A generalization of the cylinder conjecture for divisible codes

A GENERALIZATION OF THE CYLINDER CONJECTURE FOR DIVISIBLE CODES

SASCHA KURZ, SAM MATTHEUS

ABSTRACT. We extend the original cylinder conjecture on point sets in affine three-dimensional space to the more general framework of divisible linear codes overFqand their classification. Through a mix of linear programming, combinatorial techniques and computer enumeration, we investigate the structural properties of these codes. In this way, we can prove a reduction theorem for a generalization of the cylinder conjecture, show some instances where it does not hold and prove its validity for small values ofq. In particular, we correct a flawed proof for the original cylinder conjecture forq= 5and present the first proof forq= 7.

Keywords:cylinder conjecture, linear codes, divisible codes MSC:Primary 05B25; Secondary 51D20, 51E22.

1. INTRODUCTION

The cylinder conjecture, as originally formulated in [1], was motivated by direction problems in finite geometry. To be more precise, the strong version of the conjecture is formulated as follows.

Conjecture 1. LetS be a set ofp² points inAG(3, p),pprime. If S has the property that every plane is incident with0moduloppoints ofSthenSis a cylinder, i.e. the union ofpparallel lines.

The idea to classify all sets ofp²points inAG(3, p)determining few directions, is a continuation of similar results inAG(2, p), which started with the work of R´edei and Megyesi [9] and Lov´asz and Schrijver [8]. The cylinder conjecture as stated above, is of interest from a coding theoretical perspective as well. In particular, one could view this conjecture as part of a more general search for the classification of divisible codes.

A linear[n, k]q codeCis∆-divisible if all its weights are multiples of a fixed integer∆. When∆andq are coprime, the classification of∆-divisible codes overFqis almost trivial, see for example [11] for a general survey on∆-divisible codes, which contains this result. In this paper, we will focus on the case when∆is a power of the field sizeq, hence looking atq^r-divisible codes, whereris a non-negative integer. For a survey on these codes we refer to [5]. In the case of the original cylinder conjecture, a positive answer would give a classification of all[p²,4]_pcodes that arep-divisible as we will explain later on. With the classification ofq^r- divisible codes overFqin mind it thus makes sense to generalize the cylinder conjecture to higher dimensions and non-prime field characteristic. In order to state and motivate it, we require some terminology and a proper notion of cylinders in higher dimensions. We will therefore defer it to the next section.

2. PRELIMINARIES

The notation we will use throughout is the following. LetV ∼=F^vq be av-dimensional vector space over the finite fieldFq withq elements andPG(v−1, q)the projective space associated to it. By ak-space of PG(v−1, q)we mean ak-dimensional linear subspace ofV, also using the terms points, lines, planes, and hyperplanes for1-,2-,3-spaces, and(v−1)-spaces respectively. A multiset of points inPG(v−1, q)will be denoted asM, whileSrefers to a set of points.

To each multiset Mof n points inPG(v−1, q)we can assign aq-ary linear code C(M)defined by its generator matrix whosencolumns consist of representatives of thenpoints ofM. The codeC(M)is projective if and only ifMis a set. The weight of a codeword is the number of non-zero coordinates and as mentioned before, a code is calledq^r-divisible if the weight of each codeword is divisible byq^r.

Definition 2. The multisetMofnpoints inPG(v−1, q)isq^r-divisibleif and only ifC(M)is. Equivalently, Misq^r-divisible if|M ∩H| ≡ |M| (modq^r)for every hyperplaneHofPG(v−1, q).

1

For example, the set of points of a k-space is q^k−1-divisible and so is the multiset of the points of a collection ofk-spaces. The converse gives an interesting question: For which integerslis eachq^k−1divisible set of^q_q−1^k−1·lpoints the union ofl(disjoint)k-spaces? We can always choosel≥1and the maximum value forlimplies extendability results of spreads: each set ofq^k+ 1−ldisjointk-spaces inF^2kq can be extended to ak-spread. These results are mostly formulated in the language of minihypers, see e.g. [4].

An interesting property of q^r-divisible sets of points is that the divisibility is preserved (to some extent) upon intersecting with subspaces, allowing for inductive arguments.

Lemma 3. ([5, Lemma 7]) Suppose thatMis aq^r-divisible multiset ofmpoints inPG(v−1, q)andX a (v−j)-space ofPG(v−1, q)with1≤j < r. Then the restrictionM ∩X isq^r−j-divisible.

PROOF. By induction, it suffices to consider the casej = 1, i.e.X =His a hyperplane inPG(v−1, q).

The hyperplanes ofH are the (v−2)-subspaces ofPG(v−1, q)contained inH. Hence the assertion is equivalent to|M ∩U| ≡ |M| (modq^r−1)for every (v−2)-subspaceU ⊂PG(v−1, q). By assumption we have|M ∩Hi| ≡m (modq^r)for theq+ 1hyperplanesH1, . . . , Hq+1lying aboveU. This gives

(q+ 1)m≡

q+1

X

i=1

|M ∩H_i|=q· |M ∩U|+|M| ≡q· |M ∩U|+m (mod q^r)

and hencem≡ |M ∩U| (modq^r−1), as claimed.

We can now generalize the concept of a cylinder to higher dimensional vector spaces as follows.

Definition 4. Letrbe a non-negative integer. An(r+ 1)-cylinderis a multiset ofq^r+1points inPG(v−1, q) that arises as the union of the points ofq affine(r+ 1)-subspacesL1\F,. . .,Lq \F, where theLi are (r+ 1)-spaces andF is ar-space that is contained in allLi.

We remark that our definition of a2-cylinder matches the definition of a cylinder in [3] and the one stated above. By convention a1-cylinder is just a multiset ofqpoints. As the affine subspaces mentioned in the definition above will appear often, we introduce the notationA(P, F)for the affine subspacehP, Fi\F, where Pis a point andFand arbitrary subspace. Note that we havedim(A(P, F)) = dim(F) + 1. Next, we observe that(r+ 1)-cylinders can be easily constructed starting from a multiset ofqpoints.

Construction 5. Letrandv⁰be non-negative integers, and consider av⁰-spaceV⁰and a disjointr-spaceF inV =F^v

0+r

q . IfM⁰={P1, . . . , Pq}is a multiset ofqpoints inV⁰, then a(r+1)-cylinder can be constructed as the multisetMconsisting of the points ofA(Pi, F),i= 1, . . . , q.

Proposition 6. The multiset of points of an(r+ 1)-cylinder isq^r-divisible.

PROOF. We use the notation of Definition4for a given(r+ 1)-cylinder. The statement is trivial forr= 0so that we assumer≥ 1. Each hyperplaneH intersectsF either in dimensionrorr−1. In the first case we have|(Li\F)∩H| ∈ {0, q^r}. In the second case we have|(Li\F)∩H|=q^r−1for all1 ≤i≤q. Thus, we have|M ∩H| ≡0 (modq^r)for the corresponding multiset of pointsMof the(r+ 1)-cylinder.

So,(r+ 1)-cylinders yieldq^r-divisible multisets ofq^r+1points and the question arises if there are other isomorphism types. Indeed there are. Any multiset ofq(possibly equal) points with multiplicityq^reach is q^r-divisible. For that reason we will consider sets of points instead of multisets in the remaining part. We remark that studying multisets of points with restricted point multiplicity might be an interesting problem, but we will not go into this here. It will also depend on the dimension whether other isomorphism types exist.

Since each setSof points inPG(v−1, q)can be embedded inF^v

0

q forv⁰ > vwe will always assume thatS is spanning, i.e. spansPG(v−1, q).

Observation 7. In Construction5,Mis spanning if and only ifM⁰is spanning andMis a set if and only if M⁰is a set.

We call an(r+ 1)-cylinder spanning or projective if the corresponding multiset of points is.

Lemma 8. There exists a spanning projective(r+ 1)-cylinder inPG(v−1, q)if and only ifr+ 2≤v≤r+q.

PROOF. Due to the above observations it suffices to remark that2 ≤dim(hM⁰i)≤qand all dimensions in

that range can indeed be attained.

Let(v, r, q)be a triple of non-negative integers whereqis a prime power. The question of classification thus boils down to the following.

Question 9. IfSis aq^r-divisible spanning set ofq^r+1points inPG(v−1, q), does it follow thatSmust be a (r+ 1)-cylinder?

We say that the generalized cylinder conjecture is true (or false) for a triple(v, r, q)if the answer to this question is affirmative (or not). Notice that the divisibility assumption is as strong as possible and for good reason. It’s not hard to construct aq^r-divisible set ofqⁿ points inPG(v−1, q),r+ 1< n≤v−2, as the disjoint union ofq^n−r−1different(r+ 1)-cylinders, while not being a cylinder itself.

As we will see in the next section, it makes no difference if we consider point sets in affine or projective geometries. Therefore, the original cylinder conjecture asserts that the answer is affirmative for the triple (4,1, p),pprime. Remark that whenr = 0, the conjecture is trivially true, so we assumer ≥1from now on. In Section3we will prove some structural results and generalize ideas from [3], using a combination of linear programming and geometrical methods. We discuss some cases where we can determine the validity of the generalized cylinder conjecture in Section4. Our main results in these two sections can be summarised as follows. In Lemma3we have seen that the restrictionS ∩Hof aq^r-divisible set of pointsSinPG(v−1, q) to a hyperplaneH isq^r−1-divisible. We will show that if|S| = q^r+1, then there are many hyperplanesH with|S ∩H| = q^r, and hence inductive arguments can be applied. Indeed, it will turn out that everything boils down to the caser= 1, as Corollary20shows that the generalized cylinder conjecture is true for(v, r, q) if and only if it is true for(v−r+ 1,1, q). The generalized cylinder conjecture is not always true: ifqis a proper prime power, i.e., not a prime, we can construct counterexamples for suitable dimensions based on the existence of subfields, see Corollary31. For small values ofqwe find that the generalized cylinder conjecture is true wheneverq ∈ {2,3,5}, for the triples(3,1,4) and(4,1,4), but not for(5,1,4). The special case (4,1, q)is treated in detail in Section5, where the casesq = 5(fixing a flawed proof in [3]) andq = 7are fully resolved. Although our numerical data is still rather limited, we state:

Conjecture 10. The generalized cylinder conjecture is true for(r+ 3, r, q)and for(v, r, p)ifpis a prime.

Finally, we have to remark that the “right” generalization of the cylinder conjecture to non-prime field sizes is a bit unclear in the context of directions determined by a point set, but Definition4and Question9are at least reasonable in the context ofq^r-divisible sets.

3. GENERAL RESULTS AND THE REDUCTION THEOREM

The linear programming approach is based on the following three linear equations, commonly referred to as the standard equations.

Lemma 11. (See e.g.[5, Lemma 6])

LetSbe a set of points inPG(v−1, q)with|S|=n, and letaibe the number of hyperplanes inPG(v−1, q) containing exactlyipoints ofS(0≤i≤n). Then we have

n

X

i=0

a_i = q^v−1

q−1 , (1)

n

X

i=1

iai = n·q^v−1−1

q−1 , (2)

n

X

i=2

i 2

ai = n

2

·q^v−2−1

q−1 . (3)

PROOF. Double-count incidences of the tuples(H),(P1, H), and({P1, P2}, H), whereH is a hyperplane

andP16=P2are points contained inH.

The set {a_i}_i is called the spectrum of S. The setS being spanning is equivalent toa_n = 0, i.e., no hyperplane contains all points. In that case, the standard equations are equivalent to the first three MacWilliams identities for projective linear codes.

We can adapt Lemma11to our situation ofq^r-divisible sets of points.

Lemma 12. LetSbe aq^r-divisible spanning set ofq^r+1points inPG(v−1, q)and letaiq^rbe the number of hyperplanes inPG(v−1, q)containing exactlyiq^rpoints ofS(0≤i≤q). Then we have

(q−1)

q−1

X

i=0

aiq^r = q^v−1, (4)

(q−1)

q−1

X

i=0

ia_iqr = q(q^v−1−1), (5)

(q−1)

q−1

X

i=0

i(iq^r−1)aiq^r = q(q^r+1−1)(q^v−2−1). (6) PROOF. We use the equations from Lemma11. Multiplying them by q−1, usingn = q^r+1, and taking q^r-divisibility into account gives

(q−1)

q

X

i=0

aiq^r = q^v−1, (7)

(q−1)

q

X

i=0

iq^ra_iqr = q^r+1 q^v−1−1

, (8)

(q−1)

q

X

i=0

iq^r 2

aiq^r =

q^r+1 2

(q^v−2−1). (9)

Finally, dividing (8) byq^r, (9) byq^r/2, and recallinga_qr+1 = 0gives the stated result.

Lemma 13. LetS be aq^r-divisible spanning set ofq^r+1 points inPG(v−1, q). Then, the number aq^r of hyperplanes with the smallest non-zero number of points is at least ^q_q−1^v−1− q^v−r−1−q+ 1

. PROOF. Using the notation from Lemma12, Equation (6) minus2q^r−1times Equation (5) gives

(q−1)

q−1

X

i=0

q^ri(i−2)a_iqr =−q^r (q^v−1)−(q−1)q^v−r−1+ (q−1)² .

Sincei(i−2)≥0anda_iqr ≥0for all2≤i≤q−1, we conclude(q−1)aq^r ≥q^v−1−(q−1)(q^v−r−1−q+1).

In other words, almost all hyperplanes contain exactlyq^r points. For these hyperplanes we might apply induction and assume that they arer-cylinders, as we will do later on in the proof of Theorem19.

The general idea behind the proof of Lemma13is the linear programming method based on the standard equations, which is a common technique in finite geometry. To be precise, we maximize or minimize a certain a_j under the constraints of Lemma12, where we assume that alla_i∈R≥0. Bounds similar as in Lemma13 for otheraiq^r can be obtained easily.

Lemma 14. LetSbe aq^r-divisible spanning set ofq^r+1points inPG(v−1, q). Then, the numbera0of empty hyperplanes is at most(q^v−r−1−q+ 2)/2.

PROOF. From Lemma12,2q^rtimes Equation (4) minus3q^r−1times Equation (5) plus Equation (6) gives (q−1)

q−1

X

i=0

q^r(i−1)(i−2)aiq^r = (q−1)(q^v−1−q^r+1+ 2q^r).

Since(i−1)(i−2)≥0andaiq^r ≥0for all0≤i≤q−1, we conclude2q^ra0≤(q^v−1−q^r+1+ 2q^r).

Lemma 15. LetSbe aq^r-divisible spanning set ofq^r+1points inPG(v−1, q). Then, the numbera₀of empty hyperplanes is at least^qv−r−1_q−1⁻¹.

PROOF. Applying Lemma12,q^r(q−1)times Equation (4) minusq^r+1−1times Equation (5) plus Equation (6) gives

(q−1)

q−1

X

i=0

q^r(i−1)(i−q+ 1)a_iqr = (q^v−1−q^r)(q−1).

Since(i−1)(i−q+ 1)≤0anda_iqr ≥0for all1≤i≤q−1, we conclude(q−1)a₀≥q^v−r−1−1.

Corollary 16. For everyq^r-divisible spanning set ofq^r+1points inPG(v−1, q)there is at least one empty hyperplane.

PROOF. SincePG(v−1, q)containsv 1

q = q^v−1+q^v−2+· · ·+ 1points, we havev ≥ r+ 2, so that

Lemma15gives the stated result.

Thus, it makes no difference if we speak about point sets in AG(v−1, q)or PG(v −1, q). Another implication of Corollary16is that the generalized cylinder conjecture is trivially true when the dimensionvis small.

Proposition 17. Let S be aq^r-divisible spanning set of q^r+1 points inPG(v−1, q). Ifv ≤ r+ 2, then v=r+ 2andS 'AG(v−1, q).

PROOF. As in the proof of Corollary16we concludev≥r+ 2, so thatv=r+ 2. A single empty hyperplane leaves onlyq^r+1possible points, which all have be to contained inS.

We see that the generalized cylinder conjecture is trivially true for all (v, r, q), where v ≤ r+ 2. In other words, the classification ofq^r-divisible spanning sets ofq^r+1points inPG(v−1, q)ischallengingfor v≥r+ 3only.

With these auxiliary results in hand, we can prove our reduction theorem which essentially states that the validity of the generalized cylinder conjecture depends on the differencev−rand not on the values ofvand ritself. As a first step, we show that we can decreasevandrsimultaneously and preserve the truthfulness.

Proposition 18. If the generalized cylinder conjecture is true for(v+ 1, r+ 1, q), then it is true for(v, r, q).

PROOF. If the generalized cylinder conjecture is false for(v, r, q), we can apply the following construction to the corresponding counterexample and obtain a counterexample for(v+ 1, r+ 1, q).

Consider aq^r-divisible setSofq^r+1points inV := PG(v−1, q). For a pointPoutside of the ambient space we consider the new ambient spaceV⁰ :=hV, Piand setS⁰ ={A(S, P) : S∈ S}. By construction dim(V⁰) =v+ 1andS⁰is a set ofq^r+2points inV⁰. Now letH⁰by a hyperplane ofV⁰. EitherH⁰ =V or H:=H⁰∩V is a hyperplane ofV. In the first case the haveS⁰∩H⁰=S, which is of cardinalityq^r+1. In the second case we have|S ∩H| ≡0 (modq^r). IfP ≤H⁰, then|S⁰∩H⁰|=q· |S ∩H| ≡0 (modq^r+1). If Pis not contained inH⁰then each of theq^r+1affine linesA(S, P)is met byH⁰in a single point not equal to P, so that|S⁰∩H⁰|=q^r+1. Thus,S⁰isq^r+1-divisible and one can see that it is not a cylinder, ifSis not.

Theorem 19. If the generalized cylinder conjecture is true for(v,1, q), then it is true for all(v+r−1, r, q).

PROOF. Due to Proposition17we can assumev≥4. We will prove the result by induction onr. So, assume that the generalized cylinder conjecture is true for(v+r−2, r−1, q). LetS be a spanningq^r-divisible set ofq^r+1points inPG(v⁰−1, q), wherev⁰ =v+r−1andr≥2. Now letFbe the set of pointsF such that there exists a pointS∈ SwithA(S, F)⊆ S.

We will structure our proof into some intermediate results:

(1) For each pointS ∈ Sthere exists an(r−1)-spaceBsuch thatA(S, B)⊆ S.

(2) dim(hF i)∈ {r−1, r}.

(3) For eachS1, S2∈ Sthere exists an(r−1)-spaceBsuch thatA(S1, B)andA(S2, B)are both contained inS.

(4) LetS∈ Sbe a point such that there exist(r−1)-spacesB₁6=B₂withA(S, B₁), A(S, B₂)⊆ S. Then, we haveA(S,hF i)⊆ S.

(5) Sis an(r+ 1)-cylinder.

For (1) we use Lemma13to conclude that there are at mostq^v0−r−1−q+ 1hyperplanes that do not contain exactlyq^rpoints fromS. Thus, for each pointS ∈ Sat least one of the ^{qv0 −1}_q−1⁻¹ hyperplanes containingS contains exactlyq^rpoints fromS. IfHis such a hyperplane, we can apply induction forHand conclude that S ∩Hcan be partitioned into∪^q_i=1A(Si, B)for some(r−1)-spaceBandqpointsSi∈ S. Of course there exists an index1≤j≤qwithS∈A(Sj, B)and henceA(S, B) =A(Sj, B).

For (2) we note that every pointF ∈ Fis contained in every empty hyperplane, asF is the only point in A(S, F)not inS, so thathF iis contained in the intersection of all empty hyperplanes. Since there are exactly

qv0 −dim(hF i)−1

q−1 hyperplanes containinghF iwe conclude from Lemma15thatdim(hF i)≤r+ 1and moreover if equality holds then every hyperplane containinghF iis empty. SinceS 6=∅, this is only possible ifhF iis a hyperplane itself, i.e.,v⁰ =r+ 2andv= 3. Thus, we havedim(hF i)≤r. IfBis a subspace according to (1), thenB⊆ Fso thatdim(hF i)≥r−1.

For (3) we can apply the same idea as in (1). Consider the line L = hS1, S2i, then L is contained in

q^{v0 −2}−1

q−1 > q^v0−r−1−q+ 1hyperplanes (usingr≥2). So, we can use Lemma13to conclude the existence of a hyperplaneHwithL≤Hand|S ∩H|=q^r. Induction on this hyperplane then gives the existence of an (r−1)-spaceBsuch thatA(S, B)⊆ S ∩Hfor allS∈ S ∩H.

For (4) we note thatB₁, B₂⊆ Fimpliesdim(hF i)≥r, so that (2) givesdim(hF i) =r.

Ifr= 2, then take any pointFon the linehF iand a pointS∈ S. We will directly prove thatA(S, F)⊆ S.

We know that there are at mostq^v0−3−q+ 1hyperplanes not intersectingSinq²points by Lemma13, so out of theq^v0−3hyperplanes throughhS, FiintersectinghF iin onlyF, there must be at leastq−1hyperplanes containing exactlyq²points. So take one such hyperplane, apply induction and find that A(S, F) ⊆ S. It follows thatA(S,hF i)⊆ Sfor allS∈ S, i.e., (4) is valid forr= 2.

Now assume r ≥ 3. We can find a pointS ∈ S and distinct(r−1)-spacesB1, B2such thatA(S, B1) andA(S, B2)both are contained inS. Now, take a pointF ∈ hF i \ {B1, B2}, and consider any3-spaceπ throughhS, Finot intersectingB1∩B2. Then this3-spaceπintersects B1andB2each in a point, sayF1

andF₂. There are ^{qv0 −3}_q−1⁻¹ hyperplanes through this3-space, which is more thanq^v0−r−1−q+ 1ifr≥3.

We again conclude by induction that there must be a(r−1)-spaceB such thatA(S, B) ⊆ S. As F₁ and F₂must be contained inB, we conclude thatF must also be and henceA(S, F)⊆ S. It again follows that A(S,hF i)⊆ Sfor allS∈ Sand so (4) is valid forr≥3too.

For the final step (5) we can assume that there exists a pointS∈ Ssuch that there is a unique(r−1)-space BsatisfyingA(S, B)⊆ S, as otherwise by (4) we can already conclude thatSis an(r+ 1)-cylinder. Under this assumption, it follows from (3) thatA(S⁰, B)⊆ Sfor allS⁰ ∈ S, so that moduloBwe obtain a setS⁰of q²points that isq-divisible and spans a space of dimensionv⁰−r+ 1 =v. For this setS⁰we can apply the generalized cylinder conjecture for(v,1, q)and concludeS⁰ =∪^q_i=1A(S⁰_i, B⁰)for some pointsS_i⁰andB⁰. By construction, we then haveS=∪^q_i=1A(S_i⁰,hB, B⁰i)which shows thatSis an(r+ 1)-cylinder.

Combining the previous results, we find the promised reduction theorem.

Corollary 20.The generalized cylinder conjecture is true for(v, r, q)if and only if it is true for(v−r+1,1, q).

Finally, we can transfer many of the insights of [3] to our more general situation.

Lemma 21. (Cf. [3, Lemma 1])

LetSbe aq^r-divisible spanning set ofq^r+1points inPG(v−1, q). LetKbe a subspace of codimension2in PG(v−1, q). Assume that|S ∩K|=kq^r−1for some integer0< k < q. Then, any hyperplane containing Kcontains at mostkq^rpoints fromS.

PROOF. Since every hyperplane containing K contains at least kq^r−1 points, it should contain at least q^r points. Therefore, counting the number of points on hyperplanes containing K, we find at least (q+

1) q^r−kq^r−1

+kq^r−1 =q^r+1−(k−1)q^rpoints. Hence, there are(k−1)q^rpoints left, which implies

that a single hyperplane contains at mostkq^rpoints.

Corollary 22. (Cf. [3, Corollary 1])

LetSbe aq^r-divisible spanning set ofq^r+1points inPG(v−1, q). Suppose the hyperplaneH containskq^r points ofS, where0< k < q, then every hyperplaneKofH, i.e.,K≤H is a subspace of codimension2in PG(v−1, q), contains either0orlq^r−1points ofSfor somek≤l≤q.

PROOF. By Lemma3we know that|S ∩K| ≡0 (modq^r−1)holds . Suppose thatKcontainslq^r−1points for some integer0< l < k. Then, by Lemma21, every hyperplaneHinPG(v−1, q)containingKcontains

at mostlq^r< kq^rpoints, which is a contradiction.

For the next proof, denote by[x]q := (q^x−1)/(q−1)the number of hyperplanes through a codimension x-space.

Theorem 23. (Cf.[3, Theorem 2])

LetS be aq^r-divisible spanning set ofq^r+1points inPG(v−1, q). IfScontains a full affine(r+ 1)-space thenSis a(r+ 1)-cylinder.

PROOF. Due to Proposition17we can assumev≥r+ 3. Denote byAthe affine(r+ 1)-space contained inS and letA_∞be ‘its part at infinity’, i.e. the uniquer-space so thatA∪A_∞is an(r+ 1)-space ofPG(v−1, q).

Take a pointP ∈ S \Aand letHbe a hyperplane containingAandP. Byq^r-divisibility, we can assume that

|S ∩H|=kq^rwith1< k < q.

Fix a pointQ∈Aand consider the setKof(v−2)-spaces throughQ, not containingA_∞. Observe that

|K|= [v−2]q−[v−r−2]qand that every point in(S ∩H)\Ais contained ina:= [v−3]q−[v−r−3]q

of these spaces. Now double counting pairs{(K, R)|K∈ K, R∈K∩ S \A}we find (kq^r−q^r)a= X

K∈K

|(K∩ S \A)| ≥ |K|(kq^r−1−q^r−1),

by Corollary22. As|K|=aq, we see that in fact we have equality so that|K∩ S|=kq^r−1for allK∈ K.

Moreover, the pointQ∈Awas arbitrary so we conclude that every(v−2)-space not throughA_∞contains kq^r−1points.

Now retaking the point P ∈ S ∩H\A, we can consider the setK⁰ of(v−2)-spaces throughP, not containingA_∞. Again, this set has size|K⁰| = [v−2]_q −[v−r−2]_q. Denote byB the (r+ 1)-space hA_∞, Piandx =|S ∩B|. If we can show thatx =q^r, we are done. We can now double count the pairs {(K, R)|K∈ K⁰, R∈K∩ S \ {P}}. IfR∈Bthen the number of(v−2)-spaces throughP andR, not containingA_∞isb= [v−3]q−[v−r−2]q. For a point not inB, the correct count for the(v−2)-spaces is stillaand we find the following equality:

(kq^r−x)a+ (x−1)b= X

K∈K⁰

|(K∩ S \B)|=|K⁰|(kq^r−1−1).

After simplification we indeed findx=q^r, concluding the proof.

4. POSITIVE AND REGATIVE RESULTS

In this section we will discuss some triples(v, r, q)for which we can determine the validity of the general- ized cylinder conjecture. Before that, we introduce some notation to make the proofs less cumbersome.

IfKis ak-space inPG(v−1, q), thenK(S) :=|K∩ S|denotes the number of points in the intersection, which we will also refer to as the multiplicty ofK. IfKis a point, then we will say thatKis a0- or1-point, wheneverK(S) = 0or1respectively. Similarly, ifKis a line or a plane, we will say thatKis anm-line or m-plane ifK(S) =m.

First of, we will show that the generalized cylinder conjecture is true wheneverq∈ {2,3}.

Lemma 24. LetS be a spanning set ofq^r+1 points inPG(v−1, q). If every hyperplane ofPG(v−1, q) contains eitherq^ror no point fromS, thenv=r+ 2andS 'AG(v−1, q).

PROOF. Since onlya₀anda_qr are non-zero in Lemma12, we immediately findv =r+ 2, so that we can

apply Proposition17.

Corollary 25. The generalized cylinder conjecture is true for all triples(v, r,2).

Next up, we will show the generalized cylinder conjecture for(v,1,3). By Corollary20, this suffices to conclude that the generalized cylinder conjecture is true for all triples(v, r,3).

Proposition 26. LetSbe a3-divisible spanning set of9points inPG(v−1,3). ThenSis a2-cylinder.

PROOF. Assume to the contrary, thatSis not a2-cylinder, so that Proposition17impliesv ≥4. Since the maximum multiplicity of a hyperplane is6, each subspace of multiplicity6is a hyperplane. Assume thatK is a subspace of multiplicity4, so thatdim(K)≤v−2. We denote the codimensionv−dim(K)ofKby x, and hencex≥2. Since there are[x]3hyperplanes throughK, every1-point outside ofKis contained in [x−1]₃hyperplanes, and every hyperplane throughKhas multiplicity6, we have2[x]₃= 5[x−1]₃, so that 3^x−1=−3, which is impossible. Thus, no subspace can have a multiplicity of exactly4.

By Lemma24we can assume the existence of a hyperplaneHwithS(H) = 6. SinceSis spanning we have dim(H)≤6. Ifdim(H) = 6, then we can assume w.l.o.g. that the1-points inH are given byhe₁i, . . . ,he₆i, where thee_idenote the standard unit vectors. With this, the subspacehe₁, . . . , e₄iwould have multiplicity4, which is a contradiction. Now assumedim(H) = 5and that the1-points inHare given byhe1i, . . . ,he5iand a sixth pointP. Consider the subspacehe1, . . . , e4i. Since it does not containe5and there is no subspace of multiplicity four, it has to containP, which means that the fifth coordinate of the vectors inPare zero. We can repeat this argument for the subspaceh{e1, . . . , e5} \eii,i = 1, . . . ,4and conclude that thei-th coordinate of the vectors inP are zero for all1 ≤i≤5, which is a contradiction. Thus, the remaining possibilities are dim(H)∈ {3,4}, i.e.,v ∈ {4,5}. From Theorem23we conclude that the maximum line multiplicity is at most2. So, ifdim(H) = 3, thenS ∩Hwould be a set of6points inPG(2,3)with line multiplicity at most two, which does not exist. Therefore, we havedim(H) = 4 andv = 5. Using Corollary22we conclude that every planeπinH has a multiplicity in{0,2,3,5}. By the same reasoning as before, we cannot have5 points inPG(2,3)with line multiplicity at most two and henceS(π)6= 5. For the spectrum(a⁰_i)ofS ∩Hthe standard equations yield the unique solutiona⁰₀= 8,a⁰₂= 18,a⁰₃= 14. Now consider the subspaces spanned by one of the ⁶₃

= 20triples of1-points inH. As no line contains3points and every plane contains at most 3points, each triple of points spans a distinct plane, implyinga⁰₃≥20, which is a contradiction.

Corollary 27. The generalized cylinder conjecture is true for all triples(v, r,3).

An interesting implication is that the dimension v of every 3^r-divisible projective 3^r+1, v

3-code is at mostr+ 3for every positive integerr. We remark that Ward’s upper bound on the dimension of divisible codes [11, Theorem 6] givesv≤q(r+ 1)for the triple(v, r, q), and is hence not strong enough to give this result. Using the software packageLinCode[6] we have computationally checked that there is no3-divisible [9,≥5]3-code, no9-divisible[27,≥6]3-code, and no27-divisible[81,≥7]3-code. This means it might not be necessary to assume thatS is a set and not a multiset to obtain the stated upper bound for the dimension.

We remark that the truth of the cylinder conjecture for(4,1,2)and(4,1,3)was also proven in [3].

In principle it is possible to enumerate all projectiveq^r-divisible q^r+1, v

q codes and to check whether the corresponding point sets are(r+ 1)-cylinders for given finite parameters. However, given the currently available software for the exhaustive enumeration of linear codes, this approach is limited to rather small parameters. Nevertheless we report our corresponding findings here. The last step – checking whether all resulting point sets are(r+ 1)-cylinders – can be replaced by a counting argument. The numbers of projective linear codes overF5of effective lengthsn= 5ordered by their dimensionkare given by2¹3⁴4³5¹, as can be easily enumerated using the software packageLinCode[6] – even a classification by hand is possible. So, Construction5yields3¹4⁴5³6¹5-divisible projective linear codes overF5of effective lengthsn= 25, again ordered by their dimensionk. UsingLinCodewe verified that there are no further5-divisible projective linear codes overF5of effective lengthn= 25. Thus, we have computationally proven that the generalized cylinder conjecture is true for(v,1,5), where the dimensionvis arbitrary. This covers the special case(4,1,5) that we treat in the subsequent section. From Corollary20we conclude:

Corollary 28. The generalized cylinder conjecture is true for all triples(v, r,5).

Forq∈ {2,3,4}we can perform the same computation. The casesq= 2,3verify our theoretical findings for (v,1, q). The number of projective linear codes overF⁴ of effective lengths n = 4 ordered by their dimensionk are given by 2¹3²4¹. The number of 4-divisible projective linear codes over F4 of effective lengthsn = 16ordered by their dimensionkare given by3¹4²5². In other words, the generalized cylinder conjecture is true for (3,1,4) and(4,1,4) but not for (5,1,4). For v ≥ 6 there do not exist projective 4-divisible [16, v]4-codes so that the generalized cylinder conjecture is trivially true for (v,1,4) whenever v≥6. Therefore, Corollary20gives:

Corollary 29. The generalized cylinder conjecture is true for(v, r,4)if and only ifv6=r+ 4.

Of the two linear codes in dimension5, the one that does not correspond to a2-cylinder has a generator matrix given by













0110101101110000 1101100011101000 1100011111000100 1111111000000010 0111010110100001











 .

The code has weight enumeratorW(z) = 1z⁰+ 90z⁸+ 840z¹²+ 93z¹⁶ and an automorphism group of order1935360. Considered overF2the stated generator matrix gives a linearF2code with weight enumerator W(z) = 1z⁰+ 30z⁸+ 1z¹⁶, which means that the code is an affine5-space.

Computationally we also verified that the generalized cylinder conjecture is true for(5,2,4), which also follows from Theorem19. Due to the counter example for(5,1,4)there is also a counter example for(6,2,4), see Proposition18. We have computationally checked that this counter example is unique.

In order to generalize the above counter example to the generalized cylinder conjecture for(5,1,4) we remark that for each integerh≥2the fieldFqis a subfield ofFq^h, so thatF^v_qh ∼=F^vhq . Using this isomorphism we can we can embed every multiset of pointsM⁰inF^vq as a multiset of pointsM(of the same cardinality) in F^v_qh. Moreover, everyk-space inF^v_qhcorresponds to akh-space inF^vhq .

Lemma 30. Let S⁰ be a spanning projective 2h-cylinder inF^vq, where h ≥ 2. Then the corresponding embeddingSinF^v_qh ∼=F^vhq is a spanning projectiveq^h-divisible set of q^h2

points inPG v−1, q^h that is not a2-cylinder

PROOF. First we observe|S| =|S⁰| =q^2h = q^h2

. SinceS⁰is spanning and projective, the same applies toS. An arbitrary hyperplaneH inF^v_qhhas dimension(v−1)hoverFq. SinceF^vq has dimensionvoverFq, there exists a subspaceK inF^vq of dimension at leastv−hsuch that|S ∩H| = |S⁰∩K|. Note thatS⁰is q^2h−1-divisible, so that|S⁰∩K| ≡0 (modq^h)due to Lemma3. and we conclude thatSisq^h-divisible.

Now assume thatSis a2-cylinder and letLbe one of theq^h-lines. Consider two1-pointsP₁, P₂onLand denote byP₁⁰, P₂⁰ be the corresponding points inS⁰. The lineL⁰ =hP₁⁰, P₂⁰ihas multiplicity at mostqinS⁰, so thatLhas a multiplicity of at mostqinS, which is a contradiction due toh≥2.

Corollary 31. For any integer h ≥ 2, the generalized cylinder conjecture is false for v, r, q^h

whenever 2h+r≤v≤2h+r+q−2.

PROOF. From Lemma 30and Lemma8we conclude that the generalized cylinder conjecture is wrong for v,1, q^h

, where2h+ 1≤v≤2h−1 +q, so that Corollary20gives the general statement.

Ifv= 2h+r, then the point setS⁰in Lemma30is an affine geometry, so thatSis an affine subgeometry.

For the special caseh = 2one also speaks of a Baer (sub-)geometry. In general, our construction is an instance of the technique of the so-calledfield reduction, which yields a lot of non-trivial constructions and characterizations of geometric and algebraic structures, see e.g. [7]. Of course one might conjecture that every q^r-divisible set ofq^r+1points inPG(v−1, q)is either an(r+ 1)-cylinder or arises from a cylinder over a subfield.

5. THE GENERALIZED CYLINDER CONJECTURE FOR(4,1, q)

In this section, we will focus on the casev =r+ 3, and by Corollary20we can restrict ourselves to the triple(4,1, q). We will gather some more information on a possible counterexample, which leads us to be able to prove the generalized cylinder conjecture for(4,1,5)and(4,1,7). A proof of the former was also claimed in [3], but the proof contains an error which we correct here.

For the results in this section, we will often make the following assumption:

(?) Sis aq-divisible spanning set ofq²points in PG(3, q)which is not a 2-cylinder.

Sinceris fixed in this section, we will also refer to a 2-cylinder as a cylinder. Moreover, we can considerS as a set of points inAG(3, q)by Corollary16. Lastly, by Theorem23we can assume that any plane intersects Sin at mostq²−qpoints. Our general strategy is to obtain some structural results onSand find a contradiction for smallq. We will heavily rely on the standard equations for planes and points inSas stated in Lemma12, but also similar equations for lines in akq-planeHand points inS ∩H, obtained by the same double counting method. The latter gives us information on the number ofi-lines for eachi, which we will also refer to as the spectrum.

We start off by investigating the multiplicity of a line with respect toS. Summarizing the conclusions of Corollary16, Corollary22and Theorem23, we can state the following lemma.

Lemma 32. Under(?), the line multiplicities in akq-planeHare contained in{0, k, k+ 1, . . . , q−1}.

The restriction on the possible line multiplicities is quite severe. Indeed, we can investigate the existence of a setKofkqpoints inPG(2, q)admitting these line multiplicities, independently of the (generalized) cylinder conjecture. It turns out that whenkis large, such sets cannot exist and henceScannot have large intersections with planes. This idea is illustrated in the next few results.

Lemma 33. For a set ofq(q−1)points inPG(2, q), there exists a line with a multiplicity not in{0, q−1}.

PROOF. Otherwise the first two standard equations would givea0 = 1anda_q−1 =q²+q, so that the third one yields the contradiction

q−1 2

a_q−1=q(q−1)(q²−q−2)

2 <q(q−1)(q²−q−1)

2 =

q(q−1) 2

.

Corollary 34. Under(?), there can be noq(q−1)-plane.

PROOF. IfHis aq(q−1)-plane, then Lemma32and Lemma33yield a contradiction.

We can use this result to find an alternative proof of the cylinder conjecture whenq∈ {2,3}.

Corollary 35. The generalized cylinder conjecture is true for the triples(4,1,2)and(4,1,3).

PROOF. Assume thatSis not a cylinder, so that Corollary34impliesa_q(q−1)= 0. Forq= 2this means that all points ofSare contained in0-planes, which is absurd. Forq= 3solving the first two standard equations givea0= 1anda3= 39. The third implies the contradiction

3 2

a3= 117<144 = 9

2

·(3 + 1).

Remark36. The argumentation of the proof of Corollary35, i.e., the first three standard equations combined with Corollary34, would give the unique spectruma0= 7,a1= 72, anda2= 6forq= 4.

Lemma 37. For a set ofq(q−2) points inPG(2, q),q ≥ 4, there exists a line with a multiplicity not in {0, q−2, q−1}.

PROOF. Otherwise the first two standard equations would givea_q−1 = a₀(q−2)−(q−2) anda_q−2 = q²+ 2q−1−a0(q−1), so that the third yields

0 =a_q−2 q−2

2

+a_q−1 q−1

2

−

q(q−2) 2

=(q−2)(a0(q−1)−3q+ 1)

2 ,

which implies

a₀= 3q−1

q−1 = 3 + 2

q−1 ∈/ N0.

Remark38. In the proof of Lemma37the third standard equation is needed since e.g.(a0, a_q−2, a_q−1) = (2,17,2)satisfies the first two standard equations forq= 4.

Corollary 39. Under(?), there can be noq(q−2)-plane.

PROOF. Due to Corollary35we can assumeq≥4. IfH is a(q−2)q-plane, then Lemma32and Lemma37

yield a contradiction.

Corollary 40. The generalized cylinder conjecture is true for the triple(4,1,4).

PROOF. Assume that S is not a cylinder, so that Corollary34 and Corollary39 imply a_q(q−1) = 0 and a_q(q−2)= 0. Forq = 4solving the first two standard equations givea0 = 1anda4 = 84. The third implies the contradiction

4 2

a4= 504<600 = 16

2

·(4 + 1).

We might continue in the vein of Lemma37and consider sets ofq(q−3)points inPG(2, q)whose line multiplicities are contained in{0, q−3, q−2, q−1}. Due to to Corollary35and Corollary40we are only interested in the cases whereq≥5. It turns out that he unique possibility is given byq= 5and a spectrum given bya0 = 6, a2 = 15, a3 = 10, anda4 = 0. So, similar to Corollary39, we can conclude that a q(q−3)-plane does not exist unlessq = 5. This result also follows in a much shorter way from Lemma43, but we want to remark that the latter is not necessary to show that the generalized cylinder conjecture is true for the triple(4,1,5).

Proposition 41. The generalized cylinder conjecture is true for the triple(4,1,5).

PROOF. By Corollary34and Corollary39we know thata_q(q−1)= 0anda_q(q−2)= 0. With this, the standard equations forSyield the unique spectruma₀= 11,a₅ = 135, anda₁₀ = 10. Now letH be a10-plane and b_i the number of lines of multiplicityi, whereb_i = 0fori /∈ {0,2,3,4}. From the standard equations we concludeb2 = 51−6b0,b3 =−38 + 8b0, andb4 = 18−3b0, so thatb4 ≥0impliesb0 ≤6andb3 ≥0, b0 ∈N0implyb0≥19

4

= 5. Assumeb0 = 5and note that two4-lines cannot share a common1-pointP since otherwise counting points on the linesL1, . . . , L6throughPwould yield the contradiction

10 =S(H) =

6

X

i=1

S(Li)−5· S(P)≥2·4 + 4·2−5 = 11

usingS(Li)≥2for all1≤i≤6. Thus, the4-lines are pairwise disjoint, so that10 =S(H)≥a4·4 = 12, which is a contradiction. Hence, we haveb0= 6,b2= 15,b3= 10, andbi= 0otherwise.

For a lineLof multiplicity3inHdenote the other5planes byH1, . . . , H5. SinceS(Hi)∈ {5,10}for all 1≤i≤5and|S|= 25, there exists an index1≤i≤5withS(Hi) = 10. Due to1 +b3·1 = 11, there are

at least eleven10-planes, which contradictsa10= 10.

Remark42. There exists a unique projective[10,3,{6,7,8,10}]5codeCwith generator matrix





1 1 1 1 1 1 0 1 0 0 4 4 3 2 1 0 1 0 1 0 3 4 4 2 0 1 4 0 0 1



.

This code has an automorphism group of order480and weight enumeratorW_C(x) = 1x⁰+ 40x⁷+ 60x⁸+ 24x¹⁰. A geometrical interpretation for this set of points was also given in [10]. We remark that the existence of the above code was excluded in the proof of [3, Theorem 4] and so the proof is flawed. More precisely, in the last sentence of the argument showing the existence of a4-line in a10-plane where two further0-lines are constructed, it can happen that they (partially) coincide with the four0-lines found before.

Lemma 43. Under(?), the maximum line multiplicity with respect toSisq−2.

PROOF. Due to Corollary35and Corollary40we can assumeq ≥5. By Lemma32we already know that the maximum line multiplicity is at mostq−1. So assume thatLis a line of multiplicityq−1and denote byQ1, Q2the two0-points onL. LetH be an arbitrary hyperplane containingL, whereS(H) =kqfor an integer1≤k < q.

For each1-pointPonLletL1, . . . , Lq denote theqlines troughP inH that are not equal toL. Note that Lemma32impliesS(Li)≥kfor all1≤i≤q. From

kq=S(H) =S(L) +

k

X

i=1

S(Li)−q· S(P)≥q−1 +qk−q=kq−1

we conclude thatq−1of theL_i, where1≤i≤q, have multiplicitykand one has multiplicityk+ 1.

Now consider a0-pointRinHnot onLand letL⁰_ibe the lines throughRandQiinH, where1≤i≤2.

ByL⁰₃, . . . , L⁰_q+1we denote the remainingq−1lines throughRinH. Note that theL⁰_imeet the lineLin a 1-points, so thatS(L⁰_i)∈ {k, k+ 1}for3≤i≤q+ 1. From

kq=S(H) =

q+1

X

i=1

S(L⁰_i)−q· S(R)≥ S(L⁰₁) +S(L⁰₂) + (q−1)k

we concludeS(L⁰₁) +S(L⁰₂)≤k, so thatS(L⁰₁),S(L⁰₂)∈ {0, k}due to Lemma32. Thus, for1≤i≤2the q+ 1lines throughQiinH are given by the(q−1)-lineL, ^q−1_k lines of multiplicity0, and q−^q−1_k

lines of multiplicityk. Now we are ready to determine the spectrum(a_i)ofH:

a0 = 2·q−1 k

ak = (q−1)·(q−1) + 2·

q−q−1 k

ak+1 = (q−1)·1 a_q−1 = 1

andai = 0for alli /∈ {0, k, k+ 1, q−1}. Ifk =q−1ork+ 1 = q−1, then we would have to take the sum of both values, but we suppress this technical subtlety for the ease of notation. From the third standard equation we conclude

0 = k

2

a_k+ k+ 1

2

a_k+1+ q−1

2

a_q−1− kq

2

=q· (k−1)(k−q+ 1)

2 ,

so thatk∈ {1, q−1}.

So, considering theq+ 1hyperplanes throughLinPG(3, q)we conclude thatqhave multiplicityqand one has multiplicityq(q−1), where the latter contradicts Corollary34.

We remark that the preceding result can be used to simplify the proof of the cylinder conjecture forq= 5, although it is not necessary.

Lemma 44. For a set ofq(q−3) points inPG(2, q),q ≥ 7, there exists a line with a multiplicity not in {0, q−3, q−2}.

PROOF. Otherwise the first two standard equations would giveaq−2 = a0(q−3)−(q−3) andaq−3 = q²+ 2q−2−a₀(q−2)so that the third yields

0 =aq−3

q−3 2

+aq−2

q−2 2

−

q(q−3) 2

=(q−3)(a0(q−2)−4q+ 2) 2

which implies

a0= 4q−2

q−2 = 4 + 6 q−2.

Sincea0∈N0, we haveq∈ {3,4,5,8}. Due to our assumptionq≥7it remains to exclude the caseq= 8, wherea0 = 5,a5 = 48, anda6 = 20. Consider a6-lineL. The only possibility for the distribution of the multiplicities of the lines through a0-point onLis given by0²5²6⁵, so that there are3·2 = 6 0-lines. This

contradictsa₀= 5.

The assumptionq ≥ 7 in Lemma44is necessary, see Remark42for a counterexample forq = 5 and the generator matrix





1 1 0 0 1 0 1 0 1 0 0 1



 forq= 4.

Corollary 45. Under(?), there can be noq(q−3)-plane.

PROOF. Due to Corollary35, Corollary40, and Proposition41we can assumeq≥7. Lemma43implies that the maximum line multiplicity is at mostq−2. IfHis aq(q−3)-plane, then Lemma32and Lemma44yield

a contradiction.

Lemma 46. For a set ofq(q−4)points inPG(2, q),q≥7, with line multiplicities contained in{0, q−2, q− 3, q−4}, we needq= 7and spectrum(a0, aq−4, aq−3, aq−2) = (8,28,21,0).

PROOF. Solving the standard equations gives

aq−2 = (q−4)(5q−3)

2 −(q−3)(q−4)

2 a0

a_q−3 = −(q−4)(5q−2) + (q−2)(q−4)a0

aq−4 = 7q²−19q+ 6

2 −(q−2)(q−3) 2 a0, so thata_q−2≥0implies

a₀≤5q−3

q−3 = 6−q−15 q−3 anda_q−3≥0implies

a0≥ 5q−2

q−2 = 5 + 8 q−2. So, forq≥16we havea0∈/N0, which is a contradiction.

For7≤q≤13we can see thataq−3>0. So consider a(q−3)-lineL. Through each of the four0-points onLthere go at least two0-lines, since otherwise1·0 + (q−1)·(q−4) + 1·(q−3) =q(q−4) + 1>|K|.

Therefore, we havea0≥8, which is only possible forq= 7. The spectrum then follows from the equations

above.

Lemma 47. Under(?)forq= 7, the spectrum is given bya₀= 22,a₇= 357,a₁₄= 21anda₂₁= 0.

PROOF. First we will show that a21 = 0. Assume to the contrary the existence of a 21-plane H. From Lemma46we conclude that the spectrum(bi)ofS ∩H satisfiesb0= 8,b3= 28,b4= 21, andbi= 0other- wise. Consider the possible hyperplane distributions through a0-line inH:0⁵7¹21²,0⁵14²21¹,0⁴7²14¹21¹, and0³7⁴21¹. In each case there are at least three0-planes through a0-line inH, so that there are at least 3b0 = 24 0-planes in total. However, solving the standard equations for{a0, a7, a14}givesa0 = 22−a21, a₇ = 357 + 3a₂₁, anda₁₄ = 21−3a₂₁, so thata₀≤22, a contradiction. Thereforea₂₁= 0and the values

fora₀, a₇anda₁₄follow immediately as well.

Remark48. After 1671 seconds of computation time,QextNewEditionclaims that there no sets of14 points with line multiplicity at most5inPG(2,7)without lines of multiplicity1. If we additionally assume that there is no line of multiplicity5, then 908 seconds of computation time are needed. Using Lemma47this implies the truth of the cylinder conjecture for(4,1,7).

In the following we want to give an alternative, computer-free proof of the cylinder conjecture for(4,1,7).

Lemma 49. A setK of14points inPG(2,7), whose line multiplicities are contained in{0,2,3,4,5} has spectrum(a0, a2, a3, a4, a5) either(10,37,2,8,0), (11,31,10,5,0),(12,25,18,2,0), (11,30,13,2,1), or (10,36,5,5,1).

PROOF. Solving the standard equations for{a0, a2, a3}gives a₀ = 38

3 −a₅−1 3a₄ a₂ = 21 + 5a₅+ 2a₄ a₃ = 70

3 −5a₅−8 3a₄

Froma₀∈N0we concludea₄≡2 (mod 3), so that especiallya₄≥2. With this,a₃∈N0yieldsa₄≤8and a₅≤3.

In order to showa₅ ≤ 1 we consider a5-lineL. Now, letP be an arbitrary1-point onLandQbe an arbitrary0-point onL. Counting the points on the linesL, L₁, . . . , L₇throughPgives

14 =|K|=K(L) +

7

X

i=1

K(L_i)−7· K(P) =K(L₁) +

7

X

i=2

K(L_i)−2≥ K(L₁) + 6·2−2 =K(L₁) + 10, so that there is no5-line throughPbesidesL. Now assume thatM is a5-line throughQand letRbe a0-point onM not equal toQ. ByL⁰₀, . . . , L⁰₇we denote the lines throughR, where we assumeL⁰₀=M. Since five of the linesL⁰₁, . . . , L⁰₇hitLin a point they have multiplicity at least2, which yields at least5 + 5·2 = 15>14 points inK, which is a contradiction. Thus, there is also no5-line throughQbesidesL, so thata5is at most1.

To sum up, ifa5= 0, thena3∈N0impliesa4∈ {2,5,8}, and ifa5= 1, thea3∈N0impliesa4∈ {2,5}.

Plugging this into the above equations gives the five stated spectra.

Note that we have applied the same “technique” to concludea5≤1as the one used in the proof of Lemma43.

We will now rule out all possibilities from Lemma49.

Lemma 50. A setKof14points inPG(2,7), whose line multiplicities are contained in{0,2,3,4,5}, can not have5-lines.

PROOF. LetLbe a5-line andQ1,Q2, andQ3 be the0-points onL. Define a new setK⁰ as the symmetric difference ofKandL:K⁰=K \(L∩ K)∪ {Q1, Q2, Q3}. AsKdid not have any1-lines, we can see thatK⁰ is a non-trivial blocking set of size12inPG(2,7). Blocking sets of cardinality12inPG(2,7)not containing a line have been classified in [2]: besides the projective triangle there exists a unique sporadic example of non-R´edei type. Both of these constructions share the property that there exists at least one2-line through every1-point. Now considering the pointQ1, we can easily see that there are no2-lines toK⁰ through it, as

they would have to come from1-lines toK, which do not exist.

Lemma 51. No setKof14points inPG(2,7)with spectrum(a0, a2, a3, a4) = (11,31,10,5)andai = 0 otherwise exists.

PROOF. The three possible distributions of the multiplicities of the lines through a1-point are4²3¹2⁵,4¹3³2⁴, 3⁵2³and we speak of typeA1,A2, andA3, respectively.

Assume thatP is a1-point of typeA₃and letLbe an arbitrary3-line throughP. The five0-points onL are contained in another3-line besides L, by parity considerations. Thus bya₃ = 10, the two 1-points on Lthat are not equal toP are of typeA₁. SinceLwas chosen arbitrary, all ten1-points on3-lines through P that are not equal toP are of typeA₁. LetQ₁,Q₂, and Q₃ denote the three other1-points not equal to P and not forming a3-line withP. All of the five3-lines not incident withP have to consist of1-points in {Q₁, Q₂, Q₃}, which is impossible. Thus, there is no1-point of typeA₃.

Consider an arbitrary3-lineL⁰. The five0-points onL⁰are contained in another3-line besidesL⁰, for the same reason as before. Thus not all three1-points onL⁰can be of typeA2asa3= 10. So each3-line contains at least one1-point of typeA1, which is then contained in no other3-line. This means that there are at least ten 1-points of typeA1. Counting the4-lines, this givesa4≥(10·2 + 4·1)/4 = 6>5, which is a contradiction.

Proposition 52. The generalized cylinder conjecture is true for the triple(4,1,7).

PROOF. Assume the existence of a counterexampleS. Due to Lemma47, there exists a hyperplaneH so that S ∩His a set of14point whose line multiplicities are contained in{0,2,3,4,5}. From Lemma49, Lemma50 and Lemma51the spectrum ofS ∩His given by(b0, b2, b3, b4, b5) = (12,25,18,2,0). Note that everyj-line inH, wherej ≥2, is contained in(j−1)planes of multiplicity14and(9−j)planes of multiplicity7in PG(3,7). ThusScontains at leastb3+ 2b4 = 18 + 2·2 = 22planes of multiplicity14, which contradicts

a14= 21.

While our computer-free proof of the cylinder conjecture for(4,1,7)is rather lengthy, most parts are more or less systematic and might be generalized to larger field sizes. A big obstacle is that we cannot prove the truth of the observation in Remark48directly. Actually, we do not have a complete proof of this specific nonexistence result sets of14points inPG(2,7)without1-lines and just sailed around the remaining open case in the proof of Proposition52. Maybe other methods are more suitable for this kind of problems in PG(2, q). Of course, allowing computer enumerations drastically reduces the length of the argumentation.

Starting from Lemma32and Lemma32we can computationally exclude many possibilities for the restriction S ∩Hfor a hyperplaneH. In other words, the possible multiplicities for the weightsS(H)for the hyperplanes can be restricted by enumeration results for3-dimensional codes overFq. By considering a subcode of the4- dimensional projective code corresponding toSwe obtain aq²-divisible

q²−1,3

q-codeCwith a restricted set of weights that might also be enumerated computationally. For our example we remark that there are 54non-isomorphic[48,3,{21,28,35,42}]7-codes and46non-isomorphic[48,3,{28,35,42}]7-codes. More- over, the information thatSdoes not contain a full affine line restricts the possible residual codes of codewords inC. By that criterion6of of the46non-isomorphic[48,3,{28,35,42}]₇-codes can be excluded. Similar re- strictions can arise from the previously mentioned classification of projective[kq,3,{(k−1)q+ 1, . . . , k(q− 1), kq}]_q-codes. E.g., as also theoretically proven, all projective [21,3,{15,16,17,18,21}]7-codes do not contain codewords of weight15or16. So, in the residual code of a codeword of weight28inCthe weights 15and16cannot occur. This excludes12further codes. For the remaining twenty-eight[48,3,{28,35,42}]7- codes we can computationally check whether an extension to a projective[49,4,{28,35,42}]7-code exists. To this end we can utilize and ILP formulation and an ILP solver. We remark that the tightest ILP instance needed 1 238 996branch&bound nodes and 28.75 hours of computation time. At the very least, this approach gives a computational verification of Proposition52.

Let us finish with some conclusions for the generalized cylinder conjecture forq = 8. Assume thatSis an8-divisible spanning set of64points inPG(3, q)that is not a cylinder. From Corollary34, Corollary39, Corollary45and Lemma46, we conclude that the hyperplane multiplicities with respect toSare contained in {0,8,16,24}.

Solving the standard equations for the spectrum(a_i)ofSgives a0 = 29−a24

a8 = 528 + 3a24

a₁₆ = 28−3a₂₄,

so thata0 ≤29. Now assume that H is a24-plane and consider the spectrum(bi)ofS ∩H. Solving the standard equations for{b3, b5, b6}gives

b₃ = 97−5b₀−b₄ 3 b5 = −69 + 9b0−b4

b6 = 45−5b0+b4

3, so thatb₅ ≥ 0 impliesb₀ ≥ ₆₉

9

= 8. Since through every0-line inH there are at least three0-planes, a0 ≤29impliesb0≤29

3

= 9. Forb0= 8we haveb3= 62−b6,b4=−15 + 3b6, andb5= 18−3b6, so