On k-caps in PG(n, q), with q even and n \geq 3

TL;DR

This paper improves bounds on the size of k-caps in projective geometries over finite fields with even q, specifically refining the second largest complete k-cap size in PG(3, q) and deriving new bounds for higher dimensions.

Contribution

It provides improved upper bounds for the second largest complete k-cap in PG(3, q) and new bounds for maximum k-cap sizes in higher-dimensional projective spaces over even q.

Findings

Improved upper bound for m'_2(3, q) when q ≥ 8 and q even.

Derived new bounds for m_2(n, q) for n ≥ 4 and q even.

Identified gaps in previous proofs by Cao and Ou.

Abstract

Let m_2(n, q) be the maximum size of k for which there exists a k-cap in PG(n, q), and let m'_2(n, q) be the second largest value of k for which there exists a complete k-cap in PG(n, q). In this paper Chao's upper bound q^2 - q + 5 for m'_2(3, q), q even and q \geq 8, will be improved. As a corollary new bounds for m_2(n, q), q even, q\geq 8 and n \geq 4, are obtained. Cao and Ou published a better bound but there seems to be a gap in their proof.

Equations

$$m_2^{\prime }(3,q)\le q^2-q+5,q\text{even},q\ge 8.$$

$$t(t-1)\ge q(q+2-x)x.$$

$$m_2^{\prime }(3,q)<q^2-(\sqrt{5}-1)q+5,q\text{even},q\ge 8.$$

$$m_2^{\prime }(3,q)<q^2-2q+3\sqrt{q}+2,q\text{even},q\ge 2048.$$

$$m_2^{\prime }(3,q)\le q^2-q+3,q\text{even},q\ge 8.$$

$$t(t-1)\ge f(4)=f(q-2)=4q(q-2),$$

$$t\ge \frac{1+\sqrt{1+16q(q-2)}}{2}\ge 2q-\frac{7}{4}\text{for}q\ge 8.$$

$$k-1\le 2.2+(q-1)q=q^2-q+4.$$

$$\begin{gathered} s(q-1)+uq+3=q^2-q+4, \\ s+u=q. \end{gathered}$$

$$s(q-1)+uq+q-2+3=q^2-q+4,s+u=q-1.$$

$$r(q-2)+s(q-1)+uq+1=q^2-q+5,r+s+u=q+1,\text{with}r\ge 2.$$

$$r(q-2)+uq+q-1+1=q^2-q+4,r+u+1=q+1.$$

$$r(q-2)+s(q-1)+uq+1=q^2-q+4,r+s+u=q+1,r\ge 1,s\ge 1.$$

$$m_2^{\prime }(3,q)<q^2-(\sqrt{5}-1)q+5,q\text{even},q\ge 8.$$

$$m_2^{\prime }(3,4)=14.$$

$$5\le |\Pi \cap K|\le q-3.$$

$$t(t-1)\ge f(5)=f(q-3)=5q(q-3).$$

$$t\ge \frac{1+\sqrt{1+20q(q-3)}}{2}.$$

$$\sqrt{1+20q(q-3)}\ge 2\sqrt{5}q-2\alpha -1.$$

$$1+20q(q-3)\ge 20q^2+4{\alpha }^2+1-8\alpha \sqrt{5}q-4\sqrt{5}q+4\alpha ,$$

$$0\ge 4{\alpha }^2+\alpha (-8\sqrt{5}q+4)+60q-4\sqrt{5}q,$$

$$0\ge {\alpha }^2+\alpha (-2\sqrt{5}q+1)+15q-\sqrt{5}q.$$

$$k\le q^2+q+2-\sqrt{5}q+3,$$

$$k\le q^2+(1-\sqrt{5})q+5.$$

$$q^2+(1-\alpha )q+2+2\alpha >q^2+(1-\sqrt{5})q+5,$$

$$(\sqrt{5}-\alpha )q+2\alpha -3>0$$

$$\frac{q^2-q+2}{2}.(q-1)/\frac{q+2}{2}.$$

$$r(q-2)+sq+2(q-3)+1,\mathrm{with}r+s=q-1.$$

$$r(q-2)+(q-1-r)q+2(q-3)+1\le q^2-q+3,$$

Full text

Keywords: projective space, finite field, $k$-cap

1. Introduction

A *k-arc * of $\mathbf{PG}(2,q)$ is a set of $k$ points, no three of which are collinear; a k-cap of $\mathbf{PG}(n,q)$, $n\geq 3$, is a set of $k$ points, no three of which are collinear. A $k$-arc or $k$-cap is complete if it is not contained in a ($k+1$)-arc or ($k+1$)-cap. The largest value of $k$ for which a $k$-arc of $\mathbf{PG}(2,q)$, or a $k$-cap of $\mathbf{PG}(n,q)$ with $n\geq 3$, exists is denoted by $m_{2}(n,q)$. The size of the second largest complete $k$-arc of $\mathbf{PG}(2,q)$ or $k$-cap of $\mathbf{PG}(n,q)$, $n\geq 3$, is denoted by $m^{\prime}_{2}(n,q)$.

Theorem 1.1**.**

• (i)

$m_{2}(2,q)=q+2$, $q$ even **[7]**;

• (ii)

$m_{2}(3,q)=q^{2}+1$, $q$ even, $q>2$ **[6, 1, 10]**;

• (iii)

$m_{2}(n,2)=2^{n}$* [1];*

• (iv)

$m_{2}(4,4)=41$* [5];*

• (v)

$m^{\prime}_{2}(n,2)=2^{n-1}+2^{n-3}$* [4];*

• (vi)

$m^{\prime}_{2}(3,4)=14$* [8].*

Theorem 1.2** ([11, 13, 7]).**

Let $K$ be a $k$-arc of $\mathbf{PG}(2,q)$, $q$ even and $q>2$, with $q-\sqrt{q}+1<k\leq{q+1}$. Then $K$ can be uniquely extended to a ($q+2$)-arc of $\mathbf{PG}(2,q)$.

For any $k$-arc $K$ in $\mathbf{PG}(2,q)$ or $k$-cap $K$ in $\mathbf{PG}(n,q)$, $n\geq 3$, a tangent of $K$ is a line which has exactly one point in common with $K$. Let $t$ be the number of tangents of $K$ through a point $P$ of $K$ and let $\sigma_{1}(Q)$ be the number of tangents of $K$ through a point $Q\not\in K$. Then for a $k$-arc $K$ $t+k=q+2$ and for a $k$-cap $K$ $t+k=q^{n-1}+q^{n-2}+\cdots+q+2$.

Theorem 1.3** ([8]).**

If $K$ is a complete $k$-arc in $\mathbf{PG}(2,q)$, $q$ even, or a complete $k$-cap in $\mathbf{PG}(n,q)$, $n\geq 3$ and $q$ even, then $\sigma_{1}(Q)\leq t$ for each point $Q$ not on $K$.

Theorem 1.4** ([3]).**

[TABLE]

To prove Theorem 1.4 J.-M. Chao relies on the following crucial lemma.

Lemma 1.5** ([3]).**

Let $K$ be a complete $k$-cap in $\mathbf{PG}(3,q)$ with $q$ even. If $\Pi$ is a plane such that $|\Pi\cap K|=x$, then

[TABLE]

In the underlying paper the following improvement of Chao’s result will be obtained.

Theorem 1.6** (Main Theorem).**

[TABLE]

As a corollary new bounds for $m_{2}(n,q)$, $q$ even, $q\geq 8$ and $n\geq 4$, are obtained.

Combining the main theorem of [12] with Theorem 1.6, there is an immediate improvement of the upper bound for $m^{\prime}_{2}(3,q)$, $q\geq 2048$. We thank T. Szőnyi for bringing reference [12] to our attention.

Theorem 1.7**.**

[TABLE]

2. A first improvement of Chao’s bound

Theorem 2.1**.**

[TABLE]

Proof.  Let $K$ be a complete $k$-cap in $\mathbf{PG}(3,q)$, $q$ even, $q\geq 8$ and $k<q^{2}+1$.

Let $\Pi$ be a plane of $\mathbf{PG}(3,q)$ for which $4\leq|\Pi\cap K|\leq q-2$. Let $f(X)=q(q+2-X)X$. Then

[TABLE]

by Lemma 1.5. So

[TABLE]

Hence $k\leq q^{2}+q+2-2q+\frac{7}{4}=q^{2}-q+\frac{15}{4}$, and consequently $k\leq q^{2}-q+3$.

So we may assume that either $|\Pi\cap K|\leq 3$, or $|\Pi\cap K|\geq q-1$, for any plane $\Pi$ of $\mathbf{PG}(3,q)$. Let $l_{1},l_{2},...,l_{t}$ be the $t$ tangents of $K$ through the point $P\in K$. We consider three cases depending on the number of planes containing $\l_{i}$ and intersecting $K$ in at most 3 points.

(A) There exists exactly one plane $\Pi_{l_{i}}$ containing $l_{i}$ such that $|\Pi_{l_{i}}\cap K|\leq 3,i=1,2,...,t$. We will show that in this case $k\leq q^{2}-q+3$.

Assume there is exactly one plane $\Pi$ through $P$ with $|\Pi\cap K|\leq 3$. Then for $i=1,2,...,t$, $\Pi_{l_{i}}=\Pi$. Hence all tangents of $K$ containing $P$ are in $\Pi$. So $t\leq q+1$, a contradiction. Hence there are at least two planes $\Pi_{1},\Pi_{2}$ through $P$ such that $|\Pi_{i}\cap K|\leq 3,i=1,2$. Then $|\Pi_{1}\cap\Pi_{2}\cap K|=2$. Consequently $t\geq 2(q-1)$, and so $k\leq q^{2}+q+2-2q+2=q^{2}-q+4$.

Assume, by way of contradiction, that $k=q^{2}-q+4$. So $t=2(q-1)$. Then $|\Pi_{1}\cap K|=|\Pi_{2}\cap K|=3$. All tangent lines at $P$ are contained in $\Pi_{1}$ and $\Pi_{2}$. Let $l$ be a tangent of $K$ at $P$ in $\Pi_{1}$, and consider the $q+1$ planes containing $l$. The plane $\Pi_{1}$ is the only of these planes which intersects $K$ in 3 points, exactly $q-1$ planes through $l$ contain 2 tangent lines at $P$ and so intersect $K$ in a $q$-arc and the remaining plane through $l$ contains exactly one tangent line at $P$ and so intersects $K$ in a $(q+1)$-arc.

Let $\widetilde{\Pi}$ be the unique plane containing $l$ which intersects $K$ in a $(q+1)$-arc, let $\widetilde{\Pi}\cap K=O$, and let $N$ be the kernel of $O$, that is, $N$ is the unique point of $\widetilde{\Pi}$ which extends $O$ to a $(q+2)$-arc of $\widetilde{\Pi}$. Clearly $N\in l$.

If $K^{\prime}$ is a $k^{\prime}$-arc of a plane $\mathbf{PG}(2,q)$ and $P^{\prime}\in\mathbf{PG}(2,q)\setminus K^{\prime}$, then the parity of the number of tangents of $K^{\prime}$ through $P^{\prime}$ is the parity of $k^{\prime}$, see Chapter 1 of [7]. Hence, by considering $O$ and the $q-1$ $q$-arcs whose planes contain $l$, we see that the number of tangents of $K$ through $N$ is at least $q+1+q-1=2q$. As $K$ is complete we have $2q\leq t$, so $k\leq q^{2}+q+2-2q=q^{2}-q+2$, a contradiction.

Consequently $k\leq q^{2}-q+3$.

(B) Some tangent $l_{i},1\leq i\leq t$, is contained in at least two planes having at most three points in common with $K$.

First we will prove that $k\leq q^{2}-q+5$. For $k=q^{2}-q+5$ and $k=q^{2}-q+4$ a contradiction will be obtained; the case $k=q^{2}-q+4$ will be subdivided in two cases. Hence it follows that also in Case (B) we have $k\leq q^{2}-q+3$.

Counting the points of $K$ on the $q+1$ planes containing $l_{i}$ gives

[TABLE]

So $k\leq q^{2}-q+5$.

(B.1) First, assume $k=q^{2}-q+5$. Then two planes $\Pi_{1},\Pi_{2}$ containing $l_{i}$ intersect $K$ in 3 points, while the remaining planes $\Pi_{3},\Pi_{4},...,\Pi_{q+1}$ containing $l_{i}$ intersect $K$ in $q+1$ points. Let $l$ be a tangent of $K$ at $P$ in $\Pi_{1}$, distinct from $l_{i}$. Any plane $\zeta$ containing $l$, with $\zeta\not=\Pi_{1}$, intersects each $(q+1)$-arc $\Pi_{i}\cap K,i=3,4,...,q+1$, in exactly two points. Hence $|\zeta\cap K|\geq q$. Considering the lines $\zeta\cap\Pi_{2}$, we see that exactly two of the planes $\zeta$, say $\zeta_{1}$ and $\zeta_{2}$, intersect $K$ in $(q+1)$-arcs $O_{1}$ and $O_{2}$, while the $q-2$ other planes $\zeta$, say $\zeta_{3},\zeta_{4},...,\zeta_{q}$, intersect $K$ in a $q$-arc.

Let $N_{1}$ be the kernel of $O_{1}$; then $N_{1}\in l$. The number of tangents of $K$ containing $N_{1}$ is at least $q+1+q-2=2q-1$. As $K$ is complete we have $2q-1\leq t$, so $k\leq q^{2}+q+2-2q+1=q^{2}-q+3$, a contradiction.

(B.2) Next, assume $k=q^{2}-q+4$. Then, considering all planes containing $l_{i}$, there are two cases to consider.

(B.2.1) Two planes $\Pi_{1},\Pi_{2}$ containing $l_{i}$ intersect $K$ in three points, the plane $\Pi_{3}$ containing $l_{i}$ intersects K in $q$ points, and the remaining planes $\Pi_{4},\Pi_{5},...,\Pi_{q+1}$ containing $l_{i}$ intersect $K$ in $q+1$ points. Let $l$ be a tangent of $K$ at $P$ in $\Pi_{1}$, distinct from $l_{i}$. Any plane $\zeta$ containing $l$, distinct from $\Pi_{1}$, intersects each $(q+1)$-arc $\Pi_{i}\cap K,i=4,5,...,q+1$, in exactly two points; $q-1$ of these planes $\zeta$ intersect $\Pi_{3}\cap K$ in exactly two points. So for at least $q-1$ of these planes $\zeta$ we have $|\zeta\cap K|\geq q$, and for all planes $\zeta$ we have $|\zeta\cap K|\geq{q-1}$.

Assume that for all $q$ planes $\zeta$ we have $|\zeta\cap K|\geq q$. Let $s$ be the number of planes $\zeta$ for which $|\zeta\cap K|=q$ and let $u$ be the number of planes $\zeta$ for which $|\zeta\cap K|=q+1$. Then

[TABLE]

So $s(q-1)+(q-s)q+3=q^{2}-q+4$, hence $s=q-1$ and $u=1$. Let $\zeta$ be the plane which intersects $K$ in a $(q+1)$-arc $O$, and let $N\in l$ be the nucleus of $O$. The number of tangents of $K$ containing $N$ is at least $q+1+q-1=2q$, so $k\leq q^{2}+q+2-2q=q^{2}-q+2$, a contradiction.

So we may assume that for exactly $q-1$ planes $\zeta$ we have $|\zeta\cap K|\geq q$ and that for exactly one plane $\zeta$ we have $|\zeta\cap K|=q-1$. Assume that for $s$ planes $\zeta$ we have $|\zeta\cap K|=q$, and that for $u$ planes $\zeta$ we have $|\zeta\cap K|=q+1$. Then

[TABLE]

So $s(q-1)+(q-1-s)q+q+1=q^{2}-q+4$, hence $s=q-3$ and $u=2$. Let $\zeta_{1},\zeta_{2}$ be the planes containing $l$ which intersect $K$ in $(q+1)$-arcs $O_{1},O_{2}$, let $N_{1},N_{2}$ be the nuclei of $O_{1},O_{2}$, and let $\Pi_{1}\cap K=\{{P,P_{1},P_{2}}\}$. Assume first that $N_{1}\not\in P_{1}P_{2}$. Then the number of tangents of $K$ containing $N_{1}$ is at least $q+1+q-3+2=2q$, so $k\leq q^{2}-q+2$ a contradiction. Similarly if $N_{2}\not\in P_{1}P_{2}$. Hence we may assume that $N_{1}=N_{2}=P_{1}P_{2}\cap l$. Then the number of tangents of $K$ through $N_{1}$ is at least $q+1+q+q-3=3q-2$, so $k\leq q^{2}+q+2-3q+2=q^{2}-2q+4$, again a contradiction.

(B.2.2) One plane $\Pi_{1}$ containing $l_{i}$ intersects $K$ in three points, and one plane $\Pi_{2}$ containing $l_{i}$ intersects $K$ in two points. Consequently the other $q-1$ planes $\Pi_{3},\Pi_{4},...,\Pi_{q+1}$ containing $l_{i}$ intersect $K$ in $q+1$ points. Let $l$ be a tangent of $K$ at $P$ in $\Pi_{1}$, distinct from $l_{i}$. Any plane $\zeta$ containing $l$, distinct from $\Pi_{1}$, intersects each $(q+1)$-arc $\Pi_{i}\cap K$, with $i=3,4,...,q+1$, in exactly two points. As $k=q^{2}-q+4$ it easily follows that for $q-1$ of these planes $\zeta$ we have $|\zeta\cap K|=q$, while for the remaining plane $\zeta$ we have $|\zeta\cap K|=q+1$.

Let $\widetilde{\zeta}$ be the plane containing $l$ which intersects $K$ in a $(q+1)$-arc $O$, and let $N$ be the nucleus of $O$. The number of tangents of $K$ containing $N$ is at least $q+1+q-1=2q$, so $k\leq q^{2}-q+2$, again a contradiction.

(C) Some tangent $l_{i}$, with $1\leq i\leq t$, is contained in no plane having at most three points in common with $K$.

First we will prove that $k\leq q^{2}-q+5$. A contradiction will be obtained for $k\in\{q^{2}-q+5,q^{2}-q+4\}$; for $k=q^{2}-q+4$ two cases have to be considered. Hence again $k\leq q^{2}-q+3$.

Then $|\Pi_{j}\cap K|\geq q-1$ for each plane $\Pi_{j}$ containing $l_{i}$, with $j=1,2,...,q+1$. The arc $\Pi_{j}\cap K$ of $\Pi_{j}$ can be completed to a $(q+2)$-arc of $\Pi_{j}$; see Theorem 1.2. This $(q+2)$-arc meets $l_{i}$ in points $P,P_{j}$. As there are $q+1$ points $P_{j}$ and $|l_{i}\setminus\{P\}|=q$, two of the points $P_{j}$ coincide, say $P_{1}=P_{2}$. The number of tangents of $K$ containing $P_{1}$ is at least $2(q-2)+1=2q-3$, so $k\leq q^{2}-q+5$.

Now we make some observations on $(q-1)$-arcs of $\mathbf{PG}(2,q)$, $q$ even. Let $\widetilde{K}$ be any $(q-1)$-arc of $\mathbf{PG}(2,q)$, $q$ even, and let $\widetilde{l}$ be a tangent of $\widetilde{K}$ at $\widetilde{P}\in\widetilde{K}$. Let $\widetilde{C}$ be the unique $(q+2)$-arc which contains $\widetilde{K}$; see Theorem 1.2. Put $\widetilde{C}\cap\widetilde{l}=\{\widetilde{P},\widetilde{N}\}$. Then it is easy to see that exactly $q-2$ points of $\widetilde{l}\setminus\{\widetilde{P},\widetilde{N}\}$ are on exactly three tangents of $\widetilde{K}$, and that exactly one point $\widetilde{R}$ of $\widetilde{l}\setminus\{\widetilde{P},\widetilde{N}\}$ is on exactly one tangent of $\widetilde{K}$; also, $\widetilde{R}=\widetilde{l}\cap\widetilde{N}^{\prime}\widetilde{N}^{\prime\prime}$, with $\{\widetilde{N},\widetilde{N}^{\prime},\widetilde{N}^{\prime\prime}\}\cup\widetilde{K}=\widetilde{C}$.

(C.1) First, assume $k=q^{2}-q+5$. Then $\Pi_{1}\cap K$ and $\Pi_{2}\cap K$ are $(q-1)$-arcs of $\Pi_{1}$ and $\Pi_{2}$. Let $r$ be the number of $(q-1)$-arcs $\Pi_{j}\cap K$, let $s$ be the number of $q$-arcs $\Pi_{j}\cap K$ and let $u$ be the number of $(q+1)$-arcs $\Pi_{j}\cap K$. Then

[TABLE]

So $r(q-2)+s(q-1)+(q+1-r-s)q+1=q^{2}-q+5$, hence $2r+s=2q-4$, with $r\geq 2$. If $s\geq 1$, then we have an extra tangent of $K$ containing $P_{1}$, so $k\leq q^{2}-q+4$, a contradiction. Hence $s=0,r=q-2,u=3$.

As the number of tangents of $K$ containing $P_{1}$ is exactly $2q-3$, the nuclei of the three $(q+1)$-arcs $\Pi_{j}\cap K$ are distinct from $P_{1}$. Let $N$ be one of these nuclei. Also, $P_{1}$ is on exactly one tangent of each of the $q-4$ $(q-1)$-arcs $\Pi_{j}\cap K$, distinct from the $(q-1)$-arcs $\Pi_{1}\cap K,\Pi_{2}\cap K$. So $N$ is on at least three tangents of each of these $q-4$ $(q-1)$-arcs $\Pi_{j}\cap K$. Hence the number of tangents of $K$ containing $N$ is at least $2(q-4)+q+1=3q-7>2q-3$, a contradiction.

(C.2) Finally, assume that $k=q^{2}-q+4$. We have to consider two cases depending of the sizes of $\Pi_{1}\cap K$ and $\Pi_{2}\cap K$.

(C.2.1) First, assume that $\Pi_{1}\cap K$ and $\Pi_{2}\cap K$ are $(q-1)$-arcs. The tangents of $K$ containing $P_{1}$ are the tangents of $\Pi_{1}\cap K$ and $\Pi_{2}\cap K$ containing $P_{1}$, and one extra tangent $l$. Assume that $l$ is a tangent of $\Pi_{3}\cap K$. If $\Pi_{3}\cap K$ is a $(q+1)$-arc O, then $P_{1}$ is the nucleus of O, so there arise $q$ extra tangents, a contradiction; if $\Pi_{3}\cap K$ is a $(q-1)$-arc $K^{\prime}$, then $P_{1}$ is contained in at least three tangents of $K^{\prime}$, again a contradiction. Hence $\Pi_{3}\cap K$ is a $q$-arc. Also, $\Pi_{j}\cap K$, with $j=4,5,...,q+1$, cannot be a $q$-arc. Let $r$ be the number of $(q-1)$-arcs $\Pi_{j}\cap K$, and let $u$ be the number of $(q+1)$-arcs $\Pi_{j}\cap K$. Then

[TABLE]

So $r(q-2)+(q-r)q+q=q^{2}-q+4$, hence $r=q-2$ and $u=2$. Let $O_{1},O_{2}$ be the $(q+1)$-arcs $\Pi_{j}\cap K$, and let $N_{1},N_{2}$ be the nuclei of $O_{1},O_{2}$. Then $N_{i}\neq P_{1},i=1,2$. Also $P_{1}$ is contained in exactly one tangent of each of the $q-4$ $(q-1)$-arcs $\Pi_{j}\cap K$, with $j\neq 1,2$. Hence the number of tangents of $K$ containing $N_{1}$ is at least $2(q-4)+q+1=3q-7>2q-2$, clearly a contradiction.

(C.2.2) Consequently, we may assume that $\Pi_{1}\cap K$ is a $(q-1)$-arc and that $\Pi_{2}\cap K$ is a $q$-arc . Let $r$ be the number of $(q-1)$-arcs $\Pi_{j}\cap K$, let $s$ be the number of $q$-arcs $\Pi_{j}\cap K$ and let $u$ be the number of $(q+1)$-arcs $\Pi_{j}\cap K$. Then

[TABLE]

So $2r+s=2q-3,r\geq 1,s\geq 1$. Clearly, $s=1$, as otherwise we have an extra tangent containing $P_{1}$, and then $k<q^{2}-q+4$. Hence $r=q-2,s=1,u=2$. The nuclei of the two $(q+1)$-arcs $\Pi_{j}\cap K$ are distinct from $P_{1}$. Let $N$ be one of these nuclei. Also, $P_{1}$ is on exactly one tangent of each of the $q-3$ $(q-1)$-arcs $\Pi_{j}\cap K$ distinct from $\Pi_{1}\cap K$. So $N$ is on at least three tangents of each of these $q-3$ $(q-1)$-arcs $\Pi_{j}\cap K$. Consequently the number of tangents of $K$ containing $N$ is at least $2(q-3)+q+1=3q-5>2q-2$, a final contradiction. $\blacksquare$

3. Main Theorem

Theorem 3.1**.**

[TABLE]

[TABLE]

Proof  By [8] we have $m^{\prime}_{2}(3,4)=14$, and by Theorem 2.1 we have $m^{\prime}_{2}(3,8)\leq 59$, which proves Theorem 3.1 for $q=8$. So from now on we assume $q>8$.

Let $K$ be a complete $k$-cap in $\mathbf{PG}(3,q)$, $q$ even, $q>8$, and $k<q^{2}+1$. Let $\Pi$ be a plane of $\mathbf{PG}(3,q)$ for which

[TABLE]

Let $f(X)=q(q+2-X)X$. Then by Lemma 1.5 of Chao

[TABLE]

So

[TABLE]

Put $\frac{1+\sqrt{1+20q(q-3)}}{2}\geq\sqrt{5}q-\alpha$, that is,

[TABLE]

For $\alpha\leq\sqrt{5}q-(1/2)$ this is equivalent to

[TABLE]

or

[TABLE]

or

[TABLE]

Put $\alpha=3$. Then there arises $0\geq 9+3(-2\sqrt{5}q+1)+15q-\sqrt{5}q$, that is, $0\geq 12+15q-7\sqrt{5}q$. This inequality is satisfied for $q>16$.

Hence for $q>16$ we have $t\geq\sqrt{5}q-3$, and so,

[TABLE]

that is,

[TABLE]

For $q=16$ it follows from (18) that $t>32$ and so $k\leq 241$, which is equivalent to $k\leq q^{2}+(1-\sqrt{5})q+5$ with $q=16$.

**From now on suppose that either $|\Pi\cap K|\leq 4$ or $|\Pi\cap K|\geq q-2$ for any plane $\Pi$ of $\mathbf{PG}(3,q)$. Let $l_{1},l_{2},...,l_{t}$ be the $t$ tangents of $K$ containing the point $P\in K$. Assume, by way of contradiction, that $k>q^{2}+(1-\sqrt{5})q+5$. We consider three cases depending on the number of planes containing $l_{i}$ and intersecting $K$ in at most 4 points. In each case a contradiction will be obtained **.

(A) Assume, by way of contradiction, that each $l_{i}$ is contained in exactly one plane $\Pi_{l_{i}}$ for which $|\Pi_{l_{i}}\cap K|\leq 4$, with $i=1,2,...,t$.

(A.1) Assume that there is exactly one plane $\Pi$ through $P$ with $|\Pi\cap K|\leq 4$. Then for $i=1,2,...,t$ we have $\Pi_{l_{i}}=\Pi$. So $t\leq q+1$, hence $k\geq q^{2}+1$, a contradiction.

(A.2) There are at least two planes $\Pi_{1},\Pi_{2}$ through $P$ such that $|\Pi_{i}\cap K|\leq 4,i=1,2$. Then $|\Pi_{1}\cap\Pi_{2}\cap K|=2$. Consequently $t\geq 2(q-2)$, and so $k\leq q^{2}+q+2-2q+4=q^{2}-q+6$.

The plane $\Pi_{1}$ intersects $K$ in a $m$-arc, $m\leq 4$, and contains at least $q-2$ tangents of $K$ at $P$. Let $P_{1}\in(K\cap\Pi_{1})\setminus P$ and assume that $PP_{1}$ is contained in $\alpha$ planes $\Pi$ with $|\Pi\cap K|\leq 4$. Then $t\geq\alpha(q-2)$, so $k\leq q^{2}+(1-\alpha)q+2+2\alpha$. Consequently

[TABLE]

or

[TABLE]

This gives a contradiction for $\alpha>2$ with $q>8$. So $PP_{1}$ is contained in at most two planes intersecting $K$ in at most four points.

Assume, by way of contradiction, that for some plane $\Pi$ of $\mathbf{PG}(3,q)$ we have $\Pi\cap K=\{P\}$. As there are at least two planes $\Pi,\Pi^{\prime}$ through $P$ intersecting $K$ in at most four points, we have $|\Pi\cap\Pi^{\prime}\cap K|=2$ and so $|\Pi\cap K|\geq 2$, a contradiction.

Let $\mathbf{PG}(2,q)$ be a plane of $\mathbf{PG}(3,q)$ not containing $P$ and let $\sigma$ be the projection of $\mathbf{PG}(3,q)\setminus\{P\}$ from $P$ onto $\mathbf{PG}(2,q)$. Further, let $\mathcal{P}$ be the set of all images under $\sigma$ of all points of $K\setminus\{P\}$ contained in planes $\Pi$, with $P\in\Pi$, for which $|\Pi\cap K|\leq 4$, and let $\mathcal{B}$ be the set of all images under $\sigma$ of the sets $\Pi\setminus\{P\}$. Then there arises an incidence structure $(\mathcal{P},\mathcal{B})$ of points and lines for which

• (1)

$|\mathcal{B}|\geq 2$,

• (2)

any two distinct lines in $\mathcal{B}$ have exactly one point in common,

• (3)

each point is contained in at most two lines,

• (4)

each line contains at most three points and at least one point.

It follows easily that $2\leq|\mathcal{B}|\leq 4$. For each value of $|\beta|$ we will find a contradiction.

($\alpha$) $|\mathcal{B}|=4$

Then $t=4(q-2)$, so $k=q^{2}+q+2-4q+8=q^{2}-3q+10$. Hence $q^{2}-3q+10>q^{2}+(1-\sqrt{5})q+5$, or $5>(4-\sqrt{5})q$, a contradiction as $q>8$.

($\beta$) $|\mathcal{B}|=3$

If $|\mathcal{P}|=3$, then $t=3(q-1)$, so $k=q^{2}-2q+5$. Hence $q^{2}-2q+5>q^{2}+(1-\sqrt{5})q+5$, or $(3-\sqrt{5})q<0$, a contradiction.

If $|\mathcal{P}|=4$, then $t=2(q-1)+q-2$, so $k=q^{2}-2q+6$. Hence $q^{2}+(1-\sqrt{5})q+5<q^{2}-2q+6$, or $(3-\sqrt{5})q-1<0$, a contradiction.

If $|\mathcal{P}|=5$, then $t=q-1+2(q-2)$, so $k=q^{2}-2q+7$. Hence $q^{2}+(1-\sqrt{5})q+5<q^{2}-2q+7$, or $(3-\sqrt{5})q<2$, a contradiction.

If $|\mathcal{P}|=6$, then $t=3(q-2)$, so $k=q^{2}-2q+8$. Hence $q^{2}+(1-\sqrt{5})q+5<q^{2}-2q+8$, or $(3-\sqrt{5})q<3$, a contradiction.

($\gamma$) $|\mathcal{B}|=2$

By Theorem 2.1 we may assume that $k\leq q^{2}-q+3$.

If $|\mathcal{P}|=1$, then $t=2q$, so $k=q^{2}-q+2$.

If $|\mathcal{P}|=2$, then $t=2q-1$, so $k=q^{2}-q+3$.

If $|\mathcal{P}|=3$, then $t=2q-2$, so $k=q^{2}-q+4$, a contradiction.

If $|\mathcal{P}|=4$, then $t=2q-3$, so $k=q^{2}-q+5$, a contradiction.

If $|\mathcal{P}|=5$, then $t=2q-4$, so $k=q^{2}-q+6$, a contradiction.

Hence the cases $k=q^{2}-q+2$ and $k=q^{2}-q+3$ have still to be considered.

($\gamma.1$) $k=q^{2}-q+2$

On $K$ there are two points $P,P_{1}$ such that $PP_{1}$ is contained in two planes $\Pi_{1},\Pi_{2}$ intersecting $K$ in just $\{P,P_{1}\}$, and in $q-1$ planes $\Pi_{3},\Pi_{4},...,\Pi_{q+1}$ intersecting $K$ in a $(q+2)$-arc.

Let $P^{\prime}\in(\Pi_{3}\cap K)\setminus\{P,P_{1}\}$ and let $l$ be a tangent of $K$ at $P^{\prime}$. Assume, by way of contradiction, that each plane containing $l$ intersects $K$ in a $m$-arc with $m>4$, so $m\geq q-2$. These $m$-arcs $K^{\prime}_{i}$, with $i=1,2,...,q+1$, are extendable to $(q+2)$-arcs $C_{i}$. Let $C_{i}\cap l=\{N_{i},P^{\prime}\},i=1,2,...,q+1$. At least two of the points $N_{1},N_{2},...,N_{q+1}$ coincide, say $N_{1}=N_{2}$. A plane $\Pi^{\prime}$ containing $l$, but not containing $P$ nor $P_{1}$, intersects each of the $(q+2)$-arcs $\Pi_{i}\cap K$, with $i=3,4,...,q+1$, in either 0 or 2 points; so $|\Pi^{\prime}\cap K|$ is even. A plane $\Pi^{\prime}$ containing $l$ and either $P$ or $P_{1}$ intersects $K$ in $q$ points. Hence each plane containing $l$ intersects $K$ in a $m$-arc, with $m$ even. Counting tangents of $K$ containing $N_{1}$, we obtain at least $2(q-3)+1+q-1=3q-6$ tangents. So $k\leq q^{2}+q+2-3q+6=q^{2}-2q+8$, a contradiction for $q>8$. We conclude that there is a plane $\Pi^{\prime}$ containing $l$ with $|\Pi^{\prime}\cap K|\leq 4$.

Assume, by way of contradiction, that $l$ is contained in at least two planes $\Pi^{\prime},\Pi^{\prime\prime}$ with $|\Pi^{\prime}\cap K|\leq 4,|\Pi^{\prime\prime}\cap K|\leq 4$. Then, by a previous argument, these intersections have an even number of points and so $|\Pi^{\prime}\cap K|\in\{2,4\}$ and $|\Pi^{\prime\prime}\cap K|\in\{2,4\}$. Now we count the points of $K$ in planes containing $l$, and obtain $k\leq(q-1)(q-1)+7=q^{2}-2q+8$, a contradiction for $q>8$.

Hence $l$ is contained in exactly one plane $\Pi^{\prime}$ for which $|\Pi^{\prime}\cap K|\leq 4$. It follows that the roles of $P$ and $P^{\prime}$ may be interchanged.

Let $l^{\prime}$ be a second tangent of $K$ containing $P^{\prime}$, with $l^{\prime}\not\subset\Pi^{\prime}$. Let $\tilde{K}=K\cap\Pi_{3},\Pi^{\prime}\cap\tilde{K}=\{P^{\prime},P^{\prime}_{1}\}$. If $P^{\prime}_{1}\in\{P,P_{1}\}$, then $|\Pi^{\prime}\cap K|=q$, a contradiction. Hence $P^{\prime}_{1}\not\in\{P,P_{1}\}$. With $P^{\prime}$ there corresponds an incidence structure $(\mathcal{P}^{\prime},\mathcal{B}^{\prime})$ of points and lines. As $k=q^{2}-q+2$, we necessarily have $|\mathcal{P}^{\prime}|=1$ and $|\mathcal{B}^{\prime}|=2$. Hence $\Pi^{\prime}\cap K=\{P^{\prime},P^{\prime}_{1}\}$. If $\tilde{\Pi}^{\prime}$ is the unique plane containing $l^{\prime}$ and intersecting $K$ in at most 4 points, then $\tilde{\Pi}^{\prime}\cap K=\{P^{\prime},P^{\prime}_{1}\}$. Also, the roles of $P$ and $P^{\prime}_{1}$, $P^{\prime}$ and $P^{\prime}_{1}$, $P$ and $P_{1}$ can be interchanged.

Interchanging $\Pi_{3}$ and $\Pi_{i}$, $i\in\{3,4,....,q+1\}$, and interchanging $P^{\prime}$ with any point of $(\Pi_{i}\cap K)\setminus\{P,P_{1}\}$, we see that $K$ is partitioned into $(q^{2}-q+2)/2$ pairs, where each pair is contained in two planes intersecting $K$ in that pair and in $q-1$ planes intersecting $K$ in a $(q+2)$-arc. Any other plane contains either 0 or $q$ points of $K$. Each point $Q$ of $K$ is contained in $2q$ tangents; the two planes on $Q$ intersecting $K$ in two points each contain $q$ of these tangents.

Now we count the planes intersecting $K$ in a $(q+2)$-arc, and obtain

[TABLE]

Hence $q+2|(q^{2}-q+2)(q-1)$, so $q+2|24$, that is $q\in\{2,4\}$, a contradiction.

($\gamma.2$) $k=q^{2}-q+3$

Then on $K$ there are points $P,P_{1}$ such that $PP_{1}$ is contained in two planes $\Pi_{1},\Pi_{2}$ with $\Pi_{1}\cap K=\{P,P_{1}\},\Pi_{2}\cap K=\{P,P_{1},P_{2}\}$, and in $q-1$ planes $\Pi_{3},\Pi_{4},...,\Pi_{q+1}$ intersecting $K$ in a $(q+2)$-arc.

Let $P^{\prime}\in(\Pi_{3}\cap K)\setminus\{P,P_{1}\}$ and let $l$ be a tangent of $K$ at $P^{\prime}$. Assume, by way of contradiction, that each plane containing $l$ intersects $K$ in a $m$-arc with $m>4$, so $m\geq q-2$. These $m$-arcs $K^{\prime}_{i}$, with $i=1,2,...,q+1$, are extendable to $(q+2)$-arcs $C_{i}$. Let $C_{i}\cap l=\{N_{i},P^{\prime}\},i=1,2,...,q+1$. At least two of the points $N_{1},N_{2},...,N_{q+1}$ coincide, say $N_{1}=N_{2}$. A plane $\Pi^{\prime}$ containing $l$, but not containing $P$ nor $P_{1}$, intersects each of the $(q+2)$-arcs $\Pi_{i}\cap K$, with $i=3,4,...,q+1$, in either 0 or 2 points. So if $P_{2}\not\in\Pi^{\prime}$, then $|\Pi^{\prime}\cap K|$ is even. A plane $\Pi^{\prime}$ containing $l$ and either $P$ or $P_{1}$, but not $P_{2}$, intersects $K$ in $q$ points. Hence $q$ planes containing $l$ intersect $K$ in a $m$-arc, with $m$ even. Counting tangents of $K$ containing $N_{1}$, we obtain at least $2(q-3)+1+q-2=3q-7$ tangents. So $k\leq q^{2}+q+2-3q+7=q^{2}-2q+9$, a contradiction for $q>8$. We conclude that there is a plane $\Pi^{\prime}$ containing $l$ with $|\Pi^{\prime}\cap K|\leq 4$.

Assume, by way of contradiction, that $l$ is contained in at least two planes $\Pi^{\prime},\Pi^{\prime\prime}$ with $|\Pi^{\prime}\cap K|\leq 4,|\Pi^{\prime\prime}\cap K|\leq 4$. Now we count the points of $K$ in planes containing $l$, and obtain $k\leq q^{2}-2q+9$, a contradiction for $q>8$.

Hence $l$ is contained in exactly one plane $\Pi^{\prime}$ for which $|\Pi^{\prime}\cap K|\leq 4$. As all tangents of $K$ at $P_{1}$ are contained in $\Pi_{1}\cup\Pi_{2}$, it follows that each tangent of $K$ at $P_{1}$ is contained in exactly one plane intersecting $K$ in at most 4 points. Hence all points of $K\setminus\{P_{2}\}$ play the same role.

Let $l^{\prime}$ be a second tangent of $K$ containing $P^{\prime}$, with $l^{\prime}\not\subset\Pi^{\prime}$. Let $K\cap\Pi_{3}=\tilde{K},\Pi^{\prime}\cap\tilde{K}=\{P^{\prime},P^{\prime}_{1}\}$. If $P^{\prime}_{1}\in\{P,P_{1}\}$, then $|\Pi^{\prime}\cap K|\geq q$, a contradiction. Hence $P^{\prime}_{1}\not\in\{P,P_{1}\}$. With $P^{\prime}$ there corresponds an incidence structure $(\mathcal{P}^{\prime},\mathcal{B}^{\prime})$ of points and lines (see first part of (A)).

As $k=q^{2}-q+3$, we necessarily have $|\mathcal{P}^{\prime}|=2$ and $|\mathcal{B}^{\prime}|=2$. Hence $|\Pi^{\prime}\cap K|\in\{2,3\}$ and $\Pi^{\prime}\cap K\supset\{P^{\prime},P^{\prime}_{1}\}$. Let $\widetilde{\Pi}^{\prime}$ be the unique plane containing $l^{\prime}$ and intersecting $K$ in at most 4 points, and let $\widetilde{\Pi}^{\prime}\cap\widetilde{K}=\{P^{\prime},\widetilde{P}^{\prime}_{1}\}$. If $P^{\prime}_{1}\neq\widetilde{P}^{\prime}_{1}$, then by the structure of $(\mathcal{P}^{\prime},\mathcal{B}^{\prime})$ we have $\{P^{\prime}_{1},\widetilde{P}^{\prime}_{1}\}\subset\Pi^{\prime}$, clearly a contradiction. Hence $P^{\prime}_{1}=\widetilde{P}^{\prime}_{1}$, and so $\{P^{\prime},P^{\prime}_{1}\}\subset\widetilde{\Pi}^{\prime}\cap K$.

Without loss of generality we may assume that $\Pi^{\prime}\cap K=\{P^{\prime},P^{\prime}_{1},P^{\prime}_{2}\}$ and $\widetilde{\Pi}^{\prime}\cap K=\{P^{\prime},P^{\prime}_{1}\}$. As $|\Pi^{\prime}\cap K|$ is odd, the set $\Pi^{\prime}\cap K$ has to contain the point $P_{2}$. Consequently $P_{2}=P^{\prime}_{2}$.

Interchanging $\Pi_{3}$ and $\Pi_{i}$, $i\in\{3,4,\cdots,q+1\}$, and interchanging $P^{\prime}$ with any point of $(\Pi_{i}\cap K)\setminus\{P,P_{1}\}$, we see that $K\setminus\{P_{2}\}$ is partitioned into $(q^{2}-q+2)/2$ pairs, where each pair is contained in one plane intersecting $K$ in that pair, in one plane intersecting $K$ in that pair together with $P_{2}$, and in $q-1$ planes intersecting $K$ in a $(q+2)$-arc. Any other plane contains 0, 1, $q$ or $q+1$ points of $K$.

Now we count the planes intersecting $K$ in a $(q+2)$-arc and obtain

[TABLE]

Hence $q+2|(q^{2}-q+2)(q-1)$, so $q+2|24$, that is $q\in\{2,4\}$, a final contradiction.

We conclude that there is some tangent $l_{i}$ containing $P$, with $i\in\{1,2,...,t\}$, which is contained in exactly $\theta>1$ planes having at most 4 points in common with $K$.

(B) Assume, by way of contradiction, that some tangent $l$ of $K$ is contained in no plane intersecting $K$ in at most 4 points.

Hence each plane $\Pi_{i}$ containing $l$ satisfies $|\Pi_{i}\cap K|\geq q-2$, with $i=1,2,...,q+1$. By Theorem 1.2 the arc $\Pi_{i}\cap K$ can be extended to a $(q+2)$-arc $C_{i}$; let $C_{i}\cap l=\{N_{i},P\}$ with $l\cap K=\{P\}$. For at least two planes $\Pi_{i}$, say $\Pi_{1}$ and $\Pi_{2}$, we have $N_{1}=N_{2}$.

(B.1) First we prove that $N_{1}$ is on a tangent of $K$ not in $\Pi_{1}\cup\Pi_{2}$; clearly $N_{1}$ is on at least $2q-5$ tangents of $K$ contained in $\Pi_{1}\cup\Pi_{2}$. Assume the contrary. Then for any plane $\Pi_{i}\not\in\{\Pi_{1},\Pi_{2}\}$, the arc $\Pi_{i}\cap K$ must have an odd number of points. So $\Pi_{i}\cap K$ either is a $(q-1)$-arc or a $(q+1)$-arc, $i\in\{3,4,...,q+1\}$. Also, $N_{i}\neq N_{1}$ for $i=3,4,...,q-1$. If $\Pi_{i}\cap K$ is a $(q-1)$-arc and $C_{i}\setminus(\Pi_{i}\cap K)=\{N_{i},N^{\prime}_{i},N^{\prime\prime}_{i}\},i\in\{3,4,...,q+1\}$, then $N_{1}\in N^{\prime}_{i}N^{\prime\prime}_{i}$, as otherwise $N_{1}N^{\prime}_{i}$ and $N_{1}N^{\prime\prime}_{i}$ are tangents of $\Pi_{i}\cap K$.

Let $r$ be the number of planes $\Pi_{i}$, with $i\neq 1,2$, for which $\Pi_{i}\cap K$ is a $(q-1)$-arc, and let $s$ be the number of planes $\Pi_{i}$, with $i\neq 1,2$, for which $\Pi_{i}\cap K$ is a $(q+1)$-arc. The number of points of $K$ is at least

[TABLE]

As $K$ is complete, by Theorem 2.1

[TABLE]

so

[TABLE]

We may assume that $\Pi_{3}\cap K$ is a $(q-1)$-arc. The number of tangents of $K$ containing $N_{3}$ is at least

[TABLE]

Hence

[TABLE]

So

[TABLE]

a contradiction for $q>8$.

Consequently $N_{1}$ is on a tangent $l^{\prime}$ of $K$ not in $\Pi_{1}\cup\Pi_{2}$.

(B.2) **Now we consider all planes $\Pi^{\prime}_{i}$ containing the tangent $l^{\prime}$, with $i=1,2,...,q+1$. We will show that:

(a) For each plane $\Pi^{\prime}_{i}$ such that $|\Pi^{\prime}_{i}\cap K|\geq q-2$ the point $N_{1}$ does not extend the arc $\Pi^{\prime}_{i}\cap K$.

(b) For each $i$ we have $|\Pi^{\prime}_{i}\cap K|\geq q-2$**.

(a) Let $|\Pi_{1}\cap K|=\alpha,q-2\leq\alpha\leq q+1,|\Pi_{2}\cap K|=\beta,q-2\leq\beta\leq q+1$. Then $N_{1}$ is contained in at least $\alpha+\beta$ tangents of $K$. Now we consider all planes $\Pi^{\prime}_{i}$ containing the tangent $l^{\prime}$, with $i=1,2,...,q+1$. Assume, by way of contradiction, that $m=|\Pi^{\prime}_{i}\cap K|\geq q-2$ and that the $(q+2)$-arc $C^{\prime}_{i}$ extending $\Pi^{\prime}_{i}\cap K$ intersects $l^{\prime}$ in $\{N_{1},P^{\prime}\}$, with $l^{\prime}\cap K=\{P^{\prime}\},i\in\{1,2,...,q+1\}$. Then the number of tangents of $K$ containing $N_{1}$ is at least

[TABLE]

Hence

[TABLE]

So $q^{2}-2q+11>q^{2}+(1-\sqrt{5})q+5$, a contradiction. Consequently for $|\Pi^{\prime}_{i}\cap K|\geq q-2$ we have $N_{1}\not\in C^{\prime}_{i},i\in\{1,2,...,q+1\}$.

(b) Next, assume by way of contradiction that for at least one plane $\Pi^{\prime}_{i}$ containing $l^{\prime}$, say $\Pi^{\prime}_{1}$, we have $|\Pi^{\prime}_{1}\cap K|\leq 4$. Let $\Pi^{\prime}_{2}$ be the plane $ll^{\prime}$. Now we count the points of $K$ in the planes $\Pi^{\prime}_{i}$, with $i=1,2,...,q+1$. Let

$\theta_{1}$ be the number of planes $\Pi^{\prime}_{i},i\in\{3,4,...,q+1\}$, containing a tangent of $\Pi_{1}\cap K$ through $N_{1}$ and a tangent of $\Pi_{2}\cap K$ through $N_{1}$,

$\theta_{2}$ be the number of planes $\Pi^{\prime}_{i},i\in\{3,4,...,q+1\}$, containing a tangent of $\Pi_{1}\cap K$ through $N_{1}$, but no tangent of $\Pi_{2}\cap K$ through $N_{1}$,

$\theta_{3}$ be the number of planes $\Pi^{\prime}_{i},i\in\{3,4,...,q+1\}$, containing a tangent of $\Pi_{2}\cap K$ through $N_{1}$, but no tangent of $\Pi_{1}\cap K$ through $N_{1}$,

$\theta_{4}$ be the number of planes $\Pi^{\prime}_{i},i\in\{3,4,...,q+1\}$, containing no one of the tangents of $\Pi_{1}\cap K$ or $\Pi_{2}\cap K$ through $N_{1}$.

Then, as $N_{1}\notin C^{\prime}_{i}$ for $|\Pi^{\prime}_{i}\cap K|\geq q-2$, we have

[TABLE]

Hence

[TABLE]

so

[TABLE]

Now we have

$\theta_{1}+\theta_{2}\geq|\Pi_{1}\cap K|-2\geq q-4,$

$\theta_{1}+\theta_{3}\geq|\Pi_{2}\cap K|-2\geq q-4$.

Hence

[TABLE]

So $q^{2}-2q+11>q^{2}+(1-\sqrt{5})q+5$, a contradiction.

Hence no plane $\Pi^{\prime}_{i}$ containing $l^{\prime}$ intersects $K$ in a $m$-arc, with $m\leq 4,1\leq i\leq q+1$. Consequently, for each plane $\Pi^{\prime}_{i}$ containing $l^{\prime}$ we have $|\Pi^{\prime}_{i}\cap K|\geq q-2$. Also, we know that the $(q+2)$-arc $C^{\prime}_{i}$ extending $\Pi^{\prime}_{i}\cap K$ does not contain $N_{1}$, with $i=1,2,...,q+1$.

(B.3) A final contradiction will be obtained by considering the possible intersections $\Pi^{\prime}_{i}\cap K,i=1,2,...,q+1$. It is easy to see that at least $q-6$ planes $\Pi^{\prime}_{i}$ containing $l^{\prime}$ intersect $K$ in a $m$-arc having at least 3 tangents containing $N_{1}$; these planes are the planes containing $l^{\prime}$ passing through distinct tangents of $\Pi_{1}\cap K$ and $\Pi_{2}\cap K$ containing $N_{1}$. For any such plane $\Pi^{\prime}_{i}$ the arc $\Pi^{\prime}_{i}\cap K$ is either a $(q-1)$-arc or a $(q-2)$-arc. Let

$\theta^{\prime}_{1}$ be the number of planes $\Pi^{\prime}_{i}$, with $\Pi^{\prime}_{i}\not=ll^{\prime}$, containing a tangent of $\Pi_{1}\cap K$ through $N_{1}$, a tangent of $\Pi_{2}\cap K$ through $N_{1}$, where $\Pi^{\prime}_{i}\cap K$ is a $(q-1)$-arc,

$\theta^{\prime}_{2}$ be the number of planes $\Pi^{\prime}_{i}$, with $\Pi^{\prime}_{i}\not=ll^{\prime}$, containing a tangent of $\Pi_{1}\cap K$ through $N_{1}$, a tangent of $\Pi_{2}\cap K$ through $N_{1}$, where $\Pi^{\prime}_{i}\cap K$ is a $(q-2)$-arc.

Let $C^{\prime}_{i}\cap l^{\prime}=\{P^{\prime},N^{\prime}_{i}\}$, with $l^{\prime}\cap K=\{P^{\prime}\}$ and $C^{\prime}_{i}$ the $(q+2)$-arc extending $\Pi^{\prime}_{i}\cap K,i=1,2,...,q+1$. Then $N^{\prime}_{i}\not=N_{1},i=1,2,...,q+1$. We may assume that $N^{\prime}_{1}=N^{\prime}_{2}$. Assume, by way of contradiction, that $N^{\prime}_{1}=N^{\prime}_{2}=N^{\prime}_{i}$, with $i\in\{3,4,...,q+1\}$. Then $N^{\prime}_{1}$ is on at least $3(q-3)+1$ tangents of $K$. So

[TABLE]

Hence

[TABLE]

that is,

[TABLE]

clearly a contradiction. Hence we may assume that $N^{\prime}_{1}=N^{\prime}_{2},N^{\prime}_{3}=N^{\prime}_{4},N^{\prime}_{1}\not=N^{\prime}_{3},N^{\prime}_{i}\not\in\{N^{\prime}_{1},N^{\prime}_{3}\}$ for $i=5,6,...,q+1$. At least $\theta^{\prime}_{1}-4$ of the arcs $\Pi^{\prime}_{5}\cap K,\Pi^{\prime}_{6}\cap K,...,\Pi^{\prime}_{q+1}\cap K$ are $(q-1)$-arcs, say $\Pi^{\prime}_{5}\cap K,\Pi^{\prime}_{6}\cap K,...,\Pi^{\prime}_{\theta^{\prime}_{1}}\cap K$ are $(q-1)$-arcs. The number of tangents of $\Pi^{\prime}_{i}\cap K$ containing $N^{\prime}_{j}$ , with $j\in\{1,3\}$, is either 1 or 3, with $i=5,6,...,\theta^{\prime}_{1}$; if $N^{\prime}_{j}$ is contained in one tangent of $\Pi^{\prime}_{i}\cap K$, then $N^{\prime}_{u}$ is contained in 3 tangents of $\Pi^{\prime}_{i}\cap K$, with $\{j,u\}=\{1,3\}$ and $i\in\{5,6,...,\theta^{\prime}_{1}\}$. So we may assume that at least $(\theta^{\prime}_{1}-4)/2$ of the $(q-1)$-arcs $\Pi^{\prime}_{i}\cap K,i=5,6,...,\theta^{\prime}_{1}$, have 3 tangents containing $N^{\prime}_{1}$. Counting the tangents of $K$ through $N^{\prime}_{1}$, we obtain at least

[TABLE]

tangents. As $\theta^{\prime}_{1}+\theta^{\prime}_{2}\geq q-6$, this number of tangents is at least $1+q-6-6+2q-6=3q-17$. Hence

[TABLE]

So

[TABLE]

or

[TABLE]

a contradiction for $q>16$.

If at least one of the arcs $\Pi^{\prime}_{1}\cap K,\Pi^{\prime}_{2}\cap K$ is a $m$-arc with $m>q-2$, then (44) becomes

[TABLE]

which is at least $3q-15$. Hence $k\leq q^{2}-2q+17$. For $q=16$ this gives $k\leq 241$. But for $q=16$ the inequality $k>q^{2}+(1-\sqrt{5})q+5$ yields $k\geq 242$, a contradiction.

Finally we assume that $\Pi^{\prime}_{1}\cap K$ and $\Pi^{\prime}_{2}\cap K$ are $(q-2)$-arcs. Then at least $\theta^{\prime}_{1}-2$ of the arcs $\Pi^{\prime}_{i}\cap K$, with $i=5,6,...,q+1$, are $(q-1)$-arcs, say $\Pi^{\prime}_{5}\cap K,\Pi^{\prime}_{6}\cap K,...,\Pi^{\prime}_{\theta^{\prime}_{1}+2}\cap K$. So at least $(\theta^{\prime}_{1}-2)/2$ of the $(q-1)$-arcs $\Pi^{\prime}_{i}\cap K$, with $i=5,6,...,\theta^{\prime}_{1}+2$, have 3 tangents containing either $N^{\prime}_{1}$ or $N^{\prime}_{3}$. First, assume that this is the case for $N^{\prime}_{3}$. If at least one of the arcs $\Pi^{\prime}_{3}\cap K,\Pi^{\prime}_{4}\cap K$ is a $m$-arc with $m>q-2$, then the number of tangents of $K$ containing $N^{\prime}_{3}$ is at least

[TABLE]

which is at least $3q-13$. Hence $k\leq q^{2}-2q+15$, and so $q^{2}+(1-\sqrt{5})q+5<q^{2}-2q+15$, that is, $(3-\sqrt{5})q<10$, a contradiction. Hence the arcs $\Pi^{\prime}_{3}\cap K$ and $\Pi^{\prime}_{4}\cap K$ are $(q-2)$-arcs. Then the number of tangents of $K$ containing $N^{\prime}_{3}$ is at least

[TABLE]

which is at least $3q-13$. This yields again a contradiction. Consequently at least $(\theta^{\prime}_{1}-2)/2$ of the $(q-1)$-arcs $\Pi^{\prime}_{i}\cap K$, with $i=5,6,...,\theta^{\prime}_{1}+2$, have 3 tangents containing $N^{\prime}_{1}$. But then in (44) $\theta^{\prime}_{1}-4$ may be replaced by $\theta^{\prime}_{1}-2$, yielding at least $3q-15$ tangents of $K$ containing $N^{\prime}_{1}$. Hence $k\leq q^{2}-2q+17$, which is a final contradiction.

We conclude that each tangent $l$ of $K$ is contained in at least one plane intersecting $K$ in at most four points.

(C) Assume, by way of contradiction, that there is a tangent $l$ of $K$ which is contained in at least two planes $\Pi_{1},\Pi_{2}$ intersecting $K$ in a $m$-arc, with $m\leq 4$.

Assume that $l\cap K=\{P\}$ and that $\Pi_{1}\cup\Pi_{2}$ contains $2q+\delta$ tangents of $K$ through $P$. We have $-5\leq\delta\leq 1$.

(C.1) Here we will show that $2q+\delta$ is the total number of tangents of $K$ containing $P$; as a corollary it will follow that $k\in\{q^{2}-q+1,q^{2}-q+2,q^{2}-q+3\}$. Assume, by way of contradiction, that there is a tangent $l^{\prime}$ of $K$ containing $P$ with $l^{\prime}\not\subset\Pi_{1}\cup\Pi_{2}$. If $|ll^{\prime}\cap K|\leq 4$, then the number of tangents of $K$ containing $P$ is at least $2q+\delta+q-3=3q+\delta-3\geq 3q-8$, so $k\leq q^{2}+q+2-3q+8=q^{2}-2q+10$. Hence

[TABLE]

or $(3-\sqrt{5})q<5$, a contradiction. Now we consider all planes containing $l^{\prime}$. By (B) at least one of these planes intersects $K$ in a $m$-arc, with $m\leq 4$. If at least two planes containing $l^{\prime}$ intersect $K$ in at most 4 points, then $P$ is contained in at least $2q-5+2(q-5)+1=4q-14$ tangents of $K$. Hence $k\leq q^{2}+q+2-4q+14=q^{2}-3q+16$, so

[TABLE]

that is, $(4-\sqrt{5})q<11$, clearly a contradiction. Consequently exactly one plane $\Pi^{\prime}$ containing $l^{\prime}$ intersects $K$ in at most 4 points. Now we count the points of $K$ in the planes containing $l^{\prime}$. Let

$\theta_{1}$ be the number of planes , distinct from $ll^{\prime}$ and $\Pi^{\prime}$, containing $l^{\prime}$, containing a tangent of $K$ in $\Pi_{1}$ and containing a tangent of $K$ in $\Pi_{2}$,

$\theta_{2}$ be the number of planes containing $l^{\prime}$, distinct from $\Pi^{\prime}$, containing a tangent of $K$ in $\Pi_{1}$ and containing no tangent of $K$ in $\Pi_{2}$,

$\theta_{3}$ be the number of planes containing $l^{\prime}$, distinct from $\Pi^{\prime}$, containing a tangent of $K$ in $\Pi_{2}$ and containing no tangent of $K$ in $\Pi_{1}$,

$\theta_{4}$ be the number of planes , distinct from $\Pi^{\prime}$, containing $l^{\prime}$ and containing no tangent of $K$ in $\Pi_{1}$ or $\Pi_{2}$.

Then

[TABLE]

with

[TABLE]

So

[TABLE]

that is,

[TABLE]

hence

[TABLE]

Consequently

[TABLE]

or $(3-\sqrt{5})q<6$, a contradiction.

It follows that $2q+\delta$ is the total number of tangents of $K$ containing $P$ and so $k=q^{2}+q+2-2q-\delta=q^{2}-q+2-\delta$. As $k\leq q^{2}-q+3$ by Theorem 2.1, we have $-1\leq\delta\leq 1$.

(C.2) Some further properties of $K$. Let $l^{\prime\prime}$ be any tangent of $K$ not containing $P$ and let $K\cap l^{\prime\prime}=\{P^{\prime}\}$. By (B) $l^{\prime\prime}$ is contained in a plane $\Pi^{\prime\prime}$ with $|\Pi^{\prime\prime}\cap K|\leq 4$. There is a tangent $n$ of $K$ at $P^{\prime}$ not contained in $\Pi^{\prime\prime}$. The tangent $n$ is contained in a plane $\rho$ with $|\rho\cap K|\leq 4$. Let $2q+\delta^{\prime}$ be the number of tangents of $K$ at $P^{\prime}$ in $\rho\cup\Pi^{\prime\prime}$. Then $\delta^{\prime}\leq\delta$ and if $\rho\cap\Pi^{\prime\prime}$ is a tangent, then by the foregoing section we have $\delta^{\prime}=\delta$. Assume, by way of contradiction, that $\rho\cap\Pi^{\prime\prime}$ is not a tangent of $K$ and that $\delta^{\prime}<\delta$. Then there is a tangent $n^{\prime}$ of $K$ at $P^{\prime}$ not contained in $\rho\cup\Pi^{\prime\prime}$. The tangent $n^{\prime}$ is contained in a plane $\rho^{\prime}$ with $|\rho^{\prime}\cap K|\leq 4$. If $\rho\cap\rho^{\prime}$ is a tangent of $K$, then the $2q+\delta$ tangents of $K$ at $P^{\prime}$ are contained in $\rho\cup\rho^{\prime}$, a contradiction. So $\rho\cap\rho^{\prime}$ is not a tangent; similarly $\rho^{\prime}\cap\Pi^{\prime\prime}$ is not a tangent. Hence the number of tangents of $K$ at $P^{\prime}$ is at least $3(q-2)$, so $2q+\delta\geq 3q-6$, hence $\delta\geq q-6$, a contradiction. We conclude that $\delta^{\prime}=\delta$ and that all tangents of $K$ at $P^{\prime}$ are contained in $\rho\cup\Pi^{\prime\prime}$.

Hence, given any point $Q\in K$ there are two planes $\alpha_{1}$ and $\alpha_{2}$ containing all tangents of $K$ at $Q$; also $|\alpha_{1}\cap K|\leq 4$ and $|\alpha_{2}\cap K|\leq 4$. These planes are uniquely defined by $Q$, and so is $\alpha_{1}\cap\alpha_{2}$. By Section (A) the line $\alpha_{1}\cap\alpha_{2}$ is a tangent of $K$ at $Q$. Let $\widetilde{\Pi}$ be any plane containing $Q$, with $\widetilde{\Pi}\not\in\{\alpha_{1},\alpha_{2}\}$. Then $\widetilde{\Pi}\cap K$ contains at most two tangents at $Q$, so $|\widetilde{\Pi}\cap K|\geq q$. It follows that $K$ contains no $(q-2)$-arcs and no $(q-1)$-arcs.

Notice that $|\alpha_{1}\cap K|+|\alpha_{2}\cap K|+\delta=3$ and remind that $-1\leq\delta\leq 1$.

Let $\widetilde{\Pi}$ be a plane containing $Q$, with $\widetilde{\Pi}\not\in\{\alpha_{1},\alpha_{2}\}$. The arc $\widetilde{\Pi}\cap K$contains always at least one tangent of $K$ at $Q$, except when $\delta=-1,k=q^{2}-q+3,|\alpha_{1}\cap K|=|\alpha_{2}\cap K|=2$. So if $k\in\{q^{2}-q+1,q^{2}-q+2\}$ and if $k=q^{2}-q+3$ with $|\alpha_{1}\cap K|=|\alpha_{2}\cap K|+2=3$ or $|\alpha_{2}\cap K|=|\alpha_{1}\cap K|+2=3$, then $\widetilde{\Pi}\cap K$ is not a $(q+2)$-arc. If $|\alpha_{1}\cap K|=|\alpha_{2}\cap K|=2,k=q^{2}-q+3$, then there is excactly one plane $\widetilde{\Pi}$ containing $Q$ for which $\widetilde{\Pi}\cap K$ is a $(q+2)$-arc.

(C.3) $k=q^{2}-q+1$

Then $\delta=1$ and $|\Pi_{1}\cap K|=|\Pi_{2}\cap K|=1$. Let $U_{1},U_{2}\in K$, with $U_{1}\not=U_{2}$, and let $\xi_{1},\xi_{2}$ be the planes containing $U_{1}$ intersecting $K$ in at most 4 points. If $U_{2}\in\xi_{1}\cup\xi_{2}$, then $\delta\leq 0$, a contradiction. Hence $U_{2}\not\in\xi_{1}\cup\xi_{2}$. Consequently any plane containing the line $U_{1}U_{2}$ has more than 4 points in common with $K$.

Now we count the points of $K$ in planes containing the line $U_{1}U_{2}$. Let $\theta_{1}$ be the number of planes containing $U_{1}U_{2}$ intersecting $K$ in a $q$-arc, and let $\theta_{2}$ be the number of planes containing $U_{1}U_{2}$ intersecting $K$ in a $(q+1)$-arc. Then

[TABLE]

So

[TABLE]

that is $\theta_{1}=q$ and $\theta_{2}=1$.

Now we count the number of $(q+1)$-arcs on $K$, and obtain

[TABLE]

So $q+1|(q^{2}-q+1)(q-1)$, so $q+1|6$, a contradiction.

(C.4) $k=q^{2}-q+2$

Then $\delta=0$ and $\{|\Pi_{1}\cap K|,|\Pi_{2}\cap K|\}=\{1,2\}$. Let $Q$ be any point of $K$ and let $l_{Q}$ be the tangent of $K$ which is the intersection of the two planes $\alpha_{1}$ and $\alpha_{2}$ containing the $2q$ tangents of $K$ at $Q$. Let $(\alpha_{1}\cup\alpha_{2})\cap K=\{Q,Q^{\prime}\}$. Starting with $Q^{\prime}$ and $l_{Q^{\prime}}$, we find the same pair $\{Q^{\prime},Q\}$. It follows that K is partitioned into pairs of type $\{Q,Q^{\prime}\}$. Let $\mathcal{L}$ be the set of these $(q^{2}-q+2)/2$ pairs.

Let $\{Q,Q^{\prime}\}\in\mathcal{L}$, let $\alpha_{1}$ and $\alpha_{2}$ be the planes containing the $2q$ tangents of $K$ at $Q$, and assume that $Q^{\prime}\in\alpha_{1}$. Then $\alpha_{1}=l_{Q}l_{Q^{\prime}}$. Let $\Pi$ be a plane containing $QQ^{\prime}$, distinct from $\alpha_{1}$. As $\Pi$ contains a tangent of $K$ at $Q$, we have $|\Pi\cap K|\leq q+1$. Counting the points of $K$ in the planes containing $QQ^{\prime}$, we obtain $|\Pi\cap K|=q+1$. By an easy counting we see that the planes containing $l_{Q}$, but distinct from $\alpha_{1}$ and $\alpha_{2}$, intersect $K$ in $(q+1)$-arcs. This way there arise $q-1$ $(q+1)$-arcs $K_{1},K_{2},...,K_{q-1}$, having kernels $N_{1},N_{2},...,N_{q-1}$ on $l_{Q}\setminus\{Q\}$. Assume, by way of contradiction, that $N_{i}=N_{j},i\not=j$ and $i,j\in\{1,2,...,q-1\}$. Then $N_{i}$ is on at least $2q+1$ tangents of $K$, hence $k\leq q^{2}-q+1$, a contradiction. Let $l_{Q}\setminus\{N_{1},N_{2},...,N_{q-1}\}=N_{Q}$.

Assume, by way of contradiction, that $l_{Q}\cap l_{Q^{\prime}}\not=N_{Q}$. Let $l_{Q}\cap l_{Q^{\prime}}=N_{i},i\in\{1,2,...,q-1\}$, and let $R\in K_{i}\setminus\{Q\}$. Then $l_{Q^{\prime}}R\cap K$ is a $(q+1)$-arc with kernel $N_{i}$. Hence $N_{i}$ is on at least $q^{2}+2$ tangents, a contradiction. Consequently $l_{Q}\cap l_{Q^{\prime}}=N_{Q}$; similarly, $l_{Q}\cap l_{Q^{\prime}}=N_{Q^{\prime}}$.

Assume, by way of contradiction, that $l_{Q}\cap l_{S}\not=\emptyset$, with $Q\not=S$ and $\{Q,S\}\not\in\mathcal{L}$. Let $\{Q,Q^{\prime}\}$ and $\{S,S^{\prime}\}$ be elements of $\mathcal{L}$. Now we count the number of tangents of $K$ containing $l_{Q}\cap l_{S}=M$. The arc $l_{Q}l_{S}\cap K$ is a $(q+1)$-arc with kernel $M$, so $l_{Q}l_{S}$ contains $q+1$ tangents of $K$ through $M$; the arc $l_{Q}S^{\prime}\cap K$ is a $(q+1)$-arc, and as the line $MS^{\prime}$ of the plane $l_{S}S^{\prime}$ is a tangent of $K$, the point $M$ is the kernel of $l_{Q}S^{\prime}\cap K$, so $l_{Q}S^{\prime}$ contains $q+1$ tangents of $K$ through $M$; similarly the plane $l_{S}Q^{\prime}$ contains $q+1$ tangents of $K$ through $M$. Hence $M$ is contained in more than $2q$ tangents of $K$, clearly a contradiction. It follows that if $l_{Q}\cap l_{S}\not=\emptyset$, with $Q\not=S$, then $\{Q,S\}\in\mathcal{L}$.

Let $\{Q,S\}\not\in\mathcal{L}$, with $Q$ and $S$ distinct points of $K$. Then $l_{Q}\cap l_{S}=\emptyset$. Now we count the points of $K$ in the planes containing the line $QS$. Let $\theta_{1}$ be the number of planes which contain $QS$ and intersect $K$ in a $q$-arc, and let $\theta_{2}$ be the number of planes which contain $QS$ and intersect $K$ in a $(q+1)$-arc. Hence

[TABLE]

So $\theta_{1}=q-1$ and $\theta_{2}=2$. The 2 planes containing $QS$ and intersecting $K$ in a $(q+1)$-arc are the planes $l_{Q}S$ and $l_{S}Q$.

Let $\{Q,S\}\in\mathcal{L}$ and let $l_{Q}\cap l_{S}=N$. Then $N$ is kernel of no one of the $q-1$ $(q+1)$-arcs defined by planes containing the tangent $l_{Q}$ and of no one of the $q-1$ $(q+1)$-arcs defined by planes containing the tangent $l_{S}$. So for any line $n\not\in\{l_{Q},l_{S}\}$ containing $N$ we have $|n\cap K|\in\{0,2\}$. Let $n\cap K=\{U,U^{\prime}\}$.

First, assume that $\{U,U^{\prime}\}\not\in\mathcal{L}$. Then $|l_{U}U^{\prime}\cap K|=|l_{U^{\prime}}U\cap K|=q+1$. As $|l_{Q}U\cap K|=|l_{S}U\cap K|=q+1$, the planes $l_{Q}U$ and $l_{S}U$ are the two planes containing $UU^{\prime}$ and intersecting $K$ in a $(q+1)$-arc. Hence $\{l_{Q}U,l_{S}U\}=\{l_{U}U^{\prime},l_{U^{\prime}}U\}$. So we may assume that $l_{Q}U=l_{U}U^{\prime}$ and $l_{S}U=l_{U^{\prime}}U$. Consequently $l_{Q}\cap l_{U}\not=\emptyset$ and $l_{S}\cap l_{U^{\prime}}\not=\emptyset$, that is, $\{Q,U\}\in\mathcal{L}$ and $\{S,U^{\prime}\}\in\mathcal{L}$. Hence $|l_{Q}l_{U}\cap K|=|l_{S}l_{U^{\prime}}\cap K|=2$, clearly a contradiction as $Q,U,U^{\prime}\in l_{Q}l_{U}$.

It follows that $\{U,U^{\prime}\}\in\mathcal{L}$. So for any pair $\{T,T^{\prime}\}\in\mathcal{L}$, with $\{T,T^{\prime}\}\not=\{Q,S\}$, we have $N\in TT^{\prime}$. Let $n^{\prime},n^{\prime\prime}$ be distinct lines containing $N$ with $n^{\prime}\not=n\not=n^{\prime\prime}$ and $n^{\prime},n^{\prime\prime}\not\in\{l_{Q},l_{S}\}$. Assume also that $n^{\prime}\cap K=\{V,V^{\prime}\}$ and $n^{\prime\prime}\cap K=\{W,W^{\prime}\}$. Then $\{V,V^{\prime}\}\in\mathcal{L}$ and $\{W,W^{\prime}\}\in\mathcal{L}$. By the foregoing the lines $VV^{\prime},WW^{\prime},QS$ contain $N$, clearly a contradiction.

(C.5) $k=q^{2}-q+3$

Let $P$ be any point of $K$ and let $l_{P}$ be the tangent of $K$ which is the intersection of the two planes $\Pi_{1},\Pi_{2}$ containing the $2q-1$ tangents of $K$ at $P$. Two cases are considered.

(C.5.1) $\Pi_{1}\cap K=\{P,P^{\prime},P^{\prime\prime}\},\Pi_{2}\cap K=\{P\}$

Then $K$ contains no plane $(q+2)$-arcs containing $P$. Let $l$ be a tangent of $K$ at $P$, with $l$ in $\Pi_{1}$ and $l\not=l_{P}$. We count the points of $K$ in planes containing $l$. Let $\theta_{1}$ be the number of planes containing $l$ and intersecting $K$ in a $(q+1)$-arc, and let $\theta_{2}$ be the number of planes containing $l$ and intersecting $K$ in a $q$-arc. Then

[TABLE]

Hence $\theta_{1}=0$ and $\theta_{2}=q$. Let $\widetilde{\Pi}_{1},\widetilde{\Pi}_{2},...,\widetilde{\Pi}_{q}$ be the planes containing $l$ and intersecting $K$ in a $q$-arc, let $\widetilde{\Pi}_{i}\cap K=K_{i}$, let $C_{i}$ be the $(q+2)$-arc extending $K_{i}$ and let $C_{i}\cap l=\{P,N_{i}\}$, with $i=1,2,...,q$. Assume that for some $i\in\{1,2,...,q\}$ we have $N_{i}\notin P^{\prime}P^{\prime\prime}$. The number of tangents of $K$ containing $N_{i}$ is at least

[TABLE]

a contradiction. Hence $N_{1}=N_{2}=\cdots=N_{q}=l\cap P^{\prime}P^{\prime\prime}$. Then the number of tangents of $K$ containing $N_{1}$ is at least

[TABLE]

again a contradiction.

(C.5.2) $\Pi_{1}\cap K=\{P,P^{\prime}\},\Pi_{2}\cap K=\{P,P^{\prime\prime}\}$

By (C.5.1), for each point $Q\in K$ the two planes $\alpha_{1},\alpha_{2}$ through $Q$ intersecting $K$ in at most four points, intersect $K$ in exactly two points. If $\alpha_{1}\cap K=\{Q,Q^{\prime}\}$ and $\alpha_{2}\cap K=\{Q,Q^{\prime\prime}\}$, then the plane $QQ^{\prime}Q^{\prime\prime}$ is the only plane on $Q$ intersecting $K$ in a $(q+2)$-arc. Hence the $(q+2)$-arcs on $K$ partition $K$. So

[TABLE]

a contradiction.

Now the theorem is proved. $\blacksquare$

4. Corollaries

We are grateful to T. Szőnyi for bringing reference [12] to our attention which, in combination with Theorem 1.6, gives the following considerable improvement of the bound in Theorem 1.6; see also Remark 4.4.

Theorem 4.1**.**

[TABLE]

Proof.  In [12] it is proved that there does not exist a complete $k$-cap in $\mathbf{PG}(3,q)$, $q$ even, $q\geq 64$, for which

[TABLE]

where $a$ is an integer which satisfies

[TABLE]

Putting $a=3$, the desired result immediately follows from Theorem 1.6. $\blacksquare$

Theorem 4.2**.**

• (i)

$m_{2}(4,4)=41$,

• (ii)

$m_{2}(4,8)\leq 479$,

• (iii)

$m_{2}(4,q)<q^{3}-q^{2}+2\sqrt{5}q-8$, $q$ even, $q>8$.

Proof.  For $q=4$, see [5]. Assume, by way of contradiction, that $K$ is a $k$-cap of $\mathbf{PG}(4,8)$ with $k>479$, or a $k$-cap of $\mathbf{PG}(4,q)$, $q$ even and $q>8$, with $k>q^{3}-q^{2}+2\sqrt{5}q-8$. At each of its points the cap $K$ has $t=q^{3}+q^{2}+q+2-k$ tangents. Hence we assume that $t<107$ for $q=8$ and $t<2q^{2}+(1-2\sqrt{5})q+10$ for $q>8$. We obtain a contradiction in several stages.

I $K$** contains no plane $q$-arc**

Similar to the reasoning in Section I in the proof of Theorem 6.27 in [9].

II There exists no solid $\delta$ such that $q^{2}+1>|\delta\cap K|>q^{2}+(1-\sqrt{5})q+5$

Suppose $\delta$ exists. Let $\delta\cap K=K^{\prime}$. Then $K^{\prime}$ can be completed to an ovoid $O$ of $\delta$, by Theorem 3.1. Let $N\in O\setminus K^{\prime}$ and let $N^{\prime}\in K^{\prime}$. Consider the $q+1$ planes of $\delta$ through $NN^{\prime}$. Since each of these planes meets $O$ in a $(q+1)$-arc, each plane meets $K^{\prime}$ in at most a $q$-arc. By I, there is no $q$-arc on $K$; so each plane meets $K^{\prime}$ in at most a $(q-1)$-arc.

Assume, by way of contradiction, that none of these intersections is a $(q-1)$-arc. Therefore a count of the points on $K^{\prime}$ gives

[TABLE]

whence

[TABLE]

so

[TABLE]

a contradiction.

So we may assume that for one of the planes $\delta$ through $NN^{\prime}$, say $\Pi$, we have $|\Pi\cap K^{\prime}|=q-1$. Now we consider all solids of $\mathbf{PG}(4,q)$ containing the plane $\Pi$. Let $\theta$ be the number of solids $\Pi^{\prime}$ for which $|\Pi^{\prime}\cap K|>q^{2}+(1-\sqrt{5})q+5$, so $q+1-\theta$ is the number of solids $\Pi^{\prime\prime}$ for which $|\Pi^{\prime\prime}\cap K|<q^{2}+(1-\sqrt{5})q+5$. We have $\theta\geq 1$.

First, assume $\theta\geq 2$. So there are at least two solids $\Pi^{\prime}_{1},\Pi^{\prime}_{2}$ containing $\Pi$ such that $|\Pi^{\prime}_{i}\cap K|>q^{2}+(1-\sqrt{5})q+5$, with $i=1,2$. By Theorem 3.1 $\Pi^{\prime}_{i}\cap K$ can be completed to an ovoid $O_{i}$ of $\Pi^{\prime}_{i},i=1,2$. So $O_{i}\cap\Pi$ is a $(q+1)$-arc $(\Pi\cap K^{\prime})\cup\{N^{\prime}_{i},N^{\prime\prime}_{i}\},i=1,2.$ Since $\Pi\cap K^{\prime}$ can be contained in no more than three $(q+1)$-arcs, contained in a common $(q+2)$-arc, we have $|\{N^{\prime}_{1},N^{\prime\prime}_{1}\}\cap\{N^{\prime}_{2},N^{\prime\prime}_{2}\}|\geq 1$. Assume $N^{\prime}_{1}=N^{\prime}_{2}$. So the number of tangents of $K$ containing $N^{\prime}_{1}$ is at least

[TABLE]

so

[TABLE]

a contradiction.

Finally, assume that $\theta=1$. Counting the points of $K$ in the $q+1$ solids, we obtain

[TABLE]

that is,

[TABLE]

Hence, for $q>8$,

[TABLE]

so

[TABLE]

a contradiction. For $q=8$, there arises $479<479$, a contradiction.

III For a point $N$ not in $K$, there do not exist planes $\Pi_{1}$ and $\Pi_{2}$ such that $\Pi_{1}\cap\Pi_{2}=N$ and such that $\Pi_{i}\cap K$ is a $(q+1)$-arc with nucleus $N$ for $i=1,2$

Similar to the reasoning in Section III in the proof of Theorem 6.27 in [9].

IV The tangents through any point $Q$ off $K$ lie in a solid

Similar to the reasoning in Section IV in the proof of Theorem 6.27 in [9].

V The final contradiction is obtained by counting the tangents of $K$

Similar to the reasoning in Section V in the proof of Theorem 6.27 in [9]. $\blacksquare$

Theorem 4.3**.**

For $q$ even, $q>2$, $n\geq 5$,

• (i)

$m_{2}(n,4)\leq\frac{118}{3}.4^{n-4}+\frac{5}{3}$**

• (ii)

$m_{2}(n,8)\leq 478.8^{n-4}-2(8^{n-5}+\cdots+8+1)+1$,

• (iii)

$m_{2}(n,q)<q^{n-1}-q^{n-2}+2\sqrt{5}q^{n-3}-9q^{n-4}-2(q^{n-5}+\cdots+q+1)+1$, for $q>8$.

Proof  This follows directly from Theorem 1.1, Theorem 4.2 and Theorem 6.14(ii) in [9]. $\blacksquare$

Remark 4.4**.**

The bound in Theorem 4.1 leads to considerable improvements of Theorem 4.2 and Theorem 4.3. We just mention these bounds, but the proofs are the theme of a subsequent paper*.

For $q$ even, $q\geq 2048$,

[TABLE]

For $q$ even, $q\geq 2048$, $n\geq 5$,

[TABLE]

5. Remark

The bound in the MAIN THEOREM is better than the bound of Chao, see [3]. In 2014 Cao and Ou, see [2], published the bound $k<q^{2}-2q+8$ ($q$ even and $q\geq 128$), which is better than ours. I did not follow some reasoning in their proof, so I sent two mails to one of the authors explaining why I think Section 1.3 of the proof is not correct. Unfortunately I never received an answer.