How momentum is dual to the velocity vector at a point on a differentiable manifold?

The tangent space $$T_pM$$ which is a real vector space on a point $$p$$ of a differentiable manifold $$M$$, has a cotangent bundle $$T_p^*M$$ at $$p \in M$$, such that for any $$v \in T_pM$$ and for any $$w \in T_p^*M$$, we get $$w(v) = r , \quad(r \in \mathbb R)$$, Or in other notation $$\left< w,v \right> = r$$,

I am trying to realize this construction in classical mechanics,

The Lagarangian $$L$$ is a real valued function on the tangent bundle $$TM$$ (Assuming no explicit time depencence) \begin{align} L : & TM \to \mathbb R\\ &(q,\dot q) \mapsto L(q,\dot q) \end{align}

One also defines Hamiltonian $$H$$, a real valued function on the cotangent bundle $$T^*M$$, as \begin{align} H : & T^*M \to \mathbb R\\ &(q,p) \mapsto H(q,p) \end{align}

$$\dot q \in T_qM$$ and $$p \in T_q^*M$$, I am unable to see how $$\left< p,\dot q \right> = r$$?

Update:

In a differentiable manifold, only real functions I can have is of the form $$\left$$, Then the Lagrangian needed to be made out of only these objects, For a free particle that kind of Lagrangian has the form $$L = \mathbf p \cdot \dot{\mathbf q}$$
Which is not a function on $$TM$$.

The differential manifold has no other inner product defined, so I can't make $$L = \mathbf{\dot q}\cdot \mathbf{\dot q}$$. This will be valid only for Riemannian Manifolds.

Let's consider a free particle on a $$n$$ dimensional Riemannian Manifold $$(M,g)$$, The Lagrangian is given by (Summation convention is being used) \begin{align} L(\mathbf{ q},\mathbf{\dot q}) =g_{ij}(q) {\dot q}^i\cdot {\dot q^j}, \quad i = 1,\ldots,n \end{align}, The momentum $$p_i = \frac{\partial L}{\partial \dot q^i} = g_{ij}(q){\dot q^j}$$ Here $$p_i$$ are the local (in some chart $$(U,\phi)$$ of $$(M,g)$$) components of the one-form $$p$$ given by $$p(q) = p_i(q) dq^i(q) = g_{ij}(q){\dot q^j(q)} dq^i(q)$$ And the velocity vector field is locally given as $$v(q) = \dot q^i(q) \left(\frac{\partial }{\partial q^i}\right)_q$$, Now, $$dq^i(q)$$ span $$T_q^*M$$ and $$\left(\frac{\partial }{\partial q^i}\right)_q$$ span $$T_qM$$, The basis obey $$\left< dq^i(q), \left(\frac{\partial }{\partial q^j}\right)_q \right> = \delta_j^i$$ So we get, $$\left

_q = \left = g_{ij}(q){\dot q^j(q)} \dot q^i(q) = \mathbf{\dot q}(q) \cdot \mathbf{\dot q}(q) \in \mathbb R$$

, $$p$$ is a linear map and also a functional so now I can imagine this as the element of cotangent bundle that maps the vectors into the real numbers

Is this the right way of thinking?

I found something related in the question https://mathoverflow.net/questions/203138

I am not sure if this answers OP's question, but momentum can mean one of two things. It is either a "generic momentum" $$(q,p)\in T^\ast M$$, which is just a covector on $$M$$ defined at some point, or what I am calling the canonical momentum, which is actually a map $$\xi:TM\rightarrow T^\ast M$$.

Given a Lagrangian $$L:TM\rightarrow\mathbb R,\ (q,\dot q)\mapsto L(q,\dot q)$$, if the point $$q\in M$$ is fixed, the "restricted Lagrangian" $$L_q:T_qM\rightarrow\mathbb R,\ L_q(\dot q)=L(q,\dot q)$$ is a function on the single tangent space $$T_qM$$. We may take the differential of this function at $$\dot q\in T_q M$$ to get $$\mathrm dL_{q,\dot q}:T_\dot qT_qM\cong T_qM\rightarrow\mathbb R,$$ and this is a linear map, thus for fixed $$(q,\dot q)\in TM$$, $$\mathrm dL_{q,\dot q}\in T^\ast M$$, i.e. it is a covector. Then the fibre derivative of the Lagrangian is defined as $$\mathbb FL:TM\rightarrow T^\ast M,\ (q,\dot q)\mapsto \mathrm dL_{q,\dot q}.$$ Since for a vector $$\dot q$$ at $$q$$, the value $$\mathbb FL(q,\dot q)$$ is a covector at $$q$$, this is a (strict) morphism of fibre bundles, but it is not in general a morphism of vector bundles (this map is not fibrewise linear in general).

Then the canonical momentum $$p$$ (or $$(q,p)$$, depending on one's notation) corresponding to the velocity $$(q,\dot q)$$ is $$p_i=\xi_i(q,\dot q),$$ where the $$\xi_i$$ are the components of the fibre derivative $$\mathbb FL$$ in some chart.

As you said the Lagrangian is defined on the tangent bundle, whose elements, loosely speaking, are pairs of a coordinate and a derivative, e.g. $$(q, \dot{q}) = \left((q_i)_i, \; \dot{q}_j\frac{\partial}{\partial{q_j}}\right)$$ The Hamiltonian on the other hand is defined on the cotangent bundle, whose elements are pairs of a coordinate and a 1-form, e.g. $$(q, p) = \left((q_i)_i, \; p_j \text{d}q_j\right)$$ The operation $$\langle p,q \rangle$$ is then just the 1-form acting on a derivative, which per definition is $$p_i\text{d}q_i\left(\dot{q}_j\frac{\partial}{\partial{q_j}}\right) = p_i q_j \frac{\partial q_i}{\partial{q_j}} = p_i q_i \in \rm I\!R$$

• As a minor comment, I wouldn't say the derivative is part of the coordinate on the tangent bundle. Oct 5, 2020 at 21:05
• @NDewolf Now that I read your comment, I remember dimly, that tangent spaces might also be defined by means other than derivatives. Is your comment aimed at that fact or is it something else I am not considering?
– drfk
Oct 5, 2020 at 22:18
• Rather the fact that the derivatives are more like the basis vectors of the tangent spaces, not the coefficients/coordinates. Oct 6, 2020 at 7:23
• That checks out. Thanks for the answer. I know I was being somewhat mathematical incorrect. I still like to write it that way to highlight the "1-form-acting-on-derivative"-operation.
– drfk
Oct 6, 2020 at 10:11