On equation $exp [X,Y]=[exp X,exp Y]$ on a Lie group and its Lie algebra

lie-algebraslie-groups

What is a classification of all Lie groups $G$ with Lie algebra $\mathfrak{g}$ with the following equality $$exp [X,Y]=[exp X,exp Y]$$
for all $X,Y \in \mathfrak{g}$? Is there a non Abelian example?

Best Answer

Check out the BCH formula: it shows that every 2-step nilpotent Lie group satisfies this (assuming the group commutator is defined as $xyx^{-1}y^{-1}$, or as $x^{-1}y^{-1}xy$).

(Continue with the convention $[x,y]=xyx^{-1}y^{-1}$.)

Actually, a Lie group satisfies this condition ($\exp$ intertwines commutators) if and only if its Lie algebra is 2-step nilpotent.

Indeed, the BCH formula yields, writing $x=\exp(X)$ and $y=\exp(Y)$, the Taylor expansion around zero: $$\log([x,y])=[X,Y]+\frac12[X+Y,[X,Y]]+O((\|X\|+\|Y\|)^4).$$

Hence if the given equality holds, we deduce that $[X+Y,[X,Y]]$ for all $X,Y$. By homogeneity, we deduce $[X,[X,Y]]=0$ for all $X,Y$, and then writing $X=X'+Z'$ we deduce $[X,[Y,Z]]=0$ for all $X,Y,Z$, that is, the Lie algebra is 2-step nilpotent.

Certainly we reach the same conclusion with the commutator convention $x^{-1}y^{-1}xy$. While with the convention $y^{-1}x^{-1}yx$ or $yxy^{-1}x^{-1}$, we obtain by expansion at degree 2 that the given equality implies that the Lie algebra is abelian.

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