Higher analogs of simplicial and combinatorial complexity

TL;DR

This paper introduces higher analogs of simplicial and combinatorial complexity, establishing relationships with existing topological and combinatorial invariants to deepen understanding of complex structures.

Contribution

It defines higher simplicial and combinatorial complexities and connects them to topological and combinatorial invariants, expanding the theoretical framework.

Findings

Higher simplicial complexity relates to higher topological complexity.

Higher combinatorial complexity connects with the simplicial complexity of order complexes.

The paper establishes new theoretical relationships between these complexities.

Abstract

We introduce higher simplicial complexity of a simplicial complex $K$ and higher combinatorial complexity of a finite space $P$ (i.e. $P$ is a finite poset). We relate higher simplicial complexity with higher topological complexity of $|K|$ and higher combinatorial complexity with higher simplicial complexity of the order complex of $P$.

Equations

$$\pi :PX\to X\times X,\gamma \mapsto (\gamma (0),\gamma (1)).$$

$$e_n:X^{I_n}\to X^n,e_n(\alpha )=(\alpha (1_1),\alpha (1_2),\cdots ,\alpha (1_n)).$$

$$e_n^{\prime }:X^I=PX\to X^n,e_n^{\prime }(\alpha )=(\alpha (0),\alpha (\frac{1}{n-1}),\alpha (\frac{2}{n-1}),\cdots ,\alpha (1)).$$

$$cat(X)\le T{C}_n(X)\le cat(X^n)\le cat(X{)}^n.$$

$${\iota }_{{\mathrm{sd}}^{k-1}(K\times K)}:{\mathrm{sd}}^k(K\times K)\to {\mathrm{sd}}^{k-1}(K\times K),k\ge 1.$$

$${\iota }_{K\times K}^k:{\mathrm{sd}}^k(K\times K)\to K\times K,k\ge 0.$$

$$SC(K)=\underset{k\to \infty }{\lim }S{C}^k(K)=\underset{k\ge 0}{\min }\{S{C}^k(K)\}.$$

$$S{C}_n(K)=\underset{k\to \infty }{\lim }S{C}_n^k(K)=\underset{k\ge 0}{\min }\{S{C}_n^k(K)\}.$$

$$q_m:P^{J_m}\to P\times P\text{ defined by }{q}_m(\gamma )=(\gamma (0),\gamma (m)),m\ge 0$$

$$\{0,1_1,1_2,\cdots ,1_n,2_1,2_2,\cdots ,2_n,\cdots ,m_1,m_2,\cdots ,m_n\}.$$

$$0\le 1_1\ge 2_1\le \cdots \ge (\le )m_1,$$

$$0\le 1_2\ge 2_2\le \cdots \ge (\le )m_2,$$

$$\cdots$$

$$0\le 1_n\ge 2_n\le \cdots \ge (\le )m_n.$$

$$q_{n,m}:P^{J_{n,m}}\to P^n,q_{n,m}(\gamma )=(\gamma (m_1),\gamma (m_2),\cdots ,\gamma (m_n)).$$

$${\mathcal{CC}}_n(P)=\underset{m\ge 1}{\min }\{{\mathcal{CC}}_{n,m}(P)\}=\underset{m\to \infty }{\lim }{\mathcal{CC}}_{n,m}(P).$$

$$q_{n,m}^{\prime }:P^{J_{(n-1)m}}\to P^n,q_{n,m}^{\prime }(\alpha )=(\alpha (0),\alpha (m),\cdots ,\alpha ((n-1)m)).$$

$$R:J_{(n-1)(m+1)}\to J_{(n-1)m}$$

$${\mathcal{C}{\mathcal{C}}^{\prime }}_n(P)=\underset{m\ge 1}{\min }\{{\mathcal{C}{\mathcal{C}}^{\prime }}_{n,m}(P)\}=\underset{m\to \infty }{\lim }{\mathcal{C}{\mathcal{C}}^{\prime }}_{n,m}(P).$$

$$m_1\le m_1-1\ge \cdots \le 1_1\ge 0\le 1_2\ge \cdots \le m_2-1\ge m_2\le m_2-1\ge \cdots \le 1_2\ge 0\le 1_3$$

$$\ge \cdots \le m_3-1\ge m_3\le m_3-1\ge \cdots \le 1_3\ge 0\le \cdots \ge 0\le 1_n\ge \cdots \le m_n-1\ge m_n.$$

$$k\le k-1\ge k-2\le \cdots \ge (j-1)m\le (j-1)m\ge (j-1)m\le \cdots \ge (j-1)m,\text{if}j\le \frac{n}{2}$$

$$k\le k+1\ge k+2\le \cdots \ge (j-1)m\le (j-1)m\ge (j-1)m\le \cdots \ge (j-1)m\text{if}j>\frac{n}{2}.$$

$$\gamma :J_{n,k}\to P\text{ such that }{q}_{n,k}(\gamma )=(x_1,x_2,\cdots ,x_n).$$

$$\{e_{+}^0,e_{-}^0,\cdots ,e_{+}^m,e_{-}^m\},$$

$$cat(P)\le {\mathcal{CC}}_n(P)\le cat(P^n)\le cat(P{)}^n.$$

$${\mathcal{CC}}_n({\mathbb{S}}^m)=cat(({\mathbb{S}}^m{)}^n)=cat({\mathbb{S}}^m{)}^n=2^n\text{ for any }m\ge 1\text{ and }n\ge 2.$$

$${\mathcal{CC}}_n^{\infty }(P)=\underset{k\to \infty }{\lim }{\mathcal{CC}}_n^k(P)=\underset{k\ge 0}{\min }\{{\mathcal{CC}}_n^k(P)\}.$$

$$H(p_1,p_2,\cdots ,p_n,x)=s_i(p_1,p_2,\cdots ,p_n)(x),$$

$$H(p_1,p_2,\cdots ,p_n,(j-1)m)=s_i(p_1,p_2,\cdots ,p_n)((j-1)m)={\rho }_j(p_1,p_2,\cdots ,p_n),$$

Full text

1. Introduction

The *topological complexity * $TC(X)$ of a path connected space $X$ was introduced by Farber (see [3]). It is a measure of the complexity to construct a motion-planning algorithm on the space $X$. Let $I=[0,1]$ and $PX=X^{I}$ denotes the free path space. Consider the fibration

[TABLE]

Then $TC(X)$ is defined to be the least positive integer $r$ such that there exists an open cover $\{U_{1},\cdots,U_{r}\}$ of $X\times X$ with continuous section of $\pi$ over each $U_{i}$ (i.e. a continuous map $s_{i}:U_{i}\rightarrow E$ satisfying $\pi\circ s_{i}=\operatorname{Id}_{U_{i}}$ for $i=1,2,\cdots,r$). The idea was generalised by Rudyak to higher dimensions (see [7]). He introduced *n-th topological complexity * $TC_{n}(X),\ n\geq 2$ such that $TC_{2}(X)=TC(X)$. We recall the definition of higher topological complexity in the next section.

In ([4]), Gonzalez used contiguity of simplicial maps to define simplicial complexity $SC(K)$ for a simplicial complex $K$. This is a discrete analogue of topological complexity in the category of simplicial complexes. He showed that $SC(K)=TC(\mid K\mid)$ for finite $K$, where $\mid K\mid$ is the geometric realization of $K$. We introduce higher simplicial complexity $SC_{n}(K)$ of a simplicial complex $K$ and generalise the above result.

Theorem A**.**

For a finite simplicial complex $K$, $SC_{n}(K)=TC_{n}(\mid K\mid)$ for any $n\geq 2$.

(See Theorem 3.5.)

A combinatorial approach to topological complexity was introduced by Tanaka (cf. [10]). The basic idea of Tanaka’s paper is to describe topological complexity by combinatorics of finite spaces i.e. connected finite $T_{0}$ space ( see [9]). He used an analogue of the above path-space fibration for finite spaces to define combinatorial complexity $CC(P)$. It is shown that $CC(P)=TC(P)$. We introduce an analogue higher combinatorial complexity $\mathcal{CC}_{n}(P)$ and prove the above result for higher dimensions.

Theorem B**.**

For any finite space $P$, we have $\mathcal{CC}_{n}(P)=TC_{n}(P)$, for any $n\geq 2$.

(See Theorem 4.9)

Given a finite space $P$, there is a naturally associated simplicial complex, called order complex $\mathcal{K}(P)$. A finite space is equivalent to a finite poset and the $n$-simplices of $\mathcal{K}(P)$ are linearly ordered subsets of $P$ (c.f 4.7). As noted by Tanaka, $CC(P)$ is just an upper bound to to $SC(\mathcal{K}(P))$. To describe $SC(\mathcal{K}(P))$ combinatorially, he used barycentric subdivision of $P$ to define $CC^{\infty}(P)$. Finally it is shown that $CC^{\infty}(P)=SC(\mathcal{K}(P))$. Hence $CC^{\infty}(P)=TC(\mid\mathcal{K}(P)\mid)$.

We further generalise above ideas to higher combinatorial complexities $\mathcal{CC}^{\infty}_{n}(P)$ using barycentric subdivision of a finite space $P$ and prove the prove following.

Theorem C**.**

For any finite space $P$, we have $\mathcal{CC}^{\infty}_{n}(P)=SC_{n}(\mathcal{K}(P))$, for any $n\geq 2$.

(See Theorem 5.6)

The organization of the rest of paper is as follows. In Section 2 we recall basics of topological complexity and higher topological complexity of a space $X$. In Section 3 we introduce higher simplicial complexity of a simplicial complex $K$ and prove Theorem (A). In Section 4 we define higher combinatorial complexity $\mathcal{CC}_{{n}}(P)$ of a finite space $P$ and prove Theorem (B). In 5 we describe higher combinatorial complexity with barycentric subdivision $\mathcal{CC}^{\infty}_{n}(P)$ of a finite space $P$ and we prove Theorem (C).

2. Preliminaries

In this section we review basic concepts of topological complexity. For details we refer to [2, 3, 7]. We start by recalling the definition of the Schwarz genus of a fibration. Let $p:E\rightarrow B$ be a fibration. The Schwarz genus of $p$ is the minimum number $k$ such that $B$ can be cover by $k$ open subsets, $U_{1}\cup U_{2}\cup\cdots\cup U_{k}=B$ and on each $U_{i}$ there is a section of $p$. It is denoted by genus$(p)$. If no such $k$ exists then we say $genus(p)=\infty$. Then the topological complexity of $X$ is genus$(\pi)$, where $\pi$ is as in the equation 1.

Suppose $I_{n},n\geq 2$ denote the wedge of $n$ intervals $[0,1]_{j},j=1,2,\cdots,n$, where $0_{j}\in[0,1]_{j}$ are identified. Consider the mapping space $X^{I_{n}}$ and the fibration

[TABLE]

The n-th topological complexity $TC_{n}(X)$ of $X$ is defined to be genus$(e_{n})$. It can be defined alternatively as $TC_{n}(X)=\emph{genus}(e^{\prime}_{n})$, where

[TABLE]

Note that $TC_{2}(X)$ is nothing but $TC(X)$. It is proved that $TC_{n}(X)$ is homotopy invariant and $TC_{n}(X)=1$ if and only if $X$ is contractible. The sequence $TC_{n}(X)$ is an increasing sequence i.e, $TC_{n}(X)\leq TC_{n+1}(X)$. Topological complexity of a space $X$ is closely related to the Lusternik–Schnirelmann category (or L-S category) of the space $X$, which is denoted by $cat(X)$. Recall that $cat(X)$ is defined as $cat(X)=\emph{genus}(\pi_{0})$, where $\pi_{0}:P_{0}X\rightarrow X$ given by $\pi_{0}(\gamma)=\gamma(1)$ and $P_{0}X$ is the space of all paths in $X$ with a fixed starting point $x_{0}$. The topological complexity and L-S category of a space satisfy the following inequality:

[TABLE]

We will use the following Lemma to define higher simplicial complexity. It is simple generalisation of [4, Lemma 1.1 ]

2.1 Lemma**.**

The evaluation map $e^{\prime}_{n}:PX\rightarrow X^{n}$ admits a section on a subset $A$ of $X^{n}$ if and only if the composition maps $\pi_{1},\pi_{2},\cdots\pi_{n}:A\hookrightarrow X^{n}\rightarrow X$ are in same homotopy class of maps.

Proof.

Let $e^{\prime}_{n}$ admits a section $s:A\rightarrow PX$ on a subset $A$ of $X^{n}$. Let $H:A\times I\rightarrow X$ be the map given by $H(x_{1},x_{2},\cdots,x_{n},t)=s(x_{1},x_{2},\cdots x_{n})(t)$. Then clearly $H|_{A\times[\frac{j-1}{n-1},\frac{j}{n-1}]}:\pi_{j}\simeq\pi_{j+1}$ for $j=1,\cdots,n-1$. Conversely, assume that $\pi_{1},\pi_{2},\cdots\pi_{n}$ are in same homotopy class and $h^{j}_{t}:A\rightarrow X$ be a homotopy from $\pi_{j}$ to $\pi_{j+1}$ i.e, $h^{j}_{0}(x_{1},x_{2},\cdots,x_{n})=x_{j}$ and $h^{j}_{1}(x_{1},x_{2},\cdots,x_{n})=x_{j+1}$ for $j=1,2,\cdots{n-1}$. Then the concatenation of the paths $h^{1}_{t},h^{2}_{t},\cdots,h^{n-1}_{t}$ will give a section on $A$. ∎

We will use the following Proposition to relate $n$-th simplicial complexity of a simplicial complex $K$ and $n$-th topological complexity of the geometric realization of $K$. This is a simple generalisation of the result of [2, Proposition 4.12 and Remark 4.13]

2.2 Proposition**.**

Let $X$ be an ENR. Then $TC_{n}(X)=r$, where $r$ is the minimal integer such that there exist a section $s:X^{n}\rightarrow PX$ (which is not necessarily continuous) of the fibration $e^{\prime}_{n}$ and a splitting $G_{1}\cup G_{2}\cup\cdots\cup G_{r}=X^{n}$, where each $G_{i}$ is locally compact subset of $X^{n}$ and each restriction $s_{|G_{i}}:G_{i}\rightarrow PX$ is continuous for $i=1,2,\cdots,r$.

Proof.

Let us assume that $TC_{n}(X)=k$. Consider an open cover $\{U_{i}\}_{i=1}^{k}$ of $X^{n}$ such that on each open set there is a continuous section $s_{i}:U_{i}\rightarrow PX$ of $e^{\prime}_{n}$. Set $G_{1}=U_{1}$ and $G_{i}=U_{i}-(U_{1}\cup U_{2}\cup\cdots\cup U_{i-1})$ for $i\geq 2$. Then $G_{i}$’s are locally compact and cover $X^{n}$. Let $s=s_{1}\cup s_{2|_{G_{2}}}\cup s_{3|_{G_{3}}}\cdots\cup s_{k|_{G_{k}}}$. Then $s:X^{n}\rightarrow PX$ is a section of $e^{\prime}_{n}$ which is continuous on each $G_{i}$. Thus $r\leq k=TC_{n}(X)$.

Conversely, suppose that $G_{1}\cup G_{2}\cup\cdots\cup G_{r}=X^{n}$ where $\{G_{i}\}$ are locally compact subsets. Assume that $s:X^{n}\rightarrow PX$ is a section (which is not necessarily continuous) of the fibration $e^{\prime}_{n}$ such that $s|_{G_{i}}=s_{i}$ are continuous. These $s_{i}$’s are in one-to-one correspondence with homotopies $h^{i}_{t}:G_{i}\rightarrow X$, $t\in[0,1]$ such that $h^{i}_{0},h^{i}_{\frac{1}{n-1}},h^{i}_{\frac{2}{n-1}},\cdots,h^{i}_{1}:G_{i}\rightarrow X$ are the $n$ projections $X^{n}\to X$ restricted to $G_{i}$. Using the same argument as in [2, Proposition 4.12(c)], we can extend the section $s_{i}$ on an open subset containing $G_{i}$. Hence $TC_{n}(X)\leq r.$

∎

3. Higher simplicial complexity

A simplicial approach to topological complexity was introduced by Gonzalez’s ([4]). He introduced the notion of simplicial complexity $SC(K)$ for simplicial complex $K$. This was based on contiguity class of simplicial maps. It is proved in ([4]) that simplicial complexity $SC(K)$ is equal to the topological complexity $TC(\mid K\mid)$ of geometric realization of $K$, for a finite simplicial complex $K$. In this section we introduce higher analog of simplicial complexity $SC_{n}(K)$ and prove that for a finite simplicial complex $K$, $SC_{n}(K)=TC_{n}(\mid K\mid)$.

We begin by recalling the definition and some basic facts about contiguity of simplicial maps. For a positive integer $c$, two simplicial maps $\phi,\phi^{\prime}:K\rightarrow L$ are called $c$-contiguous if there is a sequence of simplicial maps $\phi=\phi_{0},\phi_{1},\phi_{2}\cdots,\phi_{c}=\phi^{\prime}:K\rightarrow L$, such that $\phi_{i-1}(\sigma)\cup\phi_{i}(\sigma)$ is a simplex of $L$ for each simplex $\sigma$ of $K$ and $i\in\{1,2,\cdots,c\}$. We write $\phi\sim\phi^{\prime}$ if $\phi$ and $\phi^{\prime}$ are $c$-contiguous for some positive integer $c$. This defines an equivalence relation on the set of simplicial maps $K\rightarrow L$ and the equivalence classes are called contiguity classes. Simplicial maps in the same contiguity class have homotopic topological realization.

We denote barycentric subdivision of $K$ by $\operatorname{sd}(K)$ and $\operatorname{sd}^{k+1}(K)=\operatorname{sd}(\operatorname{sd}^{k}(K)).$ We choose a simplicial approximation of the identity on $\mid K\mid\times\mid K\mid,$

[TABLE]

The iterated compositions are denoted by

[TABLE]

Let $\pi_{j}:\operatorname{sd}^{k}(K\times K)\rightarrow K$ denote the composition of $\iota^{k}_{K\times K}$ and $j^{th}$ projection for $j=1,2$.

Let $K$ be a finite simplicial complex. In [4] Gonzalez defined $SC^{k}(K)$ to be the smallest nonnegative integer $r$ such that there exist subcomplexes $\{L_{i}\}_{i=1}^{r}$ covering $\operatorname{sd}^{k}(K\times K)$ and the restrictions $\pi_{1},\pi_{2}:L_{i}\rightarrow K$ lie in the same contiguity class for each $i$. Then $SC^{0}\geq SC^{1}\geq SC^{2}\geq\cdots\geq 1$. The simplicial complexity $SC(K)$ is defined as the minimum of the $SC^{k}(K)$ i.e.,

[TABLE]

Now we introduce higher simplicial complexity $SC_{n}(K)$ of simplicial complex $K$. As the previous case we choose a simplicial approximation $\iota_{\operatorname{sd}^{k-1}(K^{n})}:\operatorname{sd}^{k}(K^{n})\rightarrow\operatorname{sd}^{k-1}(K^{n})$ of the identity on $\mid K\mid^{n}$ for $k\geq 0$. We denote iterated compositions by $\iota^{k}_{K^{n}}:\operatorname{sd}^{k}(K^{n})\rightarrow K^{n}$ and $\pi_{j}:\operatorname{sd}^{k}(K^{n})\rightarrow K$ denote the composition of $\iota^{k}_{K^{n}}$ and $j^{th}$ projection $K^{n}\to K$ for $j=1,2,\cdots,n$.

3.1 Definition**.**

Let $K$ be a simplicial complex. We define $SC^{k}_{n}(K)$ as the smallest nonnegative integer $r$ such that there exist subcomplexes $\{L_{i}\}_{i=1}^{r}$ covering $\operatorname{sd}^{k}(K^{n})$ and the restrictions $\pi_{j}:L_{i}\rightarrow K$, for $j=1,2,\cdots,n$ lie in the same contiguity class, for each $i$. If no such $r$ exists then we define $SC^{k}_{n}(K)$ to be $\infty$.

It is to be noted that the value $SC^{k}_{n}$ is independent of the chosen approximation $\iota^{k}_{K^{n}}:\operatorname{sd}^{k}(K^{n})\rightarrow K^{n}$ of identity on $\mid K\mid^{n}$. As in the simplicial complexity of Gonzales, we have $SC_{n}^{0}\geq SC_{n}^{1}\geq SC_{n}^{2}\geq\cdots\geq 1$ for any simplicial complex $K$. Using the Proposition (2.2) we deduce the following:

3.2 Lemma**.**

For a simplicial complex $K$, we have the following inequalities:

1. (i)

$SC_{n}^{k}(K)\geq TC_{n}(\mid K\mid)$* for any $n\geq 2$.* 2. (ii)

$SC_{n}^{k}(K)\geq SC_{n}^{k+1}(K)$* for all $n\geq 2,~{}~{}k\geq 0$.*

Proof.

(i) If we first apply the geometric realization functor of simplicial complexes, then $\mid\pi_{j}\mid$ are homotopic, for $j=1,2,\cdots,n$. Then use Lemma 2.1 to get sections $s_{i}:\mid L_{i}\mid\rightarrow P\mid K\mid$ over locally compact susbets $\mid L_{i}\mid$. If we set $G_{1}=\mid L_{1}\mid$ and $G_{i}=\mid L_{i}\mid-(\mid L_{1}\mid\cup\mid L_{2}\mid\cup\cdots\cup\mid L_{i-1}\mid)$ for $i\geq 2$, then each $G_{i}$ is also locally compact. Define $s:\mid K\mid^{n}\rightarrow P\mid K\mid$, is the union of $s_{i}$ restricted over $G_{i}$ and lastly we apply Proposition 2.2 to conclude $SC_{n}^{k}(K)\geq TC_{n}(\mid K\mid)$ for any $n\geq 2$.

(ii) Let $J$ be a subcomplex of $\operatorname{sd}^{k}(K^{n})$ on which $\pi_{1},\pi_{2},\cdots,\pi_{n}$ are in same contiguity class. Assume that $\lambda_{J}:\operatorname{sd}(J)\rightarrow J$ is an approximation of identity on $J$. Obviously $\operatorname{sd}(J)$ is a subcomplex of $\operatorname{sd}^{k+1}(K^{n})$. We will show that $\pi_{1},\pi_{2},\cdots,\pi_{n}$ are in same contiguity class on $\operatorname{sd}(J)$. The two compositions of the diagram are contiguous.

{J}$${\operatorname{sd}^{k}(K^{n})}$${\operatorname{sd}(J)}$${\operatorname{sd}^{k+1}{K^{n}}}$$\scriptstyle{\lambda}$$\scriptstyle{\iota}

So, $\pi_{1},\pi_{2},\cdots,\pi_{n}$ are in same contiguity class on $\operatorname{sd}(J)$. ∎

Since $\{SC_{n}^{k}(X)\}_{k=1}^{\infty}$ is a decreasing sequence of integers, we can make the following definition.

3.3 Definition**.**

For a finite simplicial complex $K$, the $n$-th simplicial complexity $SC_{n}(K)$ is defined as the minimum of the $SC_{n}^{k}(K)$:

[TABLE]

We now prove the main Theorem of this section. The proof is analogous to [4, Theorem 3.5]. The following result is required in the proof ([8, Chapter 3]).

3.4 Proposition**.**

Let $f_{1},f_{2},\cdots,f_{n}:\mid K\mid\rightarrow\mid L\mid$ be continuous maps, which are in same homotopic class, then there is $k_{0}\in\mathbb{N}$ such that, for each $k\geq k_{0}$ and any approximation $\phi_{1},\phi_{2},\cdots,\phi_{n}:\operatorname{sd}^{k}(K)\rightarrow L$ of $f_{1},f_{2},\cdots,f_{n}$ respectively, are belongs to same contiguity class.

3.5 Theorem**.**

For a finite simplicial complex $K$, $SC_{n}(K)=TC_{n}(\mid K\mid)$ for any $n\geq 2$.

Proof.

From the Lemma (3.2) one can say that $SC_{n}(K)\geq TC_{n}(\mid K\mid)$. We now prove the other inequality. Assume that $TC_{n}(\mid K\mid)=r$. We choose a motion planner $\{(U_{1},s_{1}),(U_{2},s_{2}),\cdots,(U_{r},s_{r})\}$ for $\mid K\mid$. Using the finiteness assumption on $K$, we choose a large positive integer $k$ so that the realization of each simplex of $\operatorname{sd}^{k}(K^{n})$ is contained in some $U_{i}~{}(0\leq i\leq r)$. For each $i\in\{1,2,\cdots,r\}$ let $L_{i}$ be the subcomplex of $\operatorname{sd}^{k}(K^{n})$ consisting of those simplices whose realization are contained in $U_{i}$. Then $L_{1},L_{2},\cdots,L_{r}$ covers $K^{n}$. Now the projections $p_{1},p_{2},\cdots,p_{n}:\mid K\mid^{n}\rightarrow\mid K\mid$ belong to same homotopy class over each $U_{i}$ and, in particular, over the realization of the corresponding subcomplex $L_{i}$. Therefore by Proposition (3.4) there is a positive integer $k^{\prime}$ such that, for each $i\in\{1,2,\cdots,r\}$ the compositions $\operatorname{sd}^{k^{\prime}}(L_{i})\hookrightarrow\operatorname{sd}^{k+k^{\prime}}(K^{n})\rightarrow K$ are belongs to same contiguity class. Hence $SC_{n}(K)\leq TC_{n}(\mid K\mid)$, and thus $SC_{n}(K)=TC_{n}(\mid K\mid)$. ∎

4. Higher combinatorial complexity

In this section we introduce the higher analogue of combinatorial complexity of a finite space, as defined by Tanaka in [10]. We refer reader to [9] for finite spaces. A finite space $P$ is a finite $T_{0}$ space. For any $x\in P$ we denote $U_{x}$ be the intersection of all open set containing $x$. Now define a partial relation on $P$ by $x\leq y$ if and only if $U_{x}\subseteq U_{y}$. So we can consider $P$ as a poset. On the other hand, given a finite poset, we have a $T_{0}$ finite set with $U_{x}=\{y:~{}~{}y\leq x\}$. Thus a finite space is equivalent to a finite poset. From now onwards we assume all our finite spaces are connected.

A map between finite spaces is continuous if and only if it preserves the partial order. Let $J_{m}$ denote the finite space consisting $m+1$ points with the zigzag order

$0\leq 1\geq 2\leq\cdots\geq(\leq)m$.

This finite space is called the finite fence with length $m$. It behaves like an interval in the category of finite spaces. An order preserving map $\gamma:J_{m}\rightarrow P$ is called a combinatorial path or simply a path. Thus a combinatorial path is just a zigzag $\gamma(0)\leq\gamma(1)\geq\gamma(2)\leq\cdots\geq(\leq)\gamma(m)$ of elements of $P$. A conneted finite space is always path connected. If $\gamma_{1}$ and $\gamma_{2}$ be two combinatorial paths in $P$ of length $m_{1}$ and $m_{2}$ respectively, satisfying $\gamma_{1}(m_{1})=\gamma_{2}(0)$. Then the concatenation of $\gamma_{1}$ and $\gamma_{2}$ is a path $\gamma_{1}\ast\gamma_{2}:J_{m}\to P$ where $m=m_{1}+m_{2}$ or $m_{1}+m_{2}+1$ according to $m_{1}$ is even or odd. It is define as:

$\gamma_{1}*\gamma_{2}(i)=\begin{cases}\gamma_{1}(i)&\mbox{ if }~{}~{}0\leq i\leq m_{1}\\ \gamma_{2}(i-m_{1})&\ \mbox{if }~{}~{}m_{1}\leq i\leq m_{1}+m_{2}.\end{cases}$                 $(m_{1}$ -even)

$\gamma_{1}*\gamma_{2}(i)=\begin{cases}\gamma_{1}(i)&\mbox{ if }~{}~{}0\leq i\leq m_{1}\\ \gamma_{1}(i-1)&\mbox{ if }~{}~{}i=m_{1}+1\\ \gamma_{2}(i-m_{1}-1)&\ \mbox{if }~{}~{}m_{1}+1\leq i\leq m_{1}+m_{2}+1.\end{cases}$ $(m_{1}$ -odd)

Two maps $f,g:P\rightarrow Q$ between two finite spaces are called homotopic if there exist $m\geq 0$ and a continuous map (or an order preserving map) $H:P\times J_{m}\rightarrow Q$ such that $H(x,0)=f(x)$ and $H(x,m)=g(x)$. The finite space of all combinatorial paths of $P$ with length $m$, equipped with the pointwise order, is denoted by $P^{J_{m}}$. As an analog of path fibration, it is equipped with the canonical order preserving map

[TABLE]

In ([10]) Tanaka defined $CC_{m}(P)$ to be the smallest nonnegative integer $r$ such that there exist an open cover of $P\times P$ consisting $r$ open sets with a section of $q_{m}$ on each open set. He proved that $CC_{m}(P)$ is decreasing sequence on $m$ and defined $CC(P)$ be the limit of $CC_{m}(P)$. Also he proved that $CC(P)=TC(P)$. In this section we will generalise the above idea.

Let $n\geq 2$ and $J_{n,m}$ be the finite set of $nm+1$ points

[TABLE]

The partial ordering on $J_{n,m}$ consists of $n$ finite fances each length $m$ as below:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Consider the space $P^{J{n,m}}$ is the space of all order preserving map $J_{n,m}\rightarrow P$. We have a canonical projection

[TABLE]

We define $\mathcal{CC}_{{n},{m}}(P)$ as the smallest positive integer $r$ such that $P^{n}$ can be cover by $r$ open sets $\{Q_{i}\}_{i=1}^{r}$ with section $s_{i}:Q_{i}\rightarrow P^{J_{n,m}}$ of $q_{n,m}$ for each $i$. If no such $r$ exist then we set $\mathcal{CC}_{{n},{m}}(P)=\infty$. The following lemma shows that $\mathcal{CC}_{{n},{m}}$ decreases as we increase $m$.

4.1 Lemma**.**

For any $m\geq 0,n\geq 2$, it holds that $\mathcal{CC}_{{n},{m+1}}(P)\leq\mathcal{CC}_{{n},{m}}(P)$.

Proof.

Let $\mathcal{CC}_{{n},{m}}(P)=r$ and $\{Q_{i}\}_{i=1}^{r}$ be an open cover of $P^{n}$ with section $s_{i}:Q_{i}\rightarrow P^{J_{n,m}}$. Consider the retraction map $R:J_{n,m+1}\rightarrow J_{n,m}$ sending each $(m+1)_{i}$ to $m_{i}$ for $1\leq i\leq n$. This is clearly an order preserving map. It will induce a map $R^{*}:P^{J_{n,m}}\rightarrow P^{J_{n,m+1}},\gamma\rightarrow\gamma\circ R$ such that the following diagram commutes:

{P^{J_{n,m}}}$${P^{J_{n,m+1}}}$${P^{n}}$$\scriptstyle{R^{*}}$$\scriptstyle{q_{n,m}}$$\scriptstyle{q_{n,m+1}}

The composition $R^{*}\circ s_{i}:Q_{i}\rightarrow P^{{n,m+1}}$ is a section of $q_{n,m+1}$ for each $i$. Thus, $\mathcal{CC}_{{n},{m+1}}(P)\leq r$. ∎

4.2 Definition**.**

For a finite space $P$ we define $\mathcal{CC}_{{n}}(P)$ is the minimum of the $\mathcal{CC}_{{n},{m}}(P)$

[TABLE]

We now give an alternative description of $\mathcal{CC}_{{n}}(P)$. Later we will use both the description interchangably. Consider the space $P^{J_{(n-1)m}}$ and the projection map

[TABLE]

Denote by $\mathcal{CC^{\prime}}_{{n},{m}}(P)$ the smallest positive integer $r^{\prime}$ such that $P^{n}$ can be cover by $r^{\prime}$ open sets $\{Q_{i}\}_{i=1}^{r^{\prime}}$ with a section $s_{i}:Q_{i}\rightarrow P^{J_{(n-1)m}}$ of $q^{\prime}_{n,m}$ for each $i$. We have a analogue of Lemma 4.1.

4.3 Lemma**.**

For any $m\geq 0,n\geq 2$, it holds that $\mathcal{CC^{\prime}}_{{n},{m+1}}(P)\leq\mathcal{CC^{\prime}}_{{n},{m}}(P)$.

Proof.

Let $q^{\prime}_{n,m}$ has a section on an open set $Q_{i}$ of $P^{n}$. We define a order preserving map

[TABLE]

which maps $i(m+1)-1$, $i(m+1)$ and $i(m+1)+1$ to $im$, if $i$ is odd otherwise it is linear. This will induce a map $R^{*}:P^{J_{(n-1)m}}\rightarrow P^{J_{(n-1)(m+1)}}$ such that the following diagram commutes:

{P^{J_{(n-1)m}}}$${P^{J_{(n-1)(m+1)}}}$${P^{n}}$$\scriptstyle{R^{*}}$$\scriptstyle{q^{\prime}_{n,m}}$$\scriptstyle{q^{\prime}_{n,m+1}}

The composition $R^{*}\circ s_{i}:Q_{i}\rightarrow P^{{(n-1)(m+1)}}$ is a section of $q^{\prime}_{n,m+1}$ for each $i$. Thus, $\mathcal{CC^{\prime}}_{{n},{m+1}}(P)\leq\mathcal{CC^{\prime}}_{{n},{m}}(P)$. ∎

4.4 Definition**.**

We define $\mathcal{CC^{\prime}}_{{n}}(P)$ is the minimum of the $\mathcal{CC^{\prime}}_{{n},{m}}(P)$

[TABLE]

We now prove that the two Definitions 4.2 and 4.4 are equivalent.

4.5 Theorem**.**

For any finite space $P$, $\mathcal{CC^{\prime}}_{{n}}(P)=\mathcal{CC}_{{n}}(P)$.

Proof.

Assume that $\mathcal{CC}_{{n}}(P)=r$. Take an open cover $\{Q_{i}\}_{i=1}^{r}$ of $P^{n}$ with order preserving section $s_{i}:Q_{i}\rightarrow P^{J_{n,m}}$ of $q_{n,m}$, for some $m$ and each $i$. Since $\mathcal{CC}_{{n},{m}}$ is decreasing with respect to $m$ by Lemma 4.1 , we can assume that $m$ is even. Define an order preserving map $f:J_{(n-1)2m}\rightarrow J_{n,m}$ by sending each element of $J_{(n-1)2m}$ to the corresponding element of the following path of $J_{n,m}$:

[TABLE]

[TABLE]

This map induces $f^{*}:P^{J_{n,m}}\rightarrow P^{J_{(n-1)2m}}$ such that the following triangle commutes.

{P^{J_{n,m}}}$${P^{J_{(n-1)2m}}}$${P^{n}}$$\scriptstyle{f^{*}}$$\scriptstyle{q_{n,m}}$$\scriptstyle{q^{\prime}_{n,2m}}

So the composition map $f^{*}\circ s_{i}:Q_{i}\rightarrow P^{J_{(n-1)2m}}$ is a section of $q^{\prime}_{n,2m}$ on $Q_{i}$ for each $i$. Thus $\mathcal{CC^{\prime}}_{{n},{2m}}(P)\leq r$ i.e. $\mathcal{CC^{\prime}}_{{n}}(P)\leq r=\mathcal{CC}_{{n}}(P)$.

Conversely, assume that $\mathcal{CC^{\prime}}_{{n}}(P)=r^{\prime}.$ Then there are sections $s_{i}:Q_{i}\rightarrow P^{J_{(n-1)m}}$ of $q^{\prime}_{n,m}$ on open subsets $Q_{i}$ of $P^{n}$ coving it, for some $m$,  $1\leq i\leq r^{\prime}$. We can take $m$ as a multiple of four, by Lemma 4.3. Let $k=\frac{(n-1)m}{2}$ so that $k$ is even. Define an order preserving retraction map $g:J_{n,k}\rightarrow J_{(n-1)m}$ by sending $0\leq 1_{j}\geq 2_{j}\leq\cdots\geq k_{j}$ to ($1\leq j\leq n$):

[TABLE]

[TABLE]

As in previous case the map $g$ map induces $g^{*}:P^{J_{(n-1)m}}\rightarrow P^{J_{n,k}}$ such that the following triangle commutes.

{P^{J_{(n-1)m}}}$${P^{J_{n,k}}}$${P^{n}}$$\scriptstyle{g^{*}}$$\scriptstyle{q^{\prime}_{n,m}}$$\scriptstyle{q_{n,k}}

So the composition map $g^{*}\circ s_{i}:Q_{i}\rightarrow P^{J_{k,n}}$ is a section of $q_{n,k}$ for each $1\leq i\leq r^{\prime}$. Thus $\mathcal{CC}_{{n},{k}}(P)\leq r^{\prime}$ i.e. $\mathcal{CC}_{{n}}(P)\leq\mathcal{CC^{\prime}}_{{n}}(P)$. Hence $\mathcal{CC^{\prime}}_{{n}}(P)=\mathcal{CC}_{{n}}(P)$. ∎

To abuse the notation we will only use $\mathcal{CC}_{{n}}(P)$ for both the descriptions. Similar to the topological setting, we have an upper bound of $\mathcal{CC}_{{n}}(P)$ in terms of $L\text{-}S$ category of $P^{n}$.

4.6 Lemma**.**

It holds that $\mathcal{CC}_{{n}}(P)\leq cat(P^{n})$ for all $n\geq 2$.

Proof.

Let $cat(P^{n})=r$. Then there exists contractible open cover $\{Q_{i}\}_{i=1}^{r}$ of $P^{n}$. For a fixed $i,$ let $Q_{i}$ be contractible to $(x_{1},x_{2},\cdots,x_{n})\in P^{n}$. Since $P$ is connected there exists a positive integer $k$ and a map

[TABLE]

Now let $(p_{1},p_{2},\cdots,p_{n})\in Q_{i}$ be an arbitrary element. Choose a contracting homotopy of $Q_{i}$ in $P^{n}$, $H:Q_{i}\times J_{l}\rightarrow P^{n}$ such that $H(p_{1},p_{2},\cdots,p_{n},0)=(x_{1},x_{2},\cdots,x_{n})$ and $H(p_{1},p_{2},\cdots,p_{n},l)=(p_{1},p_{2},\cdots,p_{n})$. Applying exponential law we get projection maps $H_{j}:Q_{i}\rightarrow P^{J_{l}}$ such that $H_{j}(p_{1},p_{2},\cdots,p_{n})(l)=p_{j}$ and $H_{j}(p_{1},p_{2},\cdots,p_{n})(0)=x_{j}$ for $j=1,2,\cdots,n$. Assume $\gamma_{j}:J_{k}\to P$ be $j$-th component of $\gamma$ which is obtained by composing $\gamma$ with $n$-different inclusions of $J_{k}$ inside $J_{n,k}$. Set $\lambda_{j}=\gamma_{j}\ast H_{j}$, concatenation of $H_{j}$ and $\gamma_{j}$. Let $\lambda:J_{n,k+l}\rightarrow P$ or $\lambda:J_{n,k+l+1}\rightarrow P$ according to $l$ even or odd, is the map whose components are $\lambda_{i}$’s. Define $s_{i}(p_{i},p_{2},\cdots,p_{n})=\lambda$, whose projection nothing but $(p_{i},p_{2},\cdots,p_{n})$. Set $m_{i}=k+l$ or $k+l+1$ according to $l$ even or odd and $m=\max\{m_{1},m_{2},\cdots,m_{r}\}$. Then $\mathcal{CC}_{{n},{m}}(P)\leq r$ and therefore $\mathcal{CC}_{{n}}(P)\leq cat(P^{n})$ for all $n\geq 2$. ∎

Now we will prove the equality between $\mathcal{CC}_{{n}}(P)$ and $TC_{n}(P)$. For this first we recall order complex $\mathcal{K}(P)$ of a finite space $P$.

4.7 Definition**.**

The order complex $\mathcal{K}(P)$ of a finite space $P$ is a simplicial complex whose $n$-simplices are linearly ordered subsets of $P$. Its geometric realization is denoted by $\mid\mathcal{K}(P)\mid$.

4.8 Example**.**

Let $\mathbb{S}^{m}$ denote the finite space consisting of $2m+2$ points

[TABLE]

with the partial order defined by $e_{p}^{k}\leq e_{q}^{l}$ if $k\leq l$ and $p,q\in\{+,-\}$. The realization of the order complex $\mid\mathcal{K}(\mathbb{S}^{m})\mid$ is homeomorphic to the sphere $S^{m}$ with dimension $m$.

4.9 Theorem**.**

For any finite space $P$, it holds that $\mathcal{CC}_{{n}}(P)=TC_{n}(P)$, $n\geq 2$.

Proof.

Let us first show that $TC_{n}(P)\geq\mathcal{CC}_{{n}}(P)$. Assume that $TC_{n}(P)=r$ with an open cover $\{Q_{i}\}_{i=1}^{n}$ of $P^{n}$ and a continuous section $Q_{i}\rightarrow P^{I}$ of $e$ for each $i$. This induces a map $f_{1}:I\rightarrow P^{Q_{i}}$ by the exponential law. Hence we obtain a order preserving map $f_{2}:J_{s}\rightarrow P^{Q_{i}}$ for some $s\geq 0$ by the homotopy theory of finite spaces. Now we can construct a order preserving map $\phi:J_{(n-1)m}\rightarrow J_{s}$ for some $m\geq 0$, such a way that $f_{1}(\frac{k}{n-1})=f_{2}\circ\phi(km),~{}~{}k=0,1,\cdots,n-1$. So we have a combinatorial section $Q_{i}\rightarrow P^{J_{(n-1)m}}$ of $q^{\prime}_{n,m}$. Hence $\mathcal{CC}_{{n}}(P)\leq r$.

For the other inequality, assume $\mathcal{CC}_{{n}}(P)=r$ with open cover $\{Q_{i}\}_{i=1}^{n}$ of $P^{n}$ and a continuous section $s_{i}:Q_{i}\rightarrow P^{J_{(n-1)m}}$ for some $m\geq 0$ of $q^{\prime}_{n,m}$, for all $i$.

Let $\alpha:[0,(n-1)m]\simeq\mid\mathcal{K}(J_{(n-1)m})\mid\rightarrow J_{(n-1)m}$ be denote the continuous map (see [6]):

$\alpha(t)=\begin{cases}2k-1&\mbox{ if }~{}~{}t=2k-1\\ 2k&\mbox{if }~{}~{}2k-1<t<2k+1.\end{cases}$

Note that, $\alpha(jm)=jm,j=0,1,\cdots,(n-1)$. Let $\beta:I\rightarrow J_{(n-1)m}$ denote the composition of $\alpha$ and the homeomorphism $I=[0,1]\simeq[0,(n-1)m]$, given by $t\rightarrow(n-1)mt$. This induces $\beta^{*}:P^{J_{(n-1)m}}\rightarrow P^{I}$ such that the diagram commutes:

{P^{J_{(n-1)m}}}$${P^{I}}$${P^{n}}$$\scriptstyle{\beta^{*}}$$\scriptstyle{q^{\prime}_{n,m}}$$\scriptstyle{e_{n}}

The composition $\beta^{*}\circ s_{i}:Q_{i}\rightarrow P^{I}$ is a continuous section for the fibration $e_{n}$. So $TC_{n}(P)\leq r$. Thus $\mathcal{CC}_{{n}}(P)=TC_{n}(P)$, $n\geq 2$. ∎

4.10 Corollary**.**

1. (a)

For any finite space $P$ we have $\mathcal{CC}_{{n}}(P)\leq\mathcal{CC}_{{n+1}}(P)$ , $n\geq 2$. 2. (b)

A finite space $P$ is contractible if and only if $\mathcal{CC}_{{n}}(P)=1$ for any $n\geq 2$. 3. (c)

The $\mathcal{CC}_{{n}}$ is homotopy invariant, i.e. $\mathcal{CC}_{{n}}(P)=\mathcal{CC}_{{n}}(Q)$ if two spaces $P$ and $Q$ are homotopy equivalent. 4. (d)

For any finite space $P$, the following inequalities hold:

[TABLE]

Proof.

This follows from above Theorem 4.9 and the corresponding inequalities about $TC_{n}(P)$.

∎

4.11 Example**.**

Let $\mathbb{S}^{m}$ denote the finite space consisting as in Example 4.8. Then

[TABLE]

Proof.

We know that $cat(\mathbb{S}^{m})=2$ (see [10, Example 3.7]) for any $m\geq 1$. So by Corollary 4.10, $\mathcal{CC}_{{n}}(\mathbb{S}^{m})\leq cat(\mathbb{S}^{m})^{n}=2^{n}$. If $\mathcal{CC}_{{n}}(\mathbb{S}^{m})<2^{n}$, then there is an open set of $(\mathbb{S}^{m})^{n}$ containing at least two distinguished maximal points of $(\mathbb{S}^{m})^{n}$. This yields a contractible open set in $\mathbb{S}^{m}$ containing $e_{+}^{m}$ and $e_{-}^{m}$, that will be nothing but the entire space $\mathbb{S}^{m}$, which is not contractible. The contradiction implies that $\mathcal{CC}_{{n}}(\mathbb{S}^{m})=2^{n}$. Thus, $\mathcal{CC}_{{n}}(\mathbb{S}^{m})=cat((\mathbb{S}^{m})^{n})=cat(\mathbb{S}^{m})^{n}=2^{n}$ for any $m\geq 1$ and $n\geq 2$. ∎

5. Higher combinatorial complexity with barycentric subdivision

The combinatorial complexity $\mathcal{CC}^{k}_{n}(P)$ does not capture the topological complexity of the naturally associated simplicial complex $\mathcal{K}(P)$. From the example of the $\mathbb{S}^{m}$ we see that $\mathcal{CC}^{k}_{n}(\mathbb{S}^{m})$ is much higher than $TC_{n}({S}^{m})$. To remedy the situation, we refine the definition of $\mathcal{CC}^{k}_{n}(P)$ using barycentric subdivision of $P$. We first recall barycentric subdivision of a finite space $P$. Then we define higher combinatorial complexity with barycentric subdivision and show it is equal to the topological complexity of $|\mathcal{K}(P)|$.

5.1 Definition**.**

The barycentric subdivision $\operatorname{sd}(P)$ of $P$ is defined as the face poset $\mathcal{X}(\mathcal{K}(P))$ of the order complex $\mathcal{K}(P)$ (4.7).

Let $\tau_{P}:\operatorname{sd}(P)\rightarrow P$ be the map sending $p_{0}\leq p_{1}\leq\cdots\leq p_{n}$ to the last element $p_{n}$. This is a weak homotopy equivalence, and the induced simplicial map $\mathcal{K}(\tau_{P}):\mathcal{K}(\operatorname{sd}(P))=\operatorname{sd}(\mathcal{K}(P))\rightarrow\mathcal{K}(P)$ is a simplicial approximation of the identity on $\mid\mathcal{K}(P)\mid$ (see [5]). For $k\geq 0$, we denote by $\tau_{P}^{k}:\operatorname{sd}^{k}(P)\rightarrow P$ the composition

$\operatorname{sd}^{k}(P)\xrightarrow{\tau_{\operatorname{sd}^{k-1}(P)}}\operatorname{sd}^{k-1}(P)\xrightarrow{\tau_{\operatorname{sd}^{k-2}(P)}}\cdots\xrightarrow{\tau_{\operatorname{sd}(P)}}\operatorname{sd}(P)\xrightarrow{\tau_{P}}P$.

5.2 Definition**.**

Let $k\geq 0$. We define $\mathcal{CC}^{k}_{n}(P)$ as the smallest nonnegative integer $r$ such that there exist an open cover $\{Q_{i}\}_{i=1}^{r}$ of $\operatorname{sd}^{k}(P^{n})$ and an positive integer $m$, with a map $s_{i}:Q_{i}\rightarrow P^{J_{n,m}}$ such that $q_{n,m}\circ s_{i}=\tau_{P^{n}}^{k}$ on $Q_{i}$ for each $i$. If no such $r$ exists, then we define $\mathcal{CC}^{k}_{n}(P)$ to be $\infty$.

If we define $\mathcal{CC^{\prime}}^{k}_{n}(P)$ as taking $P^{J_{(n-1)m}}$ instead of $P^{J_{n,m}}$ and $q^{\prime}_{n,m}$ of $q_{n,m}$ then we get same positive integer. Obviously, $\mathcal{CC}_{{n}}(P)=\mathcal{CC}^{0}_{n}(P)$ by the definition above.

5.3 Lemma**.**

For any finite space $P$ and $n\geq 2$, we have:

1. (a)

$\mathcal{CC^{\prime}}^{k}_{n}(P)=\mathcal{CC}^{k}_{n}(P)$. 2. (b)

$\mathcal{CC}^{k+1}_{n}(P)\leq\mathcal{CC}^{k}_{n}.$**

Proof.

(a)This proof is similar as the proof of the Theorem (4.5).

(b) The result for $n=2$ was proved in [10, Lemma 4.3]. In similar way we can prove this result. ∎

5.4 Definition**.**

We define the $n$-th combinatorial complexity $\mathcal{CC}^{\infty}_{n}(P)$ of $P$ to be

[TABLE]

Now we relate $\mathcal{CC}^{\infty}_{n}(P)$ to the $n$-th topological complexity $TC_{n}(\mid\mathcal{K}(P)\mid)$ of geometric realization of the order complex of $P$. For this we need the following lemma. For $k\geq 0$, let $\rho_{j}:\operatorname{sd}^{k}(P^{n})\rightarrow P$ denote the composition of $\tau^{k}_{P^{n}}:\operatorname{sd}^{k}(P^{n})\rightarrow P^{n}$ and the $j$-th projection for $j=1,2,\cdots,n$.

5.5 Lemma**.**

With notations as above, $\mathcal{CC}^{\infty}_{n}(P)\leq r$ if and only if there exist $k\geq 0$ and an open cover $\{Q_{i}\}_{i=1}^{r}$ of $\operatorname{sd}^{k}(P^{n})$ such that $\rho_{1},\rho_{2},\cdots,\rho_{n}:Q_{i}\rightarrow P$ are in same homotopy class of maps.

Proof.

Let us assume that $\mathcal{CC}^{\infty}_{n}(P)\leq r$. Then for some $k\geq 0$ there exist an open cover $\{Q_{i}\}_{i=1}^{r}$ of $\operatorname{sd}^{k}(P^{n})$ and a map $s_{i}:Q_{i}\rightarrow P^{J_{(n-1)m}}$ such that $q^{\prime}_{n,m}\circ s_{i}=\tau^{k}_{P^{n}}$ on $Q_{i}$ for each $i$.

{Q_{i}}$${P^{J_{(n-1)m}}}$${\operatorname{sd}^{k}(P^{n})}$${P^{n}}$$\scriptstyle{s_{i}}$$\scriptstyle{q^{\prime}_{n,m}}$$\scriptstyle{\tau^{k}_{P^{n}}}

We define a homotopy $H:Q_{i}\times J_{(n-1)m}\rightarrow P$ by

[TABLE]

where $(p_{1},p_{2},\cdots,p_{n})\in Q_{i},~{}~{}x\in J_{(n-1)m}.$ Then

[TABLE]

for $j\in\{1,2,\cdots,n\}$. This shows that the maps $\rho_{1},\rho_{2},\cdots,\rho_{n}:Q_{i}\rightarrow P$ are in same homotopy class of maps.

Conversely, assume that $\{Q_{i}\}_{i=1}^{r}$ is an open cover of $\operatorname{sd}^{k}(P^{n})$ such that $\rho_{1},\rho_{2},\cdots,\rho_{n}:Q_{i}\rightarrow P$ are in same homotopy class of maps for each $i$. Then there exist homotopies $H_{j}:Q_{i}\times J_{m}\rightarrow P$ for some $m$ and $j=1,2,\cdots,n-1$, such that $H_{j}$ is a homotopy between $\rho_{j}$ and $\rho_{j+1}$. Define $s_{i}:Q_{i}\rightarrow P^{J_{(n-1)m}}$ by

[TABLE]

where $\ast$ denotes concatenation and $x\in J_{(n-1)m}$. So we have

[TABLE]

Thus $q^{\prime}_{n,m}\circ s_{i}(p_{1},p_{2},\cdots,p_{n})=\tau^{k}_{P^{n}}(p_{1},p_{2},\cdots,p_{n}),~{}~{}1\leq j\leq n$. This gives a section over each $Q_{i}$ for $i=1,\cdots,r.$ Hence $\mathcal{CC}^{\infty}_{n}(P)\leq r.$

∎

We now prove the main result of the section, generalising [10, Theorem 4.9 ]

5.6 Theorem**.**

For any finite space $P$, we have $\mathcal{CC}^{\infty}_{n}(P)=SC_{n}(\mathcal{K}(P))$, $n\geq 2$.

Proof.

Assume that $\mathcal{CC}^{\infty}_{n}(P)=r$. By the Lemma (5.5) there exists $k\geq 0$ and an open cover $\{Q_{i}\}_{i=1}^{r}$ of $\operatorname{sd}^{k}(P^{n})$ such that $\rho_{1},\rho_{2},\cdots,\rho_{n}:Q_{i}\rightarrow P$ are in same homotopy class of maps. Using [1, Proposition 4.11] we can say that the maps

[TABLE]

lie in same contiguity class. The subcomplex $\mathcal{K}(Q_{i})$ form a cover of $\mathcal{K}(\operatorname{sd}^{k}(P^{n}))=\operatorname{sd}^{k}(\mathcal{K}(P^{n}))$ and $\mathcal{K}(\rho_{j})=\mathcal{K}(pr_{j}\circ\tau^{k}_{P^{n}})=\pi_{j}$, for $j=1,2,\cdots,n$. So, $SC_{n}^{k}(\mathcal{K}(P))\leq r$ and then $SC_{n}(\mathcal{K}(P))\leq r$.

Conversely, assume that $SC_{n}(\mathcal{K}(P))=r$. Then $SC_{n}^{k}(\mathcal{K}(P))=r$ for some $k\geq 0$. Let $\{L_{i}\}_{i=1}^{r}$ be a covering of $\operatorname{sd}^{k}(\mathcal{K}(P^{n}))$ and the restriction $\pi_{1},\pi_{2},\cdots,\pi_{n}:L_{i}\rightarrow\mathcal{K}(P)$ lie in same contiguity class for each $i$. The [1, Proposition 4.12] implies that $\mathcal{X}(\pi_{1}),\mathcal{X}(\pi_{2}),\cdots,\mathcal{X}(\pi_{n}):\mathcal{X}(L_{i})\rightarrow\mathcal{X}(\mathcal{K}(P))=\operatorname{sd}(P)$ are in same homotopy class of maps. The subsets $\mathcal{X}(L_{i})$ form an open cover of $\mathcal{X}(\operatorname{sd}^{k}(\mathcal{K}(P^{n})))=\operatorname{sd}^{k+1}(P^{n})$. The naturality of $\tau$ makes the following diagram commute :

{\operatorname{sd}^{k+1}(P^{n})}$${\operatorname{sd}(P^{n})}$${\operatorname{sd}(P)}$${\operatorname{sd}^{k}(P^{n})}$${P^{n}}$${P}$$\scriptstyle{\operatorname{sd}(\tau^{k}_{P^{n}})}$$\scriptstyle{\tau_{\operatorname{sd}^{k}(P^{n})}}$$\scriptstyle{\operatorname{sd}(pr_{j})}$$\scriptstyle{\tau_{P^{n}}}$$\scriptstyle{\tau_{P}}$$\scriptstyle{\tau^{k}_{P^{n}}}$$\scriptstyle{pr_{j}}

Also we have

[TABLE]

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =pr_{j}\circ\tau_{P^{n}}^{k+1}=\rho_{j}$      for $j=1,2,\cdots,n$.

So, $\mathcal{CC}^{k+1}_{n}(P)\leq r$ and then $\mathcal{CC}^{\infty}_{n}(P)\leq r$. Thus $\mathcal{CC}^{\infty}_{n}(P)=SC_{n}(\mathcal{K}(P))$, $n\geq 2$. ∎

Combining Theorem 5.6 and Theorem 3.5 we have the following corollary.

5.7 Corollary**.**

For any finite space $P$, we have $\mathcal{CC}^{\infty}_{n}(P)=TC_{n}(\mid\mathcal{K}(P)\mid)$, $n\geq 2$.