Manuscript 1 1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 Blocking sets of tangent and external lines to a hyperbolic quadric in P G(, q), q even Binod Kumar Sahoo Abstract Bikramaditya Sahu Let H be a fixed hyperbolic quadric in the three dimensional projective space P G(, q), where q is a power of. Let E (respectively; T) denote the set of all lines of P G(, q) which are external (respectively; tangent) to H. We characterize the minimum size blocking sets of P G(, q) with respect to each of the line sets T and E T. Keywords: Projective space, Blocking set, Irreducible conic, Hyperbolic quadric, Generalized quadrangle, Ovoid AMS 0 subject classification: 0B, 1E1 1 Introduction Throughout q is a prime power. Let P G(d, q) be the d-dimensional projective space over a finite field of order q and L be a subset of the line set of P G(d, q). A blocking set with respect to L (or simply, an L-blocking set) is a subset B of the point set of P G(d, q) such that every line of L contains at least one point of B. Blocking sets of P G(d, q) have been studied by several authors with respect to varying sets of lines. The first step in this regard has been to determine the smallest cardinality of a blocking set and if possible, to describe all blocking sets of that cardinality. The following fundamental result was proved by Bose and Burton in [, Theorem 1] when L is the set of all lines of P G(d, q). Theorem 1.1. [] If B is a blocking set of P G(d, q) with respect to all its lines, then B qd 1 q 1. Further, equality holds if and only if B is a hyperplane of P G(d, q). 1.1 Blocking sets of P G(, q) Let C be an irreducible conic in P G(, q). We refer to [] for the basic properties of the points and lines of P G(, q) with respect to C. There are q + 1 points in C and every line of P G(, q) meets C in at most two points. A line of P G(, q) is called external, tangent or secant to C according as it meets C in 0, 1 or points. Every point of C lies on a unique tangent line, giving exactly q + 1 tangent lines to C. If q is even, then all the q + 1 tangent lines meet in a point and this common point of intersection is called the nucleus of C. The minimum size blocking sets of P G(, q) with respect to all possible combinations of the sets of external, tangent and secant lines to C have been studied by several authors. One can refer to [1] for a brief survey of the proved results. We recall two results which are needed in this paper. 1
1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 Giulietti proved the following in [, Theorems 1.1, 1.] for the minimum size blocking sets of the external lines to C. Theorem 1.. [] Let A be a blocking set of P G(, q), q even, with respect to the external lines to C. Then A q 1, and equality holds if and only if one of the following three cases occurs: (i) A consists of all points of a tangent line, minus the tangency point in C and the nucleus of C. (ii) A consists of all points of a secant line, minus the two intersecting points with C. (iii) q is a square and A = Π \ ({N} (Π C)), where Π is a Baer subplane of P G(, q) containing the nucleus N of C and intersecting C at q + 1 points. Aguglia and Giulietti proved the following in [1, Theorem 1.] for the minimum size blocking sets of the tangent and external lines to C. Theorem 1.. [1] Let A be a blocking set of P G(, q), q even, with respect to the tangent and external lines to C. Then A q, and equality holds if and only if one of the following three cases occurs: (i) A consists of all points of a tangent line, minus the tangency point in C. (ii) A consists of all points of a secant line different from the two intersecting points with C, plus the nucleus of C. (iii) q is a square and A = Π \ (Π C), where Π is a Baer subplane of P G(, q) such that Π C is a conic in Π. 1. Blocking sets of P G(, q) Consider P G(, q) which has (q + 1)(q + 1) points and (q + 1)(q + q + 1) lines. We denote by P and L, respectively, the point set and the line set of P G(, q). Let H be a hyperbolic quadric in P G(, q), that is, a non-degenerate quadric of Witt index two. We refer to [] for the basic properties of the points, lines and planes of P G(, q) with respect to H. For l L, we have l H {0, 1,, q + 1}. A line in L is called external or secant to H according as it meets H in 0 or points. A line is called tangent to H if it meets H in 1 or q + 1 points. Tangent lines meeting H in q + 1 points (that is, which are contained in H) are also called generators of H. The quadric H consists of (q + 1) points and (q + 1) generators. Every point of H lies on two generators. Every point x P lies on q + q + 1 lines of P G(, q), and q + 1 of them are tangent to H. If x H, then the remaining q lines through x are secant to H. If x / H, then x lies on q(q + 1)/ secant lines and q(q 1)/ external lines. For any plane π of P G(, q), the intersection π H is one of the following two possibilities: (i) π H is a pair of generators of H intersecting at some point, say x. In this case, π is called a tangent plane with pole x. (ii) π H is an irreducible conic in π. In this case, π is called a secant plane and we denote by C π the conic π H in π.
1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 Let E (respectively; T, S) denote the set of all lines of P G(, q) which are external (respectively; tangent, secant) to H. Then T = (q + 1)(q + 1), S = q (q + 1) /, E = q (q 1) /. Biondi et al. studied the minimum size E-blocking sets of P G(, q) in [, Theorem 1.1] for q even and in [, Theorem.] for q odd. For q even, we shall discuss more on the E-blocking sets in Section. When q is even, the minimum size (E S)-blocking sets of P G(, q) were described in [1, Theorem 1.] using the properties of generalized quadrangles. For L {S, T S, E S}, the authors determined in [1] the minimum size L-blocking sets of P G(, q) for all q. In this paper, we characterize the minimum size T- and (T E)-blocking sets of P G(, q) when q is even. Recall that a subset O of P is called an ovoid of P G(, q) if the following two conditions are satisfied: (i) Each line of P G(, q) meets O in at most two points. (ii) For every point x O, the union of all the lines which meet O at x is a plane of P G(, q). Let O be an ovoid of P G(, q). A line of P G(, q) is called tangent to O if it meets O in one point. For x O, there are q + 1 lines through x which are tangent to O. Each of the remaining q lines through x must meet O in exactly one more point. It follows that O = q + 1 and there are (q + 1)(q + 1) lines in L which are tangent to O. We prove the following results in Sections and for the minimum size T- and (T E)- blocking sets, respectively. Proposition 1.. Let B be a T-blocking set of P G(, q), q even. Then B q + 1. Further, equality holds if and only if B is an ovoid of P G(, q) for which T is the set of tangent lines. Proposition 1.. Let B be a (T E)-blocking set of P G(, q), q even. Then B q + q and the following hold for the equality case: (i) If q =, then B = if and only if one of the following occurs: (a) B = π \ {x} for some tangent plane π with pole x H. (b) B = O {α}, where O is a an ovoid of P G(, ) for which T is the set of tangent lines and α P \ H is such that the (unique) external line through α is disjoint from O. (ii) If q, then B = q + q if and only if B = π \ {x} for some tangent plane π with pole x H. We shall adopt the following notation. For a point w P, we denote by π w the union of the q + 1 tangent lines through w. When q is even, π w is a plane and the planes of P G(, q) are precisely the π w s as w runs over all elements of P. If w H, then π w is the tangent plane with pole w. If w / H, then π w is a secant plane and the conic C w := C πw = π w H in π w consists of the q + 1 tangency points in H of the tangent lines through w, and so w is the nucleus of C w.
1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 T-blocking sets We recall few facts concerning ovoids of P G(, q). Every plane of P G(, q) meets an ovoid in a point or in an oval (that is, a set of q + 1 points no three of which are on the same line). When q >, the ovoids of P G(, q) are precisely the subsets of P of the largest possible size, no three of which are on the same line in L. This is not true in P G(, ). In this case, the complement of a plane is a subset of maximum size in which no three points are on the same line, but such a set is not an ovoid of P G(, ). For q odd, the ovoids of P G(, q) are precisely the elliptic quadrics, that is, non-degenerate quadrics in P G(, q) of Witt index one. This was proved independently by Barlotti [] and Panella []. For q even, say q = r, the known ovoids of P G(, q) are of two types: (i) the elliptic quadrics which exist for all r 1, (ii) the Tits ovoids which exist for odd r, that is, when q > is a non-square. One can refer to [, Section 1.] for more on Tits ovoids. For small values of the even prime power q, namely when r {1,,,, }, a complete classification of all ovoids of P G(, q) has been obtained. By [,,, ], we know that every ovoid of P G(, q) is an elliptic quadric if q {,, 1}, and either an elliptic quadric or a Tits ovoid if q {, }. However, for a general even prime power q, classifying all ovoids of P G(, q) is still an open problem..1 Generalized quadrangles We refer to [1] for the basics on finite generalized quadrangles. Let s and t be positive integers. Recall that a (finite) generalized quadrangle of order (s, t) is a point-line geometry X = (P, L) with point 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. (Q) Two distinct lines have at most one point in common (equivalently, two distinct points are contained in at most one line). (Q) 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) [1, 1..1]. If s = t, then X is said to have order s. If P is a subset of the point set of some projective space P G(d, q), L is a set of lines of P G(d, q) and P is the union of all lines in L, then X = (P, L) is called a projective generalized quadrangle. An ovoid of X is a set O of points with the property that each line of X contains exactly one point of O. Counting in two ways the number of point-line pairs (x, l), where x O and l is a line of X containing x, it follows that O = st + 1. The points and the lines contained in the quadric H form a generalized quadrangle of order (q, 1) and it has ovoids. The point-line geometry with point set P and line set consisting of the totally isotropic lines of P G(, q) with respect to a symplectic polarity is a generalized quadrangle of order q, denoted by W (q). It has ovoids if and only if q is even, see [1,..1,..1]. By a result of Segre [1], every ovoid of P G(, q), q even, is an ovoid of some W (q). In [1], Thas proved the converse statement that every ovoid of W (q), q even, is an ovoid of the ambient space P G(, q).
1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 Consider the point-line geometry X = (P, T) with point set P and line set T. The following result is true for q even only. Proposition.1. If q is even, then X = (P, T) is a generalized quadrangle of order q which is isomorphic to W (q). Proof. Clearly, the axioms (Q1) and (Q) are satisfied with s = q = t. We verify (Q). Let (x, l) P T be a point-line pair with x / l. We show that there exists a unique line in T which contains x and intersects l. Observe that there are four cases depending on x H or not, and l is a generator of H or not. We consider all cases together. Let π = x, l be the plane in P G(, q) generated by x and l. Since q is even, we have π = π w for some w l. If π is tangent plane, then the only tangent line containing x and intersecting l is the line through x and the pole w of π (the remaining q lines through x in π are secant to H). If π is a secant plane, then the unique tangent line through x and meeting l is the line through x and the nucleus w of the conic C π in π. This verifies the axiom (Q). [Note that, when q is odd and π is a secant plane, the uniqueness (respectively, existence) of the tangent line trough x and intersecting l fails if x is an exterior (respectively, interior) point to the conic C π in π.] Thus X = (P, T) is a projective generalized quadrangle of order q with ambient space P G(, q). Then it follows from [1,..] that X is isomorphic to W (q). Another way to see Proposition.1 is the following, see [, Theorem 1..1]: Consider the polarity ξ of P G(, q) induced by the quadric H. So ξ is an inclusion reversing bijection of order two on the set of all subspaces of P G(, q). It fixes the line set L, and interchanges P and the set of all planes of P G(, q). A subspace U of P G(, q) is called absolute with respect to ξ if it is incident with ξ(u). If q is even, then ξ is a null polarity, that is, each point (and so each plane) of P G(, q) is absolute with respect to ξ. The point set P and the line set consisting of all the absolute lines of P G(, q) with respect to ξ form a generalized quadrangle W (q) of order q. Since H is a (geometric) hyperplane of W (q), each absolute line is either contained in H or meets H at one point. Then it follows that T is precisely the set of all absolute lines of P G(, q) with respect to ξ.. Proof of Proposition 1. Here q is even. By definition, the ovoids of W (q) are precisely the minimum size blocking sets of P G(, q) with respect to the lines of W (q). If B is a T-blocking set of P G(, q), then X W (q) implies that B q + 1. Further, equality holds if and only if B is an ovoid of X, that is, if and only if B is an ovoid of P G(, q) for which T is the set of tangent lines. This completes the proof of Proposition 1.. Revisiting E-blocking sets, q even For q even, the minimum size E-blocking sets of P G(, q) were characterized in [, Theorem 1.1] with the restriction that q. More precisely, the following was proved. Proposition.1. Let B be an E-blocking set of P G(, q), q even. Then B q q. If q, then equality holds if and only if B = π \ H for some tangent plane π of P G(, q).
1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 Here we give an alternate proof of the equality case of Proposition.1 which works for all even q, in particular, for q =,. We shall use some properties of the generalized quadrangle X = (P, T) W (q) of order q. Two points of X are said to be collinear if there exists a line of X containing both of them. For a subset U of P, U denotes the set of all elements of P which are collinear with each element of U. Note that, for a P, a := {a} contains a. For two non-collinear points a, b P, we have {a, b} = q + 1 and {a, b} = q + 1. The last equality follows from the fact that all points of W (q) are regular, see [1,..1,..1]. The lines of P G(, q) are of the form {x, y} for distinct x, y P, where {x, y} is in T or S E according as y x or not. For l E (respectively, l S), we have l E (respectively, l S). Now, let B be an E-blocking set of P G(, q), q even, of minimum size q q. Then B is disjoint from H. As in the proof of [, Proposition.1], by counting in two ways the cardinality of the set {(x, l) : x B, l E, x l}, it follows that each external line contains exactly one point of B. For any secant plane π x, x P \ H, the set π x B is a blocking set of π x with respect to the external lines to the conic C x = π x H. Again, as in the proof of [, Proposition.], by counting in two ways the cardinality of the set the following hold: (i) π x B = q 1 if and only if x / B. (ii) π x B = q if and only if x B. {(y, l) : y π x B, l E, l π x, y l}, In the latter case, (π x B) \ {x} is a blocking set of π x of minimum size q 1 with respect to the external lines to C x. We denote by T the subset of T consisting of all tangent lines which are not generators of H. For l T, we denote by x l the tangency point of l in H, that is, {x l } = l H. Note that x l / B for l T, as B H is empty. Lemma.. Let l T be such that l B 1. Then every line in T through x l meets B. Proof. Suppose that there exists m T through x l such that m B =. Count the points of B contained in the planes through m. The tangent plane π xl through m contains at least one point (comes from l B). Each of the q secant planes through m contains q 1 points of B. Since m B =, it follows that B contains at least 1 + q(q 1) = q q + 1 points, a contradiction to the fact that B = q q. Lemma.. There exists a line in T containing at least two points of B. Proof. Consider a line l in E. Then l is also in E. Since every external line meets B at one point, let {a} = l B and {b} = l B. Then the tangent line m through a and b is in T and m B. Lemma.. Let l be a line in T with l B. Then l B = l \ {x l }.
1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 Proof. Clearly this holds for q =. Assume that q. We have l B l \ {x l }. Suppose that there exists a point w in l \ {x l } which is not in B. Then π w B = q 1. So one of the three cases (i) (iii) of Theorem 1. holds for the blocking set π w B of π w with respect to the external lines to C w. Since at least two points of π w B are on the same tangent line l in π w, Theorem 1.(ii) does not occur for π w B. Suppose that Theorem 1.(i) holds for π w B. Then the fact that l B implies π w B = l \ {w, x l }. Let z π w B = l B. Then (π z B) \ {z} is a blocking set of π z of minimum size q 1 with respect to the external lines to C z. So one of the three possibilities (i) (iii) of Theorem 1. holds for (π z B) \ {z}. Since the line l in π z contains q 1 points of B (which are outside C z ), it follows that Theorem 1.(i) must occur for (π z B) \ {z} and that (π z B) \ {z} = l \ {z, x l }. This gives w B, a contradiction. Now, suppose that q is a square and that Theorem 1.(iii) holds for π w B. Let π w B = Π \ ({w} (Π C w )), where Π is a Baer subplane of π w and Π C w is a conic in Π with nucleus w. Note that l contains at least three points of Π \ (Π C w ), namely, w and at least two points of π w B. This implies that l 1 = l Π is a tangent line of Π and so q 1. Thus l B = l 1 \ {w, x l } is of size q 1. Let u l B. Then (π u B) \ {u} is a blocking set of π u of minimum size q 1 with respect to the external lines to C u. Since q 1 and l B = q 1, it can be seen that Theorem 1.(iii) must hold for (π u B) \ {u}. Let (π u B) \ {u} = Π \ ({u} (Π C u )), where Π is a Baer subplane of π u and Π C u is a conic in Π with nucleus u. Since l = l Π is a tangent line of Π (as l B Π \ (Π C u )) and u l B, it follows that l B = l \ {x l } is of size q. This implies that l B = q, a final contradiction. Therefore, l B = l \ {x l }. The following lemma proves the equality case of Proposition.1. Lemma.. B = π \ H for some tangent plane π. Proof. By Lemmas. and., consider a line l T such that l B = l\{x l }. We claim that B = π xl \ H. Let l = l 1, l,, l q 1 be the q 1 lines in T through x l. Since B = q q, it is enough to show that l i B = l i \ {x l } for each i, i q 1. By Lemma., we have l i B 1. Let t = min{ l i B : i q 1}. Then 1 t q as x l / B. We show that t = q and this would complete the proof. Consider a line l k, k q 1, such that l k B = t. Let l k = {x l = x 0, x 1,, x t, x t+1,, x q }, where l k B = {x 1,, x t }. We count the points of B contained in the planes through l k. The tangent plane π xl through l k contains at least q + t(q ) points of B. Each of the t secant planes π xi, 1 i t, contains at least q t points of B different from the points of l k B. Each of the remaining q t secant planes π xi, t + 1 i q, contains at least q 1 t points of B different from that of l k B. Thus all the planes through l k together contain at least q + t(q ) + t(q t) + (q t)(q 1 t) = q t points of B and so q t B = q q. This gives t q and so t = q.
1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 (T E)-blocking sets In this section, we prove Proposition 1.. We assume throughout that q is even. The following result in P G(, q) is useful for us. Lemma.1. Let A be a blocking set of P G(, q), q even, with respect to the tangent and external lines to an irreducible conic C. If A = q, then the following hold: (a) A is disjoint from C. (b) If q, then there exists at least three secant lines through some point α of C which are disjoint from A. Proof. Part (a) directly follows from Theorem 1.. Part (b) can be seen as follows. Since A = q, one of the three cases (i) (iii) of Theorem 1. occurs. If Theorem 1.(i) holds for some tangent line l, then take α to be the tangency point of l in C. If Theorem 1.(ii) holds for some secant line l, then take α to be one of the two points in l C. Finally, suppose that q is a square and Theorem 1.(iii) holds for some Baer subplane Π of P G(, q). Then A = Π \ (Π C). Take α to be any point of C \ Π. Let β be a point in the Baer conic Π C in Π and l be the secant line (to C) through α and β. We claim that l is disjoint from A. Otherwise, l would be a Baer line which is tangent to the conic Π C in Π with tangency point β. Then the nucleus N of Π C is contained in the Baer line l. Since C and Π C share the same nucleus, it follows that the secant line l to C contains N, a contradiction. Thus the secant lines through α and meeting Π C are disjoint from A. The rest follows from the fact that Π C = q + 1. We now proceed to prove Proposition 1.. For any tangent plane π x, x H, the set B 1 = π x \ {x} is a (T E)-blocking set of size q + q. In this case, B 1 \ H = q q and B 1 H = q. For q =, we give one more example justifying the statement in Proposition 1.(i)(b). Let O be an ovoid of P G(, ) for which T is the set of tangent lines, that is, O is an ovoid of the generalized quadrangle X = (P, T) W () (Proposition.1). Then O =. We have E =. Let l 1 and l be the two lines external to H. Then P \ H = l 1 l and li = l j for {i, j} = {1, }. By [1, 1..], ( l 1 O, l O ) = (0, ) or (, 0). We may assume that ( l 1 O, l O ) = (, 0). Then, for any α l, it is clear that the set B = O {α} is a (T E)-blocking set of P G(, ) of size. Since O H is an ovoid of H, we have B H = and so B \ H =. Now, let B be a (T E)-blocking set of P G(, q). We assume that B is of minimum possible size. Then B q + q. Since B H blocks every generator of H, we have B H q + 1. Every external line meets B outside H. So B \ H is an E-blocking set of P G(, q) and hence B \ H q q by Proposition.1. Thus we have and Lemma.. For x P, the following hold: q q B \ H q 1 q + 1 B H q.
1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 (i) If x / B, then π x B q + 1. (ii) If x B \ H, then π x B q. Proof. (i) This follows, since x / B and each of the q + 1 tangent lines through x meets B. (ii) The set π x B is a blocking set of π x with respect to the tangent and external lines to the conic C x. So π x B q by Theorem 1.. Let T 0 denote the subset of T consisting of all tangent lines whose tangency points are in H \ (B H). We define T 1 = {l T 0 : l B = 1}. Lemma.. T 1 is nonempty. Proof. For every point x H, there are q 1 tangent lines in T through x. This gives T 0 = ((q + 1) B H )(q 1) (q + 1)(q 1). (1) Here the last inequality follows using the fact that B H q. Suppose that T 1 is empty. Then every line of T 0 meets B in at least two points. Consider the set Counting Z in two ways, we get Z = {(x, l) : x B, l T, x l}. B (q + 1) = Z T 0 + T \ T 0 = T 0 + T. Since B q + q and T = (q + 1)(q + 1), it follows that T 0 q 1, a contradiction to (1). So T 1 is nonempty. Lemma.. B = q + q. Proof. By Lemma., consider a line l = {x 0, x 1,, x q } of T 1. We count the points of B contained in the planes through l. We may assume that x 0 is the tangency point of l and that l B = {x q }. By Lemma., each of the planes π xi (0 i q 1) contains at least q + 1 points of B and the plane π xq contains at least q points of B. This gives q + q B = 1 + q ( π xi B 1) 1 + q + q 1 = q + q. () i=0 The first equality holds, since π xi π xj = l for 0 i j q and l B = 1. It follows from () that equality holds everywhere and so B = q + q. Using the fact that B = q + q, we have the following as a consequence of the proof of Lemma.. Corollary.. Let l T 1 and x l. Then π x B = q or q + 1 according as {x} = l B or not. In particular, the folowing hold: (i) If {x} = l B, then each of the tangent lines through x contains exactly one point of B. (ii) If {x} = l B, then the conic C x in π x is disjoint from B.
1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 Proof. Only (ii) needs a proof. The set π x B is a blocking set of π x of minimum size q with respect to the tangent and external lines to C x. So C x is disjoint from π x B (and hence from B) by Lemma.1(a). Lemma.. Let l and m be two generators of H intersecting at a point x 0. If l B = 1 and m B, then x 0 B. Proof. Suppose that x 0 / B. Let l = {x 0, x 1,, x q }. We may assume that l B = {x q }. Each of the q 1 tangent lines in T through x 0 is an element of T 0. Since m B, Corollary.(i) implies that none of them is in T 1, and so each such line meets B in at least two points. This gives that the tangent plane π x0 contains at least q + 1 points of B. By Lemma., each of the tangent planes π xi, 1 i q 1, contains at least q + 1 points of B. It follows that the planes π xi, 0 i q 1, together contain at least q + 1 + (q 1)q = q + q + 1 points of B, a contradiction to Lemma.. So x 0 B. Lemma.. Every generator of H contains one or q points of B. Proof. Let m = {x 0, x 1,, x q } be a generator of H containing at least two points of B. Fix a line l = {y 0, y 1,, y q } of T 1. We may assume that the tangency point of l is y 0 and l B = {y q }. By Corollary.(i), m is different from the two generators through y 0. Let l 1 be the generator through y 0 which intersects m. Then l 1 B = 1. We may assume that m l 1 = {x 0 }. Note that the conics C yj = π yj H, 1 j q, pairwise intersect at y 0 and each of them contains a unique point, say x j, of m \ {x 0 }. Since l B = {y q }, the conic C yq is disjoint from B by Corollary.(ii). So x q / B. We claim that m \ {x q } is contained in B. Since l 1 B = 1, m B and m l 1 = {x 0 }, Lemma. implies that x 0 B. We next show that x i B for 1 i q 1. Let m i be the tangent line through x i and y i. Since l B {y i }, m i contains exactly one point of B by Corollary.(i). If m i B {x i }, then m i T 1 as x i is the tangency point of m i. Applying Corollary.(i) again to the line m i, it follows that every tangent line through x i, and in particular, m contains exactly one point of B. This leads to a contradiction to our assumption that m B. Thus m i B = {x i } and so x i B. For two points a and b of H not lying on the same generator, we write {a, b} H = {c, d}, where c, d are the points of H such that one generator through c (respectively, d) contains a and the other one contains b. Observe that there are secant lines to H disjoint from B. Otherwise, B would be a blocking set with respect to every line of P G(, q), and then Theorem 1.1 would imply B q + q + 1. Lemma.. Let l be a secant line disjoint from B and let l H = {a, b}. If {a, b} H = {c, d}, then at least one of c and d is in B. Proof. Let π i, 0 i q, be the q + 1 planes through l, two of which are the tangent planes π c and π d. Since π i π j = l for i j and l B =, B is a disjoint union of the sets π i B. Suppose that c, d are not in B. By Lemma., each of π c and π d contains at least q + 1 points of B and each of the remaining q 1 secant planes contains at least q points of B. This gives B (q + 1) + q(q 1) = q + q +, a contradiction to Lemma..
1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 Lemma.. If q, then there exists a generator containing q points of B. Proof. By Lemma., it is enough to show that there exists a generator of H containing at least two points of B. Let π x, x P \ H, be a secant plane containing q points of B. The existence of such a plane follows from Lemma. and Corollary.. Note that x B. The set π x B is a blocking set of π x of minimum size q with respect to the tangent and external lines to the conic C x. So, for some w C x, there exists three lines l 1, l, l in π x through w which are secant to C x (and hence to H) and disjoint from π x B (and hence from B). This is possible by Lemma.1(b) as q. By Lemma., B contains a point, say z i, from li H for each i. The points z 1, z, z are pair-wise distinct and each of them lies on a generator of H through w. It follows that one of the two generators through w contains at least two points of B. Lemma.. If q, then B H = (l m) \ {x} for some generators l and m of H intersecting at x. In particular, B H = q and B \ H = q q. Proof. By Lemma., let l be a generator of H containing q points of B. Let {x} = l \ B and m be the other generator of H though x. We have m B 1. Since x / B, Lemma. implies that m B and so m B = q by Lemma.. Since B H q, it follows that B H = (l m) \ {x}. Thus B H = q and hence B \ H = q q as B = q + q..1 Proof of Proposition 1.(ii) Lemma.. If q, then B = π x \ {x} for some tangent plane π x, x H. Proof. By Lemma., let B H = (l 0 l 1 ) \ {x}, where l 0 and l 1 are two generators of H intersecting at x. We claim that B \ H = π x \ H. Let l,, l q be the other tangent lines through x. Since B \ H = q q, it is enough to show that l i B = l i \ {x} for each i q. We shall apply a similar argument as in the proof of Lemma.. We have 1 l i B q. If l j B = 1 for some j with j q, then Corollary.(i) would imply that each of the generators l 0 and l 1 meets B at exactly one point, which is not possible. So l i B. Let t = min{ l i B : i q}. We show that t = q and this would complete the proof. Consider a line l k, k q, such that l k B = t. Let l k = {x = x 0, x 1,, x t, x t+1,, x q }, where l k B = {x 1,, x t }. Now consider the planes through l k and count the points of B contained in them. The tangent plane π x0 contains at least q + t(q 1) points of B. By Lemma.(ii), each of the secant planes π xi, 1 i t, contains at least q t points of B different from the points of l k B. By Lemma.(i), each of the remaining planes π xi, t + 1 i q, contains at least q + 1 t points of B different from that of l k B. Thus the planes through l k together contain at least q + t(q 1) + t(q t) + (q t)(q + 1 t) = q + q t points of B and so q +q t B = q +q. This gives t q and so t = q. This completes the proof.
1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1. Proof of Proposition 1.(i) Here q =. We have B = and ( B \ H, B H ) = (, ) or (, ). First assume that ( B \ H, B H ) = (, ). Since q = and B H =, it is clear that there exists a generator l 0 containing exactly two points of B. Let {x} = l 0 \ B and l 1 be the other generator of H though x. We have 1 l 1 B. Since x / B, Lemma. implies that l 1 B =. So B H = (l 0 l 1 ) \ {x}. Let l be the third tangent line through x. Then 1 l B. Since l 0 B =, Corollary.(i) implies that l B =. So B \ H = l \ {x} as B \ H =. Thus B = π x \ {x}. Now assume that ( B \ H, B H ) = (, ). Then B H is an ovoid of H. There exists a unique ovoid O of X = (P, T) W () containing B H. Note that O H = B H. We show that O is contained in B. Let x O \ (B H). Suppose that x / B. Since π x B, it follows that π x B is the unique line in π x which is external to the conic C x (and hence to H). Since B =, we must have B = (B H) (π x B). Then the other external in E (through x) is disjoint from B, contradicting that B blocks every external line. Thus O B. As mentioned in the beginning of this section, exactly one of the two external lines to H, say l, is disjoint from O. Since l meets B, let α l B. Then, B = implies that B = O {α}. This completes the proof. Acknowledgement The authors wish to thank Prof. Bart De Bruyn for his comments, which helped to improve an earlier version of the article. References [1] Aguglia A and Giulietti M, Blocking sets of certain line sets related to a conic, Des. Codes Cryptogr. (00) 0 [] Barlotti A, Un estensione del teorema di Segre-Kustaanheimo, Boll. Un. Mat. Ital. (1) 0 [] Biondi P and Lo Re P M, On blocking sets of external lines to a hyperbolic quadric in P G(, q), q even, J. Geom. (00) [] Biondi P, Lo Re P M and Storme L, On minimum size blocking sets of external lines to a quadric in P G(, q), Beiträge Algebra Geom. (00) 0 1 [] Bose R C and Burton R C, A characterization of flat spaces in a finite geometry and the uniqueness of the Hamming and the MacDonald codes, J. Combinatorial Theory 1 (1) [] Fellegara G, Gli ovaloidi in uno spazio tridimensionale di Galois di ordine, Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Nat. () (1) 10 1 [] Giulietti M, Blocking sets of external lines to a conic in P G(, q), q even, European J. Combin. (00),
1 1 1 1 1 1 0 1 0 1 0 1 0 1 0 1 [] Hirschfeld J W P, Finite projective spaces of three dimensions (1) (New York: Oxford University Press) [] Moorhouse G E, Incidence geometry (00), available online at http://www.uwyo.edu/moorhouse/handouts/incidence geometry.pdf [] O Keefe C M and Penttila T, Ovoids of P G(, 1) are elliptic quadrics, J. Geom. () [] O Keefe C M and Penttila T, Ovoids of P G(, 1) are elliptic quadrics, II. J. Geom. () 10 1 [] O Keefe C M, Penttila T and Royle G F, Classification of ovoids in P G(, ), J. Geom. 0 (1) 1 10 [] Panella G, Caratterizzazione delle quadriche di uno spazio (tridimensionale) lineare sopra un corpo finito, Boll. Un. Mat. Ital. (1) 0 [1] Patra K L, Sahoo B K and Sahu B, Minimum size blocking sets of certain line sets related to a conic in P G(, q), Discrete Math. (01) 11 1 [1] Payne S E and Thas J A, Finite generalized guadrangles (00) (Zürich: European Mathematical Society) [1] Sahoo B K and Sahu B, Blocking sets of certain line sets to a hyperbolic quadric in P G(, q), communicated. [1] Sahoo B K and Sastry N S N, Binary codes of the symplectic generalized quadrangle of even order, Des. Codes Cryptogr. (01) 1 10 [1] Segre B, On complete caps and ovaloids in three-dimensional Galois spaces of characteristic two, Acta Arith. (1) 1 [1] Thas J A, Ovoidal translation planes, Arch. Math. (Basel) () 1 Address: School of Mathematical Sciences, National Institute of Science Education and Research, Bhubaneswar (HBNI), At/Po - Jatni, District - Khurda, Odisha - 00, India. Emails: bksahoo@niser.ac.in, bikram.sahu@niser.ac.in