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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?

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Aren't you supposed to derive with respect to a vector field? –  Alexei Averchenko Nov 13 '11 at 17:06
    
The exercise says a tangent vector, does that not make sense? –  user3620 Nov 13 '11 at 17:38
    
Can you prove that it either does or doesn't? Consider two vector fields $Y$ and $Z$ such that for some point $p \in S^2$ you have $Y|_p = Z|_p$. What can you say about $\nabla_Y X - \nabla_Z X$? –  Alexei Averchenko Nov 13 '11 at 19:01
    
Actually, you're right, it does. Sorry, nevermind me then :) –  Alexei Averchenko Nov 13 '11 at 19:30
    
In order to get some intuition on the geometrical meaning of Killing fields, it is usefull to know that the flow of a Killing field is by isometries. So geometrically it is kind of reasonable the field you have is Killing: it is exactly the field wich flow is by rotations around the $z$-axe of $\mathbb{R}^3$, so acts by isometries of the sphere. It is not a proof, but can help you to get some feeling. –  matgaio Apr 22 '12 at 6:02

2 Answers 2

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.

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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}

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