Abstract
This paper addresses a number of problems concerning Buekenhout-Tits unitals in \({{\,\textrm{PG}\,}}(2, q^2)\), where \(q = 2^{2e + 1}\) and \(e \ge 1\). We show that all Buekenhout-Tits unitals are equivalent under \({{\,\textrm{PGL}\,}}(3, q^2)\) [addressing an open problem in Barwick and Ebert (Unitals in Projective Planes. Springer Monographs in Mathematics. Springer, New York, 2008)], explicitly describe their stabiliser in \(\textrm{P}\Gamma \textrm{L}(3, q^2)\) [expanding Ebert’s work in Ebert (J Algebraic Comb 6(2):133–140, 1997)], and show that lines meet the feet of points not on \(\ell _{\infty }\) in at most four points. Finally, we show that feet of points not on \(\ell _{\infty }\) are not always a \(\{0, 1, 2, 4\}\)-set, in contrast to what happens for Buekenhout-Metz unitals Abarzúa et al (Adv Geom 18(2):229–236, 2018).
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1 Introduction
1.1 Background
Let \({{\,\textrm{PG}\,}}(2, q^2)\) denote the Desarguesian projective plane over the finite field with \(q^2\) elements, \({{\,\mathrm{\mathbb {F}}\,}}_{q^2}\), where q is a prime power. A unital \(U\) in \({{\,\textrm{PG}\,}}(2, q^2)\) is a set of \(q^{3} + 1\) points such that every line of \({{\,\textrm{PG}\,}}(2, q^{2})\) meets \(U\) in 1 or \(q + 1\) points. Lines meeting \(U\) in 1 point are tangent lines to \(U\), and lines meeting \(U\) in \(q + 1\) points are secant lines of U. The classical or Hermitian unital, usually denoted by \(\mathcal {H}(2,q^2)\), arises by taking the absolute points of a non-degenerate Hermitian polarity. Each point P not lying on a unital U, lies on \(q + 1\) tangent lines to \(U\); the \(q + 1\) points of \(U\) whose tangent lines contain \(P\) are called the feet of \(P\), and are denoted by \(\tau _P(U)\).
It is well-known that \({{\,\textrm{PG}\,}}(2, q^2)\) can be modelled by a Desarguesian line spread of \({{\,\textrm{PG}\,}}(3, q)\) embedded in \({{\,\textrm{PG}\,}}(4, q)\) via the André/Bruck-Bose (ABB) construction. A wide class of unitals in \({{\,\textrm{PG}\,}}(2,q^2)\), called Buekenhout unitals, arise as follows from the ABB construction; starting in \({{\,\textrm{PG}\,}}(4, q)\) fixing a hyperplane \(\Sigma \), and a Desarguesian spread of \(\Sigma \), we take any ovoidal cone \(\mathcal {C}\) such that \(\mathcal {C} \cap \Sigma \) is a spread line of \(\Sigma \). Then in \({{\,\textrm{PG}\,}}(2, q^2)\), \(\mathcal {C}\) gives rise to a unital \(U\). If the base of \(\mathcal {C}\) is an elliptic quadric, the unital is called a Buekenhout-Metz unital. The family of Buekenhout-Metz unitals contains the Hermitian unitals, but there are many non-equivalent Buekenhout-Metz unitals (see [3, 8]). If \(q = 2^{2e + 1}\), \(e \ge 1\), and the base of \(\mathcal {C}\) is a Tits ovoid, the unital is a called a Buekenhout-Tits unital. For more information on unitals and their constructions, see [4].
Unitals may be characterised based on the combinatorial properties of the feet of certain points. It is easy to see that for the classical unital \(\mathcal {H}(2,q^2)\), the feet of a point not on the unital are always collinear. Thas [13] showed the converse, namely, that a unital \(U\) is classical if and only if for all points, not on U, the feet are collinear. This was improved by Aguglia and Ebert [2] who showed that a unital \(U\) is classical if and only if there exist two tangent lines \(\ell _{1}, \ell _{2}\) such that for all points \(P \in (\ell _1 \cup \ell _2) {\setminus } U\) the feet of \(P\) are collinear. It is known (see e.g. [4]) that if \(U\) is a non-classical Buekenhout-Metz unital, the feet of a point \(P \notin U\) are collinear if and only if they lie on a distinguished tangent line \(\ell _{\infty }\) to \(U\). Furthermore, it is shown in [1] that if \(U\) is Buekenhout-Metz unital, a line meets the feet of a point \(P \notin \ell _{\infty }\) in either 0, 1, 2, or 4 points. Ebert [9] showed for a Buekenhout-Tits unital, the feet of \(P \notin U\) are collinear if and only if \(P \in \ell _{\infty }\). It is then natural to ask how a line may meet the feet of a point \(P \notin \ell _{\infty }\) for Buekenhout-Tits unitals. We will answer this question in Theorem 3.
Many characterisations of unitals make use of their stabilisers in \({{\,\textrm{PGL}\,}}\), resp. \(\textrm{P}\Gamma \textrm{L}\). In [7] it is shown that a unital is classical if its stabiliser contains a cyclic group of order \(q^{2} - q + 1\). Several other characterisations of unitals by their stabiliser group are listed in [4]. In [9], Ebert determined the stabiliser of a Buekenhout-Tits unital in \({{\,\textrm{PGL}\,}}(3, q^{2})\) (see Result 1). We will extend this work in this paper.
1.2 Summary of this paper
In this paper we present three main results:
-
1.
We show that all Buekenhout-Tits unitals are equivalent under \(\textrm{P}\Gamma \textrm{L}(3, q^2)\) (see Theorem 1). This addresses an open problem in [4], and is alluded to in [10] (see Remark 1).
-
2.
A description of the full stabiliser group of a Buekenhout-Tits unital in \(\textrm{P}\Gamma \textrm{L}(3, q^{2})\) (see Theorem 2). Ebert [9] only provides a description of stabiliser of the Buekenhout-Tits unital in \({{\,\textrm{PGL}\,}}\) (Result 1). The stabiliser of the classical unital in \(\textrm{P}\Gamma \textrm{L}(3, q^2)\) is \({\textrm{P}\Gamma \textrm{U}}(3, q^2)\), and the stabiliser of the Buekenhout-Metz unital in \(\textrm{P}\Gamma \textrm{L}(3, q^{2})\) is described in [8] for \(q\) even and [3] for \(q\) odd.
-
3.
If \(U\) is a Buekenhout-Tits unital, then a line \(\ell \) meets the feet of a point \(P \notin (\ell _{\infty } \cup U)\) in at most 4 points. Moreover, there exists a point \(P\) and line \(\ell \) such that the feet of \(P\) meet \(\ell \) in exactly three points (see Theorem 3). This highlights a difference between Buekenhout-Metz unitals and Buekenhout-Tits unitals. It also solves an open problem posed by Aguglia and Ebert [2] and later listed in [4, Chapter 8].
1.3 Coordinates for a Buekenhout-Tits unital
In [9], Ebert derives coordinates for a Buekenhout-Tits unital \(\mathcal {U}_{BT}\) in \({{\,\textrm{PG}\,}}(2, q^{2})\), \(q=2^{2e+1}\). Pick \(\epsilon \in \mathbb {F}_{q^2}\) such that \(\epsilon ^q = \epsilon + 1\), and \(\epsilon ^2 = \epsilon + \delta \) for some \(1 \ne \delta \in \mathbb {F}_q\) with absolute trace equal to one. Then the following set of points in \({{\,\textrm{PG}\,}}(2, q^2)\) is a Buekenhout-Tits unital,
where \(\sigma = 2^{e + 1}\) has the property that \(\sigma ^2\) induces the automorphism \(x\mapsto x^2\) of \({{\,\mathrm{\mathbb {F}}\,}}_{q}\). In addition, it can be verified that \(\sigma + 1\), \(\sigma + 2\), \(\sigma - 1\), and \(\sigma - 2\) all induce permutations of \(\mathbb {F}_q\) with inverses induced by \(\sigma - 1\), \(1 - \sigma /2\), \(\sigma + 1\) and \(-(\sigma /2 + 1)\) respectively.
The following theorem describes the group of projectivities (that is, elements of \({{\,\textrm{PGL}\,}}(3,q^2)\)) stabilising \(\mathcal {U}_{BT}\).
Result 1
[9, Theorem 4 and Corollary] Let \(G={{\,\textrm{PGL}\,}}(3,q^2)_{\mathcal {U}_{BT}}\), \(q=2^{2e+1}\), be the group of projectivities stabilising the Buekenhout-Tits unital \(\mathcal {U}_{BT}\). Then \(G\) is an abelian group of order \(q^{2}\), consisting of the projectivities induced by the matrices
where \(\sigma =2^{e+1}\) and matrices act on the homogeneous coordinates of points by multiplication from the right. The group \(G\) has \(q^2 - q\) orbits of length \(q^2\) on points in \({{\,\textrm{PG}\,}}(2,q^2){\setminus } (\mathcal {U}_{BT}\cup \ell _\infty ),\,{ where}\ell _\infty :x=0\).
2 On the projective equivalence of Buekenhout-Tits unitals
In this section, we show that all Buekenhout-Tits unitals are equivalent under \({{\,\textrm{PGL}\,}}(3, q^{2})\) to the unital \(\mathcal {U}_{BT}\) given in Eq. (1).
Remark 1
The authors of [10] give this result without proof and state it can be derived by the same techniques employed by Ebert in [9]. Ebert however, lists the equivalence of Buekenhout-Tits unitals as an open problem in [4] which appeared about ten years after his original paper [9].
It is easy to see that the Buekenhout-Tits unital \(\mathcal {U}_{BT}\) is tangent to the line \(\ell _{\infty }: x = 0\) at the point \(P_{\infty } = (0, 0, 1)\). From the ABB construction it follows that \(P_{\infty }\) has the following property with respect to \(\mathcal {U}_{BT}\).
Property 1
Given any unital \(U\), a point \(P \in U\) is said to have Property 1 if all secant lines through \(P\) meet \(U\) in Baer sublines.
It is shown in [5] that if two different points of \(U\) have Property 1, then \(U{ isclassical}.{ Hence},\,{ thepoint}P_\infty { istheuniquepointof}\mathcal {U}_{BT}\) admitting this property. We will count all Buekenhout-Tits unitals tangent to \(\ell _{\infty }\) at a point \(P_{\infty }\) having Property 1.
Lemma 1
There are \(q^4{(q^2 - 1)}^2\) unitals equivalent under \({{\,\textrm{PGL}\,}}(3, q^2)\) to \(\mathcal {U}_{BT}\) in \({{\,\textrm{PG}\,}}(2, q^2)\) with tangent line \(\ell _{\infty }\,:\, x = 0\) and containing the point \(P_{\infty } = (0, 0, 1)\) having Property 1.
Proof
Let \(U\) be a unital tangent to \(\ell _{\infty }\), and containing the point \(P_{\infty }\) with Property 1, that is equivalent under \({{\,\textrm{PGL}\,}}(3, q^2)\) \(\mathcal {U}_{BT}\) to \({{\,\textrm{PG}\,}}(2, q^2)\). Then, the point \(P_{\infty }\) is the unique point in \(U\) with Property 1. Thus, any projectivity mapping \(\mathcal {U}_{BT}\) to \(U\) is contained in the group \(H\) of projectivities fixing \(P_{\infty }\), and fixing \(\ell _{\infty }\) line-wise. The elements of \(H\) are induced by all matrices of the following form,
where \(x_{22} x_{33} \ne 0\) and matrices act on homogeneous coordinates by multiplication on the right. It follows that \(|H|={(q^2 - 1)}^2 q^6\). Furthermore, from the description of \(G={{\,\textrm{PGL}\,}}(3,q^2)_{\mathcal {U}_{BT}}\) in Result 1, we know that the stabiliser \(H_{\mathcal {U}_{BT}}\) in \(H\) of \(\mathcal {U}_{BT}\) coincides with \(G\). Hence, the stabiliser \(H_{\mathcal {U}_{BT}}\) has order \(q^{2}\). By the orbit-stabiliser theorem, we find that there are \({(q^2 - 1)}^2 q^4\) unitals in the orbit of \(\mathcal {U}_{BT}\) under \(H\). \(\square \)
Consider \({{\,\textrm{PG}\,}}(2, q^{2})\) modelled by the ABB construction with fixed hyperplane \(\Sigma _{\infty }\). Let \(p_{\infty }\) be the spread line corresponding to \(P_{\infty }\). Then any Buekenhout-Tits unital \(U\) tangent to \(\ell _{\infty }\) at \(P_{\infty }\) with Property 1 corresponds uniquely to an ovoidal cone \(\mathcal {C}\) meeting \(\Sigma _{\infty }\) at \(p_{\infty }\).
Lemma 2
There are \(q^{4}{(q^{2} - 1)}^{2}\) ovoidal cones \(\mathcal {C}\) in \({{\,\textrm{PG}\,}}(4, q)\) with base a Tits ovoid, such that \(\mathcal {C}\) meets \(\Sigma _{\infty }\) in the spread element \(p_{\infty }\).
Proof
Let \(V\) be a point on the line \(p_{\infty }\), and \(\Sigma \ne \Sigma _{\infty }\) a hyperplane not containing \(V\). Then, \(\Sigma \) meets \(\Sigma _{\infty }\) in a plane containing a point \(R \in p_{\infty } {\setminus } \{V\}\). Any ovoidal cone \(\mathcal {C}\) with vertex \(V\) and base a Tits ovoid, such that \(\mathcal {C}\) meets \(\Sigma _\infty \) precisely in \(p_\infty \), meets \(\Sigma \) in a Tits ovoid tangent to \(\Sigma \cap \Sigma _{\infty }\) at the point \(R\). We will count all cones of this form, for all \(V \in p_{\infty }\).
Consider the pairs of planes \(\Pi \) and Tits ovoids \(\mathcal {O}\), \((\Pi , \mathcal {O})\), where \(\Pi , \mathcal {O} \subset \Sigma \) and \(\Pi \) is tangent to \(\mathcal {O}\). On the one hand, there are \(|{{\,\textrm{PGL}\,}}(4, q)|/|\mathcal {O}_{{{\,\textrm{PGL}\,}}(4, q)}| = {(q + 1)}^2 q^4 {(q - 1)}^2 (q^2 + q + 1)\) Tits ovoids in \({{\,\textrm{PG}\,}}(3, q)\), and each has \(q^{2} + 1\) tangent planes. On the other hand, \({{\,\textrm{PGL}\,}}(4, q)\) is transitive on hyperplanes of \({{\,\textrm{PG}\,}}(3, q)\), so each plane is tangent to the same number of Tits ovoids. It thus follows, that there are
Tits ovoids tangent to \(\Sigma \cap \Sigma _{\infty }\) contained in \(\Sigma \).
Furthermore, since \({{{\,\textrm{PGL}\,}}(4, q)}_{\Sigma \cap \Sigma _{\infty }}\) is transitive on points of \(\Sigma \cap \Sigma _{\infty }\), each point of \(\Sigma \cap \Sigma _{\infty }\) is contained in the same number of Tits ovoids \(\mathcal {O}\), so it follows that the number of Tits ovoids tangent to \(\Sigma \cap \Sigma _{\infty }\) at \(R = p_{\infty } \cap \Sigma \) is \({(q - 1)}^2q^4(q + 1)\). Hence, there is an equal number of ovoidal cones with base a Tits ovoid, vertex \(V\), and meeting \(\Sigma _{\infty }\) at \(p_{\infty }\). As the choice of \(V\) was arbitrary, and there are \(q+1\) points on \(p_\infty \), there are \({(q^2 - 1)}^2q^4\) ovoidal cones with base a Tits ovoid, and meeting \(\Sigma _{\infty }\) at \(p_{\infty }\). \(\square \)
Theorem 1
All Buekenhout-Tits unitals in \({{\,\textrm{PG}\,}}(2, q^{2})\) are equivalent under \({{\,\textrm{PGL}\,}}(3, q^2)\).
Proof
From Lemmas 1 and 2, we see that the number of ovoidal cones with base a Tits ovoid, tangent to \(\Sigma _{\infty }\) at \(p_{\infty }\) is equal to the number of Buekenhout-Tits unitals that are equivalent under \({{\,\textrm{PGL}\,}}(3, q^2)\) to \(\mathcal {U}_{BT}\) and tangent to \(l_{\infty }\) at \(P_{\infty }\) with Property 1. The result follows. \(\square \)
Corollary 1
Let \(U\) be a Buekenhout-Tits unital, then the projectivity group stabilising \(U\) is isomorphic to the group \(G\) in Result 1.
Since we have shown that all Buekenhout-Tits unitals are equivalent under \({{\,\textrm{PGL}\,}}(3, q^2)\), we may use \(\mathcal {U}_{BT}\) to verify statements about general Buekenhout-Tits unitals.
3 On the stabiliser of the Buekenhout-Tits unital
We now describe the stabiliser of the Buekenhout-Tits unital \(\mathcal {U}_{BT}\) in \(\textrm{P}\Gamma \textrm{L}(3, q^{2})\).
Lemma 3
Let \(M_{u, v}, M_{s, t}\) be matrices inducing collineations of \(G\) as defined in Result 1, then \(M_{u, v} M_{s, t} = M_{u + s, t + v + su \delta }\).
Proof
Using Eq. (2), we find
Thus, we have \(M_{u,v} M_{s, t} = M_{u + s, t + v + su \delta }\). \(\square \)
Corollary 2
The order of any collineation of \(G\) induced by a matrix \(M_{u, v}\) as defined in Result 1 is four if and only if \(u \ne 0\), and two if and only if \(u = 0\) and \(v \ne 0\).
Proof
Firstly note that \(M_{0, 0} = I\). Direct calculation shows that \(M_{u,v}^2=M_{0,u^2\delta }\), \(M_{u,v}^3=M_{u,v+u^2\delta }\) and \(M_{u,v}^4=M_{0,0}\). \(\square \)
Corollary 3
The stabiliser group \(G\) as defined in Result 1 is isomorphic to \({(C_{4})}^{2e + 1}\).
Proof
Recall from Result 1 that \(|G|=q^2=2^{4e+2}{} \). From Corollary 2, we have that \(G \equiv {(C_{4})}^{k}{(C_{2})}^{l}\) for some integers \(k, l\) such that \(2^{2k + l} = |G| = 2^{4e + 2}\), and hence,
Furthermore, we see that the number of elements of order four in \(G\) is \(q^{2} - q\) as they correspond to all matrices \(M_{u,v}{} { with}u,v\in {{\,\mathrm{\mathbb {F}}\,}}_q{ and}u\ne 0\). The number of elements of order four in a group isomorphic to \({(C_{4})}^{k}{(C_{2})}^{l}\) is \((4^{k} - 2^{k})2^{l}\). Thus,
Using Eq. (3), we find that \(k = 2e + 1\), and therefore \(G \equiv {(C_{4})}^{2e + 1}\). \(\square \)
Theorem 2
Let \(q = 2^{2e + 1}\), then the stabiliser of \(\mathcal {U}_{BT}\) in \(\textrm{P}\Gamma \textrm{L}(3, q^2)\) is the order \(q^2(4e + 2)\) group \(GK\), where \(G = {{\,\textrm{PGL}\,}}(3, q^2)_{\mathcal {U}_{BT}}\) as described in Result 1, and \(K\) is a cyclic subgroup of order \(16e + 8\) generated by
(Here, \({\textbf{x}}{} { denotestherowvectorcontainingthethreehomogeneouscoordinatesofapoint},\,{ and}{\textbf{x}}^2\) denotes its elementwise power.)
Proof
From Lemma 2, the number of Buekenhout-Tits unitals tangent to \(\ell _{\infty }\,:\,x = 0\) at a point \(P_{\infty } = (0, 0, 1)\) with Property 1 is \(q^4{(q^2 - 1)}^2\). By the arguments of Lemma 1, all of these unitals are equivalent under \({{\,\textrm{PGL}\,}}(3, q^2)\) to \(\mathcal {U}_{BT}\) under the stabiliser groups \({{\,\textrm{PGL}\,}}(3, q^2)_{\{\ell _{\infty }, P_{\infty }\}}\) and \(\textrm{P}\Gamma \textrm{L}(3, q^2)_{\{\ell _{\infty }, P_{\infty }\}}\) fixing \(P_{\infty }\) and stabilising \(\ell _{\infty }\). Any collineation stabilising \(\mathcal {U}_{BT}\) must stabilise \(P_{\infty }\) and \(\ell _{\infty }\), so \(\textrm{P}\Gamma \textrm{L}(3, q^2)_{\mathcal {U}_{BT}} < \textrm{P}\Gamma \textrm{L}(3, q^2)_{\{\ell _{\infty }, P_{\infty }\}}\). Therefore, the orbit of \(\mathcal {U}_{BT}\) under \(\textrm{P}\Gamma \textrm{L}(3, q^2)_{\{\ell _{\infty }, P_{\infty }\}}\) has size \(q^4{(q^2 - 1)}^2\), that is
We can now see that \(\textrm{P}\Gamma \textrm{L}(3, q^2)_{\mathcal {U}_{BT}}\) must have order \(q^{2}(4e + 2)\).
Direct calculation shows that \(\psi \) stabilises \(\mathcal {U}_{BT}\). Because \({\textbf{x}}^{2^{4e + 2}} = {\textbf{x}}^{q^2} = {\textbf{x}}\), the collineation \(\psi ^{4e + 2}\) is a linear map stabilising \(\mathcal {U}_{BT}\), and so \(\psi ^{4e + 2} \in G\). Therefore, we deduce that \(|\psi | = (4e + 2)|\psi ^{4e + 2}|\). From Corollary 2, it follows that \(|\psi ^{4e + 2}| \in \{ 1, 2, 4\}\), with \(|\psi ^{4e + 2}| = 4\) if and only if \(\psi ^{4e + 2}\) is induced by \(M_{u, v}\) for some \(u \ne 0\). Hence, \(|\psi ^{4e + 2}| = 4\) if and only if \(\psi ^{4e + 2}(0, 1, 0) \ne (0, 1, 0)\) as \((0, 1, 0)M_{u, v} = (0,1,u + u \epsilon )\). Consider the point \((0, 1, z)\) for some arbitrary \(z \in \mathbb {F}_q\). Direct calculation shows that \(\psi (0, 1, z) = (0, 1, 1 + \mu z^2)\), where \(\mu = \frac{\delta ^{\sigma + 1}}{\delta ^{\sigma /2}(1 + \epsilon )} = \delta ^{\sigma /2}\epsilon \). Thus,
for some polynomial \(g(z)\) depending on \(k\). If \(z = 0\) and \(k = 4e + 2\) we thus find
Recall that \(\epsilon ^q=\epsilon +1,\,{ so}{{\,\textrm{Tr}\,}}_{\mathbb {F}_{q^2}/\mathbb {F}_{q}}(\epsilon )=1.{ Therefore},\,{ wehave}{{\,\textrm{Tr}\,}}_{\mathbb {F}_{q^2}/\mathbb {F}_{2}}(\delta ^{\sigma /2}\epsilon )={{\,\textrm{Tr}\,}}_{\mathbb {F}_{q}/\mathbb {F}_{2}}({{\,\textrm{Tr}\,}}_{\mathbb {F}_{q^2}/\mathbb {F}_{q}}(\delta ^{\sigma /2}\epsilon ))={{\,\textrm{Tr}\,}}_{\mathbb {F}_{q}/\mathbb {F}_{2}}(\delta ^{\sigma /2}{{\,\textrm{Tr}\,}}_{\mathbb {F}_{q^2}/\mathbb {F}_{q}}(\epsilon ))={{\,\textrm{Tr}\,}}_{\mathbb {F}_{q}/\mathbb {F}_{2}}(\delta ^{\sigma /2}) = 1.\) Hence, we see \(\psi ((0, 1, 0)) \ne (0, 1, 0)\), so \(|\psi ^{4e + 2}| = 4\) and \(|\psi | = 16e + 8\). Let \(K = \langle \psi \rangle \), because \(|K \cap G| = 4\), it follows that \(|GK| = q^{2}(4e + 2)\) and thus \(GK = \textrm{P}\Gamma \textrm{L}{(3, q^{2})}_{\mathcal {U}_{BT}}\). \(\square \)
4 On the feet of the Buekenhout-Tits unital
Recall that the feet \(\tau _P(U)\) of a point \(P\) not on a unital \(U\) is the set of all points on tangent lines to \(U\) through \(P\). The feet of the Buekenhout-Tits unital \(\mathcal {U}_{BT}\) (as coordinatised in 1) for points \(P \notin \mathcal {U}_{BT}\) are first described by Ebert in [9]. He shows that the feet of a point \(P = (1, y_1 + y_2\epsilon , z_1 + z_2\epsilon )\) is the following set of points:
If the line \(\ell \) has Equation \(\alpha x + y = 0\), where \(\alpha \in \mathbb {F}_{q^2}\), Ebert shows that \(|\ell \cap \tau _P(\mathcal {U}_{BT})| \le 1\). Otherwise, \(\ell \) has equation \((a_1 + a_2\epsilon )x + (b_1 + b_2\epsilon )y + z = 0\) and Ebert shows that \(\ell \) meets \(\tau _P(\mathcal {U}_{BT})\) in the points \(P_{r, s, t} \in \mathcal {U}_{BT}\), where \(r = s^2 + t^2\delta + st + y_1s + y_1t + y_2\delta {t} + z_1\) and \(s, t\) satisfy
We will show that for all choices of points \(P \notin \ell _{\infty }\) and lines \(\ell \), \(|\tau _{P}(\mathcal {U}_{BT}) \cap \ell | \le 4\).
Recall that the group \(G\) as described in Result 1 has \(q^2 - q\) orbits of \({{\,\textrm{PG}\,}}(2, q^2) {\setminus } (\mathcal {U}_{BT}\cup \ell _\infty )\) of size \(q^2\). Here we give a set of \(q^2 - q\) representatives for these orbits.
Lemma 4
Let \(G\) be the group of projectivities stabilising \(\mathcal {U}_{BT}\) as described in Result 1. Then, the set of \(q^2 - q\) points \(\{P_{a, b} = (1, a, b \epsilon ) \,|\, a,b \in \mathbb {F}_q,\, b \ne a^{\sigma + 2}\}\) are points from \(q^2 - q\) distinct point orbits of size \(q^2\) under \(G\).
Proof
Suppose there exists a collineation of \(G\) induced by a matrix \(M_{u, v}\) such that \(P_{a, b} M_{u, v} = P_{c, d}\). Then,
However, it is clear that \(P_{a, b}M_{u, v} = \left( 1, a + u \epsilon , v + u^{\sigma }\epsilon + a \left( u + u \epsilon \right) + b\epsilon \right) \), so \(a + u \epsilon = c\). Therefore, \(a = c\) and \(u = 0\). If \(u = 0\), then \(v + b\epsilon = d \epsilon \), and we have \(b = d\). Hence, \(P_{a, b} = P_{c, d}\) and the lemma follows. \(\square \)
There are \(q^4 - q^3 = q^{2} (q^2 - q)\) points of \({{\,\textrm{PG}\,}}(2, q^2)\) not on \(\ell _{\infty }\) or \(\mathcal {U}_{BT}\). By Lemma 4, each of these points lies in the orbit of a point of the form \((1, a, b \epsilon )\). Therefore, in order to study the feet of a point \(P\), we may assume that the point \(P = (1, y_1, z_2\epsilon )\).
The following lemma shows that the feet of a point \(P = (1, y_1, z_2\epsilon )\), with \(y_1^{\sigma + 2} \ne z_2\) meets almost all lines in at most \(2\) points.
Lemma 5
Let \(\ell :\alpha x+\beta y + z = 0{ bealinein}{{\,\textrm{PG}\,}}(2,q^2)\), where \(\alpha = a_{1} + a_{2} \epsilon \), \(\beta = b_{1} + b_{2} \epsilon \) and \(a_{1}, a_{2}, b_{1}, b_{2} \in \mathbb {F}_{q}\). Let \(P = (1, y_{1}, z_{2} \epsilon )\), with \(y_{1}, z_{2} \in \mathbb {F}_{q}\) such that \(z_{2} \ne y_{1}^{\sigma + 2}\). Unless \(b_{2} = 0\), \(y_{1} = b_{1}\) and \(a_{2} = z_{2}\), we have \(|\tau _P(\mathcal {U}_{BT}) \cap \ell | \le 2\).
Proof
From the description given in Eq. (4), we see that the points \(P_{r, s, t} \in \tau _{P}(\mathcal {U}_{BT})\) satisfy
and this equation has \(q+1\) solutions. Substituting Eqs. (8) into (5) and combining Eqs. (6) and (7), it follows that the points \(P_{r, s, t} \in \tau _P(\mathcal {U}_{BT}) \cap \ell \) have \(s, t\) satisfying
We will now count the solutions to this system, by considering the geometry of these equations in the solution space \({{\,\textrm{AG}\,}}(2, q)\) with coordinates \((s, t)\). Recall that the points \((1,s,t,s^{\sigma + 2}+t^\sigma + st),\,{ where}s,t\in {{\,\mathrm{\mathbb {F}}\,}}_q{ arethe}q^2\) affine points of a Tits ovoid in \({{\,\textrm{PG}\,}}(3, q)\) [14]. Because \(\tau _P(\mathcal {U}_{BT})\) has \(q + 1\) points, the Eq. 8 must have \(q + 1\) solutions \((s, t)\) in the solution space. Hence the \(q + 1\) points \((s, t)\) in \({{\,\textrm{AG}\,}}(2, q)\) satisfying 8 are a translation oval.
Unless \(b_{2} = 0\) and \(y_{1} = b_{1}\), Eq. (10) represents a line in the solution space \({{\,\textrm{AG}\,}}(2, q)\). A line meets the oval defined by Eq. 8 in at most two points, so we have at most two solutions to the system. If \(b_{2} = 0\), \(y_{1} = b_{1}\), and \(a_{2} \ne z_{2}\), then Eq. (10) has no solutions. \(\square \)
Remark 2
Lemma 5 is a refinement of [4, Theorem 4.33], where Barwick and Ebert rework Ebert’s earlier proof in [9] that the feet of a point \(P \notin (\ell _{\infty } \cup \mathcal {U}_{BT})\) are not collinear. This reworked proof asserts that the feet cannot be collinear because the line given by Eq. (10) and the conic from Eq. (9) cannot have \(q + 1\) common solutions. However, we can see that this logic is not complete, and leaves an interesting case to examine when Eq. (10) vanishes. Ebert’s original proof in [9] does not contain this error, instead arguing that Eqs. (9) and 8 cannot have \(q + 1\) common solutions.
It follows from Lemma 5 that the feet of a point \(P \notin (\ell _{\infty } \cup \mathcal {U}_{BT})\) is a set of \(q + 1\) points such that every line meets \(\tau _P(\mathcal {U}_{BT})\) in at most two points except for a set of \(q\) concurrent lines.
To investigate the latter case, assume that \(b_{2} = 0\), \(y_1 = b_1\) and \(a_2 = z_2\). In this case, Eq. (10) vanishes. The system describing \(\ell \cap \tau _P(\mathcal {U}_{BT})\) is thus
The lines that produce these cases are the lines with dual coordinates \([a_{1} + z_{2}\epsilon , y_{1}, 1]\). These lines are concurrent at the point \((0, 1, y_{1})\) which lies on \(\ell _{\infty }\). We will show in Corollary 4 that these latter lines meet \(\tau _P(\mathcal {U}_{BT})\) in at most four points.
Recall that an affine section of a Tits ovoid in \({{\,\textrm{PG}\,}}(3, q)\) contains \(q + 1\) points equivalent under \({{\,\textrm{PGL}\,}}(3, q^2)\) to the translation oval [14]
For a reference on translation ovals, see [11, pp. 182–186]. We require the following lemma, which adapts arguments found in [6, Lemma 2.1].
Lemma 6
Let \(\mathcal {O}\) be a translation oval in \({{\,\textrm{PG}\,}}(2, q)\) projectively equivalent to \(\mathcal {D}_{\sigma }\), and let \(\mathcal {C}\) be a non-degenerate conic. If the nucleus of \(\mathcal {O}\) is also the nucleus of \(\mathcal {C}\), then \(|\mathcal {O} \cap \mathcal {C}| \le 4\).
Proof
Without loss of generality we may take \(\mathcal {O} = \mathcal {D}_{\sigma }\), so that the nucleus of \(\mathcal {O}\) is \(N = (0, 1, 0)\). If \(N\) is also the nucleus of \(\mathcal {C}\), then \(\mathcal {C}\) is a conic of the following form,
for some \(a_1, a_2, a_3 \in \mathbb {F}_q\) with \(a_{2} \ne 0\). Suppose that \((0, 0, 1) \notin \mathcal {C}\). Then \(a_3 \ne 0\), and the point \((1, t, t^{\sigma }) \in \mathcal {C}\) if and only if \(t\) satisfies
hence,
Therefore,
and substituting Eqs. (14) into (13), we find that Eq. (13) has at most four solutions. If instead \((0, 0, 1) \in \mathcal {C}\), then \(a_3 = 0\) and arguing as above we find that Eq. (13) has at most two solutions, so \(|\mathcal {O} \cap \mathcal {C}| \le 3\). \(\square \)
Corollary 4
The feet of a point \(P \notin \left( \ell _{\infty } \cup \mathcal {U}_{BT}\right) \) meet a line \(\ell \) in at most four points.
Proof
From Lemma 5, we know we can restrict ourselves to the case \(b_2=0,y_1=b_1,a_2=z_2\) which means we are looking at the points \(P_{r, s, t} \in \tau _P(\mathcal {U}_{BT}) \cap \ell \) have \(s, t\) satisfying
where Eq. (15) represents a conic \(\mathcal {C}{} \), and Eq. (16) represents an oval \(\mathcal {O}{} \) in \({{\,\textrm{AG}\,}}(2, q)\). If the conic is degenerate, the oval and conic have at most four points in common. So we may assume that the conic is non-degenerate. The nucleus of \(\mathcal {C}{} \) is \(N = (y_1, 0, 1)\). We now show that \(N\) is the nucleus of the oval \(\mathcal {O}{} \). The line \(t = 0\) goes through \(N\) and meets the oval \(\mathcal {O}\) when \(s^{\sigma + 2} = z_{2}\), which has one solution as \(\sigma + 2\) is a permutation of \(\mathbb {F}_{q}\). The line \(s + y_{1} = 0\) through \(N\) meets the oval \(\mathcal {O}\) when \(t^{\sigma } = y^{\sigma + 2} + z_{2}\) which has one solution for \(t\). Therefore, \(N\) is the nucleus, as it is the intersection of two tangent lines to the oval. It now follows from Lemma 6 that Eqs. (15) and (16) have at most four common solutions. \(\square \)
We now show the existence of a point \(P \notin (\mathcal {U}_{BT}\cup \ell _{\infty })\) and a line \(\ell \) such that \(|\ell \cap \tau _{P}(\mathcal {U}_{BT})| = 3\), and demonstrate our bound is sharp.
Lemma 7
Consider the Equation \(s^{\sigma + 2} + t^{\sigma } + st = y_1 t + z_2\), whose solutions \((s, t)\) are a translation oval of \({{\,\textrm{AG}\,}}(2, q)\). If \(y_{1} = 0\), then the points of the oval given by Eq. (16) are
Proof
If \(y_{1} = 0\), then Eq. (16) reduces to
Using the properties of \(\sigma \) described in Sect. 1.3, one can show the point \((z_2^{1-\sigma /2}, z_{2}^{\sigma /2})\) satisfies Eq. (17). Furthermore, the points \(\overline{P_u} = (z_{2}^{1 - \sigma /2}u^{\sigma }, z_{2}^{\sigma /2}(1 + u^{\sigma }), 1 + u + u^{\sigma })\), where \(u \in \mathbb {F}_q\), are projective points satisfying the following homogeneous equation
Because \({{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(u + u^{\sigma }) = 0\), and \({{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(1) = 1\) when \(q = 2^{2e + 1}\), we have \(u^{\sigma } + u+ 1 \ne 0\) for all \(u \in \mathbb {F}_q\). Thus, normalising so \(z = 1\), the points \(\overline{P_u}\) have the form \((s, t, 1)\) where \(s\) and \(t\) satisfy Eq. (17). \(\square \)
Corollary 5
Let \(y_{1} = 0\) and consider the points \(P_u\) as described in Lemma 7. A point \(P_{u}\) lies on the conic given by Eq. (15), if and only if \(u\) is a root of the following polynomial
Proof
By directly substituting \(P_u\) into Eq. (15) we have
Raising both sides of Eq. (19) to the power of \(\sigma /2\) yields our result. \(\square \)
Theorem 3
Let \(U\) be a Buekenhout-Tits unital in \({{\,\textrm{PG}\,}}(2, q^{2})\). The feet of a point \(P \notin (\ell _{\infty } \cup U)\) meet a line \(\ell \) in at most four points. Moreover, there exists a line \(\ell \) and point \(P\) such that \(|\ell \cap \tau _{P}(U)| = k\) for each \(k \in \{0, 1, 2, 3,4\}\).
Proof
By Theorem 1 we may assume that \(U = \mathcal {U}_{BT}\). The first part of the proof comes from Corollary 4. Let \(P = (1, y_1, z_2\epsilon )\). All lines through \(P\) meet \(\tau _P(U)\) in at most one point by definition, so it is clear that there exists lines \(\ell \) such that \(|\ell \cap \tau _P(U)|\) is zero or one. Because the points of \(\tau _P(U)\) are not collinear, there exists a pair of points \(Q, R \in \tau _P(U)\) such that the line \(QR\) does not contain \((0, 1, y_1)\). Because \(QR\) does not contain \((0, 1, y_1)\) it cannot have dual coordinates of the form \([a_1 + z_2\epsilon , y_1, 1]\) for any \(a_1 \in {{\,\mathrm{\mathbb {F}}\,}}_q\), and so Lemma 5 applies to \(QR\). Hence, the line \(QR\) meets \(\tau _P(U)\) in precisely two points.
Now consider a line \(\ell \) with Equation \((\delta + \epsilon )x + z = 0\) and let \(P{ bethepoint}(1,0,\epsilon )\) (that is, \(a_1=\delta , a_2=1, b_1=b_2=y_{1} = 0, z_{2} = 1\)). The number of points of \(\ell \cap \tau _P(U)\) is the same as the number of solutions to Eqs. (11) and (12). By Lemma 7 the points \(P_u\) satisfying Eq. (12) lie on the conic determined by Eq. (11) when
Equation (20) has exactly two solutions as \(\sigma - 1\) is a permutation of \(\mathbb {F}_q\): \(u=0{ andtheuniquesolutionto}u^{\sigma -1}=\frac{1}{\delta ^{\sigma /2}}{} \). It can also be shown that \(\big (z_2^{1-\sigma /2}, z_2^{\sigma /2}\big ) = (1, 1)\) satisfies both equations. Hence, the intersection of the feet of the point \((1, 0, \epsilon )\) and \(\ell \) has exactly three points.
Finally, consider the point \(P(1,0,\frac{1}{\delta ^\sigma }\epsilon ){ andtheline}\ell { withdualcoordinates}\big [\frac{1}{\delta }+\frac{1}{\delta ^2}\epsilon ,0,1\big ]\). By Corollary 5, the number of feet of \(P{ ontheline}\ell \) is the number of roots of the polynomial (18), where \(a_1=\frac{1}{\delta }{} { and}z_2=\frac{1}{\delta ^\sigma }{} \). Substituting \(a_1 = \frac{1}{\delta }\) and \(z_2 = \frac{1}{\delta ^{\sigma }}\) yields
Since Eq. (21) describes the roots of a \({{\,\mathrm{\mathbb {F}}\,}}_2\)-linearised polynomial, and there are at most \(4\) roots, we have that the polynomial (18) has \(1,2,{ or}4{ roots}.{ Wewillshowthat},\,{ underthecondition}{{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(\delta )=1\), it has four roots. Multiplying Eq. (21) by \(\delta { yields}\delta ^{1-\sigma /2}u^\sigma +(\delta ^{\sigma -1}+1)u^2+u=0{ andnowsubstituting}a=\delta ^{\sigma -1}+1\) gives
We find that \(u = 0\) and \(u = \frac{1}{a^{1 + \sigma /2}}\) are solutions to Eq. (22). Now consider
Any solution to Eq. (23) also satisfies \((u^{\sigma } + au^2 + 1)^{\sigma / 2} + u^{\sigma } + a u^2 + 1 = 0\) which is precisely Eq. (22). Multiply Eq. (23) with \(a^{\sigma +1},\,{ thenwefind}(a^{\sigma /2+1}u)^\sigma +(a^{\sigma /2+1}u)^2+a^{\sigma +1}=0\), and letting \(z = (a^{\sigma /2 + 1}u)^2\),
which is known (see [12]) to have solutions if and only if \({{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(a^{\sigma +1})=0\). As \(z = 0\) and \(z = 1\) are not solutions of Eq. (24), no solutions of Eq. (24) correspond to the solutions \(u = 0\) or \(u = \frac{1}{a^{1 + \sigma /2}}\) of Eq. (21). Furthermore, recall that Eq. (21) has \(1,2{ or}4{ solutionsandthatwehaveassumedthat}{{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(\delta )=1.{ Since}\delta ^{\sigma -1}=a+1,\,{ itfollowsthat}\delta =(a+1)^{\sigma +1}{} { and}{{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(\delta )={{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(a^{\sigma +1}+a^\sigma +a+1)={{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(a^{\sigma +1})+{{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(1)={{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(a^{\sigma +1})+1.{ Hence},\,{ theconditions}{{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(\delta )=1{ and}{{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(a^{\sigma +1})=0\) are equivalent, and we find exactly four solutions to Eq. (21). \(\square \)
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Faulkner, J., Van de Voorde, G. On the equivalence, stabilisers, and feet of Buekenhout-Tits unitals. Des. Codes Cryptogr. (2023). https://doi.org/10.1007/s10623-023-01234-4
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DOI: https://doi.org/10.1007/s10623-023-01234-4