KelleL.Clark,LeslieD.Hatfield,JenniferD.Key*andHaroldN.Ward Dualcodesofprojectiveplanesoforder25

(de Gruyter 2003

Dual codes of projective planes of order 25

Kelle L. Clark, Leslie D. Hatfield, Jennifer D. Key* and Harold N. Ward

Dedicated to Adriano Barlotti on the occasion of his 80th birthday

Abstract.We determine improved bounds for the minimum weight of the dual code overF5

of any projective plane of order 25 and describe configurations that could give words of mini- mum weight.

1 Introduction

If Pis a projective plane of ordernand pis a prime dividingn, then the minimum weight of the dual p-ary code ofPis not, in general, known, even in the desarguesian case. It is known that when the order of the plane is a prime p, the minimum weight is 2pand words of this weight can be constructed from two distinct lines of the plane:

see, for example, [1, Chapter 6]. For the binary dual code of desarguesian planes of even orderq¼2^m the minimum weight isqþ2 and the minimum words are the incidence vectors of the hyperovals, which always exist in the desarguesian planes.

(See [10] for other results in the even case, and for when the plane has no hyperoval.

In the latter case, again the minimum weight is not known except in some particular cases.) Some other results forpodd are mentioned in Section 2. In particular, for the four planes of order 9, Key and de Resmini [11] proved that the minimum weight is 14 for the Hughes plane; and 15 for the desarguesian plane,F, the translation (Hall) plane,W, and the dual translation plane,W^D.

In this paper we concentrate on the dual code of a projective plane of order 25 and prove the following theorem:

Theorem 1.1.IfPis a projective plane of order25and C is the code ofPoverF5,then the minimum weight d^?of C^?is either42or44,or45cd^?c50.If a Baer subplane is present, then the minimum weight is either42, 44 or45.In any case,if the minimum weight is42,then a minimum-weight word has support that is the union of two projec- tive planes, p1 and p2,of order4 that are totally disjoint and the (scaled) minimum-

* This work was supported by the DoD Multidisciplinary University Research Initiative (MURI) program administered by the O‰ce of Naval Research under Grant N00014-00-1- 0565, and NSF grant #9730992.

weight word isv^p1v^p2.If the minimum weight is 44then the support of a minimum- weight word is the union of two complete22-arcs that have eleven2-secants in common.

If the minimum weight is45thenv^bv^l,wherebis a Baer subplane ofPand l is a line ofPthat is a line of the subplane,is a minimum-weight word.

(Two configurations in P are totally disjoint if they have no point or line in common.)

Corollary 1.2.The dual5-ary code of the desarguesian projective planePG2ðF25Þhas minimum weight45.

In Section 2 we give the background results, definitions and notation, and in Sec- tion 3 we prove the main theorem through a series of lemmas and propositions.

2 Background results and notation

An incidence structureD¼ ðP;B;IÞ, with point setP, block setBand incidenceI, is at-ðv;k;lÞdesign ifjPj ¼v, every blockBABis incident with preciselykpoints, and everytdistinct points are together incident with precisely lblocks. The number of blocks through a point of atd1-design is a constant, called thereplication num- ber, and denoted byr. The order of the design is defined to be rl. An incidence structure D¼ ðP;B;IÞ is a group divisible design if P is partitioned into point classessuch that two points in the same class are incident with the same numberl1of blocks, and if any two points in distinct point classes are incident with the same numberl2of blocks.

IfSis a set of points ofDand ifBis a block ofDthat meetsSinmpoints, then Bwill be called anm-secanttoS. The setSis anðn1;. . .;nrÞ-set ifShasm-secants if and only ifmAfn1;. . .;nrg. The 1-secants are thetangentstoS.

The linear code Cof the designD over the finite field F¼F_p is denotedC_pðDÞ, and is the vector space spanned by the incidence vectors of the blocks ofDoverF_p. We denote the incidence vector of any subsetSofPbyv^S. We will always takepto be a prime divisor of the order ofDwhen looking atC_pðDÞ: see [1, Theorem 2.4.1].

We viewC_pðDÞas a subspace ofF^P, the full vector space of functions fromPto F. Using the notation of functions, the value ofcAF^Pat a point XAPis denoted cðXÞ. The support set ofcis the set of points X inP for whichcðXÞ00, and the weight ofc, wtðcÞ, is the cardinality of the support set ofc. Theminimum weightof a codeC,dðCÞ ¼d, is the smallest of all the non-zero weights of the codewords ofC.

The dual or orthogonal code C^? of C is the orthogonal space with respect to the standard inner product.

LetDbe a 2-ðv;k;lÞdesign with replication numberr. LetSbe the support set of a codeword in C_p^?ðDÞ. For i¼0;. . .;jSj, let xi denote the number of i-secants to S. ForX a point inS,ziðXÞis the number ofi-secants ofSpassing throughX. It follows that x1¼0 and that z1¼0 for every point in S. From counts on the i- secantsx_i ofS, 0cick, we have

X^k

i¼0 i01

xi¼b; X^k

i¼2

ixi¼sr; X^k

i¼2

iði1Þxi¼sðs1Þl; ð1Þ

and hence

X^k

i¼3

iði2Þxi¼sððs1ÞlrÞ; ð2Þ

where the last equation is obtained from the previous two.

For a point X in S, with z_iðXÞ ¼z_i (a shorthand we shall use whenever X is obvious),

X^k

i¼2

zi¼r; X^k

i¼2

ði1Þzi¼ ðs1Þl: ð3Þ

From these two equations we obtain the useful combination X^k

i¼2

ði2Þzi¼ ðs1Þlr: ð4Þ

Since the left-hand side of this equation is nonnegative, we haveðs1Þlrd0, i.e.

sdr

lþ1 ð5Þ

for any word ofC^?.

A 2-ðn²þnþ1;nþ1;1Þdesign, fornd2, is a finite projective plane of order n.

We write PG2;1ðFqÞfor the desarguesian projective plane, i.e. the design of points and 1-dimensional subspaces of the projective space PG2ðFqÞ. Further, AG2;1ðFqÞ will denote the a‰ne desarguesian plane of orderq, i.e. the 2-design of points and 1-flats (cosets of vector subspaces of dimension one) in the a‰ne geometry AG₂ðFqÞ. Ak- arcin a plane is a set ofkpoints, no three of which are collinear. Ak-arc is said to be completeif it is not contained in aðkþ1Þ-arc in the plane.

The current state of knowledge of the minimum weights of the dual codes of finite planes is summed up in the following results. The first is a special case of the designs from finite geometries and can be found discussed in [1, Theorem 5.7.9]:

Result 2.1.Let C be the p-ary code of the desarguesian planePG2;1ðFqÞorAG2;1ðFqÞ where q¼ p^t and p is prime.Then the minimum weight d^?of C^?satisfies

ðqþpÞcd^?c2q:

Note that a similar range holds for any projective plane: ifPis a plane of ordern, pis a prime, and pjn, the minimum weightd^?ofC_p^?ðPÞsatisfies

nþ2cd^?c2n:

The lower bound is obtained by simply noticing that every one of the nþ1 lines through a point in the support set of a word of minimum weight must meet the set again, and the upper bound follows since the vectorv^lv^mis inC_p^?ðPÞ, whereland mare any two distinct lines ofP.

The next result can be found in [5, Corollary 4], and partly in [12]:

Result 2.2.Let Pbe a projective plane of odd order n,and let p be a prime such that pjn.Then the minimum weight d^?of C_p^?ðPÞsatisfies d^?d⁴₃nþ2.Further,if pd5 then d^?d³₂nþ2.

In addition the existence of a Baer subplane in a projective plane of square order gives us the following improved upper bound ford^?; see [12], [5].

Result 2.3. A projective plane of square order q² that contains a Baer subplane has words of weight2q²q in its p-ary dual code,where pjq.

In particular, this provides an upper bound for translation planes of square order;

see [6], in which improved bounds for some translation planes were obtained:

Result 2.4.Let Pbe a projective translation plane of order q^m and kernel containing F_q,where m¼2or3,q¼p^t,and p is a prime.Then the dual code of the p-ary code of Phas minimum weight at most2q^m ðq^m1þq^m2þ þqÞ.IfPis desarguesian, this also holds for m¼4.

Definition 2.5.For any vectorwAF^P_p with support setSJPandaAF_pdefine Sa¼ fXASjwðXÞ ¼ag; sa¼ jSaj

and

sðwÞ ¼ jfaAF_pjthere exists a pointY inSwithwðYÞ ¼agj:

The setSis a j-secant setifShas a 2-secant, i.e.x200, and there exists an integer jd3 such thatxi¼0 for 2<i< jandxj00.

The next result can be found in [5] and [4].

Result 2.6.LetDbe a2-ðv;k;lÞdesign with replication number r and order n.LetS be the support set of a non-zero wordwAC^?,the dual code of the p-ary code CpðDÞ, where p is an odd prime and pjn. Suppose jSj ¼sc^2r_l. Then z2¼z2ðXÞd 2r ðs1Þlfor every point X inS.Further,Sis a j-secant set for some jd3and (1) for any X inS,z₂ðXÞdlr^j1_j2l^s1_j2m

; (2) sd ^sðwÞ

sðwÞþj2 rðj1Þ

l þ1 h i

d²

j rðj1Þ

l þ1

.

Further,sðwÞis even,and if p>3and j¼3,thensðwÞd4.

In the next section we apply these results to the case whereDis a projective plane of order 25. For these parameters, the inequalities in Result 2.6 become, writingsfor sðwÞ,

z₂ðXÞd 26js25 j2

¼26 s27 j2

; ð6Þ

for anyXAS, and

sd s

sþj2ð26j25Þ: ð7Þ

Using the notation of Definition 2.5 and Result 2.6, we have the following:

Lemma 2.7.Suppose that pFr,as is the case for a projective plane.Then X

aAF_p

asa10 ðmodpÞ:

Proof.We haveP

BABv^B¼rv^P. Thus if pFr, the all-one vector||¼v^Pis inC_pðDÞ, and the congruence follows from its orthogonality to the words ofC_p^?ðDÞ. r

3 Projective planes of order 25

In what follows, let P be a projective plane of order 25 and set C¼C5ðPÞ. From Results 2.1 and 2.2, the bounds on the minimum weight d^? are 40cd^?c50; or, from Result 2.3, 40cd^?c45 for planes containing a Baer subplane. In this section we investigate the structure of a support set S of a word w in C^? having weight in this range. We first note that a constant word in C^? must have size at least ðqþ1Þðp1Þ þ1¼105. From Result 2.6,sðwÞ ¼sis either 2 or 4, and Sis a j- secant set for some jd3. We now look at the di¤erent values ofsand jto determine the possible configurations of the set S of points in P. If we fix s and take s¼ jSjc49, then by using inequality (7) we can determine the largest value of j for whichSis a j-secant set.

Lemma 3.1.Let Sbe the support set of a wordwof C^?.Suppose that s¼ jSjc49 ands¼sðwÞis as defined in Definition2.5,and supposeSis a j-secant set.Then

.

.

s 49 48 47 46 45 44 43 42 41 40

j ½4;16 ½4;12 ½4;10 ½4;8 ½4;7 ½4;6 ½4;5 ½4;5 4 4 where½4;ndenotes the range4cjcn for j.

Recall that two configurations inPare called totally disjoint if they have no point or line in common. In what follows, we often refer to a point in Sa as an a, and specify a secant by listing its point in this notation. Thus a1333(secant) is a 4-secant with one point inS1and three points inS3.

Lemma 3.2.Letwbe a word of C^?and letSbe the support set ofw.LetsðwÞ ¼4and suppose that for some X inS,z2ðXÞ ¼53s,where s¼ jSj.Then s>45.

Proof. Note first that inequality (6) implies that 53sis the smallest possible value for z2. By way of contradiction, assume that 43csc45. One has z3ðXÞ ¼s27 andziðXÞ ¼0 fori>3 from Equation (4). On scaling we may assume thatX AS1. Since the 3-secants onX have the form113or122, the only secants onX and a point ofS4 are 2-secants, ands4 ¼z2ðXÞ. LetX be ont1 113secants and t2 122secants.

Then by counting the points inS1,S2, andS3, we obtain the equations s₁¼t₁1; s₂¼2t₂; s₃¼t₁:

Thus s₃¼s₁1, and as t₁þt₂¼s27, we have s₂ ¼2s522s₁. These counts hold for all X in S1. Again as 53sis the smallest possible z₂, s_ad53s for all aAF_p. If we consider the 113 secants on a fixed3, we see that s₁ has to be even.

Similarly, the122s on a fixed2show thats₁þ1cs₂. Fors¼43 there is no set ofs_a values satisfying these conditions at all. The other possibilities are:

case s s1 s2 s3 s4

1 44 10 16 9 9

2 45 10 18 9 8

3 45 12 14 11 8

ð8Þ

Case 1 is out, because rescaling w by 3 produces another word with s4¼9 whose others_a values no longer fit the parameter lists.

Since information onz_ipossibilities fors¼45 andz₂c11 will be needed here and later, we present it now. The values not given in a row are 0, and the lists are those allowed by Equation (4):

z₂ z₃ z₄ z₅ z₆ 8 18

9 16 1

10 14 2

10 15 1

11 12 3

11 13 1 1

11 14 1

ð9Þ

IfY AS2in Case 2 of table (8), thenz2ðYÞc9 (froms3¼9), andz3ðYÞd16. ButY is on ten 122secants and at most four 244secants, which is not enough. In Case 3,

each YAS2 is on 12 122 secants, so that Y is on one further secant with another 2 and no 1. As z₂ðYÞc11, the possibilities in (9) show that this additional secant is an i-secant with i¼4;5, or 6. It is thus either a 2233, making z₂ðYÞc9; a 22344, makingz₂ðYÞc10; or a224444. Then in any event, z₃ðYÞd14, so that Y must be on at least two244s. But there are only ⁸₂ ¼28 pairs of4s for the 14 pos- sibilities for Y. Hence eachY is on exactly two244s and therefore on a224444. At this point, however, there are too many pairs of4s required, and Case 3 is also not

possible. r

Proposition 3.3. Let S be the support set of a word w of C^?. Suppose that s¼ jSjc49.Then either s¼42andSconsists of two totally disjoint projective planes of order4,or sd44.

Proof.First takes¼2 and 40csc43, so thatSis a j-secant set where jAf4;5g, by Lemma 3.1. We may scale wand assume thatS¼S1US4 without loss of gen- erality. Suppose first that S is a 4-secant set. If X is a point on a 4-secant with X AS1, we have s₄dz₂þ2, and similarly for s₁. Thus sd2l^79s₂ m

þ4, so that sd42. Ifs¼42, thens1¼s4¼21 andz2d19 for all points, whilez2ðXÞ ¼19 ifX is on a 4-secant. In this case, withX AS1, the remaining six lines throughX must have intersection with S completely in S1. Since these lines will all have to be 5- secants at least, there would have to be at least 24 more points in S1, which is too many.

Ifs¼43, thenz₂d18 and, as above, the existence of 4-secants gives that s_ad20 fora¼1;4. But by Lemma 2.7,s₁1s₄14ðmod 5Þ, so that one of thes_a must be 19 and the other 24. Thuss043.

Suppose now that S is a 5-secant set, and again s¼2, S¼S1US4. From Lemma 3.1,s¼42 or 43. Supposes¼42. Then from inequality (6),z₂d21 for any X AS. Thuss₁¼s₄ ¼21, forcingz₂¼21. Then Equations (3) give us thatz₂ðXÞ ¼ 21, z5ðXÞ ¼5, and ziðXÞ ¼0 otherwise. It follows that any two points in S1 are together on exactly one 5-secant ofS. ThusS1 is a 2-ð21;5;1Þdesign, i.e. a projec- tive plane of order 4. The setS4 is also a 2-ð21;5;1Þdesign and these designs do not share points or lines. HenceS1andS4 are a pair of totally disjoint projective planes of order 4 embedded inP.

If s¼43 in the 5-secant case, then z2d21 and sad21. Once again, Lemma 2.7 gives the contradiction that thesa are 19 and 24.

Consider now the case where s¼4, so that S is a 3-secant set and sd43, by Lemma 3.1. Ifs¼43, thenz₂ðXÞd10 for allXAS, by inequality (6). Then each s_ad10; as we cannot haves_a>10 for alla, we may scale to take s₄ ¼10, making z₂ðXÞ ¼10 forXAS1. However, Lemma 3.2 rules out this situation.

This completes all the cases for the proposition, so we havesd44. r To finish the proof of the main theorem, we need to consider the possibility that s¼44 ors¼45. We show first thats¼44 can happen only if disjoint complete 22- arcs are present, and we do this through two lemmas dealing with the di¤erent cases.

In both lemmas we haveSthe support set of a wordwinC₅^?ðPÞof weight 44, where Pis a projective plane of order 25.

Lemma 3.4.If Sis a 4-secant set of size 44 then, on scaling, s₁¼s₄¼22, and,for every point X AS,z₂¼20,z₄ ¼1,and z₅¼5.

Proof.From Lemma 3.1,s¼2, and from inequality (6), for anyX AS,z2d18. As in the 4-secant argument in Proposition 3.3, it follows that (on scaling)sad20 for a¼1;4. By Lemma 2.7, it must be thats1¼s4¼22. For a point X on a 4-secant, the only feasible solution isz2¼20,z4¼1 andz5 ¼5. The possibility of some of the points not being on 4-secants is easily ruled out by considering cases and invoking Equation (4), and so this set of parameters holds for all points ofS. r Lemma 3.5.IfShas size44,then it must be either a4-secant set of the type described in Lemma 3.4 or else the union of two disjoint complete 22-arcs that have eleven 2- secants in common. In the latter case, the parameters forSare x₀¼200,x₂ ¼x₃ ¼ 220,and x₄¼11,and,for every point ofS,z₂¼10,z₃¼15and z₄¼1.

Proof.From Lemma 3.1, if Sis not a 4-secant set, thenSis a 3, 5 or 6-secant set.

Suppose first that S is a 6-secant set. Then s¼2 from Lemma 3.1 and for any X AS,z₂d22, from inequality (6). Thusz₂¼22 for all points ofS, ands₁¼s₄ ¼ 22. For any point inS, sayXAS1, the remaining four lines that are not 2-secants must be totally inS1, so there cannot be any 6-secants.

Now suppose thatSis a 5-secant set. From Lemma 3.1,s¼2, and from inequal- ity (6), for anyX AS,z2d21. Thus we can assume that eithers1¼21 and s4¼23 or s1¼s4¼22. But Lemma 2.7 rules out the former case. If s1¼s4¼22, then if z2¼21 for some point XAS4, the remaining five lines throughX must cover one point fromS1 and 21 fromS4 excludingX. The one point fromS1 could then not havez2d21. Thus we must havez2¼22 for all points ofS. The only feasible solu- tion to this isz₅¼3 andz₁₀¼1, for all points ofS. Counting point incidences with 5-secants gives 443¼5x₅, which clearly has no solution.

Finally, suppose thatS is a 3-secant set, so that s¼4. From inequality (6), for any XAS, z₂d9. However, Lemma 3.2 excludes z₂ ¼9; thus s_ad10 for a¼ 1;2;3;4. We show that in facts_a cannot equal 10.

Suppose (by scaling) thats₄¼10; then for all points inS1,z₂¼10, and it follows that z₃¼15 and z₄¼1. IfX AS1, the 4-secant through X cannot contain a 4, so it is either 1112or 1333. If it is 1112, then on doing the point counts we arrive at s2¼362s1ands3¼s13. By the restrictions on thesa, it must be that

.

On the other hand, if the secant is1333, thens2¼322s1ands3¼s1þ2. This time there are two possibilities: one is the previous one scaled by 3, and the other is

.

Hence we may assume we have one of these two sets of values; then all points ofS1

are on the same type of 4-secant. If that secant is1112, thens₁must be divisible by 3;

so the cases₁¼13 is out. Whens₁¼10, there will be 60122secants. However, if we do the same argument for points inS4, we shall find ten2224secants. But then these two types of secants contain 60þ30¼90 pairs of2s, and yet there are only ¹²₂ ¼66 available.

We can thus takes_a¼11 for 1cac4. By Lemma 3.2,z₂09 for all points inS, and we get three possibilities for the secant counts through a point. Suppose that X AS1. ThenX is on at most ten113secants and at most five122secants. For each of the secant counts forX, we can list the possibilities for the numbers of 3-secants of the two types and see whether the remaining points can be incorporated in the needed further secants. The results are these:

case z2 z3 z4 z5 #113 #122 further secants

1 10 15 1 0 10 5 1234

2 11 13 2 0 8 5 1112,1333

3 11 14 0 1 10 4 12223

4 11 14 0 1 9 5 11233

However, in Case 3, the 3 on the 5-secant would have z₂c8 (the three 2s on the 5-secant are not on 2-secants with3), and in Case 4, the2on the 5-secant would have z₂c9. Neither of the resultingz₂values allows a 5-secant, so these two cases are out.

AsX is on five122secants in either remaining case, all 55 pairs of2s appear on these secants. But the same argument applies to all theSa. That means there can be no aaab secants at all, and z2 ¼10 is the only possibility. All the 4-secants are 1234s, and each point ofSis on exactly one of them. We havex2 ¼x3 ¼220 andx4¼11.

BothS1US4andS2US3are complete 22-arcs in the plane and the eleven 4-secants are common secants to the two arcs. This completes the proof. r Proposition 3.6.If C is the code over F₅ of a projective plane P of order25with no complete22-arcs then C^?has no word of size44.

Proof.By Lemma 3.5, ifPhas no complete 22-arcs the support setSof a word of weight 44 must have the configuration described in Lemma 3.4. LetS¼S1US4as in Lemma 3.4, and letT¼S1.

We havez₂¼20,z₄¼1, andz₅¼5 for any point ofS. Each 2-secant meetsTin one point, each 4-secant meets Tin two points, and each 5-secant meetsT in five points. FortAT, let t⁰be the other point of Ton the 4-secant throught; we have ðt⁰Þ⁰¼t. LetFbe the collection of 5-subsets ofTof the formlVT, wherel is a 5- secant meetingT. Ift;uATare distinct andt⁰0u, then the line ontandumust be a 5-secant; denote the corresponding member ofFby½t;u. It follows thatðT;FÞis a group divisible design in which the groups are the setsft;t⁰g. Moreover, ifFAF, then because each point ofF is on four other members ofF, andFand the resulting 20 members ofFare all distinct, there is a uniqueF⁰inFthat is disjoint fromF. If tAF, then ast⁰does not appear on any of these 20 members ofF,t⁰must be onF⁰. Now letT1be a set of representatives of the pairsft;t⁰g,tAT, and letF1be a set

of representatives of the pairs fF;F⁰g. Let M be the 1111 matrix with rows indexed byF1and columns indexed byT1, in which theðF;tÞentry is given by

0 if t;t⁰BF 1 if tAF 1 if t⁰AF:

Then each column of M has five nonzero entries. Suppose t and u index di¤erent columns of M, and let F¼ ½t;u, G¼ ½t⁰;u. Then F⁰¼ ½t⁰;u⁰andG⁰¼ ½t;u⁰. The rows in which both the columns indexed bytanduhave nonzero entries correspond to the pairs fF;F⁰gandfG;G⁰g. It follows that the 2 by 2 submatrix for these two rows and columns is some scaling of

1 1

1 1

:

Consequently, the columns ofMare orthogonal, andM^TM¼5I₁₁. But that would mean detM¼5¹¹⁼², which is impossible.

Thus there is no such word. r

Proposition 3.7.Letwbe a word of weight45in C^?¼C₅^?ðPÞ,wherePis a projective plane of order25.Thenwis a scalar multiple ofv^bv^l,wherebis a Baer subplane of Pand l is a line ofPthat is a line of the subplane.

Proof. Let Sbe the support set of wand let s¼sðwÞ. First we considers¼2, so thatSis a j-secant set for some jwith 4cjc7, by Lemma 3.1. Scale wto make S¼S1US4. Inequality (6) implies thatz2ðXÞd17 for anyX AS, so thatsad17.

Ass₁þs₄10ðmod 5Þands₁s₄10ðmod 5Þby Lemma 2.7, we may rescale again to assume thats₁¼25 ands₄¼20. Then forXAS4,z₂ðXÞc20. By inequality (6) again, j¼4 or 5. Suppose that j¼4. As a 4-secant meets eachSain two points, let X AS1 be on a 4-secant. ThenX is also on 17 2-secants and one more secant with a point inS4. The remaining seven secants throughX must meetSin S1 alone, and so have sizes that are multiples of 5. By Equation (4), P

ði2Þzi¼18, but these seven contribute at least 21 to the left side. Consequently j¼4 is not possible.

With j¼5, inequality (6) becomesz₂ðXÞd20. Thus for allXAS₁,z₂ðXÞ ¼20.

As above, the remaining six secants on X can only be 5-secants: z5ðXÞ ¼6 and ziðXÞ ¼0 fori02;5. IfY AS4, the secants on Y and points ofS1 are 2-secants.

Thusz2ðYÞ ¼25. The remaining secant onY must contain all the other points ofS4. That is,z20ðYÞ ¼1 andziðYÞ ¼0 fori05;20. In other words, the points ofS4are collinear; let their line bel. The 25 points ofS1along with the 5-secants now form an a‰ne plane inP. These 5-secants can meetl only in points outside ofS4. It follows easily that with the addition of the six points oflnS4 toS1 and the line l to the 5- secants, we create the Baer subplane needed in the statement of the proposition.

Supposes¼4, the value that must be ruled out. Then j¼3, by Lemma 3.1, and for anyXAS,z₂ðXÞd8. Lemma 3.2 implies thatz₂ðXÞd9, in fact, so thats_ad9 for allaAF_p. To begin with, suppose by scaling thats₄¼9, makingz₂ðXÞ ¼9 for allXAS1. Thenz₃ðXÞ ¼16 andz₄ðXÞ ¼1, from (9). As in Lemma 3.5, the possible 4-secants onX are1112and1333, and we do the point counts in the two cases, with X ont₁113secants and ont₂122secants, to find possible parameter values. If the 4- secant is1112, we have

s1¼t1þ3; s2¼2t2þ1; s3¼t1; t1þt2¼16:

If the 4-secant is1333, then

s1¼t1þ1; s2¼2t2; s3¼t1þ3; t1þt2¼16:

Sinces2is odd for1112and even for1333, allX inS1 are on the same type of 4- secant. In particular,s₁must be divisible by 3 when the secants are1112, and up to a further scaling, there is only one set of values:

.

If Y AS2, then z₂ðYÞ ¼9 also, and z₃ðYÞ ¼16. But if Y is on one of the 1112 secants, it is on at most nine122s and at most four244s, yielding too few 3-secants.

When the 4-secants are all1333,s₁þ1cs₂ from the122s on a2. Ifs₁¼9, then s2¼16 ands3¼11; scaling the word by 4 produces a parameter list withs4¼9 that no longer fits the pattern. Two possibilities remain:

.

.

If s1¼10, thenz2ðYÞ ¼10 forY AS4 (as all secants on a1 and a4are 2-secants), and z3ðYÞd14 by (9). There are ten 1333s; they use 30 pairs of 3s and leave

12

2 30¼36 pairs. Thus someY AS4is on at most four334s and so on at least ten 244s. But there are not enough4s available for this.

Similarly, ifs₁¼11, there is Y AS4 on at most five 334s. Since Y is on at most eight 244s, z₃ðYÞc13. As z₂ðYÞ ¼11, (9) shows that Y must be on a 4-secant.

There cannot be a1 on it; and as all the pairs of2s are on the122s, the only possi- bility is 3444. But z₃ðYÞd12, so that Y is on at least seven 244s; but again, too many4s are needed.

Therefores_ad10 for all aAF₅. It cannot be that s_ad11 for all a, for then three saare 11 and one is 12, and Lemma 2.7 excludes this. Scaling, we takes4¼10. Up to further scaling, Lemma 2.7 allows three possibilities:

.

.

.

Suppose thats₁¼15. IfXAS1, thenXis on at most five122s and at most ten113s,

makingz₃ðXÞc15. By (9),z₂ðXÞ ¼10 andz₃ðXÞd14. ThenX is on at least four 122s, so the 15 members ofS1require at least 60122s. As there are only 45 pairs of 2s available,s₁¼15 is ruled out.

Now let s₁¼12, and use the argument finishing the proof of Lemma 3.2: for X AS1,X is on at most five122s and at most 11113s. The possibilities are

z2 z3 z4 z5 #113 #122 further secants

9 16 1 0 11 5 1234

10 15 0 1 11 4 12223

10 15 0 1 10 5 11233

10 14 2 0 9 5 1112,1333

Regardless of the case,X is collinear with at least five pairs of2s, so that the secants on all the points ofS1 account for at least 60 pairs (each such pair appears with just one1). But there are only 55 pairs of2s; thuss1¼12 is out.

Finally, suppose thats1¼11. Begin by rescalingw by 3 to takes1¼13, s2¼10, s3¼11, ands4¼11. Again we seek to reach a contradiction by counting pairs of2s as they appear on secants with points in S1. If the secant is a122, that is the only secant this pair of2s is on. We do the same kind of secant analysis for a pointX in S1. There are quite a few, but only two in whichX is on at most three122secants:

z2 z3 z4 z5 z6 #113 #122 further secants

11 13 1 1 0 10 3 1112,12223

11 14 0 0 1 11 3 112222

As the3on the 5-secant in the first case hasz2c7, this possibility is out. In the sec- ond,Xis collinear with nine pairs of 2s, six of them on the 6-secant. Since this secant contains two1s, the count of pairs of2s from these1s is six apiece (their122s give di¤erent pairs; by (9), no1can appear on two such 6-secants). Thus regardless of the secant pattern of X, we require at least 413¼52 pairs of 2s for the secant col- linearities with points inS1, i.e. more than the 45 that are available. r Proof of theorem and corollary.The theorem is now proved, and for the corollary we note that if the planePis desarguesian then complete 22-arcs do not exist; see [3], [9].

Thus 44 is not a possibility. Furthermore,Pdoes not have subplanes of order 4; see, for example, [2]. SincePhas Baer subplanes, the minimum weight is 45. r Remarks.1) In [11] it is noted that it is easy to show that in a plane of order 9 with a word of weight 15 in its dual ternary code, the word must have the same form that is established in Proposition 3.7.

2) All the known planes of order 25 have Baer subplanes; in particular, all trans- lation planes of square order have Baer subplanes (see a new proof of this in [6]).

Thus the minimum weight is at most 45 for the known planes.

3) No plane of order 25 has been shown to contain a subplane of order 4, and some have been shown to not contain any; see [8].

4) The authors are unaware of any proofs of existence or non-existence of complete

22-arcs, except in the desarguesian case. Even if a plane does have a 22-arc, it would need to have two 22-arcs, C1 and C2, with the additional property that they share eleven secants, and of the remaining 220 secants toC1, say, 110 are external toC2and 110 are tangents toC2.

5) It seems most likely that the minimum weight is 45 for all planes of order 25.

Note that the translation planes of order 25 were classified by Czerwinski and Oak- den [7]. Most of these and some other (non-translation) planes of order 25 can be found at the web site: http://www.ces.clemson.edu/~keyj/Key/planes25

References

[1] E. F. Assmus, Jr., J. D. Key,Designs and their codes. Cambridge Univ. Press 1992.

MR 93j:51003 Zbl 0762.05001

[2] A. E. Brouwer, H. A. Wilbrink, Block designs. In: Handbook of incidence geometry, 349–382, North-Holland 1995. MR 97j:05018 Zbl 0823.51007

[3] J. M. Chao, H. Kaneta, Classical arcs in PGðr;qÞfor 23cqc29. Discrete Math.226 (2001), 377–385. MR 2002b:51005 Zbl 0977.51002

[4] K. L. Clark,Improved bounds for the minimum weight of the dual codes of some classes of designs. PhD Thesis, Clemson University 2000.

[5] K. L. Clark, J. D. Key, Geometric codes over fields of odd prime power order. In:Pro- ceedings of the Thirtieth Southeastern International Conference on Combinatorics, Graph Theory, and Computing(Boca Raton, FL,1999), volume 137, 177–186, 1999.

MR 2000k:94053 Zbl 0996.94048

[6] K. L. Clark, J. D. Key, M. J. de Resmini, Dual codes of translation planes.European J.

Combin.23(2002), 529–538. MR 1 931 937

[7] T. Czerwinski, D. Oakden, The translation planes of order twenty-five.J. Combin. Theory Ser. A59(1992), 193–217. MR 93c:51009 Zbl 0764.51004

[8] M. J. de Resmini. Private communication.

[9] G. Faina, F. Pambianco, On the spectrum of the valueskfor which a completek-cap in PGðn;qÞexists.J. Geom.62(1998), 84–98. MR 2000c:51008 Zbl 0917.51016

[10] J. D. Key, M. J. de Resmini, Small sets of even type and codewords.J. Geom.61(1998), 83–104. MR 98k:51024 Zbl 0910.51005

[11] J. D. Key, M. J. de Resmini, Ternary dual codes of the planes of order nine.J. Statist.

Plann. Inference95(2001), 229–236. MR 2002b:94031 Zbl 0978.05018

[12] H. Sachar, TheFpspan of the incidence matrix of a finite projective plane.Geom. Dedi- cata8(1979), 407–415. MR 81e:94020 Zbl 0419.51005

Received 19 December, 2002

K. L. Clark, Department of Mathematics, Queens University of Charlotte, Charlotte NC 28274, USA

Email: [email protected]

L. D. Hatfield, H. N. Ward, Department of Mathematics, University of Virginia, Charlottes- ville VA 22904, USA

Email: {ldw9e, hnw}@virginia.edu

J. D. Key, Department of Mathematical Sciences, Clemson University, Clemson SC 29634, USA

Email: [email protected]