Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I guess according to mathematical didactic, we first think of spacetime as a set and we reason about elements of its topology and then it's furthermore equipped with a metric. Appearently it is this Riemannian metric, which people consider to be the object, which induced the minimal symmetry requirements of spacetime.

1) Regarding the relation between Riemannian geometry and the Hamiltonian formalism of classical mechanisc: Does a setting for Riemannian geometry always already imply that it's possible to cook up a symplectic structre in the cotangent bundle?

2) Are there some some more natural structures which physicists might be tempted to put on spacetime, which might then also be restricting regarding the (spacetime) symmetry structures? Is constructing quantum group symmetries (of non-commutative coordinate algebras, alla Connes?) just this?

3) I'm given a solution to a differential equation which can be thought of a resulting from a Lagrangian with a set of $n$ symmetries (e.g. $n=10$ for some spacetime models). Can this solution also be the result of a Lagrangian with fewer symmetries? Here, I'm basically asking to what extend I can reconstruct the symmetries from a solution or specific sets of the soltuon. It's kind of the inverse problem of the question "are there hidden/broken symmetries?".

share|improve this question
add comment

1 Answer

The following description will be from the particle point of view, i.e., the space time manifold will refer to the configuration manifold on which a particle moves.

Remark: My wrong answer of the first question was correted following the comment by Qmechanic.

1) There is no need for a metric to define a symplectic structure on a cotangent bundle. A cotangent bundle has a canonical symplectic structure independent of any metric:

$\omega = dx^i\wedge dp_i$

However, given a metric on the configuration manifold, a cotangent bundle of a wide class of manifolds (for example compact manifolds) can be given a Kähler structure. No explicit expression is known in the general case. However, there are implicit expressions for special cases such as Lie groups, please see the example of $T^{*}SU(2)$ in Hall's lectures. The advantage of having a Kähler structure on a cotangent bundle is that enables quantization in terms of creation and anihilation operators like in the case of a flat space.

2) A natural structure that one can put on a space time manifold is a principal bundle. In this case given a metric on the base manifold and a connection on the principal bundle a Poisson structure can be defined on this principal bundle. In this case even in the case of a vanishing Hamiltonian, there will be nontrivial dynamics determined by the constraints. The classical equations of motion are the Wong equations of a colored particle in a Yang-Mills field. Please see the following work by: A. Duviryac for a clear exposition.

Regarding the second part of the question, The quantization of this system leads a quantum representation of the color group. The operator algebra of this representation has the structure of a noncommutative manifold. The most known example of this type of algebra is in the case of $SU(2)$, where this manifold is a fuzzy sphere.

3) Consider a particle moving uniformly on a great circle of a two dimensional sphere. This is a solution of a free particle on a circle whose symmetry is $U(1)$, and also a solution of a free particle on a sphere whose symmetry is $SO(3)$.

share|improve this answer
Comment to the answer(v1): If $p_j$ is supposed to transform as a co-vector under coordinate transformations $x\to x^{\prime}$, then the rhs. of the first eq. is not invariant under change of coordinates. –  Qmechanic Nov 6 '12 at 12:18
@Qmechanic Thank you –  David Bar Moshe Nov 6 '12 at 13:43
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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