A characterization of the planes meeting a hyperbolic quadric of \PG(3,q) in a conic
aa r X i v : . [ m a t h . C O ] F e b A characterization of the planes meeting a hyperbolicquadric of PG(3 , q ) in a conic
Bikramaditya SahuFebruary 9, 2021
Abstract
In this article, a combinatorial characterization of the family of planes of PG(3 , q )which meet a hyperbolic quadric in an irreducible conic, using their intersectionproperties with the points and lines of PG(3 , q ), is given.
Keywords:
Projective space, Hyperbolic quadric, Irreducible conic, Combina-torial characterization
AMS 2010 subject classification:
Let PG( n, q ) denote the n -dimensional Desarguesian projective space defined over a finitefield of order q , where q is a prime power. The quadrics of PG( n, q ) are very interestingobjects with many combinatorial properties. One of the important properties of quadricsin PG( n, q ) is that subspaces can only meet a quadric in certain ways. Therefore, we mayform families of subspaces that all meet a particular quadric in the same way and thenwe can give a characterization of that family.A characterization of the family of planes meeting a non-degenerate quadric in PG(4 , q )is given in [3]. In a series of two recent papers [1, 2], characterizations of elliptic hyper-planes and hyperbolic hyperplanes in PG(4 , q ) are given. A characterization of the familyof external lines to a hyperbolic quadric in PG(3 , q ) was given in [4] for all q (also see [6]for a different characterization in terms of a point-subset of the Klein quadric in PG(5 , q )).In a recent paper [8], a characterization of secant lines to a hyperbolic quadric is given forall odd q , q ≥
7. In this paper, we give a characterization of the family of planes meetinga hyperbolic quadric in PG(3 , q ) in an irreducible conic.Let Q be a hyperbolic quadric in PG(3 , q ), that is, a non-degenerate quadric of Wittindex two. One can refer to [5] for the basic properties of the points, lines and planes ofPG(3 , q ) with respect to Q . The quadric Q consists of ( q + 1) points and 2( q + 1) lines.Every plane of PG(3 , q ) meets Q in an irreducible conic or in two intersecting generators.In the first case we call the plane as a secant plane, otherwise, a tangent plane. There are q − q secant planes and ( q + 1) tangent planes. Each point of PG(3 , q ) \ Q lies on q secant planes and each point of Q lies on q − q secant planes. Each line of PG(3 , q ) lieson 0, q − q or q + 1 secant planes. 1n this paper, we prove the following characterization theorem. Theorem 1.1.
Let Σ be a non empty family of planes of PG(3 , q ) , for which the followingproperties are satisfied:(P1) Every point of PG(3 , q ) lies on q − q or q planes of Σ .(P2) Every line of PG(3 , q ) lies on , q − , q or q + 1 planes of Σ .Then Σ is the set of all planes of PG(3 , q ) meeting a hyperbolic quadric in an irreducibleconic. Let Σ be a nonempty set of planes of PG(3 , q ) for which the properties (P1) and (P2)stated in Theorem 1.1 hold. A point of PG(3 , q ) is said to be black or white according asit contained in q − q or q planes of Σ. Let b and w , respectively, denote the number ofblack points and the number of white points in PG(3 , q ). We have b + w = q + q + q + 1 . (1)Counting in two ways the point-plane incident pairs, { ( x, π ) | x is a point in PG(3 , q ) , π ∈ Σ and x ∈ π } , we get b ( q − q ) + wq = | Σ | ( q + q + 1) . (2)Again, counting in two ways the incident triples, { ( x, π, σ ) | x is a point in PG(3 , q ) , π, σ ∈ Σ , π = σ and x ∈ π ∩ σ } , we get b ( q − q )( q − q −
1) + wq ( q −
1) = | Σ | ( | Σ | − q + 1) . (3)If q is odd, then dividing both sides of Equation (3) by 2 to get the following equality. bq ( q − / q − q −
1) + wq ( q − q + 1) / | Σ | ( | Σ | − q + 1) / . (4) Lemma 2.1. (i) If q is even, then q + 1 divides b .(ii) If q is odd, then ( q + 1) / divides b .Proof. Since q − q − q + 11)( q −
2) + 1, we have q + 1 is co-prime to q − q −
1. Also q + 1 is co-prime to q being consecutive integers. If q is even, then q + 1 and q − q is odd,then ( q + 1) / q − / q + 1) / q are co-prime). Now (i) and (ii) follows from Equations (3) and (4), respectively. Lemma 2.2.
Every plane of Σ contains a fixed number of black points. roof. Let π be a plane in Σ. Let b π and w π , respectively, denote the number of blackand white points in π . Then w π = q + q + 1 − b π .By counting the incident point-plane pairs of the following set, { ( x, σ ) | x is a point in PG(3 , q ) , σ ∈ Σ , σ = π and x ∈ π ∩ σ } , we get b π ( q − q −
1) + w π ( q −
1) = ( | Σ | − q + 1) . Since q + 1 is co-prime to q − q − q + 1 divides b π . Let b π = ( q + 1) r π for some 0 ≤ r π ≤ q .Since w π = q + q + 1 − b π = q + q + 1 − ( q + 1) r π , by a simple calculation, we get q − qr π = | Σ | . Since | Σ | is a fixed number, r π := r is fixed. Hence b π = ( q + 1) r is a fixed number.From the proof of Lemma 2.2, q − qr = | Σ | , (5)where the number of black points in a plane in Σ is ( q + 1) r for some 0 ≤ r ≤ q . Lemma 2.3.
Any plane of
PG(3 , q ) not in Σ contains q + ( q + 1) r black points.Proof. We prove the lemma with a similar argument as that of Lemma 2.2. Let π be aplane of PG(3 , q ) not in Σ. Let b π and w π denote the number of black and white points,respectively, in π . Then w π = q + q + 1 − b π .By counting the incident point-plane pairs of the following set, { ( x, σ ) | x is a point in PG(3 , q ) , σ / ∈ Σ , σ = π and x ∈ π ∩ σ } , we get b π (2 q + 1 −
1) + w π ( q + 1 −
1) = ( q + q + q + 1 − | Σ | − q + 1) . From Equation (5,) | Σ | = q r − qr . The above equation simplifies to2 b π + w π = ( q + 1 + r )( q + 1) . Since w π = q + q + 1 − b π , we get b π = q + ( q + 1) r . Corollary 2.4. ≤ r ≤ q − .Proof. Note that 0 ≤ r ≤ q . Suppose r = 0. By Equation (5), we get | Σ | = q . FromEquation (2), it follows that b ( q −
1) + qw = q ( q + q + 1) . Putting w = q + q + q + 1 − b (Equation (1)) in the above equation to get b = q . Butthen from Equation (3), q + 1 divides q ( q − q is co-prime to q + 1, we see that q + 1 divides q − r = q . Then by Lemma 2.3, any plane of PG(3 , q ) not in Σ contains q + 2 q black points, which is a contradiction. Hence 1 ≤ r ≤ q − emma 2.5. We have r = 1 , b = ( q + 1) and | Σ | = q − q . In particular, every plane in Σ contains ( q + 1) black points.Proof. Note that, by Corollary 2.4, 1 ≤ r ≤ q −
1. If q = 2, then r = 1. Assume first that q even, q ≥ q ( q −
1) divides | Σ | ( | Σ | − | Σ | = q − qr (Equation (5)), we have q ( q −
1) divides ( q − qr )( q − qr − q − q − r ) or ( q − qr − q − r = ( q − − ( r −
1) and q − qr − q − q ) + ( q − − qr . If q − q − r ), then r = 1 or r = q . If q − q − qr −
1, then r = 0 or r = q − ≤ r ≤ q −
1, we have r = 1 or q − r = q −
1. putting | Σ | = q − qr in Equation (2) we get that b ( q −
1) + qw = ( q − q + 1)( q + q + 1) . Since w = q + q + q +1 − b , solving the above equation for b , it follows that b = q + q − q ( q + 1) − ( q − q + 1) + ( q − q ≥ r = 1 for all q even.Now assume that q is odd.By Equation (4) and Lemma 2.1(ii), we see that q ( q − / | Σ | ( | Σ | − q ( q − / q − qr )( q − qr − q − / q − r ) or ( q − qr − q − r = ( q − − ( r −
1) and q − qr − q − q ) + ( q − − qr . If( q − / q − r ), then r = 1 or r = ( q + 1) /
2. If ( q − / q − qr − r = ( q − / r = q −
1. So, r = 1 , ( q − / , ( q + 1) / q − r = q −
1. As similar to q even case, here also we get b = q + q − q + q ) − ( q −
1) + ( q − q + 1) / q − q + 1) / q + 1 −
3. This gives q = 5. So b = 5 + 5 − r = 4 and hence | Σ | = 5 − · w = 5 + 5 + 5 + 1 − b = 27. On the other hand, putting the values of b = 129 and | Σ | = 105 in Equation (3), we get that w = 552 , which is a contradiction.Suppose that r = ( q − /
2. Then | Σ | = q − q ( q − /
2. By a similar calculation asbefore using Equations (1) and (2), we get that b = ( q + 2 q − / q + 1) / q − q + 1) / q − q + 1) / q + 1 −
2. This gives q = 3and hence r = ( q − / r = ( q + 1) /
2. Again by using Equations (1) and (2), we get that b = ( q + 2 q + 4 q + 1) / q + 1) / q ( q + 1) + q . Again by Lemma 2.1(ii), ( q + 1) / q , which is a contradiction as q and q + 1 are co-prime. Hence r = 1 for all q odd.Hence for all q , we have r = 1 and | Σ | = q − q . Putting the values of | Σ | and w = q + q + q + 1 − b in Equation (2), it follows that b = ( q + 1) . Since r = 1, everyplane in Σ contains ( q + 1) black points. Corollary 2.6. (i) There are ( q + 1) planes in PG(3 , q ) which are not in Σ .(ii) Every plane in PG(3 , q ) not in Σ contains q + 1 black points.Proof. (i) follows from Lemma 2.5 and the fact that there are ( q + 1)( q + 1) planes inPG(3 , q ). Now (ii) follows from Lemma 2.3, since r = 1.4e call a plane of PG(3 , q ) tangent if it is not a plane in Σ. By Corollary 2.6, everytangent plane contains 2 q + 1 black points. Lemma 2.7.
Let l be a line of PG(3 , q ) . Then the number of tangent planes through l isequal to the number of black points contained in l .Proof. Let t and s , respectively, denote the number of tangent planes through l and thenumber of black points contained in l . We count in two different ways the point-planeincident pairs, { ( x, π ) | x ∈ l, π is a tangent plane and x ∈ π } . Hence s (2 q + 1) + ( q + 1 − s )( q + 1) = (( q + 1) − t ) · t · ( q + 1) . It follows that s = t . This proves the lemma. Corollary 2.8.
Every line of
PG(3 , q ) contains , , or q + 1 black points.Proof. From Theorem 1.1(P2), every line is contained in 0 , , q + 1 tangent planes.The proof now follows from Lemma 2.7As a consequence of Lemma 2.7, we have the following. Corollary 2.9.
Every black point is contained in some tangent plane.
Let B be the set of all black points in PG(3 , q ). We call a line black if all its q + 1 pointsare contained in B . Lemma 3.1.
Let π be a tangent plane. Then the set π ∩ B is a union of two (intersecting)black lines.Proof. By Corollary 2.6(ii), we have | π ∩ B | = 2 q + 1. Then π ∩ B is a not an arc, since anarc in any plane has at most q + 2 points. Thus there is a line l in π such that | l ∩ B | ≥ l ⊆ B .Let x be a point on l . Since | π ∩ B | = 2 q + 1 and l ⊆ B , other q points (points noton l ) of π ∩ B lie on q lines through x , different from l . If there is a line m ( = l ) through x containing two points of π ∩ B \ { x } , then | m ∩ B | ≥
3. By Corollary 2.8, m ⊆ B . So,in this case π ∩ B = l ∪ m . On the other hand, if there are two points y and z (differentfrom x ) of B which lie on two different lines (different from l ) through x , then the line m := yz intersects l at a point different from y and z . In particular | m ∩ B | ≥
3. Againby Corollary 2.8, m ⊆ B . In any case, π ∩ B = l ∪ m . This proves the lemma. Lemma 3.2.
Every black point lies on at most two black lines.Proof.
Let x be a black point. If possible, suppose that there are three distinct black lines l, m, k each of which contains x .Note that, by Lemma 2.7, each of the q + 1 planes through l as well as m are tangentplanes giving rise to 2 q + 2 tangent planes through x (with repetition allowed). Since the5angent plane h l, m i contains both l and m , each of the 2 q + 1 tangent planes through x passes through either l or m .On the other hand, again by Lemma 2.7, there are q +1 tangent planes through k . Outof these q + 1 planes through k , the planes h k, l i and h k, m i have already been countedas planes through x . Since q ≥
2, there is a tangent plane through k (and hence through x ), different from h k, l i and h k, m i . This gives a contradiction to the fact that each of thetangent plane through x is either passes through l or m . Hence, there are at most twoblack lines through through x . Lemma 3.3.
Every black point is contained in precisely two black lines.Proof.
Let x be a black point and l be a black line containing x . The existence of sucha line l follows from the facts that x is contained in a tangent plane (Corollary 2.9) andthat the set of all black points in that tangent plane is a union of two black lines (Lemma3.1). By Lemma 2.7, let π , π , . . . , π q +1 be the q + 1 tangent planes through l . For1 ≤ i ≤ q + 1, by Lemma 3.1, we have B ∩ π i = l ∪ l i for some black line l i of π i differentfrom l . Let { p i } = l ∩ l i . Lemma 3.2 implies that p i = p j for 1 ≤ i = j ≤ q + 1, and so l = { p , p , . . . , p q +1 } . Since x ∈ l , we have x = p j for some 1 ≤ j ≤ q + 1. Thus, x iscontained in precisely two black lines, namely, l and l j . We refer to [7] for the basics on finite generalized quadrangles. Let s and t be positiveintegers. A generalized quadrangle of order ( s, t ) is a point-line geometry X = ( P, L ) withpoint set P and line set L satisfying the following three axioms:(Q1) Every line contains s + 1 points and every point is contained in t + 1 lines.(Q2) Two distinct lines have at most one point in common (equivalently, two distinctpoints are contained in at most one line).(Q3) For every point-line pair ( x, l ) ∈ P × L with x / ∈ l , there exists a unique line m ∈ L containing x and intersecting l .Let X = ( P, L ) be a generalized quadrangle of order ( s, t ). Then, | P | = ( s + 1)( st + 1)and | L | = ( t + 1)( st + 1) [7, 1.2.1]. If P is a subset of the point set of some projectivespace P G ( n, q ), L is a set of lines of P G ( n, q ) and P is the union of all lines in L , then X = ( P, L ) is called a projective generalized quadrangle . The points and the lines con-tained in a hyperbolic quadric in PG(3 , q ) form a projective generalized quadrangle oforder ( q, q,
1) with ambi-ent space PG(3 , q ) is a hyperbolic quadric in PG(3 , q ), this follows from [7, 4.4.8].The following two lemmas complete the proof of Theorem 1.1.
Lemma 4.1.
The points of B together with the black lines form a hyperbolic quadric in PG(3 , q ) . roof. We have | B | = b = ( q + 1) by Lemma 2.5. It is enough to show that the points of B together with the black lines form a projective generalized quadrangle of order ( q, q + 1 points of B . By Lemma 3.3, each point of B is containedin exactly two black lines. Thus the axiom (Q1) is satisfied with s = q and t = 1. Clearly,the axiom (Q2) is satisfied.We verify the axiom (Q3). Let l = { x , x , . . . , x q +1 } be a black line and x be ablack point not contained in l . By Lemma 3.3, let l i be the second black line through x i (different from l ) for 1 ≤ i ≤ q + 1. If l i and l j intersect for i = j , then the tangentplane π generated by l i and l j contains l as well. This implies that π ∩ B contains theunion of three distinct black lines (namely, l, l i , l j ), which is not possible by Lemma 3.1.Thus the black lines l , l , . . . , l q +1 are pairwise disjoint. These q + 1 black lines contain( q + 1) black points and hence their union must be equal to B . In particular, x is a pointof l j for unique j ∈ { , , . . . , q + 1 } . Then l j is the unique black line containing x j andintersecting l .From the above, it follows that the points of B together with the black lines form aprojective generalized quadrangle of order ( q, Lemma 4.2.
The set of planes in Σ are the planes meeting B in a conic.Proof. Let π be a plane in Σ. Note that | π ∩ B | = q + 1 by Lemma 2.5. Suppose that l is a line of π containing three points of π ∩ B . Then by Corollary 2.8, l is contained in π ∩ B . Since | π ∩ B | = q + 1, we have l = π ∩ B . By Lemma 2.7, all the planes through l are tangent, which is a contradiction to the fact that π is not a tangent plane. Hence | l ∩ B | ≤