2 Unbounded block sizes

In this section we show that there is no universal bound on the block sizes needed in the block sets conjecture. In other words, for each d we give a template T and a number of colours k for which the block size cannot be at most d. We mention again that in all cases the templates we consider will be ones that we know do satisfy the block sets conjecture (all are cases that are proved in [Reference Leader, Russell and Walters8]). In fact, they will all be templates on alphabet $[3]$ .

The rough idea of our proof is as follows. We will encode (via some parts of the colouring) information not only about the ‘profile’ of a word x, meaning how many of each symbol occur in x, but also about how many times we have (say) a 3 before a 1 (not necessarily adjacent, but respecting the linear order of the coordinates). In fact, we also ask about the actual position (modulo some large fixed number) of the 1 in such a pair, so that we have several separate ‘contributions’. We make this precise below.

Proof. Let T be $1\underbrace {2\dots 2}_{d} \underbrace {3\dots 3}_{d^3}$ . In other words, T consists of a single 1, d copies of 2, and $d^3$ copies of 3. The size of T is $s=1+d(d^2+1)$ .

For the colouring, we are going to assign (as a colour) to every finite word on $[3]$ a vector of length $d^2+1$ with coordinates in $\mathbb {Z}_{d+1}$ – in other words, a point in $\mathbb {Z}_{m}^{l}$ , where $m=d+1$ and $l=d^2+1$ . The additive group notation comes from the fact that a colour will be the sum of certain ‘contributions’.

To start off, let x be a finite word on $[3]$ . For each i with $x_i=1$ , we define $a_i$ to be the number of 1’s and 2’s that occur before i when counted from the left, taken modulo $d^2+1$ . In symbols,

$$\begin{align*}a_i=|\{j<i:x_j=1\text{ or }x_j=2\}|\quad \mod d^2+1.\end{align*}$$

We define the contribution of i to be the basis vector $e_{a_i}$ in $\mathbb {Z}_m^l$ . If $x_i\neq 1$ , then the contribution of i is 0. Finally, we define the colour c of x to be the sum of all of these contributions: in other words, $\sum _{\{i:x_i=1\}}e_{a_i}$ .

We will now show that for this colouring there is no monochromatic copy of T with block sizes at most d.

Suppose for a contradiction that there exists a positive integer n, and blocks $B_1,B_2,B_3,\dots ,B_s$ all of size at most $d-1$ , and $y\in [3]^{[n]\setminus \bigcup _i B_i}$ , that yield a monochromatic copy of T.

By relabelling if necessary, we may assume that, for $i<j$ , the first coordinate of $B_i$ is less than the first coordinate of $B_j$ . For each i, let $b_i$ be the number of 1’s plus the number of 2’s that occur in y (i.e., not counting any points in the blocks) before the first coordinate of $B_i$ , counted modulo $d^2+1$ . Since we have $s=1+d(d^2+1)$ blocks and $d^2+1$ possible values for the $b_i$ , by the pigeonhole principle we must have $d+1$ blocks for which the $b_i$ have the same value, say b.

Let $B_{i_1},B_{i_2},\dots ,B_{i_{d+1}}$ be the corresponding blocks, where $i_1<i_2<\dots <i_{d+1}$ . For clarity, let $C_j=B_{i_j}$ for all $1\leq j\leq d+1$ . Since we have exactly $d+1$ such blocks, which is the number of 1’s and 2’s in our chosen template T, from now on we will only substitute the single 1 and the d 2’s in all the $C_i$ , thus forcing all the other blocks to be 3’s.

This gives us $d+1$ words $w_1,w_2,\dots ,w_{d+1}$ , given by which $C_i$ we insert the 1 into. We will derive a contradiction by showing that these $d+1$ words do not, in fact, all have the same colour.

We first observe that the colour of one of these words is the sum of the contributions of coordinates $\bigcup _i C_i$ , together with the contributions from all other 1’s, which cannot lie in any $B_i$ (so they are ‘inactive’ or fixed coordinates).

Note that the sum of contributions of the ‘inactive’ coordinates is constant since during our substitutions we never replace a 3 with a 1 or 2, or vice versa. Therefore, since we are assuming that the sum of all contributions is constant, the sum of the contributions from the coordinates of $\bigcup _i C_i$ has to be constant too, say $v\in \mathbb {Z}_m^l$ .

For each $1\leq i\leq d+1$ we denote by $p_i$ the number of 1’s and 2’s before the first coordinate of $C_i$ in any of the $w_j$ (since the $w_j$ only differ by some 1’s changing to 2’s and vice versa this is independent of which $w_j$ we choose). By construction, the $C_i$ ’s have the same number, say b, of 1’s and 2’s mod $d^2+1$ before them, when looking at y only (so not considering any of the blocks). This means that before $C_1$ we have $p_1 \equiv b\ \mod {d^2+1}$ 1’s and 2’s.

Next, $p_i$ is $b+\lambda _i\ \mod d^2+1$ , where $\lambda _i$ is the number of points of $\bigcup _j C_j$ that occur before the first point of $C_i$ .

Since all blocks have size at most $d-1$ , we get that $\lambda _i\leq (i-1)d\leq d^2$ , and so all the $p_i$ are distinct $\mod d^2+1$ . Moreover, since the first coordinate of $C_i$ occurs before the first coordinate of $C_{i+1}$ we see that $\lambda _{i+1}\geq \lambda _i+1$ .

Finally, observe that putting 1’s in the block $C_i$ contributes to the $p_i\ \mod {d^2+1}$ coordinate of the colouring. Therefore, by moving the 1 from $C_1$ to $C_i$ adds the contribution $e_{p_i}$ and subtracts the contribution $e_{p_1}$ . However, the colour v has to be the same, and thus $e_{p_i}$ has to be the contribution of one of the 1’s that were previously in $C_1$ . (This is because the other contributions from this block $C_i$ cannot cancel out the $e_{p_i}$ , as $|C_i|\le d$ .) Since this is true for any $2\leq i\leq d+1$ and all $p_i$ are distinct $\mod d^2+1$ , $C_1$ must produce at least $d+1$ contributions, thus $|C_1|\ge d+1$ , a contradiction.

We remark that the above proof is unchanged if one wishes for a more general template ${T=1\underbrace {22\dots 2}_{p}\underbrace {33\dots 3}_{q}}$ . The length of T is $s=1+p+q$ . The colouring is the same – associate to every word over $[3]$ a point in $\mathbb {Z}_m^l$ as above. Provided s is at least $pl+1$ , we still get $p+1$ blocks with the same value for $b_i$ . By taking $l>pd$ , we ensure that the $\lambda _i$ are still disjoint, and since $m>d$ we still produce at least $d+1$ contributions. Hence, if $s\geq pl+1$ , $pd<l$ and $m>d$ , the above proof goes through identically. So what we obtain is that a template with one 1, p 2’s and q 3’s does need block sizes greater than d whenever $q\geq p^2d$ .

If one considers the natural templates $123\dots m$ , it is easy to check that if the block sets conjecture holds for one such template $123\dots m$ then it also holds for all templates of length m. This is informally because identifying symbols ‘maps block sets to block sets’. Hence it also follows from Theorem 2 that for any d there is a template of the form $123\dots m$ for which one cannot guarantee that one can take blocks of size at most d.

We mention that, for Euclidean Ramsey sets, the result above has the following consequence. For any d, there exists a transitive set X that is Ramsey, and a number of colours k, such that whenever the set $(1/\alpha ) X^n$ is a k-Ramsey set for X we must have $\alpha \geq d$ . This may be read out of the equivalent form of the block sets conjecture in terms of product sets mentioned above.

3 The template $123$

As we have seen, the block sets conjecture is essentially trivial for templates over the alphabet of size 2, and even with blocks of size 1. So we now turn our attention to the first nontrivial template, namely $123$ . Here it was known that the blocks cannot be taken to be of size 1, but the block sizes that were known to work were huge. Indeed, in [Reference Kříž7] the block size is a tower, coming from an iterated application of some Ramsey theorems. And in [Reference Leader, Russell and Walters8] there is no iteration, but the proof uses van der Waerden’s theorem for k colours, so that the block size is not even constant but grows with the number of colours.

Our aim is to show that block size 2 suffices. This is best possible, in view of the remarks above. In fact, we show that a fixed pattern suffices: one can always find a block set whose pattern (the way that the set of $2\cdot 3$ coordinates is partitioned into the 3 blocks) is fixed, namely $ABCCBA$ . (Here $A,B,C$ represent the relative positions of the 2-sets $I_1,I_2,I_3$ in the fixed linear order on $[n]$ .) The key idea is in fact this palindromic aspect (the use of reversals). In contrast, the earlier proofs used patterns like $ABCABCABC\dots ABC$ , or iterated versions of that, and as we will see (see the discussion after the proof) there are reasons why this makes the proof much harder.

Proof. For simplicity and readability, we will just think of n as being ‘very large’: the actual value of n will be fixed during the proof.

Let $\theta :[3]^n\rightarrow [k]$ be a fixed k-colouring of $[3]^n$ . We must show that we can find $n-6$ fixed coordinates and 6 active ones, labelled $ABCCBA$ in this order, such that whenever we substitute $1,2,3$ (each corresponding to exactly one of A, B and C), all 6 resulting words have the same colour.

Our first step is to ensure informally that, by passing to a subset of the coordinates, ‘the positions of the 3’s do not matter’. This is a fairly standard kind of step to take in such a Ramsey argument.

Let $\mathcal A$ be the set of words of length n (on alphabet $[3])$ that contain exactly $2k+2$ 2’s and no 1’s. For example, if $k=2$ and $n=10$ , such a word could be $2322333222$ .

Let also $\chi $ be the set of words of length $2k+2$ on alphabet $[2]$ that have exactly $k+1$ 1’s and $k+1$ 2’s. For example, if $k=2$ , then words such as $111222$ , $121212$ , $1221211$ are all in $\chi $ . Clearly, the size of $\chi $ is $s=\binom {2k+2}{k+1}$ , and so let $\chi =\{w_1, w_2, \dots , w_s\}$ .

Given a word $x\in \mathcal A$ and a word $w\in \chi $ , we can generate a unique word in $[3]^n$ by replacing the 2’s in x with the word w – in other words, at the positions of the 2’s in x we insert, in order, the letters of w. Call this word $f(x,w)\in [3]^n$ . As an example, if $k=2$ , $n=10$ , $x=2322333222$ and $w=121212$ , then $f(x,w)=1321333212$ .

It is clear that f is a bijection between $\mathcal A\times \chi $ and the set of all the words of length n with exactly $k+1$ 1’s and $k+1$ 2’s. This allows us to induce a colouring $\mathcal A$ with $k^s$ colours as follows: $\Theta :\mathcal A\rightarrow [k]^s$ is given by $\Theta (x)=(\theta (f(x, w_1)),\theta (f(x,w_2)),\dots ,\theta (f(x,w_s))).$

Next, a word w in $\mathcal A$ can be seen as a subset of $[n]$ of size $2k+2$ , by selecting the coordinates at which the 2’s are. So we may also view $\Theta $ as a $k^s$ -colouring of $[n]^{(2k+2)}$ .

By Ramsey’s theorem, there exists n such that whenever $[n]^{(2k+2)}$ is $k^s$ -coloured there exists a set S of size $2k+4$ such that $S^{(2k+2)}$ is monochromatic. (This is the actual version of ‘n is very large’ that we need.)

Applying this to our colouring $\Theta $ , and assuming for clarity that $S=[2k+4]$ , we get that no matter how we insert $2k+2$ 2’s and 2 3’s in the first $2k+4$ coordinates, we get the same colour under $\Theta $ . This means that for any $w\in \chi $ , if $x_1, x_2\in \mathcal A$ have their 2’s in the first $2k+4$ coordinates, then $\theta (f(x_1,w))=\theta (f(x_2,w))$ . In other words, no matter how we insert the 2 3’s in the first $2k+4$ coordinates, the colour does not change, as long as the $12$ word is the same, namely w. In other words, informally, ‘the positions of the 3’s do not matter’.

Now consider the following $k+1$ words in $\chi $ :

$$ \begin{align*} z_1&=211212\dots12\\z_2&=122112\dots12\\&\qquad\vdots\\z_{k+1}&=121212\dots21 \end{align*} $$

In words, $z_i$ has 1’s at precisely positions $1,3,\dots , 2i-3, 2i,2i+1\dots 2k+1$ – or, simply put, it is a concatenation of $k+1$ ‘12’ blocks, with the $i^{\text {th}}$ block flipped to a ‘21’.

Let $x\in \mathcal A$ be the word that starts with $2k+2$ consecutive 2’s, the rest being all 3’s. Consider the words $f(x,z_i)$ for $1\leq i\leq k+1$ . Since we have only k colours, by the pigeonhole principle there exists $i<j$ such that the colour of $f(x,z_i)$ is the same as the colour of $f(x,z_j)$ . However, by the above choice of S, this colour does not change no matter how we reorder the first $2k+4$ coordinates of x.

To conclude the proof, we take the blocks to be $\{2i-1,2j+2\}$ , $\{2i, 2j+1\}$ and $\{2i+1, 2j\}$ , and the fixed coordinates are the $k-1$ remaining ‘12’ blocks in the first $2k+4$ coordinates and all 3’s in rest. This gives a monochromatic copy of $123$ with pattern $ABCCBA$ , as required.

To help the reader visualize the above construction, let us look at a concrete example. Suppose $k=3$ and $f(x,z_1)$ has the same colour as $f(x, z_2)$ . In other words the following receive the same colour, say a:

By moving the two 3’s around in the first word, we see that the following all have colour a:

$$ \begin{align*} \begin {array}{c} 3211231212\dots \\ 2311321212\dots \\ 2133121212\dots \end {array}\end{align*} $$

Doing the same thing for the second word, we see that the following also have colour a:

$$ \begin{align*} \begin {array}{c} 3122131212\dots \\ 1322311212\dots \\ 1233211212\dots \end {array}\end{align*} $$

And so the first 6 coordinates, under the pattern $ABCCBA$ , give a monochromatic copy of $123$ .

We remark that the above proof goes through in the exact same way more generally, for templates of the form $12\underbrace {33333}_{q}$ , again yielding block size 2. Of course, by Theorem 2 it cannot go through in the same way for templates of the form $1\underbrace {22222}_{p}\underbrace {33333}_{q}$ . However, it is possible that for each such template there could be a fixed block size that one may always take for that template (for any number of colours).

An alternative viewpoint for the second half of the above argument (after the positions of the 3’s do not matter) is to identify a word with a vector over $\mathbb {Z}$ as follows. We fix a ‘reference’ word w with t 1’s. Then, for any other word v with t 1’s, we identify v with the point in $\mathbb {Z}^t$ that represents how the 1’s have moved between w and v. In other words, if w has 1’s in positions $a_1,\dots ,a_t$ then if v has 1’s in positions $a_1+b_1,\dots ,a_t+b_t$ then it corresponds to the vector $(b_1,\dots ,b_t)$ . So we may view a colouring of the words (with t 1’s) as a colouring of $\mathbb {Z}^t$ . (Strictly, one would need to bound the coordinates of the points in $\mathbb {Z}^t$ that we are considering, as one would not want the t 1’s to overlap, for example – we ignore this point in the interests of clarity.)

In this language, the final step of the argument above is just the assertion that, whenever we k-colour $\mathbb {Z}^t$ (for fixed k and for sufficiently large t) there exist two vectors of the same colour that differ by a vector of the form $e_i-e_j$ .

It is interesting to compare this with the approach taken in [Reference Leader, Russell and Walters8], where no reversals appear. There the idea is as follows. If we used pattern $ABCABCABC\dots ABC$ (say with each block set being of size d) then we would be asking for two points x and $x+v$ of the same colour, in a k-colouring of $\mathbb {Z}^n$ with n large, such that v has d 1’s and all other coordinates zero. This is of course impossible, as we may just 2-colour by asking whether or not the coordinate sum, taken modulo $2d$ , lies in $[0,d-1]$ . So one has to consider also patterns like $AABBCCAABBCC\dots AABBCC$ , where one would be asking for v having d 2’s and all other coordinates zero, and so on. It is this (which of course works for enough candidate patterns, thanks to van der Waerden’s theorem) that means that one of these fixed patterns by itself cannot work.

To end this section, let us record that Theorem 3 has the following direct consequence for Ramsey sets. If X is a regular hexagon, then for any k the set $X^n$ , shrunk only by a factor $\sqrt {2}$ , is k-Ramsey for X. And the same would hold for any plane hexagon on which $S_3$ acts transitively – so any hexagon whose internal angles are all $120^{\circ }$ and whose sides alternate in length as $a,b,a,b,a,b$ . This follows from the equivalent formulation of the block sets conjecture in terms of product sets – see [Reference Leader, Russell and Walters8].

4 Discussion and open problems

One of the most interesting questions raised by the previous section is whether or not this phenomenon, of the block size being constant as the number of colours grows, holds for any template.

Note that this would actually imply that one can ask for the block size and the pattern of the blocks to be fixed in advance. This is because the number of patterns for a given block size is finite, so if no pattern could be guaranteed (for block size d) then a product colouring would yield no block sets of size d at all.

As mentioned above, the proof of Theorem 3 goes through in the same way for the template $12\underbrace {33333}_{q}$ , so that Conjecture 4 holds for these templates. The templates $1\underbrace {22222}_{p}\underbrace {33333}_{q}$ are the first examples of templates for which we know that the block sets conjecture does hold and yet we do not know if Conjecture 4 holds.

Of course, the most important problem on the block sets conjecture is the full conjecture itself. The simplest unknown case, in the sense of being on a small alphabet, is the template $T=112233$ . This seems to be the key ‘next step’, with the repetition of all the symbols being the crucial obstacle.

Digressing for a moment, we mention again that, for larger alphabets, the template $1234$ is known to satisfy the block sets conjecture, which may be read out of the result of Kříž [Reference Kříž7]. Indeed, Kříž showed that any set that has a group of symmetries that is soluble and acts transitively is Ramsey, and this, translated into block sets language, gives that the template $1234$ satisfies the block sets conjecture because the group $S_4$ is soluble. However, the template $12345$ is open – the fact that the group $S_5$ is not soluble means that Kříž’s result does not apply. (In fact, Kříž’s proof of his result actually always gives that the block sizes may be taken to be fixed – so that for example the template $1234$ does satisfy Conjecture 4. This is perhaps more evidence towards Conjecture 4.)

Returning to the template $112233$ , and continuing the line of thought as given in the explanation after the proof of Theorem 3, let us start by considering patterns, say with all blocks of size d, of the form $ABCDEFABCDEF \dots $ – this is not palindromic, but let us ignore that for the moment. When one ‘moves’ the 1’s from say A and B to B and C (with 2’s now moving to A), and then to C and D (with 2’s now in places A and B), then the positions of the 1’s change by 1 in the first case (as explained above), and by 2 in the second case. If we used some palindromic version, so with repeats of $FEDCBA$ as well, then that would merely change some of these $+1$ s and $+2$ into $-1$ s and $-2$ s.

So we would be asking the following question. Given k, must there exist d such that whenever we k-colour $\mathbb {Z}^n$ , for n large, we always find monochromatic $x,x+v,x+2v$ where v has d 1’s and $d -1$ s and the rest of the coordinates are zero? Sadly this is not true, for example because this v has fixed $2$ -norm, so that we would be asking for a certain nonspherical set (an arithmetic progression of length 3) to be Ramsey.

Now, if instead the pattern were a concatenation of words of the form $AABBCCDDEEFF$ (perhaps made palindromic), then these moves would be by $+2$ and $+4$ respectively (with $-2$ and $-4$ when we use the reversed word). And similarly for repeats of $AAABBBCCCDDDEEEFFF$ and so on. Hence, if we asserted that our pattern consisted of a fixed number of $ABCDEF$ words, a fixed number of $AABBCCDDEEFF$ words, and so on, then the vector representing the positions of the 1’s in the $12$ -word would take values like $x,x+v,x+2v$ , where v would have a fixed form consisting of $2t$ values, of which a given number would be $\pm 1$ s, a given number $\pm 2$ s, and so on. It turns out, however, that some similar behaviour is exhibited by those patterns where the $l_1$ norm of this v is the same – for example, the 1-norm seems also to come in when we move the 1’s to other places in the template. So it actually makes sense to ask for v to have not just fixed support size (of $2d$ ) but also fixed $1$ -norm. This means that if we have $\lambda _1$ parts of the pattern of the form $ABCDEF$ , $\lambda _2$ parts of the form $AABBCCDDEEFF$ , and so on, the we would be asking for a v of support size $2d$ whose support is made up of $\lambda _1$ values of $1$ and the same of $-1$ , and $\lambda _2$ values of $2$ and the same of $-2$ , and so one, where the sum of all the $i\cdot \lambda _i$ equals d.

Now, of course this is a very special form for a vector of 1-norm $2d$ . Nevertheless, we feel that asking the question for a general vector is a sensible one, and perhaps is relevant for making further progress. This is perhaps rather speculative, but it also seems just too nice not to be asked.

One important point is that there is a linear, not just a metric, constraint here: we are not just asking for 3 points with pairwise $l_1$ distances $d,d,2d$ . In fact, the latter is quite easy to achieve by a direct argument. Of course, in $l_2$ these pairwise distances would imply collinearity, but not in $l_1$ .

Now, for the 2-norm in place of the 1-norm Question 5 would be false, as the original argument of [Reference Erdős, Graham, Montgomery, Rothschild, Spencer and Straus2] shows - since an arithmetic progression of length 3 is not spherical. In the other direction, for the $l_\infty $ norm the result is true. As noted by Kupavskii and Sagdeev [Reference Kupavskii and Sagdeev6], this may be read out of the Hales-Jewett theorem instantly. Indeed, any monochromatic line in $[3]^n$ corresponds to an arithmetic progression with common difference having $l_\infty $ norm 1. (See that paper, and also Frankl, Kupavskii and Sagdeev [Reference Frankl, Kupavskii and Sagdeev3], for several strong bounds on the actual dimensions needed.) There is also some interesting related work on all norms $l_p$ except for $p=1$ – see Cook, Magyar and Pramanik[Reference Cook, Magyar and Pramanik1] – but the problem above remains tantalizingly open.

The above should just be the start, and there is a vast generalization that one could also ask for. If the answer to Question 5 is in the affirmative, then actually one would want longer arithmetic progressions, and indeed one would want larger parts of $l_1$ balls – where the set $x-v,x,x+v$ in Question 5 is viewed as just one very small part of an $l_1$ ball of radius d (centred at x). More usefully, one could also view this set (informally) as being the one-dimensional $l_1$ ball of radius 1 ‘generated’ by the single vector v (again centred at x). Indeed, some version of (some part of) this is needed even for the template $112233$ , since in the above we only considered the 1’s being in positions A and B or B and C or C and D – it turns out that looking at other places does correspond to taking other vectors than $x,x+v,x+2v$ and indeed these vectors do turn out to all lie on a constant radius $l_1$ sphere about some point. So again answering the question below may well prove relevant to progress on the block sets conjecture for the template $112233$ and beyond.

In general, we define the $l_1$ ball of radius r generated by $u_1,\dots ,u_t$ , where $u_1,\dots u_t$ are disjointly supported points in $\mathbb {Z}^n$ , to consist of all sums $\sum \lambda _i u_i$ , where the $\lambda _i$ are integers satisfying $\sum |\lambda _i| \leq r$ .

As explained above, the case $r=t=1$ of this is precisely Question 5. In fact, we suspect that the hard part is Question 5: if it is true then perhaps Question 6 ought to be true as well. Note that, because any finite subset of $\mathbb {Z}^n$ is contained in an $l_1$ ball, Question 6 has the attractive Ramsey flavour that the objects we are colouring and the objects that we seek a monochromatic copy of both have the same form.

Finally, we mention that, beyond the block sets conjecture itself, a key question is which (if either) of the two rival conjectures about Euclidean Ramsey sets is correct. There are explicit sets known that are spherical but do not embed into any transitive set: one example is a cyclic kite, namely the four points $(\pm 1, 0)$ and $(a, \pm \sqrt {1-a^2})$ , with a being transcendental (see [Reference Leader, Russell and Walters9]). Determining whether or not this quadrilateral is Ramsey would immediately rule out one of the two conjectures.