7
$\begingroup$

Can any properties of a ring other than being a field be captured by the geometry of its 2-dimensional free module?

Background: In his wonderful, wonderful book Geometric Algebra, Emil Artin describes the following way of putting coordinates on affine geometries:

Take a 2-dimensional Affine geometry A, that is, a set of lines and a set of points, together with the following axioms:

  1. Any two points determine a line
  2. Through any point not on a given line, there passes exactly one line parallel to the given one.
  3. There exist 3 non-collinear points.

Define a dilatation to be a map of points that sends any line to a line parallel to it (modulo some technical details). Because of the parallel postulate, dilatations are uniquely defined by the images of 2 points.

Then translations are dilatations without any fixed points (or the degenerate translation given by the identity map that has 2 fixed points and hence fixes every point). The direction of a translation τ is defined as the pencil (equivalence class of parallel lines) of the line joining the point P and τ(P). (the pencil does not depend on the choice of point P).

Translations form a group T (the identity mapping being the identity), and if there exist translations in two different directions, the group is commutative.

In the case where T is commutative, the endomorphisms of T that preserve directions form a (not necessarily commutative) ring with identity R.

Another two additional axioms, guarantee a) that R is a division ring and b) that R acts regularly on translations of the same direction, i.e. that if τ1 and τ2 have the same direction, then τ1 is sent to τ2 by some element of R.

These axioms are:

4a. For any two points P and Q, there exists a translation mapping P to Q.

4b. For any three points P, Q and R, there exists a dilatation that fixes P and sends Q to R.

Note that 4a. and 4b. are respectively equivalent to the affine formulations of Desargues' Theorem when the three lines are parallel and when the three lines intersect at a single point.

With these four axioms, it follows that choosing a point O and translations τ1 and τ2 in different directions, we can make A into an affine space over R with (0,0) corresponding corresponding to O, (1,0) to τ1(O) and (0,1) corresponding to τ2(O).

Note that R is commutative (and hence a field) if and only if A satisfies Pappus' Theorem.

The Question The above construction is also reversible and establishes a correspondence between affine geometries satisfying Desargues' theorem and division rings.

It seems to me that we can associate a 'geometry' to any ring via the 2d free module.

Are there any rings (or classes of rings) other than division rings and fields whose 'geometry' can be axiomatized similarly to affine geometries? Are any of them also uniquely determined by that `geometry'?

For example, if the ring is ℤ the geometry consists of a lattice of points in the plane, and seems to me is 'hyperbolic' in the sense that through any point not on a given line there are infinitely many parallel lines (join the given point to any point with non-integer coordinates on the line in question).

$\endgroup$
4
  • 2
    $\begingroup$ If your motivating question is "can more like this be done?" then Dembowski's Finite Geometry book is very nice. This sort of geometry uses other field like structures (near-fields) to handle the non-Desarguesian planes, and they are fairly interesting. Hall's Theory of Groups textbook has some of this. Zassenhaus's understanding of non-desarguesian planes was a very important step for finite group theorists, and was part of the path that includes Suzuki's work on exceptional characters, and the Feit-Thompson theorem. $\endgroup$ Jul 16, 2010 at 20:52
  • 2
    $\begingroup$ Instead of understanding rings R via their PGL(2,R), the group theorists understand groups G via their similarity to PGL(2,D). Just like you find the ring, you can find the group from the geometry. Even if you only have a partial knowledge of the group, it may be enough to construct the geometry, and then recognize the group. David Benson and Steven Smith has a reasonably neat book about doing this in the case of sporadic simple groups. $\endgroup$ Jul 16, 2010 at 20:56
  • $\begingroup$ I added a "group theory" tag since, to my knowledge, that is where this sort of thing is actually used. I strongly question the use of the "algebraic geometry" tag, but am not certain enough to remove it. $\endgroup$ Jul 17, 2010 at 0:35
  • $\begingroup$ @Charles Staats -- my answer is more or less what an algebraic geometer would say about this question. As you can see, I'm not sure if it's what the questioner wanted, but it's not completely off topic either. I'd leave the tag. $\endgroup$ Nov 13, 2010 at 3:10

2 Answers 2

2
$\begingroup$

I'm not sure that this is the answer you are looking for, but you might be interested in Mnev's universality theorem. Unwinding the language of semialgebraic sets, this says the following: Consider any finite set of polynomial equations and inequalities with integer coefficients, such as $$\zeta^3=1 \quad \zeta \neq 1.$$

Then there is an arrangement $A$ of points and lines such that $A$ can be realized over a commutative ring $R$ if and only if $R$ contains a solution to those equalities and inequalities.

For example, the Fano plane is representable only if $2=0$ in $R$. The statement "you can find $9$ points $x(i,j)$ labeled by $\mathbb{F}_3^2$ such that $x(i_1, j_1)$, $x(i_2, j_2)$ and $x(i_3, j_3)$ are collinear if and only if $\sum (i_s, j_s)=0$" is equivalent to the statement "$R$ contains a nontrivial cube root of unity."

But all of this is for $R$ commutative; I don't know if there is a noncommutative version of Mnev.

$\endgroup$
1
$\begingroup$

You don´t need a ring, you need a conical group, that is a group with a one-parameter of contracting automorphisms. This type of generalization of affine geometry following Artin style has been done here: Infinitesimal affine geometry of metric spaces endowed with a dilatation structure, Houston Journal of Mathematics, 36, 1 (2010), 91-136

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.