Summary of Persistent Homology

We summarize the work so far and relate it to previous results. Our input is a filtered complex K and we wish to find its kth homology H_k. In each dimension the homology of complex K^i becomes a vector space over a field, described fully by its rank \beta_k^i. (Over a field F, H_k is a F-module which is a vector space.)

We need to choose compatible bases across the filtration (compatible bases for H_k^i and H_k^{i+p}) in order to compute persistent homology for the entire filtration. Hence, we form the persistence module \mathscr{M} corresponding to K, which is a direct sum of these vector spaces (\alpha(\mathscr{M})=\bigoplus M^i). By the structure theorem, a basis exists for this module that provides compatible bases for all the vector spaces.

Specifically, each \mathcal{P}-interval (i,j) describes a basis element for the homology vector spaces starting at time i until time j-1. This element is a k-cycle e that is completed at time i, forming a new homology class. It also remains non-bounding until time j, at which time it joins the boundary group B_k^j.

A natural question is to ask when e+B_k^l is a basis element for the persistent groups H_k^{l,p}. Recall the equation \displaystyle H_k^{i,p}=Z_k^i/(B_k^{i+p}\cap Z_k^i). Since e\notin B_k^l for all l<j, hence e\notin B_k^{l+p} for l+p<j. The three inequalities \displaystyle l+p<j,\ l\geq i,\ p\geq 0 define a triangular region in the index-persistence plane, as shown in Figure below.


The triangular region gives us the values for which the k-cycle e is a basis element for H_k^{l,p}. This is known as the k-triangle Lemma:

Let \mathcal{T} be the set of triangles defined by \mathcal{P}-intervals for the k-dimensional persistence module. The rank \beta_k^{l,p} of H_k^{l,p} is the number of triangles in \mathcal{T} containing the point (l,p).

Hence, computing persistent homology over a field is equivalent to finding the corresponding set of \mathcal{P}-intervals.

Source: “Computing Persistent Homology” by Zomorodian and Carlsson

Author: mathtuition88

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