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Killing vectors

  1. Jan 9, 2010 #1
    I was wondering if you could help me with the proof of the following theorem.

    If [tex]A^{{\mu }}[/tex] and [tex]B^{{\mu }}[/tex] are Killing vectors, then so is their commutator [tex]C^{{\mu }}=[A,B]^{\mu}[/tex].

    Thanks in advance
     
  2. jcsd
  3. Jan 9, 2010 #2

    diazona

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    What have you tried?
     
  4. Jan 10, 2010 #3
    Assuming [tex]g_{\mu \nu}[/tex] as our metric, I can write

    [tex]g_{\mu \alpha}A^{\alpha}_{;\nu}=-g_{\alpha\nu}A^{\alpha}_{;\mu}[/tex] and
    [tex]g_{\mu \alpha}B^{\alpha}_{;\nu}=-g_{\alpha\nu}B^{\alpha}_{;\mu}[/tex].

    And I can correlate these two to each other, but I'm afraid about the commutator. I just need a clue as to how the commutator is written in terms of [tex]A^{\mu}[/tex] and [tex]B^{\mu}[/tex].
     
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