# Symplectic structure vs. metric structure

1. Jun 21, 2007

### jillnorth@hotmail.com

Symplectic structure vs. metric structure

A question about the relationship between the phase space of the
Hamiltonian formulation of classical mechanics and of the Lagrangian
formulation; that is, between the cotangent bundle of configuration
space, T*Q, which has a natural symplectic structure given by the
canonical symplectic form, and the tangent bundle, TQ, which has a
natural metric structure given by the Riemannian line element, viz. a
quadratic differential form of the q dots.

It seems there's a sense in which the former has *less structure*
than the latter. Hamiltonian phase space has symplectic structure
(this determines a volume element), not metric structure. And in
general, metric structure determines, or presupposes, a volume
structure, but not the other way around. A metric would then add
another level of structure to what's needed for the Hamiltonian
equations of motion. (Level of structure'' in the sense that,
starting with a set of points, we can define mathematical objects on
it, some of which presuppose others; in this sense a topological
space has more structure than a set of points, a metric space has
more structure than a topological space--a metric induces a topology--
and so on.)

It also seems the Lagrangian formulation needs this metric structure,
for the quadratic differential form is the invariant quantity of the
Lagrangian transformations, and a symplectic manifold is floppy'',
having no local notion of curvature that would distinguish one
symplectic manifold from another locally (from Darboux's theorem).
This is because of the two sets of generalized coordinates used by
the Lagrangian as opposed to the Hamiltonian formulation: for the
Hamiltonian, the canonically conjugate q's and p's, treated as
independent variables on the phase space (so that the energy function
is linear in each); for the Lagrangian, the generalized coordinates
and their first time derivatives, the generalized velocities (giving
rise to the quadratic differential form of the q dots). (Would a more
coordinate-free version of Lagrangian mechanics be able to get by
without the Riemannian metric structure on the base manifold? I
would've thought not, given the above, but I'm not sure.)

It seems this should mean that not every symplectic manifold
(similarly, not every cotangent bundle) is isomorphic to a Riemannian
manifold (tangent bundle). However, there is a natural isomorphism
between T*Q and TQ, given by the Legendre transformation. On the
other hand, the isomorphism is non-canonical (there is no basis-
independent isomorphism). More generally, then, how can we compare
the structures of these two spaces, given that they are, after all,
*different* spaces? Is there a structure-preserving map between the
two? How should one go about trying to find such a thing?

Any and all feedback or references would be extremely helpful! I am
in philosophy of physics, struggling with a paper, and having trouble
figuring out an answer on the basis of the books and online
references I've seen so far. Profuse apologies if I have simply
gotten myself rather mixed up.

2. Jun 22, 2007

### Igor Khavkine

On 2007-06-21, jillnorth@hotmail.com <jillnorth@hotmail.com> wrote:

> It seems this should mean that not every symplectic manifold
> (similarly, not every cotangent bundle) is isomorphic to a Riemannian
> manifold (tangent bundle). However, there is a natural isomorphism
> between T*Q and TQ, given by the Legendre transformation. On the
> other hand, the isomorphism is non-canonical (there is no basis-
> independent isomorphism). More generally, then, how can we compare
> the structures of these two spaces, given that they are, after all,
> *different* spaces? Is there a structure-preserving map between the
> two? How should one go about trying to find such a thing?

The Legendre transform, strictlly speaking, is not a map between tangent
and cotangent space. Rather, it is a map between functions on a vector
space and its dual (in particular the tangent space at a point, and its
dual). Specifically, it invertibly maps bounded (from below), convex
functions on the tangent space to bounded, convex functions on the
cotangent space. As such, it is independent of the choice of basis in
either vector space. The map from the cotangent space to the tangent
space is contained in one of Hamilton's equations: dq/dt = @H/@p, where
@ denotes partial derivative. The reverse map, p = @L/@(dq/dt), is
defined with the help of the Lagrangian. As long as both L and H obey
the above boundedness and convexity restrictions (as functions of dq/dt
and p, respectively), then both maps are well defined and basis
independent. However, they require (the Legendre dual) L and H to be
provided.

Note, however, that for the usual Legendre tansform to be defined, you
need to work with vector spaces. So, if your symplectic manifold is not
at least a vector bundle (where momenta live in vector space fibers
above position points), you'll have trouble getting an equivalent
Lagrangian formulation.

On the topic of comparing structures on T*Q and TQ. As you've noticed,
T*Q has a natural symplectic form on it. On the other hand TQ has a
natural prolongation structure. What I mean by this is that given a
(parametrized) curve in Q, it can be lifted to a curve in TQ simply by
attaching the curve's tangent vector to every point. While in T*Q, the
prolongation only exists for curves that solve Hamilton's equations of
motion. In a general symplectic manifold it's not even possible to
meaninfully separate the position and momentum coordinates, as they can
be freely mixed by symplectic transformations.

In the end, you can always restrict treat the phase space of a
mechanical system as the set of the solutions of its equations of
motion. Then both TQ and T*Q are just particular ways of parametrizing
it. Thus, if you have a symplectic form in any one parametrization, you
can always pull it back to different parametrization through a
diffeomorphism (like the one I described above).

> Any and all feedback or references would be extremely helpful! I am
> in philosophy of physics, struggling with a paper, and having trouble
> figuring out an answer on the basis of the books and online
> references I've seen so far. Profuse apologies if I have simply
> gotten myself rather mixed up.

For anyone trying to master mechanics from the symplectic point of view,
I would first recommend V.I. Arnold's book _Mathematical Methods of
Classical Mechanics_. Another book on this subject that I know by
reputation is Abraham's and Marsden's _Foundations of Mechanics_. I
don't know if these books will help answer your questions, but they
should contain all the basics that relate mechanics and symplectic
manifolds.

Hope this helps.

Igor

3. Jun 24, 2007

### Kay zum Felde

Hi Jill,

if M is a manifold and t(M) (t*(M)) its tangent (cotangent) bundle then the partial differentials d/dx_{j} and the differentials dx_{j} form a dual basis in the tangent- (cotangent-) vector spaces T_{x}M(T*_{x}(M)). A dual basis is a somehow orthogonal basis. That means the partial differentials and the differentials are somehow orthogonal on each other. Think of it as the dual of the electromagnetical field-tensor, where the electrical and magnetical field are simply exchanged.

I must admit, that I didn't really understand your point, when you were talking about the metric, the Lagrangian and the Poisson brackets. Are you wondering, were the metric is hidden in the Hamilton formalism ? The metric is incorporated in Einsteins variational formalism of the Hamilton function. Check any book, f.e. his original work from 1915 collected in "The Principle of Relativity" (Dover).

Regards Kay