Background: Let $(M,g)$ be a Riemannian manifold. Let $(p,v) \in TM$ and $V, W \in T_{(p,v)}TM$. We can introduce a Riemannian metric on $TM$ via $\langle V, W\rangle_{(p,v)} = \langle d\pi(V), d\pi(W) \rangle_p + \left\langle \frac{D\alpha}{dt}\!(0), \frac{D\beta}{dt}\!(0)\right\rangle_p,$ where $\alpha, \beta\colon I \to TM$ are curves in $TM$ with $\alpha(0) = \beta(0) = (p,v)$ and \alpha'(0) = V and \beta'(0) = W, and $\pi\colon TM \to M$ is the natural projection.
Given this metric on $TM$, we then call $\text{Ker}(d\pi) \subset T_{(p,v)}TM$ the vertical space, and its orthogonal complement the horizontal space. We say that a curve $\alpha\colon I \to TM$ is horizontal iff its tangent vector \alpha'(t) \in T_{\alpha(t)}TM is horizontal for all $t \in I$.
Question: How can one show that a curve $\alpha\colon I \to TM$ is horizontal if and only if $\alpha$ is parallel along its projection curve $\pi\circ \alpha$?
Source/Motivation: This is Problem 2(b) from Chapter 3 of do Carmo's "Riemannian Geometry." It was assigned as a homework problem, and the homework was collected the other day (Oct 27), but I was unable to do the forward direction $(\implies)$ of the problem.
I would especially appreciate a coordinate-free proof if possible.
My Attempt:
Suppose \alpha'(t) is horizontal, so \langle \alpha'(t), W \rangle_{\alpha(t)} = 0 for any vertical vector $W \in T_{\alpha(t)}M$. Since $d\pi(W) = 0$, we have that 0 = \langle \alpha'(t), W \rangle_{\alpha(t)} = \langle \frac{D\alpha}{dt}\!(0), \frac{D\beta}{dt}\!(0)\rangle_{(\pi\circ\alpha)(t)}, where $\beta\colon I \to TM$ is a curve with $\beta(0) = \alpha(0)$ and \beta'(0) = W.
It seems to me that if we chose $W$ cleverly enough, then we could perhaps conclude that $\frac{D\alpha}{dt}\!(0) = 0$, which is what we want to show.
Other thoughts:
- Perhaps an application of Gauss' Lemma could help somewhere?
- While attempting to prove this problem, I conjectured that if \alpha'(t) \in T_{\alpha(t)}TM is horizontal, then d\pi(\alpha'(t)) = \alpha(t). However, even if this is true -- it certainly seems likely -- I am not sure how to apply it to get a solution.