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Persistent homology groups of filtrations

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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.