Linear space is an incident structure of points and lines, such that each pair of points is contained in a unique line. We say that a linear space is embeddable in the projective plane if it can be obtained from the projective plane by deleting a certain number of points. A near pencil is a linear space in which there is one line containing v - 1 points and v - 1 lines containing two points each.
An affine plane of order n with a linear space at infinity is a finite affine plane to which up to n + 1 new points have been added. is associated with a plane parallel class, a point that is added to all lines of its class. We extend a special case of this theorem as follows: b > 1 is either a close pencil or an affine plane of order n with a. possibly degenerate) linear space at infinity. A finite linear space is a bounded set of points and a collection of subsets of points, called lines~ such that every pair of points is.
In general, v will denote the number of points and b the number of lines in a linear space. A near pencil is a linear space in which one line contains v - 1 points and the remaining lines contain two points each. A linear space satisfies b > v with equality only if the space is either a near pencil or a projective plane.
A linear space satisfying (b- v) 2 v is either an affine plane of order IV or an affine plane of order IV with. In this section we extend the classification of section 2 in the special case of v 2. 2 + c)n is either a projective plane of order n, an affine plane of order n with a linear space at infinity, or an almost pencil , where c can be taken as .147899. From the proof of Lemma 2.2, we have that the existence of a line ' of length k implies.
1 and then show that this implies that J is an affine plane of order n with a linear space at infinity. We now complete the proof of Theorem 2.4 by showing that L > n2 + 1 implies that J is an affine plane with a linear space at infinity. That is, J is an affine plane of order n with a linear space at infinity, where the space at infinity is the short lines within R.
A special case of Totten's linear space classification, Theorem 2.3, is that a linear space not near a pencil with b < n + n + l and 2 v > n 2 + 1 can be embedded in a projective plane of order n. But in the case of equality in Lemma 3.8 we also see that all points outside B have exactly degree r. 2 This is because the linear space at infinity must be a pencil or projective plane (according to Theorem 2.3), for which #points=# lines.
In other words, the double arc A' possesses only 0-secants and (n/)1)-secants (points of A become 0-secants in II'). We also have the necessary condition for a perfect )1-arc in a II , )lin (this shows that, as mentioned earlier, there is no
2, each of which lies on n/}.l lines outside AU T 1, each line of which contains 1J points of T. Equality on the lower bound implies the existence of a block design with parameters. the lower bound implies the existence of a block design with 2. J > n it establishes a fairly limited range for the number of .. points in which two perfect ]J-ares can meet. Any line containing two points from one of these sets contains ]J points from that set and no points from the other.
The points of one of these sets, together with the lines meeting only that set, form a block design on the parameters 2. The lines meeting all these sets (necessarily each in one point) and the points in . Due to the relatively small number of perfect ]J-ears known for ]J > 2 we have few examples of equality within the bounds of Theorem 3.7.
This can be shown to be the case using arguments similar to those used in the proof of construction and facts about solutions of quadratic equations over finite fields (see [21], Chapter 1). This is the maximum number of completions possible, since the exact bound for m in the proof of Theorem 3.8 yields m < 3 exactly for ~ = 2 and n= 4. We count the maximum number of ~ secants up to A\L. IT intersects the k - z points of .
We note that the derivation of the boundary implies equality in the case n > f.l()J-1) . 2 can only hold for equality in the above statement for )J full (k,2) arc in a II. 2 Then the points of II\A fall into two categories; the one on No. 1 secants, and the one on ate. Those of the second category, due to equality in the above argument, must lie on only one 2 secant and thus on k - 2 1 secant.
Consider the incidence structure, P, with Points = points of IT \A on no 1-secants and Lines = 2-secants of IT. We can now count the number of points in P. Note that any pair of points of r lies on at most one line of ~-:>. This is an improvement over Theorem 3.9 for values of ~ close to n. n These ~-secants cover a maximum number of points of are disjoint outside A. But since.
Since A is complete, each of these n points of t\ A will be covered by at least one ~ secant, and for equality in the boundary, by no more than one ~ secant. 0 is in the completion (because otherwise we would have more than one completion, one with the dot in it, the other without the dot).
