Regular pseudo-hyperovals and regular pseudo-ovals in even characteristic

0.11 MB·

by  J. A. Thas

REGULAR PSEUDO-HYPEROVALS AND REGULAR PSEUDO-OVALS IN EVEN
CHARACTERISTIC
7
1 J.A.THAS
0
2

b
e
F
ABSTRACT. S.RotteyandG.VandeVoordecharacterizedregularpseudo-ovalsof
3 PG(3n−1,q),q=2h,h>1andnprime.Hereanalternativeproofisgivenand
slightlystrongerresultsareobtained.

]
O
C 1. INTRODUCTION
.
h
Pseudo-ovalsandpseudo-hyperovalswereintroducedin[10];seealso[12]. These
t
a objectsplayakeyroleinthetheoryoftranslationgeneralizedquadrangles[6,12].
m Pseudo-hyperovals only exist in even characteristic. A characterization of regular
[ pseudo-ovalsinoddcharacteristicwasgivenin[2];seealso[12]. In[8]acharac-
terizationofregularpseudo-ovalsandregularpseudo-hyperovalsinPG(3n−1,q),
2
q even, q 6= 2 and n prime, is obtained. Here a shorter proof is given and slightly
v
6 strongerresultsareobtained.
2
1
9 2. OVALS AND HYPEROVALS
0
. Ak-arcinPG(2,q)isasetofk points,k ≥ 3,nothreeofwhicharecollinear. Any
1
0 non-singular conic ofPG(2,q)is a(q+1)-arc. If K isany k-arc of PG(2,q), then
7 k ≤q+2. Forq oddk ≤q+1andforq evena(q+1)-arcextendstoa(q+2)-arc;
1 see[3]. A(q+1)-arcisanoval;a(q+2)-arc,qeven,isacompleteovalorhyperoval.
:
v AfamoustheoremofB.Segre[9]tellsusthatforq oddeveryovalofPG(2,q)isa
i
X non-singular conic. For q even, thereare many ovals that are not conics [3]; also,
therearemanyhyperovalsthatdonotcontainaconic[3].
r
a
3. GENERALIZED OVALS AND HYPEROVALS
Arcs, ovals and hyperovals can be generalized by replacing their points with m-
dimensionalsubspacestoobtaingeneralizedk-arcs,generalizedovalsandgeneral-
izedhyperovals. Thesehavestrongconnectionstogeneralizedquadrangles,projec-
tiveplanes,circlegeometries,flocksandotherstructures. See[6,12,10,11,2,7].
Below, some basic definitions and results are formulated; for an extensive study,
manyapplicationsandopenproblems,see[12].
A generalized k-arc of PG(3n − 1,q), n ≥ 1, is a set of k (n − 1)-dimensional
subspaces of PG(3n−1,q) every three of which generate PG(3n−1,q). If q is
oddthenk ≤qn+1,ifqiseventhenk≤qn+2. Everygeneralized(qn+1)-arcof
PG(3n−1,q),q even,canbeextendedtoageneralized(qn+2)-arc.
1
2 J.A.THAS
If O is a generalized (qn + 1)-arc in PG(3n − 1,q), then it is a pseudo-oval or
generalized oval or [n−1]-oval of PG(3n−1,q). For n = 1, a [0]-oval is just an
ovalofPG(2,q). IfOisageneralized(qn+2)-arcinPG(3n−1,q),qeven,thenit
isapseudo-hyperovalorgeneralizedhyperovalor[n−1]-hyperovalofPG(3n−1,q).
Forn=1,a[0]-hyperovalisjustahyperovalofPG(2,q).
If O = {π0,π1,··· ,πqn} is a pseudo-oval of PG(3n−1,q), then πi is contained
in exactly one (2n− 1)-dimensional subspace τ of PG(3n − 1,q) which has no
i
point in common with (π0 ∪π1 ∪···∪πqn)\πi, with i = 0,1,··· ,qn; the space τi
isthetangentspaceofOat π . Forq eventheqn+1tangentspacesofOcontaina
i
common(n−1)-dimensional space πqn+1, thenucleusofO; also, O∪{πqn+1}isa
pseudo-hyperovalofPG(3n−1,q). Forqodd,thetangentspacesofapseudo-oval
Oaretheelementsofapseudo-ovalO∗ inthedualspaceofPG(3n−1,q).
4. REGULAR PSEUDO-OVALS AND PSEUDO-HYPEROVALS
In the extension PG(3n− 1,qn) of PG(3n −1,q), consider n subplanes ξ , i =
i
1,2,··· ,n,thatareconjugateintheextensionFqn ofFq andwhichspanPG(3n−
1,qn). ThismeansthattheyformanorbitoftheGaloisgroupcorrespondingtothis
extensionandspanPG(3n−1,qn).
In ξ consider an oval O = {x(1),x(1),··· ,x(1)}. Further, let x(1),x(2),··· ,x(n),
1 1 0 1 qn i i i
with i = 0,1,··· ,qn, be conjugate in Fqn over Fq. The points x(i1),x(i2),··· ,xi(n)
define an (n − 1)-dimensional subspace π over F for i = 0,1,··· ,qn. Then,
i q
O = {π0,π1,··· ,πqn} is a generalized oval of PG(3n−1,q). These objects are
theregularorelementarypseudo-ovals. IfO isreplacedbyahyperoval,andsoq is
1
even,thenthecorrespondingOisaregularorelementarypseudo-hyperoval.
Allknownpseudo-ovalsandpseudo-hyperovalsareregular.
5. CHARACTERIZATIONS
LetO={π0,π1,··· ,πqn}beapseudo-ovalinPG(3n−1,q). ThetangentspaceofO
atπ willbedenotedbyτ ,withi=0,1,··· ,qn. Chooseπ ,i∈{0,1,··· ,qn},and
i i i
letPG(2n−1,q)⊆PG(3n−1,q)beskewtoπ . Further,letτ ∩PG(2n−1,q)=η
i i i
and hπi,πji∩PG(2n−1,q) = ηj, with j 6= i. Then {η0,η1,··· ,ηqn} = ∆i is an
(n−1)-spreadofPG(2n−1,q).
Now,letqbeevenandletπbethenucleusofO. LetPG(2n−1,q)⊆PG(3n−1,q)
be skew to π. If ζj = PG(2n−1,q)∩hπ,πji, then {ζ0,ζ1,··· ,ζqn} = ∆ is an
(n−1)-spreadofPG(2n−1,q).
Next, let q be odd. Choose τ , with i ∈ {0,1,··· ,qn}. If τ ∩τ = δ , with j 6= i,
i i j j
then{δ0,δ1,··· ,δi−1,πi,δi+1,··· ,δqn}=∆⋆i isan(n−1)-spreadofτi.
Finally, let q be even and let O = {π0,π1,··· ,πqn+1} be a pseudo-hyperoval in
PG(3n−1,q). Choose π , with i ∈ {0,1,··· ,qn +1}, and let PG(2n−1,q) ⊆
i
PG(3n−1,q) be skew to π . Let hπ ,π i∩PG(2n−1,q) = η , with j 6= i. Then
i i j j
{η0,η1,··· ,ηi−1,ηi+1,··· ,ηqn+1}=∆i isan(n−1)-spreadofPG(2n−1,q).
Theorem 5.1 (Casse, Thas and Wild [2]). Consider a pseudo-oval O with q odd.
Thenatleastoneofthe(n−1)-spreads∆0,∆1,··· ,∆qn,∆⋆0,∆⋆1,··· ,∆⋆qn isregular
REGULARPSEUDO-HYPEROVALSANDREGULARPSEUDO-OVALSINEVENCHARACTERISTIC 3
ifandonlyiftheyallareregularifandonlyifthepseudo-ovalOisregular. Insucha
caseOisessentiallyaconicoverF .
qn
Theorem 5.2(RotteyandVandeVoorde[8]). Considerapseudo-ovalOinPG(3n−
1,q)withq =2h,h>1,nprime. ThenOisregularifandonlyifall(n−1)-spreads
∆0,∆1,··· ,∆qn areregular.
6. ALTERNATIVE PROOF AND IMPROVEMENTS
Theorem 6.1. Considerapseudo-hyperovalOinPG(3n−1,q),q =2h,h>1andn
prime. ThenOisregularifandonlyifall(n−1)-spreads∆ ,withi=0,1,··· ,qn+1,
i
areregular.
Proof. IfOisregular,thenclearlyall(n−1)-spreads∆ ,withi=0,1,··· ,qn+1,
i
areregular.
Conversely, assume that the (n − 1)-spreads ∆0,∆1,··· ,∆qn+1 are regular. Let
O = {π0,π1,··· ,πqn+1} and let Oˆ = {β0,β1,··· ,βqn+1} be the dual of O, with βi
beingthedualofπ .
i
Chooseβ ,i∈{0,1,··· ,qn+1},andletβ ∩β =α ,j 6=i. Then
i i j ij
(1) {αi0,αi1,··· ,αi,i−1,αi,i+1,··· ,αi,qn+1}=Γi
isan(n−1)-spreadofβ .
i
Now consider β ,β ,Γ ,Γ ,α ,j 6= i. In Γ we next consider a (n−1)-regulus γ
i j i j ij j j
containing α . The (n−1)-regulus γ is a set of maximal spaces of a Segre vari-
ij j
ety S1;n−1; see Section 4.5 in [4]. The (n−1)-regulus γj and the (n−1)-spread
Γ of β generate a regular (n −1)-spread Σ(γ ,Γ ) of PG(3n −1,q). This can
i i j i
be seen as follows. The elements of Γ intersect n lines U ,U ,··· ,U which are
i 1 2 n
conjugate in F over F , that is, they form an orbit of the Galois group corre-
qn q
spondingtothisextension. Letα ∩U ={u},withl =1,2,··· ,n. Nowconsider
ij l l
the transversals T ,T ,··· ,T of the elements of γ , with T containing u . The
1 2 n j l l
n planes T U = θ intersect all elements of γ and Γ . The (n−1)-dimensional
l l l j i
subspacesofPG(3n−1,q)intersectingθ ,θ ,··· ,θ aretheelementsoftheregu-
1 2 n
lar (n−1)-spread Σ(γ ,Γ ). The elements of this spread are the points of a plane
j i
PG(2,qn), with as lines the (2n−1)-dimensional spaces containing at least two
(andthenqn+1)elementsofthespread. Hencetheq+2elementsofOˆ containing
an element of γ , say β = β ,β ,··· ,β ,β = β , are lines of PG(2,qn).
j i i1 i2 iq+1 iq+2 j
Dualizing,theelementsπ ,π ,··· ,π arepointsofPG(2,qn).
i1 i2 iq+2
Now consider β and γ , and repeat the argument. Then there arise n planes
i2 j
θ′ intersecting all elements of γ and Γ . The (n−1)-dimensional subspaces of
l j i2
PG(3n−1,q) intersecting θ′,θ′,··· ,θ′ are the elements of the regular (n−1)-
1 2 n
spreadΣ(γ ,Γ ). TheelementsofthisspreadarethepointsofaplanePG′(2,qn),
j i2
with as lines the (2n−1)-dimensional spaces containing qn +1 elements of the
spread. Hence β ,β ,··· ,β are lines of PG′(2,qn). Dualizing, the elements
i1 i2 iq+2
π .π ,··· ,π arepointsofPG′(2,qn).
i1 i2 iq+2
First,assumethat{θ ,θ ,··· ,θ }∩{θ′,θ′,··· ,θ′}=∅. Considerπ ,π ,π ,π .
1 2 n 1 2 n i1 i2 i3 i4
The planes of PG(3n−1,qn) intersecting these four spaces constitute a set M of
maximalspacesofaSegrevarietyS2;n−1[1]. Theplanesθ1,θ2,··· ,θn,θ1′,θ2′,··· ,θn′
areelementsofM. Itfollowsthat(θ ∪θ ∪···∪θ )∩(θ′ ∪θ′ ∪···∪θ′)=∅.
1 2 n 1 2 n
4 J.A.THAS
Consider any (n−1)-dimensional subspace π ∈ {π ,π ,··· ,π } of PG(3n−
i5 i6 iq+2
1,q). Wewillshowthatπ isamaximalsubspaceofS2;n−1. Letθi∩πj ={tij},θi′∩
π = {t′ },i = 1,2,··· ,n,j = i ,i ,··· ,i . If t t ∩t t = {v },t′ t′ ∩
t′j t′ =ij{v′},with j ,j ,j ,j d1ist2inct,thqe+n2v ,vi,j1··i·j2,v aijr3eicj4onjugaite ainj1disji2m-
iliaj3rlyij4v′,v′,i··· ,v′ ar1ec2onj3uga4te. Hence hv ,v1,··2· ,v i =n hv′,v′,··· ,v′idefines
1 2 n 1 2 n 1 2 n
a(n−1)-dimensional space over Fq which intersectsθ1,θ2,··· ,θn′ (overFqn).The
points t , with j = i ,i ,··· ,i , generate a subplane of θ , and the points
ij 1 2 q+2 i
t′ , with j = i ,i ,··· ,i , generate a subplane of θ′, with i = 1,2,··· ,n. Let
ij 1 2 q+2 i
q = 2h and let F2v be the subfield of Fqn = F2hn over which these subplanes are
defined;sov|hn. Thenv <hnasotherwisethespreadsofPG(3n−1,q)definedby
θ ,θ ,··· ,θ and θ′,θ′,··· ,θ′ coincide, clearly not possible. The (n−1)-regulus
1 2 n 1 2 n
γ impliesthatthesubplanescontainaline overF ,so h|v. Asn isprimewe have
j q
v = h, so 2v = q. Hence the 2n subplanes are defined over F . It follows that
q
the q +2 elements π ,π ,··· ,π are maximal subspaces of the Segre variety
i1 i2 iq+2
S2;n−1. Hence π is a maximal subspace of S2;n−1. It follows that π1,π2,··· ,πq+2
aremaximalsubspacesofS2;n−1.
NowconsideraPG(2,q)whichintersectsπ ,π ,π ,π . The(n−1)-dimensional
i1 i2 i3 i4
spaces πi1,πi2,··· ,πiq+2 are maximal spaces of S2;n−1 which intersect PG(2,q);
theyaremaximalspacesoftheSegrevarietyS2;n−1∩PG(3n−1,q)ofPG(3n−1,q).
Considerπ andalso aPG(2n−1,q)skewtoπ . Ifweprojectπ ,π ,··· ,π
i1 i1 i2 i3 iq+2
fromπ ontoPG(2n−1,q),thenbytheforegoingparagraphtheq+1projections
i1
constitutea(n−1)-regulusofPG(2n−1,q). Similarly,ifweprojectfromπ ,sany
is
element of {1,2,··· ,q+2}. Equivalently, if s ∈ {1,2,··· ,q+2} then the spaces
β ∩β ,witht=1,2,··· ,s−1,s+1,··· ,q+2,forma(n−1)-regulusofβ .
is it is
Now assume that the condition {θ ,θ ,··· ,θ }∩{θ′,θ′,··· ,θ′} = ∅ is satisfied
1 2 n 1 2 n
foranychoiceofβ ,β ,γ ,β . Insuch acase every(n−1)-regulus containedina
i j j i2
spreadΓs definesaSegrevarietyS2;n−1overFq. Letusdefinethefollowingdesign
D. Pointsof D are theelements of Oˆ, ablock ofD is a set of q+2 elements of Oˆ,
containingatleastonespaceofa(n−1)-reguluscontainedinsomeregularspread
Γ ,andincidenceiscontainment. ThenDisa4−(qn+2,q+2,1)design. ByKantor
s
[5]thisimpliesthatq =2,acontradiction.
Consequently,wemayassumethatforatleastonequadrupleβ ,β ,γ ,β wehave
i j j i2
′ ′ ′
(2) {θ ,θ ,··· ,θ }={θ ,θ ,··· ,θ }.
1 2 n 1 2 n
Insuchacasetheqn+2elementsofOˆ arelinesoftheplanePG(2,qn). Itfollows
thatOisregular. (cid:4)
Theorem 6.2. Considerapseudo-ovalOinPG(3n−1,q),withq =2h,h>1andn
prime. ThenOisregularifandonlyifall(n−1)-spreads∆0,∆1,··· ,∆qn areregular.
Proof. IfOisregular,thenclearlyall(n−1)-spreads∆0,∆1,··· ,∆qn areregular.
Conversely, assume that the (n−1)-spreads ∆0,∆1,··· ,∆qn are regular. Let O =
{π0,π1,··· ,πqn}, let πqn+1 be the nucleus of O, let O¯ = O∪{πqn+1},let Oˆ be the
dualofO,letOˆ¯ bethedualofO¯,andletβ bethedualofπ .
i i
Chooseβ ,i∈{0,1,··· ,qn+1},andletβ ∩β =α ,j 6=i. Then
i i j ij
(3) {αi0,αi1,··· ,αi,i−1,αi,i+1,··· ,αi,qn+1}=Γi
REGULARPSEUDO-HYPEROVALSANDREGULARPSEUDO-OVALSINEVENCHARACTERISTIC 5
isan(n−1)-spreadofβ .
i
Now consider β ,β ,Γ ,Γ ,α , with j 6= i and i,j ∈ {0,1,··· ,qn}. In Γ we next
i j i j ij j
considera(n−1)-regulusγj containingαij andαj,qn+1. The(n−1)-regulusγj is
a set of maximal spaces of a Segre variety S1;n−1. The (n−1)-regulus γj and the
(n−1)-spreadΓ ofβ generatearegular(n−1)-spreadΣ(γ ,Γ )ofPG(3n−1,q).
i i j i
Such as in the proof of Theorem 6.1 we introduce the elements U ,u ,T ,θ ,l =
l l l l
1,2,··· ,n,andtheplanePG(2,qn). Theq+2elementsofOˆ¯ containinganelement
ofγj,sayβi =βi1,βi2,··· ,βiq,βj =βiq+1,βqn+1,arelinesofPG(2,qn). Dualizing,
theelementsπi1,πi2,··· ,πiq+1,πqn+1 arepointsofPG(2,qn).
Now consider β and γ , and repeat the argument. Then there arise n planes θ′
i2 j l
of PG(3n − 1,qn) intersecting all elements of γ and Γ , and a (n − 1)-spread
j i2
Σ(γ ,Γ ) of PG(3n−1,q). The elements of this spread are the points of a plane
PGj′(2,iq2n). Thespacesβi1,βi2,··· ,βiq+1,βqn+1 arelinesofPG′(2,qn). Dualizing,
theelementsπi1,πi2,··· ,πiq+1,πqn+1 arepointsofPG′(2,qn).
First,assumethat{θ ,θ ,··· ,θ }∩{θ′,θ′,··· ,θ′}=∅. Considerπ ,π ,π ,π .
1 2 n 1 2 n i1 i2 i3 i4
The planes of PG(3n−1,qn) intersecting these four spaces constitute a set M of
maximal spaces of a Segre variety S2;n−1. The planes θ1,θ2,··· ,θn,θ1′,θ2′,··· ,θn′
areelementsofM. Itfollowsthat(θ ∪θ ∪···∪θ )∩(θ′ ∪θ′ ∪···∪θ′)=∅. Let
1 2 n 1 2 n
π ∈{πi5,πi6,··· ,πiq+1,πqn+1}. AsintheproofofTheorem6.1oneshowsthatπ is
amaximalsubspaceofS2;n−1. Itfollowsthatπi1,πi2,··· ,πiq+1,πqn+1 aremaximal
subspacesofS2;n−1.
NextconsideraPG(2,q)whichintersectsπ ,π ,π ,π . The(n−1)-dimensional
i1 i2 i3 i4
spaces πi1,πi2,··· ,πiq+1,πqn+1 are maximal spaces of S2;n−1 which intersect the
planePG(2,q);theyaremaximalspacesoftheSegrevarietyS2;n−1∩PG(3n−1,q)
of PG(3n−1,q). Such as in the proof of Theorem 6.1 it follows that the spaces
βqn+1∩βit,witht=1,2,··· ,q+1,forma(n−1)-regulusofβqn+1.
Now assume that the condition {θ ,θ ,··· ,θ }∩{θ′,θ′,··· ,θ′} = ∅ is satisfied
1 2 n 1 2 n
for any choice of β ,β ,γ ,β , j 6= i and i,j ∈ {0,1,··· ,qn}. Let α ,α ,α be
i j j i2 1 2 3
distinct elements of Γqn+1. Then βi,βj,γj,βi2 can be chosen in such a way that
α ∈ β ,α ∈ β ,α ∈ γ ,β ∩β ∈ γ with α ∈ β . Hence the (n−1)-regulus
1 i 2 j 2 j i2 j j 3 i2
in βqn+1 defined by α1,α2,α3 is subset of Γqn+1. From [4] now follows that the
(n−1)-spread Γqn+1 of βqn+1 is regular. ByTheorem6.1 the pseudo-hyperovalO¯
is regular, and so O is regular. But in such a case the condition {θ ,θ ,··· ,θ }∩
1 2 n
{θ′,θ′,··· ,θ′}=∅isneversatisfied,acontradiction.
1 2 n
Consequently,wemayassumethatforatleastonequadrupleβ ,β ,γ ,β wehave
i j j i2
{θ ,θ ,··· ,θ }={θ′,θ′,··· ,θ′}. Insuchacasetheqn+2elementsofOˆ¯ arelines
1 2 n 1 2 n
oftheplanePG(2,qn). ItfollowsthatO¯,andhencealsoO,isregular. (cid:4)
Theorem 6.3. Considerapseudo-hyperovalOinPG(3n−1,q),q =2h,h>1andn
prime. ThenOisregularifandonlyifatleastqn−1elementsof{∆0,∆1,··· ,∆qn+1}
areregular.
Proof. IfOisregular,thenclearlyall(n−1)-spreads∆ ,withi=0,1,··· ,qn+1,
i
areregular.
Conversely, assume that ρ, with ρ ≥ qn−1, elements of {∆0,∆1,··· ,∆qn+1} are
regular.
6 J.A.THAS
Ifρ=qn+2,thenOisregularbyTheorem6.1;ifρ=qn+1,thenOisregularby
Theorem6.2.
Nowassumethatρ=qnandthat∆2,∆3,··· ,∆qn+1areregular. Wehavetoprove
that ∆ is regular. We use the arguments in the proof of Theorem 6.2. If one of
0
the elementsα ,α ,α , say α , in the proofof Theorem6.2 is β ∩β , thenlet γ
1 2 3 1 0 1 j
containβ ∩β ,β ∩β ,β ∩β andletβ 6=β ,withi,j ∈{2,3,··· ,qn+1}. Now
j i j 0 j 1 i2 1
seetheproofoftheprecedingtheorem.
Finally, assume that ρ = qn−1 and that ∆3,∆4,··· ,∆qn+1 are regular. We have
to prove that ∆ is regular. We use the arguments in the proof of Theorem 6.2.
0
If exactly one of the elements α ,α ,α , say α , in the proof of Theorem 6.2 is
1 2 3 1
β ∩β or β ∩β , then proceed as in the preceding paragraph with β 6= β ,β .
0 1 0 2 i2 1 2
Now assume that two of the elements α ,α ,α , say α and α , are β ∩β and
1 2 3 1 2 0 1
β ∩β . Nowconsiderall(n−1)-reguliin∆ containingα andα ,andassume,by
0 2 0 1 3
wayofcontradiction,thatnoone ofthese(n−1)-reguli containsα . Thenumber
2
of these (n−1)-reguli is qn−2, and so q = 2, a contradiction. It follows that the
q−1
(n−1)-regulus inβ definedbyα ,α ,α iscontained in∆ . Nowwe proceedas
0 1 2 3 0
intheproofofTheorem6.2. (cid:4)
7. FINAL REMARKS
7.1. Thecasesq =2andnnotprime
Forq =2ornnotprimeotherargumentshavetobedeveloped.
7.2. ImprovementofTheorem6.3
LetD=(P,B,∈)beanincidencestructuresatisfyingthefollowingconditions.
(i) |P|=qn+1,q even,q 6=2;
(ii) the elements of B are subsets of size q +1 of P and every three distinct
elementsofP arecontainedinatmostoneelementofB;
(iii) Q is a subset of size δ of P such that any triple of elements in P with at
mostoneelementinQ,iscontainedinexactlyoneelementofB;
Assumption : Any such D is a 3−(qn+1,q+1,1) design whenever δ ≤ δ with
0
δ ≤q−2.
0
Theorem 7.1. Consider a pseudo-hyperoval O in PG(3n − 1,q), q = 2h,h > 1
and n prime. Then O is regular if and only if at least qn + 1 − δ elements of
0
{∆0,∆1,··· ,∆qn+1}areregular.
Proof. SimilartotheproofofTheorem6.3. (cid:4)
7.3. Acknowledgement
WethankS.RotteyandG.VandeVoordeforseveralhelpfuldiscussions.
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