On the Girth of Groups Acting on CAT(0) Cube Complexes

It is convenient to prove the lemma in terms of the halfspaces. Suppose $({\mathfrak{a}},{\mathfrak{b}},{\mathfrak{c}})$ is the facing triple of halfspaces. Let ${\mathfrak{a}}_{1}:={\mathfrak{a}}$ and ${\mathfrak{b}}_{1}:={\mathfrak{b}}$. We start by applying an automorphism to the pair $({\mathfrak{a}}_{1},{\mathfrak{b}}_{1})$ that flips the halfspace ${\mathfrak{c}}^{*}$ to get the pair $({\mathfrak{a}}_{2},{\mathfrak{b}}_{2})$. Clearly the $\{{\mathfrak{a}}_{1},{\mathfrak{b}}_{1},{\mathfrak{a}}_{2},{\mathfrak{b}}_{2}\}$ are facing halfspaces and the pair $\{\hat{\mathfrak{a}}_{2},\hat{\mathfrak{b}}_{2}\}$ is strongly separated. By induction, suppose we have constructed the $2(n-1)$ facing halfspaces $\{{\mathfrak{a}}_{1},{\mathfrak{b}}_{1},{\mathfrak{a}}_{2},{\mathfrak{b}}_{2},\ldots,{\mathfrak{a}}_{n-1},{\mathfrak{b}}_{n-1}\}$ such that $\hat{\mathfrak{a}}_{i}$ and $\hat{\mathfrak{b}}_{i}$ are strongly separated for each $i$ where $n\geq 2$. To construct $2n$ facing hyperplanes, we apply an automorphism to the triple $({\mathfrak{a}}_{1},{\mathfrak{b}}_{1},{\mathfrak{a}}_{n-1})$ that flips ${\mathfrak{b}}^{*}_{n-1}$ to get $({\mathfrak{a}}^{\prime}_{1},{\mathfrak{b}}^{\prime}_{1},{\mathfrak{a}}^{\prime}_{n-1})$ and then apply another automorphism to the pair $({\mathfrak{a}}^{\prime}_{1},{\mathfrak{b}}^{\prime}_{1})$ that flips ${\mathfrak{a}}^{\prime*}_{n-1}$ to get another pair $({\mathfrak{a}}^{\prime\prime}_{1},{\mathfrak{b}}^{\prime\prime}_{1})$ [See figure 1]. By construction, $\{{\mathfrak{a}}^{\prime}_{1},{\mathfrak{b}}^{\prime}_{1}\}$ and $\{{\mathfrak{a}}^{\prime\prime}_{1},{\mathfrak{b}}^{\prime\prime}_{1}\}$ are both pairs of strongly separated hyperplanes. Finally, we claim that the collection $\{{\mathfrak{a}}_{1},{\mathfrak{b}}_{1},\ldots,{\mathfrak{a}}_{n-2},{\mathfrak{b}}_{n-2},{\mathfrak{a}}^{\prime}_{1},{\mathfrak{b}}^{\prime}_{1},{\mathfrak{a}}^{\prime\prime}_{1},{\mathfrak{b}}^{\prime\prime}_{1}\}$ consists of facing halfspaces [Figure 1].

First, we observe that halfspaces in the collection $\{{\mathfrak{a}}_{1}^{\prime},{\mathfrak{b}}_{1}^{\prime},{\mathfrak{a}}^{\prime\prime}_{2},{\mathfrak{b}}^{\prime\prime}_{2}\}$ are mutually disjoint. This is because ${\mathfrak{a}}_{1}^{\prime}$ and ${\mathfrak{b}}_{1}^{\prime}$ are contained in ${\mathfrak{a}}^{\prime*}_{n-1}$, and the pair $({\mathfrak{a}}^{\prime\prime}_{1},{\mathfrak{b}}^{\prime\prime}_{1})$ is obtained by applying an automorphism to the pair $({\mathfrak{a}}^{\prime}_{1},{\mathfrak{b}}^{\prime}_{1})$ that flips ${\mathfrak{a}}^{\prime*}_{n-1}$. It follows that both ${\mathfrak{a}}^{\prime\prime}_{1}$ and ${\mathfrak{b}}^{\prime\prime}_{1}$ are contained inside ${\mathfrak{a}}^{\prime}_{n-1}$. The claim follows. Next we show that both ${\mathfrak{a}}^{\prime}_{1}$ and ${\mathfrak{b}}^{\prime}_{1}$ are disjoint from the collection $\{{\mathfrak{a}}_{1},{\mathfrak{b}}_{1},\ldots,{\mathfrak{a}}_{n-2},\newline {\mathfrak{b}}_{n-2}\}$. By construction, $({\mathfrak{a}}^{\prime}_{1},{\mathfrak{b}}^{\prime}_{1})$ is obtained by applying an automorphism to the pair $({\mathfrak{a}}_{1},{\mathfrak{b}}_{1})$ that flips ${\mathfrak{b}}^{*}_{n-1}$, therefore both ${\mathfrak{a}}^{\prime}_{1}$ and ${\mathfrak{b}}^{\prime}_{1}$ are contained inside ${\mathfrak{b}}_{n-1}$. Since ${\mathfrak{b}}_{n-1}$ is disjoint from any set from the collection $\{{\mathfrak{a}}_{1},{\mathfrak{b}}_{1},\ldots,{\mathfrak{a}}_{n-2},{\mathfrak{b}}_{n-2}\}$, so are both ${\mathfrak{a}}^{\prime}_{1}$ and ${\mathfrak{b}}^{\prime}_{1}$.

Similarly, we can show both ${\mathfrak{a}}^{\prime\prime}_{1}$ and ${\mathfrak{b}}^{\prime\prime}_{1}$ are disjoint from the collection $\{{\mathfrak{a}}_{1},{\mathfrak{b}}_{1},\ldots,\newline {\mathfrak{a}}_{n-2},{\mathfrak{b}}_{n-2}\}$. First note that, ${\mathfrak{a}}^{\prime}_{n-1}\subset{\mathfrak{b}}_{n-1}$ by construction. Also, ${\mathfrak{a}}^{\prime\prime}_{1}\subset{\mathfrak{a}}^{\prime}_{n-1}$ and ${\mathfrak{b}}^{\prime\prime}_{1}\subset{\mathfrak{a}}^{\prime}_{n-1}$. Consequently both ${\mathfrak{a}}^{\prime\prime}_{1}$ and ${\mathfrak{b}}^{\prime\prime}_{1}$ are contained inside ${\mathfrak{b}}_{n-1}$ which is disjoint from any set from the collection $\{{\mathfrak{a}}_{1},{\mathfrak{b}}_{1},\ldots,{\mathfrak{a}}_{n-2},{\mathfrak{b}}_{n-2}\}$. This finishes the proof. ∎

We need to show that ${\mathfrak{b}}_{n}\cap{\mathfrak{c}}_{n}=\emptyset$ for sufficiently large $n$. By hypothesis, there exist $N$ such that ${\mathfrak{a}}_{i}\cap{\mathfrak{b}}_{i}=\emptyset$ for all $i\geq N$. Since $\{{\mathfrak{c}}_{n}\}$ is descending and intersects $\{{\mathfrak{a}}_{n}\}$, there exists $M$ such that ${\mathfrak{a}}_{N}\cap{\mathfrak{c}}_{M}\neq\emptyset$. So, either ${\mathfrak{c}}_{M}\subset{\mathfrak{a}}_{N}$, or ${\mathfrak{c}}^{*}_{M}\subset{\mathfrak{a}}_{N}$ or $\hat{\mathfrak{c}}_{M}\pitchfork\hat{\mathfrak{a}}_{N}$.

Next, we will show how to rule out the second and third cases by taking $M$ large enough. To rule out the second case, we observe that ${\mathfrak{c}}^{*}_{M}\subset{\mathfrak{c}}^{*}_{M+k}\subset{\mathfrak{a}}_{N}$ for all $k\geq 0$. However, since $X$ is finite dimensional, there can be at most finitely many hyperplanes between any two halfspaces. So, for large enough $M$, the second case does not occur. To rule out the third case, suppose $\hat{\mathfrak{c}}_{M}\pitchfork\hat{\mathfrak{a}}_{N}$ for some $M$. Since ${\mathfrak{c}}_{M}$ and ${\mathfrak{c}}_{M+1}$ are strongly separated, ${\mathfrak{a}}_{N}$ cannot be transverse to ${\mathfrak{c}}_{M+1}$. So, either ${\mathfrak{a}}_{N}\cap{\mathfrak{c}}_{M+1}=\emptyset$ or ${\mathfrak{c}}_{M+1}\subset{\mathfrak{a}}_{N}$. By hypothesis ${\mathfrak{a}}_{N}\cap{\mathfrak{c}}_{M+1}\neq\emptyset$. Therefore, we have ${\mathfrak{c}}_{M+1}\subset{\mathfrak{a}}_{N}$.

So we are left with the first possibility which is ${\mathfrak{c}}_{M}\subset{\mathfrak{a}}_{N}$ for some $M$. Since ${\mathfrak{a}}_{N}\cap{\mathfrak{b}}_{N}=\emptyset$, it follows that ${\mathfrak{c}}_{M}\cap{\mathfrak{b}}_{N}=\emptyset$. Since the sequences of halfspaces are descending, we get $\{{\mathfrak{c}}_{n}\}$ is disjoint from $\{{\mathfrak{b}}_{n}\}$. ∎

Suppose $\Gamma\leq\mathop{\mathrm{Aut}}\nolimits(X)$ does not fix a point in $X\cup\partial_{\infty}X$. We want to show that $\operatorname{Girth}(\Gamma)=\infty$.

Since $\Gamma$ does not have fixed points at infinity, $\mathop{\mathrm{Aut}}\nolimits(X)$ does not have fixed points at infinity. Furthermore, since $\mathop{\mathrm{Aut}}\nolimits(X)$ acts essentially on $X$, we can invoke [CS11, Proposition 5.1] to obtain a pair of strongly separated hyperplane $\{\hat{\mathfrak{h}}_{i},\hat{\mathfrak{d}}_{i}\}$ in each $X_{i}$.

We now apply Theorem 5 to obtain an element $\sigma\in\Gamma$ that double skewers $(\hat{\mathfrak{h}}_{i},\hat{\mathfrak{d}}_{i})$ in each $X_{i}$. Let ${\mathfrak{h}}_{i}$ be the halfspace corresponding to the hyperplane $\hat{\mathfrak{h}}_{i}$ such that $\sigma^{j+1}{\mathfrak{h}}_{i}\subset\sigma^{j}{\mathfrak{h}}_{i}$ for any $j\in\mathbb{N}$. Since $(\hat{\mathfrak{h}}_{i},\hat{\mathfrak{d}}_{i})$ are strongly separated hyperplanes, it follows that $\{\sigma^{n}{\mathfrak{h}}_{i}\}$ and $\{\sigma^{-n}{\mathfrak{h}}^{*}_{i}\}$ are two disjoint descending sequence of strongly separated halfspaces. In other words, they are the poles of $\sigma$ in $X_{i}$ (see preliminaries).

Next, we choose an arbitrary generating set $\{\gamma_{1},\gamma_{2},\ldots,\gamma_{k}\}$ of $\Gamma$. We will consider the translate of poles of $\sigma$ in each $X_{i}$ under this generating set. More precisely, we consider the following list of descending sequences of halfspaces:

where $1\leq i\leq n$ and $1\leq j\leq k$. Since isometries preserve strong separation, all the sequences in the above list consists of strongly separated halfspaces.

(Also note that, for a fix $i$, all the halfspaces in the above list may not live in $X_{i}$, because $\gamma_{j}$ can switch isomorphic factors. However, they all live in some $X_{i}$).

Now, we are going to produce another regular element $\tau\in\Gamma$ such that the poles in each $X_{i}$ are disjoint from the list ($\dagger$ ‣ 3). In particular, its poles will be disjoint from that of $\sigma$. This is where we are going to use the fact that $\Gamma$ does not fix a point in $X\cup\partial_{\infty}X$. This will ensure that each $X_{i}$ is large, in a certain sense. More precisely, we will see that each $X_{i}$ has a facing triple.

First we observe that, since each $X_{i}$ is finite dimensional and locally compact and $\mathop{\mathrm{Aut}}\nolimits(X_{i})$ acts essentially and cocompactly on $X_{i}$, by [CS11, Theorem 7.2], either $\mathop{\mathrm{Aut}}\nolimits(X_{i})$ stabilizes some Euclidean flat or there is a facing triple of hyperplanes in $X_{i}$.

Suppose $\mathop{\mathrm{Aut}}\nolimits(X_{i})$ stabilizes some Euclidean flat $F_{i}\subset X_{i}$. Since $X_{i}$ is irreducible, it follows from Lemma 9 that $F_{i}$ is R-like. Thus $\mathop{\mathrm{Aut}}\nolimits(X_{i})$ has an index two subgroup which fixes a point at infinity contradicting our assumption on $\Gamma$. Therefore, we have a facing triple of hyperplanes in each $X_{i}$. Moreover, we can choose this facing triple in a way so that there is a pair of hyperplanes in the triple that are strongly separated (see the proof of [CS11, Theorem 7.2] for details).

By Lemma 6, there exist infinitely many pairs of strongly separated hyperplanes in each $X_{i}$ that are facing each other. Given a pair of strongly separated hyperplanes taken from each $X_{i}$, we can choose an element from $\Gamma$ that double skewers each pair in the corresponding $X_{i}$ due to Theorem 5. Since all the pairs are facing each other, the poles of these group elements give us infinitely many mutually disjoint descending sequence of strongly separated hyperplanes in $X_{i}$. Of these infinitely many descending sequences of halfspaces in $X_{i}$, all but finitely many are disjoint from the list ($\dagger$ ‣ 3) by Lemma 8 because the list ($\dagger$ ‣ 3) contains a finite number of descending sequences of strongly separated halfspaces. In particular, there exists a pair of disjoint halfspaces $\{{\mathfrak{h}}^{\prime}_{i},\mathfrak{d}^{\prime}_{i}\}$ in each $X_{i}$, such that for large enough $N$, both ${\mathfrak{h}}^{\prime}_{i}$ and $\mathfrak{d}^{\prime}_{i}$ are disjoint from the following collection

Let $\tau\in\Gamma$ be an element that double skewers $(\hat{\mathfrak{h}}^{\prime}_{i},\hat{\mathfrak{d}}^{\prime}_{i})$ in each $X_{i}$. Without loss of generality, let us assume that $\{\tau^{n}{\mathfrak{h}}^{\prime}_{i}\}$ and $\{\tau^{-n}{\mathfrak{h}}^{\prime*}_{i}\}$ are the poles of $\tau$ in $X_{i}$.

We let $Y:=\sqcup_{i=1}^{n}X_{i}$. Note that $\Gamma$ acts on $Y$. We set $U_{\sigma^{\prime}}:=\bigcup_{i}(\sigma^{N}{\mathfrak{h}}_{i}\cup\sigma^{-N}{\mathfrak{h}}_{i}^{*})$ and $U_{\tau^{\prime}}:=\bigcup_{i}(\tau^{N}{\mathfrak{h}}^{\prime}_{i}\cup\tau^{-N}{\mathfrak{h}}^{\prime*}_{i})$.

Note that, the complement of $\sigma^{N}{\mathfrak{h}}_{i}\cup\sigma^{-N}{\mathfrak{h}}_{i}^{*}$ in $X_{i}$ get sent to $\sigma^{N}{\mathfrak{h}}_{i}\cup\sigma^{-N}{\mathfrak{h}}_{i}^{*}$ under high enough power of $\sigma^{\pm 1}$. Taking union over $i$, we obtain that complement of $U_{\sigma}$ in $Y$ get sent to $U_{\sigma}$ under high enough power of $\sigma^{\pm 1}$. Similarly, complement of $U_{\tau}$ in $Y$ get sent to $U_{\tau}$ under high enough power of $\tau^{\pm 1}$. We set $\sigma^{\prime}=\sigma^{M}$, $\tau^{\prime}=\tau^{M}$, where we choose $M$ so large that for all $i\in\mathbb{Z}-\{0\}$, we have

Next, we observe that, by construction, $U_{\sigma^{\prime}}$ and its translates under the generating set and their inverses do not intersect $U_{\tau^{\prime}}$. In other words we have,

Since $\gamma_{j}(U_{\sigma^{\prime}})\cap U_{\tau^{\prime}}=\emptyset$ iff $\gamma^{-1}_{j}(U_{\tau^{\prime}})\cap U_{\sigma^{\prime}}=\emptyset$ for any $\gamma_{j}$, it follows from ($2$) that $U_{\tau^{\prime}}$ and its translates under the generating set and their inverses do not intersect $U_{\sigma^{\prime}}$. In other words, we have

Furthermore, we can assume (by taking $N$ large enough) that the following set

is nonempty and for convenience, let us call this set $B$.

We claim that $\sigma^{\prime},\tau^{\prime},U_{\sigma^{\prime}},U_{\tau^{\prime}}$, and any point $x\in B$ satisfies the three properties of Lemma 4.

The first property is immediate by our choice of $x$. To check the second property, note that $x\in Y-U_{\sigma^{\prime}}$ and together with ($3$) this implies

Therefore, ($1$) gives us

which is the desired second property.

By the assumption, we can choose $p\in Y$ such that $g\cdot p\neq p$ for some $g\in G$. Now we have two cases to consider.

Suppose $p$ is a vertex in $X$. Since $G$ acts by automorphisms, translates of $p$ can only be other vertices. It follows that $k$ is at least the minimum of the distances between $p$ and other vertices. Since $X$ is locally finite, it follows that $k>0$.

If $p$ is not a vertex, then $p$ is in the interior of some cube $C$ in $X$. The infimum of distances between $p$ and all the translates of $p$ that are inside a different cube is clearly positive. Since there are only finitely many automophisms of $C$, there are only finitely many nontrivial translates of $p$ that stay inside $C$. Therefore the minimum distance from $p$ to any of its translates inside $C$ is also positive. The claim follows. ∎

Let $H$ be the index two subgroup of $G$ consisting of only the orientation preserving isometries. In particular, $H$ acts by translation on $\mathbb{R}$ and hence is a subgroup of $\mathbb{R}$. Either $H$ is trivial or $k>0$ implies that