1. Partitioning Sets to Decrease the Diameter N. Harvey U. Waterloo TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: A A AA A A A

2. Act 1: Problem Statement & History Act 2: The Solution

3. Act 1: Problem Statement & History Def: For any S ½ R d, diam(S) = sup x,y 2 S k x-y k 2 Question: For every S ½ R d, can we partition S into S 1 [ S 2 [  [ S d+1 such that diam(S i ) < diam(S) for all i=1,...,d+1 Remarks: – Only makes sense if 0 < diam(S) < 1. – For simplicity, we’ll assume S is closed. Euclidean Norm: =max

4. Our Question with d=1 Question: For every S ½ R d, can we partition S into S 1 [ S 2 [  [ S d+1 such that diam(S i ) < diam(S) for all i=1,...,d+1 0123-2 S Midpoint m = (min S + max S) / 2 = 0.5 d=1:

5. Our Question with d=1 Question: For every S ½ R d, can we partition S into S 1 [ S 2 [  [ S d+1 such that diam(S i ) < diam(S) for all i=1,...,d+1 0123-2 S Midpoint m = (min S + max S) / 2 = 0.5 Let S 1 = { x : x 2 S and x · m } Let S 2 = { x : x 2 S and x>m } Clearly diam(S i ) · diam(S) / 2

6. Our Question with d=2 Question: For every S ½ R d, can we partition S into S 1 [ S 2 [  [ S d+1 such that diam(S i ) < diam(S) for all i=1,...,d+1 S Idea: Take minimum enclosing ball Partition into 90 ± segments Can argue diam(S i ) < diam(S) S1S1 S2S2 S3S3 S4S4

7. Our Question with d=2 Question: For every S ½ R d, can we partition S into S 1 [ S 2 [  [ S d+1 such that diam(S i ) < diam(S) for all i=1,...,d+1 S Improved Idea: Take minimum enclosing ball Partition into 120 ± segments

8. Our Question with d=2 Question: For every S ½ R d, can we partition S into S 1 [ S 2 [  [ S d+1 such that diam(S i ) < diam(S) for all i=1,...,d+1 S Improved Idea: Take minimum enclosing ball Partition into 120 ± segments Is diam(S i ) < diam(S)? S1S1 S2S2 S3S3

9. Our Question with d=2 Question: For every S ½ R d, can we partition S into S 1 [ S 2 [  [ S d+1 such that diam(S i ) < diam(S) for all i=1,...,d+1 S Idea: Take minimum enclosing ball Partition into 120 ± segments Is diam(S i ) < diam(S)? Yes, unless S ¼ equi. triangle S1S1 S2S2 S3S3

10. Our Question with d=2 Question: For every S ½ R d, can we partition S into S 1 [ S 2 [  [ S d+1 such that diam(S i ) < diam(S) for all i=1,...,d+1 S Idea: Take minimum enclosing ball Partition into 120 ± segments Is diam(S i ) < diam(S)? Yes, unless S ¼ equi. triangle In this case, just rotate the partitioning. S1S1 S2S2 S3S3

11. Our Question with Parameter f Notation: f(d) = smallest integer such that every S ½ R d can be partitioned into S 1 [ S 2 [  [ S f(d) with diam(S i ) < diam(S). Our Question: Is f(d) = d+1 for all d? What do we know so far? dLower Bound on f(d)Upper Bound on f(d) 1?2 2?3

12. Lower Bound on f Notation: f(d) = smallest integer such that every S ½ R d can be partitioned into S 1 [ S 2 [  [ S f(d) with diam(S i ) < diam(S). Trivial LB: Obviously f(d) ¸ 2 for every d. Partitioning into 1 set cannot decrease diam! What do we know so far? dLower Bound on f(d)Upper Bound on f(d) 122 223 Optimal!

13. Lower Bound on f Notation: f(d) = smallest integer such that every S ½ R d can be partitioned into S 1 [ S 2 [  [ S f(d) with diam(S i ) < diam(S). Another LB: For d=2, the hard case is when S = equilateral triangle. – The 3 corner points have pairwise distance D, where D = diam(S). – To get diam(S i ) < diam(S), these 3 corner points must lie in different S i ’s. – So f(2) ¸ 3! S1S1 S2S2 S3S3 D DD

14. Lower Bound on f Notation: f(d) = smallest integer such that every S ½ R d can be partitioned into S 1 [ S 2 [  [ S f(d) with diam(S i ) < diam(S). What do we know so far? dLower Bound on f(d)Upper Bound on f(d) 122 233 Optimal!

15. Generalize to d>2? For d=2, the hard case is an equilateral triangle – Corner points have distance D, where D = diam(S) What is a d-dimensional analog? D DD Simplex (Generalized Tetrahedron) D D D D D

16. Generalize to d>2? For d=2, the hard case is an equilateral triangle – Corner points have distance D, where D = diam(S) What is a d-dimensional analog? A Simplex: The convex hull of d+1 points with pairwise distances all equal to D To get diam(S i ) < diam(S), these d+1 points must lie in different S i ’s. So f(d) ¸ d+1! D D D D D

17. Lower Bound on f Notation: f(d) = smallest integer such that every S ½ R d can be partitioned into S 1 [ S 2 [  [ S f(d) with diam(S i ) < diam(S). What do we know so far? dLower Bound on f(d)Upper Bound on f(d) 122 233 dd+1? Optimal!

18. Generalize Upper Bound to d>2? S Idea: Take minimum enclosing ball Partition using orthants Can argue diam(S i ) < diam(S) # orthants is 2 d ) f(d) · 2 d S1S1 S2S2 S3S3 S4S4 S5S5 S6S6 dLower Bound on f(d)Upper Bound on f(d) 122 233 dd+12d2d

19. History Conjecture [Borsuk ‘33]: f(d)=d+1. Karol Borsuk, 1905-1982 Field: Topology

20. History dLower Bound on f(d)Upper Bound on f(d) 122 233 [Borsuk ‘33] 344 [Perkal-Eggleston ‘47-’55] 459 [Lassak ‘82] dd+1 [Borsuk ‘33] 2 d [Knast ‘74] 1.23 d [Schramm ‘88] Special Cases: – f(d)=d+1 when S = d-dimensional sphere [Borsuk ‘33] – f(d)=d+1 for all smooth sets S [Borsuk ‘33] Conjecture [Borsuk ‘33]: f(d)=d+1. Open! Huge Gap!

21. 100101978388059 15015130603191218909 200201957243195505543289

22. History dLower Bound on f(d)Upper Bound on f(d) 122 233 [Borsuk ‘33] 344 [Perkal-Eggleston ‘47-’55] 459 [Lassak ‘82] dd+1 [Borsuk ‘33] 2 d [Knast ‘74] 1.23 d [Schramm ‘88] Theorem (Kahn-Kalai 1993): Corollary: Borsuk’s conjecture is false! Conjecture (Borsuk 1933): f(d)=d+1. [Kahn-Kalai ‘93]

24. History dLower Bound on f(d)Upper Bound on f(d) 122 233 [Borsuk ‘33] 344 [Perkal-Eggleston ‘47-’55] dd+1 [Borsuk ‘33] 2 d [Knast ‘74] 1.23 d [Schramm ‘88] Theorem [Kahn-Kalai ‘93]: Corollary: Borsuk’s conjecture is false (for d ¸ 1325). Improvements: False for d ¸ 946 [Nilli ’94],..., d ¸ 298 [Hinrichs & Richter ‘02]. Conjecture [Borsuk ‘33]: f(d)=d+1. [Kahn-Kalai ‘93]

25. Intermission A Little Story

26. Act 2: Kahn & Kalai’s Theorem Definition: f(d) = smallest integer such that every S ½ R d (with 0 < diam(S) < 1 ) can be partitioned into S 1 [ S 2 [  [ S f(d) with diam(S i ) < diam(S) 8 i. Theorem:, for large enough d. Corollary: Borsuk’s conjecture is false.

27. How could Kahn-Kalai prove this? Every smooth set satisfies f(d)=d+1, so f(d) can only be large for non-smooth sets – What’s a very non-smooth set? A finite set! Can we reduce Borsuk’s problem for finite sets to a nice combinatorics problem? Smooth Set Borsuk’s Conjecture is True Finite Set Borsuk’s Conjecture is False?

28. Natural idea: look at {0,1} d instead of R d Trick of the trade: Look at points with same # of 1’s Notation: H k = { x : x 2 {0,1} d, x has k 1’s } Each point x 2 H k corresponds to a set X ½ {1,2,...,d} with |X|=k, i.e., X = { i : x i =1 } 000 [d] Notation: H0H0 H1H1 H2H2 H3H3 100 010 001101 011 110 111

29. Combinatorial Setup H k = { x : x 2 {0,1} d, x has k 1’s } Each point x 2 H k corresponds to a set X ½ {1,2,...,d} with |X|=k, i.e., X = { i : x i =1 } Let x,y 2 H k. Let X,Y µ [d] be corresponding sets Note: We don’t care about actual distances, just whether diameter shrinks or not. It is more convenient to use “modified distances”: (this is L 1 -norm / 2) [d]

30. Combinatorial Setup Points x,y 2 H k correspond to sets X,Y µ [d], |X|=|Y|=k A set of points S = {x 1, x 2,...} µ H k corresponds to a family of sets {X 1, X 2,...}, where each X i µ [d] and |X i |=k Main Goal: find S µ H k ½ {0,1} d and p ¸ d+1 such that, letting, For all partitions S = S 1 [  [ S p some S i contains X, Y with k-|X \ Y|=D Get rid of k by setting M = k-D

31. Combinatorial Setup Points x,y 2 H k correspond to sets X,Y µ [d], |X|=|Y|=k A set of points S = {x 1, x 2,...} µ H k corresponds to a family of sets {X 1, X 2,...}, where each X i µ [d] and |X i |=k Main Goal: find S µ H k ½ {0,1} d and p ¸ d+1 such that, letting, For all partitions S = S 1 [  [ S p some S i contains X, Y with |X \ Y|=M

32. Main Goal: find S µ H k ½ {0,1} d and p ¸ d+1 such that letting, For all partitions S = S 1 [  [ S p some S i contains X, Y with |X \ Y|=M How to argue about all partitions of S? No structure... By Pigeonhole Principle, one S i is “large”: | S i | ¸ | S |/p Need a theorem saying “Every large family must contain two sets with a given intersection size” Such questions are the topic of “Extremal Set Theory” Aiming at the Goal

33. (Extremal Set) Theory “How many subsets of a finite set are there with a Property X?” Extreme (Set Theory) The Reals are Uncountable!

34. Looking for a Hammer Need a Theorem saying “Every large family must contain two sets with a given intersection size” My favorite book on Extremal Set Theory:

35. Looking for a Hammer Need a Theorem saying “Every large family must contain two sets with a given intersection size” Theorem [Frankl-Rodl]: Let n be a multiple of 4. Let B ={B 1, B 2,...} be a family of subsets of [n] with – |B i | = n/2 8 i – |B i Å B j |  n/4 8 i  j Then | B | · 1.99 n. Contrapositive: Let B ={B 1, B 2,...} be a family of subsets of [n] with each |B i |=n/2. If | B |>1.99 n then 9 B i, B j 2 B with |B i Å B j | = n/4.

36. Goal: find S µ H k ½ {0,1} d and p ¸ d+1 such that letting, For all partitions S = S 1 [  [ S p some S i contains X, Y with |X Å Y|=M Hammer: Let B ={B 1, B 2,...} be a family of subsets of [n] with |B i |=n/2. If | B |>1.99 n then 9 B i,B j 2B s.t. |B i Å B j |=n/4. How to connect Goal & Hammer? – Not obvious... – Kahn & Kalai need a clever idea... (fortunately they are very clever) Goal & Hammer

37. K n = (V,E) is complete graph; |V|=n is a multiple of 4 Let B = { B : B ½ V and |B|=n/2 } (“Bisections”) For any U µ V, let ± (U) = { uv : u 2 U and v  U } (“Cut”) Let S = { ± (B) : B 2B } (“Bisection Cuts”) For any ± (B), ± (B’) 2S, we have | ± (B) Å ± (B’)| ¸ n 2 /8 Equality holds iff |B Å B’|=n/4 a db c B = { {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, {c,d} } n = 4 U = {a,b} S = { ± ( {a,b} ), ± ( {a,c} ), ± ( {a,d} ),... } = { {ac,ad,bc,bd}, {ab,ad,bc,cd},... } ± ( {a,b} ) = { ac,ad,bc,bd } ± ( {a,c} ) = { ab,ad,bc,cd } {a,c} ± ( {a,b} ) = { ac,ad,bc,bd } ± ( {a,c} ) = { ab,ad,bc,cd } S = { ± ( {a,b} ), ± ( {a,c} ), ± ( {a,d} ),... } = { {ac,ad,bc,bd}, {ab,ad,bc,cd},... }

38. K n = (V,E) is complete graph; |V|=n is a multiple of 4 Let B = { B : B ½ V and |B|=n/2 } (“Bisections”) For any U µ V, let ± (U) = { uv : u 2 U and v  U } (“Cut”) Let S = { ± (B) : B 2B } (“Bisection Cuts”) For any ± (B), ± (B’) 2S, we have | ± (B) Å ± (B’)| ¸ n 2 /8 Equality holds iff |B Å B’|=n/4 Consider a partition S = S 1 [... [ S p and corresponding partition B = B 1 [... [ B p If p · 1.005 n then 9 i s.t. Hammer: 9 B, B’ 2 B i such that |B Å B’|=n/4 ) ± (B), ± (B’) 2S i and | ± (B) Å ± (B’)| = n 2 /8 Goal Solved:

39. Conclusion Borsuk’s Conjecture: For every S ½ R d, we can partition S into S 1 [ S 2 [  [ S d+1 such that diam(S i ) < diam(S) for all i=1,...,d+1 Kahn-Kalai: No! There exists S ½ R d and s.t. for every partition S into S 1 [ S 2 [  [ S p 9 i with diam(S i ) = diam(S). BorsukKahnKalai