I'm refering, of course, to Grothendieck's ambitious program available fully here: http://www.math.jussieu.fr/~leila/grothendieckcircle/EsquisseFr.pdf

The text is, as described in its title, a sketch, and is supposed to be written in terms one can easily follow. However, for a non-expert like me it is still, to say the least, difficult. What I've done so far is read some things that were written about it, and only glanced at "esquisse", which still strikes me as an intimidating text. My understanding is still less than preliminary.

This is going to be a series of related questions, whose answers will hopefully sketch to me the program described in "esquisse", or at least put me in better position to sketch it to myself:

  1. As I understand it, $Gal( \mathbb{Q})$ acts on $\mathbb{P}^1 _ {\bar {\mathbb{Q}}}- 0,1,\infty$ in the obvious way. Now $\mathcal{M}_{0,n}$ (the moduli space of genus $0$ $\bar {\mathbb{Q}}$-curves with $n$ marked points) can be identified with $(\mathbb{P}^1 _ {\bar {\mathbb{Q}}}- 0,1,\infty)^{n-3}-\Delta$, generalizing just $\mathbb{P}^1 _ {\bar {\mathbb{Q}}}- 0,1,\infty = \mathcal{M}_{0,4}$. So $Gal( \mathbb{Q})$ acts on $\mathcal{M} _ {0,n}$. Here I start getting confused. On the one hand it seems like the generalization that is described is an action of $Gal( \mathbb{Q})$ on all $\mathcal{M}_{g,n}$, and on the other it seems that the action described is on its universal cover, the Teichmüller space, $\mathcal{T} _{g,n}$.
  2. Okay. Now the assertion seems to be, we have a map from $Gal( \mathbb{Q})$ to each fundamental group (fundamental groupoid? I imagine that we may view $Gal( \mathbb{Q})$ as a baseless fundamental groupoid if we choose) of each $\mathcal{M}_{g,n}$. We can in fact view this as mapping simultaneously to all of them, respecting certain operations between the $\mathcal{T} _{g,n}$. What are these actions? Can you restate this last paragraph rigorously?
  3. This map (sketchily sketched in question 2) is then conjectured to be an imbedding (I think this has been proven, am I right?). According to a paper by Pierre Lochak and Leila Schneps ($\hat {GT}$ and braid groups): "A very important suggestion of Grothendieck is that the tower of Teichmüller groupoids ... should be entirely constructable from its first two levels". What does "first two levels" mean? (first two $n$'s for every $g$?) And why should that be true? (what would behoove anyone to think it's true?)
  4. How do braid groups have to do with any of this? Is the assertion that $\pi_1 ( \mathcal{M} _{0,n},basept)$ is isomorphic to $B_n$ (the braid group on $n$ generators). Is that even true? I'm confused about this point.
  5. This question is regarding notations for profinite groups, used generously anywhere related to this topic. $\hat {F_2}$, the profinite completion of the free group on two generators, is ubiquitous in such discussions because it is the fundamental group of $\mathbb{P}^1 _{\bar {\mathbb{Q}}}-0,1,\infty$. It is "generated" by two elements, meaning that it is the closure of a group generated by two elements. While defining the Grothendieck-Teichmüller group, Lochak and Schneps write "if $x, y$ are the generators of $\hat {F_2}$, we write an element of $\hat {F_2}$ as a 'profinite word' $f(x,y)$, although it is not generally a word in $x$ and $y$. This notation allows us to give a meaning to the element $f(\bar x, \bar y)$ where $\bar x$ and $\bar y$ are arbitrary elements of a profinite group". Well... then, if $f(x,y)$ is not an actual word in $x$ and $y$, then what does $f(\bar x,\bar y)$ mean?
  6. At the same place, $\hat {GT}_0$ is defined as the subgroup of $\hat{ \mathbb{Z}}^{\times} \times \hat {F_2} '$ of elements $(\lambda,f)$ such that, among other things, $f(x,y)f(y,x)=1$. What does that mean? What is $f(x,y)$, and how is it different from $f(y,x)$ in its definition? Is it that somehow I can "express" any element of $\hat {F_2}$ as a "word" in $x$ and $y$, and then $f(y,x)$ is just flipping them? How do you make that rigorous?
  7. Do I understand correctly that $\hat {GT}$ is a subgroup of the automorphism group of this tower of Teichmüller spaces, and that it is conjectured that $Gal( \mathbb{Q})$ surjects onto it (as well as imbeds into it)?
  8. Do any of these things have to do with the anabelian geometry that is sketched later in "esquisse"? If so, how? And in general, how do these sets of ideas fit into esquisse? Are they the essence, one idea of many, or a piece in a bigger picture I can't see (combining dessin d'enfants, the Teichmüller tower, and anabelian geometry)? If so, very roughly, what is that sketch?

I wasn't sure whether to make this community wiki.

  • 1
    $\begingroup$ I would suggest breaking up your string of 8 questions into several posts here. From a cursory glance it looks like 1-3 are related, 4 is more independent, 5-7 are related, and 8 is the real big picture question. $\endgroup$
    – j.c.
    May 16, 2010 at 16:37
  • 5
    $\begingroup$ 1. The point is that $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ acts on the collection of all curves defined over $\mathbb{Q}$. The connection is Belyi's theorem, that curves defined over $\mathbb{Q}$ are exactly those with a map to $P^1$ ramified only at $\{0,1,\infty\}$. 3. Grothendieck suggested that by decomposing curves into "low-complexity" curves (small dimensional moduli), it suffices to understand $M_{0,4}$, $M_{1,1}$, $M_{0,5}$, and $M_{1,2}$. 4. If your points are ordered, then $\pi_1(\mathcal{M}_{0,n})$ is the pure braid group. If unordered, then the braid group. $\endgroup$
    – Tom Church
    May 16, 2010 at 17:24
  • $\begingroup$ neverendingbooks.org/index.php/monsieur-mathieu.html and the accompanying posts might help. $\endgroup$ May 17, 2010 at 6:44
  • $\begingroup$ @Tom Church: RE point #3 - has this been proven or is it a conjecture? $\endgroup$
    – Dr Shello
    Dec 15, 2010 at 9:23

1 Answer 1


For question 5, I think the motivation for thinking of elements of $\hat{F_2}$ as "profinite words" is that $\hat{F_2}$ enjoys the same universal mapping properties as $F_2$, but in the profinite setting.

Indeed, any map from $F_2$ to a profinite group $G$ extends uniquely (at least if we require continuity) to a map from $\hat{F_2}$ to $G$. So let $f=f(x,y)\in \hat{F_2}$. Then if $\overline{x},\overline{y}$ are elements of some profinite group $G$, define $f(\overline{x},\overline{y})$ to be the image of $f$ under the unique map extending $x\mapsto \overline{x}, y\mapsto\overline{y}$.

This should also clear up number 6... the "flip" map is the unique continuous map from $\hat{F_2}$ to $\hat{F_2}$ sending $x$ to $y$, $y$ to $x$.


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