單形環面上的等譜問題

國 立 交 通 大 學 理 學 院 應 用 數 學 研 究 所

碩 士 論 文

Isospectral problem on simplicial tori

單形環面上的等譜問題

研究生：蔡志奇

Student: Chih-Chi Tsai

指導教授：康明軒

教授

Advisor: Ming-Hsuan Kang

中 華 民 國 一 百 零 三 年 一 月

March, 2014

Isospectral problem on simplicial tori

單形環面上的等譜問題

研 究 生：蔡志奇

Student: Chih-Chi Tsai

指導教授：康明軒 教授

Advisor: Professor Ming-Hsuan Kang

國 立 交 通 大 學 理 學 院 應 用 數 學 研 究 所

碩 士 論 文

A Thesis Submitted to Department of Applied Mathematics National

Chiao Tung University in partial Fulfillment of the Requirements for the

Degree of Master in Statistics

中 華 民 國 一 百 零 三 年 三 月

Isospectral problem on simplicial tori

Student: Chih-Chi Tsai

Advisor: Prof. Ming-Hsuan Kang

Department of Applied Mathematics

National Chiao Tung University

Abstract

The goal of this paper is to discuss the classical isospectral problem in graph theory for a particular type of graphs. We focus on finite regular simplicial tori arising from the Bruhat-Tits building of PGL3(Q1) and study which kinds of simplicial

單形環面上的等譜問題

國立交通大學應用數學研究所碩士班

摘要

這篇論文主要是討論特定圖型的等譜性這個古老的圖論問題。其中,

我們將目標放在從 PGL(

Q

)

的 Bruhat-Tits 結構得到的單形環面，

並從他們的 zeta 函數中所得的譜來討論哪些單形環面可以被他們的

譜所決定。

致謝

　　現在回想起來，讀數學系這件事真的是高中時想都沒想過的事。

當時高中指考完後結果非常不理想，自暴自棄的心情下選了高師大數

學系逃避，沒想到在那邊讀得不錯而重新找到了信心，研究所也順利

考到交大應數系所。雖然很可惜沒能應屆推上交大博班，再來也因為

比較現實的理由要開始考老師，學生生涯就此告一段落，不過這 6 年

來的體驗真的是學生生涯中最充實的 6 年。

　　這 6 年中，真的很感謝父母願意讓我走我想嘗試的路，感謝老師

們的鼓勵也謝謝同學們的友情支持，也非常謝謝康明軒老師願意那麼

仔細挑我論文的錯誤（挑到我自己都覺得不好意思了）。最後我想對

我自己說一句: ”恭喜！畢業了〜〜”

Contents

Abstract i

1. Introduction 1

2. Ihara zeta function 2

3. Field with one element and 1-adic field 3

4. The building of PGL3(Q1) 4

5. Simplicial tori and their isomorphism classes 5

6. Main Theorem 6

7. Geometric information encoded in the graph zeta function 9

1. Introduction

The graph zeta function is a geodesic counting function on a finite graph which was first introduced by Ihara [YI96]. Ihara’s original zeta function is defined on a discrete torsion-free co-compact subgroup Γ of PGL2(Qp) and later Serre [JPS03] reformulates it as a function on the graph obtained from the building of PGL2(Qp) quotient by Γ. The buildingB1of PGL2(Qp) is a (p + 1)-regular tree whose vertices are left PGL2(Zp)-cosets and the adjacency operator is the Hecke operator

A = PGL2(Zp) ( 1 0 0 p ) PGL2(Zp).

The group PGL2(Qp) acts on the treeB1 by left multiplication. For a torsion-free

discrete co-compact subgroup Γ of PGL2(Qp), the quotient Γ\B1is a (p+1)-regular

graph.

For PGLn(Qp), its buildingBn−1 is a (n− 1)-dimensional simplicial complex. Similar to PGL2(Qp), for a torsion-free discrete co-compact subgroup Γ of PGLn(Qp), Γ\Bn−1 is a finite complex.

One can also consider that the case that when p = 1, so that the residue field is the field with one element. In this case, the building of PGLn(Q1) becomes a single

apartment and the torsion-free discrete co-compact subgroup Γ is a free abelian group of rank n− 1. Especially, when n = 3, the building B2 of PGL3(Q1) is a

simplicial euclidean plane.

.. (0,0) . _(1,0) . (0,1) . (-1,-1)

Figure 1: The building of PGL3(Q1)

The 1-skeleton ofB2can be described as a Cayley graph onZ2with the generated

set S ={(±1, 0), (0, ±1), ±(1, 1)} shown in Figure 1.

The spectrum of a graph is the set of eigenvalues of the adjacency matrix and two graphs are isospectral if they have the same spectrum. Moreover, two graphs are isospectral if they have the same zeta function ( see section 3 ), so there are many non-isomorphic graphs with the same zeta function.

In this thesis, we would like to study if a finite quotient X ofB2(which is a torus)

can be uniquely determined by its graph zeta function ZX(u) (up to isomorphism). Let Kn be the collection of complete representatives of isomorphic classes of finite quotients ofB2with n vertices. Consider the two numbersT1(n) =|Kn| and T2(n) =|{ZX(u), X ∈ Kn}|.

By straightforward computation (using Matheamtica), we have the following re-sult n 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 T1(n) 1 2 3 2 3 3 5 4 4 3 8 4 5 6 9 4 8 5 10 T2(n) 1 2 2 2 3 3 4 4 4 3 7 4 5 6 8 4 8 5 9 n 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 T1(n) 8 7 5 15 7 8 9 13 6 14 7 15 10 10 10 20 8 T2(n) 8 7 5 14 7 8 9 12 6 14 7 14 10 10 10 19 8 n 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 T1(n) 11 12 20 9 18 9 17 16 13 9 28 12 17 14 20 10 22 T2(n) 11 12 19 9 18 9 16 16 13 9 27 12 17 14 19 10 22

From the above, we have the following conjectures onT1(n) andT2(n):

(1)T1(p) = [p+11₆ ], where p is an odd prime.

(2)T2(n) =T1(n)− δ where δ is equal to 1 if 4|n and equal to zero otherwise.

The rest of the thesis is organized as follows. In the Section two and three, we review some basic concepts in graph theory and introduce graph zeta functions. In Section four and five, we describe the group PGL3(Q1) and it’s building explicitly.

In the end, we prove (1) and a part of (2) of the conjecture. 2. Ihara zeta function

A graph X = (V, E) is an ordered pair where V is the set whose elements are called the vertices of X; E is a multi-subset of V × V , which elements are called

the oriented edges of X. Moreover, if (u, v) is an element of E, so is (v, u) and their multiplicities must be the same.

For a oriented edge e = (v1, v2), v1 is called the starting point of e, denoted by o(e); v1 is called the end point of e, denoted by t(e). A walk C on X is a sequence

of oriented edges

C = (e1, e2, . . . , en)

satisfying t(ei) = o(ei+1) for all i = 1 to n− 1. Here n is called the length of C, denoted by l(C). C has a backtrack if o(ei) = t(ei+1) for some i. C is closed if

o(e1) = t(en). When C is closed, we say C has a tail if t(e1) = o(en). Two closed walks are equivalent if one can be obtained from the other by cyclically permuting its sequence of edges. Denote the equivalence class of C by [C]. We denote by Cj the multiple of the closed walk C with j times. A closed walk is called prime if it is non-backtracking, tailless and is not a multiple of other shorter closed walk.

We denote d(v) by the number of oriented edges e with o(e) = v. Furthermore, we say X is regular if d(v) are the same for all v∈ V .

The adjacency matrix A of X is a matrix whose rows and columns are indexing by vertices. The (v, v′)-entry of A is the number of oriented edges from v to v′. Note that the trace of An _{is the number of closed walks of X of length n.}

Definition 2.1. The Ihara zeta function for X is defined to be the following function

of complex number u, with|u| sufficiently small: ZX(u) =

∏

[P ]

(1− uℓ(P ))−1

where the product is over all primes [P ] in X.

Based on Ihara’s work [YI96], Bass [BH92] prove the following theorem .

Theorem 2.1. Let X be a finite (q + 1)-regular graph. The following identities

hold ZX(u) = exp( ∞ ∑ n=1 Nn n u n_{) =} 1 (1− u2_{)|E|/2−|V |det(I− Au + qu}2₎,

where Nn is the number of non-backtracking and tailless cycles of length n; |V | is

the number of vertices of X; |E|/2 is the number of non-oriented edges of X.

Corollary 2.1. The following features of the graph X are uniquely determined by

its zeta function:

(1) The number of vertices|V |.

(2) The number of closed walks of length n in X for all n.

Proof. (1) is followed by that the degree of Z_X−1(u) =|E| = (q + 1)|V |. For (2), the number of the closed walks of X of length n equals to the trace of An_{, which is the} sum of n-th power of eigenvalues of A. On the other hand, the spectrum of A can be determined by the zeta function from the above theorem.

 3. Field with one element and 1-adic field

In the coming two sections, we call the definition of the field with one element

F1 and the building of PGLn over 1-padic field from [DK13]. The field with one element, denoted by F1 = {¯π}, is a suggestive name for an object that should

behave similarly to a finite field with a single element, if such a field could exist. The only operator on F1 is multiplication so that ¯π· ¯π = ¯π.

Now we consider the 1-adic fieldQ1={πi, i∈ Z} so that F1 is the residue field

ofQ1 and the ring of integer forQ1isZ1={πi, i≥ 0}. Since there is no addition,

the n-dimension space overQ1 is defined as

Qn 1 =Q1 ⨿ Q1 ⨿ . . .⨿Q1={πji, j = 1, 2, . . . , n, i∈ Z}. The group of automorphisms ofQn

1, denoted by GLn(Q1), consists of all

bijec-tions f onQn

1 so that f (πx) = πf (x),∀x ∈ Q

n

1. Moreover, the group PGLn(Q1) is

the quotient of GLn(Q1) by its center.

Theorem 3.1. GLn(Q1) ∼=Zno Sn, where the semi-direct product is given by the

natural permutation of Sn on n-coordinates ofZn.

Proof. For f ∈ GL(Qn

1), write f (π0j) = π aσ(j)

σ(j) for some σ ∈ Sn, aσ(j) ∈ Z and

j = 1, 2, . . . , n. Note that f is uniquely determined by the permutation σ and n-tuple of integers aσ(j).

Let ϕ : GLn(Q1)→ Zno Sn given by f7→ ((a1, a2, . . . , an), σ) which is bijective. Now for f1, f2∈ GLn(Q1), f1◦ f2(πj0) = f1(π b_σ2(j) σ2(j)) = π b_σ2(j)f 1(πσ02(j)) = π a_σ1σ2(j)+b_σ2(j) σ1σ2(j) ,

so ϕ(f1◦ f2) = ( (a1+ bσ₁−1(1), a2+ bσ₁−1(2), . . . , an+ bσ−1₁ (n), σ1σ2 ) .

On the other hand,

ϕ(f1)· ϕ(f2) = ((a1, a2, . . . , an), σ1)· ((b1, b2, . . . , bn), σ2)

= ((a1+ bσ₁−1(1), a2+ bσ₁−1(2), . . . , an+ bσ−1₁ (n)), σ1σ2).

Therefore, ϕ is a group isomorphism. 

Immediately, we have

Corollary 3.1. PGLn(Q1) ∼=Zn−1o Sn.

Remark: In the theory of the field with one element, PGLn(F1) is the group Sn, which is the Weyl group of GL(Fp). Here a similar phenomenon is occurred, so that PGLn(Q1) is isomorphic to the affine Weyl group of PGLn(Qp). [KLW10]

4. The building of PGL3(Q1)

LetZ1 be the ring of integer ofQ1. A lattice L of rank 3 in Q31 is Z1-invariant

subset so thatQ1L =Q31. This implies that the lattice L ={πji, j = 1, 2, 3, i≥ aj} for some unique a1, a2, a3 ∈ Z. Denote L by (a1, a2, a3), then we identify lattices

inQ3 1 withZ

3_.

The equivalence classes of L is

[L] ={αL, α ∈ Q1} = {(a1+ k, a2+ k, a3+ k), k∈ Z}

and denoted by [a1, a2, a3], which is an element inZ3/Z(1, 1, 1).

The building B2 of PGL3(Q1) is a 2-dimensional simplicial complex as follows.

As an abstract complex, its vertices are equivalence classes of lattices [L]; three vertices [L0], [L1], [L2] form a 2-simplex if there is a representative

L0⊇ L1⊇ L2⊇ πL0.

( The detail can be seen in [BH92], [KLW10] and [JPS03]. )

Moreover, for L = (a1, a2, a3), the adjacency vertices of [L] are{[L′] with L′ =

(a1+ b1, a2+ b2, a3+ b3) where bi ∈ {0, 1} and ∑

bi= 1 or 2. We define the i-th adjacency operator (which index is labelled by vertices of the building).

Ai[L][L′] =      1 if [L] = [a1, a2, a3] and [L′] = [a1+ b1, a2+ b2, a3+ b3] for some bi∈ {0, 1}, ∑ bi= i 0 otherwise.

Note that A1+ A2 is the adjacency operator of the underlying graph (1-skeleton)

ofB2. Observe that the setZ3/Z(1, 1, 1) has a canonical additive structure, which

is a free abelian group generated by e1= [1, 0, 0] and e2= [0, 1, 0]. In this case, the

vertices of B2 are elements in Z2 =Ze1⊕ Ze2 and the underlying graph of B2 is

the Cayley graph onZ2 _{with the generating set S ={(±1, 0), (0, ±1), ±(1, 1)}.}

The underlying space of the whole buildingB2isZ2⊗ R = R2endowed an inner

product characterized by ⟨ei, ej⟩ = { 1 if i = j −1 2 otherwise.

Moreover, PGL3(Q1) ∼=Z2oS3acts onB2as isometries as follows. With respect

to the basis e1 and e2, the subgroup S3is generated by the reflection τ = [0 11 0] and

the rotation σ =[−1 1_{−1 0}]. The groupZ2_{consists of translations so that for (a, b)∈ Z}2

and (x, y)∈ R2_{, (a, b) maps (x, y) to (x + a, y + b). Especially, the geometric ofB} 2

is relatively to Figure 1.

5. Simplicial tori and their isomorphism classes Let Γ be a finite index subgroup ofZ2_{, so that the quotient X}

Γ ofB2 by Γ is a

simplicial torus which is locally isometric toB2.

Now, the vertices of XΓare Γ−cosets in Z2and its 1-skeleton is the Cayley group

onZ2_{/Γ with the same generating set S modulo Γ. By abuse of notation, we still}

denote S modulo Γ by S.

Two simplicial tori XΓ and XΓ′ are called isomorphic, if there is an isometric

simplicial isomorphism ρ between them. It is clear that the group of automorphisms of XΓ acts transitively on vertices, so we may assume that ρ maps the vertex Γ to

the vertex Γ′. On the other hand, sinceB2 is simply connected, there is a unique

lifting ˜ρ so that ˜ρ(0, 0) = (0, 0) and the following diagram commute.

.. B..2 B2 XΓ . XΓ′ . mod Γ . ρ˜ . mod Γ′ . ρ

Since the two quotient maps and ρ are locally isometric simplicial maps so is ˜ρ. We

conclude that ˜ρ is a linear isometry onB2which maps Γ to Γ′. All linear isometries

on B2 forms the dihedral group D6 with the center Z(D6) = {I2,−I2}. We can

factor D6 as a product of S3 and Z(D6) where S3 is the symmetric group on 3

letters consisting the following elements

{I2, [0₋₁ 1−1 ] ,[−1 1_{−1 0}],[₀1−1₋₁],[−1 0_{−1 1}], [0 1 1 0]}.

Since Z(D6) fixes any translation subgroup Γ, we conclude that

Theorem 5.1. Two simplicial tori XΓ and XΓ′ are isomorphic if and only if g(Γ) = Γ′ for some g∈ S3.

Now we are able to compute the number of isomorphic classes of simplicial tori of size n. Let Λn be set of all index n subgroup ofZ2, then the group S3 defined

above acts on Λn canonically. From the above theorem, we have

T1(n) = # of isomorphic classes of finite quotients of B2 with n vertices.

= # S3-orbit of Λn. Let τ = [0 1 1 0] be a reflection and σ = [_{−1 1} −1 0 ]

be a rotation in S3. Note that there

are three reflections in τ conjugate to S3and two nontrivial rotations conjugate to σ. By Burnside’s lemma (1) # S3-orbit of Λn= 1 6(|Λn| + 2|Λ σ n| + 3|Λ τ n|). Here Λg n is a subset of Λn fixed by g.

6. Main Theorem

Recall the conjecture in Section one: let Kn be the collection of complete repre-sentatives of isomorphic classes of finite quotients of Cayley(Z2_{, S) with n vertices.}

LetT1(n) =|Kn| and T2(n) =|ZX(u), X ∈ Kn|, then (1)T1(p) = [n+11₆ ] where p is an odd prime.

(2)T2(n) =T1(n)− δ , where δ is equal to 1 if 4|n and equal to zero otherwise.

Note that every group N in Λn contains (nZ)2. Denote by Λcycn the set of all index n subgroups N ofZ2_{with N /(nZ)}2_{being cyclic. Furthermore, S}

3maps Λcycn onto Λcyc

n . For g ∈ S3, denote by Λcyc,gn the subset of Λcycn fixed by g. Then, we have the following lemma.

Lemma 6.1. For a prime p,|Λpk| =

[k 2] ∑ t=0 |Λcyc pk−2t|.

Proof. For m|n, let

Λn,m={N ∈ Λn, N /(nZ)2∼=Z/mZ × Z/(n/m)Z}.

It is clear that every element in Λn contained in some Λn,m and such m is unique provided m < n/m. Therefore, Λn = ⨿ m|n,m2_≤n Λn,m. Especially, when n = pk_{, Λ} pk= [k 2] ⨿ t=0

Λpk_,pt. On the other hand, there is a bijective

map ϕ from Λpk_,pt to Λ_pk_−2t,0= Λcyc_pk_−2t by ϕ(N ) = p−tN , hence

|Λpk| = [k 2] ∑ t=0 |Λcyc pk−2t|. 

Corollary 6.1. For a prime p,|Λg pk| = [k 2] ∑ t=0 |Λcyc,g pk−2t| for any g ∈ S3. To compute|Λcyc,g

pk−2t|, we recall the well-known Hesnel’s lemma.

Theorem 6.1 (Hensel’s Lemma). Let f (x) be a polynomial over Z.

(1) If x0 is a solution of f (x)≡ 0 mod pk such that f′(x0) ̸≡ 0 mod p, then there is a unique b ∈ {0, 1, 2, . . . , p − 1} such that x0+ pkb is a solution of f (x)≡ 0 mod pk+1_.

(2) If x1 is a solution of f (x)≡ 0 mod pk and f′(x1)≡ 0 mod p, then f(x1+ apk_{)≡ 0 mod p}k+1_{for all a∈ {0, 1, . . . , p − 1} if and only if f(x}

1)≡ 0 mod pk+1_.

The solution in (1) of the above theorem is called a non-singular solution; the solution in (2) is called a singular solution.

Recall that τ = [0 1 1 0] and σ = [_{−1 1} −1 0 ] are generators of S3.

Lemma 6.2. For all positive integer n,

(1) |Λcyc,τ

(2) |Λcyc,σ

n | = |{s ∈ Z/nZ, s2+ s + 1≡ 0 mod n}|.

Proof. Let Xcyc

n = { all cyclic subgroups of order n in (Z/nZ)2}, then there is a bijective map

η : Λcyc_n → X_ncyc by η(N) = N/(nZ)2_.

Moreover, the action of S3 on Λcycn induces an action of S3 on Xncyc given by

g(η(N )) = η(gN ) for all g∈ S3. Let Xncyc,g be the subset of Xncyc fixed by g, then

|Λcyc,g

n | = |Xncyc,g|. For G∈ Xcyc

n , G is cyclic of order n and it is generated by some (x, y)∈ (Z/nZ)

2_. If G is fixed by τ , then [ 0 1 1 0 ] [ x y ] = [ y x ] = [ sx sy ] mod n for some s∈ Z. It implies that y = s2_{y, so s}2_{≡ 1 mod n.}

Since (x, y) = (x, sx) is of order n, x has to be coprime to n. Therefore,

G =⟨(x, sx)⟩ = ⟨(1, s)⟩.

Conversely, for s satisfy s2_{≡ 1 mod n, τ fix ⟨(1, s)⟩. We conclude that for each}

solution of s2_{≡ 1 mod n, there is a unique group in X}cyc

n generated by (1, s) fixed by τ and all groups in Xcyc

n fixed by τ come from this manner. If G is fixed by σ, then [ −1 1 −1 0 ] [ x y ] = [ −x + y −x ] = [ sx sy ] mod n

for some s∈ Z. It implies that s2_{y =−(s + 1)y and s}2_{+ s + 1≡ 0 mod n.}

Since (x, y) = (−sy, y) is of order n, y has to be coprime to n. Therefore, G =⟨(−sy, y)⟩ = ⟨(−s, 1)⟩.

Conversely, for s satisfy s2_{+ s + 1≡ 0 mod n, σ fix ⟨(−s, 1)⟩. We conclude that}

for each solution of s2_{+ s + 1≡ 0 mod n, there is a unique group in X}cyc

n generated by (−s, 1) fixed by σ and all groups in Xcyc

n fixed by σ come from this manner. 

Lemma 6.3. For a prime p,

(1) |Λcyc pk | = pk+ pk−1. (2) |Λcyc,τ 2k | =      1 if k = 1 2 if k = 2 4 if k≥ 3 and |Λcyc,τ pk | = 2 if p ̸= 2. (3) |Λcyc,σ 3k | = { 1 if k = 1 0 if k≥ 2 and |Λ cyc,σ pk | = { 2 if p≡ 1 mod 3 0 if p≡ 2 mod 3.

Proof. First, for G ∈ Λcyc_pk , since G is cyclic, G = ⟨(x, y)⟩. Suppose x is coprime

to pk_{, then G =⟨(1, z)⟩ with z = x}−1_{y. In this case, G is uniquely determined by}

z and z runs through all elements inZ/pk_{Z. Therefore, we have p}k _{groups of this} type in Λcyc

pk . If x is not coprime to pk, since G is cyclic, y has to be coprime to n

and G =⟨(z, 1)⟩ with z = y−1_{x. In this case, G is uniquely determined by z and}

z runs through all elements inZ/pk_{Z not coprime to p}k_{. Therefore, we have p}k−1 groups of this type in Λcyc

pk . Hence|Λ

cyc pk| = p

k_{+ p}k_−1.

By direct computation, we have

A2={1 mod 2}, A22 ={±1 mod 22} and A₂3 ={±1, ±1 + 22 mod 23}.

Claim A2k={±1, ±1 + 2k−1 mod 2k} for k ≥ 3 which means |A2k| = 4 for k ≥ 3.

Assume A2k−1 ={±1, ±1 + 2k−2 mod 2k−1}. Note that all solutions in A2k−1

are singular solution of s2_{− 1 ≡ 0 mod 2}k−1 _and {

s2_{− 1 ≡ 0 mod 2}k _{, if s =±1}

s2− 1 ≡ 2k−1̸≡ 0 mod 2k , ifs =±1 + 2k−2.

By Theorem 6.1, we have A2k={±1, ±1+2k−1 mod 2k} and the claim is followed

by induction.

For p̸= 2, it is clear that Ap ={±1 mod p}. Since ±1 are non-singular solu-tions, by Theorem 6.1,|Apk| = |Ap| = 2, ∀k ∈ N.

Next, let us compute the cardinality of Bpk={s ∈ Z_pk, s2+ s + 1≡ 0 mod pk}.

For p = 2, B2 is empty and so is B2k for all k.

For p = 3, B3={1 mod 3}, B32 is empty and so is B₃k for all k≥ 2.

For p̸= 2 or 3,

Bp={s ∈ Zp, 4s2+ 4s + 4≡ 0 mod p} ={s ∈ Zp, (2s + 1)2≡ −3 mod p}

which cardinality is given by the Legendre symbol (−3_p ) + 1. Recall that (−3_p ) = (−1_p )(3_p) and (−1 p ) = { 1 if p≡ 1 mod 4 −1 if p ≡ 3 mod 4 and ( 3 p) = { 1 if p≡ ±1 mod 12 −1 if p ≡ ±5 mod 12, which implies |Bp| = 1 + (−3 p ) = { 2 if p≡ 1 mod 3 0 if p≡ 2 mod 3.

Note that when s2_{+ s + 1≡ 0 has two solutions, these two solutions are}

non-singular. By Theorem 6.1, we have|Bpk| = |Bp| for all p ̸= 2, 3 and all k. Together with Lemma 6.2, we complete the proof.

 Combining Corollary 6.1 and Lemma 6.3, we have

Theorem 6.2. For a prime p,

(1) |Λpk| = k ∑ t=0 pt (2) |Λτ pk| = { 2k− 1 if p = 2 k + 1 if p̸= 2 (3) |Λσ pk| =      1 if p = 3 k + 1 if p≡ 1 mod 3 0 if p≡ 2 mod 3 .

From Theorem 6.2, we can computeT1(p), where p is an odd prime. Since T1(p) = 1 6(|Λp| + 2|Λ σ p| + 3|Λ τ p|) =          1 6(3 + 0 + 3) if p = 2 1 6(4 + 2 + 6) if p = 3 1 6((p + 1) + 4 + 6) if p≡ 1 mod 3 1 6((p + 1) + 0 + 6) if p≡ 2 mod 3 and p ̸= 2 . and _{ p≡ 1 mod 6 if p≡ 1 mod 3

p≡ 5 mod 6 if p≡ 2 mod 3 and p ̸= 2. It is equivalent to the formulaT1(p) =

[p+11

6

]

, where p is an odd prime. More-over, to computeT1(n) for general n, we need the following lemma

Lemma 6.4. For g∈ S3,|Λgn| is a multiplicative function in n.

Proof. Suppose two positive integers n and m are coprime. Claim |Λmn| = |Λm| × |Λn|

Let Xn={ the subgroups of order n in (Z/nZ)2}, then there is a bijective map-ping

ψn: Λn→ Xn by N7→ N/(nZ)2.

Since the map ϕ : Xmn→ Xm× Xn by H/(mnZ)2 7→ H/(mZ)2× H/(nZ)2 is bijective ( by Chinese Remainder Theorem ) and the fact ψm, ψn, ψmnare bijective, we have|Λmn| = |Λm| × |Λn|. Moreover, we can restrict ψn to Λgn and hence|Λgn| is a multiplicative.

 From Lemma 6.4, we can computeT1(n) for general n. For the second conjecture,

we have some partial results given in next section.

7. Geometric information encoded in the graph zeta function Given a subgroup Γ ofZ2_{, let X}

Γ be the quotient of B2 by Γ which number of

vertices is V0 = V0(Γ) = [Z2 : Γ]. Let YΓ be the 1-skeleton of XΓ, which is the

Cayley graph onZ2/Γ with the generator set S ={(±1, 0), (0, ±1), ±(1, 1)}.

From Corollary 2.1, knowing the graph zeta function of YΓis equivalent to

know-ing the spectrum of the adjacent matrix A of YΓ.

On the other hand, knowing the spectrum of A is equivalent to knowing the trace of An _{for all n, which is the number of closed walks of length n in Y}

Γ.

Therefore, we would like to study if one can determine the structure of XΓ

through these numbers tr(An_).

Let Pnbe the collection of closed walks in YΓof length n starting from the vertices

Γ. Observe thatZ2 acts transitively on vertices of YΓ. Therefore, tr(An) = V0|Pn|. Now for each closed walks c in Pn, it can be uniquely lifted to a walks in YΓstarting

from (0, 0) to some element in Γ, denoted by γc. Then we can decompose Pn as

Pn = ⨿ γ∈Γ

On the other hand, one can compute Nn(γ) =|Pn(γ)| by Nn(γ) =|(s1,· · · , sn)∈ Sn, s1+· · · + sn= γ}|. Recall that S3={I2, [₀₋₁ 1−1 ] ,[−1 1_{−1 0}],[₀1−1₋₁],[−1 0_{−1 1}], [0 1 1 0]}

which acts on the set of verticesZ2 _ofB

2. We can decomposeZ2as S3-orbits as

Z2₌ ⨿

a,b_∈Z≥0

S3(a + b, b)

and we say γ inZ2_{is of type (a, b) if γ∈ S}

3(a + b, b).

Theorem 7.1. If γ̸= (0, 0) is of type (a, b), then

(1) Nn(γ) = 0 if n < a + b. (2) Na+b(γ) = (a+b a ) . (3) Na+b+1(γ) = { (a + b + 1)((a+b_b−1)+(a+b_b+1)) if ab̸= 0 (a + b + 1)(a + b) if one of a, b = 0

Proof. Note that if γ and γ′lie in the same S3-orbit, then Nn(γ) = Nn(γ′). There-fore, we may assume γ = (a+b, b). Let c be a closed walk from (0, 0) to γ of length n which uses r1,· · · , r6times of generators (1, 0), (1, 1), (0, 1), (−1, 0), (−1, −1), (0, −1)

respectively. Then we have the equations      r1+ r2− r4− r5= a + b r2+ r3− r5− r6= b r1+ r2+ . . . + r6= n

and the number of such c is given by n! r1!r2!...r6!.

Observing that its length n must be greater than or equal to a + b, and if

n = a + b, then r1= a and r2= b. Therefore, we have Nn(γ) = 0 if n < a + b and

Nn(γ) = (a+b

a )

if n = a + b.

Next we consider the case of n = a + b + 1, then we have      r1+ r2− r4− r5= a + b r2+ r3− r5− r6= b r1+ r2+ . . . + r6= a + b + 1 .

Subtracting the third equation by the first equation above, we obtain r3+ 2r4+

2r5+ r6= 1. Thus      r4= r5= 0 r1+ r2= a + b r3+ r6= 1 . If a, b ̸= 0, then we have (r1, r2, . . . , r6) = (a + 1, b− 1, 1, 0, 0, 0) or (a − 1, b + 1, 0, 0, 0, 1). Thus Na+b+1(γ) = (a + b + 1) ((a+b b₋₁ ) +(a+b_b+1)).

If a = 0 and b̸= 0, then r2+ r3− r6= r1+ r2= b. It implies that r3= r1+ r6

and r1+ 2r6 = 1. Therefore, we have (r1, r2, . . . , r6) = (1, b− 1, 1, 0, 0, 0) and Nb+1(γ) = b(b + 1).

If b = 0 and a̸= 0, then r2+r3= r6and r2+2r3= 1. It implies (r1, r2, . . . , r6) =

For an element γ of type (a, b), the graph distance of γ and the origin is a + b and we say γ is of length a + b. Observe that for a positive integer n, the set of elements of length n inZ2 _is

R(n) ={±(n, m), ±(n − m, n), ±(−m, n − m), where 0 ≤ m ≤ n}

which forms a regular hexagon with perimeter 6n. Let n be the shortest length among all nonzero elements of Γ and N0 be the number of elements in Γ of length n. N0 is an even integer since if γ is of length n so is −γ. Since Γ is a subgroup

ofZ2_{, the graph distance of any two distinct vertices in Γ is greater than or equal}

to n. Therefore, on the regular hexagon R(n) with perimeter 6n, there are at most six vertices contained in Γ. We conclude that N0 is equal to 2, 4 or 6 and when N0= 6, the six vertices in Γ∩ R(n) are

±(n, m), ±(n − m, n) and ± (m, n − m)

for some 0≤ m < n.

We define the number N0/2 to be the type of Γ so that Γ is of type 1, 2 or 3.

In the rest of the section, we study that whether the lattices of type 3 could be uniquely determined by its graph zeta function.

First, we list all Γ satisfying V0≤ 4 upto the action of S3and their corresponding

zeta functions.

Note that for any finite index subgroup Γ of Z2_{, there exist a unique basis of}

the form{(a, 0), (c, d)} with a > c ≥ 0. Therefore, we obtain the following table of S3-orbit of Γ (denoted by [Γ]S3): V0 2 [Γ]S3 {⟨(2, 0), (0, 1)⟩ , ⟨(2, 0), (1, 1)⟩} Type of Γ type 1 (ZXΓ(u)) −1 _{(1− u)(1 − 5u)(1 + 2u + 5u}2₎ V0 3 [Γ]S3 {⟨(3, 0), (0, 1)⟩ , ⟨(3, 0), (1, 1)⟩} ⟨(3, 0), (2, 1)⟩

Type of Γ type 1 type 3

(ZXΓ(u))

−1 _{(1− u)(1 − 5u)(1 + 5u}2₎2 _{(1− u)(1 − 5u)(1 + 3u + 5u}2₎2

V0 4

[Γ]S3 ⟨(2, 0), (0, 2)⟩ {⟨(2, 0), (1, 2)⟩ , ⟨(4, 0), (2, 1)⟩ , ⟨(4, 0), (3, 1)⟩}

Type of Γ type 3 type 2

(ZXΓ(u)) −1 _{(−1 + u)(−1 + 5u)(1 + 2u + 5u}2₎3 V0 4 [Γ]S3 {⟨(4, 0), (0, 1)⟩ , ⟨(4, 0), (1, 1)⟩} Type of Γ type 1 (ZXΓ(u)) −1 _{(−1 + u)(−1 + 5u)(1 − 2u + 5u}2₎2_{(1 + 2u + 5u}2₎

Note that when V0= 2 or 3, the zeta functions can uniquely determine Γ. But

when V0= 4, there are three type 2 lattice whose zeta function and the zeta function

of the type 3 lattice⟨(2, 0), (0, 2)⟩ are the same. In this rest of the section, we study

the case that V0≥ 5.

Recall that two elements v1and v2 in Γ form a reduced basis if v1 is a shortest

nonzero element in Γ and v2 is a shortest element in Γ− {Zv1}. When N0 > 2,

Γ∩ R(n) contains a reduced basis {v1, v2}. Furthermore, we can replace Γ by gΓ for some g ∈ S3 if necessary, so that we can always assume v1 = (n, m) with

0≤ m ≤n₂.

Suppose Γ is of type 3 with the shortest nonzero elements ±(n, m), ±(n − m, n) and ± (m, n − m), where 0 ≤ m ≤ n₂. In this case, the number of vertices of

YΓ is V0= n2− nm + m2. Let Tk(Γ) = 1 2 ( tr(An₎ V0 − Nk(e) ) = 1 2(|Pk| − Nk(e)) .

which can be determined by the zeta function of YΓ. We shall prove that{Tk} can be used to distinguish Γ from other subgroups.

By the above theorem, we have Tk= 0 for all k < n and

Tn(Γ) = 3 ( n m ) ; if m > 0, Tn+1= 3(n + 1) [( n m− 1 ) + ( n m + 1 )] ; if m = 0, Tn+1= 3n(n + 1).

Let Γ′ be another lattice which has the same zeta function as Γ. Therefore, we have V0(Γ) = V0(Γ′) and Tk(Γ) = Tk(Γ′) for all k. Especially, the shortest nonzero elements of Γ′ are of length n. Suppose the reduced basis of Γ′is v′₁= (n, m1) with

0≤ m1≤ n₂ and v2′ with the first coordinate of v′2is non-negative.

Case I: Γ′ is of type 3.

As we mentioned earlier, we may assume that the shortest nonzero elements of Γ′ are±(n, m′_),±(n−m′_{, n) and ±(m}′_{, n−m}′_{), where 0≤ m}′ _≤n

2. Then we have 3 ( n m ) = Tn(Γ) = Tn(Γ′) = 3 ( n m′ ) .

Since the binomial coefficient f (m) =(n m

)

is increasing when 0≤ m ≤ n

2, we have m = m′. On the other hand, Γ and Γ′ are both generated by their shortest nonzero elements. We conclude that Γ = Γ′.

Case II: Γ′ is of type 2.

In this case, v2′ = (n, m2), (m2, n) or (n− m2,−m2) for some 0 ≤ m2 ≤ n.

Observe that if v2′ = (n, m2) then v1′ − v′2 = (0, m1− m2) which is a non-zero

or ±v′

2 which is a contraction. Therefore, it is sufficient to consider two sub-cases v2′ = (m2, n) or v2′ = (n− m2,−m2).

Case II-A: v2′ = (m2, n). In this case, we have

(2) n2− nm + m2= V0(Γ) = V0(Γ′) = n2− m1m2 ⇒ m(n − m) = m1m2 and (3) 3 ( n m ) = Tn(Γ) = Tn(Γ′) = ( n m1 ) + ( n m2 ) .

If m = 0, then m1= 0 or m2= 0. Suppose m1= 0, then Eq. (3) becomes

(

n m2

) = 2.

We conclude that n = 2 and m2= 1 so that V0(Γ) = V0(Γ′) = 4.

Assume m2= 0, then similar to the case m1= 0, we have n = 2 and m1= 1 so

that V0(Γ) = V0(Γ′) = 4. The two cases contradict to our assumption.

Now suppose m > 0. If one of m1 and m2 is less than or equal to m, then by

Eq.(2), the other one is greater than or equal to n− m. Since m < n

2, in this case, we have ₍ n m1 ) + ( n m2 ) ≤ ( n m ) + ( n n− m ) < 3 ( n m )

which contradicts to Eq.(3). Therefore, we have m < m1, m2 < n− m. Next we

consider the following two sub-cases.

Case II-A-(1): Suppose there is no element of length n + 1 in Γ′. Then (4) 3(n + 1) [( n m− 1 ) + ( n m + 1 )] = Tn+1(Γ) = Tn+1(Γ′) = (n + 1) [A + B] . where A =( n m1−1 ) +(_mn 1+1 ) , B =(_mn 2−1 ) +(_mn 2+1 ) . Applying Pascard identity to 2 Eq.(3) + 1

n+1 Eq.(4), we obtain 3 ( n + 2 m + 1 ) = ( n + 2 m1+ 1 ) + ( n + 2 m2+ 1 ) . (5)

On the other hand, ( n + 2 m + 1 ) = (n + 2)(n + 1) (m + 1)(n− m + 1) ( n m ) .

Therefore, together with Eq.(3) and Eq.(5), we have (n + 2)(n + 1) (m + 1)(n− m + 1) [( n m1 ) + ( n m2 )] (6) = (n + 2)(n + 1) (m1+ 1)(n− m1+ 1) ( n m1 ) + (n + 2)(n + 1) (m2+ 1)(n− m2+ 1) ( n m2 ) .

Observe that the denominator in the above is of the form

h(x) = (x + 1)(n− x + 1) = − ( x−n 2 )2 +n 2 2 + n + 1. Thus, (7) h(x) < f (y) if x −n 2   >y −n₂ .

Therefore, h(m) < h(m1), h(m2) and the left hand side of the above equation is

Case II-A-(2): Suppose there is some element of Γ′ of length n + 1. We need the following lemma.

Lemma 7.1. The shortest non-zero elements in Γ′ − {±(n, m1),±(m2, n)} are ±(n − m2, m1− n).

Proof. Observe that the length of (n− m2, m1− n) is 2n − m1− m2. Suppose x = a(n, m1) + b(n, m2) is the desired element in Γ′− {±(n, m1),±(m2, n)}, where a, b∈ Z. If one of a and b is equal to zero or both a and b are positive, then the

length of x≥ 2n > 2n − m1− m2.

Now suppose ab < 0. Without loss of generality, we may assume that a > 0 > b and|a| ≥ |b|, then x = (a + b)(n, m1)− b(n − m2, m1− n) which length is bounded

by (a + b)n− b(n − m2).

If a + b > 0, then (a + b)n− b(n − m2)≥ n + (n − m2)≥ 2n − m1− m2.

Assume a+b = 0, then x = a(n−m2, m1−n) is length a(2n−m1−m2). Hence the

shortest non-zero elements in Γ′−{±(n, m1),±(m2, n)} are ±(n−m2, m1−n). 

In this case, the above lemma implies 2n− m1− m2 = n + 1 or equivalently m2= n−m1−1. Especially, we have (n m1 ) +(_mn 2 ) =(_mn+1 1+1 )

. The same computation in Eq.(4) as the previous case provides

(n + 2)(n + 1) (m + 1)(n− m + 1) ( n + 1 m1+ 1 ) = (n + 2)(n + 1) (n− m1)(m1+ 2) ( n m1+ 1 ) + (n + 2)(n + 1) (m1+ 1)(n− m1+ 1) ( n m1 ) + 1 n + 1 ( n + 1 m1+ 1 ) .

Here the extra term comes from the contribution of length n + 1 elements. Dividing the above equation by(n+1

m1+1 ) , we obtain (n + 2)(n + 1) (m + 1)(n− m + 1)= (n + 2) (n− m1+ 1) + (n + 2) (m1+ 2) + 1 (n + 1). On the other hand, we have m(n− m) = m1m2= m1(n− m1− 1) and

(n + 2)(n + 1) (m + 1)(n− m + 1) = (n + 2)(n + 1) m(n− m) + n + 1= (n + 2)(n + 1) m1(n− m1− 1) + n + 1 . Consider that (n + 2)(n + 1) m1(n− m1− 1) + n + 1− (n + 2) (n− m1+ 1)− (n + 2) (m1+ 2)− 1 (n + 1) = E1+ E2n + E3n 2_{+ E} 4n3+ n4 (2 + m1)(1 + n)(1− m1+ n)(1− m1− m21+ n + m1n) , where E1=−4 + 7m1+ 6m21− 2m31− m14; E2=−7 + 2m1+ 11m21+ 2m31; E3=−1 − 7m1+ m21; E4= 3− 2m1. Let (8) f (x) = E1+ E2x + E3n2+ E4x3+ x4.

then (9) f′(x) =−7 + 2m1+ 11m21+ 2m 3 1− 2x − 14m1x + 2m21x + 9x 2_{− 6m} 1x2+ 4x3. (10) f′′(x) =−2 − 14m1+ 2m21+ 18x− 12m1x + 12x2 (11) f′′′(x) = 18− 12m1+ 24x

Recall that 1 ≤ m1 ≤ n₂. Since f′′′(x) > 0,∀x ≥ 2m1, f′′(x) is increasing when x≥ 2m1. Together with

f′′(2m1) =−2 + 22m1+ 26m21> 0,

we have f′(x) is increasing when x≥ 2m1. Moreover,

f′(2m1) =−7 − 2m1+ 19m21+ 14m31> 0.

Therefore, f (x) is increasing when x≥ 2m1. On the other hand, f (2m1) =−4 − 7m1+ 6m21+ 16m

3 1+ 7m

4 1> 0,

we have f (x) > 0,∀x ≥ 2m1 and it contradicts to

(n + 2)(n + 1) (m + 1)(n− m + 1)= (n + 2) (n− m1+ 1) + (n + 2) (m1+ 2) + 1 (n + 1).

Case II-B: v′2= (n− m2,−m2). Similar to Case II-A, we have

(12) 3 ( n m ) = Tn(Γ) = Tn(Γ′) = ( n m1 ) + ( n m2 ) .

In this case, we have the same conclusion as the case II-A-(1) if we assume that there is no element of length n + 1. Therefore, it remains to consider the case that there are some elements of Γ′ of length (n + 1).

Lemma 7.2. The shortest non-zero elements in Γ′− {±(n, m1),±(n − m2,−m2)} are±(m2, m1+ m2).

Proof. When there is some element (x, y) in Γ′with length n+1, (x, y) = a(n, m1)+ b(n− m2,−m2) for some non-zero integers a and b. It implies that 0≤ (a + b)n − bm2≤ n + 1 and 0 ≤ am1− bm2≤ n + 1. Therefore, (x, y) may be (m2, m1+ m2)

or (m2− n, m1+ 2m2).

Since the length of (m2, m1+m2) is m1+m2and the length of (m2−n, m1+2m2)

is n + m1+ m2, we hence the shortest non-zero elements in Γ′− {±(n, m1),±(n −

m2,−m2)} are ±(m2, m1+ m2). 

Now we return to show that there does not exist m1 and m2 such that Eq.(12)

can be hold.

From above lemma and Eq(3), the equation Tn+1(Γ) = Tn+1(Γ′) can be written as 3(n + 2)(n + 1) (m + 1)(n− m + 1) ( n m ) = (n + 2)(n + 1) (m + 1)(n− m + 1) (( n m1 ) + ( n m2 )) = (n + 2)(n + 1) (m1+ 1)(n− m1+ 1) ( n m1 ) + (n + 2)(n + 1) (n− m2+ 1)(m2+ 1) ( n m2 ) + 1 n + 1 ( m1+ m2 m2 ) ,

where m1+ m2= n + 1. Therefore, we have (n + 2)(n + 1) (m + 1)(n− m + 1) ( n + 1 m1 ) = (n + 2)(n + 1) (m1+ 1)(n− m1+ 1) ( n m1 ) + (n + 2)(n + 1) m1(n− m2+ 2) ( n m1− 1 ) + 1 n + 1 ( n + 1 m1 ) and it implies that

(n + 2)(n + 1) m1(n− m1+ 1) + n + 1 = (n + 2) (m1+ 1) + (n + 2) (n− m2+ 2) + 1 n + 1. It is equivalent to (13) F1+ F2n + F3n 2_{+ F} 4n3 (1 + m1)(1 + n)(2− m1+ n)(m1(n− m1+ 1) + n + 1) = 0, where F1= 4 + 7m1− 6m21− 2m31+ m14; F2= 8 + 15m1− 6m21− 2m31; F3= 5 + 10m1− m21; F4= 1 + 2m1. Let (14) g(x) = F1+ F2x + F3x2+ F4x3. Therefore, (15) g′(x) = 8 + 15m1− 6m21− 2m 3 1+ 10n + 20m1x− 2m21x + 3x 2_{+ 6m} 1x2. (16) g′′(x) = 10 + 20m1− 2m21+ (6 + 12m1)x.

Recall that 1≤ m1≤n₂. Since g′′(x) is increasing and g′′(2m1) = 10 + 32m1+ 22m21> 0,

we have that g′(x) is increasing when x≥ 2m1. Moreover, since g′(2m1) = 8 + 35m1+ 46m21+ 18m

3 1> 0,

it implies that g(x) is increasing when x≥ 2m1. Together with g(2m1) = 4 + 23m1+ 44m21+ 34m

3 1+ 9m

4 1> 0,

it contradicts to Eq.(13). Hence the underlying graphs corresponding to the lattice of type 3 and type 2 can be distinguished if V0> 4.

Case III: Γ′ is of type 1.

In this case, the length of v2′ is greater than n. From Theorem 7.1, we have

(17) 3 ( n m ) = Tn(Γ) = Tn(Γ′) = ( n m1 ) ,

which implies m1≥ 1 and m1> m.

Case III-A: Suppose there is no element of length n + 1 in Γ′. If m = 0, from Theorem 7.1, we have

( n m1 ) = 3 and 3n(n + 1) = (n + 1) [( n m1− 1 ) + ( n m1+ 1 )] .

Therefore, n = 3, m1= 1 and it contradicts to 9 =

(3 0

) +(3₂).

If m̸= 0, from Theorem 7.1, we have (18) 3(n+1) [( n m− 1 ) + ( n m + 1 )] = Tn+1(Γ) = Tn+1(Γ′) = (n+1) [( n m1− 1 ) + ( n m1+ 1 )] .

Applying Pascard identity to 2 Eq.(17) + 1

n+1 Eq.(18), we have (19) 3 ( n + 2 m + 1 ) = ( n + 2 m1+ 1 ) .

Together with Eq.(17) and Eq.(19), we have (n + 2)(n + 1) (m + 1)(n− m + 1) ( n m1 ) = (n + 2)(n + 1) (m1+ 1)(n− m1+ 1) ( n m1 ) and we get a contradiction from Eq.(7) and Eq.(17).

Case III-B: Suppose there is an element of length n + 1 in Γ′.

In this case, v₂′ may be equal to (n + 1, m2), (m2, n + 1) or (n + 1− m2,−m2) for

some 0≤ m2≤ n +1. Observe that if v2′ = (n + 1, m2), then v2′− v1′ = (1, m2− m1)

which is a nonzero vector of length less than n + 1 and not equal to ±v′

1. It

contradicts to that Γ′ is of type 1 and ±v′

1 are the two shortest elements in Γ′.

Therefore, we just need to consider the case v′₂= (m2, n + 1) or (n + 1− m2,−m2).

Suppose m2= 0. If v2′ = (0, n+1), then n2+n = V0(Γ′) = V0(Γ) = n2−nm+m2

and it implies n = m(m−n), which contradicts to the assumption that 0 ≤ m ≤ n

2.

If v₂′ = (n + 1, 0), then similar to the case v₂′ = (n + 1, m2) above, we have a

contradiction.

Assume m2 = n + 1. If v′2 = (n + 1, n + 1), then v2′ − v1′ = (1, n + 1− m1),

whose length is less than or equal to n since m1≥ 1. It contradicts to that ±v′1are

the only two shortest elements in Γ′ whose lengths are n. If v′2= (0,−n − 1), then n2_{+ n = V}

0(Γ′) = V0(Γ) = n2− nm + m2 and it contradicts to our assumption

that 0≤ m ≤ n

2. Therefore, we conclude 1≤ m2≤ n.

Lemma 7.3. There is no element of length n + 1 in Γ′− {±v′

2}.

Proof. Suppose there is an element v′₃ of length n + 1 in Γ′. We may assume that the first coordinate of v₃′ is non-negative. Then v₃′ is equal to (m3, n + 1) or

(n + 1− m3,−m3) for some 1≤ m3 ≤ n. Note that both v′2 and v3′ are of length n + 1 so we can we assume m2≤ m3 and switch v2′ and v3′ if necessary.

Recall that v′2= (m2, n + 1) or (n + 1− m2,−m2) and we shall prove the lemma

holds for both cases. First, we assume v2′ = (m2, n + 1), then v3′−v2′ = (m3−m2, 0)

or (n + 1−m2−m3, m3−m2). For the case v′3−v′2= (m3−m2, 0), its length is less

or equal to n, which contradicts to that±v′

1 are the only two shortest elements in

Γ′ of length n. Now we consider v′3−v2′ = (n + 1−m2−m3, m3−m2). Observe that

if m2+ m3≤ n + 1, then the length of v′3− v2′ is max{n + 1 − m2− m3, m3− m2}.

If we assume m2+ m3 > n + 1, then the length of v′2− v3′ is 2m3− n − 1. The

length of v′₂− v′

3 in the two cases are of less than n. It contradicts to that±v1′ are

the shortest elements in Γ′ of length n. Hence there is no element of length n + 1 in Γ′− {±v′

2}. 

Case III-B-(1): v₂′ = (m2, n + 1). In this case, we have

and

length of v₁′ − v′

2= 2n− m1− m2+ 1 > n + 1

i.e.

(21) n > m1+ m2.

From Eq.(20) and Eq.(21), we have

m2− nm = n − m1m2> n− m1(n− m1) = n− m1n + m21

and it implies

(22) m2− m2₁= (m + m1)(m− m1) > n(1− m1+ m).

It contradicts to that m, m1≤ n2.

Case III-B-(2): v₂′ = (n + 1− m2,−m2). In this case, we have

3 ( n + 2 m + 1 ) = Tn+1(Γ) = Tn+1(Γ′) = ( n + 2 m1+ 1 ) + 1 n + 1 ( n + 1 m2 ) . It is equivalent to (23) (n + 2)(n + 1) (m + 1)(n− m + 1) ( n m1 ) = (n + 2)(n + 1) (m1+ 1)(n− m1+ 1) ( n m1 ) + 1 n− m2+ 1 ( n m2 ) .

Since there is no element of length less than or equal to n and (24) length of v₁′ − v′ 2= m1+ m2> n + 1, we have n + 1 n− m2+ 1 ( n m2 ) = ( n + 1 m2 ) = ( n + 1 n− m2+ 1 ) < ( n + 1 m1 ) = n + 1 n− m1+ 1 ( n m1 ) and implies (25) 1 n− m2+ 1 ( n m2 ) < 1 n− m1+ 1 ( n m1 )

On the other hand, together with the fact 0≤ m < m1≤n₂, we have

(n + 2)(n + 1) (m + 1)(n− m + 1)− (n + 2)(n + 1) (m1+ 1)(n− m1+ 1) = (n + 2)(n + 1)[m1(n− m1)− m(n − m)] (m + 1)(n− m + 1)(m1+ 1)(n− m1+ 1) > (n + 2)(n + 1)[m1(n− m1)− m(n − m1)] (m + 1)(n− m + 1)(m1+ 1)(n− m1+ 1) = (n + 2)(n + 1)(m1− m)(n − m1) (m + 1)(n− m + 1)(m1+ 1)(n− m1+ 1) > 1 (n− m1+ 1) .

Here, the last inequality is followed by the fact that (n + 2)(n + 1) > (m1+ 1)(n− m + 1), n− m1≥ m + 1 and m1− m ≥ 1.

It contradicts to the inequality (n + 2)(n + 1) (m + 1)(n− m + 1)< (n + 2)(n + 1) (m1+ 1)(n− m1+ 1) + 1 n− m1+ 1 .

Theorem 7.2. If Γ is of type 3 and V0(Γ) > 4, then YΓ can be uniquely determined by it’s zeta function.

To prove the conjecture (2), we have to compare the zeta functions of Γ of type 1 and type 2, which is more complicated. We will finish it in the future works.

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