Equal-slices measure

The equal-slices measure [math]\displaystyle{ \mu(A) }[/math] of a set [math]\displaystyle{ A \subset [3]^n }[/math] is defined by the formula

[math]\displaystyle{ \mu(A) := \sum_{(a,b,c) \in \Delta_n} \frac{|A \cap \Gamma_{a,b,c}|}{|\Gamma_{a,b,c}|} }[/math]

where [math]\displaystyle{ \Delta_n := \{ (a,b,c) \in {\Bbb Z}_+^3: a+b+c=n\} }[/math] is the triangular grid and [math]\displaystyle{ \Gamma_{a,b,c} \subset [3]^n }[/math] is the set of slices with exactly a 1s, b 2s, and c 3s. Thus every slice [math]\displaystyle{ \Gamma_{a,b,c} }[/math] has measure 1 (hence the name equal-slices), and the entire cube has measure [math]\displaystyle{ \frac{(n+1)(n+2)}{2} }[/math]. Dividing the equal slices measure by [math]\displaystyle{ \frac{(n+1)(n+2)}{2} }[/math] gives the equal-slices density.

The DHJ(3) conjecture,

DHJ(3), Version 1. For every [math]\displaystyle{ \delta \gt 0 }[/math] there exists n such that every subset [math]\displaystyle{ A \subset [3]^n }[/math] of density at least [math]\displaystyle{ \delta }[/math] contains a combinatorial line.

is equivalent to an equal slices version:

DHJ(3), Version 3. For every [math]\displaystyle{ \delta \gt 0 }[/math] there exists n such that every subset [math]\displaystyle{ A \subset [3]^n }[/math] of equal-slices density at least [math]\displaystyle{ \delta }[/math] contains a combinatorial line.

Version 3 implies Version 1

Suppose that A has density [math]\displaystyle{ \geq \delta }[/math] in the usual sense. Let m be such that every subset of [math]\displaystyle{ [3]^m }[/math] of equal-slices density [math]\displaystyle{ \geq \delta/2 }[/math] contains a combinatorial line. Now randomly embed [math]\displaystyle{ [3]^m }[/math] into [math]\displaystyle{ [3]^n }[/math] by choosing m variable coordinates and fixing the rest. We may suppose that every point in A has [math]\displaystyle{ n/3+O(\sqrt{n}) }[/math] of each coordinate value and that [math]\displaystyle{ m \ll \sqrt{n} }[/math]. Therefore, changing coordinates hardly changes the density of a slice. It follows that each point of is in approximately the same number of these random subspaces. Therefore, by averaging, there is a random subspace inside which has equal-slices-density at least [math]\displaystyle{ \geq \delta/2 }[/math], and we are done. (We could think of it Terry’s way: as we move the random subspace around, what we effectively have is a bunch of random variables, each with mean approximately [math]\displaystyle{ \delta }[/math], so by linearity of expectation we’ll get equal-slices-density at least [math]\displaystyle{ \delta/2 }[/math] at some point, whatever the measure is.)

Version 1 implies Version 3

Suppose that A has density [math]\displaystyle{ \geq \delta }[/math] in the equal-slices sense. By the first moment method, this means that A has density [math]\displaystyle{ \gg \delta }[/math] on [math]\displaystyle{ \gg \delta }[/math] on the slices.

Let m be a medium integer (much bigger than [math]\displaystyle{ 1/\delta }[/math], much less than n).

Pick (a, b, c) at random that add up to n-m. By the first moment method, we see that with probability [math]\displaystyle{ \gg \delta }[/math], A will have density on [math]\displaystyle{ \gg \delta }[/math] the of the slices [math]\displaystyle{ \Gamma_{a',b',c'} }[/math] with [math]\displaystyle{ a' = a + m/3 + O(\sqrt{m}) }[/math], [math]\displaystyle{ c' = c + m/3 + O(\sqrt{m}) }[/math], [math]\displaystyle{ c' = c + m/3 + O(\sqrt{m}) }[/math].

This implies that A has expected density on a random m-dimensional subspace generated by a 1s, b 2s, c 3s, and m independent wildcards.

Applying Version 1 to that random subspace we obtain the claim.