aa r X i v : . [ m a t h . M G ] J un CONCRETE POLYTOPES MAY NOT TILE THE SPACE
ALEXEY GARBER ⋆ AND IGOR PAK ⋄ Abstract.
Brandolini et al. conjectured in [BCRT] that all concrete lattice polytopes can multitilethe space. We disprove this conjecture in a strong form, by constructing an infinite family ofcounterexamples in R . Introduction
The study of integer points in convex polytopes is so challenging because it combines the analyticdifficulty of number theory with hardness of imagination typical to high dimensional geometry andthe computational complexity of integer programming. Consequently, whenever a new conjecture isposed it is a joyful occasion, as it suggests an order in an otherwise disordered universe. When aconjecture is occasionally disproved, this adds another layer of mystery to the subject.In this paper we study the conjecture by Brandolini et al. [BCRT, Conj. 5] that all concrete latticepolytopes can multitile the space. This conjecture was restated and further investigated in [MR,Conj. 8.6] from a different point of view. Here we disprove the conjecture by constructing a series ofexplicit counterexamples. In fact, our main result is stronger as it holds under more general notionof tileability. Our tools involve McMullen’s theory of valuations of lattice polyhedra and
Dehn’sinvariant . We conclude with final remarks and open problems.A convex polytope P ⊂ R d is called a lattice polytope if all its vertices are in Z d . Denote by ω P ( x ) := vol (cid:0) B ε ( x ) ∩ P (cid:1) vol B ε ( x )the solid angle at point x , where B ε ( x ) is a ball of radius ε centered at x , and ε > regularised ) discrete volume of P as the sum of solid angles over all integer points: χ ( P ) := X x ∈ P ∩ Z d ω P ( x ) , cf. [Bar, BR]. Pick’s theorem says that χ ( P ) = vol( P ) for all lattice polygons P ⊂ R . In an attemptto extend the theorem, Brandolini et al. [BCRT] call a lattice polytope concrete if χ ( P ) = vol( P ).They made the following curious conjecture. Conjecture 1.1 ([BCRT, MR]) . Every concrete lattice polytope P ⊂ R d multitiles R d with paralleltranslations and finitely many reflections. Here we say that P multitiles R d if there is an integer k ≥ P of congruentcopies of P such that every generic point x ∈ R d belongs to exactly k polytopes in P , see [GRS]. For k = 1 this is the usual tiling of the space, see e.g. [GS]. In the conjecture, only P ′ ∈ P are allowedif they are obtained from P with parallel translations and finitely many reflections. We disprove astronger claim in the main result of this paper: Theorem 1.2.
There is a concrete lattice polytope P ⊂ R which does not multitile R . Moreover,for all N , there is such a polytope P with more than N vertices. June 9, 2020. ⋆ School of Mathematical & Statistical Sciences, University of Texas Rio Grande Valley, Brownsville, TX, 78520.Email: [email protected] . ⋄ Department of Mathematics, UCLA, Los Angeles, CA 90095. Email: [email protected] . Brief background and the countexample idea
Let us first expound on the background and the motivation behind the conjecture. The problemof classifying polytopes which can tile (tesselate) the space is classical. It goes back to the works ofF¨edorov, Minkowski, Voronoi, Delone and Alexandrov, and was featured in
Hilbert’s 18-th Problem ,see [GS]. For tilings with parallel translations much more is known; notably that in R all suchpolytopes must be zonotopes (polytopes with centrally-symmetric faces of all dimensions). In higherdimensions, or for larger discrete groups of translations and reflections, other polytopes appear totile the space, e.g. the in R .For the lattice polytopes, the tilings are also heavily constrained and can be studied using analytictools [BCRT, GRS]. The notion of multitiling goes back to Furtw¨angler (1936), and many classicaltiling results extend to this setting [GKRS]. It is known and easy to see that if a lattice polytopemultitiles the space with parallel translations then it is concrete [BCRT] (see below). In particular,all lattice zonotopes multitile the space [GRS], and they are concrete because they can be partitionedinto parallelepipeds, see e.g. [BP, §
7] and [Zie, Ch. 7]. The conjecture can then be viewed as anattempt to say that the class of concrete lattice polytopes is very small and can be characterizedvia the large body of work towards characterization of tilings and multitilings.From this point on, we restrict ourselves to convex polytopes P ⊂ R . For the clarity, observethat ω P ( x ) = 1 for x in relative interior of P , and ω P ( x ) = for x in relative interior of a face.Similarly, ω P ( x ) = α ( e ) for x in relative interior of an edge e of P , where α ( e ) is the dihedral angleat e , and ω P ( x ) is the usual solid angle for a vertex x of P .There is a way to understand both the conjecture and our theorem as part of the same asymptoticargument. For a polytope P ⊂ R , define the ( lattice ) volume defect by δ ( P ) := χ ( P ) − vol( P ) ∈ R . Similarly, the
Dehn invariant is given by D ( P ) := X e ∈ E ( P ) ℓ ( e ) ⊗ α ( e ) ∈ R ⊗ Z (cid:0) R /π Z (cid:1) , where E ( P ) is the set of edges in P , ℓ ( e ) is the length of edge e , and α ( e ) is the dihedral angle at e ,see e.g. [Bol, Dup]. Theorem 2.1.
Let P ⊂ R be a convex polytope which multitiles the space. Then D ( P ) = 0 .Similarly, let P be a lattice convex polytope which multitiles the space with parallel translations.Then δ ( P ) = 0 , i.e. P is concrete. Versions of this result and its various generalizations have been repeatedly rediscovered, oftenwith the same asymptotic argument which goes back to Debrunner (1980) and M¨urner (1975). Werefer to [LM] for generalizations to higher dimensions and further references (see also § Proof outline.
First, suppose the multitiling is the usual tiling. Let P be the set of copies of P whichdefine the usual tiling of R , and let P R ⊂ P be the set of copies of P which intersect a ball B R ( O ) ofradius R around the origin. Denote by Γ ⊂ R the region covered with tiles in P R . On the one hand,both the volume defect and the Dehn invariant are additive, so δ (Γ) = |P R | δ ( P ) = Θ( R ) δ ( P ),and D (Γ) = |P R | D ( P ) = Θ( R ) D ( P ). On the other hand, both the volume defect and the Dehninvariant depend only on the boundary ∂ Γ, which is within a constant distance from ∂B R ( O ). Thusboth grow at most quadratically: δ (Γ), D ( P ) = O ( R ). For the defect this is clear, and for the Dehninvariant this can be made precise by using Kagan functions f (see below), which extend to ringhomomorphisms R ⊗ Z (cid:0) R /π Z (cid:1) → R , so the resulting function of R can then be viewed analytically.Comparing the lower and upper asymptotic bounds, this implies the result. For general k -tilings,the same argument work verbatim, as the changes are straightforward. (cid:3) We are implicitly using the fact that δ ( P ) = δ ( P ′ ) for all copies of P ′ ∈ P . This is not true when reflections areallowed, see § ONCRETE POLYTOPES MAY NOT TILE THE SPACE 3
Now, the idea of a counterexample to Conjecture 1.1 is very clear. We use the lattice valuationtheory to construct a lattice polytope P ⊂ R whose volume defect δ ( P ) = 0, while the Dehninvariant D ( P ) = 0. By Theorem 2.1, this implies that P cannot multitile the space.3. Minkowski additivity
Volume defect.
By definition, the volume defect δ ( P ) is a translation invariant valuation onlattice polytopes, i.e. δ ( P + x ) = δ ( P ) for all x ∈ Z , and δ ( P ∪ Q ) + δ ( P ∩ Q ) = δ ( P ) + δ ( Q ) , for all lattice polytopes P, Q ⊂ R . In particular, the volume defect is additive under disjoint union(except at the boundary). We also need the following linearity property under Minkowski addition P + Q = { x + y | x ∈ P, y ∈ Q } and expansion cP = { cx | x ∈ P } . Lemma 3.1 (see § . Let P , . . . , P k be lattice convex polytopes in R , and let t , . . . , t k ∈ N .Then δ (cid:0) t P + . . . + t k P k (cid:1) = t δ ( P ) + . . . + t k δ ( P k ) . Proof outline.
Let P ⊂ R be a lattice polytope, and let t ∈ N be a variable. Both χ ( tP ) and vol( tP )are cubic polynomials, see [M2, p. 127] (see also [BL1, Thm 7.9] and [Joc, Thm 2.1] for surveys andfurther references). Moreover, χ ( tP ) and vol( tP ) are odd cubic polynomials, see [Mac, Thm 4.8] (seealso [M1]), with the same leading coefficient. Thus, δ ( P ) = χ ( P ) − vol( P ) is linear. The polynomi-ality under Minkowski addition follows from McMullen’s homogeneous decomposition [M1] (see alsoe.g. [Joc, Thm 4.1]). Again, the cubic terms cancel, and the same argument proves multilinearityas in the lemma. (cid:3) Dehn invariant.
For the clarity of exposition, we follow [Pak, §
17] (see also [Bol, Dup]). Fixan additive function f : R −→ R , s.t. f ( a + b ) = f ( a ) + f ( b ) for all a, b ∈ R . Additive function f s.t. f ( π ) = 0 is called a Kagan function , after [Kag].For a convex polytope P ⊂ R and a Kagan function f , denote D f ( P ) := X e ∈ E ( P ) ℓ ( e ) f ( α e ) . Observe that D f is a translation invariant valuation, and that D f ( cP ) = cD f ( P ). Lemma 3.2 (see § . Let P , . . . , P k be convex polytopes in R , and let t , . . . , t k ∈ R + . Then: D f (cid:0) t P + . . . + t k P k (cid:1) = t D f ( P ) + . . . + t k D f ( P k ) , for every Kagan function f .Proof outline. For k = 2, the result follows immediately from the homogeneous decomposition againand the additivity of the Dehn invariant under disjoint union. For larger k , proceed by induction.Alternatively, recall that the Dehn invariant is a simple, translation-invariant valuation (see § D f (cid:0) t P + . . . + t k P k (cid:1) is a polynomial in t , . . . , t k of degree at most 3. Now, the restriction ofthat polynomial onto any ray from the origin is a linear function, which implies that this polynomialis linear. The details of both arguments are straightforward. (cid:3) ALEXEY GARBER AND IGOR PAK Counterexample construction
Consider the following three tetrahedra: T := conv (cid:8) (0 , , , (1 , , , (1 , , , (0 , , (cid:9) ,T := conv (cid:8) (0 , , , (2 , , − , (2 , − , , (1 , − , − (cid:9) ,T := conv (cid:8) (0 , , , (2 , , − , (3 , , − , (5 , − , − (cid:9) . In notation of [Pak, § T is regular with edge length √
2, tetrahedron T is standard with three pairwise orthogonal edges of length 3, and T is an orthoscheme (also called path simplex and Hill tetrahedron ), with three edge lengths 3.Note that six copies of T tile a cube spanned by vectors v = (2 , , − v = (1 , − , − v = (2 , − ,
2) starting at the origin O . Indeed, these six copies correspond to six permutations of { v , v , v } , and are spanned by the paths formed by these vectors. This implies that D f ( T ) = 0for every Kagan function f defined above. Proposition 4.1.
Let P := 5 T + 12 T + 19 T . Then δ ( P ) = 0 , and D f ( P ) = 0 for some Kaganfunction f .Proof. Let α = arccos , and recall that απ / ∈ Q , see e.g. [Pak, § f which satisfies f ( α ) = 0, and, moreover, f ( α ) / ∈ Q , see [Pak, Ex. 17.8].Observe that all dihedral angles of T are equal to α . Dihedral angles of T are equal to π forthe three edges at the origin, and to β := arccos √ = π − α for the three other edges. Finally, alldihedral angles of T are rational multiples of π . The values of the volume defect and the Dehninvariant for all three tetrahedra are given in Table 1 below. T T T δ ( · ) απ − − α π −
12 23 D f ( · ) √ f ( α ) − √ f ( α ) 0 Table 1.
Values of the volume defect and the Dehn invariant.Using values from the table, Lemmas 3.1 and 3.2 imply: δ ( P ) = 5 δ ( T ) + 12 δ ( T ) + 19 δ ( T ) = 0 ,D f ( P ) = 5 D f ( T ) + 12 D f ( T ) + 19 D f ( T ) = − √ f ( α ) = 0 , as desired. (cid:3) Proof of Theorem 1.2.
The polytope P ⊂ R constructed in the proof of Proposition 4.1 is concrete,but has a non-zero Dehn invariant. Thus, by Theorem 2.1, it cannot multitile the space. This provesthe first part of the theorem.For the second part, take a lattice zonotope Q ⊂ R with at least N vertices. From the resultsin §
2, we have δ ( Q ) = 0, and D f ( Q ) = 0 for all Kagan functions f . By Lemma 3.1, we have δ ( P + Q ) = δ ( P ) = 0, so ( P + Q ) is concrete. On the other hand, by Lemma 3.2, we have D f ( P + Q ) = D f ( P ) = 0. Thus, by Theorem 2.1, polytope ( P + Q ) cannot multitile R . Finally,observe that ( P + Q ) has at least N vertices, see e.g. [Zie, Prop. 7.12]. This completes the proof. (cid:3) See [Pak, § D f ( P ) = 0 without computing dihedral angles directly. ONCRETE POLYTOPES MAY NOT TILE THE SPACE 5 Final remarks and open problems R . Indeed, consider the following wedge tetrahedron and flat square pyramid : W n := conv (cid:8) (0 , , , (1 , , , (1 , , n ) , (0 , , n ) (cid:9) ,V n := conv (cid:8) (0 , , , ( n, , , (0 , n, , ( n, n, , (0 , , (cid:9) . As n → ∞ , we have: χ ( W n ) = Θ (cid:18) n (cid:19) , vol( W n ) = n , and δ ( W n ) ∼ − n . On the other hand, χ ( V n ) = n − O ( n ) , vol( V n ) = n , and δ ( V n ) ∼ n . R d which is homogeneous of degree one is Minkowski additive (see [Sch, Rem. 6.3.3]and [BL2, Cor. 32]). We include a short proof outline both for simplicity, to remain as much self-contained as possible, and as a brief guide to the literature. While Lemma 3.2 is very natural, wecould not find it stated in this form. As we explain above, its proof follows along steps similar tothe proof of Lemma 3.1.5.3. The asymptotic argument in the proof of Theorem 2.1 can also be applied in R , where it istraditionally used to show that the plane cannot be tiled with congruent convex n -gons, for n ≥ R , and for further references.5.4. One can ask if the results of this paper can be further extended. First, we can always extendTheorem 1.2 to higher dimensions d ≥
4. By the argument in the proof of Theorem 2.1, every P ⊂ R d which multitiles R d has zero Hadwiger (generalized Dehn) invariants [LM]. Take anorthogonal prism P × [0 ,
1] over the polytope P ⊂ R as in Proposition 4.1. The dihedral angles areeither π/ P . Thus the corresponding codim-2 Hadwiger invariant is non-zero,giving a counterexample in R . Proceed by induction; the details are straightforward.Going one step further, we say that a lattice polytope P ⊂ R d is super concrete , if it is concreteand scissors congruent to a d -cube. For d = 3 ,
4, by the
Sydler–Jessen theorem this is equivalentto zero Dehn invariant [Bol, Dup]. For d ≥
5, scissors congruence with a d -cube implies and isconjecturally equivalent to zero Hadwiger invariants, see e.g. [Zak]. Also, every P ⊂ R d whichmultitiles R d with translations must be super concrete, see [LM]. So in the spirit of Conjecture 1.1,one can ask whether all super concrete lattice polytopes P ⊂ R d can multitile the space.We conjecture that the answer is negative for all d ≥
3. For example, for d ≥
4, the localstructure of cones around a vertex can be constrained by a spherical Dehn invariant , see e.g. [Dup].In principle, the concrete assumption is too weak and can allow “bad cones” which would locallynot multitile the sphere S d − . It would be interesting to make this precise. The above problem iseven more interesting in R . In principle, the cones around vertices can all have nontrivial geometrygenerating a non-discrete group of symmetries, cf. [MM]. Again, it would be interesting to give anexplicit construction.5.5. Let us mention that the proof of Proposition 4.1 hinges on the following curious geometricproperty: the orthoscheme T tiles the lattice cube and thus the space, yet has δ ( T ) = 0. Inparticular, this shows that Theorem 2.1 cannot be extended to allow reflections. This non-zerovolume defect of T has to do with the fact that the remaining five orthoschemes in the tiling ofthe cube are obtained from T by reflections which do not preserve the lattice. Although congruentto T , these reflected orthoschemes have both negative and positive volume defect, giving zero intotal for the lattice cube.Note that the (primitive) lattice cubes which arise in the construction, correspond (up to paralleltranslation) to rational orthogonal matrices M ∈ O (3 , R ). These matrices are enumerated in [Cre].We conjecture that the corresponding orthoschemes have a nonzero volume defect with probability ALEXEY GARBER AND IGOR PAK at least ε >
0, as the cube edge length ℓ → ∞ . This would give further examples of polytopes as inthe proposition, all with a bounded number of vertices. Acknowledgments.
The authors are grateful to Sasha Barvinok, Alexey Glazyrin and Sinai Robinsfor interesting remarks. Special thanks to Katharina Jochemko and Monika Ludwig for help withthe valuation literature. The paper was finished when the second author was on sabbatical atthe Mittag-Leffler Institute; we are grateful for the hospitality. The second author was partiallysupported by the NSF.
References [Ale] A. D. Alexandrov, On the coverings of the plane (in Russian),
Mat. Sbornik (1937), 307–318.[Bar] A. Barvinok, Integer points in polyhedra , EMS, Z¨urich, 2008, 191 pp.[BP] A. Barvinok and J. E. Pommersheim, An algorithmic theory of lattice points in polyhedra, in
New perspec-tives in algebraic combinatorics , Cambridge Univ. Press, Cambridge, UK, 1999, 91–147.[BR] M. Beck and S. Robins,
Computing the continuous discretely (Second ed.), Springer, New York, 2015, 285 pp.[Bol] V. G. Boltyanskii,
Hilbert’s Third Problem , John Wiley, New York, 1978.[BL1] K. J. B¨or¨oczky and M. Ludwig, Valuations on lattice polytopes, in
Lecture Notes in Math. , Springer,Cham, 2017, 213–234.[BL2] K. J. B¨or¨oczky and M. Ludwig, Minkowski valuations on lattice polytopes,
Jour. EMS (2019), 163–197.[BCRT] L. Brandolini, L. Colzani, S. Robins and G. Travaglini, Convergence of multiple Fourier series and Pick’stheorem, 6 pp.; arXiv:1909.03435 .[Cre] J. Cremona, Letter to the Editor, Amer. Math. Monthly (1987), 757–758; available athttps://tinyurl.com/umoouee[Dup] J. L. Dupont, Scissors congruences, group homology and characteristic classes , World Sci., River Edge, NJ,2001, 168 pp.[GKRS] N. Gravin, M. N. Kolountzakis, S. Robins and D. Shiryaev, Structure results for multiple tilings in 3D,
Discrete Comput. Geom. (2013), 1033–1050.[GRS] N. Gravin, S. Robins and D. Shiryaev, Translational tilings by a polytope, with multiplicity, Combinator-ica (2012), 629–649.[GS] B. Gr¨unbaum and G. C. Shephard, Tilings and patterns , Freeman, New York, 1987, 700 pp.[Joc] K. Jochemko, A brief introduction to valuations on lattice polytopes, in
Algebraic and geometric combina-torics on lattice polytopes , World Sci., Hackensack, NJ, 2019, 38–55.[Kag] V. F. Kagan, ¨Uber die Transformation der Polyeder (in German),
Math. Ann. (1903), 421–424.[KPP] E. Kopczy´nski, I. Pak and P. Przytycki, Acute triangulations of polyhedra and R N , Combinatorica (2012), 85–110.[LM] J. C. Lagarias and D. Moews, Polytopes that fill R n and scissors congruence, Discrete Comput. Geom. (1995), 573–583; Acknowledgment of priority, ibid. (1995), 359–360.[MR] F. C. Machado and S. Robins, Coefficients of the solid angle and Ehrhart quasi-polynomials, 35 pp.; arXiv:1912.08017 .[Mac] I. G. Macdonald, Polynomials associated with finite cell-complexes, J. London Math. Soc. (1971), 181–192.[MM] G. A. Margulis and S. Mozes, Aperiodic tilings of the hyperbolic plane by convex polygons, Israel J.Math. (1998), 319–325.[M1] P. McMullen, Metrical and combinatorial properties of convex polytopes, in
Proc. ICM (Vancouver, BC) ,Vol. 1, 1974, 491–495.[M2] P. McMullen, Valuations and Euler-type relations on certain classes of convex polytopes,
Proc. LondonMath. Soc. (1977), 113–135.[Niv] I. Niven, Convex polygons that cannot tile the plane, Amer. Math. Monthly (1978), 785–792.[Pak] I. Pak, Lectures on discrete and polyhedral geometry
Convex bodies (Second ed.), Cambridge Univ. Press, Cambridge, UK, 2014, 736 pp.[Zak] I. Zakharevich, Perspectives on scissors congruence,