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Consider a Lagrangian $L:TQ\to \mathbb{R}$, for some smooth manifold $Q$. As explained in this answer, one can define its fiber derivative $\mathbf FL:TQ\to T^* Q$ as $$\mathbf FL(v) \equiv \mathrm d(L|_{T_x Q})(v,\bullet), \qquad v\in T_x Q.$$ More precisely, given $L|_{T_x Q}:T_x Q\to \mathbb{R}$, we're considering the differential $\mathrm d(L|_{T_x Q}):T_v(T_x Q)\to T_{L(v)}\mathbb{R}$, and exploiting the isomorphism $T_v(T_x Q)\simeq T_x Q$, we identify vertical displacements $(v,\delta v)\in T_x (T_x Q)$ with pairs of tangent vectors in $T_x Q$, and thus define $\mathbf FL(v)$ as the mapping $\delta v\mapsto \mathrm d(L|_{T_x Q})(v,\delta v)\in\mathbb{R}$.

So, as far as I can tell, formally speaking the elements in the image of $\mathbf FL$ are not functionals in the same tangent spaces that make up its domain (albeit the two things are isomorphic to one another, hence why we can identify them).

Now, people often also write the Lagrangian canonical momentum as the derivative of $L$ wrt its velocities: $p_i = \frac{\partial L}{\partial \dot q^i}$. What I'm trying to understand is whether such expressions should really be taken as a shorthand easier way to refer to the fiber derivative itself. That seems to make sense: $\mathbf FL$ sends each $v\in TQ$ to a covector, and the momentum is often said to be a covector, and the expression for $\mathbf FL$ in local coordinates also directly involves $\frac{\partial L}{\partial v^k}$.

From these observations, it would seem that I could in general define the Lagrangian canonical momentum associated to each $L:TQ\to \mathbb{R}$ and $v\in TQ$ as $p\equiv (\mathbf FL)(v)$. However, I've never seen this stated explicitly, so I'm wondering if there's some subtleties I'm not taking into account.

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glS
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  • $T_v(T_xQ)\cong T_xQ$ because $T_xQ$ is a vector space, not $T_xQ\times T_xQ$. Also, as I explain in this answer, there are a billion different things people call 'momentum'. – peek-a-boo May 16 '22 at 16:12
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    and sure, you can refer to $(\mathbf{F}L)(v)$ as the canonical momentum conjugate to the velocity $v$ (note however that it is only if $\mathbf{F}L$ is a local diffeomorphism that we can really identify the two... but in most of the basic applications this is satisfied because the Lagrangian is typically built from a Riemannian metric $g$, so the fiber derivative $\mathbf{F}L$ is nothing but the isomorphism $g^{\flat}:TQ\to T^*Q$ induced by the metric). – peek-a-boo May 16 '22 at 16:25
  • you are right, sorry. I was thinking of $TV\simeq V\times V$ for a vector space $V$ (which I think is correct, because you need one copy for basepoints and one for tangent vectors?), but fixing a basepoint that's not true, and $T_v V\simeq V$. I did read your answer there, but was still unsure whether this particular definition fits into one of those. Do you have a counterexample where the $\mathbf FL$ isn't a local diffeomorphism and thus we cannot make this identification? – glS May 16 '22 at 16:30
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    You can always refer to $(\mathbf{F}L)(v)$ as the canonical momentum associated to $v$. That's just a definition you're allowed to make. What I'm saying is that in order for you to 'identify' velocities with momenta, you need to be able to go back and forth,, i.e local diffeomorphism at the minimum. I can't give you an example or counter example; if you like it's just a definition of the loose term 'identification'. And yes, the definition given here coincides with what I wrote there, because as I said in my 2nd comment, $\mathbf{F}L=g^{\flat}$ (a diffeomorphism) in that case. – peek-a-boo May 16 '22 at 16:36

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