F_un and braid groups

Recall that an n-braid consists of n strictly descending elastic strings connecting n inputs at the top (named 1,2,…,n) to n outputs at the bottom (labeled 1,2,…,n) upto isotopy (meaning that we may pull and rearrange the strings in any way possible within 3-dimensional space). We can always change the braid slightly such that we can divide the interval between in- and output in a number of subintervals such that in each of those there is at most one crossing.

n-braids can be multiplied by putting them on top of each other and connecting the outputs of the first braid trivially to the inputs of the second. For example the 5-braid on the left can be written as $B=B_1.B_2 $ with $B_1 $ the braid on the top 3 subintervals and $B_2 $ the braid on the lower 5 subintervals.

In this way (and using our claim that there can be at most 1 crossing in each subinterval) we can write any n-braid as a word in the generators $\sigma_i $ (with $1 \leq i < n $) being the overcrossing between inputs i and i+1. Observe that the undercrossing is then the inverse $\sigma_i^{-1} $. For example, the braid on the left corresponds to the word

$\sigma_1^{-1}.\sigma_2^{-1}.\sigma_1^{-1}.\sigma_2.\sigma_3^{-1}.\sigma_4^{-1}.\sigma_3^{-1}.\sigma_4 $

Clearly there are relations among words in the generators. The easiest one we have already used implicitly namely that $\sigma_i.\sigma_i^{-1} $ is the trivial braid. Emil Artin proved in the 1930-ies that all such relations are consequences of two sets of ‘obvious’ relations. The first being commutation relations between crossings when the strings are far enough from each other. That is we have

$\sigma_i . \sigma_j = \sigma_j . \sigma_i $ whenever $|i-j| \geq 2 $


=

The second basic set of relations involves crossings using a common string

$\sigma_i.\sigma_{i+1}.\sigma_i = \sigma_{i+1}.\sigma_i.\sigma_{i+1} $


=

Starting with the 5-braid at the top, we can use these relations to reduce it to a simpler form. At each step we have outlined to region where the relations are applied


=
=
=

These beautiful braid-pictures were produced using the braid-metapost program written by Stijn Symens.

Tracing a string from an input to an output assigns to an n-braid a permutation on n letters. In the above example, the permutation is $~(1,2,4,5,3) $. As this permutation doesn’t change under applying basic reduction, this gives a group-morphism

$\mathbb{B}_n \rightarrow S_n $

from the braid group on n strings $\mathbb{B}_n $ to the symmetric group. We have seen before that the symmetric group $S_n $ has a F-un interpretation as the linear group $GL_n(\mathbb{F}_1) $ over the field with one element. Hence, we can ask whether there is also a F-un interpretation of the n-string braid group and of the above group-morphism.

Kapranov and Smirnov suggest in their paper that the n-string braid group $\mathbb{B}_n \simeq GL_n(\mathbb{F}_1[t]) $ is the general linear group over the polynomial ring $\mathbb{F}_1[t] $ over the field with one element and that the evaluation morphism (setting t=0)

$GL_n(\mathbb{F}_1[t]) \rightarrow GL_n(\mathbb{F}1) $ gives the groupmorphism $\mathbb{B}_n \rightarrow S_n $

The rationale behind this analogy is a theorem of Drinfeld‘s saying that over a finite field $\mathbb{F}_q $, the profinite completion of $GL_n(\mathbb{F}_q[t]) $ is embedded in the fundamental group of the space of q-polynomials of degree n in much the same way as the n-string braid group $\mathbb{B}_n $ is the fundamental group of the space of complex polynomials of degree n without multiple roots.

And, now that we know the basics of absolute linear algebra, we can give an absolute braid-group representation

$\mathbb{B}_n = GL_n(\mathbb{F}_1[t]) \rightarrow GL_n(\mathbb{F}_{1^n}) $

obtained by sending each generator $\sigma_i $ to the matrix over $\mathbb{F}_{1^n} $ (remember that $\mathbb{F}_{1^n} = (\mu_n)^{\bullet} $ where $\mu_n = \langle \epsilon_n \rangle $ are the n-th roots of unity)

$\sigma_i \mapsto \begin{bmatrix}
1_{i-1} & & & \\
& 0 & \epsilon_n & \\
& \epsilon_n^{-1} & 0 & \\
& & & 1_{n-1-i} \end{bmatrix} $

and it is easy to see that these matrices do indeed satisfy Artin’s defining relations for $\mathbb{B}_n $.

Bost-Connes for ringtheorists

Over the last days I’ve been staring at the Bost-Connes algebra to find a ringtheoretic way into it. Ive had some chats about it with the resident graded-guru but all we came up with so far is that it seems to be an extension of Fred’s definition of a ‘crystalline’ graded algebra. Knowing that several excellent ringtheorists keep an eye on my stumblings here, let me launch an appeal for help :

What is the most elegant ringtheoretic framework in which the Bost-Connes Hecke algebra is a motivating example?

Let us review what we know so far and extend upon it with a couple of observations that may (or may not) be helpful to you. The algebra $\mathcal{H} $ is the algebra of $\mathbb{Q} $-valued functions (under the convolution product) on the double coset-space $\Gamma_0 \backslash \Gamma / \Gamma_0 $ where

$\Gamma = { \begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix}~:~a,b \in \mathbb{Q}, a > 0 } $ and $\Gamma_0 = { \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}~:~n \in \mathbb{N}_+ } $

We have seen that a $\mathbb{Q} $-basis is given by the characteristic functions $X_{\gamma} $ (that is, such that $X_{\gamma}(\gamma’) = \delta_{\gamma,\gamma’} $) with $\gamma $ a rational point represented by the couple $~(a,b) $ (the entries in the matrix definition of a representant of $\gamma $ in $\Gamma $) lying in the fractal comb

defined by the rule that $b < \frac{1}{n} $ if $a = \frac{m}{n} $ with $m,n \in \mathbb{N}, (m,n)=1 $. Last time we have seen that the algebra $\mathcal{H} $ is generated as a $\mathbb{Q} $-algebra by the following elements (changing notation)

$\begin{cases}X_m=X_{\alpha_m} & \text{with } \alpha_m = \begin{bmatrix} 1 & 0 \\ 0 & m \end{bmatrix}~\forall m \in \mathbb{N}_+ \\
X_n^*=X_{\beta_n} & \text{with } \beta_n = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{n} \end{bmatrix}~\forall n \in \mathbb{N}_+ \\
Y_{\gamma} = X_{\gamma} & \text{with } \gamma = \begin{bmatrix} 1 & \gamma \\ 0 & 1 \end{bmatrix}~\forall \lambda \in \mathbb{Q}/\mathbb{Z} \end{cases} $

Using the tricks of last time (that is, figuring out what functions convolution products represent, knowing all double-cosets) it is not too difficult to prove the defining relations among these generators to be the following (( if someone wants the details, tell me and I’ll include a ‘technical post’ or consult the Bost-Connes original paper but note that this scanned version needs 26.8Mb ))

(1) : $X_n^* X_n = 1, \forall n \in \mathbb{N}_+$

(2) : $X_n X_m = X_{nm}, \forall m,n \in \mathbb{N}_+$

(3) : $X_n X_m^* = X_m^* X_n, \text{whenever } (m,n)=1$

(4) : $Y_{\gamma} Y_{\mu} = Y_{\gamma+\mu}, \forall \gamma,mu \in \mathbb{Q}/\mathbb{Z}$

(5) : $Y_{\gamma}X_n = X_n Y_{n \gamma},~\forall n \in \mathbb{N}_+, \gamma \in \mathbb{Q}/\mathbb{Z}$

(6) : $X_n Y_{\lambda} X_n^* = \frac{1}{n} \sum_{n \delta = \gamma} Y_{\delta},~\forall n \in \mathbb{N}_+, \gamma \in \mathbb{Q}/\mathbb{Z}$

Simple as these equations may seem, they bring us into rather uncharted ringtheoretic territories. Here a few fairly obvious ringtheoretic ingredients of the Bost-Connes Hecke algebra $\mathcal{H} $

the group-algebra of $\mathbb{Q}/\mathbb{Z} $

The equations (4) can be rephrased by saying that the subalgebra generated by the $Y_{\gamma} $ is the rational groupalgebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $ of the (additive) group $\mathbb{Q}/\mathbb{Z} $. Note however that $\mathbb{Q}/\mathbb{Z} $ is a torsion group (that is, for all $\gamma = \frac{m}{n} $ we have that $n.\gamma = (\gamma+\gamma+ \ldots + \gamma) = 0 $). Hence, the groupalgebra has LOTS of zero-divisors. In fact, this group-algebra doesn’t have any good ringtheoretic properties except for the fact that it can be realized as a limit of finite groupalgebras (semi-simple algebras)

$\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] = \underset{\rightarrow}{lim}~\mathbb{Q}[\mathbb{Z}/n \mathbb{Z}] $

and hence is a quasi-free (or formally smooth) algebra, BUT far from being finitely generated…

the grading group $\mathbb{Q}^+_{\times} $

The multiplicative group of all positive rational numbers $\mathbb{Q}^+_{\times} $ is a torsion-free Abelian ordered group and it follows from the above defining relations that $\mathcal{H} $ is graded by this group if we give

$deg(Y_{\gamma})=1,~deg(X_m)=m,~deg(X_n^*) = \frac{1}{n} $

Now, graded algebras have been studied extensively in case the grading group is torsion-free abelian ordered AND finitely generated, HOWEVER $\mathbb{Q}^+_{\times} $ is infinitely generated and not much is known about such graded algebras. Still, the ordering should allow us to use some tricks such as taking leading coefficients etc.

the endomorphisms of $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $

We would like to view the equations (5) and (6) (the latter after multiplying both sides on the left with $X_n^* $ and using (1)) as saying that $X_n $ and $X_n^* $ are normalizing elements. Unfortunately, the algebra morphisms they induce on the group algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $ are NOT isomorphisms, BUT endomorphisms. One source of algebra morphisms on the group-algebra comes from group-morphisms from $\mathbb{Q}/\mathbb{Z} $ to itself. Now, it is known that

$Hom_{grp}(\mathbb{Q}/\mathbb{Z},\mathbb{Q}/\mathbb{Z}) \simeq \hat{\mathbb{Z}} $, the profinite completion of $\mathbb{Z} $. A class of group-morphisms of interest to us are the maps given by multiplication by n on $\mathbb{Q}/\mathbb{Z} $. Observe that these maps are epimorphisms with a cyclic order n kernel. On the group-algebra level they give us the epimorphisms

$\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \longrightarrow^{\phi_n} \mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $ such that $\phi_n(Y_{\lambda}) = Y_{n \lambda} $ whence equation (5) can be rewritten as $Y_{\lambda} X_n = X_n \phi_n(Y_{\lambda}) $, which looks good until you think that $\phi_n $ is not an automorphism…

There are even other (non-unital) algebra endomorphisms such as the map $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \rightarrow^{\psi_n} R_n $ defined by $\psi_n(Y_{\lambda}) = \frac{1}{n}(Y_{\frac{\lambda}{n}} + Y_{\frac{\lambda + 1}{n}} + \ldots + Y_{\frac{\lambda + n-1}{n}}) $ and then, we can rewrite equation (6) as $Y_{\lambda} X_n^* = X_n^* \psi_n(Y_{\lambda}) $, but again, note that $\psi_n $ is NOT an automorphism.

almost strongly graded, but not quite…

Recall from last time that the characteristic function $X_a $ for any double-coset-class $a \in \Gamma_0 \backslash \Gamma / \Gamma_0 $ represented by the matrix $a=\begin{bmatrix} 1 & \lambda \\ 0 & \frac{m}{n} \end{bmatrix} $ could be written in the Hecke algebra as $X_a = n X_m Y_{n \lambda} X_n^* = n Y_{\lambda} X_m X_n^* $. That is, we can write the Bost-Connes Hecke algebra as

$\mathcal{H} = \oplus_{\frac{m}{n} \in \mathbb{Q}^+_{\times}}~\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] X_mX_n^* $

Hence, if only the morphisms $\phi_n $ and $\psi_m $ would be automorphisms, this would say that $\mathcal{H} $ is a strongly $\mathbb{Q}^+_{\times} $-algebra with part of degree one the groupalgebra of $\mathbb{Q}/\mathbb{Z} $.

However, they are not. But there is an extension of the notion of strongly graded algebras which Fred has dubbed crystalline graded algebras in which it is sufficient that the algebra maps are all epimorphisms. (maybe I’ll post about these algebras, another time). However, this is not the case for the $\psi_m $…

So, what is the most elegant ringtheoretic framework in which the algebra $\mathcal{H} $ fits??? Surely, you can do better than generalized crystalline graded algebra

Anabelian & Noncommutative Geometry 2

Last time (possibly with help from the survival guide) we have seen that the universal map from the modular group $\Gamma = PSL_2(\mathbb{Z}) $ to its profinite completion $\hat{\Gamma} = \underset{\leftarrow}{lim}~PSL_2(\mathbb{Z})/N $ (limit over all finite index normal subgroups $N $) gives an embedding of the sets of (continuous) simple finite dimensional representations

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

and based on the example $\mu_{\infty} = \mathbf{simp}_c~\hat{\mathbb{Z}} \subset \mathbf{simp}~\mathbb{Z} = \mathbb{C}^{\ast} $ we would like the above embedding to be dense in some kind of noncommutative analogon of the Zariski topology on $\mathbf{simp}~\Gamma $.

We use the Zariski topology on $\mathbf{simp}~\mathbb{C} \Gamma $ as in these two M-geometry posts (( already, I regret terminology, I should have just called it noncommutative geometry )). So, what’s this idea in this special case? Let $\mathfrak{g} $ be the vectorspace with basis the conjugacy classes of elements of $\Gamma $ (that is, the space of class functions). As explained here it is a consequence of the Artin-Procesi theorem that the linear functions $\mathfrak{g}^{\ast} $ separate finite dimensional (semi)simple representations of $\Gamma $. That is we have an embedding

$\mathbf{simp}~\Gamma \subset \mathfrak{g}^{\ast} $

and we can define closed subsets of $\mathbf{simp}~\Gamma $ as subsets of simple representations on which a set of class-functions vanish. With this definition of Zariski topology it is immediately clear that the image of $\mathbf{simp}_c~\hat{\Gamma} $ is dense. For, suppose it would be contained in a proper closed subset then there would be a class-function vanishing on all simples of $\hat{\Gamma} $ so, in particular, there should be a bound on the number of simples of finite quotients $\Gamma/N $ which clearly is not the case (just look at the quotients $PSL_2(\mathbb{F}_p) $).

But then, the same holds if we replace ‘simples of $\hat{\Gamma} $’ by ‘simple components of permutation representations of $\Gamma $’. This is the importance of Farey symbols to the representation problem of the modular group. They give us a manageable subset of simples which is nevertheless dense in the whole space. To utilize this a natural idea might be to ask what such a permutation representation can see of the modular group, or in geometric terms, what the tangent space is to $\mathbf{simp}~\Gamma $ in a permutation representation (( more precisely, in the ‘cluster’ of points making up the simple components of the representation representation )). We will call this the modular content of the permutation representation and to understand it we will have to compute the tangent quiver $\vec{t}~\mathbb{C} \Gamma $.

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.

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!