2· Proof

As discussed in the proof outline, we will use the Lovász theta function of a graph. Lovász [ Reference Lovász15 ] introduced this important graph parameter in 1979 in order to bound the Shannon capacity of a graph.

Definition 2·2 (Lovász theta function). For a graph G on vertex set [n], its Lovász number $\vartheta(G)$ is defined to be

\begin{equation*} \min_{c, U} \max_{i\in [n]} (c\cdot u_i)^{-2}, \end{equation*}

where the minimum is taken over all orthonormal representations U of G and all unit vectors $c\in V$ (where V is the Euclidean space in which the vectors $u_i$ live).

We will now verify that $f(G)=\vartheta(\overline{G})$ indeed satisfies the three key properties from the proof outline. Property (1) is well known and easy to verify.

Note that Lemma 2.3 follows from the “sandwich theorem” (see, e.g., [ Reference Knuth13 ]), stating that $\omega(G)\leq \vartheta(\overline{G})\leq \chi(G)$ for every graph G.

The next lemma verifies Property (2) in an equivalent form.

Lemma 2·4. For graphs $G_1, G_2$ on vertex set [n], let $G_1\cap G_2$ be the graph on vertex set [n] with edge set $E(G_1)\cap E(G_2)$ . The Lovász number is sub-multiplicative in the sense that

$$\vartheta(G_1\cap G_2)\leq \vartheta(G_1)\vartheta(G_2).$$

Proof. Let

\begin{align*} \vartheta(G_1) = \max_{i\in [n]} (c_1\cdot (U_1)_i)^{-2}, \\ \vartheta(G_2) = \max_{i\in [n]} (c_2\cdot (U_2)_i)^{-2}, \end{align*}

for orthonormal representations $U_1, U_2$ of $G_1, G_2$ on Euclidean spaces $V_1, V_2$ and unit vectors $c_1\in V_1, c_2\in V_2$ . Here $(U_1)_i, (U_2)_i$ denote the unit vector in $U_1, U_2$ corresponding to vertex i, respectively. Let $V = V_1\otimes V_2$ be the tensor product of $V_1$ and $V_2$ . Let U be a collection $u_1, \dots, u_n$ of unit vectors on V given by $u_i = (U_1)_i\otimes (U_2)_i$ . Notice that U is an orthonormal representation of $G_1\cap G_2$ . Indeed, if $i\neq j$ and ij is not an edge in $G_1\cap G_2$ , then we have either $ij\not \in G_1$ in which case $(U_1)_i\cdot (U_1)_j=0$ , or $ij\not \in G_2$ , in which case $(U_2)_i\cdot (U_2)_j=0$ . In both cases we have $u_i\cdot u_j=((U_1)_i\cdot (U_1)_j)((U_2)_i\cdot (U_2)_j)=0$ .

Let $c = c_1\otimes c_2$ . Then, for any $i\in [n]$ ,

\begin{equation*} c\cdot u_i = (c_1 \cdot (U_1)_i)(c_2 \cdot (U_2)_i). \end{equation*}

Hence,

\begin{align*} (c\cdot u_i)^{-2} &= (c_1 \cdot (U_1)_i)^{-2}(c_2 \cdot (U_2)_i)^{-2},\\ &\leq \vartheta(G_1)\vartheta(G_2), \end{align*}

for any $i\in [n]$ . Hence,

\begin{equation*} \vartheta(G_1\cap G_2)\leq \max_{i\in [n]} (c\cdot u_i)^{-2}\leq \vartheta(G_1)\vartheta(G_2), \end{equation*}

as desired.

Applying Lemma 2.4 with $\overline{G}_1$ and $\overline{G}_2$ in place of $G_1$ and $G_2$ , we obtain the following corollary.

Corollary 2·5. For graphs $G_1, G_2$ on vertex set [n], let $G_1\cup G_2$ be the graph on vertex set [n] with edge set $E(G_1)\cup E(G_2)$ . Then

$$\vartheta(\overline{G_1\cup G_2})\leq \vartheta(\overline{G_1})\vartheta(\overline{G_2}).$$

In [ Reference Lovász15 ], Lovász provided several equivalent descriptions of the theta function, which is one of the reasons why this function is so useful. We will need the following characterisation.

We will also need the following well-known properties of the Chebyshev polynomial of the first kind.

We are now ready to prove that the complement of the theta function has Property (3) from the proof outline. In other words, we prove that if G is an n-vertex graph that does not contain an odd cycle of length at most g, then $\vartheta(\overline{G})$ is small. Under these assumptions, Alon and Kahale [ Reference Alon and Kahale1 ] proved that $\vartheta(\overline{G})\leq 1+(n-1)^{1/g}$ . While their result is tight up to a multiplicative constant depending on g (and hence provides a good bound for constant g), it would be too weak in our setting, where g is large. Hence, we prove a different bound, using a novel argument.

Lemma 2·8. Let g be an odd positive integer and let G be a graph on vertex set [n] which does not contain any odd cycle of length at most g. Then

$$\vartheta(\overline{G})\leq 2 + ({1}/{2}) \left((2n - 2)^{1/g} - 1\right)^2.$$

Proof. We use the characterisation of $\vartheta(\overline{G})$ given by Lemma 2.6. For any orthonormal representation U of G, let $B = M(U) - I$ be the difference between its Gram matrix and the identity matrix. Note that B has vanishing diagonal, and for $i\neq j\in [n]$ , $B_{ij} = 0$ whenever ij is not an edge in G. Since G does not contain any odd cycle of length at most g, it follows that for each odd $\ell\leq g$ , G does not contain a closed walk of length $\ell$ , and therefore $\text{tr} (B^{\ell}) = 0$ . Let $\mu_n\leq \cdots\leq \mu_1$ be the eigenvalues of the real symmetric matrix B. The largest eigenvalue of M(U) is $\mu_1+1$ , so, by Lemma 2·6, it suffices to show that $\mu_1\leq 1 + ({1}/{2})\left((2n - 2)^{1/g} - 1\right)^2$ . We have

\begin{equation*} \sum_{i = 1}^n \mu_i^{\ell} = \text{tr}(B^{\ell})= 0, \end{equation*}

for any odd $\ell\leq g$ . Therefore for any odd polynomial p of degree at most g,

\begin{equation*} \sum_{i = 1}^n p(\mu_i) = 0. \end{equation*}

We take $p = T_g$ to be the gth Chebyshev polynomial of the first kind. As the Gram matrix $M(U) = B + I$ is positive semidefinite, we have $-1\leq \mu_n\leq \cdots\leq \mu_1$ . Noting that $T_g(x)\geq -1$ for all $x\geq -1$ , we have

\begin{equation*} T_g(\mu_1) = -\sum_{i = 2}^n T_g(\mu_i)\leq n - 1. \end{equation*}

We have

$$T_g(x)=\frac{1}{2}\left(\left(x-\sqrt{x^2-1}\right)^g+\left(x+\sqrt{x^2-1}\right)^g\right)\geq \frac{1}{2}\left(1+\sqrt{x^2-1}\right)^g,$$

for every $x\geq 1$ . Therefore, assuming $\mu_1\geq 1$ (otherwise we are already done),

\begin{equation*} \frac{1}{2}\left(1+\sqrt{\mu_1^2-1}\right)^g\leq T_g(\mu_1)\leq n - 1. \end{equation*}

Hence,

\begin{equation*} \mu_1\leq \sqrt{1 + \left((2n - 2)^{1/g} - 1\right)^2}\leq 1 + \frac{1}{2}\left((2n - 2)^{1/g} - 1\right)^2. \end{equation*}

As a result, $\lambda_1(M(U)) = \mu_1 + 1\leq 2 + ({1}/{2})\left((2n - 2)^{1/g} - 1\right)^2$ for any orthonormal representation U of G. Thus, $\vartheta(\overline{G})\leq 2 + ({1}/{2})\left((2n - 2)^{1/g} - 1\right)^2$ .

We are now ready to prove Theorem 1·5 in the following equivalent form.

Proof. Note that $K_n$ is the edge-union of $G_1, \dots, G_k$ . Therefore recursively applying Corollary 2·5, we have

\begin{equation*} \vartheta(\overline{K_n})\leq \vartheta(\overline{G_1})\cdots\vartheta(\overline{G_k}). \end{equation*}

By Lemma 2·8, $\vartheta(\overline{G_i})\leq 2 {+} \varepsilon_{n,g}$ for each $i = 1,\dots, k$ , where $\varepsilon_{n,g}= ({1}/{2})\left((2n - 2)^{1/g} {-} 1\right)^2$ . By Lemma 2·3, $\vartheta(\overline{K_n}) = n = (1+\delta) 2^k$ . Hence,

\begin{equation*} (1+\delta)2^k\leq (2+\varepsilon_{n,g})^k. \end{equation*}

Therefore,

\begin{equation*} 1+\delta\leq (1+\varepsilon_{n,g}/2)^k\leq \exp(k\varepsilon_{n,g}/2), \end{equation*}

so, using the estimate $\exp(\delta/2)\leq 1+\delta$ which is valid since $0\lt\delta\leq 1$ , we obtain $\delta\leq k\varepsilon_{n,g}$ . Plugging in the definition of $\varepsilon_{n,g}$ , we have

$$\delta/k\leq \frac{1}{2}\left((2n - 2)^{1/g} - 1\right)^2,$$

so

$$\sqrt{2\delta/k}\leq (2n - 2)^{1/g} - 1.$$

Noting that $2^{x/2} \leq 1 + x$ for $0\leq x\leq 2$ , we have $2^{\sqrt{\delta/2k}}\leq 1+\sqrt{2\delta/k}$ , and so

$$2^{g\sqrt{\delta/2k}}\leq 2n-2\leq 2^{2k}.$$

Hence,

$$g\sqrt{\delta/2k}\leq 2k,$$

which implies $g\leq 4k^{3/2}\delta^{-1/2}$ , as desired.

3· Concluding remarks

It is natural to wonder whether our results can be strengthened by obtaining a better bound than Lemma 2·8 for the theta function of complements of graphs of high odd girth. It turns out that no significant improvement can be obtained this way. In fact, one cannot even improve Lemma 2·8 significantly for the particular graph $C_g$ , which has odd girth g when g is odd. Indeed, it is known [ Reference Knuth13 ] that for every odd g, we have $\vartheta(\overline{C_g})=2+{\pi^2}/{2g^2}+O(g^{-4})$ , and even if we could replace the upper bound in Lemma 2·8 with $2+{\pi^2}/{2g^2}$ , this would only improve our bound in Theorem 1·4 by a polynomial factor.

The reader might also wonder whether we can choose a graph parameter f that is upper bounded by the complement of the theta function (there are natural graph parameters like this such as the Shannon capacity of the complement or the vector chromatic number) and has $f(C_g)-2\ll \vartheta(\overline{C_g})-2$ (so that the above issue does not arise), and yet it still satisfies the first two properties required in the proof overview. However, such a parameter does not exist. Indeed, for any such f we would have

(3·1) \begin{equation} g=f(K_g)\leq f(C_g)f(\overline{C_g})\leq \vartheta(\overline{C_g})\vartheta(C_g)=g, \end{equation}

where the last equality follows from the fact (see [ Reference Lovász15 ]) that $\vartheta(G)\vartheta(\overline{G})=n$ holds for every vertex-transitive n-vertex graph G. In particular, equality must hold everywhere in equation (3·1) and we have $f(C_g)=\vartheta(\overline{C_g})$ .

On the other hand, bounds on the Shannon capacity $\Theta$ for all n-vertex graphs whose complements have large odd girth directly translate to bounds on our Ramsey problem. A similar connection (between Shannon capacities of complements of odd cycles and the Ramsey problem we study) was mentioned in a paper of Zhu [ Reference Zhu17 ], building on closely related earlier observations of Erdős, McEliece and Taylor [ Reference Erdős, McEliece and Taylor10 ] and Alon and Orlitsky [ Reference Alon and Orlitsky2 ]. Indeed, consider a k-edge colouring of $K_n$ so that the colour classes define n-vertex graphs $G_1,\dots,G_k$ , none of which contain an odd cycle of length at most g. Then $\alpha(\overline{G_1}\boxtimes \overline{G_2}\boxtimes \dots \boxtimes \overline{G_k})\geq n$ , where $\boxtimes$ denotes the strong product of graphs and $\alpha$ denotes the independence number. Indeed, when each $G_i$ is identified with the graph formed by the ith colour in our k-colouring of $K_n$ , then the “diagonal” $\{(v,v,\dots,v): v\in V(K_n)\}$ is an independent set in $\overline{G_1}\boxtimes \overline{G_2}\boxtimes \dots \boxtimes \overline{G_k}$ . Now let G be the disjoint union of graphs $G_1, \dots, G_k$ . Clearly G is a graph on kn vertices which does not contain an odd cycle of length at most g. But $\overline{G_1}\boxtimes \overline{G_2}\boxtimes \dots \boxtimes \overline{G_k}$ is an induced subgraph of $\overline{G}^k$ (the k-th strong power of $\overline{G}$ ), so we have

$$\Theta(\overline{G})^k\geq \alpha(\overline{G}^k)\geq \alpha(\overline{G_1}\boxtimes \overline{G_2}\boxtimes \dots \boxtimes \overline{G_k})\geq n.$$

Hence, a sufficiently good upper bound for the Shannon capacity of kn-vertex graphs whose complements contain no short odd cycles would mean that g cannot be too large, which corresponds to an upper bound on the length of the shortest monochromatic odd cycle in any k-colouring of $K_n$ . Since the Shannon capacity of every graph is upper bounded by the theta function of the same graph, this approach could lead to an improved bound in our main results. However, the best known upper bound for the Shannon capacity of the complement of an odd cycle comes from the theta function, so we encounter the same barrier as discussed in the first paragraph of this section.