Special Issue (2003), S119–S139
Advances in Geometry
( de Gruyter 2003
On Mathon’s construction of maximal arcs in
Desarguesian planes
Frank Fiedler, Ka Hin Leung and Qing Xiang
Dedicated to Adriano Barlotti on the occasion of his 80th birthday
Abstract. We study the problem of determining the largest d of a non-Denniston maximal arc
of degree 2 d generated by a f p; 1g-map in PGð2; 2 m Þ via a recent construction of Mathon [9].
On one hand, we show that there are f p; 1g-maps that generate non-Denniston maximal arcs
of degree 2ðmþ1Þ=2 , where m d 5 is odd. Together with Mathon’s result [9] in the m even case,
this shows that there are always f p; 1g-maps generating non-Denniston maximal arcs of degree
2bðmþ2Þ=2c in PGð2; 2 m Þ. On the other hand, we prove that the largest degree of a non-Denniston
maximal arc in PGð2; 2 m Þ constructed using a f p; 1g-map is less than or equal to 2 m3 . We
conjecture that this largest degree is actually 2bðmþ2Þ=2c when m > 9.
Key words. Arc, linearized polynomial, maximal arc, quadratic form.
1 Introduction
Let PGð2; qÞ be the Desarguesian projective plane of order q, q a prime power. A set
of k points in PGð2; qÞ is called a ðk; nÞ-arc if no n þ 1 points of the set are collinear.
The number n is usually called the degree of the arc.
Let K be a ðk; nÞ-arc in PGð2; qÞ, and let P be a point in K. Then each of the q þ 1
lines through P contains at most n 1 points of K. Therefore
k c 1 þ ðq þ 1Þðn 1Þ ¼ qn þ n q:
A ðk; nÞ-arc is said to be maximal if k ¼ qn þ n q. From the above argument, it is
easily seen that any line of PGð2; qÞ that contains a point of a maximal arc K must
contain exactly n points of that arc; that is
jL V Kj ¼ 0 or n
for every line L of PGð2; qÞ. Therefore the degree n of a maximal ðqn þ n q; nÞ-arc
must divide q.
The study of arcs of degree greater than two was started by Barlotti [2]. For
q ¼ 2 m , Denniston [3] constructed maximal ðqn þ n q; nÞ-arcs in PGð2; qÞ for every
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Frank Fiedler, Ka Hin Leung and Qing Xiang
n, n j q, n < q (see also [6, p. 304]). Thas [10], [11] also gave two other constructions
of maximal arcs of certain degrees in PGð2; 2 m Þ, where m is even. For odd prime
powers q, Ball, Blokhuis and Mazzocca [1] proved that maximal arcs of degree n do
not exist in PGð2; qÞ, when n < q. Recently Mathon [9] gave a new construction of
maximal arcs in PGð2; 2 m Þ that generalizes the construction of Denniston. We give a
brief account of his construction.
Let C be the set of all conics
Fa; b; l ¼ fðx; y; zÞ A PGð2; 2 m Þ j ax 2 þ xy þ by 2 þ lz 2 ¼ 0g
where a; b A F2m and ax 2 þ x þ b is irreducible over F2 m (that is, Tr2 m =2 ðabÞ ¼ 1,
here Tr2 m =2 is the trace map from F2 m to F2 ). For l; l 0 A F2 m , l 0 l 0 we define a
composition
Fa; b; l l Fa 0 ; b 0 ; l 0 ¼ Fala 0 ; blb 0 ; lþl 0
where
a l a0 ¼
al þ a 0 l 0
l þ l0
for any a; a 0 A F2 m :
A subset F of C is said to be closed under the composition l if for any F1 ; F2 A F
with F1 0 F2 we have F1 l F2 A F. In [9] Mathon proved that the set of points of
all conics in a closed set of conics together with the common nucleus F0 ¼ Fa; b; 0 ¼
ð0; 0; 1Þ forms a maximal arc in PGð2; 2 m Þ. When all conics in a closed set of conics
come from a single pencil of conics, Mathon’s construction gives rise to Denniston
maximal arcs. In general, Mathon showed that closed sets of conics can be obtained
by using linearized polynomials over F2 m . Specifically, Mathon proved the following
theorem.
P d1
P d1
i
i
Theorem 1.1 ([9, Theorem 2.5]). Let pðxÞ ¼ i¼0
a i x 2 1 and qðxÞ ¼ i¼0
bi x 2 1
be polynomials with coe‰cients in F2 m . For an additive subgroup A of order 2 d in F2 m
let F ¼ fFpðlÞ; qðlÞ; l j l A Anf0gg H C be a set of conics with common nucleus F0 . If
Tr2 m =2 ð pðlÞqðlÞÞ ¼ 1 for every l A Anf0g, then the set of points on all conics in F
together with F0 forms a maximal ð2 mþd 2 m þ 2 d ; 2 d Þ-arc K in PGð2; 2 m Þ. If both
pðxÞ, qðxÞ have d c 2, then K is a Denniston arc.
Hamilton [4] gave the following test for when the arc K in Theorem 1.1 is a
Denniston arc.
Theorem 1.2 ([4, Theorem 2.1]). Let pðxÞ and qðxÞ be the same polynomials as given
in Theorem 1.1, let A be an additive subgroup of size 2 d in F2 m , and let K be the
maximal arc obtained in Theorem 1.1. Then K is of Denniston type if and only if for
all l; l 0 A Anf0g, l 0 l 0 , both ðpðlÞ þ pðl 0 ÞÞ=ðl þ l 0 Þ and ðqðlÞ þ qðl 0 ÞÞ=ðl þ l 0 Þ are
constant.
On Mathon’s construction of maximal arcs in Desarguesian planes
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Mathon posed several problems at the end of his paper [9]. The third problem he
posed is: What is the largest d of a non-Denniston maximal arc of degree 2 d generated by a f p; qg-map in PGð2; 2 m Þ via Theorem 1.1? When m is even, Mathon [9]
showed that there exists a non-Denniston maximal arc of degree 2 m=2þ1 generated by
a fp; 1g-map in PGð2; 2 m Þ. When m is odd, Hamilton [4] showed that there exists
a non-Denniston maximal arc of degree 8 generated by a fp; 1g-map in PGð2; 2 m Þ,
where m d 5. In this paper, we concentrate on the following restricted version of
Mathon’s problem: What is the largest d of a non-Denniston maximal arc of degree
2 d generated by a fp; 1g-map in PGð2; 2 m Þ via Theorem 1.1? In Section 2, we show
that there are fp; 1g-maps that generate non-Denniston maximal arcs of degree
2ðmþ1Þ=2 , where m d 5 is odd. Together with Mathon’s result [9, Theorem 3.2] in the
m even case, this shows that there are always f p; 1g-maps generating non-Denniston
maximal arcs of degree 2bðmþ2Þ=2c in PGð2; 2 m Þ. In Section 3 we prove that if a maximal arc generated by a f p; 1g-map via Theorem 1.1 has degree 2 m1 or 2 m2 and
m d 7, then it is a Denniston maximal arc. Hence when m d 7, the largest degree of
a non-Denniston maximal arc constructed using a f p; 1g-map via Theorem 1.1 is less
than or equal to 2 m3 . We conjecture that when m > 9, this largest degree is actually
2bðmþ2Þ=2c and provide some evidence for this conjecture.
2
Maximal arcs in PG(2, 2 m ), m odd
In this section m is always an odd positive integer, and g always denotes an element
of F2 m with Tr2 m =2 ðgÞ ¼ 1. To simplify notation, from now on, we will use Tr in place
of Tr2 m =2 if there is no confusion. We start with the following lemma.
Lemma 2.1. Let Sg ¼ fx A F2 m j Trðgx þ x 3 Þ ¼ 0g. Then there exists a choice of
g A F2 m such that Sg contains an F2 -subspace A with dimðAÞ ¼ mþ1
2 .
Proof. Let Qg ðxÞ ¼ Trðgx þ x 3 Þ and let V ¼ F2 m . The map Qg : V ! F2 is a quadratic form on V over F2 . The corresponding bilinear form B is given by Bðx; yÞ ¼
Qg ðx þ yÞ Qg ðxÞ Qg ð yÞ ¼ Trðx 2 y þ xy 2 Þ, hence
Rad V ¼ fx A V j Bðx; yÞ ¼ 0 for each y A V g
¼ fx A V j Trðx 2 y þ xy 2 Þ ¼ 0 for each y A V g
pffiffiffi
¼ fx A V j Trð yðx 2 þ x ÞÞ ¼ 0 for each y A V g
pffiffiffi
¼ fx A V j x 2 ¼ x g:
Since m is odd, we conclude that Rad V ¼ F2 . Note that in characteristic 2, the
quadratic form Qg ðxÞ is not necessarily zero on Rad V . Therefore we define
V0 ¼ fx A Rad V j Qg ðxÞ ¼ 0g:
This is an F2 -space of dimension equal to dimðRad V Þ or dimðRad V Þ 1. Since
TrðgÞ ¼ 1, we have V0 ¼ Rad V ¼ F2 . Hence rankðQg Þ ¼ m 1 is even and Qg is
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Frank Fiedler, Ka Hin Leung and Qing Xiang
either hyperbolic or elliptic. It is always possible to choose g A F2 m , with TrðgÞ ¼ 1,
such that Qg is hyperbolic on V =V0 . (This can be seen from the weight distribution
of the dual of the double-error-correcting BCH code, see [8, p. 451]). With this choice
of g, the maximum dimension of a subspace of V =V0 on which Qg vanishes is m1
2 .
Let U be such a subspace and let A ¼ U ? V0 . Then dimðAÞ ¼ mþ1
and
Q
ðxÞ
vang
2
ishes on A, hence A H Sg . This completes the proof.
r
Now let g A F2 m be chosen such that TrðgÞ ¼ 1 and Sg ¼ fx A F2 m j Trðgx þ x 3 Þ ¼
3
0g contains an F2 -subspace A of F2 m of dimension mþ1
2 . Let pðxÞ ¼ 1 þ gx þ x .
Then we have the following corollary of Theorem 1.1.
Theorem 2.2. The set of points on the conics F ¼ fFpðlÞ; 1; l j l A Anf0gg together with
the common nucleus F0 forms a maximal arc K in PGð2; 2 m Þ of degree 2ðmþ1Þ=2 . When
m d 5, the maximal arc K is non-Denniston.
Proof. Let pðlÞ ¼ 1 þ gl þ l 3 , with the choice of g as above, and let A be the mþ1
2 dimensional F2 -subspace in Sg given by Lemma 2.1. Then we have TrðpðlÞÞ ¼
Trð1Þ ¼ 1 for every l A Anf0g. By Theorem 1.1, the first part of the theorem follows.
When m d 5, the maximal arc K is non-Denniston. This can be seen as follows.
For l; l 0 A Anf0g, ðpðlÞ þ pðl 0 ÞÞ=ðl þ l 0 Þ ¼ g þ l 2 þ ll 0 þ l 02 . When jAj d 8, this
expression cannot be constant when l; l 0 , l 0 l 0 , run through Anf0g. Therefore by
Theorem 1.2, the arc K is not of Denniston type.
r
Theorem 2.2 together with Mathon’s result ([9, Theorem 3.2]) in the m even case
shows that there are always f p; 1g-maps generating non-Denniston maximal arcs of
degree 2bðmþ2Þ=2c in PGð2; 2 m Þ, when m d 5.
3
Some upper bounds on the degree of non-Denniston maximal arcs
from { p, 1}-maps
We start this section by making some remarks about Theorem 1.1. In Theorem 1.1,
Mathon restricted the degrees of the polynomials pðlÞ, qðlÞ to be less than or equal
to 2 d1 1, where the subspace A H F2 m involved has size 2 d . We will show that
there is no loss of generality in doing so.
P m1
i
Proposition 3.1. Let f ðxÞ ¼ i¼0
a i x 2 1 A F2 m ½x, and let A be an F2 -subspace in F2 m
P d1
i
of size 2 d , where d c m 1. Then there exists a polynomial f1 ðxÞ ¼ i¼0
bi x 2 1 A
F2 m ½x such that f ðlÞ ¼ f1 ðlÞ for every l A Anf0g.
Q
Proof. Let AðxÞ ¼ l A A ðx lÞ. This is a degree 2 d linearized polynomial in F2 m ½x
(see [7, p. 110], also [8, p. 119]), that is,
d
AðxÞ ¼ x 2 þ cd1 x 2
d1
þ þ c 0 x;
where ci A F2 m . Let aðxÞ ¼ x d þ cd1 x d1 þ þ c 0 . The polynomials AðxÞ and aðxÞ
are called 2-associates of each other (see [7, p. 115]). Let f ðxÞ ¼ GðxÞ=x, where GðxÞ ¼
On Mathon’s construction of maximal arcs in Desarguesian planes
P m1
i
a i x 2 , and let gðxÞ ¼
algorithm, we write
i¼0
P m1
i¼0
S123
a i x i be the 2-associate of GðxÞ. Using the division
gðxÞ ¼ kðxÞaðxÞ þ rðxÞ;
ð3:1Þ
where deg rðxÞ < deg aðxÞ ¼ d. Let KðxÞ and RðxÞ be the 2-associates of kðxÞ and
rðxÞ respectively. Turning (3.1) into linearized 2-associates, and noting that the 2associate of kðxÞaðxÞ is KðAðxÞÞ, the composition of AðxÞ with KðxÞ (cf. [7, p. 115],
Lemma 3.59), we get
GðxÞ ¼ KðAðxÞÞ þ RðxÞ;
ð3:2Þ
with deg RðxÞ c 2 d1 . With f1 ðxÞ ¼ RðxÞ=x, we see from (3.2) that f ðlÞ ¼ f1 ðlÞ for
every l A Anf0g.
r
We note that if one does not restrict the degree of the polynomials pðxÞ, qðxÞ to be
less than or equal to 2 d1 1 (where 2 d ¼ jAj), Theorem 1.1 still holds, but then it
sometimes leads to Denniston maximal arcs, which, at first sight, may not look like
Denniston. We give a couple of examples of this situation below. So by restricting the
degrees of the polynomials pðxÞ, qðxÞ to be less than or equal to 2 d1 1 in Theorem 1.1, not only is there no loss of generality (by Proposition 3.1), but also some
‘‘trivial’’ examples are avoided.
2
4
2 m1
A F2 m ½x, where Trða 0 Þ ¼ 1. Let A ¼
Example 3.2. Let pðxÞ ¼ a 0 þ xþx þx þþx
x
fx A F2 m j TrðxÞ ¼ 0g. Then we have Trð pðlÞÞ ¼ 1 for every l A Anf0g. This pðxÞ
indeed gives rise to a maximal arc of degree 2 m1 in PGð2; 2 m Þ by Mathon’s construction. But the maximal arc in this example is of Denniston type by Theorem 1.2
since for every l A Anf0g, we have pðlÞ ¼ a 0 , a constant.
P m1
i
a i x 2 1 A F2 m ½x, where Trða 0 Þ ¼ 1. We may choose
Example 3.3. Let pðxÞ ¼ i¼0
2
m1
a1 ; a2 ; . . . ; a m1 A F2 m such that A ¼ fl A F2 m j a1 l 2 þ a2 l 2 þ þ a m1 l 2 ¼ 0g
has dimension m 2 over F2 . Then we have TrðpðlÞÞ ¼ 1 for every l A Anf0g. This
pðxÞ gives rise to a maximal arc of degree 2 m2 in PGð2; 2 m Þ by Mathon’s construction. But the maximal arc in this example is again of Denniston type by Theorem 1.2
since for every l A Anf0g, pðlÞ ¼ a 0 , a constant.
Next we prove that when m d 5 the largest d of a non-Denniston maximal arc of
degree 2 d generated by a f p; 1g-map via Theorem 1.1 is less than m 1.
Theorem 3.4.PLet A be an additive subgroup of size 2 m1 in F2 m , where m d 5.
i
m2
Let pðxÞ ¼ i¼0
a i x 2 1 A F2 m ½x. If TrðpðlÞÞ ¼ 1 for all l A Anf0g, then a2 ¼
a3 ¼ ¼ a m2 ¼ 0, thus pðxÞ is linear and the maximal arc obtained via Theorem 1.1
is of Denniston type.
Proof. Every hyperplane in F2 m can be written as fx A F2 m j TrðaxÞ ¼ 0g for some
nonzero a A F2 m . By making a change of variable in pðxÞ, we may assume that A ¼
fx A F2 m j TrðxÞ ¼ 0g. We consider two cases.
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Frank Fiedler, Ka Hin Leung and Qing Xiang
Case 1: Trða 0 Þ ¼ 1. In this case, if Trð pðlÞÞ ¼ 1 for all l A Anf0g, then
P m2
P m2
i
i
a i l 2 1 Þ ¼ 0 for all l A Anf0g. Thus, ð1 þ TrðxÞÞ Trð i¼1
a i x 2 1 Þ, viewed
Trð i¼1
as a function from F2 m to itself, is identically zero. That is, in F2 m ½x, we have
ð1 þ TrðxÞÞ Tr
X
m2
aix
2 i 1
m
1 0 ðmod x 2 xÞ
ð3:3Þ
i¼1
Let tðxÞ ¼ LHS of ð3:3Þ ¼ ð1 þ x þ x 2 þ þ x 2
Claim: The coe‰cient of x 2
m
2 r þ2
m1
r
Þ Trð
P m2
i¼1
i
a i x 2 1 Þ.
r
2
2
in tðxÞ is amr
þ amrþ1
for 3 c r c m 2.
The m-bit binary representation of 2 m 2 r þ 2 is
1
...10
. . . 0 10;
|fflffl{zfflffl}
|fflffl{zfflffl}
mrd2 r2d1
which contains two blocks of 1’s (separated by 0’s). (We will always number the bits
from right to left as 0; 1; 2; . . . ; m 1.) Note that the exponents of the summands
in 1 þ TrðxÞ, written in m-bit binary representation, are 000 . . . 000,
000 . . . 001,
P m2
i
000 . . . 010; . . . ; 100 . . . 000, and the exponents of the summands in Trð i¼1
a i x 2 1 Þ
are cyclic shifts of
000 . . . 001; 000 . . . 011; 000 . . . 0111; 000 . . . 01111; . . .
and
00 11
. . 111
|fflfflfflfflffl.ffl{zfflfflfflfflffl
ffl} :
|{z}
2
m2
P m2
i
When we multiply 1 þ TrðxÞ with Trð i¼1
a i x 2 1 Þ, there are two ways to obtain
m
r
x 2 2 þ2 , namely adding the exponent of a summand in 1 þ TrðxÞ to the exponent
P m2
i
of a summand in Trð i¼1
a i x 2 1 Þ with or without carry.
Suppose that we are in the
latter case. The exponent from 1 þ TrðxÞ must be 2
P m2
i
while the exponent from Trð i¼1
a i x 2 1 Þ is a shift of 2 mr 1.
1
. . . 1 0 . . . 010 ¼ 0 . . . 010 þ 1|fflffl{zfflffl}
. . . 1 0|fflffl{zfflffl}
...0
|fflffl{zfflffl}
mrd2
mrd2
r
r
2
Thus, this case contributes the coe‰cient amr
.
Now suppose that we are in the former case. Since bit-1 of 2 m 2 r þ 2 is 1
while bit-0 is 0, the exponent 2 m 2 r þ 2 must be obtained as 2 0 added to
ð2 m 2 r Þ þ ð2 1 2 0 Þ:
1|fflffl{zfflffl}
. . . 1 0 . . . 010 ¼ 0 . . . 01 þ 1|fflffl{zfflffl}
. . . 1 0 . . . 01;
mrd2
mrd2
r
2
so this case contributes the coe‰cient amrþ1
. The claim now follows. In particular,
by (3.3), we find that a2 ¼ a3 ¼ ¼ a m2 .
On Mathon’s construction of maximal arcs in Desarguesian planes
Claim: The coe‰cient of x 2
m
4
S125
4
4
8
in tðxÞ is am2
þ am3
þ am3
.
Clearly the exponent ð2 m 4Þ ¼ 11 . . . 100 can be obtained by
11 . . . 100 ¼ 00 . . . 000 þ 11 . . . 100
4
.
This contributes the coe‰cient am2
m
Also the exponent 2 4 can bePobtained by adding a non-zero exponent in
i
m2
1 þ TrðxÞ to an exponent from Trð i¼2
a i x 2 1 Þ. Suppose that when adding the
exponents, there is no carry. We have two ways to obtain 2 m 4, namely,
11 . . . 100 ¼ 10 . . . 0 þ 011 . . . 100;
11 . . . 100 ¼ 0 . . . 0100 þ 11 . . . 1000:
4
8
This contributes the coe‰cient am3
þ am3
. Finally we note that there is no way of
m
getting 2 m 4 as a sum of exponents inducing a carry. Thus, the coe‰cient of x 2 4
in tðxÞ is as claimed. This implies a m3 ¼ 0, which yields a2 ¼ a3 ¼ ¼ a m2 ¼ 0.
Hence pðlÞ ¼ a 0 þ a1 l.
P m2
i
a i l 2 1 Þ ¼ 1 for all l A Anf0g. Hence
Case 2: Trða 0 Þ ¼ 0. PWe have Trð i¼1
i
m2
ð1 þ TrðxÞÞ ð1 þ Trð i¼1 a i x 2 1 ÞÞ, viewed as a function from F2 m to itself, is the
characteristic function of the subset f0g of F2 m . Therefore,
ð1 þ TrðxÞÞ þ ð1 þ TrðxÞÞ Tr
X
m2
ai x
2 i 1
1 1 x2
m
1
m
ðmod x 2 xÞ ð3:4Þ
i¼1
m
Note that the binary representation of the exponent of x 2 1 is 111 . . . 1 (m ones
altogether), while in the left hand side of (3.4), the binary
of the
P m2 representation
i
exponent of any term in the product ð1 þ TrðxÞÞ Trð i¼1
a i x 2 1 Þ cannot have
more than 1 þ ðm 2Þ ¼ m 1 ones. So (3.4) cannot hold. Thus, this case does not
occur. This completes our proof.
r
Remarks. (1) Theorem 3.4 is not true when m ¼ 4. In PGð2; 16Þ, there exists a degree
8 non-Denniston maximal arc (cf. Section 4.1 of [9]).
(2) It is interesting to note that when m d 5 a non-Denniston maximal arc of degree
2 m1 (i.e., the dual of a hyperoval) in PGð2; 2 m Þ can be obtained from fp; qg-maps
via Theorem 1.1, with qðxÞ 0 1. See [9, p. 362] for an example in PGð2; 32Þ. Theorem
3.4 shows that this cannot be achieved if m d 5 and qðxÞ is restricted to be 1.
The ideas in the proof of Theorem 3.4 can be further used to prove the following
theorem. The proof contains more complicated computations.
Theorem 3.5. Let P
A be an additive subgroup of size 2 m2 in F2 m , where
i
m3
m d 7. Let pðxÞ ¼ i¼0
a i x 2 1 A F2 m ½x. If TrðpðlÞÞ ¼ 1 for all l A Anf0g then
a2 ¼ a3 ¼ ¼ a m3 ¼ 0, thus pðxÞ is linear and the maximal arc obtained via Theorem 1.1 is of Denniston type.
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Frank Fiedler, Ka Hin Leung and Qing Xiang
Proof. Since A has dimension m 2 over F2 , we may assume that A ¼ fx A F2 m j
TrðxÞ ¼ 0 and TrðmxÞ ¼ 0g for some m A F2m with m 0 1. Again we consider two
cases.
Case 1: Trða 0 Þ ¼ 1. Then
ð1 þ TrðxÞÞð1 þ TrðmxÞÞ Tr
X
m3
i
a i x 2 1 1 0
m
ðmod x 2 xÞ
i¼1
ð1 þ TrðxÞ þ TrðmxÞ þ TrðxÞ TrðmxÞÞ Tr
X
m3
aix
2 i 1
ð3:5Þ
ðmod x
10
2m
xÞ
i¼1
Let rðxÞ denote
of (3.5), sðxÞ ¼ 1 þ TrðxÞ þ TrðmxÞ þ TrðxÞ TrðmxÞ, and
P m3 the2 iLHS
tðxÞ ¼ Trð i¼1
a i x 1 Þ. The exponent of each term in rðxÞ is a sum of the exponent
of a summand in sðxÞ and the exponent of some summand in tðxÞ. Similar to the
proof of Theorem 3.4, exponents of the summands in tðxÞ are 2 i 1, 1 c i c m 3,
and their cyclic shifts. Exponents from sðxÞ are 0; 2 i ; and 2 i þ 2 j , i 0 j. The terms x 0 ,
i
i
j
i
i1
i
j
x 2 , and x 2 þ2 ði 0 jÞ in sðxÞ have coe‰cients 1, 1 þ m 2 þ m 2 , and m 2 þ m 2 ,
respectively.
Claim: The coe‰cient of x ð2
m
m1
2
ð1 þ m 2
am3
m1
1Þ2 m2 2 m4
m3
þ m2
2
ðm 2
þ am4
m3
m4
in rðxÞ is
Þ þ a m4 ðm 2
þ m2
m5
m1
þ m2
Þ þ a m3 ðm 2
m1
m3
Þ
þ m2
m4
Þ:
The binary representation of the exponent of any term in rðxÞ cannot have more
than 2 þ ðm 3Þ ¼ m 1 ones. The binary expansion of ð2 m 1Þ 2 m2 2 m4 is
101011 . . . 1. This involves m 2 ones, so it can be obtained as a sum of two exponents (one from sðxÞ, the other from tðxÞ) without carry or with exactly one carry.
Assume that we are in the former case. There are only three ways to obtain ð2 m 1Þ 2 m2 2 m4 , namely,
101011 . . . 1 ¼ 001000 . . . 0 þ 100011 . . . 1
¼ 101000 . . . 0 þ 000011 . . . 1
¼ 001010 . . . 0 þ 100001 . . . 1:
m3
m4
m1
m1
m3
2
These contribute the coe‰cient ð1 þ m 2 þ m 2 Þam3
þ ðm 2 þ m 2 Þa m4 þ
2 m3
2 m5 2 m1
ð2 m 1Þ2 m2 2 m4
ðm
þm
Þam4 for x
in rðxÞ. (Here we used the assumption
4
2
that m d 7. If m ¼ 5, the coe‰cient of the term x 2 þ2 þ1 in rðxÞ is not the same
as in our claim. The reason is that, for example, 10101 ¼ 00100 þ 10001 leads to
another possibility, namely 00100 comes from a14 x 4 in tðxÞ, and 10001 comes from
0
4
0
4
ðm 2 þ m 2 Þx 2 þ2 in sðxÞ. This cannot happen if m d 7.)
On Mathon’s construction of maximal arcs in Desarguesian planes
S127
Now assume that a carry had been induced. The last carry-over must have occurred
either at bit-ðm 3Þ or bit-ðm 1Þ. The latter case cannot occur.
10101 . . . 1 ¼ 10010 . . . 0 þ 0001 . . . 1:
This contributes the coe‰cient ðm 2
we have
m1
2
am3
ð1 þ m 2
m3
m1
m
a m4 ðm 2
þ m2
m4
Þ þ a m4 ðm 2
þ m2
2
ðm 2
þ am4
Claim: The coe‰cient of x ð2
m1
m3
þ m2
m5
þ m2
m3
Þa m3 . This proves the claim. By (3.5),
m1
þ m2
Þ þ a m3 ðm 2
1Þ2 m1 2 m4
m2
m4
m1
m3
Þ
þ m2
m4
Þ¼0
ð3:6Þ
is
Þ þ a m3 ðm 2
m2
m4
þ m2
Þ:
The binary expansion of ð2 m 1Þ 2 m1 2 m4 is 011011 . . . 1. Suppose it is
obtained as a sum of exponents from sðxÞ and tðxÞ without carry. Then
01101 . . . 1 ¼ 0110 . . . 0 þ 00001 . . . 1
m2
m3
which contributes ðm 2 þ m 2 Þa m4 . (Here again we have used the assumption that
4
3
m d 7. If m ¼ 6, the coe‰cient of the term x 2 þ2 þ2þ1 in rðxÞ is not the same as in our
claim. The reason is that 011011 ¼ 011000 þ 000011 leads to another possibility,
4
3
0
0
namely 011000 comes from a28 x 2 þ2 in tðxÞ, and 000011 comes from ðm 2 þ m 2 Þx 2 þ2
in sðxÞ. This cannot happen if m d 7.)
If ð2 m 1Þ 2 m1 2 m4 is obtained as a sum of exponents from sðxÞ and tðxÞ
with a carry, the last carry-over must occur at bit-ðm 2Þ or bit-0.
01101 . . . 1 ¼ 01010 . . . 00 þ 00011 . . . 11:
This contributes the coe‰cient ðm 2
and by (3.5), we have
a m4 ðm 2
m1
2
Claim: am3
ð1 þ m 2
m3
þ m2
m2
m4
m2
þ m2
þ m2
m3
m4
Þa m3 . Therefore the claim is proved,
Þ ¼ a m3 ðm 2
m1
2
Þ þ am4
ðm 2
m3
þ m2
m2
m5
þ m2
m4
Þ:
ð3:7Þ
Þ ¼ 0.
The claim is equivalent to
a m3 ð1 þ m 2
m2
þ m2
m3
Þ þ a m4 ðm 2
m2
þ m2
m4
Þ ¼ 0:
Consider the expression
E ¼ ða m3 ð1 þ m 2
¼ a m3 ðm 2
m2
þ a m4 ðm 2
m2
þ m2
m2
þ m2
m3
þ m2
m3
Þ þ a m4 ðm 2
Þ þ a m3 ðm 2
m4
Þðm 2
m2
m2
þ m2
m2
þ m2
m3
Þ:
þ m2
m3
Þ2
m4
ÞÞðm 2
m2
þ m2
m3
Þ
S128
Frank Fiedler, Ka Hin Leung and Qing Xiang
Using (3.7), we have
E ¼ a m3 ðm 2
m2
þ m2
m3
Þ þ a m3 ðm 2
m1
þ m2
m2
Þ þ a m3 ðm 2
m2
þ m2
m4
Þ2
¼ 0:
m2
m3
Since m 0 0; 1 we have ðm 2 þ m 2 Þ 0 0 and our claim follows. In particular,
m1
m3
m1
m4
by (3.6) it implies a m4 ðm 2 þ m 2 Þ ¼ a m3 ðm 2 þ m 2 Þ. Adding this to (3.7)
we get
a m4 ðm 2
m1
þ m2
m2
Þ ¼ a m3 ðm 2
m1
þ m2
m2
Þ:
Hence a m4 ¼ a m3 . Substituting a m4 in (3.7) by a m3 , we have a m3 ¼ 0.
Claim: Let m 4 > k > 2. If aj ¼ 0 for all m 3 > j > k then ak ¼ 0.
We will use a similar argument to that in (3.7). To this end we consider the
k
m2
m3
coe‰cient of x ð2 1Þþ2 þ2
in rðxÞ. The binary expansion of its exponent is
0110 . . . 01 . . . 1. This includes 2 þ k ones. All aj with j > k are zero. The sum
of an exponent from 1 þ TrðxÞ þ TrðmxÞ þ TrðxÞ TrðmxÞ and an exponent from
Pk
i
Trð i¼0
a i x 2 1 Þ has at most 2 þ k ones. Since k > 2 there is only one way to obtain
ð2 k 1Þ þ 2 m2 þ 2 m3 , namely,
0110 . . . 0 1
. . . 1 ¼ 0110 . . . 0 þ 0 . . . 0 1|fflffl{zfflffl}
...1:
|fflffl{zfflffl}
k>2
m2
k>2
m3
It follows that ðm 2 þ m 2 Þak ¼ 0. Hence ak ¼ 0.
Since a m3 ¼ a m4 ¼ 0 we find that a3 ¼ ¼ a m4 ¼ a m3 ¼ 0 by induction.
Claim: a2 ¼ 0.
4
5
Consider the coe‰cients of x 2 þ7 and x 2 þ7 . Since for all j > 2 we have aj ¼ 0
there are only two ways to obtain each exponent.
0 . . . 0010111 ¼ 0 . . . 0010100 þ 0 . . . 0000011
¼ 0 . . . 0010001 þ 0 . . . 0000110
0 . . . 0100111 ¼ 0 . . . 0100100 þ 0 . . . 0000011
¼ 0 . . . 0100001 þ 0 . . . 0000110:
4
Hence the coe‰cient of x 2 þ7 is ðm 4 þ m 16 Þa2 þ ðm þ m 16 Þa22 and the coe‰cient of
5
x 2 þ7 is ðm 4 þ m 32 Þa2 þ ðm þ m 32 Þa22 . Adding both values we find
a2 ðm 16 þ m 32 Þ þ a22 ðm 16 þ m 32 Þ ¼ 0:
Thus, a2 is either 0 or 1. Now look at the coe‰cient of x 15 . There are only three ways
of obtaining 15 as a sum with the exponents we can use.
On Mathon’s construction of maximal arcs in Desarguesian planes
S129
0 . . . 01111 ¼ 0 . . . 01100 þ 0 . . . 00011
¼ 0 . . . 01001 þ 0 . . . 00110
¼ 0 . . . 00011 þ 0 . . . 01100:
Hence ðm 4 þ m 8 Þa2 þ ðm þ m 8 Þa22 þ ðm þ m 2 Þa24 ¼ 0. If a2 ¼ 1 then m 2 þ m 4 ¼ 0 which
is a contradiction. Thus, a2 ¼ 0.
It follows that a2 ¼ ¼ a m3 ¼ 0.
Case 2: Trða 0 Þ ¼ 0. Then Trð
P m3
i¼1
i
a i l 2 1 Þ ¼ 1 for all l A Anf0g. Hence if we view
X
m3
i
a i x 2 1
ð1 þ TrðxÞ þ TrðmxÞ þ TrðxÞ TrðmxÞÞ 1 þ Tr
i¼1
as a function from F2 m to itself, it is the characteristic function of the subset f0g of
F2 m . Therefore,
ð1 þ TrðxÞ þ TrðmxÞ þ TrðxÞ TrðmxÞÞ þ ð1 þ TrðxÞ þ TrðmxÞ þ TrðxÞ TrðmxÞÞ
X
m3
m
m
2 i 1
aix
Tr
1 1 x 2 1 ðmod x 2 xÞ:
ð3:8Þ
i¼1
m
Note that the binary representation of the exponent of x 2 1 is 111 . . . 1 (m ones
altogether), while in the left hand side of (3.8), the binary representation of the
exponent of any term in the product
ð1 þ TrðxÞ þ TrðmxÞ þ TrðxÞ TrðmxÞÞ Tr
X
m3
i
a i x 2 1
i¼1
cannot have more than 2 þ ðm 3Þ ¼ m 1 ones. So (3.8) cannot hold. Thus, this
case does not occur. This completes our proof.
r
Combining Theorem 3.4 and Theorem 3.5 with the constructive result in Section 2
and Theorem 3.2 in [9], we find that when m d 7, the largest d of a non-Denniston
maximal arc of degree 2 d in PGð2; 2 m Þ generated by a fp; 1g-map via Theorem 1.1
satisfies
mþ2
c d c m 3:
2
We have the following conjecture.
Conjecture 3.6. When m > 9, the largest d of a non-Denniston maximal
arc of degree
2 d in PGð2; 2 m Þ generated by a f p; 1g-map via Theorem 1.1 is mþ2
2 .
S130
Frank Fiedler, Ka Hin Leung and Qing Xiang
In order to prove the above conjecture, it su‰ces to prove the following. Let A
be an additive subgroup in F2 m of size 2 d , where m > 9, pðxÞ ¼ a 0 þ a1 x þ þ
d1
ad1 x 2 1 A F2 m ½x. If d d mþ2
þ 1, Trð pðlÞÞ ¼ 1 for every l A Anf0g, then a2 ¼
2
a3 ¼ ¼ ad1 ¼ 0. So far we can only prove some partial results in this direction.
Theorem 3.7. Let A be an additive subgroup in F2 m of size 2 d , where d c m 1, and
d2
let pðxÞ ¼ a 0 þ a1 x þ þ ad2 x 2 1 A F2 m ½x, with ad2 0 0. If Trð pðlÞÞ ¼ 1 for
every l A Anf0g, then d c mþ2
2 .
Proof. Assume to the contrary that d > mþ2
2 ; we will show that ad2 ¼ 0. Assume that
the defining equation for A is
ð1 þ Trðm1 xÞÞð1 þ Trðm 2 xÞÞ . . . ð1 þ Trðmmd xÞÞ ¼ 1;
where m i A F2 m , i ¼ 1; 2; . . . ; m d, are linearly independent over F2 . We consider
two cases:
Case 1: Trða 0 Þ ¼ 1. Then
ð1 þ Trðm1 xÞÞð1 þ Trðm 2 xÞÞ . . . ð1 þ Trðmmd xÞÞ
X
d2
i
m
a i x 2 1 1 0 ðmod x 2 xÞ:
Tr
ð3:9Þ
i¼1
Claim: The coe‰cient of x 1þ2þ2
2
X
þþ2 d3 þ2 d1 þ2 d þþ2 m2
is
2 m2 2 m3
2 d1
msð1Þ
msð2Þ . . . msðmdÞ
ad2 ;
s A Smd
where Smd is the symmetric group on m d letters.
The exponent of x 1þ2þ2
2
þþ2 d3 þ2 d1 þ2 d þþ2 m2
has m-bit binary representation
0 |fflfflffl{zfflfflffl}
11 . . . 1 0 11
...1:
|fflfflffl{zfflfflffl}
md
d2
Since d > mþ2
2 , we see that d 2 > m d, there is only one way to get the term
2
d3
d1
d
m2
when multiplying ð1 þ Trðm1 xÞÞð1 þ Trðm 2 xÞÞ . . .
x 1þ2þ2 þþ2 þ2 þ2 þþ2P
i
d2
ð1 þ Trðmmd xÞÞ with Trð i¼1
a i x 2 1 Þ, namely
0 11
. . . 1 0 |fflfflffl{zfflfflffl}
11 . . . 1 ¼ 0 |fflfflffl{zfflfflffl}
00 . . . 0 0 |fflfflffl{zfflfflffl}
11 . . . 1 þ 0 |fflfflffl{zfflfflffl}
11 . . . 1 0 |fflfflffl{zfflfflffl}
00 . . . 0 :
|fflfflffl{zfflfflffl}
md
d2
md
d2
Therefore the claim follows. By (3.9), we see that
md
d2
On Mathon’s construction of maximal arcs in Desarguesian planes
X
2 m2 2 m3
2 d1
msð1Þ
msð2Þ . . . msðmdÞ
ad2 ¼ 0:
S131
ð3:10Þ
s A Smd
Now set r ¼ m d. Then
X
2 m2 2 m3
msð1Þ
msð2Þ
2 d1
. . . msðmdÞ
¼
X
s A Smd
2 r1 2 r2
msð1Þ
msð2Þ
2 d1
. . . msðrÞ
:
s A Sr
Note that
0
X
r1
r2
2
2
msð1Þ
msð2Þ
. . . msðrÞ
s A Sr
m1
B m2
B
¼ detB 1
@
r1
m12
m2
m22
m22
r1
1
mr
mr2 C
C
C:
A
r1
mr2
We will use Dðm1 ; m 2 ; . . . ; m r Þ to denote this last determinant. Since m i , i ¼ 1; 2; . . . ;
r, are linearly independent over F2 , we see that Dðm1 ; m 2 ; . . . ; m r Þ 0 0 (cf. [7, p. 109]).
By (3.10), this shows that ad2 ¼ 0.
Case 2: Trða 0 Þ ¼ 0. As before, this case can be easily seen not to occur.
This completes the proof.
r
In order to extend the result in Theorem 3.7, we need to introduce more notation.
Let m1 ; m 2 ; . . . ; m r be elements in F2 m that are linearly independent over F2 . Let 0 ¼
a1 < a2 < < ar c m 1 be integers. We define
Tða1 ; a2 ; . . . ; ar Þ ¼
X
a
a
ar
2 1 2 2
2
msð1Þ
msð2Þ . . . msðrÞ
:
s A Sr
Using the above notation, we have the following lemma.
Lemma 3.8. Let m > 9 be an odd integer, let r ¼ m3
2 , and let t be an integer such that
3 c t c m1
.
Then
there
exist
0
¼
a
<
a
<
<
ar c m 1 such that
1
2
2
(i) ar c m t 3,
(ii) Tða1 ; a2 ; . . . ; ar Þ 0 0, and
(iii) the number of consecutive integers in the set fa1 ; a2 ; . . . ; ar g is less than or equal to
t 1.
We postpone the proof of this lemma to the appendix. With this lemma, we can
prove the following theorem.
S132
Frank Fiedler, Ka Hin Leung and Qing Xiang
Theorem 3.9. Let m > 9 be an odd integer, let A be an additive subgroup in F2 m of size
t
2 d , where d c m 1, and let pðxÞ ¼ a 0 þ a1 x þ a2 x 3 þ þ at x 2 1 A F2 m ½x, with
at 0 0 and t c ðd 1Þ. If 3 c t c m1
2 , and TrðpðlÞÞ ¼ 1 for every l A Anf0g, then
d c mþ1
.
2
Proof. Assume to the contrary that d > mþ1
2 ; we will show that at ¼ 0. Without loss
of generality, assume that d ¼ mþ3
,
and
let
r ¼ m d ¼ m3
2 . Assume that the defin2
ing equation for A is
ð1 þ Trðm1 xÞÞð1 þ Trðm 2 xÞÞ . . . ð1 þ Trðm r xÞÞ ¼ 1;
where m i A F2 m , i ¼ 1; 2; . . . ; r, are linearly independent over F2 . As in the proof of
Theorem 3.7, we only need to consider the case where Trða 0 Þ ¼ 1. Hence we have
ð1 þ Trðm1 xÞÞð1 þ Trðm 2 xÞÞ . . . ð1 þ Trðm r xÞÞ Tr
X
t
aix
2 i 1
10
i¼1
m
ðmod x 2 xÞ:
ð3:11Þ
By Lemma 3.8, there exist 0 ¼ a1 < a2 < < ar c m 1 such that
(i) ar c m t 3,
(ii) Tða1 ; a2 ; . . . ; ar Þ 0 0, and
(iii) the number of consecutive integers in the set fa1 ; a2 ; . . . ; ar g is less than or equal
to t 1.
a
ar
m2
m3
mt1
We will look at the coe‰cient of x 1þ2 2 þþ2 þ2 þ2 þþ2
in the left hand
side of (3.11). Note that the exponent of this monomial has the m-bit binary representation
0 11
...100
. . . 1 ffl.{zfflfflfflfflfflfflfflfflfflfflffl
. . 1 . . .ffl1} ;
|fflfflffl{zfflfflffl}
|fflfflfflfflfflfflfflfflfflfflffl
t
mt2
where at the ai th bit there is a 1, for each i ¼ 1; 2; . . . ; r.
Since the number of consecutive integers in the set fa1 ; a2 ; . . . ; ar g is less than or
equal to t 1, there is only one way to get the term
x 1þ2
a2
þþ2 ar þ2 m2 þ2 m3 þþ2 mt1
when multiplying
ð1 þ Trðm1 xÞÞð1 þ Trðm 2 xÞÞ . . . ð1 þ Trðm r xÞÞ
with
Pt
i
Trð i¼1
a i x 2 1 Þ,
namely
0 . . . 1 ffl.{zfflfflfflfflfflfflfflfflfflfflffl
. . 1 . . .ffl1} ¼ 0 |fflfflffl{zfflfflffl}
00 . . . 0 0 |fflfflfflfflfflfflfflfflfflfflffl
0 . . . 1 ffl.{zfflfflfflfflfflfflfflfflfflfflffl
. . 1 . . .ffl1} þ 0 |fflfflffl{zfflfflffl}
11 . . . 1 0 00
...0:
0 |fflfflffl{zfflfflffl}
11 . . . 1 0 |fflfflfflfflfflfflfflfflfflfflffl
|fflfflffl{zfflfflffl}
t
t
mt2
Therefore, the coe‰cient of x 1þ2
(3.11) is
a2
t
mt2
ar
þþ2 þ2
m2
þ2
m3
þþ2
mt1
mt2
in the left hand side of
On Mathon’s construction of maximal arcs in Desarguesian planes
X
S133
!
2 a1 2 a2
msð1Þ
msð2Þ
2 ar
. . . msðrÞ
at2
mt1
¼ Tða1 ; . . . ; ar Þat2
mt1
:
s A Sr
By (3.11), we see that Tða1 ; . . . ; ar Þat2
at ¼ 0. This completes the proof.
4
mt1
¼ 0. Since Tða1 ; . . . ; ar Þ 0 0, we have
r
Appendix
In this appendix, we give a proof of Lemma 3.8. First, we introduce some notation. Let x1 ; . . . ; xr be elements in F2 m that are linearly independent over F2 . For any
i
i
j
integer i, we set vi ¼ ðx12 ; . . . ; xr2 Þ. We use vi2 to denote component-wise exponentiaj
2j
2m
tion of vi by 2 . Hence vi ¼ viþj . Since xl ¼ xl for all l ¼ 1; 2; . . . ; r, we have
vm ¼ v 0 . So in what follows, the indices of vi are to be read modulo m. Now condition (ii) of Lemma 3.8 is equivalent to the vectors
va 1 ; . . . ; va r
being linearly independent over F2 m , i.e.,
0
a
x12 1
B .
det@ ..
x12
ar
a 1
xr2 1
.. C
..
. A0 0:
.
ar
xr2
Let V be the F2 m -span of v 0 ; . . . ; vm1 . By [7, Lemma 3.51], dim F2 m V ¼ r and
fvi ; viþ1 ; . . . ; viþr1 g is an F2 m -basis of V for any 0 c i c m r.
In the following, we will be considering subspaces of V spanned by some vectors
in fv 0 ; v1 ; . . . ; vm1 g. To this end, we will use binary vectors to represent subsets of
fv 0 ; . . . ; vm1 g. Let u ¼ ðu 0 ; u1 ; . . . ; ui1 Þ be a vector with entries in f0; 1g. Then the
subset of fv 0 ; v1 ; . . . ; vm1 g represented by u is
SðuÞ ¼ fvl j ul 0 0; 0 c l c i 1g:
By V ðuÞ we will denote the F2 m -span of the vectors in SðuÞ. For example, if u ¼
ð1; 1; 0; 1Þ then V ðuÞ ¼ F2 m v 0 þ F2 m v1 þ F2 m v3 . For convenience, we also allow con0
catenation of binary vectors. If u ¼ ðu 0 ; u1 ; . . . ; ui1 Þ and u 0 ¼ ðu 00 ; . . . ; uj1
Þ then the
0
concatenation of u with u is
0
Þ:
u u 0 ¼ ðu 0 ; . . . ; ui1 ; u 00 ; . . . ; uj1
Moreover u
u u is abbreviated to u l .
|fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}
l
Now we can reformulate Lemma 3.8 as follows: For every integer t such that
3 c t c m1
2 , there exists a binary vector u of length at most m ðt þ 2Þ such that
V ðuÞ ¼ V and the number of consecutive 1’s in u is at most t 1. It is this reformulation that we will prove in this appendix.
S134
Frank Fiedler, Ka Hin Leung and Qing Xiang
One final preparation before we give the proof. Given integers i and j > 0, let
I ði; jÞ denote the F2 m -span of vi , viþ1 ; . . . ; viþj1 . Given a subspace W of V, we define
t
t
t
W 2 ¼ fw 2 j w A W g, where w 2 means component-wise exponentiation of w by 2 t .
We will need the following lemma.
Lemma 4.1. Suppose that W ¼ V ðuÞ where u ¼ ðu 0 ; u1 ; . . . ; us1 Þ A F2s . If I ðs; tÞ H W
t
and W 2 V I ð0; sÞ H W then W ¼ V .
Proof. By assumption, W is spanned by a subset of fv 0 ; . . . ; vs1 g. Let vi A W , 0 c
t
i c s 1, be one of the generating vectors. If i þ t c s 1, then vi2 ¼ viþt A I ð0; sÞ V
t
t
W 2 H W . If i þ t > s 1, then vi2 ¼ viþt A I ðs; tÞ H W . Hence for any vector vi A W ,
0 c i c s 1, we have viþt A W . Extending this property to linear combinations of
the generating vectors of W, we see that I ðs þ t; tÞ H W since I ðs; tÞ H W . That is,
I ðs þ lt; tÞ H W for all l d 0. Hence vi A W for all 0 c i c m 1 and W ¼ V . r
Proof of Lemma 3.8. Write r ¼ kt þ a where 0 c a c t 1. Since r ¼ m3
2 , we have
m ¼ 2kt þ 2a þ 3. Set a ¼ ð1; 1; . . . ; 1Þ A F2a and u ¼ ð0; 1; . . . ; 1Þ A F2t . Let
V ðiÞ ¼ V ða u i Þ:
That is, V ðiÞ is the space spanned by the vectors in Sða u i Þ. Then V ðkÞ is the F2 m span of fv 0 ; v1 ; . . . ; vr1 gnfva ; vaþt ; . . . ; vaþðk1Þt g. Since fv 0 ; v1 ; . . . ; vr1 g is a basis for
V, we see that dim V ðkÞ ¼ r k. Let b be the smallest nonnegative integer such that
V ðk þ bÞ ¼ V ðk þ b þ 1Þ. In particular, V ðk þ iÞ is a proper subspace of V ðk þ i þ 1Þ
if 0 c i < b. We observe that 0 c b c k. There are three cases to consider.
Case 1: dim V ðk þ 1Þ d r k þ 2. In this case b c k 1. If V ðk þ bÞ ¼ V , then
Sða u ðkþbÞ Þ spans V. Note that a u ðkþbÞ has length a þ ðk þ bÞt c a þ ð2k 1Þt ¼
m a ðt þ 3Þ. By construction this vector does not have more than t 1 consecutive 1’s. So we are done in this case.
If V ðk þ bÞ 0 V then b c k 2. Let 0 ¼ ð0; 0; . . . ; 0Þ A F2t and a 0 ¼ ð1; 0; . . . ; 0Þ A
t
F2 . Let
wi ¼ a u ðkþbÞ 0 a 0i :
We define W ðiÞ ¼ V ðwi Þ to be the F2 m -span of the vectors in Sðwi Þ. In particular, W ð0Þ ¼ V ðk þ bÞ. Let b 0 be the smallest nonnegative integer such that
W ðb 0 Þ ¼ W ðb 0 þ 1Þ.
dim W ð0Þ ¼ dim V ðk þ bÞ
d dim V ðk þ 1Þ þ ðb 1Þ
dr k þ 2 þ b 1
¼ r ðk b 1Þ:
On Mathon’s construction of maximal arcs in Desarguesian planes
S135
Hence 0 c b 0 c k 1 b. We claim that
(i) W ðb 0 Þ I I ða þ ðb þ k þ i þ 1Þt; tÞ
t
(ii) W ðb 0 Þ I W ðb 0 Þ 2 V I ð0; a þ ðb þ k þ i þ 1ÞtÞ
for all i d 0. By Lemma 4.1, these two claims imply that W ðb 0 Þ ¼ V . The length of
wb 0 is
a þ ðk þ b þ 1 þ b 0 Þt c a þ 2kt ¼ m a 3:
Note that the last t 1 entries in wb 0 are zero. Dropping these t 1 positions we
obtain a vector of length m a ðt þ 2Þ. This vector does not have more than t 1
consecutive 1’s and it corresponds to a subset of fva1 ; . . . ; var g that spans V, hence
Lemma 3.8 is proved in this case once we prove the above two claims.
0
To prove the first claim, we recall that W ðb 0 Þ ¼ V ða u ðkþbÞ 0 a 0b Þ. Hence
vaþðkþbþ1þiÞt A W ðb 0 Þ for all i d 0 since this vector corresponds to the first position in
the i-th copy of a 0 . Now W ðb 0 Þ K V ðk þ bÞ ¼ V ðk þ b þ 1 þ iÞ for all i d 0, we also
have Sða u ðkþbþ1þiÞ Þ H W ðb 0 Þ. Thus
vaþðkþbþ1þiÞtþ1 ; . . . ; vaþðkþbþ1þiÞtþðt1Þ A W ðb 0 Þ;
since these vectors correspond to the nonzero positions in the last copy of u in
a u ðkþbþ2þiÞ . This proves our first claim.
For the second claim it su‰ces to show that Sð0 a u ðkþbÞ 0 a 0i Þ J W ðb 0 Þ.
Hence we need to show that the vectors corresponding to the ðk þ bÞ-th copy
of u and the i-th copy of a 0 , respectively, are in W ðb 0 Þ. The former is true since
W ðb 0 Þ includes Sða u ðkþbþ1Þ Þ which spans V ðk þ b þ 1Þ. The latter holds because
0
W ðb 0 Þ ¼ W ðb 0 þ 1Þ which includes the vectors in Sða u ðkþbÞ 0 a 0ðb þ1Þ ). This
proves our second claim.
Case 2: dim V ðk þ 1Þ ¼ r k þ 1 ¼ dim V ðkÞ þ 1. In this case, one of the vectors
vrþ1 ; . . . ; vrþðt1Þ does not belong to V ðkÞ. Suppose that vector is vrþj ¼ vaþktþj ,
1 c j c t 1. Then V ðk þ 1Þ ¼ V ðkÞ þ F2 m vrþj . Since any linear dependence relation translates to a linear dependence relation when both sides are raised to the
2 t th power, we get dim V ðk þ iÞ c dim V ðkÞ þ i, i d 0.
Subcase 1: V ðk þ bÞ 0 V , i.e., b < k. As seen above, all vectors vrþ1 ; . . . ; vrþðt1Þ were
linearly dependent on vectors in V ðkÞ and vrþj . Any such linear dependence translates
to a linear dependence of vrþðb1Þtþi , i 0 j, on vectors in V ðk þ b 1Þ and vrþðb1Þtþj .
Hence the vector vrþðb1Þtþj must be a vector among vrþðb1Þtþ1 ; . . . ; vrþðb1Þtþðt1Þ that
is not in V ðk þ b 1Þ. Therefore, we can replace those positions in the last copy of u
in a u ðkþb1Þ u that do not correspond to vrþðb1Þtþj by 0; we will denote the
modified vector by a u ðkþb1Þ u ð jÞ , where u ð jÞ contains only one 1. By our discussion above, we see that
V ðk þ bÞ ¼ V ða u ðkþb1Þ u ð jÞ Þ;
S136
Frank Fiedler, Ka Hin Leung and Qing Xiang
and dim V ðk þ bÞ ¼ r k þ b ¼ r ðk bÞ. Let a 0 be as defined in Case 1, and let
wi0 ¼ a u ðkþb1Þ u ð jÞ a 0i . Define W ðiÞ ¼ V ðwi0 Þ. Let b 0 be the smallest nonnegative integer such that W ðb 0 Þ ¼ W ðb 0 þ 1Þ. Then 0 c b 0 c k b. Similar to Case 1,
we have
(i) W ðb 0 Þ I I ða þ ðk þ b þ iÞt; tÞ
t
(ii) W ðb 0 Þ I W ðb 0 Þ 2 V I ð0; a þ ðk þ b þ iÞtÞ.
Thus, by Lemma 4.1 we have V ðwb0 0 Þ ¼ W ðb 0 Þ ¼ V . The length of wb0 0 is
a þ ðk þ bÞt þ b 0 t c a þ 2kt. Dropping the last t 1 zeros in wb0 0 , we get a vector
of length m a ðt þ 2Þ, which does not contain more than t 1 consecutive 1’s.
So Lemma 3.8 is proved in this subcase.
Subcase 2: V ðk þ bÞ ¼ V , i.e., b ¼ k. Since V ðk þ bÞ ¼ V ða u ðkþb1Þ u ð jÞ Þ (cf.
Subcase 1), we have
V ¼ V ða u ðkþb1Þ u ð jÞ Þ:
Note that the binary vector a u ðkþb1Þ u ð jÞ has length a þ ðk þ b 1Þt þ ð j þ 1Þ ¼
m ðt þ 2Þ ða jÞ and does not have more than t 1 consecutive 1’s. If a d j this
vector will work. So we assume that j > a. Recall that vrþ1 ; . . . ; vrþð j1Þ A V ðkÞ but
vrþj B V ðkÞ by our choice of j. Let ð0; 1; . . . ; 1Þ A F2j and w 0 ¼ a u k ð0; 1; . . . ; 1Þ.
Then V ðw 0 Þ ¼ V ðkÞ.
Let z ¼ ð1; 1; . . . ; 1 ; 0; 1; 1; . . . ; 1 Þ A F2t , and let V 0 ðiÞ be the F2 m -span of Sða z i Þ,
|fflfflfflfflfflffl{zfflfflfflfflfflffl} |fflfflfflfflfflffl{zfflfflfflfflfflffl}
tj
j1
i.e., V 0 ðiÞ ¼ V ða z i Þ. Observe that when we shift the vector a z k to the right by j
positions, we get
ð0; . . . ; 0Þ ð1; . . . ; 1Þ u ðk1Þ ð0; 1; . . . ; 1Þ :
|fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}
|fflfflfflfflfflffl{zfflfflfflfflfflffl} |fflfflfflfflfflffl{zfflfflfflfflfflffl}
j
j
tþðajÞ
The subset represented by this vector is
fvj ; vjþ1 ; . . . ; vrþj1 gnfvaþt ; vaþ2t ; . . . ; vaþkt g:
P j1
j
j
F2 m vrþi ¼ V ðkÞ. In fact, V 0 ðkÞ 2 ¼ V ðkÞ since the
Hence V 0 ðkÞ 2 J V ðkÞ þ i¼1
j
two subspaces have the same dimension r k. Similarly, V 0 ðk þ 1Þ 2 ¼ V ðk þ 1Þ.
j
Moreover, since vr2 ¼ vrþj B V ðkÞ we have vr B V 0 ðkÞ. Hence V 0 ðk þ 1Þ ¼ V 0 ðkÞ þ
F2 m vr . It follows that V 0 ðk þ iÞ ¼ V 0 ðk þ i 1Þ þ F2 m vrþit for 1 c i c k. In particular, V 0 ð2kÞ ¼ V . Since V 0 ð2kÞ ¼ V 0 ð2k 1Þ þ F2 m vrþðk1Þt , we see that actually
V 0 ð2kÞ ¼ V ða z ð2k1Þ ð1; 0; . . . ; 0 ÞÞ:
|fflfflfflfflfflffl{zfflfflfflfflfflffl}
t
a z ð2k1Þ ð1; 0; . . . ; 0 Þ
|fflfflfflfflfflffl{zfflfflfflfflfflffl}
t
m ðt þ 2Þ a. So we are also done in this subcase.
The
length
of
the
vector
is
a þ ð2k 1Þt þ 1 ¼
On Mathon’s construction of maximal arcs in Desarguesian planes
S137
Case 3: V ðkÞ ¼ V ðk þ 1Þ, i.e., dim V ðk þ 1Þ ¼ r k. In this case V is spanned by
Sða u k a 0k Þ. Thus, we have
vrþ1 ¼
r1
X
ci vi ;
i¼0
where caþnt ¼ 0 for all 0 c n c k. Let l be the largest index such that cl 0 0. We
consider three subcases.
Subcase 1: l ¼ a þ jt þ s with 2 c s c t 1 and 0 c j c k 1. Now
ðvrþ1 Þ 2
ðkj1Þt
¼ vrþ1þðkj1Þt ¼
r1
X
ðkj1Þt
ci2
viþðkj1Þt :
i¼0
Note that r t þ 2 c l þ ðk j 1Þt c r 1. Hence viþðkj1Þt A Sða u k Þ ¼ V ðkÞ
for all vi with ci 0 0. Therefore we can express vlþðkj1Þt ¼ vaþðk1Þtþs as a linear
combination of vrþ1þðkj1Þt and some vector in V ðkÞ. It follows that V is spanned by
Sða u ðk1Þ u ðlÞ a 0ðkj1Þ ð1; 1; 0; . . . ; 0Þ a 0j Þ where ð1; 1; 0; . . . ; 0Þ A F2t and
u ðlÞ ¼ ð0; 1; 1; . . . ; 1; 0; 1; 1; . . . ; 1Þ A F2t . For convenience, denote the vector a |fflfflfflfflfflffl{zfflfflfflfflfflffl}
s1
u ðk1Þ u ðlÞ a 0ðkj1Þ ð1; 1; 0; . . . ; 0Þ a 0j by z. If j > 0, then we can drop t 1
zeros from the last copy of a 0 in z to obtain a binary vector of length m a ðt þ 2Þ,
which contains no more than t 1 consecutive 1’s. If j ¼ 0 we can still drop the last
t 2 zeros from z. The resulting vector has length no more than m ðt þ 2Þ if a d 1.
Hence we only need to consider the case j ¼ 0 and a ¼ 0. In that case V is spanned
by Sðu ðk1Þ u ðlÞ a 0ðk1Þ ð1; 1ÞÞ and v 0 is not in the generating set. Thus, we can
shift every entry in u ðk1Þ u ðlÞ a 0ðk1Þ ð1; 1Þ to the left by one position. This still
is a generating vector for V which has length m ðt þ 2Þ.
Subcase 2: l ¼ a þ jt þ 1 with 0 c j c k 1. If j < k 1, the same vector z as in
Subcase 1 will suit our purpose since it does not contain more than t 1 consecutive
1’s. So we will assume j ¼ k 1. We have
2
vrþ2 ¼ vrþ1
¼
r1
X
ci2 viþ1 :
i¼0
of the viþ1 might be
Since l ¼ a þ ðk 1Þt þ 1, we have caþkt1 ¼ 0. Note that some
P k1
2
of the form vaþnt . However, since vrþ2 A V ðkÞ we must have n¼0
caþnt1
vaþnt ¼ 0.
Hence we have that vl ¼ vaþðk1Þtþ1 is a linear combination of vrþ2 and some vector
in V ðkÞ. It follows that V is spanned by Sða u ðk1Þ ð0; 1; 0; 1; . . . ; 1Þ ð1; 0; 1;
0; . . . ; 0Þ a 0ðk1Þ Þ. Denote the vector a u ðk1Þ ð0; 1; 0; 1; . . . ; 1Þ ð1; 0; 1; 0; . . . ;
0Þ a 0ðk1Þ by z 0 . We see that z 0 contains no more than t 1 consecutive 1’s. If
S138
Frank Fiedler, Ka Hin Leung and Qing Xiang
k 1 > 0 then we can drop the last t 1 zeros of z 0 and obtain a vector of length
m a ðt þ 2Þ. If k 1 ¼ 0 we can drop the last t 3 zeros of z 0 . If a d 2 this vector will have length at most m ðt þ 2Þ. We need to consider the case k ¼ 1 and
a c 1. Suppose a ¼ 0 (so r ¼ t). Then vrþ1 ¼ c1 v1 with c1 0 0. Keep squaring both
sides of this equation, we see that vrþ1þm is a nonzero scalar multiple of vrþ4 . If r > 3
then this contradicts the fact that any r consecutive vectors in the set fv 0 ; . . . ; vm1 g
are linearly independent. So r c 3. Since t d 3 and r ¼ t we have r d 3. Thus m ¼ 9.
But we assumed that m > 9, so the case a ¼ 0 cannot happen.
Now suppose a ¼ 1. Then vrþ1 ¼ c 0 v 0 þ c2 v2 and c 0 0 0, hence vrþ2 ¼ c02 v1 þ c22 v3 .
Note that since a ¼ 1, we have V ðkÞ ¼ V ð1Þ ¼ V ðð10 11
. . . 1ÞÞ. So the previous equa|fflfflffl{zfflfflffl}
t1
tion implies that vrþ2 B V ð1Þ, contradicting the assumption that V ðk þ 1Þ ¼ V ðkÞ.
Subcase 3: 0 c l c a 1. Observe that vrþ1 ; vrþ2 ; . . . ; vrþt1 A V ðkÞ as well. Note
al
2 al
that vrþ1
¼ cl2 va þ B V ðkÞ as va B V ðkÞ. It follows that a l > t 2. Since
a c t 1 this is only possible when l ¼ 0. But then vrþ1 ¼ c 0 v 0 , with c 0 0 0. This
rþ2
implies that vm ¼ v 0 ¼ c02 þ2 v1 , which is impossible. This completes the proof. r
Acknowledgements. The third author thanks the Department of Mathematics,
National University of Singapore for a research visit in January 2002, which brought
the second and the third author together to work on this subject. The research of the
third author was also partially supported by NSA grant MDA 904-01-1-0036. After
we submitted the paper, we learned (in February 2003) that an idea similar to that
given in Section 2 for constructing maximal arcs was discussed in Hamilton’s talk
in Oberwolfach in December 2001.
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Received 9 October, 2002
F. Fiedler, Q. Xiang, Department of Mathematical Sciences, University of Delaware, Newark,
DE 19716, USA
Email: {fiedler, xiang}@math.udel.edu
K. H. Leung, Department of Mathematics, National University of Singapore, Kent Ridge,
Singapore 119260
Email: matlkh@nus.edu.sg