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Derivative of the inversion operator and group identity

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Let G be a Lie group, e be its identity, and [itex] \mathfrak g [/itex] its Lie algebra. Let i be group inversion map. Show that [itex] d i_e = -\operatorname{id} [/itex].

    3. The attempt at a solution

    So this isn't terribly difficult if we have the exponentiation functor, since in that case
    [tex] e^{-\xi} = i(e^\xi) = e^{di_e\xi} [/tex]
    which implies that [itex] di_e \xi = -\xi [/itex] for all [itex] \xi \in \mathfrak g [/itex]. If the group is furthermore complex, then one could make the argument that the pushforward of the inversion map is an idempotent morphism of the adjoint representation on [itex] \mathfrak g[/itex], and use Schurr's lemma with some additional arguments to rule out the positive identity case.

    However, since the map is given explicitly, it seems to me that we should be able to show this using first principles; namely, just the properties of the exterior derivative/pushforward. However, I have not been able to compute it directly. For example, if [itex] \xi \in \mathfrak g[/itex] then
    [tex] di_e (\xi)(f) = \xi(f\circ i).[/tex]
    I am skeptical about putting a coordinate system on this, though I have tried and I do not make any forward progress. Perhaps this is obvious, but I would greatly appreciate any insight.
     
  2. jcsd
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