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Associativity of the Killing form

  1. Jun 1, 2006 #1
    This came up in an exam on Lie algebras that I had today, and it's been bugging me. How do you prove that


    The best I've managed is writing


    but I have no idea where to go from there. Hints and/or a complete proof are both appreciated :)
  2. jcsd
  3. Jun 1, 2006 #2

    George Jones

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    Expand the last commutator and use [itex]\mathrm{Trace} \left( AB \right) = \mathrm{Trace} \left( BA \right)[/itex] on one of the terms.
  4. Jun 2, 2006 #3
    Sure, that's how I would do it if the Lie bracket was defined as a commutator, but it's not in a general Lie algebra.

    Oh, hold on... since [tex]\mathrm{ad}:g\to gl(g)[/tex] does that mean that I can expand the product [ad(X),ad(Y)] as a commutator?
  5. Jun 2, 2006 #4

    George Jones

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    [ad(X) , ad(Y)] = ad(X)ad(Y) - ad(Y)ad(X),

    where the associative products are defined, since ad(X) and ad(Y) are both linear operators on g considered as a vector space.
  6. Jun 3, 2006 #5
    It's annoying that it's that simple but I couldn't do it in the exam. :(

    Still, I did much more complicated stuff so hopefully the examiners will assume I just had a dim moment - rather than assuming that I'm dim altogether!
  7. Jun 5, 2006 #6
    I thought I'd post to mention that I had to do this proof (well, variants of it) THREE times in a particle physics exam today, so thanks for clearing it up in my mind - you scored me some marks!

    Applied exams are much easier than pure ones.
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