Differentiation Problem on Lie Groups

[lavinia]

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?

 

[lavinia]

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$^{-1}$ = dg(h)θ(y)g$^{-1}$ - gθ(y)g$^{-1}$dg(h)g$^{-1}$

= dg(h)g$^{-1}$gθ(y)g$^{-1}$ - gθ(y)g$^{-1}$dg(h)g$^{-1}$

= 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?

 

[D H]

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).

 

[lavinia]

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

 

[blue_raver22]

The exterior derivative, dad(g)θ, can be computed using the Lie bracket operator, [ , ], on the Lie algebra of the Lie group G. This is because the exterior derivative of a differential form is defined as the Lie bracket of the form with the exterior derivative of the identity element. In this case, the identity element would be the tangent space of G at the identity element. Therefore, the calculation would involve taking the Lie bracket of the 1-form ad(g)θ with the exterior derivative of the identity element, which can be expressed as [d, ad(g)θ]. This will give the exterior derivative of the 1-form ad(g)θ.

For matrix Lie groups, this calculation is straightforward because the Lie bracket can be computed directly using the matrix operations. However, for abstract Lie groups, the Lie bracket may not have a direct matrix representation. In this case, the calculation would involve using the structure constants of the Lie algebra to compute the Lie bracket and then taking the exterior derivative using the same structure constants. This can be a more involved process, but it ultimately follows the same principles of using the Lie bracket to compute the exterior derivative.