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Showing that half-sum of positive roots is the sum of fundamental weights

  1. Dec 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Let L be a simple compact Lie group, and [itex] \Delta_+[/itex] is the set of positive roots. I have previously shown that if [itex]\alpha\in\Delta_+[/itex] and [itex]\alpha_i[/itex] is a simple root, then [itex]s_i\alpha\in \Delta_+[/itex] where s_i is the Weyl reflection associated with [itex]\alpha_i[/itex].

    Now, let [itex]\delta = \frac{1}{2}\sum_{\alpha\in\Delta_+}\alpha[/itex]. I want to show that
    [tex]
    s_i\delta=\delta-\alpha_i
    [/tex]


    2. Relevant equations



    3. The attempt at a solution
    It's clear that
    [tex]
    s_i\delta=\delta - \sum_{\alpha\neq \alpha_i} \frac{\alpha\cdot\alpha_i}{\alpha_i^2}\alpha_i - \alpha_i
    [/tex]

    But I have no idea how to show that [itex]\alpha\cdot\alpha_i=0\quad \forall\alpha\neq\alpha_i[/itex]. I cannot make appeal to the fact that delta might be a sum of fundamental weights because that's what I need to show later on.
     
  2. jcsd
  3. Dec 31, 2011 #2

    fzero

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    To be precise, [itex]s_i\alpha\in \Delta_+[/itex] for [itex]\alpha\neq \alpha_i[/itex]. That is, [itex]s_i[/itex] reflects [itex]\alpha_i \rightarrow -\alpha_i[/itex] but permutes the [itex]\alpha\neq \alpha_i[/itex] into one another.

    In light of the comments above, it's more straightforward to note that

    [tex]s_i \delta = \frac{1}{2}\sum_{\alpha\in\Delta_+}s_i\alpha =\frac{1}{2} \left( \sum_{\alpha\neq \alpha_i} \alpha - \alpha_i \right).[/tex]

    Restoring [itex]\delta[/itex] in an obvious way gives the required result.
     
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