An outline of a density-increment argument

Introduction

One of the proof strategies we have considered seriously is the density-increment method. The idea here is to prove that if [math]\displaystyle{ \mathcal{A} }[/math] is a subset of [math]\displaystyle{ [3]^n }[/math] of density [math]\displaystyle{ \delta }[/math] with no combinatorial line then there is a combinatorial subspace S of dimension tending to infinity with n such that the density of [math]\displaystyle{ \mathcal{A} }[/math] inside S is larger than [math]\displaystyle{ \delta }[/math] by an amount that depends on [math]\displaystyle{ \delta }[/math] only. If we can prove such a result, then we can drop down to S and repeat the argument. If the initial dimension n is large enough, then we push the density up to 1 before we run out of dimensions, and thereby obtain a contradiction.

Notation and definitions

Let x be an element of [math]\displaystyle{ [3]^n. }[/math] Write [math]\displaystyle{ U(x),V(x) }[/math] and [math]\displaystyle{ W(x) }[/math] for the sets of i such that [math]\displaystyle{ x_i=1,2 }[/math] and 3, respectively. These are called the 1-set, the 2-set and the 3-set of x. The map [math]\displaystyle{ x\mapsto(U(x),V(x),W(x)) }[/math] is a one-to-one correspondence between [math]\displaystyle{ [3]^n }[/math] and the set of all triples [math]\displaystyle{ (U,V,W) }[/math] of sets that partition [math]\displaystyle{ [n]. }[/math] We shall use the notation x and also the notation [math]\displaystyle{ (U,V,W) }[/math] and will pass rapidly between them.

Let [math]\displaystyle{ \mathcal{U} }[/math] and [math]\displaystyle{ \mathcal{V} }[/math] be collections of subsets of [math]\displaystyle{ [n]. }[/math] We shall write [math]\displaystyle{ \mathcal{U}\otimes\mathcal{V} }[/math] for the collection of all sequences [math]\displaystyle{ x }[/math] such that [math]\displaystyle{ U(x)\in\mathcal{U} }[/math] and [math]\displaystyle{ V(x)\in\mathcal{V}. }[/math] A set of the form [math]\displaystyle{ \mathcal{U}\otimes\mathcal{V} }[/math] will be called a 12-set.

In general, if X is a finite set and A and B are subsets of X, we define the density of A in B to be [math]\displaystyle{ |A\cap B|/|B|. }[/math]

A sequence fragment in [math]\displaystyle{ [3]^n }[/math] is a subset [math]\displaystyle{ Z\subset[n] }[/math] and a function [math]\displaystyle{ \sigma }[/math] from Z to [math]\displaystyle{ \{1,2,3\}. }[/math] An extension of the sequence fragment [math]\displaystyle{ (Z,\sigma) }[/math] is a sequence [math]\displaystyle{ x\in[3]^n }[/math] such that the coordinates of the restriction of x to Z are given by [math]\displaystyle{ \sigma. }[/math] For example, if n=5, then [math]\displaystyle{ §§3§2 }[/math] is a sequence fragment, and [math]\displaystyle{ 12332 }[/math] is one of its eight extensions.

A very broad overview of the argument

One of the main features of the argument is that it is local in character. In other words, it does not try to understand fully the obstructions to uniformity that cause a set to fail to have the expected number of combinatorial lines. Instead, we are content to find non-quasirandom behaviour inside small subspaces of [math]\displaystyle{ [3]^n. }[/math] This happens right from the start. A conjecture that one might hope to be true is that if a set [math]\displaystyle{ \mathcal{A} }[/math] is line-free, then there is a dense 12-set [math]\displaystyle{ \mathcal{U}\otimes\mathcal{V} }[/math] such that the density of [math]\displaystyle{ \mathcal{A} }[/math] inside [math]\displaystyle{ \mathcal{U}\otimes\mathcal{V} }[/math] is a bit larger than the density of [math]\displaystyle{ \mathcal{A} }[/math] in [math]\displaystyle{ [3]^n. }[/math] However, this is not in fact our first step. Instead, we argue that this kind of correlation happens when one restricts to an appropriate subspace. This part of the argument is given in some detail on a different page, (where what we are calling 12-sets are called special sets of complexity 1).

Once we have correlation with a 12-set (even if we have had to pass to a subspace to get it) it is natural to try to prove that a 12-set can be partitioned into combinatorial subspaces, since then by averaging we could show that [math]\displaystyle{ \mathcal{A} }[/math] correlated with one of those subspaces, and we would be done. An argument similar to the proof of the multidimensional Sperner theorem can be used to prove that a dense 12-set contains many combinatorial subspaces. However, it is not at all obvious that one can partition the 12-set into such subspaces (even if small errors are allowed).

However, a different principle comes to the rescue. It is the observation that there are two kinds of density increment that will be good enough for us. One is a density increment on a subspace, as we already know. The other is a density increment on a 12-set (obtained after passing to a subspace if necessary). Let us spell this out in more detail. Suppose that [math]\displaystyle{ \mathcal{A} }[/math] is a line-free set of density [math]\displaystyle{ \delta }[/math] and we want to obtain a contradiction. We know two things. First, we can find a 12-subset of a subspace in which [math]\displaystyle{ \mathcal{A} }[/math] has increased density [math]\displaystyle{ \delta+\eta }[/math], and secondly, if we can ever prove that there is increased density [math]\displaystyle{ \delta+\eta/2 }[/math] (say) on a subspace then we are done. However, we can add a third principle, which is that if we can find a 12-subset of a subspace on which the density goes up from [math]\displaystyle{ \delta+\eta }[/math] then we are again done. Why? Because we are in the same position as we were in before, but with a slightly higher density, so we can iterate.

Just to be explicit about this, let us briefly describe how the iteration would go. We start with a line-free set [math]\displaystyle{ \mathcal{A} }[/math] of density [math]\displaystyle{ \delta }[/math] and hope ultimately to obtain a contradiction. To do this, we adopt as a more immediate aim to find a subspace S inside which [math]\displaystyle{ \mathcal{A} }[/math] has density at least [math]\displaystyle{ \delta+\eta(\delta)/2. }[/math] As a first step, we find a 12-subset of a subspace, inside which [math]\displaystyle{ \mathcal{A} }[/math] has density at least [math]\displaystyle{ \delta+\eta. }[/math] We now focus our attention on the subproblem of trying to use this to obtain a subspace inside which [math]\displaystyle{ \mathcal{A} }[/math] has density at least [math]\displaystyle{ \delta+\eta/2. }[/math] But here again we can use the same sort of idea. We do not have to find such a subspace directly: instead, we could prove that if no such subspace exists (so the density of [math]\displaystyle{ \mathcal{A} }[/math] inside all subspaces is at most [math]\displaystyle{ \delta+\eta/2,) }[/math] then we can find a 12-subset of some further subspace, inside which [math]\displaystyle{ \mathcal{A} }[/math] has density at least [math]\displaystyle{ \delta+\eta+\theta(\delta). }[/math] And then we can try again to find a subspace where the density is at least [math]\displaystyle{ \delta+\eta/2. }[/math] Repeating this argument leads to an iteration within the main iteration, which will eventually terminate with a density increase on a subspace (since the density within a 12-set cannot exceed 1). And then we have completed another step of the main iteration, so we are done.

A brief description of the main steps

Step 1. Let [math]\displaystyle{ \mathcal{A} }[/math] be a line-free set of uniform density [math]\displaystyle{ \delta. }[/math] Then there is a subspace S of dimension tending to infinity with n and a 12-subset [math]\displaystyle{ \mathcal{U}\otimes\mathcal{V} }[/math] of S such that the equal-slices density of [math]\displaystyle{ \mathcal{A} }[/math] inside [math]\displaystyle{ \mathcal{U}\otimes\mathcal{V} }[/math] is at least [math]\displaystyle{ \delta+2\eta, }[/math] where [math]\displaystyle{ \eta=\eta(\delta). }[/math] Moreover, the equal-slices density of [math]\displaystyle{ \mathcal{U}\otimes\mathcal{V} }[/math] is at least [math]\displaystyle{ \eta(\delta). }[/math] (The proof on this wiki gives [math]\displaystyle{ \eta(\delta)=\delta^2/27.) }[/math]

Step 2. By passing to a further subspace, we can obtain the same conclusion except that now the density increase drops from [math]\displaystyle{ \eta }[/math] to [math]\displaystyle{ \eta/2 }[/math] and the density of [math]\displaystyle{ \mathcal{U}\otimes\mathcal{V} }[/math] drops to [math]\displaystyle{ \theta(\delta). }[/math] (This step is not yet checked, but should follow from a suitable averaging argument.)

Step 3. Consider the following distribution on points [math]\displaystyle{ (U,V,W)\in[3]^n. }[/math] You first choose a random pair [math]\displaystyle{ (U_0,V_0) }[/math] of disjoint subsets of [n] (uniformly from all [math]\displaystyle{ 3^n }[/math] such pairs) and some random small wildcard sets [math]\displaystyle{ Z_1,\dots,Z_m }[/math] that are disjoint from [math]\displaystyle{ U_0 }[/math] and [math]\displaystyle{ V_0. }[/math] You assign values to the wildcards, fill [math]\displaystyle{ U_0 }[/math] with 1s, [math]\displaystyle{ V_0 }[/math] with 2s and put 3s everywhere else. Then the resulting point is almost exactly uniformly distributed. (One should think of the size of each [math]\displaystyle{ Z_i }[/math] as being randomly chosen and bounded above by a large constant depending on [math]\displaystyle{ \delta. }[/math] One should also think of m as a large constant, though eventually it will replaced by a slow-growing function of n, such as [math]\displaystyle{ \log\log n. }[/math])

Step 4. Let [math]\displaystyle{ Z=Z_1\cup\dots\cup Z_m. }[/math] When we randomly choose [math]\displaystyle{ Z_1,\dots,Z_m }[/math] and the assignment of their values, we obtain a random sequence fragment [math]\displaystyle{ (Z,\sigma). }[/math] If the probability that [math]\displaystyle{ (U,V,W)\in\mathcal{A} }[/math] given [math]\displaystyle{ (Z,\sigma) }[/math] and also given that [math]\displaystyle{ U\in\mathcal{U} }[/math] and [math]\displaystyle{ V\in\mathcal{V} }[/math] is ever more than [math]\displaystyle{ \delta+\eta }[/math] (plus a tiny tiny amount that can depend on m) then we've got a density increase on a 12-set in a subspace, at least if the density of [math]\displaystyle{ \mathcal{U}\otimes\mathcal{V} }[/math] given [math]\displaystyle{ (Z,\sigma) }[/math] is not too small. This allows us to do an inner iteration (see the broad overview above). Therefore, we shall try to prove that this must happen.

Step 5. If we randomly choose [math]\displaystyle{ (U_0,V_0) }[/math] and [math]\displaystyle{ Z_1,\dots,Z_m }[/math] as in Step 3, then there is a positive probability, depending only on m and the density of [math]\displaystyle{ \mathcal{U}\otimes\mathcal{V}, }[/math] that [math]\displaystyle{ U_0+2V_0+\sum_i\eta_iZ_i\in\mathcal{U}\otimes\mathcal{V} }[/math] for every sequence [math]\displaystyle{ (\eta_1,\dots,\eta_m)\in[3]^m. }[/math] Here, [math]\displaystyle{ U_0+2V_0+\sum_i\eta_iZ_i }[/math] stands for the sequence that is 1 on [math]\displaystyle{ U_0, }[/math] 2 on [math]\displaystyle{ V_0, }[/math] [math]\displaystyle{ \eta_i }[/math] on [math]\displaystyle{ Z_i, }[/math] and 3 everywhere else. This follows from the multidimensional Sperner theorem, or possibly from very similar arguments rather than from the theorem itself. (I have no idea why that displayed expression is coming out displayed.)

Step 6. For any given choice of [math]\displaystyle{ Z_1,\dots,Z_m, }[/math] the set of all [math]\displaystyle{ (U_0,V_0) }[/math] such that [math]\displaystyle{ U_0+2V_0+\sum_i\eta_iZ_i\in\mathcal{U}\otimes\mathcal{V} }[/math] for every sequence [math]\displaystyle{ (\eta_1,\dots,\eta_m)\in[3]^m }[/math] is a 12-set, which we denote by [math]\displaystyle{ \mathcal{U}(Z_1,\dots,Z_m)\otimes\mathcal{V}(Z_1,\dots,Z_m). }[/math]

Step 7. By averaging, we can pick [math]\displaystyle{ Z_1,\dots,Z_m }[/math] such that [math]\displaystyle{ \mathcal{U}(Z_1,\dots,Z_m)\otimes\mathcal{V}(Z_1,\dots,Z_m) }[/math] is dense in [math]\displaystyle{ [3]^{[n]\setminus Z}. }[/math] Let us fix this choice of [math]\displaystyle{ Z_1,\dots,Z_m }[/math] and set [math]\displaystyle{ \mathcal{U}'=\mathcal{U}(Z_1,\dots,Z_m) }[/math] and [math]\displaystyle{ \mathcal{V}'=\mathcal{V}(Z_1,\dots,Z_m). }[/math]

Step 8. Define a bipartite graph by joining a pair [math]\displaystyle{ (U_0,V_0)\in\mathcal{U}'\otimes\mathcal{V}' }[/math] to an m-tuple [math]\displaystyle{ (\eta_1,\dots,\eta_m)\in[3]^m }[/math] if and only if [math]\displaystyle{ U_0+2V_0+\sum_i\eta_iZ_i\in\mathcal{A}. }[/math] Then if any vertex [math]\displaystyle{ (U_0,V_0) }[/math] has degree at least [math]\displaystyle{ (\delta+\eta/2)3^m, }[/math] then [math]\displaystyle{ \mathcal{A} }[/math] has density [math]\displaystyle{ \delta+\eta/2 }[/math] in the subspace [math]\displaystyle{ U_0+2V_0+\langle Z_1,\dots,Z_m\rangle, }[/math] which gives us our density increase. (Recall that eventually m is allowed to tend to infinity.)

Step 8. Therefore, we may assume that the maximum degree, and hence the average degree, of the vertices [math]\displaystyle{ (U_0,V_0) }[/math] is at most [math]\displaystyle{ (\delta+\eta/2)3^m, }[/math] which implies, by averaging, that the average degree of the [math]\displaystyle{ (\eta_1,\dots,\eta_m) }[/math] is at most [math]\displaystyle{ (\delta+\eta/2)|\mathcal{U}'\otimes\mathcal{V}'. }[/math] [math]\displaystyle{ }[/math][math]\displaystyle{ }[/math][math]\displaystyle{ }[/math][math]\displaystyle{ }[/math][math]\displaystyle{ }[/math][math]\displaystyle{ }[/math][math]\displaystyle{ }[/math][math]\displaystyle{ }[/math]