On the equivalence, stabilisers, and feet of Buekenhout-Tits unitals

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).

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,

$$\begin{aligned} \mathcal {U}_{BT}= \{(0, 0, 1)\} \cup \{P_{r, s, t} = (1, s + t \epsilon , r + (s^{\sigma + 2} + t^{\sigma } + st)\epsilon )\,|\, r, s, t \in \mathbb {F}_{q}\}, \end{aligned}$$

(1)

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

$$\begin{aligned} M_{u,v} = \left\{ \begin{bmatrix} 1 &{} u \epsilon &{} v + u^{\sigma } \epsilon \\ 0 &{} 1 &{} u + u \epsilon \\ 0 &{} 0 &{} 1 \end{bmatrix}\,\Bigg |\,u,v \in \mathbb {F}_{q}\right\} , \end{aligned}$$

(2)

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,

$$\begin{aligned} \begin{bmatrix} 1 &{}\quad x_{12} &{}\quad x_{13} \\ 0 &{}\quad x_{22} &{}\quad x_{23} \\ 0 &{}\quad 0 &{}\quad x_{33} \end{bmatrix}, \end{aligned}$$

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

$$\begin{aligned} \frac{(q +1)^2q^4{(q-1)}^2(q^2+q+1)(q^2+1) }{ q^3+q^2+q+1}= {(q-1)}^{2}q^4(q+1)(q^2+q+1) \end{aligned}$$

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

$$\begin{aligned} M_{u, v} M_{s, t} = \begin{bmatrix} 1 &{} (s + u)\epsilon &{} (t + v + su \delta ) + {(s + u)}^{\sigma } \\ 0 &{} 1 &{} (u + s) + (u + s)\epsilon \\ 0 &{} 0 &{} 1 \end{bmatrix}. \end{aligned}$$

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,

$$\begin{aligned} l = 2(2e + 1 - k). \end{aligned}$$

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,

$$\begin{aligned} (4^{k} - 2^{k})2^{l} = 4^{2e + 1} - 2^{2e + 1}. \end{aligned}$$

(3)

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

$$\begin{aligned} \psi :{\textbf{x}}\mapsto {\textbf{x}}^{2} \begin{bmatrix} 1 &{} 1 &{} \epsilon \\ 0 &{} \delta ^{\sigma /2}(1 + \epsilon ) &{} \delta ^{\sigma /2}(1 + \epsilon ) \\ 0 &{} 0 &{} \delta ^{\sigma +1} \end{bmatrix}. \end{aligned}$$

(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

$$\begin{aligned} |\textrm{P}\Gamma \textrm{L}(3, q^2)_{\mathcal {U}_{BT}}| = \frac{|\textrm{P}\Gamma \textrm{L}(3, q^2)_{\{\ell _{\infty }, P_{\infty }\}}|}{q^4{(q^2 - 1)}^2}. \end{aligned}$$

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,

$$\begin{aligned} \psi ^{k}(0, 1, z) = \left( 0, 1, \sum _{i=0}^{k}\mu ^{2^{i} - 1} + zg(z)\right) \end{aligned}$$

for some polynomial $g(z)$ depending on $k$. If $z = 0$ and $k = 4e + 2$ we thus find

$$\begin{aligned} \psi ^{4e + 2}(0, 1, 0)&= \left( 0, 1, \sum _{i=0}^{4e + 2}\mu ^{2^{i} - 1}\right) \\&= \left( 0, 1, \frac{{{\,\textrm{Tr}\,}}_{{{\,\mathrm{\mathbb {F}}\,}}_q/{{\,\mathrm{\mathbb {F}}\,}}_2}(\mu )}{\mu }\right) . \end{aligned}$$

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:

$$\begin{aligned}{} & {} \tau _P(\mathcal {U}_{BT}) = \{(1, s + t\epsilon , s^2 + t^2\delta + st + y_1s + y_1t + y_2\delta {t} + z_1 + (s^{\sigma + 2} + t^{\sigma } + st)\epsilon )\nonumber \\{} & {} |\,s, t \in \mathbb {F}_q,\, s^{\sigma + 2} + t^{\sigma } + st = y_2s + y_1t + z_2\}. \end{aligned}$$

(4)

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

$$\begin{aligned}&s^2 + \delta t^2 + st + (y_1 + b_1) s + (y_1 + y_2 \delta + b_2 \delta ) t + z_1 + a_1 = 0, \end{aligned}$$

(5)

$$\begin{aligned}&s^{\sigma + 2} + t^{\sigma } + st = b_2 s + (b_1 + b_2) t + a_2, \end{aligned}$$

(6)

$$\begin{aligned}&y_{2} s + y_{1} t + z_{2} = b_{2} s + (b_{1} + b_{2})t + a_{2}. \end{aligned}$$

(7)

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,

$$\begin{aligned} \left( 1, a, b \epsilon \right) \begin{bmatrix} 1 &{}\quad u \epsilon &{}\quad v + u^{\sigma } \epsilon \\ 0 &{}\quad 1 &{}\quad u + u \epsilon \\ 0 &{}\quad 0 &{}\quad 1 \end{bmatrix} = \left( 1, c, d \epsilon \right) . \end{aligned}$$

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

$$\begin{aligned} s^{\sigma + 2} + t^{\sigma } + st = y_{1}t + z_{2}, \end{aligned}$$

(8)

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

$$\begin{aligned} s^{\sigma + 2} + t^{\sigma } + st + y_{1}t + z_{2}&= 0. \end{aligned}$$

$$\begin{aligned} s^{2} + \delta t^{2} + st + \left( y_{1} + b_{1} \right) s + \left( y_{1} + b_{2} \delta \right) t + a_{1}&= 0 \end{aligned}$$

(9)

$$\begin{aligned} b_{2} s + \left( y_{1} + b_{1} + b_{2} \right) t + a_{2} + z_{2}&= 0 \end{aligned}$$

(10)

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

$$\begin{aligned} s^{2} + \delta t^{2} + st&= y_{1} t + a_{1} \end{aligned}$$

(11)

$$\begin{aligned} s^{\sigma + 2} + t^{\sigma } + st&= y_{1} t + z_{2}. \end{aligned}$$

(12)

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]

$$\begin{aligned} \mathcal {D}_{\sigma } = \left\{ (1, t, t^\sigma ) \,|\, t \in \mathbb {F}_q\right\} \cup \left\{ (0, 0, 1) \right\} . \end{aligned}$$

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,

$$\begin{aligned} a_1 x^2 + a_2 y^2 + a_3 z^2 + x z = 0, \end{aligned}$$

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

$$\begin{aligned} a_1 + a_2 t^2 + a_3 t^{2 \sigma } + t^\sigma = 0, \end{aligned}$$

(13)

hence,

$$\begin{aligned} 0 = {\left( a_1 + a_2 t^{2} + a_3 t^{2 \sigma } + t^{\sigma } \right) }^{\sigma /2} = a_{1}^{\sigma /2} + a_2^{\sigma /2} t^\sigma + a_3^{\sigma /2} t^2 + t. \end{aligned}$$

Therefore,

$$\begin{aligned} t^\sigma = {\left( \frac{a_3}{a_2} \right) }^{2^{e}} t^{2} + \frac{1}{a_2^{2^e}} t + {\left( \frac{a_1}{a_2}\right) }^{2^{e}}. \end{aligned}$$

(14)

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

$$\begin{aligned} s^{2} + \delta t^{2} + st&= y_{1} t + a_{1} \end{aligned}$$

(15)

$$\begin{aligned} s^{\sigma + 2} + t^{\sigma } + st&= y_{1} t + z_{2}, \end{aligned}$$

(16)

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

$$\begin{aligned} \left\{ P_{u} = \left( \frac{z_{2}^{1 - \sigma /2}u^{\sigma }}{1 + u + u^{\sigma }}, \frac{z_{2}^{\sigma /2}(1 + u^{\sigma })}{1 + u + u^{\sigma }}\right) \,\Bigg |\,u \in \mathbb {F}_q \right\} \cup \left\{ \left( z_{2}^{1 - \sigma /2}, z_{2}^{\sigma /2}\right) \right\} . \end{aligned}$$

Proof

If $y_{1} = 0$, then Eq. (16) reduces to

$$\begin{aligned} s^{\sigma + 2} + t^{\sigma } + st + z_{2} = 0. \end{aligned}$$

(17)

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

$$\begin{aligned} x^{\sigma + 2} + y^{\sigma }z^{2} + xyz^{\sigma } + z_{2}z^{\sigma + 2} = 0. \end{aligned}$$

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

$$\begin{aligned} a_{1}^{\sigma /2}u^{\sigma } + (z_{2}^{\sigma - 1} + \delta ^{\sigma /2}z_{2} + z_{2}^{\sigma /2} + a_{1}^{\sigma /2})u^{2} + z_{2}^{\sigma /2}u + \delta ^{\sigma /2} z_{2} + a_{1}^{\sigma /2}. \end{aligned}$$

(18)

Proof

By directly substituting $P_u$ into Eq. (15) we have

$$\begin{aligned} (z_{2}^{2 - \sigma } + \delta z_{2}^{\sigma } + z_{2} + a_{1})u^{2\sigma } + z_{2}u^{\sigma } + a_{1}u^{2} + (\delta z_{2}^{\sigma } + a_{1}) = 0. \end{aligned}$$

(19)

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

$$\begin{aligned} \delta ^{\sigma /2}u^{\sigma } + u = u(\delta ^{\sigma /2}u^{\sigma - 1} + 1) = 0. \end{aligned}$$

(20)

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

$$\begin{aligned} \frac{1}{\delta ^{\sigma /2}}u^{\sigma }+ \Bigg (\frac{1}{\delta ^{2-\sigma }}+\frac{1}{\delta }\Bigg )u^2+\frac{1}{\delta }u=0. \end{aligned}$$

(21)

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

$$\begin{aligned} (a^{\sigma /2}+1)u^\sigma +au^2+u=0. \end{aligned}$$

(22)

We find that $u = 0$ and $u = \frac{1}{a^{1 + \sigma /2}}$ are solutions to Eq. (22). Now consider

$$\begin{aligned} u^{\sigma } + au^2 + 1 = 0. \end{aligned}$$

(23)

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 $z = (a^{\sigma /2 + 1}u)^2$,

$$\begin{aligned} z^{\sigma /2}+z+a^{\sigma +1}=0, \end{aligned}$$

(24)

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). $\square$