This is problem 8-25 from John Lee's Introduction to Smooth Manifolds.
Prove that if $G$ is an abelian Lie group, then $Lie(G)$, the Lie algebra of the Lie group $G$ is abelian. [Hint: show that the inversion map $i:G \to G$ is a group homomorphism and use Problem 7-2.]
Problem 7-2 gives that $di_e (X_e)= -X_e$ for $X_e \in T_e G$.
So I have $(i_* X)_e = di_e (X_e)=-X_e$.
It seems like the proof suggests $[X,Y]=[-X,-Y]=[i_* X,i_* Y]=i_*[X,Y]=-[X,Y]$ and conclude that $[X,Y]=0$ for all smooth left invariant vector fields $X,Y$ in $G$.
However, this requires that $(i_* X)_g = -X_g$ for all $g\in G$. But I can't see how this can be obtained from the equality on the identity. From Problem 8-24(b) I know that $i_* X$ is right-invariant for a left invariant $X$, but not left invariant. So I cannot apply $dL_g$ on both sides to get $(i_* X)_g = -X_g$. I've been stuck on this step for a while. I would greatly appreciate any help.
Best Answer
If $G$ is abelian, then $i$ is a Lie group homomorphism. So from Teorema 8.44 in Lee's book that $i_* : LieG \to LieG$ is a Lie algebra homomorphism. So for any $X \in LieG$, $i_*X \in LieG$ is vector field that is $i$-related to $X$. So by the help of Problem 7-2, for any $X \in LieG$, $$ i_*X|_g = (di_e(X_e))^{\text{L}}|_g = d(L_g)_e (-X_e) = -X_g \implies i_*X = -X. $$ So $\forall X,Y \in LieG$, $$ -[X,Y] = i_*[X,Y] = [i_*X,i_*Y] = [-X,-Y] = [X,Y] \implies [X,Y] = 0. $$