# 'Lie correspondence' for formal power series in non-commuting indeterminates

This is related to an earlier question of mine. I would like an argument or a reference (or a missing hypothesis if needed) for the following.

Let $$\mathbb{F}\langle\langle \alpha\rangle\rangle$$ and $$\mathbb{F}\langle\langle \beta\rangle\rangle$$ be the algebras of formal power series in the given sets of non-commuting indeterminates over a field $$\mathbb{F}$$ of characteristic zero. Let $$\mathfrak{g}$$ and $$\mathfrak{h}$$ be the respective Lie subalgebras with zero constant term. Let $$G$$ and $$H$$ be the subgroups of the groups of units of the respective algebras consisting of the elements with constant term equal to 1. (So $$\log$$ maps $$G$$ bijectively onto $$\mathfrak{g}$$ and similarly for $$H$$.) Let $$\Phi:G\to H$$ be a homomorphism, and set $$\phi=\log\Phi\exp$$. Assume that $$\phi(aX)=a\phi(X)$$ for all $$a$$ and $$X$$. Then $$\phi(X+Y)=\phi(X)+\phi(Y)$$.

Let's recall the set-up in the classical case of matrix Lie groups and Lie algebras. Under suitable hypotheses there is a 1-1 correspondence between Lie group homomorphisms $$\Phi:G\to H$$ and Lie algebra homomorphisms $$\phi:\mathfrak{g}\to\mathfrak{h}$$ where $$\mathfrak{g}$$ is the Lie algebra corresponding to $$G$$, and similarly for $$\mathfrak{h}$$: they are related via $$\exp\phi=\Phi\exp$$. The proof that a Lie algebra homomorphism $$\phi$$ gives rise to a Lie group homomorphism $$\Phi$$ is fairly neat. A key step is to show that $$\log(\exp(\phi(X))\exp(\phi(Y)))=\phi(\log(\exp(X)\exp(Y)))$$, which is a consequence of the Baker-Campbell-Hausdorff formula.

To show that $$\Phi$$ gives rise to a suitable $$\phi$$, the key step this time is to show that $$\phi$$ is additive which reduces to the identity

$$\Phi\exp(\log g + \log h)=\exp(\log\Phi(g)+\log\Phi(h))$$,

but this seems to be trickier to prove in the situation I'm interested in. Even if one inverts the BCH formula (see this question and answer), the required identity is not obvious to me. (In the notation of that answer, we would need the identity $$\Phi A(g-1,h-1) = A(\Phi(g)-1,\Phi(h)-1)$$.)

The proof in Brian C. Hall's book to handle the classical case uses $$\left(e^{\frac{X}{m}}e^{\frac{Y}{m}}\right)^m$$ as an approximation to $$e^{X+Y}$$ and then involves taking $$\lim_{m\to\infty}$$, but unfortunately taking such limits doesn't seem to be valid in my situation. This could be remedied by assuming $$\mathbb{F}$$ is an ordered field and that $$\Phi$$ preserves limits in an appropriate sense, but I would like to avoid making extra assumptions if possible.

• I wouldn't discount the $\lim_{m \to \infty} \left(e^{X/m} e^{Y/m}\right)^m = e^{X+Y}$ argument all too quickly. I'm pretty sure that each single coefficient (in front of a fixed word) of $\lim_{m \to \infty} \left(e^{X/m} e^{Y/m}\right)^m$ is a rational function in $m$, and thus its limit as $m \to \infty$ is defined algebraically, by replacing $m$ by $1/m$ and substituting $0$ into the (reduced) fraction. Is this limit not equal to the corresponding coefficient of $e^{X+Y}$? (The question is not rhetorical.) Commented Jan 1, 2022 at 17:38
• @darijgrinberg I think the limit is the corresponding coefficient of $e^{X+Y}$ as you say, but I don’t see why this limit is respected by $\Phi$. Commented Jan 1, 2022 at 23:28
• Oh, good point. Commented Jan 1, 2022 at 23:58

I can provide a very partial answer to the problem in the case your field $$\mathbb{F}$$ is either the reals or the complex numbers. Then the result you are after is a consequence of the group of formal power series you are looking for actually forming an infinite-dimensional Lie group which is regular in the sense of Milnor (actually even $$C^0$$-regular). The key observation here is that the groups are character groups of suitable Hopf algebras (of non-commutative polynomials), whence the relevant Lie theory for these objects was developed in the paper Character groups of Hopf algebras as infinite-dimensional Lie groups.