# Analog of the Lie Product formula for commutators

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.

• Do you want to assume anything about the ambient Lie group in which exponentials are valued? E.g., is it compact, simply connected, …? – LSpice Aug 14 at 0:28
• @LSpice for concreteness, let's say the ambient Lie group is SU(N), although I am willing to consider any examples. – hwlin Aug 14 at 0:40
• Of course, upon replacing $X$ by $u X$ (which doesn't change anything else about the problem), it suffices to show that $e^{[X, Y]}$ lies in $G$. It might be interesting work with the root-space decomposition with respect to a torus to which $X$ is tangent (which, by conjugation, we may assume to consist of diagonal matrices if it is convenient), but I can't immediately do anything with that idea. – LSpice Aug 14 at 1:05

The formula $$e^{u[X,Y]}=\lim\limits_{N\to\infty}\left(e^{\sqrt{u/N}X}e^{\sqrt{u/N}Y}e^{-\sqrt{u/N}X}e^{-\sqrt{u/N}Y}\right)^N$$ (Godement 2017, p. 70) shows $$e^{u[X,Y]}\in G$$ always, and seems sufficiently “like” Lie’s to be what you want.
• Indeed, it's almost using the informal idea "we can compute $e^{u[X, Y]}$ for infinitesimal $u$" to compute $e^{u[X, Y]}$ for standard $u$. (Maybe some NSA can make this precise, but, at least as a conceptual explanation, it makes sense to me.) – LSpice Aug 14 at 2:01