Geodesic flow and Killing vector field

Question

Let $\gamma$ be a geodesic, $\gamma'$ its tangent vector, $X$ a Killing vector field and $X_\gamma$ the restriction of $X$ to the curve $\gamma$. Let $g$ be the metric on the manifold considered.

Prove that $g(\gamma',X_\gamma)$ is constant along $\gamma$.

The definition I have for a Killing field $X$ is that it satisfied $\mathcal{L}_Xg=0$ where $\mathcal{L}$ denotes the Lie derivative.

I thought about using the induced covariant derivative on the curve $\gamma$, but all I have is a definition and no real explanation on how to put this into pratice.

Answer 1

Hint: You will first have to understand how the fact that $X$ is a Killing field is expressed in terms of the covariant derivative $\nabla X$. This is a standard result, it can be directly derived from exapanding the equation $0=(\mathcal L_Xg)(Y,Z)$. Having done that, you can express $\tfrac{d}{dt}g(\gamma'(\gamma(t)),X(\gamma(t)))$ in terms of covariant derivatives (along the curve) and use the equations satisfied by $\gamma'$ and by $X$ to conclude that this vanishes.

Answer 2

Here's an overkill solution: the flow of $X$ consists of isometries, and so it preserves the Lagrangian $L\colon TM \to \Bbb R$ given by $L(x,v) = g_x(v,v)/2$, whose fiber derivative is $\mathbb{F}L(x,v) = g_x(v,\cdot)$. Critical points of the action functional of $L$ are geodesics, so Noether's Theorem says that along geodesics, the Noether charge $$\mathbb{F}L(x,v)X_x = g_x(v,X_x)$$is constant. So $t \mapsto g_{\gamma(t)}(\gamma'(t), X_{\gamma(t)})$ is constant if $X$ is Killing and $\gamma$ is a geodesic.