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

[demonelite123]

If X is a left invariant vector field, then $L_x \circ x_t = x_t \circ L_x$, where x_t is the flow of X and L_x 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.

 

[pat_connell]

thank you! The solution is as follows: \frac{\partial}{\partial t}(L_x \circ x_t \circ L_x^{-1}) = (dL_x \circ dx_t \circ dL_x^{-1})(\frac{\partial}{\partial t}) = dL_x \circ X \circ (L_x^{-1} \circ \gamma)′ = dL_x \circ X \circ \frac{\partial}{\partial t}(L_x^{-1}) = dL_x \circ X \circ (-dL_x^{-1})(\frac{\partial}{\partial t}) = dL_x \circ X \circ L_x^{-1}