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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?

Thanks in advance!

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  • $\begingroup$ You already know how to start based on the hints. What is your work so far? $\endgroup$ Oct 25, 2020 at 13:53
  • $\begingroup$ @SiKucing I try to use coordinate formula for Lie bracket to compute directly. But the computation is a bit lengthy and seems does not work... (Basically I just don't know how to use this hint...) $\endgroup$
    – Zorualyh
    Oct 25, 2020 at 13:57
  • $\begingroup$ Do not use coordinates for this one. I leave some sketch in my answer. $\endgroup$ Oct 25, 2020 at 14:07

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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 $X,Y \in \text{Lie}(G)$, $$ -[X,Y] = i_*[X,Y] =\cdots= [X,Y] \implies [X,Y] = 0. $$

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  • $\begingroup$ That's excellent, thanks! $\endgroup$
    – Zorualyh
    Oct 25, 2020 at 14:10
  • $\begingroup$ 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. $\endgroup$ Feb 12 at 6:38
  • $\begingroup$ @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! $\endgroup$ Feb 12 at 21:52

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