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.