I am trying to prove the following statement:
Let $(M, g)$ an Riemannian manifold an $X \in \Gamma(TM)$ a parallel vector field, i. e. $\nabla^g_Y X = 0$ for all $Y \in \Gamma(TM)$. Then all integral curves are geodesics.
My "proof" looks like that:
Let $I \subseteq \mathbb R$ and a $\gamma: I \to M$ an arbitrary integral curve, i. e. $\gamma'(t) = X(\gamma(t))$ for all $t \in I$. Since $X$ is parallel we have that
$$\nabla^g_{\partial/\partial t} \gamma ' = \nabla^g_{\partial/\partial t} X \circ \gamma = \nabla^g_{\gamma_*(\partial/\partial t)} X = \nabla^g_{\gamma'} X = 0,$$ where $\nabla^g$ denotes the Levi-Civita connection of $g$ on $TM$. Hence $\gamma$ ia a geodesic. $\square$
I am wondering if this solution is correct because I feel it is too short. I appreciate your help :)
It looks good, but I would write it as
$$ \frac{D\gamma'}{dt}(t) = \frac{D(X \circ \gamma)}{dt}(t) = (\nabla_{\gamma'(t)} X)(\gamma(t)) = 0. $$
The point is that $\gamma \colon I \rightarrow M$ is a curve and $\gamma' \colon I \rightarrow TM$ is a vector field along $\gamma$, not a vector field on $M$. To check whether $\gamma$ is a geodesic, you need to use the induced covariant derivative along $\gamma$ which I denote by $\frac{D}{dt}$. Since $\gamma$ is an integral curve of $X$, it turns out that $\gamma'$ is the "restriction" of a globally defined vector field $X$ on $M$ (that is, $\gamma' = X \circ \gamma$) and for such fields, we know that the covariant derivative along $\gamma$ is related to the regular covariant derivative of vector fields by the relation
$$ \frac{D(X \circ \gamma)}{dt}(t) = (\nabla_{\gamma'(t)} X)(\gamma(t)) $$
where the right hand side should actually be interpreted as $(\nabla_{Y_t} X)(\gamma(t))$ where $Y_t \in \mathcal{X}(M)$ is (any) vector field on $M$ for which $Y_t(\gamma(t)) = \gamma'(t)$.