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.
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Notes
In particular, the multiplicity of \([a_i,a_j)\) can be calculated using the inclusion/exclusion formula \({{\mathrm{rank}}}M(a_i \le a_{j-1}) - {{\mathrm{rank}}}M(a_i \le a_j) - {{\mathrm{rank}}}M(a_{i-1} \le a_{j-1}) + {{\mathrm{rank}}}M(a_{i-1} \le a_j)\) (Cohen-Steiner et al. 2007), which is an example of Möbius inversion (Patel 2018).
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Acknowledgements
The authors would like to that the anonymous referees for their helpful suggestions. In particular, we would like to thank the referee who contributed the proof that the enveloping distance from pointwise-finite dimensional persistence modules to q-tame persistence modules is zero. We also thank Alex Elchesen for proofreading an earlier draft of the paper. The first author would like to acknowledge the support of UFII SEED funds, ARO Research Award W911NF1810307, and the Southeast Center for Mathematics and Biology, an NSF-Simons Research Center for Mathematics of Complex Biological Systems, under National Science Foundation Grant No. DMS-1764406 and Simons Foundation Grant No. 594594.
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A The arithmetic of maps and interleavings of interval modules
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. \(\square \)
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. \(\square \)
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.
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
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 \(\sup I = \infty \).
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Bubenik, P., Vergili, T. Topological spaces of persistence modules and their properties. J Appl. and Comput. Topology 2, 233–269 (2018). https://doi.org/10.1007/s41468-018-0022-4
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DOI: https://doi.org/10.1007/s41468-018-0022-4