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Tag: topology

Manin’s geometric axis

Mumford’s drawing has a clear emphasis on the vertical direction. The set of all vertical lines corresponds to taking the fibers of the natural ‘structural morphism’ : $\pi~:~\mathbf{spec}(\mathbb{Z}[t]) \rightarrow \mathbf{spec}(\mathbb{Z}) $ coming from the inclusion $\mathbb{Z} \subset \mathbb{Z}[t] $. That is, we consider the intersection $P \cap \mathbb{Z} $ of a prime ideal $P \subset \mathbb{Z}[t] $ with the subring of constants.

Two options arise : either $P \cap \mathbb{Z} \not= 0 $, in which case the intersection is a principal prime ideal $~(p) $ for some prime number $p $ (and hence $P $ itself is bigger or equal to $p\mathbb{Z}[t] $ whence its geometric object is contained in the vertical line $\mathbb{V}((p)) $, the fiber $\pi^{-1}((p)) $ of the structural morphism over $~(p) $), or, the intersection $P \cap \mathbb{Z}[t] = 0 $ reduces to the zero ideal (in which case the extended prime ideal $P \mathbb{Q}[x] = (q(x)) $ is a principal ideal of the rational polynomial algebra $\mathbb{Q}[x] $, and hence the geometric object corresponding to $P $ is a horizontal curve in Mumford’s drawing, or is the whole arithmetic plane itself if $P=0 $).

Because we know already that any ‘point’ in Mumford’s drawing corresponds to a maximal ideal of the form $\mathfrak{m}=(p,f(x)) $ (see last time), we see that every point lies on precisely one of the set of all vertical coordinate axes corresponding to the prime numbers ${~\mathbb{V}((p)) = \mathbf{spec}(\mathbb{F}_p[x]) = \pi^{-1}((p))~} $. In particular, two different vertical lines do not intersect (or, in ringtheoretic lingo, the ‘vertical’ prime ideals $p\mathbb{Z}[x] $ and $q\mathbb{Z}[x] $ are comaximal for different prime numbers $p \not= q $).



That is, the structural morphism is a projection onto the “arithmetic axis” (which is $\mathbf{spec}(\mathbb{Z}) $) and we get the above picture. The extra vertical line to the right of the picture is there because in arithmetic geometry it is customary to include also the archimedean valuations and hence to consider the ‘compactification’ of the arithmetic axis $\mathbf{spec}(\mathbb{Z}) $ which is $\overline{\mathbf{spec}(\mathbb{Z})} = \mathbf{spec}(\mathbb{Z}) \cup { v_{\mathbb{R}} } $.

Yuri I. Manin is advocating for years the point that we should take the terminology ‘arithmetic surface’ for $\mathbf{spec}(\mathbb{Z}[x]) $ a lot more seriously. That is, there ought to be, apart from the projection onto the ‘z-axis’ (that is, the arithmetic axis $\mathbf{spec}(\mathbb{Z}) $) also a projection onto the ‘x-axis’ which he calls the ‘geometric axis’.

But then, what are the ‘points’ of this geometric axis and what are their fibers under this second projection?

We have seen above that the vertical coordinate line over the prime number $~(p) $ coincides with $\mathbf{spec}(\mathbb{F}_p[x]) $, the affine line over the finite field $\mathbb{F}_p $. But all of these different lines, for varying primes $p $, should project down onto the same geometric axis. Manin’s idea was to take therefore as the geometric axis the affine line $\mathbf{spec}(\mathbb{F}_1[x]) $, over the virtual field with one element, which should be thought of as being the limit of the finite fields $\mathbb{F}_p $ when $p $ goes to one!

How many points does $\mathbf{spec}(\mathbb{F}_1[x]) $ have? Over a virtual object one can postulate whatever one wants and hope for an a posteriori explanation. $\mathbb{F}_1 $-gurus tell us that there should be exactly one point of size n on the affine line over $\mathbb{F}_1 $, corresponding to the unique degree n field extension $\mathbb{F}_{1^n} $. However, it is difficult to explain this from the limiting perspective…

Over a genuine finite field $\mathbb{F}_p $, the number of points of thickness $n $ (that is, those for which the residue field is isomorphic to the degree n extension $\mathbb{F}_{p^n} $) is equal to the number of monic irreducible polynomials of degree n over $\mathbb{F}_p $. This number is known to be $\frac{1}{n} \sum_{d | n} \mu(\frac{n}{d}) p^d $ where $\mu(k) $ is the Moebius function. But then, the limiting number should be $\frac{1}{n} \sum_{d | n} \mu(\frac{n}{d}) = \delta_{n1} $, that is, there can only be one point of size one…

Alternatively, one might consider the zeta function counting the number $N_n $ of ideals having a quotient consisting of precisely $p^n $ elements. Then, we have for genuine finite fields $\mathbb{F}_p $ that $\zeta(\mathbb{F}_p[x]) = \sum_{n=0}^{\infty} N_n t^n = 1 + p t + p^2 t^2 + p^3 t^3 + \ldots $, whence in the limit it should become
$1+t+t^2 +t^3 + \ldots $ and there is exactly one ideal in $\mathbb{F}_1[x] $ having a quotient of cardinality n and one argues that this unique quotient should be the unique point with residue field $\mathbb{F}_{1^n} $ (though it might make more sense to view this as the unique n-fold extension of the unique size-one point $\mathbb{F}_1 $ corresponding to the quotient $\mathbb{F}_1[x]/(x^n) $…)

A perhaps more convincing reasoning goes as follows. If $\overline{\mathbb{F}_p} $ is an algebraic closure of the finite field $\mathbb{F}_p $, then the points of the affine line over $\overline{\mathbb{F}_p} $ are in one-to-one correspondence with the maximal ideals of $\overline{\mathbb{F}_p}[x] $ which are all of the form $~(x-\lambda) $ for $\lambda \in \overline{\mathbb{F}_p} $. Hence, we get the points of the affine line over the basefield $\mathbb{F}_p $ as the orbits of points over the algebraic closure under the action of the Galois group $Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p) $.

‘Common wisdom’ has it that one should identify the algebraic closure of the field with one element $\overline{\mathbb{F}_{1}} $ with the group of all roots of unity $\mathbb{\mu}_{\infty} $ and the corresponding Galois group $Gal(\overline{\mathbb{F}_{1}}/\mathbb{F}_1) $ as being generated by the power-maps $\lambda \rightarrow \lambda^n $ on the roots of unity. But then there is exactly one orbit of length n given by the n-th roots of unity $\mathbb{\mu}_n $, so there should be exactly one point of thickness n in $\mathbf{spec}(\mathbb{F}_1[x]) $ and we should then identity the corresponding residue field as $\mathbb{F}_{1^n} = \mathbb{\mu}_n $.

Whatever convinces you, let us assume that we can identify the non-generic points of $\mathbf{spec}(\mathbb{F}_1[x]) $ with the set of positive natural numbers ${ 1,2,3,\ldots } $ with $n $ denoting the unique size n point with residue field $\mathbb{F}_{1^n} $. Then, what are the fibers of the projection onto the geometric axis $\phi~:~\mathbf{spec}(\mathbb{Z}[x]) \rightarrow \mathbf{spec}(\mathbb{F}_1[x]) = { 1,2,3,\ldots } $?

These fibers should correspond to ‘horizontal’ principal prime ideals of $\mathbb{Z}[x] $. Manin proposes to consider $\phi^{-1}(n) = \mathbb{V}((\Phi_n(x))) $ where $\Phi_n(x) $ is the n-th cyclotomic polynomial. The nice thing about this proposal is that all closed points of $\mathbf{spec}(\mathbb{Z}[x]) $ lie on one of these fibers!

Indeed, the residue field at such a point (corresponding to a maximal ideal $\mathfrak{m}=(p,f(x)) $) is the finite field $\mathbb{F}_{p^n} $ and as all its elements are either zero or an $p^n-1 $-th root of unity, it does lie on the curve determined by $\Phi_{p^n-1}(x) $.

As a consequence, the localization $\mathbb{Z}[x]_{cycl} $ of the integral polynomial ring $\mathbb{Z}[x] $ at the multiplicative system generated by all cyclotomic polynomials is a principal ideal domain (as all height two primes evaporate in the localization), and, the fiber over the generic point of $\mathbf{spec}(\mathbb{F}_1[x]) $ is $\mathbf{spec}(\mathbb{Z}[x]_{cycl}) $, which should be compared to the fact that the fiber of the generic point in the projection onto the arithmetic axis is $\mathbf{spec}(\mathbb{Q}[x]) $ and $\mathbb{Q}[x] $ is the localization of $\mathbb{Z}[x] $ at the multiplicative system generated by all prime numbers).

Hence, both the vertical coordinate lines and the horizontal ‘lines’ contain all closed points of the arithmetic plane. Further, any such closed point $\mathfrak{m}=(p,f(x)) $ lies on the intersection of a vertical line $\mathbb{V}((p)) $ and a horizontal one $\mathbb{V}((\Phi_{p^n-1}(x))) $ (if $deg(f(x))=n $).
That is, these horizontal and vertical lines form a coordinate system, at least for the closed points of $\mathbf{spec}(\mathbb{Z}[x]) $.

Still, there is a noticeable difference between the two sets of coordinate lines. The vertical lines do not intersect meaning that $p\mathbb{Z}[x]+q\mathbb{Z}[x]=\mathbb{Z}[x] $ for different prime numbers p and q. However, in general the principal prime ideals corresponding to the horizontal lines $~(\Phi_n(x)) $ and $~(\Phi_m(x)) $ are not comaximal when $n \not= m $, that is, these ‘lines’ may have points in common! This will lead to an exotic new topology on the roots of unity… (to be continued).

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Mumford’s treasure map


David Mumford did receive earlier this year the 2007 AMS Leroy P. Steele Prize for Mathematical Exposition. The jury honors Mumford for “his beautiful expository accounts of a host of aspects of algebraic geometry”. Not surprisingly, the first work they mention are his mimeographed notes of the first 3 chapters of a course in algebraic geometry, usually called “Mumford’s red book” because the notes were wrapped in a red cover. In 1988, the notes were reprinted by Springer-Verlag. Unfortnately, the only red they preserved was in the title.

The AMS describes the importance of the red book as follows. “This is one of the few books that attempt to convey in pictures some of the highly abstract notions that arise in the field of algebraic geometry. In his response upon receiving the prize, Mumford recalled that some of his drawings from The Red Book were included in a collection called Five Centuries of French Mathematics. This seemed fitting, he noted: “After all, it was the French who started impressionist painting and isn’t this just an impressionist scheme for rendering geometry?””

These days it is perfectly possible to get a good grasp on difficult concepts from algebraic geometry by reading blogs, watching YouTube or plugging in equations to sophisticated math-programs. In the early seventies though, if you wanted to know what Grothendieck’s scheme-revolution was all about you had no choice but to wade through the EGA’s and SGA’s and they were notorious for being extremely user-unfriendly regarding illustrations…

So the few depictions of schemes available, drawn by people sufficiently fluent in Grothendieck’s new geometric language had no less than treasure-map-cult-status and were studied in minute detail. Mumford’s red book was a gold mine for such treasure maps. Here’s my favorite one, scanned from the original mimeographed notes (it looks somewhat tidier in the Springer-version)



It is the first depiction of $\mathbf{spec}(\mathbb{Z}[x]) $, the affine scheme of the ring $\mathbb{Z}[x] $ of all integral polynomials. Mumford calls it the”arithmetic surface” as the picture resembles the one he made before of the affine scheme $\mathbf{spec}(\mathbb{C}[x,y]) $ corresponding to the two-dimensional complex affine space $\mathbb{A}^2_{\mathbb{C}} $. Mumford adds that the arithmetic surface is ‘the first example which has a real mixing of arithmetic and geometric properties’.

Let’s have a closer look at the treasure map. It introduces some new signs which must have looked exotic at the time, but have since become standard tools to depict algebraic schemes.

For starters, recall that the underlying topological space of $\mathbf{spec}(\mathbb{Z}[x]) $ is the set of all prime ideals of the integral polynomial ring $\mathbb{Z}[x] $, so the map tries to list them all as well as their inclusions/intersections.

The doodle in the right upper corner depicts the ‘generic point’ of the scheme. That is, the geometric object corresponding to the prime ideal $~(0) $ (note that $\mathbb{Z}[x] $ is an integral domain). Because the zero ideal is contained in any other prime ideal, the algebraic/geometric mantra (“inclusions reverse when shifting between algebra and geometry”) asserts that the gemetric object corresponding to $~(0) $ should contain all other geometric objects of the arithmetic plane, so it is just the whole plane! Clearly, it is rather senseless to depict this fact by coloring the whole plane black as then we wouldn’t be able to see the finer objects. Mumford’s solution to this is to draw a hairy ball, which in this case, is sufficiently thick to include fragments going in every possible direction. In general, one should read these doodles as saying that the geometric object represented by this doodle contains all other objects seen elsewhere in the picture if the hairy-ball-doodle includes stuff pointing in the direction of the smaller object. So, in the case of the object corresponding to $~(0) $, the doodle has pointers going everywhere, saying that the geometric object contains all other objects depicted.

Let’s move over to the doodles in the lower right-hand corner. They represent the geometric object corresponding to principal prime ideals of the form $~(p(x)) $, where $p(x) $ in an irreducible polynomial over the integers, that is, a polynomial which we cannot write as the product of two smaller integral polynomials. The objects corresponding to such prime ideals should be thought of as ‘horizontal’ curves in the plane.

The doodles depicted correspond to the prime ideal $~(x) $, containing all polynomials divisible by $x $ so when we divide it out we get, as expected, a domain $\mathbb{Z}[x]/(x) \simeq \mathbb{Z} $, and the one corresponding to the ideal $~(x^2+1) $, containing all polynomials divisible by $x^2+1 $, which can be proved to be a prime ideals of $\mathbb{Z}[x] $ by observing that after factoring out we get $\mathbb{Z}[x]/(x^2+1) \simeq \mathbb{Z}[i] $, the domain of all Gaussian integers $\mathbb{Z}[i] $. The corresponding doodles (the ‘generic points’ of the curvy-objects) have a predominant horizontal component as they have the express the fact that they depict horizontal curves in the plane. It is no coincidence that the doodle of $~(x^2+1) $ is somewhat bulkier than the one of $~(x) $ as the later one must only depict the fact that all points lying on the straight line to its left belong to it, whereas the former one must claim inclusion of all points lying on the ‘quadric’ it determines.

Apart from these ‘horizontal’ curves, there are also ‘vertical’ lines corresponding to the principal prime ideals $~(p) $, containing the polynomials, all of which coefficients are divisible by the prime number $p $. These are indeed prime ideals of $\mathbb{Z}[x] $, because their quotients are
$\mathbb{Z}[x]/(p) \simeq (\mathbb{Z}/p\mathbb{Z})[x] $ are domains, being the ring of polynomials over the finite field $\mathbb{Z}/p\mathbb{Z} = \mathbb{F}_p $. The doodles corresponding to these prime ideals have a predominant vertical component (depicting the ‘vertical’ lines) and have a uniform thickness for all prime numbers $p $ as each of them only has to claim ownership of the points lying on the vertical line under them.

Right! So far we managed to depict the zero prime ideal (the whole plane) and the principal prime ideals of $\mathbb{Z}[x] $ (the horizontal curves and the vertical lines). Remains to depict the maximal ideals. These are all known to be of the form
$\mathfrak{m} = (p,f(x)) $
where $p $ is a prime number and $f(x) $ is an irreducible integral polynomial, which remains irreducible when reduced modulo $p $ (that is, if we reduce all coefficients of the integral polynomial $f(x) $ modulo $p $ we obtain an irreducible polynomial in $~\mathbb{F}_p[x] $). By the algebra/geometry mantra mentioned before, the geometric object corresponding to such a maximal ideal can be seen as the ‘intersection’ of an horizontal curve (the object corresponding to the principal prime ideal $~(f(x)) $) and a vertical line (corresponding to the prime ideal $~(p) $). Because maximal ideals do not contain any other prime ideals, there is no reason to have a doodle associated to $\mathfrak{m} $ and we can just depict it by a “point” in the plane, more precisely the intersection-point of the horizontal curve with the vertical line determined by $\mathfrak{m}=(p,f(x)) $. Still, Mumford’s treasure map doesn’t treat all “points” equally. For example, the point corresponding to the maximal ideal $\mathfrak{m}_1 = (3,x+2) $ is depicted by a solid dot $\mathbf{.} $, whereas the point corresponding to the maximal ideal $\mathfrak{m}_2 = (3,x^2+1) $ is represented by a fatter point $\circ $. The distinction between the two ‘points’ becomes evident when we look at the corresponding quotients (which we know have to be fields). We have

$\mathbb{Z}[x]/\mathfrak{m}_1 = \mathbb{Z}[x]/(3,x+2)=(\mathbb{Z}/3\mathbb{Z})[x]/(x+2) = \mathbb{Z}/3\mathbb{Z} = \mathbb{F}_3 $ whereas $\mathbb{Z}[x]/\mathfrak{m}_2 = \mathbb{Z}[x]/(3,x^2+1) = \mathbb{Z}/3\mathbb{Z}[x]/(x^2+1) = \mathbb{F}_3[x]/(x^2+1) = \mathbb{F}_{3^2} $

because the polynomial $x^2+1 $ remains irreducible over $\mathbb{F}_3 $, the quotient $\mathbb{F}_3[x]/(x^2+1) $ is no longer the prime-field $\mathbb{F}_3 $ but a quadratic field extension of it, that is, the finite field consisting of 9 elements $\mathbb{F}_{3^2} $. That is, we represent the ‘points’ lying on the vertical line corresponding to the principal prime ideal $~(p) $ by a solid dot . when their quotient (aka residue field is the prime field $~\mathbb{F}_p $, by a bigger point $\circ $ when its residue field is the finite field $~\mathbb{F}_{p^2} $, by an even fatter point $\bigcirc $ when its residue field is $~\mathbb{F}_{p^3} $ and so on, and on. The larger the residue field, the ‘fatter’ the corresponding point.

In fact, the ‘fat-point’ signs in Mumford’s treasure map are an attempt to depict the fact that an affine scheme contains a lot more information than just the set of all prime ideals. In fact, an affine scheme determines (and is determined by) a “functor of points”. That is, to every field (or even every commutative ring) the affine scheme assigns the set of its ‘points’ defined over that field (or ring). For example, the $~\mathbb{F}_p $-points of $\mathbf{spec}(\mathbb{Z}[x]) $ are the solid . points on the vertical line $~(p) $, the $~\mathbb{F}_{p^2} $-points of $\mathbf{spec}(\mathbb{Z}[x]) $ are the solid . points and the slightly bigger $\circ $ points on that vertical line, and so on.

This concludes our first attempt to decypher Mumford’s drawing, but if we delve a bit deeper, we are bound to find even more treasures… (to be continued).

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adeles and ideles

Before we can even attempt to describe the adelic description of the Bost-Connes Hecke algebra and its symmetries, we’d probably better recall the construction and properties of adeles and ideles. Let’s start with the p-adic numbers $\hat{\mathbb{Z}}_p $ and its field of fractions $\hat{\mathbb{Q}}_p $. For p a prime number we can look at the finite rings $\mathbb{Z}/p^n \mathbb{Z} $ of all integer classes modulo $p^n $. If two numbers define the same element in $\mathbb{Z}/p^n\mathbb{Z} $ (meaning that their difference is a multiple of $p^n $), then they certainly define the same class in any $\mathbb{Z}/p^k \mathbb{Z} $ when $k \leq n $, so we have a sequence of ringmorphisms between finite rings

$ \ldots \rightarrow^{\phi_{n+1}} \mathbb{Z}/p^n \mathbb{Z} \rightarrow^{\phi_n} \mathbb{Z}/p^{n-1}\mathbb{Z} \rightarrow^{\phi_{n-1}} \ldots \rightarrow^{\phi_3} \mathbb{Z}/p^2\mathbb{Z} \rightarrow^{\phi_2} \mathbb{Z}/p\mathbb{Z} $

The ring of p-adic integers $\hat{\mathbb{Z}}_p $ can now be defined as the collection of all (infinite) sequences of elements $~(\ldots,x_n,x_{n-1},\ldots,x_2,x_1) $ with $x_i \in \mathbb{Z}/p^i\mathbb{Z} $ such that
$\phi_i(x_i) = x_{i-1} $ for all natural numbers $i $. Addition and multiplication are defined componentswise and as all the maps $\phi_i $ are ringmorphisms, this produces no compatibility problems.

One can put a topology on $\hat{\mathbb{Z}}_p $ making it into a compact ring. Here’s the trick : all components $\mathbb{Z}/p^n \mathbb{Z} $ are finite so they are compact if we equip these sets with the discrete topology (all subsets are opens). But then, Tychonov’s product theorem asserts that the product-space $\prod_n \mathbb{Z}/n \mathbb{Z} $ with the product topology is again a compact topological space. As $\hat{\mathbb{Z}}_p $ is a closed subset, it is compact too.

By construction, the ring $\hat{\mathbb{Z}}_p $ is a domain and hence has a field of fraction which we will denote by $\hat{\mathbb{Q}}_p $. These rings give the p-local information of the rational numbers $\mathbb{Q} $. We will now ‘glue together’ these local data over all possible prime numbers $p $ into adeles. So, forget the above infinite product used to define the p-adics, below we will work with another infinite product, one factor for each prime number.

The adeles $\mathcal{A} $ are the restricted product of the $\hat{\mathbb{Q}}_p $ over $\hat{\mathbb{Z}}_p $ for all prime numbers p. By ‘restricted’ we mean that elements of $\mathcal{A} $ are exactly those infinite vectors $a=(a_2,a_3,a_5,a_7,a_{11},\ldots ) = (a_p)_p \in \prod_p \hat{\mathbb{Q}}_p $ such that all but finitely of the components $a_p \in \hat{\mathbb{Z}}_p $. Addition and multiplication are defined component-wise and the restriction condition is compatible with both adition and multiplication. So, $\mathcal{A} $ is the adele ring. Note that most people call this $\mathcal{A} $ the finite Adeles as we didn’t consider infinite places, i will distinguish between the two notions by writing adeles resp. Adeles for the finite resp. the full blown version. The adele ring $\mathcal{A} $ has as a subring the infinite product $\mathcal{R} = \prod_p \hat{\mathbb{Z}}_p $. If you think of $\mathcal{A} $ as a version of $\mathbb{Q} $ then $\mathcal{R} $ corresponds to $\mathbb{Z} $ (and next time we will see that there is a lot more to this analogy).

The ideles are the group of invertible elements of the ring $\mathcal{A} $, that is, $\mathcal{I} = \mathcal{A}^{\ast} $. That s, an element is an infinite vector $i = (i_2,i_3,i_5,\ldots) = (i_p)_p $ with all $i_p \in \hat{\mathbb{Q}}_p^* $ and for all but finitely many primes we have that $i_p \in \hat{\mathbb{Z}}_p^* $.

As we will have to do explicit calculations with ideles and adeles we need to recall some facts about the structure of the unit groups $\hat{\mathbb{Z}}_p^* $ and $\hat{\mathbb{Q}}_p^* $. If we denote $U = \hat{\mathbb{Z}}_p^* $, then projecting it to the unit group of each of its components we get for each natural number n an exact sequence of groups

$1 \rightarrow U_n \rightarrow U \rightarrow (\mathbb{Z}/p^n \mathbb{Z})^* \rightarrow 1 $. In particular, we have that $U/U_1 \simeq (\mathbb{Z}/p\mathbb{Z})^* \simeq \mathbb{Z}/(p-1)\mathbb{Z} $ as the group of units of the finite field $\mathbb{F}_p $ is cyclic of order p-1. But then, the induced exact sequence of finite abalian groups below splits

$1 \rightarrow U_1/U_n \rightarrow U/U_n \rightarrow \mathbb{F}_p^* \rightarrow 1 $ and as the unit group $U = \underset{\leftarrow}{lim} U/U_n $ we deduce that $U = U_1 \times V $ where $\mathbb{F}_p^* \simeq V = { x \in U | x^{p-1}=1 } $ is the specified unique subgroup of $U $ of order p-1. All that remains is to determine the structure of $U_1 $. If $p \not= 2 $, take $\alpha = 1 + p \in U_1 – U_2 $ and let $\alpha_n \in U_1/U_n $ denote the image of $\alpha $, then one verifies that $\alpha_n $ is a cyclic generator of order $p^{n-1} $ of $U_1/U_n $.

But then, if we denote the isomorphism $\theta_n~:~\mathbb{Z}/p^{n-1} \mathbb{Z} \rightarrow U_1/U_n $ between the ADDITIVE group $\mathbb{Z}/p^{n-1} \mathbb{Z} $ and the MULTIPLICATIVE group $U_1/U_n $ by the map $z \mapsto \alpha_n^z $, then we have a compatible commutative diagram

[tex]\xymatrix{\mathbb{Z}/p^n \mathbb{Z} \ar[r]^{\theta_{n+1}} \ar[d] & U_1/U_{n+1} \ar[d] \\
\mathbb{Z}/p^{n-1} \mathbb{Z} \ar[r]^{\theta_n} & U_1/U_n}[/tex]

and as $U_1 = \underset{\leftarrow}{lim}~U_1/U_n $ this gives an isomorphism between the multiplicative group $U_1 $ and the additive group of $\hat{\mathbb{Z}}_p $. In case $p=2 $ we have to start with an element $\alpha \in U_2 – U_3 $ and repeat the above trick. Summarizing we have the following structural information about the unit group of p-adic integers

$\hat{\mathbb{Z}}_p^* \simeq \begin{cases} \hat{\mathbb{Z}}_{p,+} \times \mathbb{Z}/(p-1)\mathbb{Z}~(p \not= 2) \\ \hat{\mathbb{Z}}_{2,+} \times \mathbb{Z}/2 \mathbb{Z}~(p=2) \end{cases}$

Because every unit in $\hat{\mathbb{Q}}_p^* $ can be written as $p^n u $ with $u \in \hat{\mathbb{Z}}_p^* $ we deduce from this also the structure of the unit group of the p-adic field

$\hat{\mathbb{Q}}_p^* \simeq \begin{cases} \mathbb{Z} \times \hat{\mathbb{Z}}_{p,+} \times \mathbb{Z}/(p-1)\mathbb{Z}~(p \not= 2) \\ \mathbb{Z} \times \hat{\mathbb{Z}}_{2,+} \times \mathbb{Z}/2 \mathbb{Z}~(p=2) \end{cases} $

Right, now let us start to make the connection with the apparently abstract ringtheoretical post from last time where we introduced semigroup crystalline graded rings without explaining why we wanted that level of generality.

Consider the semigroup $\mathcal{I} \cap \mathcal{R} $, that is all ideles $i = (i_p)_p $ with all $i_p = p^{n_p} u_p $ with $u_p \in \hat{\mathbb{Z}}_p^* $ and $n_p \in \mathbb{N} $ with $n_p=0 $ for all but finitely many primes p. Then, we have an exact sequence of semigroups

$1 \rightarrow \mathcal{G} \rightarrow \mathcal{I} \cap \mathcal{R} \rightarrow^{\pi} \mathbb{N}^+_{\times} \rightarrow 1 $ where the map is defined (with above notation) $\pi(i) = \prod_p p^{n_p} $ and exactness follows from the above structural results when we take $\mathcal{G} = \prod_p \hat{\mathbb{Z}}_p^* $.

This gives a glimpse of where we are heading. Last time we identified the Bost-Connes Hecke algebra $\mathcal{H} $ as a bi-crystalline group graded algebra determined by a $\mathbb{N}^+_{\times} $-semigroup crystalline graded algebra over the group algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $. Next, we will entend this construction starting from a $\mathcal{I} \cap \mathcal{R} $-semigroup crystalline graded algebra over the same group algebra. The upshot is that we will have a natural action by automorphisms of the group $\mathcal{G} $ on the Bost-Connes algebra. And… the group $\mathcal{G} = \prod_p \hat{\mathbb{Z}}_p^* $ is the Galois group of the cyclotomic field extension $\mathbb{Q}^{cyc} $!

But, in order to begin to understand this, we will need to brush up our rusty knowledge of algebraic number theory…

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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 $.

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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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M-geometry (2)

Last time we introduced the tangent quiver $\vec{t}~A $ of an affine algebra A to be a quiver on the isoclasses of simple finite dimensional representations. When $A=\mathbb{C}[X] $ is the coordinate ring of an affine variety, these vertices are just the points of the variety $X $ and this set has the extra structure of being endowed with the Zariski topology. For a general, possibly noncommutative algebra, we would like to equip the vertices of $\vec{t}~A $ also with a topology.

In the commutative case, the Zariski topology has as its closed sets the common zeroes of a set of polynomials on $X $, so we need to generalize the notion of ‘functions’ the the noncommutative world. The NC-mantra states that we should view the algebra A as the ring of functions on a (usually virtual) noncommutative space. And, face it, for a commutative variety $X $ the algebra $A=\mathbb{C}[X] $ does indeed do the job. Still, this is a red herring.

Let’s consider the easiest noncommutative case, that of the group algebra $\mathbb{C} G $ of a finite group $G $. In this case, the vertices of the tangent quiver $\vec{t}~A $ are the irreducible representations of $G $ and no sane person would consider the full group algebra to be the algebra of functions on this set. However, we do have a good alternative in this case : characters which allow us to separate the irreducibles and are a lot more manageable than the full group algebra. For example, if $G $ is the monster group then the group algebra has dimension approx $8 \times 10^{53} $ whereas there are just 194 characters to consider…

But, can we extend characters to arbitrary noncommutative algebras? and, more important, are there enough of these to separate the simple representations? The first question is easy enough to answer, after all characters are just traces so we can define for every element $a \in A $ and any finite dimensional simple A-representation $S $ the character

$\chi_a(S) = Tr(a | S) $

where $a | S $ is the matrix describing the action of a on S. But, you might say, characters are then just linear functionals on the algebra A so it is natural to view A as the function algebra, right? Wrong! Traces have the nice property that $Tr(ab)=Tr(ba) $ and so they vanish on all commutators $[a,b]=ab-ba $ of A, so characters only carry information of the quotient space

$\mathfrak{g}_A = \frac{A}{[A,A]_{vect}} $

where $[A,A]_{vect} $ is the vectorspace spanned by all commutators (and not the ideal…). If one is too focussed on commutative geometry one misses this essential simplification as clearly for $A=\mathbb{C}[X] $ being a commutative algebra,

$[\mathbb{C}[X],\mathbb{C}[X]]_{vect}=0 $ and therefore in this case $\mathfrak{g}_{\mathbb{C}[X]} = \mathbb{C}[X] $

Ok, but are there enough characters (that is, linear functionals on $\mathfrak{g}_A $, that is elements of the dual space $\mathfrak{g}_A^* $) to separate the simple representations? And, why do I (ab)use Lie-algebra notation $\mathfrak{g}_A $ to denote the vectorspace $A/[A,A]_{vect} $???

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the Manin-Marcolli cave

Yesterday, Yuri Manin and Matilde Marcolli arXived their paper
Modular shadows and the Levy-Mellin infinity-adic transform which is a
follow-up of their previous paper Continued fractions, modular symbols, and non-commutative geometry.
They motivate the title of the recent paper by :

In
[MaMar2](http://www.arxiv.org/abs/hep-th/0201036), these and similar
results were put in connection with the so called “holography”
principle in modern theoretical physics. According to this principle,
quantum field theory on a space may be faithfully reflected by an
appropriate theory on the boundary of this space. When this boundary,
rather than the interior, is interpreted as our observable
space‚Äìtime, one can proclaim that the ancient Plato’s cave metaphor
is resuscitated in this sophisticated guise. This metaphor motivated
the title of the present paper.

Here’s a layout of
Plato’s cave

Imagine prisoners, who have been chained since childhood deep inside an
cave: not only are their limbs immobilized by the chains; their heads
are chained as well, so that their gaze is fixed on a wall.
Behind
the prisoners is an enormous fire, and between the fire and the
prisoners is a raised walkway, along which statues of various animals,
plants, and other things are carried by people. The statues cast shadows
on the wall, and the prisoners watch these shadows. When one of the
statue-carriers speaks, an echo against the wall causes the prisoners to
believe that the words come from the shadows.
The prisoners
engage in what appears to us to be a game: naming the shapes as they
come by. This, however, is the only reality that they know, even though
they are seeing merely shadows of images. They are thus conditioned to
judge the quality of one another by their skill in quickly naming the
shapes and dislike those who begin to play poorly.
Suppose a
prisoner is released and compelled to stand up and turn around. At that
moment his eyes will be blinded by the firelight, and the shapes passing
will appear less real than their shadows.

Right, now how
does the Manin-Marcolli cave look? My best guess is : like this
picture, taken from Curt McMullen’s Gallery

Imagine
this as the top view of a spherical cave. M&M are imprisoned in the
cave, their heads chained preventing them from looking up and see the
ceiling (where $PSL_2(\mathbb{Z}) $ (or a cofinite subgroup of
it) is acting on the upper-half plane via
Moebius-transformations ). All they can see is the circular exit of the
cave. They want to understand the complex picture going on over their
heads from the only things they can observe, that is the action of
(subgroups of) the modular group on the cave-exit
$\mathbb{P}^1(\mathbb{R}) $. Now, the part of it consisting
of orbits of cusps
$\mathbb{P}^1(\mathbb{Q}) $ has a nice algebraic geometric
description, but orbits of irrational points cannot be handled by
algebraic geometry as the action of $PSL_2(\mathbb{Z}) $ is
highly non-discrete as illustrated by another picture from McMullen’s
gallery

depicting the ill behaved topology of the action on the bottom real
axis. Still, noncommutative _differential_ geometry is pretty good at
handling such ill behaved quotient spaces and it turns out that as a
noncommutative space, this quotient
$\mathbb{P}^1(\mathbb{R})/PSL_2(\mathbb{Z}) $ is rich enough
to recover many important aspects of the classical theory of modular
curves. Hence, they reverse the usual NCG-picture of interpreting
commutative objects as shadows of noncommutative ones. They study the
_noncommutative shadow_
$\mathbb{P}^1(\mathbb{R})/PSL_2(\mathbb{Z}) $ of a classical
commutative object, the quotient of the action of the modular group (or
a cofinite subgroup of it) on the upper half-plane.

In our
noncommutative geometry course we have already
seen this noncommutative shadow in action (though at a very basic
level). Remember that we first described the group-structure of the
modular group $PSL_2(\mathbb{Z}) = C_2 \ast C_3 $ via the
classical method of groups acting on trees. In particular, we
considered the tree

and
calculated the stabilizers of the end points of its fundamental domain
(the thick circular edge). But
later we were able to give a
much shorter proof (due to Roger Alperin) by looking only at the action
of $PSL_2(\mathbb{Z}) $ on the irrational real numbers (the
noncommutative shadow). Needless to say that the results obtained by
Manin and Marcolli from staring at their noncommutative shadow are a lot
more intriguing…

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