Dhj-lown-lower.tex
\section{Lower bounds for the density Hales-Jewett problem}\label{dhj-lower-sec}
The purpose of this section is to establish various lower bounds for $c_{n,3}$, in particular establishing Theorem \ref{dhj-lower} and the lower bound component of Theorem \ref{dhj-upper}.
As observed in the introduction, if $B \subset \Delta_{3,n}$ is a Fujimura set (i.e. a subset of $\Delta_{3,n} = \{ (a,b,c) \in \N^3: a+b+c=n\}$ which contains no upward equilateral triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$), then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a line-free subset of $[3]^n$, which gives the lower bound \begin{equation}\label{cn3}
c_{n,3} \geq |A_B| = \sum_{(a,b,c) \in B} \frac{n!}{a! b! c!}.
\end{equation} All of the lower bounds for $c_{n,3}$ in this paper will be constructed via this device.
In order to use \eqref{cn3}, one of course needs to build Fujimura sets $B$ which are ``large in the sense that the right-hand side of \eqref{cn3} is large. A fruitful starting point for this goal is the sets $$B_{j,n} := \{ (a,b,c) \in \Delta_{3,n}: a + 2b \neq j \hbox{ mod } 3 \}$$ for $j=0,1,2$. Observe that in order for a triangle $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ to lie in $B_{j,n}$, the length $r$ of the triangle must be a multiple of $3$. This already makes $B_{j,n}$ a Fujimura set for $n < 3$ (and $B_{0,n}$ a Fujimura set for $n = 3$).
When $n$ is not a multiple of $3$, the $B_{j,n}$ are all rotations of each other and give equivalent sets (of size $2 \times 3^{n-1}$). When $n$ is a multiple of $3$, the sets $B_{1,n}$ and $B_{2,n}$ are reflections of each other, but $B_{0,n}$ is not equivalent to the other two sets (in particular, it omits all three corners of $\Delta_{3,n}$); the associated set $A_{B_{0,n}}$ is slightly larger than $A_{B_{1,n}}$ and $A_{B_{2,n}}$ and thus is slightly better for constructing line-free sets.
As mentioned already, $B_{0,n}$ is a Fujimura set for $n \leq 3$, and hence $A_{B_{0,n}}$ is line-free for $n \leq 3$. Applying \eqref{cn3} one obtains the lower bounds $$ c_{0,3} \geq 1; c_{1,3} \geq 2; c_{2,3} \geq 6; c_{3,3} \geq 18.$$
For $n>3$, $B_{0,n}$ contains some triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ and so is not a Fujimura set, but one can remove points from this set to recover the Fujimura property. For instance, for $n \leq 6$, the only triangles in $B_{0,n}$ have side length $r=3$. One can ``delete these triangles by removing one vertex from each; in order to optimise the bound \eqref{cn3} it is preferable to delete vertices near the corners of $\Delta_{3,n}$ rather than near the centre. These considerations lead to the Fujimura sets \begin{align*} B_{0,4} &\backslash \{ (0,0,4), (0,4,0), (4,0,0) \}\\ B_{0,5} &\backslash \{ (0,4,1), (0,5,0), (4,0,1), (5,0,0) \}\\ B_{0,6} &\backslash \{ (0,1,5), (0,5,1), (1,0,5), (0,1,5), (1,5,0), (5,1,0) \} \end{align*} which by \eqref{cn3} gives the lower bounds $$ c_{4,3} \geq 52; c_{5,3} \geq 150; c_{6,3} \geq 450.$$ Thus we have established all the lower bounds needed for Theorem \ref{dhj-upper}.
One can of course continue this process by hand, for instance the set $$ B_{0,7} \backslash \{(0,1,6),(1,0,6),(0,5,2),(5,0,2),(1,5,1),(5,1,1),(1,6,0),(6,1,0) \}$$ gives the lower bound $c_{7,3} \geq 1302$, which we tentatively conjecture to be the correct bound.
A simplification was found when $n$ is a multiple of $3$. Observe that for $n=6$, the sets excluded from $B_{0,6}$ are all permutations of $(0,1,5)$. So the remaining sets are all the permutations of $(1,2,3)$ and $(0,2,4)$. In the same way, sets for $n=9$, $12$ and $15$ can be described as: \begin{itemize} \item $n=9$: $(2,3,4),(1,3,5),(0,4,5)$ and permutations; \item $n=12$: $(3,4,5),(2,4,6),(1,5,6),(0,2,10),(0,5,7)$ and permutations; \item $n=15$: $(4,5,6),(3,5,7),(2,6,7),(1,3,11),(1,6,8),(0,4,11),(0,7,8)$ and permutations. \end{itemize}
When $n$ is not a multiple of $3$, say $n=3m-1$ or $n=3m-2$, one first finds a solution for $n=3m$. Then for $n=3m-1$, one restricts the first digit of the $3m$ sequence to equal $1$. This leaves exactly one-third as many points for $3m-1$ as for $3m$. For $n=3m-1$, one restricts the first two digits of the $3m$ sequence to be $12$. This leaves roughly one-ninth as many points for $3m-2$ as for $3m$.
The following is an effective method to find good, though not optimal, solutions for any $n=3m$. (For $n<21$, ignore any triple with a negative entry.)
Start with thirteen groups of points in the centre, formed from adding one of the following points, or its permutation, to $M := (m,m,m)$, when $n=3m$: $$(-7,-3,+10), (-7, 0,+7),(-7,+3,+4),(-6,-4,+10),(-6,-1,+7),(-6,+2,+4)$$ $$(-5,-1,+6),(-5,+2,+3),(-4,-2,+6),(-4,+1,+3),(-3,+1,+2),(-2,0,+2),(-1,0,+1)$$ Then include eights string of points, stretching to the edges of the triangle $\Delta_n$; \begin{itemize} \item $M+(-8-2x,-6-2x,14+4x)$, $M+(-8-2x,-3-2x,11+4x)$, $M+(-8-2x,x,8+x)$, $M+(-8-2x,3+x,5+x)$ and permutations ($0\le 2x \le M-8$); \item $M+(-9-2x,-5-2x,14+4x)$, $M+(-9-2x,-2-2x,11+4x)$, $M+(-9-2x,1+x,8+x)$, $M+(-9-2x,4+x,5+x)$ and permutations ($0\le 2x \le M-9$). \end{itemize} This solution gives $O(2.7 \sqrt(\log (n)/n)3^n$ points for values of $n$ up to around $1000$. This is asymptotically smaller than the known optimum of $3^{n-O(\sqrt(\log(n)))}$. When $n=99$, it gives more than $3^n/3$ points.
The following solution gives more points for $n>1000$, but not for moderate $n$:
\begin{itemize} \item Define a sequence, of all positive numbers which, in base 3, do not contain a 1. Add 1 to all multiples of 3 in this sequence. This sequence does not contain a length-3 arithmetic progression. It starts $1,2,7,8,19,20,25,26,55, \ldots$; \item List all the $(abc)$ triples that sum to $n$, for which the larger two differ by a number from the sequence; \item Exclude the case when the smaller two differ by 1; \item Include the case when $(a,b,c)$ is a permutation of $n/3+(-1,0,1)$. \end{itemize}
An integer program was run obtain the maximum lower bound one could establish from \eqref{cn3} (see Appendix \ref{integer-sec}). The results for $1 \leq n \leq 20$ are displayed in Figure \ref{nlow}:
\begin{figure}[tb] \centerline{ \begin{tabular}{|ll|ll|} \hline $n$ & lower bound & $n$ & lower bound \\ \hline 1 & 2 &11& 96338\\ 2 & 6 & 12& 287892\\ 3 & 18 & 13& 854139\\ 4 & 52 & 14& 2537821\\ 5 & 150& 15& 7528835\\ 6 & 450& 16& 22517082\\ 7 & 1302& 17& 66944301\\ 8 & 3780&18& 198629224\\ 9 & 11340&19& 593911730\\ 10& 32864& 20& 1766894722\\ \hline \end{tabular}} \caption{Lower bounds for $c_n$ obtained by the $A_B$ construction.} \label{nlow} \end{figure}
More complete data, including the list of optimisers, can be found at \centerline{{\tt http://abel.math.umu.se/~klasm/Data/HJ/}.}
If $B$ maximises the right-hand side of \eqref{cn3}, it is easy to see that $A_B$ is a line-free set which is maximal in the sense that the addition of any further point to $A_B$ will create a combinatorial line. Thus one might conjecture that the maximal value of the right-hand side \eqref{cn3} is in fact equal to $c_{n,3}$ for all $n$; Theorem \ref{dhj-upper} asserts that this conjecture is true for $n \leq 6$.
Now we prove Theorem \ref{dhj-lower}.
\begin{proof}[Proof of Theorem \ref {dhj-lower}] Let $M$ be the circulant matrix with first row $(1,2,\ldots,k-1)$, second row $(k,1,2,\dots,k-1)$, and so on. Note that $M$ has nonzero determinant by well-known properties of circulant matrices.
Let $S$ be a subset of the interval $[-\sqrt {n}/2, \sqrt {n}/2)$ that contains no nonconstant arithmetic progressions of length k, and let $B\subset\Delta_{n, k}$ be the set
\[ B := \{(n-\sum_{i=1}^{k-1} a_i ,a_1,a_2,\dots, a_{k-1}) :
(a_1,\dots,a_{k-1})= \det(M) M^{-1}\vec{s} , \vec{s}\in S^{k-1}\}.\]
The map $(m,a_1,\dots,a_{k-1}) \mapsto M (a_1,\dots,a_{k-1})$ takes simplices to nonconstant arithmetic progressions in ${\mathbb Z}^{k-1}$, and takes $B$ to $\{det(M) \, \vec{s} \colon \vec{s} \in S^{k-1}\}$, which is a set containing no nonconstant arithmetic progressions. Thus, $B$ is a Fujimora set and so does not contain any combinatorial lines.
If all of $a_1,\ldots,a_k$ are within $C_1\sqrt{n}$ of $n/k$, then $|\Gamma_{\vec{a}}| \geq C k^n/n^{(k-1)/2}$ (where $C$ depends on $C_1$) by the central limit theorem. By our choice of $S$ and applying~\eqref{cn3}, we obtain
$$ c_ {n, k}\geq C k^n/n^{(k-1)/2} |S|^{k-1} = C k^n \left( \frac{|S|}{\sqrt{n}} \right)^{k-1}. $$
One can take $S$ to have cardinality $r_ k (\sqrt {n}) $, which from the results of O'Bryant~\cite {obryant}) satisfies (for all sufficiently large $n$, some $C>0$, and $\ell$ the largest integer satisfying $k> 2^{\ell-1}$)
$$ \frac{r_k (\sqrt{n})}{\sqrt{n}} \geq C (\log n)^{1/(2\ell)}\exp_2 (-\ell 2^{(\ell-1)/2-1/\ell} \sqrt[\ell]{\log_2 n}),$$
which completes the proof. \end {proof}
