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HELP with Killing Vectors in AdS

  1. May 26, 2013 #1
    Hi and I am sorry if you find my question naive.

    I have to find the Killing vectors of the AdS metric

    [itex]ds_{d+1}^{2} = \frac{dz^2 - dt^2 + dx^idx^i}{z^2} [/itex]

    I have found the Christoffel symbols. If I use the Killing's equation [itex]\nabla_{a}X^{b} + \nabla_{b}X^{a} = 0[/itex] I find a set of differential equations. Ok, then supposing I can solve them I get components of vectors, e.g. [itex]X_{z} = ze^{c}[/itex]. 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. jcsd
  3. May 26, 2013 #2
    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).
     
  4. May 26, 2013 #3
    Yes. I get the [itex]X_a[/itex]'s or [itex]X^a[/itex]'s. I know they are funcitons on the Manifold. The thing is, do I just say, ok, the the vector field is just [itex]X = X^a \partial_a[/itex]?
    Is it that simple?
     
  5. May 26, 2013 #4
    Yep. It's that simple.
     
  6. May 26, 2013 #5
    Ok. Thank you very much. Let's see if I can make any progress.
     
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