2·1. Setup

Fix $k \ge 11$ and $T = \binom{[k]}{2}$ . Let R be a T-coloured rainbow k-clique with colouring function $\chi_R$ and for concreteness, put $V(R)\;:\!=\;[k]$ and $\chi_R (ij) = \{i,j\}$ for all $i,j \in [k]$ .

Set

\begin{align*}a \;:\!=\; \frac{k!}{k^k-k}.\end{align*}

Our aim is to prove that $\operatorname{ind}(R) \le a$ . To this end, fix $\gamma{\,\gt\,}0$ and assume for contradiction $\operatorname{ind}(R)=a+\gamma$ . Next choose $0 {\,\lt\,} {\varepsilon} {\,\lt\,} \min\{\gamma,\operatorname{ind}(R)\}/100$ . Let $c_0$ be chosen so that $\operatorname{ind}(R, n) \le \operatorname{ind}(R)+\varepsilon$ for all $n{\,\gt\,}c_0$ . Choose $M \ge \left\lceil 2k!c_0/{\varepsilon} \right \rceil$ such that

(2·1) \begin{equation} \frac{n^k}{(n)_k} {\,\lt\,} 1 + {\varepsilon}\end{equation}

and

(2·2) \begin{equation} a \left( \frac{n^{k - 1}}{(k - 1)!} - \binom{n - 1}{k - 1} \right) {\,\lt\,} \gamma \binom{n - 1}{k - 1} - {\binom{n - 2}{k - 2}}\end{equation}

for all $n{\,\gt\,}M$ . This is possible since $\lim_{n \to \infty} n^k/(n)_k = 1$ and $n^{k - 1}/(k - 1)! - \binom{n - 1}{k - 1} = O(n^{k - 2})$ , while $\gamma \binom{n - 1}{k - 1} - {\binom{n - 2}{k - 2}} = \Omega \left( n^{k - 1} \right)$ as $n \rightarrow \infty$ . Suppose that $n{\,\gt\,}M$ is given and H is a T-coloured n-vertex graph with colouring function $\chi_H$ achieving I(R, n). This implies

\begin{align*}I(R, H) = I(R, n) = \operatorname{ind}(R,n){\binom nk},\end{align*}

where $a+\gamma =\operatorname{ind}(R) \le \operatorname{ind}(R,n) \le \operatorname{ind}(R) + \varepsilon = a+\gamma + \varepsilon$ .

The AMGM inequality yields $p(q,t) \le (q/t)^t$ and it is easy to see that

(2·3) \begin{equation} p(q, t) p(q', t') \le p(q + q', t + t')\end{equation}

for all $q, q' \ge 0$ and $t, t'{\,\gt\,}0$ (see Appendix).

For a vertex x in V(H) and $i \in [k]$ , write $d_i(x)$ for the number of copies of R containing x where x plays the role of vertex i in R. More formally, $d_i(x)$ is the number of isomorphic embeddings $\phi\;:\; R \to H$ such that $\phi(i) = x$ . Let $d(x)=\sum_i d_i(x)$ be the number of copies of R containing x. We will refer to this as the degree of x in H. Similarly, let d(x, y) be the number of copies of R containing both x and y. For $i \in [k]$ , let $N_i(x)$ be the set of $y \in V(H)\setminus \{x\}$ for which there is a copy of R in H containing both x and y in which x plays the role of vertex i in R. Note that we do not have $N_j(x) \cap N_{j'}(x) =\emptyset$ for $j \ne j'$ , but all edges between $N_j(x) \cap N_{j'}(x)$ and x have the same colour $\{j,j'\}$ . However, $N_i(x)$ has a (unique) partition $\cup_{j \ne i} N_i^j(x)$ where $N_i^j(x)$ comprises those y such that x, y lie in a copy of R with x playing the role of i and y playing the role of j. Indeed, the partition is obtained based on the colour of a vertex to x. This gives

(2·4) \begin{equation} {}d(x) = \sum_{i=1}^k d_i(x) \le \sum_{i=1}^k \prod_{j \ne i} |N_i^j(x)| \le \sum_{i=1}^k p(|N_i(x)|, k-1).\end{equation}

We partition V(H) into $V_1 \cup \cdots \cup V_k$ , where

\begin{align*}V_i = \{x \in V(H)\;:\; |N_i(x)| \ge |N_j(x)| \hbox{ for all } j \ne i\}.\end{align*}

If there is a tie, we break it arbitrarily. Set $n_i=|V_i|$ for all $i \in [k]$ .

2·2. Minimum degree

Here we show that a standard technique in extremal graph theory can be used to prove that each vertex of H lies in at least the average number of copies of R (apart from a small error term).

Proof. We write $d=b\pm c$ for the inequalities $b-c\le d \le b+c$ . Denote the average degree of H by

\begin{align*}d(H) \;:\!=\; \frac{ k\cdot I(R,H)}{n} = \operatorname{ind}(R,n){\binom{n-1}{k-1}}.\end{align*}

We claim that for every $x \in V(H)$

(2·5) \begin{equation} d(x) = d(H) \pm {\binom{n-2}{k-2}}. \end{equation}

This follows from a standard application of Zykov symmetrisation. Indeed, if the degrees of two vertices x and y differ by more than ${\binom{n - 2}{k - 2}}$ , say $d(x) {\,\gt\,}d(y) + {\binom{n - 2}{k - 2}}$ , then we can delete y and duplicate x, meaning we add a new vertex x ^′ with $\chi_H (x'z) = \chi_H (xz)$ for all other vertices z, and $\chi_H (xx')$ can be arbitrary. This transformation increases the number of copies of R by at least

\begin{align*}d(x)-d(y) -d(x,y)\ge d(x)-d(y) - {\binom{n - 2}{k - 2}} {\,\gt\,}0, \end{align*}

contradicting the maximality $I(R,H) = I(R,n)$ . Hence all degrees lie in an interval of length at most ${\binom{n - 2}{k - 2}}$ and (2·5) follows, since this interval must contain d(H). In particular, the minimum degree is at least

\begin{align*}d(H) - {\binom{n - 2}{k - 2}} = \operatorname{ind}(R,n){\binom{n - 1}{k - 1}} - {\binom{n - 2}{k - 2}}.\end{align*}

It follows from (2·2) that

\begin{align*}\operatorname{ind}(R,n){\binom{n - 1}{k - 1}} - {\binom{n - 2}{k - 2}} \ge (a+\gamma){\binom{n - 1}{k - 1}} - {\binom{n - 2}{k - 2}}{\,\gt\,} a\frac{n^{k-1}}{(k-1)!}\end{align*}

for $n {\,\gt\,} M$ , completing the proof.

2·3. Maximum colour degree

Let

\begin{align*}\alpha \;:\!=\; \frac{\max_{x,i,j} d_{\{i,j\}}(x)}{n},\end{align*}

where the maximum is taken over all vertices $x\in V(H)$ and all colours $\{i,j\} \in T$ and $d_{\{i,j\}}(x)$ is the number of edges in H incident with x in colour $\{i,j\}$ . We upper bound this value.

Proof. Let x, i, j achieve this maximum, so that $d_{\{i,j\}}(x) = \alpha n$ . Then $|N_i^j(x)| \leq \alpha n$ and $N_i(x)$ has a partition $N_{i}^j(x) \bigcup \cup_{\ell\ne j} N_i^{\ell}(x)$ where every copy of R containing x with x playing the role of i has exactly one vertex in each $N_i^{\ell}(x)$ for all $\ell \in [k]\setminus\{i\}$ . Further,

\begin{align*}\left| \bigcup_{\ell\ne j} N_i^{\ell}(x) \right| \leq n-d_{\{i,j\}}(x)\end{align*}

since a vertex incident to an edge coloured $\{i, j \}$ cannot play the role of $\ell \neq i,j$ . Consequently,

\begin{align*}d_i(x) \le |N_{i}^j(x)| \cdot p \left( n- d_{\{i,j\}}(x), k-2 \right) \le \alpha\, n \cdot \left(\frac{(1-\alpha)n}{k-2}\right)^{k-2}.\end{align*}

The same upper bound holds for $d_j(x)$ . For $\ell \not\in \{i,j\}$ , we have $N_{\ell} (x) \leq n-d_{\{i,j\}}(x)$ since x is playing the role of $\ell$ , so we cannot include an edge incident to x of colour $\{i,j\}$ since the colour must include $\ell$ . Hence

\begin{align*}d_{\ell}(x) \le p \left( n-d_{\{i,j\}}(x), k-1 \right) \le \left(\frac{(1-\alpha)n}{k-1}\right)^{k-1} \le \left(\frac{(1-\alpha)n}{k-2}\right)^{k-1}.\end{align*}

Altogether this yields

\begin{align*} d(x) \le 2 \,\alpha\, n \left(\frac{(1-\alpha)n}{k-2}\right)^{k-2} + (k-2)\left(\frac{(1-\alpha)n}{k-2}\right)^{k-1} =(1 + \alpha)\left(\frac{1 - \alpha}{k-2}\right)^{k-2}n^{k-1}.\end{align*}

Suppose for contradiction that $\alpha{\,\gt\,}0.4$ . Since $k \ge 3$ , $(1+\alpha)(1-\alpha)^{k-2}$ is a decreasing function of $\alpha$ for $\alpha \in (0.4, 1]$ , and $d(x) \geq an^{k-1}/(k-1)!$ by Lemma 2·2. Therefore

(2·6) \begin{equation} \frac{1}{k^{k-1}-1}= \frac{a}{(k-1)!} \leq \frac{d(x)}{n^{k-1}} \le 1.4 \left(\frac{0.6}{k-2}\right)^{k-2}. \end{equation}

However, this fails to hold for $k \ge 11$ (see Appendix), and we conclude that $\alpha \leq 0.4$ as desired.

2·4. The second largest neighbourhood

For a vertex $x \in V(H)$ , let Z(x) be the second largest set in $\{N_1(x), \ldots, N_k(x)\}$ and define

\begin{align*}z\;:\!=\; z_{k,n} = \max_{x \in V(H)} \frac{|Z(x)|}{n}.\end{align*}

Proof. Let x be such that $z=|Z(x)|/n$ . Let $a_i=|N_i(x)|/n$ and assume by relabeling that $a_1 \ge a_2=z \ge a_3 \ge \cdots \ge a_k$ . Since $N_j(x) \cap N_{j'}(x) \cap N_{j''}(x) = \emptyset$ for any three distinct j, j ^′, j ^′ we have $\sum a_i \le 2$ . Let $a_3+\cdots +a_k= s\le 2-(a_1+z)$ . Write $s=qz+r$ where $q \in \mathbb Z^{\ge 0}$ and $0\le r{\,\lt\,}z$ . If $x \le y$ , then $x^{k-1}+y^{k-1} {\,\lt\,} (x-\rho)^{k-1}+(y+\rho)^{k-1}$ for $0{\,\lt\,}\rho{\,\lt\,}x$ by convexity of $x^{k-1}$ so successively increasing the largest $a_i$ to z and decreasing the smallest $a_j$ to 0 or r, we obtain

\begin{align*}\sum_{i=3}^k a_i^{k-1} \le q z^{k-1} +r^{k-1}\le qz^{k-1} + \frac{r}{z} z^{k-1} = \frac{s}{z} z^{k-1} \le \frac{2-(a_1+z)}{z} z^{k-1}.\end{align*}

Consequently,

\begin{align*}\sum_{i=1}^k a_i^{k-1} = a_1^{k-1} + z^{k-1} + \sum_{i=3}^k a_i^{k-1} \le a_1^{k-1} + z^{k-1} + \frac{2-(a_1+z)}{z} z^{k-1}.\end{align*}

Since $a_1 \geq z$ , taking the derivative shows that for any z, this expression is increasing with $a_1$ . Using Lemma 2·3, we note that $a_1 + z \le 1+\alpha \le 1.4$ since

\begin{align*}|N_1(x)|+|Z(x)| =|N_1(x) \cup Z(x)|+|N_1(x) \cap Z(x)| \le n + d_{\left\{ 1,2 \right\}} (x),\end{align*}

where $d_{\left\{ 1,2 \right\}} (x) \leq \alpha n \leq 0.4 \cdot n$ . Thus $a_1 \leq 1.4-z$ and $a_1 \leq 1$ , so

\[ \sum_{i=1}^k a_i^{k-1} \leq a_1^{k-1} + z^{k-1} + \frac{2-(a_1+z)}{z} z^{k-1} \leq \min\{1.4-z,1\}^{k-1} + z^{k-1} + \frac{0.6}{z} z^{k-1}. \]

Using (2·4) and Lemma 2·2 yields

\begin{align*}\frac{1}{k^{k-1}-1} \leq \frac{d(x)}{n^{k-1}} \le \sum_{i=1}^k \left(\frac{a_i}{k-1}\right)^{k-1} \le \frac{\min\{1.4-z,1\}^{k-1} + z^{k-1} + \frac{0.6}{z} z^{k-1}}{(k-1)^{k-1}}.\end{align*}

Multiplying by $(k-1)^{k-1}$ and using the fact that

(2·7) \begin{equation} \frac{(k-1)^{k-1}}{k^{k-1}-1} \geq \frac{(k-1)^{k-1}}{k^{k-1}} = \left( 1-\frac1k \right)^{k-1}{\,\gt\,}\frac{1}{e} \end{equation}

for $k{\,\gt\,}1$ , we obtain

\begin{align*}\frac1e {\,\lt\,} (\min\{1.4-z,1\})^{k - 1} + z^{k-1} + 0.6 \cdot z^{k-2}.\end{align*}

As $z \leq a_1$ and $z + a_1 {\,\lt\,} 1.4$ , we have $z {\,\lt\,} 0.7$ . Thus, the RHS is nonincreasing with k and we may consider only the $k=11$ case. Numerical calculations show that for $z \in [0.5,0.7]$ , we have $(1.4-z)^{10} + z^{10} + 0.6z^9 {\,\lt\,} 1/e$ , so we conclude that $z {\,\lt\,} 0.5$ .

2·5. One large part

We now take care of the situation when one of the $V_i$ ’s is very large.

Proof. By contradiction, w.l.o.g. suppose that $|V_1|{\,\gt\,}(1-1/3k)n$ . If $x \in V_1$ , then $|N_1(x)|\ge |N_i(x)|$ for all $i{\,\gt\,}1$ so $|N_2(x)| \le |Z(x)| \le z n$ . Using (2·4) we have

\begin{align*}a{\binom{n}{k}} \le I(R, H) = \sum_{x \in V(H)} d_2(x) \le |V_1|p(zn, k-1)+ \frac{n}{3k} p(n,k-1)\end{align*}

and we further see that

\begin{align*}|V_1|p(zn, k-1)+ \frac{n}{3k} p(n,k-1){\,\lt\,}\left(z^{k-1}+\frac{1}{3k}\right) \frac{n^k}{(k-1)^{k-1}}.\end{align*}

Using our lower bound on M in (2·1), we get

(2·8) \begin{equation} \left( \frac{(k-1)^{k-1}}{k^k-k} \right) {\,\lt\,} (1+{\varepsilon}) \left( z^{k-1} + \frac{1}{3k} \right) . \end{equation}

This fails to hold for $k \ge 11$ (see Appendix). We conclude that $|V_i| \le (1-1/3k)n$ for all $i \in [k]$ .

2·6. Counting the copies of R in H

Here we describe the broad framework we will use to count copies of R in H. This is the same as in [ Reference Mubayi and Razborov12 ], though there are subtle differences which arise since we are in the undirected setting.

Call a copy f of R in H transversal if it includes exactly one vertex in $V_i$ for all $i \in [k]$ . We partition the copies of R in H as $H_m \cup H_g \cup H_b$ , where $H_m$ comprises those copies that lie entirely inside some $V_i$ , $H_g$ comprises those copies that intersect every $V_i$ whose edge colouring coincides with the natural one given by the vertex partition (meaning the map from R to H takes vertex i to a vertex in $V_i$ ), and $H_b$ comprises all other copies of R (including those transversal copies where some vertex is in an inappropriate $V_i$ ). Let $h_m=|H_m|, h_g=|H_g|$ and $h_b=|H_b|$ so that

\begin{align*}I(R, H) = h_m+h_g+h_b.\end{align*}

We will bound each of these three terms separately. First, note that

(2·9) \begin{equation} {}h_m = \sum_j I(R, H[V_j]) \le \sum_j I(R, n_j).\end{equation}

Next we turn to $h_g$ . Let $\Delta$ denote the number of k-sets that intersect each $V_i$ but are not counted by $h_g$ . So a k-set counted by $\Delta$ either does not form a copy of R, or forms a copy of R but its edge colouring does not coincide with the natural one given by the vertex partition $V_1 \cup \ldots \cup V_k$ . Then

(2·10) \begin{equation} {}h_g =\prod_i n_i -\Delta\end{equation}

and we need to bound $\Delta$ from below.

Note that the colour of some pair in every member of $\Delta$ does not align with the implicit one given by our partition. With this in mind, let $D_{ij}$ be the set of pairs of vertices $\{v_i, v_j\}$ where $v_i \in V_i, v_j \in V_j$ , $i\ne j$ such that $\chi_H (v_iv_j) \ne \chi_R (ij) = \{i,j\}$ . Let $\delta_{ij} =|D_{ij}|/{\binom{n}{2}}$ , $D = \cup_{ij} D_{ij}$ and $\delta= |D|/{\binom{n}{2}}$ . Let us lower bound $\Delta$ by counting the misaligned pairs from D and then choosing the remaining $k-2$ vertices, one from each of the remaining parts $V_{\ell}$ . This gives, for each $i{\,\lt\,}j$ ,

\begin{align*}\Delta \ge |D_{ij}| \prod_{\ell \ne i,j} n_{\ell}= \delta_{ij} {\binom{n}{2}} \prod_{\ell \ne i,j} n_{\ell}= \delta_{ij} {\binom{n}{2}} \frac{\prod_{\ell=1}^k n_{\ell}}{n_in_j}.\end{align*}

Since $\sum_{ij} \delta_{ij} {\binom{n}{2}} = \sum_{ij}|D_{ij}|=|D|=\delta{\binom{n}{2}},$ we obtain by summing over i, j,

\begin{align*}\Delta\left(\sum_{1\le i{\,\lt\,}j\le k} n_in_j\right) \ge \delta {\binom{n}{2}} \prod_{\ell=1}^k n_{\ell}.\end{align*}

This with along with (2·10) gives

(2·11) \begin{equation} {}h_g \le\prod_{\ell=1}^k n_{\ell}\left(1-\frac{\delta{\binom{n}{2}}}{\sum_{1\le i{\,\lt\,}j\le k}n_in_j}\right) = \prod_{\ell=1}^k n_{\ell}\left(1-\frac{\delta{\binom{n}{2}}}{{\binom{n}{2}} - \sum_i {\binom{n_i}{2}}}\right).\end{equation}

Our next task is to upper bound $h_b$ . For a vertex x and $j \in [k]$ , recall that $N_j(x) \subset V(H)$ is the set of y such that x, y lie in a copy of R with x playing the role of vertex j in R. Let us enumerate the set J of tuples (v, w, f) where $e=\{v,w\}\in D, f \in H_b$ , $e \subset f$ , and $v \in V_i$ , but $i \notin \chi_H (vw)$ . This means that v must play the role of i ^′ in f for some $i' \neq i$ , so the colours on all $k-1$ pairs (v, x) with $x \in f$ contain i ^′; in particular v is incident to $k-2$ pairs in f whose colour does not contain i. If $v\in V_i$ and $w \in V_j$ , then say that (v, w, f) is 1-sided if $|\chi_H (vw) \cap \{i,j\}|=1$ and (v, w, f) is 2-sided if $|\chi_H (vw) \cap \{i,j\}|=0$ .

Let $J_i$ be the set of i-sided tuples ( $i=1,2$ ). We consider the weighted sum

\begin{align*}S = 2|J_1| + |J_2|.\end{align*}

Observe that each $f \in H_b$ contains at least $k-2$ pairs from D. Indeed, if f is transversal, then it must contain a miscoloured vertex which yields at least $k-2$ pairs from D in f. If f is not transversal, choose $j \in [k]$ such that $|f \cap V_j|$ is largest. Set ${\mathcal{C}} \;:\!=\; f \cap V_j$ and observe that at least $|{\mathcal{C}}|-1$ of the vertices in ${\mathcal{C}}$ are miscoloured. Also, note that $2 \leq |{\mathcal{C}}| \leq k-1$ since f is not contained in one colour class and we have assumed f is not transversal.

If exactly $|{\mathcal{C}}|-1$ vertices in ${\mathcal{C}}$ are miscoloured, then every edge vw where $v\in {\mathcal{C}}$ is miscoloured and $w\in f \setminus {\mathcal{C}}$ is in D. Since $|f \setminus {\mathcal{C}}| = k - |{\mathcal{C}}|$ , this yields at least $(|{\mathcal{C}}|-1)(k-|{\mathcal{C}}|) \ge k-2$ pairs from D in f. On the other hand, if all $|{\mathcal{C}}|$ vertices in ${\mathcal{C}}$ are miscoloured, then there is a unique vertex $u \in f \setminus {\mathcal{C}}$ that plays the role of j in f. Every edge vw where $v \in {\mathcal{C}}$ and $w \in f \setminus \left( {\mathcal{C}} \cup u \right)$ is in D, so if $|{\mathcal{C}}| \leq k-2$ , this yields at least $|{\mathcal{C}}| (k - |{\mathcal{C}}| - 1) \geq k-2$ pairs from D in f. If $|{\mathcal{C}}| = k-1$ , then $f = {\mathcal{C}} \cup u$ where u plays vertex j in f but is in a different colour class, say the colour class corresponding to colour $\ell$ . There are $k-1$ edges between ${\mathcal{C}}$ and u, but only one can contain both k and $\ell$ , so at least $k-2$ edges from D are in f.

We conclude that each $f \in H_b$ contributes at least $2(k-2)$ to S since f contains at least $k-2$ pairs $e=\{v,w\}\in D$ and if (v, w, f) is 1-sided it contributes 2 to S while if it is 2-sided then it contributes 2 again since both (v, w, f) and (w, v, f) are counted with coefficient 1. This yields

(2·12) \begin{equation} S \ge 2(k-2)h_b.\end{equation}

On the other hand, we can bound S from above by first choosing $e\in D$ and then $f \in H_b$ as follows. Call $v \in e =\{v,w\} \in D$ correct in e if $v \in V_i$ , and $i \in \chi_H (vw)$ ; if v is not correct in e then $i \not\in \chi_H (vw)$ and say that v is wrong in e. The definition of D implies that every $e\in D$ has at least one wrong vertex in e (and possibly two wrong vertices). Let

\begin{align*}D_i = \{\{v, w\}\in D\;:\; \{v, w\}\ \hbox{contains exactly}\,\textit{i}\,\hbox{wrong vertices}\} \qquad \hbox{($i=1, 2$)}.\end{align*}

The crucial observation is that

(2·13) \begin{equation} ( v,w,f) \in J_i \qquad \Longrightarrow \qquad \{v,w\} \in D_i \qquad \qquad \hbox{($i=1, 2$)}.\end{equation}

To bound S from above, we use (2·13) and consider first $J_1$ and $D_1$ . We start by choosing vw in $D_1$ with wrong vertex v. Note that w is correct in vw since $vw \in D_1$ . Let $v \in V_i, w \in V_j$ . Then $\chi_H (vw) = \{j,\ell\}$ for some $\ell \ne i$ since v is wrong in e but w is correct in e. Thus for each triple $(v,w,f) \in J_1$ , vertex v plays the role of j in f or v plays the role of $\ell$ in f; thus the total number of $(v,w, f) \in J_1$ for some f is at most $p(|N_j(v)|-1,k-2) + p(|N_\ell(v)|-1,k-2)$ . Summing over all $vw \in D_1$ , we get

\[ |J_1| \leq \sum_{vw \in D_1} p(|N_j(v)|-1,k-2) + p(|N_\ell(v)|-1,k-2) \leq 2|D_1| p(zn,k-2). \]

The bound for $J_2$ is similar. Choose $vw \in D_2$ with $v \in V_i, w \in V_j$ . Let $\chi_H (vw) = \{\ell_1,\ell_2\}$ where $\{\ell_1, \ell_2\} \cap \{i,j\} = \emptyset$ . Since vw is two-sided, we see that $(v,w,f)\in J_2$ exactly when $(w,v,f) \in J_2$ . Consequently,

\begin{align*} |J_2| & \leq \sum_{vw \in D_2} p(|N_{\ell_1}(v)|-1,k-2) + p(|N_{\ell_2}(v)|-1,k-2) \\ & + p(|N_{\ell_1}(w)|-1,k-2) + p(|N_{\ell_2}(w)|-1,k-2) \\[.2 cm] & \leq 4 |D_2| p(zn,k-2).\end{align*}

This gives

(2·14) \begin{equation} {} S = 2 |J_1| + |J_2|\le 4 \, |D| \, p(zn, k-2)\le 4 \, \delta{\binom{n}{2}}\left(\frac{z}{k-2}\right)^{k-2} n^{k-2}.\end{equation}

Finally, (2·12) and (2·14) give

(2·15) \begin{equation} h_b \le \frac{S}{2(k-2)} \leq \frac{ 2 \delta{\binom{n}{2}}}{k-2}\left(\frac{z}{k-2}\right)^{k-2}n^{k-2}.\end{equation}

Using (2·9), (2·11) and (2·15) we have that

(2·16) \begin{equation} I(R, n) \leq \sum_{i} I(R, n_i) + \prod_{\ell} n_{\ell}\left(1-\frac{\delta{\binom{n}{2}}}{{\binom{n}{2}} - \sum_i {\binom{n_i}{2}}}\right) + \frac{ 2\delta{\binom{n}{2}}}{k-2}\left(\frac{z}{k-2}\right)^{k-2}n^{k-2}.\end{equation}

Our final task is to upper bound the RHS.

Since $\delta{\binom{n}{2}} \le \sum_{i \neq j}n_in_j = {\binom{n}{2}} - \sum_i {\binom{n_i}{2}}$ , we have $\delta \in I\stackrel{\rm def}{=} \left[0, 1 - \sum_i {\binom{n_i}{2}}/{\binom{n}{2}} \right]$ . Viewing (2·16) as a linear function of $\delta$ , it suffices to check the endpoints of I.

2·7. The extremal case

Proof. If $\delta=0$ , then (2·16) implies that

(2·17) \begin{equation} {}I(R, n) \le \sum_{i=1}^k I(R, n_i) + \prod_{i=1}^k n_i.\end{equation}

Let $p_i \;:\!=\; n_i/n$ . Using $\max_i p_i \le 1-1/3k$ by Lemma 2·5, convexity of $x^k$ , and $k \geq 11$ we obtain

(2·18) \begin{equation} \sum_{i=1}^k p_i^k \le \left(1-\frac{1}{3k}\right)^k + \left(\frac{1}{3k}\right)^k\le e^{-1/3} + 33^{-11} {\,\lt\,} 0.72. \end{equation}

We begin by bounding the summation in (2·17). By relabeling if necessary, let $n_1\le \cdots \le n_{\ell} \le c_0{\,\lt\,}n_{\ell+1}\le \cdots \le n_k$ where $\ell \ge 0$ . We have that

(2·19) \begin{equation} \sum_{i=1}^k I(R, n_i)\leq \ell{\binom{c_0}{k}}+\sum_{i=\ell+1}^k I(R, n_i)\le \ell{\binom{c_0}{k}}+(\operatorname{ind}(R)+\varepsilon)\sum_{i=\ell+1}^k {\binom{n_i}{k}} .\end{equation}

Observe that ${\binom{n_i}{k}}={\binom{p_in}{k}} {\,\lt\,} p_i^k{\binom{n}{k}}$ since $p_i{\,\lt\,}1$ . Dividing (2·19) by $\binom nk$ yields

(2·20) \begin{equation} \frac{1}{\binom nk} \sum_{i=1}^k I(R,n_i) \leq \ell\frac{\binom{c_0}{k}}{\binom{n}{k}} + (\operatorname{ind}(R) + {\varepsilon}) \sum_{i=\ell+1}^k p_i^k. \end{equation}

Suppose $\ell \geq 1$ . Using our bounds on ${\varepsilon}$ and M and (2·18), we can further bound

(2·21) \begin{equation} \frac{1}{\binom nk} \sum_{i=1}^k I(R,n_i) \leq \ell\frac{\binom{c_0}{k}}{\binom{n}{k}} + (\operatorname{ind}(R) + {\varepsilon}) \sum_{i=\ell+1}^k p_i^k {\,\lt\,} 0.74\operatorname{ind}(R) \end{equation}

and bound the product term

\[ \frac{1}{\binom nk} \prod_{i=1}^k n_i \leq \frac{1}{\binom nk} c_0 n^{k-1} {\,\lt\,} \frac{2k!c_0}{n} {\,\lt\,} {\varepsilon}. \]

This yields $\operatorname{ind}(R,n) \leq 0.74\operatorname{ind}(R) + {\varepsilon} {\,\lt\,} \operatorname{ind}(R)$ , a contradiction. Thus $\ell = 0$ , so using (2·20) we may rewrite (2·17) as

(2·22) \begin{equation} \operatorname{ind}(R,n) \leq (\operatorname{ind}(R) + {\varepsilon}) \sum_{i=1}^k p_i^k + \frac{1}{\binom nk} \prod_{i=1}^k n_i. \end{equation}

Isolating the product term and recalling the definition of a, as well as our lower bound on M,

\[ \frac{1}{\binom nk} \prod_{i=1}^k n_i = \frac{n^k}{\binom nk} \prod_{i=1}^k p_i \leq (a+{\varepsilon})(k^k-k) \prod_{i=1}^k p_i. \]

Plugging this into (2·22) and recalling $\operatorname{ind}(R) = a + \gamma$ ,

\[ \operatorname{ind}(R,n) \leq (a + {\varepsilon})\left( \sum_{i=1}^k p_i^k + (k^k-k) \prod_{i=1}^k p_i \right) + \gamma \sum_{i=1}^k p_i^k \leq (a + {\varepsilon} ) + 0.72\gamma. \]

The first bound $\sum p_i^k + (k^k-k) \prod p_i\le 1$ is well-known (see, e.g. (17) in [ Reference Mubayi and Razborov12 ]) and the second bound comes from (2·18). This gives the contradiction

\[ a + \gamma = \operatorname{ind}(R) \leq \operatorname{ind}(R,n) \leq a + 0.72\gamma + {\varepsilon} \]

since ${\varepsilon} {\,\lt\,} \gamma/100$ .

2·8. The absurd case

Now, we consider the other endpoint of I.

Proof. If $\delta=1 - \sum_i {\binom{n_i}{2}}/{\binom{n}{2}}$ , then (2·16) implies that

(2·23) \begin{equation} I(R, n) \le \sum_{i=1}^k I(R, n_i) +\frac{2 \sum_{i \neq j} n_in_j}{k-2}\left(\frac{z}{k-2}\right)^{k-2}n^{k-2}.\end{equation}

We first bound the second term. Dividing by $\binom nk$ and again letting $p_i \;:\!=\; n_i/n$ , we reorganise

\[ \frac{2}{\binom nk} \cdot \frac{\sum n_in_j}{k-2}\left(\frac{z}{k-2}\right)^{k-2}n^{k-2} =2 \cdot \frac{k^k-k}{(k-2)^{k-1}} \cdot \frac{n^k}{(n)_k} \cdot \left( \sum_{i \ne j} p_i p_j \right) z^{k-2}a. \]

Observe that $(k^{k-1} - 1)/(k - 2)^{k - 1}$ decreases to $e^2$ . In particular, for $k \geq 11$ , we have $(k^k-k)/(k-2)^{k-1} \leq 7.5k$ . For $n{\,\gt\,}M$ , we have $n^k/(n)_k {\,\lt\,} 1 + {\varepsilon}$ . Finally, $\sum_{i \neq j} p_ip_j = (1-\sum p_i^2) / 2 \leq (1-1/k) / 2$ as $\sum p_i^2$ is minimised when $p_i = 1/k$ for all i. Thus

\[ \frac{2}{\binom nk} \cdot \frac{\sum n_in_j}{k-2}\left(\frac{z}{k-2}\right)^{k-2}n^{k-2} \leq 7.5(1+{\varepsilon}) (k-1)z^{k-2}a {\,\lt\,} 0.25 a \]

for $k \geq 11$ as $(k-1)z^{k-2}$ is decreasing in k and $(11-1)z^{11-2} {\,\lt\,} 10 \cdot 2^{-9} {\,\lt\,} 1/50$ . Using this and (2·21) in (2·23), and $a{\,\lt\,}\operatorname{ind}(R)$ gives

\[ \operatorname{ind}(R,n) \leq 0.74 \, \operatorname{ind}(R) + 0.25a {\,\lt\,} 0.99\, \operatorname{ind}(R).\]

This contradiction completes the proof of the claim and the theorem.

3·1. Setup

Fix k and $R = ([k],E)$ a rainbow coloured graph with minimum degree at least $\eta (k-1)$ where $\eta {\,\gt\,} C \log k /(k-1)$ . We may assume that k is sufficiently large by making C sufficiently large so that the theorem is vacuous for small k. In particular, we will assume $k \geq 11$ so that we may use the same bounds as the previous section. It is notationally convenient to set $T = E \cup \{\emptyset\}$ and view R as a T-coloured complete graph $([k], \binom{[k]}{2})$ with colouring function $\chi_R$ defined as follows:

\[ \chi_R (ij) = \begin{cases} \{i,j\} & ij \in E \\ \emptyset & ij \not\in E. \end{cases} \]

Our goal is to prove that $\operatorname{ind}(R) \le a$ . To this end, fix $\gamma{\,\gt\,}0$ and assume for contradiction $\operatorname{ind}(R)=a+\gamma$ . Next choose ${\varepsilon}, c_0, M$ as in Section 2.1.

Suppose that $n{\,\gt\,}M$ is given and H is a T-coloured n-vertex graph with colouring function $\chi_H$ achieving I(R, n). This implies

\begin{align*}I(R, H) = I(R, n) = \operatorname{ind}(R,n){\binom nk},\end{align*}

where $a+\gamma =\operatorname{ind}(R) \le \operatorname{ind}(R,n) \le \operatorname{ind}(R) + \varepsilon = a+\gamma + \varepsilon$ .

Let $d_i (x)$ , d(x), d(x, y), $d_{\{i, j\}} (x)$ , $N_i (x)$ , and $N_i^j (x)$ be defined as in Section 2. Note that we do not have that all vertices in $N_j (x) \cap N_{j'}(x)$ for $j \neq j'$ have the same colour to x as it may be the case that $\chi_H (xy) = \emptyset$ and $\chi_H (xy') = \{j,j'\}$ for distinct $y, y' \in N_j (x) \cap N_{j'}(x)$ . We also do not have that $\cup_{j \ne i} N_i^j(x)$ is a partition of $N_i (x)$ as it may be the case that $y \in N_i^{j} (x) \cap N_i^{j'} (x)$ for some $j \ne j'$ satisfying $\chi_R (ij) = \chi_R (ij') = \emptyset$ and $y \in V(H)$ satisfying $\chi_H (xy) = \emptyset$ . Thus we must develop new techniques to prove a version of (2·4) from Section 2.1 to obtain bounds on $d_i(x)$ . This is the content of Section 3.2.

As in Section 2.1, we partition V(H) into $V_1 \cup \cdots \cup V_k$ , $n_i=|V_i|$ , where

\begin{align*}V_i = \{x \in V(H)\;:\; |N_i(x)| \ge |N_j(x)| \hbox{ for all } j \ne i\}.\end{align*}

If there is a tie, we break it arbitrarily.

3·2. Partitioning argument

Let the distance between two vertices v and w in a graph G, denoted $\operatorname{dist}_G (v,w)$ , be the number of edges in the shortest path between v and w in G. In our setting, a path cannot use an edge e with $\chi(e) = \emptyset$ . Then, define

\[\epsilon_G (v) \;:\!=\; \max_{w\in V(G)} \operatorname{dist}_G (v,w),\]

the eccentricity of v in G. Note that the diameter $\operatorname{diam}(G) = \max_{v \in V(G)} \epsilon_G(v)$ . For convenience, let $\epsilon (i) \;:\!=\; \epsilon_R (i)$ for all $i\in [k]$ .

Let B (x) be x in H. For $r \in \mathbb{N}$ , let $k_r(i)$ be the number of vertices in R at distance r from i. Recall that $d_{\{i,j\}}(x)$ is the number of edges in H incident with x in colour $\{i,j\}$

Lemma 3·1. Let $i,j \in [k]$ with $\{i, j\} \in T$ and $x \in V(H)$ . Then

\begin{align*} {(a)} \;\;\; & d_i(x) \le \left( \frac{|B (x)|}{k_1(i)} \right)^{k_1(i)} \left( \frac{n - |B (x)|}{k - k_1(i) - 1} \right)^{k - k_1(i) - 1} \\ {(b)} \;\;\; & d_i(x) \le \left( \frac{|N_i (x)|}{k-1} \right)^{k-1} \\ {(c)} \;\;\; & d_i(x) \le d_{\{i, j\}} (x) \cdot \left( \frac{n - d_{\{i, j\}} (x)}{k-2} \right)^{k - 2}. \end{align*}

Further, the number of copies of R in H containing vertices $x,y \in V(H)$ such that $x \in V(H)$ plays the role of vertex $i \in [k]$ in R is at most

\[ \left( \frac{|N_i (x)|}{k-2} \right)^{k-2}. \]

Proof. We start by proving the three upper bounds on $d_i (x)$ . To count the number of copies of R in H where x plays the role of i, we will recursively partition $N_i (x)$ . First, we pick $k_1 (i)$ vertices from $B(x) \cap N_i (x) \subset V(H)$ to play the role of the vertices adjacent to i in R. Notice that we may partition $B(x) \cap N_i (x)$ into $k_1 (i)$ parts based on the colour of each vertex to x as it uniquely determines its possible role in a copy of R. Set $B_1 \;:\!=\; B(x) \cap N_i (x)$ . We now recursively define $B_r$ for all $r \in [\epsilon (i)]$ . Let $2\le r \le \epsilon (i)$ , let $m \;:\!=\; k_1 (i) + \cdots + k_{r-1} (i)$ , and suppose that we have chosen $y_1, y_2, \ldots, y_m \in V(H)$ to play the roles of all vertices at distance $r-1$ or less from i in R, where $y_1,\ldots,y_{k_{r-1}(i)}$ play the roles of vertices at distance exactly $r-1$ from i. Then

\begin{align*} B_r \;:\!=\; & \, B_r (x, B_1, B_2, \ldots, B_{r-1}, y_1,\ldots, y_{k_{r-1} (i)}) \\[5pt] = & \, N_i (x) \cap \left( B (y_1) \cup \cdots \cup B (y_{k_{r-1} (i)}) \right) \setminus \left( x \cup B_1 \cup \cdots \cup B_{r-1} \right). \end{align*}

Here, $B_r$ is the set of vertices in H that can play the role of vertices at distance r from i in R, given that we have already selected all vertices at distance at most $r-1$ from i.

Note that by definition, $B_r \cap B_\ell = \emptyset$ for all $r, \ell \in [\epsilon (i)]$ and $\bigcup B_r \subseteq N_i (x)$ . For the remainder of the proof, we write $k_r \;:\!=\; k_r (i)$ for all $r \in [\epsilon (i)]$ for convenience.

Each vertex $v \in B_r$ has an edge to at least one of $y_1, \ldots, y_{k_{r-1}}$ . The colour of this edge uniquely determines the role that v may play in a copy of R, so this allows us to uniquely partition $B_{r}$ into $k_r$ parts. We note that it may be the case that v cannot legally play any role, but that only decreases the number of possible copies of R, so we may assume that this does not occur. Let $P_r \;:\!=\; P (B_r, k_r)$ be the set of tuples $\vec{y} \in B_r^{k_r}$ with one vertex from each part of $B_r$ , so

\begin{align*}|P_r| \le p(|B_r|, k_r).\end{align*}

This gives

\begin{align*} d_i (x) &\le \sum_{\vec{y}_1 \in P_1}\cdots \sum_{\vec{y}_{\epsilon (i) - 1} \in P_{\epsilon (i) - 1}} p(|B_{\epsilon (i)}|, k_{\epsilon (i)}) \\[5pt] &\le \sum_{\vec{y}_1 \in P_1}\cdots \sum_{\vec{y}_{\epsilon (i) - 1} \in P_{\epsilon (i) - 1}} p\left( |N_i (x)| - \sum_{r=1}^{\epsilon (i) - 1} |B_r|, k_{\epsilon (i)} \right). \end{align*}

Using (2·3) from Section 2.1 we see that

\begin{align*} \sum_{\vec{y}_{\epsilon (i) - 1} \in P_{\epsilon (i) - 1}} \; p&\left( |N_i (x)| - \sum_{r=1}^{\epsilon (i) - 1} |B_r|, k_{\epsilon (i)} \right) \\[5pt] &\le p \left( |B_{\epsilon (i) - 1}|, k_{\epsilon (i) - 1} \right) \cdot p\left( |N_i (x)| - \sum_{r=1}^{\epsilon (i) - 1} |B_r| , k_{\epsilon (i)} \right) \\[5pt] &\le p\left( |N_i (x)| - \sum_{r=1}^{\epsilon (i) - 2} |B_r| , k_{\epsilon (i) - 1} (i) + k_{\epsilon (i)} \right). \end{align*}

Using $\sum_{r = 1}^{\epsilon (i)} k_r = k-1$ , we obtain

\begin{align*}d_i(x) &\le \sum_{\vec{y}_1 \in P_1}\cdots \sum_{\vec{y}_{\epsilon (i) - 2} \in P_{\epsilon (i) - 2}} \; p\left( |N_i (x)| - \sum_{r=1}^{\epsilon (i) - 2} |B_r| , k_{\epsilon (i) - 1} + k_{\epsilon (i)} \right)\\&= \sum_{\vec{y}_1 \in P_1}\cdots \sum_{\vec{y}_{\epsilon (i) - 2} \in P_{\epsilon (i) - 2}} \; p\left( |N_i (x)| - \sum_{r=1}^{\epsilon (i) - 2} |B_r| , (k-1)-\sum_{r=1}^{\epsilon(i)-2} k_r \right).\end{align*}

Continuing this process, we obtain, for each $1 \le \ell \le \epsilon(i)-1$ ,

\begin{align*} d_i(x) \le \sum_{\vec{y}_1 \in P_1}\cdots \sum_{\vec{y}_{\ell} \in P_\ell} p\left( |N_i (x)| - \sum_{r=1}^{\ell} |B_r| , (k-1) - \sum_{r=1}^{\ell} k_r \right). \end{align*}

When $\ell=1$ this becomes

(3·1) \begin{align} d_i (x) &\le \sum_{\vec{y}_1 \in P_1} p\left( |N_i (x)| - |B_1|, k -1- k_1 \right)\nonumber \\ &\le p(|B_1 |, k_1) \cdot p\left( |N_i (x)| - |B_1 |, k - k_1 - 1 \right). \end{align}

As $|B(x)| \ge |B_1|$ and $n - |B(x)| \ge |N_i (x)| - |B_1 |$ for all $i \in [k]$ by definition,

\[ p(|B_1|, k_1) \cdot p\left( |N_i (x)| - |B_1 |, k - k_1 - 1 \right) \le \left( \frac{|B (x)|}{k_1} \right)^{k_1} \left( \frac{n - |B (x)|}{k - k_1 - 1} \right)^{k - k_1 - 1}, \]

so (a) holds. Alternatively, (2·3) also yields

\[ p(|B (x)|, k_1) \cdot p\left( |N_i (x)| - |B (x)|, k - k_1 - 1 \right) \leq p(|N_i(x)|, k-1) \leq \left( \frac{|N_i (x)|}{k-1} \right)^{k-1}, \]

so (b) holds.

For (c), let $j \in [k]$ such that $ij \in E$ . We bound $d_i (x)$ as before, but we choose the vertex y that plays role j separately. We see that

\[ |P_1| \le d_{\{i, j\}} (x) \cdot p (|B_1| - d_{\{i,j\}}(x), k_1 -1). \]

This combined with (3·1) and (2·3) gives

\begin{align*} d_i (x) &\le \sum_{\vec{y}_1 \in P_1} p\left( |N_i (x)| - |B_1|, k - k_1 - 1 \right) \\ &\le d_{\{i, j\}} (x) \cdot p (|B_1| - d_{\{i, j\}} (x), k_1 -1) \cdot p\left( |N_i (x)| - |B_1|, k - k_1 - 1 \right) \\ &\le d_{\{i, j\}} (x) \cdot p (n - d_{\{i, j\}} (x), k -2) \\ &\le d_{\{i, j\}} (x) \cdot \left( \frac{n - d_{\{i, j\}} (x)}{k-2} \right)^{k - 2}. \end{align*}

It remains to prove the last sentence of the lemma. We proceed as before except that, for $\ell \in [\epsilon (i)]$ such that $y \in B_\ell$ , we require that y is chosen. This means that instead of choosing $k_{\ell} (i)$ vertices from $B_{\ell}$ , we only need to choose $k_{\ell} (i)-1$ vertices from $B_{\ell}$ as we have already chosen y. Following the same procedure as before, we see that

\begin{align*} d_i (x)& \le \sum_{\vec{y}_1 \in P_1}\cdots \sum_{\vec{y}_{\ell} \in P_{\ell}} p \left( |B_{\ell}|, k_{\ell} - 1 \right) \cdot p\left( |N_i (x)| - \sum_{r=1}^{\ell + 1} |B_r| , (k-1) - \sum_{r=1}^{\ell + 1} k_r \right) \\ &\le \sum_{\vec{y}_1 \in P_1}\cdots \sum_{\vec{y}_{\ell} \in P_\ell} p\left( |N_i (x)| - \sum_{r=1}^{\ell} |B_r| , (k-2) - \sum_{r=1}^{\ell} k_r \right) \\ &\vdots \\ &\le \sum_{\vec{y}_1 \in P_1} p\left( |N_i (x)| - |B_1|, k - 2-k_1 \right) \\ &\le p(|B_1|, k_1) \cdot p(|N_i(x)| - |B_1|, k - 2 - k_1) \\ &\le p(|N_i (x)|, k - 2) \\ &\le \left( \frac{|N_i (x)|}{k-2} \right)^{k-2}. \end{align*}

This completes the proof.

3·3. Minimum degree

As in Section 2.2, we wish to show that each vertex of H lies in approximately the average number of copies of R.

This follows from an identical Zykov symmetrisation argument as used in the proof of Lemma 2·2. Note that we have assumed the same inequalities for M as we did in Section 2.1.

3·4. Maximum colour and non-edge degrees

The following two claims are used in the proof of Lemma 3·5 to bound the size of the second largest neighborhood.

Let

\begin{align*}\alpha \;:\!=\; \frac{\max_{x,i,j} d_{\{i,j\}}(x)}{n}\end{align*}

where the maximum is taken over all vertices $x \in V(H)$ and all colours $\{i,j\} \in T$ . We upper bound this value.

Proof. Let x achieve this maximum, so that $d_{\{i,j\}}(x) = \alpha n$ for some $\{i,j\} \in T$ . By Lemma 3·1(c) and $\alpha \leq 1$ , we get

\begin{align*} \max\{d_i(x), d_j(x)\} \leq \alpha n \left( \frac{(1-\alpha)n}{k-2} \right)^{k-2} \leq n^{k-1} \left( \frac{1-\alpha}{k-2} \right)^{k-2}. \end{align*}

For any other vertex $\ell \ne i,j$ , we have $|N_\ell(x)| \leq (1-\alpha)n$ , since $\ell$ is adjacent to no edges of colour $\{i,j\}$ in R. Thus by Lemma 3·1(b), we get

\[ d_\ell(x) \leq \left( \frac{|N_\ell (x)|}{k-1} \right)^{k-1} \leq \left( \frac{(1-\alpha)n}{k-1} \right)^{k-1} \leq n^{k-1} \left( \frac{1-\alpha}{k-2} \right)^{k-2}. \]

The last inequality comes as decreasing the denominator increases the fraction, and the base is less than 1, so decreasing the exponent increases the result. Summing over all indices in [k] and using Lemma 3·2, we get

\begin{equation*} \frac{1}{k^{k-1}-1} = \frac{a}{(k-1)!} \leq \frac{d(x)}{n^{k-1}} \leq k \left( \frac{1-\alpha}{k-2} \right)^{k-2}. \end{equation*}

Rearranging yields

\[ \frac{1}{k} \cdot \frac{(k-2)^{k-2}}{k^{k-1}-1} \le (1-\alpha)^{k-2}. \]

We see that

\[ \frac{(k-2)^{k-2}}{k^{k-1}-1} \geq \frac{(k-2)^{k-2}}{k^{k-1}} = \frac1k \left( 1-\frac2k \right)^{k-2} {\,\gt\,} \frac1{e^2k}, \]

so

\[ \frac{1}{e^2k^2} {\,\lt\,} (1-\alpha)^{k-2} \leq \exp(-(k-2)\alpha). \]

Assume for contradiction that $\alpha \geq \eta/4 {\,\gt\,} C\log k/(4(k-1))$ . Then

\[ \frac{1}{e^2k^2} {\,\lt\,} k^{-C(1-1/(k-1))/4} {\,\lt\,} k^{-0.9C/4}, \]

since $k \geq 11$ . For $C {\,\gt\,} 10$ , this gives a contradiction for sufficiently large k.

Let

\begin{align*}\beta \;:\!=\; \frac{\max_{x} d_\emptyset(x)}{n},\end{align*}

where the maximum is taken over all vertices $x \in V(H)$ and $d_\emptyset(x)$ is the number of edges in H incident with x in colour $\emptyset$ (non-edges). Note that we may assume that R has at least one non-edge, since otherwise the proof from Section 2 suffices. Thus we may also assume that H has at least one non-edge, so $\beta {\,\gt\,}0$ . We upper bound $\beta$ .

Proof. Assume for contradiction that $\beta \geq 1 - \eta/2$ . Let x achieve this maximum so that $d_\emptyset (x) = \beta n$ . This implies that $B(x) = (1 - \beta) n$ . For any $i \in V(R)$ , Lemma 3·1(a) gives

\begin{align*} d_i (x) &\le \left( \frac{B (x)}{k_1(i)} \right)^{k_1(i)} \left( \frac{n - B (x)}{k - k_1(i) - 1} \right)^{k - k_1(i) - 1} \\ &= \left( \frac{(1- \beta) n}{k_1(i)} \right)^{k_1(i)} \left( \frac{\beta n}{k - k_1(i) - 1} \right)^{k - k_1(i) - 1} \\ &= n^{k-1} \frac1{k_1 (i)^{k_1 (i)}} (1-\beta)^{k_1 (i)}\beta^{k- k_1 (i) -1} \left( \frac{1}{k-1-k_1 (i)} \right)^{k-1-k_1 (i)}. \end{align*}

For this section, we take the convention $0^0 = 1$ to handle the case that $k_1(i) = k-1$ .

Let $q = k_1(i)/(k-1) \in (0,1]$ . Then

(3·2) \begin{equation} d_i (x) \le \left( \frac{n}{k-1} \right)^{k-1} \left( \frac{(1-\beta)^q\beta^{1-q}}{q^q(1-q)^{1-q}} \right)^{k-1}. \end{equation}

We will first bound the term

(3·3) \begin{equation} \frac{(1-\beta)^q \beta^{1-q}}{q^q (1-q)^{1-q}}. \end{equation}

Regarding (3·3) as a function of $\beta$ , we see that the derivative

\[ \frac{\partial}{\partial \beta}\left( \frac{(1-\beta)^q \beta^{1-q}}{q^q (1-q)^{1-q}} \right) = \frac{(1-\beta)^{q-1}\beta^{-q}}{q^q(1-q)^{1-q}}((1-q)-\beta) \]

is negative for $\beta {\,\gt\,} 1-q$ since the fraction is nonnegative. Recall that $k_1(i) = \deg_R(i) \geq \eta (k-1)$ by assumption, so $q {\,\gt\,} \eta$ . We have also assumed for contradiction that $\beta \geq 1-\eta/2 {\,\gt\,} 1-\eta {\,\gt\,} 1-q$ . Thus decreasing $\beta$ to $1 - \eta/2$ will only increase (3·3), i.e.

(3·4) \begin{equation} \frac{(1-\beta)^q \beta^{1-q}}{q^q (1-q)^{1-q}} \leq \frac{(\eta/2)^q (1-\eta/2)^{1-q}}{q^q (1-q)^{1-q}}. \end{equation}

We now have a function purely of q. Taking the derivative, we get

\[ \frac{\partial}{\partial q} \left( \frac{(\eta/2)^q (1-\eta/2)^{1-q}}{q^q (1-q)^{1-q}} \right) = \frac{1}{2} (1 - q)^{q - 1} q^{-q} (2 - \eta)^{1-q} \eta^q \log \left( \frac{\eta (1-q)}{q (2 - \eta)} \right), \]

where all terms are positive except the logarithm, which is negative for $q {\,\gt\,} \eta/2$ . Thus (3·4) is decreasing with q for $q {\,\gt\,} \eta/2$ , so we may take the further upper bound

\begin{equation*} \frac{(1-\beta)^q \beta^{1-q}}{q^q (1-q)^{1-q}} \leq \frac{(\eta/2)^\eta (1-\eta/2)^{1-\eta}}{\eta^\eta (1-\eta)^{1-\eta}} = 2^{-\eta} \left( 1+\frac{\eta}{2(1-\eta)} \right)^{1-\eta}, \end{equation*}

where $2^{-\eta} \left( 1+\frac{\eta}{2(1-\eta)} \right)^{1-\eta} \leq \exp(-(\!\log 2 - 1/2)\eta)$ .

We now have an appropriate upper bound. Substituting into (3·2) and recalling that $\eta {\,\gt\,} C \log k / k$ , we see that

\[ d_i(x) \leq \left( \frac{n}{k-1} \right)^{k-1} \exp(-(\!\log 2 - 1/2)(k-1)\eta) {\,\lt\,} \left( \frac{n}{k-1} \right)^{k-1} k^{-C(\!\log 2-1/2)}. \]

Using Lemma 3·2, we get

\[ \frac{1}{k^{k-1}-1} = \frac{a}{(k-1)!} \leq \frac{d(x)}{n^{k-1}} \leq k \left( \frac{1}{k-1} \right)^{k-1} k^{-C(\!\log 2-1/2)}. \]

Rearranging terms and using the standard inequality (2·7) yields

\[ \frac{1}{ek} \leq k^{-C(\!\log 2-1/2)}. \]

For $C {\,\gt\,} 1/(\!\log 2-1/2) \approx 5.18$ , this yields a contradiction for sufficiently large k.

3·5. The second largest neighborhood

For a vertex $x \in V(H)$ , let Z(x) be the second largest set in $\{N_1(x), \ldots, N_k(x)\}$ and define

\begin{align*}z\;:\!=\; z_{k,n} = \max_{x \in V(H)} \frac{|Z(x)|}{n}.\end{align*}

Proof. Let $x \in V(H)$ such that $z=|Z(x)|/n$ . Suppose $x \in V_i$ and $Z(x) = N_j(x)$ for distinct $i,j \in [k]$ . Then we want to bound $|N_i(x) \cap Z(x)|$ . Suppose $y \in N_i(x) \cap Z(x)$ . If $xy \in E$ , then $i \in \chi_H(xy)$ and $j \in \chi_H(xy)$ , so $\chi_H(xy) = \{i,j\}$ . Thus $|N_1(x)\cap Z(x)| \leq d_{\{i,j\}}(x) + d_\emptyset(x)$ . It follows that

\begin{align*}|N_i(x)|+|Z(x)| =|N_i(x) \cup Z(x)|+|N_i(x) \cap Z(x)| \le n + d_{\{i,j\}} (x) + d_\emptyset(x),\end{align*}

where $d_{\{i,j\}} (x) \leq \alpha n$ and $d_\emptyset(x) \leq \beta n$ . Thus, $|N_i(x)| + |Z(x)| {\,\lt\,} (2-\eta/4)n$ by Claims 3·3 and 3·4 Since $|Z(x)| \leq |N_i(x)|$ , this gives $z {\,\lt\,} 1-\eta/8$ .

3·6. One large part

We now take care of the situation when one of the $V_i$ ’s is very large.

Proof. By contradiction, w.l.o.g. suppose that $|V_1|{\,\gt\,}(1-1/3k)n$ . If $x \in V_1$ , then $|N_1(x)|\ge |N_i(x)|$ for all $i{\,\gt\,}1$ so $|N_2(x)| \le |Z(x)| \le z n$ . Applying Lemma 3·1(b) to $d_2 (x)$ gives

\begin{align*}a{\binom{n}{k}} \le I(R, H) = \sum_{x \in V(H)} d_2(x) \le |V_1| \left( \frac{zn}{k-1} \right)^{k-1} + \frac{n}{3k} \left( \frac{n}{k-1} \right)^{k-1}\end{align*}

and we further see that

\begin{align*}|V_1| \left( \frac{zn}{k-1} \right)^{k-1} + \frac{n}{3k} \left( \frac{n}{k-1} \right)^{k-1} \leq \left(z^{k-1}+\frac{1}{3k}\right) \frac{n^k}{(k-1)^{k-1}}.\end{align*}

Rearranging and using $M^k/(M)_k {\,\lt\,} 1.01$ as assumed in (2·1), we get

\begin{equation*} \frac{(k-1)^{k-1}}{k^k-k} \leq 1.01 \left( z^{k-1} + \frac{1}{3k} \right). \end{equation*}

Using the standard inequality (2·7) and then Lemma 3·5 gives

\[ \left( \frac1{1.01e} - \frac13 \right)\frac1k {\,\lt\,} z^{k-1} {\,\lt\,} \left( 1-\frac\eta8 \right)^{k-1} \leq \exp(\!-C\log k/8 ) = k^{-C/8}. \]

For any $C{\,\gt\,}8$ this fails to hold for sufficiently large k.

3·7. Counting the copies of R in H

The way we count copies of R in H is very similar to the previous section and to [ Reference Mubayi and Razborov12 ]. While we do not have as much information in this case, without a focus on optimising for small k, we allow ourselves to be less strict with the counting arguments.

Call a copy f of R in H transversal if it includes exactly one vertex in $V_i$ for all $i \in [k]$ . We partition the copies of R in H as $H_m \cup H_g \cup H_b$ where $H_m$ comprises those copies that lie entirely inside some $V_i$ , $H_g$ comprises those copies that intersect every $V_i$ whose edge colouring coincides with the natural one given by the vertex partition (meaning the map from R to H takes vertex i to a vertex in $V_i$ ), and $H_b$ comprises all other copies of R (including those transversal copies where some vertex is in an inappropriate $V_i$ ). Thus a transversal copy f is in $H_b$ if and only if the unique map $\phi \;:\; [k] \to f$ with $\phi(i) \in V_i$ for all i is not a graph isomorphism from $R \to H[f]$ . Let $h_m=|H_m|, h_g=|H_g|$ and $h_b=|H_b|$ so that

\begin{align*}I(R, H) = h_m+h_g+h_b.\end{align*}

We will bound each of these three terms separately. As in Section 2.6, let D be the set of all pairs $\{v,w\}$ such that $v \in V_i, w \in V_j,$ and $\chi_H(vw) \ne \chi_R(ij)$ where $i \ne j$ . Let $\delta \;:\!=\; |D|/\binom n2$ . The identical reasoning as in Section 2.6 gives the first two bounds

(3·5) \begin{equation} {}h_m = \sum_{j=1}^k I(R, H[V_j]) \le \sum_{j=1}^k I(R, n_j)\end{equation}

and

(3·6) \begin{equation} {}h_g \le\prod_{\ell=1}^k n_{\ell}\left(1-\frac{\delta{\binom{n}{2}}}{\sum_{1\le i{\,\lt\,}j\le k}n_in_j}\right) = \prod_{\ell=1}^k n_{\ell}\left(1-\frac{\delta{\binom{n}{2}}}{{\binom{n}{2}} - \sum_i {\binom{n_i}{2}}}\right).\end{equation}

Our next task is to upper bound $h_b$ . This argument must be carried out differently. For a vertex $x \in V(H)$ and $j \in [k]$ , recall that $N_j(x) \subset V(H)$ is the set of y such that x, y lie in a copy of R with x playing the role of vertex j in R. Let us enumerate the set J of ordered pairs (e, f) where $e\in D, f \in H_b$ , and $e \subset f$ . To simplify the argument and notation, we count pairs instead of triples as in Section 2.

We must show that each $f \in H_b$ contains an edge in D. If f is transversal, then as we have noted, the natural map is not a graph isomorphism. Thus there is some incorrectly coloured edge which is in D. If f is not transversal, there is some $i \in [k]$ such that $|f \cap V_i|\ge 2$ . Note that $f \notin H_m$ , so $|f \cap V_i| {\,\lt\,} k$ . As R is connected, there exist $v \in V_i,\;u \in V_j$ for some $j \ne i$ such that vu is an edge in f. Since $|f \cap V_i| \geq 2$ , choose also $w \in f \cap V_i$ with $w \ne v$ . If $\chi_R (ij) = \emptyset$ then $vu \in D$ . If $\chi_R (ij) = \{ i, j\},$ then as $\chi_H (vu) = \chi_H (wu) = \{ i, j\}$ would contradict that f is a copy of R in H, we must have that uv or uw in D. This gives us that

\[ h_b \le |J| .\]

To bound $|J|$ from above, we start by choosing some bad edge $vw \in D$ . Let $f \subset V(H)$ such that $(vw,f) \in J$ . Either $v \in V_i$ does not play the role of i or $w \in V_j$ does not play the role of j in f. Then $f \subset N_\ell (v) \cup \{ v \}$ for some $\ell \neq i$ or $f \subset N_\ell(w) \cup \{w\}$ for some $\ell \neq j$ . We have $|N_\ell(v)|,|N_\ell(w)| \leq zn$ by the definition of z and the partition $V_1 \cup \cdots \cup V_k = V(H)$ . By the final statement of Lemma 3·1,

\[ |J|\le \sum_{vw \in D} \left( \sum_{\ell \ne i} \left( \frac{|N_\ell(v)|}{k-2} \right)^{k-2} + \sum_{\ell \ne j} \left( \frac{|N_\ell(w)|}{k-2} \right)^{k-2} \right) \leq 2|D| (k-1) \left( \frac{zn}{k-2} \right)^{k-2} . \]

Thus, recalling that $\delta \;:\!=\; |D|/\binom n2$ , we obtain

(3·7) \begin{equation} h_b \leq 2\delta (k-1) \binom{n}{2} \left( \frac{zn}{k-2} \right)^{k-2}.\end{equation}

Using (3·5), (3·6), and (3·7) we obtain

(3·8) \begin{equation} I(R, n) \leq \sum_{i} I(R, n_i) + \prod_{\ell} n_{\ell}\left(1-\frac{\delta{\binom{n}{2}}}{{\binom{n}{2}} - \sum_i {\binom{n_i}{2}}}\right) + 2\delta (k-1) \binom{n}{2} \left( \frac{zn}{k-2} \right)^{k-2}.\end{equation}

Our final task is to upper bound the RHS.

As in Section 3.7, we see that $\delta \in I\stackrel{\rm def}{=} \left[0, 1 - \sum_i {\binom{n_i}{2}}/{\binom{n}{2}} \right]$ . Viewing (3·8) as a linear function of $\delta$ , it again suffices to check the endpoints of I.

3·8. The extremal case

Proof. If $\delta=0$ , then (3·8) implies that

\begin{equation*} I(R, n) \le \sum_{i=1}^k I(R, n_i) + \prod_{i=1}^k n_i.\end{equation*}

This is the same equation as (2·17), and we have all the same assumptions. The same argument as in Section 2.7 derives a contradiction.

3·9. The absurd case

Now, we consider the other endpoint of I.

Proof. If $\delta=1 - \sum_i {\binom{n_i}{2}}/{\binom{n}{2}}$ , then (3·8) implies that

(3·9) \begin{equation} I(R, n) \le \sum_{i=1}^k I(R, n_i) + 2(k-1) \sum_{i \neq j} n_in_j \left(\frac{z}{k-2}\right)^{k-2}n^{k-2}.\end{equation}

This is similar to (2·23) with an extra factor of approximately $k^2$ in the second term. We can bound the first sum using the same techniques as in Section 2.7, giving (2·23):

(3·10) \begin{equation} \frac{1}{\binom nk} \sum_{i=1}^k I(R,n_i) \leq \ell\frac{\binom{c_0}{k}}{\binom{n}{k}} + (\!\operatorname{ind}(R) + {\varepsilon}) \sum_{i=\ell+1}^k p_i^k {\,\lt\,} 0.74 \operatorname{ind}(R). \end{equation}

We now bound the second term. Dividing by $\binom nk$ , we reorganize

\[ \frac{2}{\binom nk} (k-1) \left( \sum_{i \ne j} n_in_j \right)\left(\frac{z}{k-2}\right)^{k-2}n^{k-2} = \frac{2 (k-1)(k^k-k) n^k}{(k-2)^{k-2} (n)_k}\left( \sum_{i \ne j} p_i p_j \right) z^{k-2}a. \]

We first relax $(k-1)(k^k-k) {\,\lt\,} k^{k+1}$ . Observe that $k^{k-2}/(k-2)^{k-2} \leq e^2$ . Thus this first quotient is at most $e^2k^3$ . For $n{\,\gt\,}M$ , we have $n^k/(n)_k {\,\lt\,} 1 + {\varepsilon} \leq 1.01$ . Finally, $\sum_{i \neq j} p_ip_j = (1-\sum p_i^2) / 2 \leq (1-1/k) / 2 \leq 1/2$ as $\sum p_i^2$ is minimised when $p_i = 1/k$ for all i. Thus

\[ \frac{2}{\binom nk} (k-1) \left( \sum_{i \ne j} n_in_j \right)\left(\frac{z}{k-2}\right)^{k-2}n^{k-2} \leq 2 e^2 k^3 \cdot 1.01 \cdot \frac{1}{2} \cdot z^{k-2}a = 1.01e^2k^3 z^{k-2} a. \]

By Lemma 3·5, we know that $z^{k-2} \leq \left( 1-\eta/8 \right)^{k-2}$ , and we further see that

\[ \left( 1-\frac{\eta}{8} \right)^{k-2} \leq \exp\left(\! -\frac{(k-2)\eta}8 \right) \leq \exp\left(\! -\frac C8 \left( 1 - \frac{1}{k-1} \right) \log k \right) {\,\lt\,} k^{-0.9C/8}. \]

We again used $k \geq 11$ here. Thus for $C {\,\gt\,} 24/0.9 \approx 26.67$ and $k \geq 11$ , we have $1.01e^2k^3z^{k-2}a {\,\lt\,} 0.25a$ . Recalling that $a{\,\lt\,}\operatorname{ind}(R)$ , plugging this and (3·10) into (3·9) gives

\[ \operatorname{ind}(R,n) \leq 0.74 \operatorname{ind}(R) + 0.25 \, a {\,\lt\,} 0.99\operatorname{ind}(R).\]

This contradiction completes the proof of the claim and the theorem.