$f:M\rightarrow M$ is an isometry and $ker(Id - d_{\gamma(t)}f) = \mathbb{R} \dot{\gamma}(t)$, show that $\gamma$ is a geodesic

Question

Let M be a Riemannian manifold, $\gamma: I \rightarrow M$ be a curve which is parameterized with constant speed and $f:M\rightarrow M$ be an isometry with $f \circ \gamma = \gamma$. Furthermore $\ker(Id - d_{\gamma(t)}f) = \mathbb{R} \dot{\gamma}(t), \forall t$. Show that $\gamma$ is a geodesic then.

I need some help with this. I don't know how f plays into this. We know that $d_{\gamma(t)}f(a\dot{\gamma}(t)) = a\dot{\gamma}(t)$ for any $a \in \mathbb{R}$ , then for $ \mathbb{R} \ni c = \langle \dot{\gamma(t)}, \dot{\gamma(t)} \rangle = \langle d_{\gamma(t)}f(\dot{\gamma}(t)), d_{\gamma(t)}f(\dot{\gamma}(t)) \rangle$.

I don't see how any of this is supposed to be related to $\nabla_{\dot{\gamma}}\dot{\gamma}$.

Why is it important that f is an isometry?

Answer 1

This is a consequence of the fact that, under your given assumptions, $\gamma$ constitutes a local parameterization of an immersed submanifold of $M$ which is fixed by an isometry. But such immersed submanifolds are totally-geodesic. In this instance, the immersed submanifold is of dimension 1 and an immersed, totally-geodesic submanifold of dimension 1 is just a geodesic.

Answer 2

I will give an answer without using the fact that the fixed point set of an isometry is a totally geodesic submanifold.

Instead we use that isometry preserves Levi-Civite connection. Hence we have (for any $t\in I$)

$$\tag{1} df_{\gamma(t)}( \nabla _{\gamma'(t)} \gamma'(t) )= \nabla _{df_{\gamma(t) } \ \gamma'(t)}df_{\gamma(t) } \gamma'(t) = \nabla _{\gamma'(t)} \gamma'(t).$$

Since $\langle \nabla _{\gamma'(t)} \gamma'(t) , \gamma'(t)\rangle =0$ as $\gamma$ is of constant speed, together with (1) and $\ker (df_{\gamma(t)} - \operatorname{id}) = \mathbb R \gamma'(t)$ one has $\nabla _{\gamma'(t)} \gamma'(t) = 0$. That is, $\gamma$ is a geodesic.