The Topology of Configuration Spaces of Coverings Shuchi Agrawal, - - PowerPoint PPT Presentation

The Topology of Configuration Spaces of Coverings

Shuchi Agrawal, Daniel Barg, Derek Levinson

Summer@ICERM

November 5, 2015

Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 1 / 30

Overview

1

Introduction

2

k-coverings of the unit interval

3

Excess 0 coverings of S1

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Introduction

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General Question

Question

Given a metric space Y , a radius r and n closed balls of this radius, what is the topology of the configuration space of the balls (i.e. their centers) such that every point in Y is covered by (at least) one ball?

Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 4 / 30

Configuration Spaces

Given n balls, label these balls 1, 2, . . . , n. Suppose Y ⊂ Rd, and consider all vectors in Y n ⊂ Rdn of the form x = ( x1, x2, . . . , xn), where ball i has center xi.

Definition (Configuration Space)

The configuration space of coverings of Y is all x ∈ Y n such that Y is covered, i.e. Covn(r, Y ) = { x ∈ Y n | ∀y ∈ Y ∃ 1 ≤ i ≤ n s.t. d(y, xi) ≤ r}

Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 5 / 30

Single coverings of the interval

Suppose we now consider coverings of the unit interval I = [0, 1].

1 8 3 8 5 8 7 8

x1 x2 x3 x4

Figure: The configuration above corresponds to the point ( 1

8, 3 8, 5 8, 7 8) ∈ R4

1 8 3 8 5 8 7 8

x2, x5 x4, x6 x3 x1

Figure: The configuration above corresponds to the point ( 7

8, 1 8, 5 8, 3 8, 1 8, 3 8) ∈ R6

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“Excess”

Definition (Excess)

The excess given a radius r is defined as the largest number m for which it is still possible to cover the interval with (n − m) r-balls.

1 8 3 8 5 8 7 8

x1 x2 x3 x4

Figure: Excess 0, as with 4 balls of radius 1

8, we can just cover I.

1 8 3 8 5 8 7 8

x2, x5 x4, x6 x3 x1

Figure: Excess 2, as we can cover the interval with 4 balls of radius 1

8.

Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 7 / 30

Background and Goal

Theorem (Baryshnikov)

Covn(r, I) ∼ Skelm(Pn), where m is the excess, Skelm is an m-skeleton, and Pn is the permutahedron on n vertices.

Example (n = 3; the 3-permutahedron is a 2-dimensional hexagon)

for 0 ≤ 2r < 1

3, cannot cover, so Cov3(r, I) ∼

= ∅ for 1

3 ≤ 2r < 1 2, m = 0, Cov3(r, I) ∼ vertices of hexagon (0-sk.)

for 1

2 ≤ 2r < 1, m = 1, Cov3(r, I) ∼ 1-sk. of hexagon ∼ S1

for 1 ≤ 2r, m = 2, contractible

Our Goal

Find an analogue for the case of k-covering I, where k is arbitrary.

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Indices

Definition

Define “indices” as points in the unit interval of the form ij = 2j−1

2n

for 1 ≤ j ≤ n. i1 = 1

8

i2 = 3

8

i3 = 5

8

i4 = 7

8

Suppose we have balls of radius r =

1

• 2n. Then if we are k-covering I, kn

balls will cover I, and then the excess m =(# of balls)−kn.

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k-coverings of the unit interval

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The Space of Double Coverings: Excess 1

Suppose now that we want to double-cover every point in I, so that ∀ y ∈ I, ∃ 1 ≤ j = k ≤ 2n + 1 s.t. max(d(y, xj), d(y, xk)) ≤ r

Definition

Let 2-Cov2n+1(r, I) be the configuration space of double coverings of the interval with 2n + 1 balls, with

1 2n ≤ r < 1 2(n−1), so the excess is 1. 1 4 3 4

x1, x2, x3 x4, x5 1

1 4 1 2 3 4

x1 x2 x3 x4, x5 1

Figure: Two configurations with 5 balls of radius 1

4 which double-cover I,

corresponding to the the points ( 1

4, 1 4, 1 4, 3 4, 3 4) and ( 1 4, 1 8, 3 8, 3 4, 3 4), respectively, in

2-Cov5( 1

4, I).

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The n = 5 case and the Desargues Graph

Theorem

For 1

4 ≤ r < 1 2 (excess 1),

2-Cov5(r, I) ∼ =h G(10, 3), the “Desargues Graph”- the bipartite double cover of the Petersen Graph G(5, 2).

1 4 2 5 3 1 3 2 4 5 1 3 2 5 4 1 2 3 4 5 1 3 2 4 5 2 1 3 4 5 1 2 3 5 4 1 2 4 3 5 1 2 3 4 5 2 1 4 3 5 2 1 3 5 4 1 5 2 3 4 2 1 3 5 4 2 1 5 3 4 1 2 4 3 5 3 1 4 2 5 2 1 4 3 5 3 1 5 2 4 3 1 4 2 5 4 5 1 2 3

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Theorem 1: Our space is homotopic to a graph

Theorem

For r =

1 2n (excess 1), 2-Cov2n+1(r, I) ∼ G, for G a graph, i.e. a

1-dimensional simplicial complex.

Definition

Let G2,n ⊂ 2-Cov2n+1(r, I) be the following graph. For x ∈-Cov2n+1(r, I) to be in G2,n we first of all require that ∀ 1 ≤ j ≤ n, ∃ 1 ≤ p = q ≤ 2n + 1 s.t. ij = xp = xq. Thus, any point on this graph has at least 2 balls centered at each index ij. A vertex of this graph also has one index with 3 balls centered at it. An edge of this graph has exactly 2 balls centered at each index, and one ball centered in an interval of the form (ij, ij+1) = ( 2j−1

2n , 2(j+1)−1 2n

) for 1 ≤ j ≤ n − 1.

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Some Lemmas

Theorem

For any double-covering in 2-Cov2n+1( 1

2n, I) ⊂ I 2n+1, every index must

have at least 1 ball centered at it, that is: ∀ 1 ≤ j ≤ n, ∃ 1 ≤ k ≤ 2n + 1 s.t. ij = xk.

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Some Lemmas

Theorem

For any double covering, at most 1 ball can be centered in any interval of the form (ij, ij+1) for 1 ≤ j ≤ n − 1. ij−1 ij ij+1 ij+2 xi−1 xi xi+1 xi+2 1

Figure: The above cannot happen.

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Some Lemmas

Theorem

Suppose a double-covering has no balls centered in (0, i1) or (in, 1). Suppose the balls with centers xI1, xI2, . . . , xIp (re-labeled in ascending

• rder) are not centered at indices, for 1 ≤ Ij ≤ 2n + 1 and 1 ≤ p ≤ n + 1.

Then xI1 ∈ (ij, ij+1), . . . , xIp ∈ (ij+p−1, ij+p) for 1 ≤ j ≤ n. ij−1 ij ij+1 ij+2 xI1 xI2 xI3 xI4 1

Figure: The above cannot happen.

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Flow for the Space of Double Coverings

n = 3, excess 1; 7 balls of radius 1

6 1 6 1 2 5 6

x3, x4 x5 x6, x7 x1 x2 1

x1 x2

1 6 1 2 5 6 1 6 1 2 5 6

Figure:

1 6 ≤ x1 ≤ 1 2 ∩ 1 2 ≤ x2 ≤ 5 6 ∩ x2 − x1 ≤ 1 3

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Flow for the Space of Double Coverings

n = 4, excess 1; 9 balls of radius 1

8 1 8 3 8 5 8 7 8

x4, x5 x6 x7 x8, x9 x1 x2 x3 1 x1 x2 x3 ( 1

8, 3 8, 5 8)

( 3

8, 3 8, 5 8)

( 3

8, 5 8, 5 8)

( 3

8, 5 8, 7 8)

( 1

2, 1 2, 1 2)

Figure:

1 8 ≤ x1 ≤ 3 8 ∩ 3 8 ≤ x2 ≤ 5 8 ∩ 5 8 ≤ x3 ≤ 7 8 ∩ x2 − x1 ≤ 1 4 ∩ x3 − x2 ≤ 1 4

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Main Theorem

We calculated the vertex and edge counts of Gk,n: V (Gk,n) = k

• kn+1

k k k k ... k+1

• =

k(kn+1)! (k!)n−1·(k+1)

E(Gk,n) =

2 n ·V ·(k+1)+ n−2 n ·V ·2(k+1)

2

= V (n−1)(k+1)

n

= k(kn+1)!(n−1)

n(k!)n−1

Theorem (See, for example: [Katok, 2006])

Suppose X and Y are 1-dimensional simplicial complexes, i.e. graphs. Then X ∼ Y ↔ χ(X) = χ(Y ), where χ(X) = V (X) − E(X), where χ(X) is called the “Euler Characteristic” of X.

Theorem (Main Theorem)

k-Covkn+1( 1

2n, I) ∼

=h Gk,n, with the above vertex and edge counts.

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Future Work: Extending to higher excesses

Conjecture

k-Cov2n+m( 1

2n, I) ∼ m-dimensional simplicial complex.

Need extension of excess 1 flows.

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Future Work: Non-smooth Morse Theory

Theorem (Milnor, Classical Morse Theory)

Let f : M → R be smooth. Let a < b. If f −1[a, b] is compact, and contains no points where ▽f = 0, then Ma = f −1[−∞, a] is homotopy equivalent to Mb = f −1[−∞, b].

Definition (Tautological Function)

Define τ : Covn(r, Y ) → R by τ( x = (x1, . . . , xn)) = max

y∈Y min 1≤i≤n d(xi, y)

τ is only piece-wise smooth; must use techniques such as in [Agrachev, 1997].

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Future Work: Non-smooth Morse Theory

Definition (Tautological Function for Double-covering)

Define τ : 2-Covn(r, Y ) → R by τ( x) = max

y∈Y

min

1≤i<j≤n max{d(xi, y), d(xj, y)}

Conjecture

The only critical points of τ occur when the excess changes.

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Excess 0 coverings of S1

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Space of Single Coverings of the Circle

n balls of radius

1 2n, total length n · 1 n = 1

1 2 3 Permutations 123, 312, 231 equivalent up to rotation. 1 3 2 Permutations 132, 321, 213 equivalent up to rotation.

Theorem

Covn( 1

2n, S1) ∼

=

(n−1)!

• i=1

S1

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Space of Double Coverings of the Circle, for odd n

n balls of radius 1

n, total length n · 1 2n = 2

1 2 3

Theorem (same reasoning as single-covering case)

2-Covn( 1

n, S1) ∼

=

(n−1)!

• i=1

S1

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Space of Double Coverings for n = 2, r = 1

2

Both balls cover the entire circle, can be moved independently of each

• ther.

Theorem

2-Cov2( 1

2, S1) ∼

= S1 × S1 ∼ = T 2

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Space of Double Coverings for n ≥ 4, n even

1, 2 3, 4 1 3 2 4 1 2 3 4

Theorem

2-Cov4( 1

4, S1) ∼

= 3 tori, glued as below. In general, for n even, 2-Covn( 1

n, S1) ∼

= 2(n−1)!

n

tori; each torus glued to n

2 · (2

n−2 2 − 1) other tori.

1, 2 3, 4 1, 3 2, 4 1, 4 2, 3

π1(2 − Cov4( 1

4 , S1)) =

Z × F3

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Generalization to k-covering, Future Work

Small Result: k-Covk( 1

2, S1) ∼

= T k.

Conjecture

k-Covn( k

n, S1) ∼

=

(n−1)!

• i=1

S1 if k ∤ n. Generalize to higher k, and look at higher excess.

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References (selection)

John Milnor (1963) Morse Theory Princeton University Press. Han Wang (2014) On the Topology of the Spaces of Coverings PhD Thesis University of Illinois at Urbana-Champaign. Anatole Katok and Alexey Sossinsky (2006) Introduction to Modern Topology and Geometry Lecture Notes Penn State University 44-50. Yuliy Baryshnikov, Peter Bubenik and Matthew Kahle (2014) Min-Type Morse Theory for Configuration Spaces of Hard Spheres University of Illinois at Urbana-Champaign.

• A. A. Agrachev, D. Pallaschke and S. Scholtes (2014)

On Morse Theory For Piecewise Smooth Functions Journal of Dynamical and Control Systems 449-469.

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Acknowledgements

Thank you to Yuliy for inventing the problem, and for being an incredibly helpful advisor. Thank you to Tarik for many enlightening conversations. Thank you to Stefan for his willingness to listen and help. Thank you to ICERM for the opportunity, for the facilities, and for the sustenance (shout out to Danielle!).

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