Odd Order

Write $b_r(q)=\frac r{r+1}q(q+1)$ for the bound of Theorem A and $\Delta_r(q)=\kappa_r(q)-b_r(q)$ for its deficiency.

6.1 Strictness

Theorem 6.1. Let $q$ be odd and $1\le r\le q-2$. Then $\kappa_r(q)>b_r(q)$. Consequently $\kappa_r(q)\ge\lceil b_r(q)\rceil$, and $\kappa_r(q)\ge b_r(q)+1$ whenever $(r+1)\mid q(q+1)$.

Proof. This is Corollary 4.3: equality would require a maximal arc of degree $r+1$ with $2\le r+1\le q-1$ in $PG(2,q)$, $q$ odd, which Theorem 2.6 excludes. The integrality statements follow. ∎

The divisibility condition holds for $r=1$ (always), for $r=2$ iff $q\equiv0,2\pmod3$, and in general depends on the residue of $q$ modulo $r+1$. For $q=5$, $r=2$ it holds and $\kappa_2(5)=21=b+1$; for $q=7$, $r=2$ it fails, $\lceil b\rceil=38$, and $\kappa_2(7)=40$. So the deficiency is sometimes exactly what arithmetic forces and sometimes more. The only proof I have of the strict inequality is through the Ball–Blokhuis–Mazzocca theorem, even for $r=2$, where the statement is that a set of $2(q+1)$ points with two on each line through $O$ cannot have all its other secants of length exactly three.

6.2 The deficiency near the top

Proposition 6.2. For every $q$ and every $1\le s\le D(q)-2$,

$$\Delta_{q-s}(q)=\frac{s(s-1)}{q+1-s}.$$

Proof. By Theorem 5.4, $\kappa_{q-s}(q)=q^2-s$, and

$$q^2-s-\frac{(q-s)q(q+1)}{q+1-s}=\frac{(q^2-s)(q+1-s)-(q-s)q(q+1)}{q+1-s}=\frac{s^2-s}{q+1-s},$$

since both products expand to $q^3+q^2-sq^2-sq$ plus, in the first, $s^2-s$. ∎

For $s=1$ this is $0$, as it must be ($r=q-1$ attains the bound). For $s\ge2$ it is positive, consistent with Corollary 4.3 (a maximal arc of degree $q-s+1$ with $2\le s\le q-1$ would need $q-s+1\mid q$, impossible), and it tends to $0$ as $q\to\infty$ with $s$ fixed. So in the regime $r=q-s$ the counting bound is asymptotically sharp for every $q$ but attained only at $s=1$. The extremal configuration of Theorem E(i) misses only the $s$ collinear points of $T$, and its multiplicities are $q-s$ or $q-s+1$ at the points off the line $M$ carrying $T$ and $q$ at the $q-s$ points of $M\setminus T$; the deficiency $s(s-1)/(q+1-s)$ is the variance of this distribution. The exact values for $q=7$ illustrate the formula: $\Delta_6=0$, $\Delta_5=\frac26=\frac13$, $\Delta_4=\frac65$, all confirmed by the search.

6.3 The deficiency for fixed $r$

The opposite regime, $r$ fixed and $q\to\infty$ through odd values, is where the problem is open. For $r=1$ the Blokhuis–Mazzocca theorem gives $\Delta_1(q)=\frac{q-1}2$: linear in $q$. The computed values for $r\ge2$ are few:

(For $q=9$, $r=2$ the search of Appendix A.1 did not finish; it found a configuration of $65$ points, and Theorem 6.1 gives $61$ as the lower bound. The $r=3,4$ entries are lower bounds from Theorem 6.1.)

In terms of the deviations $d_P=m_P-r-1$ of Chapter 3, $\Delta_r(q)=\frac1{(r+1)^2}\sum_Pd_P^2$, so a linear lower bound $\Delta_r(q)\ge c_rq$ would say that a set with $r$ points on each line of a pencil, in a plane of odd order, is forced to have at least $c_r(r+1)^2q$ units of "squared irregularity" among its other secants. For $r=1$ that is the Blokhuis–Mazzocca bound in its dual form: a $(q+2)$-set with an internal nucleus (a point of the set through which every line meets it in exactly one further point) in a plane of odd order meets at least $\binom{q+2}2+\frac{q-1}2$ lines (Blokhuis and Mazzocca 2008, Proposition 8), i.e. $\Delta_1(q)\ge\frac{q-1}2$. Their proof uses Segre's lemma of tangents, with which Bichara and Korchmáros (1982) had shown that such a set has at most two internal nuclei, together with the Jamison–Brouwer–Schrijver bound on affine blocking sets. For $r\ge2$ nothing of the kind seems to be known, and I state the question as the principal open problem of the dissertation.

Question 6.3. For fixed $r\ge2$, does $\Delta_r(q)\to\infty$ as $q\to\infty$ through odd prime powers? Is $\Delta_r(q)\ge c_rq$ for some $c_r>0$?

Two remarks bear on it. First, the two extremal configurations for $q=7$, $r=2$ found by the search and by the conic-pencil construction (§7.3) both have deficiency $\frac83$, and the pencil construction has deficiency $\approx\frac{q^2}{12}$ for large $q$ (§7.3), far more than linear; a linear-deficiency family, if one exists, must be of a different kind. Second, the near-extremal configurations of Chapter 3 (those with $|U|=\lceil b_r(q)\rceil$) have all multiplicities in $\{r,r+1,r+2\}$ in the cases computed, so the question is equivalent, for such configurations, to bounding below the number of points of multiplicity $\ne r+1$.

6.4 Spectra of the extremal configurations

The multiplicity spectra of the minimal configurations found for odd $q$ are:

In every case the deviations lie in $\{-2,-1,0,+1\}$, with $+1$ the only positive value: the extremal configurations have no point of multiplicity above $r+2$, and their deficiency is carried mostly by points of multiplicity $r$ and $r+2$. Lemma 3.3(b) says that on each line the deviations sum to zero, so each point of multiplicity $r+2$ on a line is balanced on that line by points of multiplicity $r$ or $r-1$. Whether extremal configurations always have deviations bounded independently of $q$ is a natural sub-question of Question 6.3; the $r=1$ case says yes ($\{-1,0,+1\}$ for the conic construction).