On Hermitian cones in $\mathrm{P}\mathrm{G}(3,q^2)$

1. Introduction

Algebraic varieties in a finite projective space $\mathrm{P}\mathrm{G}(r,q)$ have few finite intersection sizes with all the members of one (or more than one) prescribed family of subspaces. Thus, it is natural to ask if it is possible to reconstruct their structure starting from these intersection numbers and possibly other additional arithmetic and/or geoemetric conditions. A first example for such a characterization problem is the famous theorem of B. Segre (1954) [9] which characterizes the non–degenerate conics of $\mathrm{P}\mathrm{G}(2,q)$, $q$ odd, as sets of $(q+1)$–points intersected by any line in at most two points. There is a wide literature devoted to this problem, mostly when there are only two possibilities for the intersection sizes (cf e.g. [2,5,7,8,10] and the references cited therein). Whereas less is known, as regards sets with more than two intersection sizes, (cf e.g. [3,4,10]).

Recently, some papers have been published concerning characterizations of cones with a Baer subplane or an Hermitian arc as base (director) curve in $3$–dimensional finite projective spaces, as sets of points of $\mathrm{P}\mathrm{G}(3,q^2)$ with three intersection numbers with respect to the planes [1,6,4,3]. In [1], using a combination of combinatorial methods and algebraic geometry over finite fields, the author obtains a characterization of the Hermitian cone in $\mathrm{P}\mathrm{G}(3,q^2)$ as a surface of degree $q+1$ and with intersection sizes with the planes as those of an Hermitian cone and using this result on cones she gets also a characterization of (nonsingular) Hermitian surfaces in $\mathrm{P}\mathrm{G}(4,q^2)$.

In this paper, we present a combinatorial characterization of Hermitian cones of $\mathrm{P}\mathrm{G}(3,q^2)$ starting from slightly weaker conditions than those assumed in [1].

Before to state our result, we recall some definitions to which we refer throughout the paper.

Let $\mathbb{P}=\mathrm{P}\mathrm{G}(n,q)$ be the $n$–dimensional (desarguesian) projective pace of order $q$, and $m_1,\dots ,m_s$ be $s$ integers such that $0\le m_1<\cdots <m_s$ and for any integer $h$, $1\le h\le n-1$, let ${\mathcal{P}}_h$ denote the family of all the $h$–dimensional subspaces of $\mathbb{P}$. A subset $\mathcal{K}$ of $k$ points of $\mathbb{P}$ is a $k$–set of class $[m_1,\dots ,m_s{]}_h$ if $|\mathcal{K}\cap \pi |\in \{m_1,\dots ,m_s\}$ for every $\pi \in {\mathcal{P}}_h$. Furthermore, if for every $m_j\in \{m_1,\dots ,m_s\}$ there is at least one subspace $\pi \in {\mathcal{P}}_h$ such that $|\mathcal{K}\cap \pi |=m_j$ $\mathcal{K}$ is of type $(m_1,\dots ,m_s{)}_h$. The integers $m_1,\dots ,m_s$ are the intersection numbers of $\mathcal{K}$ (with respect to the dimension $h$). If $h=1,2$ one says that $\mathcal{K}$ is of line (plane)–class (type) $[m_1,\dots ,m_s{]}_h$ (resp. $(m_1,\dots ,m_s{)}_h$).

Let $\mathcal{K}$ be a subset of points of $\mathbb{P}$, a line (plane) intersecting $\mathcal{K}$ in exactly $i$ points is called $i$–line (plane). If $i=1$ a $1$-line (plane) is called tangent. An external line is a $0$–line.

An Hermitian arc (or unital of $PG(2,q^2)$ is a set of points of $\mathrm{P}\mathrm{G}(2,q^2)$ of size $q^3+1$ and of line–type $(1,q+1{)}_1$.

Let $\pi$ be a plane of $\mathrm{P}\mathrm{G}(2,q^2)$ and $\mathcal{H}$ be a Hermitian arc in $\pi$, and $V$ a point not in $\pi$. A Hermitian cone is the set of points of $\mathrm{P}\mathrm{G}(3,q^2)$ which is the union of the lines through $V$ and any point of $\mathcal{H}$.

We are going to prove the following result.

Let us end this section by recalling the statement of the result in [1] referred to above.

2. The proof

Throughout this section, the number of $i$–lines, ($1\le i\le q^2+1$), in a plane is denoted with $b_i$. For $i\in \{q^2+1,q^3+1,q^3+q^2+1\}$ the number of all $i$–planes of $\mathrm{P}\mathrm{G}(3,q)$ and the number of all $i$–planes passing through a line $\ell$ of the space are denoted with $c_i$ and $c_i(\ell )$, respectively.

The following Lemmata give the proof of Theorem 1.

In particular, $(q^2+1)$–lines do exist and a $(q^3+q^2+1)$–plane $\pi$ contains no external line.

Hence, $\mathcal{K}$ is a blocking set². Thus, $(q^2+1)$–planes intersects $\mathcal{K}$ in exactly one $(q^2+1)$–line.

If there is no $(q+1)$–line, let $\pi$ be a $(q^3+q^2+1)$–plane, then counting its number of points via the lines on a point of $\pi$ not in $\mathcal{K}$ gives $q^3+q^2+1\le (q^2+1)q$, a contradiction. Thus, $(q+1)$–lines do exist.

By the above proof, it follows that all the planes through a $q$–line are $(q^3+1)$–planes.

The basic equations for $k$–sets of points of $\mathrm{P}\mathrm{G}(3,q^2)$ of plane–type $(q^2+1,q^3+1,q^3+q^2+1{)}_2$ are

$$\begin{array}{rl}{c}_{q^2+1}+c_{q^3+1}+c_{q^3+q^2+1}=& (q^2+1)(q^4+1),\\ (q^2+1)c_{q^2+1}+(q^3+1)c_{q^3+1}+(q^3+q^2+1)c_{q^3+q^2+1}=& k(q^4+q^2+1),\\ q^2(q^2+1)c_{q^2+1}+q^3(q^3+1)c_{q^3+1}+(q^3+q^2)(q^3+q^2+1)c_{q^3+q^2+1}=& k(k-1)(q^2+1).\end{array}$$

Being $k=q^5+q^2+1$ it follows that $c_{q^2+1}=q^3+1$, $c_{q^3+1}=q^6$ and $c_{q^3+q^2+1}=q^4-q^3+q^2$.

Let $\ell$ be a $(q^2+1)$–line, and let $x$ and $y$ be the number of $(q^2+1)$–planes and of $(q^3+1)$–planes on $\ell$, respectively. Counting the number of points of $\mathcal{K}$ via the lines on $\ell$ gives

$$\begin{array}{rl}{q}^5+q^2+1=& q^2+1+y(q^3-q^2)+(q^2+1-x-y)q^3,\\ q^3=& x{q}^3+y{q}^2,\end{array}$$ that is $q=xq+y$. So, either $x=0$ and $y=q$ or $x=1$ and $y=0$. It follows that on a $(q^2+1)$–line there is at most one $(q^2+1)$–plane. Let $\mathcal{L}$ be the set of all the $(q^2+1)$–lines contained into exactly one $(q^2+1)$–plane. Double countings give $|\mathcal{L}|=c_{q^2+1}=q^3+1$.

Since a $(q^3+1)$–plane with no $(q^2+1)$–line intersects $\mathcal{K}$ in an Hermitian arc, it follows that there are no $q$–lines.

Let $\pi$ be a $(q^3+q^2+1)$–plane, and let $\ell$ be a $(q^2+1)$–line contained in $\pi$. Let $p$ be a point of $\pi \cap \mathcal{K}\setminus \ell$ and let ${\ell }_1$ a $(q^2+1)$–line of $\pi$ on $p$. Let $p_0$ be the point $\ell \cap {\ell }_1$. Let $x_0(\le q^2-1)$ be the number of $(q+1)$–lines of $\pi$ on $p_0$. Assume that there is no tangent line on $p_0$. Counting the number of points of $\pi \cap \mathcal{K}$ via the lines passing through $p_0$ gives

$$q^3+q^2+1=1+x_{0}q+(q^2+1-x_0)q^2,$$ that is $$x_0(q-1)=q^2(q-1),$$ and so $x_0=q^2$, a contradiction since $x_0\le q^2-1$.

Thus in $\pi$, and so in every $(q^3+q^2+1)$–plane there is at least one tangent line. So in every $(q^3+q^2+1)$–plane all the $(q^2+1)$–lines are concurrent in a point on which there are all the tangent lines to $\mathcal{K}$ of the plane, too. If $\ell$ is a $(q+1)$–line of a $(q^3+q^2+1)$–plane, since on each of its point there is a $(q^2+1)$–line and since these lines have to be concurrent we have that the number of all the $(q^2+1)$–lines of a $(q^3+q^2+1)$ plane is $(q+1)$.

So, in every $(q^3+q^2+1)$–plane, the $(q^2+1)$–lines are part of a pencil containing no $(q+1)$–line.

Let $\ell$ be a $(q^2+1)$–line and ${\pi }_0$ be a $(q^3+1)$–plane. Let $p$ be the point of $\mathcal{K}$ in which $\ell$ intersects ${\pi }_0$. Through $\ell$ there are exactly one $(q^2+1)$–plane, say ${\pi }_1$ and $q^2$ $(q^3+q^2+1)$–planes ${\pi }_i$, $i=2,\dots ,q^2+1$. The plane ${\pi }_1$ intersects ${\pi }_0\cap \mathcal{K}$ in the only tangent line at $p$, the planes ${\pi }_i$ intersects ${\pi }_0\cap \mathcal{K}$ in the remaining $q^2$ $(q+1)$–lines.

Any other $(q^2+1)$–line, say ${\ell }_1$ intersects $\ell$, otherwise there is no $(q^2+1)$–plane on ${\ell }_1$ and so counting the number of points of $\mathcal{K}$ through the planes on ${\ell }_1$ gives $q^5+q^2+1=k\ge q^2+1+(q^2+1)q^3$.

All the $(q^2+1)$–lines different from $\ell$ are concurrent in a point $V$ of $\ell$, otherwise let ${\ell }_1$ and ${\ell }_2$ be two $(q^2+1)$–lines intersecting $\ell$ in two different points. Then ${\ell }_1$ and ${\ell }_2$ are disjoint and so skew. Hence on each of them there pass only $(q^3+q^2+1)$ planes, and so counting the number of points of $\mathcal{K}$ via the planes on one of the ${\ell }_i$, $i=1,2$, gives a number bigger than $q^5+q^2+1$, a contradiction. Thus, there is a point $V$ on $\ell$ such that on it there are $(q^3+1)$ $(q^2+1)$–lines and all the other lines are tangent ones.

Hence, $\mathcal{K}$ is the Hermitian cone with vertex $V$ and base $\pi \cap \mathcal{K}$, and the proof is complete.