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APPEARED IN BULLETIN OF THE

arXiv:math/9210222v1 [math.MG] 1 Oct 1992APPEARED IN BULLETIN OF THEAMERICAN MATHEMATICAL SOCIETYVolume 27, Number 2, October 1992, Pages 279-283KELLER’S CUBE-TILING CONJECTUREIS FALSE IN HIGH DIMENSIONSJeffrey C. Lagarias and Peter W. ShorAbstract. O. H. Keller conjectured in 1930 that in any tiling of Rn by unit n-cubesthere exist two of them having a complete facet in common.

O. Perron proved thisconjecture for n ≤6. We show that for all n ≥10 there exists a tiling of Rn by unitn-cubes such that no two n-cubes have a complete facet in common.1.

IntroductionIn 1907 Minkowski [5] conjectured that all the extremal lattices for the supre-mum norm were of a certain simple form and observed that this conjecture had ageometric interpretation: in any lattice tiling of Rn with unit n-cubes there mustexist two cubes having a complete facet ((n −1)-face) in common. He proved thisfor n = 2 and 3.

In studying this question, Keller [4] generalized it to conjecturethat any tiling of Rn by unit n-cubes contains two cubes having a complete facet incommon. In 1940 Perron [6] proved Keller’s conjecture for dimensions n ≤6.

Soonafter, Haj´os [2] proved that Minkowski’s original conjecture is true in all dimensions.Keller’s stronger conjecture remained open. Haj´os [3] later gave a combinatorialproblem concerning factorization of abelian groups, which he proved was equiva-lent to Keller’s conjecture.

Stein [7] gave a survey of these results and other relatedtiling problems.More recently, Szab´o [8] showed that if Keller’s conjecture is false in Rn, thenthere exists a counterexample tiling in some Rm (with m ≥n) having the followingextra properties: the centers of all cubes are in 12Zm, and the tiling is periodic withperiod lattice containing 2Zm. Corr´adi and Szab´o [1] studied a graph-theoretic ver-sion of this latter problem, showing directly that there are no such counterexamplesfor m ≤5.We explicitly construct a counterexample tiling of Szab´o’s type in R10.

Keller’sconjecture is then false for all n ≥10 because a counterexample tiling in Rn givesone in Rn+1 by “stacking” layers of this tiling with suitable translations madebetween adjacent layers.2. Main resultWe prove the following result.1991 Mathematics Subject Classification.

Primary 52C22 Secondary 05B45.Received by the editors March 5, 1992c⃝1992 American Mathematical Society0273-0979/92 $1.00 + $.25 per page1

2J. C. LAGARIAS AND P. W. SHORTheorem A.

For n = 10 and 12 there exists a tiling of Rn by unit cubes suchthat(1) The centers of all cubes are in 12Zn;(2) The tiling is periodic with period lattice 2Zn;(3) No two cubes have a complete facet in common.Before giving the constructions, we describe Corr´adi and Szab´o’s equivalentgraph-theoretic criterion for such a tiling to exist in Rn.Scale everything up by a factor of 2 to consider tilings of Rn by translates of thecubeC = {(x1, . .

. , xn) : −1 ≤xi ≤1 for all i}of side 2 centered at the origin, such that the centers of all cubes are in Zn, andthe tiling is periodic with period lattice 4Zn.

There are 2n equivalence classes ofcubes m + C + 4Zn in such a tiling, and each equivalence class contains a uniquecube with center(1)m = (m1, . .

. , mn) ∈Zn,0 ≤mi ≤3 .The collection S of these 2n vectors describes the tiling.Now form two graphs Gn and G∗n, each of which has 4n vertices labeled by the4n vectors in Zn of form (1), as follows.

Consider the conditions:(a) m and m′ have some |mi −m′i| = 2. (b) m and m′ differ in two coordinate directions.Gn has an edge between vertices m and m′ if (a) holds, while G∗n has an edgebetween m and m′ if (a) and (b) both hold.

Condition (a) says that all translatesunder 4Zn of cubes C centered at m and m′ have disjoint interiors, while (a) and(b) together say that all translates under 4Zn of such cubes also do not have acomplete facet in common.A set S of 2n vectors satisfying (1) yields a 4Zn-periodic cube tiling if and onlyif S forms a clique in Gn and it yields a 4Zn-periodic cube tiling with no two cubeshaving a complete facet in common if and only if S forms a clique in G∗n. Thisgives the Corr´adi-Szab´o criterion that a Szab´o-type counterexample exists in Rn ifand only if G∗n contains a clique of size 2n.The graph Gn is the complement of the product graph C4 ⊗C4 ⊗· · · ⊗C4 ofn copies of the 4-cycle C4.

It has maximal clique size equal to the independentset number of C4 ⊗· · · ⊗C4, which is α(C4)n = 2n since C4 is a perfect graphand α(C4) = 2. In fact, Gn has an enormous number of maximal cliques, and theproblem is whether or not any of them remain a clique in G∗n.Note also that the graphs Gn and G∗n have large groups of automorphisms.

Onboth graphs one can relabel the vertices m = (m1, . .

. , mn) by relabeling the ithcoordinate using the group generated by the cyclic permutations (0123) and the2-cycle (13), and one also can permute coordinates.

This generates a group of 8nn!automorphisms.Proof of Theorem A. We give the easier 12-dimensional construction first.

It startswith the set T of vectors given in Table 1, which (ignoring the primes on somezeros) is a clique of size 8 in G3. It is very nearly a clique for G∗3, in that it omits

KELLER’S CUBE-TILING CONJECTURE3Table 1. Clique T in G3.00020112001220′3320′0′32222only three edges, namely, 201–20′3, 120–320′, 012–0′32.

If somehow 0′ were distinctfrom 0, while it was still the case that 2 −0′ = 2, then this would be a 23-clique forG∗3 and would give a counterexample.We use a block substitution construction that in effect accomplishes this in R3kfor suitable k.Assign to each of 0, 0′, 1, 2, 3 sets S0, S′0, S1, S2, S3 of vectors in{0, 1, 2, 3}k having the following properties:(i) Each of S0, S′0, S1, S2, S3 is a clique in G∗k. (ii) No two of the sets S0, S′0, S1, S2, S3 have a common vector.

(iii) S0 ∪S2, S′0 ∪S2, and S1 ∪S3 are each a clique in Gk.Assuming (i), (ii), the last condition (iii) says, e.g., for S0 ∪S2 each element of S0differs from each element of S2 by 2 (mod 4) in some coordinate.Call the vectors in the sets Si blocks. Form the set S of all vectors in {0, 1, 2, 3}3kthat can be formed by taking any vector (m1, m2, m3) ∈T and for each mi substi-tuting any block in the corresponding Smi, independently for each i.Claim.

S is a clique in G∗3k.To prove this, let v, v′ be distinct elements of S constructed from m = (m1, m2, m3)and m′ = (m′1, m′2, m′3) in T , respectively. If m = m′ then v, v′ have some blockw, w′ ∈Smi where they differ, and condition (i) forces an edge between v and v′in G∗3k.

If m ̸= m′ then m and m′ differ by 2 in some coordinate (here 0′ and 2are considered to differ by 2), which carries over to v and v′ by condition (iii), andm and m′ also differ in another coordinate, where 0 is treated as distinct from 0′,and this carries over to v and v′ by condition (ii), proving the claim.If one can choose |S0| = |S′0| = a, |S1| = b, |S2| = c, and |S3| = d with a+c = 2k,b + d = 2k, then |S| = a3 + 3abc + 3acd + c3 = 23k will be a clique in G∗3k, thusgiving a counterexample.We achieve this with k = 4, a = b = 12, c = d = 4, with the sets S0, S′0, S1, S2, S3given in Table 2 below. The sets S0, S′0, S2 were obtained from a 28-clique for G∗5given in [1, Table 2].

Examining the first column of this 28-clique, one finds twelvevectors each having value 0 and 2 and four having value 1. Deleting this column andgrouping the resulting Z4-vectors as S0, S′0, S2, the G∗5-clique property guaranteesthat all the conditions (i), (ii), (iii) that concern only S0, S′0, S2, automaticallyhold.

Next we apply a suitable automorphism of G∗4 to S0, S2 to obtain S1, S3.For any automorphism conditions (i) and (iii) will automatically hold for S1, S3obtained this way. Thus we need only to find an automorphism where (ii) holds.The automorphism that cyclically permutes the labels of the first coordinate 0 →1 →2 →3 →0 gives suitable S1, S3, as listed in Table 2.

4J. C. LAGARIAS AND P. W. SHORTable 2.

Blocks used in constructions.S0S′0S2S1S′1S300000303021110001303121100121011113210122011213202131113230312132113330302301130302012302130002003321323133223231020133120202331210022113100321121123001311200012220302232200022230131033301010323223223332202233132323101320231The conditions (i), (ii), (iii) can be verified directly for S0, S′0, S1, S2, S3 by handcalculation. Aside from the distinctness of all elements, the automorphism sending(S0, S2) to (S1, S3) means that one need only check properties for S0, S′0, S2.

Thecalculation can be further reduced by observing that there is an automorphismof G∗4 that fixes S2 and sends S0 to S′0. This automorphism cyclically permutesthe labels of the first coordinate 0 →1 →2 →3 →0 and the last coordinate0 →3 →2 →1 →0, and then exchanges these coordinates.

Thus one need onlyverify that S0 and S2 are G∗4-cliques and S0 ∪S2 is a G4-clique.The 10-dimensional construction is similar in nature and is based on the factthat the set eT = S0 ∪S2 from Table 2 is a clique of size 24 in G4, which is verynearly a 24-clique for G∗4. In G∗4 it omits only the four edges 0213–0211, 3132–1132,2301–2303, 1020–3020.

Now regard S2 as being021′111′32230′330′20where we want 0 ̸= 0′ and 1 ̸= 1′.Assign to 0, 0′, 1, 1′, 2, 3 the sets of blocksS0, S′0, S1, S′1, S2, S3 in Table 2, where S′1 is constructed from S1 similarly to S′0from S0. These sets satisfy:(i) Each of S0, S′0, S1, S′1, S2, S3 is a clique in G∗4.

(ii) No two of these sets have a common vector. (iii) S0 ∪S2, S′0 ∪S2, S1 ∪S3, and S′1 ∪S3 are each a clique in G4.Apply the block substitution construction to the second and third columns onlyon eT to obtain a set eS of 210 10-vectors.

This is a clique in G∗10, as required. Notethat only the second and third columns need to be expanded in blocks, because theprimed elements in S2 above appear only in these columns.□3.

DiscussionThe failure of Keller’s Conjecture in high dimensions illustrates the general phe-nomenon that Euclidean space allows more freedom of movement in high dimen-

KELLER’S CUBE-TILING CONJECTURE5sions than in low ones. It is interesting that the critical dimension where Keller’sConjecture first fails, which is at least 7, is as high as it is.It may be a difficult matter to determine exactly the critical dimension.

Ex-haustive search for Szab´o-type counterexamples already seems infeasible for G∗7;the maximum clique problem is a well-known NP-complete problem, which is alsocomputationally hard in practice. The authors ruled out the existence of any 27-clique in G∗7 that is invariant under a cyclic permutation of coordinates by computersearch.

It is conceivable that there exist Szab´o-type counterexamples in dimension7, 8, or 9, which are all so structureless that they will be hard to find. In any casewe have so far found no variant of the constructions of Theorem A that work inthese dimensions.A natural extension of Keller’s conjecture is to determine the largest integerKn such that every tiling of Rn by unit cubes contains two cubes that have acommon face of at least dimension Kn.

For a Szab´o-type tiling, two cubes havingcoordinates (m1, . .

. , mn) and (m′1, .

. .

, m′n) in G∗n have a k-dimensional face incommon if |mi −m′i| = 0 or 2 for all i, and exactly k values |mi −m′i| = 0. The10-dimensional and 12-dimensional cube tilings eS and S constructed in TheoremA each contain two cubes sharing a common face of codimension 2, so they implyonly K10 ≤8 and K12 ≤10.

We have found a different 10-dimensional cube tiling(using a similar construction) which shows that K10 ≤7. We also can show thatn −Kn →∞as n →∞; details will appear elsewhere.AcknowledgmentWe thank Victor Klee for bringing this problem and the work of Corr´adi andSzab´o to our attention.References1.

K. Corr´adi and S. Szab´o, A combinatorial approach for Keller’s conjecture, Period. Math.Hungar.

21 (1990), 91–100.2. G. Haj´os, ¨Uber einfache und mehrfache Bedeckung des n-dimensionalen Raumes mit einenW¨urfelgitter, Math.

Z. 47 (1942), 427–467.3., Sur la factorisation des groupes abelians, ˇCasopis Pˇest.

Mat. Fys.

74 (1950),157–162.4. O. H. Keller, ¨Uber die l¨uckenlose Einf¨ullung des Raumes mit W¨urfeln, J. Reine Angew,Math.

163 (1930), 231–248.5. H. Minkowski, Diophantische Approximationen, Teubner, Leipzig, 1907.

(Reprint: 1961Physica-Verlag, W¨urzberg.) [see Chapter 2, §4 and Chapter 3, §7.

Minkowski’s Conjectureappears on p. 28 and its geometric interpretation on p. 74.]6.

O. Perron, ¨Uber l¨uckenlose Ausf¨ullung des n-dimensionalen Raumes durch kongruenteW¨urfel, I, II, Math. Z.

46 (1940), 1–26, 161–180.7. S. K. Stein, Algebraic tiling, Amer.

Math. Monthly 81 (1974), 445–462.8.

S. Szab´o, A reduction of Keller’s conjecture, Period. Math.

Hungar. 17 (1986), 265–277.AT&T Bell Laboratories, Murray Hill, New Jersey 07974

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