Edge connectivity of simplicial polytopes

Let $n_{X}:=\#(X\cap\mathcal{V}(D))$ and let $\delta:=\delta(G)$. If $n_{X}\leq d-1$, then removing these $n_{X}$ vertices does not disconnect the graph. Thus $X=X\cap\mathcal{V}(D)$. For any $x\in X$, $x$ is incident with at least $\delta-n_{X}+1$ edges in $D$ (as $\deg(x)\geq\delta$). Since $\#D\leq\delta$, we have $n_{X}(\delta-n_{X}+1)\leq\delta$, and so $\delta\leq n_{X}\leq d-1$, a contradiction to the fact that $\delta\geq d$. Hence $n_{X}\geq d$. The same reasoning yields that $\#(\overline{X}\cap\mathcal{V}(D))\geq d$. ∎

Let $G$ be a plane triangulation, and let $D:=\mathcal{E}(X,\overline{X})$ be a minimum edge cut for some $X\subseteq\mathcal{V}(G)$. By Euler’s formula, the minimum degree of a plane graph is at most five. Thus $\#D\leq 5$. Suppose, by way of contradiction, that $D$ is nontrivial. It is a standard result of graph theory that a plane triangulation is 3-connected; this also follows from Balinski’s theorem (1). Let $n_{X}:=\#(X\cap\mathcal{V}(D))$ and $n_{\overline{X}}:=\#(\overline{X}\cap\mathcal{V}(D))$. Then, Lemma 4 ensures that $n_{X}\geq 3$ and $n_{\overline{X}}\geq 3$.

There is a vertex $u\in{X}\cap\mathcal{V}(D)$ with a unique neighbour $v$ in $\overline{X}\cap\mathcal{V}(D)$; if every vertex in ${X}\cap\mathcal{V}(D)$ was incident with at least two edges from $D$, then $\#D\geq 6$, a contradiction. Moreover, since $D$ is nontrivial, the vertex $u$ has at least one neighbour in $X$, say $u_{1}$. It follows that every $u_{1}-v$ path in $G$ passes through an edge of $D$. Thus, the edge cut $D$ must separate $u_{1}$ from $v$ in the link of $u$. Let $D_{u}:=D\cap\mathcal{E}(\operatorname{link}(u))$. The set $D_{u}$ contains precisely the edges incident with $v$ in $\mathcal{E}(\operatorname{link}(u))$; otherwise either $D_{u}$ consists of the edges incident with $u_{1}$ or there would be at least two nonadjacent edges in $D_{u}$, both cases implying the existence of at least two neighbours of $u$ in $\overline{X}$. Because $\operatorname{link}(u)$ is a cycle (see also Proposition 7), we conclude that $v$ is incident with at least three edges in $D$, including $uv$.

Again, since $\#D\leq 5$ and $v$ is incident with at least three edges in $D$, there is a vertex $v^{\prime}\in\overline{X}\cap\mathcal{V}(D)$ with a unique neighbour $u^{\prime}$ in ${X}\cap\mathcal{V}(D)$. Following the same line of reasoning as before, the vertex $u^{\prime}$ is incident with $u^{\prime}v^{\prime}$ and exactly two edges in $D\cap\operatorname{link}(v^{\prime})$, one of these two edges is $u^{\prime}v$. By the 3-connectivity of $G$, the vertices $u^{\prime},v$ cannot disconnect the graph. Thus, there is at least one vertex in $(X\cap\mathcal{V}(D))\setminus\{u,u^{\prime}\}$ adjacent to a vertex in $\overline{X}\setminus\{v^{\prime},v\}$. Since this means that $\#D\geq 6$, we arrive at a contradiction. The proof of the theorem is complete. ∎

Let $P$ be a simplicial $d$-polytope, let $G$ be its graph, let $\delta$ be the minimum degree of $G$, and let $D:=\mathcal{E}(X,\overline{X})$ be a minimum edge cut of $G$ for some $X\subseteq\mathcal{V}(G)$. By our hypothesis, $\#D\leq 4d-7$. The theorem holds for $d=3$ by Theorem 5, and so assume that $d\geq 4$. Let $n_{X}:=\#(X\cap\mathcal{V}(D))$ and $n_{\overline{X}}:=\#(\overline{X}\cap\mathcal{V}(D))$. Suppose, by way of contradiction, that $D$ is nontrivial. By Balinski’s theorem (1), $G$ is $d$-connected, in which case Lemma 4 ensures that $n_{X}\geq d$ and $n_{\overline{X}}\geq d$. We need two simple claims.

Consider a vertex $v\in\mathcal{N}_{\overline{X}}(w)$. As the graph of $\operatorname{link}(w)$ is isomorphic to the graph of a simplicial $(d-1)$-polytope (Proposition 7), the vertex $v$ has degree at least $d-1$ in $\operatorname{link}(w)$. Of the neighbours of $v$ in $\operatorname{link}(w)$, at most $\#\mathcal{N}_{\overline{X}}(w)-1$ of them are in $\operatorname{link}(w)\cap\overline{X}$, since $\mathcal{V}(\operatorname{link}(w))=\mathcal{N}_{X}(w)\cup\mathcal{N}_{\overline{X}}(w)$. It follows that, of the neighbours of $v$ in $\operatorname{link}(w)$, at least $d-1-(\#\mathcal{N}_{\overline{X}}(w)-1)$ are in $X$; in other words, $v$ is incident with at least $d-\mathcal{N}_{\overline{X}}(w)$ edges in $D\cap\mathcal{E}(\operatorname{link}(w))$. If we also count the edge $wv\notin\mathcal{E}(\operatorname{link}(kw))$, then we get the desired number of edges incident with $v$. The statement about the vertex $z\in\overline{X}\cap\mathcal{V}(D)$ is proved analogously. $\square$

Let $k:=\min(\{\#\mathcal{N}_{\overline{X}}(x):x\in X\cap\mathcal{V}(D)\}\cup\{\#\mathcal{N}_{{X}}(x):x\in\overline{X}\cap\mathcal{V}(D)\})$, and let $u\in\mathcal{V}(D)$ be a vertex such that $\#\mathcal{N}_{\overline{X}}(u)=k$. Without loss of generality, assume that $u\in X\cap\mathcal{V}(D)$. Then, by Claim 1 we have that $\#\mathcal{N}_{\overline{X}}(u)\leq d-2$, and by Claim 2 every vertex $v$ in $\mathcal{N}_{\overline{X}}(u)$ is incident with $d+1-k$ edges of $D$. Every vertex in $\mathcal{N}_{\overline{X}}(u)$ is in $\mathcal{V}(D)\cap\overline{X}$, and so there are at least $d-k$ vertices in $(\mathcal{V}(D)\cap\overline{X})\setminus\operatorname{link}(u)$ (as $n_{\overline{X}}\geq d$). Because every vertex in $\mathcal{V}(D)$ is incident with at least $k$ edges from $D$, for $k\geq 2$ we get that

a contradiction. Therefore $k=1$. Denote by $v$ the unique vertex in $\mathcal{N}_{\overline{X}}(u)$. By Claim 2 we have that $\#\mathcal{N}_{X}(v)\geq d$.

If every vertex in $\overline{X}\cap\mathcal{V}(D)$ other than $v$ was incident with at least three edges in $D$, then we would have $\#D\geq d+3(d-1)=4d-3$, and so there exists a vertex $v^{\prime}\in\overline{X}\cap\mathcal{V}(D)$ such that $\#\mathcal{N}_{X}(v^{\prime})\leq 2$. We consider two cases according to the cardinality of $\mathcal{N}_{X}(v^{\prime})$.

First suppose that $\#\mathcal{N}_{X}(v^{\prime})=1$. Then the unique vertex $u^{\prime}\in\mathcal{N}_{X}(v^{\prime})$ is incident with at least $d$ edges from $D$ by Claim 2. Since $G$ is $d$-connected by Balinski’s theorem (1), $G$ would remain $(d-2)$-connected after removing $u^{\prime}$ and $v$. By Menger’s theorem (8), there are $(d-2)$ pairwise internally disjoint $u-v^{\prime}$ paths in $G-\{u^{\prime},v\}$, and so $D$ contains $d-2$ pairwise disjoint edges not containing $u^{\prime}$ or $v$, say $u_{1}v_{1},\ldots,u_{d-2}v_{d-2}$ with $u_{i}\in X$ and $v_{i}\in\overline{X}$. It is not possible that each $u_{i}$ is incident with at least three edges in $D$, as otherwise Claim 2 would ensure that every vertex $v_{i}$ ($i=1,\ldots,d-2$) would be incident with at least $d-2$ edges in $D$. Thus, counting the edges in $D$ incident with $v$, $v^{\prime}$, and $v_{1},\ldots,v_{d-2}$ we get that $\#D\geq d+1+3(d-2)=4d-5$. So there is a vertex $u_{i}\in X$ adjacent to at most two vertices in $\overline{X}$, one of which is $v_{i}$. By Claim 2, this implies that $\#\mathcal{N}_{X}(v_{i})\geq d+1-2$. Of the edges in $D$ not incident with any vertex in $\{v,v^{\prime},v_{i}\}$, $d-3$ are edges $u_{j}v_{j}$ for $j\neq i$, and another $d-3$ of them are edges incident with $u^{\prime}$. Therefore,

another contradiction. Thus, $\#\mathcal{N}_{X}(v^{\prime})=2$.

The proof of the case $\#\mathcal{N}_{X}(v^{\prime})=2$ is analogous to the proof of the case $\#\mathcal{N}_{X}(v^{\prime})=1$, but we provide the details for the sake of completeness. Let $\{u^{\prime},u^{\prime\prime}\}:=\mathcal{N}_{X}(v^{\prime})$. Then Claim 2 yields that both $u^{\prime}$ and $u^{\prime\prime}$ are incident with $d-1$ edges from $D$. Removing $u^{\prime}$, $u^{\prime\prime}$, and $v$ would make $G$ $(d-3)$-edge-connected by Balinski’s theorem (1). So $D$ contains $d-3$ pairwise disjoint edges not containing $u^{\prime}$, $u^{\prime\prime}$, or $v$, say $u_{1}v_{1},\ldots,u_{d-3}v_{d-3}$ with $u_{i}\in X$ and $v_{i}\in\overline{X}$. It is not possible that each $u_{i}$ is incident with at least three edges in $D$; otherwise Claim 2 would ensure that every vertex $v_{i},i=1,\ldots,d-3$ would be incident with at least $d-2$ edges in $D$, which yields that $\#D\geq d+2+3(d-2)=4d-4>4d-7$ by counting the edges of $D$ incident with $v$, $v^{\prime}$, and $v_{1},\ldots,v_{d-3}$. So there is a vertex $u_{i}\in X$ adjacent to at most two vertices in $\overline{X}$, one of which is $v_{i}$. By Claim 2, this implies that $\#\mathcal{N}_{X}(v_{i})\geq d-1$. Of the edges in $D$ not incident with any vertex in $\{v,v^{\prime},v_{i}\}$, $d-4$ are edges $u_{j}v_{j}$ for $j\neq i$, and another $2d-2-6$ of them are edges incident with $u^{\prime}$ or $u^{\prime\prime}$. Therefore, for $d\geq 5$ we have that

another contradiction. In the case $d=4$ we get $\#D=5d-11$. This means that $\#\mathcal{N}_{X}(v)=4$, $\#\mathcal{N}_{X}(v_{i})=3$, and $\#\mathcal{N}_{X}(v^{\prime})=2$. But, since $n_{\overline{X}}\geq d$, there should exist a vertex $v^{\prime\prime}$ in $\overline{X}\cap\mathcal{V}(D)$ other than $\{v,v_{i},v^{\prime}\}$; this gives an edge that has not been accounted for. This final contradiction completes the proof of the theorem. ∎