I was not able to do the following exercise from Manfredo's Riemannian Geometry:
Chapter 3, Exercise 5 Let $M$ be a Riemannian manifold and $X\in \mathfrak{X}(M)$. Let $p\in M$ and let $U\subset M$ be a neighborhood of $p$. Let $\varphi:(-\varepsilon,\varepsilon)\times U\to M$ be a differentiable mapping such that for any $q\in U$ the curve $t\mapsto \varphi(t,q)$ is a trajectory of $X$ passing through $q$ at $t=0$ ($U$ and $\varphi$ are given by the fundamental theorem for ordinary differential equations, Cf. Theorem 2.2). $X$ is called a Killing field (or an infinitesimal isometry) if, for each $t_0\in (-\varepsilon,\varepsilon)$, the mapping $\varphi(t_0,\,):U\to M$ is an isometry. Prove that:
(a) (…)
(d) $X$ is a Killing field $\iff \langle \nabla_YX,Z\rangle+\langle \nabla_ZX,Y\rangle=0$, for all $Y,Z\in \mathfrak{X}(M)$ (the equation above is called the Killing equation).
Then Manfredo gives a hint for proving $(\Rightarrow)$ and therefore I suppose that $(\Leftarrow)$ must be the "easy" part. However, I could not prove it. In $(\Leftarrow)$, I want to show that
$$\langle d(\varphi_{t_0})_q\cdot u,d(\varphi_{t_0})_q\cdot v\rangle=\langle u,v\rangle,$$
for all $t_0\in (-\varepsilon,\varepsilon)$, $q\in U$ and $u,v\in T_qM$. But I am not being able to relate this with the Killing equation. I know that $\frac{d}{dt}\varphi(t,q)=X(\varphi(t,q))$ but $d(\varphi_t)_q$ is not exactly $\frac{d}{dt}\varphi(t,q)$…
Best Answer
A Killing vector field is one satisfying $\mathcal{L}_X g=0.$ Since \begin{align} (\mathcal{L}_Xg)(Y,Z)&=Xg(Y,Z)-g([X,Y],Z)-g(Y,[X,Z]) \\ &=g(\nabla_XY,Z)+g(Y,\nabla_XZ) -g([X,Y],Z)-g(Y,[X,Z]) \\ &=g(\nabla_YX,Z)+g(Y,\nabla_ZX), \end{align} where in the second line we used the fact that the Levi-Civita connection is Riemannian and in the third that it is torsion free, we have your equivalence.