TILING ABELIAN GROUPS WITH A SINGLE TILE

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1 TILING ABELIAN GROUPS WITH A SINGLE TILE S.J. EIGEN AND V.S. PRASAD Abstract. Suppose G is an infinite Abelian group that factorizes as the direct sum G = A B: i.e., the B-translates of the single tile A, evenly tile the group G (B is called the tile set). In this note we consider conditions for another set C G to tile G with the same tile set B. In an earlier paper we answered a question of Sands regarding such tilings of G when A is a finite tile. We consider extensions of Sands s question when A is infinite. We offer two approaches to this question. The first approach involves a combinatorial condition used by Tijdeman and Sands. This condition completely characterizes when a set C can tile G with the tile set B; the condition is applied to simplify the proofs and extend some results of Sands [8]. The second approach is measure theoretic and follows Eigen, Hajian, Ito s work on exhaustive weakly wandering sets for ergodic infinite measure preserving transformations. 1. Introduction Two subsets A and B of an abelian group G induce a tiling (or factorization) of G, if every element of G has a unique representation as a sum a + b with a A and b B, and we write G = A B. Note that this is the same as saying that the sets A and B have the following two properties: (1) The B-translates of the tile A are disjoint; i.e., the sets {A + b : b B} are disjoint. Borrowing terminology from the ergodic theory of infinite measure preserving transformations [1], we say A is weakly wandering under B. (2) The B-translates of A fill up G; i.e., b B (A + b) = G. Again borrowing from ergodic theory, we say that A is exhaustive under B. Throughout this paper we assume that A and B are subsets of G so that our group G = A B. We normalize by requiring that 0 A B. We refer to this as a tiling of G by the single tile A with tile set B. In [8] A.D. Sands asked the following question: suppose the tile A is finite and G is tiled by A with tile set B, and suppose C is any other finite set of the same size as A which satisfies the condition (1) that the B-translates of C are disjoint (i.e., C is weakly wandering under B), must then G be tiled by C with tile set B (i.e., do the B-translates of C exhaust all of G)? Note that if G is finite then the answer to Sands s question is yes and the proof is a straightforward counting argument. A much more careful counting argument provides a positive answer to Sands s question for infinite groups G in [2]. Mike Keane has asked us what happens when the tile A is infinite? We address this question here. Date: January 31, Mathematics Subject Classification. 20K01(primary), 37A05, 28D05(secondary). Key words and phrases. tilings, exhaustive weakly wandering, ergodic infinite measure preserving transformation To appear in Discrete and Continuous Dynamical Systems. 1

2 2 S.J. EIGEN AND V.S. PRASAD Note that if A is an infinite tile which tiles G with tile set B, then simply removing an element from A would provide another infinite tile C of the same size as A which is weakly wandering under B, but clearly not exhaustive under B, so cardinality is not the required condition for infinite tiles A. In this note we give two approaches to Keane s question when A need not be a finite tile. First, in Theorem 2.1, we consider a combinatorial condition due to Tijdemann [10] and obtain a necessary and sufficient condition for C to tile with tile set B. In Corollary 2.5 we use this characterization to simplify some proofs of Sands in [8]. A second approach to Keane s tiling question (following Eigen, Hajian and Ito [1]) provides a solution to a measure theoretic version of Sands s question. We conclude this introduction by noting that problems on the factorization of abelian groups have long been of interest starting with Minkowski s 1907 work on diophantine approximation and the Minkowski conjecture [3] (and the related Keller conjecture [4]) on lattice tilings of Euclidean space with the unit cube. In 1941, the lattice tiling conjecture was shown by Hajós [3] to be equivalent to a group theoretic factorization theorem which Hajós then proved (see also simplifications due to Rédei [6], [7], and recent work of Kolountzakis [5] the latter using a spectral theory approach to show various equivalent forms of Hajós s theorem) An important part of Hajós s work involves replacing a factor A (for a factorization G = A B) by another factor C, with special properties (so that G = C B). A beautiful exposition of the Hajós theorem and many other current tiling problems can be found in Szabo and Stein s Carus Monograph (volume 25), [9]. Factorizations of abelian groups have been considered by many mathematicians including Coven, de Bruijn, Fary, Fuchs, Hajós, Long, Meyerowitz, Rédei, Sands, Stein, Swenson, Szabo, Szele, Tijdeman. An approach to factorizations of Z using the ergodic theory of infinite measure preserving transformations has been developed by Eigen, Hajian, Ito and Kakutani (see for example [1]). 2. Tilings: A combinatorial approach Our first approach involves a condition of Tijdeman [10]. Since G = A B we normalize our sets so that A B = {0}. Each element g G is uniquely written as g A + g B, where g A A and g B B are called the A and B names of g respectively. Sands showed [8, Thm 1] that when A is finite and G = A B, then for each g G: {(g + a) A : a A}. = A where. = means not only are the two sets equal, but also that every element of the left side of. = is uniquely represented as an element of the right side. Suppose G = A B. The next result characterizes sets C which tile G with the same tile set B. It provides our first answer to Keane s question. Theorem 2.1. Suppose the Abelian group G = A B. Let C be any subset of G. Then the following two conditions are equivalent: (1) G = C B (2) For each g G, {(g c) A : c C}. = A This theorem is an immediate consequence of the following Lemmas 2.2 and 2.3 which characterize sets C that are respectively exhaustive and weakly wandering under the tile set B.

3 TILING ABELIAN GROUPS 3 Lemma 2.2. Suppose that G = A B. Then the tile C is exhaustive with tile set B (i.e., G = C + B) if and only if for each g G, the collection of A-names {(g c) A : c C} = A. Proof. Suppose for each g G, the collection of A-names {(g c) A : c C} = A. Given g G, since 0 A, there is a c C such that (g c) A = 0. Thus g c = (g c) A + (g c) B = 0 + b where b B, and so g = c + b. Thus G = C + B. Conversely, given g G and a A, if G = C + B, then there are elements c C and b B such that g a = c + b g c = a + b. Thus (g c) A = a and we have for each g G, {(g c) A : c C} = A. Lemma 2.3. Suppose that G = A B. Then the tile C is weakly wandering with tile set B (i.e., the B-translates of C are disjoint) if and only if for each g G, the collection of A-names {(g c) A : c C} are distinct. Proof. Assume that the B-translates of C are disjoint. Then (C C) (B B) = {0}. Suppose for some g G, that there are elements c 1, c 2 C so that (g c 1 ) A = (g c 2 ) A. Then g c 1 = (g c 1 ) A + b 1 g c 2 = (g c 2 ) A + b 2. where b 1, b 2 B. Therefore subtracting these two equations, we get c 2 c 1 = b 1 b 2 (C C) (B B). But, since C is weakly wandering under B, c 1 = c 2. Thus the A-names are distinct. Conversely, suppose for each g G, the A-names {(g c) A : c C} are distinct. If, c 1, c 2 C and b 1, b 2 B satisfy then there is a A and b B so that Consequently, c 1 + b 1 = c 2 + b 2 c 1 b 1 = c 2 b 2 = a + b b c 1 = a + b 1 b c 2 = a + b 2. This implies that ( b c 1 ) A = ( b c 2 ) A. The uniqueness of A-names of {( b c) A : c C} implies c 1 = c 2. It then follows that b 1 = b 2 as well. Note that if we apply this condition to C = A we get that for each g G, {(g a) A : a A}. = A. Note it is the A-names, (g C) A that are the object of interest, and not the A-names, (g + C) A, as the following example shows. Example 2.4. Consider the following subsets of the integers: A = SF S{2 2n+1 : n = 0, 1, 2,...} where this means that A is the set of all Sums of Finite Subsets of 2 raised to odd powers (note the sum over the empty subset implies 0 A), and B = SF S{2 2n : n = 0, 1, 2,...}. The following facts are true: Z = A B; if C = A, then C does not tile Z with tile set B, since C B = N, the nonpositive

4 4 S.J. EIGEN AND V.S. PRASAD integers. However the remark before the example implies that for each g Z, {(g + c) A : c C}. = A. Our next result contains Theorems 1 and 2 of Sands [8] and extends those theorems when A need not be a finite tile. Corollary 2.5. Let G = A B. Then the following four conditions are equivalent. (a) G = A ( B) (b) G = ( A) B. (c) {g + a : a A} A = A. (d) {g + b : b B} B = B Furthermore, they are all implied by (e) A <. Proof. (a) (b) Multiply either statement by 1 to obtain the other (and note that neither condition is equivalent to A B = G). (b) (c) Let C = A in our Theorem 2.1. (d) (a) Again this just uses our Theorem 2.1 with C = B and tile set A. (e) (c) Since all elements (g + a) A are distinct as a A varies, then we can state that {(g + a) A : a A}. = A when A is a finite set. This is just Sands s Theorem 1 from [8]. 3. Tilings: A measure theoretic approach We end this note with a measure theoretic approach to Keane s question, thereby providing an answer to an extension of Sands s problem to infinite sets. This method follows the work of Eigen, Hajian, Ito and Kakutani [1] on the ergodic theory of infinite measure preserving transformations (see also [2]). Suppose now that our group G has a translation invariant measure µ defined on it. Note that in this section all statements are assumed to be true modulo a set of µ-measure zero. Here is a measure theoretic version of Sands s question. Suppose that G = A B, where A and B are measurable subsets such that 0 < µ(a) <, for some translation invariant measure µ. If C is a measurable subset with µ(c) = µ(a) and disjoint B-translates, then must G be tiled by C with tile set B? This is clearly true if µ(g) <, so we assume that µ(g) =. The following result is a direct generalization of our main Theorem from [2] and answers this extension of Sands s question. Indeed the proof of this theorem just replaces counting measure in [2] with the translation invariant measure µ and the argument remains exactly the same. Theorem 3.1. Let G be an abelian group with µ a translation invariant measure on G. Let G = A B be a factorization of the abelian group G with measurable subsets A and B where 0 < µ(a) <. Consider the three conditions on a measurable set C G. (1) µ(c) = µ(a). (2) C is weakly wandering under B (i.e., the B-translates of C are disjoint). (3) C is exhaustive under B (i.e., the union of the B-translates of C is G). Then, whenever a subset C satisfies any two of the above conditions it must satisfy the remaining condition. Note, we remind the reader again, that all conclusions and hypotheses are assumed to hold modulo sets of µ-measure 0.

5 TILING ABELIAN GROUPS 5 Proof. The argument in [2] used counting measure which is a translation invariant measure. If we replace cardinality in that proof by µ( ) throughout, then the same arguments, prove this theorem. As in [2], we note that the implication that (ii) and (iii) (i) does not require that µ(a) <. This approach to Sands s questions has its roots in the ergodic theory of infinite measure preserving transformations [1]. Let (X, µ) be an infinite measure space. Let T be an invertible measure preserving transformation of (X, µ), T : (X, µ) (X, µ). A subset W X is said to be weakly wandering for a sequence B of the integers if the collection of sets {T b (W ) : b B} is pairwise disjoint. The set W is called exhaustive for the sequence B if X = b B T b (W ). The following theorem on exhaustive weakly wandering sets was proved by Eigen, Hajian and Ito [1]. Theorem 3.2 ((Eigen Hajian Ito) [1]). Let T be an ergodic infinite measure preserving transformation of the sigma finite measure space (X, µ). Let W X be an exhaustive and weakly wandering set for the sequence of integers B. Suppose µ(w ) <. Consider the following three conditions on some V X (1) µ(v ) = µ(w ). (2) V is weakly wandering for B. (3) V is exhaustive for the set B. Then, when a subset V satisfies any two of the above conditions, it satisfies the third. Furthermore, the implication that the latter two conditions imply the first does not require the assumption that µ(w ) <. Our proof above is just the Eigen-Hajian-Ito proof for the G-action {T g : g G}, when X = G, µ is a translation invariant measure, and G acts on itself by T g (x) = x + g for x G, W = A and V = C. References [1] Eigen, S., Hajian, A., and Ito, Y., Ergodic measure preserving transformations of finite type Tokyo J. Math. 11 (1988), no. 2, [2] Eigen, S., and Prasad, V.S., Solution to a problem of Sands on the factorization of groups, Indagationes Mathematicae (New Series) 14 (2003) [3] Hajós, G., Uber einfache und mehrfache Bedeckung des n-dimensionalen Raumes mit einem Wurfelgitter, Math. Z. 47 (1941), [4] Hajós, G., Sur la factorisation des groupes abéliens, Casopis Pest. Mat. Fys. 74 (1949), [5] Kolountzakis, M., Lattice tilings by cubes: whole, notched and extended, Electronic J. Combin. 5 (1998), Research paper 14, 11p. [6] Rédei, L. Vereinfachter Beweis des Satzes von Minkowski-Hajós, Acta Univ. Szeged. Sci. Math. 13 (1949) [7] Rédei, L. Kurzer Beweis des gruppentheoretischen Satzes von Hajós, Comment. Math. Helv. 23 (1949) [8] A. D. Sands, Replacement of factors by subgroups in the factorization of abelian groups Bull. London Math. Soc. 32 (2000) [9] Stein, S. K. and Szabó, S., Algebra and tiling. Homomorphisms in the service of geometry, Carus Mathematical Monographs, 25, published by Mathematical Association of America, Washington, DC,(1994) [10] R. Tijdeman, Decomposition of the integers as a direct sum of two subsets, Number theory, (ed. S. David), (Cambridge University Press, 1995), Northeastern University address: eigen@neu.edu

6 6 S.J. EIGEN AND V.S. PRASAD University of Massachusetts Lowell address: vidhu