Quantitative Simplification of Filtered Simplicial Complexes

Abstract

We introduce a new invariant defined on the vertices of a given filtered simplicial complex, called codensity, which controls the impact of removing vertices on the persistent homology of this filtered complex. We achieve this control through the use of an interleaving type of distance between filtered simplicial complexes. We study the special case of Vietoris–Rips filtrations and show that our bounds offer a significant improvement over the immediate bounds coming from considerations related to the Gromov–Hausdorff distance. Based on these ideas we give an iterative method for the practical simplification of filtered simplicial complexes. As a byproduct of our analysis we identify a notion of core of a filtered simplicial complex which admits the interpretation as a minimalistic simplicial filtration which retains all the persistent homology information.

Appendix

Proof of Proposition 2.8

Let R be a correspondence between M, N (i.e. $R \subseteq M \times N$ and $\pi _M(R)=M$, $\pi _N(R)=N$). Note that R can be considered as a tripod between $X^*,Y^*$. By Remark 2.7, it is enough to show that the distortion of R as a metric correspondence between M, N is same with the distortion of R as a tripod between $X^*,Y^*$. Let us denote the first one by $\mathrm {dis}^\mathrm{met}(R)$ and the second one by $\mathrm {dis}^\mathrm{tri}(R)$.

Claim

$\mathrm {dis}^\mathrm{tri}(R) \ge \mathrm {dis}^\mathrm{met}(R)$.

Proof

By Remark 2.3, the size functions of $X^*,Y^*$ are given by the diameter. Hence we have:

$$\begin{aligned} \mathrm {dis}^\mathrm{tri}(R)&\ge \max _{(x,y),(x',y') \in R} |\mathrm {diam}_M(x,x')-\mathrm {diam}_N(y,y')|\\&= \max _{(x,y),(x',y') \in R} |d_M(x,x')-d_N(y,y')|\\&=\mathrm {dis}^\mathrm{met}(R). \end{aligned}$$

$\square$

Claim

$\mathrm {dis}^\mathrm{tri}(R) \le \mathrm {dis}^\mathrm{met}(R)$.

Proof

Let $\alpha \in \mathrm {P}(R)$. Let $\alpha \in \mathrm {P}(R)$ such that

$$\begin{aligned} \mathrm {dis}^\mathrm{tri}(R)=|\mathrm {diam}_M(\pi _M(\alpha ))-\mathrm {diam}_N(\pi _N(\alpha ))|. \end{aligned}$$

Without loss of generality, we can assume that

$$\begin{aligned} \mathrm {diam}_M(\pi _M(\alpha )) \ge \mathrm {diam}_N(\pi _N(\alpha )). \end{aligned}$$

Let $x,x'$ be points in $\pi _M(\alpha )$ so that

$$\begin{aligned} \mathrm {diam}_M(\pi _M(\alpha ))=d_M(x,x'). \end{aligned}$$

There exists points $y,y'$ in N such that $(x,y),(x',y') \in \alpha$. Then we have

$$\begin{aligned} \mathrm {dis}^\mathrm{tri}(R)&= \mathrm {diam}_M(\pi _M(\alpha ))-\mathrm {diam}_N(\pi _N(\alpha )) \\&= d_M(x,x')-\mathrm {diam}_N(\pi _N(\alpha ))\\&\le d_M(x,x')-d_N(y,y')\\&\le \mathrm {dis}^\mathrm{met}(R). \end{aligned}$$

$\square$

Proof of Proposition 2.9

Non-negativity and symmetry properties follows from the definition. $d_\mathrm {GH}(X^*,X^*)=0$ since the distortion of the identity tripod on the vertex set of X is 0. Let us show the triangle inequality. Let $(Z,p,p')$ be a tripod between $X^*,X'^*$ and $(Z',q',q'')$ be a tripod between $X'^*,X''^*$. Let $Z''$ be the fiber product

$$\begin{aligned} Z''=Z {_{p'}\times _{q'}} Z' . \end{aligned}$$

Then $(Z'',p \circ \pi _Z,q'' \circ \pi _{Z'})$ is a tripod between $X^*,X''^*$. Given $\alpha \in \mathrm {P}(Z'')$, we have

$$\begin{aligned} |\mathrm {D}_X(p \circ \pi _Z(\alpha ))- \mathrm {D}_{X''}(q'' \circ \pi _{Z'})|&\le |\mathrm {D}_X(p \circ \pi _Z(\alpha ))-\mathrm {D}_{X'}(p' \circ \pi _Z(\alpha ))| \\&\quad +|\mathrm {D}_{X'}(p' \circ \pi _Z(\alpha )) - \mathrm {D}_{X''}(q'' \circ \pi _{Z'}(\alpha ))| \\&=|\mathrm {D}_X(p \circ \pi _Z(\alpha ))-\mathrm {D}_{X'}(p' \circ \pi _Z(\alpha ))| \\&\quad +|\mathrm {D}_{X'}(q' \circ \pi _{Z'}(\alpha )) - \mathrm {D}_{X''}(q'' \circ \pi _{Z'}(\alpha ))| \\&\le \mathrm {dis}(Z)+\mathrm {dis}(Z'). \end{aligned}$$

Since $\alpha \in \mathrm {P}(Z'')$ was arbitrary, $\mathrm {dis}(Z'') \le \mathrm {dis}(Z')+\mathrm {dis}(Z'')$. Since the tripods $Z,Z'$ were arbitrary, $d_\mathrm {GH}(X^*,X''^*) \le d_\mathrm {GH}(X^*,X'^*)+d_\mathrm {GH}(X'^*,X''^*)$. $\square$

Proof of Lemma 2.23

Let $p_X$ (resp. $p_Y$) be the projection map from R to the vertex set of $X^*$ (resp. $Y^*)$. Let $\alpha$ be a non-empty subset of the vertex set of $X^*$. Note that

$$\begin{aligned} f(\alpha ) \subseteq p_Y(p_X^{-1}(\alpha )). \end{aligned}$$

Let $\varepsilon :=\mathrm {dis}(R)$. We have

$$\begin{aligned} \mathrm {D}_{Y^*}(f(\alpha ))&\le \mathrm {D}_{Y^*}(p_Y(p_X^{-1}(\alpha )))\\&\le \mathrm {D}_{X^*}(p_X(p_X^{-1}(\alpha )))+\varepsilon \\&=\mathrm {D}_{X^*}(\alpha )+\varepsilon . \end{aligned}$$

Hence $\mathrm {deg}(f) \le \varepsilon$. Similarly $\mathrm {deg}(g) \le \varepsilon$.

Note that $p_X(p_Y^{-1}(f(\alpha )))$ contains both $\alpha$ and $g \circ f(\alpha )$. Hence

$$\begin{aligned} \mathrm {D}_{X^*}(g \circ f (\alpha ) \cup \alpha )&\le \mathrm {D}_{X^*}(p_X(p_Y^{-1}(f(\alpha ))))\\&\le \mathrm {D}_{Y^*}(p_Y(p_Y^{-1}(f(\alpha ))))+\varepsilon \\&= \mathrm {D}_{Y^*}(f(\alpha ))+\varepsilon \\&\le \mathrm {D}_{X^*}(\alpha )+2\varepsilon . \end{aligned}$$

This shows that

$$\begin{aligned} \mathrm {codeg}^\infty (g \circ f, \mathrm {id}_{X^*})\le \mathrm {codeg}(g \circ f, \mathrm {id}_{X^*}) \le 2\varepsilon . \end{aligned}$$

Similarly,

$$\begin{aligned} \mathrm {codeg}^\infty (f \circ g,\mathrm {id}_{Y^*}) \le 2\varepsilon . \end{aligned}$$

This completes the proof. $\square$

Proof of Proposition 4.6

Let us start by fixing some notation. Let A be a finite subset of X. We denote the homology maps induced by the inclusion $A \hookrightarrow X$ by $\iota _A:\mathrm {H}_k(\mathrm {VR}^r(A)) \rightarrow \mathrm {H}_k(\mathrm {VR}^r(X))$. If B is another finite subset of X such that $A \subseteq B$, we denote the homology map induced by this inclusion by $\iota _{A,B}:\mathrm {H}_k(\mathrm {VR}^r(A)) \rightarrow \mathrm {H}_k(\mathrm {VR}^r(B))$. If z is chain in $\mathrm {VR}^r(X)$ such that the vertices of z is contained in A, we denote the homology class it represents in $\mathrm {H}_k(\mathrm {VR}^r(A))$ by $[z]_A$.

The homology maps $\iota _A:\mathrm {H}_k(\mathrm {VR}^r(A)) \rightarrow \mathrm {H}_k(\mathrm {VR}^r(X))$ shows that $\mathrm {H}_k(\mathrm {VR}^r(X))$ is a cocone for J. Let us show that it is universal. Let M be another cocone for J with morphisms $\phi _A:\mathrm {H}_k(\mathrm {VR}^r(A)) \rightarrow M$. Define a map $u:\mathrm {H}_k(\mathrm {VR}^r(X)) \rightarrow M$ as follows. Given a homology class c in $\mathrm {H}_k(\mathrm {VR}^r(X))$, let z be a cycle representing it and let A be a finite subset containing the vertices of z. Define $u(c):= \phi _A([z]_A)$. Let us show that this map is well defined. Let $z'$ be another cycle representing c and $A'$ be a finite subset containing the vertices of $z'$. Note that $z-z'$ is a boundary in $\mathrm {VR}^r(X)$, let w be a chain in $\mathrm {VR}^r(X)$ so that $\partial w = z-z'$. Let B a finite subspace of X containing $A,A'$ and vertices of w. Note that $[z]_B=[z']_B$ as w is contained in B. We have

$$\begin{aligned} \phi _A([z]_A)= & {} \phi _B \circ \iota _{A,B}([z]_A)=\phi _B([z]_B)=\phi _B([z']_B)\\= & {} \phi _B \circ \iota _{A',B}([z']_B)=\phi _{A'}([z']_{A'}). \end{aligned}$$

This shows that $u:\mathrm {H}_k(\mathrm {VR}^r(X)) \rightarrow M$ is well defined. It is an R-module homomorphism since if $c,c'$ are homology classes in $\mathrm {H}_k(\mathrm {VR}^r(X))$ represented by cycles $z,z'$ whose vertices are contained in a finite subspace A, then for any $\lambda \in R$ we have

$$\begin{aligned} u(c + \lambda \, c')= & {} \phi _A([z]_A + \lambda \, [z']_A)\\= & {} \phi _A([z]_A)+\,\lambda \, \phi _A([z']_A) =u(c)+\lambda \, u(c'). \end{aligned}$$

Given a homology class c in $\mathrm {H}_k(\mathrm {VR}^r(A))$ which is represented by a cycle z, we have

$$\begin{aligned} u \circ \iota _A (c)= \phi _A([z]_A)=\phi _A(c), \end{aligned}$$

hence u commutes with structure maps. The uniqueness of such u follows from the fact that for every homology class c in $\mathrm {H}_k(\mathrm {VR}^r(X))$ there exists a finite subset A such that c is contained in the image of $\iota _A:\mathrm {H}_k(\mathrm {VR}^r(A)) \rightarrow \mathrm {H}_k(\mathrm {VR}^r(X))$. $\square$

Proof of Proposition 4.7

Let $p \in X$, $q \in Y$ be the chosen points for the wedge sum. Note that $\mathrm {VR}^r(X) \vee \mathrm {VR}^r(Y)$ is contained in $\mathrm {VR}^r(X \vee Y)$ for each $r \ge 0$. Let us show that this inclusion induces an isomorphism between homology groups. Note that this is enough for our proof as these inclusions commutes with the structure maps of both filtered simplicial complexes.

Order the disjoint union of $X \vee Y$ and denote the smallest element of a finite subset $\sigma$ by $\min (\sigma )$. It is known that for a simplicial complex S with ordered vertices, the map $B(S) \rightarrow S$ from the barycentric subdivision B(S) of S to S defined by $\sigma \mapsto \min (\sigma )$ is simplicial and induces an isomorphism between homology groups [28, pp. 166, 167].

Consider the map $f:B\big (\mathrm {VR}^r(X \vee Y) \big ) \rightarrow \mathrm {VR}^r(X) \vee \mathrm {VR}^r(Y)$ defined as follows: $f(\sigma )=\min (\sigma )$ if $\sigma$ is contained in X or Y, $f(\sigma )=p=q$ else. Let us see that this map is simplicial. Take a simplex $S=(\sigma _1 \subset \sigma _2 \subset \dots \subset \sigma _n)$ in $B\big (\mathrm {VR}^r(X \vee Y) \big )$. Without loss of generality, assume that k is the maximal integer such that for $i \le k$, $\sigma _i \subseteq X$ and furthermore assume that for $i>k$, $\sigma _i$ is neither contained in X nor in Y. If $k=n$, then $f(S) \subseteq \sigma _n$, hence it is a simplex in $\mathrm {VR}^r(X) \vee \mathrm {VR}^r(Y)$. If $k \ne n$, then $\sigma _n$ contains elements from both X and Y, hence $\sigma _n \cup \{p\}$ is a simplex in in $\mathrm {VR}^r(X \vee Y)$, which in turn implies that $\sigma _k \cup \{p\}$ is a simplex in $\mathrm {VR}^r(X)$. Hence, $f(S) \subseteq \sigma _k \cup \{p\}$ is a simplex in $\mathrm {VR}^r(X) \vee \mathrm {VR}^r(Y)$. Therefore, f is simplicial.

Consider the following (non-commutative) diagram:

Here, i, j are inclusions and $\varphi ,\psi$ are maps defined by $\sigma \mapsto \min (\sigma )$. This diagram is non-commutative only because $i \circ f$ is not equal to $\psi$. All other commutativity relations hold. Let us show that $i \circ f$ is contiguous to $\psi$, hence the non-commutativity disappears when we pass to homology.

As above, take a simplex $S=(\sigma _1 \subset \sigma _2 \subset \dots \subset \sigma _n)$ in $B\big (\mathrm {VR}^r(X \vee Y) \big )$. Without loss of generality, assume that k is the maximal integer such that for $i \le k$, $\sigma _i \subseteq X$ and furthermore assume that for $i>k$, $\sigma _i$ is neither contained in X nor in Y. If $k=n$, then S is in the image of j, therefore $i \circ f(S) = \psi (S)$. If $k<n$, then we have

$$\begin{aligned} i \circ f(S) \cup \psi (S) \subseteq \sigma _n \cup \{p \} \in \mathrm {VR}^r(X\vee Y). \end{aligned}$$

This shows the contiguity. Now we have the following commutative diagram:

Now, the induced map $f_*$ is surjective since $\varphi _*$ is surjective and $f_*$ is injective since $\psi _*$ is injective. Hence $f_*$ is an isomorphism, which implies that $i_*=\psi _* \circ f_*^{-1}$ is an isomorphism. $\square$