How does the tautological one-form convert a velocity to a momentum?

Question

The Wikipedia page on the "tautological one-form" $\theta$ says that it

is used to create a correspondence between the velocity of a point in a mechanical system and its momentum, thus providing a bridge between Lagrangian mechanics and Hamiltonian mechanics

and that

velocities are appropriate for the Lagrangian formulation of classical mechanics, but in the Hamiltonian formulation, one works with momenta, and not velocities; the tautological one-form is a device that converts velocities $\dot{q}$ into momenta $p$.

I certainly understand why this device is physically useful, but unfortunately, the Wikipedia doesn't explicitly explain how $\theta$ maps $\dot{q}$ to $p$.

In order to make sure that I understand the tautological one-form correctly, I'd like to explain it in very concrete detail with a minimum of mathematical formalism; if anything below is incorrect, then please let me know. As I understand it, $\theta$ is constructed via the following steps:

1. We start with the configuration space, which is represented by an $n$-dimensional smooth manifold $Q$. The velocities $v := \dot{q}$ live in the tangent space $TQ$.
2. The cotangent bundle $T^*Q$ is a $2n$-dimensional smooth manifold that can (loosely) be thought of as the set of all ordered pairs $(q, p|_q)$, where $q \in Q$ and $p|_q$ is a one-form that linearly maps the tangent space $T_qQ$ at the point $q$ to $\mathbb{R}$.
3. $q$ and $p|_q$ both have $n$ degrees of freedom, so the cotangent space $T^*Q$ is a $2n$-dimensional smooth manifold. It can therefore be locally parameterized by local coordinate charts $(U \subset T^*Q) \to \mathbb{R}^{2n}$, which we can split up into $2n$ different coordinate charts $(U \subset T^*Q) \to \mathbb{R}$ that we'll call $q^i,\ i = 1, \dots, n$ and $p_j,\ j = 1, \dots, n$, where we've used the cotangent bundle geometry to distinguish the $q^i$ from the $p_j$. Specifically, we separate the coordinate charts such that the $n$ charts $q^i$ only depend nontrivially on the first argument $q$ in each $(q,p|_q) \in T^*Q$, while the other $n$ coordinate charts $p_j$ can depend nontrivially on both $q$ and $p|_q$. These are just coordinate charts $(U \subset T^*Q) \to \mathbb{R}$, not vectors or one-forms or tensors of any kind.
4. Fix an ordered pair $m = (q, p|_q) \in T^*Q$, where $q \in Q$ and $p|_q \in T^*_q$ map to a fixed set of $2n$ real numbers $q^i(m)$ and $p_i(m)$.
5. The tautological one-form corresponding to the element $m \in T^*Q$ is the fixed linear functional $$\theta_m := \sum_{i=1}^n p_i(m)\ dq^i(m).$$ Despite the notation, the sum on $i$ is not a tensor contraction, but just a plain old sum.

If we now let $m$ vary over $T^*Q$, then we get a one-form $$\theta \in T^*T^*Q := \{(m, \theta_m) | m \in T^*Q\}$$ over the whole cotangent bundle manifold $T^*Q$.

For a given $i$, $q^i$ is a real-valued function of (an open subset of) the $2n$-dimensional cotangent bundle smooth manifold $T^*Q$. The one-form $\theta$ over $T^*Q$ is therefore technically $2n$-dimensional. But as mentioned above, we used the contangent-bundle structure of $T^*Q$ to separate out the coordinate charts $q^i$ and $p_i$ so that the $n$ functions $q^i$ only depend nontrivially on the $q \in Q$ in the element $(q, p|_q) \in T^*Q$. But each summand (with fixed $i$) in the definition of $\theta$ is a one-form that points along one of the coordinate basis directions $dq^i$, and none of the summands point along a $dp_j$ direction. So while the linear functional $\theta_m$ technically acts on a $2n$-dimensional space, it is only nonzero within the $n$-dimensional subspace spanned by the $dq^i$. The defining map $m \to \theta_m$ technically maps the $2n$-dimensional cotangent bundle $T^*Q$ to $T^*_mT^*Q$, the $2n$-dimensional dual space spanned by the $dq^i$ and $dp_j$. But the image of this map lies entirely with the $n$-dimensional dual subspace spanned by the $dq^i$, so without loss of generality we can restrict the target space down to that dual subspace and think of the defining map $m \to \theta_m$ as a map from the $2n$-dimensional cotangent bundle $T^*Q$ to the only $n$-dimensional dual space $T^*_mQ$.

Is everything above correct (or at least correct enough for a physics level of rigor)?

If so, I still don't see exactly what is the map from velocities $v \in TQ$ to momenta $p \in T^*Q$. Given an explicit position $q \in Q$ and a list of coefficients $\dot{q}^i$ in the expansion $v = \dot{q}^i(v) \frac{\partial}{\partial q^i}$, how exactly do we use the tautological one-form $\theta$ to figure out the corresponding momentum $(q,p|_q) \in T^*Q$?

Answer 1

Your statements are correct for me. However your final claim is not true, you need to be careful which manifolds are being identified.

The identifying map it gives is between $T(T^*M)$ and $T^*(T^*M)$. This is done by the symplectic structure. Taking the exterior derivative of the tautological one from, you get a two form on the cotangent bundle: $$ \omega=d\theta $$ The non degeneracy of $\omega$ induces at every $m\in T^*M$ an isomorphism between $T_m(T^*M)$ and $T_m^*(T^*M)$ given by: $$ \alpha \in T_m(T^*M) \to \omega_m(\alpha,\cdot)\in T_m^*(T^*M) $$ which extends to the entire manifolds $T(T^*M)$ and $T^*(T^*M)$.

However, this does not give an identification from $TM$ to $T^*M$. For this, you will need a Lagrangian (or conversely a Hamiltonian). Explicitly, given a Lagrangian $L$ defined on $TM$, its “partial derivative with respect to velocity” gives the map. Its inverse is constructed by constructing the Legendre transform of $L$, called the Hamiltonian, and taking the derivative with respect to conjugate momentum. Intuitively, you need to find a way to raise/lower the indices, which cannot be done only using the differential structure of $M$, you need something more.

Hope this helps.