Linearity of Generalized Cactus Groups

The notion of cactus group can be generalized by replacing symmetric groups with general Coxeter groups. Geometrically, it is a special case of the group $A$ introduced by David, Januszkiewicz and Scott as the group of all lifts of the $W$-action on the blow-up $\Sigma_{\#}$ of a reflection tiling $\Sigma$ to its universal cover in [DJS03].

Let $(W,S)$ be a finitely generated Coxeter group and $(m_{st})_{s,t\in S}$ be its associated Coxeter matrix. A spherical subgroup of $W$ is a finite subgroup $W_{I}$ generated by some $I\subset S.$ We denote its longest element by $w_{I}$ following [Hum90]. A nonempty subset $I\subset S$ is connected if $W_{I}$ is finite and there is no way to write $I=J\cup K$ such that $J$ and $K$ are nonempty and $W_{I}=W_{J}\times W_{K}.$ Let $\mathcal{F}(S)\subset\mathcal{P}(S)$ denote the family of all connected subsets of $S.$

Since $w_{J}$ is the longest element of $W_{J}$, it follows that $w_{J}(I)$ is still a subset of $S$ for $I\subset J.$ In particular if $W_{I\cup J}=W_{I}\times W_{J},$ then $w_{J}(I)=I$ and $\gamma_{I}$ commutes with $\gamma_{J}.$

Similar to the case of cactus group, there is a homomorphism $g_{W}:C_{W}\to W$ given by $\gamma_{I}\mapsto w_{I}.$ Indeed, $w_{I}^{2}=1$ since it is the longest element in $W_{I}$. If $W_{I}\times W_{J}=W_{I\cup J},$ then $w_{I}$ commutes with $w_{J}$. When $I\subset J,$ there is $w_{J}(I)\subset J$ and the longest element in $W_{w_{J}(I)}=w_{J}(W_{I})$ is $w_{w_{J}(I)}=w_{J}w_{I}w_{J}.$ Call the kernel of $g_{W}$ the generalized pure cactus group $PC_{W}.$

Our main theorem in this section is that the generalized pure cactus group can be embedded into a right-angled Coxeter group, generalizing the approach taken by Mostovoy in [[]Proposition 2]Mos:

We set up a technical lemma before giving the proof. Let $(W,S)$ be a right-angled Coxeter group and $G$ a subgroup of $\operatorname{Aut}(W,S).$ Let $I$ be a subset of $S$ and $\{g_{i}\}_{i\in I}$ a family of involutions in $G$ satisfying

1. (1)

   $g_{i}(s)\in S$ for all $i\in I$ and $s\in S.$

2. (2)

   Given $i,j\in I$ with $ij=ji,$ either

   $$g_{i}(j)\in I,g_{j}(i)=i,\text{ and }g_{g_{i}(j)}=g_{i}g_{j}g_{i},$$

   or

   $$g_{j}(i)\in I,g_{i}(j)=j,\text{ and }g_{g_{j}(i)}=g_{j}g_{i}g_{j}.$$

3. (3)

   $g_{i}(i)=i$ for all $i\in I.$

Let $H$ be the subgroup of $W\rtimes G$ generated by $\{ig_{i}\}_{i\in I}.$ Let $L$ be the group generated by $l_{i}$ for all $i\in I$ subject to relations

1. (1)

   $l_{i}l_{j}=l_{g_{i}(j)}l_{i}$ when $ij=ji,g_{j}(i)=i,g_{i}(j)\in I,$ and $g_{g_{i}(j)}=g_{i}g_{j}g_{i},$

2. (2)

   $l_{i}^{2}=1$ for all $i\in I.$

Now we give the proof of Theorem 2.3:

As an example, the usual cactus group $J_{n}$ defined on the Coxeter group $S_{n}$ coincides with the corresponding generalized cactus group $C_{n}.$ It follows from Theorem 2.3 that $J_{n}$ embeds into $D_{n}\rtimes S_{n}$ where $D_{n}$ is the diagram group and $C_{n}$ embeds into $\mathbb{W}\rtimes\operatorname{Aut}(\mathbb{W},\mathbb{S})$ for an appropriate Coxeter system $(\mathbb{W},\mathbb{S}),$ see [Mos19]. Furthermore, there is a bijection $f$ from $\mathbb{S}$ to the set of subsets $I\subset\{1,2,\cdots,n\}$ with $|I|\geq 2$ given by

$$W_{\{s_{i},s_{i+1},\cdots,s_{j}\}}\mapsto\{i,i+1,\cdots,j+1\}$$

and

$$wW_{I}w^{-1}\mapsto w(f(W_{I})).$$

It follows that there is a group isomorphism $\mathbb{W}\simeq D_{n}$ and a commutative diagram

Let $(W,S)$ be a Coxeter system and $M$ its associated Coxeter matrix. A representation of $W\rtimes\operatorname{Aut}(W,S)$ can be constructed as follows. Let $E=\mathbb{R}^{S}$ with standard basis $\{\varepsilon_{s}\}_{s\in S}.$ Define a bilinear form $B_{t}$ on $E$ by

$$B_{t}(\varepsilon_{s},\varepsilon_{v})=\begin{cases}-\cos\left(\pi/m_{s,v}\right)&m_{s,v}<\infty\\ -t&m_{s,v}=\infty\end{cases}$$

For any $t>0,$ there is a representation $\pi$ of $W$ (in particular the Tits representation when $t=1$) generated by $s\mapsto\sigma_{s}$ where

$$\sigma_{s}(x)=x-2B_{t}(x,\varepsilon_{s})\varepsilon_{s}.$$

To show $\pi$ is a representation, it suffices to show $\sigma_{s}\sigma_{v}$ has order $m_{s,v}$ for $s,v\in S.$ Clearly $\sigma_{s}$ preserves the bilinear form $B_{t}.$ Let $\lambda=B_{t}(e_{s},e_{v}).$ Then

$$(\sigma_{s}\sigma_{v})(e_{s})=\sigma_{s}(e_{s}-2\lambda e_{v})=-e_{s}-2\lambda(e_{v}-2\lambda e_{s})=(4\lambda^{2}-1)e_{s}-2\lambda e_{v}$$

and $(\sigma_{s}\sigma_{v})(e_{v})=-e_{v}+2\lambda e_{s}.$ If $m_{s,v}=\infty,$ then $\lambda=-t.$ For all $t\geq 1,$ the equation $k^{2}+2\lambda k+1=0$ has a positive real root $k=k_{0}>0.$ Set $r=4\lambda^{2}+2k_{0}\lambda-1$ and assume $r\neq 0$ (this is true for all but finitely many $t$). Then

$$(\sigma_{s}\sigma_{v})(e_{s}+k_{0}e_{v})=(4\lambda^{2}+2\lambda k_{0}-1)e_{s}-(2\lambda+k_{0})e_{v}=r(e_{s}+k_{0}e_{v}).$$

Hence using the second equation which can be rewritten into $(\sigma_{s}\sigma_{v})(e_{v})=2\lambda(e_{s}+k_{0}e_{v})-(2\lambda k_{0}+1)e_{v},$ we can obtain

$$(\sigma_{s}\sigma_{v})^{m}(e_{v})=a_{m}(e_{s}+k_{0}e_{v})+(-2\lambda k_{0}-1)^{m}e_{v}$$

where $a_{1}=2\lambda$ and $a_{m+1}=ra_{m}+2\lambda(-2\lambda k_{0}-1)^{m}.$ If $t>1,$ then $-2\lambda k_{0}-1=k_{0}^{2}>1$ and $(\sigma_{s}\sigma_{v})^{m}$ can never be the identity. If $t=1,$ then $a_{m}=2m$ and $(\sigma_{s}\sigma_{v})^{m}$ cannot be the identity, either. Thus $\sigma_{s}\sigma_{v}$ has order $\infty.$

When $m_{s,v}<\infty,$ we use the identification $\text{span}(e_{s},e_{v})\approx\mathbb{C}$ by identifying $e_{s}$ to 1 and $e_{v}$ to $-e^{-i\theta},$ where $\theta=\pi/m_{s,v}.$ Then $\sigma_{s}\sigma_{v}$ acts by a rotation of $2\pi/m_{s,v}$ and thus has order $m_{s,v}.$ Hence $\pi$ is a representation.