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Mumford’s treasure map

Dec 13th

Brave New Geometries

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 $\wis{spec}(\Z[x])$ , 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 $\wis{spec}(\C[x,y])$ corresponding to the two-dimensional complex affine space $\mathbb{A}^2_{\C}$ . 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 $\wis{spec}(\Z[x])$ 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 so when we divide it out we get, as expected, a domain $\Z[x]/(x) \simeq \Z$ , 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 $\Z[x]/(x^2+1) \simeq \Z[i]$ , 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 . These are indeed prime ideals of $\Z[x]$ , because their quotients are $\Z[x]/(p) \simeq (\Z/p\Z)[x]$ are domains, being the ring of polynomials over the finite field $\Z/p\Z = \mathbb{F}_p$ . 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 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 $\mathfrak{m} = (p,f(x))$ where is a prime number and f(x) is an irreducible integral polynomial, which remains irreducible when reduced modulo (that is, if we reduce all coefficients of the integral polynomial f(x) modulo 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

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

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 $\wis{spec}(\Z[x])$ are the solid . points on the vertical line ~(p) , the $~\mathbb{F}_{p^2}$ -points of $\wis{spec}(\Z[x])$ 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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geometry, Grothendieck, Manin, modular

the Bost-Connes coset space

Jan 17th

Posted by lievenlb in featured

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Noncommutative geometry and the Riemann zeta function

By now, everyone remotely interested in Connes’ approach to the Riemann hypothesis, knows the one line mantra

one can use noncommutative geometry to extend Weil’s proof of the Riemann-hypothesis in the function field case to that of number fields

But, can one go beyond this sound-bite in a series of blog posts? A few days ago, I was rather optimistic, but now, after reading-up on the Connes-Consani-Marcolli project, I feel overwhelmed by the sheer volume of their work (and by my own ignorance of key tools in the approach). The most recent account takes up half of the 700+ pages of the book Noncommutative Geometry, Quantum Fields and Motives by Alain Connes and Matilde Marcolli…

So let us set a more modest goal and try to understand one of the first papers Alain Connes wrote about the RH : Noncommutative geometry and the Riemann zeta function. It is only 24 pages long and relatively readable. But even then, the reader needs to know about class field theory, the classification of AF-algebras, Hecke algebras, etc. etc. Most of these theories take a book to explain. For example, the first result he mentions is the main result of local class field theory which appears only towards the end of the 200+ pages of Jean-Pierre Serre’s Local Fields, itself a somewhat harder read than the average blogpost…

Anyway, we will see how far we can get. Here’s the plan : I’ll take the heart-bit of their approach : the Bost-Connes system, and will try to understand it from an algebraist’s viewpoint. Today we will introduce the groups involved and describe their cosets.

For any commutative ring let us consider the group of triangular $2 \times 2$ matrices of the form

$P_R = \{ \begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix}~|~b \in R, a \in R^* \}$

(that is, in an invertible element in the ring ). This is really an affine group scheme defined over the integers, that is, the coordinate ring

$\mathbb{Z}[P] = \mathbb{Z}[x,x^{-1},y]$ becomes a Hopf algebra with comultiplication encoding the group-multiplication. Because

$\begin{bmatrix} 1 & b_1 \\ 0 & a_1 \end{bmatrix} \begin{bmatrix} 1 & b_2 \\ 0 & a_2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \times b_2 + b_1 \times a_2 \\ 0 & a_1 \times a_2 \end{bmatrix}$

we have $\Delta(x) = x \otimes x$ and $\Delta(y) = 1 \otimes y + y \otimes x$ , or is a group-like element whereas is a skew-primitive. If $R \subset \mathbb{R}$ is a subring of the real numbers, we denote by P_R^+ the subgroup of P_R consisting of all matrices with a > 0 . For example,

$\Gamma_0 = P_{\mathbb{Z}}^+ = \{ \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}~|~n \in \mathbb{Z} \}$

which is a subgroup of $\Gamma = P_{\mathbb{Q}}^+$ and our first job is to describe the cosets.

The left cosets $\Gamma / \Gamma_0$ are the subsets $\gamma \Gamma_0$ with $\gamma \in \Gamma$ . But,

$\begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix} \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & b+n \\ 0 & a \end{bmatrix}$

so if we represent the matrix $\gamma = \begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix}$ by the point ~(a,b) in the right halfplane, then for a given positive rational number the different cosets are represented by all $b \in [0,1) \cap \mathbb{Q} = \mathbb{Q}/\mathbb{Z}$ . Hence, the left cosets are all the rational points in the region between the red and green horizontal lines. For fixed the cosets correspond to the rational points in the green interval (such as over $\frac{2}{3}$ in the picture on the left.

Similarly, the right cosets $\Gamma_0 \backslash \Gamma$ are the subsets $\Gamma_0 \gamma$ and as

$\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix} = \begin{bmatrix} 1 & b+na \\ 0 & a \end{bmatrix}$

we see similarly that the different cosets are precisely the rational points in the region between the lower red horizontal and the blue diagonal line. So, for fixed they correspond to rational points in the blue interval (such as over $\frac{3}{2}$ ) $[0,a) \cap \mathbb{Q}$ . But now, let us look at the double coset space $\Gamma_0 \backslash \Gamma / \Gamma_0$ . That is, we want to study the orbits of the action of $\Gamma_0$ , acting on the right, on the left-cosets $\Gamma / \Gamma_0$ , or equivalently, of the action of $\Gamma_0$ acting on the left on the right-cosets $\Gamma_0 \backslash \Gamma$ . The crucial observation to make is that these actions have finite orbits, or equivalently, that $\Gamma_0$ is an almost normal subgroup of $\Gamma$ meaning that $\Gamma_0 \cap \gamma \Gamma_0 \gamma^{-1}$ has finite index in $\Gamma_0$ for all $\gamma \in \Gamma$ . This follows from

$\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix} \begin{bmatrix} 1 & m \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & b+m+an \\ 0 & a \end{bmatrix}$

and if varies then takes only finitely many values modulo $\mahbb{Z}$ and their number depends only on the denominator of . In the picture above, the blue dots lying on the line over $\frac{2}{3}$ represent the double coset

$\Gamma_0 \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & \frac{2}{3} \end{bmatrix}$ and we see that these dots split the left-cosets with fixed value $a=\frac{2}{3}$ (that is, the green line-segment) into three chunks (3 being the denominator of a) and split the right-cosets (the line-segment under the blue diagonal) into two subsegments (2 being the numerator of a). Similarly, the blue dots on the line over $\frac{3}{2}$ divide the left-cosets in two parts and the right cosets into three parts.

This shows that the $\Gamma_0$ -orbits of the right action on the left cosets $\Gamma/\Gamma_0$ for each matrix $\gamma \in \Gamma$ with $a=\frac{2}{3}$ consist of exactly three points, and we denote this by writing $L(\gamma) = 3$ . Similarly, all $\Gamma_0$ -orbits of the left action on the right cosets $\Gamma_0 \backslash \Gamma$ with this value of a consist of two points, and we write this as $R(\gamma) = 2$ .

For example, on the above picture, the black dots on the line over $\frac{2}{3}$ give the matrices in the double coset of the matrix

$\gamma = \begin{bmatrix} 1 & \frac{1}{7} \\ 0 & \frac{2}{3} \end{bmatrix}$

and the gray dots on the line over $\frac{3}{2}$ determine the elements of the double coset of

$\gamma^{-1} = \begin{bmatrix} 1 & -\frac{3}{14} \\ 0 & \frac{3}{2} \end{bmatrix}$

and one notices (in general) that $L(\gamma) = R(\gamma^{-1})$ . But then, the double cosets with $a=\frac{2}{3}$ are represented by the rational b’s in the interval $[0,\frac{1}{3})$ and those with $a=\frac{3}{2}$ by the rational b’s in the interval $\frac{1}{2}$ . In general, the double cosets of matrices with fixed $a = \frac{r}{s}$ with ~(r,s)=1 are the rational points in the line-segment over with $b \in [0,\frac{1}{s})$ .

That is, the Bost-Connes double coset space $\Gamma_0 \backslash \Gamma / \Gamma_0$ are the rational points in a horrible fractal comb. Below we have drawn only the part of the dyadic values, that is when $a = \frac{r}{2^t}$ in the unit inverval

and of course we have to super-impose on it similar pictures for rationals with other powers as their denominators. Fortunately, NCG excels in describing such fractal beasts…

UPDATE : here is a slightly beter picture of the coset space, drawing the part over all rational numbers contained in the 15-th Farey sequence. The blue segments of length one are at 1,2,3,…

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Connes, geometry, groups, Marcolli, noncommutative, Riemann

Conway’s puzzle M(13)

Jun 16th

Posted by lievenlb in featured

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Mathieu-games

Recently, I’ve been playing with the idea of writing a book for the general public. Its title is still unclear to me (though an idea might be “The disposable science”, better suggestions are of course wellcome) but I’ve fixed the subtitle as “Mathematics’ puzzling fall from grace”. The book’s concept is simple : I would consider the mathematical puzzles creating an hype over the last three centuries : the 14-15 puzzle for the 19th century, Rubik’s cube for the 20th century and, of course, Sudoku for the present century.

For each puzzle, I would describe its origin, the mathematics involved and how it can be used to solve the puzzle and, finally, what the differing quality of these puzzles tells us about mathematics’ changing standing in society over the period. Needless to say, the subtitle already gives away my point of view. The final part of the book would then be more optimistic. What kind of puzzles should we promote for mathematical thinking to have a fighting chance to survive in the near future?

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15-puzzle, apple, arxiv, Conway, Elkies, Mathieu, puzzle, sudoku

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