Let $X$ and $Y$ be two Killing fields, prove that $[X,Y]$ is a Killing Field I want to prove that the Lie derivative of two Killing fields is a Killing field. The exercise is stated as following:
Let (M,g) be a riemannian mannifold with $\nabla$ the Levi-Civita connexion. $X\in \mathfrak{X}(M)$ is a Killing field if it satisfies
$$
Xg(Y,Z)=g([X,Y],Z)+g(Y,[X,Z]) \ \forall Y,Z\in \mathfrak{X}(M)
$$
The first part of the exercise is to prove that this condition is equivalent to
$$
g(\nabla_YX,Z)+g(\nabla_ZX,Y)=0
$$
which I have done successfully. 
Using one of those relations I have to prove that given $X,Y\in \mathfrak{X}(M)$ two Killing fields $[X,Y]$ is also a Killing field. The Jacobi relation $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$ is given as a hint but I don't know if it is really usefull or it is a trap. I have tried a lot of calculations but haven't got anything yet. 
Thank you for your help
 A: The Lie derivative $[X,Y]$ acts on functions $f\in C^\infty(M)$ as
$$ [X,Y](f) = X(Y(f))-Y(X(f)).$$
Then it's a straightforward calculation:
\begin{align}
[X,Y]g(V,W)&=X\left(Yg(V,W)\right)-Y\left(Xg(V,W)\right)\\[1em]
&\qquad\qquad\text{Now, $X$ and $Y$ are Killing fields. Use twice:}\\[1em]
&=X(g([Y,V],W)+g(V,[Y,W]))-Y(g([X,V],W)+g(V,[X,W]))\\[1em]
&=g([X,[Y,V]],W) + g([Y,V],[X,W]) + g([X,V],[Y,W])+g(V,[X,[Y,W]])\\
&\phantom{=}-\left\{g([Y,[X,V]],W) + g([X,V],[Y,W]) + g([Y,V],[X,W])+g(V,[Y,[X,W]])\right\}\\[1em]
&=g([X,[Y,V]],W)+g(V,[X,[Y,W]])-g([Y,[X,V]],W)-g(V,[Y,[X,W]])\\[1em]
&=g\left(\left[X,\left[Y,V\right]\right] - \left[Y,\left[X,V\right]\right],W
\right)+g\left(V,\left[X,\left[Y,W\right]\right] - \left[Y,\left[X,W\right]\right]
\right)\\[1em]
&\qquad\qquad \text{use anticommutativity of Lie bracket}\\[1em]
&=g\left(\left[X,\left[Y,V\right]\right] + \left[Y,\left[V,X\right]\right],W
\right)+g\left(V,\left[X,\left[Y,W\right]\right] +\left[Y,\left[W,X\right]\right]
\right)\\[1em]
&\qquad\qquad\text{Use the Jacobi identity}\\[1em]
&=g(-[V,[X,Y]],W)+g(V,- [W,[X,Y]])\\[1em]
&=g([[X,Y],V],W)+g(V,[[X,Y],W])
\end{align}
And with this, you're done.
