Topological spaces of persistence modules and their properties

Abstract

Persistence modules are a central algebraic object arising in topological data analysis. The notion of interleaving provides a natural way to measure distances between persistence modules. We consider various classes of persistence modules, including many of those that have been previously studied, and describe the relationships between them. In the cases where these classes are sets, interleaving distance induces a topology. We undertake a systematic study the resulting topological spaces and their basic topological properties.

A The arithmetic of maps and interleavings of interval modules

In this appendix, we give some basic results on interval modules, maps of interval modules, interleavings of interval modules, and neighborhoods of interval modules.

1.1 A.1 Some relations between intervals

First we define some relations between intervals that will be useful in the following sections and describe some of their properties.

Recall that $I \subset {\mathbb {R}}$ is an interval if $a, c \in I$ and $a \le b \le c$ then $b \in I$. It follows that the intersection of two intervals is an interval.

Definition 9

For $A,B \subset {\mathbb {R}}$, define the relation $A \le B$ if

• 1. for all $a \in A$ there is a $b \in B$ such that $a \le b$, and

• 2. for all $b \in B$ there is an $a \in A$ such that $a \le b$.

Lemma 18

This relation defines a partial order on intervals.

Proof

Let A, B, and C be intervals. $A \le A$ since for all $a \in A$, $a \le a$. Assume $A \le B$ and $B \le A$. Let $a \in A$. Then by Definition 9 (1), there is $b \in B$ with $a \le b$, and by Definition 9 (2), there is $b' \in B$ with $b' \le a$. Since B is an interval $a \in B$. Thus $A \subset B$. Similarly $B \subset A$.

Finally assume $A \le B$ and $B \le C$. For all $a \in A$ there is a $b \in B$ with $a \le b$ and $c \in C$ with $b \le c$. Thus $a \le c$. For all $c \in C$ there is a $b \in B$ with $b \le c$ and $a \in A$ with $a \le b$. Thus $a \le c$. Therefore $A \le C$. $\square$

Let us define another relation.

Definition 10

For $A,B \subset {\mathbb {R}}$, define $A \prec B$ if for all $a \in A$ and $b \in B$, $a \le b$.

Lemma 19

Let I and J be disjoint, nonempty intervals. Then $J \le I$ iff $J \prec I$.

Proof

See Fig. 8. Let $j \in J$. Then either condition implies that there is an $i \in I$ with $j \le i$. The negation of either condition implies that there is an $i \in I$ with $i<j$. Since I is an interval, this would imply that $j \in I$ which is a contradiction. $\square$

Fig. 8

Two disjoint nonempty intervals

Lemma 20

If J and I are intervals with $J \le I$ then $J \setminus (I \cap J) = J \setminus I$ is an interval and $I \setminus (I \cap J) = I \setminus J$ is an interval.

Proof

Let $a,c \in J \setminus (I \cap J)$ and $a \le b \le c$. Since J is an interval, $b \in J$. Since $c \in J$ there is a $d \in I$ with $c \le d$. Since $c \not \in I$ and I is an interval, $b \not \in I$. Thus $b \in J \setminus (I \cap J)$.

Let $a,c \in I \setminus (I \cap J)$ and $a \le b \le c$. Since I is an interval $b \in I$. Since $a \in I$ there is a $x \in J$ with $x \le a$. Since $a \not \in J$ and J is an interval, $b \not \in J$. Thus $b \in I \setminus (I \cap J)$. $\square$

Lemma 21

Let I and J be intervals with $J \le I$. Then $J \setminus (I \cap J) \prec (I \cap J)$, and $(I \cap J) \prec I \setminus (I \cap J)$.

Proof

See Fig. 9. First note that if either A or B is empty then $A \prec B$. Suppose $j \in J \setminus (I \cap J)$ and $i \in I \cap J$ with $i < j$. Since $J \le I$, there is an $i' \in I$ with $j \le i'$. Since I is an interval, $j \in I$, which is a contradiction. Thus, for all $j \in J \setminus (I \cap J)$ and for all $i \in I \cap J$, $j \le i$. That is, $J \setminus (I \cap J) \prec (I \cap J)$. Similarly, let $j \in I \cap J$ and $i \in I \setminus (I \cap J)$ with $i < j$. Again, since $J \le I$, there is a $j' \in J$ with $j'\le i$. Since J is an interval, $i \in J$, which is a contradiction. $\square$

Fig. 9

The interval modules in Lemmas 20, 21, 22, Proposition 16 and Corollary 11

1.2 A.2 Nonzero maps of interval modules

In this section we characterize nonzero maps of interval modules.

Proposition 16

Let I and J be nonempty intervals. There is a nonzero map of persistence modules $f:I \rightarrow J$ if and only if $J \le I$ and $I \cap J \ne \emptyset$.

Proof

$(\Rightarrow )$ Assume $f \ne 0$. Then there is an $a \in {\mathbb {R}}$ such that $0 \ne f_a:I(a) \rightarrow J(a)$. Without loss of generality, assume that $f_a = 1$. Thus $a \in I$ and $a \in J$. We need to check the conditions in Definition 9.

(1) For all $i \in I$ with $a \le i$, the condition is satisfied by $a \in J$. For all $i \in I$ with $i \le a$, we have the following commutative diagram,

which implies that $i \in J$, and thus $J(i\le a) = 1$, and therefore $f_i= 1$. (2) For all $j \in J$ with $j \le a$, the condition is satisfied by $a \in I$. For all $j \in J$ with $a \le j$, we have the following commutative diagram,

which implies that $j \in I$, $I(a \le j) = 1$, and $f_j= 1$.

$(\Leftarrow )$ Define $f:I \rightarrow J$ by $f_a=1$ if $a \in I \cap J$, and $f_a=0$ otherwise. We claim that f is a natural transformation. For $a \le b$, we need to check that the diagram

commutes. There are four cases to check. If $a,b \in I \cap J$, then all four maps are the identity and thus the diagram commutes. If $a,b \not \in I \cap J$ then both vertical maps are zero and thus the diagram commutes.

If $a \in I \cap J$ and $b \not \in I \cap J$ then by definition the left map is the identity and the right map is zero. If $b \in J$ then $b \not \in I$, which implies, since I is an interval, that for all $c \ge b$, $c \not \in I$. But this contradicts Definition 9 (1). Therefore $b \not \in J$. Thus $J(b) = 0$ and hence the diagram commutes.

If $a \not \in I \cap J$ and $b \in I \cap J$, then $f_a=0$ and without loss of generality $f_b=1$. Again $a \in I$ implies $a \not \in J$, which implies that for all $c \le a$, $c \not \in J$, which is a contradiction. Therefore $a \not \in I$ which implies that $I(a) = 0$ and thus the diagram commutes. $\square$

Lemma 22

Assume there is a nonzero map $f: I \rightarrow J$ of interval modules. Then (up to isomorphism) $f_a = 1$ if $a \in I \cap J$ and $f_a = 0$ otherwise.

Proof

Assume $f \ne 0$. The there is a $b \in I \cap J$ such that $f_b$ is nonzero. Without loss of generality, we may assume that $f_b = 1$. Let $a \le b \le c \in I \cap J$. We have the following commutative diagram,

which implies that $f_a = 1$ and $f_c=1$. Thus $f_a=1$ for all $a \in I \cap J$.

If $a \not \in I \cap J$ then either $I(a)=0$ or $J(a)=0$, which implies that $f_a=0$. $\square$

Corollary 11

Let $f:I \rightarrow J$ be a nonzero map of interval modules. Then the image of f is $I \cap J$, the kernel of f is $I \setminus (I \cap J)$, the cokernel of f is $J \setminus (I \cap J)$, and f factors as follows.

1.3 A.2 Interleavings of interval modules

In this section we characterize interleavings of interval modules.

Definition 11

Let I be an interval and $\varepsilon \in {\mathbb {R}}$. Define the shifted interval$I[\varepsilon ]$ by $x \in I[\varepsilon ]$ if and only if $x+\varepsilon \in I$. For example, $[a,b)[\varepsilon ] = [a-\varepsilon ,b-\varepsilon )$.

The next lemma follows immediately from the definitions.

Lemma 23

If I is a nonempty interval and $\varepsilon \ge 0$, then $I[\varepsilon ] \le I$.

Definition 12

Let M be a persistence module and let $\varepsilon \in {\mathbb {R}}$. We define the shifted persistence module$M[\varepsilon ]$ by $M[\varepsilon ](a) = M(a + \varepsilon )$ and $M[\varepsilon ](a \le b) = M(a + \varepsilon \le b + \varepsilon )$. That is, $M[\varepsilon ] = MT_{\varepsilon }$.

We remark that these two definitions are compatible. If I is an interval module and $\varepsilon \in {\mathbb {R}}$, then the shifted persistence module $I[\varepsilon ]$ is the interval module on the interval $I[\varepsilon ]$. Also note that $0[\varepsilon ] = 0$.

Let I be an interval and $\varepsilon \ge 0$. If $I \cap I[\varepsilon ] \ne \emptyset$, we denote the corresponding nonzero map from Proposition 16 by $I^{(\varepsilon )}:I \rightarrow I[\varepsilon ]$. If I and $I[\varepsilon ]$ are disjoint, we denote the zero map by $I^{(\varepsilon )}:I \rightarrow I[\varepsilon ]$. In either case, $I^{(\varepsilon )} = I\eta _{\varepsilon }$.

Definition 13

Given a map of persistence modules $\alpha : M \rightarrow N$ and $\varepsilon \in {\mathbb {R}}$, define $\alpha [\varepsilon ]: M[\varepsilon ] \rightarrow N[\varepsilon ]$ by $\alpha [\varepsilon ]_a = \alpha _{a+\varepsilon }$. That is, $\alpha [\varepsilon ] = \alpha T_{\varepsilon }$.

As a special case of Definition 1, we have the following.

Definition 14

Let I and J be interval modules and $\varepsilon \ge 0$. Then I and J are $\varepsilon$-interleaved if there exist maps $\varphi : I \rightarrow J[\varepsilon ]$ and $\psi : J \rightarrow I[\varepsilon ]$ such that $\psi [\varepsilon ]\varphi = I^{(2\varepsilon )}$ and $\varphi [\varepsilon ]\psi = J^{(2\varepsilon )}$.

Lemma 24

If intervals satisfy $K \le J \le I$ then $I \cap K \subset J$.

Proof

Let $x \in I \cap K$. Then there is a $j \in J$ such that $x \le j$. Also, there is a $j' \in J$ with $j' \le x$. Since J is an interval, $x \in J$. $\square$

Lemma 25

If $K \le J \le I$ then $I \cap K = (I \cap J) \cap (J \cap K)$.

Proof

One direction is easy: $(I \cap J) \cap (J \cap K) = I \cap J \cap K \subset I \cap K$. The other direction follows from Lemma 24. $\square$

Proposition 17

Let I and J be interval modules and $\varepsilon \ge 0$. If $J[\varepsilon ] \le I$ and $I[\varepsilon ] \le J$ then I and J are $\varepsilon$-interleaved.

Proof

Define $\varphi :I \rightarrow J[\varepsilon ]$ by $\varphi _x = 1$ if $x \in I \cap J[\varepsilon ]$ and $\varphi _x = 0$ otherwise. Similarly define $\psi : J \rightarrow I[\varepsilon ]$ by $\psi _x = 1$ if $x \in J \cap I[\varepsilon ]$ and $\psi _x = 0$ otherwise. We claim that these provide the desired $\varepsilon$-interleaving.

First, $I^{(2\varepsilon )}: I \rightarrow I[2\varepsilon ]$ is given by $I^{(2\varepsilon )}_x = 1$ if $x \in I \cap I[2\varepsilon ]$ and $I^{(2\varepsilon )}_x = 0$ otherwise. Next, $\psi [\varepsilon ]\varphi _x = 1$ if $x \in (I \cap J[\varepsilon ]) \cap (J[\varepsilon ]\cap I[2\varepsilon ])$ and $\psi [\varepsilon ]\varphi _x = 0$ otherwise. By Lemma 25, these maps are equal. Similarly, $\varphi [\varepsilon ]\psi = J^{(2\varepsilon )}$. $\square$

Next, we define the erosion of a persistence module. Compare with (Patel 2018) and (Puuska 2017).

Definition 15

Let I be an interval or an interval module and $\varepsilon \ge 0$. We define the $\varepsilon$-erosion of I to be $I^{-\varepsilon } = I[\varepsilon ] \cap I[-\varepsilon ]$.

Note that $I[\varepsilon ] \le I^{-\varepsilon } \le I[-\varepsilon ]$. See Fig. 10.

Fig. 10

An interval module and its erosion

Corollary 12

If $J[\varepsilon ] \le I$ and $I[\varepsilon ] \le J$ then $I^{-\varepsilon } \subset J$ and $J^{-\varepsilon } \subset I$.

Proof

It follows from the assumptions that we also have $J \le I[-\varepsilon ]$ and $I \le J[-\varepsilon ]$. So we have $J[\varepsilon ] \le I \le J[-\varepsilon ]$ and $I[\varepsilon ] \le J \le I[-\varepsilon ]$. The result follows from Lemma 24. $\square$

Theorem 10

Let I and J be interval modules and $\varepsilon \ge 0$. Then I and J are $\varepsilon$-interleaved if only if $I^{-\varepsilon } \subset J$ and $J^{-\varepsilon } \subset I$.

Proof

$(\Rightarrow )$ Let $\varphi$ and $\psi$ be an $\varepsilon$-interleaving. If either $\varphi$ or $\psi$ are zero, then from Definition 14, $I^{(2\varepsilon )}$ and $J^{(2\varepsilon )}$ are zero. It follows that $I^{-\varepsilon }$ and $J^{-\varepsilon }$ are both empty and the condition is satisfied. If both $\varphi$ and $\psi$ are nonzero, then by Proposition 16, $J[\varepsilon ] \le I$ and $I[\varepsilon ] \le J$. The result follows from Corollary 12.

$(\Leftarrow )$ We need to check four cases. (1) $I^{-\varepsilon }$ and $J^{-\varepsilon }$ are both empty. Then I and J are $\varepsilon$-interleaved by $\varphi =0$ and $\psi =0$.

(2) $I^{-\varepsilon }$ and $J^{-\varepsilon }$ are both nonempty. Let $a \in I[\varepsilon ]$. Then there is an element $b \in I^{-\varepsilon } \subset J$ with $a \le b$. Let $b \in I$. Then there is an element $a \in I^{-\varepsilon }[\varepsilon ] \subset J[\varepsilon ]$ with $a \le b$. Let $a \in J[\varepsilon ]$. Then there is an element $b \in J^{-\varepsilon } \subset I$ with $a \le b$. Let $b \in J$. Then there is an element $a \in J^{-\varepsilon }[\varepsilon ] \subset I[\varepsilon ]$ with $a \le b$. The result follows from Proposition 17.

(3) $I^{-\varepsilon }$ is nonempty and $J^{-\varepsilon }$ is empty. Let $a \in I[\varepsilon ]$. Then there is $b \in I^{-\varepsilon } \subset J$ with $a\le b$. Since J is shorter than I, it follows that $I[\varepsilon ] \le J$. Let $b \in I$. Then there is $a \in I^{-\varepsilon }[\varepsilon ] \subset J[\varepsilon ]$ with $a \le b$. Since J is shorter than I, it follows that $J[\varepsilon ] \le I$. The result follows from Proposition 17.

(4) is the same as the third case. $\square$

1.4 A.4 Neighborhoods of interval modules

Using Theorem 10, one obtains a complete characterization of the interval modules within distance $\varepsilon$ of an interval module.

Example 9

Consider the interval module [a, b) and let $\varepsilon \in [0,\frac{b-a}{2})$. Then an interval module I is $\varepsilon$-interleaved with [a, b) if and only if $[a+\varepsilon ,b-\varepsilon ) \subset I \subset [a-\varepsilon ,b+\varepsilon )$. Furthermore $B_\varepsilon ([a,b))$ consists of those interval modules I satisfying

$$\begin{aligned} a-\varepsilon< \inf I< a+\varepsilon \quad \text {and} \quad b-\varepsilon< \sup I < b+\varepsilon . \end{aligned}$$

Example 10

Consider the interval module [a, b) and let $\varepsilon \ge \frac{b-a}{2}$. Then an interval I is $\varepsilon$-interleaved with [a, b) if and only if either $I \subset [a-\varepsilon ,b+\varepsilon )$ or if for no $x \in {\mathbb {R}}$ do we have $[x-\varepsilon ,x+\varepsilon ] \subset I$. Furthermore, $B_{\varepsilon }([a,b))$ consists of those interval modules I with either $a-\varepsilon < \inf I$ and $\sup I < b+\varepsilon$ or ${{\mathrm{diam}}}I < \varepsilon$.

Example 11

Consider the interval module $[a,\infty )$ and let $\varepsilon \ge 0$. Then an interval module I is $\varepsilon$-interleaved with $[a,\infty )$ if and only if $[a+\varepsilon ,\infty ) \subset I \subset [a-\varepsilon ,\infty )$. Furthermore $B_{\varepsilon }([a,\infty ))$ consists of interval modules I satisfying $a-\varepsilon< \inf I < a + \varepsilon$ and $\sup I = \infty$.