Infinite Reduced Words, Lattice Property And Braid Graph of Affine Weyl Groups

TL;DR

This paper explores the structure of infinite reduced words in affine Weyl groups, establishing a bijection with biclosed sets, and proves the connectedness of their braid graphs, extending classical results to an infinite setting.

Contribution

It introduces a bijection between infinite reduced words and biclosed sets, and proves the lattice and connectivity properties of affine Weyl groups' braid graphs.

Findings

Bijection between infinite reduced words and biclosed sets.

Biclosed sets form a complete algebraic ortholattice in rank 3 affine Weyl groups.

Braid graphs of these groups are connected, extending finite case results.

Abstract

In this paper, we establish a bijection between the infinite reduced words of an affine Weyl group and certain biclosed sets of its positive system and determine all finitely generated biclosed sets in the positive system of an affine Weyl group. Using these results, we show first that the biclosed sets in the standard positive system of rank 3 affine Weyl groups when ordered by inclusion form a complete algebraic ortholattice and secondly that the (generalized) braid graphs of those Coxeter groups are connected, which can be thought of as an infinite version of Tit's solution to the word problem.

Equations

$$x\cdot \Gamma :=({\Phi }_x\backslash x(-\Gamma ))\cup (x(\Gamma )\backslash (-{\Phi }_x)).$$

$$\alpha \prec s_{\alpha }(\beta )\prec s_{\alpha }{s}_{\beta }(\alpha )\prec \cdots \prec s_{\beta }(\alpha )\prec \beta$$

$${\Phi }_{{\Delta }^{{}^{\prime }},{\Delta }^{{}^{\prime \prime }}}^{+}=({\Phi }^{+}\backslash {\Phi }_{{\Delta }^{\prime }})\cup {\Phi }_{{\Delta }^{\prime \prime }}$$

$$s_1{s}_2{s}_3\cdots$$

$${\Phi }_{s_1}\subset {\Phi }_{s_1{s}_2}\subset {\Phi }_{s_1{s}_2{s}_3}\subset \dots .$$

$${\Phi }_{s_1{s}_2{s}_3\cdots }=\bigcup _{i=1}^{\infty }{\Phi }_{s_1{s}_2\cdots s_i}$$

$${\Phi }_{s_1{s}_2{s}_3\cdots }^{\prime }={\Phi }^{+}\backslash {\Phi }_{s_1{s}_2{s}_3\cdots }.$$

$$s_1(\cdots (s_{k-1}(s_{k}w))\cdots )$$

$$b_1\le b_1\vee b_2\le b_1\vee b_2\vee b_3\le \dots .$$

$$b\wedge s_1\le b\wedge s_1{s}_2\le b\wedge s_1{s}_2{s}_3\le \dots .$$

$$W_1<W_2<W_3<\dots$$

$$w_0(W_1)\le w_0(W_2)\le w_0(W_3)\le \dots .$$

$$s_{1,1}{s}_{1,2}{s}_{1,3}\cdots$$

$$s_{2,1}{s}_{2,2}{s}_{2,3}\cdots$$

$$s_{3,1}{s}_{3,2}{s}_{3,3}\cdots$$

$$\cdots$$

$$b_1\le b_1\vee b_2\le b_1\vee b_2\vee b_2\le \dots .$$

$$b_1\le b_1\vee b_2\le b_1\vee b_2\vee b_2\le \dots$$

$$h_L:{\Phi }^{+}\to \mathbb{N}.$$

$$d_L:{\Phi }^{+}\to \mathbb{N}.$$

$$d_L(\alpha )=\max \{n|\alpha \text{can be written as the sum of}n\text{elements in}{\Phi }_{L,\varnothing }^{+}\}.$$

$$({\alpha }^{\vee },\gamma )\ge 0,$$

$$({\beta }^{\vee },\gamma )\ge 0,$$

$$(\alpha +\beta +\gamma ,-{\alpha }^{\vee })\ge 0,$$

$$(\alpha +\beta +\gamma ,-{\beta }^{\vee })\ge 0.$$

$$(\beta ,{\alpha }^{\vee })\le -2,$$

$$(\alpha ,{\beta }^{\vee })\le -2.$$

$$(\alpha ,\beta {)}^2\ge (\alpha ,\alpha )(\beta ,\beta ).$$

$$\alpha +\beta =(\alpha +{\beta }_{m+1})+({\beta }_1)+({\beta }_2)+\cdots +({\beta }_m).$$

$$\alpha +\beta =(\alpha +{\beta }_1+{\beta }_2+\cdots +{\beta }_m)+{\beta }_{m+1}.$$

Full text

Introduction

Given an arbitrary Coxeter system $(W,S)$ one can attach a root system to it in a canonical way. A particular total ordering on the set of the positive roots called a reflection order is the notion analogous to the reduced expression of the longest element of a finite Coxeter group. Such orderings encapsulate important combinatorics of the Iwahori-Hecke algebra, Kazhdan-Lusztig polynomials and $R$-polynomials. For example, reflection orders can be used to generalize the symmetry of the structure constants of the Iwahori-Hecke algebra associated to an infinite Coxeter group by playing the role of the reduced expressions of the longest element of a finite Coxeter group. See for example [4], [5]. Reflection orders can also be used to provide nonrecursive combinatorial formulae for Kazhdan-Lusztig polynomials and $R$-polynomails. See [1] Chapter 5.

However, the structure of the reflection orders of an infinite Coxeter group are in general not well understood. Several questions about the reflection orders and their initial sections are still open. Among them are two long-standing conjectures of Dyer, the first of which relates the initial sections of the reflection orders to the subsets of the positive system that are both closed and coclosed under a certain natural closure operator (these sets will be called biclosed sets). The first conjecture asserts that any maximal chain of biclosed sets arises from the set of all initial sections of some reflection order and that a set is biclosed if and only if it is an initial section of some reflection order. The second conjecture asserts that the initial sections of all reflection orders when partially ordered by inclusion form a complete lattice. These two conjectures can be motivated by the analogy between the reflection orders and the basic properties of the reduced expressions by considering the initial sections as elements of possibly “infinite length”. With this point of view, the second conjecture is a natural generalization to the set of initial sections of the classical result that the weak order of a finite Coxeter group is a lattice and the first conjecture is analogous to the fact that every element has a reduced expression. The conjectures are true for all finite Coxeter groups (where they reduce to well-known properties of weak order on $W$) and can be easily verified for the infinite dihedral case ($\widetilde{A}_{1}$).

In [6], Dyer proves the first conjecture in the case of affine Weyl groups. Provided that the first conjecture holds, the second conjecture can be restated as follows: the set of all biclosed sets when partially ordered by inclusion forms a complete lattice. Such a conjecture stimulates many studies. In [8], some evidence in support of the conjectures is given and the conjectural join and meet of two biclosed sets are described in terms of the closure operator. In [19], the author uses a slightly different convex closure operator and proves a variant of the conjecture in rank three cases. In [14], the authors characterize the existence of the join of a set of elements in $W$ by the corresponding inversion sets and their position relative to the imaginary cone. In [23], the author studies the conjecture using his theory of projective valued digraphs.

In this paper, we study the second conjecture in the case of rank three affine Weyl groups. To do this, we first consider the poset of those biclosed sets that are inversion sets of $W$ and the infinite reduced expressions. We show that such a poset is a complete meet semilattice. Then we investigate the infinite reduced words of an affine Weyl group and characterize the biclosed sets which are the inversion sets of infinite reduced words. An immediate consequence of this characterization is that for rank three affine Weyl groups any biclosed set is either an inversion set of some (finite or infinite) word or the complement of an inversion set. This special property makes the conjecture for rank three affine Weyl groups more accessible. To be precise, in rank three affine cases the poset of the biclosed sets splits into a lower half (consisting of the inversion sets of finite or infinite words) and a top half (consisting of the complements of the inversion sets of finite or infinite words). The key ingredient to show the conjecture is to prove that in the lower half, a family of biclosed sets is either bounded or its orthogonal family is bounded. We prove this property by some case studies. In this way we show that for rank three affine Weyl groups Dyer’s conjecture holds. We also classify the finitely generated biclosed sets in the positive system of an affine root system and this allows us to conclude that the lattices under study are algebraic. The paper is organized as follows.

In section 2, we consider the set $\overline{W}$ consisting of the infinite reduced words and the elements in the group $W$. The weak order on $W$ is naturally extended to $\overline{W}$. We study such an order and show that $\overline{W}$ is a complete meet semilattice which is a complete lattice if and only if $W$ is the weak direct product of countably many finite or locally finite Coxeter groups. We also show that $\overline{W}$ has a Join Orthogonality Property and admits maximal elements provided the rank of $W$ is at most infinite countable.

In section 3, We specialize to the case of affine Weyl groups and establish a ($W-$equivariant) bijection between the infinite reduced words and the certain biclosed sets in the positive system. By using this bijection we give a characterization of affine Weyl groups: a finite rank irreducible infinite Coxeter system is affine if and only if $\overline{W}$ admits finitely many maximal elements.

A consequence of the classification obtained in section 3 is that for rank three affine positive systems, the poset of the biclosed sets splits into two halves where the lower half consists of the inversion sets of the (finite or infinite) words. Section 4 is devoted to proving that the lower half of the poset has the following favorable property: a family of its elements is either bounded or its orthogonal family is bounded. By using this property we show the existence of the join of two biclosed sets and therefore prove that the poset of biclosed sets is a lattice. We also prove that in the positive system of an affine root system if the closure of the union of a family of finite biclosed sets is finite then it is biclosed.

In section 5, we study those biclosed sets in the positive system of an affine root system that are finitely generated. In rank three cases they are the compact elements of the lattice of biclosed sets. Our results show that those lattices are algebraic.

In section 6, we show that the (generalized) braid graphs (defined using reflection orders) of rank three affine Weyl groups are all connected, which generalizes the fact that the reduced expressions of a (finite) word are connected by braid moves. In this infinite setting, a biclosed set of the standard positive system can be thought of as an element of “infinite length”. The reflection order is a substitute for reduced expression. A braid move is characterized by the reversal of a dihedral string. The proof makes use of the explicit description of the biclosed sets as either inversion sets or their complements in affine rank three cases.

1. Preliminaries

A partially ordered set $L$ is a meet semilattice if any two elements $x,y\in L$ admit a greatest lower bound (meet), denoted by $x\wedge y$. A partially ordered set $L$ is called a complete meet semilattice if any subset $A\subset L$ admits a greatest lower bound (meet), denoted by $\bigwedge A.$ A partially ordered set $L$ is a join semilattice if any two elements $x,y\in L$ admit a least upper bound (join), denoted by $x\vee y$. A partially ordered set $L$ is called a complete join semilattice if any subset $A\subset L$ admits a least upper bound (join), denoted by $\bigvee A.$ If a partially ordered set is both a meet semilattice and a join semilattice, it is called a lattice. If a partially ordered set is both a complete meet semilattice and a complete join semilattice, it is called a complete lattice. An ortholattice $L$ is a lattice $L$ such that it has a minimum element, denoted by 0, and a maximum element, denoted by 1 and there exists an order-reversing involution on $L$, denoted by $x\mapsto x^{oc}$ such that $x\wedge x^{oc}=0$ and $x\vee x^{oc}=1$ for all $x\in L$.

An element $x$ in a complete lattice $L$ is called compact if $x\leq\bigvee Y$ for some $Y\subset L$ implies that $x\leq\bigvee Y_{0}$ for some finite $Y_{0}\subset Y.$ A complete lattice is called algebraic if every element of it is a join of a set of compact elements.

We record the following easy lemma that will be used frequently in the paper.

Lemma 1.1**.**

Let $(P,\leq)$ be a complete meet semilattice. Suppose that $A$ is a bounded subset of $P$. Then the join of $A$ exists in $P$.

For a totally ordered set $T$, a set $I\subset T$ is called an initial section if for all $x\in I,y\in T\backslash I$ we have that $x\leq y$. A set $F\subset T$ is called a final section if for all $x\in F,y\in T\backslash F$ we have that $x\geq y.$

Let $(P,\leq)$ be a partially ordered set. A subset $Q$ of $P$ is said to be a totally ordered subset (chain) of $P$ if the restriction of $\leq$ to $Q$ is a total order. A totally ordered subset $Q$ of $P$ is called a maximal totally ordered subset if for any $x\in P\backslash Q$, $Q\cup\{x\}$ is not a totally ordered subset. A poset $(P,\leq)$ is said to be chain complete if any chain of it has a least upper bound and a greatest lower bound.

Let $(W,S)$ be a Coxeter system with root system $\Phi$, see Chapter 5 of [16] for example. Note that in this paper we allow $S$ to be infinite. The set of positive roots and negative roots are denoted by $\Phi^{+}$ and $\Phi^{-}$ respectively. The set of simple roots is denoted by $\Pi$. An effective way to visualize the infinite root system in lower rank is to consider the projective representation of $W$. The idea is to project all the roots in the standard reflection representation to an affine hyperplane which is transverse to $\Phi^{+}.$ For the projective representation see Section 2 of [10], Section 2 of [15] and Section 2.3 of [19]. Given $w\in W$ define the inversion set of $w^{-1}$ to be $\Phi_{w}:=\{\alpha\in\Phi^{+}|w^{-1}(\alpha)\in\Phi^{-}\}.$ Let $T:=\{wsw^{-1}|w\in W,s\in S\}.$ The set $T$ is called the set of reflections. There exists a canonical bijection between $T$ and $\Phi^{+}$ and denote by $s_{\alpha}$ the reflection corresponding to a positive root $\alpha$. Let $W^{\prime}$ be a subgroup of $W$ generated by a subset of $T$. Then $W^{\prime}$ is called a reflection subgroup. It can be shown that $(W^{\prime},\{t\in T\cap W^{\prime}|l(t^{\prime}t)>l(t),\forall t^{\prime}\in T\cap W^{\prime}\backslash\{t\}\})$ is a Coxeter system. For reflection subgroups see Section 3 of [7].

The (right) weak order $\leq$ on $W$ is defined as $x\leq y$ if and only if $\Phi_{x}\subset\Phi_{y}$ for all $x,y\in W$. The poset $(W,\leq)$ is a complete meet semilattice. It is a complete lattice if and only if $W$ is finite. See Chapter 3 of [1].

A set $\Gamma\subset\Phi^{+}$ is called closed (in $\Phi^{+}$) if for all $\alpha,\beta\in\Gamma$, $k_{1}\alpha+k_{2}\alpha\in\Phi^{+}$ with $k_{1},k_{2}\in\mathbb{R}_{\geq 0}$ implies that $k_{1}\alpha+k_{2}\alpha\in\Gamma$. A subset $\Gamma$ is called coclosed (in $\Phi^{+}$) if $\Phi^{+}\backslash\Gamma$ is closed (in $\Phi^{+}$). A set $\Gamma\subset\Phi^{+}$ is called biclosed (in $\Phi^{+}$) if $\Gamma$ and $\Phi^{+}\backslash\Gamma$ are both closed (in $\Phi^{+}$). Replacing $\Phi^{+}$ with $\Phi$ in the above definition, we define the closed sets, coclosed sets and biclosed sets in $\Phi$. We usually denote by $\overline{X}$ the closure of a set $X$. Such a closure operator is well-defined (since the intersection of closed sets stays closed).

It is known that the finite biclosed sets in $\Phi^{+}$ are precisely the inversion sets of the elements in the Coxeter group (See [8] Lemma 4.1(d)).

Denote by $\mathscr{B}(\Phi^{+})$ the set of all biclosed sets in $\Phi^{+}$, regarded as a partially ordered set under set inclusion. Two biclosed sets $B_{1},B_{2}$ of this poset are said to be in the same block if they differ by finitely many roots.

For a set $\Gamma$ biclosed in $\Phi^{+}$ and $x\in W$, define

[TABLE]

Then $x\cdot\Gamma$ is again a biclosed set and this is a group action of $W$ on the set of biclosed sets in $\Phi^{+}.$ See [8] Lemma 4.1(a)(c).

A total order $\prec$ on $\Phi^{+}$ is called a reflection order if for all $\alpha,\beta\in\Phi^{+},\alpha\prec\beta$ and $a,b\in\mathbb{R}_{\geq 0}$ such that $a\alpha+b\beta\in\Phi^{+}$ we have that $\alpha\prec a\alpha+b\beta\prec\beta.$ Suppose that $(W,S)$ is a dihedral Coxeter system with $S=\{s_{\alpha},s_{\beta}\}$. Then

[TABLE]

is a reflection order.

A locally finite Coxeter system $(W,S)$ is a Coxeter system such that $W$ is infinite and each parabolic subgroup $W_{J}$ of it is finite if $J$ is finite. Irreducible locally finite Coxeter systems are precisely the ones of type $A_{\infty},A_{\infty,\infty},B_{\infty},D_{\infty}.$ See Exercise 4.14 of [18].

Now let $\Phi$ be an irreducible crystallographic root system of a Weyl group $W$ and be contained in the Euclidean space $V$ as in [2] Chapter VI, §1. A subset $\Gamma\subset\Phi$ is called $\mathbb{Z}-$closed if for all $\alpha,\beta\in\Gamma$ such that $\alpha+\beta\in\Phi$, we have that $\alpha+\beta\in\Gamma.$

Let $\Delta$ be a simple system of $\Phi$ with the corresponding positive system $\Phi^{+}$. Let $\gamma$ be the highest root. Suppose that $\Delta^{{}^{\prime}}$ and $\Delta^{{}^{\prime\prime}}$ are two subsets of $\Delta$ which are orthogonal (i.e. $(\alpha,\beta)=0,\forall\alpha\in\Delta^{{}^{\prime}},\beta\in\Delta^{{}^{\prime\prime}}$). Define

[TABLE]

where $\Phi_{\Delta^{\prime}}$ is the root subsystem of $\Phi$ generated by $\Delta^{\prime}.$

In [6] (Theorem 1.15) and in [12] (Theorem 4) all biclosed sets in $\Phi$ are classified:

Theorem 1.2**.**

Let $\Gamma\subset\Phi$. The set $\Gamma$ is biclosed in $\Phi$ if and only if there exists a simple system $\Delta$ with the corresponding positive system $\Psi^{+}$ and orthogonal subsets $\Delta^{{}^{\prime}},\Delta^{{}^{\prime\prime}}\subset\Delta$ such that $\Gamma=\Psi^{+}_{\Delta^{{}^{\prime}},\Delta^{{}^{\prime\prime}}}$.

Define a real vector space $V^{\prime}=V\oplus\mathbb{R}\delta$ and extend the inner product on $V$ to $V^{\prime}$ by requiring $(\delta,V^{\prime})=0.$ For $\alpha\in\Phi^{+}$, we define $\widehat{\alpha}=\{\alpha+n\delta|n\in\mathbb{Z}_{\geq 0}\}\subset V^{\prime}$. For $\alpha\in\Phi^{-}$, we define $\widehat{\alpha}=\{\alpha+(n+1)\delta|n\in\mathbb{Z}_{\geq 0}\}\subset V^{\prime}$. For a set $\Gamma\subset\Phi$, define $\widehat{\Gamma}=\bigcup_{\alpha\in\Gamma}\widehat{\alpha}\subset V^{\prime}$.

Denote by $\widetilde{W}$ the (irreducible) affine Weyl group corresponding to $W$. Then $\widetilde{W}$ has $\widehat{\Phi}\cup-\widehat{\Phi}$ as root system with $\widehat{\Phi}(=\widetilde{\Phi}^{+})$ as the set of positive roots. The set of simple roots $\widetilde{\Delta}=\Delta\bigcup\{\delta-\gamma\}$. For this construction, see Chapter 6 of [18] or Section 3.3 of [11]. It is known that $\widetilde{W}=W\ltimes Q$ where $Q$ is the coroot lattice. Denote by $\pi$ the natural epimorphism from $\widetilde{W}$ to $W$.

In [6] (Proposition 5.11, Corollary 5.12) all biclosed sets in $\widehat{\Phi}$ are classified:

Theorem 1.3**.**

(1) Let $\Gamma\subset\widehat{\Phi}$. The set $\Gamma$ is biclosed in $\widehat{\Phi}$ if and only if $\Gamma=w\cdot\widehat{\Lambda}$ where $\Lambda$ is biclosed in $\Phi$ and $w\in\widetilde{W}$.

(2) The biclosed sets in $\widehat{\Phi}$ that are in the same block as $\widehat{\Psi^{+}_{\Delta_{1},\Delta_{2}}}$ are those of the form $w\cdot\widehat{\Psi^{+}_{\Delta_{1},\Delta_{2}}}$ for some $w\in W^{\prime}<\widetilde{W}$ where $W^{\prime}$ is the reflection subgroup generated by $\{s_{\alpha}|\alpha\in\widehat{\Psi_{\Delta_{1}\cup\Delta_{2}}}\}$. In particular $w\cdot\widehat{\Psi^{+}_{\Delta_{1},\Delta_{2}}}=\widehat{\Psi^{+}_{\Delta_{1},\Delta_{2}}}+(\Phi_{W^{\prime}}^{+}\cap w(-\Phi_{W^{\prime}}^{+}))$ where $+$ denotes the set symmetric difference and $\Phi_{W^{\prime}}^{+}\subset\widehat{\Phi}$ is the positive system of $W^{\prime}$.

It is worth noting that in (2) of the above Theorem $\widehat{\Psi_{\Delta_{1}}}$ and $\widehat{\Psi_{\Delta_{2}}}$ are orthogonal. So, letting $W_{i},i=1,2$ be the reflection subgroup generated by $\widehat{\Psi_{\Delta_{i}}},i=1,2$, one has $W^{\prime}=W_{1}\times W_{2}$.

We also need the following results which are proved in [6] (Lemma 5.10(f), Proposition 5.20, Proposition 5.22).

Theorem 1.4**.**

(1) Any maximal totally ordered subset of $\mathscr{B}(\widehat{\Phi})$ is the set of all initial sections of some reflection order of $\widehat{\Phi}$ and any biclosed set in $\widehat{\Phi}$ is an initial section of some reflection order.

(2) Any maximal totally ordered subset of $\mathscr{B}(\widehat{\Phi})$ is also a maximal totally ordered subset of the poset $(\mathcal{P}(\widehat{\Phi}),\subset)$, the poset of all subsets of $\widehat{\Phi}$ under inclusion.

(3) Denote by $\pi$ the canonical epimorphism from $\widetilde{W}$ to $W$. Suppose that $\Psi^{+}$ is a positive system in $\Phi$ and $M,N$ are orthogonal subsets of its simple system. Let $w\in\widetilde{W}$. Then the biclosed sets $w\cdot\widehat{\Psi^{+}_{M,N}}$ and $\widehat{\pi(w)(\Psi^{+}_{M,N})}=\widehat{\Lambda^{+}_{M^{\prime},N^{\prime}}}$ (where $\Lambda^{+}=\pi(w)(\Psi^{+})$, $M^{\prime}=\pi(w)M$, $N^{\prime}=\pi(w)N$) are in the same block in the poset $\mathscr{B}(\widehat{\Phi})$.

In this paper we denote the disjoint union by $\uplus$ and we denote the cardinality of a countably infinite set by $\aleph_{0}$.

2. infinite reduced words of a Coxeter System

Let $(W,S)$ be a Coxeter system with $W$ infinite.

Definition 2.1**.**

An infinite reduced expression is a sequence

[TABLE]

with $s_{i}\in S$ such that for all $k\geq 1$, $s_{1}s_{2}\cdots s_{k}$ is a reduced expression.

If $s_{1}s_{2}s_{3}\cdots$ is an infinite reduced expression then immediately we have that

[TABLE]

Definition 2.2**.**

Let $s_{1}s_{2}s_{3}\cdots$ be an infinite reduced expression. Define

[TABLE]

and

[TABLE]

It is clear from the definition that $\Phi_{s_{1}s_{2}s_{3}\cdots}$ is infinite and biclosed. Given two infinite reduced expressions $s_{1}s_{2}s_{3}\cdots$ and $r_{1}r_{2}r_{3}\cdots$. Write $s_{1}s_{2}s_{3}\cdots\thicksim r_{1}r_{2}r_{3}\cdots$ if $\Phi_{s_{1}s_{2}s_{3}\cdots}=\Phi_{r_{1}r_{2}r_{3}\cdots}$. It is clear that this is an equivalence relation.

Definition 2.3**.**

Let $L$ be the set of all infinite reduced expressions. Define $W_{l}=L/\thicksim$. An element in $W_{l}$ is called an infinite reduced word of $(W,S)$. Denote $\overline{W}=W_{l}\uplus W.$ Given $w=[s_{1}s_{2}s_{3}\cdots]\in W_{l}$, $s_{1}s_{2}s_{3}\cdots$ is called a reduced expression of $w$.

If no confusion is possible, we write $w=s_{1}s_{2}s_{3}\cdots$ if $w=[s_{1}s_{2}s_{3}\cdots]$.

Definition 2.4**.**

Given $x,y\in\overline{W}$, define $x\leq y$ if and only if $\Phi_{x}\subset\Phi_{y}$. Call this order the weak order on $\overline{W}$. Let $x\in W_{l}$. An element $w\in W$ is called a prefix of $x$ if $w\leq x$. Say two elements $x,y\in\overline{W}$ are orthogonal, denoted by $x\perp y$ if $\Phi_{x}\cap\Phi_{y}=\emptyset.$

In the above definition we extend the weak order on $W$ to $\overline{W}$ in an obvious way. Such an order has been studied by various authors, for example see [14], [20] and [21].

Remark*.*

It follows from the definition that we can define the weak order in terms of braid moves. Let $w_{1},w_{2}$ be two infinite reduced words. Suppose that the reduced expression $s_{1}s_{2}\cdots s_{k}$ (denoted by $\underline{u_{k}}$) appear as the first $k$ letters in a reduced expression of $w_{1}$. Then $w_{1}\leq w_{2}$ if and only if for any reduced expression of $w_{2}$ and any $k$, one can apply braid moves finitely many times to convert such reduced expression to a reduced expression starting with $\underline{u_{k}}$.

Example*.*

Consider the Coxeter group $W$ of type $\widetilde{A}_{2}$, i.e. the group with presentation $\langle s_{1},s_{2},s_{3}|s_{1}^{2}=s_{2}^{2}=s_{3}^{2}=(s_{1}s_{2})^{3}=(s_{1}s_{3})^{2}=(s_{2}s_{3})^{2}=e\rangle$. Let $w_{1}=s_{1}s_{2}s_{3}s_{1}s_{2}s_{3}s_{1}s_{2}s_{3}\cdots$ and $w_{2}=s_{1}s_{2}s_{1}s_{3}s_{1}s_{2}s_{1}s_{3}s_{1}s_{2}s_{1}s_{3}\cdots$. Then $w_{1}\lvertneqq w_{2}$. To see this, one notes that any (finite) prefix of $w_{1}$ is also a (finite) prefix of $w_{2}$. But the prefix $s_{1}s_{2}s_{1}s_{3}$ of $w_{2}$ is not a prefix of $w_{1}.$

Definition 2.5**.**

Let $s\in S$ and $w$ be an infinite reduced word of $W$. We define $sw$ in the following way:

Case I: if $s\perp w$, we define $sw$ to be the word obtained by concatenating $s$ with a reduced expression of $w$.

Case II: if $s\not\perp w$ (i.e. $s\in\Phi_{w}$), choose one reduced expression $s_{1}s_{2}s_{3}\cdots$ of $w$ and one can find $u=s_{1}s_{2}\cdots s_{k}$ such that $l(su)=l(u)-1$. We define $sw$ to be the word obtained by concatenating one reduced expression of $su$ with the word $s_{k+1}s_{k+2}\cdots$.

This multiplication is well-defined thanks to Lemma 2.7 below.

Definition 2.6**.**

Let $u\in W$ and $w$ be an infinite reduced word. Let $s_{1}s_{2}\cdots s_{k}$ be any (reduced or non-reduced) expression of $u$. Define $uw$ to be

[TABLE]

Again Lemma 2.7 below ensures that this multiplication is well-defined.

We summarize the basic properties of infinite reduced words in the following lemmas. Their proofs reduce easily to the analogous statements of Coxeter groups.

Lemma 2.7**.**

(a) Let $w_{1}=s_{1}s_{2}s_{3}\cdots$ and $w_{2}=r_{1}r_{2}r_{3}\cdots$ be two infinite reduced words of $W$. Then $w_{1}\leq w_{2}$ if and only if any prefix of $w_{1}$ is a prefix of $w_{2}$.

(b) Let $x,y\in\overline{W}$. Then $x\perp y$ if and only if for any prefix $u$ of $x$ and any prefix $v$ of $y$ we have that $l(u^{-1}v)=l(u)+l(v)$.

(c) The multiplications in Definition 2.5 and 2.6 are all well-defined, i.e. in 2.5 case I the product $sw$ is independent of the choice of the reduced expression of $w$, and in 2.5 case II the product $sw$ is independent of the choice of reduced expression of $w$, the choice of $u$ and the choice of the reduced expression of $su$, and in 2.6 the product $uw$ does not depend on the choice of the expression of $u$.

(d) Let $x,y\in W$ and $z\in\overline{W}$. Then $x\cdot\Phi_{z}=\Phi_{xz}$, $x\cdot\Phi_{z}^{\prime}=\Phi_{xz}^{\prime}$ and $(xy)z=x(yz).$

Proof.

(a) and (b) follow easily from the definition and Proposition 3.1.3 of [1]. Now let $s_{1}s_{2}s_{3}\cdots$ and $r_{1}r_{2}r_{3}\cdots$ be two reduced expressions of an infinite reduced word $w$ and suppose that $s\perp w$. It follows that $\Phi_{ss_{1}s_{2}s_{3}}=\bigcup_{i=1}^{\infty}\Phi_{ss_{1}s_{2}\cdots s_{i}}=\bigcup_{i=1}^{\infty}(\Phi_{s}\cup s\Phi_{s_{1}s_{2}\cdots s_{i}})=\Phi_{s}\cup s(\bigcup_{i=1}^{\infty}\Phi_{s_{1}s_{2}\cdots s_{i}})=\Phi_{s}\cup s(\bigcup_{i=1}^{\infty}\Phi_{r_{1}r_{2}\cdots r_{i}})=\bigcup_{i=1}^{\infty}\Phi_{sr_{1}r_{2}\cdots r_{i}}=\Phi_{sr_{1}r_{2}\cdots}$. Now instead assume that $s\not\perp w$. Let $u=s_{1}s_{2}\cdots s_{k}$ and $l(su)=k-1$. By the basic properties of the reduced expressions of an element in a Coxeter group, one has $l(sus_{k+1}s_{k+2}\cdots s_{j})=j-1$ for any $j\geq k+1$. Therefore concatenating a reduced expression of $su$ (denoted by $\underline{su}$) with $s_{k+1}s_{k+2}\cdots$ gives rise to a valid infinite reduced word. It also follows that $\Phi_{\underline{su}s_{k+1}s_{k+2}\cdots}=\Phi_{su}\cup su\Phi_{s_{k+1}s_{k+2}\cdots}$. Therefore the product is independent of the choice of the reduced expression of $su$. Also note that $\Phi_{su}\cup su\Phi_{s_{k+1}s_{k+2}\cdots}=(s\Phi_{u}\backslash(-\Phi_{s}))\cup s(u\Phi_{s_{k+1}s_{k+2}\cdots})=s(\Phi_{w})\backslash(-\Phi_{s})$. This shows that the product is independent of the choice of the reduced expression of $w$ and the prefix $u$. Therefore the multiplication in Definition 2.5 is well defined.

Now let $s\in S,w\in\overline{W}$. One checks in two cases: $s\perp w$ and $s\not\perp w$ (by using the formula $x\cdot\Gamma:=(\Phi_{x}\backslash x(-\Gamma))\cup(x(\Gamma)\backslash(-\Phi_{x}))$) that $s\cdot\Phi_{w}=\Phi_{sw}$. This proves that the multiplication in Definition 2.6 is well defined. It also follows that $x\cdot\Phi_{z}=\Phi_{xz}$ and $x(yz)=(xy)z$ in (d). Finally $x\cdot\Phi_{z}^{\prime}=\Phi_{xz}^{\prime}$ follows from the general fact that $x\cdot(\Phi^{+}\backslash\Gamma)=\Phi^{+}\backslash(x\cdot\Gamma)$ for $\Gamma$ biclosed in $\Phi^{+}$. ∎

Now we study the weak order on $\overline{W}$. First we prove an easy but useful lemma.

Lemma 2.8**.**

Let $A\subset W$ and $|A|<\infty$. Suppose that $A$ is bounded in $\overline{W}$. Then $A$ is bounded in $W$ and $\bigvee A$ exists in $W$.

Proof.

Suppose that $A$ is bounded by $s_{1}s_{2}s_{3}\cdots$. Then every element of $A$ is bounded by some $s_{1}s_{2}\cdots s_{k}\in W$. Since $A$ is finite, we can take the longest of these bounds and it is a bound in $W$ of $A$. The second assertion follows from the fact that $W$ is a complete meet semilattice and Lemma 1.1. ∎

Theorem 2.9**.**

The poset $(\overline{W},\leq)$ is a complete meet semilattice.

Proof.

We first show

Claim: Let $w$ be an infinite reduced word. Then the set of its prefixes is countably infinite.

Let $s_{1}s_{2}s_{3}\cdots$ be a reduced expression of $w$. Then the set of the prefixes of $w$ is $\bigcup_{i=1}^{\infty}\{u\in W|u\leq s_{1}s_{2}\cdots s_{i}\}$ and each set in this union is finite. Therefore the claim is true.

We first show that any non-empty subset of $W_{l}$ has a meet. Take $\emptyset\neq A\subset W_{l}$. As a consequence of the claim, the set $B:=\{w\in W|w\leq a,\forall a\in A\}$ is countable. So we let $B=\{b_{1},b_{2},\dots\}$. Then we obtain an ascending chain under weak order:

[TABLE]

Each term in this chain exists in $W$ by Lemma 2.8. Therefore this chain gives rise to a well-defined infinite reduced word or an element in $W$. Denote it by $v$. We shall show that $v$ is the meet of $A$.

First we note that it is immediate that $v\geq b,\forall b\in B$. Then take some $w\in\overline{W}$ which is a lower bound of $A$. So any prefix of $w$ is in $B$ and therefore is bounded by $v$. This shows that $w\leq v$. So $v$ is the meet of $A$.

Since $W$ is a complete meet semilattice, any subset of $W$ admits a meet in $W$ and it is clear that such meet remains to be the meet in $\overline{W}$.

Now we consider $A\subset\overline{W}$ with $A\cap W_{l}\neq\emptyset$ and $A\cap W\neq\emptyset$. Let $a=\bigwedge(A\cap W_{l})$ and $b=\bigwedge(A\cap W)$. Fix the reduced expressions $s_{1}s_{2}s_{3}\cdots$ (finite if $a\in W$) for $a$. Then we have a chain:

[TABLE]

This chain defines an infinite reduced word or an element in the group. Denote it by $y$. One readily sees that all lower bounds of $A$ are less than or equal to $y$. ∎

The poset $\overline{W}$ and its semilattice property was considered by various authors, for example in [20] Theorem 4.10. We incorporate this proof for general Coxeter groups and emphasize that this meet semilattice is complete.

It is natural to ask when $\overline{W}$ is a complete lattice. This question is answered in the following Theorem.

Theorem 2.10**.**

The poset $\overline{W}$ is a complete lattice if and only if $W\simeq{}^{w}\Pi_{i=1}^{\infty}W_{i}$ where each $W_{i}$ is either a finite or locally finite irreducible Coxeter group and ${}^{w}\Pi$ denotes the weak direct product.

Proof.

We first prove the “only if” part. A Coxeter group $W$ can always be written as the weak direct product of a family of irreducible Coxeter groups, called its components. Since any two elements of $W$ have an upper bound in $\overline{W}$, they have an upper bound in $W$ by Lemma 2.8. So each group of that family must either be locally finite or finite. If the family consists of uncountably many non-trivial Coxeter groups, then we take one generator from each of these Coxeter groups. Clearly their join cannot possibly be an infinite reduced word or an element in $W$.

To prove the converse, we first note that for a finite Coxeter group $W$ there exists a longest element $w_{0}(W)$. For a locally finite Coxeter group $W$, one can find a sequence of parabolic subgroups

[TABLE]

such that each $W_{i}$ is a finite and $\bigcup_{i=1}^{\infty}W_{i}=W$ by the classification of locally finite Coxeter groups. Thus we have a chain

[TABLE]

This chain defines an infinite reduced word $w_{0}(W)$ which is the unique maximal element in $(\overline{W},\leq)$. Now suppose that $W\simeq{}^{w}\Pi_{i=1}^{\infty}W_{i}$ and that $w_{0}(W_{i})=s_{i,1}s_{i,2}s_{i,3}\cdots.$ Listing these elements

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and then enumerating $s_{i,j}$’s in a zig-zag way (i.e. $s_{1,1}s_{1,2}s_{2,1}s_{3,1}s_{2,2}s_{1,3}s_{1,4}s_{2,3}\cdots$) give rise to a well-defined infinite reduced word which is maximal under the weak order. The existence of such an element makes the complete semilattice a complete lattice. ∎

Remark*.*

It can be shown that for a locally finite Coxeter group any biclosed set in $\Phi^{+}$ is the inversion set of a (finite or infinite) word. Thus the above theorem proves that the poset of biclosed sets in $\Phi^{+}$ is a lattice for those types of Coxeter groups.

Next we show that $\overline{W}$ has the following favorable property and we refer to it as Join Orthogonal Property.

Theorem 2.11**.**

Suppose that $A\subset\overline{W}$ and $A$ admits a join $w\in\overline{W}.$ Let $v\in\overline{W}$ such that $a\perp v$ for all $a\in A$. Then $w\perp v$.

Proof.

Suppose that $w$ is in $W$ (then $A\subset W$). If $v\in W$, the assertion follows from Lemma 4.2(b) in [8]. If $v\in W_{l}$, then every prefix of $v$ is orthogonal to $w$ by the above argument and thus $v\perp w$. Suppose that $w$ is in $W_{l}$. Consider $B=\{b\in W|\exists a\in A,b\leq a\}$. Clearly $\bigvee B=\bigvee A=w$ and the set $B$ is contained in the set of prefixes of $w$ and hence countable by the claim in the proof of Theorem 2.9. Then write $B=\{b_{1},b_{2},\dots\}$. We also have that $b_{i}\perp v$ for all $i$. Therefore any prefix of $w$ must be smaller than $\bigvee_{i=1}^{n}b_{i}$ for some $n$ and thereby orthogonal to $v$. ∎

The following proposition will be used in proving a characterization of affine Weyl groups in the next section. Now suppose that $W$ is countably generated (i.e. $|S|\leq\aleph_{0}$). As a consequence $W$ is countable.

Proposition 2.12**.**

The poset $(\overline{W},\leq)$ contains maximal elements. Every element $w\in\overline{W}$ is less than or equal to some maximal element.

Proof.

To prove the first assertion it suffices to show that the condition of Zorn’s Lemma is satisfied. Suppose that we have an ascending chain in $\overline{W}$: $\{w_{i}\}_{i\in I}$. Then the set $B=\{b\in W|b\leq w_{i}$ for some $i\}$ is countable since $W$ is countable. Let $B=\{b_{1},b_{2},b_{3},\dots\}$. For $b_{1},b_{2},\dots,b_{k}$, one can find some $w_{j}$ which is their upper bound. Hence $b_{1},b_{2},\dots,b_{k}$ admit a join in $W$ by Lemma 2.8. So we have an ascending chain of elements in $W$:

[TABLE]

This defines an element in $\overline{W}$ (denoted by $v$) which will be shown to be the join of the original chain $\{w_{i}\}_{i\in I}$. Take any $w_{i}$ in the chain. Since any finite prefix of $w_{i}$ is $b_{l}$ for some $l$, it is less than $v$. Therefore we have that $w_{i}\leq v$. Let $u$ be any upper bound of the chain $\{w_{i}\}_{i\in I}$. Since $\bigvee_{i=1}^{k}b_{i}$ is bounded by some $w_{j}$, $\bigvee_{i=1}^{k}b_{i}\leq w_{j}\leq u$. Therefore $v\leq u$. Hence we conclude that $v=u.$ The second assertion follows by the same argument applied to $\{x\in\overline{W}|x\geq w\}$. ∎

Another consequence of the above proof is the following Lemma.

Lemma 2.13**.**

Let $W$ be countably generated and $\{w_{i}\}_{i\in I}$ be an ascending chain in $\overline{W}$ under the weak order. Then the join of $\{w_{i}\}_{i\in I}$ exists in $\overline{W}$ and $\Phi_{\vee_{i}w_{i}}=\cup_{i\in I}\Phi_{w_{i}}$.

Proof.

Such a join (denoted by $v$) is constructed in the proof of Proposition 2.12. We only need to show that $\Phi_{v}=\cup_{i\in I}\Phi_{w_{i}}.$ Take a finite prefix $b$ of $w_{i}$ for some $i\in I$. By the construction of $v$, $\Phi_{b}\subset\Phi_{v}$. Therefore $\Phi_{w_{i}}\subset\Phi_{v}$ for any $i$. Hence $\Phi_{v}\supset\cup_{i\in I}\Phi_{w_{i}}.$ Conversely $v$ is the word defined by the ascending chain:

[TABLE]

where $\{b_{1},b_{2},b_{3},\dots\}$ is the set of finite prefixes of $\{w_{i}\}_{i\in I}$. Hence the inversion set of each term in this ascending chain is contained in $\cup_{i\in I}\Phi_{w_{i}}$. Therefore we see the reverse inclusion. ∎

3. infinite reduced words And Biclosed Sets In an Affine Root

System

The main aim of this section is to obtain a description of $\Phi_{w}$ for $w\in\overline{W}$ in the affine case. Related descriptions have been studied by other authors, for example in [20] Section 4. We characterize those biclosed sets arising from infinite reduced words by using the description of affine root systems as introduced in Section 1 and the action of $W$ on the set of biclosed sets. Our proof is different and uses an interesting property of the finite root systems (Proposition 3.2) which has not been observed previously to the knowledge of the author and this property will be used again later in the paper (Theorem 5.2). A proof of the classification without using Proposition 3.2 is also possible and is given in a subsequent paper [24].

Let $\Phi$ be an irreducible crystallographic root system of a Weyl group $W$. Let $\Pi$ be a simple system of $\Phi$ and $L\varsubsetneqq\Pi.$ Denote the positive system corresponding to $\Pi$ by $\Phi^{+}$.

We begin by defining the function

[TABLE]

Suppose that $\Phi^{+}\ni\alpha=\sum_{\alpha_{i}\in\Pi}k_{i}\alpha_{i}$. Let $h_{L}(\alpha):=\sum_{\alpha_{i}\not\in L}k_{i}$. (Note that each $k_{i}$ is an integer and that $h_{L}$ is clearly additive.)

We define another function

[TABLE]

If $\alpha\in\Phi^{+}_{L,\emptyset}$,

[TABLE]

If $\alpha\not\in\Phi^{+}_{L,\emptyset}$ (that is $\alpha\in\Phi_{L}$), then $d_{L}(\alpha)=0.$

When $L$ is fixed (as we do in the following discussion), we denote the functions $d_{L}$ and $h_{L}$ by $d$ and $h$ respectively for convenience.

In order to establish the relation between these two functions, we need the following lemma which is suggested and proved by Matthew Dyer. (This has been generalized to sets of $n$ roots in [22] Theorem 3.7.)

Lemma 3.1**.**

Let $\alpha,\beta,\gamma\in\Phi^{+}$. Suppose that $\alpha+\beta$ and $\alpha+\beta+\gamma$ are both in $\Phi^{+}$. Then either $\alpha+\gamma$ or $\beta+\gamma$ is in $\Phi^{+}$.

Proof.

Assume that to the contrary, neither $\alpha+\gamma$ nor $\beta+\gamma$ is a (positive) root. Since the root system is crystallographic, we have that $(\alpha,\gamma)\geq 0$ and that $(\beta,\gamma)\geq 0.$ Also since neither $(\alpha+\beta+\gamma)+(-\beta)$ nor $(\alpha+\beta+\gamma)+(-\alpha)$ is a (positive) root, we have that $(\alpha+\beta+\gamma,-\beta)\geq 0$ and that $(\alpha+\beta+\gamma,-\alpha)\geq 0.$ Let $v^{\vee}=\frac{2v}{(v,v)}$. Then we conclude that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

This gives that

[TABLE]

[TABLE]

This yields that

[TABLE]

But by Cauchy-Schwarz inequality this forces that $\alpha=\beta$, which again contradicts the assumption that $\alpha+\beta$ is a root. ∎

Proposition 3.2**.**

Let functions $d,h$ be defined as above. Then for any $\alpha\in\Phi^{+}$, $d(\alpha)=h(\alpha)$.

Proof.

(M. Dyer) If $\alpha\not\in\Phi^{+}_{L,\emptyset},$ then the both sides are [math]. So we assume that $\alpha\in\Phi^{+}_{L,\emptyset}$. (This implies that $\alpha\not\in\Phi_{L}$.)

Suppose that $\alpha$ can be written as a sum of $n$ elements in $\Phi^{+}_{L,\emptyset}.$ When writing such an element as the positive linear combination of simple roots, there must be at least one simple root outside $L$ appearing in the sum. Hence we see that $d(\alpha)\leq h(\alpha)$.

Now we prove that $h(\alpha)\leq d(\alpha)$. For simple roots we clearly have that $h(\alpha)=d(\alpha)$. Also note that any positive root $\alpha$ can be written as $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{k}$ where each $\alpha_{i}$ is simple and $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{i}$ is a root. So it suffices to show that $d(\alpha+\beta)\geq d(\alpha)+d(\beta)$ for $\alpha,\beta,\alpha+\beta$ all positive roots.

If $d(\alpha)=d(\beta)=0$, the inequality is trivial. If $d(\alpha)>0$ and $d(\beta)>0$, then the inequality follows from the definition of the function $d$. So it suffices to prove the case where $d(\alpha)=0$ and $d(\beta)>0.$ We prove this by induction on $d(\beta)$. The case where $d(\beta)=1$ is trivial. Suppose that for $d(\beta)\leq m$, $d(\alpha)+d(\beta)\leq d(\alpha+\beta)$.

Assume that $d(\beta)=m+1$ and $\beta=\beta_{1}+\beta_{2}+\cdots+\beta_{m+1}$ with each $\beta_{i}\not\in\Phi_{L}$. We can even assume that $\beta_{1}+\beta_{2}+\cdots+\beta_{i}$ is a root for all $i$ by Chapter VI §1 Proposition 19 in [2]. By Lemma 3.1, either $\alpha+\beta_{m+1}$ or $\alpha+\beta_{1}+\beta_{2}+\cdots+\beta_{m}$ is a (positive) root. In the first case

[TABLE]

So we have that $d(\alpha+\beta)\geq m+1$. In the second case,

[TABLE]

We have that $d(\alpha+\beta_{1}+\beta_{2}+\cdots+\beta_{m})>0$ and so by the previous discussion $d(\alpha+\beta)\geq d(\alpha+\beta_{1}+\beta_{2}+\cdots+\beta_{m})+1$. But by induction $d(\alpha+\beta_{1}+\beta_{2}+\cdots+\beta_{m})\geq m$ and we are done. ∎

Corollary 3.3**.**

If $\alpha=k_{1}\beta+k_{2}\gamma$ with $\alpha,\beta,\gamma\in\Phi^{+}$ and $k_{1},k_{2}\in\mathbb{R}_{\geq 0}$, then $d(\alpha)=k_{1}d(\beta)+k_{2}d(\gamma)$.

Proof.

By Proposition 3.2 it suffices to show that $h(\alpha)=k_{1}h(\beta)+k_{2}h(\gamma)$. But this is clear from the definition of the $h$ function. ∎

From now on we consider the associated affine root system $\widetilde{\Phi}$ with the positive system $\widehat{\Phi}$. The corresponding (irreducible) affine Weyl group is denoted by $\widetilde{W}$.

Definition 3.4**.**

Denote the set

[TABLE]

by $(\Phi^{+}_{L,\emptyset})_{n}$.

Lemma 3.5**.**

The identity

[TABLE]

holds.

Proof.

Let $0\leq k\leq nd(\alpha)$. Then we can find $0\leq a_{1},a_{2},\dots,a_{d(\alpha)}\leq n$ such that

[TABLE]

By definition of the function $d$, we have that $\alpha=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{d(\alpha)}$ where $\alpha_{i}\in\Phi^{+}_{L,\emptyset}$. Therefore

[TABLE]

This combined with Chapter VI §1 Proposition 19 in [2] shows that $\alpha+k\delta\in\overline{(\Phi^{+}_{L,\emptyset})_{n}}.$

Conversely we just need to prove that the set on the right is closed, which is enough since it lies between the set and its closure. Assume that $\alpha_{1},\alpha_{2}\in\Phi^{+}_{L,\emptyset}$, $0\leq k_{1}\leq nd(\alpha_{1}),0\leq k_{2}\leq nd(\alpha_{2})$, $m_{1},m_{2}\geq 0$ ($k_{1},k_{2}\in\mathbb{Z}$, $m_{1},m_{2}\in\mathbb{R}$) and $m_{1}(\alpha_{1}+k_{1}\delta)+m_{2}(\alpha_{2}+k_{2}\delta)$ is a root in $\widehat{\Phi}$. Then $m_{1}\alpha_{1}+m_{2}\alpha_{2}\in\Phi.$ Since $h(\alpha_{i})>0$ for $i=1,2$ and $m_{1},m_{2}$ cannot both be zero so we have that $h(m_{1}\alpha_{1}+m_{2}\alpha_{2})>0$ and therefore $m_{1}\alpha_{1}+m_{2}\alpha_{2}\in\Phi^{+}_{L,\emptyset}$. By Corollary 3.3, $d(m_{1}\alpha_{1}+m_{2}\alpha_{2})=m_{1}d(\alpha_{1})+m_{2}d(\alpha_{2})\geq m_{1}\frac{k_{1}}{n}+m_{2}\frac{k_{2}}{n}$. So

[TABLE]

This implies that $m_{1}\alpha_{1}+m_{2}\alpha_{2}+(m_{1}k_{1}+m_{2}k_{2})\delta$ is still in $\{\alpha+k\delta|\alpha\in\Phi^{+}_{L,\emptyset},0\leq k\leq nd(\alpha)\}$. ∎

An immediate consequence of this lemma is

Lemma 3.6**.**

The set $\overline{(\Phi^{+}_{L,\emptyset})_{n}}$ is finite.

We also have

Lemma 3.7**.**

The set $\overline{(\Phi^{+}_{L,\emptyset})_{n}}$ is biclosed.

Proof.

We only need to show that the set is coclosed.

Case I:

Consider $\alpha+k_{1}\delta,\beta+k_{2}\delta$ where $\alpha,\beta\in\Phi^{+}_{L,\emptyset}$ and $k_{1}>nd(\alpha),k_{2}>nd(\beta).$ Suppose that $t_{1}\alpha+t_{2}\beta+(t_{1}k_{1}+t_{2}k_{2})\delta,t_{1},t_{2}\geq 0$ is a root of $\widehat{\Phi}$. So $d(t_{1}\alpha+t_{2}\beta)=t_{1}d(\alpha)+t_{2}d(\beta)<t_{1}\frac{k_{1}}{n}+t_{2}\frac{k_{2}}{n}$. Therefore

[TABLE]

That implies that $t_{1}\alpha+t_{2}\beta+(t_{1}k_{1}+t_{2}k_{2})\delta\not\in\overline{(\Phi^{+}_{L,\emptyset})_{n}}.$

Case II:

Consider $\alpha+k_{1}\delta,\beta+k_{2}\delta$ where $\alpha,\beta\not\in\Phi^{+}_{L,\emptyset}$. Suppose that $t_{1}\alpha+t_{2}\beta+(t_{1}k_{1}+t_{2}k_{2})\delta,t_{1},t_{2}\geq 0$ is a root of $\widehat{\Phi}$. Then $t_{1}\alpha+t_{2}\beta\not\in\Phi^{+}_{L,\emptyset}$ as $\Phi^{+}_{L,\emptyset}$ is a biclosed set in $\Phi.$

Case III:

Consider $\alpha+k_{1}\delta,\beta+k_{2}\delta$ where $\alpha\not\in\Phi^{+}_{L,\emptyset}$ and $\beta\in\Phi^{+}_{L,\emptyset}$ and $k_{2}>nd(\beta)$. Suppose that $t_{1}\alpha+t_{2}\beta+(t_{1}k_{1}+t_{2}k_{2})\delta,t_{1},t_{2}\geq 0$ is a root of $\widehat{\Phi}$. We only need to examine the case where $t_{1}\alpha+t_{2}\beta\in\Phi^{+}_{L,\emptyset}$. In this case $d(t_{1}\alpha+t_{2}\beta)\leq t_{2}d(\beta)<t_{2}\frac{k_{2}}{n}$(If $\alpha$ is positive then the first $\leq$ is $=$). Therefore

[TABLE]

That implies that $t_{1}\alpha+t_{2}\beta+(t_{1}k_{1}+t_{2}k_{2})\delta\not\in\overline{(\Phi^{+}_{L,\emptyset})_{n}}.$ ∎

Corollary 3.8**.**

The biclosed set $\widehat{\Phi^{+}_{L,\emptyset}}$ is of the form $\Phi_{w}$ where $w$ is an infinite reduced word (in $\overline{\widetilde{W}}$).

Proof.

This follows from the above Lemma 3.6 and Lemma 3.7 and

[TABLE]

∎

Lemma 3.9**.**

Let $\Psi^{+}$ be a positive system of $\Phi$, $\Delta$ be its simple system, $N,M\subset\Delta$ with $N$ orthogonal to $M$ and $M\neq\emptyset$. Then $\widehat{\Psi_{N,M}^{+}}$ is not of the form $\Phi_{w}$ where $w$ is an infinite reduced word.

Proof.

Let $\alpha\in M.$ Then by definition $\alpha+p\delta$ and $-\alpha+q\delta$ are both in $\widehat{\Psi_{N,M}^{+}}$ for all $p,q\in\mathbb{Z}_{\geq 1}$. Suppose that to the contrary $\widehat{\Psi_{N,M}^{+}}=\Phi_{w}$ where $w$ is an infinite reduced word. Then they must both be contained in some finite biclosed set $\Phi_{u},u\in\widetilde{W}$. Since $\Phi_{u}$ is biclosed so we have that $2(\alpha+p\delta)+(-\alpha+q\delta)=\alpha+(2p+q)\delta\in\Phi_{u},3(\alpha+p\delta)+2(-\alpha+q\delta)=\alpha+(3p+2q)\delta,\dots$.

This implies that $\Phi_{u}$ is infinite and this is a contradiction. ∎

The proof of the above lemma also implies the following useful result that will be repeatedly used later in the paper.

Lemma 3.10**.**

Let $C$ be a closed set in $\widehat{\Phi}$ and $\alpha\in\Phi$. If there exist $\alpha+n\delta,-\alpha+m\delta,m,n\in\mathbb{Z}_{\geq 1}$ both in $C$, then $C$ is infinite.

Lemma 3.11**.**

One has the following equality of sets:

$\{v\cdot\widehat{\Psi^{+}_{M,N}}|v\in\widetilde{W},\Psi^{+}\,\text{is a positive system of }\,\Phi,M,N\,\text{are orthogonal subsets}\newline \text{of the simple system of }\Psi^{+}\}$* $=\{w\cdot\widehat{\Phi^{+}_{L,K}}|w\in\widetilde{W},L,K\subset\Pi,L\,\text{and}\,K\,\text{are othogonal}\}$. Furthermore, if $v\cdot\widehat{\Psi^{+}_{M,N}}=w\cdot\widehat{\Phi^{+}_{L,K}}$, then $|M|=|L|$ and $|N|=|K|.$*

Proof.

It is clear that we only need to show that the left hand side is contained in the right hand side. Furthermore it suffices to show that $\widehat{\Psi^{+}_{M,N}}$ is contained in the right hand side for $\Psi^{+}$ a positive system of $\Phi$ and $M,N$ two orthogonal subsets of the simple system of $\Psi^{+}$.

There exists $z\in W$ such that $z\Psi^{+}=\Phi^{+}.$ Then $z\Psi^{+}_{M,N}=\Phi^{+}_{z(M),z(N)}$ is a biclosed set in $\Phi$. Now by Theorem 1.4 (2) take $u\in\widetilde{W}$ such that $\pi(u)=z^{-1}$ (where $\pi$ is the canonical surjection from $\widetilde{W}$ to $W$), we have that $\widehat{\Psi^{+}_{M,N}}$ and $u\cdot\widehat{\Phi^{+}_{z(M),z(N)}}$ differ only by finitely many roots. Now by Theorem 1.3 (2), there exists $u^{\prime}\in W^{\prime}$ where $W^{\prime}$ is the reflection subgroup generated by $\{s_{\alpha}|\alpha\in\widehat{\Psi_{M\cup N}}\}$ such that $u^{\prime}\cdot(u\cdot\widehat{\Phi^{+}_{z(M),z(N)}})=\widehat{\Psi^{+}_{M,N}}$. So the equality of the two sets is proved.

For the second assertion, one notes that if $v\cdot\widehat{\Psi^{+}_{M,N}}=w\cdot\widehat{\Phi^{+}_{L,K}}$ one has that $\pi(w^{-1}v)(\Psi^{+})=\Phi^{+},\pi(w^{-1}v)(M)=L$ and $\pi(w^{-1}v)(N)=K.$ Therefore $|M|=|L|$ and $|N|=|K|.$ ∎

Theorem 3.12**.**

The map

[TABLE]

gives

(1) a bijection between $\widetilde{W}_{l}$ and the following subset of $\mathscr{B}(\widehat{\Phi})$

[TABLE]

(2) a bijection between $\overline{\widetilde{W}}$ and the following subset of $\mathscr{B}(\widehat{\Phi})$

[TABLE]

Proof.

Suppose that $u\in\overline{\widetilde{W}}$. In view of Theorem 1.2, 1.3 and Lemma 3.11, we need to examine whether a set $\widehat{\Phi^{+}_{M,N}},M,N\subset\Pi$ is in the image of the given map. Corollary 3.8 shows that $\widehat{\Phi_{L,\emptyset}^{+}}$ is in the image of the map $u\mapsto\Phi_{u}$. By Lemma 2.7 (d) $w\cdot\widehat{\Phi_{L,\emptyset}^{+}}$ is also in the image. Lemma 3.9 shows that $\widehat{\Phi_{N,M}^{+}}$ where $M\neq\emptyset$ is not in the image. Then neither is $w\cdot\widehat{\Phi_{N,M}^{+}}$ by Lemma 2.7 (d). Finally we see that $w\cdot\widehat{\Phi_{\Pi,\emptyset}^{+}}$ is finite and thus not in the image if $u\in\widetilde{W}_{l}$. ∎

Remark*.*

It is worth noting that the case $L=\Pi$ gives the finite biclosed sets.

Corollary 3.13**.**

Let $\Psi^{+}$ be a positive system in $\Phi$ and $\Delta$ be the simple system of it. Suppose that $M\subset\Delta$ and that $v\in\widetilde{W}$. Then $v\cdot\widehat{\Psi^{+}_{M,\emptyset}}$ is equal to $\Phi_{x}$ for some $x\in\overline{\widetilde{W}}.$

Proof.

The assertion follows immediately from Lemma 3.11, Theorem 3.12 and Lemma 2.7 (d). ∎

Example*.*

Let $W$ be of type $A_{2}$. Denote by $\alpha,\beta$ the two simple roots of $\Phi^{+}$, the standard positive system of $\Phi$. The biclosed set $\widehat{\Phi^{+}_{\{\beta\},\emptyset}}=\widehat{\{\alpha,\alpha+\beta\}}$ is an inversion set of an infinite reduced word by the above Theorem. In fact it is equal to $\Phi_{s_{\alpha}s_{\beta}s_{\delta-\alpha-\beta}s_{\alpha}s_{\beta}s_{\delta-\alpha-\beta}s_{\alpha}s_{\beta}s_{\delta-\alpha-\beta}\cdots}$

In the subsequent work [24], for arbitrary infinite Coxeter groups a conjectural characterization of biclosed sets from infinite words by the semilattice property of the associated “twisted weak order” is given.

As a consequence of the bijection discussed above, we provide a characterization of affine Weyl groups. Let $B$ be a subset of $\widehat{\Phi}$. Define $I_{B}=\{\alpha\in\Phi||\widehat{\alpha}\bigcap B|=\aleph_{0}\}$. Define $A_{B}=\{\alpha\in\Phi|\widehat{\alpha}\bigcap B\neq\emptyset\}$.

Lemma 3.14**.**

Let $\Gamma\subset\Phi$ be $\mathbb{Z}-$closed. If $\Gamma\cap-\Gamma=\emptyset$ then $\Gamma$ is contained in a positive system of $\Phi$.

Proof.

This is [2] Chapter IV, $\S 1$, Proposition 22. ∎

Lemma 3.15**.**

Let $\Gamma$ be a closed subset of $\widetilde{\Phi}^{+}$. Then $A_{\Gamma}$ is $\mathbb{Z}-$closed in $\Phi$.

Proof.

Let $\alpha,\beta\in A_{\Gamma}$ such that $\alpha+\beta$ is a root. This implies that $\alpha+m\delta$ and $\beta+n\delta$ are in $\Gamma$ for some $m,n\in\mathbb{Z}_{\geq 0}$. Then $m+n\in\mathbb{Z}_{\geq 0}$. If $\alpha+\beta\in\Phi^{-}$, then at least one root among $\alpha$ and $\beta$ is in $\Phi^{-}$, in which case $m+n\in\mathbb{Z}_{>0}$. So $\alpha+\beta+(m+n)\delta\in\widetilde{\Phi}^{+}$, and therefore in $\Gamma$. ∎

Theorem 3.16**.**

Let $(W,S)$ be a finite rank, irreducible, infinite Coxeter system. Then $W$ is affine if and only if $\overline{W}$ admits finitely many maximal elements.

Proof.

The existence of such maximal elements is guaranteed by Proposition 2.12. For only if, we show that the maximal elements in $\overline{W}$ where $W$ is an affine Weyl group correspond to $\widehat{\Psi^{+}}$ where $\Psi^{+}$ is a positive system of $\Phi$ (and therefore there are only finitely many of them). First from Corollary 3.13 every $\widehat{\Psi^{+}}$ is indeed the inversion set of an infinite reduced word. Lemma 3.10 shows that any biclosed set properly containing $\widehat{\Psi^{+}}$ is not the inversion set of an element in $\overline{W}$ (such biclosed sets must contain some $\alpha+m\delta$ and $-\alpha+n\delta$ for some $\alpha\in\Phi$). Now it suffices to show that any other element $u$ in $\overline{W}$ is not maximal. By Lemma 3.15, $A_{\Phi_{u}}$ is $\mathbb{Z}$-closed. We claim that $A_{\Phi_{u}}\cap-A_{\Phi_{u}}=\emptyset$. To see the claim suppose that $\alpha\in A_{\Phi_{u}}\cap-A_{\Phi_{u}}$. So some $\alpha+m\delta$ and $-\alpha+n\delta$ are both in $\Phi_{u}$. Then Lemma 3.10 shows that this is not possible. Hence by Lemma 3.14, $A_{\Phi_{u}}\subset\Psi^{+}$ where $\Psi^{+}$ is a positive system in $\Phi$ and $\Phi_{u}\subset\widehat{\Psi^{+}}$.

Conversely, by [13] Theorem 2.7.2, for $W$ finite rank, irreducible, infinite and non-affine, there exists a universal reflection subgroup of arbitrarily large rank. Suppose that $\overline{W}$ has $k$ maximal elements. Choose a rank $k+1$ universal reflection subgroup generated by $\{t_{1},t_{2},\dots,t_{k+1}\}$ (as canonical generators). These reflections correspond to a set of positive roots $\{\alpha_{1},\alpha_{2},\dots,\alpha_{k+1}\}$. There must be a maximal element $w$ such that $\Phi_{w}$ contains $\alpha_{i}$ and $\alpha_{j},i\neq j$ by pigeonhole. But $\{k_{1}\alpha_{i}+k_{2}\alpha_{j}|k_{1},k_{2}\in\mathbb{R}_{\geq 0}\}\cap\Phi$ is infinite and therefore they cannot be contained in any finite biclosed set. This is a contradiction. ∎

Example*.*

Let $W$ be of type $A_{2}$. Then $\overline{\widetilde{W}}$ has 6 maximal elements. They are in bijection with the 6 positive systems of $\Phi$. Denote by $\alpha,\beta$ the two simple roots of $\Phi^{+}$.

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4. A Lattice of Biclosed Sets in Rank 3 Affine Root System

Let $W$ be a Weyl group and $\Phi$ be its irreducible crystallographic root system. Let $\Phi^{+}$ be a positive system of $\Phi$ with $\Pi$ the corresponding simple system. Denote by $\widetilde{W}$ the corresponding (irreducible) affine Weyl group. Then $\widetilde{\Phi}:=\widehat{\Phi}\bigcup-\widehat{\Phi}$ is the root system of $\widetilde{W}$ and $\widetilde{\Phi}^{+}=\widehat{\Phi}$. For $\alpha\in\Phi^{+}$, define $\alpha_{0}=\alpha$. For $\alpha\in\Phi^{-}$, define $\alpha_{0}=\alpha+\delta.$ The following lemma in fact reveals the so called dominance order in an affine root system.

Lemma 4.1**.**

Let $x\in\widetilde{W}$. If $\alpha\in\Phi$ and $\alpha_{0}+k\delta\in\Phi_{x}$ with $k\geq 0$, then $\alpha_{0},\alpha_{0}+\delta,\dots,\alpha_{0}+k\delta$ are all contained in $\Phi_{x}$.

Proof.

Suppose that $x^{-1}(\alpha_{0})=\beta+m\delta$ for some $\beta\in\Phi$. One has that $x^{-1}(\alpha_{0}+k\delta)=\beta+(k+m)\delta\in\widetilde{\Phi}^{-}$. If $\beta\in\Phi^{+}$ then $k+m\leq-1$ i.e. $m\leq-k-1.$ Then for $0\leq p\leq k$, $x^{-1}(\alpha_{0}+p\delta)=\beta+(m+p)\delta$. Note that $m+p\leq m+k\leq-1$. So $x^{-1}(\alpha_{0}+p\delta)\in\widetilde{\Phi}^{-}.$ If $\beta\in\Phi^{-}$ then $k+m\leq 0$ i.e. $m\leq-k.$ Then for $0\leq p\leq k$, $x^{-1}(\alpha_{0}+p\delta)=\beta+(m+p)\delta$. Note that $m+p\leq m+k\leq 0$. So $x^{-1}(\alpha_{0}+p\delta)\in\widetilde{\Phi}^{-}.$ ∎

The following theorem answers a question from [8] in the case of irreducible affine Weyl groups (and easily generalized to affine Coxeter groups). See the remark after Theorem 1.5 of [8].

Theorem 4.2**.**

Let $X\subset\widetilde{W}$. If $\overline{\bigcup_{x\in X}\Phi_{x}}$ is finite, then it is biclosed and the join of $X$ exists.

Proof.

We note that it could not happen that $\alpha+k\delta$ and $-\alpha+m\delta$ are both in the finite set $\overline{\bigcup_{x\in X}\Phi_{x}}$ thanks to Lemma 3.10. Therefore by Lemma 3.15 all $\alpha\in\Phi$ with $\alpha+p\delta\in\overline{\bigcup_{x\in X}\Phi_{x}}$ for some $p$ form a $\mathbb{Z}-$closed set and this set has empty intersection with its negative. So by Lemma 3.14 they belong to some positive system $\Xi^{+}$ of $\Phi$. So we have that $\overline{\bigcup_{x\in X}\Phi_{x}}\subset\widehat{\Xi^{+}}$. Since $\overline{\bigcup_{x\in X}\Phi_{x}}$ is finite, it is contained in $\Phi_{u}$ where $u$ is a finite prefix of the infinite reduced word corresponding to $\widehat{\Xi^{+}}$. So $X$ has a bound in $\widetilde{W}$ and the assertion follows from Theorem 1.5(a) of [8]. ∎

Lemma 4.3**.**

Let $\Gamma$ be a closed subset of $\widetilde{\Phi}^{+}$ and $\alpha\in\Phi.$ If $\alpha$ is in $I_{\Gamma}$ and $\alpha+N\delta\in\Gamma$ then $\alpha+n\delta$ is in $\Gamma$ for all $n\geq N$.

Proof.

There must be some $M>N$ such that $\alpha+M\delta\in\Gamma$ since $\alpha\in I_{\Gamma}$. Then

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∎

Lemma 4.4**.**

Let $x,y\in\overline{\widetilde{W}}$. Then it cannot happen that $\Phi_{x}^{\prime}\subsetneq\Phi_{y}.$

Proof.

By Lemma 3.9, Corollary 3.13 and Lemma 2.7(d) the biclosed sets $w\cdot\widehat{\Psi^{+}_{\emptyset,M}}$ where $w\in\widetilde{W}$ are precisely those $\Phi_{x}^{\prime},x\in\overline{\widetilde{W}}$. Suppose that $M\neq\emptyset$. Then by Theorem 1.3 (2) and Theorem 1.4 (2), those biclosed sets of the form $w\cdot\widehat{\Psi^{+}_{\emptyset,M}}$ must contain some infinite $\delta$ chain through some $\alpha$ and $-\alpha$ (where $\alpha\in\Phi$) as $w\cdot\widehat{\Psi^{+}_{\emptyset,M}}$ differs from $\widehat{\pi(w)(\Psi^{+}_{\emptyset,M})}$ by finitely many roots. This implies that these sets cannot be possibly contained in a biclosed set coming from $\Phi_{y},y\in\overline{\widetilde{W}}$ by the proof of Lemma 3.9. Now suppose that $M=\emptyset.$ Then $w\cdot\widehat{\Psi^{+}_{\emptyset,\emptyset}}=\widehat{\Xi^{+}}$ for some positive system $\Xi^{+}$ in $\Phi$. Clearly if there is a biclosed set properly containing $\widehat{\Xi^{+}}$, again it must have infinite $\delta$-chains through some $\alpha,-\alpha\in\Phi$, which implies that it is not of the form $\Phi_{y},y\in\overline{\widetilde{W}}$. ∎

Remark*.*

Suppose that $\Phi_{x}^{\prime}=\Phi_{y}$. Again $\Phi_{x}^{\prime}=w\cdot\widehat{\Psi^{+}_{\emptyset,N}}$ for some $w\in\widetilde{W}$. The same reasoning as in the proof of the above Lemma implies that $N=\emptyset$. Therefore $\Phi_{x}^{\prime}=\widehat{\Xi^{+}}$ for some positive system $\Xi^{+}=\pi(w)\Psi^{+}$ in $\Phi$. Therefore we conclude that if $\Phi_{x}^{\prime}=\Phi_{y}$ then $x,y$ are both maximal in $\overline{\widetilde{W}}$ (by the first paragraph of the proof of Theorem 3.16). Conversely if $x$ is maximal in $\overline{\widetilde{W}}$, then $\Phi_{x}=\widehat{\Xi^{+}}$ for some positive system $\Xi^{+}$ and consequently $\Phi_{x}^{\prime}=\widehat{\Xi^{-}}$. Take the infinite reduced word $y$ such that $\Phi_{y}=\widehat{\Xi^{-}}$. Then we have that $\Phi_{x}^{\prime}=\Phi_{y}$.

From now on, let $\widetilde{W}$ be of type $\widetilde{A}_{2}$ or $\widetilde{B}_{2}=\widetilde{C}_{2}$ or $\widetilde{G}_{2}$.

Lemma 4.5**.**

Suppose that $x,y\in\widetilde{W}$ and that $x\wedge y=e$. Let $\Gamma=\overline{\Phi_{x}\cup\Phi_{y}}$ and assume that $I_{\Gamma}=A_{\Gamma}=\Psi^{+}_{\emptyset,\{\alpha\}}$ for some positive system $\Psi^{+}$ in $\Phi$ and a simple root $\alpha$ of $\Psi^{+}$. Then $\Gamma=\widehat{\Psi^{+}_{\emptyset,\{\alpha\}}}$.

Proof.

The proof will be a case-by-case analysis. But before that we make a few case-free observations.

(1) The sets $\widehat{\alpha}$ and $\widehat{-\alpha}$ are both contained in $\Gamma$.

Proof of (1). Note that $I_{\Gamma}=A_{\Gamma}=\Psi^{+}_{\emptyset,\{\alpha\}}$. We claim that both $\alpha$ and $-\alpha$ are contained in $A_{\Phi_{x}\cup\Phi_{y}}$. If $-\alpha\not\in A_{\Phi_{x}\cup\Phi_{y}}$, then $A_{\Phi_{x}\cup\Phi_{y}}\subset\Psi^{+}$ (note that $A_{\Phi_{x}\cup\Phi_{y}}\subset A_{\Gamma}$). Consequently $\widehat{-\alpha}\cap\Gamma=\emptyset$ (as one easily sees that $A_{\overline{S}}\subset\overline{A_{S}}$ for any $S\subset\widehat{\Phi}$), which is a contradiction. The same reasoning works for $\alpha$ in place of $-\alpha$. Therefore $\alpha_{0}$ and $(-\alpha)_{0}$ are in $\Phi_{x}\bigcup\Phi_{y}$ by Lemma 4.1. By Lemma 4.3 we get the assertion.

(2) There is at least one simple root $\rho_{0}\in\widetilde{\Phi}^{+}$ such that $\rho\neq\pm\alpha$ and $\widehat{\rho}$ is contained in $\Gamma.$

Proof of (2). Because $\alpha_{0}$ and $(-\alpha)_{0}$ cannot both be simple roots in $\widetilde{\Phi}^{+}$ (the simple roots of $\widetilde{\Phi}^{+}$ are $\gamma,\gamma$ simple in $\Phi^{+}$ and $-\eta+\delta$ where $\eta$ is the highest root in $\Phi^{+}$) and we assume that $x$ and $y$ have trivial meet, there must be $\rho_{0}\in\Phi_{x}\bigcup\Phi_{y},\rho_{0}\neq\alpha_{0},(-\alpha)_{0}$ and $\rho_{0}$ is simple in $\widetilde{\Phi}^{+}$. Since $\rho\in A_{\Gamma}=I_{\Gamma}$, this will cause $\widehat{\rho}$ to be contained in $\Gamma$ by Lemma 4.3.

(3) If neither $\alpha_{0}$ nor $(-\alpha)_{0}$ is simple in $\widetilde{\Phi}^{+}$, there are two simple roots $\rho_{0},\gamma_{0}\in\widetilde{\Phi}^{+}$ such that $\widehat{\rho}$ and $\widehat{\gamma}$ are both contained in $\Gamma.$

Proof of (3). Again because $x$ and $y$ have trivial meet we must have at least two simple roots of $\widetilde{\Phi}^{+}$ in $\Phi_{x}\cup\Phi_{y}$. Suppose that these two simple roots are $\rho_{0},\gamma_{0}$ (with $\gamma,\rho\in\Phi$). Then Lemma 4.3 forces the assertion.

Now we carry out the case study.

$\mathbf{A}_{2}:$

The two simple roots of $\Phi^{+}$ are denoted by $\alpha,\beta.$ First we consider the case

$I_{\Gamma}=A_{\Gamma}=\{\alpha,\beta,\alpha+\beta,-\alpha\}$.

By the above fact (1), the sets $\widehat{\alpha}$ and $\widehat{-\alpha}$ have to be contained in $\Gamma$. Then by the above fact (2), the set $\widehat{\beta}$ also has to be contained in $\Gamma$. Then $\widehat{\alpha+\beta}$ can clearly be generated by $\widehat{\alpha}\cup\widehat{\beta}$ and is in $\Gamma$ (since $\alpha+\beta+k\delta=(\alpha)+(\beta+k\delta)$).

Any permutation of the simple roots $\alpha,\beta,-\alpha-\beta+\delta$ induces an isomorphism of the affine root system $\widetilde{\Phi}$ while preserving the positive system $\widetilde{\Phi}^{+}$. By making use of this symmetry of $\widetilde{A}_{2}$, it is sufficient to treat the above case.

$\mathbf{B}_{2}:$

The short simple root is denoted by $\alpha$ and the long simple root is denoted by $\beta.$

(1) $I_{\Gamma}=A_{\Gamma}=\{\alpha,\beta,\alpha+\beta,2\alpha+\beta,-\alpha\}$.

By the above fact (1) the sets $\widehat{\alpha}$ and $\widehat{-\alpha}$ have to be contained in $\Gamma$. By the above fact (2) we also have that $\widehat{\beta}\subset\Gamma$. Then $\widehat{\alpha+\beta}$ and $\widehat{2\alpha+\beta}$ can clearly be generated by $\widehat{\{\alpha,-\alpha,\beta\}}$ and are thus in $\Gamma$ (since $\alpha+\beta+k\delta=(\alpha)+(\beta+k\delta)$ and $2\alpha+\beta+k\delta=2(\alpha)+(\beta+k\delta)$).

A permutation of the simple roots $\beta,\delta-2\alpha-\beta$ induces an isomorphism of the affine root system $\widetilde{\Phi}$ while preserving the positive system $\widetilde{\Phi}^{+}$. By making use of this symmetry of $\widetilde{B}_{2}$, the case where $I_{\Gamma}=A_{\Gamma}=\{-2\alpha-\beta,-\alpha,-\beta,-\alpha-\beta,\alpha\}$ can be proved in the same way.

(2) $I_{\Gamma}=A_{\Gamma}=\{-2\alpha-\beta,\beta,\alpha+\beta,2\alpha+\beta,-\alpha\}$.

By the above fact (1) the sets $\widehat{2\alpha+\beta}$ and $\widehat{-2\alpha-\beta}$ have to be contained in $\Gamma$. By the above fact (2) the set $\widehat{\beta}$ is also contained in $\Gamma$. Then note that $\frac{1}{2}(2\alpha+\beta)+\frac{1}{2}\beta=\alpha+\beta,-\alpha+\delta=\frac{1}{2}(-2\alpha-\beta+\delta)+\frac{1}{2}(\beta+\delta)$. Therefore $\widehat{\alpha+\beta}$ and $\widehat{-\alpha}$ are in $\Gamma$ by Lemma 4.3.

Again a permutation of the simple roots $\beta,\delta-2\alpha-\beta$ induces an isomorphism of the affine root system $\widetilde{\Phi}$ while preserving the positive system $\widetilde{\Phi}^{+}$. By making use of this symmetry of $\widetilde{B}_{2}$, the case where $I_{\Gamma}=A_{\Gamma}=\{-2\alpha-\beta,\beta,-\beta,-\alpha-\beta,-\alpha\}$ can be proved in the same way.

(3) $I_{\Gamma}=A_{\Gamma}=\{-2\alpha-\beta,\beta,\alpha+\beta,-\alpha-\beta,-\alpha\}$.

By the above fact (1), the sets $\widehat{\alpha+\beta}$ and $\widehat{-\alpha-\beta}$ are contained in $\Gamma$. Since neither $\alpha+\beta$ nor $\delta-\alpha-\beta$ is simple, by the above fact (3), the sets $\widehat{\beta},\widehat{-2\alpha-\beta}$ have to be contained in $\Gamma$. Then note that $-\alpha+\delta=(-\alpha-\beta+\delta)+\beta$. So the set $\widehat{-\alpha}$ is in $\Gamma$ by Lemma 4.3.

(4) $I_{\Gamma}=A_{\Gamma}=\{-2\alpha-\beta,2\alpha+\beta,-\beta,-\alpha-\beta,\alpha\}$.

By the above fact (1) the sets $\widehat{2\alpha+\beta}$ and $\widehat{-2\alpha-\beta}$ have to be contained in $\Gamma$. By the above fact (2) we also have that $\widehat{\alpha}\subset\Gamma$. Then note that $-\alpha-\beta+\delta=(-2\alpha-\beta+\delta)+\alpha$ and $-\beta+\delta=(-\alpha-\beta+\delta)+\alpha$. So $\widehat{-\alpha-\beta}$ and $\widehat{-\beta}$ are in $\Gamma$ by Lemma 4.3.

Again a permutation of the simple roots $\beta,\delta-2\alpha-\beta$ induces an isomorphism of the affine root system $\widetilde{\Phi}$ while preserving the positive system $\widetilde{\Phi}^{+}$. By making use of this symmetry of $\widetilde{B}_{2}$, the case where $I_{\Gamma}=A_{\Gamma}=\{\alpha+\beta,2\alpha+\beta,-\beta,\beta,\alpha\}$ can be proved in the same way.

(5) $I_{\Gamma}=A_{\Gamma}=\{\alpha+\beta,2\alpha+\beta,-\beta,-\alpha-\beta,\alpha\}$.

By the above fact (3) this situation cannot happen as there will at most be one simple root in $\Gamma.$

$\textbf{G}_{2}$:

The long simple root is denoted by $\beta$ and the short simple root is denoted by $\alpha.$

(1) $I_{\Gamma}=A_{\Gamma}=\{\alpha,\beta,\alpha+\beta,2\alpha+\beta,3\alpha+\beta,3\alpha+2\beta,-\alpha\}$.

By the above fact (1) the sets $\widehat{\alpha}$ and $\widehat{-\alpha}$ have to be contained in $\Gamma$. By the above fact (2) we also have that $\widehat{\beta}\subset\Gamma$. Then the sets $\widehat{\alpha+\beta}$, $\widehat{2\alpha+\beta}$, $\widehat{3\alpha+\beta}$ and $\widehat{3\alpha+2\beta}$ can clearly be generated and are all in $\Gamma$ thanks to Lemma 4.3.

(2) $I_{\Gamma}=A_{\Gamma}=\{-3\alpha-\beta,\beta,\alpha+\beta,2\alpha+\beta,3\alpha+\beta,3\alpha+2\beta,-\alpha\}$.

By the above fact (3) this situation cannot happen as there will at most be one simple root in $\Gamma.$

(3) $I_{\Gamma}=A_{\Gamma}=\{-3\alpha-\beta,\beta,\alpha+\beta,2\alpha+\beta,-2\alpha-\beta,3\alpha+2\beta,-\alpha\}$.

By the above fact (3) this situation cannot happen as there will at most be one simple root in $\Gamma.$

(4) $I_{\Gamma}=A_{\Gamma}=\{-3\alpha-\beta,\beta,\alpha+\beta,-3\alpha-2\beta,-2\alpha-\beta,3\alpha+2\beta,-\alpha\}$.

By the above fact (1) the sets $\widehat{3\alpha+2\beta}$ and $\widehat{-3\alpha-2\beta}$ are contained in $\Gamma$. By the above fact (2) the set $\widehat{\beta}$ also has to be contained in $\Gamma$. Then note that $\alpha+\beta=\frac{1}{3}(3\alpha+2\beta)+\frac{1}{3}\beta$, $-\alpha+\delta=\frac{1}{3}(-3\alpha-2\beta+\delta)+\frac{2}{3}(\beta+\delta)$, $-3\alpha-\beta+\delta=(-3\alpha-2\beta+\delta)+\beta$ and $-2\alpha-\beta+\delta=(-3\alpha-2\beta+\delta)+(\alpha+\beta).$ Hence $\widehat{\alpha+\beta},\widehat{-\alpha},\widehat{-3\alpha-\beta}$ and $\widehat{-2\alpha-\beta}$ are all in $\Gamma$ thanks to Lemma 4.3.

(5) $I_{\Gamma}=A_{\Gamma}=\{-3\alpha-\beta,\beta,\alpha+\beta,-3\alpha-2\beta,-2\alpha-\beta,-\alpha-\beta,-\alpha\}$.

By the above fact (1) the sets $\widehat{\alpha+\beta}$ and $\widehat{-\alpha-\beta}$ are contained in $\Gamma$. Since neither $\alpha+\beta$ nor $\delta-\alpha-\beta$ is simple, by the above fact (3) the sets $\widehat{\beta}$ and $\widehat{-3\alpha-2\beta}$ are also contained in $\Gamma$. Then note that $-\alpha+\delta=\frac{1}{3}(-3\alpha-2\beta+\delta)+\frac{2}{3}(\beta+\delta)$, $-3\alpha-\beta+\delta=(-3\alpha-2\beta+\delta)+\beta$ and $-2\alpha-\beta+\delta=(-3\alpha-2\beta+\delta)+(\alpha+\beta).$ Hence $\widehat{-\alpha},\widehat{-3\alpha-\beta}$ and $\widehat{-2\alpha-\beta}$ are all in $\Gamma$ thanks to Lemma 4.3.

(6) $I_{\Gamma}=A_{\Gamma}=\{-3\alpha-\beta,\beta,-\beta,-3\alpha-2\beta,-2\alpha-\beta,-\alpha-\beta,-\alpha\}$.

By the above fact (1) the sets $\widehat{\beta}$ and $\widehat{-\beta}$ have to be contained in $\Gamma$. By the above fact (2) we also have that $\widehat{-3\alpha-2\beta}\subset\Gamma.$ Then note that $-\alpha+\delta=\frac{1}{3}(-3\alpha-2\beta+\delta)+\frac{2}{3}(\beta+\delta)$, $-3\alpha-\beta+\delta=(-3\alpha-2\beta+\delta)+\beta$, $-2\alpha-\beta+\delta=\frac{2}{3}(-3\alpha-2\beta+\delta)+\frac{1}{3}(\beta+\delta)$ and $\frac{1}{3}(-3\alpha-2\beta+2\delta)+\frac{1}{3}(-\beta+\delta)=-\alpha-\beta+\delta$. Hence $\widehat{-\alpha},\widehat{-3\alpha-\beta},\widehat{-2\alpha-\beta}$ and $\widehat{-\alpha-\beta}$ are all in $\Gamma$ thanks to Lemma 4.3.

(7) $I_{\Gamma}=A_{\Gamma}=\{-3\alpha-\beta,\alpha,-\beta,-3\alpha-2\beta,-2\alpha-\beta,-\alpha-\beta,-\alpha\}$.

By the above fact (1) the sets $\widehat{\alpha}$ and $\widehat{-\alpha}$ have to be contained in $\Gamma$. By the above fact (2) we also have that $\widehat{-3\alpha-2\beta}\subset\Gamma.$ Then note that $-\alpha-\beta+\delta=\frac{1}{2}(-3\alpha-2\beta+\delta)+\frac{1}{2}(\alpha+\delta)$, $-\beta+\delta=(-\alpha-\beta+\delta)+\alpha,-2\alpha-\beta+\delta=\frac{1}{2}(-3\alpha-2\beta+\delta)+\frac{1}{2}(-\alpha+\delta)$. Hence $\widehat{-\alpha-\beta},\widehat{-\beta}$ and $\widehat{-2\alpha-\beta}$ are all in $\Gamma$ thanks to Lemma 4.3. Finally we note that in this case the two elements $x,y$ are of the form $s_{\alpha}\cdots$ and $s_{\delta-3\alpha-2\beta}\cdots$. Because $\Gamma$ is infinite they could not be $s_{\alpha}$ and $s_{\delta-3\alpha-2\beta}$ (note that $s_{\alpha}$ and $s_{\delta-3\alpha-2\beta}$ commute to each other). Because the two elements have trivial meet we cannot have $s_{\alpha}s_{\delta-3\alpha-2\beta}\cdots$ or $s_{\delta-3\alpha-2\beta}s_{\alpha}$. Finally if one of the elements is $s_{\alpha}s_{\beta}\cdots$ then we will have that $3\alpha+\beta\in A_{\Gamma}$, which is a contradiction. So we must have one element being equal to $s_{\delta-3\alpha-2\beta}s_{\beta}\cdots$. Therefore we have that $-3\alpha-\beta+\delta\in\Phi_{x}\cup\Phi_{y}$. So we conclude that $\widehat{-3\alpha-\beta}\subset\Gamma$ thanks to Lemma 4.3.

(8) $I_{\Gamma}=A_{\Gamma}=\{-3\alpha-\beta,\alpha,-\beta,-3\alpha-2\beta,-2\alpha-\beta,-\alpha-\beta,3\alpha+\beta\}$.

By the above fact (1) the sets $\widehat{-3\alpha-\beta}$ and $\widehat{3\alpha+\beta}$ are contained in $\Gamma$. Since neither $-3\alpha-\beta+\delta$ nor $3\alpha+\beta$ is simple, by the above fact (3) we also have that $\widehat{\alpha}\subset\Gamma$ and $\widehat{-3\alpha-2\beta}\subset\Gamma.$ Then note that $-2\alpha-\beta+\delta=(-3\alpha-\beta+\delta)+\alpha$, $-\alpha-\beta+\delta=(-2\alpha-\beta+\delta)+\alpha,$ and $-\beta+\delta=(-\alpha-\beta+\delta)+\alpha$. Hence $\widehat{-2\alpha-\beta},\widehat{-\alpha-\beta}$ and $\widehat{-\beta}$ are all in $\Gamma$ thanks to Lemma 4.3.

(9) $I_{\Gamma}=A_{\Gamma}=\{2\alpha+\beta,\alpha,-\beta,-3\alpha-2\beta,-2\alpha-\beta,-\alpha-\beta,3\alpha+\beta\}$.

By the above fact (1) the sets $\widehat{-2\alpha-\beta}$ and $\widehat{2\alpha+\beta}$ are contained in $\Gamma$. Since neither $-2\alpha-\beta+\delta$ nor $2\alpha+\beta$ is simple, by the above fact (3) the sets $\widehat{\alpha}$ and $\widehat{-3\alpha-2\beta}$ have to be in $\Gamma.$ Then note that $3\alpha+\beta=(2\alpha+\beta)+\beta$, $-\alpha-\beta+\delta=(-2\alpha-\beta+\delta)+\alpha,$