# DHJ(k) implies multidimensional DHJ(k)

## Introduction

This is a result that will be needed if the proof of DHJ(3) is correct and we want to push it through for DHJ(k).

## The proof

Let $\mathcal{A}$ be a density-$\delta$ subset of $[k]^n$ and let M be large enough so that every subset of $[k]^M$ of density at least $\theta$ contains a combinatorial line. Now split $[k]^n$ up into $[k]^M\times[k]^{n-M}.$ For a proportion at least $\delta/2$ of the points y in $[k]^{n-M}$ the set of $x\in[k]^M$ such that $(x,y)\in\mathcal{A}$ has density at least $\delta/2.$ Therefore, by DHJ(k) (with $\theta=\delta/2$) we have a combinatorial line. Since there are fewer than $(k+1)^M$ combinatorial lines to choose from, by the pigeonhole principle we can find a combinatorial line $L\subset[k]^M$ and a set $\mathcal{A}_1$ of density $\delta/2(k+1)^M$ in $[k]^{n-M}$ such that $(x,y)\in\mathcal{A}$ whenever $x\in L$ and $y\in\mathcal{A}_1.$ And now by induction we can find an (m-1)-dimensional subspace in $\mathcal{A}_1$ and we're done.

## A weak multidimensional DHJ(k) implies DHJ(k)

It is also true that a weak multidimensional DHJ(k) implies DHJ(k). We will show that the following statement is equivalent to DHJ(k):

“There is a constant, c < 1 that for every d there is an n that any c-dense subset of $[k]^n$ contains a d-dimensional subspace.”

We should show that the statement above implies DHJ(k). As before, write $[k]^n$ as $[k]^r\times[k]^s$ , where s is much bigger than r. For each $y\in [k]^s$ , define $\mathcal{A}_y$ to be $\{x\in[k]^r:(x,y)\in\mathcal{A}\}$ . Let Y denote the set of $y\in [k]^s$ such that $\mathcal{A}_y$ is empty. Suppose that $\mathcal{A}$ is large, line-free, and its density is $\delta =\Delta-\epsilon$ where $\Delta$ is the limit of density of line-free sets and $\epsilon \lt (1-c)\Delta$ . We can also suppose that no $\mathcal{A}_y$ has density much larger than $\Delta$ as that would guarantee a combinatorial line. But then the density of Y is at most 1-c, so there is a c-dense set, $Z=[k]^s-Y$, such that any element is a tail of some elements of $\mathcal{A}$ . For every $y \in Z$ choose an $x\in [k]^r:(x,y)\in\mathcal{A}$ . This x will be the colour of y. It gives a $[k]^r$ colouring on Z. By the initial condition Z contains arbitrary large subspaces, so by HJ(k) we get a line in $\mathcal{A}$ .