ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 21 (2021) #P1.04 / 45–55
https://doi.org/10.26493/1855-3974.2358.3c9
(Also available at http://amc-journal.eu)
On Hermitian varieties in PG(6, q2)
Angela Aguglia *
Dipartimento di Meccanica, Matematica e Management, Politecnico di Bari,
Via Orabona 4, I-70125 Bari, Italy
Luca Giuzzi †
DICATAM, University of Brescia, Via Branze 53, I-25123 Brescia, Italy
Masaaki Homma
Department of Mathematics and Physics, Kanagawa University,
Hiratsuka 259-1293, Japan
Received 8 June 2020, accepted 15 February 2021, published online 10 August 2021
Abstract
In this paper we characterize the non-singular Hermitian variety H(6, q2) of PG(6, q2),
q ̸= 2 among the irreducible hypersurfaces of degree q + 1 in PG(6, q2) not containing
solids by the number of its points and the existence of a solid S meeting it in q4 + q2 + 1
points.
Keywords: Unital, Hermitian variety, algebraic hypersurface.
Math. Subj. Class. (2020): 51E21, 51E15, 51E20
1 Introduction
The set of all absolute points of a non-degenerate unitary polarity in PG(r, q2) determines
the Hermitian variety H(r, q2). This is a non-singular algebraic hypersurface of degree
q + 1 in PG(r, q2) with a number of remarkable properties, both from the geometrical and
the combinatorial point of view; see [6, 16]. In particular, H(r, q2) is a 2-character set with
respect to the hyperplanes of PG(r, q2) and 3-character blocking set with respect to the
*Corresponding author. The author was supported by the Italian National Group for Algebraic and Geometric
Structures and their Applications (GNSAGA - INdAM).
†The author was supported by the Italian National Group for Algebraic and Geometric Structures and their
Applications (GNSAGA - INdAM).
E-mail addresses: angela.aguglia@poliba.it (Angela Aguglia), luca.giuzzi@unibs.it (Luca Giuzzi),
homma@kanagawa-u.ac.jp (Masaaki Homma)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
46 Ars Math. Contemp. 21 (2021) #P1.04 / 45–55
lines of PG(r, q2) for r > 2. An interesting and widely investigated problem is to provide
combinatorial descriptions of H(r, q2).
First, we observe that a condition on the number of points and the intersection numbers
with hyperplanes is not in general sufficient to characterize Hermitian varieties; see [1,
2]. On the other hand, it is enough to consider in addition the intersection numbers with
codimension 2 subspaces in order to get a complete description; see [7].
In general, a hypersurface H of PG(r, q) is viewed as a hypersurface over the algebraic
closure of GF(q) and a point of PG(r, qi) in H is called a GF(qi)-point. A GF(q)-point
of H is also said to be a rational point of H. Throughout this paper, the number of GF(qi)-
points of H will be denoted by Nqi(H). For simplicity, we shall also use the convention
|H| = Nq(H).
In the present paper, we shall investigate a combinatorial characterization of the Her-
mitian hypersurface H(6, q2) in PG(6, q2) among all hypersurfaces of the same degree
having also the same number of GF(q2)-rational points.
More in detail, in [12, 13] it has been proved that if X is a hypersurface of degree q+1
in PG(r, q2), r ≥ 3 odd, with |X | = |H(r, q2)| = (qr+1 + (−1)r)(qr − (−1)r)/(q2 − 1)
GF(q2)-rational points, not containing linear subspaces of dimension greater than r−12 ,
then X is a non-singular Hermitian variety of PG(r, q2). This result generalizes the char-
acterization of [8] for the Hermitian curve of PG(2, q2), q ̸= 2.
The case where r > 4 is even is, in general, currently open. A starting point for
a characterization in arbitrary even dimension can be found in [3] where the case of a
hypersurface X of degree q + 1 in PG(4, q2), q > 3 is considered. There, it is shown that
when X has the same number of rational points as H(4, q2), does not contain any subspaces
of dimension greater than 1 and meets at least one plane π in q2+1GF(q2)-rational points,
then X is a Hermitian variety.
In this article we deal with hypersurfaces of degree q + 1 in PG(6, q2) and we prove
that a characterization similar to that of [3] holds also in dimension 6. We conjecture that
this can be extended to arbitrary even dimension.
Theorem 1.1. Let S be a hypersurface of PG(6, q2), q > 2, defined over GF(q2), not
containing solids. If the degree of S is q + 1 and the number of its rational points is
q11 + q9 + q7 + q4 + q2 + 1, then every solid of PG(6, q2) meets S in at least q4 + q2 + 1
rational points. If there is at least a solid Σ3 such that |Σ3 ∩ S| = q4 + q2 + 1, then S is a
non-singular Hermitian variety of PG(6, q2).
Furthermore, we also extend the result of [3] to the case q = 3.
2 Preliminaries and notation
In this section we collect some useful information and results that will be crucial to our
proof.
A Hermitian variety in PG(r, q2) is the algebraic variety of PG(r, q2) whose points ⟨v⟩
satisfy the equation η(v, v) = 0 where η is a sesquilinear form GF(q2)r+1×GF(q2)r+1 →
GF(q2). The radical of the form η is the vector subspace of GF(q2)r+1 given by
Rad(η) := {w ∈ GF(q2)r+1 : ∀v ∈ GF(q2)r+1, η(v, w) = 0}.
The form η is non-degenerate if Rad(η) = {0}. If the form η is non-degenerate, then the
corresponding Hermitian variety is denoted by H(r, q2) and it is a non-singular algebraic
A. Aguglia et al.: On Hermitian varieties in PG(6, q2) 47
variety, of degree q + 1 containing
(qr+1 + (−1)r)(qr − (−1)r)/(q2 − 1)
GF(q2)-rational points. When η is degenerate we shall call vertex Rt of the degenerate Her-
mitian variety associated to η the projective subspace Rt := PG(Rad(η)) := {⟨w⟩ : w ∈
Rad(η)} of PG(r, q2). A degenerate Hermitian variety can always be described as a cone
of vertex Rt and basis a non-degenerate Hermitian variety H(r − t, q2) disjoint from Rt
where t = dim(Rad(η)) is the vector dimension of the radical of η. In this case we shall
write the corresponding variety as RtH(r − t, q2). Indeed,
RtH(r − t, q2) := {X ∈ ⟨P,Q⟩ : P ∈ Rt, Q ∈ H(r − t, q2)}.
Any line of PG(r, q2) meets a Hermitian variety (either degenerate or not) in either
1, q + 1 or q2 + 1 points (the latter value only for r > 2). The maximal dimension of
projective subspaces contained in the non-degenerate Hermitian variety H(r, q2) is (r −
2)/2, if r is even, or (r − 1)/2, if r is odd. These subspaces of maximal dimension are
called generators of H(r, q2) and the generators of H(r, q2) through a point P of H(r, q2)
span a hyperplane P⊥ of PG(r, q2), the tangent hyperplane at P .
It is well known that this hyperplane meets H(r, q2) in a degenerate Hermitian variety
PH(r − 2, q2), that is in a Hermitian cone having as vertex the point P and as base a
non-singular Hermitian variety of Θ ∼= PG(r − 2, q2) contained in P⊥ with P ̸∈ Θ.
Every hyperplane of PG(r, q2) that is not tangent meets H(r, q2) in a non-singular
Hermitian variety H(r−1, q2), and is called a secant hyperplane of H(r, q2). In particular,
a tangent hyperplane contains
1 + q2(qr−1 + (−1)r)(qr−2 − (−1)r)/(q2 − 1)
GF(q2)-rational points of H(r, q2), whereas a secant hyperplane contains
(qr + (−1)r−1)(qr−1 − (−1)r−1)/(q2 − 1)
GF(q2)-rational points of H(r, q2).
We now recall several results which shall be used in the course of this paper.
Lemma 2.1 ([15]). Let d be an integer with 1 ≤ d ≤ q + 1 and let C be a curve of degree
d in PG(2, q) defined over GF(q), which may have GF(q)-linear components. Then the
number of its rational points is at most dq + 1 and Nq(C) = dq + 1 if and only if C is a
pencil of d lines of PG(2, q).
Lemma 2.2 ([10]). Let d be an integer with 2 ≤ d ≤ q + 2, and C a curve of degree d
in PG(2, q) defined over GF(q) without any GF(q)-linear components. Then Nq(C) ≤
(d− 1)q+ 1, except for a class of plane curves of degree 4 over GF(4) having 14 rational
points.
Lemma 2.3 ([11]). Let S be a surface of degree d in PG(3, q) over GF(q). Then
Nq(S) ≤ dq2 + q + 1
Lemma 2.4 ([8]). Suppose q ̸= 2. Let C be a plane curve over GF(q2) of degree q + 1
without GF(q2)-linear components. If C has q3 + 1 rational points, then C is a Hermitian
curve.
48 Ars Math. Contemp. 21 (2021) #P1.04 / 45–55
Lemma 2.5 ([7]). A subset of points of PG(r, q2) having the same intersection numbers
with respect to hyperplanes and spaces of codimension 2 as non-singular Hermitian vari-
eties is a non-singular Hermitian variety of PG(r, q2).
From [9, Theorem 23.5.1, Theorem 23.5.3] we have the following.
Lemma 2.6. If W is a set of q7 + q4 + q2 + 1 points of PG(4, q2), q > 2 such that every
line of PG(4, q2) meets W in 1, q + 1 or q2 + 1 points, then W is a Hermitian cone with
vertex a line and base a unital.
Finally, we recall that a blocking set with respect to lines of PG(r, q) is a point set
which blocks all the lines, i.e., intersects each line of PG(r, q) in at least one point.
3 Proof of Theorem 1.1
We first provide an estimate on the number of points of a curve of degree q+1 in PG(2, q2),
where q is any prime power.
Lemma 3.1. Let C be a plane curve over GF(q2), without GF(q2)-lines as components
and of degree q + 1. If the number of GF(q2)-rational points of C is N < q3 + 1, then
N ≤
 q
3 − (q2 − 2) if q > 3
24 if q = 3
8 if q = 2.
(3.1)
Proof. We distinguish the following three cases:
(a) C has two or more GF(q2)-components;
(b) C is irreducible over GF(q2), but not absolutely irreducible;
(c) C is absolutely irreducible.
Suppose first q ̸= 2.
Case (a) Suppose C = C1 ∪ C2. Let di be the degree of Ci, for each i = 1, 2. Hence
d1 + d2 = q + 1. By Lemma 2.2,
N ≤ Nq2(C1) +Nq2(C2) ≤ [(q + 1)− 2]q2 + 2 = q3 − (q2 − 2)
Case (b) Let C′ be an irreducible component of C over the algebraic closure of GF(q2). Let
GF(q2t) be the minimum defining field of C′ and σ be the Frobenius morphism of GF(q2t)
over GF(q2). Then
C = C′ ∪ C′σ ∪ C′σ
2
∪ . . . ∪ C′σ
t−1
,
and the degree of C′, say e, satisfies q + 1 = te with e > 1. Hence any GF(q2)-rational
point of C is contained in ∩t−1i=0C′σ
i
. In particular, N ≤ e2 ≤ ( q+12 )
2 by Bezout’s Theorem
and ( q+12 )
2 < q3 − (q2 − 2).
Case (c) Let C be an absolutely irreducible curve over GF(q2) of degree q + 1. Either C
has a singular point or not.
In general, an absolutely irreducible plane curve M over GF(q2) is q2-Frobenius non-
classical if for a general point P (x0, x1, x2) of M the point P q
2
= P q
2
(xq
2
0 , x
q2
1 , x
q2
2 ) is
A. Aguglia et al.: On Hermitian varieties in PG(6, q2) 49
on the tangent line to M at the point P . Otherwise, the curve M is said to be Frobenius
classical. A lower bound of the number of GF(q2)-points for q2-Frobenius non-classical
curves is given by [4, Corollary 1.4]: for a q2-Frobenius non-classical curve C′ of degree
d, we have Nq2(C′) ≥ d(q2 − d + 2). In particular, if d = q + 1, the lower bound is just
q3 + 1.
Going back to our original curve C, we know that C is Frobenius classical because
N < q3 + 1. Let F (x, y, z) = 0 be an equation of C over GF(q2). We consider the curve
D defined by ∂F∂x x
q2 + ∂F∂y y
q2 + ∂F∂z z
q2 = 0. Then C is not a component of D because C is
Frobenius classical. Furthermore, any GF(q2)-point P lies on C ∩ D and the intersection
multiplicity of C and D at P is at least 2 by Euler’s theorem for homogeneous polynomials.
Hence by Bézout’s theorem, 2N ≤ (q + 1)(q2 + q). Hence
N ≤ 1
2
q(q + 1)2.
This argument is due to Stöhr and Voloch [18, Theorem 1.1]. This Stöhr and Voloch’s
bound is lower than the estimate for N in case (a) for q > 4 and it is the same for q = 4.
When q = 3 the bound in case (a) is smaller than the Stöhr and Voloch’s bound.
Finally, we consider the case q = 2. Under this assumption, C is a cubic curve and neither
case (a) nor case (b) might occur. For a degree 3 curve over GF(q2) the Stöhr and Voloch’s
bound is loose, thus we need to change our argument. If C has a singular point, then C is a
rational curve with a unique singular point. Since the degree of C is 3, singular points are
either cusps or ordinary double points. Hence N ∈ {4, 5, 6}. If C is nonsingular, then it is
an elliptic curve and, by the Hasse-Weil bound, see [19], N ∈ I where I = {1, 2, . . . , 9}
and for each number N belonging to I there is an elliptic curve over GF(4) with N points,
from [14, Theorem 4.2]. This completes the proof.
Henceforth, we shall always suppose q > 2 and we denote by S an algebraic hypersur-
face of PG(6, q2) satisfying the following hypotheses of Theorem 1.1:
(S1) S is an algebraic hypersurface of degree q + 1 defined over GF(q2);
(S2) |S| = q11 + q9 + q7 + q4 + q2 + 1;
(S3) S does not contain projective 3-spaces (solids);
(S4) there exists a solid Σ3 such that |S ∩ Σ3| = q4 + q2 + 1.
We first consider the behavior of S with respect to the lines.
Lemma 3.2. An algebraic hypersurface T of degree q + 1 in PG(r, q2), q ̸= 2, with
|T | = |H(r, q2)| is a blocking set with respect to lines of PG(r, q2)
Proof. Suppose on the contrary that there is a line ℓ of PG(r, q2) which is disjoint from
T . Let α be a plane containing ℓ. The algebraic plane curve C = α ∩ T of degree q + 1
cannot have GF(q2)-linear components and hence it has at most q3 + 1 points because of
Lemma 2.2. If C had q3 +1 rational points, then from Lemma 2.4, C would be a Hermitian
curve with an external line, a contradiction since Hermitian curves are blocking sets. Thus
Nq2(C) ≤ q3. Since q > 2, by Lemma 3.1, Nq2(C) < q3−1 and hence every plane through
r meets T in at most q3−1 rational points. Consequently, by considering all planes through
r, we can bound the number of rational points of T by Nq2(T ) ≤ (q3 − 1) q
2r−4−1
q2−1 =
50 Ars Math. Contemp. 21 (2021) #P1.04 / 45–55
q2r−3 + · · · < |H(r, q2)|, which is a contradiction. Therefore there are no external lines to
T and so T is a blocking set w.r.t. lines of PG(r, q2).
Remark 3.3. The proof of [3, Lemma 3.1] would work perfectly well here under the
assumption q > 3. The alternative argument of Lemma 3.2 is simpler and also holds for
q = 3.
By the previous Lemma and assumptions (S1) and (S2), S is a blocking set for the
lines of PG(6, q2) In particular, the intersection of S with any 3-dimensional subspace Σ
of PG(6, q2) is also a blocking set with respect to lines of Σ and hence it contains at least
q4 + q2 + 1 GF(q2)-rational points; see [5].
Lemma 3.4. Let Σ3 be a solid of PG(6, q2) satisfying condition (S4), that is Σ3 meets S
in exactly q4 + q2 + 1 points. Then, Π := S ∩ Σ3 is a plane.
Proof. S∩Σ3 must be a blocking set for the lines of PG(3, q2); also it has size q4+q2+1.
It follows from [5] that Π := S ∩ Σ3 is a plane.
Lemma 3.5. Let Σ3 be a solid of satisfying condition (S4). Then, any 4-dimensional
projective space Σ4 through Σ3 meets S in a Hermitian cone with vertex a line and basis a
Hermitian curve.
Proof. Consider all of the q6 + q4 + q2 + 1 subspaces Σ3 of dimension 3 in PG(6, q2)
containing Π = S ∩ Σ3.
From Lemma 2.3 and condition (S3) we have |Σ3 ∩ S| ≤ q5 + q4 + q2 + 1. Hence,
|S| = (q7 + 1)(q4 + q2 + 1) ≤ (q6 + q4 + q2)q5 + q4 + q2 + 1 = |S|.
Consequently, |Σ3 ∩ S| = q5 + q4 + q2 + 1 for all Σ3 ̸= Σ3 such that Π ⊂ Σ3.
Let C := Σ4 ∩ S . Counting the number of rational points of C by considering the
intersections with the q2+1 subspaces Σ′3 of dimension 3 in Σ4 containing the plane Π we
get
|C| = q2 · q5 + q4 + q2 + 1 = q7 + q4 + q2 + 1.
In particular, C ∩Σ′3 is a maximal surface of degree q+ 1; so it must split in q+ 1 distinct
planes through a line of Π; see [17]. So C consists of q3 + 1 distinct planes belonging to
distinct q2 pencils, all containing Π ; denote by L the family of these planes. Also for each
Σ′3 ̸= Σ3, there is a line ℓ′ such that all the planes of L in Σ′3 pass through ℓ′. It is now
straightforward to see that any line contained in C must necessarily belong to one of the
planes of L and no plane not in L is contained in C.
In order to get the result it is now enough to show that a line of Σ4 meets C in either 1,
q + 1 or q2 + 1 points. To this purpose, let ℓ be a line of Σ4 and suppose ℓ ̸⊆ C. Then, by
Bezout’s theorem,
1 ≤ |ℓ ∩ C| ≤ q + 1.
Assume |ℓ ∩ C| > 1. Then we can distinguish two cases:
1. ℓ∩Π ̸= ∅. If ℓ and Π are incident, then we can consider the 3-dimensional subspace
Σ′3 := ⟨ℓ,Π⟩. Then ℓ must meet each plane of L in Σ′3 in different points (otherwise
ℓ passes through the intersection of these planes and then |ℓ ∩ C| = 1). As there are
q + 1 planes of L in Σ′3, we have |ℓ ∩ C| = q + 1.
A. Aguglia et al.: On Hermitian varieties in PG(6, q2) 51
2. ℓ ∩ Π = ∅. Consider the plane Λ generated by a point P ∈ Π and ℓ. Clearly
Λ ̸∈ L. The curve Λ∩S has degree q+1 by construction, does not contain lines (for
otherwise Λ ∈ L) and has q3 + 1 GF(q2)-rational points (by a counting argument).
So from Lemma 2.4 it is a Hermitian curve . It follows that ℓ is a q + 1 secant.
We can now apply Lemma 2.6 to see that C is a Hermitian cone with vertex a line.
Lemma 3.6. Let Σ3 be a space satisfying condition (S4) and take Σ5 to be a 5-dimensional
projective space with Σ3 ⊆ Σ5. Then S ∩ Σ5 is a Hermitian cone with vertex a point and
basis a Hermitian hypersurface H(4, q2).
Proof. Let
Σ4 := Σ
1
4,Σ
2
4, . . . ,Σ
q2+1
4
be the 4-spaces through Σ3 contained in Σ5. Put Ci := Σi4 ∩ S , for all i ∈ {1, . . . , q2 +1}
and Π = Σ3 ∩ C1. From Lemma 3.5 Ci is a Hermitian cone with vertex a line, say ℓi.
Furthermore Π ⊆ Σ3 ⊆ Σi4 where Π is a plane. Choose a plane Π′ ⊆ Σ14 such that
m := Π′ ∩ C1 is a line m incident with Π but not contained in it. Let P1 := m ∩ Π. It is
straightforward to see that in Σ14 there are exactly 1 plane through m which is a (q
4+q2+1)-
secant, q4 planes which are (q3 + q2 + 1)-secant and q2 planes which are (q2 + 1)-secant.
Also P1 belongs to the line ℓ1. There are now two cases to consider:
(a) There is a plane Π′′ ̸= Π′ not contained in Σi4 for all i = 1, . . . , q2+1 with m ⊆ Π′′ ⊆
S ∩ Σ5.
We first show that the vertices of the cones Ci are all concurrent. Consider mi :=
Π′′ ∩ Σi4. Then {mi : i = 1, . . . , q2 + 1} consists of q2 + 1 lines (including m) all
through P1. Observe that for all i, the line mi meets the vertex ℓi of the cone Ci in
Pi ∈ Π. This forces P1 = P2 = · · · = Pq2+1. So P1 ∈ ℓ1, . . . , ℓq2+1.
Now let Σ4 be a 4-dimensional space in Σ5 with P1 ̸∈ Σ4; in particular Π ̸⊆ Σ4. Put
also Σ3 := Σ14 ∩ Σ4. Clearly, r := Σ3 ∩ Π is a line and P1 ̸∈ r. So Σ3 ∩ S cannot be
the union of q + 1 planes, since if this were to be the case, these planes would have to
pass through the vertex ℓ1. It follows that Σ3∩S must be a Hermitian cone with vertex
a point and basis a Hermitian curve. Let W := Σ4 ∩ S . The intersection W ∩ Σi4, as i
varies, is a Hermitian cone with basis a Hermitian curve, so, the points of W are
|W| = (q2 + 1)q5 + q2 + 1 = (q2 + 1)(q5 + 1);
in particular, W is a hypersurface of Σ4 of degree q + 1 such that there exists a plane
of Σ4 meeting W in just one line (such planes exist in Σ3). Also suppose W to contain
planes and let Π′′′ ⊆ W be such a plane. Since Σi4∩W does not contain planes, all Σi4
meet Π′′′ in a line ti. Also Π′′′ must be contained in
⋃q2+1
i=1 ti. This implies that the set
{ti}i=1,...,q2+1 consists of q2 + 1 lines through a point P ∈ Π \ {P1}.
Furthermore each line ti passing through P must meet the radical line ℓi of the Hermi-
tian cone S ∩Σi4 and this forces P to coincide with P1, a contradiction. It follows that
W does not contain planes.
So by the characterization of H(4, q2) of [3] we have that W is a Hermitian variety
H(4, q2).
52 Ars Math. Contemp. 21 (2021) #P1.04 / 45–55
We also have that |S∩Σ5| = |P1H(4, q2)|. Let now r be any line of H(4, q2) = S∩Σ4
and let Θ be the plane ⟨r, P1⟩. The plane Θ meets Σi4 in a line qi ⊆ S for each
i = 1, . . . , q2 + 1 and these lines are concurrent in P1. It follows that all the points of
Θ are in S. This completes the proof for the current case and shows that S ∩ Σ5 is a
Hermitian cone P1H(4, q2).
(b) All planes Π′′ with m ⊆ Π′′ ⊆ S ∩Σ5 are contained in Σi4 for some i = 1, . . . , q2+1.
We claim that this case cannot happen. We can suppose without loss of generality
m ∩ ℓ1 = P1 and P1 ̸∈ ℓi for all i = 2, . . . , q2 + 1. Since the intersection of the
subspaces Σi4 is Σ3, there is exactly one plane through m in Σ5 which is (q
4+ q2+1)-
secant, namely the plane ⟨ℓ1,m⟩. Furthermore, in Σ14 there are q4 planes through m
which are (q3+q2+1)-secant and q2 planes which are (q2+1)-secant. We can provide
an upper bound to the points of S ∩ Σ5 by counting the number of points of S ∩ Σ5
on planes in Σ5 through m and observing that a plane through m not in Σ5 and not
contained in S has at most q3 + q2 + 1 points in common with S ∩ Σ5. So
|S ∩ Σ5| ≤ q6 · q3 + q7 + q4 + q2 + 1.
As |S∩Σ5| = q9+q7+q4+q2+1, all planes through m which are neither (q4+q2+1)-
secant nor (q2+1)-secant are (q3+q2+1)-secant. That is to say that all of these planes
meet S in a curve of degree q + 1 which must split into q + 1 lines through a point
because of Lemma 2.1.
Take now P2 ∈ Σ24 ∩ S and consider the plane Ξ := ⟨m,P2⟩. The line ⟨P1, P2⟩ is
contained in Σ24; so it must be a (q+1)-secant, as it does not meet the vertex line ℓ2 of
C2 in Σ24. Now, Ξ meets every of Σ
i
4 for i = 2, . . . , q
2 + 1 in a line through P1 which
is either a 1-secant or a q + 1-secant; so
|S ∩ Ξ| ≤ q2(q) + q2 + 1 = q3 + q2 + 1.
It follows that |S ∩ Ξ| = q3 + q2 + 1 and S ∩ Ξ is a set of q + 1 lines all through the
point P1. This contradicts our previous construction.
Lemma 3.7. Every hyperplane of PG(6, q2) meets S either in a non-singular Hermitian
variety H(5, q2) or in a cone with vertex a point over a Hermitian hypersurface H(4, q2).
Proof. Let Σ3 be a solid satisfying condition (S4). Denote by Λ a hyperplane of PG(6, q2).
If Λ contains Σ3 then, from Lemma 3.6 it follows that Λ∩S is a Hermitian cone PH(4, q2).
Now assume that Λ does not contain Σ3. Denote by S
j
5 , with j = 1, . . . , q
2 + 1
the q2 + 1 hyperplanes through Σ14, where as before, Σ
1
4 is a 4-space containing Σ3. By
Lemma 3.6 again we get that Sj5 ∩ S = P jH(4, q2). We count the number of rational
points of Λ∩ S by studying the intersections of Sj5 ∩ S with Λ for all j ∈ {1, . . . , q2 + 1}.
Setting Wj := Sj5 ∩ S ∩ Λ, Ω := Σ14 ∩ S ∩ Λ then
|S ∩ Λ| =
∑
j
|Wj \ Ω|+ |Ω|.
If Π is a plane of Λ then Ω consists of q + 1 planes of a pencil. Otherwise let m be the
line in which Λ meets the plane Π. Then Ω is either a Hermitian cone P0H(2, q2), or q+1
A. Aguglia et al.: On Hermitian varieties in PG(6, q2) 53
planes of a pencil, according as the vertex P j ∈ Π is an external point with respect to m or
not.
In the former case Wj is a non singular Hermitian variety H(4, q2) and thus |S ∩Λ| =
(q2 + 1)(q7) + q5 + q2 + 1 = q9 + q7 + q5 + q2 + 1.
In the case in which Ω consists of q+1 planes of a pencil then Wj is either a P0H(3, q2)
or a Hermitian cone with vertex a line ℓ and basis a Hermitian curve H(2, q2).
If there is at least one index j such that Wj = ℓH(2, q2), then there must be a 3-
dimensional space Σ′3 of S
j
5 ∩Λ meeting S in a generator. Hence, from Lemma 3.6 we get
that S ∩ Λ is a Hermitian cone P ′H(4, q2).
Assume that for all j ∈ {1, . . . , q2 + 1}, Wj is a P0H(3, q2). In this case
|S ∩ Λ| = (q2 + 1)q7 + (q + 1)q4 + q2 + 1 = q9 + q7 + q5 + q4 + q2 + 1 = |H(5, q2)|.
We are going to prove that the intersection numbers of S with hyperplanes are only two
that is q9 + q7 + q5 + q4 + q2 + 1 or q9 + q7 + q4 + q2 + 1.
Denote by xi the number of hyperplanes meeting S in i rational points with i ∈ {q9 +
q7 + q4 + q2 + 1, q9 + q7 + q5 + q2 + 1, q9 + q7 + q5 + q4 + q2 + 1}. Double counting
arguments give the following equations for the integers xi:
∑
i xi = q
12 + q10 + q8 + q6 + q4 + q2 + 1∑
i ixi = |S|(q10 + q8 + q6 + q4 + q2 + 1)∑
i=1 i(i− 1)xi = |S|(|S| − 1)(q8 + q6 + q4 + q2 + 1).
(3.2)
Solving (3.2) we obtain xq9+q7+q5+q2+1 = 0. In the case in which |S ∩ Λ| = |H(5, q2)|,
since S ∩ Λ is an algebraic hypersurface of degree q + 1 not containing 3-spaces, from
[19, Theorem 4.1] we get that S ∩Λ is a Hermitian variety H(5, q2) and this completes the
proof.
Proof of Theorem 1.1. The first part of Theorem 1.1 follows from Lemma 3.4. From
Lemma 3.7, S has the same intersection numbers with respect to hyperplanes and 4-spaces
as a non-singular Hermitian variety of PG(6, q2), hence Lemma 2.5 applies and S turns
out to be a H(6, q2).
Remark 3.8. The characterization of the non-singular Hermitian variety H(4, q2) given
in [3] is based on the property that a given hypersurface is a blocking set with respect to
lines of PG(4, q2), see [3, Lemma 3.1]. This lemma holds when q > 3. Since Lemma 3.2
extends the same property to the case q = 3 it follows that the result stated in [3] is also
valid in PG(4, 32).
4 Conjecture
We propose a conjecture for the general 2n-dimensional case.
Let S be a hypersurface of PG(2d, q2), q > 2, defined over GF(q2), not containing d-
dimensional projective subspaces. If the degree of S is q+1 and the number of its rational
points is |H(2d, q2)|, then every d-dimensional subspace of PG(2d, q2) meets S in at least
θq2(d − 1) := (q2d−2 − 1)/(q2 − 1) rational points. If there is at least a d-dimensional
54 Ars Math. Contemp. 21 (2021) #P1.04 / 45–55
subspace Σd such that |Σd ∩ S| = |PG(d − 1, q2)|, then S is a non-singular Hermitian
variety of PG(2d, q2).
Lemma 3.1 and Lemma 3.2 can be a starting point for the proof of this conjecture since
from them we get that S is a blocking set with respect to lines of PG(2d, q2).
ORCID iDs
Angela Aguglia https://orcid.org/0000-0001-6854-8679
Luca Giuzzi https://orcid.org/0000-0003-3975-7281
Masaaki Homma https://orcid.org/0000-0003-4568-6408
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