Potential Energy in Geometric Mechanics I am reading through Montgomery's A Tour of Subriemannian Geometries, Their Geodesics and Applications. In the appendix section on Geometric Mechanics, the potential function $V$ satisfies the following properties:
Given a Riemannian manifold $(Q,\langle \ , \rangle)$, where $Q$ represents the configuration space, suppose some local coordinates on $Q$ are given by $(q^i)$, and those on $TQ$ by $(q^i, v^i)$, and the Levi-Civita connection is $\nabla$. Then the potential energy $V:Q \to \mathbb{R}$ satisfies $$ \nabla_{\dot{q}} \dot{q} = - \nabla V. $$
I presume there is some abuse of notation here. Clearly, we can always uniquely extend $V$ to $\hat{V}:TQ \to \mathbb{R}$, but I am unclear as to the rest of the notation, as clearly the two sides of the equation above do not seem to match up in terms of units/objects. Does this just mean we can slot in any tangent vector we like on the right hand side?
 A: The potential $V$ isn't given by the equation you wrote; $V$ is just an arbitrary differentiable function $V \colon Q \to \mathbb{R}$ specified in advance. You can take its differential $dV$ and raise the index using the metric to get the gradient $\nabla V$, which is a vector field on $Q$. 
(So $\nabla$ on the right-hand side denotes the covariant derivative of the function (scalar field) $V$, where $\nabla_X V$ is simply the directional derivative of $V$ in the direction of the vector $X$.)
The equation the you wrote is an ODE, “Newton's law”, for a particle (of mass $m=1$) moving in the force field $-\nabla V$  on the manifold $Q$.
That it, you seek a curve $q(t)$, with tangent/velocity vector $\dot q(t)$, such that for each $t$, the vector $\nabla_{\dot q} \dot q$ (the covariant derivative of the vector field $\dot q$ in the direction of the curve, evaluated at the point $q(t)$) equals the vector $-\nabla V(q(t))$. So the equation expresses an equality between two vectors in the tangent space $T_{q(t)}Q$ (for each $t$).
So on the left-hand side, $\nabla_X$ denotes the covariant derivative of a vector field on $V$, in the $X$ direction. Now, $\dot q$ isn't really a vector field on $Q$, since it's only defined along the curve $q(t)$,
but the covariant derivative of a vector field $Y$ in the $X$ direction only depends on the values of $Y$ along any curve with tangent vector $X$, so $\dot q$ is “sufficiently defined” to be differentiated along its own direction.
A: The phrasing "the potential energy is given by" is very strange. I don't think $V$ is meant to be extended: I believe $\nabla_{\dot q} \dot q = -\nabla V$ is to be interpreted as the equation of motion for a particle with position $q(t)$ at time $t$. Thus all three relevant objects live in $TQ$:


*

*$\dot q(t) \in T_{q(t)} Q \subset TQ$ is the velocity at time $t$

*$\nabla_{\dot q(t)} q \in T_{q(t)} Q$ is the covariant acceleration at time $t$

*$\nabla V|_{q(t)} \in T_{q(t)} Q$ is the gradient vector of the function $V:Q \to \mathbb R$ at the position $q(t)$.

