# Cotangent bundle

This may be a poorly phrased question - please let me know of it - but what is the correct way to think of the cotangent bundle? It seems odd to think of it as the dual of the tangent bundle (I am finding it odd to reconcile the notions of "maps to the ground field" with this object).

One fruitful way to think about it, if you have physics background, is as phase space. Your manifold is the configuration space for some system of particles, and the cotangent bundle is then the phases, so the cotangent directions are velocities. This is helpful also with the symplectic structure on $T^*M$.

• Your manifold, in fact, is the configuration space for a particle traveling on your manifold! Jul 28, 2010 at 0:33
• Well, of course there's that! But often the manifold is the configuration space of something nontrivial. Though these manifolds are often open... Jul 28, 2010 at 0:36
• Finally I have enough rep to comment! This is a subtle point: the tangent bundle is the (velocity)-phase space as you've described it. The Cotangent bundle is the momentum-phase space. Now, one often has the equation p = mv, so often one can be sloppy about this distinction. However, if, say, a magnetic field is present, then the magnetic field doesn't carry any velocity, but it DOES carry momentum. Jul 31, 2010 at 20:17
• Fair enough, but for intuition sake, we can get away with conflating velocity and momentum for a massive particle. Jul 31, 2010 at 20:43

You might be interested in this MathOverflow post: https://mathoverflow.net/questions/17325/why-is-cotangent-more-canonical-than-tangent

(Sorry, I'd leave this as a comment but I just joined this site and don't have enough reputation.)

I'm not completely sure what you mean by this: "It seems odd to think of it as the dual of the tangent bundle (I am finding it odd to reconcile the notions of "maps to the ground field" with this object)," but maybe the following will help you see why it is natural to consider the dual space of the tangent bundle.

Given a function f on our manifold, we want to associate something like the gradient of f. Well, in calculus, what characterized the gradient of a function? Its the vector field such that when we take its dot product with a vector v at some point p, we get the directional derivative, at p, of f along v. In a general manifold we don't have a dot product (which is a metric) but we can form a covector field (something which gives an element of the cotangent bundle at any point) such that, when applied to a vector v, we get the directional derivative of f along v. This covector field is denoted df and is called the exterior derivative of f.

• $A^2=B^2-C^2$ ghjj May 18, 2020 at 7:04