Killing vector field of the sphere Given a tangent vector field $X(x,y,z) = y\frac{\partial}{\partial x} -x\frac{\partial}{\partial y}$ of the sphere $S^2 \subset \mathbb{R}^3$.
Compute the Levi-Civita covariant derivative $\nabla_{v_p}X$ of any tangent vector $v_p$.
Secondly, show that this is a Killing vector field for the sphere.
I am having trouble with the first part, computing the covariant derivative.
Is the easiest way to compute it to use the ambient covariant derivative?
 A: For the first part, you can use the formula 
$$2\langle\nabla_{X}y, Z\rangle = X\langle Y, Z\rangle + Y\langle Z, X\rangle - Z\langle X, Y\rangle + \langle[X, Y], Z\rangle + \langle[Z, X], Y\rangle - \langle[Y, Z], X\rangle
$$
This formula probably occurs in every differential geometry book. You can just complete $v_p$ to a rotational vector field in the obvious way. 
For the second part, I think matgaio's comment is good enough, because as far as I know, having isometric flow is the definition of Killing field (see Peter Petersen's book for example). However, you can also check from its usual formula 
$$
\mathcal{L}_X g (U, V) = \langle\nabla_U X, V\rangle + \langle U, \nabla_V X\rangle
$$
For this, the computation for the first part should come in handy.
A: for the first part
for an arbitary vector field $Y=y_1\dfrac{\partial}{\partial x} +y_2\dfrac{\partial}{\partial y}+y_3\dfrac{\partial}{\partial z}$ we have (at point p):
\begin{equation}
\nabla_{Y_p}X = y_1 (\dfrac{\partial}{\partial x} .y)\dfrac{\partial}{\partial x} -y_1(\dfrac{1}{\partial x}.x)\dfrac{1}{\partial y}+y_2 (\dfrac{\partial}{\partial y} .y)\dfrac{\partial}{\partial x} -y_2(\dfrac{\partial}{\partial y} .x)\dfrac{1}{\partial y}+y_3(\dfrac{\partial}{\partial z} .y)\dfrac{\partial}{\partial x} -y_3(\dfrac{\partial}{\partial z} .x)\dfrac{\partial}{\partial y}
\end{equation}
then
\begin{equation}
\nabla_{Y_p}X= y_2\dfrac{1}{\partial x}-y_1 \dfrac{1}{\partial y}
\end{equation}
for the second part, we have:
\begin{equation}
L_Xg(Y,X)=g(\nabla _Y X,Z)+g(\nabla _ZX,Y)=(y_2z_1-y_1z_2)+(z_2y_1-z_1y_2)=0
\end{equation}
