CHARACTERIZATION OF A CLASS OF PLANAR SELF-AFFINE TILE DIGIT SETS
LI-XIANG AN AND KA-SING LAU
Abstract. We call a finite setD ⊂Zsa(self-affine) tile digit setwith respect to an expanding integral matrixAif the self-affine setT(A,D) is a tile inRs. It has been a widely open problem to characterize the tile digit sets for a givenA. While there are substantial investigations onR, there is no result onRsother than the case where |detA|=pwith pa prime. In this paper, we make an initiation to study a basic case A=pI2 in R2. We characterize the tile digit sets by making use of the zeros of the mask polynomial of D associated with a tile criterion of Kenyon [K], together with a recent result of Iosevichet alon factorization of sets in Zp×Zp [IMP].
Contents
1. Introduction 1
2. Preliminaries 4
3. Directional projection of D 8
4. Decomposition of D 15
5. Proof of the main theorem 18
6. Remarks 24
References 27
1. Introduction
Let A be an s×s expanding matrix (i.e., all eigenvalues have moduli >1) with integral entries; let D={0=d0,d1,· · ·db−1} ⊂Zsbe a finite set, and call it a digit set. Theaffine pair(A,D) defines an iterated function system {ϕi}b−1i=0 withϕi(x) = A−1(x+di). It follows that there exists a unique compact set T :=T(A,D)⊂ Rs (self-affine set) satisfying the set-valued relation
AT =T +D. (1.1)
2010 Mathematics Subject Classification. Primary 11B75, 52C22; Secondary 11A63, 28A80 . Key words and phrases. Digit sets, direct summands, mask polynomials, modulus, prime num- ber, spectral sets, self-affine tiles, tile digit sets,zeros.
The research is supported in part by the HKRGC grant and the NNSF of China (no. 11371382 and 11601175).
1
It is well-known that when T has non-empty interior, then T is a translational tile (Bandt [B]), i.e., there exists a discrete set J such that
T +J =Rs and (T +t1)o∩(T +t2)o=∅, t1 ̸=t2 ∈ J.
We call such T a self-affine tile, and D a tile digit set with respect to A. In this case, #D=|detA|.
The self-affine tile has its origin in the study of finite automata and recurrent sets ([K], [Th], [R]). The development was strongly motivated by wavelet theory, fractal geometry and geometry of numbers ([AB1,2], [B], [BS], [CT], [GaY], [HL1,2], [KiL1,2], [LgW1-4], [LL], [RWY], [SW]), and the connection with the Fuglede’s conjecture on tiles and spectral sets ([F], [T], [Ko], [KoM], [FHL]). Despite the large literature on self-affine tiles, there is a fundamental question that is still widely open:
Question. For a given expanding integral matrix A, characterize the tile digit sets D of A, i.e., the digit sets D such that T(A,D) is a self-affine tile.
If D is a complete residue set modulus A, then T(A,D) is a self-affine tile [B].
Kenyon [K] proved that on R, the converse is also true if A = [p] is a prime;
it was extended to Rs with a mild assumption that D spans Rs [HL1] (see also [LgW3] for another more restricted condition). For the non-prime case, Lagarias and Wang [LgW3] enlarged the class of tile digit sets by generalizing a direct sum setup (Odlyzko [O]) to the class of product-form digit sets:
D=E0⊕Aℓ1E1· · · ⊕AℓkEk (mod Aℓk+1), where
E =E0⊕ E1· · · ⊕ Ek,
and E is a complete residue set modulusA. More recently Lai, Rao and one of the author further extended the above form to certainmodulo product-forms andhigher order modulo product-forms inR ([LR], [LLR1,2]). These classes cover all tile digit sets with respect to A = [pl] and [prq], where p, q are primes, and it is speculated that the higher order modulo product-form may cover all tile digit sets in R. The techniques used are based on the cyclotomic polynomials, and factorization of cyclic groups on R. As there is no suitable extension to higher dimension, we know very little about the structure of tile digit sets in higher dimensional spaces.
In this paper, our main purpose is to initiate a study of the tile digit sets D in R2 with respect to A = pI2, where p is prime and I2 is the identity matrix on R2; necessarily, #D =p2. For simplicity, we write T(p,D) :=T(pI2,D). The structure of tile digit set D with respect to pI2 has closed connection with the decomposition of sets in Z2p, which was studied in detail by Iosevich, Mayeli and Pakianatha [IMP].
We call a digit setW aZ2p-summandif there is a setE such thatW ⊕E is a complete residue set modulus pI2, and callE a complementary summand ofW. (In [IMP],W
2
is called a Z2p-tile, we use an alternative name to avoid confusion with the tile digit set.) Note that E is not uniquely defined (see Section 2). We let
FW ={E :E is a complementary summand of W}. Our main theorem is:
Theorem 1.1. Assume A = pI2, and the affine pair (A,D) is primitive. Then D is a tile digit set with respect to A if and only if there is an invertible matrix B ∈M2(Z) with |detB|= 1 and a Z2p-summand W ={wi}pi=0−1 such that
BD ≡
p∪−1
i=0
(wi+AkEi) (mod Ak+1), (1.2) for some k ≥ 0, where Ei ∈ FW. In particular, when k = 0, then the Ei’s are all identical, and D is a complete residue set modulus A.
Since W is a Z2p-summand, the wi’s are in different coset modulo A, hence the expression in (1.2) is a disjoint union. When allEi is the same, sayE, then the digit set
D ≡ W ⊕AkE (mod Ak+1),
which is a product-form. For the case p = 2 or 3, the characterization of the tile digit sets is particularly simple: for p= 2, they are complete residue sets with respect to 2I2; for p = 3, they are product-forms. These will be proved separately in the remark section.
The sufficiency of Theorem 1.1 is straight forward (Proposition 2.5). The main part is on the necessity. Forv ∈Z2\{0}, we define a projection ofDin the direction of v by
πv(D) ={⟨d,v⟩:d∈ D}, and let mD(ξ) = ∑
d∈De2πi⟨d,ξ⟩ be the mask polynomial of D. We will use the fol- lowing relation of πv(D) and the zeros of mD(ξ) throughout the paper (Proposition 3.2).
Proposition 1.2. Let (A,D) be as the above, then mD(A−(k+1)v) = 0 if and only if
πv(D)≡ {0 =η0, η1,· · · , ηp−1}+pk{0,1,· · ·, p−1} (mod pk+1), (1.3) where 0≤ηi ≤pk−1.
3
By using the Kenyon’s criterion ([K]) of tile digit sets through the zeros of mD(ξ) (Theorem 2.4), and by multiplying an invertible matrix B with |detB|= 1 to D if necessary, we have (Proposition 3.11)
πe1(D)≡ {0, α1,· · ·, αp−1}+pk1{0,1,· · · , p−1} (mod pk1+1), πe2(D)≡ {0, β1,· · · , βp−1}+pk2{0,1,· · · , p−1} (mod pk2+1),
for some k1, k2 ≥ 0, where e1 = (1,0)t,e2 = (0,1)t. The case that k1 = k2 = 0 is equivalent to D being a complete residue set (Corollary 3.10). Our main effort is to consider the case that k1, k2 ≥ 1 and prove that k1 = k2. For this we use the projection on e1 to decompose the tile digit set D (Lemma 4.1, Definition 4.2), and show that in this decomposition, the second coordinate match up with the projection on e2 (Section 5). This decomposition eventually yields the expression (1.2) in Theorem 1.1.
For the organization of the paper, in Section 2, we summarize some of the direct sum decomposition of Z2p in [IMP], and prove the sufficiency of Theorem 1.1. In Section 3, we consider the πv(D), and lay down the connection with the zeros of the mask polynomial of D. We carry out the decomposition of D in Section 4, and prove the necessity of the theorem in Section 5. In Section 6, we prove the stronger conclusion for the case p = 2,3, and also give some remarks and outlooks of the related problems.
2. Preliminaries
In this section, we will introduce some standard notations and basic facts that are needed. Letpbe a prime number, and letZ2p be the quotient group ofZ2p :=Z2/pZ2. Definition 2.1. We say that W ⊂ Z2p is a Z2p-summand if there exists E such that W ⊕ E = Z2p. Note that E is also a Z2p-summand, and call it a complementary summand of W.
It follows that a non-trivial Z2p-summand has cardinality p. If we let W′,E′ ⊂Z2 be representations of the above W and E respectively, then we have
W′⊕ E′⊕pZ2 =Z2.
Hence W′ tiles Z2 with a tiling set E′ ⊕pZ2. With a slight abuse of notation for convenience, we use the same W to mean a set in the quotient group Z2p, or its representation in Z2.
Let W be aZ2p-summand with #W =p, and let mW(ξ) =∑
w∈We2πi⟨w,ξ⟩ be the mask function of W. Let Z(g) denote the zeros of a function g. It follows that
Z(mW)∪ Z(mE) =Z(mZ2p) =p−1Z2\Z2. (2.1)
4
Therefore there is non-zero v ∈ Z2 \pZ2 such that mW(p−1v) = 0. According to [IMP, Section 5], a set W is a Z2p-summand if and only if W is a graph, i.e.,
W ={xe1+f(x)e2 :x∈Zp}, (2.2) where e1,e2 is a basis for Z2p and f : Zp → Zp is a function. The basis can be chosen from a zero v of mW(p−1ξ) in two cases: (i) if ⟨v,v⟩ ̸∈ pZ, let e1 = v, e2
orthogonal to e1; (ii) if ⟨v,v⟩ ∈pZ, let e2 =v and let e1 be an element off the line generated bye2 and ⟨e1,e2⟩ ∈1 +pZ. Also it follows from the expression ofW that {⟨v,w⟩:w ∈ W} ≡ {0,1,· · · , p−1}(modp).Hence we can obtain a complementary summand by takingE ⊂Z2pto be a line passes through the origin and perpendiculars to v. The following example is an illustration of this construction, and shows that such E is not unique.
Example 2.2. Let W ={ 0,
[ 3 1
] ,
[ 4 0
] ,
[ 4 3
] ,
[ 4 1
]}, E1 ={0,1,2,3,4}[
1 1
]
, E2 ={0,1,2,3,4}[
1 3
]
Then W is aZ25-summand, and bothE1,E2 are complementary summands ofW, and FW ={E1,E2}.
Proof: We observe that mW(5−1(1,4)t) = 0. Let e1 = (1,4)t, and let e2 = (1,1)t (as in case (i) of (2.2)). Then {e1,e2}forms an orthogonal basis for Z25, and
W ={xe1+f(x)e2 :x∈ {0,1,2,3,4}}, with
(f(0), f(1), f(2), f(3), f(4)) = (0,2,2,1,0).
Therefore by the aboveWis aZ25-summand. Alternatively, note thatmW(5−1(2,1)t) = 0 and ⟨(2,1)t,(2,1)t⟩= 5, as in case (ii) of (2.2), we can lete2 = (2,1)t,e1 = (3,0)t, and express W in terms of this basis.
LetE1andE2be as given, they are lines through the origin, perpendicular to (1,4)t and (2,1)t respectively, they are complementary summands of W. To show that FW ={E1,E2}, we let E ∈ FW, then W ⊕ E =Z25 and E ={ye′1+g(y)e′2 :y∈Z5}. To determine e′1 and e′2, we note that Z = {(1,0)t,(1,1)t,(1,2)t,(0,1)t} is a zero set of mE(5−1ξ) (by (2.1)). Similar to the above, we can choose e′1 = (1,0)t and e′2 = (0,1)t to be a basis of Z25, and for each v ∈Z,
{⟨ye′1+g(y)e′2,v⟩:y∈Z5} ≡ {0,1,2,3,4} (mod 5).
From 0 ∈ E and E ∩ W = ∅, we have g(0) = 0 and g(4) = 2 or 4. By a direct check, we have for y = 1,2,3, g(y) = 3y (when g(4) = 2) or y (when g(4) = 4).
This implies either E = E1 or E2. We can check the three other elements in Z (for e′1 = (1,2)t, use (ii) in (2.2), let e′2 = (1,0)t), and they end up to be the sameE1 or E2. 2
5
Remark 1. Note that E1 is also a Z25-summand with W as a complementary summand; butW is not a line inZ25. Therefore the above method of orthogonal line does not determine the class FE1 of all complementary summands.
Remark 2. For anyZ2p-summand, there are at mostpp−1complementary summand- s. In fact, if we letW =
{ 0,
[ 1 0
] ,
[ 2 0
] ,· · ·[
p−1 0
]}
, then it has pp−1 complemen- tary summands of the form E =
{ 0,
[ n1 1
] ,
[ n2 2
] ,· · ·[
np−1 p−1
]}
, 0≤ni ≤p−1.
Let A∈M2(Z), ford,d′ ∈Z2, we write d≡d′ (mod A) to mean d−d′ ∈AZ2. For D,D′ ⊂ Z2, we define D ≡ D′(mod A) similarly; note that in this case D and D′ may have different cardinalities. We also use Dres (or D(mod A)) to mean the residue set of DinZ2∩A[0,1)2. For the special caseA=pI2, we will write (modp) to replace (mod pI2).
For an affine pair (A,D), there is a smallestA-invariant sublatticeZ(A,D) of Zs that contains the difference set D − D. We call a digit setD primitiveif Z(A,D) = Zs, and also call the associated tile T(A,D) is primitive tile. The investigation in [LgW3] shows that every self-affine tile is a primitive tile in essence: there is an invertible matrix B and an affine pair (A,e De) such that Z(A,e De) =Zs and
T(A,D) = BT(A,e De).
Without loss of generality, we make the assumption thatD is primitive with respect toA throughout the paper. The following simple fact will be used frequently in the sequel.
Lemma 2.3. Let A =pI2 and D a primitive tile digit set with respect to A. Then for any invertible matrix B with |detB|= 1, BD is again a primitive tile digit set with respect to A.
Proof. It follows from a direct check of (1.1) and A and B are commutative. 2 For D ≡ D′ (mod pk), and d∈ D, write d=d′+pkn with d′ ∈ D′,n∈Z2, then
mD(p−kv) = ∑
d∈De2πi⟨d′,p−kv⟩e2πi⟨n,v⟩ =mD′(p−kv), v ∈Z2. (2.3) Similarly, if v≡v′(mod pk), then mD(p−kv) = mD(p−kv′).
For any integral expanding matrix A, let T := T(A,D) be a self-affine set in R2, and let χT be the characteristic function of T. It follows from the functional equation AT =T +D that
χT(x) = ∑
d∈D
χT(Ax−d), x∈R2.
6
By taking Fourier transform of χT formally, we have b
χT(ξ) =mD((AT)−1ξ)χbA−1T(ξ) =
∏∞ k=1
mD((AT)−kξ).
The following theorem, due to Kenyon [K], is a basic criterion for a digit set to define a self-affine tile.
Theorem 2.4. (Kenyon [K]) The self-affine set T(A,D) is a tile if and only if for any 0̸=v ∈Zs, there exists an integer k≥1 such that mD((AT)−kv) = 0.
Let A=pI2, then the condition in the theorem reduces to : for any 0̸=v ∈Z2, there exists a k ≥ 0 such that mD(p−(k+1)v) = 0 (equivalently mp−(k+1)D(v) = 0).
The follow proposition is the sufficiency of Theorem 1.1.
Proposition 2.5. Let p be a prime integer. Suppose W ={0=w0,w1,· · · ,wp−1} is a Z2p-summand, let
D ≡ ∪p−1
i=0(wi+pkEi) (mod pk+1)
with Ei ∈ FW. Then T(p,D) is a self-affine tile if k ≥ 1, or if k = 0, and the Ei’s are identical.
Proof. Fork= 0 in the second case, letE denote the identicalEi’s, thenD=W ⊕E. Hence D is a complete residue set modulus pI2, and is a tile digit set.
To prove the case k ≥ 1, let Ej ∈ FW, then Dj :=W ⊕ Ej is a complete residue set modulus pI2, i.e., (Dj)res ={0,· · · , p−1} × {0,· · · , p−1}. Then by (2.3)
m(Dj)res(p−1v) =mDj(p−1v) = mW(p−1v)mEj(p−1v), ∀ v ∈Z2. Note that Z(m(Dj)res) =p−1Z2\Z2, therefore for anyEj ∈ FW,
Z(mW)∪ Z(mEj) = p−1Z2\Z2. (2.4) Now, we check the Kenyon criterion (Theorem 2.4) holds on D. In view of (2.3), we can assume without loss of generality, that D = ∪p−1
j=0(wj +pkEj). For any v ∈ Z2 \ {0}, let n ≥ 0 be such that v′ = p−nv ∈ Z2\pZ2. Then by (2.4), either p−1v′ ∈ Z(mW) or ∈ Z(mEj) for all Ej ∈ FW. In the first case,p−(n+1)v ∈ Z(mW), then
mD(p−(n+1)v) = ∑p−1
j=0e2πi⟨wj, p−1v′⟩mpkEj(p−1v′)
= ∑p−1
j=0e2πi⟨wj, p−1v′⟩mEj(pk−1v′)
= p∑p−1
j=0e2πi⟨wj, p−1v′⟩ =pmW(p−1v′) = 0
7
(we need k ≥ 1 in the first identity of the last line). In the second case, p−1v′ ∈ Z(mEj) for all Ej ∈ FW. Then p−(n+k+1)v ∈ Z(mpkEj), and hence
mD(p−(n+k+1)v) = ∑p−1
j=0e2πi⟨wj,p−(n+k+1)v⟩mpkEj(p−(n+k+1)v) = 0.
By Theorem 2.4, T(p,D) is a self-affine tile. 2
3. Directional projection of D
In this section, we will set up some preparation work for the proof of the necessity of Theorem 1.1. We assume that0∈ D, andDis primitive with respect toA=pI2. For a digit set S ⊂Z+, denote PS(x) = ∑
s∈Sxs.
LetZ[x] (Z+[x]) denote the set of polynomials with integer (non-negative integer, respectively) coefficients. We use a |b to denotea divides b, and a-b means a does not divide b; the notations apply to both integers and polynomials. Let Φd(x) be the d-th cyclotomic polynomial, the minimal polynomial of the primitive d-th root of unity, i.e., Φd(e2πi/d) = 0. Then for a prime p, Φp(x) = 1 +x+· · ·+xp−1, and Φps(x) = 1 +xps−1 +· · ·+xps−1(p−1).
Lemma 3.1. Let p be a prime. Suppose f(x)∈ Z+[x] and has degree less than ps. If Φps(x)|f(x), then there exists a polynomialQ(x)∈Z+[x] such that
f(x) = Φps(x)Q(x).
Proof. We only need to show that Q(x) has non-negative coefficients. Since by assumption f(x) has degree ≤ ps−1 and Φps(x) has degree ps−1(p−1), Q(x) has degree ≤ps−1−1. Write
f(x) =
p∑s−1
n=0
anxn, Q(x) =
ps∑−1−1
n=0
bnxn, an∈Z+, bn ∈Z. Then
p∑s−1
n=0
anxn = Φps(x)Q(x) =
p−1
∑
i=0 ps−1∑−1
n=0
bnxn+ips−1.
Note that for any 0 ≤n1, n2 ≤ps−1−1, 0≤i1, i2 ≤p−1, we have n1+i1ps−1 =n2+i2ps−1 ⇐⇒ n1 =n2, i1 =i2.
It follows that an+ips−1 =bn∈Z+. This proves Q(x)∈Z+[x]. 2 Assume that D ⊂Z2 (and 0∈ D by convention). For anyv ∈R2, denote
πv(D) := {⟨d,v⟩:d∈ D},
8
and call it the projection of D in the v-direction, or in short, thev-projection of D. If v =e1 = (1,0)t, then we denote πv(D) by π1(D).
Proposition 3.2. Let D ⊂ Z2 be a primitive digit set with respect to pI2, then mD(p−(k+1)v) = 0 if and only if
πv(D)≡ {0 =η0, η1,· · ·, ηs−1}+pk{0,1,· · ·, p−1}(mod pk+1), (3.1)
where s ≥1 and 0≤ηi ≤pk−1.
Remark. Note that the ⟨d,v⟩ can be positive or negative, but the entries on the right side are always nonnegative; also note that the ηi’s can be identical.
Proof. Let A ≡πv(D) (mod pk+1) with A ⊂ {0,1,· · · , pk+1−1}. Then PA(e2πi/pk+1) = ∑
d∈A
e2πip−(k+1)d = ∑
d∈πv(D)
e2πip−(k+1)d=mD(p−(k+1)v) = 0.
This implies Φp(k+1)(x)|PA(x). By Proposition 3.1, there is a polynomial Q(x) ∈ Z+[x] such that
PA(x) = Φpk+1(x)Q(x).
Since s := Q(1) is a positive integer, we can write Q(x) = ∑s−1
i=0xηi, η0 = 0 (ηi
may be repeated); furthermore 0 ≤ ηi ≤ pk−1 (as PA has degree ≤ pk+1−1 and Φpk+1(x) =∑p−1
j=0xjpk). We conclude that
A={0, η1,· · · , ηs−1}+pk{0,1,· · · , p−1}.
For the sufficiency, we observe that the expression of πv(D) in (3.1) implies that PA(x) = Φpk+1(x)Q(x) where Q(x) =∑s−1
i=0xηi. It follows that p−(k+1) is a root of
PA(x), so that p−(k+1)v is a root of mD(ξ). 2
Note that Proposition 3.2 implies that if the zero set of mD(ξ) is nonempty, then p|#D. Let v = (n, m)t be such that mD(p−(k+1)v) = 0, and without loss of generality, we assume that #D = ps where s is as in (3.1). Let B =
[ n m 0 1
] . Consider BD, it changes the first coordinate of D, but keep the second coordinate unchanged, and
π1(BD) = πv(D)≡ {0 =η0, η1,· · · , ηs−1}+pk{0,1,· · · , p−1}(mod pk+1).
Corollary 3.3. Let D ⊂Z2 be as in Proposition 3.2. Suppose there is v = (1, m)t such that mD(p−(k+1)v) = 0 for some k ≥1, then ηi ̸= 0 for somei.
9
Proof. Let B =
[ 1 m 0 1
]
be the matrix defined by v. By assumption, D is a primitive digit set with respect to pI2. Since detB = 1,BDis also a primitive digit set with respect to pI2. If ηi = 0 for alli, then we have
π1(BD)≡pk{0,1,· · · , p−1} (mod pk+1) with k ≥1.
Hence the first coordinate of BD is contained in pkZ. This implies thatBD cannot be a primitive digit set with respect to pI2, a contradiction. 2 By Proposition 3.2 and its proof, we obtain a preliminary decomposition of BD through its first coordinate.
Corollary 3.4. Let D ⊂ Z2 be as in Proposition 3.2, and let B be defined by v = (n, m)t as the above. Assume that mD(p−(k+1)v) = 0, k ≥0. Then BD admits a decomposition BD=∪s−1
i=1Di where Di ={[ ηi+pk+1n0
di,0
] ,
[ ηi+pk+pk+1n1
di,1
] ,· · · ,
[ ηi+pk(p−1) +pk+1np−1
di,p−1
]},
and mDi(p−(k+1)e1) = 0.
Let
V ={(n, m)t:n ∈Z\pZ, m ∈Z}, V1 ={(1, m)t :m∈Z}. Clearly, the condition in Kenyon’s criterion is equivalent toV∪Ve ⊂∪
k≥0Z(mp−(k+1)D) for A =pI2, whereVe ={(m, n)t: (n, m)t∈V}.
Lemma 3.5. For v ∈V, p−(k+1)v is a root of mD(ξ) if and only if p−(k+1)v1 is a root mD(ξ) for some v1 ∈V1.
Proof. We only need to prove the necessity. For v = (n, m)t ∈ V, since n and pk+1 are co-prime, there exist integers t1, t2 such that nt1 +pk+1t2 = m; hence p−(k+1)(0, nt1−m)∈Z2. For v1 = (1, t1)t,
mD(p−(k+1)nv1) = mD(p−(k+1)v+p−(k+1)(0, nt1−m)t)
= mD(p−(k+1)v)
= 0.
Since n is not a factor ofp, we see that p−(k+1)v1 is a root of mD(ξ) as well. 2 Hence to consider the zeros of mD(ξ), we can use V and V1 interchangeably. We define the following condition on mD(ξ) which fulfils part of the condition in the Kenyon criterion.
(∗) For every v ∈V1, there exists k > 0 such that mD(p−(k+1)v) = 0.
10
We use kv to denote the smallest k such that mD(p−(k+1)v) = 0. Note thatkv = 0 for all v ∈V1 is equivalent tomD(p−1V1) = 0. Let
C ={ 0,
[ 1 0
] ,· · ·,
[ p−1 0
]}.
ThenmC(p−1V1) = 0. In the following, we will discuss the relation ofmD(p−1V1) = 0 and C is a direct summand in Dres (recall that Dres is defined as D(mod p) ⊂ Z2 ∩p[0,1)2).
Proposition 3.6. Let D ⊂ Z2 be a primitive digit set with respect to pI2, and by translation, assume that 0 ∈ Dres has the largest number of repetition. Then the following are equivalent:
(i) kv = 0 for all v ∈V1 (or V);
(ii) D satisfies condition (∗), and C ⊂ Dres;
(iii) there are 0≤di ≤p−1 (they may be equal) such that Dres=
{ 0,
[ 0 d1
]· · ·[
0 ds−1
]}
+C. (3.2)
Remark. The condition on Dres has no essential significance. Without that, we change the Dres in (ii), (iii) by a shift of the digits.
Proof. (i) ⇒ (ii) We need only show the second part of (ii). Denote D0 :={d∈ D :⟨d,(1, m)t⟩ ≡0 (mod p) for some 0≤m≤p−1}. It is easy to check by definition that
(D0)res ={0} ∪ {[
x y
]∈ Dres :y ̸= 0}.
Let ℓdenote the number of repetition of 0∈ Dres, and is largest by assumption. As mD(p−1v) = 0, Proposition 3.2 implies that
πv(D)≡ {| {z }0,· · · ,0}
s
+{0,1,· · ·, p−1} (mod p).
Therefore, each 0 ≤m≤p−1 corresponding to exactly selements inD0: ℓ of them
≡0 (modp), and s−ℓ of them do not. Hence #D0 =ℓ+ (s−ℓ)p.LetD0c=D \ D0, then
π2((Dc0)res) = {0}, #Dc0 =sp−(ℓ+ (s−ℓ)p) = ℓ(p−1)>0. (3.3) If p = 2, then it follows from the first part of (3.3) that (Dc0)res =
{[ 1 0
]}
, and hence C =
{ 0,
[ 1 0
]} ⊂ Dres. For the casep > 2, we divide our consideration into two cases:
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Case 1. ℓ =s, thenD0 ≡ {0} (mod p). Hence
π2(D) = π2(D0)∪π2(Dc0)≡ {0} (mod p).
As ke1 = 0 (i.e., mD(p−1e1) = 0), we have
π1(D)(mod p) =π1(Dres) ={| {z }0,· · · ,0}
s
+{0,1,· · · , p−1}. The two identities imply Dres={|0,· · ·{z ,0}}
s
+C.
Case 2. ℓ < s. Assume the contrary, C * Dres, we prove that there is a d ∈ D such that D −d contains at least ℓ+ 1 elements ≡0 (mod p), a contradiction.
As C * Dres, there is an x0 ∈ {1,· · · , p−1} and (x0,0)t ̸∈ Dres. This together with (3.3) yield
(D0c)res ⊂{[
x 0
]
:x∈ {1,· · · , p−1} \ {x0}} .
As #Dc0 =ℓ(p−1) and p > 2 by assumption, the pigeon hole principle implies that there must be an x ∈ {1,· · · , p−1} such that D0c contains at least ℓ+ 1 elements
≡(x,0)t(mod p). Choose one from them, sayd. ThenD −dcontains at leastℓ+ 1 elements ≡0 (mod p), and completes the proof.
(ii) ⇒ (iii) We show that Dres has the form in (3.2) by induction on s. It is trivial for s = 1. Suppose the statement is true for k ≤ s−1. For the case s. By assumption, there is a subset D1 of D satisfies (D1)res =C. Denote D2 := D \ D1, then #D2 = (s−1)p. For any v ∈V1,
0 = mD(p−1v)
= mD1(p−1v) +mD2(p−1v)
= mC(p−1v) +mD2(p−1v)
= mD2(p−1v).
By translating ad∈ D2 with 0∈ D2−dand by induction, we have mD2−d(p−1v) = 0. By induction,
(D2)res={[ 0
d1
] ,· · ·[
0 ds−1
]}+C, for some di ∈ {0,1,· · · , p−1}, and
Dres = (D1)res∪(D2)res ={ 0,
[ 0 d1
] ,· · ·[
0 ds−1
]}+C.
(iii) ⇒(i) It follows from a direct check. 2
In Proposition 3.8, we show that the condition in Proposition 3.6(i) is implied by ke1 = 0 where e1 = (1,0)t. We need a lemma.
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Lemma 3.7. SupposeD satisfies condition (∗), and #D=p, then Dres=C if and only if ke1 = 0.
Proof. The necessity follows directly check of mC(p−1e1). For the sufficiency, we observe that if #D=p, then by condition (∗) and Proposition 3.2, we have for any v ∈V, there is k >0 such that
πv(D)≡pk{0,1,· · · , p−1} (mod pk+1).
Note that #πv(D) = p. In particular, for ke1 = 0, we have π1(D)≡ {0,1,· · · , p−1} (mod p).
If Dres ̸= C, then there is 0 ≤ j ≤ p−1 such that [ j
0
] ̸∈ Dres. Therefore there
is [ d1
d2
] ∈ D with d1 = j +pn and d2 ∈ Z\pZ. Then v = (d2,−d1)t ∈ V and
⟨v, [ d1
d2
]⟩= 0. Hence #πv(D)< p, a contradiction. Hence Dres=C. 2
Proposition 3.8. SupposeDwith#D=sp < p2 satisfies property (∗), andke1 = 0, then condition (i) in Proposition 3.6 is satisfied, i.e., kv = 0 for all v ∈V1.
Proof. In view of Proposition 3.6, it suffices to prove C ⊂ Dres here. When s = 1, the conclusion follows from Lemma 3.7. For s >1, we prove that if C *Dres, then there is a De satisfies the conditions in Lemma 3.7, andC *Deres. It is impossible.
By Proposition 3.6, if C * Dres, then there exists v0 = (1, m0) such that k0 :=
kv0 ≥ 1. Corollary 3.3 implies that there exist i such that η := η(vi 0) ̸= 0. Let B =
[ 1 m0
0 1
]
and considerBD, then we can use
π1(BD) =πv0(D)≡ {0 =η0, η1,· · · , ηs−1}+pk0{0,1,· · · , p−1} (mod pk0+1) to partition BD according to its first coordinate (see Corollary 3.4):
BD =
[ pk0 0
0 1
]D0′ ∪ · · · ∪([
ηs−1
0
] +
[ pk0 0
0 1
]Ds′−1
)
, (3.4)
and mD′
i(p−1e1) = 0. Let D′ be the union of those D′i with ηi = 0. Then #D′ =s′p with s′ < s. Then we have
(i) D′ satisfies condition (∗). Indeed for any v = (1, m) ∈ V1, let v′ = (1, m0 + pk0m)t. Because v′ ≡v0 (mod pk0), we have, fork ≤k0,
mD(p−kv′) = mD(p−kv0)̸= 0
(by the statement after (2.3)). This implies kv′ ≥k0. In view of Corollary 3.4 and the expression of Di′, we have mD′(p−(kv′−k0+1)v) = 0.
(ii) C *D′res. For otherwise, C ⊆ D′res. Consider B−1
[ pk0 0
0 1
]C ⊂ B−1
[ pk0 0
0 1
]D′res.
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