A Strange Ultrametric Geometry

Abstract

The main purpose of this paper is to answer a question raised by Dence in [2]. In that article, he defines a certain kind of metric, called an ultrametric, on the rationals Q and questions whether it can be extended to Q x Q, while still maintaining its special properties. We do just that in this article, creating thereby an unusual geometry that is very close to a model of Euclidean geometry in the sense that most of the Euclidean axioms are satisfied. However, it is quite different from the ordinary model of Euclidean geometry in many ways. Thus it shows that when attempting proofs in axiomatic geometries, one must not rely too much on 'common sense', but rather on straightforward deductions form the axioms and previous theorems. To construct our model, we need to introduce the concept of an ultrametric distance. Recall that a metric on S is a function d: SxS->[O,oo) such that for x, y, zES, (1) d(x,y)=O iff x=y; (2) d(x, y) = d(y, x); and (3) d(x, z) _ d(x, y) + d(y, z). Hence, on the real line R, ordinary distance (i.e., d(a, b) = la b 1) is a metric. To get an ultrametric distance, we replace (3) by the condition (3') d(x, z) _ max(d(x, y), d(y, z)). The reader will easily notice that (3') is a strictly stronger condition than (3) and that the ordinary distance function on R is not an ultrametric. We now describe a particular ultrametric distance function on the set Q of rational numbers. Given x E Q, X7 0, write x = 2k(a/b) where a and b are odd integers. Then the exponent k is uniquely determined. We then define Ix 12 = (1/2)k and d(x, y) = x y 12. If x = 0, then IX = 0. For example, if x = 5/12 = 2-2(5/3), then [x 12 = (1/2)-2 = 4. The verification that this function satisfies the ultrametric inequality can be found in [1]. (There is nothing special about the number 2 in this definition. Any prime number p will work: if x = pk(alb), where a and b are relatively prime to p, then we can define Ix |P = (1/p)k. And even this is a specific case in the general theory of ultrametric spaces. A nice discussion of some of these general abstract ideas can be found in [2].) Dence raised the question as to whether or not we can non-trivially extend this ultrametric to two dimensions. Let us first examine a few possible ways of extending, all of which fail. If A = (a,, a2), B = (b,, b2) E Q X Q, define D1(A, B) = d(al, b1) + d(a2, b2) and D2(A, B) = d(al, b1) (where d is the ultrametric distance on Q described above). Then both of the above functions fail to satisfy the ultrametric distance inequality; in fact D2 is not even metric. A more familiar way to extend a metric is exhibited by D3(P1, P2) = (d(pi, p2)2 + d(ql, q2)2)12. The reader will recognize that this is the same procedure as is used in extending the ordinary distance function on R to the Cartesian plane. Unfortunately, though D3 is a metric, it also does not satisfy the ultrametric inequality. To see this, let A = (1/2,1/2), B = (0, 1), C = (1/3, 2/3); then