffeos on the configuration space, so canonical transformations - aka symplectomorphisms - are more general than that (though not too much more general if you look at things locally, but that is a different story to think about).

$\endgroup$
0
$\begingroup$

I invite you to read this book, especially chapter 4 and 5: Jean-Louis Koszul, Introduction to Symplectic Geometry https://link.springer.com/book/10.1007%2F978-981-13-3987-5

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.