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Is the Lie Algebra of a connected abelian group abelian? I guess that this should be true, but how do you prove it?

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  • $\begingroup$ "connected" is only relevant for the converse: if a connected group has abelian Lie algebra then that group is abelian $\endgroup$ – user399601 Feb 2 '17 at 18:18
  • $\begingroup$ How can you show that the converse is also true? $\endgroup$ – koch Feb 2 '17 at 18:28
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    $\begingroup$ @koch If $G$ is connected then it is generated by $exp(\mathfrak{g})$, and $e^X e^Y e^{-X} = exp(e^{ad X} Y) = exp(Y)$ for all $X,Y$ since $ad = 0$. $\endgroup$ – user399601 Feb 2 '17 at 18:33
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    $\begingroup$ In what sense is this question "off-topic"? $\endgroup$ – geometricK Jan 18 '20 at 3:20
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Yes, and connectedness is not necessary. I know three proofs:

Proof 1

When $G$ is abelian, the inverse map $$i:G\to G,\quad g\mapsto g^{-1}$$ is a group homomorphism. Hence, its differential at $1\in G$ $$di_1:\mathfrak{g}\to\mathfrak{g},\quad X\mapsto -X$$ is a Lie algebra homomorphism. But then $$-[X,Y]=di_1([X,Y])=[di_1(X),di_1(Y)]=[-X,-Y]=[X,Y],$$ so $[X,Y]=0$.

Proof 2

For any Lie group $G$, the differential at $1$ of the map $\mathrm{Ad}:G\to GL(\mathfrak{g})$ is $\mathrm{ad}:\mathfrak{g}\to\mathrm{End}(\mathfrak{g})$ where $\mathrm{ad}(X)(Y)=[X,Y]$. But when $G$ is abelian, $\mathrm{Ad}$ is the constant map to the identity (since $\mathrm{Ad}(g)$ is the differential of the map $G\to G,a\mapsto gag^{-1}$ which is the identity when $G$ is abelian), so $\mathrm{ad}=0$.

Proof 3

For any Lie group $G$ we have that for $X,Y\in\mathfrak{g}$, $$\exp(sX)\exp(tY)=\exp(tY)\exp(sX),\forall s,t\in\mathbb{R}\quad\iff[X,Y]=0.$$ If $G$ is abelian, the left-hand side always hold, so $[X,Y]=0$ for all $X,Y\in\mathfrak{g}$.

Remark about the converse

The last proof can be used to prove the converse when $G$ is connected. This is because $\exp$ restricts to a diffeomorphism from a neighborhood of $0$ in $\mathfrak{g}$ to a neighborhood of the identity in $G$ and a connected group is generated by any neighborhood of the identity.

However, connectedness is necessary for the converse. For example, if $T$ is any abelian connected Lie group and $H$ is any non-abelian finite group, then $G=T\times H$ is a non-abelian Lie group with abelian Lie algebra.

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  • $\begingroup$ How do you show that the converse is true? $\endgroup$ – koch Feb 2 '17 at 18:28
  • $\begingroup$ @koch See the remark. $\endgroup$ – Spenser Feb 2 '17 at 18:35
  • $\begingroup$ @Spenser Hi, hope it's not too late to ask. In Proof 1, how did you calculate the differential of the inverse map for an arbitrary Lie group G? $\endgroup$ – Khal Jul 19 '20 at 14:51

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