A simplification of "Effective quasimorphisms on right-angled Artin groups"

TL;DR

This paper simplifies the proof of a key result in the study of quasimorphisms on RAAG-like groups acting on CAT(0) cube complexes, making the argument more accessible and concise.

Contribution

It provides a shorter, more straightforward proof that quasimorphisms on RAAG-like groups have a uniform defect and specific value properties, removing the need for complex tools.

Findings

Quasimorphisms on RAAG-like groups have a uniform defect of 12.

Hyperbolic elements in RAAG-like groups have stable commutator length at least 1/24.

The simplified proof avoids technical tools like essential characteristic sets and equivariant Euclidean embeddings.

Abstract

This paper presents a simplification of the main argument in "Effective quasimorphisms on right-angled Artin groups" by Fern\'os, Forester and Tao. Their article introduces a family of quasimorphisms on a certain class of groups (called RAAG-like) acting on CAT(0) cube complexes. They show that these have uniform defect of $12$ and take the value of at least $1$ on a chosen element. With the Bavard Duality they conclude that hyperbolic elements of RAAG-like groups have stable commutator length of at least $1/24$. Their proof that the quasimorphisms take the value $1$ is quite technical and relies on some tools they introduce: the essential characteristic set and equivariant Euclidean embeddings. Here it is shown that this is unnecessary and a much shorter proof is given. This was originally part of my master thesis "Stable commutator length gap in RAAG-like groups" supervised by Alessandra Iozzi at ETH Z\"urich.

Equations

$$[x,y]:=\left\{\Phi \in \mathcal{H}:x\notin \Phi \text{ and }y\in \Phi \right\}.$$

$$\delta (g)=\underset{x\in X^{(0)}}{\min }{d}^c(x,gx).$$

$$\text{scl}(g)=\underset{n\to \infty }{\lim }\frac{\text{cl}(g^n)}{n}.$$

$$D(\varphi ):=\underset{g,h\in G}{\sup }|\varphi (gh)-\varphi (g)-\varphi (h)|.$$

$$\mathrm{scl}(g)=\underset{\overline{\varphi }\in Q(G)}{\sup }\frac{\overline{\varphi }(g)}{2D(\overline{\varphi }))}.$$

$$\left|{\omega }_{\gamma }(x,y)-{\omega }_{\gamma }(x,m)-{\omega }_{\gamma }(m,y)\right|<2$$

$$\begin{array}{rl}& \left|{\omega }_{\gamma }(x,y)-{\omega }_{\gamma }(x,m)-{\omega }_{\gamma }(m,y)\right|\\ & \le \left|c_{\gamma }(x,y)-c_{\gamma }(x,m)-c_{\gamma }(m,y)\right|+\left|c_{\bar{\gamma }}(x,y)-c_{\bar{\gamma }}(x,m)-c_{\bar{\gamma }}(m,y)\right|\\ & \le 2\end{array}$$

$$\left|{\omega }_{\gamma }(x,y)+{\omega }_{\gamma }(y,z)+{\omega }_{\gamma }(z,x)\right|\le 6$$

$$\begin{array}{rl}\left|{\omega }_{\gamma }(x,y)+{\omega }_{\gamma }(y,z)+{\omega }_{\gamma }(z,x)\right|\le & |{\omega }_{\gamma }(x,m)+{\omega }_{\gamma }(m,y)+{\omega }_{\gamma }(y,m)\\ & +{\omega }_{\gamma }(m,z)+{\omega }_{\gamma }(z,m)+{\omega }_{\gamma }(m,x)|+6\\ =& 6,\end{array}$$

$$\begin{array}{rl}|\delta {\varphi }_{\gamma }(g,h)|& =|{\varphi }_{\gamma }(gh)-{\varphi }_{\gamma }(g)-{\varphi }_{\gamma }(h)|\\ & =|{\omega }_{\gamma }(x_0,gh{x}_0)-{\omega }_{\gamma }(x_0,g{x}_0)-{\omega }_{\gamma }(x_0,h{x}_0)|\\ & =|{\omega }_{\gamma }(x_0,gh{x}_0)-{\omega }_{\gamma }(x_0,g{x}_0)-{\omega }_{\gamma }(g{x}_0,gh{x}_0)|\\ & =|{\omega }_{\gamma }(x_0,gh{x}_0)+{\omega }_{\gamma }(g{x}_0,x_0)+{\omega }_{\gamma }(gh{x}_0,g{x}_0)|\\ & \le 6,\end{array}$$

$${\overline{\varphi }}_{\gamma }(g)=\underset{n\to \infty }{\lim }\frac{{\varphi }_{\gamma }(g^n)}{n}=\underset{n\to \infty }{\lim }\frac{{\omega }_{\gamma }(o,g^{n}o)}{n}\ge \frac{n}{n}=1$$

Full text

Their article introduces a family of quasimorphisms on a certain class of groups (called RAAG-like) acting on CAT(0) cube complexes. They show that these have uniform defect of $12$ and take the value of at least $1$ on a chosen element. With the Bavard Duality they conclude that hyperbolic elements of RAAG-like groups have stable commutator length of at least $1/24$.

Their proof that the quasimorphisms take the value $1$ is quite technical and relies on some tools they introduce: the essential characteristic set and equivariant Euclidean embeddings. Here it is shown that this is unnecessary and a much shorter proof is given.

This was originally part of my master thesis “Stable commutator length gap in RAAG-like groups” supervised by Alessandra Iozzi at ETH Zürich.

1. 1 Introduction

2. 2 Premliminaries

3. 2.1 Halfspaces in CAT(0) cube complexes

4. 2.2 Haglund’s combinatorial axis

5. 2.3 The Bavard Duality

6. 3 RAAG-like actions

7. 4 The quasimorphisms and their defect

8. 5 Effectiveness

1. Introduction

In [FFT16], a special class of group actions on CAT(0) cube complexes is defined by the behaviour of halfspaces under the action (see Definition 3 in this paper). Since they generalise right-angled Artin groups (RAAGs) acting on the associated RAAG-complex (the universal cover of the associated Salvetti complex), they are called RAAG-like actions.

The authors define quasimorphisms on all groups acting non-transversally 111A group acts non-transversally, if orbits of halfspaces are nested. In particular, RAAG-like actions are non-transverse. on CAT(0) cube complexes, generalising Fujiwara–Epstein counting quasimorphisms on free groups (introduced in [EF97]). The quasimorphisms in [FFT16] are shown to have defect of at most $6$ and, hence, their homogenisation at most $12$. This is done using the median property of CAT(0) spaces.

The authors then prove effectiveness of these quasimorphisms, that is, for every element $g$ acting hyperbolically on the CAT(0) cube complex, one of their homogeneous quasimorphisms $\overline{\phi}$ satisfies $\overline{\phi}(g)\geq 1$. For this, using Haglunds combinatorial axis ([Hag07]), they isolate a subcomplex with respect to $g$ called ‘essential set’ and construct a $g$-equivariant embedding of this into a Euclidean space. They then use a rather intricate series of technical lemmas to prove that $\overline{\phi}(g)\geq 1$.

The purpose of this paper is to show that these technical arguments can be avoided and replaced by a short proof that also uses the full properties of RAAG-like actions.

Finally, we conclude as in [FFT16] that hyperbolic elements of RAAG-like actions have stable commutator length at least $1/24$ using the Bavard Duality. In particular, RAAGs have a stable commutator length gap of $1/24$. Note that, using a different class of quasimorphisms, Heuer [Heu19] has already proved a better (and optimal) bound of $1/2$ for the stable commutator length in RAAGs.

2. Premliminaries

2.1. Halfspaces in CAT(0) cube complexes

Let $X$ be a CAT(0) cube complex. Denote by $\mathcal{H}(X)$ the set of half-spaces and by $\overline{\Phi}$ the complement of a halfspace $\Phi$. Two halfspaces $\Phi,\Psi\in\mathcal{H}$ are said to be nested, if either $\Phi\subseteq\Psi$, $\overline{\Phi}\subseteq\Psi$, $\Phi\subseteq\overline{\Psi}$ or $\overline{\Phi}\subseteq\overline{\Psi}$. Otherwise, they are transverse. Two distinct $\Phi,\Psi\in\mathcal{H}$ are tightly nested, if they are nested and there is no $\Phi^{\prime}\in\mathcal{H}$ with $\Phi\subsetneq\Phi^{\prime}\subsetneq\Psi$ or $\overline{\Phi}\subsetneq\Phi^{\prime}\subsetneq\Phi$ etc..

For every oriented edge $E$ in $X$, there is exactly one $\Phi\in\mathcal{H}$ such that the beginning vertex of $E$ is in $\Phi$ and the end vertex is in $\overline{\Phi}$. We say that $\Phi$ and $E$ are dual to each other.

Given $x,y\in\mathcal{H}$ the interval between $x$ and $y$ is

[TABLE]

Given vertices $x,y,z\in X$, there is a unique vertex $m(x,y,z)$ called median with the property $[a,b]=[a,m]\cup[m,b]$ for any distinct pair $a,b$ in ${x,y,z}$ (see Preliminaries in [FFT16] for a simple proof).

2.2. Haglund’s combinatorial axis

Let $X$ be a CAT(0) cube complex. We can introduce a metric called combinatorial distance $d^{c}$ on the set of vertices $X^{(0)}$, by defining $d^{c}(x,y)$ as the minimal number of edges in an edge path from $x$ to $y$. A combinatorial geodesic is an optimal (with respect to $d^{c}$) oriented edge path.

The translation distance of an automorphism $g$ of $X$ is the natural number

[TABLE]

An automorphism $g$ of a CAT(0) cube complex $X$ is hyperbolic, if it fixes no vertex of $X$, i.e. $\delta(g)>0$. Otherwise, $g$ is called elliptic. A combinatorial axis is an infinite combinatorial geodesic on which $g$ acts as a shift. According to Haglund in [Hag07], if $g$ is a hyperbolic automorphism all of whose powers act without inversion (that is, there are no $\Phi\in\mathcal{H}$ and $n\in\mathbb{Z}$ with $g^{n}\Phi=\overline{\Phi}$), then every vertex in $X^{(0)}$ on which $g$ attains its translation distance is contained in some combinatorial axis.

As in [FFT16], let $A^{+}_{g}$ denote all halfspaces dual to an oriented edge in some combinatorial axis (indeed, a halfspace dual to some combinatorial axis is also dual to all other ones according to [Hag07]). Clearly, $A^{+}_{g}=\bigcup_{n\in\mathbb{N}}[g^{n}o,g^{n+1}o]$ for any vertex $o$ where $g$ attains its translation distance. An important fact is that for $\Phi,\Psi\in A^{+}_{g}$ either $\Phi\subseteq\Psi$, $\Phi\supseteq\Psi$ or $\Phi$ and $\Psi$ are transverse, which follows because a combinatorial geodesic may never leave a halfspace after entering.

2.3. The Bavard Duality

Given a group $G$, the commutator length cl is a function $\textrm{cl}:[G,G]\rightarrow\mathbb{N}$, where $[G,G]$ is the commutator subgroup. For $g\in[G,G]$, it is defined as the minimal number of commutators whose product is $g$. The stable commutator length is the well defined limit

[TABLE]

Bounds of scl can be estimated using homogeneous quasimorphisms and the Bavard Duality. A quasimorphism is a function $\phi:G\rightarrow\mathbb{R}$ with a bounded defect

[TABLE]

The quasimorphism $\phi$ is called homogeneous, if $\phi(g^{n})=n\phi(g)$ for all $g$ and $n\in\mathbb{Z}$. Denote by $Q(G)$ the space of homogeneous quasimorphisms on $G$. Every quasimorphism yields a homogeneous quasimorphism called homogenisation $\overline{\phi}(g):=\lim_{n\to\infty}\frac{\phi(g^{n})}{n}$ with the following property:

Lemma 1**.**

Let $\phi$ be a quasimorphism. Then its homogenisation satisfies $D(\overline{\phi})\leq 2D(\phi)$.

The Bavard Duality states that:

Theorem 2** ([Bav91]).**

For any $g\in[G,G]$

[TABLE]

Therefore, to prove that $\textrm{scl}(g)$ is bounded from below by some constant, it suffices to find a homogeneous quasimorphism which has low enough defect (we say it is efficient) and at the same time does not vanish on $g$ (that is, it is effective).222See [Cal09, Chapter 2] for more detail and proofs on quasimorphisms and scl.

3. RAAG-like actions

Let us reproduce the definition of RAAG-like actions given in [FFT16, Chapter 7]:

Definition 3**.**

Let $G$ be a group acting on a CAT(0) cube complex $X$ with halfspaces $\mathcal{H}(X)$. The action is called RAAG-like if the following are satisfied:

1. (i)

There are no $\Phi\in\mathcal{H}(X)$ and $h\in G$ with $h\overline{\Phi}=\Phi$ (“no inversions”) 2. (ii)

there are no $\Phi\in\mathcal{H}(X)$ and $h\in G$ with $\Phi$ and $h\Phi$ transverse (“non-transverse”), 3. (iii)

there are no tightly nested $\Phi,\Phi^{\prime}\in\mathcal{H}(X)$ and $h\in G$ with $\Phi$ and $h\Phi^{\prime}$ transverse, 4. (iv)

there are no $\Phi\in\mathcal{H}(X)$ and $h\in G$ with $\Phi\subset h\overline{\Phi}$ tightly.

A group is called RAAG-like, if it has a faithful RAAG-like action on some CAT(0) cube complex.

Remark*.*

If $G$ acts on $X$ freely, then RAAG-likeness of the action is equivalent to $X/G$ being a A-special (often simply called special) cube complex in the sense of Haglund and Wise [HW08, Definition 3.2]. In particular, we have the following correspondences:

1. (i)

corresponds to all hyperplanes in $X/G$ being two-sided, 2. (ii)

corresponds to no hyperplane in $X/G$ intersecting itself, 3. (iii)

corresponds to no pair of hyperplanes in $X/G$ inter-osculating and 4. (iv)

corresponds to no pair of hyperplanes in $X/G$ directly self-osculating.

Hence $G$ is the fundamental group of an A-special cube complex and conversely the fundamental group of an A-special cube complex acts RAAG-like and freely on its universal cover. In particular, RAAGs are RAAG-like, as they are the fundamental group of an A-special cube complex (their Salvetti complex).

Lemma 4**.**

Every non-trivial element of a RAAG-like action is hyperbolic.

Proof.

Suppose $h\in G$ is elliptic, i.e. $h$ has at least one fixed vertex, and acts non-trivially. If for some fixed vertex of $h$ in $X$, every incident edge is fixed, then all neighbouring vertices of $v$ are also fixed. Therefore, there must be some fixed vertex $v$ with an incident edge which is not fixed, or else every single vertex of $X$ would be fixed. Let $E$ be adjacent to $v$ and mapped to some other edge $F$ adjacent to $v$. If $E$ and $F$ bound a square, then $h$ is transverse, as the halfspace $\Phi$ dual to $E$ is transverse to $h\Phi$, the halfspace dual to $F$. If they do not bound a square, then $\Phi$ and $h\Phi$ are tightly nested if they are not transverse. ∎

4. The quasimorphisms and their defect

From now on, let $G$ be a group with a non-transverse (not necessarily RAAG-like) action on a CAT(0) cube complex $X$.

We recall the quasimorphisms defined in [FFT16, Chapter 4] and, for completeness, the proof that their defect is bounded by $12$.

Definition 5**.**

A segment is a series of half-spaces $\gamma=\left\{\Phi_{0},...,\Phi_{r}\right\}$ such that $\Phi_{i}\supsetneq\Phi_{i+1}$ tightly for $0\leq i<r$.

The reverse segment of $\gamma$ is $\overline{\gamma}=\left\{\overline{\Phi}_{r},...,\overline{\Phi}_{0}\right\}$.

Definition 6**.**

Two segments $\gamma_{1},\gamma_{2}\in\mathcal{H}(X)$ are said to overlap, if there is $\Phi_{1}\in\gamma_{1}$ and $\Phi_{2}\in\gamma_{2}$ that are equal or transverse. A set of segments is called non-overlapping, if no two of its segments overlap.

Remark*.*

If $\gamma=\left\{\Phi_{0},...,\Phi_{r}\right\}$ and $\gamma^{\prime}=\left\{\Psi_{0},...,\Psi_{s}\right\}$ are non-overlapping segments, then one of the following holds:

• $\Phi_{0}\supset...\supset\Phi_{r}\supset\Psi_{0}\supset...\supset\Psi_{s}$
• $\Psi_{0}\supset...\supset\Psi_{s}\supset\Phi_{0}\supset...\supset\Phi_{r}$
• $\overline{\Phi}_{r}\supset...\supset\overline{\Phi}_{0}\supset\Psi_{0}\supset...\supset\Psi_{s}$
• $\Psi_{0}\supset...\supset\Psi_{s}\supset\overline{\Phi}_{r}\supset...\supset\overline{\Phi}_{0}$

Otherwise, it can easily be seen, that tight nestedness of the segments is contradicted. We write $\gamma>\gamma^{\prime}$, $\gamma^{\prime}>\gamma$, $\overline{\gamma}>\gamma^{\prime}$ and $\gamma^{\prime}>\overline{\gamma}$ respectively in these cases.

Remark*.*

If $S$ is a set of non-overlapping segments, then for any $\gamma_{1},\gamma_{2}\in S$ either $\gamma_{1}>\gamma_{2}$ or $\gamma_{2}>\gamma_{1}$. Thus, if $S$ is finite, it must contain a maximal segment that contains every other segment in $S$, and a minimal segment that is contained by every other segment in $S$, respectively.

Definition 7**.**

Given a segment $\gamma$, let $G\gamma=\{g\gamma:g\in G\}$ denote the set of copies of $\gamma$. The function $c_{\gamma}:X^{2}\rightarrow\mathbb{R}$ is defined sucht that $c_{\gamma}(x,y)$ is the cardinality of the largest non-overlapping subset of $G\gamma$ in $[x,y]$.

Furthermore, define $\omega_{\gamma}:X^{2}\rightarrow\mathbb{R}$ by $\omega_{\gamma}(x,y):=c_{\gamma}(x,y)-c_{\bar{\gamma}}(x,y)$.

Remark*.*

$\omega_{\gamma}(\cdot,\cdot)$ is $G$-invariant, i.e. $\omega_{\gamma}(x,y)=\omega_{\gamma}(gx,gy)$ for any $x,y\in X$ and $g\in G$, since any non-overlapping subset of $G\gamma$ in $[x,y]$ can be pushed by $g$ to one in $[gx,gy]$, and vice versa.

Furthermore, $\omega_{\gamma}(\cdot,\cdot)$ is antisymmetric, since if $g\gamma\in[x,y]$, then $g\overline{\gamma}\in[y,x]$, and vice versa.

The following lemmas show that $\omega_{\gamma}(o,go)$ as a function of $g$ (where $o$ is any vertex of $X$) is a quasimorphism.

Lemma 8**.**

For $x,m,y\in X$ with $m=m(x,m,z)$,

[TABLE]

holds.

Proof.

Let us first prove $c_{\gamma}(x,y)\geq c_{\gamma}(x,m)+c_{\gamma}(m,y)-1$. Let $S_{1}$ and $S_{2}$ be maximal non-overlapping sets of copies of $\gamma=\left\{\Phi_{0},...,\Phi_{r}\right\}$ in $[x,m]$ and $[m,y]$, respectively. Let $g\gamma$ be the minimal element of $S_{1}$. We have $a\gamma>b\gamma$ for any $a\gamma\in S_{1}\setminus\{g\gamma\}$ and $b\gamma\in S_{2}$, because for any $a\Phi_{k}\in a\gamma$ and $b\Phi_{l}\in b\gamma$ we have $a\Phi_{k}\subsetneq g\Phi_{0}\subsetneq b\Phi_{0}\subsetneq b\Phi_{l}$. Thus, $S_{1}\setminus\{g\gamma\}$ and $S_{2}$ do not overlap, whence the inequality follows.

Let us now prove $c_{\gamma}(x,z)\leq c_{\gamma}(x,m)+c_{\gamma}(m,y)+1$. Let $S$ be a maximal set of copies of $\gamma$ in $[x,m]$. There can be at most one copy $g\gamma\in S$ containing halfspaces $\Phi$ and $\Psi$ such that $y\in\Phi$ and $y\notin\Psi$ since all other copies of $\gamma$ in $S$ either elementwise contain $\Phi$ or are contained elementwise in $\Psi$. The remaining $|S|-1$ copies can be assigned to sets $S_{1}$ and $S_{2}$ contained in $[x,y]$ and $[m,y]$, respectively, which proves the inequality.

This proves $|c_{\gamma}(x,y)-c_{\gamma}(x,m)-c_{\gamma}(m,y)|<1$ and therefore the lemma, as

[TABLE]

∎

Lemma 9**.**

For any $x,y,z\in X$

[TABLE]

holds.

Proof.

Let $m$ be the median of $x,y,z$. By the last lemma, $\left|\omega_{\gamma}(x,y)-\omega_{\gamma}(x,m)-\omega_{\gamma}(m,y)\right|<2$ holds, and analogous inequalities after replacing $x$ or $y$ by $z$. Therefore,

[TABLE]

where antisymmetry of $\omega_{\gamma}$ was used on the last line. ∎

Lemma 10**.**

Given a segment $\gamma$ and a vertex $x_{0}\in X$, the map $\phi_{\gamma}:G\rightarrow\mathbb{R}$ given by $\phi_{\gamma}(g)=\omega_{\gamma}(x_{0},gx_{0})$ is a quasimorphism with defect bounded by $6$.

As a consequence, its homogenisation $\overline{\phi}_{\gamma}$ has defect bounded by $12$.

Proof.

[TABLE]

where $\omega_{\gamma}(x_{0},hx_{0})=\omega_{\gamma}(gx_{0},ghx_{0})$, antisymmetry of $\omega_{\gamma}$ and Lemma 9 were used in this order. ∎

5. Effectiveness

From now on let $G$ be a group with a RAAG-like action on $X$. Let $g\in G$ be a hyperbolic element and let $o\in X$ denote a vertex where $g$ attains its translation distance.

The aim is to find a segment $\gamma$ in $[o,go]$ such that for $m\in\mathbb{N}$ the interval $[g^{m}o,g^{m+1}o]$ contains at least one copy of $\gamma$ and no copies of $\overline{\gamma}$. This will guarantee $\overline{\phi_{\gamma}}(g)\geq 1$.

The following are the segments we need:

Definition 11**.**

A segment $\gamma=\left\{\Phi_{0},...,\Phi_{r}\right\}$ in $[o,go]$ is called $g$-nested if $\gamma>g\gamma$. It is a maximal $g$-nested segment, if it is not contained in any other $g$-nested segment in $[o,go]$

Remark*.*

A maximal $g$-nested segment always exists, since a single halfspace inn $[o,go]$ is a $g$-nested segment by non-transversality.

The $g$-nestedness is to guarantee, that $[g^{m}o,g^{m+1}o]$ contains a copy of $\gamma$ for every $m\in\mathbb{N}$, while the maximality will be crucial to ensure that no copies of $\overline{\gamma}$ occur in these intervals.

Here is a useful characterisation of maximality:

Lemma 12**.**

Let $\gamma=\left\{\Phi_{0},...,\Phi_{r}\right\}$ be maximal $g$-nested in $[o,go]$. Then:

1. (i)

Any $\Psi\in[o,go]$ with $\Psi\supsetneq\Phi_{0}$ is transverse to $g^{-1}\Phi_{r}$. 2. (ii)

Any $\Psi\in[o,go]$ with $\Phi_{r}\supsetneq\Psi$ is transverse to $g\Phi_{0}$.

Proof.

Let $\Psi\in[o,go]$ with $\Psi\supsetneq\Phi_{0}$ and suppose by contradiction that $\Psi$ is not transverse to $g^{-1}\Phi_{r}$. Since $o\notin\Psi$, we have $g^{-1}\Phi_{r}\supsetneq\Psi$ as $\Psi\supseteq g^{-1}\Phi_{r}$ would imply $o\notin g^{-1}\Phi_{r}$. Let $\Psi^{\prime}$ be a halfspace with $\Psi^{\prime}\supsetneq\Phi_{0}$ tightly and $\Psi\supseteq\Psi^{\prime}$. Clearly, $g^{-1}\Phi_{r}\supsetneq\Psi^{\prime}$. Therefore, $\{g^{-1}\Psi^{\prime}\}\cup g^{-1}\gamma>\{\Psi^{\prime}\}\cup\gamma$. Applying $g$ yields $\{\Psi^{\prime}\}\cup\gamma>\{g\Psi^{\prime}\}\cup g\gamma$ which means $\{\Psi^{\prime}\}\cup\gamma$ is $g$-nested and thereby $\gamma$ not maximal $g$-nested.

The proof of the second part is symmetric. ∎

The following lemma, overlooked in [FFT16], will be the key:

Lemma 13**.**

Let $\alpha=\left\{\Phi_{0},...,\Phi_{r}\right\}$ be a segment in $A^{+}_{g}$ and $h\in G$ such that $h\overline{\alpha}\subset A^{+}_{g}$. Then either $h\overline{\alpha}>\alpha$ or $\alpha>h\overline{\alpha}$.

Proof.

Note that for $0\leq k\leq r$, exactly one of $\Phi_{k}\supsetneq h\overline{\Phi}_{k}$ or $h\overline{\Phi}_{k}\supsetneq\Phi_{k}$ must hold, because they are nested ($h$ is non-transverse) and equality would amount to an inversion ($\Phi_{k}\supsetneq h\Phi_{k}$ and $h\Phi_{k}\supsetneq\Phi_{k}$ are impossible, as $\alpha,h\overline{\alpha}\subset A^{+}_{g}$).

We may assume $0<r$ as the case $0=r$ is trivial. If the Lemma were false, then $\Phi_{0}\supsetneq h\overline{\Phi}_{0}$ and $h\overline{\Phi}_{r}\supsetneq\Phi_{r}$. There must be some $0\leq i<r$ such that $\Phi_{i}\supsetneq h\overline{\Phi}_{i}$ and $h\overline{\Phi}_{i+1}\supsetneq\Phi_{i+1}$. Let us show that all four possible relations between $h\overline{\Phi}_{i+1}$ and $\Phi_{i}$ are impossible, leading to a contradiction:

1. (a)

$h\overline{\Phi}_{i+1}=\Phi_{i}$: By RAAG-like (iv) (see Definition 3), this is impossible. 2. (b)

$h\overline{\Phi}_{i+1}\supsetneq\Phi_{i}$: Then $h\overline{\Phi}_{i+1}\supsetneq\Phi_{i}\supsetneq h\overline{\Phi}_{i}$, which contradicts that $h\overline{\Phi}_{i+1}$ and $h\overline{\Phi}_{i}$ are tightly nested. 3. (c)

$\Phi_{i}\supsetneq h\overline{\Phi}_{i+1}$: Then $\Phi_{i}\supsetneq h\overline{\Phi}_{i+1}\supsetneq\Phi_{i+1}$, which contradicts that $\Phi_{i}$ and $\Phi_{i+1}$ are tightly nested. 4. (d)

$h\overline{\Phi}_{i+1}$ and $\Phi_{i}$ are transverse: Then RAAG-like (iii) is violated.

∎

And we are ready to show, that no reverse copies of maximal $g$-nested segments occur:

Lemma 14**.**

Let $\gamma=\left\{\Phi_{0},...,\Phi_{r}\right\}$ in $A^{+}_{g}$ maximal $g$-nested. Then there is no $h\in G$ such that $h\overline{\gamma}\subset A^{+}_{g}$ and $h\overline{\Phi}_{r}\in[o,go]$.

Proof.

By Lemma 13 there are two cases: either (1) $h\overline{\gamma}>\gamma$ or (2) $\gamma>h\overline{\gamma}$.

Case (1): If $h\overline{\gamma}>g^{-1}\gamma$, then $o\notin h\overline{\Phi}_{r}$ clearly cannot hold. If $g^{-1}\gamma>h\overline{\gamma}$, then $h\overline{\Phi}_{r}$ is in $[o,go]$ and contains $\Phi_{0}$, but is not transverse to $g^{-1}\Phi_{r}$, in contradiction to Lemma 12.

Case (2): If $g\gamma>h\overline{\gamma}$, then $go\in h\overline{\Phi}_{r}$ clearly cannot hold. If $h\overline{\gamma}>g\gamma$, then $h\overline{\Phi}_{r}$ is in $[o,go]$ and contained in $\Phi_{r}$, but is not transverse to $g\Phi_{0}$, in contradiction to Lemma 12. ∎

Tying things together gives us the following:

Theorem 15**.**

Let $\gamma$ be maximal $g$-nested in $[o,go]$. Then $\overline{\phi}_{\gamma}(g)\geq 1$.

Proof.

Let $\gamma=\left\{\Phi_{0},...,\Phi_{r}\right\}$. For $n>0$, $\left\{g^{0}\gamma,...,g^{n}\gamma\right\}$ is non-overlapping and in $[o,go]$. Therefore, $c_{\gamma}(g^{n})\geq n$.

On the other hand, if $h\overline{\gamma}\subset[o,g^{n}o]$, then $g^{-m}h\overline{\Phi}_{r}\in[o,go]$ (and $g^{-m}h\overline{\gamma}\subset A^{+}$) for some $m\in\mathbb{N}$. But this contradicts Lemma 14. Hence, $c_{\overline{\gamma}}(g^{n})=0$.

Now

[TABLE]

∎

An application of the Bavard Duality yields the main result:

Corollary 16**.**

Let $G$ be a group with a RAAG-like action on a CAT(0) cube complex. Then any element acting non-trivially has $\mathrm{scl}(g)\geq\frac{1}{24}$. In particular RAAG-like groups have a stable commutator length gap of $1/24$.

Proof.

By Lemma 4 every element of $G$ is hyperbolic. By the Theorem 15 there is a quasimorphism $\overline{\phi}_{\gamma}$ with $\overline{\phi}_{\gamma}(g)\geq 1$ that has defect $\leq 12$ by Lemma 10. By the Bavard Duality (Theorem 2) the corollary follows. ∎