# Symplectic manifold question

1. Jul 13, 2007

### 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 pi is clearly a cotangent vector.

To me it seems like what pi 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?

2. Jul 13, 2007

### George Jones

Staff Emeritus
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,

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

then

$$\frac{\partial L}{\partial \dot{q}^j} = g_{jk} \dot{q}^k.$$

Last edited: Jul 13, 2007
3. Jul 13, 2007

### 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.