Coordinate rings of topological Klingenberg planes I: The affine perspective

Abstract

Abstract. Although the coordinate ternary field of a topological affine plane is topological, the converse does not hold. However, an affine plane is topological precisely when its coordinate biternary fields are topological. We extend this result to topological biternary rings and their topological affine Klingenberg planes. Then we examine the locally compact situation. Finally, following the ideas of Knarr and Weigand, we show that in certain circumstances, the continuity of the ternary operators is sufficient to ensure that the biternary ring is topological. This facilitates the construction of locally compact, locally connected affine Klingenberg planes.Mathematics Subject Classifications (1991): 51E 15, 54H 13.

Coordinate rings of topological Klingenberg planes II: the algebraic foundation for a projective theory

“…However, as examples of Eisele [6,7,8] confirm, a topological ternary field in the sense of Salzmann (i.e., the ternary operator and its associated inverses are continuous [14]) does not necessarily produce a topological plane. Surprisingly, the explanation of this phenomenon becomes evident in the more general theory of topological Klingenberg planes as we exhibited in [3], for the affine case.…”

“…In the affine case, incomplete biternary rings are appropriate for coordinatizing AK-planes, but not topological AK-planes [3]. There, we actually needed a biternary ring and we showed that every incomplete biternary ring can be extended to a biternary ring in a unique way.…”

“…First, we note that R; T * 1 , T * 3 is an incomplete biternary ring, so by [3], it has a unique biternary ring extension R; T 1 , T 3 , in which T 1 = T * 1 and…”

“…Now P = P p 1 ∪ P p 2 ∪ P p 3 and P = ∪ 3 i=1 P i . If P ∈ P 3 (i.e., P ≈ p 3 in P), then we define the affine coordinates of P in P p 3 to be x = P 3 E ∧P 2 P , y = P 3 E ∧P 1 P , and we write P = (x, y) 3 . We note that P = P 2 x ∧P 1 y and (x, y) 3 is the point (x, y, 1) of P.…”

“…Suppose that (x, y) 3 = (u, v) 1 . Then P 3 (x, y) 3 ∧ EP 2 = (1, u) 3 and (x, y) 3 I P 3 (1, u) 3 = [u, 1, 0], so y = T 1 (x, u, 0) = x · u. In addition, (1, v) …”