## Saturday, January 7, 2012

### Constructing the Grothendieck-Teichmuller Group

So for the past six or seven months I've been trying to get a copy of the paper On quasitriangular Quasi-Hopf algebras and a group closely connected with ${\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})}$'' by V.G. Drinfel'd. My online search uncovered only a Russian copy (which, unfortunately, I don't read). My library searches were equally unfruitful: the British library's copy is unavailable'' and I was unable to find a copy at another UK institution. My supervisor was sure he had a copy, but when he discovered that it mysteriously disappeared he encouraged me to request an inter-library loan with my university. I did, and my university said they found a copy in Leeds, but the last ten pages are missing. So - if any readers have a copy of the aforementioned paper: I'd greatly appreciate your lending me it!

In the meanwhile, I've been reading On Associators and the Grothendieck-Teichmuller Group'' by Dror Bar-Natan. Among other things, that paper described the construction of said group via a Parenthesized Braid Category''. I'll discuss that construction in this post.

## 1. Braids and Categories

I really don't want to give a precise definition of a braid (I've not yet seen one that doesn't make my head hurt), but hopefully I won't be too wrong if I propose the following:

Def1 1 A braid on ${n}$ strands is a set of ${n}$ oriented curves with distinct starting and ending points where the starting points lie on the same line, each endpoint lies on a separate line, and no curve contains a loop.

For example, the following is a braid:

Clearly we can enumerate the starting and ending points, and treat a braid as a permutation on them. In this way, braids form a category where the braid itself is the morphism and the objects are simply sets to be permuted.

We're going to change our class of objects slightly - we're going to consider parenthesizations of sets. Id est, pretend that the sets have some underlying multiplicative structure. We want to order the sets and denote the order of multiplication. Exempli gratia, the ordering ${(1 \, 2 \, 3)}$ has two parenthesizations: ${((1 \, 2) \, 3)}$ and ${(1 \, (2 \, 3))}$. Our braid morphisms are allowed to act on any parenthesizations.

We aren't done complicated things yet: we're going to modify the morphisms themselves so that they become formal sums:

$\displaystyle \sum_{i=1}^{k} \beta_i B_i$

Where each ${B_i}$ is a braid on ${n}$ strands, all ${B_i}$'s have the same effective permutation, and ${\beta_i}$ belongs to some ${\mathbb{Q}}$-algebra, usually ${\mathbb{Q}}$ or ${\mathbb{C}}$. We're forming the algebriod'' over ${\mathbb{Q}}$ or ${\mathbb{C}}$ over the set of braids on ${n}$ strands, with composition defined to be a bilinear map in the same way you would do for a group algebra.

This category is our Parenthesized Braid Category ${\textbf{PaB}}$.

## 2. Fibred Linear Categories

Our category is fibred. To see this, consider the category ${\textbf{PaP}}$ of parenthesized objects as before, but with simple permutations as the morphisms. Then consider the functor ${\textbf{S} : \textbf{PaB} \rightarrow \textbf{PaP}}$ that maps each object to itself, but takes each formal sum of braids to that one effective permutation. Clearly the set of morphisms in ${\text{PaB}}$ from ${O_1}$ to ${O_2}$ is:

$\displaystyle \text{Hom}_\textbf{PaB}(O_1, O_2) = \textbf{S}^{-1} \left( \text{Hom}_\textbf{PaP}(O_1, O_2) \right)$

Where P is the effective permutation.

One might also note that for a fixed permutation ${P}$, the fibre of morphisms ${\textbf{S}^{-1}(P)}$ is a vector space (actually, its an algebra, because we can compose morphisms as our product). This allows us to define some very linear'' notions, for instance:

We can define subcategories: a category is a subcategory of ${\textbf{PaB}}$ if it shares the same collection of objects and each ${\text{Hom}(O_1, O_2)}$ is a subspace of the corresponding ${\text{Hom}_\textbf{PaB}(O_1, O_2)}$.

A subcategory is an ideal if whenever one of two composable ${B_1}$ or ${B_2}$ belongs to it then the composition belongs to it, too. Exempli gratia, let ${\textbf{I}}$ be the ideal where all morphisms ${\sum \beta_i B_i}$ have ${\sum \beta_i = 0}$. Then we can define the quotient ${\textbf{PaB}/\textbf{I}}$ in an obvious way: each set of morphisms ${\text{Hom}(O_1, O_2)}$ is the corresponding quotient in the bigger set of morphisms.

Additionally, we can define idealic powers: ${\textbf{I}^n}$ has morphisms that can be written as compositions of ${n}$ morphisms in ${\textbf{I}}$. We can continue this and take an inverse systems with an inverse limits, in fact, we can even do the ${\textbf{I}}$-${adic}$ completion:

$\displaystyle \widehat{\textbf{PaB}} = \varprojlim_n \textbf{PaB}/\textbf{I}^n$

Finally, we can do a tensor product of ${\textbf{PaB}}$ with itself (in the same way we would do tensor powers of an algebra). For ${\textbf{PaB}^{2\otimes}}$ we define the morphism as the disjoint union:

$\displaystyle \text{Hom}_{\textbf{PaB}^{2\otimes}}(O_1,O_2) = \coprod_{P \in \text{Hom}_{\textbf{PaB}}(O_1, O_2)} \textbf{S}^{-1}(P) \otimes \textbf{S}^{-1}(P)$

Now we can even do a coproduct ${\Delta: \textbf{PaB} \rightarrow \textbf{PaB}^{2\otimes}}$ by ${B \mapsto B \otimes B}$. This coproduct is a functor. (I suspect that there is a bialgebra/hopf algebra structure hidden somewhere in here but I've not thought about it enough yet. Not for this post.)

## 3. Some Functors

In addition to ${\textbf{S} : \textbf{PaB} \rightarrow \textbf{PaP}}$, we also have a few functors ${\textbf{PaB} \rightarrow \textbf{PaB}}$. We can describe these by their action on a single braid and one can expand them as a linear action on the formal sums we use. First, we have the extension functors ${d_0}$ and ${d_{n+1}}$ that add a single straight strand to the left or right of the braid, respectively.

Second, we have cabling functors ${d_i}$ for ${1 \leq i \leq n}$ on a braid of n strands. This functor simply doubles the ${i}$-th strand.

Finally, we have a strand-removal functor ${s_i}$, which, as you might guess, removes the ${i}$-th strand.

We can now define the Grothendieck-Teichmuller group ${\widehat{\text{GT}}}$. This is the group of all invertible functors ${a: \textbf{PaB} \rightarrow \textbf{PaB}}$ such that ${\textbf{S} \circ a = S}$, ${d_i \circ a = a \circ d_i}$, ${s_i \circ a = a \circ s_i}$, ${\Delta \circ a = a \otimes a \circ \Delta}$. We also require that $a$ leaves the braid constant:

Right! There we are. Sooner or later I actually do some maths, but for now I'm stuck understanding trying to understand definitions, so hence this post.

## 4. Source

This is all just a reiteration of the material in the aforementioned paper by Bar-Natan.