When ρ is composite, the design has more structure: this aspect is discussed here.
Following Carmichael, split the lines of the main geometry into 3 classes:
The points can likewise be put into 3 classes, as follows:
The next bits of information refer to the numbers of each of one class of point on each class of line, and vice versa, using the matrix layout of Greig. The row borders refer to the point classes, and the column borders to the line classes. The entries in the body comprise a pair of numbers: the first is the number of points on a line, in the given categories, and the second gives the number of lines on a point, in the given categories.
| q2ν+qν+1 | (q2ν+qν+1)(qρν-qν) | q3ν(q(ρ-1)ν-1)(q(ρ-2)ν-1) | ||
|---|---|---|---|---|
| s | t | x | ||
| q2ν+qν+1 | C0 | qν+1,qν+1 | 1,qρν-qν | 0,0 |
| (q2ν+qν+1)(qρν-qν) | C1 | qρν-qν,1 | q2ν,q2ν | q2ν+qν+1,qρν-q2ν |
| q3ν(q(ρ-1)ν-1)(q(ρ-2)ν-1) | C2 | 0,0 | qρν-q2ν,q2ν+qν+1 | qρν-q2ν-qν,qρν-q2ν-qν |
The complete design above is a BIBD with λ=1.
An informal proof of the above is straightforward (and follows Carmichael). The number of secants and of points in C0 is given by the parameters of the sub-geometry, as are the number of points of the sub-geometry in those secants and their replication there. The number of C1 points (resp tangents), is the number of points (lines) in the sub-geometry times the difference in the replication (block size) between the full geometry and the sub-geometry. There is obviously just one point of the sub-geometry in a tangent, and none in the exterior lines. The number of exterior lines, and of C2 points, is given by difference. There are no C2 points on the secants, by definition, and the number of C1 points there is given by the difference in block sizes. These points can occur once only in the secants, otherwise two lines would have two points in common, as they already have one point in common in the sub-geometry.
So that gives numbers of all the constituent elements, and the first row and column of the matrix proper. The key to the rest of the matrix lies in the central element. Each point of the sub-geometry occurs (qρν-qν) times in the tangents, and has already occurred with all the other sub-geometry points and with (qν+1)(qρν-qν) points of C1. So there are q2ν(qρν-qν) such points left, giving q2ν per line. This is the maximum, since each tangent must have one point in common with each of the secants; qν+1 of these will have the same point of the sub-geometry as the given tangent, leaving q2ν points from C1.
Conversely, each point of C1 must occur with the q2ν points of the sub-geometry not in the secant it occurred in.
The remaining entries in the matrix are then obtained by difference, though they are all of some interest in their own right.
Thus, consider the number of points of C1 in an exterior line: each exterior line must have one point in common with each of the secants, in particular with the C1 points of the secants (since these are the only points in common). So there must be q2ν+qν+1 of them. Note that each of these transversals can have at most one point in common with each of the partial transversals represented by the C1 points of the tangents.
Given then the number of C2 points in the tangents and exterior lines, what is the relationship between them? For the moment, forget the C1 points, and suppose that the C2 points of the exterior line are S,T,...,U. These points cannot occur together in the tangents, and each occur q2ν+qν+1 times in the tangents: so that accounts for (q2ν+qν+1)(qρν-q2ν-q2ν) lines. This leaves (q2ν+qν+1)q2ν lines with which our given exterior line must have a point in common, not any of S,T,...,U. So this is where the C1 points must come back in: each of the (q2ν+qν+1) C1 points occurs q2ν times in the tangents, thus accounting for the remaining lines. In other words, given any exterior line, the tangents can be partitioned into sets according to the point held in common with the given exterior line.
A similar argument applies the other way round, obviously.
A completely different set of PBDs can be obtained by taking the points of C0 and C1, or C0 and C2, or C1 or C2 individually (with all the appropriate lines). These can then be augmented, as was the PBD above, by adding back single points, or spiking to include several collinear points. Many of these are given in the applet here: this does not consider whether ρ is composite.
Last Updated on 11/07/2004
By D.H.Rees