Let $(M,g)$ be a Riemannian manifold. Show that if $Y$ is a Killing vector field and $\gamma : (a,b) \rightarrow M$ is a geodesic, then the function $g(\dot{\gamma},Y)$ is conserved along the geodesic.
I have seen a proof using the Einstein notation in GR. Let $X = \dot{\gamma}$,
\begin{align}
\frac{d}{dt}X_a Y^a
&= X^c \nabla_c(X_aY^a)\\
&= X^cY^a\nabla_cX_a + X^cX_a\nabla_cY^a\\
&= 0
\end{align}
by geodesic equation $\nabla_X X = 0$ and antisymmetry of $\nabla_cY^a$. However, how can one prove rigorously without restricting to a coordinate chart? Here is my attempt:
We know that $\mathcal{L}_Yg = 0$ and the Lagrangian squared $L^2 = g(X,X)$ is conserved along the geodesic, so
\begin{align}
\mathcal{L}_X(g(X,Y))
&= \mathcal{L}_X g(X,Y) + g(\underbrace{\mathcal{L}_X X}_{=0},Y) + g(X,\mathcal{L}_X Y)\\
&= \mathcal{L}_X g(X,Y) + g(X,-\mathcal{L}_Y X)\\
&= \mathcal{L}_X g(X,Y) – \frac{1}{2}\mathcal{L}_Y(\underbrace{g(X,X)}_{=\text{const}}) + \frac{1}{2} \underbrace{\mathcal{L}_Y g}_{\text{Killing field}, =0}(X,X)\\
&= \mathcal{L}_X g(X,Y)
\end{align}
The question is how can we show that this last term is $0$?
Best Answer
The vector field $X=\dot \gamma$ is only defined along the geodesic $\gamma$, not on an open set of $M$, so it does not make sense to calculate the Lie derivative $\mathcal{L}_X (g(X,Y))$. In order to show that $g(X,Y)$ is constant along $\gamma$, we need to show that $\nabla_X g(X,Y)$ vanishes. We get $$ \begin{align*} \nabla_X g(X,Y) = g(\nabla_X X, Y) + g(X, \nabla_X Y) = 0 \end{align*} $$ In the first equality we used that $\nabla g = 0$. The first term is zero since $\gamma$ is a geodesic. The second term vanishes because $Y$ is Killing and $g(X,X)$ is constant: $$ \begin{align*} 0 = (\mathcal{L}_Y g)(X,X) &= \mathcal{L}_Y (g(X,X)) - 2g(\mathcal{L}_Y X, X) \\ &= 2 g([X,Y], X) = 2 g(\nabla_X Y-\nabla_Y X, X) \\ &= 2 g(\nabla_X Y, X)- Yg(X,X) = 2 g(\nabla_X Y,X). \end{align*} $$