This is a very half-baked question relating to something in my PhD research. I am trying to apply these ideas to a different situation, but this is a starter case that I am trying to think about. Therefore I'd like to try to set things up quite generally, although I think this is still incredibly vague. Apologies in advance!

Let's suppose I have some commutative $\mathbb{Q}$-algebra $A$ (think: subring of the complex numbers) and let $B = A[\alpha, \alpha^{-1}]$ for some element $\alpha$ (thought of as a complex number not in $A$).

Let $A\langle\langle x_0, x_1\rangle\rangle$ denote the ring of formal power series in noncommuting variables $x_0, x_1$ with coefficients in $A$. That is, let $X$ be the free monoid on $\left\{ x_0, x_1\right\}$. Then an element of $A\langle\langle x_0, x_1\rangle\rangle$ is a formal series $$S = \sum_{w \in X}S_w w, \quad S_w\in A.$$

The ring $A\langle\langle x_0, x_1\rangle\rangle$ is a Hopf algebra with the coproduct for which $x_0$ and $x_1$ are primitive. Let $\text{GrL}(A\langle\langle x_0, x_1\rangle\rangle)$ denote the set of grouplike elements of this Hopf algebra.

I am interested in the following map: set $\alpha = 2\pi i$, and let $f: \text{GrL}(A\langle\langle x_0, x_1\rangle\rangle)\to \text{GrL}(B\langle\langle x_0, x_1\rangle\rangle)$ be the conjugation map

$$S\mapsto S \exp(2 \pi i x_1) S^{-1}=S(1+2\pi i x_1 + \frac{(2\pi i)^2}{2}x_1 x_1 + \dots)S^{-1}.$$

Question: I have heard that it is possible to recover the (space of) coefficients $S_w$ of $S$ from the "twisted series" $f(S)$. How can one go about doing this? Is it a direct computation or is there an abstract way to see it? I have heard that this is discussed in Deligne-Goncharov's paper Groupes fondamentaux motiviques de Tate mixte but unfortunately my French isn't very good and the exposition seems fairly complicated.

The motivation for this comes from what happens to the Drinfeld associator $S = \Phi(x_0, x_1)$ under monodromy around the punctured point $1\in\mathbb{P}^1\backslash\left\{0,1,\infty\right\}$. More abstractly (and unfortunately much less certainly in my mind), let $_{0}\Pi_{1}$ denote the (de Rham) torsor of paths from the tangential basepoint $\vec{1}_0$ at $0$ to the tangential basepoint $-\vec{1}_1$ at $1$. I believe that we have $_{0}\Pi_{1}\cong \text{Spec}(\mathbb{Q}\langle x_0, x_1\rangle)$, the spectrum of the shuffle algebra. (I am actually unsure on whether this is correct because $_{0}\Pi_{1}$ is a torsor under the de Rham fundamental group - which is also this same scheme - but it is only isomorphic after picking a point. Perhaps this isomorphism above is not natural?)

Set $_{0}\Pi_0$ (resp. $_1\Pi_1$) to be the de Rham fundamental group at the tangential basepoint $\vec{1}_0$ (resp. $-\vec{1}_1$). Then for $K$ a $\mathbb{Q}$-algebra, the $K$-points of each of these schemes $_a\Pi_b$ should be

$$_a\Pi_b (K) \cong \text{GrL}(K\langle\langle x_0, x_1\rangle\rangle),$$

although these isomorphisms may be different in each case. The Drinfeld associator lives in $_0\Pi_1 (\mathbb{R})\hookrightarrow _0\Pi_1 (\mathbb{C})$, and can be thought of as the "straight line path" $\text{dch}$ from $\vec{1}_0$ to $-\vec{1}_1$. Then by picking the element $\exp(2\pi i x_1)\in _1\Pi_1 (\mathbb{C})$ (thought of as a small loop around the puncture at $1$) we first go along $\text{dch}$, then go around the loop around $1$, and then go along the reverse path $\text{dch}^{-1}$. This results in a map

$$_0\Pi_1 (\mathbb{C}) \to _{0}\Pi_0 (\mathbb{C}), \quad S\mapsto S \exp(2\pi i x_1) S^{-1}.$$

The idea is that one should be able to recover the original "path" from $\vec{1}_0$ to $-\vec{1}_1$ from this new "loop". I am trying to understand this case before moving onto a much more complicated twisting map arising from the case of a punctured elliptic curve.

  • $\begingroup$ Do you think it would be better/also beneficial to post this question in MathOverflow? As I understand things, that setting might bear more fruit for a research-related question like this. $\endgroup$ May 12, 2017 at 4:16
  • $\begingroup$ And if you're needing to read math in French, here's a great set of links and answers for algebraic geometry (including motives and all things Deligne/Grothendieck, Serre, etc.) in particular: mathoverflow.net/questions/117622/math-french-words $\endgroup$ May 12, 2017 at 4:18
  • $\begingroup$ Also possibly check out Milne's PDF on motives: jmilne.org/math/xnotes/mot.html $\endgroup$ May 12, 2017 at 4:22
  • $\begingroup$ Dear @TannerStrunk, thank you for these helpful comments! It's great to have a set of links for French algebraic geometry. Thanks for the advice about posting on MO as well. I will have a think about how to reformulate the question to make it more suitable there and then I will probably repost! $\endgroup$
    – Alex Saad
    May 12, 2017 at 13:14

1 Answer 1


Almost. You recover $S$ only up to right multiplication by a$$\text{exp}(\text{multiple of }e_1).$$The point is that the Hopf algebra—with a grain of salt: completed tensor product needed—is the universal enveloping algebra of the free Lie algebra in $x_1$ and $x_2$, and that in this Lie algebra, the centralizer of $x_1$ is reduced to the multiples of $x_1$.

Model: In a group $G$, given $x$ with centralizer $Z(x)$, the conjugate $gxg^{-1}$ determines $g$ up to right multiplication by a $z$ in $Z(x)$.

  • $\begingroup$ Thank you so much for this answer - this is very helpful! Before I accept it, would you be able to provide either some more details or a reference as to how you recover the coset $g Z(x)$ from $g x g^{-1}$? I'm sure this is standard group theory but I would like to see how it is done. $\endgroup$
    – Alex Saad
    May 14, 2017 at 11:13
  • $\begingroup$ Sorry, I was being stupid and confusing myself but I understand now what you mean by "up to a right multiple of $Z(x)$": from $b=g x g^{-1}$ we uniquely recover $g$ as $g = b g x^{-1}$, but if $g$ were replaced by $g h$ for any $h\in Z(x)$ we would only be able to recover $g$ and not $gh$ from $b = (gh)x(gh)^{-1}$ via this process. $\endgroup$
    – Alex Saad
    May 15, 2017 at 11:54

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