An identity about Killing vector field I am studying Kähler geometry and I started reading these lecture notes and this is the first exercise (p.2). 
If $X$ is a Killing vector field on a Riemannian manifold and $Y,Z$ be two vector fields, then 
\begin{equation}
 \nabla^2X(Y,Z)+R(X,Y)Z=0
\end{equation}
I thought by writing everything in local coordinates would work but I really want some kind of neat proof. Can someone give some suggestion? 
Thank you in advance. 
 A: I was talking to my advisor these days and he told me this little gem of a proof: $$\begin{align}(\mathcal{L}_X\nabla)_YZ &= \mathcal{L}_X\nabla_YZ - \nabla_{\mathcal{L}_XY}Z - \nabla_Y\mathcal{L}_XZ \\ &= [X,\nabla_YZ] - \nabla_{[X,Y]}Z - \nabla_Y[X,Z] \\ &= \nabla_X\nabla_YZ - \nabla_{\nabla_YZ}X - \nabla_{[X,Y]}Z - \nabla_Y\nabla_XZ+\nabla_Y\nabla_ZX \\ &= R(X,Y)Z + \nabla^2X(Y,Z). \end{align}$$If $X$ is Killing, then the flow of $X$ consists of isometries. Isometries preserve the Levi-Civita connection $\nabla$, so $\mathcal{L}_X\nabla = 0$. Done.
A: That's called Killing's identity and basically says that a Killing vector field is uniquely determined by its value and the value of its derivative at a point.  Suppose $\xi$ is a vector field on $(M,g)$.  Let $A : TM \to TM$ be defined by $A X = -\nabla_X \xi$.  Then $\xi$ is Killing if and only if $A \in \mathfrak{so}(TM)$; that is, $g(AX,Y)= - g(X,AY)$.  Killing's identity then says that $\nabla_X A = R(X,\xi)$ as sections of $\mathfrak{so}(TM)$.  This is the identity that you are asking about.  The proof is calculational: just compute the derivative of $A$.  It is simple to do in indices (either abstract or relative to a local chart) but you can also prove it in a coordinate-free fashion.  Let me write it as a sequence of exercises for you.


*

*Show that
$$(\nabla_X A)(Y) = - \nabla_X \nabla_Y \xi + \nabla_{\nabla_X Y} \xi$$

*Show that
$$(\nabla_X A)(Y) - (\nabla_Y A)(X) = - R(Y,\xi)(X) + R(X,\xi)(Y)$$

*Define $S(X,Y,Z) := g((\nabla_X A - R(X,\xi))(Y),Z)$ and notice that the identity you are after is $S(X,Y,Z) = 0$.  Show that
$$ S(X,Y,Z) = S(Y,X,Z) $$
and
$$S(X,Y,Z) = - S(Y,Z,X) $$
and hence that $S(X,Y,Z) = 0$.

A: Here's a solution (up to your curvature sign conventions): since $X$ is Killing, you have that $$g(\nabla_YX,Z) + g(Y, \nabla_ZX) = 0$$ for all $Y$ and $Z$. Take the derivative of that in the direction of another field $W$ and look at cyclic permutations of $Y,Z,W$: $$\begin{align}  
g(\nabla_W\nabla_YX,Z) + g(\nabla_YX,\nabla_WZ) + g(\nabla_WY,\nabla_ZX) + g(Y,\nabla_W\nabla_ZX) &= 0 \\ g(\nabla_Y\nabla_ZX,W) + g(\nabla_ZX,\nabla_YW) + g(\nabla_YZ,\nabla_WX) + g(Z,\nabla_Y\nabla_WX) &= 0 \\ g(\nabla_Z\nabla_WX,Y) + g(\nabla_WX,\nabla_ZY) + g(\nabla_ZW,\nabla_YX) + g(Y,\nabla_Z\nabla_YX) &= 0\end{align}$$Now make $-(1)+(2)+(3) = 0$. This simplifies to $$R(Z,W,X,Y) + R(Y,W,X,Z) + g(\nabla_Y\nabla_ZX+\nabla_Z\nabla_YX,W)+g(\nabla_YZ+\nabla_ZY, \nabla_WX)=0.$$Using the definition of $\nabla^2X$ we get $$g(\nabla_Y(\nabla X)Z + \nabla_Z(\nabla X)Y, W) = R(Y,X,Z,W) + R(Z,X,Y,W).$$This implies that $$\nabla_Y(\nabla X)Z + \nabla_Z(\nabla X)Y = R(Y,X)Z + R(Z,X)Y.$$But from the definition, we also get the Ricci identity$$\nabla_Z(\nabla X)Y = \nabla_Y(\nabla X)Z - R(Y,Z)X.$$Plug that in the previous equality and apply Bianchi to get $$2\nabla_Y(\nabla X)Z = R(Y,Z)X + R(Z,X)Y + R(Y,Z)X = 2R(Y,X)Z,$$and so $\nabla_Y(\nabla X)Z = R(Y,X)Z$. Now you rewrite that as $\nabla^2X(Y,Z) + R(X,Y)Z = 0$, as wanted.
