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 \(\mathrm {PG}(3,q^n)\).
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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
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
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:
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
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 \(\forall \; u,v \in V\):
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
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:
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:
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:
-
i.1)
-
i)
-
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,
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
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.
-
2)
-
a)
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:
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
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
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 \(\lbrace x_1=x_2=x_3=0 \rbrace \).\(\square \)
Data sharing is not applicable to this article as no new data were created or analyzed in this study.
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Durante, N., Grimaldi, G.G. Absolute points of correlations of \(PG(4,q^n)\). J Algebr Comb 56, 873–887 (2022). https://doi.org/10.1007/s10801-022-01135-0
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DOI: https://doi.org/10.1007/s10801-022-01135-0