Stability Theorem Xinyi Kong, 1281909 Eindhoven, university of - - PowerPoint PPT Presentation

Stability Theorem

Xinyi Kong, 1281909

Eindhoven, university of Technology

31 May 2018

Overview

◮ Motivation ◮ Persistence Diagram ◮ Main Theory ◮ Hausdorff Stability ◮ Bottleneck Stability

Motivation

Motivation

Motivation

Stability means that if the input changes a tiny bit, the output should not change by much either.

Recap

Let f : R → R be a smooth function. We call that x is a critical point and f (x) id a critical value of f if f

′(x) = 0. A critical

point x is non-degenerate if f

′′(x) = 0.

Example

Homological critical value

Definition: Let X be a topological space and f : X → R be a

• function. A homological critical value of f is a real number b for

which there exists an integer k such that for all sufficiently small ε > 0 the map Hk(f −1(−∞, b − ε]) → Hk(f −1(−∞, b + ε]) is not an isomorphism.

A tame function

Definition: A function f : X → R is tame if it has a finite number

• f homological critical values and the homology groups

Hk(f −1(−∞, a]) are finite-dimensional for all k ∈ Z and a ∈ R.

Figure 1: tame Figure 2: non-tame

Persistence Diagram

How to draw a persistence diagram of f ?

◮ Pair the critical points of f . ◮ Map each pair to the corresponding point in persistence

diagram.

Persistence Diagram

Pair the critical points of f : When we pass a local maximum and merge two components, we pair the maximum with the higher (younger) of the two local minima that represent the two components. The other minimum is now the representative of the component resulting from the merger.

Example

Persistence Diagram of a tame function

Let f : X → R be a tame function, (ai)i=1...n its homological critical values, and (bi)i=0...n an interleaved sequence, namely bi−1 < ai < bi for all i. We set b−1 = a0 = −∞ and bn+1 = an = +∞. We consider the corresponding sequence of homology groups, 0 = Hk(Xb−1) → Hk(Xb0) → ... → Hk(Xbn+1) = Hk(X) and the maps between them. We define the multiplicity of the pair (ai, aj) by µj

i = βbj bi−1 − βbj bi + βbj−1 bi

− βbj−1

bi−1

Persistence Diagram of a tame function

Definition: The persistence diagram D(f ) ⊂ ¯ R2 of f is the set of points (ai, aj), counted with multiplicity µj

i for 0 ≤ i < j ≤ n + 1,

union all points on the diagonal, counted with infinite multiplicity.

Example

♯(A)

The total multiplicity of a multiset A is written as ♯(A)

example

The total multiplicity of the persistence diagram without the diagonal is written as follow: ♯(D(f ) − ∆) =

i<j µj i

L∞ − norm

For points p = (p1, p2, ..., pn) and q = (q1, q2, ..., qn) in ¯ Rn, p − q∞ := max(|p1 − q1|, |p2 − q2|, ..., |pn − qn|). For function f and g, f − g∞ = supx|f (x) − g(x)|.

Figure 3: p − q∞ Figure 4: f − g∞

Hausdorff distance and bottleneck distance

Definition: Let X and Y be multisets of points. The Hausdorff distance and bottleneck distance between X and Y are dH(X, Y ) = max

• sup

x inf y x − y∞ , sup y inf x y − x∞

• dB(X, Y ) = inf

γ sup x x − γ(x)∞

where x ∈ X and y ∈ Y range over all points and γ ranges over all bijections from X to Y .

Hausdorff distance and bottleneck distance

dH(X, Y ) = max

• sup

x inf y x − y∞ , sup y inf x y − x∞

• Figure 5: sup

x inf y x − y∞

Figure 6: sup

y inf x y − x∞

It is the greatest of all the distances from a point in one set to the closest point in the other set.

Hausdorff distance and bottleneck distance

dB(X, Y ) = inf

γ sup x x − γ(x)∞

Figure 7: sup

x x − γ1(x)∞

Figure 8: sup

x x − γ2(x)∞

Hausdorff distance and bottleneck distance

dH(X, Y ) ≤ dB(X, Y ) Bottleneck distance has one more constrain which makes it cannot map all the points in one set to the closest point in the other set.

Figure 9: sup

x inf y x − y∞

Figure 10: sup

y inf x y − x∞

Figure 11: sup

x x − γ1(x)∞

Figure 12: sup

x x − γ2(x)∞

Main Theory

Main Theory: Let X be a triangulable space with continuous tame functions f ,g : X → R. Then the persistence diagrams satisfy dB(D(f ), D(g)) ≤ f − g∞.

To proof the main theorem: I will first show what we can get from dH(D(f ), D(g)) ≤ f − g∞ Then strengthen the result to proof dB(D(f ), D(g)) ≤ f − g∞

Hausdorff Stability

If the inequality dH(D(f ), D(g)) ≤ f − g∞ = ε is true, we can find a point p(x, y) ∈ D(f ), then there must be a point of D(g) at the distance less than or equal to ε from p(x, y). That means there must be a point q of D(g) inside the square [x − ε, x + ε] × [y − ε, y + ε].

Box lemma

For a < b < c < d, let R = [a, b] × [c, d] be a box in ¯ R2 and let Rε = [a + ε, b − ε] × [c + ε, d − ε] be the box obtained by shrinking R at all four sides. Box Lemma: ♯(D(f ) ∩ Rε) ≤ ♯(D(g) ∩ R)

Recall

Main Theory: Let X be a triangulable space with continuous tame functions f ,g : X → R. Then the persistence diagrams satisfy dB(D(f ), D(g)) ≤ f − g∞.

Bottleneck stability

Let’s start with a special case. Given a tame function f : X → R, we consider the minimum distance between off-diagonal points or between ans off-diagonal point and the diagonal: δf = min {p − q∞ |(D(f ) − ∆) ∋ p = q ∈ D(f )}

Bottleneck Stability

δf = min {p − q∞ |(D(f ) − ∆) ∋ p = q ∈ D(f )} We get Figure 1 by drawing squares of radius r = δf /2 around the points of D(f ).

Figure 13: D(f )

Bottleneck Stability

δf = min {p − q∞ |(D(f ) − ∆) ∋ p = q ∈ D(f )} Then we add another tame function g : X → R which is very close to f . That means f and g satisfy f − g∞ ≤ δf /2

Figure 14: D(f ) Figure 15: D(f ) and D(g)

Bottleneck Stability

δf = min {p − q∞ |(D(f ) − ∆) ∋ p = q ∈ D(f )} f and g satisfy f − g∞ ≤ δf /2 Writing µ for the multiplicity of the point p ∈ (D(f ) − ∆) and ε for the square with center p and radius ε = f − g∞

Bottleneck Stability

Recall the box lemma

For a < b < c < d, let R = [a, b] × [c, d] be a box in ¯ R2 and let Rε = [a + ε, b − ε] × [c + ε, d − ε] be the box obtained by shrinking R at all four sides. Box Lemma: ♯(D(f ) ∩ Rε) ≤ ♯(D(g) ∩ R) From the box lemma we get: µ ≤ ♯(D(g) ∩ ε) ≤ ♯(D(f ) ∩ 2ε)

Bottleneck Stability

From the box lemma we get: µ ≤ ♯(D(g) ∩ ε) ≤ ♯(D(f ) ∩ 2ε) Since 2ε ≤ δf , p is the only point of D(f ) in 2ε, which implies µ = ♯(D(g) ∩ ε).

Bottleneck Stability

Since 2ε ≤ δf , p is the only point of D(f ) in 2ε, which implies µ = ♯(D(g) ∩ ε). Therefore, we can map all points of D(g) in ε to p. And the rest of point will be mapped to the nearest point on the diagonal, because dH(D(f ), D(g)) ≤ ε

Bottleneck Stability

Easy Bijection lemma: Let f , g : X → R be tame functions and g very close to f . then the persistence diagrams satisfy dB(D(f ), D(g)) ≤ f − g∞

Summary

◮ Persistence diagram of tame functions ◮ Hausdorff distance and bottleneck distance of two persistence

diagrams

◮ Use the box lemma to proof the bottleneck distance ◮ If a function change a little bit, its persistence diagram is

stable.

Source

◮ D. Cohen-Steiner, H. Edelsbrunner, J. Harer, Stability of

Persistence Diagrams, Disc. Comp. Geom. 37: 103–120, 2007.

◮ H. Edelsbrunner, J. L. Harer, Persistent homology – a Survey,

Surveys on discrete and computational geometry, 257–282,

• Contemp. Math., 453, Amer. Math. Soc., Providence, RI,

2008.