# How is this (Tannakian) de Rham fundamental group calculated?

Edit: I originally asked this question on MSE, but migrated it to MO after a long period of inactivity and a recommendation from another user.

Let $X$ be a complex elliptic curve and $e$ the identity element of $X$. Let $E^\times$ denote the punctured curve $X\backslash\left\{e\right\}$. In their 2007 paper Towards Multiple Elliptic Polylogarithms, Levin and Racinet compute the de Rham fundamental group $\pi_1 ^{dR} (E^\times, e)$. To do this they prove that there is an equivalence of categories (Theorem 1)

$$R\mathsf{-NMod}\xrightarrow{\sim} \mathsf{NConn}(X;e)$$

where:

• $R = \mathbb{C}\langle\langle \mathbf{t}, A\rangle\rangle$ is the Hopf algebra of noncommutative power series in two formal variables $\mathbf{t}$ and $A$. $R$ is also the completed universal enveloping algebra of the free Lie algebra $\mathbb{L}(\mathbf{t}, A)$ generated by $\mathbf{t}$ and $A$. The coproduct in $R$ is given by

$$\Delta(\mathbf{t}) = \mathbf{t}\otimes 1 + 1\otimes\mathbf{t},\quad \Delta(A) = A\otimes 1 + 1\otimes A;$$

• $R-\mathsf{NMod}$ is the category of (finitely generated) "nilpotent left $R$-modules", which I take to mean (as they define) finite-dimensional complex vector spaces $V$ with two nilpotent endomorphisms $\mathbf{t}, A\in\text{End}(V)$;
• $\mathsf{NConn}(X;e)$ is the category of "nilpotent vector bundles" on $X$ with log-poles at $e$, which I take to mean unipotent vector bundles i.e. those $\mathcal{V}$ possessing a finite filtration by subbundles (with compatible connections) such that the quotients are all isomorphic to the trivial bundle with trivial connection $(\mathcal{O}_X, d)$. This category is often called $\mathsf{Un} (X;e)$ or something to that effect (e.g. in the papers of M. Kim).

By definition the de Rham fundamental group $\pi_1^{dR} (E^\times;b)$ is the Tannakian fundamental group of the category $\mathsf{Un}(X;e)$, with fiber functor $\omega$ sending a bundle to the stalk over a basepoint $b$ of $X$ (or for $e$, a suitable notion of "tangential basepoint"). By the above equivalence, it follows that there is an isomorphism between $\pi_1^{dR} (E^\times;e)$ and the Tannakian fundamental group of $R-\mathsf{NMod}$ (with the forgetful functor as fiber functor).

In Corollary 2.2.7, Levin and Racinet claim that this Tannakian fundamental group is

$$\pi_1^{dR} (E^\times; e)\cong \exp \mathbb{L}(\mathbf{t}, A).$$

I don't understand how this is calculated as the Tannakian fundamental group of $R-\mathsf{NMod}$. Would anyone be able to help me or point to how this is calculated? Many thanks!

• You should try Mathoverflow for this question. – Moishe Kohan Apr 19 '17 at 13:18

First, a correction, I believe "simultaneously nilpotent" in this context means that any sequence of $t$s and $A$s eventually multiplies to zero, or in other words there is a filtration such that both $t$ and $A$ send elements to elements of lower degree. This makes sense as we want to get the same filtration on the vector bundle. It's equivalent to say that every element of the Lie algebra, not just the generators, acts nil potently.

To calculate the Tannakian group, we can use the alternate description of the Tannakian fundamental group, that it is the unique pro-algebraic group whose category of representations (w/ Tannakian structure) is equivalent to your Tannakian category.

For any Lie algebra, there is an equivalence of categories between the representations of the Lie algebra and the representations of the universal enveloping algebra.

There is also an equivalence of categories between representations of a simply-connected Lie group and its Lie algebra. In the algebraic group setting, however, we have to be careful, because not all representations produced this way are algebraic.

Instead, note that the Lie algebra in question is an inverse limit of nilpotent Lie algebras (the quotients by the ideal generated by the $n$th order commutators, say). Hence its exponential is an inverse limit of unipotent Lie groups. Algebraic representations of a unipotent Lie group are exactly the representations of its Lie algebra sending every representation to a unipotent group.

To check this, note that any unipotent subgroup of $GL_n$ is contained in the upper triangular subgroup, and take the logarithm. Conversely, use the power series formula for exponentials, and note that it is algebraic when the object being exponentiated is nilpotent.

It follows that there is a series of equivalences of categories between nilpotent modules for the universal enveloping algebra, representations of the pro-Lie algebra that make every element nilpotent, and algebraic representations of the pro-unipotent pro-algebraic exponential group. Because these equivalences agree with the tensor product and the fiber functor, this must be the Tannakian group.