# How we compute killing vector for two-sphere

1. Sep 29, 2009

### off-diagonal

The metric on $S^2$ is given by,
$\displaystyle ds^2=d\theta^2 + sin^2\theta d\phi^2$

$\displaystyle \xi ^{\mu}_{(1)}\partial _{\mu} = \partial_{\phi}$

$\displaystyle \xi^{\mu}_{(2)}\partial_{\mu} = \ -(cos\phi \partial_{\theta} - cot\theta sin\phi \partial_{\phi})$

$\displaystyle \xi^{\mu}_{(3)}\partial_{\mu} = sin\phi \partial_{\theta} + cot\theta cos\phi \partial_{\phi}$

from Black Hole Physics: Basic Concepts and New Development by Frolov & Novikov
Appendix B

Anyone can explain me how to compute this 3 Killing vector?

2. Sep 30, 2009

### javierR

You'll need to pull off the $$g_{\mu\nu}$$ from the line element and then easily calculate the inverses $$g^{\mu\nu}$$. Then you have to calculate the Christoffel symbols from the metric components (looking hard at the equations for these, can you see which symbols will be non-vanishing ahead of time?). Then you know how the covariant derivative acts (See standard equations for GR). Then the Killing vector equation is $$\nabla_{\mu}\zeta_{\nu} + \nabla_{\nu}\zeta_{\mu}=0$$. After you've carefully written these things out, you can do some integrals to pull out the Killing vectors from the above equation.

3. Oct 4, 2009

### off-diagonal

but after I perform calculation. I found a problem that may cause from my misunderstand about Killing vector equation.

Here's the detail
from these metric
$\displaystyle ds^2=d\theta^2 + sin^2\theta d\phi^2$ I can read out metric tensor component as $\displaystyle g_{\theta}_{\theta}= 1 g_{\phi}_{\phi}=sin^2\theta$
so I can compute Christoffel symbol , there are two component in $S^2$ case
$\displaystyle \Gamma^{\theta}_{\phi\phi}=-sin\theta cos\theta$
$\displaystyle \Gamma^{\phi}_{\phi \theta}=cot\theta$

and from Killing equation I've got three equation

$\displaystyle \mu=\nu=\theta$
$\displaystyle \partial_{\theta}\xi_{\theta}=0$ since there are no $\displaystyle \Gamma^{i}_{\theta\theta}$

$\displaystyle \mu=\nu=\phi$
$\displaystyle \partial_{\phi}\xi_{\phi}+sin\theta cos\theta \xi_{\theta}= 0$

$\displaystyle \mu=\theta,\nu=\phi$
$\displaystyle \partial_{\theta}\xi_{\phi}+\partial_{\phi}\xi_{\theta}-2cot\theta \xi_{\phi}= 0$

What should I do next? to find all ${\xi}$
PS. since my answer has a term which depend on$\phi$ but until my last calculation I found no $\phi$ terms appear so am I miscalculate somewhere above?

4. Oct 14, 2009

### javierR

Sorry, I didn't see your message till now. If you still need it: Now you can solve for the $$\xi_{\mu}$$ by integration for each of the three equations. Try finding the form of the theta one first and then proceed to the next one using that result, and so on. You'll have to do some analysis to get the explicit form of the functions, and don't forget the presence of integration constants (constant with respect to one or both variables that is, since we're dealing with partial derivatives). Once you get the functions for theta and phi, you can easily invert them with the form of the metric you wrote down. Then the independent Killing vectors are $$\xi^{\theta}\partial_{\theta}$$ and $$\xi^{\phi}\partial_{\phi}$$, and a general Killing vector is a linear combination of these. (You can also show that you can write such a Killing vector split into three parts corresponding to generators of rotations on the sphere).