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If X is a left invariant vector field, then L_x o x_t = x_t o L_x

oblixps

Member
May 20, 2012
38
Let [tex] L_x: G \rightarrow G [/tex] be the left multiplication map that sends any point y to xy. Let [tex] x_t [/tex] be the flow of the left invariant vector field X. I want to show that [tex] L_x \circ x_t = x_t \circ L_x [/tex].

I let q be any point in G, and let [tex]x_{t}(q) = y [/tex]. Also, [tex] L_{x}(y) = xy [/tex]. So I have [tex] L_x \circ x_{t} (q) = xy [/tex] and i want to show that [tex] x_t \circ L_{x} (q) = x_{t}(xq) = xy [/tex].

i'm having trouble showing the last statement that [tex] x_{t}(xq) = xy [/tex]. Since X is a left invariant vector field, i know that [tex] dL_{x}X_{q} = X_{xq} [/tex] and [tex] \frac{\partial x_t(q)}{\partial t} = X_{q} [/tex] since x_t is the flow of X. i can't figure out how to use these to get to my desired result however.

would someone mind offering hints on how to proceed?
 

oblixps

Member
May 20, 2012
38
I came across the hint that equivalently, i should show that [tex] x_t = L_x \circ x_t \circ L_{x}^{-1} [/tex] by showing that [tex] L_x \circ x_t \circ L_{x}^{-1} [/tex] is the flow of [tex] dL_x \circ X \circ L_{x}^{-1} [/tex]. So i want to show that [tex] \frac{\partial}{\partial t} (L_x \circ x_t \circ L_{x}^{-1}) = dL_x \circ X \circ L_{x}^{-1} [/tex].

I am having a little trouble with the chain rule and was wondering if someone could help out. so far, [tex] \frac{\partial}{\partial t} (L_x \circ x_t \circ L_{x}^{-1}) = d(L_x \circ x_t \circ L_{x}^{-1})(\frac{\partial}{\partial t}) = (dL_x \circ dx_t \circ dL_{x}^{-1})(\frac{\partial}{\partial t}) [/tex] but i am not sure how to simplify this.

i have tried [tex] (dL_{x}^{-1})(\frac{\partial}{\partial t}) = (L_{x}^{-1} \circ \gamma)' [/tex] where [tex] \gamma [/tex] is the curve that goes through the vector [tex] \frac{\partial}{\partial t} [/tex]. i also tried using the fact that [tex] (dL_{x}^{-1})(\frac{\partial}{\partial t})f = \frac{\partial}{\partial t}(f \circ L_{x}^{-1}) [/tex] so [tex] (dL_x \circ dx_t \circ dL_{x}^{-1})(\frac{\partial}{\partial t}) = dL_x \circ X \circ (f \circ L_{x}^{-1}) [/tex] but now i have an f in there i can't get rid of.

i am still a beginner on differential geometry so these manipulations are relatively new to me. any help on this would be greatly appreciated.