3-torsion in the homology of complexes of graphs

of bounded degree

Jakob Jonsson

August 30, 2008

Abstract

For δ ≥ 1 and n ≥ 1, we examine the simplicial complex of graphs onn vertices in which each vertex has degree at most δ; we identify a givengraph with its edge set and admit one loop at each vertex. δ = 1 yieldsthe matching complex, and it is known that there is 3-torsion in degree dof the homology of this complex whenever n−4

3≤ d ≤ n−6

2. We establish

similar bounds for δ ≥ 2. Specifically, for δ = 2, there is 3-torsion indegree d of the homology whenever 5n−8

6≤ d ≤ n− 3. For δ ≥ 3, there is

3-torsion in degree d whenever (3δ−1)n−86

≤ d ≤ δn−δ−62

and n ≥ 6δ. Thesituation for other pairs (d, n) remains unknown in general. To detecttorsion, we construct an explicit cycle z that is easily seen to have theproperty that 3z is a boundary. Defining a homomorphism that sends zto a non-boundary element in the chain complex of a certain matchingcomplex, we obtain that z itself is a non-boundary. In particular, thehomology class of z has order 3.

1 Introduction

Our aim is to examine the integral homology of certain simplicial complexesdefined in terms of degree bounds in graphs. Specifically, each face in a givencomplex corresponds to a graph such that the degree of each vertex is boundedfrom above by a certain fixed value. In the case that the value is 1 for eachvertex, we obtain the well-studied matching complex. The infinite part of thehomology has been computed [9], but not very much is known about the finitepart. We make some progress on the latter problem, detecting 3-torsion in thehomology for various choices of parameters.

Let us formulate the problem more precisely, starting with basic graph-theoretic definitions. A graph G consists of a vertex set V and a set E oftwo-sets vw = {v, w} such that v, w ∈ V . We restrict our attention to simplegraphs admitting loops; a graph being simple means that no two-set appearsmore than once in E, and loops are edges of the form vv.

The degree degG(v) of a vertex v in G is the number of occurrences of v inE; we adopt the convention that v occurs twice in the loop vv. For example,

1

in the graph G with edge set {aa, ab, ac, bc, bd}, we have that degG(a) = 4,degG(b) = 3, degG(c) = 2, and degG(d) = 1.

Let n ≥ 1 and let λ = (λ1, . . . , λn) be an arbitrary sequence of integers.Define BDλ

n to be the family of graphs G on the vertex set [n] = {1, . . . , n} suchthat degG(i) ≤ λi for each i ∈ [n]. For a graph G on the vertex set [n], letBDλ

n(G) be the subfamily of BDλn obtained by restricting to subgraphs of G. We

write BD(δ,...,δ)n = BDδ

n. For δ = 1, we obtain the matching complex Mn = BD1n.

Identifying each graph with its edge set, we may interpret BDλn as an ab-

stract simplicial complex. This complex is our main object of study; we maysummarize our results as follows.

• For n1, n2 ≥ 0, the group Hd(BD2n21n1

n1+n2; Z) contains 3-torsion whenever

2n1 + 5n2 − 86

≤ d ≤ n1 + 2n2 − 62

,

where 2n21n1 denotes the sequence consisting of n2 copies of 2 and n1

copies of 1. In particular, Hd(BD2n; Z) contains 3-torsion whenever

5n− 86

≤ d ≤ n− 3.

• For δ ≥ 3, the group Hd(BDδn; Z) contains 3-torsion whenever

(3δ − 1)n− 86

≤ d ≤ δn− δ − 62

and n ≥ 6δ.

With n2 = 0 in the first result, we recover a previous result [4, §11.2.3] thatHd(Mn; Z) contains 3-torsion whenever

n− 43

≤ d ≤ n− 62

.

For δ = 3, the inequality n ≥ 6δ is a consequence of the other inequalities. Wehave some hope that this condition can be dropped for all δ. Given the situationfor δ = 2, it is also plausible that one may provide a slight improvement to theupper bound on d. We are more skeptical about the possibility of improvingthe lower bound. See Section 5.3 for more discussion.

One may also consider the subcomplex of BDλn obtained by restricting to

loopless graphs. The reason for focusing on the variant admitting loops is thatthis variant appears naturally in algebra. Specifically, one may express thehomology of certain Lie algebras [8, 11] and also the minimal free resolutionof certain semigroup algebras [9, 3] in terms of the homology of BDλ

n. Theconstruction in Section 5.1 would apply to the subcomplex of loopless graphs,but the other constructions only apply to the full complex BDλ

n.

2

2 Simplicial chain complexes

2.1 Notation

Most material in this section is standard, but we present a fairly detailedoverview of the subject to avoid ambiguity in later sections.

Let ∆ be a simplicial complex and let F be the ring of integers or a field.For d ≥ −1, let Cd(∆; F) be the free F-module with one basis element, denotedas s1 ∧ · · · ∧ sd+1, for each d-dimensional face {s1, . . . , sd+1} of ∆. We refer tos1∧· · ·∧sd+1 as an oriented simplex. Let Sn be the symmetric group on the set[n]. For any permutation π ∈ Sd+1 and any face σ = {s1, . . . , sd+1}, we define

sπ(1) ∧ sπ(2) ∧ · · · ∧ sπ(d+1) = sgn(π) · s1 ∧ s2 ∧ · · · ∧ sd+1. (1)

We will find it convenient to write

[σ] = s1 ∧ s2 ∧ . . . ∧ sd+1,

implicitly assuming that we have a fixed linear order on the 0-cells in ∆.We extend the definition of s1∧· · ·∧sd+1 to arbitrary sequences (s1, . . . , sd+1)

by defining s1 ∧ · · · ∧ sd+1 = 0 if si = sj for some i 6= j. Note that (1) impliesthat 2 · s1 ∧ · · · ∧ sd+1 = 0 for such a sequence.

The boundary map ∂d : Cd(∆; F) → Cd−1(∆; F) is the homomorphism de-fined by

∂d(s1 ∧ . . . ∧ sd+1) =d+1∑i=1

(−1)i−1s1 ∧ . . . ∧ si−1 ∧ si+1 ∧ . . . ∧ sd+1.

Combining all ∂d, we obtain an operator ∂ on the direct sum C(∆; F) of allCd(∆; F). It is well-known and easy to see that ∂2 = 0.

For the chain complex (C(∆; F), ∂) on the simplicial complex ∆, we refer toelements in ∂−1({0}) as cycles and elements in ∂(C(∆; F)) as boundaries. Definethe dth reduced homology group of ∆ with coefficients in F as the quotient F-module

Hd(∆; F) := ∂−1d ({0})/∂d+1(Cd+1(∆; F)) = ker ∂d/im ∂d+1.

2.2 Some useful constructions

Whenever σ = {s1, . . . , sa} and τ = {t1, . . . , tb} are faces such that σ∪τ ∈ ∆, wedefine the product of the oriented simplices [σ] = s1∧· · ·∧sa and [τ ] = t1∧· · ·∧tbto be the element

[σ] ∧ [τ ] = s1 ∧ · · · ∧ sa ∧ t1 ∧ · · · ∧ tb. (2)

Note that [σ]∧ [τ ] is zero whenever σ ∩ τ is nonempty, because this means thatsi = tj for some i and j.

3

Let ∆1 and ∆2 be subcomplexes of ∆ such that σ1 ∪ σ2 ∈ ∆ wheneverσ1 ∈ ∆1 and σ2 ∈ ∆2. Given elements ci ∈ Cdi−1(∆i; F) for i = 1, 2, we definethe product c1 ∧ c2 ∈ Cd1+d2−1(∆; F) by extending the product (2) bilinearly.We have that

∂(c1 ∧ c2) = ∂(c1) ∧ c2 + (−1)d1c1 ∧ ∂(c2). (3)

In particular, if c1 and c2 are cycles, then so is c1 ∧ c2.For a face σ, let lk∆(σ) be the complex {τ : τ ∪ σ ∈ ∆, τ ∩ σ = ∅} and let

fdel∆(σ) be the complex {τ : τ ∈ ∆, τ $ σ}.Let σ = {s1, . . . , sr} ∈ ∆ and let c ∈ Cd−1(∆; F). There is a unique decom-

position of c asc = s1 ∧ · · · ∧ sr ∧ c′ + x,

where c′ ∈ Cd−r−1(lk∆(σ); F) and x ∈ Cd−r−1(fdel∆(σ); F). We write c′ =lkc([σ]) and x = fdelc([σ]); thus

c = [σ] ∧ lkc([σ]) + fdelc([σ]).

Since

∂(c) = ∂([σ]) ∧ lkc([σ]) + (−1)r · [σ] ∧ ∂(lkc([σ])) + ∂(fdelc([σ])),

we have that lk∂(c)([σ]) = (−1)r ·∂(lkc([σ])) and fdel∂(c)([σ]) = ∂([σ])∧lkc([σ])+∂(fdelc([σ])). In particular, lkc([σ]) is a cycle whenever c is a cycle.

Let ∆1, . . . ,∆k be subcomplexes of ∆ such that⋃k

i=1 σi ∈ ∆ wheneverσi ∈ ∆i for each i. Suppose that we are given an element c = c1∧· · ·∧ck, whereci is an element in Cdi−1(∆i; F) for each i.

Lemma 2.1 Let σ be a face of ∆. We have that

[σ] ∧ lkc([σ]) =∑

(τ1,...,τk)

[τ1] ∧ lkc1([τ1]) ∧ · · · ∧ [τk] ∧ lkck([τk]),

where the sum is over all ordered partitions (τ1, . . . , τk) of σ such that τi ∈ ∆i.

Proof. By linearity, we need only prove the lemma in the case that each ci

coincides with an oriented simplex [ρi]. For any τi ⊆ ρi, we have that [τi] ∧lk[ρi]([τi]) = [ρi]. Moreover, if τi 6⊆ ρi, then [τi] ∧ lk[ρi]([τi]) = 0. In particular,each summand in the right-hand side is either c or 0. As a consequence, if someelement appears in both ρi and ρj for some i 6= j, meaning that c = 0, then theright-hand side is zero. Clearly, the left-hand side is also zero in this case.

Assume that ρ1, . . . , ρk are pairwise disjoint and write ρ = ρ1 ∪ . . .∪ ρk. If ρdoes not contain σ, then both sides in the lemma are zero. Assume that ρ doescontain σ. Then [σ] ∧ lkc([σ]) = c. Moreover,

c = [ρ1] ∧ · · · ∧ [ρk]= [σ ∩ ρ1] ∧ lk[ρ1]([σ ∩ ρ1]) ∧ · · · ∧ [σ ∩ ρk] ∧ lk[ρk]([σ ∩ ρk]).

The latter expression coincides with the right-hand side in the lemma, because(σ ∩ ρ1, . . . , σ ∩ ρk) is the only partition (τ1, . . . , τk) of σ such that lk[ρi]([τi]) isnonzero for each i. �

4

3 Basic properties of cycle products in BDλn

Let A = {a1, . . . , aq−1} and B = {b1, . . . , bq} be subsets of [n], not necessarilydisjoint, such that |B| − 1 = |A| ≥ 0. Define

φA,B =∑

π∈Sq

sgn(π) · a1bπ(1) ∧ · · · ∧ aq−1bπ(q−1).

φA,B is a cycle in the chain complex of BDλn, where λi equals 2, 1, or 0 depending

on whether i belongs to both sets A and B, one of the sets, or neither of the sets.For example, φ[k],[2k+1]\[k] is a cycle in the chain complex of BD1

2k+1 = M2k+1.We refer to φA,B as an (|A|, |B|)-cycle. Note that φ{a},{b,c} = ab− ac and thatφ∅,{b} = [∅] for any b; the latter cycle is the generator of C−1(M{b}; Z) ∼= Z.

The following result is due to Bouc [2] and Shareshian and Wachs [10].

Theorem 3.1 Let η ∈ {0, 1, 2} and α ≥ 0 and let X be a set of size n =3α + 2η + 1. Let X =

⋃αi=0(Ai ∪ Bi) be a partition of X such that |Ai| = 1

and |Bi| = 2 for 1 ≤ i ≤ α and such that |A0| = η and |B0| = η + 1. Then thehomology class of the cycle

z =α∧

i=0

φAi,Bi

is nonzero and generates the group

Hα+η−1(MX ; Z) ∼= Hα+η−1(Mn; Z).

Moreover, this group is an elementary 3-group for n ≥ 15 and n ∈ {7, 10, 12, 13},finite of order divisible by 3 for n = 14, and infinite for n ∈ {1, 3, 4, 5, 6, 8, 9, 11}.

The group in the theorem is the bottom nonvanishing homology group of Mn

[2, 10]. For n = 14, the order of the group is in fact divisible by 15 [5].Let k ≥ 1. For 1 ≤ i ≤ k, let

λi = (λi1, . . . , λ

in)

be a sequence of nonnegative integers and let E be a set of edges on the vertex set[n]. Let di ≥ 0 and γi ∈ Cdi−1(BDλi

n (E); Z). Write λ =∑k

i=1 λi, d =∑k

i=1 di,and

γ = γ1 ∧ · · · ∧ γk.

For an edge set X, define µ(X) to be the sequence (µ1, . . . , µn), where µi is thenumber of occurrences of the vertex i in the edge set X.

Lemma 3.2 γ is an element in Cd−1(BDλn(E); Z). If each γi is a cycle, then

so is γ. Moreover, the exponent of the homology class of γ in Hd−1(BDλn(E); Z)

divides the exponent of the homology class of γi in Hdi−1(BDλi

n (E); Z) for 1 ≤i ≤ k.

5

Proof. By construction, if e1 ∧ · · · ∧ ed appears in the expansion of γ1 ∧ · · · ∧ γk,then µ({e1, . . . , ed}) is bounded by

∑i λi = λ. As a consequence, γ is indeed

an element in Cd−1(BDλn(E); Z). (3) and a straightforward induction argument

yield that γ is a cycle whenever each γi is a cycle. Finally, if the homology classof, say, γ1 has exponent a, then there is an element c ∈ Cd1(BDλ1

n (E); Z) suchthat ∂(c) = a · γ1. Since c ∧ γ2 ∧ · · · ∧ γk belongs to Cd(BDλ

n(E); Z) and

∂(c ∧ γ2 ∧ · · · ∧ γk) = (aγ1) ∧ γ2 ∧ · · · ∧ γk = a · γ1 ∧ · · · ∧ γk,

it follows that the exponent of the homology class of γ1 ∧ · · · ∧ γk divides a. Bysymmetry, the same is true for γi instead of γ1 for each i ∈ {2, . . . , k}. �

From now on, assume that each γi is a cycle. We will make repeated use ofthe following result.

Lemma 3.3 Suppose that either of the following holds for some index i.

(1) We have that

γi = φ{a1},{b1,c1} ∧ φ{a2},{b2,c2} ∧ φ∅,{x},

and the seven elements in the vertex set V = {a1, b1, c1, a2, b2, c2, x} areall distinct.

(2) We have that

γi = φ{a1},{b1,c1} ∧ φ{a2},{b2,c2} ∧ φ{a3},{b3,c3} ∧ φ{a4},{b4,c4},

and the twelve elements in the vertex set V = {ar, br, cr : r ∈ {1, 2, 3, 4}}are all distinct.

(3) We have that

γi = φ{a1},{b1,c1} ∧ φ{a2},{b2,c2} ∧ φ{a3},{b3,c3} ∧ φ{v,w},{x,y,z},

and the 14 elements in the vertex set V = {ar, br, cr : r ∈ {1, 2, 3}} ∪{v, w, x, y, z}. are all distinct.

Suppose that E contains all edges between vertices in V . In cases (1) and (2),the exponent of the homology class of γ in Hd−1(BDλ

n(E); Z) divides 3. In case(3), the exponent of the homology class of γ divides a nonzero multiple of 3.

Proof. For simplicity, assume that i = 1. Write γ = γ1 ∧ γ′, where γ′ =γ2 ∧ · · · ∧ γk. It is clear that γ1 is a cycle in the chain complex of MV andthat γ′ is a cycle in the chain complex of BDλ′

n (E), where λ′ is obtained fromλ by subtracting one from each v ∈ V . Theorem 3.1 yields that the homologyclass of γ1 in the chain complex of MV has exponent 3 in cases (1) and (2) andexponent a nonzero multiple of 3 in case (3). By Lemma 3.2, we are done. �

6

Suppose that we are given pairwise disjoint faces σi ∈ BDλi

n (E). Writeσ =

⋃ki=1 σi. Note that γ′i = lkγi([σi]) is a cycle in the chain complex of

lkBDλi

n (E)(σi) = BDλi−µ(σi)

n (E \ σi).

Lemma 3.4 With σ as above, suppose that the following condition is satisfied:

• If σ is the disjoint union of the sets τ1, . . . , τk and lkγi([τi]) is nonzero forall i, then τi = σi for all i.

Thenlkγ([σ]) = ±lkγ1([σ1]) ∧ · · · ∧ lkγk

([σk]), (4)

and the exponent of the homology class of lkγ([σ]) in Hd−|σ|−1(BDλ−µ(σ)n (E \

σ); Z) divides the exponent of the homology class of γ in Hd−1(BDλn(E); Z).

Proof. By Lemma 2.1 and the assumption in the present lemma,

[σ] ∧ lkγ([σ]) = [σ1] ∧ lkγ1([σ1]) ∧ · · · ∧ [σk] ∧ lkγk([σk]).

Thus (4) follows immediately.Suppose that c is an element in Cd(BDλ

n(E); Z) such that

∂(c) = a · γ1 ∧ γ2 ∧ · · · ∧ γk = a · γ.

Then∂(lkc([σ])) = ±lk∂(c)([σ]) = ±lka·γ([σ]) = ±a · lkγ([σ]).

Since we may view lkc([σ]) as an element in the chain complex of BDλ−µ(σ)n (E \

σ), we are done. �

Assume that lkγi([σi]) is nonzero for 1 ≤ i ≤ k. Note that if the condition inLemma 3.4 is satisfied, then lkγi([σi]) does not contain any edge from σ in itsexpansion for 1 ≤ i ≤ k. Namely, suppose e ∈ σj appears in lkγi([σi]) for somej 6= i. Then each of lkγi([σi∪{x}]) and lkγj ([σj \{x}]) is nonzero, contradictingthe uniqueness of the partition (σ1, . . . , σk).

Recall that our goal is to detect 3-torsion in the homology of BDδn for various

values of n and δ. To achieve this, we will build a cycle of the form

z = φA1,B1 ∧ · · · ∧ φAk,Bk

and apply Lemma 3.3 to conclude that the exponent of the homology class ofz in the chain complex of BDδ

n divides 3. To prove that the exponent is indeed3 and not 1, we will construct a set σ such that Lemma 3.4 applies: There isa unique partition σ =

⋃ki=1 σ such that lkφAi,Bi

([σi]) is nonzero for all i. Inparticular,

lkz([σ]) = ±lkφA1,B1([σ1]) ∧ · · · ∧ lkφAk,Bk

([σk]).

By Lemma 3.4, it then suffices to show that the homology class of lkz([σ]) isnonzero in the chain complex of BD(δ,...,δ)−µ(σ)

n (En \ σ), where En is the edge

7

set of the complete graph on n vertices. In fact, it suffices to show that this istrue in the chain complex of the larger complex BD(δ,...,δ)−µ(σ)

n .We conclude this section with a useful lemma about cycles in matching

complexes.

Lemma 3.5 Let X be a nonempty vertex set and let X =⋃α

i=1(Ai ∪ Bi) bea partition of X such that |Bi| = |Ai| + 1 for each i. If at most one Ai

is empty, then the homology class of the cycle z =∧α

i=1 φAi,Bi is nonzero inH |X|−α

2 −1(MX ; Z).

Proof. First, assume that one of the sets Ai is empty and the others have size 1.Equivalently, z is a product consisting of one (0, 1)-cycle and α−1 (1, 2)-cycles.Theorem 3.1 yields that the homology class of z is nonzero.

Next, assume that we do not have the situation just examined. Let i havethe property that |Ai| is maximal and let a ∈ Ai and b ∈ Bi. By Lemma 3.4with σ = {ab}, we obtain that the exponent of the homology class of lkz(ab) in

H |X\{a,b}|−α2 −1

(MX\{a,b}; Z)

divides the exponent of the homology class of z in H |X|−α2 −1

(MX ; Z). Usinginduction on |X|, we deduce that the homology class of z is nonzero. Namely,lkφAi,Bi

(ab) = ±φAi\{a},Bi\{b}. �

4 All degrees at most two

We start by examining BDλn in the case that all λi are at most two. Let k ≥ 0

and consider n = 6k + 10. Define

(A,B) = ({2, 4, 6, . . . , 6k + 6} ∪ {6k + 8, 6k + 9, 6k + 10},{1, 3, 5, 7, . . . , 6k + 7} ∪ {6k + 8, 6k + 9, 6k + 10});

(C1, D1) = ({1}, {2, 3});(C2, D2) = ({4}, {5, 6});

· · ·(C2k+2, D2k+2) = ({6k + 4}, {6k + 5, 6k + 6});

(E,F ) = (∅, {6k + 7}).

To facilitate comparison between the different pairs, we illustrate the sets intable form in Table 1.

Each vertex is contained in exactly two sets and |A|+ |C1|+ · · ·+ |C2k+2|+|E| = 5k + 8, which implies that the cycle

z = φA,B ∧2k+2∧i=1

φCi,Di ∧ φE,F

8

Table 1: The pairs (A,B), (Ci, Di), (E,F ).

1 2 3 4 5 6 · · · 6k − 5 6k − 4 6k − 3 6k − 2 6k − 1 6k

B A B A B A · · · B A B A B AC1 D1 D1 C2 D2 D2 · · · C2k−1 D2k−1 D2k−1 C2k D2k D2k

6k + 1 6k + 2 6k + 3 6k + 4 6k + 5 6k + 6 6k + 7 6k + 8 6k + 9 6k + 10

B A B A B A B A B AC2k+1 D2k+1 D2k+1 C2k+2 D2k+2 D2k+2 F B A B

Table 2: A and B extended to {v1, . . . , vm}.

v1 v2 v3 · · · vm−1 vm

A A A · · · A AB B B · · · B B

is an element in C7+5k(BD210+6k; Z). Observe that φE,F is the empty cycle [∅].

Note that φA,B is a (3k+6, 3k+7)-cycle, φCi,Di a (1, 2)-cycle for i ∈ [2k+2],and (E,F ) a (0, 1)-cycle. Since we have at least two (1, 2)-cycles and one (0, 1)-cycle defined on disjoint vertex sets, we may apply Lemma 3.3 to conclude thatthe exponent of the homology class of z in H7+5k(BD2

10+6k; Z) divides 3.Note that φA,B is defined on the edge set AB = {ab : a ∈ A, b ∈ B} and

similarly for φCi,Di for all i and for the empty cycle φE,F . Define

σ = {i(6k + 11− i) : i ∈ [3k + 5]}.

One easily checks that the edges in σ are all present in AB and not present inCiDi for i ∈ [2k + 2]. For the latter, note that (x + y) mod 3 ∈ {0, 1} wheneverxy ∈ CiDi, whereas (x + y) mod 3 = 2 for all xy ∈ σ. We conclude that σ0 = σand σi = ∅ for i ∈ [2k +2] is the only way to partition σ as σ0 ∪σ1 ∪ · · · ∪σ2k+2

such that σ0 ⊆ AB and σi ⊆ CiDi for i ∈ [2k + 2]. It follows that

lkz([σ]) = ±lkφA,B([σ]) ∧

2k+2∧i=1

φCi,Di ∧ φE,F

= ±φ{6k+9},{6k+8,6k+10} ∧2k+2∧i=1

φCi,Di∧ φE,F

By Lemma 3.5, the cycle lkz([σ]) forms a nonzero homology class in the chaincomplex of M6k+10; thus Lemma 3.4 implies that the same is true for z in thechain complex of BD2

6k+10. We conclude that the exponent of the homologyclass of z in H7+5k(BD2

10+6k; Z) is equal to 3.We may extend the construction further by adding yet a set of vertices

{v1, . . . , vm}, m ≥ 0 and extending A and B according to Table 2.

9

Table 3: t ≥ κ pairs removed.

1 2 3 4 5 6 · · · 6k − 5 6k − 4 6k − 3 6k − 2 6k − 1 6k

C1 D1 D1 C2 D2 D2 · · · C2k−1 D2k−1 D2k−1 C2k D2k D2k

6k + 1 6k + 2 6k + 3 6k + 4 6k + 5 6k + 6 6k + 7 6k + 8 6k + 9 6k + 10

C2k+1 D2k+1 D2k+1 C2k+2 D2k+2 D2k+2 F B A B

v1 v2 · · · v2t′′−1 v2t′′ v2t′′+1 v2t′′+2 · · · vm−1 vm

A − · · · A − A A · · · A A− B · · · − B B B · · · B B

Adding the loops vivi to σ and arguing as before, we obtain that lkz([σ])remains the same cycle as before, implying that the exponent of the homologyclass of z in H7+5k+m(BD2

12+6k+m; Z) remains 3.To summarize, we have the following result:

Theorem 4.1 There is 3-torsion in H7+5k+m(BD210+6k+m; Z) whenever k ≥ 0

and m ≥ 0. Equivalently, Hd(BD2n; Z) contains 3-torsion whenever

5n− 86

≤ d ≤ n− 3.

To see that we indeed have equivalence, note that the transformation(nd

)=

(107

)+

(6 15 1

) (km

)is invertible over Z.

So far, we only considered the case that all degree bounds are 2. For a moregeneral construction, start as before by defining the pairs (A,B) and so on asin Tables 1 and 2. Write κ = 3k +5; note that 2κ+m is the number of vertices.Let m ≥ 0 and 0 ≤ t ≤ (2κ + m)/2 = κ + m/2. Decompose t as t = t′ + t′′,where t′ = min{t, κ} and t′′ = max{0, t − κ}. Note that t′′ ≤ m/2. Remove2t = t + t elements from the sets A and B in the following manner.

• For κ− t′ +1 ≤ i ≤ κ+ t′, remove i from A whenever i is even and removei from B whenever i is odd.

• For 1 ≤ i ≤ 2t′′, remove vi from A whenever i is even and remove vi fromB whenever i is odd.

Thus if t ≥ κ, then we obtain the situation in Table 3. If t < κ, say t = κ− 5,then we obtain the situation in Table 4.

Note that

z = φA,B ∧2k+2∧i=1

φCi,Di ∧ φE,F

10

Table 4: t = κ− 5 pairs removed.

1 2 3 4 5 6 · · · 6k − 5 6k − 4 6k − 3 6k − 2 6k − 1 6k

B A B A B − · · · − − − − − −C1 D1 D1 C2 D2 D2 · · · C2k−1 D2k−1 D2k−1 C2k D2k D2k

6k + 1 6k + 2 6k + 3 6k + 4 6k + 5 6k + 6 6k + 7 6k + 8 6k + 9 6k + 10

− − − − − A B A B AC2k+1 D2k+1 D2k+1 C2k+2 D2k+2 D2k+2 F B A B

v1 v2 v3 · · · vm−1 vm

A A A · · · A AB B B · · · B B

is a cycle of degree

(κ + 1 + m− t′ − t′′) + (2k + 2)− 1 = 7 + 5k + m− t

in the chain complex of a complex isomorphic to BD22κ+m−2t12t

2κ+m . As before, zcontains at least two (1, 2)-cycles and one (0, 1)-cycle defined on disjoint vertexsets, which implies that the exponent of the homology class of z divides 3; useLemma 3.3.

Define σ to be the union of the sets

{i(2κ + 1− i) : i ∈ [κ− t′]}.

and{v2j−1v2j : j ∈ [t′]} ∪ {vivi : 2t′ + 1 ≤ i ≤ m}.

Using the same argument as before, we obtain that all edges in σ belong to{ab : a ∈ A, b ∈ B} and to no other set. Hence lkz([σ]) is again equal to±φ{6k+9},{6k+8,6k+10} ∧

∧2k+2i=1 φCi,Di ∧ φE,F , which forms a nonzero homology

class in the chain complex of M6k+10; apply Lemma 3.5. By Lemma 3.4, weobtain that the exponent of the homology class of z in Hd(BD2n21n1

n1+n2; Z) is equal

to 3, where n1

n2

d

=

0107

+

2 0 0−2 6 1−1 5 1

tkm

⇐⇒

tkm

=

0−38

+

1/2 0 01/2 1 −1−2 −5 6

n1

n2

d

.

Note that the inequalities k ≥ 0, m ≥ 0, and 0 ≤ t ≤ 3k+5+m/2 are equivalentto

2n1 + 5n2 − 86

≤ d ≤ n1 + 2n2 − 62

(5)

11

Table 5: Three new vertices w1, w2, w3 added.

w1 w2 w3

− B AG H H

orw1 w2 w3

G H H

and n1, n2 ≥ 0. Moreover, k, m, t are integers if and only if n1 is an even integerand n2 and d are integers.

To obtain results for odd n1, for any given k, m, t with properties as above,start as before and add three new vertices w1, w2, w3. Introduce a new pair(G, H) and possibly extend (A,B) to {w2, w3} according to one of the twotables in Table 5. Choosing the table on the left, we obtain n1 = 2t + 1 andn2 = 2κ + m − 2t + 2. Choosing the table on the right, we obtain n1 = 2t + 3and n2 = 2κ + m − 2t. Adding one to t in the latter case, we get n1 = 2t + 1and n2 = 2κ + m− 2t + 2 in both cases. This tweak yields the modified bounds0 ≤ t ≤ 1 + κ + m/2 on t.

Taking into account the above modification to t, the element

z = φA,B ∧2k+2∧i=1

φCi,Di ∧ φE,F ∧ φG,H

is a cycle of degree 9+5k +m− t in the chain complex of a complex isomorphicto BD22+2κ+m−2t11+2t

3+2κ+m . Adding w2w3 to σ if w2 ∈ B and w3 ∈ A and arguing asbefore, we obtain that the homology class of z in Hd(BD2n21n1

n1+n2; Z) has exponent

3, where n1

n2

d

=

1129

+

2 0 0−2 6 1−1 5 1

tkm

⇐⇒

tkm

=

−1/2−7/2

8

+

1/2 0 01/2 1 −1−2 −5 6

n1

n2

d

.

The inequalities k ≥ 0, m ≥ 0, and 0 ≤ t ≤ 3k + 6 + m/2 are equivalent to

2n1 + 5n2 − 86

≤ d ≤ n1 + 2n2 − 72

, (6)

n1 ≥ 1, and n2 ≥ 0. Since n1 is odd, we may replace the second inequality in(6) with d ≤ n1+2n2−6

2 and hence obtain exactly the same inequalities as in (5).To summarize, we have the following result.

Theorem 4.2 For n1, n2 ≥ 0, Hd(BD2n21n1

n1+n2; Z) contains 3-torsion whenever

2n1 + 5n2 − 86

≤ d ≤ n1 + 2n2 − 62

.

In particular, BD2n21n1

n1+n2is not d 2n1+5n2−8

6 e-connected if either n1 +n2 ≥ 10 andn1 is even or n1 + n2 ≥ 13 and n1 is odd.

12

We have performed computer calculations [7] for all n1 and n2 satisfying n1 +2n2 ≤ 16, and for these values it turns out that there is 3-torsion in the groupHd(BD2n21n1

n1+n2; Z) if and only if either of the following is true.

(a) The inequalities in Theorem 4.2 are satisfied.

(b) d = n1+2n2−52 and n1 ≥ 7, where n1 is odd.

We conjecture that this is true in general. The general situation in case (b)remains unknown. Even in the special case n2 = 0, it is not known whetherHn−5

2(Mn; Z) = Hn−5

2(BD1

n; Z) contains 3-torsion for odd n ≥ 17.Note that (a) and (b) combined are not equivalent to 2n1+5n2−8

6 ≤ d ≤n1+2n2−5

2 . Namely, the latter inequalities include the case that d = n1+2n2−52

and n1 < 7 ≤ n1 + n2, which is not included in (b).There is 5-torsion in H4(BD2r114−2r

14−r ; Z) for 0 ≤ r ≤ 7 [5, 7]. In particular,there is nonvanishing homology below the lower bound on d in Theorem 4.2 for(n1, n2) ∈ {(0, 7), (2, 6), (4, 5)}.

5 All degrees equal to δ ≥ 3

In the previous section, we dealt with two parameters k and m. For fixed n1

and n2, the degree d was minimal when k was maximal and m was minimal.Increasing m and decreasing k, we obtained larger values of d. One may expressthis as saying that k yielded 3-torsion in lower-degree homology groups, whereasm yielded 3-torsion in higher-degree homology groups.

We use a similar approach in this section when dealing with the complexBDδ

n for δ ≥ 3. Since the construction is complicated, we proceed in steps,restricting to the parameter m and an auxiliary parameter ` in the first stepand introducing the parameter k in terms of ` in a later step.

5.1 Even number of vertices

Let m, ` ≥ 0 and δ ≥ 3. The construction to be described easily extends to anydegree δ ≥ 1, but we already examined δ = 1, 2 in the previous section. Let thevertex set be

V = {r+i , r−i : 1 ≤ i ≤ δ} ∪ {s+

i , s−i : 1 ≤ i ≤ δ + `} ∪ {t+i , t−i : 1 ≤ i ≤ δ + m}.

The size of this vertex set is always even; we will provide a modified constructionfor vertex sets of odd size in Section 5.2.

For 1 ≤ p ≤ δ, we define sets Ap, Bp, Xp in the manner described in Table 6.Write X =

⋃δp=1 Xp. Using table notation as before, we have the situation

in Table 7. For ` = 0, note that the assignments for s±i are identical to thosefor r±i . For m = 0, the same is true for t±i , except that the roles of Ap and Bp

are switched.

13

Table 6: Definition of the sets Ap, Bp, Xp.

Elements contained inAp Bp Xp

r+i , s+

i , t−i r−i , s−i , t+i for 1 ≤ i ≤ p− 1;t−p r−p , s−p r+

p , s+p , t+p

r−i , s−i , t+i r+i , s+

i , t−i for p + 1 ≤ i ≤ δ;s−i s+

i (p < δ) for δ + 1 ≤ i ≤ δ + `;s+

i , s−i (p = δ)t+i t−i for δ + 1 ≤ i ≤ δ + m;

Table 7: The pairs (Ap, Bp) and the set X.

r+1 r+

2 · · · r+δ−1 r+

δ

X B1 · · · B1 B1

A2 X · · · B2 B2

· · · · · · · · · · · · · · ·Aδ−1 Aδ−1 · · · X Bδ−1

Aδ Aδ · · · Aδ X

r−1 r−2 · · · r−δ−1 r−δB1 A1 · · · A1 A1

B2 B2 · · · A2 A2

· · · · · · · · · · · · · · ·Bδ−1 Bδ−1 · · · Bδ−1 Aδ−1

Bδ Bδ · · · Bδ Bδ

s+1 s+

2 · · · s+δ−1 s+

δ s+δ+1 · · · s+

δ+`

X B1 · · · B1 B1 B1 · · · B1

A2 X · · · B2 B2 B2 · · · B2

· · · · · · · · · · · · · · · · · · · · · · · ·Aδ−1 Aδ−1 · · · X Bδ−1 Bδ−1 · · · Bδ−1

Aδ Aδ · · · Aδ X X · · · X

s−1 s−2 · · · s−δ−1 s−δ s−δ+1 · · · s−δ+`

B1 A1 · · · A1 A1 A1 · · · A1

B2 B2 · · · A2 A2 A2 · · · A2

· · · · · · · · · · · · · · · · · · · · · · · ·Bδ−1 Bδ−1 · · · Bδ−1 Aδ−1 Aδ−1 · · · Aδ−1

Bδ Bδ · · · Bδ Bδ X · · · X

t+1 t+2 · · · t+δ−1 t+δ t+δ+1 · · · t+δ+m

X A1 · · · A1 A1 A1 · · · A1

B2 X · · · A2 A2 A2 · · · A2

· · · · · · · · · · · · · · · · · · · · · · · ·Bδ−1 Bδ−1 · · · X Aδ−1 Aδ−1 · · · Aδ−1

Bδ Bδ · · · Bδ X Aδ · · · Aδ

t−1 t−2 · · · t−δ−1 t−δ t−δ+1 · · · t−δ+m

A1 B1 · · · B1 B1 B1 · · · B1

A2 A2 · · · B2 B2 B2 · · · B2

· · · · · · · · · · · · · · · · · · · · · · · ·Aδ−1 Aδ−1 · · · Aδ−1 Bδ−1 Bδ−1 · · · Bδ−1

Aδ Aδ · · · Aδ Aδ Bδ · · · Bδ

14

Table 8: The pairs (A′p, B

′p) along with the set X.

r+1 r+

2 · · · r+δ−1 r+

δ r−1 r−2 · · · r−δ−1 r−δX X · · · X X B′

1 B′2 · · · B′

δ−1 B′δ

s+1 s+

2 · · · s+δ−1 s+

δ s+δ+1 · · · s+

δ+`

X X · · · X X X · · · X

s−1 s−2 · · · s−δ−1 s−δ s−δ+1 · · · s−δ+`

B′1 B′

2 · · · B′δ−1 B′

δ X · · · X

t+1 t+2 · · · t+δ−1 t+δ t+δ+1 · · · t+δ+m

X X · · · X X − · · · −

t−1 t−2 · · · t−δ−1 t−δ t−δ+1 · · · t−δ+m

A′1 A′

2 · · · A′δ−1 A′

δ − · · · −

For given ε ≥ 0, let w denote a cycle of degree ε− 1 in the chain complex ofMX ; hence

w ∈ Cε−1(MX ; Z). (7)

We will define w later on. Note that |X| = 3δ + 2`. Let

z =δ∧

p=1

φAp,Bp ∧ w. (8)

Since φAp,Bp is a (3δ + ` + m − 2, 3δ + ` + m − 1)-cycle for 1 ≤ p ≤ δ − 1 andφAδ,Bδ

is a (3δ + m− 2, 3δ + m− 1)-cycle, we obtain that

z ∈ C3δ2−2δ−1+(δ−1)`+δm+ε(BDδ6δ+2`+2m; Z). (9)

5.1.1 Defining σ

We want to define a set σ such that

lkz([σ]) = ±δ∧

p=1

φA′p,B′

p∧ w,

where the pairs (A′p, B

′p) are defined according to Table 8, in which we also

indicate the set X. It is clear that lkz([σ]) belongs to Cδ+ε(M6δ+2`; Z). If thiscycle is nonzero, then Lemma 3.4 yields that the exponent of the homology classof z is nontrivial. In particular, if w is such that Lemma 3.3 applies to z, thenthe exponent is equal to, or a nonzero multiple of, 3.

It remains to construct σ and specify w. First, we focus on σ. For p ∈ [δ],define

σRp = {r+

i r−j : i, j ∈ [δ], (i + j) mod (δ + 1) = p};σS

p = {s+i s−j : i, j ∈ [δ + `], (i + j) mod (δ + ` + 1) = p, i + j 6= 2δ + ` + 1};

σTp = {t+i t−j : i, j ∈ [δ + m], (i + j) mod (δ + m + 1) = p};

15

Table 9: Pairings in σRp , σS

p , and σTp .

p r+1 r+

2 r+3 · · · r+

δ−1 r+δ

1 − r−δ r−δ−1 · · · r−3 r−22 r−1 − r−δ · · · r−4 r−33 r−2 r−1 − · · · r−5 r−4· · · · · · · · · · · · · · · · · · · · ·

δ − 1 r−δ−2 r−δ−3 r−δ−4 · · · − r−δδ r−δ−1 r−δ−2 r−δ−3 · · · r−1 −

p s+1 s+

2 s+3 · · · s+

δ−1 s+δ s+

δ+1 · · · s+δ+`

1 − s−δ+` s−δ−1+` · · · s−3+` s−2+` s−1+` · · · s−22 s−1 − s−δ+` · · · s−4+` s−3+` s−2+` · · · s−33 s−2 s−1 − · · · s−5+` s−4+` s−3+` · · · s−4· · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

δ − 1 s−δ−2 s−δ−3 s−δ−4 · · · − s−δ+` s−δ−1+` · · · s−δδ s−δ−1 s−δ−2 s−δ−3 · · · s−1 − − · · · −

p t+1 t+2 t+3 · · · t+δ−1 t+δ t+δ+1 · · · t+δ+m

1 − t−δ+m t−δ−1+m · · · t−3+m t−2+m t−1+m · · · t−22 t−1 − t−δ+m · · · t−4+m t−3+m t−2+m · · · t−33 t−2 t−1 − · · · t−5+m t−4+m t−3+m · · · t−4· · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

δ − 1 t−δ−2 t−δ−3 t−δ−4 · · · − t−δ+m t−δ−1+m · · · t−δδ t−δ−1 t−δ−2 t−δ−3 · · · t−1 − t−δ+m · · · t−δ+1

16

see Table 9 for an overview. Note that the condition i + j 6= 2δ + ` + 1 forσS

p only applies to p = δ and means that we exclude edges containing s±i forδ + 1 ≤ i ≤ δ + `.

Let σR =⋃δ

p=1 σRp and define σS and σT analogously; note that

σR = {r+i r−j : i, j ∈ [δ], i + j mod (δ + 1) ∈ [δ]};

σS = {s+i s−j : i, j ∈ [δ + `], (i + j) mod (δ + ` + 1) ∈ [δ],

i + j 6= 2δ + ` + 1};σT = {t+i t−j : i, j ∈ [δ + m], (i + j) mod (δ + m + 1) ∈ [δ]}.

We have that the vertices r±i , s±i , and t±i all appear exactly δ − 1 times in σR,σS , and σT , respectively, with the exception that t±i appears exactly δ times inσT for δ + 1 ≤ i ≤ δ + m.

Write σp = σRp ∪ σS

p ∪ σTp and σ = σR ∪ σS ∪ σT =

⋃δp=1 σp. Later in

this section, we will define w such that Lemma 3.3 applies to z and such thatthe edges in σ are not used in w and hence only appear in the cycles φAp,Bp .In particular, given such a w, to apply Lemma 3.4, it suffices to show thatσ =

⋃δp=1 σp is the only partition σ =

⋃δp=1 τp such that lkφAp,Bp

([τp]) is nonzerofor each p ∈ [δ]. We prove this by examining each of σR, σS , and σT separately.

Lemma 5.1 σR =⋃δ

p=1 σRp is the unique partition σR =

⋃δp=1 τp such that

lkφAp,Bp([τp]) is nonzero for all p, and the analogous properties hold for σS

p andσT

p .

Proof. We start by considering the set σT . Suppose that we have a partitionσT =

⋃δp=1 τp such that lkφAp,Bp

([τp]) is nonzero for all p. Write M = δ+m+1.We use induction on a parameter y to show that τp = σT

p for all p. The inductionassumption is that t+i t−j belongs to τ(i+j) mod M whenever j < y and t+i t−j ∈ σT ;this is trivially true for y = 1.

First, assume that y ≤ δ. We want to prove that t+p−y+M t−y belongs to τp

for 1 ≤ p ≤ y − 1 and that t+p−yt−y belongs to τp for y + 1 ≤ p ≤ δ.Let 1 ≤ p ≤ y − 1. By induction, t+p−y+M t−y−(p−q) belongs to τq for 1 ≤ q ≤

p − 1. Moreover, if t+p−y+M t−y belongs to {ab : a ∈ Aq, b ∈ Bq} for some q ≥ y,then

p− y + M ≤ q − 1 ⇐⇒ y − (M + p− q) ≥ 1.

By induction, t+p−y+M t−y−(M+p−q) ∈ τq. As a consequence, t+p−y+M t−y mustbelong to τp′ for some p′ satisfying p ≤ p′ ≤ y − 1. Clearly, the only possibilityis that t+p−y+M t−y belongs to τp for 1 ≤ p ≤ y − 1.

Next, let y + 1 ≤ p ≤ δ. From what we just concluded, t+p−yt−y belongs to aset τp′ such that p′ ≥ y. For t+p−yt−y to belong to {ab : a ∈ Ap′ , b ∈ Bp′}, we alsoneed p′ ≥ p− y + 1. Now, by induction, t+p−yt−y−i ∈ τp−i for 1 ≤ i ≤ y− 1. Thisimplies that t+p−yt−y belongs to τp′ for some p′ ≥ p. Again, the only possibilityis that t+p−yt−y belongs to τp for y + 1 ≤ p ≤ δ.

17

Next, assume that δ + 1 ≤ y ≤ δ + m. We want to prove that t+p−y+M t−ybelongs to τp for 1 ≤ p ≤ δ. By induction, t+p−y+M t−y−(p−q) belongs to τq for1 ≤ q ≤ p − 1. As a consequence, t+p−y+M t−y must belong to τp′ for some p′

satisfying p ≤ p′ ≤ δ. Once again, we deduce that t+p−y+M t−y belongs to τp for1 ≤ p ≤ δ.

By symmetry, the proof for σR is identical; restrict to m = 0 and switch theroles of Ap and Bp.

It remains to consider σS . The proof is almost exactly the same as for σT

but with ` instead of m. The only change is that the range of p is 1 ≤ p ≤ δ− 1when δ + 1 ≤ y ≤ δ + `. �

5.1.2 Defining the cycle w

Our next goal is to define the cycle w. We do this using two different approaches.The first approach yields 3-torsion in higher-degree homology groups, whereasthe second approach yields 3-torsion in lower-degree homology groups.

Higher-degree homology. In the first approach, we consider the case that` = 0 and δ ≥ 4; it suffices to consider the second approach when δ = 3. Notethat X has size 3δ and that σ contains no edges with both endpoints in X.

First, assume that δ is even. Let 2 ≤ α ≤ δ/2 and partition X as

X =2α⋃i=1

Xi,

where each set Xi has an odd number of elements, |X1| = 1, |X2| = |X3| = 3,and |Xi| ≥ 3 for 4 ≤ i ≤ α. Partition each Xi as Xi = Ci ∪ Di, where|Di| = |Ci|+ 1. Define

w =2α∧i=1

φCi,Di .

We have that ε = 3δ2 −α, where ε−1 is the degree of w as specified in (7). Using

(9), we obtain that z as defined in (8) is an element in Cd(BDδn; Z), where(

nd + 1

)=

(6δ + 2m

3δ2 − 2δ + δm + 3δ2 − α

)=

(6δ + 2m

3δ2 − δ2 − α + δm

)=

(δ2 + α

) (6

3δ − 1

)+ (2m + 3δ − 6α)

(1δ2

).

By Lemma 3.3, the exponent of the homology class of z divides 3. By Lemma 3.5,the homology class of w is nonzero; hence Lemma 3.4 yields that the homologyclass of z equals 3.

Write k = δ/2 + α and m′ = 2m + 3δ − 6α. Since 2 ≤ α ≤ δ/2 and m ≥ 0,we have that δ/2 + 2 ≤ k ≤ δ and m′ ≥ 3δ − 6α = 6δ − 6k. Writing m insteadof m′ for simplicity, we obtain the following result.

18

Lemma 5.2 Let δ ≥ 4 be even. The group H(3δ−1)k+ δ2 m−1(BDδ

6k+m; Z) con-tains 3-torsion whenever δ/2 + 2 ≤ k ≤ δ and 6k + m ≥ 6δ with m even.

Note that the transformation (k, m) 7→ (n, d) is invertible over Z;(nd

)=

(0−1

)+

(6 1

3δ − 1 δ2

) (km

)⇐⇒

(km

)=

(−16

)+

(δ2 −1

−3δ + 1 6

) (nd

).

Next, assume that δ is odd and δ ≥ 5. Let 2 ≤ α ≤ (δ − 1)/2 and partitionX as before as

X =2α+1⋃i=1

(Ci ∪Di),

where again |Di| = |Ci| + 1, |C1| = 0, |C2| = |C3| = 1, and |Ci| ≥ 1 for4 ≤ i ≤ 2α + 1. Define

w =2α+1∧i=1

φCi,Di.

This time, ε = 3δ−12 − α. Using (9), we conclude that z as defined in (8) is an

element in Cd(BDδn; Z), where(

nd + 1

)=

(6δ + 2m

3δ2 − 2δ + δm + 3δ−12 − α

)=

(6δ + 2m

3δ2 − δ+12 − α + δm

)= (δ + 1 + 2α)

(3

3δ−12

)+

(m + 3δ−3

2 − 3α) (

2δ

).

As above, we obtain that the homology class of z equals 3.Write k = δ + 1 + 2α and m′ = m + 3δ−3

2 − 3α. Since 2 ≤ α ≤ (δ− 1)/2 andm ≥ 0, we have that δ + 5 ≤ k ≤ 2δ and m′ ≥ 3δ−3

2 − 3α = 3δ − 3k2 . Again,

write m instead of m′.

Lemma 5.3 Let δ ≥ 5 be odd. The group H 3δ−12 k+δm−1(BDδ

3k+2m; Z) contains3-torsion whenever δ + 5 ≤ k ≤ 2δ with k even and 3k + 2m ≥ 6δ.

Again, the transformation (k, m) 7→ (n, d) is invertible over Z;(nd

)=

(0−1

)+

(3 2

3δ−12 δ

) (km

)⇐⇒

(km

)=

(−23

)+

(δ −2

−3δ+12 3

) (nd

).

19

Lower-degree homology. In the second approach, we consider arbitrary ` ≥0 and δ ≥ 3. Divide X into two sets X1 and X2 such that |X2| is a multiple of3 (possibly zero) and such that X1 contains the set {s+

i : 1 ≤ i ≤ δ + `} andX2 contains the set {s−i : δ + 1 ≤ i ≤ δ + `}. Since there are 2δ ≥ 6 additionalvertices r+

i and t+i in X, these conditions are easy to achieve.Note that there are no edges in σ with both endpoints in X1 or with both

endpoints in X2; this is because we put all s+i in X1 and all s−i in X2. Write

|X1| = 3a + (2η + 1), where η ∈ {0, 1, 2}, and |X2| = 3b; we have that 3δ + 2` =3(a + b) + 2η + 1. Partition X1 as

X1 =a⋃

i=1

(Ci ∪Di) ∪ (E ∪ F )

such that |Ci| = 1 and |Di| = 2 for all i and such that |E| = η and |F | = η + 1.Similarly, partition X2 as

X2 =a+b⋃

i=a+1

(Ci ∪Di)

such that |Ci| = 1 and Di = 2 for all i. Define

w =a+b∧i=1

φCi,Di∧ φE,F . (10)

Note that w consists of a+ b (1, 2)-cycles and one (η, η +1)-cycle (thus a+ b+1(1, 2)-cycles if η = 1). It follows that lkz([σ]) consists of δ + a + b (1, 2)-cyclesand one (η, η+1)-cycle. In particular, the homology class of lkz([σ]) in the chaincomplex of M3(δ+a+b)+2η+1 = M3δ+2` is nonzero; use Theorem 3.1. ApplyingLemma 3.3 and Lemma 3.4, we conclude that the homology class of z hasexponent 3 whenever |X| = 3δ + 2` ≥ 7 and |X| /∈ {8, 9, 11, 14} and that thehomology class of z has exponent a finite multiple of 3 whenever |X| = 14. Fora fixed δ, we are in either situation for values of ` as specified in the followingtable.

δ `3 ≥ 2≥ 4 ≥ 0

We obtain one case for each possible congruence class of ` modulo 3.

• If ` = 3k, then w consists of δ +2k (1, 2)-cycles. Using (9), we obtain thatz is a cycle in Cd(BDδ

n; Z), where(n

d + 1

)=

(6δ

3δ2 − δ

)+ k ·

(6

3δ − 1

)+ m ·

(2δ

)= (k + δ) ·

(6

3δ − 1

)+ m ·

(2δ

).

For δ = 3, we need k ≥ 1 to get 3-torsion, whereas any k ≥ 0 yields3-torsion for δ ≥ 4.

20

• If ` = 3k+1, then w consists of δ+2k−1 (1, 2)-cycles and one (2, 3)-cycle;hence z is a cycle in Cd(BDδ

n; Z), where(n

d + 1

)=

(6δ + 23δ2

)+ k ·

(6

3δ − 1

)+ m ·

(2δ

)= (k + δ) ·

(6

3δ − 1

)+ (m + 1) ·

(2δ

).

For δ = 3, we need k ≥ 1 to get 3-torsion, whereas any k ≥ 0 yields3-torsion for δ ≥ 4.

• If ` = 3k+2, then w consists of δ+2k+1 (1, 2)-cycles and one (0, 1)-cycle;hence z is a cycle in Cd(BDδ

n; Z), where(n

d + 1

)=

(6δ + 4

3δ2 + δ − 1

)+ k ·

(6

3δ − 1

)+ m ·

(2δ

)= (k + δ + 1) ·

(6

3δ − 1

)+ (m− 1) ·

(2δ

)We get 3-torsion whenever k ≥ 0 for all δ ≥ 3.

Lemma 5.4

• For δ = 3, the group H4k+3m−1(BD33k+2m; Z) contains 3-torsion whenever

k ≥ 8 with k even and m ≥ −1.

• For δ ≥ 4 and δ even, the group H(3δ−1)k+ 12 δm−1(BDδ

6k+m; Z) contains3-torsion whenever k ≥ δ +1 and m ≥ −2 with m even and also wheneverk = δ and m ≥ 0, again with m even.

• For δ ≥ 5 and δ odd, the group H 3δ−12 k+δm−1(BDδ

3k+2m; Z) contains 3-torsion whenever k ≥ 2δ + 2 and m ≥ −1 with k even and also wheneverk = 2δ and m ≥ 0.

Proof. By the above discussion, for δ ≥ 4, we obtain 3-torsion for any linearcombination (

nd + 1

)= k′ ·

(6

3δ − 1

)+ m′ ·

(2δ

)(11)

such that k′ ≥ δ + 1 and m′ ≥ −1 or such that k′ ≥ δ and m′ ≥ 0. Forδ = 3, we obtain 3-torsion whenever k′ ≥ 4 and m′ ≥ −1. The substitutions(k, m) = (2k′,m′) for odd δ and (k, m) = (k′, 2m′) for even δ yield the lemma;these substitutions are for alignment with Lemmas 5.2 and 5.3. �

Combining Lemma 5.4 with Lemmas 5.2 and 5.3, we obtain the followingsummary of the situation for vertex sets of even size.

Lemma 5.5

21

Table 10: Modification to Table 7 with one additional vertex r−δ+1 in the casethat δ is even.

r−1 r−2 r−3 r−4 r−5 · · · r−δ r−δ+1

B1 A2 A1 A1 A1 · · · A1 A1

B2 B2 B2 A2 A2 · · · A2 A2

B3 B3 B3 A4 A3 · · · A3 A3

B4 B4 B4 B4 B4 · · · A4 A4

· · · · · · · · · · · · · · · · · · · · · · · ·Bδ−1 Bδ−1 Bδ−1 Bδ−1 Bδ−1 · · · Aδ Aδ−1

Bδ Bδ Bδ Bδ Bδ · · · Bδ Bδ

• For δ = 3, the group H4k+3m−1(BD33k+2m; Z) contains 3-torsion whenever

k ≥ 8 with k even and m ≥ −1.

• For δ ≥ 4 and δ even, the group H(3δ−1)k+ 12 δm−1(BDδ

6k+m; Z) contains3-torsion whenever k ≥ δ/2 + 2, m ≥ −2 with m even, and 6k + m ≥ 6δ.

• For δ ≥ 5 and δ odd, the group H 3δ−12 k+δm−1(BDδ

3k+2m; Z) contains 3-torsion whenever k ≥ δ + 5 with k even, m ≥ −1, and 3k + 2m ≥ 6δ.

5.2 Odd number of vertices

As promised, we now introduce a modified construction with an odd number ofvertices, and we do it by adding a vertex r−δ+1. The construction depends onthe parity of δ. For δ even, we modify the definitions of Ai and Bi according toTable 10, adding the diagonal of elements in boxes and moving everything onthe right of this diagonal one step to the right. The construction for odd δ issimilar, except that we add the element r−δ+1 to the set X; see Table 11.

For even δ, this modification yields a cycle z

z ∈ C3δ2− 32 δ−1+(δ−1)`+δm+ε(BDδ

6δ+1+2`+2m; Z). (12)

For odd δ, we need to redefine w, as we have added one element r−δ+1 to the setX; w is now a cycle of degree ε′ − 1 in the chain complex of MX for some ε′.This yields a cycle

z ∈ C3δ2− 32 δ− 3

2+(δ−1)`+δm+ε′(BDδ6δ+1+2`+2m; Z). (13)

We modify σ by redefining

σR = {r+i r−j : i, j ∈ [δ], i + j mod (δ + 2) ∈ [δ]} ∪ {r−2jr

−2j+1 : 1 ≤ j ≤ b δ

2c}.

When constructing w in Section 5.2.2 in the case that δ is odd, we will make surethat there are no edges of the form r+

i r−δ+1 in w. Assuming that w indeed has thisproperty, we can prove the following result, which is analogous to Lemma 5.1.

22

Table 11: Modification to Table 7 with one additional vertex r−δ+1 in the casethat δ is odd.

r−1 r−2 r−3 r−4 r−5 · · · r−δ−1 r−δ r−δ+1

B1 A2 A1 A1 A1 · · · A1 A1 A1

B2 B2 B2 A2 A2 · · · A2 A2 A2

B3 B3 B3 A4 A3 · · · A3 A3 A3

B4 B4 B4 B4 B4 · · · A4 A4 A4

· · · · · · · · · · · · · · · · · · · · · · · · · · ·Bδ−2 Bδ−2 Bδ−2 Bδ−2 Bδ−2 · · · Aδ−1 Aδ−2 Aδ−2

Bδ−1 Bδ−1 Bδ−1 Bδ−1 Bδ−1 · · · Aδ−1 Bδ−1 Aδ−1

Bδ Bδ Bδ Bδ Bδ · · · Bδ Bδ X

Table 12: Pairings in σRp \ Ep.

p r+1 r+

2 r+3 r+

4 · · · r+δ−1 r+

δ r+δ+1

1 − − r−δ r−δ−1 · · · r−4 r−3 r−22 r−1 − − r−δ · · · r−5 r−4 r−33 r−2 r−1 − − · · · r−6 r−5 r−4· · · · · · · · · · · · · · · · · · · · · · · · · · ·

δ − 1 r−δ−2 r−δ−3 r−δ−4 r−δ−5 · · · − − r−δδ r−δ−1 r−δ−2 r−δ−3 r−δ−4 · · · r−1 − −

23

Lemma 5.6 For p ∈ [δ], define

σRp = {r+

i r−j : i, j ∈ [δ], (i + j) mod (δ + 2) = p} ∪ Ep,

where Ep = {r−p r−p+1} if p is even and Ep = ∅ otherwise. Then σR =⋃δ

p=1 σRp

is the unique partition σR =⋃δ

p=1 τp such that lkφAp,Bp([τp]) is nonzero for all

p.

Remark. See Table 12 for an illustration of σRp \ Ep.

Proof. Suppose that we have a partition σR =⋃δ

p=1 τp such that lkφAp,Bp([τp])

is nonzero for all p. We want to prove that τp = σRp for all p. First, note that

the edge r−2jr−2j+1 only appears in φA2j ,B2j and in no other cycle φAi,Bi ; hence

we must have that r−2jr−2j+1 ∈ τ2j for 1 ≤ j ≤ bδ/2c. In particular, for the

remainder of the proof, we may forget about all elements in boxes in Tables 10and 11.

From now on, the proof is almost identical to the proof of Lemma 5.1.Specifically, we use induction on a parameter y to show that τp = σR

p for allp. Writing M = δ + 2, we want to prove that r+

p−y+Mr−y belongs to τp for1 ≤ p ≤ y − 2 and that r+

p−yr−y belongs to τp for y + 1 ≤ p ≤ δ.Let 1 ≤ p ≤ y − 2. By induction, r+

p−y+Mr−y−(p−q) belongs to τq for 1 ≤ q ≤p − 1, which implies that r+

p−y+Mr−y does not belong to τq for any such q. Ify is even, then r−y is not contained in Ay−1 ∪ By−1. If y is odd, then r−y−1r

−y

belongs to τy−1. In either case, r+p−y+Mr−y cannot belong to τy−1. Moreover, if

r+p−y+Mr−y belongs to {ab : a ∈ Aq, b ∈ Bq} for some q ≥ y, then

p− y + M ≤ q − 1 ⇐⇒ y − (M + p− q) ≥ 1.

By induction, r+p−y+Mr−y−(M+p−q) ∈ τq, which implies that r+

p−y+Mr−y does notbelong to τq.

We conclude that r+p−y+Mr−y must belong to τp′ for some p′ satisfying p ≤

p′ ≤ y − 2. Clearly, the only possibility is that r+p−y+Mr−y belongs to τp for

1 ≤ p ≤ y − 2.Next, let y + 1 ≤ p ≤ δ. From what we just concluded, r+

p−yr−y belongs to aset τp′ such that p′ ≥ y. For r+

p−yr−y to belong to {ab : a ∈ Ap′ , b ∈ Bp′}, we alsoneed p′ ≥ p− y + 1. Now, by induction, r+

p−yr−y−i ∈ τp−i for 1 ≤ i ≤ y− 1. Thisimplies that r+

p−yr−y belongs to τp′ for some p′ ≥ p. Again, the only possibilityis that r+

p−yr−y belongs to τp for y + 1 ≤ p ≤ δ. �

5.2.1 Even δ

For even δ ≥ 4, the set X remains the same as for even vertex sets and we mayhence define w as before. In the higher-degree case with ` = 0 as in Section 5.1.2,

24

we thus have that ε = 3δ2 − α, where 2 ≤ α ≤ δ/2. It follows that z as defined

in (12) is an element in Cd(BDδn; Z), where(

nd + 1

)=

(6δ + 1 + 2m

3δ2 − 3δ2 + δm + 3δ

2 − α

)=

(6δ + 1 + 2m3δ2 − α + δm

)=

(δ2 + α

) (6

3δ − 1

)+ (2m + 3δ + 1− 6α)

(1δ2

).

By Lemma 3.3, the exponent of the homology class of z divides 3. By Lemma 3.5,the homology class of w is nonzero; hence Lemma 3.4 yields that the homologyclass of z equals 3. Using the same approach as in the proof of Lemma 5.2, wededuce the following.

Lemma 5.7 Let δ ≥ 4 be even. The group H(3δ−1)k+ δ2 m−1(BDδ

6k+m; Z) con-tains 3-torsion whenever δ/2 + 2 ≤ k ≤ δ and 6k + m ≥ 6δ with m odd.

For the last inequality, use the fact that 6k + m ≥ 6δ + 1 and m is odd.Next, consider the lower-degree case with ` ≥ 0 as in Section 5.1.2. Define

w as in (10). We have the following situation for the three possible congruenceclasses of ` modulo 3; apply (12). In all three cases, we get 3-torsion for allk ≥ 0.

• If ` = 3k, then w consists of δ + 2k (1, 2)-cycles; hence z is a cycle inCd(BDδ

n; Z), where(n

d + 1

)=

(6δ + 1

3δ2 − 12δ

)+ k ·

(6

3δ − 1

)+ m ·

(2δ

)= (k + δ) ·

(6

3δ − 1

)+ (2m + 1) ·

(112δ

).

• If ` = 3k+1, then w consists of δ+2k−1 (1, 2)-cycles and one (2, 3)-cycle;hence z is a cycle in Cd(BDδ

n; Z), where(n

d + 1

)=

(6δ + 3

3δ2 + 12δ

)+ k ·

(6

3δ − 1

)+ m ·

(2δ

)= (k + δ) ·

(6

3δ − 1

)+ (2m + 3) ·

(112δ

).

• If ` = 3k+2, then w consists of δ+2k+1 (1, 2)-cycles and one (0, 1)-cycle;hence z is a cycle in Cd(BDδ

n; Z), where(n

d + 1

)=

(6δ + 5

3δ2 + 32δ − 1

)+ k ·

(6

3δ − 1

)+ m ·

(2δ

)= (k + δ + 1) ·

(6

3δ − 1

)+ (2m− 1) ·

(112δ

).

25

Lemma 5.8 For δ ≥ 4 and δ even, the group H(3δ−1)k+ δ2 m−1(BDδ

6k+m; Z) con-tains 3-torsion whenever k ≥ δ + 1 and m ≥ −1 with m odd and also wheneverk = δ and m ≥ 1, again with m odd.

Proof. The proof is almost identical to that of Lemma 5.4. �

Combining Lemma 5.8 with Lemma 5.7, we obtain the following summaryof the situation for vertex sets of odd size in the case of even degree.

Lemma 5.9 For δ ≥ 4 and δ even, the group H(3δ−1)k+ δ2 m−1(BDδ

6k+m; Z) con-tains 3-torsion whenever k ≥ δ/2 + 2, m ≥ −1 with m odd, and 6k + m ≥ 6δ.

5.2.2 Odd δ

Next, assume that δ is odd and δ ≥ 3. The set X no longer remains the same,as we have added the element r−δ+1 to X. First, consider the higher-degree casewith ` = 0 and δ ≥ 5 as in Section 5.1.2.

The size of X is 3δ + 1. Let 2 ≤ α ≤ (δ + 1)/2 and partition X as

X =2α⋃i=1

(Ci ∪Di),

where |Di| = |Ci| + 1, C1 = ∅, D1 = {r−δ+1}, |C2| = |C3| = 1, and |Ci| ≥ 1 for4 ≤ i ≤ 2α. Define

w =2α∧i=1

φCi,Di.

Since r−δ+1 is used for an empty cycle, there are no edges from σ in w. Weget that ε′ = 3δ+1

2 − α and hence that z as defined in (13) is an element inCd(BDδ

n; Z), where(n

d + 1

)=

(6δ + 1 + 2m

3δ2 − 32δ − 1

2 + δm + 3δ+12 − α

)=

(6δ + 1 + 2m3δ2 − α + δm

)= (δ + 2α)

(3

3δ−12

)+

(m +

3δ + 12

− 3α

) (2δ

).

As above, we obtain that the homology class of z equals 3.

Lemma 5.10 For δ ≥ 5 and δ odd, the group H 3δ−12 k+δm−1(BDδ

3k+2m; Z) con-tains 3-torsion whenever δ + 4 ≤ k ≤ 2δ + 1 with k odd, m ≥ −1, and3k + 2m ≥ 6δ.

For the last inequality, use the fact that 3k + 2m ≥ 6δ + 1 and k is odd.Next, consider the lower-degree case in Section 5.1.2 with ` ≥ 0 and δ ≥ 3.

Since δ is odd, the set X has size 3δ + 2` + 1. Partition X into two sets X1 and

26

X2 as before with |X2| a multiple of 3 but with the further assumption that allelements r+

i appear in X1 along with s+1 , . . . , s+

δ+` and the single element r−δ+1

appears in X2 along with s−δ+1, . . . , s−δ+`. Again, there are no edges in σ with

both endpoints in X1 or with both endpoints in X2. Defining w as in (10), weobtain 3-torsion for ` ≥ 0 for all odd δ ≥ 3. We have the following situationfor the three possible congruence classes of ` modulo 3; apply (13). In all threecases, we get 3-torsion for all k ≥ 0.

• If ` = 3k, then w consists of δ +2k (1, 2)-cycles and one (0, 1)-cycle; hencez is a cycle in Cd(BDδ

n; Z), where(n

d + 1

)=

(6δ + 1

3δ2 − 12δ − 1

2

)+ k ·

(6

3δ − 1

)+ m ·

(2δ

)= (k + δ + 1

2 ) ·(

63δ − 1

)+ (m− 1) ·

(2δ

).

• If ` = 3k + 1, then w consists of δ + 2k + 1 (1, 2)-cycles; hence z is a cyclein Cd(BDδ

n; Z), where(n

d + 1

)=

(6δ + 3

3δ2 + 12δ − 1

2

)+ k ·

(6

3δ − 1

)+ m ·

(2δ

)= (k + δ + 1

2 ) ·(

63δ − 1

)+ m ·

(2δ

).

• If ` = 3k + 2, then w consists of δ + 2k (1, 2)-cycles and one (2, 3)-cycle;hence z is a cycle in Cd(BDδ

n; Z), where(n

d + 1

)=

(6δ + 5

3δ2 + 32δ − 1

2

)+ k ·

(6

3δ − 1

)+ m ·

(2δ

)= (k + δ + 1

2 ) ·(

63δ − 1

)+ (m + 1) ·

(2δ

).

Lemma 5.11 For δ ≥ 3 and δ odd, the group H 3δ−12 k+δm−1(BDδ

3k+2m; Z) con-tains 3-torsion whenever k ≥ 2δ + 1 with k odd and m ≥ −1.

Proof. The proof is almost identical to that of Lemma 5.4. �

Combining Lemma 5.11 with Lemma 5.10, we obtain the following summaryof the situation for vertex sets of odd size in the case of odd degree.

Lemma 5.12

• For δ = 3, the group H4k+3m−1(BDδ3k+2m; Z) contains 3-torsion whenever

k ≥ 7 with k odd and m ≥ −1.

• For δ ≥ 5 and δ odd, the group H 3δ−12 k+δm−1(BDδ

3k+2m; Z) contains 3-torsion whenever k ≥ δ + 4 with k odd, m ≥ −1, and 3k + 2m ≥ 6δ.

27

5.3 Summary

Combining Lemma 5.5 and Lemma 5.9, we obtain the following summary of thesituation for even δ ≥ 4.

Theorem 5.13 For δ ≥ 4 and δ even, the group H(3δ−1)k+ δ2 m−1(BDδ

6k+m; Z)contains 3-torsion whenever k ≥ δ/2 + 2, m ≥ −2, and 6k + m ≥ 6δ. Equiva-lently, Hd(BDδ

n; Z) contains 3-torsion whenever

(3δ − 1)n− 86

≤ d ≤ δn− δ − 62

and n ≥ 6δ.

Combining Lemma 5.5 and Lemma 5.12, we obtain the following summaryof the situation for odd δ ≥ 5.

Theorem 5.14 For δ ≥ 5 and δ odd, the group H 3δ−12 k+δm−1(BDδ

3k+2m; Z)contains 3-torsion whenever k ≥ δ+4, m ≥ −1, and 3k+2m ≥ 6δ. Equivalently,Hd(BDδ

n; Z) contains 3-torsion whenever

(3δ − 1)n− 86

≤ d ≤ δn− δ − 62

and n ≥ 6δ.

Finally, we summarize the situation for δ = 3 using the same lemmas.

Theorem 5.15 For δ = 3, the group H4k+3m−1(BD33k+2m; Z) contains 3-torsion

whenever k ≥ 7 and m ≥ −1. Equivalently, Hd(BD3n; Z) contains 3-torsion

whenever8n− 8

6≤ d ≤ 3n− 9

2.

Note that the conclusion in all three theorems is the same: Hd(BDδn; Z)

contains 3-torsion whenever

(3δ − 1)n− 86

≤ d ≤ δn− δ − 62

and n ≥ 6δ. For the case δ = 3, this is true because the inequalities for d implythat n ≥ 19 > 6δ.

We have some hope that the following stronger result holds.

Conjecture 5.16 For δ ≥ 1, the group Hd(BDδn; Z) contains 3-torsion when-

ever(3δ − 1)n− 8

6≤ d ≤ δn− δ − 4

2.

Note that the inequalities in the conjecture imply that n ≥ 3δ + 4. For δ ≥ 3,this is less than the lower bound 6δ on n in the above theorems.

By Theorem 4.1, the conjecture is true for δ = 2. For δ = 1, the remainingcase to consider is d = n−5

2 and n ≥ 7 for n odd. Now, 3-torsion is known toexist for n ∈ {7, 9, 11, 13, 15} [2, 10, 7]; thus one need only consider odd n ≥ 17.For δ ≥ 3, the following cases remain.

28

• (3δ−1)n−86 ≤ d ≤ δn−δ−6

2 and 3δ + 10 ≤ n < 6δ.

• d = δn−δ−42 and n ≥ 3δ + 4. Note that δ is even or n is odd.

• d = δn−δ−52 , and n ≥ 3δ + 7. Note that δ is odd and n is even.

References

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[2] S. Bouc, Homologie de certains ensembles de 2-sous-groupes des groupessymetriques, J. Algebra 150 (1992), 187–205.

[3] X. Dong and M. L. Wachs, Combinatorial Laplacian of the matching com-plex, Electronic J. Combin. 9 (2002), no. 1, R17.

[4] J. Jonsson, Simplicial Complexes of Graphs, Lecture Notes in Mathematics1928, Springer, 2008.

[5] J. Jonsson, Five-torsion in the homology of the complex on 14 vertices, J.Algebraic Combin., Accepted.

[6] J. Jonsson, Exact sequences for the homology of the matching complex, J.Combin. Theory, Ser. A, Accepted.

[7] J. Jonsson, More torsion in the homology of the matching complex,Preprint, 2008.

[8] T. Jozefiak and J. Weyman. Representation-theoretic interpretation of aformula of D. E. Littlewood. Math. Proc. Cambridge Philos. Soc. 103(1988), 193–196.

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