TITLE: $X$ is a Killing field $\iff \langle \nabla_YX,Z \rangle+\langle \nabla _ZX,Y\rangle=0$
QUESTION [9 upvotes]: 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 $\frac{d}{dt}\varphi(t,q)$...

REPLY [2 votes]: The metric in this case is given by
$$\left<(d\phi(t)) u, (d\phi(t)) v\right>=\left<(d\phi(t_0)) u, (d\phi(t_0)) v\right>
+O(t^2)$$
and hence
$$\lim_{t\to 0}\frac{\left<(d\phi(t)) u, (d\phi(t)) v\right>-\left<(d\phi(t_0)) u, (d\phi(t_0)) v\right>}{t}=0$$
i.e.
$$\left({\frak{L}}_Xg\right)(u,v)=0$$
Then you can use the answer of Aloizio Macedo.