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Differentiation Problem on Lie Groups

  1. Sep 20, 2013 #1

    lavinia

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    Suppose θ is a differential 1 form defined on a manifold and with values in the Lie algebra of a Lie group,G.

    On MxG define the 1 form, ad(g)θ ,where θ is extended by letting it be zero on the tangent space to G

    How do you compute the exterior derivative, dad(g)θ ?

    BTW: For matrix Lie groups this is straightforward. What is the abstract calculation?
     
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  3. Oct 7, 2013 #2

    lavinia

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    Here is some computation for matrix groups

    dad(g)θ(x + h, y + k) = (x + h).ad(g)θ(y + k) - (y + k).ad(g)θ(x + h) -ad(g)θ[x + h, y + k]

    where Y and X are tangent to the manifold and h and K are left invariant vector fields.

    Computing:

    dad(g)θ(x + h, y + k) = (x + h).ad(g)θ(y) - (y + k).ad(g)θ(x) -ad(g)θ[x, y]

    = ad(g)dθ(x,y) + h.ad(g)θ(y) - k.ad(g)θ(x)


    h.ad(g)θ(y) = h. gθ(y)g[itex]^{-1}[/itex] = dg(h)θ(y)g[itex]^{-1}[/itex] - gθ(y)g[itex]^{-1}[/itex]dg(h)g[itex]^{-1}[/itex]

    = dg(h)g[itex]^{-1}[/itex]gθ(y)g[itex]^{-1}[/itex] - gθ(y)g[itex]^{-1}[/itex]dg(h)g[itex]^{-1}[/itex]

    = ω(h)ad(g)θ(y) - ad(g)θ(y)ω(h) where ω is the right invariant Maurer-Cartan form.

    So h.ad(g)θ(y) - k.ad(g)θ(x) = ω(h)ad(g)θ(y) - ad(g)θ(y)ω(h) - ω(k)ad(g)θ(x) + ad(g)θ(x)ω(k)

    Simplify?
     
    Last edited: Oct 7, 2013
  4. Oct 7, 2013 #3

    D H

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    It looks like you are writing about the Baker–Campbell–Hausdorff formula, which does have a somewhat simplified representation in the case of SO(n).
     
  5. Oct 7, 2013 #4

    lavinia

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    This doesn't simplify anything but succinctly rewrites the formula as

    ad(g)dθ +[ω,ad(g)θ] where the second term is the Lie bracket of the two differential forms ω and ad(g)θ
     
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