# Why do horizontal curves have zero covariant derivative along their projection?

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.
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It's been a while since I did this, but could you remind me what $\frac{D}{dt}$ means? I vaguely recall it has something to do with covariant differentiation of a vector field along a curve. – Zhen Lin Oct 30 '11 at 11:24
A note on your second "other thoughts": that $\alpha$ is a parallel vector field along the curve $\pi\circ\alpha$ is independent of parametrisation. On the other hand, $\alpha'$ is dependent on parametrisation. So the expression $d\pi(\alpha'(t)) = \alpha(t)$ cannot hold for arbitrary parallel vector fields (and hence for arbitrary horizontal $\alpha'$). In particular, if you let $\alpha(t) = \alpha(0)$ the constant curve, you have $\alpha' = 0$ and $d\pi(\alpha'(t)) \neq \alpha(t)$ – Willie Wong Nov 1 '11 at 10:36

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 $D_t\alpha(0) = 0$.

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Warning: There follows my answer, I do not know if I have been enough clear, but anyway you can find the original argument on page 343 in Sasaki's paper The differential geometry of the tangent bundles of Riemannian manifolds. However any feedback is highly appreciated.

Notations: Let $x$ be a coordinate system on $M$, and let us denote by $(x,y)$ and $((x,y),(u,v))$ the associated standard coordinate systems respectively on $TM$ and $T(TM)$ .

The part(a) of do Carmo's problem requires to prove that the displayed scalar product on the fibers of $\tau_{T(TM)}:T(TM)\to TM$ is well-defined.
In order to show this, you have to prove, for any $(x,y)\in TM$, and $(u,v)\in T_{(x,y)}(TM)$, that $\frac{D\alpha}{dt}\!(0)\in T_xM$ is independent by $\alpha$, for $\alpha$ varying among the curves on $TM$ such that $\alpha(0)=(x,y)$, and $\alpha'(0)=(u,v)$.
Infact, preserving the previous notations, you should have proved that $$\frac{D\alpha}{dt}\!(0)=(x,v+\Gamma[x](y,u)).$$ (Where $\Gamma$ is the bilinear map whose components are the Christoffel symbols.)
This holds because the expression for the covariant derivative along an arbitrary curve $t\in I\to\alpha(t)=(x(t),y(t))\in TM$ is $\frac{D\alpha}{dt}\!(t)=(x(t),y'(t)+\Gamma[x(t)](x'(t),y(t))$, as reported in formula (1) on page 51 in do Carmo.

The same identical argument implies that:

a curve $\alpha(t)=(x(t),y(t))$ is autoparallel, i.e. $\frac{D\alpha}{dt}(t)\equiv 0$, iff $y'(t)+\Gamma[x(t)](y(t),x'(t))\equiv 0\ (\ast)$.

If the metric on $M$ is $g_{ij}dx^i\otimes dx^j$ then the metric on $TM$ is $$g_{ij}dx^i\otimes dx^j+g_{ij}\left(dy^i+\Gamma^i_{kl}y^k dx^l\right)\otimes\left(dy^j+\Gamma^j_{mn}y^m dx^n\right).$$

Now we find that:

an element $(u,v)$ in $T_{(x,y)}M$ is horizontal iff $v+\Gamma[x](y,u)=0\ (\ast\ast)$.

This follows from the coordinate expression of the Sasaki metric and because an element $(u,v)$ of $T_{(x,y)}M$ is vertical iff $u=0$.

From $(\ast)$ and $(\ast\ast)$, we conclude that:

a curve $\alpha(t)=(x(t),y(t))$ is autoparallel, iff $\alpha'(t)=(x'(t),y'(t))$ is horizontal for all $t$.

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