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Let $k$ be a field of characteristic $0$ and let $\mathfrak{g}$ be a finite dimensional Lie algebra over $k$. $\mathfrak{g}$ corresponds to a formal group scheme $\mathcal{G} = \text{Spf} (U(\mathfrak{g})^{*})$. In fact, there is an equivalence of categories between finite-dimensional Lie algebras and infinitesimal formal group schemes over $k$ with finite-dimensional tangent space. This is elaborated on here by Akhil Mathew.

There should be an exponential map $\text{exp} : \mathfrak{g} \rightarrow \mathcal{G}$. This is mentioned at the end of the blog post above, but I can't figure out how to construct it explicitly.

Any help on this would be much appreciated.

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    $\begingroup$ You didn't specify that your field k has characteristic zero, but to get this equivalence, it is essential that your field k have characteristic zero. Part of the whole point of formal groups is that, over fields of positive characteristic, there are more formal groups than Lie algebras, so the formal group of an abelian variety remembers more structure than the Lie algebra! $\endgroup$ – user126192 Jul 25 '18 at 14:08
  • $\begingroup$ Oh- that's right. I see I overlooked that requirement. $\endgroup$ – Dean Young Jul 25 '18 at 22:38
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$\newcommand{\g}{\mathfrak{g}}$ In a way this is tautological, in the sense that $\mathcal G$ can be seen as a formal exponentation of $\g$, although I don't think there is an actual exponential map from one to the other in general. Rather, there is a map (in fact an isomorphism of formal schemes) from the formal completion of $\g$ at the origin to $\mathcal G$.

Namely, the dual $U(\g)^*$ can be identified as a (topological) algebra with the algebra $\widehat S(\g^*)$ of formal power series on $\g^*$, ie with the formal completion of $\mathcal O(\g)$ w.r.t to the ideal of function vanishing at the origin.

This identification, should indeed be thought as pulling back functions through the exponential map: indeed, the induced coproduct on $\widehat S(\g^*)$ is expressed through the Baker-Campbell-Hausdorff formula $$\forall x,y \in \g, \Delta(f)(x\otimes y)=f(BCH(x,y))$$ where BCH is the Lie formal power series defined on formal variables $a,b$ by $$BCH(a,b):=\log(e^ae^b).$$

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  • $\begingroup$ Thanks. What is $S(-)$, here? $\endgroup$ – Dean Young Jul 25 '18 at 13:20
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    $\begingroup$ Sorry, I meant the symmetric/polynomial algebra. $\endgroup$ – Adrien Jul 25 '18 at 13:21
  • $\begingroup$ If it's not too much extra, how do I make the identification between the space of formal exponentials and the space $\text{Spf} (U(\mathfrak{g})^*)$? $\endgroup$ – Dean Young Jul 25 '18 at 22:37
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    $\begingroup$ Sorry, that part was somewhat wrong, and the other confusing, I edited my answer. Hope it's clearer now. $\endgroup$ – Adrien Jul 26 '18 at 14:27
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    $\begingroup$ Indeed, $U(\mathfrak g)$ is a cocommutative Hopf algebra, hence its full linear dual is a commutative, topological Hopf algebra, the multiplication being the transpose of the coproduct. $\endgroup$ – Adrien Jul 26 '18 at 14:55

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