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

Solution 1:

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 $D_t\alpha(0) = 0$.

Solution 2:

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 $t$.