Does $e^Xe^Y = e^Ye^X$ iff $[X,Y]=0$ hold once we are sufficiently close to the identity of a Lie group?

Question

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Must there exist a neighbourhood $U \subseteq \mathfrak{g}$ of $0$ such that, for all $X,Y \in U$, the implication $$e^Xe^Y = e^Ye^X \Rightarrow [X,Y]=0$$ is valid?

I don't have any formal education in Lie groups and algebras, self-administered or otherwise, so, while I often have guesses as to what sorts of things should be true, I rarely know how to go about proving them. My only idea went something like this:

• Word on the street is that, locally, the product structure of Lie group can be described by analytic functions on the Lie algebra.
• We should be able to view $\{(X,Y): e^Xe^Y=e^Ye^X\}$ as the zero set of some analytic function.
• The zero set of a (multivariable) analytic function probably can't look that bad
• ....?

Answer 1

Yes, it's true. Indeed, we can locally write the BCH formula as $$\log(e^xe^y)=x+y+\frac12[x,y]+Q(x,y,[x,y]).$$ Here $Q$ is defined as follows: the remainder (sum of terms of order $\ge 3$ in BCH) is written as $\sum_i \lambda_iP_i(x,y)([x,y])$, where $P_i$ is some nonempty finite product of the operators $\mathrm{ad}(x)$ and $\mathrm{ad}(y)$, and $Q(x,y,z)=\sum \lambda_iP_i(x,y)(z)$. In particular we can bound around zero $\|Q(x,y,z)\|\le C\max(\|x\|,\|y\|)\|z\|$ for a suitable constant (and choice of norms).

If $e^xe^y=e^ye^x$ for small $x,y$, this yields
$$x+y+\frac12[x,y]+Q(x,y,[x,y])=y+x+\frac12[y,x]+Q(y,x,[y,x]),$$ that is, $$[x,y]=Q(y,x,-[x,y])-Q(x,y,[x,y]).$$

This yields $\|[x,y]\|\le 2C\max(\|x\|,\|y\|)\|[x,y]\|$. If $[x,y]\neq 0$, this yields $1\le 2C\max(\|x\|,\|y\|)$, which is a contradiction if $(x,y)$ is too close to $(0,0)$.

Answer 2

It is enough to prove the result for $X$, $Y$ $n\times n$ complex matrices.

Now, we have the following formula:

$$e^X e^Y e^{-X}= \exp( Ad(e^X)(Y))$$ where

$$Ad(e^X)(Y) = \exp( ad(X)) (Y)$$ where again the operator $ad(X)$ acts on $n\times n$ matrices by $$Z \mapsto ad(X) Z \colon = [X,Z]$$ and $\exp ad(X)$ is the exponential of this linear map from $M_n(\mathbb{C})$ to itself.

Now we have to show that $\exp (T)(Y) = Y$ implies $T Y = 0$, if $T$ is small enough. In fact we have the following:

Let $T$ be a linear map on a finite dimensional (complex) vector space $V$ and $v$ in $V$ so that $\exp T v = v$. Assume moreover that $T$ has no eigenvalues $\lambda = 2 k \pi i$, $k \ne 0$. Then $T v = 0$. Hint: consider the Jordan canonical form of $T$.

$\bf{Added:}$ You can also show the last statement is true for $T$ small enough using the formula

$$\log ( I + ( e^T - I)) = T$$ where $$\log ( I + S) = S - \frac{S^2}{2} + \frac{S^3}{3} + \cdots$$