The Borsuk-Ulam Theorem If f : S n R n n = 2 is continuous then - PowerPoint PPT Presentation

Using the Borsuk-Ulam Theorem G Eric Moorhouse based on Matou šek’s book ...

The Borsuk-Ulam Theorem If f : S n → R n n = 2 is continuous then there exists x 2 S n such that f ( – x ) = f ( x ) . T = 69.154 ° C T = 69.154 ° C P = 102.79 kPa P = 102.79 kPa

In a deflated sphere, there is a point directly above its antipode.

Brouwer Fixed-Point Theorem If f : B n → B n then there exists x 2 B n such that f ( x ) = x . n = 2

The Ham Sandwich Theorem Given n mass distributions in R n , there exists a hyperplane dividing each of the masses. n = 3 ham, cheese, bread

The Necklace Theorem n = 3 Every open necklace with n types of stones can be divided between two thieves using no more than n cuts. There is a version for several thieves.

The Necklace Theorem n = 3 Every open necklace with n types of stones can be divided between two thieves using no more than n cuts. All known proofs are topological

Tucker’s Lemma Consider a triangulation +2 of B n with vertices labeled +2 +1 ± 1, ±2, ..., ± n , such that +2 +1 the labeling is antipodal +2 on the boundary. Then – 2 there exists an edge (1- – 1 simplex) whose endpoints – 2 – 1 have opposite labels i , – i . – 1 – 2 n = 2

Borsuk-Ulam Theorem Tucker’s Ham Sandwich Lemma Theorem Brouwer Fixed- Point Theorem Necklace Theorem

Versions of the Borsuk-Ulam Theorem (1) ( Borsuk 1933) If f : S n → R n is continuous then- -----there exists x 2 S n such that f ( – x ) = f ( x ) . (2) If f : S n → R n is antipodal , i.e. f ( – x ) = – f ( x ), -----then there exists x 2 S n such that f ( x ) = 0 . (3) There is no antipodal map S n → S n – 1 . (4) ( Lyusternik- Schnirel’man 1930) If { A 1 , A 2 ,..., A n +1 } -----is a closed cover of S n , then some A i contains a pair of - -----antipodal points. (5) generalizing (4), each A i is either open or closed (Henceforth all maps are continuous functions.)

(1) If f : S n → R n is continuous then there exists ----- x 2 S n such that f ( – x ) = f ( x ) . * + (2) If f : S n → R n is antipodal , i.e. f ( – x ) = – f ( x ), -----then there exists x 2 S n such that f ( x ) = 0 . Let f : S n → R n be antipodal. There exists x 2 S n such that f ( x ) = f ( – x ) = – f ( x ). So f ( x ) = 0. Let f : S n → R n and define g ( x ) = f ( x ) – f ( – x ). Since g is antipodal, there exists x 2 S n such that g ( x ) = 0. So f ( – x ) = f ( x ) .

(1) ( Borsuk 1933) If f : S n → R n is continuous then- -----there exists x 2 S n such that f ( – x ) = f ( x ) . + (4) ( Lyusternik- Schnirel’man 1930) If { A 1 , A 2 ,..., A n +1 } -----is a closed cover of S n , then some A i contains a pair of - -----antipodal points. Define f : S n → R n , x 7! (dist( x , A 1 ), ..., dist( x , A n )). There exists x 2 S n such that f ( – x ) = f ( x ) = y , say. If y i = 0 ( i ≤ n ) then x, – x 2 A i . Otherwise x, – x 2 A n +1 .

Radon’s Theorem Let n ≥1 . Every set of n +2 points in R n can be partitioned as A 1 [ A 2 such that conv ( A 1 ) ∩ conv ( A 2 ) ≠ Ø. n = 2

Radon’s Theorem Let n ≥1 . Every set of n +2 points in R n can be partitioned as A 1 [ A 2 such that conv ( A 1 ) ∩ conv ( A 2 ) ≠ Ø. n = 2

Radon’s Theorem (Alternative Formulation) Let σ n +1 be an n +1-simplex where n ≥1 and let f : σ n +1 → R n be affine linear. There exist two complementary sub-simplices α , β of σ n +1 such that f ( α) ∩ f ( β) ≠ Ø. R 2 σ 3 ½ R 3 n = 2

Radon’s Theorem (Alternative Formulation) Let σ n +1 be an n +1-simplex where n ≥1 and let f : σ n +1 → R n be affine linear. There exist two complementary sub-simplices α , β of σ n +1 such that f ( α) ∩ f ( β) ≠ Ø. R 2 σ 3 ½ R 3 n = 2

Radon’s Theorem (Alternative Formulation) Let σ n +1 be an n +1-simplex where n ≥1 and let f : σ n +1 → R n be affine linear. There exist two complementary sub-simplices α , β of σ n +1 such that f ( α) ∩ f ( β) ≠ Ø. R 2 σ 3 ½ R 3 n = 2

Topological Radon Theorem Let σ n +1 be an n +1-simplex where n ≥1 and let f : σ n +1 → R n be continuous . There exist two complementary sub-simplices α , β of σ n +1 such that f ( α) ∩ f ( β) ≠ Ø. R 2 σ 3 ½ R 3 n = 2

Topological Radon Theorem Let σ n +1 be an n +1-simplex where n ≥1 and let f : σ n +1 → R n be continuous . There exist two complementary sub-simplices α , β of σ n +1 such that f ( α) ∩ f ( β) ≠ Ø. R 2 σ 3 ½ R 3 n = 2

Tverberg’s Theorem Let n ≥1, r ≥2. Every set of nr + r – n points in R n can be partitioned as A 1 [ A 2 [ ... [ A r such that conv ( A 1 ) ∩ conv ( A 2 ) ∩ ... ∩ conv ( A r ) ≠ Ø. n = 2, r = 3

Tverberg’s Theorem Let n ≥1, r ≥2. Every set of nr + r – n points in R n can be partitioned as A 1 [ A 2 [ ... [ A r such that conv ( A 1 ) ∩ conv ( A 2 ) ∩ ... ∩ conv ( A r ) ≠ Ø. This generalization of Radon’s Theorem also has a valid topological version. n = 2, r = 3

Lovász-Kneser Theorem n Kneser Graph KG n,k has ( ) vertices k A µ {1,2,..., n }, | A | = k . 12 Here 1 ≤ k ≤ ( n +1)/2. Vertices A , B are adjacent iff A ∩ B = Ø. 35 34 13 25 45 24 14 23 15 χ ( KG 5,2 ) = 3

Lovász-Kneser Theorem n Kneser Graph KG n,k has ( ) vertices k A µ {1,2,..., n }, | A | = k . 12 Here 1 ≤ k ≤ ( n +1)/2. Vertices A , B are adjacent iff A ∩ B = Ø. 35 34 13 25 45 Kneser Conjecture (1955) χ ( KG n,k ) = n – 2 k +2. Proved by Lovász (1978) using the 24 14 Borsuk-Ulam Theorem. The fractional chromatic number 23 15 gives the very weak lower bound χ ( KG 5,2 ) = 3 χ ( KG n,k ) ≥ n / k .

Lovász-Kneser Theorem n Kneser Graph KG n,k has ( ) vertices k A µ {1,2,..., n }, | A | = k . Here 1 ≤ k ≤ ( n +1)/2. Vertices A , B are adjacent iff A ∩ B = Ø. A proper colouring of KG n,k with colours 1,2,..., n – 2 k +2: min A ∩ {1,2,..., n – 2 k +1}, if this intersection is nonempty; A is coloured: n – 2 k +2 otherwise, i.e. A µ { n – 2 k +2, ..., n }.

Proof of Lovász-Kneser Theorem Vertices of KG n,k : k -subsets of an n -set X ½ S d , d = n – 2 k +1. WLOG points of X are in general position (no d +1 points on any hyperplane through 0). Each x 2 S d gives a partition R d +1 = H ( x ) [ x ? [ H ( – x ). Suppose there is a proper colouring of ( ) using colours 1,2,..., d . X k Define the point sets A 1 , A 2 , ..., A d µ S d : A i is the set of all x 2 S d for which some k -set B µ H ( x ) has colour i . A d +1 = S d – ( A 1 [ A 2 [ ... [ A d ). A 1 , A 2 , ..., A d are open; A d +1 is closed. So some A i contains a pair of antipodal points x , – x . Case i ≤ d : we get k -tuples A µ H ( x ), B µ H ( – x ) of colour i . No! Case i = d +1: H ( x ) contains at most k – 1 points of X . So does H ( – x ). So x ? contains at least n – 2( k – 1) = d +1 points of X . No!

Similar techniques yield lower bounds for chromatic numbers for more general graphs using Z 2 -indices ...

Sequence of spheres S n = {( x 0 , x 1 ,..., x n ) 2 R n +1 : Σ x i 2 = 1} ... → → → → → S 0 S 1 S 2 S 3 S 4 Antipodal maps S n → S n +1 i.e. f ( – x ) = – f ( x ) but S n +1 → S n /

The n -simplex σ n 3 2 2 0 0 1 0 0 1 1 ... || σ 0 || || σ 1 || || σ 2 || || σ 3 || 0123 {0,1,2} {0,1} 012 013 023 123 {0} {0,1} {0,2} {1,2} {0} {1} 01 02 12 03 13 23 {} {0} {1} {2} {} 0 1 2 3 {} {}

Its geometric realization || σ n || ½ R n 3 2 2 0 0 1 0 0 1 1 ... || σ 0 || || σ 1 || || σ 2 || || σ 3 || || σ 0 || || σ 1 || || σ 2 || || σ 3 || 0123 {0, 1,2 } { 0,1 } 012 013 023 123 { 0 } { 0,1 } { 0,2 } { 1,2 } { 0 } {1} 01 02 12 03 13 23 {} { 0 } { 2 } {1} 0 1 2 3 {} The n -simplex σ n {} {}

A Simplicial Complex e.g. || K || = geometric realization of K K {1,2,3} 3 {0,1} {1,2} {1,3} {2,3} 0 2 {0} {1} {2} {3} 1 {}

e.g. K = σ 2 Skeletons K ≤ 1 K {0,1,2} = the 1-skeleton of K K ≤ 0 = the 0-skeleton of K {0,1} {0,2} {1,2} {0,1} {0,2} {1,2} {0} {1} {2} {0} {1} {2} {0} {1} {2} {} {} {} 2 2 2 || K ≤ 1 || = || K ≤ 0 || = || K || = 0 1 0 1 0 1

|| σ n || = B n , || ( σ n ) ≤ n – 1 || = S n – 1 , ||( σ n ) ≤ 1 || = K n +1 e.g. 2 2 ||( σ 2 ) ≤ 1 || = = K 3 = S 2 || σ 2 || = = B 2 0 0 1 1

Topological join S n * S m = S n + m +1 = = e.g. S 0 * S 0 = S 1 S 0 S 0 S 1 = = S 0 * S 1 = S 2 S 0 S 1 S 2 In particular S n = ( S 0 ) *( n +1) = S 0 * S 0 * ... * S 0

Join ( σ 1 ) *2 = σ 1 * σ 1 = σ 3 σ 1 σ 1 More generally, ( σ n ) *2 = σ 2 n +1 , ||( σ n ) *2 || = B 2 n +1 . Deleted Join ||( σ 1 ) *2 || = S 1 = Δ σ 1 σ 1 S 1 More generally, ||( σ n ) *2 || = S n . Δ

Z 2 -action on a topological space X : a homeomorphism X → X , x 7! x ’ such that ( x ’ ) ’ = x (not necessarily fixed-point-free). Denote – x = x ’ . S n and R n have natural Z 2 -actions. The first is free, the second is not. Let X and Y be topological Z 2 -spaces. Write X → Y if there exists a Z 2 -equivariant map f : X → Y , i.e. f ( – x ) = – f ( x ). If not, write X → Y . / Thus S n → S n +1 , S n +1 → S n . / If X → Y and Y → W , then X → W . So ‘ → ’ defines a partial order.