# Definition of conjugate momentum on a manifold

I have trouble understanding this definition:

Let $$Q$$ be some manifold and $$L: TQ \to \mathbb{R}$$ a smooth function. Then for some local coordinates $$(q, \dot{q})$$ on $$TQ$$ the conjugated momentum is defined as $$\frac{\partial L}{\partial \dot{q}}$$, which is an element of the co-tangential bundle $$T^{*}Q$$.

How is the expression $$\frac{\partial L}{\partial \dot{q}}$$ to be interpreted? If one simply expresses $$L$$ in local coordinates by $$L \circ(q^{-1}, \dot{q}^{-1}): \mathbb{R}^{2n} \to \mathbb{R}$$ and differentiates it with respect to the second variable one gets a function $$\mathbb{R}^n \to \mathbb{R}$$ and not an element of the co-tangential bundle $$T^{*}Q$$. Is the correct expression $$\partial_2 ( L \circ(q^{-1}, \dot{q}^{-1}))\circ (q, \dot{q}) \in T^{*}Q\ ?$$

• yes, the issue is that the derivatives you get away from $0 \in \Bbb R^n$ are not actually independent of the choice of local coordinates in any reasonable sense. The only invariant data is what you're writing down along the 0-section. – user98602 Dec 31 '18 at 0:39

The notation $$\partial L/\partial\dot{q}$$ is no more than a symbol that denotes the Frechet derivative of $$L$$ with respect to $$\dot{q}$$. More precisely, let $$L:Q\times TQ\to\mathbb{R}.$$ Then for all $$x\in Q$$, $$T_xQ$$ is a vector space equipped with the norm induced by the metric on $$Q$$. In this sense, one may define a bounded linear operator $$A:TQ\to\mathbb{R}$$, such that $$\lim_{Y\to 0}\frac{\left\|L(x,X+Y)-L(x,X)-A(Y)\right\|_{\mathbb{R}}}{\left\|Y\right\|_{T_xQ}}=0$$ for a given $$x\in Q$$ and a given $$X\in T_xQ$$. This linear operator $$A$$, if it exists, is called the partial derivative of $$L$$ with respect to $$X$$, also denoted by $$\partial L/\partial X$$ for intuition. Further, if one considers a trajectory $$q:\mathbb{R}\to Q$$, we have $$\dot{q}:\mathbb{R}\to T_{q}Q$$. Hence when considering the special form $$L(q,\dot{q})$$, we also take $$\partial L/\partial\dot{q}$$ instead of $$\partial L/\partial X$$, again, for intuition.
Finally, whatever the notation is assigned to the operator defined above, either $$A$$ or $$\partial L/\partial X$$ or $$\partial L/\partial\dot{q}$$, it is (1) linear and is (2) from $$TQ$$ to $$\mathbb{R}$$. On the other hand, the collection of all linear operators from $$TQ$$ to $$\mathbb{R}$$ is definitely $$T^*Q$$. Therefore, it follows that $$A\in T^*Q,\quad\text{or}\quad\frac{\partial L}{\partial X}\in T^*Q,\quad\text{or}\quad\frac{\partial L}{\partial\dot{q}}\in T^*Q.$$