Why are generalized momenta cotangent vectors in symplectic manifolds?

[StatusX]

I've been reading about the abstract formulation of dynamics in terms of symplectic manifolds, and it's amazing how naturally everything falls out of it. But one thing I can't see is why the generalized momenta should be cotangent vectors. I can see why generalized velocities are tangent vectors, making the Lagrangian a function on the tangent bundle of configuration space. But then the book I'm reading claims that since:

p_i = \frac{\partial L}{\partial \dot q_i}

it follows that p_i is clearly a cotangent vector.

To me it seems like what p_i is is a function assigning numbers to vectors in the "tangent space of the tangent space of a point in the manifold". Note that by what's in the quotes, I don't mean "the tangent space of the tangent bundle" (which would put the momentum in T^*(TM) ), because it doesn't depend on the change in configuration.

But it also shouldn't be in T^*M unless the Lagrangian is a linear function of velocity, so that its partial derivative with respect to velocity doesn't depend on your location in the tangent space, and so can be taken as a uniform linear functional on the tangent space. What am I missing here?

 

[George Jones]

The \dot{q}^i are, at the point q of M labeled by the generalized coordinates q^i, components of the tangent vector

v = \dot{q}^i \partial_{q^i}

of a curve in configuration space M.

The p_i are, at q, components of a particular covector \omega with respect to a particular basis of covectors. This covector \omega is the covector naturally associated with the tangent vector v by the metric g that comes from the kinetic energy

T = \frac{1}{2} g_{ij} \dot{q}^i \dot{q}^j.

In other words,

\omega \left( u \right) = g \left(v,u\right) = g_{ij} \dot{q}^i u^j[/itex]<br /> <br /> for all tangent vectors u at q.<br /> <br /> Setting \omega = p_i dq^i gives that \omega_j = g_{ij} \dot{q}^i. But, if<br /> <br /> L = \frac{1}{2} g_{jk} \dot{q}^j \dot{q}^k - U \left( q \right),<br /> <br /> then<br /> <br /> \frac{\partial L}{\partial \dot{q}^j} = g_{jk} \dot{q}^k.

 

[StatusX]

Ok, my problem was that, unless the dependence of L on velocity is linear, the covector depends on \dot q. An ordinary covector field is normally defined at each point on the manifold, not on the tangent bundle. But in fact with the momentum you do want a covector field on the entire tangent bundle, or more specifically, a map from the tangent bundle to the cotangent bundle, allowing you to replace the tangent manifold from Lagrangian mechanics with the cotangent manifold from Hamiltonian mechanics. I think I have it now, but please correct me if that sounds wrong. Thanks.