TITLE: Analog of the Lie Product formula for commutators
QUESTION [6 upvotes]: Let $X, Y$ be elements of a Lie algebra. Consider the group $G$ generated by (limits of) arbitrary products of the elements
$$ G = \langle{e^{tX},e^{sY}\rangle}$$
for all $t,s$. The Lie product formula tells us that $e^{u(X+Y)} \in G$. What about the commutator? E.g., for all $u$, what conditions are necessary so that
$$ e^{u [X,Y]} \in G$$
Clearly we can achieve this for ``infinitesimal'' $u$ by considering something like $e^{t X} e^{t Y} e^{-t X} e^{-t Y}$ for infinitesimal $t$. But what about for finite $u$? Is there a formula like
$$ e^{u [X,Y]} = \mathcal{P} e^{\int X t' + Y s' d\tau  }$$
for some $t(\tau), s(\tau)$? If so, is there an explicit formula for $t$ and $s$ in terms of $u$? I know of such a formula for SU(2), but I do not know how general group.

REPLY [7 votes]: The formula
$$
e^{u[X,Y]}=\lim\limits_{n\to\infty}\left(e^{\sqrt{u/n}\,X}e^{\sqrt{u/n}\,Y}e^{-\sqrt{u/n}\,X}e^{-\sqrt{u/n}\,Y}\right)^n
\tag{$*$}
$$
(Godement 2017, p. 70) shows that $e^{u[X,Y]}\in G$ always, and seems sufficiently “like” “Lie’s” to be what you want.
Note added: The origin of this formula was left unresolved before. Howe (1983) and Godement (originally 1982) point to von Neumann (1929, §II.3) who proves $e^{u[X,Y]}\in G$, but apparently not $(*)$. Formula $(*)$ is explicit in e.g. Chorin et al. (1978, p. 207), Goldstein (1970), Nelson (1969, p. 111), Goto (1969, p. 159), Hausner-Schwartz (1968, p. 78), Cohn (1957, p. 112), and somewhat implicit in Gluškov (1957, p. 137), Yamabe (1950, p. 14), Yosida (1936, Remark p. 470). Almost none of whom cite each other!