Given a right and a left invariant vector field $R_A$ and $L_A$ with $L_A(e)=A=R_A$ on a Lie group $G$, do the vector fields generally also agree on other points of the Lie group?
In particular, I'm thinking of the 1-parameter subgroup $e^{t A}$.
- For matrices, I believe that we have $R_A(e^{tA})=e^{tA}A$ and $L_A(e^{tA})=Ae^{tA}$, implying $R_A(e^{tA})=L_A(e^{tA})$ due to $e^{tA}A=Ae^{tA}$.
Is this true?
Suppose that $G$ is a Lie group whose Lie algebra is ${\cal G}$, for every $g\in G, A\in {\cal G}$, we denote by $l_g(x)=gx, r_g(x)=xg$, the left-invariant vector field $L_A$ defined by $A$ is the vector field such that $L_A(e)=A$ and for every $g\in G, l_g^*(L_A)(x)={dl_g}_{g^{-1}(x)}L_A(g^{-1}(x))=L_A(x)$. In particular if $x=g$, we obtain ${dl_g}_e(A)=L_A(g)$. Similarly, we have ${dr_g}_e(A)=R_A(g)$.
Suppose that $L_A(g)=R_A(g)$, this is equivalent to saying that ${dl_g}_e(A)={drg}_e(A)$, we deduce that $Ad(g)_e(A)=A$.