A characterization of the family of secant lines to a hyperbolic quadric in PG(3,q), q odd

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Discrete Mathematics

journal homepage:www.elsevier.com/locate/disc

A characterization of the family of secant lines to a hyperbolic quadric in PG(3 , ^{q), q odd}

Puspendu Pradhan

^a,b,1

, Bikramaditya Sahu

^c,d,2

aSchool of Mathematical Sciences, National Institute of Science Education and Research (NISER), Bhubaneswar, P.O.- Jatni, District- Khurda, Odisha 752050, India

bHomi Bhabha National Institute (HBNI), Training School Complex, Anushakti Nagar, Mumbai 400094, India

cDepartment of Mathematics, Indian Institute of Science, Bangalore 560012, India

dDepartment of Mathematics, National Institute of Technology, Rourkela, Odisha 769008, India

a r t i c l e i n f o

Article history:

Received 15 July 2019

Received in revised form 25 June 2020 Accepted 28 June 2020

Available online 18 July 2020

Keywords:

Projective space Hyperbolic quadric Secant line

Combinatorial characterization

a b s t r a c t

We give a combinatorial characterization of the family of lines of PG(3,^q),q≥7 odd, which meet a hyperbolic quadric in two points (the so called secant lines) using their intersection properties with the points and planes of PG(3,^q).

©2020 Elsevier B.V. All rights reserved.

1. Introduction

Throughout,qis a prime power and PG(d

,

q) denotes thed-dimensional Desarguesian projective space defined over a finite field of orderq. Characterizations of the family of external or secant lines to an ovoid/quadric in PG(3

,

^{q) with} respect to certain combinatorial properties have been given by several authors. A characterization of the family of secant lines to an ovoid in PG(3

,

q) was obtained in [7] forqodd and in [11] forq

>

2 even, which was further improved in [6]

for allq

>

2. A characterization of the family of external lines to a hyperbolic quadric in PG(3

,

q) was given in [5] for allq(also see [9] for a different characterization in terms of a point-subset of the Klein quadric in PG(5

,

q)) and to an ovoid in PG(3

,

q) was obtained in [6] for allq

>

2. One can refer to [1,2,12,13] for characterizations of external lines in PG(3

,

q) with respect to quadric cone, oval cone and hyperoval cone. Here we give a characterization of the secant lines with respect to a hyperbolic quadric in PG(3

,

^{q) for oddq}

≥

7.

LetQ be a hyperbolic quadric in PG(3

,

q), that is, a non-degenerate quadric of Witt index two. One can refer to [8]

for the basic properties of the points, lines and planes of PG(3

,

q) with respect toQ. Every line of PG(3

,

^{q) meetsQin} 0

,

¹

,

^{2 orq}

+

1 points. A line of PG(3

,

q) is calledsecantwith respect toQif it meetsQin 2 points. The lines of PG(3

,

^q) that meetQinq

+

1 points are calledgeneratorsofQ. Each point ofQlies on two generators. The quadricQconsists of (q

+

1)²points and 2(q

+

1) generators.

E-mail addresses: puspendu.pradhan@niser.ac.in(P. Pradhan),bikramadityas@iisc.ac.in,sahuba@nitrkl.ac.in(B. Sahu).

1 The author was supported by the Council of Scientific and Industrial Research, India grant No. 09/1002(0040)/2017-EMR-I, Ministry of Human Resource Development, Government of India.

2 The author was supported by Department of Science and Technology Grant No. EMR/2016/006624, by the UGC Center for Advanced Studies and by National Board for Higher Mathematics Grant No. 0204/18/2019/R&D-II/10462, Department of Atomic Energy, Government of India.

https://doi.org/10.1016/j.disc.2020.112044 0012-365X/©2020 Elsevier B.V. All rights reserved.

The total number of secant lines of PG(3

,

q) with respect toQis^q2(q₂⁺¹⁾². We recall the distribution of secant lines with respect to points and planes of PG(3

,

q) which plays an important role in this paper. Each point ofQlies onq²secant lines. Each point of PG(3

,

^q)

\

Qlies on ^q(q₂⁺¹⁾ secant lines. Each plane of PG(3

,

q) containsq²or ^q(q₂⁺¹⁾ secant lines. If a plane containsq²secant lines, then every pencil of lines in that plane contains 0 orqsecant lines. Ifqis odd and a plane contains ^q(q₂⁺¹⁾ secant lines, then every pencil of lines in that plane contains^q−₂¹

,

^q+₂^{1 orq}secant lines.

We prove the following result in this paper.

Theorem 1.1. LetSbe a family of lines of PG(3

,

^{q), q}

≥

7odd, for which the following properties are satisfied:

(P1) There are ^q(q₂⁺¹⁾ or q²lines of Sthrough a given point of PG(3

,

q). Further, there exists a point which is contained in

q(q+₁₎

2 lines ofSand a point which is contained in q²lines ofS.

(P2) Every plane

π

^{of PG(3}

,

^q)contains at least one line ofSand one of the following two cases occurs:

(P2a) every pencil of lines in

π

^contains0or q lines ofS. (P2b) every pencil of lines in

π

^{contains q−}₂^1,q+₂¹ or q lines ofS.

Then eitherSis the set of all secant lines with respect to a hyperbolic quadric in PG(3

,

q), or the set of points each of which is contained in q²lines ofSform a line l of PG(3

,

^q)andSis a hypothetical family of ^q4+q3₂^+2q2 lines of PG(3

,

^q)not containing l.

Letm_i, 1

≤

i

≤

k, bekintegers with 0

≤

m₁

<

^m2

< · · · <

^mk

≤

q

+

1. A setK of points of PG(2

,

q) is said to be of class

[

m₁

,

^m2

, . . . ,

^mk

]

if every line of PG(2

,

^{q) meetsK inm}i points for somei

∈ {

1

,

²

, . . . ,

^k

}

. We need the following result which was proved in [4, Theorem 4.6] for oddq

≥

7.

Proposition 1.2([4]).Let K be a set of class[

1

,

^q+₂¹

,

^q+₂³]

in PG(2

,

q), where q

≥

7is odd. Then K consists of an irreducible conic in PG(2

,

^q)and its interior points.

2. Combinatorial results

LetSbe a family of lines of PG(3

,

q) for which the properties stated inTheorem 1.1hold. A plane of PG(3

,

^{q) is said} to betangentorsecantaccording as it satisfies the property (P2a) or (P2b). For a given plane

π

^{of PG(3}

,

q), we denote by S_π the set of lines ofSwhich are contained in

π

^.

2.1. Tangent planes

Lemma 2.1. If

π

is a tangent plane, then

|

S_π

| =

q².

Proof. Fix a linelofS_π. Every line ofS_π meetsl. By property (P2a), every point oflis contained inqlines ofS_π. Then the numbers of lines ofS_π is 1

+

(q

+

1)(q

−

1)

=

q². □

Lemma 2.2. Let

π

be a tangent plane. Then there are q²

+

q points of

π

, each of which is contained in q lines ofS_π. Equivalently, there is only one point of

π

which is contained in no lines ofS_π.

Proof. LetA_π(respectively,B_π) be the set of points of

π

each of which is contained in no lines (respectively,qlines) of S_π. Then

|

A_π

| + |

B_π

| =

q²

+

q

+

1. Consider the following set of point-line pairs:

X

= {

(x

,

^l)

:

x

∈

B_π

,

^l

∈

S_π

,

^x

∈

l

} .

ByLemma 2.1,

|

S_π

| =

q². Counting

|

X

|

in two ways, we get

|

B_π

| ·

q

= |

X

| =

q²

·

(q

+

1). This gives

|

B_π

| =

q²

+

qand hence

|

A_π

| =

1. _□

For a tangent plane

π

, we denote byp_π the unique point of

π

which is contained in no lines ofS_π (Lemma 2.2) and call it thepoleof

π

^.

Corollary 2.3. Let

π

be a tangent plane. Then the q

+

1lines of

π

not contained inS_π are precisely the lines of

π

^through the pole p_π.

2.2. Secant planes

In the rest of this paper,we assume that q is odd and that q

≥

7. As an application ofProposition 1.2, we have the following.

Lemma 2.4. Let K be a set of class[_q−₁ 2

,

^q+₂¹

,

^q]

in PG(2

,

q). Then K consists of the exterior points of an irreducible conic in PG(2

,

^q).

Proof. The complementKofKin PG(2

,

q) is a set of class[

1

,

^q+₂¹

,

^q+₂³]

. Sinceq

≥

7,Kconsists of an irreducible conicC and its interior points byProposition 1.2. ThenKconsists of the exterior points ofC. □

Lemma 2.5. Let

π

be a secant plane. ThenS_π consists of the secant lines of an irreducible conic in

π

^{and so}

|

S_π

| =

^q(q₂⁺¹⁾. Proof. Each point of

π

is contained in (q

−

1)

/

^{2, (q}

+

1)

/

^{2 orqlines ofS}π. Therefore, in the dual plane

π

^Dof

π

^,Sπ is a set of points of class[_q−₁

2

,

^q+₂¹

,

^q]

. Sinceq

≥

7, byLemma 2.4,S_π consists of the exterior points of an irreducible conic C^Din

π

^D. The dual ofC^Dis an irreducible conicCin

π

and the lines of

π

contained inC^Dare precisely the tangent lines ofC. Since exterior points ofC^Din

π

^Dcorrespond to the secant lines ofC in

π

, it follows thatS_π is precisely the set of secant lines ofCin

π

^{and hence}

|

S_π

| =

^q(q+1)

2 . _□

For a secant plane

π

^{of PG(3}

,

q), we denote by

γ

⁽

π

) the irreducible conic obtained inLemma 2.5with respect to which S_π is the set of secant lines in

π

. The sets of exterior and interior points of

γ

⁽

π

^{) in}

π

are denoted by

α

⁽

π

^{) and}

β

⁽

π

^), respectively. Then we have

| γ

⁽

π

⁾

| =

q

+

1,

| α

⁽

π

⁾

| =

^q2+q

2 and

| β

⁽

π

⁾

| =

^q2−q

2 . Note that

α

⁽

π

^),

β

⁽

π

^{) and}

γ

⁽

π

) are precisely the sets of points of

π

which are contained in ^q−₂¹, ^q+₂¹ andqlines ofS_π, respectively.

3. Black points

Recall that every point of PG(3

,

q) is contained inq²or ^q(q₂⁺¹⁾ lines ofSby property (P1). We call a point of PG(3

,

^q) blackif it is contained inq²lines ofS.

3.1. Existence of tangent and secant planes

We show that both tangent and secant planes exist for which the following result is needed.

Lemma 3.1. Let l be a line of PG(3

,

q). Then the number of tangent planes through l is equal to the number of black points contained in l.

Proof. Lettandb, respectively, denote the number of tangent planes throughland the number of black points contained inl. We count in two different ways the total number of lines ofS

\{

l

}

meetingl. We have

|

S_π

| =

q²or^q(q₂⁺¹⁾byLemmas 2.1 and2.5for a plane

π

. Any line ofSmeetinglis contained in some plane throughl. Ifl

∈

S, then we get

t(q²

−

1)

+

(q

+

1

−

t)

(q(q

+

1)

2

−

1

)

=

b(q²

−

1)

+

(q

+

1

−

b)

(q(q

+

1)

2

−

1

)

.

Ifl

∈ /

S, then we get

tq²

+

(q

+

1

−

t)q(q

+

1)

2

=

bq²

+

(q

+

1

−

b)q(q

+

1)

2

.

In both cases, it follows that (t

−

b)^q2₂^−q

=

0 and hencet

=

b. _□ Corollary 3.2. Both tangent and secant planes exist.

Proof. By property (P1), letx(respectively,y) be a point of PG(3

,

q) which is contained inq²(respectively, ^q(q₂⁺¹⁾) lines ofS. Takinglto be the line throughxandy, the corollary follows fromLemma 3.1using the facts thatxis a black point butyis not a black point. □

Corollary 3.3. Every line of a tangent plane contains at least one black point.

Corollary 3.4. Every black point is contained in some tangent plane.

3.2. Black points in secant planes

Lemma 3.5. If

π

is a secant plane, then the set of black points in

π

is contained in the conic

γ

⁽

π

). In particular, there are exactly q lines ofS_π through a black point in

π

^.

Proof. Letxbe a black point in

π

. Suppose thatxis not contained in

γ

⁽

π

). Fix a linelofS_π throughxand consider the q

+

1 planes of PG(3

,

^{q) through}l. There areq²lines ofSthroughxand each of them is contained in some plane through l. Sincex

∈ / γ

⁽

π

), the plane

π

contains at most ^q+₂¹lines ofSthroughx. Each of the remainingqplanes throughlcontains at mostqlines ofSthroughx. This implies that there are at most^q+₂¹

+

q(q

−

1) lines ofSthroughx. This is not possible, as ^q+₂¹

+

q(q

−

1)

<

^{q2. Sox}

∈ γ

⁽

π

^). □

Lemma 3.6. The number of black points in a given secant plane is independent of that plane.

Proof. Let

π

be a secant plane and

λ

_πdenote the number of black points in

π

. We count the total number of lines ofS. The lines ofSare divided into two types:

(I) the ^q(q₂⁺¹⁾ lines ofSwhich are contained in

π

^, (II) those lines ofSwhich meet

π

in a singleton.

Let

θ

be the number of type (II) lines ofS. In order to calculate

θ

, we divide the points of

π

into four groups:

(a) The

λ

π black points contained in

π

: These points are contained in

γ

⁽

π

^{) by}Lemma 3.5. Out of theq²lines of S through such a point,qof them are contained in

π

^.

(b) The

| γ

⁽

π

⁾

| − λ

π points of

γ

⁽

π

) which are not black: Out of the ^q(q₂⁺¹⁾ lines ofSthrough such a point,qof them are contained in

π

^.

(c) The points of

α

⁽

π

): Out of the ^q(q₂⁺¹⁾ lines ofSthrough such a point, ^q−₂¹ of them are contained in

π

^. (d) The points of

β

⁽

π

): Out of the ^q(q₂⁺¹⁾ lines ofSthrough such a point, ^q+₂¹ of them are contained in

π

^. Since

| γ

⁽

π

⁾

| =

q

+

1,

| α

⁽

π

⁾

| =

^q2+q

2 and

| β

⁽

π

⁾

| =

^q2−q

2 , we get

θ = λ

π(q²

−

q)

+

(q

+

1

− λ

π)

(q(q

+

1)

2

−

q

)

+ | α

⁽

π

⁾

|

(q(q

+

1)

2

−

^q

−

1 2

)

+ | β

⁽

π

⁾

|

(q(q

+

1)

2

−

^q

+

1 2

)

= λ

π

(q²

−

q 2

)

+

^q

3(q

+

1)

2

.

Then

|

S

| = θ +

^q(q

+

1) 2

= λ

π

(q²

−

q 2

)

+

^q

4

+

q³

+

q²

+

q

2 . Since

|

S

|

is a fixed number, it follows that

λ

πis independent of the secant plane

π

^. □

ByLemma 3.6, we denote by

λ

the number of black points in a secant plane. From the proof ofLemma 3.6, we thus have the following equation involving

λ

^and

|

S

|

:

λ

(q²

−

q

2 )

+

^q

4

+

q³

+

q²

+

q

2

= |

S

| .

⁽¹⁾

As a consequence ofLemma 3.5, we have Corollary 3.7.

λ ≤

q

+

1.

3.3. Black points in tangent planes

Lemma 3.8. The number of black points in a given tangent plane is independent of that plane.

Proof. Let

π

be a tangent plane with polep_π and

µ

π be the number of black points in

π

. We shall apply a similar argument as in the proof ofLemma 3.6by calculating

|

S

|

. The lines ofSare divided into two types: (I) theq²lines ofS which are contained in

π

, and (II) those lines ofSwhich meet

π

in a singleton. Let

θ

be the number of type (II) lines of S. In order to calculate

θ

, we divide the points of

π

into two groups:

(a) The

µ

π black points contained in

π

^,

(b) Theq²

+

q

+

1

− µ

π points of

π

which are not black.

Ifxis a point of

π

which is different fromp_π, thenLemma 2.2implies that the number of lines ofSthroughxwhich are not contained in

π

^isq2

−

qor ^q(q₂⁺¹⁾

−

qaccording asxis a black point or not. We consider two cases depending onp_π is a black point or not.

Case-1:p_π is a black point. In this case,Lemma 2.2implies that none of theq²lines ofSthroughp_π is contained in

π

^. Then

θ =

q²

+

(

µ

π

−

1)(q²

−

q)

+

(q²

+

q

+

1

− µ

π)

(q(q

+

1)

2

−

q

)

= µ

π

(q²

−

q 2

)

+

^q

4

+

q 2

.

Case-2:p_πis not a black point. In this case, none of the^q(q₂⁺¹⁾ lines ofSthroughp_πis contained in

π

^byLemma 2.2. Then

θ = µ

π(q²

−

q)

+

^q(q

+

1)

2

+

(q²

+

q

− µ

π)

(q(q

+

1)

2

−

q

)

= µ

π

(q²

−

q 2

)

+

^q

4

+

q 2

.

In both cases,

|

S

| = θ +

q²

= µ

π

(q²

−

q 2

)

+

^q

4

+

2q²

+

q

2 . Since

|

S

|

is a fixed number, it follows that

µ

π is independent of the tangent plane

π

^. □

ByLemma 3.8, we denote by

µ

the number of black points in a tangent plane. From the proof ofLemma 3.8, we thus have the following equation involving

µ

^and

|

S

|

:

µ

(q²

−

q

2 )

+

^q

4

+

2q²

+

q

2

= |

S

|

(2)

From Eqs.(1)and(2), we have

µ = λ +

q

.

⁽³⁾

3.4. Black points on a line Lemma 3.9. The following hold:

(i) Every line of PG(3

,

^q)contains0

,

¹

,

^{2or q}

+

1black points.

(ii) If a line of PG(3

,

^q)contains exactly two black points, then it is a line ofS.

Proof. Letlbe a line of PG(3

,

^{q) andb}be the number of black points contained inl. Assume thatb

>

2. If there exists a secant plane

π

^{throughl, thenLemma 3.5}implies that the linelcontainsb

≥

3 number of points of the conic

γ

⁽

π

^{) in}

π

^, which is not possible. So all planes throughlare tangent planes. Then all theq

+

1 points oflare black byLemma 3.1.

This proves (i).

Ifb

=

2, thenLemma 3.1implies that there exists a secant plane

π

^{throughl. ByLemma 3.5,l}is a secant line of the conic

γ

⁽

π

^{) in}

π

and hence a line ofS_π. This proves (ii). □

4. Proof ofTheorem 1.1

We shall continue with the notation used in the previous sections and the assumption thatq

≥

7. We denote byH the set of all black points of PG(3

,

^{q), and byH}π the set of black points of PG(3

,

q) which are contained in a given plane

π

^.

Lemma 4.1.

|

H

| = λ

^(q

+

1). In particular,

|

H

| ≤

(q

+

1)².

Proof. Fix a secant plane

π

^{. Letl}be a line of

π

which is external to the conic

γ

⁽

π

^{). By}Lemma 3.5, none of the points of lis black. Then, byLemma 3.1, each plane throughlis a secant plane. The number of black points contained in a secant plane is

λ

. Counting all the black points contained in theq

+

1 planes throughl, we get

|

H

| = λ

^(q

+

1). Since

λ ≤

q

+

1 byCorollary 3.7, we have

|

H

| ≤

(q

+

1)². □

The following result was proved by Bose and Burton in [3, Theorem 1]. We need it in the plane case.

Proposition 4.2([3]).Let B be a set of points of PG(d

,

^q)such that every line of PG(d

,

^q)meets B. Then

|

B

| ≥

(q^d

−

1)

/

^(q

−

1), and equality holds if and only if B is a hyperplane of PG(d

,

^q).

Lemma 4.3. If

π

be a tangent plane, thenH_π contains a line.

Proof. ByCorollary 3.3, every line of

π

^meetsHπ. ByProposition 4.2(takingn

=

2), we then have

|

H_π

| ≥

q

+

1, and equality holds if and only ifH_π itself is a line of

π

^.

Therefore, assume that

|

H_π

| >

^q

+

1. Sinceqis odd, the maximum size of an arc in

π

^isq

+

1. SoH_π cannot be an arc and hence there exists a linelof

π

which contains at least three points ofH_π. Then all points oflare black by Lemma 3.9(i) and solis contained inH_π. □

Lemma 4.4. Let

π

be a tangent plane. ThenH_π is either a line or union of two (intersecting) lines.

Proof. UsingLemmas 3.9(i) and4.3, observe that there are only four possibilities for the setH_π:

(1) H_π is a line.

(2) H_π is the union of a lineland a point of

π

not contained inl.

(3) H_π is the union of two (intersecting) lines.

(4) H_π is the whole plane

π

^.

We show that the possibilities(2)and (4)do not occur. IfH_π is the whole plane

π

^{, then}

µ =

q²

+

q

+

1 and so

λ =

q²

+

1 by Eq.(3), which is not possible byCorollary 3.7.

Now suppose thatH_πis the union of a lineland a pointxnot onl. If the polep_πof

π

is different fromx, then taketto be the line throughp_πandx(note thatp_π may or may not be onl). Ifp_π

=

x, then taket to be any line throughp_π

=

x.

Since

π

is a tangent plane,tis not a line ofS_π byCorollary 2.3and hence is not a line ofS. On the other hand, sincet contains only two black points (namely, the pointxand the intersection point oflandt),tis a line ofSbyLemma 3.9(ii).

This leads to a contradiction. □

Lemma 4.5. Let

π

be a tangent plane. IfH_πis a line of PG(3

,

q), then the following hold:

(i)H_π is not a line ofS. (ii) H_π

=

H.

(iii) Sis a set of q⁴

+

q³

+

2q²

2 lines of PG(3

,

^q)not containing the lineH.

Proof.(i) Suppose thatH_π is a line ofS. Then, byCorollary 2.3, the polep_πof

π

must be a point of

π \

H_π. Fix a linem of

π

^throughpπ. Note thatm

∈ /

Sagain byCorollary 2.3. Letxbe the point of intersection ofmandH_π. Sincemcontains only one black point (which isx),Lemma 3.1implies thatmis contained in one tangent plane (namely,

π

^{) andqsecant} planes. Sincex

̸=

p_π, by Lemma 2.2, there areqlines of

π

^throughxwhich are contained inS. In each of theqsecant planes throughm,xbeing a black point, there areqlines throughxwhich are contained inSbyLemma 3.5. Sincem

∈ /

S, we getq(q

+

1)

=

q²

+

qlines ofSthroughx, which is not possible by property (P1).

(ii) Suppose thatxis a black point which is not contained inH_π. Let

π

^′be the plane generated by the lineH_π and the pointx. We have

π ̸= π

^′asxis not a black point of

π

. Each of the planes through the lineH_π is a tangent plane by Lemma 3.1. In particular,

π

^′is a tangent plane. Note that

π

^containsq

+

1 black points, whereas

π

^′contains at leastq

+

2 black points. This contradictsLemma 3.8.

(iii) The lineHis not contained inSby (i) and (ii). Since the tangent plane

π

^{contains q}

+

1 black points, we have

µ =

q

+

1. Then Eq.(2)gives that

|

S

| =

^q

4

+

q³

+

2q²

2 . _□

Lemma 4.5proves the second possibility as mentioned inTheorem 1.1for the familyS. However, the existence of a familySof lines with

|

S

| =

^q4+q3+2q2

2 and satisfying the conditions ofTheorem 1.1is yet to be explored.

In the rest of this section, we assume thatH_π is the union of two (intersecting) lines for every tangent plane

π

^{. So}

µ =

2q

+

1 and then Eq.(3)gives that

λ =

q

+

1. From Eq.(2)andLemma 4.1, we get

|

S

| =

^q

2(q

+

1)²

2 and

|

H

| =

(q

+

1)²

.

⁽⁴⁾

Lemma 4.6. Let

π

be a tangent plane. IfH_π is the union of the lines l and l^′of

π

, then the pole p_π of

π

is the intersection point of l and l^′.

Proof. Letxbe the intersection point oflandl^′. Suppose thatp_π

̸=

x. Lett be a line of

π

^throughpπ which does not containx(note thatp_πmay or may not be contained inl

∪

l^′). Since

π

is a tangent plane,tis not a line ofS_πbyCorollary 2.3 and hence is not a line ofS. On the other hand, sincetcontains two black points (namely, the two intersection points of twithlandl^′), it is a line ofSbyLemma 3.9(ii). This leads to a contradiction. _□

We call a line of PG(3

,

^q)blackif it is contained inH.

Lemma 4.7. Every black point is contained in at most two black lines.

Proof. Letxbe a black point. If possible, suppose that there are three distinct black linesl

,

^l1

,

^l2each of which contains x. Let

π

(respectively,

π

^′) be the plane generated byl

,

^l1(respectively,l

,

^l2). Each plane throughlis a tangent plane by Lemma 3.1. So

π

^and

π

^′are tangent planes. SinceH_π

=

l

∪

l₁andH_π^′

=

l

∪

l₂, it follows that

π ̸= π

^{′. ByLemma 4.6,} xis the pole of both

π

^and

π

^′. So the lines throughxwhich are contained in

π

^or

π

^′are not lines ofSbyCorollary 2.3.

Thus each line ofSthroughxis contained in some plane throughlwhich is different from both

π

^and

π

^′. It follows that the number of lines ofSthroughxis at mostq(q

−

1), which contradicts to the fact that there areq²lines ofSthrough x(being a black point). □

Lemma 4.8. Every black point is contained in precisely two black lines.

Proof. Letxbe a black point andlbe a black line containingx. The existence of such a linelfollows from the facts that xis contained in a tangent plane (Corollary 3.4) and that the set of all black points in that tangent plane is a union of two black lines. ByLemma 3.1, let

π

1

, π

2

, . . . , π

q+₁be theq

+

1 tangent planes throughl. For 1

≤

i

≤

q

+

1, we have H_πi

=

l

∪

l_ifor some black linel_iof

π

idifferent froml. Let

{

p_i

} =

l

∩

l_i.Lemma 4.7implies thatp_i

̸=

p_jfor 1

≤

i

̸=

j

≤

q

+

1, and sol

= {

p₁

,

^p2

, . . . ,

^pq+₁

}

. Sincex

∈

l, we havex

=

p_jfor some 1

≤

j

≤

q

+

1. Thus, applyingLemma 4.7again, it follows thatxis contained in precisely two black lines, namely,landl_j. □

We refer to [10] for the basics on finite generalized quadrangles. Let s and t be positive integers. A generalized quadrangle of order (s

,

^t) is a point-line geometryX

=

(P

,

L) with point set P and line set Lsatisfying the following three axioms:

(Q1) Every line containss

+

1 points and every point is contained int

+

1 lines.

(Q2) Two distinct lines have at most one point in common (equivalently, two distinct points are contained in at most one line).

(Q3) For every point-line pair (x

,

^l)

∈

P

×

Lwithx

∈ /

l, there exists a unique linem

∈

Lcontainingxand intersectingl.

LetX

=

(P

,

L) be a generalized quadrangle of order (s

,

^{t). Then,}

|

P

| =

(s

+

1)(st

+

1) and

|

L

| =

(t

+

1)(st

+

1) [10, 1.2.1]. If Pis a subset of the point set of some projective space PG(d

,

^q),Lis a set of lines of PG(d

,

^{q) andP}is the union of all lines in L, thenX

=

(P

,

L) is called aprojective generalized quadrangle. The points and the lines contained in a hyperbolic quadric in PG(3

,

q) form a projective generalized quadrangle of order (q

,

1). Conversely, any projective generalized quadrangle of order (q

,

1) with ambient space PG(3

,

q) is a hyperbolic quadric in PG(3

,

q), this follows from [10, 4.4.8].

The following two lemmas complete the proof ofTheorem 1.1.

Lemma 4.9. The points ofHtogether with the black lines form a hyperbolic quadric in PG(3

,

^q).

Proof. We have

|

H

| =

(q

+

1)² by(4). It is enough to show that the points ofHtogether with the black lines form a projective generalized quadrangle of order (q

,

^1).

Each black line containsq

+

1 points ofH. ByLemma 4.8, each point ofHis contained in exactly two black lines. Thus the axiom (Q1) is satisfied withs

=

qandt

=

1. Clearly, the axiom (Q2) is satisfied.

We verify the axiom (Q3). Let l

= {

x₁

,

^x2

, . . . ,

^xq+₁

}

be a black line and x be a black point not contained inl. By Lemma 4.8, letl_ibe the second black line throughx_i(different froml) for 1

≤

i

≤

q

+

1. Ifl_iandl_jintersect fori

̸=

j, then the tangent plane

π

generated byl_i andl_jcontains las well. This implies thatH_π contains the union of three distinct black lines (namely,l

,

^li

,

^lj), which is not possible. Thus the black linesl₁

,

^l2

, . . . ,

^lq+₁ are pairwise disjoint. Theseq

+

1 black lines contain (q

+

1)² black points and hence their union must be equal toH. In particular,xis a point of l_jfor uniquej

∈ {

1

,

²

, . . . ,

^q

+

1

}

. Thenl_jis the unique black line containingx_jand intersectingl.

From the above two paragraphs, it follows that the points of H together with the black lines form a projective generalized quadrangle of order (q

,

1). This completes the proof. □

Lemma 4.10. The lines ofSare precisely the secant lines to the hyperbolic quadricH.

Proof. By(4), we have

|

S

| =

^q

2(q

+

1)²

2 , which is equal to the number of secant lines toH. It is enough to show that every secant line toHis a line ofS. This follows fromLemma 3.9(ii), as every secant line toHcontains exactly two black points. □

Declaration of competing interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

Acknowledgments

The authors wish to thank Dr. Binod Kumar Sahoo for his comments, which helped improve an earlier version of the article. The authors also thank the anonymous referees for their helpful comments and for suggesting to use certain results from the paper [4].

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