Norm of geodesic velocity vector

Question

The definition of affine geodesic is clear: a curve with covariant derivative respective to Levi-Civita connection $\nabla$ of velocity vector $\dot{\gamma}$ respective to vector field $\dot{\gamma}$, for arbitrary instant $t \in [0, 1]$ equal to zero. Locally, we may interchange the term "geodesic curve" with exponential map exp${}_p(s v)$.

My question is: is the norm $\langle \dot{\gamma}, \dot{\gamma} \rangle$ along a geodesic constant?

My best answer is "No.". My best argument is "Let some arbitrary positive scalar $\varepsilon$ such that following inequality $0 < \varepsilon < 1$ and a geodesic curve such that initial values are $(p, \, v)$ and the vector norm $\langle v, \, v\rangle$ is equal scalar $\ell$. The velocity vector $\dot{\gamma}(\varepsilon)$ at instant $\varepsilon$. Therefore, the velocity vector norm at $\dot{\gamma}(\varepsilon)$ is equal to $\varepsilon \ell$ since we may define thee exponential map exp$_{\gamma(\varepsilon)} \dot{\gamma}(\varepsilon)$."

It contradicts the fact of covariant derivatives as a parallel transport operation i.e. that the vector norm is constant along the geodesic curve.

Answer 1

The norm of a geodesic curve's velocity vector is constant and can be seen using the following argument.

Let $\gamma: I\to M$ be a geodesic, i.e. it satisfies $\nabla_{\dot{\gamma}}\dot{\gamma}=0$. We then have $$\frac{d}{dt} ||\dot{\gamma}(t)||^2=\frac{d}{dt}\langle \dot{\gamma}(t), \dot{\gamma}(t)\rangle=(\nabla_{\dot{\gamma}} g)(\dot{\gamma}(t),\dot{\gamma}(t))+\langle \nabla_{\dot{\gamma}}\dot{\gamma}(t),\dot{\gamma}(t)\rangle+\langle\dot{\gamma}(t), \nabla_{\dot\gamma}\dot\gamma\rangle.$$ By metric compatibility $\nabla_{\dot\gamma}g=0$ and the geodesic equation tells us that $\nabla_{\dot\gamma}\dot\gamma=0$, hence $$\frac{d}{dt}||\dot\gamma||^2=0.$$