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New dimensional dual hyperovals, which are not quotients of the classical dual hyperovals Hiroaki Taniguchi National Institute of Technology, Kagawa College, 355, Chokushicho, Takamatsu city, Kagawa, 761-8058, Japan

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Article history: Received 24 September 2013 Received in revised form 9 August 2014 Accepted 9 August 2014

Keywords: Dual hyperoval

abstract Let d ≥ 2. If d-dimensional dual hyperoval exists in V (n, 2) (n-dimensional vector space over GF (2)), then 2d + 1 ≤ n ≤ (d + 1)(d + 2)/2 + 2 (Yoshiara [15], 2004), and it is conjectured that n ≤ (d + 1)(d + 2)/2. In V ((d + 1)(d + 2)/2, 2), there are four known non-isomorphic d-dimensional dual hyperovals. These are the Huybrechts dual hyperoval (Huybrechts [5], 2002), the Buratti–Del Fra dual hyperoval (Buratti–Del Fra [1], 2003), (Del Fra and Yoshiara [2], 2005), (Taniguchi and Yoshiara [13], 2012), the Veronesean dual hyperoval (Thas and van Maldeghem [14], 2004, Yoshiara [15], 2004) and the deformation of Veronesean dual hyperoval (Taniguchi [9], 2009). Many of the known dual hyperovals in V (n, 2) for 2d + 2 < n < (d + 1)(d + 2)/2 are quotients of the four examples in V ((d + 1)(d + 2)/2, 2). Very recently, we constructed examples of d-dimensional dual hyperovals for n = 3d + 3 with d ≥ 3, n = 4d − 2 with d ≥ 4, and n = 3d + 1 with 3 ≤ d ≤ 13 satisfying some conditions (Taniguchi [11], 2013), which are not quotients of the four examples in V ((d + 1)(d + 2)/2, 2). In this paper, we modify the method of constructing the Buratti–Del Fra dual hyperoval (Taniguchi and Yoshiara [13], 2012) combined with Example 3.4 of Dempwolff [3] (2013), and construct many non-isomorphic symmetric bilinear dual hyperovals in V (((1/r )d2 + 3d + 2)/2, 2) for d = lr ≥ 4, which are not quotients of the four dual hyperovals in V ((d + 1)(d + 2)/2, 2). © 2014 Elsevier B.V. All rights reserved.

1. Introduction Let d, n be integers with d ≥ 2 and n > d. Let V (n, 2) be an n-dimensional vector space over the field GF (2). Definition 1. A family S of (d + 1)-dimensional vector subspaces of V (n, 2) is a d-dimensional dual hyperoval in V (n, 2) if it satisfies the following conditions: (d1) (d2) (d3) (d4)

any two distinct members of S intersect in one-dimensional subspace, any three mutually distinct members of S intersect trivially, the members of S generate V (n, 2), and there are exactly 2d+1 members of S .

The definition of higher dimensional dual hyperovals was given by C. Huybrechts and A. Pasini in [6]. We call V (n, 2) of (d3) the ambient space of the dual hyperoval S and it is denoted by ⟨S ⟩. For d-dimensional dual hyperovals S1 and S2 in V (n, 2), we say that S1 is isomorphic to S2 by the mapping φ , if φ is a linear automorphism of V (n, 2) which sends the members of S1 onto the members of S2 . If S := S1 = S2 , we say that φ is an automorphism of S . The automorphism group Aut(S ) of a dual hyperoval S is the group consisting of the automorphisms of S . Let Si be a d-dimensional dual hyperoval

E-mail address: [email protected] http://dx.doi.org/10.1016/j.disc.2014.08.004 0012-365X/© 2014 Elsevier B.V. All rights reserved.

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in V (ni , 2), i = 1, 2. We say S2 is a quotient of S1 , or S1 is a cover of S2 if there is a surjective GF (2)-linear mapping (a covering map) π : V (n1 , 2) → V (n2 , 2) such that π (S1 ) = S2 . It is proved in [15] that, if a d-dimensional dual hyperoval exists in V (n, 2), then 2d + 1 ≤ n ≤ (d + 1)(d + 2)/2 + 2, and it is conjectured n ≤ (d + 1)(d + 2)/2. Many non-isomorphic d-dimensional dual hyperovals are known in V (n, 2) for n = 2d + 1 and n = 2d + 2 (see for example [8,15,16]). There are four known non-isomorphic examples for n = (d + 1)(d + 2)/2 (see [1,5,9,14]). On the other hand, many of the known dual hyperovals in V (n, 2) for 2d + 2 < n < (d + 1)(d + 2)/2 are quotients of the four examples in V ((d + 1)(d + 2)/2, 2). Very recently, we constructed examples of d-dimensional dual hyperovals for n = 3d + 3 with d ≥ 3, n = 4d − 2 with d ≥ 4, and n = 3d + 1 with 3 ≤ d ≤ 13 satisfying some conditions [11], which are not quotients of the four examples in V ((d + 1)(d + 2)/2, 2). In this paper, we modify the method of constructing the Buratti–Del Fra dual hyperoval [13] combined with Example 3.4 of [3], and present many non-isomorphic examples for n = ((1/r )d2 + 3d + 2)/2 with d = lr ≥ 4, which are not quotients of the four examples in V ((d + 1)(d + 2)/2, 2). Let V be a (d + 1)-dimensional GF (2)-vector space, and W an l-dimensional GF (2)-vector space. A dual hyperoval S := {X (e) | e ∈ V } in V ⊕ W = V (d + l + 1, 2) is said to be a bilinear dual hyperoval if there is a GF (2)-bilinear mapping B : V ⊕ V → W such that X (e) := {(x, B(x, e)) | x ∈ V } ⊂ V ⊕ W for any e ∈ V . We recall that B(x, b) = 0 must have only one non-zero solution x ∈ V for any non-zero b ∈ V and B(a, y) = 0 must have only one non-zero solution y ∈ V for any non-zero a ∈ V , since S is a dual hyperoval. (These two conditions are independent.) A bilinear dual hyperoval has a translation group T := {ta | a ∈ V }, where X (t )ta = X (t + a) defined by ta : V ⊕ W ∋ (x, y) → (x, y + B(x, a)) ∈ V ⊕ W . Translation groups of bilinear dual hyperovals are characterized and studied in Section 3 of [4]. There are examples of dual hyperovals which have more than one translation groups (see Section 6 of [4]), which are conjugate in their automorphism groups (Theorem 3.11 of [4]). We call a bilinear dual hyperoval symmetric if B is symmetric, that is, B(x, y) = B(y, x) for x, y ∈ V . (See [4] or [8] for more details.) There are many bilinear dual hyperovals, such as the Huybrechts dual hyperoval [5], the APN dual hyperovals [16], the Buratti–Del Fra dual hyperoval [1], the Yoshiara dual hyperovals [15]. However, there exist also non-bilinear dual hyperovals, such as the Veronesean dual hyperoval [14], and the deformation of Veronesean dual hyperoval [9]. d-dimensional dual hyperovals constructed in this paper are symmetric, d = lr ≥ 4, and the ambient space of dimension ((1/r )d2 + 3d + 2)/2. Let us recall our construction of the Buratti–Del Fra dual hyperoval in [13]. We will modify this construction in Section 2 to construct new dual hyperovals. Denote by I the set of integers between 0 and d and set I0 = I \ {0}. Let V be a (d + 1)dimensional vector space over GF (2) with a basis ei (i ∈ I). Inside V ⊗ V , let W be a GF (2)-vector subspace generated by

(ei ⊗ ej ) + (ej ⊗ ei ) for all i, j ∈ I with i < j, e0 ⊗ e0 and (ei ⊗ ei ) + (e0 ⊗ ei ) for all i ∈ I0 . We denote by v the image v + W of a vector v ∈ V ⊗ V under the canonical projection of V ⊗ V onto (V ⊗ V )/W . Then x ⊗ y = y ⊗ x and (x1 + x2 ) ⊗ y = x1 ⊗ y + x2 ⊗ y for any x, x1 , x2 , y ∈ V . Furthermore, the following vectors form a basis for (V ⊗ V )/W (of dimension (d + 1)2 − (d + 1)(d + 2)/2 = d(d + 1)/2): ei ⊗ ej

for all i, j ∈ I0 with i < j,

and e0 ⊗ ei

for all i ∈ I0 .

We notice that, for non-zero x, y ∈ V , we have x ⊗ y = 0 if and only if x = y + e0 in case y ̸= e0 and x = e0 in case y = e0 (see Lemma 2 of [13]). We consider the direct sum A := V ⊕ ((V ⊗ V )/W ) := {(x, v) | x ∈ V , v ∈ V ⊗ V }. The dimension of A is (d + 1) + d(d + 1)/2 = (d + 1)(d + 2)/2. For each t ∈ V , we define a subset X (t ) of A by X (t ) := {(x, x ⊗ t ) | x ∈ V }. Then X (t ) is a subspace of A of dimension d + 1. Fact 2 (Proposition 3 of [13]). SB := {X (t ) | t ∈ V } is a d-dimensional dual hyperoval over GF (2) with ambient space A, which is isomorphic to the Buratti–Del Fra dual hyperoval constructed originally in [1]. Remark 3. From Fact 2, we see that Buratti–Del Fra dual hyperoval is a bilinear dual hyperoval with the bilinear mapping B(x, t ) = x ⊗ t. 2. A dual hyperoval Sc for c ∈ GF (2r ) with Tr(c ) = 1 Let l ≥ 1 and r ≥ 1 be integers with d = lr ≥ 4, and GF (2r ) a finite field of 2r elements. From now on, the letter I is used to denote the set of integers i with 0 ≤ i ≤ l, and we set I0 := I \ {0}. Let V1 be an l-dimensional vector space over GF (2r ) with a basis {ei | i ∈ I0 } and V2 an (l + 1)-dimensional vector space over GF (2r ) with a basis {ei | i ∈ I }. Let V ⊂ V2 be an (rl + 1)-dimensional GF (2)-vector space generated by V1 and e0 . (V = V1 ⊕ ⟨e0 ⟩ as GF (2)-vector space.) Let c ∈ GF (2r ) be a non-zero element such that the equation x2 + (x/c ) + 1 = 0 has no solution in GF (2r ). Remark 4. If there exists x ∈ GF (2r ) with x2 + (x/c ) + 1 = 0, then, if we put y := cx, we have y2 + y + c 2 = 0, hence the absolute trace Tr(c ) = 0. Conversely, if Tr(c ) = 0, there exists y ∈ GF (2r ) with y2 + y + c 2 = 0, hence there exists x ∈ GF (2r ) with x2 + (x/c ) + 1 = 0, by additive form of Hilbert’s Theorem 90. Therefore, for c ∈ GF (2r ), we have x2 + (x/c ) + 1 ̸= 0 for any x ∈ GF (2r ) if and only if Tr(c ) = 1 in GF (2).

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Set V2 ⊗ V2 = V2 ⊗GF (2r ) V2 . We define the action of a ∈ GF (2r ) on V2 ⊗ V2 naturally, as a(x ⊗ y) = (ax) ⊗ y = x ⊗ (ay). Inside V2 ⊗ V2 , let Wc (c is defined above) be the GF (2r )-vector subspace generated by

(ei ⊗ ej ) + (ej ⊗ ei ) for all i, j ∈ I with i < j, e0 ⊗ e0 and c (ei ⊗ ei ) + (e0 ⊗ ei ) for all i ∈ I0 . We denote by x ⊗c y the image x ⊗ y + Wc of a vector x ⊗ y ∈ V2 ⊗ V2 under the canonical projection of V2 ⊗ V2 onto (V2 ⊗ V2 )/Wc . Then x ⊗c y = y ⊗c x and (x1 + x2 ) ⊗c y = x1 ⊗c y + x2 ⊗c y for any x, x1 , x2 , y ∈ V2 . We notice that x ⊗c e0 = e0 ⊗c x = (cx) ⊗c x = x ⊗c (cx) for any x ∈ V1 and e0 ⊗c e0 = 0 in (V2 ⊗ V2 )/Wc . Hence {ei ⊗c ej | i, j ∈ I0 , i ≤ j} generate (V2 ⊗ V2 )/Wc as a GF (2r )-vector space. Next consider the symmetric tensor space Sym(V1 ) := (V1 ⊗ V1 )/W over GF (2r ), where W ⊂ V1 ⊗ V1 is a GF (2r )-vector subspace defined by W := ⟨(ei ⊗ ej ) + (ej ⊗ ei ) | i, j ∈ I0 with i < j⟩. We regard V1 ⊗ V1 ⊂ V2 ⊗ V2 naturally, by the GF (2r )-linear mapping defined by V1 ⊗ V1 = ⟨ei ⊗ ej | i, j ∈ I0 ⟩ ∋ ei ⊗ ej → ei ⊗ ej ∈ V2 ⊗ V2 = ⟨ei ⊗ ej | i, j ∈ I ⟩ for i, j ∈ I0 . Lemma 5. Sym(V1 ) ≃ (V2 ⊗ V2 )/Wc . Hence the dimension of (V2 ⊗ V2 )/Wc is l(l + 1)/2 over GF (2r ), and rl(l + 1)/2 over GF (2). Proof. Recall {ei ⊗c ej | i, j ∈ I0 , i ≤ j} generate (V2 ⊗ V2 )/Wc . Hence dimGF (2r ) (V2 ⊗ V2 )/Wc ≤ l(l + 1)/2. We have ⟨Wc , V1 ⊗ V1 ⟩/(V1 ⊗ V1 ) ≃ ⟨e0 ⊗ e0 ⟩ ⊕ ⟨c (ei ⊗ ei ) + (e0 ⊗ ei ) | i ∈ I0 ⟩ ⊕ ⟨ei ⊗ e0 + e0 ⊗ ei | i ∈ I0 ⟩ ≃ Wc /W as GF (2r )-vector spaces, and since ⟨Wc , V1 ⊗ V1 ⟩/(V1 ⊗ V1 ) ≃ Wc /Wc ∩ (V1 ⊗ V1 ), we must have Wc ∩ (V1 ⊗ V1 ) = W . Hence the natural injection V1 ⊗ V1 → V2 ⊗ V2 induces an isomorphism Sym(V1 ) ≃ (V2 ⊗ V2 )/Wc . Therefore {ei ⊗c ej | i, j ∈ I0 , i ≤ j} is a basis of (V2 ⊗ V2 )/Wc as a GF (2r )-vector space. Thus the dimension of the vector space (V2 ⊗ V2 )/Wc over GF (2r ) is l(l + 1)/2, and rl(l + 1)/2 over GF (2). Lemma 6. Let x, y ∈ V1 . If x ⊗c y + c (x + y) ⊗c (x + y) = 0, then x = y = 0. Proof. Let us express x, y ∈ V1 as x =

i∈I0

xi ei and y =

i∈I0

yi ei with xi , yi ∈ GF (2r ). Then, by easy calculations, we have

2 2 (xi + (xi yi /c ) + yi )(ei ⊗c ei ) = 0. (xi yj + xj yi )(ei ⊗c ej ) + c

x ⊗c y + c (x + y) ⊗c (x + y) =

i

i∈I0

Recall {ei ⊗c ej | i, j ∈ I0 , i ≤ j} is a basis of (V2 ⊗ V2 )/Wc as a GF (2 )-vector space. Hence, if x ⊗c y + c (x + y) ⊗c (x + y) = 0, then xi 2 +(xi yi /c )+yi 2 = 0 for any i ∈ I0 . If (xi , yi ) ̸= (0, 0) for some i ∈ I0 , for example yi ̸= 0, then (xi /yi )2 +(xi /cyi )+1 = 0 gives a solution xi /yi ∈ GF (2r ) for the equation x2 + (x/c ) + 1 = 0, which contradicts our assumption on c ∈ GF (2r ). Hence (xi , yi ) = (0, 0) for any i ∈ I0 , thus x = y = 0. r

Lemma 7. Let x, y ∈ V1 and y ̸= 0. If x ⊗c y = 0, then x = 0. r Proof. Let x, y ∈ V1 and y ̸= 0. Let x = i∈I0 xi ei and y = i∈I0 yi ei with xi , yi ∈ GF (2 ). Then we have i

Lemma 8. Let x, y ∈ V with x, y ̸∈ V1 . If x ⊗c y = 0, then x = y = e0 . Proof. Recall that V = V1 ⊕ ⟨e0 ⟩ ⊂ V2 as a GF (2)-vector space. Since x, y ̸∈ V1 , x and y have expressions x = x0 + e0 and y = y0 + e0 with x0 , y0 ∈ V1 . Assume x ⊗c y = 0. Then, since e0 ⊗c e0 = 0, we have x0 ⊗c y0 + e0 ⊗c (x0 + y0 ) = x0 ⊗c y0 + c (x0 + y0 ) ⊗c (x0 + y0 ) = 0. Hence x0 = y0 = 0 by Lemma 6, and x = y = e0 . Lemma 9. Let x, y ∈ V with x ̸∈ V1 and y ∈ V1 \ {0}. Then x ⊗c y = 0 if and only if x = cy + e0 . r Proof. Let y ∈ V1 \ {0} with y = i∈I0 yi ei , yi ∈ GF (2 ). From cei ⊗c ei + e0 ⊗c ei = 0 for all i ∈ I0 , we see that cy ⊗c y + e0 ⊗c y = 0. Hence (cy + e0 ) ⊗c y = 0. Now, let x ⊗c y = 0 with x ∈ V \ {0}. We must have x = x0 + e0 for some x0 ∈ V1 by assumption. Adding these equations, we have (x0 + cy) ⊗c y = 0 with x0 + cy ∈ V1 , hence x0 + cy = 0 by Lemma 7, and we have x = x0 + e0 = cy + e0 .

If we exchange x and y in Lemma 9, we have the following lemma. Lemma 10. Let x, y ∈ V with x ∈ V1 \ {0} and y ̸∈ V1 . Then x ⊗c y = 0 if and only if x = c −1 (y + e0 ). From these lemmas, we have the following proposition. Proposition 11. For non-zero x, y ∈ V , we have x ⊗c y = 0 if and only if x = cy + e0 ̸∈ V1 in case y ∈ V1 , x = c −1 (y + e0 ) ∈ V1 in case y ̸∈ V1 with y ̸= e0 , and x = e0 in case y = e0 .

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We set d := rl. The dimension of V ⊕ ((V2 ⊗ V2 )/Wc ) over GF (2) is (rl + 1) + rl(l + 1)/2 = (rl2 + 3rl + 2)/2 = (1/2r )d2 + (3/2)d + 1 = ((1/r )d2 + 3d + 2)/2. Inside the vector space V (((1/r )d2 + 3d + 2)/2, 2) := V ⊕ ((V2 ⊗ V2 )/Wc ), for each t ∈ V , let us define a (d + 1)-dimensional vector subspace Xc (t ) by Xc (t ) := {(x, x ⊗c t ) | x ∈ V }. Theorem 12. Sc := {Xc (t ) | t ∈ V } is a d-dimensional symmetric bilinear dual hyperoval in V (((1/r )d2 + 3d + 2)/2, 2). Proof. As {x ⊗ y + Wc | x, y ∈ V1 } generates (V2 ⊗ V2 )/Wc and V1 ⊂ V , V (((1/r )d2 + 3d + 2)/2, 2) = V ⊕ ((V2 ⊗ V2 )/Wc ) is generated by {(x, 0) | x ∈ V } = Xc (0) and {(x, x ⊗c t ) | x ∈ V } = Xc (t ) for t ∈ V \ {0}. Hence the ambient space of Sc is V (((1/r )d2 + 3d + 2)/2, 2). For s ̸= t ∈ V , let Xc (s) ∩ Xc (t ) ∋ (x, x ⊗c s) = (x, x ⊗c t ) with x ̸= 0. Then we have x ⊗c s = x ⊗c t, i.e., x ⊗c (s + t ) = 0. From x ⊗c (s + t ) = 0 and Proposition 11, we have x = c (s + t ) + e0 ̸∈ V1 if s + t ∈ V1 , x = c −1 (s + t + e0 ) ∈ V1 if s + t ̸∈ V1 with s + t ̸= e0 , and x = e0 if s + t = e0 . Hence any two members Xc (s) and Xc (t ) with s ̸= t intersect in one dimensional subspace. Obviously, for mutually distinct elements s, t1 , t2 ∈ V , we have Xc (s)∩ Xc (t1 ) ̸= Xc (s)∩ Xc (t2 ) from the above result. Hence Xc (s)∩ Xc (t1 )∩ Xc (t2 ) = {0}. There are |V | = 2d+1 members in Sc (|V | is the cardinality of V ). By the definition of Wc , we have B(x, y) := x ⊗c y is symmetric (i.e., x ⊗c y = y ⊗c x) and bilinear (i.e., x ⊗c (y1 + y2 ) = x ⊗c y1 + x ⊗c y2 ). Therefore, Sc is a symmetric bilinear dual hyperoval in V (((1/r )d2 + 3d + 2)/2, 2). Remark 13. We see that, if r = 1, then GF (2r ) = GF (2), c = 1 and ((1/r )d2 + 3d + 2)/2 = (d + 1)(d + 2)/2, and our dual hyperoval Sc is exactly the Buratti–Del Fra dual hyperoval in [13]. If l = 1, the dimension of the ambient space ((1/r )d2 + 3d + 2)/2 = 2r + 1 = 2d + 1, and our dual hyperoval Sc is the dual hyperoval constructed by U. Dempwolff in Example 3.4 of [3]. In the notation there, Sβ with c replaced by 1/c correspond to Sc with l = 1. 3. Facts and propositions In this section, I follow the conventions of [4]. So, in this section, we will write, for example, the image of a set S under a mapping π by S π (instead of S π or π (S )). Let S be a d-dimensional dual hyperoval over GF (2) in the ambient space U. Then there exists a bijective map X : V → S , such that

S = {X (t ) | t ∈ V }. Define a mapping a := aS : V × V → U by a(s, s) = 0 for all s ∈ V and for s ̸= t, we define a(s, t ) by X (t ) ∩ X (s) = ⟨a(s, t )⟩. Note that a(s, t ) = a(t , s). We say that the dual hyperoval has property (B), if there exists a basis {e0 , . . . , ed } of V such that: (B1) {a(ei , 0) | 0 ≤ i ≤ d} ∪ {a(ei , ej ) | 0 ≤ i < j ≤ d} is a basis of U (i.e., dim U = (d + 1)(d + 2)/2). (B2) X (t ) = ⟨a(e, t ) | e ∈ B0 ⟩ where B0 = {0, e0 , . . . , ed }. All classical dual hyperovals in V ((d + 1)(d + 2)/2, 2), i.e., the Huybrechts dual hyperoval, the Buratti–Del Fra dual hyperoval, the Veronesean dual hyperoval and the deformation of Veronesean dual hyperoval, satisfy the property (B). Remark 14. For the Huybrechts dual hyperoval and the Buratti–Del Fra dual hyperoval, it is immediate to see that they satisfy the property (B). See for example [10]. As for the Veronesean dual hyperoval and the deformation of Veronesean dual hyperoval, to see that they satisfy the property (B), we need to change the suffix of the (d + 1)-dimensional subspace X (∞) of [7,9] by X (0), and ⟨a(s, s)⟩ = X (s) ∩ X (∞) by ⟨a(s, 0)⟩ = X (s) ∩ X (0) for s ̸= 0. Theorem 15. Let S be a d-dimensional dual hyperoval over GF (2) with property (B) and let π : U → W be an epimorphism, such that T = S π is a d-dimensional dual hyperoval with ambient space W . Let φ be an automorphism of T . Then there exists an automorphism φ of S , such that for all t ∈ V X (t )π φ = X (t )φπ . In particular Aut(S ) contains a subgroup isomorphic to Aut(T ). Proof. There exists a permutation ρ of V , such that X (t )π φ = X (t ρ)π for all t ∈ V . As π restricted to ∪X ∈S X is a bijection, we have ⟨a(s, t )π φ⟩ = X (s)π φ ∩ X (t )π φ = ⟨a(sρ, t ρ)π ⟩, i.e. a(s, t )π φ = a(sρ, t ρ)π for all s, t ∈ V . Using (B1), we define φ ∈ GL(U ) by a(u, v)φ = a(uρ, vρ) for u, v ∈ B0 . We want to prove the following claim: Claim: X (t )φπ = X (t )π φ

for all t ∈ V .

(1)

In particular φ is an automorphism of S . For each t ∈ V we have X (t ) = {a(s, t ) | s ∈ V } by the definition of a dual hyperoval. So for a(s1 , t ), a(s2 , t ) ∈ X (t ), there exists a unique ut (s1 , s2 ) ∈ V , such that a(s1 , t ) + a(s2 , t ) = a(ut (s1 , s2 ), t ).

H. Taniguchi / Discrete Mathematics 337 (2014) 65–75

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Then a(ut (s1 , s2 )ρ, t ρ)π = a(ut (s1 , s2 ), t )π φ

= = = =

(a(s1 , t ) + a(s2 , t ))π φ (a(s1 , t )π + a(s2 , t )π )φ (a(s1 ρ, t ρ) + a(s2 ρ, t ρ))π a(ut ρ (s1 ρ, s2 ρ)t ρ)π .

Hence (as π restricted to ∪X ∈S X is a bijection) a(ut (s1 , s2 )ρ, t ρ) = a(ut ρ (s1 ρ, s2 ρ), t ρ). By (B2) we deduce, that for each s, t ∈ V , we can write a(s, t ) =

m

a(wi , t )

i=0

with w1 , . . . , wm ∈ B0 . We set m(s, t ) = m, if m is minimal in such a representation (in particular m(t , t ) = 0). We denote by l(s) the number of nontrivial coefficients in the linear combination of s by the ei ’s. We show by induction on (l(t ), m(s, t )) that a(s, t )φ = a(sρ, t ρ).

(2)

For t = 0 (l(t ) = 0) or t = ei (l(t ) = 1) and m(s, t ) = 1 (i.e. a(s, t ) is a basis element of U) (2) holds by the definition of φ . Assume, (2) holds for all a(s, t ) with l(t ) < l0 and m(s, t ) < m0 . Assume, l(t ) = l0 , m(s, t ) = m = m0 > 1, and that m−1 m ′ ′ ′ ′ a(s, t ) = i=0 a(wi , t ). Define s by a(s , t ) = i=0 a(wi , t ), i.e., m(s , t ) < m(s, t ) and a(s, t ) = a(s , t ) + a(wm , t ). Hence ′ ′ ′ ′ a(s, t ) = a(ut (s , wm ), t ), i.e., s = ut (s , wm ). Then by induction a(s , t )φ = a(s ρ, t ρ), a(wm , t )φ = a(wm ρ, t ρ) and a(s, t )φ = a(s′ , t )φ + a(wm , t )φ

= = = =

a(s′ ρ, t ρ) + a(wm ρ, t ρ) a(ut ρ (s′ ρ, em ρ), t ρ) a(ut (s′ , wm )ρ, t ρ) a(sρ, t ρ).

Thus, by induction (2) holds for every a(s, t ) with l(t ) = l0 and arbitrary m(s, t ). Suppose l(t ) = l0 + 1 > l0 ≥ 1. By induction (and as a(s, t ) = a(t , s)) (2) holds if l(s) ≤ l0 , in particular if m(s, t ) ≤ 1. Now we can repeat the argument from above and (2) follows. Clearly, (2) implies (1). Thus φ is an automorphism of S and as the representation of the automorphism as a permutation on the dual hyperoval is faithful, we see, that the inverse of π (considered as a map of the set T onto the set S ) defines an equivalence map of permutation groups. The proof is complete. We say that a dual hyperoval S splits with respect to a subspace Y ⊂ ⟨S ⟩ if ⟨S ⟩ = X ⊕ Y for any X ∈ S . All known dual hyperovals split with respect to some subspaces. Definition 16 (Section 2 of [4]). We call T ⊂ Aut(S ) a translation group of Aut(S ) if T acts regularly on the elements (i.e., (d + 1)-subspaces) of S , and S splits with respect to C⟨S ⟩ (T ) := {u ∈ ⟨S ⟩ | uτ = u for any τ ∈ T }. Fact 17 (Theorem 3.2 of [4]). For a dual hyperoval S , Aut(S ) has a translation group T if and only if S is a bilinear dual hyperoval. For more details, see [4]. Fact 18 (Proposition 10 of [2]). The Huybrechts or the Buratti–Del Fra dual hyperoval S have only one translation group T ≃ 2d+1 , such that T is normal in Aut(S ). Moreover Aut(S ) = T : GL(V , 2) (semi-direct product of T and GL(V )) for the Huybrechts dual hyperoval, and Aut(S ) = T : H, where H < GL(V , 2) consists of the elements of GL(V , 2) which fix e0 for the Buratti–Del Fra dual hyperoval. Let us identify T and V as elementary abelian groups of size |T | = |V | = 2d+1 . Then any element (l, A) ∈ T : GL(V , 2) or (l, A) ∈ T : H above for the Huybrechts or the Buratti–Del Fra dual hyperoval acts on S as X (t ) → X (tA + l). The translation group T of the Huybrechts or the Buratti–Del Fra dual hyperoval consists of the elements (l, 1) for l ∈ V (1 is the identity element) which acts on S as X (t ) → X (t + l). Definition 19 (Section 2 of [4]). Let V = V (d + 1, 2) and let S = {X (t ) | t ∈ V } be a bilinear dual hyperoval with X (t ) = {(x, B(x, t )) | x ∈ V }, where B(x, t ) is a GF (2)-bilinear mapping. We call the set N := {ω | B(xω, y) = B(x, yω) for any x, y ∈ V } of GF (2)-linear automorphisms of V the nucleus of the dual hyperoval S . We say that a dual hyperoval has a trivial nucleus if N = {1}.

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An automorphism of a bilinear dual hyperoval S in V ⊕ W of Section 1 fixing V and W , i.e., V ⊕ W ∋ (x, y) → (xλ, yρ) ∈ V ⊕ W for some λ ∈ GL(V ) and ρ ∈ GL(W ), is called an autotopism. It is easy to see that there exists µ ∈ GL(V ) such that B(x, e)ρ = B(xλ, eµ) for any x, e ∈ V . So, it is convenient to denote an autotopism by a triple (λ, µ, ρ). We say that this autotopism is special if λ = µ. (See [4].) Lemma 20. Let S be a bilinear dual hyperoval as in Definition 19. Assume moreover that S is symmetric and has a trivial nucleus. Let (λ, µ, ρ) be any autotopism of S . Then λ = µ. That is, any autotopism is special. Proof. Using symmetry of B(x, y), we have B(xλ, yµ) = B(x, y)ρ = B(y, x)ρ = B(yλ, xµ) = B(xµ, yλ). Substituting xµ and yµ by x and y, we have B(xω, y) = B(x, yω), where ω = µ−1 λ. Since S has a trivial nucleus, we have ω = 1, hence λ = µ. Lemma 21. The Huybrechts dual hyperoval, the Buratti–Del Fra dual hyperoval, and the dual hyperoval Sc constructed in Section 2 have trivial nuclei. Proof. For theHuybrechts dual hyperoval, we have B(x, t ) = x ∧ t (see [10]). Let (xω) ∧ y = x ∧ (yω) for any x, y ∈ V . Write ei ω = 0≤j≤d aij ej for 0 ≤ i ≤ d with aij ∈ GF (2) for 0 ≤ j ≤ d. Expand the equation (ei ω) ∧ ek = ei ∧ (ek ω) for 0 ≤ i < k ≤ d. Then we obtain on both sides GF (2)-linear combinations in es ∧ et for 0 ≤ s < t ≤ d and a comparison of coefficients shows aij = 0 for 0 ≤ i, j ≤ d with i ̸= j and aii = 1 for 0 ≤ i ≤ d. Hence we have ω = 1. For the Buratti–Del Fra dual hyperoval, we have B(x, t ) = x ⊗ y (see Remark 3 and [13]). Write ei ω = 0≤j≤d aij ej for 0 ≤ i ≤ d with aij ∈ GF (2) for 0 ≤ j ≤ d. Expand the equation (ei ω) ⊗ ek = ei ⊗ (ek ω) for 0 ≤ i, k ≤ d with i ̸= k, and replace terms of the form ej ⊗ e0 for 0 ≤ j ≤ d by ej ⊗ ej . Then, by the same method as above, we have aij = 0 for 0 ≤ i, j ≤ d with i ̸= j and aii = 1 for 0 ≤ i ≤ d, and ω = 1. For the dual hyperoval Sc constructed in Section 2, we have B(x, t ) = x ⊗c y. Write ei ω = 0≤j≤l aij ej for i ∈ I with r ai0 ∈ GF (2) and aij ∈ GF (2 ) for j ∈ I0 . Expand the equation (ei ω) ⊗c ek = ei ⊗ (ek ω) for i, k ∈ I with i ̸= k, and replace terms of the form a(ej ⊗c e0 ), j ∈ I0 for some a ∈ GF (2r ) by ca(ej ⊗c ej ). Then as above, aij = 0 for i, j ∈ I with i ̸= j and aii = 1 for i ∈ I. Hence we have ω = 1. Theorem 22. Let V = V (d + 1, 2) (recall d + 1 > 4 by assumption), and let Bi : V ⊕ V → Wi , i = 1, 2, define bilinear dual hyperovals Si ⊆ Ui = V ⊕ Wi of the form Si = {Xi (v) | v ∈ V }, Xi (v) = {(u, Bi (u, v)) | u ∈ V }. Assume, that S1 is symmetric and that this dual hyperoval has a trivial nucleus. Assume further, that π : U1 → U2 is an epimorphism, which defines a covering map from S1 to S2 and that there is an isomorphism T ∋ t → t ∈ T ≤ Aut(S1 ), T a translation group of S2 , such that Xπ t = Xtπ ,

X ∈ S1 , t ∈ T .

Then T is a translation group of S1 . Proof. First we show: (i) We may assume, that T is the standard translation group of S2 . (ii) We may assume, that T normalizes the standard translation group of S1 . (iii) We may assume, that X1 (0)π = X2 (0). As translation groups of a dual hyperoval are conjugate in the automorphism group [4, Theorem 3.11], there is a

γ ∈ Aut(S2 ), such that T γ is the standard translation group. Set π ′ = π ◦ γ . Then π ′ is a covering map, such that X π ′ t γ = X π γ γ −1 t γ = X π t γ = X t π ′ ,

so that we may assume, that (i) holds. Let S be a Sylow 2-subgroup of Aut(S1 ) containing T and δ ∈ Aut(S1 ), such that S δ contains the standard translation group of S1 . Set π ′ = δ −1 ◦ π . Then π ′ is a covering map, such that δ

X δ t π ′ = X t π = X π t = X δπ ′ t , δ

so replacing t → t by t → t , we may assume, that T ≤ S δ . By [4, Theorem 3.11] S δ contains only one translation group (see also the last line of the proof of Theorem 3.11), i.e., the standard translation group, so that T normalizes this translation group and (ii) follows. Finally, choose τ ∈ T , such that X1 (0)π τ = X2 (0) and set π ′ = π ◦ τ . Then X1 (0)π ′ = X2 (0) and the computation for (i) shows X π ′t = X π ′t τ = X t π ′, so that we also get (iii). We now identify Xi (0), i = 1, 2, with V = V (d + 1, 2) and choosing a suitable basis (for instance of X2 (0)), we may assume, that the restriction of π to V is the identity. Let T1 be the standard translation group of S1 . Since T normalizes

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T1 , and since the normalizer NAut(S1 ) (T1 ) = T1 A, where A is a group consisting of autotopisms by [4, Lemma 2.5(b)], we deduce that for tv ∈ T with v ∈ V which acts as tv : V ⊕ W2 ∋ (x, y) → (x, y + B2 (x, v)) ∈ V ⊕ W2 , there exists a τv a ∈ T1 with v a ∈ V (which acts as τv a : V ⊕ W1 ∋ (x, y) → (x, y + B1 (x, v a )) ∈ V ⊕ W1 ) and an autotopism α(v) : V ⊕ W1 ∋ (x, y) → (xλ(v), yρ(v)) with B1 (x, w)ρ(v) = B1 (xλ(v), wµ(v)) (conventions as in [4]), such that t v = α(v)τv a . Since B1 is symmetric and has a trivial nucleus, we see by Lemma 20, that autotopisms are special, i.e., µ(v) = λ(v). Since X1 (t ) = {(x, B1 (x, t )) | x ∈ V } we have X1 (t )t v = X1 (t )α(v)τv a = {(xλ(v), B1 (x, t )ρ(v)) + (0, B(xλ, v a )) | x ∈ V } = {(xλ(v), B1 (xλ(v), t µ(v) + v a )) | x ∈ V }, hence X1 (t )t v = X1 (t µ(v) + v a ). From the regular action of T we deduce, that v → v a is a permutation on V , which fixes 0. Using t v+w = t v t w , we deduce

µ(v + w) = µ(v)µ(w) and (v + w)a = v a µ(w) + wa . Let W ′ be the inverse image of W2 with respect to π . Since S2 splits over W2 and as π restricted to W ′ ∩ X1 (v) for v ∈ V is injective, we conclude, that W ′ ∩ X1 (v) = {0} for v ∈ V and that S1 splits over W ′ . Hence there exist A ∈ Hom(W1 , V ) and an isomorphism B : W1 → W ′ , such that V ⊕W1 ≃ V ⊕W ′ by V ⊕W1 ∋ (x, y) → (x+yA, yB) ∈ V ⊕W ′ . Now, for (x, 0) ∈ V ⊕W1 we have (x, 0)t v π = (x, 0)π tv = (x, 0)tv = (x, B2 (x, v)) by (iii), while (x, 0)t v = (xµ(v)+ B1 (xµ(v), v a )A, B1 (xµ(v), v a )B) in V ⊕ W ′ , we get B2 (x, v) = B1 (xµ(v), v a )Bπ .

(3)

As T is a 2-group, there exists by a basic result about p-groups 0 ̸= x0 ∈ V , such that x0 µ(v) = x0 for all v ∈ V , i.e., B2 (x0 , v) = B1 (x0 , v a )Bπ

for all v.

(4)

The equation B2 (x0 , v + w) = B2 (x0 , v) + B2 (x0 , w) leads together with (4) to B1 (x0 , v a (µ(w) + 1))Bπ = 0 and using (3) we get B2 (x0 , (v a (µ(w) + 1))a

−1

) = 0 for all v, w ∈ V .

By the basic properties of a dual hyperoval, there exists a permutation f of V ∗ = V \ {0}, such that B2 (x, f (x)) = 0. −1

Thus (v a (µ(w) + 1))a ∈ ⟨f (x0 )⟩ or equivalently v(µ(w) + 1) ∈ ⟨z0 ⟩, z0 = f (x0 )a for all v, w ∈ V . Hence there exists α(v, w) ∈ GF (2), such that

vµ(w) = v + α(v, w)z0 . Clearly, α is additive in both arguments. We have by (3) B2 (x, v + w) = B1 (x + α(x, v + w)z0 , (v + w)a )Bπ

= B1 (x + α(x, v + w)z0 , v a + α(v a , w)z0 + wa )Bπ and B2 (x, v) + B2 (x, w) = B1 (x + α(x, v)z0 , v a )Bπ + B1 (x + α(x, w)z0 , w a )Bπ . As B2 is additive, we obtain

α(v a , w)B1 (x, z0 )Bπ = (gB1 (z0 , v a ) + hB1 (z0 , wa ) + kB1 (z0 , z0 ))Bπ for functions g , h, k in the variables x, v, w, z0 taking values in GF (2). Suppose, α(v a , w) ̸= 0 for some v, w ∈ V . Then the image of B1 (x, z0 )Bπ , which has dimension d (we use (3) again), lies in ⟨B1 (z0 , v a ), B1 (z0 , w a ), B1 (z0 , z0 )⟩Bπ , i.e., d ≤ 3, a contradiction. Thus α(v, w) = 0 for all v, w , i.e., µ(v) = 1 for all v . But then of course ρ(v) = 1, i.e., t v = τv a . The assertion follows. Remark 23. For the Huybrechts or the Buratti–Del Fra dual hyperoval S , there exists an elementary abelian subgroup E < Aut(S ) with |E | = 2d+1 which acts regularly on S such that E ̸= T (T is the translation group of S ). In case d + 1 = 5, for example, let V = V (5, 2) and put e0 := (0, 0, 0, 0, 1). For t = (t1 , t2 , t3 , t4 , t5 ), l = (l1 , l2 , l3 , l4 , l5 ) ∈ V , let α(t , l) := l4 t1 + l3 t2 + l2 t3 + l1 t4 and define A(l) ∈ H ≤ GL(V ) for H in Fact 18 by tA(l) := t + α(t , l)e0 for any t ∈ V (5, 2). Let E = {(l, A(l)) | l ∈ V } ≤ T : H. Then E acts on S by X (t ) → X (tA(l) + l). Hence we see that E is an elementary abelian subgroup of Aut(S ) = T : H (for the Buratti–Del Fra dual hyperoval) or T : GL(V ) (for the Huybrechts dual hyperoval) which acts regularly on S . However E is not conjugate to T in Aut(S ) by Fact 18. Proposition 24. Under the same assumptions as in Theorem 22, there exist two linear isomorphisms λ, µ : V → V , and a linear surjection ρ : W1 → W2 such that B2 (xλ, t µ) = B1 (x, t )ρ .

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Proof. Let π : U1 ⊃ S1 → S2 ⊂ U2 be a covering map from S1 to S2 . By Theorem 15, Aut(S2 ) is isomorphic to a subgroup of Aut(S1 ), hence T2 in Theorem 22, for the standard translation group T2 of S2 , is a subgroup of Aut(S1 ). As in the proof of Theorem 22, there exists σ ∈ Aut(S1 ) such that X1 (0)σ π = X2 (0) and T2 = T1σ , the standard translation group of S1 . Hence we have W1 σ π = (CV ⊕W1 (T1 ))σ π = CV ⊕W2 (T2 ) = W2 . Now σ π : V ⊕ W1 → V ⊕ W2 induces X1 (0) = V ⊕ {0} → V ⊕ {0} = X2 (0) and {0} ⊕ W1 → {0} ⊕ W2 . We define a linear bijection λ : V → V by σ π : X1 (0) ∋ (x, 0) → (xλ, 0) ∈ X2 (0), and a bijection µ : V → V by σ π : S1 ∋ X1 (t ) → X2 (t µ) ∈ S2 . As X1 (0)σ π = X2 (0), we have 0µ = 0. Also we can check X1 (t + a)σ π = X2 (t µ + aµ), hence (t + a)µ = t µ + aµ and the mapping µ is linear. Now we have (x, 0)σ π = (xλ, 0) and (x, B1 (x, t ))σ π = (xλ, B2 (xλ, t µ)). Since W1 σ π = W2 , there exists ρ = σ π|W1 : B1 (x, t ) → B2 (xλ, t µ) such that (x, B1 (x, t ))σ π = (xλ, B1 (x, t )ρ). Therefore we have B1 (x, t )ρ = B2 (xλ, t µ). Proposition 25. Let the assumptions be given as in Theorem 22 and let S2 be a dual hyperoval Sc constructed in Section 2. Then we have λ = µ in Proposition 24. Proof. Using symmetry of B1 (x, y) for S1 and B2 (x, y) = x ⊗c y for S2 = Sc in Proposition 24, one gets (xλ) ⊗c (yµ) = (xµ) ⊗c (yλ) and substituting xµ and yµ by x and y one gets (xω) ⊗c y = x ⊗c (yω), where ω = µ−1 λ. By Lemma 21, we have ω = 1. Hence we have λ = µ. 4. Sc is not a quotient of Buratti–Del Fra DHO SB if c ̸= 1 Let V and V1 be as in Section 2. We use the expression of Buratti–Del Fra dual hyperoval SB described in Fact 2, that is, SB := {XB (t ) | t ∈ V } where XB (t ) = {(x, x ⊗ t ) | x ∈ V } ⊂ V ⊕ ((V ⊗ V )/W ) for any t ∈ V . We recall, for non-zero x, y ∈ V , x ⊗ y = 0 if and only if x = y + e0 or x = y = e0 .

Remark 26. For c = 1 and r is odd (which means Tr(c ) = Tr(1) = 1), Sc is a quotient of Buratti–Del Fra dual hyperoval SB by Lemma 4. (3) of [13]. In this section, we will show the following: Theorem 27. The dual hyperoval Sc is not a quotient of SB if c ̸= 1. We assume that Sc is a quotient of SB , and deduce a contradiction. Since Sc is a quotient of SB by assumption, from Proposition 24, there exist two GF (2)-isomorphisms λ, µ : V → V and a GF (2)-surjection ρ : (V ⊗ V )/W → (V2 ⊗ V2 )/Wc , such that

(x ⊗ y)ρ = xλ ⊗c yµ for any x, y ∈ V . We recall that the Buratti–Del Fra dual hyperoval satisfies the property (B), and has trivial nucleus by Lemma 21, hence satisfies the assumption of Proposition 25, therefore λ = µ. Then we have eλ0 = e0 because 0 = e0 ⊗ e0 = eλ0 ⊗c eλ0 and by Proposition 11. Proof of Theorem 27. We have (x ⊗ y)ρ = xλ ⊗c yλ for x, y ∈ V by Proposition 25. However, if x ̸= 0 and xλ ∈ V1 , we have (x ⊗ x)ρ = (e0 ⊗ x)ρ = eλ0 ⊗c xλ = e0 ⊗c xλ = c (xλ ⊗ xλ ). As c ̸= 1, we have a contradiction. 5. Sc is not a quotient of Huybrechts DHO, Veronesean DHO and the deformation of Veronesean DHO Huybrechts dual hyperoval is defined by SH := {XH (t ) | t ∈ V }, where XH (t ) := {(x, x ∧ t ) | x ∈ V } ⊂ V ⊕(V ∧ V ) for any t ∈ V . (SH is originally constructed in [5]. As for the expression above, see [10].) Firstly, we prove the following proposition in this section. Proposition 28. Sc is not a quotient of the Huybrechts dual hyperoval. Proof. We recall that the Huybrechts dual hyperoval satisfies the property (B), and has trivial nucleus by Lemma 21, hence satisfies the assumption of Proposition 25, therefore we have (x ∧ y)ρ = xλ ⊗c yλ for x, y ∈ V by Proposition 25 with some λ and ρ . However, if x ̸= 0 and xλ ∈ V1 , we have 0 = (x ∧ x)ρ = xλ ⊗c xλ ̸= 0, a contradiction. We now use the description of bilinear dual hyperovals and translation groups given in Section 1. Recall that the translation subgroup T = {ta | a ∈ V } ⊂ Aut(S ) of any bilinear dual hyperoval S acts regularly on the members of S . Proposition 29. The bilinear dual hyperoval Sc is not a quotient of the Veronesean dual hyperoval SV , or the deformation of Veronesean dual hyperoval SD . Proof. Assume Sc is a quotient of the Veronesean dual hyperoval SV , or a quotient of the deformation of Veronesean dual hyperoval SD , respectively. Since SV and SD satisfy property (B), we see Aut(SV ) or Aut(SD ) has a subgroup G isomorphic to Aut(Sc ) by Theorem 15. Also the action of G on SV and Aut(SD ) respectively is permutation equivalent to the action of Aut(Sc ) on Sc , a contradiction, since Aut(Sc ) is transitive, whereas Aut(SV ) and Aut(SD ) are not transitive (see [7,12]).

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6. Sc1 ≃ Sc2 if and only if c2 = c1σ for some σ ∈ Gal(GF (2r )/GF (2)) Recall our dual hyperoval Sc is a d = rl-dimensional dual hyperoval with c ∈ GF (2r ). Let r = r1 and l = l1 and c1 ∈ GF (2r1 ) for a dual hyperoval Sc1 , and r = r2 and l = l2 and c2 ∈ GF (2r2 ) for a dual hyperoval Sc2 . We note that, if Sc1 is isomorphic to Sc2 , then r1 = r2 and l1 = l2 . (Since the dimensions of the dual hyperovals are equal, we have d = r1 l1 = r2 l2 , and since the dimensions of the ambient spaces are equal, we have (r1 l21 + 3r1 l1 + 2)/2 = (r2 l22 + 3r2 l2 + 2)/2. From these equations, we easily see r1 = r2 and l1 = l2 .) Let d = nr ≥ 4. We assume l = l1 = l2 and r = r1 = r2 for the dual hyperovals Sc1 and Sc2 . We will show the following theorem. Theorem 30. Let c1 and c2 be two elements in GF (2r ) such that the absolute traces are Tr(c1 ) = 1 and Tr(c2 ) = 1. Then the dual hyperoval Sc1 is isomorphic to Sc2 if and only if there exists an automorphism σ of GF (2r ) over GF (2) such that c2 = c1σ . We use the following result of Dempwolff and Edel [4]. Fact 31 (Proposition 3.13 of [4]). Let d ≥ 3. Let Si := {Xi (e) | e ∈ V }, where Xi (e) := {(x, Bi (x, e)) | x ∈ V } ⊂ V ⊕ W be d-dimensional bilinear dual hyperovals in V ⊕ W with GF (2)-bilinear mappings Bi : V ⊕ V → W for i = 1, 2. Then S1 is isomorphic to S2 if and only if there exist two GF (2)-isomorphisms λ, µ : V → V and a GF (2)-isomorphism ρ : W → W , such that (B1 (x, y))ρ = B2 (xλ , yµ ) for any x, y ∈ V . From Remark 26 and Theorem 27, we see that Sc is isomorphic to S1 if and only if c = 1 (if both Sc and S1 are defined). So, from now on, we assume that c1 ̸= 1 and c2 ̸= 1. Firstly, we assume that Sc1 is isomorphic to Sc2 . Then, from Fact 31, there exist two GF (2)-isomorphisms λ, µ : V → V and a GF (2)-isomorphism ρ : (V ⊗ V )/Wc1 → (V ⊗ V )/Wc2 , such that

(x ⊗c1 y)ρ = xλ ⊗c2 yµ for any x, y ∈ V . µ µ µ µ Lemma 32. (a) We have eλ0 = e0 if and only if e0 = e0 .(b) We have eλ0 + e0 ̸∈ V1 if eλ0 = ̸ e0 , or if e0 = ̸ e0 .(c) eλ0 + e0 ̸= 0 µ implies eλ0 + e0 ̸∈ V1 . µ

µ

µ

µ

Proof. (a) Since e0 ⊗c1 e0 = 0, we have eλ0 ⊗c2 e0 = e0 ⊗c2 eλ0 = 0. If eλ0 = e0 , then e0 ⊗c2 e0 = 0, hence e0 = e0 by µ Proposition 11. By the same way, if e0 = e0 , we have eλ0 ⊗c2 e0 = 0, hence eλ0 = e0 by Proposition 11. (b) Assume eλ0 ∈ V1 µ µ µ λ λ and e0 ∈ V1 . From (ce0 + e0 ) ⊗c2 e0 = 0 by Proposition 11 and e0 ⊗c2 eλ0 = 0, we have V1 ∋ e0 = ceλ0 + e0 ̸∈ V1 , a µ µ λ λ λ −1 λ contradiction. Next, if e0 ̸∈ V1 and e0 ̸∈ V1 with e0 ̸= e0 and e0 ̸= e0 , then since c (e0 + e0 ) ⊗c2 e0 = 0 by Proposition 11 µ µ µ and e0 ⊗c2 eλ0 = 0, we also have a contradiction V1 ̸∋ e0 = c −1 (eλ0 + e0 ) ∈ V1 . Hence, if eλ0 ̸= e0 or e0 ̸= e0 , we must have µ µ µ µ λ λ λ λ e0 ∈ V1 and e0 ̸∈ V1 , or e0 ̸∈ V1 and e0 ∈ V1 from (a). Therefore, we have e0 + e0 ̸∈ V1 . (c) If e0 + e0 ̸= 0, then eλ0 ̸= e0 and µ µ e0 ̸= e0 by (a), hence eλ0 + e0 ̸∈ V1 by (b). Proof of Theorem 27. We recall V = V1 ⊕ ⟨e0 ⟩ as GF (2)-vector space, and λ, µ : V → V are GF (2)-linear isomorphisms. µ−1

Let Vλµ be a vector space defined by Vλµ := V1 ∩ V1λ ∩ V1 . Then, since d = rn ≥ 4 by assumption, the dimension dimGF (2) Vλµ ≥ 2. Let x be a non-zero element of Vλµ . From the equations (x ⊗c1 (c1 x + e0 ))ρ = xλ ⊗c2 (c1 x + e0 )µ = 0 and xλ ⊗c2 (c2 xλ + e0 ) = 0 by Proposition 11, we have −1

µ

c2 xλ + e0 = (c1 x)µ + e0 .

(5)

From the equations ((c1 x + e0 ) ⊗c1 x)ρ = (c1 x + e0 )λ ⊗c2 xµ = 0 and (c2 xµ + e0 ) ⊗c2 xµ = 0 by Proposition 11, we have c2 xµ + e0 = (c1 x)λ + eλ0 .

(6)

We add these equations, then we have µ

c2 (xλ + xµ ) = ((c1 x)λ + (c1 x)µ ) + (eλ0 + e0 ). λ

µ

(7) λ

λ

µ

µ

Since Vλµ ∋ x → c2 (x + x ) ∈ V1 and Vλµ ∋ x → (c1 x) + (c1 x) ∈ V1 are linear mappings, e0 + e0 must be 0. (Here, we µ use the assumption that |Vλµ | ≥ 4.) Thus, eλ0 = e0 = e0 by Lemma 32. µ µ Now, for x ∈ V1 \ {0} with x ∈ V1 , since c2 x + e0 = (c1 x)λ + eλ0 as we obtained above, and since eλ0 = e0 , we must have c2 xµ = (c1 x)λ

if xµ ∈ V1 .

(8) µ

µ

µ

We show now, that V1 is invariant under λ and µ. Assume V1 ̸= V1 . As |V1 − (V1 ∩ V1 )| = 2d − 2d−1 (V1 is a hyperplane µ of V ), there exists x, y ∈ V1 with x ̸= y such that xµ , yµ ̸∈ V1 . Let x ∈ V1 \ {0} and xµ ̸∈ V1 . Then, xµ ̸= e0 since e0 = e0 −1 µ ρ λ µ µ and µ is a bijection. From the equations ((c1 x + e0 ) ⊗c1 x) = (c1 x + e0 ) ⊗c2 x = 0 and c2 (x + e0 ) ⊗c2 x = 0 by Proposition 11 (note that xµ ̸∈ V1 ), we have c2−1 (xµ + e0 ) = (c1 x)λ + eλ0 = (c1 x)λ + e0 .

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If there exist x, y ∈ V1 with x ̸= y such that xµ ̸∈ V1 and yµ ̸∈ V1 , then c2−1 (xµ + e0 ) = (c1 x)λ + e0 and c2−1 (yµ + e0 ) = (c1 y)λ + e0 , by the same reason as above. Adding these equations, we have c2−1 (x + y)µ = (c1 (x + y))λ . However, since (x + y)µ = xµ + yµ ∈ V1 (we obtain xµ + yµ ∈ V1 from xµ ̸∈ V1 and yµ ̸∈ V1 , and V = V1 ⊕ ⟨e0 ⟩), we have c2 (x + y)µ = (c1 (x + y))λ as we see in (8). Thus we must have c2−1 (x + y)µ = c2 (x + y)µ . µ

Since c2 ̸= 1 by assumption and (x + y)µ ̸= 0, we have a contradiction. Hence V1 = V1 , and by symmetry, V1λ = V1 . µ

Since x ∈ V1 and x ∈ V1 for x ∈ V1 , we have c2 x + e0 = (c1 x) + e0 and c2 xµ + e0 = (c1 x)λ + eλ0 as we see before, µ and from eλ0 = e0 = e0 , we have µ

c2 xµ = (c1 x)λ

λ

and

λ

c2 xλ = (c1 x)µ

µ

for any x ∈ V1 .

(9)

Eq. (9) shows µLc2 = Lc1 λ and λLc2 = Lc1 µ, where Lc denotes the left multiplication by c (consider as a GF (2)-linear map). Thus µLc2 λ−1 = Lc1 and λLc2 µ−1 = Lc1 , so that

µL2c2 µ−1 = µLc2 λ−1 λLc2 µ−1 = L2c1 . Since multiplicative groups of finite fields of characteristic 2 have odd order, we get µLkc2 µ−1 = Lkc1 for all k. Thus

σ : x → µ−1 xµ, which is a ring homomorphism, induces a field isomorphism of GF (2)[Lc1 ] onto GF (2)[Lc2 ]. Hence we may regard σ ∈ Gal(GF (2r )/GF (2)), the Galois group of GF (2r ) over GF (2). Thus, we have c1σ = c2 as desired. Conversely, let us assume c1σ = c2 for some σ ∈ Gal(GF (2r )/GF (2)). Recall V1 is an l-dimensional GF (2r )-vector space with a basis {e1 , . . . , el }, V2 is an (l + 1)-dimensional GF (2r )-vector space with a basis {e0 , e1 , . . . , el }, and a symmetric tensor space Sym(V1 ) is an l(l + 1)/2-dimensional vector space over GF (2r ) with a basis {ei ⊗ ej | 1 ≤ i ≤ j ≤ l} (see Section 2). Let us define a GF (2)-linear mapping ρ : Sym(V1 ) → Sym(V1 ) by ρ = xσij ei ⊗ ej for xij ∈ GF (2r ) with 1 ≤ i ≤ j ≤ l. xij ei ⊗ ej For x = x1 e1 + · · · + xl el ∈ V1 with x1 , . . . , xl ∈ GF (2r ), we define a GF (2)-linear mapping σ : V1 → V1 by xσ := xσ1 e1 + · · · + xσl el . Then, for x, y ∈ V1 , we can easily check that

(x ⊗ y)ρ = xσ ⊗ yσ ∈ Sym(V1 ). Since Sym(V1 ) ≃ (V2 ⊗ V2 )/Wci for i = 1, 2 as GF (2r )-vector spaces induced by a natural injection V1 ⊗ V1 → V2 ⊗ V2 (see Section 2), we are able to regard ρ as a GF (2)-linear bijection ρ : (V2 ⊗ V2 )/Wc1 → (V2 ⊗ V2 )/Wc2 by ρ : (V2 ⊗ V2 )/Wc1 ≃ Sym(V1 ) → Sym(V1 ) ≃ (V2 ⊗ V2 )/Wc2 . Now recall that, for x, y ∈ V1 , the natural injection V1 ⊗ V1 → V2 ⊗ V2 induces the mappings Sym(V1 ) ∋ x ⊗ y → x ⊗c1 y ∈ (V2 ⊗ V2 )/Wc1 and Sym(V1 ) ∋ x ⊗ y → x ⊗c2 y ∈ (V2 ⊗ V2 )/Wc2 , and that we have e0 ⊗c1 x = (c1 x) ⊗c1 x in (V2 ⊗ V2 )/Wc1 and (c2 xσ ) ⊗c2 xσ = e0 ⊗c2 xσ in (V2 ⊗ V2 )/Wc2 . Then we see that ρ satisfies the following equations, for x , y ∈ V1 :

(x ⊗c1 y)ρ = (x ⊗ y)ρ = xσ ⊗ yσ = xσ ⊗c2 yσ , and (e0 ⊗c1 x)ρ = ((c1 x) ⊗c1 x)ρ = (c1σ xσ ) ⊗ xσ = (c2 xσ ) ⊗c2 xσ = e0 ⊗c2 xσ . Thus, by linearity, and since e0 ⊗c1 e0 = e0 ⊗c2 e0 = 0, ρ satisfies

((x0 + α e0 ) ⊗c1 (y0 + β e0 ))ρ = (xσ0 + α e0 ) ⊗c2 (yσ0 + β e0 )

(10)

for x0 , y0 ∈ V1 and α, β ∈ GF (2). Let us define a GF (2)-linear bijection λ : V = V1 ⊕ ⟨e0 ⟩ ∋ x1 e1 + · · · + xl el + α e0 → xσ1 e1 + · · · + xσl el + α e0 ∈ V = V1 ⊕ ⟨e0 ⟩ for α ∈ GF (2) and x1 , . . . , xl ∈ GF (2r ). Then, for x, y ∈ V , we have

(x ⊗c1 y)ρ = xλ ⊗c2 yλ from (10). Therefore, by Fact 31, Sc1 is isomorphic to Sc2 .

Acknowledgments I would like to thank the referees for their helpful comments and suggestions. This work was supported by JSPS KAKENHI Grant No. 26400029.

H. Taniguchi / Discrete Mathematics 337 (2014) 65–75

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