Prove $\exp(x)\exp(y)=\exp(y)\exp(x)$ if $[x,y]=0$.

Question

How to prove the following statement

Prove that $\exp(x)\exp(y)=\exp(y)\exp(x)$ if $x,y\in \mathfrak{g}$ and $[x,y]=0$.

by using the following results:

1. $\mathrm{ad} (x)\cdot y=[x,y]$;
2. $\mathrm{Ad} (\exp(x))=\exp(\mathrm{ad}x)$.

where $\mathrm{Ad}:G\rightarrow \mathrm{Gl}(\mathfrak{g})$ and $\mathrm{ad}:=\mathrm{Ad}_{\ast}:~\mathfrak{g}\rightarrow \mathfrak{gl}(\mathfrak{g})$.

(Without using the Campbell-Hausdorff formula...)

---

Attempt:

If $[x,y]=0$, then $\mathrm{ad}(x)\mathrm{ad}(y)=\mathrm{ad}(y)\mathrm{ad}(x)$.

Also, we have $\mathrm{Ad}(\exp(xy))=1$. But I cannot move forward.

Answer 1

I use $e^x = \exp(x)$. Note that

$$e^x e^y = e^y e^x$$

is the same as $c(e^y)(e^x) = e^x$, where $c(g) : G \to G$ is the conjugation $c(g) h = g^{-1} hg$. Note that by definition $Ad(g) = c(g)_* :\mathfrak g \to \mathfrak g$.

Note from (1) and (2)

\begin{equation} \begin{split} Ad(e^y) (x) &= \operatorname{Exp}(ad (y)) (x)\\ &= x + ad(y)x + \frac{1}{2} ad(y)^2 x + \frac{1}{3!} ad(y)^3 x + \cdots \\ &= x \end{split} \end{equation}

since $ad(y) x = 0$.

Write $g = e^y$. From $Ad(g) x = x$ we want to integrate to say that $g^{-1} e^x g = e^x$. Consider

$$\gamma(t) = g^{-1} e^{tx} g,$$

Note that by definition of $Ad$, $\gamma'(0) = Ad(g)x = x$.

Also $\gamma(0) = e_G$ and $\gamma(t_1+ t_2) = \gamma(t_1)\gamma(t_2)$. In particular, $$\gamma'(t) = \gamma(t)_* \gamma'(0) = \gamma(t)_* x.$$ Thus $\gamma(t)$ and $e^{tx}$ are both the integral curve of the vector fields $X (g)= g_* x$ (just the left invariant vector fields defined by $x\in \mathfrak g$) which start at the identity $e_G$. Thus $$ \gamma(t) = g^{-1} e^{tx} g = e^{tx}$$

for all $t$ and in particular $g, e^x$ commutes.