1. Independence Complexes of Certain Families of Graphs
Rickard Fors
rifo02@kth.se
Master thesis in Mathematics at KTH
Presented August 19, 2011
Supervisor: Jakob Jonsson
2.
3. Abstract
The focus of this report will be on independence complexes constructed from independent
sets of members of sequences of graphs. For such independence complexes, we will study
generating 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 recursive
methods will be used instead. The generating functions, bounded formulae and recursive
equations will be compared to known number sequences, and where possible bijections
to other problems will be established. For the independence complexes of each graph
sequence, formulae will be given for where the homology groups are nonzero, as well as
in some cases formulae for the exact dimension of the homology groups for each complex.
6. Chapter 1
Introduction
In this text we study independence complexes of members of sequences of graphs. Other
studies of independence complexes include [1], [3], [7] and [9], the interested reader is
encouraged to read these for even more references. This chapter contains some prereq-
uisite knowledge, as well as definitions and theorems used throughout this report. In
chapters 2 and 3 ˜f-polynomials, Euler characteristics and homology groups are studied
for independence complexes of members of different sequences of complete graphs.
A geometric simplicial complex is a geometric object in Euclidean space. It can consist
of 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-dimensional
counterparts. These pieces are put together following certain rules. An abstract simpli-
cial complex is a combinatorial description of a geometric simplicial complex. It consists
of 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 ∆ is
the 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} form
line 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 as
sets with 1 element, edges as sets with 2 elements, and more generally, n-dimensional
simplices as sets with n + 1 elements. If ∆ is a simplicial complex, then dim(∆) is the
the maximal dimension of all faces of ∆. A face over the vertices {v0, v1, · · · , vk} will be
denoted 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 edge
between each other. In other words, if you consider an independent set I ⊆ G, then for
4
7. all vertices x, y ∈ I, x and y are not neighbors.
Taking a graph G, we construct the independence complex Ind(G) from the independent
sets 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 sets
of G.
Example.
In the left graph, the light red and green dots form an independent set, the dark red
and blue dots another.
Figure 1.1: From the graph in the left figure, we construct the independence complex in
the 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 dimension
the complex consists of. In this report, we will write the ˜f-polynomial of a simplicial
complex ∆ as ˜f(∆, t). The tilde sign above the f is used to indicate that we include the
empty 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)i
ai
where n is the dimension of the complex (the highest dimension of a face of ∆), and ai
is the number of faces of ∆ with dimension i. We see that the Euler characteristic of
5
8. the complex in Figure 1.1 is −1 + 6 − 8 + 2 = −1. It also follows directly from the above
formula 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 in
abstract algebra [4].
Given a field F, a vector space (V, F, +, ·) over F is a set V together with the two binary
operators addition + : V × V → V , and scalar multiplication · : F × V → V , that satisfy
the 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 all
v ∈ 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 is
thus 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-modules
over 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 ∈ 3Z
and any r in Z, the product rx ∈ 3Z. Thus 3Z is a Z-module and a submodule of Z.
6
9. 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 a
homomorphism 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 the
kernel 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 all
elements 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 have
im 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 → N
that 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-module
structure 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 this
module 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 an
equivalence relation ∼ on Z:
x ∼ y ⇔ y − x ∈ B.
7
10. We define Z/B to be the set of equivalence classes of this relation. This means that
every 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-module
structure 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 that
maps 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 quotient
Z/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 which
consists of only even numbers. For x − y to be even, either both x and y are even, or both
of them are odd, so one equivalence class will consist of all even numbers, and the other one
of 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, we
form 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 degree
n. We will write ˜Cn(∆) or ˜Cn for short.
For oriented simplices, we will write v0 ∧ v1∧· · · ∧vn since it is more convenient than
ev0,v1,...,vn . The ∧ is used to denote exterior product and satisfies
b ∧ a = −a ∧ b
a ∧ a = 0.
The boundary map δn: ˜Cn(∆) → ˜Cn−1(∆), is an F-module homomorphism that takes an
element 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)r
v0∧· · · vr−1 ∧ ˆvr ∧ vr+1∧· · · ∧vn
where ˆvr denotes the exclusion of vr.
8
11. 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 boundary
maps:
C: · · ·
δn+2
−−−→ ˜Cn+1
δn+1
−−−→ ˜Cn
δn
−→ ˜Cn−1
δn−1
−−−→ · · ·
C is finite if finitely many chain groups are nonzero. For each simplicial complex ∆, we
may 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 see
Section 1.7. Since all members of Zn(C) are cycles, and all members of Bn(C) are
boundaries, we have Bn(C) ⊆ Zn(C), since all boundaries are also cycles. We can now
define the homology in degree n of C, ˜Hn(C):
˜Hn(C) =
Zn(C)
Bn(C)
=
ker δn
im δn+1
For the definition of quotients, see Section 1.6. From here on, when we refer to the
simplicial homology ˜Hn(∆, F) of a simplicial complex ∆, we are really talking about the
homology of the associated simplicial chain complex C(∆, F). The tilde above the H is
used 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:
9
12. Using the definition above, let us calculate the homology of degree 1 of this complex, which
we 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 the
two 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 this
report.
Proposition 1.9.1 Let ∆ be a simplicial complex, then we have that
˜χ(∆) =
n≥−1
(−1)n
rank ˜Hn(∆).
For proof of this, see Theorem 22.2 in [8].
Proposition 1.9.2 Let ∆ and Γ be simplicial complexes. The Mayer-Vietoris sequence
for 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 in
the 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 is
thus 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 and
δ0(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
10
13. ˜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 when
dim(∆) ≤ 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 number
of 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 means
that B1(∆, F) = 0. Thus we have that ˜H1(∆, F) = Z1(∆, F). Theorem 1.9.5 in [2] gives
us 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 as
the 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 and
components, so 0 = e − v + k follows. Let us assume we have i cycles in ∆ and that
i = e − v + k. If we add an edge between two different components of ∆, we have an
extra edge but one less component, so the result still holds. Let us thus assume that we
add an edge between two vertices of the same component of ∆. If the new edge goes
between v0 and vk, these vertices were not neighbors before (since the edge would not
be new in that case), and since v0 and vk already were in the same component, there
is some path {v0, v1, ..., vk} in ∆. This means that adding the edge {v0, vk} creates a
new cycle {v0, v1, ..., vk, v0} ∈ ∆, and since we have the same number of vertices and
components, we get i + 1 = (e + 1) − v + k. By subtracting 1 on both sides the result
follows.
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(Gx) + t · ˜f(GN[x]), where x is any vertex ∈ ∆.
Proof. Ind(Gx) consists of all faces that do not contain x, which means the remaining
faces of Ind(G) all contain x. Let Σ be the family of all faces σ ∈ Ind(G). Then σ ∈ Σ
if and only if σx ∈ Ind(GN(x)). As a consequence
σ∈Σ
t|σ|
= t ˜f(GN[x])
and the result follows.
11
14. 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(GN[a])
j∗
n
−→ ˜Hn(Ga)
i∗
n
−→ ˜Hn(G)
t∗
n
−→ ˜Hn−1(GN[a])
j∗
n−1
−−−→ ˜Hn−1(Ga)
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. Then
there is the following long exact sequence:
· · ·
f∗
n+1
−−−→ ˜Hn+1(Ga) ⊕ ˜Hn+1(Gb)
g∗
n+1
−−−→ ˜Hn+1(G)
h∗
n+1
−−−→ ˜Hn(G{a, b})
f∗
n
−→ ˜Hn(Ga) ⊕ ˜Hn(Gb)
g∗
n
−→ ˜Hn(G)
h∗
n
−→ ˜Hn−1(G{a, b})
f∗
n−1
−−−→ ˜Hn−1(Ga) ⊕ ˜Hn−1(Gb)
g∗
n−1
−−−→ · · ·
Proof. Note that
Ind(Ga) ∩ Ind(Gb) = Ind(G{a, b})
Ind(Ga) ∪ Ind(Gb) = 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 neighbor
is a. Then
˜Hn(G) ∼= ˜Hn−1(GN[a])
for all n.
Proof. We note that b is an isolated vertex in Ga, which implies that ˜H(Ga) = 0 for
all 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 triangle
such that N(c) = {a, b}. Then
˜Hn(G) ∼= ˜Hn−1(GN[a]) ⊕ ˜Hn−1(GN[b])
for all n.
To prove this, one applies Proposition 1.9.6 to conclude that ˜Hn(G) ∼= ˜H(Ga)⊕ ˜H(Gb),
and then applies Corollary 1.9.7 to each of Ga and Gb. 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.
12
15. 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 isolated
in G {a, b, c}. We also have that the union ∆ ∪ Γ = Ind(G), since no independent set
contains 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.
13
16. 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 an
edge, then another vertex is added with edges to both of the previous ones, forming
G1. Next, another vertex comes with two more edges, as shown in Figure 2.1, forming
a second triangle. Here the index of the graph corresponds directly to the number of
3-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(Gkx) +
t · ˜f(GkN[x]) (see Section 1.9), where x is any vertex and N[x] = N(x) ∪ x (the union
of 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
14
17. As stated in section 1.2, the first 1 is the empty set, this is a constant coefficient. The
coefficient in front of t is the number of vertices in the graph. Let ˜χ(Gk) represent
˜χ(Ind(Gk)), using the f-vectors we can quickly calculate the Euler characteristics as
well 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(Gkx) + t · ˜f(GkN[x]),
choosing x as the bottom right vertex:
Figure 2.2: We see that G4x = G3, and that G4N[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(Gkx) + t · ˜f(GkN[x]), using the top right vertex as
x in Figure 2.4:
15
18. Figure 2.4: The red parts disappear when we remove x respectively x ∪ N(x).
We note that Gk{x} = Gk−1 and that GkN[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 for
members 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 by
setting 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). Note
that 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
+t2
x2
+
∞
k=3
akxk
−x−2tx−x2
−3tx2
−
∞
k=3
ak−1xk
−
∞
k=3
tak−3xk
=
= 1 + 2t + tx + tx2
+ t2
x2
+
∞
k=3
(akxk
− ak−1xk
− tak−3xk
) =
= 1 + 2t + tx + tx2
+ t2
x2
.
This yields:
f(x) =
1 + 2t + tx + tx2 + t2x2
1 − x − tx3
.
16
19. The generating function for the Euler characteristic is found by setting t = −1 and then
multiplying 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 the
recursive 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 values
of 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 ak
will 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 (see
Section 1.9)
˜Hn(G, Z) ∼= ˜Hn−1(GN[y], Z) ⊕ ˜Hn−1(GN[z], Z)
17
20. 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: GkN[y], the red part disappears from Gk.
We see that GkN[y] = Gk−4. Using the same technique, we also get that GkN[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, we
conclude 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 independence
complexes 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 ∆ (see
Section 1.9), we get
18
21. ˜H0(G0, Z) ∼= Z
˜H0(G1, Z) ∼= Z2
˜H0(G2, Z) ∼= Z2
˜H0(G3, Z) ∼= Z
˜H0(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), and
˜Hn(Gk, Z) ∼= ˜Hn−1(GkN[y], Z) ⊕ ˜Hn−1(GkN[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 complete
graph 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+1
i + k+1
i−1 + k+1
i−2 and a
b = 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 show
that:
x =
k + 1
i
+
k + 1
i − 1
+
k + 1
i − 2
.
If we assume the above, we have the inductive step:
19
22. x =
k
i
+
k
i − 1
+
k
i − 2
+
k
i − 1
+
k
i − 2
+
k
i − 3
.
That this is equal to k+1
i + k+1
i−1 + k+1
i−2 follows from using the formula
k+1
i = k
i + k
i−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 = 0 in theorem 2.1.2. We see that x = 0 if 0
≤ i ≤ k+3 (this follows directly from our definition of a
b ).
2.2 Another K3 Graph Sequence
Let us look at a different K3 graph sequence. This time each K3 graph shares a vertex
with 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.
2.2.1 ˜f-Polynomials and Euler Characteristics
Let ˜f(Gk) again represent ˜f(Ind(Gk), t) and ˜χ(Gk) represent ˜χ(Ind(Gk)). As we did for
the previous sequence, we use the formula ˜f(Gk) = ˜f(Gkx)+t· ˜f(GkN[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 Euler
characteristics are found using ˜χ(∆) = − ˜f(∆, −1) :
20
23. ˜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(Gkx) + t · ˜f(GkN[x]). Choosing the bottom right vertex as x, we get the graphs
in Figure 2.9
Figure 2.9: The red parts disappear when we remove x and N[x].
We see that Gkx = Gk−1 and GkN[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 odd
indices 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
24. 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+3t
and 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 indices
next:
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 + 2t2
x +
∞
k=2
bkxk
− x − 2tx − tx − 2t2
x −
∞
k=2
bk−1xk
−
∞
k=2
tbk−2xk
=
= 1 + 2t + tx +
∞
k=2
(bkxk
− (1 + t)bk−1xk
− 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 and
multiplying 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
25. a2k = a2k−1 + ta2k−3
However, since both a2k−1 and a2k−3 are odd, this can not either be used as is. Using
our 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 the
one 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 recursive
equation:
bk + bk−2 = 0.
So we get the characteristic polynomial
P(x) = x2
+ 1 = 0.
23
26. 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 same
characteristic polynomial, all we need to do to find the closed formula for it is to solve
the 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
27. 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
k
Using 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
28. 2.2.4 Homology Groups
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(GN[y], Z) ⊕ ˜Hn−1(GN[z], Z)
to calculate the homology groups for independence complexes of members with even
indices 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 the
left of x and z the one above to the left of x. Then N[y] will contain five vertices, and
N[z] will contain three. The index k corresponds directly to the number of vertices in
Gk, and the formula follows. Since this is very similar to what we did in section 2.1.4, a
more 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(GN[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 disappear
from 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 odd
values 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:
˜H−1(Gk, Z) ∼= 0 for all Gk since no member of the sequence is the empty complex.
We use ˜H0(∆, Z) ∼= Zk−1, where k is the number of connected components of ∆ (see
Section 1.9), to get
26
29. ˜H0(G0, Z) ∼= 0
˜H0(G1, Z) ∼= Z
˜H0(G2, Z) ∼= Z2
˜H0(G3, Z) ∼= Z
˜H0(G4, Z) ∼= Z
˜H0(G5, Z) ∼= 0
Using the fact that ˜H1(∆, Z) ∼= Ze−v+k, for graphs, where e is the number of edges, v
vertices 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) ∼= Z
˜H1(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 when
k = 1 (all those homology groups are listed earlier). We assume that the theorem is true
for 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 and
˜H1(G10, Z) ∼= ˜H0(G7, Z) ⊕ ˜H0(G5, Z). ˜H0(G5, Z) ∼= 0 and ˜H0(G7, Z) ∼= ˜H−1(G3, Z) ∼= 0
since G3 = ∅.
27
30. Next we assume that the theorem is true for k when i > 4. We now need to prove
that ˜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 = 2
x = 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 odd
since 4k is always even. So i = 1 or i = 3, which means that ˜Hk(G4k+1, Z) ∼= Z and
˜Hk(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
31. The above equalities use the proof for odd indices, and Theorem 2.2.2. This concludes
our proof.
Note that the only nonzero homology groups not included by this theorem are those for
k = 0. Those are listed explicitly earlier in this section.
29
32. 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, where
K4 is the complete graph on 4 vertices. Each K4 graph shares two vertices and one edge
with the previous one, and the index of a member of the sequence corresponds directly
to the number of K4 graphs connected this way.
Figure 3.1: The first four graphs of the sequence: G0, G1, G2 and G3.
3.1.1 ˜f-Polynomials and Euler Characteristics
As we did for the K3 graph sequences in the previous chapter, let ˜f(Gk) represent
˜f(Ind(Gk), t) and ˜χ(Gk) represent ˜χ(Ind(Gk)). We again make use of the formula
˜f(Gk) = ˜f(Gkx) + t · ˜f(GkN[x]) (see Section 1.9), to calculate the f-polynomials for
G0 to G3:
˜f(G0) = 1 + 2t
˜f(G1) = 1 + 4t
˜f(G2) = 1 + 6t + 4t2
˜f(G3) = 1 + 8t + 12t2
30
33. Since G0 and G1 are complete graphs (K2 and K4 respectively), their corresponding
independence complexes lack edges, and thus their t2 terms are 0. We use the formula
˜χ(∆) = − ˜f(∆, −1) to calculate the Euler characteristics for the first four members of
the 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(Gkx) + t · ˜f(GkN[x]) with
the top right vertex as x, we get:
Figure 3.3: Red parts disappear when we remove x and N[x].
We note that GkN[x] = Gk−2, and we apply the same formula to Gk{x}, choosing the
bottom right corner as y:
Figure 3.4: Red parts disappear when we remove y and N[y].
31
34. We see that (Gk{x}){y} = Gk−1, and that (Gk{x})N[y] = Gk−2. Putting it all
together 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 using
˜χ(∆) = − ˜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 + 2t
and 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−1xk
− 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 Euler
characteristic
f(x) =
1 + 2x
1 − x + 2x2
.
32
35. 3.1.3 Closed Formula
To find the closed formula for the ˜f-polynomial, let ak = ˜f(Ind(Gk)). We have the
recursive 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
2
x2 =
1 −
√
1 + 8t
2
Thus the closed formula for ak is
ak = A
1 +
√
1 + 8t
2
k
+ B
1 −
√
1 + 8t
2
k
where A and B are constants. We can find the values of A and B by solving the following
system of equations:
a0 = A + B = 1 + 2t
a1 = A 1+
√
1+8t
2 + B 1−
√
1+8t
2 = 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
36. These numbers correspond to the number of unique ways to cover a 3×n rectangle with
blocks 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 x
blocks of size 2 × 2. So a0 = 1 + 2t says that given a 3 × 2 rectangle, we can cover it in
blocks of only size 1 × 1 in 1 way (just placing them at the six available places), and in
two 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 tiles
at the bottom, or at the bottom leaving room for the 1 × 1 tiles at the top. Let us look
at another example:
Figure 3.5: The top left corner of G2 marked red, it corresponds to placing the red 2 × 2
block 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 rightmost
vertices (since all the other vertices are neighbors of the red). This corresponds to us
having two possible positions to place another 2 × 2 block next to the red one, either at
the same height to the right, or slightly below to the right. In general, when placing a
2 × 2 block in a 3 × n rectangle, we can never place another one directly below it, this
corresponds to two vertices in the same column in Gn−2 never belonging to the same
independent 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 choosing
two such vertices would correspond to trying to place a 2 × 2 block halfway on top of
another. 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
37. 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
k
Since a0 = 1 and a1 = 3 we get the following system of equations:
a0 = A + B = 1
a1 = A 1+i
√
7
2 + B 1−i
√
7
2 = 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
k
Since both roots contain imaginary numbers, this sequence is also erratic. Its first entries
are:
1, 3, 1, −5, −7, 3, 17, 11, −23, −45, ...
3.1.4 Homology Groups
Let ˜Hn(G, Z) denote the homology of degree n of Ind(G). To calculate the homology
groups for the independence complexes of members of this sequence, we use the formula
(see Section 1.9):
˜Hn(G, Z) ∼= ˜Hn−1(GN[y], Z) ⊕ ˜Hn−1(GN[z], Z) ⊕ ˜Hn−1(GN[w], Z)
where there exists some vertex x such that N(x) = {w, y, z}, and w, y, z are adjacent to
each other.
We see that GkN[y] = Gk−2, GkN[z] = GkN[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
38. Figure 3.6: x is red, y is light blue, z light green and w dark blue in the left graph. The
red parts disappear when we remove N[y] in the right.
Using formulae from section 1.9 we calculate the homology groups for the independence
complexes of the first four members of the sequence:
˜H−1(Gk, Z) ∼= 0 for all Gk since no member of the sequence is the empty complex.
˜H0(G0, Z) ∼= Z
˜H0(G1, Z) ∼= Z3
˜H0(G2, Z) ∼= Z2
˜H0(G3, Z) ∼= 0
˜H1(G0, Z) ∼= 0
˜H1(G1, Z) ∼= 0
˜H1(G2, Z) ∼= Z
˜H1(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+1
i 2i + k+1
i−1 2i−1 and a
b = 0 if b > a
or b < 0.
Proof. We begin by checking a base case: ˜H0(G0, Z) ∼= Zx, where x = 0+1
0 20 +
0+1
0−1 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
˜Hn(Gk, Z) ∼= ˜Hn−1(Gk−2, Z) ⊕ ˜Hn−1(Gk−3, Z) ⊕ ˜Hn−1(Gk−3, Z)
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
39. 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 + 2
similarly 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, as
show 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
40. 3.2.1 Homology Groups
As before let ˜Hn(G, Z) denote the homology of degree n of Ind(G). We use the same
formula as in Section 3.1.4 (see also Section 1.9):
˜Hn(G, Z) ∼= ˜Hn−1(GN[y], Z) ⊕ ˜Hn−1(GN[z], Z) ⊕ ˜Hn−1(GN[w], Z)
where there exists some vertex x such that N(x) = {w, y, z}. Let us say we pick the top
right vertex as x in G3 and let the vertex to the left of x be denoted y. If we remove
N[y] we get the graph in Figure 3.8.
Figure 3.8: The graph GN[y].
This is however not a graph in our sequence, but using
˜Hn(G) ∼= ˜Hn−1(GN[y])
where there is a vertex x such that y is the only neighbor of x (see Section 1.9), we get
the graph in Figure 3.9.
Figure 3.9: The resulting graph when choosing x as the rightmost vertex and removing
N[y] in Figure 3.8.
We see that if Gk was the original graph, this graph corresponds to Gk−3. If we let z
be the vertex below x in the original graph, we get GkN[z] = Gk−2. Finally, letting w
denote the vertex diagonally below to the left of x, we get GkN[w] = Gk−3. We omit
the details of the proofs of these last two equalities due to similarities with the proofs in
earlier sections. Putting it all together, we get
38
41. ˜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 independence
complexes of the first four members of the sequence:
˜H−1(Gk, Z) ∼= 0 for all Gk since no member of the sequence is the empty complex.
˜H0(G0, Z) ∼= Z
˜H0(G1, Z) ∼= Z2
˜H0(G2, Z) ∼= Z
˜H0(G3, Z) ∼= 0
˜H1(G0, Z) ∼= 0
˜H1(G1, Z) ∼= Z
˜H1(G2, Z) ∼= Z2
˜H1(G3, Z) ∼= Z3
Theorem 3.2.1
For even k, ˜Hk(G3k/2+i, Z) = 0 if and only if 0 ≤ i ≤ 3k/2 + 2.
For odd k, ˜Hk(G(3k−1)/2+i, Z) = 0 if and only if 0 ≤ i ≤ 3k/2 + 5/2.
Proof. We will use the formula
˜Hn(Gk, Z) ∼= ˜Hn−1(Gk−2, Z) ⊕ ˜Hn−1(Gk−3, Z) ⊕ ˜Hn−2(Gk−3, Z)
and mathematical induction. The base cases are easily verified by using the homology
groups 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
42. If we assume the theorem is true for some k, then
1. ˜Hm(G(3m−1)/2+i, Z) = 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) = 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) = 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 when
i < 0. For the upper bound, term 2 is nonzero when i = 3k/2 + 2 and all terms are zero
when 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) = 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) = 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) = 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 nonzero
when i = 3k/2 + 5/2 and all terms are zero when i > 3k/2 + 5/2, this concludes the
proof.
Finding an exact formula for the dimensions of the homology groups of the independence
complexes of this graph sequence proved to take too much time, and is therefore omitted
in this report. The table in Figure 3.10 shows these dimensions for ˜Hn(Gk). These are
to be read from left to right, that is, the numbers 1, 2, 1 on the top row correspond to
the dimension of ˜H0 for G0, G1 and G2.
We see that the rightmost column is filled with ones, and that the next column is a
sequence of integers all equal to k + 2 on the corresponding row. The column one step
further 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 between
the numbers increases. It is left as an open problem to find an exact formula for the
dimensions of these homology groups.
40
43. dim ˜Hn(Gk) n range of k
1 2 1 0 0 − 2
1 2 3 3 1 1 1 − 5
2 5 6 6 4 1 2 3 − 8
1 4 10 14 13 10 5 1 3 4 − 11
2 10 20 30 31 24 15 6 1 4 6 − 14
1 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 − 20
1 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 from
our sequence.
3.2.2 Euler Characteristics
We can calculate a recursive equation for the Euler characteristic for this sequence by
using the formulae
˜Hn(Gk, Z) ∼= ˜Hn−1(Gk−2, Z) ⊕ ˜Hn−1(Gk−3, Z) ⊕ ˜Hn−2(Gk−3, Z)
and
˜χ(∆) =
n≥−1
(−1)n
rank ˜Hn(∆)
(see Sections 3.2.1 and 1.9). We get
n≥−1
(−1)n
rank ˜Hn(Gk) =
n≥−1
(−1)n
rank ˜Hn−1(Gk−2) +
+
n≥−1
(−1)n
rank ˜Hn−1(Gk−3) +
n≥−1
(−1)n
rank ˜Hn−2(Gk−3).
We look at the terms on the right side:
n≥−1
(−1)n
rank ˜Hn−1(Gk−2) = [m = n − 1] =
m≥−2
(−1)m+1
rank ˜Hm(Gk−2)
n≥−1
(−1)n
rank ˜Hn−1(Gk−3) = [m = n − 1] =
m≥−2
(−1)m+1
rank ˜Hm(Gk−3)
n≥−1
(−1)n
rank ˜Hn−1(Gk−3) = [m = n − 2] =
m≥−3
(−1)m+2
rank ˜Hm(Gk−3)
Note that there are no faces of dim < −1, thus we have
41
44. m≥−2
(−1)m+1
rank ˜Hm(Gk−2) = −
m≥−1
(−1)m
rank ˜Hm(Gk−2) = −˜χ(Gk−2)
m≥−2
(−1)m+1
rank ˜Hm(Gk−3) = −
m≥−1
(−1)m
rank ˜Hm(Gk−3) = −˜χ(Gk−3)
m≥−3
(−1)m+2
rank ˜Hm(Gk−3) =
m≥−1
(−1)m
rank ˜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 we
omit the calculations for the generating function and bounded formula. Calculating
˜χ(G0) = 1 and ˜χ(G1) = 1 we can find all Euler characteristics for this sequence, they
are:
1, 1, −1, −1, 1, 1, −1, −1, 1, ...
42
45. Chapter 4
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43
