Absolute points of correlations of $PG(4,q^n)$

Abstract

The sets of the absolute points of (possibly degenerate) polarities of a projective space are well known. The sets of the absolute points of (possibly degenerate) correlations, different from polarities, of $\mathrm {PG}(2,q^n)$, have been completely determined by B.C. Kestenband in 11 papers from 1990 to 2014, for non-degenerate correlations and by D’haeseleer and Durante (Electron J Combin 27(2):2–32, 2020) for degenerate correlations. The sets of the absolute points of degenerate correlations, different from degenerate polarities, of a projective space $\mathrm {PG}(3,q^n)$ have been classified in (Donati and Durante in J Algebr Comb 54:109–133, 2021). In this paper, we consider the four dimensional case and completely determine the sets of the absolute points of degenerate correlations, different from degenerate polarities, of a projective space $\mathrm {PG}(4,q^n).$ As an application, we show that some of these sets are related to the Kantor’s ovoid and to the Tits’ ovoid of $Q(4,q^n)$ and hence also to the Tits’ ovoid of $\mathrm {PG}(3,q^n)$.

1 Introduction and preliminary results

1.1 Sesquilinear forms and correlations

Let V and W be two ${\mathbb {F}}$-vector spaces, where ${\mathbb {F}}$ is a field. A map $f:V\longrightarrow W$ is called semilinear or $\sigma$-linear if there exists an automorphism $\sigma$ of ${\mathbb {F}}$ such that

$$\begin{aligned}f(v+v') = f(v)+f(v') \, \, \, \, \text{ and } \, \, \, \, f(a v) = a^\sigma f(v) \end{aligned}$$

for all vectors $v\in V$ and all scalars $a\in {\mathbb {F}}$. If $\sigma$ is the identity map, then f is a usual linear map.Let V be an ${\mathbb {F}}$-vector space with finite dimension d. A map

$$\begin{aligned} \langle , \rangle : (v,v') \in V\times V \longrightarrow \langle v , v' \rangle \in {\mathbb {F}}\end{aligned}$$

is a sesquilinear form or a semibilinear form on V if it is a linear map on the first argument, and it is a $\sigma$-linear map on the second argument, that is:

$$\begin{aligned} \langle v+v', v'' \rangle= & {} \langle v , v'' \rangle + \langle v', v''\rangle ,\\ \langle v, v'+v'' \rangle= & {} \langle v , v' \rangle + \langle v, v''\rangle ,\\ \langle a v, v' \rangle= & {} a \langle v, v' \rangle , \, \, \, \, \langle v, a v' \rangle = a^\sigma \langle v, v' \rangle , \end{aligned}$$

for all $v,v',v''\in V$, $a\in {\mathbb {F}}$ and $\sigma$ an automorphism of ${\mathbb {F}}$. If $\sigma$ is the identity map, then $\langle , \rangle$ is a usual bilinear form. If ${{\mathcal {B}}} = (e_1,e_2,\ldots , e_{d+1})$ is an ordered basis of V, then for $x,y \in V$ we have $\langle x,y \rangle = X^t A Y^\sigma$, where $A=(\langle e_i, e_j \rangle )$ is the associated matrix to the sesquilinear form with respect to the ordered basis ${{\mathcal {B}}}$; X and Y are the columns of the coordinates of x, y w.r.t. ${{\mathcal {B}}}$. The term sesqui comes from the Latin, and it means one and a half. For every subspace S of V, put

$$\begin{aligned} S^{T} := & {} \{y \in V: \langle x, y \rangle =0 \,\,\, \forall x \in S \},\\ S^\perp := & {} \{y \in V: \langle y, x \rangle =0 \,\,\, \forall x \in S \}. \end{aligned}$$

Both $S^{T}$ and $S^\perp$ are subspaces of V. The subspaces $V^{T}$ and $V^\perp$ are called the right and the left radical of $\langle , \rangle$ and will be also denoted by $\mathrm {Rad}_r(V)$ and $\mathrm {Rad}_l(V)$, respectively.

Proposition 1.1

The right and the left radical of a sesquilinear form of a vector space V has the same dimension.

A non-degenerate sesquilinear form $\langle , \rangle$ has ${V}^\perp ={V}^{T}=\{0\}.$

Definition 1.2

A $\sigma$-sesquilinear form is reflexive if $\forall \; u,v \in V$:

$$\begin{aligned} \langle u, v \rangle =0 \iff \langle v, u \rangle =0. \end{aligned}$$

Definition 1.3

Let V be an ${\mathbb {F}}$-vector space of dimension greater than two. A bijection $g:\mathrm {PG}(V) \longrightarrow \mathrm {PG}(V)$ is a collineation if g, together with $g^{-1}$, maps k-dimensional subspaces into k-dimensional subspaces. If V has dimension two, then a collineation is a map $\langle v\rangle \in \mathrm {PG}(V) \longrightarrow \langle f(v)\rangle \in \mathrm {PG}(V)$, induced by a bijective semilinear map $f:V\longrightarrow V$.

Theorem 1.4

(Fundamental Theorem) Let V be an ${\mathbb {F}}$-vector space. Every collineation of $\mathrm {PG}(V)$ is induced by a bijective semilinear map $f: V \longrightarrow V$.

In the sequel if S is a vector subspace of V, we will denote with the same symbol S the associated projective subspace of $\mathrm {PG}(V)$. If v is a nonzero vector of V, we denote by $\langle v \rangle$ a point of $\mathrm {PG}(V)$.

Definition 1.5

Let $f:V\longrightarrow V$ be a semilinear map, with $\mathrm {Ker} f \ne \{0\}$. The map

$$\begin{aligned} \langle v\rangle \in \mathrm {PG}(V) \setminus \mathrm {Ker} f \longrightarrow \langle f(v)\rangle \in \mathrm{{PG}}(V), \end{aligned}$$

will be called a degenerate collineation of $\mathrm {PG}(V)$.

Definition 1.6

A (degenerate) correlation or (degenerate) duality of $\mathrm {PG}(d,{\mathbb {F}})$ is a (degenerate) collineation between $\mathrm {PG}(d,{\mathbb {F}})$ and its dual space $\mathrm {PG}(d,{\mathbb {F}})^*$.

Remark 1.7

A correlation of $\mathrm {PG}(d,{\mathbb {F}})$ can be seen as a bijective map of $\mathrm {PG}(d,{\mathbb {F}})$ that maps k-dimensional subspaces into $(d-1-k)$-dimensional subspaces reversing inclusion and preserving incidence.

A correlation of $\mathrm {PG}(d,{\mathbb {F}})$ applied twice gives a collineation of $\mathrm {PG}(d,{\mathbb {F}})$.

Theorem 1.8

Any (possibly degenerate) correlation of $\mathrm {PG}(d,{\mathbb {F}})$, $d>1$, is induced by a $\sigma$-sesquilinear form of the underlying vector space ${\mathbb {F}}^{d+1}$. Conversely, every $\sigma$-sesquilinear form of ${\mathbb {F}}^{d+1}$ induces two (possibly degenerate) correlations of $\mathrm {PG}(d,{\mathbb {F}})$. The two correlations coincide if and only if the form $\langle , \rangle$ is reflexive.

Remark 1.9

A (possibly degenerate) correlation induced by a $\sigma$-sesquilinear form will be also called a $\sigma$-correlation. Sometimes a (degenerate) correlation whose associated form is bilinear is called linear.

Definition 1.10

A (degenerate) polarity is a (degenerate) correlation whose square is the identity.

If $\perp$ is a (possibly degenerate) polarity, then for every pair of points P and R the following holds:

$$\begin{aligned} P \in R^\perp \iff R\in P^\perp . \end{aligned}$$

Proposition 1.11

A (degenerate) correlation is a (degenerate) polarity if and only if the induced sesquilinear form is reflexive.

The non-degenerate, reflexive $\sigma$-sesquilinear forms of a $(d+1)$-dimensional ${\mathbb {F}}$-vector space V have been classified (for a proof see, e.g., Theorem 3.6 in [1] or Theorem 6.3 and Proposition 6.4 in [2]).

In this paper, we will focus on degenerate non-reflexive $\sigma$-sesquilinear form of a five dimensional vector space over a finite field ${\mathbb {F}}_{q^n}$.

2 The $\sigma$-quadrics of $\mathrm {PG}(d,q^n)$

Let V be an ${\mathbb {F}}$-vector space. If V is equipped with a sesquilinear form $\langle , \rangle$ we may consider in $\mathrm {PG}(V)$ the set $\Gamma$ of absolute points of the associated correlation that is the points X such that $X \in X^{\perp }$ (or equivalently $X \in X^{T}$). If A is the associated matrix to the $\sigma$-sesquilinear form $\langle , \rangle$ w.r.t. an ordered basis of V, then the set $\Gamma$ has equation $X^tAX^{\sigma }=0$.

The definition of $\sigma$-quadrics of $\mathrm {PG}(d,q^n)$ has been first given in [10] (see also [4, 9]).

Definition 2.1

A $\sigma$-quadric of $d,q^n)$ is the set of the absolute points of a (possibly degenerate) $\sigma$-correlation, $\sigma \ne 1$, of $\mathrm {PG}(d,q^n)$. A $\sigma$-quadric of $\mathrm {PG}(2,q^n)$ will be called a $\sigma$-conic.

Proposition 2.2

Let $\Gamma :X^t A X^\sigma =0$ be a $\sigma$-quadric of $\mathrm {PG}(d,q^n)$. Every subspace S intersects $\Gamma$ either in a $\sigma$-quadric of S, or it is contained in $\Gamma$.

Proof

Let S be an h-dimensional subspace of $\mathrm {PG}(d,q^n)$. We may assume, w.l.o.g. that $S:x_{h+2}=0, \ldots , x_{d+1}=0$. Let $A'$ be the submatrix of A obtained by deleting the last $d-h$ rows and columns of A. If $A'=0$, then $S\subset \Gamma$. If $A'\ne 0$, then $S\cap \Gamma$ is a $\sigma$-quadric of S.$\square$

Regarding subspaces contained in $\sigma$-quadrics, in [9] the following has been proved.

Proposition 2.3

If $S_h$ is an h-dimensional subspace of $\mathrm {PG}(d,q^n)$ contained in a $\sigma$-quadric $\Gamma$ with equation $X^t A X^\sigma =0$, then $h\le \left\lfloor {d-\frac{{{\,\mathrm{\mathrm {rank}}\,}}(A)}{2}}\right\rfloor$. Moreover, there exists a $\sigma$-quadric with equation $X^t A X^\sigma =0$, containing a subspace with dimension $\left\lfloor {d-\frac{{{\,\mathrm{\mathrm {rank}}\,}}(A)}{2}}\right\rfloor .$

We recall that $\sigma$-quadrics have been completely classified in $\mathrm {PG}(d,q^n)$ for $d\in \{1,2\}$ (see [4]) and partially classified for $d=3$ (see [9]). Here we will deal with the four dimensional case. As in [9], we will divide the $\sigma$-quadrics of $\mathrm {PG}(4,q^n)$ according to the rank of the associated matrix. We start with the rank 4 case.

For what follows, we can assume $\sigma \ne 1$. Let $V={\mathbb {F}}_{q^n}^{5}$, let $\langle , \rangle$ be a degenerate $\sigma$-sesquilinear form with associated (degenerate) correlations $\perp$, $T$ and let $\Gamma :X^t A X^\sigma =0$ be the associated $\sigma$-quadric. We will denote by $L=V^{\perp }$ and $R=V^{T}$, the left and right radicals of $\langle , \rangle$, respectively, seen as subspaces of $\mathrm {PG}(4,q^n)$ that will be called the vertices of $\Gamma$.

Proposition 2.4

[9, Proposition 2.3] Let $\Gamma :X^tAX^{\sigma }=0$ be a $\sigma$-quadric of $\mathrm {PG}(d,q^n)$.

• For every point $Y \in \Gamma \setminus R$, the hyperplane $Y^{T}$ is the union of lines through Y either contained or 1-secant or 2-secant to $\Gamma$.

• For every point $Y \in \Gamma \setminus L$, the hyperplane $Y^{\perp }$ is the union of lines through Y either contained or 1-secant or 2-secant to $\Gamma$.

Corollary 2.5

[9, Corollary 2.4] Let $\Gamma :X^tAX^{\sigma }=0$ be a $\sigma$-quadric of $\mathrm {PG}(d,q^n)$ and let $L=V^{\perp }$, $R=V^{T}$.

• For every point $Y \in L$, the set $Y^{T} \cap \Gamma$ is the union of lines through Y.

• For every point $Y \in R$, the set $Y^{\perp } \cap \Gamma$ is the union of lines through Y.

Remark 2.6

Let $\Gamma :X^tAX^{\sigma }=0$ be a $\sigma$-quadric of $\mathrm {PG}(d,q^n)$, $n \ge 2$, and let $L=V^{\perp }$ and $R=V^{T}$ be its vertices. For every point $Y \in L$, the hyperplane $Y^{T}$ does not contain $(q+1)$-secant lines through Y to $\Gamma$, and for every point $Y \in R$, the hyperplane $Y^{\perp }$ does not contain $(q+1)$-secant lines through Y to $\Gamma$. Indeed, let $Y \in L$ and let Z be a point of $Y^{T}$. The line YZ has equation $X=\lambda Y+ \mu Z$, $(\lambda , \mu ) \in \mathrm {PG}(1,q^n)$; hence, $Y^{T} \cap \Gamma$ is determined by the solutions in $(\lambda , \mu )$ of the following equation:

$$\begin{aligned} Y^tAY^{\sigma }\lambda ^{\sigma +1}+Y^tAZ^{\sigma }\lambda \mu ^{\sigma }+Z^tAY^{\sigma }\lambda ^{\sigma }\mu +Z^tAZ^{\sigma }\mu ^{\sigma +1}=0. \end{aligned}$$

(1)

In the previous equation, it is $Y^tAY^{\sigma }=Y^tAZ^{\sigma }=0$, since $Y \in L$ and $Z^tAY^{\sigma }=0$, since $Z \in Y^{T}$. Hence, Eq. (1) becomes $Z^tAZ^{\sigma }\mu =0$ and two cases occur:

• If $Z^tAZ^{\sigma }=0$, then the line YZ is contained in $\Gamma$.

• If $Z^tAZ^{\sigma } \ne 0$, then the line YZ intersects $\Gamma$ exactly at the point Y.

If $Y \in R$, the result follows in a similar way.

3 $\sigma$-Quadrics of rank 4 in $\mathrm {PG}(4,q^n)$

Let $\Gamma : X^tAX^{\sigma }=0$ be a $\sigma$-quadric of $\mathrm {PG}(4,q^n)$ associated with a $\sigma$-sesquilinear form $\langle$ ,  $\rangle$. In this section, we assume that $\mathrm {rk}(A)=4$. Therefore, the radicals $V^{\bot }$ and $V^{\top }$ are one-dimensional vector subspaces of V, so they are points of $\mathrm {PG}(4,q^n)$. We distinguish several cases:

• 1) $V^{\bot } \ne V^{\top }.$ We may assume w.l.o.g. that the point $R=(1,0,0,0,0)$ is the right radical and the point $L=(0,0,0,0,1)$ is the left radical. It follows that

  $$\begin{aligned} A= \left( \begin{array}{lllll} 0 &{} a_{12} &{} a_{13} &{} a_{14} &{} a_{15} \\ 0 &{} a_{22} &{} a_{23} &{} a_{24} &{} a_{25} \\ 0 &{} a_{32} &{} a_{33} &{} a_{34} &{} a_{35} \\ 0 &{} a_{42} &{} a_{43} &{} a_{44} &{} a_{45} \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 \end{array}\right) , \end{aligned}$$

  and $\Gamma :(a_{12}x_1+a_{22}x_2+a_{32}x_3+a_{43}x_4)x_2^{\sigma }+(a_{13}x_1+a_{23}x_2+a_{33}x_3+a_{43}x_4)x_3^{\sigma }$ $+(a_{14}x_1+a_{24}x_2+a_{34}x_3+a_{44}x_4)x_4^{\sigma }+(a_{15}x_1+a_{25}x_2+a_{35}x_3+a_{45}x_4)x_5^{\sigma }=0.$ The degenerate collineation

  $$\begin{aligned} \top : Y \in \mathrm {PG}(4,q^n) \setminus \lbrace R \rbrace \mapsto X^tAY^{\sigma }=0 \in \mathrm {PG}(4,q^n)^{*} \end{aligned}$$

  associated with the sesquilinear form maps points into hyperplanes through the point L. Points that are on a common line through R are mapped into the same hyperplanes through L. Therefore, $\top$ induces a collineation $\Phi : {\mathcal {S}}_R \longrightarrow {\mathcal {S}}_L^{*}$. Let

  $$\begin{aligned} {\mathcal {S}}_R= \lbrace \textit{l}_{\alpha ,\beta ,\gamma ,\delta }: (\alpha ,\beta ,\gamma ,\delta ) \in \mathrm {PG}(3,q^n) \rbrace , \end{aligned}$$

  where

  $$\begin{aligned} \textit{l}_{\alpha ,\beta ,\gamma ,\delta }:{\left\{ \begin{array}{ll} x_1=\lambda \\ x_2=\mu \alpha \\ x_3=\mu \beta \\ x_4=\mu \gamma \\ x_5=\mu \delta \end{array}\right. }, (\lambda , \mu ) \in \mathrm {PG}(1,q^n) \end{aligned}$$

  and

  $$\begin{aligned} {\mathcal {S}}_L^{*} = \lbrace \Sigma _{\alpha ',\beta ',\gamma ',\delta '} : (\alpha ',\beta ',\gamma ',\delta ') \in \mathrm {PG}(3,q^n) \rbrace , \end{aligned}$$

  where

  $$\begin{aligned} \Sigma _{\alpha ',\beta ',\gamma ',\delta '}: \alpha ' x_1 + \beta ' x_2 + \gamma ' x_3 + \delta ' x_4 = 0. \end{aligned}$$

  The collineation $\Phi$ is given by $\Phi (\textit{l}_{\alpha ,\beta ,\gamma ,\delta }) = \Sigma _{\alpha ',\beta ',\gamma ',\delta '}$, with

  $$\begin{aligned} (\alpha ',\beta ',\gamma ',\delta ')^t=A'{\left[ (\alpha ,\beta ,\gamma ,\delta )^t\right] }^{\sigma }, \end{aligned}$$

  where $A'$ is the matrix obtained by A by deleting the last row and the first column. Note that $|A'| \ne 0$ since $\mathrm {rk}(A)=3$. It is easy to see that $\Gamma$ is the set of points of intersection of corresponding elements under the collineation $\Phi$. If $Y=(y_1,y_2,y_3,y_4,y_5)$ is a point of $\Gamma \setminus \lbrace R \rbrace$, then the tangent hyperplane to $\Gamma$ at the point Y is the hyperplane $\Sigma _{Y}=Y^{\top }$ with equation $X^tAY^{\sigma }=0$. It follows that for every point Y of $\Gamma \setminus \lbrace R \rbrace$ the hyperplane $\Sigma _Y$ contains the point L. The tangent hyperplane $\Sigma _{L}=L^{\top }$ to $\Gamma$ at the point L is the hyperplane with equation $X^tAL^{\sigma }=0$, that is:

  $$\begin{aligned} \Sigma _L: a_{15}x_1 + a_{25}x_2 + a_{35}x_3 + a_{45}x_4=0. \end{aligned}$$

  We again distinguish some cases.

  • i) First assume that $\Sigma _L$ contains the line RL. It follows that, w.l.o.g., we may put $\Sigma _L : x_4=0$. Hence, $a_{15}=a_{25}=a_{35}=0$ and we can put $a_{45} =1,$ obtaining $\Gamma :(a_{12}x_1+a_{22}x_2+a_{32}x_3+a_{42}x_4)x_2^{\sigma }+(a_{13}x_1+a_{23}x_2+a_{33}x_3+a_{43}x_4)x_3^{\sigma }$ $+(a_{14}x_1+a_{24}x_2+a_{34}x_3+a_{44}x_4)x_4^{\sigma }+x_4 x_5^{\sigma }=0.$ With this assumption, the collineation $\Phi$ maps the line RL into the hyperplane $\Sigma _L$. Consider now the star ${\mathcal {S}}_{R,\Sigma _{L}}$ of lines through R in $\Sigma _L$. We distinguish two cases.

    • i.1) Suppose that $\Phi$ maps the lines of ${\mathcal {S}}_{R,\Sigma _{L}}$ into the hyperplanes through the line RL. In this case, we can assume that $\Phi$ maps the line $x_3=x_4=x_5=0$ into the hyperplane $x_2=0$, the line $x_2=x_4=x_5=0$ into the hyperplane $x_3=0$ and the line $x_2=x_3=x_5=0$ into the hyperplane $x_1=0$ obtaining

      $$\begin{aligned} \Gamma : ax_1 x_4^{\sigma } + bx_2^{\sigma +1} - cx_3^{\sigma + 1} + x_4 x_5^{\sigma } =0. \end{aligned}$$

      We can assume that $\Gamma$ contains the points $(0,1,0,1,-1)$ and $(1,0,0,1,-1)$ obtaining $a=b=1$. By Corollary 2.5, we know that $R^{\perp } \cap \Gamma$ is the union of lines through R. Since $R^{\perp }$ has equation $x_4=0$, then a line $l_{\alpha , \beta , \gamma , \delta }$ through R is contained in $\Gamma$ if, and only if, $\alpha ^{\sigma +1}-c\beta ^{\sigma +1}=0$. Hence, the number of lines through R contained in $\Gamma$, different from the line RL, depends on the cardinality of the set $\lbrace l_{x,1,0,y} \in {\mathcal {S}}_R : x^{\sigma + 1} = c \rbrace$, and this is given by $q^n \vert \{ x \in {\mathbb {F}}_{q^n} : x^{\sigma + 1} = c \} \vert$. Moreover, the number of lines through L contained in $\Gamma$, different from the line RL, is equal to $q^n \vert \{ x \in {\mathbb {F}}_{q^n} : x^{\sigma + 1} = c \} \vert$. Indeed, let ${\mathcal {S}}_L=\{ t_{\alpha ,\beta ,\gamma ,\delta } : (\alpha ,\beta ,\gamma ,\delta ) \in \mathrm {PG}(3,q^n)\}$ be the set of lines through L, where

      $$\begin{aligned} \textit{t}_{\alpha ,\beta ,\gamma ,\delta }:{\left\{ \begin{array}{ll} x_1=\lambda \alpha \\ x_2=\lambda \beta \\ x_3=\lambda \gamma \\ x_4=\lambda \delta \\ x_5=\mu \end{array}\right. }, (\lambda , \mu ) \in \mathrm {PG}(1,q^n). \end{aligned}$$

      By Corollary 2.5, $\Sigma _L \cap \Gamma$ is the union of lines through L. Then a line $t_{\alpha ,\beta ,\gamma ,\delta }$ is contained in $\Gamma$ if, and only if, $\beta ^{\sigma +1}-c\gamma ^{\sigma +1}=0$. This yields that the number of lines through L contained in $\Gamma$, different from the line RL, depends on the cardinality of the set $\{ t_{x,y,1,0} \in {\mathcal {S}}_L : y^{\sigma +1}=c \}$. The number of solutions of the equation $x^{\sigma + 1}=c$ is either 0, 1, 2 or $q+1$ depending upon q even or odd and n even or odd. We distinguish several cases:

• If q is even and n is even, then there are either 0 or 1 or $q+1$ solutions giving either 0 or $q^n$ or $(q+1)q^n$ lines through R (and hence through L) contained in $\Gamma$.

• If q is even and n is odd, then there is a unique solution of the equation giving $q^n$ lines through R (and through L) contained in $\Gamma$.

• If q is odd and n is even, then there are either 0 or $q+1$ solutions of the equation giving either 0 or $(q+1)q^n$ lines through R (and through L) contained in $\Gamma$.

• If q is odd and n is odd, then there are either 0 or 2 solutions of the equation giving either 0 or $2q^n$ lines through R (and through L) contained in $\Gamma$.

In these cases, we will call the set $\Gamma$ either an elliptic or a $q^n$-parabolic or a $2q^n$-hyperbolic or a $(q+1)q^n$-hyperbolic $\sigma$-quadric with collinear vertex points R and L according to the number of lines through R (different from the line RL) contained in $\Gamma$ is either 0 or $q^n$ or $2q^n$ or $(q+1)q^n$. Now, let l a line through R. If $l \notin {\mathcal {S}}_{R, \Sigma _L}$ then $\Phi (l) \not \supset RL$ and so $l \cap \Phi (l)$ is a point. If $l \in {\mathcal {S}}_{R, \Sigma _L}$ then $l \cap \Phi (l)$ is either the point R or the line l. Recalling that $\Gamma$ contains the line RL, we get $|\Gamma |= q^{3n}+q^n \cdot q^n | \{ x \in {\mathbb {F}}_{q^n} : x^{\sigma +1}=c \}| + q^n+1$. If q is even and n is even, put $d=(q^n-1, q^m+1)$.

Theorem 3.1

Let $\Gamma$ be a degenerate elliptic $\sigma$-quadric of $\mathrm {PG}(4,q^n)$ with collinear vertex points R and L. Then, $\Gamma$ has canonical equation $\Gamma : x_1 x_4^{\sigma } + x_2^{\sigma +1} - cx_3^{\sigma + 1} + x_4 x_5^{\sigma } =0$, with c a nonsquare if q is odd and $c^{(q^n-1)/d} \ne 1$ if q is even and n is even. Moreover, $|\Gamma | = q^{3n}+q^n+1$ and $\Gamma$ contains only the line RL.

Theorem 3.2

Let $\Gamma$ be a degenerate $q^n$-parabolic $\sigma$-quadric of $\mathrm {PG}(4,q^n)$ with collinear vertex points R and L. Then, q is even and $\Gamma$ has canonical equation $\Gamma : x_1 x_4^{\sigma } + x_2^{\sigma +1} - cx_3^{\sigma + 1} + x_4 x_5^{\sigma } =0$, where the equation $x^{\sigma +1} = c$ has a unique solution. Moreover, $|\Gamma | = q^{3n}+q^{2n}+q^n+1$ and $\Gamma$ contains $q^n$ lines through R and $q^n$ lines through L (beside RL).

Theorem 3.3

Let $\Gamma$ be a degenerate $2q^n$-hyperbolic $\sigma$-quadric of $\mathrm {PG}(4,q^n)$ with collinear vertex points R and L. Then q and n are odd and $\Gamma$ has canonical equation $\Gamma : x_1 x_4^{\sigma } + x_2^{\sigma +1} - cx_3^{\sigma + 1} + x_4 x_5^{\sigma } =0$, where $x^{\sigma +1} = c$ has exactly two solutions. Moreover, $|\Gamma | = q^{3n}+2q^{2n}+q^n+1$ and $\Gamma$ contains $2q^n$ lines through R and $2q^n$ lines through L (beside RL).

Theorem 3.4

Let $\Gamma$ be a degenerate $(q+1)q^n$-hyperbolic $\sigma$-quadric of $\mathrm {PG}(4,q^n)$ with collinear vertex points R and L. Then, n is even and $\Gamma$ has canonical equation $\Gamma : x_1 x_4^{\sigma } + x_2^{\sigma +1} - cx_3^{\sigma + 1} + x_4 x_5^{\sigma } =0$, $x^{\sigma +1} = c$ has exactly $q+1$ solutions. Moreover, $|\Gamma | = q^{3n}+(q+1)q^{2n}+q^n+1$ and $\Gamma$ contains $(q+1)q^n$ lines through R and $(q+1)q^n$ lines through L (beside RL).

i.2) Now, suppose that $\Phi$ does not map the lines of ${\mathcal {S}}_{R, \Sigma _L}$ into the hyperplanes through the line RL. In this case, there exists a hyperplane $\Sigma$ containing RL such that the lines of the star ${\mathcal {S}}_{R, \Sigma }$ are mapped, under $\Phi$, into the hyperplanes through RL. Hence, there is another line through R (together with RL) contained in $\Gamma$. In this case, we may assume that $\Sigma : x_3=0$ and $\Phi$ maps the line $x_3=x_4=x_5=0$ into the hyperplane $x_2=0$, the line $x_2=x_4=x_5=0$ into the hyperplane $x_1=0$, and the line $x_2=x_3=x_5=0$ into the hyperplane $x_3=0$. Hence,

$$\begin{aligned} \Gamma : -ax_2^{\sigma + 1} + bx_1x_3^{\sigma }+cx_3x_4^{\sigma }+x_4 x_5^{\sigma }=0. \end{aligned}$$

Assuming that $\Gamma$ contains the points (0, 1, 0, 1, 1), $(0,0,-1,1,1)$, $(-1,0,1,1,0)$, we get $a= b=c=1$. Since $R^{\perp }$ has equation $x_3=0$; then, a line $l_{\alpha ,\beta ,\gamma ,\delta }$ through R is contained in $\Gamma$ if, and only if, $\alpha ^{\sigma +1}=\gamma \delta ^{\sigma }$. Observe that if $\alpha =0$, then either $\gamma =0$, which gives the line RL, or $\delta =0$, which gives the line $l_{0,0,1,0}$. So, the number of lines through R contained in $\Gamma$, different from the lines RL and $l_{0,0,1,0}$, depends on the cardinality of the set $\{ l_{1,0,x,y} \in {\mathcal {S}}_R: xy^{\sigma }=1 \}$. A pair (x, y) is a solution of $xy^{\sigma }=1$ if, and only if, $y=x^{-\sigma ^{-1}}.$ Hence, there are $q^n-1$ solutions of the equation giving $q^n-1$ lines through R contained in $\Gamma$. We will call the set $\Gamma$ a non-degenerate parabolic $\sigma$-quadric with collinear vertex points R and L. The following holds:

Theorem 3.5

Let $\Gamma$ be a non-degenerate parabolic $\sigma$-quadric of $\mathrm {PG}(4,q^n)$ with collinear vertex points R and L. Then, $\Gamma$ has canonical equation $\Gamma : -x_2^{\sigma + 1} +x_1x_3^{\sigma }+x_3x_4^{\sigma }+x_4 x_5^{\sigma }=0$. Moreover, $|\Gamma | = q^{3n}+q^{2n}+q^n+1$ and $\Gamma$ contains $q^n$ lines through R and $q^n$ lines through L (beside RL).

ii) Next, assume that $\Sigma _L$ does not contain the line RL (or equivalently $\Phi$ does not map the line RL into a hyperplane through the line RL). W.l.o.g. we may put $\Sigma _L : x_1=0$. In this case, there is a hyperplane through R (not containing L), say $\Sigma _R$, such that the star of lines through R in $\Sigma _R$ is mapped, under $\Phi$, into the hyperplanes through RL. We may assume that $\Sigma _R: x_5=0$. Hence, $\Phi$ maps the lines $l_{\alpha , \beta , \gamma , 0}$ into the hyperplanes $\Sigma _{0, \beta ', \gamma ', \delta '}$ so we may assume that $\Phi$ maps the line $l_{1,0,0,0}$ into the hyperplanes $\Sigma _{0,1,0,0}$, the line $l_{0,1,0,0}$ into the hyperplanes $\Sigma _{0,0,1,0}$, and the line $l_{0,0,1,0}$ into the hyperplanes $\Sigma _{0,0,0,1}$. Hence, the points of $\Gamma$ satisfy the equation

$$\begin{aligned} \Gamma : ax_2^{\sigma + 1} + b x_3^{\sigma + 1} + cx_4^{\sigma + 1} + x_1x_5^{\sigma }=0. \end{aligned}$$

Assuming, w.l.o.g., that the point $(-1,1,0,0,1)$ belongs to $\Gamma$ we obtain $a=1$. Since $R^{\perp }$ has equation $x_5=0$, then a line $l_{\alpha ,\beta ,\gamma ,\delta }$ through R is contained in $\Gamma$ if, and only if, $\alpha ^{\sigma + 1}+b \beta ^{\sigma +1}+c \gamma ^{\sigma +1}=0$. Hence, the numbers of lines through R contained in $\Gamma$ depend on the number of points of the Kestenband $\sigma$-conic of $\mathrm {PG}(2,q^n)$ (see [11,12,13,14,15,16,17,18,19,20,21]) given by the equation $x^{\sigma +1} + by^{\sigma +1} + cz^{\sigma +1}=0$. We distinguish several cases:

• If q is odd and n is odd, then there are $q^n+1$ points giving $q^n+1$ lines through R (and through L) contained in $\Gamma$.

• If q is even and n is odd, then there are $q^n+1$ points giving $q^n+1$ lines through R (and through L) contained in $\Gamma$.

• If n is even, then there are either $q^n+1+(-q)^{n/2+1}(q-1)$ or $q^n+1+(-q)^{n/2}(q-1)$ or $q^n+1-2(-q)^{n/2}$ points giving either $q^n+1+(-q)^{n/2+1}(q-1)$ or $q^n+1+(-q)^{n/2}(q-1)$ or $q^n+1-2(-q)^{n/2}$ lines through R (and through L) contained in $\Gamma$.

In these cases, we will call the set $\Gamma$ either of type 1 or of type 2 or of type 3 or of type 4 with vertex points R and L according to the number of lines through R contained in $\Gamma$ is either $q^n+1$ or $q^n+1+(-q)^{n/2+1}(q-1)$ or $q^n+1+(-q)^{n/2}(q-1)$ or $q^n+1-2(-q)^{n/2}.$ Now, let l a line through R. If l is not contained in $\Sigma _R$, then $\Phi (l) \not \supset RL$ and so $l \cap \Phi (l)$ is a point. If l is contained in $\Sigma _R$, then $l \cap \Phi (l)$ is either the point R or the line l. Recalling that $\Gamma$ does not contain the line RL, we get $|\Gamma |= q^{3n}+q^n | \{ (x,y,z) \in 2,q^n) : x^{\sigma +1}+by^{\sigma +1}+cz^{\sigma +1}=0 \}| +1$.

Theorem 3.6

Let $\Gamma$ be a non-degenerate $\sigma$-quadric of type 1 of $\mathrm {PG}(4,q^n)$ with vertex points R and L. Then, q is odd and n is either odd or even and $\Gamma$ has canonical equation $\Gamma : x_2^{\sigma + 1} + x_3^{\sigma + 1} + x_4^{\sigma + 1} + x_1x_5^{\sigma } = 0$, where the Kestenband $\sigma$-conic of $\mathrm {PG}(2,q^n)$ given by the equation $x^{\sigma +1}+y^{\sigma +1}+z^{\sigma +1}=0$ has $q^n+1$ points. Moreover, $|\Gamma | = q^{3n}+q^{2n}+q^n+1$ and $\Gamma$ contains exactly $q^n+1$ lines through R and exactly $q^n+1$ lines through L.

Theorem 3.7

Let $\Gamma$ be a non-degenerate $\sigma$-quadric of type 2 of $\mathrm {PG}(4,q^n)$ with vertex points R and L. Then, n is even and $\Gamma$ has canonical equation $\Gamma : x_2^{\sigma + 1} + x_3^{\sigma + 1} + x_4^{\sigma + 1} + x_1x_5^{\sigma } = 0$, where the Kestenband $\sigma$-conic of $\mathrm {PG}(2,q^n)$ given by the equation $x^{\sigma +1}+y^{\sigma +1}+z^{\sigma +1}=0$ has $q^n+1+(-q)^{n/2+1}(q-1)$ points. Moreover, $|\Gamma | = q^{3n} + q^{2n} + q^n + q^n(-q)^{n/2+1}(q-1)+1$ and $\Gamma$ contains exactly $q^n+1+(-q)^{n/2+1}(q-1)$ lines through R and exactly $q^n+1+(-q)^{n/2+1}(q-1)$ lines through L.

Theorem 3.8

Let $\Gamma$ be a non-degenerate $\sigma$-quadric of type 3 of $\mathrm {PG}(4,q^n)$ with vertex points R and L. Then, n is even and $\Gamma$ has canonical equation $\Gamma : x_2^{\sigma + 1} + x_3^{\sigma + 1} + cx_4^{\sigma + 1} + x_1x_5^{\sigma }= 0$, where the Kestenband $\sigma$-conic of $\mathrm {PG}(2,q^n)$ given by the equation $x^{\sigma +1}+y^{\sigma +1}+cz^{\sigma +1}=0$ has $q^n+1+(-q)^{n/2}(q-1)$ points, with $c \notin \lbrace x^{q+1} : x \in {\mathbb {F}}_{q^n} \rbrace .$ Moreover, $|\Gamma |= q^{3n} + q^{2n} + q^n + q^n(-q)^{n/2}(q-1)+1$ and $\Gamma$ contains exactly $q^n+1+(-q)^{n/2}(q-1)$ lines through R and exactly $q^n+1+(-q)^{n/2}(q-1)$ lines through L.

Theorem 3.9

Let $\Gamma$ be a non-degenerate $\sigma$-quadric of type 4 of $\mathrm {PG}(4,q^n)$ with vertex points R and L. Then, n is even and $\Gamma$ has canonical equation $\Gamma : x_2^{\sigma + 1} + bx_3^{\sigma + 1} + cx_4^{\sigma + 1} + x_1x_5^{\sigma }= 0$, where the Kestenband $\sigma$-conic of $\mathrm {PG}(2,q^n)$ given by the equation $x^{\sigma +1}+by^{\sigma +1}+cz^{\sigma +1}=0$ has $q^n+1-2(-q)^{n/2}$ points, with b, c, $b/c \notin \lbrace x^{q+1} : x \in {\mathbb {F}}_{q^n} \rbrace .$ Moreover, $|\Gamma |= q^{3n} + q^{2n} +q^n -2q^n(-q)^{n/2} + 1$ and $\Gamma$ contains exactly $q^n+1-2(-q)^{n/2}$ lines through R and exactly $q^n+1-2(-q)^{n/2}$ lines through L.

2) $V^{\bot } = V^{\top }.$ We may assume w.l.o.g. that the point $R=L=(1,0,0,0,0)$ is both the left radical and the right radicals. It follows that, in this case, $\Gamma$ is a cone with vertex the point R projecting a $\sigma$-quadric of rank 4 in a hyperplane not through R. Indeed, since the matrix A has rank four with first column and last row equal to 0, by choosing a hyperplane not through the point R, e.g., $\Sigma : x_1=0$, we get that the set $\Gamma \cap \Sigma$ is a $\sigma$-quadric of the hyperplane $\Sigma$ with associated matrix of rank 4.

Proposition 3.10

Let $\Gamma : X^tAX^{\sigma }=0$ be a $\sigma$-quadric of $\mathrm {PG}(4,q^n)$, with $\mathrm {rk}(A)= 4$. The set $\Gamma$ is one of the following:

• a degenerate either elliptic or $q^n$-parabolic or $2q^n$-hyperbolic or $(q+1)q^n$-hyperbolic $\sigma$-quadric with two collinear vertex points;

• a non-degenerate parabolic $\sigma$-quadric with two collinear vertex points;

• a non-degenerate $\sigma$-quadric either of type 1 or of type 2 or of type 3 or of type 4 with two vertex points;

• a cone with vertex a point V projecting a $\sigma$-quadric of rank 4 in a hyperplane $\Sigma$, with $V \notin \Sigma$.

4 $\sigma$-Quadrics of rank 3 in $\mathrm {PG}(4,q^n)$

In this section, a $\sigma$-quadric $\Gamma$ of $\mathrm {PG}(4,q^n)$ will have equation $X^tAX^{\sigma }=0$ with $\mathrm {rk}(A)=3$. Hence, $\mathrm {dim}V^{\bot }=\mathrm {dim}V^{\top }=2$ so right and left radicals in $\mathrm {PG}(4,q^n)$ are two lines r and l. We distinguish three cases:

• 1) $r \cap l = \emptyset .$ We may assume w.l.o.g. that $r:x_3=x_4=x_5=0$ and $l:x_1=x_2=x_3=0$. Then:

  $$\begin{aligned}&\Gamma :(a_{13}x_1+a_{23}x_2+a_{33}x_3)x_3^{\sigma }+(a_{14}x_1+a_{24}x_2+a_{34}x_3)x_4^{\sigma }\\&\quad +(a_{15}x_1+a_{25}x_2+a_{35}x_3)x_5^{\sigma }=0. \end{aligned}$$

  Let

  $$\begin{aligned} {\mathcal {P}}_r = \lbrace \pi _{a,b,c} : (a,b,c) \in \mathrm {PG}(2,q^n) \rbrace , \end{aligned}$$

  where

  $$\begin{aligned} \pi _{a,b,c} = {\left\{ \begin{array}{ll} x_1=\lambda \\ x_2 = \mu \\ x_3= \gamma a \\ x_4=\gamma b \\ x_5= \gamma c \end{array}\right. }, (\lambda , \mu , \gamma ) \in \mathrm {PG}(2,q^n) \end{aligned}$$

  and

  $$\begin{aligned} {\mathcal {S}}_{l} = \lbrace \Sigma _{a,b,c} : (a,b,c) \in \mathrm {PG}(2, q^n) \rbrace , \hbox { where }\Sigma _{a,b,c} : ax_1 + bx_2 + cx_3 = 0. \end{aligned}$$

  The $\sigma$-quadric $\Gamma$ is the set of points of $\mathrm {PG}(4,q^n)$ of intersection of corresponding elements under a collineation $\Phi : {\mathcal {P}}_r \longrightarrow {\mathcal {S}}_l$. Let $\Sigma _{rl}$ be the hyperplane spanned by the lines r and l, it follows that:

  $$\begin{aligned} \Sigma _{rl} : x_3=0. \end{aligned}$$

  Let $\pi$ the plane through r s.t. $\Sigma _{rl} = \Phi (\pi )$. We distinguish two cases.

  • a) First assume that $\Sigma _{rl}$ contains the plane $\pi$. W.l.o.g., we may put $\pi : x_3=x_4=0$. Then, $\Phi (\pi _{0,0,1})=\Sigma _{0,0,1}$. We may assume that $\Phi$ maps the plane $\pi _{1,0,0}$ into the hyperplane $\Sigma _{1,0,0}$, the plane $\pi _{0,1,0}$ into the hyperplane $\Sigma _{0,1,0}$ and the plane $\pi _{1,1,1}$ into the hyperplane $\Sigma _{1,1,1}$, and hence, a canonical equation of $\Gamma$ in this case is given by

    $$\begin{aligned} \Gamma : x_1x_3^{\sigma }+x_2x_4^{\sigma }+x_3x_5^{\sigma }=0. \end{aligned}$$

    The set $\Gamma$ is the union of the plane $\pi$ and $q^{2n}+q^n$ lines. Let P be a point of the plane $\pi _{a,b,c} \in {\mathcal {P}}_r$, then $P=(\lambda , \mu , \gamma a, \gamma b, \gamma c).$ Observe that P belongs to l if, and only if, $\lambda = \mu = a=0$ and $\gamma \ne 0$. It follows that the plane $\pi _{a,b,c}$ is skew with l if, and only if, $a \ne 0$. Therefore, $\Gamma$ contains $q^n$ lines which are transversal with r and l, and $q^{2n}$ which are incident with r and skew with l. Hence, $\Gamma$ has $q^{3n}+2q^{2n}+q^n+1$ points. We will call this set a degenerate hyperbolic $\sigma$-quadric with skew vertex lines r and l.

  • b) Now assume that $\Sigma _{rl}$ does not contain the plane $\pi$. It follows that, w.l.o.g., we may put $\pi : x_4=x_5=0$. Then, $\Phi (\pi _{1,0,0})=\Sigma _{0,0,1}$. We may assume that $\Phi$ maps the plane $\pi _{0,1,0}$ into the hyperplane $\Sigma _{0,1,0}$, the plane $\pi _{0,0,1}$ into the hyperplane $\Sigma _{1,0,0}$ and the plane $\pi _{1,1,1}$ into the hyperplane $\Sigma _{1,1,1}$, and hence, a canonical equation of $\Gamma$ in this case is given by

    $$\begin{aligned} \Gamma : x_3^{\sigma + 1}+ x_2x_4^{\sigma }+x_1x_5^{\sigma } = 0. \end{aligned}$$

    The set $\Gamma$ is the union of $q^{2n}+q^n+1$ lines whose $q^n+1$ lines are transversal with r and l, and $q^{2n}$ lines are incident with r and skew with l. Observing that $\pi \cap \Sigma _{rl} = r$, it follows that $\Gamma$ has $q^{3n}+q^{2n}+q^n+1$ points. We will call this set a non-degenerate parabolic $\sigma$-quadric with skew vertex lines r and l.

    • 2) $r \cap l = \lbrace V \rbrace$ is a point. We may assume w.l.o.g. that $r:x_3=x_4=x_5=0$ and $l: x_2=x_3=x_4=0$. In this case, the $\sigma$-quadric $\Gamma$ is a cone with vertex the point V. Since the matrix A has rank three with first two columns and first and last rows equal to 0, by choosing a hyperplane not through the point V, e.g., $\Sigma :x_1=0$, we get that the set $\Gamma \cap \Sigma$ is a $\sigma$-quadric of the hyperplane $\Sigma$ with associated matrix of rank three and two (collinear or not) vertex points given by $R=r \cap \Sigma$ and $L=l \cap \Sigma$. It follows that $\Gamma$ is a cone with vertex the point V projecting a $\sigma$-quadric of rank 3 in a hyperplane not through V with two (collinear or not) vertex points.

    • 3) $r = l.$ We may assume w.l.o.g. that $r= l : x_3=x_4=x_5=0$. It follows that, in this case, $\Gamma$ is a cone with vertex the line r. Since the matrix A has rank three with first two columns and first two rows equal to 0, by choosing a plane not through the line r, e.g., $\pi : x_1=x_2=0$, we get that the set $\Gamma \cap \pi$ is a $\sigma$-conic of the plane $\pi$ with associated matrix of rank three. Hence, it is a Kestenband $\sigma$-conic of $\pi$. It follows that $\Gamma$ is a cone with vertex the line r projecting a Kestenband $\sigma$-conic in a plane not through r. In particular, if $n=2$ and $\sigma ^2=1$, then $\Gamma$ is a Hermitian cone with vertex the line r.

Proposition 4.1

Let $\Gamma : X^tAX^{\sigma }=0$ be a $\sigma$-quadric of $\mathrm {PG}(4,q^n)$, with $\mathrm {rk}(A)= 3$. The set $\Gamma$ is one of the following:

• a degenerate hyperbolic $\sigma$-quadric with skew vertex lines r and l;

• a non-degenerate parabolic $\sigma$-quadric with skew vertex lines r and l;

• a cone with vertex a point V projecting a $\sigma$-quadric of rank 3 in a hyperplane $\Sigma$ with two (collinear or not) vertex points, with $V \notin \Sigma$;

• a cone with vertex a line v projecting a Kestenband $\sigma$-conic of a plane $\pi$ not through v.

5 $\sigma$-Quadrics of rank 2 in $\mathrm {PG}(4,q^n)$

In this section, a $\sigma$-quadric $\Gamma$ of $\mathrm {PG}(4,q^n)$ will have equation $X^tAX^{\sigma }=0$ with $\mathrm {rk}(A)=2$. Hence, $\mathrm {dim}V^{\bot }=\mathrm {dim}V^{\top }=3$ so right and left radicals in $\mathrm {PG}(4,q^n)$ are two planes $\pi _R$ and $\pi _L$. We distinguish three cases:

• 1) $\pi _R \cap \pi _L = \lbrace V \rbrace$ is a point. We may assume w.l.o.g. that $\pi _R : x_4 = x_5=0$ and $\pi _L: x_1=x_2=0$. It follows that, in this case, $\Gamma$ is a cone with vertex the point V. Since the matrix A has rank two with first three columns and last three rows equal to 0, by choosing a hyperplane not through the point V, e.g., $\Sigma : x_3=0$, we get that the set $\Gamma \cap \Sigma$ is $\sigma$-quadric of the hyperplane $\Sigma$ with associated matrix of rank 2 and vertices the lines

  $$\begin{aligned} r:x_3=x_4=x_5=0 \hbox { and }l:x_1=x_2=x_3=0. \end{aligned}$$

  Hence, it is a $\sigma$-quadric of pseudoregulus type of $\Sigma$ with skew vertex lines r and l (see [7, 9]). It follows that $\Gamma$ is a cone with vertex the point V projecting a $\sigma$-quadric of pseudoregulus type in a hyperplane not through V with skew vertex lines.

• 2) $\pi _R \cap \pi _L = t$ is a line. We may assume w.l.o.g. that $\pi _R : x_4 = x_5 = 0$, $\pi _L : x_3 = x_4 = 0$. In this case, the $\sigma$-quadric $\Gamma$ is a cone with vertex the line t projecting a (degenerate or not) $C_F^m$-set (see [5, 6, 8]) in a plane not through t. Indeed, let $\pi$ a plane not through the line t and let $A = \pi _R \cap \pi$, $B = \pi _L \cap \pi$. It follows that $\Gamma \cap \pi$ is a set of points of $\pi$ generated by a collineation between the pencils of lines of $\pi$ with center the points A and B induced by the collineation between the pencils of hyperplanes ${\mathcal {P}}_{\pi _R}$ and ${\mathcal {P}}_{\pi _L}$ that is associated with $\Gamma$.

• 3) $\pi _R = \pi _L$. We may assume w.l.o.g. that $\pi _R = \pi _L : x_4 = x_5=0$. In this case, the $\sigma$-quadric is a cone with vertex the plane $\pi _R$ over a $\sigma$-quadric of a line skew with $\pi _R$. That is, $\Gamma$ is either just the plane $\pi _R$ or a hyperplane through $\pi _R$ or a pair of distinct hyperplanes through $\pi _R$ or $q+1$ hyperplanes through $\pi _R$ forming an ${\mathbb {F}}_q$-subpencil of hyperplanes through $\pi _R$.

Proposition 5.1

Let $\Gamma : X^tAX^{\sigma }=0$ be a $\sigma$-quadric of $\mathrm {PG}(4,q^n)$, with $\mathrm {rk}(A)= 2$. The set $\Gamma$ is one of the following:

• a cone with vertex a point V projecting a $\sigma$-quadric of pseudoregulus type in a hyperplane $\Sigma$ with skew vertex lines, with $V \notin \Sigma$;

• a cone with vertex a line v projecting a (possibly degenerate) $C_F^m$-set of a plane $\pi$, with $v \cap \pi = \emptyset$;

• a cone with vertex a plane $\pi$ projecting a $\sigma$-quadric of a line v, with $v \cap \pi = \emptyset$ (hence either just the plane $\pi$ or one, two or $q+1$ hyperplanes through $\pi$).

6 $\sigma$-Quadrics of rank 1 in $\mathrm {PG}(4,q^n)$

In this section, a $\sigma$-quadric $\Gamma$ of $\mathrm {PG}(4,q^n)$ will have equation $X^tAX^{\sigma }=0$ with $\mathrm {rk}(A)=1$. Hence, $\mathrm {dim}V^{\bot }=\mathrm {dim}V^{\top }=4$ so left and right radicals in $\mathrm {PG}(4,q^n)$ are hyperplanes. We distinguish two cases:

• $V^{\bot } \ne V^{\top }.$ We may assume that $\Sigma _R : x_5=0$ is the right radical and $\Sigma _L : x_1=0$ is the left radical. Hence, $\Gamma : x_1 x_5^{\sigma } = 0$ that is the union of two different hyperplanes.

• $V^{\bot } = V^{\top }.$ We may assume that $\Sigma _R = \Sigma _L : x_5=0$ is both the left and the right radical. Hence, $\Gamma : x_5^{\sigma + 1} = 0$ that is a hyperplane of $\mathrm {PG}(4,q^n)$.

7 $\sigma$-Quadrics of $\mathrm {PG}(4,q)$ and ovoids of Q(4, q)

In this final section, we will show, as an application of $\sigma$-quadrics of $\mathrm {PG}(4,q)$, that two of the known ovoids of Q(4, q) can be obtained as intersection of a suitable $\sigma$-quadric with Q(4, q).

An ovoid of Q(4, q) is a set of $q^{2}+1$ points no two collinear on the quadric. Let $Q(4,q) : x_1x_5 +x_2x_4 + x_3^2 = 0$. Any ovoid of Q(4, q) can be written in the following way:

$$\begin{aligned} {\mathcal {O}}(f) = \lbrace (0,0,0,0,1) \rbrace \cup \lbrace (1,x,y, f(x,y), -y^2 - xf(x,y)) : x,y \in {\mathbb {F}}_{q} \rbrace \end{aligned}$$

for some function f(x, y). They are rare objects and, beside the classical example given by an elliptic quadric, only three classes are known for q odd, one class for q even and a sporadic example for $q=3^5$. They have been studied since the end of the 1980s also because of their connections with many other important and well studied objects such as semifield flocks of a three-dimensional quadratic cone, ovoids of $\mathrm {PG}(3,q)$, eggs of finite projective spaces, translation generalized quadrangles, rank 2 commutative semifields, etc. Here we present the two classes of ovoids related to $\sigma$-quadrics of $\mathrm {PG}(4,q)$. Let n be a non-square of ${\mathbb {F}}_q$, $q=p^h$, q odd and $h > 1$, and let $\sigma \ne 1$ be an automorphism of ${\mathbb {F}}_q$, then the set ${\mathcal {O}}(f_1)$ with $f_1(x,y)=-nx^{\sigma }$ is an ovoid of Q(4, q) and it is called Kantor ovoid. If $q=2^{2h+1}$ and $\sigma =2^{h+1}$, then ${\mathcal {O}}(f_2)$ with $f_2(x,y)=x^{\sigma +1} +y^\sigma$ is an ovoid of Q(4, q), and it is called Tits ovoid. The following holds:

Proposition 7.1

Let n be a non-square of ${\mathbb {F}}_q$, $q=p^h$, q odd and $h > 1$, and let $\sigma \ne 1$ be an automorphism of ${\mathbb {F}}_q$. The $\sigma$-quadric $\Gamma$ of rank 2 of $4,q)$ given by the equation $x_4x_1^{\sigma } + n x_1x_2^{\sigma } = 0$ meets the quadric $Q(4,q) : x_1x_5 +x_2x_4 + x_3^2 = 0$ in the union of a Kantor ovoid and a quadratic cone contained in a hyperplane of $\mathrm {PG}(4,q)$.

Proof

Start observing that $\Gamma = \lbrace x_1=0 \rbrace \cup \lbrace (1,x,y, -nx^{\sigma },z) : x,y,z \in {\mathbb {F}}_q \rbrace .$

First let $P=(1,x,y, -nx^{\sigma },z)$ be a point in $\Gamma \setminus \lbrace x_1=0 \rbrace .$ It follows that P belongs to Q(4, q) if, and only if, $z=-y^2 + nx^{\sigma + 1}.$

Now observe that the intersection $\lbrace x_1=0 \rbrace \cap Q(4,q)$ is given by

$$\begin{aligned} {\left\{ \begin{array}{ll} x_1=0 \\ x_2x_4+x_3^2=0, \end{array}\right. } \end{aligned}$$

that is a quadratic cone of the hyperplane $x_1=0$.

Proposition 7.2

Let $q=2^{2h+1}$ and let $\sigma =2^{h+1}$. The $\sigma$-quadric $\Gamma$ of rank 3 of $\mathrm {PG}(4,q)$ given by the equation $x_4x_1^{\sigma } + x_2^{\sigma +1}+x_1x_3^{\sigma } = 0$ meets the quadric $Q(4,q) : x_1x_5 +x_2x_4 + x_3^2 = 0$ in the union of a Tits ovoid and the line $x_1=x_2=x_3=0$.

Proof

Start observing that

$$\begin{aligned} \Gamma = \lbrace x_1=x_2=x_3=0 \rbrace \cup \lbrace (1,x,y, x^{\sigma +1}+y^\sigma ,z) : x,y,z \in {\mathbb {F}}_q \rbrace . \end{aligned}$$

First let $P=(1,x,y, x^{\sigma +1}+y^\sigma ,z)$ be a point in $\Gamma \setminus \lbrace x_1=x_2=x_3=0 \rbrace .$ It follows that P belongs to Q(4, q) if, and only if, $z=y^2 + x^{\sigma + 2}+xy^{\sigma }.$ Now observe that the quadric Q(4, q) contains the lines $\lbrace x_1=x_2=x_3=0 \rbrace$.$\square$

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