Weighted product of point clouds and simplicial complexes

Abstract

This paper extends the concept of weighted point clouds and weighted simplicial complexes by introducing product point clouds and product simplicial complexes within a commutative ring with unity. Within an integral domain, the introduction of a weighted product chain group, along with the induced product weighted homomorphism and weighted product boundary maps, leads to significant outcomes and findings. To explore the algebraic characteristics of a weighted product structure, we introduce the concept of weighted product homology. This homology considers the relationship of weights assigned to elements within the structure and their impact on the structure’s underlying algebraic properties.

1 Introduction

Even after investing decades of effort into unveiling the inherent features like shape and structure within data sets, it becomes evident that noteworthy advancements were predominantly accomplished through the development of the potent mathematical tool known as topological data analysis. Since the approach of homeomorphism requires a more discriminating tool for identifying objects more precisely, a more effective technique called homology was developed in topological data analysis with the help of algebraic tools such as groups, rings, etc. One of the main benefits of homology is its ability to classify objects by identifying higher-dimensional holes and connected components. Regarding homology, $0-$ dimensional holes can be analogous to connected components, $1-$ dimensional holes can be thought of as tunnels, and $2-$dimensional holes can be compared to cavities. Furthermore, it provides a bridge between geometry and topology, allowing for the study of continuous transformations and deformations while maintaining a focus on essential topological properties. Apart from the comparative study of distinct spaces, homology serves practical purposes in a range of diverse domains, including image processing [6, 8, 21] and digital image analysis [1, 3, 10]. This encompasses activities such as identifying components within circuit diagrams [9, 15], detecting tumors in X-ray images [13], as well as applications in geoscience [22, 23] and fluid dynamics [11, 12, 14].

Simplicial complexes serve as the primary data structure for representing topological spaces, constituting one of the numerous approaches to representing a topological space by separating it down into basic components [5, 24, 26]. In addition, detecting holes in objects using homology through the simplicial approach is characterized by its simplicity and efficiency, owing to its computational ease and effectiveness in managing data storage efficiently. Betti numbers are numerical invariants associated with these spaces that provide information about their connectivity as well as the presence of holes, voids, or high-dimensional features [18]. Due to the limitations of unweighted simplicial complexes in representing higher-order interactions, the use of weighted simplicial complexes enables a remedy. Specifically, weighted simplicial complexes possess the capability to illustrate higher-order networks without any loss of information, facilitating the simultaneous depiction of both the weighted topology of data and its relationships. The exploration of weighted homology within simplicial complexes was initially undertaken by Robert J. Dawson [4], followed by a broader extension of this concept by Ren et al. [19, 20]. By introducing appropriate weights, weighted homology demonstrates the capacity to differentiate between simplicial complexes that exhibit homotopy equivalence. Also when all weights are equal and non-zero, the weighted homology maintains an isomorphic correspondence with the standard simplicial homology. In the study of Wu et al., they applied Forman’s discrete Morse theory within the framework of weighted homology [25]. Their work involves the development of weighted extensions of standard theorems from discrete Morse theory.

In Dowson’s study on weighted simplicial complexes, a significant advancement was made by introducing weight values that are restricted to nonnegative integers [4]. His study went beyond the mere introduction of weighted simplicial complexes; it delved deeply into the categorical properties of these structures. This categorical analysis laid the groundwork for uncovering significant insights into the interplay between weights and topology, enabling us to grasp the intricate ways in which different weights impact the structure of the complex. The serious consideration of categorical properties in Dowson’s work not only established a solid theoretical basis for weighted simplicial homology but also opened avenues for practical applications in various domains. The study conducted by Ren et al. [19] represents a significant extension and generalization of the concepts surrounding weighted simplicial complexes and their associated homology. Unlike previous work that restricted weights to nonnegative integers, Ren et al. [19] expanded the scope by considering weights from an integral domain R, thereby introducing a broader framework for studying the topological properties of weighted simplicial complexes. This generalization to integral domains R allows for a more flexible and versatile representation of weights. Integral domains encompass a wide range of algebraic structures, including familiar examples like the integers and various polynomial rings [17].

The Cartesian product of point clouds involves a fundamental mathematical operation where each point in one cloud is paired with every point in another cloud, resulting in a new space of combined points. When considering weighted simplicial complexes in this scenario, one can extend the integral domain-based weight assignment to this product construction. This presents a novel and powerful means of associating weights to the combined simplices, potentially yielding deeper insights into the structural relationships between the original point clouds. This paper is poised to contribute significantly to the understanding of weighted simplicial complexes and their behavior under Cartesian product operations, particularly when weights are drawn from an integral domain. The results of this study have the potential to yield profound insights into the topological characteristics of complex data sets and point clouds, offering a valuable resource for everyone. Also, we can find weighted product homology emerges within the Cartesian product by introducing the concept of a weighted product simplicial map and a weighted product boundary operator.

Section 2 of the paper contains the preliminary definitions, results, theorems, and notations. Section 3 of the paper includes several significant outcomes, which form the core contributions of the study. This section delves into the exploration of weighted product point clouds, weighted product simplicial complexes, weighted product homology, and their associated results. To illustrate and substantiate these concepts, some of examples are provided, offering practical insights into the application of these ideas. Our study extends beyond the analysis of Cartesian products between two point clouds to a broader context involving finitely many point clouds.

2 Preliminaries

Definition 1

[5, 7, 24] A simplex is a generalization of the notion of a triangle or tetrahedron to arbitrary dimensions. Specifically, a s simplex is a s -dimensional polytope which is the convex hull of $s+1$ affinely independent points.

Definition 2

[5, 7, 24] A simplicial complex S is a collection of simplices such that

• 1. If S contains a simplex $s_{1}$, then S also contains every face of $s_{1}$

• 2. If two simplices in S intersect,then their intersection is a face of each of them.

Definition 3

[5, 7, 24] Let S be a simplicial complex and n a dimension. A $n-$ chain is a formal sum of $n-$simplices in S. The standard notation is $c=\sum a_{i}\sigma _{i}$, where the $\sigma _{i}$ are the $p-$ simplices and the a_i are the integer coefficients. The set of all n-chains form a group $C_{n}$ under addition.

Definition 4

[5, 7, 24] The the boundary of a n-simplex is the sum of its $(n-1)$-dimensional faces. If $\sigma = [u_{0}, u_{1},..., u_{n}]$ for the simplex spanned by the listed vertices, then its boundary is $\partial _{n}\sigma =\sum _{j=0}^{n} (-1)^{j}[u_{0},..., \hat{u_{j}},..., u_{p}]$, where the hat indicates that $u_{j}$ is omitted

Definition 5

[5, 7, 24] A $n-$ cycle c is a $n-$chain with empty boundary, $\partial _{n} c = 0$. And the set of all n- cycles will form a subgroup of chain group $C_{n}$.

Definition 6

[5, 7, 24] A n-boundary b is a n-chain that is the boundary of a $(n + 1)$-chain, $b = \partial d$ with $d \in C_{n+1}$

Definition 7

[5, 7, 24] Let $C_{n}$ be a chain group whose elements are the N chains and $\partial _{n}:C_{n}\rightarrow C_{n-1}$ maps each n-chain to the sum of the $(n-1)$dimensional faces of its n cells which is a $(n-1)$ chain.

Writing the groups and maps in sequence, we get the chain complex

$$\begin{aligned} \cdots \xrightarrow []{\partial _{p+2}} C_{p+1} \xrightarrow []{\partial _{p+1}} C_{p} \xrightarrow []{\partial _{p}} C_{p-1}\xrightarrow []{\partial _{p-1}}\cdots \end{aligned}$$

Then the n-th homology group is defined as

$$\begin{aligned} H_{n}={Ker(\partial _{n})}/{Im(\partial _{n+1})} \end{aligned}$$

Definition 8

[2, 16] Category, $\mathcal {C}$, is formed by two sorts of objects, the objects of the category $\mathcal {C}_{0}$ and for any $\mathcal {X}_{1},\mathcal {X}_{2}\in \mathcal {C}_{0}$, set of morphisms, $\mathcal {C}(\mathcal {X}_{1},\mathcal {X}_{2})$, which related to two objects called source and target of the morphism. For any $\mathcal {X}_{1},\mathcal {X}_{2},\mathcal {X}_{3} \in \mathcal {C}_{0}$ the mapping of sets

$$\begin{aligned} \mathcal {C}(\mathcal {X}_{2},\mathcal {X}_{3})\times \mathcal {C}(\mathcal {X}_{1},\mathcal {X}_{1})\rightarrow \mathcal {C}(\mathcal {X}_{1},\mathcal {X}_{3}) \end{aligned}$$

is called composition and composition should be associative. Also for all $\mathcal {W} \in \mathcal {C}_{0}$ there exist an identity morphism $Id_{\mathcal {W}}: \mathcal {W}\rightarrow \mathcal {W}$ satisfying $Id_{\mathcal {W}}\mathcal {F}=\mathcal {F}$ and $\mathcal {G}Id_{\mathcal {W}}=\mathcal {G}$ for all morphisms $\mathcal {F}:{\mathcal {U}}\rightarrow \mathcal {W}$ and $\mathcal {G}:{\mathcal {W}}\rightarrow \mathcal {V}$.

Definition 9

[7] Let S and $S'$ be simplicial complexes. Then the Cartesian product $S\times S'$ is a simplicial complex such that

• 1. $F_{0}\left( S\times S'\right)$ is the Cartesian product $F_{0}(S) \times F_{0}\left( S'\right)$

• 2. $\sigma \in F_{k}\left( S\times S'\right)$ if and only if $\pi _{S}(\sigma ) \in F_{k}(S)$ and $\pi _{S}(\sigma ) \in F_{k}\left( S'\right) ,$ where for $\sigma = \left( \left( v_{1}, v_{1}'\right) ,\ldots ,\left( v_{n},v_{n}'\right) \right) \in S\times S',~ \pi _{S}: S\times S' \rightarrow S$ which maps $\sigma \rightarrow (v_{1}, \ldots , v_{n})$ and $~\pi _{S'}: S\times S' \rightarrow S'$ which maps $\sigma \rightarrow \left( v_{1}',\ldots ,v_{n}'\right)$ where $F_{i}(S) = \{\sigma \in S: dim(\sigma ) = i\}.$

Remark 1

It is vital to understand the differences between the simplicial complexes join and the Cartesian product. The simplicial complex resulting from the Cartesian product of two simplicial complexes is formed by taking the Cartesian product of their vertex sets and subsequently constructing simplices through the Cartesian product of the simplices within each respective simplicial complex. Indeed, a simplicial complex join is formed by combining the vertex sets of the two complexes via a disjoint union and creating connections between each simplex in the first complex and each simplex in the second.

Definition 10

[19] let $\mathcal {R}$ to be a commutative ring with multiplicative identity. A nonzero element $a\in \mathcal {R}$ is said to divide an element $b\in \mathcal {R}$ (denoted a|b) if there exists $x\in \mathcal {R}$ such that $ax = b$. A nonzero element a in a ring $\mathcal {R}$ is called a zero divisor if there exists a nonzero $x \in \mathcal {R}$ such that $ax = 0.$

A commutative ring $\mathcal {R}$ with $1_{R}\ne 0$ and no zero divisors is called an integral domain.

Proposition 1

[19] Let $\mathcal {R}$ be an integral domain. Let Sbe the set of all nonzero elements in $\mathcal {R}$.Then the map $\phi _{s}: \mathcal {R} \rightarrow S^{-1}\mathcal {R}$ from $\mathcal {R}$ with to the quotient field $S^{-1}\mathcal {R}$, given by $r\rightarrow rs/s$ (for any $s\in S$) is a monomorphism. Hence, the integral domain $\mathcal {R}$ can be embedded in its quotient field.

Definition 11

[19] Let n be a positive integer. The point cloud data $\mathcal {X}$ in $\mathbb {R}^{n}$ is a finite subset of $\mathbb {R}^{n}$. Given some point cloud data $\mathcal {X},$ a weight on $\mathcal {X}$ is a function $\mathfrak {w}_{0}: \mathcal {X} \rightarrow \mathcal {R}$, where $\mathcal {R}$ is a commutative ring. The pair $(\mathcal {X}, \mathfrak {w}_{0})$ is called weighted point cloud data, or WPCD for short.

Definition 12

[19] A weighted simplicial complex (or WSC for short) is a pair $(K, \mathfrak {w})$ consisting of a simplicial complex K and a function $\mathfrak {w}: K \rightarrow \mathcal {R}$, where$\mathcal {R}$ is a commutative ring, such that for any $\sigma _{1}, \sigma _{2}\in K$ with $\sigma _{1} \subseteq \sigma _{2}$, we have $\mathfrak {w}(\sigma _{1}) |\mathfrak {w}(\sigma _{2})$.

Definition 13

[19] Let $(\mathcal {X}, \mathfrak {w}_{0})$ be a weighted point cloud data,with weight function $\mathfrak {w}_{0}: \mathcal {X} \rightarrow \mathcal {R}$. Let K be a simplicial complex whose set of vertices is $\mathcal {X}$. We define a weight function $\mathfrak {w}_{k}: K \rightarrow \mathcal {R}$ by $\mathfrak {w}_{k}(\sigma )=\prod _{1=0} ^{k} \mathfrak {w}_{0}(vi).$ where $\sigma = [v_{0}, v_{1},..., v_{k}]$ is a $k-$ simplex of K. We call $\mathfrak {w},$ defined as such, the product weighting.

3 Results and discussion

The core of this section are the weighted product point clouds, where the integral weights from an integral domain enriches the structure of the Cartesian product of point clouds. The paper defines and characterizes this novel construct, showing how integral domain-based weights influence the integrated point clouds and contribute to their topological representation.

Definition 14

Let $(\mathcal {X},\mathfrak {w}_{{x}_{0}})$ be a weighted point cloud data in $\mathbb {R}^{m_{1}}$ with weight function $\mathfrak {w}_{{x}_{0}}:\mathcal {X}\rightarrow \mathcal {R}$ and $(\mathcal {Y},\mathfrak {w}_{{y}_{0}})$ be a weighted point cloud data in $\mathbb {R}^{m_{2}}$ with weight function $\mathfrak {w}_{{y}_{0}}:\mathcal {Y}\rightarrow \mathcal {R}$ where $m_{1},m_{2}\in \mathbb {Z^{+}}$ and $\mathcal {R}$ be a commutative ring with unity. Then the pair $(\mathcal {X}\times \mathcal {Y},(\mathfrak {w}_{x\times y})_{0})$ is called weighted product point cloud in $\mathbb {R}^{m_{1}+m_{2}}$ where $(\mathfrak {w}_{x}\times \mathfrak {w}_{y})_{0}$ is a function from $\mathcal {X}\times \mathcal {Y}\rightarrow \mathcal {R}$ defined by

$$\begin{aligned} (\mathfrak {w}_{x\times y})_{0}(v_{x},v_{y})=\mathfrak {w}_{{x}_{0}}(v_{x}).\mathfrak {w}_{{y}_{0}}(v_{y}) \end{aligned}$$

for all $(v_{x},v_{y})\in \mathbb {R}^{m_{1}+m_{2}}$

Definition 15

Let $(\mathcal {X}_{i},\mathfrak {w}_{{x_{i}}_{0}})$ be weighted point cloud in $\mathbb {R}^{n_{i}}$ with weight function $\mathfrak {w}_{{x_{i}}_{0}}:\mathcal {X}_{i}\rightarrow \mathcal {R}$ for $i=1,2,\ldots , m$ where $n_{1},n_{2},\ldots , n_{m}$ are positive integers and $\mathcal {R}$ be a commutative ring with unity. Then $(\mathcal {X}_{1}\times \mathcal {X}_{2}\times \ldots \times \mathcal {X}_{m},(\mathfrak {w}_{x_{1}\times x_{2}\times \cdots \times x_{m}})_{0})$is called weighted m product point cloud data in $\mathbb {R}^{n_{1}+n_{2}+\ldots n_{m}},$ where

$$\begin{aligned} (\mathfrak {w}_{x_{1}\times x_{2}\times \cdots \times x_{m}})_{0}(v_{x_{1}},v_{x_{2}}\ldots v_{x_{m}})=\displaystyle \prod _{i=1}^{m}\mathfrak {w}_{x_{i_{0}}}(v_{x_{i}}) \end{aligned}$$

for all $(v_{x_{1}},v_{x_{2}},\ldots , v_{x_{m}})\in \mathbb {R}^{n_{1}+n_{2}+\ldots +n_{m}}$

Definition 16

Let $\left( K_{1},\mathfrak {w}_{k_{1}}\right) \left( K_{2},\mathfrak {w}_{k_{2}}\right)$ are weighted simplicial complexes with weight functions $\mathfrak {w}_{k_{1}}:K_{1}\rightarrow {\mathcal {R}}$ and $\mathfrak {w}_{k_{2}}:K_{2}\rightarrow {\mathcal {R}}$, $\mathcal {R}$ be a commutative ring with unity. A weighted product simplicial complex is a pair $\left( K_{1}\times K_{2}, \mathfrak {w}_{k_{1}\times k_{2}}\right)$ of product of simplicial complexes $K_{1}$ and $K_{2}$ with a weight function $\mathfrak {w}_{k_{1}\times k_{2}}:K_{1}\times K_{2}\rightarrow \mathcal {R}$ such that for any simplices $\sigma _{k_{1}\times k_{2}}$ and ${\sigma '_{k_{1}\times k_{2}}} \in K_{1}\times K_{2}$ with $\sigma _{k_{1}\times k_{2}}\subseteq {\sigma '_{k_{1}\times k_{2}}}$, $\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) /\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma '_{k_{1}\times k_{2}}\right) .$

Definition 17

Let $\left( K_{1},\mathfrak {w}_{k_{1}}\right) \left( K_{2},\mathfrak {w}_{k_{2}}\right) \ldots \left( K_{m},\mathfrak {w}_{k_{m}}\right)$ are weighted simplicial complexes and $\mathcal {R}$ be a commutative ring with unity. A weighted m product simplicial complex is a pair $(K_{1}\times K_{2}\times \ldots \times K_{m}, \mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}})$ with weight function $\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}:K_{1}\times K_{2}\times \ldots \times K_{m}\rightarrow \mathcal {R}$ such that for any $\sigma _{(k_{1}\times k_{2}\times \ldots \times k_{m})}$ and ${\sigma '_{(k_{1}\times k_{2}\times \ldots \times k_{m})}} \in K_{1}\times K_{2}\times \ldots \times K_{m}$ with $\sigma _{(k_{1}\times k_{2}\times \ldots \times k_{m})}\subseteq {\sigma '_{(k_{1}\times k_{2}\times \ldots \times k_{m})}}$, $\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}(\sigma _{(k_{1}\times k_{2}\times \ldots \times k_{m})}/\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}(\sigma '_{(k_{1}\times k_{2}\times \ldots \times k_{m}))}.$

The next two definitions are extending the Definitions 14 and 15 to simplices with higher dimensions, further extending the concept of weighted product point clouds and weighted m product point clouds. We also discuss some efficient results which are related to weighted product simplicial complexes and weighted m product simplicial complexes. The results concerning the preimage of product weight functions with values in ideals are significant in the following results. Additionally, finding a relation between the inverse image of weight functions for product weights within ideals and the inverse image of individual weight functions within ideals.

Definition 18

Let $\left( \mathcal {X},\mathfrak {w}_{{x}_{0}}\right)$ be a weighted point cloud data in $\mathbb {R}^{m_{1}}$ with weight function $\mathfrak {w}_{{x}_{0}}:\mathcal {X}\rightarrow \mathcal {R}$ and $\left( \mathcal {Y},\mathfrak {w}_{{y}_{0}}\right)$ be a weighted point cloud data in $\mathbb {R}^{m_{2}}$ with weight function $\mathfrak {w}_{{y}_{0}}:\mathcal {Y}\rightarrow \mathcal {R}$ where $m_{1},m_{2}\in \mathbb {Z^{+}}$ and $\mathcal {R}$ be a commutative ring with unity. Let $K_{1}$ be a simplicial complex of $\left( \mathcal {X},\mathfrak {w}_{{x}_{0}}\right)$ and $K_{2}$ be a simplicial complex of $(\mathcal {Y},\mathfrak {w}_{{y}_{0}})$. Consider the product $K_{1}\times K_{2}$ and the weight function $\mathfrak {w}_{k_{1}\times k_{2}}: K_{1}\times K_{2}\rightarrow \mathcal {R}$ defined by

$$\begin{aligned} \mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})=\displaystyle \prod _{i=0}^{l} \mathfrak {w}_{x_{0}}(v_{x_{i}}).\mathfrak {w}_{y_{0}}(v_{y_{i}}) \end{aligned}$$

where $\sigma _{k_{1}\times k_{2}}=[(v_{x_{0}},v_{y_{0}}),(v_{x_{1}},v_{y_{1}}),\ldots ,(v_{x_{l}},v_{y_{l}})]$ be a l simplex. Then $\mathfrak {w}_{k_{1}\times k_{2}}$ is called the product weighing of product of two simplicial complexes.

Definition 19

Let $\left( \mathcal {X}_{i},\mathfrak {w}_{{x_{i}}_{0}}\right)$ are weighted point clouds with weight function $\mathfrak {w}_{{x_{i}}_{0}}:X_{i}\rightarrow \mathcal {R}$ for $i=1,2,\ldots ,m$ and $\mathcal {R}$ be a commutative ring with unity. Let $K_{i}$ be the simplicial complex whose vertices are from $\mathcal {X}_{i}$ for $i=1,2,\ldots ,m$. Consider the product $K_{1}\times K_{2}\times \ldots \times K_{m}$ with weight function $\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}:K_{1}\times K_{2}\times \ldots \times K_{m}\rightarrow \mathcal {R}$ defined by

$$\begin{aligned} \mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}(\sigma _{k_{1}\times k_{2}\times \ldots \times k_{m}})=\displaystyle \prod _{i=0}^{l}\mathfrak {w}_{{x_{1}}_{0}}(v_{x_{1}})_{i}.\mathfrak {w}_{{x_{2}}_{0}}(v_{x_{2}})_{i}\ldots \mathfrak {w}_{{x_{m}}_{0}}(v_{x_{m}})_{i}, \end{aligned}$$

where, $\sigma _{(k_{1}\times k_{2}\times \ldots \times k_{m}})=[(v_{x_{1_{0}}},v_{x_{2_{0}}},\ldots ,v_{x_{m_{0}}}),(v_{x_{1_{1}}},v_{x_{2_{1}}},\ldots ,v_{x_{m_{1}}}), \ldots ,(v_{x_{1_{l}}},v_{x_{2_{l}}},\ldots ,v_{x_{m_{l}}})]$. $\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}$ called product weighing of product of m simplicial complexes.

Result 1

For any l simplex $\sigma _{k_{1}\times k_{2}}$ in $K_{1}\times K_{2}$, we can find $l_{1}$ simplex $\sigma _{k_{1_{l_{1}}}}$ in $K_{1}$ and $l_{2}$ simplex $\sigma _{k_{2_{l_{2}}}}$ in $K_{2}$ where $l_{1}\le l$ and $l_{2}\le l$ such that $\mathfrak {w}_{k_{1}}\left( \sigma _{k_{1_{l_{1}}}}\right) /\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right)$ and $\mathfrak {w}_{k_{2}}\left( \sigma _{k_{2_{l_{2}}}}\right) /\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right)$.

Proof

Any l simplex in $K_{1}\times K_{2}$ can be written as

$$\begin{aligned} \sigma _{k_{1}\times k_{2}}=[(v_{x_{0}},v_{y_{0}}),(v_{x_{1}},v_{y_{1}}),\ldots ,(v_{x_{l}},v_{y_{l}})]. \end{aligned}$$

Here $\sigma _{k_{1}} =[v_{x_{0}},v_{x_{1}},\ldots v_{x_{l}}]$ be $l_{1}$ simplex in $K_{1}$ for some $l_{1}\le l$ and $\sigma _{k_{2}}=[v_{y_{0}},v_{y_{1}},\ldots v_{y_{l}}]$ be $l_{2}$ simplex in $K_{2}$ for some $l_{2}\le l$.

Without loss of generality take

$$\begin{aligned} \sigma _{k_{1}}=[v_{x_{0}},v_{x_{1}},\ldots ,v_{x_{l_{1}}},\ldots , v_{x_{l}}] \end{aligned}$$

where $v_{x_{0}},v_{x_{1}},\ldots ,v_{x_{l_{1}}}$ are distinct vertices and $v_{x_{l_{1}+1}},v_{x_{l_{1}+2}},\ldots ,v_{x_{l}}$ are repeating vertices in $K_{1}$ and

$$\begin{aligned} \sigma _{k_{2}}=[v_{y_{0}},v_{y_{1}},\ldots ,v_{y_{l_{2}}},\ldots , v_{y_{l}}] \end{aligned}$$

where $v_{y_{0}},v_{y_{1}},\ldots ,v_{y_{l_{2}}}$ are distinct vertices and $v_{y_{l_{2}+1}},v_{y_{l_{2}+2}},\ldots ,v_{y_{l}}$ are repeating vertices in $K_{2}$. Let $\sigma _{k_{1_{l_{1}}}}=[v_{x_{0}},v_{x_{1}},\ldots ,v_{x_{l_{1}}}]$ and $\sigma _{k_{2_{l_{2}}}}=[v_{y_{0}},v_{y_{1}},\ldots ,v_{y_{l_{2}}}]$. Then,

$$\begin{aligned} \mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right)&= \displaystyle \prod _{i=0}^{l} \mathfrak {w}_{x_{0}}(v_{x_{i}}) .\mathfrak {w}_{y_{0}}(v_{y_{i}})\\&= \mathfrak {w}_{k_{1}}\left( \sigma _{k_{1_{l_{1}}}}\right) \displaystyle \prod _{i=l_{1}+1}^{l} \mathfrak {w}_{x_{0}}(v_{x_{i}}).\mathfrak {w}_{k_{2}}\left( \sigma _{k_{2_{l_{2}}}}\right) \displaystyle \prod _{i=l_{2}+1}^{l} \mathfrak {w}_{y_{0}}(v_{y_{i}}) \end{aligned}$$

Hence, $\mathfrak {w}_{k_{1}}\left( \sigma _{k_{1_{l_{1}}}}\right) /\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right)$ and $\mathfrak {w}_{k_{2}}\left( \sigma _{k_{2_{l_{2}}}}\right) /\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right)$. $\square$

Corollary 1

For any l simplex $\sigma _{k_{1}\times k_{2}\times \ldots \times k_{m}}$ in $K_{1}\times K_{2}\times \ldots \times K_{m}$, we can find $l_{i}$ simplex $\sigma _{k_{i_{l_{i}}}}$ in $K_{i}$ where $l_{i}\le l \text { for } i=1,2,\ldots ,m$ such that $\mathfrak {w}_{k_{i}}\left( \sigma _{k_{i_{l_{i}}}}\right) /\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}\left( \sigma _{k_{1}\times k_{2}\times \ldots \times k_{m}}\right)$ for all $i=1,2,\ldots ,m$.

Result 2

Let $\sigma _{k_{1}\times k_{2}}=[(v_{x_{0}},v_{y_{0}}),(v_{x_{1}},v_{y_{1}}),\ldots ,(v_{x_{l}},v_{y_{l}})]$ be a $l-$ simplex in $K_{1}\times K_{2}.$ If $\sigma _{k_{1}}=[v_{x_{0}},v_{x_{1}},\ldots ,v_{x_{l}}]$ be a $l-$ simplex in $K_{1}$ and $\sigma _{k_{2}}=[v_{y_{0}},v_{y_{1}},\ldots ,v_{y_{l}}]$ be a $l-$ simplex in $K_{2}$ then

$$\begin{aligned} \mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) =\mathfrak {w}_{k_{1}}\left( \sigma _{k_{1}}\right) .\mathfrak {w}_{k_{2}}\left( \sigma _{k_{2}}\right) . \end{aligned}$$

Proof

$\sigma _{k_{1}}$ is a $l-$ simplex in $K_{1}$ implies all $v_{x_{i}} \text { for } i=0,1,\ldots ,l$ are distinct vertices of $K_{1}$ and $\sigma _{k_{2}}$ is a $l-$ simplex in $K_{2}$ implies all $v_{y_{i}} \text { for } i=0,1,\ldots ,l$ are distinct vertices of $K_{2}$. Hence

$$\begin{aligned} \mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right)&=\displaystyle \prod _{i=0}^{l} \mathfrak {w}_{x_{0}}(v_{x_{i}})\displaystyle \prod _{i=0}^{l} \mathfrak {w}_{y_{0}}(v_{y_{i}}) \\&=\mathfrak {w}_{k_{1}}\left( \sigma _{k_{1}}\right) .\mathfrak {w}_{k_{2}}\left( \sigma _{k_{2}}\right) . \end{aligned}$$

$\square$

Proposition 2

Let $\left( \mathcal {X}\times \mathcal {Y},(\mathfrak {w}_{x\times y})_{0}\right)$ be a weighted product point cloud data. Also let $\mathfrak {w}_{k_{1}\times k_{2}}$ be the product weighing defined on the product $K_{1}\times K_{2}$ of weighted simplices $K_{1}$ and $K_{2}$. Then

• (i) The restriction of $\mathfrak {w}_{k_{1}\times k_{2}}$ to the vertices of $K_{1}\times K_{2}$ is $(\mathfrak {w}_{x\times y})_{0}$

• (ii) For any $\sigma _{k_{1}\times k_{2}}\subseteq {\sigma '_{k_{1}\times k_{2}}}$, $\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) /\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma '_{k_{1}\times k_{2}}\right)$

Proof

(i) We have the vertex set of $K_{1}\times K_{2}$ will be of the form $\left( v_{x_{i}},v_{y_{i}}\right)$ for some positive integer i. Then from the definition of product weighing

$$\begin{aligned} \mathfrak {w}_{k_{1}\times k_{2}}\left( v_{x_{i}},v_{y_{i}}\right)&=w_{x_{0}}\left( v_{x_{i}}\right) .w_{y_{0}}\left( v_{y_{i}}\right) \\&= (\mathfrak {w}_{x\times y})_{0}\left( v_{x_{i}},v_{y_{i}}\right) . \end{aligned}$$

(ii) Let $\sigma _{k_{1}\times k_{2}}=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\left( v_{x_{1}},v_{y_{1}}\right) ,\ldots ,\left( v_{x_{l}},v_{y_{l}}\right) \right]$ be l simplex, and

$\sigma '_{k_{1}\times k_{2}}=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\left( v_{x_{1}},v_{y_{1}}\right) , \ldots ,\left( v_{x_{p}},v_{y_{p}}\right) \right]$ be a p simplex with $\sigma _{k_{1}\times k_{2}}\subseteq \sigma '_{k_{1}\times k_{2}}$.

Clearly $l\le p$, and

$$\begin{aligned} \mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma '_{k_{1}\times k_{2}}\right)&= \displaystyle \prod _{i=0}^{p} w_{x_{0}}\left( v_{x_{i}}\right) w_{y_{0}}\left( v_{y_{i}}\right) \\&= \displaystyle \prod _{i=0}^{l} w_{x_{0}}\left( v_{x_{i}}\right) w_{y_{0}}\left( v_{y_{i}}\right) .\displaystyle \prod _{i=l+1}^{p} w_{x_{0}}\left( v_{x_{i}}\right) w_{y_{0}}\left( v_{y_{i}}\right) \\&=\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) .\displaystyle \prod _{i=l+1}^{p} w_{x_{0}}\left( v_{x_{i}}\right) w_{y_{0}}\left( v_{y_{i}}\right) \end{aligned}$$

since $\displaystyle \prod _{i=l+1}^{p} w_{x_{0}}\left( v_{x_{i}}\right) w_{y_{0}}\left( v_{y_{i}}\right) \in \mathcal {R}$, $\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) /\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma '_{k_{1}\times k_{2}}\right) .$ $\square$

Corollary 2

Let $\left( \mathcal {X}_{1}\times \mathcal {X}_{2}\times \ldots \times \mathcal {X}_{m},(\mathfrak {w}_{x_{1}\times x_{2}\times \ldots \times x_{m}})_{0}\right)$ be a weighted m product point cloud in $\mathbb {R}^{n}$. Also let $\mathfrak {w}_{k_{1}\times k_{2}\times \ldots \times k_{m}}$ be the product weighing defined on the product $K_{1}\times K_{2}\times \ldots \times K_{m}$. Then,

• (i) The restriction of $\mathfrak {w}_{k_{1}\times k_{2}\times \ldots \times k_{m}}$ to the vertices of $K_{1}\times K_{2}\times \ldots \times K_{m}$ is $\left( \mathfrak {w}_{x_{1}\times x_{2}\times \ldots \times x_{m}}\right) _{0}$

• (ii) For any $\sigma _{k_{1}\times k_{2}\times \ldots \times k_{m}}\subseteq {\sigma '_{k_{1}\times k_{2}\times \ldots \times k_{m}}}$,

  $$\begin{aligned} \mathfrak {w}_{k_{1}\times k_{2}\times \ldots \times k_{m}}\left( \sigma _{k_{1}\times k_{2}\times \ldots \times k_{m}}\right) /\mathfrak {w}_{k_{1}\times k_{2}\times \ldots \times k_{m}}\left( \sigma '_{k_{1}\times k_{2}\times \ldots \times k_{m}}\right) \end{aligned}$$

Lemma 1

Let $\mathfrak {I}$ be an ideal of a commutative ring $\mathcal {R}$ and $\left( K_{1}\times K_{2},\mathfrak {w}_{k_{1}\times k_{2}}\right)$ be weighted product simplicial complex. Let $\left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I})$ denote the preimage of $\mathfrak {I}$ under $\left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}.$ If $\sigma _{k_{1}\times k_{2}}\in \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I})$, then, for all simplices $\tau _{k_{1}\times k_{2}}$ containing $\sigma _{k_{1}\times k_{2}}$ we have $\tau _{k_{1}\times k_{2}}\in \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I})$.

Proof

Let $\mathfrak {I}$ be an ideal of a commutative ring $\mathcal {R}$. Assume $\sigma _{k_{1}\times k_{2}}\in (\mathfrak {w}_{k_{1}\times k_{2}})^{-1}(\mathfrak {I})$ then $\mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})\in \mathfrak {I}$. Since $\sigma _{k_{1}\times k_{2}}\subseteq \tau _{k_{1}\times k_{2}}$, from the above proposition

$$\begin{aligned} \mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) /\mathfrak {w}_{k_{1}\times k_{2}}\left( \tau _{k_{1}\times k_{2}}\right) . \end{aligned}$$

So,

$$\begin{aligned} \mathfrak {w}_{k_{1}\times k_{2}}\left( \tau _{k_{1}\times k_{2}}\right) = r~ \mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \end{aligned}$$

for some $r\in \mathcal {R}$. That is,

$$\begin{aligned} \tau _{k_{1}\times k_{2}}\in \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I}) \end{aligned}$$

. $\square$

Corollary 3

Let $\mathfrak {J}$ be an ideal of a commutative ring $\mathcal {R}$ and $\left( K_{1}\times K_{2}\times \ldots \times K_{m},\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}\right)$ be weighted m product simplicial complex. Let $\left( \mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}\right) ^{-1}(\mathfrak {J})$ denote the preimage of $\mathfrak {J}$ under $\left( \mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}\right) ^{-1}$. If $\sigma _{k_{1}\times k_{2}\times \ldots \times k_{m}}\in \left( \mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}\right) ^{-1}(\mathfrak {J})$, the for all simplices $\tau _{k_{1}\times k_{2}\times \ldots \times K_{m}}$ containing $\sigma _{k_{1}\times k_{2}\times \ldots \times k_{m}}$ we have $\tau _{k_{1}\times k_{2}\times \ldots \times k_{m}}\in \left( \mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}\right) ^{-1}(\mathfrak {J})$.

Theorem 3

Let $\mathfrak {P}$ be a prime ideal of a commutative ring $\mathcal {R}$. Then for any l simplex $\sigma _{k_{1}\times k_{2}}=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\left( v_{x_{1}},v_{y_{1}}\right) ,\ldots ,\left( v_{x_{l}},v_{y_{l}}\right) \right]$ in $\left( K_{1}\times K_{2}, \mathfrak {w}_{k_{1}\times k_{2}}\right)$ with $\sigma _{k_{1}\times k_{2}}\in \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {P})$, then there will be atleast one vertex $v_{x_{i}}$ or $v_{y_{i}}$ such that either $v_{x_{i}}\in \left( \mathfrak {w}_{k_{1}}\right) ^{-1}\left( \mathfrak {P}\right)$ or $v_{y_{i}}\in \left( \mathfrak {w}_{k_{2}}\right) ^{-1}(\mathfrak {P})$ for some $i=0,1,2,\ldots ,l$.

Proof

Let $\sigma _{k_{1}\times k_{2}}=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\left( v_{x_{1}},v_{y_{1}}\right) ,\ldots ,\left( v_{x_{l}},v_{y_{l}}\right) \right]$ be a l simplex in $\left( K_{1}\times K_{2},\mathfrak {w}_{k_{1}\times k_{2}}\right)$. If $\sigma _{k_{1}\times k_{2}}\in \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {P})$ then $\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \in \mathfrak {P}$.

$$\begin{aligned}&\implies \displaystyle \prod _{i=0}^{p} w_{x_{0}}(v_{x_{i}})w_{y_{0}}(v_{y_{i}})\in \mathfrak {P} \\&\implies \text {~either~} \displaystyle \prod _{i=0}^{p} w_{x_{0}}(v_{x_{i}})\in \mathfrak {P} \text { or } \displaystyle \prod _{i=0}^{p} w_{y_{0}}(v_{y_{i}})\in \mathfrak {P}. \end{aligned}$$

Without loss of generality assume $\displaystyle \prod _{i=0}^{p} w_{x_{0}}(v_{x_{i}})\in \mathfrak {P}$

$$\begin{aligned}&\implies w_{x_{0}}(v_{x_{0}}) \displaystyle \prod _{i=1}^{p} w_{x_{0}}(v_{x_{i}})\in \mathfrak {P}\\&\implies \text { either } w_{x_{0}}(v_{x_{0}})\in \mathfrak {P} \text { or } \displaystyle \prod _{i=1}^{p} w_{x_{0}}(v_{x_{i}})\in \mathfrak {P}. \end{aligned}$$

If $\mathfrak {w}_{x_{0}}(v_{x_{0}})\in \mathfrak {P}$ we are done. If $\displaystyle \prod _{i=1}^{p} \mathfrak {w}_{x_{0}}(v_{x_{i}})\in \mathfrak {P}$

$$\begin{aligned} \implies \text {either~} \mathfrak {w}_{x_{0}}(v_{x_{1}})\in \mathfrak {P} \text { or } \displaystyle \prod _{i=2}^{p} \mathfrak {w}_{x_{0}}(v_{x_{i}})\in \mathfrak {P}. \end{aligned}$$

If $\mathfrak {w}_{x_{0}}(v_{x_{1}})\in \mathfrak {P}$ we are done. Continuing like this we will get atleast one vertex $v_{x_{i}}$ such that $v_{x_{i}}\in (\mathfrak {w}_{k_{1}})^{-1}(\mathfrak {P})$. $\square$

Corollary 4

Let $\mathfrak {P}$ be a prime ideal of a commutative ring $\mathcal {R}$. Then for any l simplex $\sigma _{(k_{1}\times k_{2}\times \ldots \times k_{m})}=\left[ \left( v_{x_{1_{0}}},v_{x_{2_{0}}},\ldots ,v_{x_{m_{0}}}\right) ,\left( v_{x_{1_{1}}},v_{x_{2_{1}}},\ldots ,v_{x_{m_{1}}}\right) ,\ldots ,((v_{x_{1}})_{l},(v_{x_{2}})_{l},\ldots , (v_{x_{m}})_{l})\right]$ in $\left( K_{1}\times K_{2}\times \ldots \times K_{m}, \mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}} \right)$ with $\sigma _{(k_{1}\times k_{2}\times \ldots \times k_{m})}\in \left( \mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}\right) ^{-1}(\mathfrak {P})$, then there will be atleast one vertex $v_{x_{i_{j}}} \in (w_{k_{i}})^{-1}(\mathfrak {P})$ for some $i,j=0,1,2,\ldots ,l$.

Result 4

Let $\mathfrak {P}$ be a prime ideal of a commutative ring $\mathcal {R}$ and $\sigma _{k_{1}\times k_{2}}=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\left( v_{x_{1}},v_{y_{1}}\right) , \ldots ,\left( v_{x_{l}},v_{y_{l}}\right) \right]$ be a l simplex in $(K_{1}\times K_{2}, \mathfrak {w}_{k_{1}\times k_{2}})$. Let $\sigma _{k_{1_{l_{1}}}}$ be the largest simplex of order $l_{1}$ in $K_{1}$ spanned by the vertices $\{v_{x_{0}},v_{x_{1}},\ldots ,v_{x_{l}}\}$ and $\sigma _{k_{2_{l_{2}}}}$ be the largest simplex of order $l_{2}$ in $K_{2}$ spanned by the vertices $\{v_{y_{0}},v_{y_{1}},\ldots ,v_{y_{l}} \}\text { where } l_{1},l_{2}\le l.$ If $\sigma _{k_{1}\times k_{2}}\in \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {P})$ then either $\sigma _{k_{1_{l_{1}}}}\in (\mathfrak {w}_{k_{1}})^{-1}(\mathfrak {P})$ or $\sigma _{k_{2_{l_{2}}}}\in (\mathfrak {w}_{k_{2}})^{-1}(\mathfrak {P}).$

Proof

Let $\mathfrak {P}$ be a prime ideal of a commutative ring $\mathcal {R}.$ If $\sigma _{k_{1}\times k_{2}}\in \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {P})$, the above theorem there will be atleast one vertex $v_{x_{i}}$ or $v_{y_{i}}$ such that either $v_{x_{i}}\in (\mathfrak {w}_{k_{1}})^{-1}(\mathfrak {P})$ or $v_{y_{i}}\in (\mathfrak {w}_{k_{2}})^{-1}(\mathfrak {P})$ for some $i=0,1,2,\ldots ,l$.

Assume that $v_{x_{q}}\in (\mathfrak {w}_{k_{1}})^{-1}(\mathfrak {P})$ for $0\le q \le l.$ Then $\sigma _{k_{1_{l_{1}}}}$ is a $l_{1}$ simplex in $K_{1}$ which contains $v_{x_{q}}$. Hence, $\sigma _{k_{1_{l_{1}}}}\in (\mathfrak {w}_{k_{1}})^{-1}(\mathfrak {P})$ by Lemma 1. If $v_{y_{q}}\in (\mathfrak {w}_{k_{2}})^{-1}(\mathfrak {P})$ for $0\le q \le l.$ Then $\sigma _{k_{2_{l_{2}}}}$ is a $l_{2}$ simplex in $K_{2}$ which contains $v_{y_{q}}$. Hence, $\sigma _{k_{2_{l_{2}}}}\in (\mathfrak {w}_{k_{2}})^{-1}(\mathfrak {P})$ by Lemma 1. $\square$

Corollary 5

Let $\mathfrak {P}$ be a prime ideal of a commutative ring $\mathcal {R}.$ Let $\sigma _{(k_{1}\times k_{2}\times \ldots \times k_{m})}=\left[ \left( v_{x_{1_{0}}},v_{x_{2_{0}}},\ldots ,v_{x_{m_{0}}}\right) , \left( v_{x_{1_{1}}},v_{x_{2_{1}}},\ldots ,v_{x_{m_{1}}}\right) ,\ldots ,\left( v_{x_{1_{l}}},v_{x_{2_{l}}},\ldots ,v_{x_{m_{l}}}\right) \right]$ be a l simplex in $\left( K_{1}\times K_{2}\times \ldots \times K_{m}, \mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}} \right)$ with $\sigma _{(k_{1}\times k_{2}\times \ldots \times k_{m})}^{-1}(\mathfrak {P}).$ If $\sigma _{k_{i_{l_{i}}}}$ be the largest simplex of order $l_{i}$ in $K_{i}$ spanned by the vertices $\{v_{x_{i_0}},v_{x_{i_1}},\ldots ,v_{x_{i_l}}\} \text { where } l_{i}\le l \text { for } i=1,2,\ldots ,m$ then $\sigma _{k_{i_{l_{i}}}}\in (\mathfrak {w}_{k_{i}})^{-1}(\mathfrak {P})$ for atleast one i.

Theorem 5

Let $\mathfrak {I}$ be an ideal of a commutative ring $\mathcal {R}.$ Also let $\left( K_{1}\times K_{2},\mathfrak {w}_{k_{1}\times k_{2}}\right)$ be a weighted simplicial complex with weight function $\mathfrak {w}_{k_{1}\times k_{2}}:K_{1}\times K_{2}\rightarrow \mathcal {R}$ is a weight function. Then $K_{1}\times K_{2}{\setminus }\left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I})$ is a simplicial subcomplex of $K_{1}\times K_{2}$.

Proof

If $K_{1}\times K_{2}{\setminus }\left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I})=\phi ,$ then it is the empty subcomplex of $K_{1}\times K_{2}$. Otherwise let $\tau _{k_{1}\times k_{2}}\in K_{1}\times K_{2}{\setminus }\left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I}).$ Let $\sigma _{k_{1}\times k_{2}}$ be any nonempty subset of $\tau _{k_{1}\times k_{2}}$. Suppose $\sigma _{k_{1}\times k_{2}}\in \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I}).$ Then by Lemma 1, $\tau _{k_{1}\times k_{2}}\in (\mathfrak {w}_{k_{1}\times k_{2}})^{-1}(\mathfrak {I})$. Which is a contradiction. So $\sigma _{k_{1}\times k_{2}}\in K_{1}\times K_{2}{\setminus }\left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I})$. Hence $K_{1}\times K_{2}{\setminus }\left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I})$ is a simplicial subcomplex of $K_{1}\times K_{2}$. $\square$

Corollary 6

Let $\mathfrak {I}$ be an ideal of a commutative ring $\mathcal {R}.$ Also let $\left( K_{1}\times K_{2}\times \cdots \times K_{m},\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}\right)$ be a weighted simplicial complex with weight function $\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}:K_{1}\times K_{2}\times \cdots \times K_{m}\rightarrow \mathcal {R}$ is a weight function. Then $K_{1}\times K_{2}\times \cdots \times K_{m}{\setminus }\left( \mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}\right) ^{-1}(\mathfrak {I})$ is a simplicial subcomplex of $K_{1}\times K_{2}\times \cdots \times K_{m}$.

Result 6

Let $\mathfrak {I}$ be an ideal of a commutative ring $\mathcal {R}.$ Let $\left( K_{1},\mathfrak {w}_{k_{1}}\right) , \left( K_{2},\mathfrak {w}_{k_{2}}\right)$ are weighted simplicial complexes and $\left( K_{1}\times K_{2},\mathfrak {w}_{k_{1}\times k_{2}}\right)$ be the weighted product of simplicial complex. Then $K_{1}{\setminus } \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})\times K_{2}{\setminus } \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I})$ is a sub complex of $K_{1}\times K_{2}.$

Proof

For any ideal $\mathfrak {I}$, $K_{1}{\setminus } \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})$ is a sub complex of $K_{1}$ and $K_{2}\setminus \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I})$ is a sub complex of $K_{2}$. Hence $K_{1}{\setminus } \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})\times K_{2}{\setminus } \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I})$ is a sub complex of $K_{1}\times K_{2}.$ $\square$

Corollary 7

Let $\mathfrak {I}$ be an ideal of a commutative ring $\mathcal {R}.$ Let $\left( K_{1},\mathfrak {w}_{k_{1}}\right) , \left( K_{2},\mathfrak {w}_{k_{2}}\right) ,\ldots ,\left( K_{m}, \mathfrak {w}_{k_{m}}\right)$ are m weighted simplicial complexes and $\left( K_{1}\times K_{2}\times \cdots \times K_{m},\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}\right)$ be the weighted m product of simplicial complexes. Then $K_{1}{\setminus } \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})\times K_{2}{\setminus } \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I})\times \cdots \times K_{m}{\setminus } \mathfrak {w}_{k_{m}}^{-1}(\mathfrak {I})$ is a sub complex of $K_{1}\times K_{2}\times \cdots \times K_{m}.$

Theorem 7

Let $\mathfrak {I}$ be an ideal of a commutative ring $\mathcal {R}.$ Let $\left( K_{1},\mathfrak {w}_{k_{1}}\right) , \left( K_{2},\mathfrak {w}_{k_{2}}\right)$ are weighted simplicial complexes and $\left( K_{1}\times K_{2},\mathfrak {w}_{k_{1}\times k_{2}}\right)$ be the weighted product of simplicial complex. Then $K_{1}\times K_{2}{\setminus }(\mathfrak {w}_{k_{1}\times k_{2}})^{-1}(\mathfrak {I}) \subseteq K_{1}{\setminus } \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})\times K_{2}{\setminus } \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I}).$

Proof

Let ${\sigma _{{k_{1}\times k_{2}}}}\in K_{1}\times K_{2}{\setminus }(\mathfrak {w}_{k_{1}\times k_{2}})^{-1}(\mathfrak {I})$ with

$$\begin{aligned} \sigma _{k_{1}\times k_{2}}=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\left( v_{x_{1}},v_{y_{1}}\right) ,\ldots ,\left( v_{x_{l}},v_{y_{l}}\right) \right] \end{aligned}$$

where $\sigma _{k_{1}} =\left[ v_{x_{0}},v_{x_{1}},\ldots v_{x_{l}}\right]$ be $l_{1}$ simplex in $K_{1}$ for some $l_{1}\le l$ and $\sigma _{k_{2}}=\left[ v_{y_{0}},v_{y_{1}},\ldots v_{y_{l}}\right]$ be $l_{2}$ simplex in $K_{2}$ for some $l_{2}\le l$. Without loss of generality take

$$\begin{aligned} \sigma _{k_{1}}=\left[ v_{x_{0}},v_{x_{1}},\ldots ,v_{x_{l_{1}}},\ldots , v_{x_{l}}\right] \end{aligned}$$

where $v_{x_{0}},v_{x_{1}},\ldots ,v_{x_{l_{1}}}$ are distinct vertices $v_{x_{l_{1}+1}},v_{x_{l_{1}+2}},\ldots ,v_{x_{l}}$ are repeating vertices in $K_{1}$ and

$$\begin{aligned} \sigma _{k_{2}}=\left[ v_{y_{0}},v_{y_{1}},\ldots ,v_{y_{l_{1}}},\ldots , v_{y_{l}}\right] \end{aligned}$$

where $v_{y_{0}},v_{y_{1}},\ldots ,v_{y_{l_{2}}}$ are distinct vertices and $v_{y_{l_{2}+1}},v_{y_{l_{2}+2}},\ldots ,v_{y_{l}}$ are repeating vertices in $K_{2}$. Let

$$\begin{aligned} \sigma _{k_{1_{l_{1}}}}=\left[ v_{x_{0}},v_{x_{1}},\ldots ,v_{x_{l_{1}}}\right] \text { and } \sigma _{k_{2_{l_{2}}}}=\left[ v_{y_{0}},v_{y_{1}},\ldots ,v_{y_{l_{1}}}\right] . \end{aligned}$$

We have ${\sigma _{{k_{1}\times k_{2}}}}\in K_{1}\times K_{2}{\setminus }(\mathfrak {w}_{k_{1}\times k_{2}})^{-1}(\mathfrak {I})$

$$\begin{aligned}&\implies \mathfrak {w}_{k_{1}\times k_{2}}( {\sigma _{{k_{1}\times k_{2}}}})\notin \mathfrak {I}\\&\implies \displaystyle \prod _{i=0}^{l}\mathfrak {w}_{x_{0}}(v_{x_{i}})\prod _{i=0}^{l}\mathfrak {w}_{y_{0}}(v_{y_{i}})\notin \mathfrak {I}\\&\implies \displaystyle \prod _{i=0}^{l_{1}}\mathfrak {w}_{x_{0}}(v_{x_{i}}) \displaystyle \prod _{i=l_{1}+1}^{l} \mathfrak {w}_{x_{0}}(v_{x_{i}}) \displaystyle \prod _{i=0}^{l_{2}}\mathfrak {w}_{y_{0}}(v_{y_{i}}) \displaystyle \prod _{i=l_{2}+1}^{l}\mathfrak {w}_{y_{0}}(v_{y_{i}})\notin \mathfrak {I}\\&\implies \mathfrak {w}_{x_{0}}(v_{x_{i}}),\mathfrak {w}_{y_{0}}(v_{y_{i}}) \notin \mathfrak {I} \text { for all } i=0,1,...,l\\&\implies \sigma _{k_{1}\times k_{2}}\in K_{1}\setminus \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})\times K_{2}\setminus \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I}) \end{aligned}$$

Hence,

$$\begin{aligned} K_{1}\times K_{2}\setminus (\mathfrak {w}_{k_{1}\times k_{2}})^{-1}(\mathfrak {I}) \subseteq K_{1}\setminus \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})\times K_{2}\setminus \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I}). \end{aligned}$$

$\square$

The converse of the theorem stated above might not hold in all cases. Example 8 will demonstrate this point (Fig. 1).

Example 8

Let $\mathcal {R}=\mathbb {Z}$ and $\mathfrak {I}=4\mathbb {Z}$. Also let $(X_{0},\mathfrak {w}_{x_{0}})=\{(u_{0},2),(u_{1},4)\}$ and $(Y_{0},\mathfrak {w}_{y_{0}})=\{(v_{0},2),(v_{1},4)\}$. $L_{1}$ be the weighted simplicial complex of weighted point cloud $(X_{0},\mathfrak {w}_{x_{0}})$ and $L_{2}$ be the weighted simplicial complex of weighted point cloud $(Y_{0},\mathfrak {w}_{y_{0}})$ and $L_{1}\times L_{2}.$

$$\begin{aligned}{} & {} L_{1}\setminus \mathfrak {w}_{l_{1}}^{-1}(\mathfrak {I})=[u'_{0}(2)], L_{2}\setminus \mathfrak {w}_{l_{2}}^{-1}(\mathfrak {I})=[v'_{0}(2)]\\{} & {} \quad \implies L_{1}\setminus \mathfrak {w}_{l_{1}}^{-1}(\mathfrak {I})\times L_{2}\setminus \mathfrak {w}_{l_{2}}^{-1}(\mathfrak {I})=[(u'_{0},v'_{0})(4)]. \end{aligned}$$

But we have,

$$\begin{aligned} L_{1}\times L_{2}\setminus {\mathfrak {w}_{l_{1}}\times \mathfrak {w}_{l_{2}}}^{-1}(\mathfrak {I})=\phi \end{aligned}$$

Fig. 1

Weighted product of weighted simplices $L_1$ and $L_2$

Corollary 8

Let $\mathfrak {I}$ be an ideal of a commutative ring $\mathcal {R}.$ Let $\left( K_{1},\mathfrak {w}_{k_{1}}\right) , \left( K_{2},\mathfrak {w}_{k_{2}}\right) ,\ldots , \left( K_{m},\mathfrak {w}_{k_{m}}\right)$ are weighted simplicial complexes and $\left( K_{1}\times K_{2}\times \cdots \times K_{m},\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}\right)$ be the weighted m product of simplicial complex. Then,

$K_{1}\times K_{2}\times \cdots \times K_{m}{\setminus }(\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}})^{-1}(\mathfrak {I}) \subseteq K_{1}{\setminus } \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})\times K_{2}{\setminus } \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I})\times \cdots \times K_{1}{\setminus } \mathfrak {w}_{k_{m}}^{-1}(\mathfrak {I}).$

Proposition 3

Let $\mathfrak {I}$ and $\mathfrak {I'}$ are ideals ideal of a commutative ring $\mathcal {R}$ and $(K_{1}\times K_{2},\mathfrak {w}_{k_{1}\times k_{2}})$ be weighted product simplicial complex. Then,

$$\begin{aligned}{} & {} K_{1}\times K_{2}\setminus \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I}\cap \mathfrak {I'})\\{} & {} \quad = \{K_{1}\times K_{2}\setminus \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I})\}\cup \{K_{1}\times K_{2}\setminus \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I'})\}. \end{aligned}$$

Proof

Let $\sigma _{k_{1}\times k_{2}}\in K_{1}\times K_{2}{\setminus }(\mathfrak {w}_{k_{1}\times k_{2}})^{-1}(\mathfrak {I}\cap \mathfrak {I'})$

$$\begin{aligned}&\iff \mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \notin \mathfrak {I}\cap \mathfrak {I'}\\&\iff \mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \notin \mathfrak {I} \text { or } \mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \notin \mathfrak {I'} \end{aligned}$$

That is,

$$\begin{aligned}{} & {} K_{1}\times K_{2}\setminus \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I}\cap \mathfrak {I'})\\{} & {} \quad = \{K_{1}\times K_{2}\setminus \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I})\}\cup \{K_{1}\times K_{2}\setminus \left( \mathfrak {w}_{k_{1}\times k_{2}}\right) ^{-1}(\mathfrak {I'})\}. \end{aligned}$$

$\square$

Corollary 9

Let $\mathfrak {I}$ and $\mathfrak {I'}$ are ideals ideal of a commutative ring $\mathcal {R}$ and Let $\left( K_{1},\mathfrak {w}_{k_{1}}\right) , \left( K_{2},\mathfrak {w}_{k_{2}}\right) ,\ldots , \left( K_{m},\mathfrak {w}_{k_{m}}\right)$ are weighted simplicial complexes and $\left( K_{1}\times K_{2}\times \cdots \times K_{m},\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}\right)$ be the weighted m product of simplicial complex. Then,

$$\begin{aligned}&K_{1}\times K_{2}\times \cdots \times K_{m}\setminus (\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}})^{-1}(\mathfrak {I}\cap \mathfrak {I'})\\&\quad = K_{1}\times K_{2}\times \cdots \times K_{m}\setminus (\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}})^{-1}(\mathfrak {I})\\&\quad \cup K_{1}\times K_{2}\times \cdots \times K_{m} \setminus (\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}})^{-1}(\mathfrak {I'}) \end{aligned}$$

Result 9

Let $\mathfrak {I}$ and $\mathfrak {J}$ are ideals ideal of a commutative ring $\mathcal {R}$ and $\left( K_{1}\times K_{2},\mathfrak {w}_{k_{1}\times k_{2}}\right)$ be weighted product simplicial complex. Then,

$$\begin{aligned} K_{1}\times K_{2}\setminus (\mathfrak {w}_{k_{1}\times k_{2}})^{-1}(\mathfrak {I}\cap \mathfrak {I'})\subseteq \left\{ K_{1}\setminus \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})\cap {\mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I'})}\right\} \times \left\{ K_{2}\setminus \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I})\cap {\mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I'})}\right\} \end{aligned}$$

Proof

$$\begin{aligned}&K_{1}\times K_{2}\setminus (\mathfrak {w}_{k_{1}\times k_{2}})^{-1}\left( \mathfrak {I}\cap \mathfrak {I'}\right) \\&\quad = \left\{ K_{1}\times K_{2}\setminus (\mathfrak {w}_{k_{1}\times k_{2}})^{-1}(\mathfrak {I})\right\} \cup \left\{ K_{1}\times K_{2}\setminus (\mathfrak {w}_{k_{1}\times k_{2}})^{-1}(\mathfrak {I'})\right\} \\&\quad \subseteq \left\{ K_{1}\setminus \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})\times K_{2}\setminus \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I})\right\} \cup \left\{ K_{1}\setminus \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I'})\times K_{2}\setminus \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I'})\right\} \\&\quad \subseteq \left\{ K_{1}\setminus \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})\cup K_{1}\setminus \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I'})\right\} \times \left\{ K_{2}\setminus \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I})\cup K_{2}\setminus \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I'})\right\} \\&\quad \subseteq \left\{ K_{1}\setminus {\mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})\cap \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I'})}\right\} \times \left\{ K_{2}\setminus {\mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I})\cap \mathfrak {w}_{k_{2}}^{-1}(\mathfrak {I'})}\right\} \end{aligned}$$

$\square$

Corollary 10

Let $\mathfrak {I}$ and $\mathfrak {I'}$ are ideals ideal of a commutative ring $\mathcal {R}$ and Let $(K_{1},\mathfrak {w}_{k_{1}}), (K_{2},\mathfrak {w}_{k_{2}}),\ldots , (K_{m},\mathfrak {w}_{k_{m}})$ are weighted simplicial complexes and $(K_{1}\times K_{2}\times \cdots \times K_{m},\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}})$ be the weighted m product of simplicial complex. Then,

$$\begin{aligned} \begin{aligned}&K_{1}\times K_{2}\times \cdots \times K_{m}\setminus (\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}})^{-1}(\mathfrak {I}\cap \mathfrak {I'})\\&\quad \subseteq K_{1}\setminus \mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I})\cap {\mathfrak {w}_{k_{1}}^{-1}(\mathfrak {I'})}\times \cdots \times \\&\qquad K_{m}\setminus \mathfrak {w}_{k_{m}}^{-1}(\mathfrak {I})\cap {\mathfrak {w}_{k_{m}}^{-1}(\mathfrak {I'})} \end{aligned} \end{aligned}$$

Definition 20

Let $(K_{1},\mathfrak {w}_{k_{1}}), (K_{2},\mathfrak {w}_{k_{2}}), (P_{1},\mathfrak {w}_{p_{1}})$ and $(P_{2},\mathfrak {w}_{p_{2}})$ are weighted simplicial complexes of weighted point clouds $(X_{0},\mathfrak {w}_{x_{0}}), (Y_{0},\mathfrak {w}_{y_{0}}), (X'_{0},\mathfrak {w}_{x'_{0}})$ and $(Y'_{0},\mathfrak {w}_{y'_{0}})$ respectively. If $f_{1}:K_{1}\rightarrow P_{1}$ is a weighted simplicial map from $K_{1}$ to $P_{1}$ and $f_{2}:K_{2}\rightarrow P_{2}$ is a weighted simplicial map from $K_{2}$ to $P_{2}$, then the function $F_{f_{1}\times f_{2}}:K_{1}\times K_{2} \rightarrow P_{1}\times P_{2}$ defined by $F_{f_{1}\times f_{2}}=(f_{1},f_{2})$ which maps vertices of $K_{1}\times K_{2}$ to vertices of $P_{1}\times P_{2}$ is called weighted simplicial map induced by $f_{1}$ and $f_{2}$. That is, if $\left\{ \left( v_{x_{0}},v_{y_{0}}\right) , \left( v_{x_{1}},v_{y_{1}}\right) ,\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right\}$ span a simplex of $K_{1}\times K_{2},$ $\left\{ F_{f_{1}\times f_{2}}\left( v_{x_{0}},v_{y_{0}}\right) , F_{f_{1}\times f_{2}}\left( v_{x_{1}},v_{y_{1}}\right) ,\ldots , F_{f_{1}\times f_{2}}\left( v_{x_{n}},v_{y_{n}}\right) \right\}$(not necessarily distinct) will span a simplex of $P_{1}\times P_{2}$.

Given two simplicial maps that operate from one simplicial complex to another, it is possible to construct a simplicial map induced by these mappings, which is defined as the product of the corresponding simplices. But the converse need not be true as illustrated in the following example. That is, All simplicial maps from $K_{1}\times K_{2}$ to $P_{1}\times P_{2}$ need not be induced by the simplicial maps from $K_{1}$ to $P_{1}$ and $K_{2}$ to $P_{2}.$

Example 10

Let $K_{1}=[u],K_{2}=[u_{1},u_{2}],P_{1}=[v_{1},v_{2}]$ and $P_{2}=[v]$ then,$K_{1}\times K_{2}=[(u,u_{1}),(u,u_{2})]$ and $P_{1}\times P_{2}=[(v,v_{1}),(v,v_{2})]$ as in Fig. 2.

Consider the map $F: K_{1}\times K_{2} \rightarrow P_{1}\times P_{2}$ defined by $F(u,u_{1})=(v,v_{1}), F(u,u_{2})=(v,v_{2})$ Clearly it is a simplicial map. But it is not induced by any simplicial map from $K_{1}$ to $P_{1}$ and $K_{2}$ to $P_{2}.$ If it is induced by any simplicial map $f_{1}:K_{1}\rightarrow P_{1}$ and $f_{2}:K_{2}\rightarrow P_{2}$, $f_{1}(u)$ should be a unique vertex in $P_{1}.$ Any simplicial map $F: K_{1}\times K_{2} \rightarrow P_{1}\times P_{2}$ induced by simplicial maps $f_{1}:K_{1}\rightarrow P_{1}$ and $f_{2}:K_{2}\rightarrow P_{2}$ will be either $F(u,u_{1})=(v,v_{1}),F(u,u_{2})=(v,v_{1})$ or $F(u,u_{1})=(v,v_{2}),F(u,u_{2})=(v,v_{2}).$

Fig. 2

Weighted product of weighted simplices $K_1,K_2$ and $P_1,P_2$

Definition 21

Let $(K_{i},\mathfrak {w}_{k_{i}}), (P_{i},\mathfrak {w}_{p_{i}})$ and are simplicial complexes of point clouds $(X_{{i}_{0}},\mathfrak {w}_{x_{i_{0}}})$ and $(Y_{{i}_{0}},\mathfrak {w}_{y_{i_{0}}})$respectively for $i=1,2,\ldots ,m$. If $f_{i}:K_{i}\rightarrow P_{i}$ is a simplicial map from $K_{i}$ to $P_{i}$ for each $i=1,2,\ldots ,m$, then the function $F_{f_{1}\times f_{2}\times \cdots \times f_{m}}:K_{1}\times K_{2}\times \cdots \times K_{m} \rightarrow P_{1}\times P_{2}\times \cdots \times P_{m}$ defined by $F_{f_{1}\times f_{2}\times \cdots \times f_{m}}=(f_{1},f_{2},\ldots , f_{m})$ which maps vertices of $K_{1}\times K_{2}\times \cdots \times K_{m}$ to vertices of $P_{1}\times P_{2}\times \cdots \times P_{m}$ is called simplicial map induced by $f_{1}, f_{2},\ldots ,f_{m}$.

Result 11

If $f_{1}:K_{1}\rightarrow P_{1}$ is a morphism from $K_{1}$ to $P_{1}$ and $f_{2}:K_{2}\rightarrow P_{2}$ is a morphism from $K_{2}$ to $P_{2}$, then the function $F_{f_{1}\times f_{2}}:K_{1}\times K_{2} \rightarrow P_{1}\times P_{2}$ induced by $f_{1}$ and $f_{1}$ is a morphism.

Proof

Since $f_{1}$ and $f_{2}$ are two morphisms then $\mathfrak {w}_{p_{1}}\left( f_{1}\left( \sigma _{p_{1}}\right) \right) /\mathfrak {w}_{k_{1}}(\sigma _{k_{1}})$ for all $\sigma _{k_{1}}\in K_{1}.$

$$\begin{aligned} \implies \mathfrak {w}_{x'_{0}}\left( f_{1}\left( v_{x_{i}}\right) \right) /\mathfrak {w}_{x_{0}}(v_{x_{i}}), i=1,2,\ldots ,m_{1} \end{aligned}$$

and $\mathfrak {w}_{p_{2}}\left( f_{2}\left( \sigma _{p_{2}}\right) \right) /\mathfrak {w}_{k_{2}}(\sigma _{k_{2}})$ for all $\sigma _{k_{2}}\in K_{2}.$

$$\begin{aligned} \implies \mathfrak {w}_{y'_{0}}\left( f_{2}\left( v_{y_{i}}\right) \right) /\mathfrak {w}_{y_{0}}(v_{y_{i}}), i=1,2,\ldots ,m_{2} \end{aligned}$$

where $m_{1}$ be the number of vertices in $K_{1}$ and $m_{2}$ be the number of vertices in $K_{2}.$

Let

$$\begin{aligned} \sigma _{k_{1}\times k_{2}}=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\left( v_{x_{1}},v_{y_{1}}\right) ,\ldots ,\left( v_{x_{l}},v_{y_{l}}\right) \right] \end{aligned}$$

be a l simplex in $K_{1}\times K_{2}.$ We have

$$\begin{aligned} \mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})=\displaystyle \prod _{i=0}^{l}\mathfrak {w}_{x_{0}}(v_{x_{i}})\mathfrak {w}_{y_{0}}(v_{y_{i}}). \end{aligned}$$

From the definition of simplicial map induced by $f_{1}$ and $f_{2}$,

$$\begin{aligned} F_{f_{1}\times f_{2}}=\left[ \left( f_{1}\left( v_{x_{0}}\right) ,f_{2}\left( v_{y_{0}}\right) \right) ,\left( f_{1}\left( v_{x_{1}}\right) ,f_{2}\left( v_{y_{1}}\right) \right) ,\ldots , \left( f_{1}\left( v_{x_{l}}\right) ,f_{2}\left( v_{y_{l}}\right) \right) \right] . \end{aligned}$$

Clearly $F_{f_{1}\times f_{2}}(\sigma _{k_{1}\times k_{2}}))$ is a q simplex in $L_{1}\times L_{2}$ with $q \le l.$

Let $\left( f_{1}\left( v_{x_{0}}\right) ,f_{2}\left( v_{y_{0}}\right) \right) ,\left( f_{1}\left( v_{x_{1}}\right) ,f_{2}\left( v_{y_{1}}\right) \right) ,\ldots ,\left( f_{1}\left( v_{x_{q}}\right) ,f_{2}\left( v_{y_{q}}\right) \right)$ are distinct vertices, and

$\left( f_{1}\left( v_{x_{q+1}}\right) ,f_{2}\left( v_{y_{q+1}}\right) \right) ,\left( f_{1}\left( v_{x_{p+2}}\right) ,f_{2}\left( v_{y_{p+2}}\right) \right) ,\ldots ,\left( f_{1}\left( v_{x_{l}}\right) ,f_{2}\left( v_{y_{l}}\right) \right)$ are repeating vertices. Then,

$$\begin{aligned} F_{f_{1}\times f_{2}}&=\left[ \left( f_{1}\left( v_{x_{0}}\right) ,f_{2}\left( v_{y_{0}}\right) \right) ,\left( f_{1}\left( v_{x_{1}}\right) ,f_{2}\left( v_{y_{1}}\right) \right) ,\ldots , \left( f_{1}\left( v_{x_{q}}\right) ,f_{2}\left( v_{y_{q}}\right) \right) \right] \\&\implies \mathfrak {w}_{p_{1}\times p_{2}}\left( F_{f_{1}\times f_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \right) = \displaystyle \prod _{i=0}^{q}\mathfrak {w}_{x'_{0}}\left( f_{1}\left( v_{x_{i}}\right) \right) \mathfrak {w}_{y'_{0}}f_{2}((v_{y_{i}})).\\&\implies \mathfrak {w}_{p_{1}\times p_{2}}\left( F_{f_{1}\times f_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \right) /\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \end{aligned}$$

Hence $F_{f_{1}\times f_{2}}:K_{1}\times K_{2} \rightarrow P_{1}\times P_{2}$ induced by $f_{1}$ and $f_{1}$ is a morphism. $\square$

Corollary 11

If $f_{i}:K_{i}\rightarrow P_{i} \text { for } i=1,2,\ldots ,m$ is a morphism from $K_{i}$ to $P_{i}$, then the function $F_{f_{1}\times f_{2}\times \cdots \times f_{m}}:K_{1}\times K_{2} \times \cdots \times K_{m}\rightarrow P_{1}\times P_{2}\times \cdots \times P_{m}$ induced by $f_{1},f_{1},\ldots ,f_{m}$ is a morphism.

Before defining the weighted product homology group, it is essential to define the notions of the weighted product chain group, the induced product weighted homomorphism, and the weighted product boundary maps. All of these terminologies are being introduced with the aid of induced maps. From now onward, $\mathcal {R}$ has to be considered as an integral domain.

Definition 22

Let $C_{n}(K_{1}\times K_{2})$ be the $\mathcal {R}$ module with basis the n simplices of $K_{1}\times K_{2}$ with nonzero weight. Elements of $C_{n}(K_{1}\times K_{2}),$ called n chains, are finite formal sums $\sum _{\alpha }n_{\alpha }\sigma _{(k_{1}\times k_{2})\alpha }$ with coefficients $n_{\alpha }\in \mathcal {R}$ and $\sigma _{(k_{1}\times k_{2})\alpha }\in K_{1}\times K_{2}.$

Definition 23

Let $C_{n}(K_{1}\times K_{2}\times \cdots \times K_{m})$ be the $\mathcal {R}$ module with basis the n simplices of $K_{1}\times K_{2}\times \cdots \times K_{m}$ with nonzero weight. Elements of $C_{n}(K_{1}\times K_{2}\times \cdots \times K_{m}),$ called n chains, are finite formal sums $\sum _{\alpha }n_{\alpha }\sigma _{(k_{1}\times k_{2}\times \cdots \times K_{m})\alpha }$ with coefficients $n_{\alpha }\in \mathcal {R}$ and $\sigma _{(k_{1}\times k_{2}\times \cdots \times K_{m})\alpha }\in K_{1}\times K_{2}\times \cdots \times K_{m}.$

Definition 24

Given a simplicial map $F_{f_{1}\times f_{2}}:K_{1}\times K_{2} \rightarrow P_{1}\times P_{2}$ which is induced by $f_{1}$ and $f_{1}$, the induced homomorphism $F_{(f_{1}\times f_{2})\alpha }: C_{n}(K_{1}\times K_{2})\rightarrow C_{n}(P_{1}\times P_{2})$ is defined on the on the generators of $C_{n}(K_{1}\times K_{2})$ as follows. For $\sigma _{k_{1}\times k_{2}}=[(v_{x_{0}},v_{y_{0}}),(v_{x_{1}},v_{y_{1}}),\ldots ,(v_{x_{n}},v_{y_{n}})]\in C_{n}(K_{1}\times K_{2}),$ we define

$$\begin{aligned}{} & {} F_{{\#}(f_{1}\times f_{2})\alpha }\\{} & {} \quad ={\left\{ \begin{array}{ll} \frac{\mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})}{\mathfrak {w}_{p_{1}\times p_{2}}\left( F_{f_{1}\times f_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \right) }F_{f_{1}\times f_{2}}(\sigma _{k_{1}\times k_{2}}) &{}\quad \text {if}~ \left( f_{1}\left( v_{x_{0}}\right) ,f_{2}\left( v_{y_{0}}\right) \right) ,\ldots , \left( f_{1}\left( v_{x_{n}}\right) ,f_{2}\left( v_{y_{n}}\right) \right) \\ &{} \quad \text {are distinct}, \\ 0 &{} \quad \text {otherwise } \end{array}\right. } \end{aligned}$$

Clearly it is well defined since, if $\mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})\ne 0,$ then $\mathfrak {w}_{p_{1}\times p_{2}}( F_{f_{1}\times f_{2}}(\sigma _{k_{1}\times k_{2}}))/\mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})$ implies $\mathfrak {w}_{p_{1}\times p_{2}}\left( F_{f_{1}\times f_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \right) \ne 0$.

Definition 25

Given a simplicial map $F_{f_{1}\times f_{2}\times \cdots \times f_{m}}:K_{1}\times K_{2}\times \cdots \times K_{m} \rightarrow P_{1}\times P_{2}\times \cdots \times P_{m}$ which is induced by $f_{1},f_{2},\ldots ,f_{m}$, the induced homomorphism $F_{(f_{1}\times f_{2}\times \cdots \times f_{m})\alpha }: C_{n}(K_{1}\times K_{2}\times \cdots \times K_{m})\rightarrow C_{n}( P_{1}\times P_{2}\times \cdots \times P_{m})$ is defined on the on the generators of $C_{n}(K_{1}\times K_{2}\times \cdots \times K_{m})$ as follows. For

$$\begin{aligned} \sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}}=\left[ \left( v_{x_{1_{0}}},v_{x_{2_{0}}},\ldots ,v_{x_{m_{0}}}\right) ,\ldots ,\left( v_{x_{1_{n}}},v_{x_{2_{n}}},\ldots ,v_{x_{m_{n}}}\right) \right] \in C_{n}(K_{1}\times K_{2}), \end{aligned}$$

we define

$$\begin{aligned}&F_{{\#}(f_{1}\times f_{2}\times \cdots \times f_{m})\alpha }(\sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}})\\&\quad ={\left\{ \begin{array}{ll} \frac{\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}(\sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}})}{\mathfrak {w}_{p_{1}\times p_{2}}( F_{f_{1}\times f_{2}}(\sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}}))}F_{f_{1}\times f_{2}}(\sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}}) &{}\quad \text {if all ~} ({f_{1}(v_{x_{1_{i}}})},\ldots ,{f_{m}(v_{x_{m_{i}}})})\\ &{} \quad \text {are distinct for all }i=0,\ldots ,m, \\ 0 &{} \quad \text {otherwise } \end{array}\right. } \end{aligned}$$

Definition 26

The weighted boundary map of cartesian product of two weighted simplicial complexes is a map $\partial _{(k_{1}\times k_{2})n}: C_{n}(K_{1}\times K_{2})\rightarrow C_{n-1}(K_{1}\times K_{2})$ defined by

$$\begin{aligned}\partial _{(k_{1}\times k_{2})n}(\sigma _{k_{1}\times k_{2}})=\sum _{i=0}^{n}\frac{\mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})}{\mathfrak {w}_{k_{1}\times k_{2}}(d_{i}(\sigma _{k_{1}\times k_{2}}))}(-1)^{i}d_{i}(\sigma _{k_{1}\times k_{2}}) \end{aligned}$$

where

$$\begin{aligned}{} & {} d_{i}(\sigma _{k_{1}\times k_{2}})\\{} & {} \quad =\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\ldots ,\widehat{\left( v_{x_{i}},v_{y_{i}}\right) },\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right] \text { (deleting the vertex } (v_{x_{i}},v_{y_{i}})) \end{aligned}$$

for any $n-$ simplex $\sigma _{k_{1}\times k_{2}}=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\left( v_{x_{1}},v_{y_{1}}\right) ,\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right]$

If $\mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})\ne 0,$ then $\mathfrak {w}_{k_{1}\times k_{2}}(d_{i}(\sigma _{k_{1}\times k_{2}}))\ne 0.$ So, $\partial _{(k_{1}\times k_{2})n}$ is well defined.

Definition 27

The weighted boundary map of cartesian m product of weighted simplicial complexes is a map $\partial _{(k_{1}\times k_{2}\times \cdots \times k_{m})n}: C_{n}(K_{1}\times K_{2}\times \cdots \times K_{m})\rightarrow C_{n-1}(K_{1}\times K_{2}\times \cdots \times K_{m})$ defined by

$$\begin{aligned}{} & {} \partial _{(k_{1}\times k_{2}\times \cdots \times k_{m})n}(\sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}})\\{} & {} \quad =\sum _{i=0}^{n}\frac{\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}(\sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}})}{\mathfrak {w}_{k_{1}\times k_{2}\times \cdots \times k_{m}}(d_{i}(\sigma _{k_{1}\times k_{2}\times \cdots \times k_{m} }))}(-1)^{i}d_{i}(\sigma _{k_{1}\times k_{2}\times \cdots \times k_{m} }) \end{aligned}$$

where

$$\begin{aligned}{} & {} d_{i}(\sigma _{k_{1}\times k_{2}\times \cdots \times k_{m} })\\{} & {} \quad =\left[ \left( v_{x_{1_{0}}},v_{x_{2_{0}}},\ldots ,v_{x_{m_{0}}}\right) ,\ldots ,\widehat{\left( v_{x_{1_{i}}},v_{x_{2_{i}}},\ldots ,v_{x_{m_{i}}}\right) },\ldots ,\left( v_{x_{1_{n}}},v_{x_{2_{n}}},\ldots ,v_{x_{m_{n}}}\right) \right] \end{aligned}$$

$\left( \text { deleting the vertex } \left( v_{x_{1_{i}}},v_{x_{2_{i}}},\ldots ,v_{x_{m_{i}}}\right) \right)$ for any $n-$ simplex

$$\begin{aligned}{} & {} \sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}}\\{} & {} \quad =\left[ \left( v_{x_{1_{0}}},v_{x_{2_{0}}},\ldots ,v_{x_{m_{0}}}\right) ,\left( v_{x_{1_{2}}},v_{x_{2_{2}}},\ldots ,v_{x_{m_{2}}}\right) ,\ldots ,\left( v_{x_{1_{n}}},v_{x_{2_{n}}},\ldots ,v_{x_{m_{n}}}\right) \right] \end{aligned}$$

Result 12

Let $\mathcal {R}$ is an integral domain, $K_{1}$ be the simplicial complex of weigted point cloud $(X_{0},\mathfrak {w}_{{x_{0}})})$ and $K_{2}$ be the simplicial complex of weigted point cloud $(Y_{0},\mathfrak {w}_{y_{0})})$. If $\partial _{(k_{1}\times k_{2})n}: C_{n}(K_{1}\times K_{2})\rightarrow C_{n-1}(K_{1}\times K_{2})$ is a weighted boundary map of cartesian product of $K_{1}$ and $K_{2}$, then

$$\begin{aligned} \partial _{(k_{1}\times k_{2})n}(\sigma _{k_{1}\times k_{2}})=\sum _{i=0}^{n}\mathfrak {w}_{x_{0}}(v_{x_{i}}) \mathfrak {w}_{y_{0}}(v_{y_{i}})(-1)^{i}d_{i}(\sigma _{k_{1}\times k_{2}}) \end{aligned}$$

for all $\sigma _{k_{1}\times k_{2}}\in K_{1}\times K_{2}.$

Proof

Let

$$\begin{aligned} \sigma _{k_{1}\times k_{2}})=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\ldots ,\widehat{\left( v_{x_{i}},v_{y_{i}}\right) },\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right] \in K_{1}\times K_{2} \end{aligned}$$

we have

$$\begin{aligned} d_{i}(\sigma _{k_{1}\times k_{2}})&=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\ldots ,\widehat{\left( v_{x_{i}},v_{y_{i}}\right) },\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right] \\&=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\ldots ,\left( v_{x_{i-1}},v_{y_{i-1}}\right) ,\left( v_{x_{i+1}},v_{y_{i+1}}\right) ,\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right] \end{aligned}$$

also

$$\begin{aligned} \mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})&=\prod _{j=0}^{n}\mathfrak {w}_{x_{0}}(v_{x_{j}})\mathfrak {w}_{y_{0}}(v_{y_{j}})\\ \mathfrak {w}_{k_{1}\times k_{2}}(d_{i}(\sigma _{k_{1}\times k_{2}}))&=\prod _{j=0}^{i-1}\mathfrak {w}_{x_{0}}(v_{x_{j}})\mathfrak {w}_{y_{0}}(v_{y_{j}})\prod _{j=i+1}^{n}\mathfrak {w}_{x_{0}}(v_{x_{j}})\mathfrak {w}_{y_{0}}(v_{y_{j}}) \end{aligned}$$

sine $\mathcal {R}$ is an integral domain,

$$\begin{aligned} \frac{\mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})}{\mathfrak {w}_{k_{1}\times k_{2}}(d_{i}(\sigma _{k_{1}\times k_{2}}))}= \mathfrak {w}_{x_{0}}(v_{x_{i}})\mathfrak {w}_{y_{0}}(v_{y_{i}}). \end{aligned}$$

This is true for all $\sigma _{k_{1}\times k_{2}}\in K_{1}\times K_{2}.$ Hence

$$\begin{aligned} \partial _{(k_{1}\times k_{2})n}(\sigma _{k_{1}\times k_{2}})=\sum _{i=0}^{n}\mathfrak {w}_{x_{0}}(v_{x_{i}}) \mathfrak {w}_{y_{0}}(v_{y_{i}})(-1)^{i}d_{i}(\sigma _{k_{1}\times k_{2}}) \end{aligned}$$

$\square$

Corollary 12

Let $\mathcal {R}$ be an integral domain,$K_{i}$ be the simplicial complex of weigted point cloud $(X_{i_{0}},\mathfrak {w}_{{x_{i_{0}}}}) \text { for all }i=1,2,\ldots ,m$. If $\partial _{(k_{1}\times k_{2}\times \cdots \times k_{m})n}: C_{n}(K_{1}\times K_{2}\times \cdots \times K_{m})\rightarrow C_{n-1}(K_{1}\times K_{2}\times \cdots \times K_{m})$ is a weighted boundary map of cartesian $m-$ product of weighted simplicial complexes $K_{i}$, then

$$\begin{aligned}{} & {} \partial _{(k_{1}\times k_{2}\times \cdots \times k_{m})n}(\sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}})\\{} & {} \quad =\sum _{i=0}^{n}\mathfrak {w}_{x_{1_{0}}}(v_{x_{1_{i}}})\mathfrak {w}_{x_{2_{0}}}(v_{x_{2_{i}}})...\mathfrak {w}_{x_{m_{0}}}(v_{x_{m_{i}}}) (-1)^{i}d_{i}(\sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}}) \end{aligned}$$

for all $\sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}}\in K_{1}\times K_{2}\times \cdots \times K_{m}.$

Example 13

Let $\mathcal {R}=\mathbb {Z}, (X_{0},\mathfrak {w}_{x_{0}})=\{(u_{0},2),(u_{1},3)\}$ and $(Y_{0},\mathfrak {w}_{y_{0}})=\{(v_{0},5),(v_{1},2)\}$. $K_{1}$ be the simplicial complex of weighted point cloud $(X_{0},\mathfrak {w}_{x_{0}})$ and $K_{2}$ be the simplicial complex of weighted point cloud $(Y_{0},\mathfrak {w}_{y_{0}})$ and $K_{1}\times K_{2}$ be the Cartesian product of $K_{1}$ and $K_{2}$ as in Fig. 3.

Let $\sigma _{k_{1}\times k_{2}}=[(u_{0},v_{0}),(u_{0},v_{1})]$ be a $2-$ simplex in $K_{1}\times K_{2}.$ We have

$$\begin{aligned}&\partial _{k_{1}\times k_{2}}1[(u_{0},v_{0}),(u_{0},v_{1})]\\&\quad =\frac{\mathfrak {w}_{k_{1}\times k_{2}}[(u_{0},v_{0}),(u_{0},v_{1})]}{\mathfrak {w}_{k_{1} \times k_{2}}(u_{0},v_{0})}(u_{0},v_{0})\\&\qquad -\frac{\mathfrak {w}_{k_{1}\times k_{2}}[(u_{0},v_{0}),(u_{0},v_{1})]}{\mathfrak {w}_{k_{1} \times k_{2}}(u_{0},v_{1})}(u_{0},v_{1})\\&\quad =4(u_{0},v_{0})-10 (u_{0},v_{1}) \end{aligned}$$

also

$$\begin{aligned} \mathfrak {w}_{x_{0}}({u_{0}})\mathfrak {w}_{y_{0}}({v_{1}})(u_{0},v_{0})-\mathfrak {w}_{x_{0}}({u_{0}})\mathfrak {w}_{y_{0}}({v_{0}})(u_{0},v_{1})= 4(u_{0},v_{0})-10 (u_{0},v_{1}) \end{aligned}$$

Fig. 3

Weighted product of weighted simplices $K_1$ and $K_2$

Proposition 4

$\partial _{k_{1}\times k_{2}}^{2}=0.$ That is the composition $C_{n}(K_{1}\times K_{2})\xrightarrow []{\partial _{(k_{1}\times k_{2})n}} C_{n-1}(K_{1}\times K_{2})\xrightarrow []{\partial _{(k_{1}\times k_{2})n-1}} C_{n-2}(K_{1}\times K_{2})$ is the zero map.

Proof

Let

$$\begin{aligned} \sigma _{k_{1}\times k_{2}}=[(v_{x_{0}},v_{y_{0}}),(v_{x_{1}},v_{y_{1}}),\ldots ,(v_{x_{n}},v_{y_{n}})] \end{aligned}$$

be an $n-$ simplex in $K_{1}\times K_{2}.$ We have

$$\begin{aligned}&\partial _{(k_{1}\times k_{2})n}(\sigma _{k_{1}\times k_{2}}) =\sum _{i=0}^{n}\frac{\mathfrak {w}_{k_{1}\times {k_{2}}}(\sigma _{k_{1}\times k_{2}})}{\mathfrak {w}_{k_{1}\times {k_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}}))}(-1)^{i}d_{i}(\sigma _{k_{1}\times k_{2}})\\&\quad \implies \partial _{(k_{1}\times k_{2})n-1} \partial _{(k_{1}\times k_{2})n}(\sigma _{k_{1}\times k_{2}}) = \partial _{n-1}\left\{ \sum _{i=0}^{n}\frac{\mathfrak {w}_{k_{1}\times {k_{2}}}(\sigma _{k_{1}\times k_{2}})}{\mathfrak {w}_{k_{1}\times {k_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}}))}(-1)^{i}d_{i}(\sigma _{k_{1}\times k_{2}})\right\} \\&\quad =\sum _{j<i}\frac{\mathfrak {w}_{k_{1}\times {k_{2}}}(\sigma _{k_{1}\times k_{2}})}{\mathfrak {w}_{k_{1}\times {k_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}}))}(-1)^{i}\\&\qquad \frac{\mathfrak {w}_{k_{1}\times {k_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}}))}{\mathfrak {w}_{k_{1}\times {k_{2}}}\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\ldots , \widehat{\left( v_{x_{j}},v_{y_{j}}\right) },\ldots , \widehat{\left( v_{x_{i}},v_{y_{i}}\right) },\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right] }\\&\qquad (-1)^{j}\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\ldots , \widehat{\left( v_{x_{j}},v_{y_{j}}\right) },\ldots , \widehat{\left( v_{x_{i}},v_{y_{i}}\right) },\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right] \\&\qquad +\sum _{i<j}\frac{\mathfrak {w}_{k_{1}\times {k_{2}}}(\sigma _{k_{1}\times k_{2}})}{\mathfrak {w}_{k_{1}\times {k_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}}))}(-1)^{i}\\&\qquad \frac{\mathfrak {w}_{k_{1}\times {k_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}}))}{\mathfrak {w}_{k_{1}\times {k_{2}}}\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\ldots , \widehat{\left( v_{x_{i}},v_{y_{i}}\right) },\ldots , \widehat{\left( v_{x_{j}},v_{y_{j}}\right) },\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right] }\\&\qquad (-1)^{j-1}\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\ldots , \widehat{\left( v_{x_{i}},v_{y_{i}}\right) },\ldots , \widehat{\left( v_{x_{j}},v_{y_{j}}\right) },\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right] \\&\quad =\sum _{j<i}\frac{\mathfrak {w}_{k_{1}\times {k_{2}}}(\sigma _{k_{1}\times k_{2}})}{\mathfrak {w}_{k_{1}\times {k_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}}))}(-1)^{i+j}\\&\qquad \frac{\mathfrak {w}_{k_{1}\times {k_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}}))}{\mathfrak {w}_{k_{1}\times {k_{2}}}\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\ldots , \widehat{\left( v_{x_{j}},v_{y_{j}}\right) },\ldots , \widehat{\left( v_{x_{i}},v_{y_{i}}\right) },\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right] }\\&\qquad \left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\ldots , \widehat{\left( v_{x_{j}},v_{y_{j}}\right) },\ldots , \widehat{\left( v_{x_{i}},v_{y_{i}}\right) },\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right] \\&\qquad +\sum _{i<j}\frac{\mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})}{\mathfrak {w}_{k_{1}\times {k_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}}))}(-1)^{i+j-1}\\&\qquad \frac{\mathfrak {w}_{k_{1}\times {k_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}}))}{\mathfrak {w}_{k_{1}\times {k_{2}}}[(v_{x_{0}},v_{y_{0}}),\ldots ,\widehat{(v_{x_{i}},v_{y_{i}})},\ldots ,\widehat{(v_{x_{j}},v_{y_{j}})},\ldots ,(v_{x_{n}},v_{y_{n}})]}\\&\qquad [(v_{x_{0}},v_{y_{0}}),\ldots ,\widehat{(v_{x_{i}},v_{y_{i}})},\ldots ,\widehat{(v_{x_{j}},v_{y_{j}})},\ldots ,(v_{x_{n}},v_{y_{n}})]\\&\quad =0 \end{aligned}$$

$\square$

Corollary 13

$\partial _{k_{1}\times k_{2}\times \cdots \times k_{m}}^{2}=0.$ The composition $C_{n}(K_{1}\times K_{2}\times \cdots \times K_{m})\xrightarrow []{\partial _{(k_{1}\times k_{2}\times \cdots \times k_{m})n}} C_{n-1}(K_{1}\times K_{2}\times \cdots \times K_{m})\xrightarrow []{\partial _{(k_{1}\times k_{2}\times \cdots \times k_{m})n-1}} C_{n-2}(K_{1}\times K_{2}\times \cdots \times K_{m})$ is the zero map.

Lemma 2

Let $F_{f_{1}\times f_{2}}:K_{1}\times K_{2}\rightarrow P_{1}\times P_{2}$ be a simplicial map induced by $f_{1}$ and $f_{2}, d_{i}$ be the $i^{th}$ face map. Then

$$\begin{aligned} d_{i}\left( F_{f_{1}\times f_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \right) =F_{f_{1}\times f_{2}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) \end{aligned}$$

$\text {for all } \sigma _{k_{1}\times k_{2}}=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\left( v_{x_{1}},v_{y_{1}}\right) ,\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right] \in K_{1}\times K_{2}$ with $F_{f_{1}\times f_{2}}(v_{x_{0}},v_{y_{0}}), F_{f_{1}\times f_{2}}(v_{x_{1}},v_{y_{1}}), \ldots , F_{f_{1}\times f_{2}}(v_{x_{n}},v_{y_{n}}) \text { are distinct.}$

Proof

Let $\sigma _{k_{1}\times k_{2}}=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\left( v_{x_{1}},v_{y_{1}}\right) ,\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right]$ be an n simplex in $K_{1}\times K_{2}.$ Then, we have

$$\begin{aligned} d_{i}\left( F_{f_{1}\times f_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \right)&=d_{i}\left[ F_{f_{1}\times f_{2}}\left( v_{x_{0}},v_{y_{0}}\right) , F_{f_{1}\times f_{2}}\left( v_{x_{1}},v_{y_{1}}\right) ,\ldots ,F_{f_{1}\times f_{2}}\left( v_{x_{n}},v_{y_{n}}\right) \right] \\&=d_{i}\left[ \left( f_{1}\left( v_{x_{0}}\right) ,f_{2}\left( v_{y_{0}}\right) \right) , \left( f_{1}\left( v_{x_{1}}\right) ,f_{2}\left( v_{y_{1}}\right) \right) ,\ldots ,\left( f_{1}\left( v_{x_{n}}\right) ,f_{2}\left( v_{y_{n}}\right) \right) \right] \\&=\left[ \left( f_{1}\left( v_{x_{0}}\right) ,f_{2}\left( v_{y_{0}}\right) \right) ,\ldots ,\hat{\left( f_{1}\left( v_{x_{i}}\right) ,f_{2}\left( v_{y_{i}}\right) \right) },\ldots ,\left( f_{1}\left( v_{x_{n}}\right) ,f_{2}\left( v_{y_{n}}\right) \right) \right] \\&=F_{f_{1}\times f_{2}}\left[ \left( v_{x_{0}},v_{y_{0}}\right) , \left( v_{x_{1}},v_{y_{1}}\right) ,\ldots ,\hat{\left( v_{x_{i}},v_{y_{i}}\right) },\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right] \\&=F_{f_{1}\times f_{2}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) \end{aligned}$$

$\square$

Corollary 14

Let $F_{f_{1}\times f_{2}\times \cdots \times f_{m}}:K_{1}\times K_{2}\times \cdots \times K_{m}\rightarrow P_{1}\times P_{2}\times \cdots \times P_{m}$ be a simplicial map induced by $f_{i} \text { for } i=1,2,\ldots ,m, d_{i}$ be the $i^{th}$ face map. Then

$$\begin{aligned} d_{i}\left( F_{f_{1}\times f_{2}\times \cdots \times f_{m}}\left( \sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}}\right) \right) =F_{ f_{1}\times \cdots \times f_{m}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}}\right) \right) \end{aligned}$$

$\text {for all } \sigma _{k_{1}\times k_{2}\times \cdots \times k_{m}}=\left[ \left( v_{x_{1_{0}}},v_{x_{2_{0}}},\ldots ,v_{x_{m_{0}}}\right) ,\left( v_{x_{1_{1}}},v_{x_{2_{1}}},\ldots ,v_{x_{m_{1}}}\right) ,\ldots ,\left( v_{x_{1_{n}}},v_{x_{2_{n}}},\ldots ,v_{x_{m_{n}}}\right) \right]$

with $F_{{f_{1}\times f_{2}\times \cdots \times f_{m}}}\left( v_{x_{1_{0}}},v_{x_{2_{0}}},\ldots ,v_{x_{m_{0}}}\right) , \ldots , F_{{f_{1}\times f_{2}\times \cdots \times f_{m}}}\left( v_{x_{m_{n}}},v_{x_{m_{n}}},\ldots ,v_{x_{m_{n}}}\right) \text { are distinct.}$

Proposition 5

Let $F_{f_{1}\times f_{2}}:K_{1}\times K_{2}\rightarrow P_{1}\times P_{2}$ be a simplicial map induced by $f_{1}$ and $f_{2}.$ Then,

$$\begin{aligned} F_{{\#}{f_{1}\times f_{2}}}\partial _{k_{1}\times k_{2}}=\partial _{k_{1}\times k_{2}}F_{{\#}{f_{1}\times f_{2}}} \end{aligned}$$

Proof

Let $\sigma _{k_{1}\times k_{2}}=\left[ \left( v_{x_{0}},v_{y_{0}}\right) ,\left( v_{x_{1}},v_{y_{1}}\right) ,\ldots ,\left( v_{x_{n}},v_{y_{n}}\right) \right]$ be an n simplex in $K_{1}\times K_{2}.$ Also let $\tau _{k_{1}\times k_{2}}$ be a simplex of $P_{1}\times P_{2}$ which is spanned by $F_{f_{1}\times f_{2}}(v_{x_{0}},v_{y_{0}}),F_{f_{1}\times f_{2}}(v_{x_{1}},v_{y_{1}}),\ldots , F_{f_{1}\times f_{2}}(v_{x_{n}},v_{y_{n}})$

Case 1. If $\dim \tau _{k_{1}\times k_{2}}=n,$ then $F_{f_{1}\times f_{2}}(v_{x_{0}},v_{y_{0}}),F_{f_{1}\times f_{2}}(v_{x_{1}},v_{y_{1}}),\ldots ,F_{f_{1}\times f_{2}}(v_{x_{n}},v_{y_{n}})$ are distinct.

$$\begin{aligned}&F_{{\#}{f_{1}\times f_{2}}}\partial _{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})\\&\quad =F_{{\#}{f_{1}\times f_{2}}}\sum _{i=0}^{n}\frac{\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) }{\mathfrak {w}_{k_{1}\times k_{2}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) }(-1)^{i}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) \\&\quad = \sum _{i=0}^{n}\frac{\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) }{\mathfrak {w}_{k_{1}\times k_{2}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) } (-1)^{i} F_{{\#}{f_{1}\times f_{2}}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) \\&\quad = \sum _{i=0}^{n}\frac{\mathfrak {w}_{k_{1}\times k_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) }{\mathfrak {w}_{k_{1}\times k_{2}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) } (-1)^{i}\frac{\mathfrak {w}_{k_{1}\times k_{2}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) }{\mathfrak {w}_{k_{1}\times k_{2}}( F_{f_{1}\times f_{2}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) }F_{f_{1}\times f_{2}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) \end{aligned}$$

$$\begin{aligned}&\partial _{k_{1}\times k_{2}}F_{{\#}{f_{1}\times f_{2}}}(\sigma _{k_{1}\times k_{2}})\\&\quad =\partial _{k_{1}\times k_{2}}\left\{ \frac{\mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}}))}{\mathfrak {w}_{k_{1}\times k_{2}}( F_{f_{1}\times f_{2}}(\sigma _{k_{1}\times k_{2}}))} (-1)^{i}F_{f_{1}\times f_{2}}(\sigma _{k_{1}\times k_{2}})\right\} \\&\quad =\sum _{i=0}^{n}\frac{\mathfrak {w}_{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})}{\mathfrak {w}_{k_{1}\times k_{2}}\left( F_{f_{1}\times f_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \right) }\frac{\mathfrak {w}_{k_{1}\times k_{2}}\left( F_{f_{1}\times f_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \right) }{\mathfrak {w}_{k_{1}\times k_{2}}\left( d_{i}(F_{f_{1}\times f_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \right) }(-1)^{i}d_{i}\left( F_{f_{1}\times f_{2}}\left( \sigma _{k_{1}\times k_{2}}\right) \right) \\&\quad = F_{{\#}{f_{1}\times f_{2}}}\partial _{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}}) \end{aligned}$$

Case 2. $\dim \tau _{k_{1}\times k_{2}}\le n-2.$ $F_{{\#}{f_{1}\times f_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}}))=0$ for all i. So $F_{{\#}{f_{1}\times f_{2}}}\partial _{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})$ will be zero. Since $F_{{\#}{f_{1}\times f_{2}}}(\sigma _{k_{1}\times k_{2}})=0$ then, $\partial _{k_{1}\times k_{2}}F_{{\#}{f_{1}\times f_{2}}}(\sigma _{k_{1}\times k_{2}})=0$

Case 3. $\dim \tau _{k_{1}\times k_{2}}= n-1$. Without loss of generality, assume that $F_{f_{1}\times f_{2}}(v_{x_{0}},v_{y_{0}})=F_{f_{1}\times f_{2}}(v_{x_{1}},v_{y_{1}})$ and $F_{f_{1}\times f_{2}}(v_{x_{1}},v_{y_{1}}),\ldots ,F_{f_{1}\times f_{2}}(v_{x_{n}},v_{y_{n}})$ are distinct. Clearly

$$\begin{aligned} \partial _{k_{1}\times k_{2}}F_{{\#}{f_{1}\times f_{2}}}(\sigma _{k_{1}\times k_{2}})=0. \end{aligned}$$

We have,

$$\begin{aligned} F_{{\#}{f_{1}\times f_{2}}}\partial _{k_{1}\times k_{2}}(\sigma _{k_{1}\times k_{2}})= \sum _{i=0}^{n}\frac{\mathfrak {w}_{k_{1}\times {k_{2}}}(\sigma _{k_{1}\times k_{2}})}{\mathfrak {w}_{k_{1}\times {k_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}}))} (-1)^{i} F_{{\#}{f_{1}\times f_{2}}}(d_{i}(\sigma _{k_{1}\times k_{2}})). \end{aligned}$$

This sum will contain only two nonzero terms. Terms are

$$\begin{aligned}&\sum _{i=0}^{1}\frac{\mathfrak {w}_{k_{1}\times {k_{2}}}\left( \sigma _{k_{1}\times k_{2}}\right) }{\mathfrak {w}_{k_{1}\times {k_{2}}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) } (-1)^{i} F_{{\#}{f_{1}\times f_{2}}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) \\&\quad = \sum _{i=0}^{1}\frac{\mathfrak {w}_{k_{1}\times {k_{2}}}\left( \sigma _{k_{1}\times k_{2}}\right) }{\mathfrak {w}_{k_{1}\times {k_{2}}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) } (-1)^{i}\frac{\mathfrak {w}_{k_{1}\times {k_{2}}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) }{\mathfrak {w}_{k_{1}\times {k_{2}}}\left( F_{f_{1}\times f_{2}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) \right) }F_{f_{1}\times f_{2}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) \\&\quad =\sum _{i=0}^{1}\frac{\mathfrak {w}_{k_{1}\times {k_{2}}}\left( \sigma _{k_{1}\times k_{2}}\right) }{\mathfrak {w}_{k_{1}\times {k_{2}}}\left( F_{f_{1}\times f_{2}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) \right) } (-1)^{i}F_{f_{1}\times f_{2}}\left( d_{i}\left( \sigma _{k_{1}\times k_{2}}\right) \right) \\&\quad =0, \text { since } F_{f_{1}\times f_{2}}(v_{x_{0}},v_{y_{0}})=F_{f_{1}\times f_{2}}(v_{x_{1}},v_{y_{1}}). \end{aligned}$$

$\square$

Definition 28

The Homology group of the weighted product of simplicial complexes is defined as

$$\begin{aligned} H_{n}(K_{1}\times K_{2})=\text {ker}(\partial _{(k_{1}\times k_{2})n})/\text {Im}(\partial _{(k_{1}\times k_{2})n+1}) \end{aligned}$$

where $\partial _{(k_{1}\times k_{2})n}$is the weighted boundary map of the Cartesian product of two weighted simplicial complexes.

Definition 29

The Homology group of weighted m product of simplicial complexes is defined as

$$\begin{aligned} H_{n}(K_{1}\times K_{2}\times \cdots \times K_{m})=\text {ker}(\partial _{(k_{1}\times k_{2}\times \cdots \times k_{m})n})/\text {Im}(\partial _{(k_{1}\times k_{2}\times \cdots \times k_{m})n+1}) \end{aligned}$$

where $\partial _{(k_{1}\times k_{2}\times \cdots \times k_{m})n}$is the weighted boundary map of Cartesian product of m weighted simplicial complexes (Fig. 4).

Example 14

Consider Fig. 3. Let $\mathcal {R}=\mathbb {Z}, (X'_{0},\mathfrak {w}_{x'_{0}})=\{(x_{0},1),(x_{1},1)\}$ and $(Y'_{0},\mathfrak {w}_{y'_{0}})=\{(y_{0},1),(y_{1},n); n\in \mathbb {Z^{+}}\}$. $K'_{1}$ be the simplicial complex of weighted point cloud $(X'_{0},\mathfrak {w}_{x'_{0}})$ and $K'_{2}$ be the simplicial complex of weighted point cloud $(Y'_{0},\mathfrak {w}_{y'_{0}})$ and $K'_{1}\times K'_{2}$ be the Cartesian product of $K'_{1}$ and $K'_{2}.$

$$\begin{aligned} \partial _{(K'_{1}\times K'_{2})1}[(x_{0},y_{0}),(x_{0},y_{1})]&=n(x_{0},y_{0})-(x_{0},y_{1})\\ \partial _{(K'_{1}\times K'_{2})1}[(x_{0},y_{1}),(x_{1},y_{1})]&=n(x_{0},y_{1})-n(x_{1},y_{1})\\ \partial _{(K'_{1}\times K'_{2})1}[(x_{1},y_{0}),(x_{1},y_{1})]&=n(x_{1},y_{1})-(x_{1},y_{0})\\ \partial _{(K'_{1}\times K'_{2})1}[(x_{1},y_{0}),(x_{0},y_{0})]&=(x_{1},y_{0})-(x_{0},y_{0}) \end{aligned}$$

We have,

$$\begin{aligned} H_{0}(K_{1}\times K_{2})&=\text {ker}(\partial _{(k_{1}\times k_{2})0})/\text {Im}(\partial _{(k_{1}\times k_{2})1})\\&=\langle (x_{0},y_{0}),(x_{0},y_{1}),(x_{1},y_{1}),(x_{1},y_{0})~| ~n(x_{0},y_{0})-(x_{0},y_{1})=0,\\&\qquad n(x_{0},y_{1})-n(x_{1},y_{1})=0, n(x_{1},y_{1})-(x_{1},y_{0})\\&=0, (x_{1},y_{0})-(x_{0},y_{0})=0\rangle \\&=\mathbb {Z}\oplus \mathbb {Z}_{n} \end{aligned}$$

Fig. 4

Weighted product of weighted simplices $K'_1$ and $K'_2$

4 Conclusion and directions of future works

In the study, the weighted product point clouds and weighted product simplicial complexes are defined, where the weights are taken from a commutative ring with unity. We have introduced the concept of weighted product chain group, the induced product weighted homomorphism, and the weighted product boundary maps on integral domain. Applying these concepts, we developed the concept of weighted product homology with examples. For future studies, it is recommended to delve into the exploration of weighted product persistence homology and its associated concepts. As weighted persistence homology discerns filtrations that ordinary persistent homology cannot differentiate, its application within the field of product space would be an innovation.