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Killing fields

  1. May 26, 2006 #1
    A question on a General Relativity exam that I have asks how many linearly independent Killing fields there can be in an n-dimensional manifold. I'm sure I've seen this question before and I think that the answer is n(n+1)/2, but I can't remember why!

    Any help?
     
    Last edited: May 26, 2006
  2. jcsd
  3. May 27, 2006 #2

    George Jones

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    The reason might might be something like this.

    Killing's equation is anti-symmetric in 2 indices, which gives

    [tex]
    \left(
    \begin{array}{cc}
    n\\
    2
    \end{array}
    \right)
    [/tex]

    degrees of freedom, plus [itex]n[/itex] initial degrees of freedom, one for each component of a killing vector.

    Regards,
    George
     
  4. May 27, 2006 #3
    Another way of understanding this number is considering the symmetries belonging to the Killing fields.

    The simplest example of a maximally symmetric (n dim.) manifold is just Euclidian [tex]\mathbb{R}^n[/tex] (or Minkowski space in case of Lorentzian metric). Independent symmetries are given by the translations and rotations. There are n independent translations and you can convince yourself that there are n(n-1)/2 independent rotations (the dimension of SO(4)). So the total number of independent symmetries is [tex]n+n(n-1)/2=n(n+1)/2[/tex].

    So any (Riemannian) manifold can have at most n(n+1)/2 symmetries, and at most the same number of independent Killing fields.

    -- Timbuqtu
     
  5. May 27, 2006 #4

    selfAdjoint

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    This is excellent. My only gripe is using the term "rotation" for an element of SO(n). Isometry would be better.
     
  6. May 28, 2006 #5
    Okay, thanks for the help guys. I suppose the question to ask now is: Why do Killing vectors have a one-to-one correspondence to spacetime symmetries?
     
  7. May 28, 2006 #6
    The killing equation [tex]\nabla_{(\mu}k_{\nu)}=0[/tex] is equivalent to the vanishing of the Lie-derivative of the metric along k, [tex]\mathcal{L}_k g_{\mu\nu}=0[/tex]. This last equation actually just tells you that the metric does not change if you pull it along the vector field k, so k represents a symmetry of the metric (and therefore a symmetry of the space time).

    -- Timbuqtu
     
  8. May 28, 2006 #7
    Ah, fantastic. Cheers. :)
     
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