ARS MATHEMATICA CONTEMPORANEA
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 14 (2018) 251-265
https://doi.org/10.26493/1855-3974.1285.f3c
(Also available at http://amc-journal.eu)
Inherited unitals in Moulton planes
*
Gábor Korchmáros , Angelo Sonnino
Dipartimento di Matematica, Informática ed Economia Universita degli Studi della Basilicata Viale dell'Ateneo Lucano 10, 85100 Potenza, Italy
Tamás Szonyi
ELTE Eotvos Loránd University, Institute of Mathematics and MTA-ELTE Geometric and Algebraic Combinatorics Research Group H-1117 Budapest, Pázmany P. s. 1/c, Hungary
Received 11 January 2017, accepted 24 July 2017, published online 4 September 2017
We prove that every Moulton plane of odd order—by duality every generalised Andre plane—contains a unital. We conjecture that such unitals are non-classical, that is, they are not isomorphic, as designs, to the Hermitian unital. We prove our conjecture for Moulton planes which differ from PG(2, q2) by a relatively small number of point-line incidences. Up to duality, our results extend previous analogous results—due to Barwick and Gruning—concerning inherited unitals in Hall planes.
Keywords: Unitals, Moulton planes. Math. Subj. Class.: 51E20, 05B25
1 Introduction
A unital is a set of q3+1 points together with a family of subsets, each of size q+1, such that every pair of distinct points are contained in exactly one subset of the family. Such subsets are usually called blocks so that unitals are block-designs 2-(q3+1, q +1, 1). The classical example of a unital arises from the unitary polarity in the Desarguesian projective plane PG(2, q2) where the points are the absolute points, and the blocks are the non-absolute lines of the unitary polarity. The name of "Hermitian unital" is commonly used for the
* This research was carried out within the activities of the GNSAGA of the Italian INdAM. E-mail address: gabor.korchmaros@unibas.it (Gabor Korchmaros), angelo.sonnino@unibas.it (Angelo Sonnino), szonyi@cs.elte.hu (Tamas Szonyi)
Abstract
©® This work is licensed under http://creativecommons.org/licenses/by/3.0/
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classical example since the absolute points of the unitary polarity are the points of the Hermitian curve defined over GF(q2).
A unital U is embedded in a projective plane n of order q2, if its points are points of n and its blocks are intersections with lines. As usual, we adopt the term "chord" to indicate a block of U. Aline I of n is either a tangent or a (q +1)-secantto U according as nU| = 1 or nU| = q + 1, and in the latter case I n U is a chord. Examples of unitals embedded in PG(2, q2) other than the Hermitian ones are known to exist.
A unital is classical if it is isomorphic, as a block-design, to a Hermitian unital. Classical unitals contain no O'Nan configurations, and it has been conjectured that any non-classical unital embedded in PG(2, q2) must contain a O'Nan configuration.
In several families of non-desarguesian planes, the problem of constructing and characterizing unitals has also been investigated; see [1, 2, 5, 6, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 27, 28]. Apart from the examples of unitals arising from a unitary polarity in a commutative semifield plane, the known examples are inherited unitals from the Hermitian unital. In a non-desarguesian plane n of order q2 arising from PG(2, q2) by altering some of the point-line incidences, the adjective "inherited" is used for those pointsets of PG(2, q2) which keep their intersection properties with lines when moving from PG(2, q2) to n.
In this paper we construct inherited unitals in Moulton planes of odd order q2, and, by duality, in generalised Andre planes of the same order; see Theorem 3.1. We also investigate the problem whether these unitals are classical; see Theorems 3.5 and 3.6. We show that if such a plane differs from PG(2, q2) by a relatively small number of incidences only, then the inherited unital is non-classical. Also, we exhibit non-classical inherited unitals in case of many point-line incidence alterations. Such unitals appear to be of interest in coding theory; see [25].
What emerges from our work leads us to conjecture that the inherited unitals constructed in our paper are all non-classical. It should be noticed that our results extend previous analogous results due to Barwick and Griming concerning inherited unitals in Hall planes which are very special Andre planes; see [8, 16] and Remark 3.4. The methods used in [8] are mostly geometric and involve Baer subplanes and blocking sets. In this paper, we adopt a more algebraic approach that allows us to exploit results on the number of solutions of systems of polynomial equations over a finite field.
2 Two new results on the Hermitian unital
We establish and prove two theorems on Hermitian unitals that will play a role in our study on unitals in Moulton planes.
Up to a change of the homogeneous coordinate system (X1, X2, X3) in PG(2, q2), the points of the classical unital U are those satisfying the equation
+ X2q+1 + X3q+1 = 0.	(2.1)
In the affine plane AG(2, q2) arising from PG(2, q2) with respect to the line X3 = 0, we use the coordinates (X, Y) where X = X1/X3 and Y = X2/X3). Then the points of U in AG(2, q2) are the solutions of the equation
X 9+1 + Y9+1 + 1=0.	(2.2)
Since GF(q2) is the quadratic extension of GF(q) by adjunction of a root i of the polynomial X2 — s with a non-square element s of GF(q), every element u of GF(q2) can
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uniquely be written as u = u + iu2 with u^u2 € GF(q). Then uq = u — iu2 and ||u|| = uq+1 = ui — su2. Therefore, the points P(x, y) € U lying in AG(2, q2) are those satisfying the equation
x2 — sx2 + y2 — sy2 + 1=0.	(2.3)
For a subset T C GF(q) \ {0}, let St denote the set of points { (x, y) | ||x|| = t € T }. Hence the pointset St n U comprises all points P(x, y) such that both x2 — sx2 = t and (2.3) hold. Therefore, a point P (x, y) € AG(2,q2) is in St nU if and only if P1(x1,x2) € AG(2, q) lies on the non-degenerate conic C1 : X2 — sY2 — t = 0 while P2(y1 ,y2) € AG(2, q) does lie on the conic C2 : X2 — sY2 + 1 + t = 0. This shows that St n U has size (q + 1)2 apart from the case t = —1 when it consists of the q + 1 points of U lying on the X -axis.
Lemma 2.1. Let i be a non-vertical line in AG(2, q2). Then |i nUnSt| € {0,1,2, q +1} for every t € T .If q +1 occurs then i is either a horizontal line, or it passes through the origin.
Proof. The points P (x, 0) with ||x|| = t form a Baer subline. As U is classical, i nU is a Baer subline of i, and hence the projection of inU on the X-axis from is a Baer-subline, as well. Since two distinct Baer sublines have at most two common points, the first assertion follows. To prove the second one, we need some computation. If i has equation Y = Xm+ b, we have to count the roots x of the polynomial f (X) = Xq+1 + (Xm + b)q+1 + 1 whose norm ||x|| is equal to t. If ||x|| = t, then f (x) = bmqxq + bqmx +1(1 + mq+1) + bq+1 + 1 and hence
xf (x) = bq mx2 + (t(1 + mq+1) + bq+1 + 1)x + bmq t.
If we have at least three such roots x then either m = 0 and t + 1 = —bq+1, or b = 0 and
t(1 + mq+1) = —1.	□
Take any two distinct non-tangent lines i1 and i2 of U. We are interested in the intersection of the projection of i1 n U from P on i2 with i2 n U. For any point P outside both i1 and i2, the projection of i1 to i2 from P takes the chord i1 n U to a Baer subline of i2. Since two Baer sublines of i2 intersect in 0,1, 2 or q +1 points, one may want to determine the size of the sets (i = 0,1, 2, q +1) consisting of all points P for which this intersection number is equal to i. The points in are called elliptic, parabolic, hyperbolic, or full with respect to the pair (i1, i2), according as i = 0, i = 1, i = 2, or i = q + 1, respectively; see [21].
We go on to compute the size of n U. Since the linear collineation group G = PGU(3, q) of PG(2, q2) preserving U acts transitively on the points outside U, we may assume that = i1 n i2. The stabiliser of in G acts on the pencil with center in
as the general projective group PGL(2, q) on the projective line PG(1, q2). Therefore, it has two orbits, one consisting of all tangents the other of all chords to U through YTO. This allows us to assume without loss of generality that i1 is the line at infinity. Since i2 is not a tangent to U, its equation is of the form X = c with cq+1 + 1=0. Therefore, cq+1 + 1 is either a non-zero square or a non-square element of GF(q). These two cases occur depending upon whether a linear collineation 7 € PGL(2, q) taking i1 to i2 is in the subgroup isomorphic to the special projective group PSL(2, q) or not. Accordingly, {i1 , i2 } is called a special pair or a general pair. Further, since P is a point outside i1 and i2, it is an affine point P = (a, b) with a = c.
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¿2 \
Y = 0
X = 0 X = c
Figure 1: The initial configuration.
Let P = (a, b) denote a point of U, that is,
aq+i + bq+i + 1 = 0.
(2.4)
Takealine r of equation Y = m(X - a) + b through P = (a, b). A necessary and sufficient condition for r to meet both ¿4 and ¿2 in U is the existence of a solution t g GF(q2) of the system consisting of (2.4) together with
Cq+1 + t q+l + 1 =
mq+1 + 1 = 0.
(2.5)
(2.6)
In fact, Q(c, t) with t = m(c — a) + b is the point of r on ¿2. Then (2.5) holds if and only if Q G U. Furthermore, (2.6) is the necessary and sufficient condition for the infinite point of r to be in U; see Figure 1.
The above discussion also shows how to count lines through P meeting both ¿1 n U and ¿2 n U. Essentially, one has to find the number of solutions in the indeterminate t of the system consisting of the equations (2.4), (2.5), and (2.6). Observe that (2.4), (2.5), (2.6) are equivalent to
a2 - sa2 + b2 - sb2 + 1 = 0, c2 - sc2 + t2 - sr22 + 1 = 0, biTi - sb2T2 + aici - sa2C2 + 1 = 0.
(2.7)
(2.8) (2.9)
¿1 — ¿CO
0
From this the following result is obtained.
Proposition 2.2. The number of lines through P meeting both ¿1 nU and ¿2 nU equals the number of solutions (t1, t2), with T1, t2 g GF(q), of the system consisting of (2.7), (2.8), (2.9).
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In investigating the above system, two cases are distinguished according as (61,62) is (0,0) or not.
In the former case, Equations (2.7) and (2.9) read a2 - sa2 + 1 = 0 and a1c1 - sa2c2 + 1 = 0. Geometrically in AG(2, q), the point U = (a1, a2) is the intersection of the ellipse E, with equation X2 — sY2 + 1 = 0, and the line v with equation c1X — sc2Y + 1 = 0. Since cq+1 + 1 = c2 — sc2 + 1 is a non-zero element of GF(q), v must be either a secant, or an external line to E and this occurs according as c2 — sd^ + 1 is a non-zero square or non-square element in GF(q). In fact, from (2.7) and (2.9),
sc2&2 — 1	—sc2 ± ic1\Jc2 — sc2 + 1
a1 = -, a2 = -T2-^-.
c1	s ( c12 — sc22 )
Therefore, if P is on the X-axis, then P is elliptic in general, apart from the case where cq+1 +1 = c2 — sc2 + 1 is a non-square element in GF(q) and P is one of the two common points of C and v, namely P = P(a, 0) where
. —1 ± V1+ c«+1
a = a1 + «a2 = -.
cq
Further, in the exceptional case, P is a full point as for any c1, c2 G GF(q) with c1 — sc2 + 1 = 0, Equation (2.8) always has q +1 solutions (t1, t2) with t1, t2 g GF(q).
In the latter case, either 61 or 62 is not zero. If 61 = 0, retrieving t1 from (2.9) and putting it in (2.8) gives a quadratic equation in the indeterminate t2, namely
(s2 62 — s62)t22 — 2s62(a1c1 — sa2c2 + 1)t2 +
(a1c1 — sa2c2 + 1)2 + 62(1 + c2 — sc2) = 0, (2.10)
whose discriminant is A1 = s62 A with
A = (62 — s62)(1 + c2 — sc2) + (a1c1 — sa2c2 + 1)2 which can also be written by (2.7) as
A = —(1 + c2 — sc2)(a2 — sa2 + 1) + (a^ — sa2c2 + 1)2.
For 62 = 0, retrieving t2 from (2.9) and putting it in (2.8) gives the following quadratic equation in the indeterminate t1:
(—61 + 62)tx2 + 2a161c1T1 —
a1 — s2a2c2 — 62 c1 + s62c2 + 2sa2c2 — 62 — 1 = 0 (2.11)
with discriminant A2 = s362A. Since A1 and A2 are simultaneously zero, or a square, or anon-square in GF(q), each of the equations (2.10) and (2.11) has 2,1 or zero solutions in GF(q), depending upon whether A is a square element, zero, or a non-square element of GF(q), respectively. This leads to the study of the zeroes of the polynomial
F (X,Y,Z) = —(1 + c1 — sc2)(X2 — sY2 + 1) + (c1X — sc2Y + 1)2 — Z2. (2.12)
Geometrically, F(X, Y, Z) = 0 is the equation of a quadric Q in AG(3, q). Actually, Q
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Table 1: Elliptic, parabolic, hyperbolic and full points.
	P(a, 0)		P(a,b), b = 0	
	1± l|c|| G □	1± l|c||G □	1± ||c|| G □	1± |c|G □
N£	i ±1	i — 1	—3 — 9q ± q2 ± q3	—3 — 5q — q2 ± q3
			2	2
NP	0	0	2q — 1	0
NH	0	0	(i —21)2 (q ±1)	(q ±21)2 (q— 1)
Nf	0	2	0	0
is a cone. In fact, the system FX = FY = FZ = 0 has a (unique) solution (c1,c2,0) and hence the point V(c1,c2,0) is the vertex of Q. In particular, the intersection of Q with the plane Z = 0 splits into two lines over GF(q) or its quadratic extension GF(q2), and this occurs according as the infinite points of the conic with equation
-(1 + c2 - sc2)(X2 - sY2) + (c1X - sc2Y)2 = 0
lie in PG(2, q) or in PG(2, q2) \ PG(2, q). By a direct computation, this condition only depends on cq+1, namely whether 1 + cq+1 is a square or a non-square element of GF(q). Therefore, Q contains either 2q-1 or 1 points in the plane Z = 0, and this occurs according as the pair [£1, l2} is special or general. Also, in the former case there are exactly 2q - 1 parabolic points P but in the latter case no point P is parabolic. Therefore, the following result holds.
Theorem 2.3. Let i1, t2 be any two distinct non-tangent lines of the classical unital U in PG(2, q2) whose common point is off U. The number NE,Np,N%,NS, of elliptic, parabolic, hyperbolic and full points of U with respect to the pair {£1,£2} is given in Table 1.
We state a corollary of Theorem 2.3 that will be used in Section 3. For i = 1,2 let Aj be a subset of ^ n U such that | A11 = | A21 = A.
Theorem 2.4. If
x>yj (i±M	(2.13)
there exists a non-degenerate quadrangle AiBiA2B2 with vertices Ai, Bi G Aj for i = 1,2 such that its diagonal point AiB2 n BiA2 lies in U.
Proof. We prove the existence of a hyperbolic point D in U such that the projection of Ai from D on l2 share two points with A2. From Theorem 2.3, we have at least 2 (q— 1)2 (q±1) hyperbolic points in U. We omit those hyperbolic points projecting Ai = (£i n U) \ Ai to a pointset of l2 meeting l2 n U nontrivially. The number of such hyperbolic points is
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A(q - 1)(q + 1) with A = q + 1 - A. Similarly we omit all hyperbolic points projecting A2 = (¿2 nM) \ A2 to apointset of ¿1 meeting ¿4 n U nontrivially. Therefore, the total number of omitted hyperbolic points is 2A(q2-1)-A2(q-1) = (q -1)A(2q+2-A(q-1)). From Theorem 2.3, this number is smaller than the total number of hyperbolic points as far as (2.13) holds.	□
To state the other new result on the classical unital a couple of ad hoc notation in AG(2, q2) will be useful: For a non-vertical line ¿ with equation Y = Xm + b, A denotes the non-vertical line with equation Y = Xmq + b. Given a point P(a, b) outside U, two lines ¿1 and ¿2 are said to be a good line-pair whenever the lines ¿1 and ¿2 meet in a point of U. Our goal is to show that if a = 0 then there exist many good pairs.
For i = 1, 2, write the equations of ¿ in the form Y = (X - a)m¿ + b. Then ¿i has equation Y = Xm? - am¿ + b. Hence P(x, y) = ¿A n ¿2 where
a(mi - m.2)
and hence
a(mi - m2) q
-q-q— mi — ami + b.
Note that
lldl = xq+1 = aq+1
1
(mi — m2)q 1
q+i
(mi — m-2 )q'
2- = llaH = 0.
The condition for P(x, y) to lie in W is
xq+i + yq+i + 1 = aq+i + a'
q+i + yq+i + 1 = aq+i + aq+^ i^l_mi
V (mi — m2)q
q+1
m? — mi +— ) +1=0. 1a
Let
Then the last equation reads
e = —
aq+i + 1
aq+i
e GF(q).
(mi — m2) q
--^-mi
(mi — m2)q
q	b q+1
mi — mi +— = e.
1a
(2.14)
Henceforth we assume that With
Ml = —1.
mi = a + ¿ß, m2 = 7 + ¿5, — = u + ¿v,
a
(2.14) reads
(a — 7)+ i(ß — 5) (a — y) — i(ß — 5)
(a — ¿ß) — (a + ¿ß) + u + ¿v
q+1
e,
qq
mm
1
2
y
1
2
a
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whence
(ua — «y — svfi + svJ)2 — s(2fiY — 2aJ — u(fi — J) + v(a — y))2
- C((a — y)2 — s(fi — J)2) = 0,
that is,
(u(a — y ) — sv(fi — J))2 — s(2fiY — 2aJ — u(fi — J) + v(a — y))2
— ^((a — Y)2 — s(fi — J)2)=0. (2.15)
With	_
Y = a — Y, J = fi — J,
Equation (2.15) becomes
(«Y — svJ)2 — s(—2fiY — 2aJ — «J + vY )2 — C(Y2 — sJ2) = 0, (2.16) which can be viewed as a quadratic form in Y and J:
F (Y, J) = («2 — v2s + 4vfis — 4fi2s — £) Y2 + 2(—2ufis + 2vas — 4afis) yJ
+ (—u2 s — 4«as + v2 s2 — 4a2 s + s£) J2 (2.17)
with discriminant
A = —u4s + 2u2v2s2 + 2u2s£ — v4s3 — 2v2s2£ — s£2
+ (—4u3s + 4«v2s2 + 4us£) a + (— 4«2vs2 + 4v3s3 + 4vs2£) fi
— 8uvs2 afi + (—4u2s + 4s£) a2 + (—4v2s3 — 4s2£) fi2.
Note that P(x, y) € U if and only if A = A2 for some A G GF(q). This leads us to consider the quadric Q in AG(3, q) of equation
aoo + aoi X + a^Y + a^XY + an! 2 + a22Y2 — Z2 =0,
a00 = —u4s + 2u2v2s2 + 2u2s£ — v4s3 — 2v2s2£ — s£2,
aoi = —4«3s + 4uv2s2 + 4us£,
a02 = —4u2vs2 + 4v3 s3 + 4vs2£,
ai2 = —8uvs2,
aii = —4u2s + 4s£,
a22 = —4v2s3 — 4s2£.
The above coefficients are related by the following equations:
(i)	aoo — 2(èaoi« — ao2v) = s£(u2 — sv2 — £);
(ii)	2aoi — è(aii« — èai2v) = 0;
where
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(iii) 2a02 - 1 (2ai2U - 022«) = 0.
Therefore, the determinant D of the 4 x 4 matrix associated with Q is equal to -sC(u2 -sv2 - C) multiplied by the determinant of the cofactor of a00. The latter determinant
ana22 - 4a22 is equal to
Do = s3C(u2 - sv2 - C) = s3(aq+1 + bq+1 + 1)(aq+1 + 1).	(2.18)
It turns out that
D = -(s2C(u2 - sv2 - C))2. Observe that C = 0 if and only if aq+1 = -1, while
2 ^ bq+1 aq+1 + 1 aq+1 + bq+1 + 1 u - sv - C = —tt +--n— =-n-
vanishes only for P(a, b) G U. Therefore, Q is non-degenerate. More precisely, the quadric Q is either elliptic or hyperbolic according as q = -1 (mod 4) or q = 1 (mod 4). The plane at infinity cuts out from Q a conic C with homogeneous equation a11X2 + a12XY + a22 Y2 - Z2 =0. Observe that C is non-degenerate by D0 = 0. Thus, the number of points of Q in AG(3, q) is q2 ± q with q = ±1 (mod 4). Furthermore, the point at infinity on the Z-axis does not lie on Q, and it is an external point or an internal point to C according as -D0 is a non-zero square or a non-square in GF(q). Therefore, the number of tangents to Q through in AG(3, q) is equal to q - 1 or q +1 according as -D0 is a (non-zero) square or a non-square in GF(q). From the above discussion, the numbers Ns and Nt of secants and tangents to Q through are those given in the following lemma:
Lemma 2.5. For q = -1 (mod 4), either Nt = q + 1, Ns = 1 (q - 1)2, or Nt = q -1, Ns = 1 (q2 - 2q -1), according as D0 is a (non-zero) square or a non-square in GF(q). For q = 1 (mod 4), either Nt = q - 1, Ns = 1 (q2 + 1), or Nt = q +1, Ns = 1 (q2 - 1), according as D0 is a (non-zero) square or a non-square in GF(q).
Going back to the discriminant A, we see that A vanishes for Ns + Nt ordered pairs (a, that is, Ns + Nt is the number of lines ¿4 through P(a, b) for which there exists a line ¿2 such that (¿1, ¿2) is a good line-pair. For each counted in Nt (resp. Ns), we have q - 1 (resp. 2(q - 1)) such lines ¿2, since if (2.17) has a non-trivial solution (7, J) in GF(q) x GF(q) then it has exactly q - 1 solutions, the multiples of (7,7) by the non-zero elements of GF(q).
If we do not count the q +1 tangents to U through P(a, b), each of the lines through P(a, b) counted in Ns is in at least 2(q - 1) - (q + 1) = q - 3 good line-pairs. Therefore Lemma 2.5 has the following corollary.
Theorem 2.6. Let P(a, b) be a point of AG(2, q2) outside U. If a = 0, ||a|| = -1 and q > 3, then there exist at least two non-tangent lines , ¿2 of U through P, such that the non-tangent lines ¿1 and ¿2 meet in a point of U. Further, if q > 5 then and ¿2 may be chosen among the lines through P(a, b) other than the horizontal lines and those passing through the origin.
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3 Unitals in Moulton planes
Let T be a non-empty subset of the multiplicative group of GF(q). The (affine) Moulton plane MT (q2) which is considered in our paper is the affine plane coordinatized by the left quasifield GF(q2)(+, o) where
Jxy if ||x||G T, x o y = <
[xyq if ||x|| G T,
with ||x|| = xq+1 being the norm of x G GF(q2) over GF(q). Geometrically, MT(q2) is constructed on AG(2, q2) by replacing the non-vertical lines with the graphs of the functions
Y = X o m + b.	(3.1)
This also shows that to the non-vertical line £ of equation Y = Xm + b there corresponds the line of equation £ of equation Y = X o m + b in MT (q2), and viceversa. It is useful to look at the partition of the points outside the Y-axis into q - 1 subsets Si, called stripes, where P(x, y) G Si if and only if ||x|| = wi with w a fixed primitive element of GF(q). Such stripes were already defined in Section 2; here we just abbreviate the subscript wi by i. In fact, moving to MT(q2) the point-line incidences P G £ in AG(2, q2) do not alter as long as P G Si with wi G T. The projective Moulton plane is the projective closure of Mt(q2) and it has the same points at infinity as AG(2, q2). For a similar description of Moulton planes see also [3, 4, 26].
The dual of the Moulton plane is the Andre plane AT (q2 ) coordinatized by the right quasifield GF(q2)(+, *) where
x * y
xy if | x| G T, xqy if ||x|| G T.
In this duality, the correspondence occurs between the point (u, v) of MT(q2) and the line of equation Y = u * X - v, as well as between the line of equation Y = X o m + b and the point (m, -b) of AT(q2). The correspondence between points at infinity and lines through Yto, and viceversa, is the same as the canonical duality between PG(2, q2) and its dual plane PG* (2, q2). If T consists of just one element, then the arising Andre planes are pairwise isomorphic and they are also known as Hall planes.
Let U be the classical unital in PG(2, q2) given in its canonical form (2.1). We prove that U is an inherited unital in the Moulton plane, that is, the point-set of U is a unital in Mt(q2) as well.
Theorem 3.1. Let U be the classical unital in PG(2, q2) given in its canonical form (2.1). Then, for any T, U is a unital in the projective Moulton plane MT (q2) as well.
Proof. In the very special case T = {-1}, the proof is straightforward. It is enough to show that if a non-vertical line I of equation Y = Xm + b meets U in a point P(x, y) with ||x|| = —1 then y = 0 and x = —b/m with (—b/m)q+1 = 1. In fact, the corresponding line ¿ in Mt(q2) has the same property: if P(x, y) G í n U then y = 0 and x = (—b/mq)q+1. Since (-b/m)q+1 = (-b/mq)q+1, the assertion follows for T = {-1}.
In the general case, it suffices to exhibit a bijective map from I n U to í n U for every line í of AG(2, q2). We may limit ourselves to non-vertical lines with non zero slopes. Let
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Y = Xm + b be the equation of such a line t and take any point P (x, y) lying in t nU. Then m = 0 and x = (y — b)m-1. Define the map p: t ^ t by
Obviously, p(P) = P whenever ||x|| G T.
Since p is bijective, it suffices to show that P gM yields p(P) G U, and the converse also holds. P(x, y) = ((y - 6)m-1, y) G U if and only if
((y - 6)m-1)q+1 + yq+1 - 1 = (y - 6)q+1(m-1)q+1 + yq+1 - 1 = 0.
By (mq)q+1 = mq+1, the latter equation is equivalent to
Theorem 3.1 and its proof also show that if i is a tangent to U in AG(2, q2) then the corresponding line i is a tangent to U in the projective Moulton plane, and the converse also holds. In particular, the tangent to U at a point outside the X-axis is the line i of equation Y = X(-cd-1)q - d-q with tangency point P(c, d). Therefore, the corresponding line i of equation Y = X o (-cd-1)q - d-q is a tangent to U at the point p(P) = P(c, d) with c = c or c = c(cd-1)q-1 according as ||c|| G T or ||c|| G T. Since ||c|| = ||c||, the tangency points of i and i lie in the same stripe. The tangents of U with tangency point at infinity contain the origin and each of them has equation Y = Xm with mq+1 + 1 = 0. By the proof of Theorem 3.1, the corresponding lines Y = X o m are the tangents of U in the projective Moulton plane.
Now look at dual plane of the projective Moulton plane MT (q2) which is the projective Andre plane AT (q2). In this duality, the tangent line i of U with equation Y = X o (-cd-1)q - d-q corresponds to the point P*(w*,v*) G (q2) where u* = -(-cd-1)q and v* = d-q. Since ((-cd-1)q)q+1 + (d-q)q+1 + 1 = 0, we have u*q+1 + v*q+1 + 1 = 0. Similarly, the tangent line i of U with equation Y = X o m, mq+1 + 1=0, corresponds to the point P* (u* ,v*) G AT(q2) where u* = u and v* = 0. Therefore u*q+1 + v*q+1 + 1 = 0. In terms of PG* (2, q2), the Desarguesian plane which gives rise to the projective Andre plane AT (q2), the points P*(u*, v*) lie on the classical unital U* given in its canonical form. This shows that U* can be viewed as an inherited unital in the projective Andre plane AT (q2).
Remark 3.2. If T = {-1} then the unique stripe where incidence are altered meets U in q +1 points lying on the X-axis. The unital U* in the Hall plane is the Gruning unital [16] while for T = {¿} with wJ = -1, U* in the Hall plane is the Barwick unital [7].
A O'Nan configuration of a unital consists of four blocks b1, b2, 63 and 64 intersecting in six points P1, P2, P3, P4, P5 and P6 as in Figure 2. As mentioned in the introduction, the Hermitian unital contains no O'Nan configuration. This fundamental result due to O'Nan dates back to 1972, see [22] and [9, Section 4.2].
p(P)
P((y — b)m-1,y) for ||x|| £ T, P((y — b)m-q, y) for ||x|| G T.
((y — b)q+1(m-q )q+1 + yq+1 — 1 = ((y — b)m-q )q+1 + yq+1 — 1 = 0,
whence the claim follows.
□
Lemma 3.3. If T = { — 1} then the unitalU of (q2) is non-classical.
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Proof. We show that the unital U in (q2) with T = {-1} contains a O'Nan configuration. Take a e GF(q2) such that ||a|| = -1. The line ¿1 of equation Y = X - a meets U in Q(a, 0) and q more points. Take m e GF(q2) such that mq-1 = -1. The line i2 of equation Y = Xm + am meets U in R (-a, 0) and q more points. Further, the common point of l1 and l2 is
S =
-a(m + 1) -2am
(m - 1) (m - 1)
Since
-a(m + 1)
(m - 1)
^q+1 (m +1)q+1 = (m - 1)q+1
mq+1 + mq + m +1 mq+1 - mq - m +1


-	m2 - m + m + 1
-	m2 + m - m + 1
1,
the point S is outside U. Further, in the Moulton plane (q2) with T = {-1}, the corresponding lines ¿i and ¿2 meet in Q(a, 0) which is a point of U.
To show that U is not a classical unital in our Moulton plane (q2), it suffices to exhibit a O'Nan configuration {P0, P1, P2, P3, P4, P5} lying in U. The idea is to start off with P0 = Q(a, 0), and to find four more affine points P1, P2 e ¿i and P3, P4 e ¿2 each lying in U, so that U also contains one of the two diagonal points P5 of the quadrangle P1P2P3P4 that are different from P0. First we show that P1 e ¿1. Let P1 = P1 (x1, y1). Then, ||x1| = -1. In fact, otherwise, we would have yq+1 = 0 and hence y1 = 0, contradicting P0 = P1. Similarly, P2 e and P3,P4 e ¿2. Now we use a counting argument in PG(2, q2) to show that the quadrangle P1P2P3P4 can be chosen in such a way that P5 e U. Since S = ¿1 n ¿2 is outside U, the lines of U joining a point of ¿^ with a point of ¿"2 cover (q + 1)2(q - 1) points of U other than those lying in ¿^ U ¿2 From (q +1)2 (q - 1) > q3 + 1 - 2q, there exists a quadrangle F1P2P3P4 in PG(2, q2) such that
P1, P2 e ¿1 n U, P3, P4 e ¿2 n U, P5 = P1P3 n P2P4 e U.
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Since (q +1)2 (q -1) > q3 +1 - 2q + (q +1) we may also assume that either P5 g nU, or P5 = (x5,y5) with ||x5|| = -1. In particular, P5 is not on the X-axis.
If Pi, P2 = Q and P3, P4 = R then P5 remains a diagonal point of the quadrangle PiP2P3P4 in (q2), and we are done.
Otherwise, take the cyclic subgroup G of PGU(3, q) of order q +1 fixing the point S and preserving each line through S. Since |G| > 4, G contains an element g such that Q G {g(Pi), g(P2)} and R G {g(P3), g(P4)}. Then g takes the quadrangle P1P2P3P4 to another one, whose vertices are different from both Q and R. The image g(P5) is on the line r through S and P5. Since r n U has at most one point on the X-axis, there exists at most one g G G such that g(P5) lies on the X-axis. Therefore, if |G| > 5, some g G G also takes P5 either to a point of infinity or a point (x5, y5) with ||x51| = -1. In the Moulton plane (q2), the O'Nan configuration P0, g(P1), g(P2), g(P3), g(P4), g(P5) arising from the quadrangle g(P^1)g(P2)g(P3)g(P4) lying in U has also two diagonal points, namely P0 and g(P5), belonging to U.	□
Remark 3.4. Lemma 3.3 can also be obtained from Griining's work. In fact, if T = {-1} then U is isomorphic to its dual, see [16, Theorem 4.2], and the dual of U contains some O'Nan configuration, see [16, Lemma 5.4c].
We conjecture that Lemma 3.3 holds true for any T. Theorem 3.5 proves this as long as T is small enough. On the other end, Theorem 3.6 provides Moulton planes with large T for which the conjecture holds.
Theorem 3.5. If q > 5 and
then U in the Moulton plane MT (q2) is a non-classical unital.
Proof. As in the proof of Lemma 3.3, we show the existence of a O'Nan-configuration {Po, Pi, P2, P3, P4, P5} lying in U. For a point P(a, b) G AG(2, q2) with a = 0 and ||«y G T \ { —1}, Theorem 2.6 ensures the existence of two non-vertical lines G and through P such that
(i)	neither G nor G is horizontal or passes through the origin,
(ii)	P0 = G n G g U.
From Lemma 2.1, there exist at least q +1 - 2|T| points P(x,y) lying on G nU such that ||x|| G T, and the same holds for G n U. Therefore, Theorem 2.4 applies with A = q +1 - 2|T| showing that if (3.2) is assumed, then the unital U in MT(q2) contains a O'Nan configuration.	□
Theorem 3.6. If q > 5, then there exists a T with |T | > q — 4 such that U is a non-classical unital in MT(q2).
Proof. From the proof of Theorem 3.5, some Moulton plane MT (q2 ) contains O'Nan configurations lying in U .If {P0, P^ P2, P3, P4, P5} one of them, add each non-zero element s G GF(q) to T which satisfies the condition s = ||xj|| for Pj = Pj(xj,yj) with 1 < i < 5. Then T expands and its size becomes at least q—4. In the resulting Moulton plane MT (q2), the above hexagon {P0, P1, P2, P3, P4, P5} is still a O'Nan configuration lying in the unital W.	□
(3.2)
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