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

by demonelite123
Tags: field, invariant, vector
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Nov22-12, 05:46 PM
P: 219
If X is a left invariant vector field, then [itex] L_x \circ x_t = x_t \circ L_x [/itex], where xt is the flow of X and Lx is the left translation map of the lie group G.

In order to show this, I am trying to show that [itex] x_t = L_x \circ x_t \circ L_x^{-1} [/itex] by showing that [itex] L_x \circ x_t \circ L_x^{-1} [/itex] is the flow of [itex] dL_x \circ X \circ L_x^{-1} [/itex]. So i want to show that [itex] \frac{\partial}{\partial t} (L_x \circ x_t \circ L_x^{-1}) = dL_x \circ X \circ L_x^{-1}[/itex].

I am having some trouble with the chain rule and was wondering if someone could help me out. so far, i have [itex] \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}) [/itex] but i am not sure how to simplify this.

i know of 2 different ways to manipulate tangent vectors. i have tried [itex] (dL_x^{-1})(\frac{\partial}{\partial t})=(L_x^{-1} \circ \gamma)′ [/itex] where γ is the curve that goes through the vector [itex] \frac{\partial}{\partial t}[/itex], but that doesn't seem to get me to where i want. i also tried using the fact that [itex] (dL_x^{-1})(\frac{\partial}{\partial t})f = \frac{\partial}{\partial t}(f \circ L_x^{-1})[/itex] so [itex](dL_x \circ dx_t \circ dL_x^{-1})(\frac{\partial}{\partial t}) = dL_x \circ X \circ (f \circ L_x^{-1}) [/itex]but now i have an f in there i can't get rid of.

any help on this would be greatly appreciated.
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