Definition of Simplicial Homology group 📂Topological Data Analysis

Definition of Simplicial Homology group

Buildup

Despite the complexity of the content, I made sure to leave detailed calculations and explanations to make it as understandable as possible. If you’re interested in homology, I highly recommend reading this.

Consider a topological space $X$ of interest, represented through a $\Delta$-complex structure according to a specific simplicial complex. As a small example, in the image on the right, the torus represents $X$, and the left side corresponds to the simplicial complex.

Definition of a simplex:
The convex hull of $v_{0}, v_{1} , \cdots , v_{n} \in \mathbb{R}^{n+1}$, which are affinely independent, is called an $n$-simplex $\Delta^{n}$, and the vectors $v_{k}$ are called vertices. Mathematically, it is expressed as follows. $$ \Delta^{n} := \left\{ \sum_{k} t_{k} v_{k} : v_{k} \in \mathbb{R}^{n+1} , t_{k} \ge 0 , \sum_{k} t_{k} = 1 \right\} $$
An $n-1$-simplex $\Delta^{n-1}$ created by removing a vertex from $\Delta^{n}$ is called a face of $\Delta^{n}$. The union of all faces of $\Delta^{n}$ is called the boundary of $\Delta^{n}$ and is denoted as $\partial \Delta^{n}$.
The interior of a simplex $\left( \Delta^{n} \right)^{\circ} := \Delta^{n} \setminus \partial \Delta^{n}$ is called an open simplex.

Let’s say a simplicial complex is a complex made up of simplices, specifically forming a CW complex as follows:

The definition of $n$:
$D^{n} \subset \mathbb{R}^{n}$ defined as follows is called an $n$-unit disk. $$ D^{n} := \left\{ \mathbf{x} \in \mathbb{R}^{n} : \left\| \mathbf{x} \right\| \le 1 \right\} $$
A subset $e^{n}$ that is homeomorphic to $D^{n} \setminus \partial D^{n}$ is called an $n$-cell.

Definition of CW Complex:
A discrete set $X^{0} \ne \emptyset$ is considered as 0-cells.
An $n$-skeleton $X^{n}$ is made by attaching $n$-cells $e_{\alpha}^{n}$ to $X^{n-1}$ using the maps $\phi_{\alpha} : S^{n-1} \to X^{n-1}$.
$X := \bigcup_{n \in \mathbb{N}} X^{n}$ becomes a topological space with a weak topology, then $X$ is called a cell complex.

Definition

Consider a topological space $X$ with a $\Delta$-complex structure.

Let’s denote the free Abelian group with a basis of open $n$-simplices, or $n$-cells $e_{\alpha}^{n}$ in $X$, as $\Delta_{n} (X)$. Elements of $\Delta_{n} (X)$ are called $n$-chains and are represented as formal sums with coefficients $k_{\alpha} \in \mathbb{Z}$ as follows. $$ \sum_{\alpha} k_{\alpha} e_{\alpha}^{n} $$ Each $n$-cell $e_{\alpha}^{n}$ corresponds to a characteristic map $\sigma_{\alpha} : \Delta^{n} \to X$, allowing representation as follows. $$ \sum_{\alpha} k_{\alpha} \sigma_{\alpha} $$
The boundary homomorphism $\partial_{n} : \Delta_{n} (X) \to \Delta_{n-1} (X)$ is defined as follows, where $\sigma_{\alpha} | \left[ v_{1} , \cdots , \hat{v}_{i} , \cdots , v_{n} \right]$ indicates the restriction of $\sigma_{\alpha}$ to an $n-1$-simplex in $X$. $$ \partial _{n} \left( \sigma_{\alpha} \right) := \sum_{i=0}^{n} \left( -1 \right)^{i} \sigma_{\alpha} | \left[ v_{1} , \cdots , \hat{v}_{i} \cdots , v_{n} \right] $$
The quotient group $\ker \partial_{n} / \operatorname{Im} \partial_{n+1}$ is denoted as $H_{n}^{\Delta}$, and since $H_{n}^{\Delta}$ is a homology group, it is called the $n$th simplicial homology group of $X$.

The group $0$ is a magma defined on ${ 0 }$, essentially an empty algebraic structure.
The homomorphism $\partial^{2} = 0$ is a zero morphism.
$\operatorname{Im}$ refers to the image.
$\ker$ refers to the kernel.
In a set, the notation $\hat{v}_{i}$ means excluding $v_{i}$, as follows: $$ { v_{1} , \cdots , \hat{v}_{i} , \cdots , v_{n} } := { v_{1} , \cdots , v_{n} } \setminus { v_{i} } $$

Explanation

The definition section might be overwhelming with its dense text. It’s normal if it’s not immediately clear. The explanation aims to be thorough and accessible, addressing points that were confusing during my study.

Why are elements of $\Delta_{n} (X)$ called chains?

Considering the notation $\sigma_{\alpha} : \Delta^{n} \to X$, we can abstract away whether $e_{\alpha}^{n}$ is an element of $\Delta^{n}$ or $X$. For $n=2$ and all coefficients $k_{\alpha} = 1$, the geometric representation can be imagined as the figure on the right, denoted as $\sum_{i=1}^{7} \sigma_{i}$.

The term “chain” might make sense now, but it’s not crucial for understanding. What’s important is that the collection of $n$-chains in $\Delta_{n} (X)$ forms a chain complex.

Is $\Delta_{n} (X)$ really a group?

It’s crucial to note that the “formal sum” used to describe chains is not an algebraic operation within $\Delta_{n} (X)$. This notation is merely symbolic. For example, the expression

2😀 + 💎 - 3🍌

has no mathematical meaning as it's unclear what "twice 😀 plus 💎 minus three 🍌" would entail. This confusion is similar to the uncertainty in $\sum\_{\alpha} k\_{\alpha} e\_{\alpha} \simeq \sum\_{\alpha} k\_{\alpha} \sigma\_{\alpha}$ regarding - The addition of open simplices $e\_{\alpha}^{n}$, which is undefined - The interpretation of $\sigma\_{\alpha}$, which is a function - The meaning of operations like $-3 e\_{1}^{n} + 7 e\_{2}^{n} \simeq -3 \sigma\_{1} + 7 \sigma\_{2}$

Thankfully, these concerns are irrelevant to $\Delta_{n} (X)$. If we define

$\sigma=$2😀 + 💎 - 3🍌

as an $n$-chain in $\Delta\_{n} (X)$, its inverse can be defined using the inverses of coefficients $k\_{\alpha} \in (\mathbb{Z}, +)$, resulting in

$-\sigma=$ (-2)😀 + (-1)💎 + 3🍌

This definition is sufficient regardless of the specific structure of $\Delta_{n} (X)$. The identity element of $\Delta_{n} (X)$ can be defined as $0 := \sigma + (-\sigma)$, and since $\mathbb{Z}$ is an Abelian group, so is $\Delta_{n} (X)$. The operation $+$ in $(\Delta_{n} (X), +)$ is induced from $(\mathbb{Z}, +)$ but is distinct, and $\Delta_{n} (X)$ is a free Abelian group, with $\sum_{\alpha} k_{\alpha} \sigma_{\alpha}$ now being an algebraic sum.

In summary:

The initial definition’s appearance of addition in $\sum_{\alpha} k_{\alpha} \sigma_{\alpha}$ was merely notational, not an operation.
The $+$ in $(\Delta_{n} (X), +)$ is derived from $(\mathbb{Z}, +)$ but is not the same.
$(\Delta_{n} (X), +)$ is a free Abelian group, and $\sum_{\alpha} k_{\alpha} \sigma_{\alpha}$ is now an algebraic sum.

Why is $\partial$ called the boundary?

The definition of $\partial_{n}$ may seem abstract, but the following illustration clarifies its meaning.

For example, for $\partial_{2}$, we can perform the following calculation. $$ \begin{align*} & \partial _{2} \left[ v_{0} ,v_{1}, v_{2} \right] \\ =& \sum_{i=0}^{2} (-1)^{i} \left[ v_{0} ,v_{1}, v_{2} \right] \setminus \left[ v_{i} \right] \\ =& (-1)^{0} \left[ v_{1}, v_{2} \right] + (-1)^{1} \left[ v_{0}, v_{2} \right] + (-1)^{2} \left[ v_{0}, v_{1} \right] \\ =& \left[ v_{1}, v_{2} \right] - \left[ v_{0}, v_{2} \right] + \left[ v_{0}, v_{1} \right] \end{align*} $$

If you’re studying homology, it’s generally accepted that the boundary of a triangle $\left[ v_{0} ,v_{1}, v_{2} \right]$ consists of the segments $\left[ v_{1}, v_{2} \right], \left[ v_{0}, v_{2} \right], \left[ v_{0} , v_{1} \right]$. The real challenge is understanding what $\left[ v_{1}, v_{2} \right] - \left[ v_{0}, v_{2} \right]$ means. How can segments be subtracted? And how about operations on 2-simplices like triangles?

These questions miss the point. Refocusing, $\partial_{2} \left[ v_{0} ,v_{1}, v_{2} \right] \in \Delta_{1} (X)$ is simply a formal sum of the three elements $\left[ v_{1}, v_{2} \right], \left[ v_{0}, v_{2} \right], \left[ v_{0} , v_{1} \right]$. $$ (+1) \left[ v_{1}, v_{2} \right] + (-1) \left[ v_{0}, v_{2} \right] + (+1) \left[ v_{0}, v_{1} \right] $$

Denoting these as $$ \begin{align*} a := \left[ v_{1}, v_{2} \right] \ b:= \left[ v_{0}, v_{2} \right] \ c:= \left[ v_{0} , v_{1} \right] \end{align*} $$ reveals the nature of $\Delta_{1} (X)$. For example, a $1$-chain $x \in \Delta_{1} (X)$ can be represented with coefficients $k_{a} , k_{b} , k_{c} \in \mathbb{Z}$ as $$ x = k_{a} a + k_{b} b + k_{c} c $$

Viewing from the perspective of $a,b,c$, the free group $\Delta_{1} (X) := F[{ a,b,c }]$ is constructed, essentially equivalent to $\mathbb{Z}^{3} \simeq \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$.

This shift in perspective is crucial for understanding subsequent examples. We must think algebraically rather than geometrically.

Examples

Consider the following scenario: $$ \begin{align*} \\ \partial_{n} :& \Delta_{n} (X) \to \Delta_{n-1} (X) \\ H_{n}^{\Delta} (X) =& \ker \partial_{n} / \operatorname{Im} \partial_{n+1} \end{align*} $$

For $n = 0$, $\partial_{0} : \Delta_{0} (X) \to 0$ implies $\ker \partial_{0} = \Delta_{0} (X)$.

Circle $S^{1}$

For a circle $X = S^{1}$, there’s one 0-simplex (vertex $v$), one 1-simplex (edge $e$), and no $n$-simplices for $n \ge 2$. The chain complex is structured as follows: $$ \cdots \longrightarrow 0 \longrightarrow \Delta_{1}\left( S^{1} \right) \overset{\partial_{1}}{\longrightarrow} \Delta_{0}\left( S^{1} \right) \overset{\partial_{0}}{\longrightarrow} 0 $$

$\Delta_{1}(S^{1})$, being generated solely by $e$, is isomorphic to $\mathbb{Z}$, and similarly, $\Delta_{0}(S^{1})$ is isomorphic to $\mathbb{Z}$ due to being generated by $v$ alone. Since $\partial_{1}$ is a zero morphism: $$ \partial e = v - v = 0 $$

For $n = 0$, $\ker \partial_{0} = \Delta_{0} (S^{1})$, and since $\partial_{1}$ is a zero morphism, its image is ${ 0 }$, leading to: $$ \begin{align*} H_{0}^{\Delta} \left( S^{1} \right) =& \ker \partial_{0} / \operatorname{Im} \partial_{1} \\ \simeq& \Delta_{0} \left( S^{1} \right) / \left\{ 0 \right\} \\ \simeq& \mathbb{Z} \end{align*} $$

For $n = 1$, $\operatorname{Im} \partial_{2} = { 0 }$ since $\partial_{1}$ is a zero morphism, and $\ker \partial_{1} = \Delta_{1} (S^{1})$, resulting in: $$ \begin{align*} H_{1}^{\Delta} \left( S^{1} \right) =& \ker \partial_{1} / \operatorname{Im} \partial_{2} \\ \simeq& \Delta_{1} \left( S^{1} \right) / \left\{ 0 \right\} \\ \simeq& \mathbb{Z} \end{align*} $$

For $n \ge 2$, $H_{n}^{\Delta} (S_{1}) \simeq 0$, summarizing as: $$ H_{n}^{\Delta} \left( S_{1} \right) \simeq \begin{cases} \mathbb{Z} & , \text{if } n = 0, 1 \\ 0 & , \text{if } n \ge 2 \end{cases} $$

Torus $T^{2}$

Considering a torus $T^{2}$ as in the image, there’s one 0-simplex (vertex $v$), three 1-simplices (edges $a$, $b$, $c$), two 2-simplices ($U$, $L$), and no $n$-simplices for $n \ge 3$. The chain complex is organized as follows: $$ \cdots \longrightarrow 0 \longrightarrow \Delta_{2}\left( T \right) \overset{\partial_{2}}{\longrightarrow} \Delta_{1}\left( T \right) \overset{\partial_{1}}{\longrightarrow} \Delta_{0}\left( T \right) \overset{\partial_{0}}{\longrightarrow} 0 $$

Hence, the free groups $\Delta_{n} (T)$ are: $$ \Delta_{n} \left( T \right) \simeq \begin{cases} \mathbb{Z}^{1} & , \text{if } n = 0 \\ \mathbb{Z}^{3} & , \text{if } n = 1 \\ \mathbb{Z}^{2} & , \text{if } n = 2 \\ 0 & , \text{if } n \ge 3 \end{cases} $$

Since the edges $a$, $b$, $c$ connect to vertex $v$ at both ends: $$ \begin{align*} \partial a =& v - v = 0 \\ \partial b =& v - v = 0 \\ \partial c =& v - v = 0 \end{align*} $$ and $\partial_{1}$ is a zero morphism, similar to the circle case.

For $n = 0$, the situation mirrors that of the circle: $$ \begin{align*} H_{0}^{\Delta} \left( T \right) =& \ker \partial_{0} / \operatorname{Im} \partial_{1} \\ \simeq& \Delta_{0} \left( T \right) / \left\{ 0 \right\} \\ \simeq& \mathbb{Z} \end{align*} $$

For $n = 1$, since $\partial_{1}$ is a zero morphism, $\ker \partial_{1} = \Delta_{1} (T)$. The boundary homomorphism $\partial_{2} : \Delta_{2}(T) \to \Delta_{1}(T)$ yields: $$ \partial_{2} U = a + b - c = \partial_{2} L $$ and since ${ a, b, a + b - c }$ is a basis for $\Delta_{1}(T)$, $H_{1}^{\Delta}$ is isomorphic to the free group generated by $a$ and $b$, resulting in: $$ H_{1}^{\Delta} \left( T \right) \simeq \mathbb{Z} \oplus \mathbb{Z} $$

For $n = 2$, $\operatorname{Im} \partial_{3} = { 0 }$ and considering the dimensions of $\Delta_{2}(T)$ and $\Delta_{1}(T)$, we get: $$ \begin{align*} H_{2}^{\Delta} \left( T \right) =& \ker \partial_{2} / \operatorname{Im} \partial_{3} \\ \simeq& \mathbb{Z}^{3-2} / \left\{ 0 \right\} \\ \simeq& \mathbb{Z} \end{align*} $$

For $n \ge 3$, $H_{n}^{\Delta} (T) \simeq 0$, summarizing as: $$ H_{n}^{\Delta} \left( T \right) \simeq \begin{cases} \mathbb{Z} & , \text{if } n = 0 \\ \mathbb{Z} \oplus \mathbb{Z} & , \text{if } n = 1 \\ \mathbb{Z} & , \text{if } n = 2 \\ 0 & , \text{if } n \ge 3 \end{cases} $$

Theorem

$H_{n}^{\Delta}$ is a homology group

Definition of a homology group:
Let $n \in \mathbb{N}_{0}$. A sequence of Abelian groups $C_{n}$ and homomorphisms $\partial_{n} : C_{n} \longrightarrow C_{n-1}$ forming a chain $$ \cdots \longrightarrow C_{n+1} \overset{\partial_{n+1}}{\longrightarrow} C_{n} \overset{\partial_{n}}{\longrightarrow} C_{n-1} \longrightarrow \cdots \longrightarrow C_{1} \overset{\partial_{1}}{\longrightarrow} C_{0} \overset{\partial_{0}}{\longrightarrow} 0 $$ that satisfies $$ \partial_{n} \circ \partial_{n+1} = 0 $$ for all $n$ is called a chain complex.
The quotient group $H_{n} := \ker \partial_{n} / \operatorname{Im} \partial_{n+1}$ is called the $n$th homology group of the complex.
The homomorphism $\partial_{n} : C_{n} \longrightarrow C_{n-1}$ is called the boundary or differential operator.

For the chain complex ${ (\Delta_{n} (X), \partial_{n}) }_{n=0}^{\infty}$, $H_{n}^{\Delta} := \ker \partial_{n} / \operatorname{Im} \partial_{n+1}$ is a homology group. That is, $\partial_{n} \circ \partial_{n+1}$ is a zero morphism for all $n \in \mathbb{N}$.

Proof

Applying $\partial_{n-1} \circ \partial_{n}$ to $\sigma \in \Delta_{n}$ yields: $$ \begin{align*} & \left( \partial_{n-1} \circ \partial_{n} \right) \left( \sigma \right) \\ =& \partial_{n-1} \left( \partial_{n} \left( \sigma \right) \right) \\ =& \partial_{n-1} \left( \sum_{i=0}^{n} \left( -1 \right)^{i} \sigma_{\alpha} | \left[ v_{1} , \cdots , \hat{v}_{i} , \cdots , v_{n} \right] \right) \\ =& \sum_{j < i} \left( -1 \right)^{i} \left( -1 \right)^{j} \sigma_{\alpha} | \left[ v_{1} , \cdots , \hat{v}_{i} , \cdots , \hat{v}_{j} , \cdots , v_{n} \right] \\ & + \left( -1 \right) \sum_{j >i} \left( -1 \right)^{i} \left( -1 \right)^{j} \sigma_{\alpha} | \left[ v_{1} , \cdots , \hat{v}_{i} , \cdots , \hat{v}_{j} , \cdots , v_{n} \right] \\ =& 0 \end{align*} $$

Such proofs are often more illuminating with specific examples rather than generalizations. $$ \begin{align*} & \partial_{1} \left( \partial_{2} \left[ v_{0}, v_{1} , v_{2} \right] \right) \\ =& \partial_{1} \left( \left[ v_{1} , v_{2} \right] - \left[ v_{0}, v_{2} \right] + \left[ v_{0}, v_{1} \right] \right) \\ =& \partial_{1} \left[ v_{1} , v_{2} \right] - \partial_{1} \left[ v_{0}, v_{2} \right] + \partial_{1} \left[ v_{0}, v_{1} \right] \\ =& \left[ v_{2} \right] - \left[ v_{1} \right] - \left( \left[ v_{2} \right] - \left[ v_{0} \right] \right) + \left[ v_{1} \right] - \left[ v_{0} \right] \\ =& 0 \end{align*} $$

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