Erd s-Hajnal Sets and Semigroup Decompositions

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Erdős-Hajnal Sets and
Semigroup Decompositions
Joshua N. Cooper
Courant Institute
Suppose we have a collection of lines in 3D…
stack(L)=3
What is the largest stacked subset?
Denote
A set ofthe
lines
cardinality
is stacked
of the
if it is
largest
linearly
stacked
ordered
subset
by
“passing
ofthe
a family
L ofover”
linesrelation.
stack(L).
There is a natural tournament on L:
Define
stack(n)
= min stack(L)
.
Stacked
subsets
correspond
to transitive
|L| = n
subtournaments.
Ramsey’s Theorem implies stack(n) >> log n.
For a tournament T, define trans(T) to be the size of the largest
transitive subtournament.
trans(T) is notoriously sensitive to randomness.
And line configurations are decidedly not random.
z
ℓ2
ℓ1
(a,b,1)
(a′,b ′,1)
z=1
y
z = -1
(c′,d ′,-1)
(c,d,-1)
ℓ1 passes over ℓ2 iff g(a-a′, b-b′, c-c′, d-d ′) > 0,
where g(x, y, z, w) = (x-z)(xw-yz).
V = {v1,…,vn}
∩
Re-define a line configuration tournament:
R4
E = {(v,w) : g(v-w) > 0}
Theorem. [Milnor '64, Thom '65] Let K be the number of ±1-vectors in the set
{(r1(x) ,…, rm(x)) : x Rm}
where rj = sgn ○ fj, 1 ≤ i ≤ m, and each fj is a polynomial of degree at most d in
k variables. Then
R ≤ (4edm/k) k.
d = 3, k = 4n, m = n(n-1)/2
R = 2O(n) << 2n(n-1)/2
There are very few line configurations compared to the number of tournaments.
There is some T which cannot be a subtournament of a line configuration
tournament. So what?
Definition. A homogeneous subset in a graph is a clique or an independent set.
Conjecture [Erdős-Hajnal]. For every graph H, there is an ε >
0 so that every
graph on n vertices which has no induced copy of H contains a homogeneous
subset of size nε.
Alon, Pach, and Solymosi showed that the following is equivalent:
Conjecture. For every tournament T, there is an ε >
0 so that every tournament
on n vertices which has contains no copy of T as a subtournament contains a
transitive subset of size nε.
So, if you believe Erdős-Hajnal, every line configuration tournament contains a
large transitive subtournament.
f ), for a k-variable polynomial f, by
V = {v1,…,vn}
∩
Define the digraph G(
Rk
E = {(v,w) : f(v-w) > 0}
Theorem. [Alon, Pach, Pinchasi, Radoičić, Sharir ‘04] Every digraph G( f )
contains a large transitive subtournament, independent set, or complete graph.
(Actually, they proved a lot more.)
V = {v1,…,vn}
∩
Generalize: Define G(S), for a subset S of Rk, to be the digraph
These definitions agree when
Rk
E = {(v,w) : v-w
S = f -1(R≥0).
S}
Call a subset S of Rk Erdős-Hajnal if G(S) must contain a large transitive
subtournament, independent set, or clique.
Which other sets are Erdős-Hajnal ?
Theorem. Any bounded set S such that 0
∂S is Erdős-Hajnal.
Either n1/2 points fall in a single square, or lots of points fall into n1/2 squares.
Rk is a semigroup if x, y
∩
Theorem. Any semigroup S
S implies x+y
S.
Rk is Erdős-Hajnal.
∩
A set S
b
c
a
b–a
S
&
c–b
S
c–a
S
Then G(S) is a quasiorder. Define [a] = {x : a←x←a}, so G(S)
induces a partial order on the equivalent classes.
Apply Dilworth’s Theorem : Either (A) there is an antichain of size n1/3, or a
chain of size n2/3. In the latter case, there is either (B) an equivalence class
of size n1/3 or (C) a chain of size n1/3 among elements of equiv. classes.
(A) = Independent Set, (B) = Clique, (C) = Transitive Subtournament.
T are Erdős-Hajnal, then S ∩ T, S T and S c are, too.
∩
Proposition. If S,
A semialgebraic set is a subset of Rk defined by polynomial inequalities.
The Alon-Pach-Pinchasi-Radoičić-Sharir Theorem means, in particular, that
all semialgebraic sets are Erdős-Hajnal.
Might it already be true that all semialgebraic sets belong to the set algebra
generated by semigroups?
Theorem. When k = 1 or 2, yes.
Conjecture. When k = 3, yes. When k > 3, no.
The “proof”…
The green area is a semigroup.
Parting Questions:
1. Do all 3-dimensional semialgebraic sets belong to the set
algebra generated by semigroups?
2. What other sets are Erdős-Hajnal? Positive sets of
Chebyshev systems?
3. What is the right exponent for stacked subsets?
1/6 ≤ ε ≤ 0.565 ≈ log37
4. Is anyone here hiring this year?
Thank you!
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