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SOME TOPICS m FmITE GEOMETRICAL STRUCTURES
A. Barlotti
University of North Carolina
(on leave from the University of Florence, Italy)
Institute of Statistics Mimeo Series No. 439
August 1965
This research was supported by the National Science
Foundation Grant No. GP-3792.
DEPARTMENT OF STATISTICS
UNIVERSITY OF NORTH CAROLINA
Chapel Hill, N. C.
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TABLE OF C<ETENTS
CHAPTER
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II
PAGE
FINITE WEAK AFFINE SPACES • • • • • • • • • • • • • • • • •
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Review of finite projective and affine spaces ••
Definition of finite weak affine spa'ces • • • • •
Examples of S-spaces. • • • • • • • • • • • • • •
An open question. • • • • • • • • • • • • • • • •
A more general geometric structure of R.C. Bose •
I-l
I-3
I-6
I-l4
I-l5
·. .....
II-l
k-ARCS IN A PLANE • • • • • • • • • • • • • •
...
k-arcs in a projective plane. • • • • •
II-l
Some properties of k-arcs • • • • • • • • • • • •
II-l.
Some properties of (q+l)-arcs, in a plane
of order q( q odd) • • • • • • • '. '. • • • • • • 1I-3
On q-arc of PG(2, q) with q odd• • • • • • • • • •
1I-9
II-12
Complete k-arcs other than conics • '. '• • • • • •
II-l4
(q+2) -arcs with q even. • • • • • • • '. • • • • •
II-l6
Diophantine system relative to a k-arc'. • • • • •
II-19
Index and rank of a k-arc • • • • '. • • • • • • •
Examples of k-arc other than ovals. • • • • • • •
1I-22
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III
(3.l)
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A PROJECTIVE PrANE• • • • • • • • • • • •
III-l
Definitions • • • • • • • • • • • • • • • • • • •
III-l
THE (k,n)-ARCS
(3.2) An upper bound for the number of points of
(3.3)
(3.4)
(3.5)
(3.6)
a (k,n)-arc• • • • • • • • • • • • • • • • • •
Some properties of (k,n)-arcs with
q == 0 (mod n) • • • • • • • • • • • • • • • • •
Same examples of (k,n)-arcs • • • • • • • • • • •
On the non-existence of certain (k;n)-arcs. • • •
An open question on the relationship between
certain (k,n)-arcs and algebraic curves. • • •
v
(5.3)
(5.4)
(5.5)
III-ll
IV-l
k·arcs in PG(r, q) • • • • • • • • • • • • • • • •
(k,n)-arcs in PG(r,q) • • • • • • • • • • • • • •
IV-l
IV-2
k-CAPS IN PG(r,q), r ~ 3• • • • • • • • • • • • • • • • • •
(5.1)
(5.2)
1II-4
1II-6
1II-7
>3 • • • • • • • • • •
k-ARCS AND (k,n)-ARCS IN PG(r,q), r
(4.l)
(4.2)
III-l
Definitions • • • • • • • • • • • • • • • • • • •
Sore results on maximal k-caps in PG(3, q) • • • •
Upper bounds on k-caps contained in a quadric • •
other results on k-caps contained in a
quadric in PG(3, q) • • • • • • • • • • • • • •
Upper bounds for the number of p'oint'sin
a k-cap in PG(r,q) • • • • • • • • • • • • • •
(cont~ued)
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V-13
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':CABLE OF CONTENTS (continued)
Lower bounds for the number of points 'in
a maximal k-cap in PG(r,q), r ~ 4• • • • • • •
Exarrq;>les of complete k-caps other than
ova1oids • • • • • • • • • • • • • • • • •
The Segre-Tits ova1oid. • • • • • • • • • • • • •
VI
k-SETS OF KIND s • • • • • • • • • • • • • • • • • • • •
(6.1)
(6.2)
(6.3)
(6.4)
(6.5)
APPENDIX•• • •
.
REFERENCES. • • •
·· •• ••
. .
····•
·· · · · · · · ·
• . . · . . · • ·
·• • • • ·• • ·
.... ·..• • • • • •
Definitions • • • • • • • • • • • • • • • •
Embedding a K(s) in a set of lower kind • •
An upper bound for the number of points
in a k-set •
• • • • • • • •
• •
Lower bounds for the number of points
in a maximal k-set •
• • • •
• •
The Bose-Chaudhuri sets
• • •
•
•
•
• • • • • •
.....• • •
·.
V-27
V-29
V-37
VI-I
VI-I
VI-I
VI-6
VI-8
VI-9
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ACKNOWLEDGMENT
The author wishes to acknowledge with sincere thanks the help
received from Mr. T. A. Dowling in the preparation of these lecture-notes.
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CHAPl'ER I
FnTITE WEAK AFF:rnE SPACES
(1.1)
1.
Review of finite ;Erojective and aff':ine spaces.
Axioms for the finite projective space.
We silB.1l calsider tlD
classes of objects, P and.t , at least one of wl1l.ch is non empty, and which
do not have any element in
in
P x
'!'he elements of P
are called "paints"
are called "lines:' We luppose given a relation "incidence",
and those at!
E ,
CannDl.
,~
, such that a given point and a given line are either inciThe geometrical structure ( P , ,~
dent or not incideDt.
to be a f1nite projective SPace,
1,
,
E )
is then said
if the following axioms ue satisfiedJ
(Pl) There il exactly one line which is incident with each of' t10
distinct paints.
(P2) (Veblen)
If a line intersects two sides at a triangle (not at
their intersection), it also intersects the third side.
(P3) (Fane)
Ivery line oontains at least three points.
(p4) '!'here is at least one line, the number of points incident with
which is finite.
It is easy to see that
Pl, P2, P3 and p4 imply that every line oon-
tains the same number of points.
If' this number is s+l, the integer
s
is
called the order of the space.
We make the following definiticns:
Definition (1.1.1),. "fiat" is a set of points of
1
such that, if' it
contains two distinct points A and B, it also ccntains all paints on the
line joining A and B.
Definition (1.1.2>' 1
is finite-d1menaiCllal i f there is a finite set of
points such tbat any nat which Caltains them contains the whole space.
If
I
N+ 1
is the slllLllest number of points for which this is true, then N is
caJJ.ed the dimension of
2t •
Every flat of a finite dimensional space is also a finite dimensial8.l
space, whose dimension is defined be the dimensicn of the flat.
Planes and
h;yperplanes are flats of d.inension two and N-l respectively.
2. The finite affine space.
Starting from a projective space an
*
~-l'
affine space,J: , can be obtained by deleting a hyperplane,
points and lines in it.
and aU
The dimension and the order of this affine space are
defined to be the d1mensiQl and the order of the original projective space.
In addition to the incidence relation, wbich renains unchanged, we
have a new relation of parallelism.
betl'Teen flats of any dimension.
In general parallelism can be defined
We Shall, mwever, require only the definition
for lines:
J: are called ''parallel'' it the lines
Definition (1.1.3)Two lines of
of
2t
*
"B-1
from which they come, are incident in
In every affine space the following properties ho1d:
i) There is exactly one line which is incident "lith each of two
points.
ii) Every line contains exactly
s
points, where
s
is the order
of the space.
iii) Parallelism is an equivalence relation.
iV) Given a point
M and a line b, there is exactly me line, m ,
which is incident with
M and parallel to
v) The number of points of J: is
sN.
vi) Through everJ point there pass exactly
N
s - 1
s - 1
lines.
1_2
b.
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3. Theorem of Desarques.
We sha.ll use the notation (BC,B'C') to denote the intersectim of the
lines
BC and B'C'.
Desargues' theorem state s :
trian~es
such that the lines
if ABC
AA', BB', CC'
and A'B'C'
are two (ncn degenerate)
pass thrcngh the same point 0,
tllen (BC,BtCt), (CA,CtAt), (AB,AtB t ) lie on the same line.
It is well known that v4rlle there exist projective and affine planes
in l'liuch the theorem of Desargues is not valid, a projective or an affine space
of dimension iri 3 is necessarily desarguesian.
(1.2) Definiticm of finite weak aftine spaces.
1. Introduction.
Recently E. Sperner looked at the possibility of
constructing geometric stroctures (weak. af:f'1ne spaces) which are ncn-desarguesian
and, in sane cases, can be considered as having dimension greater tl'an two (1).
The cJ.ass of these structures incJ11des:
a) the affine spaces of an,y dimension iri 3;
b) the affine place s associated with a ternary ring (2);
c) many new classes of structures 'Which can be considered as nondesarguesian spaces.
In the finite case the struc1llres indicated in c) provide examples of
resolvable designs which are mt isomorphic to any a.ffine space EG(R,pn) (3).
(1) See: Sperner E.: [1], [2], [3].
(2) See: Hall M. Jr. [I], 20,:5; Pickert, G. [1], 1.5; or the appendix
by L. Lombardo Radice in Segre, B. [13], A2., A3.
(3) See also, for resolvable designs non isomorphic to any EG(n,pn),
Bose, R.C. [3].
I-.3
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2. Axioms far the ,-reak aftine space of Sperner
We shall consider t'WO classes of objects which do not have any element
in canmon:
a) A class, P , of "points" which we shall indicate with capital
letters:
A, B,C, ... ;
b) a class, .s= , of "lines": a,b,c, . . . . .
We suppose given:
I) a relation"1ncidence" in
P x
, such that a given point and
:
a given line are either incident or not incident (we will use respectively
the symbols A
E
b
and A _ b).
II) a relation "para.Uelism" in S. x !.
are parallel or not (in symbols:
The
a
II
b
such tlBt two given lines
or a}t b).
geometrical structure S. ( P, : ,
E,
II)
is then said to form
a weak affine space of Sperner, which we will call an S-space, if the following
axioms are satisfied.
(Al)
There is eJCactl;y one line which is incident
~th
each of two
distinct points.
(A2) Each line ccntains eDct]y s<!...g) points.
The cardinal
s
of the points of a line can of course be infinite,
but in what follows we will alNS suppose tmt
s
s
is fin! tee
The integer
is called the order of the S-space.
(A3)
Parallelism is an equivalence relation.
(A4) To each line, a, and each point, B1 there exists one 8Di ally
me line b, such that B
E b
and a
L.l.!..
Using (A') and (A4) we find that
an
b
a
II
b
means either a • b
or
=¢ •
~:
The
axioms (Al) - (A4) are the same as properties (i) - (iv) tor
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affine spaces in (1.1).
Thus every ordinary affine space is an S-lpace.
In what tollows we shall smw that there ext,st other classes of S-spaces.
3.
The dimension of S-BP!ces
ot finite order.
From (A4) it to1lmrn that through every point in an S-space of erder
s
there pass the same nwnber of lines.
Let this number be
k.
There are
tlIree possibilities:
1)
k
is tinite and of the torm
2)
k
is finite but not of the
3)
k is infinite.
If It
is
n.
is ot the form 1), we shall say that the e:ti.m!nsion of the space
Note that from property vi) in (1.1) the number of lines thrcllgh a
points in an ordinary finite affine space of dimension n
If
k
is c£ thia form.
is at the form 2), the space is called irregular, with
respect to its dimension, and we shall not attempt to define a c1.1nension.
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If k
is infinite, the dimension of the space is infinite.
An S-space is called finite if
k is finite.
the space IIIlst have a finite number of points.
if k
is finite,
the order
munber of points on a line b
s
is finite.
I f 1-;:
is finite, then
Also from(AJ.) we see that
s < k, fer the
In particular
is no greater tlB.n the number of liDes through
a point P not on b. Since one at the
k lines thrmgh P is parallel to
b, there can be no more than k-l points on b.
The number of points in a finite S-space of order
s
is
since thrmgb any point P pass k liDes, each of wtxI.ch contains
(s-l)k + 1,
a-1 points
other than P.
i'heorem (1.2.1) Every S-space of dimension two is an aff'ine plane.
Proof:
It auf'f'1ces to prove tlat if two lines 1B.ve no points in
1-5
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conanon, they are parallel.
be a point on b.
all the
a
and
b
are t'\iO such lines.
Let p
Since the d:imensia1 of the space is t'\«>, the number of
(s2-1)/s-1 = s+l.
lines through P is
wi th the points of
Suppose
By (Al)
of these lines jaln P
s
a, and one of them is parallel to
s+l lines thrcugh P, it follows that
A s:hn11 ar resul.t
Since th! se are
and b
are parallel.
to Theorem 1.2.1 for S-spaces at dimens:la:1 greater
~les
than two is not valid.
a
a.
of S-spaces of dimension three whtch are not
affine spaces are given in the following section.
For eDMPles of S-spaces with irregular dimensiCl1, see R.C. :Bose
[1], [4].
An exatlq)le of an i.n:f'in1te-dimensicna1 S-space is givm in Bar10tti
[5] n.ll.
parallism defined in this space, is isormorphic to an aff'ine k-flat, then
this set is defined as an a:f'f1ne k-flat in the B-space.
Ex!.DE1es of S-spaces.
:l!ix:anJ?le 1 (Sperner [1], § 3)
The construction of thi.s first e:x.&nJ?le of an
S-space is related to the ortllographic projections of
ge~try.
Monge :in
descriptive
It is possible to obtain such a ccnstruction for the n-eUmensicneJ.
case (Barlotti [5], n.10), but far simplicity we shall consider only the
tbree-d1mensiCll&1 case here.
Let CX:J.'
choose a line
~,
~
and
in
ex:;
be three affine planes of the same order s.
CX:J. and a line
k2 in
~.
Because Cl:L and
~
We
are of
the same order, it is possible to set up a (1-1) correspondence between the
points of
~
and
~.
B',r identi1'ying these correspmding points, we can
join tIle t'\«) planes so that they intersect in the
1-6
Canmal
line
k=~=~.
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I f a set of points and lines in an S-space, with incidence and
(1.3)
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a
i
(~=1,2)
we choose a set of parallel lines Li
If ~ (1-1,2) is a. point in
to k.
passes tl1rcugh Qi'
we define
ai' and
"i
is the ]jne of L
Z(~) to be the intersect:1al of "i
,
An ordered pair of points (~,~) such that
Qi c a i (i-.1.,2) and
i
wh1ch
and
It.
Z(~,>.Z(~)
is defined to be a point of S-space.
Ther e are two different kinds of lines in the S-space.
We.b&1.l refer
to these as lines of the first kind aDd Un•• of the seccnd kind.
A line of the first kind is an ordered pair of l:Ines (gl'~) luch tlBt
gi c a i (i=1,2) and gi ~ Li (i-1,2).
Incidence between points and lines of the first ldnd il defined as
A point (\,~) is inci,dent with a line (gl'~) if an,d only if ~
follOl'1S:
We write (~,~) c
is incident with gi in a i (i=l.,2).
(gl'~).
Two l.ines (gl'~) and (~,~) of the first kind are parallel i f ~
only if
~i
is parallel. to
hi
in a i (i-l.,2).
we
write (gl.'~) II(~,~).
The set of points (~,~,) and lines (gl.'~) thus far described do not
have the structure of an S-space.
We still need to define the line joining
tlro points (~,~) and (Q{,~) with
Z(~) = Z(~,>.
These lines w:Ul be
those of the second kind.
To define lines of tIE second kind, we censider a fixed point Z en
k and the set
plane
~,
l.et
any point in
In
a
i
~
Hz of all points
~
and
~
(~,~) such tha.t
Z(\)
be two sets of parallel l.ines.
as the intersection of a line r l. C
~
III
Z.
two (l.-l.) correspondences ep'i
and a line
(i=l.,2) between the llnes of Ki
I.
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1-7
In the tl11rd
We can represent
(i=l.,2) we consider the set of l.:1nes K parallel. to l~.
i
Ri (see figure l..3.1).
I
which &1"e nat parallel
r 2 c R2 •
We can make
and those of
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k
Z, lre ccnsider a point (~,~) in the set
For a fixed point
To tIn s point we can associate a unique point of
the line of Ki
<1t CPi
€
Ri • The lines
unique point (Ql li'l' ~q>2) of ~.
the points of
Nz
Z
a
gives us a point AI
of
Z
al =
Z
CAIZ
I A e a)
in~.
Mz.
as follows.
Let
Then to the line
Nz.
<l:i
be
<l:i e Ki
ClfPi (i=1,2) intersect in a
We shall denote tI1i.s correspondence bett-reen
and the points of
Consider a line
I
~ (i=1,2).
which passes, through
corresponds a line
~
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I
~
by
-1
IZ •
To each point A of a
the correspondence
We define the set of points
to be a line of the second ldlld.
The incidence relaticn between points and line s of the secood kind
is clear from the above definition.
Two llnes of the second kind
h I Z• are parallel if and only if g
II
h in
g I
Z
and
a,.
It is easy to verify that all the axioms of an S..space ere satisfied
by the set of points and tlro sets of lines wbi ch mve been described.
c;,.
~,
and
a,
If
are desarguesian plAnes, the S..space is ord:tnary aff:Jne space
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of three d1mens:kms.
desar~esien,
at least ale of these p1azles is ncz-
If, however,
then we have an exaq>le of an S-space wili ch is not an ordinary
at:f'i.ne space.
~,,~, ~
To see this, let
and suppose that
(Qi, Z( Q{»
~
be three non-collinear points of <X:L
is non-desarguesian.
and (Qi, z(~) )
(~, z( Q,..L) ),
Then the t l1:t'ee points
are non-collinear and defme
a 2-flat in the S-space which 1s isorDDrphic to Cl:J. and is therefore nondesarguesian.
Note that the dimension of this S-space is three, since thrrogh a
fixed point there pass
s2 lines of the first kind and s+l lines of the
sS_l
second kind, for a total of s2+s+l = - 1 •
sIn connection with this example, we have the following:
Theorem 1.,.1.
Through every point of this S-space there pass at
least 2s+l planes.
We shall prove that through any point there pass at least
Proof&
s
planes isomorphic to C%.I.' at least
one plane isamrphic to
s
other planes isomorphic to
~,
and
~.
Let (P ,P ) be a fixed point.
l 2
Nz(P ).
1
We have smwn that there is a (1-1) correspondence betl-leel1 the po:ints of this
set and the points of
~,
Consider the set of points
and this correspondence, in vie'''' of the definition
of a line of the second kind, gives us an iSOIOOrphism.
isomorphic to
ex,.
To :f'1nd a plane isomorphic to
'1.'
plAne through
(pl' P2)
P b~t not in ~.
2
~E~.
calsider a. fixed line
a 2 thrc:ngh
Let (a2 ) denate the set of all points (~,~) with
'rhe lines joininG any two points of (a2 ) with different ~ are all
lines of the kind
t\·70
Thus there is one
points of
(a1''2) ,·r.l.th ~
('2) with the same
a:-bitrary in
~
1-9
a.t.
The 1inl!s joining en.y
are obtained as follows:
Let
(~,~)
I
and (Q{,~) be two such points.
The line joining these tw points is
line in
a,
joining the points
-1
( ~,~ ) tz(~)
r 2 IZ(~)' '\-'here
and
r
is the
2
( . ) -1
~,~ IZ(~) (see figure 1.3.2)
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Let us denote these two sets of lines by 11 and 11 , respectively.
2
1
Every two lines 'Jf ,the set 111 U N2 which are not parallel intersect
in a point of the set
(~).
The, set of points
(~)
and the set of lines
11 U N2 therefore give us a plane. This plane is iSOIlX)rphic to CX:I., the
1
isoIOOrphism being defined by the correspondence
There are
through P2.
s
different
~s
Thus there are at least
in which we ccnld choose the line
s
~
planes thrcugh (P ,P ) isomorphic
1 2
to ~.. Simi1.a.rly there are at least s planes through (P ,P2 ) isomorphic
1
to ~. Therefore they are at least 2s + 1 planes through a point of the
space.
We state the fol.l.ow:i.ng theorem without proof:
TlEorem 1.'.2.
If there are JOOre than
I-10
2s + 1 planes through a
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point of tIlLs S-space, then
Let 1: ( P, l
E?!atIi>le 2
d
>2
and order
s
~
and ~ are isomrphic (See Barlotti [5]n.7).
II)
,6 ,
be an ordinary aft.1.ne space of cUmena:lon
3. Ccmsider a plane a in 1:
a. Starling with
on
c;.
1:, we can obtain an
and tl«> points
A and B
S-space S( P , l , c'
cl1aneing the incidence relation in the following way:
,II)
by
The incidence v1ll be
the same except tor the lines wtxl.ch lie in a and pass through A and mt
through B or PUll thrCllgh B and not through A.
J'or a line of a
A (not AB), the incidence is changed by :replacing A with
B.
through
S1mUarly far
a ]Jne of a through B.
It is easy to see that the tour postulates for S-spaces are satisfied..
a
Clearly the plane
remains desarguesian after tm ch8l1ge, but the space is
no J.onBer an ordinary att1ne space since the Veblen axiom is not satisfied..
!o see th1 s, let
r
be a line in a
through A but not through B (when
reGarded as a line of 1:), and let u be a line not in a
through A.
~
Then r
lines of
r
and
u.
r
and u.
~
and
u
determine a plane
~
in 1:.
which paslles
Let
a
and b be
not throueJ,1 A wtnch intersect in e, point belonging to neither
In the S-space, the lines
a
and
b int.ersect, but not the lines
1'hus the VebJ.en axtom for an ordinary aff'.i.ne space is not
sa.tisfied.. (see figure 1.3.3).
L
---
I-ll
I
lxa!!()le 3
A third eDJIIille of an S-space can be Calstructed by an algebraic
(see Sperner [3], § 1).
method based on a finite near-fiel4/. A ••t of elements K(al~,11".) il a
finite near-field if tbere are defined two binary operations e&lled &dditl.an
and multiplication which satisfy the following axioms.
1)
J[
i. an abelian group with respect to addit:lon.
(The identity
of this group i. denoted by 0.)
2)
Th. non-zero element. of
K
form a group with respect to multi-
plication.
a.
3) 'or every
K,
o.a. a.o •
0 •
4) a (~ + ,) • a ~+ a 1
Using tb11 finite near-field, we can construct an S-space in the
following way:
Consider the vector space V of all n-tuples
(<;, ~I".,a~),
a
• K (i==1,2, ... ,n). Additim of two vectors is defined in the uau&1 way.
i
For any vector g • V and any scalar ~. J[, we define right scalar multiplication as
g~.(<;,A,~~,.",~~) •
The points of our S-space are vectors belonging to
two distinct points g and
J!
v.
~
line joining
any
is the set of points of the form
g + (! - 9 ) ~,
where ~ c
J[
•
We JDI18t show tlat every tllO pointe of the line determine the same set
of points.
Let Z and! be two other points on the line joining
Then
z • g +(! -
g) ~.
! • g +(! - g) \
'
and therefore
! -
z..
•
(! -
9>
~ 2 - (! - g) ~ 1
(~- g) (~2 - ~)
I-12
g
and
! .
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z + (~+z)p.
[g + (~ - g)'Al ] +
(2 - g) \- ~)p
•
g + (~ - g) [ \+( ~ - \ ) p ]
•
g + (~ - g)
say,
J.1
g
and therefore is on the line joining
every point on the line joining g and
and~.
~
Similarly we can sbow that
is on the line joining
Z md ! .
Thus oo.r definition of a line is invariant under different choices of the
t1'10 initial points.
It is apparent from this proof that axiom (A]J for S-spaces is satisfied.
Also axiom ~ is satisfied, for from
g+
it follows that
(~ -
z) 'A. g +(~ - z)
'A. J.1e
J.1
,
Therefore there are s points on a lme, where s
is the order of K.
g
The line joining
and! and the line joining Z and
pare.llel if and ally if there eJdsts an element
'A
E
K ('A
Ia
~
are called
0) such that
(~ - g) 'A. (A - z) •
-
Since tins relation is reflexive, ByDDDetric, and transitive, it follows tbi,t
&Xiam (A.3J is satisfied.
We need only to show tlBt for each line
there is exactly one line parallel to
g
the line joining g and 1!, then the line h
through Z
and is parallel to
g.
g
and each point
and passing through
joining
Z and 1
Z. If g is
+<.~
assume that the line joining 9 and! and the line joining Z and
It foll.aws that
or
winch implies that
.§ •
! •
Z' V
- g) puses
To show that this parallel is unique,
-Z +(2z -
(2 - g) 'A • .§ -
I.
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Z and ! is of the torm
Then every point on the line joining
g)'A ,
h •
I-13
!
are parallel.
I
We have shown" th&t all the axioms are valid; and there fore this set
of paints and lines with the relations as defined calstitutes a Sperner space.
If' we joing a fixed
The number of points in the space is clearly s.
point
P to each of the other sn_1 points of the space" each line is replicated
s-l times" since there are s-l points on each line in addition to P.
It follows
n
that the mnnber of lines through P
dimension of the space 1s
is
a -1
s-l
" and therefore that the
n.
Note that if we regard points as treatments and lines as blocks" a
finite Sperner space is equivalent to a resolvable baJ.anced incoq>lete block
design ,,71th
1. 4)
~..
1.
An ORen queaticm.
Through every point of a :r1n:lte Sperner space pass a certain number
(possibly zero) of planes.
Consider the set of number (nc) where nc .. the
nwnber of planes pass1ng through the point C.
bound of the numbers in
(1~3)
{n
c
} and
m denote the lower
M the upper bCWld.
Example 2 1n sectiCll
showed that even if M is the same as for ordinary affine space" the
8perner space my not be affine.
through every point
in
Let
In fact" in this exanple we can show that
C on the lineAB .(C
Ia
A" C .; B) pass as many planes as
r.
We state witbout proof the following theorem concerning the lower
bound m (see Barlotti [5]" n.5).
Theorem 1.4.1
If'the d1mensiCX1 of an S-space of order
s
is three,
and if m > s2+8 -l, the space is an ordinary affine space.
An open question is to determine whether or not it is possible to
improve this result.
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(1.5)
A I!K)re general pometric structure of R. C. Bose.
R. C. Bose has studied a more general geometric structure.
considers a set of points, a set of lines and an incidence relation, 1I1ich
satisfy the following a.x:l.OI:lS:
Bl.
Two points
are .joined by at mst one line.
B2.
There are
r
lines incident with every point.
B3.
There are k
points incident with every line.
B4. Given a line J
are exactl,y t
lines
(t
il;
and a point
P
1) passing tl'xrough
(See R. C. Bose [6]).
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Be
I-15
not incident with', there
P and intersecting
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CHAPTER II
k-ARCS Dr A POOlE
(~~.l)
l,:-arcs in a projective plane
In 1Tda.t follows we shall demtc b:,' :n:{q)
q , While
order
q.
He
He
I)G{2, q)
::;ilall use t11e symbol
be(~in
any projective plane of order
only for a desarcuesian plane of
''1i ttl two definitions:
Definition (2.1.1)
A k-arc ·01' :n:{q) is a set of
k
points of :n:{q)
such that no tllree of them are collinear.
Definition (2.1.2J
no k -arc, Hi tjl
l
(2.2)
1:
1
>k
A k-arc in said to be complete if tJlere e:d sts
, which conta.ins it.
Some properties of l:-arcs
A line of the plane will meet a Civen k-arc in hro, one, or zero
points.
Tilesc lines are called
secants~
tangents, and external lines, respect-
ively, to the C;iven k-arc.
Theorcia (2.2.1)
throuGll a roint
Proof:
P
In
:n:{q)
tile number of tangents of a l:-arc passing
not on tIle arc ,las
SUP110se there are
11
t~le
same parity as
k.
secants passing through
P.
Tile number
of points of t.ile l:-arc Which are inciCi.cnt llith these secants passinG
P
is 211.
T,le lines joining the other ::-211 points of the arc ,lit'll
thrOUt~_l
P
must
be tanGent::;.
We lrill consider the question of determing the maxill1U1:l value of
for w11ich at-arc "rill exist.
k
In tile ::.:;enera.l case, we 118.ve the follolTing
theorem:
Tileorcm (2.2.21 (R. C. Bose [2]0.
If there exists a k-arc in :n:(q),
then if q is odd, k'::::: q + 1; if q is even, k:S q
of
2.
I
Proof:
Consider a point
The secants tllrouc;l1
P
points of the k-o.rc.
P
on a Ie-arc.
o
are obtained b:,r joinine;
o
Through
P
0
P
pass q+l lines.
0
"Ti th tile
All the remaininG lines are tangents.
1:-1 other
Tile nm:lber of
tl1ese is
q + 1 - (k-l)
But tl1is number is nonneeativee
=q
- k + 2
Tilcrefore
k:S:q+2.
If
q
i;:; even, it is not possible to improve this result.
clearly exists in
A (Q+2,)-arc
PG(2,2), since (me of t,le axioms for a projective plane
states that tl1ere exist four point::::._ no t·,J.ree of Which are collinear.
Since tile number of tangents t,11':Jn[:;11 a point on a I,-arc is
q-I:+2,
a (q+2}-arc ~lo.S no tangents.
If
('
is odd, q+2
is odd, o.nc1
1);)r
number of tancents passing tl1rOuGh a point
contradicts tlle fact tha.t a (Q+2)-arc
a. (q+2) -arc
(l.OCS
dllS
Tneorem (2.2.1) tl1ere are an odd
P
not on tile (q+2)-arc.
no tangents.
Tl1erefare if
But this
q
is odd,
not exist, and so :L:1 tJ ds ca.se
Ie
:s:
q + 1 •
Tne e::::i.stence of (q+l)-arcs i:1 tile (lcsarcuesian planes fol1011:::: from
Tl1eorc.1 (2.2.3)
In
PG(2,':l), un irreducible conic is a h+l) -arc.
1"01' l)rJofs of this tl1eorer'1: Gee R. C. Bose [5], and B. SeCre [h], page
121, or [1:;] n. 1'(3.
'inet,er or not tl1ere exist «(.1.+1) -arCG in evelJr projective :plane
is still all 02.)cn qucstjon.
desarcuesin.l1
l~lalles
Example;::;
or
(11-:-1) arcs in some rmrt:i.cEln.:·.'
1f( q)
ll·:m.-
are given in A. Ui..10lCl' [1] and G. Rodxignes [1].
A cJ.llGf.lification of tile (q+l)-urcs (ovals) in finite non-dcsm~cuesian
planes is.ivcn b~r T. G. Ostrom (scc T. G. Ostrom [1]).
II-2
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Defil1ition (2.2.l)
according as
q
(2.3)
The k-arcs of ~(q) having q+l or q+2 points,
is odd or even, are called ovals.
Some properties of (c;+l)-arcs, in a plane of order
In ~(q), if
Dleorem (2.3.1).
q
(<t odd).
is odd, no three tangents of a
q
(q+l)-arc pass tIlrouCh the same point of the plane.
Proof:
Suppose there exists a point
tyro tangents of tIle oval.
Let
t
sects tIle oval in a single point
points of tlle oval other tl1&n
points on
t
pass tl1rougil
besides
through which pass more than
lJe one of these tangents.
t
inter-
T;le tangents passing throuGh t:le
Q.
Q :',Ulst intersect
t.
But tilen
t
q
Since there are
P, tilen there is at least one point A on
(~+l
Tl1en
Q, it fol101'TS that if tyro or nDre of t'ilese
pass none of t'jleSe tangents.
and since
P
t
q
q
tangents
tilrouell whicIl
is the only tangent tilroUc;ll
A,
is even, tIlis contl~a(!.icts Theorem (2.2.1).
Thus the points which do not belong to an oval fall into t110 classes:
those incident lritil two tangents of tile oval, and those incident -r;rith no
tane:ents.
Tllese points are called external points and internal points,
respectivel;y:.
Since ever<J t't'ID distinct tanc;ents of tIle oval determine an e::cternal
point, the l1'l1nber of external
..
po~nts ~s
clearly
of internal points is obtained by subtraction as
Lemma (2.3.2).
any tI1ree points
S",
(Lemma of
~2and ~3
tanGen~s;
( q+1) g,{q+l)
2
=2:
•
Tile number
q2+ q+l_(q+l)_ q(q+l) = q(g,-l)
.22
B. Segr e [2], [3]).
If in PG(2,q)
of a l:-arc arc chosen as the f'unda~ntal points
(1,0,0), (0,1,0), and (0,0,1) of the s:,rstem of reference, and the q-1:+2
tangents to t',1e l:-arc at t1'lese tI1ree points are written in the forn
(") An anal,vtical criterian iThic11 i'Till Show Whether a point is all
external or internal point to a given conic in PG(2,q), q Odd, is in Qvist [1],
n. 9.
II-3
(-)
I
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(i, ;],1,1
= 1,2, ••••••• ,
q-k+2), t.,lCl1
n
1{
r,s
rs
=-1.
For the prDof Df this lemma, see B. Sccre [13], n. 17 (*) •
~leorem
2.3.3
(B. Segre[l])
D1
PG(2.q), if
q
is odd, ~len every
(q+1)-arc (oval) is an irreducible conic.
The first proof of this theoren can be found in B. SeC;re [1] or [2J;
see also B. SeGre [13], n.174.
Proof:
We s}1O.11 cive an alternative proof (A. Cossu[l]).
Consider the Galois field of order
q
= pn,
and denotc its non-
b~l
zero elements
c l ,c2 ,···,c q _l •
Clearly for
i
I
(j fixed), ue imve
j
i - cj
C
If
i
I
h, and
j
I
0 •
is fixed,
ilnve
l'Te
c. - c.
J
n
and
- c .•
J
The product of all the non-zero elements of the field is -1, i.e.,
= -1
(*) For
t.:uc study of questions rclated to k-arcG it is ver:: iUl!ortant
to have a hl0"lTlecice of this and otilcr proofs [in particular those of
(2.3.3), (2.4.3), (2.5.5) (2.6.5)] uhich vie omit because they can
be fOtmd in Chapter 17 of the bo01: h:/ B. Sec;re [13].
~1eorems
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Therefore
q-1
n
-c.
J i=l
(c -c.)=
i J
-1
Consider a (q+1)-arc in the plane_
\ole can choose two points of the
arc as
= (1,0,0)
Al
A2 = (0,1,0)
~
The point
at A
and A -
1
= (0, 0,1)
Every point
2
coordinates non-zero_
we can take as the intersection of tile tangents
a of the arc other than
A
1
and
A
2
has all three
Therefore we can write the coordinates of such a point
as
a = (~ , Yrj/ 1) The tangent at
Cf,
will be called
nle coefficient t a
then
than
A
1
s
and A
ext3
tllan Al or
=
Tile
s<.xp
its equation beine;
,
GF(q) , for if
and hence will not be a taneent_
\x = a ,
We will
to:.
have the form
2
- Yex
up
x p - Xo:
x
13
p = (x.~ ,"';y' /?/1)
and
Note that
A 2
were true, sa:;r
I
secants of the arc "rllich join ex to the points of tIle arc other
q-2
v
where
~
a will also pass through
The
i
is a non-zero element of
't
find the value of
T
is another point of the arc other
1-'
X
t3
F X cx
)
Yt3
= Xo: ' tllen tIle points
are all non-zero, and for :[,ixed
s,
!" ,
l_~
I
1I-5
F Yex
0:"
' for if one of these equalities
t3, and A2 lmuld be collinear.
0:,
and
f3
F ., ,
I
for if
s Gp ~
the tlu-ee points
a, f3 , and
S
C.
'Y
"lOuld be collinear.
'Y
The tangent
T
J
is
""
different fron all the secants, and hence
\( ~ scp·
Therefore
\,
q-2
and the
I ..
Sex ~
are the q-1 non-zero elements of GF (q),
0
and so
q-1
t
ex
Yf3 - Ye:.
n
xf3 -
~=l
= -1
Xc;
~fc.
Usinr, (2.3.1),
we get
q-l
n (v 13 -,,)
"ex
- ya
oJ
~=1
~,ta
= -1
- xex
Dividinr.;, we have
Yex
-~
q-1
Yl? - Ya
=1
x~ - X a:
n
f3=U.
~;a.
Comparinr; this with
(2.3.2),
t
a
i'Te
see that
-"
tJ ,.,
-
=
•
,-"
'"
·c~
The line joining
~
with
~
has tIle equation
Y - mx
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where
Ya
m =x
Q
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Let
Q.
and
R be the intersections of the line joining Al
(z = 0) ,dtll ~(L and
To:, respectively.
Yc;
Q
= (1, x'
c;
R
= (1,
and A
2
Then
0)
-Y
Tnen
~,A2,Q,R
- acx ,
oX
0)
are four points 118.v1ng cross ratio -1. (1)
In symbols,
(~,A2,Q,R) = -1.
We have also proved the following projective property:
(p)
Let
B
l
and
B
2
be two distinct points of an oval, and let
denote tile intersection of
oval other than
B
I
and
~leir
tangents.
Let
B:3
a be a point of the
B , and denote its tangent by
2
To:.
the points
QI :=
R
I
BIB'")
...
n 13-.0;
)
-
Then
Fic::ure 1
(l}por a definition of this concept, see B. Segre, [1.3], n.100.
1I-7
Define
I
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Note tl1at we l1ave not yet used tIle condition that q is odd. If q
Ya
Y
is even , then t c; = - -x
= -xa
• Tilen the tangent at a passes tl1rough
a
a
A7. , and therefore all the tane;ents of tIle arc pass through the
SaIne
J
Hence the arc is not complete, since lTe can adjoin
NOvT sUJ7.Pose
is odd.
q
c ,AI and A2 , the tan..r:;cnt
'Hllere
t,..,
1-'
=
"t"~
If
=(
=
*
t.o the intersection of
Consider the line joinine A2
a13 in a point
*
R.
(a, 13, Q-x-, R*)
=-
By property
1.
_+-
~
A,,!
Ficure 2
If
* = 1:3 ,
Q
tl1en
x_
u;
y
13
- X
'r
13 "a
(q+2)-arc.
=0
I1-8
AI
I
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I
I
-.
xc!13 - x13Ya
xa - x
13
0,
3
is a point of the arc different from
in the point
Q
It intersects
to obtain a
al3 has tIle equation
;/ - Ya
A.,J,
3
13 has tIle equation
at
TIle l:in e
anG. intersects
(J
A
point A •
(p),
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and
have
'\'Te
= t~
\}
*
R_€ Al~ •
and
Consider the conic
~A;
and
~ A
3
follmvs that
~
Ta
T
let
and
C
2
which is tangent at Al
and A
2
, respectively, and which passes thrOuel1 a.
belongs to
13
C •
q> 3, (2) and 13
If
2
be the intersections of AI-3 with
to the lines
From (2.3.3) it
does not lie on
aA '
3
a and T13 ' respectively.
T
Then
Ta = (0,
2ya '
= (0,
T~
1)
2Y13 ' 1).
Note from property (p) that if the four points
from
B
3
onto
T
B ,B , Q',R'.
l 2
a
B ,B ,Q',R'
l 2
are projected
' the four points obtained have the same cross-ratio as
If instead of
a we consider the tangent at A2 , this property
T
s:i1m-Ts that
But this implies
Since
a
is a fixed point and f3
is an arbitrary point on tile arc, we see
til8.t all points of the arc satisfy
'XY
"fl1icj] is a conic.
(2.4)
= xa
Ya
This cor.rpletes tl1e proof of Theorem (2.3.3)
On q-arc of
PG(2,q) with q odd.
vle ha.ve two lemmas:
Lemr:Ja (2.4.1) (B. Segre [3], p... 365). Let
r, ~2, ~3, ~o, ~
be five distinct points
of PG(2, q), no three of which are collinear.
(2) If q=3, the (q+l)-arc is a conic since four points no
of lTilic;l are collinear deterT.line a. non-degenerate conic.
II-9
tl~ee
I
Consider the homoGraphy
accordinG to whicl1
~
11
o
whicl1 fixes the points
"i' (-l)
ho(~2)
~l (~)
0
no(so) =
on
~
This line iTill clearly depend
•
-:>
S1 , S)
and on the four pointG
and
That is,
Consider t'ile line
~
~2, S3
= ~1
= f
= ~3
'0 :,
Let
o
~.
corresponds to
~l,
~
3
,
~
0
, but it does not depend on
t~le
order in iii'lich these last four points are taken (i. e., on the choice of tile
tlJree fixed poi.nts of thellomocrapl1y within the set of rom' points
~
.
';,
)
r,..2
, s..0
.,
Pj,'ool:
)
•
The line
::"r:'
*
~"O
will be denoted 1)y
lve can choose .the s:-Gtem of reference so that tile coordinates of the
fOUl' point s
0:: 1
are
~2(0,1,O)
(1,0,0)
Tllen the coordinates of
~(al,a2'~)
are all non-zero and unequal,
::dncc no t"ilree points in tillS set of fj,ve are collinear.
He find that the coordinates of
* ("
ai ,
~o
Consider t:i.le homoGraphy
~3
and transforms
into
11
3
s.
S*
are
a 22 ' ~2) •
i'll1icll leaves the points
fixed
He find that it transforns
S into t11e point
Similar1J' we can ShOrT that t:i.le points
~:.
and S2
tile line
and
obta.ined from tlle;lo:lOcrap"Jlies
S~ 0
*
, resPectively, lie on
•
Let
six pointG in
11*
(i
= 0,1, ••• ,4)
ro(2, q), no ti1I'ee of which are collinear.
II-10
and
S be
Then the lines
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I
('\iilere
i , i , i3' i
4
l 2
is
aJ.'1~i'
set of four of the indices 0,1, ... , l~) cannot
all coincide.
of tIle lines
We aGain can cll00se the s:;rsten of reference as in Lemma (2.4.1), and let tilC
coordil1ates of
S and g40 be
~(a,.,a2'a.,)
;4o( cl' c 2 ' c 3 ) •
,..
s ( S...1
If' the line
~2 ,;;
_3 , s
,.. 0) ,
,.~
w:iiI
1 C 1 joins the points
~k..,
2)
So(a
,a..., ,B.§
, coincides Hit:1 t he line
~
1 2
g (~,c~,~)
contain the point
vnl:tc~l fixes
c\:~: ~
2c3 a.l ~ - c l ~
c - c
l
3
If tile three points
F
*'
'~l;>o
1:.
0 into g.
-
and
~
':l
alc l - a 2c 2 + (a2 -al )c
3
Sinilarly if the line
3
_(--1
S
-;, , 'c,7 4 , ,,0':> , ',~oo)
;;,
ss*
in the homoeraplly
He find that
2c3a2~ - c2~ - c3a.§
a.5
c3
r::'
S
"iilicll corresponds to
Sl , ~, g40 and transforms
=
( S,S,S,,:>
1 ... 2 ,. 4 eO)
, tins line "Till
c
3
- c
2
.
•
0.
2
3
are collinear, tllen "Te find tha.t
=0
coincides "Tith either
S(S4o,S2,S3,~0) or
•."'" fO~nd t',et
J.....
,,,~
- alc l + (~-~)C2 + ~c3
=0
It follovTs that if all threc of these above equations hold, tllen
c
l
= c
2
= c
3
' wlucll implies that
F. 4 =
~o.
II-11
Thus '\ore have a contradiction.
I
The preceding two ler.1lnas are used in proving the following theorem:
Tlleorem (2.4.3) (B. Secre [3])
If
q
is odd, eveT"./ q-w:,c of
is contained in a conic, uniquely determined if
q
PG(2,q)
2: 5.
For tIle proof of tili s theorem see B. See;re [13], n. 175.
COliJP1ete k-arcs other tllan conics
If tIle number of points in n k-arc is SlMll in cOY:1p3.rison 1'r.i. th the
nLt: lbcr
of points on a. line in·t;:le plane,
to
arc.
t~le
,.,e
can often adjoin additional points
TilC follmdnc; theore11. provides a necessarJr condition for a 1;:-a1'c in a.
<1 to be cOJIp1ete.
pla.ne of order
T'Jlco:.t.'el.l (2.5.1)
A k-arc in a plane
(I
~
If
fJ.
P1'oo:::':
1f(q)
is corrq:>letc onl~r if
2
:is a power of a PyLIC, the inequal i ty is strict.
A l:-nrc is
cOllI.Ple·i~e
onJ.:/ if throngh ever;! point of the plane tiler0
}!!:'.cs ()nc of t1le k(h-l)/2 cecants.
of tIle arc.
Consider a tancent
Tile number of sccants not passing throuGh
t
tilrough a point
P
Since at J.east one of thesc llRtSt pass tl1rouc;h each of the
ot~ler tilB.l.l
q
P
(l\:-l~ (k-2)
t
points of
p, it follows tllat
(1;:-1) (k-2)
2
If
is
q
1f(q), 'Hitil
q
odd, the lines joininc two by two
points c1l05cn arbi tra.ril~r fron an ·::>val
ph"1JIC e::cep~l'
•
is a power of a prime, it is obvious that the equality cannot
ThcorcI.l (2.5.2}. In
the other points of
<1;5
C pass throur.;h all points of the
C.
II-12
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.-
< (k-l)(k-2)
-
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Proof:
Let
C*
denote a subset of
C consistinG of
CJb5
"-
is a point not on
C throuGh irllich pass no secants of
at lea:::;t
secants of
(tJ.-l)/2
C pass tlu-o\l811
C*.
n+l _ '1-1
2
B~{ Tl1eorem (2.3.1)
C.,·
Since both points of alV
=
C which
q+3
2
If
is odd, and
q
K
is a l~-arc of
in an:! oval, then the rna.xir.Iur1 munber of points comr.lon to
q+7
n.::>t '-'
r-reater than
2:J.
Suppose there is an oval
belol1c;inC to
K
rc('1)
K
C.
and an oval is
Denote
K.
Then b JT Theorem (2.5.2), throuGh a point
*
but not to
not contained
C containil1G (q+5¥2 points of
*
tiLts set of common points l)~T
P
C there "Till pass a secant of
have a contradiction, since tIlere will then be three points of
*
C.
K
Thus 11e
which are
collinear.
Tile following theorem is valid for deSarguesian planes onlJ':
Tlleorem(2.5.4).
If
q
~ 7 ('1
odd), there are
~
- arcs not contained in
D.
conic.
Proof:
Let
secants of
P
Since there are ('1+5)/2 points in C*, we have a contradiction.
Corollary (2.5.3).
Proof':
Suppose
Thus there are at most
'1
points in C*.
points.
* there are at least ('1-1)/2 points of
C,
c;iven secant eannot belone to
cannot belong to
P.
..
C be a conic and
C passing tllrouch
P
P.
an external point.
TIlere are
'1-1
I")
c...
If we consider the set of points obtained
b:; includinc only one of eacll pair of points of C through which one of these
secants pass, together with the two points in '-1hic11 tile ta..'1Cents tbrOuc;ll
intersect tlle conic, and
P
itself, then we obtain a
II-J3
q;5 - arc.
P
Since tilis
I
al'C
~2 points of C, it cannot lie in a conic, for two conics
contains
ca.nnot have rore than foul' points in common.
Since
q
2:
7, ~3 > 4 , and
the tileorem is proved.
In (3Cneral, it is not l:ncmn whether the q;5 - arcs obtained in this
'H'E'.;lr
are complete.
The proof of the following theorem , 110irever, shows tllat
in some cases they are.
Tneorem (2.5.5)
(L.
Lombardo Radice [1]).
If
q
= 4h
+3, there exist complete
q~5 _ arcs in PG(2,q).
c;.
For the construction of such arcs, see B. Segre [13], n.176.
(2.6)
(q + 2) - arcs with
When
q
q
even.
is even, instead of Theorem (2.3.1) we have the following
theorem.
Theorem (2.6.1) (Qv.ist [1])
In ~(q), if
q
is even, all the tangents of
a (cl+1)-arc are concurrent.
Proof:
By Theorem (2.2.1) througll every point of a secant of the (q+l)-arc
there passes at least one tangent.
But since there are q+l
tangents it
follows tllat through everJ point on a secant there passes exactly one tancent.
TIlis means that no secant can pass through the point of intersection of 1;\°10
tWlcents.
in th
the
Therefore the lines joining the intersection of any two
q+1
ta~ents
points of the arc are all the tangents of tOile arc.
The point in whic"ll all the tangents of the (q+l)-arc meet is called
tile nucleus of the (q+l)-arc.
Corollary (2.6.2).
L~
~(q),
if
q
is even, every (q+l)-arc is
incomplete and can be uniquely completed to form a (q+2)-arc.
Proof: The points of the (q+l)-arc, together with the nucleus, obviously
form a (q+2)-arc.
11-14
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Theorem (2.6.3)
~
-I-
In n(q), with
q
even, the lines joining, in pairs,
2 points c110sen arbitrarily from a (q+2)-arc ) C, pass tl1rough all points
of t.Ile plane except the other points of
C.
Proof: Similar to ~le proof of Theorem (2.5.2).
Corollary (2.6.4).
If
q
is even, and
K is a k-arc of n(q) not
contained in any oval, then the maximum number of points common to
anJ'
K and
( q+2) -arc is not greater than ~ + 1 •
Proof:
In
See proof of Corollary (2.5.3)
h
PG (2, 2 )
we can obtain a (q+2) -arc by adj oinill[j the nucleus to a
conic (see also Theorem (2.2.3». By deleting a po:illt otrler than the nucleus
from this (q+2)-arc we obtain a (q+1)-arc which is not a conic provided
q > 5.
For if tIlis arc is a conic, by the theorem of Bezout it cannot have more then
fOl).r points in connnon with tile original conic.
Since it clearly has
q-l
points in common, the result follows.
~1e
question arises as to vnlether or not there exist (q+2)-arcs whidl
do not contain a conic.
Ti.leorem (2.6.5)
t~lere
exist
TJle answer is given by the fol101'Ting theorem:
(B. Sec;re [5])
In
h
PG(2,2 ) for
11
=5
or h
2:
7 ,
(q+2)-arcs i·tiuc}) do not contain a conic.
TIle proof of Theorem (2.6.2) is Biven in B. Segre [13], n. 178.
He s'jlall mention here onl~r t.hat such (q+2)-arcs are obta:L.'1ed by adjoininr:
tl1e nucleus to the (q+l)-arcs c;iven by the algebraic curve
C: y = x
i·l11ere
r = 213 , and
g
r
is an integer satisfying
II-15
2:5 g :5 11 - 2,
(11,13) = 1.
I
(2.7)
Diophantine syste.lIl relative to a k-arc (see B. Segre [11]" n.10).
In this section
K be a k-arc in PG(2,q).
Let
Definition (2.7.1).
relative to K
If
can be either odd or even.
q
j
A
point P of the plane is said to llave index i
if exactly i
secants of K pass thrOUGh it.
is the nwnber of tangents to K passing through P, then
clearl~r
2i + j =
k
Hence
j
k
-
(mod 2)
[see T11eorem (2.l.1)J.
Also we have
:s 1!. = (~J
o :s i
We denote by c i
(~O)
•
= l,2, ••• ,1!.)
(i
plane not belonging to K l1aving index i.
comnlete
if and only if
•
c0
=0
the number of points of the
It is obvious that a k-arc is
•
co' cl ' ••• , c1!. are projective characters of K, i.e.,
are invariant under projectivities. They satisfy tJle following
Tile integers
tjle~r
relations:
1!.
(i)
(ii)
~
=1
+ Cl + q"""" - Ie
1
E
iC i
= (~)
(~)
c.1
i=O
(iii)
c.
i=O
1!.
IE
i=O
Relation (i)
(q-l)
= :3 d~)
foll~rn
easily by noting that all tIle points of the plane
not belonr;ing to the k-arc fall into disjoint sets determined by their indices.
Tile nwnber of secants passing through K is
II-16
(~), each of which
.1
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contains
q-l
to
of secants
1-
(2) (q-l).
points not belonsing to
and points n::>t on
K
K.
Therefore the number of pairs
but belonging to these secants is
K
The number of pairs of secants and points of index
to tilese secants is
ic .•
pairs of secants to
K
Smnming over
J.
i
belonging
i, we get acain tIle total number of
and points not on
belonginc to these secants.
K
Tllis establishes relation (ii).
Tile number of pairs of secants which intersect in a point not on
cen be cletermined as follmrs.
each point of index
index
i.
i )and summing over
such pairs passing tllroueh
There are
Hlutiplyinr; b~r
c
(the nwnber of points having
i
i, "re get the total number of such pairs, wilic·;)
is tile left hand side of relation (iii).
b~r
K
This same nW-:lber can be obtained
cotU1tinc the number of distinct sets of four points of
K, and rml1.tipl:ling
b: r three, since the secants iiilicll join two-by-two eaC!1 set. of four points
:.lCct in ti.lree points not on
Let us number the
2'
= 1,2, ••• ,
of index
TIlis establishe s relation (iii).
K.
of
( k) sec''''J.lts
"
2
i = 1,2, ••• ,£ ,
l~rine
i
~re
K '1n an axb·.1ll.~·a.l·~;·
define
on the r-th secant to
K.
i"JaY.
For
as t"le number of points
These neil clJa.racters of the
1:-crc satisfy the follOiTin:::, relations:
(1\r)
t
I:
i=l
(~~)
(v)
I:
1'=1
(\~)
£
I:
i=2
d:
J.
= q-l
r = 1
,c., .•• , (1-:)
2
0
d: = ic.
i = 1,2, ••• ,
(i-1)d~ =(k;)
l'
J.
1
= 1,2, ••• ,
t
]-
(~)
Relations (1v) follolW easily by noting that both sides represent t,;c
nurriber oi' points not
be10nCil1£.~
to
K lyinr; on a secant.
II-17
I
Bac11 side of relations(v) ey,;presses tIle number of pairs of secants
o.nc1 points llavinc index
i
lyinc on these secants.
The number of secants lThicl! meet the NIl secant in a point not on
1(-2)
is clcarl:,r ( 2
•
The sane number is obtained b:,'
Ie
tIle number of
COtU1tin~
:::ccants distinct from tile r-th secant which meet the :Pot!l secant in a point
of index
1, (i-l)d: , anel sur;lJ:Un/3 over
i.
J.
By s\mmUnc; each side of
(iv)
over
r
This proves relations (vi).
and eaCD side of
(v) over
i,
ire can equate the right-hand,-sides to obtain relation (ii).
For fixed
r, we can add side b:,r side equations
(iv)
and (vi) to
obtain
t
~
~
(vii)
i=l
1 ~
~• q-1
-- «(-~)
0
r
J.°di
r = 1,2, ... ,
(~)
c:.
Siuilarly by addinC side-by-side equation (ii) to tlTice equation (iii)
lre obtain
t
E
(viii)
=
1:(::-1)
(t-2 ) (k-3)
+ (~) (q-l)
i=l
Tnis last relation could also be obtained by SUP.ll".dne each side of (vii)
over
r
and using relation
Definition (2. 7.2).
Ol
r
for
(v).
r
d.J.
A k-arc is called symmetric if the
i = 1,2, ••• ,£
are independent
.
•
Fron relation (v) it follo"'TS that a necessary condition for a k-arc to
be G:r:r::letric is that
kO:-l) divid.e 2ic.
it Y.ll1.St divide every
c., end so by
Similarl:/ if
J.
J.
(i),
k
i.
Hence if
must divide
k
is a. prime,
q2 + q + 1.
k-l is prime, it must divide every co and therefore divide
J.
(12+ 0
".1
for all
~.
1I-18
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(2.0)
Index and rank of.' a l~-arc (B. Segre [11], n.ll).
-
Definition (2. 8. l). Let
and let b
a
~lel1
a
bc the smallest subscript SUC'll that
is called the index
Obviously a
k-arc
and b
>
is called the
>0
,
a
for a given k-arc.
0
of the k-arc.
~
is complete if and only if its index is positive.
iTc llave the follovr.i.ng tlleorem:
Theorem (2.3.1). The necessary and sufficient condition for the existence in
PG(2, q) of "a k-arc having a=b
Proof:
is that
is even and
k = q+2 •
(See proof of ~leorem 2.2.2». ~lerefore eve~J point not on ~le
Uu1Ccnts.
arc lies on
q,;2
secants.
To prove t"he necessity, note that
a.nd all i Fa.
d
q
The sufficiency follmrs from tIle fact that a (q+2)-arc cannot ImYe
r = q-l
a
a=b
implies
c. =
J.
0.: = 0
J.
for all
From relation (iv) in section (2. 7),it follm'Ts tllat
for all
r.
I
I
Relations (v), (Vi), and (i) tllen inply tI1e folloninG:
aC
a
k
= (q-l) (2)
(a-l)(q-l) = (k~2)
c
I
I
I
I
Ie
I
I
~
be tIle lare;est sv.bscript such that
c
El~ninating
a
and
c
a
a
=q,2+ q +l_k
from these three eqnations, 'ue e;et
(q-k+2) [2(q,-k+l)q + (k-l) (k-2)] = 0
If
C~Ul0t
1::
is a nonnegative intee;er less than
be satisfied.
Tileorem(2.8.2)~
If k
~lis
q+2, the above equation
contradiction establishes
F '1+2,
then the index
satisfy
II-19
a
tl~
proof.
and the ra.nl~ b
of a k-arc
r
I
(i)
a.
<
•
4b(q2+ Cl+1-1:) - 2(q-1)k(k-1)
Proof:
.-I
2(b-l)(q,-1)1~(k-l) - k(k-1)(1I:-2)(k-3)
By the definition of index and rank,
b
(i-a)(i-b) c. < 0
E
~
i=a
-
Tl1is eives
a
b c
{b i=aE
i
E J
b
i=a
-
iC i
< {b
-
~
~
ic. ~
i=a
i=a
i
2C.} .
~
UsinG relations (i), (ii), and (iii) of Section~.~, the desired
ineq,uality follows.
~leorem~.G. 3). Suppose there eY~sts a k-arc with
1;: ~ q+2
whose rank
b
satisfies
(ii)
b<
. k(k-l)(k-2)(k-3)
•
Tilcn the k-arc is incomplete.
Proof:
If (ii) holds, ti.len it follows from (i) that
iJe a l1on.nec;ative integer,
= :) ,
a
< 1.
Since
a
Inlst
and hence the k-arc is incomplete.
TileOl·en(g.3.4). A k-arc in PG(2,CJ.) is incomplete if
to
a
k
is small in relation
CJ.; more precisely, if
1:(L-3) + 2
for k even
2
~:(l:-3)
for Ie odd
2
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The first inequality follous innnediately from Tlleorem~.5.l). He
Proof:
1;;1.0.11 prove the theorem for
q
1:
odd by Sllowing that (ii) must hold if
2; k(k-3) /2 •
Since
If 1:
b
< A
is odd, 2.t
1-
= [-12],
= k-l
(ii) is clearly satisfied if
a..!1cl 't'le can divide
k-l, obtaininc
Since each term of the left hand side is even, their StU,l is even and therefore not greater than -2.
TiTuS 'tre can add 2 to tIle left-iland side and replace
tile strict inequality b:>r a sirrrple inequality, whici1 can be written in the
forn
(q-It+i) {1;:(}~-5)-2CJ.J ~ 0
Since
CJ.
is odd,
•
q ~ k + 1 and tIle result follows since
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both sides by
II-2l
q
2:
k(k-3 )/2.
I
EXB.11Rles of k-arc other than ovals
(2.9)
We have already seen some examples of complete
other examples are given 8Jld discussed in
K-arCS
(section (2.5».
C. DiComite [1], L. Lune11! - M. Sce [1],
[2], M. Scafati [1], M. See, [3], B. Segre [3] n.21, [5] § V, [11] n.15.
We shall give here an exanp1e of a complete k-arc in a particular nondesarguesian plane of order 9, which was originally found by Veblen and
Wedderburn [1], which is also known as the Hughes plane of order 9.
In the determination of this arc we shall use the representation of
the plane by a complete set of mutually orthogonal latin square s (R. C. Bose
[0] or [5 ] (1. 4) )•
The plane is given by the following set of squares
(G. Pickert [1], p. 293):
L(l)
0
2
1
1
0
2
2
1
0
6 7 8
8 6 7
7 8 6
3 4 5
5 3 4
4 5 3
4 5 6 7 8
4 8 6 7
3
5
4 5 3 7 8 6
0 1 2 3 4 5
2 0 1 5 3 4
1 2 0 4 5 3
6 7 8 0 1 2
8 6 7 2 0 1
7 8 6 1 2 0
0
1
2
3
4
5
6
7
8
0
1
2
3
4
0
0
8 6
2
1
7
6 4 5 3
3
0
3
4
5
4 5 6
5 3 7
3 4 8
7 8 0
8 6 1
6 7 2
4 5 6
5 3 7
3 4 8
7 8 0 1 2 3
8 6 1 2 0 4
6 7 2 0 1 5
1 2 3 4
8 6 7 2 0
4 5 3 7 8
5 3 4 1 2
1 2 0 6 7
6 7 8 5 3
7 8 6 4 5
3 4 5 0 1
2 0 1 8 6
6 7 8
2 3 4 5
8 0 1 2
7 5 3 4
5
4
1
2
0
1
7 8
8 6
6 7
2
0
1
4 5
5 3
3 4
1
2
0
L(4)
L(3)
6 7 8 0 1
3 4 5 6 7
1 2 0 8 6
7 8 6 5 3
4 5 3 2 0
2 0 1 7 8
8 6 7 4 5
5 3 4 1 2
1
2
0
1
2
0
7 8 6
5 6 7
1 5 3
6 1 2
0 8 6
8 4 5
4 0 1
3
2
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0
8
4
0
7
3
2
1
7 8 6
7 3 4 5
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1I-22
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4)
Not more than two of the points
X, Y, L , ~, ••• , L are in
1
8
s.
The k-arc is complete if it is not possible to adjoin a point to S
without losing one of these properties.
Following are examples of complete
k-arcs in this plane with k = 7,8,9;
k =
k
k
7:
= 8:
= 9:
(0,0) , (0,1), (1,0), (1,1), (2,3)
(3,6), (3,8).
(0,0),
(3,2),
(0,1), (1,0), (1,1), (2,3), (2,7),
(0,7),
(7,3),
(0,8), (1,1), (1,8), (3,7), (7,0)
(8,0), (8,1).
L4
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The existence of a complete 9-arc in this plane shows that a result
similar to Theorem (2.4.3) does not hold in general for non-desarguesian planes.
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1I-24
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CHAPTER
THE (k, n) -ARCS
(3.1)
m A PROJECTIVE
PrANE
Definitions
Definition (3.1.1).
that
III
n
A
(k,n)-arc of n(q) is a set of k points ofn(q) such
is the largest number of them which are collinear.
Note that the k-arcs of Chapter II are (k,2)-arcs in this notation.
Definition (3.1.2).
A (k,n)-arc is said to be complete it there exists
no (k,n)-arc, with k > k, which contains it.
1
Definition (3.1.3).
Relative to a (k,n)-arc
of the plane fall into three classes:
A secant to
K
r
of
n(q) the lines
secants, tangents, and external lines.
is a line which contains
A tangent of order
K
n
points of K.
(1 ~ r ~ n-l) to K is a line which contains r
points of K.
An external line is one which contains no points of K.
Definition (3.1.4).
A point of n(q) is an interior point relative
to a (k,n)-arc, K)if it is not contained in K and does not lie on a tangent
to K.
A point not in K which is not an interior point is called an exterior
point.
(3.2)
An upper bound for the number of points of a ~.
Consider a point of a (k,n)-arc in n(q).
q + 1 lines.
the arc.
Every such line can contain no more than n-l other points of
Therefore
k < (n-l)(q+l)+ 1
(i)
If
Through this point pass
n
= q+1,
= (n-l)q
+ n
this bound cannot be improved since all the points of
III-l
I
the plane form a {q2+q+l , q+l)-arc.
If n
= q, we can obtain a {q2, q)-arc
by cancelling a line of the plane together with all its points.
the best upper bounds are given in Theorem (2.2.2).
If n
= 2,
We shall assume in what
follows, therefore, that
3
:s n :s q
- 1.
We shall show that (i) can be improved if
(ii)
q
Since n
:s q -
+
0 (mod n) •
1, there is a point of n{q) not in a {k,n)-arc
K.
Through this point there will pass at least one tangent, since
{n-l)q
+0 (mod n).
Therefore by an argument similar to the one which pre-
cedes (i), starting from a point of K which lies on a tangent, we see that
k
:s (q+l)(n-l)
•
Suppose there exists a {k,n)-arc with k
point of the arc.
There are
= (q+l){n-l). Let P be a
q+1 lines through P; q of which are secants and
the other a tangent of order n-l.
Since there are (q+l){n-l) points, and
one tangent of order n-l through each point; the total number of tangents
to the arc is q+1.
Suppose (ii) holds and k = (q+l)(n-l).
Then through every exterior
point to the (k,n)-arc there pass at least two tangents.
For if only one
tangent passes through an exterior point Q, the q{n-l) points of the arc not
lying on this tangent must lie on secants passing through Q.
implies q{n-l)
5
But this
0 (mod n), which is a contradiction in view of (ii).
There are q-n+l points not belonging to the arc Which lie on a given
secant.
Since there are q+l tangents, exactly one through each point of the
arc, the number of tangents passing through points of the arc not lying on
III-2
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the given secant is also
q-n+l.
Since we have shown that at least two
tangents pass through every exterior point, it follows that there do exist
interior points to this arc.
All the lines through these interior points
which meet the arc must be secants.
Therefore (q+l)(n-l) == 0 (md n).t
which implies
q + 1 == 0 (rood n).
(iii)
Let
i+l
denote the number of tangents through an exterior point Q.
The number of points of the arc which lie on secants passing through Q is then
= (n-l)(q-i).
(n-l)(q+l) - (n-l)(i+l)
This implies
(iv)
(n-l)(q-i) == 0 (mod n)
Adding (iii) and (iv), we have
i + 1 == 0 (md n) •
Thus the number of tangents through an exterior point must be a mUltiple of n.
Let
is q-n+2.
t
be a tangent to the arc.
The number of exterior points on T
At least n-l tangents in addition to
of these points.
t
must pass through each
Since the total number of tangents to the arc is q+l, we
mst have
(v)
If
(n-l)(q-n+2) Sq.
n > 2, inequality
(v)
can be satisfied only if n ~ q + 1.
is a contradiction, since we suppose n S q - 1.
cannot exist a
Theorem <:~.2.l)
the number
k
((q+l)(n-l), n)- arc.
(A. Barlotti [3]).
And so if q
4: 0
But this
(mod n), t11ere
We have proved the following theorem:
If
q
of points in a (k,n)-arc in
4: 0
(rood n) and 2 < n < q , then
n(q) satisfies
k S (n-l)q+n-2
I.
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It is not possible to lower this bound without further restrictions
on n or q, as is demonstrated by the following exam;ples.
III-3
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Example 1.
In
PG(2,4) we can obtain a (9,3)-arc by cancelling all the
points lying on three lines which do not meet in a conunon point.
q
= 4,
n
= 3,
Example 2.
Here
so that the bound is obtained.
In PG(2,5), an (1l,3)-arc is obtained in a similar way by
cancelling all the points lying on fixe lines, no three of which pass through
the same point.
Here
q
= 5,
= 3,
n
and the bound is attained.
By :fUrther restricting n, the bound in Theorem (3.2.1) can be improved
:fUrther:
Theorem (3.2.21.
or
If
k
+0 (ood n),
> 4, then k
and if
~ (n-l)q + (n-3)
(1)
n
(2)
n ~ 9, then k ~ (n-l)q + (n-4)
For the proof of this theorem see L. Lunelli e M. Sce [3].
It has been conjectured (L. Lunelli e M. Sce [3]) that the least upper
bound for
k
when q" 0 (ood n) is (n-l)q + 1.
That this bound cannot be
decreased :fUrther is shown by the following theorem:
Theorem (3.2.3).
In
PG(2, p
2m
), the curve
I:
with the equation
m
m
m
~ +1 + ~ +1 + ~ +1 = 0
is a (p3m+ 1, pm + l)-arc.
For the proof of this result see R. C. Bose [4], n.4.
(3.3)
Some ;properties of (kzn)-arcs with q;5 0 (ood n).
If
q;5 0 (IOOd n), in 3t(q)
we have the following theorems concerning
(k,n)-arcs:
Theorem (3.3.1). (A. Cossu [2]).
(n-l)q+n, n)- arc with
q
=0
If there exists in PG(2, q) an
(mod n), then there also exists an
111-4
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((n.f -l)q + n", n")- arc, where n"
Proof:
Let
counting tIle other points of
a paint of K.
therefore
•
K be the given (n-l)q+n, n) -arc.
There cannot be any tangents to
K, a fact which is easily seen by
K which can lie on the q+l lines passing through
The number of secants passing through an external point is
q+l-n' , so that the number of external lines passing through eve ry
external point is n'.
The total number of secants to
K
is
(q+l-n' )(q+l)
where
k
= (n-l)q+n.
The total number of external lines is therefore (n' -l)q+n' •
therefore have a set of
(n' -l)q+n"
which pass through the same poi. nt.
We
lines of PG(2, q), not rore than n' of
The dual of this set of lines is the
required arc.
Theorem
complete
(3.3.2).
If q= 0 (mod n), then there do not exist
(n-l)(q+l);n)- arcs in ~(q).
The proof is similar to that of Corollary (2.6.2), which is
Proof:
the special case n
= 2.
Suppose an
The number of tangents to
Section
(3.2».
Since
K
(n-l)(q+l); n} - arc
Section
(3.2)
K exists in
q = 0 (rod n), (n-l)(q+l)
+ 0 (rod n),
so that at
It vms shown in
that exactly one tangent passes through each point of
Consider a secant to
K.
~(q).
is q+l, all of which are of order n-l (see
least one tangent passes through every point not on K.
least one tangent.
I.
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= qJn
K.
Through every point of the secant there passes at
But since the total number of tangents is equal to the
total number of points on the secant, exactly one tangent passes through each
point of the secant.
Thus through every point of intersection of two tangents
III-5
I
there cannot pass any secants.
through a cOmJOOn point.
It follows that all
q+l tangents must pass
Clearly this point could be adjoined to
other properties of
K.
{(n-l)q+n; n) - arcs with q == 0 (rod n) in
PG(2, q) can be found in A. Cossu [2].
(3.4)
Some examples of (k,n) - arcs.
Exa.m;ple 1.
the set of
3'
in
Let C be an oval in 3({ q) (q odd).
~(q-l) internal points of
i(q+l) points, a secant to
tangent to C does not meet 3'
H· q{q-l),
i(q+l)} -arc in 3({q).
in
in
3'
in any points.
i(q-l) points, and a
Therefbre the set 3'
is a
It is not complete, however, for no line
in
i
(q+l) points.
to obtain a complete
We can therefore
(i q{q-l)+l,
i(q+l) }-arc
3({q).
Exa.m;ple 2.
q
An external line to C meets
C meets
through a point of C will meet 3'
adjoin a point of C to 3'
C.
We denote by 3'
(k,q)-arc with
k < cf.
Let
relative to a conic C in
(i q(q+l), q }- arc,
to e some, but not
Example 3.
We shall give another example of a complete
e
(a)
be the set of
3({q) (q odd).
i
q{q+l) external points
It is easily seen that
but it is not complete.
e is a
It my be completed by adjoining
all, of the internal point of C.
Let C be an oval in 3({ q) (q odd), and 3'
internal points relative to C.
Which is complete.
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We noted in Section (3.2) that the affine plane of order
(cf , q) -arc.
is a complete
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The set C + 3'
is a
the set of
(i (cr+q+2),
i( q+3)} -arc
There are two kinds of points in the set C + 3':
Points through which pass tangents of order one only.
the points of C.
III-6
These are
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(b)
Points through which pass tangents of order
~-<q+l) only.
These
are the points of 3'.
Example 4
subplane
~., (m)
~(q)
Let
of order
be a projective plane which contains a proper
m for which the following property holds:
Through
every point of ~(q) there passes at least one line containing a line of
~'(m).
~(q)
Then it can be shown that through every point of
not on ~I(m)
there passes exactly one line containing a line of ~., (m) and that
(Dembowski [1], 2.4).
q
= m2
The points of the subplane obviously constitute a
(IJi2+m+l; m+l)-arc in ~(m2).
Ex.a.tIi?le 2,.
and
~.,
~
Let
be the Hughes plane of order 9 [see Section (2.9)],
its desarguesian subp1ane of order 3 (D. H. Hughes [1]).
a line of ~ I ,let
be the set of 9 points of ~'-r'.
C1'
which we obtain by adjoining to
of ~
containing r'.
C1'
Let
K
three points of r-r', where
Then K is a complete
struction in any Hughes plane of order
q2
(J2, 3) -arc.
If
r' is
be the set
r
is the line
The same con-
will give us a complete
(<f+q, q) -arc.
Example
6.
(A.
Cossu [2]).
This example illustrates the existence
.
of complete ( 2h-l - 1 ) 2 h + 2h-l ,2h-l) -arc ~n
PG (2h)
2,
•
.
S~nce
a
h
(2h + 2, 2) - arc exists in PG(2,2 ), Theorem (3.3.1) gives us immediately the
existence of a
(3.5) On
( 2 h-l -1 ) 2 h + 2h-l , 2h-l) - arc.
the non-existence of certain (k;n)-arcs
We have the following theorems:
Theorem
(3.5.11.
The necessary condition for the existence in
~(q)
(q = ph) of' an «n-l)q+l;n) - arc having the property that through each of its
points there passes exactly one tangent of order one and no other tangents,
III-7
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is that either
q == 0 (mod n)
q ==
or
P.
1 (mod n).
*
Proof: Suppose q 0 (mod n). Let t be a tangent to the arc at
Through a point on t" not belonging to the arc
there will pass at least one other tangent; for if not, then we must have
(n-l)q == 0 (IOOd n), contrary to hypothesis.
q points of t
Therefore, through each of the
other than P there pass tangents other than t.
Let
i
denote the minimum number of these other tangents passing through one of the
q
points.
Then i
satisfies
a < i < n-l
,
the second inequality following from the fact that if
i > n-l, the total
number of tangents is greater than (n-l)q+l, which is impossible.
i < n-l.
Suppose
The number of points of the arc which lie on tangents passing
through a common point not on the arc must be in the same residue class
(mod n) as the number lying on tangents passing through a different point
not on the arc.
Therefore we can let
through which pass
which pass
i
other tangents,
x be the number of points of t
y be the number of points through
i+n other tangents, z be the number of points through which pass
i0f2n other tangents, etc.
Then the following two relations hold:
ix + (i+n)y + (i+2n)z + •••
ix +
iy
+
iz
= (n-l)q
+ ••• =
iq
Subtracting, we get
= (n-i-l)q
ny + 2nz + •••
The left-hand side is divisible by n.
fore also be divisible by n.
If
The right-hand side must there-
i < n-l, it follows that we must have
q == 0 (mod n), which is a contradiction.
III-8
Hence
i
= n-l.
Therefore
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through every point of
follows that
point of
t
n
t
not on the arc there pass
n
tangents.
It
points of the arc nRlst lie on tangents passing through a
and the others nRlst lie on secants passing through this point.
This implies that (n-1)q+1
Theorem (3.5.2).
existence in
Let
7t(q) of an
n
!!:
0 (md n), i.e., that q== 1 (mod n).
be a prime.
A necessary condition for too
{(n-1)q+1, n )-arc
K having the property that
through everyone of its points there pass n-1 tangents of order n-1 and no
others, is that
either
q == 1 (IOOd n)
or
q == -2 (IOOd n)
Proof:
Thro~h
every point of K
pass n-1 tangents of order n-1.
total number of points in K is (n-1)q+1.
tangents to
t
be a tangent to
of K lying on
t
Therefore the total number of
K is
(n-1) {(n-1)q+l)
n-1
Let
The
t.
K and let
Denote by
passing through a point of
i
t
=
(n-l)q + 1
•
Pl'P2 , ... , P n 1
be the points
the mininn.un number of tangents other than
other than
Pl'P2 , ... , Pn - l •
The total
number of tangents Which do not pass through one of Pl'P2 , ••• , P n - l is
(n-1)q - (n-1)(n-2)
= (n-1)(q-n+2)
and the number of points other than Pl' P2' ••• , Pn-l on
Then clearly
i
u
and
v
is
q-n+2.
satisfies
o~
Let
t
i ~ n-1
be the number of points of
ing through any two points of
t
K which belong to tangents pass-
other than
III-9
Pl' P2' ••• , P n-1.
Then
I
(i)
There are
point of
than t
t
i
other than Pl'P2 , ... , P - l •
n
passing through at least one
If there are
passing through another point of t
i.e.
is prime to
n.
j
j
tangents other
(rod n)
(nx:>d n)
As in the proof of the previous theorm, let
be the number of points of t
not on K through which pass
i
x
tangents
other than t, y be the number through which pass i+n other tangents, etc.
Then the following relations hold:
ix + (i+n)y + (i+2n)z + •••
= (n-l)(q-n+2)
ix +
=i
iy
+
iz
+ •••
(q-n+2)
= (n-i-l)(q-n+2)
The left-hand side is divisible by n.
Since
n
is prime, if it
divides the right-hand side it must divide one of the factors.
let
Q be a point of t
If
i = n-l,
not on K through which pass n-l other tangents.
The number of points of K which belong to tangents passing through Q is
n(n-l).
Therefore the number of points of K which belong to secants pass-
ing through
Q
is
(n-l)q + 1 - n(n-l)
This number must be divisible by n, which implies
(n-l)q + 1
= 0 (mod n)
i.e.
q = 1 (mod n) •
I f n divides q-n+2 we have
q + 2 - 0 (rod n)
111-10
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By subtraction, we have
ny + 2nz + ••••
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not on K, then by (i)
= (j+l)(n-l)
i -
n-l
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(lOOd n)
tangents other than t
(i+l)(n-l)
since
=v
u
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i. e.
=
Theorem (3.5.2) for the case
-2 (mod n).
n
=3
(3.6)
An open question on the relationship betl~en certain (k,n)-arcs
and algebraic curves.
We have seen [Theorem (3.2.3)] that in PG(2,p2m), the algebraic
curve
xim+1 + { m+1 + ~m+1 = 0
is a
(p2m+ 1, pm+ l)-arc.
This arc has the property that through every one
of its points there passes one tangent of order one and no others. (See
R. C. Bose [4], n.5).
The question of whether, with a suitable choice of
m
the system of reference, every (p3 + 1, pm+ l)-arc in PG(2, p2m) having
this property can be written in the form (3.6.1) is as yet unanswered.
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was first proved by B. d'Orgeval
(see B. d'Orgeva1 [1]).
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CHAPTER IV
k-ARCS .AND (k,n)-ARCS TIl PG{r, q), r ~ 3
(4.1)
k-arcs in PG{r,q)
Definition (4.1.1)
A k-arc in PG{r,q) is a set of k points such
that: if k :5. r, they are independent; if k
>r
+ 1, no r + 1 of them are
dependent.
Definition (4.1.2)
A k-arc in PG{r,q) is complete if it is not con-
tained in any (k+l)-arc of PG{r, q).
Definition (4.1. 3)
A normal rational curve of PG{r, q) is an irredu-
cible algebraic variety of dimension one and order r, not belonging to any
subspace PG{r', q) of dimension r' < r.
Theorem (4.1.1) (B. Segre [3])
In PG{r,q), q odd, every normal
rational curve is a (q + 1) -arc.
For the proof of this theorem and the remaining ones in this section
see B. Segre [13],
§§ 186, 187.
If q + 1 > r + 2, then an (r + 3)-arc
Theorem (4.1.2) (B. Segre[3])
of PG{r, q) determines a normal rational curve which contains it.
In addition,
any two normal rational curves are homographic.
Theorem (4.1.3)
If q ~ 5 is odd, every (q+l)-arc
(B. Segre[3])
of PG{3,q) is a skew cubic.
Theorem (4.1. 4) (B. Segre [3 ] )
If
q ~ 5 is odd, the ma.xi.mum value
of k for which there exist k-arcs in PG{3,q) is q+1 •
Theorem (4.1.5) (B. Segre [3])
If r ~ 4 and q ~ r + 2 is odd, the
ma.xinn.un value of k for '\'mich k-arcs exist in PG(r, q) can not exceed
(q + r - 3).
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(4.2)
(k,n}-arcs in PG{r,q)
A (k,n)-arc in PG(r,q), with k > n
Definition (4.2.1)
set of k points of PG(r,q) such that no hyperplane contains
~
r, is a
~e
than n
points of the set.
Definition (4.2.2)
arc
(~
A (k,n)-arc is complete if there is no (kl,n)-
> k) which contains it.
Note that if n = r, a (k,r)-arc of PG{r,q) is called a k-arc of
PG{r, q).
Theorem (4.2.1)
An upper bound for the value of K in a (k,n)-arc
in PG{3, q) is given by
k ~ (n-2)(q+1) + 2 •
Proof:
Consider a line joining two points of the arc.
Through this
line there pass q+l planes, each of which can contain at most n-2 other
points of the arc.
Note that if there is a line in PG(3,q) which contains h points of
the are, the upper bound for K can be reduced to
k
Theorem (4.2.2)
and n > 3, then
Proof:
~
(n-h){q+l) + h •
If a (k,n)-arc exists in PG{3,q) with q
F0
(mod{n-1»
k ~ (n-2)(q+l) •
If we project onto a plane fran a point of the are, then we
obtain a (k-1,n-l)-arc in the plane.
Applying Theorem (3.2.1) the result
follows.
Theorem (4.2.3)
If a (k,n)-arc exists in PG(r,q) such that n > r,
upper bounds for the value of k are given by the following formulae:
(i)
(ii)
k ~ {n-r+l)q + n-2 if q ~ 0 (mod n-r+2)
k
~
(n-r+l)q + n
if q
E
0 (mod n-r+2) •
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Proof:
It is possible to choose r-2 independent POints
~,~, ••• ,A
_
r 2
of the arc; for if not, then all k points of the arc would lie in PG(h, q)
with h < r.
They would therefore not constitute a (k,n)-arc in PG(r, q)
according to Definition (4.2.1).
Consider the {r-3)-nat
1t
be a plane which is skew to 1:.
of the arc other than
1: determined by ~,A2' ... , A _
r 2 ' and let
Suppose first that 1: contains no points
~,~, ... '~-2.
Then the projections from
E onto
1t
of the k-r+2 points of the arc not in I: constitute a {k-r+2,n-r+2)-arc
in
1t.
Then (i) follows from Theorem (3.2.l) and (ii) from (i), section (3.2).
If 1: contains r+h-2 points of the arc, h
from 1: onto
1t
n-r-h+2)-arc in
> 0, then the projections
of the k-r-h+2 points not in 1: constitute a {k-r-h+2 ,
1t.
From (i), section (3.2),·we have
k ~ (n-r+l)q + (n-r+2) - h(q+l)
which is a lower bound than (i) or (ii).
IV-3
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C~V
k-CAPS IN PG(r,q), r ~ 3
(5.1) Definitions
Definition
(5.1.1)
A set of k distinct points of PG(r,q) (r >
no three of which are cOllinear, is called a k-cap.
Clearly the intersection of a k-cap with any t-flat is a kl-cap (k
l
~
k)
according to this definition.
Definition
(5.1.2)
In a given PG(r,q) a k-cap is complete if there is
no kl-cap , k1 > k, which contains it.
Definition
(5.1.3)
In PG(r,q) a k-cap for which k is maximal is
(5.1.4)
A line in PG(r,q) is a secant, tangent, or external
called an ovaloid.
Definition
line with respect to a k-cap according as it intersects the cap is t\\U, one,
or zero points, respectively.
Through each point of a k-arc pass the same number (k-l) of secants
and tl1e same number
r 1
(3....:.:!::1 - k + 1) of tangents.
q-
(5.2) Some results on IDBJdmal k-caps in PG(3,q)
Theorem (5.2.1) (Bose [2], Seiden [1], Qvist [1])
A k-cap in PG(3, q), q > 2, contains at most q2 + 1 poi nts.
Proof:
(a)
q odd.
Consider the q+l planes passing through a secant.
Each of them intersects the cap in at most q+l points, since q is odd
(Theorem (2.2.2».
Since the two points belonging to the secant are common
to aJ.l of them, we have
k < (q+l)(q-l) + 2
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3),
(b)
q even, (q > 2).
= q2
+1 •
The proof requires several steps.
V-J.
We shall
I
first prove that
k.:s q2 + q + 2.
cf + q + 1 lines passing through
There are
a point P of the cap.
Each of these lines can contain at most one point of
the cap other than P.
Thus
k:S q2 + q + 2 •
Suppose there exists a (cf + q + 2)-cap.
that there are no tangents.
It is clear from the above
Every plane which intersects the cap must there-
fore have at least q + 2 points in common with it.
On the other hand, no
plane can contain more than q + 2 points of the cap {Theorem (2.2.2».
Thus
a plane intersects the cap in either q+e points or in no point.
The number of secants to the cap is (q2+q+2) (cf+q+1)/2, and the number
of lines in PG(3,q) is (~+1)(cf+q+1). Since q
lines which do not intersect the cap.
one of these external lines.
> 1,
we see that there exist
Consider the q+1 planes passing through
Each of these planes intersects the cap in
either q+2 points or in no points.
Therefore
q+2Icf+q+2
which implies
q + 21 q2 + 4(q+2) = {q+2)2 + 4 ,
q + 21
i.e.
Since
q
> 2,
Therefore a (q2+q+e)-cap does not
exist in PG(3,q) if q > 2 •
We now suppose that there exists a maximal k-cap in ro(3, q) with
k = ~ + a , Where 1 < a < q + 2.
P
Let P
be a point on this cap.
Through
there pass q2 + a - 1 secants, and there remain
cf+q+1 - (q2+ a-1)
lines through P which are tangents.
= q-a+2 > 0
Let
t
be one of these tangents.
None
of the q+1 planes passing through t can intersect the cap:in a {q+2)-arc
since such an arc has no tangents.
At least one plane through t must inter-
sect the cap in a {q+l)-arc, for if not then the number of points in the cap
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this is impossible.
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is less than q2+l, which contradicts the assumption on a •
Let
1t
be a plane which intersects the cap in a (q2+l ) arc.
tangents to this arc meet in a point Q on
(Theorem (2.6.1».
1t.
All the
However,
aLl q2+q+l lines through Q. cannot be tangents to the cap, for if this were so
then we could adjoin Q. to the cap, contradicting the assumption that it is
maximal.
1t
Let s be a secant tl1rough Q..
A plane passing through s intersects
in a line passing through Q. Which also is tangent to the k-cap.
Thus the
intersection of this plane and the cap has a tangent and therefore cannot be
a (q+2)-arc.
Therefore by counting the number of point
of the cap lying
in the different plane through s we see that
k ~ (q-l)(q+l) + 2
= q2
+ 1
This is again a contradiction since we assume a
> 1, so the theorem is proved.
The bound given in Theorem (5.2.1) cannot be improved (:R. C. Bose [2],
§
5, 5), as the following theorem shows.
Theorem (5.2.2)
.An elliptic quadric in PG(3, q) is a set of q2+l
points, no three of which are collinear.
For the proof see, for example, :R. C. Bose [5] theorem (2.3.6),
Primrose [1], B. Segre [11], §I, 1.
The converse to Theorem (5.2.2) is also valid if q is odd
Theorem (5.2.3) (Barlotti [1])
Every (q2+l)-cap in PG(3,q) (q odd)
is an elliptic quadric.
Proof:
Let
K be a (q2+l)-cap in PG(3,q).
Then K has the following
properties:
a)
No plane intersects K in more than q+l points.
This follows from Theorem (2.2.2) since q is odd.
b)
If a plane contains t"TO points of K, then it contains q+1 points
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of K; and these points form a conic.
If a plane contains two points of K but less than q+l, then by counting
the points of K lying on the q+l planes through these two points we have
q2+1
Which is a contradiction.
< (q-l)(q+l)+2 = q2+1
By Theorem (2.3.3), a (q+l)-arc in PG(2,q) (q odd),
is a conic.
c)
If
P
€
K, there is exactly one plane through P which does not
pass through any other points of K
(tangent plane).
A plane passing through P and another point of the cap intersects
in a (q+l)-arc, by b).
Let
t
be the tangent to this arc at P.
K
If there
is a plane passing through P which does not intersect K in any other points,
it must pass through t.
Iifow q of the planes through t exhaust all the
points of K, so that there remains one plane which passes through no points
of K other than P.
Let ~ be a plane which cuts K in a (q+l)-arc, Which by (b) is a
r.
conic
We first suppose q
Let a
conic.
K in a conic
~.
~
5.
Let A ,A , ••• ,A be five points of this
l 2
5
be another plane through Al and ~.
r 2 • Let
BI,B2'~
~ cuts
be three points on r 2 different from Al and
and the tangent plane to K at ~ (at ~).
AI'~'
Let
cuts K in a conic
et:3
o:t or ~.
~
Then
(at ~) is given
and the tangent plane to K at ~ (at A ).
2
Denote by I'* the intersection between Q and
•
be a third plane pass-
r 3 , and again the tangent to r 3 at
Let Q be the quadric \<lhich passes through
=:. r 3
ex,
and a point C which is in K but not in
by the intersection of
I'*
Again by (b),
The tangent line to r 2 at Al ( at ~) is given by the intersection of
ing through
et:3
2
~
0:.,.
Al,~,~,~,A5,BI,B2,B3'C.
We want to prove that
The tangent plane to Q at Al(at A ) is determined by the tangent
2
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lines to
rl
r2
and
at Al (at
.t\),
and so it is the same as the tangent plane
to K at ~ (at ~).
Therefore the tangents to I'* at A and ~ are given by
l
the intersection of
~
and the tangent planes to Q at
~ and~,
respectively.
Since these tangent planes are also the tangent planes to K at Al and A ,
2
these two tangent lines coincide with the tangents to r:3 at A and ~. It
l
follows that r:3 coincides 'With I'* , and thus r.3 lies on the quadric Q.
Let P be any point of K.
P is different from A , since
l
through P and
r l ,r2 ,
or
r.3
~
A:r. €
Q by definition.
and which does not contain
at~.
ot11er than AI.
We will prove that P
Then ex
cuts
r l ,r2 ,
r.3
of ex and the tangent plane to K at AI.
tangent plane to Q at
€
Let ex be a plane passing
in three distinct points
The intersection of ex and K is the conic determined by these
three points, by A , and by the tangent line at
l
proves that P
We can suppose
Q.
or any of the tangent lines to
~
and
€
~,
~
which is the intersection
But this tangent plane is also the
and therefore this conic also lies in Q.
This
Q •
We need now to prove that Q contains no points other than those of K •
Suppose there is a point which belongs to Q but not to K.
Then Q contains
nore than q2+1 points, and therefore it is either a cone (containing q2+q+1
points) or a hyperbolic quadric (containing q2+2q+l points).
then it contains q+l lines; since q
least six points.
~
If Q is a cone,
5, each of these lines contains at
It is clear that there is no way to obtain a {<f+l)-cap
by deleting q of these points, since we will always have at least one line
remaining which passes through three points of the set.
Similarly if Q is
a hyperbolic quadric, we cannot obtain a (q2+1)-cap by deleting 2q points
(see also section (5 • .3».
If q = 3, let r be a line Which does not contain any points of K.
V-5
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There are four planes passing through r.
~
see that two of these planes,
t"\'1O,
and
~
Since K has ten points, by (b) we
cut K in four points and the other
ct.? and C\' are tangent planes.
The nine points of K belonging to
al'~'
and
0:,
determine a quadric Q.
It can be readily verified that a set of nine points belonging to a cone or a
hyperbolic quadric must have at least three collinear.
Therefore Q is an
We need to prove that the tenth point R of Q is the point
elliptic quadric.
P of K which lies on
C\
Let P' be the point of K belonging to
ct.?
I f P lies on the tangent
plane to Q at P', then this plane must contain two other points of K.
These
two other points must also belong to Q, and so we have a contradiction since
a tangent plane cannot contain three points of Q.
Therefore P does not lie
on the tangent plane to Q at P', and so the line PP' must be a secant to Q.
It cannot pass through any of the four points of K belonging to
four points of K belonging t,o
on
C\'
and it is therefore P.
Theorem (5.2.4)
(q = 2 h , h
> 1),
Proof:
' \ or the
a2 ; and so the intersection of P 'P and
Q must lie
This completes the proof of the theorem.
I f P is a point on a (q2+l)-cap K in PG(:3,q),
then the q+l lines tangent to K at P all lie in a plane.
Let
t
be one of the q+l
tangents through P.
There are
q+l
planes passing through t, none of Which can intersect K in a (q+2)-arc; since
such an arc has no tangents.
However, by counting points of K belonging to
the planes passing through t, we see that at least one of these planes must
intersect K in a (q+l)-arc.
Let a be such a plane.
Then there is a point Q
on a through which pass all the tangents to this arc, since q is even.
Theorem (5.2.1) at least one secant to K must pass through Q.
one of these secants.
v-6
Let
r
By
be
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None of the planes through r can intersect K in a (q+2)-arc.
this, let 13 be a plane through r.
through Q.
By
This 13 intersects
ex
To see
in a line passing
This line must be a tangent to Ie, but a (q+2)-arc has no tangents.
counting the points of K belonging to planes passing through r, we see
that every such plane cuts K in a (q+l)-arc.
A line passing through Q which does not lie in
For suppose
8
Ie in a (q+l)-arc •
A line passing through Q which does not lie in
in
a. secant to K.
is a tangent to K which passes through Q but does not lie in
ex. The plane containing r and s cuts
K.
ex must be
ex mst be a secant to
For suppose s is a tangent to K which passes through Q but does not lie
ex. The plane containing r and s cuts
lt in a (q+l)-arc which has two
tangents passing through Q (one is s and the other lies in
ex). This is
impossible since r is a secant to the (q+l)-arc which alSo passes through Q
[Theorem (2.6.1) J.
I f a plane through t coo.tains a point P1 of K different from P, then
by the above P1Q is a secant; and thus the plane cuts K in a (q+l)-arc.
Then clearly q of the planes through t exhaust the
P.
r:f
points of Ie other than
It follows that there is one plane through t which cuts K only in P.
Corollary (5.2.5)
h
No plane cuts a (q2+l)-cap in PGC3,q), (q=2 ,h > 1),
in a (q+2)-arc.
The proof follows immediately from the existence of a tangent plane
at every point.
Corollary (5.2.6)
A plane in PG(3,q) intersects a (q2+l)-cap in
either one or q+l points.
Proof:
The result folloWS by counting the secant planes, which are
q3 + q in number.
Together with the q2+l tangent planes, these are all the
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planes of PG(3, q).
Theorem (5.2.7)
If a (q2+l)-cap lC exists in PG(3,q), q ~ 4 even,
such that its intersection with every secant plane is a conic, then lC is an
elliptic quadric.
Proof:
Using the result of Theorem (5.2.4), we can repeat the proof
of Theorem (5.2.3).
We shall see later (Theorem (5.8.3» that it is not necessary to require
that every plane section of K be a conic in order that the conclusion of
Theorem (5.2.7) be valid.
Theorem (5.2.81
Let H be a point not belonging to a (q2+l)-cap
K in PG(3,q), (q = 2h , h > 1).
Then there are q+l tangents to K which pass
through H, and they all lie in a plane.
Proof:
Since a (q2+l)-cap in PG(3,q) is complete, there must be at
least one secant h to K passing through H.
.very plane through h intersects
Let r 1 and r 2 be two of these arcs, and let t l and t be
2
the tangents to r 1 and r , respectively, Which pass through H [see proof of
2
Theorem (2.6.1)]. The plane containing t l and t 2 passes through a secant to
K and therefore cuts 1C in a (q+1)-arc. Since two tangents to this arc pass
K in a (q+l)-arc.
through h, all the q+l tangents to this arc pass through H.
As in the proof
of Theorem (5.2.4), we see that there are no other tangents to K passing
through H.
Theorem (5.2.9)
Let H be a point not belonging to a (q2+1)-cap K
h
in PG(3,q), (q = 2 , h > 1). Then there are q+l tangent planes to K passing
through H.
Proof:
To every tangent plane to 1C which passes through H there
corresponds a unique tangent line contained in the plane and pUling through
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On
the other band, through every tangent line there passes exactly one
tangent plane.
Thus the number of tangent planes through H is q+1 by Theorem
(5.2.6).
Theorem (5.2.lOl
Not mare than two tangent planes to a (q2+l)-cap K
in PG(3, q) can pass through the same line.
Proof:
Two tangent planes to K intersect in an external line.
By
counting the points of K belonging to the q+1 planes passing through an
external line, we see that q-l of them must cut K in a (q+l)-arc.
The two
remaining planes are tangent.
h
Consider an ovaloid K of PG(3,q), with q = 2
2: 4,
and any plane
1t
From Corollary (5.2.6) there are only two possible cases to be
of PG(3,q).
distinguished:
(1)
1t
is a tangent plane to K at some point P;
(2)
1t
cuts K in a (q+l)-arc.
(q, +1) -arc meet in a point P in
In this case, all the tangents to the
1t •
In either case, given the plane
1C ,
the point P is uniquely determined
as the intersection of all the q+l tangent lines to K lying in
called the ;eole of
1t
1C ,
and is
with respect to K.
Theorem (5.2.11) (B. Segre [12])
The correspondence associating to
every plane of PG(3,q) its pole with respect to a (q2+l)-cap K is a null
polarity.
Proof:
pole P.
We have shown that to each plane
1t
of ro(3, q) corresponds its
We now show that to a given point P, there is exactly one plane
having P as its pole, i.e. the correspondence
1C ~
P is one-to-one.
This
result folloWS from Theorem (5.2.4) if P belongs to K, and from Theorem
(5.2.8) if P is not on K.
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Now consider the point P corresponding to a plane n:.
We shall show
that the poles of all the planes passing through P are all the points of
For let ex be one of these planes.
The result is trivial if ex = n:.
ex 1= n: , then since ex passes through P, the line ex
n
n:
n:.
Ii'
is a tangent to K.
Since the pole of ex is the intersection of all the tangents of K which lie
in ex , it follows that the pole of ex lies in
1f.
But there are as many planes
through P as there are points in n: , and since the poles of distinct planes are
distinct, we have the result.
n: is called the EQla.r plane of the point P. This
completes the proof of Theorem (5.2.ll).
Some properties of' this null polarity
are discussed in the following paragraphs.
Notice that if B is a point on the polar plane ex of a point A, then A
is on the polar plane 13 of B.
Let
be any line of PG(:5,q).
r
We shall prove that the poles of all
the planes passing through r lie on a line r', and that the poles of all the
planes through r' lie on r.
their polar planes.
Let A and B be two points of r, and ex and 13
Since the poles of all planes through A lie on ex , and
since those of B lie on 13 , the poles of all planes through r • AB, lie on
r
t
= ex n 13.
If A' and B t are two points of r', then since the polar planes
of each of A' and B' contain both A and B, the result follows in the same wa;y
as above.
r
t
r
t
is called the 12.olar line of r, and vice versa.
We observe that
is the intersection of all the polar planes of points on r.
If r is a tangent line, then r is its own polar line, for by definition
r contains the poles of all planes which pass through it.
then r
t
is an external line.
If r is a secant,
To see this, let A and B be the two points of K
which lie on r.
Then the polar planes of A and B are the tangent planes at
and
those points, reErpectiveJ:y,/r t is the intersection of these two tangent planes,
line
which is an external line. The polar line of an external/is a secant since
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an ovaJ..oid has the same number of secants and external lines, as can be seen
by subtracting the number of tangents, (q+l){q2+l), and the number of secants,
q2+l
( 2)' from (q2+l){q2+q+l), the total number of lines in ro(3,q).
When a point describes a line r in ro(3, q), its polar plane turns about
the polar line r' of r.
of r and planes
We now prove that this correspondence between points
through r
I
is hOlWgraphic.
This result is trivially true if
r is not a tangent, since in that case points and polar planes are incident.
Assume then that r is tangent; then r' = r.
not tangent to the ova.loid.
Let P be a point of r,
of A.
1t
Let
s
be a line skew to r and
Its polar line, s', has these same properties.
its polar plane, A
= 1t n
s, and a
the polar plane
As we have seen earlier, the point P is incident with
incident with
1t ,
the polar plane of P.
a since A is
Denoting by (P) the set of points
of r, and by (1t), (A), and (a) the sets of the corresponding elements,1t , A ,
a , 'ire have the following homographies:
(1t) 7\
(A)
(A) 7\
(a)
(a) 7\
(p)
It follows by the transitive properties of the projectivity that (1t) 7\ (p) •
An inmediate consequence of Theorem (5.2.11) is the following:
Theorem (5.2.12)
The tangent planes of any given ovaloid constitute
the dual of an ovaloid.
(5.3) umzer bounds on k-caps contained in a g,uadrig
If a quadric does not contain lines, then it is well known that the
quadric is an elliptic quadric and the dimension of the space is three.
quadric has cf+l points which form a (q2+l)-cap.
V-ll
The
Any subset of k of these
I
points form a k-cap contained in this quadric.
incomplete.
If k <
cf
+ 1, the cap is
The only complete cap contained in this quadric is the quadric
itself.
In this section we shall derive some upper bounds for the number of
points in a k-cap which is contained in a ruled quadric, .i.e. a quadric which
contains lines.
used:
The following method of obtaining
thes~
u.pper bounds is
Suppose that through each point of the quadric there pas.
contained in the quadric.
a
lines
Then the number of pairs of points of the cap and
lines of the quadric which pass through them is Gk.
If the quadric contains
<B lines, then since no line can contain more than t'WO points of the cap, the
number of such pairs is not greater than aB.
Hence
2<B
k < -
or
-
To determine the values of G and
as
a
we distinguish between three types of
nondegenerate ruled quadrics (*) , and we obtain the following theorem:
Theorem
(5.,.1) (B.
Segre [ll],§III, 20)
If
K is a k-cap contained
in a nondegenerate ruled quadric Q in PG(r, q), then an upper bound for k is
given below
(1)
r
= 2j
(j
> 1)
(In this case Q contains linear flats of dimension j - 1)
k < 2(qj+l) (qj-l)
cf (2)
1
r = 2j - 1, (j > 1), Where j-l is the dimension of the linear
flat of highest d1mensicn contained in Q.
*
() The appropriate values of G and
Bose . [5], section (2.4) §4, §5.
as
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in each case are &1.so given in
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k < 2(qj-l+l)(qj - 1)
ef r = 2j + 1, (j
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> 1), where j - 1 is the dimensim of the linear
flat of highest diIllmsion contained in Q.
k < 2(qj+l+ 1) (qj - 1)
ef -
1
In a similar way upper bounds for k can be obtained when the quadric
is degenerate.
We shall require in the sequel only the particular cases (a) r=3, j=2,
k ~ 2(q+l) by (2); and (b) r
= 4,
k ~ 2(q2 + 1) by (1).
It is easy to verify
directly thet the first bound holds if K is a cone in PG(3, q) and the latter
bound holds if Q is a cone in PG(4,q), q > 2, having as vertex a point •
(5.4)
other results on k-cags contained in a quadric in PG(;;,9,).
Theorem (5.4.1) (A. Barlotti [2])
In PG(3,q), q
> 7 odd, a k-cap
K having
k~ef-q+7
is contained in an elliptic quadric.
Proof:
If k
fare suppose k ~
ef
= q2
+ 1, the result follows from Theorem (5.2.3).
and let h = q2 - k.
Then 0 ~ h ~ q - 7.
There-
Since lC is a
k-cap and k = ef-h satisfies the inequality, the following properties hold:
a)
In PG(3,q), there is no plane which contains IIX)re than q+l points
b)
I f a plane contains two points of lC, it contains at least q-h
of K.
points of K.
c)
At least q-h of the planes passing through a given secant cut
K in a (q+l)-arc.
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Property (a) is obvious, and (b) and (c) follow immediately by counting the points of K belonging to planes which pass through a secant.
Since q-h ~
7, by Property (c) at least seven planes cutting K in a
(q+1) -arc pass through every secant.
Let A and B be two points of K.
containing A and B.
and B.
Let r and s be two different (q+l)-arcs
Choose a point C on r and Dons, both different from A
Consider a plane thrcugh CD which does not contain either A or B or
the tangent to r at C or to s at D and which intersects K.
Denote the otlEr
points of intersection of t with r and s by E and F, respectively.
The set of
nine different points consisting of A,B.C,D,E,F, and three other points of K,
chosen one each from r,s, and t, determine a quadric
r
which contain r,s, and t.
Choose a point M on t different from the five previously considered.
Consider a plane which passes through AM and cuts K in a (q+l)-arc u, but
which does not contain any of B,C,D,E, and F.
Then
three other points of r , one each from r, s, and t.
all belong to r , it follmTs that
u
belongs to
u
contains A and M and
Since these five points
r •
In the same way, starting from another point N of t different from the
six previously considered, 'He can find a (q+l)-arc v which contains A and N
and belongs to both K and
Now given
any
r.
point X
€
K, we want to prove that X
to r, s, u, or v, this result follows at once.
€
r.
If X belongs
Therefore let X be a point of
K which does not lie on r, s, u, or v.
The four (q+l)-arcs r, s, u, and v intersect in at DX>st
at her than A.
(~) = 6
points
Therefore we can find a plane passirg through AX which does
not contain any of these points of intersection, and which cuts K in a
(q+l)-arc w.
Then w contains A and four other points of
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r,
ale each on
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r, s, u, and v.
If
It follows that w belongs to
r , and therefore
r is a cone or a hyperbolic quadric, then
than 2q+2 points (see section (5.3.».
rf is false.
For q ~
X e
r •
K cannot contain more
7, the inequality
q + 7 ~ 2q + 2
Therefore we have reached a contradiction and so
r must
be an
elliptic quadric.
The following theorem gives an a.na.logous result for the remaining odd
values of q.
Theorem (5.4.2)
If
q = 3 or q = 5, every q2_cap K in PG{3, q) is
contained in an elliptic quadric.
Proof:
If
q = 3, q2 = 9 and so there is a quadric containing IC.
If
this quadric is not elliptic, among these nine points we can find at least
three collinear.
If q
= 5,
let P be a point on K.
tangent lines passing through P.
There are (q2+q+l) - (q2_l)
= q+2
Suppose no three of these are coplanar.
Then these tangents intersect a plane not passing through P in q+2 points, no
three of which are collinear.
Since q = 5 is odd, this is a contradiction.
But a plane which contains three tangents cannot contain another point of K,
since then it would contain q points of K (Property{b) of Theorem (5.4.1)
holds for q = 5), and a q-arc cannot have three tangents at the same point.
It follows that this plane contains q+1 of the tangents.
We will call this
plane the tangent plane to K at P.
Let A and B be two points of K and ex and
A and B, respectively.
Let
r
~
the tangent planes to K at
be a plane passing through A and B and cutting
K in a {q+l)-arc u (There are four such planes).
through A and belonging to ex but not to 7; and by
V-15
Denote by G a. line passing
0
a plane
passing through
I
a
but not containing B, and 'Which cuts K in a (q+I) -arc v.
Let
.-I
C be the
intersection of u and v other than A, and D and E two points of v other than
A and C.
r passing through A,B,C,D,E and
There is a quadric
as tangent planes at A and B, respectively.
u and v.
If
It is easily seen that
and
~
r contains
r contains lines, there is at least one line of r belonging to
This line intersects 8
in a point of v, which is therefore a point of K.
contradicts the fact that
ing through
having a
a,
~
is a tangent plane to K.
which cuts K in a (q+l)-arc w.
~.
This
Consider any plane pass-
It is easily seen that w belongs
to r .
We have proved that all the points of K which lie on q-l of the planes
through A and B belong to
r.
the remaining two planes,
1t
point on
1t
1
1
We need only
to show that this is true for
and 1t2 ' passing through A and B.
and suppose X does not belong to
r.
Let
be a
X
We want to show that
X
cannot be long to K.
Suppose we delete the following points of r : a) All points belonging
; b) All points belonging to lines joining X with
l
points belonging to tangents to r which pass through X.
to
1t
four points Which belong to both
r and
1t
2
;
c) All
There remain at least
The se renaming points can be
K.
divided into pairs, each pair being collinear with X.
belong to K.
r n
Therefore X cannot
It follows that r contains all points of K.
Theorem (5.4.3)
(B. Segre [12])
an irreducible quadric Q of PG(3, q),
Let
K be a k-cap contained in
q?: 4 even or odd, and suppose that
k~i<q2+q+4) •
Then Q is elliptic, and every cap containing K lies entirely in Q.
Proof:
It is inmediately seen that Q must be elliptic, for otherwise
k < 2q + 2, which contradicts the hypothesis.
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Suppose that there is a cap containing K which does nat lie entirely
in Q.
are q+1 tangents to Q passing through 0 since Q is a quadric.
joining 0 to K can meet K in only one point.
The
k
There
lines
Therefore there are at least
k-(q+l) ~ i<q2 - q + 2) lines through 0 which are secants to Q.
By
counting
the points of Q belonging to lines through 0, we find that Q contains at
least q2 + 3 points, which is a contradiction.
Note that Theorem (5.4.3) remains true if we replace "irreducible
quadric" by "ovaloid" (Fellegara [1]).
(5.5)
U,pper bounds for the number of points in a k-cap in PG(r, g)
We shall denote by ~(r+l,q) the maximum number of points in PG(r,q)
such that no
'Dle known values, of ~ (r+1, q) are
h of them are dependent.
given below (see Theorem (2.2.2), Theorem (5.2.2) and R.C. Bose [2]).
(1)
~(2,q)
= 2
~(3,q)
= q + 1
= q + 2
{
~(4,q)
= q2
m,(r+l,2)
= 2r
The values of
(q odd)
(q even)
+ 1 (any q)
~ (r+l, q) for
r
> 3 and q > 2 are
unknown.
'Dle
results of this section provide same bounds on ~(r+l,q) in those cases.
Theorem (5.5.1)
(Tallini[l]).
In PG(r,q) with r
> 3 and q > 2, we
have
~ ( r+1,q)
Proof:
I.
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Then there is at least one point, 0 say, of this cap outside Q.
r-1
<q
+ 1
Let us suppose that (qr-l+ 1)-cap K exists in PG(r, q).
obtain a contradiction.
We shaJ.l
I
A plane passing through a secant cuts
IC in a t-arc, Where
if q odd,
if q even.
The first step wU1 show that the value
if q is even, so that
t = qi2 cannot be attained
t:s q+1 in either case.
Suppose there exists a plane
r-3
1C which cuts IC in a {qi2)-arc.
There are ~ 3 = Z qJ 3-spaces which
J-o
contain 1C; let d i (i = 1,2'''.'~_3) be the munber of points of 1t lying
in the ith 3-space but not in 1C. Then
~-3
Z
(2)
i=l
d
=
qr-l+ 1 - (q+2)
=
qr-l - q - l
i
We know that D) (~, q) = ef + 1 by the previous section.
d
i
Therefore
:s q2 - q - 1
Adding, we get
Z~-3
i=l
d :s
i
~_3{q2_q-l)
r-l
= q
This contradicts (2).
-q~-3
r-l
<q
- q - 1
Therefore no plane can cut K in JOOre than q+l points.
The second step is to prove that any plane passing through a secant
cuts K in a {q+l)-arc. The number of planes passing through a secant is
r-2 J .
~-2 =
Z q • If one of these cuts IC in less than q+l points, then the
t-=O
number of points in K satisfies
This contradiction shows that every plane passing through a secant cuts K
in a {q+l)-arc.
Let
S 1 be a hyperplane passing through a secant s. The intersection
r-
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of Sr_l and lC is a ( qr-2 + 1 ) -cap, as can easily be seen by counting the points
belonging to planes of Sr_l passing through S.
Next we will show that all the tangents to K at a point lie in a
hyperplane which contains no secants. Let
r-l
r-l J
r-l
there pass q
secants and
~ q - q
t
be a point of K. Through T
r-2 J,
= ~ q tangents. Let t l and
, /,--0
/,=()
be two tangents to K at T. There cannot be any secants lying in the plane
2
determined by t
T
and. t , for we 'WOuld then have a (q+1) -arc having two tangents
l
2
at the same point T.
d
~
-
Consider a hyperplane ~ which does not contain T. The set
of
r-2
2 =
~ qJ, points which are the intersections of ~ and the tangents
J=O
through T has the property that if two points P and P belong t o ; , then
l
2
all points on the line P P2 do also. It follows that ,,) is a linear subspace(*),
1
and since.J has
~-2 points,
it is an (r-2)-flat.
Joining T to
J , we get
a hyperplane which contains all the tangents and no secants.
r-2
A hyperplane must intersect K in either 0,1, or q
+ 1 points.
Let
T and T be two points of K, and denote by 1'1 and 1'2 the tangent hyperplanes
2
l
at these two points.
Then
contains no points of K.
1'1 and
* which
1'2 intersect in an (r-2)-flat ~-2
Let t be the number of tangent hyperplanes through
*
E r-2' other than 1'1 and 1'2' and let s be the number of secant hyperplanes
through <-2.
Then by counting the points of K other than Tl and T2 which
belong to hyperplanes through
qr-l + 1
=t
*
~r-2'
we have
21)
+ s (qr - +
) (r-2
< q-l + (
q-l
q
+ 1)
The inequality implies that
qr-3
<2 ,
(*)This result can easily b~ proved by induction on the dimension of
the space joining the points of ~!/.
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>3
which is a contradiction since r
and q
> 2.
This completes the prex..c.
Improvements on the upper bound on m,(r+l,q) given in Theorem (5.~.1)
were obtained by Barlotti [4], B. Segre [11], and Bose and Srivastava [1].
We give below these results for the cases in which they are the best known.
Theorem (5.5.• 2) (B. Segre [11] n. 24)(*)
> 7 odd, then
If a k-cap exists in PG(4, q), q
k
:s q3
- cf + 8q - 14 •
Alternatively:
for q ;:; 7 odd.
Proof:
Suppose there exists a k-cap X in PG(4, q) with k ::: q3 - q2 + 8q - 13.
We wiD. prove that this leads to a contradiction.
A plane which cuts K in a (q+l)-arc will be called a plane of the
A hyperplane 'Which cuts X in at least cf - q + 7 points will be
first kind.
called a hyperplane of the first kind.
belong to an elliptic quadric.
By Theorem (5.4.1), these points
The following two lemmas are required for
the proof.
Lemma (5.5.3)
Through any secant of K there pass at least 9q-14
planes of the first ..J.nd.
I f not, then by count: ng the points on the different q2 + q + 1
planes through a secant, we cet
k
:s 2
+ (9q-15)(~,1) + [(cf+q+l)-(9q-15)](q-2)
= q3 _
q2 +
8q _
=
15
Throl.gh any plane of the first kind there pass at
(*>This result is actl.ally a small improvement on Se'gre IS, obtained
by replacing Segre IS Theorem I (in [11])' Theorem (5.4.1).
by
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least five hyperplanes of the first kind.
If not, then by counting the points on the hyperplanes passing through
a plane of the first kind, we have
k ~ (q+l) + 4 [q2+l-(q+l} ] + (q-3}[q2_ q+6-(q+l)]
=
q3 _ q2+ 8q _ 14
We now proceed to the proof of Theorem (5.5.2).
Let ~ be a plane of
the first kind and let S' and S" be hyperplanes of the first kind passing
through
~.
Denote the conic in which
~
cuts K by C, and by Q' and Q" the
elliptic quadrics containing the points of lC in S' and S", respectively.
Obviously Q' and Q" cut
~
in C.
Let A be a point on C and P a point of K not contained in either S' or
S" (The existence of P is easily shown by counting points).
least (9q-14}-(q+l)
which do not cut
= 8q-15
~
There are at
planes of the first kind through the secant AP
in a line.
Let
p
be one of these planes.
Of the q+l
hyperplanes passing through p , there is at mst one tangent to Qr at A.
same statement holds for Q".
The
By Lemma (5. 5.4) , there are at least three
hyperplanes of the first kind passing through p which are not tangent to
either Q' or Q" at A.
Let
S be one of these hyperplanes, let Q be the
elliptic quadric which contains the points of lC in S, and denote by
~"
the intersections of S with S' and S", respectively.
different from
~
since
does not belong to S.
~
For if
~'and~"
~ €
S, then
~,
and
are
~
would
intersect p in a line.
The plane
~'is
irreducible conic C'.
not tangent to Q r, and therefore intersects Q' in an
There are at mst (q2+l}-(q2-q+7) = q-6 points in QI
which do not belong to K.
of K on C'.
Thus there are at least (q+l}-(q-6) = 7 points
But these 7 points belong to Q, and therefore Q contains C'.
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Similarly Q contains CIt, the intersection of lC and
are different from
1t,
1t". Since
1t'
and
1t"
C' is different from C".
There is exactly one hyperquadric V in ro(4, q) which contains Q', Q",
Q cuts V in C', C", and P, and so Q belongs to V and is the inter-
and P.
section of V with S.
Since Q, Q', and Q" are non-degenerate quadrics, V is
either non-degenerate or else a cone having a single point as vertex.
In the
latter case, since Q is an elliptic quadric, V is a cone having a point as
vertex and which projects an unruled quadric.
We will show that V contains
aJ.l points of le.
All the points of K belonging to S, S', S" lie in V.
point of K not belonging to S, S', or S".
to V.
S' or SIt.
through
1t
in a line.
Clearly
p* does not belong to either
There are at least five hyperplanes of the first kind passing
P*.
Of these, two can be tangent to Q' and two others tangent to
There is at least one hyperplane S* which is tangent to neither of
these quadrics.
1t
We want to prove that P* belongs
Let p* be a plane of the first kind passing through the secant PP* and
which does not intersect
Q".
Let P* be any
s* cannot contain
1t
since
P*, which is in S*, does not cut
in a line.
Let Q* be the elliptic quadric which contains the points of lC in S*,
and let
r' and r" be the intersections of S* with Q' and Q", respectively.
Then, .as we saw before,
which belong to Q*.
to V.
r'
and
r"
must be different irreducible conics
But Q* intersects V in
r', r",
But p* belongs to Q*, and thus P* belongs to V.
contains K.
Then by section (5.3), k:S 2(~+1).
qS _
ef
qS _
3cr + 8q -
+ 8q-13
:s k :s 2(q2+l)
15 :s 0
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and P, and so Q* belongs
It follows that V
We have
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The last inequality is impossible for q
~
3.
This contradiction completes
the proof.
Theorem (5.5.5)
(Segre [11], n. 24)(*)
If a k-cap K exists in PG(r, q) with r
k
< qr-1
- q
r4
+
Sr-3
q
-
> 4, q
~ 7 odd, then
6 r~.
~
J.
~ q
-
S
i=O
Alternatively,
for r
> 4, q
Proof:
Case 1:
~ 7 odd.
We have two cases.
There is a 3-fiat which contains at least cf-q points of le.
Then by
counting the number of points of K which lie on 4-flats passing through this
3-flat and using Theorem (5.5.2) we have
(i)
k
:s
(q2. q ) +
(qr~+
'. .. + q + 1)[(q3 _ q2 + Sq-14).(q2_ q )]
r-l
r-2
S r-3
r-4 i
=q
-q
+ q
-6E q -8
i=O
Case 2:
No 3-flat contains q2_q points of K.
The number of pairs of points
of K and 3-flats passing through these points is
l(
k •
( qr -1 qr-l - l)(qr-2 -1 )
(q3_ l ) (q2_l) (q-l)
•
(See R. C. Bose [5], Theorem (2.1.2) or B. Segre [13] §167).
On
the other
habd, since every 3-flat contains less than cf-q points of K, the number of
pairs is not greater than
(*) The improvement of Segre t s result for ro( 4, q) given by Theorem
(5.5.2) leads to the improvements of his bounds for PG(r, q), r > 4, given by
this theorem.
V-23
I
(r+1
q
-1 )( qr -1 )( qr-1 -l}(qr-2 -1)
(q4_1 ) (q3_1 ) (q2_1) (q-1)
.'I
•
Combining these results, we have
q
(ii)
r+1 - 1
q4 _ 1
.An upper bound for k in the general. case is given by the larger of
the expressions on the right-hand sides of (i) and (ii).
It is easily shown
that (i) is larger than (ii).
If q =
7 the bound for k given by the last theorem can be improved.
Theorem (5.5.6).
If a k-cap K exists inPG(r,q) with r
> 4, q
= 7,
r-3 i
k ~ qr-1 _ ~ q •
i=l
.,
Alternatively,
In; (r+1, 7) =s
T-1
r-3
.
~ 7~
-
1=1
for
r
Proof:
Case 1:
>4 •
There are three cases to consider.
There is a secant to K such that no plane through this secant cuts
K in a (q+1)-arc.
By counting the points of Ie on the planes through this
secant we have
k ~ {-l _
(i)
Case 2:
r-2 .
~ q~
i=l
There is a plane ex 'Which cuts Ie in a (q+l)-arc but such that
no '-flat through ex cuts K in a (q2+l)-cap.
these 3-flats through ex , we have
(ii)
k
=s qr-l -
r-3
~
q
i
i=l
v-24
then
I
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I
I
By counting points of K on
I
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Case 3:
There is at least one 3-flat S Which cuts K in a (q2+l)-cap.
counting points of K on the 4-flats thrOugh S we obtain, using Theorem (5.5.2),
(iii)
r-4 .
k < q2 + 1 + [q3_~+8q_14_(q2+l)] Z q~
i~
r-4
= qr-l_ qr-2 + 7qr-3 _ 8 Z qi+ q _ 14
i=l
Comparing (i), (ii), and (iii), we see that the upper bound given by
(ii) is largest.
Hence in the general case this is an upper bound for k.
Theorem (5.5.7)
If a k-cap K exists in PG(4,q), q
k
Alternatively,
Proof:
~ (5,5)
::s
= 5,
then
q3 - 1
::s 124.
Let A and B be t\'/O points of K.
We distinguish two cases according
as there is or is not a 3-flat passing through AB which contains less than
q2 points of K.
AB.
Suppose first there is such a 3-flat
Z passing through
There are q2 planes passing through AB which do not lie in Z.
Each
of these planes cannot contain more than q-l points of K other than A and
B.
By counting points of K on these planes and on Z , we have
< 2 + q2(q-l) + (cf - 2)
k
Consider now the case in which any 3-flat passing through AB contains
at least ~ points of K.
no more than q - 1
=4
If there is a plane passing through AB containing
points of K, then by counting points of K on planes
through AB, we have
k
I.
I
I
By
::s 2
+ (~+q)(q-l) + (q-3)
3
= q
- 1
V-25
I
Suppose now that every plane through AB cuts K in at least q points.
Not all of these planes can cut in q+l points, for this would contradict
Theorem (5.5.1) as is easily seen again by counting points of K on planes
through AB.
Therefore there exists a plane
r.
These five points determine a conic
conic.
1C
which cuts K in q
=5
points.
Let C be the sixth point of this
This means it is impossible to
We can assume that K is complete.
adjoin C to K; hence there are two points D and E of K such that C lies on
'the secant DE.
But
1C
and DE determine a 3-flat through AB, which by hypo-
thesis contains at least q2 points of K.
contained in an elliptic quadric Q.
By Theorem (5.4.2), these points are
But the intersection of Q and
conic containing the five points of K in
that C lies on Q.
1C ,
and therefore is
r.
1C
is a
It follows
But since Q is elliptic, it cannot contain the three
collinear points C, D, E.
This contradiction completes the proof of the
theorem.
Theorem (5.5.8).
If a k-cap exists in PG(r,q), r
k
::s {-I
r-5
- 2q
> 4,
q
= 5,
then
If q
> 2,
.
~ q~ - 1
i=O
Alternatively,
r-l
~ ( r+1,5 ) < 5
- 10
r-5.
~ 5~ - 1
i=O
for
r
> 4.
The proof is the same as for Theorem (5.5.6).
Theorem (5.5.9).
r
2: .3,
(R. C. Bose and J. N. Srivastava [1]).
then the ma.ximu.m possible number of points in any k-cap in PG(r, q)
cannot exceed the positive root of the quadratic equation
f(x)
= (q2_ q_l)x2
r.
r i
- [(q2-2q-l)+(q-2) ~ q~]x - 2 ~ q
i=O
Hence if q > 2, r
2: .3:
v-26
i=O
=0
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I.
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where
~
i
r
=
i~
q
•
For the proof see R. C. Bose and J. N. Srivastava [1], § 2-5 •
Using Theorem (5.5.9), we obtain improvements over previous results in
the following special cases (Ibid.§ 6):
~(r+1,3) ~
3r +l + 23
10
~(5,q) ~ q3 - 1
(iii)
~(r+1, q)
,
r
>4
,
q
>2 ,
even •
2
r
i
q -2q-l+(q-2){i~ q )
< ------..,;;;....;;.-- +
q2 _ q _ 1
1,
q
> 2, even.
The last bound is improved by the following theorem..
Theorem (5.5.l0).
In PG{r, q), r
> 4, q > 2 even, no k-cap can contain
more than
points.
Alternatively
~(r+1,q)~qr-1_2{
for r
> 4,
q
>2
r-4 i
1:
q)-l
i=O
even.
The proof is similar to that of Theorem (5. 5. 6), using the bound
~ (5, q) ~ q3 - 1 given by the previous theorem.
We must consider the addi-
tional case in which a plane cuts the k-cap in a (q+2)-arc, however.
(5.6) Lower bounds for the nwnber of points in a maximal k-cap in FG{r, q),
r ~
4.
In this section we obtain some lower bounds for
bounds are obtained by constructing caps.
V-27
~ (r+1, q).
The
I
Theorem (5.6.1)
(B. Segre [ll]).
2: 4
For r
and q arbitrary, there
exists a k-cap in PG(r,q) with
k
= q2
m,(r-2,q) + 1 •
EquivaJ.ently,
m,(r+l,q)
Proof:
to
1r.
Consider a plane
1r
2: q2
m,(r-2,q) + 1 •
and an (r-3) -flat
I: 3 in FG(r, q) which is skew
r-
There exists a max:inal cap lC* in I:r _ with m, (r-2, q) points.
3
m, (r-2, q) 3-flats 8 i determined by a point of JC* and
Let P be a fixed point of
is tangent to
7C
at P.
1r ,
are all distinct.
and denote by Q an elliptic quadric in 8 which
i
i
Ther.e are clearly
belonging to these quadrics.
1r
The
rf
m,(r-2,q) + 1 distinct points
We shall show that the set K of these points
form:; a cap.
Consider two points A and B belonging to Qi'
The line AB belongs to
Si and can therefore intersect Sj (j 1= i) in a point of
only point of
7C
7C.
Since P is the
which belongs to K, AB cannot contain another point of K.
Now let A and B be, points of Si and Sj' respectively (i 1= j), which
are both different from P.
S. and will intersect
J
I:
~ containing Si and B will contain
The 4-flat
~
r-~
in a line containing two points of K*.
Since
this line cannot cootain another point of lC*, the line AB cannot meet another
of the 3-flats S.t (.I 1= i,j).
This proves that K is a cap.
The preceding theorem is the best lower bound known in the general
case.
It can be improved upon in certain particu1.ar cases, however, as in
the following theorem.
Theorem (5.6.2)
(B. Segre [11]).
If
q = ph where
p = 7 (mod. 8), then there exists a k-cap in PG(4, q) with
k
_
-
5q2
-
2q + 1
12
V-28
h
is odd and
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The proof may be found in B. Segre [11].
The following theorem gives explicit lower bounds for
Theorem (5.6.3)
(B. Segre [11]).
2h
m,(3h+2,q) ~ q
+
m.,. (3h+3 q)
~
,
q-1
q2_ 1
2h+2
m;(3 h +3,q) > q2h + q
- 1
q-1
m, (3h+4, q)
Proof:
_
(q odd.)
2h
q(q
- 1)
q2_1
(q even)
2h+4
>
q
-
We can suppose q
for q = 2.
(any q)
q2 _ 1
> q2h+2- 1 _ q(q2h- 1)
-
> 0 we have
For any h
q2h+2 _ 1
- 1
q2 _ 1
(any q)
> 2 since the exact vaJ.ue of m;(r+1,q) is
••I
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I
I
but are not ma.x:i.mal..
I.
I
I
known
For h = 0, we get the results given by (1), section (5.5).
proof is completed by induction on h using Theorem (5.6.1) •
(5.7)
m; (r, q).
The
Examples of complete k-caps other than ovaloids.
In this section we shall give examples of k-caps which are complete
Theorem (5.7.1) (B. Segre [11]).
Let
r
2: 3 and
q be odd.
The set
r
of k = 2 -1 points of PG(r, q) other than the unit point whose hanogeneous
coordinates consist entirely of +1 's and -1 's constitute a k-cap.
If q
Proof:
= 3,
Let P'
we obtain a complete k-cap by adjoining the unit point.
= p'(x~,
Xi, ...,
distinct points of the set.
xp and P"
respectively.
xl, ..., x;>
be tl«>
Let a', a" denote the number of +1 's among the
coordinates of pI, P" respectively.
r-a '+1, r-a"+l,
= P"(xo,
Then the number of -l's in pI, P" are
Since the coordinates are homogeneoo.s, we
can take
V-29
I
(l)
.-I
a' > r - a' + 1
a" > r - a" + 1 •
Denote by
b
the number of subscripts I
the number for which
= x" = -1.
x'
for which
x'I -- x"I -- +1,
and by
c
Then we have
o :::
b ::: min{a ',a")
o :::
c ::: min{r-a'+l, r-a"+l)
and
(2)
a'+a"-b+c = r+l •
From (l) it follows that
b = 0 implies c
= O.
a'+a"
2:
r+l, and therefore from (2) we see that
Since we suppose P' and P" are distinct, this is
impossible so that b
> O.
AIry other point A on the line passing through P' and P" has coordinates
of the form
xI = x'I + Ax"I
where
A
€
GF{q) and A/:
o.
The coordinates of A fall into four groups as
follows:
Value of
XI
I'umber of
x.e
1 + A
b
-1 - A
c
having this value
a'-b
1 - A
-1 + A
r+l-a'-c
We wi.l1 prove that at least three of these groups are non-empty, which will
imply that the coordinates of A have at least three distinct values and hence
that A cannot belong to the set.
a' -b = O.
We have seen that b > O.
Suppose c
= 0,
Then by (2) we have a" = r+l which is impossible since the unit
point is excluded.
If c = 0, r-c-a'+l = 0, then again we have the contradiction
V-30
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a' = r+1.
Suppose a'-b = 0, r-c-a'+l = 0, then b+c = r+1 which implies
that P' and P" are the same point.
It is easily proved that if any two of the e1em:!nts l+A, -l-A, 1-A,
-l+A are equal, it follows that A = ± 1 and that two of the elements are zero,
while the other two are distinct and non-zero.
belong to the set, and hence that it constitutes a k-cap.
If q = 3, then we can adjoin the unit point to the cap.
case,
"
can assume only the values 1 and -1.
For in this
Since two distinct points of
the cap l'ID.lst differ in at least one coordinate, but not in all, we get at
least one zero coordinate by taking the linear combinations with
" = -1.
,,= 1 and
Therefore the unit point cannot be collinear with two points of
the cap•
To prove that the cap obtained by adjoining the unit point When q
=3
is complete, it is sufficient to show that any point A containing zero coordinates can be represented as a linear combination of two points of the cap.
can represent
A
We
as the sum of the two points P1 and P2 of the cap which
satisfy the following conditions:
If A has a zero in the ith coordinate, then
one of P1 and P2 has a 1 and the other has a -1 in the i th coordinate; if A
has a 1 in the i th coordinate, then both P1 and P2 have -1' s in the i th
coordinate; and if A has a -1, then both P and P have +l's in the ith2
1
coordinate, for i
= O,l, ••• ,r.
r
This example shows that a complete 2 -cap exists in PG(r,3); but it is
not an ovaJ.oid since for any r, 2
r
is less than the lower bounds for
~(r+1,3) given by Theore~ (5.6.3).
Theorem (5.7.2)
I.
I
I
This proves that A cannot
(B.
Segre [11]).
element of GF(q) different from 0,1,-1.
Suppose r ~ 3, q ~
For n
= 1,2, ••• ,
4
and let i be an
r, consider the
points of PG(r, q) having n (horoogeneous) coordinates equal to 1 and the
V-31
I
remaining r-n+l equal to i.
The set of these 2
r+l
- 2 points is a k-cap.
If
q = 4, the cap is complete.
Proof:
= p'
Let p'
(xO'
x;., ..., x;)
distinct points of the set.
and pll
= pll{XO'
xl, ..., x;)
be two
Denote by n' and nil the number of coordinates
which are 1 in p' and pll, respectively; and by a the number of coordinates in
Then 1 :s n', nil < r and
which both p' and p" have 1 's.
O:s a:s min (n', nil).
A third point A on the line passing through p' and p" has coordinates
of the form
Xl = xj + ~x.t, Where A is a non-zero element of GF{q).
Then we
have the following table as in the previous theorem:
Number of Xl having this value
Value of Xl
1 + Ai
i +
nil - a
1 + A
a
i + Ai
r - n' - nil + a + 1
To show that A cannot belong to the set, we distinguish several cases.
If n' - a = 0 and nil - a = 0, then n' = nil = a and it follows that p'
and p" would be the same point.
n'
=0
(n"
= 0)
r - n' - n" + a + 1 = 0, then
hypothesis.
Howerer, if a
If a = 0 and n' - a = 0 (or nil - a .. 0), then
which contradicts the hypothesis.
we
have nil • r+ 1
If n r
-
a
=0
and
which again contradicts the
Hence no two groups can be empty, except possibly the last two.
=0
and r-n' -n"+a+l
=0
, then P and p" have no common
coordinates, and any other point of the line p'p" has as coordinates 1+ Ai
and i + A.
To show that this point cannot belong to the set, it is
sufficient to show that the ratio (i+A)/ (l+Ai) cannot equal i or 1/i.
In
either case, we obtain i 2 = 1, Which is a contradiction since i is different
V-32
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I
I
I
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I
••I
n' - a
~
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I
I
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I.
I
I
from 1 and -1.
Since
,,1= 0, i 1= 1, if two of the values l+fd, i+1\, 1+1\, i+M are
equal, then we must have either 'N=l or "
= -1.
In either case, the other two
values are distinct from the cormnon value of the first two and also from each
Therefore we have three distinct values, and hence A cannot belong
other.
to the set.
If' q
= 4,
let 0, 1, i, i 2
be any point of PG(r,4).
= i+l
denote the elements of GF(4).
By a suitable permutation of coordinates, we can
write the coordinates of 0 in the abbreviated form (0
where for any "
€
If'
nO
, 1 , i , i2 )
nl
~
n,
GF(4), the symbol ~ represents a row vector having n
coordinates all of which are equal to ".
the cap.
Let 0
Suppose that 0 does not belong to
no = 0, then all of n ,n2 ,n are non-zero, for otherwise 0
l
3
would belong to the cap.
If'
no 1= 0, at least one of nl'~'n, is non-zero.
In either case consider the two distinct points P' and P" of the cap woose
coordinates are
P' :
(in' ln ' 1 , i ).
o
1
n2
n,
o belongs to the secant line P'P" since the coordinates of 0 can be obtained
P":
by adding the coordinates of P I to the coordinates of P" nmltiplied by i 2 •
Hence the cap is complete, but it is not maximal by Theorem (5.6.3).
The following lemma is required to construct a complete 12-cap in
PG(3,4):
Lemma. (5. 7. 3) (B. Segre [12]).
If C and C' are any two ovals of
PG(2,4) with no common points, then no line of PG(2,4) can be external to
both C and C'.
V-33
I
Proof:
Both C and CI consist of q+e = 6 points.
Through each point of C
there pass q+l=5 lines, one through each of the rema.i.ning 5 points of C.
an oval has no tangents,
three of these five lines will be secants to C' and
two will be externa.l to CI .
6· 2
~ =
to both C and C',
Since
£..:.2.
2
Hence there are
= 9 lines which are secants
6 lines Which are secants to C and external to CI ,
and 6 lines which are secants to C' and external to C.
lines having some point in cammon with C U C'
(B. Segre [12]).
9+6+6
is therefore
Let ~
ftl
and
= 21,
be two distinct
planes in PGC3, 4) and let r be their line of intersection.
oval C in
~
and an oval C in
l
Then the set C U C
l
r.
Proof:
~l'
or
~l
on
~
Consider an
both ovals having no point in camnon with
is a complete 12-cap in PG(3,4).
It is clear that no thr,ee points of C U C can be collinear, so we
l
need only to prove completeness.
Since C and C are ovals, no point of
l
~
can be adjoined to C U C • Suppose there is a point 0 outside ~ U ~l
l
which is not collinear with any two points of C U C • Then if we project C
l
I
from 0, we get two ovals in
external to r.
~
which have no common points and are both
By the previous lenma, this is impossible.
It follows that
C U Cl is a complete cap.,
Theorem (5.7.5) (B. Segre [12]).
of PG(3,q), q = 2
ble conic C in
h
~
Let ~ and
1C
, and r their line of intersection.
and an irreducible conic C in
I
~l'
l
be two distinct planes
Consider an irreduci-
both conics touching
r at the same point T and having the same nucleus 0 (necessarily on rand
distinct from T).
by
~
Denote by A any of the q points of C distinct from T and
any of the q points of C distinct from T.
l
Then every point
intersection of two of the ,q2 lines AA , and not on either
l
on exactly q of these lines.
The points
V-34
~
I
I
The total number of
which is the total number of lines in PG(2, 4).
Theorem (5.7.4)
.-I
1C
or
~
of
~l' lies
are q in number, lie on a plane
I
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.,
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,.
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I
I.
I
I
1t
2
1t ) which contains r, and together with T they
l
(distinct from 1t and
consti. tute the points of an irreducible conic C which touches r at T and
2
has 0 as its nucleus.
Let~, A'Ai
Proof:
A, A' are in C
and~,
be two of the q2 lines defined in the theorem, where
Ai
point ~ not on either
are in C •
l
"'1' so that A, A', ~, Ai are all distinct.
1t or
~
Then the projection of C on1t from
l
T, A, A' and having nucleus O.
at these tlu'ee points:
Suppose these two lines meet in a
is a conic Ci containing the points
Therefore C and Ci have the same tangents
TO, AO A' 0 and so they coincide.
It follows that
~
must lie on exactly q of the q2 lines defined in the theorem.
Conversely, fix arbitrarily one of these lines,
be a point of r distinct from T and O.
intersection with C and Cl , respectively.
~
~
outside 1t and
is the intersection of AA
intersection of
AAi
Ai to be
The lines
result exactly q of these lines pass through
If
say, and let R
The lines AR and A1R are secants
to C and C respectively, and so we can take A' and
l
and hence meet in a point
~
1tl •
~
~
their other
and A
coplanar
Therefore by the above
•
and A'Ai as above, and
l
'Ai are
A2
is the
and A'Ai, then the points R, ~, and.A2 are the diagonal
points of the quadrangle
AA'Ai~
; and
thus are collinear since the field
has characteristic two.
Since there are
exactly one point
points.
~
rf
lines
AI\,
and since on each of these there is
through which pass q of these lines, there are q such
Also the line joining any of these points with a point of C other
than T meets C in a point distinct fran T. Let ~'.A2 be any two of these q
l
meeting points, and let A be any point of C other than T. Then the lines
~
and
A.A2
meet C in points
l
~
and
Ai,
V-35
respectively, distinct from T.
Let
R be the intersection of .A:LAi with r.
Then R is distinct from 0 and T and
so the line .AR in" meets C in a second point A' distinct from T.
points
A2,
~,
R are the diagonal points of the quadrangle AA 'Ai~
are collinear.
Then the
and hence
Thus the line joining any two of the q meeting points
"2
r, and so it follows that these q points are all in a plane
~
meets
containing r.
These q points can therefore be obtained by projecting from any fixed point
A(I: T) of C the points ~
on a conic C of
2
"2'
(/=
T) on C onto".
l
Consequently, they all lie
which touches r at T and has 0 as nucleus.
This completes
the proof.
We note that the relation among three conics C, C ' C is symmetric.
l
2
Starting from any two of them, we obtain the third one as indicated in the
theorem.
Theorem (5.7.6) (B. Segre [12]).
PG(3,q), q = 2
h
•
With the notation of the previous theorem, the set C U C U 0 is a
l
Proof:
(2q+2)-cap in PG{3,q).
"I'
There exist complete (3q+2)-caps in
Since
C U 0 is an oval in " , and C U 0 an oval in
l
it is clear that no other point of
1f
or
1f
l
can be adjoined to the set.
On the other hand, each line ~ (joining a point A
Al
/=
T of C to a point
T of C ' and thus containing a point ~ of C ) has exactly (q+l)-3
l
2
points outside the planes
in a point outside
1f, 1f
/=
l
, and
PG(3,Q)
side
1f
2
1f, 1f
l
1f, 1f
l
, or
, which is the
1f
2
,
1f
l
, and
1f
2
since no two of' the q2 lines
;
easi~
seen to be the total number of points of
It follows that through each point out-
there passes exactly one secant.
points can be adjoined to
Hence none of these
C UC U O.
I
We shall show that the IIBXimum number of points of
is Cl.
..v,. meet
, these lines contain <f(Cl-2) points outsd.de
outside these three planes.
1f, 1f
2
= q-2
2 we can adjoin
1f
It wiJ.J. then follow that the {3q+2)-cap so obtained is complete.
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It is clear from the proof of the previous theorem that no point of C or of
2
r can be adjoined to the cap.
Since T and 0 are points of
1t
2
in the (2q+2)-
cap, we clearly cannot adjoin more than (q+2)-2 = q other points of ~.
We
can obtain q additional points by considering the pencil of conics determined
by C and the line r counted twice.
2
Any two nondegenerate conics of this
pencil will intersect only at T and will have the same nucleus 0 (*).
We can
choose any one of them different from C and adjoin its q points other than
2
T to the cap.
This completes the proof.
other examples of k-ca.ps are given in B. Segre [11] n. 23 and
G. Ta.llini [11].
(5.8)
The Segre-Tits Ova.loid
We have seen that if q is odd, an ovaloid in PG(3,q) is an elliptic
quadric (Theorem (5.2.3».
This result is not true if q is even, as the
examples of this section will illustrate.
We begin with som:! remarks on the plane sections of an ovaloid K in
PG(3,q), q
= 2h •
not tangent to K.
Let
1t
be one of the q3 + q planes of PG(3,q) which are
Then by Corollary (5.2.6)
from which we can obtain an oval
ruP
1t
cuts K in a (q+l)-arc, r say,
by adjoining the pole P of
1t.
For
r we have three possibilities:
<*)ThiS can be proved a.na.lytica.lly by choosing the system of references
in such a way that the pencil of conics is given by
x2
+ yz +
"if
= 0
V-37
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(1)
r
(2)
r can be obtained from a conic r' c r u P by replacing one of
is a conic (and then P is its nucleus)
r' by its nucleus O.
the points P of
We then say that
r is a pointed conic,
having as nucleus the point 0, and we note that the tangent of
r
at 0 is the
line OP.
(3)
The oval
r
U P cannot be obtained from a conic by adjoining the
nucleus to it.
If K is a quadric, only case (1) arises, and conversely (Theorem 5.2.7).
We shall say that K is singular or regular according as some or none of the
It is known (B. Segre [5]) that every
plane sections of lC are of case (3).
ovaloid is regular for q
(5.8.1)
Theorem
K in PG(3,q), q
= 2h ,
=4
and q
= 8.
(B. Segre [12])
All the pointed conics of an oValoid
which have as nucleus a given point 0, have the same
tangent at O.
Proof:
Let
1(, 1(1 be two distinct planes of PG(3, q) containing
r, r l
K in two pointed conics
say, both having nucleus O.
° and cutting
Let r
= 1( n
1(1
and suppose that r intersects K at a point T distinct from 0; so that the
poles (with respect to K) P of 1( and PI of 1(1 do not lie on r.
On
1(, 1(1
we have two conics C, C touching r at T and having 0 as nucleus, such that
l
r u
Denote by
P
=C U
0,
r1
U P1
= Cl
U 0 •
H, HI the quadric cones projecting C, C from PI' P respectively.
l
From Theorem (5.7. 5) and the proof of Theorem (5.7.6) we see that C, C
l
define a plane
1(2
passing through r, and that the lines joining points of
r - (0 U T) = C -(p U T)
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inth the points of
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fill up all the points of the space not on
and therefore all these points belong to secants of
r u rl
•
Since K is a
cap, it follows that the points of
must lie on
1[2 U H U~.
The set of points
1[2 U H U
11.
cannot contain more
than 3q points of K - (ru r ), however.
l
is (q2+l ) - 2q
>
But the number of points in K-(rU r )
l
3q, since the existence of pointed conics implies q ~ 8. This
I
contradiction completes the proof.
Corollary (5.8.2).
Any point of an ovaloid of PG(3, q) is the nucleus
of at most q of its pointed conics.
Theorem (5.8.3) (B. Segre [12]).
which contains
Proof:
i< q3 _q2+2q)
An ovaloid K of PG(3,q), q ~ 8,
conics is an elliptic quadric.
We start by showing that we can find two points A, B on K such that
the following two conditions are satisfied:
f
(1)
there exist
+ 1 distinct conics in K passing through A
(2)
there exist ~ + 1 distinct conics in Ie passing through both A and
B.
If we could not find an A for which (1) holds, then every point of K
lies on no mre than
%
conics'.
By counting in two different ways the number
of pairs formed by a conic of K and one of its points, we obtain
(q+l)
which cannot hold.
If, given A, we cannot find another point B of Ie for
which (2) is true, then A and any one of the q2 points of K distinct from A
would both lie on no more than ~ conics of K.
V-39
Again by counting in two
I
different ways the number of pairs fOI'JOOd by a conic of K containing A and
one of its q points different from A, and using (1), we obtain
(-f + 1)
2
q
S
q2.
~
which likewise does not hold.
If A,B are two points of K for which (1) and (2) hold, we denote by
a,~ the tangent planes to K at A,B, by CO,C l ' ••• , CqJ2
conics of K passing through both A and B, and by
through A but not B.
Then the plane of
i)
tangent to at most one of tIle CIS at A.
sects the planes of Cl , ••• , CqJ2
~
~ + 1 distinct
a conic of K passing
will meet a
in a line which is
~
We can therefore suppose that
PqJ2
in secants of K; if P , ••• ,
l
inter-
denote
the points of K different from A lying on these secants, then P. is a cammon
~
(i = 1, ••• ,
point of C. and ,
~
qJ2).
Let Q be the quadric which contains C , C , and P •
2
l
3
Then Q contains
the four points A, PI' P2' P3 of ~; and a , the tangent plane to Q at A,
contains the tangent to tJ at A.
conic
C. (i
~
= 3,
It follows that Q contains~.
••• , qJ2) belongs to Q, since Q contains three points
A, B, P. of C.; and the tangent planes to Q at A and B,
~
Also the
~
a and
~
, contain
the tangents of C. at A and B.
~
From (1) and (2) we see that there exist conics of K which pass through
A but not B and which do not touch Co at A.
for
~
On substituting such a conic
in the previous argument, we see that Co must lie on a quadric which
contains at least
~ - 1 of the conics C ' ••• , CqJ2 ; hence Q also contains
l
Co •
The number of points in the set Co U C U ••• U CqJ2 U ~ is at least
l
2 +
(qJ2
+ l)(q-l) +
(qJ2 -
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1)
>
t
(q2 + q + 4) •
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Since the 'Points lie both on K and Q, from Theorem (5.4.3) it follows that
K is contained in Q, and hence K = Q since they both have q2+l points.
Corollary (5.8.4).
If K is a regular ovaloid of PG(3,q) which is not
contained in a quadric, then there exists a point of K which is the nucleus
of at least qj2 + 1 of the pointed conics of K.
Proof:
If every point of K is the nucleus of no more than qj2 pointed conics,
then K could contain at most (q2+l)qj2 pointed conics, and so the number of
conics of K is at least
Which contrasts with Theorem (5.8.3) since K is not contained in a quadric.
Lemma (5.8.5) (B. Segre [12]).
PG(3,Q) (q
= 2h ),
Let 1( and
1(1 be t:wo planes of
and let r be their line of intersection.
Consider an
irreducible conic C in 1( touching r at a point T, and an irreducible conic
C in 1(1 touching r at a point T 1= T, the two conics having the same nucleus
l
l
o (lying on r). Then there exists a third plane passing through r, such that
the points of w-r are all the points of the space not on
1( U 1(1 which lie
on exactly one line meeting both C and C •
l
Proof:
We can choose the system of reference so that
T
T
1
= (1, 0, 0, 0),
= (0, 1, 0, 0),
o=
(1, 1,
0, 0),
and the equations of C and C are of the form:
l
C:
x4 = 0
Cl : ~ = 0
The coordinates of points A,
(~+~)~+~=O
(~+~)~ + ~= 0 •
~
of C, Cl ' different from T, Tl respectively,
are
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where
A:
'1. = A2
Al :
x l =1J
+ A, ~
,
~ = 1,
,~=Jf-+IJ,
~ = 0,
A, IJ are arbitrary elenents of GF (q).
P of PG(3,q) not contained in
Xl = a,
P:
where
=A
c
I
~
1C
U
= b,
1C
l
.-I
The coordinates of any point
can be written in the f'orm
~ =
c,
x4 = 1
0 •
A:t,
Then A,
and P are collinear if' and only if'
a = C(A2 +A) + IJ
b = c
A + (~+IJ)
Eliminating IJ , we obtain
c2A4 + C(C+l)A2 + (a+b+a2
)
= 0 ;
and this equation has just one root in GF(q) if' and only if'
if and only if' P lies on the plane
~+x4
= 0,
c+l = 0, i. e.
which we take as w.
This
completes the proof'.
We shall now indicate the construction given by B. Segre [12] of' a
reguJ.ar ovaloid in PG(3, q) which is not a quadric.
PG(3, q), q = 2
h
2: 8,
which contains
1- + 2
2:
2
Let K be an ovaloid in
pointed conics
having the same nucleus 0 and denote by w the tangent plane of' K at O.
the planes
1C, 1C1'
Xl' ••• ' Xl-
0 f'
Then
these pointed conics must all contain a
f'ixed line r tangent to K at 0 by Theorem (5.8.1); in addition, the poles
of' these planes Dnlst be distinct points of' r by Theorem (5.2.9).
If
J,
> 0,
the points 0, T, T , U1' ••• , U1- correspond homographically to the planes
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w,
~, ~l'
Xl' ••• , XI.
In
any
case, we obtain the existence of 1+2 conics
C, Cl , ~l' ••• ,
'I
~'~l'
Xl ' ••• , XI ' touching r at T, Tl , Ul , ... , UI respectively,
and having as nucleus 0; these conics being defined by
lying in
r
r1
= (C U 0)
= (C1U: 0)
-,.• •
- T
- T1
•
b.
.t
= ('.I U 0)
- U.t
The plane through r associated--fram Lemma (5.8.5)--with any two of the conics
C, Cl,'l' ••• , ~ ~
cannot intersect K outside of 0, and therefore coincides
with w.
h
We recall that an elenent a of GF(2 ) is said to be of the first or of
the second category according as the equation
p2+ p + a =0
h
has tl-IO roots or no roots in GF(2 ), and that the sum of two elements of the
same or opposite categories is of the first or second category, respectively
(see B. Segre [13 ], § 80).
We now have
Theorem (5.8.6) (B. Segre [12]). Let al'~' ••• , a q_2 denote all the
h
elements of GF (q), q = 2 , different from 0 and 1 taken in any order. Then in
ro(3, q) there exist s an ovaloid containing q pointed conic s w.i. th the same
nucleus if it is possible to choose q-2 (not necessarily distinct) ele:roonts
b , b , ... , b _ of the second category of GF(q), such that the determinants
q 2
l 2
of the submatrices consisting of any two distinct columns of the matrix
[
•••••
a
•••••
b
V-43
q-2
q-2
J
I
are of the second category.
Proof:
l
With the notation given immediately before this theorem, and with
= q-2,
we can choose a system of reference for PG(3,q) such that the line
r has the equation
~
r:
= x4 = 0
•
The q+1 points of r are then
0(1000)
T(OlOO)
T (1100)
l
U ( ~i 100)
= 1,2,
i
i
••• , q-2
and they correspond hanographically to the planes
passing through r.
w:
~
=0
1(:
x4
=0
1(1
x4
=~
~
~ =
.fai
~
••• , q-2
The conics
C:
Cl :
x4
=0
x4 = ~
=0
xi + X2~
+ ~ = 0
1(, 1(1'
xt'
+
X2~
+
ai~
+ bi
x3 = 0
touch r at the points T, T , U respectively,
l
i
and each of them has nucleus O.
The theorem will be proved if we show that on choosing the b r s in the
required manner, we obtain a cap by adjoining the point 0 to the q2 points
which lie on the conics defined above but not on r.
to prove that any three of these
never collinear.
rf
To do this, we need only
points taken from three different conics are
For brevity, we omit the lengthy calculations required to
v-44
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~ + X2~
~l: x4 = .fai~ ~
lie on the planes
= 1,2,
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prove this.
See B. Segre [12].
Theorem (5.8.7)
(B. Segre [12]). It is possible to satisfy the condi-
tions in Theorem (5.8.6) if q
Proof:
For q
= 8,
a
l
=x
a2
=x
let
= 8, but not if q = 16.
x be a primitive root of GF(23 ) and let
~=x2
+ 1
a4
=x2
a
+ 1
5
a6
= x2
+ x
= x2
+ x + 1 •
It is easily verified that
0, aI'
a3 , a 5
are of the first category, and
are of the second category, and that the conditions of Theorem (5.8.6) are
satisfied if we take
bl
= b 6 = a4
b2
= b 3 = a6
4 = b5 = a2
For q = 16, see B. Segre [12].
•
J. Tits [1] has proved that if
q
b
=
2
2rn+l
(m
~
1), then there exists
in PG(3,q) a class of ovaloids which are not elliptic quadrics.
The example
of B. Segre (Th. (5.8.6) and (5.8.7) is a particular case of this general
class (Fellegara [1]).
It has been verified with the aid of a computer that
in PG(3,8) all ovaloids which are not elliptic quadrics belong to this class
(Fellegara [1]).
other results on this class of ovaloids are given in J. Tits [2], [3],
[4].
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CHAPmR VI
k-SETS OF KIND s
(6.1)
DefinitionS
;::
Definition (6.1.1)
of
k
points such that any
s+l
distinct points are linearly independent,
but there is at least one sUbset of
In other words, there is no
s-+e
linearly dependent points.
(s-l)-flat of PG(r, q) containing
s+l
points of K(s), but there is an s-flat which contains s+e of these points.
Clearly 2:S s
-Note:
.:s r.
No subset of A < s + 1 points of a K(s) can be dependent.
-
Vie see that a k-set of l<;:ind two is a k-cap, but the converse is not
necessarily true.
Also a l<;:-set of' kind r in PG(r, q) is a k-arc in that
space.
Definition (6.1.2)
A k-set of kind
s
is incomplete or COmplete
according as it is or is not a subset of a (k+l)-set of kind s.
The definitions of secants, tangents, and external lines to k-sets
of kind s are the same as those for k-caps.
(6.2)
Embedding a K(s) in a set of lower kind.
We note that since a K(s) contains
s+2
dependent points, it is
impossible to embed a k-set of kind s in a set of higher l<;:ind.
We shall
examine the problem of embedding a K(s) in a set of lower kind.
Theorem (6.2.1)
(Tallini [7])
is a proper subset of a set of kind
Proof
I.
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A k-set of kind s, K(s), in PG(r,q) is a set
Every k-set of kind s, with s
> 4,
t < s-2 •
Let S 2 denote the (s-2)-flat determined by s-l independent
spoints of K(s). Then S 2 contains no other points of K(s). Let P be a
spoint of Ss_2 not belonging to one of the (S~l)secants of Ss_2 n K(S).
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Such a point exists, since s
~
4.
Then no secants of K(s) pass through P,
for if P lies on a secant joining two points of K(s) - S
of this secant and S
s-
2' then the join
s-
2 is a (S-l)-flat containing s+l points of K(s).
can adjoin P to K(s) to obtain a (k+l)-set.
dependent points (p and those of Ss_2
Corollary (6.2.2)
n
K(s», the set if of kind t
We
s
Since this set contains
~
s-2.
A k-set of kind s is contained in a complete set
of ldnd two or three.
Theorem (6.2.3) (Tallini [7])
Every incomplete K(s) is contained
in a complete set of kind two.
The theorem is trivial if s
= 2,
complete set of kind s containing K(s).
let 1\ and
~
so suppose s
distinct (see ~ following Definition 6.1.1).
to
H(s).
P~
Then, fori = 1,2,
Let
H(s)
be a
Let P be a point of H(s)_ K(s) and
be two distinct points of K(s).
,·rhich lie on the lines Pl\ and
> 2.
The lines
Let
N
l
P~
and
P~
are
and N be two points
2
, respectively, and which do not belong
BM
i
is the only line through N which coni
tains t"ro points of H(s), for otherwise we would have four point s of H(s)
coplanar.
The line N N is external to K(s) for the same reason. If we
1 2
adjoin N and N to K(s), ,re get a set of k+2 points, no three of which
2
l
are collinear, but at least four coplanar, i. e. a (k+2)-set of kind two.
If this set is incomplete, then it is contained in a complete set.
Some examples given later show that it is possible to have k-sets of
kind three which cannot be embedded in a set of kind two.
The following
theorem gives a necessary condition for the existence of such sets.
Ti1eorem (6.2.4) (Tallini
[7])
A necessary condition for the
existence in PG(r, q) of a complete K(3) not contained in a set of kind tyro
is that
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k
=
q-3 +
J 8{+1 + q2_6q+l
2(q-l)
be an integer and that the equation
r-l J,
+ (~-k-2 1: q)
= (2k+q-4)x
qx2
=0
/'--0
have t'tro distinct positive integer roots h
l
and~.
If such a K(3) exists,
the hyperplanes of mer, q) can be divided into two non-empty disjoint systems,
according as they have h
Proof
l
or
~
points in common with K(3).
~3) denote
two. ~1en ~3)
Let
in a set of kind
a k-set of kind three which is not contained
is complete by Theorem (6.2.3).
each point of PG(r,q) which does not belong to
~3)
Through
there passes exactly
one secant of ~3); since there is at least one by the condition that /<..(3)
is not contained in a K(2), and not more than one by the fact that a K(3)
cannot have four points coplanar.
Let Q
the
(~)
r
=
secants of
r+l 1
q
q-
i
pc3 )
denote the number of points of PG(r, q).
each contain q-l of the
~-k points
Since
of PG(r,q)- ?«3),
and since no point of PG(r, q) - /(,3) can belong to two secants, we have
k
(2)(q-l) = ~-k
(q-l)k"2 - (q-3}k - 2~
(1)
or
=0
Solving this quadratic equation for k, we obtain
J
k _ q-3 + 8qr+l+ cl
2(q-l)
-
(2)
-
6q + 1
(The other root is negative since q ~ 2).
of ;t(3) , the right-hand
Let
D~note
by
~
Since k is the number of points
side of (2) must be a positive integer.
Sr_l be any (r-l)-flat of PG(r,q) and let
the number of points in Hl •
VI-3
I1. = ~3) n Sr_l
•
The number of points of S 1 which
r-
I
do not lie on secants of
~
••I
is
h
l
~-l - ( 2 )(q-l)-~
•
Through each of these points there passes exactly one secant of
J«3)_~.
Since the number of these secants is
we have the following relation:
T11US
~
is a root of
Since (3) has one real positive root, and since ~-k-~_l
follows that the equation has two real positive roots.
>0
by (1), it
Both of these roots
:must be integers; for if (3) has only the integer root h
' it follows from
l
the above argument that every hyperplane meets ')( (3) in the same number h
l
of points.
This implies that every
number of points.
3<.
To see this, consider an (r-2)-flat
(3) in n points.
through
S 2
r-
in the same
which meets
Then by counting points of )<, (3) on hyperplanes
S 2' we haVB the equality
r-
which determines
for a given l, 1
points.
:K(:n
(r-2)-flat meets
n
S
uniquely.
l
S r-l,
He could proceed inductively to show that
every l-flat meets
~ (3)
in the same number of
But this leads to a contradiction, since there exist secant and
tangcnt lines to
9< (3).
It follows that if (3) has one positive integraJ.
root, it has two such roots.
This completes the proof of the theorem.
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The conditions of the theorem are satisfied for
r=3,
q~
and r=4,
q=3, being k=5, h =3, ~ =1 and k=ll, h1 =5, h =2 respectively. The follmTing
2
1
two examples prove that these K(3) 's which are not contained in K(2), s do
exist.
A 5-set of kind three in PG(3,2) is given by the vertices of
the system of reference and the unity point.
These can be seen to be the
points of the quadric
~~ + xJ.x, + x1x4 + ~x, + ~x, + ~x4 + x, x4 =
°·
An example of an 11-set of kind 3 in PG(4,3) is given by the following points:
A (J.,O,O,O,O),
1
~(O,l,O,O,O),
A4 (O,0,0,1,0),
A (O,0,0,0,1)
5
B (l,0,-l,l,-1),
2
B (O,1,-1,-1,1),
1
~ (0,0, 1,0, 0),
u(l, 1, 1, 1, 1),
13 (J., 1,0, -1, -1),
B4 (l,-J., 1,0,1),
B (J., -J.,1, -1, 0).
5
These 11 points belong to the following five quadric cones:
~x, - ~x4 - ~x5 + x,x4 - x,x5 + x4~ =
~x, - ~x4 - ~x5 - x,x4 + x,~ + x4x 5 =
x4~ - x4~ - ~x5 + x1~ - x1~ + ~~
=
°
°
°
°
x4X, - x4~ - x4~ - x,~ + x,~ + ~x1 = °
~~
- ~~ -.
~x, + x5~ - x 5x, - XlX, =
It has been proved (Makowski[l]) that for q = 2,3,4, there do not exist any
other k-sets of kind three which are not contained in a k-set of kind two.
Values of r and q for vThich there exist K(3), s in PG(r, q) not contained
in any K(2) are rather exceptional, as is shown by the following theorem.
Theorem (6.2.5) (Tallini[7]) For any fixed r
integer
<l:r
such that, for every
q
>~ ,
in a set of kind two.
The proof is given in Ta.llini [7].
VI-5
any
> 3, we can find an
K(3) in PG(r, q) is contained
I
Theorem (6.2.6) (TaJ.lini [7])
Let
Q
be a point of K(3) in PG(r,q),
and let H denote the set of points obtained by adjoining to K(3) -Q any k-l
points outside K(3) but
~ing
a 2(k-l)-set of kind two.
If
one each on the secants through Q.
Then H is
K(3) is not contained in a set of kind two,
H is complete.
Proof
If three points of H are collinear, the lines joining these
points with Q are distinct and belong to the same plane.
The plane will
contain four points of K(3), Which is contradiction.
If
K(3) is not contained in a set of kind
t~,
then throo.gh each
point of the space not belonging to K(3) there passes exactly one secant of
K(3), which is also a secant of H.
Therefore H is complete.
It follows that the existence of an ll-set of kind three in PG(4,3)
implies the existence in PG(4,3) of a complete 20-set of kind two.
(6.3)
An uwer bound for the number of points in a k-set
We now establish a general inequality for the number of points belonging to a k-set
of kind s
Theorem (6.3.1)
(*)$
(C. R. Rao [1]) The number of points k in a k-set
of kind s in PG(r, q) satisfies the folJowing inequalities:
r+l
(i)
q
(ii)
q
k
> 1 + (1)
(q-l)
-
+ ••• + (k)(q_l)u
u
if s
= 2u-l
k
r+l > 1 + (1)
+ (k-l)(q_l)u+l
(q-l) + ••• + (k)(q_l)u
u
u
if s = 2u
Proof
Clearly it is
The following proof is due to R. C. Bose [2].
sufficient to prove (i) and (ii) When the k-set is ma.xima.J..
Let
m denote
the number of points in such a set.
Case I
s = 2u-l
Now m is the maximum number of points it is possible to choose in
{*)The exact value of this number is known only in particular cases
(see Tallini [6]).
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m(r" q) such that no
°o"01"
(iii)
Let
I,
2u
:s -u.
of the points are dependent.
••• ,
°m-1
Let the points be
•
Points which lie on any (I,-l)-flat determined by any
I,
of the
points (iii)" but not lying on arty' (1,-2)-flat determined by 1,-1 of them" may
be called I,-th
themselves.
stage points.
Thus we have m first stage points, vis the points(iii)
Again any two of them say 0i and OJ determine q-l second stage
Points" namely the points on the line 0iOj other than 0i and OJ.
Proceeding
in this manner we see that any I, of the points (iii) determine {q_l)l,-l
There are (~) ways of chosing
I,-th stage points.
(iii).
I,
points out of the points
Corresponding to any choice we get (q_l)l,-ll,_th stage points.
The
I,-th stage points corresponding to two different choices cannot have any
point common if
I, ~
u, since no 2u of the pal.nts (iii) are dependent.
Hence
counting all points up to the u-th stage, the number of distinct points
obtained is
(iv)
This must not exceed the total number of pd nts in the space, vis.
H-l
.9...-• From this we at once get (i).
q-1
Case II.
Let
s = 2u.
In this case
m is the maximum number of points that it is possible
to choose in FG(r, q) so that no (2u+l) should be dependent.
If these
points are
••• ,
°m-l
then as before the number of distinct points obtained up to the u-th stage
is given by (iv).
Choose any u of the points 01" 02" ••• " 0m_l ' and add to
VI-7
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these the joint 0
o
•
These
u + 1
points determine a corresponding set of
(q_l)u , (u+l)-th stage points, and it is seen as before that the (u+l)-th
stage points corresponding to two different ways of choosing have no point
in connnon.
We thus get in all
distinct points.
Since this must not exceed the total number of points in
PG(r,q) we get (ii).
(6. 1.!-)
(See also R. C. Bose [4.1]).
wwer bounds for the number of points in a maximal k-set
Consider a matrix M "Those rows consist of all
vectors over GF(q).
r+l
q
- 1 non-null (r+l)-
We choose a set of these rows as follows:
any row of M, and then choose a
~
2
Let
a
l
be
fran among the rows which are not multiples
of al'
The third row
is chosen from among those rows not linearly depend-
ent on a
and a ; and in general the (i+l}:th row a + is selected fran anxmg
l
2
i l
those rows not linearly dependent on s or few of
al,~, ••• ,ai'
This is
possible as long as all linear combinations of s or fewer of a ,a , ••• ,a
l 2
i
do not exhaust the rows of M.
combinations are distinct.
In the worst possible case, all the se linear
Since we can choose the non-zero coefficients
of any row in q-l ways, the number of linear combinations of s or fewer rows
•
If this is less than c{+l_ 1, the number of rows in M, then we can choose
another row a i +1'
We have the following theorem:
Theorem (6.4.1)
(Varshamov [1])
If k-l is the maximum value of the integer x for which
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then there exists a k-set of l:".ind s in PG(r, q).
Another
result on lower bounds for k in a maximal k-set is given
by Theorem (6.5.4).
The lower bound for the number of points in a max:l.nal k-set given by
Theorem (6.4.1) can be increased in sane particular cases by using BoseChaudhuri sets.
(6.5)
The Bose-Chaudhuri sets
A particular class of k-sets has been found by R. C. Bose and D. K.
Ray-Chaudhuri [1], [2], for the case q
= 2,
and their I1Ethod has been general-
ized to arbitrary q by W. W. Peterson [2].
We have the following theorem:
."..
~eorem
(6.5.1)
Let
s
and
m be positive integers.
Then in
PG(ms-l, q) we can find qm- 1 points such that no s are dependent.
~:
Let
x
The result is obvious if
s
2:
qm-1.
Therefore suppose s < qm-1.
be a primitive element of the Galois field <IF ( qm).
Each element of
GF (qm) can be written as a polynomial
m-l
a o +a1x+ ••• +am- IX
(i)
of degree m-l or less with coefficients from GF(q).
Denote by V (q) the
m
vector space of m-vectors with elements from GF(q).
We can institute a 1-1 correspondence between the elements of Vm(q)
and GF(qm) by making ex
nomial (i).
= (ao,aI'
••• , a
1) of V (q) correspond to the poly-
m-
m
In this correspondence the null vector of V ('I) corresponds to
m
the null element of GF(~), and the sum of any two vectors corresponds to the
sum of the corresponding elements of GF(qm).
VI-9
So we can identify a vector of
I
vm(q) and the corresponding element in G.F(qm).
We can therefore define a
multiplication in V (a) by putting
m
Q13
=.,
a(x)f3(x) =
if
r(x)
in GF(qm) and a(x), f3(x), r(x), correspond respectively to a,f3,r.
Consider the matrix
=
M
1
1
• • •
1
x
;i!-
;i!-
• • •
X
(x2)2
x3
• • •
(x2)3
• • •
•
•
•
•
•
m
• qm-2 •
{;i!-)q -2
x
• • •
s
(XS)2
{XS)3
•
•
• m
(xs)q -2
m
The q -1 rows of M can be interpreted as s-vectors over GF(qm) or as msvectors over GF{q) using the correspondence described above.
If we can
prove that no s rows of' M axe independent over GF{qm), it would f'ollow that
no s axe dependent over the subfield GF{q).
Now the determinant of the
submatrix consisting of the il-th, i -th, ••• , i -th rows of M is
2
5
i
(x2) 1
i
(~) 2
•
••
• i
(;i!-) s
IX:
1
x 2
,.
Ui
s
which is non-zero since
x
• • •
• • •
• • •
is a primitive element of GF(qm).
that no s rows of' M axe dependent.
This shows
Writing the rows of' M as ms-vectors over
GF{q), i.e. as points of PG(ms-l,q), we have the result.
VI-10
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Theorem (6.5.2) (R. C. Bose and D. K. Ra.y Chaudhuri [1] and [2])
Let
t
and
m be positive integers.
Then in PG(nrt-l,2) we can find
2r:l_l points such that no 2t are dependent.
Proof
The result follows from the above theorem, since any linear
relation between the rows of
1
1
•
• • •
··•
•
•
•••
•
•
• m
-2
· . .•
·..
• m
(cP)2 -2 •
eP
,,.•.
••
•
• •
implies the same linear relation a.nx>ng the rows of M (with s • 2t, q
because of the auton:orphism x
= 2),
m
-+ ,(! of G.F(2 ) •
The result of Theorem (6.5.2) is obtained from the previous theorem
by omitting dependent columns from the matrix M without changing its property
of having no 2t or fewer dependent rows.
If we denote the rank of M in the
general case by R, then we can choose R independent columns and drop the
remaining ones.
The resulting matrix will still have the property that no
s or fewer rows are dependent.
Theorem (6.5.3)
The following theorem is due to W. W. Peterson:
The rank of M is equal to the number of distinct
residue classes n:odule (qm-1) among the ms integers.
qju
u
j
= 1,2, ••• s
= O,l, ... ,m-l
•
See W. W. Peterson [1] and [2] Chapter 8.
It follows that there exists a (qm_J)-set of kind s in PG(R-l, q).
The correspondence used in the proof of Theorem (6.5.1) leads immediately
to the following result:
VI-ll
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Theorem (6.5.4)
The existence of k points in PG(r_l,qn), no s
of which are dependent, implies the existence of k points in PG(nr-l, q), no
s of which are dependent.
In other words
m (nr,q) > m (r,qn)
s
- s
with the notation of section (5.5).
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APPENDIX
Bibliographical. note
on the study of algebraic varieties in ro(r, q)
The geometry of quadric hyper surfaces in PG(r, q) has been studied by
Primrose [1], B. Segre [11], and Ray-Chaudhuri [1].
Geometrical characteri-
zations of quadrics has been given by A. Barlotti [1], G. Panella [1], and
u.
Tallini [1], [2].
Properties of algebraic varieties of higher order are investigated
in the following Plpers:
R. C. Bose [4], R. C. Bose and I. M. Chakravarti
[1], M. Cicchese [1], S. Lang and A. Weil [1], L. B. Nisnevic [1], L. A.
Rosati [1], [2], B. Segre [3], [4], [11], [ 13], [12.4], [15], [16], [17],
G. Tallini [4], [5], (8!, [9].
other references on this topic are given in B. Segre [9].
Sections 2-3
of this paper deal with problems encountered in studying these varieties.
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OSTROM, T. G.:
[1]
Ovals, dualities and Desargues IS theorem.
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Canad, J. Math. 7 (1955),
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PETERSON,
Caratterizzazione delle quadriche di uno spazio (tridimensimale)
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The Me I. T. Press, Cambridge, Mass., 1961.
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RAY-CHAUDHURI, D.K.:
[1]
Some results on quadrics in finite projective geometries based on
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Factorial experimmts derivable from combinatorial arrangements
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RODRIQUEZ, G.:
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Un esempio di ovale che non e una quasi-conica.
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Boll. Un. Mat.
ROSATI, L. A.:
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Sul numero dei punti di una superficie cubica in uno spazio lineare
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L r equazione delle 27 rette della superficie cubica generale in un
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612-626; (2,) 13 (1958), 84-99.
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SCAFATI, M.:
[1]
Sui 6-archi completi di un piano lineare S2 8. Convegno internazionale:
Reticoli e geometrie proiettive (Palermo, ~ssina 1957) pp. 128-132.
Cremonese, Rome, 1958.
SCE, M.:
[1]
Sui kq-archi di indice h. Convegno internazionale:Reticoli e
geometrie proiettive(Palermo, Messina 1957) pp. 133-135.
Cremonese, Rome, 1958.
[2]
Sulia completezza degli archi nei piani proiettivi finiti. Atti
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43-51.
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[3]
Preliminari ad una teoria aritmetico-gruppale dei k-archi.
Mat. e Appl. (5) 19 (1960), 241-291.
Rend.
SEGRE, B.:
[1]
Suile ovali nei piani lineari finitie Atti Accad. Naz. Lincei Rend.
C1. Sci. Fis. Mat. Natur. (~) 17 (1954), 141-142.
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Ovals in a finite projective plane.
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Curve razionall normali e k-archi negli spazi finitie Ann. Mat.
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Canad. J. Math. 7 (1955),
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Sui k-archi nei piani fin!ti di caratteristica 2.
Pures Appl. 2 (1957), 289-300.
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di caratteristica p ~ 0,
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Ist. Mat. Univ. 1957.
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Corso C. I. M. E., Rome,
Sulle geametrie proiettive finite. Convegno internazionale:
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Cremnese, Rome, 1958.
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Rev. MB.th.
Archimede 10
Proc. Internat. Congress Math. 1958 pp. 488Cambridge Univ. Press, New York, 1960.
On Galois geanetrieS'.
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Le geometrie di Galois.
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On complete caps and ovaloids in three-dimensional Galois spaces
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Sulla teoria delle equazioni e delle congruenze algebriche (I and II).
Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur. (8) 27
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Le geometrie di Galois. Arch! ed ovali; calotte ed ovaloidi.
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Sul numero delle soluzioni di un qualsiasi si'Stema di equazioni
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Lectures on Modern Geometry, with an appendix by Lucio LombardoRaidice. Crenx:mese, Rome, 1961.
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Atti
(i§b21,
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Geometry and Algebra in Galois spaces.
Hamburg 25 (1962), 129-139.
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Abh. Math. Sem. Univ.
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Intorno ad una congettura di Lang e Wei1. Atti Accad. Naz. Lincei
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Arithmetische Eigenschaften von Galois-RHumen. I.
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SEmEN E.:
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A theorem in finite projective geometry and an application to
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Affine RHume mit schwacher Inzidenz und zugel18rige algebraische
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Verallgemeinerte affine RHume und ihre algebraische Darste11ung.
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Ann. Mat. Pura
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Caratterizzazione grafica delle quadriche ellittiche neg1i spazi
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Sui q-archi di un piano lineare finito di caratteristica p = 2.
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(1957), 242-245.
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Una proprieta grafica caratteristica delle superficie di Veronese
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Caratterizzazione grafica di certe superficie' cubiche di s3 q
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Natur. (8) 26 (1959), 484-489 and 644-648•
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Le geometrie di Galois e le loro applicazioni alla statistica e
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Acta Arith.
Sulle ipersuperficie irriducibi1i d'ordine minimo che contengono
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Intorno a11e fome di uno spazio di Galois ed a;gli spazi subordinati
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Un' app1icazione delle Geometrie di Galois a questioni di Statistica.
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Le groupes simples de Suzuki et de Ree.
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Ovoides a translations.
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Non-Desarguesian and non-Pascalian geometries.
Soc. 8 (1907), 379-388.
Trans. Amer. Math.
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