Let be a differentiable manifold. Analogously to the construction of the tangent bundle, we can make the set of covectors on a given manifold into a vector bundle over , denoted and called the cotangent bundle of .
To make this definition precise it is convenient to use the classical definition of a manifold (http://planetmath.org/NotesOnTheClassicalDefinitionOfAManifold). Let be an -dimensional differentiable manifold, let (each is an open subset of ) be an atlas of with transition functions .
As an atlas for , we may take . We may construct transition functions as follows:
For these to be valid transition functions, they must satisfy the three criteria. For a verification that these criteria are satisfied, please see the attachment.
The cotangent bundle is a vector bundle over the manifold . To substantiate this claim, we must specify a projection map onto the manifold and local trivializations and transition functions and verify that they satisfies the defining properties of a bundle. In terms of the local coordinates used above, it is easy to describe the projection map :
The local trivializations are also somewhat trivial:
Finally, the transition functions are given as follows:
For a verification that satisfies the three criteria for a bundle, please see the attachment.
The cotangent bundle is the vector bundle dual to the tangent bundle . On any differentiable manifold, (for example, by the existence of a Riemannian metric), but this identification is by no means canonical, and thus it is useful to distinguish between these two objects.
The cotangent bundle to any manifold has a natural symplectic structure given in terms of the Poincaré 1-form, which is in some sense unique. This is not true of the tangent bundle. The existence of a symplectic structure implies that the cotangent bundle is always orientable, even if the original manifold is not.
|Date of creation||2013-03-22 13:59:02|
|Last modified on||2013-03-22 13:59:02|
|Last modified by||rspuzio (6075)|