I stumbled upon this while reading about Legendre Transforms today. So consider an n-particle system. The Lagrangian is a function of $ q_i$'s and $\dot q_i$'s. If you consider the manifold $M$ where the $q_i$'s are constrained to (in this case it is simply $\mathbb{R}^n$), then the $(\dot q_1, \dot q_2,... \dot q_n) $'s lie in the tangent space of $M$ at the $(q_1,q_2,...q_n)$'s. So Lagrangian can be considered as a function on the tangent bundle $T_M$. The transformation from a Lagrangian to a Hamiltonian converts $\dot q_i$ dependence into $p_i = \frac{\partial L}{\partial \dot q_i} $ dependence.
Here comes my confusion. The transformation from a Lagrangian to a Hamiltonian converts $\dot q_i$ dependence into $p_i = \frac{\partial L}{\partial \dot q_i} $ dependence. If we consider $\dot q_i$s as components of the velocity vector $\vec{v}$, then the $p_i$'s are components of a vector in the DUAL space, and the Hamiltonian is a function on the cotangent bundle $T^*_M$ . However, the conventional definition of momentum in classical mechanics is $\vec{p} = m\vec{v} $ which implies that $p$ is a vector. So it seems that canonical momentum conjugates are dual to the conventional momentum. What do I make of that? This distinction is not necessary in $\mathbb{R}^n$ since vectors and one-forms have the same components. But what happens in a general manifold? Should I attach any significance to the fact that canonical momentum conjugates are one-forms?
