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 Introduction 1 k -coverings of the unit interval 2 Excess 0 coverings of S 1 3 Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 2 / 30

Introduction Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 3 / 30

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 ⊂ R d , and consider all vectors in Y n ⊂ R dn of the form � x = ( � x 1 , � x 2 , . . . , � x n ), where ball i has center � x i . Definition (Configuration Space) x ∈ Y n such that Y is The configuration space of coverings of Y is all � covered, i.e. x ∈ Y n | ∀ y ∈ Y ∃ 1 ≤ i ≤ n s.t. d ( y , x i ) ≤ r } Cov n ( r , Y ) = { � 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]. x 1 x 2 x 3 x 4 1 3 5 7 8 8 8 8 Figure: The configuration above corresponds to the point ( 1 8 , 3 8 , 5 8 , 7 8 ) ∈ R 4 x 2 , x 5 x 4 , x 6 x 3 x 1 1 3 5 7 8 8 8 8 Figure: The configuration above corresponds to the point ( 7 8 , 1 8 , 5 8 , 3 8 , 1 8 , 3 8 ) ∈ R 6 Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 6 / 30

“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. x 1 x 2 x 3 x 4 1 3 5 7 8 8 8 8 Figure: Excess 0, as with 4 balls of radius 1 8 , we can just cover I . x 2 , x 5 x 4 , x 6 x 3 x 1 1 3 5 7 8 8 8 8 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) Cov n ( r , I ) ∼ Skel m ( P n ) , where m is the excess, Skel m is an m-skeleton, and P n is the permutahedron on n vertices. Example ( n = 3; the 3-permutahedron is a 2-dimensional hexagon) 3 , cannot cover, so Cov 3 ( r , I ) ∼ for 0 ≤ 2 r < 1 = ∅ for 1 3 ≤ 2 r < 1 2 , m = 0, Cov 3 ( r , I ) ∼ vertices of hexagon (0-sk.) for 1 2 ≤ 2 r < 1, m = 1, Cov 3 ( r , I ) ∼ 1-sk. of hexagon ∼ S 1 for 1 ≤ 2 r , m = 2, contractible Our Goal Find an analogue for the case of k -covering I , where k is arbitrary. Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 8 / 30

Indices Definition Define “indices” as points in the unit interval of the form i j = 2 j − 1 for 2 n 1 ≤ j ≤ n . i 1 = 1 i 2 = 3 i 3 = 5 i 4 = 7 8 8 8 8 1 Suppose we have balls of radius r = 2 n . Then if we are k -covering I , kn balls will cover I , and then the excess m =(# of balls) − kn . Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 9 / 30

k -coverings of the unit interval Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 10 / 30

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 ≤ 2 n + 1 s.t. max( d ( y , x j ) , d ( y , x k )) ≤ r Definition Let 2-Cov 2 n +1 ( r , I ) be the configuration space of double coverings of the 1 1 interval with 2 n + 1 balls, with 2 n ≤ r < 2( n − 1) , so the excess is 1. x 1 , x 2 , x 3 x 4 , x 5 x 4 , x 5 x 1 x 2 x 3 0 1 0 1 1 3 1 1 3 4 4 4 2 4 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-Cov 5 ( 1 4 , I ). Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 11 / 30

The n = 5 case and the Desargues Graph 1 4 1 3 2 5 2 4 3 5 Theorem 1 3 1 2 2 5 3 4 For 1 4 ≤ r < 1 4 5 2 (excess 1 ), 2 -Cov 5 ( r , I ) ∼ = h G (10 , 3) , the 1 3 2 1 2 4 3 4 5 5 “Desargues Graph”- the bipartite 1 2 1 2 3 5 4 3 double cover of the Petersen 4 5 Graph G (5 , 2) . 1 2 2 1 3 4 4 3 5 5 2 1 1 5 3 5 2 3 4 4 2 1 2 1 3 5 5 3 4 4 1 2 3 1 4 3 4 2 5 5 2 1 3 1 4 3 5 2 5 4 3 1 4 5 4 2 1 2 5 3 Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 12 / 30

Theorem 1: Our space is homotopic to a graph Theorem 1 For r = 2 n (excess 1 ), 2 -Cov 2 n +1 ( r , I ) ∼ G, for G a graph, i.e. a 1 -dimensional simplicial complex. Definition Let G 2 , n ⊂ 2-Cov 2 n +1 ( r , I ) be the following graph. For � x ∈ -Cov 2 n +1 ( r , I ) to be in G 2 , n we first of all require that ∀ 1 ≤ j ≤ n , ∃ 1 ≤ p � = q ≤ 2 n + 1 s.t. i j = x p = x q . Thus, any point on this graph has at least 2 balls centered at each index i j . 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 ( i j , i j +1 ) = ( 2 j − 1 2 n , 2( j +1) − 1 ) for 1 ≤ j ≤ n − 1. 2 n Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 13 / 30

Some Lemmas Theorem For any double-covering in 2 -Cov 2 n +1 ( 1 2 n , I ) ⊂ I 2 n +1 , every index must have at least 1 ball centered at it, that is: ∀ 1 ≤ j ≤ n , ∃ 1 ≤ k ≤ 2 n + 1 s.t. i j = x k . Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 14 / 30

Some Lemmas Theorem For any double covering, at most 1 ball can be centered in any interval of the form ( i j , i j +1 ) for 1 ≤ j ≤ n − 1 . x i − 1 x i x i +1 x i +2 0 1 i j − 1 i j i j +1 i j +2 Figure: The above cannot happen. Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 15 / 30

Some Lemmas Theorem Suppose a double-covering has no balls centered in (0 , i 1 ) or ( i n , 1) . Suppose the balls with centers x I 1 , x I 2 , . . . , x I p (re-labeled in ascending order) are not centered at indices, for 1 ≤ I j ≤ 2 n + 1 and 1 ≤ p ≤ n + 1 . Then x I 1 ∈ ( i j , i j +1 ) , . . . , x I p ∈ ( i j + p − 1 , i j + p ) for 1 ≤ j ≤ n. x I 2 x I 3 x I 1 x I 4 0 1 i j − 1 i j i j +1 i j +2 Figure: The above cannot happen. Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 16 / 30

Flow for the Space of Double Coverings n = 3, excess 1; 7 balls of radius 1 6 x 2 x 3 , x 4 x 5 x 6 , x 7 x 1 x 2 5 0 1 6 1 1 5 6 2 6 1 2 1 6 x 1 1 1 5 6 2 6 1 6 ≤ x 1 ≤ 1 2 ∩ 1 2 ≤ x 2 ≤ 5 6 ∩ x 2 − x 1 ≤ 1 Figure: 3 Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 17 / 30

Flow for the Space of Double Coverings n = 4, excess 1; 9 balls of radius 1 8 ( 3 8 , 5 8 , 7 8 ) x 4 , x 5 x 6 x 7 x 8 , x 9 x 1 x 2 x 3 0 1 1 3 5 7 ( 3 8 , 5 8 , 5 8 ) 8 8 8 8 ( 1 8 , 3 8 , 5 ( 3 8 , 3 8 , 5 8 ) 8 ) ( 1 2 , 1 2 , 1 2 ) x 3 x 2 x 1 8 ≤ x 1 ≤ 3 1 8 ∩ 3 8 ≤ x 2 ≤ 5 8 ∩ 5 8 ≤ x 3 ≤ 7 8 ∩ x 2 − x 1 ≤ 1 4 ∩ x 3 − x 2 ≤ 1 Figure: 4 Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 18 / 30

Main Theorem We calculated the vertex and edge counts of G k , n : kn +1 k ( kn +1)! � � V ( G k , n ) = k = ( k !) n − 1 · ( k +1) k k k k ... k +1 2 n · V · ( k +1)+ n − 2 n · V · 2( k +1) = V ( n − 1)( k +1) = k ( kn +1)!( n − 1) E ( G k , n ) = n ( k !) n − 1 2 n 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) 2 n , I ) ∼ k-Cov kn +1 ( 1 = h G k , n , with the above vertex and edge counts. Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 19 / 30

Future Work: Extending to higher excesses Conjecture k -Cov 2 n + m ( 1 2 n , I ) ∼ m -dimensional simplicial complex. Need extension of excess 1 flows. Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 20 / 30

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 M a = f − 1 [ −∞ , a ] is homotopy equivalent to M b = f − 1 [ −∞ , b ] . Definition (Tautological Function) Define τ : Cov n ( r , Y ) → R by τ ( � x = ( x 1 , . . . , x n )) = max y ∈ Y min 1 ≤ i ≤ n d ( x i , y ) τ is only piece-wise smooth ; must use techniques such as in [Agrachev, 1997]. Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 21 / 30

Future Work: Non-smooth Morse Theory Definition (Tautological Function for Double-covering) Define τ : 2-Cov n ( r , Y ) → R by τ ( � x ) = max 1 ≤ i < j ≤ n max { d ( x i , y ) , d ( x j , y ) } min y ∈ Y Conjecture The only critical points of τ occur when the excess changes. Agrawal, Barg, Levinson (ICERM) Topology of Coverings November 5, 2015 22 / 30