Transparencies (preprints) available at http://faculty.uml.edu/vprasad/reprints
Tiling Abelian groups with a
single tile and the ergodic
theory of transformations
preserving an infinite measure
S. Eigen and V.S. Prasad CRM/CMI
Tiles and Tile sets
Suppose abelian group factorizes G = A ⊕ B: i.e., {A + b : b ∈ B} are disjoint: A is weakly wandering under B
and
∪b∈B(A + b) = G : A is exhaustive under B A is called a tile and B a tile set
Sands’s Question
Problem. (Sands, Bull. LMS ’00) Suppose abelian group G = A⊕B and |A| < ∞. Suppose C ⊂ G so that |A| = |C| and is weakly wander-ing under B. Does it follow that G = C + B? The answer is yes and more generally
Theorem. (Eigen-P. Indag ’04) For a finite tile A with tile set B, (G = A ⊕ B). Consider conditions on a set C ⊂ G.
i) |C| = |A|.
ii) C is weakly wandering under B iii) C is exhaustive under B.
If C satisfies any two of the above conditions it must satisfy the remaining condition.
So i) and ii) implies iii) answers Sands’s ques-tion.
When |A| = ∞, answer to Sands’s question is false. Only the implication iii) and ii) ⇒ i) does not require the finiteness of A.
Eigen-Hajian-Ito Theorem (Tokyo J. Math. 88)
Let T : (X, µ) → (X, µ) ergodic infinite measure preserving transformation of the sigma finite measure space. Let W ⊂ X be an exhaustive and weakly wandering set for the sequence of integers B (i.e., w.w. for B: Tb(W ) for b ∈ B are disjoint collection; exhaustive for B if those images cover X). Suppose µ(W ) < ∞. Con-sider the following three conditions on some V ⊂ X
1. µ(V ) = µ(W ).
Then, when a subset V satisfies any two of the above conditions, it satisfies the third. Fur-thermore, the implication that the latter two conditions imply the first does not require the assumption that µ(W ) < ∞.
Tijdemann’s Condition
G = A ⊕ B and normalize so A ∩ B = {0}
so for g ∈ G, g = gA + gB are the A and B names of g.
Sands showed when finite set A tiles G with tile set B, then for each g ∈ G
{(g + a)A : a ∈ A} $ A
“$” means: sets are equal and each element of LHS is uniquely represented as element of RHS
Tiles with tile set B
Theorem. (E-P, D&CDS) Suppose the Abelian group G = A ⊕ B. Let C be any subset of G,
then
G = C ⊕ B ⇔
Exhaustive
Lemma 1 Let G = A ⊕ B. Then the tile C is exhaustive with tile set B (i.e., G = C + B, not necessarily disjoint translates) 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.
Weakly Wandering
Lemma 2 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 c1, c2 ∈ C so that (g − c1)A = (g − c2)A. Then
g − c1 = (g − c1)A + b1 g − c2 = (g − c2)A + b2.
where b1, b2 ∈ B. Therefore subtracting these two equations, we get c2 − c1 = b1 − b2 ∈ (C − C) ∩ (B − B). But, since C is weakly wan-dering under B, c1 = c2. Thus the A-names are distinct.
Conversely, suppose for each g ∈ G, the A-names {(g − c)A : c ∈ C} are distinct. If, c1, c2 ∈ C and b1, b2 ∈ B satisfy
c1 + b1 = c2 + b2
then there is a ∈ A and b ∈ B so that −c1 − b1 = −c2 − b2 = a + b Consequently,
−b − c1 = a + b1 −b − c2 = a + b2.
This implies that (−b − c1)A = (−b − c2)A. The uniqueness of A-names of {(−b − c)A : c ∈ C} implies c1 = c2. It then follows that b1 = b2 as well.
Cor and Generalized Sands
This extends Sands’s Theorem 1 and 2 from (Bull LMS 2000)
Cor Let G = A ⊕ B. Then the following four conditions are equivalent.
(a) G = A ⊕ (−B) (b) G = (−A) ⊕ B
(c) ∀g ∈ G, {g + a, a ∈ A}A $ A (d) ∀g ∈ G, {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 condi-tion is equivalent to A ⊕ B = G).
(b) ⇔ (c) Let C = −A in our Theorem.
(d) ⇔ (a) Again this just uses our Theorem with C = −B and tile set A.
(e) ⇒ (c) Since all elements (g + a)A are dis-tinct 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 (Bull LMS 2000)
Measure Theoretic Tilings
µ some translation invariant measure. All state-ments are modulo µ-null sets.
Thm 2 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 re-maining condition.
E-P proof in Indag 04 works when counting measure replaced by µ translation invariant mea-sure.
Good Pairs of Integers S = {0, 1, 4, 5, 16, 17, 20, 21, · · · }
i.e., sums of finite subsets of the even powers of 2.
Fact: Z = S 2S
That is, each integer is obtained, and it is ob-tained uniquely.
•∀n ∈ Z, ∃si ∈ S such that n = s1 − 2s2
• s1 − 2s2 = t1 − 2t2 ⇒ s1 = t1 and s2 = t2
Definition. (de Bruijn, 1950, 1964) A pair (a, b) of positive odd integers is good if Z =
A Little Ergodic Theory
In 1970, Hajian and Kakutani presented an ex-ample of an Ergodic Infinite Measure Preserv-ing Transformation with an explicit Exhaustive Weakly Wandering set and sequence.
Definition. B a sequence of integers is Ex-haustive Weakly Wandering Sequence for the transformation T if there exists a set W of positive measure satisfying
1: µ(TbW ∩ Tb0W ) = 0, b 6= b0, b, b0 ∈ B. 2: µ(X\ ∪b∈B TbW ) = 0.
The set W is then an Exhaustive Weakly Wandering Set for T under B.
Consider the point 0 ∈ [0, 1) = W . It’s return sequence
While the exhaustive weakly wandering sequence for the set W = [0, 1) is 2S.
A similar transformation and skyscraper con-struction corresponds to any good pair of in-tegers.
Back to Good Pairs S = {0, 1, 4, 5, 16, 17, 20, 21, · · · }
(a, b) positive odd integers is good if Z = aS 2bS. Otherwise the pair is called Bad.
Examples: (1, 1), (1, 7), and (7, 13).
Example: (1, 22k+1 − 1) good for all k ≥ 0. Quick Facts from de Bruijn
(a, b) good requires gcd(a, b) = 1 (a, b) good iff (b, a) good.
(a, b) good requires a ≡ b Mod 6
de Bruijn (1950) listed good pairs (a, b) for 1 ≤ a ≤ b ≤ 100 obtained by “pencil and paper” and “shuffling four strips of paper”
Universally Bad
DEF: A positive odd integer u is Universally Bad (U.B.) if (ua, b) is bad for all pairs of positive odd integers a and b.
THM. (De Bruijn) u = 2k + 1 are U.B.
THM. u = φpk(4) are U.B., where p is prime and φn(x) is the n’th cyclotomic polynomial.
The first two U.B. integers are 3 And 5.
• We will see they are bad in slightly different ways.
THM 3, 22k+1 + 1, and φpk(4), p > 2 prime
will all be de Bruijn U.B.
THM 5, and 22k + 1 will be U.B. but not de Bruijn U.B.
2-Adics Z2
The 2-adic integers is the completion of the nonnegative integers in the 2-adic norm.
Z2 = ( z = X i≥0 zi2i : zi ∈ {0, 1} ) Write 0 < n = 2k · m, m odd.
DEF. The 2-adic order of n > 0 is defined by ord(n) = ord2(n) = k, if k is the highest power of 2 which divides n.
DEF. The 2-adic norm is |n| = |n|2 = 2−ord(n) = 2−k.
Identify Z2 with {0, 1}N,
i.e. z = P
zi2i ↔ (z0, z1, z2, · · · ). Extend ord to Z2 by setting
ord(z) = i, coordinate of first nonzero zi where z = P
zi2i = (z0z1 · · · ) for z ∈ Z2. Extend the 2-adic norm |z| = 2−ord(z).
Thus we get all integers represented in Z2 Positive integers end in all 0’s
Negative integers end in all 1’s
Addition in this representation is coordinate-wise from left to right with “carry” to the right.
• A sequence z(n) ∈ Z2 converges if and only if
ord(z(n) − z(m)) → ∞ as n, m → ∞.
Lemma (Geometric Series) If ord(x) > 1, (i.e. |x|2 < 1) then ∞ X 0 xi = 1 1 − x in the 2-adic integers.
Illustrations −1 = (¯1) = P∞ 0 2i = 1 1−2. Repeating pattern (10) = P∞ 0 4i = 1 1−4 = −13, and in general, P∞ i=0 4in = 1 1−4n.
S denotes the closure of S in the 2-adic norm.
Fact: S and its closure S have 0’s in all odd coordinate places; i.e., α = (α0α1 · · · ) ∈ S if and only if α2i+1 = 0 for all i.
Fundamental Theorem: For all pairs (a, b) of positive odd integers, the 2-adic integers can be written as
Z2 = aS 2bS = aS 2bS.
Idea. The Fundamental Theorem clarifies how it is possible for Z 6= aS 2bS.
THM. The pair of positive odd integers (a, b) is bad, i.e. Z 6= aS 2bS, if and only if there is an integer n such that
n = aσ − 2bτ where σ or τ ∈ S\S.
Illustration of why 3 is Universally Bad
Let (a, b) be any pair of odd positive integers; we need to show that (3a, b) is bad.
The fraction 1
3 = (10) belongs to S \ S.
Putting σ = −13 and τ = 0
we have −a = 3aσ − 2b · 0 and so (3a, b) is bad.
DEF. An odd positive integer u is a de Bruijn universally bad integer if there is some σ ∈ S \ S such that uσ ∈ Z.
Example
The integer u = 85 is de Bruijn universally bad. To see this, observe that the fraction
−21/255 = (10101000) = 1 + 4 + 4
2
1 − 44 ∈ S \ S, and 85 · −21255 = −7.
Example
The integer u = 341 is de Bruijn universally bad.
In this case, the fraction
−102381 = (1000101000) = 1 + 4
2 + 43 5
The integer 341 is a new universally bad inte-ger not on de Bruijn’s list. It is a product of 11 and 31 neither of which is universally bad.
The integer 85 is on de Bruijn’s list since it is a multiple of 5 = 22 + 1 and 17 = 24 + 1, which are universally bad, though neither is a de Bruijn universally bad integer.
These illustrate
THM. A positive odd integer u is a de Bruijn universally bad integer if and only if there exists a fraction of the form
σ =
PR−1
i=0 δi4i
1 − 4R ∈ S \ S
Illustration of why 5 is Universally Bad 5 is not de Bruijn universally bad . But it is still Universally Bad.
The two numbers −1 3 and − 1 15 = (10001000) = X 42i are both in S \ S.
Multiply by 5: −53 and −13 are both in 5(S \S). Now multiplying by a (assuming it is not a mul-tiple of 3) the set of fractional parts of the two numbers −3a and −5a3 will be {1/3, 2/3}.
On the other hand, since τ = −1/3 ∈ S \ S it follows that −2bτ has a fractional part of 1/3 or 2/3.
Back to Proof of Fund. Thm
Fundamental Theorem: For all pairs (a, b) of positive odd integers, the 2-adic integers can be written as
Z2 = aS 2bS = aS 2bS.
This is not true for arbitrary subsets of Z2.
Def A 2-adic integer z ∈ Z2 is of even order if ord(z) = 2i and of odd order if ord(z) = 2i+1.
Language Issue: All odd integers a, (posi-tive and nega(posi-tive) are of even order: in fact, ord(a) = 0 since the highest power of 2 which divides an odd integer a is 20.
Even integers may be either odd order or even order. By convention, 0 is considered both even order and odd order, and is the only such number.
It is easy to see
Lemma. Multiplication by an odd integer is ord-preserving.
The theorem is proved through a series of el-ementary lemmas (which extends to p-adics). Lemmas
(a) s is of even order for all s ∈ S.
(b) s − s0 is of even order for all s, s0 ∈ S Let a be a positive odd integer. Then,
(c) aσ is of even order for all σ ∈ S and ord(aσ) = ord(σ).
(d) aσ − aσ0 is of even order for all σ, σ0 ∈ S and ord(aσ − aσ0) = ord(σ − σ0).
(e) ord(bτ ) = ord(τ )
and each element in 2S is of odd order. (f ) ord(bτ − bτ0) = ord(τ − τ0)
Some Proofs
Proof: of (1) aS − 2bS = aS 2bS
We want to show that each difference is unique. Suppose aσ − 2bτ = aσ0 − 2bτ0.
Rearranging the terms gives a(σ − σ0) = b(2τ − 2τ0).
The left hand side has even order and the right hand side has odd order. The only number with both odd and even order is 0.
Proof: of (2) aS − 2bS = aS − 2bS.
One containment, aS − 2bS ⊃ aS − 2bS, is obvi-ous.
The other containment follows from
Lemma. Suppose as(n) − 2bt(n) → z with se-quences s(n), t(n) ∈ S (and a, b are positive odd integers). Then s(n) and t(n) each converge. The latter follows by rewriting the Cauchy dif-ference
as(n) − 2bt(n) − (as(m) − 2bt(m))
ord(x ± y) = min(ord(x), ord(y)).
Apply this to x = [as(n) − as(m)] which has even order and y = b[2(t(n) − 2(t(m))] which has odd order.
Some Questions
Question If u is a de Bruijn universally bad integer, is it divisible by 3 or φpk(4) for some p and k?
Question The integers 4k + 1 (and their mul-tiples) are universally bad but not de Bruijn universally bad. What other integers are uni-versally bad but not de Bruijn uniuni-versally bad? Question What is the (upper and lower) den-sity of the universally bad integers? Assuming the density exists, it is certainly more than 1/3. Question Given an a which has at least one b with which it is a good pair, are there infinitely many b’s with which it is a good pair? de Bruijn shows that this is true for a = 1.
References
N. G. de Bruijn, On bases for the set of inte-gers, Publ. Math. Debrecen 1 (1950), 232– 242.
N. G. de Bruijn, Some direct decompositions of the set of integers, Math. Comp. 18 (1964), 537–546.
Presentation based upon: Universally Bad In-tegers and the 2-Adics, S. Eigen, Y. Ito and V.S. Prasad. J. Num. Theory 107 (2004), pg