The Lie exponential map and commuting elements

Question

By Baker–Campbell–Hausdorff formula, if $[v,w]=0$ for $v,w$ in the Lie algebra $\mathfrak g$ of a Lie group $G$ then $\exp(v)$ and $\exp(w)$ commute in $G$.

Does anyone know a reference or a method of proof of the following partial inverse: For $v,w$ sufficiently close to $0$ in $\mathfrak g$, $\exp(v)\exp(w)=\exp(w)\exp(v)$ implies $[v,w]=0$?

Being sufficiently close to zero is of course necessary here.

Answer 1

There is something similar: For connected Lie group G,

$$X, Y \in \mathfrak g \text{ commute if and only if } \exp(tX) \exp(sY) = \exp(sY) \exp(tX) \text{ for all $s,t \in \mathbb R$.} $$

I do not know direct proof of the converse, but it follow directly from a theorem in smooth manifold theory, which says that smooth vector fields commute iff their flows commute. You can read about it in Lee's Introduction to smooth manifolds.

Answer 2

Let $W$ denote a neighbourhood of $0$ such that for $X \in W$ the logarithm of $e^{\mathrm{ad}X}$ is well-defined, then $e^{\mathrm{ad}X}Y=Y$ implies $[X, Y] = 0$. Let $U$ be a neighbourhood of $0$ such that $\mathrm{exp}$ is injective on $U$. Let $V$ be a neighbourhood of $0$ such that for $X, Y \in V$ we have $\mathrm{Ad}(\mathrm{exp}(X))(Y) \in U$ and $Y \in U$. Then $V \cap W$ has the wanted property.

Because: $\mathrm{exp}(Y) = \mathrm{exp}(X)\mathrm{exp}(Y)\mathrm{exp}(X)^{-1} = \mathrm{exp}(\mathrm{Ad}(\mathrm{exp}(X))Y)$. Which implies $Y = \mathrm{Ad}(\mathrm{exp}(X))Y = e^{\mathrm{ad}X}Y$.

Answer 3

There exists the following partial inverse, though this is not as strong as the statement you require. This is related to exercise 9.1 in the book of Fulton and Harris on representation theory. Let $G$ be a Lie group and $\mathfrak{g}$ its Lie algebra. If $X \in \mathfrak{g}$ is small enough and satisfies $[X,Y] = 0$ for all $Y \in \mathfrak{g}$, then $\exp(X)$ commutes with all elements in a neighbourhood of the identity in $G$ by the BCH-formula. On the other hand, if $\exp(X)$ commutes with all elements of $G$, then $1 = Ad(\exp(X))$. Furthermore by naturality of the exponential map we then must have $1 = Ad(\exp(X)) = \exp(ad(X))$. However, if $X$ is small enough this implies $ad(X) = 0$, i.e. $[X,Y] = 0$ for all $Y \in \mathfrak{g}$.