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Maurer-Cartan form involved in Lie bracket

  1. Jul 31, 2014 #1
    The Maurer-Cartan one-form ##\Theta = g^{-1} dg## is though of as a lie algebra valued form.
    It arises in connection with Yang-Mill's theory where the gauge potential transforms as
    $$A \mapsto g Ag^{-1} - g^{-1} dg.$$

    However, one also defines for lie-algebra valued differential forms ##\alpha, \beta \in \Omega_p(M,\mathfrak g)##, the Lie bracket
    $$[\alpha, \beta] = [\xi_k, \xi_l] \alpha^k \wedge \beta^l.$$

    The question then arise, what does one mean by the lie-bracket when ##g^{-1} dg## is involved?
    For example, how would one compute
    $$[g^{-1} dg, g\alpha g^{-1}],$$
    for a lie algebra valued form ##\alpha##?
     
  2. jcsd
  3. Aug 12, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
     
  4. Sep 9, 2014 #3
    Hi,

    An exterior product exists on the exterior algebra of vector space valued differential forms only if the vectorspace carries in addition the structure of an algebra.

    Let G be a lie group and [tex]\varOmega^{*}:=\bigoplus_{r=0}^{dim(G)}\Omega^{r}(G,\mathfrak{g})[/tex]
    the exterior algebra of the associated lie algebra valued differential forms. If so, the exterior product is defined by

    [tex][\alpha,\beta](v_{1},\ldots,v_{p},v_{p+1},\ldots,v_{p+q}):=\frac{1}{p!q!}\sum_{\sigma\in S_{p+q}}sign(\sigma)[\alpha(v_{\sigma(1)},\ldots,v_{\sigma(p)}),\beta(v_{\sigma(p+1)},\ldots,v_{\sigma(p+q)})],[/tex]
    So the bracket is not a lie bracket, the bracket denotes the product of the algebra.
     
    Last edited: Sep 9, 2014
  5. Sep 15, 2014 #4

    lavinia

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    Gold Member

    Correct. So what is your question?
     
  6. Sep 16, 2014 #5
    The question was asked in the first post.
     
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