maximal antichains on two levels of the boolean latticetkalinow/presentations/mfac.pdf · maximum...

Maximal antichains on two levels of theBoolean Lattice

Thomas Kalinowski

Institut fur MathematikUniversitat Rostock

Seminar Plzen9 May 2012

joint work with

Martin Gruttmuller Sven Hartmann

Uwe Leck Ian Roberts

Outline

1 Introduction

2 Maximal antichains and extremal graph theory

3 Maximal regular antichains

4 Open problems

The Boolean lattice Bn

Subsets of [n] = {1, . . . ,n} ordered by inclusion

B3B4

Set families – Subsets of Bn

F = {13,23,24,124,1234}

Antichains and completely separating systems

DefinitionAn antichain is a set family A,such that for A 6= B ∈ A alwaysA 6⊆ B.

Example:{12,13,14,23,24,34} 1 1 0 0

1 0 1 01 0 0 10 1 1 00 1 0 10 0 1 1

DefinitionA completely separatingsystem (CSS) is a family C,such that for every two pointsx 6= y there is a block B ∈ Cwith x ∈ B and y 6∈ B.

Example:{123,145,246,356}( 1 1 1 0 0 0

1 0 0 1 1 00 1 0 1 0 10 0 1 0 1 1

)

Flat antichains and fair CSSs

DefinitionAn antichain is called flat if, forsome k , it consists completelyof k - and (k + 1)-subsets.

Example:{125,126,127,345,346,347,

13,14,23,24,56,57,67}

DefinitionA CSS is called fair if, for somek , every point occurs k or k + 1times.

Example:{12378,1239a,45679,4568a,

4bc,5bd ,6cd}

Maximum flat antichains of minimum size

Problem

Given k < n/2, what is the minimum size of a maximal antichain onlevels k and k + 1?

Theorem (Gruttmuller, Hartmann, K, Leck, Roberts 2009)

The minimum size of a maximal antichain containing only 2- and3-sets is

(n2

)− b(n + 1)2/8c.

For n 6≡ 1 (mod 4) there is a unique extremal antichain.

For n ≡ 1 (mod 4) there are two different maximal antichainshaving the same minimum size.

More general problem

Given 2 6 k < l < n, what is the minimum size of a maximal antichainon levels k and l?

Extremal graph problem

With an antichain A = A2 ∪ A3 we associate a graphG(A) = (V ,E):

V = [n], E =([n]2

)\ A2.

{125,126,127,345,346,347,

13,14,23,24,56,57,67}

A antichain

The elements of A3 are triangles in G(A).

A maximal

Every edge of G(A) is contained in an element of A3, i.e. ina triangle.Every triangle of G(A) is an element of A.

The case (k , l) = (2,3)

A is a maximal antichain iff

Every edge of G(A) is contained in a triangle.A3 equals T , the set of triangles in G(A).

|A| = |T |+((n

2

)− |E |

)→ min

Extremal graph problem

Maximize |E | − |T |

subject to the condition that every edge is contained in atriangle.

The case (k , l) = (2,3)

Theorem (Gruttmuller, Hartmann, K, Leck, Roberts 2009)

In any graph G with the property that every edge is contained in atriangle we have

|E | − |T | 6⌊

(n + 1)2

8

⌋.

For n 6≡ 1 (mod 4) there is a unique extremal graph, and for n ≡ 1(mod 4) there are two extremal graphs.

T = {125,126,127,345,346,347}([n]

2

)\ E = {13,14,23,24,56,57,67}

A model proof

Theorem (Mantel 1908)

A triangle-free graph on n vertices has at most n2/4 edges.

local structure: d(x) + d(y) 6 n + t(xy) for every edge xysumming over the edges and using Cauchy-Schwarz:

n|E |+ 3|T | >∑

xy∈E

(d(x) + d(y)) =∑x∈V

d(x)2

>1n

(∑x∈V

d(x)

)2

=4|E |2

n

=⇒ |T | >(4|E | − n2) |E |

3n.

The (2,3)-proof

x y

z

Nx Ny

Nz

Nxy

NyzNxz

Nxyz

Local structure

d(x) + d(y) + d(z)

− t(xy)− t(yz)− t(xz) 6 n

Sum over triangles

∑x∈V

d(x)t(x)−∑

xy∈E

t(xy)2 6 n · |T |

. . . rather painful manipulations . . .

|E | − |T | 6 18

n(n + 1)2

Hypergraph formulation

A ⊆([n]

k

)∪([n]

l

)k -uniform hypergraph (k-graph): Edges are k -subsets ofthe elements of Al .k -graph G = (V ,E) with vertex set V = [n] and such thatevery edge is contained in an l-clique. → (k , l)-graphsDenoting the number of i-cliques by ei we get

Extremal problem

Maximize |E | − el ,

subject to the condition that every edge is contained in anl-clique.

An asymptotic upper bound

exk (n, l) maximal number of edges in a k -graph on nvertices that does not contain an l-clique.

t(k , l) Turan number: limn→∞

exk (n, l)/(n

k

)TheoremFor (k , l)-graphs we have for n→∞,

|E | − el 6

(( lk

)− 2( l

k

)− 1

t(k , l) + o(1)

)(nk

).

Proof: hypergraph removal lemma

A construction for (k , l) = (2,4)

Assume n = 4t .E = [1,2t ]× [2t + 1,4t ] ∪ {(2i − 1,2i) : i = 1,2, . . . ,2t}

|E | − e4 = 3n2/16 + n/2

Conjecture

In any (2,4)-graph we have |E | − e4 6 (3/16 + o(1)) n2.

general theorem: |E | − e4 6 (4/15 + o(1))n2.

Improving the bound for (k , l) = (2,4)

TheoremIn any (2,4)-graph we have

|E | − e4 6

2(

39 +√

21)

375+ o(1)

n2 < (0.233 + o(1)) n2.

Proof by averaging arguments similar to the (2,3)-case.

Regularity

We now require that every element of the ground setappears in the same number r of blocks.

ProblemWhat is the smallest possible regularity for a maximal antichainin([n]

2

)∪([n]

3

)?

Graph formulation

What is the maximal r ′ = r ′(n) such that there is a graph G onthe vertex set [n] with the following properties?

Every edge is contained in a triangle.For every vertex x , we have d(x)− t(x) = r ′, where

d(x) is the degree of vertex x , andt(x) is the number of triangles containing x .

A lower bound

TheoremFor n→∞, we have

r ′(n) 6 (1/10− o(1))n.

Corollary

For n→∞, the minimal regularity of a maximal (2,3)-antichainis

r(n) > (9/10− o(1))n.

Proof: combination of averaging and removal lemmaarguments

A construction

Let Γ be an abelian group of size m.Let A ⊆ Γ be a set without 3-term arithmetic progressions.Let G be a tripartite graph with vertex set Γ× [3] and

(g,1) and (h,2) are adjacent iff h − g ∈ A,(g,2) and (h,3) are adjacent iff h − g ∈ A,(g,1) and (h,3) are adjacent iff h − g ∈ 2A.

d(x) = 2|A|, t(x) = |A| for every vertex x .

TheoremFor n ≡ 0 (mod 3), there are maximal (2,3)-antichains ofregularity

n

(1− c log1/4(n)

22√

2 log2 n

).

We can do better!

TheoremFor n ≡ 0 (mod 16), there is a maximal (2,3)-antichain whichis (15n/16− 1)-regular.

Construction (1/2)

A B

000

001

010

011

100

101

110

111

000

001

010

011

100

101

110

111

n = 16t

vertex set{A,B} × [t ]× {0,1}3

matchings inside A and B:{(i ,A, x , y ,0), (i , x , y ,1)}

crossing edges 1:{(i ,A, x , y , z), (j ,B, x , y ′, y)}

so far: (2t + 1)-regular, onlyedges in B uncovered

Construction (2/2)A B

000

001

010

011

100

101

110

111

000

001

010

011

100

101

110

111

A B

000

001

010

011

100

101

110

111

000

001

010

011

100

101

110

111

crossing edges 2: {(i ,A, x , y , z), (j ,B,1− x , z, z ′)}d(v) = 4t + 1 for all vertices vv ∈ A: 2t triangles of type AAB and t triangles of type BBAv ∈ B: 2t triangles of type BBA and t triangles of type AAB

=⇒ d(v)− t(v) = t + 1 = n/16 + 1

Problems

Conjecture

In any (2,4)-graph we have |E | − e4 6 (3/16 + o(1)) n2.

ProblemDetermine the maximal (k , l)-antichains of minimum size.Interesting special cases:

k = 2l = k + 1

Conjecture

For n→∞, the minimal regularity of a maximal (2,3)-antichainis (15/16 + o(1))n.