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Semisimple algebra (Killing form)

  1. May 9, 2010 #1
    I have started learning Lie algebra and I can't understand one example given in the notes.
    Given:
    [tex] [h_{\alpha},e_{\alpha}] = 2 e_{\alpha} [/tex]
    [tex] [h_{\alpha},f_{\alpha}] = -2 f_{\alpha} [/tex]
    [tex] [e_{\alpha},f_{\alpha}] = h_{\alpha} [/tex]

    and that
    [tex]
    [x,y] = K(x,y) t_{\alpha}
    [/tex]
    if [tex]\alpha[/tex] is a root and [tex] x \in L_{\alpha}, y \in L_{-\alpha} [/tex]
    Now, the example is application of the theorem to [tex]A_2[/tex].
    Generators are
    [tex] h_{\alpha} = E_{11} -E_{22} [/tex]
    [tex] h_{\beta} = E_{22} -E_{33} [/tex]
    [tex] e_{\alpha} = E_{12} [/tex]
    [tex] e_{\beta} = E_{23}[/tex]
    [tex] e_{-\alpha} = E_{21} [/tex]
    [tex]e_{-\beta} = E_{32} [/tex]
    and Postive roots are {[tex] \alpha, \beta, \alpha+\beta [/tex]}.

    I am meant to check that
    1.[tex]\alpha(h_{\alpha}) = \beta(h_{\beta}) =2 [/tex]
    2.[tex] \alpha(h_{\beta}) = \beta(h_{\alpha}) =-1 [/tex]

    I can't do part (2). Part (1) seems simple:
    [tex] \alpha(h_{\alpha}) = K(t_{\alpha},h_{\alpha} )= K(t_{\alpha},2\frac{t_{\alpha}}{K(t_{\alpha},t_{\alpha})}) = 2 [/tex]
    My problem is with finding [tex]t_{\alpha}[/tex] and [tex] t_{\beta}[/tex] to calculate [tex]K(t_{\alpha},t_{\alpha})[/tex]. How would one go about doing it?
    Because
    [tex] \alpha(h_{\beta}) = K(t_{\alpha},h_{\beta} )= K(t_{\alpha},2\frac{t_{\beta}}{K(t_{\beta},t_{\beta})}) = \frac{2}{K(t_{\beta},t_{\beta})} K(t_{\alpha},t_{\beta}) [/tex]
    Thanks.

    EDIT: I hope I had posted in the right thread. Should I have posted this in HW help?
     
    Last edited: May 9, 2010
  2. jcsd
  3. May 9, 2010 #2
    I have got it. It's just calculating the Cartan integers from the basis.
     
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