HELP with Killing Vectors in AdS

  • Thread starter Thread starter llorgos
  • Start date Start date
  • Tags Tags
    Vectors
llorgos
Messages
18
Reaction score
0
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.
 
Physics news on Phys.org
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).
 
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?
 
Yep. It's that simple.
 
Ok. Thank you very much. Let's see if I can make any progress.
 
Back
Top