Let S=(P, B, I) be a generalized quadrangle of order (q, q 2 ), q>1, and assume that S satisfies Property (G) at the flag (x, L). If q is odd then S is the dual of a flock generalized quadrangle. This solves (a stronger version of ) a ten-year-old conjecture. We emphasize that this is a powerful the
Generalized Quadrangles of Order (s, s2), II
β Scribed by J.A Thas
- Publisher
- Elsevier Science
- Year
- 1997
- Tongue
- English
- Weight
- 514 KB
- Volume
- 79
- Category
- Article
- ISSN
- 0097-3165
No coin nor oath required. For personal study only.
β¦ Synopsis
ac.be. * . The set of all points r i * is denoted by B; we have |B| =q 2n . Further, let A be the set of all intersections of PG(n+1, q n ) with the tangent lines of the conics C i at s 1 . The tangent line U i of C i at s 1 belongs to {Γ , hence
of B and just one point of A. Let ? be the plane containing R i and r j * , r j * Γ R i . If we now take for 8 1 the space (R i , s 1 , r j *), then the base B of the pencil P consists of two conics C i and C j and (C i _ C j )&[s 1 ] is the set of all points r d in 8 1 . It immediately follows that ? contains the line R j , and that ? contains no line R l # R with i{l{j; also, ? & B consists of all points of
Suppose that r i * and r j * are points of B not on a common line of R. Considering all planes of PG(n+1, q n ), the previous section shows that r i * and r j * are on the same number of lines of R. Now let r i * and r j *, i{j, be points of B on a common line of R. Let t u +1 be the number of lines of R containing r u * . By way of contradiction we assume that t i {t j . Let r l * be a point of B not on r i *r j *. The plane
not in A, is on t j +1 lines of R. Let r u * # B be not on r i *r j * and not on r i *r l * . Assume, by way of contradiction that r u *r j * # R. As r i *r u * Γ R, the point r u * is on t i +1 lines of R. For any point r c * # r i *r j *&[r j *, r i *] the line r u *r c * does not belong to R, and so r u * is on t j +1 lines of R. Hence t i =t j , a contradiction. Consequently r u *r j * Γ R. Interchanging roles of r u * and r l * , we see that necessarily r u *r i * # R. It follows that all lines of R contain r i * . Hence t j =0. So (t 1 +1)(q n &1)+1= |B| =q 2n , that is, t 1 =q n . Consequently there are q n +1 conics in V containing s 1 , and they all share a point r i . In PG(n&1, q n ) we now consider a point s 2 {s 1 on an element of V. Let C be a conic of V containing r i and s 2 . Choose on C a point r w different from r i . Then there is a conic C$ in V containing r w and s 1 , and so C$ contains r i . Consequently V contains two distinct conics C and C$ through r i and r w , a contradiction. We conclude that the number of lines of R through any point r i * # B is independent of the index i. The constant number of lines of R which contains r i * # B will be denoted by t+1.
Fix a line R # R and count on two ways the number of ordered pairs (r i *, R$), with r i * # B, r i * Γ R, R$ & R{< and r i * # R$. We obtain q n t(q n &1)+;q n =q 2n &q n ,
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