Independence Complexes

Independence Complexes of Certain Families of Graphs

Rickard [email protected]

Master thesis in Mathematics at KTHPresented August 19, 2011

Supervisor: Jakob Jonsson

Abstract

The focus of this report will be on independence complexes constructed from independentsets of members of sequences of graphs. For such independence complexes, we will studygenerating functions and closed formulae for the Euler characteristics and f -polynomials,as well as homology groups of different degrees. All of these can be computed by hand,although this quickly becomes tedious as well as really difficult to do, hence recursivemethods will be used instead. The generating functions, bounded formulae and recursiveequations will be compared to known number sequences, and where possible bijectionsto other problems will be established. For the independence complexes of each graphsequence, formulae will be given for where the homology groups are nonzero, as well asin some cases formulae for the exact dimension of the homology groups for each complex.

Contents

1 Introduction 41.1 Independent Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 f -polynomial and Euler Characteristic . . . . . . . . . . . . . . . . . . . . 51.3 Vector Spaces and Modules . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Homomorphisms and Isomorphisms . . . . . . . . . . . . . . . . . . . . . . 71.5 Direct Sums and Free Modules . . . . . . . . . . . . . . . . . . . . . . . . 71.6 Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.7 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.8 Simplicial Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.9 Useful Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.9.1 Homology in Low Degrees . . . . . . . . . . . . . . . . . . . . . . . 101.9.2 Independence Complex Formulae . . . . . . . . . . . . . . . . . . . 11

2 K3 Graph Sequences 142.1 A K3 Graph Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.1.1 f -Polynomials and Euler Characteristics . . . . . . . . . . . . . . . 142.1.2 Generating Functions . . . . . . . . . . . . . . . . . . . . . . . . . 162.1.3 Closed Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.1.4 Homology Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Another K3 Graph Sequence . . . . . . . . . . . . . . . . . . . . . . . . . 202.2.1 f -Polynomials and Euler Characteristics . . . . . . . . . . . . . . . 202.2.2 Generating Functions . . . . . . . . . . . . . . . . . . . . . . . . . 222.2.3 Closed Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.2.4 Homology Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3 K4 Graph Sequences 303.1 A K4 Graph Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.1.1 f -Polynomials and Euler Characteristics . . . . . . . . . . . . . . . 303.1.2 Generating Function . . . . . . . . . . . . . . . . . . . . . . . . . . 323.1.3 Closed Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.1.4 Homology Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.2 Another K4 Graph Sequence . . . . . . . . . . . . . . . . . . . . . . . . . 373.2.1 Homology Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2

3.2.2 Euler Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4 References 43

3

Chapter 1

Introduction

In this text we study independence complexes of members of sequences of graphs. Otherstudies of independence complexes include [1], [3], [7] and [9], the interested reader isencouraged to read these for even more references. This chapter contains some prereq-uisite knowledge, as well as definitions and theorems used throughout this report. Inchapters 2 and 3 f -polynomials, Euler characteristics and homology groups are studiedfor independence complexes of members of different sequences of complete graphs.

A geometric simplicial complex is a geometric object in Euclidean space. It can consistof points and line segments like regular graphs, but on top of that it may also con-tain simplices of higher dimensions such as triangles, tetrahedra and their n-dimensionalcounterparts. These pieces are put together following certain rules. An abstract simpli-cial complex is a combinatorial description of a geometric simplicial complex. It consistsof a family of sets that is closed under deletion of elements. This means that if we let∆ be a simplicial complex and τ ∈ ∆, then π ⊆ τ ⇒ π ∈ ∆. For example, if ∆ isthe simplicial complex containing the sets {a, b, c}, {a, b}, {a, c}, {b, c}, {a}, {b}, {c} and∅, we may view ∆ geometrically: {a, b, c} forms a triangle, {a, b}, {a, c} and {b, c} formline segments, {a}, {b} and {c} are vertices, and ∅ is the empty set.

The members of a simplicial complex are called faces. If τ is a face of a simplicial com-plex, then the dimension of τ , dim(τ) = |τ |−1. For the empty set we have dim(∅) = −1,points have dimension 0, line segments dimension 1 and so forth. We consider points assets with 1 element, edges as sets with 2 elements, and more generally, n-dimensionalsimplices as sets with n + 1 elements. If ∆ is a simplicial complex, then dim(∆) is thethe maximal dimension of all faces of ∆. A face over the vertices {v0, v1, · · · , vk} will bedenoted v0v1· · · vk for convenience.

1.1 Independent Sets

An independent set of a graph G consists of vertices of G that do not have an edgebetween each other. In other words, if you consider an independent set I ⊆ G, then for

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all vertices x, y ∈ I, x and y are not neighbors.

Taking a graph G, we construct the independence complex Ind(G) from the independentsets of G, by relating each independent set I ⊆ G to a simplex of dimension |I| − 1,where |I| is the number of vertices in I. Ind(G) is thus the family of independent setsof G.

Example.

In the left graph, the light red and green dots form an independent set, the dark redand blue dots another.

Figure 1.1: From the graph in the left figure, we construct the independence complex inthe right. The 2-dimensional simplices (triangles) are filled with grey for clarity.

1.2 f-polynomial and Euler Characteristic

The f -polynomial of a simplicial complex tells us how many faces of each dimensionthe complex consists of. In this report, we will write the f -polynomial of a simplicialcomplex ∆ as f(∆, t). The tilde sign above the f is used to indicate that we include theempty set in our calculations. We define the f -polynomial as:

f(∆, t) =∑τ∈∆

tdim(τ)+1,

where τ is a face of ∆ and dim(τ) is the dimension of τ .

Example.In Figure 1.1 the f -polynomial of the independence complex is: f(∆, t) = 1+6t+8t2 +2t3.Here the 1 corresponds to the empty set, 6 is the number of vertices, 8 the number of edges,and 2 the number of triangles. There are no simplices of higher dimensions in the complex.

The Euler characteristic of a simplicial complex ∆ is defined as

χ(∆) =

n∑i=−1

(−1)iai

where n is the dimension of the complex (the highest dimension of a face of ∆), and aiis the number of faces of ∆ with dimension i. We see that the Euler characteristic of

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the complex in Figure 1.1 is −1 + 6− 8 + 2 = −1. It also follows directly from the aboveformula that:

χ(∆) = −f(∆,−1).

1.3 Vector Spaces and Modules

This report assumes knowledge of basic theory of groups, rings and fields. For expla-nations about these concepts, the reader is referred to Fraleigh’s book A first course inabstract algebra [4].

Given a field F, a vector space (V,F,+, ·) over F is a set V together with the two binaryoperators addition + : V ×V → V , and scalar multiplication · : F×V → V , that satisfythe following axioms:

1. u+ (v + w) = (u+ v) + w for all u, v, w ∈ V.

2. u+ v = v + u for all u, v ∈ V.

3. There exists an element 0 ∈ V , such that v + 0 = v for all v ∈ V.

4. There exists an additive inverse of v, denoted −v, such that v + (−v) = 0 for allv ∈ V.

5. a(u+ v) = au+ av for all a ∈ F and all u, v ∈ V.

6. (a+ b)u = au+ bu for all a, b ∈ F and all u ∈ V.

7. a(bu) = (ab)u for all a, b ∈ F and all u ∈ V.

8. 1u = u, where 1 denotes the multiplicative identity in F, for all u ∈ V

The set of integers Z does not form a field, since it lacks multiplicative inverses and isthus not closed under division. This means that we cannot define vector spaces over Z.However, given any commutative ring R that satisfy the axioms above when R = F, V =(V,R,+, ·) is called an R-module. Since Z is a commutative ring, we can define Z-modulesover Z, which are the modules that we will be focusing on in this report. For exam-ple, using the axioms one can verify that every abelian group is a module over the ring Z.

Let M be an R-Module and S a subgroup of M . Then S is an R-submodule of M if,for any s ∈ S and any r ∈ R, rs ∈ S.

Example. Consider the Z-module of integer multiples of 3, we will call it 3Z. For any x ∈ 3Zand any r in Z, the product rx ∈ 3Z. Thus 3Z is a Z-module and a submodule of Z.

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1.4 Homomorphisms and Isomorphisms

A homomorphism is a map between two algebraic structures that preserves their struc-ture. In the case of modules, if M and N are R-modules, then a map f : M → N is ahomomorphism of R-modules if, for any r, s ∈ R and any m,n ∈M , the following holds:

f(rm+ sn) = rf(m) + sf(n).

Given an R-module homomorphism f : M → N , an important submodule of M is thekernel of f , which consists of all elements of M that f maps to zero:

ker f = {m ∈M : f(m) = 0}.

There is also an important submodule of N , called the image of f , which consists of allelements in N that at least one element of M is mapped to under f :

im f = {n ∈ N : n = f(m) for some m ∈M}.

Example. Consider the homomorphism f : Z → Z15, where f(x) = 3x mod 15. We haveim f = {0, 3, 6, 9, 12} and ker f = {15z, where z ∈ Z}.

A bijective homomorphism is an isomorphism. If there exists a map f : M → Nthat is an isomorphism, we say that M and N are isomorphic and we write M ∼= N .

1.5 Direct Sums and Free Modules

Let {Mi : 1 ≤ i ≤ k} be a family of R-modules. We define the direct sum

M = M1 ⊕M2 ⊕ · · · ⊕Mk

as the set of all vectors (m1,m2, . . . ,mk) such that mi ∈Mi for 1 ≤ i ≤ k. The R-modulestructure is inherited by component-wise addition and scalar multiplication:

(m1, . . . ,mk) + (n1, . . . , nk) = (m1 + n1, . . . ,mk + nk),

r(m1, . . . ,mk) = (rm1, . . . , rmk).

Using the direct sum of n copies of R, we form the R-module Rn. We refer to thismodule as the free R-module of rank n. For more discussion on free modules, see [4].

1.6 Quotients

Given two R-modules Z and B, with B being a submodule of Z, we can define anequivalence relation ∼ on Z:

x ∼ y ⇔ y − x ∈ B.

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We define Z/B to be the set of equivalence classes of this relation. This means thatevery element of Z/B has the form

[x] = {x+ b : b ∈ B} = {y : y − x ∈ B}.

For convenience, we can denote the equivalence classes as [x] = x + B. The R-modulestructure can be inherited by Z/B by defining

λ[x] + µ[y] = [λx+ µy].

The projection map p : Z → Z/B defined by p(z) = z + B is a homomorphism thatmaps each element z ∈ Z to the corresponding equivalence class z +B ∈ Z/B.

Example. Let Z = Z and B = 2Z, where 2Z consists of all even numbers. The quotientZ/2Z consists of only two elements, one for the odd numbers and one for the even numbers.To see this, remember that x ∼ y if and only if y − x ∈ B, and our B here is 2Z whichconsists of only even numbers. For x− y to be even, either both x and y are even, or bothof them are odd, so one equivalence class will consist of all even numbers, and the other oneof all the odd numbers. Using the projection map, when you add z ∈ Z with any b ∈ 2Z,the result is always odd if z is odd, and always even if z is even, since all b ∈ 2Z are even.

1.7 Definitions

Let F be a commutative ring, and let ∆ be a simplicial complex. For each n ≥ −1, weform a free F-module Cn(∆,F) with a basis indexed by the n-dimensional faces of ∆.This means that to every face v0v1· · · vn ∈ ∆, we associate a basis element ev0,v1,...,vn .We call this element an oriented simplex, and we call Cn(∆,F) the chain group of degreen. We will write Cn(∆) or Cn for short.

For oriented simplices, we will write v0 ∧ v1∧· · · ∧vn since it is more convenient thanev0,v1,...,vn . The ∧ is used to denote exterior product and satisfies

b ∧ a = −a ∧ ba ∧ a = 0.

The boundary map δn: Cn(∆)→ Cn−1(∆), is an F-module homomorphism that takes anelement x ∈ Cn(∆) and associates it to its boundary, which is an element in Cn−1(∆).Formally, if we have an oriented simplex v0 ∧ v1∧· · · ∧vn, then

δn(v0 ∧ v1∧· · · ∧vn) =n∑r=0

(−1)rv0∧· · · vr−1 ∧ vr ∧ vr+1∧· · · ∧vn

where vr denotes the exclusion of vr.

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Example. δ3(a ∧ b ∧ c ∧ d) = b ∧ c ∧ d− a ∧ c ∧ d+ a ∧ b ∧ d− a ∧ b ∧ c.

The boundary maps satisfy δn−1 ◦δn = 0, that is, if we take the boundary of a boundary,we get 0. For proof of this, see [6].

We define a chain complex C as a sequence of chain groups together with boundarymaps:

C: · · · δn+2−−−→ Cn+1δn+1−−−→ Cn

δn−→ Cn−1δn−1−−−→ · · ·

C is finite if finitely many chain groups are nonzero. For each simplicial complex ∆, wemay associate the simplicial chain complex C(∆;F).

1.8 Simplicial Homology

We define Zn(C) as

Zn(C) = ker δn = {z ∈ Cn : δn(z) = 0}

and we define Bn(C) as

Bn(C) = im δn+1 = {z ∈ Cn : z = δn+1(x) for some x ∈ Cn+1}

For definitions of ker and im , see Section 1.4, and for definitions of C, δn and Cn seeSection 1.7. Since all members of Zn(C) are cycles, and all members of Bn(C) areboundaries, we have Bn(C) ⊆ Zn(C), since all boundaries are also cycles. We can nowdefine the homology in degree n of C, Hn(C):

Hn(C) =Zn(C)

Bn(C)=

ker δnim δn+1

For the definition of quotients, see Section 1.6. From here on, when we refer to thesimplicial homology Hn(∆,F) of a simplicial complex ∆, we are really talking about thehomology of the associated simplicial chain complex C(∆,F). The tilde above the H isused to indicate that we are considering the reduced homology. In the reduced homology,the chain group of degree−1 is defined to be F, instead of 0 as in the unreduced homology.

Example. Let us again look at the simplicial complex from Section 1.1:

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Using the definition above, let us calculate the homology of degree 1 of this complex, whichwe will call ∆. The members of Z1(∆,F) are easy to see in the figure: the triangles{red, light green, dark green} and {brown, light blue, dark blue}, and the tetra-angle{red, dark green, light blue, brown}. The members B1(∆,F) are the boundaries of thetwo triangles in the figure, thus we have:

H1(∆,F) =Z1(∆,F)

B1(∆,F)=

ker δ1

im δ2

∼=F3

F2∼= F

1.9 Useful Formulae

In this section we will look at formulae that will be used throughout the rest of thisreport.

Proposition 1.9.1 Let ∆ be a simplicial complex, then we have that

χ(∆) =∑n≥−1

(−1)nrank Hn(∆).

For proof of this, see Theorem 22.2 in [8].

Proposition 1.9.2 Let ∆ and Γ be simplicial complexes. The Mayer-Vietoris sequencefor the pair (∆,Γ) is the following long exact sequence:

· · ·f∗n+1−−−→ Hn+1(∆)⊕ Hn+1(Γ)

g∗n+1−−−→ Hn+1(∆ ∪ Γ)h∗n+1−−−→ Hn(∆ ∩ Γ)

f∗n−→ Hn(∆)⊕ Hn(Γ)g∗n−→ Hn(∆ ∪ Γ)

h∗n−→ Hn−1(∆ ∩ Γ)f∗n−1−−−→ Hn−1(∆) ∪ Hn−1(Γ)

g∗n−1−−−→ · · ·

The image of each map in an exact sequence is equal to the kernel of the next map inthe sequence. For more on this sequence, see Section 2.2 in [6].

1.9.1 Homology in Low Degrees

If ∆ is any other simplicial complex than the void simplicial complex, ∅, Z−1(∆,F) =F · e∅. Note that the empty simplicial complex, {∅}, contains the empty set, and isthus not the same as the void simplicial complex, which contains nothing. If ∆ = {∅}we have B−1(∆,F) = 0, so H−1(∆,F) ∼= F. Otherwise ∆ contains some vertex a andδ0(a) = e∅. This in turn means that B−1(∆,F) = Z−1(∆,F), so we have H−1(∆,F) ∼= 0.To summarize,

H−1(∆,F) ∼= F if ∆ = {∅}H−1(∆,F) ∼= 0 otherwise

For (reduced) homology in degree 0, we have

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H0(∆,F) ∼= Fk−1

where k is the number of connected components of ∆. See [6] for proof of this formula.

When it comes to homology in degree 1, there is a formula that applies only whendim(∆) ≤ 1, or in other words, when ∆ is a graph:

Proposition 1.9.3 H1(∆,F) ∼= Fe−v+k, where e is the number of edges, v the numberof vertices, and k the number of components of ∆, and dim(∆) ≤ 1.

Proof. Since ∆ is a graph, there is no face τ ⊆ ∆ such that dim(τ) > 1, which meansthat B1(∆,F) = 0. Thus we have that H1(∆,F) = Z1(∆,F). Theorem 1.9.5 in [2] givesus that Z1(∆,F) is the cycle space of ∆, and that the rank of Z1(∆,F) is e− v + k . �

We claim that the number of cycles in ∆ is e− v+k. The number of cycles is defined asthe least number of edges we need to remove to make ∆ acyclic (a graph without cycles).When we have 0 edges, we obviously have 0 cycles and the same number of vertices andcomponents, so 0 = e − v + k follows. Let us assume we have i cycles in ∆ and thati = e − v + k. If we add an edge between two different components of ∆, we have anextra edge but one less component, so the result still holds. Let us thus assume that weadd an edge between two vertices of the same component of ∆. If the new edge goesbetween v0 and vk, these vertices were not neighbors before (since the edge would notbe new in that case), and since v0 and vk already were in the same component, thereis some path {v0, v1, ..., vk} in ∆. This means that adding the edge {v0, vk} creates anew cycle {v0, v1, ..., vk, v0} ∈ ∆, and since we have the same number of vertices andcomponents, we get i + 1 = (e + 1) − v + k. By subtracting 1 on both sides the resultfollows.

1.9.2 Independence Complex Formulae

In this section, we let Hn(G) denote Hn(Ind(G),F).

Proposition 1.9.4 Let G be a graph and let f(G) represent f(Ind(G), t). Then f(G) =f(G\x) + t · f(G\N [x]), where x is any vertex ∈ ∆.

Proof. Ind(G\x) consists of all faces that do not contain x, which means the remainingfaces of Ind(G) all contain x. Let Σ be the family of all faces σ ∈ Ind(G). Then σ ∈ Σif and only if σ\x ∈ Ind(G\N(x)). As a consequence∑

σ∈Σ

t|σ| = tf(G\N [x])

and the result follows. �

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Proposition 1.9.5 Let G be a graph and let a be a vertex in G. Then there is the fol-lowing long exact sequence:

· · ·i∗n+1−−−→ Hn+1(G)

t∗n+1−−−→ Hn(G\N [a])j∗n−→ Hn(G\a)

i∗n−→ Hn(G)t∗n−→ Hn−1(G\N [a])

j∗n−1−−−→ Hn−1(G\a)i∗n−1−−−→ Hn−1(G)

t∗n−1−−−→ · · ·

See Theorem 3.5.1 in [6] for more on this sequence.

Proposition 1.9.6 Let G be a graph and let a and b be adjacent vertices in G. Thenthere is the following long exact sequence:

· · ·f∗n+1−−−→ Hn+1(G\a)⊕ Hn+1(G\b)

g∗n+1−−−→ Hn+1(G)h∗n+1−−−→ Hn(G\{a, b})

f∗n−→ Hn(G\a)⊕ Hn(G\b) g∗n−→ Hn(G)h∗n−→ Hn−1(G\{a, b})

f∗n−1−−−→ Hn−1(G\a)⊕ Hn−1(G\b)g∗n−1−−−→ · · ·

Proof. Note that

Ind(G\a) ∩ Ind(G\b) = Ind(G\{a, b})Ind(G\a) ∪ Ind(G\b) = Ind(G)

The result then follows from applying Proposition 1.9.2. �

Corollary 1.9.7 Let G be a graph, and let a and b be vertices such that b’s only neighboris a. Then

Hn(G) ∼= Hn−1(G\N [a])

for all n.

Proof. We note that b is an isolated vertex in G\a, which implies that H(G\a) = 0 forall n. Applying Proposition 1.9.5 the result follows. �

Proposition 1.9.8 Let G be a graph, and let a, b and c be vertices forming a trianglesuch that N(c) = {a, b}. Then

Hn(G) ∼= Hn−1(G\N [a])⊕ Hn−1(G\N [b])

for all n.

To prove this, one applies Proposition 1.9.6 to conclude that Hn(G) ∼= H(G\a)⊕H(G\b),and then applies Corollary 1.9.7 to each of G\a and G\b. See Section 3.5 in [2] for details.

Proposition 1.9.9 Let G be a graph, and let a, b, c, and d be vertices forming a K4

graph such that N(d) = {a, b, c}. Then

Hn(G) ∼= Hn−1(G−N [a])⊕ Hn−1(G−N [b])⊕ Hn−1(G−N [c])

for all n.

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Proof. Let

∆ = Ind(G \ {a})Γ = Ind(G \ {b, c})

Then ∆ ∩ Γ = Ind(G \ {a, b, c}). This complex has zero homology, since d is isolatedin G \ {a, b, c}. We also have that the union ∆ ∪ Γ = Ind(G), since no independent setcontains more than one of the elements a, b and c.

Proposition 1.9.2 yields that

Hn(G) = Hn(G \ {a})⊕ Hn(G \ {b, c}).

The graf G \ {a} looks like the one in Corollary 1.9.8, but with b, c and d instead of a,b and c. This yields that

Hn(G \ {a}) = Hn−1(G \ {a} \N [b])⊕ Hn−1(G \ {a} \N [c]) =

= Hn−1(G \N [b])⊕ Hn−1(G \N [c])

since a ∈ N(b) and a ∈ N(c).

The graf G \ {b, c} looks like the one in Corollary 1.9.7 but with a, d instead of a,b. This yields

Hn(G \ {b, c}) = Hn−1(G \ {b, c} \N [a]) = Hn−1(G \N [a])

since b and c are in N [a]. The result follows from this.�

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Chapter 2

K3 Graph Sequences

2.1 A K3 Graph Sequence

Let K3 denote the complete graph with 3 vertices, also known as a 3-clique. The con-struction of the sequence starts with G0 which is just two vertices connected with anedge, then another vertex is added with edges to both of the previous ones, formingG1. Next, another vertex comes with two more edges, as shown in Figure 2.1, forminga second triangle. Here the index of the graph corresponds directly to the number of3-cliques (triangles) in it.

Figure 2.1: The first four graphs of the sequence: G0, G1, G2 and G3.

2.1.1 f-Polynomials and Euler Characteristics

Let f(Gk) represent f(Ind(Gk), t), that is, the f -polynomial for the independence com-plex of the (k+ 1):th member of the sequence. We use the formula f(Gk) = f(Gk\x) +t · f(Gk\N [x]) (see Section 1.9), where x is any vertex and N [x] = N(x) ∪ x (the unionof the neighborhood of x and x itself), to calculate the f -polynomials for 0 ≤ k ≤ 4 :

f(G0) = 1 + 2t

f(G1) = 1 + 3t

f(G2) = 1 + 4t+ t2

f(G3) = 1 + 5t+ 3t2

f(G4) = 1 + 6t+ 6t2

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As stated in section 1.2, the first 1 is the empty set, this is a constant coefficient. Thecoefficient in front of t is the number of vertices in the graph. Let χ(Gk) representχ(Ind(Gk)), using the f -vectors we can quickly calculate the Euler characteristics aswell using χ(∆) = −f(∆,−1):

χ(Ind(G0)) = −1 + 2 = 1

χ(Ind(G1)) = −1 + 3 = 2

χ(Ind(G2)) = −1 + 4− 1 = 2

χ(Ind(G3)) = −1 + 5− 3 = 1

χ(Ind(G4)) = −1 + 6− 6 = −1

Let us see what happens when we use the formula f(Gk) = f(Gk\x) + t · f(Gk\N [x]),choosing x as the bottom right vertex:

Figure 2.2: We see that G4\x = G3, and that G4\N [x] = G1.

So f(G4) = f(G3) + t · f(G1) = 1 + 5t+ 3t2 + t(1 + 3t) = 1 + 6t+ 6t2.This can be generalized to:

Theorem 2.1.1 For the f -polynomial we have

f(Gk) = f(Gk−1) + t · f(Gk−3).

Proof. Figure 2.3 shows the end to a graph in the sequence (that is, the rightmost part):

Figure 2.3: The rightmost part of a graph in the sequence.

We apply the formula f(Gk) = f(Gk\x) + t · f(Gk\N [x]), using the top right vertex asx in Figure 2.4:

15

Figure 2.4: The red parts disappear when we remove x respectively x ∪N(x).

We note that Gk\{x} = Gk−1 and that Gk\N [x] = Gk−3, from which f(Gk) = f(Gk−1)+t · f(Gk−3) follows. This concludes the proof (note that the situation is identical formembers with even and odd indices of the sequence). �

As said in Section 1.2 the equation for the Euler Characteristic χ(Gk) can be found bysetting t = −1 in Theorem 2.1.1. This yields that

χ(Gk) = χ(Gk−1)− χ(Gk−3).

2.1.2 Generating Functions

To find the generating function for the f -polynomial, we let ak = f(Ind(Gk), t). Notethat a0 = 1 + 2t, a1 = 1 + 3t and a2 = 1 + 4t+ t2. By Theorem 2.1.1:

ak − ak−1 − tak−3 = 0.

Now let f(x) =∑∞

k=0 akxk. We get

(1− x− tx3)f(x) =

∞∑k=0

akxk −

∞∑k=0

akxk+1 −

∞∑k=0

takxk+3

We shift indexes next:

(1− x− tx3)f(x) =

∞∑k=0

akxk −

∞∑k=1

ak−1xk −

∞∑k=3

tak−3xk =

= 1+2t+x+3tx+x2+4tx2+t2x2+

∞∑k=3

akxk−x−2tx−x2−3tx2−

∞∑k=3

ak−1xk−

∞∑k=3

tak−3xk =

= 1 + 2t+ tx+ tx2 + t2x2 +∞∑k=3

(akxk − ak−1x

k − tak−3xk) =

= 1 + 2t+ tx+ tx2 + t2x2.

This yields:

f(x) =1 + 2t+ tx+ tx2 + t2x2

1− x− tx3.

16

The generating function for the Euler characteristic is found by setting t = −1 and thenmultiplying f(x) with −1; this renders:

f(x) =1 + x

1− x+ x3.

2.1.3 Closed Formula

To find a closed formula for the Euler characteristic, let ak = χ(Ind(Gk)). We have therecursive equation:

ak − ak−1 + ak−3 = 0.

The characteristic polynomial for the Euler characteristic is then:

P (x) = x3 − x2 + 1 = 0.

Using an equation solving calculator, we get the approximate values:

x1 ≈ −0.755

x2 ≈ 0.877 + i0.745

x3 ≈ 0.877− i0.745

Thus the closed formula for ak is:

ak ≈ A(−0.755)k +B(0.877 + i0.745)k + C(0.877− i0.745)k

where A, B and C are constants.

The absolute value of the real root x1 is approximately 0.755, and the absolute valuesof x2 and x3 are approximately:√

0.8772 + 0.7452 ≈ 1.151

As the imaginary roots have a higher absolute value than the real root, the sequence akwill have an erratic behavior. Indeed, its first numbers are:

1, 2, 2, 1,−1,−3,−4,−3, 0, 4, 7, 7, 3,−4,−11,−14, ...

2.1.4 Homology Groups

Let Hn(G,Z) denote the homology of degree n of Ind(G). We use the formula (seeSection 1.9)

Hn(G,Z) ∼= Hn−1(G\N [y],Z)⊕ Hn−1(G\N [z],Z)

17

where y and z are vertices such that N(x) = {y, z} for some x, y ∈ N(z), and N [x] =N(x) ∪ x.

Let us again consider a member of Gk when k is even:

Figure 2.5: The rightmost part of a graph in the sequence.

Choosing the bottom right vertex as x, we get:

Figure 2.6: Our x marked in red, y and z in blue.

Let y be the top right corner, and look at N [y]:

Figure 2.7: Gk\N [y], the red part disappears from Gk.

We see that Gk\N [y] = Gk−4. Using the same technique, we also get that Gk\N [z] =Gk−5, where z is the neighbor to the left of x. So for even Gk we have:

Hn(Gk,Z) ∼= Hn−1(Gk−4,Z)⊕ Hn−1(Gk−5,Z).

The case for odd members of Gk is identical to the one with even members. Thus, weconclude that for k ≥ 5, k ∈ N:

Hn(Gk,Z) ∼= Hn−1(Gk−4,Z)⊕ Hn−1(Gk−5,Z).

Let us compute the homology groups for the first five members of the independencecomplexes of Gk:

H−1(Gk,Z) ∼= 0 for all Gk since no member of the sequence is the empty complex.

Using H0(∆,Z) ∼= Zk−1, where k is the number of connected components of ∆ (seeSection 1.9), we get

18

H0(G0,Z) ∼= ZH0(G1,Z) ∼= Z2

H0(G2,Z) ∼= Z2

H0(G3,Z) ∼= ZH0(G4,Z) ∼= 0

Using H1(∆,Z) ∼= Ze−v+k, for graphs (that is, simplicial complexes of degree 1 or lower),where e is the number of edges, v vertices and k components (see Section 1.9), andHn(Gk,Z) ∼= Hn−1(Gk\N [y],Z)⊕ Hn−1(Gk\N [z],Z), we get:

H1(G0,Z) ∼= 0

H1(G1,Z) ∼= 0

H1(G2,Z) ∼= 0

H1(G3,Z) ∼= 0

H1(G4,Z) ∼= Z

Example. H1(G4,Z) ∼= H0(G0,Z) ⊕ H0(K1,Z) ∼= Z ⊕ 0 ∼= Z, where K1 is the completegraph with one vertex.

There are no nonzero homology groups of higher degrees in these first five indepen-dence complexes.

Theorem 2.1.2 Hk(G4k+i,Z) ∼= Zx, where x =(k+1i

)+(k+1i−1

)+(k+1i−2

)and

(ab

)= 0 if

b > a or b < 0.

Proof. We use the formula Hn(Gk,Z) ∼= Hn−1(Gk−4,Z) ⊕ Hn−1(Gk−5,Z) and mathe-

matical induction. Adjusting the values we get:

Hk(G4k+i,Z) ∼= Hk−1(G4(k−1)+i,Z)⊕ Hk−1(G4(k−1)+i−1,Z)

The base cases have already been computed earlier in this section. We want to showthat:

x =

(k + 1

i

)+

(k + 1

i− 1

)+

(k + 1

i− 2

).

If we assume the above, we have the inductive step:

19

x =

(k

i

)+

(k

i− 1

)+

(k

i− 2

)+

(k

i− 1

)+

(k

i− 2

)+

(k

i− 3

).

That this is equal to(k+1i

)+(k+1i−1

)+(k+1i−2

)follows from using the formula(

k+1i

)=(ki

)+(ki−1

)three times. �

Corollary 2.1.3 Hk(G4k+i) is nonzero for 0 ≤ i ≤ k + 3

Proof. We are looking for values where x 6= 0 in theorem 2.1.2. We see that x 6= 0 if 0≤ i ≤ k+3 (this follows directly from our definition of

(ab

)). �

2.2 Another K3 Graph Sequence

Let us look at a different K3 graph sequence. This time each K3 graph shares a vertexwith the previous one. The index k of a member of the sequence, Gk, is equal to |Gk|+1.Unlike the previous sequence, this one contains ”half” K3 graphs as well.

Figure 2.8: The first four graphs of this sequence: G0, G1, G2 and G3.


Let f(Gk) again represent f(Ind(Gk), t) and χ(Gk) represent χ(Ind(Gk)). As we did forthe previous sequence, we use the formula f(Gk) = f(Gk\x)+t · f(Gk\N [x]) (reference),to calculate the f -polynomials for G0 to G3:

f(G0) = 1 + t

f(G1) = 1 + 2t

f(G2) = 1 + 3t

f(G3) = 1 + 4t+ 2t2

An example of the above calculations can be found in section 2.1. Now the Eulercharacteristics are found using χ(∆) = −f(∆,−1) :

20

χ(G0) = −1 + 1 = 0

χ(G1) = −1 + 2 = 1

χ(G2) = −1 + 3 = 2

χ(G3) = −1 + 4− 2 = 1

Theorem 2.2.1 The recursive formulae for the f -polynomials for this sequence are

f(Gk) = f(Gk−1) + t · f(Gk−2)

for members with odd indices and

f(Gk) = f(Gk−1) + t · f(Gk−3)

for members with even indices.

Proof. Let us look at the members with even indices. We again use the formula f(Gk)= f(Gk\x) + t · f(Gk\N [x]). Choosing the bottom right vertex as x, we get the graphsin Figure 2.9

Figure 2.9: The red parts disappear when we remove x and N [x].

We see that Gk\x = Gk−1 and Gk\N [x] = Gk−3 by counting the number of red vertices.f(Gk) = f(Gk−1)+ t · f(Gk−3) follows for members with even indices. The proof for oddindices is almost identical, and therefore we omit it here.

�

As before we find the recursive formulae for the Euler characteristics by using χ(∆) =−f(∆,−1); this gives us

χ(Gk) = χ(Gk−1)− χ(Gk−2).

for members with odd indices and

χ(Gk) = χ(Gk−1)− χ(Gk−3).

for members with even indices.

21

2.2.2 Generating Functions

Let ak = f(Ind(Gk), t). In section 2.2.1 we found that a0 = 1+t, a1 = 1+2t, a2 = 1+3tand a3 = 1 + 4t+ 2t2. For odd indices we have:

a2k+1 = a2k + ta2k−1

We can however not use this as is since 2k is even, so we use the formula for even indicesnext:

a2k = a2k−1 + ta2k−3

which gives us:a2k+1 = (1 + t)a2k−1 + ta2k−3

Now let bk = a2k+1, bk−1 = a2k−1 and so on, and let f(x) =∑∞

k=0 bkxk. We get

(1− (1 + t)x− tx2)f(x) =

∞∑k=0

bkxk −

∞∑k=0

(1 + t)bkxk+1 −

∞∑k=0

tbkxk+2

then we shift indices:

(1− (1 + t)x2 − tx4)f(x) =∞∑k=0

bkxk −

∞∑k=1

(1 + t)bk−1xk −

∞∑k=2

tbkxk =

= 1 + 2t+ x+ 4tx+ 2t2x+∞∑k=2

bkxk − x− 2tx− tx− 2t2x−

∞∑k=2

bk−1xk −

∞∑k=2

tbk−2xk =

= 1 + 2t+ tx+∞∑k=2

(bkxk − (1 + t)bk−1x

k − tbk−2xk) =

= 1 + 2t+ tx

Thus we have:

f(x) =1 + 2t+ tx

1− (1 + t)x− tx2.

The generating function for the Euler characteristic is found by setting t = −1 andmultiplying f(x) with −1:

f(x) =1 + x

1 + x2.

To find the generating function for the f -vector for even indices first look at our formula:

22

a2k = a2k−1 + ta2k−3

However, since both a2k−1 and a2k−3 are odd, this can not either be used as is. Usingour formula for odd indices we get:

a2k−1 = (1 + t)a2k−3 + ta2k−5

a2k−3 = (1 + t)a2k−5 + ta2k−7

We also have:

a2k−2 = a2k−3 + ta2k−5

a2k−4 = a2k−5 + ta2k−7

Putting the above together we get:

a2k = (1 + t)a2k−2 + ta2k−4

We omit the calculation for the generating function here, since it is very similar to theone for odd indices. It is:

f(x) =1 + t+ tx− t2x

1− (1 + t)x− tx2.

where

f(x) =∞∑k=0

a2kxk.

So for the Euler characteristic, we have:

f(x) =2x

1 + x2.

2.2.3 Closed Formulae

Let bk = a2k+1 as before, and ak = χ(Ind(Gk)). For odd indices, we have the recursiveequation:

bk + bk−2 = 0.

So we get the characteristic polynomial

P (x) = x2 + 1 = 0.

23

P (x) has the roots

x1 = i

x2 = −i

Thus our closed formula for the Euler characteristic for odd indices is

bk = A(i)k +B(−i)k

We can find the values of A and B by using b0 = a1 = 1 and b1 = a3 = 1:

b0 = A+B = 1

b1 = Ai−Bi = 1

Solving the above system of equations yields:

bk =

(1− i

2

)(i)k +

(1 + i

2

)(−i)k

Since the recursive equation for the Euler characteristic for even indices has the samecharacteristic polynomial, all we need to do to find the closed formula for it is to solvethe following system of equations, where ck = a2k, ck−1 = a2k−2 and so forth:

c0 = C +D = 0

c1 = Ci−Di = 2

This yields:ck = (−i)(i)k + (i)(−i)k.

The sequence of Euler characteristics is periodic, its values are:

0, 1, 2, 1, 0,−1,−2,−1, 0, 1, 2, 1, 0,−1,−2,−1, 0, ...

Now we will look at closed formulae for the f -vectors. Let ak = f(Ind(Gk)), bk = a2k+1

still, and ck = a2k. bk and ck again both have the same characteristic polynomial, P (x)

P (x) = x2 − (1 + t)x− t = 0.

This polynomial has the roots

24

x1 =1 + t+

√1 + 6t+ t2

2

x2 =1 + t−

√1 + 6t+ t2

2

So we have

bk = A

(1 + t+

√1 + 6t+ t2

2

)k+B

(1 + t−

√1 + 6t+ t2

2

)k

ck = C

(1 + t+

√1 + 6t+ t2

2

)k+D

(1 + t−

√1 + 6t+ t2

2

)kUsing the values c0 = a0 = 1 + t, c1 = a2 = 1 + 3t, b0 = a1 = 1 + 2t and b1 = a3 =1 + 4t+ 2t2, we get the following systems of equations:

A+B = 1 + 2t

A

(1 + t+

√1 + 6t+ t2

2

)+B

(1 + t−

√1 + 6t+ t2

2

)= 1 + 4t+ 2t2

C +D = 1 + t

C

(1 + t+

√1 + 6t+ t2

2

)+D

(1 + t−

√1 + 6t+ t2

2

)= 1 + 3t

Solving these yields

A =1 + 5t+ 2t2 + (1 + 2t)

√1 + 6t+ t2

2√

1 + 6t+ t2

B =−1− 5t− 2t2 + (1 + 2t)

√1 + 6t+ t2

2√

1 + 6t+ t2

C =1 + 4t− t2 + (1 + t)

√1 + 6t+ t2

2√

1 + 6t+ t2

D =−1− 4t+ t2 + (1 + t)

√1 + 6t+ t2

2√

1 + 6t+ t2

25


As before, let Hn(G,Z) denote the homology of degree n of Ind(G). We use the formula(see Section 1.9)

Hn(G,Z) ∼= Hn−1(G\N [y],Z)⊕ Hn−1(G\N [z],Z)

to calculate the homology groups for independence complexes of members with evenindices in the sequence. We get the following recursive formula:

Hn(Gk,Z) ∼= Hn−1(Gk−3,Z)⊕ Hn−1(Gk−5,Z)

This follows from picking the rightmost vertex as x, let y denote the neighbor to theleft of x and z the one above to the left of x. Then N [y] will contain five vertices, andN [z] will contain three. The index k corresponds directly to the number of vertices inGk, and the formula follows. Since this is very similar to what we did in section 2.1.4, amore extensive proof is omitted here. To calculate the homology groups when k is odd,we use the formula (see Section 1.9)

Hn(G,Z) ∼= Hn−1(G\N [y],Z)

where y is the only neighbor of some vertex x.

Figure 2.10: x marked with red, y with blue in the left graph. The red parts disappearfrom the right when we remove N [y].

In the right graph in figure 2.10 we see that N [y] contains four vertices. Thus for oddvalues of k we have:

Hn(Gk,Z) ∼= Hn−1(Gk−4,Z).

We proceed to calculate the homology groups for the first six members of the sequence:


We use H0(∆,Z) ∼= Zk−1, where k is the number of connected components of ∆ (seeSection 1.9), to get

26

H0(G0,Z) ∼= 0

H0(G1,Z) ∼= ZH0(G2,Z) ∼= Z2

H0(G3,Z) ∼= ZH0(G4,Z) ∼= ZH0(G5,Z) ∼= 0

Using the fact that H1(∆,Z) ∼= Ze−v+k, for graphs, where e is the number of edges, vvertices and k components (see Section 1.9), we get

H1(G0,Z) ∼= 0

H1(G1,Z) ∼= 0

H1(G2,Z) ∼= 0

H1(G3,Z) ∼= 0

H1(G4,Z) ∼= ZH1(G5,Z) ∼= Z

There are no nonzero homology groups of higher degrees in these independence com-plexes.

Theorem 2.2.2 Hk(G4k+i,Z) ∼= 0 for k > 0, if i < 0 or i > 4.

Proof. Let us begin by looking at i < 0, we have the base case Hk(G4k+i,Z) ∼= 0 whenk = 1 (all those homology groups are listed earlier). We assume that the theorem is truefor k when i < 0. We then need to prove that Hk+1(G4(k+1)+i,Z) ∼= 0.

Assume i < 0 and i odd. Hk+1(G4(k+1)+i,Z) ∼= Hk(G4k+i,Z) ∼= 0 by assumption.

Assume i < 0 and i even. This means that i < −1. We obtain that

Hk+1(G4(k+1)+i,Z) ∼= Hk(G4k+1+i,Z)⊕ Hk(G4k−1+i,Z) ∼= 0⊕ 0 ∼= 0

(note that 1 + i < 0 since i < −1).

Now we look at i > 4, for base cases with k = 1 we have: H1(G9,Z) ∼= H0(G5,Z) ∼= 0 andH1(G10,Z) ∼= H0(G7,Z) ⊕ H0(G5,Z). H0(G5,Z) ∼= 0 and H0(G7,Z) ∼= H−1(G3,Z) ∼= 0since G3 6= ∅.

27

Next we assume that the theorem is true for k when i > 4. We now need to provethat Hk+1(G4(k+1)+i,Z) ∼= 0.

Assume i > 4 and i odd. Hk+1(G4(k+1)+i,Z) ∼= Hk(G4k+i,Z) ∼= 0 by assumption.

Assume i > 4 and i even. This means that i > 5. We obtain that

Hk+1(G4(k+1)+i,Z) ∼= Hk(G4k+1+i,Z)⊕ Hk(G4k−1+i,Z) ∼= 0⊕ 0 ∼= 0

(note that −1 + i > 4 since i > 5). This concludes the proof.�

Theorem 2.2.3 Hk(G4k+i,Z) ∼= Zx, where 0 ≤ i ≤ 4, k > 0 and{x = 2 if i = 2x = 1 otherwise.

Proof. We begin by proving the theorem for odd indices, by using the formula Hn(Gk,Z) ∼=Hn−1(Gk−4,Z) and mathematical induction. For the base cases, we find that H1(G5,Z) ∼=Z and H1(G7,Z) ∼= Z.

Now we assume that Hk(G4k+i,Z) ∼= Z for odd indices, which means that i is oddsince 4k is always even. So i = 1 or i = 3, which means that Hk(G4k+1,Z) ∼= Z andHk(G4k+3,Z) ∼= Z.

We need to prove that Hk+1(G4(k+1)+i,Z) ∼= Z. Note that

4(k + 1) + i = 4k + 4 + i. So 4(k + 1) + i = 4k + 5 or 4k + 7.

This gives us:

1. Hk+1(G4k+5,Z) ∼= Hk(G4k+1,Z) ∼= Z

2. Hk+1(G4k+7,Z) ∼= Hk(G4k+3,Z) ∼= Z

which concludes the proof for odd indices.

Let us now look at even indices. We will use the formula Hn(Gk,Z) ∼= Hn−1(Gk−3,Z)⊕Hn−1(Gk−5,Z). This time the base case is H1(G4,Z) ∼= Z. We have:

1. Hk+1(G4k+4,Z) ∼= Hk(G4k+1,Z)⊕ Hk(G4k−1,Z) ∼= Z⊕ 0 ∼= Z.

2. Hk+1(G4k+6,Z) ∼= Hk(G4k+3,Z)⊕ Hk(G4k+1,Z) ∼= Z⊕ Z ∼= Z2.

3. Hk+1(G4k+8,Z) ∼= Hk(G4k+5,Z)⊕ Hk(G4k+3,Z) ∼= 0⊕ Z ∼= Z.

28

The above equalities use the proof for odd indices, and Theorem 2.2.2. This concludesour proof.

�

Note that the only nonzero homology groups not included by this theorem are those fork = 0. Those are listed explicitly earlier in this section.

29

Chapter 3

K4 Graph Sequences

3.1 A K4 Graph Sequence

In this section we look at the independence complexes of a sequence of K4 graphs, whereK4 is the complete graph on 4 vertices. Each K4 graph shares two vertices and one edgewith the previous one, and the index of a member of the sequence corresponds directlyto the number of K4 graphs connected this way.

Figure 3.1: The first four graphs of the sequence: G0, G1, G2 and G3.


As we did for the K3 graph sequences in the previous chapter, let f(Gk) representf(Ind(Gk), t) and χ(Gk) represent χ(Ind(Gk)). We again make use of the formulaf(Gk) = f(Gk\x) + t · f(Gk\N [x]) (see Section 1.9), to calculate the f -polynomials forG0 to G3:

f(G0) = 1 + 2t

f(G1) = 1 + 4t

f(G2) = 1 + 6t+ 4t2

f(G3) = 1 + 8t+ 12t2

30

Since G0 and G1 are complete graphs (K2 and K4 respectively), their correspondingindependence complexes lack edges, and thus their t2 terms are 0. We use the formulaχ(∆) = −f(∆,−1) to calculate the Euler characteristics for the first four members ofthe sequence:

χ(G0) = −1 + 2 = 1

χ(G1) = −1 + 4 = 3

χ(G2) = −1 + 6− 4 = 1

χ(G3) = −1 + 8− 12 = −5

Theorem 3.1.1 The recursive formula for the f -polynomials for this sequence is

f(Gk) = f(Gk−1) + 2t · f(Gk−2)

Proof. We begin by looking at the rightmost parts of a graph in the sequence:

Figure 3.2: Rightmost part of K4 sequence.

Let f(Ind(Gk), t) = f(Gk). Using the formula f(Gk) = f(Gk\x) + t · f(Gk\N [x]) withthe top right vertex as x, we get:

Figure 3.3: Red parts disappear when we remove x and N [x].

We note that Gk\N [x] = Gk−2, and we apply the same formula to Gk\{x}, choosing thebottom right corner as y:

Figure 3.4: Red parts disappear when we remove y and N [y].

31

We see that (Gk\{x})\{y} = Gk−1, and that (Gk\{x})\N [y] = Gk−2. Putting it alltogether we get

f(Gk) = t · f(Gk−2) + f(Gk−1) + t · f(Gk−2) = f(Gk−1) + 2t · f(Gk−2).

�

From this we can also directly find the formula for the Euler characteristic by usingχ(∆) = −f(∆,−1), this gives us

χ(Gk) = χ(Gk−1)− 2 · χ(Gk−2).

3.1.2 Generating Function

Let ak = f(Ind(Gk), t) as in section 2.1.2. In section 3.1.1 we found that a0 = 1 + 2tand a1 = 1 + 4t. We have:

ak − ak−1 − 2tak−2 = 0.

Let f(x) =∑∞

k=0 akxk. We get

(1− x− 2tx2)f(x) =∞∑k=0

akxk −

∞∑k=0

akxk+1 −

∞∑k=0

2takxk+2

Shifting indices yields

(1− x− 2tx2)f(x) =∞∑k=0

akxk −

∞∑k=1

ak−1xk −

∞∑k=2

2tak−2xk =

= 1 + 2t+ x+ 4tx+∞∑k=2

akxk − x− 2tx−

∞∑k=2

ak−1xk −

∞∑k=2

2tak−2xk =

= 1 + 2t+ 2tx+∞∑k=2

(akxk − ak−1x

k − 2tak−2xk) =

= 1 + 2t+ 2tx.

Thus we have

f(x) =1 + 2t+ 2tx

1− x− 2tx2.

Setting t = −1 and multiplying f(x) with −1 gives the generating function for the Eulercharacteristic

f(x) =1 + 2x

1− x+ 2x2.

32

3.1.3 Closed Formula

To find the closed formula for the f -polynomial, let ak = f(Ind(Gk)). We have therecursive equation:

ak − ak−1 − 2tak−2 = 0.

This means we have the following characteristic polynomial:

P (x) = x2 − x− 2t = 0.

This polynomial has the roots

x1 =1 +√

1 + 8t

2x2 =

1−√

1 + 8t

2

Thus the closed formula for ak is

ak = A

(1 +√

1 + 8t

2

)k+B

(1−√

1 + 8t

2

)kwhere A and B are constants. We can find the values of A and B by solving the followingsystem of equations:

a0 = A+B = 1 + 2t

a1 = A(

1+√

1+8t2

)+B

(1−√

1+8t2

)= 1 + 4t

After some calculations we get

A =

√1 + 8t+ 2t

√1 + 8t+ 6t+ 1

2√

1 + 8t

B =

√1 + 8t+ 2t

√1 + 8t− 6t− 1

2√

1 + 8t

We have

a0 = 1 + 2t

a1 = 1 + 4t

a2 = 1 + 6t+ 4t2

a3 = 1 + 8t+ 12t2

a4 = ...

33

These numbers correspond to the number of unique ways to cover a 3×n rectangle withblocks of size 2× 2 and 1× 1 as follows:

an corresponds to the rectangle of size 3 × (n + 2), and tx says that we are using xblocks of size 2× 2. So a0 = 1 + 2t says that given a 3× 2 rectangle, we can cover it inblocks of only size 1× 1 in 1 way (just placing them at the six available places), and intwo ways if we use 1 block of size 2× 2. The latter follows from that given a 2× 2 block,we may either place it at the top of the rectangle, which leaves room for two 1× 1 tilesat the bottom, or at the bottom leaving room for the 1× 1 tiles at the top. Let us lookat another example:

Figure 3.5: The top left corner of G2 marked red, it corresponds to placing the red 2× 2block as shown in the right figure.

In Figure 3.5 we can see that to form an independent set with two vertices from G2,where one of them is the red vertex, we can only choose from either of the two rightmostvertices (since all the other vertices are neighbors of the red). This corresponds to ushaving two possible positions to place another 2× 2 block next to the red one, either atthe same height to the right, or slightly below to the right. In general, when placing a2 × 2 block in a 3 × n rectangle, we can never place another one directly below it, thiscorresponds to two vertices in the same column in Gn−2 never belonging to the sameindependent set (since they are always neighbors). By the same argument, vertices im-mediately to the right or left of each other are always neighbors in Gn−2, and choosingtwo such vertices would correspond to trying to place a 2 × 2 block halfway on top ofanother. Since there is only one way to place 1× 1 blocks (one on each remaining tile),the result follows. See also [10].

Now let ak = χ(Ind(Gk)). For the Euler Characteristic we have

ak − ak−1 + 2ak−2 = 0.

So the characteristic polynomial is

P (x) = x2 − x+ 2 = 0,

34

which has the roots

x1 =1 + i

√7

2

x2 =1− i

√7

2

This time the closed formula for ak is

ak = A

(1 + i

√7

2

)k+B

(1− i

√7

2

)kSince a0 = 1 and a1 = 3 we get the following system of equations:

a0 = A+B = 1

a1 = A(

1+i√

72

)+B

(1−i√

72

)= 3

Solving for A and B we get

ak =

(1

2+

5

2i√

7

)(1 + i

√7

2

)k+

(1

2− 5

2i√

7

)(1− i

√7

2

)kSince both roots contain imaginary numbers, this sequence is also erratic. Its first entriesare:

1, 3, 1,−5,−7, 3, 17, 11,−23,−45, ...


Let Hn(G,Z) denote the homology of degree n of Ind(G). To calculate the homologygroups for the independence complexes of members of this sequence, we use the formula(see Section 1.9):

Hn(G,Z) ∼= Hn−1(G\N [y],Z)⊕ Hn−1(G\N [z],Z)⊕ Hn−1(G\N [w],Z)

where there exists some vertex x such that N(x) = {w, y, z}, and w, y, z are adjacent toeach other.We see that Gk\N [y] = Gk−2, Gk\N [z] = Gk\N [w] = Gk−3. Thus we have

Hn(Gk,Z) ∼= Hn−1(Gk−2,Z)⊕ Hn−1(Gk−3,Z)⊕ Hn−1(Gk−3,Z)

35

Figure 3.6: x is red, y is light blue, z light green and w dark blue in the left graph. Thered parts disappear when we remove N [y] in the right.

Using formulae from section 1.9 we calculate the homology groups for the independencecomplexes of the first four members of the sequence:


H0(G0,Z) ∼= ZH0(G1,Z) ∼= Z3

H0(G2,Z) ∼= Z2

H0(G3,Z) ∼= 0

H1(G0,Z) ∼= 0

H1(G1,Z) ∼= 0

H1(G2,Z) ∼= ZH1(G3,Z) ∼= Z5

There are no nonzero homology groups of higher degrees in these independence com-plexes.

Theorem 3.1.2 Hk(G2k+i,Z) ∼= Zx, where x =(k+1i

)2i+

(k+1i−1

)2i−1 and

(ab

)= 0 if b > a

or b < 0.

Proof. We begin by checking a base case: H0(G0,Z) ∼= Zx, where x =(

0+10

)20 +(

0+10−1

)20−1 = 1 + 0 = 1. So H0(G0,Z) ∼= Z and the base case holds.

Next we assume that the theorem is true for k. We use the formula


to get:

Hk(G2k+i,Z) ∼= Hk−1(G2(k−1)+i,Z)⊕ Hk−1(G2(k−1)+i−1,Z)⊕ Hk−1(G2(k−1)+i−1,Z).

36

We want to show that

x =

(k + 1

i

)2i +

(k + 1

i− 1

)2i−1.

Our assumption that the theorem is true for k gives us

x =

(k

i

)2i +

(k

i− 1

)2i−1 +

(k

i− 1

)2i−1 +

(k

i− 2

)2i−2 +

(k

i− 1

)2i−1 +

(k

i− 2

)2i−2 =

=

(k

i

)2i +

(k

i− 1

)2i +

(k

i− 2

)2i−1 +

(k

i− 1

)2i−1 =

=

(k + 1

i

)2i +

(k + 1

i− 1

)2i−1

which concludes the proof. �

Corollary 3.1.3 Hk(G2k+i,Z) is nonzero if and only if 0 ≤ i ≤ k + 2.

Proof. Setting i < 0 in Theorem 3.1.2 gives b < 0 in both terms. Setting i > k + 2similarly yields b > a in both terms and the corollary follows. �

3.2 Another K4 Graph Sequence

Let us see what happens when we add another level of K4 graphs on top of the first, asshow in Figure 3.7.

Figure 3.7: G0, G1, G2 and G3 of this sequence.

Here the index k of a graph Gk is equal to the number of columns of K4 graphs.

37


As before let Hn(G,Z) denote the homology of degree n of Ind(G). We use the sameformula as in Section 3.1.4 (see also Section 1.9):

Hn(G,Z) ∼= Hn−1(G\N [y],Z)⊕ Hn−1(G\N [z],Z)⊕ Hn−1(G\N [w],Z)

where there exists some vertex x such that N(x) = {w, y, z}. Let us say we pick the topright vertex as x in G3 and let the vertex to the left of x be denoted y. If we removeN [y] we get the graph in Figure 3.8.

Figure 3.8: The graph G\N [y].

This is however not a graph in our sequence, but using

Hn(G) ∼= Hn−1(G\N [y])

where there is a vertex x such that y is the only neighbor of x (see Section 1.9), we getthe graph in Figure 3.9.

Figure 3.9: The resulting graph when choosing x as the rightmost vertex and removingN [y] in Figure 3.8.

We see that if Gk was the original graph, this graph corresponds to Gk−3. If we let zbe the vertex below x in the original graph, we get Gk\N [z] = Gk−2. Finally, letting wdenote the vertex diagonally below to the left of x, we get Gk\N [w] = Gk−3. We omitthe details of the proofs of these last two equalities due to similarities with the proofs inearlier sections. Putting it all together, we get

38

Hn(Gk,Z) ∼= Hn−1(Gk−2,Z)⊕ Hn−1(Gk−3,Z)⊕ Hn−2(Gk−3,Z).

Using formulae from Section 1.9 we calculate the homology groups for the independencecomplexes of the first four members of the sequence:


H0(G0,Z) ∼= ZH0(G1,Z) ∼= Z2

H0(G2,Z) ∼= ZH0(G3,Z) ∼= 0

H1(G0,Z) ∼= 0

H1(G1,Z) ∼= ZH1(G2,Z) ∼= Z2

H1(G3,Z) ∼= Z3

Theorem 3.2.1

For even k, Hk(G3k/2+i,Z) 6= 0 if and only if 0 ≤ i ≤ 3k/2 + 2.

For odd k, Hk(G(3k−1)/2+i,Z) 6= 0 if and only if 0 ≤ i ≤ 3k/2 + 5/2.

Proof. We will use the formula


and mathematical induction. The base cases are easily verified by using the homologygroups listed earlier in this section. We begin by looking at the case when k is even:

Hk(G3k/2+i,Z) ∼= Hk−1(G3k/2+i−2,Z)⊕ Hk−1(G3k/2+i−3,Z)⊕ Hk−2(G3k/2+i−3,Z).

We proceed to look at each of the terms on the right hand side separately:

1. Hk−1(G3k/2+i−2,Z) = [m = k − 1] = Hm(G3(m+1)/2+i−2,Z) = Hm(G(3m−1)/2+i,Z)

2. Hk−1(G3k/2+i−3,Z) = [m = k−1] = Hm(G3(m+1)/2+i−3,Z) = Hm(G(3m−1)/2−1+i,Z)

3. Hk−2(G3k/2+i−3,Z) = [m = k − 2] = Hm(G3(m+2)/2+i−3,Z) = Hm(G3m/2+i,Z)

39

If we assume the theorem is true for some k, then

1. Hm(G(3m−1)/2+i,Z) 6= 0 if and only if 0 ≤ i ≤ 3m/2 + 5/2 = 3(k − 1) + 5/2 =3k/2 + 1 (since k is even m = k − 1 is odd)

2. Hm(G(3m−1)/2−1+i,Z) 6= 0 if and only if 0 ≤ i − 1 ≤ 3m/2 + 5/2 ⇒ 1 ≤ i ≤3m/2 + 7/2 = 3(k − 1)/2 + 7/2 = 3k/2 + 2

3. Hm(G3m/2−1+i,Z) 6= 0 if and only if 0 ≤ i ≤ 3m/2+2 = 3(k−2)/2+2 = 3k/2−1.

For the lower bound, terms 1 and 3 are nonzero when i = 0 and all terms are zero wheni < 0. For the upper bound, term 2 is nonzero when i = 3k/2 + 2 and all terms are zerowhen i > 3k/2 + 2, and the result follows. Now let us assume that k is odd:

1. Hk−1(G(3k−1)/2+i−2,Z) = [m = k−1] = Hm(G(3(m+1)−1)/2+i−2,Z) = Hm(G3m/2−1+i,Z)

2. Hk−1(G(3k−1)/2+i−3,Z) = [m = k−1] = Hm(G(3(m+1)−1)/2+i−3,Z) = Hm(G3m/2−2+i,Z)

3. Hk−2(G(3k−1)/2+i−3,Z) = [m = k−2] = Hm(G(3(m+2)−1)/2+i−3,Z) = Hm(G(3m−1)/2+i,Z)

Again assume the theorem is true for some k to get

1. Hm(G3m/2−1+i,Z) 6= 0 if and only if 0 ≤ i− 1 ≤ 3m/2 + 2⇒ 1 ≤ i ≤ 3m/2 + 3 =3(k − 1)/2 + 3 = 3k/2 + 3/2

2. Hm(G3m/2−2+i,Z) 6= 0 if and only if 0 ≤ i− 2 ≤ 3m/2 + 2⇒ 2 ≤ i ≤ 3m/2 + 4 =3(k − 1)/2 + 4 = 3k/2 + 5/2

3. Hm(G(3m−1)/2+i,Z) 6= 0 if and only if 0 ≤ i ≤ 3m/2 + 5/2 = 3(k − 2)/2 + 5/2 =3k/2− 1/2.

This time term 3 is nonzero if i = 0 and all terms are zero when i < 0. Term 2 is nonzerowhen i = 3k/2 + 5/2 and all terms are zero when i > 3k/2 + 5/2, this concludes theproof.

�

Finding an exact formula for the dimensions of the homology groups of the independencecomplexes of this graph sequence proved to take too much time, and is therefore omittedin this report. The table in Figure 3.10 shows these dimensions for Hn(Gk). These areto be read from left to right, that is, the numbers 1, 2, 1 on the top row correspond tothe dimension of H0 for G0, G1 and G2.

We see that the rightmost column is filled with ones, and that the next column is asequence of integers all equal to k + 2 on the corresponding row. The column one stepfurther to the left has a value equal to

(k+2

2

)(where the value of k again is taken from

the same row). Every time we move to the left, the complexity of the relation betweenthe numbers increases. It is left as an open problem to find an exact formula for thedimensions of these homology groups.

40

dim Hn(Gk) n range of k

1 2 1 0 0− 21 2 3 3 1 1 1− 5

2 5 6 6 4 1 2 3− 81 4 10 14 13 10 5 1 3 4− 11

2 10 20 30 31 24 15 6 1 4 6− 141 6 22 44 63 71 60 40 21 7 1 5 7− 17

3 17 48 96 138 158 146 106 62 28 8 1 6 9− 201 9 42 109 207 305 356 344 273 175 91 36 9 1 7 10− 23

.

Figure 3.10: Dimensions of homology groups of independence complexes of graphs fromour sequence.

3.2.2 Euler Characteristics

We can calculate a recursive equation for the Euler characteristic for this sequence byusing the formulae


andχ(∆) =

∑n≥−1

(−1)nrank Hn(∆)

(see Sections 3.2.1 and 1.9). We get

∑n≥−1

(−1)nrank Hn(Gk) =∑n≥−1

(−1)nrank Hn−1(Gk−2) +

+∑n≥−1

(−1)nrank Hn−1(Gk−3) +∑n≥−1

(−1)nrank Hn−2(Gk−3).

We look at the terms on the right side:∑n≥−1

(−1)nrank Hn−1(Gk−2) = [m = n− 1] =∑m≥−2

(−1)m+1rank Hm(Gk−2)

∑n≥−1

(−1)nrank Hn−1(Gk−3) = [m = n− 1] =∑m≥−2


∑n≥−1

(−1)nrank Hn−1(Gk−3) = [m = n− 2] =∑m≥−3


Note that there are no faces of dim < −1, thus we have

41

∑m≥−2

(−1)m+1rank Hm(Gk−2) = −∑m≥−1

(−1)mrank Hm(Gk−2) = −χ(Gk−2)

∑m≥−2

(−1)m+1rank Hm(Gk−3) = −∑m≥−1

(−1)mrank Hm(Gk−3) = −χ(Gk−3)

∑m≥−3

(−1)m+2rank Hm(Gk−3) =∑m≥−1

(−1)mrank Hm(Gk−3) = χ(Gk−3)

which gives us

χ(Gk) = −χ(Gk−2)− χ(Gk−3) + χ(Gk−3) = −χ(Gk−2).

This is however the same formula as the one for odd indices in Section 2.2.3, thus weomit the calculations for the generating function and bounded formula. Calculatingχ(G0) = 1 and χ(G1) = 1 we can find all Euler characteristics for this sequence, theyare:

1, 1,−1,−1, 1, 1,−1,−1, 1, ...

42

Chapter 4

References

[1] M. Bousquet-Melou, S. Linusson, E. Nevo, On the independence complex of squaregrids, J. Algebraic Combin. 27 (2008), 423–450

[2] R. Diestel, Graph Theory, http://diestel-graph-theory.com/basic.html

[3] A. Engstrom, Upper bounds on the Witten index for supersymmetric lattice modelsby discrete Morse theory, European J. Combin. 30 (2009), no. 2, 429-438

[4] John B. Fraleigh, A first course in abstract algebra, Addison-Wesley 1989

[5] A. Hatcher, Algebraic Topology, Cambridge University Press in 2002

[6] J. Jonsson, Introduction to simplicial homologyNotes available at http://www.math.kth.se/∼jakobj/homology.html

[7] J. Jonsson, Certain homology cycles of the independence complex of grid graphs,Disc. Comp. Geom. 43 (2010), No. 4, 927–950

[8] J. R. Munkres, Elements of Algebraic Topology, Addison-Wesley 1984

[9] J. Skwarski, Operations on a graph G that shift the homology of the independencecomplex of G 2010

[10] The On-Line Encyclopedia of Integer Sequences, http://oeis.org/A128099

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Independence Complexes