DHJ(2.6)
From Polymath1Wiki
DHJ(2.6) is the statement that if
has density δ, and n is sufficiently large depending on δ, then there exists r such that for each ij=01,12,20, there exists a combinatorial line w0,w1,w2 with r wildcards whose i and j elements are in A. This is weaker than DHJ(2.7) and DHJ(3), but implies DHJ(2.5).
First proof
To prove DHJ(2.6), we use the Version 5 formulation, thus we now have Bernoulli variables
of probability at least δ, and we will find a combinatorial line w0,w1,w2 such that any two of the events
jointly hold with positive probability.
We can color each line w0,w1,w2 in one of eight colours depending on whether
intersect with positive probability for ij=01,12,20. Applying the Graham-Rothschild theorem, we can pass to a large subspace on which all lines have the same color. But by (the Version 5 formulation of) DHJ(2.5),
has to have positive probability for at least one line, hence for all lines. The claim follows.
Second proof
One can tone the Ramsey theory in the above proof down a notch, by replacing the Graham-Rothschild theorem by the simpler Folkman's theorem (this is basically because of the permutation symmetry of the random subcube ensemble). Indeed, given a dense set A in [3]n we can colour [n] in eight colours, colouring a shift r depending on whether there exists a combinatorial line with r wildcards whose first two (or last two, or first and last) vertices lie in A. By Folkman's theorem, we can find a monochromatic m-dimensional pinned cube Q in [n] (actually it is convenient to place this cube in a much smaller set, e.g. [n0.1]). If the monochromatic colour is such that all three pairs of a combinatorial line with r wildcards can occur in A, we are done, so suppose that there is no line with r wildcards with r in Q with (say) the first two elements lying in A.
Now consider a random copy of [3]m in [3]n, using the m generators of Q to determine how many times each of the m wildcards that define this copy will get. The expected density of A in this cube is about δ, so by DHJ(2.5) at least one of the copies is going to get a line whose first two elements lie in A, which gives the contradiction.
A Ramsey-free proof?
Here is a potential angle for a Ramsey-free proof of DHJ(2.6). The idea would be to soup up the Sperner-based proof of DHJ(2.5) and show that the set D of (Hamming) distances where we can find a 01-half-line in A is “large” and “structured”.
So again, let be x a random string in [3]n and condition on the location of x’s 2’s. There must be some set of locations such that A has density at least δ on the induced 01 hypercube (which will likely have dimension around 2n/3). So fix 2’s into these locations and pass to the 01 hypercube. Our goal is now to show that if A is a subset of 0,1n' of density then the set of Hamming distances where has Sperner pairs is large/structured, where
.
Almost all the action in {0,1}n' is around the
middle slices. Let’s simplify slightly (although this is not much of a cheat) by pretending that has A density δ on the union of the middle slices *and* that these slices have “equal weight”. Index the slices by
, with the ith slice actually being the
slice.
Let Ai denote the intersection of A with the ith slice, and let μi denote the relative density of Ai within its slice.
Here is a slightly wasteful but simplifying step: Since the average of the μi’s is δ, a simple Markov-type argument shows that
for at least a
fraction of the i’s. Call such an i “marked”.
Now whenever we have, say, 3 / δ marked i’s, it means we have a portion of A with a number of points at least times the number of points in a single middle slice. Hence Sperner's theorem tells us we get two comparable points from among these slices.
Thus we have reduced to the following setup:
There is a graph G=(V,E), where
(the “marked i’s”) has density at least δ / 2 and the edge set has the following density property: For every collection
with
, there is at least one edge among the vertices in V'.
Let D be the set of “distances” in this graph, where an edge (i,j) has “distance” |i-j|. Show that this set of distances must be large and/or have some useful arithmetic structure.
