# 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
 P: 219 If X is a left invariant vector field, then $L_x \circ x_t = x_t \circ L_x$, 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 $x_t = L_x \circ x_t \circ L_x^{-1}$ by showing that $L_x \circ x_t \circ L_x^{-1}$ is the flow of $dL_x \circ X \circ L_x^{-1}$. So i want to show that $\frac{\partial}{\partial t} (L_x \circ x_t \circ L_x^{-1}) = dL_x \circ X \circ L_x^{-1}$. I am having some trouble with the chain rule and was wondering if someone could help me out. so far, i have $\frac{\partial}{\partial t}(L_x \circ x_t \circ L_x^{-1}) = d(L_x \circ x_t \circ L_x^{-1})(\frac{\partial}{\partial t}) = (dL_x \circ dx_t \circ dL_x^{-1})(\frac{\partial}{\partial t})$ but i am not sure how to simplify this. i know of 2 different ways to manipulate tangent vectors. i have tried $(dL_x^{-1})(\frac{\partial}{\partial t})=(L_x^{-1} \circ \gamma)′$ where γ is the curve that goes through the vector $\frac{\partial}{\partial t}$, but that doesn't seem to get me to where i want. i also tried using the fact that $(dL_x^{-1})(\frac{\partial}{\partial t})f = \frac{\partial}{\partial t}(f \circ L_x^{-1})$ so $(dL_x \circ dx_t \circ dL_x^{-1})(\frac{\partial}{\partial t}) = dL_x \circ X \circ (f \circ L_x^{-1})$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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