Arcs, linear sets and hermitian curves in the finite projective plane

Universitat Polit`ecnica de Catalunya Facultat de Matem`atiques i Estad´ıstica Degree in Mathematics Bachelor’s Degree Thesis Arcs, linear sets and hermitian curves in the finite projective plane. Jordi Jofre Senciales Supervised by Simeon Ball September, 2018 Thanks to Simeon Ball. Abstract The main object of the study of this thesis are arcs in PG(2, q2). An arc in PG(2, q2) is set of points with the property that every line intersects with the arc in at most 2 points. One can prove that an arc in PG(2, q2) has at most q+ 2 points and one would like to find arcs which contain a lot of points. An arc is equivalent to a maximum distance separable code. The larger the arc, the longer the code and the greater error-correcting properties the arc will have. In Section 3we study arcs of size q+ 1 that are the set of zeros of a quadratic form. In Section 4we will study an arc constructed as the intersection of two hermitian curves. This arc is of size q−√q+ 1 and is not contained in a conic. In the last section we study the intersection between a linear set and a hermitian curve. Firstly we calculate some examples in GAP. Then we prove some results about hermitian curves that will help us to interpret the computational results. We will prove that the intersection between a linear set over scattered space and non-degenerate hermitian curve is an arc. Keywords Arcs, MDS codes, Hermitian curves, Linear sets, Scattered subspaces 1 Contents 1 Introduction to projective geometry 4 1.1 Projective Space ........................................ 4 1.2 Projective subspaces ...................................... 4 1.3 Projective references ...................................... 5 1.4 Equations in projective spaces ................................. 5 1.5 Dual projective spaces ..................................... 6 2 Finite geometry 8 2.1 Finite Projective Spaces .................................... 8 2.2 Projective planes ........................................ 9 2.3 Maximum distance separable codes and arcs ......................... 10 3 Ovals 13 3.1 Ovals .............................................. 13 3.2 Segre’s theorem ........................................ 14 4 Kestenband 16 4.1 Hermitian curves ........................................ 16 4.2 The arc of Kestenband .................................... 19 5 Baer lines and linear sets 20 5.1 Regulus ............................................ 20 5.2 Baer lines ........................................... 21 5.3 Linear sets ........................................... 22 6 Scatared spaces 24 6.1 Spreads ............................................ 24 6.2 Scattered spaces with respect to Desarguesian spreads ................... 25 7 Trying to construct an arc 29 7.1 Programming in GAP ..................................... 29 7.2 Intersection between a linear set and a hermitian curve ................... 29 7.3 More about hermitian curves ................................. 32 7.4 Interpreting the results .................................... 34 8 Bibliography 41 A Codes of GAP 42 2 A.1 Calculus of function sigma .................................. 42 A.2 Construction of random linear set ............................... 42 A.3 Hermitian curve ........................................ 44 A.4 Secant distribution ...................................... 45 A.5 Intersection between Hand B(U).............................. 46 3 Finite geometry 1. Introduction to projective geometry 1.1 Projective Space We start with a short introduction to projective geometry. Let Pbe a set. Let Kbe a field. Let Ebe a vector space over K. A projective space is the triplet (P,E,π) where π:E−{0} −→ Psuch that: •πis surjective. •π(u) = π(v)⇐⇒ ∃λ∈K∗such that u=λv. The dimension of Pis rank(E)−1. We will denote the elements of Pas p=π(u) = [u] = [u]π. In this thesis we will use dimension for the dimension of a projective space and rank for the dimension of vector space. Definition 1.1. Let Ea vector space over Kof dimension n+1. Let P(E) = {subspaces of Eof dimension 1}.π:E−{0} −→ P(E) defined by π(u) =<u>. Then (P(E), E,π) is a projective space of dimension n. 1.2 Projective subspaces Definition 1.2. A projective subspace is a subset Lof Psuch that L=π(F− {0}) where F is a vector subspace of E. We will denote L= [F]. Lemma 1.3. Let L = [F]projective subspace of P. Then we have: 1. π−1(L) = F−{0}. 2. F =π−1(L)∪{0}. 3. Moreover, if we have p ∈L and u ∈E such that p =π(u)then u ∈F . Proof. We know that L=π(F−{0}), taking π−1in each part we have π−1(L) = π−1(π(F−{0})) = F− {0}. To prove the last equality, it’s clear that F− {0} ⊂ π−1(π(F− {0})). To prove the other inclusion: v∈π−1(π(F−{0})) ⇒π(v)∈π(F−{0})⇒ ∃u∈F−{0}s.t.π(u) = π(v)⇒u=λv,λ∈K∗⇒v∈F−{0} 2 is an immediate consequence of 1. For 3 we have p∈Land u∈Esuch that p=π(u) using this π(u)∈L⇒u∈π−1(L) = F−{0} ⇒ u∈ F. Corollary 1.4. Let L1= [F1]and L2= [F2]be projectives subspaces then we have: 1. L1⊂L2⇔F1⊂F2 2. L1⊂L2⇒dim(L1)≤dim(L2) 4 3. L1⊂L2and dim(L1)≤dim(L2)then L1=L2 We can define the following operations on projective subspaces L1= [F1] and [L2]=[F2], the intersection and the sum: •Sum: The sum of two projective subspaces is the smallest projective subspace which contain them. We can check easily the formula L1+L2= [F1+F2]. •Intersection: The intersection of two projective subspaces is a projective subspace. We can prove it by the following equality: L1∪L2= [F1∪F2]. Definition 1.5. Let A={Pi}i∈Ibe a family of points of a projective plane Pwhere Pi= [vi]. We will say the points of Aare independent if {vi}i∈Iare linearly independent. 1.3 Projective references Let Ebe a vector space over the field K. Let P= (P,E,π) a projective space of dimension n. We want to find a bijection with good characteristics between Pand P(Kn+1). For this reason we introduce projective references. Definition 1.6. Let R={p0, ..., pn;U}be a family of points of Pwhere any subfamily of size n+ 1 is independent. We will say that Ris a projective reference. Definition 1.7. Let e0, ..., enbe a basis of E. I will say this basis is adapted to the projective reference R if we have the following relation: pi= [ei] and U= [e0+ ... + en]. There always exists an adapted basis over R. We can assume that pi= [vi] and {vi}iform a basis over E because {p0, ..., pn}is a independent family. If we have that U= [u] we can write uas a linear combination of {vi}:u=λ0v0+ ... + λnvn. If we take ei=λivithen {ei}is an adapted basis over R. Reciprocally, if we take {e0, ..., en}a basis of Ethen it is adapted to the reference R={[e0], ..., [en]; [e0+ ... + en]}. Definition 1.8. Let Rbe a projective reference. Let Bbe an adapted basis over R. Let p= [v]∈P. If the coordinates of vwith respect to the basis Bare (v0, ..., vn), then we will say that the coordinates of p with respect to the reference Ris the object [(v0, ..., vn)] ∈P(Kn+1). 1.4 Equations in projective spaces Let P= (P,E,π) be a projective space. Let Rbe a projective reference and {e0, ..., en}be an adapted basis. Let Vedenote the coordinates in this basis of a vector v. Let La projective subspace of Pwhere L= [v0, ..., vd]. Let p= [x]∈Pthen we have p∈L⇔x∈< v0, ..., vn>⇔Xe∈<V0,e, ..., Vn,e>. The last expression helps us to find an equation of the projective subspace in the reference R. Now we are going to introduce the conics. Let S2(E) = {bilinear forms of E}. We will define quadrics using the objects of P(S2(E)) in the following way. (I will refer to these objects as conics when Phave dimension 2). We define Q= [φ], a quadric, as follows. The set of points of Qis {p∈P|p= [x], φ(x,x) = 0}. Note that the condition p∈Qdoes not depend on the representative of p. Let us deduce the equation of a quadric. Let Q= [φ] be a quadric and M=Me(φ) be the matrix of φwith the respect to the adapted base {e0, ...en}. Then we define the matrix of Qon the reference Ras 5 Finite geometry the hyperplane. Suppose that αr= 0. Then [(0, ..., 0, 1)] is in the hyperplane but the equation α0+α1t+ ... + αr−1tr−1 has at most r−1 solutions. The next proposition is about the size of arcs, a very important issue to study. Proposition 2.20. Let A be an arc of PG(r,q)then |A| ≤ q+r. Proof. Take a subset Sof Awith r−1 points. Sis a subset of PG(r,q) which spans a projective subspace Lof dimension r−2. If not, we can add points of Aso that we have r+ 1 points in the same hyperplane and we will have a contradiction because Ais an arc. We can count the hyperplanes through Lusing proposition 2.4 with d=n−2, k=n−1. It give us that there are q+ 1 hyperplanes through L. Each hyperplane that contains Scontains at most one other point of A−S, and each point of A−Sis contained in a hyperplane which contains S. Thus we have |A| ≤ r−1 + q+ 1 = r+q. 12 3. Ovals In this section we will study the ovals of a projective plane seen as an incidence structure. And then we will see that ovals in PG(2, q), qodd, are given by a quadratic equation, this is the main theorem of this section, Segre’s theorem. 3.1 Ovals Definition 3.1. Let Γ = (P,L) be a projective plane of order q. We will say that a subset A⊂Pis an oval if it has size q+ 1 and every line of Γ is incident at most to two points of A. We can take a point Pof an oval. This point is incident with q+ 1 lines, qof them are incident with another point of the oval. The last line is only incident with P. We will call it the tangent to P. An oval has q+ 1 tangents, one for each point. In PG(2, q) it is equivalent the definition of an oval or an arc of size q+ 1. In PG(2, q) we have the example of the conics. Recall that a conic is a subset of PG(2, q) defined by a quadratic equation. For each non-degenerate conic we can find a projective reference so that the equation is x2=yz. C={[(x,y,z)]|x2=yz}={[(0, 1, 0)]}∪{[(t,t2, 1)]|t∈Fq} It is easy to see that Chas q+ 1 points. Since Cis defined by a quadratic equation, each line of PG(2, q) is incident with Cat most 2 times. I will give an other example in PG(2, q): Proposition 3.2. The set A={[(1, t,t2i)]|t∈Fq}∪{[(0, 0, 1)]} is an oval in PG(2, q), q= 2hif mcd(i,h)=1. Proof. A has q+ 1 points. We only have to check that every line in PG(2, 2h) is incident at most two points of A: There are qlines incident to [(0, 0, 1)] of the form y=ax. Each line is incident to [(1, a,a2i)]. It is clear that the last line incident to [(0, 0, 1)] is x= 0 and it is the tangent of this point. We have counted all lines incident to this point. Now I am going to study the lines of the form z=ax +by. Fix a line L: z=ax +by. Firstly note that exponents in the field Fqare defined in Z/Zq−1. I will see that if we have two distinct points incident to Lof the form [(1, u,u2i)] and [(1, v,v2i)] then we have u−v=bmwhere m= (2i−1)−1. It means that the difference of the parameter of two points of Aincident with Lonly depends on L. This tells us that there are at most two points in Athat are incident to L. Suppose that [(1, u,u2i)] and [(1, v,v2i)] are in L. Then we have u2i=a+bu and v2i=a+bv. Since the field has characteristic 2 we can write (u−v)2i=u2i−v2i=b(u−v) and u−v6= 0 because the points are different. The equation is (u−v)2i−1=b. By the hypothesis mcd(i,h)=1⇒mcd(2i−1, q−1) = 1 and using Bezout we get that ∃m,n∈Zsuch that m(2i−1) + n(q−1) = 1 that in Z/Zq−1is m(2i−1) = 1. Then using this: bm= (u−v)(2i−1)m=u−v. We are going to prove a very beautiful result which says if we have an oval then each point of the plane is incident to zero or two tangents. 13 Finite geometry Proposition 3.3. Let Obe an oval in a projective plane of order q, q odd. Then every point of the projective plane not in Ois incident with zero or two tangents of O. Proof. Let xibe the number of points not in Osuch that are incident with itangents of O. Fix a point p/∈ O. Suppose that pis incident with an odd number of tangents. Then the number of points of Ois odd, since the other lines incident with pare incident with an even number of points of O. However, q+ 1 is even, so pis incident with an even number of tangents, so xi= 0 if iis odd. We are going to count the pairs (p,l) where lis a tangent of Oand p∈l−O in two different ways: Take a tangent lhave qpoints not in O. There are q+ 1 tangents in O, consequently the number of pairs is q(q+ 1). If we calculate joining the points on the number of tangents we get that the number of pairs is Pi≥2ixi. So the equation is Pi≥2ixi=q(q+ 1). Now, we are going to count the triples (p,l,m) such that land mare different tangents of Oand p∈l∩m in two different ways: Note that if p∈l∩mthen p/∈ O because there is only one tangent incident with each point of the oval. Taking p/∈ O such that is in itangents then we can choose i(i−1) triples. Then we have the number of triples is Pi≥2(i−1)xi. Fix ltangent of O, then all the other qtangents intersect lonce. Since there are q+ 1 tangents the number of triples is q(q+ 1). So we can get that Pi≥2i(i−1)xi=q(q+ 1). Combining the two last equations we have: X i≥2 i(i−2)xi= 0 All terms of the sum are non-negative, so xi= 0 if i6= 0, 2. 3.2 Segre’s theorem Now we are going to prove Segre’s theorem, which was proven by Beniamino Segre in 1955 [5]. The proof given here is adapted from [2]. It is the main theorem of this section because it says that when qis odd, in PG(2, q), an oval and a conic are the same thing. Theorem 3.4. An oval in PG(2, q), q odd, is a conic. Proof. We are going to prove it building a quadratic equation that contains all points of the oval. Then the set defined by it will be the oval because both have q+ 1 points. Let [x], [y], [z] be three different points of O. We take the basis x,y,zof V(3, q). It is a basis because Ois an oval. The tangents to the points [x], [y], [z] are α21X2+α31X3= 0, α12X1+α32X3= 0, α13X1+α23X2= 0 respectively. Let [s]∈ O − {[x], [y], [z]}where s= (s1,s2,s3) on the basis {x,y,z}. Then we can consider the line joining [s] and [z]: s2X1−s1X2= 0. Because the property of the ovals all lines are different, so the following set has all no null elements of Fq: s2 s1|[s]∈ O−{[x], [y], [z]}∪−α13 α23  Using it we have: −α13 α32 Y [s]∈O−{[x],[y],[z]} s2 s1 =−1 (1) 14 I define the following linear maps on basis {x,y,z}:Tx(X) = α21X2+α31X3,Ty(X) = α12X1+α32X3, Tz(X) = α13X1+α23X2. It is easy to see that Tz(x) = α31 and Tz(y) = α32. Using (1) this implies Tz(x)Qs2=Tz(y)Qs1. Similarly we can obtain Tx(y)Qs3=Tx(z)Qs2and Ty(z)Qs1=Ty(x)Qs3. Combining this we get: Tx(y)Ty(z)Tz(x) = Tx(z)Ty(x)Tz(y) (2) Let [v], [u], [w] three different points of O −{[x]}. By interpolation, we can verify the following equation evaluating in X=vand X=ubecause in both sides are polynomials of degree 1: Tx(X) = Tx(u)det(X,u,x) det(u,v,x)+Tx(v)det(X,u,x) det(v,u,x) Evaluating this in X=wwe get: Tx(w) det(u,v,x) + Tx(v) det(w,u,x) + Tx(u) det(v,w,x) = 0 (3) Changing the rolls of x,u,v,wwe can obtain the following three equations: Tu(w) det(x,v,u) + Tu(v) det(w,x,u) + Tu(x) det(v,w,u) = 0 Tv(w) det(x,u,v) + Tv(u) det(w,x,v) + Tv(x) det(u,w,v)=0 Tw(u) det(x,v,w) + Tw(v) det(u,x,w) + Tw(x) det(v,u,w) = 0 Using (2) we get the following relation: Tw(x) Tx(w)=Tw(u)Tu(x) Tx(u)Tu(w)=Tw(v)Tv(x) Tx(v)Tv(w) Multiplying it by (3) then we have the following equation: Tw(x) det(u,v,x) + Tv(x)Tw(v) Tv(w)det(w,v,x) + Tu(x)Tw(u) Tu(w)det(v,w,x) = 0 Using the three equations following (3) we can change Tw(x), Tv(x) and Tu(x) from the last one and we obtain: det(u,v,x)(Tw(u) det(x,v,w) + Tw(v) det(u,x,w)) +Tw(v) Tv(w)det(w,u,x)(Tv(w) det(x,u,v) + Tv(u) det(w,x,v)) −Tw(v) Tu(w)det(v,w,x)(Tu(w) det(x,v,u) + Tu(v) det(w,x,u)) = 0 and rearranging the last coefficient using (2): 2Tw(u) det(u,v,x) det(x,v,w)+2Tw(v) det(u,v,x) det(u,x,w)+2Tv(u)Tw(v) Tv(w)det(w,u,x) det(w,x,v)=0 Now with the basis {u,v,w}we have that an arbitrary point [x]∈ O satisfies the equation: 2Tw(u)x3x1+ 2Tw(v)x3x2+ 2Tv(u)Tw(v) Tv(w)x2x1= 0 We have to check that this equation is a non degenerate quadratic form. All the coefficients are different from 0 because the characteristic of the field isn’t 2 and there are no other point of the oval in a tangent. There we use the hypothesis that qis odd. 15 Finite geometry 4. Kestenband In this section we are going to see the construction of an arc of size q−√q+ 1 in PG(2, q) that it is not contained in a conic. This arc was found by Kestenband [3]. In this section firstly we will study the hermitian curves which we will use in the construction. 4.1 Hermitian curves Definition 4.1. Let βbe a map in V=V(n,q) such that β:V×V−→ Fqand let σbe automorphism of Fqsuch that σ2= 1 and σ6= 1. We will say that βis an hermitian form if it have the following properties: 1. β(u+w,v) = β(u,v) + β(w,v) 2. β(u,v+w) = β(u,v) + β(u,w) 3. β(au,bv) = abσβ(u,v) 4. β(u,v) = β(v,u)σ Note that the existence of σimplies that qmust be a square. Then because of Galois theory σis unique and xσ=x√qand the fixed field of σis F√q. For this reason we can say that ∀u∈Fq,β(u,u)∈F√q. We will say βis degenerate if there exists w6= 0 such that β(w,u)=0∀u∈V(n,q). A vector uis isotropic to β, a non-degenerate hermitian form, if β(u,u) = 0 and a pair {u,v}is said hyperbolic if β(u,v) = 1 and u,vare isotropic. A subspace Uis anisotropic if β(u,u)6= 0, ∀u∈U−0. We prove some lemmas that will help us to determinate the equation of an hermitian curve. These lemmas can be found in [1]. Lemma 4.2. Suppose that L is a subspace of rank 2 of V (n,q)that contains an isotropic vector to β, a hermitian form. Then βrestricted to L is degenerate or has an hyperbolic pair {u,v}such that L=<v,u>. Proof. If βrestricted to Lis non-degenerate then there exists w∈V(n,q) such that β(u,w)6= 0. If β(w,w) = 0 we have finished the proof by scaling wappropriately. Suppose that β(w,w)6= 0. Take d such that dσ6=−d. There exists such a dsince the change of sign is not a morphism if the characteristic is different of 2 and if it is 2, it is the identity. We can define c:= dσβ(w,w) d+dσ. Then c+cσ=β(w,w). Define a:= β(w,w) and take v=−a−1−σcu +a−σw. Lemma 4.3. Let βbe a non-degenerate hermitian form of V (n,q). Let W be a maximal totally isotropic subspace and suppose that it has rank r. Then exists a basis {ei|i= 1, ..., r}∪{fi|i= 1, ..., r}of a subspace X ⊂V such that W =<e1, ..., er> and V =X⊕U. Moreover (ei,fi)is an hyperbolic pair. Proof. If we have r= 0 then we have finished. Suppose that r>0, then exists e1∈Wthat is isotropic. Since βis non-degenerate, there exists v∈V such that β(e1,v)6= 0 and βrestricted to <e1,v>is non-degenerate. By Lemma 4.2 we have that there exists a hyperbolic pair {e1,f1}such that V=<e1,f1>⊕V1where V1=<e1,f1>⊥. Define W1=W∩V1. We can calculate the rank of W1by Grassman: rank(W1) = rank(W∩V1) = rank(W) + rank(V1)−rank(W⊕V) = r+ (n−2) −(n−1) = r−1 16 It is not difficult to check that rank(W⊕V1) = n−1. If βrestricted to V1is degenerate then there exists u∈V1−{0}such that β(u,w)=0∀w∈V1. Since β(u,v)=0∀v∈<e1,f1>then βwill be degenerate in V. This implies that βis non-degenerate in V1. Then applying the same to the spaces V1and W1, which is totally isotropic, we are finished by induction. The next lemma tell us that if the rank of subspace is more than 2 then the maximal isotropic space is non-trivial. This is important because it allows us to apply the last lemma to find a basis such that the form has an easy expression. Lemma 4.4. Let βbe a non-degenerate hermitian form of V (n,q) = V . If V has at least rank 2, V will have an isotropic vector. Proof. Suppose that vis not an isotropic vector and define b=β(v,v)∈F√q. Take u∈<v>⊥and consider β(u+av,u+av) = β(u,u) + aσ+1β(v,v). Since −β(u,u) b∈F√qand the map f:Fq−→ F√q such that f(x) = xσ+1 is surjective. Using it we can take a∈Fqso that aσ+1 =−β(u,u) b. Then the vector u+av is isotropic. We only have to check that fis surjective. Firstly we can see that |f−1(a)| ≤ √q+ 1 because this set is defined by polynomial of degree √q+ 1. Suppose that exists a∈F√qsuch that f−1(a) = ∅. Then we can get the following contradiction: q=|Fq|=|∪c∈F√q−{a}f−1(c)|=X c∈F√q−{a}|f−1(c)| ≤ (√q−1)(√q+ 1) = q−1 Let βbe a non-degenerate hermitian curve. By Lemma 4.4 we know that an anisotropic space has rank 1 or 0. Suppose that Vhas rank n. By the lemma 4.3, we can find a basis such that β(u,v) = u1vσ 2+u2vσ 1+...+un−1vσ n+unvσ n−1if nis even and β(u,v) = u1vσ 2+u2vσ 1+...+un−2vσ n−1+un−1vσ n−2+unvσ n if nis odd. Let H={[x]∈PG(n,q)|β(x,x)=0}where βis a hermitian form in PG(n,q). We will call it a hermitian surface. Now I am interested to know how many points it has. We will say that His nondegenerate if βis non-degenerate. Proposition 4.5. Let H be a non-degenerate hermitian curve of PG(2, q). Let L be a line. If H ∩L has at least two points then it has √q+ 1 points. Proof. The two points, that the intersection has, are [x], [y]. Let hbe the hermitian form associate to the curve. All the points of the line are {[x]}∪[x+λy]}λ∈Fq. We can define a map such that φ:Fq−→ F√q where φ(λ) = h(x+λy,x+λy) = λh(x,y) + λ√qh(x,y). It is clear that all points of the intersection are [y], which are associated with one λ∈φ−1(0). Suppose that h(x,y)6= 0. For each a∈F√q,φ−1(a) has at most √qelements because it is defined by a polynomial of degree √q. Now we are going to see that it is exactly √q: q=|Fq|=|∪a∈F√qφ−1(a)|=X a∈F√q |φ−1(a)| ≤ X a∈F√q √q=√q√q=q This equation implies that all must be equal. Then we have that |φ−1(0)|=√q. We have seen that the intersection has √q+ 1 points. 17 Finite geometry Suppose that h(x,y) = 0. This implies that all points of the line are in H. Since Lhas dimension 1, this contradicts Lemma 4.3. Now, by counting, we can calculate the number of points of a hermitian curve in PG(2, q). To prove it we only need an extra lemma, which says that for every point of an hermitian curve has only one tangent (line that only intersects in one point to the curve). We will prove this later. Corollary 4.6. A non-degenerate hermitian curve in PG(2, q)has q√q+ 1 points. Proof. Let Hbe the curve. We are going to prove it by double counting. We count pairs (p,l), where p∈Hand lis a line of PG(2, q) for which l3p, in two ways. Firstly fix the point. Each point is incident with q+ 1 lines. The number of pairs (p,l) is |H|(q+ 1). Finally fix the line. Let lbe a line of PG(2, q). This line intersects Hin 1 or √q+ 1 points. In PG(2, q) there are q2+q+ 1 lines. There are |H|tangents, which only intersect Hin one point, and there are q2+q+ 1 − |H|lines which intersect Hin √q+ 1 points. This implies that the number of pairs is (q2+q+ 1 −|H|)(√q+ 1) + |H|. Combining these two counts, we get that |H|=q√q+ 1. Lemma 4.7. Let H be a non-degenerate hermitian curve of PG(2, q)represented by h. Let L = [W] =< u,v>be a line such that u,v are isotropic vectors of h. Then h|Wis non-degenerate. Proof. Suppose that h|Wis degenerate. This implies that exists a vector w∈Wsuch that h(w,w0) = 0 for all w0∈W. Because of Whas rank 2 we can suppose that W=<u,w>and expressing h|Win this basis we have that is the 0 form. This implies that H∩Lhas q+1 points, contradicting Proposition 4.5. Lemma 4.8. Let H be a non-degenerate hermitian curve in PG(2, q)defined by the form h. For each p∈H then ∃!tangent of H through p. Proof. Let p= [x] and [y]/∈H. The points of the line defined by [x], [y] are {[y]}∪{[x+λy]}λ∈Fq. Suppose that h(x,y)6= 0. We can define a map φ(λ) = h(x+λy,x+λy) and proceeding similarly proposition 4.5 we obtain that this line has √q+ 1 points in H. Thus, it is not a tangent. Suppose that h(x,y) = 0. We have the equation 0 = h(x+λy,x+λy) = h(x,x)+λh(x,y)+λ√qh(x,y)+ λ√q+1h(y,y). Since [y]/∈Hthe only point of Hon the line joining [x] and [y] is [x]. We have proved that if h(x,y) = 0 and [y]/∈Hthe line defined by [x], [y] is a tangent through [x]. But all the points such that 0 = h(x,y) = h(y,x) = y1xσ 2+y2xσ 1+y3xσ 3are incident with the same line. This line is the unique tangent to Hthrough p. We are using that the curve is non-degenerate, so it will be useful to have a lemma that help us know when a hermitian form is non-degenerate: Lemma 4.9. Let M be a matrix such that Mt=M√q, we will call a matrix hermitian with this property. Then the form defined by β(x,y) = xtMy√qis a non-degenerate hermitian form if and only if det(M)6= 0. Proof. Define f:V(n,q)−→ V(n,q) such that f(x) =    β(x,e1) . . . β(x,en)   where e1, ..., enis the canonical basis of V(n,q). This is a linear map with matrix Min the canonical basis. And we have: det M= 0 ⇔ker(f)6={0}⇔∃v6= 0 such that ∀u∈V(n,q)β(v,u) = 0 ⇔βis degenerate 18 4.2 The arc of Kestenband The following theorem concerns the construction of an arc of size q−√q+ 1 that it is not contained in a conic: Theorem 4.10. Let q >9be a square. Let I be the identity matrix and let H be a hermitian matrix. For any hermitian matrix M we define the following set: V(M) = {[x]∈PG(2, q)|xtMxq= 0} If the characteristic polynomial of H is irreducible over Fq, then the set S =V(I)∩V(H)is an arc of PG(2, q)that is not contained in a conic. Proof. Since the characteristic polynomial of His irreducible over Fq, we have that det(H−λI)6= 0 for every λ∈Fq. Then the set V(H+λI) will be a non degenerate hermitian curve. Take the curves Vλ=V(H+λI) for all λ∈F√qand V∞=V(I). If [x] is at least in two curves Va, then making linear combinations of the equations of these curves we find that [x] is in all curves Va. Suppose that [x]/∈V(I) then we have xtHx√q=a∈F√qand xtIx√q=−bwhere b∈(F√q)∗. Multiplying the second equation by a band adding to the first we found that [x]∈V(H+a bI). Using this we know that every point is in all curves Vaor only in one. Now we use double counting to calculate the size of the set. We will count the pairs (p,V) where p∈PG(2, q) and Vis one of previous curves such that p∈V. Fixing Vwe have that the number of pairs is P|V|= # hermitian curves V# points in hermitian curve = (√q+ 1)(q√q+ 1). Now fixing p, if p∈Sthen pis in all curves Vand if p/∈Sthen pis in exactly one curve V. So the number of pairs (p,V) is |S|(√q+ 1) + q2+q+ 1 −|S|. Using the two ways of counting we get that |S|=q−√q+ 1. Now we will see that Sis an arc. Let lbe a line incident with r≥2 points of S. Then lintersects each curve Vin √q+ 1 points, √q+ 1 −rof which are not in S. The number of points that are in lbut not in Sis q+ 1 −r. Since each point of l−Sis incident with exactly one curve V, q+ 1 −r= (√q+ 1)(√q+ 1 −r) Hence, r= 2. This proves that Sis an arc. We want to see that Sis not a subset of a conic. We will use Bezout’s theorem. Firstly we have to check that a conic and a hermitian curve don’t share any component. We can assume that the conic is y2=xz and we take a general non-degenerate hermitian curve h(x,y,z). We know that the conic is irreducible, this implies that if they share a component then the conic divides the hermitian curve. This says us that the polynomial his 0 in Fq[X,Y,Z] <Y2−XZ>. Using that a basis of this space is the residues of {YXiZj,XiZj}i,j≥0 we get that the hermitian form is 0. A contradiction since it is non-degenerate. Using Bezout’s theorem we get that Shave at most 2√q+ 2 points of the conic but q−√q+ 1 >2√q+ 2 for q>9 . 19 Finite geometry 5. Baer lines and linear sets 5.1 Regulus Definition 5.1. Let Σ = PG(3, q). A regulus is a set Rof q+ 1 disjoint lines of Σ such that every line that intersects at least with 3 lines of Rintersects all q+ 1 lines of R. Lemma 5.2. For every three disjoint lines l1, l2, l3of PG(n,q)there are at most q + 1 lines such that intersects l1, l2and l3. Equivalently #{m line of PG(n,q)|m∩li6=∅for i = 1, 2, 3} ≤ q+ 1. Proof. We have to check for every p∈l3there is at most one line mthrough psuch that m∩l26=∅and m∩l36=∅because l3has q+ 1 points. Assume that ∃m1,m2different lines with the previous hypothesis. We will call pi=m1∩liand qi=m2∩lifor i= 1, 2. The previous points are all different because l1and l2are disjoint and m1and m2are different lines. Let πbe the plane defined by m1and m2. This implies that p1,p2,q1,q2∈π. Because of l1=p1∧q1and l2=p2∧q2we know that l1,l2∈π. Every two lines in a projective plane have non-empty intersection. Hence l1∩l26=∅. We get a contradiction, since l1and l2are disjoint. This proposition will be very useful: Proposition 5.3. Given 3disjoint lines of Σ = PG(3, q)then ∃!regulus such that contain these lines. Proof. We will call these three lines l0,l1,l∞. We have to check that we can take a basis such that these lines have the following expression: l0=<(0, 0, 0, 1), (0, 0, 1, 0) >,l1=<(1, 0, 0, 1), (0, 1, 1, 0) >,l∞=<(1, 0, 0, 0), (0, 1, 0, 0) > Suppose that l0=<v1,v2>,l1=<u1,u2>,l∞=<w1,w2>. Because of l0and l∞are disjoint {v1,v2,w1,w2}is a basis. Hence we can write u1,u2as a linear combination of these four vectors: u1=λ1v1+λ2v2+λ3w1+λ4w2 u2=µ1v1+µ2v2+µ3w1+µ4w2 Claim: The set {n1=λ1v1+λ2v2,n2=λ3w1+λ4w4,n3=µ1v1+µ2v2,n4=µ3w1+µ4w2}is a basis. We have to check that they are linearly independent: γ1n1+γ2n2+γ3n3+γ4n4= 0 Substituting the expressions of niand using that v1,v2,w1,w2is basis we get the following linear system where the variables are γi: λ1γ1+µ1γ3= 0 λ2γ1+µ2γ3= 0 λ3γ2+µ3γ4= 0 λ4γ2+µ4γ4= 0 To find γi= 0 we need that det(A) = (µ1λ2−µ2λ1)(λ3µ4−λ4µ3)6= 0. Firstly we will check that µ1λ2−µ2λ16= 0. Suppose µ1λ2−µ2λ1= 0. Then we have µ1λ2=µ2λ1. Using this we get µ2u1−λ2u2∈ 20 l1∩l∞. A contradiction because they are disjoint. A similar argument verifies that λ3µ4−µ3λ4= 0. Taking {n2,n4,n3,n1}as a basis we get the expressions of l0,l1,l∞that we want. We define the following lines: ma:= [<(a, 0, 0, 1), (0, a, 1, 0) >], m∞= [<(0, 0, 0, 1), (1, 0, 0, 0) >], la= [<(a, 0, 0, 1), (0, a, 1, 0) >] for all a∈Fq. It is easy to check that lines of the sets {la}a∈Fq∪{l∞}and {ma}a∈Fq∪{m∞}are disjoint. Now we check that the set R={la}a∈Fq∪{l∞}is a regulus. We know that the lines are disjoint. Rhas the property that each line lafor all a∈Fq∪ {∞} intersects with all mb, we only have to calculate a determinant for each mband la. Suppose that there is a line Ssuch that intersects with three lines of R. Using the last lemma we know that there are at most q+ 1 lines that intersects these three, and we know that this q+ 1 lines are the mb. This implies that Sis one of mb. This implies Sintersects all la. We have proved existence. Now we will see that Ris the unique regulus such that contains l0,l1,l∞. Suppose that R0is a regulus that contains l0,l1,l∞. Because of m0,m1,m∞intersect with this three lines of R0then they have to intersect with all lines of R0. But for the lemma 5.2 there are at most q+ 1 lines that intersects with m0,m1,m∞ and we know that this q+ 1 lines are the lines of R. This implies R=R0. Corollary 5.4. The q + 1 transversal lines of a regulus R form another regulus. We will denote it by opp(R). Proof. Applying the same argument of the existence in the set {mb}b∈Fq∪ {m∞}we get that it is a regulus. 5.2 Baer lines In this subsection we will need to talk about field extensions to be able to define what is a Baer line. We will use the definition of regulus to define a Baer line and we will use some properties of them to study Baer lines. Lemma 5.5. Let L/K be a finite extension field of degree m. Let V a vector space over L of rank n. Then V is a vector space over K of rank nm. Proof. It is easy to check the axioms of a vector space over Kusing the axioms of Vas a vector space over L. To check the multiplicity of the rank: let {e1, ..., en}be a basis of V as vector space over Land let {l1, ..., lm}be a basis of the extension L/K. It is easy to check that {ljei}i,jis a basis of Vas a vector space over K. Let q≥2 a power of a prime. Consider the following map: ψ: subspaces of PG(2, q2)π1 ←→ subspaces of V(3, q2)i −→ subspaces of V(6, q)π2 ←→ subspaces of PG(5, q) where π1,π2are correspondences between the projective subspaces and linear subspaces and iis the identification of the lemma 5.5 between subspaces of V(3, q2) and V(6, q). Note that if Whas dimension n−1 then ψ(W) has dimension 2n−1. When we will want to calculate this map, it will depend on the basis we will chose for the isomorphism between V(6, q) and V(3, q2) seen as a vector field over Fq. This fact is not important for the following definitions. 21 Finite geometry w,u∈Q∩<T,wm+1 >. This tells us that Tis contained in a space of rank m+ 2 if wm+1 /∈ ∪Q∈S|Q∩T6={0}<Q,T>. Firstly we have to count the number of non zero vectors of ∪Q∈S|Q∩T6={0}<Q,T>. Take Q∈Ssuch that Q∩T6= {0}. This implies that Q∩Thas rank 1 because Tis scattered. The rank of <Q,T>is t+musing Grassmann. This implies that there are qt+m−qmvectors in <Q,T>−T. Since Tis scattered, there exists Q∈Ssuch that Q∩H6={0}for each point of T. This implies that there are Θm(q) = qm+ ... + 1 Qof this type. Then in ∪Q∈S|Q∩T6={0}<Q,T>there are (qm+t−qm)(qm+ ... + q+ 1) + qm+1 −1 non null vectors. But we need that this number will be smaller than the number of non null vectors in V(rt,q) that it is qrt −1. We need that: qrt >(qm+t−qm)(qm+ ... + q+ 1) + qm+1 Supposing m<rt−t 2suffices. The following proposition proves the existence of maximum scattered spaces with respect to Desarguesian 1-spreads of PG(2r−1, q). This is all we need because in this thesis we work always with a line spread. Proposition 6.9. Let S be a Desarguesian 1-spread of PG(2r−1, q). Then the dimension of maximum scattered space with respect to S will be r −1. Proof. This is an immediate consequence of Lemma 6.8 and Proposition 6.7. 28 7. Trying to construct an arc 7.1 Programming in GAP In this section we will see how I have found examples of linear sets with the help of the computer to try construct an arc. I use GAP to program the calculations. The programs are included in the annex. Firstly, for the calculations we choose ∈Fq2such that Fq2=Fq(). Since we have an extension of degree 2, 2=a+bfor some a,b∈Fq. We have to calculate ψ:PG(2, q2)−→ lines of PG(5, q). Let [v] be a point of PG(2, q2): [v]−→ {λv|λ∈Fq2} −→ {(λ1+λ2)v|λ1,λ2∈Fq} −→ [<v,v>] where [<v,v>] is a line of PG(5, q). Take a basis v1,v2,v3of V(3, q2) then we can construct a basis of V(6, q) taking v1,v2,v3,v1,v2,v3. Let v=xv1+yv2+zv3where x,y,z∈Fq2and x=x1+x2,y= y1+y2,z=z1+z2where x1,x2,y1,y2,z1,z2∈Fq. Now we have to found the expression of v,von the basis of V(6, q): v=xv1+yv2+zv3= (x1+x2)v1+(y1+y2)v2+(z1+z2)v3=x1v1+y1v2+z1v3+x2v1+y2v2+z2v3 v=xv1+yv2+zv3= (x1+2x2)v1+ (y1+2y2)v2+ (z1+2z2)v3= = (x1+ (a+b)x2)v1+ (y1+ (a+b)y2)v2+ (z1+ (a+b)z2)v3 =x2bv1+y2bv2+z2bv3+ (x1+x2a)v1+ (y1+y2a)v2+ (z1+z2a)v3 Choosing vi=eiwe have that ψsends [(x,y,z)] to [<(x1,y1,z1,x2,y2,z2), (x2b,y2b,z2b,x1+x2a,y1+ y2a,z1+z2a)>]. Now the problem is given a x∈Fq2found efficiently x1,x2∈Fqsuch that x=x1+x2. To solve it at the start of the code I build an ordered vector of tuples where each tuple has the form (x,x1,x2). It is easy to calculate taking each x1,x2∈Fqand using that each x∈Fq2has a unique expression of this type. Since the vector is ordered it is efficient to ask for any x∈Fq2. To calculate a random linear set of rank n, firstly we choose nrandom vectors of V(6, q) and then for each point p∈PG(2, q2) asks if u1, ..., un,ψ(p) has maximum rank. Since we choose a random linear set we can fix the hermitian curve xyq+yxq+zq+1 = 0 of PG(2, q2). Then we will be interested to know the secant distribution of the intersection of these sets. The secant distribution of a set Sis the sequence of numbers {Ti}i≥0where Tiis the number of lines which intersect Sin ipoints. If we have Ti= 0 for all i≥3 then the set will be an arc. 7.2 Intersection between a linear set and a hermitian curve When we intersect a linear set and a hermitian curve with gap we get the following results. Let Sbe the intersection: For q= 3 we found 4 distinct tangent distributions: Table 1. For q= 7 we found 4 distinct tangent distributions: Table 2. Proposition 7.1. Let U be a projective subspace of PG(5, q). Let L be a line of PG(2, q2). Let H be a hermitian curve of PG(2, q2). If dim(U∩ψ(L)) ≥2and L is not tangent to H then L ∩H∩B(U)is a Baer line. 29 Finite geometry |S|T0T1T2T4 13 18 30 36 7 4 54 36 0 1 10 27 34 27 3 16 15 16 48 12 Table 1: Tangent distribution S=H∩B(U), q= 3 |S|T0T1T2T8 57 882 378 1176 15 50 1050 338 1057 6 8 2058 392 0 1 64 763 320 1344 24 Table 2: Tangent distribution S=H∩B(U), q= 7 Proof. Firstly we will see that L⊂B(U). Take a point p∈Lthen ψ(p)⊂ψ(L). We know that ψ(L) has dimension 3 and ψ(L)∩Uhas dimension 2. Using this and Green’s formula we get that dim(ψ(p)∩U)≥0 which implies that ψ(p)∩U6=∅. By the definition of B(U) we have p∈B(U). We have L⊂B(U)⇒L∩B(U)∩H=L∩Hand it is clear that it is a Baer line since there are at least two points. Then next proposition gives a structure to the linear sets of B(U) when Uhas dimension 3. Proposition 7.2. Let U be a projective subspace of PG(5, q)of dimension 3. Then we have two cases for B(U): •There exists only one P ∈PG(2, q2)such that ψ(P)⊂U then the set of lines contained in B(U)is a bear line in PG(2, q2)∗. •There exists at least two P1,P2∈PG(2, q2)such that ψ(P1), ψ(P2)⊂U. Let L be the line defined by P1,P2. Then we have B(U) = L. Proof. There always exists a P∈PG(2, q2) such that ψ(P)⊂Usince Proposition 6.7 implies that U can’t be scattered with respect to D3,2,qsince Uhas rank 4. If it has two points P1,P2such that ψ(P1), ψ(P2)⊂Uthen ψ(P1) + ψ(P2)⊆U, which implies that ψ(L)⊆Uand they have the same dimension. This implies that ψ(L) = Uand since distinct lines have disjoint images we get L=B(U). Suppose that there exists only one point P= [u]∈PG(2, q2) such that ψ(P)⊂U. Then we can express U=<u,u,v1,v2>such that {u,v1,v2}is a basis of V(3, q2). We have to check this, let λi∈Fq2: λ0u+λ1v1+λ2v2= 0 ⇒λ01u+λ02u+λ11v1+λ12v1+λ21v2+λ22v2= 0 ⇒ λ01u+λ02u+λ11v1+λ21v2=−λ12v1−λ22v2 30 Where λij ∈Fq. If we have λ01u+λ02u+λ11v1+λ21v2=−λ12v1−λ22v2= 0 this implies that λij = 0 because {v,v,v1,v2}and {v1,v2}are linearly independent in V(6, q). And we have that {u,v1,v2}are linearly independent in V(3, q2). Suppose that λ01u+λ02u+λ11v1+λ21v2=−λ12v1−λ22v26= 0. Let Q= [−λ12v1−λ22v2] be a point of PG(2, q2). Then we have ψ(Q)⊂Usince −λ12v1−λ22v2∈Uand (−λ12v1−λ22v2) = −λ12v1− λ22v2=λ01u+λ02u+λ11v1+λ21v2∈U. This implies that Q=Psince we supposed that Pis the unique point such that ψ(P)⊂U. This implies the following equality: λ12v1+λ22v2=µu=µ1u+µ2u. All coefficients are in Fqand these vectors are linearly independent in V(6, q). This implies that all coefficients have to be 0. This is a contradiction because it is a representation of a point. We define the following set of lines: A={L∈PG(2, q2)∗|P∈Land dim(ψ(L)∩U)=2} We want to prove that ∪L∈AL=B(U). Suppose that Lis a line of PG(2, q2) such that contains P. Since P∈Lwe know that ψ(L)∩Uhas at least dimension 1. Suppose that ψ(L)∩Uhas dimension 3. Then ψ(L)⊂Uand this is the case one. We have proved that ψ(L)∩Uhas dimension 1 or 2. If we have dim(ψ(L)∩U) = 2 then Lwill be a subset of B(U). If dim(ψ(L)∩U) = 1 then ψ(L)∩U=ψ(P) and this implies that L∩B(U) = {P}. Firstly we will prove the following inclusion: ∪L∈AL⊂B(U). It is clear because for each L∈ A has the property that ψ(L)∩Uhas dimension 2 and this implies that L⊂B(U). Now we have to prove the other inclusion. Let Q∈B(U). If Q=Pthen it is clear that Q∈ ∪L∈A. Suppose that Q6=P. Then exists a line Lsuch that P,Q∈L. This implies that ψ(L)∩Uhas dimension 2. Now we have that L∈ A. We have proved the equality. Now we want to prove that the set Ais a bear line in PG(2, q2)∗. Firstly we will see the following equality: A={La}a∈Fq∪ {L∞}where La=<u,v1+av2>and L∞=<u,v2>. We know that U=<u,u,v1,v2>where {u,v1,v2}is a basis of V(3, q2). Let Lbe a line of the set Athen L=<u,s> and ψ(L) =<u,u,s,s>. To impose that ψ(L)∩Uwill have 3 we have to impose that ψ(L) + Uhas at most rank 5. This is equivalent to the existence of a non-trivial linear combination with coefficients in V(6, q) of the vectors s,s,u,u,v1,v2: λ1s+λ2s+λ3u+λ4u+λ5v1+λ6v2= 0 ⇒ (λ1+λ2)s=λ1s+λ2s=−λ3u−λ4u−λ5v1−λ6v2 Since u,u,s,sare linearly independent in V(6, q) we have that (λ5,λ6)6= (0, 0). Using the last equality we get: L=<u,s>=<u, (λ1+λ2)s>=<u,−λ3u−λ4u−λ5v1−λ6v2>= =<u,λ5v1+λ6v2> It is clear that Lis one of the Lawith a∈Fqor L∞. This proves one inclusion and the other is trivial. Now we only have to see that the set {La}a∈Fq∪{L∞}is a bear line in PG(2, q2)∗. Take {w0,w1,w2}the basis of V(3, q2)∗associated with {u,v1,v2}. We will take the basis w1,−w2of P∗the line of PG(2, q2)∗ associated to P. Then La= [aw1−w2] = [(a, 1)] and L∞= [w1] = [(1, 0)], which implies that this set is a bear line. 31 Finite geometry Figure 1: Example of the linear set of first case for q= 3 7.3 More about hermitian curves The objective of this section is prove that in PG(2, q2) the set of lines Lsuch that Lis incident with fix P and is tangent to a fix hermitian curve then this set of lines is a bear line on the dual projective space. Lemma 7.3. Let H a non-degenerate hermitian surface of PG(n,q2)represented by h. Let PG(n,q2)∗ the dual projective space of PG(n,q2). Then for each [w]∈PG(n,q2)∗exists a unique [x]∈PG(n,q2) such that [w]=[h(., x)]. Proof. Existence: Let v1, ..., vn+1 a basis of V(n+ 1, q2) so that the hermitian form is h(x,y) = x1yq 2+ x2yq 1+ ... + xnyn+1 +xn+1yq n(suppose that n is odd, if n is pair is the same procedure). Take w1, ...wn+1 the dual basis associate to the basis v1, ..., vn+1. Let w=λ1w1+ ... + λn+1wn+1 ∈V(n+ 1, q2)∗we have to found a vector x=x1v1+ ... + xn+1vn+1 ∈V(n+ 1, q2) such that for all vector y=y1v1+ ... + yn+1 ∈ V(n+ 1, q2) we have w(y) = y1λ1+ ... + yn+1λn+1 =y1xq 2+ ... + yn+1xq n. Because of an automorfism on a field is bijective we can take x1, ..., xnso that the equality will be true for each y∈V(n,q2). It is clear that exists a unique [x]∈PG(n,q2) because in the equation we can take yi= 0 for i6=jand yj= 1. We can see that there are there are a correspondence between V(n,q2) and V(n,q2)∗sending xto h(., x). The following proposition will help us to identify the hyper-planes such that are tangent to an hermitian curve. The idea is the same as in dimension 2. Proposition 7.4. Let W ∈PG(n,q2)∗and let H be a non-degenerate hermitian curve of PG(n,q2). Then W is tangent to H if and only if W = [h(., y)] for some [y]∈H. Proof. Firstly we see that if [y]∈Hthen we have that [h(., y)] is tangent to H. Suppose that exists a point [x]∈Hsuch that h(x,y) = 0. Then this implies that all points of L=<x,y>will be in H. It means that His degenerate, which is a contradiction. 32 Finally we see that if [h(., y)] is tangent to Hthen we have that [y]∈H. If we see that there is only one hyperplane tangent to Hthrough [y] we will have finished. Suppose that there is a hyperplane Mtangent to [y] that is not [h(., y)]. Then we have that there exists [x]∈M−Hsuch that h(x,y)6= 0. Doing a similar proof for dimension 2 we know that the line L=<x,y>has q+ 1 points in common with H. This implies that Mis not a tangent because it contains L, a contradiction. We have proved that [h(., y)] is the unique hyperplane tangent to Hthrough [y]. The following theorem is the main result of this subsection since the objective of this subsection will be a corollary of this. But firstly we observe that we can extend easily the definition of hermitian curve to the dual space and we have the same properties since V(n,q2)∗≡V(n,q2). Theorem 7.5. Let H be a non-degenerate hermitian curve of PG(n,q2). Define: H∗={W∈PG(n,q2)∗|W is tangent to H} is a subset of the dual space. Then H∗is a non-degenerate hermitian curve of PG(n,q2)∗. Proof. Define β:V(n+ 1, q2)∗−→ Fq2on the following way: β(h(., x), h(., y)) = h(y,x) It is clear that βis a non-degenerate hermitian form on the dual because his a non-degenerate hermitian form. Let Bthe hermitian curve associate to bon the dual projective space. Because of the definition we have: [h(., x)] ∈B⇔[x]∈H⇔[h(., x)] ∈H∗ The following corollary is an immediately consequence of the theorem. Corollary 7.6. Let H be a non-degenerate hermitian curve of PG(2, q2). Take a point p ∈PG(2, q2). Then if the set of lines B={L∈PG(2, q2)|p∈L and L is tangent to H} has at least two lines then it is a bear line in PG(2, q2)∗. Proof. We know that the set of lines through a point in the dual space is a line. Using Theorem 7.5 we know that the set of lines tangents to His a non-degenerate hermitian curve. Bis the intersection between these last two sets. If it has at least two lines it will be a bear line. The following proposition will help us to understand why we don’t get an arc. Proposition 7.7. Let H be a non-degenerate hermitian curve represented by h. Let A be a set of points of H. We define the following set of PG(2, q2)∗: B={L∈PG(2, q2)∗|L is tangent to H through a point in A} B is the set of tangents of H through a point in A. Then we have that A is a bear line in PG(2, q2)if and only if B is a bear line in PG(2, q2)∗. 33 Finite geometry Proof. Using the above propositions it is clear that we can express Bin the following form: B={[h(., v)] ∈PG(2, q2)∗|[v]∈A} Firstly suppose that Ais a bear line. This means that exists v1,v2∈V(3, q2) such that: A={[av1+v2]|a∈Fq}∪{[v1]} Take the elements w1=h(., v1) and w2=h(., v2) of V(3, q2)∗. We know that the elements of Bare h(., v) for each [v] in A. h(., v) = h(., av1+v2) = ah(., v1) + h(., v2) = aw1+w2 We have used that aq=abecause a∈Fq. This says us that B={[(a, 1)]}a∈Fq∪{[(1, 0)]}in the basis w1,w2. It is clear that Bis a bear line. Now suppose that Bis bear line in PG(2, q2)∗. This implies that exists w1,w2∈V(3, q2)∗such that: B={[aw1+w2]|a∈Fq}∪{[w1]} Using Proposition 7.4 we know that wi=h(., vi) for some vi∈V(3, q2)∗ aw1+w2=ah(., v1) + h(., v2) = h(., av1+v2). This implies that the points of Aare [av1+v2] for a∈Fqand v1, since the lines of Bare tangent to these points. This implies that Ais a bear line. Another way to prove the last proposition is using the corollary 7.6 Let {pi}i=0,...,qwhere pi= [vi] a subset of Hsuch that is a bear line. This is equivalent to the existence of a line L= [h(., v)] ∈PG(2, q2)∗ such that L∩H={pi}i=0,...,q. This is equivalent to have that h(v,vi) = h(vi,v) = 0 for all i= 0, ..., q because 0q= 0. This is the same that the tangents of Hthrough piare concurrent in [v]. Using the corollary 7.6 we have that the tangents of Hthrough piare concurrent to [v] if and only if this tangents are a bear line in PG(2, q2)∗. 7.4 Interpreting the results In this section we will use the above sections in order to understand the results obtained by computer. The first issue is to know why we didn’t get an arc. Proposition 7.8. Let U be a subspace of PG(5, q)of dimension 3. Let H be a non-degenerate hermitian curve of PG(2, q2). Then B(U)∩H isn’t an arc. Proof. We know that there are two possible cases for B(U) using Proposition 7.2. Suppose that B(U) is a line. Then B(U)∩His a bear line or a point. This implies that it isn’t an arc. Suppose that B(U) is a bear line in PG(2, q2)∗. Suppose that point Pis the point where all q+ 1 lines of B(U) are concurrent. If P∈Hthen at least qlines of B(U) are not tangent to H, since for each point of Hthere exists a unique tangent. This implies that B(U)∩Hhas at least qbear lines. Now suppose P/∈Hthen we have that the lines that are tangents to Hwhich are concurrent to the same point not in His a bear line in PG(2, q2)∗. This implies that we have 0, 1, 2, q+ 1 lines of B(U) tangent to H. If we have a no tangent line then B(U)∩Hwill be an arc. Suppose that all q+ 1 lines of B(U) are tangent to Hin points p0, ..., pq. This implies that B(U)∩H={p0, ..., pq}. Because of lines contained in B(U) is a bear line in PG(2, q2)∗and his lines are tangent to pi, using Proposition 7.7 the points piform a bear line. This implies that this case isn’t an arc. 34 Now we will see the different cases that we can get with the computer for B(U)∩Hwhere Uis a subspace of PG(5, q) of dimension 3 and His a non-degenerate hermitian curve of PG(2, q2). To be able to better understand the results it is clear that all lines which intersects in q+ 1 points to B(U)∩Hare bear lines because H∩Lis at most a bear line for any line L. There are two possible cases for B(U). If B(U) is a line then B(U)∩His a bear line or a point. Suppose that B(U) = ∪i=0,...,qLiis a bear line in PG(2, q2) where P∈PG(2, q2) is the point that the q+ 1 lines of B(U) are concurrent. Suppose that P∈Hwe have two cases: exists Lisuch that is tangent to Hthrough Por all Liare not tangents to H. If exists Litangent to Hthen we have that Lj∩His a bear line for j6=i. This implies that B(U)∩Hhas at least qbear lines. There are not more bear lines: suppose that exists a bear line that no provide by Lj∩H. This implies that exists a line L6=Ljfor all jsuch that L∩B(U)∩His a bear line. Lhas to intersect for each jto Lj∩Hin one different point. But Li∩H={P}and P∈Lj∩H for all j. We get a contradiction because Lhas to intersect to Lj∩Hto unique different point for each j. This implies that this case has qbear lines. It is easy to see that this case B(U)∩Hhas q2+ 1 points (Figure 4). Suppose that does not exists any Ljsuch that is tangent to H. Then we have at least q+ 1 bear lines which are Lj∩Hfor each j. This case we have q2+q+ 1 points (Figure 5). Now suppose that P/∈H.B(U) and the set of lines which are concurrent to Psuch that are tangent to H are bear lines in PG(2, q2)∗. For this reason we have that the intersection in PG(2, q2)∗of this two sets has 0, 1, 2, q+ 1 lines. This implies that B(U) has 0, 1, 2, q+ 1 lines Ljtangents to H. Suppose that we have q+ 1 tangents. Following the proof of the proposition 7.8 we know that this case B(U)∩His a bear line (Figure 2). Suppose that we have 2 lines Li,Lktangent to H. This case has q−1 lines Ljsuch that Lj∩H is a bear line. We have to see that this bear lines are all the bear lines of B(U)∩H. Suppose that exists a line L6=Ljfor all jsuch that L∩B(u)∩His a bear line. Then we have that Lhas to intersect in only one different point to each Lj∩H. We will call this points pj. We know that the set {pj}j=0,...,qis a bear line. The tangent to Hthrough pjwe will call Sj. Then the set of PG(2, q2)∗formed by Sjis a bear line. We know that Si=Li,Sk=Lkthis implies that the point where all line Sjare concurrent is P. This implies that for each Sjshare the points Pand pjwith Lj. This implies that each Sj=Lj. This is a contradiction because we have that some Ljare not tangents. In this case we have q−1 bear lines. It is easy to count the points. It has (q−1)(q+ 1) + 2 = q2+ 1 points. Suppose that we have only Lisuch that are tangent to H. This implies that we have qlines Ljsuch that Lj∩His a bear line. This case has at least qbear lines. We can count q2+q+1 points. The last case is when all Ljare not tangent to H. We have the q+1 bear lines Lj∩H. This implies that at least there are q+1 bear lines. This case has (q+1)2points (Figure 3). The following proposition should help us to count the tangent distribution of H∩B(U). In the following, Uwill be a subspace of dimension 3 of PG(5, q) such that there exists only one P∈PG(2, q2) such that ψ(P)⊂U. Proposition 7.9. Define a map in the following way: ϕ:B(U)−{P} −→ U−ψ(P)such that ϕ(Q) = ψ(Q)∩U. Then we have that ϕdefines a bijection between B(U)−{P}and U −ψ(P). Proof. Firstly we have to check that ϕis well defined. Let Q∈B(U) we have that ψ(Q)∩U6=∅. This implies that ψ(Q)∩Uhas dimension 0 or 1. But the only point Q∈B(U) such that ψ(Q)∩Uhas dimension 1 is P. This implies that for all Q∈B(U)−{P}we have that ψ(Q)∩Uhas dimension 0. This implies that it is a point of U−ψ(P) because the images of ψare disjoint. Now we will see that ϕis surjective. Let Sa point of U−ψ(P). We know that ψ(PG(2, q2)) is a partition of PG(5, q). This implies that exists a point Q∈PG(2, q2) such that {S} ⊂ ψ(Q). We know that S∈U 35 Finite geometry Figure 2: The union of black lines is B(U). The point Pis red. The blue points are points of the hermitian curve. The purple line is which contains the bear line. This is a representation for q= 3 Figure 3: The union of the black lines is B(U). The point Pis red. The blue points are points of the hermitian curve. This is a representation for q= 3 this implies that ψ(Q)∩U6=∅. This implies that Q∈B(U) and Q6=Pbecause S/∈ψ(P). For this reason we have that ϕ(Q) = S 36 Figure 4: The union of the four lines is B(U). The purple line is the tangent line to H. The blue points are in H. This is a representation for q= 3 Figure 5: The union of the four lines is B(U). The blue points are in H. This is a representation for q= 3 Now we have to check that ϕis injective. It is clear because the images of ψare disjoint. Now the objective is define a map between PG(5, q)×PG(5, q) to Fq=Fq∪ {∞}. This map will help us to identify which lines intersect B(U)∩Hin a bear line. Firstly we observe that for each hermitian form h:V(3, q2)×V(3, q2)−→ Fq2can be seen in the following way: h:V(6, q)×V(6, q)−→ Fq2, since V(3, q2) and V(6, q) are isomorphic as vector spaces. Moreover, his bilinear over V(6, q). Let p,q be two arbitrary points of PG(5, q) represented by v,u∈V(6, q). Definition 7.10. We define the map w:PG(5, q)×PG(5, q)−→ Fq. We know that h(v,u) = c+d where c,d∈Fq. If d= 0 then w(p,q) = ∞else w(p,q) = c d. We have to check that wis well defined. Suppose that p= [v1]=[v2], q= [u1]=[u2]∈PG(2, q2). 37 Finite geometry AddSet(B,p); fi; return B; end; Firstly we choose a random vector subspace Uof rank 4. Then for all points of PG(2, q2) we check if ψ(P) + Uhas rank at most 5. A.3 Hermitian curve HermitianCurve:=function(q) local S,H, s,i,j,x1,x2; s:=RootInt(q); H:=[[0,1,0],[1,0,0],[0,0,1]]; H:= One(GF(q))*H; S:=[]; for x1 in GF(q) do for x2 in GF(q) do if [Z(q)^0,x1,x2]*H*[Z(q)^0,x1^s,x2^s]=0*Z(q) then AddSet(S,[Z(q)^0,x1,x2]); fi; od; od; for x1 in GF(q) do if [Z(q)*0,Z(q)^0,x1]*H*[Z(q)*0,Z(q)^0,x1^s]=0*Z(q) then AddSet(S,[Z(q)*0,Z(q)^0,x1]); fi; od; if [Z(q)*0,Z(q)*0,Z(q)^0]*H*[Z(q)*0,Z(q)*0,Z(q)^0]=0*Z(q) then AddSet(S,[Z(q)*0,Z(q)*0,Z(q)^0]); fi; return S; end; This function return the set of points of the hermitian curve xyq+yxq+zq+1 = 0. This function calculate it for each point if PG(2, q2) is in H. 44 A.4 Secant distribution SecantDistribution:=function(S,q) local x1,x2,j,tau,a,sec; tau:=[]; for j in [1..(q+1)] do tau[j]:=0; od; for x1 in GF(q) do for x2 in GF(q) do sec:=0; for a in [1..Size(S)] do if (x1*S[a][1]+x2*S[a][2]+Z(q)^0*S[a][3])=0*Z(q) then sec:=sec+1; fi; od; if sec<>0 then tau[sec]:=tau[sec]+1; fi; od; od; for x1 in GF(q) do sec:=0; for a in [1..Size(S)] do if x1*S[a][1]+Z(q)^0*S[a][2]=0*Z(q) then sec:=sec+1; fi; od; if sec<>0 then tau[sec]:=tau[sec]+1; fi; od; sec:=0; for a in [1..Size(S)] do if S[a][1]=0*Z(q) then sec:=sec+1; fi; od; if sec<>0 then tau[sec]:=tau[sec]+1; fi; return(tau); end; 45 Finite geometry This function returns a vector with the secant distribution of the set S. It calculate the secant distribution counting the points that each line intersects with S. A.5 Intersection between Hand B(U) q = 11^2; s = RootInt(q); SET:= Set(GF(q));; #taula per escriure els numeros en funci´o del del cos petit. Coords:=[];; for i in GF(s) do for j in GF(s) do AddSet(Coords,[i*Z(q)+j,i,j]);; od; od; U:=HermitianCurve(q);; for b in [1..100] do T:=Burbuja(q); S:=Intersection(T,U);; tau:=SecantDistribution(S,q);; Print(Size(S)," ",tau," ","\n"); od; This is the main code. It calculates the intersection between a Hand B(U). It calculate the table coords that is vector of vectors (x,x1,x2). It is useful to calculate the factorization very fast. 46