# Abelian Lie group implies abelian Lie algebra

Here is the exercise from Lee's Introduction to smooth manifold 8-25

Prove that if $$G$$ is an abelian Lie group, then $$Lie(G)$$ is abelian. [Hint: show that the inversion map $$i:G\rightarrow G$$ is a group homomorphism, and use $$di_e: T_eG\rightarrow T_eG$$ is given by $$di_e(X)=-X$$.]

where $$Lie(G)$$ is defined as all left-invariant vector fields. I don't know how to get started on this. Does anyone know why the hint helps?

Since $$G$$ abelian, then $$i$$ is a Lie group homomorphism and hence $$i_* : \text{Lie}(G) \to \text{Lie}(G)$$ is a Lie algebra homomorphism, where for each $$X \in \text{Lie}(G)$$, $$i_*X \in \text{Lie}(G)$$ is a $$i$$-related vector field to $$X$$. By the second hint, we can show that for any $$X \in \text{Lie}(G)$$,
$$i_*X = -X.$$ Hence for any $$X,Y \in \text{Lie}(G)$$, $$-[X,Y] = i_*[X,Y] =\cdots= [X,Y] \implies [X,Y] = 0.$$
• Why is $i_*X = -X$? I can't figure out how to show this from $(i_*X)_e = -X_e$ which is what we get from using the hint. Is $i_*X$ left invariant as well? 8-24(b) shows that it is right invariant, so I'm not sure if it is left invariant as well. Feb 12 at 6:38
• @nomadicmathematician I haven't touch math for loong time. But fortunately I've still have my notes with me. Seems like I used Theorem 8.44, that is use the fact that $i_*X$ is $i-$related to X. So you can start with this $i_*X|_g = (di_e(X_e))^{\text{L}}|_g$. Cheers! Feb 12 at 21:52