More on lines in Euclidean Ramsey theory

Let $\ell_m$ be a sequence of $m$ points on a line with consecutive points at distance one. Answering a question raised by Fox and the first author and independently by Arman and Tsaturian, we show that there is a natural number $m$ and a red/blue-colouring of $\mathbb{E}^n$ for every $n$ that contains no red copy of $\ell_3$ and no blue copy of $\ell_m$.


Introduction
Let E n denote n-dimensional Euclidean space, that is, R n equipped with the Euclidean metric.Given two sets X 1 , X 2 ⊂ E n , we write E n → (X 1 , X 2 ) if every red/blue-coloring of E n contains either a red copy of X 1 or a blue copy of X 2 , where a copy for us will always mean an isometric copy.Conversely, E n (X 1 , X 2 ) means that there is some red/blue-coloring of E n which contains neither a red copy of X 1 nor a blue copy of X 2 .
The study of which sets X 1 , X 2 ⊂ E n satisfy E n → (X 1 , X 2 ) is a particular case of the Euclidean Ramsey problem, which has a long history going back to a series of seminal papers [6,7,8] of Erdős, Graham, Montgomery, Rothschild, Spencer and Straus in the 1970s.Despite the vintage of the problem, surprisingly little progress has been made since these foundational papers (though see [9,12] for some important positive results).For instance, it is an open problem, going back to the papers of Erdős et al. [7], as to whether, for every n, there is m such that E n (X, X) for every X ⊂ E n with |X| = m.
Write ℓ m for the set consisting of m points on a line with consecutive points at distance one.Perhaps because it is a little more accessible than the general problem, the question of determining which n and X satisfy the relation E n → (ℓ 2 , X) has received considerable attention.For instance, it is known [11,14] that E 2 → (ℓ 2 , X) for every four-point set X ⊂ E 2 and that E 2 → (ℓ 2 , ℓ 5 ).On the other hand [5], there is a set X of 8 points in the plane, namely, a regular heptagon with its center, such that E 2 (ℓ 2 , X).In higher dimensions, by combining results of Szlam [13] and Frankl and Wilson [10], it was observed by Fox and the first author [4] that E n → (ℓ 2 , ℓ m ) provided m ≤ 2 cn for some positive constant c (see also [1,2] for some better bounds in low dimensions).Our concern here will be with a question raised independently by Fox and the first author [4] and also by Arman and Tsaturian [2], namely, as to whether an analogous result holds with ℓ 2 replaced by ℓ 3 .That is, for every natural number m, is there a natural number n such that E n → (ℓ 3 , ℓ m )?We answer this question in the negative.
Theorem 1.1.There exists a natural number m such that E n (ℓ 3 , ℓ m ) for all n.
Before our work, the best result that was known in this direction was a 50-year-old result of Erdős et al. [6], who showed that E n (ℓ 6 , ℓ 6 ) for all n.Their proof uses a spherical colouring, where all points at the same distance from the origin receive the same colour.We will also use a spherical colouring, though, unlike the colouring in [6], which is entirely explicit, our colouring will be partly random.
Research supported by NSF Award DMS-2054452.

Preliminaries
In this short section, we note two key lemmas that will be needed in our proof.The first says that certain real-valued quadratic polynomials are reasonably well-distributed modulo a prime q.
Lemma 2.1.Let p(x) = x 2 + αx + β, where α and β are real numbers, and let q be a prime number.Then, for m = q 3 , the set {p(i)} m i=1 overlaps with at least q/6 of the intervals [j, j + 1) with 0 ≤ j ≤ q − 1 when considered mod q.
Proof.By a standard argument using the pigeonhole principle, there exists some k ≤ q 2 such that |kα| ≤ 1/q mod q.We split into two cases, depending on whether k is a multiple of q or not.
Suppose first that k ≡ 0 mod q and consider the set of values {p(ki)} q i=1 .Note first that {i 2 } q i=1 is a set of (q + 1)/2 distinct integers mod q, so, since k is not a multiple of q, the same is also true of the set {k 2 i 2 } q i=1 .Hence, letting p 1 (x) = x 2 + β, we see that the set {p 1 (ki)} q i=1 overlaps with at least q/2 of the intervals [j, j + 1) with 0 ≤ j ≤ q − 1 when considered mod q.But |kiα| ≤ 1 mod q for all 1 ≤ i ≤ q, so that |p(ki) − p 1 (ki)| ≤ 1 for all 1 ≤ i ≤ q.Therefore, since exactly three different intervals are within distance one of any particular interval, the set {p(ki)} q i=1 overlaps with at least q/6 of the intervals [j, j + 1) mod q.
Suppose now that k = sq for some s ≤ q.Then sqα = rq + ǫ for some |ǫ| ≤ 1/q, which implies that α = r s + ǫ ′ , where |ǫ ′ | ≤ 1/q 2 .Without loss of generality, we may assume that r and s have no common factors.Consider now the polynomial p 2 (x) = x 2 + r s x and the set {p 2 (si)} q i=1 .Since p 2 (si) = s 2 i 2 +ri, it is easy to check that p 2 (si) ≡ p 2 (sj) mod q if and only if s 2 (i+j)+r ≡ 0 mod q.Since r and s are coprime, this implies that the set {p 2 (si)} q i=1 takes at least q/2 values mod q.Hence, letting p 3 (x) = x 2 + r s x + β, we see that the set {p 3 (si)} q i=1 overlaps with at least q/2 of the intervals [j, j + 1) with 0 ≤ j ≤ q − 1 when considered mod q.But, since |α − r/s| ≤ 1/q 2 , we have that |p(si) − p 3 (si)| = |α − r s |si ≤ 1, so that, as above, the set {p(si)} q i=1 overlaps with at least q/6 of the intervals [j, j + 1) mod q.
Given M real polynomials p 1 , . . ., p M in N variables, a vector σ ∈ {−1, 0, 1} M is called a sign pattern of p 1 , . . ., p M if there exists some x ∈ R N such that the sign of p i (x) is σ i for all 1 ≤ i ≤ M .The second result we need is the Oleinik-Petrovsky-Thom-Milnor theorem (see, for example, [3]), which, for N fixed, gives a polynomial bound for the number of sign patterns.Lemma 2.2.For M ≥ N ≥ 2, the number of sign patterns of M real polynomials in N variables, each of degree at most D, is at most 50DM N N .

Proof of Theorem
If the points are at distances x 1 , x 2 and x 3 , respectively, from the origin o and the angle a 1 a 2 o is θ, then we have Similarly, if a 1 , a 2 , . . ., a m ∈ R n form a copy of ℓ m with |a i − a i+1 | = 1 for all i = 1, 2, . . ., m − 1, then, again writing x i for the distance of a i from the origin, we have Given these observations, our aim will be to colour R ≥0 so that there is no red solution to y 1 + y 3 = 2y 2 + 2 and no blue solution to the system y i−1 + y i+1 = 2y i + 2 with i = 2, . . ., m − 1.Assuming that we have such a colouring χ, we can simply colour a point a ∈ R n by χ(|a| 2 ) and it is easy to check that there is no red copy of ℓ 3 and no blue copy of ℓ m .
We have therefore moved our problem to one of finding a natural number m and a colouring χ of R ≥0 with no red solution to y 1 +y 3 = 2y 2 +2 and no blue solution to the system y i−1 +y i+1 = 2y i +2 with i = 2, . . ., m − 1.Let q be a prime number.We will take m = q 3 and define χ by choosing an appropriate colouring χ ′ of Z q and then setting χ(y) = χ ′ (⌊y⌋ mod q) for all y ∈ R ≥0 .Our aim now is to show that there is a suitable choice for χ ′ .For this, we consider a random red/blue-colouring χ ′ of Z q and show that, for q sufficiently large, the probability that χ contains either of the banned configurations is small.
Concretely, suppose that Z q is coloured randomly in red and blue with each element of Z q coloured red with probability p = q −3/4 and blue with probability 1 − p.With this choice, the expected number of solutions in red to any of the equations y 1 + y 3 = 2y 2 + c with c ∈ {1, 2, 3} is at most where we used that there are at most 3q solutions to any of our 3 equations with two of the variables {y 1 , y 2 , y 3 } being equal and that q is sufficiently large.Note that if there are indeed no red solutions to these three equations over Z q , then there is no red solution to y 1 + y 3 = 2y 2 + 2 in the colouring χ of R. Indeed, if y i = n i + ǫ i with 0 ≤ ǫ i < 1, then n i is coloured red in χ ′ and for c ∈ {1, 2, 3}.However, we know that there are no red solutions to any of these equations in the colouring χ ′ , so there is no red solution to y 1 + y 3 = 2y 2 + 2 in the colouring χ.
Our aim now is to apply Lemma 2.2 to count the number of different ways in which a set of solutions (y 1 , y 2 , . . ., y m ) to our system of equations can overlap the collection of intervals [j, j + 1) mod q.Without loss of generality, we may assume that 0 ≤ a, d < q.Since, under this assumption, any set of solutions over R to our system of equations is contained in the interval [0, 2m 2 ), it will suffice to count the number of feasible overlaps with the intervals [j, j + 1) with 0 ≤ j ≤ 2m 2 − 1.Since we need to check at most two linear inequalities in the two variables a and d to check whether each of the m points are placed in each of the 2m 2 intervals, we can apply Lemma 2.2 with N = 2, D = 1 and M = 2 • m • 2m 2 = 4m 3 to conclude that the points y 1 , . . ., y m overlap the intervals [j, j + 1) with 0 ≤ j ≤ 2m 2 − 1 in at most (100m 3 ) 2 = 10 4 m 6 different ways.But now, since at least q/6 of the y i must always be in distinct intervals, a union bound implies that the probability we have a blue solution to our system of equations is at most 10 4 m 6 (1 − q −3/4 ) q/6 < 1 2 for m sufficiently large.Combined with our earlier estimate for the probability of a red solution to y 1 + y 3 = 2y 2 + 2, we see that for m sufficiently large (m = 10 50 will suffice) there exists a colouring with no red ℓ 3 and no blue ℓ m , as required.

Concluding remarks
We say that a set X ⊂ E d is Ramsey if for every natural number r there exists n such that every r-colouring of E n contains a monochromatic copy of X.In [4], it was shown that a set X is