Two-character sets, Veronese varieties and Grassmannian of lines

Abstract

We show that, for any q, the pointset covered by the conic planes of a quadric Veronesean of $\textrm{PG}(3,q)$ is a two character set with respect to hyperplanes of $\textrm{PG}(9,q)$.

1 Introduction

A two–character set in the projective space $\textrm{PG}(d,q)$ is a set ${\mathcal {T}}$ of n points with the property that intersection number with any hyperplane only takes two values, say $n-w_1$ and $n-w_2$. The positive integers $w_1$ and $w_2$ are called the constants of ${\mathcal {T}}$. Two–character sets with respect to hyperplanes are strictly connected to strongly regular graphs. Embed $\textrm{PG}(d,q)$ as a hyperplane $\Pi$ in $\textrm{PG}(d+1,q)$. The linear representation graph $\Gamma _d^*({{\mathcal {T}}})$ is the graph having as vertices the points of $\textrm{PG}(d+1,q)\setminus \Pi$ and two vertices are adjacent whenever the line defined by them meets ${\mathcal {T}}$. Then $\Gamma _d^*({{\mathcal {T}}})$ has $q^{d+1}$ vertices and valence $k=(q-1)n$. Delsarte [5] showed that this graph is strongly regular due to the fact that ${\mathcal {T}}$ is a two–character set. The other parameters of the graph are $\lambda =k-1+(k-qw_1+1)(k-qw_2+1)$ and $\mu =k+(k-qw_1)(k-qw_2)$. Regarding the coordinates of the elements of ${\mathcal {T}}$ as columns of the generator matrix of a code ${\mathcal {L}}$ of length n and dimension $d+1$, the two–character set property implies that the code ${\mathcal {L}}$ has two weights [1]. Such a code is called a projective two–weight code. In [3, Proposition 2.1] we proved that the secant variety ${\mathcal {M}}_4^3$ of a Veronese surface of $\textrm{PG}(5,q)$ is a two–character set with respect to hyperplanes. The Veronese surface contains $q^2+q+1$ plane conics mutually intersecting in one point. The planes of these conics are called conic planes and it can be proved that the points covered by them is exactly ${\mathcal {M}}_4^3$. Since any quadric Veronesean contains plane conics that are either disjoint or mutually intersect in one point, it is natural to ask if the pointset covered by them is always a two–character set with respect to hyperplanes. In this paper we show that the answer is affirmative for the quadric Veronesean of $\textrm{PG}(3,q)$, any q, but this property breaks down for $\textrm{PG}(4,q)$. Also, in Sect. 3, we observe that the Grassmannian of lines of $\textrm{PG}(4,q)$ and the secant variety of the Veronese variety of $\textrm{PG}(3,q)$, any q, lie in the same projective space $\textrm{PG}(9,q)$, have the same size and the same characters with respect to hyperplanes.

In the last section, we construct another two-character set of $\textrm{PG}(5,q)$ having the same characters as $\mathcal {M}_4^3$ but with a different automorphism group.

2 The quadric Veronesean ${\mathcal {V}}_n$ of $\textrm{PG}\left( \frac{n(n+3)}{2},q\right)$

Let q be a fixed prime power. For any integer k, denote by $\textrm{PG}(k,q)$ the k–dimensional projective space over the Galois field $\mathbb {F}_q$. Let $n\ge 1$ be an arbitrary integer. We choose homogeneous projective coordinates in $\mathbb {P}:=\textrm{PG}(n,q)$ and in $\Sigma :=\textrm{PG}(n(n+3)/2,q)$. The Veronesean map, say $\nu$, sends a point of $\textrm{PG}(n,q)$ with coordinates $(x_0,\dots ,x_n)$ onto the points of $\Sigma$ with coordinates

$$\begin{aligned} \left( {x_{0}^{2} ,x_{1}^{2} , \ldots ,x_{n}^{2} ,x_{0} x_{1} , \ldots ,x_{0} x_{n} , \ldots ,x_{{n - 1}} x_{n} } \right) . \end{aligned}$$

The quadric Veronesean or the Veronese variety ${\mathcal {V}}_n^{2^n}$, or, for short ${\mathcal {V}}_n$, is the image of the Veronesean map. It turns out that the Veronesean map is a bijection between points of $\mathbb {P}$ and points of ${\mathcal {V}}_n$. Also, $\mathcal {V}_n$ is a cap of $\Sigma$, cf. [6, Theorem 25.1.8].

Consider the quadric Veronesean ${\mathcal {V}}_n$ in $\Sigma$ and the correspondiing Veronesean map. The image of an arbitrary plane of $\mathbb {P}$ under the Veronesean map is a quadric Veronesean ${\mathcal {V}}_2$ and the subspace of $\Sigma$ generated by it has dimension 5. Such a subspace is called a ${\mathcal {V}}_2$–subspace of $\Sigma$. The image of a line of $\mathbb {P}$ is a conic and the plane generated by it is called a conic plane.

The automorphism group G of ${\mathcal {V}}_n$ contains $\textrm{PGL}(n+1,q)$ and acts transitively on conic planes of ${\mathcal {V}}_n$, cf. [6, Theorem 25.1.10].

Let $q>2$. The quadrics and the lines of $\mathbb {P}$ are in $1-1$ correspondence (under $\nu$) with the hyperplanes and the conics of the Veronese variety $\mathcal {V}_n$, respectively (see [6, Theorems 4.3, 4.10 and 4.12]). Let $\mathcal {M}:=\mathcal {M}_2(\mathcal {V}_n)$ be the secant variety of $\mathcal {V}_{n}$, i.e. the set of points on the lines joining two points of $\mathcal {V}_n$. Since through two distinct points of $\mathcal {V}_n$ there passes a unique conic of $\mathcal {V}_n$ then the variety $\mathcal {M}$ turns out to be the union of the conic planes of $\Sigma$.

In this setting the following preliminary result can be easily proven.

Proposition 2.1

Let $H=\nu (\mathcal {Q})$ be a hyperplane of $\Sigma$ corresponding to a quadric $\mathcal {Q}$ of $\mathbb {P}$. Let $\ell$ be a line of $\mathbb {P}$ and let $\pi _\ell \subset \mathcal {M}$ be the corresponding conic plane of the nondegenerate conic $\nu (\ell )$ of $\mathcal {V}_n$. We have the following cases:

• (1) If $\ell \subset \mathcal {Q}$ then $|\pi _\ell \cap (H{\setminus }\mathcal {V}_n)|=q^2$.

• (2) If $|\ell \cap \mathcal {Q}|=2$ then $|\pi _\ell \cap (H{\setminus }\mathcal {V}_n)|=q-1$.

• (3) If $|\ell \cap \mathcal {Q}|=1$ then $|\pi _\ell \cap (H{\setminus }\mathcal {V}_n)|=q$.

• (4) If $|\ell \cap \mathcal {Q}|=0$ then $|\pi _\ell \cap (H{\setminus }\mathcal {V}_n)|=q+1$.

Proof

The result follows from the fact that:

• (1) If $\ell \subset \mathcal {Q}$ then $\pi _\ell \subset H$ and $\pi _\ell \cap \mathcal {V}_n=\nu (\ell )$.

• (2) If $|\ell \cap \mathcal {Q}|=2$ then $\pi _\ell \cap H$ is a line with exactly two points on $\mathcal {V}_n$.

• (3) If $|\ell \cap \mathcal {Q}|=1$ then $\pi _\ell \cap H$ is a line with exactly one point on $\mathcal {V}_n$.

• (4) If $|\ell \cap \mathcal {Q}|=0$ then $\pi _\ell \cap H$ is a line with no point on $\mathcal {V}_n$.

$\square$

In this geometric setting let $x_s,x_0,x_1,x_2$ denote the number of lines contained, external, tangent and secant to $\mathcal {Q}$, respectively. Also, let $\Theta _m$ be the number of points of a projective space isomorphic to $\textrm{PG}(m,q)$. We have the following result.

Proposition 2.2

$|H\cap \mathcal {M}|=|\mathcal {Q}|+x_sq^2+x_0(q+1)+x_1q+x_2(q-1)$.

Proof

We first observe that $|\nu (\mathcal {Q})|=|H\cap \mathcal {V}_n|=|\mathcal {Q}|$ and the conics contained in $H\cap \mathcal {V}_n$ correspond (under $\nu ^{-1}$) to the lines of $\mathbb {P}$ contained in $\mathcal {Q}$. Hence, for any line $\ell \subset \mathcal {Q}$, the corresponding conic plane $\pi _\ell$ is contained in $H\cap \mathcal {M}$ and intersects $\mathcal {M}\setminus \mathcal {V}_n$ in $q^2$ points. Any other line, say m, of $\mathbb {P}$ is not contained in $\mathcal {Q}$ and the corresponding conic plane $\pi _m\subset \mathcal {M}$ of $\Sigma$ intersects H in a line meeting $\mathcal {V}_n$ in 0, 1 or 2 points according to the fact that m is external, tangent or secant to $\mathcal {Q}$. Also, since two conic planes are disjoint or they meet in a point of $\mathcal {V}_n$, the result follows. $\square$

Proposition 2.3

Let $q>2$ and $n\ge 4$. The secant variety $\mathcal {M}$ of $\Sigma$ has at least three intersection characters with respect to the hyperplanes, more precisely: if n is even

$$\begin{aligned} T1&:= \frac{{\left( {q^{n} - 1} \right) \left( {q^{{n + 1}} + q^{{n + 2}} - q^{n} - 1} \right) }}{{\left( {q + 1} \right) \left( {q - 1} \right) ^{2} }};\\ T2&:= \frac{{\left( {q^{4} + q - 1} \right) q^{{2n - 2}} - \left( {q + 1} \right) q^{{n + 1}} + 1}}{{\left( {q - 1} \right) ^{2} \left( {q + 1} \right) }};\\ T3&:= \frac{{\left( {q^{n} - 1} \right) \left( {q^{{n + 2}} - 1} \right) }}{{\left( {q - 1} \right) ^{2} \left( {q + 1} \right) }}; \end{aligned}$$

and if $n=2m-1$ is odd

$$\begin{aligned} T1,\ T2 \text { and } T4:=\frac{(q^m+1)(q^{m-1}+q^{2m} -q^{2m+1}+q^{3m-1}-q^{3m-3}+q^3-q-1)}{q-1}. \end{aligned}$$

Also $T_1>T_2>T_3$ and $T_4>T_1>T_2$.

Proof

Consider the following cases.

Case 1: $\mathcal {Q}$ is an $(n-1)$-dimensional projective space of $\mathbb {P}$. In such a case any line of $\mathbb {P}$ is either contained in $\mathcal {Q}$ or intersects $\mathcal {Q}$ in exactly one point. Straightforward computations show that

$$\begin{aligned} |\mathcal {Q}|= & {} \frac{q^n-1}{q-1};\\ x_s= & {} \frac{(q^n-1)(q^n-q)}{(q^2-1)(q^2-q)};\\ x_1= & {} \frac{(q^n-1)q^{n-1}}{q-1}. \end{aligned}$$

From Proposition 2.2 it follows

$$\begin{aligned} |H\cap \mathcal {M}|= \frac{q^n-1}{q-1}+q^2 \frac{(q^n-1)(q^n-q)}{(q^2-1)(q^2-q)} +q\frac{(q^n-1)q^{n-1}}{q-1} \end{aligned}$$

and hence

$$\begin{aligned} T_1=|H\cap \mathcal {M}|= \frac{(q^n-1) (q^{n+1}+q^{n+2}-q^n-1)}{(q-1)^2(q+1)}. \end{aligned}$$

(1)

Case 2: $\mathcal {Q}$ is a quadratic cone of $\mathbb {P}$ with vertex an $(n-3)$-dimensional projective space, say V, and base a non-degenerate conic $\mathcal {C}$. Then

$$\begin{aligned} |\mathcal {Q}|=q^{n-1}+q^{n-2}+\frac{q^{n-2}-1}{q-1}. \end{aligned}$$

Also, the number of lines contained in $\mathcal {Q}$ is

$$\begin{aligned} x_s=\frac{ q(q^{n-1}-1)(q^{n-1}-q)-q(q^{n-2}-1) (q^{n-2}-q)+(q^{n-1}-1)(q^{n-1}-q)}{(q^2-1)(q^2-q)}. \end{aligned}$$

If P is a point of the vertex V then through P there pass $q^{n-1}$ tangent lines to $\mathcal {Q}$, whereas if $P\notin V$, then through it there are $q^{n-2}$ tangents to $\mathcal {Q}$. It follows that

$$\begin{aligned} x_1=\Theta _{n-3}q^{n-1}+(|\mathcal {Q}|-\Theta _{n-3})q^{n-2}=q^{2n-4} +\frac{q^{n-1}(q^{n-1}-1)}{q-1}. \end{aligned}$$

Since through any point of $\mathcal {Q}\setminus V$ there pass $q^{n-1}$ secant lines to $\mathcal {Q}$, then

$$\begin{aligned} x_2=\frac{(|\mathcal {Q}|-\Theta _{n-3})q^{n-1}}{2}=\frac{q^{2n-3}(q+1)}{2}. \end{aligned}$$

Finally, the number of external lines to $\mathcal {Q}$ is

$$\begin{aligned} x_0=\frac{q^{2n-3}(q-1)}{2}. \end{aligned}$$

From Proposition 2.2, it follows that

$$\begin{aligned} T_2=|H\cap \mathcal {M}|= \frac{(q^4+q-1)q^{2n-2}-(q+1)q^{n+1}+1}{(q-1)^2(q+1)}. \end{aligned}$$

(2)

Case 3: $n=2m$ is even and $\mathcal {Q}$ is a parabolic quadric Q(n, q) of $\mathbb {P}$. Then

$$\begin{aligned} |\mathcal {Q}|=\Theta _{n-1}. \end{aligned}$$

From [2, p.451] (see also [6, pp. 23 and 25]) the number of lines contained, external and secant to $\mathcal {Q}$ is:

$$\begin{aligned} x_s= & {} \frac{(q^m-1)(q^{m-1}-1)(q^m+1)(q^{m-1}+1)}{(q-1)(q^2-1)};\\ x_0= & {} \frac{q^{n-1}(q^n-1)}{2*(q+1)};\\ x_2= & {} \frac{(q^m+1)(q^m-1)q^{n-1}}{2*(q-1)}. \end{aligned}$$

Hence, the number of tangent line to $\mathcal {Q}$ is

$$\begin{aligned} x_1=\frac{(q^{n+1}-1)(q^{n+1}-q)}{(q^2-1)(q^2-q)}-x_s-x_0-x_2. \end{aligned}$$

From Proposition 2.2, it follows

$$\begin{aligned} T_3=|H\cap \mathcal {M}|= \frac{(q^n-1)(q^{n+2}-1)}{(q-1)^2(q+1)}. \end{aligned}$$

(3)

Case 4: $n=2m-1$ is odd and $\mathcal {Q}$ is an elliptic quadric $Q^-(n,q)$ of $\mathbb {P}$. Then

$$\begin{aligned} |\mathcal {Q}|=\frac{(q^m+1)(q^{m-1}-1)}{q-1}. \end{aligned}$$

From [2, p.451] the number of lines contained, external and secant to $\mathcal {Q}$ is:

$$\begin{aligned} x_s= & {} (q^m+1)(q^{m-1}+1)(q^{m-1}-1)(q^{m-2}-1);\\ x_0= & {} \frac{(q^m+1)(q^{m-1}+1)q^{2m-2}}{2(q+1)};\\ x_2= & {} \frac{(q^m+1)(q^{m-1}-1)q^{2m-2}}{2(q-1)}. \end{aligned}$$

Hence, the number of tangent lines to $\mathcal {Q}$ is

$$\begin{aligned} x_1=\frac{(q^{n+1}-1)(q^{n+1}-q)}{(q^2-1)(q^2-q)}-x_s-x_0-x_2. \end{aligned}$$

From Proposition 2.2, it follows

$$\begin{aligned} T_4=|H\cap \mathcal {M}|= \frac{(q^m+1)(q^{m-1}+q^{2m} -q^{2m+1}+q^{3m-1}-q^{3m-3}+q^3-q-1)}{q-1}. \end{aligned}$$

(4)

Comparing the numbers $T_1, T_2,T_3$ and $T_1, T_2,T_4$ it can be seen that they are different. More precisely, direct computations show that:

$$\begin{aligned} T_1-T_2=q^{2n-2}, \quad T_2-T_3=\frac{q^{n-2}(q^n-q^2)}{q^2-1} \end{aligned}$$

and, since $q>2$,

$$\begin{aligned} T_4>\frac{q^{4m+1}-q^{4m}}{(q-1)^2(q+1)}>\frac{q^{4m}+q^{4m-1}}{(q-1)^2(q+1)} =\frac{q^{2n+2}+q^{2n+1}}{(q-1)^2(q+1)}>T1. \end{aligned}$$

$\square$

Remark 2.4

Note that if $\mathcal {Q}$ is either the union of two hyperplanes of $\textrm{PG}(n,q)$ or an $(n-2)$-dimensional projective subspace of $\mathbb {P}$, direct computations show that the size of $H\cap \mathcal {M}$ always equals $T_1$. Also, if $n=2m+1$ and $\mathcal {Q}$ is a hyperbolic quadric $Q^+(n,q)$ of $\mathbb {P}$ direct computations show that the size of $H\cap \mathcal {M}$ equals $T_2$.

2.1 The case $n=3$

In this section, using the previous results, we prove our main theorem.

Theorem 2.5

The secant variety $\mathcal {M}:=\mathcal {M}_2(\mathcal {V}_3)$ of $\Sigma =\textrm{PG}(9,q)$ is a two-character set with respect to the hyperplanes. The two characters are:

$$\begin{aligned} q^5+2q^4+2q^3+2q^2+q+1 \text { and }q^5+q^4+2q^3+2q^2+q+1 \end{aligned}$$

Proof

We have to consider all the possibilities for a quadric of $\mathbb {P}:=\textrm{PG}(3,q)$.

Case 1: $\mathcal {Q}$ is plane of $\mathbb {P}$. From (1), we get

$$\begin{aligned} |H\cap \mathcal {M}|=q^5+2q^4+2q^3+2q^2+q+1. \end{aligned}$$

Case 2: $\mathcal {Q}$ is a pair of planes of $\mathbb {P}=\textrm{PG}(3,q)$ intersecting in a line. From Remark 2.4 we have again

$$\begin{aligned} |H\cap \mathcal {M}|=q^5+2q^4+2q^3+2q^2+q+1. \end{aligned}$$

Case 3: $\mathcal {Q}$ is a quadratic cone of $\mathbb {P}$ with vertex a point and base a non-degenerate conic. From (2), we get

$$\begin{aligned} |H\cap \mathcal {M}|=q^5+q^4+2q^3+2q^2+q+1. \end{aligned}$$

Case 4: $\mathcal {Q}$ is a hyperbolic quadric of $\mathbb {P}$. From Remark 2.4 we have again

$$\begin{aligned} |H\cap \mathcal {M}|=q^5+q^4+2q^3+2q^2+q+1. \end{aligned}$$

Case 5: $\mathcal {Q}$ is an elliptic quadric of $\mathbb {P}$. From (4) we get again

$$\begin{aligned} |H\cap \mathcal {M}|=q^5+q^4+2q^3+2q^2+q+1. \end{aligned}$$

Case 6: $\mathcal {Q}$ is a line of $\mathbb {P}$. From Remark 2.4 we have

$$\begin{aligned} |H\cap \mathcal {M}|=q^5+2q^4+2q^3+2q^2+q+1. \end{aligned}$$

$\square$

2.2 The Case $n=4$

In this section we completely determine the characters of the secant variety $\mathcal {M}_2(\mathcal {V}_4)$.

Theorem 2.6

The secant variety $\mathcal {M}=\mathcal {M}_2(\mathcal {V}_4)$ of $\Sigma =\textrm{PG}(14,q)$ is a three-character set with respect to the hyperplanes. The three characters are:

$$\begin{aligned}{} & {} q^7+2q^6+2q^5+3q^4+2q^3+2q^2+q+1,\\{} & {} q^7 + q^6 + 2 q^5 + 3 q^4 + 2 q^3 + 2 q^2 + q + 1,\\{} & {} q^7 + 2 q^6 + 2 q^5 + 3 q^4 + 2 q^3 + 2 q^2 + q + 1. \end{aligned}$$

Proof

We have to consider all the possibilities for a quadric of $\mathbb {P}=\textrm{PG}(4,q)$.

Case 1: $\mathcal {Q}$ is a 3-dimensional projective space of $\mathbb {P}$. From (1), we get

$$\begin{aligned} |H\cap \mathcal {M}|=q^7+2q^6+2q^5+3q^4+2q^3+2q^2+q+1. \end{aligned}$$

Case 2: $\mathcal {Q}$ is pair of hyperplanes of $\mathbb {P}$ intersecting in a plane. From Remark 2.4 we have again

$$\begin{aligned} |H\cap \mathcal {M}|=q^7+2q^6+2q^5+3q^4+2q^3+2q^2+q+1. \end{aligned}$$

Case 3: $\mathcal {Q}$ is a quadratic cone of $\mathbb {P}$ with vertex a line and base a non-degenerate conic. From (2), we get

$$\begin{aligned} |H\cap \mathcal {M}|=q^7 + q^6 + 2 q^5 + 3 q^4 + 2 q^3 + 2 q^2 + q + 1. \end{aligned}$$

Case 4: $\mathcal {Q}$ is plane of $\mathbb {P}$. From Remark 2.4 we have

$$\begin{aligned} |H\cap \mathcal {M}|=q^7 + 2 q^6 + 2 q^5 + 3 q^4 + 2 q^3 + 2 q^2 + q + 1. \end{aligned}$$

Case 5: $\mathcal {Q}$ is parabolic quadric Q(4, q) of $\mathbb {P}$. From (3) we have

$$\begin{aligned} |H\cap \mathcal {M}|=q^7 + q^6 + 2 q^5 + 2 q^4 + 2 q^3 + 2 q^2 + q + 1. \end{aligned}$$

Case 6: $\mathcal {Q}$ is quadratic cone with vertex a point P and base a hyperbolic quadric $Q^+(3,q)$ of $\mathbb {P}$. In such a case $|\mathcal {Q}|=q(q+1)^2+1$ and $x_s=2q^3+3q^2+2q+1$. Through any point of $\mathcal {Q}$ different from P there pass $q^3$ secant lines to $\mathcal {Q}$. Hence $x_2=(|\mathcal {Q}|-1)/2=q^4(q+1)^2/2$. Also, the number of tangents through P is $q^3-q$ and through any other point of $\mathcal {Q}$ is $q^2-q$. Then $x_1=q^5+q^4-q^2-q$. Finally, $x_0=q^4(q^2-1)/2$. From Proposition 2.2, direct computations show that

$$\begin{aligned} |H\cap \mathcal {M}|=q^7+q^6+2q^5+3q^4+2q^3+2q^2+q+1. \end{aligned}$$

Case 7: $\mathcal {Q}$ is quadratic cone with vertex a point P and base an elliptic quadric $Q^-(3,q)$ of $\mathbb {P}$. In such a case $|\mathcal {Q}|=q(q^2+1)+1$ and $x_s=q^2+1$. Through any point of $\mathcal {Q}$ different from P there pass $q^3$ secant lines to $\mathcal {Q}$. Hence $x_2=(|\mathcal {Q}|-1)/2=q^4(q^2+1)/2$. Also, the number of tangents through P is $q^3+q$ and through any other point of $\mathcal {Q}$ is $q^2+q$. Then $x_1=q^5+q^4+2q^3+q^2+q$. Finally, $x_0=q^4(q^2+1)/2$. From Proposition 2.2, direct computations show that

$$\begin{aligned} |H\cap \mathcal {M}|=q^7 + q^6 + 2 q^5 + 3 q^4 + 2 q^3 + 2 q^2 + q + 1. \end{aligned}$$

$\square$

3 Grassmannians

Let ${\mathcal {G}}(1,n,q)$ denote the set of two–dimensional subspaces of the vector space $\mathbb {F}_q^{n+1}$. A two–dimensional subspace of $\mathbb {F}_q^{n+1}$ corresponds to a line of the associated projective space $\textrm{PG}(n,q)$ so that ${\mathcal {G}}(1,n,q)$ can be viewed as the set of such lines. Denote by $(\dots ,X_{ij},\dots )$, with $0\le i <j\le n$, the homogeneous coordinates of $\textrm{PG}(N,q)=PG(\Lambda ^2(\mathbb {F}_q^{n+1}))$, where $N=n(n+1)/2-1$. They are called the Plücker coordinates on ${\mathcal {G}}(1,n,q)$. If $\ell$ is a line of $\textrm{PG}(n,q)$ defined by two points, say $P_1=P(X_1)$ and $P_2=P(X_2)$, we can associate with $\ell$ the point $\rho (\ell )=P(X_1\wedge X_2)\in \textrm{PG}(N,q)$. The map of sets given by

$$\begin{aligned}{} & {} \rho : {\mathcal {G}}(1,n,q)\rightarrow PG(N,q),\\{} & {} \ell \mapsto \rho (\ell ) \end{aligned}$$

is a well–defined map called the Plücker embedding of ${\mathcal {G}}(1,n,q)$. The subset ${\mathcal {G}}(1,n,q)$ is an algebraic variety called the Grassmannian of lines of $\textrm{PG}(n,q)$. It is known that ${\mathcal {G}}(1,n,q)$ contains two systems of maximal subspaces. The first system consists of the $(n-1)$–spaces $\Pi _{n-1}$ with $\rho ^{-1}(\Pi _{n-1})$ the set of all lines through a common point; this system is called Latin system and its elements are called Latins spaces. The second system consists of planes $\Pi _2$ with $\rho ^{-1}(\Pi _2)$ the set of lines contained in a common plane; this system is called the Greek system and its elements are called Greek planes.

A linear complex ${\mathcal {L}}$ of $\textrm{PG}(n,q)$ is a set of lines whose Plu$\ddot{\textrm{c}}$ker coordinates satisfy a linear equation $\sum _{i<j}a_{ij}X_{ij}=0$. There exists a one-to-one correspondence between linear complexes of $\textrm{PG}(n,q)$ and hyperplane sections of ${\mathcal {G}}(1,n,q)$. Let $H:\sum _{i<j}a_{ij}X_{ij}=0$ be a hyperplane of $\textrm{PG}(N,q)$. We associate with H the $(n+1)\times (n+1)$ skew-symmetric matrix

$$\begin{aligned} A_H= \left( \begin{array}{cccc} 0 &{} a_{01} &{} \dots &{} a_{0n} \\ -a_{01} &{} 0 &{} \dots &{} a_{1n} \\ \vdots &{} \vdots &{} \vdots &{} \vdots \\ -a_{0n} &{} -a_{1n} &{} \dots &{} 0 \\ \end{array}\right) . \end{aligned}$$

up to a non–zero scalar in $\mathbb {F}_q\setminus \{0\}$. The linear complex ${\mathcal {L}}$ is said to be of type r if the corresponding matrix $A_H$ has rank r. A linear complex of type $n+1$ is said to be non-singular. It is known that the rank of a skew–symmetric matrix is an even number. Hence, in $\textrm{PG}(n,q)$, n even, all linear complexes are singular.

Lemma 3.1

Let $\ell =\langle P_0,P_1\rangle$ be a line of $\textrm{PG}(n,q)$, with $P_0=P(X)$ and $P_1=P(Y)$ two distinct points of $\textrm{PG}(n,K)$. The point $\rho (\ell )$ lies on the hyperplane H if and only if $XA_HY^t=0$.

Proof

Let $H:\sum _{i<j}a_{ij}X_{ij}=0$, $X=(X_i)_{0\le i\le n}$, $Y=(Y_i)_{0\le i\le n}$ and $\rho (\ell )=(p_{ij})_{0\le i<j\le n}$, where $p_{ij}=X_iY_j-X_jY_i$. Since

$$\begin{aligned} \sum _{i<j}a_{ij}p_{ij}=\sum _{i<j}a_{ij}x_iy_j -\sum _{i<j}a_{ij}x_jy_i=0, \end{aligned}$$

we see that $XA_HY^t=0$ if and only if $\rho (\ell )\in H$. $\square$

From now on we assume that $n=4$ and hence $N=9$.

Lemma 3.2

[6, Theorem 24.2.15] The automorphism group G of ${\mathcal {G}}(1,4,q)$ in $\textrm{PGL}(10,q)$ is isomorphic to $\textrm{PGL}(5,q)$.

There are two possibilities for a hyperplane section of ${\mathcal {G}}(1,4,q)$ since a skew-symmetric matrix of order 5 has type 2 or 4. The corresponding singular linear complexes correspond to:

• (type 2) lines of $\textrm{PG}(4,q)$ incident to or contained in a plane. In this case ${\mathcal {L}}$ consists of $(q^2+q+1)(q^3+q^2+1)$ lines;

• (type 4) lines of $\textrm{PG}(4,q)$ incident to or contained in a parabolic quadric Q(4, q). In this case ${\mathcal {L}}$ consists of $(q+1)^2(q+1)$ lines.

Theorem 3.3

The Grassmannian of lines of $\textrm{PG}(4,q)$ is a two–character set of $\textrm{PG}(9,q)$ with respect to hyperplanes. It has size $(q^2+1)(q^4+q^3+q^2+q+1)$ and the characters are $(q^2+q+1)(q^3+q^2+1)$ and $(q+1)^2(q+1)$.

Remark 3.4

The secant variety $\mathcal {M}(\mathcal {V}_n)$ and the Grassmannian of lines of $\textrm{PG}(n+1,q)$ lie in the same projective ambient $\textrm{PG}\left( \frac{n(n+3)}{2},q\right)$ and, for any n, they have the same number of points, i.e. $(q^n-1)(q^{n+1}-1)q^2/(q-1)(q^2-1)$. Also, as we have shown for $n=2$ and $n=3$, they have the same characters (with respect to hyperplanes). Is this just numerology or there is a combinatorial connection between the two algebraic varieties?

4 A derivation technique

In this section we construct another two-character set of $\textrm{PG}(5,q)$ with respect to hyperplanes having the same characters as ${\mathcal {M}}_4^3:=\mathcal {M}(\mathcal {V}_2)$ but not isomorphic to ${\mathcal {M}}_4^3$.

Let $X_{11},X_{12},X_{13},X_{22},X_{23},X_{33}$ be projective homogeneous coordinates in $\textrm{PG}(5,q).$ The Veronese surface ${\mathcal {V}}_2$ in $\textrm{PG}(5,q)$ has parametric equations

$$\begin{aligned} (a^2,ab,ac,b^2,bc,c^2), \end{aligned}$$

where $a,b,c\in \mathbb {F}_q$ and $(a,b,c)\ne (0,0,0)$ and the secant variety $\mathcal {M}_3^4$ has equation

$$\begin{aligned} X_{11}X_{22}X_{33}+2X_{12}X_{23}X_{13}-X_{12}^2X_{33} -X_{13}^2X_{22}-X_{23}^2X_{11}=0. \end{aligned}$$

Consider the 3-dimensional projective space $T:X_{13}=X_{33}=0$. It is easy to see that $T\cap \mathcal {M}_{3}^4$ consists of two planes, i.e., $\pi _t:X_{11}=X_{13}=X_{33}=0$ and $\pi _c:X_{23}=X_{13}=X_{33}=0$, which are a tangent plane and a conic plane of $\mathcal {V}_2$, respectively.

Also, let $\ell =\pi _t\cap \pi _c$. In T there are, apart from $\pi _t$ and $\pi _c$, another $q-1$ planes through $\ell$.

Lemma 4.1

If $\Pi$ is a hyperplane intersecting $\mathcal {V}_2$ in exactly one point P, then $\Pi$ contains the tangent plane to $\mathcal {V}_2$ in P.

Proof

Let $P=\nu (p)$, with $p\in \textrm{PG}(2,q)$. Embed $\textrm{PG}(2,q)$ in $\textrm{PG}(2,q^2)$. Let G be the stabilizer of $\textrm{PG}(2,q)$ in $\textrm{PGL}(3,q^2)$. Then G acts transitively on the points of $\textrm{PG}(2,q)$ and the stabilizer of p in G acts transitively on the imaginary lines through p. Hence, without loss of generality, we can always choose a hyperplane of $\textrm{PG}(5,q)$ intersecting $\mathcal {V}_2$ in a point.

Assume q is odd and let $\Pi : X_{22}=\alpha X_{33}$, with $\alpha$ a nonzero square in $\mathbb {F}_q$. Then $\Pi$ intersects $\mathcal {V}_2$ in the point $P:=(1,0,0,0,0,0)$ and contains the tangent plane to $\mathcal {V}_2$ in P with equation $X_{22}=X_{23}=X_{33}=0$.

Suppose q even and let $\Pi :\ X_{22}+X_{23}+\alpha X_{33}$, with $\textrm{Tr}_{q/2}(\alpha )=1$, where $\textrm{Tr}_{q/2}$ denotes the absolute trace from $\mathbb {F}_q$ over $\mathbb {F}_2$. Then $\Pi$ intersects $\mathcal {V}_2$ in the point $P:=(1,0,0,0,0,0)$ and contains the tangent plane to $\mathcal {V}_2$ in P with equation $X_{22}=X_{23}=X_{33}=0$. $\square$

Lemma 4.2

If a hyperplane $\Pi$ contains a tangent plane to $\mathcal {V}_2$ then $\Pi$ cannot intersect $\mathcal {V}_2$ in a quartic normal rational curve.

Proof

By way of contradiction suppose that the intersection $\Pi \cap \mathcal {V}_2$ is a quartic normal rational curve. Then the projective subspace tangent to the curve is a line. $\square$

Let $\pi$ be a plane through $\ell$ distinct from $\pi _t$ and $\pi _c$. The following result generalizes [4, Theorem 2].

Theorem 4.3

The sets $({\mathcal {M}}_4^3\setminus \pi _c)\cup \pi$ and $({\mathcal {M}}_4^3\setminus \pi _c)\cup \pi$ are two-character sets with respect to hyperplanes. The two characters are:

$$\begin{aligned} q^3+q^2+q+1 \text { and } q^3+2q^2+q+1. \end{aligned}$$

Proof

From [3] a hyperplane $\Pi$ of $\textrm{PG}(5,q)$ meets ${\mathcal {M}}_4^3$ in either $q^3+2q^2+q+1$ points or $q^3+q^2+q+1$ points. The first case occurs when the intersection between $\Pi$ and $\mathcal {V}_2$ is either a point or a conic or a pair of conics. The latter case occurs when the intersection between $\Pi$ and $\mathcal {V}_2$ is a quartic normal rational curve.

We first suppose that ${\mathcal {M}}'=({\mathcal {M}}_4^3{\setminus } \pi _c)\cup \pi$.

Let $\Pi$ be a hyperplane of $\textrm{PG}(5,q)$ and let $P=\ell \cap \mathcal {V}_2$. If $\Pi$ contains T there is nothing to prove. Otherwise, $\Pi$ meets T in a plane $\sigma$.

Four possibilities occur.

$\boxed {\text {Case 1}.}$ If $\sigma$ coincides with $\pi$ then $\Pi$ cannot contain a conic $\mathcal {C}$ of $\mathcal {V}_2$. In such a case $\Pi$ would contain the tangent line to such a conic $\mathcal {C}$ through P. Such a line is different from $\ell$ (otherwise $\pi _c\subset \Pi$ and so $T\subset \Pi$) and hence $\pi _t\subset \Pi$, i.e. $T\subset \Pi$, a contradiction. By Lemma 4.1, if $\Pi \cap \mathcal {V}_2=\{P\}$ then $\pi _t\subset \Pi$, again a contradiction. Then $\Pi$ intersects $\mathcal {V}_2$ in a quartic normal rational curve and hence $|\Pi \cap \mathcal {M}_3^4|=q^3+q^2+q+1$ and hence $|\Pi \cap \mathcal {M}'|=q^3+2q^2+q+1$.

$\boxed {\text {Case 2}.}$ If $\sigma$ coincides with $\pi _c$ then, by Lemma 4.2, $\Pi$ meets ${\mathcal {M}}_4^3$ in $q^3+2q^2+q+1$ points and hence ${\mathcal {M}}'$ in $q^3+q^2+q+1$ points.

$\boxed {\text {Case 3}.}$ If $\sigma$ coincides with $\pi _t$ then $\Pi$ meets both ${\mathcal {M}}_4^3$ and ${\mathcal {M}}'$ in $q^3+2q^2+q+1$ points.

$\boxed {\text {Case 4}.}$ The last possibility is when $\sigma$ meets both planes $\pi _c$ and $\pi$ in two distinct lines. In this case the line $\sigma \cap \pi _P$ is replaced by $\sigma \cap \pi$ and hence $\Pi$ meets ${\mathcal {M}}_4^3$ and ${\mathcal {M}}'$ in the same number of points that is either $q^3+2q^2+q+1$ or $q^3+q^2+q+1$.

If ${\mathcal {M}}':=({\mathcal {M}}_4^3\setminus \pi _t)\cup \pi$ the cases 2 and 3 interchange. $\square$

Proposition 4.4

${\mathcal {M}}'$ admits an automorphism group of order $q^3(q-1)$.

Proof

The group of ${\mathcal {M}}_4^3$ in $\textrm{PSL}_6(q)$ is isomorphic to $H:=\textrm{PSL}_3(q)$ and it acts transitively on the $(q+1)(q^2+q+1)$ tangent lines to ${\mathcal {V}}_2$. Then the stabilizer of $\ell$ in H has order $q^3(q-1)^2$ and permutes the $q-1$ planes through $\ell$ in T distinct from $\pi _t$ and $\pi _c$. Choosing one among these $q-1$ planes we get the result. $\square$