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.
Best Answer
You are almost there.
Observe the following: if $W$ is vertical, $d\pi(W) = 0$. Using the (assumed) knowledge that the metric on the tangent bundle $TTM$ is a Riemannian metric and hence non-degenerate, we have that for any non-vanishing vertical $W$, the corresponding map $$ \langle \cdot, \frac{D\beta_W}{dt}(0) \rangle_p \in T^*_pM $$ is non-zero (since its action on $D_t\beta_W(0)$ is non-zero).
Now applying the rank nullity theorem to the map $d\pi$ (we are working with finite dimensional Riemannian manifolds, right?) you have that the vertical space of $T_{p,v}TM$ has the same dimension as $T_pM$. So applying the rank nullity theorem again to the linear map $$ W\mapsto \langle \cdot, \frac{D\beta_W}{dt}(0)\rangle_p $$ you see that it is invertible.
So what you have shown, that for any vertical $W$, $\langle D_t\alpha(0), D_t\beta_W(0)\rangle = 0$ implies that $D_t\alpha(0) = 0$.