# In the context of Lie algebras, why does the “diagonal map” commute with the exponential?

I'm referring to Serre's lecture notes "Lie Groups and Lie Algebras".

Let $k$ be a field of characteristic 0. Let $X$ be a set (of generators), and let $Ass_X$ be the free associative algebra on $X$. This is just the tensor algebra of the vector space with basis $X$. Let $Ass^n_X$ be the $n$th graded summand consisting of the degree $n$ tensors.

He defines $\widehat{Ass_X}$ as the product $\widehat{Ass_X} := \prod_{n=0}^\infty Ass_X^n$, and defines the completed tensor product as $$\widehat{Ass_X}\hat{\otimes}\widehat{Ass_X} := \prod_{p,q}Ass_X^p\otimes Ass_X^q$$

There is a diagonal map $\Delta : Ass_X\rightarrow Ass_X\otimes Ass_X$ coming from the identification of $Ass_X$ as the universal algebra of the free lie algebra $L_X$ on $X$, which is induced by the usual diagonal map $L_X\rightarrow L_X\oplus L_X$ given by $x\mapsto (x,x)$, and $L_X$ can be identified with the Lie subalgebra of $Ass_X$ given by the elements $a\in Ass_X$ on which $\Delta(a) = a\otimes 1 + 1\otimes a$.

Let $\hat{\mathfrak{m}}$ be the ideal of $\widehat{Ass_X}$ generated by $X$.

In the proof of Corollary 7.3 in Chapter IV (Free Lie Algebras), Serre says that the exponential map given by $x\mapsto \sum_{n\ge 0}\frac{x^n}{n!}$ commutes with $\Delta$, at least when restricted to $\hat{\mathfrak{m}}\cap L_X$, but doesn't give a proof.

How would one argue that $\Delta$ commutes with $\exp$?

• If I understand your question correctly: The map $\Delta$ is an algebra homomorphism, and thus commutes with any polynomial. If it is also continuous, then it commutes with power series as well, and $\exp$ is power series. I'll let you figure out what "continuous" means here. – darij grinberg Sep 3 '17 at 19:56
• @darijgrinberg Ah! Of course, that's obvious now that I think about it. I'm still rather new to this stuff so I think I just got a bit dazed from the large number of new objects being introduced. – oxeimon Sep 4 '17 at 0:47