HELP with Killing Vectors in AdS

1. May 26, 2013

llorgos

Hi and I am sorry if you find my question naive.

I have to find the Killing vectors of the AdS metric

$ds_{d+1}^{2} = \frac{dz^2 - dt^2 + dx^idx^i}{z^2}$

I have found the Christoffel symbols. If I use the Killing's equation $\nabla_{a}X^{b} + \nabla_{b}X^{a} = 0$ I find a set of differential equations. Ok, then supposing I can solve them I get components of vectors, e.g. $X_{z} = ze^{c}$. So this is a component of the Killing vector?

I am quite confused and I would appreciate if someone could explain in simple steps how to proceed.

Thank you very much for your help and patience.

2. May 26, 2013

ForMyThunder

The metric you have, the $ds_{d+1}^2$, gives you the components of the metric, $g_{ab}$, which you can just read off. Feeding this into the Christoffel symbols and the Killing equation gives a system of differential equations which you solve for $X^a$. I think you've got this far.

I think you may be confused because the $X^a$'s are functions? Correct me if I'm wrong.

These $X^a$ should be functions on the manifold, since they correspond to the components of a vector field on it. Thus, the Killing vector field is just (locally, that is, in the coordinate system specified) $X=X^a\partial_a$, where $\partial_a$ is the coordinate frame (I'm not sure how physicists do their notation).

3. May 26, 2013

llorgos

Yes. I get the $X_a$'s or $X^a$'s. I know they are funcitons on the Manifold. The thing is, do I just say, ok, the the vector field is just $X = X^a \partial_a$?
Is it that simple?

4. May 26, 2013

ForMyThunder

Yep. It's that simple.

5. May 26, 2013

llorgos

Ok. Thank you very much. Let's see if I can make any progress.