A Characterisation of the Lines Externaltoan Oval Cone in PG(3,q), q Even

J. Geom. 93 (2009), 21–27 c 2009 Birkh¨  auser Verlag Basel/Switzerland 0047-2468/010021-7, published online 15 April 2009 DOI 10.1007/s00022-009-1966-2

Journal of Geometry

A Characterisation of the Lines External to an Oval Cone in PG(3, q), q Even S. G. Barwick and David K. Butler Abstract. In this article, the lines not meeting an oval cone in PG(3, q) (q even) are characterised by their intersection properties with points and planes. Mathematics Subject Classification (2000). 51E20. Keywords. Projective space, oval cone, lines, characterisation.

1. Introduction Recently, Di Gennaro, Durante and Olanda [1], and Durante and Olanda [2] have characterised the lines external to each of the two non-singular quadrics in PG(3, q). Durante and Olanda proved this result for the elliptic quadric: Theorem 1.1 ([2]). Let L be a set of lines in PG(3, q), q > 2 such that: (I) every point lies on 0 or 12 q(q + 1) lines of L , (II) every plane contains q 2 or 12 q(q − 1) lines of L . Then L is the set of external lines to an ovoid in PG(3, q). In their proof, they investigate the set of points K not lying on any of the lines of L , and show that K is an ovoid. In the q odd case, K is an elliptic quadric. Following this success for the elliptic quadric, Di Gennaro, Durante and Olanda used similar techniques to prove the following characterisation for lines external to the hyperbolic quadric: Theorem 1.2 ([1]). Let L be a set of lines in PG(3, q), q odd, such that: (I) every point lies on 0 or 12 q(q − 1) lines of L , (II) every plane contains 0 or 12 q(q − 1) lines of L , (III) in every plane, there are 0, 12 (q − 1) or 12 (q + 1) lines of L through any point.

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Then there are these possibilities for L : • L is the set of lines external to a hyperbolic quadric, • L is contained in the set of lines not meeting a fixed pair of skew lines, • L is contained in the set of lines not meeting a single fixed line. Given these two results, it is reasonable to ask whether a similar result may be obtained for the irreducible, singular quadric PG(3, q), that is, for the quadric cone. In this article, we prove such a result for the q even case, namely we prove the following characterisation: Theorem 1.3. Let L be a non-empty set of lines in PG(3, q), q even, such that: (I) every plane contains 0, q 2 or 12 q(q − 1) lines of L , (II) every point is on 0 or 12 q 2 lines of L . Then L is the set of lines external to a hyperoval cone C . Further, L is the set of lines external to each of the q + 2 different oval cones contained in C .

2. The proof of Theorem 1.3 We prove Theorem 1.3 by a series of lemmas. Let L be a set of lines as described in the theorem. In order to make the argument clearer, we will define some terminology: • A point on 0 lines of L will be called a black point; a point on 12 q 2 lines of L will be called a white point. • A plane containing 0 lines of L will be called a 0-plane. • A plane containing q 2 lines of L will be called a V-plane. • A plane containing 12 q(q − 1) lines of L will be called a secant plane. Lemma 2.1. There are q secant planes and one V-plane through any line of L . Proof. Let  be a line of L and let v be the number of V-planes through . We will count the number of lines of L meeting (but not equal to)  in two ways; let this number be L . We first count L by considering the lines through each point on . Each point on  is a white point, and so is on 12 q 2 lines of L (including ). Counting this way, we have included  itself q + 1 times. So we have: L =

1 2 q (q + 1) − (q + 1) . 2

On the other hand, we can count L by considering the lines in each plane through . There are v V-planes through , and no 0-planes (since  is a line of L ). This leaves (q + 1 − v ) secant planes through . Each V-plane contains q 2 lines of L

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(including ) and each secant plane contains 12 q(q − 1) lines of L (including ). Again we have counted  itself q + 1 times, so we have: 1 L = q 2 · v + q(q − 1) · (q + 1 − v ) − (q + 1) 2 1 1 1 = q(q + 1)v + q 2 (q + 1) − q(q + 1) − (q + 1) . 2 2 2 Hence, 1 2 1 1 1 q (q + 1) − (q + 1) = q(q + 1)v + q 2 (q + 1) − q(q + 1) − (q + 1) , 2 2 2 2 and so v = 1. That is, there is one V-plane through . Hence there are q secant planes through .  Note that the above lemma ensures the existence of both secant planes and Vplanes as L is non-empty. Lemma 2.2. There is at most one black point in a V-plane Proof. Let π be a V-plane and suppose P and Q are two black points in π. Now black points are on no lines of L , so none of the 2q + 1 lines through these two points can be lines of L . This leaves q 2 − q other lines in π. So π has at most q 2 − q lines of L . But π is a V-plane and contains q 2 lines of L . This is a contradiction, so there can be at most one black point in π.  Lemma 2.3. The number of black points in each type of plane is constant. That is, all secant planes have the same number of black points, all V-planes have the same number of black points, and all 0-planes have the same number of black points. Proof. Let π be a plane containing Lπ lines of L . Let the number of black points in π be Bπ . Thus there are (q 2 + q + 1 − Bπ ) white points in π. We will count the size of L by considering the lines of L through each point in π. Through each black point there are no lines of L , and through each white point there are 12 q 2 lines of L . However, if we count in this way, each line of L in π has been counted q + 1 times – once for each white point on it. So we must subtract q times the number of lines of L in π. Thus: 1 |L | = 0 · Bπ + q 2 · (q 2 + q + 1 − Bπ ) − qLπ 2 1 = q 2 (q 2 + q + 1 − Bπ ) − qLπ . 2 In the above equation, |L | is a constant, so Bπ is only dependent on Lπ – the number of lines of L in π. In other words, two planes of the same type have the same number of black points.  Lemma 2.4. Every 0-plane contains 2q + 1 black points, every V-plane contains 1 black point, and every secant plane contains q + 2 black points. Also, |L | = 1 3 2 q (q − 1).

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Proof. Let B0 be the number of black points in a 0-plane, BV the number of black points in a V-plane, and Bs the number of black points in a secant plane. From the proof of Lemma 2.3, if π is a plane containing Bπ black points and Lπ lines of L , then: 1 (∗) |L | = q 2 (q 2 + q + 1 − Bπ ) − qLπ . 2 2 If we choose π to be a V-plane, then Lπ = q and equation (∗) becomes: 1 2 2 q (q + q + 1 − BV ) − q 3 . 2 Now, by Lemma 2.2, there is at most one black point in a V-plane, so BV = 0 or 1. When BV = 0, |L | = 12 q 2 (q 2 + q + 1) − q 3 = 12 q 2 (q 2 − q + 1) and when BV = 1, |L | = 12 q 2 (q 2 + q) − q 3 = 12 q 3 (q − 1). |L | =

Let X = {(, π) |  is a line of L , π is a V-plane,  ∈ π}. We will count |X| in two ways: 1. Count  first, then π: There are |L | choices for , and then 1 choice for π through  by Lemma 2.1. Thus |X| = |L |. 2. Count π first, then : Let v be the total number of V-planes. Then we have v choices for π. After this we have q 2 choices for  within π. Thus |X| = v · q 2 . Hence: |L | = q 2 v, and so v =

|L | q2 .

Now if |L | = 12 q 2 (q 2 − q + 1), then v = 12 (q 2 − q + 1). But q 2 − q + 1 is an odd number, which means that v cannot be a whole number. This is a contradiction, since v is the number of V-planes, so |L | = 12 q 3 (q − 1), and hence also BV = 1. If we choose our original π to be a secant plane, then Lπ = 12 q(q − 1). Substituting this and the value of |L | into equation (∗) gives 12 q 3 (q − 1) = 12 q 2 (q 2 + q + 1 − Bs ) − 12 q 2 (q − 1), and so Bs = q + 2. Finally, if we choose our original π to be a 0-plane, then Lπ = 0. Substituting this and the value of |L | into equation (∗) gives 12 q 3 (q − 1) = 12 q 2 (q 2 + q + 1 − B0 ),  and so B0 = 2q + 1. Lemma 2.5. There are (q + 1)2 black points in total Proof. Let  be a line of L and consider the planes about . All the black points must be contained in these planes. Now  has one V-plane through it and q secant planes. The V-plane has one black point, and the secant planes have q + 2. Thus  the number of black points is 1 · 1 + q · (q + 2) = (q + 1)2 . Lemma 2.6. There are q 3 secant planes, V-planes.

1 2 (q

+ 1)(q + 2) 0-planes, and

1 2 q(q

− 1)

Proof. Let s be the number of secant planes, v the number of V-planes, and n the 1 3 | 2 q (q−1) = 12 q(q −1). number of 0-planes. From the proof of Lemma 2.4, v = |L q2 = q2

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To find s, let X = {(, π) |  is a line of L , π is a secant plane,  ∈ π}. We will count X in two ways: 1. Count  first, then π: There are 12 q 3 (q − 1) choices for  and q choices for π through . So |X| = 12 q 4 (q − 1). 2. Count π first, then : There are s choices for π and 12 q(q − 1) choices for  in π. So |X| = s · 12 q(q − 1). Thus 12 q(q − 1)s = 12 q 4 (q − 1), and so s = q 3 . Finally, there are q 3 + q 2 + q + 1 planes in total, so n = q 3 + q 2 + q + 1 − v − s = 1  2 (q + 1)(q + 2). Lemma 2.7. There exists a single point V through which pass all V-planes and 0-planes. The secant planes are precisely those planes not through V . Proof. Let π be a V-plane. By Lemma 2.4, π contains a single black point, V . The q + 1 lines through V are clearly not lines of L , so the remaining q 2 lines in π must be the q 2 lines of L in π. That is, every line not through V is a line of L . Now by Lemma 2.1, a line of L is contained in exactly one V-plane. For the lines not through V in π, this plane is π itself. So, any other V-planes must pass through V . (Note that if π is the only V-plane, then it clearly passes through V ). Also, a line of L is contained in no 0-planes. So the lines of π not through V are contained in no 0-planes. Hence, any 0-planes must pass through V . The total number of V-planes and 0-planes is 12 q(q − 1) + 12 (q + 1)(q + 2) = q 2 + q + 1. This is the total number of planes through V . So there are no secant planes through V . There are q 3 secant planes in total, and q 3 planes not through V , so the planes not through V are precisely the secant planes.  Lemma 2.8. A line through V either has V as its only black point, or consists entirely of black points. A line not through V is either a line of L (containing no black points) or contains exactly two black points. Proof. Let m be a line through V and suppose m contains a further black point, P . Consider the planes through m. Any plane through m contains V , and so is not a secant plane. Also, any plane through m has at least two black points (P and V ), so it is not a V-plane. Thus all planes through m are 0-planes. Since there are no lines of L in any 0-plane, there can be no lines of L meeting m. Thus m must consist entirely of black points. So a line through V either has V as its only black point, or consists entirely of black points. Now let m be a line not through V and suppose m is not a line of L . Let Bm be the number of black points on m. We will count the total number of all black points by considering the black points in the planes about m. Now exactly one plane passes through both m and V . If this plane were a V-plane, then m would be a line of L (as noted in the proof of Lemma 2.7). Thus this plane is a 0-plane and contains 2q + 1 black points by Lemma 2.4. The other q planes

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through m are secant planes and each contains q + 2 black points. Counting this way, the black points on m have been counted q + 1 times, once for each plane through m. Thus (q + 1)2 = 2q + 1 + q(q + 2) − qBm , and so Bm = 2. Hence a line not through V is either a line of L or contains exactly 2 black points.  Lemma 2.9. The q + 2 black points in a secant plane form a hyperoval. Proof. Let π be a secant plane and let m be a line of π. Then m cannot contain the point V , since no secant plane passes through V . Hence, by Lemma 2.8, every line of π is either a line of L or contains exactly two black points. In particular, no three black points of π are contained in a line. Hence, the set of black points in π form an arc, which is a hyperoval because it has q + 2 points.  Lemma 2.10. The set of black points C is a hyperoval cone and L is its set of external lines. Proof. Let π be a secant plane and O the hyperoval of black points in π. Consider the q + 2 lines joining V to a point of O. Now each of these lines passes through V and contains at least two black points, so by Lemma 2.8, each of these lines must consist entirely of black points. The number of points on these q + 2 lines is: q(q + 2) + 1 = q 2 + 2q + 1 = (q + 1)2 , and this is the size of C . Thus, C is the set of points on the lines joining V to the points of a hyperoval O. That is, C is a hyperoval cone. By Lemma 2.8, a line not in L must contain a black point, so the lines of L are precisely the lines not meeting C .  Note that if g is a generator of the hyperoval cone C , then the set of points obtained by removing all the points of g except V is an oval cone. Since C has q + 2 generators, there are q + 2 different oval cones contained in C . It remains to show that each of these cones has L as its set of external lines. Lemma 2.11. Each of the oval cones contained in C has L as its set of external lines. Proof. Let g be a generator of C and let C  be the oval cone obtained by removing the points of g except V . Now the lines of L are external to C , so they must also be external to C  , since C  is contained in C . So we must show that there are no further external lines to C  . Let  be a line not in L . If  contains V , then it meets C  . If  does not contain V , then by Lemma 2.8, it must contain exactly two points of C . These two points are on different generators, so when we remove the points of g,  still contains at least one point of C  . Hence, a line not in L must meet C  . So, L is the set of external lines to C  .  This completes the proof of Theorem 1.3.

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References [1] R. Di Gennaro, N. Durante and D. Olanda, A characterization of the family of lines external to a hyperbolic quadric of PG(3, q), J. Geom. 80 (2004) 65–74. [2] N. Durante and D. Olanda, A characterization of the family of secant or external lines of an ovoid of PG(3, q), Bull. Belg. Math. Soc. Simon Stevin 12 (2005) 1–4. S. G. Barwick and David K. Butler School of Pure Mathematics The University of Adelaide Adelaide, 5005 Australia e-mail: [email protected] [email protected] Received: 29 November 2005.