AbstractWe generalize and complete Ferri's characterization of the finite quadric Veronesean V24 by showing that Ferri's assumptions also characterize the quadric Veroneseans in spaces of even characteristic

Thas, J.A.

Van Maldeghem, H.

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On Ferri's characterization of the finite quadric Veronesean V24 

Journal of Combinatorial Theory, Series A 110 (2005) 217–221
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On Ferri’s characterization of the ﬁnite quadric
Veronesean V42
J.A. Thas, H. Van Maldeghem
Galglaan 2, B-9000 Gent, Belgium
Received 1 May 2003
Communicated by Francis Buekenhout
Available online 11 January 2005
Abstract
Wegeneralize and complete Ferri’s characterization of the ﬁnite quadricVeroneseanV42 by showing
that Ferri’s assumptions also characterize the quadric Veroneseans in spaces of even characteristic.
© 2004 Elsevier Inc. All rights reserved.
MSC: 51E20; 51B99; 51E25
Keywords:Veronesean cap; Quadric Veronesean
1. Introduction
Let q be a ﬁxed prime power. For any integer k, denote by PG(k, q) the k-dimensional
projective space over the ﬁnite (Galois) ﬁeldGF(q) of q elements.We choose coordinates in
PG(2, q) and in PG(5, q). TheVeronesean mapmaps a point of PG(2, q)with coordinates
(x0, x1, x2) onto the point of PG(5, q) with coordinates
(x20 , x
2
1 , x
2
2 , x0x1, x0x2, x1x2).
The quadric Veronesean V42 is the image of the Veronesean map. The set V42 is a cap of
PG(5, q) and has a lot of other nice geometric and combinatorial properties, summarized in
[2].We also refer to [2] for characterizations of this cap, sometimes called aVeronesean cap.
In particular, there exists a characterization of V42 in terms of the intersection numbers of a
E-mail addresses: jat@cage.ugent.be (J.A. Thas), hvm@cage.ugent.be (H. Van Maldeghem).
0097-3165/$ - see front matter © 2004 Elsevier Inc. All rights reserved.
doi:10.1016/j.jcta.2004.10.009
218 J.A. Thas, H. Van Maldeghem / Journal of Combinatorial Theory, Series A 110 (2005) 217–221
hyperplane which is valid for q odd. It was ﬁrst considered and proved by Ferri [1]; the proof
in [2] is much shorter because Hirschfeld and Thas make use of the other characterizations.
Also, the proof of Ferri did not work for q = 3; see [1]. Recently, the authors proved a new
characterization of the ﬁnite quadric Veroneseans, and they will use it here to generalize
Ferri’s result to all q.
We now prepare the statement of our Main result.
2. Main result
Recall from [2] that the quadric Veronesean V42 is a cap K in PG(5, q) satisfying the
following two properties:
(VC1) For every hyperplane  of PG(5, q), we have | ∩ K| = 1, q + 1 or 2q + 1, and
there exists some hyperplane  such that | ∩K| = 2q + 1.
(VC2) Any plane of PG(5, q) with four points in K has at least q + 1 points in K.
It is also proved in [2] that these two properties characterize V42 for all odd q; Ferri [1] had
proved this for all odd q = 3. In the present paper we will prove this for all q. In fact, we
will be able to copy the proof in [2] for the general case (now relying on the Main results
of [4]) except for q = 4, for which we produce a separate argument.
So we obtain the following general characterization:
Theorem 2.1. Let K be a set of points of PG(5, q), q > 2, satisfying (VC1) and (VC2).
Then K is projectively equivalent with the quadric Veronesean V42 in PG(5, q). For q = 2,
a set of points in PG(5, 2) satisfying (VC1) and (VC2) is either a quadric Veronesean or
an elliptic quadric in some subspace PG(3, 2).
3. Proof of the main result
We now prove Theorem 2.1.
Let K be a set of points of PG(5, q), q > 2, satisfying (VC1) and (VC2) (see above).
We ﬁrst prove that K is a (q2 + q + 1)-cap. This follows from the results in [2] if q = 4.
So we ﬁrst deal with the case q = 4.
In the next three lemmas, we assume that q = 4 and that K satisﬁes (VC1) and (VC2).
We adopt the terminology of [2]: a solid is a 3-dimensional subspace of PG(5, 4), while a
prime is a 4-dimensional subspace of PG(5, 4).
Lemma 3.1. K generates PG(5, 4).
Proof. By (VC1) the set K does not generate a line. Assume that K generates a plane 2.
By Lemma 25.3.5 of [2] there is a line L of 2 with |L ∩ K| ∈ {2, 3}. Let 4 be a prime
which contains L but not 2. Then |4 ∩ K| ∈ {2, 3}, contradicting (VC1). Next, assume
that K generates a solid 3. Then |K| = 9 and each plane of 3 has one or ﬁve points in K.
Let p and p′ be distinct points of K. Suppose that the line pp′ = L has b2 points in K.
Counting the points ofK in the planes of 3 through the line L, we obtain 5(5−b)+b = 9,
J.A. Thas, H. Van Maldeghem / Journal of Combinatorial Theory, Series A 110 (2005) 217–221 219
whence b = 4. Let L ∩K = {p, p′, p′′, p′′′} and let 2 ∩K = {p, p′, p′′, p′′′, r}, with 2
some plane of 3 through L. Then the line rp has only 2 = b points in K, a contradiction.
Finally, assume that K generates a prime 4. By (VC1) we have again |K| = 9 and each
solid 3 of 4 has one or ﬁve points inK. Let L be a line having at least 2 points inK, and let
2 be a plane of 4 containing L. Further, let |L ∩K| = a and |2 ∩K| = b. Counting the
points of K in the solids of 4 containing 2, we obtain 5(5− b)+ b = 9, whence b = 4.
Counting the points of K in the planes of 4 containing L, we obtain 21(4 − a) + a = 9.
Consequently a = 15/4, a contradiction. The lemma is proved. 
Lemma 3.2. K is a cap.
Proof. Let L be a line. By Lemma 25.3.2 of [2] we have either L ⊆ K or |L ∩K|3.
First assume that L∩K = {p, p′, p′′}. Choose points r1, r2, r3 onK \ {p, p′, p′′} so that
〈L, r1, r2, r3〉 is a prime 4. Then |4 ∩ K| = 9. Necessarily 〈L, ri〉 contains ﬁve points
of K, i = 1, 2, 3 (use (VC2)). The solid 〈L, r1, r2〉 contains either seven or eight points. If
〈L, r1, r2〉 contains eight points, then it contains the three planes 〈L, ri〉, i = 1, 2, 3, so it
contains nine points, a contradiction. Hence |K ∩ 〈L, r1, r2〉| = 7. Considering the primes
containing 〈L, r1, r2〉 there arises |K| = 17. Now we project K \ L from L onto a solid 3
skew to L. There arises a set K′ of size 7 in 3 which intersects each plane of 3 in either
one or three points. By [3] such a set K′ does not exist.
Next, assume that K contains a line L. Choose points r1, r2, r3 ∈ K \ L such that
〈L, r1, r2, r3〉 generates a prime 4. Then |4 ∩K| = 9. Let (K∩ 4) \L = {r1, r2, r3, r4}.
By the preceding paragraph r4 /∈ 〈L, ri〉, i = 1, 2, 3, as otherwise there is a line containing
exactly three points of K. Now we project K \ L from L onto a solid 3 skew to L. There
arises a set K′ which intersects each plane of 3 in either one or four points. By [3] such a
set K′ does not exist.
The lemma is proved. 
Lemma 3.3. The cap K contains exactly 21 points.
Proof. Put |K| = k. Let 14,24, . . . be the primes of PG(5, 4), and let si be the number
of points of K in i4. Counting in two ways the number of ordered pairs (p,i4), with
p ∈ K ∩ i4, we obtain
1365∑
i=1
si = 341k.
Counting in two ways the number of ordered triples (p, p′,i4), with p, p′ ∈ K ∩ i4, and
p = p′, we obtain
1365∑
i=1
si(si − 1) = 85k(k − 1).
220 J.A. Thas, H. Van Maldeghem / Journal of Combinatorial Theory, Series A 110 (2005) 217–221
The setK is a cap; so counting in two ways the number of ordered 4-tuples (p, p′, p′′,i4),
with p, p′, p′′ ∈ K ∩ i4, and p = p′ = p′′ = p, we obtain
1365∑
i=1
si(si − 1)(si − 2) = 21k(k − 1)(k − 2).
Since si ∈ {1, 5, 9} for all i, we have
1365∑
i=1
(si − 1)(si − 5)(si − 9) = 0.
Hence
1365∑
i=1
si(si − 1)(si − 2)− 12
1365∑
i=1
si(si − 1)+ 45
1365∑
i=1
si − 61425 = 0.
We obtain, substituting the previous equalities,
21k(k − 1)(k − 2)− 1020k(k − 1)+ 15345k − 61425 = 0.
Hence 7k3 − 361k2 + 5469k − 20475 = 0. It follows that k = 21 or k = 25.
Assume that k = 25. If 3 is a solid which contains a6 points of K, then |K| = 25 =
a + 5(9− a), so a = 5, a contradiction. If 2 is a plane which contains at least four points
of K, then 2 contains at least ﬁve points of K (by (VC2)), so there exists a solid which
contains at least six points of K, a contradiction. Hence any four points of K are linearly
independent.
Let p be a ﬁxed point of K. Let ci be the number of primes of PG(5, 4) which contain p
and intersect K in i points, i = 1, 5, 9. Counting pairs {p′,4} with p′ ∈ K, p = p′, with
4 a prime and p, p′ ∈ 4, we obtain 4c5+ 8c9 = 2040. Counting triples {p′, p′′,4} with
p′, p′′ ∈ K,p = p′ = p′′ = p, with4 a prime andp, p′, p′′ ∈ 4, we obtain 6c5+28c9 =
5796. Counting quadruples {p′, p′′, p′′′,4} with p′, p′′, p′′′ ∈ K, p, p′, p′′, p′′′ distinct,
4 a prime and p, p′, p′′, p′′′ ∈ 4, we obtain 4c5 + 56c9 = 10120, clearly contradicting
the previous equalities.
So we conclude that k = 21 and the lemma is proved. 
Now it is clear that Lemmas 25.3.10–25.3.13 of [2] hold for all q3. In particular, this
means that there are exactly q2 + q + 1 planes of PG(5, q) meeting K in an oval (which is
a (q + 1)-arc), and every pair of points of K is contained in exactly one such plane. Also,
two such planes meet in exactly one point, which belongs toK. LetK be as in Theorem 2.1
and suppose q > 2. By the proof of Theorem 25.3.14 of [2], we now also have that every
three planes of PG(5, q) that intersect K in an oval generate PG(5, q). By Theorem 1.3 of
[4],K either is the quadricVeronesean V42 or q = 4 andK is the unique 2-dimensional dual
hyperoval of PG(5, 4). As in the latter case (VC2) is not satisﬁed, we proved Theorem 2.1
for all q > 2.
Finally suppose q = 2. We use similar terminology as before. Let 4 be a prime of
PG(5, 2) containing 5 points of K. If these ﬁve points generate 4, then, considering the
J.A. Thas, H. Van Maldeghem / Journal of Combinatorial Theory, Series A 110 (2005) 217–221 221
three primes through a solid contained in 4 and itself containing four points of K, it is
easily seen that |K| = 7 and every six points of K generate PG(5, 2). In this case K is a
skeleton and hence isomorphic to the quadricVeronesean V42 . So we may assume that these
ﬁve points do not generate 4. Clearly this implies |K| = 5. It is now an easy exercise to
see that K generates a solid and is an elliptic quadric in that solid (because every plane of
that solid contains either one or three points of K).
The proof of Theorem 2.1 is complete.
References
[1] O. Ferri, Su di una caratterizzazione graﬁca della superﬁcie di Veronese di un S5,q , Atti Accad. Naz. Lincei
Rend. 61 (1976) 603–610.
[2] J.W.P. Hirschfeld, J.A. Thas, General Galois Geometries, Oxford University Press, Oxford, 1991.
[3] J.A. Thas, A combinatorial problem, Geom. Dedicata 1 (1973) 236–240.
[4] J.A. Thas, H.VanMaldeghem, Characterizations of the ﬁnite quadricVeroneseansV2nn , Quart. J. Math. Oxford
Ser. 55 (2004) 99–113.


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On Ferri's characterization of the finite quadric Veronesean V24

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