Metric Thickenings of Euclidean Submanifolds

Transcription

1 Metric Thickenings of Euclidean Submanifolds Advisor: Dr. Henry Adams Committe: Dr. Chris Peterson, Dr. Daniel Cooley Joshua Mirth Masters Thesis Defense October 3, 2017

2 Introduction

3 Motivation Can we recover the object on the right from the one on the left? 2

4 Motivation Can we recover the object on the right from the one on the left? They share essentially no topological properties (connectedness, loops, dimension). 2

5 Motivation Can we recover the object on the right from the one on the left? They share essentially no topological properties (connectedness, loops, dimension). A reconstruction method should work given a perfect sample. 2

6 Background

7 Simplicial Complexes Definition Let V be a set, called the set of vertices. An abstract simplicial complex K on vertex set V is a subset of the power set of V with the property that if σ K, then all subsets of σ are in K. 3

8 Simplicial Complexes Definition Let V be a set, called the set of vertices. An abstract simplicial complex K on vertex set V is a subset of the power set of V with the property that if σ K, then all subsets of σ are in K. For example: V = {a, b, c, d, e} K = {[abc], [ac], [bc], [ac], [ad], [cd], [a], [b], [c], [d], [e]} 3

9 Simplicial Complexes Every simplicial complex has a geometric realization: V = {a, b, c, d, e} K = {[abc], [ac], [bc], [ac], [ad], [cd], [a], [b], [c], [d], [e]} b c a e d 4

10 Simplicial Complexes Every simplicial complex has a geometric realization: V = {a, b, c, d, e} K = {[abc], [ac], [bc], [ac], [ad], [cd], [a], [b], [c], [d], [e]} b c a e d 4

11 Simplicial Complexes Every simplicial complex has a geometric realization: V = {a, b, c, d, e} K = {[abc], [ac], [bc], [ac], [ad], [cd], [a], [b], [c], [d], [e]} b c a e d 4

12 Simplicial Complexes Every simplicial complex has a geometric realization: V = {a, b, c, d, e} K = {[abc], [ac], [bc], [ac], [ad], [cd], [a], [b], [c], [d], [e]} b c a e d 4

13 Simplicial Complexes Every simplicial complex has a geometric realization: b c a e d The topology on a finite simplicial complex is the subspace topology of its geometric realization in R n. 4

14 The Vietoris Rips Complex Definition Let X be a metric space and r > 0 a scale parameter. The Vietoris Rips complex, VR (X; r), of X, has vertex set X and a simplex for every finite subset σ X such that diam(σ) r. b c a e d 5

15 The Vietoris Rips Complex Definition Let X be a metric space and r > 0 a scale parameter. The Vietoris Rips complex, VR (X; r), of X, has vertex set X and a simplex for every finite subset σ X such that diam(σ) r. b c a e d 5

16 The Vietoris Rips Complex Definition Let X be a metric space and r > 0 a scale parameter. The Vietoris Rips complex, VR (X; r), of X, has vertex set X and a simplex for every finite subset σ X such that diam(σ) r. b c a e d 5

17 The Vietoris Rips Complex Definition Let X be a metric space and r > 0 a scale parameter. The Vietoris Rips complex, VR (X; r), of X, has vertex set X and a simplex for every finite subset σ X such that diam(σ) r. b c a e d 5

18 The Vietoris Rips Complex Definition Let X be a metric space and r > 0 a scale parameter. The Vietoris Rips complex, VR (X; r), of X, has vertex set X and a simplex for every finite subset σ X such that diam(σ) r. b c a e d 5

19 The Čech Complex Definition Let X Y be a submetric space and r > 0 a scale parameter. The Čech complex Č (X, Y ; r), of X, has vertex set X and a simplex for every finite subset σ X such that. x i σ B(x i, r/2) 6

20 The Čech Complex b c a e d 7

21 The Čech Complex b c a e d 7

22 The Čech Complex b c a e d 7

23 The Čech Complex b c a e d 7

24 The Čech Complex b c a e d 7

25 The Čech Complex b c a e d 7

26 Homotopy Equivalence Definition Let f : X Y and g : X Y be continuous maps. Then f is homotopic to g, denoted f g, if there exists a continuous function H : X [0, 1] Y such that H(x, 0) = f(x), H(x, 1) = g(x). 8

27 Homotopy Equivalence Definition Let f : X Y and g : X Y be continuous maps. Then f is homotopic to g, denoted f g, if there exists a continuous function H : X [0, 1] Y such that H(x, 0) = f(x), H(x, 1) = g(x). Definition Let X and Y be topological spaces. Then X is homotopy equivalent to Y, written X Y, if there exists a pair of continuous functions f : X Y and g : Y X such that g f id X and f g id Y. 8

28 Homotopy equivalence permits stretching and bending in a way that allows the dimension to change: 9

29 Homotopy equivalence permits stretching and bending in a way that allows the dimension to change: 9

30 Homotopy equivalence permits stretching and bending in a way that allows the dimension to change: 9

31 Nerve Lemma Lemma (Nerve Lemma: Convex Version) Let U α for α A an index set be convex subsets of R n. Then N ({U α }) α A U α. b c a e d 10

32 Nerve Lemma Lemma (Nerve Lemma: Convex Version) Let U α for α A an index set be convex subsets of R n. Then N ({U α }) α A U α. b c a e d The Čech complex is the nerve of balls of radius r/2, so it is homotopy equivalent to the underlying space for a good cover. 10

33 Hausmann s Theorem Theorem Let M be a compact Riemannian manifold and r > 0 be sufficiently small. Then VR(M; r) M [4]. 11

34 Hausmann s Theorem Theorem Let M be a compact Riemannian manifold and r > 0 be sufficiently small. Then VR(M; r) M [4]. The bound on r depends upon the curvature of M. 11

35 Hausmann s Theorem Theorem Let M be a compact Riemannian manifold and r > 0 be sufficiently small. Then VR(M; r) M [4]. The bound on r depends upon the curvature of M. VR(M; r) does not inherit the metric of M. 11

36 Hausmann s Theorem Theorem Let M be a compact Riemannian manifold and r > 0 be sufficiently small. Then VR(M; r) M [4]. The bound on r depends upon the curvature of M. VR(M; r) does not inherit the metric of M. Thus: Hausmann s proof only gives a map T : VR(M; r) M, and proves the equivalence using algebraic techniques. 11

37 Hausmann s Theorem Theorem Let M be a compact Riemannian manifold and r > 0 be sufficiently small. Then VR(M; r) M [4]. The bound on r depends upon the curvature of M. VR(M; r) does not inherit the metric of M. Thus: Hausmann s proof only gives a map T : VR(M; r) M, and proves the equivalence using algebraic techniques. T depends upon a total order of the points in M. 11

38 Hausmann s Theorem Theorem Let M be a compact Riemannian manifold and r > 0 be sufficiently small. Then VR(M; r) M [4]. The bound on r depends upon the curvature of M. VR(M; r) does not inherit the metric of M. Thus: Hausmann s proof only gives a map T : VR(M; r) M, and proves the equivalence using algebraic techniques. T depends upon a total order of the points in M. In particular, the inclusion ι: M VR(M; r) does not provide the inverse (in fact, ι is not even continuous.) 11

39 Metric Thickenings

40 Metric Thickenings The Metric thickening of a simplicial complex was first introduced by Adamaszek, Adams, and Frick [1]. It puts the 1-Wasserstein metric on the geometric realization of a simplicial complex. This lets us use the theory of metric spaces to prove reults analogous to Hausmann and the Nerve Lemma. 12

41 Metric Vietoris Rips Thickenings Definition (Adamaszek, Adams, Frick) For a metric space X and r 0, the Vietoris Rips thickening VR m (X; r) is the set { k } VR m (X; r) = λ ix i k N, x i X, and diam({x 0,..., x k }) r, i=0 where λ i 0 and i λ i = 1, equipped with the 1-Wasserstein metric.[1] 13

42 Metric Vietoris Rips Thickenings Definition (Adamaszek, Adams, Frick) For a metric space X and r 0, the Vietoris Rips thickening VR m (X; r) is the set { k } VR m (X; r) = λ ix i k N, x i X, and diam({x 0,..., x k }) r, i=0 where λ i 0 and i λ i = 1, equipped with the 1-Wasserstein metric.[1] As a set this is identical to the geometric realization of VR(X; r), but the topology is different. 13

43 Metric Vietoris Rips Thickenings Definition (Adamaszek, Adams, Frick) For a metric space X and r 0, the Vietoris Rips thickening VR m (X; r) is the set { k } VR m (X; r) = λ ix i k N, x i X, and diam({x 0,..., x k }) r, i=0 where λ i 0 and i λ i = 1, equipped with the 1-Wasserstein metric.[1] As a set this is identical to the geometric realization of VR(X; r), but the topology is different. By identifying x X with δ x P(X), we can view VR m (X; r) as a subset of P(X), the set of all Radon probability measures on X. 13

44 Metric Vietoris Rips Thickenings Definition (Adamaszek, Adams, Frick) For a metric space X and r 0, the Vietoris Rips thickening VR m (X; r) is the set { k } VR m (X; r) = λ ix i k N, x i X, and diam({x 0,..., x k }) r, i=0 where λ i 0 and i λ i = 1, equipped with the 1-Wasserstein metric.[1] As a set this is identical to the geometric realization of VR(X; r), but the topology is different. By identifying x X with δ x P(X), we can view VR m (X; r) as a subset of P(X), the set of all Radon probability measures on X. This makes VR m (X; r) a (metric) thickening of X. 13

45 Wasserstein Metric Let x, x VR m (X; r) with x = k i=0 λ ix i and x = k i=0 λ i x i. Define a matching p between x and x to be any collection of non-negative real numbers {p i,j } such that k j=0 p i,j = λ i and k i=0 p i,j = λ j. Define the cost of the matching p to be cost(p) = i,j p i,jd(x i, x j ). Definition The 1-Wasserstein metric on VR m (X; r) is the distance d W defined by d W (x, x ) = inf { cost(p) p is a matching between x and x }. 14

46 Euclidean Submanifolds

47 Sets of Positive Reach We will prove an analogue of Hausmann s theorem in the context of subsets of Euclidean space, rather than Riemannian manifolds. This is a natural setting for data analysis. Positive reach was first introduced by Federer [3]. In particular, any C k submanifold of R n has positive reach, for k 2, so sets of positive reach include many potentially interesting objects. 15

48 Sets of Positive Reach The medial axis of X R n is the closure, Y, of Y = {y R n x 1 x 2 M with d(y, x 1 ) = d(y, x 2 ) = d(y, X)}. The reach, τ, of X is the minimal distance τ = d(x, Y ) between X and its medial axis. τ Y X 16

49 Sets of Positive Reach r τ τ = 0 τ = r Sets with corners have zero reach. 17

50 Sets of Positive Reach r τ τ = 0 τ = r Sets with corners have zero reach. Smooth manifolds embedded in R n have positive reach. 17

51 Sets of Positive Reach r τ τ = 0 τ = r Sets with corners have zero reach. Smooth manifolds embedded in R n have positive reach. Reach is half the distance between non-connected components. 17

52 Nearest Point Projection Define the α-offset of X R n : Tub α = {x R n d(x, X) < α} = x X B(x, α). If X has reach τ, then π : Tub τ X where x maps to its nearest point in X is well-defined and continuous [3]. τ Y Tub α 18

53 Proposition (Niyogi, Smale, Weinberger) Let X R n have reach τ > 0. Let p X and suppose x Tub τ \ X satisfies π(x) = p. If c = p + τ x p x p, then B(c, τ) X =. Proof. For any 0 < t < τ, let y t = p + t x p x p. Since y t Tub τ, we have B(y t, t) X = {p} and d(y t, p) = t, so B(c, t) X =. Note that B(c, τ) = 0<t<τ B(y t, t). Indeed, to see the inclusion, suppose that z B(c, τ), so that d(z, c) = τ ɛ for some ɛ > 0. Let t = τ ɛ 3. By the triangle inequality, d(y t, z) d(y t, c) + d(c, z) = τ 2ɛ 3 < t, giving z B(y t, t). The reverse inclusion is straightforward. It follows that B(c, τ) X =. 19

54 Results

55 Main Theorem Theorem (Metric Hausmann) Let X R n and suppose the reach τ of X is positive. Then for all r < τ, the metric Vietoris Rips thickening VR m (X; r) is homotopy equivalent to X. 20

56 Main Theorem Theorem (Metric Hausmann) Let X R n and suppose the reach τ of X is positive. Then for all r < τ, the metric Vietoris Rips thickening VR m (X; r) is homotopy equivalent to X. f τ Y X with Tub τ π i VR m (X; r) X 20

57 Theorem (Metric Nerve Theorem) Let X be a subset of Euclidean space R n, equipped with the Euclidean metric, and suppose the reach τ of X is positive. Then for all r < τ, the metric Čech thickening Čm (X; 2r) is homotopy equivalent to X. f τ Y X with Tub τ π i Č m (X; 2r) X 21

58 Lemmas Lemma For X R n and r > 0, the linear projection map f : VR m (X; r) R n has its image contained in Tub r. 22

59 Lemmas Lemma For X R n and r > 0, the linear projection map f : VR m (X; r) R n has its image contained in Tub r. Proof. Let x = k i=0 λ ix i VR m (X; r); we have diam(conv{x 0,..., x k }) = diam([x 0,..., x k ]) r. Since f(x) conv{x 0,..., x k }, it follows that d(f(x), X) d(f(x), x 0 ) r, and so f(x) Tub r. 22

60 Lemmas Lemma For X R n and r > 0, the linear projection map f : VR m (X; r) R n has its image contained in Tub r. Proof. Let x = k i=0 λ ix i VR m (X; r); we have diam(conv{x 0,..., x k }) = diam([x 0,..., x k ]) r. Since f(x) conv{x 0,..., x k }, it follows that d(f(x), X) d(f(x), x 0 ) r, and so f(x) Tub r. Lemma Let x 0,..., x k R n, let y conv{x 0,..., x k }, and let C be a convex set with y / C. Then there is at least one x i with x i / C. 22

61 Lemma Let X R n have positive reach τ, let [x 0,... x k ] be a simplex in VR(X; r) with r < τ, let x = λ i x i VR m (X; r), and let p = π(f(x)). Then the simplex [x 0,..., x k, p] is in VR(X; r). Proof. T p H x0 B(c, τ) c B(x 0, r) f(x) x 0 p 23

62 Main Result We are now prepared to prove our main result. Theorem Let X be a subset of Euclidean space R n, equipped with the Euclidean metric, and suppose the reach τ of X is positive. Then for all r < τ, the metric Vietoris Rips thickening VR m (X; r) is homotopy equivalent to X. 24

63 Proof. By [1, Lemma 5.2], map f : VR m (X; r) R n is 1-Lipschitz and hence continuous. It follows from Lemma 12 that the image of f is a subset of Tub τ. Let i: X VR m (X; r) be the inclusion map. Note that π f i = id X. f τ Y X with Tub τ π i VR m (X; r) X 25

64 Proof. Consider H : VR m (X; r) I VR m (X; r) defined by H(x, t) = t id VR m (X;r) + (1 t)i π f. H is well-defined by Lemma 14, and continuous by [1, Lemma 3.8].It follows that H is a homotopy equivalence from i π f to id VR m (X;r). f τ Y X with Tub τ π i VR m (X; r) X 25

65 Čech Result Theorem Let X be a subset of Euclidean space R n, equipped with the Euclidean metric, and suppose the reach τ of X is positive. Then for all r < τ, the metric Čech thickening Čm (X; 2r) is homotopy equivalent to X. Proof. The proof uses similar techniques to that of Theorem 15 26

66 Conclusion

67 Conclusions Metric analogue of Hausmann in Euclidean space. 27

68 Conclusions Metric analogue of Hausmann in Euclidean space. For a Riemannian version see [1]. Or: Corollary If N is a smooth, compact, Riemannian manifold, there exists a τ > 0 such that VR m (N; r) N for all 0 < r < τ. Proof. This follows from the Nash Embedding theorem [7]. 27

69 Conclusions Metric analogue of Hausmann in Euclidean space. For a Riemannian version see [1]. Or: Corollary If N is a smooth, compact, Riemannian manifold, there exists a τ > 0 such that VR m (N; r) N for all 0 < r < τ. Proof. This follows from the Nash Embedding theorem [7]. The same techniques hold for metric Čech thickenings. 27

70 Conclusions Metric analogue of Hausmann in Euclidean space. For a Riemannian version see [1]. Or: Corollary If N is a smooth, compact, Riemannian manifold, there exists a τ > 0 such that VR m (N; r) N for all 0 < r < τ. Proof. This follows from the Nash Embedding theorem [7]. The same techniques hold for metric Čech thickenings. Worth considering version for dense-samplings [6, 2]. 27

71 References [1] M. Adamaszek, H. Adams, and F. Frick, Metric reconstruction via optimal transport. Preprint, arxiv/ [2] F. Chazal and S. Oudot, Towards persistence-based reconstruction in Euclidean spaces, in Proceedings of the 24th Annual Symposium on Computational Geometry, ACM, 2008, pp [3] H. Federer, Curvature measures, Transactions of the American Mathematical Society, 93 (1959), pp [4] J.-C. Hausmann, On the Vietoris Rips complexes and a cohomology theory for metric spaces, in Prospects In Topology, F. Quinn, ed., vol. 138 of Annals of Mathematics Studies, Princeton University Press, 1995, pp [5] H. Karcher, Riemannian center of mass and mollifier smoothing, Communications on pure and applied mathematics, 30 (1977), pp [6] J. Latschev, Vietoris Rips complexes of metric spaces near a closed Riemannian manifold, Archiv der Mathematik, 77 (2001), pp [7] J. Nash, The imbedding problem for Riemannian manifolds, Annals of Mathematics, 63 (1956), pp [8] P. Niyogi, S. Smale, and S. Weinberger, Finding the homology of submanifolds with high confidence from random samples, Discrete Computational Geometry, 39 (2008), pp