1 Introduction

The Lunelli–Sce hyperoval was discovered [16] in 1958 by computer search. Later with the help of a computer Hall [12] described a number of automorphisms of the Lunelli–Sce hyperoval, but did not show that they generate the full automorphism group of this hyperoval. Payne and Conklin [18] showed that the Hall automorphisms form the full automorphism group. Korchmáros [14] independently gave a constructive proof of this result and also showed that the Lunelli–Sce hyperoval is the unique irregular hyperoval in PG(2, 16) admitting a transitive automorphism group. O’Keefe and Penttila [17] reproved Hall’s classification result without the use of a computer. Brown and Cherowitzo [4] provided a group-theoretic construction of the Lunelli–Sce hyperoval. Automorphism group of the Lunelli–Sce hyperoval has order 144 and is isomorphic to \(((C_3 \times C_3) \rtimes C_8) \times C_2\). Cherowitzo, Penttila, Pinneri and Royle  [8] showed that the Lunelli–Sce hyperoval is the first non-trivial member of the Subiaco family, and Cherowitzo, O’Keefe and Penttila [9] showed that the hyperoval is the first non-trivial member of the Adelaide family.

All known representations of the Lunelli–Sce hyperoval are complicated. We would like to give a new representation of the hyperoval. The advantage of this approach is that the new description of the Lunelli–Sce hyperoval is very simple and natural, and its automorphism group can be easily described in the new terms. This construction also immediately explains why the Lunelli–Sce hyperoval is a member of both Subiaco and Adelaide families.

2 Preliminary considerations and notation

In this section we recall some definitions and notation [5,6,7, 10, 13, 15]. Let \(F={\mathbb {F}}_{2^m}\) be a finite field of order \(q=2^m\). Consider points of a projective plane PG(2, q) in homogeneous coordinates as triples (x, y, z), where \(x, y, z \in F\), \((x,y,z)\ne (0,0,0)\), and we identify (x, y, z) with \((\lambda x, \lambda y, \lambda z)\), \(\lambda \in F^*\). Then points of PG(2, q) are

$$\begin{aligned} \{ (x, y, 1) \mid x \in F, \ y \in F \} \cup \{ (x,1,0) \mid x\in F\} \cup \{ (1,0,0) \}. \end{aligned}$$

We shall call points of the form (x, y, 0) the points at infinity. Removing points at infinity from PG(2, q), we will get an affine plane

$$\begin{aligned} AG(2,q) = \{ (x, y, 1) \mid x, y \in F \}. \end{aligned}$$

Associating (x, y, 1) with (x, y) we can identify points of the affine plane AG(2, q) with the elements of the vector space

$$\begin{aligned} V(2,q) = \{ (x,y) \mid x, y \in F \}, \end{aligned}$$

and we will write \(AG(2,q)=V(2,q)\).

Let K be a quadratic extension of the field \(F={\mathbb {F}}_{2^m}\). For \(x \in K\), let \(T(x)=Tr_{K/F}(x)\) be the trace function with respect to a finite field extension K / F. The conjugate of \(x\in K\) over F is

$$\begin{aligned} {\bar{x}} =x^q. \end{aligned}$$

Then the trace and the norm maps from K to F are

$$\begin{aligned} T(x)= & {} Tr_{K/F}= x + {\bar{x}}= x + x^q,\\ N(x)= & {} N_{K/F} (x) = x {\bar{x}} = x^{1+q}. \end{aligned}$$

The unit circle of K is the set of elements of norm 1:

$$\begin{aligned} S= \{ u\in K : u{\bar{u}} =1 \}. \end{aligned}$$

Therefore, S is the multiplicative group of \((q+1)\)st roots of unity in K. Since \(F\cap S = \{ 1\}\), each non-zero element of K has a unique polar coordinate representation

$$\begin{aligned} x=\lambda u \end{aligned}$$

with \(\lambda \in F^*\) and \(u \in S\). For any \(x\in K^*\) we have \(\lambda = \sqrt{x {\bar{x}}}\), \(u= \sqrt{x/ {\bar{x}}}\).

Following [11], consider an element \(i\in K\) with property \(T(i)=i+i^q=1\). Then \(K=F(i)\) and i is a root of a quadratic equation

$$\begin{aligned} z^2+z+\delta =0, \end{aligned}$$

where \(\delta = N(i) \in F\). Any element \(z\in K\) can be represented as \(z=x+yi\), where \(x, y \in F\). It is easy to see that \(x= z+ (z+{\bar{z}})i\), \(y= z + {\bar{z}}\).

Since K is a two dimensional vector space over F, we can introduce new representation of \(AG(2,q)=V(2,q)\) using the field K, and write \(AG(2,q)=K\). If \(z\in K=x+yi\) then \(z\in AG(2,q)\) corresponds to the element \((x,y,1) \in PG(2,q)\).

An oval in PG(2, q) is a set of \(q+1\) points, no three of which are collinear. Any line of the plane meets the oval \({\mathcal {O}}\) at either 0, 1 or 2 points and is called exterior, tangent or secant, respectively. All the tangent lines to the oval \({\mathcal {O}}\) concur  [13] at the same point N, called the nucleus of \({\mathcal {O}}\). The set \({\mathcal {H}}={\mathcal {O}} \cup \{N\}\) becomes a hyperoval, that is a set of \(q+2\) points, no three of which are collinear. Conversely, by removing any point from hyperoval one gets an oval.

For \(m=4\), the Lunelli–Sce hyperoval in PG(2, 16) can be described by

$$\begin{aligned} {\mathcal {D}} (f)= \{( t, f(t), 1) \mid t\in F\} \cup \{ (1,0,0) \} \cup \{ (0,1,0) \}, \end{aligned}$$

where

$$\begin{aligned} f(x) = x^{12} + x^{10} + \eta ^{11} x^8 + x^6 + \eta ^2 x^4 + \eta ^9 x^2 \end{aligned}$$

is an o-polynomial, and \(\eta \) is a primitive element of GF(16) satisfying

$$\begin{aligned} \eta ^4 = \eta + 1. \end{aligned}$$

There is an equivalent representation  [1, 2] of this hyperoval in PG(2, q) using the field K, namely, we can write the hyperoval \({\mathcal {D}} (f)\) in the form

$$\begin{aligned} {\mathcal {H}} (g) = \left\{ \frac{u}{g(u)} \mid u \in S\right\} \cup \{ 0 \} \subseteq K \end{aligned}$$

for a function \(g: S \rightarrow F\), where \(g(u)=1+u^5+{\bar{u}}^5\), \(m=4\).

For arbitrary m, if \(g(u)=1+u^5+{\bar{u}}^5\) then \({\mathcal {H}} (g)\) is one of the Subiaco hyperovals, and if \(g(u)=1+u^{(q-1)/3}+{\bar{u}}^{(q-1)/3}\), m even, then \({\mathcal {H}} (g)\) is the Adelaide hyperoval [1, 2]. So one can immediately see that the Lunelli–Sce hyperoval is a particular case of the Subiaco and Adelaide hyperovals for \(m=4\). (If \(g(u)=0\) for some \(u\in S\), then we assume that \( \frac{u}{g(u)} =u_{\infty }\), an element at infinity in direction u. This could happen only for the Subiaco hyperovals with odd m.)

3 Automorphism groups

For \(z\in K=AG(2,q)\), we define maps:

$$\begin{aligned} \sigma (z)= & {} z^2, \quad \sigma \in Gal(K/ {\mathbb {F}}_2),\\ \tau _1(z)= & {} \frac{1+z}{1+z+{\bar{z}}}, \\ \tau= & {} \sigma ^2 \tau _1, \quad \tau (z)= \frac{1+z^4}{1+z^4+{\bar{z}}^4}, \\ \rho (z)= & {} \frac{(\eta ^{14}+h^3i)+(\eta ^{10}+i)z +(\eta ^9+\eta ^4i){\bar{z}}}{1+(\eta ^7+\eta ^2i)z +(\eta ^{12}+\eta ^2i){\bar{z}}}. \end{aligned}$$

We note that in case of the Subiaco and Adelaide hyperovals (\(m>4\)) the map \(\sigma \) is an automorphism of the hyperoval (in case of the Adelaide hyperoval it even generates the entire automorphism group), and the map \(\tau _1\) is a collineation which permutes points \(0\in K\) and \(1\in K\), but not an automorphism. We also have \(\tau _1^2(z)={\bar{z}}\).

Theorem 3.1

 

  1. (a)

    \({\mathcal {H}} (g)\) is the Lunelli–Sce hyperoval for \(m=4\), and the full automorphism group \(Aut({\mathcal {H}})\) of \({\mathcal {H}} (g)\) is generated by elements \(\sigma , \tau , \rho \).

  2. (b)

    \(N= \langle \rho , \rho ^{\sigma } \rangle \cong C_3 \times C_3\) is a normal elementary abelian 3-group in \(Aut({\mathcal {H}})\).

  3. (c)

    \(\tau \) is a central element in \(Aut({\mathcal {H}})\) of order 2.

  4. (d)

    \(Aut({\mathcal {H}}) = (N \rtimes \langle \sigma \rangle ) \times \langle \tau \rangle \cong ((C_3 \times C_3) \rtimes C_8) \times C_2\).

  5. (e)

    \(Aut({\mathcal {H}}) = (N \times \langle \tau \rangle ) \rtimes \langle \sigma \rangle \cong ((C_3 \times C_3) \times C_2 ) \rtimes C_8\), where \(N \times \langle \tau \rangle \cong (C_3 \times C_3) \times C_2\) is a subgroup of order 18 acting regularly on the hyperoval \({\mathcal {H}} (g)\), and the subgroup \(\langle \sigma \rangle = Gal(K/ {\mathbb {F}}_2)\) acts naturally on \({\mathcal {H}} (g)\).

Proof

It is clear that \(\sigma \) is a collineation of AG(2, q) since it is a semilinear map:

$$\begin{aligned} \sigma (\lambda z_1 + z_2) = \lambda ^2 \sigma (z_1) + \sigma (z_2) \end{aligned}$$

for all \(\lambda \in F\), \(z_1, z_2 \in K\).

Let \(z=x+yi\), \(x, y \in F\). Then \(x= z+ (z+{\bar{z}})i\), \(y= z + {\bar{z}}\). Consider a collineation generated by the matrix

$$\begin{aligned} A=\left( \begin{array}{ccc} 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 1 \\ 1 &{}\quad 0 &{}\quad 1 \end{array} \right) \end{aligned}$$

We show that it corresponds to the map \(\tau _1\). Indeed, we have

$$\begin{aligned} (x,y,1) A= (x+1, y, y+1)= & {} (z+ (z+{\bar{z}})i+1, z + {\bar{z}}, z + {\bar{z}} +1) \\= & {} \left( \frac{z+ (z+{\bar{z}})i+1}{1+z + {\bar{z}}}, \frac{z + {\bar{z}}}{1+z + {\bar{z}}}, 1\right) . \end{aligned}$$

Therefore, this collineation maps element \(z\in K\) to the element

$$\begin{aligned} \frac{z+ (z+{\bar{z}})i+1}{1+z + {\bar{z}}} + \frac{z + {\bar{z}}}{1+z + {\bar{z}}}i = \frac{1+z}{1+z + {\bar{z}}}. \end{aligned}$$

We have \(\sigma \tau _1 = \tau _1 \sigma \) and \(\tau _1^2=\sigma ^4\). Then \(\tau = \sigma ^2 \tau _1\) and \(\tau ^2 = \mathrm{id}\).

Now we will show that the collineation \(\rho \) is generated by the matrix

$$\begin{aligned} B=\left( \begin{array}{ccc} \eta ^{13} &{}\quad \eta &{}\quad \eta ^2 \\ \eta ^{14} &{}\quad \eta ^6 &{}\quad \eta ^{12} \\ \eta ^{14} &{}\quad \eta ^3 &{}\quad 1 \end{array} \right) \end{aligned}$$

Indeed,

$$\begin{aligned} (x,y,1) B= & {} (\eta ^{13}x+\eta ^{14}y+\eta ^{14}, \eta x+\eta ^6 y+\eta ^3, \eta ^2 x+\eta ^{12}y+1) \\= & {} (\eta ^{13}[z+ (z+{\bar{z}})i]+\eta ^{14}[z + {\bar{z}}]+\eta ^{14} , \eta [z+ (z+{\bar{z}})i]+\eta ^6 [z + {\bar{z}}]\\&+\,\eta ^3, \eta ^2 [z+ (z+{\bar{z}})i]+\eta ^{12}[z + {\bar{z}}]+1)\\= & {} \big ((\eta ^2 + \eta ^{13}i)z + (\eta ^{14} +\eta ^{13}i){\bar{z}} + \eta ^{14}, (\eta ^{11} + \eta i)z + (\eta ^6 +\eta i){\bar{z}} \\&+\, \eta ^3, (\eta ^7 + \eta ^2 i)z + (\eta ^{12} +\eta ^2 i){\bar{z}} + 1 \big ) \\= & {} \left( \frac{(\eta ^2 + \eta ^{13}i)z + (\eta ^{14} +\eta ^{13}i){\bar{z}} + \eta ^{14}}{(\eta ^7 + \eta ^2 i)z + (\eta ^{12} +\eta ^2 i){\bar{z}} + 1}, \frac{(\eta ^{11} + \eta i)z + (\eta ^6 +\eta i){\bar{z}} + \eta ^3}{(\eta ^7 + \eta ^2 i)z + (\eta ^{12} +\eta ^2 i){\bar{z}} + 1}, 1\right) . \end{aligned}$$

Therefore, element \(z\in K\) is mapped to the element

$$\begin{aligned}&\frac{(\eta ^2 + \eta ^{13}i)z + (\eta ^{14} +\eta ^{13}i){\bar{z}} + \eta ^{14}}{(\eta ^7 + \eta ^2 i)z + (\eta ^{12} +\eta ^2 i){\bar{z}} + 1} + \frac{(\eta ^{11} + \eta i)z + (\eta ^6 +\eta i){\bar{z}} + \eta ^3}{(\eta ^7 + \eta ^2 i)z + (\eta ^{12} +\eta ^2 i){\bar{z}} + 1} i \\&\quad = \frac{(\eta ^{14}+h^3i)+(\eta ^{10}+i)z +(\eta ^9+\eta ^4i){\bar{z}}}{1+(\eta ^7+\eta ^2i)z +(\eta ^{12}+\eta ^2i){\bar{z}}}, \end{aligned}$$

which shows that the collineation \(\rho \) is generated by the matrix B.

We have \(\rho ^3 = \mathrm{id}\) since \(B^3=\eta ^3 I\), where I is the identity matrix.

Let us numerate elements of the hyperoval \({\mathcal {H}}(g)\) in the following way: the elements \(w^i/g(w^i)\) are denoted by i, where \(1\le i\le 17\), and the element \(0\in K\) is denoted by 18. The corresponding points of the Lunelli–Sce hyperoval \({\mathcal {H}} (g)\) in PG(2, 16) are given in Table 1.

Table 1 Points of the Lunelli–Sce Hyperoval
Full size table

Then one has an injective homomorphism

$$\begin{aligned} \pi : Aut({\mathcal {H}}) \rightarrow S_{18} \end{aligned}$$

where \(S_{18}\) is the symmetric group on a set of 18 elements. Elementary Magma [3] calculations (but it can be done also by hand) show that

$$\begin{aligned} \pi (\sigma )= & {} (1, 2, 4, 8, 16, 15, 13, 9)(3, 6, 12, 7, 14, 11, 5, 10),\\ \pi (\tau )= & {} (1, 6)(2, 12)(3, 9)(4, 7)(5, 15)(8, 14)(10, 13)(11, 16)(17, 18),\\ \pi (\rho )= & {} (1, 13, 15)(2, 4, 16)(3, 14, 18)(5, 6, 10)(7, 11, 12)(8, 17, 9),\\ \pi (\rho ^\sigma )= & {} (1, 16, 17)(2, 9, 13)(3, 10, 12)(4, 8, 15)(5, 7, 14)(6, 11, 18), \end{aligned}$$

and

$$\begin{aligned} \pi (N)= & {} \{(1, 13, 15)(2, 4, 16)(3, 14, 18)(5, 6, 10)(7, 11, 12)(8, 17, 9), \\&(1, 15, 13)(2, 16, 4)(3, 18, 14)(5, 10, 6)(7, 12, 11)(8, 9, 17), \\&(1, 16, 17)(2, 9, 13)(3, 10, 12)(4, 8, 15)(5, 7, 14)(6, 11, 18), \\&(1, 17, 16)(2, 13, 9)(3, 12, 10)(4, 15, 8)(5, 14, 7)(6, 18, 11), \\&(1, 9, 4)(2, 15, 17)(3, 7, 6)(5, 18, 12)(8, 16, 13)(10, 14, 11), \\&(1, 4, 9)(2, 17, 15)(3, 6, 7)(5, 12, 18)(8, 13, 16)(10, 11, 14), \\&(1, 2, 8)(3, 5, 11)(4, 17, 13)(6, 12, 14)(7, 18, 10)(9, 15, 16), \\&(1, 8, 2)(3, 11, 5)(4, 13, 17)(6, 14, 12)(7, 10, 18)(9, 16, 15), e \}. \end{aligned}$$

It is clear that \(N \times \langle \tau \rangle \) acts regularly on \({\mathcal {H}}(g)\). \(\square \)

Note that [1, 2] if \(g(u)=1+u^{(q-1)/3}+{\bar{u}}^{(q-1)/3}\), \(m\ge 6\), m even, then the automorphism group of the Adelaide hyperoval \({\mathcal {H}} (g)\) has order 2m and is equal to \(Gal(K/{\mathbb {F}}_2) = \langle \sigma \rangle \). If \(g(u)=1+u^5+{\bar{u}}^5\), \(m\ge 5\) and \(m\not \equiv 2 \pmod {4}\), then the automorphism group of the Subiaco hyperoval \({\mathcal {H}} (g)\) has order 2m and is equal to \(Gal(K/{\mathbb {F}}_2) = \langle \sigma \rangle \). If \(g(u)=1+u^5+{\bar{u}}^5\), \(m\ge 6\) and \(m\equiv 2 \pmod {4}\), then the automorphism group of the Subiaco hyperoval \({\mathcal {H}} (g)\) has order 10m and is equal to the semidirect product \(\langle \varphi _{v} \rangle \langle \sigma \rangle \), where \(\varphi _{v}(z)=vz\), \(v=w^{(q+1)/5}\), \(S=\langle w\rangle \).

4 Conclusion

All known representations of the Lunelli–Sce hyperoval are complicated. We gave a new representation of the hyperoval. The advantage of this approach is that the new description of the Lunelli–Sce hyperoval is very simple and natural, and its automorphism group can be easily described in the new terms. This construction also explains why the Lunelli–Sce hyperoval is a member of both Subiaco and Adelaide families.