# Inversion Sets and Quotient Root Systems

Ivan Dimitrov  Department of Mathematics and Statistics, Queen’s University
<EMAIL_ADDRESS>,  Cole Gigliotti  Department of Mathematics, University
of British Columbia<EMAIL_ADDRESS>,  Etan Ossip  Faculty of Engineering
and Applied Science, Queen’s University<EMAIL_ADDRESS>,  Charles
Paquette  Department of Mathematics and Computer Science, Royal Military
College of Canada<EMAIL_ADDRESS>and  David Wehlau
Department of Mathematics and Computer Science, Royal Military College of
Canada<EMAIL_ADDRESS>
###### Abstract.

We provide a recursive description of all decompositions of the positive roots
$R^{+}$ of a quotient root system $R$ into disjoint unions of inversion sets.
Our description is type-independent and generalizes the analogous result for
type $\mathbb{A}$ root systems in [USRA]. The main tool is the notion of an
inflation of a subset of a quotient root system. This new notion allows us to
treat all root systems (and their quotients) uniformly. We also obtain some
numerical results about the number of special decompositions. The new
sequences we obtain may be considered as extensions of Catalan numbers.

Keywords: Root system, Quotient root system, Inversion set.

###### 2020 Mathematics Subject Classification:

Primary 17B22; Secondary 17B20, 17B25, 22F30.

## Introduction

Let $\Delta$ be a root system with Weyl group $W$ and set of positive roots
$\Delta^{+}$. The inversion set of $w\in W$ is the set

$\Phi(w):=\\{\alpha\in\Delta^{+}\,|\,w(\alpha)\in\Delta^{-}\\}=\Delta^{+}\cap
w^{-1}(\Delta^{-})\ ,$

where $\Delta^{-}=-\Delta^{+}$. If $\Delta$ is of type $\mathbb{A}_{n}$, then
$\Delta^{+}$ can be identified with the set

$\\{(i,j)\in\mathbb{Z}\times\mathbb{Z}\,|\,1\leqslant i<j\leqslant n+1\\}\ ,$

$W=S_{n+1}$ is the symmetric group on $n+1$ elements, and the inversion set of
$\sigma\in S_{n+1}$ is the set

$\Phi(\sigma)=\\{(i,j)\in\Delta^{+}\,|\,\sigma(i)>\sigma(j)\\}\ .$

The problem of describing the decompositions of $\Delta^{+}$ as a disjoint
union of inversion sets arrises in connection with studying the Littlewood-
Richardson cone, see [USRA] and the references therein for details. In [USRA],
a complete description of such decompositions was obtained in terms of
inflations of permutations. The results for type $\mathbb{A}$ are also carried
over to types $\mathbb{B}$ and $\mathbb{C}$ by exploiting a realization of
each of the latter root systems as a root system of type $\mathbb{A}$ with
additional symmetries. Unfortunately, the methods developed there do not apply
to root systems of type $\mathbb{D}$ and to exceptional root systems. The goal
of this paper with to provide a uniform approach to studying decompositions of
$\Delta^{+}$ into a disjoint union of inversion sets.

The uniform approach to inversion sets and inflations requires expanding the
class of root systems to include quotient root systems (QRSs for short).
Roughly speaking, if $\Delta$ is a root system with a base $\Sigma$ and $I$ is
a subset of $\Sigma$, the QRS $\Delta/I$ is the image of $\Delta$ under the
natural projection
$\pi_{I}:\mathrm{span\,}\Delta\to\mathrm{span\,}\Delta/\mathrm{span\,}I$. QRSs
have been studied extensively in connection with Lie theory, [K], simplicial
hyperplane arrangements, [Cu] and the references therein, and 3-fold flopping
contractions, [IW], to mention a few. The notions of positive roots, bases,
etc. extend to QRSs and the counterparts of Weyl groups are certain groupoids
called Weyl groupoids.

Let $R$ be a QRS with positive roots $R^{+}$. By definition, a subset
$\Phi\subset R^{+}$ is an inversion set if $\Phi$ is both closed and coclosed.
Of course, for root systems, this definition is equivalent to the usual
definition of an inversion set. The main result of this paper is Theorem 5.9
which describes recursively all decompositions of $R^{+}$ into a disjoint
union of inversion sets. The main tool in our description is the notion of
inflation of a subset of a quotient of $R^{+}$, see Section 3. In preparation
for Theorem 5.9, we prove that every subset of $R^{+}$ (not necessarily an
inversion set) can be represented in a canonical way as the inflation of the
empty subset, a primitive subset, or the whole of a quotient of $R^{+}$, see
Theorem 3.14.

If $R$ is a root system of type $\mathbb{A}$, the canonical form of an
inversion set and the decomposition provided by Theorem 5.9 coincide with the
ones from [USRA]. The reason for this is that every quotient of a root system
of type $\mathbb{A}$ is equivalent to a root system of type $\mathbb{A}$ of
smaller rank. For root systems of types $\mathbb{B}$ and $\mathbb{C}$ the
situation is a bit more complicated since the quotients do not remain
equivalent to root systems of the same type; however, as far as inversion set
are concerned these new QRS behave exactly as root systems of types
$\mathbb{B}$ and $\mathbb{C}$, see Remark 1.4. The situation is dramatically
different for root systems of type $\mathbb{D}$ and for exceptional root
systems, which explains why these were beyond reach with the methods developed
in [USRA].

The paper is organized as follows: In Section 1 we provide the necessary
background on QRSs and inversion sets. In Section 2 we introduce and study
paths of roots in QRSs. In Section 3 we define inflations, study their
properties, and prove Theorem 3.14 which proves that every subset of $R^{+}$
admits a canonical representation as an inflation. In Section 4 we introduce
and study the properties of the main tool for proving the main theorem, namely
the graph $G_{\Phi}^{\Phi^{c}}$ associated to an inversion set $\Phi\subset
R^{+}$. In Section 5 we prove the main theorem of the paper, Theorem 5.9. In
Section 6 we evaluate the number of fine decompositions of $R^{+}$ into
inversion sets. Note that root systems of type $\mathbb{A}$ give rise to
Catalan numbers, so the sequences we obtain in Section 6 may be considered as
analogs and extensions of Catalan numbers.

##  1\. Preliminaries

###  1.1. Quotient Root Systems

In this section, after briefly recalling the notion of classical root system,
we define quotient root systems. The idea is that quotient root systems are
general enough to allow nice reductive arguments to understand root systems,
but still close enough to classical root systems to share many similarities
with them.

Let us now fix some notation and recall the notion of a root system. The
reader is referred to [B] and [H] for basic properties of root systems. We fix
a finite dimensional real vector space $E$ equipped with the standard bilinear
form $\langle\cdot,\cdot\rangle$, called a Euclidean space. A _root system_
$\Delta$ is a finite subset of non-zero vectors in $E$, called _roots_ ,
satisfying the following properties with respect to the bilinear form:

  1. (1)

The roots span $E$;

  2. (2)

If $\alpha\in\Delta$ and $r$ is a non-zero integer with $r\alpha$ a root, then
$r=\pm 1$;

  3. (3)

For $\alpha\in\Delta$, the reflection with respect to the hyperplane
$H_{\alpha}$ of $E$ perpendicular to $\alpha$ sends a root to a root;

  4. (4)

For $\alpha,\beta\in\Delta$, the number
$2\frac{\langle\alpha,\beta\rangle}{\langle\alpha,\alpha\rangle}$ is an
integer.

We let $\Sigma\subset\Delta$ be a fixed _base_ , that is, the set $\Sigma$
forms a basis of $E$ and each root of $\Delta$ can be written as an integral
combination of the elements from $\Sigma$ with either all coefficients non-
negative, or all coefficients non-positive. When a base is given, a root such
that all of its coefficient in that base are non-negative is called a
_positive_ root (with respect to the given base). The other roots are called
_negative_ roots. We let $\Delta^{+}$ denote the set of positive roots with
respect to $\Sigma$ and $\Delta^{-}$ that of negative roots. We have a
disjoint union decomposition $\Delta=\Delta^{+}\cup\Delta^{-}$. We say that
$\Delta$ is _irreducible_ when $\Delta$ cannot be partitioned into two subsets
of mutually orthogonal roots.

Next we introduce and briefly discuss quotient root systems. For more details
on quotient root systems, we refer the reader to [K] and [DR]. For a subset
$I\subset\Sigma$, we let $\Delta_{I}$ denote the subsystem
$\Delta_{I}=\Delta\cap\mathrm{span\,}I$. Clearly, $\Delta_{I}$ is a root
system (not necessarily irreducible) with a base $I$ and a fixed set of
positive roots $\Delta_{I}^{+}=\Delta_{I}\cap\Delta^{+}$.

Now, we would like to endow the quotient space $E/\mathrm{span\,}I$ with a
bilinear form that makes the projection of the roots into a system with rich
combinatorics. In order to do this, consider the canonical projection
$\pi_{I}:E\to E/\mathrm{span\,}I$. We identify the quotient
$E/\mathrm{span\,}I$ with the orthogonal complement of $\mathrm{span\,}I$.
Hence, $E/\mathrm{span\,}I$ is endowed with a corresponding bilinear form that
we also denote $\langle\cdot,\cdot\rangle$. We let $\Delta/I$ denote the non-
zero images of elements of $\Delta$. In other words, we set

$\Delta/I:=\\{\pi_{I}(\beta)\,|\,\beta\in\Delta\backslash\mathrm{span\,}I\\}.$

We call the set $\Delta/I$ a _quotient root system_ or QRS for short while a
given element from $\Delta/I$ is called a _root_. As for classical root
systems, a _base_ $\Sigma^{\prime}$ of such a QRS is a subset of the set of
roots that forms a basis of the underlying vector space $E/\mathrm{span\,}I$,
and such that every root can be written as an integral combination of
$\Sigma^{\prime}$, and where the coefficients are either all non-negative (the
root is then called positive with respect to $\Sigma^{\prime}$) or all non-
positive (the root is then called negative). It is easy to see that

$\Sigma/I:=\\{\pi_{I}(\beta)\,|\,\beta\in\Sigma\backslash I\\}$

is a base of $\Delta/I$. This gives rise to the corresponding sets
$(\Delta/I)^{+}$ of positive roots and $(\Delta/I)^{-}$ of negative roots. We
observe that we have the disjoint union decomposition
$\Delta/I=(\Delta/I)^{+}\cup(\Delta/I)^{-}$ and where
$(\Delta/I)^{+}=\pi_{I}(\Delta^{+})\backslash\\{0\\}$ and
$(\Delta/I)^{-}=\pi_{I}(\Delta^{-})\backslash\\{0\\}$. The _rank_ or a QRS is
just the dimension of the ambient vector space, or equivalently, the number of
roots in a base.

######  Remark 1.1.

  1. (i)

Note that when $I=\emptyset$, the QRS $\Delta/I$ is canonically identified
with the root system $\Delta$. When considering QRSs, we will assume that
$I\neq\Sigma$, so that the vector space $E/\mathrm{span\,}I$ is not trivial.

  2. (ii)

Unlike root systems, QRSs allow for multiples of roots which are also roots,
i.e., a QRS may contain both $\alpha$ and $2\alpha$. A root $\alpha$ for which
$\alpha/r$ is not a root for all $r$ with $|r|>1$ is called primitive.

We will usually denote a QRS by $R$, with a given fixed base by $\Sigma$. With
respect to this base, $R^{+}$ denotes the set of positive roots and $R^{-}$
that of negative roots. The underlying root system giving rise to the QRS $R$
will be denoted $\Delta$.

####  1.1.1. Coefficient vectors, partial order.

Let $R$ be a QRS with base $\Sigma=\\{\theta_{1},\ldots,\theta_{r}\\}$. We can
identify any root by its _coefficient vector_ , which is just the vector in
$\mathbb{Z}^{r}$ collecting the coefficients of the root when writing it as an
integral combination of the elements of the base. In other words, for a root
$\alpha$, we consider the expression $\alpha=\sum_{i=1}^{r}d_{i}\theta_{i}$
and define $d(\alpha)=(d_{1},\ldots,d_{r})$ to be the coefficient vector of
$\alpha$, with respect to the base $\Sigma$. By definition of a base, any
coefficient vector is in $(\mathbb{Z}_{\geqslant 0})^{r}$ or in
$(\mathbb{Z}_{\leqslant 0})^{r}$.

We can use these to define a partial order on roots (which again depends on
$\Sigma$). For two roots $\alpha,\beta$, we write $\alpha\leqslant\beta$ if
$d(\alpha)\leqslant d(\beta)$, where for two vectors
$\vec{u},\vec{v}\in\mathbb{Z}^{r}$, we write $\vec{u}\leqslant\vec{v}$ when
$u_{i}\leqslant v_{i}$ for $i=1,2,\ldots,r$.

####  1.1.2. Adjacency diagram and support.

Let $R$ be a QRS with base $\Sigma=\\{\theta_{1},\ldots,\theta_{r}\\}$. The
_adjacency diagram_ of $R$ with respect to $\Sigma$ is the graph on vertices
$\Sigma$ and such that there is an edge connecting $\theta_{i}$ with
$\theta_{j}$ if and only if $\langle\theta_{i},\theta_{j}\rangle<0$. We note
that a QRS is connected precisely when its adjacency diagram is connected.

The _support_ of a root $\alpha$, with respect to a given base $\Sigma$, is

$\operatorname{supp}(\alpha)=\\{\theta_{i}\in\Sigma\mid d(\alpha)_{i}\neq
0\\}.$

If $S$ is a set of roots, then we define its support $\operatorname{supp}(S)$
to be the union of the supports of the roots in $S$. It is not hard to see
that the full subgraph of the adjacency diagram corresponding to the support
of a root is always connected. This follows from the similar fact on classical
root systems.

###  1.2. Inversion Sets

In this subsection, we fix a QRS $R$ with base $\Sigma$. For a subset $\Phi$
of $R^{+}$, we denote by $\Phi^{c}$ its complement in $R^{+}$.

######  Definition 1.2.

Let $\Phi\subseteq R^{+}$. We say that

  1. (1)

$\Phi$ is _closed_ if $\alpha+\beta\in\Phi$ whenever $\alpha,\beta\in\Phi$.

  2. (2)

$\Phi$ is _co-closed_ if $\alpha+\beta\in\Phi^{c}$ whenever
$\alpha,\beta\in\Phi^{c}$.

  3. (3)

$\Phi$ is an _inversion set_ if it is both closed and co-closed.

######  Remark 1.3.

It is easy to see that intersections of closed sets are closed and unions of
co-closed sets are co-closed.

For classical roots systems, the notion of inversion set takes its origin from
the fact that $\Phi$ is an inversion set precisely when there is an element
$w$ in the Weyl group of the root system such that

$\Phi=\\{\alpha\in R^{+}\mid w(\alpha)\in R^{-}\\}.$

As we will be interested in decomposing inversion sets, we need the following
definition.

######  Remark 1.4.

A set $\Phi\subset R^{+}$ is an inversion set if and only if it can be
realized as the set of elements of $R^{+}$ lying on one side of some
hyperplane of $E$. In particular, if two QRSs $R_{1}$ and $R_{2}$ have the
same set of primitive elements, cf. Remark 1.1, then there is a natural
bijection between the inversion sets in $R_{1}^{+}$ and $R_{2}^{+}$.

######  Definition 1.5.

Let $\Phi\subseteq R^{+}$ be a non-empty inversion set. We say that $\Phi$ is
_irreducible_ if, whenever $\Phi=\Phi_{1}\cup\Phi_{2}$ as a disjoint union of
inversion sets, then $\Phi_{1}=\Phi$ or $\Phi_{2}=\Phi$.

###  1.3. Decomposition of inversion sets

We will be interested in decomposing an inversion set into disjoint union of
inversion sets. We have the following definition.

######  Definition 1.6.

Let $\Phi$ be an inversion set in $R^{+}$. A _decomposition_ of $\Phi$ is an
expression

$\Phi=\Phi_{1}\sqcup\cdots\sqcup\Phi_{r}$

where all $\Phi_{i}$ are non-empty pairwise disjoint inversion sets. Such a
decomposition is _fine_ if $r$ is the rank of $R$.

######  Proposition 1.7.

Let $\Phi_{1}\sqcup\cdots\sqcup\Phi_{r}=\Phi$ be a decomposition of an
inversion set $\Phi$. Then for each $1\leqslant k\leqslant r$,
$\Phi_{1}\sqcup\cdots\sqcup\Phi_{k}$ is an inversion set.

###### Proof.

It suffices to prove that $\Phi_{1}\sqcup\Phi_{2}$ is closed and co-closed. It
is clear that $\Phi_{1}\sqcup\Phi_{2}$ is co-closed as the union of co-closed
sets. Further, $\Phi_{1}\sqcup\Phi_{2}$ is closed as the complement of
$\Phi_{3}\sqcup\cdots\sqcup\Phi_{r}\sqcup\Phi^{c}$, which is co-closed. ∎

##  2\. Paths in QRSs

One difficulty of working with QRSs (as well as in root systems themselves) is
that sums of roots are not in general roots, and it is unclear whether given a
system $R$, roots $\alpha,\beta\in R$, and a set $K\subseteq R$ with
$\beta-\alpha\in\mathrm{span\,}K$, there is some way to add roots in $K$ to
get from $\alpha$ to $\beta$ without leaving $R$. This question motivates the
definition of paths in QRSs and is answered in the affirmative in Proposition
2.7.

######  Definition 2.1.

A path between two roots $\alpha$ and $\beta$ is a collection of roots
$\kappa_{1},\kappa_{2},\ldots,\kappa_{n}$ such that
$\alpha+\kappa_{1}+\cdots+\kappa_{n}=\beta$ and
$\alpha+\kappa_{1}+\cdots+\kappa_{i}\in R$ for all $i$. We denote this path
$[\alpha;\kappa_{1},\ldots,\kappa_{n};\beta]$ and call each $\kappa_{i}$ a
step.

###  2.1. Existence of Paths

Seeing as the key condition in a path is that the partial sums are roots, it
will be essential to have some sufficient condition for the sum of two roots
to be a root. The following proposition is part of Theorem 2.3 in [K].

######  Proposition 2.2.

Let $\alpha,\beta\in R$. If $\left\langle\alpha,\beta\right\rangle<0$, then
$\alpha+\beta\in R$. If $\left\langle\alpha,\beta\right\rangle>0$, then
$\alpha-\beta\in R$.

######  Proposition 2.3.

Let $\alpha,\beta,\kappa_{1},\ldots,\kappa_{n}\in R$. If
$\beta=\alpha+\kappa_{1}+\cdots+\kappa_{n}$ where
$\kappa_{1}+\kappa_{2}+\cdots+\kappa_{n}\notin R$, then $\alpha+\kappa_{i}\in
R$ for some $1\leqslant i\leqslant n$.

###### Proof.

Since $\beta$ is a root but $\kappa_{1}+\kappa_{2}+\cdots+\kappa_{n}$ is not,
$\alpha\neq 0$. Next, because $\beta-\alpha\notin R$,
$0\geqslant\left\langle\alpha,\beta\right\rangle=\left\langle\alpha,\alpha\right\rangle+\left\langle\alpha,\kappa_{1}\right\rangle+\cdots+\left\langle\alpha,\kappa_{n}\right\rangle$.
Since $\left\langle\alpha,\alpha\right\rangle>0$, there must be at least one
$\kappa_{i}$ such that $\left\langle\alpha,\kappa_{i}\right\rangle<0$, so that
$\alpha+\kappa_{i}\in R$. ∎

######  Lemma 2.4.

Let $\alpha,\beta,\gamma$ be roots such that $\alpha+\beta+\gamma\in R$ but
$\beta+\gamma\notin R$. If $\alpha+\gamma\neq 0$, then $\alpha+\beta\in R$. If
$\alpha+\beta\neq 0$, then $\alpha+\gamma\in R$.

###### Proof.

We assume that $\alpha+\gamma\neq 0$ and show that $\alpha+\beta\in R$. It
suffices to show that either $\left\langle\alpha,\beta\right\rangle<0$ or
$\left\langle\alpha+\beta+\gamma,\gamma\right\rangle>0$. For that reason, we
will assume $\left\langle\alpha,\beta\right\rangle\geqslant 0$ and show that
$\left\langle\alpha+\beta+\gamma,\gamma\right\rangle>0$. Note that because
$\beta+\gamma\notin R$, both $\left\langle\beta,\gamma\right\rangle\geqslant
0$ and $\left\langle\alpha+\beta+\gamma,\alpha\right\rangle\leqslant 0$. We
extract from these inequalities that
$\left\langle\alpha+\gamma,\alpha\right\rangle\leqslant-\left\langle\beta,\alpha\right\rangle\leqslant
0$. Because $\left\langle\alpha+\gamma,\alpha+\gamma\right\rangle>0$ and
$\left\langle\alpha+\gamma,\alpha\right\rangle\leqslant 0$, we now have that
$\left\langle\alpha+\gamma,\gamma\right\rangle>0$. Finally, we note that
$\left\langle\alpha+\beta+\gamma,\gamma\right\rangle=\left\langle\alpha+\gamma,\gamma\right\rangle+\left\langle\beta,\gamma\right\rangle>0$,
and so we conclude that $\alpha+\beta\in R$. By the symmetry of the statement,
$\alpha+\gamma$ is a root provided $\alpha+\beta\neq 0$. ∎

######  Remark 2.5.

In cases where $\alpha+\beta\neq 0$ and $\alpha+\gamma\neq 0$, for example
when $\alpha,\beta,\gamma>0$, the lemma amounts to the statement that if
$\alpha,\beta,\gamma,$ and $\alpha+\beta+\gamma$ are roots, then at least two
of $\alpha+\beta$, $\alpha+\gamma$, and $\beta+\gamma$ are roots.

######  Proposition 2.6.

Let $\alpha,\beta,\kappa_{1},\ldots,\kappa_{n}\in R$. If
$\beta=\alpha+\kappa_{1}+\cdots+\kappa_{n}$ where
$\mu:=\kappa_{1}+\kappa_{2}+\cdots+\kappa_{n}\in R$ and $\mu\neq 0$, then
there exists some $\kappa_{i}$ such that $\alpha+\kappa_{i}\in R$ or
$\alpha+(\mu-\kappa_{i})\in R$.

###### Proof.

Since
$\left\langle\mu,\kappa_{1}\right\rangle+\left\langle\mu,\kappa_{2}\right\rangle+\cdots+\left\langle\mu,\kappa_{n}\right\rangle=\left\langle\mu,\mu\right\rangle>0$,
there is some $\kappa_{i}$ such that
$\left\langle\mu,\kappa_{i}\right\rangle>0$ and $\mu-\kappa_{i}\in R$. We then
apply Lemma 2.4 to the expression $\beta=(\mu-\kappa_{i})+\alpha+\kappa_{i}$
to conclude that if $\alpha+\kappa_{i}\notin R$, then
$\mu-\kappa_{i}+\alpha\in R$, provided that $(\mu-\kappa_{i})+\kappa_{i}\neq
0$, which is true by assumption. ∎

######  Proposition 2.7.

Let $\beta=\alpha+\kappa_{1}+\cdots+\kappa_{n}$ for roots $\alpha,\beta,$ and
$\kappa_{i}$, where no subcollection of
$\kappa_{1},\kappa_{2},\ldots,\kappa_{n}$ sums to 0. There is a path
$[\alpha;\mu_{1},\mu_{2},\ldots,\mu_{n};\beta]$ from $\alpha$ to $\beta$ for
some permutation $\mu_{1},\mu_{2},\ldots,\mu_{n}$ of
$\kappa_{1},\kappa_{2},\ldots,\kappa_{n}$.

###### Proof.

We prove the sufficient claim that, given any expression
$\beta^{\prime}=\alpha^{\prime}+\kappa_{1}^{\prime}+\cdots+\kappa_{n}^{\prime}$
for roots
$\alpha^{\prime},\beta^{\prime},\kappa_{1}^{\prime},\ldots,\kappa_{n}^{\prime}$,
there is some $1\leqslant i\leqslant n$ such that
$\alpha^{\prime}+\kappa_{i}^{\prime}\in R$. We prove this by induction on $n$,
the length of the expression, with the base case being trivial. In the
inductive step, we write
$\beta^{\prime}=\alpha^{\prime}+\kappa_{1}^{\prime}+\cdots+\kappa_{n}^{\prime}$
and let
$\mu:=\kappa_{1}^{\prime}+\kappa_{2}^{\prime}+\cdots+\kappa_{n}^{\prime}$. If
$\mu\notin R$, then by Proposition 2.3, there exists some
$\kappa_{i}^{\prime}$ such that $\alpha^{\prime}+\kappa_{i}^{\prime}\in R$. If
$\mu\in R$, we use Proposition 2.6 to select some $\kappa_{i}^{\prime}$ such
that either $\alpha^{\prime}+\kappa_{i}^{\prime}\in R$ or
$\alpha^{\prime}+(\mu-\kappa_{i}^{\prime})\in R$. If
$\alpha^{\prime}+\kappa_{i}^{\prime}\in R$, we are done. Otherwise,
$\alpha^{\prime}+(\mu-\kappa_{i}^{\prime})\in R$ is an expression with one
fewer term than $\alpha^{\prime}+\mu$, so we obtain our result by our
inductive hypothesis. ∎

###  2.2. Reduced Paths

While Proposition 2.7 describes the existence of paths between roots, there
are often multiple ways to rearrange the steps in a path without breaking the
condition that partial sums be roots. Having the freedom to permute the steps
in a path may allow us to ensure that all partial sums are positive, for
example, or stay within an inversion set. We therefore introduce the notion of
a reduced path, and show in Proposition 2.12 that, under some conditions,
every permutation of the steps of a reduced path defines another reduced path.

######  Definition 2.8.

A path $[\alpha;\kappa_{1},\ldots,\kappa_{n};\beta]$ is reduced if
$\kappa_{i}+\kappa_{j}\notin R$ for any $1\leqslant i\neq j\leqslant n$.

######  Proposition 2.9.

No non-trivial sum of steps in a reduced path is a root. More strongly, if
$\kappa_{1},\kappa_{2},\ldots,\kappa_{n}\in R$ with
$\kappa_{1}+\kappa_{2}+\cdots+\kappa_{n}\in R$ and $n>1$, then for any
$i\leqslant n$, there exists some $i\neq j\leqslant n$ such that
$\kappa_{i}+\kappa_{j}\in R$.

###### Proof.

This follows from Proposition 2.7, by forming a path from $\kappa_{i}$ to
$\kappa_{1}+\kappa_{2}+\cdots+\kappa_{n}$ with steps in $\\{\kappa_{j}\mid
j\neq i\\}$. ∎

######  Proposition 2.10.

Let $[\alpha;\kappa_{1},\ldots,\kappa_{n};\beta]$ be a path. There exists a
reduced path $[\alpha;\mu_{1},\ldots,\mu_{m};\beta]$ between $\alpha$ and
$\beta$ with each $\mu_{i}$ being a sum of a subcollection of the steps
$\kappa_{1},\kappa_{2},\ldots,\kappa_{n}$ of the original path.

###### Proof.

We define an iterative process on the collection of roots
$\kappa_{1},\kappa_{2},\ldots,\kappa_{n}$. At each step, we replace two roots
whose sum is a root with that sum, decreasing the size of the collection but
preserving its sum. When this process terminates, which it necessarily does
since it decreases the length of the collection at each step, we are left with
some $\mu_{1},\mu_{2},\ldots,\mu_{m}$ with $m\leqslant n$ and $\mu_{i}\in R$
for all $i$. By Proposition 2.7, we can order these $\mu_{i}$ as steps to form
a path from $\alpha$ to $\beta$, avoiding the condition that any subset of
them sum to 0 through Proposition 2.9. This is our desired reduced path. ∎

######  Remark 2.11.

Given some initial path, the procedure outlined in the above proof does not
produce a unique output. There may be more than one way to reduce a path.

######  Proposition 2.12.

Let $[\alpha;\kappa_{1},\ldots,\kappa_{n};\beta]$ be a reduced path with
$\alpha\neq-\kappa_{i}$ for any $1\leqslant i\leqslant n$. For any permutation
$\mu_{1},\mu_{2},\ldots,\mu_{n}$ of $\kappa_{1},\kappa_{2},\ldots,\kappa_{n}$,
$[\alpha;\mu_{1},\ldots,\mu_{n};\beta]$ is a reduced path.

###### Proof.

We prove that any adjacent transposition of steps in
$[\alpha;\kappa_{1},\ldots,\kappa_{n};\beta]$ results in another reduced path,
from which the desired result follows. To that end, we take some $1\leqslant
i<n$ and consider swapping the positions of $\kappa_{i}$ and $\kappa_{i+1}$.
The resulting sequence of partial sums is the same as that of the original
path, except $\alpha+\kappa_{1}+\cdots+\kappa_{i-1}+\kappa_{i}$ is replaced by
$\alpha+\kappa_{1}+\cdots+\kappa_{i-1}+\kappa_{i+1}$. We therefore need only
to show that this is a root. To that end, we apply Lemma 2.4 to the expression
$(\alpha+\kappa_{1}+\cdots+\kappa_{i-1})+\kappa_{i}+\kappa_{i+1}$ to conclude
that $\alpha+\kappa_{1}+\cdots+\kappa_{i-1}+\kappa_{i+1}\in R$ provided
$\alpha+\kappa_{1}+\cdots+\kappa_{i-1}+\kappa_{i}\neq 0$. If $i=1$, this is
guaranteed by the condition that $\alpha\neq-\kappa_{j}$ for any $1\leqslant
j\leqslant n$. If $i>1$, then we would have a non-trivial sum of steps in a
reduced path equal to $-\alpha$, which is impossible by Proposition 2.9. We
therefore conclude that
$[\alpha;\kappa_{1},\ldots,\kappa_{i+1},\kappa_{i},\ldots,\kappa_{n};\beta]$
is another reduced path. ∎

######  Example 2.13.

In the root system $D_{5}$ we have the following equation,

$010^{\displaystyle 0}_{\displaystyle 0}-000^{\displaystyle 0}_{\displaystyle
1}+001^{\displaystyle 0}_{\displaystyle 1}+001^{\displaystyle
1}_{\displaystyle 1}+100^{\displaystyle 0}_{\displaystyle
0}=112^{\displaystyle 1}_{\displaystyle 1}$

and indeed these roots can be arranged to form the path

$\left[010^{\displaystyle 0}_{\displaystyle 0};100^{\displaystyle
0}_{\displaystyle 0},001^{\displaystyle 1}_{\displaystyle
1},-000^{\displaystyle 0}_{\displaystyle 1},001^{\displaystyle
0}_{\displaystyle 1};112^{\displaystyle 1}_{\displaystyle 1}\right].$

We can reduce this path, following the procedure outlined in Proposition 2.10,
obtaining

$\left[010^{\displaystyle 0}_{\displaystyle 0};100^{\displaystyle
0}_{\displaystyle 0},001^{\displaystyle 1}_{\displaystyle
0},001^{\displaystyle 0}_{\displaystyle 1};112^{\displaystyle
1}_{\displaystyle 1}\right],$

which is reduced. To verify Proposition 2.12, we check that each permutation
of the steps in the above path properly defines a path.

$\begin{array}[]{ccc}\left[010^{\displaystyle 0}_{\displaystyle
0};100^{\displaystyle 0}_{\displaystyle 0},001^{\displaystyle
1}_{\displaystyle 0},001^{\displaystyle 0}_{\displaystyle
1};112^{\displaystyle 1}_{\displaystyle 1}\right]&\left[010^{\displaystyle
0}_{\displaystyle 0};100^{\displaystyle 0}_{\displaystyle
0},001^{\displaystyle 0}_{\displaystyle 1},001^{\displaystyle
1}_{\displaystyle 0};112^{\displaystyle 1}_{\displaystyle
1}\right]&\left[010^{\displaystyle 0}_{\displaystyle 0};001^{\displaystyle
1}_{\displaystyle 0},100^{\displaystyle 0}_{\displaystyle
0},001^{\displaystyle 0}_{\displaystyle 1};112^{\displaystyle
1}_{\displaystyle 1}\right]\\\\[8.0pt] \left[010^{\displaystyle
0}_{\displaystyle 0};001^{\displaystyle 1}_{\displaystyle
0},001^{\displaystyle 0}_{\displaystyle 1},100^{\displaystyle
0}_{\displaystyle 0};112^{\displaystyle 1}_{\displaystyle
1}\right]&\left[010^{\displaystyle 0}_{\displaystyle 0};001^{\displaystyle
0}_{\displaystyle 1},100^{\displaystyle 0}_{\displaystyle
0},001^{\displaystyle 1}_{\displaystyle 0};112^{\displaystyle
1}_{\displaystyle 1}\right]&\left[010^{\displaystyle 0}_{\displaystyle
0};001^{\displaystyle 0}_{\displaystyle 1},001^{\displaystyle
1}_{\displaystyle 0},100^{\displaystyle 0}_{\displaystyle
0};112^{\displaystyle 1}_{\displaystyle 1}\right]\end{array}$

It is not too difficult to see that every partial sum along each of these
paths is a root in $D_{5}$.

The following proposition will be useful in Section 5. Its proof uses some
properties of reduced paths that we just proved.

######  Proposition 2.14.

If $\Phi$ is a co-closed subset of $R^{+}$, then
$\mathrm{span}_{\mathbb{Z}}\Phi=\mathrm{span}_{\mathbb{Z}}\operatorname{supp}\Phi$.

###### Proof.

It follows from the definition of $\operatorname{supp}\Phi$ that
$\mathrm{span}_{\mathbb{Z}}\Phi\subseteq\mathrm{span}_{\mathbb{Z}}\operatorname{supp}\Phi$.
To demonstrate the reverse inclusion, we take
$\theta\in\operatorname{supp}\Phi$ and show that it is in
$\mathrm{span}_{\mathbb{Z}}\Phi$. This is obvious if $\theta\in\Phi$, so we
consider the case where $\theta\notin\Phi$. Since
$\theta\in\operatorname{supp}\Phi$, there is some $\alpha\in\Phi$ such that
$\theta<\alpha$. We may therefore, appealing to Proposition 2.7, form a path
from $\theta$ to $\alpha$ with positive simple roots as steps. By Proposition
2.10, we may reduce this path into
$[\theta;\kappa_{1},\ldots,\kappa_{n};\alpha]$ with positive steps. Using
Proposition 2.12, we take the ordering of the $\kappa_{i}$’s to be as follows,
with those in $\Phi$ first, and later those in $\Phi^{c}$.

$[\theta;\underbrace{\kappa_{1},\ldots,\kappa_{m}}_{\kappa_{i}\in\Phi},\underbrace{\kappa_{m+1},\ldots,\kappa_{n}}_{\kappa_{i}\in\Phi^{c}};\alpha]$

Co-closure implies that $\theta+\kappa_{1}+\ldots+\kappa_{m}\in\Phi$, from
which it follows that $\theta\in\mathrm{span}_{\mathbb{Z}}\Phi$. ∎

##  3\. Inflations

######  Definition 3.1.

Let $I\subseteq S$ and $\Phi\subseteq R^{+}$. We say that $\Phi$ is an
inflation from $I$ of $\Psi\subseteq(R/I)^{+}$ by $X\subseteq R_{I}^{+}$ if
$\Phi=\pi_{I}^{-1}(\Psi)\cup X$ where $\pi_{I}$ is the canonical projection
from $V$ onto $V/\mathrm{span\,}I$. We write
$\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$.

######  Definition 3.2.

Let $\alpha,\beta\in R^{+}$ and $\Phi\subseteq R^{+}$. Then we write
$\alpha\sim_{\Phi}\beta$ if either both $\alpha,\beta\in\Phi$ or both
$\alpha,\beta\notin\Phi$.

######  Remark 3.3.

We will often use the following simple properties of inflations without
explicit reference.

  1. (i)

Every $\Phi\subseteq R^{+}$ is an inflation
$\Phi=\mathrm{inf}_{S}^{S}(\emptyset,\Phi)$ as well as
$\Phi=\mathrm{inf}_{\emptyset}^{S}(\Phi,\emptyset)$.

  2. (ii)

Let $I\subseteq S$. If
$\mathrm{inf}_{I}^{S}(\Psi_{1},X_{1})=\mathrm{inf}_{I}^{S}(\Psi_{2},X_{2})$,
then $\Psi_{1}=\Psi_{2}$ and $X_{1}=X_{2}$.

  3. (iii)

If $\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$, then
$\Phi^{c}=\mathrm{inf}_{I}^{S}(\Psi^{c},X^{c})$.

  4. (iv)

$\mathrm{inf}_{I}^{S}(\Psi_{1},X_{1})\cup\mathrm{inf}_{I}^{S}(\Psi_{2},X_{2})=\mathrm{inf}_{I}^{S}(\Psi_{1}\cup\Psi_{2},X_{1}\cup
X_{2})$

The following two propositions are easily verified.

######  Proposition 3.4.

Let $\Phi\subseteq R^{+}$ and $I\subseteq S$. The following are equivalent:

  1. (i)

$\Phi$ is inflated from $I$.

  2. (ii)

For any $\alpha,\beta\in R^{+}$ such that $\pi_{I}(\alpha)=\pi_{I}(\beta)\neq
0$, $\alpha\sim_{\Phi}\beta$.

  3. (iii)

For any $\alpha\in R^{+}$ and $\theta\in\pm I$ such that $\pi_{I}(\alpha)\neq
0$ and $\alpha+\theta\in R^{+}$, $\alpha+\theta\sim_{\Phi}\alpha$.

  4. (iv)

For any $\overline{\alpha}\in R/I$, $\pi^{-1}(\overline{\alpha})\subseteq\Phi$
or $\pi^{-1}(\overline{\alpha})\subseteq\Phi^{c}$.

  5. (v)

$\pi_{I}(\Phi)\cap\pi_{I}(\Phi^{c})\subseteq\\{0\\}$.

######  Proposition 3.5.

Let $I\subseteq J\subseteq S$ and let $X\subseteq R_{I}^{+}$, $T\subseteq
R_{J/I}^{+}$, and $\Psi\subseteq R_{S/J}^{+}$. Then

(1)
$\mathrm{inf}_{I}^{S}(\mathrm{inf}_{J/I}^{S/I}(\Psi,T),X)=\mathrm{inf}_{J}^{S}(\Psi,\mathrm{inf}_{I}^{J}(T,X))$

Here we use the natural isomorphism $(S/I)/(J/I)\cong S/J$ to make sense of
$\mathrm{inf}_{J/I}^{S/I}(\Psi,T)$.

######  Proposition 3.6.

Assume $I\subseteq J$ and
$\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)=\mathrm{inf}_{J}^{S}(\Theta,Y)$. Then
$\Psi=\mathrm{inf}_{J/I}^{S/I}(\Theta,Z)$ for some $Z\subseteq R_{J/I}$.

###### Proof.

Note that $Y\subset\mathrm{span\,}J\subset\mathrm{span\,}S$. Since
$Y=\Phi\cap\mathrm{span\,}J$, using Proposition 3.4 (ii), one verifies easily
that $Y=\mathrm{inf}_{I}^{J}(Z,X)$ for some $Z\subset\mathrm{span\,}J/I$. Then
(1) above implies

$\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)=\mathrm{inf}_{J}^{S}(\Theta,Y)=\mathrm{inf}_{J}^{S}(\Theta,\mathrm{inf}_{I}^{J}(Z,X))=\mathrm{inf}_{I}^{S}(\mathrm{inf}_{J/I}^{S/I}(\Theta,Z),X)\
,$

proving that $\Psi=\mathrm{inf}_{J/I}^{S/I}(\Theta,Z)$. ∎

######  Lemma 3.7.

Let $I\subseteq S$ and $\overline{\alpha},\overline{\beta},\overline{\gamma}$
be non-zero elements in $R/I$ such that
$\overline{\gamma}=\overline{\alpha}+\overline{\beta}$. Then

  1. (i)

If $\gamma\in\pi_{I}^{-1}(\overline{\gamma})$, then there are
$\alpha\in\pi_{I}^{-1}(\overline{\alpha})$ and
$\beta\in\pi_{I}^{-1}(\overline{\beta})$ with $\gamma=\alpha+\beta$.

  2. (ii)

If $\alpha\in\pi_{I}^{-1}(\overline{\alpha})$, then there are
$\beta\in\pi_{I}^{-1}(\overline{\beta})$ and
$\gamma\in\pi_{I}^{-1}(\overline{\gamma})$ with $\gamma=\alpha+\beta$.

###### Proof.

Statement (i) follows from [K, Theorems 1.9 and 2.3] and statement (ii)
follows from (i) by considering the elements
$\overline{\alpha}=\overline{\gamma}+(-\overline{\beta})$. ∎

######  Proposition 3.8.

Let $\Phi\subseteq R^{+}$. If $\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$. $\Phi$ is
an inversion set if and only if both $\Psi$ and $X$ are inversion sets.

###  3.1. The set $\mathrm{Gen}(\Phi)$

######  Definition 3.9.

Given $\Phi\subseteq R^{+}$, define

$\mathrm{Gen}(\Phi):=\\{K\subseteq
S\,|\,\Phi=\mathrm{inf}_{K}^{S}(\Theta,Y)\text{ for some }\Theta\text{ and
some }Y\\}\ ,$

the sets from which $\Phi$ is inflated.

######  Remark 3.10.

The following properties of $\mathrm{Gen}(\Phi)$ are obvious:

  1. (i)

$\emptyset,S\in\mathrm{Gen}(\Phi)$

  2. (ii)

$\mathrm{Gen}(\Phi)=\\{\emptyset,S\\}$ if and only if $\Phi$ is primitive.

  3. (iii)

$\mathrm{Gen}(\Phi)=2^{S}$ if and only if $\Phi=\emptyset$ or $\Phi=R^{+}$.

  4. (iv)

$\mathrm{Gen}(\Phi)=\mathrm{Gen}(\Phi^{c})$.

######  Proposition 3.11.

Let $\Phi\subseteq R^{+}$, and $I,J\in\mathrm{Gen}(\Phi)$. Then $I\cup
J\in\mathrm{Gen}(\Phi)$ and $I\cap J\in\mathrm{Gen}(\Phi)$.

###### Proof.

The statement about $I\cap J$ is straightforward. The proof about $I\cup J$ is
an immediate consequence of the fact that, for any $K\subset S$ and any non-
zero $\nu\in R/K$, the set $\pi_{K}^{-1}(\nu)$ is connected by elements of
$K$, which is a consequence of Proposition 2.7. ∎

###  3.2. Canonical Form

The aim of this subsection is to provide a canonical way to describe a subset
of $R^{+}$ as an inflation.

######  Definition 3.12.

$\Phi\subset R^{+}$ is primitive if $\Phi\neq\emptyset,\Phi\neq R^{+}$, and
$\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$ implies that either $I=\emptyset$ or
$I=S$.

######  Remark 3.13.

$\Phi\subset R^{+}$ is primitive if and only if $\Phi^{c}$ is primitive.

######  Theorem 3.14.

Let $R$ be a connected QRS, and let $\Phi\subseteq R^{+}$. Then
$\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$ where $I$ is a proper subset of $S$ and
one of the following mutually exclusive alternatives holds:

  1. (i)

$\Psi$ is primitive

  2. or

  3. (ii)

$\Psi=\emptyset$ or $\Psi=(R/I)^{+}$ and $I$ is smallest with this property.

Moreover, this decomposition is unique.

We call the expression $\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$ described in
Theorem 3.14 the canonical form of $\Phi$. Note that in [USRA], the notion of
a “simple form of a permutation” was studied. The canonical form of an
inversion set in type $\mathbb{A}$ is the counterpart of the simple form of
the corresponding permutation.

Before proving the theorem, we establish two useful results.

######  Lemma 3.15.

Let $\Phi\subseteq R^{+}$. There exists a unique set $I\subseteq S$ with the
following property: $\Phi=\mathrm{inf}_{K}^{S}(\emptyset,X)$ for some set $X$
if and only if $I\subseteq K$.

###### Proof.

Set $I:=\operatorname{supp}\Phi$. Clearly, $\Phi\subset\mathrm{span\,}I$ and
$\Phi=\mathrm{inf}_{I}^{S}(\emptyset,\Phi)$. Hence, for any $I\subset K\subset
S$, we have
$\Phi=\mathrm{inf}_{I}^{S}(\emptyset,\Phi)=\mathrm{inf}_{K}^{S}(\emptyset,\Phi)$.
Conversely, if $\Phi=\mathrm{inf}_{K}^{S}(\emptyset,X)$ for some
$X\subset\mathrm{span\,}K$, then $\Phi=X\subset\mathrm{span\,}K$. In
particular, $\operatorname{supp}(\alpha)\subset K$ for every $\alpha\in\Phi$,
proving that $I\subset K$. ∎

######  Lemma 3.16.

Assume that $I\cap J=\emptyset$ and $I\cup J=S$. Define

$I^{\prime}:=\\{\theta\in I\,|\,\theta\perp\lambda{\text{ for any }}\lambda\in
J\\}\quad{\text{and}}\quad J^{\prime}:=\\{\theta\in
J\,|\,\theta\perp\lambda{\text{ for any }}\lambda\in I\\}\ .$

If $\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)=\mathrm{inf}_{J}^{S}(\Theta,Y)$, then
$\Phi=\mathrm{inf}_{I^{\prime}\cup J^{\prime}}^{S}(\Xi,Z)$, where
$\Xi=\emptyset$ or $\Xi=(R/(I^{\prime}\cup J^{\prime}))^{+}$.

###### Proof.

Let $\tau$ denote the highest root of $R^{+}$. If $\gamma\in
R^{+}\backslash(R_{I}\cup R_{J})$, then $\tau$ and $\gamma$ are linked by a
sequence of roots, none of which belongs to $R_{I}\cup R_{J}$ and the
difference between any two consecutive roots in the sequence is an element of
$S=I\cup J$. Hence $\gamma\sim_{\Phi}\tau$.

Next we prove that if $\gamma\in R^{+}\backslash(R_{I^{\prime}}\cup
R_{J^{\prime}})$, then $\gamma\sim_{\Phi}\tau$. Without loss of generality we
may assume that $\gamma\in R_{I}\backslash R_{I^{\prime}}$. The definition of
$I^{\prime}$ implies that there is $\beta\in J$ which is connected to the
support of $\gamma$. Then $\langle\gamma,\beta\rangle<0$ and hence
$\gamma+\beta\in R^{+}$. Since $\pi_{J}(\gamma)=\pi_{J}(\gamma+\beta)\neq 0$,
we conclude that $\gamma\sim_{\Phi}\gamma+\beta\sim_{\Phi}\tau$. Since the
support of a root is connected, we get $R_{I^{\prime}\cup
J^{\prime}}=R_{I^{\prime}}\cup R_{J^{\prime}}$, which proves that $\Phi$ is an
inflation from $I^{\prime}\cup J^{\prime}$ of $(R/{I^{\prime}\cup
J^{\prime}})^{+}$ or $\emptyset$ depending on whether $\tau$ belongs to $\Phi$
or not. ∎

###### Proof of Theorem 3.14.

Existence. If $\Phi$ is primitive, $\Phi=\emptyset$, or $\Phi=R^{+}$, then we
observe that $\Phi=\mathrm{inf}_{\emptyset}^{S}(\Phi,\emptyset)$ and we are
done. If $\Phi$ is not primitive, then $\Phi=\mathrm{inf}_{K}^{S}(\Xi,Y)$ for
some proper non-empty subset $K$ of $S$. If $\Xi$ is primitive,
$\Xi=\emptyset$, or $\Xi=(R/K)^{+}$, we are done (invoking Lemma 3.15 if
necessary). If not, Proposition 3.5 implies that either $\Phi$ is an inflation
of from a proper subset $K^{\prime}$ of $S$ which contains $K$ as a proper
subset. This process is finite since, for every $L\subset S$ with
$\\#L={\mathrm{rk\,}}R-1$, every subset of $(R/L)^{+}$ is either primitive,
the empty set, or $(R/L)^{+}$ itself, proving the existence part of the
statement.

Uniqueness. If $\Phi=\mathrm{inf}_{I}^{S}(\emptyset,X)$ for some $X$, then the
highest root of $R^{+}$ does not belong to $\Phi$ in contrast to the case when
$\Phi$ is an inflation of $(R/I)^{+}$. This proves that $\Phi$ cannot be an
inflation of both $\emptyset$ and $(R/I)^{+}$.

Now assume that
$\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)=\mathrm{inf}_{J}^{S}(\Theta,Y)$, where
$\Psi$ is primitive and $\Theta$ is either primitive or equals $\emptyset$ or
$(R/J)^{+}$. By Proposition 3.11, we have that $\Phi=\mathrm{inf}_{I\cup
J}^{S}(\Gamma,Z)$ for some $\Gamma$ and $Z$. Now, since
$\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)=\mathrm{inf}_{I\cup J}^{S}(\Gamma,Z)$, by
Proposition 3.6, it follows that $\Psi=\mathrm{inf}_{I\cup
J/I}^{S/I}(\Gamma,T)$ for some $T$. Since $\Psi$ is primitive, we have $I\cup
J=I$ or $I\cup J=S$. In the first case, uniqueness follows from Proposition
3.6. Hence, we may assume that $I\cup J=S$. Moreover, if $I\cap
J\neq\emptyset$, then $\Phi=\mathrm{inf}_{I\cap J}^{S}(\Xi,Z)$, implying that
$\Xi$ is an inflation of $\Psi$ and $\Theta$ which proves the uniqueness by
induction on ${\mathrm{rk\,}}R$.

It remains to deal with the case $I\cap J=\emptyset$, $I\cup J=S$. Lemma 3.16
implies that $\Phi$ is an inflation from $I^{\prime}\cup J^{\prime}$. If
$J^{\prime}$ is not empty, then $\Phi$ is also an inflation from $I\cup
J^{\prime}$, contradicting the assumption that $\Psi$ is primitive. If both
$I^{\prime}$ and $J^{\prime}$ are empty, Lemma 3.16 implies that
$\Phi=\emptyset$ or $\Phi=R^{+}$, again contradicting the assumption that
$\Psi$ is primitive.

Finally we consider the case when $I^{\prime}\neq\emptyset$,
$J^{\prime}=\emptyset$, $\Theta=\emptyset$ (the case $\Theta=(R/J)^{+}$
follows because of Remark 3.13). Lemma 3.16 implies that $\Phi$ is an
inflation of the empty set from $I^{\prime}$ and $J$ and Lemma 3.15 implies
that $\Phi=\emptyset$, completing the proof of the uniqueness. ∎

######  Remark 3.17.

If $\Phi$ is canonically inflated from $I$, then so is $\Phi^{c}$.

######  Proposition 3.18.

In the language of Theorem 3.14,

  1. (i)

$\Psi$ is primitive if and only if
$\operatorname{supp}\Phi=\operatorname{supp}\Phi^{c}=S$;

  2. (ii)

$\Psi=\emptyset$ if and only if $\operatorname{supp}\Phi$ is a proper subset
of $S$ and $\operatorname{supp}\Phi^{c}=S$. Respectively, $\Psi=(R/I)^{+}$ if
and only if $\operatorname{supp}\Phi^{c}$ is a proper subset of $S$ and
$\operatorname{supp}\Phi=S$.

###### Proof.

If $\Phi=\mathrm{inf}_{I}^{S}(\emptyset,X)$, then $\operatorname{supp}\Phi=I$.
Conversely, if $\operatorname{supp}\Phi=J$ is a proper subset of $S$, then
$\Phi\subset R_{J}^{+}$ and $\Phi=\mathrm{inf}_{J}^{S}(\emptyset,\Phi)$. This
proves that $\operatorname{supp}\Phi=S$ unless
$\Phi=\mathrm{inf}_{I}^{S}(\emptyset,X)$. ∎

We complete this section by describing the set $\mathrm{Gen}(\Phi)$ when in
the canonical form $\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$, $\Psi$ is primitive.
When $\Psi$ above is not primitive, the description of $\mathrm{Gen}(\Phi)$ is
more complicated and we omit it.

######  Proposition 3.19.

Let $\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$ be the canonical form of
$\Phi\subseteq R^{+}$ where $\Psi$ is primitive. Then
$\mathrm{Gen}(\Phi)=\mathrm{Gen}(X)\cup\\{S\\}$.

###### Proof.

First we prove that $\mathrm{Gen}(\Phi)\backslash\\{S\\}$ contains a unique
maximal element. Assume, by way of contradiction, that
$\mathrm{Gen}(\Phi)\backslash\\{S\\}$ contains two distinct maximal elements
$K$ and $L$. Then $K\cap L$ and $K\cup L$ are both elements of
$\mathrm{Gen}(\Phi)$ by Proposition 3.11 and $K\cup L=S$ by the maximality of
$K$ and $L$. Write

$\Phi=\mathrm{inf}_{K}^{S}(\Psi^{\prime},X^{\prime})=\mathrm{inf}_{L}^{S}(\Psi^{\prime\prime},X^{\prime\prime})=\mathrm{inf}_{K\cap
L}^{S}(\Gamma,Y)\ .$

We define $K^{\prime\prime}:=K/(K\cap L)$ and $L^{\prime\prime}:=L/(K\cap L)$
and $S^{\prime\prime}:=S/(K\cap L)$. Then $K^{\prime\prime}\cap
L^{\prime\prime}=\emptyset$ and $K^{\prime\prime}\cup
L^{\prime\prime}=S^{\prime\prime}$. By Proposition 3.6, there exists
$\theta^{\prime},\theta^{\prime\prime},Z^{\prime},Z^{\prime\prime}$ such that

$\Gamma=\mathrm{inf}_{K^{\prime\prime}}^{S^{\prime\prime}}(\theta^{\prime},Z^{\prime})=\mathrm{inf}_{L^{\prime\prime}}^{S^{\prime\prime}}(\theta^{\prime\prime},Z^{\prime\prime})\
.$

Applying Lemma 3.16, we define
$K^{\prime}:=K^{\prime\prime}\cap(L^{\prime\prime})^{\perp}\subset
S^{\prime\prime}$ and
$L^{\prime}:=L^{\prime\prime}\cap(K^{\prime\prime})^{\perp}\subset
S^{\prime\prime}$ to get $\Gamma=\mathrm{inf}_{K^{\prime}\cup
L^{\prime}}^{S^{\prime\prime}}(\Xi,W)$ where $\Xi=\emptyset$ or
$\Xi=R/(K^{\prime}\cup L^{\prime})$.

Note that $K^{\prime}\cup L^{\prime}\neq S^{\prime\prime}$. To see this,
recall that $S^{\prime\prime}$ is connected and thus there exists $\alpha\in
K^{\prime\prime}$ and $\beta\in L^{\prime\prime}$ with $\alpha$ and $\beta$
not perpendicular. Thus both $\alpha$ and $\beta$ lie in
$S^{\prime\prime}\setminus(K^{\prime}\cup L^{\prime})$.

Therefore

$\Phi=\mathrm{inf}_{K\cap L}^{S}(\Gamma,Y)=\mathrm{inf}_{K\cap
L}^{S}(\mathrm{inf}_{K^{\prime}\cup
L^{\prime}}^{S^{\prime\prime}}(\Xi,W),Y)=\mathrm{inf}_{K^{\prime}\cup
L^{\prime}}^{S}(\Xi,U)$

for some $U$ where $\Xi=\emptyset$ or $\Xi=R/(K^{\prime}\cup L^{\prime})$. But
this contradicts the hypothesis that $\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$ is
the canonical form of $\Phi$ where $\Psi$ is primitive. This contradiction
shows that $\mathrm{Gen}(\Phi)\backslash\\{S\\}$ has a unique maximal element.

Next we note that this maximal element is $I$. Suppose, to the contrary, that
$\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)=\mathrm{inf}_{K}^{S}(\Psi^{\prime},X^{\prime})$
where $I\subsetneq K\subsetneq S$. Then by Proposition 3.6, we have
$\Psi=\mathrm{inf}_{K/I}^{S/I}(\Psi^{\prime},Z)$ for some
$Z\subset\text{span}(R/I)$. This contradicts the assumption that $\Psi$ is
primitive, proving that $I$ is the unique maximal element of
$\mathrm{Gen}(\Phi)\backslash\\{S\\}$.

To complete the proof that $\mathrm{Gen}(\Phi)=\mathrm{Gen}(X)\cup\\{S\\}$,
assume that $J\in\mathrm{Gen}(\Phi)\backslash\\{S\\}$. Then $J\subset I$ and

$\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)=\mathrm{inf}_{J}^{S}(K,Y)\ .$

By Proposition 3.6, $K=\mathrm{inf}_{I/J}^{S/J}(\theta,Z)$ and

$\Phi=\mathrm{inf}_{J}^{S}(K,Y)=\mathrm{inf}_{J}^{S}(\mathrm{inf}_{I/J}^{S/J}(\theta,Z),Y)=\mathrm{inf}_{I}(\theta,\mathrm{inf}_{I}^{J}(Z,Y))\
.$

Comparing the above with $\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$, we conclude that
$\theta=\Psi$ and $X=\mathrm{inf}_{J}^{I}(Z,Y)$, proving that
$J\in\mathrm{Gen}(X)$. This shows that
$\mathrm{Gen}(\Phi)\subset\mathrm{Gen}(X)\cup\\{S\\}$. The converse inclusion
is obvious. ∎

##  4\. The graph $G_{\Phi}^{\Phi^{c}}$

For the rest of the paper, the symbol $\Phi$ will denote an inversion set of
$R$, and in particular, a

While Section 2 deals with the existence and properties of paths between roots
in $R$, it is often useful to restrict the possible partial sums and steps
associated with a path. To that end, we introduce the family of graphs
$G_{A}^{B}$ and in particular the graph $G_{\Phi}^{\Phi^{c}}$, the structure
of whose components gives important information about the properties of
$\Phi$.

######  Definition 4.1.

For $A,B\subseteq R$, we define $G_{A}^{B}$ to be the graph whose vertices are
roots in $A$, where $\alpha$ and $\alpha^{\prime}$ are connected by an edge if
$\alpha-\alpha^{\prime}\in\pm B$.

Many statements about paths can be translated into statements about these
graphs. The statement that there is a path between any two roots with simple
roots as steps, for example, is the statement that $G_{R}^{S}$ is connected,
while the statement that
$\mathrm{span\,}_{\mathbb{Z}}\Phi=\mathrm{span\,}_{\mathbb{Z}}\operatorname{supp}\Phi$
can be understood to mean that the components of $G_{R}^{\Phi}$ are inflated
from $\operatorname{supp}\Phi$.

######  Proposition 4.2.

Let $\Phi=\Phi_{1}\sqcup\Phi_{2}$. If $C$ is a component of
$G_{\Phi}^{\Phi^{c}}$, then $C\subseteq\Phi_{1}$ or $C\subseteq\Phi_{2}$.

###### Proof.

Suppose to the contrary that neither $C\cap\Phi_{1}$ nor $C\cap\Phi_{2}$ is
empty. By definition we would have $\alpha+\kappa=\beta$ for some
$\alpha\in\Phi_{1},\beta\in\Phi_{2},$ and $\kappa\in\pm\Phi^{c}$. Without loss
of generality we may take $\kappa>0$, from which we obtain the contradiction
that $\beta\notin\Phi_{2}$ by co-closure. ∎

######  Remark 4.3.

Let $\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$ and $C$ be a component of
$G_{\Phi}^{\Phi^{c}}$. Then $C\subseteq\mathrm{inf}_{I}^{S}(\Psi,\emptyset)$
or $C\subseteq X$.

Proposition 4.2 hints that when studying decompositions of an inversion set
$\Phi$, the relevant objects to examine are not arbitrary subsets of $\Phi$,
but rather collections of components of $G_{\Phi}^{\Phi^{c}}$. For that
reason, understanding the structure of the set of these components is the
focus of the following subsections.

###  4.1. Addition of Components

Somewhat surprisingly, the components of $G_{\Phi}^{\Phi^{c}}$ admit an
approximately additive structure whose properties we investigate.

######  Lemma 4.4.

Let $\Phi\subseteq R^{+}$ and $A,B,C$ be components of $G_{\Phi}^{\Phi^{c}}$
with $\alpha,\in A,\beta\in B$, and $\alpha+\beta\in C$.

  1. (i)

If $A\neq C$, then for all $\alpha^{\prime}\in A$ there exists
$\beta^{\prime}\in B$ such that $\alpha^{\prime}+\beta^{\prime}\in C$

  2. (ii)

If $A\neq C$ and $B\neq C$, then for all $\gamma^{\prime}\in C$, there exist
$\alpha^{\prime}\in A$ and $\beta^{\prime}\in B$ such that
$\gamma^{\prime}=\alpha^{\prime}+\beta^{\prime}$.

###### Proof.

First we prove (i). It suffices to verify (i) for every neighbour
$\alpha^{\prime}\in A$ of $\alpha$. Write $\alpha^{\prime}=\alpha-\kappa$
where $\kappa\in\pm\Phi^{c}$. Then
$\alpha^{\prime}+\kappa+\beta=\alpha+\beta$. Since the right hand side is a
root, either $\alpha^{\prime}+\beta\in R$ or $\beta+\kappa\in R$.

If $\alpha^{\prime}+\beta$ is a root then it lies in $\Phi$ by the closure of
$\Phi$. Since $(\alpha^{\prime}+\beta)-(\alpha+\beta)=-\kappa\in\pm\Phi^{c}$
we see that $\alpha^{\prime}+\beta\in C$ verifying (i).

Conversely suppose that $\beta+\kappa\in R$. If $\beta+\kappa\in\pm\Phi^{c}$
then $(\alpha+\beta)-\alpha^{\prime}\in\pm\Phi^{c}$ and thus $C=A$. This
contradiction shows that $\beta+\kappa\notin\pm\Phi^{c}$. If
$\beta+\kappa\in-\Phi$ then $-\kappa=\beta+(-\beta-\kappa)\in\Phi$. This
contradiction shows that $\beta+\kappa\notin-\Phi$. Therefore
$\beta+\kappa\in\Phi$ and thus $\beta+\kappa\in B$. Then
$\alpha^{\prime}+(\beta+\kappa)=\alpha+\beta\in C$ again verifying (i).

To prove (ii), it suffices show that (ii) holds for every neighbour $\gamma$
of $\gamma^{\prime}$. Write $\gamma=\gamma^{\prime}+\kappa$ where
$\kappa\in\pm\Phi^{c}$. Consider the equation
$\alpha^{\prime}+\beta^{\prime}+\kappa=\gamma$. Since the right hand side is a
root either $\alpha^{\prime}+\kappa\in R$ or $\beta^{\prime}+\kappa\in R$.
Without loss of generality we may suppose that $\alpha^{\prime}+\kappa\in R$.

If $\alpha^{\prime}+\kappa\in\pm\Phi^{c}$ then
$\gamma-\beta^{\prime}=\alpha^{\prime}+\kappa\in\pm\Phi^{c}$ and thus $C=B$.
This contradiction shows that $\alpha^{\prime}+\kappa\notin\pm\Phi^{c}$. If
$\alpha^{\prime}+\kappa\in-\Phi$ then
$-\kappa=\alpha+(-\alpha^{\prime}-\kappa)\in\Phi$. This contradiction shows
that $\alpha^{\prime}+\kappa\notin-\Phi$. Therefore
$\alpha^{\prime}+\kappa\in\Phi$ and thus $\alpha^{\prime}+\kappa\in A$. Then
$(\alpha^{\prime}+\kappa)+\beta^{\prime}=\gamma$ as required. ∎

######  Lemma 4.5.

Suppose that $A,B,C,D$ are components of $G^{\Phi^{c}}_{\Phi}$ with $B\neq C$,
and $D\neq C$, and $A\neq D$. Further suppose that $n$ is minimal such that
exists $\alpha^{\prime}\in A$, and $\beta^{\prime},\beta^{\prime\prime}\in B$
such that the $\alpha^{\prime}+\beta^{\prime}\in C$ and
$\alpha^{\prime}+\beta^{\prime\prime}\in D$ and there exists a path
$[\beta^{\prime};\kappa^{\prime}_{1},\kappa^{\prime}_{2},\dots,\kappa^{\prime}_{n};\beta^{\prime\prime}]$
of length $n$ in $G^{\Phi^{c}}_{\Phi}$. Then

  1. (1)

$\alpha^{\prime}+\beta^{\prime}+\kappa^{\prime}_{1}$ is not a root,

  2. (2)

$\alpha^{\prime}+\beta^{\prime\prime}-\kappa^{\prime}_{1}$ is not a root,

  3. (3)

$\alpha^{\prime}_{1}:=\alpha^{\prime}-\kappa_{1}^{\prime}\in A$

  4. (4)

$[\beta^{\prime}+\kappa^{\prime}_{1};\kappa^{\prime}_{2},\dots,\kappa^{\prime}_{n},\kappa^{\prime}_{1};\beta^{\prime\prime}+\kappa^{\prime}_{1}]$
is a path in $B$ of length $n$, and
$\alpha^{\prime}_{1}+\beta^{\prime}+\kappa^{\prime}_{1}=\alpha^{\prime}+\beta\in
C$ and
$\alpha^{\prime}_{1}+\beta^{\prime\prime}+\kappa_{1}=\alpha^{\prime}+\beta^{\prime\prime}\in
D$.

###### Proof.

Since
$[\beta^{\prime};\kappa_{1},\kappa_{2},\dots,\kappa_{n};\beta^{\prime\prime}]$
is a path in $G^{\Phi^{c}}_{\Phi}$, each difference
$\kappa^{\prime}_{i}\in\pm\Phi^{c}$. Define
$\beta^{\prime}_{0}=\beta^{\prime}$ and
$\beta^{\prime}_{i}=\beta^{\prime}_{i-1}+\kappa_{i}$ for $i=1,2,\dots,n$. Then
each $\beta^{\prime}_{i}\in B$.

Consider the equation:

$\beta^{\prime}_{1}+\alpha^{\prime}-\kappa^{\prime}_{1}=\beta^{\prime}+\alpha^{\prime}$

Since the right hand side is a root, at least one of
$\beta^{\prime}_{1}+\alpha^{\prime}$ or $\alpha^{\prime}-\kappa^{\prime}_{1}$
is a root.

Assume by way of contradiction that $\beta^{\prime}_{1}+\alpha^{\prime}$ is a
root. Then $\beta^{\prime}_{1}+\alpha^{\prime}\in\Phi$ by the closure of
$\Phi$. Then
$\beta^{\prime}_{1}+\alpha^{\prime}=(\beta^{\prime}+\alpha)+\kappa_{1}$ lies
in $C$. This violates the minimality of $n$ as the path
$[\beta^{\prime}_{1};\kappa^{\prime}_{2},\kappa_{3}\dots,\kappa^{\prime}_{n};\beta^{\prime\prime}]$
in $G^{\Phi^{c}}_{\Phi}$ shows. Therefore $\beta^{\prime}_{1}+\alpha^{\prime}$
is not a root which proves (1). Furthermore, it implies that
$\alpha^{\prime}_{1}:=\alpha^{\prime}-\kappa^{\prime}_{1}$ is a root which
proves ( 3).

If
$\alpha^{\prime}-\kappa^{\prime}_{1}=(\alpha^{\prime}+\beta^{\prime})-\beta^{\prime}_{1}\in\pm\Phi^{c}$
then $C=B$. This contradiction shows
$\alpha^{\prime}-\kappa^{\prime}_{1}\notin\pm\Phi^{c}$. If
$\alpha^{\prime}-\kappa^{\prime}_{1}\in-\Phi$ then
$\kappa^{\prime}_{1}=\alpha^{\prime}+(-\alpha^{\prime}+\kappa^{\prime}_{1})\in\Phi$.
This contradiction shows $\alpha^{\prime}-\kappa^{\prime}_{1}\notin-\Phi$.
Therefore $\alpha^{\prime}-\kappa^{\prime}_{1}\in\Phi$ and thus
$\alpha^{\prime}_{1}:=\alpha^{\prime}-\kappa^{\prime}_{1}\in A$.

Consider the equation:

$\beta^{\prime\prime}+\alpha^{\prime}_{1}+\kappa^{\prime}_{1}=\beta^{\prime\prime}+\alpha^{\prime}$

Since the right hand side is a root at least one of
$\beta^{\prime\prime}+\alpha^{\prime}_{1}$ or
$\beta^{\prime\prime}+\kappa^{\prime}_{1}$ is a root.

Assume by way of contradiction that $\beta^{\prime\prime}+\alpha^{\prime}_{1}$
is a root. Then $\beta^{\prime\prime}+\alpha^{\prime}_{1}\in\Phi$ by the
closure of $\Phi$ and
$\beta^{\prime\prime}+\alpha^{\prime}_{1}=(\beta^{\prime\prime}+\alpha^{\prime})-\kappa_{1}$
lies in $D$. Hence we have a path
$[\beta^{\prime}+\kappa^{\prime}_{1};\kappa^{\prime}_{2},\kappa^{\prime}_{3},\dots,\kappa^{\prime}_{n};\beta^{\prime\prime}]$
of length $n-1$ in $G_{\Phi}^{\Phi^{c}}$ where
$\alpha^{\prime}_{1}+(\beta^{\prime}+\kappa^{\prime}_{1})\in C$ and
$\alpha^{\prime}_{1}+\beta^{\prime\prime}\in D$. This contradicts the
minimality of $n$ and thus $\beta^{\prime\prime}+\alpha^{\prime}_{1}$ cannot
be a root. This proves (2). Furthermore it implies that
$\beta^{\prime\prime}+\kappa^{\prime}_{1}$ must be a root.

If $\beta^{\prime\prime}+\kappa^{\prime}_{1}\in\pm\Phi^{c}$ then
$(\alpha^{\prime}+\beta^{\prime\prime})-(\alpha^{\prime}-\kappa^{\prime}_{1})=\beta^{\prime\prime}+\kappa^{\prime}_{1}\in\pm\Phi^{c}$
which implies that $\alpha^{\prime}+\beta^{\prime\prime}$ and
$\alpha^{\prime}-\kappa^{\prime}_{1}$ lie in the same component, and thus
$D=A$. This contradiction shows
$\beta^{\prime\prime}+\kappa^{\prime}_{1}\notin\pm\Phi^{c}$. If
$\beta^{\prime\prime}+\kappa^{\prime}_{1}\in-\Phi$ then
$-\kappa^{\prime}_{1}=\beta^{\prime\prime}+(-\beta^{\prime\prime}-\kappa^{\prime}_{1})\in\Phi$.
This contradiction shows
$\beta^{\prime\prime}+\kappa^{\prime}_{1}\notin-\Phi$. Therefore
$\beta^{\prime\prime}+\kappa^{\prime}_{1}\in\Phi$. Hence
$\beta^{\prime}_{n+1}:=\beta^{\prime\prime}+\kappa^{\prime}_{1}\in B$.

Thus we have the path
$[\beta^{\prime}_{1};\kappa^{\prime}_{2},\kappa^{\prime}_{3},\dots,\kappa^{\prime}_{n},\kappa^{\prime}_{1};\beta^{\prime}_{n+1}]$
of length $n$ in $G^{\Phi^{c}}_{\Phi}$ such that
$\alpha^{\prime}_{1}+\beta^{\prime}_{1}=\alpha^{\prime}-\kappa^{\prime}_{1}+\beta^{\prime}+\kappa^{\prime}_{1}+\in
C$ and
$\alpha^{\prime}_{1}+\beta^{\prime}_{n+1}=\alpha^{\prime}-\kappa^{\prime}_{1}+\beta^{\prime\prime}+\kappa^{\prime}_{1}\in
D$ which proves (4).

∎

######  Proposition 4.6.

Let $A,B,C,D$ be components of $G_{\Phi}^{\Phi^{c}}$. Suppose that $\alpha\in
A$ and $\beta,\beta^{\prime}\in B$ with $\alpha+\beta\in C$ and
$\alpha+\beta^{\prime}\in D$. Further suppose that either $A\neq B$ or that
$A=B\notin\\{C,D\\}$. Then $C=D$.

###### Proof.

We assume that $C\neq D$. The hypotheses guarantee that we may assume that
$B\neq C$ and $A\neq D$. There exists a path
$[\beta;\kappa_{1},\kappa_{2},\dots,\kappa_{n};\beta^{\prime}]$ in $B$. Each
of the differences $\kappa_{i}\in\pm\Phi^{c}$. Put $\beta_{0}:=\beta$ and
$\beta_{i}=\beta_{i-1}+\kappa_{i}$ for $i=1,2,\dots,n$. We are required to
prove that $\alpha+\beta$ and $\alpha+\beta^{\prime}$ are connected by a path
in the graph $G_{\Phi}^{\Phi^{c}}$.

We proceed by induction on $n$, the length of the path in
$G_{\Phi}^{\Phi^{c}}$ joining $\beta$ to $\beta^{\prime}$. If $n=0$, then
$\beta=\beta^{\prime}$ and the result is true.

The induction hypothesis is that $C=D$ whenever there exists
$\alpha^{\prime}\in A$, $\beta^{\prime\prime},\beta^{\prime\prime\prime}\in B$
with $\alpha^{\prime}+\beta^{\prime\prime}\in C$, and
$\alpha^{\prime}+\beta^{\prime\prime\prime}\in D$ such that there is a path in
$G_{\Phi}^{\Phi^{c}}$ of length less than $n$ joining $\beta^{\prime\prime}$
to $\beta^{\prime\prime\prime}\in B$.

If there exists a path in $G_{\Phi}^{\Phi^{c}}$ of length less than $n$
joining $\beta$ to $\beta^{\prime}$. then by the induction hypothesis we are
done. We assume by way of contradiction that no such path exists. This means
that Lemma 4.5 applies.

Thus both $\alpha+\beta+\kappa_{1}$ and $\alpha+\beta^{\prime}-\kappa_{1}$ are
not roots. Furthermore
$[\beta+\kappa_{1};\kappa_{2},\kappa_{3},\dots,\kappa_{n},\kappa_{1};\beta^{\prime}+\kappa_{1}]$
is a path in $B$ of length $n$ where $\alpha_{1}+\beta+\kappa_{1}\in C$ and
$\alpha_{1}+\beta^{\prime}+\kappa_{1}\in D$ with
$\alpha_{1}:=\alpha-\kappa_{1}\in A$.

We may apply Lemma 4.22 to this new path. Thus
$\alpha_{1}+(\beta+\kappa_{1})+\kappa_{2}=\alpha-\kappa_{1}+\beta+\kappa_{1}+\kappa_{2}=\alpha+\beta+\kappa_{2}$
is not a root. Similarly $\alpha+\beta^{\prime}-\kappa_{2}$ is not a root.
Furthermore
$[\beta+\kappa_{1}+\kappa_{2};\kappa_{3},\dots,\kappa_{n}.\kappa_{1},\kappa_{2};\beta^{\prime}+\kappa_{1}+\kappa_{2}]$
is a path in $B$ of length $n$ where
$\alpha_{2}+\beta+\kappa_{1}+\kappa_{2}\in C$ and
$\alpha_{2}+\beta^{\prime}+\kappa_{1}+\kappa_{2}\in D$ where
$\alpha_{2}:=\alpha_{1}-\kappa_{2}=\alpha-\kappa_{1}-\kappa_{2}\in A$.

Applying the Lemma 4.22 successively $n$ times we discover that both
$\alpha+\beta+\kappa_{i}$ and $\alpha+\beta^{\prime}-\kappa_{i}$ are not roots
for all $i=1,2,\dots,n$.

Since
$(\alpha+\beta)+\kappa_{1}+\kappa_{2}+\dots+\kappa_{n}=\alpha+\beta^{\prime}$
is a root, Proposition 2.3 implies that
$\kappa_{1}+\kappa_{2}+\dots+\kappa_{n}$ is a root. Hence Proposition 2.6
applies. This implies that there is some $i$ such that either
$\alpha+\beta+\kappa_{i}$ or $\alpha+\beta^{\prime}-\kappa_{i}$ is a root.
This contradiction shows that a path satisfying the induction hypothesis must
exist and thus $C=D$. ∎

######  Proposition 4.7.

Let $A,B$ be components of $G_{\Phi}^{\Phi^{c}}$. Let
$\alpha,\alpha^{\prime}\in A$, $\beta,\beta^{\prime}\in B$ with
$\alpha+\beta,\alpha^{\prime}+\beta^{\prime}\in R$. If $A\neq B$ or $A=B$ but
$\alpha+\beta,\alpha^{\prime}+\beta^{\prime}\notin A$, then $\alpha+\beta$ and
$\alpha^{\prime}+\beta^{\prime}$ belong to the same component of
$G_{\Phi}^{\Phi^{c}}$.

###### Proof.

Suppose that $A$ and $B$ are components of $G_{\Phi}^{\Phi^{c}}$ and
$\alpha,\alpha^{\prime}\in A$, $\beta,\beta^{\prime}\in B$, with both
$\alpha+\beta,\alpha^{\prime}+\beta^{\prime}\in R$. Let $D$ denote the
component of $G_{\Phi}^{\Phi^{c}}$ which contains
$\alpha^{\prime}+\beta^{\prime}$. By Lemma 4.4 there is a root
$\beta^{\prime\prime}\in B$ such that $\alpha+\beta^{\prime\prime}\in D$. Let
$C$ denote the component of $G_{\Phi}^{\Phi^{c}}$ which contains
$\alpha+\beta$. Then $\alpha+\beta\in C$ and $\alpha+\beta^{\prime\prime}\in
D$. We can apply Proposition 4.6 to conclude $C=D$ since $A\neq D$ and $B\neq
D$. ∎

We are now ready to define addition of components of $G_{\Phi}^{\Phi^{c}}$.

######  Definition 4.8.

Consider two components $A,B$ of $G_{\Phi}^{\Phi^{c}}$ and the set of roots
$Z:=\\{\alpha+\beta\mid\alpha\in A,\beta\in B\\}\cap R$.

  1. $(1)$

If $Z\neq\emptyset$ and is contained in a single component $C$, then we define
$A+B=C$. Note that this includes the cases where $A\neq B$.

  2. $(2)$

If $Z\neq\emptyset$ and intersects more than one component, then $B=A$, and
$Z$ meets exactly two components, which are $A,C$ (see Proposition 4.7). In
this case, we define $A+A=C$.

  3. $(3)$

If $Z=\emptyset$, then we leave $A+B$ undefined.

######  Remark 4.9.

In the above definition, we have no example where case (2) occurs. Although we
suspect it never occurs, we do not yet know how to exclude this possibility.

######  Remark 4.10.

Let $A,B,C$ be components of $G_{\Phi}^{\Phi^{c}}$. If $\alpha+\beta=\gamma$
for $\alpha\in A,\beta\in B$, and $\gamma\in C$, then $A+B=C$ or $A=B=C$ (or
both).

######  Example 4.11.

Here is an inversion set $\Phi$ in $\mathbb{D}_{7}$ which is the inflation of
a primitive $\Psi$. It is given by $\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$ where
the graph $G_{\Phi}^{\Phi^{c}}$ has two components outside $X$. Here

$\Psi=\\{100,001,111,012,112\\}$ and $I$ is the set consisting of the last $4$
simple roots, namely, $\\{\theta_{4},\theta_{5},\theta_{6},\theta_{7}\\}$.

$\Phi=\left\\{\begin{array}[]{ccccc}10000^{\displaystyle 0}_{\displaystyle
0},&00100^{\displaystyle 0}_{\displaystyle 0},&00010^{\displaystyle
0}_{\displaystyle 0},&00001^{\displaystyle 0}_{\displaystyle
0},&00110^{\displaystyle 0}_{\displaystyle 0}\\\\[6.0pt] 00011^{\displaystyle
0}_{\displaystyle 0},&00001^{\displaystyle 0}_{\displaystyle
1},&11100^{\displaystyle 0}_{\displaystyle 0},&00111^{\displaystyle
0}_{\displaystyle 0},&00011^{\displaystyle 0}_{\displaystyle 1},\\\\[6.0pt]
11110^{\displaystyle 0}_{\displaystyle 0},&00111^{\displaystyle
1}_{\displaystyle 0},&00111^{\displaystyle 0}_{\displaystyle
1},&11111^{\displaystyle 0}_{\displaystyle 0},&00111^{\displaystyle
1}_{\displaystyle 1},\\\\[6.0pt] 11111^{\displaystyle 1}_{\displaystyle
0},&11111^{\displaystyle 0}_{\displaystyle 1},&00112^{\displaystyle
1}_{\displaystyle 1},&11111^{\displaystyle 1}_{\displaystyle
1},&00122^{\displaystyle 1}_{\displaystyle 1},\\\\[6.0pt] 11112^{\displaystyle
1}_{\displaystyle 1},&11122^{\displaystyle 1}_{\displaystyle
1},&01222^{\displaystyle 1}_{\displaystyle 1},&11222^{\displaystyle
1}_{\displaystyle 1}\end{array}\right\\}$

######  Example 4.12.

Here is an inversion set $\Phi$ in $\mathbb{E}_{6}$ which is
$\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$ where
$I=\\{\theta_{2},\theta_{4},\theta_{5},\theta_{6}\\}$,
$\Psi=\mathbb{C}_{2}=\\{10,01,11,12\\}$ (which is not primitive) and
$X=\left\\{{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip 1pt}}&&0&&\hfil\hskip
1.0\\\\[-4.0pt] 0&0&1&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&0&0&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&1&1\hfil\hskip
1.0\end{array}}\right\\}$.
The graph $G_{\Phi}^{\Phi^{c}}$ has five components of which the first four
are outside $X$. Here

$\Phi=\left\\{\begin{array}[]{ccccc}{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 1&0&0&0&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&0&0&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&0&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&0&0&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&0&0&0\hfil\hskip 1.0\end{array}},\\\
{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip 1pt}}&&1&&\hfil\hskip
1.0\\\\[-4.0pt] 0&0&1&0&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&0&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&0&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&0&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&0\hfil\hskip 1.0\end{array}},\\\
{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip 1pt}}&&0&&\hfil\hskip
1.0\\\\[-4.0pt] 0&0&1&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&0&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&1&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&1\hfil\hskip 1.0\end{array}},\\\
{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip 1pt}}&&1&&\hfil\hskip
1.0\\\\[-4.0pt] 1&1&1&1&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&2&1&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&2&1&0\hfil\hskip 1.0\end{array}},\\\
{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip 1pt}}&&1&&\hfil\hskip
1.0\\\\[-4.0pt] 1&1&1&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&2&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&2&1&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&2&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&2&2&1\hfil\hskip 1.0\end{array}},\\\
{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip 1pt}}&&1&&\hfil\hskip
1.0\\\\[-4.0pt] 1&2&2&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&2&2&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&2&2&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&3&2&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&2&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&3&2&1\hfil\hskip
1.0\end{array}}.\end{array}\right\\}$

The first four components are:
$C_{1}=\left\\{{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 1&0&0&0&0\hfil\hskip
1.0\end{array}}\right\\}$,
$C_{2}=\left\\{{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&2&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&0&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&2&2&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&2&1&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&0\hfil\hskip
1.0\end{array}}\right\\}$,
$C_{3}=\left\\{{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&2&1&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&0&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&2&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&2&2&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&1&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&1&0\hfil\hskip
1.0\end{array}}\right\\}$, and
$C_{4}=\left\\{{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&2&1&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&2&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&2&2&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&3&2&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&2&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&3&2&1\hfil\hskip
1.0\end{array}}\right\\}$
The fifth component is $C_{5}=X$. The addition table is given by

$\begin{array}[]{l|ccccc}&1&2&3&4&5\\\ \hline\cr 1&-&3&-&-&-\\\ 2&3&-&4&-&2\\\
3&-&4&-&-&3\\\ 4&-&-&-&-&4\\\ 5&-&2&3&4&-\end{array}$

######  Example 4.13.

Here is an inversion set $\Phi$ in $\mathbb{E}_{6}$ which is
$\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$ where
$I=\\{\theta_{1},\theta_{3},\theta_{4},\theta_{5}\\}$,
$\Psi=\mathbb{B}_{2}=\\{10,01,11,21\\}$ (which is not primitive) and
$X=\Big{\\{}{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&0&1&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&1&0\hfil\hskip
1.0\end{array}}\Big{\\}}$.
The graph $G_{\Phi}^{\Phi^{c}}$ has seven components of which the first four
are outside $X$. Here

$\Phi=\left\\{\begin{array}[]{ccccc}{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&0&0&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&0&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&0&1&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&0&0&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&0&0\hfil\hskip 1.0\end{array}},\\\
{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip 1pt}}&&0&&\hfil\hskip
1.0\\\\[-4.0pt] 0&0&1&1&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&0&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&0&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&1&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&1&1\hfil\hskip 1.0\end{array}},\\\
{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip 1pt}}&&1&&\hfil\hskip
1.0\\\\[-4.0pt] 1&1&1&0&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&1&0\hfil\hskip 1.0\end{array}},\\\
{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip 1pt}}&&0&&\hfil\hskip
1.0\\\\[-4.0pt] 1&1&1&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&2&1&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&2&1&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&1&1\hfil\hskip 1.0\end{array}},\\\
{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip 1pt}}&&1&&\hfil\hskip
1.0\\\\[-4.0pt] 0&1&2&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&2&1&0\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&2&1&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&2&2&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&2&1&1\hfil\hskip 1.0\end{array}},\\\
{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip 1pt}}&&1&&\hfil\hskip
1.0\\\\[-4.0pt] 1&1&2&2&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&2&2&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&3&2&1\hfil\hskip
1.0\end{array}},&{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&2&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&3&2&1\hfil\hskip
1.0\end{array}}\end{array}\right\\}$

The first four components are:
$C_{1}=\left\\{{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&2&1&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&0&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&2&1&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&2&1&0\hfil\hskip 1.0\end{array}},\\\
{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip 1pt}}&&1&&\hfil\hskip
1.0\\\\[-4.0pt] 0&1&1&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&1&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&0&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&0\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&1&0\hfil\hskip
1.0\end{array}}\right\\}$,
$C_{2}=\left\\{{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&0&0&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&0&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&1&1\hfil\hskip
1.0\end{array}}\right\\}$,
$C_{3}=\left\\{{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&2&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&2&2&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&2&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&2&2&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&2&2&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&3&2&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&1&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 1&1&1&1&1\hfil\hskip
1.0\end{array}},{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&1&&\hfil\hskip 1.0\\\\[-4.0pt] 0&1&2&1&1\hfil\hskip
1.0\end{array}}\right\\}$, and
$C_{4}=\left\\{{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&2&&\hfil\hskip 1.0\\\\[-4.0pt] 1&2&3&2&1\hfil\hskip
1.0\end{array}}\right\\}$.

The last three components are
$C_{5}=\left\\{{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&0&0\hfil\hskip
1.0\end{array}}\right\\}$,
$C_{6}=\left\\{{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&0&1&0\hfil\hskip
1.0\end{array}}\right\\}$,
$C_{7}=\left\\{{\begin{array}[]{@{}c@{}c@{}c@{}c@{}c@{\hskip
1pt}}&&0&&\hfil\hskip 1.0\\\\[-4.0pt] 0&0&1&1&0\hfil\hskip
1.0\end{array}}\right\\}$ where $X=C_{5}\sqcup C_{6}\sqcup C_{7}$.

The addition table is given by

$\begin{array}[]{l|ccccccc}&1&2&3&4&5&6&7\\\ \hline\cr 1&-&3&4&-&1&1&1\\\
2&3&-&-&-&2&2&2\\\ 3&4&-&-&-&3&3&3\\\ 4&-&-&-&-&-&-&-\\\ 5&1&2&3&-&-&7&-\\\
6&1&2&3&-&7&-&-\\\ 7&1&2&3&-&-&-&-\\\ \end{array}$

######  Example 4.14.

Here is an inversion set $\Phi$ in $\mathbb{B}_{5}$ which is
$\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$ where $I=\\{\theta_{4},\theta_{5}\\}$,
$\Psi=\\{100,001,111,012,112\\}$, which is primitive, and
$X=\\{00010,00001,00011,00012\\}$.
The graph $G_{\Phi}^{\Phi^{c}}$ has six components of which the first two are
outside $X$. Here

$\Phi=\left\\{\begin{array}[]{ccccc}10000,&00100,&00010,&00001,&00110,\\\
00011,&11100,&00111,&00012,&11110,\\\ 00112,&11111,&00122,&11112,&11122,\\\
01222,&11222\end{array}\right\\}.$

The first two components are:
$C_{1}=\left\\{10000,00100,00110,11100,00111,11110,\\\
00112,11111,00122,11112,11122,01222\right\\}$, and
$C_{2}=\\{11222\\}$.

The last four components are
$C_{3}=\\{00010\\}$ $C_{4}=\\{00001\\}$, $C_{5}=\\{00011\\}$,
$C_{6}=\\{00012\\}$ where $X=C_{3}\sqcup C_{4}\sqcup C_{5}\sqcup C_{6}$.

The addition table is given by

$\begin{array}[]{l|cccccc}&1&2&3&4&5&6\\\ \hline\cr 1&2&-&1&1&1&1\\\
2&-&-&-&-&-&-\\\ 3&1&-&-&5&-&-\\\ 4&1&-&5&-&6&-\\\ 5&1&-&-&6&-&-\\\
6&1&-&-&-&-&-\\\ \end{array}$

###  4.2. Partial Order of Components

In this subsection, we define a partial order for the components of
$G_{\Phi}^{\Phi^{c}}$ in two different ways, one using the notion of addition
from Subsection 4.1.

The following definition explains our convention for adding multiple
components and omitting bracketing.

######  Definition 4.15.

Given components $C_{1},C_{2},\ldots,C_{n}$, a bracketed sum of
$C_{1},C_{2},\ldots,C_{n}$ is a sum in which the components appear in this
order and which contains $n-2$ pairs of parentheses and these pairs of
parentheses are properly nested. Moreover every subsum of components in this
expression must be defined. The standard sum of $C_{1},C_{2},\ldots,C_{n}$ is
the bracketed sum $(\ldots((C_{1}+C_{2})+C_{3})+\ldots)+C_{n}$. We denote the
standard sum by $C_{1}+C_{2}+\ldots+C_{n}$.

As usual we denote the sum $A+A+\dots+A$ of $k$ copies of $A$ by $kA$, when it
is defined.

######  Definition 4.16.

Let $A,B$ be components of $G_{\Phi}^{\Phi^{c}}$ for some $\Phi$. We say
$A\leqslant B$ if there exists a (possibly empty) collection of components
$C_{1},C_{2},\ldots,C_{n}$ such that $A+C_{1}+\ldots+C_{n}=B$.

######  Proposition 4.17.

$\leqslant$ is a partial order on the components of $G_{\Phi}^{\Phi^{c}}$.

###### Proof.

Reflexivity and transitivity are clear from the definition. To see that
$\leqslant$ is antisymmetric, suppose $A<B$ and $B<A$. We may then write
$B=A+C_{1}+\ldots+C_{m}$ and $A=B+C_{1}^{\prime}+\ldots+C_{n}^{\prime}$ with
$n,m>0$. Without loss of generality, we may take $A+C_{1}+\ldots+C_{i}\neq
A+C_{1}+\ldots+C_{i+1}$ and $B+C_{1}^{\prime}+\ldots+C_{i}^{\prime}\neq
B+C_{1}^{\prime}+\ldots+C_{i+1}^{\prime}$ for all $i$. Starting with any root
$\alpha\in A$, Lemma 4.4(i) allows us to find a root
$\alpha+\gamma_{1}+\ldots+\gamma_{m}\in B$ where $\gamma_{i}\in C_{i}$.
Similarly, for any root $\beta\in B$, we find a root
$\beta+\gamma_{1}^{\prime}+\ldots+\gamma_{n}^{\prime}\in A$ with
$\gamma^{\prime}_{i}\in C_{i}^{\prime}$. Given a root $\alpha^{\prime}\in A$,
we can thus find roots $\beta^{\prime}\in B$ with
$\beta^{\prime}>\alpha^{\prime}$ and $\alpha^{\prime\prime}\in A$ with
$\alpha^{\prime\prime}>\beta^{\prime}$. There are thus no maximal roots in
$A$, contradicting the fact that $R$ is finite. We conclude that if
$A\leqslant B$ and $B\leqslant A$, then $A=B$. ∎

######  Example 4.18.

Below are Hasse diagrams that correspond to the posets of components in
Examples 4.12, 4.13, and 4.14, respectively.

${{C_{4}}}$${{C_{2}}}$${{C_{4}}}$${{C_{3}}}$${{C_{1}}}$${{C_{3}}}$${{C_{1}}}$${{C_{2}}}$${{C_{6}}}$${{C_{1}}}$${{C_{2}}}$${{C_{7}}}$${{C_{6}}}$${{C_{5}}}$${{C_{5}}}$${{C_{6}}}$${{C_{3}}}$${{C_{4}}}$

######  Proposition 4.19.

Let $A,B,C$ be components of $G_{\Phi}^{\Phi^{c}}$ with $A+B=C$. Then
$\operatorname{supp}C=\operatorname{supp}A\cup\operatorname{supp}B$.

###### Proof.

First we show that $\operatorname{supp}A\subseteq\operatorname{supp}C$. If
$A=C$, the result is immediate. Otherwise, for any
$\theta\in\operatorname{supp}A$, we pick $\alpha\in A$ supported on $\theta$.
By Lemma 4.4 (i), there exists a $\beta\in B$ such that $\alpha+\beta\in C$.
Since $\theta\in\operatorname{supp}(\alpha+\beta)$,
$\theta\in\operatorname{supp}C$ as well, so
$\operatorname{supp}A\subseteq\operatorname{supp}C$. By symmetry, this also
shows that $\operatorname{supp}B\subseteq\operatorname{supp}C$, so that
$\operatorname{supp}A\cup\operatorname{supp}B\subseteq\operatorname{supp}C$.

To show that
$\operatorname{supp}C\subseteq\operatorname{supp}A\cup\operatorname{supp}B$ we
note that the statement is trivial if $C=A$ or $C=B$. Otherwise, by Lemma 4.4
(ii), each $\gamma\in C$ can be written as $\alpha+\beta$ for $\alpha\in A$
and $\beta\in B$, so that if $\theta\in\operatorname{supp}\gamma$, then
$\theta\in\operatorname{supp}\alpha$ or $\theta\in\operatorname{supp}\beta$,
which implies that $\theta\in\operatorname{supp}A\cup\operatorname{supp}B$ and
concludes the argument. ∎

######  Proposition 4.20.

Let $A,B$ be components of $G_{\Phi}^{\Phi^{c}}$ with $A\leqslant B$. Then
$\operatorname{supp}A\subseteq\operatorname{supp}B$.

###### Proof.

This follows immediately from Proposition 4.19. ∎

######  Proposition 4.21.

$A\leqslant B$ if and only if there exist $\alpha\in A,\beta\in B$ with
$\alpha\leqslant\beta$ in the usual sense.

###### Proof.

Suppose that $A\leqslant B$ and $\alpha\in A$. We construct a root $\beta\in
B$ such that $\alpha\leqslant\beta$. Writing $A+C_{1}+\ldots+C_{n}=B$ with
$A+C_{1}+\ldots+C_{i}\neq A+C_{1}+\ldots+C_{i+1}$ for all $i$, we find through
Lemma 4.4(i) a root $\beta:=\alpha+\gamma_{1}+\ldots+\gamma_{n}$ with
$\gamma_{i}\in C_{i}$. We observe that $\beta\in B$ and
$\alpha\leqslant\beta$. In the trivial case where $A=B$ this process simply
chooses $\beta=\alpha$. Conversely, we suppose that there is a pair $\alpha\in
A$ and $\beta\in B$ with $\alpha\leqslant\beta$. By definition, this means
that there is some expression $\beta=\alpha+\theta_{1}+\ldots+\theta_{n}$ with
$\theta_{i}\in S$. By Proposition 2.7, we may form a path
$[\alpha;\kappa_{1},\ldots,\kappa_{n};\beta]$ with
$\kappa_{1},\kappa_{2},\ldots,\kappa_{n}$ being a permutation of
$\theta_{1},\theta_{2},\ldots,\theta_{n}$. By Proposition 2.10, we may reduce
this path into a reduced path of the form
$[\alpha;\mu_{1},\ldots,\mu_{m};\beta]$, with $\mu_{i}>0$. Since Proposition
2.12 guarantees that any rearrangement of the steps in this path gives another
valid path, we may first put steps in $\Phi$ and then those in $\Phi^{c}$,
obtaining a path like the one below.

$[\alpha;\underbrace{\nu_{1},\ldots,\nu_{k}}_{\nu_{i}\in\Phi},\underbrace{\nu_{k+1},\ldots,\nu_{m}}_{\nu_{i}\in\Phi^{c}};\beta]$

First by closure, then by co-closure, the partial sums
$\alpha+\nu_{1}+\ldots+\nu_{i}$ are in $\Phi$. Let
$\beta^{\prime}=\alpha+\nu_{1}+\cdots+\nu_{k}$. Observe that
$\beta^{\prime},\beta$ lie in the same component. Hence we have a reduced path
of elements in $\Phi$ of length $k$ from a root in $A$ to a root in $B$. We
proceed by induction on $k$. If $k=0$, then $A=B$ and we are done. Assume that
$k\geqslant 1$. Let $\gamma=\alpha+\nu_{1}+\cdots+\nu_{k-1}$, which is a root
in $\Phi$. Let this root be in a component $D$. By induction, there are
components $C_{1},\ldots C_{s}$ such that $D=A+C_{1}+\cdots+C_{s}$. We may
further assume that $A+C_{1}+\cdots+C_{i}\neq A+C_{1}+\cdots+C_{i}+C_{i+1}$
for all $i$. If $D=B$, then we are done. Otherwise, consider the component $C$
of $G_{\Phi}^{\Phi^{c}}$ containing $\nu_{k}$. Since $\gamma+\nu_{k}$ is a
root in $B$ and $D\neq B$, this means that $D+C=B$. That means that
$A+C_{1}+\cdots+C_{s}+C=B$. Hence, $A\leqslant B$ as required. ∎

###  4.3. Further properties of Component Addition

The addition of components is not associative, since one can find components
$A,B,C$ such that $A+B$ and $(A+B)+C$ is defined, but $B+C$ is not defined.
However, we will prove below that weaker forms of associativity hold, cf.
Proposition 4.23.

We will use the following Lemma in the proof of Proposition 4.24.

######  Lemma 4.22.

Suppose that $A,B,C,D$ where $A\neq D$ and $B\neq D$ are components of
$G^{\Phi^{c}}_{\Phi}$ Further suppose that $n$ is minimal such that exist
$\alpha^{\prime}\in A$, and $\beta^{\prime}\in B$, and $\gamma^{\prime}\in C$
and both $\alpha^{\prime}+\beta^{\prime}\in D$ and
$\alpha^{\prime}+\gamma^{\prime}\in D$ and a path
$[\alpha^{\prime}+\beta^{\prime};\kappa_{1},\kappa_{2},\dots,\kappa_{n};\alpha^{\prime}+\gamma^{\prime}]$
of length $n$ in $G^{\Phi^{c}}_{\Phi}$.

Then

  1. (1)

$\beta^{\prime}+\kappa^{\prime}_{1}$ is not a root,

  2. (2)

$\gamma^{\prime}-\kappa^{\prime}_{1}$ is not a root,

  3. (3)

$\alpha^{\prime}_{1}:=\alpha^{\prime}+\kappa_{1}^{\prime}\in A$

  4. (4)

$[\alpha^{\prime}_{1}+\beta^{\prime};\kappa^{\prime}_{2},\dots,\kappa^{\prime}_{n},\kappa^{\prime}_{1};\alpha^{\prime}_{1}+\gamma^{\prime}]$
is a path in $G^{\Phi^{c}}_{\Phi}$ of length $n$ and
$\alpha^{\prime}_{1}+\beta^{\prime},\alpha^{\prime}_{1}+\gamma^{\prime}\in D$.

###### Proof.

Since
$[\alpha^{\prime}+\beta^{\prime};\kappa_{1},\kappa_{2},\dots,\kappa_{n};\alpha^{\prime}+\gamma^{\prime}]$
is a path in $G^{\Phi^{c}}_{\Phi}$, each difference
$\kappa^{\prime}_{i}\in\pm\Phi^{c}$. Define
$\delta^{\prime}_{0}=\alpha^{\prime}+\beta^{\prime}$ and
$\delta^{\prime}_{i}=\delta^{\prime}_{i-1}+\kappa_{i}$ for $i=1,2,\dots,n$.
Then each $\delta^{\prime}_{i}\in D$.

Consider the equation:

$\delta^{\prime}_{1}-\alpha^{\prime}-\kappa^{\prime}_{1}=\beta^{\prime}$

Since the right hand side is a root, at least one of
$\delta^{\prime}_{1}-\alpha^{\prime}$ or $\alpha^{\prime}+\kappa^{\prime}_{1}$
is a root.

Assume by way of contradiction that $\delta^{\prime}_{1}-\alpha^{\prime}$ is a
root. If $\delta^{\prime}_{1}-\alpha^{\prime}\in\pm\phi^{c}$ then
$\delta^{\prime}_{1}$ and $\alpha^{\prime}$ are in the same component. This
contradiction shows $\delta^{\prime}_{1}-\alpha^{\prime}\notin\pm\Phi^{c}$. If
$\delta^{\prime}_{1}-\alpha^{\prime}=\beta^{\prime}+\kappa^{\prime}_{1}\in-\Phi$
then
$-\kappa^{\prime}_{1}=(-\kappa^{\prime}_{1}-\beta^{\prime})+\beta^{\prime}\in\Phi$
by the closure of $\Phi$. This contradiction shows
$\delta^{\prime}_{1}-\alpha^{\prime}\notin-\Phi^{c}$. If
$\delta^{\prime}_{1}-\alpha^{\prime}=\beta^{\prime}+\kappa^{\prime}_{1}\in\Phi$
then it lies in the component $B$. Then we have a path
$[(\alpha^{\prime}+(\beta^{\prime}+\kappa^{\prime}_{1});\kappa^{\prime}_{2},\kappa^{\prime}_{3},\dots,\kappa^{\prime}_{n};\delta^{\prime}_{n}=\gamma^{\prime}+\alpha^{\prime}]$
of length $n-1$ in $D$. By the minimality of $n$, this cannot happen and thus
$\delta^{\prime}_{1}-\alpha^{\prime}=\beta^{\prime}+\kappa^{\prime}_{1}$ is
not a root. This proves (1). Moreover it implies that
$\alpha^{\prime}_{1}:=\alpha^{\prime}+\kappa^{\prime}_{1}$ is a root.

If
$\alpha^{\prime}+\kappa^{\prime}_{1}=\delta^{\prime}_{1}-\beta^{\prime}\in\pm\phi^{c}$
then we find that $\delta^{\prime}_{1}$ and $\beta^{\prime}$ lie in the same
component. This contradiction shows
$\alpha^{\prime}+\kappa^{\prime}_{1}\notin\pm\Phi^{c}$. If
$\alpha^{\prime}+\kappa^{\prime}_{1}\in-\Phi$ then
$-\kappa^{\prime}_{1}=(-\kappa^{\prime}_{1}-\alpha^{\prime})+\alpha^{\prime}\in\phi$
by the closure of $\Phi$. This contradiction shows
$\alpha^{\prime}+\kappa^{\prime}_{1}\notin-\Phi$. Hence we must have
$\alpha^{\prime}_{1}=\alpha^{\prime}+\kappa^{\prime}_{1}\in\Phi$ and thus
$\alpha^{\prime}_{1}\in A$ proving (3).

Consider the equation:

$\delta^{\prime}_{n}-\alpha^{\prime}_{1}+\kappa^{\prime}_{1}=\gamma^{\prime}$

Since the right hand side is a root at least one of
$\delta^{\prime}_{n}-\alpha^{\prime}_{1}$ or
$\delta^{\prime}_{n}+\kappa^{\prime}_{1}$ is a root.

Assume by way of contradiction that $\delta^{\prime}_{n}-\alpha^{\prime}_{1}$
is a root. If $\delta^{\prime}_{n}-\alpha^{\prime}_{1}\in\pm\Phi^{c}$ then
$\delta^{\prime}_{n}$ and $\alpha^{\prime}_{1}$ are in the same component.
This contradiction shows
$\delta^{\prime}_{n}-\alpha^{\prime}_{1}\notin\pm\Phi^{c}$. If
$\delta^{\prime}_{n}-\alpha^{\prime}_{1}=\gamma^{\prime}-\kappa^{\prime}_{1}\in-\Phi$
then
$\kappa^{\prime}_{1}=(\kappa^{\prime}_{1}-\gamma^{\prime})+\gamma^{\prime}\in\Phi$
by the closure of $\Phi$. This contradiction shows
$\delta^{\prime}_{n}-\alpha^{\prime}_{1}\notin-\Phi^{c}$. . If
$\delta^{\prime}_{n}-\alpha^{\prime}_{1}=\gamma^{\prime}-\kappa^{\prime}_{1}\in\Phi$
then it lies in the component $C$. We now have a path
$[\alpha^{\prime}_{1}+\beta^{\prime}=\delta^{\prime}_{1};\kappa^{\prime}_{2},\kappa^{\prime}_{3},\dots,\kappa^{\prime}_{n};\delta^{\prime}_{n}=\alpha^{\prime}_{1}+(\gamma^{\prime}-\kappa^{\prime}_{1})]$
of length $n-1$ in $D$. By the minimality of $n$, this cannot happen and thus
$\delta^{\prime}_{n}-\alpha^{\prime}_{1}=\gamma^{\prime}-\kappa^{\prime}_{1}$
is not a root. This proves (2). Moreover it implies that
$\delta^{\prime}_{n}+\kappa^{\prime}_{1}$ is a root.

The root
$\delta^{\prime}_{n}+\kappa^{\prime}_{1}=\gamma^{\prime}+\alpha^{\prime}_{1}$
lies in $\Phi$ by the closure of $\Phi$. Moreover it is an element of $D$.

We now have a path
$[\alpha^{\prime}_{1}+\beta^{\prime}=\delta^{\prime}_{1};\kappa^{\prime}_{2},\kappa^{\prime}_{3},\dots,\kappa^{\prime}_{n};\delta^{\prime}_{n+1}=\gamma^{\prime}+\alpha^{\prime}_{1}]$
in $D$ of length $n$, with differences
$\kappa_{2},\kappa_{3},\dots,\kappa_{n},\kappa_{1}$ which proves (4). ∎

As noted above, addition of components is not associative either but the
following weaker statements hold.

######  Proposition 4.23.

Let $A,C_{1},C_{2},\ldots,C_{n}$ be components of $G_{\Phi}^{\Phi^{c}}$.

  1. (i)

Suppose that $C_{1}+C_{2}+C_{3}$ is defined. Then there exist distinct $r,s$
with $\\{r,s\\}\neq\\{1,2\\}$ such that $C_{r}+C_{s}$ is defined. Furthermore,
we can pick such $r,s$ with the property that
$(C_{r}+C_{s})+C_{t}=C_{1}+C_{2}+C_{3}$ where
$t\in\\{1,2,3\\}\backslash\\{r,s\\}$.

  2. (ii)

Suppose a bracketed sum of components $C_{1},C_{2},\ldots,C_{n}$ exists and
equals $A$. Then there is a permutation $\\{i_{1},i_{2},\ldots,i_{n}\\}$ of
$\\{1,2,\ldots,n\\}$ such that $A=C_{i_{1}}+C_{i_{2}}+\ldots+C_{i_{n}}$.

###### Proof.

First we prove (i). Let $\alpha_{i}\in C_{i}$ for $i=1,2,3$ be such that both
$\alpha_{1}+\alpha_{2}\in C_{1}+C_{2}$ and
$\delta=(\alpha_{1}+\alpha_{2})+\alpha_{3}\in C_{1}+C_{2}+C_{3}$ are roots. By
Lemma 2.4 there exists at least two distinct unordered pairs
$\\{i,j\\},\\{j,k\\}\subset\\{1,2,3\\}$ such that both $\alpha_{i}+\alpha_{j}$
and $\alpha_{j}+\alpha_{k}$ are roots. Thus both $C_{i}+C_{j}$ and
$C_{j}+C_{k}$ are defined. The remaining of the proof of (i) considers several
cases and subcases.

Case A. Assume first that the $C_{i}$ are pairwise distinct. Since
$(C_{1}+C_{2})+C_{3}$ is defined, there are roots $\alpha_{i}\in C_{i}$ with
$\alpha_{1}+\alpha_{2}\in C_{1}+C_{2}$ and
$\alpha_{1}+\alpha_{2}+\alpha_{3}\in(C_{1}+C_{2})+C_{3}$. That includes the
case where $C_{1}+C_{2}=C_{3}$ and $C_{3}+C_{3}=C_{3}$. Now, either
$\alpha_{1}+\alpha_{3}$ or $\alpha_{2}+\alpha_{3}$ is a root. Without loss of
generality, assume the first case happens. Then $C_{1}+C_{3}$ is defined.
Since $C_{1}\neq C_{3}$, we know that $\alpha_{1}+\alpha_{3}\in C_{1}+C_{3}$.
The expression that $(\alpha_{1}+\alpha_{3})+\alpha_{2}$ is a root says that
$(C_{1}+C_{3})+C_{2}$ is defined. If $C_{1}+C_{3}\neq C_{2}$, then it is clear
that we get $(C_{1}+C_{3})+C_{2}=C_{1}+C_{2}+C_{3}$, as required. Assume that
$C_{1}+C_{3}=C_{2}$. In this case, $C_{1}+C_{2}\neq C_{3}$ as it would
otherwise yield $C_{2}\leq C_{3}\leq C_{2}$, a contradiction. We take
$\beta_{2}$ maximal in $C_{2}$ such that $\beta_{1}\in C_{1}$ is such that
$\beta_{1}+\beta_{2}\in C_{1}+C_{2}$, and then we pick $\beta_{3}\in C_{3}$
such that $\beta_{1}+\beta_{2}+\beta_{3}\in C_{1}+C_{2}+C_{3}$.

Subcase A.1. If $\beta_{1}+\beta_{3}$ is a root, then
$\beta_{1}+\beta_{3}\in(C_{1}+C_{3})=C_{2}$. However, we know that
$(\beta_{1}+\beta_{3})+\beta_{2}$ is defined and by maximality of $\beta_{2}$,
we know that $\mathrm{C}_{2}+C_{2}\neq C_{2}$. In this case, we get
$(C_{1}+C_{3})+C_{2}=C_{1}+C_{2}+C_{3}$.

Subcase A.2. If $\beta_{1}+\beta_{3}$ is not a root, then
$\beta_{2}+\beta_{3}$ is a root. In this case, $C_{2}+C_{3}$ is defined. As
above, the conclusion follows if $C_{2}+C_{3}\neq C_{1}$. Otherwise, we get
$C_{1}\leq C_{2}\leq C_{1}$, a contradiction.

Now, assume that two of the $C_{i}$ coincide. The conclusion is clear if they
are all the same. We call these two components $A$ and $B$ where $A$ is the
component that is repeated twice. We consider the cases where the sums are
$A+A+B,A+B+A$ or $B+A+A$.

Case B. Consider first the case $A+A+B$.

Subcase B.1. Assume that $A+A=A$. The expression $A+B$ is defined, and we need
to prove that $A+A+B=A+(A+B)$. If $A+B=A$, then the result follows. Otherwise,
we can find a root of the form $\alpha_{1}+\alpha_{2}$ in $A+A$ and a root
$\beta$ in $B$ such that $\alpha_{1}+\alpha_{2}+\beta$ is a root. Now, one of
$\alpha_{1}+\beta,\alpha_{2}+\beta$ is a root. In the first case,
$(\alpha_{1}+\beta)+\alpha_{2}$ being a root and $A+B\neq A$ shows what we
want. Similarly in the other case.

Subcase B.2. Assume that $A+A=2A\neq A$. We can pick $\alpha_{1},\alpha_{2}$
and $\beta$ such that $\alpha_{1}+\alpha_{2}\in 2A$ and
$\alpha_{1}+\alpha_{2}+\beta\in A+A+B$. That includes the case where $B=2A$.
The 2 out of 3 rule shows that $A+B$ is defined and $\alpha_{1}+\beta$ or
$\alpha_{2}+\beta$ lies in $A+B$. Assume the first case. The only problematic
case is when $A+B=A$. But then $(\alpha_{1}+\beta)+\alpha_{2}$ is a sum of two
roots in $A$. If it is not in $2A$, then it is in $A$ but this would yield
$2A<A$, a contradiction.

Case C. We have $B+A+A$. This case is the same as case B up to relabeling.

Case D. The only remaining case is $A+B+A$. Obviously, we have that $B+A$ is
defined and $(B+A)+A=(A+B)+A=A+B+A$, as wanted.

This finishes the proof of (i).

Now we prove (ii). Let $A=X+Y$, with $X$ being a bracketed sum of the
components $C_{1},\ldots,C_{k}$ and $Y$ being a bracketed sum of the
components $C_{k+1},\ldots,C_{n}$. Induction on $n$, $n=1$ being trivial.
Assume the statement is true for fewer than $n$ components. Induction on
$j:=\min(k,n-k)$. The case $j=1$ follows from the induction on $n$. Assume
$j>1$ and, without loss of generality, $j=k$. Then
$X=X^{\prime}+X^{\prime\prime}$ and, by (i) we have

$X+Y=(X^{\prime}+X^{\prime\prime})+Y=(X^{\prime}+Y)+X^{\prime\prime}{\text{ or
}}(X^{\prime\prime}+Y)+X^{\prime}\ .$

Since both $X^{\prime}$ and $X^{\prime\prime}$ involve fewer components than
$X$, the statement follows by induction on $j$. ∎

The addition of components lacks some of the usual properties of addition. For
example, $A+B=A+C$ does not necessarily imply that $B=C$. In Example 4.13,
$C_{1}+C_{5}=C_{1}+C_{6}=C_{1}+C_{7}=C_{1}$. However, the following weaker
cancellation rules apply:

######  Proposition 4.24.

Let $A,B,C$ be components of $G_{\Phi}^{\Phi^{c}}$.

  1. (i)

If $A+B=A+C\neq A$, then $B=C$.

  2. (ii)

If $A+B+C=A+B$, then $A+C=A$ or $B+C=B$.

  3. (iii)

If $A+B+C=B+C$, then $A+B=B$ or $B+C=C$.

###### Proof.

First we prove (i). Let $D:=A+B=A+C\neq A$. If $B=C$ we are done and thus
without loss of generality we may assume $B\neq D$. Fix $\alpha\in A$. By
Lemma 4.4 there exists $\beta\in B$ and $\gamma\in C$ such that both
$\alpha+\beta$, and $\alpha+\gamma$ are elements of $D$. Moreover there exists
a path $[\alpha+\beta;\kappa_{1},\kappa_{2},\dots,\kappa_{n};\alpha+\gamma]$
in $D$. Each of the differences $\kappa_{i}\in\pm\Phi^{c}$. Put
$\delta_{0}:=\alpha+\beta$ and $\delta_{i}=\delta_{i-1}+\kappa_{i}$ for
$i=1,2,\dots,n$. We are required to prove that $\beta$ and $\gamma$ are
connected by a path in the graph $G_{\Phi}^{\Phi^{c}}$.

We proceed by induction on $n$, the length of the path in
$G_{\Phi}^{\Phi^{c}}$ joining $\alpha+\beta$ to $\alpha+\gamma$. If $n=0$,
then $\beta=\gamma$ and the result is true.

The induction hypothesis is that $B=C$ whenever there exists
$\alpha^{\prime}\in A$, $\beta^{\prime}\in B$ and $\gamma^{\prime}\in C$ such
that there is a path in $G_{\Phi}^{\Phi^{c}}$ of length less than $n$ joining
$\alpha^{\prime}+\beta^{\prime}\in D$ to $\gamma^{\prime}+\delta^{\prime}\in
D$.

If there exists $\alpha^{\prime}\in A$, $\beta^{\prime}\in B$ and
$\gamma^{\prime}\in C$ such that there is a path in $G_{\Phi}^{\Phi^{c}}$ of
length less than $n$ joining $\alpha^{\prime}+\beta^{\prime}\in D$ to
$\gamma^{\prime}+\delta^{\prime}\in D$ then by the induction hypothesis we are
done. We assume by way of contradiction that no such path exists. This means
that Lemma 4.22 applies.

Thus both $\beta+\kappa_{1}$ and $\gamma-\kappa_{1}$ are not roots.
Furthermore
$[\beta+\alpha_{1};\kappa_{2},\kappa_{3},\dots,\kappa_{n},\kappa_{1};\gamma+\alpha_{1}]$
is a path in $D$ of length $n$ where $\alpha_{1}:=\alpha+\kappa_{1}$.

We may apply Lemma 4.22 to this new path. Thus both $\beta+\kappa_{2}$ and
$\gamma-\kappa_{2}$ are not roots and
$[\beta+\alpha_{2};\kappa_{2},\kappa_{3},\dots,\kappa_{n},\kappa_{1},\kappa_{2};\gamma+\alpha_{2}]$
is a path in $D$ of length $n$, where
$\alpha_{2}:=\alpha_{1}+\kappa_{2}=\alpha+\kappa_{1}+\kappa_{2}$.

Applying the Lemma 4.22 successively $n$ times we discover that both
$\beta+\kappa_{i}$ and $\gamma-\kappa_{i}$ are not roots for all
$i=1,2,\dots,n$.

Since $\alpha+\beta+\kappa_{1}+\kappa_{2}+\dots+\kappa_{n}=\alpha+\gamma$ we
have $\beta+\kappa_{1}+\kappa_{2}+\dots+\kappa_{n}=\gamma$. By Proposition 2.3
we must have that $\kappa_{1}+\kappa_{2}+\dots+\kappa_{n}$ is a root. Hence
Proposition 2.6 applies. This implies that there is some $i$ such that either
$\beta+\kappa_{i}$ or $\gamma-\kappa_{i}$ is a root. This contradiction shows
that a path satisfying the induction hypothesis must exist and thus $B=C$.

We now prove (ii). If $A+B=A$ this becomes $A+C=A$. Similarly if $A+B=B$ this
becomes $B+C=B$. Therefore we may suppose that $A+B\neq A$ and $A+B\neq B$. By
Proposition 4.23 (i), either $(A+C)+B=A+B$ or $A+(B+C)=A+B$. By (i) we can
cancel $B$ in the former case and $A$ in the latter to obtain either $A+C=A$
or $B+C=B$ as required.

Now we prove the third assertion (iii). Suppose that $A+B+C=B+C$. If $B+C\neq
C$ then we can use part (i) to cancel $C$ from this equation to get $A+B=B$ as
required.

∎

###  4.4. Simple Components

######  Definition 4.25.

A component $C$ of $G_{\Phi}^{\Phi^{c}}$ is simple if, whenever $A+B=C$, $A=C$
or $B=C$.

The next proposition tells us that the notion of simple for component addition
is compatible with the same notion for their underlying QRS.

######  Proposition 4.26.

A component $C$ of $G_{\Phi}^{\Phi^{c}}$ is simple if and only if it contains
a simple root.

###### Proof.

$\left(\implies\right)$: Suppose $C$ is a simple component. Let $\gamma\in C$
be non-simple. Since $\gamma$ is not simple, there exist positive roots
$\alpha,\beta$ such that $\alpha+\beta=\gamma$. If, one of $\alpha,\beta$ is
in $\phi^{c}$, we would have that the other is in $\phi$, and in particular,
in $C$. However, if $\alpha,\beta$, are both in $\phi$, they belong to their
own components $A,B$ respectively, such that $A+B=C$. Since $C$ is simple,
either $A$ or $B$ is $C$. In both cases, there is a root in $C$ strictly
smaller than $\gamma$. By induction, $C$ contains a simple root.

$\left(\impliedby\right)$: Suppose that $C$ is a component which contains a
simple root, but is not simple. let $\theta\in C$ be a simple root and $A,B$
be components not equal to $C$ such that $A+B=C$. By Lemma 4.4(ii), there
exists $\alpha\in A$ and $\beta\in B$ such that $\alpha+\beta=\theta$, a
contradiction. ∎

######  Proposition 4.27.

There is at most one simple component $C$ of $G_{\Phi}^{\Phi^{c}}$ with
$\operatorname{supp}C=\operatorname{supp}\Phi$.

###### Proof.

Suppose $C_{1}$ and $C_{2}$ are simple components in $G_{\Phi}^{\Phi^{c}}$
with
$\operatorname{supp}C_{1}=\operatorname{supp}C_{2}=\operatorname{supp}\Phi$.
By Proposition 4.26, there is a simple root $\theta\in C_{1}$, and since
$C_{2}$ is of full support in $\Phi$, there exists a root $\alpha\in C_{2}$
supported on $\theta$, that is $\theta\leqslant\alpha$. This gives us by
Proposition 4.21 that $C_{1}\leqslant C_{2}$. Symmetrically, $C_{2}\leqslant
C_{1}$, and we conclude that $C_{1}=C_{2}$. ∎

######  Proposition 4.28.

Any component $A$ can be written as a standard sum of simple components, (cf.
Definition 4.15).

###### Proof.

If $A$ is simple, we are done. Otherwise there exist components $B$ and $C$
such that $A=B+C$ with $B<A$ and $C<A$. Using induction and Proposition 4.23
(ii), the result is clear.

∎

##  5\. Proof of Main Theorem

###  5.1. Primitive implies Irreducible

######  Proposition 5.1.

Let $\Phi$ be an inversion set. If $C$ is a component of
$G_{\Phi}^{\Phi^{c}}$, then $\operatorname{supp}C\in\mathrm{Gen}(\Phi)$.

###### Proof.

Let $I=\operatorname{supp}C$ and $D=\bigcup_{C^{\prime}\leqslant
C}C^{\prime}$. It follows from Proposition 4.20 that $\operatorname{supp}D=I$.
Next, suppose $\alpha+\beta\in D$ for some $\alpha,\beta\in R^{+}$. We will
show that either $\alpha$ or $\beta$ is in $D$ to prove that $D$ is co-closed.
Since $C\subseteq\Phi$ and $\Phi$ is co-closed, either $\alpha$ or $\beta$ is
in $\Phi$, so without loss of generality we let $\alpha\in\Phi$. We then give
the labels $C_{\alpha+\beta}$ and $C_{\alpha}$ to the components respectively
containing $\alpha+\beta$ and $\alpha$. Since $\alpha\leqslant\alpha+\beta$,
we have by Proposition 4.21 that $C_{\alpha}\leqslant
C_{\alpha+\beta}\leqslant C$, so $C_{\alpha}\subseteq D$, $\alpha\in D$, and
$D$ is co-closed.

Having constructed the co-closed set $D$, we use it to prove our result. We
suppose toward a contradiction that there exist $\gamma,\delta\in R^{+}$ with
$\pi_{I}(\gamma)=\pi_{I}(\delta)\neq 0$, $\gamma\in\Phi$, and
$\delta\in\Phi^{c}$. By Proposition 2.14,
$\delta-\gamma\in\mathrm{span}_{\mathbb{Z}}(D)$, and so Proposition 2.7
guarantees a path $[\gamma;\kappa_{1},\ldots,\kappa_{n};\delta]$ with
$\kappa_{i}\in\pm D$. Since $\gamma\in\Phi$, $\delta\in\Phi^{c}$, and all
partial sums are positive, there will be some $i$ such that the $i$-th partial
sum $\gamma_{i}$ is in $\Phi$ while the $(i+1)$-th partial sum $\gamma_{i+1}$
is in $\Phi^{c}$. We would then have $\gamma_{i}+\kappa_{i+1}=\gamma_{i+1}$.
We note that $\kappa_{i+1}$ must be negative since since $\Phi$ is closed and
$\gamma_{i+1}\in\Phi^{c}$. But then $\gamma_{i}-(-\kappa_{i+1})=\gamma_{i+1}$,
so that $\gamma_{i}$ and $-\kappa_{i+1}$ are linked in $G_{\Phi}^{\Phi^{c}}$.
Since $-\kappa_{i+1}\in D$ and $D$ is a union of components of
$G_{\Phi}^{\Phi^{c}}$, this implies $\gamma^{\prime}\in D$, which is not
possible because $\pi_{I}(\gamma^{\prime})\notin\pi_{I}(D)=\\{0\\}$. ∎

######  Theorem 5.2.

Let $\Phi$ be an inversion set. The following are equivalent:

  1. (i)

$\Phi$ is irreducible.

  2. (ii)

Every component $C$ of $G_{\Phi}^{\Phi^{c}}$ has
$\operatorname{supp}C=\operatorname{supp}\Phi$.

  3. (iii)

$G_{\Phi}^{\Phi^{c}}$ has a unique simple component.

  4. (iv)

There exists a component $A$ of $G_{\Phi}^{\Phi^{c}}$ such that for all other
components $C$ of $G_{\Phi}^{\Phi^{c}}$, $C=kA$ for some $k\geqslant 1$.

###### Proof.

We show that $(i)\implies(ii)$ by contrapositive. If there were a component
$C$ with $\operatorname{supp}C\subsetneq\operatorname{supp}\Phi$, then
$\Phi=\mathrm{inf}_{\operatorname{supp}C}^{S}(\Psi,X)$ by Proposition 5.1.
Because $\operatorname{supp}C\neq\operatorname{supp}\Phi$, $\Psi$ is non-
empty. Since $C\subset X$, $X$ is also non-empty, and thus $\Phi$ is
reducible. That $(ii)\implies(iii)$ follows immediately from Proposition 4.27.
$(iii)\implies(iv)$ because by Proposition 4.28 every component can be written
as the sum of simple components, and by $(iii)$ there is a unique simple
component. To see that $(iv)\implies(i)$, suppose that
$\Phi=\Phi_{1}\sqcup\Phi_{2}$ and, without loss of generality, let
$A\subseteq\Phi_{1}$. By closure of $\Phi_{1}$, every other component also
belongs to $\Phi_{1}$, so $\Phi_{1}=\Phi$, $\Phi_{2}=\emptyset$, and $\Phi$ is
irreducible. ∎

######  Corollary 5.3.

Primitive inversion sets are irreducible.

###  5.2. Proof of the Main Theorem

######  Proposition 5.4.

Let $A,B$ be components of $G_{\Phi}^{\Phi^{c}}$ such that $A+B$ is defined.
If $I\in\mathrm{Gen}(A)\cap\mathrm{Gen}(B)$, then $I\in\mathrm{Gen}(A+B)$.

###### Proof.

If $A+B$ is equal to $A$ or $B$ the result is immediate. Otherwise, for any
$\gamma\in A+B$ such that $\pi_{I}(\gamma)\neq 0$ and $\theta\in I$, we show
that $\gamma\pm\theta\in A+B$ given that $\gamma\pm\theta\in R$. To that end,
we apply Lemma 4.4 (ii) and write $\gamma=\alpha+\beta$ for $\alpha\in
A,\beta\in B$. Then $\gamma\pm\theta=\alpha+\beta\pm\theta$. Using Lemma 2.4,
we conclude that either $\beta\pm\theta$ is a root or $\alpha\pm\theta$ is a
root, provided that $\alpha+\beta\neq 0$, which is true because both
$\alpha,\beta>0$. We can therefore take $\alpha\pm\theta\in R$ without loss of
generality. Note that $\pi_{I}(\alpha)\neq 0$, since otherwise $\beta$ and
$\alpha+\beta$ would be in the same fibre, and since $I\in\mathrm{Gen}(B)$, we
would have $A+B=B$. This means that $\alpha\pm\theta\in A$, since
$I\in\mathrm{Gen}(A)$, and therefore $(\alpha\pm\theta)+\beta\in A+B$. We thus
conclude that $I\in\mathrm{Gen}(A+B)$. ∎

######  Proposition 5.5.

Let $\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$ expressed canonically with $\Psi$
primitive. There is a unique simple component of full support in
$G_{\Phi}^{\Phi^{c}}$, while all other simple components are contained in $X$.

###### Proof.

Pick a simple root $\theta\in\mathrm{inf}_{I}^{S}(\Psi,\emptyset)$ and let $C$
be the component containing $\theta$ in $G_{\Phi}^{\Phi^{c}}$. By Proposition
5.1, $\operatorname{supp}C\in\mathrm{Gen}(\Phi)$, and by Proposition 3.19,
either $\operatorname{supp}C\subseteq I$ or $\operatorname{supp}C=S$. Since
$\theta\in\operatorname{supp}C$ but $\theta\notin I$, we conclude that $C$ is
of full support. Because we chose $\theta$ arbitrarily, it follows that any
component containing a simple root of $\mathrm{inf}_{I}^{S}(\Psi,\emptyset)$
is of full support, and by Proposition 4.27 there is only one such component.
All other simple components therefore must be contained in $X$. ∎

######  Proposition 5.6.

Let $\Phi$ be an inversion set canonically inflated from $I$. Then all
components of $G_{\Phi}^{\Phi^{c}}$ are inflated from $I$.

###### Proof.

We write $\Phi=\mathrm{inf}_{I}^{S}(\Psi,X)$ and split the argument into three
mutually exclusive cases.

  1. (1)

$\Psi=\emptyset$:

Here the result is trivial, since
$\operatorname{supp}C\subseteq\operatorname{supp}\Phi=I$ for every component
$C$ of $G_{\Phi}^{\Phi^{c}}$, so that $I\in\mathrm{Gen}(C)$.

  2. (2)

$\Psi=(R/I)^{+}$:

Since $I$ is canonical, $\operatorname{supp}\Phi^{c}=I$. This implies that
each component of $G_{\Phi}^{\Phi^{c}}$ is contained in a single fibre of
$\pi_{I}$. To show that each is inflated from $I$ we need only to show that
those components that are contained in a fibre over a non-zero root of $R/I$
are exactly the fibre that contains them. This is a consequence of
Propositions 2.14 and 2.7, that state respectively that roots in $\Phi^{c}$
span the difference between any two roots $\alpha,\beta$ in the same fibre,
and that roots of $\Phi^{c}$ can therefore be used to construct a path between
$\alpha$ and $\beta$. This path stays in the same fibre and therefore in
$\Phi$, so that components of $G_{\Phi}^{\Phi^{c}}$ are either fibres of
$\pi_{I}$ over non-zero elements or contained in the zero fibre. In both
cases, they are inflated from $I$.

  3. (3)

$\Psi$ is primitive:

Let $C$ be the unique simple component of full support defined in Proposition
5.5. Since all other simple components are contained in $X$, we label them
$X_{1},X_{2},\ldots,X_{n}$. Each $X_{i}$ contains a simple root $\theta_{i}$
such that $C$ is supported at $\theta_{i}$. It follows that $X_{i}\leqslant C$
for all $i$, and so $X_{i}+A_{1}+\ldots+A_{n}=C$ for some components
$A_{1},A_{2},\ldots A_{n}$. If all $A_{j}\subseteq X$, then
$\operatorname{supp}C\subseteq I$, which we know is not the case. Thus, there
is some $A_{k}$ that is not contained in $X$. It is clear that $A_{k}\leqslant
C$, but because $C$ is minimal among components not contained in $X$, we must
have that $A_{k}=C$. We consider now the sum $X_{i}+A_{1}+\ldots+A_{k-1}+C$.
Once again, it is clear that $C\leqslant X_{i}+A_{1}+\ldots+A_{k-1}+C$, but
because $(X_{i}+A_{1}+\ldots+A_{k-1}+C)+A_{k+1}+\ldots+A_{n}=C$, we also have
that $X_{i}+A_{1}+\ldots+A_{k-1}+C\leqslant C$, for the conclusion that
$X_{i}+A_{1}+\ldots+A_{k-1}+C=C$.

We are given $X_{i}+A_{1}+\cdots+A_{k-1}+C=C$ where $X_{i}\subseteq X$ for all
$i$. We may assume that for each $j>1$, $X_{i}+A_{1}+\cdots+A_{j-1}\neq
X_{i}+A_{1}+\cdots+A_{j}$. It follows from Lemma 4.4 that there exist
$\kappa_{i}\in\mathrm{A}_{i}$ and $\gamma\in C$ such that we have a path
$[\alpha;\kappa_{1},\ldots,\kappa_{k-1},\gamma;\beta]$ where $\alpha\in X_{i}$
and $\beta\in C$. We proceed by induction on $k$. If $k=2$, then
$X_{i}+A_{1}=C$ forces $A_{1}$ to have support outside of $I$. Hence
$A_{1}=tC$ for some $t$. If $A_{1}\neq C$, this would yield $tC=A_{1}<C$, a
contradiction. Hence, $X_{i}+C=C$. Now, assume $k\geqslant 3$. In particular,
we have

$(\alpha+\kappa_{1}+\cdots+\kappa_{k-2})+\kappa_{k-1}+\gamma=\beta$

where all three terms on the left-hand side are roots. It follows that either
$\kappa_{k-1}+\gamma$ is a root or that
$(\alpha+\kappa_{1}+\cdots+\kappa_{k-2})+\gamma$ is a root. In the first case,
as $\kappa_{k-1}+\gamma$ belongs to a component larger than or equal to $C$,
hence to $tC$ for some $t\geqslant 1$. Given that adding
$\alpha+\kappa_{1}+\cdots+\kappa_{k-2}$ gives a root in $C$, this yields
$tC\leq C$, so that $tC=C$. Hence, $\kappa_{k-1}+\gamma=\gamma^{\prime}$ for
some $\gamma^{\prime}\in C$. We proceed by induction. Assume the second case,
that is, that $(\alpha+\kappa_{1}+\cdots+\kappa_{k-2})+\gamma$ is a root.
Again, this root has to be in a component of the form $tC$ for some
$t\geqslant 1$, but we get $tC\leq C$ showing that $tC=C$. Hence, we can use
induction as well.

With this knowledge, we show that $I\in\mathrm{Gen}(C)$. To that end, let
$\gamma\in C$ and $\theta\in\pm I$ such that $\gamma+\theta\in R$. First, we
note that $C\subseteq\mathrm{inf}_{I}^{S}(\Psi,\emptyset)\subseteq\Phi$, so
that $\gamma+\theta\in\Phi$. If $\theta\in\pm\Phi^{c}$, then $\gamma+\theta$
is adjacent to $\gamma$ in $G_{\Phi}^{\Phi^{c}}$, so $\gamma+\theta\in C$. If
$\theta\in\Phi$, then the component $X_{\theta}$ of $G_{\Phi}^{\Phi^{c}}$
containing $\theta$ is simple and contained in $X$, so that $\gamma+\theta\in
X_{\theta}+C=C$. Otherwise, if $\theta\in-\Phi$,
$\gamma+\theta\leqslant\gamma$ and so the component containing it is lesser or
equal to $C$. Since $C$ is minimal among components not contained in $X$ and
$\pi_{I}(\gamma+\theta)=\pi_{I}(\gamma)\neq 0$, $\gamma+\theta\in C$. We
therefore find that $I\in\mathrm{Gen}(C)$. Note that this means that all
simple components are inflated from $I$, since $I\in\mathrm{Gen}(C)$ and all
other simple components are contained in $X\subseteq R_{I}^{+}$.

Since all components are sums of simple components by Proposition 4.28 and
sums of inflations from $I$ are inflations from $I$ by Proposition 5.4, every
component is inflated from $I$.

Since the three preceding cases cover all possibilities for $\Psi$ in a
canonical expression, the proof is complete. ∎

######  Corollary 5.7.

Suppose $\mathrm{inf}_{I}^{S}(\Psi,X)=\Phi_{1}\sqcup\Phi_{2}$ with $I$
canonical. Then $I\in\mathrm{Gen}(\Phi_{1})\cap\mathrm{Gen}(\Phi_{2})$.

######  Remark 5.8.

Note that if $R$ is a QRS with connected components
$R_{1},R_{2},\ldots,R_{l}$, every decomposition of $R^{+}$ into inversion sets
yields a decomposition of each $R_{i}$ into inversion sets and vice-versa. In
particular, Theorem 5.9 provides a recursive description of all decompositions
of a QRS into inversion sets.

######  Theorem 5.9.

Let $R$ be connected and
$R^{+}=\Phi_{1}\sqcup\Phi_{2}\sqcup\cdots\sqcup\Phi_{n}$ for inversion sets
$\Phi_{1},\Phi_{2},\ldots,\Phi_{n}$ where $\Phi_{1}$ contains the highest root
of $R$. If $\Phi_{1}=\mathrm{inf}_{I}^{S}(\Psi,X_{1})$ when expressed
canonically, then up to a relabelling,
$\Phi_{2}=\mathrm{inf}_{I}^{S}(\Psi^{c},X_{2})$ and
$\Phi_{i}=\mathrm{inf}_{I}^{S}(\emptyset,X_{i})$ for $i>2$.

###### Proof.

We note that $\Psi\neq\emptyset$, since in that case
$\operatorname{supp}\Phi_{1}=I\subsetneq S$, and $\Phi_{1}$ would contain no
roots of full support. We therefore split our argument into the two remaining
cases:

  1. (i)

$\Psi=(R/I)^{+}$:

In this case the result is clear, since
$\Phi_{i}\subseteq\Phi_{1}^{c}=\mathrm{inf}_{I}^{S}(\emptyset,X_{1}^{c})$ for
each $i\geqslant 2$.

  2. (ii)

$\Psi$ is primitive:

Note that
$\mathrm{inf}_{I}^{S}(\Psi^{c},X_{1}^{c})=\Phi_{1}^{c}=\Phi_{2}\sqcup\Phi_{3}\sqcup\ldots\sqcup\Phi_{n}$,
so by Corollary 5.7, $I\in\mathrm{Gen}(\Phi_{i})$ for each $i\geqslant 2$.
Writing $\Phi_{i}=\mathrm{inf}_{I}^{S}(\Xi_{i},X_{i})$, we find that
$\Psi^{c}=\Xi_{2}\sqcup\Xi_{3}\sqcup\ldots\sqcup\Xi_{n}$. Since $\Psi^{c}$ is
primitive and therefore irreducible by Corollary 5.1, all but one $\Xi_{i}$ is
empty. Labelling the non-empty set $\Xi_{2}$, we find that
$\Phi_{2}=\mathrm{inf}_{I}^{S}(\Psi^{c},X_{2})$ and
$\Phi_{i}=\mathrm{inf}_{I}^{S}(\emptyset,X_{i})$ for $i>2$.

∎

##  6\. Fine decompositions of root systems

Recall that a decomposition
$R^{+}=\Phi_{1}\sqcup\Phi_{2}\sqcup\cdots\sqcup\Phi_{n}$ of the positive roots
of a QRS $R$ is said to be fine if $n={\mathrm{rk\,}}R$ and each inversion set
is non-empty.

The aim of this section is to use Theorem 5.9 to count the number of fine
decompositions of the classical root systems. The paper [USRA], derived
formulae for the number of fine decompositions of $\mathbb{A}_{n}$ and
$\mathbb{B}_{n}$ but not for $\mathbb{D}_{n}$. With the hindsight of the more
natural definition of inflation we see that part of the difficulty arises from
the fact that there are multiple types of rank 2 quotients of
$\mathbb{D}_{n}$, which have different numbers of primitive inversion sets. As
we will see, Theorem 5.9 allows us to treat each root system with a uniform
approach. We begin by introducing a few relevant propositions.

######  Proposition 6.1.

Let $R^{+}=\Phi_{1}\sqcup\Phi_{2}\sqcup\cdots\sqcup\Phi_{n}$ be a fine
decomposition of a QRS $R$ with base $S$. Then:

  1. (i)

For each $1\leqslant i\leqslant n$, $|\Phi_{i}\cap S|=1$.

  2. (ii)

There exists an $I\subseteq S$ such that for all $1\leqslant i\leqslant n$,
$\Phi_{i}=\mathrm{inf}_{I}^{S}(\Psi_{i},\emptyset)$ or
$\Phi_{i}=\mathrm{inf}_{I}^{S}(\emptyset,X_{i})$ for some
$\Psi_{i}\subseteq(R/I)^{+}$ or $X_{i}\subseteq R_{I}^{+}$. Further,
${\mathrm{rk\,}}(R/I)=1$ or ${\mathrm{rk\,}}(R/I)=2$.

###### Proof.

$(i)$ By co-closure, $|\Phi_{i}\cap S|\geqslant 1$ for each $i$. The $n$
simple roots of $S$ are divided into $n$ pairwise disjoint inversion sets in
such a way that each inversion set contains at least one simple root. This
forces $|\Phi_{i}\cap S|=1$ for all $i$. $(ii)$ By Theorem 5.9, there exists
some $I\in\bigcap_{1\leqslant i\leqslant n}\mathrm{Gen}(\Phi_{i})$, so we may
write $\Phi_{i}=\mathrm{inf}_{I}^{S}(\Psi_{i},X_{i})$. By $(i)$, however,
either $\Psi_{i}$ or $X_{i}$ does not contain a simple root and is thus empty.
Theorem 5.9 also implies that $\Psi_{1}\sqcup\Psi_{2}=(R/I)^{+}$, but because
both $\Psi_{1}$ and $\Psi_{2}$ contain at most one simple root,
${\mathrm{rk\,}}(R/I)\leqslant 2$. ∎

######  Remark 6.2.

The converse to (i) also holds, since if each inversion set in a decomposition
contains exactly 1 simple root, there must be $n={\mathrm{rk\,}}R$ inversion
sets in the decomposition.

Proposition 6.1 allows us to count fine decompositions of a QRS $R$ with rank
$n$ as follows. We first take $I\subseteq S$ where $|I|=n-1$ or $|I|=n-2$. If
$|I|=n-1$, we let $\Phi_{1}=\mathrm{inf}_{I}^{S}((R/I)^{+},\emptyset)$ and
$\Phi_{2},\Phi_{3},\ldots,\Phi_{n}$ be a fine decomposition of $R_{I}$. If
$|I|=n-2$, we choose a primitive inversion set $\Psi\subset(R/I)^{+}$ and let
$\Phi_{1}=\mathrm{inf}_{I}^{S}(\Psi,\emptyset)$ and
$\Phi_{2}=\mathrm{inf}_{I}^{S}(\Psi^{c},\emptyset)$. Once again we let
$\Phi_{3},\Phi_{4},\ldots,\Phi_{n}$ be a fine decomposition of $R_{I}$. It is
clear that in both cases $\Phi_{1},\Phi_{2},\ldots,\Phi_{n}$ is a fine
decomposition of $R$, and by Proposition 6.1, this characterizes all fine
decompositions. Introducing some notation, we let
$S=\\{\theta_{1},\theta_{2},\ldots,\theta_{n}\\}$ be a base of the QRS $R$ and
let $F(R)$ denote the number of fine decompositions of $R$. Then

(2)
$F(R)=\sum_{i=1}^{n}F(R_{\\{\theta_{i}\\}^{c}})+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\Pi(R/\\{\theta_{i},\theta_{j}\\}^{c})F(R_{\\{\theta_{i},\theta_{j}\\}^{c}})$

where $\Pi(R):=\dfrac{1}{2}\left|\\{\Psi\subset R^{+}\mid\Psi\text{ is a
primitive inversion set}\\}\right|$. We introduce one more proposition before
counting the fine decompositions of the classical root systems.

Recall that if $R_{1},R_{2}$ are QRSs with ambient Euclidean spaces
$E_{1},E_{2}$, then we have a QRS $R_{1}\times R_{2}$ with ambient Euclidean
space $\mathrm{E}_{1}\oplus E_{2}$. Its roots are identified with $R_{1}\cup
R_{2}$. It is realized as a quotient of the root system
$\Delta_{1}\times\Delta_{2}$ where $R_{i}$ is a quotient of $\Delta_{i}$ for
$i=1,2$.

In the following, we write $\mathbb{A}_{0}=\mathbb{B}_{0}=\mathbb{D}_{0}$ to
denote the rank zero root system with trivial ambient Euclidean space. We also
use the convention that $\mathbb{D}_{1}=\mathbb{B}_{1}=\mathbb{A}_{1}$, and
$\mathbb{D}_{2}=\mathbb{A}_{1}\times\mathbb{A}_{1}$ and
$\mathbb{D}_{3}=\mathbb{A}_{3}$. The proof of the following follows from
Remark 5.8.

######  Proposition 6.3.

Suppose $R_{1}$ and $R_{2}$ are QRSs. Then $F(R_{1}\times
R_{2})=F(R_{1})F(R_{2})$.

###  6.1. Type $\mathbb{A}_{n}$.

We take some $n\geqslant 1$ and consider the fine decompositions of the root
system $\mathbb{A}_{n}$. To do this, we first consider the quotients and
subsystems of $\mathbb{A}_{n}$. It is not hard to see that all quotients of
$\mathbb{A}_{n}$ are also of type $\mathbb{A}_{k}$ for some $k\leqslant n$.
Moreover, if $I=\\{\theta_{i_{1}},\theta_{i_{2}},\ldots,\theta_{i_{m}}\\}$ are
simple roots for some $1\leqslant i_{1}<i_{2}<\ldots<i_{m}\leqslant n$, then
$(\mathbb{A}_{n})_{I}=\mathbb{A}_{i_{1}-1}\times\mathbb{A}_{i_{2}-i_{1}-1}\times\cdots\times\mathbb{A}_{n-i_{m}}$.
Defining $a_{n}:=F(\mathbb{A}_{n})$ to be the number of fine decompositions of
$\mathbb{A}_{n}$, we apply Equation (2) to find that

$a_{n}=\sum_{i=1}^{n}a_{i-1}a_{n-i}+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\Pi(\mathbb{A}_{2})a_{i-1}a_{j-i-1}a_{n-j}$

It is quickly verified that there are no primitive inversion sets in
$\mathbb{A}_{2}$, so $\Pi(\mathbb{A}_{2})=0$. Thus,

$a_{n}=\sum_{i=1}^{n}a_{i-1}a_{n-i}$

along with the fact that $a_{0}=1$, this recurrence relation implies that
$a_{n}=\binom{2n}{n}/(n+1)$, the $n$-th Catalan number.

###  6.2. Type $\mathbb{B}_{n}$.

We proceed similarly for $\mathbb{B}_{n}$. Unlike type $\mathbb{A}$, the
quotient of root systems of type $\mathbb{B}$ are not necessarily of type
$\mathbb{B}$. However, the primitive elements in a quotient of a root system
of type $\mathbb{B}$ form a root system of type $\mathbb{B}$ (see [DR]). Thus
Remark 1.4 implies that counting the inversion sets (and the respective
decompositions) in a quotient of a root system of type $\mathbb{B}$ is
equivalent to counting the inversion sets (and the respective decompositions)
in the corresponding root system of type $\mathbb{B}$.

Taking $n\geqslant 1$, if
$I=\\{\theta_{i_{1}},\theta_{i_{2}},\ldots,\theta_{i_{m}}\\}$ are simple roots
for some $1\leqslant i_{1}<i_{2}<\ldots<i_{m}\leqslant n$, then
$(\mathbb{B}_{n})_{I}\cong\mathbb{A}_{i_{1}-1}\times\mathbb{A}_{i_{2}-i_{1}-1}\times\cdots\times\mathbb{B}_{n-i_{m}}$.
Unlike in $\mathbb{A}_{2}$, there are two primitive inversion sets in
$\mathbb{B}_{2}$, so $\Pi(\mathbb{B}_{2})=1$. Once again applying Equation
(2), we define $b_{n}:=F(\mathbb{B}_{n})$ to be the number of fine
decompositions of $\mathbb{B}_{n}$ and find

$\displaystyle b_{n}$
$\displaystyle=\sum_{i=1}^{n}a_{i-1}b_{n-i}+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}a_{i-1}a_{j-i-1}b_{n-j}$
$\displaystyle=\sum_{i=1}^{n}a_{i-1}b_{n-i}+\sum_{\begin{subarray}{c}k+\ell+m=n-2\\\
k,\ell,m\geqslant 0\end{subarray}}a_{k}a_{\ell}b_{m}$
$\displaystyle=\sum_{i=1}^{n}a_{i-1}b_{n-i}+\sum_{i=2}^{n}\sum_{j=1}^{i-1}a_{i-j-1}a_{j-1}b_{n-i}$
$\displaystyle=\sum_{i=1}^{n}a_{i-1}b_{n-i}+\sum_{i=2}^{n}b_{n-i}\sum_{j=1}^{i-1}a_{i-j-1}a_{j-1}$
$\displaystyle=\sum_{i=1}^{n}a_{i-1}b_{n-i}+\sum_{i=2}^{n}b_{n-i}a_{i-1}$
$\displaystyle=b_{n-1}+2\sum_{i=2}^{n}a_{i-1}b_{n-i}$

where the second-to-last step follows from the recurrence relation
$a_{i-1}=\sum_{j=1}^{i-1}a_{i-j-1}a_{j-1}$ from previous subsection. The first
few values of the sequence $(b_{n})_{n\geqslant 1}$ are as follows:
$1,3,9,29,97,333,1165$.

###  6.3. Type $\mathbb{D}_{n}$.

Let $I:=\\{\theta_{i},\theta_{j}\\}^{c}$, where $i<j$. Then for
$i,j\in\\{1,n-1,n\\}$, the QRS is of type $\mathbb{A}_{2}$, which has no
primitive inversion sets. Whereas in the other cases, the quotient has the
same number of primitives as $\mathbb{B}_{2}$. The remainder of the simple
roots fall into a root system isomorphic to

$\mathbb{D}_{n}^{(i,j)}:=\begin{cases}\mathbb{A}_{i-1}\times\mathbb{A}_{j-i-1}\times\mathbb{D}_{n-j}&\textnormal{
if $j<n-1$}\\\ \mathbb{A}_{i-1}\times\mathbb{A}_{n-i-1}&\textnormal{ if
$j=n-1$ or $j=n$}\end{cases}$

If instead $I=\\{\theta_{i}\\}^{c}$, the remainder of the roots fall into a
root system isomorphic to

$\mathbb{D}_{n}^{(i)}:=\begin{cases}\mathbb{A}_{i-1}\times\mathbb{D}_{n-i}&\textnormal{
if $i<n-1$}\\\ \mathbb{A}_{n-1}&\textnormal{ if $i=n-1$ or $i=n$}\end{cases}$

We now let $d_{n}:=F(\mathbb{D}_{n})$ for $n\geqslant 4$ and
$d_{0}=d_{1}=d_{2}=1$ and $d_{3}=a_{3}=5$. Then for $n\geqslant 3$ we have

$\displaystyle d_{n}=$
$\displaystyle\sum_{i=1}^{n}F(\mathbb{D}_{n}^{(i)})+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\Pi(\mathbb{D}_{n}/\\{\theta_{i},\theta_{j}\\}^{c})F(\mathbb{D}_{n}^{(i,j)})$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{n-2}a_{i-1}d_{n-i}+2a_{n-1}+\sum_{i=1}^{n-3}\sum_{j=i+1}^{n-2}\Pi(\mathbb{D}_{n}/\\{\theta_{i},\theta_{j}\\}^{c})F(\mathbb{D}_{n}^{(i,j)})$
$\displaystyle+\sum_{i=1}^{n-2}\Pi(\mathbb{D}_{n}/\\{\theta_{i},\theta_{n-1}\\}^{c})F(\mathbb{D}_{n}^{(i,n-1)})+\sum_{i=1}^{n-1}\Pi(\mathbb{D}_{n}/\\{\theta_{i},\theta_{n}\\}^{c})F(\mathbb{D}_{n}^{(i,n)})$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{n-2}a_{i-1}d_{n-i}+2a_{n-1}+\sum_{i=1}^{n-3}\sum_{j=i+1}^{n-2}a_{i-1}a_{j-i-1}d_{n-j}$
$\displaystyle+\sum_{i=2}^{n-2}\Pi(\mathbb{D}_{n}/\\{\theta_{i},\theta_{n-1}\\}^{c})F(\mathbb{D}_{n}^{(i,n-1)})+\sum_{i=2}^{n-2}\Pi(\mathbb{D}_{n}/\\{\theta_{i},\theta_{n}\\}^{c})F(\mathbb{D}_{n}^{(i,n)})$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{n-2}a_{i-1}d_{n-i}+2a_{n-1}+\sum_{i=1}^{n-3}\sum_{j=i+1}^{n-2}a_{i-1}a_{j-i-1}d_{n-j}$
$\displaystyle+\sum_{i=2}^{n-2}a_{i-1}a_{n-i-1}+\sum_{i=2}^{n-2}a_{i-1}a_{n-i-1}\
.$

Now

$\displaystyle\sum_{i=1}^{n-3}$
$\displaystyle\sum_{j=i+1}^{n-2}a_{i-1}a_{j-i-1}d_{n-j}=\sum_{j=2}^{n-2}\sum_{i=1}^{j-1}a_{i-1}a_{j-i-1}d_{n-j}$
$\displaystyle=\sum_{j=2}^{n-2}\sum_{i=0}^{j-2}a_{i}a_{j-i-2}d_{n-j}=\sum_{j=0}^{n-4}\sum_{i=0}^{j}a_{i}a_{j-i}d_{n-j-2}$
$\displaystyle=\sum_{j=0}^{n-4}\left(\sum_{i=0}^{j}a_{i}a_{j-i}\right)d_{n-j-2}=\sum_{j=0}^{n-4}a_{j+1}d_{n-j-2}\
.$

Therefore

$\displaystyle d_{n}=$
$\displaystyle\sum_{i=1}^{n-2}a_{i-1}d_{n-i}+2a_{n-1}+\sum_{i=1}^{n-3}\sum_{j=i+1}^{n-2}a_{i-1}a_{j-i-1}d_{n-j}$
$\displaystyle+\sum_{i=2}^{n-2}a_{i-1}a_{n-i-1}+\sum_{i=2}^{n-2}a_{i-1}a_{n-i-1}$
$\displaystyle=$ $\displaystyle
a_{0}d_{n-1}+\sum_{i=2}^{n-2}a_{i-1}d_{n-i}+2a_{n-1}+\sum_{j=0}^{n-4}a_{j+1}d_{n-j-2}+2\sum_{i=2}^{n-2}a_{i-1}a_{n-i-1}$
$\displaystyle=$ $\displaystyle
d_{n-1}+\sum_{i=0}^{n-4}a_{i+1}d_{n-i-2}+2a_{n-1}+\sum_{j=0}^{n-4}a_{j+1}d_{n-j-2}+2\sum_{i=2}^{n-2}a_{i-1}a_{n-i-1}$
$\displaystyle=$ $\displaystyle
d_{n-1}+2\sum_{i=0}^{n-4}a_{i+1}d_{n-i-2}+2a_{n-1}+2\sum_{i=1}^{n-3}a_{i}a_{n-i-2}$
$\displaystyle=$ $\displaystyle
d_{n-1}-2a_{n-2}d_{1}-2a_{n-1}d_{0}+2\sum_{i=1}^{n-1}a_{i}d_{n-i-1}+2a_{n-1}$
$\displaystyle-2a_{0}a_{n-2}-2a_{n-2}a_{0}+2\sum_{i=0}^{n-2}a_{i}a_{n-i-2}$
$\displaystyle=$ $\displaystyle
d_{n-1}+2\sum_{i=0}^{n-2}a_{i+1}d_{n-i-2}-6a_{n-2}+2\sum_{i=0}^{n-2}a_{i}a_{n-i-2}$
$\displaystyle=$ $\displaystyle
d_{n-1}+2\sum_{i=0}^{n-2}a_{i+1}d_{n-i-2}+2a_{n-1}-6a_{n-2}\ .$

The first few values of the sequence $(d_{n})_{n\geqslant 1}$ are: 1, 1, 5,