How to show $~\nabla _TT=0$?

Question

$M$ is a Riemann manifold. $$ \begin{align} \alpha : & [a,b]\times(-\varepsilon,\varepsilon) \rightarrow M \\ &(t,s) \rightarrow \alpha(t,s) \end{align} $$ For any given $s\in [a,b]$, $\alpha(t,s)$ is geodesic.Let $$ T=\alpha_*(\frac{\partial}{\partial t}) ~~,~~V=\alpha_*(\frac{\partial}{\partial s}) $$ How to show $~\nabla _TT=0$ ? I always got stuck when it contain pushforwards or pullbacks.

Answer 1

The $s$ variable doesn't really appear in this question. If $\gamma$ is a geodesic then by definition, $\nabla_{\gamma '} \gamma' =0$. Also, by definition $\gamma' =\gamma_* \frac{d}{dt}$. Let $\alpha_s$ be the map $t \mapsto \alpha(t,s)$ . It is the composition of $\alpha$ with an inclusion map. Letting $\gamma= \alpha_s$ and noting that $ {\alpha_s} _* \frac{d}{dt}$ $ = \alpha_* \circ i_*\frac{d}{dt} =\alpha_* \frac{\partial}{\partial t}$ we get the answer.