2.1 The construction

For some positive integer $p$ , let

\begin{align*} 1 \leq r_1 \leq r_2 \leq \cdots \leq r_p \end{align*}

be fixed positive integers such that $r_d | r_{d+1}$ for each $d=1,\ldots,p-1$ . These $r_i$ will be called the parameters of our edge-colouring.

For any $\alpha \geq r_p$ , let $n=2^{\alpha }$ , and associate each vertex of the complete graph $K_n$ with its own unique binary string of length $\alpha$ . For each $d=1,\ldots,p$ , let $\alpha = a_dr_d+b_d$ for positive integers $a_d,b_d$ such that $1 \leq b_d \leq r_d$ . For each string $x \in \{0,1\}^{\alpha }$ , we let

\begin{align*} x=\left (x_1^{(d)},x_2^{(d)},\ldots,x_{a_d+1}^{(d)}\right ) \end{align*}

where $x_i^{(d)}$ denotes a binary string in $\{0,1\}^{r_d}$ for each $i=1,\ldots,a_d$ and $x_{a_d+1}^{(d)}$ denotes a binary string from $\{0,1\}^{b_d}$ . We will call these substrings $r_d$ -blocks of $x$ , including the final one which may or may not actually have length equal to $r_d$ .

In the following definitions, we let $r_0=1$ and $r_{p+1}=\alpha$ . First, we define a function $\eta _d$ for any $d=0,\ldots,p$ on domain $\{0,1\}^{\beta } \times \{0,1\}^{\beta }$ where $\beta$ is any positive integer as

\begin{align*} \eta _d(x,y) = \left \{ \begin{array}{ll} \left (i,\{x_i^{(d)},y_i^{(d)}\}\right ) & \quad x \neq y\\ 0 & \quad x=y \end{array} \right . \end{align*}

where $i$ denotes the minimum index for which $x_i^{(d)} \neq y_i^{(d)}$ .

For $x,y \in \{0,1\}^{\alpha }$ and $0 \leq d \leq p$ , let

\begin{align*} \xi _d(x,y) = \left (\eta _d\left (x_{1}^{(d+1)},y_{1}^{(d+1)}\right ),\ldots,\eta _d\left (x_{a_{d+1}+1}^{(d+1)},y_{a_{d+1}+1}^{(d+1)}\right )\right ). \end{align*}

And let

\begin{align*} c_p(x,y) = \left (\xi _p(x,y),\ldots,\xi _0(x,y)\right ). \end{align*}

Next, we assume that the binary strings of $\{0,1\}^{\beta }$ are lexicographically ordered for every positive integer $\beta$ . For $1 \leq i \leq a_p+1$ and binary strings $x\lt y$ , define

\begin{align*} \delta _{p,i}(x,y) = \left \{ \begin{array}{ll} +1 & \quad \text{if }\; x_i^{(p)} \leq y_i^{(p)}\\ -1 & \quad \text{if }\; x_i^{(p)} \gt y_i^{(p)}. \end{array} \right . \end{align*}

Let

\begin{align*} \Delta _p(x,y) = \left (\delta _{p,1}(x,y),\ldots,\delta _{p,a_p+1}(x,y)\right ). \end{align*}

Finally, let

\begin{align*} \psi _p(x,y) = \left (c_p(x,y),\Delta _p(x,y)\right ). \end{align*}

2.2 Number of colours

For any positive integer $n$ , let $\beta$ be the positive integer for which

\begin{align*} 2^{(\beta -1)^{p+1}}\lt n \leq 2^{\beta ^{p+1}}. \end{align*}

For each $d=1,\ldots,p+1$ , let $r_d=\beta ^d$ in the construction of $\psi _p$ . Specifically, this means we are constructing the colouring on the complete graph with vertex set $\{0,1\}^{\alpha }$ where $\alpha =\beta ^{p+1}$ . We can apply this colouring to $K_n$ by arbitrarily associating each vertex of $K_n$ with a unique binary string from $\{0,1\}^{\alpha }$ and taking the induced colouring.

As shown in [Reference Conlon, Fox, Lee and Sudakov3], for these choices of parameters $r_d$ , the colouring $c_p$ uses at most $2^{4(p+1)\beta ^p\log _2 \beta }$ colours. On the other hand, $\Delta _p$ uses

\begin{align*} 2^{a_p+1} \leq 2^{\beta } \end{align*}

colours. So all together, $\psi _p$ uses at most $2^{4(p+1)\beta ^p\log _2 \beta +\beta }$ colours, where

\begin{align*} (\log _2n)^{1/(p+1)} \leq \beta \lt (\log _2n)^{1/(p+1)} +1. \end{align*}

Thus, for any fixed $p$ , $\psi _p$ uses a total of $n^{o(1)}$ colours.

2.3 Refinement of functions

Before we prove Lemma 2.1, it will be helpful to give the following definition and results about refinement of functions. The definition and Lemma 2.3 are paraphrased from [Reference Conlon, Fox, Lee and Sudakov3].

Lemma 2.4. Let $f,g$ be functions on domain $A$ . If $f$ refines $g$ and $S \subseteq A$ is a finite subset for which $|f(S)|=|g(S)|$ , then

\begin{align*} f(s_1)=f(s_2) \iff g(s_1)=g(s_2) \end{align*}

for all $s_1,s_2 \in S$ .

In particular, Lemma 2.4 implies that if some edge-colouring of a clique $S$ is refined by another edge-colouring, but $S$ contains the same number of colours under each, then the edge-colourings must be isomorphic.

2.4 Proof of Lemma 2.1

Let $S \subseteq \{0,1\}^{\alpha }$ be a set of $|S| \leq p+3$ vertices which contains exactly $|S|-1$ distinct edge colours under $c_p$ . We will prove that $S$ either has a leftover structure or contains a striped $K_4$ by induction on $\alpha$ , similar to the proof of Theorem 2.2 from [Reference Conlon, Fox, Lee and Sudakov3].

For the base case, consider $\alpha \leq r_p$ . Then for any $x,y \in S$ , the first component of $c_p(x,y)$ is

\begin{align*} \xi _p(x,y)=\left (\eta _p(x,y)\right )=\left ((1,\{x,y\})\right ). \end{align*}

Therefore, all of the edges of $S$ receive distinct colours. So it must be that $|S|-1=\binom{|S|}{2}$ , which happens only when $|S|=1,2$ . In either case, $S$ trivially has a leftover structure.

Now assume that $\alpha \gt r_p$ and that the statement is true for shorter binary strings. For each $d=1,\ldots,p$ , let $\alpha _d$ be the largest integer strictly less than $\alpha$ that is divisible by $r_d$ . For any $x \in S$ , let $x=(x'_{\!\!d},x''_{\!\!\!d})$ for $x'_{\!\!d} \in \{0,1\}^{\alpha _d}$ and $x''_{\!\!\!d} \in \{0,1\}^{\alpha -\alpha _d}$ .

Let $S_d$ denote the set of $\alpha _d$ -prefixes of $S$ ,

\begin{align*} S_d=\left \{x'_{\!\!d} \in \{0,1\}^{\alpha _d} | \exists x \in S, x=(x'_{\!\!d},x''_{\!\!\!d}) \right \}. \end{align*}

For each $x'_{\!\!d} \in S_d$ , let

\begin{align*} T_{x'_{\!\!d}} = \{x \in S | x=(x'_{\!\!d},x''_{\!\!\!d})\}. \end{align*}

Let $\Lambda _I^{(d)}$ be the set of colours contained in $S$ found on edges that go between vertices from two distinct $T$ -sets,

\begin{align*} \Lambda _I^{(d)}=\{c_p(x,y) | x,y \in S;\; x'_{\!\!d} \neq y'_{\!\!d}\}. \end{align*}

Similarly, let $\Lambda _E^{(d)}$ denote the set of colours contained in $S$ found on edges between vertices from the same $T$ -set,

\begin{align*} \Lambda _E^{(d)} = \{c_p(x,y)| x,y \in S;\; x \neq y;\; x'_{\!\!d} = y'_{\!\!d}\}. \end{align*}

Note that these sets of colours, $\Lambda _I^{(d)}$ and $\Lambda _E^{(d)}$ , partition all of the colours contained in $S$ . Therefore,

\begin{align*} |S|-1=|\Lambda _I^{(d)}|+|\Lambda _E^{(d)}|. \end{align*}

Next, define

\begin{align*} C_I^{(d)} &= \left \{(c_p(x'_{\!\!d},y'_{\!\!d}),\eta _{d-1}(x''_{\!\!\!d},y''_{\!\!\!d}))|x,y \in S;\; x'_{\!\!d} \neq y'_{\!\!d}\right \}\\ C_E^{(d)} &= \left \{\{x''_{\!\!\!d},y''_{\!\!\!d}\}|x,y \in S;\; x \neq y;\; x'_{\!\!d} = y'_{\!\!d}\right \}.\\ \end{align*}

It is shown in [Reference Conlon, Fox, Lee and Sudakov3] that $|\Lambda _I^{(d)}| \geq |C_I^{(d)}|$ and that $|\Lambda _E^{(d)}| \geq |C_E^{(d)}|$ . The second inequality is easier to see since any distinct $x,y \in S$ for which $x'_{\!\!d} = y'_{\!\!d}$ give $\xi _d=\left (0,\ldots,0,(i,\{x''_{\!\!\!d},y''_{\!\!\!d}\})\right )$ as the appropriate component of $c_p(x,y)$ . Although the first inequality seems intuitively true, its proof is a bit more subtle. The following fact (proved in [Reference Conlon, Fox, Lee and Sudakov3]) together with Lemma 2.3 gives us the desired inequality.

Fact 2.1 (Lemma 4.3 from [Reference Conlon, Fox, Lee and Sudakov3]). For $x,y \in \{0,1\}^{\alpha }$ , let

\begin{align*} \gamma _d(x,y)=(c_p(x'_{\!\!d},y'_{\!\!d}),\eta _{d-1}(x''_{\!\!\!d},y''_{\!\!\!d})). \end{align*}

Then $c_p$ refines $\gamma _d$ as functions on domain $\{0,1\}^{\alpha } \times \{0,1\}^{\alpha }$ .

We will also use the following fact which is proven in [Reference Conlon, Fox, Lee and Sudakov3], although not stated as a claim or lemma that can be easily cited. (See the final sentence of the second-to-final paragraph on page 11.)

Fact 2.2 (proved in [Reference Conlon, Fox, Lee and Sudakov3]). There exists an integer $1 \leq d \leq p$ for which

\begin{align*} |C_I^{(d)}|+|C_E^{(d)}| \geq |S|-1. \end{align*}

Therefore,

\begin{align*} |S|-1=|\Lambda _I^{(d)}|+|\Lambda _E^{(d)}| \geq |C_I^{(d)}|+|C_E^{(d)}| \geq |S|-1, \end{align*}

which implies that

\begin{align*} |S|-1=|\Lambda _I^{(d)}|+|\Lambda _E^{(d)}| = |C_I^{(d)}|+|C_E^{(d)}|. \end{align*}

Let

\begin{align*} \tilde{c}_p(x,y) = \left \{ \begin{array}{ll} (c_p(x'_{\!\!d},y'_{\!\!d}),\eta _{d-1}(x''_{\!\!\!d},y''_{\!\!\!d})) & \quad \text{if }\; x'_{\!\!d} \neq y'_{\!\!d}\\ \{x''_{\!\!\!d},y''_{\!\!\!d}\} & \quad \text{otherwise}. \end{array} \right . \end{align*}

Then by Fact 2.1, we know that $\tilde{c}_p$ refines $c_p$ . And since $|\Lambda _I^{(d)}|+|\Lambda _E^{(d)}| = |C_I^{(d)}|+|C_E^{(d)}|$ , then by Lemma 2.4, we know that the structure of $S$ under $\tilde{c}_p$ must be the same as the structure of $S$ under $c_p$ . Therefore, we need only show that $S$ either has a leftover structure or contains a striped $K_4$ under $\tilde{c}_p$ to complete the proof. We consider two cases: either there exists some $\omega \in C_E^{(d)}$ that appears more than once in $S$ under $\tilde{c}_p$ or each $\omega \in C_E^{(d)}$ appears exactly once in $S$ under $\tilde{c}_p$ .

Case 1: Let $\omega \in C_E^{(d)}$ appear on at least two edges in $S$ . This implies that $\omega =\{x''_{\!\!\!d},y''_{\!\!\!d}\}$ and so there must exist $a,b,c,e \in S$ such that $a=(x'_{\!\!d},x''_{\!\!\!d})$ , $b=(x'_{\!\!d},y''_{\!\!\!d})$ , $c=(y'_{\!\!d},x''_{\!\!\!d})$ , and $e=(y'_{\!\!d},y''_{\!\!\!d})$ for some $x'_{\!\!d} \neq y'_{\!\!d}$ . Therefore,

\begin{align*} \tilde{c}_p(a,b)=\tilde{c}_p(c,e)&=\{x''_{\!\!\!d},y''_{\!\!\!d}\}\\ \tilde{c}_p(a,c)=\tilde{c}_p(b,e)&=(c_p(x'_{\!\!d},y'_{\!\!d}),0)\\ \tilde{c}_p(a,e)=\tilde{c}_p(b,c)&=(c_p(x'_{\!\!d},y'_{\!\!d}),\eta _{d-1}(x''_{\!\!\!d},y''_{\!\!\!d})), \end{align*}

and all three colours are distinct. Hence, $S$ contains a striped $K_4$ under $\tilde{c}_p$ .

Case 2: If each $\omega \in C_E^{(d)}$ appears exactly once in $S$ under $\tilde{c}_p$ , then we know that

\begin{align*} |C_E^{(d)}| = \sum _{x'_{\!\!d} \in S_d}\binom{|T_{x'_{\!\!d}}|}{2} \end{align*}

since each edge within a given $T$ -set receives a unique colour. Moreover, if we let

\begin{align*} C_B^{(d)} = \{c_p(x'_{\!\!d},y'_{\!\!d}) | x'_{\!\!d},y'_{\!\!d} \in S_d\}, \end{align*}

then we know that

\begin{align*} |C_I^{(d)}| \geq |C_B^{(d)}| \geq |S_d|-1. \end{align*}

Therefore,

\begin{align*} |S_d|-1+\sum _{x'_{\!\!d} \in S_d}\binom{|T_{x'_{\!\!d}}|}{2} &\leq |S|-1\\[4pt] \sum _{x'_{\!\!d} \in S_d}\binom{|T_{x'_{\!\!d}}|}{2} &\leq |S|-|S_d|\\[4pt] \sum _{x'_{\!\!d} \in S_d}\binom{|T_{x'_{\!\!d}}|}{2} &\leq \sum _{x'_{\!\!d} \in S_d} (|T_{x'_{\!\!d}}|-1)\\[4pt] \sum _{x'_{\!\!d} \in S_d}(|T_{x'_{\!\!d}}|-1)(|T_{x'_{\!\!d}}|-2) &\leq 0. \end{align*}

Hence, we have $|T_{x'_{\!\!d}}|=1,2$ for each $x'_{\!\!d} \in S_d$ . This implies that $|C_E^{(d)}|=\sum _{x'_{\!\!d} \in S_d} (|T_{x'_{\!\!d}}|-1)$ and $|C_I^{(d)}|=|C_B^{(d)}|=|S_d|-1$ . So by induction, $S_d$ either has a leftover structure or contains a striped $K_4$ under $c_p$ . Furthermore, the colouring defined by

\begin{align*} c'_{\!\!p}(x,y) = \left \{ \begin{array}{ll} c_p(x'_{\!\!d},y'_{\!\!d}) & \quad \text{if }\; x'_{\!\!d} \neq y'_{\!\!d}\\ \{x''_{\!\!\!d},y''_{\!\!\!d}\} & \quad \text{otherwise} \end{array} \right . \end{align*}

is refined by $\tilde{c}_p$ , and $S$ contains exactly $|S|-1$ colours under both $c'_{\!\!p}$ and $\tilde{c}_p$ . So by Lemma 2.4, the edge-colouring of $S$ under $\tilde{c}_p$ is isomorphic to the one under $c'_{\!\!p}$ , and hence it is sufficient to show that $S$ has either a leftover structure or contains a striped $K_4$ under $c'_{\!\!p}$ .

If $S_d$ has a leftover structure under $c_p$ , then we see that $S$ also has a leftover structure under $c'_{\!\!p}$ since we can form $S$ under $c'_{\!\!p}$ from $S_d$ under $c_p$ by a sequence of splits as described in the definition of a leftover structure. That is, for each $x'_{\!\!d} \in S_d$ for which $|T_{x'_{\!\!d}}| = 2$ , we replace $x'_{\!\!d}$ with two vertices with a new edge colour between them, and the same edge colours that $x'_{\!\!d}$ already had to the rest of the vertices.

On the other hand, if $S_d$ contains a striped $K_4$ under $c_p$ , then $S$ must contain a striped $K_4$ under $c'_{\!\!p}$ with colours entirely from $C_B^{(d)}$ . This concludes the proof.

2.5 Proof of Lemma 2.2

Let $a,b,c,d \in \{0,1\}^{\alpha }$ be four distinct vertices, and assume towards a contradiction that they form a striped $K_4$ under $\psi _p$ . Specifically, assume that $\psi _p(a,b)=\psi _p(c,d)$ , $\psi _p(a,c)=\psi _p(b,d)$ , and $\psi _p(a,d)=\psi _p(b,c)$ .

Without loss of generality, we may assume the following: that $a$ is the minimum element of the four under the lexicographic ordering of $\{0,1\}^{\alpha }$ ; that for some $i \leq j,k$ ,

\begin{align*} \eta _p(a,b) = \eta _p(c,d) &= (i,\{x,y\})\\ \eta _p(a,c) = \eta _p(b,d) &= (j,\{z,w\})\\ \eta _p(a,d) = \eta _p(b,c) &= (k,\{s,t\}); \end{align*}

and that $a_i^{(p)}=c_i^{(p)}=x$ while $b_i^{(p)}=d_i^{(p)}=y$ . It follows from the ordering that $x \lt y$ and that $a\lt c\lt b,d$ . Furthermore, we have $i\lt j$ since $a$ and $c$ do not differ in the $i^{th}$ block. Similarly, we see that $(k,\{s,t\})=(i,\{x,y\})$ . Without loss of generality, we may assume $a_j^{(p)}=b_j^{(p)}=z$ and $c_j^{(p)}=d_j^{(p)}=w$ . Therefore, $z\lt w$ and $a\lt c\lt b\lt d$ .

Now, it follows that $\delta _j(a,d) = +1$ and that $\delta _j(c,b)=-1$ , contradicting our assumption that $\psi _p(a,d)=\psi _p(c,b)$ .

3.1 The construction

Let ${\mathbb{F}}_q^*$ denote the nonzero elements of the finite field with $q$ elements, and let $({\mathbb{F}}_q^*)^d$ denote the set of ordered $d$ -tuples of elements from ${\mathbb{F}}_q^*$ . In other words, $({\mathbb{F}}_q^*)^d$ is the set of $d$ -dimensional vectors over the field ${\mathbb{F}}_q$ without zero components. In what follows, we will assume that the set ${\mathbb{F}}_q^*$ is endowed with a linear order which can be arbitrarily chosen. We then order the set $({\mathbb{F}}_q^*)^d$ with lexicographic ordering based on the order applied to ${\mathbb{F}}_q^*$ .

Define a set of colours $C_d$ as the disjoint union

\begin{align*} C_d = \text{DOT} \sqcup \text{ZERO} \sqcup \text{UP} \sqcup \text{DOWN}, \end{align*}

where $\text{DOT}={\mathbb{F}}_q^*$ , and ZERO, UP, and DOWN are each copies of the set $\{1,\ldots,d\} \times{\mathbb{F}}_q$ . Let

\begin{align*} \varphi _d\;:\;\binom{({\mathbb{F}}_q^*)^d}{2} \rightarrow C_d \end{align*}

be a colouring function of pairs of distinct vectors, $x\lt y$ , defined by

\begin{align*} \varphi _d\left (x,y\right ) = \left \{ \begin{array}{ll} (i,x_i+y_i)_{\text{ZERO}} & \quad \text{if }\; x \cdot y=0\\ (i,x_i+y_i)_{\text{UP}} & \quad \text{if }\; x \cdot y \neq 0 \text{ and } x \cdot y = x \cdot x\\ (i,x_i+y_i)_{\text{DOWN}} & \quad \text{if }\; x \cdot y \not \in \{0,x \cdot x\} \text{ and } x \cdot y = y \cdot y\\ x \cdot y & \quad \text{otherwise}\\ \end{array} \right . \end{align*}

where $i$ is the first coordinate for which $x=(x_1,\ldots,x_d)$ differs from $y=(y_1,\ldots,y_d)$ and $x \cdot y$ denotes the standard inner product (dot product).

3.2 Number of colours

Let $n$ be a positive integer. Let $q$ be the smallest odd prime power for which $n \leq (q-1)^d$ . Then we can colour the edges of $K_n$ by arbitrarily associating each vertex with a unique vector from $({\mathbb{F}}_q^*)^d$ and taking the colouring induced by $\varphi _d$ . By Bertrand’s Postulate, $q \leq 2(n^{1/d}+1)$ . Therefore, the number of colours used by $\varphi _d$ on $K_n$ is at most

\begin{align*} (3d+1)q-1 \leq (6d+2)n^{1/d}+(6d+1). \end{align*}

3.3 Proof of Theorem 1.2

Lemma 3.1. Let $\{s_1,\ldots,s_t\} \subseteq ({\mathbb{F}}_q^*)^d$ be a set of linearly independent vectors and let $a,b \in ({\mathbb{F}}_q^*)^d$ such that

\begin{align*} \varphi _d\left (a,b\right )=\varphi _d\left (a,s_i\right )&=\alpha \\ \varphi _d\left (b,s_i\right )&=\beta \end{align*}

for some $\alpha,\beta \in C_d$ and for each $1\leq i \leq t$ . Then $s_1,\ldots, s_{t},b$ are linearly independent.

Proof. Assume towards a contradiction that $b=\sum _{j=1}^{t} \lambda _j s_j$ for some scalars $\lambda _1,\ldots,\lambda _{t} \in{\mathbb{F}}_q$ . We will first show that $\sum _{j=1}^{t}\lambda _j=1$ .

If $\alpha \in$ DOT, then $b=\sum _{j=1}^{t}\lambda _j s_j$ implies that

\begin{align*} \alpha =a \cdot b = \sum _{j=1}^{t}\lambda _j(a\cdot s_j)=\sum _{j=1}^{t}\lambda _j\alpha . \end{align*}

Therefore, $\sum _{j=1}^{t}\lambda _j=1$ since $\alpha \notin$ ZERO.

If $\alpha \notin$ DOT, then

\begin{align*} a_i+b_i=a_i+s_{1,i}=\cdots =a_i+s_{t,i} \end{align*}

where $i$ is the first index of difference between $a$ and $b$ . Thus, $s_{j,i}=b_i$ for all $1\leq j\leq t$ . But then $b=\sum _{j=1}^{t}\lambda _j s_j$ implies that

\begin{align*} b_i=\sum _{j=1}^{t}\lambda _j s_{j,i}=\sum _{j=1}^{t}\lambda _j b_i. \end{align*}

Hence, $\sum _{j=1}^{t}\lambda _j=1$ since $b_i \neq 0$ . Therefore, for any $\alpha \in C_d$ we have $\sum _{j=1}^{t} \lambda _j=1$ .

Now, if $\beta$ has the dot property, then let $\beta '$ denote $b \cdot s_j$ for all $j=1,\ldots,t$ . We have

\begin{align*} b \cdot b =\sum _{j=1}^{t}\lambda _j(b\cdot s_j)=\sum _{j=1}^{t}\lambda _j\beta '=\beta '. \end{align*}

But this implies that $\beta \in \text{UP} \cup \text{DOWN}$ , contradicting that $\beta$ has the dot property.

So we must assume that $\beta$ does not have the dot property. It follows that

\begin{align*} b_k+s_{1,k}=\cdots =b_k+s_{t,k} \end{align*}

where $k$ is the first index of difference between $b$ and $s_1$ . Therefore, $s_{1,k}=\cdots =s_{t,k}$ , and so

\begin{align*} b_k = \sum _{j=1}^t \lambda _j s_{j,k} = \sum _{j=1}^t \lambda _j s_{1,k} = s_{1,k}, \end{align*}

contradicting our choice of $k$ .

Since we reach a contradiction for all colours $\beta$ , it must be the case that $s_1,\ldots, s_{t},b$ are linearly independent vectors, as desired.

We now define a particular instance of leftover structure that will be useful in our arguments.

Definition 3.3. We call the set of vectors $S=\{s_1,\ldots,s_t\} \subseteq ({\mathbb{F}}_q^*)^d$ a $t$ -falling star under the colouring $\varphi _d$ if $\varphi _d\left (s_i,s_j\right )=\alpha _i$ for all $1\leq j \lt i\leq t$ , as shown in Figure 2. For any set of vectors $T \subseteq ({\mathbb{F}}_q^*)^d$ under $\varphi _d$ , let $FS(T)$ denote the maximum $t$ such that $T$ contains a $t$ -falling star.

Figure 2. A $t$ -falling star.

The following result about these falling stars is an easy consequence of Lemma 3.1 which can be shown by induction on the number of vectors.

Corollary 3.2. Let $S=\{s_1,\ldots,s_t\} \subseteq ({\mathbb{F}}_q^*)^d$ be a $t$ -falling star under $\varphi _d$ . Then the vectors $s_1, \ldots, s_{t-1}$ are linearly independent. Consequently, for any subset $T \subseteq ({\mathbb{F}}_q^*)^d$ ,

\begin{align*}{\text{rk}}(T) \geq FS(T)-1. \end{align*}

Moreover, if $T$ is contained within a monochromatic neighbourhood of some other vector, then

\begin{align*}{\text{rk}}(T) \geq FS(T). \end{align*}

Lemma 3.3. Let $A,B \subseteq{\mathbb{F}}_q^d$ be disjoint sets of vectors such that $A$ confines $B$ . Then

\begin{align*}{\text{af}}(B) \leq d-{\text{rk}}(A). \end{align*}

Proof. Let $t={\textrm{rk}}(A)$ , and let $a_1,\ldots, a_t$ be linearly independent vectors from $A$ . Since $A$ confines $B$ , then for each $a_i$ , there exists an $\alpha _i \in{\mathbb{F}}_q$ such that $a_i \cdot b = \alpha _i$ for all $b \in B$ . Therefore, $B$ is a subset of the solution space for the matrix equation,

\begin{align*} \left (\begin{array}{c} - a_1 - \\ \vdots \\ - a_t - \end{array}\right ) \left (\begin{array}{c} \mid \\ x \\ \mid \end{array}\right ) = \left (\begin{array}{c} \alpha _1\\ \vdots \\ \alpha _t\end{array}\right ). \end{align*}

Since $a_1,\ldots, a_t$ are linearly independent, the matrix of these $t$ vectors has full rank, and hence, the solution set is an affine space of dimension $d-t$ , as desired.

Lemma 3.5. Let $t \geq 2$ be an integer. An affine subspace of ${\mathbb{F}}_q^d$ of dimension $t-2$ will contain no $t$ -falling stars of $({\mathbb{F}}_q^*)^d$ under $\varphi _d$ . Therefore,

\begin{align*}{\text{af}}(S) \geq FS(S)-1 \end{align*}

for any subset of vectors $S \subseteq ({\mathbb{F}}_q^*)^d$ .

Proof. We will proceed by induction on $t$ . The base case $t=2$ is trivial since an affine subspace of dimension 0 is just one vector while a $2$ -falling star contains two distinct vectors.

So assume that $t \geq 3$ and that the statement is true for $t-1$ . Let $s_1,\ldots,s_t$ be $t$ distinct vectors that form a $t$ -falling star. That is, let $\alpha _1,\ldots,\alpha _{t-1} \in C_d$ and let $\varphi _d\left (s_i,s_j\right ) = \alpha _i$ when $1 \leq i \lt j \leq t$ . Assume towards a contradiction that these vectors are contained inside an affine subspace of dimension $t-2$ . Then there exist scalars $\lambda _1,\ldots,\lambda _{t-1} \in{\mathbb{F}}_q$ such that $s_t=\sum _{j=1}^{t-1} \lambda _j s_j$ and $\sum _{j=1}^{t-1} \lambda _j=1$ .

First, note that if $\lambda _1=0$ , then the vectors $s_2,\ldots,s_t$ form a $(t-1)$ -falling star and are contained in an affine subspace of dimension $t-3$ , a contradiction of the inductive hypothesis. So we must assume in what follows that $\lambda _1 \neq 0$ .

Now, we consider two cases: either $\alpha _1 \in \text{DOT}$ or $\alpha _1 \not \in \text{DOT}$ . If $\alpha _1 \in \text{DOT}$ , then

\begin{align*} \alpha _1=s_1 \cdot s_t = s_1 \cdot \sum _{j=1}^{t-1} \lambda _j s_j = \lambda _1(s_1 \cdot s_1)+\alpha _1 \sum _{j=2}^{t-1} \lambda _j = \lambda _1(s_1 \cdot s_1)+\alpha _1(1-\lambda _1). \end{align*}

Therefore, $\lambda _1(s_1 \cdot s_1 - \alpha _1)=0$ . Since $\lambda _1 \neq 0$ , it follows that

\begin{align*} s_1 \cdot s_1 = \alpha _1 = s_1 \cdot s_2, \end{align*}

which that implies $\alpha _1 \not \in \text{DOT}$ , a contradiction.

So assume that $\alpha _1 \not \in \text{DOT}$ , and let $i$ denote the index of the first component where $s_1$ differs from the other vectors. In this case,

\begin{align*} s_{1,i}+s_{2,i}=\cdots =s_{1,i}+s_{t,i}, \end{align*}

and hence $s_{2,i}=\cdots =s_{t,i}$ . Therefore,

\begin{align*} s_{t,i}=\sum _{j=1}^{t-1}\lambda _j s_{j,i} = \lambda _1 s_{1,i} + s_{t,i} \sum _{j=2}^{t-1} \lambda _j = \lambda _1 s_{1,i}+s_{t,i}(1-\lambda _1). \end{align*}

So $\lambda _1(s_{1,i}-s_{t,i})=0$ . Since $\lambda _1 \neq 0$ , we have $s_{1,i}=s_{t,i}$ , a contradiction of our choice of $i$ .

Lemma 3.6. Let $S \subseteq ({\mathbb{F}}_q^*)^d$ be a set of $p \geq 1$ vectors with a leftover structure under the colouring $\varphi _d$ . Then

\begin{align*} FS(S) \geq \left \lceil \log _2{p} \right \rceil + 1. \end{align*}

Proof. We will prove this by induction on $p$ . The base case when $p=1$ is trivial, so assume that $S$ has $p \geq 2$ vectors. Then $S$ has an initial bipartition, $S=A \cup B$ , and we note that

\begin{align*} FS(S) \geq 1+ \max \left (FS(A),FS(B)\right ). \end{align*}

Since $|A|,|B|\lt p$ , then by induction $FS(T) \geq \left \lceil \log _2(|T|) \right \rceil + 1$ for $T=A,B$ . Thus, we have

\begin{align*} FS(S) \geq \left \lceil \log _2\left (\max (|A|,|B|)\right ) \right \rceil + 2, \end{align*}

and since $\max (|A|,|B|) \geq \left \lceil \frac{p}{2} \right \rceil$ , then

\begin{align*} FS(S)\geq \left \lceil \log _2\left (\left \lceil \frac{p}{2} \right \rceil \right ) \right \rceil + 2 = \left \lceil \log _2{p} \right \rceil + 1. \end{align*}

Proof. We will prove this by induction on $p$ . For the base case, let $p=2$ . Let $x_1=1$ be the entire sequence. Then the first two conditions hold trivially since the sum of the sequence is 1, and since

\begin{align*} \left \lceil \log _2(1) \right \rceil + \left \lceil \log _2(1) \right \rceil =0 \leq d-1 \end{align*}

for any $d \geq 1$ . For the third condition, since $\left \lceil \log _2(1) \right \rceil = 0$ , it suffices to show that $T+1 \leq d$ . This follows from Corollary 3.2, since $S \cup \{a_1,\ldots,a_T\}$ forms a $(T+2)$ -falling star, and hence $d \geq{\textrm{rk}}\left (S \cup \{a_1,\ldots,a_T\}\right ) \geq T+1$ .

So assume that $S$ is a set of $p$ vertices for $p \geq 3$ and that the statement is true for smaller sets. Let the initial bipartition of $S$ be $S=A \cup B$ . By Lemma 3.4, we may assume without loss of generality that $A$ confines $B$ . Therefore, ${\textrm{af}}(B) \leq d-{\textrm{rk}}(A)$ by Lemma 3.3. By Corollary 3.2, we know that ${\textrm{rk}}(A) \geq FS(A)$ since $A$ is in a monochromatic neighbourhood of any vector from $B$ . And by Lemma 3.5, we know that ${\textrm{af}}(B) \geq FS(B)-1$ . Thus, $FS(A)+FS(B)-1 \leq d$ . So by Lemma 3.6, we can conclude that

\begin{align*} \left \lceil \log _2(|A|) \right \rceil + \left \lceil \log _2(|B|) \right \rceil \leq d-1. \end{align*}

Therefore, setting $x_1=\min \{|A|,|B|\}$ guarantees that $1 \leq x_1 \leq \left \lfloor \frac{p}{2} \right \rfloor$ and that

\begin{align*} \left \lceil \log _2(x_1) \right \rceil + \left \lceil \log _2(p-x_1) \right \rceil \leq d-1. \end{align*}

This gives us a positive integer $x_1$ which satisfies the first two conditions. Moreover, by Corollary 3.2 and Lemma 3.6,

\begin{align*} d \geq{\textrm{rk}}\left (S \cup \{a_1,\ldots,a_T\}\right ) &\geq FS(S\cup \{a_1,\ldots, a_T\})-1\\ & \geq (T+1+\max\!\left (FS(A),FS(B)\right ))-1\\ &\geq T + \left \lceil \log _2(p-x_1) \right \rceil +1. \end{align*}

Thus, $x_1$ also satisfies the third condition.

Let $S'$ denote the larger of the two parts $A$ and $B$ , and let $a_{T+1}$ denote an arbitrary vector from $S \setminus S'$ . Then $S'$ contains $p-x_1\lt p$ vectors and has a leftover structure under $\varphi _d$ . Moreover, $S'$ and $a_1,\ldots,a_{T},a_{T+1}$ satisfy the monochromatic neighbourhood conditions of the hypothesis. Hence, by induction there exists a sequence of positive integers $x_1',\ldots,x'_{\!\!t^{\prime}}$ such that $\sum _{i=1}^{t'}x'_{\!\!i}=p-x_1-1$ and for each $i=1,\ldots,t'$ , the following three conditions hold:

1. (1) $1 \leq x'_{\!\!i} \leq \left \lfloor \frac{p-x_1-s'_{\!\!i}}{2} \right \rfloor$ ;

2. (2) $\left \lceil \log _2(x'_{\!\!i}) \right \rceil + \left \lceil \log _2(p-x_1-s'_{\!\!i}-x'_{\!\!i}) \right \rceil \leq d-1$ ;

3. (3) $\left \lceil \log _2(p-x_1-s'_{\!\!i}-x'_{\!\!i}) \right \rceil \leq d-i-(T+1)$ ,

where $s'_{\!\!i}=0$ if $i=1$ and $s'_{\!\!i}=\sum _{j=1}^{i-1}x'_{\!\!j}$ otherwise.

Let $x_i=x'_{i-1}$ for $2\leq i \leq t'+1$ and let $t=t'+1$ to get a sequence $x_1,\ldots,x_t$ for which

\begin{align*} \sum _{i=1}^t x_i=x_1+\sum _{i=1}^{t'}x_i'=x_1+p-x_1-1=p-1. \end{align*}

For each $i=2,\ldots,t$ , the first two conditions are satisfied since $x_1+s'_{\!\!i}=s_{i+1}$ , and the third condition is satisfied since $d-i-(T+1)=d-(i+1)-T$ .

Proof. If such a set exists, then by Lemma 3.7 with $T=0$ , a positive integer $x_1$ exists such that $1 \leq x_1 \leq 3$ and

\begin{align*} \left \lceil \log _2(x_1) \right \rceil + \left \lceil \log _2(6-x_1) \right \rceil \leq 2. \end{align*}

It is simple to check that no such integer exists.

Proof. If such a set exists, then by Lemma 3.7 with $T=0$ , we must be able to find a sequence of positive integers $x_1,x_2,\ldots,x_t$ that satisfy the conditions given in the Lemma. In particular, $1 \leq x_1 \leq 4$ and

\begin{align*} \left \lceil \log _2(x_1) \right \rceil + \left \lceil \log _2(8-x_1) \right \rceil \leq 3. \end{align*}

We can check and find that $x_1=1$ is the only possibility. Therefore, $1 \leq x_2 \leq 3$ such that

\begin{align*} \left \lceil \log _2(7-x_2) \right \rceil &\leq 2\\ \left \lceil \log _2(x_2) \right \rceil + \left \lceil \log _2(7-x_2) \right \rceil &\leq 3.\\ \end{align*}

A quick check reveals that no such integer exists.

Theorem 1.2 follows from Theorem 1.1 and Corollaries 3.8 and 3.9.