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Prove Killing vector is an affine collineation

  1. Mar 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Given a manifold M with metric gab and associated derivative operator a, let ξ a be a Killing vector on M. Prove that ξ a is an affine collineation for a.

    2. Relevant equations

    (a) For a vector ξ a to be an affine collineation for a derivative operator a, it must satisfy
    ab ξ c = Rmabc ξ m.​

    (b) We are told that ξ a is a Killing vector, so
    (a ξ b) = 0.​

    (c) The action of the Riemann tensor on a vector is
    Rabcd ξ c = –2 ∇[ab] ξ d.​

    3. The attempt at a solution

    I've gone down one blind alley after another. Below I reproduce what seemed like the most promising path I've explored so far. Can somebody help me with the next step, or suggest an alternative path that could be more fruitful?

    Start with the right-hand side of (a):
    Rmabc ξ m = gcp Rmabp ξ m.​
    It's a symmetry property of the Riemann tensor that I can switch the first pair of indices with the second:
    gcp Rmabp ξ m = gcp Rbpma ξ m = gcp gaq Rbpmq ξ m.​
    We've now finagled this into the form of (c):
    gcp gaq Rbpmq ξ m = –2 gcp gaq [bp] ξ q = –2 gcp[bp] ξ a.​
    Performing the antisymmetrization,
    –2 gcp[bp] ξ a = – gcp (∇bp ξ a – ∇pb ξ a) .​
    Now it's easy to use (b) on the first of these two terms:
    gcp (∇bp ξ a – ∇pb ξ a) = + gcp (∇ba ξ p + ∇pb ξ a).​
    But I still have that second term sitting there! What do I do about it?
     
  2. jcsd
  3. Mar 17, 2014 #2

    WannabeNewton

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    Let me start you off on an easier path. We have by definition ##\nabla_a \nabla_b \xi_c - \nabla_b \nabla_a \xi_c = R_{abc}{}{}^{d}\xi_d## which we can rewrite using ##\nabla_{(a}\xi_{b)} = 0## as ##\nabla_a \nabla_b \xi_c + \nabla_b \nabla_c \xi_a = R_{abc}{}{}^{d}\xi_d##. Now perform cyclic permutations of the indices ##(abc)## and combine the resulting equations in the appropriate manner in order to get the result ##\nabla_a\nabla_b \xi_c = -R_{bca}{}{}^{d}\xi_d##. Your desired result is then immediate.
     
  4. Mar 18, 2014 #3
    So as you say, I have the 3 equations:[tex]\left\{ \begin{array}{l}
    R_{abc}{}^{d}\xi_d = \nabla_a\nabla_b\xi_c+\nabla_b\nabla_c\xi_a \\
    R_{cab}{}^{d}\xi_d = \nabla_c\nabla_a\xi_b+\nabla_a\nabla_b\xi_c \\
    R_{bca}{}^{d}\xi_d = \nabla_b\nabla_c\xi_a+\nabla_c\nabla_a\xi_b.
    \end{array}\right.[/tex]If I add together the first two equations and subtract the third, I get
    [tex](R_{abc}{}^{d}+R_{cab}{}^{d}-R_{bca}{}^{d})\xi_d = 2\nabla_a\nabla_b\xi_c.[/tex]The Riemann tensor antisymmetrizes to 0 in its first 3 indices:[tex]R_{[abc]}{}^{d} = \tfrac13(R_{abc}{}^{d}+R_{bca}{}^{d}+R_{cab}{}^{d})=0.[/tex]So the [itex]R_{abc}{}^{d}+R_{cab}{}^{d}[/itex] in my equation can be replaced by a [itex]-R_{bca}{}^{d}[/itex]:[tex]-2R_{bca}{}^{d}\xi_d = 2\nabla_a\nabla_b\xi_c.[/tex]I can exchange the first pair of indices in the Riemann tensor with the second pair for free:[tex]-R_{ambc}\xi^m = \nabla_a\nabla_b\xi_c.[/tex]But swapping the order of the first two costs a minus sign:[tex]+R_{mabc}\xi^m = \nabla_a\nabla_b\xi_c.[/tex]And finally raising c on both sides completes the proof.

    Cool. Cool cool cool. Thanks for the help!
     
  5. Mar 20, 2014 #4

    WannabeNewton

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    Science Advisor

    No problem! The utilization of ##\epsilon_{abcd}##, ##R_{[abc]d}## (usually in conjunction with the former) and index permutation constitute very standard techniques that you will make use of over and over when performing these kinds of calculations so it's good to internalize them.
     
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