This site is devoted to mathematics and its applications. Created and run by Peter Saveliev.

# Persistence of homology classes in filtrations

Redirect page

## Introduction

While constructing a simplicial complex from a point cloud we gradually add cells of various dimensions in order to discover the topology of the object.

For example, this is what a point cloud of a circle might look like:

But how do we know which one of these complexes is the "true" representation without looking at it?

Or, more narrowly, which one captures the hole of the circle, without us looking at it to confirm?

To answer this question we look at the whole sequence and how it is developing.

To evaluate the topology one has to take into account the robustness of the topological features in such a sequence. Recall, a filtration is a sequence of "nested" cell complexes, where the arrows are inclusions: $$K^1↪K^2↪K^3↪...↪K^s.$$

The measurement for this robustness is called persistence.

## Example

Let's consider a simpler example:

In the top row we see the complexes of the filtration. In each "frame", the new cells are marked in red.

In the bottom row, the new cycles are given in various colors.

• The brown $0$-cycle appears in the 1st frame and dies in the 2nd.
• The blue $0$-cycle appears in the 3rd frame and dies in the 5th.
• The green $1$-cycle appears in the 3rd frame and dies in the 4th.
• The orange $1$-cycle appears in the 4th frame and dies in the 6th.

The persistence a homology class can be evaluated as its life-span: how long does it take for these cycles to disappear.

Therefore, the persistences of the above cycles are 1, 2, 1, 2, respectively.

Now, in order to settle on a particular topology we choose a threshold for persistence. If the choice is 1, then the "persistence homology" of the filtration is generated by two $0$-cycles and two $1$-cycles. If, however, the choice is $2$, there is only one of each.

Which is the "correct" one, is still a judgement call.

Why is this a fruitful idea? Because this way we deal with ambiguity in a purely topological way.

## Definition

Now, what about homology groups?

Let's look at the homology groups of the elements of the filtration.

Dimension $0$ homology groups: $${\bf Z}\times {\bf Z}\rightarrow {\bf Z} \rightarrow {\bf Z}\times {\bf Z} \rightarrow {\bf Z}\times {\bf Z} \rightarrow {\bf Z} \rightarrow {\bf Z}.$$ These tell the whole story; cycles appear and disappear and the homology groups change accordingly.

Dimension $1$ homology groups: $$0\rightarrow 0 \rightarrow {\bf Z} \rightarrow {\bf Z} \rightarrow {\bf Z} \rightarrow 0.$$ These don't tell the whole story; it is impossible to deduce that one cycle appeared and another disappeared at the same time! The persistence information is lost.

Recall that the $k$th homology group of $$K^{i}$$ is defined as $$H_{k}(K^{i})=Z_{k}^{i}/B_{k}^{i},$$ where

• $$Z_{k}^{i}$$ is the group of k-cycles in $$K^{i}$$ and
• $$B_{k}^{i}$$ is the group of k-boundaries in $$K^{i}$$.

Now, we are trying to capture the cycles that appear, as cycles, in $i$th complex and were killed, as boundaries, in the $(i+p)$th.

Given a positive integer $p$, the $p$-persistence $k$th homology group of $\{K^{i}\}$ is $$H_{k}^{i,p}(K^{i})=Z_{k}^{i}/(B_{k}^{i+p}∩Z_{k}^{i}).$$

The persistent Betti numbers are the ranks of the persistent homology groups.

For a better analysis one will need homology maps of the inclusions.

## Experiment with a dataset

Below is an example of how we determine the persistent Betti numbers of a point cloud computed with the homology software called JPlex.

Specifically, noise has been added to a point cloud of the plane in space.

In these diagrams we provide the lifespans of the cycles called barcodes:

We can see that there is one cycle in dimension $0$ that has a much longer lifespan than others. So the set probably connected!

In dimension $1$, no single feature persists significantly. So it's probably simply connected!

Image analysis software called Pixcavator finds persistent Betti numbers of 2D gray scale images with respect to the filtration created by thresholding and, in this case,

persistence = contrast.