On rigidity of Coxeter systems up to finite twists and separations of Coxeter generating sets

Tetsuya Hosaka Department of Mathematics, Faculty of Science, Shizuoka University, 836 Ohya, Suruga-ku, Shizuoka, 422-8529, Japan hosaka.tetsuya@shizuoka.ac.jp

(Date: June 20, 2023)

Abstract.

In this paper, we study the twist-conjecture for Coxeter systems and rigidity of Coxeter systems up to finite twists. For Coxeter systems $(W,R)$ and $(W,S)$ , under the untangle-condition for conjugate subsets, we investigate separations and type(I) and type(II) subsets of $R$ and $S$ and give an equivalent condition of $R$ and $S$ that are conjugate up to finite twists. We provide one direction of approach to solving the twist-conjecture and the isomorphism problem for Coxeter groups of finite ranks.

Key words and phrases:

Coxeter groups; Coxeter systems; the isomorphism problem for Coxeter groups; rigidity of Coxeter systems; twists of Coxeter systems

2010 Mathematics Subject Classification:

20F55

This work was partly supported by JSPS KAKENHI Grant Number JP18K03273.

1. Introduction and preliminaries

The purpose of this paper is to investigate the twist-conjecture for Coxeter systems and rigidity of Coxeter systems up to finite twists. Definitions and details of Coxeter groups and Coxeter systems are found in [1] and [11]. In this paper, we suppose that all Coxeter groups are of finite ranks. The isomorphism problem for Coxeter groups is open.

Problem 1.1 (The isomorphism problem for Coxeter groups).

For given Coxeter systems $(W,S)$ and $(W^{\prime},S^{\prime})$ , find an algorithm to determine whether the Coxeter groups $W$ and $W^{\prime}$ are isomorphic or not.

For a Coxeter system $(W,S)$ and a subset $T$ of $S$ , $W_{T}$ is defined as the subgroup generated by $T$ in $W$ . It is well known that $(W_{T},T)$ also becomes a Coxeter system. A subset $T$ of $S$ is said to be spherical, if the subgroup $W_{T}$ is finite. If $T$ is a spherical subset of $S$ , then it is known that there exists a unique longest length element in $W_{T}$ that is denoted by $w_{T}$ . For a subset $T$ of $S$ , we denote $T^{\perp}:=\{s\in S:st=ts\ \text{for all}\ t\in T\}$ . For a Coxeter group $W$ , a subset $R$ of $W$ is called a Coxeter generating set for $W$ , if $(W,R)$ is a Coxeter system.

Recent research on the isomorphism problem for Coxeter groups and rigidity of Coxeter groups and Coxeter systems can be found in [2], [4], [5], [10], [12], [14], [15], [16], [17] and [18].

A Coxeter system $(W,S)$ is said to be reflection-rigid, if it is determined by the Coxeter group $W$ and the set of reflections

{\mathcal{R}}_{S}:=\{wsw^{-1}\in W:w\in W,\ s\in S\}=S^{W}

up to isomorphisms; that is, for any Coxeter generating set $S^{\prime}$ for $W$ such that the reflection sets ${\mathcal{R}}_{S}$ and ${\mathcal{R}}_{S^{\prime}}$ are equal, two Coxeter systems $(W,S)$ and $(W,S^{\prime})$ are isomorphic.

Also a Coxeter system $(W,S)$ is said to be strongly-reflection-rigid, if $S$ is determined by the Coxeter group $W$ and the reflection set ${\mathcal{R}}_{S}$ up to conjugate; that is, for any Coxeter generating set $S^{\prime}$ for $W$ such that the reflection sets ${\mathcal{R}}_{S}$ and ${\mathcal{R}}_{S^{\prime}}$ are equal, two Coxeter generating sets $S$ and $S^{\prime}$ are conjugate in $W$ .

Let $W$ be a Coxeter group. Two Coxeter generating sets $S$ and $S^{\prime}$ for $W$ are said to be angle-compatible [5] (or sharp-angled [12]), if for any $\{a,b\}\subset S$ as $o(ab)<\infty$ there exists $\{a^{\prime},b^{\prime}\}\subset S^{\prime}$ as $o(a^{\prime}b^{\prime})<\infty$ such that $\{a,b\}$ and $\{a^{\prime},b^{\prime}\}$ are conjugate in $W$ . Here for $w\in W$ , $o(w)$ is the order of $w$ in the group $W$ . We say that a Coxeter system $(W,S)$ is angle-rigid, if for any Coxeter generating set $S^{\prime}$ for $W$ such that $S$ and $S^{\prime}$ are angle-compatible, two Coxeter systems $(W,S)$ and $(W,S^{\prime})$ are isomorphic. Also a Coxeter system $(W,S)$ is said to be strongly-angle-rigid, if for any Coxeter generating set $S^{\prime}$ for $W$ such that $S$ and $S^{\prime}$ are angle-compatible, two Coxeter generating sets $S$ and $S^{\prime}$ are conjugate in $W$ .

It is known that if two Coxeter generating sets $S$ and $S^{\prime}$ for $W$ are angle-compatible then the reflection sets ${\mathcal{R}}_{S}$ and ${\mathcal{R}}_{S^{\prime}}$ are equal (see [5]). Hence if a Coxeter system $(W,S)$ is (strongly-)reflection-rigid, then it is (strongly-)angle-rigid.

Recently, Marquis and Mühlherr have proved the following.

Theorem 1.2 ([12, Corollary 1.1]).

The isomorphism problem for Coxeter groups is solved as soon as the following problem is solved.

Problem 1.3 ([12]).

Let $(W,R)$ be a Coxeter system. Find all Coxeter generating sets $S\subset R^{W}$ such that $S$ is sharp-angled with respect to $R$ .

Thus if for a given Coxeter system $(W,R)$ we can output all Coxeter generating sets $S$ for $W$ such that $R$ and $S$ are angle-compatible, then the isomorphism problem for Coxeter groups is solved.

Caprace and Przytycki have given the following.

Theorem 1.4 ([5, Theorem 1.1]).

Let $S$ and $R$ be angle-compatible Coxeter generating sets for a group $W$ . If $S$ is twist-rigid, then $S$ and $R$ are conjugate.

Hence, every twist-rigid Coxeter system is strongly-angle-rigid.

Here $(W,S)$ (and $S$ ) is said to be twist-rigid, if $(W,S)$ has no elementary twist [5]. We give a definition and detail of a “twist” later.

Definition 1.5.

Let $(W,S)$ be a Coxeter system and let $U$ be a subset of $S$ . We consider the irreducible decomposition

W_{U}=W_{U_{1}}\times\cdots\times W_{U_{n}},

where $U=U_{1}\cup\cdots\cup U_{n}$ is a disjoint union and each $(W_{U_{i}},U_{i})$ is an irreducible Coxeter system. Here each $W_{U_{i}}$ is either finite or infinite. We define

	$\displaystyle U_{\sigma}:=\bigcup\{U_{i}:\text{$W_{U_{i}}$ is finite}\}\ \text{and}$
	$\displaystyle U_{\nu}:=\bigcup\{U_{i}:\text{$W_{U_{i}}$ is infinite}\}.$

Then $U=U_{\sigma}\cup U_{\nu}$ is a disjoint union, $W_{U_{\sigma}}$ is finite, $W_{U_{\nu}}$ is infinite (if $U_{\nu}$ is non-empty), and $W_{U}=W_{U_{\sigma}}\times W_{U_{\nu}}$ .

Let $(W,S)$ be a Coxeter system. We say that a subset $U$ of $S$ is a spherical-product subset of $S$ , if $U$ is non-empty and $U\subset\sigma\cup\sigma^{\perp}$ for some non-empty spherical subset $\sigma$ of $S$ . Here if $U$ is a spherical-product subset of $S$ , then either

(1)

$U_{\sigma}=U$ and $U_{\nu}=\emptyset$ (where $U$ is spherical),
(2)

$U_{\sigma}\neq\emptyset$ and $U_{\nu}\neq\emptyset$ (where $U=U_{\sigma}\cup U_{\nu}\subset U_{\sigma}\cup(U_{\sigma})^{\perp}$ ), or
(3)

$U_{\sigma}=\emptyset$ and $U_{\nu}=U$ (where for a spherical subset $\sigma$ of $S$ such that $U\subset\sigma\cup\sigma^{\perp}$ , if we put $\overline{U}:=\sigma\cup U$ then $\overline{U}_{\sigma}=\sigma$ and $\overline{U}_{\nu}=U$ ).

A subset $T$ of $S$ is said to be connected, if for any $a,b\in T$ as $a\neq b$ , there exists a sequence $a=t_{1},t_{2},\ldots,t_{n}=b$ in $T$ such that $o(t_{i}t_{i+1})$ is finite for any $i=1,\ldots,n-1$ (that is, the nerve of $(W_{T},T)$ is connected). Also we often say that a spherical-product subset $U$ separates $S$ , if $S$ is connected and $S-U$ is not connected (that is, the nerve $L_{U}$ of $(W_{U},U)$ separates the connected nerve $L$ of $(W,S)$ to some at least two components).

If a spherical-product subset $U$ of $S$ separates $S$ , then for some non-empty subsets $X$ and $Y$ of $S$ ,

(1)

$S-U=X\cup Y$ that is a disjoint union and
(2)

$o(xy)=\infty$ for any $x\in X$ and $y\in Y$ .

There is a spherical subset $\sigma$ of $S$ such that $U\subset\sigma\cup\sigma^{\perp}$ . Here if $U_{\sigma}$ is non-empty then we take $\sigma:=U_{\sigma}$ . Then $w_{\sigma}Uw_{\sigma}=U$ and we obtain a Coxeter generating set for $W$ as

S^{\prime}:=X\cup U\cup(w_{\sigma}Yw_{\sigma})

that is an elementary twist ([2], [3], [5], [13], [14]). Also in the case that $S$ is not connected, we can consider that $U=\emptyset$ separates $S$ . If $S=X\cup Y$ that is a disjoint union and $o(xy)=\infty$ for any $x\in X$ and $y\in Y$ , then for any $w\in W_{X}$ we obtain a Coxeter generating set $S^{\prime}:=X\cup(wYw^{-1})$ for $W$ that is also an elementary twist ([2], [3], [5], [6], [14]).

Hence, $(W,S)$ is twist-rigid if and only if $S$ is connected and any spherical-product subset $U$ does not separate $S$ .

More generally, let $U$ be a spherical-product subset of $S$ that separates $S$ and let $X$ and $Y$ be non-empty subsets of $S$ such that

(1)

$S-U=X\cup Y$ that is a disjoint union,
(2)

$o(xy)=\infty$ for any $x\in X$ and $y\in Y$ and
(3)

$wUw^{-1}=U$ for some $w\in W_{X\cup U}$ .

Then we obtain a Coxeter generating set for $W$ as

S^{\prime}:=X\cup U\cup(wYw^{-1})

that is called a $($ general- $)$ twist. Here there is a possibility that every twist can be denoted by some elementary-twists. The author does not know whether this always holds.

In this paper, we often say that a Coxeter generating set $S^{\prime}$ is obtained from $S$ by some finite twists, if there exists a sequence $S=S_{1},S_{2},\ldots,S_{n}=S^{\prime}$ of Coxeter generating sets for $W$ such that each $S_{i+1}$ is obtained from $S_{i}$ by some twist. We also say that Coxeter generating sets $R$ and $S$ for $W$ are conjugate up to finite twists, if there exists a Coxeter generating set $R^{\prime}$ for $W$ such that $R^{\prime}$ is obtained from $R$ by some finite twists and $R^{\prime}$ is conjugate to $S$ . Here Coxeter generating sets $R$ and $S$ for $W$ are conjugate up to finite twists if and only if there exist Coxeter generating sets $R^{\prime}$ and $S^{\prime}$ for $W$ such that $R^{\prime}$ and $S^{\prime}$ are obtained from $R$ and $S$ by some finite twists respectively and $R^{\prime}$ is conjugate to $S^{\prime}$ .

Now we define new concepts “separations” and “untangle-conjugate” on Coxeter systems. These definitions are technical. The purpose of this paper is to give a further reduction of the isomorphism problem for Coxeter groups. These definitions are designed and constructed as the main results in this paper hold and as two Coxeter generating sets with some conditions using them are conjugate up to finite twists.

First, the following lemma is known.

Lemma 1.6 (cf. [2]).

Let $(W,S)$ and $(W,S^{\prime})$ be Coxeter systems. For each maximal spherical subset $T$ of $S$ , there exists a unique maximal spherical subset $T^{\prime}$ of $S^{\prime}$ such that $W_{T}$ and $W_{T^{\prime}}$ are conjugate in $W$ .

Here we say that two Coxeter systems $(W,S)$ and $(W,S^{\prime})$ are maximal-spherical-subset-compatible, if for each maximal spherical subset $T$ of $S$ , there exists a maximal spherical subset $T^{\prime}$ of $S^{\prime}$ such that $T$ and $T^{\prime}$ are conjugate in $W$ (where $T^{\prime}$ is unique by Lemma 1.6). If two Coxeter systems are maximal-spherical-subset-compatible then they are angle-compatible. Indeed for each $\{a,b\}\subset S$ as $o(ab)<\infty$ , there exists a maximal spherical subset $T$ of $S$ containing $\{a,b\}$ . If $T^{\prime}$ is a maximal spherical subset of $S^{\prime}$ such that $T$ and $T^{\prime}$ are conjugate in $W$ then some subset $\{a^{\prime},b^{\prime}\}\subset T^{\prime}$ is conjugate to $\{a,b\}$ .

For a Coxeter system $(W,S)$ as $S$ is connected and for $A\subset S$ , we say that $A$ is a twist-rigid subset of $S$ , if $A$ is connected and if there does not exist a spherical-product subset $U$ of $S$ such that $U$ separates $S$ and $U\cap A$ separates $A$ . Here we note that $(W_{A},A)$ need not be a twist-rigid Coxeter system in general (see examples in Section 2).

Let $(W,S)$ be a Coxeter system. We suppose that $S$ is connected. Let ${\mathcal{A}}$ be a set of subsets of $S$ such that

(1)

$S=\bigcup_{A\in\mathcal{A}}A$ ,
(2)

each $A\in{\mathcal{A}}$ is connected and a union of some maximal twist-rigid subsets of $S$ , and
(3)

for each maximal twist-rigid subset $A_{0}$ of $S$ , there exists a unique element $A\in{\mathcal{A}}$ such that $A_{0}\subset A$ .

A subset $U$ of $S$ is called a separator of ${\mathcal{A}}$ , if the following conditions (i)–(v) hold:

(i)

$U$ is a spherical-product subset of $S$ .
(ii)

$U$ separates $S$ .
(iii)

For any $A\in{\mathcal{A}}$ , there exists a unique $j\in\{1,\ldots,t\}$ such that $A\subset\overline{X}_{j}$ . Here $S-U=X_{1}\cup\cdots\cup X_{t}$ and $X_{1},\ldots,X_{t}$ are the connected components of $S-U$ and we define $\overline{X}_{j}:=\bigcup\{A\in{\mathcal{A}}:A\subset X_{j}\cup U\}$ for each $j=1,\ldots,t$ .
(iv)

There exist $A_{1},A_{2}\in{\mathcal{A}}$ such that $A_{1}\cap A_{2}=U$ , $A_{1}\subset\overline{X}_{j_{1}}$ and $A_{2}\subset\overline{X}_{j_{2}}$ for some $j_{1},j_{2}\in\{1,\ldots,t\}$ as $j_{1}\neq j_{2}$ .
(v)

Let $j\in\{1,\ldots,t\}$ . For any distinct elements $A_{1},\ldots,A_{n}\in{\mathcal{A}}$ such that $U\cap\overline{X}_{j}\subset A_{1}\cup\cdots\cup A_{n}\subset\overline{X}_{j}$ , if $(A_{1}\cup\cdots\cup A_{i})\cap A_{i+1}$ is maximal in

$\{(A_{1}\cup\cdots\cup A_{i})\cap A:A\in{\mathcal{A}}-\{A_{1},\ldots,A_{i}\}\}$

for any $i=1,\ldots,n-1$ , then $U\cap\overline{X}_{j}\subset A_{i}$ for some $i\in\{1,\ldots,n\}$ .

A set ${\mathcal{A}}$ of subsets of $S$ is called a separation of $S$ , if the following conditions (1)–(4) hold:

(1)

$S=\bigcup_{A\in\mathcal{A}}A$ .
(2)

Each $A\in{\mathcal{A}}$ is connected and a union of some maximal twist-rigid subsets of $S$ .
(3)

For each maximal twist-rigid subset $A_{0}$ of $S$ , there exists a unique element $A\in{\mathcal{A}}$ such that $A_{0}\subset A$ .
(4)
For distinct elements $A_{1},A_{2},\ldots,A_{n}\in{\mathcal{A}}$ , if $(A_{1}\cup\cdots\cup A_{i})\cap A_{i+1}$ is maximal in

$\{(A_{1}\cup\cdots\cup A_{i})\cap A:A\in{\mathcal{A}}-\{A_{1},\ldots,A_{i}\}\}$

for any $i=1,\ldots,n-1$ , then for each $i=1,\ldots,n-1$ ,
1. (a)
  
  $U_{i}:=(A_{1}\cup\cdots\cup A_{i})\cap A_{i+1}$ is a separator of ${\mathcal{A}}$ , and
2. (b)
  
  $A_{1}\cup\cdots\cup A_{i}\subset X_{j_{1}}\cup U_{i}$ and $A_{i+1}\subset X_{j_{2}}\cup U_{i}$ for some $j_{1},j_{2}\in\{1,\ldots,t\}$ as $j_{1}\neq j_{2}$ , where $S-U_{i}=X_{1}\cup\cdots\cup X_{t}$ and $X_{1},\ldots,X_{t}$ are the connected components of $S-U_{i}$ .

In the case that $S$ is not connected, a set ${\mathcal{A}}$ of subsets of $S$ is called a separation of $S$ , if for the connected components $S_{1},\ldots,S_{k}$ of $S$ (that is, each $S_{i}$ is a maximal connected subset of $S$ and $S=S_{1}\cup\cdots\cup S_{k}$ is a disjoint union), ${\mathcal{A}}_{i}:=\{A\in{\mathcal{A}}:A\subset S_{i}\}$ is a separation of $S_{i}$ for any $i=1,\ldots,k$ .

For two sets ${\mathcal{A}}_{1}$ and ${\mathcal{A}}_{2}$ of subsets of $S$ , we denote ${\mathcal{A}}_{1}\preceq{\mathcal{A}}_{2}$ , if for any $A_{1}\in{\mathcal{A}}_{1}$ , $A_{1}\subset A_{2}$ for some $A_{2}\in{\mathcal{A}}_{2}$ .

A separation ${\mathcal{A}}$ of $S$ is said to be minimal, if

there does not exist a separation ${\mathcal{A}}^{\prime}$ of $S$ such that ${\mathcal{A}}^{\prime}\neq{\mathcal{A}}$ and any $A^{\prime}\in{\mathcal{A}}^{\prime}$ is contained in some $A\in{\mathcal{A}}$ .

A minimal separation is “minimal” with respect to the partial order “ $\preceq$ ”.

Let $(W,S)$ be a Coxeter system and let ${\mathcal{A}}$ be a set of subsets of $S$ . We consider a twist

S^{\prime}:=X\cup U\cup(wYw^{-1})

where $U$ is a spherical-product subset of $S$ and $w\in W_{X\cup U}$ such that $wUw^{-1}=U$ , $U$ separates $S$ as $S-U=X\cup Y$ that is a disjoint union and $o(xy)=\infty$ for any $x\in X$ and $y\in Y$ . A twist $S^{\prime}$ is said to be preserving ${\mathcal{A}}$ , if for any $A\in{\mathcal{A}}$ , $A\subset X\cup U$ or $A\subset Y\cup U$ (that is, there does not exist $A\in{\mathcal{A}}$ such that $A\cap X\neq\emptyset$ and $A\cap Y\neq\emptyset$ ). If the twist $S^{\prime}$ is preserving ${\mathcal{A}}$ then

{\mathcal{A}}^{\prime}:=\{A:A\in{\mathcal{A}},\ A\subset X\cup U\}\cup\{wAw^{-1}:A\in{\mathcal{A}},\ A\not\subset X\cup U\}

is a set of subsets of $S^{\prime}$ and ${\mathcal{A}}^{\prime}$ is called the set induced by ${\mathcal{A}}$ and the twist.

Here if ${\mathcal{A}}$ is a separation of $S$ , then the induced set ${\mathcal{A}}^{\prime}$ is a separation of $S^{\prime}$ and it is called the separation induced by ${\mathcal{A}}$ and the twist.

Let ${\mathcal{A}}$ be a set of subsets of $S$ . We often say that $S^{\prime\prime}$ is obtained from $S$ by some finite twists preserving ${\mathcal{A}}$ , if there exists a sequence $S=S_{1},S_{2},\ldots,S_{n}=S^{\prime\prime}$ of Coxeter generating sets for $W$ such that $S_{i+1}$ is a twist of $S_{i}$ preserving ${\mathcal{A}}_{i}$ for each $i=1,\ldots,n-1$ where ${\mathcal{A}}_{1}:={\mathcal{A}}$ and ${\mathcal{A}}_{i}$ is the set of subsets of $S_{i}$ induced by ${\mathcal{A}}_{i-1}$ and the twist for each $i=2,\ldots,n-1$ .

Definition 1.7.

Let $(W,R)$ and $(W,S)$ be Coxeter systems and let ${\mathcal{A}}$ and ${\mathcal{B}}$ be separations of $R$ and $S$ respectively. We say that $(W,R)$ and $(W,S)$ are compatible on separations ${\mathcal{A}}$ and ${\mathcal{B}}$ , if

(i)

each $A\in{\mathcal{A}}$ is conjugate to some unique $B\in{\mathcal{B}}$ and
(ii)

each $B\in{\mathcal{B}}$ is conjugate to some unique $A\in{\mathcal{A}}$ .

We define that two Coxeter systems $(W,R)$ and $(W,S)$ are some-separation-compatible, if $(W,R)$ and $(W,S)$ are compatible on some separations ${\mathcal{A}}$ and ${\mathcal{B}}$ of $R$ and $S$ respectively.

Every spherical subset $T$ of $S$ is not separated by any spherical(-product) subset $\sigma$ of $S$ . Hence $T$ is a twist-rigid subset of $S$ and $T\subset A$ for some maximal twist-rigid subset $A$ of $S$ .

If two Coxeter systems are some-separation-compatible, then they are maximal-twist-rigid-subset-compatible and maximal-spherical-subset-compatible, hence they are angle-compatible.

We define and investigate type(I)-type(II)-compatible later.

In this paper, we study when are Coxeter generating sets conjugate up to finite twists under “the untangle-condition” on conjugate subsets.

Let $(W,S)$ be a Coxeter system. We define “untangle-conjugate”.

Definition 1.8.

For spherical subsets $\sigma$ , $\tau$ and $T$ of $S$ as $\sigma\cup\tau\subset T$ , we denote $\displaystyle\sigma\mathop{\simeq}_{w_{T}}\tau$ , if $w_{T}\sigma w_{T}=\tau$ .

Let $U$ and $U^{\prime}$ be non-empty subsets of $S$ . They are uniquely denoted by $U=U_{\sigma}\cup U_{\nu}$ and $U^{\prime}=U^{\prime}_{\sigma}\cup U^{\prime}_{\nu}$ . For a spherical subset $T$ of $S$ as $U_{\sigma}\cup U^{\prime}_{\sigma}\subset T$ , we denote $\displaystyle U\mathop{\simeq}_{w_{T}}U^{\prime}$ , if $w_{T}U_{\sigma}w_{T}=U^{\prime}_{\sigma}$ , $U_{\nu}=U^{\prime}_{\nu}$ and $st=ts$ for any $s\in U_{\nu}$ and $t\in T$ .

Here if $\displaystyle U\mathop{\simeq}_{w_{T}}U^{\prime}$ , then $U$ and $U^{\prime}$ are conjugate and

	$\displaystyle w_{T}Uw_{T}$	$\displaystyle=w_{T}(U_{\sigma}\cup U_{\nu})w_{T}=(w_{T}U_{\sigma}w_{T})\cup(w_{T}U_{\nu}w_{T})$
		$\displaystyle=U^{\prime}_{\sigma}\cup U_{\nu}=U^{\prime}.$

Also if

\displaystyle U=U_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}U_{2}\mathop{\simeq}_{\,\,w_{T_{2}}}\cdots{}\mathop{\simeq}_{\ \ w_{T_{q-1}}}U_{q}=U^{\prime},

then $U$ and $U^{\prime}$ are conjugate and for $w:=w_{T_{q-1}}\cdots w_{T_{2}}w_{T_{1}}$ ,

	$\displaystyle wUw^{-1}$	$\displaystyle=w_{T_{q-1}}\cdots w_{T_{2}}w_{T_{1}}U_{1}w_{T_{1}}w_{T_{2}}\cdots w_{T_{q-1}}$
		$\displaystyle=w_{T_{q-1}}\cdots w_{T_{2}}U_{2}w_{T_{2}}\cdots w_{T_{q-1}}$
		$\displaystyle=\cdots$
		$\displaystyle=w_{T_{q-1}}U_{q-1}w_{T_{q-1}}$
		$\displaystyle=U_{q}=U^{\prime}.$

We say that conjugate subsets $U$ and $U^{\prime}$ of $S$ are untangle, if the following statements (1) and (2) hold:

(1)

In the case that $U\neq U^{\prime}$ , there exist a sequence $U_{1},\cdots,U_{q}$ of subsets of $S$ and a sequence $T_{1},\cdots,T_{q-1}$ of spherical subsets of $S$ such that

\displaystyle U=U_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}U_{2}\mathop{\simeq}_{\,\,w_{T_{2}}}\cdots{}\mathop{\simeq}_{\ \ w_{T_{q-1}}}U_{q}=U^{\prime}.

(2)

In the case that $U=U^{\prime}$ , for $w\in W$ as $wUw^{-1}=U$ , we define the bijective map $f_{w}:U\to U$ by $f_{w}(a)=waw^{-1}$ for any $a\in U$ . For any $w\in W$ as $wUw^{-1}=U$ , $f_{w}=f_{1}={\rm id}_{U}$ for $1\in W$ , or, there exist a sequence $U_{1},\cdots,U_{q}$ of subsets of $S$ and a sequence $T_{1},\cdots,T_{q-1}$ of spherical subsets of $S$ such that

$\displaystyle U=U_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}U_{2}\mathop{\simeq}_{\,\,w_{T_{2}}}\cdots{}\mathop{\simeq}_{\ \ w_{T_{q-1}}}U_{q}=U$

and $f_{w}=f_{w_{0}}$ for $w_{0}:=w_{T_{q-1}}\cdots w_{T_{2}}w_{T_{1}}$ .

Now we define “the untangle-conjugate-condition” and “the untangle-condition”.

Definition 1.9.

We say that a Coxeter system $(W,S)$ has the untangle-conjugate-condition, if all conjugate subsets $U$ and $U^{\prime}$ of $S$ are untangle and the following holds:

( $*$ )

Let $R$ be a Coxeter generating set for $W$ , let $S_{1},\ldots,S_{n}$ be the connected components of $S$ and let $R_{1},\ldots,R_{n}$ be the connected components of $R$ (that is, $W=W_{S_{1}}*\cdots*W_{S_{n}}=W_{R_{1}}*\cdots*W_{R_{n}}$ ). If $S_{i}$ and $R_{i}$ are conjugate in $W$ for any $i=1,\ldots,n$ , then there exists a Coxeter generating set $R^{\prime}$ for $W$ such that $R^{\prime}$ is obtained from $R$ by some finite twists and $S$ and $R^{\prime}$ are conjugate.

Also we say that a Coxeter system $(W,S)$ has the untangle-condition, if all Coxeter systems obtained from $(W,S)$ by some finite twists have the untangle-conjugate-condition.

Example 1.10.

We consider a Coxeter system $(W,S)$ defined by Figure 1 (i) or (ii).

Refer to caption — Figure 1. Example 1.10

Here $S$ is the vertex set of the figure and $o(ss)=1$ for any $s\in S$ . If two vertices $s,t\in S$ as $s\neq t$ do not span any edge in the figure then we define $o(st)=\infty$ . If two vertices $s,t\in S$ as $s\neq t$ span an edge numbering $m$ then define $o(st)=m$ .

Let $\sigma=\sigma_{1}:=\{a_{1},b_{1}\}$ , $\sigma_{2}:=\{a_{2},b_{1}\}$ , $\sigma_{3}:=\{a_{2},b_{2}\}$ , $\sigma_{4}:=\{a_{2},b_{3}\}$ and $\tau=\sigma_{5}:=\{a_{3},b_{3}\}$ . Also let $T_{1}:=\{a_{1},a_{2},b_{1}\}$ , $T_{2}:=\{a_{2},b_{1},b_{2}\}$ , $T_{3}:=\{a_{2},b_{2},b_{3}\}$ and $T_{4}:=\{a_{2},a_{3},b_{3}\}$ .

Then $\sigma$ and $\tau$ are untangle-conjugate and

\displaystyle\sigma=\sigma_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}\sigma_{2}\mathop{\simeq}_{\,\,w_{T_{2}}}\sigma_{3}\mathop{\simeq}_{\,\,w_{T_{3}}}\sigma_{4}\mathop{\simeq}_{\,\,w_{T_{4}}}\sigma_{5}=\tau

in both cases (i) and (ii).

In the case (ii), for $T_{0}:=\sigma=\{a_{1},b_{1}\}$ ,

\displaystyle\sigma\mathop{\simeq}_{\,\,w_{T_{0}}}\sigma.

Then $w_{T_{0}}\sigma w_{T_{0}}=\sigma$ . Here $w_{T_{0}}a_{1}w_{T_{0}}=b_{1}$ and $w_{T_{0}}b_{1}w_{T_{0}}=a_{1}$ . Hence $f_{w_{T_{0}}}:\sigma\to\sigma$ is the bijective map such that $f_{w_{T_{0}}}(a_{1})=b_{1}$ and $f_{w_{T_{0}}}(b_{1})=a_{1}$ .

We can see [3] and [8] on conjugate spherical subsets.

Example 1.11.

We consider a Coxeter system $(W,S)$ defined by Figure 2.

Here we identify $a_{1}=b_{5}$ and $b_{1}=a_{5}$ . (The nerve of $(W,S)$ is a triangulation of the Möbius band.)

Let $\sigma=\sigma_{1}:=\{a_{1},b_{1}\}=\{b_{5},a_{5}\}$ , $\sigma_{2}:=\{a_{2},b_{1}\}$ , $\sigma_{3}:=\{a_{2},b_{2}\}$ , $\sigma_{4}:=\{a_{3},b_{2}\}$ , $\sigma_{5}:=\{a_{3},b_{3}\}$ , $\sigma_{6}:=\{a_{4},b_{3}\}$ , $\sigma_{7}:=\{a_{4},b_{4}\}$ and $\sigma_{8}:=\{a_{5},b_{4}\}$ . Also let $T_{1}:=\{a_{1},a_{2},b_{1}\}$ , $T_{2}:=\{a_{2},b_{1},b_{2}\}$ , $T_{3}:=\{a_{2},a_{3},b_{2}\}$ , $T_{4}:=\{a_{3},b_{2},b_{3}\}$ , $T_{5}:=\{a_{3},a_{4},b_{3}\}$ , $T_{6}:=\{a_{4},b_{3},b_{4}\}$ , $T_{7}:=\{a_{4},a_{5},b_{4}\}$ and $T_{8}:=\{a_{5},b_{4},b_{5}\}$ .

Then

\displaystyle\sigma=\sigma_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}\sigma_{2}\mathop{\simeq}_{\,\,w_{T_{2}}}\sigma_{3}\mathop{\simeq}_{\,\,w_{T_{3}}}\sigma_{4}\mathop{\simeq}_{\,\,w_{T_{4}}}\sigma_{5}\mathop{\simeq}_{\,\,w_{T_{5}}}\sigma_{6}\mathop{\simeq}_{\,\,w_{T_{6}}}\sigma_{7}\mathop{\simeq}_{\,\,w_{T_{7}}}\sigma_{8}\mathop{\simeq}_{\,\,w_{T_{8}}}\sigma_{1}=\sigma.

For $w_{0}:=w_{T_{8}}w_{T_{7}}w_{T_{6}}w_{T_{5}}w_{T_{4}}w_{T_{3}}w_{T_{2}}w_{T_{1}}$ , we have that $w_{0}\sigma w_{0}^{-1}=\sigma$ . Here $w_{0}a_{1}w_{0}^{-1}=b_{1}$ and $w_{0}b_{1}w_{0}^{-1}=a_{1}$ .

For example, if $w=1$ , $w=a_{1}$ , $w=b_{1}$ , $w=w_{\sigma}=a_{1}b_{1}$ , $w=w_{0}$ , $w=w_{0}^{2}$ , or $w=a_{1}w_{0}a_{1}b_{1}w_{0}$ , then $w\sigma w^{-1}=\sigma$ .

For any $w\in W$ as $w\sigma w^{-1}=\sigma$ , if [ $wa_{1}w^{-1}=a_{1}$ and $wb_{1}w^{-1}=b_{1}$ ] then $f_{w}=f_{1}$ for $1\in W$ . Also if [ $wa_{1}w^{-1}=b_{1}$ and $wb_{1}w^{-1}=a_{1}$ ] then $f_{w}=f_{w_{0}}$ as above.

The author does not know whether there is an example of non-untangle-conjugate subsets of a Coxeter generating set.

After some preliminaries in Sections 2 and 3, we prove the following theorem in Section 4.

Theorem 1.12.

Let $(W,R)$ and $(W,S)$ be Coxeter systems with the untangle-condition. If $(W,R)$ and $(W,S)$ are some-separation-compatible, then $R$ and $S$ are conjugate up to finite twists.

We obtain a corollary from Theorem 1.12.

Corollary 1.13.

For Coxeter systems $(W,R)$ and $(W,S)$ with the untangle-condition, the following statements are equivalent $:$

(i)

$R$ and $S$ are conjugate up to finite twists.
(ii)

$(W,R_{0})$ and $(W,S)$ are some-separation-compatible for some Coxeter generating set $R_{0}$ obtained from $R$ by finite twists.

Now we define “type(I)” and “type(II)” subsets and “the standard-separation” of a Coxeter generating set.

Definition 1.14.

Let $(W,S)$ be a Coxeter system. For each minimal separation ${\mathcal{A}}$ of $S$ , we define ${\mathcal{U}}_{\mathcal{A}}$ as the set of separators of ${\mathcal{A}}$ . Let

\overline{\mathcal{U}}:=\bigcap\{{\mathcal{U}}_{\mathcal{A}}:\text{${\mathcal{A}}$ is a minimal separation of $S$}\}.

Then we define the separation $\widetilde{\mathcal{A}}_{S}$ of $S$ by $\overline{\mathcal{U}}$ as follows.

Let ${\mathcal{A}}_{0}$ be the set of maximal twist-rigid subsets of $S$ . For $A,A^{\prime}\in{\mathcal{A}}_{0}$ , we denote $A\sim A^{\prime}$ if any $U\in\overline{\mathcal{U}}$ does not separate $A$ and $A^{\prime}$ ; that is, there exist a minimal separation ${\mathcal{A}}$ of $S$ and an element $\overline{A}\in{\mathcal{A}}$ such that $A\cup A^{\prime}\subset\overline{A}$ . Then “ $\sim$ ” is an equivalence relation on the set ${\mathcal{A}}_{0}$ . Let $[A]:=\{A^{\prime}\in{\mathcal{A}}_{0}:A^{\prime}\sim A\}$ that is the equivalence class for $A\in{\mathcal{A}}_{0}$ . Here

{\mathcal{A}}_{0}/\!\!\sim\;=\{[A]:A\in{\mathcal{A}}_{0}\}=\{[A_{1}],\ldots,[A_{n}]\}

for some $A_{1},\ldots,A_{n}\in{\mathcal{A}}_{0}$ as $[A_{i}]\neq[A_{j}]$ if $i\neq j$ . Let

\overline{A}_{i}:=\bigcup[A_{i}]=\bigcup\{A\in{\mathcal{A}}_{0}:A\sim A_{i}\}

for each $i=1,\ldots,n$ . Then we define

\widetilde{\mathcal{A}}_{S}:=\{\overline{A}_{1},\ldots,\overline{A}_{n}\}.

In Section 5, we show that $\widetilde{\mathcal{A}}_{S}$ is a separation of $S$ . We say that $\widetilde{\mathcal{A}}_{S}$ is the standard separation of $S$ .

A subset $A$ of $S$ is said to be type(I), if $A\in{\mathcal{A}}$ holds for any minimal separation ${\mathcal{A}}$ of $S$ . Let $\widetilde{\mathcal{A}}_{S}^{\rm(I)}$ be the set of type(I) subsets of $S$ ; that is,

\widetilde{\mathcal{A}}_{S}^{\rm(I)}=\{A\in\widetilde{\mathcal{A}}_{S}:A\in{\mathcal{A}}\ \text{for any minimal separation ${\mathcal{A}}$ of $S$}\}.

We also define

\widetilde{\mathcal{A}}_{S}^{\rm(II)}:=\widetilde{\mathcal{A}}_{S}-\widetilde{\mathcal{A}}_{S}^{\rm(I)}

and every element $A\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ is called a type(II) subset of $S$ .

Then $\widetilde{\mathcal{A}}_{S}=\widetilde{\mathcal{A}}_{S}^{\rm(I)}\cup\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ is a disjoint union. Here $\widetilde{\mathcal{A}}_{S}$ , $\widetilde{\mathcal{A}}_{S}^{\rm(I)}$ and $\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ are uniquely determined by the Coxeter system $(W,S)$ .

Let $(W,S)$ be a Coxeter system. We say that for a subset $A$ of $S$ , $A^{\prime}$ is a twist of $A$ that induces some twist of $S$ , if there exist a spherical-product subset $U$ of $S$ and $w\in W$ such that $U\subset A$ , $U$ separates $A$ , $A^{\prime}$ is a twist of $A$ obtained by $U$ and $w$ , $U$ separates $S$ and some twist $S^{\prime}$ of $S$ is obtained by $U$ and $w$ . We also say that $A^{\prime\prime}$ is obtained from $A$ by some finite twists that induces some twist of $S$ , if there exist a sequence $A=A_{1},A_{2},\ldots,A_{n}=A^{\prime\prime}$ and a sequence $S=S_{1},S_{2},\ldots,S_{n}=S^{\prime\prime}$ of Coxeter generating sets for $W$ such that $A_{i}\subset S_{i}$ for any $i=1,\ldots,n$ and $A_{i+1}$ is a twist of $A_{i}$ that induces a twist $S_{i+1}$ of $S_{i}$ for each $i=1,\ldots,n-1$ .

We define “type(II)-compatible”.

Definition 1.15.

Let $(W,R)$ and $(W,S)$ be Coxeter systems. We say that $A\in\widetilde{\mathcal{A}}_{R}^{\rm(II)}$ and $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ are type(II)-compatible, if

there exists $A_{0}$ obtained from $A$ by some finite twists that induce some twists of $R$ preserving $\widetilde{\mathcal{A}}_{R}-\{A\}$ such that $A_{0}$ and $B$ are conjugate in $W$ .

This means that the type(II) subsets $A\in\widetilde{\mathcal{A}}_{R}^{\rm(II)}$ and $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ are conjugate up to finite twists that induce some twists of $R$ preserving $\widetilde{\mathcal{A}}_{R}-\{A\}$ .

Now we define “type(I)-type(II)-compatible”.

Definition 1.16.

Two Coxeter systems $(W,R)$ and $(W,S)$ are said to be type(I)-type(II)-compatible, if for the standard separations $\widetilde{\mathcal{A}}_{R}$ and $\widetilde{\mathcal{A}}_{S}$ of $R$ and $S$ respectively,

(i)

each $A\in\widetilde{\mathcal{A}}_{R}^{\rm(I)}$ is conjugate to some unique $B\in\widetilde{\mathcal{A}}_{S}^{\rm(I)}$ ,
(ii)

each $B\in\widetilde{\mathcal{A}}_{S}^{\rm(I)}$ is conjugate to some unique $A\in\widetilde{\mathcal{A}}_{R}^{\rm(I)}$ ,
(iii)

for each $A\in\widetilde{\mathcal{A}}_{R}^{\rm(II)}$ , there exists a unique $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ such that $A$ and $B$ are type(II)-compatible, and
(iv)

for each $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ , there exists a unique $A\in\widetilde{\mathcal{A}}_{R}^{\rm(II)}$ such that $B$ and $A$ are type(II)-compatible.

In Section 6, we show that if two Coxeter systems with the untangle-condition are type(I)-type(II)-compatible, then they are some-separation-compatible up to finite twists. Thus, if two Coxeter systems $(W,R)$ and $(W,S)$ with the untangle-condition are type(I)-type(II)-compatible, then $R$ and $S$ are conjugate up to finite twists by Theorem 1.12.

We show the following theorem in Section 6.

Theorem 1.17.

For Coxeter systems $(W,R)$ and $(W,S)$ with the untangle-condition, the following two statements are equivalent $:$

(i)

$R$ and $S$ are conjugate up to finite twists.
(ii)

$(W,R)$ and $(W,S)$ are type(I)-type(II)-compatible.

For given Coxeter systems $(W,R)$ and $(W,S)$ , it seems that to consider whether

(a)

$(W,R)$ and $(W,S)$ are type(I)-type(II)-compatible

is more simple than to consider whether

(b)

$(W,R_{0})$ and $(W,S)$ are some-separation-compatible for some Coxeter generating set $R_{0}$ obtained from $R$ by finite twists.

In Theorem 1.17, we use two conditions as “the untangle-condition” and “type(I)-type(II)-compatible” for Coxeter systems.

Problem 1.18.

The untangle-conjugate-condition will always hold for all Coxeter systems $(W,S)$ ?

Problem 1.19.

Angle-compatible Coxeter systems $(W,R)$ and $(W,S)$ will be always type(I)-type(II)-compatible?

If there exist counter-examples of angle-compatible Coxeter systems $(W,R)$ and $(W,S)$ such that any $R_{0}$ and $S_{0}$ obtained from $R$ and $S$ by some finite twists respectively are not type(I)-type(II)-compatible, or a Coxeter system $(W,S)$ that does not have the untangle-conjugate-condition, then they are meaningful examples.

If Problems 1.18 and 1.19 both can be solved affirmatively, then the twist-conjecture for Coxeter systems can be solved affirmatively from Theorem 1.17. Also if Problem 1.3 can be solved from the twist-conjecture for Coxeter systems affirmatively, then the isomorphism problem for Coxeter groups of finite ranks can be solved by Theorem 1.2.

2. Remarks and examples on separations

We introduce some remarks and examples on separations of Coxeter generating sets.

Remark 2.1.

Let $(W,S)$ be a Coxeter system as $S$ is connected. Let ${\mathcal{A}}$ be a separation of $S$ and let $n:=|{\mathcal{A}}|$ . Then we may denote ${\mathcal{A}}=\{A_{1},A_{2},\ldots,A_{n}\}$ such that for each $i=1,\ldots,n-1$ , $U_{i}:=(A_{1}\cup\cdots\cup A_{i})\cap A_{i+1}$ is maximal in

\{(A_{1}\cup\cdots\cup A_{i})\cap A_{j}:j=i+1,\ldots,n\}.

Here by the definition of a separation of $S$ , $U_{i}$ is a separator of ${\mathcal{A}}$ and it is a spherical-product subset of $S$ that separates $S$ for any $i=1,\ldots,n-1$ . Hence $W$ has a structure as

W=(\cdots((W_{A_{1}}*_{W_{U_{1}}}W_{A_{2}})*_{W_{U_{2}}}W_{A_{3}})*_{W_{U_{3}}}\cdots)*_{W_{U_{n-1}}}W_{A_{n}}.

Remark 2.2.

Let $(W,S)$ be a Coxeter system as $S$ is connected and let $U$ be a spherical-product subset of $S$ that separates $S$ . Let ${\mathcal{A}}_{0}$ be the set of maximal twist-rigid subsets of $S$ . Suppose that $X_{1},\ldots,X_{t}$ are the connected components of $S-U$ and $S-U=X_{1}\cup\cdots\cup X_{t}$ is a disjoint union. Let

		$\displaystyle\overline{X}_{i}:=\bigcup\{A\in{\mathcal{A}}_{0}:A\subset X_{i}\cup U\ \text{and}\ A\not\subset U\}$
for each $i=1,\ldots,t$ and let
		$\displaystyle\overline{Y}:=\bigcup\{A\in{\mathcal{A}}_{0}:A\subset U\}.$

Here each $\overline{X}_{i}$ (and $\overline{Y}$ ) is connected and it is a union of some maximal twist-rigid subsets of $S$ (if $\overline{Y}$ is non-empty). Let ${\mathcal{A}}:=\{\overline{X}_{1},\ldots,\overline{X}_{t},\overline{Y}\}$ if $\overline{Y}$ is non-empty, and let ${\mathcal{A}}:=\{\overline{X}_{1},\ldots,\overline{X}_{t}\}$ if $\overline{Y}$ is empty. Then ${\mathcal{A}}$ is a separation of $S$ . We say that ${\mathcal{A}}$ is the induced separation of $S$ by $U$ .

Let ${\mathcal{A}}^{\prime}$ be a minimal separation of $S$ such that ${\mathcal{A}}^{\prime}\preceq{\mathcal{A}}$ . Then $U$ does not separate any $A\in{\mathcal{A}}^{\prime}$ . Thus we can obtain a minimal separation ${\mathcal{A}}^{\prime}$ of $S$ from a spherical-product subset $U$ of $S$ that separates $S$ such that $U$ does not separate any $A\in{\mathcal{A}}^{\prime}$ .

Let $\{U_{j}\}$ be the set of spherical-product subsets of $S$ that separate $S$ . By the above argument, we obtain the separation ${\mathcal{A}}_{j}$ of $S$ induced by each $U_{j}$ . If $\{U_{j}\}$ is non-empty (that is, if $(W,S)$ is not a twist-rigid Coxeter system), then for any minimal separation ${\mathcal{A}}^{\prime}$ of $S$ , ${\mathcal{A}}^{\prime}\preceq{\mathcal{A}}_{j}$ for some $j$ . Hence, from considering the set $\{U_{j}\}$ of spherical-product subsets that separate $S$ and the induced separations ${\mathcal{A}}_{j}$ , we can obtain all minimal separations of $S$ .

We give some examples.

Example 2.3.

We consider Coxeter systems $(W_{i},S_{i})$ ( $i=1,2,3,4$ ) defined by Figure 3.

Here $S_{1}$ , $S_{2}$ , $S_{3}$ and $S_{4}$ are the vertex sets of the figures $(1)$ , $(2)$ , $(3)$ and $(4)$ respectively in Figure 3. Let $i\in\{1,2,3,4\}$ . We define $o(ss)=1$ for any $s\in S_{i}$ . If two vertices $s,t\in S_{i}$ as $s\neq t$ do not span any edge in the figure then we define $o(st)=\infty$ . If two vertices $s,t\in S_{i}$ as $s\neq t$ span an edge then we consider that $o(st)=o(ts)$ has some (arbitrary) finite number at least $2$ . Then a Coxeter system $(W_{i},S_{i})$ is obtained.

(1) Let $U:=\{a_{1},a_{2}\}$ , $X_{1}:=\{x_{1},x_{2}\}$ and $X_{2}:=\{y_{1},y_{2}\}$ in $(W_{1},S_{1})$ . Then $U$ is a spherical-product subset of $S_{1}$ that separates $S_{1}$ and $S_{1}-U=X_{1}\cup X_{2}$ where $X_{1}$ and $X_{2}$ are the connected components of $S_{1}-U$ . Then $\overline{X}_{1}=\{x_{1},x_{2},a_{1},a_{2}\}$ and $\overline{X}_{2}=\{y_{1},y_{2},a_{1},a_{2}\}$ . Here ${\mathcal{A}}:=\{\overline{X}_{1},\overline{X}_{2}\}$ is the separation of $S_{1}$ induced by $U$ , and $U$ is a separator of ${\mathcal{A}}$ .

(2) Let $U:=\{a_{1},a_{2}\}$ , $X_{1}:=\{x_{1},x_{2}\}$ and $X_{2}:=\{y_{1},y_{2},y_{3}\}$ in $(W_{2},S_{2})$ . Then $U$ is a spherical-product subset of $S_{2}$ that separates $S_{2}$ and $S_{2}-U=X_{1}\cup X_{2}$ where $X_{1}$ and $X_{2}$ are the connected components of $S_{2}-U$ . Then $\overline{X}_{1}=\{x_{1},x_{2},a_{1},a_{2}\}$ and $\overline{X}_{2}=\{y_{1},y_{2},y_{3},a_{1}\}$ . Here ${\mathcal{A}}:=\{\overline{X}_{1},\overline{X}_{2}\}$ is the separation of $S_{2}$ induced by $U$ , and $U^{\prime}:=\{a_{1}\}$ is a separator of ${\mathcal{A}}$ (here $U=\{a_{1},a_{2}\}$ is not a separator of ${\mathcal{A}}$ ).

(3) Let $U:=\{a_{1}\}$ , $X_{1}:=\{x\}$ , $X_{2}:=\{y\}$ and $X_{3}:=\{z\}$ in $(W_{3},S_{3})$ . Then $U$ is a spherical-product subset of $S_{3}$ that separates $S_{3}$ and $S_{3}-U=X_{1}\cup X_{2}\cup X_{3}$ where $X_{1}$ , $X_{2}$ and $X_{3}$ are the connected components of $S_{3}-U$ . Then $\overline{X}_{1}=\{x,a_{1}\}$ , $\overline{X}_{2}=\{y,a_{1}\}$ and $\overline{X}_{3}=\{z,a_{1}\}$ . Here ${\mathcal{A}}:=\{\overline{X}_{1},\overline{X}_{2},\overline{X}_{3}\}$ is the separation of $S_{3}$ induced by $U$ , and $U$ is a separator of ${\mathcal{A}}$ .

(4) Let $U:=\{a_{1},a_{2}\}$ , $X_{1}:=\{x\}$ and $X_{2}:=\{y\}$ in $(W_{4},S_{4})$ . Then $U$ is a spherical-product subset of $S_{4}$ that separates $S_{4}$ and $S_{4}-U=X_{1}\cup X_{2}$ where $X_{1}$ and $X_{2}$ are the connected components of $S_{4}-U$ . Then $\overline{X}_{1}=\{x,a_{1}\}$ , $\overline{X}_{2}=\{y,a_{1}\}$ and $\overline{Y}=\{a_{1},a_{2}\}$ . Here ${\mathcal{A}}:=\{\overline{X}_{1},\overline{X}_{2},\overline{Y}\}$ is the separation of $S_{4}$ induced by $U$ , and $U^{\prime}:=\{a_{1}\}$ is a separator of ${\mathcal{A}}$ (here $U=\{a_{1},a_{2}\}$ is not a separator of ${\mathcal{A}}$ ).

Example 2.4.

We consider a Coxeter system $(W,S)$ defined by Figure 4.

Here $S$ is the vertex set of the figure. Let $o(ss)=1$ for any $s\in S$ . If two vertices $s,t\in S$ as $s\neq t$ do not span any edge in the figure then we define $o(st)=\infty$ . If two vertices $s,t\in S$ as $s\neq t$ span an edge numbering $2$ then define $o(st)=2$ . Also if two vertices $s,t\in S$ as $s\neq t$ span an edge with no numbering then we consider that $o(st)=o(ts)$ has some (arbitrary) finite number at least $2$ . Then a Coxeter system $(W,S)$ is obtained.

Let $A_{i}$ be the set of $4$ -vertices around “ $A_{i}$ ” in the figure and let $B_{i}$ be the set of $4$ -vertices around “ $B_{i}$ ” in the figure for each $i=1,2,3,4$ .

(a)

The set of maximal twist-rigid subsets of $S$ is

$\{A_{1},\ A_{2},\ A_{3},\ A_{4},\ B_{1},\ B_{2},\ B_{3},\ B_{4}\}.$

Here

$D:=\{d_{1},\,d_{2},\,d_{3},\,d_{4},\,d_{5},\,d_{6},\,d_{7},\,d_{8}\}$

is not a twist-rigid subset of $S$ . Indeed for example $U:=\{d_{2},\,d_{0},\,d_{6}\}$ is a spherical-product subset of $S$ that separates $S$ and $U\cap D=\{d_{2},\,d_{6}\}$ separates $D$ . We also note that $(W_{D},D)$ is a twist-rigid Coxeter system.
(b)
The minimal separations of $S$ are the 4-sets as
1. (1)
  
  $\{A_{1},\ A_{2},\ A_{3},\ A_{4},\ B_{1}\cup B_{2},\ B_{3}\cup B_{4}\}$ ,
2. (2)
  
  $\{A_{1},\ A_{2},\ A_{3},\ A_{4},\ B_{1}\cup B_{4},\ B_{2}\cup B_{3}\}$ ,
3. (3)
  
  $\{A_{1},\ A_{2},\ A_{3},\ A_{4},\ B_{1}\cup B_{3},\ B_{2},\ B_{4}\}$ and
4. (4)
  
  $\{A_{1},\ A_{2},\ A_{3},\ A_{4},\ B_{1},\ B_{3},\ B_{2}\cup B_{4}\}$ .
(c)

The set of type(I) subsets of $S$ is

$\widetilde{\mathcal{A}}_{S}^{\rm(I)}=\{A_{1},\ A_{2},\ A_{3},\ A_{4}\}.$
(d)

The set of type(II) subsets of $S$ is

$\widetilde{\mathcal{A}}_{S}^{\rm(II)}=\{B_{1}\cup B_{2}\cup B_{3}\cup B_{4}\}.$

(e)

The standard separation of $S$ is

	$\displaystyle\widetilde{\mathcal{A}}_{S}$	$\displaystyle=\widetilde{\mathcal{A}}_{S}^{\rm(I)}\cup\widetilde{\mathcal{A}}_{S}^{\rm(II)}$
		$\displaystyle=\{A_{1},\ A_{2},\ A_{3},\ A_{4},\ B_{1}\cup B_{2}\cup B_{3}\cup B_{4}\}.$

Example 2.5.

We consider a Coxeter system $(W,S)$ defined by Figure 5.

Here $S$ is the vertex set of the figure and $o(st)$ is defined as in Example 2.4.

Let $A_{i}$ be the set of $4$ -vertices around “ $A_{i}$ ” in the figure for each $i=1,2,3,4,5,6,7,8$ .

(a)

The set of maximal twist-rigid subsets of $S$ is

$\{A_{1},\ A_{4},\ A_{5},\ A_{8},\ A_{2}\cup A_{6},\ A_{3}\cup A_{7}\}.$
(b)
The minimal separations of $S$ are the 9-sets as
1. (1)
  
  $\{A_{1}\cup A_{5},\ A_{2}\cup A_{6},\ A_{3}\cup A_{7},\ A_{4}\cup A_{8}\}$ ,
2. (2)
  
  $\{A_{1}\cup A_{5},\ A_{2}\cup A_{6},\ A_{3}\cup A_{7}\cup A_{8},\ A_{4}\}$ ,
3. (3)
  
  $\{A_{1}\cup A_{5},\ A_{2}\cup A_{6},\ A_{3}\cup A_{4}\cup A_{7},\ A_{8}\}$ ,
4. (4)
  
  $\{A_{1},\ A_{2}\cup A_{5}\cup A_{6},\ A_{3}\cup A_{7},\ A_{4}\cup A_{8}\}$ ,
5. (5)
  
  $\{A_{1},\ A_{2}\cup A_{5}\cup A_{6},\ A_{3}\cup A_{7}\cup A_{8},\ A_{4}\}$ ,
6. (6)
  
  $\{A_{1},\ A_{2}\cup A_{5}\cup A_{6},\ A_{3}\cup A_{4}\cup A_{7},\ A_{8}\}$ ,
7. (7)
  
  $\{A_{1}\cup A_{2}\cup A_{6},\ A_{5},\ A_{3}\cup A_{7},\ A_{4}\cup A_{8}\}$ ,
8. (8)
  
  $\{A_{1}\cup A_{2}\cup A_{6},\ A_{5},\ A_{3}\cup A_{7}\cup A_{8},\ A_{4}\}$ and
9. (9)
  
  $\{A_{1}\cup A_{2}\cup A_{6},\ A_{5},\ A_{3}\cup A_{4}\cup A_{7},\ A_{8}\}$ .
(c)

The set $\widetilde{\mathcal{A}}_{S}^{\rm(I)}$ of type(I) subsets of $S$ is empty.
(d)

The set of type(II) subsets of $S$ is

$\widetilde{\mathcal{A}}_{S}^{\rm(II)}=\{B_{1},\ B_{2}\}$

where $B_{1}:=A_{1}\cup A_{2}\cup A_{5}\cup A_{6}$ and $B_{2}:=A_{3}\cup A_{4}\cup A_{7}\cup A_{8}$ .
(e)

The standard separation of $S$ is

$\widetilde{\mathcal{A}}_{S}=\{B_{1},\ B_{2}\}.$

Example 2.6.

We consider the Coxeter system $(W,S)$ defined by Figure 6. Here $S:=\{a,\,b_{1},\,b_{2},\,b_{3},\,c\}$ .

Figure 6. Example 2.6

Let $A_{1}:=\{a,\,b_{1},\,c\}$ , $A_{2}:=\{a,\,b_{2},\,c\}$ , $A_{3}:=\{a,\,b_{3},\,c\}$ , $B_{1}:=\{a,\,b_{1},\,b_{2},\,b_{3}\}$ and $B_{2}:=\{b_{1},\,b_{2},\,b_{3},\,c\}$ .

(a)

The set of maximal twist-rigid subsets of $S$ is

$\{\,\{a,\,b_{1}\},\,\{a,\,b_{2}\},\,\{a,\,b_{3}\},\,\{b_{1},\,c\},\,\{b_{2},\,c\},\,\{b_{3},\,c\}\,\}.$
(b)
The minimal separations of $S$ are the 2-sets as
1. (1)
  
  $\{A_{1},\ A_{2},\ A_{3}\}$ and
2. (2)
  
  $\{B_{1},\ B_{2}\}$ .
(c)

The set $\widetilde{\mathcal{A}}_{S}^{\rm(I)}$ of type(I) subsets of $S$ is empty.
(d)

The set of type(II) subsets of $S$ is $\widetilde{\mathcal{A}}_{S}^{\rm(II)}=\{S\}$ .
(e)

The standard separation of $S$ is $\widetilde{\mathcal{A}}_{S}=\{S\}$ .

Example 2.7.

We consider a Coxeter system $(W,S)$ defined by Figure 7. Here $S$ is the vertex set of the figure; that is,

S:=\{a,\,b,\,c,\,d,\,e,\,f,\,g,\,h,\,i,\,j,\,k,\,l\},

and we define $o(ek)\geq 3$ .

Let

	$\displaystyle A_{1}$	$\displaystyle:=\{a,\,b,\,d,\,e\},$
	$\displaystyle A_{2}$	$\displaystyle:=\{b,\,d,\,e,\,g,\,i,\,j,\,k,\,l\},$
	$\displaystyle A_{3}$	$\displaystyle:=\{d,\,e,\,f,\,g\}\ \text{and}$
	$\displaystyle A_{4}$	$\displaystyle:=\{b,\,c,\,e,\,g,\,h\}.$

Then

{\mathcal{A}}:=\{A_{1},\ A_{2},\ A_{3},\ A_{4}\}

is a separation of $S$ . For example,

	$\displaystyle U_{1}:=A_{1}\cap A_{2}\ \text{is maximal in}\ \{A_{1}\cap A_{j}:j=2,3,4\},$
	$\displaystyle U_{2}:=(A_{1}\cup A_{2})\cap A_{3}\ \text{is maximal in}\ \{(A_{1}\cup A_{2})\cap A_{j}:j=3,4\}\ \text{and}$
	$\displaystyle U_{3}:=(A_{1}\cup A_{2}\cup A_{3})\cap A_{4}\ \text{is maximal in}\ \{(A_{1}\cup A_{2}\cup A_{3})\cap A_{j}:j=4\}.$

Also $U_{1}$ , $U_{2}$ and $U_{3}$ are separators of ${\mathcal{A}}$ and they are spherical-product subsets of $S$ that separate $S$ . Here we note that $A_{1}\cap A_{3}$ , $A_{1}\cap A_{4}$ and $A_{3}\cap A_{4}$ are spherical-product subsets that do not separate $S$ .

The spherical-product subsets of $S$ that separate $S$ are the 3-sets as

U_{1}=\{b,\,d,\,e\},\ U_{2}=\{d,\,e,\,g\}\ \text{and}\ U_{3}=\{b,\,e,\,g\}.

Each $U_{i}$ induces the separation ${\mathcal{A}}_{i}$ of $S$ as in Remark 2.2. Then for any minimal separation ${\mathcal{A}}^{\prime}$ of $S$ , ${\mathcal{A}}^{\prime}\preceq{\mathcal{A}}_{i}$ for some $i=1,2,3$ . Here ${\mathcal{A}}\preceq{\mathcal{A}}_{i}$ holds for any $i=1,2,3$ , where ${\mathcal{A}}=\{A_{1},A_{2},A_{3},A_{4}\}$ . Hence ${\mathcal{A}}$ is the unique minimal separation of $S$ . Thus, there are no type(II) subsets of $S$ and

\widetilde{\mathcal{A}}_{S}=\widetilde{\mathcal{A}}_{S}^{\rm(I)}=\{A_{1},\ A_{2},\ A_{3},\ A_{4}\}.

Example 2.8.

We consider a Coxeter system $(W^{\prime},S^{\prime})$ defined by Figure 8. Here $S^{\prime}$ is the vertex set of the figure; that is,

S^{\prime}:=\{a,\,b,\,c,\,d,\,e,\,f,\,g,\,h\}.

Then $W^{\prime}$ can be considered as a standard subgroup of $(W,S)$ in Example 2.7 generated by $S^{\prime}$ .

Let

	$\displaystyle A_{1}$	$\displaystyle:=\{a,\,b,\,d,\,e\},$
	$\displaystyle A_{3}$	$\displaystyle:=\{d,\,e,\,f,\,g\}\ \text{and}$
	$\displaystyle A_{4}$	$\displaystyle:=\{b,\,c,\,e,\,g,\,h\}.$

Then the spherical-product subsets of $S^{\prime}$ that separate $S^{\prime}$ are the 3-sets as

U_{1}=\{b,\,d,\,e\},\ U_{2}=\{d,\,e,\,g\}\ \text{and}\ U_{3}=\{b,\,e,\,g\}.

Each $U_{i}$ induces the separation ${\mathcal{A}}^{\prime}_{i}$ of $S^{\prime}$ as in Remark 2.2. Then for any minimal separation ${\mathcal{A}}^{\prime}$ of $S^{\prime}$ , ${\mathcal{A}}^{\prime}\preceq{\mathcal{A}}^{\prime}_{i}$ for some $i=1,2,3$ . Here we note that the set $\{A_{1},A_{3},A_{4}\}$ is not a separation of $S^{\prime}$ . Then the minimal separations of $S^{\prime}$ are the 3-sets as

	$\displaystyle\{A_{1}\cup A_{3},\ A_{4}\},$
	$\displaystyle\{A_{1},\ A_{3}\cup A_{4}\}\ \text{and}$
	$\displaystyle\{A_{1}\cup A_{4},\ A_{3}\}.$

Hence, there are no type(I) subsets of $S^{\prime}$ and

\widetilde{\mathcal{A}}_{S^{\prime}}=\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(II)}=\{S^{\prime}\}.

Example 2.9.

We consider Coxeter systems $(W,S)$ and $(W,S^{\prime})$ defined by Figure 9. Here $S$ and $S^{\prime}$ are the vertex sets of the corresponding figures and we define $o(a_{1}a_{2})=3$ .

Let $\sigma:=\{a_{1},a_{2}\}$ , $U:=\{a_{1},a_{2},a_{3},a_{4}\}$ , $X:=\{x_{1},x_{2},x_{3}\}$ , $Y:=\{y_{1},y_{2},y_{3}\}$ and $Y^{\prime}:=\{y^{\prime}_{1},y^{\prime}_{2},y^{\prime}_{3}\}$ . Here $U_{\sigma}=\sigma$ , $U_{\nu}=\{a_{3},a_{4}\}$ and $w_{\sigma}U=w_{\sigma}=U$ . Then $S^{\prime}=X\cup U\cup w_{\sigma}Yw_{\sigma}$ is a twist of $S$ by $U$ and $w_{\sigma}$ (where $y^{\prime}_{i}=w_{\sigma}y_{i}w_{\sigma}$ for $i=1,2,3,4$ ).

In $(W,S)$ , let $A_{i}$ be the set of $4$ -vertices around “ $A_{i}$ ” in the figure for each $i=1,2,3,4$ . Then the minimal separations of $S$ are the 9-sets as

(1)

$\{U,\ A_{1}\cup A_{2},\ A_{3}\cup A_{4}\}$ ,
(2)

$\{U,\ A_{1},\ A_{4},\ A_{2}\cup A_{3}\}$ ,
(3)

$\{U,\ A_{1}\cup A_{4},\ A_{2},\ A_{3}\}$ ,
(4)

$\{U\cup A_{1},\ A_{2},\ A_{3}\cup A_{4}\}$ ,
(5)

$\{U\cup A_{2},\ A_{1},\ A_{3}\cup A_{4}\}$ ,
(6)

$\{A_{1}\cup A_{2},\ U\cup A_{3},\ A_{4}\}$ ,
(7)

$\{A_{1}\cup A_{2},\ U\cup A_{4},\ A_{3}\}$ ,
(8)

$\{U\cup A_{1}\cup A_{3},\ A_{2},\ A_{4}\}$ and
(9)

$\{U\cup A_{2}\cup A_{4},\ A_{1},\ A_{3}\}$ .

The Coxeter system $(W,S^{\prime})$ is denoted by Figure 10.

In $(W,S^{\prime})$ , let $A_{i}$ and $B_{i}$ be the sets of $4$ -vertices around “ $A_{i}$ ” and “ $B_{i}$ ” in the figure for each $i=1,2$ respectively. Then the minimal separations of $S$ are the 9-sets as

(1)

$\{A_{1}\cup A_{2},\ U,\ B_{1}\cup B_{2}\}$ ,
(2)

$\{A_{1}\cup A_{2},\ U\cup B_{1},\ B_{2}\}$ ,
(3)

$\{A_{1}\cup A_{2},\ U\cup B_{2},\ B_{1}\}$ ,
(4)

$\{A_{1},\ A_{2}\cup U,\ B_{1}\cup B_{2}\}$ ,
(5)

$\{A_{1},\ A_{2}\cup U\cup B_{1},\ B_{2}\}$ ,
(6)

$\{A_{1},\ A_{2}\cup U\cup B_{2},\ B_{1}\}$ ,
(7)

$\{A_{2},\ A_{1}\cup U,\ B_{1}\cup B_{2}\}$ ,
(8)

$\{A_{2},\ A_{1}\cup U\cup B_{1},\ B_{2}\}$ and
(9)

$\{A_{2},\ A_{1}\cup U\cup B_{2},\ B_{1}\}$ .

Here $U$ is a spherical-product subset that separates $S$ and $S^{\prime}$ both. We consider the separations ${\mathcal{A}}$ and ${\mathcal{A}}^{\prime}$ of $S$ and $S^{\prime}$ induced by $U$ respectively. Then $(W,S)$ and $(W,S^{\prime})$ are compatible on the separations ${\mathcal{A}}$ and ${\mathcal{A}}^{\prime}$ . Thus $(W,S)$ and $(W,S^{\prime})$ are some-separation-compatible.

In $(W,S)$ , there are no type(I) subsets of $S$ and

\widetilde{\mathcal{A}}_{S}=\widetilde{\mathcal{A}}_{S}^{\rm(II)}=\{S\}.

The Coxeter system $(W,S^{\prime})$ is denoted by Figure 10. In $(W,S^{\prime})$ , there are no type(I) subsets of $S^{\prime}$ and

\widetilde{\mathcal{A}}_{S^{\prime}}=\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(II)}=\{S^{\prime}\}.

Then the type(II) subsets $S\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ and $S^{\prime}\in\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(II)}$ are type(II)-compatible. Hence $(W,S)$ and $(W,S^{\prime})$ are type(I)-type(II)-compatible. Here $(W,S)$ and $(W,S^{\prime})$ are not compatible on their standard separations.

3. On separations of Coxeter generating sets and compatible Coxeter systems

Let $(W,S)$ be a Coxeter system. A maximal spherical subset $\sigma$ of $S$ is not separated by any subset of $\sigma$ . Hence every maximal spherical subset $\sigma$ is a twist-rigid subset of $S$ . Also for a maximal spherical subset $\sigma$ of $S$ , there exists a maximal twist-rigid subset $A$ of $S$ such that $\sigma\subset A$ .

We consider that two Coxeter systems $(W,R)$ and $(W,S)$ are maximal-twist-rigid-subset-compatible, if each maximal twist-rigid subset $A$ of $R$ is conjugate to some unique maximal twist-rigid subset $B$ of $S$ and each maximal twist-rigid subset $B$ of $S$ is conjugate to some unique maximal twist-rigid subset $A$ of $R$ .

By the above argument, if two Coxeter systems are maximal-twist-rigid-subset-compatible, then they are maximal-spherical-subset-compatible. Also by the argument in Section 1, if two Coxeter systems are maximal-spherical-subset-compatible, then they are angle-compatible.

Question 3.1.

(i)

Angle-compatible Coxeter systems $(W,R)$ and $(W,S)$ will be maximal-spherical-subset-compatible?
(ii)

Maximal-spherical-subset-compatible Coxeter systems $(W,R)$ and $(W,S)$ will be maximal-twist-rigid-subset-compatible?

The following technical lemma and remark are used in the proof of the main theorem.

Lemma 3.2.

Let $(W,S)$ be a Coxeter system and let ${\mathcal{A}}$ be a separation of $S$ . Suppose that the following statements (1)–(4) hold:

(1)

$A_{1},\ldots,A_{n}\in{\mathcal{A}}$ .
(2)

$(A_{1}\cup\cdots\cup A_{i})\cap A_{i+1}$ is maximal in

$\{(A_{1}\cup\cdots\cup A_{i})\cap C:C\in{\mathcal{A}}-\{A_{1},\ldots,A_{i}\}\}$

for any $i=1,\ldots,n-1$ .
(3)

$\overline{A}_{0}:=A_{1}\cup\cdots\cup A_{n}$ .
(4)

$A\in{\mathcal{A}}-\{A_{1},\ldots,A_{n}\}$ .

Then there exists a subset $U_{0}$ of $\overline{A}_{0}$ such that $U_{0}$ is a separator of ${\mathcal{A}}$ and $U_{0}$ separates $\overline{A}_{0}$ and $A$ ; that is, for the connected components $X_{1},\dots,X_{t}$ of $S-U_{0}$ (where $S-U_{0}=X_{1}\cup\cdots\cup X_{t}$ is a disjoint union), $\overline{A}_{0}\subset X_{i}\cup U_{0}$ and $A\subset X_{j}\cup U_{0}$ for some $i,j\in\{1,\ldots,t\}$ as $i\neq j$ .

Proof.

We suppose that the statements (1)–(4) hold.

Let $A_{n+1}\in{\mathcal{A}}-\{A_{1},\ldots,A_{n}\}$ such that $U:=(A_{1}\cup\cdots\cup A_{n})\cap A_{n+1}$ is maximal in $\{(A_{1}\cup\cdots\cup A_{n})\cap C:C\in{\mathcal{A}}-\{A_{1},\ldots,A_{n}\}\}$ .

If $U$ separates $\overline{A}_{0}$ and $A$ , then we obtain $U_{0}:=U$ and this lemma is proved, since $U$ is a subset of $\overline{A}_{0}$ and $U$ is a separator of ${\mathcal{A}}$ by the definition of a separation.

We suppose that $U$ does not separate $\overline{A}_{0}$ and $A$ .

Let $A_{n+1},\ldots,A_{p}\in{\mathcal{A}}-\{A_{1},\ldots,A_{n}\}$ such that $(A_{1}\cup\cdots\cup A_{i})\cap A_{i+1}$ is maximal in

\{(A_{1}\cup\cdots\cup A_{i})\cap C:C\in{\mathcal{A}}-\{A_{1},\ldots,A_{i}\}\}

for any $i=1,\ldots,n-1,n,\ldots,p-1$ and $(A_{1}\cup\cdots\cup A_{p-1})\cap A_{p}$ separates $\overline{A}_{0}$ and $A$ . Here we may suppose that $p$ is the minimum number as $(A_{1}\cup\cdots\cup A_{p-1})\cap A_{p}$ separates $\overline{A}_{0}$ and $A$ ; that is, $(A_{1}\cup\cdots\cup A_{i-1})\cap A_{i}$ does not separate $\overline{A}_{0}$ and $A$ for any $i=n+1,\ldots,p-1$ .

Let $U:=(A_{1}\cup\cdots\cup A_{p-1})\cap A_{p}$ that is a separator of ${\mathcal{A}}$ and separates $\overline{A}_{0}$ and $A$ . Here if $U\subset\overline{A}_{0}$ then we obtain $U_{0}:=U$ and this lemma is proved.

We suppose that $U\not\subset\overline{A}_{0}$ .

We consider the connected components $Y_{1},\dots,Y_{r}$ of $S-U$ where $S-U=Y_{1}\cup\cdots\cup Y_{r}$ is a disjoint union. We may suppose that $\overline{A}_{0}\subset Y_{1}\cup U$ and $A\subset Y_{2}\cup U$ . By the definition of a separator, there exists $A_{i_{0}}$ ( $1\leq i_{0}\leq p-1$ ) such that $U\subset A_{i_{0}}\subset Y_{1}\cup U$ . Since $U\not\subset\overline{A}_{0}$ , we have that $n+1\leq i_{0}\leq p-1$ .

(a) Since $U$ separates $\overline{A}_{0}$ and $A$ , for any $a\in\overline{A}_{0}$ and $b\in A$ , any path $[a,b]$ form $a$ to $b$ in the nerve of $(W,S)$ intersects $U$ .

(b) Let $V:=(A_{1}\cup\cdots\cup A_{i_{0}-1})\cap A_{i_{0}}$ that is a separator of ${\mathcal{A}}$ and does not separate $\overline{A}_{0}$ and $A$ . We consider the connected components $Z_{1},\dots,Z_{q}$ of $S-V$ where $S-V=Z_{1}\cup\cdots\cup Z_{q}$ is a disjoint union. We may suppose that $\overline{A}_{0}\cup A\subset Z_{1}\cup V$ and $A_{i_{0}}\subset Z_{2}\cup V$ . Here $\overline{A}_{0}\not\subset V$ and $A\not\subset V$ . Hence there exist $a_{1}\in Z_{1}\cap\overline{A}_{0}$ and $b_{1}\in Z_{1}\cap A$ . Since $Z_{1}$ is connected, there exists a path $[a_{1},b_{1}]$ from $a_{1}$ to $b_{1}$ in the nerve of $(W_{Z_{1}},Z_{1})$ . Here $Z_{1}\cap A_{i_{0}}=\emptyset$ , since $A_{i_{0}}\subset Z_{2}\cup V$ . Thus the path $[a_{1},b_{1}]$ does not intersect $A_{i_{0}}$ . In particular, the path $[a_{1},b_{1}]$ does not intersect $U$ , because $U\subset A_{i_{0}}$ . This contradicts (a).

Thus, $U\subset\overline{A}_{0}$ and we obtain $U_{0}:=U$ . This lemma is proved. ∎

The following remark is obtained from Lemma 3.2.

Remark 3.3.

Let $(W,S)$ be a Coxeter system with the untangle-conjugate-condition and let ${\mathcal{A}}$ be a separation of $S$ . We suppose that the following statements (1)–(8) hold:

(1)

$A_{1},\ldots,A_{n}\in{\mathcal{A}}$ .
(2)

$(A_{1}\cup\cdots\cup A_{i})\cap A_{i+1}$ is maximal in

$\{(A_{1}\cup\cdots\cup A_{i})\cap C:C\in{\mathcal{A}}-\{A_{1},\ldots,A_{i}\}\}$

for any $i=1,\ldots,n-1$ .
(3)

$\overline{A}_{0}:=A_{1}\cup\cdots\cup A_{n}$ .
(4)

$A\in{\mathcal{A}}-\{A_{1},\ldots,A_{n}\}$ .
(5)

$U$ is a spherical-product subset of $S$ such that $U_{\sigma}$ is non-empty and $U\subsetneq\overline{A}_{0}$ .
(6)

$U^{\prime}$ is a spherical-product subset of $S$ such that $U^{\prime}\subsetneq A$ .
(7)

$U$ and $U^{\prime}$ are conjugate and $U\neq U^{\prime}$ .
(8)

$|U|\geq\max\{\,|\overline{A}_{0}\cap C|\,:C\in{\mathcal{A}},\ C\not\subset\overline{A}_{0}\}$ .

Then by (7) and the untangle-conjugate-condition, there exist a sequence $U_{1},\cdots,U_{q}$ of subsets of $S$ and a sequence $T_{1},\cdots,T_{q-1}$ of spherical subsets of $S$ such that

\displaystyle U=U_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}U_{2}\mathop{\simeq}_{\,\,w_{T_{2}}}\cdots{}\mathop{\simeq}_{\ \ w_{T_{q-1}}}U_{q}=U^{\prime}.

Here $|U|=|U_{1}|=\cdots=|U_{q}|=|U^{\prime}|$ .

Since $U\neq U^{\prime}$ by (7), $q\geq 2$ . We suppose that $U_{i}\not\subset\overline{A}_{0}$ for any $i\in\{2,\ldots,q\}$ and $T_{i}\not\subset\overline{A}_{0}$ for any $i\in\{1,\ldots,q-1\}$ .

By Lemma 3.2, there exists a subset $U_{0}$ of $\overline{A}_{0}$ such that $U_{0}$ is a separator of ${\mathcal{A}}$ and $U_{0}$ separates $\overline{A}_{0}$ and $A$ .

By the above untangle-conjugate sequence, $F:=T_{1}\cup\cdots\cup T_{q-1}$ is connected. Here $(U_{i})_{\sigma}\cup(U_{i+1})_{\sigma}\subset T_{i}$ for each $i=1,\ldots,q-1$ . Also $U_{1}=U\subsetneq\overline{A}_{0}$ by (5) and $U_{q}=U^{\prime}\subsetneq A$ by (6). Then $U\subset\overline{A}_{0}\cap F$ and $U^{\prime}\subset A\cap F$ . Also $U_{\nu}=U^{\prime}_{\nu}\subset\overline{A}_{0}\cap A$ .

For any $a\in U_{\sigma}$ , $F-(U-\{a\})=(F-U)\cup\{a\}$ is connected and there exists a path from $a$ to some point of $A$ in $F-(U-\{a\})$ , because each $T_{i}$ is spherical and $o(st)<\infty$ for any $s,t\in T_{i}$ .

Hence $U\subset U_{0}$ , since $U_{0}$ separates $\overline{A}_{0}$ and $A$ .

We show that $U_{0}=\overline{A}_{0}\cap C$ for some $C\in{\mathcal{A}}$ as $C\not\subset\overline{A}_{0}$ . Let $X_{1},\ldots,X_{t}$ be the connected components of $S-U_{0}$ and let $\overline{X}_{j}:=\bigcup\{C\in{\mathcal{A}}:C\subset X_{j}\cup U_{0}\}$ for each $j=1,\ldots,t$ . Here $\overline{X}_{j}\cap\overline{X}_{j^{\prime}}\subset U_{0}$ if $j\neq j^{\prime}$ . Since $U_{0}$ is a separator of ${\mathcal{A}}$ , there exist $C_{1},C_{2}\in{\mathcal{A}}$ such that $U_{0}=C_{1}\cap C_{2}$ , $C_{1}\subset\overline{X}_{j_{1}}$ and $C_{2}\subset\overline{X}_{j_{2}}$ for some $j_{1},j_{2}\in\{1,\ldots,t\}$ as $j_{1}\neq j_{2}$ by (iv) in the definition of a separator. Here $\overline{A}_{0}\subset\overline{X}_{j_{0}}$ for some $j_{0}\in\{1,\ldots,t\}$ . Then $j_{0}\neq j_{1}$ or $j_{0}\neq j_{2}$ , since $j_{1}\neq j_{2}$ . We suppose that $j_{0}\neq j_{1}$ . Then $C_{1}\not\subset\overline{A}_{0}$ and $U_{0}=\overline{A}_{0}\cap C_{1}$ , because $U_{0}\subset\overline{A}_{0}$ and $U_{0}=C_{1}\cap C_{2}\subset C_{1}$ (here $\overline{X}_{j_{0}}\cap\overline{X}_{j_{1}}\subset U_{0}$ ). Hence $U_{0}=\overline{A}_{0}\cap C$ for some $C\in{\mathcal{A}}$ as $C\not\subset\overline{A}_{0}$ .

Thus $|U|\geq|U_{0}|$ by (8). Hence we have that $U=U_{0}$ , because $U\subset U_{0}$ .

Then $U$ separates $S$ , and $V:=U\cup T_{1}=T_{1}\cup U_{\nu}$ separates $S$ . Here $st=ts$ for any $s\in U_{\nu}$ and $t\in T_{1}$ .

Let $X$ and $Y$ be subsets of $S$ such that $S-V=X\cup Y$ is a disjoint union, $\overline{A}_{0}\subset X\cup V$ , $A\subset Y\cup V$ and $o(xy)=\infty$ for any $x\in X$ and $y\in Y$ . Then we can obtain a twist

S^{\prime}=X\cup V\cup w_{T_{1}}Yw_{T_{1}}.

Here for each $C\in{\mathcal{A}}$ , $C\subset X\cup V$ or $C\subset Y\cup V$ . Hence the twist $S^{\prime}$ is preserving ${\mathcal{A}}$ and fixing $\overline{A}_{0}$ . Let ${\mathcal{A}}^{\prime}$ be the induced separation of $S^{\prime}$ by ${\mathcal{A}}$ and the twist. Then $w_{T_{1}}U_{2}w_{T_{1}}=U_{1}=U$ . Let $A^{\prime}:=w_{T_{1}}Aw_{T_{1}}$ , let $U^{\prime}_{i}:=w_{T_{1}}U_{i}w_{T_{1}}$ for $i=2,\ldots,q$ and let $T^{\prime}_{i}:=w_{T_{1}}T_{i}w_{T_{1}}$ for $i=2,\ldots,q-1$ . Then in $S^{\prime}$ , we obtain that $U^{\prime}_{2}\subset\overline{A}_{0}$ and $U^{\prime}_{q}\subset A^{\prime}$ . Also

\displaystyle U^{\prime}_{2}\mathop{\simeq}_{\,\,w_{T^{\prime}_{2}}}U^{\prime}_{3}\mathop{\simeq}_{\,\,w_{T^{\prime}_{3}}}\cdots{}\mathop{\simeq}_{\ \ w_{T^{\prime}_{q-1}}}U^{\prime}_{q}

in $S^{\prime}$ . Here we can iterate this argument.

4. Some-separation-compatible Coxeter generating sets are conjugate up to finite twists

We prove that two some-separation-compatible Coxeter generating sets are conjugate up to finite twists.

Theorem 4.1.

Let $(W,R)$ and $(W,S)$ be Coxeter systems with the untangle-condition. If $(W,R)$ and $(W,S)$ are some-separation-compatible, then $R$ and $S$ are conjugate up to finite twists.

Proof.

We suppose that $(W,R)$ and $(W,S)$ are some-separation-compatible and suppose that $R$ and $S$ are connected. Then there exist separations ${\mathcal{A}}_{R}$ and ${\mathcal{A}}_{S}$ of $R$ and $S$ respectively such that $(W,R)$ and $(W,S)$ are compatible on the separations ${\mathcal{A}}_{R}$ and ${\mathcal{A}}_{S}$ ; that is, each $A\in{\mathcal{A}}_{R}$ is conjugate to some unique $B\in{\mathcal{A}}_{S}$ and each $B\in{\mathcal{A}}_{S}$ is conjugate to some unique $A\in{\mathcal{A}}_{R}$ .

We note that if $R^{\prime}$ and $S^{\prime}$ are Coxeter generating sets for $W$ that are obtained from $R$ and $S$ by some finite twists preserving ${\mathcal{A}}_{R}$ and ${\mathcal{A}}_{S}$ respectively, then also each $A\in{\mathcal{A}}_{R^{\prime}}$ is conjugate to some unique $B\in{\mathcal{A}}_{S^{\prime}}$ and each $B\in{\mathcal{A}}_{S^{\prime}}$ is conjugate to some unique $A\in{\mathcal{A}}_{R^{\prime}}$ .

Now we show that the following statement $({\rm P}_{i})$ holds for each $i=1,\ldots,n$ , where $n:=|{\mathcal{A}}_{R}|=|{\mathcal{A}}_{S}|$ .

$({\rm P}_{i})$

There exist Coxeter generating sets $R_{i}$ and $S_{i}$ for $W$ such that $R_{i}$ and $S_{i}$ are obtained from $R$ and $S$ by some finite twists preserving ${\mathcal{A}}_{R}$ and ${\mathcal{A}}_{S}$ respectively and $\overline{A}_{i}$ and $\overline{B}_{i}$ are conjugate in $W$ , where $\overline{A}_{i}$ is a union of some $i$ -th elements $C_{1},\ldots,C_{i}$ of ${\mathcal{A}}_{R_{i}}$ in $R_{i}$ as $(C_{1}\cup\cdots\cup C_{k})\cap C_{k+1}$ is maximal in

$\{(C_{1}\cup\cdots\cup C_{k})\cap C:C\in{\mathcal{A}}_{R_{i}}-\{C_{1},\ldots,C_{k}\}\}$

for any $k=1,\ldots,i-1$ , and $\overline{B}_{i}$ is a union of some $i$ -th elements $D_{1},\ldots,D_{i}$ of ${\mathcal{A}}_{S_{i}}$ in $S_{i}$ as $(D_{1}\cup\cdots\cup D_{k})\cap D_{k+1}$ is maximal in

$\{(D_{1}\cup\cdots\cup D_{k})\cap D:D\in{\mathcal{A}}_{S_{i}}-\{D_{1},\ldots,D_{k}\}\}$

for any $k=1,\ldots,i-1$ .

We prove $({\rm P}_{i})$ by induction on $i$ .

Let $R_{1}:=R$ and $S_{1}:=S$ . For $A_{1}\in{\mathcal{A}}_{R_{1}}$ , there exists a unique $B_{1}\in{\mathcal{A}}_{S_{1}}$ such that $A_{1}$ and $B_{1}$ are conjugate. Let $\overline{A}_{1}:=A_{1}$ and $\overline{B}_{1}:=B_{1}$ . Then $({\rm P}_{1})$ holds.

Let $i_{0}\geq 1$ . We suppose that $({\rm P}_{i_{0}})$ holds; that is, there exist Coxeter generating sets $R_{i_{0}}$ and $S_{i_{0}}$ for $W$ such that $R_{i_{0}}$ and $S_{i_{0}}$ are obtained from $R$ and $S$ by some finite twists preserving ${\mathcal{A}}_{R}$ and ${\mathcal{A}}_{S}$ respectively and $\overline{A}_{i_{0}}$ and $\overline{B}_{i_{0}}$ are conjugate, where $\overline{A}_{i_{0}}$ is a union of some $i_{0}$ -th elements $C_{1},\ldots,C_{i_{0}}$ of ${\mathcal{A}}_{R_{i_{0}}}$ in $R_{i_{0}}$ as $(C_{1}\cup\cdots\cup C_{k})\cap C_{k+1}$ is maximal in

\{(C_{1}\cup\cdots\cup C_{k})\cap C:C\in{\mathcal{A}}_{R_{i_{0}}}-\{C_{1},\ldots,C_{k}\}\}

for any $k=1,\ldots,i_{0}-1$ , and $\overline{B}_{i_{0}}$ is a union of some $i_{0}$ -th elements $D_{1},\ldots,D_{i_{0}}$ of ${\mathcal{A}}_{S_{i_{0}}}$ in $S_{i_{0}}$ as $(D_{1}\cup\cdots\cup D_{k})\cap D_{k+1}$ is maximal in

\{(D_{1}\cup\cdots\cup D_{k})\cap D:D\in{\mathcal{A}}_{S_{i_{0}}}-\{D_{1},\ldots,D_{k}\}\}

for any $k=1,\ldots,i_{0}-1$ .

Then we show that $({\rm P}_{i_{0}+1})$ holds.

Let

	$\displaystyle d_{1}:=\max\{\;\|\overline{A}_{i_{0}}\cap C\|\,:C\in{\mathcal{A}}_{R_{i_{0}}},\ C\not\subset\overline{A}_{i_{0}}\}\text{ and}$
	$\displaystyle d_{2}:=\max\{\;\|\overline{B}_{i_{0}}\cap D\|\,:D\in{\mathcal{A}}_{S_{i_{0}}},\ D\not\subset\overline{B}_{i_{0}}\}.$

Since $R$ and $S$ are connected, $R_{i_{0}}$ and $S_{i_{0}}$ are connected. Hence $d_{1}>0$ and $d_{2}>0$ .

Now we suppose that $d_{1}\leq d_{2}$ .

Then there exists $B\in{\mathcal{A}}_{S_{i_{0}}}$ such that $B\not\subset\overline{B}_{i_{0}}$ and $d_{2}=|\overline{B}_{i_{0}}\cap B|$ . Let $S_{i_{0}+1}:=S_{i_{0}}$ and let $\overline{B}_{i_{0}+1}:=\overline{B}_{i_{0}}\cup B$ . Here $S_{i_{0}+1}$ is preserving ${\mathcal{A}}_{S_{i_{0}}}$ and fixing $\overline{B}_{i_{0}}$ . Also $S_{i_{0}+1}$ is obtained from $S$ by some finite twists preserving ${\mathcal{A}}_{S}$ . We also note that $\overline{B}_{i_{0}+1}$ is a union of some $(i_{0}+1)$ -th elements of ${\mathcal{A}}_{S_{i_{0}+1}}$ in $S_{i_{0}+1}$ .

Let $A$ be the element of ${\mathcal{A}}_{R_{i_{0}}}$ such that $A$ and $B$ are conjugate in $W$ . Let $U:=\overline{B}_{i_{0}}\cap B$ . Then since $|U|=d_{2}$ , $U$ is a spherical-product subset of $S_{i_{0}+1}$ and $U$ separates $S_{i_{0}+1}$ by the definition of a separation, because ${\mathcal{A}}_{S_{i_{0}+1}}$ is a separation of $S_{i_{0}+1}$ .

Since $\overline{A}_{i_{0}}$ and $\overline{B}_{i_{0}}$ are conjugate in $W$ , $x\overline{B}_{i_{0}}x^{-1}=\overline{A}_{i_{0}}$ for some $x\in W$ . Then $U\subset\overline{B}_{i_{0}}$ and $U^{\prime}:=xUx^{-1}\subset\overline{A}_{i_{0}}$ . Also since $A$ and $B$ are conjugate in $W$ , $yBy^{-1}=A$ for some $y\in W$ . Then $U\subset B$ and $U^{\prime\prime}:=yUy^{-1}\subset A$ . Here $U^{\prime}$ and $U^{\prime\prime}$ are conjugate spherical-product subsets of $R_{i_{0}}$ in $W$ . Hence they are untangle by the untangle-condition. Here if $U^{\prime}_{\sigma}$ is empty then $U^{\prime}=U^{\prime}_{\nu}=U^{\prime\prime}_{\nu}=U^{\prime\prime}$ .

(a) We consider the case that $U^{\prime}=U^{\prime\prime}$ .

Then $\overline{A}_{i_{0}}\cap A=U^{\prime}=U^{\prime\prime}$ and $d_{1}=d_{2}$ , since $|U^{\prime}|=|U|=d_{2}$ and $d_{1}\leq d_{2}$ . Hence $|U^{\prime}|=d_{1}$ , $U^{\prime}=\overline{A}_{i_{0}}\cap A$ is maximal in

\{\overline{A}_{i_{0}}\cap C:C\in{\mathcal{A}}_{R_{i_{0}}},\ C\not\subset\overline{A}_{i_{0}}\}

and $U^{\prime}=U^{\prime\prime}$ separates $R_{i_{0}}$ .

Here $xUx^{-1}=U^{\prime}$ and $yUy^{-1}=U^{\prime\prime}=U^{\prime}$ . Hence

xy^{-1}U^{\prime}yx^{-1}=xUx^{-1}=U^{\prime}.

We define the bijective map $f_{xy^{-1}}:U^{\prime}\to U^{\prime}$ by $f_{xy^{-1}}(a)=(xy^{-1})a(yx^{-1})$ for any $a\in U^{\prime}$ .

By the untangle-conjugate-condition, $f_{xy^{-1}}=f_{1}$ for $1\in W$ , or, there exist a sequence $U_{1},\cdots,U_{q}$ of subsets of $R_{i_{0}}$ and a sequence $T_{1},\cdots,T_{q-1}$ of spherical subsets of $R_{i_{0}}$ such that

\displaystyle U^{\prime}=U_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}U_{2}\mathop{\simeq}_{\,\,w_{T_{2}}}\cdots{}\mathop{\simeq}_{\ \ w_{T_{q-1}}}U_{q}=U^{\prime},

$(U_{i})_{\sigma}\cup(U_{i+1})_{\sigma}\subset T_{i}$ for any $i=1,\ldots,q-1$ and $f_{xy^{-1}}=f_{w_{0}}$ for $w_{0}:=w_{T_{q-1}}\cdots w_{T_{2}}w_{T_{1}}$ .

(a-1) Suppose that $f_{xy^{-1}}=f_{1}$ . Then $\overline{A}_{i_{0}+1}=\overline{A}_{i_{0}}\cup A$ in $R_{i_{0}}$ is conjugate to $\overline{B}_{i_{0}+1}=\overline{B}_{i_{0}}\cup B$ in $S_{i_{0}}$ where $\overline{A}_{i_{0}+1}$ and $\overline{B}_{i_{0}+1}$ are unions of some $(i_{0}+1)$ -th elements of ${\mathcal{A}}_{R_{i_{0}}}$ and ${\mathcal{A}}_{S_{i_{0}}}$ respectively. Let $R_{i_{0}+1}:=R_{i_{0}}$ and $S_{i_{0}+1}:=S_{i_{0}}$ . Here $\overline{A}_{i_{0}}\cap A$ is maximal in

\{\overline{A}_{i_{0}}\cap C:C\in{\mathcal{A}}_{R_{i_{0}+1}},\ C\not\subset\overline{A}_{i_{0}}\}

and $\overline{B}_{i_{0}}\cap B$ is maximal in

\{\overline{B}_{i_{0}}\cap D:D\in{\mathcal{A}}_{S_{i_{0}+1}},\ D\not\subset\overline{B}_{i_{0}}\},

because $|\overline{A}_{i_{0}}\cap A|=|\overline{B}_{i_{0}}\cap B|=d_{1}=d_{2}$ .

(a-2) Suppose that there exist a sequence $U_{1},\cdots,U_{q}$ of subsets of $R_{i_{0}}$ and a sequence $T_{1},\cdots,T_{q-1}$ of spherical subsets of $R_{i_{0}}$ such that

\displaystyle U^{\prime}=U_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}U_{2}\mathop{\simeq}_{\,\,w_{T_{2}}}\cdots{}\mathop{\simeq}_{\ \ w_{T_{q-1}}}U_{q}=U^{\prime},

$(U_{i})_{\sigma}\cup(U_{i+1})_{\sigma}\subset T_{i}$ for any $i=1,\ldots,q-1$ and $f_{xy^{-1}}=f_{w_{0}}$ for $w_{0}:=w_{T_{q-1}}\cdots w_{T_{2}}w_{T_{1}}$ .

Then we will attach $A$ to $\overline{A}_{i_{0}}$ by gluing $U^{\prime}$ and $U^{\prime\prime}=U^{\prime}$ by some finite twists of $R_{i_{0}}$ (preserving ${\mathcal{A}}_{R_{i_{0}}}$ and fixing $\overline{A}_{i_{0}}$ ) induced by the above untangle-conjugate sequence from $U^{\prime}$ to $U^{\prime\prime}=U^{\prime}$ .

Since $U^{\prime}=\overline{A}_{i_{0}}\cap A$ separates $R_{i_{0}}$ , $V:=U^{\prime}\cup T_{1}=T_{1}\cup U^{\prime}_{\nu}$ separates $R_{i_{0}}$ . For some two subsets $X$ and $Y$ of $R_{i_{0}}$ ,

(1)

$R_{i_{0}}-V=X\cup Y$ that is a disjoint union,
(2)

$o(xy)=\infty$ for any $x\in X$ and $y\in Y$ ,
(3)

$\overline{A}_{i_{0}}\cup(U_{1}\cup\cdots\cup U_{q})\subset X\cup V$ and
(4)

$A\subset Y\cup V$ .

Here for any $C\in{\mathcal{A}}_{R_{i_{0}}}$ , $C\subset X\cup V$ or $C\subset Y\cup V$ , since $U^{\prime}=\overline{A}_{i_{0}}\cap A$ is a separator of ${\mathcal{A}}_{R_{i_{0}}}$ and $V=U^{\prime}\cup T_{1}$ .

Then $R^{\prime}_{i_{0}}:=X\cup V\cup(w_{T_{1}}Yw_{T_{1}})$ is a twist of $R_{i_{0}}$ preserving ${\mathcal{A}}_{R_{i_{0}}}$ and fixing $\overline{A}_{i_{0}}$ . Let ${\mathcal{A}}_{R^{\prime}_{i_{0}}}$ be the separation of $R^{\prime}_{i_{0}}$ induced by ${\mathcal{A}}_{R_{i_{0}}}$ and the twist.

Let $A^{\prime}:=w_{T_{1}}Aw_{T_{1}}\in{\mathcal{A}}_{R^{\prime}_{i_{0}}}$ that is conjugate to $A$ and $B$ . Then

A^{\prime}=w_{T_{1}}Aw_{T_{1}}\supset w_{T_{1}}U^{\prime}w_{T_{1}}=w_{T_{1}}U_{1}w_{T_{1}}=U_{2},

$U^{\prime}=U^{\prime\prime}=U_{q}\subset\overline{A}_{i_{0}}$ and

\displaystyle U_{2}\mathop{\simeq}_{\,\,w_{T_{2}}}U_{3}\mathop{\simeq}_{\,\,w_{T_{3}}}\cdots{}\mathop{\simeq}_{\ \ w_{T_{q-1}}}U_{q}=U^{\prime}.

Since $U^{\prime}=U_{1}$ separates $R_{i_{0}}$ and separates $\overline{A}_{i_{0}}$ and $A$ , we have that $U_{2}$ separates $R^{\prime}_{i_{0}}$ and $U_{2}$ separates $\overline{A}_{i_{0}}$ and $A^{\prime}$ in $R^{\prime}_{i_{0}}$ . Hence $T_{2}\cup U^{\prime}_{\nu}$ separates $R^{\prime}_{i_{0}}$ and $T_{2}\cup U^{\prime}_{\nu}$ separates $\overline{A}_{i_{0}}$ and $A^{\prime}$ in $R^{\prime}_{i_{0}}$ .

We iterate this argument for $T_{2},\ldots,T_{q-1}$ . Then we obtain a Coxeter generating set $R^{\prime\prime}_{i_{0}}$ from $R_{i_{0}}$ by some finite twists preserving ${\mathcal{A}}_{R_{i_{0}}}$ and fixing $\overline{A}_{i_{0}}$ such that for $w_{0}=w_{T_{q-1}}\cdots w_{T_{1}}$ and $A^{\prime\prime}:=w_{0}Aw_{0}\in{\mathcal{A}}_{R^{\prime\prime}_{i_{0}}}$ that is conjugate to $A$ and $B$ ,

A^{\prime\prime}=w_{0}Aw_{0}\supset w_{0}U^{\prime}w_{0}=w_{T_{q-1}}\cdots w_{T_{1}}U_{1}w_{T_{1}}\cdots w_{T_{q-1}}=U_{q}=U^{\prime},

where ${\mathcal{A}}_{R^{\prime\prime}_{i_{0}}}$ is the separation of $R^{\prime\prime}_{i_{0}}$ induced by ${\mathcal{A}}_{R_{i_{0}}}$ and the twists.

Let $R_{i_{0}+1}:=R^{\prime\prime}_{i_{0}}$ and $S_{i_{0}+1}:=S_{i_{0}}$ . Here $\overline{A}_{i_{0}}\cap A^{\prime\prime}$ is maximal in

\{\overline{A}_{i_{0}}\cap C:C\in{\mathcal{A}}_{R_{i_{0}+1}},\ C\not\subset\overline{A}_{i_{0}}\}

and $\overline{B}_{i_{0}}\cap B$ is maximal in

\{\overline{B}_{i_{0}}\cap D:D\in{\mathcal{A}}_{S_{i_{0}+1}},\ D\not\subset\overline{B}_{i_{0}}\},

because $|\overline{A}_{i_{0}}\cap A^{\prime\prime}|=|\overline{B}_{i_{0}}\cap B|=d_{1}=d_{2}$ . Here $\overline{A}_{i_{0}}\cup A^{\prime\prime}$ in $R_{i_{0}+1}$ is conjugate to $\overline{B}_{i_{0}}\cup B$ in $S_{i_{0}+1}$ where $A^{\prime\prime}=w_{0}Aw_{0}$ is the element of ${\mathcal{A}}_{R_{i_{0}+1}}$ that is conjugate to $A$ and $B$ . Thus we obtain that $({\rm P}_{i_{0}+1})$ holds.

(b) We consider the case that $U^{\prime}\neq U^{\prime\prime}$ .

Here $U^{\prime}_{\sigma}$ and $U^{\prime\prime}_{\sigma}$ are non-empty. Then there exists a sequence of spherical-product subsets

\displaystyle U^{\prime}=U_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}U_{2}\mathop{\simeq}_{\,\,w_{T_{2}}}\cdots{}\mathop{\simeq}_{\ \ w_{T_{q-1}}}U_{q}=U^{\prime\prime}

in $R_{i_{0}}$ . Here $T_{1},\cdots,T_{q-1}$ are spherical subsets of $R_{i_{0}}$ such that $(U_{j})_{\sigma}\cup(U_{j+1})_{\sigma}\subset T_{j}$ for $j=1,\ldots,q-1$ , $(U_{j})_{\nu}=U^{\prime}_{\nu}$ for $j=1,\ldots,q$ , and $st=ts$ for any $s\in U^{\prime}_{\nu}$ and $t\in T_{1}\cup\cdots\cup T_{q-1}$ . Also $U^{\prime}=U_{1}\subset\overline{A}_{i_{0}}$ and $U^{\prime\prime}=U_{q}\subset A$ .

Let $\sigma_{j}:=(U_{j})_{\sigma}$ for each $j=1,\ldots,q$ . Then

\displaystyle U^{\prime}_{\sigma}=\sigma_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}\sigma_{2}\mathop{\simeq}_{\,\,w_{T_{2}}}\cdots{}\mathop{\simeq}_{\ \ w_{T_{q-1}}}\sigma_{q}=U^{\prime\prime}_{\sigma}.

We will attach $A$ to $\overline{A}_{i_{0}}$ by gluing $U^{\prime}$ and $U^{\prime\prime}$ by some finite twists of $R_{i_{0}}$ (preserving ${\mathcal{A}}_{R_{i_{0}}}$ and fixing $\overline{A}_{i_{0}}$ ) induced by the untangle-conjugate sequence from $U^{\prime}$ to $U^{\prime\prime}$ .

We first suppose that $U_{q}\not\subset\overline{A}_{i_{0}}$ (that is, $\sigma_{q}\not\subset\overline{A}_{i_{0}}$ ). Let $j_{0}\in\{1,\ldots,q-1\}$ be the number as $\sigma_{j_{0}}\subset\overline{A}_{i_{0}}$ and $\sigma_{j}\not\subset\overline{A}_{i_{0}}$ for any $j=j_{0}+1,\ldots,q$ . For $T:=T_{j_{0}}$ , $w_{T}\sigma_{j_{0}+1}w_{T}=\sigma_{j_{0}}$ and $T\not\subset\overline{A}_{i_{0}}$ .

Let $V:=T\cup U^{\prime}_{\nu}=U_{j_{0}}\cup T$ . Here $st=ts$ for any $s\in U^{\prime}_{\nu}$ and $t\in T$ . Then $T$ is a spherical subset and $V$ is a spherical-product subset of $R_{i_{0}}$ . Also $\overline{A}_{i_{0}}\cap V=U_{j_{0}}$ and $|U_{j_{0}}|=|U^{\prime}|=|U|=d_{2}\geq d_{1}$ .

Then by Remark 3.3, $U_{j_{0}}$ separates $R_{i_{0}}$ and $U_{j_{0}}$ separates $\overline{A}_{i_{0}}$ and $A$ in $R_{i_{0}}$ .

Thus $V=U_{j_{0}}\cup T$ separates $R_{i_{0}}$ and $V$ separates $\overline{A}_{i_{0}}$ and $A$ . For some two subsets $X$ and $Y$ of $R_{i_{0}}$ ,

(1)

$R_{i_{0}}-V=X\cup Y$ that is a disjoint union,
(2)

$o(xy)=\infty$ for any $x\in X$ and $y\in Y$ ,
(3)

$\overline{A}_{i_{0}}\subset X\cup V$ and
(4)

$A\subset Y\cup V$ .

Here for any $C\in{\mathcal{A}}_{R_{i_{0}}}$ , $C\subset X\cup V$ or $C\subset Y\cup V$ , since $U_{j_{0}}=\overline{A}_{i_{0}}\cap A^{\prime\prime}$ and $V=U_{j_{0}}\cup T$ . Then $R^{\prime}_{i_{0}}:=X\cup V\cup(w_{T}Yw_{T})$ is a twist of $R_{i_{0}}$ preserving ${\mathcal{A}}_{R_{i_{0}}}$ and fixing $\overline{A}_{i_{0}}$ . Let ${\mathcal{A}}_{R^{\prime}_{i_{0}}}$ be the separation of $R^{\prime}_{i_{0}}$ induced by ${\mathcal{A}}_{R_{i_{0}}}$ and the twist.

Then $w_{T}\sigma_{j_{0}+1}w_{T}=\sigma_{j_{0}}$ and $w_{T}U_{j_{0}+1}w_{T}=U_{j_{0}}$ . Hence $U^{\prime\prime}$ moves one step toward $U^{\prime}$ by this twist. Let $\sigma^{\prime}_{k}:=w_{T}\sigma_{k}w_{T}$ and $U^{\prime}_{k}:=w_{T}U_{k}w_{T}$ for each $k=j_{0}+1,\ldots,q$ and let $T^{\prime}_{k}:=w_{T}T_{k}w_{T}$ for $k=j_{0}+1,\ldots,q-1$ (that are the corresponding subsets of $R^{\prime}_{i_{0}}$ to $\sigma_{k}$ , $U_{k}$ and $T_{k}$ in $R_{i_{0}}$ respectively). Then

\displaystyle U^{\prime}_{\sigma}=\sigma_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}\cdots{}\mathop{\simeq}_{\ \ w_{T_{j_{0}-1}}}\sigma_{j_{0}}=\sigma^{\prime}_{j_{0}+1}{}\mathop{\simeq}_{\ \ w_{T^{\prime}_{j_{0}+1}}}\sigma^{\prime}_{j_{0}+2}{}\mathop{\simeq}_{\ \ w_{T^{\prime}_{j_{0}+2}}}\cdots{}\mathop{\simeq}_{\ \ w_{T^{\prime}_{q-1}}}\sigma^{\prime}_{q}=(\overline{U}^{\prime\prime})_{\sigma}

and

\displaystyle U^{\prime}=U_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}\cdots{}\mathop{\simeq}_{\ \ w_{T_{j_{0}-1}}}U_{j_{0}}=U^{\prime}_{j_{0}+1}{}\mathop{\simeq}_{\ \ w_{T^{\prime}_{j_{0}+1}}}U^{\prime}_{j_{0}+2}{}\mathop{\simeq}_{\ \ w_{T^{\prime}_{j_{0}+2}}}\cdots{}\mathop{\simeq}_{\ \ w_{T^{\prime}_{q-1}}}U^{\prime}_{q}=\overline{U}^{\prime\prime}

in $R^{\prime}_{i_{0}}$ where $\overline{U}^{\prime\prime}:=w_{T}U^{\prime\prime}w_{T}$ . Here $A^{\prime}:=w_{T}Aw_{T}\in{\mathcal{A}}_{R^{\prime}_{i_{0}}}$ is conjugate to $A$ and $B$ , and $\overline{U}^{\prime\prime}\subset A^{\prime}$ holds.

We iterate this argument. Then we obtain a Coxeter generating set $R^{\prime\prime}_{i_{0}}$ from $R_{i_{0}}$ by some finite twists preserving ${\mathcal{A}}_{R_{i_{0}}}$ and fixing $\overline{A}_{i_{0}}$ such that the above sequence of spherical-product subsets transforms to

\displaystyle U^{\prime}=U_{1}\mathop{\simeq}_{\,\,w_{T_{1}}}U_{2}\mathop{\simeq}_{\,\,w_{T_{2}}}\cdots{}\mathop{\simeq}_{\ \ w_{T_{j_{0}-1}}}U_{j_{0}}=U^{\prime\prime\prime}

in $R^{\prime\prime}_{i_{0}}$ where ${\mathcal{A}}_{R^{\prime\prime}_{i_{0}}}$ is the separation of $R^{\prime\prime}_{i_{0}}$ induced by ${\mathcal{A}}_{R_{i_{0}}}$ and the twist, and $U^{\prime\prime\prime}\subset A^{\prime\prime}$ for $A^{\prime\prime}\in{\mathcal{A}}_{R^{\prime\prime}_{i_{0}}}$ that is conjugate to $A$ and $B$ . Here $A^{\prime\prime}\not\subset\overline{A}_{i_{0}}$ and $U_{j_{0}}\subset\overline{A}_{i_{0}}$ by the assumption and the definition of the number $j_{0}$ .

Then $U_{j_{0}}=\overline{A}_{i_{0}}\cap A^{\prime\prime}$ and $|U_{j_{0}}|=d_{1}=d_{2}$ . Hence $U_{j_{0}}$ separates $R^{\prime\prime}_{i_{0}}$ . Thus if $j_{0}\geq 2$ then for $T_{0}:=T_{j_{0}-1}$ , $V_{0}:=T_{0}\cup U^{\prime}_{\nu}=U_{j_{0}}\cup T_{0}$ separates $R^{\prime\prime}_{i_{0}}$ and we have a twist $R^{\prime\prime\prime}_{i_{0}}$ of $R^{\prime\prime}_{i_{0}}$ by $V_{0}$ and $w_{T_{0}}$ preserving ${\mathcal{A}}_{R^{\prime\prime}_{i_{0}}}$ and fixing $\overline{A}_{i_{0}}$ . Then $U^{\prime\prime\prime}$ moves one step toward $U^{\prime}$ by this twist. Here $A^{\prime\prime\prime}:=w_{T_{0}}A^{\prime\prime}w_{T_{0}}$ is conjugate to $A$ and $B$ , and $U_{j_{0}-1}=w_{T_{0}}U_{j_{0}}w_{T_{0}}\subset A^{\prime\prime\prime}$ in $R^{\prime\prime\prime}_{i_{0}}$ .

Since $U_{j_{0}}$ separates $R^{\prime\prime}_{i_{0}}$ and separates $\overline{A}_{i_{0}}$ and $A^{\prime\prime}$ in $R^{\prime\prime}_{i_{0}}$ , we have that $U_{j_{0}-1}$ separates $R^{\prime\prime\prime}_{i_{0}}$ and $U_{j_{0}-1}$ separates $\overline{A}_{i_{0}}$ and $A^{\prime\prime\prime}$ in $R^{\prime\prime\prime}_{i_{0}}$ . Hence $T_{j_{0}-1}\cup U^{\prime}_{\nu}$ separates $R^{\prime\prime\prime}_{i_{0}}$ and $T_{j_{0}-1}\cup U^{\prime}_{\nu}$ separates $\overline{A}_{i_{0}}$ and $A^{\prime\prime\prime}$ in $R^{\prime\prime\prime}_{i_{0}}$ .

We can iterate this argument for $T_{j_{0}-1},\ldots,T_{1}$ .

By iterating this argument and by (a), we obtain a Coxeter generating set $R_{i_{0}+1}$ from $R_{i_{0}}$ by some finite twists preserving ${\mathcal{A}}_{R_{i_{0}}}$ and fixing $\overline{A}_{i_{0}}$ such that $\overline{A}_{i_{0}}\cup A^{\prime}$ in $R_{i_{0}+1}$ is conjugate to $\overline{B}_{i_{0}}\cup B$ in $S_{i_{0}+1}$ where $A^{\prime}$ is the element of ${\mathcal{A}}_{R_{i_{0}+1}}$ that is conjugate to $A$ and $B$ . Hence, we obtain that $({\rm P}_{i_{0}+1})$ holds.

Thus $({\rm P}_{i})$ holds for any $i=1,\ldots,n$ .

Then $({\rm P}_{n})$ implies that there exist Coxeter generating sets $R_{n}$ and $S_{n}$ for $W$ such that $R_{n}$ and $S_{n}$ are obtained from $R$ and $S$ by some finite twists respectively and $R_{n}$ and $S_{n}$ are conjugate, because the unions of the $n$ -th elements of ${\mathcal{A}}_{R_{n}}$ in $R_{n}$ and ${\mathcal{A}}_{S_{n}}$ in $S_{n}$ are just $R_{n}$ and $S_{n}$ respectively.

Therefore, $R$ and $S$ are conjugate up to finite twists.

In the case that $R$ and $S$ are connected, we showed that if $(W,R)$ and $(W,S)$ are some-separation-compatible, then $R$ and $S$ are conjugate up to finite twists.

We suppose that $R$ and $S$ are not connected.

Let $R_{1},\ldots,R_{n}$ be the connected components of $R$ and let $S_{1},\ldots,S_{n}$ be the connected components of $S$ ; that is,

W=W_{S_{1}}*\cdots*W_{S_{n}}=W_{R_{1}}*\cdots*W_{R_{n}},

where the numbers of the connected components of $R$ and $S$ are equal.

By the same argument as above (in the case that $R$ and $S$ are connected), we can obtain Coxeter generating sets $R^{\prime}$ and $S^{\prime}$ for $W$ such that $R^{\prime}$ and $S^{\prime}$ are obtained from $R$ and $S$ by some finite twists respectively and $R^{\prime}_{i}$ and $S^{\prime}_{i}$ are conjugate for any $i=1,\ldots,n$ , where $R^{\prime}_{1},\ldots,R^{\prime}_{n}$ are the connected components of $R^{\prime}$ and $S^{\prime}_{1},\ldots,S^{\prime}_{n}$ are the connected components of $S^{\prime}$ .

Then by the untangle-condition, there exists a Coxeter generating set $R^{\prime\prime}$ for $W$ such that $R^{\prime\prime}$ is obtained from $R^{\prime}$ by some finite twists and $S^{\prime}$ and $R^{\prime\prime}$ are conjugate.

Therefore, $R$ and $S$ are conjugate up to finite twists. ∎

We obtain the following corollary from Theorem 4.1.

Corollary 4.2.

For Coxeter systems $(W,R)$ and $(W,S)$ with the untangle-condition, the following two statements are equivalent:

(i)

$R$ and $S$ are conjugate up to finite twists.
(ii)

$(W,R_{0})$ and $(W,S)$ are some-separation-compatible for some Coxeter generating set $R_{0}$ obtained from $R$ by finite twists.

5. $\widetilde{\mathcal{A}}_{S}$ is a separation of $S$

In this section, we prove the following proposition.

Proposition 5.1.

Let $(W,S)$ be a Coxeter system and let $\widetilde{\mathcal{A}}_{S}$ be the set of subsets of $S$ in Definition 1.14. Then $\widetilde{\mathcal{A}}_{S}$ is a separation of $S$ .

Let $(W,S)$ be a Coxeter system and let ${\mathcal{A}}_{0}$ be the set of maximal twist-rigid subsets of $S$ . We suppose that $S$ is connected. For $\widetilde{\mathcal{A}}_{S}$ , we show the statements (1)–(4) of the definition of a separation (in Section 1) hold.

(1) and (3): For each $A_{0}\in{\mathcal{A}}_{0}$ , there exists a unique $B\in\widetilde{\mathcal{A}}_{S}$ such that $A_{0}\subset B$ . (Here we can denote $B=\bigcup[A_{0}]$ by Definition 1.14.) Hence $\bigcup\widetilde{\mathcal{A}}_{S}=\bigcup{\mathcal{A}}_{0}=S$ .

(2) Let $B\in\widetilde{\mathcal{A}}_{S}$ . Then $B=\bigcup[A_{0}]$ for some $A_{0}\in{\mathcal{A}}_{0}$ by Definition 1.14. Hence $B$ is a union of some maximal twist-rigid subsets of $S$ . Also $B=\bigcup[A_{0}]$ is connected by the definition of the equivalence relation “ $\sim$ ” on ${\mathcal{A}}_{0}$ .

To prove (4) in the definition of a separation for $\widetilde{\mathcal{A}}_{S}$ , we show some lemmas.

Let ${\mathcal{A}}$ be a minimal separation of $S$ (and fix ${\mathcal{A}}$ ).

Lemma 5.2.

Proof.

We can denote $B_{0}=\bigcup[A^{\prime}]$ for some $A^{\prime}\in{\mathcal{A}}_{0}$ . Let $[A^{\prime}]=\{A^{\prime}_{1},\ldots,A^{\prime}_{t}\}$ (where $A^{\prime}_{j}\in{\mathcal{A}}_{0}$ as $A^{\prime}_{j}\sim A^{\prime}$ ). For each $j=1,\ldots,t$ , $A^{\prime}_{j}\subset A^{\prime\prime}_{j}$ for some unique $A^{\prime\prime}_{j}\in{\mathcal{A}}$ , since ${\mathcal{A}}$ is a separation of $S$ . Here $A^{\prime\prime}_{1},\ldots,A^{\prime\prime}_{t}$ need not be different all together.

We show that $A^{\prime\prime}_{j}\subset B_{0}$ for any $j\in\{1,\ldots,t\}$ .

Let $j\in\{1,\ldots,t\}$ . Here $A^{\prime}_{j}\subset B_{0}=\bigcup[A^{\prime}]$ and $A^{\prime}_{j}\in{\mathcal{A}}_{0}$ . Hence $A^{\prime}_{j}\sim A^{\prime}$ . Let $A^{\prime\prime}\in{\mathcal{A}}_{0}$ as $A^{\prime\prime}\subset A^{\prime\prime}_{j}$ . Then $A^{\prime\prime}\sim A^{\prime}_{j}$ by the definition of the equivalence relation “ $\sim$ ” on ${\mathcal{A}}_{0}$ , since $A^{\prime\prime},A^{\prime}_{j}\in{\mathcal{A}}_{0}$ and $A^{\prime\prime}\cup A^{\prime}_{j}\subset A^{\prime\prime}_{j}\in{\mathcal{A}}$ . Thus $A^{\prime\prime}\sim A^{\prime}_{j}\sim A^{\prime}$ . We obtain that $A^{\prime\prime}\subset\bigcup[A^{\prime}]=B_{0}$ . Hence

A^{\prime\prime}_{j}=\bigcup\{A^{\prime\prime}\in{\mathcal{A}}_{0}:A^{\prime\prime}\subset A^{\prime\prime}_{j}\}\subset B_{0}.

Thus $A^{\prime\prime}_{j}\subset B_{0}$ for any $j\in\{1,\ldots,t\}$ .

Then $\{A^{\prime\prime}_{1},\ldots,A^{\prime\prime}_{t}\}\subset\{A_{1},\ldots,A_{l}\}$ (in fact, the equality holds) and

\textstyle B_{0}=\bigcup[A^{\prime}]=\bigcup_{j=1}^{t}A^{\prime}_{j}\subset\bigcup_{j=1}^{t}A^{\prime\prime}_{j}\subset\bigcup_{i=1}^{l}A_{i}.

Also obviously $\bigcup_{i=1}^{l}A_{i}\subset B_{0}$ holds. Thus $B_{0}=\bigcup_{i=1}^{l}A_{i}$ . ∎

Now we say that a sequence $A_{1},\ldots,A_{l}\in{\mathcal{A}}$ satisfies the condition $(*)$ in ${\mathcal{A}}$ , if

$(*)$

$(A_{1}\cup\cdots\cup A_{i})\cap A_{i+1}$ is maximal in

$\{(A_{1}\cup\cdots\cup A_{i})\cap A:A\in{\mathcal{A}}-\{A_{1},\ldots,A_{i}\}\}$

for any $i=1,\ldots,l-1$ .

We next show the following.

Lemma 5.3.

Let $B_{0}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ . Let $A_{1},\ldots,A_{l}$ be the distinct elements of ${\mathcal{A}}$ such that $A_{i}\subset B_{0}$ for any $i=1,\ldots,l$ . Then there exists a bijective map $f:\{1,\ldots,l\}\to\{1,\ldots,l\}$ such that the sequence $A_{f(1)},\ldots,A_{f(l)}$ satisfies the condition $(*)$ in ${\mathcal{A}}$ .

Proof.

Let $f(1)\in\{1,\ldots,l\}$ that is arbitrary. Let $f(2)\in\{1,\ldots,l\}-\{f(1)\}$ such that $A_{f(1)}\cap A_{f(2)}$ is maximal in $\{A_{f(1)}\cap A:A\in\{A_{1},\ldots,A_{l}\}-\{A_{f(1)}\}\}$ . Then we show that $A_{f(1)}\cap A_{f(2)}$ is maximal in $\{A_{f(1)}\cap A:A\in{\mathcal{A}}-\{A_{f(1)}\}\}$ . Indeed if this does not hold, then there exists $A^{\prime}_{2}\in{\mathcal{A}}-\{A_{f(1)}\}$ such that $A_{f(1)}\cap A^{\prime}_{2}$ is maximal in $\{A_{f(1)}\cap A:A\in{\mathcal{A}}-\{A_{f(1)}\}\}$ and $A_{f(1)}\cap A_{f(2)}\subsetneq A_{f(1)}\cap A^{\prime}_{2}$ . Here $A^{\prime}_{2}\not\in\{A_{1},\ldots,A_{l}\}$ and $A^{\prime}_{2}\not\subset B_{0}$ . Hence some $\overline{U}_{1}\in\overline{\mathcal{U}}$ separates $B_{0}$ and $A^{\prime}_{2}$ , where $\overline{\mathcal{U}}$ is the separators set as in Definition 1.14. Here “ $\overline{U}_{1}$ separates $B_{0}$ and $A^{\prime}_{2}$ ” means that $B_{0}\subset\overline{X}_{j_{1}}$ and $A^{\prime}_{2}\subset\overline{X}_{j_{2}}$ for some $j_{1},j_{2}\in\{1,\ldots,t\}$ as $j_{1}\neq j_{2}$ , where $X_{1},\ldots,X_{t}$ are the connected components of $S-U_{1}$ and $S-U_{1}=X_{1}\cup\cdots\cup X_{t}$ .

Then $\overline{U}_{1}$ separates $A_{f(1)}$ and $A^{\prime}_{2}$ . Thus $\overline{U}_{1}$ separates $A_{f(1)}$ and $A_{f(2)}$ , because $A_{f(1)}\cap A_{f(2)}\subsetneq A_{f(1)}\cap A^{\prime}_{2}\subset\overline{U}_{1}$ . This contradicts that $A_{f(1)}\cup A_{f(2)}\subset B_{0}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ . Thus $A_{f(1)}\cap A_{f(2)}$ is maximal in $\{A_{f(1)}\cap A:A\in{\mathcal{A}}-\{A_{f(1)}\}\}$ .

Let $f(3)\in\{1,\ldots,l\}-\{f(1),f(2)\}$ such that $(A_{f(1)}\cup A_{f(2)})\cap A_{f(3)}$ is maximal in

\{(A_{f(1)}\cup A_{f(2)})\cap A:A\in\{A_{1},\ldots,A_{l}\}-\{A_{f(1)},A_{f(2)}\}\}.

Then we show that $(A_{f(1)}\cup A_{f(2)})\cap A_{f(3)}$ is maximal in

\{(A_{f(1)}\cup A_{f(2)})\cap A:A\in{\mathcal{A}}-\{A_{f(1)},A_{f(2)}\}\}.

Indeed if this does not hold, then there exists $A^{\prime}_{3}\in{\mathcal{A}}-\{A_{f(1)},A_{f(2)}\}$ such that $(A_{f(1)}\cup A_{f(2)})\cap A^{\prime}_{3}$ is maximal in

\{(A_{f(1)}\cup A_{f(2)})\cap A:A\in{\mathcal{A}}-\{A_{f(1)},A_{f(2)}\}\}

and

(A_{f(1)}\cup A_{f(2)})\cap A_{f(3)}\subsetneq(A_{f(1)}\cup A_{f(2)})\cap A^{\prime}_{3}.

Here $A^{\prime}_{3}\not\in\{A_{1},\ldots,A_{l}\}$ and $A^{\prime}_{3}\not\subset B_{0}$ . Hence some $\overline{U}_{2}\in\overline{\mathcal{U}}$ separates $B_{0}$ and $A^{\prime}_{3}$ , and $\overline{U}_{2}$ separates $A_{f(1)}\cup A_{f(2)}$ and $A^{\prime}_{3}$ . Then $\overline{U}_{2}$ separates $A_{f(1)}\cup A_{f(2)}$ and $A_{f(3)}$ , because $(A_{f(1)}\cup A_{f(2)})\cap A_{f(3)}\subsetneq(A_{f(1)}\cup A_{f(2)})\cap A^{\prime}_{3}\subset\overline{U}_{2}$ . This contradicts that $A_{f(1)}\cup A_{f(2)}\cup A_{f(3)}\subset B_{0}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ .

By iterating this argument, we obtain a sequence $A_{f(1)},\ldots,A_{f(l)}$ that satisfies the condition $(*)$ in ${\mathcal{A}}$ . ∎

Lemma 5.4.

Let $B_{1},\ldots,B_{n}\in\widetilde{\mathcal{A}}_{S}$ such that the sequence $B_{1},\ldots,B_{n}$ satisfies the condition $(*)$ in $\widetilde{\mathcal{A}}_{S}$ . Then there exists a sequence

A_{1},\ldots,A_{l_{1}},A_{l_{1}+1},\ldots,A_{l_{2}},\ldots,A_{l_{n-1}+1},\ldots,A_{l_{n}}\;\in{\mathcal{A}}

satisfying the condition $(*)$ in ${\mathcal{A}}$ such that

\textstyle B_{1}=\bigcup_{i=1}^{l_{1}}A_{i},\ B_{2}=\bigcup_{i=l_{1}+1}^{l_{2}}A_{i},\ \ldots,\ B_{n}=\bigcup_{i=l_{n-1}+1}^{l_{n}}A_{i}.

Also $(B_{1}\cup\cdots\cup B_{i})\cap B_{i+1}=(B_{1}\cup\cdots\cup B_{i})\cap A_{l_{i}+1}$ for any $i=1,\ldots,n-1$ .

Proof.

In the case that $n=1$ , by Lemma 5.3, there exists a sequence $A_{1},\ldots,A_{l_{1}}\in{\mathcal{A}}$ satisfying the condition $(*)$ in ${\mathcal{A}}$ such that $B_{1}=\bigcup_{i=1}^{l_{1}}A_{i}$ .

Let $n\geq 2$ . Suppose that $B_{1},\ldots,B_{n-1},B_{n}\in\widetilde{\mathcal{A}}_{S}$ satisfies the condition $(*)$ in $\widetilde{\mathcal{A}}_{S}$ , a sequence

A_{1},\ldots,A_{l_{1}},A_{l_{1}+1},\ldots,A_{l_{2}},\ldots,A_{l_{n-2}+1},\ldots,A_{l_{n-1}}\;\in{\mathcal{A}}

satisfies the condition $(*)$ in ${\mathcal{A}}$ and

\textstyle B_{1}=\bigcup_{i=1}^{l_{1}}A_{i},\ \;B_{2}=\bigcup_{i=l_{1}+1}^{l_{2}}A_{i},\ \;\ldots,\ \;B_{n-1}=\bigcup_{i=l_{n-2}+1}^{l_{n-1}}A_{i}.

Here $B_{n}\in\widetilde{\mathcal{A}}_{S}$ and $(B_{1}\cup\cdots\cup B_{n-1})\cap B_{n}$ is maximal in

\{(B_{1}\cup\cdots\cup B_{n-1})\cap B:B\in\widetilde{\mathcal{A}}_{S}-\{B_{1},\ldots,B_{n-1}\}\}.

Since $B_{n}\not\in\{B_{1},\ldots,B_{n-1}\}$ , by the definition of $\widetilde{\mathcal{A}}_{S}$ , there exists $\overline{U}\in\overline{\mathcal{U}}$ such that $B_{n}\subset\overline{X}_{1}$ and $B_{1}\cup\cdots\cup B_{n-1}\subset\overline{X}_{2}\cup\cdots\cup\overline{X}_{t}$ where $S-\overline{U}=X_{1}\cup\cdots\cup X_{t}$ is a disjoint union, $X_{1},\ldots,X_{t}$ are the connected components of $S-\overline{U}$ and $\overline{X}_{j}=\bigcup\{A\in{\mathcal{A}}:A\subset X_{j}\cup\overline{U}\}$ for $j=1,\ldots,t$ . Here $\overline{U}$ is a separator of ${\mathcal{A}}$ . For $U_{0}:=\overline{U}\cap\overline{X}_{1}$ , by (v) in the definition of a separator, there exists $A^{\prime}_{0}\in{\mathcal{A}}$ such that $U_{0}\subset A^{\prime}_{0}\subset\overline{X}_{1}$ . Then by (iv) in the definition of a separator, the following equation holds;

	$\displaystyle U_{0}$	$\displaystyle=(\overline{X}_{2}\cup\cdots\cup\overline{X}_{t})\cap A^{\prime}_{0}=\overline{U}\cap A^{\prime}_{0}$
		$\displaystyle=(\overline{X}_{2}\cup\cdots\cup\overline{X}_{t})\cap\overline{X}_{1}=\overline{U}\cap\overline{X}_{1}.$

Let $U^{\prime}_{0}:=(B_{1}\cup\cdots\cup B_{n-1})\cap B_{n}$ . Then

U^{\prime}_{0}=(B_{1}\cup\cdots\cup B_{n-1})\cap B_{n}\subset(\overline{X}_{2}\cup\cdots\cup\overline{X}_{t})\cap\overline{X}_{1}=U_{0}.

Hence $U^{\prime}_{0}\subset U_{0}\subset A^{\prime}_{0}\subset\overline{X}_{1}$ . Thus

	$\displaystyle U^{\prime}_{0}$	$\displaystyle=(B_{1}\cup\cdots\cup B_{n-1})\cap A^{\prime}_{0}$
		$\displaystyle=(B_{1}\cup\cdots\cup B_{n-1})\cap\overline{X}_{1}.$

Here $U^{\prime}_{0}=(B_{1}\cup\cdots\cup B_{n-1})\cap B_{n}$ is maximal in

\{(B_{1}\cup\cdots\cup B_{n-1})\cap B:B\in\widetilde{\mathcal{A}}_{S}-\{B_{1},\ldots,B_{n-1}\}\}.

By the above argument, $U^{\prime}_{0}=(B_{1}\cup\cdots\cup B_{n-1})\cap A^{\prime}_{0}$ is maximal in

\{(B_{1}\cup\cdots\cup B_{n-1})\cap A:A\in{\mathcal{A}}-\{A_{1},\ldots,A_{l_{n-1}}\}\}

and it is maximal in

\{(A_{1}\cup\cdots\cup A_{l_{n-1}})\cap A:A\in{\mathcal{A}}-\{A_{1},\ldots,A_{l_{n-1}}\}\}.

Hence the sequence

A_{1},\ldots,A_{l_{1}},A_{l_{1}+1},\ldots,A_{l_{2}},\ldots,A_{l_{n-2}+1},\ldots,A_{l_{n-1}},\,A^{\prime}_{0}\;\in{\mathcal{A}}

satisfies the condition $(*)$ in ${\mathcal{A}}$ . Here $U^{\prime}_{0}$ is a separator of ${\mathcal{A}}$ .

Let $A^{\prime}_{1},\ldots,A^{\prime}_{l^{\prime}}$ be the set of ${\mathcal{A}}$ such that $A^{\prime}_{i}\subset B_{n}$ for any $i=1,\ldots,l^{\prime}$ . By Lemma 5.2, $B_{n}=A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{l^{\prime}}$ . By Lemma 5.3, we may suppose that $A^{\prime}_{1}=A^{\prime}_{0}$ and the sequence $A^{\prime}_{1},\ldots,A^{\prime}_{l^{\prime}}$ satisfies the condition $(*)$ in ${\mathcal{A}}$ . Then

(B_{1}\cup\cdots\cup B_{n-1})\cap B_{n}=U^{\prime}_{0}=(B_{1}\cup\cdots\cup B_{n-1})\cap A^{\prime}_{1}.

We show that

((B_{1}\cup\cdots\cup B_{n-1})\cup A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A^{\prime}_{i+1}=(A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A^{\prime}_{i+1}

for any $i=1,\ldots,l^{\prime}-1$ . Let $U_{i+1}:=((B_{1}\cup\cdots\cup B_{n-1})\cup A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A^{\prime}_{i+1}$ . Then

	$\displaystyle U_{i+1}$	$\displaystyle=((B_{1}\cup\cdots\cup B_{n-1})\cup A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A^{\prime}_{i+1}$
		$\displaystyle\subset(\overline{X}_{2}\cup\cdots\cup\overline{X}_{t})\cup A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A^{\prime}_{i+1}$
		$\displaystyle=((\overline{X}_{2}\cup\cdots\cup\overline{X}_{t})\cap A^{\prime}_{i+1})\cup((A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A^{\prime}_{i+1})$
		$\displaystyle=((\overline{X}_{2}\cup\cdots\cup\overline{X}_{t})\cap\overline{X}_{1}\cap A^{\prime}_{i+1})\cup((A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A^{\prime}_{i+1})$
		$\displaystyle=((U_{0}\cap A^{\prime}_{i+1})\cup((A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A^{\prime}_{i+1})$
		$\displaystyle=(U_{0}\cup A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A^{\prime}_{i+1}$
		$\displaystyle=(A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A^{\prime}_{i+1},$

where $A^{\prime}_{i+1}\subset\overline{X}_{1}$ and $A^{\prime}_{i+1}=\overline{X}_{1}\cap A^{\prime}_{i+1}$ . Also here $(\overline{X}_{2}\cup\cdots\cup\overline{X}_{t})\cap\overline{X}_{1}=U_{0}$ and $U_{0}\subset A^{\prime}_{0}=A^{\prime}_{1}$ . Obviously $(A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A^{\prime}_{i+1}\subset U_{i+1}$ holds.

Then $U_{i+1}$ is maximal in

\{(A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A:A\in{\mathcal{A}}-\{A^{\prime}_{1},\ldots,A^{\prime}_{i}\}\}

for any $i=1,\ldots,l^{\prime}-1$ . Hence $U_{i+1}$ is maximal in

\{((B_{1}\cup\cdots\cup B_{n-1})\cup A^{\prime}_{1}\cup\cdots\cup A^{\prime}_{i})\cap A:\\ A\in{\mathcal{A}}-\{A_{1},\ldots,A_{l_{1}},A_{l_{1}+1},\ldots,A_{l_{2}},\ldots,A_{l_{n-2}+1},\ldots,A_{l_{n-1}},A^{\prime}_{1},\ldots,A^{\prime}_{i}\}\}

for any $i=1,\ldots,l^{\prime}-1$ .

Thus the sequence

A_{1},\ldots,A_{l_{1}},A_{l_{1}+1},\ldots,A_{l_{2}},\ldots,A_{l_{n-2}+1},\ldots,A_{l_{n-1}},A^{\prime}_{1}\ldots,A^{\prime}_{l^{\prime}}\;\in{\mathcal{A}}

satisfies the condition $(*)$ in ${\mathcal{A}}$ . ∎

(4) Now we show that (4) in the definition of a separation of $S$ holds for $\widetilde{\mathcal{A}}_{S}$ .

Suppose that $B_{1},\ldots,B_{n}\in\widetilde{\mathcal{A}}_{S}$ is a sequence satisfying the condition $(*)$ in $\widetilde{\mathcal{A}}_{S}$ . Then we show that $U_{i}:=(B_{1}\cup\cdots\cup B_{i})\cap B_{i+1}$ is a separator of $\widetilde{\mathcal{A}}_{S}$ for any $i=1,\ldots,n-1$ .

Let $i\in\{1,\ldots,n-1\}$ and let $U:=U_{i}=(B_{1}\cup\cdots\cup B_{i})\cap B_{i+1}$ . By Lemma 5.4, $U$ is a separator of ${\mathcal{A}}$ .

(i) and (ii): Then $U$ is a spherical-product subset of $S$ and $U$ separates $S$ .

(iii) Suppose that $X-U=X_{1}\cup\cdots\cup X_{t}$ is a disjoint union and $X_{1},\ldots,X_{t}$ are the connected components of $S-U$ . Let

	$\displaystyle\overline{X}_{j}:=\bigcup\{A\in{\mathcal{A}}:A\subset X_{j}\cup U\}\ \text{and}$
	$\displaystyle\overline{X}^{\prime}_{j}:=\bigcup\{B\in\widetilde{\mathcal{A}}_{S}:B\subset X_{j}\cup U\}$

for each $j=1,\ldots,t$ .

By the definition of $\widetilde{\mathcal{A}}_{S}$ , each $B\in\widetilde{\mathcal{A}}_{S}$ is not separated by any separator of a minimal separation of $S$ . Hence for any $B\in\widetilde{\mathcal{A}}_{S}$ , $U$ does not separate $B$ and $B-U\subset X_{j_{0}}$ for some unique $j_{0}\in\{1,\ldots,t\}$ . Then $B\subset\overline{X}_{j_{0}}$ and $B\subset\overline{X}^{\prime}_{j_{0}}$ hold.

Here $\overline{X}_{j}=\overline{X}^{\prime}_{j}$ holds for any $j=1,\ldots,t$ , because each $B\in\widetilde{\mathcal{A}}_{S}$ is denoted by $B=\bigcup_{i=1}^{l}A_{i}$ (where $A_{1},\ldots,A_{l}$ are the elements of ${\mathcal{A}}$ such that $A_{i}\subset B$ ) and each $A\in{\mathcal{A}}$ is contained in some unique $B\in\widetilde{\mathcal{A}}_{S}$ .

(iv) Since $U$ is a separator of ${\mathcal{A}}$ , there exist $A_{1},A_{2}\in{\mathcal{A}}$ such that $A_{1}\cap A_{2}=U$ , $A_{1}\subset\overline{X}_{j_{1}}$ and $A_{2}\subset\overline{X}_{j_{2}}$ for some $j_{1},j_{2}\in\{1,\ldots,t\}$ as $j_{1}\neq j_{2}$ . Then $A_{1}\subset B^{\prime}_{1}$ and $A_{2}\subset B^{\prime}_{2}$ for some unique $B^{\prime}_{1},B^{\prime}_{2}\in\widetilde{\mathcal{A}}_{S}$ . Here $B^{\prime}_{1}\subset\overline{X}_{j_{1}}=\overline{X}^{\prime}_{j_{1}}$ and $B^{\prime}_{2}\subset\overline{X}_{j_{2}}=\overline{X}^{\prime}_{j_{2}}$ . Then $B^{\prime}_{1}\cap B^{\prime}_{2}=U$ holds.

(v) Let $j\in\{1,\ldots,t\}$ . Suppose that $B^{\prime}_{1},\ldots,B^{\prime}_{k}\in\widetilde{\mathcal{A}}_{S}$ satisfies the condition $(*)$ in $\widetilde{\mathcal{A}}_{S}$ and $U\cap\overline{X}^{\prime}_{j}\subset B^{\prime}_{1}\cup\cdots\cup B^{\prime}_{k}\subset\overline{X}^{\prime}_{j}$ . By Lemma 5.4, there exists a sequence

A^{\prime}_{1},\ldots,A^{\prime}_{p_{1}},\ldots,A^{\prime}_{p_{k-1}+1},\ldots,A^{\prime}_{p_{k}}\;\in{\mathcal{A}}

satisfying the condition $(*)$ in ${\mathcal{A}}$ such that

\textstyle B^{\prime}_{1}=\bigcup_{i=1}^{p_{1}}A^{\prime}_{i},\ \ldots,\ B^{\prime}_{k}=\bigcup_{i=p_{k-1}+1}^{p_{k}}A^{\prime}_{i}.

Since $U$ is a separator of ${\mathcal{A}}$ , we have that $U\cap\overline{X}^{\prime}_{j}\subset A^{\prime}_{i_{0}}$ for some $i_{0}\in\{1,\ldots,p_{k}\}$ . Here $A^{\prime}_{i_{0}}\subset B^{\prime}_{i_{1}}$ for some $i_{1}\in\{1,\ldots,k\}$ . Then $U\cap\overline{X}^{\prime}_{j}\subset A^{\prime}_{i_{0}}\subset B^{\prime}_{i_{1}}$ .

Thus from the above (i)–(v), $U$ is a separator of $\widetilde{\mathcal{A}}_{S}$ .

We show (b) on (4) in the definition of a separation for $\widetilde{\mathcal{A}}_{S}$ ; that is, $B_{1}\cup\cdots\cup B_{i}\subset X_{j_{1}}\cup U$ and $B_{i+1}\subset X_{j_{2}}\cup U$ for some $j_{1},j_{2}\in\{1,\ldots,t\}$ as $j_{1}\neq j_{2}$ .

Since $\widetilde{\mathcal{A}}$ is a separation, by Lemma 5.4, we obtain that $B_{1}\cup\cdots\cup B_{i}\subset X_{j_{1}}\cup U$ and $A_{l_{i}+1}\subset X_{j_{2}}\cup U$ for some $j_{1},j_{2}\in\{1,\ldots,t\}$ as $j_{1}\neq j_{2}$ . Since $U$ is a separator of $\widetilde{\mathcal{A}}_{S}$ , $U$ does not separate $B_{i+1}=A_{l_{i}+1}\cup\cdots\cup A_{l_{i+1}}\in\widetilde{\mathcal{A}}_{S}$ . Hence $B_{i+1}\subset X_{j_{2}}\cup U$ .

Thus (4) in the definition of a separation of $S$ holds for $\widetilde{\mathcal{A}}_{S}$ .

Therefore $\widetilde{\mathcal{A}}_{S}$ is a separation of $S$ .

6. Type(I)-type(II)-compatible and conjugate up to finite twists

We investigate on type(I)-type(II)-compatible Coxeter systems and conjugate up to finite twists. We show that for Coxeter systems $(W,R)$ and $(W,S)$ with the untangle-condition, $R$ and $S$ are conjugate up to finite twists if and only if $(W,R)$ and $(W,S)$ are type(I)-type(II)-compatible.

We show a lemma on conjugate spherical-product subsets.

Lemma 6.1.

Let $(W,S)$ be a Coxeter system and let $U$ be a spherical-product subset of $S$ . If $wUw^{-1}=U$ for some $w\in W$ , then $wU_{\sigma}w^{-1}=U_{\sigma}$ and $wU_{\nu}w^{-1}=U_{\nu}$ .

Proof.

Suppose that $wUw^{-1}=U$ for some $w\in W$ . Then

	$\displaystyle W_{U}=W_{U_{\sigma}}\times W_{U_{\nu}}\ \text{and}$
	$\displaystyle W_{wUw^{-1}}=wW_{U}w^{-1}=wW_{U_{\sigma}}w^{-1}\times wW_{U_{\nu}}w^{-1}$
	$\displaystyle\hskip 40.97194pt=W_{wU_{\sigma}w^{-1}}\times W_{wU_{\nu}w^{-1}}.$

Here $W_{U_{\nu}}$ and $W_{wU_{\nu}w^{-1}}$ are the minimal standard (parabolic) subgroups of finite index in $W_{U}$ and $W_{wUw^{-1}}$ respectively (see [9]). Since $W_{U}=W_{wUw^{-1}}$ and $U=wUw^{-1}$ , we have that $W_{U_{\nu}}=W_{wU_{\nu}w^{-1}}$ . Hence $U_{\nu}=wU_{\nu}w^{-1}$ . Thus $wU_{\nu}w^{-1}=U_{\nu}$ and $wU_{\sigma}w^{-1}=U_{\sigma}$ . ∎

We show a lemma on spherical subsets and type(II) subsets.

Lemma 6.2.

Let $(W,S)$ be a Coxeter system. If $\sigma$ is a spherical subset of $S$ that separates $S$ , then $\sigma$ does not separate any type(II) subset of $S$ .

Proof.

Let $\sigma$ be a spherical subset of $S$ that separates $S$ . We can denote $S-\sigma=X_{1}\cup\cdots\cup X_{t}$ where $X_{1},\ldots,X_{t}$ are the connected components of $S-\sigma$ .

Then for any maximal twist-rigid subset $A$ of $S$ , $\sigma$ does not separate $A$ and $A\subset X_{i}\cup\sigma$ for some $i\in\{1,\ldots,t\}$ . Also for any minimal separation ${\mathcal{A}}$ of $S$ and for any $B\in{\mathcal{A}}$ , $\sigma$ does not separate $B$ and $B\subset X_{i}\cup\sigma$ for some $i\in\{1,\ldots,t\}$ . (Indeed if $\sigma$ separates some $B\in{\mathcal{A}}$ , then there exists a separation ${\mathcal{A}}^{\prime}$ of $S$ induced by $\sigma$ from ${\mathcal{A}}$ such that ${\mathcal{A}}^{\prime}\prec{\mathcal{A}}$ . This contradicts that ${\mathcal{A}}$ is minimal.)

Hence $\sigma$ does not separate any type(II) subset of $S$ . ∎

Now we show the following.

Lemma 6.3.

Let $(W,S)$ be a Coxeter system with the untangle-condition and let $(W,S^{\prime})$ be a Coxeter system obtained from $(W,S)$ by some twist. Then $(W,S)$ and $(W,S^{\prime})$ are type(I)-type(II)-compatible.

Proof.

It is sufficient to show this lemma in the case that $(W,S)$ and $(W,S^{\prime})$ are connected. Now we suppose this.

Let $U$ be a spherical-product subset of $S$ and let $w\in W$ such that $U$ separates $S$ , $U=wUw^{-1}$ and $S^{\prime}$ is obtained from $S$ by some twist of $U$ and $w$ . There exist non-empty subsets $X$ and $Y$ of $S$ such that $S-U=X\cup Y$ , $X\cap Y=\emptyset$ , $o(xy)=\infty$ for any $x\in X$ and $y\in Y$ , and $S^{\prime}=X\cup U\cup(wYw^{-1})$ .

Let ${\mathcal{A}}_{0}$ and ${\mathcal{A}}^{\prime}_{0}$ be the sets of maximal twist-rigid subsets of $S$ and $S^{\prime}$ respectively. Here for each $C\in{\mathcal{A}}_{0}$ , if we put $C^{\prime}:=C$ (if $C\subset X\cup U$ ) and $C^{\prime}:=w^{-1}Cw$ (if $C\not\subset X\cup U$ ), then $C^{\prime}\in{\mathcal{A}}^{\prime}_{0}$ . Hence

\begin{split}{\mathcal{A}}^{\prime}_{0}=&\{C:C\in{\mathcal{A}}_{0},\ C\subset X\cup U\}\\ &\ \hskip 34.1433pt\cup\{wCw^{-1}:C\in{\mathcal{A}}_{0},\ C\not\subset X\cup U\}.\end{split}

By the definition of type(I) subsets and Remark 2.2, each $A\in\widetilde{\mathcal{A}}_{S}^{\rm(I)}$ is not separated by any spherical-product subset that separates $S$ .

Let $A\in\widetilde{\mathcal{A}}_{S}^{\rm(I)}$ . Here $A$ is not separated by $U$ . We consider the subset $A^{\prime}$ of $S^{\prime}$ as $A^{\prime}=A$ (if $A\subset X\cup U$ ) and $A^{\prime}=wAw^{-1}$ (if $A\not\subset X\cup U$ ). Then since $U$ separates $S^{\prime}$ , there is an induced separation ${\mathcal{A}}^{\prime}$ of $S^{\prime}$ by $U$ as in Remark 2.2. Then $A^{\prime}$ is not separated by any spherical-product subset $U^{\prime}$ that separates $S^{\prime}$ . Indeed, suppose that $A^{\prime}$ is separated by some spherical-product subset $U^{\prime}_{0}$ that separates $S^{\prime}$ . Here either $U$ does not separate any $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ or $U$ separates some $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ . In both cases, $U^{\prime}_{0}\subset X\cup U$ or $U^{\prime}_{0}\subset wYw^{-1}\cup U$ . For $U_{0}:=U^{\prime}_{0}$ (if $U^{\prime}_{0}\subset X\cup U$ ) and $U_{0}:=w^{-1}U^{\prime}_{0}w$ (if $U^{\prime}_{0}\not\subset X\cup U$ ), $U_{0}$ is a spherical-product subset of $S$ that separates $S$ and $U_{0}$ has to separate $A$ that is a contradiction. Hence $A^{\prime}\in\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(I)}$ .

Also by the same argument, for each $A^{\prime}\in\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(I)}$ , if $A$ is the subset of $S$ defined by $A:=A^{\prime}$ (if $A^{\prime}\subset X\cup U$ ) and $A:=w^{-1}A^{\prime}w$ (if $A^{\prime}\not\subset X\cup U$ ) then $A\in\widetilde{\mathcal{A}}_{S}^{\rm(I)}$ .

Thus

\begin{split}\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(I)}=&\{A:A\in\widetilde{\mathcal{A}}_{S}^{\rm(I)},\ A\subset X\cup U\}\\ &\ \hskip 34.1433pt\cup\{wAw^{-1}:A\in\widetilde{\mathcal{A}}_{S}^{\rm(I)},\ A\not\subset X\cup U\},\end{split}

and $|\widetilde{\mathcal{A}}_{S}^{\rm(I)}|=|\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(I)}|$ .

(a) We first suppose that $U$ does not separate any $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ .

Then for any minimal separation ${\mathcal{A}}$ of $S$ , $S^{\prime}$ is a twist of $S$ that is preserving ${\mathcal{A}}$ and the separation ${\mathcal{A}}^{\prime}$ of $S^{\prime}$ induced by ${\mathcal{A}}$ and the twist is minimal. Also for any minimal separation ${\mathcal{A}}^{\prime}$ of $S^{\prime}$ , the separation ${\mathcal{A}}$ of $S$ induced by ${\mathcal{A}}^{\prime}$ and the twist is minimal. Hence

\begin{split}\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(II)}=&\{B:B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)},\ B\subset X\cup U\}\\ &\ \hskip 34.1433pt\cup\{wBw^{-1}:B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)},\ B\not\subset X\cup U\}.\end{split}

Thus $(W,S)$ and $(W,S^{\prime})$ are type(I)-type(II)-compatible.

(b) We suppose that $U$ separates some unique $B_{0}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ . Here $U$ does not separate any $A\in\widetilde{\mathcal{A}}_{S}-\{B_{0}\}$ .

Let

	$\displaystyle\overline{\mathcal{U}}_{S}:=\bigcap\{{\mathcal{U}}_{\mathcal{A}}:\text{${\mathcal{A}}$ is a minimal separation of $S$}\}\ \text{and}$
	$\displaystyle\overline{\mathcal{U}}_{S^{\prime}}:=\bigcap\{{\mathcal{U}}_{{\mathcal{A}}^{\prime}}:\text{${\mathcal{A}}^{\prime}$ is a minimal separation of $S^{\prime}$}\},$

where ${\mathcal{U}}_{\mathcal{A}}$ and ${\mathcal{U}}_{{\mathcal{A}}^{\prime}}$ are the sets of separators of minimal separations ${\mathcal{A}}$ and ${\mathcal{A}}^{\prime}$ of $S$ and $S^{\prime}$ respectively. The standard separations $\widetilde{\mathcal{A}}_{S}$ and $\widetilde{\mathcal{A}}_{S^{\prime}}$ are defined by the separator sets $\overline{\mathcal{U}}_{S}$ and $\overline{\mathcal{U}}_{S^{\prime}}$ respectively.

Then for any minimal separation ${\mathcal{A}}$ of $S$ , $S^{\prime}$ is a twist of $S$ that is preserving $\{A\in{\mathcal{A}}:A\not\subset B_{0}\}$ ; that is, $A\subset X\cup U$ or $A\subset wYw^{-1}\cup U$ for each $A\in{\mathcal{A}}$ as $A\not\subset B_{0}$ . Also for any minimal separation ${\mathcal{A}}^{\prime}$ of $S^{\prime}$ , $S$ is a twist of $S^{\prime}$ that is preserving $\{A^{\prime}\in{\mathcal{A}}^{\prime}:A^{\prime}\not\subset B^{\prime}_{0}\}$ where $B^{\prime}_{0}:=(B_{0}\cap(X\cup U))\cup w(B_{0}\cap Y)w^{-1}$ ; that is, $A^{\prime}\subset X\cup U$ or $A^{\prime}\subset w^{-1}Yw\cup U$ for each $A^{\prime}\in{\mathcal{A}}^{\prime}$ as $A^{\prime}\not\subset B^{\prime}_{0}$ .

Hence for any $\overline{U}\in{\mathcal{U}}_{\mathcal{A}}$ as $\overline{U}\subset X\cup U$ , $\overline{U}\in{\mathcal{U}}_{{\mathcal{A}}^{\prime}}$ . Also for any $\overline{U}\in{\mathcal{U}}_{\mathcal{A}}$ as $\overline{U}\subset Y\cup U$ , $w\overline{U}w^{-1}\in{\mathcal{U}}_{{\mathcal{A}}^{\prime}}$ .

Let $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}-\{B_{0}\}$ . We consider the subset $B^{\prime}$ of $S^{\prime}$ as $B^{\prime}=B$ (if $B\subset X\cup U$ ) and $B^{\prime}=wBw^{-1}$ (if $B\not\subset X\cup U$ ). Then $B^{\prime}\in\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(II)}$ by the above argument. Here $B$ and $B^{\prime}$ are type(II)-compatible, since $B$ and $B^{\prime}$ are conjugate. Hence

\begin{split}\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(II)}\supset&\{B:B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}-\{B_{0}\},\ B\subset X\cup U\}\\ &\ \hskip 34.1433pt\cup\{wBw^{-1}:B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}-\{B_{0}\},\ B\not\subset X\cup U\}.\end{split}

Thus $|\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(II)}|\geq|\widetilde{\mathcal{A}}_{S}^{\rm(II)}|$ , because $|\widetilde{\mathcal{A}}_{S}^{\rm(II)}-\{B_{0}\}|=|\widetilde{\mathcal{A}}_{S}^{\rm(II)}|-1$ and there is a possibility that the corresponding subset $B^{\prime}_{0}=(B_{0}\cap(X\cup U))\cup w(B_{0}\cap Y)w^{-1}$ of $S^{\prime}$ to $B_{0}$ in $S$ is a union of some type(II) subsets. Also since $S$ is a twist of $S^{\prime}$ , by the same argument, $|\widetilde{\mathcal{A}}_{S}^{\rm(II)}|\geq|\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(II)}|$ . Hence $|\widetilde{\mathcal{A}}_{S}^{\rm(II)}|=|\widetilde{\mathcal{A}}_{S^{\prime}}^{\rm(II)}|$ . This implies that

B^{\prime}_{0}=(B_{0}\cap(X\cup U))\cup w(B_{0}\cap Y)w^{-1}

is a type(II) subset of $S^{\prime}$ . Here $B^{\prime}_{0}$ is a twist of $B_{0}$ that induces the twist of $S$ .

Thus, if $U$ separates some unique $B_{0}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ then $(W,S)$ and $(W,S^{\prime})$ are type(I)-type(II)-compatible.

(c) We suppose that $U$ separates some $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ that is not unique. Let $B_{1},\ldots,B_{l}$ be the type(II) subsets of $S$ such that $U$ (also $U\cap B_{i}$ ) separates $B_{i}$ for $i=1,\ldots,l$ .

For each $i=1,\ldots,l$ , we show that $U\cap B_{i}$ separates unique $B_{i}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ . We suppose that $U\cap B_{i}$ separates some $B_{j}$ as $j\neq i$ . Then $U\cap B_{i}\cap B_{j}$ separates $B_{j}$ . The standard separation $\widetilde{\mathcal{A}}_{S}$ of $S$ is defined by the separator-set $\overline{\mathcal{U}}_{S}$ . Here $B_{i}\cap B_{j}\subset U_{0}$ for some $U_{0}\in\overline{\mathcal{U}}_{S}$ . Then $U\cap B_{i}\cap B_{j}\subset U\cap U_{0}\subset U_{0}$ . Since $U\cap B_{i}\cap B_{j}$ separates $B_{j}$ , we have that $U_{0}$ separates $B_{j}$ . This contradicts that $U_{0}\in\overline{\mathcal{U}}_{S}$ does not separate any $B\in\widetilde{\mathcal{A}}_{S}$ .

Now $S$ is connected. Let $Y_{1},\ldots,Y_{t}$ be the connected components of $Y$ . Then $Y=Y_{1}\cup\cdots\cup Y_{t}$ and $S-U=X\cup Y=X\cup Y_{1}\cup\cdots\cup Y_{t}$ that are disjoint unions.

Let ${\mathcal{A}}_{0}$ be the maximal twist-rigid subset of $S$ , let ${\mathcal{A}}$ be the induced separation of $S$ by $U$ as Remark 2.2, and let ${\mathcal{A}}^{\prime}$ be a minimal separation of $S$ such that ${\mathcal{A}}^{\prime}\preceq{\mathcal{A}}$ .

Let $i\in\{1,\ldots,t\}$ and let

\overline{Y}_{i}:=\bigcup\{A\in{\mathcal{A}}_{0}:A\subset Y_{i}\cup U\ \text{and}\ A\not\subset U\}.

Then

\overline{Y}_{i}=\bigcup\{A\in{\mathcal{A}}^{\prime}:A\subset Y_{i}\cup U\ \text{and}\ A\not\subset U\}.

Since ${\mathcal{A}}^{\prime}$ is a separation of $S$ , there exists $A^{\prime}_{i}\in{\mathcal{A}}^{\prime}$ such that $\overline{Y}_{i}\cap A^{\prime}_{i}$ is a separator of ${\mathcal{A}}^{\prime}$ and $A^{\prime}_{i}\subset B_{j_{i}}$ for some $j_{i}\in\{1,\ldots,l\}$ . Then $\overline{Y}_{i}\cap A^{\prime}_{i}\subset U\cap B_{j_{i}}$ separates $B_{j_{i}}$ . (Also $\overline{Y}_{i}\cap A^{\prime}_{i}\subset U\cap B_{j_{i}}$ separates $S$ .) Here $B_{j_{i}}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ is uniquely determined.

Hence $t=l$ and the map $g(i)=j_{i}$ is bijective on the set $\{1,\ldots,l\}$ . We may suppose that $g(i)=i$ for any $i=1,\ldots,l$ . Then for each $i=1,\ldots,l$ , $\overline{Y}_{i}\cap A^{\prime}_{i}$ is a separator of ${\mathcal{A}}^{\prime}$ and $A^{\prime}_{i}\subset B_{i}$ .

Let $U_{i}:=(U\cap B_{i})\cup U_{\sigma}$ for each $i=1,\ldots,l$ . Since $U\cap B_{i}$ separates $S$ , $U_{i}$ separates $S$ . Here $U_{i}=U_{\sigma}\cup(U_{\nu}\cap U_{i})$ holds. Then $wU_{\sigma}w^{-1}=U_{\sigma}$ and $wU_{\nu}w^{-1}=U_{\nu}$ by Lemma 6.1. Hence $waw^{-1}=a$ for any $a\in U_{\nu}$ by the untangle-condition. Thus

	$\displaystyle wU_{i}w^{-1}$	$\displaystyle=(wU_{\sigma}w^{-1})\cup(w(U_{\nu}\cap U_{i})w^{-1})$
		$\displaystyle=U_{\sigma}\cup(U_{\nu}\cap U_{i})=U_{i},$

for any $i=1,\ldots,l$ .

Here $U_{i}$ separates $B_{i}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ . Indeed

B_{i}-U_{i}=B_{i}-(U_{i}\cap B_{i})=B_{i}-(U\cap B_{i})=B_{i}-U.

We show that $U_{i}$ separates the unique element $B_{i}$ of $\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ . We suppose that $U_{i}$ separates $B_{j}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ as $j\neq i$ . Then

B_{j}\cap U_{i}=(U\cap B_{i}\cap B_{j})\cup(U_{\sigma}\cap B_{j})

separates $B_{j}$ . Since the spherical subset $U_{\sigma}\cap B_{j}$ does not separate the type(II) subset $B_{j}$ by Lemma 6.2, $U\cap B_{i}\cap B_{j}\neq\emptyset$ and $B_{i}\cap B_{j}\neq\emptyset$ . Hence $B_{i}\cup B_{j}$ is connected. There exists the separation ${\mathcal{A}}_{1}$ induced by $U_{i}$ as in Remark 2.2 and we can obtain a minimal separation ${\mathcal{A}}^{\prime}_{1}$ as ${\mathcal{A}}^{\prime}_{1}\preceq{\mathcal{A}}_{1}$ . Here the spherical-product subset $U_{i}=(U\cap B_{i})\cup U_{\sigma}$ separates $B_{i}$ and $U_{i}$ separates $B_{j}$ by hypothesis. This contradicts that $B_{i}$ and $B_{j}$ are distinct type(II) subsets of $S$ . Thus $U_{i}$ separates unique $B_{i}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ .

For each $i=1,\ldots,l$ and for $X_{i}:=S-(U_{i}\cup Y_{i})$ , $S-U_{i}=X_{i}\cup Y_{i}$ that is a disjoint union and $o(xy)=\infty$ for any $x\in X_{i}$ and $y\in Y_{i}$ . Here

S=X\cup U\cup(Y_{1}\cup\cdots\cup Y_{l})

is a disjoint union and $wU_{i}w^{-1}=U_{i}$ as above.

Let $S_{1}:=X_{1}\cup U_{1}\cup wY_{1}w^{-1}$ that is a twist of $S$ by $U_{1}$ and $w$ . Then $U_{1}$ separates unique $B_{1}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ by the above argument. Hence $S$ and $S_{1}$ are type(I)-type(II)-compatible by the above argument (b). Then

S_{1}=X\cup U\cup(wY_{1}w^{-1}\cup Y_{2}\cup Y_{3}\cup\cdots\cup Y_{l}).

For the corresponding subset $B^{\prime}_{2}$ of $S_{1}$ to the subset $B_{2}$ of $S$ , $U_{2}$ separates unique $B^{\prime}_{2}\in\widetilde{\mathcal{A}}_{S_{1}}^{\rm(II)}$ in $S_{1}$ . Also for the corresponding subset $X^{\prime}_{2}$ of $S_{1}$ to the subset $X_{2}$ of $S$ , $S_{1}-U_{2}=X^{\prime}_{2}\cup Y_{2}$ is a disjoint union.

Let $S_{2}:=X^{\prime}_{2}\cup U_{2}\cup wY_{2}w^{-1}$ that is a twist of $S_{1}$ by $U_{2}$ and $w$ . Then $U_{2}$ separates unique $B^{\prime}_{2}\in\widetilde{\mathcal{A}}_{S_{2}}^{\rm(II)}$ in $S_{2}$ . Hence $S_{1}$ and $S_{2}$ are type(I)-type(II)-compatible by the above argument (b). Here

S_{2}=X\cup U\cup(wY_{1}w^{-1}\cup wY_{2}w^{-1}\cup Y_{3}\cup\cdots\cup Y_{l}).

By iterating this argument, we obtain a sequence of Coxeter generating sets $S=S_{0},S_{1},S_{2},\ldots,S_{l}$ such that $S_{i}$ is some twist of $S_{i-1}$ by $U_{i}$ and $w$ for each $i=1,\ldots,l$ . Here $U_{i}$ separates unique $B^{\prime}_{i}\in\widetilde{\mathcal{A}}_{S_{i}}^{\rm(II)}$ in $(W,S_{i})$ for any $i=1,\ldots,l$ , where $B^{\prime}_{i}$ is the type(II) subset of $S_{i}$ corresponding to $B_{i}$ in $S$ . Then $(W,S_{i})$ and $(W,S_{i+1})$ are type(I)-type(II)-compatible for each $i=1,\ldots,l$ by the above argument (b). Here

	$\displaystyle S=S_{0}=X\cup U\cup(Y_{1}\cup Y_{2}\cup Y_{3}\cup\cdots\cup Y_{l}),$
	$\displaystyle S_{1}=X\cup U\cup(wY_{1}w^{-1}\cup Y_{2}\cup Y_{3}\cup\cdots\cup Y_{l}),$
	$\displaystyle S_{2}=X\cup U\cup(wY_{1}w^{-1}\cup wY_{2}w^{-1}\cup Y_{3}\cup\cdots\cup Y_{l}),$
	$\displaystyle\ \cdots$
	$\displaystyle S_{l}=X\cup U\cup(wY_{1}w^{-1}\cup wY_{2}w^{-1}\cup\cdots\cup wY_{l}w^{-1})=S^{\prime}.$

Thus, we obtain that $(W,S)$ and $(W,S^{\prime})$ are type(I)-type(II)-compatible. ∎

Remark 6.4.

Let $(W,S)$ be a Coxeter system. Let $U_{1}$ and $U_{2}$ be spherical-product subsets of $S$ that separate $S$ such that for each $i=1,2$ ,

$\text{(a)}_{i}$

$U_{i}$ does not separate any $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ , or
$\text{(b)}_{i}$

$U_{i}$ separates some unique $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ .

We consider a twist $S^{\prime}=X\cup U_{1}\cup wYw^{-1}$ of $S$ by $U_{1}$ and $w\in W$ . Here we suppose that $wU_{1}w^{-1}=U_{1}$ , $S-U_{1}=X\cup Y$ is a disjoint union and $o(xy)=\infty$ for any $x\in X$ and $y\in Y$ .

Then we investigate a spherical-product subset $U^{\prime}_{2}$ of $S^{\prime}$ corresponding to $U_{2}$ in $S$ from the proof of Lemma 6.3.

$\text{(a)}_{1}$

$U_{1}$ does not separate any $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ ,

then for $U^{\prime}_{2}:=U_{2}$ (if $U_{2}\subset X\cup U_{1}$ ) and $U^{\prime}_{2}:=wU_{2}w^{-1}$ (if $U_{2}\not\subset X\cup U_{1}$ ), $U^{\prime}_{2}$ is the spherical-product subset of $S^{\prime}$ corresponding to $U_{2}$ in $S$ .

We suppose that

$\text{(b)}_{1}$

$U_{1}$ separates some unique $B_{1}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ .

$\text{(a)}_{2}$

$U_{2}$ does not separate any $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ ,

$\text{(b)}_{2}$

$U_{2}$ separates some unique $B_{2}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$

and if $B_{1}\neq B_{2}$ , then for $U^{\prime}_{2}:=U_{2}$ (if $U_{2}\subset X\cup U_{1}$ ) and $U^{\prime}_{2}:=wU_{2}w^{-1}$ (if $U_{2}\not\subset X\cup U_{1}$ ), $U^{\prime}_{2}$ is the spherical-product subset of $S^{\prime}$ corresponding to $U_{2}$ in $S$ .

Suppose that

$\text{(b)}_{1}$

$U_{1}$ separates some unique $B_{1}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ and
$\text{(b)}_{2}$

$U_{2}$ separates some unique $B_{2}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$

and suppose that $B_{1}=B_{2}$ . In this case, $U_{1}$ and $U_{2}$ both separate the unique type(II) subset $B_{1}=B_{2}$ .

We obtain the following theorem from the proof of Lemma 6.3 and Remark 6.4.

Theorem 6.5.

Let $(W,R)$ be a Coxeter system with the untangle-condition and let $S$ be a Coxeter generating set for $W$ obtained from $R$ by some finite twists. Then $(W,R)$ and $(W,S)$ are type(I)-type(II)-compatible.

Proof.

There exists a sequence $R=S_{1},S_{2},\ldots,S_{n}=S$ of Coxeter generating sets for $W$ such that $S_{i+1}$ is obtained from $S_{i}$ by some twist for any $i=1,\ldots,n-1$ . We suppose that the twist $S_{i+1}$ of $S_{i}$ is obtained by a spherical-product subset $U_{i}$ of $S_{i}$ and $w_{i}\in W$ as $w_{i}U_{i}w_{i}^{-1}=U_{i}$ for each $i=1,\ldots,n-1$ . Here by the proof of Lemma 6.3, for each $i=1,\ldots,n-1$ , either

(a)

$U_{i}$ does not separate any $B\in\widetilde{\mathcal{A}}_{S_{i}}^{\rm(II)}$ ,
(b)

$U_{i}$ separates some unique $B\in\widetilde{\mathcal{A}}_{S_{i}}^{\rm(II)}$ , or
(c)

$U_{i}$ separates some $B\in\widetilde{\mathcal{A}}_{S_{i}}^{\rm(II)}$ that is not unique.

Here if (c) $U_{i}$ separates some $B\in\widetilde{\mathcal{A}}_{S_{i}}^{\rm(II)}$ that is not unique, then by the proof of Lemma 6.3, there exist a sequence $U_{i}^{1},\ldots,U_{i}^{t}$ of spherical-product subsets and a sequence $S_{i}=S_{i}^{1},S_{i}^{2},\ldots,S_{i}^{t}=S_{i+1}$ of Coxeter generating set for $W$ such that each $S_{i}^{j+1}$ is obtained from $S_{i}^{j}$ by some twist induced by $U_{i}^{j}$ and $w_{i}$ as $w_{i}U_{i}^{j}w_{i}^{-1}=U_{i}^{j}$ and $U_{i}^{j}$ separates some unique $B_{i}^{j}\in\widetilde{\mathcal{A}}_{S_{i}^{j}}^{\rm(II)}$ .

Thus there exists a sequence $R=S_{1},S_{2},\ldots,S_{k}=S$ of Coxeter generating sets for $W$ such that for any $i=1,\ldots,k-1$ , $S_{i+1}$ is obtained from $S_{i}$ by some twist of some spherical-product subset $U_{i}$ of $S_{i}$ and $w_{i}\in W$ as $w_{i}U_{i}w_{i}^{-1}=U_{i}$ and either

(a)

$U_{i}$ does not separate any $B\in\widetilde{\mathcal{A}}_{S_{i}}^{\rm(II)}$ , or
(b)

$U_{i}$ separates some unique $B\in\widetilde{\mathcal{A}}_{S_{i}}^{\rm(II)}$ .

By the proof of Lemma 6.3 and Remark 6.4, we can obtain that $(W,R)=(W,S_{1})$ and $(W,S)=(W,S_{k})$ are type(I)-type(II)-compatible. ∎

We show the following theorem.

Theorem 6.6.

Let $(W,R)$ and $(W,S)$ be Coxeter systems with the untangle-condition. If $(W,R)$ and $(W,S)$ are type(I)-type(II)-compatible, then there exists a Coxeter generating set $\overline{R}$ obtained from $R$ by some finite twists such that $(W,\overline{R})$ and $(W,S)$ are compatible on the standard separations $\widetilde{\mathcal{A}}_{\overline{R}}$ and $\widetilde{\mathcal{A}}_{S}$ (hence they are some-separation-compatible).

Proof.

Let $A_{1}\in\widetilde{\mathcal{A}}_{R}^{\rm(II)}$ and $B_{1}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ such that $A_{1}$ and $B_{1}$ are type(II)-compatible. There exists $A^{\prime}_{1}$ obtained from $A_{1}$ by some finite twists that induce some twists of $R$ preserving $\widetilde{\mathcal{A}}_{R}-\{A_{1}\}$ such that $A^{\prime}_{1}$ and $B_{1}$ are conjugate in $W$ . Let $R_{1}$ be the Coxeter generating set for $W$ obtained from $R$ by the induced finite twists preserving $\widetilde{\mathcal{A}}_{R}-\{A_{1}\}$ such that $A^{\prime}_{1}$ in $R_{1}$ is conjugate to $B_{1}$ in $S$ .

Then by the proofs of Lemma 6.3 and Theorem 6.5, $(W,R)$ and $(W,R_{1})$ are type(I)-type(II)-compatible, $A^{\prime}_{1}$ is a type(II) subset of $R_{1}$ and

\widetilde{\mathcal{A}}_{R_{1}}=(\widetilde{\mathcal{A}}_{R}-\{A_{1}\})^{\prime}\cup\{A^{\prime}_{1}\},

where $(\widetilde{\mathcal{A}}_{R}-\{A_{1}\})^{\prime}$ is the set of the corresponding conjugate subsets of $R_{1}$ to the elements of $\widetilde{\mathcal{A}}_{R}-\{A_{1}\}$ in $R$ .

Hence

(1)

$(W,R_{1})$ and $(W,S)$ are type(I)-type(II)-compatible,
(2)

$A^{\prime}_{1}\in\widetilde{\mathcal{A}}_{R_{1}}^{\rm(II)}$ is conjugate to $B_{1}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ and
(3)

$\widetilde{\mathcal{A}}_{R_{1}}-\{A^{\prime}_{1}\}=(\widetilde{\mathcal{A}}_{R}-\{A_{1}\})^{\prime}$ .

Let $A^{\prime}_{2}\in\widetilde{\mathcal{A}}_{R_{1}}^{\rm(II)}-\{A_{1}^{\prime}\}$ and $B_{2}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}-\{B_{1}\}$ such that $A^{\prime}_{2}$ and $B_{2}$ are type(II)-compatible. Let $A_{2}\in\widetilde{\mathcal{A}}_{R}^{\rm(II)}$ be the corresponding conjugate subset of $R$ to $A^{\prime}_{2}\in\widetilde{\mathcal{A}}_{R_{1}}^{\rm(II)}$ in $R_{1}$ .

There exists $A^{\prime\prime}_{2}$ obtained from $A^{\prime}_{2}$ by some finite twists that induce some twists of $R_{1}$ preserving $\widetilde{\mathcal{A}}_{R_{1}}-\{A^{\prime}_{2}\}$ such that $A^{\prime\prime}_{2}$ and $B_{2}$ are conjugate. Let $R_{2}$ be the induced Coxeter generating set.

Then by the proofs of Lemma 6.3 and Theorem 6.5, $(W,R_{1})$ and $(W,R_{2})$ are type(I)-type(II)-compatible, $A^{\prime\prime}_{2}$ is a type(II) subset of $R_{2}$ and

\widetilde{\mathcal{A}}_{R_{2}}=(\widetilde{\mathcal{A}}_{R_{1}}-\{A^{\prime}_{2}\})^{\prime}\cup\{A^{\prime\prime}_{2}\}=(\widetilde{\mathcal{A}}_{R}-\{A_{1},\,A_{2}\})^{\prime\prime}\cup\{A^{\prime\prime}_{1},\,A^{\prime\prime}_{2}\},

where $A^{\prime\prime}_{1}$ is the corresponding conjugate subset of $R_{2}$ to $A^{\prime}_{1}$ in $R_{1}$ , and $(\widetilde{\mathcal{A}}_{R_{1}}-\{A^{\prime}_{2}\})^{\prime}$ (and $(\widetilde{\mathcal{A}}_{R}-\{A_{1},\,A_{2}\})^{\prime\prime}$ ) is the set of the corresponding conjugate subsets of $R_{2}$ to the elements of $\widetilde{\mathcal{A}}_{R_{1}}-\{A^{\prime}_{2}\}$ in $R_{1}$ (and $\widetilde{\mathcal{A}}_{R}-\{A_{1},\,A_{2}\}$ in $R$ respectively).

Hence

(1)

$(W,R_{2})$ and $(W,S)$ are type(I)-type(II)-compatible,
(2)

$A^{\prime\prime}_{1}\in\widetilde{\mathcal{A}}_{R_{2}}^{\rm(II)}$ is conjugate to $B_{1}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ ,
(3)

$A^{\prime\prime}_{2}\in\widetilde{\mathcal{A}}_{R_{2}}^{\rm(II)}$ is conjugate to $B_{2}\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ and
(4)

$\widetilde{\mathcal{A}}_{R_{2}}-\{A^{\prime\prime}_{1},\,A^{\prime\prime}_{2}\}=(\widetilde{\mathcal{A}}_{R}-\{A_{1},\,A_{2}\})^{\prime\prime}$ .

By iterating this argument, we obtain a Coxeter generating set $\overline{R}$ for $W$ from $R$ by some finite twists preserving $\widetilde{\mathcal{A}}_{R}^{\rm(I)}$ such that for each $A\in\widetilde{\mathcal{A}}_{R}^{\rm(II)}$ , the corresponding subset $A^{\prime}\in\widetilde{\mathcal{A}}_{\overline{R}}^{\rm(II)}$ is conjugate to some unique $B\in\widetilde{\mathcal{A}}_{S}^{\rm(II)}$ .

Also by the definition of type(I)-type(II)-compatible and the above argument, each $A\in\widetilde{\mathcal{A}}_{\overline{R}}^{\rm(I)}$ is conjugate to some unique $B\in\widetilde{\mathcal{A}}_{S}^{\rm(I)}$ .

Thus, $(W,\overline{R})$ and $(W,S)$ are compatible on the standard separations $\widetilde{\mathcal{A}}_{\overline{R}}$ and $\widetilde{\mathcal{A}}_{S}$ . ∎

We obtain the following theorem from Theorems 4.1, 6.5 and 6.6.

Theorem 6.7.

Let $(W,R)$ and $(W,S)$ be Coxeter systems with the untangle-condition. Then the following two statements are equivalent $:$

(i)

$R$ and $S$ are conjugate up to finite twists.
(ii)

$(W,R)$ and $(W,S)$ are type(I)-type(II)-compatible.

Proof.

By Theorem 6.5, if $R$ and $S$ are conjugate up to finite twists, then $(W,R)$ and $(W,S)$ are type(I)-type(II)-compatible.

If $(W,R)$ and $(W,S)$ are type(I)-type(II)-compatible, then there exists a Coxeter generating set $\overline{R}$ obtained from $R$ by some finite twists such that $(W,\overline{R})$ and $(W,S)$ are compatible on the standard separations $\widetilde{\mathcal{A}}_{\overline{R}}$ and $\widetilde{\mathcal{A}}_{S}$ by Theorem 6.6. Thus by Theorem 4.1, $R$ and $S$ are conjugate up to finite twists. ∎

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On rigidity of Coxeter systems up to finite twists and separations of Coxeter generating sets

Abstract.

Key words and phrases:

2010 Mathematics Subject Classification:

1. Introduction and preliminaries

Problem 1.1 (The isomorphism problem for Coxeter groups).

Theorem 1.2 ([12, Corollary 1.1]).

Problem 1.3 ([12]).

Theorem 1.4 ([5, Theorem 1.1]).

Definition 1.5.

Lemma 1.6 (cf. [2]).

Definition 1.7.

Definition 1.8.

Definition 1.9.

Example 1.10.

Example 1.11.

Theorem 1.12.

Corollary 1.13.

Definition 1.14.

Definition 1.15.

Definition 1.16.

Theorem 1.17.

Problem 1.18.

Problem 1.19.

2. Remarks and examples on separations

Remark 2.1.

Remark 2.2.

Example 2.3.

Example 2.4.

Example 2.5.

Example 2.6.

Example 2.7.

Example 2.8.

Example 2.9.

3. On separations of Coxeter generating sets and compatible Coxeter systems

Question 3.1.

Lemma 3.2.

Proof.

Remark 3.3.

4. Some-separation-compatible Coxeter generating sets are conjugate up to finite twists

Theorem 4.1.

Proof.

Corollary 4.2.

5. 𝒜~S\widetilde{\mathcal{A}}_{S} is a separation of SS

Proposition 5.1.

Lemma 5.2.

Proof.

Lemma 5.3.

Proof.

Lemma 5.4.

Proof.

6. Type(I)-type(II)-compatible and conjugate up to finite twists

Lemma 6.1.

Proof.

Lemma 6.2.

Proof.

Lemma 6.3.

Proof.

Remark 6.4.

Theorem 6.5.

Proof.

Theorem 6.6.

Proof.

Theorem 6.7.

Proof.

References

5. $\widetilde{\mathcal{A}}_{S}$ is a separation of $S$