Structural properties of Toeplitz graphs

For each $i\in[t]$, we let $V_{i}=\{i+st\mid s=0,1,\ldots,\left\lfloor(n-i)/t\right\rfloor\}$. Then $|V_{i}|=\lfloor(n-i)/t\rfloor+1$ for each $i\in[t]$. By Lemma 2.2, $H_{i}:=G[V_{i}]$ is a component of $G$ for each $i\in[t]$. Fix $i\in[t]$ and let $f$ be a function from $V(G_{\lfloor(n-i)/t\rfloor+1}\langle 1,2,\ldots,k\rangle)$ to $V_{i}$ defined by $f(s+1)=i+st$ for each $s\in\{0,1,\ldots,\lfloor(n-i)/t\rfloor\}$. It is easy to see that $f$ is a bijection. Then $u$ and $v$ are adjacent in $G_{\lfloor(n-i)/t\rfloor+1}\langle 1,2,\ldots,k\rangle$ if and only if $|u-v|\in\{1,\ldots,k\}$, which is equivalent to $|(i+(u-1)t)-(i+(v-1)t)|\in\{t,\ldots,kt\}$, that is, $f(u)$ and $f(v)$ are adjacent in $H_{i}$. ∎

Let $S_{1}=\{i_{1},i_{2},\ldots,i_{\ell}\}$ be a maximum clique in $G$ and let $i_{1}=\min S_{1}$. Then, by Proposition 2.1, $|i_{u}-i_{v}|\in\{t_{1},\ldots,t_{k}\}$ for each $\{u,v\}\subset[l]$. If $i_{1}=1$, then we are done. If $i_{1}>1$, then the vertices $1,i_{2}-i_{1}+1,i_{3}-i_{1}+1,\ldots,i_{\ell}-i_{1}+1$ form a clique of size $\ell$ in $G$ by Proposition 2.1. ∎

Let $N(1)$ be the set of neighbors of $1$ in $G$. Then $N(1)=\{t_{1}+1,t_{2}+1,\ldots,t_{k}+1\}$. Now $G$ is $K_{q}$-free if and only if $1$ does not belong to a clique of size $q$ (by Lemma 2.4), equivalently, any subset of $N(1)$ with size $q-1$ contains a pair $t_{i}+1$, $t_{j}+1$ such that $|t_{i}-t_{j}|=|(t_{i}+1)-(t_{j}+1)|\notin\{t_{1},\ldots,t_{k}\}$.

If any subset $S\subseteq[k]$ with size $q-1$ contains a pair of distinct integers $a,b\in S$ such that $|t_{a}-t_{b}|\notin\{t_{1},\ldots,t_{k}\}$, then any subset of $N(1)$ with size $q-1$ contains a pair $t_{i}+1$, $t_{j}+1$ such that $|t_{i}-t_{j}|\notin\{t_{1},\ldots,t_{k}\}$. Suppose that any subset of $N(1)$ with size $q-1$ contains a pair $t_{i}+1$, $t_{j}+1$ such that $|t_{i}-t_{j}|\notin\{t_{1},\ldots,t_{k}\}$. Let $S$ be a subset of $[k]$ with size $q-1$. Then $\{t_{i}+1\mid i\in S\}$ is a subset of $N(1)$ with size $q-1$, so there is a pair $t_{i}+1$, $t_{j}+1$ such that $|t_{i}-t_{j}|\notin\{t_{1},\ldots,t_{k}\}$ by the last equivalence above. Thus there is a pair of distinct integers $a,b\in S$ such that $|t_{a}-t_{b}|\notin\{t_{1},\ldots,t_{k}\}$.∎

We can easily check the ‘if’ part. To show the ‘only if’ part, suppose that $|t_{i}-t_{j}|\in B(G)$ for every $\{t_{i},t_{j}\}\subset B(G)$ with $i\neq j$. Then $t_{2}-t_{1}\in B(G)$. Since $t_{2}-t_{1}<t_{2}$, $t_{2}-t_{1}=t_{1}$. Therefore $t_{2}=2t_{1}$. By the supposition again, $t_{3}-t_{1}\in B(G)$. Since $t_{3}-t_{1}<t_{3}$, $t_{3}-t_{1}=t_{2}$ or $t_{3}-t_{1}=t_{1}$. If $t_{3}-t_{1}=t_{1}$, then $t_{3}=t_{2}$, a contradiction. Thus $t_{3}-t_{1}=t_{2}$ and consequently $t_{3}=3t_{1}$. By repeating this procedure, we conclude that $t_{i}=it_{1}$ for each $i\in[k]$ and thus $B(G)=\{t_{1},2t_{1},\ldots,kt_{1}\}$. ∎

By Lemma 2.4, there is a maximum clique that contains the vertex $1$. Then the clique is contained in the closed neighborhood of $1$. Since $\deg(1)=k$, the inequality holds. Now we show the equality part. If $B(G)=\{t_{1},2t_{1},\ldots,kt_{1}\}$, then $\{1,1+t_{1},\ldots,1+kt_{1}\}$ is a clique of size $k+1$ in $G$ and so $\omega(G)=k+1$.

Suppose that $\omega(G)=k+1$. Then, by Lemma 2.4, there is a maximum clique of size $k+1$ containing $1$. Therefore $\{1,1+t_{1},\ldots,1+t_{k}\}$ is a clique. Now, for each $i,j\in[k]$, $1+t_{i}$ and $1+t_{j}$ are adjacent since they belong to the same clique. Then $|(1+t_{j})-(1+t_{i})|=|t_{j}-t_{i}|\in B(G)$ by Proposition 2.1, so $B(G)=\{t_{1},2t_{1},\ldots,kt_{1}\}$ by Lemma 2.7.∎

Suppose that $n-kt\leq t$. Then $n\leq(k+1)t$. By Proposition 2.3, $G$ has $t$ components and so $\theta_{E}(G)\geq t$. Again, by Proposition 2.3, each component is isomorphic to $G_{\lfloor(n-i)/t\rfloor+1}\langle 1,2,\ldots,k\rangle$ for each $i\in[t]$. Since $n\leq(k+1)t$, the number of vertices in a component is at most $k+1$ and so $G_{\lfloor(n-i)/t\rfloor+1}\langle 1,2,\ldots,k\rangle$ for each $i\in[t]$ is a complete graph by Proposition 2.1. Thus $\theta_{E}(G)=t$.

Now, suppose that $t<n-kt$. Then $n>(k+1)t$. Let $C_{i}=\{i,i+t,\ldots,i+kt\}$ for each $i\in\{1,\ldots,n-kt\}$. Then, for each $i\in\{1,\ldots,n-kt\}$, an element in $C_{i}$ is at least $i\geq 1$ and at most $i+kt\leq(n-kt)+kt=n$, so $C_{i}$ is a vertex set of $G$ . In addition, $C_{i}$ is a clique for each $i\in\{1,\ldots,n-kt\}$ by definition. Take an edge $uv\in E(G)$ such that $u<v$. If $kt<v$, then $C_{v-kt}=\{v-kt,v-(k-1)t,\ldots,v\}$ contains $u$ and $v$. Suppose that $v\leq kt$. Then there exists an integer $r$ such that $r\equiv v\pmod{t}$ with $1\leq r\leq t$. Obviously, $r\leq v$ and $v-r$ is a multiple of $t$. Since $v\leq kt$, $v-r\leq kt$ and so $v-r\in\{t,2t,\ldots,kt\}$. Thus $C_{r}=\{r,r+t,\ldots,r+kt\}$ contains $v$. Thus $\{C_{i}\mid 1\leq i\leq n-kt\}$ is an edge-clique cover of $G$ and so $\theta_{E}(G)\leq n-kt$.

Now, let $F=\{i\ i+kt\mid 1\leq i\leq n-kt\}$. Take edges $i\ i+kt$ and $j\ j+kt$ for some $1\leq i<j\leq n-kt$. Since $j+kt-i>kt$, $i$ and $j+kt$ are not adjacent by Proposition 2.1 and so $i\ i+kt$ and $j\ j+kt$ do not belong to the same clique. Thus $\theta_{E}(G)\geq|F|=n-kt$ and hence $\theta_{E}(G_{n})=n-kt$.

Let $H:=H_{n}\langle s_{1},s_{2},\ldots,s_{k}\rangle$ be a Toeplitz graph with $s_{1}=t$ and $s_{k}=kt$. Note that $G=H$ if $k\leq 2$, so $\theta_{E}(G)=\theta_{E}(H)$. We assume that $k\geq 3$. Suppose that $kt<n\leq(k+1)t$. Let $I=\{i\ i+s_{1}\mid 1\leq i\leq s_{1}\}$. Since $k\geq 3$, $2s_{1}=2t<kt<n$ and so, by Proposition 2.1, $I\subset E(H)$. Take edges $i\ i+s_{1}$ and $j\ j+s_{1}$ for some $1\leq i<j\leq s_{1}$. Since $j-i<s_{1}$, $i$ and $j$ are not adjacent by Proposition 2.1 and so $i\ i+s_{1}$ and $j\ j+s_{1}$ do not belong to the same clique. Thus we have shown that $\theta_{E}(H)\geq|I|=s_{1}=t=\theta_{E}(G)$ if $kt<n\leq(k+1)t$. Suppose that $n>(k+1)t$. Since $s_{k}=kt$, $F\subset E(H)$ by Proposition 2.1 and so $\theta_{E}(H)\geq|F|=n-kt$. Therefore $\theta_{E}(G)\leq\theta_{E}(H)$ if $(k+1)t<n$. Thus the “moreover” part is true. ∎

Since $s\equiv n\pmod{t}$ and $1\leq s\leq t$, it follows from Proposition 2.3 that $G$ has $t$ components, each of the first $s$ components $H_{1},\ldots,H_{s}$ is isomorphic to $G_{\lceil n/t\rceil}\langle 1,2,\ldots,k\rangle$, and each of the other $t-s$ components $H_{s+1},\ldots,H_{t}$ is isomorphic to $G_{\lfloor n/t\rfloor}\langle 1,2,\ldots,k\rangle$.

Since $H_{i}\simeq G_{\lceil n/t\rceil}\langle 1,2,\ldots,k\rangle$ for each $i\in[s]$ and $k+1$ consecutive vertices of $\{1,2,\ldots,\lceil\frac{n}{t}\rceil\}$ form a clique in $G_{\lceil n/t\rceil}\langle 1,2,\ldots,k\rangle$, we have $\theta_{v}(H_{i})\leq\left\lceil\frac{\lceil n/t\rceil}{k+1}\right\rceil$ for each $i\in[s]$. Now, by Theorem 2.8, $\omega(H_{i})=k+1$ for each $i\in[s]$, so we can conclude that $\theta_{v}(H_{i})=\left\lceil\frac{\lceil n/t\rceil}{k+1}\right\rceil$ for each $i\in[s]$. Similarly, we can show that $\theta_{v}(H_{i})=\left\lceil\frac{\lfloor n/t\rfloor}{k+1}\right\rceil$ for each $i\in\{s+1,\ldots,t\}$. Therefore

By Proposition 2.3, it suffices to show that $G:=G_{m}\langle 1,2,\ldots,k\rangle$ is chordal for each integer $m$, $m>k$. To reach a contradiction, suppose that $G$ contains a hole $C:=v_{1}v_{2}\ldots v_{\ell}v_{1}$ for some integer $\ell\geq 4$. We identify $v_{\ell+1}$ with $v_{1}$. Since the sequence $C$ cannot strictly increase or decrease, either $v_{i-1}<v_{i}$ and $v_{i}>v_{i+1}$ or $v_{i-1}>v_{i}$ and $v_{i}<v_{i+1}$ for some $i$, $2\leq i\leq\ell$. Assume the former. Then $v_{i}-v_{i-1}=|v_{i}-v_{i-1}|=a$ and $v_{i}-v_{i+1}=|v_{i}-v_{i+1}|=b$ for some $a,b\in\{1,2,\ldots,k\}$ by Proposition 2.1. Since $\ell\geq 4$, $a\neq b$ and so $|v_{i-1}-v_{i+1}|=|a-b|\in\{1,2,\ldots,k-1\}$. Therefore $v_{i-1}v_{i+1}$ is a chord of $C$ and we reach a contradiction. We can similarly show that $C$ also has a chord in the latter case. Thus $G$ is chordal. ∎

For each $i\in\{1,\ldots,t_{1}\}$, let $P_{i}$ be the path such that

Then $P_{1},\ldots,P_{t_{1}}$ are $t_{1}$ disjoint paths which contain all the vertices of $G$.

Now we construct a cycle in the following way. We start from the vertex $1$ and consider the edge $1\ 1+t_{2}$. Then $1+t_{2}$ is on the path $P_{j}$ for $j\equiv 1+t_{2}\pmod{t_{1}}$. We denote by $P_{1}^{\prime}$ the $(1+t_{2},j)$-section of ${P_{j}}^{-1}$. Since $j\leq t_{1}$, $j+t_{2}\leq t_{1}+t_{2}\leq n$ and so $j+t_{2}$ is a vertex of $G$. Then we take the edge $j\ j+t_{2}$ and the path $P_{k}$ where $k\equiv j+t_{2}\equiv 1+2t_{2}\pmod{t_{1}}$. We denote by $P_{2}^{\prime}$ the $(j+t_{2},k)$-section of ${P_{k}}^{-1}$. Then $1$ $P_{1}^{\prime}$ $P_{2}^{\prime}$ is a $(1,k)$-path. Noting that $1\equiv 1+xt_{2}\pmod{t_{1}}$ has a solution $s:=t_{1}/d$ where $d=\gcd(t_{1},t_{2})$, we may conclude that $1$ $P_{1}^{\prime}$ $P_{2}^{\prime}$ $\cdots$ $P_{s}^{\prime}$ is a cycle in $G$. By construction, the length of this cycle is the minimum of sum of two positive integers $x$ and $y$ satisfying $yt_{2}=xt_{1}$, that is, $\min\{x+y\mid yt_{2}-xt_{1}=0,x,y\in\mathbb{Z}^{+}\}=(t_{1}+t_{2})/d$.

To show that the cycle has no chord, take two vertices $u$ and $v$ with $u<v$ on the cycle. Then $u=1+u_{1}t_{2}-v_{1}t_{1}$ and $v=1+u_{2}t_{2}-v_{2}t_{1}$ for some $\{u_{1},u_{2}\}\subset\{1,\ldots,t_{1}/d\}$ and $\{v_{1},v_{2}\}\subset\{1,\ldots,t_{2}/d\}$. If $u$ and $v$ are adjacent, $v-u$ is either $t_{1}$ or $t_{2}$. Suppose $v-u=t_{1}$. Then $(u_{2}-u_{1})t_{2}-(v_{2}-v_{1}+1)t_{1}=0$. However, $|u_{2}-u_{1}|<t_{1}/d$ and so $u_{2}=u_{1}$ and $|v_{2}-v_{1}|=1$, which implies that $u$ and $v$ are consecutive on the cycle. Similarly one can show that if $v-u=t_{2}$, then $u$ and $v$ are also consecutive. Thus the cycle has no chord. Since $t_{2}\neq 2t_{1}$, $(t_{1}+t_{2})/d>3$ and so the cycle is a hole. ∎

$\mbox{(i)}\Rightarrow\mbox{(ii)}$ is obvious. By Theorem 2.8, $\mbox{(iii)}\Leftrightarrow\mbox{(iv)}$. To complete the proof, we shall show that $\mbox{(ii)}\Rightarrow\mbox{(iii)}$ and $\mbox{(iii)}\Rightarrow\mbox{(i)}$. To show $\mbox{(ii)}\Rightarrow\mbox{(iii)}$, we denote by $C_{4}(t_{i},t_{j})$ the $4$-cycle

for each $\{i,j\}\subset[k]$ with $i<j$ and $t_{i}+t_{j}<n$. Since $G$ is chordal, $C_{4}(t_{i},t_{j})$ has a chord for each $\{i,j\}\subset[k]$ with $i<j$ and $t_{i}+t_{j}<n$. Thus, for each $\{i,j\}\subset[k]$ with $i<j$ and $t_{j}+t_{i}<n$,

Suppose that $k=2$. If $t_{2}\neq 2t_{1}$, then $G$ has a hole of length $(t_{1}+t_{2})/\gcd(t_{1},t_{2})$ by Lemma 3.2 and we reach a contradiction. Thus $t_{2}=2t_{1}$. Now we suppose that $k=3$. Since $t_{1}+t_{2}<t_{1}+t_{3}<t_{2}+t_{3}\leq n$, $C_{4}(t_{1},t_{2})$ and $C_{4}(t_{1},t_{3})$ exist. Therefore we have $t_{2}-t_{1}=t_{1}$ or $t_{1}+t_{2}=t_{3}$ from $C_{4}(t_{1},t_{2})$ and $t_{3}-t_{1}\in\{t_{1},t_{2}\}$ from $C_{4}(t_{1},t_{3})$. If $t_{2}-t_{1}=t_{1}$, then $t_{3}-t_{1}=t_{2}$, so $t_{2}=2t_{1}$ and $t_{3}=3t_{1}$. Assume that $t_{1}+t_{2}=t_{3}$. To reach a contradiction, suppose that $t_{2}\neq 2t_{1}$. By definition, $H=G_{t_{1}+t_{2}}\langle t_{1},t_{2}\rangle$ is a subgraph of $G$ and $H$ contains a hole $C$ by Lemma 3.2. Since $G$ is chordal, $C$ has a chord in $G$. Then the difference of two ends of a chord is $t_{3}$, for otherwise the chord also exists in $H$. However, since $C$ is a subgraph of $H$, the difference of any pair of vertices on $C$ is at most $t_{1}+t_{2}-1=t_{3}-1$ and we reach a contradiction. Thus $t_{2}=2t_{1}$ and $t_{3}=t_{1}+t_{2}=3t_{1}$.