Tag: profinite

Bost-Connes for ringtheorists

Published January 23, 2008 by lievenlb

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…

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Anabelian & Noncommutative Geometry 2

Published December 19, 2007 by lievenlb

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

Published December 14, 2007 by lievenlb

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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