Vector field with constant length

Question

Is it correct, that for some pseudo-Riemannian manifold $M$, $X \in \mathfrak{X}(M)$, if $g(X,X)=1$, then the integral curves for $X$ are geodesics? I have the following explanation: Let $\gamma$ be a curve, s.t. $\gamma'(t)=X_{\gamma(t)}$. Then

$g(\gamma', \nabla_{\gamma'} \gamma')=\gamma'g(\gamma', \gamma')- g(\gamma', \nabla_{\gamma'} \gamma')$ which implies that $g(\gamma', \nabla_{\gamma'} \gamma')=0$ and therefore, since the metric is non-degenerate and $\gamma' \neq 0$ ,that $\nabla_{\gamma'} \gamma'=0$.

I'm pretty sure that this can't be true but at the moment I don't see why it does not hold..

Answer 1

$g(\gamma',\nabla_{\gamma'}\gamma')=0$ means that the velocity and acceleration vectors are orthogonal, this does not necessarily implies $\nabla_{\gamma'}\gamma'=0$. A simple example is the punctured plane $\mathbb R^2\setminus\{0\}$. In polar coordinates, take the smooth vector field $X=(-\sin\theta,\cos\theta)$. The integral curves are circles, which are not geodesics.

Answer 2

The result is true if you assume also that $X$ is a Killing field (i.e., that the flow of $X$ consists of isometries, which is equivalent to $\nabla X$ being skew). If $g(X,X)$ is constant, for all $Y$ we have that $Yg(X,X) = 0$, so $g(\nabla_YX,X) = 0$. But Killing's equation $$g(\nabla_YX, X) + g(Y, \nabla_XX) = 0$$reduces to $g(Y, \nabla_XX) = 0$. Since $Y$ is arbitrary, now non-degeneracy kicks in to give $\nabla_XX = 0$. This means that the integral curves of $X$ are geodesics.