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On maximal cliques of polar graphs
Finite Fields and Their Applications 47 (2017) 276–285
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Finite Fields and Their Applications www.elsevier.com/locate/ffa
On maximal cliques of polar graphs ✩ Antonio Cossidente a , Giuseppe Marino b , Francesco Pavese c,∗ a
Dipartimento di Matematica Informatica ed Economia, Università della Basilicata, Contrada Macchia Romana, I-85100 Potenza, Italy b Dipartimento di Matematica e Fisica, Università degli Studi della Campania “Luigi Vanvitelli”, Viale Lincoln, 5, I-81100 Caserta, Italy c Dipartimento di Meccanica, Matematica e Management, Politecnico di Bari, Via Orabona, 4, I-70125 Bari, Italy
a r t i c l e
i n f o
Article history: Received 2 September 2015 Received in revised form 15 June 2017 Accepted 23 June 2017 Available online xxxx Communicated by L. Storme
a b s t r a c t The maximal cliques of the graph NU(4, q 2 ) related to the Hermitian surface of PG(3, q 2 ) and of the graph NO± (2n + 2, q), q even, n ≥ 1, are classified. © 2017 Elsevier Inc. All rights reserved.
MSC: 05E30 05C99 Keywords: Strongly regular graph Maximal clique Hermitian surface Hyperbolic quadric Elliptic quadric
✩ The research was supported by Ministry for Education, University and Research of Italy MIUR (Project PRIN 2012 “Geometrie di Galois e strutture di incidenza”) and by the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA-INdAM). * Corresponding author. E-mail addresses: [email protected] (A. Cossidente), [email protected] (G. Marino), [email protected] (F. Pavese).
http://dx.doi.org/10.1016/j.ffa.2017.06.014 1071-5797/© 2017 Elsevier Inc. All rights reserved.
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1. Introduction Let Q± (2n + 1, q) be a non-degenerate quadric of PG(2n + 1, q), n ≥ 1. Let NO± (2n + 2, q) be the graph whose vertices are the points of PG(2n + 1, q) \ Q± (2n + 1, q) and two vertices P1 , P2 are adjacent if the line joining P1 and P2 contains exactly one point of Q± (2n + 1, q) (i.e., it is a line tangent to Q± (2n + 1, q)). The graph NO± (2n + 2, q) is a strongly regular graph if and only if q = 2 [1, Table 9.9, p. 145]. Let H(n, q 2 ) be a non-degenerate Hermitian variety of PG(n, q 2 ), n ≥ 2. Let NU(n + 1, q 2 ) be the graph whose vertices are the points of PG(n, q 2 ) \ H(n, q 2 ) and two vertices P1 , P2 are adjacent if the line joining P1 and P2 contains exactly one point of H(n, q 2 ) (i.e., it is a line tangent to H(n, q 2 )). The graph NU(n + 1, q 2 ) is a strongly regular graph for any q [1, Table 9.9, p. 145]. The maximal cliques of the graphs NU(3, q 2 ) were classified in [2, Corollary 3]. In particular the authors showed that there are exactly two types of cliques of size q 2 and q + 2, respectively. In the first case, it corresponds to the points on a tangent line with the tangent point excluded. In the second case, it corresponds to q + 1 points forming a Baer subline of a tangent line , and the other point not on . In this paper we will classify the maximal cliques of the graph NU(4, q 2 ) related to the Hermitian surface of PG(3, q 2 ) and of the graph NO± (2n + 2, q), q even, n ≥ 1 (cf. Theorem 3.1). In particular we will prove the following classification theorem. Theorem 1.1. Let C be a maximal clique of NU(4, q 2 ), then either 1. |C| = q 2 , and C consists of the points on a tangent line without the tangent point itself; or C spans the whole space and one of the following cases occur: 2. |C| = q + 4, q > 2, and C consists of a Baer subline and a triangle; 3. |C| = 2q + 2, q ≡ 1 (mod 3) and C consists of two Baer sublines on two skew lines; 4. |C| = 2q + 3, q ≡ 1 (mod 3) and C consists of two Baer sublines on two skew lines and a further point not on the lines; 5. C is an arc of PG(3, q 2 ). 2. The maximal cliques of NU(4, q 2 ) The graph NU(4, q 2 ) is a strongly regular graph with parameters v = q 3 (q 2 +1)(q −1), k = (q 2 − 1)(q 3 + 1), λ = q 4 + q 2 − 2 and μ = q(q 2 − 1)(q + 1). We first recall that an O’Nan configuration consists of six points in a projective plane, namely the four vertices of a non-degenerate quadrangle together with two of their three diagonal points. In [14] it has been proved that no O’Nan configuration lies on a Hermitian curve. By using the polarity of the Hermitian curve, we have the following remark.
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Remark 2.1. Let H(2, q 2 ) be a Hermitian curve in PG(2, q 2 ). There do not exist four points not on H(2, q 2 ) with the property that no three of them are collinear and such that the line joining any two of them is tangent to H(2, q 2 ). Let H(3, q 2 ) be the Hermitian surface of PG(3, q 2 ) with equation X1q+1 + X2q+1 − X3q+1 + X4q+1 = 0. Let G be the subgroup of PΓL(4, q 2 ) fixing H(3, q 2 ). Definition 2.2. Let Z be a point set of PG(3, q 2 ) \ H(3, q 2 ). We say that a point R of PG(3, q 2 ) \ H(3, q 2 ) is compatible with Z, if each of the lines joining R with a point of Z is tangent to H(3, q 2 ). We need the following result. Proposition 2.3. Let C be a maximal clique of NU(4, q 2 ) which is neither an arc of PG(3, q 2 ) (i.e., there are no four points of C in a plane) nor contained in a line. Then C is contained in the union of a plane and a line. Proof. Assume that C is not an arc and that it is not contained in a line. Then we can find a set A of four non-collinear non-isotropic points in a plane π, such that the line joining any two of them is tangent to H(3, q 2 ). Note that in this case π is a plane secant to H(3, q 3 ). Indeed, since in a tangent plane to H(3, q 2 ) all tangent lines pass through the tangent point, if there were at least two non-isotropic points in a plane tangent to H(3, q 2 ) such that the line joining any two of them is tangent to H(3, q 2 ), then they must lie on a tangent line and so they must be collinear. Without loss of generality, we may assume that π is the plane having equation X4 = 0 [10, Theorem 24.4.2]. From Remark 2.1, we have that three points of A are necessarily collinear, i.e., they lie on a tangent line, say t. Then, the further point of A, say U1 , does not lie on t. The stabilizer of π in the group G, say Gπ , is transitive on the set of non-isotropic points of π [10, Theorem 24.4.2]. Therefore, we may assume that U1 is the point (1, 0, 0, 0). From [7,12], the stabilizer of U1 in the group Gπ , say M , induces on π a subgroup of order (q + 1)(q 3 − q) that is the direct product of a homology group M1 of order q + 1 and of a group M2 of order q 3 − q isomorphic to PGL(2, q). The group M stabilizes the line : X1 = X4 = 0. Every element in M1 fixes all points on and all lines through the point U1 , while M2 acts in its natural representation on the q + 1 isotropic points of . Since M2 permutes in a single orbit the non-isotropic points of and M1 permutes in a single orbit the q + 1 isotropic points on secant lines through U1 , we have that the group M acts transitively on the q 3 − q isotropic points of π \ . It follows that M is transitive on the set of q 3 − q tangent lines of π not passing through U1 . Then we may choose t as the line tangent to H(3, q 2 ) at the point P = (1, 0, 1, 0) X4 = 0 . By intersecting t with the q + 1 tangent lines of π passing having equations X1 = X3
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through U1 , we obtain a set, say A , consisting of q + 1 non-isotropic points of π forming a Baer subline. In particular, the three points of A distinct from U1 , say P1 , P2 , P3 , are contained in A . The stabilizer of t in M induces on π a homology group, say Mt , of order q + 1, having as axis the line X2 = X4 = 0 and center the ⎧ point ⎫ ⎛ U2 = (0, ⎞ 1, 0, 0). Hence, it fixes all ⎪ ⎪ 1 0 0 ⎨ ⎬ ⎜ ⎟ 2 q+1 lines through the point U2 . In particular Mt = ⎝ 0 z 0 ⎠ : z ∈ GF (q ), z =1 ⎪ ⎪ ⎩ 0 0 1 ⎭ permutes in a single orbit the q + 1 points of A . Hence, we may assume that P1 is the point (1, 1, 1, 0). A straightforward calculation shows that the points P2 , P3 are of the form (1, αi , 1, 0), where αiq+1 = 1, αi = 1, i = 1, 2 and α1 = α2 . If R ∈ π \ t is compatible with A, then R lies on a tangent line through U1 . By Remark 2.1, R ∈ t and hence R ∈ A . It turns out that the set of points of π compatible with A coincides with A \ A. Let R = (X1 , X2 , X3 , X4 ) be a point of PG(3, q 2 ) \ H(3, q 2 ) that is compatible with A. From [10, Lemma 23.2.1 ii)], we have that ⎧ q+1 ⎪ − X3q+1 + X4q+1 = 0 X ⎪ ⎪ 2 ⎪ ⎨ X q+1 − 2X q+1 + X X q + X q X − X X q − X q X + X X q + X q X = 0 1 3 1 2 2 3 4 3 1 3 1 2 2 3 ⎪ X4q+1 − 2X3q+1 + X1 X3q + X1q X3 − αi X1 X2q − αiq X1q X2 + +αiq X2 X3q + αi X2q X3 = 0, ⎪ ⎪ ⎪ ⎩ i = 1, 2. By subtracting the third equation from the second for i = 1, 2, we get
(α1 − 1)X1 X2q + (α1q − 1)X1q X2 + (1 − α1q )X2 X3q + (1 − α1 )X2q X3 = 0 (α2 − 1)X1 X2q + (α2q − 1)X1q X2 + (1 − α2q )X2 X3q + (1 − α2 )X2q X3 = 0
(1)
Equations (1) are two distinct Hermitian forms of rank two, say H1 , H2 , and in π both represent a set of lines through P . Assume, by way of contradiction, that H1 = λH2 , for some λ ∈ GF (q 2 )∗ . Then we have that α1 = λα2 +(1 −λ) with λ ∈ GF (q)∗ . In particular λ = 1, otherwise α1 = α2 . Since α1q+1 = 1, it follows that α2 + α2q = 2 and taking into account that α2q+1 = 1, we have α2 = 1, a contradiction. From a geometric point of view, the forms in (1) represent in PG(3, q 2 ) q + 1 planes passing through the line joining P with the point U4 = (0, 0, 0, 1). Also, they cannot share more than two planes, since they are distinct. It can be easily seen that they share exactly the two planes π1 and π2 , with equations X2 = 0 and X1 = X3 , respectively. Note that the plane π2 is tangent to H(3, q 2 ) at the point P . Therefore all the points of π2 that are compatible with A, lie on t and hence must be points of A . It follows that if there exist non-isotropic points compatible with A that are not in π, then they must lie on π1 , and so they are contained in π1 ∪ t. 2
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Let C be a maximal clique of NU(4, q 2 ) which is neither an arc of PG(3, q 2 ) nor contained in a line. With the notations introduced in the previous proof, it turns out that C ⊆ A ∪π1 , with π1 : X2 = 0. In particular, the points of C ∩π1 satisfy the following equations: ⎧ q+1 q+1 ⎪ ⎨ X4 − X3 = 0 (2) X4q+1 − 2X3q+1 + X1 X3q + X1q X3 = 0 ⎪ ⎩X = 0 2 Lemma 2.4. A maximal clique containing A must contain A ∪ {U1 }. Proof. We keep the notation of Proposition 2.3. The points of A ∪ {U1 } are pairwise compatible. Hence, we only need to check that they are compatible with the points of π1. If R is a point of π1 compatible with A, then R ∈ / π ∩ π1 (since π ∩ π1 is a (q + 1)-secant to H(3, q 2 ) containing U1 ). Furthermore the set of points of π1 that are compatible with {R, P1 , P2 , P3 } are compatible with all points of A ∪{U1 } as well. Observe that the plane R, t meets H(3, q 2 ) in a Hermitian curve H(2, q 2 ) since t is a tangent line not through R (and hence the tangent lines are not concurrent). Then, we consider the projection of the isotropic points of the polar line of R (with respect to H(2, q 2 )) onto the line t. Since the points P1 , P2 , P3 define a unique Baer subline we are done. 2 The only point satisfying Equations (2) having X3 = 0 is the point U1 ∈ A. The further points of π1 , satisfying Equations (2) are of the form (a, 0, 1, b), where a, b ∈ GF (q 2 ), with a +aq = 1 and bq+1 = 1. They form a set of q 2 +q points, say B . Set B := B ∪{U1 }. We have the following result. Proposition 2.5. The set B has size q 2 + q + 1 and it is of type (0, 1, 2, q + 1) with respect to lines of π1 . Also the following properties hold true: (a) Each line of π1 that is tangent to H(3, q 2 ) and passing through U1 contains q points of B . (b) There exist q lines of π1 passing through U4 containing q + 1 points of B . Among them there are 0, 1 or 2 lines that are tangent to H(3, q 2 ), according as q ≡ 1, 0 or −1 (mod 3), respectively. Q (c) For each point Q ∈ B there are q−1 tangent lines to H(3, q 2 ) ∩π1 , say mQ 1 , . . . , mq−1 , passing through Q and containing exactly one further point of B . (d) Let Q ∈ B such that the line QU4 is not tangent to H(3, q 2 ). Then, any two distinct points of RQ := {mQ i ∩ B : i = 1, . . . , q − 1} \ {Q} are not collinear with U1 . (e) Let Q ∈ B such that the line QU4 is tangent to H(3, q 2 ). If Q is compatible with two points distinct from U1 on a tangent line r through U1 not containing Q, then Q is compatible with r ∩ B. (f ) Let q ≡ 1 (mod 3). Let Q ∈ B such that Q U4 is not tangent to H(3, q 2 ). Then there are one or two points Q ∈ B such that U1 ∈ / QQ and with the property that
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the lines QU4 , QQ are tangents to H(3, q 2 ) according as q ≡ 0 or −1 (mod 3), respectively. Moreover, if Q is a point fulfilling these conditions then Q is compatible with Q U1 ∩ B. Proof. The set B can be seen as the intersection of a Hermitian curve H contained in the plane π1 with a Baer pencil represented by Equations (2), i.e., a cone whose vertex is a point on H and whose base is a Baer subline. In [5] and [6], it is proved that every such set is of type (0, 1, 2, q + 1) with respect to lines of π1 and that every (q + 1)-secant line is a Baer subline. The number of (q + 1)-secant lines to B is 2q + 1 [5, p. 227]. There are q + 1 different (q + 1)-secants to B passing through U1 and they all happen to be tangent lines to H(3, q 2 ), whereas the remaining q lines pass through U4 and necessarily contain q + 1 points of B . This proves Statement (a) and the first part of Statement (b). Note that each intersection point between a (q + 1)-secant line to B passing through U1 and a (q + 1)-secant line to B passing through U4 , belongs to B . The q lines having q +1 points of B are those joining U4 with a point Qλ = (λ, 0, 1, 0), where λ ∈ GF (q 2 ), λ + λq = 1. These are all (q + 1)-secants to B through U4 . Indeed, each of them contains the q + 1 points (λ, 0, 1, b), bq+1 = 1 of B . Since is the polar line of U4 with respect to the polarity of the Hermitian curve H(3, q 2 ) ∩ π1 , each of these lines is tangent to H(3, q 2 ) if and only if the point Qλ belongs to H(3, q 2 ) if and only if λq+1 = 1. Consider the following system:
X q+1 = 1 X q + X = 1.
(3)
We have three cases, [8, Section 1.5, (xi), (xii), (xiii)]: i) if q ≡ 1 (mod 3), there are no solutions of (3) in GF (q 2 ); ii) if q ≡ 0 (mod 3), the unique solution is −1; iii) if q ≡ −1 (mod 3), there exist two distinct solutions of (3) in GF (q 2 ) \ GF (q). Therefore, among the previously mentioned q lines passing through U4 , there are 0, 1 or 2 lines that are tangent to H(3, q 2 ), according as q ≡ 1, 0 or −1 (mod 3), respectively. This proves the second part of Statement (b). Let C = (1, 0, 0, c) be a point of the line joining U1 with U4 and Q = (a, 0, 1, b) a generic point of B , hence a +aq = 1 and bq+1 = 1. The line joining Q with the point C is tangent to H(3, q 2 ) if and only if the following equation is satisfied [10, Lemma 23.2.1 ii)]: (1 − aq+1 )cq+1 + aq bcq + abq c = 0.
(4)
If aq+1 = 1, there are q + 1 solutions for c, see [11, Lemma 2.44], one being c = 0, and so C = U1 . If aq+1 = 1, there are q solutions for c. In this case, both lines QU1 , QU4 , are tangent to H(3, q 2 ). Let c be a solution of (4), with c = 0 if aq+1 = 1, and
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c = 0, −b/(1 − aq+1 ) if aq+1 = 1. A generic point R of the line QC, distinct from C, has coordinates (a+λ, 0, 1, b+λc). Then R ∈ B if and only if λ = 0 or λ = (bq c−bcq )/cq+1 and R = ((acq+1 +bq c −bcq )/cq+1 , 0, 1, bq c/cq ). Finally, note that λ = 0 otherwise, if aq+1 = 1, then bcq = 0, that is a contradiction, whereas if aq+1 = 1, then c = −b/(1 − aq+1 ) and we get again a contradiction. It follows that R = Q. This proves Statement (c). Let Q be a point of B such that QU4 is not tangent to H(3, q 2 ) (i.e., aq+1 = 1). We show that the q − 1 points of RQ lie on distinct tangent lines passing through U1 . To this aim, we only need to prove that bq c/cq = bq c /c q , for distinct solutions c, c of (4). Assume the contrary holds true. Then cc q = c cq or cc q ∈ GF (q)∗ , which implies that c = dc, for some d ∈ GF (q)∗ . Then, Equation (4) is satisfied if and only if (d − 1)(1 − aq+1 ) = 0, i.e., if and only if d = 1, a contradiction. This proves Statement (d). Let Q be a point of B such that QU4 is tangent to H(3, q 2 ). Assume that Q is compatible with two points distinct from U1 on a tangent line r through U1 not containing Q, say Y and Z. Then, we consider the projection of the isotropic points of the polar line of Q (with respect to H(3, q 2 ) ∩ π1 ) onto the line r. Since the points U1 , Y, Z define a unique Baer subline, we have proved Statement (e). Let Q = (a, 0, 1, b) ∈ B such that Q U4 is not tangent to H(3, q 2 ), i.e., aq+1 = 1. By [10, Lemma 23.2.1 ii)], direct calculations show that a point Q of type (x0 , 0, 1, b/x0 ), where x0 is a solution of (3), satisfies the conditions of Statement (f ). The first part of (f ) now follows from Statement (b) and from Remark 2.1. Since the point Q U1 ∩ QU4 belongs to B , by applying Statement (e), we find that the point Q is compatible with Q U1 ∩ B. This proves the last part of Statement (f ). 2 Now, we are able to prove Theorem 1.1. Proof. If C consists of the q 2 points on a tangent line t to H(3, q 2 ) without the tangent point itself, then C is maximal. By way of contradiction, assume that there exists a further point P compatible with C, then the plane σ := t, P cannot be neither a secant plane nor a tangent plane to H(3, q 2 ). Indeed, in the first case there would be q 2 tangent lines of σ through P and in the second case there are non-concurrent tangent lines of σ. A contradiction. Assume that C is not as in cases 1 and 5. From Propositions 2.3 and 2.5, using the same notation, we have A ∪ {U1 } ⊆ C ⊆ A ∪ B ∪ {U1 }, where A = {(1, αi , 1, 0) : αi ∈ GF (q 2 ), αiq+1 = 1}, U1 = (1, 0, 0, 0), B = {(a, 0, 1, b) : a, b ∈ GF (q 2 ), a + aq = 1, bq+1 = 1}. Let H(2, q 2 ) be the Hermitian curve H(3, q 2 ) ∩ π1 . By the maximality of C, we can certainly choose a point X ∈ C belonging to B . Let X = (a, 0, 1, b) and let tb := XU1 .
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We first assume that the line XU4 is not tangent to H(2, q 2 ). Observe that this assumption forces q to be bigger than 2. Indeed, from Proposition 2.5 (b), if q = 2, it is not possible to find a point of B not lying on a tangent to H(2, q 2 ) through U4 . A further point, say X , compatible with A ∪ {X} must lies on a tangent through X. Hence, it belongs either to tb = XU1 or to RX (cf. Proposition 2.5 (c), (d)). Suppose that X ∈ RX and X U4 is not tangent to H(2, q 2 ). Then, the clique C = A ∪ {U1 , X, X } so obtained is maximal and has size q + 4. Indeed, if there was a further point X compatible with C then, by Remark 2.1, since XX is secant to B , then X belongs either to XU1 or to X U1 and in both cases, we have a contradiction from Proposition 2.5 (d). This means that C is a maximal clique as in case 2. Suppose that X ∈ RX and X U4 is tangent to H(2, q 2 ) (hence q ≡ 1 (mod 3)). By Remark 2.1, since XX is secant to B , a further point X of B can be added to C if and only if it belongs either to tb or to X U1 . The latter case cannot happen by Proposition 2.5 (d). This means that necessarily X ∈ tb and X is compatible with all q + 1 points of B ∩ tb by Proposition 2.5 (e). Note that, by Proposition 2.5 (f ), this happens if and only if X = (x0 , 0, 1, b/x0 ) ∈ B , where x0 is a solution of (3). Then C = A ∪ (B ∩ tb ) ∪ {X } is a maximal clique as in case 4. Suppose that X ∈ tb . Note that a further point X ∈ B \ tb , compatible with A ∪{U1 , X, X } such that X U4 is not tangent to H(2, q 2 ) cannot exist by Proposition 2.5 (d). Hence, if q ≡ 1 (mod 3) by Proposition 2.5 (b), C = A ∪ (tb ∩ B) is a maximal clique as in case 3. If q ≡ 1 (mod 3), the situation reduces to the situation in the first part of this case (a point in RX ) and a maximal clique as in case 4 arises. We now assume that the line XU4 is tangent to H(2, q 2 ) and hence q ≡ 1 (mod 3) by Proposition 2.5 (b). Let X ∈ B \ {X} be compatible with A ∪ {X}. We may assume that X U4 is a tangent line for any X we are adding, for otherwise we can reduce the situation to the previous case. Suppose that X ∈ tb . A further point X ∈ B \ tb , compatible with A ∪ {U1 , X, X } such that X U4 is tangent to H(2, q 2 ), can be added and by Proposition 2.5 (b), (e) and (f ) necessarily q ≡ −1 mod 3 and X = (x0 , 0, 1, b/x0 ) ∈ B , where x0 is a solution of (3). Then C = A ∪ (tb ∩ B) ∪ {X } is a maximal clique as in case 4. / tb and X ∈ / XU4 . Then q ≡ −1 (mod 3) by Proposition 2.5 Suppose that X ∈ (b). A further point X ∈ B \ {X, X } compatible with A ∪ {U1 , X, X } lies either on tb or on X U1 . By Proposition 2.5 (e), (f ), we have that C = A ∪ (tb ∩ B) ∪ {X } or C = A ∪ (X U1 ∩ B) ∪ {X}, according as X ∈ tb or X ∈ X U1 , respectively. In both cases C is a maximal clique as in case 4. Finally, suppose that X ∈ XU4 . A further point X ∈ B \ {X, X } compatible with A ∪ {U1 , X, X } lies either on tb or on X U1 or on XU4 . Since XU4 is tangent to H(2, q 2 ), from Proposition 2.5 (b), we have that X = (x0 , 0, 1, b), X = (x0 , 0, 1, ¯b) and XU4 : X2 = 0, X1 = x0 X3 , where x0 is a solution of (3) and bq+1 = ¯bq+1 = 1, b = ¯b. Note that U1 is compatible with all points of XU4 ∩ B and C = A ∪ (XU4 ∩ B ) ∪ {U1 } is a maximal clique as in case 4. If X ∈ tb or X ∈ X U1 , then, by Proposition 2.5 (e), (f ), necessarily ¯b = b/x0 or b = ¯b/x0 , and we have that C = A ∪ (tb ∩ B) ∪ {X }
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or C = A ∪ (X U1 ∩ B) ∪ {X}, respectively. In both cases C is a maximal clique as in case 4. 2 Remark 2.6. With the aid of MAGMA [3] we investigated the cliques of NU(4, q 2 ), q = 2, 3 in more details. q=2 The graph NU(4, 4) has 90 maximal cliques of size 4 corresponding to points on tangent lines and 2880 projectively equivalent maximal cliques of size 7 as in Theorem 1.1 case 4. q=3 The graph NU(4, 9) has 1088640 projectively equivalent maximal cliques of size 7 as in Theorem 1.1 case 2, 409920 maximal cliques of size 9 of which 1680 correspond to points on tangents and the remaining cliques are as in Theorem 1.1 case 4. There are 77760 projectively equivalent maximal cliques of size 8 all corresponding to complete 8-arcs in PG(3, 9) and admitting an absolutely irreducible subgroup G PSL(2, 7) as an automorphism group. The group G is a subgroup of the maximal subgroups A7 and PSL(3, 4) of PSU(4, 9) [13]. It is interesting to note that the group G has two orbits on generators of the Hermitian surface H(3, 9), say S1 and S2 , both of size 7 and each of them is a complete partial spread of H(3, 9), i.e., a set of mutually disjoint generators of H(3, 9) which is maximal with respect to set-theoretic inclusion. Also there is an orbit of size 42 that together with S1 and S2 gives rise to the unique hemisystem of H(3, 9) associated to the Gewirtz graph [4,15]. The remaining 56 generators are partitioned in two orbits of equal size and they form the complementary hemisystem. The group G has also two orbits of size 28 on lines of PG(3, 9) that are not generators and each of them consists of mutually disjoint lines. The point sets covered by the lines of each of these orbits of size 28 are subsets of PG(3, 9) of size 280 forming quasi-Hermitian surfaces, i.e., a subset of PG(3, 9) having the same intersection numbers with respect to planes as H(3, 9). We suspect that when q 3 ≡ 1 mod (7), the group G has an orbit of size 8 on non-singular points giving rise to a (possibly not maximal) clique of NU(4, q 2 ). Finally NU(4, 9) has 2612736 projectively equivalent maximal cliques of size 5 corresponding to 5-arcs of PG(3, 9) that can be completed to a twisted cubic of PG(3, 9) and admitting a cyclic subgroup of order 5 of PGU(4, 9) as an automorphism group. Remark 2.7. In [9], the authors show that NU(4, q 2 ), q odd, has a maximal clique of size 2q + 2 consisting of points on two skew tangent lines 1 and 2 [9, Theorem 2.5]. From our main result this is not completely true. Indeed, such a clique can always be extended to a maximal clique of size 2q + 3 if q ≡ 0 (mod 3) or q ≡ −1 (mod 3). 3. Maximal cliques of NO± (2n + 2, q), q even In this section we classify the maximal cliques of the graph NO± (2n + 2, q), q even, n ≥ 1.
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Theorem 3.1. Every maximal clique of NO± (2n + 2, q), q even, n ≥ 1, is an affine space isomorphic to AG(n, q). Proof. Since q is even, the polarity induced by Q± (2n + 1, q) gives rise to a symplectic polar space W(2n +1, q). In particular every generator of W(2n +1, q) either is contained in Q+ (2n + 1, q) or meets Q± (2n + 1, q) in a projective subspace of dimension n − 1. Since the number of generators of Q+ (2n + 1, q) is less than the number of generators of W(2n + 1, q) the second possibility actually occurs. A tangent line r to Q± (2n + 1, q) at a point R is contained in the polar hyperplane of R with respect to the symplectic polarity and hence every tangent line to Q± (2n + 1, q) is a totally isotropic line of W(2n + 1, q). Let S be a nonempty subset of non-singular points of PG(2n + 1, q) pairwise joined by a tangent line to Q± (2n + 1, q). Since points in S are pairwise conjugate with respect to the symplectic polarity associated to W(2n + 1, q), we have that S is a totally isotropic subspace of W(2n + 1, q) not contained in Q± (2n + 1, q). It follows that S is maximal (as clique of NO± (2n + 1)) if and only if M := S is a generator of W(2n + 1, q) with M not contained in Q± (2n + 1, q) and S consists of the q n non-singular points of M , i.e., those points of M not on Q± (2n + 1, q). 2 Acknowledgments The authors would like to thank one of the anonymous referees for helpful comments and suggestions improving the first version of the paper. References [1] A.E. Brouwer, W.H. Haemers, Spectra of Graphs, Universitext, Springer, New York, 2012. [2] A.A. Bruen, J.C. Fisher, An observation on certain point-line configurations in classical planes, Discrete Math. 106/107 (1992) 93–96. [3] J. Cannon, C. Playoust, An Introduction to MAGMA, University of Sydney Press, Sydney, 1993. [4] A. Cossidente, T. Penttila, Hemisystems on the Hermitian surface, J. Lond. Math. Soc. (2) 72 (3) (2005) 731–741. [5] M. Crismale, (q 2 + q + 1)-sets of type (0, 1, 2, q + 1) in translation planes of order q 2 , Ann. Discrete Math. 18 (1983) 225–228. [6] G. Donati, N. Durante, Some subsets of the Hermitian curve, Eur. J. Comb. 24 (2003) 211–218. [7] R.W. Hartley, Determination of the ternary collineation groups whose coefficients lie in the GF (2n ), Ann. Math. 27 (1925/26) 140–158. [8] J.W.P. Hirschfeld, Projective Geometries over Finite Fields, Oxford Mathematical Monographs, Oxford Science Publications, The Clarendon Press, Oxford University Press, New York, 1998. [9] J.W.P. Hirschfeld, G. Kiss, Tangent sets in finite spaces, Discrete Math. 155 (1–3) (1996) 107–119. [10] J.W.P. Hirschfeld, J.A. Thas, General Galois Geometries, Oxford Mathematical Monographs, Oxford Science Publications, The Clarendon Press, Oxford University Press, New York, 1991. [11] D.R. Hughes, F.C. Piper, Projective Planes, Springer-Verlag, New York–Berlin, 1973. [12] H.H. Mitchell, Determination of the ordinary and modular ternary linear groups, Trans. Am. Math. Soc. 12 (1911) 207–242. [13] B. Mwene, On some subgroups of the group PSL(4, q), q odd, Geom. Dedic. 12 (1982) 189–199. [14] M.E. O’Nan, Automorphism of unitary block designs, J. Algebra 20 (1972) 492–511. [15] B. Segre, Forme e geometrie hermitiane, con particolare riguardo al caso finito, Ann. Mat. Pura Appl. (4) 70 (1965) 1–201.
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Finite Fields and Their Applications www.elsevier.com/locate/ffa
On maximal cliques of polar graphs ✩ Antonio Cossidente a , Giuseppe Marino b , Francesco Pavese c,∗ a
Dipartimento di Matematica Informatica ed Economia, Università della Basilicata, Contrada Macchia Romana, I-85100 Potenza, Italy b Dipartimento di Matematica e Fisica, Università degli Studi della Campania “Luigi Vanvitelli”, Viale Lincoln, 5, I-81100 Caserta, Italy c Dipartimento di Meccanica, Matematica e Management, Politecnico di Bari, Via Orabona, 4, I-70125 Bari, Italy
a r t i c l e
i n f o
Article history: Received 2 September 2015 Received in revised form 15 June 2017 Accepted 23 June 2017 Available online xxxx Communicated by L. Storme
a b s t r a c t The maximal cliques of the graph NU(4, q 2 ) related to the Hermitian surface of PG(3, q 2 ) and of the graph NO± (2n + 2, q), q even, n ≥ 1, are classified. © 2017 Elsevier Inc. All rights reserved.
MSC: 05E30 05C99 Keywords: Strongly regular graph Maximal clique Hermitian surface Hyperbolic quadric Elliptic quadric
✩ The research was supported by Ministry for Education, University and Research of Italy MIUR (Project PRIN 2012 “Geometrie di Galois e strutture di incidenza”) and by the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA-INdAM). * Corresponding author. E-mail addresses: [email protected] (A. Cossidente), [email protected] (G. Marino), [email protected] (F. Pavese).
http://dx.doi.org/10.1016/j.ffa.2017.06.014 1071-5797/© 2017 Elsevier Inc. All rights reserved.
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1. Introduction Let Q± (2n + 1, q) be a non-degenerate quadric of PG(2n + 1, q), n ≥ 1. Let NO± (2n + 2, q) be the graph whose vertices are the points of PG(2n + 1, q) \ Q± (2n + 1, q) and two vertices P1 , P2 are adjacent if the line joining P1 and P2 contains exactly one point of Q± (2n + 1, q) (i.e., it is a line tangent to Q± (2n + 1, q)). The graph NO± (2n + 2, q) is a strongly regular graph if and only if q = 2 [1, Table 9.9, p. 145]. Let H(n, q 2 ) be a non-degenerate Hermitian variety of PG(n, q 2 ), n ≥ 2. Let NU(n + 1, q 2 ) be the graph whose vertices are the points of PG(n, q 2 ) \ H(n, q 2 ) and two vertices P1 , P2 are adjacent if the line joining P1 and P2 contains exactly one point of H(n, q 2 ) (i.e., it is a line tangent to H(n, q 2 )). The graph NU(n + 1, q 2 ) is a strongly regular graph for any q [1, Table 9.9, p. 145]. The maximal cliques of the graphs NU(3, q 2 ) were classified in [2, Corollary 3]. In particular the authors showed that there are exactly two types of cliques of size q 2 and q + 2, respectively. In the first case, it corresponds to the points on a tangent line with the tangent point excluded. In the second case, it corresponds to q + 1 points forming a Baer subline of a tangent line , and the other point not on . In this paper we will classify the maximal cliques of the graph NU(4, q 2 ) related to the Hermitian surface of PG(3, q 2 ) and of the graph NO± (2n + 2, q), q even, n ≥ 1 (cf. Theorem 3.1). In particular we will prove the following classification theorem. Theorem 1.1. Let C be a maximal clique of NU(4, q 2 ), then either 1. |C| = q 2 , and C consists of the points on a tangent line without the tangent point itself; or C spans the whole space and one of the following cases occur: 2. |C| = q + 4, q > 2, and C consists of a Baer subline and a triangle; 3. |C| = 2q + 2, q ≡ 1 (mod 3) and C consists of two Baer sublines on two skew lines; 4. |C| = 2q + 3, q ≡ 1 (mod 3) and C consists of two Baer sublines on two skew lines and a further point not on the lines; 5. C is an arc of PG(3, q 2 ). 2. The maximal cliques of NU(4, q 2 ) The graph NU(4, q 2 ) is a strongly regular graph with parameters v = q 3 (q 2 +1)(q −1), k = (q 2 − 1)(q 3 + 1), λ = q 4 + q 2 − 2 and μ = q(q 2 − 1)(q + 1). We first recall that an O’Nan configuration consists of six points in a projective plane, namely the four vertices of a non-degenerate quadrangle together with two of their three diagonal points. In [14] it has been proved that no O’Nan configuration lies on a Hermitian curve. By using the polarity of the Hermitian curve, we have the following remark.
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Remark 2.1. Let H(2, q 2 ) be a Hermitian curve in PG(2, q 2 ). There do not exist four points not on H(2, q 2 ) with the property that no three of them are collinear and such that the line joining any two of them is tangent to H(2, q 2 ). Let H(3, q 2 ) be the Hermitian surface of PG(3, q 2 ) with equation X1q+1 + X2q+1 − X3q+1 + X4q+1 = 0. Let G be the subgroup of PΓL(4, q 2 ) fixing H(3, q 2 ). Definition 2.2. Let Z be a point set of PG(3, q 2 ) \ H(3, q 2 ). We say that a point R of PG(3, q 2 ) \ H(3, q 2 ) is compatible with Z, if each of the lines joining R with a point of Z is tangent to H(3, q 2 ). We need the following result. Proposition 2.3. Let C be a maximal clique of NU(4, q 2 ) which is neither an arc of PG(3, q 2 ) (i.e., there are no four points of C in a plane) nor contained in a line. Then C is contained in the union of a plane and a line. Proof. Assume that C is not an arc and that it is not contained in a line. Then we can find a set A of four non-collinear non-isotropic points in a plane π, such that the line joining any two of them is tangent to H(3, q 2 ). Note that in this case π is a plane secant to H(3, q 3 ). Indeed, since in a tangent plane to H(3, q 2 ) all tangent lines pass through the tangent point, if there were at least two non-isotropic points in a plane tangent to H(3, q 2 ) such that the line joining any two of them is tangent to H(3, q 2 ), then they must lie on a tangent line and so they must be collinear. Without loss of generality, we may assume that π is the plane having equation X4 = 0 [10, Theorem 24.4.2]. From Remark 2.1, we have that three points of A are necessarily collinear, i.e., they lie on a tangent line, say t. Then, the further point of A, say U1 , does not lie on t. The stabilizer of π in the group G, say Gπ , is transitive on the set of non-isotropic points of π [10, Theorem 24.4.2]. Therefore, we may assume that U1 is the point (1, 0, 0, 0). From [7,12], the stabilizer of U1 in the group Gπ , say M , induces on π a subgroup of order (q + 1)(q 3 − q) that is the direct product of a homology group M1 of order q + 1 and of a group M2 of order q 3 − q isomorphic to PGL(2, q). The group M stabilizes the line : X1 = X4 = 0. Every element in M1 fixes all points on and all lines through the point U1 , while M2 acts in its natural representation on the q + 1 isotropic points of . Since M2 permutes in a single orbit the non-isotropic points of and M1 permutes in a single orbit the q + 1 isotropic points on secant lines through U1 , we have that the group M acts transitively on the q 3 − q isotropic points of π \ . It follows that M is transitive on the set of q 3 − q tangent lines of π not passing through U1 . Then we may choose t as the line tangent to H(3, q 2 ) at the point P = (1, 0, 1, 0) X4 = 0 . By intersecting t with the q + 1 tangent lines of π passing having equations X1 = X3
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through U1 , we obtain a set, say A , consisting of q + 1 non-isotropic points of π forming a Baer subline. In particular, the three points of A distinct from U1 , say P1 , P2 , P3 , are contained in A . The stabilizer of t in M induces on π a homology group, say Mt , of order q + 1, having as axis the line X2 = X4 = 0 and center the ⎧ point ⎫ ⎛ U2 = (0, ⎞ 1, 0, 0). Hence, it fixes all ⎪ ⎪ 1 0 0 ⎨ ⎬ ⎜ ⎟ 2 q+1 lines through the point U2 . In particular Mt = ⎝ 0 z 0 ⎠ : z ∈ GF (q ), z =1 ⎪ ⎪ ⎩ 0 0 1 ⎭ permutes in a single orbit the q + 1 points of A . Hence, we may assume that P1 is the point (1, 1, 1, 0). A straightforward calculation shows that the points P2 , P3 are of the form (1, αi , 1, 0), where αiq+1 = 1, αi = 1, i = 1, 2 and α1 = α2 . If R ∈ π \ t is compatible with A, then R lies on a tangent line through U1 . By Remark 2.1, R ∈ t and hence R ∈ A . It turns out that the set of points of π compatible with A coincides with A \ A. Let R = (X1 , X2 , X3 , X4 ) be a point of PG(3, q 2 ) \ H(3, q 2 ) that is compatible with A. From [10, Lemma 23.2.1 ii)], we have that ⎧ q+1 ⎪ − X3q+1 + X4q+1 = 0 X ⎪ ⎪ 2 ⎪ ⎨ X q+1 − 2X q+1 + X X q + X q X − X X q − X q X + X X q + X q X = 0 1 3 1 2 2 3 4 3 1 3 1 2 2 3 ⎪ X4q+1 − 2X3q+1 + X1 X3q + X1q X3 − αi X1 X2q − αiq X1q X2 + +αiq X2 X3q + αi X2q X3 = 0, ⎪ ⎪ ⎪ ⎩ i = 1, 2. By subtracting the third equation from the second for i = 1, 2, we get
(α1 − 1)X1 X2q + (α1q − 1)X1q X2 + (1 − α1q )X2 X3q + (1 − α1 )X2q X3 = 0 (α2 − 1)X1 X2q + (α2q − 1)X1q X2 + (1 − α2q )X2 X3q + (1 − α2 )X2q X3 = 0
(1)
Equations (1) are two distinct Hermitian forms of rank two, say H1 , H2 , and in π both represent a set of lines through P . Assume, by way of contradiction, that H1 = λH2 , for some λ ∈ GF (q 2 )∗ . Then we have that α1 = λα2 +(1 −λ) with λ ∈ GF (q)∗ . In particular λ = 1, otherwise α1 = α2 . Since α1q+1 = 1, it follows that α2 + α2q = 2 and taking into account that α2q+1 = 1, we have α2 = 1, a contradiction. From a geometric point of view, the forms in (1) represent in PG(3, q 2 ) q + 1 planes passing through the line joining P with the point U4 = (0, 0, 0, 1). Also, they cannot share more than two planes, since they are distinct. It can be easily seen that they share exactly the two planes π1 and π2 , with equations X2 = 0 and X1 = X3 , respectively. Note that the plane π2 is tangent to H(3, q 2 ) at the point P . Therefore all the points of π2 that are compatible with A, lie on t and hence must be points of A . It follows that if there exist non-isotropic points compatible with A that are not in π, then they must lie on π1 , and so they are contained in π1 ∪ t. 2
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Let C be a maximal clique of NU(4, q 2 ) which is neither an arc of PG(3, q 2 ) nor contained in a line. With the notations introduced in the previous proof, it turns out that C ⊆ A ∪π1 , with π1 : X2 = 0. In particular, the points of C ∩π1 satisfy the following equations: ⎧ q+1 q+1 ⎪ ⎨ X4 − X3 = 0 (2) X4q+1 − 2X3q+1 + X1 X3q + X1q X3 = 0 ⎪ ⎩X = 0 2 Lemma 2.4. A maximal clique containing A must contain A ∪ {U1 }. Proof. We keep the notation of Proposition 2.3. The points of A ∪ {U1 } are pairwise compatible. Hence, we only need to check that they are compatible with the points of π1. If R is a point of π1 compatible with A, then R ∈ / π ∩ π1 (since π ∩ π1 is a (q + 1)-secant to H(3, q 2 ) containing U1 ). Furthermore the set of points of π1 that are compatible with {R, P1 , P2 , P3 } are compatible with all points of A ∪{U1 } as well. Observe that the plane R, t meets H(3, q 2 ) in a Hermitian curve H(2, q 2 ) since t is a tangent line not through R (and hence the tangent lines are not concurrent). Then, we consider the projection of the isotropic points of the polar line of R (with respect to H(2, q 2 )) onto the line t. Since the points P1 , P2 , P3 define a unique Baer subline we are done. 2 The only point satisfying Equations (2) having X3 = 0 is the point U1 ∈ A. The further points of π1 , satisfying Equations (2) are of the form (a, 0, 1, b), where a, b ∈ GF (q 2 ), with a +aq = 1 and bq+1 = 1. They form a set of q 2 +q points, say B . Set B := B ∪{U1 }. We have the following result. Proposition 2.5. The set B has size q 2 + q + 1 and it is of type (0, 1, 2, q + 1) with respect to lines of π1 . Also the following properties hold true: (a) Each line of π1 that is tangent to H(3, q 2 ) and passing through U1 contains q points of B . (b) There exist q lines of π1 passing through U4 containing q + 1 points of B . Among them there are 0, 1 or 2 lines that are tangent to H(3, q 2 ), according as q ≡ 1, 0 or −1 (mod 3), respectively. Q (c) For each point Q ∈ B there are q−1 tangent lines to H(3, q 2 ) ∩π1 , say mQ 1 , . . . , mq−1 , passing through Q and containing exactly one further point of B . (d) Let Q ∈ B such that the line QU4 is not tangent to H(3, q 2 ). Then, any two distinct points of RQ := {mQ i ∩ B : i = 1, . . . , q − 1} \ {Q} are not collinear with U1 . (e) Let Q ∈ B such that the line QU4 is tangent to H(3, q 2 ). If Q is compatible with two points distinct from U1 on a tangent line r through U1 not containing Q, then Q is compatible with r ∩ B. (f ) Let q ≡ 1 (mod 3). Let Q ∈ B such that Q U4 is not tangent to H(3, q 2 ). Then there are one or two points Q ∈ B such that U1 ∈ / QQ and with the property that
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the lines QU4 , QQ are tangents to H(3, q 2 ) according as q ≡ 0 or −1 (mod 3), respectively. Moreover, if Q is a point fulfilling these conditions then Q is compatible with Q U1 ∩ B. Proof. The set B can be seen as the intersection of a Hermitian curve H contained in the plane π1 with a Baer pencil represented by Equations (2), i.e., a cone whose vertex is a point on H and whose base is a Baer subline. In [5] and [6], it is proved that every such set is of type (0, 1, 2, q + 1) with respect to lines of π1 and that every (q + 1)-secant line is a Baer subline. The number of (q + 1)-secant lines to B is 2q + 1 [5, p. 227]. There are q + 1 different (q + 1)-secants to B passing through U1 and they all happen to be tangent lines to H(3, q 2 ), whereas the remaining q lines pass through U4 and necessarily contain q + 1 points of B . This proves Statement (a) and the first part of Statement (b). Note that each intersection point between a (q + 1)-secant line to B passing through U1 and a (q + 1)-secant line to B passing through U4 , belongs to B . The q lines having q +1 points of B are those joining U4 with a point Qλ = (λ, 0, 1, 0), where λ ∈ GF (q 2 ), λ + λq = 1. These are all (q + 1)-secants to B through U4 . Indeed, each of them contains the q + 1 points (λ, 0, 1, b), bq+1 = 1 of B . Since is the polar line of U4 with respect to the polarity of the Hermitian curve H(3, q 2 ) ∩ π1 , each of these lines is tangent to H(3, q 2 ) if and only if the point Qλ belongs to H(3, q 2 ) if and only if λq+1 = 1. Consider the following system:
X q+1 = 1 X q + X = 1.
(3)
We have three cases, [8, Section 1.5, (xi), (xii), (xiii)]: i) if q ≡ 1 (mod 3), there are no solutions of (3) in GF (q 2 ); ii) if q ≡ 0 (mod 3), the unique solution is −1; iii) if q ≡ −1 (mod 3), there exist two distinct solutions of (3) in GF (q 2 ) \ GF (q). Therefore, among the previously mentioned q lines passing through U4 , there are 0, 1 or 2 lines that are tangent to H(3, q 2 ), according as q ≡ 1, 0 or −1 (mod 3), respectively. This proves the second part of Statement (b). Let C = (1, 0, 0, c) be a point of the line joining U1 with U4 and Q = (a, 0, 1, b) a generic point of B , hence a +aq = 1 and bq+1 = 1. The line joining Q with the point C is tangent to H(3, q 2 ) if and only if the following equation is satisfied [10, Lemma 23.2.1 ii)]: (1 − aq+1 )cq+1 + aq bcq + abq c = 0.
(4)
If aq+1 = 1, there are q + 1 solutions for c, see [11, Lemma 2.44], one being c = 0, and so C = U1 . If aq+1 = 1, there are q solutions for c. In this case, both lines QU1 , QU4 , are tangent to H(3, q 2 ). Let c be a solution of (4), with c = 0 if aq+1 = 1, and
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c = 0, −b/(1 − aq+1 ) if aq+1 = 1. A generic point R of the line QC, distinct from C, has coordinates (a+λ, 0, 1, b+λc). Then R ∈ B if and only if λ = 0 or λ = (bq c−bcq )/cq+1 and R = ((acq+1 +bq c −bcq )/cq+1 , 0, 1, bq c/cq ). Finally, note that λ = 0 otherwise, if aq+1 = 1, then bcq = 0, that is a contradiction, whereas if aq+1 = 1, then c = −b/(1 − aq+1 ) and we get again a contradiction. It follows that R = Q. This proves Statement (c). Let Q be a point of B such that QU4 is not tangent to H(3, q 2 ) (i.e., aq+1 = 1). We show that the q − 1 points of RQ lie on distinct tangent lines passing through U1 . To this aim, we only need to prove that bq c/cq = bq c /c q , for distinct solutions c, c of (4). Assume the contrary holds true. Then cc q = c cq or cc q ∈ GF (q)∗ , which implies that c = dc, for some d ∈ GF (q)∗ . Then, Equation (4) is satisfied if and only if (d − 1)(1 − aq+1 ) = 0, i.e., if and only if d = 1, a contradiction. This proves Statement (d). Let Q be a point of B such that QU4 is tangent to H(3, q 2 ). Assume that Q is compatible with two points distinct from U1 on a tangent line r through U1 not containing Q, say Y and Z. Then, we consider the projection of the isotropic points of the polar line of Q (with respect to H(3, q 2 ) ∩ π1 ) onto the line r. Since the points U1 , Y, Z define a unique Baer subline, we have proved Statement (e). Let Q = (a, 0, 1, b) ∈ B such that Q U4 is not tangent to H(3, q 2 ), i.e., aq+1 = 1. By [10, Lemma 23.2.1 ii)], direct calculations show that a point Q of type (x0 , 0, 1, b/x0 ), where x0 is a solution of (3), satisfies the conditions of Statement (f ). The first part of (f ) now follows from Statement (b) and from Remark 2.1. Since the point Q U1 ∩ QU4 belongs to B , by applying Statement (e), we find that the point Q is compatible with Q U1 ∩ B. This proves the last part of Statement (f ). 2 Now, we are able to prove Theorem 1.1. Proof. If C consists of the q 2 points on a tangent line t to H(3, q 2 ) without the tangent point itself, then C is maximal. By way of contradiction, assume that there exists a further point P compatible with C, then the plane σ := t, P cannot be neither a secant plane nor a tangent plane to H(3, q 2 ). Indeed, in the first case there would be q 2 tangent lines of σ through P and in the second case there are non-concurrent tangent lines of σ. A contradiction. Assume that C is not as in cases 1 and 5. From Propositions 2.3 and 2.5, using the same notation, we have A ∪ {U1 } ⊆ C ⊆ A ∪ B ∪ {U1 }, where A = {(1, αi , 1, 0) : αi ∈ GF (q 2 ), αiq+1 = 1}, U1 = (1, 0, 0, 0), B = {(a, 0, 1, b) : a, b ∈ GF (q 2 ), a + aq = 1, bq+1 = 1}. Let H(2, q 2 ) be the Hermitian curve H(3, q 2 ) ∩ π1 . By the maximality of C, we can certainly choose a point X ∈ C belonging to B . Let X = (a, 0, 1, b) and let tb := XU1 .
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We first assume that the line XU4 is not tangent to H(2, q 2 ). Observe that this assumption forces q to be bigger than 2. Indeed, from Proposition 2.5 (b), if q = 2, it is not possible to find a point of B not lying on a tangent to H(2, q 2 ) through U4 . A further point, say X , compatible with A ∪ {X} must lies on a tangent through X. Hence, it belongs either to tb = XU1 or to RX (cf. Proposition 2.5 (c), (d)). Suppose that X ∈ RX and X U4 is not tangent to H(2, q 2 ). Then, the clique C = A ∪ {U1 , X, X } so obtained is maximal and has size q + 4. Indeed, if there was a further point X compatible with C then, by Remark 2.1, since XX is secant to B , then X belongs either to XU1 or to X U1 and in both cases, we have a contradiction from Proposition 2.5 (d). This means that C is a maximal clique as in case 2. Suppose that X ∈ RX and X U4 is tangent to H(2, q 2 ) (hence q ≡ 1 (mod 3)). By Remark 2.1, since XX is secant to B , a further point X of B can be added to C if and only if it belongs either to tb or to X U1 . The latter case cannot happen by Proposition 2.5 (d). This means that necessarily X ∈ tb and X is compatible with all q + 1 points of B ∩ tb by Proposition 2.5 (e). Note that, by Proposition 2.5 (f ), this happens if and only if X = (x0 , 0, 1, b/x0 ) ∈ B , where x0 is a solution of (3). Then C = A ∪ (B ∩ tb ) ∪ {X } is a maximal clique as in case 4. Suppose that X ∈ tb . Note that a further point X ∈ B \ tb , compatible with A ∪{U1 , X, X } such that X U4 is not tangent to H(2, q 2 ) cannot exist by Proposition 2.5 (d). Hence, if q ≡ 1 (mod 3) by Proposition 2.5 (b), C = A ∪ (tb ∩ B) is a maximal clique as in case 3. If q ≡ 1 (mod 3), the situation reduces to the situation in the first part of this case (a point in RX ) and a maximal clique as in case 4 arises. We now assume that the line XU4 is tangent to H(2, q 2 ) and hence q ≡ 1 (mod 3) by Proposition 2.5 (b). Let X ∈ B \ {X} be compatible with A ∪ {X}. We may assume that X U4 is a tangent line for any X we are adding, for otherwise we can reduce the situation to the previous case. Suppose that X ∈ tb . A further point X ∈ B \ tb , compatible with A ∪ {U1 , X, X } such that X U4 is tangent to H(2, q 2 ), can be added and by Proposition 2.5 (b), (e) and (f ) necessarily q ≡ −1 mod 3 and X = (x0 , 0, 1, b/x0 ) ∈ B , where x0 is a solution of (3). Then C = A ∪ (tb ∩ B) ∪ {X } is a maximal clique as in case 4. / tb and X ∈ / XU4 . Then q ≡ −1 (mod 3) by Proposition 2.5 Suppose that X ∈ (b). A further point X ∈ B \ {X, X } compatible with A ∪ {U1 , X, X } lies either on tb or on X U1 . By Proposition 2.5 (e), (f ), we have that C = A ∪ (tb ∩ B) ∪ {X } or C = A ∪ (X U1 ∩ B) ∪ {X}, according as X ∈ tb or X ∈ X U1 , respectively. In both cases C is a maximal clique as in case 4. Finally, suppose that X ∈ XU4 . A further point X ∈ B \ {X, X } compatible with A ∪ {U1 , X, X } lies either on tb or on X U1 or on XU4 . Since XU4 is tangent to H(2, q 2 ), from Proposition 2.5 (b), we have that X = (x0 , 0, 1, b), X = (x0 , 0, 1, ¯b) and XU4 : X2 = 0, X1 = x0 X3 , where x0 is a solution of (3) and bq+1 = ¯bq+1 = 1, b = ¯b. Note that U1 is compatible with all points of XU4 ∩ B and C = A ∪ (XU4 ∩ B ) ∪ {U1 } is a maximal clique as in case 4. If X ∈ tb or X ∈ X U1 , then, by Proposition 2.5 (e), (f ), necessarily ¯b = b/x0 or b = ¯b/x0 , and we have that C = A ∪ (tb ∩ B) ∪ {X }
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or C = A ∪ (X U1 ∩ B) ∪ {X}, respectively. In both cases C is a maximal clique as in case 4. 2 Remark 2.6. With the aid of MAGMA [3] we investigated the cliques of NU(4, q 2 ), q = 2, 3 in more details. q=2 The graph NU(4, 4) has 90 maximal cliques of size 4 corresponding to points on tangent lines and 2880 projectively equivalent maximal cliques of size 7 as in Theorem 1.1 case 4. q=3 The graph NU(4, 9) has 1088640 projectively equivalent maximal cliques of size 7 as in Theorem 1.1 case 2, 409920 maximal cliques of size 9 of which 1680 correspond to points on tangents and the remaining cliques are as in Theorem 1.1 case 4. There are 77760 projectively equivalent maximal cliques of size 8 all corresponding to complete 8-arcs in PG(3, 9) and admitting an absolutely irreducible subgroup G PSL(2, 7) as an automorphism group. The group G is a subgroup of the maximal subgroups A7 and PSL(3, 4) of PSU(4, 9) [13]. It is interesting to note that the group G has two orbits on generators of the Hermitian surface H(3, 9), say S1 and S2 , both of size 7 and each of them is a complete partial spread of H(3, 9), i.e., a set of mutually disjoint generators of H(3, 9) which is maximal with respect to set-theoretic inclusion. Also there is an orbit of size 42 that together with S1 and S2 gives rise to the unique hemisystem of H(3, 9) associated to the Gewirtz graph [4,15]. The remaining 56 generators are partitioned in two orbits of equal size and they form the complementary hemisystem. The group G has also two orbits of size 28 on lines of PG(3, 9) that are not generators and each of them consists of mutually disjoint lines. The point sets covered by the lines of each of these orbits of size 28 are subsets of PG(3, 9) of size 280 forming quasi-Hermitian surfaces, i.e., a subset of PG(3, 9) having the same intersection numbers with respect to planes as H(3, 9). We suspect that when q 3 ≡ 1 mod (7), the group G has an orbit of size 8 on non-singular points giving rise to a (possibly not maximal) clique of NU(4, q 2 ). Finally NU(4, 9) has 2612736 projectively equivalent maximal cliques of size 5 corresponding to 5-arcs of PG(3, 9) that can be completed to a twisted cubic of PG(3, 9) and admitting a cyclic subgroup of order 5 of PGU(4, 9) as an automorphism group. Remark 2.7. In [9], the authors show that NU(4, q 2 ), q odd, has a maximal clique of size 2q + 2 consisting of points on two skew tangent lines 1 and 2 [9, Theorem 2.5]. From our main result this is not completely true. Indeed, such a clique can always be extended to a maximal clique of size 2q + 3 if q ≡ 0 (mod 3) or q ≡ −1 (mod 3). 3. Maximal cliques of NO± (2n + 2, q), q even In this section we classify the maximal cliques of the graph NO± (2n + 2, q), q even, n ≥ 1.
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Theorem 3.1. Every maximal clique of NO± (2n + 2, q), q even, n ≥ 1, is an affine space isomorphic to AG(n, q). Proof. Since q is even, the polarity induced by Q± (2n + 1, q) gives rise to a symplectic polar space W(2n +1, q). In particular every generator of W(2n +1, q) either is contained in Q+ (2n + 1, q) or meets Q± (2n + 1, q) in a projective subspace of dimension n − 1. Since the number of generators of Q+ (2n + 1, q) is less than the number of generators of W(2n + 1, q) the second possibility actually occurs. A tangent line r to Q± (2n + 1, q) at a point R is contained in the polar hyperplane of R with respect to the symplectic polarity and hence every tangent line to Q± (2n + 1, q) is a totally isotropic line of W(2n + 1, q). Let S be a nonempty subset of non-singular points of PG(2n + 1, q) pairwise joined by a tangent line to Q± (2n + 1, q). Since points in S are pairwise conjugate with respect to the symplectic polarity associated to W(2n + 1, q), we have that S is a totally isotropic subspace of W(2n + 1, q) not contained in Q± (2n + 1, q). It follows that S is maximal (as clique of NO± (2n + 1)) if and only if M := S is a generator of W(2n + 1, q) with M not contained in Q± (2n + 1, q) and S consists of the q n non-singular points of M , i.e., those points of M not on Q± (2n + 1, q). 2 Acknowledgments The authors would like to thank one of the anonymous referees for helpful comments and suggestions improving the first version of the paper. References [1] A.E. Brouwer, W.H. Haemers, Spectra of Graphs, Universitext, Springer, New York, 2012. [2] A.A. Bruen, J.C. Fisher, An observation on certain point-line configurations in classical planes, Discrete Math. 106/107 (1992) 93–96. [3] J. Cannon, C. Playoust, An Introduction to MAGMA, University of Sydney Press, Sydney, 1993. [4] A. Cossidente, T. Penttila, Hemisystems on the Hermitian surface, J. Lond. Math. Soc. (2) 72 (3) (2005) 731–741. [5] M. Crismale, (q 2 + q + 1)-sets of type (0, 1, 2, q + 1) in translation planes of order q 2 , Ann. Discrete Math. 18 (1983) 225–228. [6] G. Donati, N. Durante, Some subsets of the Hermitian curve, Eur. J. Comb. 24 (2003) 211–218. [7] R.W. Hartley, Determination of the ternary collineation groups whose coefficients lie in the GF (2n ), Ann. Math. 27 (1925/26) 140–158. [8] J.W.P. Hirschfeld, Projective Geometries over Finite Fields, Oxford Mathematical Monographs, Oxford Science Publications, The Clarendon Press, Oxford University Press, New York, 1998. [9] J.W.P. Hirschfeld, G. Kiss, Tangent sets in finite spaces, Discrete Math. 155 (1–3) (1996) 107–119. [10] J.W.P. Hirschfeld, J.A. Thas, General Galois Geometries, Oxford Mathematical Monographs, Oxford Science Publications, The Clarendon Press, Oxford University Press, New York, 1991. [11] D.R. Hughes, F.C. Piper, Projective Planes, Springer-Verlag, New York–Berlin, 1973. [12] H.H. Mitchell, Determination of the ordinary and modular ternary linear groups, Trans. Am. Math. Soc. 12 (1911) 207–242. [13] B. Mwene, On some subgroups of the group PSL(4, q), q odd, Geom. Dedic. 12 (1982) 189–199. [14] M.E. O’Nan, Automorphism of unitary block designs, J. Algebra 20 (1972) 492–511. [15] B. Segre, Forme e geometrie hermitiane, con particolare riguardo al caso finito, Ann. Mat. Pura Appl. (4) 70 (1965) 1–201.