AI have the following questions about the Homogenity of geodesic: Let $\gamma:(-\delta,\delta)\to M$, where $t\to\gamma(t,q,v)$ is a geodesic, then $\gamma:\left(-\dfrac{\delta}{a},\dfrac{\delta}{a}\right)\to M$, where $t\to\gamma(t,q,av)$, $a\in\mathbb{R}$ is a geodesic and $\gamma(t,q,av)=\gamma(at,q,v)$.
The proof is the follow, define by $h(t)=\left(-\dfrac{\delta}{a},\dfrac{\delta}{a}\right)\to M$ the curve $t\to h(t)=\gamma(at,q,v)$, this implies that $h'(t)=a\gamma'(at,q,v)$, $h(0)=q$ and $h'(0)=av$. Take the conection $$\dfrac{D}{dt}\left(\dfrac{d\gamma}{dt}\right)=\nabla_{\frac{dh}{dt}}{\dfrac{dh}{dt}}=a^2\nabla_{\gamma'(at,q,v)}{\gamma'(at,q,v)}=0$$ Therefore, $h$ is a geodesic and we're done. But I don't see why $$ \nabla_{\gamma'(at,q,v)}{\gamma'(at,q,v)}=0, $$ from the hypothesis we have $$ \nabla_{\gamma’(t,q,v)}{\gamma’(t,q,v)}=0 $$ Is this enough to imply that $\gamma(at,q,v)$ is a geodesic? It seems intuitive that the rate of change of $\gamma(at,q,v)$ in the direction of $\gamma(at,q,v)$ is zero but I am missing the details. As a side question, is it also true $\nabla_{X}{X}=0$ for an arbitrary vector field $X$? (I guess not...)
You have to pay attention to where your $t$ is or lives. At this point of the proof $q$ and $v$ don't matter in the notation, so let's just write $\gamma(\cdot)=\gamma(\cdot,q,v)$. By assumptiom $\nabla_{\gamma'(\tau)}\gamma'(\tau)=0$ for all $\tau \in (-\delta, \delta)$.But now when $a$ is fixed and $t \in (-\frac{\delta}{a},\frac{\delta}{a})$, $at$ is in the $\delta$-Intervall, so you can apply the assumption.