Confusion about role of covectors in behavior of momentum This question has been beaten to death on math stackexchange, so I'm a little embarassed I have to ask again, but I am really confused about how/why momentum is a covector. I read Arnold but that didn't really help, since he doesn't explain it much. I will try to ask a specific question.
If we have a "coordinate function" $q : \mathbb{R}^n \to \mathbb{R}$, the gradient gives us a tangent vector $$\nabla q(x)|_a = \left[\frac{\partial}{\partial x_1} q(a), ..., \frac{\partial}{\partial x_n} q(a)\right],$$ and we can take a specific component ($x_i$ component of the velocity, at a point $x=a$) $$\nabla q(x)|_a\cdot e_i = \frac{\partial}{\partial x_i} q(x)|_a = \dot{q}_{a,i}.$$
Its usually claimed that $p$ is a linear functional of $\dot{q}$, since $$p_i = \frac{\partial L}{\partial \dot{q}_i}$$ and this function acts linearly on velocities. Here $L$ is the lagrangian density. Its also true that we can say $p(\dot{q})=m\dot{q}$ which is a scalar. So far so good, but here is what I don't understand, specifically:

*

*Usually a covector is motivated as "eating a tangent vector and returning a number" or by related statements using differentials and dual bases. I have no idea, none at all, how this maps onto the momentum function. My guess it has to do with the fact that we look at specific components of the gradient, maybe we can use $dq(\mathbf{e}_i) = \dot{q}_i$, but I really don't know.

And a sort of sub question:


*How would it even be written in terms of the cotangent bundle coordinates? Using $\sum p_i dq_i$? In that case $p_i$ is the coefficient of the basis $dq_i$ and corresponds to $\frac{\partial q}{\partial x_i}$... right?

Basically theres a huge mental disconnect between the math and the physics I've read. As you can see I'm badly confused, so maybe you can simply recommend a good resource to wrap my head around it.
 A: The word "momentum" gets thrown around more often than candy during Halloween. Asking why is momentum a covector/coordinate function in some circumstances while it is a vector in other circumstances doesn't do much to clarify the situation. What is useful I think is to clearly write down all the objects involved, and to clarify which space (manifold) everything lives in. In physics we often don't clarify which space we're working on, and as a result several things get mixed up. This by itself may not be a bad thing (because ultimately it is possible to nicely "identify" things), but it becomes confusing for a student (myself included) when we don't explicitly mention how things are being identified.
Fix an $n$-dimensional smooth manifold $Q$ ("the manifold of all possible configurations"). Let $TQ,T^*Q$ be its tangent and cotangent bundle respectively. Let us adopt the following notation:

*

*$\pi_{TQ}:TQ\to Q$ the tangent bundle projection

*$\pi_{T^*Q}:T^*Q\to Q$ the cotangent bundle projection.

*$(TU, (q,\dot{q})=(q^1,\dots, q^n, \dot{q}^1,\dots, \dot{q}^n))$ an adapted coordinate chart for $TQ$

*$(T^*U, (z,p)= (z^1,\dots, z^n, p_1,\dots, p_n))$ an adapted coordinate chart for $T^*Q$.

Now, here's a proposition:

Theorem 1.
For every smooth function $f:TQ\to\Bbb{R}$, there is a smooth 1-form $\mu_f$ on $TQ$ (recall this means a smooth section $\mu_f:TQ\to T^*(TQ)$) such that for every adapted coordinate chart $(TU, (q^i,\dot{q}^i))$ of $TQ$, we have
\begin{align}
\mu_f&=\frac{\partial f}{\partial \dot{q}^i}\,dq^i
\end{align}
(the equality is on the open set $TU$, so to be really strict I should write $\mu_f|_{TU}$ on the LHS).

Notice what type of object everything is. $f$ is a smooth function on $TQ$, and $(q^i,\dot{q}^i)$ are coordinate functions on $TU$, so $\frac{\partial f}{\partial \dot{q}^i}$ is a smooth function on $TU$ and $dq^i$ is a $1$-form on $TU$, so their product and summation over $i$ is again a 1-form on $TU$. We are strictly on the tangent bundle, so forget about momentum and cotangent bundles temporarily.

Proof Outline of Theorem 1.
Consider two adapted coordinate charts $(TU, (q,\dot{q}))$ and $(TV, (r,\dot{r}))$ on $TQ$. It is a good exercise for you to prove that on the intersection $TU\cap TV$, we have
\begin{align}
\frac{\partial r^i}{\partial \dot{q}^j}&= 0 \quad \text{and} \quad \frac{\partial \dot{r}^i}{\partial \dot{q}^j}=\frac{\partial r^i}{\partial q^j}.\tag{$*$}
\end{align}
Therefore,
\begin{align}
\frac{\partial f}{\partial \dot{r}^i}\,dr^i&=
\left(\frac{\partial f}{\partial q^j}\frac{\partial q^j}{\partial \dot{r}^i}+\frac{\partial f}{\partial \dot{q}^j}\frac{\partial \dot{q}^j}{\partial \dot{r}^i}\right)
\cdot 
\left(\frac{\partial r^i}{\partial q^k}\,dq^k+\frac{\partial r^i}{\partial \dot{q}^k}\,d\dot{q}^k\right) \\
&=\left(0+\frac{\partial f}{\partial \dot{q}^j}\frac{\partial q^j}{\partial r^i}\right)\cdot \left(\frac{\partial r^i}{\partial q^k}\,dq^k+0\right)\\
&=\frac{\partial f}{\partial \dot{q}^j}\delta^j_k\,dq^k\\
&=\frac{\partial f}{\partial \dot{q}^i}\,dq^i
\end{align}
(in the last line I resolved the Kronecker delta summation and I renamed indices). Since both expressions agree in the overlap region, we can unambiguously define the 1-form $\mu_f$ on all of $TQ$.

If we want to give names, then we can call $\mu_f$ the "canonical momentum $1$-form of the smooth function $f:TQ\to\Bbb{R}$". Again, don't get hung up on the word "momentum", because it's just a word. Likewise, we can refer to the components $\frac{\partial f}{\partial \dot{q}^i}$ as "The canonical momentum of the function $f$ with respect to $q^i$ where $(TU, (q,\dot{q}))$ is an adapted coordinate chart of $TQ$". Or for short, the canonical momentum of $f$ conjugate to the $i^{th}$ position coordinate, or even shorter, simply "the $i^{th}$ component of momentum". As I said, these are just names, so it doesn't really matter what you call it, as long as you're aware of the object you're talking about (in this case a $1$-form on $TQ$ specially constructed from a smooth function $f:TQ\to\Bbb{R}$).
To make some contact with Physics, let us now suppose $Q$ is equipped with a Riemannian metric $g$. This sets up a vector bundle isomorphism $g^{\flat}:TQ\to T^*Q$ and its inverse $g^{\sharp}:T^*Q\to TQ$ (the so called "musical isomorphisms", or in GR called index lowering/raising). Then, given a smooth function $\psi:Q\to\Bbb{R}$ we can consider a smooth function $L:TQ\to\Bbb{R}$ as
\begin{align}
L(\xi)=\frac{1}{2}g_{\pi_{TQ}(\xi)}(\xi,\xi)-\psi(\pi_{TQ}(\xi))
\end{align}
Or in terms of an adapted coordinate chart $(TU, (q,\dot{q}))$, we have
\begin{align}
L&=\frac{1}{2}(g_{ij}\circ \pi_{TQ})\dot{q}^i\dot{q}^j-\psi\circ \pi_{TQ}
\end{align}
In words, $\psi$ is the potential energy (the fact that it's a function on $Q$ reflects that potential energy depends on position only rather than position and velocity). So, this $L$ is quadratic in the velocity coordinates and we're subtracting off a potential energy term. So, this is precisely what we encounter in elementary Lagrangian mechanics.
Now, due to the symmetry of $g_{ij}$, one can easily calculate that
\begin{align}
\frac{\partial L}{\partial \dot{q}^i}=(g_{ij}\circ \pi_{TQ})\,\, \dot{q}^j
\end{align}
So, the canonical $1$-form associated with $L$ is
\begin{align}
\mu_L&=\frac{\partial L}{\partial \dot{q}^i}\,dq^i=(g_{ij}\circ \pi_{TQ})\,\dot{q}^j\,\, dq^j.
\end{align}

The word "momentum" comes up because if we think of $Q=\Bbb{R}^3$, with the Riemannian metric $g=m(dx\otimes dx+dy\otimes dy+dz\otimes dz)$ for some $m>0$, then $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$ and likewise with $y$ and $z$, so it coincides with what we have all learnt since kids was momentum. Therefore, by this analogy, we extend our terminology and thus I defined $\mu_f$ above to be  "the momentum $1$-form of $f$".

I hope this addresses part of why sometimes we like to think of momentum as a $1$-form on $TQ$.
Now, you may wonder what does the cotangent bundle has to do with anything, and why the coordinates are labeled by a $p$ in $(T^*U, (z,p)= (z^1,\dots, z^n, p_1,\dots, p_n))$. A-priori there is no relationship; the use of the letter $p$ is for now just an accident. I would suggest you now read the following answer of mine: pulling back tautological one-form using Riemannian metric. What happens is that on $T^*Q$ there is a very nice $1$-form $\theta=p_i\,dz^i$ (one can show is well-defined using coordinates; or one can also give an intrinsic definition).
Using the Riemannian metric and the corresponding vector bundle isomorphism $g^{\flat}:TQ\to T^*Q$, one can pull back the form $\theta$ to a $1$-form $\theta_g:= (g^{\flat})^*\theta$ on $TQ$. In my linked answer, I prove that
\begin{align}
\theta_g&= (g^{\flat})^*(p_i\,dz^i)\\
&= (p_i\circ g^{\flat})\, dq^i\\
&= (g_{ij}\circ \pi_{TQ})\,\, \dot{q}^j\,dq^i
\end{align}
But this is none other than the $1$-form $\mu_L$ for the specific quadratic-in-velocity Lagrangian $L$ I defined above. So, because we have already agreed above that the functions $\frac{\partial L}{\partial \dot{q}^i}= (g_{ij}\circ \pi_{TQ})\, \dot{q}^j$ generalize our elementary notion of momentum, it follows that $p_i\circ g^{\flat}$ also ought to be called the $i^{th}$ momentum function. The isomorphism $g^{\flat}$ simply allows us to convert between the tangent and cotangent bundles, so by extension of terminology, we may also decide to call the coordinate functions $p_i$ on $T^*Q$ as momentum.
Thus to summarize, from every smooth function $f$ on $TQ$ we obtain a $1$-form $\mu_f$. From the special case of having a Riemannian metric, we can define a function $L$ which is quadratic in velocities. Then, the components of the $1$-form $\mu_L$ in a hyper special case coincide with our elementary notion of momentum. Finally, we can show that $\theta_g:= (g^{\flat})^*\theta=\mu_L$, and thus by extension, the coordinate functions $p_i$ also deserve to be called momentum (and note that since the Riemannian metric $g$ relates the velocity coordinates $\dot{q}^i$ with the momentum coordinates $p_i$, we may sometimes physically interpret a Riemannian metric as a generalized mass matrix/moment of inertia of a system).
