# 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.

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