This site is being phased out.

# Persistence via homology operators

Redirect to:

The stability of topology in the presence of noise is called persistence.

The analysis is based on the concept of filtration, which is a sequence of "nested" cell complexes, where the arrows are inclusions: $$K^1↪K^2↪K^3↪...↪K^s.$$ Then we have a sequence of homology groups connected by homomorphisms: $$H_{∗}(K¹)→H_{∗}(K²)→… →H_{∗}(K^{s})→0.$$

Previously we defined persistent homology on the chain level. It is easier to understand on the homology level, how long does it take for the homology operators to "kill" the given homology class?

## Example

Previously we have considered persistence of homology classes in filtrations for this example, on the chain level:

This time we look at the homology classes specifically. The persistence of a homology class can be evaluated as its *life-span*: how many applications of the homology operators of the inclusions does it take for the homology class to become $0$.

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

Since we don't want to look at one class at a time, let's look at the homology groups (over ${\bf R}$) and the homology operators of the whole filtration.

- Dimension $0$:
- homology groups: $${\bf R}\times {\bf R}\rightarrow {\bf R} \rightarrow {\bf R}\times {\bf R} \rightarrow {\bf R}\times {\bf R} \rightarrow {\bf R} \rightarrow {\bf R}. $$
- homology maps: $$[1,1], [1,0]^T, Id, [1,1], Id.$$

- Dimension $1$:
- homology groups: $$0\rightarrow 0 \rightarrow {\bf R} \rightarrow {\bf R} \rightarrow {\bf R} \rightarrow 0. $$
- homology maps: $$0, 0, [0,1]^T, Id, 0.$$

Now, the definition of the persistent homology of a complex $K^i$ in the filtration is clear:

- the $0$-persistence homology is simply its homology group;
- the $1$-persistence group is the orthogonal complement of the kernel of the homology map of the inclusion of the complex into the next one;
- the $2$-persistence group is the orthogonal complement of the kernel of the homology map of the composition of the two inclusions;
- etc.

**Exercise.** Compute the persistent homology groups of this filtration.

Later we define a single group that captures all information about the homology of the filtration.

## Another example

It is tempting to conclude that each topological feature in the space being discretized is captured by at least one of the elements of the filtration. The example below shows that this isn't always the case:

Here we discretize a V-shaped figure (left) with $3\times 3$, $2\times 2$, and $1\times 1$ grids. In each case the result is topologically the same:

Of course, there is no hole in the original!

It is clear that improving the resolution won't change the result. In fact the last image on the right shows the left bottom pixel zoomed in and a disconnected pixel (hole) is still present.

Now, these three images -- going from right to left -- form a filtration.

The bad news is that the homology, $H_1(K^i),i=1,2,3$, produced by this filtration are identical and the hole might appear -- incorrectly -- persistent.

The good news that with a help of the homology maps the issue is resolved. At every step, the hole goes to zero under the homology operator of the inclusion. Therefore the persistence of each of the holes is equal to 1!

## Experiments

Below is an example of how we determine the persistent Betti numbers of a point cloud computed with JPlex. To be precise, these are relative Betti numbers, i.e., the edges are identified (see quotient spaces).

Below noise has been added to a plane in space.

In addition to computing the Betti numbers as the global topological characteristics of the complex, homology theory can also provide local information. For every small patch $P$ of $K$ we compute the *relative homology* $H_{\ast}(K,K\backslash P)$ by, essentially, collapsing the complement of $P$ to a single point. Then, if $K$ is a manifold, its *dimension* is equal to $n$ provided:

- $H_{n}(K,K\backslash P)\neq0$ and
- $H_{k}(K,K\backslash P)=0$ for $k\neq n.$

In the above barcodes, we see the following.

- In dimension $0$, one feature persists longer that other dimension $0$ features.
- In dimension $1$, no single feature has a particularly long lifespan compared to other dimension $1$ features.
- In dimension $2$, one features has a significant lifespan.

By analyzing the barcodes of relative Betti numbers, we determine the dimension of our data set. In this example, the persistent Betti numbers are:
$$B_0 = 1, B_1 = 0, B_2 = 0, …$$
Thus the data set constitutes a single part, and has no other topological features. Next, the persistent relative Betti numbers are:
$$B_0 = 1, B_1 = 0, B_2 = 1, B_3 = 0, …$$
The result confirms that the data set is *two-dimensional*.

**Project:** Create a digital geometric figure with no symmetries, create the dataset of all photos (or, alternatively, the projections) from all possible directions, evaluate the homology of the dataset with JPlex. The homology produced this way is supposed to coincide with

- either the homology of the sphere ${\bf S}^2$:

$$H_1({\bf S}^2)=0,H_2({\bf S}^2)={\bf Z};$$

- or the homology of the rotation group $SO(3)$:

$$H_1(SO(3);{\bf Z})={\bf Z}_2.$$