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# Persistent homology groups of filtrations

In the general context of filtrations the measure of importance of a homology class is its persistence which is the length of its lifespan within the direct system of homology of the filtration.

Recall that a one-parameter filtration is
\begin{equation*}
K^{1}\hookrightarrow K^{2}\hookrightarrow K^{3}\hookrightarrow \ldots \
\hookrightarrow K^{s},
\end{equation*}
where $K^{1},K^{2},\ldots ,K^{s}$ are cell complexes and the arrows represent the inclusions $i^{n,n+1}:K^{n}\hookrightarrow K^{n+1}$. We denote the filtration by $\{K^{n}\}.$ The homology generates a "direct system":
\begin{equation*}
H_{\ast }(K^{1})\rightarrow H_{\ast }(K^{2})\rightarrow \ldots \ \rightarrow
H_{\ast }(K^{s})\longrightarrow 0.
\end{equation*}
denoted by $\{H_{\ast }(K^{n})\}.$ Then the *homology group of filtration* $\{K^{n}\}$ is defined as the product of the kernels of the inclusions:
\begin{equation*}
H_{\ast }(\{K^{n}\})=\ker i_{\ast }^{1,2}\oplus \ker \,i_{\ast }^{2,3}\oplus
\ldots \oplus \ker i_{\ast }^{s,s+1}.
\end{equation*}

Given filtration $\{K^{n}\},$ we say that *the persistence* $P(x)$ of $x\in H_{\ast}(K^{n})$ *is equal to* $p$ if

- $i_{\ast}^{n,n+p}(x)=0$ and
- $i_{\ast}^{n,n+p-1}(x)\neq 0.$

Our interest is in the "robust" homology classes, i.e., the ones with high persistence. However, the collection of these classes is not a group as it doesn't even contain $0$. Therefore we have to deal with "noise" first.

Given a positive integer $p,$ *the $p$-noise (homology) group $N_{\ast}^{p}(K^{n})$ of $\{K^{n}\}$* is the group of all elements of $K^{n}$ with persistence less than $p.$

Alternatively, we can define these groups via the kernels of the homomorphisms of the inclusions: $$N_{\ast }^{p}(K^{n})=\ker \,i_{\ast }^{n,n+p}.$$

**Proposition.** $N_{\ast}^{p+1}(K^{n})\subset N_{\ast}^{p}(K^{n}).$

Next, we "remove" the noise from the homology group. *The $p$-persistent (homology) group} of $K^{n}$ with respect to the filtration $\{K^{n}\}$* is defined as
\begin{equation*}
H_{\ast }^{p}(K^{n})=H_{\ast }(K^{n})/N_{\ast }^{p}(K^{n}).
\end{equation*}
The point of this definition is that, given a threshold for noise, *if the difference between two homology classes is noise, they should be equivalent*.

Next, just as in the case of noise-less analysis, we define a single structure to capture all (robust) homology classes.

Let $p$ be a positive integer. Suppose $x\in \ker i_{\ast }^{k,k+p}$ and let $y=$ $i_{\ast}^{k,k+1}(x).$ Then \begin{eqnarray*} i_{\ast }^{k+1,k+1+p}(y) &=&i_{\ast }^{k+1,k+1+p}(i_{\ast }^{k,k+1}(x)) \\ &=&i_{\ast }^{k,k+1+p}(x)=i_{\ast }^{k+p,k+p+1}(i_{\ast }^{k,k+p}(x)) \\ &=&i_{\ast }^{k+p,k+p+1}(0)=0. \end{eqnarray*} Hence $y\in \ker i_{\ast }^{k+1,k+1+p}.$ We have proved that \begin{equation*} i_{\ast }^{k,k+1}(\ker i_{\ast }^{k,k+p})\subset \ker i_{\ast }^{k+1,k+1+p}. \end{equation*} It follows that the homomorphism $$i_{\ast }^{k,k+1}:\ker i_{\ast}^{k,k+p}\rightarrow \ker i_{\ast }^{k+1,k+1+p}$$ generated by the inclusion is well-defined.

Next, we use these homomorphisms to define *the $p$-noise (homology) group $N_{\ast }^{p}(\{K^{n}\})$ of filtration $\{K^{n}\}$ * as
\begin{equation*}
N_{\ast }^{p}(\{K^{n}\})=\ker i_{\ast }^{1,2}\oplus \ldots \oplus \ker
i_{\ast }^{s,s+1}.
\end{equation*}
Observe that the formula is the same as the one in the definition of $H_{\ast}^{p}(\{K^{n}\}).$

Since $i_{\ast }^{k,k+1}:\ker i_{\ast}^{k,k+p}\rightarrow \ker i_{\ast }^{k+1,k+1+p}$ is a restriction of $$i_{\ast }^{k,k+1}:H_{\ast }^{p}(K^{k})\rightarrow H_{\ast }^{p}(K^{k+1}),$$ each term in the above definition is a subgroup of the corresponding term in the definition of $H_{\ast }(\{K^{n}\}).$ The proposition below follows.

**Proposition.** $N_{\ast }^{p}(\{K^{n}\})\subset H_{\ast }(\{K^{n}\}).$

Finally, *the $p$-persistent (homology) group of filtration $\{K^{n}\}$* is
\begin{equation*}
H_{\ast }^{p}(\{K^{n}\})=H_{\ast }(\{K^{n}\})/N_{\ast }^{p}(\{K^{n}\}).
\end{equation*}

The results about $H_{\ast }^{p}(\{K^{n}\})$ analogous to the ones about $H_{\ast }(\{K^{n}\})$ also hold.

**Exercise:** State and prove these propositions.