Discrete Differential Calculus on (Co-)Simplicial Complexes and Generalized
(Co-)Homology

Shiquan Ren

> ###### Abstract
>
> Let $V$ be a finite set. Let $\mathcal{K}$ be a simplicial complex with its
> vertices in $V$. In this paper, we discuss some differential calculus on
> $V$. We construct some generalized homology groups of $\mathcal{K}$ by using
> the differential calculus on $V$. Moreover, we define a co-simplicial
> complex to be the complement of a simplicial complex in the complete
> hypergraph on $V$. Let $\mathcal{L}$ be a co-simplicial complex with its
> vertices in $V$. We construct some generalized cohomology groups of
> $\mathcal{L}$ by using the differential calculus on $V$.
>
> 2010 Mathematics Subject Classification. Primary 55U10, 55U15, Secondary
> 53A45, 08A50
>
> Keywords and Phrases. simplicial complexes, hypergraphs, (co-)chain
> complexes, (co-)homology, differential calculus

##  1 Introduction

Simplicial complexes play an important and fundamental role in algebraic
topology. So far, topologists have developed the homology and cohomology
theory for simplicial complexes. We refer to [23, Chapter 1] and [19, Section
2.1] for a systematic introduction to the simplicial homology theory. We also
refer to [23, Section 42, Chapter 5] and [19, Section 3.1 and Section 3.2] for
an introduction to the simplicial cohomology theory. On the other hand, since
1950’s, topologists have developed the simplicial homotopy theory (for
example, we may refer to [10, 11, 12, 22, 30]), which has been found to have
significant applications in various topics in algebraic and geometric topology
(for example, we refer to [5, 20, 25] for some of such applications). In
simplicial homotopy theory, simplicial complexes are the fundamental models
for simplicial sets.

The notion of hypergraphs is a higher dimensional generalization of the notion
of graphs (cf. [1, 24]). In a graph, an edge consists of two vertices while in
a hypergraph, a hyperedge is allowed to be consisted of $n$-vertices for any
$n\geq 1$. From a topological point of view, a hypergraph can be obtained by
deleting some non-maximal faces in a simplicial complex (cf. [3, 24]) while a
simplicial complex is a special hypergraph with no non-maximal faces missing.
The complete hypergraph $\Delta[V]$ on a finite set $V$ has its set of the
hyperedges as all the non-empty subsets of $V$ (cf. Definition 5). A
simplicial complex with all of its vertices in $V$ has its set of the
simplices as a subset of $\Delta[V]$. The complement of the set of the
simplices in $\Delta[V]$ is called a co-simplicial complex (cf. Definition 8
and Proposition 2.1).

Differential calculus is an important tool in (co)homology theory. In some
textbooks in algebraic topology (for example, [2, 21]), the methods of
differential calculus have been applied to the (co)homology theory of
differentiable manifolds and fibre bundles. During the 1990s, A. Dimakis and
F. Müller-Hoissen [7, 8, 9] initiated the study of discrete differential
calculus on discrete sets with a motivation from theoretical physics. During
the 2010s, based on the study of A. Dimakis and F. Müller-Hoissen [7, 8, 9],
Alexander Grigor’yan, Yong Lin and Shing-Tung Yau [13], Alexander Grigor’yan,
Yong Lin, Yuri Muranov and Shing-Tung Yau [14, 15, 16] and Alexander
Grigor’yan, Yuri Muranov and Shing-Tung Yau [17, 18] developed the discrete
differential calculus methods on discrete sets and applied the methods to the
study of digraphs.

In this paper, we apply the method of the (discrete) differential calculus and
give some generalized homology for simplicial complexes as well as generalized
cohomology for co-simplicial complexes.

Let $V$ be a finite set. Let $\mathcal{K}$ be a simplicial complex whose the
set of the vertices is a subset of $V$. Let $n\geq 0$. Let $v_{0}v_{1}\ldots
v_{n}$ be an $n$-simplex of $\mathcal{K}$. The usual boundary operator (cf.
[19, p. 105], [23, p. 28]) is given by

$\displaystyle\partial_{n}(v_{0}v_{1}\ldots
v_{n})=\sum_{i=0}^{n}(-1)^{i}v_{0}\ldots\widehat{v_{i}}\ldots v_{n}.$ (1.1)

We generalize the usual boundary operator and define a weighted boundary
operator

$\displaystyle\frac{\partial}{\partial v}(v_{0}v_{1}\ldots
v_{n})=\sum_{i=0}^{n}(-1)^{i}\delta(v,v_{i})v_{0}\ldots\widehat{v_{i}}\ldots
v_{n}$

with respect to any fixed vertex $v\in V$. Note that

$\displaystyle\partial_{n}=\sum_{v\in V}\frac{\partial}{\partial v}.$

We take the exterior algebra $T_{*}(V)$ generated by the
$\frac{\partial}{\partial v}$’s for all $v\in V$. We prove in Subsection 4.2
that for any $t\geq 0$ and any $\alpha\in T_{2t+1}(V)$, there is a generalized
homology group of $\mathcal{K}$ with respect to $\alpha$. Moreover, we prove
in Theorem 4.6 that for any $s\geq 0$ and any $\beta\in T_{2s}(V)$, the
element $\beta$ induces a homomorphism between the corresponding generalized
homology groups.

We point out that the generalized homology groups which will be investigated
in Subsection 4.2 are generalizations of the weighted homology groups
investigated by Robert. J. MacG. Dawson[6], Shiquan Ren, Chengyuan Wu and Jie
Wu in [27, 28] and Shiquan Ren, Chengyuan Wu, Jie Wu and Kelin Xia [29] for
weighted simplicial complexes. Let $f$ be a real function on $V$. We take
$t=1$ and

$\displaystyle\alpha=\sum_{v\in V}f(v)\frac{\partial}{\partial v}$

in Definition 21, Subsection 4.2. Then the generalized homology groups of the
simplicial complex $\mathcal{K}$ with respect to $\alpha$, which will be
investigated in Subsection 4.2, give the weighted homology groups of the
weighted simplicial complex $(\mathcal{K},f)$ which have been investigated in
[27, 28, 29].

On the other hand, Let $\mathcal{L}$ be a co-simplicial complex whose the set
of the vertices is a subset of $V$. For any $v\in V$, we consider the adjoint
linear map $dv$ of the element $\frac{\partial}{\partial v}$ in $T_{*}(V)$. We
define $T^{*}(V)$ as the exterior algebra generated by the $dv$’s for all
$v\in V$. We prove in Subsection 4.3 that for any $t\geq 0$ and any $\omega\in
T^{2t+1}(V)$, there is a generalized cohomology group of $\mathcal{L}$ with
respect to $\omega$. Moreover, we prove in Theorem 4.8 that for any $s\geq 0$
and any $\mu\in T^{2s}(V)$, the element $\mu$ induces a homomorphism between
the generalized cohomology groups.

The remaining part of this paper is organized as follows. In Section 2, we
introduce the definitions of hypergraphs, simplicial complexes and co-
simplicial complexes. In Section 3, as a preparation for Section 4, we discuss
some differential calculus for paths on discrete sets. In Section 4, we define
the generalized homology groups for simplicial complexes in Definition 21 and
define the generalized cohomology groups for co-simplicial complexes in
Definition 22. We prove Theorem 4.6 and Theorem 4.8. Finally, in Section 5, we
give some examples for Section 4.

##  2 Hypergraphs, Simplicial Complexes and Co-Simplicial Complexes

In this section, we review the definitions of hypergraphs and simplicial
complexes. We give the definition of co-simplicial complexes. We explain the
topological meanings of these definitions by introducing the geometric
realizations of hypergraphs, simplicial complexes and co-simplicial complexes.

###  2.1 Combinatorial Models of Hypergraphs, Simplicial Complexes, and Co-
Simplicial Complexes

Let $V$ be a discrete set whose elements are called vertices. Suppose $V$ has
a total order $\prec$. Let $n\geq 0$ be a non-negative integer.

######  Definition 1.

An $n$-hyperedge on $V$ is a sequence

$\displaystyle\sigma^{(n)}=v_{0}v_{1}\ldots v_{n}$ (2.1)

where $v_{0}\prec v_{1}\prec\cdots\prec v_{n}$ are vertices in $V$. For
simplicity, an $n$-hyperedge is also called a hyperedge and $\sigma^{(n)}$ in
(2.1) is also denoted as $\sigma$.

######  Remark 1:

By Definition 1, a $0$-hyperedge on $V$ is just a single vertex $v_{0}$ in $V$
and a $1$-hyperedge on $V$ is just an edge $v_{0}v_{1}$ in the complete graph
on $V$.

######  Definition 2.

The complete $n$-uniform hypergraph $\Delta_{n}(V)$ on $V$ is the collection
of all the possible $n$-hyperedges on $V$. In other words, $\Delta[V]$
consists of all the subsets of $V$ with $n$-vertices:

$\displaystyle\Delta_{n}(V)=\\{v_{0}v_{1}\ldots v_{n}\mid
v_{0},v_{1},\cdots,v_{n}\in V{\rm~{}and~{}}v_{0}\prec v_{1}\prec\cdots\prec
v_{n}\\}.$

######  Remark 2:

In particular, let $n=2$ in Definition 2. Then the complete $2$-uniform
hypergraph $\Delta_{2}(V)$ is just the complete graph on $V$.

######  Definition 3.

An $n$-uniform hypergraph  $\mathcal{H}^{(n)}$ on $V$ is a collection of some
of the $n$-hyperedges on $V$. In other words, $\mathcal{H}^{(n)}$ consists of
some of the subsets of $V$ with $n$-vertices:

$\displaystyle\mathcal{H}^{(n)}\subseteq\\{v_{0}v_{1}\ldots v_{n}\mid
v_{0},v_{1},\cdots,v_{n}\in V{\rm~{}and~{}}v_{0}\prec v_{1}\prec\cdots\prec
v_{n}\\}.$

######  Definition 4.

A hypergraph on $V$ is a disjoint union

$\displaystyle\mathcal{H}=\bigcup_{n\geq 0}\mathcal{H}^{(n)}$ (2.2)

where $\mathcal{H}^{(n)}$ is an $n$-uniform hypergraph on $V$ for each $n\geq
0$.

######  Definition 5.

The complete hypergraph $\Delta[V]$ on $V$ is the collection of all the
possible hyperedges on $V$. In other words, $\Delta[V]$ consists of all the
non-empty finite subsets of $V$.

######  Remark 3:

It is direct that we have a disjoint union

$\displaystyle\Delta[V]=\bigcup_{n\geq 0}\Delta_{n}(V).$

######  Definition 6.

Let $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ be two hypergraphs on $V$. The
complement of $\mathcal{H}_{1}$ in $\mathcal{H}_{2}$ is defined to be a
hypergraph $\mathcal{H}_{2}\setminus\mathcal{H}_{1}$ on $V$ by

$\displaystyle\mathcal{H}_{2}\setminus\mathcal{H}_{1}=\\{\sigma{\rm~{}is~{}a~{}hyperedge~{}on~{}}V\mid\sigma\in\mathcal{H}_{2}{\rm~{}and~{}}\sigma\notin\mathcal{H}_{1}\\}.$

######  Definition 7.

A simplicial complex (pl. simplicial complexes) $\mathcal{K}$ on $V$ is a
hypergraph on $V$ such that for any hyperedge $\sigma\in\mathcal{K}$ and any
non-empty subset $\tau\subseteq\sigma$, we always have $\tau\in\mathcal{K}$. A
hyperedge in a simplicial complex is also called a simplex (pl. simplices).

######  Definition 8.

A co-simplicial complex (pl. co-simplicial complexes) $\mathcal{L}$ on $V$ is
a hypergraph on $V$ such that for any hyperedge $\sigma\in\mathcal{L}$ and any
hyperedge $\tau$ on $V$ satisfying $\sigma\subseteq\tau$, we always have
$\tau\in\mathcal{L}$. A hyperedge in a co-simplicial complex is also called a
co-simplex (pl. co-simplices).

######  Remark 4:

From Definition 3, Definition 7 and Definition 8, it is direct that

  * •

for any $n\geq 1$, an $n$-uniform hypergraph is not a simplicial complex;

  * •

for any $n\leq\\#V-1$ where $\\#V$ is the cardinality of $V$ (here $\\#V$ can
be either finite or infinite), an $n$-uniform hypergraph is not a co-
simplicial complex.

######  Remark 5:

From Definition 5, Definition 7 and Definition 8, it is direct that

  * •

the complete hypergraph $\Delta[V]$ is a simplicial complex on $V$;

  * •

the complete hypergraph $\Delta[V]$ is a co-simplicial complex on $V$.

######  Proposition 2.1.

Let $\Delta[V]$ be the complete hypergraph on $V$. Let $\mathcal{K}$ be a
simplicial complex on $V$. Let $\mathcal{L}$ be a co-simplicial complex on
$V$. Then both the followings are satisfied:

  1. (i).

$\Delta[V]\setminus\mathcal{K}$ is a co-simplicial complex on $V$;

  2. (ii).

$\Delta[V]\setminus\mathcal{L}$ is a simplicial complex on $V$.

###### Proof.

(i). Let $\sigma\in\Delta[V]\setminus\mathcal{K}$. Let $\tau$ be a hyperedge
on $V$ such that $\sigma\subseteq\tau$. In order to prove that
$\Delta[V]\setminus\mathcal{K}$ is a co-simplicial complex, it suffices to
prove $\tau\in\Delta[V]\setminus\mathcal{K}$. Suppose to the contrary,
$\tau\notin\Delta[V]\setminus\mathcal{K}$. Then $\tau\in\mathcal{K}$. Since
$\mathcal{K}$ is a simplicial complex and $\sigma\subseteq\tau$, we have
$\sigma\in\mathcal{K}$. This contradicts
$\sigma\in\Delta[V]\setminus\mathcal{K}$. Therefore,
$\tau\in\Delta[V]\setminus\mathcal{K}$, which implies that
$\Delta[V]\setminus\mathcal{K}$ is a co-simplicial complex.

(ii). Let $\sigma\in\Delta[V]\setminus\mathcal{L}$. Let $\tau$ be a hyperedge
on $V$ such that $\tau\subseteq\sigma$. In order to prove that
$\Delta[V]\setminus\mathcal{L}$ is a simplicial complex, it suffices to prove
$\tau\in\Delta[V]\setminus\mathcal{L}$. Suppose to the contrary,
$\tau\notin\Delta[V]\setminus\mathcal{L}$. Then $\tau\in\mathcal{L}$. Since
$\mathcal{L}$ is a co-simplicial complex and $\tau\subseteq\sigma$, we have
$\sigma\in\mathcal{L}$. This contradicts
$\sigma\in\Delta[V]\setminus\mathcal{L}$. Therefore,
$\tau\in\Delta[V]\setminus\mathcal{L}$, which implies that
$\Delta[V]\setminus\mathcal{L}$ is a co-simplicial complex. ∎

######  Example 2.2.

Consider the set $V=\\{v_{0},v_{1},v_{2},v_{3},v_{4},v_{5}\\}$. Then

  1. (i).

$\sigma^{(3)}=v_{0}v_{2}v_{4}v_{5}$ is $3$-hyperedge on $V$;

  2. (ii).

$\mathcal{H}^{(2)}=\\{v_{0}v_{2}v_{3},v_{1}v_{2}v_{3},v_{1}v_{3}v_{5},v_{2}v_{4}v_{5}\\}$
is a $2$-uniform hypergraph on $V$;

  3. (iii).

$\mathcal{H}=\\{v_{0},v_{0}v_{1},v_{4}v_{5},v_{0}v_{1}v_{2},v_{2}v_{3}v_{4}v_{5}\\}$
is a hypergraph on $V$;

  4. (iv).

$\Delta[V]=\\{v_{i}\mid 0\leq i\leq 5\\}\cup\\{v_{i}v_{j}\mid 0\leq i<j\leq
5\\}\cup\\{v_{i}v_{j}v_{k}\mid 0\leq i<j<k\leq
5\\}\cup\\{v_{i}v_{j}v_{k}v_{l}\mid 0\leq i<j<k<l\leq
5\\}\cup\\{v_{i}v_{j}v_{k}v_{l}v_{s}\mid 0\leq i<j<k<l<s\leq
5\\}\cup\\{v_{0}v_{1}v_{2}v_{3}v_{4}v_{5}\\}$;

  5. (v).

$\mathcal{K}=\\{v_{0},v_{0}v_{1},v_{0}v_{2},v_{1}v_{2},v_{0}v_{1}v_{2}\\}$ is
a simplicial complex on $V$;

  6. (vi).

$\mathcal{L}=\\{v_{0}v_{1}v_{2}v_{4},v_{0}v_{1}v_{2}v_{3}v_{5},v_{0}v_{1}v_{2}v_{3}v_{4},v_{0}v_{1}v_{2}v_{4}v_{5},v_{0}v_{1}v_{2}v_{3}v_{4}v_{5}\\}$
is a co-simplicial complex on $V$.

######  Example 2.3.

Consider the set $V=\mathbb{Z}$ of all the integers. Then

  1. (i).

for any $p\in\mathbb{Z}$ and any $q\geq 0$, the sequence $p(p+1)\ldots(p+q)$
of subsequent integers is a $q$-hyperedge on $V$;

  2. (ii).

for any $q\geq 0$, the collection $\mathcal{H}^{(q)}=\\{p(p+1)\ldots(p+q)\mid
p\equiv 1~{}({\rm mod~{}}3)\\}$ of sequences of subsequent integers is a
$q$-uniform hypergraph on $V$;

  3. (iii).

the collection $\mathcal{H}=\\{p(p+1)\ldots(p+q)\mid p\equiv 1~{}({\rm
mod~{}}3){\rm~{}and~{}}2\leq q\leq 5\\}$ of sequences of subsequent integers
is a hypergraph on $V$;

  4. (iv).

$\Delta[V]=\\{i_{0}\in\mathbb{Z}\\}\cup\\{i_{0}i_{1}\in\mathbb{Z}\times\mathbb{Z}\mid
i_{0}<i_{1}\\}\cup\\{i_{0}i_{1}i_{2}\in\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}\mid
i_{0}<i_{1}<i_{2}\\}\cup\cdots$;

  5. (v).

the collection $\mathcal{K}=\\{p(p+1)\ldots(p+q)\mid
p\in\mathbb{Z}{\rm~{}and~{}}0\leq q\leq 5\\}$ of sequences of subsequent
integers is a simplicial complex on $V$;

  6. (vi).

the collection $\mathcal{L}=\\{p(p+1)\ldots(p+q)\mid
p\in\mathbb{Z}{\rm~{}and~{}}q>5\\}$ of sequences is a co-simplicial complex on
$V$.

###  2.2 Geometric Realizations of Hypergraphs, Simplicial Complexes, and Co-
Simplicial Complexes

Throughout this subsection, we suppose $V$ is a finite set. Let $N=\\#V$ be
the number of the vertices in $V$. We enumerate the vertices in $V$ as
$v_{0},v_{1},\ldots,v_{N}$ such that $v_{0}\prec v_{1}\prec\cdots\prec v_{N}$.

Consider an Euclidean space $\mathbb{R}^{N}$ with an orthonormal basis
$e_{1}$, $e_{2}$, $\ldots$, $e_{N}$. For each $1\leq i\leq N$, take the point

$\displaystyle p_{i}=(0,\cdots,0,1,0,\cdots,0)$

in $\mathbb{R}^{N}$ where the $i$-th coordinate is $1$ and all the other
coordinates are $0$. Then the vector from $0$ to $p_{i}$ in $\mathbb{R}^{N}$
is the unit vector $e_{i}$. Define a map

$\displaystyle\rho:~{}~{}~{}V\longrightarrow\mathbb{R}^{N}$

by $\rho(v_{i})=p_{i}$ for $i=1,2,\ldots,N$.

######  Definition 9.

Let $0\leq k_{0}\leq N$. Let $\sigma^{(0)}=v_{k_{0}}$ be a $0$-hyperedge
(which is just a vertex) on $V$. The geometric realization of $\sigma$,
denoted by $|\sigma|$, is the point $\rho(v_{k_{0}})$ in $\mathbb{R}^{N}$.

######  Definition 10.

Let $1\leq n\leq N-1$. Let $0\leq k_{0}<k_{1}<\cdots<k_{n}\leq N$. Let
$\sigma^{(n)}=v_{k_{0}}v_{k_{1}}\ldots v_{k_{n}}$ be an $n$-hyperedge on $V$.
The geometric realization of $\sigma$, denoted by $|\sigma|$, is a subset of
$\mathbb{R}^{N}$ given by

$\displaystyle|\sigma|$ $\displaystyle=$
$\displaystyle\Big{\\{}(x_{1},x_{2},\ldots,x_{N})\in\mathbb{R}^{N}\mid{\rm
there~{}exist~{}}0<t_{0},t_{1},\ldots,t_{n}<1$
$\displaystyle{\rm~{}such~{}that~{}}\sum_{i=0}^{n}t_{i}=1{\rm~{}and~{}}(x_{1},x_{2},\ldots,x_{N})=\sum_{i=0}^{n}t_{i}\rho(v_{k_{i}})\Big{\\}}.$

######  Lemma 2.4.

Let $V$ be a finite set consisting of $N$ vertices. Let $1\leq n\leq N-1$. Let
$\sigma$ be an $n$-hyperedge on $V$. Then the geometric realization $|\sigma|$
is an open subset of an $n$-hyperplane in $\mathbb{R}^{N}$.

###### Proof.

Suppose $\sigma^{(n)}=v_{k_{0}}v_{k_{1}}\ldots v_{k_{n}}$. Let
$P(\rho(\sigma))$ be the $n$-dimensional hyperplane in $\mathbb{R}^{N}$
affinely spanned by the $(n+1)$-points $\rho(v_{k_{0}})$, $\rho(v_{k_{1}})$,
$\ldots$, $\rho(v_{k_{n}})$. We observe that the geometric realization
$|\sigma|$ is the interior of the convex hull of the subset

$\displaystyle\rho(\sigma)=\\{\rho(v_{k_{0}}),\rho(v_{k_{1}}),\ldots,\rho(v_{k_{n}})\\}$
(2.3)

of $P(\rho(\sigma))$. Thus $|\sigma|$ is an open subset of the $n$-hyperplane
$P(\rho(\sigma))$ in $\mathbb{R}^{N}$. ∎

######  Lemma 2.5.

Let $V$ be a finite set consisting of $N$ vertices. Let $\sigma$ and $\tau$ be
two distinct hyperedges on $V$. Then $|\sigma|\cap|\tau|=\emptyset$.

###### Proof.

Suppose $\sigma=v_{k_{0}}v_{k_{1}}\ldots v_{k_{n}}$ and
$\tau=u_{l_{0}}u_{l_{1}}\ldots u_{l_{m}}$. Let $p\in|\sigma|$ and
$q\in|\tau|$. Then the coordinate of $p$ in $\mathbb{R}^{N}$ is

$\displaystyle(0,\cdots,0,t_{0},0,\cdots,0,t_{1},0,\cdots)$ (2.4)

where

  * •

for each $0\leq i\leq n$, the $k_{i}$-th coordinate is $t_{i}>0$ such that
$t_{0}+t_{1}+\cdots+t_{n}=1$;

  * •

all the other coordinates are $0$.

Similarly, the coordinate of $q$ in $\mathbb{R}^{N}$ is

$\displaystyle(0,\cdots,0,s_{0},0,\cdots,0,s_{1},0,\cdots)$ (2.5)

where

  * •

for each $0\leq j\leq m$, the $l_{j}$-th coordinate is $s_{j}>0$ such that
$s_{0}+s_{1}+\cdots+s_{m}=1$;

  * •

all the other coordinates are $0$.

Since $\sigma\neq\tau$, we have
$(k_{0},k_{1},\ldots,k_{n})\neq(l_{0},l_{1},\ldots,l_{m})$ 111We write
$(k_{0},k_{1},\ldots,k_{n})=(l_{0},l_{1},\ldots,l_{m})$ if $n=m$ and
$k_{i}=l_{i}$ for each $0\leq i\leq m$. We write
$(k_{0},k_{1},\ldots,k_{n})\neq(l_{0},l_{1},\ldots,l_{m})$ otherwise. .
Consequently, the coordinates of $p$ given by (2.4) do not equal to the
coordinates of $q$ given by (2.5), which implies $p\neq q$. Therefore,
$|\sigma|\cap|\tau|=\emptyset$. ∎

######  Definition 11.

Let $\mathcal{H}$ be a hypergraph on $V$. The geometric realization of
$\mathcal{H}$, denoted by $|\mathcal{H}|$, is a subset of $\mathbb{R}^{N}$
given by

$\displaystyle|\mathcal{H}|=\bigcup_{\sigma\in\mathcal{H}}|\sigma|.$ (2.6)

That is, the geometric realization $|\mathcal{H}|$ is the union of the
geometric realizations $|\sigma|$ for all the hyperedges
$\sigma\in\mathcal{H}$.

######  Remark 6:

By Lemma 2.5, the union in (2.6) is a disjoint union.

######  Lemma 2.6.

Let $V$ be a finite set consisting of $N$ vertices. Let $\mathcal{H}_{1}$ and
$\mathcal{H}_{2}$ be two hypergraphs on $V$. Let
$\mathcal{H}_{2}\setminus\mathcal{H}_{1}$ be the complement hypergraph of
$\mathcal{H}_{1}$ in $\mathcal{H}_{2}$. Then the geometric realizations of
$\mathcal{H}_{1}$, $\mathcal{H}_{2}$ and
$\mathcal{H}_{2}\setminus\mathcal{H}_{1}$ satisfy

$\displaystyle|\mathcal{H}_{2}\setminus\mathcal{H}_{1}|=|\mathcal{H}_{2}|\setminus|\mathcal{H}_{1}|.$

###### Proof.

It is direct to verify

$\displaystyle|\mathcal{H}_{2}\setminus\mathcal{H}_{1}|$ $\displaystyle=$
$\displaystyle\bigcup_{\sigma\in\mathcal{H}_{2}\setminus\mathcal{H}_{1}}|\sigma|$
$\displaystyle=$
$\displaystyle\Big{(}\bigcup_{\sigma\in\mathcal{H}_{2}}|\sigma|\Big{)}\setminus\Big{(}\bigcup_{\sigma\in\mathcal{H}_{1}}|\sigma|\Big{)}$
$\displaystyle=$ $\displaystyle|\mathcal{H}_{2}|\setminus|\mathcal{H}_{1}|.$

Here the second equality follows from Lemma 2.5 (or Remark 6) while the first
and the third equalities follow from Definition 11. ∎

######  Proposition 2.7.

Let $V$ be a finite set consisting of $N$ vertices. Let $\mathcal{K}$ be a
simplicial complex on $V$. Then the geometric realization $|\mathcal{K}|$ is a
closed subset of $\mathbb{R}^{N}$.

###### Proof.

Let $\mathcal{K}$ be a simplicial complex. To prove that $|\mathcal{K}|$ is a
closed subset of $\mathbb{R}^{N}$, we take $x$ to be a point in the boundary
of $|\mathcal{K}|$. It suffices to prove $x\in|\mathcal{K}|$. Since $x$ is in
the boundary of $|\mathcal{K}|$, it follows that there exists
$\sigma\in\mathcal{K}$ such that $x$ is in the boundary of $|\sigma|$.

Case 1. $\sigma$ is a $0$-simplex.

Then $x$ is the single point in $|\sigma|$. Thus $x\in|\sigma|$, which implies
$x\in|\mathcal{K}|$.

Case 2. $\sigma$ is an $n$-simplex for $n\geq 1$.

Then $x$ is in the boundary of the convex hull of $\rho(\sigma)$ (recall
$\rho(\sigma)$ is given by (2.3) in the proof of Lemma 2.4). Thus there exists
a non-empty proper subset $\tau$ of $\sigma$ such that $x$ is in the convex
hull of $\rho(\tau)$. Note that for any two such subsets $\tau$ and
$\tau^{\prime}$ of $\sigma$, the intersection $\tau\cap\tau^{\prime}$ is still
such a subset of $\sigma$. Hence without loss of generality, we assume $\tau$
is the smallest subset of $\sigma$ such that $x$ is in the convex hull of
$\rho(\tau)$. Since $\mathcal{K}$ is a simplicial complex, we have that $\tau$
is a simplex in $\mathcal{K}$.

Subcase 2.1. $\tau$ is a $0$-simplex.

Then $x$ is the single point in $|\tau|$, which implies $x\in|\mathcal{K}|$.

Subcase 2.2. $\tau$ is an $m$-simplex for some $1\leq m\leq n-1$.

Then $x$ is in the convex hull of $\rho(\tau)$. Moreover, for any non-empty
proper subset $\delta$ of $\tau$, since $\tau$ is assumed to be smallest, we
have that $x$ is not in the convex hull of $\rho(\delta)$. It follows that $x$
is in the interior of the the convex hull of $\rho(\tau)$. Therefore,
$x\in|\tau|$, which implies $x\in|\mathcal{K}|$.

Summarizing the above cases, we have $x\in|\mathcal{K}|$. Thus $|\mathcal{K}|$
is a closed subset of $\mathbb{R}^{N}$. ∎

######  Corollary 2.8.

Let $V$ be a finite set consisting of $N$ vertices. Let $\mathcal{L}$ be a co-
simplicial complex on $V$. Then the geometric realization $|\mathcal{L}|$ is
an open subset of the geometric realization $|\Delta[V]|$.

###### Proof.

Since $\Delta[V]$ is a simplicial complex (cf. Remark 5), it follows from
Proposition 2.7 that $|\Delta[V]|$ is a closed subset of $\mathbb{R}^{N}$. By
Proposition 2.1, the complement hypergraph $\Delta[V]\setminus\mathcal{L}$ is
a simplicial complex on $V$. Thus also by Proposition 2.7, the geometric
realization $|\Delta[V]\setminus\mathcal{L}|$ is a closed subset of
$\mathbb{R}^{N}$ as well. In addition, since each simplex of
$\Delta[V]\setminus\mathcal{L}$ is a simplex of $\Delta[V]$, it follows that
$|\Delta[V]\setminus\mathcal{L}|$ is a closed subset of $|\Delta[V]|$. Note
that $\Delta[V]$ is the disjoint union of $\mathcal{L}$ and
$\Delta[V]\setminus\mathcal{L}$. Consequently, it follows from Lemma 2.6 that
$|\mathcal{L}|$ is the complement of $|\Delta[V]\setminus\mathcal{L}|$ in
$|\Delta[V]|$. Therefore, $|\mathcal{L}|$ is an open subset of $|\Delta[V]|$.
∎

######  Example 2.9.

Let $V=\\{v_{0},v_{1},v_{2}\\}$.

  1. (i).

The complete hypergraph on $V$ is

$\displaystyle\Delta[V]=\\{v_{0},v_{1},v_{2},v_{0}v_{1},v_{1}v_{2},v_{0}v_{2},v_{0}v_{1}v_{2}\\}.$

The geometric realization $|\Delta[V]|$ is in the following figure:

$v_{0}$$v_{1}$$v_{2}$

  2. (ii).

Let the hypergraph $\mathcal{H}=\\{v_{1},v_{2},v_{0}v_{1},v_{0}v_{1}v_{2}\\}$.
The geometric realization $|\mathcal{H}|$ is in the following figure:

$v_{0}$$v_{1}$$v_{2}$

  3. (iii).

Let the simplicial complex
$\mathcal{K}=\\{v_{0},v_{1},v_{2},v_{0}v_{1},v_{0}v_{2},v_{1}v_{2}\\}$. Then
$\Delta[V]\setminus\mathcal{K}=\\{v_{0}v_{1}v_{2}\\}$ is a co-simplicial
complex. The geometric realizations $|\mathcal{K}|$ and
$|\Delta[V]\setminus\mathcal{K}|$ are in the following figure:

$v_{0}$$v_{1}$$v_{2}$$v_{0}$$v_{1}$$v_{2}$

From the figure we can see that $|\mathcal{K}|$ is closed in $|\Delta[V]|$,
$|\Delta[V]\setminus\mathcal{K}|$ is open in $|\Delta[V]|$, and $|\Delta[V]|$
is the disjoint union of $|\mathcal{K}|$ and
$|\Delta[V]\setminus\mathcal{K}|$;

  4. (iv).

Let the co-simplicial complex
$\mathcal{L}=\\{v_{0}v_{1},v_{0}v_{2},v_{0}v_{1}v_{2}\\}$. Then
$\Delta[V]\setminus\mathcal{L}=\\{v_{0},v_{1},v_{2},v_{1}v_{2}\\}$ is a
simplicial complex. The geometric realizations $|\mathcal{L}|$ and
$|\Delta[V]\setminus\mathcal{L}|$ are in the following figure:

$v_{0}$$v_{1}$$v_{2}$$v_{0}$$v_{1}$$v_{2}$

From the figure we can see that $|\mathcal{L}|$ is open in $|\Delta[V]|$,
$|\Delta[V]\setminus\mathcal{L}|$ is closed in $|\Delta[V]|$, and
$|\Delta[V]|$ is the disjoint union of $|\mathcal{L}|$ and
$|\Delta[V]\setminus\mathcal{L}|$.

##  3 Differential Calculus for Paths on Discrete Sets

In this section, we review the definitions of the paths and the elementary
paths on a discrete set (cf. [14]). By applying some discrete differential
calculus, we construct certain chain complexes and co-chain complexes for the
space of paths on a discrete set.

###  3.1 Paths on Discrete Sets

Throughout this section, we let $V$ be a discrete set. Let $n\geq 0$ be a non-
negative integer.

######  Definition 12.

(cf. [14, Definition 2.1]). An elementary $n$-path on $V$ is an ordered
sequence $v_{0}v_{1}\ldots v_{n}$ of $n+1$ vertices in $V$. Here for any
$0\leq i<j\leq n$, we do not require $v_{i}\prec v_{j}$, $v_{j}\prec v_{i}$ or
$v_{i}\neq v_{j}$.

######  Definition 13.

(cf. [14, Definition 2.2]). A formal linear combination of elementary
$n$-paths on $V$ with coefficients in the real numbers $\mathbb{R}$ is called
an $n$-path on $V$.

######  Notation 1.

(cf. [14, Subsection 2.1]). Denote by $\Lambda_{n}(V)$ the vector space of all
$n$-paths. Then any element in $\Lambda_{n}(V)$ is of the form

$\displaystyle\sum_{v_{0},v_{1},\ldots,v_{n}\in V}{r_{v_{0}v_{1}\ldots
v_{n}}v_{0}v_{1}\ldots v_{n}},~{}~{}~{}r_{v_{0}v_{1}\ldots
v_{n}}\in\mathbb{R}.$

######  Notation 2.

Letting $n$ run over all non-negative integers, we have a graded vector space

$\displaystyle\Lambda_{*}(V)=\bigoplus_{n=0}^{\infty}\Lambda_{n}(V).$

######  Notation 3.

For each $n\geq 0$, we have a canonical inner product

$\displaystyle\langle~{},~{}\rangle:~{}~{}~{}\Lambda_{n}(V)\times\Lambda_{n}(V)\longrightarrow\mathbb{R}$

on $\Lambda_{n}(V)$ by

$\displaystyle\langle u_{0}u_{1}\ldots u_{n},v_{0}v_{1}\ldots
v_{n}\rangle=\prod_{i=0}^{n}\delta(u_{i},v_{i}).$ (3.1)

######  Remark 7:

It follows from (3.1) that

  * •

if $u_{0}u_{1}\ldots u_{n}$ and $v_{0}v_{1}\ldots v_{n}$ are identically the
same elementary $n$-path, then

$\displaystyle\langle u_{0}u_{1}\ldots u_{n},v_{0}v_{1}\ldots v_{n}\rangle=1;$

  * •

if $u_{0}u_{1}\ldots u_{n}$ and $v_{0}v_{1}\ldots v_{n}$ are not the same
elementary $n$-path, then

$\displaystyle\langle u_{0}u_{1}\ldots u_{n},v_{0}v_{1}\ldots v_{n}\rangle=0.$

###  3.2 Partial Derivatives on Path Spaces

######  Definition 14.

For any $v\in V$, we define the partial derivative of $\Lambda_{*}(V)$ with
respect to $v$ to be a sequence of linear maps

$\displaystyle\frac{\partial}{\partial
v}:~{}~{}~{}\Lambda_{n}(V)\longrightarrow\Lambda_{n-1}(V),~{}~{}~{}n\geq 0$

by letting

$\displaystyle\frac{\partial}{\partial v}(v_{0}v_{1}\ldots
v_{n})=\sum_{i=0}^{n}(-1)^{i}\delta(v,v_{i})v_{0}\ldots\widehat{v_{i}}\ldots
v_{n}.$ (3.2)

Here in (3.2), for any vertices $u,v\in V$, we use the notation
$\delta(u,v)=1$ if $u=v$ and $\delta(u,v)=0$ if $u\neq v$. We extend (3.2)
linearly over $\mathbb{R}$.

######  Remark 8:

By Definition 14, for any distinct vertices $v_{0},v_{1},\ldots,v_{n}$ in $V$
we have the followings:

  * •

if $v_{i}=v$ for some $0\leq i\leq n$, then

$\displaystyle\frac{\partial}{\partial v}(v_{0}v_{1}\ldots
v_{n})=(-1)^{i}v_{0}\ldots\widehat{v_{i}}\ldots v_{n};$

  * •

if $v_{i}\neq v$ for any $0\leq i\leq n$, then

$\displaystyle\frac{\partial}{\partial v}(v_{0}v_{1},\ldots v_{n})=0.$

######  Lemma 3.1.

([26, Lemma 2.7]). For any $u,v\in V$, we have

$\displaystyle\frac{\partial}{\partial u}\circ\frac{\partial}{\partial
v}=-\frac{\partial}{\partial v}\circ\frac{\partial}{\partial u}.$ (3.3)

###### Proof.

Since both $\frac{\partial}{\partial u}$ and $\frac{\partial}{\partial v}$ are
linear, it follows that both $\frac{\partial}{\partial
u}\circ\frac{\partial}{\partial v}$ and $\frac{\partial}{\partial
v}\circ\frac{\partial}{\partial u}$ are linear as well. Hence in order to
prove the identity (3.3) as linear maps from $\Lambda_{n}(V;R)$ to
$\Lambda_{n-1}(V;R)$, we only need to verify the identity (3.3) on an
elementary $n$-path $v_{0}v_{1}\ldots v_{n}$. By the definition (3.2), we have

$\displaystyle\frac{\partial}{\partial u}\circ\frac{\partial}{\partial
v}(v_{0}v_{1}\ldots v_{n})$ $\displaystyle=$
$\displaystyle\frac{\partial}{\partial
u}\Big{(}\sum_{j=0}^{n}(-1)^{j}\delta(v,v_{j})v_{0}\ldots\widehat{v_{j}}\ldots
v_{n}\Big{)}$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{n}(-1)^{j}\delta(v,v_{j})\frac{\partial}{\partial
u}(v_{0}\ldots\widehat{v_{j}}\ldots v_{n})$ $\displaystyle=$
$\displaystyle\sum_{j=0}^{n}(-1)^{j}\delta(v,v_{j})\sum_{i=0}^{j-1}(-1)^{i}\delta(u,v_{i})(v_{0}\ldots\widehat{v_{i}}\ldots\widehat{v_{j}}\ldots
v_{n})$
$\displaystyle+\sum_{j=0}^{n}(-1)^{j}\delta(v,v_{j})\sum_{i=j+1}^{n}(-1)^{i-1}\delta(u,v_{i})(v_{0}\ldots\widehat{v_{i}}\ldots\widehat{v_{j}}\ldots
v_{n})$ $\displaystyle=$ $\displaystyle\sum_{0\leq i<j\leq
n}(-1)^{i+j}\delta(u,v_{i})\delta(v,v_{j})(v_{0}\ldots\widehat{v_{i}}\ldots\widehat{v_{j}}\ldots
v_{n})$ $\displaystyle+\sum_{0\leq j<i\leq
n}(-1)^{i+j-1}\delta(u,v_{i})\delta(v,v_{j})(v_{0}\ldots\widehat{v_{j}}\ldots\widehat{v_{i}}\ldots
v_{n}).$

Similarly,

$\displaystyle\frac{\partial}{\partial v}\circ\frac{\partial}{\partial
u}(v_{0}v_{1}\ldots v_{n})$ $\displaystyle=$ $\displaystyle\sum_{0\leq j<i\leq
n}(-1)^{i+j}\delta(u,v_{i})\delta(v,v_{j})(v_{0}\ldots\widehat{v_{i}}\ldots\widehat{v_{j}}\ldots
v_{n})$ $\displaystyle+\sum_{0\leq i<j\leq
n}(-1)^{i+j-1}\delta(u,v_{i})\delta(v,v_{j})(v_{0}\ldots\widehat{v_{j}}\ldots\widehat{v_{i}}\ldots
v_{n}).$

Therefore, for any elementary $n$-path $v_{0}v_{1}\ldots v_{n}$ on $V$, we
have

$\displaystyle\frac{\partial}{\partial u}\circ\frac{\partial}{\partial
v}(v_{0}v_{1}\ldots v_{n})+\frac{\partial}{\partial
v}\circ\frac{\partial}{\partial u}(v_{0}v_{1}\ldots v_{n})=0.$

Consequently, by the linear property of $\frac{\partial}{\partial
u}\circ\frac{\partial}{\partial v}$ and $\frac{\partial}{\partial
v}\circ\frac{\partial}{\partial u}$, we obtain (3.3). ∎

######  Notation 4.

We denote $\frac{\partial}{\partial v}\circ\frac{\partial}{\partial u}$ as
$\frac{\partial}{\partial v}\wedge\frac{\partial}{\partial u}$ for any $u,v\in
V$.

######  Definition 15.

We consider the exterior algebra

$\displaystyle T_{*}(V)=\wedge\Big{(}\frac{\partial}{\partial v}\mid v\in
V\Big{)}$

and call it the differential algebra on $V$.

We have the following observations:

  * •

The differential algebra $T_{*}(V)$ is a direct sum

$\displaystyle T_{*}(V)=\bigoplus_{k=0}^{\infty}T_{k}(V);$

  * •

$T_{0}(V)=\mathbb{R}$ while for each $k\geq 1$, the space $T_{k}(V)$ is the
vector space spanned by all the following elements

$\displaystyle\frac{\partial}{\partial v_{1}}\wedge\frac{\partial}{\partial
v_{2}}\wedge\cdots\wedge\frac{\partial}{\partial
v_{k}},~{}~{}~{}v_{1},v_{2},\ldots,v_{k}\in V$

modulo the relation

$\displaystyle\frac{\partial}{\partial
v_{1}}\wedge\cdots\wedge\frac{\partial}{\partial
v_{i}}\wedge\frac{\partial}{\partial
v_{i+1}}\wedge\cdots\wedge\frac{\partial}{\partial
v_{k}}=-\frac{\partial}{\partial
v_{1}}\wedge\cdots\wedge\frac{\partial}{\partial
v_{i+1}}\wedge\frac{\partial}{\partial
v_{i}}\wedge\cdots\wedge\frac{\partial}{\partial v_{k}}$

for any $1\leq i\leq k-1$;

  * •

The exterior product

$\displaystyle\wedge:~{}~{}~{}T_{k}(V)\times T_{l}(V)\longrightarrow
T_{k+l}(V),~{}~{}~{}k,l\geq 1,$

is the composition of linear maps. It is given by

$\displaystyle(\frac{\partial}{\partial
v_{1}}\wedge\cdots\wedge\frac{\partial}{\partial
v_{k}})\wedge(\frac{\partial}{\partial
u_{1}}\wedge\cdots\wedge\frac{\partial}{\partial
u_{l}})=\frac{\partial}{\partial
v_{1}}\wedge\cdots\wedge\frac{\partial}{\partial
v_{k}}\wedge\frac{\partial}{\partial
u_{1}}\wedge\cdots\wedge\frac{\partial}{\partial u_{l}}$

which extends bi-linearly over $\mathbb{R}$.

  * •

For any $k\geq 1$ and any $\alpha\in T_{k}(V)$, we have that $\alpha$ gives a
sequence of linear maps

$\displaystyle\alpha_{n}:~{}~{}~{}\Lambda_{n}(V)\longrightarrow\Lambda_{n-k}(V),~{}~{}~{}n\geq
0.$ (3.4)

Here we adopt the notation that $\Lambda_{-n}(V)=0$ for any $n\geq 0$.
Precisely, if we write

$\displaystyle\alpha=\sum_{v_{1},v_{2},\ldots,v_{k}\in
V}r_{v_{1},v_{2},\ldots,v_{k}}\frac{\partial}{\partial
v_{1}}\wedge\frac{\partial}{\partial
v_{2}}\wedge\cdots\wedge\frac{\partial}{\partial
v_{k}},~{}~{}~{}r_{v_{1},v_{2},\ldots,v_{k}}\in\mathbb{R},$

then for any elementary $n$-path $u_{0}u_{1}\ldots u_{n}$ on $V$ with $n\geq
k$, we have 222The expression of $\alpha(u_{0}u_{1}\ldots u_{n})$ follows from
the following two observations:  (i). for any $0\leq
i_{1}<i_{2}<\ldots<i_{k}\leq n$, by applying (3.2) for $k$-times, we have
$\displaystyle\frac{\partial}{\partial
u_{i_{1}}}\wedge\frac{\partial}{\partial
u_{i_{2}}}\wedge\cdots\wedge\frac{\partial}{\partial
u_{i_{k}}}(u_{0}u_{1}\ldots
u_{n})=(-1)^{i_{1}+i_{2}+\cdots+i_{k}}u_{0}\ldots\widehat{u_{i_{1}}}\ldots\widehat{u_{i_{2}}}\ldots\widehat{u_{i_{k}}}\ldots
u_{n};$ (ii). for any $\sigma\in S_{k}$, by applying (3.3) iteratively, we
have $\displaystyle\frac{\partial}{\partial
u_{i_{\sigma(1)}}}\wedge\frac{\partial}{\partial
u_{i_{\sigma(2)}}}\wedge\cdots\wedge\frac{\partial}{\partial
u_{i_{\sigma(k)}}}={\rm sgn}(\sigma)\frac{\partial}{\partial
u_{i_{1}}}\wedge\frac{\partial}{\partial
u_{i_{2}}}\wedge\cdots\wedge\frac{\partial}{\partial u_{i_{k}}}.$

$\displaystyle\alpha(u_{0}u_{1}\ldots u_{n})$ $\displaystyle=$
$\displaystyle\sum_{v_{1},v_{2},\ldots,v_{k}\in
V}r_{v_{1},v_{2},\ldots,v_{k}}\frac{\partial}{\partial
v_{1}}\wedge\frac{\partial}{\partial
v_{2}}\wedge\cdots\wedge\frac{\partial}{\partial v_{k}}(u_{0}u_{1}\ldots
u_{n})$ $\displaystyle=$ $\displaystyle\sum_{0\leq
i_{1}<i_{2}<\ldots<i_{k}\leq n~{}}\sum_{\sigma\in
S_{k}}r_{u_{i_{\sigma(1)}},u_{i_{\sigma(2)}},\ldots,u_{i_{\sigma(k)}}}{\rm
sgn}(\sigma)$
$\displaystyle{\rm~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}}(-1)^{i_{1}+i_{2}+\cdots+i_{k}}u_{0}\ldots\widehat{u_{i_{1}}}\ldots\widehat{u_{i_{2}}}\ldots\widehat{u_{i_{k}}}\ldots
u_{n}.$

Here $S_{k}$ is the permutation group on $k$-letters and for any permutation
$\sigma\in S_{k}$, we use ${\rm sgn}(\sigma)$ to denote the signature of
$\sigma$.

###  3.3 Partial Differentiations on Path Spaces

######  Definition 16.

For any $v\in V$, we define the partial differentiation $dv$ with respect to
$v$ to be a sequence of linear maps

$\displaystyle
dv:~{}~{}~{}\Lambda_{n}(V)\longrightarrow\Lambda_{n+1}(V),~{}~{}~{}n\geq 0,$

such that $dv$ is the adjoint linear map of $\frac{\partial}{\partial v}$ for
each $n\geq 0$. Precisely, for any $n\geq 0$, any $\xi\in\Lambda_{n}(V)$, and
any $\eta\in\Lambda_{n+1}(V)$, we have

$\displaystyle\langle\frac{\partial}{\partial
v}(\eta),\xi\rangle=\langle\eta,dv(\xi)\rangle.$ (3.5)

The next lemma gives an explicit formula for $dv$.

######  Lemma 3.2.

([26, Lemma 2.10]). For any $n\geq 1$, any $v\in V$, and any elementary
$(n-1)$-path $u_{0}u_{1}\ldots u_{n-1}$ on $V$, we have

$\displaystyle dv(u_{0}u_{1}\ldots
u_{n-1})=\sum_{i=0}^{n}(-1)^{i}u_{0}u_{1}\ldots u_{i-1}vu_{i}u_{i+1}\ldots
u_{n-1}.$ (3.6)

###### Proof.

In (3.5), we take $\eta$ to be an elementary $n$-path $v_{0}v_{1}\ldots
v_{n}\in\Lambda_{n}(V)$ and take $\xi$ to be an elementary $(n-1)$-path
$u_{0}u_{1}\ldots u_{n-1}\in\Lambda_{n-1}(V)$. Then

$\displaystyle\langle v_{0}v_{1}\ldots v_{n},dv(u_{0}u_{1}\ldots
u_{n-1})\rangle$ $\displaystyle=$
$\displaystyle\langle\frac{\partial}{\partial v}(v_{0}v_{1}\ldots
v_{n}),u_{0}u_{1}\ldots u_{n-1}\rangle$ $\displaystyle=$
$\displaystyle\langle\sum_{i=0}^{n}(-1)^{i}\delta(v,v_{i})v_{0}\ldots\widehat{v_{i}}\ldots
v_{n},u_{0}u_{1}\ldots u_{n-1}\rangle$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{n}(-1)^{i}\delta(v,v_{i})\prod_{j=0}^{i-1}\delta(v_{j},u_{j})\prod_{j=i}^{n-1}\delta(v_{j+1},u_{j}).$

Consequently, we have

$\displaystyle dv(u_{0}u_{1}\ldots u_{n-1})$ $\displaystyle=$
$\displaystyle\sum_{v_{0},v_{1},\ldots,v_{n}\in V}\langle v_{0}v_{1}\ldots
v_{n},dv(u_{0}u_{1}\ldots u_{n-1})\rangle v_{0}v_{1}\ldots v_{n}$ (3.7)
$\displaystyle=$ $\displaystyle\sum_{v_{0},v_{1},\ldots,v_{n}\in
V}\Big{(}\sum_{i=0}^{n}(-1)^{i}\delta(v,v_{i})\prod_{j=0}^{i-1}\delta(v_{j},u_{j})\prod_{j=i}^{n-1}\delta(v_{j+1},u_{j})\Big{)}v_{0}v_{1}\ldots
v_{n}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{n}(-1)^{i}\Big{(}\sum_{v_{0},v_{1},\ldots,v_{n}\in
V}\delta(v,v_{i})\prod_{j=0}^{i-1}\delta(v_{j},u_{j})\prod_{j=i}^{n-1}\delta(v_{j+1},u_{j})\Big{)}v_{0}v_{1}\ldots
v_{n}$ $\displaystyle=$ $\displaystyle\sum_{i=0}^{n}(-1)^{i}u_{0}u_{1}\ldots
u_{i-1}vu_{i}u_{i+1}\ldots u_{n-1}.$

We obtain (3.6). ∎

The next corollary gives the case $n=1$ in Lemma 3.2.

######  Corollary 3.3.

For any $u,v\in V$ we have $dv(u)=vu-uv$. ∎

Similar with the proof of Lemma 3.1, it is direct to verify the next lemma.

######  Lemma 3.4.

([26, Lemma 2.7]). For any $u,v\in V$ we have

$\displaystyle du\circ dv=-dv\circ du.$ (3.8)

###### Proof.

333An alternative proof for Lemma 3.4 follows from Lemma 3.1 directly: Let
$u,v\in V$. For any any $n\geq 0$, any $\xi\in\Lambda_{n}(V)$ and any
$\eta\in\Lambda_{n+2}(V)$, we have  $\displaystyle\langle\eta,du\wedge
dv(\xi)\rangle=\langle\frac{\partial}{\partial
u}(\eta),dv(\xi)\rangle=\langle\frac{\partial}{\partial
v}\wedge\frac{\partial}{\partial u}(\eta),\xi\rangle$ $\displaystyle=$
$\displaystyle-\langle\frac{\partial}{\partial
u}\wedge\frac{\partial}{\partial
v}(\eta),\xi\rangle=-\langle\frac{\partial}{\partial
v}(\eta),du(\xi)\rangle=-\langle\eta,dv\wedge du(\xi)\rangle.$ This implies
(3.8). Nevertheless, the proof for Lemma 3.4 in the main-body consolidates
(3.6) in Lemma 3.2.

For any $n\geq 0$ and any elementary $n$-path $v_{0}v_{1}\ldots
v_{n}\in\Lambda_{n}(V)$, by (3.6), we have

$\displaystyle du\circ dv(v_{0}v_{1}\ldots v_{n})$ $\displaystyle=$
$\displaystyle du\Big{(}\sum_{i=0}^{n+1}(-1)^{i}v_{0}\ldots
v_{i-1}vv_{i}\ldots v_{n}\Big{)}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{n+1}(-1)^{i}du(v_{0}\ldots v_{i-1}vv_{i}\ldots
v_{n})$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{n+1}(-1)^{i}\Big{(}\sum_{j=0}^{i-1}(-1)^{j}v_{0}\ldots
v_{j-1}uv_{j}\ldots v_{i-1}vv_{i}\ldots v_{n}$
$\displaystyle+(-1)^{i}v_{0}\ldots v_{i-1}uvv_{i}\ldots
v_{n}+(-1)^{i+1}v_{0}\ldots v_{i-1}vuv_{i}\ldots v_{n}$
$\displaystyle+\sum_{j=i+1}^{n+1}(-1)^{j+1}v_{0}\ldots v_{i-1}vv_{i}\ldots
v_{j-1}uv_{j}\ldots v_{n}\Big{)}$

while

$\displaystyle dv\circ du(v_{0}v_{1}\ldots v_{n})$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{n+1}(-1)^{i}\Big{(}\sum_{j=0}^{i-1}(-1)^{j}v_{0}\ldots
v_{j-1}vv_{j}\ldots v_{i-1}uv_{i}\ldots v_{n}$
$\displaystyle+(-1)^{i}v_{0}\ldots v_{i-1}vuv_{i}\ldots
v_{n}+(-1)^{i+1}v_{0}\ldots v_{i-1}uvv_{i}\ldots v_{n}$
$\displaystyle+\sum_{j=i+1}^{n+1}(-1)^{j+1}v_{0}\ldots v_{i-1}uv_{i}\ldots
v_{j-1}vv_{j}\ldots v_{n}\Big{)}.$

Thus

$\displaystyle du\circ dv(v_{0}v_{1}\ldots v_{n})=-dv\circ du(v_{0}v_{1}\ldots
v_{n})$

for any $n\geq 0$ and any elementary $n$-path $v_{0}v_{1}\ldots
v_{n}\in\Lambda_{n}(V)$. Consequently, we obtain (3.8). ∎

######  Notation 5.

We denote $du\circ dv$ as $du\wedge dv$ for any $u,v\in V$.

######  Definition 17.

We consider the exterior algebra

$\displaystyle T^{*}(V)=\wedge\Big{(}dv\mid v\in V\Big{)}$

and call it the co-differential algebra on $V$.

We have the following observations:

  * •

$T^{*}(V)$ is a direct sum

$\displaystyle T^{*}(V)=\bigoplus_{k=0}^{\infty}T^{k}(V).$

  * •

$T^{0}(V)=\mathbb{R}$ while for each $k\geq 1$, the space $T^{k}(V)$ is
spanned by

$\displaystyle dv_{1}\wedge dv_{2}\wedge\cdots\wedge
dv_{k},~{}~{}~{}v_{1},v_{2},\ldots,v_{k}\in V$

modulo the relation

$\displaystyle dv_{1}\wedge\cdots\wedge dv_{i}\wedge
dv_{i+1}\wedge\cdots\wedge dv_{k}=-dv_{1}\wedge\cdots\wedge dv_{i+1}\wedge
dv_{i}\wedge\cdots\wedge dv_{k}$

for any $1\leq i\leq k-1$.

  * •

For any $k\geq 1$ and any $\omega\in T^{k}(V)$, we have that $\omega$ gives a
sequence of linear maps

$\displaystyle\omega_{n}:~{}~{}~{}\Lambda_{n}(V)\longrightarrow\Lambda_{n+k}(V),~{}~{}~{}n\geq
0.$ (3.9)

######  Definition 18.

Let $k\geq 1$, $\alpha\in T_{k}(V)$ and $\omega\in T^{k}(V)$. We say that
$\alpha$ and $\omega$ are adjoint to each other if for any $n\geq 0$, any
$\xi\in\Lambda_{n}(V)$ and any $\eta\in\Lambda_{n+k}(V)$, the identity

$\displaystyle\langle\alpha(\eta),\xi\rangle=\langle\eta,\omega(\xi)\rangle$

is satisfied.

######  Proposition 3.5.

Let $k\geq 1$ be any positive integer. Let $\alpha\in T_{k}(V)$ be given by

$\displaystyle\alpha=\sum_{v_{1},v_{2},\ldots,v_{k}\in
V}r_{v_{1},v_{2},\ldots,v_{k}}\frac{\partial}{\partial
v_{1}}\wedge\frac{\partial}{\partial
v_{2}}\wedge\cdots\wedge\frac{\partial}{\partial
v_{k}},~{}~{}~{}r_{v_{1},v_{2},\ldots,v_{k}}\in\mathbb{R}.$

Suppose $\omega\in T^{k}(V)$ is adjoint to $\alpha$. Then $\omega$ is given by

$\displaystyle\omega={\rm sgn}(k)\sum_{v_{1},v_{2},\ldots,v_{k}\in
V}r_{v_{1},v_{2},\ldots,v_{k}}dv_{1}\wedge dv_{2}\wedge\cdots\wedge
dv_{k},~{}~{}~{}r_{v_{1},v_{2},\ldots,v_{k}}\in\mathbb{R}$ (3.10)

where ${\rm sgn}(k)=1$ if $k\equiv 0,1$ modulo $4$ and ${\rm sgn}(k)=-1$ if
$k\equiv 2,3$ modulo $4$.

###### Proof.

Let $n\geq 0$, $\xi\in\Lambda_{n}(V)$, and $\eta\in\Lambda_{n+k}(V)$. Then we
have

$\displaystyle\langle\alpha(\eta),\xi\rangle$ $\displaystyle=$
$\displaystyle\sum_{v_{1},v_{2},\ldots,v_{k}\in
V}r_{v_{1},v_{2},\ldots,v_{k}}\langle\frac{\partial}{\partial
v_{1}}\wedge\frac{\partial}{\partial
v_{2}}\wedge\cdots\wedge\frac{\partial}{\partial v_{k}}(\eta),\xi\rangle$
$\displaystyle=$ $\displaystyle\sum_{v_{1},v_{2},\ldots,v_{k}\in
V}r_{v_{1},v_{2},\ldots,v_{k}}\langle\frac{\partial}{\partial
v_{2}}\wedge\cdots\wedge\frac{\partial}{\partial
v_{k}}(\eta),dv_{1}(\xi)\rangle$ $\displaystyle=$
$\displaystyle\sum_{v_{1},v_{2},\ldots,v_{k}\in
V}r_{v_{1},v_{2},\ldots,v_{k}}\langle\frac{\partial}{\partial
v_{3}}\wedge\cdots\wedge\frac{\partial}{\partial v_{k}}(\eta),dv_{2}\wedge
dv_{1}(\xi)\rangle$ $\displaystyle=$ $\displaystyle\cdots$ $\displaystyle=$
$\displaystyle\sum_{v_{1},v_{2},\ldots,v_{k}\in
V}r_{v_{1},v_{2},\ldots,v_{k}}\langle\eta,dv_{k}\wedge
dv_{k-1}\wedge\cdots\wedge dv_{1}(\xi)\rangle$ $\displaystyle=$
$\displaystyle\sum_{v_{1},v_{2},\ldots,v_{k}\in
V}r_{v_{1},v_{2},\ldots,v_{k}}{\rm sgn}(k)\langle\eta,dv_{1}\wedge
dv_{2}\wedge\cdots\wedge dv_{k}(\xi)\rangle.$

The last equality follows from the fact that the permutation
$(k,k-1,\ldots,1)$ of $(1,2,\ldots,k)$ has the signature

$\displaystyle{\rm
sgn}{{1,2,\ldots,k}\choose{k,k-1,\ldots,1}}=(-1)^{(k-1)+(k-2)+\cdots+1}=(-1)^{\frac{k(k-1)}{2}}$

for any $k\geq 2$ and the permutation $(k,k-1,\ldots,1)$ of $(1,2,\ldots,k)$
has the signature $1$ for $k=1$. In other words, the permutation
$(k,k-1,\ldots,1)$ of $(1,2,\ldots,k)$ has the signature $1$ for $k\equiv 0,1$
(mod $4$) and has the signature $-1$ for $k\equiv 2,3$ (mod $4$). Therefore,
we have that the $\omega$ given by (3.10) is adjoint to $\alpha$. The
proposition follows. ∎

###  3.4 Some Chain Complexes and Co-Chain Complexes on Path Spaces

######  Proposition 3.6.

Let $t\geq 0$ be a non-negative integer. Let $\alpha\in T_{2t+1}(V)$ and
$\omega\in T^{2t+1}(V)$. Then for any $0\leq q\leq 2t$, we have a chain
complex

and a co-chain complex

###### Proof.

Let $t\geq 0$. Let $\alpha\in T_{2t+1}(V)$ and $\omega\in T^{2t+1}(V)$. Let
$0\leq q\leq 2t$. Note that for each $n\geq 0$, the maps

$\displaystyle\alpha:~{}~{}~{}\Lambda_{n(2t+1)+q}(V)\longrightarrow\Lambda_{(n-1)(2t+1)+q}(V)$

and

$\displaystyle\omega:~{}~{}~{}\Lambda_{n(2t+1)+q}(V)\longrightarrow\Lambda_{(n+1)(2t+1)+q}(V)$

are well-defined. By the anti-symmetric property of exterior algebras, we have

$\displaystyle\alpha\wedge\alpha=(-1)^{(2t+1)^{2}}\alpha\wedge\alpha,~{}~{}~{}~{}~{}~{}\omega\wedge\omega=(-1)^{(2t+1)^{2}}\omega\wedge\omega.$

Since $(2t+1)^{2}$ is odd, we have

$\displaystyle\alpha\circ\alpha=\alpha\wedge\alpha=0,~{}~{}~{}~{}~{}~{}\omega\circ\omega=\omega\wedge\omega=0.$

Thus for any $0\leq q\leq 2t$, we have the chain complex as well as the co-
chain complex as given in the proposition. ∎

######  Notation 6.

Let $0\leq q\leq 2t$. Let $\alpha\in T_{2t+1}(V)$ and $\omega\in T^{2t+1}(V)$.
We adopt the following notations:

  1. (i).

denote the chain complex in Proposition 3.6 as

$\displaystyle\Lambda_{*}(V,\alpha,q)=\\{\Lambda_{n(2t+1)+q}(V),\alpha\\}_{n\geq
0};$

  2. (ii).

denote the co-chain complex in Proposition 3.6 as

$\displaystyle\Lambda^{*}(V,\omega,q)=\\{\Lambda_{n(2t+1)+q}(V),\omega\\}_{n\geq
0}.$

######  Notation 7.

For any integer $m$, there is a unique integer $\lambda$ (not necessarily non-
negative) and a unique integer $0\leq q\leq 2t$ such that $m=\lambda(2t+1)+q$.
We adopt the following notations:

  1. (i).

denote the chain complex

as

$\displaystyle\Lambda_{*}(V,\alpha,m)=\\{\Lambda_{(n+\lambda)(2t+1)+q}(V),\alpha\\}_{n\geq
0};$

  2. (ii).

denote the co-chain complex

as

$\displaystyle\Lambda^{*}(V,\omega,m)=\\{\Lambda_{(n+\lambda)(2t+1)+q}(V),\omega\\}_{n\geq
0}.$

Here in both (i) and (ii), we use the notation $\Lambda_{k}(V)=0$ for $k<0$.

######  Proposition 3.7.

Let $t,s\geq 0$ be non-negative integers. Let $m\in\mathbb{Z}$. Let $\alpha\in
T_{2t+1}(V)$ and $\omega\in T^{2t+1}(V)$. Let $\beta\in T_{2s}(V)$ and $\mu\in
T^{2s}(V)$. Then $\beta$ gives a chain map

$\displaystyle\beta:~{}~{}~{}\Lambda_{*}(V,\alpha,m)\longrightarrow\Lambda_{*}(V,\alpha,m-2s)$

and $\mu$ gives a co-chain map

$\displaystyle\mu:~{}~{}~{}\Lambda^{*}(V,\omega,m)\longrightarrow\Lambda^{*}(V,\omega,m+2s).$

###### Proof.

Note that as linear maps,

$\displaystyle\beta:~{}~{}~{}\Lambda_{(n+\lambda)(2t+1)+q}(V)\longrightarrow\Lambda_{(n+\lambda)(2t+1)+q-2s}(V)$

and

$\displaystyle\mu:~{}~{}~{}\Lambda_{(n+\lambda)(2t+1)+q}(V)\longrightarrow\Lambda_{(n+\lambda)(2t+1)+q+2s}(V)$

are well-defined. By the anti-symmetric property of exterior algebras, we have
(cf. [4, p, 53, Anticommutative Law])

$\displaystyle\alpha\wedge\beta=(-1)^{2s(2t+1)}\beta\wedge\alpha=\beta\wedge\alpha.$

That is,

$\displaystyle\alpha\circ\beta=\beta\circ\alpha.$

Thus $\beta$ is a chain map from $\Lambda_{*}(V,\alpha,m)$ to
$\Lambda_{*}(V,\alpha,m-2s)$. Moreover, we also have (cf. [4, p, 53,
Anticommutative Law])

$\displaystyle\omega\wedge\mu=(-1)^{2s(2t+1)}\mu\wedge\omega=\mu\wedge\omega.$

That is,

$\displaystyle\omega\circ\mu=\mu\circ\omega.$

Thus $\mu$ is a co-chain map from $\Lambda^{*}(V,\omega,m)$ to
$\Lambda^{*}(V,\omega,m+2s)$. The proposition follows. ∎

##  4 Generalized (Co-)Homology for (Co-)Simplicial Complexes

In this section, we define the generalized homology groups for simplicial
complexes and the generalized cohomology groups for co-simplicial complexes.
We prove that any element $\beta\in T_{2s}(V)$, where $s\geq 0$, induces
homomorphisms between the generalized homology groups for the simplicial
complexes on $V$. We also prove that any element $\mu\in T^{2s}(V)$, where
$s\geq 0$, induces homomorphisms between the generalized cohomology groups for
the co-simplicial complexes on $V$.

###  4.1 Some Auxiliaries

Throughout this section, we $V$ be a finite set. Let $\Delta[V]$ be the
complete hypergraph on $V$. For each $n\geq 0$, let

$\displaystyle C_{n}(\Delta[V];\mathbb{R})={\rm
Span}_{\mathbb{R}}\\{\sigma^{(n)}\in\Delta[V]\\}$

be the vector space consisting of all the linear combinations of the
$n$-hyperedges on $V$. Consider the direct sum

$\displaystyle C_{*}(\Delta[V];\mathbb{R})=\bigoplus_{n\geq
0}C_{n}(\Delta[V];\mathbb{R}).$ (4.1)

Note that since $V$ is assumed to be a finite set, the direct sum in the
right-hand side of (4.1) is a finite sum.

######  Lemma 4.1.

Let $t\geq 0$. Let $m\in\mathbb{Z}$. Suppose $m=\lambda(2t+1)+q$ where
$\lambda\in\mathbb{Z}$ and $0\leq q\leq 2t$. Then for any $\alpha\in
T_{2t+1}(V)$, the graded vector space

$\displaystyle
C_{(n+\lambda)(2t+1)+q}(\Delta[V];\mathbb{R}),~{}~{}~{}~{}~{}~{}n\geq 0$ (4.2)

equipped with the boundary map $\alpha$ gives a sub-chain complex of
$\Lambda_{*}(V,\alpha,m)$, which will be denoted as
$C_{*}(\Delta[V],\alpha,m)$.

###### Proof.

For each $n\geq 0$, the vector space
$C_{(n+\lambda)(2t+1)+q}(\Delta[V];\mathbb{R})$ is a subspace of the vector
space $\Lambda_{(n+\lambda)(2t+1)+q}(V)$. Hence in order to prove that (4.2)
equipped with $\alpha$ is a sub-chain complex of $\Lambda_{*}(V,\alpha,m)$, it
suffices to prove that the map

$\displaystyle\alpha:~{}~{}~{}C_{(n+\lambda)(2t+1)+q}(\Delta[V];\mathbb{R})\longrightarrow
C_{(n-1+\lambda)(2t+1)+q}(\Delta[V];\mathbb{R})$ (4.3)

is well-defined for each $n\geq 0$. This follows from the observation that for
any $[(n+\lambda)(2t+1)+q]$-simplex

$\displaystyle v_{0}v_{1}\ldots v_{(n+\lambda)(2t+1)+q}\in
C_{(n+\lambda)(2t+1)+q}(\Delta[V];R)$

and any

$\displaystyle\alpha=\frac{\partial}{\partial
u_{1}}\wedge\frac{\partial}{\partial
u_{2}}\wedge\cdots\wedge\frac{\partial}{\partial u_{2t+1}}$

where $u_{1},u_{2},\ldots,u_{2t+1}\in V$ and $u_{1}\prec u_{2}\prec\cdots\prec
u_{2t+1}$, we have

$\displaystyle\alpha(v_{0}v_{1}\ldots v_{(n+\lambda)(2t+1)+q})\in
C_{(n-1+\lambda)(2t+1)+q}(\Delta[V];R).$

By a calculation of linear combinations, it follows that the map (4.3) is
well-defined. Therefore, the graded vector space (4.2) equipped with $\alpha$
is a sub-chain complex of $\Lambda_{*}(V,\alpha,m)$. ∎

######  Definition 19.

For any $n\geq 0$ and any elementary $n$-path $v_{0}v_{1}\ldots v_{n}$ on $V$,
we call $v_{0}v_{1}\ldots v_{n}$ a cyclic elementary $n$-path if there exists
$0\leq i<j\leq n$ such that $v_{i}=v_{j}$.

######  Definition 20.

Let $\mathcal{O}_{n}(V)$ be the vector space spanned by all the cyclic
elementary $n$-paths on $V$. Then $\mathcal{O}_{n}(V)$ consists of all the
linear combinations of the cyclic elementary $n$-paths on $V$. We call an
element in $\mathcal{O}_{n}(V)$ a cyclic $n$-path on $V$.

######  Lemma 4.2.

Let $t\geq 0$. Let $m\in\mathbb{Z}$. Suppose $m=\lambda(2t+1)+q$ where
$\lambda\in\mathbb{Z}$ and $0\leq q\leq 2t$. Then for any $\omega\in
T^{2t+1}(V)$, the graded vector space

$\displaystyle\mathcal{O}_{(n+\lambda)(2t+1)+q}(V),~{}~{}~{}~{}~{}~{}n\geq 0$
(4.4)

equipped with the co-boundary map $\omega$ gives a sub-co-chain complex of
$\Lambda^{*}(V,\omega,m)$, which will be denoted as
$\mathcal{O}^{*}(V,\omega,m)$.

###### Proof.

It suffices to verify that the map

$\displaystyle\omega:~{}~{}~{}\mathcal{O}_{(n+\lambda)(2t+1)+q}(V)\longrightarrow\mathcal{O}_{(n+1+\lambda)(2t+1)+q}(V)$
(4.5)

is well-defined for each $n\geq 0$. This follows from the observation that for
any cyclic elementary $(n+\lambda)(2t+1)+q$-path

$\displaystyle v_{0}v_{1}\ldots
v_{(n+\lambda)(2t+1)+q}\in\mathcal{O}_{(n+\lambda)(2t+1)+q}(V)$

and any

$\displaystyle\omega=du_{1}\wedge du_{2}\wedge\cdots\wedge du_{2t+1}$

where $u_{1},u_{2},\ldots,u_{2t+1}\in V$ and $u_{1}\prec u_{2}\prec\cdots\prec
u_{2t+1}$, we have

$\displaystyle\omega(v_{0}v_{1}\ldots
v_{(n+\lambda)(2t+1)+q})\in\mathcal{O}_{(n+1+\lambda)(2t+1)+q}(V).$

By a calculation of linear combinations, it follows that the map (4.5) is
well-defined. Therefore, the graded vector space (4.4) equipped with $\omega$
is a sub-co-chain complex of $\Lambda^{*}(V,\omega,m)$. ∎

######  Lemma 4.3.

Let $t\geq 0$. Let $m\in\mathbb{Z}$. Suppose $m=\lambda(2t+1)+q$ where
$\lambda\in\mathbb{Z}$ and $0\leq q\leq 2t$. Then for any $\omega\in
T^{2t+1}(V)$, the graded vector space

$\displaystyle
C_{(n+\lambda)(2t+1)+q}(\Delta[V];\mathbb{R}),~{}~{}~{}~{}~{}~{}n\geq 0$ (4.6)

equipped with the co-boundary map $\omega$ gives a quotient co-chain complex
$\Lambda^{*}(V,\omega,m)/\mathcal{O}^{*}(V,\omega,m)$ which will be denoted as
$C^{*}(\Delta[V],\omega,m)$.

###### Proof.

Note that the canonical inclusion of the sub-co-chain complex
$\mathcal{O}^{*}(V,\omega,m)$ into the co-chain complex
$\Lambda^{*}(V,\omega,m)$ gives a quotient co-chain complex
$\Lambda^{*}(V,\omega,m)/\mathcal{O}^{*}(V,\omega,m)$. On the other hand, for
each $n\geq 0$, the quotient vector space

$\displaystyle\Lambda_{(n+\lambda)(2t+1)+q}(V)/\mathcal{O}_{(n+\lambda)(2t+1)+q}(V)$

is canonically isomorphic to the vector space
$C_{(n+\lambda)(2t+1)+q}(\Delta[V];\mathbb{R})$. Therefore, the quotient co-
chain complex $\Lambda^{*}(V,\omega,m)/\mathcal{O}^{*}(V,\omega,m)$ is given
by the graded vector space (4.6) equipped with the co-boundary map $\omega$.
The lemma follows. ∎

With the help of Proposition 3.7, the next proposition follows.

######  Proposition 4.4.

Let $t,s\geq 0$ be non-negative integers. Let $m\in\mathbb{Z}$. Let $\alpha\in
T_{2t+1}(V)$ and $\omega\in T^{2t+1}(V)$. Let $\beta\in T_{2s}(V)$ and $\mu\in
T^{2s}(V)$. Then $\beta$ is a chain map

$\displaystyle\beta:~{}~{}~{}C_{*}(\Delta[V],\alpha,m)\longrightarrow
C_{*}(\Delta[V],\alpha,m-2s)$ (4.7)

and $\mu$ is a co-chain map

$\displaystyle\mu:~{}~{}~{}C^{*}(\Delta[V],\omega,m)\longrightarrow
C^{*}(\Delta[V],\omega,m+2s).$ (4.8)

###### Proof.

By a similar argument in the proof of Lemma 4.1, it can be verified that as a
linear map, $\beta$ in (4.7) is well-defined. Thus by Proposition 3.7 and
Lemma 4.1, it follows that $\beta$ in (4.7) is a chain map. On the other hand,
by a similar argument in the proof of Lemma 4.3, it can be verified that as a
linear map, $\mu$ in (4.8) is well-defined. Thus by Proposition 3.7 and Lemma
4.3, it follows that $\omega$ in (4.8) is a co-chain map. ∎

###  4.2 Generalized Homology for Simplicial Complexes

Let $\mathcal{K}$ be a simplicial complex with its vertices in $V$.

######  Notation 8.

For each $n\geq 0$, let $C_{n}(\mathcal{K};\mathbb{R})$ be the (real) vector
space consisting of all the linear combinations of the $n$-simplices in
$\mathcal{K}$.

######  Theorem 4.5.

Let $t,s\geq 0$. Let $m\in\mathbb{Z}$. Suppose $m=\lambda(2t+1)+q$ where
$\lambda\in\mathbb{Z}$ and $0\leq q\leq 2t$. Then

  1. (i).

for any $\alpha\in T_{2t+1}(V)$, the graded vector space

$\displaystyle
C_{(n+\lambda)(2t+1)+q}(\mathcal{K};\mathbb{R}),~{}~{}~{}~{}~{}~{}n\geq 0$
(4.9)

equipped with the chain map $\alpha$ gives a sub-chain complex of
$C_{*}(\Delta[V],\alpha,m)$, which will be denoted as
$C_{*}(\mathcal{K},\alpha,m)$;

  2. (ii).

for any $\beta\in T_{2s}(V)$, there is an induced chain map

$\displaystyle\beta:~{}~{}~{}C_{*}(\mathcal{K},\alpha,m)\longrightarrow
C_{*}(\mathcal{K},\alpha,m-2s).$ (4.10)

###### Proof.

We prove (i) and (ii) subsequently.

(i). For each $n\geq 0$, the vector space
$C_{(n+\lambda)(2t+1)+q}(\mathcal{K};\mathbb{R})$ is a subspace of the vector
space $C_{(n+\lambda)(2t+1)+q}(\Delta[V];\mathbb{R})$. Hence in order to prove
that the graded vector space (4.9) equipped with the chain map $\alpha$ is a
sub-chain complex of $C_{*}(\Delta[V],\alpha,m)$, it suffices to prove that
the map

$\displaystyle\alpha:~{}~{}~{}C_{(n+\lambda)(2t+1)+q}(\mathcal{K};\mathbb{R})\longrightarrow
C_{(n-1+\lambda)(2t+1)+q}(\mathcal{K};\mathbb{R})$ (4.11)

is well-defined for each $n\geq 0$. This follows from the observation that for
any $[(n+\lambda)(2t+1)+q]$-simplex

$\displaystyle v_{0}v_{1}\ldots v_{(n+\lambda)(2t+1)+q}\in
C_{(n+\lambda)(2t+1)+q}(\mathcal{K};R)$

and any

$\displaystyle\alpha=\frac{\partial}{\partial
u_{1}}\wedge\frac{\partial}{\partial
u_{2}}\wedge\cdots\wedge\frac{\partial}{\partial u_{2t+1}}$

where $u_{1},u_{2},\ldots,u_{2t+1}\in V$ and $u_{1}\prec u_{2}\prec\cdots\prec
u_{2t+1}$, we have

$\displaystyle\alpha(v_{0}v_{1}\ldots v_{(n+\lambda)(2t+1)+q})\in
C_{(n-1+\lambda)(2t+1)+q}(\mathcal{K};R).$

By a calculation of linear combinations, it follows that the map (4.11) is
well-defined. Therefore, the graded vector space (4.9) equipped with the chain
map $\alpha$ is a sub-chain complex of $C_{*}(\Delta[V],\alpha,m)$.

(ii). Similar with the verification that the map $\alpha$ in (4.11) is well-
defined for each $n\geq 0$, we can prove that the map

$\displaystyle\beta:~{}~{}~{}C_{(n+\lambda)(2t+1)+q}(\mathcal{K};\mathbb{R})\longrightarrow
C_{(n+\lambda)(2t+1)+q-2s}(\mathcal{K};\mathbb{R})$

is well-defined for each $n\geq 0$. Therefore, with the help of (4.7) in
Proposition 4.4, we have that $\beta$ gives a chain map in (4.10). ∎

######  Definition 21.

Let $t\geq 0$. Let $\alpha\in T_{2t+1}(V)$. Let $m\in\mathbb{Z}$. Suppose
$m=\lambda(2t+1)+q$ where $\lambda\in\mathbb{Z}$ and $0\leq q\leq 2t$. Let
$\mathcal{K}$ be a simplicial complex with its vertices in $V$. For each
$n\geq 0$, we define the $n$-th generalized homology group
$H_{n}(\mathcal{K},\alpha,m)$ of $\mathcal{K}$ with respect to $\alpha$ and
$m$ to be the $n$-th homology group

$\displaystyle H_{n}(\mathcal{K},\alpha,m):$ $\displaystyle=$ $\displaystyle
H_{n}(C_{*}(\mathcal{K},\alpha,m))$ $\displaystyle=$ $\displaystyle\frac{{\rm
Ker}\Big{(}\alpha:C_{(n+\lambda)(2t+1)+q}(\mathcal{K};\mathbb{R})\longrightarrow
C_{(n-1+\lambda)(2t+1)+q}(\mathcal{K};\mathbb{R})\Big{)}}{{\rm
Im}\Big{(}\alpha:C_{(n+1+\lambda)(2t+1)+q}(\mathcal{K};\mathbb{R})\longrightarrow
C_{(n+\lambda)(2t+1)+q}(\mathcal{K};\mathbb{R})\Big{)}}$

of the chain complex $C_{*}(\mathcal{K},\alpha,m)$.

The next theorem follows from Theorem 4.5 and Definition 21 immediately.

######  Theorem 4.6 (Main Result I).

Let $t,s\geq 0$. Let $m\in\mathbb{Z}$. Suppose $m=\lambda(2t+1)+q$ where
$\lambda\in\mathbb{Z}$ and $0\leq q\leq 2t$. Then for any $\alpha\in
T_{2t+1}(V)$ and $\beta\in T_{2s}(V)$, there is an induced homomorphism

$\displaystyle\beta_{*}:~{}~{}~{}H_{n}(\mathcal{K},\alpha,m)\longrightarrow
H_{n}(\mathcal{K},\alpha,m-2s),~{}~{}~{}~{}~{}~{}n\geq 0$ (4.12)

of the generalized homology groups.

###### Proof.

Apply the homology functor to the chain complex in Theorem 4.5 (i) and the
chain map in Theorem 4.5 (ii). We obtain the homomorphism $\beta_{*}$ of the
generalized homology groups in (4.12). ∎

###  4.3 Generalized Cohomology for Co-Simplicial Complexes

Let $\mathcal{L}$ be a co-simplicial complex with its vertices in $V$.

######  Notation 9.

For each $n\geq 0$, let $C_{n}(\mathcal{L};\mathbb{R})$ be the (real) vector
space consisting of all the linear combinations of the $n$-co-simplices in
$\mathcal{L}$.

######  Theorem 4.7.

Let $t,s\geq 0$. Let $m\in\mathbb{Z}$. Suppose $m=\lambda(2t+1)+q$ where
$\lambda\in\mathbb{Z}$ and $0\leq q\leq 2t$. Then

  1. (i).

for any $\omega\in T^{2t+1}(V)$, the graded vector space

$\displaystyle
C_{(n+\lambda)(2t+1)+q}(\mathcal{L};\mathbb{R}),~{}~{}~{}~{}~{}~{}n\geq 0$
(4.13)

equipped with the co-boundary map $\omega$ gives a sub-co-chain complex of
$C^{*}(\Delta[V],\omega,m)$, which will be denoted as
$C^{*}(\mathcal{L},\omega,m)$;

  2. (ii).

for any $\mu\in T^{2s}(V)$, there is an induced co-chain map

$\displaystyle\mu:~{}~{}~{}C^{*}(\mathcal{L},\omega,m)\longrightarrow
C^{*}(\mathcal{L},\omega,m+2s).$ (4.14)

###### Proof.

We prove (i) and (ii) subsequently.

(i). For each $n\geq 0$, the vector space
$C_{(n+\lambda)(2t+1)+q}(\mathcal{L};\mathbb{R})$ is a subspace of the vector
space $C_{(n+\lambda)(2t+1)+q}(\Delta[V];\mathbb{R})$. Hence in order to prove
that the graded vector space (4.13) equipped with the co-boundary map $\omega$
is a sub-co-chain complex of $C^{*}(\Delta[V],\omega,m)$, it suffices to prove
that the map

$\displaystyle\omega:~{}~{}~{}C_{(n+\lambda)(2t+1)+q}(\mathcal{L};\mathbb{R})\longrightarrow
C_{(n+1+\lambda)(2t+1)+q}(\mathcal{L};\mathbb{R})$ (4.15)

is well-defined for each $n\geq 0$. This follows from the observation that for
any $[(n+\lambda)(2t+1)+q]$-co-simplex

$\displaystyle v_{0}v_{1}\ldots v_{(n+\lambda)(2t+1)+q}\in
C_{(n+\lambda)(2t+1)+q}(\mathcal{L};R)$

and any

$\displaystyle\omega=du_{1}\wedge du_{2}\wedge\cdots\wedge du_{2t+1}$

where $u_{1},u_{2},\ldots,u_{2t+1}\in V$ and $u_{1}\prec u_{2}\prec\cdots\prec
u_{2t+1}$, we have

$\displaystyle\omega(v_{0}v_{1}\ldots v_{(n+\lambda)(2t+1)+q})\in
C_{(n+1+\lambda)(2t+1)+q}(\mathcal{L};R).$

By a calculation of linear combinations, it follows that the map (4.15) is
well-defined. Therefore, the graded vector space (4.13) equipped with the co-
boundary map $\omega$ is a sub-co-chain complex of
$C^{*}(\Delta[V],\omega,m)$.

(ii). Similar with the verification that the map $\omega$ in (4.15) is well-
defined for each $n\geq 0$, we can prove that the map

$\displaystyle\mu:~{}~{}~{}C_{(n+\lambda)(2t+1)+q}(\mathcal{L};\mathbb{R})\longrightarrow
C_{(n+\lambda)(2t+1)+q+2s}(\mathcal{L};\mathbb{R})$

is well-defined for each $n\geq 0$. Therefore, with the help of (4.8) in
Proposition 4.4, we have that $\mu$ gives a co-chain map in (4.14). ∎

######  Definition 22.

Let $t\geq 0$. Let $\omega\in T^{2t+1}(V)$. Let $m\in\mathbb{Z}$. Suppose
$m=\lambda(2t+1)+q$ where $\lambda\in\mathbb{Z}$ and $0\leq q\leq 2t$. Let
$\mathcal{L}$ be a co-simplicial complex with its vertices in $V$. For each
$n\geq 0$, we define the $n$-th generalized cohomology group
$H^{n}(\mathcal{L},\omega,m)$ of $\mathcal{L}$ with respect to $\omega$ and
$m$ to be the cohomology group

$\displaystyle H^{n}(\mathcal{L},\omega,m):$ $\displaystyle=$ $\displaystyle
H^{n}(C^{*}(\mathcal{L},\omega,m))$ $\displaystyle=$ $\displaystyle\frac{{\rm
Ker}\Big{(}\omega:C_{(n+\lambda)(2t+1)+q}(\mathcal{L};\mathbb{R})\longrightarrow
C_{(n+1+\lambda)(2t+1)+q}(\mathcal{L};\mathbb{R})\Big{)}}{{\rm
Im}\Big{(}\omega:C_{(n-1+\lambda)(2t+1)+q}(\mathcal{L};\mathbb{R})\longrightarrow
C_{(n+\lambda)(2t+1)+q}(\mathcal{L};\mathbb{R})\Big{)}}$

of the co-chain complex $C^{*}(\mathcal{L},\omega,m)$.

The next theorem follows from Theorem 4.7 and Definition 22 immediately.

######  Theorem 4.8 (Main Result II).

Let $t,s\geq 0$. Let $m\in\mathbb{Z}$. Suppose $m=\lambda(2t+1)+q$ where
$\lambda\in\mathbb{Z}$ and $0\leq q\leq 2t$. Then for any $\omega\in
T^{2t+1}(V)$ and $\mu\in T^{2s}(V)$, there is an induced homomorphism

$\displaystyle\mu_{*}:~{}~{}~{}H^{n}(\mathcal{L},\omega,m)\longrightarrow
H^{n}(\mathcal{L},\omega,m+2s),~{}~{}~{}~{}~{}~{}n\geq 0$ (4.16)

of the generalized cohomology groups.

###### Proof.

Apply the cohomology functor to the co-chain complex in Theorem 4.7 (i) and
the co-chain map in Theorem 4.7 (ii). We obtain the homomorphism $\mu_{*}$ of
the generalized cohomology groups in (4.16). ∎

##  5 Examples

We give some examples for Theorem 4.5, Theorem 4.6, Theorem 4.7 and Theorem
4.8.

######  Example 5.1.

Let $V$ be any finite set. Then we have the followings.

  1. (i).

Any element $\alpha\in T_{1}(V)$ can be expressed as

$\displaystyle\alpha=\sum_{v\in V}f(v)\frac{\partial}{\partial v}$ (5.1)

for some function $f:V\longrightarrow\mathbb{R}$. Let $\mathcal{K}$ be a
simplicial complex with its vertices in $V$. Then for any $n\geq 0$ and any
$n$-simplex $v_{0}v_{1}\ldots v_{n}$ in $\mathcal{K}$, we have

$\displaystyle\alpha(v_{0}v_{1}\ldots v_{n})$ $\displaystyle=$
$\displaystyle\sum_{v\in V}f(v)\frac{\partial}{\partial v}(v_{0}v_{1}\ldots
v_{n})$ $\displaystyle=$ $\displaystyle\sum_{v\in
V}f(v)\sum_{i=0}^{n}(-1)^{i}\delta(v,v_{i})v_{0}\ldots\widehat{v_{i}}\ldots
v_{n}$ $\displaystyle=$ $\displaystyle\sum_{i=0}^{n}(-1)^{i}\Big{(}\sum_{v\in
V}\delta(v,v_{i})f(v)\Big{)}v_{0}\ldots\widehat{v_{i}}\ldots v_{n}$
$\displaystyle=$
$\displaystyle\sum_{i=0}^{n}(-1)^{i}f(v_{i})v_{0}\ldots\widehat{v_{i}}\ldots
v_{n}.$

In [27, 28, 29], the $\alpha$ given in (5.1) is called the $f$-weighted
boundary operator on $\mathcal{K}$ and the $(\alpha,0)$-homology of
$\mathcal{K}$ is denoted as the weighted homology $H_{*}(\mathcal{K},f)$ of
the weighted simplicial complex $(\mathcal{K},f)$. Particularly, if $f$ takes
the constant value $1$ for all $v\in V$, then $\alpha$ is the usual boundary
operator $\partial_{*}$ given in (1.1) and $H_{*}(\mathcal{K},f)$ is the usual
homology $H_{*}(\mathcal{K})$ (cf. [23, Chapter 1] and [19, Section 2.1]) of
$\mathcal{K}$.

  2. (ii).

Any element $\omega\in T^{1}(V)$ can be expressed as

$\displaystyle\omega=\sum_{v\in V}f(v)dv$ (5.2)

for some function $f:V\longrightarrow\mathbb{R}$. Let $\mathcal{L}$ be a co-
simplicial complex with its vertices in $V$. Then for any $n\geq 0$ and any
$n$-co-simplex $v_{0}v_{1}\ldots v_{n}$ in $\mathcal{L}$, we have

$\displaystyle\omega(v_{0}v_{1}\ldots v_{n})$ $\displaystyle=$
$\displaystyle\sum_{v\in V}f(v)dv(v_{0}v_{1}\ldots v_{n})$ $\displaystyle=$
$\displaystyle\sum_{v\in V}f(v)\sum_{i=0}^{n+1}(-1)^{i}v_{0}v_{1}\ldots
v_{i-1}vv_{i}v_{i+1}\ldots v_{n}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{n+1}(-1)^{i}\Big{(}\sum_{v\in
V}f(v)v_{0}\ldots\ldots v_{i-1}vv_{i}\ldots v_{n}\Big{)}.$

Similar with (i), we call the $\omega$ given in (5.2) the $f$-weighted co-
boundary operator on $\mathcal{L}$ and denote the $(\omega,0)$-cohomology of
$\mathcal{L}$ as $H^{*}(\mathcal{L},f)$. Particularly, if $f$ takes the
constant value $1$ for all $v\in V$, then we denote the $\omega$ as $d_{*}$
denote the $H^{*}(\mathcal{L},f)$ as $H^{*}(\mathcal{L})$.

######  Example 5.2.

Let $V=\\{v_{0},v_{1},v_{2}\\}$. Let $f:V\longrightarrow\mathbb{R}$ be a
function on $V$.

  1. (i).

Let

$\displaystyle\mathcal{K}=\\{v_{0},v_{1},v_{2},v_{0}v_{1},v_{0}v_{2},v_{1}v_{2}\\}$

be a simplicial complex with its vertices in $V$. Then we have

$\displaystyle C_{0}(\mathcal{K};\mathbb{R})$ $\displaystyle=$
$\displaystyle{\rm Span}_{\mathbb{R}}\\{v_{0},v_{1},v_{2}\\},$ $\displaystyle
C_{1}(\mathcal{K};\mathbb{R})$ $\displaystyle=$ $\displaystyle{\rm
Span}_{\mathbb{R}}\\{v_{0}v_{1},v_{0}v_{2},v_{1}v_{2}\\},$ $\displaystyle
C_{n}(\mathcal{K};\mathbb{R})$ $\displaystyle=$ $\displaystyle
0{\rm~{}~{}~{}~{}for~{}all~{}}n\geq 2.$

     * •

Let $t=1$. Let

$\displaystyle\alpha$ $\displaystyle=$ $\displaystyle\sum_{v\in
V}f(v)\frac{\partial}{\partial v}$ $\displaystyle=$ $\displaystyle
f(v_{0})\frac{\partial}{\partial v_{0}}+f(v_{1})\frac{\partial}{\partial
v_{1}}+f(v_{2})\frac{\partial}{\partial v_{2}}.$

With the help of Example 5.1 (i), we have

$\displaystyle\alpha(v_{0})=\alpha(v_{1})=\alpha(v_{2})=0,$
$\displaystyle\alpha(v_{0}v_{1})=f(v_{0})v_{1}-f(v_{1})v_{0},$
$\displaystyle\alpha(v_{0}v_{2})=f(v_{0})v_{2}-f(v_{2})v_{0},$
$\displaystyle\alpha(v_{1}v_{2})=f(v_{1})v_{2}-f(v_{2})v_{1},$
$\displaystyle\alpha(v_{0}v_{1}v_{2})=f(v_{0})v_{1}v_{2}-f(v_{1})v_{0}v_{2}+f(v_{2})v_{0}v_{1}.$

Note that

$\displaystyle\dim{\rm
Ker}\Big{(}\alpha:C_{0}(\mathcal{K};\mathbb{R})\longrightarrow 0\Big{)}=3$

and

$\displaystyle\dim{\rm
Im}\Big{(}\alpha:C_{1}(\mathcal{K};\mathbb{R})\longrightarrow
C_{0}(\mathcal{K};\mathbb{R})\Big{)}=\begin{cases}2,&{\rm~{}if~{}}f(v_{i})\neq
0{\rm~{}for~{}some~{}}i=0,1,2;\\\
0,&{\rm~{}if~{}}f(v_{0})=f(v_{1})=f(v_{2})=0.\end{cases}$

Thus

$\displaystyle
H_{0}(\mathcal{K},f)=H_{0}(\mathcal{K},\alpha,0)=\begin{cases}\mathbb{R},&{\rm~{}if~{}}f(v_{i})\neq
0{\rm~{}for~{}some~{}}i=0,1,2;\\\
\mathbb{R}^{3},&{\rm~{}if~{}}f(v_{0})=f(v_{1})=f(v_{2})=0.\end{cases}$

Note that

$\displaystyle\dim{\rm
Ker}\Big{(}\alpha:C_{1}(\mathcal{K};\mathbb{R})\longrightarrow
C_{0}(\mathcal{K};\mathbb{R})\Big{)}=\begin{cases}1,&{\rm~{}if~{}}f(v_{i})\neq
0{\rm~{}for~{}some~{}}i=0,1,2;\\\
3,&{\rm~{}if~{}}f(v_{0})=f(v_{1})=f(v_{2})=0\end{cases}$

and

$\displaystyle\dim{\rm
Im}\Big{(}\alpha:C_{2}(\mathcal{K};\mathbb{R})\longrightarrow
C_{1}(\mathcal{K};\mathbb{R})\Big{)}=0.$

Thus

$\displaystyle
H_{1}(\mathcal{K},f)=H_{1}(\mathcal{K},\alpha,0)=\begin{cases}\mathbb{R},&{\rm~{}if~{}}f(v_{i})\neq
0{\rm~{}for~{}some~{}}i=0,1,2;\\\
\mathbb{R}^{3},&{\rm~{}if~{}}f(v_{0})=f(v_{1})=f(v_{2})=0.\end{cases}$

     * •

Let $s=1$. Let

$\displaystyle\beta=b_{01}\frac{\partial}{\partial
v_{0}}\wedge\frac{\partial}{\partial v_{1}}+b_{02}\frac{\partial}{\partial
v_{0}}\wedge\frac{\partial}{\partial v_{2}}+b_{12}\frac{\partial}{\partial
v_{1}}\wedge\frac{\partial}{\partial v_{2}}.$

Then

$\displaystyle\beta(v_{i})=0~{}~{}~{}~{}~{}~{}{\rm~{}for~{}}0\leq i\leq 2$

and

$\displaystyle\beta(v_{i}v_{j})=0~{}~{}~{}~{}~{}~{}{\rm~{}for~{}}0\leq i<j\leq
2.$

Thus the induced homomorphism $\beta_{*}$ between the homology groups is
identically zero.

  2. (ii).

Let

$\displaystyle\mathcal{L}=\\{v_{0}v_{1},v_{0}v_{2},v_{0}v_{1}v_{2}\\}$

be a co-simplicial complex with its vertices in $V$. Then we have

$\displaystyle C_{0}(\mathcal{L};\mathbb{R})$ $\displaystyle=$ $\displaystyle
0,$ $\displaystyle C_{1}(\mathcal{L};\mathbb{R})$ $\displaystyle=$
$\displaystyle{\rm Span}_{\mathbb{R}}\\{v_{0}v_{1},v_{0}v_{2}\\},$
$\displaystyle C_{2}(\mathcal{L};\mathbb{R})$ $\displaystyle=$
$\displaystyle{\rm Span}_{\mathbb{R}}\\{v_{0}v_{1}v_{2}\\},$ $\displaystyle
C_{n}(\mathcal{L};\mathbb{R})$ $\displaystyle=$ $\displaystyle
0{\rm~{}~{}~{}~{}for~{}all~{}}n\geq 3.$

     * •

Let $t=1$. Let

$\displaystyle\omega$ $\displaystyle=$ $\displaystyle\sum_{v\in V}f(v)dv$
$\displaystyle=$ $\displaystyle
f(v_{0})dv_{0}+f(v_{1})dv_{1}+f(v_{2})dv_{2}+f(v_{3})dv_{3}.$

With the help of Example 5.1 (ii), we have

$\displaystyle\omega(v_{0}v_{1})=f(v_{2})v_{0}v_{1}v_{2},$
$\displaystyle\omega(v_{0}v_{2})=-f(v_{1})v_{0}v_{1}v_{2},$
$\displaystyle\omega(v_{0}v_{1}v_{2})=0.$

Note that

$\displaystyle\dim{\rm
Ker}\Big{(}\omega:C_{1}(\mathcal{L};\mathbb{R})\longrightarrow
C_{2}(\mathcal{L};\mathbb{R})\Big{)}=\begin{cases}1,&{\rm~{}if~{}}f(v_{i})\neq
0{\rm~{}for~{}some~{}}i=1,2;\\\
2,&{\rm~{}if~{}}f(v_{1})=f(v_{2})=0\end{cases}$

or equivalently,

$\displaystyle\dim{\rm
Im}\Big{(}\omega:C_{1}(\mathcal{L};\mathbb{R})\longrightarrow
C_{2}(\mathcal{L};\mathbb{R})\Big{)}=\begin{cases}1,&{\rm~{}if~{}}f(v_{i})\neq
0{\rm~{}for~{}some~{}}i=1,2;\\\
0,&{\rm~{}if~{}}f(v_{1})=f(v_{2})=0.\end{cases}$

Thus

$\displaystyle
H^{1}(\mathcal{L},f)=H^{1}(\mathcal{L},\omega,0)=\begin{cases}\mathbb{R},&{\rm~{}if~{}}f(v_{i})\neq
0{\rm~{}for~{}some~{}}i=1,2;\\\
\mathbb{R}^{2},&{\rm~{}if~{}}f(v_{1})=f(v_{2})=0\end{cases}$

and

$\displaystyle
H^{2}(\mathcal{L},f)=H^{2}(\mathcal{L},\omega,0)=\begin{cases}0,&{\rm~{}if~{}}f(v_{i})\neq
0{\rm~{}for~{}some~{}}i=1,2;\\\
\mathbb{R},&{\rm~{}if~{}}f(v_{1})=f(v_{2})=0.\end{cases}$

Moreover,

$\displaystyle H^{n}(\mathcal{L},f)=H^{n}(\mathcal{L},\omega,0)=0$

for any $n\neq 1,2$.

     * •

Let $s=1$. Let

$\displaystyle\mu=u_{01}dv_{0}\wedge dv_{1}+u_{02}dv_{0}\wedge
dv_{2}+u_{12}dv_{1}\wedge dv_{2}.$

Then

$\displaystyle\mu(v_{0}v_{1}v_{2})=mu(v_{0}v_{1})=\mu(v_{0}v_{2})=0.$

Thus the induced homomorphism $\mu_{*}$ between the cohomology groups is
identically zero.

######  Example 5.3.

Let $V=\\{v_{0},v_{1},v_{2},v_{3}\\}$. Let $t=1$. Then any $\alpha\in
T_{3}(V)$ can be expressed as

$\displaystyle\alpha$ $\displaystyle=$ $\displaystyle
f(v_{0},v_{1},v_{2})\frac{\partial}{\partial
v_{0}}\wedge\frac{\partial}{\partial v_{1}}\wedge\frac{\partial}{\partial
v_{2}}+f(v_{0},v_{1},v_{3})\frac{\partial}{\partial
v_{0}}\wedge\frac{\partial}{\partial v_{1}}\wedge\frac{\partial}{\partial
v_{3}}$ $\displaystyle+f(v_{0},v_{2},v_{3})\frac{\partial}{\partial
v_{0}}\wedge\frac{\partial}{\partial v_{2}}\wedge\frac{\partial}{\partial
v_{3}}+f(v_{1},v_{2},v_{3})\frac{\partial}{\partial
v_{1}}\wedge\frac{\partial}{\partial v_{2}}\wedge\frac{\partial}{\partial
v_{3}}$

where

$\displaystyle f:~{}~{}~{}V\times V\times V\longrightarrow\mathbb{R}$

is a real function on the $3$-fold Cartesian product of $V$. By Proposition
3.5, the adjoint $\omega\in T^{3}(V)$ of $\alpha$ is given by

$\displaystyle\omega$ $\displaystyle=$
$\displaystyle-f(v_{0},v_{1},v_{2})dv_{0}\wedge dv_{1}\wedge
dv_{2}-f(v_{0},v_{1},v_{3})dv_{0}\wedge dv_{1}\wedge dv_{3}$
$\displaystyle-f(v_{0},v_{2},v_{3})dv_{0}\wedge dv_{2}\wedge
dv_{3}-f(v_{1},v_{2},v_{3})dv_{1}\wedge dv_{2}\wedge dv_{3}.$

Let $s=1$. Then any $\beta\in T_{2}(V)$ can be expressed as

$\displaystyle\beta$ $\displaystyle=$ $\displaystyle
g(v_{0},v_{1})\frac{\partial}{\partial v_{0}}\wedge\frac{\partial}{\partial
v_{1}}+g(v_{0},v_{2})\frac{\partial}{\partial
v_{0}}\wedge\frac{\partial}{\partial
v_{2}}+g(v_{0},v_{3})\frac{\partial}{\partial
v_{0}}\wedge\frac{\partial}{\partial v_{3}}$
$\displaystyle+g(v_{1},v_{2})\frac{\partial}{\partial
v_{1}}\wedge\frac{\partial}{\partial
v_{2}}+g(v_{1},v_{3})\frac{\partial}{\partial
v_{1}}\wedge\frac{\partial}{\partial
v_{3}}+g(v_{2},v_{3})\frac{\partial}{\partial
v_{2}}\wedge\frac{\partial}{\partial v_{3}}$

where

$\displaystyle g:~{}~{}~{}V\times V\longrightarrow\mathbb{R}$

is a real function on the $2$-fold Cartesian product of $V$. By Proposition
3.5, the adjoint $\mu\in T^{2}(V)$ of $\beta$ is given by

$\displaystyle\mu$ $\displaystyle=$ $\displaystyle-g(v_{0},v_{1})dv_{0}\wedge
dv_{1}-g(v_{0},v_{2})dv_{0}\wedge dv_{2}-g(v_{0},v_{3})dv_{0}\wedge dv_{3}$
$\displaystyle-g(v_{1},v_{2})dv_{1}\wedge dv_{2}-g(v_{1},v_{3})dv_{1}\wedge
dv_{3}-g(v_{2},v_{3})dv_{2}\wedge dv_{3}.$

Consider the complete hypergraph

$\displaystyle\Delta[V]$ $\displaystyle=$
$\displaystyle\\{v_{0},v_{1},v_{2},v_{3},v_{0}v_{1},v_{0}v_{2},v_{0}v_{3},v_{1}v_{2},v_{1}v_{3},v_{2}v_{3},$
$\displaystyle
v_{0}v_{1}v_{2},v_{0}v_{1}v_{3},v_{0}v_{2}v_{3},v_{1}v_{2}v_{3},v_{0}v_{1}v_{2}v_{3}\\}.$

Then $\Delta[V]$ is a simplicial complex and is also a co-simplicial complex.

  * •

By a direct calculation,

$\displaystyle\alpha(v_{i})$ $\displaystyle=$ $\displaystyle
0,~{}~{}~{}i=0,1,2,3,$ $\displaystyle\alpha(v_{i}v_{j})$ $\displaystyle=$
$\displaystyle 0,~{}~{}~{}0\leq i<j\leq 3,$
$\displaystyle\alpha(v_{i}v_{j}v_{k})$ $\displaystyle=$ $\displaystyle
0,~{}~{}~{}0\leq i<j<k\leq 3,$ $\displaystyle\alpha(v_{0}v_{1}v_{2}v_{3})$
$\displaystyle=$
$\displaystyle(-1)^{0+1+2}f(v_{0},v_{1},v_{2})v_{3}+(-1)^{0+1+3}f(v_{0},v_{1},v_{3})v_{2}$
$\displaystyle+(-1)^{0+2+3}f(v_{0},v_{2},v_{3})v_{1}+(-1)^{1+2+3}f(v_{1},v_{2},v_{3})v_{0}$
$\displaystyle=$
$\displaystyle-f(v_{0},v_{1},v_{2})v_{3}+f(v_{0},v_{1},v_{3})v_{2}-f(v_{0},v_{2},v_{3})v_{1}$
$\displaystyle+f(v_{1},v_{2},v_{3})v_{0}.$

It follows that

$\displaystyle\dim{\rm
Im}\Big{(}\alpha:C_{3}(\Delta[V];\mathbb{R})\longrightarrow
C_{0}(\Delta[V];\mathbb{R})\Big{)}$ $\displaystyle=$
$\displaystyle\begin{cases}1,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k}),0\leq
i<j<k\leq 3,{\rm~{}are~{}not~{}all~{}zero};\\\
0,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k})=0{\rm~{}for~{}any~{}}0\leq i<j<k\leq
3\end{cases}$

or equivalently,

$\displaystyle\dim{\rm
Ker}\Big{(}\alpha:C_{3}(\Delta[V];\mathbb{R})\longrightarrow
C_{0}(\Delta[V];\mathbb{R})\Big{)}$ $\displaystyle=$
$\displaystyle\begin{cases}0,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k}),0\leq
i<j<k\leq 3,{\rm~{}are~{}not~{}all~{}zero};\\\
1,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k})=0{\rm~{}for~{}any~{}}0\leq i<j<k\leq
3.\end{cases}$

Consequently,

$\displaystyle
H_{0}(\Delta[V],\alpha,0)=\begin{cases}3,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k}),0\leq
i<j<k\leq 3,{\rm~{}are~{}not~{}all~{}zero};\\\
4,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k})=0{\rm~{}for~{}any~{}}0\leq i<j<k\leq
3\end{cases}$

and

$\displaystyle
H_{3}(\Delta[V],\alpha,0)=\begin{cases}0,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k}),0\leq
i<j<k\leq 3,{\rm~{}are~{}not~{}all~{}zero};\\\
1,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k})=0{\rm~{}for~{}any~{}}0\leq i<j<k\leq
3.\end{cases}$

By a similar calculation, we have

$\displaystyle
H_{1}(\Delta[V],\alpha,0)=\mathbb{R}^{6},~{}~{}~{}~{}~{}~{}H_{2}(\Delta[V],\alpha,0)=\mathbb{R}^{4}.$

Moreover,

$\displaystyle H_{n}(\Delta[V],\alpha,0)=0$

for any $n\neq 0,1,2,3$.

  * •

It is direct that

$\displaystyle\beta\circ\alpha(v_{i})=0$

for any $0\leq i\leq 3$,

$\displaystyle\beta\circ\alpha(v_{i}v_{j})=0$

for any $0\leq i<j\leq 3$,

$\displaystyle\beta\circ\alpha(v_{i}v_{j}v_{k})=0$

for any $0\leq i<j<k\leq 3$, and

$\displaystyle\beta\circ\alpha(v_{0}v_{1}v_{2}v_{3})=0.$

Therefore, the induced homomorphism $\beta_{*}$ between the homology groups is
the zero map.

  * •

By a direct calculation,

$\displaystyle\omega(v_{0})$ $\displaystyle=$
$\displaystyle-f(v_{1},v_{2},v_{3})dv_{1}\wedge dv_{2}\wedge dv_{3}(v_{0})$
$\displaystyle=$ $\displaystyle f(v_{1},v_{2},v_{3})v_{0}v_{1}v_{2}v_{3},$
$\displaystyle\omega(v_{1})$ $\displaystyle=$
$\displaystyle-f(v_{0},v_{2},v_{3})dv_{0}\wedge dv_{2}\wedge dv_{3}(v_{1})$
$\displaystyle=$ $\displaystyle-f(v_{0},v_{2},v_{3})v_{0}v_{1}v_{2}v_{3},$
$\displaystyle\omega(v_{2})$ $\displaystyle=$
$\displaystyle-f(v_{0},v_{1},v_{3})dv_{0}\wedge dv_{1}\wedge dv_{3}(v_{2})$
$\displaystyle=$ $\displaystyle f(v_{0},v_{1},v_{3})v_{0}v_{1}v_{2}v_{3},$
$\displaystyle\omega(v_{3})$ $\displaystyle=$
$\displaystyle-f(v_{0},v_{1},v_{2})dv_{0}\wedge dv_{1}\wedge dv_{2}(v_{3})$
$\displaystyle=$ $\displaystyle-f(v_{0},v_{1},v_{2})v_{0}v_{1}v_{2}v_{3},$
$\displaystyle\omega(v_{i}v_{j})$ $\displaystyle=$ $\displaystyle
0,~{}~{}~{}0\leq i<j\leq 3,$ $\displaystyle\omega(v_{i}v_{j}v_{k})$
$\displaystyle=$ $\displaystyle 0,~{}~{}~{}0\leq i<j<k\leq 3,$
$\displaystyle\omega(v_{0}v_{1}v_{2}v_{3})$ $\displaystyle=$ $\displaystyle
0.$

It follows that

$\displaystyle\dim{\rm
Im}\Big{(}\omega:C_{0}(\Delta[V];\mathbb{R})\longrightarrow
C_{3}(\Delta[V];\mathbb{R})\Big{)}$ $\displaystyle=$
$\displaystyle\begin{cases}1,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k}),0\leq
i<j<k\leq 3,{\rm~{}are~{}not~{}all~{}zero};\\\
0,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k})=0{\rm~{}for~{}any~{}}0\leq i<j<k\leq
3\end{cases}$

or equivalently,

$\displaystyle\dim{\rm
Ker}\Big{(}\omega:C_{0}(\Delta[V];\mathbb{R})\longrightarrow
C_{3}(\Delta[V];\mathbb{R})\Big{)}$ $\displaystyle=$
$\displaystyle\begin{cases}3,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k}),0\leq
i<j<k\leq 3,{\rm~{}are~{}not~{}all~{}zero};\\\
4,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k})=0{\rm~{}for~{}any~{}}0\leq i<j<k\leq
3.\end{cases}$

Consequently,

$\displaystyle
H^{0}(\Delta[V],\omega,0)=\begin{cases}3,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k}),0\leq
i<j<k\leq 3,{\rm~{}are~{}not~{}all~{}zero};\\\
4,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k})=0{\rm~{}for~{}any~{}}0\leq i<j<k\leq
3\end{cases}$

and

$\displaystyle
H^{3}(\Delta[V],\omega,0)=\begin{cases}0,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k}),0\leq
i<j<k\leq 3,{\rm~{}are~{}not~{}all~{}zero};\\\
1,&{\rm~{}if~{}}f(v_{i},v_{j},v_{k})=0{\rm~{}for~{}any~{}}0\leq i<j<k\leq
3.\end{cases}$

By a similar calculation, we have

$\displaystyle
H^{1}(\Delta[V],\omega,0)=\mathbb{R}^{6},~{}~{}~{}~{}~{}~{}H^{2}(\Delta[V],\omega,0)=\mathbb{R}^{4}.$

Moreover,

$\displaystyle H^{n}(\Delta[V],\omega,0)=0$

for any $n\neq 0,1,2,3$.

  * •

It is direct to see that the induced homomorphism $\mu_{*}$ between the
cohomology groups is the zero map.

## Acknowledgement

The author would like to express his deep gratitude to the referee for the
careful reading of the manuscript.

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Shiquan Ren

Address: School of Mathematics and Statistics, Henan University, Kaifeng
475004, China.

E-mail<EMAIL_ADDRESS>