Subplanes of PG( 2, q 3 ) and the Ruled Varieties V 2 5 of PG( 6,q )

Abstract

In this note we study subplanes of order q of the projective plane Π=PG( 2, q 3 ) and the ruled varieties V 2 5 of Σ=PG( 6,q ) using the spatial representation of Π in Σ, by fixing a hyperplane Σ with a regular spread of planes. First are shown some configurations of the affine q-subplanes. Then to prove that a variety V 2 5 of Σ represents a non-affine subplane of order q of Π, after having shown basic incidence properties of it, such a variety V 2 5 is constructed by choosing appropriately the two directrix curves in two complementary subspaces of Σ. The result can be translated into further incidence properties of the affine points of V 2 5 . Then a maximal bundle of varieties V 2 5 having in common one directrix cubic curve is constructed.

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Vincenti, R. (2024) Subplanes of PG( 2, q 3 ) and the Ruled Varieties V 2 5 of PG( 6,q ). Open Journal of Discrete Mathematics, 14, 16-27. doi: 10.4236/ojdm.2024.142003.

1. Introduction

It is known that a projective translation plane Π of order q r and kernel F = G F ( q ) can be represented by a 2r-dimensional projective space Σ = P G ( 2 r , q ) over F, fixing a hyperplane Σ = P G ( 2 r 1, q ) and a spread (partition) S of Σ with ( r 1 ) -dimensional subspaces (cf. [1] and [2] ).

The points of Π are represented by 1) the points of Σ \ Σ (the affine points) and by 2) the elements of S (the points at infinity). The lines of Π are represented by 1) the r-dimensional subspaces S of Σ \ Σ such that S Σ belongs to S and by 2) the spread S . The translation line l of Π (the line at infinity) is represented by S (cf. Lemma 2.6).

If a subplane of Π meets l in a subline, then such a subplane is affine, if it meets l in one point is non-affine.

An affine subplane A of order q is represented by every transversal plane α to S , that is, by a plane α Σ \ Σ such that the line t = α Σ meets q + 1 elements of S , t is a transversal line to S . In such a way l is a line of the projective completion of A . Of course all that holds also in case Π is the Desarguesian plane P G ( 2, q r ) when S is a regular spread (cf. [1] [2] [3] [4] [5] for r = 2 ).

Fix r = 3 so that Π = P G ( 2, q 3 ) , Σ = P G ( 6, q ) , Σ = P G ( 5, q ) and S is a regular spread of planes.

About the affine subplanes of Π = P G ( 2, q 3 ) of order q (having the same subline at infinity) we prove there exist q 2 + q + 1 through one fixed affine point, while q 4 partition the affine points of Π (cf. Proposition 2.11, Theorem 2.12).

A variety V 2 5 of Σ is a ruled variety of P G ( 6, q ) with the minimum order directrix a conic and a maximum order directrix a skew cubic in a 3-dimensional subspace, the two curves lying in two complementary spaces (cf. [6] , Capter 13, 8., 9.). The variety can be obtained by joining points of the two directrix curves corresponding via a projectivity.

We choose a conic in a plane π S and a cubic in a 3-dimensional subspace S 0 Σ \ Σ with π 0 = S 0 Σ S and π 0 π . Some fundamental incidence properties of V 2 5 are shown (cf. Paragraph 3.1). Then, if q 1 , ( mod 3 ) , we can prove that the variety V 2 5 represents a non-affine subplane Πq of order q of P G ( 2, q 3 ) (cf. Theorem 3.6).

The properties of Πq of being a plane, translate into further incidence properties of the affine points of V 2 5 (cf. Theorem 3.8).

By fixing the 3-subspace S 0 with the chosen directrix cubic curve C 0 , a maximal bundle of varieties V 2 5 having in common C 0 is constructed (cf. Theorem 3.9).

At the end is formulated the conjecture that a ruled variety V 2 2 r 1 of P G ( 2 r , q ) represents a non affine subplane of order q of P G ( 2, q r ) , via the spatial representation.

2. Preliminary Notes and Results

Let F = G F ( q ) be a finite field, q = p s , p an odd prime. Denote F r + 1 the ( r + 1 ) -dimensional vector space over F, P G ( r , q ) = P r F r + 1 the r-dimensional projective space contraction of F r + 1 over F. Let F ¯ be the algebraic closure of the field F = G F ( q ) .

The geometry P G ( r , q ) is considered a sub-geometry of P G ( r , q ) ¯ , the projective geometry over F ¯ . We refer to the points of P G ( r , q ) as the rational points of P G ( r , q ) ¯ .

Denote S h or h-space with 1 h r 1 a subspace of P G ( r , q ) of dimension h. A hyperplane S r 1 will be denoted also by H, a plane by π. If A , B , C , are subspaces denote A + B + C + = A , B , C , , the subspace generated by them. More simply, when A , B are points, AB denote the line defined by them.

Definition 2.1 A variety V u v of dimension u and of order v of P G ( r , q ) is the set of the rational points of a projective variety V ¯ u v of P G ( r , q ) ¯ defined by a finite set of polynomials with coefficients in the field F.

From [6] , pp. 290, 7., for r 4 follows.

Lemma 2.2 The ruled variety V 2 r 1 of P G ( r , q ) is generated by the lines joining the corresponding points of two birationally (projectively) equivalent curves of order m and r 1 m , respectively, lying in two complementary subspaces of the same dimensions. As the directrix curves have no point in common, then the number of points of V 2 r 1 is ( q + 1 ) 2 and the order is the sum of the orders of the curves.

Let Σ be the projective space P G ( 2 r 1, q ) over the field F = G F ( q ) , r > 1 an integer.

Definition 2.3 A spread of Σ is a partition S with ( r 1 ) -dimensional subspaces (that is, every point of Σ lies in one element of S ). A regulus R of S is a collection of subspaces of S such that:

1) R contains at least 3 elements,

2) Every line meeting 3 elements of R , a transversal line, meets every element of R ,

3) Every point of a transversal line to R lies in one element of R ,

4) Every plane through a transversal line is a transversal plane to S .

Any three pairwise disjoint ( r 1 ) -dimensional subspaces of S lie in a unique regulus. Any two distinct transversal lines are skew.

A spread is regular if for any three distinct elements of S , all the members of the unique regulus determined by them are in S .

Regular spreads represent Desarguesian planes P G ( 2, q r ) (cf. [2] , pp. 162-163).

The construction of a regular spread can be described as follows.

Choose a coordinate system in Σ so that for a point P of Σ , P ( x , y ) = ( x 1 , x 2 , , x r ; y 1 , y 2 , , y r ) = F * ( x , y ) , F * = F \ { 0 } .

A regular spread of Σ is given by the set { J = ( 0 , y ) | y F r } { J m = ( x , x m ) | x , m F r } where y = x m is the multiplication in the field F r . Such a multiplication can be represented also by y = x M with M a r × r matrix over F so that x M = x m . The set M = { M | x M = x m } is a field isomorphic to ( F r ) 2 , acting strictly transitively over F r .

In case of a projective plane over a skew-field, a spread can be constructed in the same way. The set of matrices is not a field, anyway it operates strictly transitively over F r (cf. [1] and [2] ).

Let Σ be the projective geometry P G ( 6, q ) .

Denote π and S0 a plane and a 3-space, respectively, of Σ. Assume π and S0 are complementary. Let C 2 be an irreducible conic of π , C 3 a skew cubic curve of S0. The curves C 2 and C 3 are projectively equivalent so that their points can be connected by a projectivity.

Lemma 2.4 A variety V 2 5 of Σ is obtained by joining the corresponding points of C 2 and of C 3 .

Proof. See [6] , p. 291.

Corollary 2.5 The variety V 2 5 consists of ( q + 1 ) 2 points of the q + 1 generatrix lines including the points of the minimum order directrix C 2 and of the maximum order directrix C 3 .

Note that the set of the q + 1 generatrix lines partition the variety.

Choose a hyperplane Σ = P G ( 5, q ) of Σ as the hyperplane at infinity.

Fix a coordinate system in Σ so that it is a coordinate system also for Σ ¯ . Denote a point P ( x , y , t ) = ( x 1 , x 2 , x 3 , y 1 , y 2 , y 3 , t ) : = F ¯ * ( x , y , t ) , F ¯ * = F ¯ \ { 0 } . Let t = 0 be the equation of Σ .

P is a rational point if there exists ( x , y , t ) F 7 such that P ( x , y , t ) .

A variety V of Σ is the set of the rational points of Σ ¯ solutions of a finite set of polynomials of F [ x , y , t ] .

Let Π = P G ( 2, q 3 ) be the Desarguesian plane over the field G F ( q 3 ) . Denote l the line at infinity of Π. Represent Π in Σ = P G ( 6, q ) by a regular spread S of planes of Σ , with | S | = q 3 + 1 .

More precisely define the following incidence structure Π = ( P , L , I ) (points, lines, incidence, respectively) where

P = { P Σ \ Σ } { π S } ,

L = { L 0 = { S 3 Σ \ Σ | S 3 Σ S } } { l = S } ,

I is defined as follows

if P Σ \ Σ , l L 0 then P I l P l , no point of Σ \ Σ incides l , π I l for all π S , π I l where l L 0 l Σ = π .

From [1] [2] [3] and [4] pp. 38-39 follows.

Lemma 2.6 Π Π .

In short, the affine points of Π are represented by the q 6 affine points of Σ \ Σ , the points at infinity by the q 3 + 1 planes of S . The affine lines of Π are represented by the 3-spaces S of Σ \ Σ such that the plane S Σ belongs to S , the line at infinity l by the spread S .

Definition 2.7 A subplane π = ( P , L , I ) of a plane π = ( P , L , I ) is a subgeometry of π, that is, an incidence substructure for which P P , for each line l L , there exists a line l L such that l L and I = I .

Definition 2.8 A subplane of Π = P G ( 2, q 3 ) of order q is affine if it meets the line l of Π in a subline consisting of q + 1 points, it is non-affine if it meets the line l in one point.

Let r be any transversal line to S . As S is regular, the q + 1 planes that r meets form a regulus R S (cf. [2] , Lemma 12.2). Choose and fix a transversal plane α through the line r.

It is easy to prove the following.

Proposition 2.9 The plane α is isomorphic to a subplane π P G ( 2, q ) of Π whose points at infinity are represented by the q + 1 planes of R , the lines of π being represented by the sublines intersections of α with the 3-spaces of Σ through the planes of R . As the line at infinity of π is a subline of the infinity line of Π, then π is an affine subplane.

To construct transversal lines to R in Σ in a synthetic way the procedure is similar to the one used for dimension 3.

Proposition 2.10 The transversal lines to R are q 2 + q + 1 .

Proof. Denote π 1 , π 2 , π 3 three planes of the regulus R . Fix a point P π 1 and denote S = P , π 2 the 3-space of Σ direct sum of P and π 2 and S = P , π 3 the 3-space of Σ direct sum of P and π 3 . Lying in a 5-dimensional subspace, then S S = r is a line. As a line of S, r meets π 2 in a point, as a line of S , r meets π 3 in a point. Therefore r is a transversal line to the planes π 1 , π 2 , π 3 . As π 1 , π 2 , π 3 belong to the regulus R , the line r meets each of the q + 1 elements of R . In such a way one can construct a transversal line for every point P chosen in π 1 , that is, q 2 + q + 1 .

Proposition 2.11 The cardinality of the set of all affine subplanes of Π isomorphic to P G ( 2, q ) having the same subline of q + 1 points at infinity and containing one affine point is q 2 + q + 1 .

Proof. Let r 0 be a transversal line to R , α 0 r 0 a transversal plane, O an affine point of α 0 . Denote { r i | i = 0 , , q 2 + q } the transversal lines to the regulus R . Each of the q 2 + q + 1 planes α i = O , r i represents an affine subplane π i of Π, π i P G ( 2, q ) (cf. Proposition 2.9).

Choose and fix a transversal line r. Consider the bundle (r) of the planes of Σ \ Σ having the line r as axis. Each plane α ( r ) is isomorphic to P G ( 2, q ) (cf. Proposition 2.9) and it is an affine subplane of Π having the same subline of q + 1 points at infinity.

Theorem 2.12 The planes of (r) partition the q 6 affine points of Π.

Proof. The planes of (r) through the transversal line r are parallel, therefore they have no affine point in common otherwise they would coincide. Each such a plane contains q 2 affine points.

To prove the statement it is appropriate to calculate the number of the planes of Σ \ Σ of the bundle through the line r.

It is known that the number of k-subspaces in a n-space (vector notation) is

[ n | k ] = ( q n 1 ) ( q n q ) ( q n q k 1 ) ( q k 1 ) ( q k q ) ( q k q k 1 ) .

The number h of the planes through a line in P G ( 6, q ) \ P G ( 5, q ) equals the number of 3-spaces in a 4-space, minus the number of the planes through a line in P G ( 5, q ) , which equals the number of the planes in a 3-space.

Therefore, after simplification of the two ratios, follows

h = q 5 1 q 1 q 4 1 q 1 = q 4 .

More simply, as a line and an independent point define a plane, fixed the line r, there are q 6 choices for a point in Σ \ Σ to get the plane r , P Σ \ Σ , this number to be divided by q 2 , which equals the choices of an affine point on a same plane, hence again h = q 4 .

As each plane of (r) contains q 2 points of Σ \ Σ and | ( r ) | = q 4 , then the total number of points of Σ \ Σ covered by them is q 6 , which are all the affine points of Π.

3. Main Results

3.1. The Variety V 2 5 and Its Sections

Denote Σ = P G ( 6 , q ) , Σ = P G ( 5, q ) a hyperplane of Σ, S a regular spread of Σ . Choose and fix a plane π S and a 3-space S 0 in Σ \ Σ such that S 0 Σ = π 0 S \ π . Then choose and fix:

1) A non-degenerate conic C 2 π ,

2) A skew cubic curve C 3 in S 0 such that C 3 π 0 = (cf. [7] , p. 234, Corollary 4, N 5 ).

Let Λ : C 2 C 3 be a projectivity. Represent C 2 = { G i | i = 1 , , q + 1 } , C 3 = { G i = Λ G i | i = 1 , , q + 1 } . Denote V 2 5 the variety arising by connecting corresponding points of C 2 and C 3 via Λ (cf. [6] , p. 291). The curves C 2 and C 3 are directrix curves of V 2 5 , the set G = { g i = G i G i | i = 1 , , q + 1 } is the set of the generatrix lines of V 2 5 . The set G partitions the variety.

Let H be any hyperplane. In a suitable complexification of Σ, H V 2 5 is a curve of order 5 (cf. [6] , p. 288, 5.).

Proposition 3.1 The variety V 2 5 consists of q + 1 skew affine generatrix lines and of q 2 + q affine points.

1) A directrix curve C C 2 cut by a hyperplane on V 2 5 cannot lie in a plane. The conic C 2 is the unique minimum order 2 directrix.

2) The 3-space joining two generatrix lines cannot contain the plane π .

3) The 3-space joining one generatrix line and the plane π meets no other generatrix.

4) Three generatrices g 1 , g 2 , g 3 are joint by a hyperplane H that contains the plane π , so that H V 2 5 = { g 1 , g 2 , g 3 } C 2 .

5) A hyperplane contains neither a fixed directrix, nor a fixed generatrix.

Proof. Let g , g be two generatrix lines. Denote G = g π , G = g π , with G , G C 2 , G = g C 3 , G = g C 3 , G , G S 0 . Let r be the line G G , r π , r the line G G , r S 0 . Assume g , g meet in a point. Hence they define a plane τ = r , r so that r r = P with P π S 0 , a contradiction.

As C 3 has no points in π 0 , then all the q + 1 generatrices are affine and the points of them are affine except the points of C 2 , therefore the affine points of V 2 5 are q ( q + 1 ) = q 2 + q .

1) Assume a hyperplane H meets V 2 5 in a directrix curve C C 2 lying in a plane τ . Then V 2 5 is contained at most in the 5-space generated by τ and π , a contradiction. The conic C 2 is the unique minimum order 2 directrix, otherwise the variety generated by the two conics would have order at most 4.

2) Assume there exists a 3-dimensional subspace S containing π and two generatrix lines, g 1 , g 2 . Denote G i = g i C 3 , i = 1,2 and G 1 G 2 Σ = G . The line G 1 G 2 belongs to S 0 and to S, so that the point G is a common point of π and π 0 , a contradiction.

3) Let S = g , π , with g G , be a 3-space. If S g with g g , g G , then g S , so that S contains two generatrix lines and the plane π , a contradiction to (2).

4) Assume three generatrices g 1 , g 2 , g 3 are joint by a 4-space S . As S contains the three points G 1 , G 2 , G 3 C 2 , G i = g i C 2 , i = 1,2,3 , then π S and C 2 S . As S cannot contain V 2 5 , a hyperplane H S and through a further point P V 2 5 \ S should contain also the generatrix g P through P. Hence H would meet V 2 5 in 4 generatrix lines and a conic, that is, in a variety of order 6, a contradiction (cf. [6] , p. 288, 5.). Therefore a hyperplane H containing three generatrices g 1 , g 2 , g 3 , contains the non collinear points G 1 , G 2 , G 3 hence the whole plane π and then the conic directrix C 2 . Therefore H V 2 5 = { g 1 , g 2 , g 3 } C 2 that is a curve of order 5 (and H contains no further point of V 2 5 ).

5) Let g , g two generatrices of V 2 5 . Denote S the 3-space containing them. Let H be a generic hyperplane with H S and assume H contains a fix directrix C . Let P be a point of V 2 5 , P H \ C . Denote S 4 = S , P . Then every hyperplane containing S 4 and S 4 itself, would contain the generatrix g P through P, so that g , g , g P S 4 , a contradiction to (4). An analogous contradiction is reached if we assume a generic hyperplane containing g , g contained a fix generatrix (cf. [6] , 6., pp. 289-290).

The following propositions are a rereading in the current case of [6] , pp. 287-290.

Proposition 3.2 A hyperplane H containing two generatrices, contains a residual cubic curve C lying in a 3-space S H , S skew to π . C is irreducible and is a directrix.

Proof. In [6] , 3., p. 287, the 2nd paragraph, is affirmed that a hyperplane H meets V 2 5 in a rational normal curve of order 5 (as it lives in a 5-space) or in a curve of order m < 5 met by all the generatrix lines and in 5 m generatrices. In our case it is m = 2 or m = 3 . Set m = 2 . If a hyperplane H contains the unique conic directrix C 2 (cf. 1), Proposition 3.1), then it must contain 5 m = 3 generatrix lines and viceversa (cf. 4), Proposition 3.1).

Set m = 3 . If a hyperplane contains 5 m = 2 generatrix lines, then it meets V 2 5 in a residual cubic curve C and viceversa.

Assume the cubic C exists in a plane π. Let H be a hyperplane containing π and three further points P , Q , R V 2 5 \ π . Then H contains also the 3 generatrix lines g P , g Q , g R as all the generatrix meet C . In such a way H V 2 5 { C , g P , g Q , g R } , that is a curve of order 6, a contradiction. Hence C lies in a 3-space S.

If such a space S met π , then a hyperplane H S , π would contain V 2 5 , a contradiction, therefore S π = .

Assume a cubic curve C is reducible. Of course, the unique possibility is that C consists of at most 3 generatrix lines. In such a case C would meet the conic C 2 and then the plane π , a contradiction with S π = .

Each irreducible curve of order 3 lying in V 2 5 , meets each generatrix lines (as they partition V 2 5 ) that is, it is a directrix curve.

Corollary 3.3 All the directrix cubic curves are obtaining by cutting V 2 5 with the hyperplanes through any two generatrix lines. The maximal hyperplane section of V 2 5 consists 4 q + 1 points.

Proof. An irreducible cubic curve C V 2 5 is a rational normal curve that is, it lies in a 3-space S (cf. Proposition 3.2). If C V 2 5 is a cubic curve, for any two generatrix lines g , g G is uniquely defined the hyperplane H = g , g , S .

Let H be a hyperplane. If H V 2 5 were an irreducible curve of order 5, then | H V 2 5 | = q + 1 . If H V 2 5 = { g 1 , g 2 , C 3 } , g 1 , g 2 G , then | H V 2 5 | = 3 q + 1 . If H V 2 5 = { g 1 , g 2 , g 3 , C 2 } , g 1 , g 2 , g 3 G , then | H V 2 5 | = 4 q + 1 .

Proposition 3.4 1) No two directrix cubic curves belong to a same 3-space.

2) Two directrix cubic curves meet in at most one point.

Proof. 1) Assume two directrix cubic curves C , C V 2 5 belong to a same 3-space S. Then any hyperplane H S meets V 2 5 in a curve of order at least 6, a contradiction.

2) Let C and C be two cubic curves with C S , C S , where S , S are 3-spaces, S S from (1). Assume the curves have at least 2 points in common, P , Q C C , P Q . Then S S P Q so that the hyperplane H = S , S meets V 2 5 in a curve of order 6, a contradiction.

3.2. Bundles of Cubics on V 2 5 and a Non-Affine Subplane

Denote F = G F ( q ) , Σ = P G ( 6, q ) . Let Σ = P G ( 5, q ) be a hyperplane of Σ, S a regular spread of Σ .

Choose a coordinate system ( x , y , t ) = ( x 1 , x 2 , x 3 , y 1 , y 2 , y 3 , t ) in Σ so that t = 0 represents Σ , ( x , y ) are internal coordinates for Σ and for a point P Σ \ Σ , P ( x , y , t ) = ( x 1 , x 2 , x 3 , y 1 , y 2 , y 3 , t ) = F * ( x , y , t ) , F * = F \ { 0 } .

The spread S can be represented as follows

S = { π = ( 0, y ) | y F 3 } { π m = ( x , x m ) | x , m F 3 }

where y = x m is the multiplication in the field F 3 . Such a multiplication can be represented also by y = x M with M a 3 × 3 matrix over F so that x M = x m . The set M = { M | x M = x m } is a field isomorphic to ( F 3 ) 2 , strictly transitive over F 3 .

Denote R = { π , π k | k F } the regulus of S represented by the scalar matrices k I .

From now on we choose q 1 ( mod 3 ) , so that in F = G F ( q ) q 1 3 elements are cubes, while the remaining ones are non cubes.

Let C be an irreducible conic of π and C 0 a skew cubic curve in the 3-space S 0 of Σ \ Σ through π 0 so that C 0 π 0 = (cf. [7] , p. 234, Corollary 4, N 5 ):

C = { ( 0,0,0,1, λ , λ 2 ,0 ) | λ G F ( q ) } { O = ( 0,0,0,0,0,1,0 ) }

C 0 = { ( 1 , λ , λ 2 , 0 , 0 , 0 , s λ 3 ) | λ G F ( q ) } { O = ( 0 , 0 , 0 , 0 , 0 , 0 , 1 ) } ,

where s G F ( q ) is a non cube.

The two curves are referred through a projectivity Λ : C 0 C represented by having inserted the same parameter λ for which it is agreed that the points are considered corresponding to each other, plus Λ ( O ) = O .

Denote V the ruled variety V 2 5 defined by C and C 0 . The curves C and C 0 are directrix curves of V , the set G of the lines connecting corresponding points are the generatrix lines of V .

Let us consider the affinity φ of Σ represented by the following 6 × 6 matrix in 3 × 3 blocks

M φ = ( I k I 0 I )

so that the extended projectivity φ ¯ is represented by the 7 × 7 matrix M φ ¯ obtained from M φ by adding the vector ( 0,0,0,0,0,0,1 ) as the 7th column and the 7th row.

Theorem 3.5 1) For every point P O O \ { O } there exists a bundle C P of q cubic curves on the variety V = V 2 5 having the point P in common, each curve lying in one 3-space intersecting a plane of R \ π . Each bundle cover the q 2 points of V \ O O .

2) Such cubic curves are q 2 .

Proof. 1) For each point A = ( 0 , a , 0 ) = ( 0 , 0 , 0 , a 1 , a 2 , a 3 , 0 ) π it is φ ¯ ( A ) : = ( A ) M φ ¯ = A , that is, π is pointwise fixed. For a point B = ( b , 0 , 0 ) = ( b 1 , b 2 , b 3 , 0 , 0 , 0 , 0 ) π 0 it is φ ¯ ( B ) : = ( B ) M φ ¯ = ( b , k b , 0 ) , that is, φ ¯ ( π 0 ) = π k , and φ ¯ ( O ) = O . Hence φ ¯ ( S 0 ) is a 3-space S k through O with S k Σ = π k . The cubic C 0 S 0 is mapped onto a cubic C k S k with O C k and C k π k = . Therefore there exists a bundle C0 of q cubic curves through O collecting the q 2 points of V \ O O .

Let P = ( 0,0,0,0,0, h ,1 ) be a point of O O \ { O , O } and denote τ h the associated translation. Therefore τ h ( O ) = P and τ h ( C 0 ) = C P .

2) The cardinality of { C P | P O O \ { O } } is q 2 as for each point P O O \ { O } it is | C P | = q and the points of O O \ { O } are q.

Note that, chosen π and S 0 , the variety V 2 5 selections in the spread S the regulus R to which π and π 0 belong.

Denote Π the projective plane P G ( 2, q 3 ) . Represent Π in Σ, Π = ( P , L , I ) as in Lemma 2.6.

Denote V the set of the q 2 affine points of V = V 2 5 .

Let Π q = ( P , L , I ) be the incidence substructure of Π defined as follows:

P = { P V } { π } ,

L = { C C P | P O O \ { O } } G ,

I is defined as follows

I = I restricted to the affine points and lines, π I g for all g G .

Theorem 3.6 Πq is a non-affine subplane of Π of order q.

Proof. It is known from [8] [9] pp. 160-161 and [4] pp. 40-41, that if in an incidence structure the following four properties hold

( 1 3 2 6 2 )

where

1: the number of the points is q 2 + q + 1 ,

2: the number of the lines is q 2 + q + 1 ,

3: each line contains q + 1 points,

62: two lines meet in at most one point,

then the structure is a projective plane of order q.

The affine points of V are the affine points of the q + 1 generatrix lines of G , that is, they are q ( q + 1 ) = q 2 + q to which the point at infinity π has to be added. Hence | P | = q 2 + q + 1 , that is, 1 - holds.

From Theorem 3.5 follows | { C C P | P O O \ { O } } | = q 2 . As | G | = q + 1 then | L | = q 2 + q + 1 , that is, 2 - holds.

Each cubic curve of CP has as many points as C 0 has, that is q + 1 . Each generatrix line g G has q affine points and the point ad infinity π , hence 3 - holds.

From Proposition 3.4, (2) follows that two cubic curves meet in at most one point. Moreover each cubic curve, being a directrix, meets a generatrix line in one point. Two generatrix lines meet only in the point π . Hence 62 holds.

To end proving that Πq is a subplane of Π it needs to verify that Πq is a subgeometry of Π (cf. Definition 2.7). Its set of points is clearly a subset of the points of Π. Moreover, every line g G is contained in a unique 3-space S = g , π which meets no other generatrix (cf. Proposition 3.1, 3)) and every cubic of CP lies in a unique 3-space (cf. Proposition 3.4, 1)) meeting Σ in a plane of R (cf. Theorem 3.5, 1)).

From Theorem 3.6 follows.

Corollary 3.7 Let P, Q be two points of V . If PQ is not a generatrix line then P, Q belong to one directrix cubic, if P V and Q = π then the line PQ of Πq is the generatrix g P .

The properties of Πq of being a plane can be translated into further incidence properties of the affine points of V 2 5 .

Theorem 3.8 Let P, Q be two points of V 2 5 \ C . Then P, Q are joined by one generatrix line or by one directrix cubic C S , where S is a 3-space with S Σ = π k R \ π . Every two directrix cubic curves of V 2 5 \ C meet in one point.

In Theorem 3.5, 1), is shown that the variety V selects a regulus R to which both π , π 0 belong. Denote π : = π 1 , C : = C 1 . R : = R 1 . Fix the directrix cubic curve C 0 S 0 .

Theorem 3.9 There exists a bundle B of varieties V 2 5 with the cubic C 0 as directrix, any to varieties having C 0 in common, | B | = q 2 .

Proof. The construction is done step by step, by choosing at each step a plane of the spread S out the regulus identified by the variety of the previous step, and a directrix conic in it.

Step 1 - Construct the variety V 1 = V 2 5 starting from the conic C 1 and the cubic C 0 S 0 . In S \ R 1 are q 3 q possible choices for the next step.

Step 2 - Choose a plane π 2 S \ R 1 . Fix a conic C 2 in it and construct the variety V 2 = V 2 5 starting from the conic C 2 and the cubic C 0 S 0 . Let R 2 be the regulus of S to which π 2 and π 0 belong. In S \ { R 1 , R 2 } are q 3 2 q possible choices for the next step.

Step 3 - Choose a plane π 3 S \ { R 1 , R 2 } . Fix a conic C 3 in it and construct the variety V 3 = V 2 5 starting from the conic C 3 and C 0 S 0 . Let R 3 be the regulus of S to which π 3 and π 0 belong. In S \ { R 1 , R 2 , R 3 } are q 3 3 q possible choices for the next step. And so on.

The procedure ends evidently at the q2-th step. Therefore B = { V i | i = 1,2, , q 2 } and | B | = q 2 .

Conjecture - A variety V 2 2 r 1 of P G ( 2 r , q ) represents a non-affine subplane of order q of P G ( 2, q r ) via the spatial representation. This is partially addressed in Theorem 11 of the following paper: M. Lavrauw, C. Zanella, Subspaces intersecting each element of a regulus in one point, Andr-Bruck-Bose Representation and Clubs, Electron. J. Combin. 23 (2016), Paper 1.37, pp. 1-11.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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