# Tag: profinite

It is perhaps surprising that Alain Connes and Katia Consani, two icons of noncommutative geometry, restrict themselves to define commutative algebraic geometry over $\mathbb{F}_1$, the field with one element.

My guess of why they stop there is as good as anyone’s. Perhaps they felt that there is already enough noncommutativity in Soule’s gadget-approach (the algebra $\mathcal{A}_X$ as in this post may very well be noncommutative). Perhaps they were only interested in the Bost-Connes system which can be entirely encoded in their commutative $\mathbb{F}_1$-geometry. Perhaps they felt unsure as to what the noncommutative scheme of an affine noncommutative algebra might be. Perhaps …

Remains the fact that their approach screams for a noncommutative extension. Their basic object is a covariant functor

$N~:~\mathbf{abelian} \rightarrow \mathbf{sets} \qquad A \mapsto N(A)$

from finite abelian groups to sets, together with additional data to the effect that there is a unique minimal integral scheme associated to $N$. In a series of posts on the Connes-Consani paper (starting here) I took some care of getting rid of all scheme-lingo and rephrasing everything entirely into algebras. But then, this set-up can be extended verbatim to noncommuative $\mathbb{F}_1$-geometry, which should start from a covariant functor

$N~:~\mathbf{groups} \rightarrow \mathbf{sets}$

from all finite groups to sets. Let’s recall quickly what the additional info should be making this functor a noncommutative (affine) F_un scheme :

There should be a finitely generated $\mathbb{C}$-algebra $R$ together with a natural transformation (the ‘evaluation’)

$e~:~N \rightarrow \mathbf{maxi}(R) \qquad N(G) \mapsto Hom_{\mathbb{C}-alg}(R, \mathbb{C} G)$

(both $R$ and the group-algebra $\mathbb{C} G$ may be noncommutative). The pair $(N, \mathbf{maxi}(R))$ is then called a gadget and there is an obvious notion of ‘morphism’ between gadgets.

The crucial extra ingredient is an affine $\mathbb{Z}$-algebra (possibly noncommutative) $S$
such that $N$ is a subfunctor of $\mathbf{mini}(S)~:~G \mapsto Hom_{\mathbb{Z}-alg}(S,\mathbb{Z} G)$ together with the following universal property :

any affine $\mathbb{Z}$-algebra $T$ having a gadget-morphism $~(N,\mathbf{maxi}(R)) \rightarrow (\mathbf{mini}(T),\mathbf{maxi}(T \otimes_{\mathbb{Z}} \mathbb{C}))$ comes from a $\mathbb{Z}$-algebra morphism $T \rightarrow S$. (If this sounds too cryptic for you, please read the series on C-C mentioned before).

So, there is no problem in defining noncommutative affine F_un-schemes. However, as with any generalization, this only makes sense provided (a) we get something new and (b) we have interesting examples, not covered by the restricted theory.

At first sight we do not get something new as in the only example we did in the C-C-series (the forgetful functor) it is easy to prove (using the same proof as given in this post) that the forgetful-functor $\mathbf{groups} \rightarrow \mathbf{sets}$ still has as its integral form the integral torus $\mathbb{Z}[x,x^{-1}]$. However, both theories quickly diverge beyond this example.

For example, consider the functor

$\mathbf{groups} \rightarrow \mathbf{sets} \qquad G \mapsto G \times G$

Then, if we restrict to abelian finite groups $\mathbf{abelian}$ it is easy to see (again by a similar argument) that the two-dimensional integer torus $\mathbb{Z}[x,y,x^{-1},y^{-1}]$ is the correct integral form. However, this algebra cannot be the correct form for the functor on the category of all finite groups as any $\mathbb{Z}$-algebra map $\phi~:~\mathbb{Z}[x,y,x^{-1},y^{-1}] \rightarrow \mathbb{Z} G$ determines (and is determined by) a pair of commuting units in $\mathbb{Z} G$, so the above functor can not be a subfunctor if we allow non-Abelian groups.

But then, perhaps there isn’t a minimal integral $\mathbb{Z}$-form for this functor? Well, yes there is. Take the free group in two letters (that is, all words in noncommuting $x,y,x^{-1}$ and $y^{-1}$ satisfying only the trivial cancellation laws between a letter and its inverse), then the corresponding integral group-algebra $\mathbb{Z} \mathcal{F}_2$ does the trick.

Again, the proof-strategy is the same. Given a gadget-morphism we have an algebra map $f~:~T \mapsto \mathbb{C} \mathcal{F}_2$ and we have to show, using the universal property that the image of $T$ is contained in the integral group-algebra $\mathbb{Z} \mathcal{F}_2$. Take a generator
$z$ of $T$ then the degree of the image $f(z)$ is bounded say by $d$ and we can always find a subgroup $H \subset \mathcal{F}_2$ such that $\mathcal{F}_2/H$ is a fnite group and the quotient map $\mathbb{C} \mathcal{F}_2 \rightarrow \mathbb{C} \mathcal{F}_2/H$ is injective on the subspace spanned by all words of degree strictly less than $d+1$. Then, the usual diagram-chase finishes the proof.

What makes this work is that the free group $\mathcal{F}_2$ has ‘enough’ subgroups of finite index, a property it shares with many interesting discrete groups. Whence the blurb-message :

if the integers $\mathbb{Z}$ see a discrete group $\Gamma$, then the field $\mathbb{F}_1$ sees its profinite completion $\hat{\Gamma} = \underset{\leftarrow}{lim}~\Gamma/ H$

So, yes, we get something new by extending the Connes-Consani approach to the noncommutative world, but do we have interesting examples? As “interesting” is a subjective qualification, we’d better invoke the authority-argument.

Alexander Grothendieck (sitting on the right, manifestly not disputing a vacant chair with Jean-Pierre Serre, drinking on the left (a marvelous picture taken by F. Hirzebruch in 1958)) was pushing the idea that profinite completions of arithmetical groups were useful in the study of the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, via his theory of dessins d’enfants (children;s drawings).

In a previous life, I’ve written a series of posts on dessins d’enfants, so I’ll restrict here to the basics. A smooth projective $\overline{\mathbb{Q}}$-curve $X$ has a Belyi-map $X \rightarrow \mathbb{P}^1_{\overline{\mathbb{Q}}}$ ramified only in three points ${ 0,1,\infty }$. The “drawing” corresponding to $X$ is a bipartite graph, drawn on the Riemann surface $X_{\mathbb{C}}$ obtained by lifting the unit interval $[0,1]$ to $X$. As the absolute Galois group acts on all such curves (and hence on their corresponding drawings), the action of it on these dessins d’enfants may give us a way into the multiple mysteries of the absolute Galois group.

In his “Esquisse d’un programme” (Sketch of a program if you prefer to read it in English) he writes :

“C’est ainsi que mon attention s’est portée vers ce que j’ai appelé depuis la “géométrie algêbrique anabélienne”, dont le point de départ est justement une étude (pour le moment limitée à la caractéristique zéro) de l’action de groupe de Galois “absolus” (notamment les groupes $Gal(\overline{K}/K)$, ou $K$ est une extension de type fini du corps premier) sur des groupes fondamentaux géométriques (profinis) de variétés algébriques (définies sur $K$), et plus particulièrement (rompant avec une tradition bien enracinée) des groupes fondamentaux qui sont trés éloignés des groupes abéliens (et que pour cette raison je nomme “anabéliens”). Parmi ces groupes, et trés proche du groupe $\hat{\pi}_{0,3}$, il y a le compactifié profini du groupe modulaire $SL_2(\mathbb{Z})$, dont le quotient par le centre $\pm 1$ contient le précédent comme sous-groupe de congruence mod 2, et peut s’interpréter d’ailleurs comme groupe “cartographique” orienté, savoir celui qui classifie les cartes orientées triangulées (i.e. celles dont les faces des triangles ou des monogones).”

and a bit further, he writes :

“L’élément de structure de $SL_2(\mathbb{Z})$ qui me fascine avant tout, est bien sur l’action extérieure du groupe de Galois $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ sur le compactifié profini. Par le théorème de Bielyi, prenant les compactifiés profinis de sous-groupes d’indice fini de $SL_2(\mathbb{Z})$, et l’action extérieure induite (quitte à passer également à un sous-groupe overt de $Gal(\overline{\mathbb{Q}},\mathbb{Q})$), on trouve essentiellement les groupes fondamentaux de toutes les courbes algébriques définis sur des corps de nombres $K$, et l’action extérieure de $Gal(\overline{K}/K)$ dessus.”

So, is there a noncommutative affine variety over $\mathbb{F}_1$ of which the unique minimal integral model is the integral group algebra of the modular group $\mathbb{Z} \Gamma$ (with $\Gamma = PSL_2(\mathbb{Z})$? Yes, here it is

$N_{\Gamma}~:~\mathbf{groups} \rightarrow \mathbf{sets} \qquad G \mapsto G_2 \times G_3$

where $G_n$ is the set of all elements of order $n$ in $G$. The reason behind this is that the modular group is the free group product $C_2 \ast C_3$.

Fine, you may say, but all this is just algebra. Where is the noncommutative complex variety or the noncommutative integral scheme in all this? Well, we can introduce them too but as this post is already 1300 words long, I’ll better leave this for another time. In case you cannot stop thinking about it, here’s the short answer.

The complex noncommutative variety has as its ‘points’ all finite dimensional simple complex representations of the modular group, and the ‘points’ of the noncommutative $\mathbb{F}_1$-scheme are exactly the (modular) dessins d’enfants…

A quick recap of last time. We are trying to make sense of affine varieties over the elusive field with one element $\mathbb{F}_1$, which by Grothendieck’s scheme-philosophy should determine a functor

$\mathbf{nano}(N)~:~\mathbf{abelian} \rightarrow \mathbf{sets} \qquad A \mapsto N(A)$

from finite Abelian groups to sets, typically giving pretty small sets $N(A)$. Using the F_un mantra that $\mathbb{Z}$ should be an algebra over $\mathbb{F}_1$ any $\mathbb{F}_1$-variety determines an integral scheme by extension of scalars, as well as a complex variety (by extending further to $\mathbb{C}$). We have already connected the complex variety with the original functor into a gadget that is a couple $~(\mathbf{nano}(N),\mathbf{maxi}(R))$ where $R$ is the coordinate ring of a complex affine variety $X_R$ having the property that every element of $N(A)$ can be realized as a $\mathbb{C} A$-point of $X_R$. Ringtheoretically this simply means that to every element $x \in N(A)$ there is an algebra map $N_x~:~R \rightarrow \mathbb{C} A$.

Today we will determine which gadgets determine an integral scheme, and do so uniquely, and call them the sought for affine schemes over $\mathbb{F}_1$.

Let’s begin with our example : $\mathbf{nano}(N) = \underline{\mathbb{G}}_m$ being the forgetful functor, that is $N(A)=A$ for every finite Abelian group, then the complex algebra $R= \mathbb{C}[x,x^{-1}]$ partners up to form a gadget because to every element $a \in N(A)=A$ there is a natural algebra map $N_a~:~\mathbb{C}[x,x^{-1}] \rightarrow \mathbb{C} A$ defined by sending $x \mapsto e_a$. Clearly, there is an obvious integral form of this complex algebra, namely $\mathbb{Z}[x,x^{-1}]$ but we have already seen that this algebra represents the mini-functor

$\mathbf{min}(\mathbb{Z}[x,x^{-1}])~:~\mathbf{abelian} \rightarrow \mathbf{sets} \qquad A \mapsto (\mathbb{Z} A)^*$

and that the group of units $(\mathbb{Z} A)^*$ of the integral group ring $\mathbb{Z} A$ usually is a lot bigger than $N(A)=A$. So, perhaps there is another less obvious $\mathbb{Z}$-algebra $S$ doing a much better job at approximating $N$? That is, if we can formulate this more precisely…

In general, every $\mathbb{Z}$-algebra $S$ defines a gadget $\mathbf{gadget}(S) = (\mathbf{mini}(S),\mathbf{maxi}(S \otimes_{\mathbb{Z}} \mathbb{C}))$ with the obvious (that is, extension of scalars) evaluation map

$\mathbf{mini}(S)(A) = Hom_{\mathbb{Z}-alg}(S, \mathbb{Z} A) \rightarrow Hom_{\mathbb{C}-alg}(S \otimes_{\mathbb{Z}} \mathbb{C}, \mathbb{C} A) = \mathbf{maxi}(S \otimes_{\mathbb{Z}} \mathbb{C})(A)$

Right, so how might one express the fact that the integral affine scheme $X_T$ with integral algebra $T$ is the ‘best’ integral approximation of a gadget $~(\mathbf{nano}(N),\mathbf{maxi}(R))$. Well, to begin its representing functor should at least contain the information given by $N$, that is, $\mathbf{nano}(N)$ is a sub-functor of $\mathbf{mini}(T)$ (meaning that for every finite Abelian group $A$ we have a natural inclusion $N(A) \subset Hom_{\mathbb{Z}-alg}(T, \mathbb{Z} A)$). As to the “best”-part, we must express that all other candidates factor through $T$. That is, suppose we have an integral algebra $S$ and a morphism of gadgets (as defined last time)

$f~:~(\mathbf{nano}(N),\mathbf{maxi}(R)) \rightarrow \mathbf{gadget}(S) = (\mathbf{mini}(S),\mathbf{maxi}(S \otimes_{\mathbb{Z}} \mathbb{C}))$

then there ought to be $\mathbb{Z}$-algebra morphism $T \rightarrow S$ such that the above map $f$ factors through an induced gadget-map $\mathbf{gadget}(T) \rightarrow \mathbf{gadget}(S)$.

Fine, but is this definition good enough in our trivial example? In other words, is the “obvious” integral ring $\mathbb{Z}[x,x^{-1}]$ the best integral choice for approximating the forgetful functor $N=\underline{\mathbb{G}}_m$? Well, take any finitely generated integral algebra $S$, then saying that there is a morphism of gadgets from $~(\underline{\mathbb{G}}_m,\mathbf{maxi}(\mathbb{C}[x,x^{-1}])$ to $\mathbf{gadget}(S)$ means that there is a $\mathbb{C}$-algebra map $\psi~:~S \otimes_{\mathbb{Z}} \mathbb{C} \rightarrow \mathbb{C}[x,x^{-1}]$ such that for every finite Abelian group $A$ we have a commuting diagram

$\xymatrix{A \ar[rr] \ar[d]_e & & Hom_{\mathbb{Z}-alg}(S, \mathbb{Z} A) \ar[d] \\ Hom_{\mathbb{C}-alg}(\mathbb{C}[x,x^{-1}],\mathbb{C} A) \ar[rr]^{- \circ \psi} & & Hom_{\mathbb{C}-alg}(S \otimes_{\mathbb{Z}} \mathbb{C}, \mathbb{C} A)}$

Here, $e$ is the natural evaluation map defined before sending a group-element $a \in A$ to the algebra map defined by $x \mapsto e_a$ and the vertical map on the right-hand side is extensions by scalars. From this data we must be able to show that the image of the algebra map

$\xymatrix{S \ar[r]^{i} & S \otimes_{\mathbb{Z}} \mathbb{C} \ar[r]^{\psi} & \mathbb{C}[x,x^{-1}]}$

is contained in the integral subalgebra $\mathbb{Z}[x,x^{-1}]$. So, take any generator $z$ of $S$ then its image $\psi(z) \in \mathbb{C}[x,x^{-1}]$ is a Laurent polynomial of degree say $d$ (that is, $\psi(z) = c_{-d} x^{-d} + \ldots c_{-1} x^{-1} + c_0 + c_1 x + \ldots + c_d x^d$ with all coefficients a priori in $\mathbb{C}$ and we need to talk them into $\mathbb{Z}$).

Now comes the basic trick : take a cyclic group $A=C_N$ of order $N > d$, then the above commuting diagram applied to the generator of $C_N$ (the evaluation of which is the natural projection map $\pi~:~\mathbb{C}[x.x^{-1}] \rightarrow \mathbb{C}[x,x^{-1}]/(x^N-1) = \mathbb{C} C_N$) gives us the commuting diagram

$\xymatrix{S \ar[r] \ar[d] & S \otimes_{\mathbb{Z}} \mathbb{C} \ar[r]^{\psi} & \mathbb{C}[x,x^{-1}] \ar[d]^{\pi} \\ \mathbb{Z} C_n = \frac{\mathbb{Z}[x,x^{-1}]}{(x^N-1)} \ar[rr]^j & & \frac{\mathbb{C}[x,x^{-1}]}{(x^N-1)}}$

where the horizontal map $j$ is the natural inclusion map. Tracing $z \in S$ along the diagram we see that indeed all coefficients of $\psi(z)$ have to be integers! Applying the same argument to the other generators of $S$ (possibly for varying values of N) we see that , indeed, $\psi(S) \subset \mathbb{Z}[x,x^{-1}]$ and hence that $\mathbb{Z}[x,x^{-1}]$ is the best integral approximation for $\underline{\mathbb{G}}_m$.

That is, we have our first example of an affine variety over the field with one element $\mathbb{F}_1$ : $~(\underline{\mathbb{G}}_m,\mathbf{maxi}(\mathbb{C}[x,x^{-1}]) \rightarrow \mathbf{gadget}(\mathbb{Z}[x,x^{-1}])$.

What makes this example work is that the infinite group $\mathbb{Z}$ (of which the complex group-algebra is the algebra $\mathbb{C}[x,x^{-1}]$) has enough finite Abelian group-quotients. In other words, $\mathbb{F}_1$ doesn’t see $\mathbb{Z}$ but rather its profinite completion $\hat{\mathbb{Z}} = \underset{\leftarrow} \mathbb{Z}/N\mathbb{Z}$… (to be continued when we’ll consider noncommutative $\mathbb{F}_1$-schemes)

In general, an affine $\mathbb{F}_1$-scheme is a gadget with morphism of gadgets
$~(\mathbf{nano}(N),\mathbf{maxi}(R)) \rightarrow \mathbf{gadget}(S)$ provided that the integral algebra $S$ is the best integral approximation in the sense made explicit before. This rounds up our first attempt to understand the Connes-Consani approach to define geometry over $\mathbb{F}_1$ apart from one important omission : we have only considered functors to $\mathbf{sets}$, whereas it is crucial in the Connes-Consani paper to consider more generally functors to graded sets. In the final part of this series we’ll explain what that’s all about.

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 $\mathbb{B}_n$.