Covariant derivative of Killing vector field is antisymmetric Let $M$ be a Riemannian manifold with Levi-Civita connection $\nabla$. A vector field $X\in \mathfrak{X}(M)$ is said to be a Killing vector field if, for every $t$, its flow $X_{t}\colon M_{t}\to M_{t}$ is an isometry. I'm trying to prove that the covariant derivative $(\nabla X)_{p}\colon T_{p}(M)\to T_{p}(M)$ is an antisymmetric linear map with respect to the metric.
As far as I know, this statement is easy to prove by using Lie derivatives, but I'd like to see a proof where we do not make explicit use of that. So far, I've attempted the following:
Let $v,w\in T_{p}(M)$. I need to see that $\langle\nabla_{v}X,w\rangle+\langle v,\nabla_{w}X\rangle=0$. It seems natural to me to consider the function $g(t)=\langle (X_{t})_{*p}(v),(X_{t})_{*p}(w) \rangle=\langle v,w\rangle$ and to try to convert the equation $g'(0)=0$ into either the equation above or something similar. Am I on the right track or should I try a different approach?
Thank you in advance!
EDIT
I've tried to give a proof of the result by using coordinates (I'm not too fond of it, but at least I think it gets the job done).
Suppose $X(p)\neq 0$. Then, by the Tubular Neighborhood Lemma, we can find a coordinate chart $(U,(x^{1},...,x^{n}))$ where $U$ corresponds to the cube $(-\varepsilon,\varepsilon)^{n}$, and $X=\partial_{1}$. To prove the result, it suffices to see that $\langle \nabla_{\partial_{i}}\partial_{1},\partial_{j} \rangle=-\langle \partial_{i},\nabla_{\partial_{j}}\partial_{1}\rangle$.
On one hand:
$$
\langle \nabla_{\partial_{i}}\partial_{1},\partial_{j} \rangle_{p}=\langle \Gamma_{i1}^{k}\partial_{k},\partial_{j}\rangle_{p}=\Gamma_{i1}^{k}(p)g_{kj}(p).
$$
On the other hand, since $X_{t}$ is now given in coordinates by $(u^{1},...,u^{n})\mapsto(u^{1}+t,...,u^{n})$ and is an isometry, it follows that the functions $g_{ij}$ are independent of $x^{1}$. That is: $\partial_{1}g_{ij}=0$.
Now, by using the Koszul formula
$$
\Gamma_{i 1}^{k} g_{k j}=\frac{1}{2}\left(\partial_{i} g_{1 j}+\partial_{1} g_{i j}-\partial_{j} g_{i 1}\right)=\frac{1}{2}\left(\partial_{i} g_{1 j}-\partial_{j} g_{1 i}\right)=-\Gamma_{j 1}^{k} g_{k i}
$$
so we get the desired result.
Now, by continuity, we deduce that $\nabla X$ is antisymmetric at every $p\in \operatorname{Supp}(X)$.
Finally, if $p\notin \operatorname{Supp}(X)$, then $X=0$ in a neighborhood of $p$, which implies $(\nabla X)_{p}=0$ (which is clearly antisymmetric).
Is this proof correct? Also, is there a way of seeing this without appealing to coordinates?
 A: I have found a coordinate free proof of the fact. I'm sharing it here in case somebody stumbles upon this question in the future.
Since $X\in \mathfrak{X}(M)$ is a Killing vector field, all of its flows are isometries. This means that, if $g$ denotes the Riemannian metric on $M$, then $X_{t}^{*}g=g$ for all $t\in \mathbb{R}$. Therefore, for any $p\in M$, we get that
$$
0=\dfrac{d}{dt}\Big|_{t=0}g_{p}=\dfrac{d}{dt}\Big|_{t=0}(X_{t}^{*}g)_{p}=:(\mathscr{L}_{X}g)_{p}.
$$
Notice two things: the first one is that the limit is taken in the vector space $T_{p}^{*}(M)\otimes T_{p}^{*}(M)$ of all $(0,2)$-tensors over $T_{p}(M)$. The tensor $\mathscr{L}_{X}g$ is known as the Lie derivative of $g$ with respect to $X$. We get that if $X$ is a Killing field then $\mathscr{L}_{X}g=0$ (in fact, these two conditions are equivalent).
Now, there is a known expression for $\mathscr{L}_{X}g$ (c.f. "Introduction to Smooth Manifolds" (2nd ed.) by Lee, Corollary 12.33) given by
$$
\mathscr{L}_{X}g(Y,Z)=X\langle Y,Z \rangle-\langle [X,Y],Z \rangle-\langle Y,[X,Z] \rangle, \quad X,Y,Z\in \mathfrak{X}(M).
$$
Since the Lie derivative is identically zero, we get
$$
0=X\langle Y,Z \rangle-\langle [X,Y],Z \rangle-\langle Y,[X,Z] \rangle=\langle \nabla_{X}Y-\nabla_{X}Y+\nabla_{Y}X,Z \rangle+\langle Y,\nabla_{X}Z-\nabla_{X}Z+\nabla_{Z}X \rangle.
$$
We simplify to get $\langle \nabla_{Y}X,Z \rangle=-\langle Y,\nabla_{Z}X \rangle$, so $\nabla X$ is antisymmetric.
