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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.

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  • $\begingroup$ Do you want to assume anything about the ambient Lie group in which exponentials are valued? E.g., is it compact, simply connected, …? $\endgroup$
    – LSpice
    Commented Aug 14, 2019 at 0:28
  • $\begingroup$ @LSpice for concreteness, let's say the ambient Lie group is SU(N), although I am willing to consider any examples. $\endgroup$
    – hwlin
    Commented Aug 14, 2019 at 0:40
  • $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented Aug 14, 2019 at 1:05
  • $\begingroup$ Yamabe's theorem: every path connected subgroup of a Lie group is an immersed Lie subgroup. This implies that the commutator is in the Lie algebra of that subgroup, but doesn't give the formula. $\endgroup$
    – Ben McKay
    Commented Oct 9, 2019 at 17:55

1 Answer 1

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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!

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    $\begingroup$ 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.) $\endgroup$
    – LSpice
    Commented Aug 14, 2019 at 2:01

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