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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?
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?