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daddy wasn’t impressed

A first year-first semester course on group theory has its hilarious moments. Whereas they can relate the two other pure math courses (linear algebra and analysis) _somewhat_ to what they’ve learned before, with group theory they appear to enter an entirely new and strange world. So, it is best to give them concrete examples : symmetry groups of regular polygons and Platonic solids, the symmetric group etc. One of the lesser traditional examples I like to give is Nim addition and its relation to combinatorial games.

For their first test they had (among other things) to find a winning move for the position below in the Lenstra’s turtle turning game. At each move a player must put one turtle on its back and may also turn over any single turtle to the left of it. This second turtle, unlike the first, may be turned either onto its feet or onto its back. The player wins who turns the last turtle upside-down.

So, all they needed to see was that one turtle on its feet at place n is equivalent to a Nim-heap of height n and use the fact that all elements have order two to show that any zero-move in the sum game can indeed be played by using the second-turtle alternative. (( for the curious : the answer is turning both 9 and 4 on their back ))

A week later, one of the girls asked at the start of the lecture :

Are there real-life applications of group-theory? I mean, my father asked me what I was learning at school and I told him we were playing games turning turtles. I have to say that he was not impressed at all!.

She may have had an hidden agenda to slow me down because I spend an hour talking about a lot of things ranging from codes to cryptography and from representations to elementary particles…

For test three (on group-actions) I asked them to prove (among other things) Wilson’s theorem that is

$~(p-1)! \equiv -1~\text{mod}~p $

for any prime number $p $. The hint being : take the trivial action of $S_p $ on a one-element set and use the orbit theorem. (they know the number of elements in an $S_n $-conjugacy class)

Fingers crossed, hopefully daddy approved…

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profinite groups survival guide

Even if you don’t know the formal definition of a profinte group, you know at least one example which explains the concept : the Galois group of the algebraic numbers $Gal = Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ aka the absolute Galois group. By definition it is the group of all $\mathbb{Q} $-isomorphisms of the algebraic closure $\overline{\mathbb{Q}} $. Clearly, it is an object of fundamental importance for mathematics but in spite of this very little is known about it. For example, it obviously is an infinite group but, apart from the complex conjugation, try to give one (1!) other nontrivial element… On the other hand we know lots of finite quotients of $Gal $. For, take any finite Galois extension $\mathbb{Q} \subset K $, then its Galois group $G_K = Gal(K/\mathbb{Q}) $ is a finite group and there is a natural onto morphism $\pi_K~:~Gal \rightarrow G_K $ obtained by dividing out all $K $-automorphisms of $\overline{\mathbb{Q}} $. Moreover, all these projections fit together nicely. If we take a larger Galois extension $K \subset L $ then classical Galois theory tells us that there is a projection $\pi_{LK}~:~G_L \rightarrow G_K $ by dividing out the normal subgroup of all $K $-automorphisms of $L $ and these finite maps are compatible with those from the absolute Galois group, that is, for all such finite Galois extensions, the diagram below is commutative

[tex]\xymatrix{Gal \ar[rr]^{\pi_L} \ar[rd]_{\pi_K} & & G_L \ar[ld]^{\pi_{LK}} \\
& G_K &}[/tex]

By going to larger and larger finite Galois extensions $L $ we get closer and closer to the algebraic closure $\overline{Q} $ and hence a better and better finite approximation $G_L $ of the absolute Galois group $Gal $. Still with me? Congratulations, you just rediscovered the notion of a profinite group! Indeed, the Galois group is the projective limit

$Gal = \underset{\leftarrow}{lim}~G_L $

over all finite Galois extensions $L/\mathbb{Q} $. If the term ‘projective limit’ scares you off, it just means that all the projections $\pi_{KL} $ coming from finite Galois theory are compatible with those coming from the big Galois group as before. That’s it : profinite groups are just projective limits of finite groups.

These groups come equipped with a natural topology : the Krull topology. Again, this notion is best clarified by considering the absolute Galois group. Now that we have $Gal $ we would like to extend the classical Galois correspondence between subgroups and subfields $\mathbb{Q} \subset K \subset \overline{\mathbb{Q}} $ and between normal subgroups and Galois subfields. For each finite Galois extension $K/\mathbb{Q} $ we have a normal subgroup of finite index, the kernel $U_K=Ker(\pi_K) $ of the projection map above. Let us take the set of all $U_K $ as a fundamental system of neighborhoods of the identity element in $Gal $. This defines a topology on $Gal $ and this is the Krull topology. As every open subgroup has finite index it is clear that this turns $Gal $ into a compact topological group. Its purpose is that we can now extend the finite Galois correspondence to Krull’s Galois theorem :

There is a bijective lattice inverting Galois correspondence between the set of all closed subgroups of $Gal $ and the set of all subfields $\mathbb{Q} \subset F \subset \overline{\mathbb{Q}} $. Finite field extensions correspond in this bijection to open subgroups and the usual normal subgroup and factor group correspondences hold!

So far we had a mysterious group such as $Gal $ and reconstructed it from all its finite quotients as a projective limit. Now we can reverse the situation : suppose we have a wellknown group such as the modular group $\Gamma = PSL_2(\mathbb{Z}) $, then we can look at the set of all its normal subgroups $U $ of finite index. For each of those we have a quotient map to a finite group $\pi_U~:~\Gamma \rightarrow G_U $ and clearly if $U \subset V $ we have a quotient map of finite groups $\pi_{UV}~:~G_U \rightarrow G_V $ compatible with the quotient maps from $\Gamma $

[tex]\xymatrix{\Gamma \ar[rr]^{\pi_U} \ar[rd]_{\pi_V} & & G_U \ar[ld]^{\pi_{UV}} \\
& G_V &}[/tex]

For the family of finite groups $G_U $ and groupmorphisms $\pi_{UV} $ we can ask for the ‘best’ group mapping to each of the $G_U $ compatible with the groupmaps $G_{UV} $. By ‘best’ we mean that any other group with this property will have a morphism to the best-one such that all quotient maps are compatible. This ‘best-one’ is called the projective limit

$\hat{\Gamma} = \underset{\leftarrow}{lim}~G_U $

and as a profinite group it has again a Krull topology making it into a compact group. Because the modular group $\Gamma $ had quotient maps to all the $G_U $ we know that there must be a groupmorphism to the best-one
$\phi~:~\Gamma \rightarrow \hat{\Gamma} $ and therefore we call $\hat{\Gamma} $ the profinite compactification (or profinite completion) of the modular group.

A final remark about finite dimensional representations. Every continuous complex representation of a profinite group like the absolute Galois group $Gal \rightarrow GL_n(\mathbb{C}) $ has finite image and this is why they are of little use for people studying the Galois group as it conjecturally reduces the study of these representations to ‘just’ all representations of all finite groups. Instead they consider representations to other topological fields such as p-adic numbers $Gal \rightarrow GL_n(\mathbb{Q}_p) $ and call these Galois representations.

For people interested in Grothendieck’s dessins d’enfants, however, continuous complex representations of the profinite compactification $\hat{\Gamma} $ is exactly their object of study and via the universal map $\phi~:~\Gamma \rightarrow \hat{\Gamma} $ above we have an embedding

$\mathbf{rep}_c~\hat{\Gamma} \rightarrow \mathbf{rep}~\Gamma $

of them in all finite dimensional representations of the modular group (
and we have a similar map restricted to simple representations). I hope this clarifies a bit obscure terms in the previous post. If not, drop a comment.

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Anabelian vs. Noncommutative Geometry

This is how my attention was drawn to what I have since termed
anabelian algebraic geometry, whose starting point was exactly a study
(limited for the moment to characteristic zero) of the action of absolute
Galois groups (particularly the groups $Gal(\overline{K}/K) $, where K is an extension of finite type of the prime field) on (profinite) geometric fundamental
groups of algebraic varieties (defined over K), and more particularly (breaking with a well-established tradition) fundamental groups which are very far
from abelian groups (and which for this reason I call anabelian). Among
these groups, and very close to the group $\hat{\pi}_{0,3} $ , there is the profinite compactification of the modular group $SL_2(\mathbb{Z}) $, whose quotient by its centre
$\{ \pm 1 \} $ contains the former as congruence subgroup mod 2, and can also be
interpreted as an oriented cartographic group, namely the one classifying triangulated oriented maps (i.e. those whose faces are all triangles or
monogons).

The above text is taken from Alexander Grothendieck‘s visionary text Sketch of a Programme. He was interested in the permutation representations of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ as they correspond via Belyi-maps and his own notion of dessins d’enfants to smooth projective curves defined over $\overline{\mathbb{Q}} $. One can now study the action of the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q}) $ on these curves and their associated dessins. Because every permutation representation of $\Gamma $ factors over a finite quotient this gives an action of the absolute Galois group as automorphisms on the profinite compactification

$\hat{\Gamma} = \underset{\leftarrow}{lim}~\Gamma/N $

where the limit is taken over all finite index normal subgroups $N \triangleleft PSL_2(\mathbb{Z}) $. In this way one realizes the absolute Galois group as a subgroup of the outer automorphism group of the profinite group $\hat{\Gamma} $. As a profinite group is a compact topological group one should study its continuous finite dimensional representations which are precisely those factoring through a finite quotient. In the case of $\hat{\Gamma} $ the simple continuous representations $\mathbf{simp}_c~\hat{\Gamma} $ are precisely the components of the permutation representations of the modular group. So in a sense, anabelian geometry is the study of these continuous simples together wirth the action of the absolute Galois group on it.

In noncommutative geometry we are interested in a related representation theoretic problem. We would love to know the simple finite dimensional representations $\mathbf{simp}~\Gamma $ of the modular group as this would give us all simples of the three string braid group $B_3 $. So a natural question presents itself : how are these two ‘geometrical’ objects $\mathbf{simp}_c~\hat{\Gamma} $ (anabelian) and $\mathbf{simp}~\Gamma $ (noncommutative) related and can we use one to get information about the other?

This is all rather vague so far, so let us work out a trivial case to get some intuition. Consider the profinite completion of the infinite Abelian group

$\hat{\mathbb{Z}} = \underset{\leftarrow}{lim}~\mathbb{Z}/n\mathbb{Z} = \prod_p \hat{\mathbb{Z}}_p $

As all simple representations of an Abelian group are one-dimensional and because all continuous ones factor through a finite quotient $\mathbb{Z}/n\mathbb{Z} $ we see that in this case

$\mathbf{simp}_c~\hat{\mathbb{Z}} = \mu_{\infty} $

is the set of all roots of unity. On the other hand, the simple representations of $\mathbb{Z} $ are also one-dimensional and are determined by the image of the generator so

$\mathbf{simp}~\mathbb{Z} = \mathbb{C} – { 0 } = \mathbb{C}^* $

Clearly we have an embedding $\mu_{\infty} \subset \mathbb{C}^* $ and the roots of unity are even dense in the Zariski topology. This might look a bit strange at first because clearly all roots of unity lie on the unit circle which ‘should be’ their closure in the complex plane, but that’s because we have a real-analytic intuition. Remember that the Zariski topology of $\mathbb{C}^*$ is just the cofinite topology, so any closed set containing the infinitely many roots of unity should be the whole space!

Let me give a pedantic alternative proof of this (but one which makes it almost trivial that a similar result should be true for most profinite completions…). If $c $ is the generator of $\mathbb{Z} $ then the different conjugacy classes are precisely the singletons $c^n $. Now suppose that there is a polynomial $a_0+a_1x+\ldots+a_mx^m $ vanishing on all the continuous simples of $\hat{\mathbb{Z}} $ then this means that the dimensions of the character-spaces of all finite quotients $\mathbb{Z}/n\mathbb{Z} $ should be bounded by $m $ (for consider $x $ as the character of $c $), which is clearly absurd.

Hence, whenever we have a finitely generated group $G $ for which there is no bound on the number of irreducibles for finite quotients, then morally the continuous simple space for the profinite completion

$\mathbf{simp}_c~\hat{G} \subset \mathbf{simp}~G $

should be dense in the Zariski topology on the noncommutative space of simple finite dimensional representations of $G $. In particular, this should be the case for the modular group $PSL_2(\mathbb{Z}) $.

There is just one tiny problem : unlike the case of $\mathbb{Z} $ for which this space is an ordinary (ie. commutative) affine variety $\mathbb{C}^* $, what do we mean by the “Zariski topology” on the noncommutative space $\mathbf{simp}~PSL_2(\mathbb{Z}) $ ? Next time we will clarify what this might be and show that indeed in this case the subset

$\mathbf{simp}_c~\hat{\Gamma} \subset \mathbf{simp}~\Gamma $

will be a Zariski closed subset!

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