Connectedness of friends-and-strangers graphs of complete bipartite graphs and others

Assume that $c_{1}c_{2}\cdots c_{n}c_{1}$ is a $C_{n}$. For any bijection $\sigma:V(C_{n})\mapsto V(K_{k,n-k})$, we can map $\sigma$ to a cyclic ordering $\rho(\sigma)=\sigma(c_{1})\sigma(c_{2})\cdots\sigma(c_{n})\sigma(c_{1})$ of $V(K_{k,n-k})$. This cyclic ordering $\rho(\sigma)$ can be divided uniquely to a pair of cyclic orderings $(\rho_{A}(\sigma),\rho_{B}(\sigma))$ of $A$ and $B$. So we obtain a map that sends any bijection $\sigma$ to a pair of cyclic orderings $(\rho_{A}(\sigma),\rho_{B}(\sigma))$ of $A$ and $B$.

Because any vertices $a\in A$ and $b\in B$ are adjacent in $K_{k,n-k}$, two bijections are in the same component of $\mathsf{FS}(C_{n},K_{k,n-k})$ if and only if they are mapped to the same pair of cyclic orderings. Thus, each component of $\mathsf{FS}(C_{n},K_{k,n-k})$ corresponds to a pair of cyclic orderings of $A$ and $B$. Because a set $S$ has precisely $|S|!/|S|=(|S|-1)!$ cyclic orderings, the graph $\mathsf{FS}(C_{n},K_{k,n-k})$ has precisely $(k-1)!(n-k-1)!$ components. $\blacksquare$

Fix such a bijection $\sigma$ and let $C_{t}=c_{1}c_{2}\cdots c_{t}c_{1}$ satisfying $c_{1}=\sigma^{-1}(1),c_{2}=\sigma^{-1}(2)$. Let $S$ be the sequence of swaps along the edges $c_{2}c_{3},c_{3}c_{4},\dots,$ $c_{t-1}c_{t},$ $c_{1}c_{2},c_{t}c_{1}$. Then the sequence $S$ transforms $\sigma$ into a new bijection $\sigma_{1}=S(\sigma)$ satisfying $\sigma_{1}(c_{2})=\sigma(c_{1}),$ $\sigma_{1}(c_{1})=\sigma(c_{2}),$ $\sigma_{1}(c_{3})=\sigma(c_{4}),$ $\sigma_{1}(c_{4})=\sigma(c_{5}),$ $\dots,\sigma_{1}(c_{t-1})=\sigma(c_{t}),\sigma_{1}(c_{t})=\sigma(c_{3})$. Set $\sigma_{i+1}=S(\sigma_{i})$, see Table 1. It is easy to verify that $\sigma_{t-2}=(1\ 2)\circ\sigma$ since $t-2$ is an odd integer, which implies that $\sigma$ and $(1\ 2)\circ\sigma$ lie in the same component of $\mathsf{FS}(C_{t},K_{2,t-2})$. $\blacksquare$

If $Y$ is disconnected, Lemma 2.3 implies that $\mathsf{FS}(K_{k,n-k},Y)$ is disconnected. If $Y$ is bipartite, Lemma 2.4 implies that $\mathsf{FS}(K_{k,n-k},Y)$ is disconnected since $K_{k,n-k}$ is also bipartite. If $Y$ contains a non-trivial $k$-bridge, then Lemma 2.5 implies that $\mathsf{FS}(K_{k,n-k},Y)$ is disconnected since $K_{k,n-k}$ is not $(k+1)$-connected. If $Y$ is a cycle, then Lemma 2.7 implies that $\mathsf{FS}(K_{k,n-k},Y)$ is disconnected. $\blacksquare$

It is easy to see that the disconnectedness part of Theorem 1.4 follows from Theorem 1.3, just taking $k=2$.

For the connectedness part of Theorem 1.4, we need the following.

To make the arguments easier to follow, we postpone the proof of Proposition 3.1 until the end of this section. Note that $K_{k,n-k}=C_{4}$ if $n=4$ and so Theorem 1.4 holds by the connectedness of $\mathsf{FS}(C_{n},Y)$ as shown by Defant and Kravitz [4]. If $n\geq 5$, then Theorem 1.4 follows from Proposition 3.1.

Therefore, we complete the proof of Theorem 1.4. $\blacksquare$

Since $X$ is non-bipartite, there exists an odd cycle $C$ in $X$. Fix a bijection $\sigma$ satisfying $\sigma^{-1}(1)\sigma^{-1}(2)\in E(X)$.

If both $\sigma^{-1}(1)$ and $\sigma^{-1}(2)$ are in $V(C)$, then Lemma 2.10 implies directly that $1,2\in V(K_{2,n-2})$ are $(X,K_{2,n-2})$-exchangeable from $\sigma$ by considering swaps along edges in $E(C)$.

If exactly one of $\sigma^{-1}(1)$ and $\sigma^{-1}(2)$ is in $V(C)$, assume that $\sigma^{-1}(1)\in V(C)$. Let $c\in N(\sigma^{-1}(1))\cap V(C)$ be one of the two neighbors of $\sigma^{-1}(1)$ in $V(C)$ and swap along the edges $\sigma^{-1}(1)c,\sigma^{-1}(2)c$, which results a new bijection $\tau$ satisfying $\tau^{-1}(1),\tau^{-1}(2)\in V(C)$ and $\tau^{-1}(1)\tau^{-1}(2)\in E(C)$. Applying Lemma 2.10 to $\tau$ obtains a sequence of swaps that transforms $\tau$ to $(1\ 2)\circ\tau$. Then we can swap along the edges $\sigma^{-1}(2)c,\sigma^{-1}(1)c$ to get the desired $(1\ 2)\circ\sigma$.

If none of $\sigma^{-1}(1)$ and $\sigma^{-1}(2)$ are in $V(C)$, let $x_{1}x_{2}\cdots x_{p}$ be a shortest path in $X$ between the two vertex sets $\{\sigma^{-1}(1),\sigma^{-1}(2)\}$ and $V(C)$. Assume without loss of generality that $\sigma^{-1}(1)=x_{1}$ and denote $\sigma^{-1}(2)$ by $x_{0}$. Denote by $S$ the sequence of swaps along the edges $x_{1}x_{2},\dots,x_{p-1}x_{p}$, $x_{0}x_{1}$, $x_{1}x_{2},\dots,x_{p-2}x_{p-1}$. The sequence $S$ transforms $\sigma$ into a new bijection $\tau$ satisfying $\tau^{-1}(1)\in V(C)$, $\tau^{-1}(2)\notin V(C)$ and $\tau^{-1}(1)\tau^{-1}(2)\in E(X)$. Similar to the case when exactly one of $\sigma^{-1}(1)$ and $\sigma^{-1}(2)$ is in $V(C)$, there is a sequence of swaps that transforms $\tau$ into $(1\ 2)\circ\tau$. Then, the sequence $S^{-1}$ transforms $(1\ 2)\circ\tau$ into the desired $(1\ 2)\circ\sigma$. $\blacksquare$

By Lemma 2.8, it suffices to show that any two vertices $u,v\in V(K_{2,n-2})$ are $(X,K_{2,n-2})$-exchangeable from any bijection $\sigma:V(X)\mapsto V(K_{2,n-2})$ satisfying $\sigma^{-1}(u)\sigma^{-1}(v)\in E(X)$. If $\{u,v\}=\{1,2\}$, then Lemma 3.2 implies directly that $u,v$ are $(X,K_{2,n-2})$-exchangeable from $\sigma$. If $|\{u,v\}\cap\{1,2\}|=1$, then we can swap along the edge $\sigma^{-1}(u)\sigma^{-1}(v)$ to obtain directly the bijection $(u\ v)\circ\sigma$, i.e., $u,v$ are $(X,K_{2,n-2})$-exchangeable from $\sigma$. So we are left to consider the case $\{u,v\}\cap\{1,2\}=\varnothing$.

Firstly, we prove that there exists a sequence of swaps only involving edges in $K_{2,n-2}|_{[n]\setminus\{u,v\}}$ that transforms $\sigma$ into a bijection $\tau$ such that $X|_{\tau^{-1}(\{u,v,1,2\})}$ is connected and $\tau^{-1}(u)\tau^{-1}(v)\in E(X)$. If there are no edges between $\{\sigma^{-1}(1),\sigma^{-1}(2)\}$ and $\{\sigma^{-1}(u),\sigma^{-1}(v)\}$, let $x_{1}x_{2}\cdots x_{q}$ be a shortest path between the two vertex sets $\{\sigma^{-1}(1),\sigma^{-1}(2)\}$ and $\{\sigma^{-1}(u),\sigma^{-1}(v)\}$. Assume without loss of generality that $\sigma^{-1}(1)=x_{1}$. Denote by $S$ the sequence of swaps along the edges $x_{1}x_{2},\dots,x_{q-2}x_{q-1}$, which can transform $\sigma$ into a bijection $\tau^{\prime}$ such that $X|_{\tau^{\prime-1}(\{u,v,1\})}$ is connected and $\tau^{\prime-1}(u)\tau^{\prime-1}(v)\in E(X)$. For the same reason, if $\tau^{\prime-1}(2)$ is not adjacent to any vertex in $\{\tau^{\prime-1}(u),\tau^{\prime-1}(v)$, $\tau^{\prime-1}(1)\}$, then there exists a sequence of swaps transforming $\tau^{\prime}$ into a bijection $\tau$ satisfying that $X|_{\tau^{-1}(\{u,v,1,2\})}$ is connected and $\tau^{-1}(u)\tau^{-1}(v)\in E(X)$. So, by Lemma 2.11, $u,v$ are $(X,K_{2,n-2})$-exchangeable from $\tau$ if and only if $u,v$ are $(X,Y)$-exchangeable from $\sigma$. So, we assume that $X|_{\sigma^{-1}(\{u,v,1,2\})}$ is connected.

If one of $\sigma^{-1}(1)$ and $\sigma^{-1}(2)$ has no neighbor in $\{\sigma^{-1}(u),\sigma^{-1}(v)\}$, assume that it is $\sigma^{-1}(1)$ and that $\sigma^{-1}(2)\sigma^{-1}(v)\in E(X)$. Denote by $S_{0}$ the sequence of swaps along the edges $\sigma^{-1}(2)\sigma^{-1}(v)$, $\sigma^{-1}(v)\sigma^{-1}(u)$. The sequence $S_{0}$ transforms $\sigma$ into a new bijection $\eta$ such that both $\eta^{-1}(1)$ and $\eta^{-1}(2)$ have neighbors in $\{\eta^{-1}(u),\eta^{-1}(v)\}$ and $\eta^{-1}(u)\eta^{-1}(v)\in E(X)$. If $u,v$ are $(X,K_{2,n-2})$-exchangeable from $\eta$, then there will exist a sequence $S_{1}$ of swaps that transforms $\eta$ to $(u\ v)\circ\eta$. Thus, the sequences $S_{0},S_{1},S^{-1}_{0}$ will transform $\sigma$ to $(u\ v)\circ\sigma$. So, to complete the proof, it suffices to show that $u,v$ are $(X,K_{2,n-2})$-exchangeable from $\sigma$ under the assumption that both $\sigma^{-1}(1)$ and $\sigma^{-1}(2)$ have neighbors in $\{\sigma^{-1}(u),\sigma^{-1}(v)\}$.

Assume that $\sigma^{-1}(u)$ is a common neighbor. Swapping along the edges $\sigma^{-1}(1)\sigma^{-1}(u)$, $\sigma^{-1}(u)\sigma^{-1}(v)$, $\sigma^{-1}(2)\sigma^{-1}(u)$ transforms $\sigma$ into a new bijection $\tau$ with $\tau^{-1}(2)=\sigma^{-1}(u)$, $\tau^{-1}(1)=\sigma^{-1}(v)$ and $\tau^{-1}(1)\tau^{-1}(2)\in E(X)$, see Figure 2. Apply Lemma 3.2 to $\tau$, we obtain a sequence of swaps that transforms $\tau$ to $(1\ 2)\circ\tau$. Then swapping along the edges $\sigma^{-1}(1)\sigma^{-1}(u)$, $\sigma^{-1}(v)\sigma^{-1}(u)$, $\sigma^{-1}(2)\sigma^{-1}(u)$ transforms $(1\ 2)\circ\tau$ into the desired bijection $(u\ v)\circ\sigma$.

Since $\sigma^{-1}(u)\sigma^{-1}(v)$ is not a trivial cut edge and $X$ contains no non-trivial cut edge, there exist cycles in $X$ that contain the edge $\sigma^{-1}(u)\sigma^{-1}(v)$. Let $C$ be such a shortest one. The length of $C$ is strictly less than $n$ since $X$ is not a cycle. Assume without loss of generality that $\sigma^{-1}(1)\sigma^{-1}(u),\sigma^{-1}(2)\sigma^{-1}(v)$ are edges in $X$.

Assume that it is $\sigma^{-1}(1)$. Let $C=c_{1}c_{2}\cdots c_{r}c_{1}$ with $c_{1}=\sigma^{-1}(u)$, $c_{2}=\sigma^{-1}(1)$, $c_{r}=\sigma^{-1}(v)$.

Denote by $S$ the sequence of swaps along the edges $c_{2}c_{3}$, $c_{3}c_{4},\dots,c_{r-2}c_{r-1}$. The sequence $S$ transforms $\sigma$ into a new bijection $\tau$ such that $\tau^{-1}(1)$ and $\tau^{-1}(2)$ are adjacent to a common vertex $\tau^{-1}(v)$, see Figure $3$. The same as Case $1$, we obtain a sequence of swaps that transforms $\tau$ to $(u\ v)\circ\tau$. Then the sequence of swaps $S^{-1}$ transforms $(u\ v)\circ\tau$ into the desired $(u\ v)\circ\sigma$.