Differentiation Problem on Lie Groups

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  • #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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  • #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?
 
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  • #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).
 
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  • #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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