Quadratic Sets 4.5 Elliptic, Parabolic and Hyperbolic Quadratic Sets

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1 Quadratic Sets 4.5 Elliptic, Parabolic and Hyperbolic Quadratic Sets

2 Definitions Def: Let Q be a nondegenerate quadratic set in a d-dimensional projective space P (not assumed to be finite). If d is even and the index of Q is ½d then Q is called parabolic. If d is odd and the index of Q is ½(d-1) then Q is called elliptic. Finally, if d is odd and the index of Q is ½(d+1) then Q is called hyperbolic.

3 Examples a) Theorem can be reformulated as: Any nonempty nondegenerate quadratic set of a finite projective space is elliptic, parabolic or hyperbolic. b) The parabolic quadratic sets of a projective plane are the ovals. c) In a 3-dimensional projective space, the elliptic quadratic sets are the ovoids and the hyperbolic quadratic sets are the hyperboloids.

4 A More General Definition of a Cone Def: Let H be a hyperplane of a projective space P, and let V be a point outside of H. If Q* is a nondegenerate quadratic set of H the quadratic set Q = (VX), X Q* is called a cone with vertex V over Q*.

5 Parabolic Quadratic Sets Theorem 4.5.1: Let Q be a parabolic quadratic set in a 2tdimensional projective space P with t > 1. a) Let H = Q P be a tangent hyperplane. Then Q' = Q H is a cone over a parabolic quadratic set with vertex P. b) Let H* be a hyperplane that is not a tangent hyperplane. Then Q* = Q H* is an elliptic or hyperbolic quadratic set. Pf: a) Let W be a complement of P in H, and define Q'' = Q W. By Lemma 4.1.4, Q'' is a nondegenerate quadratic set. If U denotes a maximal Q-subspace through P, then U'' = U W has dimension dim(u) 1. By Theorem 4.2.4, U'' is a maximal Q''-subspace, therefore Q'' is parabolic. Since,by Lemma 4.1.3, the radical of Q' consists of just one point, namely P, Q' is a cone over Q''.

6 Parabolic Quadratic Sets Theorem 4.5.1: Let Q be a parabolic quadratic set in a 2tdimensional projective space P with t > 1. a) Let H = Q P be a tangent hyperplane. Then Q' = Q H is a cone over a parabolic quadratic set with vertex P. b) Let H* be a hyperplane that is not a tangent hyperplane. Then Q* = Q H* is an elliptic or hyperbolic quadratic set. Pf (cont): b) Since H* is not a tangent hyperplane, Q* is nondegenerate (see 4.1.4). A maximal Q-subspace (which is a subspace of dimension t-1) intersects H* in a subspace of dimension t-1 or t-2. This means that Q* is elliptic or hyperbolic.

7 Finite Parabolic Quadratic Sets Corollary 4.5.2: Let Q be a nonempty, nondegenerate quadratic set in P = PG(4,q). Then Q induces in any tangent hyperplane a cone, and in any other hyperplane an ovoid or a hyperboloid. Furthermore, Q consists of exactly q 3 + q 2 + q + 1 points, the number of hyperplanes in which Q induces an ovoid is ½q 2 (q 2-1), and the number of hyperplanes in which Q induces a hyperboloid is ½q 2 (q 2 +1). Pf: We know that Q is parabolic. From this the first assertion follows from the Theorem. In particular, we get that for each point P in Q, the quadratic set induced in Q P is a cone with vertex P. Thus the number a of Q-lines through P is q+1. So, by Thm we have Q = 1 + q 3 + aq = 1 + q 3 + (q+1)q.

8 Finite Parabolic Quadratic Sets Corollary 4.5.2: Let Q be a nonempty, nondegenerate quadratic set in P = PG(4,q). Then Q induces in any tangent hyperplane a cone, and in any other hyperplane an ovoid or a hyperboloid. Furthermore, Q consists of exactly q 3 + q 2 + q + 1 points, the number of hyperplanes in which Q induces an ovoid is ½q 2 (q 2-1), and the number of hyperplanes in which Q induces a hyperboloid is ½q 2 (q 2 +1). Pf(cont.): Let t = Q = q 3 + q 2 + q + 1, be the number of tangent hyperplanes, h the number of hyperboloid hyperplanes and e the number of ovoid hyperplanes. Clearly, t + h + e = q 4 + q 3 + q 2 + q + 1. So, h + e = q 4. Now a tangent hyperplane contains exactly q 2 +q+1 points of Q, while a hyperboloid hyperplane contains (q+1) 2 and an ovoid hyperplane contains q Any point of Q is on exactly q 3 + q 2 + q + 1 hyperplanes, thus we have:

9 Finite Parabolic Quadratic Sets Corollary 4.5.2: Let Q be a nonempty, nondegenerate quadratic set in P = PG(4,q). Then Q induces in any tangent hyperplane a cone, and in any other hyperplane an ovoid or a hyperboloid. Furthermore, Q consists of exactly q 3 + q 2 + q + 1 points, the number of hyperplanes in which Q induces an ovoid is ½q 2 (q 2-1), and the number of hyperplanes in which Q induces a hyperboloid is ½q 2 (q 2 +1). Pf(cont.): t(q 2 + q + 1) + h(q+1) 2 + e(q 2 + 1) = Q (q 3 + q 2 + q + 1). Which implies, h(q+1) 2 + e(q 2 + 1) = Q (q 3 + q 2 + q + 1) - t(q 2 + q + 1) = q 3 (q 3 + q 2 + q + 1). So, h(q+1) 2 + (q 4 - h)(q 2 + 1) = q 3 (q 3 + q 2 + q + 1) or 2qh = q 3 (q 2 + 1).

10 Hyperbolic Quadratic Sets Theorem 4.5.3: Let Q be a hyperbolic quadratic set of a (2t+1)- dimensional projective space P with t > 1. a) If H = Q P is a tangent hyperplane then Q' = Q H is a cone over a hyperbolic quadratic set with vertex P. b) If H* is a hyperplane which is not a tangent hyperplane then Q* = Q H* is a parabolic quadratic set. Pf: a) Let W be a complement of P in H, and define Q'' = Q W. By Lemma 4.1.4, Q'' is a nondegenerate quadratic set. If U denotes a maximal Q-subspace through P, then U'' = U W has dimension dim(u) 1. By Theorem 4.2.4, U'' is a maximal Q''-subspace, therefore Q'' is hyperbolic. Since,by Lemma 4.1.3, the radical of Q' consists of just one point, namely P, Q' is a cone over Q''. b) follows directly from and

11 A Finite Hyperbolic Quadratic Set Theorem 4.5.4: Let Q be a hyperbolic quadratic set in P = PG(5,q). Then a = (q+1) 2 and Q = q 4 + q 3 + 2q 2 + q + 1 = (q 2 + q + 1)(q 2 + 1). Pf: The number of Q lines through a point P ( = a) is the number of lines through the vertex P in Q P, which equals the number of points in the hyperbolic quadratic set of Theorem This hyperbolic quadratic set lies in a 3 dimensional space, and so, has (q+1) 2 points. By b) we obtain: Q = 1 + q 4 + (q+1) 2 q.

12 Equivalence Def: Let Q be a hyperbolic quadratic set of a 5-dimensional projective space P. We say that two Q-planes π 1 and π 2 are equivalent (written π 1 ~ π 2 ) if π 1 and π 2 are equal or intersect in precisely one point. Lemma 4.5.5: Let Q be a hyperbolic quadratic set of a 5- dimensional projective space P. Then the relation ~ is an equivalence relation. Pf: The relation is obviously reflexive and symmetric, so we need only show that it is transitive. Let π 1, π 2 and π 3 be three Q-planes such that π 1 and π 2 meet at a point P and π 2 and π 3 meet at a point R. Since all lines through P in π 1 and π 2 are tangent lines, it follows that Q P = <π 1,π 2 >. Similarly, Q R = <π 2,π 3 >.

13 Equivalence Lemma 4.5.5: Let Q be a hyperbolic quadratic set of a 5-dimensional projective space P. Then the relation ~ is an equivalence relation. Pf(cont.): Case I: P = R. We must show that π 1 and π 3 have no additional points in common. Let W be a complement of P in H = Q P and define Q' = Q W. Let g i = π i W (i = 1,2,3). By we know that Q' is a hyperboloid. Since g 1 and g 2 are skew lines, they belong to the same class of Q' (both in the regulus or both in the opposite regulus). Similarly, g 2 and g 3 are skew, so they are in the same class. Thus, both g 1 and g 3 are in the same class as g 2 and so are skew. Therefore, π 1 π 3 = P.

14 Equivalence Lemma 4.5.5: Let Q be a hyperbolic quadratic set of a 5-dimensional projective space P. Then the relation ~ is an equivalence relation. Pf(cont.): Case II: P R. In this case π 3 is not contained in Q P. Thus, π 3 intersects the hyperplane Q P in a line g 3. Let W be a complement of P in Q P containing g 3, and define Q' = Q W. Then Q' is a hyperboloid, and g 1 = π 1 W and g 2 = π 2 W are lines of the same class of Q'. Since g 3 intersects g 2 in the point R, g 3 belongs to the other class. This implies that g 3 and g 1 also intersect in some point S. Thus, π 1 π 3 = S.

15 Equivalence Classes Theorem 4.5.6: Let Q be a hyperbolic quadratic set of a 5- dimensional projective space P. Then the set of all Q-planes is partitioned into exactly two equivalence classes with respect to ~. Pf: Let π 1 and π 2 be two Q-planes that intersect in a line g (these exist by the proof of Theorem 4.2.3). These belong to different equivalence classes. We have to show that every other Q-plane π belongs to one of these two classes. The subspace V = <π 1,π 2 > has dimension 3 and the quadratic set induced by Q in V consists only of the points on these two planes (if it contained another point then V would be a Q-space... but its dimension is too large). Now let π be any Q-plane different from π 1 and π 2. The intersection of π with V must be contained in π 1 π 2 and by the dimension formula must be a point or a line since π is not contained in V (if empty the dimension of <π,v> would be 6).

16 Equivalence Classes Theorem 4.5.6: Let Q be a hyperbolic quadratic set of a 5-dimensional projective space P. Then the set of all Q-planes is partitioned into exactly two equivalence classes with respect to ~. Pf (cont.): Assume π intersects V in the points of g. Consider a point P on g. By 4.5.3, Q induces in any complement W of P in Q P a hyperboloid. However a point other than P in which π and g intersect is on three Q-planes, thus in W this point is on three distinct Q-lines, a contradiction. So π can not meet g in more than one point. Hence if the intersection is a line, π intersects either one of the planes π 1 or π 2 in a line and the other in a point (the point of intersection of the line with g). On the other hand, if the intersection is a point, the point could not be on g, since that implies that π is equivalent to both π 1 and π 2 (which are not equivalent to each other), so the intersection point is in just one of the planes. In either case π intersects only one of the two planes in a point and so is in the equivalence class of that plane.

17 Generalized Quadrangles Def: A generalized quadrangle is a rank 2 geometry consisting of points and lines such that a) Any two distinct points are on at most one line. b) All lines are incident with the same number of points; all points are incident with the same number of lines. c) If P is a point outside a line g, then there is precisely one line through P intersecting g.

18 Structure Theorem Theorem 4.5.8: The geometry consisting of the points, lines and planes of a hyperbolic quadratic set of a 5-dimensional projective space has the following diagram. Pf: The points, lines and planes of a hyperbolic quadratic set Q form a rank 3 geometry. The residue of a Q-plane (all Q-points and Q-lines incident with a plane) clearly form a projective plane, hence the first link. The residue of a Q-line consists of the points on the line and the planes containing the line, and since each of the former is incident with each of the latter, this is the trivial geometry. Finally, consider the residue of a Q-point, all Q-lines and Q-planes containing the point.

19 Structure Theorem Theorem 4.5.8: The geometry consisting of the points, lines and planes of a hyperbolic quadratic set of a 5-dimensional projective space has the following diagram. Pf: These lines and planes all lie in the tangent hyperplane at that point. Furthermore, by Theorem (a), the intersection of Q with this hyperplane is a cone over a hyperboloid in 3-dimensional space. The lines of this cone can be identified with the points of the hyperboloid and the planes with the lines on the hyperboloid. That is, the Q-planes, thought of as lines of the residue form a regulus and its opposite regulus. This structure is the 3-dimensional grid, a generalized quadrangle.