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Any biinvariant metric proportional to Killing metric

  1. Jul 14, 2014 #1
    The killing form on a lie algebra is defined as
    $$B(X,Y) = \text{Tr}ad_X \circ ad_Y$$
    where ##ad_X: \mathfrak{g} \to \mathfrak{g}## is given by ##ad_X(Y) = [X,Y]##, where the latter is the lie bracket between X and Y in ##\mathfrak{g}##. Expressed in terms of components on a basis on ##\mathfrak{g}## we have
    $$B_{ij} = c_{il}^{\ \ k} c_{jk}^{\ \ l}$$.

    The killing form can serve as a biinvariant metric on the lie group G, and I've seen it stated several times that, if the group G is simple, then all other biinvariant metrics are proportional to the killing form. Especially the formula
    $$B(X,Y) \sim \text{Tr}(\rho(X)\rho(Y))$$
    where ##\rho## is a representation of ##\mathfrak{g}## is thrown a lot around.

    So I wonder how this statement is proved?
  2. jcsd
  3. Jul 14, 2014 #2
    It's essentially an application of Schur's Lemma in the context of Lie algebras which I will just give a quick sketch and leave out all the minor details.

    Let [itex] B,B'[/itex] be two invariant symmetric bilinear forms. Use the fact that [itex] \mathfrak{g} [/itex] is simple to show [itex] B,B'[/itex] are both either zero or nondegenerate (use the properties of B to show [itex] \left\{ x\in \mathfrak{g} : B(x,y)=0~\forall y\in \mathfrak{g} \right\} [/itex] is an ideal.) We may as well assume each is nondegenerate since otherwise the statement is trivial. We have two different isomorphisms between [itex] \mathfrak{g} [/itex] and [itex] \mathfrak{g^*} [/itex], say [itex] \phi,\psi [/itex] corresponding to the two different forms. Apply Schur's lemma to conclude [itex] \psi^{-1}\circ\phi [/itex] is a scalar multiple of the identity and show this implies the forms [itex] B,B' [/itex] are themselves scalar multiples of each other.
  4. Aug 1, 2014 #3
    Can the statement be proved without referring to Schur's Lemma? I'm a phycisist, so I have not
    had the time to go deep into representation theory. However, I have read up on ideals, simple and semisimple lie algebras.
  5. Aug 1, 2014 #4


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    Schur's lemma is really elementary. You can probably understand it if you spend 15 minutes at it.
  6. Aug 2, 2014 #5


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    Why don't you look into it and get back to us with any question you may have.
  7. Aug 2, 2014 #6
    Any suggestions where to read up on it? Can it be done and still avoiding "modules"?
  8. Aug 2, 2014 #7
    I have a deadline for something in which it would be nice have an explanation of the Killing form proportionality. Do you have any suggestions where I could read up on it without having to learn additional material?
  9. Aug 2, 2014 #8


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    I don't know many books, but Liebeck and James, it comes with solutions. BTW, representations can be seen as modules, and modules are somewhat like vector spaces, but use rings instead of fields. See the section for Schur's lemma http://www.mathematik.uni-kl.de/~taylor/PDF/jamesliebeck.pdf [Broken] The book is broken into small, self-contained sections.
    Last edited by a moderator: May 6, 2017
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