Partial Flocks of Quadratic Cones with a Point Vertex in PG ( n , q ), n Odd

(1)

P1: VTL/SRK P2: VTL/PMR/ASH P3: PMR/ASH QC:

Journal of Algebraic Combinatorics KL472-04-Okeefe July 30, 1997 11:0

Journal of Algebraic Combinatorics 6 (1997), 377–392

°c 1997 Kluwer Academic Publishers. Manufactured in The Netherlands.

Partial Flocks of Quadratic Cones with a Point Vertex in PG ( n , q ), n Odd

CHRISTINE M. O’KEEFE^∗ [email protected]

Department of Mathematics, The University of Adelaide, Adelaide, 5005, Australia

J.A. THAS [email protected]

Department of Pure Mathematics and Computer Algebra, University of Gent, Krijgslaan 281, B-9000 Gent, Belgium

Received December 15, 1995; Revised June 24, 1996

Abstract. We generalise the definition and many properties of flocks of quadratic cones in PG(3,q)to partial flocks of quadratic cones with vertex a point in PG(n,q), for n≥3 odd.

Keywords: Galois geometry, flock, cone, ovoid, cap

1. Introduction

For information on the properties of quadrics in projective spaces, see [4, Section 5.1], [5, Chapter 16] and especially [8, Chapter 22]. In the following, we always assume that n≥3 is odd.

In PG(n,q),n≥3 odd, letK=vQbe a cone with vertex the pointvand baseQ, where Qis a non-singular (parabolic) quadric in a hyperplane PG(n−1,q)not onv.

A partial flock ofKof size k is a set of hyperplanesπ1, . . . , πkof PG(n,q), each not on v, such that for each i,j ∈ {1, . . . ,k}with i 6= j the(n−2)-dimensional spaceπi∩πj

meetsKin a non-singular elliptic quadric. The set of (non-singular, parabolic) quadrics πi∩Kfor i =1, . . . ,k is also called a partial flock ofK.

In the case n=3, since an elliptic quadric in PG(1,q)has no points, the above definition coincides with the existing definition of a partial flock of a quadratic cone in PG(3,q). 2. The size of a partial flock, q even

It is easy to see that a partial flock of a quadratic cone in PG(3,q),q odd or even, has size at most q,since the conics in the flock are disjoint. In this section we use Lemma 1 (a generalisation of [12, 1.5.2]) to show that this bound also holds for odd n ≥ 5 and q even. Our proof is also valid in the case n=3.

∗This work was supported by the Australian Research Council, Department of Pure Mathematics and Computer Algebra of the University of Gent and the National Fund for Scientific Research of Belgium.

(2)

378 O’KEEFE AND THAS

Lemma 1 In PG(n,q),where n ≥ 3 is odd and q is even,letF = {π1, . . . , πk}be a partial flock of the coneK=vQ.Let u be the nucleus ofQin the subspace PG(n−1,q) of PG(n,q). Then each spaceπi∩πj,i 6= j,is disjoint from the linevu.

Proof: Suppose, to the contrary, that there exist i 6= j such thatπi∩πj ∩vu =u⁰,say.

Then u⁰is the nucleus of the (parabolic) quadricK∩πi, soπi∩πj ∩Kis parabolic, a

contradiction. 2

Theorem 2 In PG(n,q),where n≥3 is odd and q is even,a partial flock of a quadratic cone has size at most q.

Proof: LetF be a partial flock of the coneK =vQ.Let u be the nucleus ofQin the subspace PG(n−1,q)of PG(n,q). By Lemma 1, no two elements ofFcan meet on the linevu.Since each element ofFmust meetvu\{v}, we have k≤q. 2

3. Generalising known results

In this section we generalise some results which are well-known for flocks of quadratic cones in PG(3,q). In particular, the dual setting for q even generalises [12, 1.5.3], the al- gebraic condition generalises [12, 1.5.5], the existence of the partial ovoid ofQ⁺(n+2,q) generalises [12, 1.3], the process of derivation for q odd generalises [1] and the construction of herds of caps for q even generalises [2, Theorem 1] (see also [11, Theorem 2.1]).

4. The dual setting

Case (1) q odd: First suppose that q is odd. In PG(n,q), let F = {π1, . . . , πk} be a partial flock of the cone K = vQ.We apply a duality to PG(n,q). The point v is mapped to a hyperplane V of PG(n,q)and the set of lines of K on v is mapped to the set of all tangent hyperplanes to a non-singular quadricQ⁰of V . The hyperplanes π1, . . . , πkofFare mapped to points p1, . . . ,pkof PG(n,q)\V.For i6= j the(n−2)- dimensional space πi ∩πj meets K in the points of a non-singular elliptic quadric Q⁻(n−2,q); so the hyperplanehπi ∩πj, vi,generated byπi ∩πj andv,contains exactly the lines ofvQon the conevQ⁻(n−2,q).It follows that the line pipj meets V in a point pi j on exactly the tangent hyperplanes ofQ⁰which correspond under the duality to the lines ofvQ⁻(n−2,q); so the tangent points of these hyperplanes are the points of a non-singular elliptic quadricQˆ⁻(n−2,q)onQ⁰.Hence pi j is an interior point ofQ⁰.

Thus, for n and q odd, a dual partial flock of a non-singular quadricQ⁰of a hyperplane PG(n−1,q)of PG(n,q)is a set of points of PG(n,q)\PG(n−1,q)such that the line joining any two of them meets PG(n−1,q)in a point interior toQ⁰. It is clear that a partial flock gives rise to a dual partial flock and conversely.

Case (2) q even: Now suppose that q is even.

(3)

PARTIAL FLOCKS OF QUADRATIC CONES 379

We use the following notation, introduced in [7]. LetQ be a non-singular quadric in PG(n,q),let PG(n−1,q)be a hyperplane and let Q be a point of PG(n,q)\PG(n−1,q) not lying onQand distinct from its nucleus. The projection ofQfrom Q onto PG(n−1,q) is the setR= {PQ∩PG(n−1,q): P ∈Q}.If n is odd andQis hyperbolic then we write R=R⁺while ifQis elliptic then we writeR=R⁻.We note, see [7], that a setRhas type(1,q/2+1,q+1)with respect to lines, that a setR⁺contains a unique hyperplane PG(n−2,q)such that(PG(n−1,q)\R⁺)∪PG(n−2,q)is a setR⁻and that a setR⁻ contains a unique hyperplane PG(n−2,q)such that(PG(n−1,q)\R⁻)∪PG(n−2,q) is a setR⁺.

In PG(n,q),for odd n≥5,letF= {π1, . . . , πk}be a partial flock of the coneK=vQ.

Again, we apply a duality to PG(n,q). The pointvis mapped to a hyperplane V=PG(n− 1,q)of PG(n,q). LetGbe the set of generators (((n−3)/2)-dimensional subspaces) lying onQ.A((n−1)/2)-dimensional subspacevG,G ∈ G,is mapped by the duality to an ((n−1)/2)-dimensional subspace of V,and we denote byRthe union of the points lying on such((n−1)/2)-dimensional subspaces of V . The setRcontains the subspace PG(n−2,q) of V which is the dual of the line uv,with u the nucleus ofQ.It can be shown thatRhas type (1,q/2+1,q+1)with respect to lines, by showing that an(n−2)-dimensional subspace of PG(n,q)onvlies in exactly 1,q/2+1 or q+1 hyperplanes containing an elementvG,G∈ G.Then, sinceRcontains((n−1)/2)-dimensional subspaces not in PG(n−2,q),it follows thatRis a setR⁺in V (this also follows from|R| =qⁿ⁻¹/2+qⁿ⁻²+· · ·+q+1+q⁽ⁿ⁻¹^)/²/2 and [7]). The hyperplanesπ1, . . . , πkofFare mapped to points p₁, . . . ,p_kof PG(n,q)\V. For i 6= j the(n−2)-dimensional spaceπi ∩ πj does not meet the line uvand meetsK in exactly the points of a non-singular elliptic quadricQ⁻(n−2,q);hence the hyperplane hπi ∩πj, vidoes not contain any element ofG.So the line pipj meets V in a point of V\R⁺=R⁻\PG(n−2,q).

For n odd and q even a dual partial flock of a setR⁺of type(1,q/2+1,q+1)in a hyperplane PG(n−1,q)of PG(n,q)is a set of points of PG(n,q)\PG(n−1,q)such that the line joining any two of them meets PG(n−1,q)in a point of PG(n−1,q)\R⁺.It is clear that a partial flock gives rise to a dual partial flock and conversely.

We remark that the results of this last section also hold in the case n=3 (see [12]); here a setR⁺is the set of points of a dual regular hyperoval.

4.1. The algebraic conditions

For q=2^h, the map trace is defined by trace: GF(q)→GF(2), x7→

h−1

X

i=0

x²ⁱ.

Theorem 3 In PG(n,q)for n ≥ 3 odd,letK = vQbe a quadratic cone with vertex the pointvand baseQ,whereQis a non-singular quadric in a hyperplane not onv,and letF = {π1, . . . , πk}be a set of hyperplanes not onv. Without loss of generality,we can suppose that the quadratic coneK=vQhas equation x₀x₁+x₂x₃+· · ·+x_n₋₃x_n₋₂=x_n²₋₁, so thatv=(0, . . . ,0,1)andQhas equation x0x1+x2x3+ · · · +xn−3xn−2 =x_n²₋₁in the

(4)

hyperplane PG(n−1,q)with equation xn = 0. For i = 1, . . . ,k the hyperplaneπi has equation a⁽₀ⁱ⁾x0+ · · · +a⁽_nⁱ₋⁾₁xn−1+xn = 0 for some a⁽_jⁱ⁾ ∈ GF(q). If q is odd,F is a partial flock ofKif and only if

−4¡

a₀⁽ⁱ⁾−a₀⁽^j⁾¢¡

a⁽₁ⁱ⁾−a⁽₁^j⁾¢

− · · ·

−4¡

a_n⁽ⁱ₋⁾₃−a_n⁽₋^j⁾₃¢¡

a⁽_nⁱ₋⁾₂−a_n⁽₋^j⁾₂¢ +¡

a_n⁽ⁱ₋⁾₁−a⁽_n₋^j⁾₁¢2

is a non-square in GF(q)for all i,j ∈ {1, . . . ,k},i 6= j.If q is even,F is a partial flock ofKif and only if a_n⁽ⁱ₋⁾₁−a_n⁽₋^j⁾₁6=0 and

trace

Ã¡a₀⁽ⁱ⁾−a₀⁽^j⁾¢¡

a⁽₁ⁱ⁾−a⁽₁^j⁾¢

+ · · · +¡

a_n⁽ⁱ₋⁾₃−a⁽_n₋^j⁾₃¢¡

a_n⁽ⁱ₋⁾₂−a⁽_n₋^j⁾₂¢

¡a_n⁽ⁱ₋⁾₁−a⁽_n^j₋⁾₁¢2

!

=1

for all i,j ∈ {1, . . . ,k},i6= j .

Proof: For i,j∈ {1, . . . ,k},i 6= j , the hyperplanehπi∩πj, vimeetsK∩PG(n−1,q)= Qin the quadricQ⁰with equations

¡a₀⁽ⁱ⁾−a₀⁽^j⁾¢

x₀+ · · · +¡

a_n⁽ⁱ₋⁾₁−a⁽_n₋^j⁾₁¢

x_n₋₁=0,

(1) x₀x₁+x₂x₃+ · · · +x_n₋₃x_n₋₂=x_n²₋₁.

At least one of(a₀⁽ⁱ⁾−a₀⁽^j⁾), . . . , (a⁽_nⁱ₋⁾₂−a_n⁽₋^j⁾₂)is not zero, for otherwisehπi ∩πj, vimeets Kin a hyperbolic quadratic cone with vertexv, soπi∩πjmeetsKin a hyperbolic quadric, contrary to the definition of partial flock. Therefore, without loss of generality, we suppose that a⁽₀ⁱ⁾6=a₀⁽^j⁾. The quadricQ⁰is the intersection of the cone

¡a₀⁽^j⁾−a₀⁽ⁱ⁾¢₋1¡¡

a⁽₁ⁱ⁾−a⁽₁^j⁾¢

x1+ · · · +¡

a_n⁽ⁱ₋⁾₁−a_n⁽^j₋⁾₁¢ xn−1

¢x1

+x₂x₃+ · · · +x_n₋₃x_n₋₂=x_n²₋₁,

that is,

¡a₁⁽ⁱ⁾−a₁⁽^j⁾¢ x²₁+¡

a₀⁽ⁱ⁾−a₀⁽^j⁾¢

x_n²₋₁+¡

a₂⁽ⁱ⁾−a₂⁽^j⁾¢

x1x2+ · · · +¡

a_n⁽ⁱ₋⁾₁−a_n⁽^j₋⁾₁¢

x1xn−1+¡

a₀⁽^j⁾−a₀⁽ⁱ⁾¢

x2x3+¡

a₀⁽^j⁾−a⁽₀ⁱ⁾¢

x4x5+ · · · +¡

a₀⁽^j⁾−a₀⁽ⁱ⁾¢

xn−3xn−2=0, (2)

with the hyperplane (1) not through its vertex. We determine exactly when the quadric Q⁰ is non-singular and elliptic. Let the matrix A = [a_{i j}]_i_,_j₌₁_,...,_n₋₁, where a_{i i} is twice the coefficient of x_i² in (2) and for i<j ai j=aj i is the coefficient of xixj in (2).

(5)

Then A is





 2³

a₁⁽ⁱ⁾−a₁⁽^j⁾´ ³

a⁽₂ⁱ⁾−a⁽₂^j⁾´ ³

a⁽₃ⁱ⁾−a₃⁽^j⁾´

. . . . ³

a⁽_nⁱ₋⁾₃−a⁽_n₋^j⁾₃´ ³

a⁽_nⁱ₋⁾₂−a_n⁽₋^j⁾₂´ ³

a_n⁽ⁱ₋⁾₁−a_n⁽₋^j⁾₁´

³ a₂⁽ⁱ⁾−a₂⁽^j⁾

´ 0

³ a⁽₀^j⁾−a⁽ⁱ⁾₀

´

0 . . . 0 0 0

³

a₃⁽ⁱ⁾−a₃⁽^j⁾´ ³

a⁽₀^j⁾−a₀⁽ⁱ⁾´

0 0 . . . 0 0 0

³

a₄⁽ⁱ⁾−a₄^(j⁾´

0 0 ... ... 0 0 0

... ... ... ... ... ... ... ...

³

a_n−3⁽ⁱ⁾ −a⁽_n−3^j)´

0 0 0 . . . 0 ³

a⁽₀^j)−a₀⁽ⁱ⁾´

³ 0

a_n⁽ⁱ⁾₋₂−a⁽_n^j₋⁾₂´

0 0 0 . . . ³

a⁽₀^j⁾−a₀⁽ⁱ⁾´

0 0

³

a_n−1⁽ⁱ⁾ −a⁽_n−1^j)´

0 0 0 . . . 0 0 2³

a⁽₀ⁱ⁾−a₀⁽^j)´







with determinant (expanding by the last row; then expanding the two resulting subdetermi- nants by the last column and first row respectively)

|A| =(−1)⁽ⁿ⁻³^)/²¡

a₀⁽ⁱ⁾−a₀⁽^j⁾¢n−3¡ 4¡¡

a₀⁽ⁱ⁾−a₀⁽^j⁾¢¡

a₁⁽ⁱ⁾−a₁⁽^j⁾¢ +¡

a₂⁽ⁱ⁾−a⁽₂^j⁾¢

×¡

a⁽₃ⁱ⁾−a₃⁽^j⁾¢

+ · · · +¡

a_n⁽ⁱ₋⁾₃−a⁽_n₋^j⁾₃¢¡

a_n⁽ⁱ₋⁾₂−a_n⁽₋^j⁾₂¢¢

−¡

a_n⁽ⁱ₋⁾₁−a⁽_n₋^j⁾₁¢2¢ . If q is odd, by [8, 22.2.1], the quadric Q⁰ is non-singular and elliptic if and only if (−1)⁽ⁿ⁻¹^)/²|A|is a non-square in GF(q), which is if and only if

−4¡

a₀⁽ⁱ⁾−a⁽₀^j⁾¢¡

a₁⁽ⁱ⁾−a₁⁽^j⁾¢

− · · · −4¡

a⁽_nⁱ₋⁾₃−a⁽_n₋^j⁾₃¢¡

a_n⁽ⁱ₋⁾₂−a_n⁽^j₋⁾₂¢ +¡

a_n⁽ⁱ₋⁾₁−a_n⁽^j₋⁾₁¢2

is a non-square in GF(q).

For q even, by [8, 22.2.1], the quadricQ⁰is non-singular if and only if|A| 6=0,that is, if and only if a_n⁽ⁱ₋⁾₁−a_n⁽^j₋⁾₁ 6=0. Further, the non-singular quadricQ⁰is elliptic if and only if trace((|B| −(−1)⁽ⁿ⁻¹^)/²|A|)/(4|B|)) = 1, where the matrix B =[bi j]i,j=1,...,n−1

has bii=0 and bj i = −bi j= −ai jfor i < j.(The formula(|B| −(−1)⁽ⁿ⁻¹^)/²|A|)/(4|B|) should be interpreted as follows: the terms ai j are replaced by indeterminates zi j, the formula is evaluated as a rational function over the integers Z , and then zi jis specialized to ai jto give the result.) Thus B is







0 ³

a₂⁽ⁱ⁾−a₂⁽^j⁾´ ³

a₃⁽ⁱ⁾−a₃⁽^j⁾´

. . . . ³

a_n⁽ⁱ⁾₋₃−a⁽_n^j₋⁾₃´ ³

a_n⁽ⁱ⁾₋₂−a_n⁽^j₋⁾₂´ ³

a⁽ⁱ_n₋⁾₁−a_n⁽₋^j⁾₁´

−³

a₂⁽ⁱ⁾−a₂⁽^j)´

0 ³

a₀⁽^j)−a⁽₀ⁱ⁾´

0 . . . 0 0 0

−³

a₃⁽ⁱ⁾−a₃⁽^j⁾´

−³

a₀⁽^j⁾−a₀⁽ⁱ⁾´

0 0 . . . 0 0 0

−³

a₄⁽ⁱ⁾−a₄⁽^j⁾´

0 0 ... ... 0 0 0

... ... ... ... ... ... ... ...

−³

a_n⁽ⁱ₋⁾₃−a_n⁽^j₋⁾₃´

0 0 0 . . . 0 ³

a₀⁽^j⁾−a⁽₀ⁱ⁾´

0

−³

a_n−2⁽ⁱ⁾ −a_n−2^(j⁾

´

0 0 0 . . . −³

a⁽₀^j)−a₀⁽ⁱ⁾

´

0 0

−³

a_n⁽ⁱ₋⁾₁−a_n⁽^j₋⁾₁´

0 0 0 . . . 0 0 0







(6)

and|B| =(a₀⁽ⁱ⁾−a₀⁽^j⁾)ⁿ⁻³(a_n⁽ⁱ₋⁾₁−a⁽_n₋^j⁾₁)².Thus, the non-singular quadricQ⁰is elliptic if and only if

trace

Ã¡a₀⁽ⁱ⁾−a₀⁽^j⁾¢¡

a₁⁽ⁱ⁾−a₁⁽^j⁾¢

+ · · · +¡

a⁽_nⁱ₋⁾₃−a_n⁽^j₋⁾₃¢¡

a_n⁽ⁱ₋⁾₂−a_n⁽^j₋⁾₂¢

¡a⁽_nⁱ₋⁾₁−a_n⁽₋^j⁾₁¢2

!

=1. 2

4.2. The corresponding partial ovoid ofQ⁺(n+2,q)

Theorem 4 In PG(n,q),n ≥ 3 odd,letFbe a partial flock of size k of the quadratic coneK = vQ. Then there exists a partial ovoid of the non-singular hyperbolic quadric Q⁺(n+2,q)of size kq +1 comprising k conics mutually tangent at a common point.

Conversely,given any such partial ovoid there exists a partial flockFofK.

Proof: EmbedKin a non-singular hyperbolic quadricQ⁺in PG(n+2,q)and let⊥denote the polarity determined byQ⁺. LetF= {π1, . . . , πk}.First, since PG(n,q)∩Q⁺=vQ, the line L=PG(n,q)^⊥meetsQ⁺in the single pointv.For i =1, . . . ,k,πi^⊥is a plane on L meetingQ⁺in a (non-singular) conicCionv.Since, for i,j ∈ {1, . . . ,k},i 6= j, πi∩πj

meetsKand hence alsoQ⁺in a non-singular elliptic quadric, it follows thathπi^⊥, π^⊥j ialso meetsQ⁺in a non-singular elliptic quadric. Hence no two points ofCi∪Cjare collinear onQ⁺, soC1∪ · · · ∪Ckis a partial ovoid ofQ⁺of size kq+1.The converse is immediate

as the polarity is bijective and involutory. 2

Corollary 5 Let q be even. A partial ovoid ofQ⁺(n+2,q)which is a union of conics mutually tangent at a common point has size at most q²+1.

Proof: Theorems 2 and 4. 2

The construction in Theorem 4 gives a bound on the size of a partial flock. If n>3 and q is even, this is not as good as the bound in Theorem 2.

Theorem 6 In PG(n,q),n ≥ 3 odd,letFbe a partial flock of size k of the quadratic coneK=vQin PG(n,q).Then k≤q⁽ⁿ⁻¹^)/².

Proof: GivenF, by Theorem 4 there exists a partial ovoidOof size kq+1 ofQ⁺(n+2,q). ThusO≤q⁽ⁿ⁺¹^)/²+1 ([8, A VI]) and the result follows. 2 We remark that in the case n=3, the bound is best possible as there exist partial flocks of size q of a quadratic cone in PG(3,q), called flocks, associated with certain ovoids of Q⁺(5,q).

LetF= {π1, . . . , πk}be a partial flock ofK=vQin PG(n,q),n odd. If the elements of the partial flock contain a common m-dimensional subspaceξ, then the corresponding partial ovoid ofQ⁺(n+2,q)is contained in an(n+1−m)-dimensional subspace. In particular, if m=n−3 and ifξ∩Kis non-singular then the corresponding partial ovoid is

(7)

contained in a quadricQ(4,q). If, further, q is odd then there corresponds a partial spread of size kq+1 of the generalized quadrangle W(q). If k=q then this is a spread and there arises a translation plane.

4.3. Derivation of a partial flock ofK, q odd

Let Q(n +1,q) be the non-singular quadric of PG(n +1,q) defined by the equation x0x1+x2x3+· · ·+xn−3xn−2−x_n²₋₁+xnxn+1=0 and let⊥denote the polarity determined byQ(n+1,q).The tangent hyperplane H0ofQ(n+1,q)at the point p0 =(0, . . . ,0,1,0) has equation xn+1 =0 and intersectsQ(n+1,q)in the quadratic coneK0with equation x0x1+x2x3+ · · · +xn−3xn−2−x_n²₋₁=xn+1=0 and vertex p0.

LetF0be a partial flock of size k ofK0, where for i=1, . . . ,k the elementπiofF0has equations a⁽₀ⁱ⁾x0+ · · · +a⁽_nⁱ₋⁾₁xn−1+xn =xn+1 =0.For i =1, . . . ,k, we define the line Li =πi^⊥,and note that LimeetsQ(n+1,q)in p0and the further point

pi = µ

a⁽₁ⁱ⁾,a⁽₀ⁱ⁾,a⁽₃ⁱ⁾,a⁽₂ⁱ⁾, . . . ,a_n⁽ⁱ₋⁾₂,a⁽_nⁱ₋⁾₃,−1 2 a_n⁽ⁱ₋⁾₁,1

4

¡a_n⁽ⁱ₋⁾₁¢2

−a⁽₀ⁱ⁾a₁⁽ⁱ⁾

−a⁽₂ⁱ⁾a₃⁽ⁱ⁾− · · · −a_n⁽ⁱ₋⁾₃a_n⁽ⁱ₋⁾₂,1

¶ .

Since pi ∈Q(n+1,q), it follows that the hyperplane Hi =p^⊥_i with equation a₀⁽ⁱ⁾x0+a⁽₁ⁱ⁾x1+ · · · +a⁽_nⁱ₋⁾₁xn−1+xn+a⁽_nⁱ₊⁾₁xn+1=0,

where

a_n⁽ⁱ₊⁾₁=1/4¡ a⁽_nⁱ₋⁾₁¢2

−a₀⁽ⁱ⁾a⁽₁ⁱ⁾−a⁽₂ⁱ⁾a₃⁽ⁱ⁾− · · · −a⁽_nⁱ₋⁾₃a_n⁽ⁱ₋⁾₂, (3) meetsQ(n+1,q)in a quadratic coneKi.For each i,j ∈ {1, . . . ,k}with i 6= j , define the (n −1)-dimensional spaceπi j = Hi ∩Hj. For each j ∈ {1, . . . ,k}letπj j be the (n−1)-dimensional spaceπj.

Theorem 7 With the notation introduced above,for any j∈ {1, . . . ,k},the set Fj = {πi j: i=1, . . . ,k}is a partial flock of the quadratic coneKj in H_j.

Proof: We use the notation and definitions made in this subsection. Let the collineation σ of PG(n+1,q)be defined by

σ:(x0,x1, . . . ,xn+1)7→

µ

x0−a₁⁽^j⁾xn+1,x1−a₀⁽^j⁾xn+1, . . . ,xn−3−a_n⁽^j₋⁾₂xn+1, xn−2−a_n⁽^j₋⁾₃xn+1,xn−1+1

2a⁽_n₋^j⁾₁xn+1,xn+a⁽₀^j⁾x0

+a₁⁽^j⁾x1+ · · · +a_n⁽^j₋⁾₁xn−1+a_n⁽^j₊⁾₁xn+1,xn+1

¶ .