Homotopy Type of Independence Complexes of Certain Families of Graphs

TL;DR

This paper investigates the topological structure of independence complexes for specific graph families, showing they are homotopy equivalent to wedges of spheres or circles, and explores how graph perturbations affect this topology.

Contribution

It characterizes the homotopy types of independence complexes for generalized Mycielskian and categorical product graphs, and analyzes the effects of graph perturbations.

Findings

Independence complexes of generalized Mycielskian graphs are wedge sums of spheres.

Independence complexes of categorical product graphs are wedge sums of circles.

Perturbing a graph can make its independence complex homotopy equivalent to a suspension.

Abstract

We show that the independence complexes of generalised Mycielskian of complete graphs are homotopy equivalent to a wedge sum of spheres, and determine the number of copies and the dimensions of these spheres. We also prove that the independence complexes of categorical product of complete graphs are wedge sum of circles, upto homotopy. Further, we show that if we perturb a graph $G$ in a certain way, then the independence complex of this new graph is homotopy equivalent to the suspension of the independence complex of $G$.

Tables

Table 1. Table 1 : Betti numbers of Ind ( K 2 × K 3 × K n {\rm{Ind}}(K_{2}\times K_{3}\times K_{n} )

$n$	${\beta }_3$	${\beta }_i$, $i\ne 3$
2	$4$	$0$
3	$14$	$0$
$4$	$30$	$0$
$5$	$52$	$0$
$6$	$80$	$0$

Equations

$$l{k}_K(v):=\{\sigma \in K|v\notin \sigma \text{and}\sigma \cup \{v\}\in K\}.$$

$$s{t}_K(\sigma ):=\{\tau \in K|\sigma \cup \tau \in K\}.$$

$$S{C}_K(\sigma ):=\bigcup _{u\in \sigma }s{t}_K(\{u\}).$$

$${\rm{Ind}}(G)\simeq\Sigma(st_{{\rm{Ind}}(G)}(\{v\})\cap SC_{{\rm{Ind}}(G)}\big{(}N(v))).$$

$$\mathrm{Ind}(G)=\mathrm{Ind}(G_1\bigsqcup G_2)\simeq \mathrm{Ind}(G_1)\ast \mathrm{Ind}(G_2),$$

$$\mu (a_1)\succ a_1\prec \mu (a_2)\succ a_2\prec \mu (a_3)\succ a_3\dots \mu (a_t)\succ a_t\prec \mu (a_1),t\ge 2.$$

$$A_{x_i}=\{\sigma \in A_{x_{i-1}}^{\prime }|x_i\notin \sigma ,\text{and}\sigma \cup \{x_i\}\in A_{x_{i-1}}^{\prime }\},\text{where}{A}_{x_0}^{\prime }=P,$$

$${\mu }_{x_i}:A_{x_i}\to A_{x_{i-1}}^{\prime }\setminus A_{x_i}\text{by}{\mu }_{x_i}(\sigma )=\sigma \cup \{x_i\}\text{and}$$

$$A_{x_i}^{\prime }=A_{x_{i-1}}^{\prime }\setminus {\mathfrak{S}}_{x_i},\text{where}{\mathfrak{S}}_{x_i}=A_{x_i}\cup {\mu }_{x_i}(A_{x_i}).$$

$$\text{Ind}(K_m\times K_n)\simeq \underset{(m-1)(n-1)}{\bigvee }{\mathbb{S}}^1.$$

$$(a_1,b_1)<\dots <(a_1,b_n)<(a_2,b_1)<(a_3,b_1)<\dots <(a_m,b_1).$$

$$\mathrm{Ind}(\underset{(r-1)\text{-copies}}{\underbrace{K_2\times \dots \times K_2}}\times K_n)\simeq \underset{(n-1{)}^{2^{r-2}}}{\bigvee }{\mathbb{S}}^{2^{r-1}-1}.$$

$$\mathrm{Ind}(K_2\times K_3\times K_n)\simeq \underset{(n-1)(3n-2)}{\bigvee }{\mathbb{S}}^3.$$

$$\mathrm{Ind}(G)\simeq \mathrm{Ind}(G\setminus u^{\prime }).$$

$$\mathrm{Ind}(G)\simeq \underset{i=1}{\overset{k}{\bigvee }}\Sigma (\mathrm{Ind}(G-N[w_i])).$$

$$\Big{(}Y\bigsqcup Z\Big{)}/\sim,$$

$$\mathrm{Ind}(K_n\times L_0)\cong \mathrm{Ind}(K_n)=\underset{n-1}{\bigvee }{\mathbb{S}}^0.$$

$$\mathrm{Ind}(K_n\times L_2)\simeq \mathrm{Ind}(K_2\times K_n)\simeq \underset{n-1}{\bigvee }{\mathbb{S}}^1.$$

$$\mathrm{Ind}(K_n\times L_{3s+t})$$

$$\simeq \mathrm{Ind}(K_2\times K_n)\ast \mathrm{Ind}(K_n\times L_{3s+t-3})$$

$$\displaystyle\simeq\big{(}\bigvee\limits_{n-1}\mathbb{S}^{1}\big{)}\ast\big{(}{\rm{Ind}}(K_{n}\times L_{3(s-1)+t})\big{)}.$$

$$\mathrm{Ind}(K_n\times L_{3s+t})$$

$$\simeq \{\begin{array}{ll}\underset{(n-1{)}^{s+1}}{\bigvee }{\mathbb{S}}^{2s}& \text{if }t=0,\\ \{\text{point}\}& \text{if}t=1,\\ \underset{(n-1{)}^{s+1}}{\bigvee }{\mathbb{S}}^{2s+1}& \text{if }t=2.\end{array}$$

$$\mathrm{Ind}(I_{1,t})$$

$$\mathrm{Ind}(I_{1,2})\simeq \underset{n-1}{\bigvee }\Sigma (\mathrm{Ind}(K_2))\simeq \underset{n-1}{\bigvee }\Sigma ({\mathbb{S}}^0)\simeq \underset{n-1}{\bigvee }{\mathbb{S}}^1.$$

$$\text{Ind}(I_{1,t-2})$$

$$\displaystyle\textstyle{\rm{Ind}}(I_{1,t-2})\simeq\bigvee\limits_{(n-1)}\Sigma\Big{(}{\rm{Ind}}(I_{1,t-5}\sqcup K_{2})\Big{)}\simeq\bigvee\limits_{(n-1)}\Sigma^{2}\Big{(}{\rm{Ind}}(I_{1,t-5}\big{)}\Big{)}.$$

$$\mathrm{Ind}(I_{1,t-2})\simeq \underset{(n-1{)}^{k-1}}{\bigvee }{\Sigma }^{2(k-1)}(\mathrm{Ind}(I_{1,1}))\simeq \underset{(n-1{)}^{k-1}}{\bigvee }{\Sigma }^{2(k-1)}(\underset{n-1}{\bigvee }{\mathbb{S}}^0)\simeq \underset{(n-1{)}^k}{\bigvee }{\mathbb{S}}^{2(k-1)}.$$

$$\mathrm{Ind}(I_{1,t-2})\simeq \underset{(n-1{)}^{k-1}}{\bigvee }{\Sigma }^{2(k-1)}(\mathrm{Ind}(I_{1,2}))\simeq \underset{(n-1{)}^{k-1}}{\bigvee }{\Sigma }^{2(k-1)}(\underset{n-1}{\bigvee }{\mathbb{S}}^1)\simeq \underset{(n-1{)}^k}{\bigvee }{\mathbb{S}}^{2(k-1)+1}.$$

$$\displaystyle{\rm{Ind}}(M_{r}(K_{n}))\simeq\Sigma\big{(}st_{{\rm{Ind}}(M_{r}(K_{n}))}(w_{r})\cap SC_{{\rm{Ind}}(M_{r}(K_{n}))}(\{(1,r-1),\ldots,(n,r-1)\})\big{)}.$$

$$\mathrm{Ind}(M_r(K_n))\simeq \Sigma (K)\simeq \underset{n(n-1{)}^k}{\bigvee }{\mathbb{S}}^{2k}.$$

$$\mathrm{Ind}(M_r(K_n))\simeq \underset{(n-1{)}^{k+1}}{\bigvee }{\mathbb{S}}^{2k+1}.$$

$$\mathrm{Ind}(M_r(K_n))\simeq \Sigma (\mathrm{Ind}(K_n\times L_{r-3}))\simeq \Sigma (\underset{(n-1{)}^k}{\bigvee }{\mathbb{S}}^{2(k-1)})\simeq \underset{(n-1{)}^k}{\bigvee }{\mathbb{S}}^{2k-1},$$

Full text

Keywords : Independence complexes, generalised Mycielskian, discrete Morse theory

2010 Mathematics Subject Classification: primary 05C69, secondary 55P15

1 Introduction

A subset $I$ of vertex set of a graph $G$ is called independent, if the induced subgraph of $G$ on $I$ is a collection of isolated vertices. The independence complex, ${\rm{Ind}}(G)$, of a simple graph $G$ is the simplicial complex whose simplices are the independent sets of $G$. In last few years a lot of attention has been drawn towards the study of independence complexes of graphs.

In [2], Babson and Kozlov used the topology of independence complexes of cycles to prove a conjecture by Lovász. Meshulam, in [17], gave a connection between the domination number of a graph $G$ and certain homological properties of ${\rm{Ind}}(G)$; and their application to Hall-type theorems for coloured independent sets. Properties of independence complexes have also been used to study the Tverberg graphs [11] and the independent system of representatives [1]. For more on these complexes, interested reader is referred to [3, 4, 5, 7, 8, 9, 10, 14].

There are only a few classes of graphs for which a closed form formula for the homotopy type of independence complexes is known. For instance, see [5] for stabe Kneser graphs, [13] for forests, [16] for cycle graphs, and [18] for a family of regular bipartite graphs. In this article, we compute the homotopy type of independence complex of a few families of graphs, and give the exact formula for the same. We also give results about the homotopy type of such complexes when a graph has a certain local structure.

The layout of this article is as follows: In Section 3, we analyze the independence complexes of product of complete graphs, and show that it is homotopy equivalent to a wedge of circles (cf. Proposition 3.2). Section 4 is devoted towards computation of independence complexes of generalised Mycielskian (see Definition 2.2) of complete graphs (cf. Theorem 4.12) which turns out to be a wedge sum of spheres. In Section 5, we show that if we perturb a graph $G$ (locally; by removing some edges, and adding new edges and vertices) to obtain a new graph $H$, in a certain manner (refer to Figure 2), then Ind$(H)$, is homotopy equivalent to the suspension of Ind$(G)$ (cf. Theorem 5.1). As an application of Theorem 5.1, we determine the homotopy type of independence complexes of cycles with certain type of subdivisions.

2 Preliminaries

A graph is an ordered pair $G=(V,E)$ where $V$ is called the set of vertices and $E\subseteq V\times V$, the set of unordered edges of $G$. The vertices $v_{1},v_{2}\in V$ are said to be adjacent, if $(v_{1},v_{2})\in E$. This is also denoted by $v_{1}\sim v_{2}$, and if $v_{1}=v_{2}$, then $v_{1}$ is said to be a looped vertex. If $v$ is a vertex of $G$, then the set of its neighbours in $G$ is $\{x\in V(G):x\sim v\}$, and is denoted by $N(v)$. A graph $H$ with $V(H)\subseteq V(G)$ and $E(H)\subseteq E(G)$ is called a subgraph of the graph $G$. For a nonempty subset $U$ of $V(G)$, the induced subgraph $G[U]$, is the subgraph of $G$ with vertices $V(G[U])=U$ and $E(G[U])=\{(a,b)\in E(G)\ |\ a,b\in U\}$. In this article, $G[V(G)\setminus A]$ will be denoted by $G-A$ for $A\subsetneq V(G)$.

The complete graph on $n$ vertices is a graph where any two distinct vertices are adjacent, and it is denoted by $K_{n}$. For $n\geq 3$, the cycle graph $C_{n}$ is the graph with $V(C_{n})=\{1,\ldots,n\}$ and $E(C_{n})=\{(i,i+1):1\leq i\leq n-1\}\cup\{(1,n)\}$.

Definition 2.1**.**

The categorical product of two graphs $G$ and $H$, denoted by $G\times H$ is the graph where $V(G\times H)=V(G)\times V(H)$ and $(g,h)\sim(g^{\prime},h^{\prime})$ in $G\times H$, if and only if $g\sim g^{\prime}$ in $G$ and $h\sim h^{\prime}$ in $H$.

For $r\geq 1$, let $L_{r}$ denote the path graph of length $r$ with loop at one end, i.e., it is a graph with vertex set $V(L_{r})=\{0,\ldots,r\}$ and edge set $E(L_{r})=\{(i,i+1)\ |\ 0\leq i\leq r-1\}\cup\{(0,0)\}$.

Definition 2.2**.**

Let $G$ be a graph and $r\geq 1$. The $r$-th generalised Mycielskian, $M_{r}(G)$, of $G$ is the graph $(G\times L_{r})/\sim_{r}$, where $\sim_{r}$ is the equivalence which identifies all those vertices whose second coordinate is $r$. The graph $M_{2}(G)$ is called the Mycielskian of $G$.

An (abstract) simplicial complex $K$ is a collection of finite sets such that if $\tau\in K$ and $\sigma\subset\tau$, then $\sigma\in K$. The elements of $K$ are called the simplices of $K$. If $\sigma\in K$ and $|\sigma|=k+1$, then $\sigma$ is said to be $k$-dimensional. The set of [math]-dimensional simplices of $K$ is denoted by $V(K)$, and its elements are called vertices of $K$. A subcomplex of a simplicial complex $K$ is a simplicial complex whose simplices are contained in $K$. In this article, we always assume empty set as a simplex of any simplicial complex.

The link of a vertex $v\in V(K)$ is the subcomplex of $K$ defined as

[TABLE]

The star of a simplex $\sigma\in K$ is the subcomplex of $K$ defined as

[TABLE]

In this article, we consider any simplicial complex as a topological space, namely its geometric realization. For the definition of geometric realization, we refer to book [15] by Kozlov.

Definition 2.3** ([3]).**

Let $K$ be a simplicial complex and $\sigma\in K$. The star cluster of $\sigma$ in $K$ is a subcomplex of $K$ defined as

[TABLE]

The following results by Barmak will be used in this article.

Lemma 2.4** (Lemma $3.2$, [3]).**

The star cluster of a simplex in independence complex is contractible.

Lemma 2.5** (Lemma $3.3$, [3]).**

Let $K_{1}$ and $K_{2}$ be two contractible subcomplexes of a simplicial complex $K$ such that $K=K_{1}\cup K_{2}$. Then $K\simeq\ \Sigma(K_{1}\cap K_{2})$, where $\Sigma(X)$ denotes the suspension of space $X$.

Lemma 2.6** (Theorem $3.6$, [3]).**

Let $G$ be a graph and $v$ be a non-isolated vertex of $G$ which is contained in no triangle. Then $N(v)$ is a simplex of ${\rm{Ind}}(G)$, and

[TABLE]

The following observation directly follows from the definition of independence complexes of graphs.

Lemma 2.7**.**

Let $G$ be a graph obtained by taking disjoint union of two graphs $G_{1}$ and $G_{2}$. Then,

[TABLE]

where $\ast$ denotes the join operation.

Now we discuss some tools needed from discrete Morse theory ([12]).

Definition 2.8** (Definition $11.1$, [15]).**

A partial matching on a poset $P$ is a subset $M\subseteq P\times P$ such that

• (i)

$(a,b)\in M$ implies $b\succ a;$ i.e., $a<b$ and no $c$ satisfies $a<c<b$, and

• (ii)

each $a\in P$ belong to at most one element in $M$.

Note that, $\mathcal{M}$ is a partial matching on a poset $P$ if and only if there exists $\mathcal{A}\subset P$ and an injective map $\mu:\mathcal{A}\rightarrow P\setminus\mathcal{A}$ such that $\mu(a)\succ a$ for all $a\in\mathcal{A}$.

Definition 2.9**.**

An acyclic matching is a partial matching $\mathcal{M}$ on the poset $P$ such that there does not exist a cycle

[TABLE]

For an acyclic partial matching on $P$, those elements of $P$ which do not belong to the matching are said to be *critical *.

Theorem 2.10** (Theorem $11.13$, [15]).**

Let $\Delta$ be a simplicial complex and $M$ be an acyclic matching on the face poset of $\Delta$. Let $c_{i}$ denote the number of critical $i$-dimensional cells of $\Delta$ with respect to the matching $M$. Then $\Delta$ is homotopy equivalent to a cell complex $\Delta_{c}$ with $c_{i}$ cells of dimension $i$ for each $i\geq 0$, plus a single [math]-dimensional cell in the case where the empty set is also paired in the matching.

Following can be inferred from Theorem 2.10.

Remark 1**.**

If an acyclic matching has critical cells only in a fixed dimension $i$, then $\Delta$ is homotopy equivalent to a wedge of $i$-dimensional spheres.

3 Independence complex of $K_{m}\times K_{n}$

In this section, we compute the independence complex of $K_{m}\times K_{n}$ for $m,n\geq 2$. We first start by defining an acyclic matching on the face poset of a general simplicial complex; and then use a special case of this matching to prove the result for ${\rm{Ind}}(K_{m}\times K_{n})$.

Let $K$ be a simplicial complex and let $X=\{x_{1},\ldots,x_{n}\}\subseteq V(K)$. The elements of $X$ are ordered as; $x_{1}<x_{2}<\ldots<x_{n}.$

Let $P$ be the face poset of $(K,\subseteq)$. For $1\leq i\leq n,$ define

[TABLE]

We note that by construction, $A_{x_{i}}\cap A_{x_{j}}=\emptyset$ whenever $i\neq j$. Let $A=\bigcup\limits_{i=1}^{n}A_{x_{i}}$ and $\mu_{K}^{X}:A\rightarrow P\setminus A$ be defined by $\mu_{K}^{X}(\sigma)=\mu_{x_{i}}(\sigma)$, where $x_{i}$ is the unique element such that $\sigma\in A_{x_{i}}$.

Clearly, $\mu_{K}^{X}$ is injective and is therefore, a well defined partial matching on $P$. It follows from [18, Proposition 3.1] that $\mu_{K}^{X}$ is an acyclic matching. For the sake of completeness, we give a proof here as well.

Proposition 3.1**.**

$\mu_{K}^{X}$* is an acyclic matching on $P$.*

Proof.

Let there exist distinct cells $\sigma_{1},\ldots,\sigma_{t}\in A$ such that $\mu_{K}^{X}(\sigma_{i})\succ\sigma_{i+1\ (\text{mod}\ t)},1\leq i\leq t$.

Let $x\in X$ be the least element such that $\{\sigma_{1},\ldots,\sigma_{t}\}\cap A_{x}\neq\emptyset$. Without loss of generality, assume that $\sigma_{1}\in A_{x}$, i.e., $x\notin\sigma_{1}$ and $\mu_{K}^{X}(\sigma_{1})=\sigma_{1}\cup\{x\}$. $\mu_{K}^{X}(\sigma_{1})\succ\sigma_{2}$ and $\sigma_{1}\neq\sigma_{2}$ implies that there exists $x^{\prime}\in\mu_{1}(\sigma_{1}),x^{\prime}\neq x$ such that $\sigma_{2}=\mu_{K}^{X}(\sigma_{1})\setminus\{x^{\prime}\}$. We now have the following two possibilities:

$x\in\sigma_{t}$.

$\sigma_{1}\in A_{x}$ implies that $x\notin\sigma_{1}$. $x\in\sigma_{t}$ implies that $x\in\mu_{K}^{X}(\sigma_{t})$. Therefore, $\sigma_{1}=\mu_{K}^{X}(\sigma_{t})\setminus\{x\}$ which implies that $\mu_{K}^{X}(\sigma_{1})=\mu(\sigma_{t})$ a contradiction, since $\sigma_{1}\neq\sigma_{t}$. 2. 2.

$x\notin\sigma_{t}$, i.e., there exists a least $l\in\{2,\ldots,t\}$ such that $x\notin\sigma_{l}$.

$x\in\mu_{K}^{X}(\sigma_{l-1})$ and $x\notin\sigma_{l}$ implies that $\sigma_{l}=\mu_{K}^{X}(\sigma_{l-1})\setminus\{x\}$ i.e., $\mu_{K}^{X}(\sigma_{l-1})=\sigma_{l}\cup\{x\}$. Since $\sigma_{l}$ and $\mu_{K}^{X}(\sigma_{l-1})\notin A_{i}\cup\mu_{x_{i}}(x_{i})\ \forall\ i<x$, from the definition $\sigma_{l}\in A_{x}$. This implies that $\mu_{K}^{X}(\sigma_{l})=\sigma_{l}\cup\{x\}=\mu_{K}^{X}(\sigma_{l-1})$, which implies that $\sigma_{l}=\sigma_{l-1}$, a contradiction.

Therefore, $\mu_{K}^{X}$ is an acyclic matching on $P$. ∎

Let $m,n\geq 2$ and $V(K_{m})=\{a_{1},\ldots,a_{m}\}$, $V(K_{n})=\{b_{1},\ldots,b_{n}\}$.

Remark 2**.**

Observe that the maximal simplices of ${\rm{Ind}}(K_{m}\times K_{n})$ are only of the following two types:

sets of the form $\{(a_{i},b_{j})\ |\ j\in[n]\}$, where $i\in[m]$, and 2. 2.

sets of the form $\{(a_{i},b_{j})\ |\ i\in[m]\}$, where $j\in[n]$.

Using the above classification of simplices of ${\rm{Ind}}(K_{m}\times K_{n})$, we prove the following result.

Proposition 3.2**.**

Let $m,n\geq 2$. Then

[TABLE]

Proof.

Let $I:={\rm{Ind}}(K_{m}\times K_{n})$ and let $J=\{(a_{1},b_{i})~{}|~{}1\leq i\leq n\}\cup\{(a_{i},b_{1})~{}|\ 2\leq i\leq m\}\subseteq V(I)$. Further, let $P_{m,n}$ be the face poset of $(I,\subseteq)$. We define the ordering on $J$ as follows:

[TABLE]

Let $\mu_{I}^{J}$ be the matching defined as in the beginning of this section with respect to the ordering of elements of $J$ given as above. From Proposition 3.1 $\mu_{I}^{J}$ is an acyclic matching. Let $C$ be the set of critical cells for the matching $\mu_{I}^{J}$.

Claim 1**.**

C= $\{\{(a_{i},b_{1}),(a_{i},b_{j})\}\ |\ 2\leq i\leq m,2\leq j\leq n\}$.

Proof of Claim 1.

In this proof, for the convenience of notation, we denote $\mu_{I}^{J}$ by $\mu$.

Here, we first show that every element of $C$ is critical. Let $i\in\{2,\ldots,m\}$ and $j\in\{2,\ldots,n\}$. First observe that $\mu(\{(a_{i},b_{j})\})=\{(a_{1},b_{j}),(a_{i},b_{j})\}$. Since $i,j\geq 2$, it follows from the definition of $\mu$ that $\{(a_{i},b_{1}),(a_{i},b_{j})\}$ is a critical cell.

Now, let $\sigma\in I$ be a critical cell. Note that $\mu(\{\emptyset\})=\{(a_{1},b_{1})\}$, therefore $\sigma\neq\{(a_{1},b_{1})\}$. Since for each $j\geq 2,\ \mu(\{(a_{1},b_{j})\})=\{(a_{1},b_{j}),(a_{1},b_{1})\}$; and for each $i\geq 2$ and $k\geq 1$, $\mu(\{(a_{i},b_{k})\})=\{(a_{i},b_{k}),(a_{1},b_{k})\}$, we thus conclude that $\sigma$ has at least two elements. From Remark 2, either $\sigma=\{(a_{i_{1}},b_{j}),\ldots,(a_{i_{t}},b_{j})\}$ for some fixed $j\in[n]$ and $t\geq 2$ or $\sigma=\{(a_{i},b_{j_{1}}),\ldots,(a_{i},b_{j_{l}})\}$ for some fixed $i\in[m]$ and $l\geq 2$.

Suppose $\sigma=\{(a_{i_{1}},b_{j}),\ldots,(a_{i_{t}},b_{j})\}$ for some $j\in[n]$ and $t\geq 2$. If $(a_{1},b_{j})\notin\sigma$, then $\mu(\sigma)=\sigma\cup\{(a_{1},b_{j})\}$; and if $(a_{1},b_{j})\in\sigma$, then $\sigma=\mu(\sigma\setminus\{(a_{1},b_{j})\})$, which contradicts that $\sigma$ is a critical cell. Therefore, $\sigma=\{(a_{i},b_{j_{1}}),\ldots,(a_{i},b_{j_{l}})\}$ for some $i\in[m]$ and $l\geq 2$.

Note that, if $(a_{i},b_{1})\notin\sigma$ then $\mu(\sigma)=\sigma\cup\{(a_{1},b_{1})\}$, which is again a contradiction. Therefore, $(a_{i},b_{1})\in\sigma$. Further, if $i=1$, then $\sigma=\mu(\sigma\setminus\{(a_{1},b_{1})\})$. Therefore, $\sigma=\{(a_{i},b_{j_{1}}),\ldots,(a_{i},b_{j_{l}})\}$ for some $i\in\{2,\ldots,m\}$, $l\geq 2$ and $(a_{i},b_{1})\in\sigma$.

To prove Claim 1, it now suffices to show that $|\sigma|=2$. Suppose $|\sigma|\geq 3$. Since $|\sigma\setminus\{(a_{i},b_{1})\}|\geq 2$ and $i\geq 2$, by definition of $\mu$, we have $\mu(\sigma\setminus\{(a_{i},b_{1})\})=\sigma$, which is a contradiction to the fact that $\sigma$ is critical and therefore result follows. ∎

From Claim 1, all the critical cells for matching $\mu$ are of the same dimension, i.e., one dimensional. Moreover, the cardinality of the set $C$ is $(m-1)(n-1)$. Therefore result follows from Remark 1. ∎

Remark 3**.**

Observe that the graph $\underbrace{K_{2}\times\ldots\times K_{2}}_{(r-1)\text{-copies}}\times K_{n}$ is isomorphic to $2^{r-2}$ disjoint copies of $K_{2}\times K_{n}$. Therefore, using Lemma 2.7, we get

[TABLE]

It is thus natural to ask if one can generalise the Proposition 3.2 to $r$-fold product of complete graphs for $r\geq 3$, i.e., if the independence complexes of $r$-fold product of complete graphs are homotopy equivalent to wedge sum of spheres. We strongly believe that the independence complexes of $r$-fold product of complete graphs are homotopy equivalent to wedge of spheres of dimension $2^{r-1}-1$.

In support of our intuition, we present our computer based computations for the Betti numbers, denoted $\beta_{i}$, of the independence complexes of $K_{2}\times K_{3}\times K_{n}$ in Table 1.

Based on our computations, we propose the following conjecture

Conjecture 1**.**

For $n\geq 2$,

[TABLE]

4 Independence complex of $M_{r}(K_{n})$

This section is devoted to the computation of independence complexes of Mycielskian of graphs. To start with, we compute ${\rm{Ind}}(M_{2}(G))$ for any graph $G$. We then focus on the generalised Mycielskian of graphs, and determine the homotopy type of Ind$(M_{r}(K_{n}))$ for any $n$ and $r\geq 2$.

Theorem 4.1**.**

For any graph $G$, Ind$(M_{2}(G))\simeq\Sigma({\rm{Ind}}(G))$.

Proof.

Let $V(G)=\{v_{1},\ldots,v_{n}\}$ and let $w=(v_{1},2)=\ldots=(v_{n},2)$. Let $K=st_{{\rm{Ind}}(M_{2}(G))}(w)\cap SC_{{\rm{Ind}}(M_{2}(G))}(\{(v_{1},1),\ldots,(v_{n},1)\})$. Since $N(w)=\{(v_{1},1),\ldots,(v_{n},1)\}\in{\rm{Ind}}(M_{2}(G))$, Lemma 2.6 implies that

${\rm{Ind}}(M_{2}(G))\simeq\Sigma(K).$

Let $H$ be the subgraph of $M_{2}(G)$ induced by $\{(v_{1},0),\ldots,(v_{n},0)\}$. Clearly, $H\cong G$ and therefore ${\rm{Ind}}(H)\cong{\rm{Ind}}(G)$. We now show that $K={\rm{Ind}}(H)$.

Let $\sigma\in{\rm{Ind}}(H)$ and $(v_{i},0)\in\sigma$, then $N((v_{i},0))\cap\sigma=\emptyset$. Since $N((v_{i},1))\subseteq N((v_{i},0))\cup\{w\}$, $N((v_{i},1))\cap\sigma=\emptyset$ thereby implying that $\sigma\in$ $st_{{\rm{Ind}}(M_{2}(G))}(\{(v_{i},1)\})$. Since $\sigma\subseteq V(H)$, we see that $\sigma\cup\{w\}\in{\rm{Ind}}(M_{2}(G))$. Therefore, $\sigma\in st_{{\rm{Ind}}(M_{2}(G))}(\{w\})\cap SC_{{\rm{Ind}}(M_{2}(G))}(\{(v_{1},1),\ldots,(v_{n},1)\})$ and hence ${\rm{Ind}}(H)\subseteq K$.

Now suppose that $\sigma\in K$. For each $i$, $w$ is adjacent to $(v_{i},1)$ in $M_{2}(G)$, therefore $\sigma\cap\{w,(v_{i},1)\}=\emptyset$ for all $i$, and hence $K\subseteq{\rm{Ind}}(H)$. ∎

We now fix some notations and list a few results which will be used in this section for the computation of the independence complex of the generalised Mycielskian of complete graphs.

For a vertex $v$ of a graph $G$, $N[v]:=N(v)\cup\{v\}$. Also, if $A\subseteq V(G)$, then $N(A):=\bigcup\limits_{v\in A}N(v)$ and $N[A]:=\bigcup\limits_{v\in A}N[v]$.

Lemma 4.2** (Lemma $3.4$, [9]).**

Let $G$ be a graph and $u,u^{\prime}\in V(G)$ such that $N(u)\subseteq N(u^{\prime})$. Then,

[TABLE]

Lemma 4.3** (Proposition $2.10$, [19]).**

Let $G$ be a graph and let $\{a,b\}\in{\rm{Ind}}(G)$. If ${\rm{Ind}}(G-N[\{a,b\}])$ is collapsible, then ${\rm{Ind}}(G)$ collapses onto ${\rm{Ind}}(\tilde{G})$, where $V(\tilde{G})=V(G)$ and $E(\tilde{G})=E(G)\cup\{(a,b)\}$. In particular, ${\rm{Ind}}(G)\simeq{\rm{Ind}}(\tilde{G})$.

Lemma 4.4** (Lemma $2.1$, [14]).**

Let $G$ be graph and $v$ be a simplicial vertex444A vertex $v$ of $G$ is called simplicial if the subgraph induced by $N(v)$ is a complete graph. of $G$. Let $N(v)=\{w_{1},w_{2},\ldots,w_{k}\}$. Then

[TABLE]

Definition 4.5**.**

Let $p:X\to Y$ and $q:X\to Z$ be two continuous maps. The pushout of the diagram $Y\xleftarrow{p}X\xrightarrow{q}Z$ is the space

[TABLE]

where $\sim$ denotes the equivalence relation $p(x)\sim q(x)$ for $x\in X$.

The homotopy pushout of $Y\xleftarrow{p}X\xrightarrow{q}Z$ is the space \big{(}Y\sqcup(X\times I)\sqcup Z\big{)}/\sim, where $\sim$ denotes the equivalence relation $(x,0)\sim p(x)$, and $(x,1)\sim q(x)$ for $x\in X$. It can be shown that homotopy pushouts of any two homotopy equivalent diagrams are homotopy equivalent.

Remark 4**.**

If spaces are CW complexes and maps are subcomplex inclusions, then their homotopy pushout and pushout spaces are equivalent up to homotopy. For elaborate discussion of these results, we refer interested readers to [6, Chapter 7].

Lemma 4.6**.**

Let $X$ be a simplicial complex and $v\in V(X)$. Let $Y=\{\sigma\in X\ |\ v\notin\sigma\}$ be a subcomplex of $X$. If $lk_{X}(v)$ is contractible, then $X\simeq Y$.

Proof.

Let $A=lk_{X}(v)$ and let $Z$ be the homotopy pushout of the diagram $A\xleftarrow{=}A\xhookrightarrow{}Y$. Since $A\times I$ is homotopy equivalent to $A$, $Y\simeq Z$. Also, contractibility of $A$ implies that $Z$ is of the same homotopy type as $Z/(A\times\{1\})$. Therefore, $Y\simeq Z\simeq Z/(A\times\{1\})$ which is homeomorphic to $X$. ∎

Lemma 4.7**.**

Let $n\geq 2$ and $X_{1},X_{2},\ldots,X_{n}$ be simplicial complexes. If each $X_{i}$ is contractible and for each $j\in\{2,3,\ldots,n\}$, \big{(}\bigcup\limits_{i=1}^{j-1}X_{i}\big{)}\cap X_{j}\simeq\bigvee\limits_{r}\mathbb{S}^{k}, then $X_{1}\cup X_{2}\cup\ldots\cup X_{n}\simeq\bigvee\limits_{(n-1)r}\mathbb{S}^{k+1}$.

Proof.

Observe that $X_{1}\cup X_{2}$ is the pushout of the diagram $X_{1}\xhookleftarrow{}X_{1}\cap X_{2}\xhookrightarrow{}X_{2}$, where $\hookrightarrow$ denotes inclusion maps. From Remark 4, the homotopy pushout and pushout of $X_{1}\xhookleftarrow{}X_{1}\cap X_{2}\xhookrightarrow{}X_{2}$ are homotopy equivalent to each other. Further, the homotopy pushout of $X_{1}\xhookleftarrow{}X_{1}\cap X_{2}\xhookrightarrow{}X_{2}$ is homotopy equivalent to the homotopy pushout of $\{\text{point}\}\longleftarrow{}\bigvee\limits_{r}\mathbb{S}^{k}\longrightarrow{}\{\text{point}\}$ (since $X_{1}$ and $X_{2}$ are contractible and $X_{1}\cap X_{2}\simeq\bigvee\limits_{r}\mathbb{S}^{k}$). Moreover, homotopy pushout of $\{\text{point}\}\longleftarrow{}\bigvee\limits_{r}\mathbb{S}^{k}\longrightarrow{}\{\text{point}\}$ is homotopy equivalent to \Sigma\big{(}\bigvee\limits_{r}\mathbb{S}^{k}\big{)}. Therefore, $X_{1}\cup X_{2}\simeq\bigvee\limits_{r}\mathbb{S}^{k+1}$.

Let $n\geq 3$. Inductively assume that for any $2\leq t<n$, $\bigcup\limits_{i=1}^{t}X_{i}\simeq\bigvee\limits_{(t-1)r}\mathbb{S}^{k+1}$. In particular, $\bigcup\limits_{i=1}^{n-1}X_{i}\simeq\bigvee\limits_{(n-2)r}\mathbb{S}^{k+1}$. Further, the pushout of the diagram \bigcup\limits_{i=1}^{n-1}X_{i}\xhookleftarrow{}\big{(}\bigcup\limits_{i=1}^{n-1}X_{i}\big{)}\cap X_{n}\xhookrightarrow{}X_{n} is the space $\bigcup\limits_{i=1}^{n}X_{i}$. Thus, from Remark 4, $\bigcup\limits_{i=1}^{n}X_{i}$ is homotopy equivalent to the homotopy pushout of the diagram $\bigvee\limits_{(n-2)r}\mathbb{S}^{k+1}\longleftarrow{}\bigvee\limits_{r}\mathbb{S}^{k}\longrightarrow{}\{\text{point}\}$ which is homotopy equivalent to $\bigvee\limits_{(n-2)r+r}\mathbb{S}^{k+1}$.

∎

Lemma 4.8**.**

Let $n\geq 2$ and $X_{1},X_{2},\ldots,X_{n}$ be simplicial complexes. If for each $i\in\{1,2,\ldots,n\}$, $X_{i}\simeq\bigvee\limits_{r}\mathbb{S}^{k}$ and for each $j\in\{2,3,\ldots,n\}$, \big{(}\bigcup\limits_{i=1}^{j-1}X_{i}\big{)}\cap X_{j} is contractible, then $X_{1}\cup X_{2}\cup\ldots\cup X_{n}\simeq\bigvee\limits_{nr}\mathbb{S}^{k}$.

Proof.

Using similar arguments as in the proof of Lemma 4.7, we get that $X_{1}\cup X_{2}$ is homotopy equivalent to the homotopy pushout of $\bigvee\limits_{r}\mathbb{S}^{k}\xhookleftarrow{}\{\text{point}\}\xhookrightarrow{}\bigvee\limits_{r}\mathbb{S}^{k}$ (since $X_{1}\simeq\bigvee\limits_{r}\mathbb{S}^{k}\simeq X_{2}$ and $X_{1}\cap X_{2}$ is contractible).

Further, homotopy pushout of $\bigvee\limits_{r}\mathbb{S}^{k}\xhookleftarrow{}\{\text{point}\}\xhookrightarrow{}\bigvee\limits_{r}\mathbb{S}^{k}$ is homotopy equivalent to $\bigvee\limits_{r+r}\mathbb{S}^{k}$. Thus, $X_{1}\cup X_{2}\simeq\bigvee\limits_{2r}\mathbb{S}^{k}$. As before, the result now follows from induction. ∎

Proposition 4.9**.**

Let $r\geq 0$ and $n\geq 2$. Then

${\rm{Ind}}(K_{n}\times L_{r})\simeq\begin{cases}\bigvee\limits_{(n-1)^{k+1}}\mathbb{S}^{2k}&\text{if }r=3k,\\ \hskip 8.5359pt\{\rm{point}\}&\text{if}\ r=3k+1,\\ \bigvee\limits_{(n-1)^{k+1}}\mathbb{S}^{2k+1}&\text{if }r=3k+2.\\ \end{cases}$**

Proof.

Let $r=3k+t$ for some $t\in\{0,1,2\}$ and $k\geq 0$. We prove this result by induction on $k$.

To prove the base step, let $k=0$. We show that the result holds for $t\in\{0,1,2\}$. If $t=0$, then $K_{n}\times L_{0}$ isomorphic to $K_{n}$ implies

[TABLE]

For $t=1$, let $H_{1}$ be the induced subgraph of $K_{n}\times L_{1}$ with vertex set $\{(i,1)\ |\ 1\leq i\leq n\}$. Since $H_{1}$ does not have any edge, ${\rm{Ind}}(H_{1})\simeq\{\text{point}\}$. Observe that, in $K_{n}\times L_{1}$, $N((i,1))\subseteq N((i,0))$ for each $i\in\{1,2,\ldots,n\}$. Now repeated use of Lemma 4.2 for each $i$ gives us ${\rm{Ind}}(K_{n}\times L_{1})\simeq{\rm{Ind}}(H_{1})\simeq\{\text{point}\}.$ This proves the result for $k=0$ and $t=1$.

Finally, if $t=2$, let $H_{2}$ be the induced subgraph of $K_{n}\times L_{2}$ with vertex set $\{(i,j)\ |\ 1\leq i\leq n,\ 1\leq j\leq 2)\}$. Clearly, $H_{2}\cong K_{2}\times K_{n}$. We observe that in $K_{n}\times L_{2}$, $N((i,2))\subseteq N((i,0))$ for each $i\in\{1,2,\ldots,n\}$. By Lemma 4.2 and Proposition 3.2, we conclude that

[TABLE]

Inductively assume that the result is true for $k<s$ and $t\in\{0,1,2\}$. We now show that the result holds for $k=s>0$ and every $t\in\{0,1,2\}$.

Let $H$ be the induced subgraph of $K_{n}\times L_{3s+t}$ with vertex set $V(K_{n}\times L_{3s+t})\setminus\{(i,3s+t-2)\ |\ 1\leq i\leq n\}$. Further, in $K_{n}\times L_{3s+t}$, $N((i,3s+t))\subseteq N((i,3s+t-2))$. Thus, using Lemma 4.2, we have that ${\rm{Ind}}(K_{n}\times L_{3s+t})\simeq{\rm{Ind}}(H)$. Now observe that, $H\cong(K_{2}\times K_{n})\bigsqcup(K_{n}\times L_{3s+t-3})$. Using Lemma 2.7 and Proposition 3.2, we get

[TABLE]

By induction hypothesis, we get

[TABLE]

This completes the proof of Proposition 4.9. ∎

We fix a natural number $n\geq 3$, and define a few notations that would be used in rest of this section.

Let $V(K_{n})=\{1,2,\dots,n\}$. 2. 2.

In $M_{r}(K_{n})$, the equivalence class of vertices with second coordinate as $r$ is denoted by $w_{r}$. 3. 3.

For $r\geq 3$, let $i\in\{1,\ldots,n\}$ and $j\in\{1,\ldots,r-1\}$. We define $I_{i,j}^{n}$ to be the subgraph of $M_{r}(K_{n})$ induced by $V(M_{j}(K_{n}))\setminus\{w_{j}\}\cup\{(i,j)\}$. Also, let $I_{i,0}^{n}$ be the subgraph of $M_{r}(K_{n})$ induced by the vertex $(i,0)$.

Example: $I_{3,2}^{3}$ is the induced subgraph of $M_{r}(K_{3})$ on the vertex set $\{(s,t)\ :\ 1\leq s\leq 3,\ 0\leq t\leq 1\}\cup\{(3,2)\}$ (see Figure 1).

Since $n\geq 3$ is fixed for the rest of this section, we would write $I_{i,j}$ to denote $I_{i,j}^{n}$ for the simplicity of notation.

Lemma 4.10**.**

Let $I_{1,t}$ be as defined above, then

[TABLE]

Proof.

In $I_{1,1}$, $N((1,0))=\{(2,0),\ldots,(n,0)\}=N((1,1))$ and therefore by Lemma 4.2, ${\rm{Ind}}(I_{1,1})\simeq{\rm{Ind}}(I_{1,1}\setminus\{(1,1)\})\cong{\rm{Ind}}(K_{n})\simeq\bigvee\limits_{n-1}\mathbb{S}^{0}$.

Recall that a vertex $v$ is simplicial if the subgraph induced by $N(v)$ is a complete graph. In $I_{1,2}$, $N((1,1))=\{(2,0),\ldots,(n,0)\}$ and therefore $(1,1)$ is a simplicial vertex. Moreover, for each $2\leq i\leq n$, $V(I_{1,2})\setminus N[(i,0)]=\{(1,2),(i,1)\}$, implying that $I_{1,2}-N[(i,0)]\cong K_{2}$. Therefore, using Lemma 4.4, we get that

[TABLE]

Since the graph $I_{1,3}-N[\{(1,1),(2,1)\}]$ contains an isolated vertex $(1,3)$, ${\rm{Ind}}(I_{1,3}-N[\{(1,1),(2,1)\}])$ is a cone by Lemma 2.7, and hence collapsible. Using Lemma 4.3, ${\rm{Ind}}(I_{1,3})\simeq{\rm{Ind}}(I_{1,3}^{\prime})$, where $I_{1,3}^{\prime}$ is the graph with $V(I_{1,3}^{\prime})=V(I_{1,3})$ and $E(I_{1,3}^{\prime})=E(I_{1,3})\cup\{((1,1),(2,1))\}$. We repeat this process for all pair of vertices $((i,1),(j,1))$ $1\leq i\neq j\leq n$ and apply Lemma 4.3, which thereby implies that ${\rm{Ind}}(I_{1,3})\simeq{\rm{Ind}}(\tilde{I}_{1,3})$, where $V(\tilde{I}_{1,3})=V(I_{1,3})$ and $E(\tilde{I}_{1,3})=E(I_{1,3})\cup\{((i,1),(j,1))\ :\ 1\leq i\neq j\leq n\}$. For each $1\leq i\leq n$, $\tilde{I}_{1,3}-N[\{(i,1)\}]$ contains an isolated vertex $(i,0)$ and therefore ${\rm{Ind}}(\tilde{I}_{1,3}-N[\{(i,1)\}])$ is collapsible by Lemma 2.7. Now, using the fact that $(1,2)$ is a simplicial vertex in $\tilde{I}_{1,3}$ and ${\rm{Ind}}(\tilde{I}_{1,3}-N[\{(i,1)\}])\cong{\rm{Ind}}(\tilde{I}_{1,3}-N[\{(j,1)\}])$ for all $2\leq i\neq j\leq n$, by Lemma 4.4, ${\rm{Ind}}(\tilde{I}_{1,3})$ is contractible. Hence, ${\rm{Ind}}(I_{1,3})$ is contractible. ∎

We now generalise Lemma 4.10 and compute the homotopy type of independence complex of $I_{1,t}$, for any natural number $t$.

Lemma 4.11**.**

Let $t\geq 6$. Then

[TABLE]

Proof.

To prove this, we first construct a graph $\tilde{I}_{1,t-2}$ that contains $I_{1,t-2}$ as a subgraph such that $(1,t-3)$ is a simplicial vertex in $\tilde{I}_{1,t-2}$ and ${\rm{Ind}}(I_{1,t-2})\simeq{\rm{Ind}}(\tilde{I}_{1,t-2})$. Observe that the vertex $(1,t-2)$ is an isolated vertex in $I_{1,t-2}-N[\{(1,t-4),(2,t-4)\}]$ and hence ${\rm{Ind}}(I_{1,t-2}-N[\{(1,t-4),(2,t-4)\}])$ is collapsible. By Lemma 4.3, ${\rm{Ind}}(I_{1,t-2})\simeq{\rm{Ind}}(H)$, where $V(H)=V(I_{1,t-2})$ and $E(H)=E(I_{1,t-2})\cup\{((1,t-4),(2,t-4))\}$. By repeating this process for all pairs $(i,j),\ 1\leq i\neq j\leq n$, we get the graph $\tilde{I}_{1,t-2}$ such that ${\rm{Ind}}(I_{1,t-2})\simeq{\rm{Ind}}(\tilde{I}_{1,t-2})$, where $V(\tilde{I}_{1,t-2})=V(I_{1,t-2})$ and $E(\tilde{I}_{1,t-2})=E(I_{1,t-2})\cup\{((i,t-4),(j,t-4))\ :\ 1\leq i\neq j\leq n\}$. Since $N((1,t-3))=\{(2,t-4),\ldots,(n,t-4)\}$, the vertex $(1,t-3)$ is a simplicial vertex in $\tilde{I}_{1,t-2}$ . From Lemma 4.4, we have

{\rm{Ind}}(I_{1,t-2})\simeq{\rm{Ind}}(\tilde{I}_{1,t-2})\simeq\bigvee\limits_{i=2}^{n}\Sigma\big{(}{\rm{Ind}}(\tilde{I}_{1,t-2}-N[\{(i,t-4)\}])\big{)}.

Observe that for each $2\leq i\leq n$, the graph $\tilde{I}_{1,t-2}-N[\{(i,t-4)\}]$ is isomorphic to $I_{i,t-5}\sqcup K_{2}$, where $K_{2}$ appears because of the edge $((1,t-2),(i,t-3))$ in $\tilde{I}_{1,t-2}-N[\{(i,t-4)\}]$. We note that for any $j$, $I_{i,j}\cong I_{l,j}$ and therefore

[TABLE]

We consider the following three cases.

Case 1. $t=3k$.

In this case, since $t-2=3(k-1)+1$, using Equation (4.1) and Lemma 4.10, we conclude that

[TABLE]

Case 2. $t=3k+1$.

In this case, $t-2=3(k-1)+2$. Again by Lemma 4.10 and Equation (4.1), we have

[TABLE]

Case 3. $t=3k+2$.

In this case, since $t-2=3(k-1)+3$, ${\rm{Ind}}(I_{1,t-2})\simeq\bigvee\limits_{(n-1)^{k-1}}\Sigma^{2(k-1)}({\rm{Ind}}(I_{1,3}))$. By Lemma 4.10, ${\rm{Ind}}(I_{1,3})$ is contractible and so is ${\rm{Ind}}(I_{1,t-2})$. ∎

We are now ready to prove the main result of this section. Firstly, note that $M_{r}(K_{2})$ is isomorphic to odd cycle $C_{2r+1}$, for which independence complex has been computed by Kozlov in [16]. Here, we determine the homotopy type of ${\rm{Ind}}(M_{r}(K_{n}))$ for $n>2$ and any $r$.

Theorem 4.12**.**

Let $r\geq 2$ be a positive integer. Then,

${\rm{Ind}}(M_{r}(K_{n}))\simeq\begin{cases}\bigvee\limits_{(n-1)^{k}}\mathbb{S}^{2k-1}&\text{if }r=3k,\\ \bigvee\limits_{n(n-1)^{k}}\mathbb{S}^{2k}&\text{if }r=3k+1,\\ \bigvee\limits_{(n-1)^{(k+1)}}\mathbb{S}^{2k+1}&\text{if }r=3k+2.\\ \end{cases}$**

Proof.

If $r=2$, then Theorem 4.1 implies that ${\rm{Ind}}(M_{r}(K_{n}))\simeq\Sigma({\rm{Ind}}(K_{n}))$, and hence the result follows. So we assume that $r\geq 3$. In $M_{r}(K_{n})$, $N(w_{r})=\{(1,r-1),\ldots,(n,r-1)\}$ and there is no edge among the vertices of $N(w_{r})$. Therefore, from Lemma 2.6

[TABLE]

Let $K$ denote the complex $st_{{\rm{Ind}}(M_{r}(K_{n}))}(w_{r})\cap SC_{{\rm{Ind}}(M_{r}(K_{n}))}(\{(1,r-1),\ldots,(n,r-1)\})$.

Claim 2**.**

$K=\bigcup\limits_{i=1}^{n}\emph{{\rm{Ind}}}(I_{i,r-2}).$**

Let $\sigma\in K$. If $\sigma\cap\{(1,r-2),\ldots,(n,r-2)\}=\emptyset$, then $\sigma\in\bigcup\limits_{i=1}^{n}{\rm{Ind}}(I_{i,r-2})$. On the other hand if $\sigma\cap\{(1,r-2),\ldots,(n,r-2)\}\neq\emptyset$, then there exists a unique $i$ such that $(i,r-2)\in\sigma$ and in this case $\sigma\in{\rm{Ind}}(I_{i,r-2})$. Hence, $K\subseteq\bigcup\limits_{i=1}^{n}{\rm{Ind}}(I_{i,r-2})$. If $\tau\in\bigcup\limits_{i=1}^{n}{\rm{Ind}}(I_{i,r-2})$, then $\tau\in{\rm{Ind}}(I_{i,r-2})$ for some $i$, therefore $\tau\in st_{{\rm{Ind}}(M_{r}(K_{n}))}((i,r-1))\cap st_{{\rm{Ind}}(M_{r}(K_{n}))}(w_{r})$. This completes the proof of Claim 2.

For $1\leq i\neq l\leq n$, observe that $I_{i,j}\cap I_{l,j}\cong K_{n}\times L_{j-1}$, therefore ${\rm{Ind}}(I_{i,j}\cap I_{l,j})\simeq{\rm{Ind}}(K_{n}\times L_{j-1})$. Also, for any arbitrary graph $G$ and $A,B\subseteq V(G)$, ${\rm{Ind}}(G[A\cap B])={\rm{Ind}}(G[A])\cap{\rm{Ind}}(G[B])$, and hence ${\rm{Ind}}(I_{i,j}\cap I_{l,j})={\rm{Ind}}(I_{i,j})\cap{\rm{Ind}}(I_{l,j})$. Further, \big{(}\bigcup\limits_{s=1}^{i}{\rm{Ind}}(I_{s,j})\big{)}\bigcap{\rm{Ind}}(I_{i+1,j})={\rm{Ind}}(I_{i,j})\cap{\rm{Ind}}(I_{i+1,j})\simeq{\rm{Ind}}(K_{n}\times L_{j-1}) for all $1\leq i\leq n-1$.

Case 1: $r=3k+1$.

From Lemmas 4.10 and 4.11, ${\rm{Ind}}(I_{i,r-2})\simeq\bigvee\limits_{(n-1)^{k}}\mathbb{S}^{2(k-1)+1}$. Proposition 4.9 implies that ${\rm{Ind}}(K_{n}\times L_{r-3})$ is contractible and therefore by using Claim 2 and Lemma 4.8, we conclude that $K\simeq\bigvee\limits_{n(n-1)^{k}}\mathbb{S}^{2(k-1)+1}$. Thus, from Equation (4.2),

[TABLE]

Case 2 : $r=3k+2$.

In this case, ${\rm{Ind}}(I_{i,r-2})$ is contractible from Lemmas 4.10 and 4.11. Since, $r-3=3(k-1)+2$, ${\rm{Ind}}(K_{n}\times L_{r-3})\simeq\bigvee\limits_{(n-1)^{k}}\mathbb{S}^{2k-1}$ from Lemma 4.9. Hence by using Lemma 4.7 and Claim 2, we conclude that $K\simeq\bigvee\limits_{(n-1)(n-1)^{k}}\mathbb{S}^{2k}$. Thus,

[TABLE]

Case 3. $r=3k$.

First we show that $K\simeq{\rm{Ind}}(K_{n}\times L_{r-3})$. Observe that $lk_{K}((1,r-2))={\rm{Ind}}(I_{1,r-3})$. Using Lemmas 4.10 and 4.11, we get that ${\rm{Ind}}(I_{1,r-3})$ is contractible. Hence Lemma 4.6 implies that $K\simeq K^{1}:=K\setminus\{\sigma\in K\ :\ (1,r-3)\in\sigma\}$. Now $lk_{K^{1}}((2,r-2))={\rm{Ind}}(I_{2,r-3})\cong{\rm{Ind}}(I_{1,r-3})$ and therefore $K^{1}\simeq K^{2}:=K^{1}\setminus\{\sigma\in K^{1}:(2,r-3)\in\sigma\}$. Repeating this process for all $(i,r-2),3\leq i\leq n$, we conclude that $K\simeq K^{n}:=\{\sigma\in K\ :\ (1,r-2),\ldots,(n,r-2)\notin\sigma\}$. It is easy to check that $K^{n}\cong{\rm{Ind}}(K_{n}\times L_{r-3})$. From Proposition 4.9, we conclude that

[TABLE]

and hence the theorem. ∎

5 Independence complexes of graphs with a specific local structure

Let $G$ be a graph that contains a crossing of two edge as in Figure 2(a). In this section we prove that if we replace a crossing in graph $G$ with the structure as in Figure 2(b), then the independence complex of $H$ is the suspension of the independence complex of $G$.

Theorem 5.1**.**

Let $G$ be a graph that contains graph depicted in Figure 2(a) as a subgraph. If $H$ is the graph with $V(H)=V(G)\sqcup\{a,b,c,d\}$, $E(H)=(E(G)\setminus\{(1,4),(2,3)\})\cup\{(1,a),(a,b),(b,3),(2,c),(c,d),(d,4),(a,c),(b,d)\}$ obtained from $G$ as shown in Figure 2(b), then

[TABLE]

Proof.

Observe that $\{a,d\}$ and $\{b,c\}$ are simplices of Ind$(H)$. Let $K=\text{Ind}(H)$, $K_{1}=SC_{K}(\{a,d\})$ and $K_{2}=SC_{K}(\{b,c\})$. From Lemma 2.4, $K_{1}$ and $K_{2}$ are contractible subcomplexes of $K$.

We first show that $K=K_{1}\cup K_{2}.$ Clearly, $K_{1}\cup K_{2}$ is a subcomplex of $K$. Let $\sigma\in K$. If $\{a,b,c,d\}\cap\sigma\neq\emptyset$, then by definition $\sigma\in K_{1}$ or $K_{2}$. Therefore we assume that $\{a,b,c,d\}\cap\sigma=\emptyset$. If $1\notin\sigma$, then $\sigma\in st_{K}(\{a\})$. If $1\in\sigma$, then $2\notin\sigma$ and thereby implying that $\sigma\in st_{K}(\{c\})$.

Claim 3**.**

$K_{1}\cap K_{2}={\rm{Ind}}(G).$**

Let $\sigma\in K_{1}\cap K_{2}$. Clearly, $\{a,b,c,d\}\cap\sigma=\emptyset$. To show that $\sigma\in\text{Ind}(G)$, it is enough to show that $\{1,4\}\nsubseteq\sigma$ and $\{2,3\}\nsubseteq\sigma$. However, $\{1,4\}\notin K_{1}$ implies that $\{1,4\}\nsubseteq\sigma$. Similarly, $\{2,3\}\notin K_{2}$ implies that $\{2,3\}\nsubseteq\sigma$.

Now, let $\sigma\in\text{Ind}(G)$. Since $\{1,4\}\nsubseteq\sigma$, $\sigma$ is in $st_{K}(\{a\})$ or $st_{K}(\{d\})$. Moreover, $\{2,3\}\nsubseteq\sigma$ implies that either $\sigma$ is in $st_{K}(\{b\})$ or is in $st_{K}(\{c\})$. Therefore, $\sigma\in K_{1}\cap K_{2}.$

Thus result follows from Lemma 2.5. ∎

We note that the proof of Theorem 5.1 also holds if we assume that the vertices $3$ and $4$ are same (cf. Figure 3), i.e., $3=4$. We, therefore, have the following result as a special case.

Theorem 5.2** (Section $3.3.1$, [19]).**

Let $G$ be a graph that contains triangle depicted in Figure 3(a) as a subgraph. If $H$ is the graph with $V(H)=V(G)\sqcup\{a,b,c,d\}$, $E(H)=(E(G)\setminus\{(1,3),(2,3)\})\cup\{(1,a),(a,b),(b,3),(2,c),(c,d),(d,3),(a,c),(b,d)\}$ obtained from $G$ as shown in Figure 3(b), then

[TABLE]

Before proceeding further, we would like to point out that Skwarski has considered a similar construction as in Figure 3 in [19, Section 3.3.1].

We now record a straight forward observation that follows from Theorem 5.2.

Corollary 5.3**.**

Let $G$ be a graph with Figure 3(b) as an induced subgraph. Then the independence complex of $G$ has the homotopy type of a suspension.

As an application of Theorem 5.1 and Theorem 5.2, we compute the homotopy type of the independence complexes of a particular family of graphs.

Let $C_{n}^{0}\equiv C_{n}$ be the cycle graph on the vertex set $\{1,2,\ldots,n\}$. Let $C_{n}^{1}$ be the graph obtained from $C_{n}$ by subdividing the edges adjacent to $1$ and adding an edge between the newly created vertices. Let $x_{1},y_{1}$ be the vertices of $V(C_{n}^{1})\setminus V(C_{n}^{0})$. We iteratively define the graph $C_{n}^{j}$ to be the graph obtained from $C_{n}^{j-1}$ as per the above construction. We note that $V(C_{n}^{j})=\{1,2,\dots,n\}\cup\{x_{1},x_{2},\dots,x_{j}\}\cup\{y_{1},y_{2},\dots,y_{j}\}$, i.e., $|V(C_{n}^{j})|=n+2j$ and $E(C_{n}^{j})=(E(C_{n}^{0})\setminus\{(1,2),(1,n)\})\cup\{(x_{i},y_{i}):1\leq i\leq j\}\cup\{(x_{i},x_{i+1}),(y_{i},y_{i+1}):1\leq i\leq j-1\}\cup\{(1,x_{j}),(1,y_{j}),(2,x_{1}),(n,y_{1})\}$.

For example, Figure 4 shows $C_{3}^{2}$ and $C_{5}^{3}$.

Corollary 5.4**.**

Let $i,r\geq 1$, then

${\rm{Ind}}(C_{3r}^{i})\simeq\begin{cases}\mathbb{S}^{r-1+j}\bigvee\mathbb{S}^{r-1+j}&\text{if }i=2j,\\ \text{\rm\{\text{point}\}}&\text{otherwise},\\ \end{cases}$**

${\rm{Ind}}(C_{3r+1}^{i})\simeq\begin{cases}\mathbb{S}^{r-1+j}\bigvee\mathbb{S}^{r-1+j}&\text{if }i=2j-1,\\ \text{\rm\{point\}}&\text{otherwise},\\ \end{cases}$**

and

${\rm{Ind}}(C_{3r+2}^{i})\simeq\begin{cases}\mathbb{S}^{r}\bigvee\mathbb{S}^{r}&\text{if }i=1,\\ \mathbb{S}^{r+j}\bigvee\mathbb{S}^{r+j}&\text{if }i=2j\text{ or }2j+1,\ j\geq 1.\\ \end{cases}$**

Proof.

Let $P_{n}$ be the path graph on $n$ vertices with $n-1$ edges. From [15, Proposition 11.16], we know that

[TABLE]

We give the proof by induction on $i$. Observe that in $C_{n}^{i}$, vertex $1$ is a simplicial vertex for each $i\geq 1$. Therefore, using Lemma 4.4 and Equation 5.3 we get

[TABLE]

Similarly, using Lemma 4.4 and Equation 5.3 we get

[TABLE]

Now using Theorem 5.2, we observe that for any $i\geq 3$, ${\rm{Ind}}(C_{n}^{i})=\Sigma({\rm{Ind}}(C_{n}^{i-2}))$ and therefore by induction on $i$, the result follows from ${\rm{Ind}}(C_{n}^{1})$ and ${\rm{Ind}}(C_{n}^{2})$. ∎

Acknowledgements

We are thankful to Priyavrat Deshpande and Dheeraj Kulkarni for inviting us to the workshop “Young Topologists’ Meet" held at Chennai Mathematical Institute in 2018, where the work of Section 5 was done. The first author was partially supported by IRCC, IIT Bombay and the third author was partially supported by a grant from Infosys Foundation.