Tag: topology

Mazur’s knotty dictionary

Published December 27, 2008 by lieven

In the previous posts, we have depicted the ‘arithmetic line’, that is the prime numbers, as a ‘line’ and individual primes as ‘points’.

However, sometime in the roaring 60-ties, Barry Mazur launched the crazy idea of viewing the affine spectrum of the integers, $spec (Z)$ , as a 3-dimensional manifold and prime numbers themselves as knots in this 3-manifold…

After a long silence, this idea was taken up recently by Mikhail Kapranov and Alexander Reznikov (1960-2003) in a talk at the MPI-Bonn in august 1996. Pieter Moree tells the story in his recollections about Alexander (Sacha) Reznikov in Sipping Tea with Sacha : “Sasha’s paper is closely related to his paper where the analogy of covers of three-manifolds and class field theory plays a big role (an analogy that was apparently first noticed by B. Mazur). Sasha and Mikhail Kapranov (at the time also at the institute) were both very interested in this analogy. Eventually, in August 1996, Kapranov and Reznikov both lectured on this (and I explained in about 10 minutes my contribution to Reznikov’s proof). I was pleased to learn some time ago that this lecture series even made it into the literature, see Morishita’s ‘On certain analogies between knots and primes’ J. reine angew. Math 550 (2002) 141-167.”

Here’s a part of what is now called the Kapranov-Reznikov-Mazur dictionary :

What is the rationale behind this dictionary? Well, it all has to do with trying to make sense of the (algebraic) fundamental group $π_{1}^{a l g} (X)$ of a general scheme $X$ . Recall that for a manifold $M$ there are two different ways to define its fundamental group $π_{1} (M)$ : either as the closed loops in a given basepoint upto homotopy or as the automorphism group of the universal cover $\tilde{M}$ of $M$ .

For an arbitrary scheme the first definition doesn’t make sense but we can use the second one as we have a good notion of a (finite) cover : an etale morphism $Y \to X$ of the scheme $X$ . As they form an inverse system, we can take their finite automorphism groups $A u t_{X} (Y)$ and take their projective limit along the system and call this the algebraic fundamental group $π_{1}^{a l g} (X)$ .

Hendrik Lenstra has written beautiful course notes on ‘Galois theory for schemes’ on all of this starting from scratch. Besides, there are also two video-lectures available on this at the MSRI-website : Etale fundamental groups 1 by H.W. Lenstra and Etale fundamental groups 2 by F. Pop.

But, what is the connection with the ‘usual’ fundamental group in case both of them can be defined? Well, by construction the algebraic fundamental group is always a profinite group and in the case of manifolds it coincides with the profinite completion of the standard fundamental group, that is,
$π_{1}^{a l g} (M) ≃ \hat{π_{1} (M)}$ (recall that the cofinite completion is the projective limit of all finite group quotients).

Right, so all we have to do to find a topological equivalent of an algebraic scheme is to compute its algebraic fundamental group and find an existing topological space of which the profinite completion of its standard fundamental group coincides with our algebraic fundamental group. An example : a prime number $p$ (as a ‘point’ in $spec (Z)$ ) is the closed subscheme $spec (F_{p})$ corresponding to the finite field $F_{p} = Z / p Z$ . For any affine scheme of a field $K$ , the algebraic fundamental group coincides with the absolute Galois group $G a l (\overset{―}{K} / K)$ . In the case of $F_{p}$ we all know that this abslute Galois group is isomorphic with the profinite integers $\hat{Z}$ . Now, what is the first topological space coming to mind having the integers as its fundamental group? Right, the circle $S^{1}$ . Hence, in arithmetic topology we view prime numbers as topological circles, that is, as knots in some bigger space.

But then, what is this bigger space? That is, what is the topological equivalent of $spec (Z)$ ? For this we have to go back to Mazur’s original paper Notes on etale cohomology of number fields in which he gives an Artin-Verdier type duality theorem for the affine spectrum $X = spec (D)$ of the ring of integers $D$ in a number field. More precisely, there is a non-degenerate pairing $H_{e t}^{r} (X, F) \times E x t_{X}^{3 - r} (F, G_{m}) \to H_{e t}^{3} (X, F) ≃ Q / Z$ for any constructible abelian sheaf $F$ . This may not tell you much, but it is a ‘sort of’ Poincare-duality result one would have for a compact three dimensional manifold.

Ok, so in particular $spec (Z)$ should be thought of as a 3-dimensional compact manifold, but which one? For this we have to compute the algebraic fundamental group. Fortunately, this group is trivial as there are no (non-split) etale covers of $spec (Z)$ , so the corresponding 3-manifold should be simple connected… but wenow know that this has to imply that the manifold must be $S^{3}$ , the 3-sphere! Summarizing : in arithmetic topology, prime numbers are knots in the 3-sphere!

More generally (by the same arguments) the affine spectrum $spec (D)$ of a ring of integers can be thought of as corresponding to a closed oriented 3-dimensional manifold $M$ (which is a cover of $S^{3}$ ) and a prime ideal $p ◃ D$ corresponds to a knot in $M$ .

But then, what is an ideal $a ◃ D$ ? Well, we have unique factorization of ideals in $D$ , that is, $a = p_{1}^{n_{1}} \dots p_{k}^{n_{k}}$ and therefore $a$ corresponds to a link in $M$ of which the constituent knots are the ones corresponding to the prime ideals $p_{i}$ .

And we can go on like this. What should be an element $w \in D$ ? Well, it will be an embedded surface $S \to M$ , possibly with a boundary, the boundary being the link corresponding to the ideal $a = D w$ and Seifert’s algorithm tells us how we can produce surfaces having any prescribed link as its boundary. But then, in particular, a unit $w \in D^{*}$ should correspond to a closed surface in $M$ .

And all these analogies carry much further : for example the class group of the ring of integers $C l (D)$ then corresponds to the torsion part $H_{1} (M, Z)_{t o r}$ because principal ideals $D w$ are trivial in the class group, just as boundaries of surfaces $\partial S$ vanish in $H_{1} (M, Z)$ . Similarly, one may identify the unit group $D^{*}$ with $H_{2} (M, Z)$ … and so on, and on, and on…

More links to papers on arithmetic topology can be found in John Baez’ week 257 or via here.

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Manin’s geometric axis

Published December 23, 2008 by lieven

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’ : $π : spec (Z [t]) \to spec (Z)$ coming from the inclusion $Z \subset Z [t]$ . That is, we consider the intersection $P \cap Z$ of a prime ideal $P \subset Z [t]$ with the subring of constants.

Two options arise : either $P \cap Z \neq 0$ , 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 $p Z [t]$ whence its geometric object is contained in the vertical line $V ((p))$ , the fiber $π^{- 1} ((p))$ of the structural morphism over $(p)$ ), or, the intersection $P \cap Z [t] = 0$ reduces to the zero ideal (in which case the extended prime ideal $P Q [x] = (q (x))$ is a principal ideal of the rational polynomial algebra $Q [x]$ , 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 $m = (p, f (x))$ (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 $V ((p)) = spec (F_{p} [x]) = π^{- 1} ((p))$ . In particular, two different vertical lines do not intersect (or, in ringtheoretic lingo, the ‘vertical’ prime ideals $p Z [x]$ and $q Z [x]$ are comaximal for different prime numbers $p \neq q$ ).

That is, the structural morphism is a projection onto the “arithmetic axis” (which is $spec (Z)$ ) 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 $spec (Z)$ which is $\overset{―}{spec (Z)} = spec (Z) \cup v_{R}$ .

Yuri I. Manin is advocating for years the point that we should take the terminology ‘arithmetic surface’ for $spec (Z [x])$ a lot more seriously. That is, there ought to be, apart from the projection onto the ‘z-axis’ (that is, the arithmetic axis $spec (Z)$ ) 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 $spec (F_{p} [x])$ , the affine line over the finite field $F_{p}$ . 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 $spec (F_{1} [x])$ , over the virtual field with one element, which should be thought of as being the limit of the finite fields $F_{p}$ when $p$ goes to one!

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

Over a genuine finite field $F_{p}$ , the number of points of thickness $n$ (that is, those for which the residue field is isomorphic to the degree n extension $F_{p^{n}}$ ) is equal to the number of monic irreducible polynomials of degree n over $F_{p}$ . This number is known to be $\frac{1}{n} \sum_{d | n} μ (\frac{n}{d}) p^{d}$ where $μ (k)$ is the Moebius function. But then, the limiting number should be $\frac{1}{n} \sum_{d | n} μ (\frac{n}{d}) = δ_{n 1}$ , 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 $F_{p}$ that $ζ (F_{p} [x]) = \sum_{n = 0}^{\infty} N_{n} t^{n} = 1 + p t + p^{2} t^{2} + p^{3} t^{3} + \dots$ , whence in the limit it should become
$1 + t + t^{2} + t^{3} + \dots$ and there is exactly one ideal in $F_{1} [x]$ having a quotient of cardinality n and one argues that this unique quotient should be the unique point with residue field $F_{1^{n}}$ (though it might make more sense to view this as the unique n-fold extension of the unique size-one point $F_{1}$ corresponding to the quotient $F_{1} [x] / (x^{n})$ …)

A perhaps more convincing reasoning goes as follows. If $\overset{―}{F_{p}}$ is an algebraic closure of the finite field $F_{p}$ , then the points of the affine line over $\overset{―}{F_{p}}$ are in one-to-one correspondence with the maximal ideals of $\overset{―}{F_{p}} [x]$ which are all of the form $(x - λ)$ for $λ \in \overset{―}{F_{p}}$ . Hence, we get the points of the affine line over the basefield $F_{p}$ as the orbits of points over the algebraic closure under the action of the Galois group $G a l (\overset{―}{F_{p}} / F_{p})$ .

‘Common wisdom’ has it that one should identify the algebraic closure of the field with one element $\overset{―}{F_{1}}$ with the group of all roots of unity $μ_{\infty}$ and the corresponding Galois group $G a l (\overset{―}{F_{1}} / F_{1})$ as being generated by the power-maps $λ \to λ^{n}$ on the roots of unity. But then there is exactly one orbit of length n given by the n-th roots of unity $μ_{n}$ , so there should be exactly one point of thickness n in $spec (F_{1} [x])$ and we should then identity the corresponding residue field as $F_{1^{n}} = μ_{n}$ .

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

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

Indeed, the residue field at such a point (corresponding to a maximal ideal $m = (p, f (x))$ ) is the finite field $F_{p^{n}}$ 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 $Φ_{p^{n} - 1} (x)$ .

As a consequence, the localization $Z [x]_{c y c l}$ of the integral polynomial ring $Z [x]$ 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 $spec (F_{1} [x])$ is $spec (Z [x]_{c y c l})$ , which should be compared to the fact that the fiber of the generic point in the projection onto the arithmetic axis is $spec (Q [x])$ and $Q [x]$ is the localization of $Z [x]$ 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 $m = (p, f (x))$ lies on the intersection of a vertical line $V ((p))$ and a horizontal one $V ((Φ_{p^{n} - 1} (x)))$ (if $d e g (f (x)) = n$ ).
That is, these horizontal and vertical lines form a coordinate system, at least for the closed points of $spec (Z [x])$ .

Still, there is a noticeable difference between the two sets of coordinate lines. The vertical lines do not intersect meaning that $p Z [x] + q Z [x] = Z [x]$ for different prime numbers p and q. However, in general the principal prime ideals corresponding to the horizontal lines $(Φ_{n} (x))$ and $(Φ_{m} (x))$ are not comaximal when $n \neq m$ , 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

Published December 13, 2008 by lieven

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 $spec (Z [x])$ , the affine scheme of the ring $Z [x]$ of all integral polynomials. Mumford calls it the”arithmetic surface” as the picture resembles the one he made before of the affine scheme $spec (C [x, y])$ corresponding to the two-dimensional complex affine space $A_{C}^{2}$ . 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 $spec (Z [x])$ is the set of all prime ideals of the integral polynomial ring $Z [x]$ , 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 $Z [x]$ 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 $Z [x] / (x) ≃ Z$ , 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 $Z [x]$ by observing that after factoring out we get $Z [x] / (x^{2} + 1) ≃ Z [i]$ , the domain of all Gaussian integers $Z [i]$ . 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 $Z [x]$ , because their quotients are
$Z [x] / (p) ≃ (Z / p Z) [x]$ are domains, being the ring of polynomials over the finite field $Z / p Z = F_{p}$ . 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 $Z [x]$ (the horizontal curves and the vertical lines). Remains to depict the maximal ideals. These are all known to be of the form
$m = (p, f (x))$
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 $F_{p} [x]$ ). 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 $m$ 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 $m = (p, f (x))$ . Still, Mumford’s treasure map doesn’t treat all “points” equally. For example, the point corresponding to the maximal ideal $m_{1} = (3, x + 2)$ is depicted by a solid dot $.$ , whereas the point corresponding to the maximal ideal $m_{2} = (3, x^{2} + 1)$ 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

$Z [x] / m_{1} = Z [x] / (3, x + 2) = (Z / 3 Z) [x] / (x + 2) = Z / 3 Z = F_{3}$ whereas $Z [x] / m_{2} = Z [x] / (3, x^{2} + 1) = Z / 3 Z [x] / (x^{2} + 1) = F_{3} [x] / (x^{2} + 1) = F_{3^{2}}$

because the polynomial $x^{2} + 1$ remains irreducible over $F_{3}$ , the quotient $F_{3} [x] / (x^{2} + 1)$ is no longer the prime-field $F_{3}$ but a quadratic field extension of it, that is, the finite field consisting of 9 elements $F_{3^{2}}$ . 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 $F_{p}$ , by a bigger point $\circ$ when its residue field is the finite field $F_{p^{2}}$ , by an even fatter point $◯$ when its residue field is $F_{p^{3}}$ 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 $F_{p}$ -points of $spec (Z [x])$ are the solid . points on the vertical line $(p)$ , the $F_{p^{2}}$ -points of $spec (Z [x])$ 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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