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Complete Arcs in Planes of Square Order
Annab of Discrete Mathematics 30 (1986) 243-250 8 Elsevier Science Publishers B.V.(North-HoUand)
243
COMPLETE ARCS IN PLANES OF SQUARE ORDER J.C. Fisher1, J.W.P. Hirschfeld2 and J . A . Thas3 'Department of Mathematics, University of Regina, Regina, Canada, S4S OA2. 2Mathematics Division, University of Sussex, Brighton, U.K. BN1 9QH. 3Seminar of Geometry and Combinatorics, University of Ghent, 9000 Gent, Belgium. Large arcs in cyclic planes of square order are constructed as orbits of a subgroup of a group whose generator acts as a single cycle. In the Desarguesian plane of even square order, this gives an example of an arc achieving the upper bound for complete arcs other than ovals. 1.
INTRODUCTION
Our aim is to demonstrate the existence of complete (q2 - q + 1)-arcs in a 2 2 cyclic projective plane II(q ) of order q . The only such plane known is
PG(2,q2),
the plane over the field GF(q2)
.
These arcs were found incidentally
by Kestenband [S], using different methods, as one of the possible types of intersection of two Hermitian curves in PG(2,q2) . The importance of these arcs, not observed in [S], is Segre's result that for q e en, a complete m-arc in 1 . Thus, this example of a PG(2,q) with m < q + 2 satisfies m 5 q - Jq +
complete arc attains the upper bound f o r q even As a by-product of the investiis the disjoint union of gation, it is shown that a Hermitian curve in PG 2,q') q + l of these arcs. 2.
NOTATION
Let
n
= n(q
2
)
be a cyclic projective plane of order 9'.
identify its points with the elements i of
ZV, v = q4
+
q2
+
cyclic group is generated by the automorphism u with o(i) = i [ 3 ] , 5 4 . 2 . The lines are obtained from a perfect difference set lo =
j = O,l,.. ., v
Ido,dl,., . dq2} as the sets u J ( l o ) , Let b = q2 + q + 1
and k = q
2
- q
are relatively prime, Zv= Zb x Zk. i = (1,s)
where i
In this notation u(i) = (r + 1, s
+
+
For
1;
i
-
l),
+
1, i
E
Zv,
1.
then v = bk.
Since b
and k
in Zv, we write
r(mod b), i L s(mod k)
taken modulo b and the second modulo k . to any arithmetical operation in Zv.
One can so that the
1,
.
where the sum of the first component is The notation extends in a natural way
J.C. Fischer, J . W.P.Hirschfeld and J.A. Thar
244
By the multiplier theorem of Hall [2], q 3 is a multiplier of II ; this 3 means that the mapping J, given by $(i) = q i is an automorphism of Il. Since q6 F 1 (mod v) , so J, is an involution. Indeed, J1 is a Baer involution since it fixes all b points of kZv = [(r,O) If we define r ( q 3 - 1) E 0 (mod b)
:
.
= I(r,s) : r
B
E
r
3
Zb1; this is because q r - r
E
Zbl for s
O,l, ..., k
=
-
=
1,
then o(Bs) = BS+l and the q2 - q + 1 Baer subplanes Bs partition Il. A l i n e of a Baer subplane Bs is a line of Il meeting Bs in q t 1 points. Similarly define = {(r,s) : s
K
E
iZk 1 for r = 0,1,. . . , b - 1 ,
whence o(Kr) = K r+l and the Kr also partition Il. Thus i = ( r , s ) = B n K s r It will turn out that Kr is a complete (q2 - q + 1)-arc. 3.
COMPLETE k-ARCS LEMMA 3 . 1 :
$(i)
,
=
(1,
k
-
s)
Proof:
i = (r,s) i n
For each
3
q3s F - s s k
For any l i n e
IK Proof:
izb
Zk, we have that
x
of i
,
q s + s = s ( q + l)k
LEMMA 3.2:
=
fixes the first component r
It was noted in 52 that
Now, for each s in iZk
whence
izv
.
n
~~1
.t
-
3
0 (mod k) ,
s (mod k )
.
0
BS , w i t h
of the Baer subpZane
is
odd if even
(r,s)
E
if ( r , s ) #
By lemma 3.1, the involution
$J
(r,s)
=
Bs n Kr,
K p..
fixes exactly one point of Kr
3
namely the point (r,O) where it meets B o ; the other points of Kr are interwhich implies changed in pairs. If K is a line of Bo it is fixed by $ , that the number of points of .t n Kr 1.t
of
n Krl Bs,
outside Bo
is even. Thus the parity of
varies as L n Kr n Bo is empty o r the point (r,O) . For a line .t apply the same argument to o-’(.t) , which is a line in Bo. 0
245
Complete Arcs in Planes of Square Order Let
LEMMA 3 . 3 :
be an automorphism group that a c t s regularly on t h e
S
p o i n t s of some p r o j e c t i v e p h n e
of order
n(n)
n,
and suppose t h a t
VO,V1, ...,Vt are the orbits of t h e p o i n t s under t h e a c t i o n of a subgroup s . If .t i s a Zine of n(n) and A . = 19.. n v.1 , then j=1 To each of the
n2
+
n
of
3
1
Proof:
G
A.(A. J
J
-
1) = I G I
of
elements y
1.
-
S\{l}
there corresponds
a unique pair of points P ,
Q of 9. for which y ( P ) = Q ; in fact, n 2 and Q = 2 n y(L) . If there was another such pair on 2 S would not act regularly on the lines of n(n) . Now we count the set
P
1
= y- ( 2 )
in two ways. First, each y
IJI
whence
= IGI - 1 .
then
,
other than the identity gives a unique pair
(P%Q),
Second, 9. is a disjoint union o f the sets 9. n V. , 3
and to each pair (P,Q) , P # Q , in 9. n V . there is a unique y in G such J that y ( P ) = Q ; hence IJI = 1 4 . (A. - 1) and so I J I = A . (A. - 1 ) . 0 Aj>l J J j=1 J J We are now ready to prove the main result.
In 14, an alternative proof is
provided that makes use of the properties of perfect difference sets. For
THEOREM 3.4:
k
=
q2
lie i n
-
q + 1 in
B
q
n(q2) . q + 1
are t h e
2,
>
each o r b i t
Kr
is a complete k-arc w i t h
Furthennore, the l i n e s through tangents t o
Kr
at
(r,s)
.
Bs n Kr = ( r , s ) t h a t
Proof: Fix a Baer subplane B and let II be one of its lines. For each orbit K r j ( j = 0 , 1 , . . . , q ) that meets .9. n Bs, set C I . + 1 = 12 n K I ; J rj for the remaining orbits, set @ . = n K 1 , j = q + 1, q + 2 , ... , b - 1 . 1 j' By lemma 3.2 both a . and B . are even. I
3
By definition,
By lemma 3.3, b-1
1
j =q+1
Bj (Bj - 1)
+
j=1
( a . + 1) a J. = q J
whence subtraction yields b-1 j =O
2
- 9,
J.C. Fischer, J. W.P.Hirschfeld and J.A. Thas
246
Consequently
B. I
E
{0,2}
for
j 2 q
Summarily, f o r any l i n e
(rj,s)
(i)
Krj
E
a t the point
1.
II o f t h e s u b p l a n e B s ,
so t h a t
II n Kr j
t
aj
= 0,
(!2 n
K
rj
1
either
= 1 and
is t a n g e n t t o
II
i = (r.,s) 1
or (ii)
.t n K,.
1
Bs = 0 and II meets
n
in
Krj
or
0
points.
2
i s a l i n e of e x a c t l y one o f t h e s u b p l a n e s Bs, it f o l l o w s i n more t h a n two p o i n t s ; t h a t i s , Krj is a t h a t no l i n e meets K r j (q2 q + 1 ) - a r c . From ( i ) it i s c l e a r t h a t , f o r each p o i n t ( r j , s ) o f K,. , I t h e q + 1 l i n e s of Bs through ( r j , s ) a r e t h e q + 1 t a n g e n t s o f K r j at t h i s point. S i n c e each l i n e of
il
-
For q
a s i m p l e c o u n t i n g argument suffices t o show t h a t t h e k - a r c
2 4 ,
Kr
i s complete. Assume t h e c o n t r a r y . Then t h e r e i s a p o i n t P through which p a s s q 2 - q t 1 t a n g e n t s o f Kr , one from each of i t s p o i n t s . S i n c e P E K r , for
some
r’ # r ,
q 2 - q + l
i t f o l l o w s t h a t through each p o i n t o f
tangents of
t h e g r o u p g e n e r a t e d by
(because
K
br a )
.
Since
Krl
l i n e s i s counted more t h a n t w i c e , whence tangents.
But a s
i ( q 2 - q + 1)’ When
L
Kr
(q
q = 3,
t
has e x a c t l y l)(q2 - q
+
and
Krl
is i t s e l f a k - a r c , none of t h e s e t a n g e n t 2 - q + l )2 has a t l e a s t ;(q
Kr
(q + l ) ( q 1)
,
- q
t
t a n g e n t s , we have
1)
q
a contradiction for
it must f i r s t b e observed t h a t
p l a n e of o r d e r 9 , Bruck [l].
there are
Krl
are o r b i t s u n d e r t h e a c t i o n o f
Kr
Then t h e o n l y 7 - a r c o f
2
4.
i s t h e unique c y c l i c
PG(2.9)
whose automorphism
PG(2,9)
group c o n t a i n s an element o f o r d e r 7 i s a complete arc, [ 3 ] , 514.7. q = 2
The c a s e
i s a genuine e x c e p t i o n : a 3 - a r c i s never complete. 0 Remark A theorem o f Segre [ 3 ] , 510.3, s t a t e s t h a t a complete m-arc i n
q
even, i s e i t h e r an o v a l , t h a t i s a (q + 2 ) - a r c , o r
m
5
q
-
PG(2,q),
Jq + 1 .
So, f o r
q
247
Complete Arcs ira Planes of Square Order an even s q u a r e , theorem 3 . 4 g i v e s an example of a complete (q - Jq + 1 ) - a r c and shows t h a t S e g r e ' s theorem cannot be improved i n t h i s c a s e . is e i t h e r a c o n i c , t h a t i s a (q + 1 ) - a r c , o r
m
odd
q ,
shows t h a t
PG(2,9)
q
odd,
This r e s u l t
However, t h e e x i s t e n c e o f a
iq + 1
i s n o t t h e b e s t bound f o r a l l
514.7.
[3],
LINES IN lI(q2)
4.
(i)
q
+
1 p o i n t s of the form
D
difference set (ii)
d ( k - 1) = j = 1,2,..
('12
.,
A line
one p o i n t .
Each o f i t s l i n e s
Bo.
L.
(rj, k - j)
L
S i n ce
of
j
Zk\{O}
i s f i x e d by
$,
r
Zb
of
j
r. # D
I t remains t o show t h a t
t h a n twice among t h e p o i n t s o f
3
L
.
, s # 0,
k - 1 differences
Bs.
l i n e of
i s a t an g e n t t o t h o s e
the other =
Kt
Bs ,
common w i t h i t .
in
q
Bs
+
D
generates 1
for
Zb
.
of t he p a r t i t i o n i n exact l y
o c c ur s a s t h e second component o f (I
interchanges
it f o l lo w s t h a t both
( r j , j ) and j
and
k - j
and t h a t no
r.
I
in
Zb\D
are
can ap p ear more
T h is f o l l o w s from t h e fact t h a t t h e p o i n t s of
Zv : each of t h e
t ( ( r j , j ) - (rj, k
that lie in
os(Bo)
therefore contains
ok
k - 1 differences
Bo
-
j))
.0
g iv e n i n t h e theorem i s e s s e n t i a l l y t h e
is a k - a r c whose t a n g e n t s are t h e l i n e s through
Kr
a l t e r n a t i v e proof t h a t
since
q,
must o c c u r e x a c t l y once, and t h e s e are accounted
The d e s c r i p t i o n of a l i n e of
Bo
for
.
c o n s t i t u t e a p e r f e c t d i f f e r e n c e set f o r i = (0,s)
II
Lemma 3 . 1 shows t h a t
p a i r e d with t h e same element
f o r by t h e
i s an element o f a p e r f e c t
meets any o t h e r subplane
Bo
of
e x a c t l y one p o i n t o f
of t h e form
d
i s an element o f a p e r f e c t d i f f e r e n c e s e t
d
Thus each element
.
, where
is i t s e l f a c y c l i c p l a n e of o r d e r
where
L
(d,O)
Zb;
for
3
a c y c l i c group f o r (d,O)
c o n s i s t s of
pairs of p o i n t s of the form ( r j , j ) and ( r j , k - j ) $ ( k - 1) , w i t h the r . d i s t i n c t elements of Zb\D .
Bo
Proof: elements
Bo
Any l i n e of
THEOREM 4 . 1 :
a
-
q
odd, [ 3 ] ,
PC(2,q),
q - Jq/4 + 7 / 4 .
5
has been s l i g h t l y improved by t h e t h i r d a u t h o r . complete 8 - a r c i n
q
For
§10.4, t h e comparable theorem s t a t e s t h a t a complete m-arc i n
(r,s)
The d e s c r i p t i o n of p o i n t s of t y p e ( i ) and ( i i ) shows t h a t any 0
or
2
points.
Kt
t h a t meet i t i n a p o i n t o f
Bo;
it meets
The proof i s completed by n o t i n g t h a t
which e i t h e r c o i n c i d e s with
Bo
o r h as no p o i n t s o r l i n e s i n
248
J.C. Fischer. J. W.P. Hirschfeld and J.A. Thas HERMITIAN CURVES
5.
The only known cyclic planes are the Desarguesian ones and, in this section, we restrict our attention to P G ( 2 , q 2 )
.
.
R . = o-’(Ro) 1
element of
Then
Lo.
and R .
i
1
In particular, L
theorem 4.1; now, D
Lo
of Bo and define are incident exactly when i + j (mod v)
It is convenient to distinguish one line n
B0
= { i = (d,O) : d
0 is a distinguished, perfect difference set f o r
iil The s e t
THEOREM 5 . 2 :
H = {
(d/2, s) : d
Hermitian curve and is t h e d i s j o i n t union of the
q
6
+
D, s
is an
as in (i) of
DI
E
.
‘b
Zk} i s a
E
1 compZete k-arcs
Kd/2 ’
Bs is a conic or a l i n e of Bs according a s q is odd or even, whence tI is a disjoint union o f k subconics or k s u b l i n e s
liii H
n
accordingly.
Define the correlations $ : i a . and p : ( r , s ) c-f 2 (r,-s) . is an ordinary polarity for q odd and a pseudo polarity for q even, [ 3 ] , g 8 . 3 . Thus, with J , as in 12, we have that p = $0 = $ 9 . In fact, p is Pro0f:
Then
+-f
$
a Hermitian polarity since the self-conjugate points of p are the q3 + 1 points (r,s) satisfying ( r , s ) + (r,-s) = (d,O) for d in D , From this ( i ) follows. In Bo in Zb. So
the points are
Bo
(r,O)
while the lines are R
is self-polar with respect to
self-conjugate points of the polarity
@
p
(r,O)
and meets H
induced on
B0
by
p
’
.
both with r in the q + 1 These self-
conjugate points form a subconic when q is odd and a subline when q is even. there exists s ‘ such that bs’ t s (mod k) since b and k are Given s , coprime. Thus H n Bs = 0bs’ (13 n Bo) = I(d/2, s) : d 6 D} is a conic or a line of Bs according as q is odd o r even, and the last part of (ii) follows. 0 THEOREM 5 . 2 : q
The tangents t o any complete
( q 2 - q + 1)-arc
in
2 PG(2,q )
,
even, form a dual Hermitian arc. Proof:
See Thas [ 6 ] .
THEOREM 5.3: The tangents t o any o f t h e complete 2 PG(2,q ) form a dual H e m i t i a n curve i f and only i f q
Proof:
Let
chosen s o that 21) = D and each multiplier of
q
(q2 - q + 1)-arcs
Kr
in
i s even.
be even and consider the arc K O , where D has been (which is always possible since 2 is a Hall multiplier Bo fixes at least one line of Bo) . Then the tangents
249
Complete Arcs in Planes of Square Order to
II II s
(0,O) ,
at
KO
with
(d, 0 )
to
(d,O)
namely t h e l i n e s o f
in
l i n e s of t h e p o l a r i t y
D = 2D
determined by
p
c o i n c i d e s with t h e s e t o f t a n g e n t s t o q
b e odd.
q'
Hence t h e t a n g e n t s t o
,
to
{I. : j
have t h e form
,
(0,s) =
takes
(d,s),
d
U,
6
J these lines a r e the self-conjugate
Thus t h e s e t of t a n g e n t s t o
H.
KO
H.
- q
and so i s odd, t h i s number i s never
1
+
not i n
P
do n o t form a d u a l H e r m i t i a n a r c . 0
Kr
Each of t h e
Tb'EOREM 5 . 4 :
is
KO
(0,O)
(0,O)
S i n c e t h e number o f t a n g e n t s from a p o i n t
h a s t h e p a r i t y of
Kr
to
Kr
q + 1 .
which t a k e s
t h e set of t a n g e n t s t o
(d,-s) ' From t h e assumption t h a t
Now l e t
containing
Bo
obs' ,
Since
D .
9,
Zk}.
E
d
(q2 - q
+
i s t h e intersection of t u o
Kr
1)-arcs
Wermitian curves. First, let
Proof: contained i n a p(H)
=
H
q
be even.
H,
Hermitian c u r v e
and l e t
H*
by t h e t a n g e n t s t o Now, l e t
q
Kr
Then a s i n theorem 5 . 1 , t h e a r c
which d e t e r m i n e s a p o l a r i t y
Then, a s i n theorem 5 . E
D,
E
D, s
s c izk
Hence Hr = { ( d / 2 + r , s ) : d
E
iZk
In f a c t , s i n c e t h e r e e x i s t s r ' kr' we have t h a t Hr = o (Ho). Since D
i s a l s o a Hermitian c u r v e .
,
Hrl
n Hr2
and
d2
= Kt
in
D
,
so
Zb,
Let
(b)
If
H
r2,
Also
s i n c e t h e r e e x i s t unique
dl
0 , q
even, and l e t
then
q > 2 ,
Suppose m
theorem 1 0 . 3 . 3 , c o r o l l a r y 2 ) , (q2 + 2 ) - a r c .
+
is a perfect
r1 # r 2 .
ti.
m = 4 if q = 2 . 2 m = q - q + 1 and (i)
f o r any
k
such t h a t
be a Hemitian curve in PG(2,q')
(m+1)-'zw in H , 2 m = q - q + l if q > 2 ;
Pro0f:
=
rl = i d
+
If there is no (a)
(ii)
n Hr21
d l - d 2 E 2 ( r 2 - r l ) (nod b ) .
be an m-izric contained in
(i)
IHrl
t = ad
where such t h a t
THEOREM 5.5: K
is
Kr
Let
b e t h e d u a l H e r m i t i a n c u r v e o f theorem 5 . 3 t h a t is formed Then p(Kr) = H* n h , whence K = p(H*) n H .
be even o r odd.
is a Hermitian c u r v e .
difference set in
.
.
H = Ho = { ( d / 2 , s ) : d
k r ' : r (mod b)
p
>
K
Now, count t h e p a i r s
q'
then -
K
q + 1.
is compLete. Then by S e g r e ' s theorem ( [ 3 ] ,
is c o n t a i n e d i n an o v a l (P,Q)
such t h a t
P
E
t h a t is a
0, K,
Q
E
0 ,
P
# 9 and
250
J. C.Fischer. J. W.P. Hirschfeld and J.A. Thas
PQ is tangent to H .
There are at most two points P f o r a given Q ,
since
three would be collinear. So
Hence 3m
2q2
5
+
4,
and 3q2
2 ( q 2 + 2 - m)
m
.C
-
3q
+
3
c
2q2
+
. 4
implies that q = 2 .
This
gives the result. (ii)
Suppose K is not complete, then the same argument as (i)
gives q ' whence q2
-
3q
-
1
5 0 ;
-
q
+ 1 5
2(q
that is, q = 2 .
+
l),
U
For q odd, the points of Bo together with the q2 + q + 1 Remark: conics Cr = { (6d + r, 0) : d E D} , r E Zb , form a plane of order q This
.
plane is isomorphic to PG(2,q) via the isomorphism 6 given by 8(x,O) = (gx, 0) For all q , this configuration of conics also appears as the section
.
by a plane
TI
in PG(3,q),
of the q2 + q + 1 quadric surfaces through a twisted cubic T where 71 is skew to T ; see [4], theorem 21.4.5.
REFERENCES [l]
Bruck, R.H., Quadratic extensions of cyclic planes, Proc. Sympos. AppZ.
Math. 10 (1960), 15-44. [2]
Hall, M., Cyclic projective planes, Duke Math. J. 14 (1947), 1079-1090.
[3]
Hirschfeld, J.W.P.
Projective Geometries over Finite Fields (Oxford
University Press, Oxford, 1979). [4]
Hirschfeld, J.W.P., Finite Projective Spaces of Three Dimensions (Oxford University Press, Oxford, to appear).
[5]
Kestenband, B., Unital intersections in finite projective planes, G e m . Dedicata 11 (1981), 107-117.
[6]
Thas, J.A., Elementary proofs of two fundamental theorems of B. Segre without using the Hasse-Weil theorem, J . Combin. Theory Ser. A . 34 (1983), 381-384.
243
COMPLETE ARCS IN PLANES OF SQUARE ORDER J.C. Fisher1, J.W.P. Hirschfeld2 and J . A . Thas3 'Department of Mathematics, University of Regina, Regina, Canada, S4S OA2. 2Mathematics Division, University of Sussex, Brighton, U.K. BN1 9QH. 3Seminar of Geometry and Combinatorics, University of Ghent, 9000 Gent, Belgium. Large arcs in cyclic planes of square order are constructed as orbits of a subgroup of a group whose generator acts as a single cycle. In the Desarguesian plane of even square order, this gives an example of an arc achieving the upper bound for complete arcs other than ovals. 1.
INTRODUCTION
Our aim is to demonstrate the existence of complete (q2 - q + 1)-arcs in a 2 2 cyclic projective plane II(q ) of order q . The only such plane known is
PG(2,q2),
the plane over the field GF(q2)
.
These arcs were found incidentally
by Kestenband [S], using different methods, as one of the possible types of intersection of two Hermitian curves in PG(2,q2) . The importance of these arcs, not observed in [S], is Segre's result that for q e en, a complete m-arc in 1 . Thus, this example of a PG(2,q) with m < q + 2 satisfies m 5 q - Jq +
complete arc attains the upper bound f o r q even As a by-product of the investiis the disjoint union of gation, it is shown that a Hermitian curve in PG 2,q') q + l of these arcs. 2.
NOTATION
Let
n
= n(q
2
)
be a cyclic projective plane of order 9'.
identify its points with the elements i of
ZV, v = q4
+
q2
+
cyclic group is generated by the automorphism u with o(i) = i [ 3 ] , 5 4 . 2 . The lines are obtained from a perfect difference set lo =
j = O,l,.. ., v
Ido,dl,., . dq2} as the sets u J ( l o ) , Let b = q2 + q + 1
and k = q
2
- q
are relatively prime, Zv= Zb x Zk. i = (1,s)
where i
In this notation u(i) = (r + 1, s
+
+
For
1;
i
-
l),
+
1, i
E
Zv,
1.
then v = bk.
Since b
and k
in Zv, we write
r(mod b), i L s(mod k)
taken modulo b and the second modulo k . to any arithmetical operation in Zv.
One can so that the
1,
.
where the sum of the first component is The notation extends in a natural way
J.C. Fischer, J . W.P.Hirschfeld and J.A. Thar
244
By the multiplier theorem of Hall [2], q 3 is a multiplier of II ; this 3 means that the mapping J, given by $(i) = q i is an automorphism of Il. Since q6 F 1 (mod v) , so J, is an involution. Indeed, J1 is a Baer involution since it fixes all b points of kZv = [(r,O) If we define r ( q 3 - 1) E 0 (mod b)
:
.
= I(r,s) : r
B
E
r
3
Zb1; this is because q r - r
E
Zbl for s
O,l, ..., k
=
-
=
1,
then o(Bs) = BS+l and the q2 - q + 1 Baer subplanes Bs partition Il. A l i n e of a Baer subplane Bs is a line of Il meeting Bs in q t 1 points. Similarly define = {(r,s) : s
K
E
iZk 1 for r = 0,1,. . . , b - 1 ,
whence o(Kr) = K r+l and the Kr also partition Il. Thus i = ( r , s ) = B n K s r It will turn out that Kr is a complete (q2 - q + 1)-arc. 3.
COMPLETE k-ARCS LEMMA 3 . 1 :
$(i)
,
=
(1,
k
-
s)
Proof:
i = (r,s) i n
For each
3
q3s F - s s k
For any l i n e
IK Proof:
izb
Zk, we have that
x
of i
,
q s + s = s ( q + l)k
LEMMA 3.2:
=
fixes the first component r
It was noted in 52 that
Now, for each s in iZk
whence
izv
.
n
~~1
.t
-
3
0 (mod k) ,
s (mod k )
.
0
BS , w i t h
of the Baer subpZane
is
odd if even
(r,s)
E
if ( r , s ) #
By lemma 3.1, the involution
$J
(r,s)
=
Bs n Kr,
K p..
fixes exactly one point of Kr
3
namely the point (r,O) where it meets B o ; the other points of Kr are interwhich implies changed in pairs. If K is a line of Bo it is fixed by $ , that the number of points of .t n Kr 1.t
of
n Krl Bs,
outside Bo
is even. Thus the parity of
varies as L n Kr n Bo is empty o r the point (r,O) . For a line .t apply the same argument to o-’(.t) , which is a line in Bo. 0
245
Complete Arcs in Planes of Square Order Let
LEMMA 3 . 3 :
be an automorphism group that a c t s regularly on t h e
S
p o i n t s of some p r o j e c t i v e p h n e
of order
n(n)
n,
and suppose t h a t
VO,V1, ...,Vt are the orbits of t h e p o i n t s under t h e a c t i o n of a subgroup s . If .t i s a Zine of n(n) and A . = 19.. n v.1 , then j=1 To each of the
n2
+
n
of
3
1
Proof:
G
A.(A. J
J
-
1) = I G I
of
elements y
1.
-
S\{l}
there corresponds
a unique pair of points P ,
Q of 9. for which y ( P ) = Q ; in fact, n 2 and Q = 2 n y(L) . If there was another such pair on 2 S would not act regularly on the lines of n(n) . Now we count the set
P
1
= y- ( 2 )
in two ways. First, each y
IJI
whence
= IGI - 1 .
then
,
other than the identity gives a unique pair
(P%Q),
Second, 9. is a disjoint union o f the sets 9. n V. , 3
and to each pair (P,Q) , P # Q , in 9. n V . there is a unique y in G such J that y ( P ) = Q ; hence IJI = 1 4 . (A. - 1) and so I J I = A . (A. - 1 ) . 0 Aj>l J J j=1 J J We are now ready to prove the main result.
In 14, an alternative proof is
provided that makes use of the properties of perfect difference sets. For
THEOREM 3.4:
k
=
q2
lie i n
-
q + 1 in
B
q
n(q2) . q + 1
are t h e
2,
>
each o r b i t
Kr
is a complete k-arc w i t h
Furthennore, the l i n e s through tangents t o
Kr
at
(r,s)
.
Bs n Kr = ( r , s ) t h a t
Proof: Fix a Baer subplane B and let II be one of its lines. For each orbit K r j ( j = 0 , 1 , . . . , q ) that meets .9. n Bs, set C I . + 1 = 12 n K I ; J rj for the remaining orbits, set @ . = n K 1 , j = q + 1, q + 2 , ... , b - 1 . 1 j' By lemma 3.2 both a . and B . are even. I
3
By definition,
By lemma 3.3, b-1
1
j =q+1
Bj (Bj - 1)
+
j=1
( a . + 1) a J. = q J
whence subtraction yields b-1 j =O
2
- 9,
J.C. Fischer, J. W.P.Hirschfeld and J.A. Thas
246
Consequently
B. I
E
{0,2}
for
j 2 q
Summarily, f o r any l i n e
(rj,s)
(i)
Krj
E
a t the point
1.
II o f t h e s u b p l a n e B s ,
so t h a t
II n Kr j
t
aj
= 0,
(!2 n
K
rj
1
either
= 1 and
is t a n g e n t t o
II
i = (r.,s) 1
or (ii)
.t n K,.
1
Bs = 0 and II meets
n
in
Krj
or
0
points.
2
i s a l i n e of e x a c t l y one o f t h e s u b p l a n e s Bs, it f o l l o w s i n more t h a n two p o i n t s ; t h a t i s , Krj is a t h a t no l i n e meets K r j (q2 q + 1 ) - a r c . From ( i ) it i s c l e a r t h a t , f o r each p o i n t ( r j , s ) o f K,. , I t h e q + 1 l i n e s of Bs through ( r j , s ) a r e t h e q + 1 t a n g e n t s o f K r j at t h i s point. S i n c e each l i n e of
il
-
For q
a s i m p l e c o u n t i n g argument suffices t o show t h a t t h e k - a r c
2 4 ,
Kr
i s complete. Assume t h e c o n t r a r y . Then t h e r e i s a p o i n t P through which p a s s q 2 - q t 1 t a n g e n t s o f Kr , one from each of i t s p o i n t s . S i n c e P E K r , for
some
r’ # r ,
q 2 - q + l
i t f o l l o w s t h a t through each p o i n t o f
tangents of
t h e g r o u p g e n e r a t e d by
(because
K
br a )
.
Since
Krl
l i n e s i s counted more t h a n t w i c e , whence tangents.
But a s
i ( q 2 - q + 1)’ When
L
Kr
(q
q = 3,
t
has e x a c t l y l)(q2 - q
+
and
Krl
is i t s e l f a k - a r c , none of t h e s e t a n g e n t 2 - q + l )2 has a t l e a s t ;(q
Kr
(q + l ) ( q 1)
,
- q
t
t a n g e n t s , we have
1)
q
a contradiction for
it must f i r s t b e observed t h a t
p l a n e of o r d e r 9 , Bruck [l].
there are
Krl
are o r b i t s u n d e r t h e a c t i o n o f
Kr
Then t h e o n l y 7 - a r c o f
2
4.
i s t h e unique c y c l i c
PG(2.9)
whose automorphism
PG(2,9)
group c o n t a i n s an element o f o r d e r 7 i s a complete arc, [ 3 ] , 514.7. q = 2
The c a s e
i s a genuine e x c e p t i o n : a 3 - a r c i s never complete. 0 Remark A theorem o f Segre [ 3 ] , 510.3, s t a t e s t h a t a complete m-arc i n
q
even, i s e i t h e r an o v a l , t h a t i s a (q + 2 ) - a r c , o r
m
5
q
-
PG(2,q),
Jq + 1 .
So, f o r
q
247
Complete Arcs ira Planes of Square Order an even s q u a r e , theorem 3 . 4 g i v e s an example of a complete (q - Jq + 1 ) - a r c and shows t h a t S e g r e ' s theorem cannot be improved i n t h i s c a s e . is e i t h e r a c o n i c , t h a t i s a (q + 1 ) - a r c , o r
m
odd
q ,
shows t h a t
PG(2,9)
q
odd,
This r e s u l t
However, t h e e x i s t e n c e o f a
iq + 1
i s n o t t h e b e s t bound f o r a l l
514.7.
[3],
LINES IN lI(q2)
4.
(i)
q
+
1 p o i n t s of the form
D
difference set (ii)
d ( k - 1) = j = 1,2,..
('12
.,
A line
one p o i n t .
Each o f i t s l i n e s
Bo.
L.
(rj, k - j)
L
S i n ce
of
j
Zk\{O}
i s f i x e d by
$,
r
Zb
of
j
r. # D
I t remains t o show t h a t
t h a n twice among t h e p o i n t s o f
3
L
.
, s # 0,
k - 1 differences
Bs.
l i n e of
i s a t an g e n t t o t h o s e
the other =
Kt
Bs ,
common w i t h i t .
in
q
Bs
+
D
generates 1
for
Zb
.
of t he p a r t i t i o n i n exact l y
o c c ur s a s t h e second component o f (I
interchanges
it f o l lo w s t h a t both
( r j , j ) and j
and
k - j
and t h a t no
r.
I
in
Zb\D
are
can ap p ear more
T h is f o l l o w s from t h e fact t h a t t h e p o i n t s of
Zv : each of t h e
t ( ( r j , j ) - (rj, k
that lie in
os(Bo)
therefore contains
ok
k - 1 differences
Bo
-
j))
.0
g iv e n i n t h e theorem i s e s s e n t i a l l y t h e
is a k - a r c whose t a n g e n t s are t h e l i n e s through
Kr
a l t e r n a t i v e proof t h a t
since
q,
must o c c u r e x a c t l y once, and t h e s e are accounted
The d e s c r i p t i o n of a l i n e of
Bo
for
.
c o n s t i t u t e a p e r f e c t d i f f e r e n c e set f o r i = (0,s)
II
Lemma 3 . 1 shows t h a t
p a i r e d with t h e same element
f o r by t h e
i s an element o f a p e r f e c t
meets any o t h e r subplane
Bo
of
e x a c t l y one p o i n t o f
of t h e form
d
i s an element o f a p e r f e c t d i f f e r e n c e s e t
d
Thus each element
.
, where
is i t s e l f a c y c l i c p l a n e of o r d e r
where
L
(d,O)
Zb;
for
3
a c y c l i c group f o r (d,O)
c o n s i s t s of
pairs of p o i n t s of the form ( r j , j ) and ( r j , k - j ) $ ( k - 1) , w i t h the r . d i s t i n c t elements of Zb\D .
Bo
Proof: elements
Bo
Any l i n e of
THEOREM 4 . 1 :
a
-
q
odd, [ 3 ] ,
PC(2,q),
q - Jq/4 + 7 / 4 .
5
has been s l i g h t l y improved by t h e t h i r d a u t h o r . complete 8 - a r c i n
q
For
§10.4, t h e comparable theorem s t a t e s t h a t a complete m-arc i n
(r,s)
The d e s c r i p t i o n of p o i n t s of t y p e ( i ) and ( i i ) shows t h a t any 0
or
2
points.
Kt
t h a t meet i t i n a p o i n t o f
Bo;
it meets
The proof i s completed by n o t i n g t h a t
which e i t h e r c o i n c i d e s with
Bo
o r h as no p o i n t s o r l i n e s i n
248
J.C. Fischer. J. W.P. Hirschfeld and J.A. Thas HERMITIAN CURVES
5.
The only known cyclic planes are the Desarguesian ones and, in this section, we restrict our attention to P G ( 2 , q 2 )
.
.
R . = o-’(Ro) 1
element of
Then
Lo.
and R .
i
1
In particular, L
theorem 4.1; now, D
Lo
of Bo and define are incident exactly when i + j (mod v)
It is convenient to distinguish one line n
B0
= { i = (d,O) : d
0 is a distinguished, perfect difference set f o r
iil The s e t
THEOREM 5 . 2 :
H = {
(d/2, s) : d
Hermitian curve and is t h e d i s j o i n t union of the
q
6
+
D, s
is an
as in (i) of
DI
E
.
‘b
Zk} i s a
E
1 compZete k-arcs
Kd/2 ’
Bs is a conic or a l i n e of Bs according a s q is odd or even, whence tI is a disjoint union o f k subconics or k s u b l i n e s
liii H
n
accordingly.
Define the correlations $ : i a . and p : ( r , s ) c-f 2 (r,-s) . is an ordinary polarity for q odd and a pseudo polarity for q even, [ 3 ] , g 8 . 3 . Thus, with J , as in 12, we have that p = $0 = $ 9 . In fact, p is Pro0f:
Then
+-f
$
a Hermitian polarity since the self-conjugate points of p are the q3 + 1 points (r,s) satisfying ( r , s ) + (r,-s) = (d,O) for d in D , From this ( i ) follows. In Bo in Zb. So
the points are
Bo
(r,O)
while the lines are R
is self-polar with respect to
self-conjugate points of the polarity
@
p
(r,O)
and meets H
induced on
B0
by
p
’
.
both with r in the q + 1 These self-
conjugate points form a subconic when q is odd and a subline when q is even. there exists s ‘ such that bs’ t s (mod k) since b and k are Given s , coprime. Thus H n Bs = 0bs’ (13 n Bo) = I(d/2, s) : d 6 D} is a conic or a line of Bs according as q is odd o r even, and the last part of (ii) follows. 0 THEOREM 5 . 2 : q
The tangents t o any complete
( q 2 - q + 1)-arc
in
2 PG(2,q )
,
even, form a dual Hermitian arc. Proof:
See Thas [ 6 ] .
THEOREM 5.3: The tangents t o any o f t h e complete 2 PG(2,q ) form a dual H e m i t i a n curve i f and only i f q
Proof:
Let
chosen s o that 21) = D and each multiplier of
q
(q2 - q + 1)-arcs
Kr
in
i s even.
be even and consider the arc K O , where D has been (which is always possible since 2 is a Hall multiplier Bo fixes at least one line of Bo) . Then the tangents
249
Complete Arcs in Planes of Square Order to
II II s
(0,O) ,
at
KO
with
(d, 0 )
to
(d,O)
namely t h e l i n e s o f
in
l i n e s of t h e p o l a r i t y
D = 2D
determined by
p
c o i n c i d e s with t h e s e t o f t a n g e n t s t o q
b e odd.
q'
Hence t h e t a n g e n t s t o
,
to
{I. : j
have t h e form
,
(0,s) =
takes
(d,s),
d
U,
6
J these lines a r e the self-conjugate
Thus t h e s e t of t a n g e n t s t o
H.
KO
H.
- q
and so i s odd, t h i s number i s never
1
+
not i n
P
do n o t form a d u a l H e r m i t i a n a r c . 0
Kr
Each of t h e
Tb'EOREM 5 . 4 :
is
KO
(0,O)
(0,O)
S i n c e t h e number o f t a n g e n t s from a p o i n t
h a s t h e p a r i t y of
Kr
to
Kr
q + 1 .
which t a k e s
t h e set of t a n g e n t s t o
(d,-s) ' From t h e assumption t h a t
Now l e t
containing
Bo
obs' ,
Since
D .
9,
Zk}.
E
d
(q2 - q
+
i s t h e intersection of t u o
Kr
1)-arcs
Wermitian curves. First, let
Proof: contained i n a p(H)
=
H
q
be even.
H,
Hermitian c u r v e
and l e t
H*
by t h e t a n g e n t s t o Now, l e t
q
Kr
Then a s i n theorem 5 . 1 , t h e a r c
which d e t e r m i n e s a p o l a r i t y
Then, a s i n theorem 5 . E
D,
E
D, s
s c izk
Hence Hr = { ( d / 2 + r , s ) : d
E
iZk
In f a c t , s i n c e t h e r e e x i s t s r ' kr' we have t h a t Hr = o (Ho). Since D
i s a l s o a Hermitian c u r v e .
,
Hrl
n Hr2
and
d2
= Kt
in
D
,
so
Zb,
Let
(b)
If
H
r2,
Also
s i n c e t h e r e e x i s t unique
dl
0 , q
even, and l e t
then
q > 2 ,
Suppose m
theorem 1 0 . 3 . 3 , c o r o l l a r y 2 ) , (q2 + 2 ) - a r c .
+
is a perfect
r1 # r 2 .
ti.
m = 4 if q = 2 . 2 m = q - q + 1 and (i)
f o r any
k
such t h a t
be a Hemitian curve in PG(2,q')
(m+1)-'zw in H , 2 m = q - q + l if q > 2 ;
Pro0f:
=
rl = i d
+
If there is no (a)
(ii)
n Hr21
d l - d 2 E 2 ( r 2 - r l ) (nod b ) .
be an m-izric contained in
(i)
IHrl
t = ad
where such t h a t
THEOREM 5.5: K
is
Kr
Let
b e t h e d u a l H e r m i t i a n c u r v e o f theorem 5 . 3 t h a t is formed Then p(Kr) = H* n h , whence K = p(H*) n H .
be even o r odd.
is a Hermitian c u r v e .
difference set in
.
.
H = Ho = { ( d / 2 , s ) : d
k r ' : r (mod b)
p
>
K
Now, count t h e p a i r s
q'
then -
K
q + 1.
is compLete. Then by S e g r e ' s theorem ( [ 3 ] ,
is c o n t a i n e d i n an o v a l (P,Q)
such t h a t
P
E
t h a t is a
0, K,
Q
E
0 ,
P
# 9 and
250
J. C.Fischer. J. W.P. Hirschfeld and J.A. Thas
PQ is tangent to H .
There are at most two points P f o r a given Q ,
since
three would be collinear. So
Hence 3m
2q2
5
+
4,
and 3q2
2 ( q 2 + 2 - m)
m
.C
-
3q
+
3
c
2q2
+
. 4
implies that q = 2 .
This
gives the result. (ii)
Suppose K is not complete, then the same argument as (i)
gives q ' whence q2
-
3q
-
1
5 0 ;
-
q
+ 1 5
2(q
that is, q = 2 .
+
l),
U
For q odd, the points of Bo together with the q2 + q + 1 Remark: conics Cr = { (6d + r, 0) : d E D} , r E Zb , form a plane of order q This
.
plane is isomorphic to PG(2,q) via the isomorphism 6 given by 8(x,O) = (gx, 0) For all q , this configuration of conics also appears as the section
.
by a plane
TI
in PG(3,q),
of the q2 + q + 1 quadric surfaces through a twisted cubic T where 71 is skew to T ; see [4], theorem 21.4.5.
REFERENCES [l]
Bruck, R.H., Quadratic extensions of cyclic planes, Proc. Sympos. AppZ.
Math. 10 (1960), 15-44. [2]
Hall, M., Cyclic projective planes, Duke Math. J. 14 (1947), 1079-1090.
[3]
Hirschfeld, J.W.P.
Projective Geometries over Finite Fields (Oxford
University Press, Oxford, 1979). [4]
Hirschfeld, J.W.P., Finite Projective Spaces of Three Dimensions (Oxford University Press, Oxford, to appear).
[5]
Kestenband, B., Unital intersections in finite projective planes, G e m . Dedicata 11 (1981), 107-117.
[6]
Thas, J.A., Elementary proofs of two fundamental theorems of B. Segre without using the Hasse-Weil theorem, J . Combin. Theory Ser. A . 34 (1983), 381-384.