Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The reason might might be something like this.

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


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

  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


    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    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. :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook