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Let $G$ be a Lie group, $\mathfrak{g}$ it's Lie algebra. Assume $[x,y]=0 \, \, \forall x,y \in \mathfrak{g}$. Is it true that $G$ is abelian?

Remarks: (1) The other direction ($G$ abelian $\Rightarrow \mathfrak{g}$ abelian) is a basic result.

(2) For matrix groups, the commutator of the Lie algebra is just the usual matrix commutator. So I think the other direction holds for such groups. (Am I right?)

What about general Lie groups?

An attempted proof:

Maybe we can claim that $exp(x+y)=exp(x)\cdot exp(y)$, so the multiplication in $\text{Image}(exp)$ is commutative.

The image of $exp$ contains an open neighbourhood $U$ of the identity $e $, so there is an open neighbourhood $V$ of $e$ such that $V \cdot V \subseteq U$. Since $V$ generates $G$ we are done.

Is this true?

Dietrich Burde
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Asaf Shachar
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    You need to assume that $G$ is connected, or this is false: $O(2,\mathbb R)$, is a natural example. Or, more artificially, the product of any abelian Lie group and a non-abelian finite group. – paul garrett Nov 06 '15 at 13:58

1 Answers1

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Think about the Lie algebra as invariant vector fields on the group and recall that flowing by $t$ along $X$ corresponds to multiplying by $e^{tX}$. The condition $[X,Y]=0$ means that the flows of $X,Y$ commute, so we have $e^X e^Y = e^Y e^X$. Thus $\exp(\mathfrak g)$ is Abelian, and the rest of your argument shows that the connected component of the identity is Abelian. If the group is not connected then $V$ does not generate $G$.

Anthony Carapetis
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