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DOI 10.1515 / ADVGEOM.2009.030 de Gruyter 2010

Nearly flag-transitive affine planes

William M. Kantor and Michael E. Williams

Dedicated to Tim Penttila on the occasion of his 50th birthday (Communicated by T. Grundh¨ofer)

Abstract. Spreads of orthogonal vector spaces are used to construct many translation planes of even orderqm, for oddm >1, having a collineation with a(qm−1)-cycle on the line at infinity and on each of two affine lines.

Key words. Affine plane, translation plane, spread, flag, symplectic geometry, orthogonal geome- try.

2000 Mathematics Subject Classification. Primary 51A40; Secondary 05E20, 51A35, 51A50

1 Introduction

In [13, 14] we used the relationship between symplectic and orthogonal geometries in characteristic 2 in order to construct new affine planes: flag-transitive planes in [13], and semifield planes in [14]. In this paper we continue those papers by proving the following theorem (whereρ(m)denotes the number of prime factors ofm, counting multiplicities, and logarithms are always to the base 2):

Theorem 1.1. Letq≥4be a power of2, and letm >1be an odd integer. Then there are more thanq3ρ(m)−2 pairwise nonisomorphic translation planes of orderqm, with kernel of orderq, for which there is a collineation of orderqm−1having a(qm−1)-cycle on the line at infinity and on each of two affine lines.

For better estimates on the number of planes in the theorem, see Theorem 9.2 and Corollary 9.3. These planes are constructed using explicitly definedprequasifields (cf.

(4.2)). They arenearly flag-transitive affine planes: their collineation groups have 2 or 3 flag-orbits. (In [6, p. 794] these are called “triangle-transitive planes”, which suggests even more transitivity than we obtain.)

This research was supported in part by NSF grants DMS 9301308 and DMS 0753640.

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The proof combines methods in [8] and [18]: these planes arise using symplectic and orthogonal spreads together with changing from fields to proper subfields, keeping track of these field changes using kernels of the associated planes. We settle the isomorphism problem for these planes using an elementary Sylow argument (Proposition 5.2 and The- orem 8.5). A similar argument is used in [13, Proposition 5.1] (and in [9, III.C]) for flag-transitive planes; but no such argument is possible for the semifield planes in [14].

We also relate the present planes to those in [13]:

Theorem 1.2. There is a natural bijection between the set of isomorphism classes of flag- transitive planes in[13]and a subset of the isomorphism classes of nearly flag-transitive planes in Theorem1.1; each plane in[13]is a Baer subplane of the corresponding plane in Theorem1.1.

Our prequasifields all have the form(F,+,∗)withFa finite field andx∗y=yL(xy) for an additive mapL:F →Fsuch thatx→xL(x)is a permutation ofF. Some of these mapsLare essentially in [8, II p. 312], and also appear in [1]. In view of Theorem 1.1, one of our goals is to produce large numbers of such mapsL.

The planes studied in [7, 8, 10–14] and here are symplectic translation planes. Re- markable results of Maschietti [16] use line ovals to distinguish such planes among all translation planes of characteristic 2.

2 Up and down: from symplectic to orthogonal spreads and back Throughout this paper, all fields are finite of characteristic 2. We briefly review some of the background required from [3] and [8, 14]. The best background source is probably [11], with the coding-theoretic aspects discarded.

Spreads. Consider a 2m-dimensional vector spaceW overK = GF(q). Aspread in W is a familySofqm+1 subspaces of dimensionmthat partition the nonzero vectors.

The corresponding affine translation planeA = A(S)has as points the vectors of V and as lines all cosets of members ofS [3, pp. 131–133]. The collineations fixing 0 and stabilizing every member ofSgenerate a fieldK(A), called thekernelof the plane. This is the largest field over which the members ofScan be viewed as subspaces.

The full collineation group ofAis

AutA=Wo(AutA)0=WoΓL(W)S,

whereΓL(W)S = (AutA)0denotes the setwise stabilizer ofS, and is a group of semi- linear transformations overK(A). More generally, any isomorphism between translation planes sends the kernel of one to the kernel of the other.

One standard way translation planes are constructed is through the use of coordinatiz- ingquasifieldsorprequasifields[3, p. 129]. We will see these starting in Section 3.

Symplectic spreads. If there is also a nondegenerate alternating bilinear form(,)onW such that eachX ∈ Sis totally isotropic (i.e.,(X, X) =0), thenSis called asymplectic spreadandA(S)is asymplectic translation plane.

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Orthogonal spreads. Consider a 2m+2-dimensional vector spaceV overK= GF(q), equipped with a nondegenerate quadratic formQof maximal Witt index. This means that there arem+1-spacesX that aretotally singular(i.e.,Q(X) = 0). Moreover, V has (qm+1−1)(qm+1)nonzero singular vectors.

Anorthogonal spread ofV is a familyΣofqm+1 totally singularm+1-spaces that partition the set of nonzero singular vectors. Note that an orthogonal spread is not a spread in the sense of the earlier definition. We recall that there are two types of totally singularm+1-spaces such that totally singularm+1-spacesX andY have the same type if and only ifdimX ∩Y ≡m+1 (mod 2) [17, pp. 170–172]. Thus,ifV has an orthogonal spread thenmmust be odd.

Down: from orthogonal spreads to symplectic spreads. There is also a symplectic structure on the orthogonal vector spaceV, determined by the nondegenerate alternating bilinear form(u, v) :=Q(u+v)−Q(u)−Q(v). Ifzisanynonsingular point ofV, then z/z inherits the nondegenerate alternating bilinear form (u+z, v+z) := (u, v)for u, v∈z. IfX∈ΣthenhX∩z, zi/zis anm-dimensional totally isotropic subspace.

Moreover,

Σ/z:={hX∩z, zi/z|X∈Σ} (2.1) is a symplectic spread ofz/z.

Up: from symplectic spreads to orthogonal spreads. This process can be reversed:

any symplectic spreadSofz/zcan be lifted to an essentially unique orthogonal spread ΣSofV such thatΣS/z=S(see [4,5] and [8, I]). We will exhibit such a lifting explicitly in (3.7).

The simplest example of this lifting process was studied at length in [8, I]. It arises from the orthogonal spreadΣthat determines the desarguesian planeA(S), and hence is called the (orthogonal)desarguesian spread.

Down: from symplectic spreads to symplectic spreads. Given a vector spaceV over a fieldF, with associated nondegenerate alternating bilinear form(,), ifKis a subfield ofF andT: F → Kis the trace map, then T(,)defines a nondegenerate alternating K-bilinear form on theK-spaceV. IfSis a symplectic spread of theF-spaceV thenS is also a symplectic spread of theK-spaceV.

Scions. LetSbe a symplectic spread. Suppose thatS0is another symplectic spread aris- ing via a (repeated) up and down process of passing between symplectic and orthogonal geometries, or passing to subfields, as above. Then we callS0ascionofS[18]. IfS0is a scion ofSthenA(S0)will be called ascionofA(S).

Instances of this notion are crucial for Theorem 1.1.

Groups. For a symplectic vector space W, letΓSp(W)(or ΓSp(WK)if we need to specify the underlying fieldK) denote the subgroup ofΓL(W)consisting of thoseg ∈ ΓL(W) that preserve the symplectic structure, so that (ug, vg) = k(u, v)α for some k ∈ K, someα ∈ Aut(K)and all u, v ∈ W. Similarly, for an orthogonal vector

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spaceV, letΓO+(V)(orΓO+(VK)) denote the group of allg∈ΓL(V)that preserve the orthogonal structure determined by the quadratic formQ, so thatQ(vg) = kQ(v)αfor somek∈K, someα∈Aut(K)and allv∈V.

Equivalences. Theautomorphism groupof an orthogonal spreadΣis the set-stabilizer ΓO+(V)Σ of ΣinΓO+(V). Two orthogonal spreads areequivalent if some element of ΓO+(V)sends one to the other. According to [8, Theorem 3.5 and Corollary 3.6]

(cf. [12]), equivalences among orthogonal spreads are closely related to isomorphisms among the affine planes they spawn:

Theorem 2.2. LetSi be a symplectic spread in the symplecticK-spaceWi,i = 1,2, such thatA(S1)andA(S2)are isomorphic.

(i) There is aK-semilinear mapg: W1 → W2 sending S1 toS2 and preserving the symplectic structure.

(ii) Suppose that there are orthogonal spreadsΣiin an orthogonal vector spaceV, and nonsingular pointsziofV, such thatSi = Σi/zifori=1,2. Then some element ofΓO+(V)sendsΣ1toΣ2andz1toz2.

Moreover, AutA(S1) = Wo K(A(S1))ΓSp(W)S1

in (i) [8, Theorem 3.5 and Corollary 3.6].

Methodological remarks. The common thread in [2, 7, 8, 10, 13, 14] and the present paper is the use of scions of desarguesian planes. In those references and here, the specific up and down process employed is designed to preserve subgroups ofSL(2, qm)that fix nonsingular points in the 2m+2-dimensional orthogonal space obtained by lifting the desarguesian spread. A start in this direction can be seen in [8], using subgroups of orderqm+1,qmorqm−1 in order to obtain flag-transitive, semifield and nearly flag- transitive planes, respectively. These groups also appear in [13], in [14] and here, but using arbitrarily long chains of subfields and hence of up and down moves. Moreover, up to conjugacy (the normalizers of) these groups are precisely the stabilizers inSL(2, qm) of nonsingular points in the space underlying the (orthogonal) desarguesian spread; no further planes can be obtained in this up and down manner that are preserved by other subgroups ofSL(2, qm).

Slight variations on these constructions are used in [7, 14] to obtain affine planes whose full collineation groups are unusually small.

At the moment,everyknown symplectic plane in characteristic 2 having odd dimen- sion over its kernel is a scion of a desarguesian plane. There must be many others, but we have no idea where to look for them.

3 Prequasifields

Consider finite fieldsF ⊇Kof characteristic 2, with corresponding trace mapT:F → K, where[F:K]is odd.

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Hypothesis 3.1. P = (F,∗,+)satisfies the following for allx, y, z ∈ F, allk ∈ K, and somel∈F.

(i) (x+y)∗z=x∗z+y∗z, (ii) xy=0⇒x∗y=0,

(iii) x∗y=x∗z⇒x(y+z) =0, (iv) k(x∗y) = (kx)∗y,

(v) T(x(x∗y)) =T(lxy)2, and (vi) z(x∗y) = (z−1x)∗(zy)ifz6=0.

Here (i)–(iii) are precisely the definition of aprequasifield. (This is not a quasifield since it does not necessarily have an identity element.) The associated spread

S:={S[s]|s∈F∪ ∞}

of theK-vector spaceF2consists of the followingK-subspaces (cf. (iv)):

S[s] :={(x, x∗s)|x∈F}, s∈F, and S[∞] :={(0, y)|y∈F}. (3.2) The kernelK(A(S))containsKsinceSconsists ofK-subspaces.

Conditions (v) and (vi) are of special interest. Namely, (v) implies thatSis a sym- plectic spread with respect to the alternatingK-bilinear form

(x, y),(v, w)

:=T(xw+yv), (3.3)

while (vi) produces collineations of the affine planeA(S), and automorphisms of the orthogonal spreadΣ, discussed in the next proposition.

Remark 3.4. A slight modification of the multiplication∗would allow us to assume that l=1 (namely, usex∗0y:=x∗(ly)). Instead we will chooselso that the formula for the binary operation∗is as nice as possible.

Equip theK-vector spaceV :=F ⊕K⊕F⊕Kwith the nondegenerate quadratic formQ(x, a, y, b) :=T(xy)+ab. We can identify the symplecticK-spacesh0,1,0,1i/ h0,1,0,1iandF2. Defineφζ:F2→F2andϕζ:V →V,ζ∈F, by

(x, y)φζ = (ζ−1x, ζy) and (x, a, y, b)ϕζ = (ζ−1x, a, ζy, b), respectively. (3.5) Proposition 3.6. Suppose thatP(F,∗,+)satisfies Hypothesis3.1.

(i) The symplectic spreadSis invariant under the cyclic groupG:={φζ |ζ∈F} of symplectic isometries ofF2. Moreover, Ginduces a group of collineations of the affine planeA(S)that regularly permutes qm−1 points at infinity, fixes the remaining two points, and regularly permutesqm−1points of each of the two lines joining either of these two points to0.

(ii) Slifts to an orthogonal spreadΣ={Σ[s]|s∈F∪ {∞}}ofV, where Σ[∞] =0⊕0⊕F⊕K

Σ[s] =

x, a, x∗s+ls(a+T(lxs)), T(lxs)

|x∈F, a∈K , s∈F.

(3.7)

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HereΣ/h0,1,0,1iisS. The groupGˆ :={ϕζ | ζ ∈F}consists of isometries ofV, and acts onΣby stabilizingΣ[0]andΣ[∞]and regularly permuting the remaining members ofΣ.

(iii) CV( ˆG) = {(0, a,0, b) | a, b ∈ K}, and the nonsingular points fixed byGˆ are h0, λ,0,1i,λ∈K.

(iv) For eachλ∈K, the symplectic spreadΣ/h0, λ2,0,1iis coordinatized by a pre- quasifieldP= (F,◦,+), defined by

x◦y:=x∗y+lyT(lxy) +lλyT(lλxy) =x∗y+ly(1+λ2)T(lxy), that satisfies Hypothesis3.1withlλin place ofl.

Proof. These are straightforward calculations, most of which are given in [14, Theo- rem 2.18]. Hypothesis 3.1(v) is proved as follows:

z(x◦y) =z(x∗y) + (1+λ2)lzyT(lxy)

= (z−1x)∗(zy) + (1+λ2)l(zy)T l(z−1x)(zy)

= (z−1x)◦(zy).

For (iv), note that

x, λ2T(lxs), x∗s+ls(λ2T(lxs) +T(lxs)), T(lxs)

= x,0, x∗s+ls(1+λ2)T(lxs)),0)

+T(lxs)(0, λ2,0,1 . 2

We will need some elementary properties of trace functions.

Lemma 3.8. Suppose thatF ⊇F0 ⊇Kare fields with[F:K]odd and corresponding trace mapsT:F →KandT0:F →F0. Ifx∈Fandu∈F0, then

(i) T T0(x) =T(x), (ii) T(ux) =T(uT0(x)), (iii) T uxT0(x)

=T(ux2), and (iv) T0(u) =uandT(1) =1.

Proof. See [14, Lemma 2.14] for assertions (i), (ii), (iv). For (iii), use (i):T uxT0(x)

= T T0 uxT0(x)

=T uT0 xT0(x)

=T uT0(x)T0(x)

=T T0(ux2) =T(ux2). 2

4 Nearly flag-transitive planes

Nearly flag-transitive planes were defined in Section 1. We now give examples that are scions of desarguesian planes (cf. Section 2).

LetF =F0 ⊃ · · · ⊃ Fn be a chain of fields, with[F:Fn]odd and corresponding trace mapsTi: F →Fi. If 1 ≤i≤nletVi be theFi-vector spaceF ⊕Fi⊕F⊕Fi, equipped with the quadratic formQi(x, a, y, b) :=Ti(xy) +ab.

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Proposition 4.1. For1 ≤i≤nletλi ∈Fi; setλ0 =1andci :=Qi

j=0λj whenever 0≤i≤n. If

x∗y:=xy2+

n

X

i=1

ci−1yTi(ci−1xy) +ciyTi(cixy)

, (4.2)

thenP(F,∗,+)satisfies Hypothesis3.1withl=cn. Furthermore,A(S)is a symplec- tic nearly flag-transitive scion of a desarguesian plane.

Thus,ci is any element ofF1 such thatci/ci−1 = λi ∈ Fi when i ≥ 1; but (4.2) makes it clear that we will want to require thatci−16=ci. We begin with some observa- tions concerning the sum in (4.2):

Lemma 4.3. (i) x∗y=y xy+f1(xy)

, wheref1(u) :=Pn

i=1(1+λ2i)ci−1Ti(ci−1u) is inF1.

(ii) Tj

x2+Pj i=1

ci−1xTi(ci−1x) +cixTi(cix)

=Tj(cjx)2wheneverx∈F and 1≤j≤n.

(iii) Ifj < i≤nandx, y∈F, thenTj(ci−1x)Ti(ci−1y) +Tj(cix)Ti(ciy) = [ci−1/cj](1+λ2i)Tj(cjx)Ti(ci−1y).

Proof. (i)x∗y=xy2+yPn

i=1 ci−1Ti(ci−1xy) +λici−1Tiici−1xy)

i∈Fi. (ii) Usex2=c0xT0(c0x)and Lemma 3.8(iii):

Tj x2+

j

X

i=1

ci−1xTi(ci−1x) +cixTi(cix)

=

j−1

X

i=0

Tj(cixTi+1(cix)) +Tj(cixTi(cix))

+Tj cjxTj(cjx)

=

j−1

X

i=0

Tj (cix)2

+Tj (cix)2

+Tj (cjx)2

=Tj(cjx)2.

(iii) Sinceciici−1andci−1/cj =Qi−1

l=j+1λl∈Fj+1⊂Fj,

Tj(ci−1x)Ti(ci−1y) +Tj(cix)Ti(ciy) = (1+λ2i)Tj(ci−1x)Ti(ci−1y)

= [ci−1/cj](1+λ2i)Tj(cjx)Ti(ci−1y). 2 We now give two proofs thatPsatisfies Hypothesis 3.1.

First proof of Proposition4.1 (geometric). We apply the up and down process along the chain(Fi)n0 of fields, beginning with the desarguesian spread. DefinePj = (F,∗j,+) by

x∗jy=xy2+

j

X

i=1

ci−1yTi(ci−1xy) +ciyTi(cixy) .

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Clearly,x∗0y=xy2coordinatizes the desarguesian plane. Sincex(x∗0y) = (xy)2,P0 satisfies Hypothesis 3.1, withF =K,T =1 andl=c0 =1.

Suppose that 0≤j−1 ≤n−1 andPj−1 = (F,∗j−1,+)satisfies Hypothesis 3.1 withK =Fj−1,T =Tj−1andl=cj−1. ThenPj−1also satisfies Hypothesis 3.1 with K =Fj,T =Tj andl =cj−1, sinceTjTj−1 =Tj by Lemma 3.8(i). In the preceding section we saw that this implies thatPj−1defines a symplectic spreadSj−1inF2, and hence also an orthogonal spreadΣj−1in the orthogonal spaceVj−1=F⊕Fj⊕F⊕Fj admitting the groupGˆ in Proposition 3.6(ii). By Proposition 3.6(iv),Σj/h0, λ2j,0,1iis (equivalent to) the symplectic spread of theFj-vector spaceF2 coordinatized by Pj, andPj satisfies Hypothesis 3.1 withK=Fj,T =Tjandl=λjcj−1=cj.

Hence, the desired result holds by induction. 2

Second proof of Proposition4.1 (algebraic). For completeness, as in [14, p. 908] we give a direct proof thatPsatisfies Hypothesis 3.1. Parts (i), (ii), (iv) and (vi) are straightfor- ward calculations, and (v) holds by Lemma 4.3(ii) withl=cj, so we focus on part (iii):

we assume thatx∗y1 = x∗y2, and deduce thatx(y1 +y2) = 0. Let z = xy1 and w=xy2. By (4.2),x(x∗y1) =x(x∗y2)becomes

z2+

n

X

i=1

ci−1zTi(ci−1z) +cizTi(ciz)

=w2+

n

X

i=1

ci−1wTi(ci−1w) +ciwTi(ciw) . (4.4)

We use backwards induction to show that

Tj(cjz) =Tj(cjw) whenever 0≤j≤n. (4.5) ApplyingTn to (4.4), by Lemma 4.3(ii) we obtainTn(cnz)2 =Tn(cnw)2, so that (4.5) holds whenj=n.

Now suppose that, for some j such that 0 ≤ j < n, wheneverj < l ≤ n we haveTl(clz) = Tl(clw)(and then alsoTl(cl−1z) = Tl(cl−1w)sincecl = λlcl−1 with λl ∈ Fl). We must show thatTj(cjz) = Tj(cjw). By our induction hypothesis, (4.4) states that

n z2+

j

X

i=1

ci−1zTi(ci−1z) +cizTi(ciz)o +

n

X

i=j+1

ci−1zTi(ci−1z) +cizTi(ciz)

=n w2+

j

X

i=1

ci−1wTi(ci−1w)+ciwTi ciw)o +

n

X

i=j+1

ci−1wTi(ci−1w)+ciwTi(ciw) .

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ApplyTj, using Lemma 4.3(ii) and the fact thatTi(ci−1z), Ti(ciz)∈Fjfori≥j+1:

Tj(cjz)2+

n

X

i=j+1

Tj(ci−1z)Ti(ci−1z) +Tj(ciz)Ti(ciz)

=Tj cjw)2+

n

X

i=j+1

Tj(ci−1w)Ti(ci−1w) +Tj(ciw)Ti(ciw) .

By Lemma 4.3(iii),

Tj(cjz)2+

n

X

i=j+1

[ci−1/cj](1+λ2i)Tj(cjz)Ti(ci−1z)

=Tj(cjw)2+

n

X

i=j+1

[ci−1/cj](1+λ2i)Tj(cjw)Ti(ci−1w).

By induction,Tiici−1z) =Tiici−1w)ifi≥j+1, so that Tj(cjz) +Tj(cjw)2

=

Tj(cjz) +Tj(cjw)

n

X

i=j+1

[ci−1/cj](1+λ2i)Ti(ci−1z).

IfTj(cjz)6=Tj(cjw), then Tj(cjz) +Tj(cjw) =

n

X

i=j+1

[ci−1/cj](1+λ2i)Ti(ci−1z)∈Fj+1

sinceci−1/cj, λi ∈Fj+1fori≥j+1. SinceTj+1j+1cjz) =Tj+1j+1cjw)by our inductive hypothesis, from Lemma 3.8(i),(iv) we obtain

Tj(cjz) +Tj(cjw) =Tj+1 Tj(cjz) +Tj(cjw)

=Tj+1(cjz) +Tj+1(cjw) =0, which contradicts the fact thatTj(cjz)6=Tj(cjw).

By induction, this proves (4.5). In particular,xy1 =z=T0(c0z) =T0(c0w) =w=

xy2, as required. 2

Definition 4.6. Let(Fi)n0 be a chain of distinct fields with[F0:Fn]odd, and let(λi)n0 be a sequence of elements withλ0=1 andλi∈Fi, 1≤i≤n. Then we call (Fi)n0,(λi)n0 adefining pairfor various objects obtainedalong the chain(Fi)n0:

the prequasifieldP (Fi)n0,(λi)n0

in (4.2),

the symplectic spreadS (Fi)n0,(λi)n0

in (3.2), and

the orthogonal spreadΣ (Fi)n0,(λi)n0

in (3.7).

It is areduceddefining pair ifλi6=1 for alli≥1.

Each defining pair (Fi)n0,(λi)n0

determines a reduced defining pair (Fi0)n00,(λ0i)n00 obtained by deleting all entriesFj andλj withj ≥1 andλj =1. Then (Fi)n0,(λi)n0

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and (Fi0)n00,(λ0i)n00

determine thesameprequasifield by (4.2) or Lemma 4.3(i). Hence, we will only consider reduced defining pairs. Sinceλn 6=1 for a reduced pair, we have

|Fn|>2:each of our nearly flag-transitive scions of the desarguesian plane has kernel properly containingGF(2).

Whenn=1, a reduced pair is(F,1), and produces the desarguesian plane by (4.2).

In Lemma 4.3(i) we introduced the functionf1(x). Later we will need a slightly more general function:

Lemma 4.7. Let (Fi)n0,(λi)n0

be a reduced defining pair. For1 ≤j ≤nand for all x∈F set

fj(x) := (1+λ2j)Tj(cj−1x) +λj

n

X

i=j+1

[ci−1/cj](1+λ2i)Ti ci−1x .

ThenFjis the subfield ofF generated byfj(F).

Proof. Sinceλj 6=1, x → (1+λ2j)Tj(cj−1x)maps ontoFj. This proves the lemma ifj = n. If j < n thenPn

i=j+1[ci−1/cj](1+λ2i)Ti ci−1x

∈ Fj+1 sinceci−1/cj, λi ∈ Fj+1. Thus,fj maps ontoFj

jFj+1). Since[Fj: Fj+1] ≥3, this implies that the subfield ofFjgenerated byfj(F)has size>|Fj|1/2and hence isFj. 2

5 Isomorphisms between nearly flag-transitive planes

We digress for a general though elementary observation concerning symplectic nearly flag-transitive planes that permits us to decrease the amount of calculation used in the proof of Theorem 1.1 (cf. [9, III.C]). LetF be a finite field of characteristic 2, letK, T,F2andGbe as in Section 3 (but here we can allowmto be even), and consider the G-invariant nondegenerate alternatingK-bilinear form (3.3) on theK-spaceF2. ThenG leaves invariant the totally isotropic subspacesX:=F⊕0 andY :=0⊕F.

Assume that we do not have both|F|=64and|K|=2.By Zsigmondy’s Theorem [19], there is a Sylowp-subgroupP of Gthat acts irreducibly on bothX andY. By checking orders we find thatPis a Sylow subgroup ofΓSp(F2, K).

Lemma 5.1. (i) CΓL(F2,K)(P) ={(x, y)→(ax, by)|a, b∈F}, and

(ii) NΓSp(F2,K)(P) ={(x, y)→(kaxα, ka−1yα)|k∈K, a∈F, α∈Aut(F)}hθi whereθ: (x, y)→(y, x).

Proof. By Schur’s Lemma,CΓL(F2,K)(P|X)is the multiplicative group of a field of size

|X|, and hence is isomorphic toF. The same holds forCΓL(F2,K)(P|Y). No semilinear transformation can centralizeP and interchangeX andY (note thatθinvertsP). This proves the first assertion.

For the second one, note that all scalars are inCΓSp(F2,K)(P), so that we can view CΓSp(F2,K)(P)asG×K. All field automorphisms are present inNΓSp(F2,K)(P). Any g∈NΓSp(F2,K)(P)acts on the pair{X, Y}and hence has the form(x, y)→(axα, byα)

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or(x, y) → (ayα, bxα)for somea, b, α. By (3.3),T(abzα) = kT(z)β for someβ ∈ Aut(F),k∈K, and allz∈F. Thus,ab=k, as asserted. 2

The following consequence is analogous to [13, Proposition 5.1]:

Proposition 5.2. LetS and S0 beG-invariant symplectic spreads in the K-space F2, both containing theG-invariant subspacesF ⊕0and0⊕F, such thatGis transitive on the remaining members of bothS andS0. IfA(S)andA(S0)are isomorphic, then S0=SαorSθαfor someα∈Aut(F), whereθis as above.

Proof. By Theorem 2.2(i), we may assume that the given isomorphism is induced by a symplectic transformation f of the K-space F2. LetP be the Sylow subgroup ofG used above. ThenP andPf are Sylow subgroups of(AutA(S0))∩Sp(F2, K), so that Pf h = P for someh ∈ (AutA(S0))∩Sp(F2, K). Now f h ∈ ΓSp(F2, K)is an isomorphismA(S) → A(S0) that normalizesP. SinceK andGleave both spreads invariant, the second part of the preceding lemma concludes the proof. 2

6 Kernels

Our first use of the preceding section is to calculate the kernels of our nearly flag-transitive scions of desarguesian planes. See [13, Section 6] for an argument based on the same idea.

Theorem 6.1. If (Fi)n0,(λi)n0

is a reduced defining pair, thenFn is the kernel of the associated planeA (Fi)n0,(λi)n0

.

Proof. By (4.2), the kernelKofA (Fi)n0,(λi)n0

containsFn. Since|Fn| >2, we can apply Section 5 (withK=Fn). LetP be as in that section. ThenPnormalizesK, hence induces semilinear transformations on theK-spaceF2, and hence centralizesKin view of|P|. By Lemma 5.1(i), each element ofKhas the formh: (x, y)→(ax, by)for some a, b∈F. We must show thata=b∈Fn.

In the notation of (3.2),S[s]g =S[s]for eachs. By Lemma 4.3(i), (ax)s2+sf1(axs) = (ax)∗s=b(x∗s) =b xs2+sf1(xs)

for allx∈F, s∈ F. Then(a−b)xs=−f1(axs) +bf1(xs). Varyxin order to see that the left side produces either 0 orF. However, the right side lies in the 2-dimensional F1-subspaceF1+bF1ofF, which is not all ofF since[F:F1]≥3. Hence,(a−b)xs must be 0, so thata=b.

Thus,f1(ax) =af1(x)for allx∈F.

Suppose thata /∈Fn, and choosej≤nsuch thata∈Fj−1−Fj. By Lemma 4.3(i),

n

X

i=1

(1+λ2i)ci−1Ti(ci−1ax) =a

n

X

i=1

(1+λ2i)ci−1Ti(ci−1x)

for allx∈F, and hencefj(ax) =afj(x)in the notation of Lemma 4.7. By that lemma, fj(F)generatesFj, so thatfj(x)6=0 and hencea=fj(ax)/fj(x)∈Fjfor somex.

This contradiction proves thata=b∈Fn, as required. 2

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7 InterchangingX andY

In the next section we will determine all isomorphisms among the nearly flag-transitive planesA (Fi)n0,(λi)n0

. We first need to see what happens when X andY are inter- changed using the symplectic mapθ: (x, y)→(y, x).

Theorem 7.1. θ is an isomorphismA (Fi)n0,(λi)n0

→ A (Fi)n0,(λ−1i )n0

for any re- duced defining pair (Fi)n0,(λi)n0

.

We will prove this by induction onn, using the following inductive step:

Lemma 7.2. In Proposition3.6, assume thatθsendsStoS0, where(F,+,∗0)satisfies Hypothesis3.1(i)–(iv),(vi)andT(x(x∗0y)) = T(l−1xy)for allx, y∈F. Definex◦y as in Proposition3.6(iv), so thatS= Σ/h0, λ2,0,1i; and similarly define

x◦0y:=x∗0y+ (1+λ−2)l−1yT(l−1xy), so thatS0 = Σ0/h0, λ−2,0,1i. ThenSθ=S0.

Moreover, ifθsends the liney=x∗stoy=x∗s0for a permutations→s0ofF, then it also sendsy=x◦stoy=x◦0s0.

Proof. By Proposition 3.6(ii),SandS0 lift to unique orthogonal spreads Σ andΣ0 (respectively) ofV = F ⊕K⊕F ⊕KcontainingXb := F ⊕K⊕0⊕0 andYb :=

0⊕0⊕F⊕K. For the present proof it is convenient to writeSc:= ΣandSc0 := Σ0. Note that θ lifts to a unique orthogonal map θˆ: (x, a, y, b) → (y, b, x, a) sending Xb ↔ Yb and fixingh0,1,0,1i. For, θ lifts to a unique isometry ofh0,1,0,1i fixing h0,1,0,1i, and hence to a unique isometry ofV sendingXbto the subspaceYbof the same type.

EachZ∈ Slifts to the subspaceZˆof singular points of the hyperplane ofh0,1,0,1i that containsh0,1,0,1i and projects ontoZ in h0,1,0,1i/h0,1,0,1i; and then lifts further to the unique totally singular subspaceZbofV that containsZˆand has the same type asXb andYb.

Similarly,Zθ∈ Sθlifts to(Zθ)ˆand then toZcθ. SinceθsendsZtoZθ,θˆsends lifts to lifts:(Z)b θˆ=Zcθ. Consequently, by hypothesis,cSθˆ=Scθ=Sc0.

Sinceλ∈K,θˆsendsh0, λ2,0,1itoh0,1,0, λ2i=h0, λ−2,0,1iandh0, λ2,0,1ito h0, λ−2,0,1i, and hence alsoSc/h0,1,0, λ2i=S toSc0/h0, λ−2,0,1i=S0, using the multiplications◦and◦0 of the prequasifields in Proposition 3.6(iv) and the present lemma. This proves the first assertion of the lemma.

Ifs∈FandZ =S[s]as in (3.2), thenZθ=S0[s0]by the definition ofs0, so that Zˆθˆ=Zcθ = Σ0[s0]. Then Zˆθˆ∩ h0, λ−2,0,1i

/h0, λ−2,0,1i= Σ0[s0], which is also Zˆ∩ h0, λ2,0,1i

/h0, λ2,0,1iθˆ

= Σ[s]θˆ. Ifx∈Fandb∈K, then

(x, λ2b, y, b) = (x,0, y,0) +b(0, λ2,0,1) (y, b, x, λ2b) = (y,0, x,0) +λ2b(0, λ−2,0,1).

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Thus, the maph0, λ2,0,1i/h0, λ2,0,1i → h0, λ−2,0,1i/h0, λ−2,0,1iinduced byθˆ behaves likeθ, and sendsΣ[s]toΣ0[s0], as required. 2 Proof of Theorem7.1. We use induction. Whenn=0,x∗y =xy2 =x∗0y,λ=1 = λ−1, andθis a collineation of the corresponding desarguesian plane interchanging the linesy=xsandy=xs−1fors6=0.

Suppose that the theorem holds for somen−1≥0. By the lemma (withl=cn−1and λ=λn as in the first proof of Proposition 4.1), sinceSθ =S0 we also haveSθ =S0. The element “l” for∗0 isc−1n−1 by our inductive hypothesis. The definition of◦0 shows that the element “λ” for◦0in Proposition 3.6(iv) isλ−1n . Hence, that proposition implies that “l” for◦0isλ−1n c−1n−1=c−1n , so thatS0is the spread forA (Fi)n0,(λ−1i )n0

. 2

Remark 7.3. We conclude with computational remarks concerning the preceding results that are not needed for Theorem 1.1.

(1) As in Hypothesis 3.1(vi), we havez(x∗0 y) = (z−1x)∗0 (zy)if z 6= 0. For, (x, y)→(z−1x, zy)is a collineation of the plane determined by∗and hence also of the plane determined by∗0.

(2) Sincey=x∗simplies thatx=y∗0s0, for allx∈F we have x= (x∗s)∗0s0.

(3) There is an elementc ∈ F such thats0 =cs−1. For, by (1), (2) and Hypothe- sis 3.1(v), ift6=0 then

{(t−1x)∗(ts)} ∗0(ts)0=t−1x

=t−1{(x∗s)∗0s0}

={t(x∗s)} ∗0(t−1s0)

={(t−1x)∗(ts)} ∗0(t−1s0).

Thus,(ts)0=t−1s0. Uses=1 in order to obtaint0=t−1cwithc=10, as required.

(4) For the planes in Theorem 7.1, s0 = s−1. Namely, this holds for desarguesian planes (whenn=0), and hence it holds by induction using the final part of the lemma.

(5) By the lemma together with Proposition 3.6(iv) and (2), x◦y=x∗y+µlyT(lxy) x◦0y=x∗0y+µ0l0yT(l0xy)

x= (x∗y)∗0y0

whereµ=1+λ20=1+λ−2andl0 =l−1. Hence, again by (2), x= (x◦y)◦0y0

={x∗y+µlyT(lxy)} ◦0y0

={x∗y+µlyT(lxy)} ∗0y00l0y0T(l0y0{x∗y+µlyT(lxy)})

=x+µT(lxy)·(ly)∗0y00l0y0T(l0y0(x∗y)) +µ0l0y0µT(lxy)T(l0y0ly).

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By (3),yy0 =cfor some constantc. By (1) and Hypothesis 3.1(vi), (ly)∗0y0 =y−1(l∗0c) =l−1y−1(1∗0(cl))

y0(x∗y) = (xy0−1)∗(y0y) = (xyc−1)∗c=c((xy)∗1).

Also,v→T(l−1c(v∗1))is a linear functionalF →K, so for a uniquea∈Fwe have T(l−1c(v∗1)) =T(av)for allv∈F. It follows that

0=

µT(lxy)l−1y−1(1∗0(cl)) +l−1µ0cy−1T(l−1c((xy)∗1)) +µ0µl−1cy−1T(lxy)T(c) = µT(lxy)l−1y−1(1∗0(cl)) +cµ0l−1y−1T(axy) +µ0µcl−1y−1T(lxy)T(c).

Hence, for allv∈F,

0=µ(1∗0(cl))T(lv) +cµ0T(av) +cµ0µT(lv)T(c) T(av) ={µ(1∗0(cl)) +cµ0µT(c)}(cµ0)−1T(lv).

Since the left side is inK, it follows that

a={µ(1∗0(cl)) +cµ0µT(c)}(cµ0)−1l.

Thus, we have proved that the parameterscandaare related as follows:

a(cµ0) =µl(1∗0(cl)) +clµ0µT(c)

T(av) =T(cl−1(v∗1)). (7.4) These parameters evidently depend somehow on∗andλ.

(6) These parameters can, however, be determined in the situation of Theorem 7.1. We assume that we are passing from the casen−1 ton, just as in the proof of the theorem.

We havec =1 by (4), and we useµ =µn,T = Tn andl = cn−1. By Lemma 4.3(i) and (7.4),

Tn(av) =Tn c−1n−1 v+

n−1

X

i=1

(1+λ2i)ci−1Ti(ci−1v)

=Tn(c−1n−1v) +

n−1

X

i=1

Tn (1+λ2i)c−1n−1ci−1Ti(ci−1v)

=Tn(c−1n−1v) +

n−1

X

i=1

TnTi (1+λ2i)c−1n−1ci−1ci−1v

=Tn(c−1n−1v) +

n

X

i=1

Tn(c−1n−1c2i−1v) +

n

X

1

Tn(c−1n−1c2iv)

=Tn(c−1n−1v) +Tn(c−1n−1v) +Tn(c−1n−1c2n−1v),

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so thata=cn−1. Also, sincec0i−1cn−1=c−1i−1cn−1∈Fiandcl=cn, µnl(1∗0(cl)) =µncn−1

c2n−1+

n−1

X

i=1

(1+λ02i)c0i−1cn−1Ti(c0i−1cn−1)

ncn−1 c2n−1+

n−1

X

i=1

(1+λ02i)c0i−1cn−1c0i−1cn−1

ncn−1 c2n−1+

n−1

X

i=1

c02i−1c2n−1+

n−1

X

i=1

c02ic2n−1

ncn−1{c2n−1+c2n−1+c−2n−1c2n−1}

ncn−1, while

a(cµ0) +clµ0nµnT(c) =cn−1µ0n+cn−1µ0nµn

=cn−1µ0n(1+µn)

=cn−1µn, as required in (7.4).

8 Isomorphisms among nearly flag-transitive scions

Our goal in this section is Theorem 8.5, which completely solves the isomorphism prob- lem for the planesA (Fi)n0,(λi)n0

. As in [13, Lemma 5.3] and [14, Proposition 3.38], we first need to know when two of our spreads coincide. While the required result offers no surprises, proving it appears to be harder than one might expect, in fact slightly harder than in the preceding two references:

Proposition 8.1. Let (Fi)n0,(λi)n0

and (Fi0)n00,(λ0i)n00

be reduced defining pairs with F0=F =F00. They determine the same prequasifield(i.e., the exact same multiplication) if and only ifn=n0,Fi=Fi0andλi0iwhenever1≤i≤n.

Proof. We may assume without loss of generality thatn0≥n. We will prove that Fj=Fj0 and λj−10j−1 whenever 1≤j≤n.

For this purpose we use induction to prove that, for eachjwith 1≤j≤n,

Fl=Fl0 whenever 0≤l≤j, and λl0l whenever 0≤l < j. (8.2) Whenj = 1, we haveλ0 = 1 = λ00 by definition. We must show that F1 = F10. By hypothesis,x∗y=x◦yfor allx, y∈F. Then Lemma 4.3(i) implies thatyf1(z) =yf10(z) for allz ∈ F (wherez =xy, andf1 andf10 are as in that lemma). Thus,F1 = F10 by Lemma 4.7, and (8.2) holds whenj=1.

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Now assume that (8.2) holds for somej with 1 ≤ j < n. Thenc0k = Qk i=0λ0i = Qk

i=0λi=ckwhenever 0≤k < j.

We first show thatλj0j.By Lemma 4.3(i), ifx, y∈Fthen

xy2+

j−1

X

i=1

(1+λ2i)ci−1yTi(ci−1xy)

+cj−1h

(1+λ2j)yTj(cj−1xy) +

n

X

i=j+1

(1+λ2i)[ci−1/cj−1]yTi(ci−1xy)i

=xy2+

j−1

X

i=1

(1+λ2i)ci−1yTi(ci−1xy)

+cj−1h

(1+λ0j2)yTj(cj−1xy) +

n0

X

i=j+1

(1+λ0i2)[c0i−1/cj−1]yTi0(c0i−1xy)i .

By our inductive hypothesis,

(1+λ2j)Tj(cj−1xy) +

n

X

i=j+1

(1+λ2i)[ci−1/cj−1]Ti(ci−1xy)

= (1+λ0j2)Tj(cj−1xy) +

n0

X

i=j+1

(1+λ0i2)[c0i−1/cj−1]Ti0(c0i−1xy).

(8.3)

Letx=c−1j−1. By Lemma 3.8(iv), ify∈Fjthen (1+λ2j)y+λj

n

X

i=j+1

(1+λ2i)[ci−1/cj]Ti([ci−1/cj−1]y)

= (1+λ0j2)y+λ0j

n0

X

i=j+1

(1+λ0i2)[c0i−1/cj]Ti0([c0i−1/cj−1]y),

where 1+λ2i, ci−1/cj =Qi−1

j+1λj ∈Fj+1and 1+λ02i, c0i−1/cj ∈ Fj+10 fori ≥j+1.

Then

j0j)2y=λjg(y) +λ0jg0(y) for ally∈Fj,

for additive mapsg: Fj → Fj+1 andg0: Fj → Fj+10 . Since[Fj: Fj+1] ≥ 3 and [Fj: Fj+10 ] ≥ 3, we have|kerg| ≥ (2/3)|Fj|and|kerg0| ≥ (2/3)|Fj|, so that there is somey6=0 inkerg∩kerg0. Then(λj0j)2y=0, so thatλj0j, as claimed.

Next we show thatFj+1=Fj+10 .By induction and the preceding paragraph, we have Fi =Fi0andλi0i wheneveri≤j. Now (8.3) states thatcjfj+1(xy) =cjfj+10 (xy) for allx, y∈F, in the notation of Lemma 4.7. ThenFj+1=Fj+10 by that lemma.

By induction, this proves (8.2).

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