Are integral curves of a vector field $X$ such that $\nabla_X X = 0$ geodesics? Let $(M,g)$ be a riemannian manifold, $X$ a vector field of $M$ and $\gamma \colon M \to \mathbb{R}$ an integral curve of $X$ that is $\gamma'(t) = X(\gamma(t))$. Assume also that the Levi-Civita connection $\nabla_X X = 0$ where 

Are the integral curves of a $X$ geodesics?

I believe is true. A curve $\gamma$ of $M$ is a geodesic if and only if the covariant derivative $\nabla_{\gamma'(t)}\gamma'(t) = 0$. But since $\gamma$ is an integral curve of $X$, just plugging $X(\gamma(t))$ in the connection and using the hypothesis that $\nabla_X X = 0$ gives the result. 
Is this correct?
 A: $X(X,X)=0$ so that $|X|$ is constant on an integral curve $c(t)$. Since $c$ is a constant speed, so by an assumption, $c$ is a geodesic.
A: The answer is yes. I believe we can prove it with the pullback connection. On any curve $\alpha$ we have the induced pullback connection $\alpha^\ast \nabla$. The geodesic equation becomes with it $$\alpha^\ast \nabla_{d/ds} \alpha' = 0$$
We show this is satisfied. Let $\gamma$ be integral curve. Thus we have that $X \circ \gamma = \gamma'$. It turns out that this means that $\gamma' = \gamma^\ast X$, so that it is a pullback section. Thus  $$\gamma^\ast \nabla_{d/ds} \gamma' =\gamma^\ast \nabla_{d/ds} \gamma^\ast X, $$
and by definition of the pullback connection $$\gamma^\ast \nabla_{d/ds} \gamma' =\gamma^\ast \nabla_{\gamma_\ast d/ds} X, $$
but $\gamma_\ast d/ds =  X \circ \gamma$ and since the connection is $C^\infty$ linear on the below slot we have $$\gamma^\ast \nabla_{d/ds} \gamma' =\gamma^\ast \nabla_{X} X, $$
so when your condition is satisfied the curve is a geodesic.
