characterizations of buekenhout-metz unitals

Geometriae Dedicata 59: 29--42, 1996. 29 © 1996 KluwerAcademic Publishers. Printed in the Netherlands.

Characterizations of Buekenhout-Metz Unitals

L. R. A. CASSE 1, CHRISTINE M. O'KEEFE 1 and TIM PENTTILA 2 1Department of Pure Mathematics, The University of Adelaide, Adelaide, S.A. 5005, Australia 2Department of Mathematics, The University of Western Australia, Nedlands, W.A. 6009, Australia

(Received: 8 January 1993)

Abstract. We give a characterization of the Buekenhout-Metz unitals in PG(2, q2), in the cases that q is even or q = 3, in terms of the secant lines through a single point of the unital. With the addition of extra conditions, we obtain further characterizations of Buekenhout-Metz unitals in PG(2, q2), for all q. As an application, we show that the dual of a Buekenhout-Metz unital in PG(2, q2) is a Buekenhout-Metz unital.

Mathematics Subject Classification (1991): 51E20.

Key words: Buekenhout-Metz unitals, classical unitals, Baer sublines.

1. Introduction

We denote by PG(2, q2), where q = ph and p is a prime, the Desarguesian projective plane of square order q2. A unital (or Hermitian arc) H in PG(2, q2) is a set of q3 + 1 points of PG(2, q2) such that each line meets it in either 1 or q + 1 points. A line is a tangent or a secant line of H if it contains 1 or q -4- 1 points of H, respectively. A point P E H lies on one tangent and q2 secant lines, while a point Q ~ H lies on q + 1 tangents and q2 _ q secant lines.

An example of a unital in PG(2, q2) is given by the set of absolute points of a unitary polarity. This is called a classical unital (or Hermitian curve), and any classical unital is the image under an element of PFL(3, qZ) of the set of points (x0, zl , z2) satisfying the following equation

+ + = o.

In 1976, Buekenhout [9] proved the existence of unitals in every translation plane of square order q2 with kernel containing GF(q). In particular, he noted that his construction gave a family of non-classical unitals in PG(2, q2) for q > 2 even and not a square. Metz [22], in 1979, extended this observation to the case of q even and square, and q odd; hence, for any prime power q > 2, there exist non-classical unitals in PG(2, q2).

A Buekenhout-Metz unital in PG(2, q2) is a unital which arises by the construction due to Buekenhout [9, §4, Remark (4)]. Since the classical unital can be constructed in this way, it is included in the class of Buekenhout-Metz unitals.

30 L.R. A, CASSE ET AL.

We note that every unital in PG(2, 22) is classical and hence is a Buekenhout- Metz unital. Consequently, in the following we will consider unitals in PG(2, q2) where q > 2. Further, a recent computer search by Penttila and Royle [25] con- cludes that every unital in PG(2, 32) is a Buekenhout-Metz unital.

The classes of classical and Buekenhout-Metz unitals have been characterized in the class of unitals in PG(2, qZ) for q > 2 by Faina and Korchmgtros [12] and by Lef6vre-Percsy [19] and [20], as follows. Note that a Baer subplane of PG(2, q2) is a subplane of order q, and any line of PG(2, q2) meets a Baer subplane in 1 or q + 1 points. A set of q + 1 points which is the intersection of a line with a Baer subplane is a Baer subline.

(1.1) THEOREM [12], [20]. Let ~ be a unital in PG(2, q2), where q > 2. I f each line of the plane which is secant to bl intersects it in a Baer subline, then bt is a classical unital.

(1.2) THEOREM [19]. Let U be a unital in PG(2, q2) where q > 2 and let ~ be some tangent line to lg. I f all Baer sublines having a point on g intersect Lt in O, 1,2 or q + 1 points, then lg is a Buekenhout-Metz unital.

In fact (1.2) is a necessary and sufficient condition, and also holds for parabolic unitals in any translation plane of order q2 whose kernel contains GF(q), see [19] and recall that a parabolic unital in a translation plane is a unital for which the translation line is a tangent.

The characterization of Buekenhout-Metz unitals in__PG(2, q2) described in (1.2) is equivalent to the following. Let T be the point of / , /on the tangent g. If the points of / , /on lines through T form Baer sublines and if each Baer subline meeting g \ { T } intersects/ t in at most two points, then b / i s a Buekenhout-Metz unital. (This follows from the fact that two Baer sublines of a given line in PG(2, q2) meet in 0, 1 , 2o r q + 1 points, see [10].)

In this paper we improve the characterization of Lef6vre-Percsy in PG(2, q2) in the cases q even and q = 3 by weakening the hypotheses to apply only to the lines through one point T of/,/. Our result is:

(1.3) THEOREM. Let ~ be a unital in PG(2, q2), where q > 2 is even or q = 3. Then lg is a Buekenhout-Metz unital if and only if there exists a point T ofl.t such that the points o f ~ on each of the q2 secant lines to ~ through T form a Baer subline.

We can obtain a characterization of Buekenhout-Metz unitals in PG(2, q2) which holds for all q > 2 with the addition of an extra hypothesis. This extra hypothesis is either a condition on the stabilizer of the unital; namely the assumption that the unital is fixed by a certain type of automorphism of PG(2, q2), or a hypothesis of a synthetic nature on some of the tangents of the unital.

For the condition on the stabilizer, recall (see [10]) that a point P is a centre of an automorphism 9 of PG(2, q2) if each line through P is fixed by 9 and a line

CHARACTERIZATIONS OF BUEKENHOUT-METZ UNITALS 31

g is an axis of 9 if each point on g is fixed by g. An elation of PG(2, q2) is an automorphism which has an incident centre-axis pair.

Our characterizations are given in the next two theorems:

(1.4) THEOREM. Let U be a unital in PG(2, q2), where q > 2. Then ~ is a Buekenhout-Metz unital i f and only i f

(i) there exists a point T o f ~ such that the points o f ~ on each of the q2 secant lines to bl through T form a Baer subIine, and

(ii) there is a non-trivial elation with centre T fixing bt.

(1.5) THEOREM. Let U be a unital in PG(2, q2), where q > 2. Then U is a Buekenhout-Metz unital i f and only i f

(i) there exists a point T o f ~ such that the points o f U on each of the q2 secant lines to bl through T form a Baer subline, and

(ii) there exists a point P on go~ \{T} such that the points o f tangency o f the q + 1 tangents to bl on P lie on a Baer subline n (through T).

There are very few known constructions of unitals in PG(2, q2). In particular, constructions are known for the classical unitals, the Buekenhout-Metz unitals and the Buekenhout hyperbolic unitals (arising from [9, Th. 4]). Let PG(2, q2)a denote the dual plane of PG(2, q2), with points the lines of PG(2, q2) and lines the points of PG(2, q2). The dual plane PG(2, q2)d is a projective plane isomorphic

to PG(2, q2). Now the dual ~a in PG(2, q2)d of a unital ~ in Pa(2, q2) has as

its points the tangents of 9 . It is straightforward to check that ~a is a unital in PG(2, q2)a, which is therefore isomorphic to a unital in PG(2, q2). The only unitals in PG(2, q2) known to the authors are those arising from the above constructions, and the duals of such unitals.

It is well known that the dual of a classical unital in PG(2, q2) is classical. Further, a Buekenhout hyperbolic unital in PG(2, q2) is classical, see [6]. Since classical unitals are Buekenhout-Metz unitals, in order to show that all the known unitals in PG(2, q2) are Buekenhout-Metz unitals we need only show that:

(1.6) THEOREM. The dual o f a Buekenhout-Metz unital in PG(2, q2) is a Buekenhout-Metz unital.

When q is odd, Baker and Ebert [5] have shown that this is true, in fact each Buekenhout-Metz unital in PG(2, q2), q odd, is self-dual. Further, the same result is true for those Buekenhout-Metz unitals in PG(2, qa), q even, for which the underlying ovoid is an elliptic quadric [11]. Our proof is valid for any q and any underlying ovoid.

32 L.R.A. CASSE ET AL.

2. Preliminary Results

To discuss Buekenhout's construction of unitals and to prove our characterization, we use the following representation of PG(2, q2) in PG(4, q) due to Andr6 [3] and Bruck and Bose [7] and [8]. The construction is discussed in [15, §17.7].

Let ~ = PG(3, q) be embedded as a hyperplane in PG(4, q). Let S be a regular spread of Zoo. We define a new incidence structure: the points are the points of PG(4, q ) \ Z ~ , the lines are the planes of PG(4, q) which do not lie in E ~ but which meet Zoo in a line of S and incidence is natural. This incidence structure is a Desarguesian affine plane AG(2, q2) and can be completed to a projective plane PG(2, q2) by the addition of an ideal line g~ whose points are the elements of the spread S.

Under a slight abuse of notation, in the following work the phrase a subspace of PG(4, q ) \ E ~ will mean a subspace of PG(4, q) which is not contained in E ~ .

Under this representation of PG(2, q2) in PG(4, q), a plane of PG(4, q ) \ E ~ which does not contain any line of S corresponds to a Baer subplane of PG(2, qa) with q + 1 points on g~ , (see [27, Prop. 2]). It follows that a line of PG(4, q)\Eoo corresponds to a Baer subline of PG(2, q2) with a point on £~. The converse of each of these results is true, (see [27, Prop. 1]).

Let V and W be points on a line t of S, and let O be an ovoid in some 3- dimensional subspace of PG(4~ q) not containing V and such that O meets E ~ in exactly the point W. Let 9" be the cone with vertex V projecting (_9, that is, H* is the union of the lines joining V to each point of O. Equivalently, let V be a point on a line t of S and let (9 be an ovoid of the quotient space with respect to V which meets ~ in exactly t (so O is a set of q2 + 1 lines through V, no three of which lie in a plane through V, and O contains t). In either case, let H* be the union of the lines so constructed. Then the set H of points of PG(2__, q 2) corresponding to H* is a Buekenhout-Metz unital. Note that gc~ is tangent to H at a point T.

All known ovoids in PG(3, q) are either elliptic quadrics or Tits ovoids (see [15, 16.4]). However the classification of ovoids in PG(3, q) is complete only for all q odd and for q even with q _< 32, (see [15], [23], [24]). Every ovoid in PG(3, q), q odd and q = 4, 16 is an elliptic quadric while every ovoid in PG(3, q) for q = 8, 32 is either an elliptic quadric or a Tits ovoid.

Note that we include in the class of Buekenhout-Metz unitals the unitals constructed using any ovoid O in the above construction. This agrees with the usage of the term Buekenhout-Metz unital found in [19]. However in the definition of Buekenhout-Metz unital given in [5] and [11] the ovoid O is assumed to be an elliptic quadric. Hence the unitals arising from the Tits ovoids (and any further ovoids, if such exist) are excluded from the class of Buekenhout-Metz unitals considered in [11].

The following lemma is fundamental to our arguments. As it originally appeared as a corollary of a more general result, we provide a direct proof.


(2.1) LEMMA [14, Cor. 3.3]. Let H be a unital and let B be a Baer subplane o f PG(2, q2). I f bl denotes the number o f lines o f B which, when extended to PG(2, q2), are tangents o f H and 8 denotes the number o f points in B N H then

bl + 8 = 2(q + 1).

Proof We count in two ways the ordered pairs (P, ~) where P is a point of H, is a line of B (extended to PG(2, q2)) and P E g. We obtain:

s(q + 1) + (q3 + 1 - s).l = bl.1 + (q2 + q + 1 - b l ) ( q + 1)

and the result follows. []

3. Characterization of Buekenhout-Metz Unitals in PG(2, q2)

Let H be a unital in PG(2, q2) where q > 2. Suppose there exists a point T E such that the points of H on lines of PG(2, q2) through T, and distinct from the tangent £~ at T, form Baer sublines.

In the representation of PG(2, q2) in PG(4, q) given in Section 2, and with the notation introduced there, the unital H corresponds to a set of points H* in PG(4, q) and we let H* -- H * \ ~ . The above hypothesis is equivalent to the hypothesis that H* is t together with a union of q2 lines gl,. • •, gq2 of PG(4, q), each meeting t but pairwise having no common point in PG(4, q ) \E~.

We proceed via a sequence of Lemmas. Let AG(2, q2) = PG(2, q2)\£~ and let u :

(3.1) LEMMA. Let c~ be a plane of PG( 4, q) \ E oo passing through t. Then ~ meets bt m t together wtth a umque hne gi o fH .

Proof First, ~ corresponds to a line of PG(2, q2) through the point T, distinct from goo. This line meets H in q + 1 points (including T) which comprise a Baer subline, by hypothesis. Hence ~ fq N* is t together with q points on a line gi of PG(4, q ) \ P ~ and the result follows. []

(3.2) LEMMA. I f for distinct i, j E 1, . . . , q2, two lines gi and gj o f H-* meet, then the plane they span contains no further point o f H*.

Proof The plane/3 = (gi, ~j) does not contain t, for otherwise it would correspond to a line ofPG(2, q2) containing more than q + 1 points of H. Thus/3 meets P, oo in a line not belonging to S, and hence corresponds to a Baer subplane B of PG(2, q2) with q + 1 points on £oo. At least one line o r b is a tangent of H, namely ~ , and by Lemma 2.1, B meets/.4 in at most 2q + 1 points. Thus B = B\g~o meets H in at most 2q points. Since/3 contains 2q points of R* which are exactly the 2q points of (~i U gj)\Eoo, it follows that/3 contains no further point of H-*. []

(3.3) LEMMA. Suppose there exist distinct i , j E 1 , . . . ,q2 such that gi meets gj in a point V and let/3 = (gi, ~.j). Then the 3-dimensional subspace P, = (/3, t) meets H* in a cone projecting an oval from the point V C t.


Proof First we show that q of the lines ~1, . . . , gq2 pass through V. By (3.1), the

q planes ofPG(4, q ) \ E ~ in E passing through t each contain a distinct line of H*. These q lines all pass through V, for if one such line does not pass through V then it meets/3 in a point not on gi or gj, contradicting (3.2). Note that if P, contains a point not on t which lies on one of the lines ~1 , . . . , gq2 then it must contain the

whole line, so that ~ meets ~* in a union of t together with q of the lines £1 , . . . , gq2. It is enough now to show that no plane through V contains more than two of these lines. By (3.1) a plane through t contains at most one line from ~1, . . . , gq2 and by (3.2) a plane not through t contains at most two lines from ~1,. • •, gq2. []

(3.4) LEMMA. Suppose there exist distinct i , j E 1 , . . . , q2 such that gi meets ~j in a point V. Let/3 = (gi,gj) and E = (/3, t). By (3.3), E fq ~* is a cone. I f a further line gk of H* passes through V then all of gl, . . . , gq2 pass through V.

Proof Without loss of generality, suppose that t, gl ,...,/~q are the lines of the

cone E N ~* and that £q+1 also passes through the point V. For each i = 1 , . . . , q,

the space (gq+l, gi, t) is a 3-dimensional space meeting ~* in a cone with vertex V, by (3.3). By varying i, we see that q(q - 1) + 1 = q2 _ q + 1 of the lines ~1,. • •, £q: pass through V, leaving q - 1 further lines. The remainder of the argument is perhaps more straightforward in a quotient space.

Let Q be the quotient space PG(4, q) with respect to t, so that planes on t are represented as points of Q. A plane of PG(4, q) \~ ,~ through t, and the corresponding point of Q, are called full if the plane contains a line of U* through V. The space E ~ corresponds to a line Eoo of Q, none of whose points are full. The space E corresponds to a line E of Q, all of whose points apart from P = E N Zoo are full. The plane 7- = (t, gq+l) corresponds to the full point 7- in Q. The preceding paragraph has shown that each point of Q is full, apart from the points of E ~ and the points of the line PT- in Q. The q - 1 points of PT-\{P, 7-) correspond to exactly the q - 1 planes left out in the argument in the preceding paragraph. By (3.3), if two points of Q\Eco are full then every point of Q\Zc~ on the line joining them is full. Thus the set of full points of Q is an affine subspace of Q and hence is either empty, a single point, the points of Q \ Z ~ on a line of Q distinct from E ~ or the whole of Q\Eo~. Therefore the set of full points of Q must be the whole of Q \ E ~ , so every plane of PG(4, q ) \Eec meets H* in a line ~k through V, for some k. Thus we have shown that the q2 lines g l , . . . , £q2 all pass through V. []

(3.5) LEMMA. Suppose that all the lines ~.1 , . . . , gq2 pass through the same point

oft. Then bl is a Buekenhout-Metz unital. Proof The hypothesis implies that the set ~* is t together with q2 lines

~1, • . . , gq2 of PG(4, q)\ ~oo all passing through a point V of t. Note that 9" meets Zoo in exactly the points of the line t.

If a plane through V contains a point not on t which lies on one of the lines gl,.. . ,/~q2 then it contains the whole line. By (3.1) and (3.2), no plane through


V meets ~* in more than two lines. Since there are q2 + 1 lines, they form an ovoid in the quotient space with respect to V, noting that q > 2. Thus L/ is a Buekenhout-Metz unital, as it is constructed by the method given in Section 2. []

(3.6) LEMMA. Suppose that not all the lines ~ 1 , . . . , ~q2 pass through the same point o f t. Then the lines gl , . . . , ~fl 2 fal l into q cones C1 , . . . , Cq in 3-dimensional spaces E1 , . . . , Eq and with distinct vertices VI, • • •, Vq on t. Further, the common intersection o f the spaces Y;,w, E l , . . . , Eq is a plane 7r tangent to each cone Ci in

Proof By (3.3), if two of the lines g l , . . . , gq2 pass through a point of t then q of them do, forming a cone. By (3.4), no more than q of the lines £1,- . . , £q2 pass

through a point of t. Therefore the q2 lines e l , . . . , gq2 form q cones, with distinct vertices on t. Finally, consider the plane 7r = E ~ N Ei for some i. Since there is no point of/./* in E ~ , it follows that 7r is a tangent plane to each cone C1,. . . ,Cq in E 1 , . . . , E q ont . []

In the cases q even and q --- 3 we are able to show that the configuration suggested by (3.6) cannot be a unital. The arguments appear in (3.7) and (3.8), respectively.

(3.7) LEMMA. l f q is even, it is impossible that the lines ~1, . . . , g.q2 form q cones C1, • . . , Cq with distinct vertices V1, . . . , Vq on t.

Proof Since q is even, each cone C/has a nuclear line ni where the nuclear line of a cone is the line which, in the quotient space, is the nucleus of the conic comprising the lines of the cone. Since 7r is the tangent plane to each cone Ci on t, each nuclear line ni lies in 7r. Thus 7r contains exactly q nuclear lines n l , . . . , nq, and there is a point P of 7r\t on no nuclear line. Let m be the element of the spread S o n P .

A plane w on ra and not in E ~ meets each solid E~ in a line g on P. Since ~ is not contained in E ~ , it follows that w N E ~ = ra and hence ~ fq 7r = {P) . Since all the tangent planes to Ci in Ei contain the nuclear line n~, and since g fq ni is empty, g lies in no such tangent plane. Thus I£ N C~[ = 0 or 2, hence

I nWl

is even. But w corresponds to a line of PG(2, q2) not through T, hence

Iw ML/* I = 1 o r q + 1,

which is odd. This contradiction proves our claim. []

(3.8) LEMMA. I f q = 3, it is impossible that the lines ~ 1 , . . . ,£q2 form q cones C1 , . . . , Cq with distinct vertices V1, . . . , Vq on t.

Proof First we introduce some additional notation. For a point P E 7r\t, let m be the element of S through P. For each i = 1 , . . . , q there are two tangent planes


to the cone Ci on the line ~ P , namely 7r and a further plane meeting Ci in a line ViQ i for some point Q i E Ci. We say that ~ P and ViQ i are paired for Ci. Note that distinct lines ~ P are paired with distinct lines ViQi for C~. We observe that a plane about ra intersects Ei in a line through P; thus such a plane intersects the cone Ci in 0, 1 or 2 points. A plane is called j-secant to Ci if it meets Ci in j points, for j = 0, 1 or 2. It follows that if R E ViQi then the plane (m, R) is 1-secant to Ci while if R E Ci\{ViQi, t) then (m, R) is 2-secant to Ci.

We assume that the unital U is not a Buekenhout-Metz unital, and by (1.2) there is a_Baer subline ofPG(2, q2) meeting t ~ \ { T } and containing exactly three points of///. Thus in PG(4, q) there is a line gl meeting three lines £1, g2, £3 say, among the lines gl, • • •, gq2. The lines gl, g2, Ca lie in different cones C1, C2, C3 respectively, for otherwise the line 91 lies in one of the spaces Ei and hence cannot meet three lines of U*.

Now £1, £2~ £3 lie in the 3-dimensional subspace (91, t); so they lie in a unique regulus T¢ whose opposite regulus ~opp contains Yl and t. Let Z be the point of t which is not the vertex of any cone Ci~ and consider the line r of T¢ on Z. Since r meets 91 in E ~ and meets t in E~ , r lies in E ~ , but (r, t) is not 7r (since 91 ~ Ei for any i so that gl N r ( r ) .

Let t, raa, ra2, m3 be the elements of S through the points o f t , where ml passes through the point gl fq r. Since S is regular, t, ral, m2, m3 form a regulus, which we denote by ~t . Let g2, g3 be the lines of ~opp through ra2 fq r and ra3 t3 r, respectively.

Now 7r meets ~ in the line t, so there is a line 8 E 7r (distinct from r as r 7~ 70 of (~,)opp meeting t. Without loss of generality, suppose that t N s = V3. Let P~ = mi N s f o r / = 1 ,2 ,3 and let Qij = 9i [q £j for i , j = 1,2,3.

The plane (ga, ral) corresponds to a line of PG(2, 9) meeting g~ \{T} which contains exactly four points of L/; hence (91, ml) contains exactly four points of ~*. We continue the argument in three cases:

Case 1. Suppose that (91, ral) is 2-secant to C1; so it is 1-secant to C2 and C3. Since Q12 c V2Q12 c ~2 and (Q12, ml) = (gl, ml ) is 1-secant toC2, it follows

that V2P1 and £2 are paired for C2. Similarly, V3P1 and £3 are paired for C3. Consider the plane (92, m2). Now V3P2 and £3 are paired for C3, so the plane

(92, m2) is 1-secant to C3. Also V2P2 and £2 are not paired for C2 (as V2P1 and £2 are paired for C2) so (92, rn2) is 2-secant to C2. Hence (92, m2} is 1-secant to C1 (as /4 is a unital) and so VIP2 and el are paired for C1.

It follows that V1P3 and £1 are not paired for C1 and that V2P3 and £2 are not paired for C2. Thus (93, ra3) is 2-secant to C1, 2-secant to C2 and at least 1-secant to C3. This is a contradiction as then (93, ra3) represents a line in PG(2, 9) with more than four points of a unital.

Case 2. Suppose that (91, ral) is 2-secant to C2. The arguments in this case are analogous to those used in Case 1, except that the roles of C1 and C2 are interchanged.


Case 3. Suppose that (gl, ml ) is 2-secant to C3. Then (gl, ml) is 1-secant to each of C1 and C2. It follows that V1 P1 and gl are paired for C1 and that V2P1 and g2 are paired for C2. Thus (g2, m2) is 2-secant to each of C1 and C2, and at least 1-secant to C3, which is impossible. []

We have now proved the following result.

(1.3) THEOREM. Let U be a unital in PG(2, q2), where q > 2 is even or q = 3. Then Lt is a Buekenhout-Metz unital i f and only if there exists a point T ofbt such that the points of ~ on each of the q2 secant lines to U through T form a Baer subline.

Proof Suppose that there exists a point T of b / such that the points o f / , / o n each of the q2 secant lines to ~ through T form a Baer subline. Then (3.1) to (3.8) show that/ . / is a Buekenhout-Metz unital.

The converse is immediate. []

The authors conjecture that in fact (1.3) holds for all values of q. We now prove the result for all values of q with an extra condition. This extra condition could be either a condition on the stabiliser of the unital, see (1.4), or a condition on some tangents of the unital, see (1.5). We first prove some preliminary lemmas.

(3.9) LEMMA. Let C be a cone in PG(3, q),for q > 2. I f there is an elation with centre on C and fixing C then the centre of the elation is the vertex of C.

Proof Suppose the centre C of the elation e is not the vertex o f t . Since q > 2, there is a line g on C which does not lie in the axis of the elation, which does not lie on the cone and which meets C at C and at a unique further point B. Now g is fixed by e; and since e also fixes C then e must fix B, contrary to the fact that the elation e fixes no point not on its axis ([10, p. 119]). []

(3.10) LEMMA. Let ~ be a Buekenhout-Metz unital in PG(2, q2), where q > 2. Then there exists a point T o f ~ such that the points o f ~ on each of the q2 secant lines to lg through T form a Baer subline.

Proof Let the point T of/ , /and the tangent go~ to/./at T be as in the construction in Section 2; so PG(2, q2) is represented in PG(4, q), with gc¢ corresponding to Noo. With the usual notation, U corresponds to a set ~ in PG(4, q) which is the line t of the spread S together with q2 lines of PG(4, q ) \ N ~ through a point V of t. The result follows immediately since a line in PG(4, q ) \ E ~ corresponds to a Baer subline in PG(2, q2). []

(3.11) LEMMA. Let U be a Buekenhout-Metz unital in PG(2, q2). Let T be a point o f ~ such that the points o f ~ on each of the q2 secant lines to U on T form a Baer subline. Let g~ be the (unique) tangent to bt on T. Then for any point P on


g~ \ {T}, the points of tangency of the q + 1 tangents to bt on P are coUinear and form a Baer subline.

Proof A tangent line other than g~ on P corresponds in PG(4, q) to a plane r, on the line p of the spread S corresponding to P and containing a single point Q of 9*. The 3-dimensional space (~-, V 1 meets 9" in exactly the points of a line ei for some i E {1 , . . . , q2}, for if (r, V / contains a further point o fg* then it contains a further line on V which must meet T, necessarily at Q. Each of the q planes in (r, V) on p not in ~o~ contains a unique point of 9" (namely, the intersection of the plane with gi) and therefore corresponds to a tangent line to 9 in PG(2, q2). The points of tangency on the tangents on P are the points corresponding to the points of gi \gc~ together with T, and form a Baer subline with a point on g~. n

(1.4) THEOREM. Let 9 be a unital in PG(2, q2), where q > 2. Then 9 is a Buekenhout-Metz unital if and only if

(i) there exists a point T of 9 such that the points of 9 on each of the q2 secant lines to bl through T form a Baer subline, and

(ii) there is a non-trivial elation ofPG(2, q2) with centre T fixing 9.

Proof Recall that an elation of PG(2, q2) with axis g~ induces an elation of PG(4, q) with axis ~ o , and conversely (see [21]), where an elation of PG(4, q) with axis ~ is an automorphism of PG(4, q) which fixes each point of ~ and fixes each line through a given point of ~ (called the centre of the elation.)

Suppose that 9 is a Buekenhout-Metz unital in PG(2, q2). Property (i) follows from (3.10). Let g~ be the tangent to 9 at the point T and let PG(2, q2) be represented in PG(4, q) as above, with g~ corresponding to ~ . With the usual notation, 9 corresponds to a set 9* in PG(4, q) which is the line t of the spread S together with q2 lines of PG(4, q)\Eoo through a point V of t. Then, (ii) follows since any elation of PG(4, q) with centre V and axis ~ induces an elation of PG(2, q2) with centre T which fixes 9 .

Now suppose that b/is a unital such that (i) and (ii) hold. By (3.1)-(3.6), (i) implies that either 9 is a Buekenhout-Metz unital or else it corresponds to the set 9" of t together with the union of q2 lines in PG(4, q)\P,~o, which fall into q cones C1,. . . , Cq in 3-dimensional spaces P,1,... , P'q with distinct vertices V1,. . . , Vq on t. We now show that this second possibility cannot occur.

By (ii), there is a non-identity elation 9 of PG(2, q2) with centre T E 9 which fixes b/. The axis of 9 is the unique tangent go~ to b/at T, for if 9 has axis a secant ra of 9 then let R be a point of 9 on m but distinct from the centre T ofg. Then 9 permutes the q2 lines on R distinct from the axis, so fixes the unique tangent to b/ on R. But this is impossible as all fixed lines of 9 pass through the centre.

Thus, to 9 there corresponds an elation 9* of PG(4, q), with axis ~ and centre some point of t, which fixes 9*. Certainly, 9* fixes t and each line through its centre, hence it also fixes each subspace on t. In particular, for i = 1 , . . . , q we have that 9* fixes each subspace Ni and fixes the cone Ci in ~i. Consider the


restriction 9* ofg* onto Ei; it is clear that 9* is an elation with the same centre as 9* and axis Eoo N Ei. By (3.9), the centre of 9~', and hence the centre ofg, is ~ for i = 1 , . . . , q, a contradiction. As we have seen above, the only possibility is that b/ is a Buekenhout-Metz unital. []

(1.5) THEOREM. Let U be a unital in PG(2, q2), where q > 2. Then ~ is a Buekenhout-Metz unital i f and only i f

(i) there exists a point T o f ~ such that the points o f ~ on each of the q2 secant lines to bt through T form a Baer subline, and

(ii) there exists a point P on ~oo \ {T) such that the points o f tangency o f the q + 1 tangents to bl on P are collinear.

Proof First, suppose that b/is a Buekenhout-Metz unital. Then (i) follows from (3.10) and (ii) follows from (3.11).

Now suppose that b/ is a unital such that (i) and (ii) hold, but that L/is not a Buekenhout-Metz unital. By (1.3), q is odd and q > 3. By (i) and (3.1)-(3.6), in the usual notation, the lines ~1, . . . , gqz fall into q cones C1 , . . . , Cq in 3-dimensional spaces E l , . - . , Eq with distinct vertices V1,. . . , Vq on t. Let p be the element of the spread S corresponding to P, and choose i with 1 < i < q. The point p N rc lies on two tangent planes to Ci in Ei, namely 7r and one other, ~ri. Let £i = Ci N ~ri and consider the 3-dimensional space E = (p, gi). We count the points of U* in E, that is, the points of ~* in E \ Eoo, in two different ways. Let a be the number of planes 7r' in E and on p such that 17d fq H*[ = 1. Then ]E fq U*l = a + (q - a)(q + 1), as each plane :r / in E \ E ~ corresponds to a line of PG(2, q2) on P, distinct from £oo. If a of these lines are tangents then the remaining q - a are secants. Now

For each 1 _< j _< q with j # i, the plane E N ~j meets ~* in a conic meeting E ~ in ~ ; while E N Ei meets 9 " in the line gi. Hence I~ N b/* I = (q - 1)q + q = q2. Comparing the two expressions for I E N///'1, we see that a = 1. In other words, there is a unique point Xi of gi such that (p, X~) corresponds to a tangent line to b/in PG(2, q2). Hence, in PG(2, q2), each of q lines on T, other than goo, contains exactly one point lying on a tangent through P. Thus no two points of tangency to L/of tangents on P are collinear with T; contrary to (ii). []

Of course, in (1.5)(ii) we need only assume that T and two further points of tangency on tangents to L/on P are collinear.

Also, we note that the hypothesis in (ii) is closely related to work of Hirschfeld, Storme, Thas and Voloch [16] and Thas [26].

4 0 L.R.A. CASSE ET AL.

Remarks. In fact any translation plane 7r of order q2 with kemel containing GF(q) has the above representation in PG(4, q) using a spread S in a hyperplane PG(3, q) of PG(4, q), where the spread S is not assumed to be regular (see [3], [7], [8]). However our current method does not extend to prove (1.3) for unitals in these non-Desarguesian planes, for we rely on the fact that the Baer sublines in PG(2, q2) with a point on the line at infinity correspond exactly to the lines of PG(4, q)\PG(3, q). This result need not hold in a non-Desarguesian translation plane; in particular results of Vincenti [27, Prop. 2] and Freeman [13, Th. 4.2] provide counter-examples in certain translation planes of fourth power order, which we now exhibit.

Let 7r be a non-Desarguesian translation plane of order q2 and kemel GF(q) constructed from a spread S in Eo~ embedded as a hyperplane in PG(4, q). Suppose that q is a square and that there exists a subgeometry PG(4, v~) such that S induces a spread on the subgeometry PG(3, x/r~) = PG(4, x/q) N E~. The points of PG(4, x/~)\PG(3, x/~) represent an affine Baer subplane of 7r ([27, Prop. 2]). A plane in PG(4, x/~) which meets PG(3, x/~) in exactly a line of S N PG(3, v/-~) is clearly not a line in PG(4, q) but represents a Baer subline of 7r, since it is the intersection of the sets of points corresponding to a line and a Baer subplane of 71".

The existence of such subgeometry and spread configurations is guaranteed by results of Freeman, see [13, Th. 4.2].

We note that in the course of the proof of a related characterization, Larato [18] appears to have neglected these examples of Baer sublines in non-Desarguesian planes.

Larato's result provides an alternative characterization of Buekenhout-Metz unitals in PG(2, q2) for q odd, using hypotheses which are slightly stronger than those of our (1.3). In fact Larato's result follows from (3.1)-(3.6) and the observation that two Baer sublines which each contain the point T lie in an affine Baer subplane if and only if the corresponding lines in PG(4, q) meet Eo~ in the same point and are not coplanar with the spread line they meet. Also, (1.3) extends Larato's result to the case of q even while weakening the hypotheses.

4. The Known Unitals in PG(2, q2)

As in the introduction, each unital in PG(2, q2) known to the authors is either a classical unital, a Buekenhout-Metz unital, a Buekenhout hyperbolic unital or is the dual of one of these unitals. We note, in particular, that the unitals investigated by Baker and Ebert [4], [5], by Ebert [11] and by Hirschfeld and SzSnyi [17] are shown, in the respective papers, to be Buekenhout-Metz unitals. These unitals have been characterized, in terms of their groups, by Abatangelo [1] and by Abatangelo and Larato [2].

It is well known that the dual of a classical unital in PG(2, q2) is classical. Fur- ther, a Buekenhout hyperbolic unital in PG(2, q2) is classical, see [6]. Since classical


unitals are Buekenhout-Metz unitals, in order to show that all the known unitals are Buekenhout-Metz unitals we need only show that the dual of a Buekenhout-Metz unital is again a Buekenhout-Metz unital.

By [5, Cor. 2] and [11, Cor. 6], we only need prove the result in the case of q even and (_9 not an elliptic quadric. However we give a proof valid in all cases.

(1.6) THEOREM. The dual of a Buekenhout-Metz unital is also a Buekenhout- Metz unital.

Proof Let H be a Buekenhout-Metz unital in PG(2, q2) and let ~d be its dual in PG(2, q2)g. Let T and g~ be as in the statement of (3.11). We will apply (1.4), first checking conditions (i) and (ii).

First, consider the point g~ of H d, and let m be a secant line in PG(2, q2)d on

~ . The points of ~d on m correspond in PG(2, q2) to the set of tangent lines to the unital H passing through a point M on the tangent go~. By (3.11), the points of tangency on these lines form a Baer subline; so the tangents on M form a Baer

subpencil of the pencil of lines on M. In PG(2, q2)a, therefore, the points of ~a on m form a Baer subline. Thus (i) holds.

By (1.4), there is a non-trivial elation 9 with centre T fixing H. As in the proof of (1.4), the axis of g is ~o~. Since a unital has a unique tangent at each of its points, an automorphism of PG(2, q2) fixing a unital is an automorphism of the dual plane

fixing the dual unital. Thus 9 is an automorphism of PG(2, q2)a fixing Ha, and it is clear that g is an elation with centre go~ and axis T.

By (1.4), the dual unital ~a is a Buekenhout-Metz unital in the dual plane, hence is a Buekenhout-Metz unital in PG(2, q2). []

Acknowledgements

The authors wish to thank Catherine Quinn and the referee for pointing out an error in the original version of this paper.

The authors acknowledge the support of the Australian Research Council.

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N o t e added in proof: T h e o r e m 1.3 has n o w been p roved for q > 3 by Cather ine

T. Qu inn and R e y Casse, 'Conce rn ing a character isat ion o f B u e k e n h o u t - M e t z

uni ta ls ' , J. Geom. 52 (1995), 159-167 .

characterizations of buekenhout-metz unitals

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