I am trying to solve Exercise 3.2(c) of Do Carmo's Riemannian Geometry.The definition is:
Moreover:
A vector at $(p,v)\in TM$ that is orthogonal (for the metric above) to the fiber $\pi^{-1}(p)\approx T_pM$ is called a horizontal vector.
A coordinates chart of $p\in M$ induces a coordinates chart of $(p,v)\in TM$,for convenience set $M=R^n$,$TM=R^n\times R^n$ and the Christoffel symbols are $\Gamma_{ij}^k$.
Prove that the geodesic field is a horizontal vector field(i.e.,it is horizontal at every point).
My approach is:
pick up $(p,v)\in TM=R^n\times R^n$,by definition,the geodesic field at this point is $(v,s),s_k=-\Gamma_{ij}^kv_iv_j.$ it is easy to see vector $(a,\alpha)$ at $(p,v)$ is horizontal iff $\forall k(\alpha_k+\Gamma_{ij}^ka_i\alpha_j=0)$.Thus $(v,s)$ is horizontal iff $(-\Gamma_{i_1j_1}^kv_{i_1}v_{j_1}+\Gamma_{i_2j_2}^kv_{i_2}\Gamma_{rs}^{j_2}v_rv_s=0)$.it suffices to prove $(\Gamma_{ij}^kv_{i}(-v_{j}+\Gamma_{rs}^{j}v_rv_s)=0)$
Does the following really hold?or I missed anything?
$\Gamma_{ij}^kv_{i}(-v_{j}+\Gamma_{rs}^{j}v_rv_s)=0$
I know it could be done by just choosing a geodesic $(\gamma(t),\gamma'(t))$ passing $(p,v)$,then use $\frac{\mathrm{D} }{\mathrm{d} t}\gamma'(t)=0$ and use the following to conclude the proof:
$\langle V,W \rangle_{(p,v)} = \left\langle d\pi(V), d\pi(W) \right\rangle_{p} + \left\langle \frac{Dv}{dt}(0) , \frac{Dw}{ds}(0) \right \rangle_{p}$
But I also want to know if the first way works.Any help would be appreciated.
