📂Topological Data Analysis

Definition of Persistent Homology groups

Build Up

In the Free group obtained from an Oriented Simplex $k$-Simplex $K$, the boundary operator $\partial_{k} : \mathsf{C}_{k} \to \mathsf{C}_{k-1}$ that satisfies $\partial_{k} \circ \partial_{k+1} = 0$ constitutes a Chain Complex. The Homology group defined as the quotient group of the Cycle group $\mathsf{Z}_{k} := \ker \partial_{k}$ and the Boundary group $\mathsf{B}_{k} := \operatorname{Im} \partial_{k+1}$ is called the $k$th Homology group. $$ \mathsf{H}_{k} := \mathsf{Z}_{k} / \mathsf{B}_{k} $$ Meanwhile, let’s say $K$ has a filtration like the following in a Filtered Complex. $$ K^{0} \subset \cdots \subset K^{i-1} \subset K^{i} \subset \cdots \subset K^{i+p} \subset K^{i+p+1} \subset \cdots \subset K $$ Therefore, since $K^{i}$ are all Simplicial Complexes, we can consider the corresponding boundary operators $\partial_{k}^{i}$ and $\mathsf{C}_{k}^{i}, \mathsf{Z}_{k}^{i}, \mathsf{B}_{k}^{i}$ for each index $i$.

Definition ¹

The following group is called the $k$th $p$-persistent homology group of $K^{i}$. $$ \mathsf{H}_{k}^{i,p} := \mathsf{Z}_{k}^{i} / \left( \mathsf{B}_{k}^{i+p} \cap \mathsf{Z}_{k}^{i} \right) $$ The rank $\beta_{k}^{i,p}$ of $\mathsf{H}_{k}^{i,p}$ is called the $k$th $p$-persistent Betti Number of $K^{i}$.

Explanation

If $p = 0$, then $\mathsf{B}_{k}^{i+0} \cap \mathsf{Z}_{k}^{i} = \mathsf{B}_{k}^{i}$, which is consistent with the original definition of homology groups.

The plethora of subscripts might be dizzying, but what is important is only the conceptual meaning given by being $p$-persistent. $$ \cdots \subset K^{i} \subset \cdots \subset K^{i+p} \subset \cdots $$ In reality, although there are simplicial complexes between $K^{i}$ and $K^{i+p}$ upon looking at the filtration, the definition of the $p$-persistent homology groups does not at all mention them. Conversely, it can be seen as not included in the definition because the intermediate parts are not of interest, considering the same from $i$ to $i+p$. If we interpret this as ‘$K^{i}$ does not algebraically change up to $K^{i+p}$’, then the expression ‘$p$-persistent’ should now be intuitively understood.