A family of translation planes

An infinite family of non-Desarguesian translation planes of order q with kernel GF(q) is constructed, for any odd prime power q. The collineation group of each plane has orbits of lengths 1, q, and q − q on the translation line. The method used for the construction is net replacement, starting from a Dickson-Knuth semifield plane. The planes are constructed via spreads and the spreads via spread sets. A spread set is a set S of q matrices, each 2 by 2, with entries in GF(q) such that the difference of any two of them is non-singular. Given a spread set S, a spread πS of PG(3, q) arises. (The elements of πS in GF(q) 4 are the line x1 = x2 = 0 and the lines {(x1, x2, x3, x4) | (x1, x2)A = (x3, x4)}, for A ∈ S.) Applying the Andre/Bruck-Bose construction to a spread of PG(3, q) gives a translation plane of order q with kernel containing GF(q) [1]. Let F = GF(q) > GF(q) = K, q odd, with corresponding automorphism x → x = x̄ defining the conjugate x̄ of x, and the corresponding norm x → x = xx̄, x ∈ F . Fix η ∈ F, η / ∈ K such that η̄ = −η; in general if ᾱ = −α, then α is skew-symmetric. Remark 1 If α, β ∈ F , with ᾱ = −α, then α± β = 0 implies α = β = 0. (This follows from adding the equation to its conjugate.) We start with two copies of a Dickson-Knuth semifield plane. Lemma 1 The following sets of matrices are spread sets closed under addition. Δ = {( s t s̄+ t̄ ηs̄+ t̄ ) | s, t ∈ F } ; Δ− = {( s t s̄− t̄ ηs̄+ t̄ ) | s, t ∈ F } . 54 A. HUDSON AND T. PENTTILA Proof. Since Δ and Δ− are closed under addition, we need only check that their non-zero elements are nonsingular. Since η̄ = −η, the determinants of the generic elements of Δ and Δ− are the sum of a skew symmetric part (ηss̄+ (st̄− s̄t)) and ±tν , which by the remark is non-zero if s and t are non-zero. Fix c ∈ F ∗ to be a non-square, and let δ1,2 denote the entry at location (1, 2) of any matrix M ∈ Δ ∪Δ−. Then M is considered square or non-square according to whether δ1,2 = θ , θ ∈ F ∗ or δ1,2 = cθ, θ ∈ F ∗, respectively. Let Sq(Δ±) and NSq(Δ±) be respectively the partial spread sets of all square and non-square matrices in Δ±. We claim that the partial subspread associated withNSq(Δ) ⊂ Δ and NSq(Δ−) ⊂ Δ− are replacements of each other. This will follow by noting the following. Lemma 2 If A ∈ NSq(Δ), B ∈ Sq(Δ−), then det(B − A) = 0. Proof. The difference of distinct typical elements may be expressed (using the notation σ = s− s′ ) as ( s θ s̄− θ̄ ηs̄+ θ̄ ) − ( s′ cφ s̄+ ̄ cφ2 ηs+ ̄ cφ2 ) = ( σ θ − cφ σ̄ − (θ2 + cφ2) ησ̄ + (θ2 − cφ2) ) . Equating the determinant to zero, ησσ̄ + (σ(θ2 − cφ2)− σ̄(θ − cφ)) + (θ2 + cφ2)((θ − cφ)) = 0. Now ησσ̄ + (σ(θ2 − cφ2)− σ̄(θ − cφ)) is skew-symmetric. Expanding (θ2 + cφ2)((θ − cφ)) gives (θθ̄ − cc̄φ2φ̄2)− cφθ̄ − θcφ2, and, again, −cφ2θ̄2 − θcφ2 is skew-symmetric. Thus adding the determinant to its conjugate gives (θθ̄ − cc̄φφ̄) = 0. Thus c is a square inK, contrary to c being a non-square in F , as ν is surjective. Corollary 1 Δ := Sq(Δ) ∪NSq(Δ−) is a spread set, and the associated spread πΔ is a replacement of the Knuth semifield spread πΔ+ obtained by replacing the partial spread associated with NSq(Δ) by the partial spread associated with NSq(Δ−), of the Knuth spread Δ−. A FAMILY OF TRANSLATION PLANES 55 Let m = (q + 2q − 1)/2. Then t = −tq for non-square t and similarly t = t for square t. Then Δ = { Ss,t = ( s t s + t ηs + t ) | s, t ∈ GF(q) } . Let π be the plane (of order q) arising from the spread πΔ via the Andre/Bruck-Bose construction. Theorem 1 The plane π is non-Desarguesian, and the collineation group Autπ of π has orbits of lengths 1, q and q − q on the translation line. Proof. If S is any spread set then the additive group Σ = {A ∈ S : S + A ⊂ S} corresponds to the y-axis elations. So a fixed matrix A = Ss1,t1 corresponds to a y-axis elation, if and only if S +A ⊂ S, if and only if t − t1 = (t − t1) for all t, which implies t1 = 0. Thus the full elation group Σ with axis the y-axis has order q, from which it follows that π is neither Desarguesian nor a semifield. We claim that Σ consists of all elations in the translation complement C of π. If not, by the Hering-Ostrom theorem [1] Δ admits SL(2, q), so, by the Schaeffer-Walker theorem [1] π is either a Hall plane (but this cannot have an elation group of order greater than 2) or a Hering plane (but this cannot have a kernel of square order). So Σ is a normal subgroup of C. Now π admits the homology group HY = {Diag(x2, x̄, 1, 1) : x ∈ F}, of order (q − 1)/2, and the homology group HX = {Diag(1, 1, y, y) : y ∈ F}, of order (q − 1)/2. Moreover, the orbits of G = 〈Σ, HX , HY 〉 on the translation line of π have lengths 1, q, q − q 2 , q − q 2 . Let O be the orbit of G of length q. There is an element δ of ΓL(4, q) fixing Δ, O and the orbit of G of length 1, and interchanging the two remaining orbits of G, namely that induced by the map A → X−1 1 ĀX2, where X1 = ( 1 0 0 −1 ) , and X2 = ( 1 0 η2+1 η η )