# Decomposing inversion sets of permutations and applications to faces of the Littlewood–Richardson cone

## Abstract

If \(\alpha \in S_n\) is a permutation of \(\{1, 2,\ldots ,n\}\), the inversion set of \(\alpha \) is \(\Phi (\alpha ) = \{ (i, j) \,| \, 1 \leqslant i < j \leqslant n, \alpha (i) > \alpha (j)\}\). We describe all *r*-tuples \(\alpha _1, \alpha _2, \ldots , \alpha _r \in S_n\) such that \(\Delta _n^+ = \{ (i, j) \, | \, 1 \leqslant i < j \leqslant n\}\) is the disjoint union of \(\Phi (\alpha _1), \Phi (\alpha _2), \ldots , \Phi (\alpha _r)\). Using this description, we prove that certain faces of the Littlewood–Richardson cone are simplicial and provide an algorithm for writing down their sets of generating rays. We also discuss analogous problems for the Weyl groups of root systems of types *B*, *C* and *D* providing solutions for types *B* and *C*. Finally, we provide some enumerative results and introduce a useful tool for visualizing inversion sets.

### Keywords

Inversion set Simple permutation Littlewood–Richardson cone Catalan numbers### Mathematics Subject Classification

05E15 05A05 05E10 52B20## 1 Introduction

*n*, we set

*positive roots*, the element (1,

*n*) is the

*highest root*, and the elements \((i,i+1)\) with \(1\leqslant i \leqslant n-1\) are the

*simple roots*.

*the inversion set of*\(\alpha \), \(\Phi (\alpha )\), by

*n*! subsets of \(\Delta _n^{+}\) which are inversion sets.

Throughout the paper we use the following notational conventions. We use the symbol \(\sqcup \) to denote a disjoint union. Often we will write \({ I}\), Open image in new window, \(\Delta ^+\), etc. instead of \({ I}_n\), Open image in new window, \(\Delta ^+_n\), etc. when the value of *n* is clear from the context.

### Definition 1.1

*decomposition*of an inversion set \({\Phi ({\alpha })}\) is a set of disjoint inversion sets \({\Phi ({\alpha _1})},{\Phi ({\alpha _2})},\dots ,{\Phi ({\alpha _r})}\) such that

*trivial*if \({\Phi ({\alpha })}={\Phi ({\alpha _a})}\) for some

*a*with \(1 \leqslant a \leqslant r\), and hence that \(\alpha _i={ I}\) for all \(i\ne a\).

We say that an element \(\alpha \in S_n\) (and its inversion set \({\Phi ({\alpha })}\)) is *reducible* if there exists a non-trivial decomposition of \({\Phi ({\alpha })}\). Otherwise we say that \(\alpha \) (and \({\Phi ({\alpha })}\)) is *irreducible*.^{1} We call a decomposition as above an *irreducible decomposition* if each \({\Phi ({\alpha _i})}\) is irreducible.

Solving the following problem was the motivation for this article.

### Problem 1.2

- (i)
determining the regular codimension

*n*faces of the Littlewood–Richardson cone; - (ii)
studying the cup product of the cohomology of line bundles on homogeneous varieties.

*The Littlewood–Richardson cone*If

*A*is a Hermitian matrix, denote by \(\lambda = (\lambda _1 \geqslant \lambda _2 \geqslant \ldots \geqslant \lambda _n) \in {\mathbb R}^n\) its eigenvalues and let \({\mathbb R}^{3n}_+ = \{(\lambda , \mu , \nu ) \, | \, \lambda _i \geqslant \lambda _{i+1}, \mu _i \geqslant \mu _{i+1}, \nu _i \geqslant \nu _{i+1} { \text{ for } } 1 \leqslant i \leqslant n-1\}\). In 1912 H. Weyl posed the following question: For which triples \((\lambda , \mu , \nu ) \in {\mathbb R}^{3n}_+\) do there exist Hermitian matrices

*A*,

*B*,

*C*such that \(C = A+ B\) and whose eigenvalues are \(\lambda , \mu ,\nu \) respectively. In 1962 A. Horn proved that the set of such triples is a polyhedral cone \(\mathcal {C'}\) and conjectured inequalities determining \(\mathcal {C'}\). Horn’s conjecture was proved in the 1990’s by Klyachko and Knutson and Tao, see [6] for a nice exposition on Horn’s conjecture. It is worth mentioning that the lattice points of \(\mathcal {C'}\) are exactly the triples \((\lambda , \mu , \nu )\) for which the corresponding Littlewood–Richardson coefficient \(c_{\lambda , \mu }^\nu \) is nonzero. It is often convenient to study the cone \(\mathcal {C''}\) corresponding to the relation \(A+B+C = 0\) instead of \(\mathcal {C'}\) corresponding to \(C = A+B\) thus symmetrizing the roles of \(\lambda , \mu \), and \(\nu \). Ressayre [11] described all

*regular faces of*\(\mathcal {C''}\), i.e., faces that intersect the interior of \({\mathbb R}^{3n}_+\). The regular faces of codimension

*n*are in a bijection with triples \(\alpha _1, \alpha _2, \alpha _3\) of elements of \(S_n\) with the property that \(\Delta _n^+ = {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup {\Phi ({\alpha _3})}\): each regular face of codimension

*n*is the intersection of \({\mathbb R}^{3n}_+\) with the subspace of codimension

*n*defined by

*Cup products of line bundles on homogeneous varieties*Let \(G = GL_n({\mathbb C})\), let \(B \subset G\) be a Borel subgroup, and let \(X = G/B\). The Picard group of

*G*-equivariant line bundles on

*X*is isomorphic to \({\mathbb Z}^n\). We denote by \(\mathcal {L}_\lambda \) the line bundle on

*X*which corresponds to the

*B*-character \(-\lambda \). We call \(\lambda \in {\mathbb Z}^n\)

*dominant*if \(\lambda _1 \geqslant \lambda _2 \geqslant \ldots \geqslant \lambda _n\) and

*strictly dominant*if \(\lambda _1> \lambda _2> \ldots > \lambda _n\). Let \(\rho = (n-1, n-2, \ldots , 0) \in {\mathbb Z}^n\). We call \(\lambda \in {\mathbb Z}^n\)

*regular*if there exists \(\alpha \in S_n\) such that \(\alpha \cdot \lambda := \alpha (\lambda + \rho ) - \rho \) is dominant. Such an element \(\alpha \) is uniquely determined by \(\lambda \) and we denote it by \(\alpha _\lambda \). The celebrated Borel-Weil-Bott theorem calculates the cohomology groups \(H^q(X, \mathcal {L}_\lambda )\). In particular, it states that \(H^q(X, \mathcal {L}_\lambda )\) is zero unless \(\lambda \) is regular and

*q*equals the length of \(\alpha _\lambda \). In this case, \(H^q(X, \mathcal {L}_\lambda ) \cong V(\alpha _\lambda \cdot \lambda )^*\), where for any dominant weight \(\mu \), \(V(\mu )\) denotes the irreducible

*G*-module with highest weight \(\mu \). In [5] the following question was studied: For which pairs \(\lambda , \mu \in {\mathbb Z}^n\) is the cup product map

Both of these motivating problems have versions involving an arbitrary number of factors, (i.e., the sum of *r* matrices, or the cup product of *r* cohomology groups), and their solutions are similarly expressed as decompositions of \(\Delta _{n}\) with \(r+1\) factors. We were thus led to consider Problem 1.2.

1.3. Before we state the main results of the paper, we introduce some concepts and state background results.

### Definition 1.3

An *interval* (of size *t*) is a set of consecutive integers \(\{i,i+1,i+2,\dots ,i+t-1\}\). For a permutation \(\alpha \in S_n\), a *block* (of size *t*) of \(\alpha \) is an interval \(\{i,i+1,i+2,\dots ,i+t-1\}\) of size *t* such that the set \(\{\alpha (i),\alpha (i+1),\dots ,\alpha (i+t-1)\}\) is also an interval (of size *t*). Every permutation in \(S_n\) has *n* blocks of size 1 and a block of size of *n*. If \(\alpha \in S_n\) has no blocks of size *t* for all \(1< t < n\), then we say that \(\alpha \) is *simple*.^{2}

### Example 1.4

The permutation \((9,7,1,5,3,4,6,8,2)\in S_9\) has a block of size 8 corresponding to the interval \(\{2,3,4,5,6,7,8,9\}\) and a block of size 4 corresponding to the interval \(\{4,5,6,7\}\). The permutation \((5,2,6,1,4,7,3)\in S_7\) has no non-trivial blocks and so is simple.

To state our results, we need to introduce an inflation procedure to describe permutations inductively. We describe this procedure heuristically as follows. We consider a permutation on *n* letters as a shuffling of a deck of *n* cards. To shuffle, we first cut the deck into *m* piles of sizes \(z_1,z_2,\dots ,z_m\), respectively. Shuffle each of these piles according to a permutation \(\beta _i \in S_{z_i}\). Finally, reassemble the piles in an order determined by a permutation \(\sigma \in S_m\). The resulting permutation in \(S_n\) is denoted by \(\sigma [\beta _1,\beta _2,\dots ,\beta _m]\) and is called an *inflation of *\(\sigma \)*by*\(\beta _1, \beta _2, \ldots , \beta _m\).

### Example 1.5

*i*-th block to get . Finally, we permute each of the blocks as specified by \(\sigma =(3,1,4,2)\) to get . The resulting permutation sends 1 to the 4th position, 2 to the 5th position, 3 to the 6th position, and so on, finally sending 8 to the 2nd position. In function notation, this is the permutation \(\alpha =(4,5,6,1,7,8,3,2)\).

For a formal characterization of inflation see Sect. 3; for a history of the inflation procedure and a discussion of a number of applications, we refer the reader to the survey article of Brignall [4]. The definition of inflation in Brignall’s article [4, §1.1] is equivalent to the characterization in Sect. 3, and to the “shuffling cards” description above. A graphical description of inflation in terms of inversion sets appears in appendix section “Inflation”.

Note that a permutation \(\alpha \in S_n\) is simple if and only if \(\alpha \) cannot be expressed as an inflation \(\alpha =\sigma [\beta _1,\beta _2,\dots ,\beta _m]\) with \(\sigma \in S_m\) and \(2 \leqslant m \leqslant n-1\).

### Definition 1.6

A permutation \(\alpha \in S_n\) is called *plus-decomposable* if \(\alpha \) may be written in the form \(\alpha = { I}_2[\beta _1,\beta _2]\). Otherwise \(\alpha \) is *plus-indecomposable*. Similarly, \(\alpha \in S_n\) is called *minus-decomposable* if \(\alpha \) may be written in the form Open image in new window. Otherwise \(\alpha \) is *minus-indecomposable*.

We follow [2] in using the terms “plus-indecomposable” and “minus-indecomposable”. These are called “sum indecomposable” and “skew indecomposable”, respectively, in [4]. It is not difficult to verify that \(\alpha \in S_n\) cannot be both plus-decomposable and minus-decomposable. On the other hand, there are permutations which are both plus-indecomposable and minus-indecomposable, e.g. every simple \(\alpha \in S_n\) with \(n>2\).

The following theorem of Albert, Atkinson and Klazar illustrates the importance of simple permutations and the inflation procedure.

### Theorem 1.7

[2, Theorem 1] Let \(n \geqslant 2\). For every permutation \(\alpha \in S_n\) there exists a simple permutation \(\sigma \in S_m\) and permutations \(\beta _1,\beta _2,\dots ,\beta _m\) such that \(\alpha = \sigma [\beta _1,\beta _2,\dots ,\beta _m]\). Moreover if \(\sigma \ne { I}_2\) and Open image in new window then \(\beta _1,\beta _2,\dots ,\beta _m\) and \(\sigma \) are unique. If \(\sigma = { I}_2\), then \(\beta _1,\beta _2\) and \(\sigma \) are unique if we add the additional condition that \(\beta _1\) is plus-indecomposable. Similarly, if Open image in new window then \(\beta _1,\beta _2\) and \(\sigma \) are unique if we add the additional condition that \(\beta _1\) is minus-indecomposable. \(\square \)

For our purposes, we modify the statement of the above theorem as follows.

### Theorem 1.8

Let \(n \geqslant 2\). For every permutation \(\alpha \in S_n\) there exists a permutation \(\sigma \in S_m\) and permutations \(\beta _1,\beta _2,\dots ,\beta _m\) such that \(\alpha = \sigma [\beta _1,\beta _2,\dots ,\beta _m]\) where either \(\sigma \) is simple and \(m \geqslant 4\) or \(\sigma = { I}_m\) or Open image in new window. Furthermore, this expression for \(\alpha \) is unique if we require that *m* be maximal when \(\sigma ={ I}_m\) or Open image in new window, i.e., that each \(\beta _b\) is plus-indecomposable when \(\sigma ={ I}\) and each \(\beta _b\) is minus-indecomposable when Open image in new window. \(\square \)

### Corollary 1.9

In the notation of Theorem 1.8 above \(\sigma = I_m\), \(\sigma =J_m\) or \(\sigma \) is simple and \(m \geqslant 4\), if and only if \(\alpha \) is plus-decomposable, \(\alpha \) is minus-decomposable, or \(\alpha \) is both plus-indecomposable and minus-indecomposable, respectively.

### Definition 1.10

We say that \(\alpha \) is expressed in *simple form* when we write \(\alpha = \sigma [\beta _1,\beta _2,\dots ,\beta _m]\) in the form guaranteed by Theorem 1.8, i.e, when \(\sigma \) is simple with \(m \geqslant 4\) or Open image in new window or \(\sigma = { I}_m\) with *m* maximal.

1.4.

*m*intervals of lengths \(z_1, z_2, \ldots , z_m\) and let \(\sigma _a \in S_m\), and \(\beta _{ab} \in S_{z_b}\) (\(1 \leqslant a \leqslant r\), \( 1 \leqslant b \leqslant m\)) be such that

### Theorem 1.11

*m*intervals of lengths \(z_1, z_2, \ldots , z_m\). Then, up to reordering of \(\alpha _2, \alpha _3, \ldots , \alpha _r\), there exists a unique set of elements \(\sigma _a\in S_m\) and \(\beta _{ab}\in S_{z_b}\) such that \(\alpha _a = \sigma _a[\beta _{a1},\beta _{a2},\dots ,\beta _{am}]\), for \(a=2,\dots ,r\), \(b=1,2,\dots ,m\) and

- (i)$$\begin{aligned}\begin{array}{rcl} \Delta _m^+ &{}=&{} \Phi (\sigma _1) \sqcup \Phi (\sigma _2) \sqcup \ldots \sqcup \Phi (\sigma _r), \\ &{}&{}\\ \Delta ^+_{z_1} &{} = &{} {\Phi ({\beta _{11}})} \sqcup {\Phi ({\beta _{21}})} \sqcup \dots \sqcup {\Phi ({\beta _{r1}})}, \\ \Delta ^+_{z_2} &{} = &{} {\Phi ({\beta _{12}})} \sqcup {\Phi ({\beta _{22}})} \sqcup \dots \sqcup {\Phi ({\beta _{r2}})},\\ &{}\vdots &{}\\ \Delta ^+_{z_m} &{} = &{} {\Phi ({\beta _{1m}})} \sqcup {\Phi ({\beta _{2m}})} \sqcup \dots \sqcup {\Phi ({\beta _{rm}})}; \end{array} \end{aligned}$$
- (ii)
if \(\alpha _1\) is minus-decomposable then Open image in new window and \(\sigma _2=\sigma _3=\dots =\sigma _r={ I}_m\);

- (iii)
if \(\alpha _1\) is minus-indecomposable then \(\sigma _1\) is simple and Open image in new window, and \(\sigma _3=\sigma _4=\dots =\sigma _r={ I}_m\).

Let *q* denote the number of \(\sigma _a\) which are not \({ I}_m\), i.e., \(q := {\left\{ \begin{array}{ll} 1, &{}\text {if }\alpha _1 \text { is minus-decomposable};\\ 2, &{}\text {if }\alpha _1 \text { is minus-indecomposable}. \end{array}\right. }\)

- (i)
each of the decompositions \(\Delta ^+_{z_b} = {\Phi ({\beta _{1b}})} \sqcup {\Phi ({\beta _{2b}})} \sqcup \dots \sqcup {\Phi ({\beta _{rb}})}\) is irreducible;

- (ii)
exactly one of \(\beta _{a1},\beta _{a2},\dots ,\beta _{am}\) is not equal to the identity for \(a=q+1,\dots , r\);

- (iii)
\(\beta _{ab} = { I}_{z_b}\) for \(a=1,\dots , q\) and \(b=1,\dots , m\);

- (iv)
\(m = 2\) if \(\alpha _1\) is minus-decomposable.

### Example 1.12

The recursive form of this theorem allows us to inductively solve many problems concerning decompositions. For example, in Sect. 6 we exploit this recursiveness to obtain a number of results enumerating various solutions to the main problem. In Sect. 8 we use the form to prove a result about the decompositions which yields an algorithm producing all generating rays on a given regular codimension *n* face of the Littlewood–Richardson cone.

1.5. The problem discussed above has a Lie theoretic background and a natural generalization. We recommend the book by Fulton and Harris, [7] as a general Lie Theory reference. Let \(\Delta \) be a root system with corresponding Weyl group \({\mathcal W}\). Fix a splitting \(\Delta = \Delta ^+ \sqcup \Delta ^-\) of \(\Delta \) into positive and negative roots. For \(\alpha \in {\mathcal W}\), the inversion set of \(\alpha \), \({\Phi ({\alpha })}\) is defined by \({\Phi ({\alpha })} := \{ v \in \Delta ^+ \mid \alpha \cdot v \in \Delta ^-\}\). We are concerned with ways to express the positive roots as a disjoint union of inversion sets: \(\Delta ^+ = {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup \dots \sqcup {\Phi ({\alpha _r})}\) where \(\alpha _1,\alpha _2,\dots ,\alpha _r \in {\mathcal W}\).

Problem 1.2, which is solved by Theorem 1.11, is the \(A_{n-1}\)-case of the more general problem for arbitrary root systems. (Both of the motivating problems also have versions for arbitrary root systems, and their solution is again in terms of decompositions of the positive roots into inversion sets, i.e., the more general problem.) It is natural to attempt to solve the general problem for all root systems. Section 5 is devoted to studying the root systems of types *B*, *C* and *D*. We provide a solution for root systems of types *B* and *C*. Root systems of types *B* and *C* have isomorphic Weyl groups and so yield identical answers to our questions; nevertheless, we consider them separately because this gives us two different ways of looking at the same problem. Root systems of type *D* are more complicated and we only provide a brief discussion of the difficulties we encountered when attempting to deal with them. The exceptional root systems are also interesting, but our methods are unlikely to yield any results. The solution of Problem 1.2 for root systems of type \(G_2\) is elementary: all non-trivial decompositions are of the form \({\Phi ({\alpha })}\sqcup {\Phi ({J\alpha })}\), where \(J\in {\mathcal W}(G_2)\) is the longest element of \({\mathcal W}(G_2)\). The root system \(F_4\) is probably easily treated by direct computations (possibly aided by a computer). Root systems of type *E*, especially \(E_8\), may be too complicated to treat even by computer computations.

*B*and

*C*. We then turn to applications of the main theorem. In Sect. 6 we give some enumerative results deduced from the theorems. In Sect. 7 we give an algorithm to decompose a single inversion set. In Sect. 8 we use the main theorem to parameterize regular codimension

*n*faces of the Littlewood–Richardson cone. Finally, in appendix, we describe

*sign diagrams*, a visual method of displaying inversion sets which has proved useful to us in thinking about these problems.

## 2 Preliminaries on inversion sets

- (i)
If \((i,j), (j,k) \in \Phi \) then \((i,k) \in \Phi \) (

*closed condition*). - (ii)
If \((i,j), (j,k) \notin \Phi \) then \((i,k) \notin \Phi \) (

*co-closed condition*).

### Proposition 2.1

[10, Proposition 5.10] A set \(\Phi \subset \Delta _n^{+}\) is an inversion set if and only if \(\Phi \) is both closed and co-closed, i.e., both \(\Phi \) and its complement \(\Delta ^{+}_n\setminus \Phi \) satisfy the closed condition.

The following simple result is often useful.

### Lemma 2.2

Every non-empty inversion set \({\Phi ({\alpha })}\) contains at least one simple root.

### Proof

This follows from the fact that if \(\alpha (i) < \alpha (i+1)\) for all \(i=1,2,\dots ,n-1\) then \(\alpha ={ I}_n\). \(\square \)

### Definition 2.3

The graph of a permutation \(\alpha \in S_n\) is the set of *n* lattice points \(\{(i,\alpha (i))\mid i=1,2,\dots ,n\}\) considered as a subset of \([1,n]\times [1,n] \subset {\mathbb R}^2\).

We have already noted that Open image in new window. The following two lemmas give further indication of the importance of Open image in new window.

### Lemma 2.4

Let \(\alpha \in S_n\). The permutation \(\alpha \) is simple if and only if Open image in new window is simple.

### Proof

A block of size \(t+1\) for the permutation \(\alpha \) corresponds to a \(t \times t\) closed square in \([1,n]\times [1,n]\) which contains \(t+1\) points of the graph of \(\alpha \). Hence \(\alpha \) is simple if there does not exist a \(t\times t\) closed square in \([1,n]\times [1,n]\) containing \(t+1\) points of the graph of \(\alpha \) with \(2\leqslant t \leqslant n-1\). If the graph of \(\alpha \) satisfies this condition, then so does the graph of \(J\alpha \), which is obtained from that of \(\alpha \) by reflecting in the horizontal line \(y=n/2\). Thus \(\alpha \) is simple if and only if \(J\alpha \) is simple. \(\square \)

To each element \((i,j)\in \Delta _n^{+}\) we associate the line segment joining the points \((i,\alpha (i))\) and \((j,\alpha (j))\) on the graph of \(\alpha \). We note that \((i,j)\in {\Phi ({\alpha })}\) if and only if the corresponding line segment has negative slope.

### Lemma 2.5

Let \(\alpha \in S_n\). Then Open image in new window, or equivalently, Open image in new window.

### Proof

The graph of Open image in new window is obtained from the graph of \(\alpha \) by reflecting in the horizontal line \(y=n/2\). Using the characterization of \({\Phi ({\alpha })}\) as those positive roots whose corresponding line segment has negative slope completes the proof of the lemma. \(\square \)

### Corollary 2.6

The element Open image in new window is reducible for \(m\geqslant 3\) and irreducible for \(m=2\).

### Proof

By Lemma 2.5 any Open image in new window gives a non-trivial decomposition Open image in new window, and the set Open image in new window is non-empty if \(m\geqslant 3\). Conversely, Open image in new window is clearly irreducible. \(\square \)

Next we discuss some basic properties of decompositions.

As the following proposition shows, the union of an arbitrary collection of inversion sets appearing in a decomposition is again the inversion set of a permutation.

### Proposition 2.7

Suppose \(\Delta _n^+ = {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup \dots \sqcup {\Phi ({\alpha _r})}\) is a decomposition and let *A* be any subset of \(\{1,2,\dots ,r\}\). Then there exists \(\alpha \in S_n\) such that \({\Phi ({\alpha })}= \sqcup _{a\in A} {\Phi ({\alpha _a})}\).

### Proof

Clearly it suffices to prove the assertion for doubleton sets \(A=\{p,q\}\). Thus it suffices to show that \({\Phi ({\alpha _p})} \sqcup {\Phi ({\alpha _q})}\) is both closed and co-closed. For ease of notation, we will assume \(A = \{1,2\}\). First we show that \({\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})}\) is co-closed. Suppose that \((i,j),(j,k) \notin {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})}\). Then for \(b=1,2\) we have \((i,k) \notin {\Phi ({\alpha _b})}\) since \({\Phi ({\alpha _b})}\) is co-closed. Thus \((i,k) \notin {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})}\) which shows \({\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})}\) is co-closed.

To see that \({\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})}\) is closed, suppose that \((i,j), (j,k) \in {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})}\). Then for \(a=3,4,\dots ,r\) we have \((i,j),(j,k) \notin {\Phi ({\alpha _a})}\) and thus \((i,k) \notin {\Phi ({\alpha _a})}\) since \({\Phi ({\alpha _a})}\) is co-closed. Hence \((i,k) \notin \sqcup _{a=3}^r {\Phi ({\alpha _a})}\) which implies that \((i,k) \in {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})}\). This shows that \({\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})}\) is closed and completes the proof of proposition. \(\square \)

Note that some hypothesis of the type \(\Delta _n^+ = {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup \dots \sqcup {\Phi ({\alpha _r})}\) is necessary in the above proposition; arbitrary unions of inversion sets need not be inversion sets. For example, consider \(n=3\), \(\alpha _1 = (2,1,3)\) and \(\alpha _2 = (1,3,2)\). Then \({\Phi ({\alpha _1})} = \{(1,2)\}\), \({\Phi ({\alpha _2})} = \{(2,3)\}\) and \({\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})}\) is not closed and so is not an inversion set.

### Corollary 2.8

If \({\Phi ({\alpha })}= {\Phi ({\alpha _1})} \,\sqcup \, {\Phi ({\alpha _2})}\, \sqcup \, \cdots \, \sqcup \, {\Phi ({\alpha _r})}\) and *A* is any subset of \(\{1,2,\dots ,r\}\) then there exists \(\alpha _A\in S_n\) with \({\Phi ({\alpha _A})} = \sqcup _{a\in A} {\Phi ({\alpha _a})}\).

### Proof

Set Open image in new window. By Lemma 2.5 we then have the decomposition \(\Delta _n^{+}={\Phi ({\alpha _1})}\sqcup \cdots \sqcup {\Phi ({\alpha _{r+1}})}\). The corollary then follows from Proposition 2.7 applied to this decomposition. \(\square \)

Recall that an element \(\alpha \in S_n\) is called reducible if there exists a non-trivial decomposition \({\Phi ({\alpha })}= {\Phi ({\alpha _1})}\sqcup \cdots \sqcup {\Phi ({\alpha _r})}\). By Corollary 2.8 if \(\alpha \) is reducible there exists such a non-trivial decomposition with \(r=2\).

Given a decomposition \(\Delta _{n}^{+}={\Phi ({\alpha _1})}\sqcup \cdots \sqcup {\Phi ({\alpha _r})}\) we may obtain finer and coarser decompositions as follows. We obtain a coarser decomposition by choosing one or more disjoint subsets \(A_i \subset \{1,2,\dots ,r\}\) and replacing \(\sqcup _{a \in A_i} {\Phi ({\alpha _a})}\) by the single inversion set \({\Phi ({\alpha _{A_i}})}\) where \(\alpha _{A_i}\) is the element whose existence is guaranteed by Corollary 2.8. We obtain a finer decomposition by further decomposing one or more of the \({\Phi ({\alpha _i})}\) into smaller inversion sets.

We may continue refining a decomposition until we arrive at an *irreducible decomposition*, i.e., a decomposition \(\Delta ^+_n = {\Phi ({\gamma _1})} \sqcup {\Phi ({\gamma _2})} \sqcup \dots \sqcup {\Phi ({\gamma _s})}\) where each \({\Phi ({\gamma _i})}\) is irreducible.

Clearly every decomposition of \(\Delta _n^+\) may be obtained by applying the coarsening operation to some irreducible decomposition. For this reason, studying and classifying the irreducible decompositions is of particular interest. Accordingly we characterize the irreducible decompositions in our main theorem, Theorem 1.11, as well as in the analogous theorems for root systems of type B and C.

## 3 Restriction maps and proof of the main theorem

Given a subset \({\mathcal F}\subseteq \{1,2,\ldots , n\}\) and an element \(\alpha \in S_n\) we obtain a permutation in \(S_m\), with \(m=|{\mathcal F}|\), by noting how \(\alpha \) changes the relative order of elements of \({\mathcal F}\). This procedure gives rise to a map of sets \(\theta _{\mathcal F}:S_n\longrightarrow S_m\) called a *restriction map*. Although not homomorphisms, the maps \(\theta _{{\mathcal F}}\) are useful in making inductive arguments on inversion sets. In this section we use restriction maps to establish several results on simple and irreducible permutations, culminating in a proof of the main theorem (Theorem 1.11).

We start by giving formal descriptions of the restriction maps and the inflation procedure.

### Definitions 3.1

- (a)
Two sequences \(x_1,x_2,\dots ,x_m\) and \(y_1,y_2,\dots ,y_m\) each comprised of

*m*distinct real numbers are*order isomorphic*if \(x_i > x_j\) if and only if \(y_i > y_j\). - (b)
Suppose \({{\mathcal F}}\) is a subset of \(\{1,2,\dots ,n\}\) of size

*m*, and write \({\mathcal F}=\{i_1,i_2,\dots ,i_m\}\) where \(i_1< i_2< \dots < i_m\). For any \(\alpha \in S_n\) restricting \(\alpha \) to \({\mathcal F}\) yields a sequence \(\alpha (i_1),\alpha (i_2),\dots ,\alpha (i_m)\) which is order isomorphic to the sequence \(\mu (1),\mu (2),\dots ,\mu (m)\) corresponding to a unique element \(\mu \in S_m\). We denote this element \(\mu \) by \(\mu = \theta _{\mathcal F}(\alpha )\) and use \(\theta _{{\mathcal F}}:S_n\longrightarrow S_m\) for the corresponding map of sets. - (c)
For \({\mathcal F}\subseteq \{1,2,\dots ,n\}\) we write \(\Delta ^+_{\mathcal F}\) to denote the set \(\Delta ^+_{\mathcal F}:= \{(i,j) \in \Delta ^+_n \mid i,j \in {\mathcal F}\}\).

- (d)
A decomposition of \(\{1,\ldots , n\}\) into an

*ordered disjoint union of intervals*is a decomposition \(\{1,2,\dots ,n\} = U_1 \sqcup U_2 \sqcup \dots \sqcup U_m\) where each \(U_i\) is an interval and where for each \(1\leqslant i < j\leqslant m\), we have \(a<b\) for each \(a\in U_i\) and \(b\in U_j\). - (e)
Given a decomposition of \(\{1,\ldots , n\}\) into an ordered disjoint union of intervals as above, a subset \({\mathcal F}\subset \{1,2,\dots ,n\}\) is

*admissible*if \(|{\mathcal F}\cap U_i|=1\) for all \(i=1,2,\dots ,m\).

Note that the condition of being admissible in (*e*) depends on the choice of decomposition into ordered disjoint intervals. In every case we use this term we will be careful to make the choice of decomposition explicit.

*m*. In addition to the description by shuffling cards given in Sect. 1.3, the inflation \(\alpha := \sigma [\beta _1,\beta _2,\dots ,\beta _m] \in S_n\), \(n=\sum |U_i|\), is characterized by the following two conditions:

- (1)
\(\theta _{U_i}(\alpha ) = \beta _i\) for all \(i=1,2,\dots ,m\).

- (2)
\(\theta _{\mathcal F}(\alpha )=\sigma \) for any admissible \({\mathcal F}\).

The following lemma, which computes the inversion set of \(\alpha =\sigma [\beta _1,\ldots , \beta _m]\) from those of its components, follows from either of the descriptions of the inflation procedure. The reader may find an illustration of this lemma and the arguments for its proof in appendix section “Inflation”.

### Lemma 3.2

If a permutation \(\sigma \) is reducible it is clear that any inflation \(\sigma [{ I}_{z_1},\ldots , { I}_{z_m}]\) (for any positive integers \(z_1\),..., \(z_m\)) is also reducible: one simply takes a decomposition of \({\Phi ({\sigma })}\) and inflates the permutations which appear. However, it is not immediately clear that an inflation of an irreducible element remains irreducible; *a priori* it seems that there could be decompositions of the inflation which do not respect the inflation structure, and therefore do not come from decompositions of the original \(\sigma \). That this can never happen is a consequence of the following more precise statement.

### Lemma 3.3

For any \(\sigma \in S_m\), and any positive integers \(z_1\),..., \(z_m\), inflation of the decompositions of \({\Phi ({\sigma })}\) gives a one-to-one correspondence between the decompositions of \({\Phi ({\sigma })}\) and the decompositions of \({\Phi ({\sigma [{ I}_{z_1},\ldots , { I}_{z_m}]})}\).

### Proof

Set \(\alpha =\sigma [{ I}_{z_1},\ldots , { I}_{z_m}]\) and \(n=\sum z_i\). We must show that for any decomposition \({\Phi ({\alpha })}={\Phi ({\alpha _1})}\, \sqcup \, \cdots \, \sqcup \, {\Phi ({\alpha _r})}\) there are unique \(\sigma _1\),..., \(\sigma _r\in S_m\) such that \(\alpha _k=\sigma _k[{ I}_{z_1},\ldots , { I}_{z_m}]\) for \(k=1\),..., *r*. Lemma 3.2 then implies that \({\Phi ({\sigma })}={\Phi ({\sigma _1})}\,\sqcup \, \cdots \, \sqcup \, {\Phi ({\sigma _r})}\).

Let \(\{1,2,\dots ,n\} = U_1\, \sqcup \, U_2\, \sqcup \, \cdots \, \sqcup \, U_m\) be the decomposition into ordered disjoint intervals corresponding to the inflation \(\sigma [{ I}_{z_1},{ I}_{z_2},\dots ,{ I}_{z_m}]\). By Lemma 3.2, we have \({\Phi ({\alpha })} = \{(a,b) \in \Delta _n^+ \mid a \in U_{i}, b \in U_{j}, (i,j) \in {\Phi ({\sigma })}\}\).

Choose any root \((i,j)\in {\Phi ({\sigma })}\). The fact that no root \((a,a')\) with \(a,a'\in U_i\) is in \({\Phi ({\alpha })}\) means that no such root is in \({\Phi ({\alpha _1})}\),..., \({\Phi ({\alpha _r})}\), and so each of \(\alpha _1\),..., \(\alpha _r\) preserves the relative order of the elements in \(U_i\). Similarly each of \(\alpha _1\), ..., \(\alpha _r\) preserves the relative order of elements in \(U_j\).

Let \(a_0\) be the smallest element in \(U_i\) and \(b_1\) the largest element in \(U_j\). The root \((a_0,b_1)\) is in \({\Phi ({\alpha })}\) and so must be contained in one of \({\Phi ({\alpha _1})}\),..., \({\Phi ({\alpha _r})}\). Suppose that \((a_0,b_1)\in {\Phi ({\alpha _k})}\), i.e., that \(\alpha _k(b_1) < \alpha _k(a_0)\). Then the fact that \(\alpha _k\) preserves the relative order of the elements in \(U_i\) and \(U_j\) now implies that \(U_i\times U_j:=\{(a,b)\mid a\in U_i, b\in U_j\} \subseteq {\Phi ({\alpha _k})}\). Since the decomposition of \({\Phi ({\alpha })}\) is into disjoint subsets, we therefore have \((U_i\times U_j)\bigcap {\Phi ({\alpha _{\ell }})}=\emptyset \) if \(\ell \ne k\).

For each \(k=1\),..., *r* set \(T_k=\{(i,j)\in {\Phi ({\sigma })}\mid U_i\times U_j\subseteq {\Phi ({\alpha _k})}\}\). We have just shown that for any \((i,j)\in {\Phi ({\sigma })}\) there is a unique *k* such that \((U_i\times U_j)\bigcap {\Phi ({\alpha _k})}\ne \emptyset \), and for that *k* we have \(U_i\times U_j\subseteq {\Phi ({\alpha _k})}\). From this we conclude first that \({\Phi ({\sigma })}=T_1\sqcup \cdots \sqcup T_r\), and second, since \({\Phi ({\alpha })}=\bigcup _{(i,j)\in {\Phi ({\sigma })}} U_i\times U_j\), that \({\Phi ({\alpha _k})}=\bigcup _{(i,j)\in T_k} U_i\times U_j\) for each *k*.

The fact that \({\Phi ({\alpha _k})}\) is both closed and co-closed implies that the same holds for \(T_k\), and thus there is a unique permutation \(\sigma _k\in S_m\) such that \(T_k={\Phi ({\sigma _k})}\). Lemma 3.2 then says that \({\Phi ({\sigma _k[{ I}_{z_1},\ldots , { I}_{z_k}]})}= \bigcup _{(i,j)\in T_k} U_i\times U_j\). Since the inversion set uniquely determines the permutation, we therefore have \(\alpha _k= \sigma _k[{ I}_{z_1},\ldots , { I}_{z_k}]\) for each \(k=1\),..., *r*. \(\square \)

### Corollary 3.4

Let \(\sigma \in S_m\), and let \(z_1,z_2,\dots ,z_m\) be positive integers. Then the permutation \(\alpha :=\sigma [{ I}_{z_1},{ I}_{z_2},\dots ,{ I}_{z_m}] \in S_n\) is irreducible if and only if \(\sigma \) is irreducible.

### Corollary 3.5

Let \(\alpha =\sigma [\beta _1,\beta _2,\dots ,\beta _m] \ne { I}\) where \(\beta _i \in S_{|U_i|}\) for \(i=1,2,\dots ,m\). Then \(\alpha \) is irreducible if and only if exactly one of the permutations \(\sigma ,\beta _1,\beta _2,\dots ,\beta _m\) is a non-identity permutation and that non-identity permutation is itself irreducible. In particular, if \(\alpha \) is irreducible with \(\sigma \ne { I}\) then \(\alpha =\sigma [{ I},{ I},\dots ,{ I}]\) where \(\sigma \) is irreducible.

### Proof

*m*. If \(\alpha \) is irreducible all then all but one of the inversion sets in the decomposition on the right are empty, and hence all but one of the corresponding elements are the identity. Conversely, if more than one of the inversion sets in this decomposition of \({\Phi ({\alpha })}\) is non-empty, then we have a non-trivial decomposition of \(\alpha \). Furthermore, if \(\sigma \) is reducible then Lemma 3.3 shows that \(\alpha \) is reducible too. Similarly if some \(\beta _i\) is reducible then the order preserving bijection \(\Phi _i\) from Lemma 3.2 induces a decomposition of \({\Phi ({{ I}_m[{ I}_{z_1},\ldots , { I}_{z_{i-1}}, \beta _i,{ I}_{z_{i+1}},\ldots , { I}_{z_m}]})}\), showing again that \(\alpha \) is reducible. \(\square \)

### Definition 3.6

*connected*if

- (i)
\(\theta _{\mathcal F}(\alpha )\) and \(\theta _{{\mathcal F}'}(\alpha )\) are irreducible;

- (ii)
\({\Phi ({\alpha })}\cap \Delta ^+_{\mathcal F}\cap \Delta ^+_{{\mathcal F}'} \ne \emptyset \).

The following two results will be used several times.

### Lemma 3.7

- (1)
If \(\mu \) is irreducible then there exists a unique index \(\delta ({\mathcal F})\) with \(1 \leqslant \delta ({\mathcal F}) \leqslant r\) such that \({\Phi ({\alpha })}\cap \Delta ^+_{\mathcal F}\subseteq {\Phi ({\alpha _{\delta ({\mathcal F})}})}\), and hence \({\Phi ({\alpha })}\cap \Delta ^+_{\mathcal F}\cap {\Phi ({\alpha _{i}})}=\emptyset \) for all \(i\ne \delta ({\mathcal F})\).

- (2)
If \({\mathcal F}\) and \({\mathcal F}'\) are \(\alpha \)-connected then \(\delta ({\mathcal F})=\delta ({\mathcal F}')\).

### Proof

Let \(m=|{\mathcal F}|\) and suppose that \(\mu \) is irreducible. Put \(\mu _a = \theta _{{\mathcal F}}(\alpha _a)\) for \(a=1,2,\dots ,r\). There exists an order preserving bijection \(\Psi : {\mathcal F}\rightarrow \{1,2,\dots ,m\}\). It is easy to see that \((i,j) \in {\Phi ({\alpha })}\cap \Delta ^+_{\mathcal F}\) if and only if \((\Psi (i),\Psi (j)) \in {\Phi ({\theta _{\mathcal F}(\alpha )})}\). Thus \(\Psi \) identifies \({\Phi ({\alpha })}\cap \Delta ^+_{\mathcal F}\) with \({\Phi ({\theta _{\mathcal F}(\alpha )})}\). Intersecting \({\Phi ({\alpha })}= {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup \dots \sqcup {\Phi ({\alpha _r})}\) with \(\Delta ^+_{\mathcal F}\) and using this identification we get \({\Phi ({\mu })}= {\Phi ({\mu _1})} \sqcup {\Phi ({\mu _2})} \sqcup \dots \sqcup {\Phi ({\mu _r})}\). Since \(\mu \) is irreducible, there exists a unique \(\delta ({\mathcal F})\) such that \({\Phi ({\mu })}= {\Phi ({\mu _{\delta ({\mathcal F})}})}\), furthermore \({\Phi ({\mu _i})}=\emptyset \) for all \(i\ne \delta ({\mathcal F})\). Therefore Open image in new window and \({\Phi ({\alpha })}\cap \Delta ^+_{\mathcal F}\cap {\Phi ({\alpha _{i}})}=\emptyset \) for all \(i\ne \delta ({\mathcal F})\).

For the second assertion, recall that \(\mu \) and \(\mu '\) are irreducible by definition. By the above, \({\Phi ({\alpha })} \cap \Delta ^+_{{\mathcal F}} \subseteq {\Phi ({\alpha _{\delta ({\mathcal F})}})}\) and \({\Phi ({\alpha })} \cap \Delta ^+_{{\mathcal F}'} \subseteq {\Phi ({\alpha _{\delta ({\mathcal F}')}})}\). Since \({\mathcal F}\) and \({\mathcal F}'\) are \(\alpha \)-connected there exists \((i,j) \in {\Phi ({\alpha })}\cap \Delta ^+_{\mathcal F}\cap \Delta ^+_{{\mathcal F}'}\). Thus \((i,j) \in {\Phi ({\alpha _{\delta ({\mathcal F})}})} \cap {\Phi ({\alpha _{\delta ({\mathcal F}')}})}\). Hence \(\delta ({\mathcal F})=\delta ({\mathcal F}')\). \(\square \)

### Corollary 3.8

Suppose \({\mathcal F}_1,{\mathcal F}_2,\dots ,{\mathcal F}_s \subset \{1,2,\dots ,n\}\) where \({\mathcal F}_i\) and \({\mathcal F}_{i+1}\) are \(\alpha \)-connected for all \(1\leqslant i\leqslant s-1\). (In particular, \(\theta _{{\mathcal F}_i}(\alpha )\) is irreducible for all \(i=1,2,\dots ,s\).) Assume further that Open image in new window. Then \(\alpha \) is irreducible.

### Proof

Suppose that \({\Phi ({\alpha })} = {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup \dots \sqcup {\Phi ({\alpha _r})}\). By Lemma 3.7, we have \(j = \delta ({\mathcal F}_1) = \delta ({\mathcal F}_2) = \dots = \delta ({\mathcal F}_r)\) with \({\Phi ({\alpha })} \cap \Delta _{{\mathcal F}_i}^+ \subseteq {\Phi ({\alpha _j})}\). Therefore Open image in new window and thus the decomposition \({\Phi ({\alpha })}= {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup \dots \sqcup {\Phi ({\alpha _r})}\) is trivial. \(\square \)

The basic objects for describing a permutation by inflation are the simple permutations, while in describing decompositions the basic objects are the irreducible permutations. In our recursive method of describing decompositions by inflations it is therefore natural to choose the basic object to be those permutations which are both simple and irreducible.

It turns out that simple permutations are automatically irreducible (and thus our basic building blocks are again the simple permutations). To prove this and a related useful fact we need an additional definition. Recall that a permutation \(\alpha \in S_n\) is simple if it has no blocks of length *t* for \(2\leqslant t\leqslant n-1\).

### Definition 3.9

A permutation \(\alpha \in S_n\) is *two-block simple* if \(\alpha (1) \ne 1\), \(\alpha (i+1) \ne \alpha (i) + 1\) for \(1 \leqslant i \leqslant n-1\), and \(\alpha (n) \ne n\).

The name is somewhat inaccurate: the condition that \(\alpha \) has no blocks of length two is that \(\alpha (i+1) \ne \alpha (i) \pm 1\) for all *i*, whereas we are only asking that \(\alpha (i+1) \ne \alpha (i)+1\) for all *i*, and imposing the additional conditions that \(\alpha (1)\ne 1\) and \(\alpha (n)\ne n\). Nonetheless, we continue to use this name since it gives an indication of the defining conditions. In Proposition 3.15 below we will show that the property of being simple is equivalent to the property of being both irreducible and two-block simple. From this equivalence and Lemma 2.4 we will deduce that if \(\alpha \) is simple then \(J\alpha \) is irreducible.

Our method of proving the equivalence is inductive. The base cases of the induction are a particular family of permutations previously appearing in the literature.

### Definition 3.10

*exceptional*if it is one of the following permutations

- (1)
\(\alpha = (2,4,6,\dots ,2m-2,2m,1,3,5,\dots ,2m-3,2m-1)\),

- (2)
\(\alpha =(m+1,1, m+2, 2, m+3, 3,\dots , 2m-1, m-1, 2m, m)\),

- (3)
\(\alpha = (2m-1,2m-3,2m-5,\dots ,3,1,2m,2m-2,2m-4,\dots ,4,2)\),

- (4)
\(\alpha =(m, 2m, m-1, 2m-1,m-2,2m-2,\dots ,2, m+2, 1, m+1)\).

Here are the graphs of the exceptional permutations in the case \(n=10\).

These patterns, which hold for all even *n*, allow for a quick verification of the fact that the exceptional permutations are “simple and do not have simple one-point deletions”. Moreover this property characterizes exceptional permutations (see Definition 3.12 and Theorem 3.13 below).

### Lemma 3.11

Let \(\alpha \in S_n\) be exceptional. Then \(\alpha \) is irreducible and two-block simple.

### Proof

It is easily seen that all of these permutations are two-block simple.

(1) Suppose \(\alpha = (2,4,6,\dots ,2m-2,2m,1,3,5,\dots ,2m-3,2m-1)\). Then \({\Phi ({\alpha })}\) has only one simple root, \((m,m+1)\), and so is irreducible.

For the remaining cases, we proceed by induction using Corollary 3.8 repeatedly.

We now turn to the reduction step and then the inductive proof of Proposition 3.15.

### Definition 3.12

Let \(\alpha \in S_n\). Choose *k* with \(1 \leqslant k \leqslant n\) and put \({\mathcal F}= \{1,2,\dots ,n\} \setminus \{k\}\). The permutation \(\alpha ^\circ =\theta _{\mathcal F}(\alpha ) \in S_{n-1}\) is called a *one point deletion* of \(\alpha \). We say that \(\alpha ^\circ \) is obtained from \(\alpha \) by deleting \((k,\alpha (k))\).

The following theorem, expressed in the language of posets, was first proved by Schmerl and Trotter [13]. The version below in terms of permutations appears as [1, Theorem 5].

### Theorem 3.13

Let \(n \geqslant 2\) and suppose \(\alpha \in S_n\) is simple but not exceptional. Then \(\alpha \) has a one-point deletion \(\alpha ^\circ \) which is simple.

### Lemma 3.14

Suppose that \(\alpha \in S_n\) is reducible and has a one-point deletion which is irreducible. Then \(\alpha \) is not simple.

### Proof

Let \({\Phi ({\alpha })}= {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})}\) be a non-trivial decomposition, and let \(\alpha ^\circ \) be an irreducible one-point deletion obtained from \(\alpha \) by deleting \((k,\alpha (k))\). Since \(\alpha ^\circ \) is irreducible, applying Lemma 3.7(1) with \({\mathcal F}=\{1,\ldots , n\}\setminus \{k\}\) gives that either \({\Phi ({\alpha })}\cap \Delta _{{\mathcal F}}^{+}\subset {\Phi ({\alpha _1})}\) or \({\Phi ({\alpha })}\cap \Delta _{{\mathcal F}}^{+}\subset {\Phi ({\alpha _2})}\) . By relabeling we may assume that \({\Phi ({\alpha })}\cap \Delta _{{\mathcal F}}^{+}\subset {\Phi ({\alpha _1})}\). Concretely this means that all roots of the form \((i,j)\in {\Phi ({\alpha })}\) with either \(i\ne k\) or \(j\ne k\) are in \({\Phi ({\alpha _1})}\). The remaining roots in \({\Phi ({\alpha })}\), those of the form (*i*, *k*) or (*k*, *j*) may appear in either \({\Phi ({\alpha _1})}\) or \({\Phi ({\alpha _2})}\), and \({\Phi ({\alpha _2})}\) only has roots of this form. Furthermore, since the decomposition is assumed non-trivial, there is at least one root in \({\Phi ({\alpha _2})}\).

Write \(|{\Phi ({\alpha _2})}| = p+q\) where *p* of the elements of \({\Phi ({\alpha _2})}\) are of the form (*i*, *k*) with \(1 \leqslant i < k\) and *q* of the elements of \({\Phi ({\alpha _2})}\) are of the form (*k*, *j*) with \(k < j \leqslant n\). Suppose both *p* and *q* are nonzero. Then there exist \(i < k\) and \(j > k\) with \((i,k), (k,j) \in {\Phi ({\alpha _2})}\). Since \({\Phi ({\alpha _2})}\) is closed this means that the root \((i,j)\in {\Phi ({\alpha _2})}\) contrary to the description above. Thus only one of *p* and *q* is nonzero.

Assume first that \(p\ne 0\) and \(q=0\). By the form of the roots in \({\Phi ({\alpha _2})}\) the only simple root in \({\Phi ({\alpha _2})}\) is \((k-1,k)\). This implies that \(\alpha _2\) preserves the relative order of the elements in \(\{1,2,\ldots , k-1\}\) and the relative order of the elements in \(\{k,k+1,\ldots , n\}\). Along with the fact that the only roots in \({\Phi ({\alpha _2})}\) are of the form (*i*, *k*), this implies that \({\Phi ({\alpha _2})}= \{ (k-p,k), (k-p+1,k), \dots , (k-1,k) \}\).

Set \(s= \max \{\alpha (k-i) \mid 1\leqslant i \leqslant p\} - \alpha (k)\), and let *R* be the \(p\times s\) rectangle \(R := [k-p,k] \times [\alpha (k),\alpha (k)+s]\). There are \(p+1\) vertical lattice lines and \(s+1\) horizontal lattice lines through *R*. Exactly \(p+1\) points of the graph of \(\alpha \) lie inside the rectangle *R*: \((k-p,\alpha (k-p)), (k-p+1,\alpha (k-p+1)), \dots , (k-1,\alpha (k-1))\) and \((k,\alpha (k))\). Since \(\alpha \) is a permutation (and hence injective), no two points of its graph may lie on the same horizontal line, and thus \(s\geqslant p\). We will now show that \(s\leqslant p\) and hence \(s=p\).

We first claim that there are no points of the graph of \(\alpha \) strictly to the left of *R*, i.e., a point \((x,\alpha (x))\) with \(x < k-p\) and \(\alpha (k)< \alpha (x) < \alpha (k)+s\). Assume to the contrary that \((x, \alpha (x))\) is such a point. A potential graph of such an \(\alpha \) is shown below

(although, as part of the proof will show, certain features of the graph are incorrect). Let \({\mathcal F}= \{x,z,k\}\) with \(z := \alpha ^{-1}(\alpha (k)+s)\). By this choice of *x*, the slope between \((x,\alpha (x))\) and \((k,\alpha (k))\) is negative, and so (*x*, *k*) is a root of \({\Phi ({\alpha })}\). This root is not contained in \({\Phi ({\alpha _2})}\) since if \((i,k)\in {\Phi ({\alpha _2})}\) then \((i,\alpha (i))\) is in *R*, and we have chosen \((x,\alpha (x))\) outside of *R*. Thus \((x,k)\in {\Phi ({\alpha _1})}\), and hence \({\Phi ({\theta _{{\mathcal F}}(\alpha _1)})}\ne \emptyset \). On the other hand, the slope between \((z,\alpha (k)+s)\) and \((k,\alpha (k))\) is also negative, and thus \((z,k)\in {\Phi ({\alpha })}\). Since \((z,\alpha (k))\) is in *R*, this root is in \({\Phi ({\alpha _2})}\), and hence \({\Phi ({\theta _{{\mathcal F}}(\alpha _2)})}\ne \emptyset \). Thus applying \(\theta _{\mathcal F}\) to the decomposition of \({\Phi ({\alpha })}\) produces a non-trivial decomposition of \(\theta _{\mathcal F}(\alpha )=(2,3,1)\). However the permutation (2, 3, 1) is irreducible, and this contradiction establishes the claim.

We similarly claim that there are no points of the graph of \(\alpha \) strictly to the right of *R*. Assume such a point \((y,\alpha (y))\) exists with \(k < y\) and \(\alpha (k)< \alpha (y) < \alpha (k)+s\). Set \({\mathcal F}= \{z,k,y\}\) with \(z = \alpha ^{-1}(\alpha (k)+s)\) as above. As before, we deduce that \((z,y)\in {\Phi ({\alpha _1})}\) and so \({\Phi ({\theta _{{\mathcal F}}(\alpha _1})})\ne \emptyset \), that \((z,k)\in {\Phi ({\alpha _2})}\) and so \({\Phi ({\theta _{{\mathcal F}}(\alpha _2})})\ne \emptyset \), and hence that the decomposition of \({\Phi ({\alpha })}\) induces a non-trivial decomposition of the irreducible element \(\theta _{\mathcal F}(\alpha )=(3,1,2)\). Thus there are no points on the graph of \(\alpha \) to the right of *R* either.

Since \(\alpha \) is a permutation (and hence surjective) there is a point of its graph on each of the \(s+1\) horizontal lattice lines through *R*. We have just shown that none of these points lie outside of *R*, and hence all are in *R*. Each of these points lies on a different vertical lattice line of which there are \(p+1\), and so \(s\leqslant p\). We conclude that \(p=s\), that *R* is a square, and that the graph of \(\alpha \) contains \(p+1\) points in *R*. If \(p+1 < n\) then \(\alpha \) is not simple because it has the block of size \(p+1\) corresponding to *R*. If \(p+1 = n\) then \((k,\alpha (k))=(n,1)\) and \(\alpha \) is not simple because it has the block \(\{1, 2, \ldots , n-1\}\) of size \(n-1\).

A similar argument, with a rectangle of the form \(R=[k,k+q]\times [\alpha (k),\alpha (k)-s]\), handles the case \(p=0\) and \(q\ne 0\). \(\square \)

We can now prove our characterization of simple permutations.

### Proposition 3.15

Let \(n \geqslant 2\) and let \(\alpha \in S_n\) with \(\alpha \ne { I}_2\). Then \(\alpha \in S_n\) is simple if and only if it is irreducible and two-block simple.

### Proof

First suppose \(\alpha \) is irreducible and two-block simple, and express \(\alpha \) in simple form: \(\alpha =\sigma [\beta _1,\beta _2,\dots ,\beta _m]\) with \(\beta _b \in S_{z_b}\) for \(b=1,2,\dots ,m\). By Corollary 3.5, exactly one of the permutations \(\sigma ,\beta _1,\beta _2,\dots ,\beta _m\) is a non-identity permutation. Since \(\alpha \) is two-block simple we have \(\alpha (1)\ne 1\) and \(\alpha (n)\ne n\). If \(\sigma ={ I}_m\) then this would imply that \(\beta _1 \ne { I}\) and \(\beta _m \ne { I}\), which is a contradiction. Thus \(\sigma \ne { I}\) and \(\alpha =\sigma [{ I}_{z_1},{ I}_{z_2},\dots ,{ I}_{z_m}]\). Since \(\alpha \) is two-block simple, \(z_b=1\) for all *b* and hence \(\alpha =\sigma \). Now either \(\sigma \) is simple or Open image in new window. In the former case, \(\alpha =\sigma \) is simple. Otherwise, if Open image in new window, then we must have \(m=2\) by Corollary 2.6 and thus Open image in new window is simple.

Next we suppose that \(\alpha \) is simple. We proceed by induction on *n*. For \(n=2\), we must have Open image in new window which is simple, two-block simple and irreducible. For \(n=3\), no elements are simple. For \(n \geqslant 4\), it follows immediately from the definitions that \(\alpha \) is two-block simple. It remains for us to prove, by induction, that simple implies irreducible.

For \(n=4\) the only two simple permutations are (2, 4, 1, 3) and (3, 1, 4, 2), both of which are exceptional (they appear as (1) and (2) on the list with \(m=2\)) and hence irreducible by Lemma 3.11.

Suppose \(n \geqslant 5\) and that \(\alpha \in S_n\) is simple. If \(\alpha \) is exceptional then the result follows from Lemma 3.11. If not, then by Theorem 3.13, \(\alpha \) has a one-point deletion \(\alpha ^{\circ }\) which is also simple, and hence irreducible by the inductive hypothesis. But then \(\alpha \) must also be irreducible. If not, then Lemma 3.14 would apply to show that \(\alpha \) is not simple, contrary to assumption. \(\square \)

### Remark 3.16

The permutation \(\alpha =(2,4,5,1,3) \in S_5\) is irreducible since its inversion set contains only one simple root. However \(\alpha \) is not two-block simple (and hence also not simple). I.e., although the condition of being simple implies that of being irreducible, the reverse implication does not hold.

### Corollary 3.17

Suppose that \(\alpha \in S_n\) is simple. Then Open image in new window is irreducible.

### Proof

By Lemma 2.4 the fact that \(\alpha \) is simple implies that Open image in new window is simple. But then Open image in new window is irreducible (and also two-block simple) by Proposition 3.15. \(\square \)

This corollary is required for the proof of the main theorem and was one of the motivations for proving Proposition 3.15.

We will also need the following lemma in the proof of the main theorem.

### Lemma 3.18

Let \(\alpha \in S_n\) be simple with \(n\geqslant 4\). Fix *k* with \(1 \leqslant k \leqslant n\). Then there exists \((i,j) \in {\Phi ({\alpha })}\) with \(i \ne k\) and \(j \ne k\).

### Proof

Define \(a := \alpha ^{-1}(1)\) and \(b := \alpha ^{-1}(n)\). Then \(a \ne 1\), \(a \ne n\), \(b\ne 1\) and \(b \ne n\) since \(\alpha \) is simple. Clearly \((1,a), (b,n) \in {\Phi ({\alpha })}\). The conditions \(1 \ne n\), \(a \ne b\), \(b \ne 1\) and \(a \ne n\) imply that \(\{1,a\} \cap \{b,n\} = \emptyset \). Thus either \(k \notin \{1,a\}\) or \(k \notin \{b,n\}\). Hence one of (1, *a*) or (*b*, *n*) may be used as the required root (*i*, *j*). \(\square \)

We now prove the main theorem.

### Proof of Theorem 1.11

Suppose that \(\Delta _n^+ = {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup \dots \sqcup {\Phi ({\alpha _r})}\) is a decomposition with \({\Phi ({\alpha _s})} \ne \emptyset \) for all *s*, and express \(\alpha _1 = \sigma _1[\beta _{11},\beta _{12},\dots ,\beta _{1m}]\) in simple form. Recall that by assumption the highest root (1, *n*) is an element of \({\Phi ({\alpha _1})}\). Let \(\{1,2,\dots ,n\} = U_1\, \sqcup \, U_2\, \sqcup \, \dots \,\sqcup \, U_m\) be the intervals corresponding to the simple form \(\alpha _1 = \sigma _1[\beta _{11},\beta _{12},\dots ,\beta _{1m}]\) with \(z_b=|U_b|\). Throughout this proof “admissible” refers to the intervals \(U_1, \ldots , U_m\).

The assumption that \((1,n) \in {\Phi ({\alpha _1})}\) means that \(\sigma _1 \ne { I}_m\). Corollary 1.9 then implies that either \(\sigma _1\in S_m\) is simple with \(m\geqslant 4\) or that Open image in new window (These two possibilities correspond to the cases that \(\alpha _1\) is minus-indecomposible or is minus-decomposible, respectively).

We first consider the case that \(\sigma _1\in S_m\) is simple with \(m\geqslant 4\). We next show that there is an \(s\geqslant 2\) such that \(\alpha _s\) is of the form Open image in new window. Let \({\mathcal F}\) be an admissible set. Then \(\theta _{\mathcal F}(\alpha _1)=\sigma _1\) and \({\Phi ({\sigma _1})} \sqcup {\Phi ({\theta _{\mathcal F}(\alpha _2)})} \sqcup \dots \sqcup {\Phi ({\theta _{\mathcal F}(\alpha _r)})} = \Delta _{\mathcal F}^+\). Thus Open image in new window. The element Open image in new window is irreducible by Corollary 3.17 and thus there exists \(\delta ({\mathcal F})\geqslant 2\) such that Open image in new window, and \({\Phi ({\theta _{\mathcal F}(\alpha _s)})}=\emptyset \) for all \(s\ne \delta ({\mathcal F}), s\geqslant 2\). We claim that the number \(\delta ({\mathcal F})\) is independent of the choice of admissible set \({\mathcal F}\).

Recall that an admissible set is the choice of a single element from each of the sets \(U_1\), \(U_2\), ..., \(U_m\), and thus the admissible sets are in one to one correspondence with the points of \(U_1\times U_2\times \cdots \times U_m\). Given any two admissible subsets \({\mathcal F}\) and \({\mathcal F}'\) we may find a sequence of admissible subsets \({\mathcal F}={\mathcal F}_0\), \({\mathcal F}_1\), ..., \({\mathcal F}_{l-1}\), \({\mathcal F}_l={\mathcal F}'\) such that each \({\mathcal F}_i\) and \({\mathcal F}_{i+1}\) differ by only a single element (i.e, under the correspondence with elements of \(U_1\times \cdots \times U_m\), differ in only a single coordinate). To prove that \(\delta ({\mathcal F})\) is independent of the choice of admissible set we may thus reduce to the case that \({\mathcal F}\) and \({\mathcal F}'\) differ by a single element.

Suppose that \({\mathcal F}'\) is obtained from \({\mathcal F}\) by replacing \(u_k\in U_k\) with \(u'_k\in U_k\) for some \(1\leqslant k\leqslant m\). There is a root (*i*, *j*) in Open image in new window with \(i\ne k\) and \(j\ne k\) by Lemma 3.18. Since \({\mathcal F}\) and \({\mathcal F}'\) differ only in the element in \(U_k\), the elements they choose from \(U_i\) (respectively \(U_j\)) are the same. Set \(a=\Delta _{{\mathcal F}}\cap U_i = \Delta _{{\mathcal F}'}\cap U_i\) and \(b=\Delta _{{\mathcal F}}\cap U_j = \Delta _{{\mathcal F}'}\cap U_j\). Then \((i,j)\in {\Phi ({\theta _{{\mathcal F}}(\alpha _s)})}\) if and only if \((a,b)\in {\Phi ({\alpha _s})}\), and similarly \((i,j) \in {\Phi ({\theta _{F'}(\alpha _s)})}\) if and only if \((a,b) \in {\Phi ({\alpha _s})}\). We have seen above that \({\Phi ({\theta _{\mathcal F}(\alpha _s)})}=\emptyset \) (respectively \({\Phi ({\theta _{{\mathcal F}'}(\alpha _s)})}=\emptyset \)) for all \(s\ne \delta ({\mathcal F})\) (respectively, \(s\ne \delta ({\mathcal F}')\)), \(s\geqslant 2\). Thus \(\delta ({\mathcal F})=\delta ({\mathcal F}')\), and so \(\delta ({\mathcal F})\) is constant for all admissible sets \({\mathcal F}\). By reordering the elements \(\alpha _2, \ldots , \alpha _r\), we may assume that this constant value is 2.

The statement we have just proved, that Open image in new window for all admissible sets \({\mathcal F}\), is equivalent to the statement that for any \((i,j)\in \Delta _m\), and any \(a\in U_i\), \(b\in U_j\), \((a,b)\in {\Phi ({\alpha _2})}\) if and only if Open image in new window. This implies that \(\alpha _2\) permutes the intervals \(U_1\),..., \(U_m\), in the manner specified by Open image in new window and thus can be written as an inflation Open image in new window. Specifically, \(\beta _{2t} = \theta _{U_t}(\alpha _2)\) for \(t=1\),..., *m*.

For the remaining \(\alpha _s\), \(s=3\),..., *r*, we have \(\alpha _s(a)<\alpha _s(b)\) for all \(a\in U_i\), \(b\in U_j\) and \(1\leqslant i<j\leqslant m\) since the roots (*a*, *b*) are all contained in \({\Phi ({\alpha _1})}\sqcup {\Phi ({\alpha _2})}\). This implies that for each such \(\alpha _s\) we have \(\alpha _s(U_i)=U_i\) for each *i* and hence that \(\alpha _s={ I}_m[\beta _{s1},\ldots , \beta _{sm}]\) with \(\beta _{si}=\theta _{U_i}(\alpha _s)\) for \(i=1\),..., *m*.

On the other hand, if \(\alpha _1\) is minus-decomposable then (by definition) the simple form of \(\alpha _1\) is Open image in new window. For the remaining *s*, \(s=2\),..., *r*, we again have that \(\alpha _s(a)<\alpha _s(b)\) whenever \(a\in U_i\), \(b\in U_j\), and \(1\leqslant i<j\leqslant m\) since the roots (*a*, *b*) are all contained in \({\Phi ({\alpha _1})}\). We conclude as in the first case that each \(\alpha _s\), \(s\geqslant 2\) is of the form \({ I}_m[\beta _{s1},\ldots , \beta _{sm}]\) with \(\beta _{si}=\theta _{U_i}(\alpha _s)\) for \(i=1\),..., *m*.

This proves the theorem in the case of a general (possibly reducible) decomposition.

Next we consider irreducible decompositions. Corollary 3.5, shows the necessity of conditions (i), (ii) and (iii) of the final assertion. The element \(J_m\) is irreducible if and only if \(m=2\) and this shows the necessity of condition (iv). Thus these four conditions hold for irreducible decompositions. Finally if these four conditions hold it is clear that each of the inversion sets \({\Phi ({\alpha _a})}\) is irreducible by Corollary 3.5. \(\square \)

## 4 Symmetric permutations

The aim of this section is to extend the results obtained so far to a special class of permutations. The results will then be used in the next section to study root systems of types *B*, *C* and *D*.

A permutation \(\alpha \in S_N\) is *symmetric* if Open image in new window. Equivalently, \(\alpha \in S_N\) is symmetric if the graph of \(\alpha \) is symmetric under rotation by \(\pi \) radians about the point \((\frac{N+1}{2}, \frac{N+1}{2})\).

### Proposition 4.1

Let \(\alpha \in S_N\) and write \(\alpha \) in simple form: \(\alpha =\sigma [\beta _1,\beta _2,\dots ,\beta _s]\). If \(\alpha \) is symmetric then \(\sigma \) is necessarily symmetric and Open image in new window for all \(b=1,2,\dots ,s\). Consequently, if *N* is odd then \(s = 2m+1\) is odd, \(z_{m+1}\) is odd and \(\beta _{m+1} \in S_{z_{m+1}}\) is symmetric. If *N* is even then then *s* may be even or odd; if, in addition, \(s = 2m+1\) is odd then \(z_{m+1}\) is even and \(\beta _{m+1} \in S_{z_{m+1}}\) is symmetric.

### Proof

Next we define an inflation operation which produces symmetric permutations. Let \(0 \leqslant p \leqslant n\) and let \(\{1,2,\dots ,n-p\} = U_1\, \sqcup \, U_2\, \sqcup \, \cdots \, \sqcup \, U_m\) be a decomposition into intervals. Put \(z_b = |U_b|\) for \(b=1,2,\dots ,m\). Suppose that \(\beta _b \in S_{z_b}\) for \(b=1,2,\dots ,m\) and \(\beta _{m+1} \in S_{2p+1}\) or \(\beta _{m+1} \in S_{2p}\). (For uniformity of notation we allow considering \(S_{2p}\) for \(p=0\); we will use \(\varnothing \) to denote the “phantom” element of \(S_0\)). Let \(\sigma \in \left\{ \begin{array}{lcl} S_{2m+1} &{} { \text{ if } } &{} \beta _{m+1} \ne \varnothing \\ S_{2m} &{} { \text{ if } } &{} \beta _{m+1} = \varnothing \end{array} \right. \),

*symmetric inflation*and denote it by

### Corollary 4.2

Finally, Theorem 1.8 implies the existence of a *simple symmetric form expression* of a symmetric element \(\alpha \in S_N\).

### Proposition 4.3

*M*be maximal when \(\sigma = I_M\) or Open image in new window. \(\square \)

It also natural to discuss decomposing \(\Delta _N^+\) into symmetric inversion sets. Theorem 1.11 and Proposition 4.1 imply in a straightforward manner the following theorem.

### Theorem 4.4

- (i)$$\begin{aligned}\begin{array}{rcl} \Delta ^+_M &{}=&{} {\Phi ({\sigma _1})} \sqcup {\Phi ({\sigma _2})} \sqcup \dots \sqcup {\Phi ({\sigma _r})}, \\ &{}&{}\\ \Delta ^+_{z_1} &{} = &{} {\Phi ({\beta _{11}})} \sqcup {\Phi ({\beta _{21}})} \sqcup \dots \sqcup {\Phi ({\beta _{r1}})}, \\ \Delta ^+_{z_2} &{} = &{} {\Phi ({\beta _{12}})} \sqcup {\Phi ({\beta _{22}})} \sqcup \dots \sqcup {\Phi ({\beta _{r2}})},\\ &{}\vdots &{}\\ \Delta ^+_{z_m} &{} = &{} {\Phi ({\beta _{1m}})} \sqcup {\Phi ({\beta _{2m}})} \sqcup \dots \sqcup {\Phi ({\beta _{rm}})},\\ &{}&{}\\ \Delta ^+_P &{} = &{} {\Phi ({\beta _{1(m+1)}})} \sqcup {\Phi ({\beta _{2(m+1)}})} \sqcup \dots \sqcup {\Phi ({\beta _{r(m+1)}})}; \end{array} \end{aligned}$$
- (ii)
if \(\alpha _1\) is minus-decomposable then Open image in new window and \(\sigma _2=\sigma _3=\dots =\sigma _r={ I}\);

- (iii)
if \(\alpha _1\) is minus-indecomposable then \(\sigma _1\) is simple and, after relabeling \(\alpha _2, \ldots , \alpha _r\), we have Open image in new window, and \(\sigma _3=\sigma _4=\dots =\sigma _r={ I}\).

Let *q* denote the number of \(\sigma _a\) which are not \({ I}\), i.e., \(q := {\left\{ \begin{array}{ll} 1, &{}\text {if }\alpha _1 \text { is minus-decomposable};\\ 2, &{}\text {if }\alpha _1 \text { is minus-indecomposable}. \end{array}\right. }\) Again, after relabeling \(\alpha _2, \ldots , \alpha _r\), we assume that \(\sigma _{q+1} = \ldots = \sigma _r = I\).

- (i)
each of the decompositions listed in (i) above is irreducible;

- (ii)
exactly one of \(\beta _{a1},\beta _{a2},\dots ,\beta _{am}\) is not equal to the identity for \(a=q+1,\dots , r\);

- (iii)
\(\beta _{ab} = { I}\) for \(a=1,\dots , q\) and \(b=1,\dots , m+1\);

- (iv)
\(m = 1\) if \(\alpha _1\) is minus-decomposable. \(\square \)

## 5 Decompositions of types B, C and D

*B*,

*C*and

*D*. We introduce some notation related to these root systems; our exposition is limited only to the minimum that we need. For a reference on root systems, see [8, Chap. III]. We will compare the root systems of types

*B*,

*C*and

*D*with the root systems of type

*A*. We take \(\{e_1,e_2,\dots ,e_{n+1}\}\) as a standard basis for \({\mathbb R}^{n+1}\) and consider the Weyl group \({\mathcal W}(A_n) \cong S_{n+1}\) as the group of all permutations of this basis. With this notation, the positive roots are

*m*-dimensional vector space. The weights of these respective natural representations are \(\{\pm \varepsilon _i\}_{i=1}^{n}\) for \(C_n\) or \(D_n\) and \(\{0\} \cup \{\pm \varepsilon _i\}_{i=1}^{n}\) for \(B_n\). We thus obtain the following faithful permutation representations of the Weyl groups.

- The Weyl group \({\mathcal W}(B_n)\) is the set of sign-compatible permutations \(\alpha \) ofThat is, permutations obeying the condition that \(\alpha (-\varepsilon _i)=-\alpha (\varepsilon _i)\) for all \(1 \leqslant i \leqslant n\) and \(\alpha (0)=0\). Abstractly, \({\mathcal W}(B_n) \cong S_n \rtimes ({\mathbb Z}/2{\mathbb Z})^{n}\).$$\begin{aligned} \{\varepsilon _1,\varepsilon _2,\dots ,\varepsilon _n,0,-\varepsilon _n,\dots ,-\varepsilon _2,-\varepsilon _1 \}. \end{aligned}$$
- The Weyl group \({\mathcal W}(C_n)\) is the set of sign-compatible permutations of the setAbstractly, \({\mathcal W}(C_n) \cong S_n \rtimes ({\mathbb Z}/2{\mathbb Z})^{n}\).$$\begin{aligned} \{\varepsilon _1,\varepsilon _2,\dots ,\varepsilon _n,-\varepsilon _n,\dots ,-\varepsilon _2,-\varepsilon _1 \}. \end{aligned}$$
- The Weyl group \({\mathcal W}(D_n)\) is the set of sign-compatible permutations of the setinvolving an even number of sign changes, i.e., permutations \(\alpha \) for which \(\alpha (\varepsilon _i) = - \varepsilon _j\) for an even number of indices \(1 \leqslant i \leqslant n\). Abstractly, \({\mathcal W}(D_n) \cong S_n \rtimes ({\mathbb Z}/2{\mathbb Z})^{n-1}\).$$\begin{aligned} \{\varepsilon _1,\varepsilon _2,\dots ,\varepsilon _n,-\varepsilon _n,\dots ,-\varepsilon _2,-\varepsilon _1 \} \end{aligned}$$

In order to treat the roots systems of types *B*, *C* and *D* and their Weyl groups uniformly, we introduce some notation. Instead of discussing separately the root systems \(B_n, C_n\) or \(D_n\) we will sometimes discuss the root system \(X_n\) understanding that *X* stands for one of *A*, *B*, *C* or *D*. For uniformity of notation below, when considering \(X_n\), we allow all values of \(n \geqslant 0\): for instance, \(\Delta _{X_0}^+ = \emptyset \).

For \(\alpha \in {\mathcal W}(X_n)\) we define \({\Phi ({\alpha })} := \{v \in \Delta _{X_n}^+ \mid \alpha (v) \not \in \Delta _{X_n}^+\}\). It is clear that this definition of inversion set agrees with our previous definition when \(X =A\). Let Open image in new window denote the element of \({\mathcal W}(X_n)\) such that Open image in new window. As in the type *A* case, Open image in new window is called the *longest element* of the corresponding Weyl group.

Most of the contents of Sect. 2 transfer to the cases when \(X = B, C\) or *D*. For instance, call a set \(\Phi \subset \Delta _{X_n}^+\)*closed* if \(\alpha _1 + \alpha _2 \in \Phi \) whenever \(\alpha _1, \alpha _2 \in \Phi \) and \(\alpha _1 + \alpha _2 \in \Delta _{X_n}^+\). Proposition 2.1 still holds: \(\Phi \subset \Delta _{X_n}^+\) is an inversion set if and only if both \(\Phi \) and \(\Delta _{X_n}^+ \backslash \Phi \) are closed. Similarly, the obvious analogs of Lemmas 2.2 and 2.5, Proposition 2.7 and Corollary 2.8 hold in general.

We now turn to the basic problem of this paper, that of decomposing inversion sets, in the \(B_n\), \(C_n\), and \(D_n\) cases. Our idea is to relate it to the type *A* case by using the permutation representations of \({\mathcal W}(B_n)\), \({\mathcal W}(C_n)\) and \({\mathcal W}(D_n)\) listed above. Specifically, using the orders \(\varepsilon _1,\varepsilon _2,\ldots , \varepsilon _n,0,-\varepsilon _n,\ldots , -\varepsilon _2,-\varepsilon _1\) for \(B_n\), and \(\varepsilon _1,\varepsilon _2,\ldots , \varepsilon _n, -\varepsilon _n,\ldots , -\varepsilon _2,-\varepsilon _1\) for \(C_n\) and \(D_n\), these permutation representations define embeddings of groups \(\iota :{\mathcal W}(B_n)\hookrightarrow S_{2n+1}\), \(\iota :{\mathcal W}(C_n)\hookrightarrow S_{2n}\) and \(\iota :{\mathcal W}(D_n)\hookrightarrow S_{2n}\) (We use the same name for each of these homomorphisms, trusting that the name of the domain will make the homomorphism clear).

Note that for \(X = B\) or *C* we have Open image in new window, while for \(X = D_n\) this is true if and only if *n* is even.

The following proposition establishes the behavior of inversion sets under the maps \(\iota \) and \(\rho \) above. Its proof is straightforward and is left to the reader.

### Proposition 5.1

*D*.

- (i)
For any \(\alpha \in {\mathcal W}(X_n)\), \(\iota (\alpha ) \in S_{\tilde{n}}\) is symmetric. If \(X = B\) or

*C*, the image of \(\iota \) consists of all symmetric permutations in \(S_{\tilde{n}}\); if \(X = D\), the image of \(\iota \) consists of all symmetric permutations \(\beta \in S_{\tilde{n}}\) such that an even number of the elements \(\beta (1), \ldots , \beta (n)\) are greater than*n*. - (ii)
The map \(\rho \) is surjective. More precisely, each element of \(\hat{\Delta }_{X_n}^+\) of the form \(2\varepsilon _i\) has a unique preimage in \(\Delta _{\tilde{n}}^+\) and each of the remaining elements of \(\hat{\Delta }_{X_n}^+\) has exactly two preimages in \(\Delta _{\tilde{n}}^+\).

- (iii)
If \(\beta \in S_{\tilde{n}}\) is symmetric then \(\rho ({\Phi ({\beta })}) \cap \Delta _{X_n}^+ \subseteq \Delta _{X_n}^+\) is an inversion set.

- (iv)
If \(\alpha \in {\mathcal W}(X_n)\) then \({\Phi ({\iota (\alpha )})} = \rho ^{-1}({\Phi ({\alpha })})\).

- (v)
Let \(\alpha \in {\mathcal W}(X_n)\). If \(X = B\) or

*C*then \(\beta = \iota (\alpha )\) is the unique element of \(S_{\tilde{n}}\) such that \({\Phi ({\alpha })} = \rho ({\Phi ({\beta })}) \cap \Delta _{X_n}^+\). If \(X = D\) there are exactly two elements \(\beta = \iota (\alpha )\) and \(\beta ' \not \in \iota ({\mathcal W}(X_n))\) such that \({\Phi ({\alpha })} = \rho ({\Phi ({\beta })}) \cap \Delta _{X_n}^+ = \rho ({\Phi ({\beta '})}) \cap \Delta _{X_n}^+\). \(\square \)

Proposition 5.1 implies that the map \(\iota \) interacts well with decompositions into inversion sets. More precisely, the following statements follow immediately from Proposition 5.1.

### Corollary 5.2

- (i)Assume that \(\alpha _1, \alpha _2 \in {\mathcal W}(X_n)\) satisfy \({\Phi ({\iota (\alpha _1)})} \cap {\Phi ({\iota (\alpha _2)})} = \emptyset \). Then$$\begin{aligned} \rho ({\Phi ({\iota (\alpha _1)})} \sqcup {\Phi ({\iota (\alpha _2)})}) = \rho ({\Phi ({\iota (\alpha _1)})}) \sqcup \rho ({\Phi ({\iota (\alpha _2)})}). \end{aligned}$$
- (ii)Let \(\alpha _1,\alpha _2,\dots ,\alpha _r \in {\mathcal W}(X_n)\). Then
- (iii)
An element \(\alpha \in {\mathcal W}(X_n)\) is irreducible if and only if \(\iota (\alpha ) \in S_{\tilde{n}}\) is irreducible. \(\square \)

Corollary 5.2 suggests that one can approach studying decompositions of \(\Delta _{X_n}^+\) inversion sets by studying decompositions of \(\Delta _{\tilde{n}}^+\) into symmetric inversion sets. Indeed, this approach can be carried out successfully in the cases when \(X = B\) and *C*. Unfortunately, the ambiguity in Proposition 5.1 (i), (iv) prevented us from obtaining results for \(X=D\). The first step is to define (or attempt to define) an inflation operation for the Weyl groups of types *B*, *C* and *D*. Proposition 5.1 (i) allows us to transfer the inflation operation for symmetric permutations to an inflation operation for the Weyl groups of types *B* and *C* but not *D*.

*B*and

*C*. Let \(X =B\) or

*C*. Let \(0 \leqslant p \leqslant n\) and let \(\{1,2,\dots ,n-p\} = U_1 \sqcup U_2 \sqcup \dots \sqcup U_m\) be a decomposition into intervals. Put \(z_b = |U_b|\) for \(b=1,2,\dots ,m\). Suppose that \(\sigma \in \left\{ \begin{array}{lcl} {\mathcal W}(B_m) &{} { \text{ if } } &{} X_p \ne C_0\\ {\mathcal W}(C_m) &{} { \text{ if } } &{} X_p = C_0 \end{array} \right. \), \(\beta _{m+1} \in {\mathcal W}(X_p)\) and \(\beta _b \in S_{z_b}\) for \(b=1,2,\dots ,m\). We form the inflation

An element \(\alpha \in {\mathcal W}(X_n)\) which cannot be realized as such an inflation in \({\mathcal W}(X_n)\) except with \(m=0\) or \(m=n\) is said to be *simple in*\({\mathcal W}(X_n)\). Propositions 4.1 and 5.1(i) imply immediately the following statement.

### Proposition 5.3

Let \(X = B\) or *C* and let \(\alpha \in {\mathcal W}(X_n)\). Then \(\alpha \) is simple in \({\mathcal W}(X_n)\) if and only if \(\iota (\alpha )\) is simple in \(S_{\tilde{n}}\). \(\square \)

*simple form expression for*\(\alpha \in {\mathcal W}(X_n)\) ifis the simple form expression for \(\iota (\alpha )\) in \(S_{\tilde{n}}\).

*D*. On one hand, the element \(\tilde{\alpha }\) defined above may not belong to the image of \(\iota \) and, on the other hand, for \(\alpha \in {\mathcal W}(D_n)\) the element \(\iota (\alpha )\) may be an inflationwhere \(\tilde{\sigma }\) and \(\tilde{\beta }_{m+1}\) are symmetric but not necessarily in the image of \(\iota \).

### Theorem 5.4

*C*. Suppose \(\alpha _1, \alpha _2,\dots ,\alpha _r \in {\mathcal W}(X_n)\) and

- (i)$$\begin{aligned}\begin{array}{rcl} \Delta ^+_{X_m} &{}=&{} {\Phi ({\sigma _1})} \sqcup {\Phi ({\sigma _2})} \sqcup \dots \sqcup {\Phi ({\sigma _r})}, \\ &{}&{}\\ \Delta ^+_{A_{z_1-1}} &{} = &{} {\Phi ({\beta _{11}})} \sqcup {\Phi ({\beta _{21}})} \sqcup \dots \sqcup {\Phi ({\beta _{r1}})}, \\ \Delta ^+_{A_{z_2-1}} &{} = &{} {\Phi ({\beta _{12}})} \sqcup {\Phi ({\beta _{22}})} \sqcup \dots \sqcup {\Phi ({\beta _{r2}})},\\ &{}\vdots &{}\\ \Delta ^+_{A_{z_m-1}} &{} = &{} {\Phi ({\beta _{1m}})} \sqcup {\Phi ({\beta _{2m}})} \sqcup \dots \sqcup {\Phi ({\beta _{rm}})},\\ &{}&{}\\ \Delta ^+_{X_p} &{} = &{} {\Phi ({\beta _{1(m+1)}})} \sqcup {\Phi ({\beta _{2(m+1)}})} \sqcup \dots \sqcup {\Phi ({\beta _{r(m+1)}})}; \end{array} \end{aligned}$$
- (ii)
if \(\alpha _1\) is minus-decomposable then Open image in new window and \(\sigma _2=\sigma _3=\dots =\sigma _r={ I}\);

- (iii)
if \(\alpha _1\) is minus-indecomposable then \(\sigma _1\) is simple (in \({\mathcal W}(B_m)\) or in \({\mathcal W}(C_m)\)) and, after relabeling \(\alpha _2, \ldots , \alpha _r\), we have Open image in new window, and \(\sigma _3=\sigma _4=\dots =\sigma _r={ I}\).

Let *q* denote the number of \(\sigma _a\) which are not \({ I}\), i.e., \(q := {\left\{ \begin{array}{ll} 1, &{}\text {if }\alpha _1 \text { is minus-decomposable};\\ 2, &{}\text {if }\alpha _1 \text { is minus-indecomposable}. \end{array}\right. }\) Again, after relabeling \(\alpha _2, \ldots , \alpha _r\), we assume that \(\sigma _{q+1} = \ldots = \sigma _r = I\).

- (i)
each of the decompositions listed in (i) above is irreducible;

- (ii)
exactly one of of \(\beta _{a1},\beta _{a2},\dots ,\beta _{am}\) is not equal to the identity for \(a=q+1,\dots , r\);

- (iii)
\(\beta _{ab} = { I}\) for \(a=1,\dots , q\) and \(b=1,\dots , m+1\);

- (iv)
\(m = 1\) if \(\alpha _1\) is minus-decomposable.

### Proof

This result follows directly from Theorem 1.11 and the results of this section. Only two additional observations are needed. The first is that Open image in new window is irreducible if and only if \(m=1\). The second is that the assumption \(e_1 - e_{\tilde{n}} \in {\Phi ({\iota (\alpha _1)})}\) implies that \(\iota (\sigma _1)\) is not the identity. \(\square \)

We conclude this section with a few remarks about decomposing \(\Delta _{D_n}^+\) into inversion sets. As we already mentioned, it is not clear how to define the inflation operation for type *D*. Another possible approach to decomposing \(\Delta _{D_n}^+\) may be to use the fact that \(\Delta _{D_n}^+\) embeds naturally into \(\Delta _{C_n}^+\). Indeed, one can show that every decomposition of \(\Delta _{C_n}^+\) into inversion sets produces a unique decomposition of \(\Delta _{D_n}^+\) into inversion sets. We do not know, however, whether the converse is true.

## 6 Enumerative results

The inductive description for a decomposition provided by Theorems 1.11 and 5.4 allows us to use generating series or recursion to enumerate many different types of decompositions. We give a few examples.

Let \(s_n\) be the number of simple pairs in \(S_n\), i.e., the number of sets Open image in new window with \(\alpha \in S_{n}\) and both \(\alpha \) and Open image in new window simple (note that by Lemma 2.4\(\alpha \) is simple if and only if Open image in new window is simple). Let \(S_A(z) = \sum _{n\geqslant 0} s_n z^n = z^2 + z^4 + 3z^5+\cdots \) be the corresponding generating function. By [2, page 5] we have the following description of *S*(*z*). Let \(F(z) = \sum _{n\geqslant 1} n! z^{n}\) and \(G(z) = \sum _{n\geqslant 1} g_n z^n\) its functional inverse, i.e., the function defined by the relation \(G(F(z))=z\). Then \(s_1=0\), \(s_2=1\), and \(s_n = -g_n/2 - (-1)^n\) for \(n\geqslant 3\).

*Number of decompositions into irreducibles*Let \(a_n\) be the number of decompositions \(\Delta _n^{+} = {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup \dots \sqcup {\Phi ({\alpha _r})}\) into non-empty inversion sets, where each \(\alpha _k\in S_n\) is irreducible, and where we ignore the order in the decomposition. Let \(A(z) = \sum _{n\geqslant 1} a_n z^n\) be the generating series. Theorem 1.11 leads to the relation \(A(z) = S_A(A(z)) + z\), which recursively determines the coefficients \(a_n\). Here are the low order terms of

*A*(

*z*):

*Decompositions of maximal length* If \(\alpha \ne { I}\) then the inversion set \({\Phi ({\alpha })}\) must contain at least one simple root. Since there are only \(n-1\) simple roots, any decomposition \(\Delta _n^{+} = {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup \dots \sqcup {\Phi ({\alpha _r})}\), with no \(\alpha _s={ I}\) must satisfy \(r\leqslant n-1\). Let \({\text {Cat}_\text {A}(n-1)}\) denote the number of decompositions of \(\Delta _n^+\) into exactly \(n-1\) non-empty inversion sets (Thus each inversion set appearing in the decomposition must contain exactly one simple root).

### Lemma 6.1

\({\text {Cat}_\text {A}(n)}= \frac{1}{n+1}\genfrac(){0.0pt}{}{2n}{n}\), the \(n^\text {th}\) Catalan number.

### Proof

We consider decompositions of the form \(\Delta _n^{+} = {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup \dots \sqcup {\Phi ({\alpha _{n-1}})}\) and compute \({\text {Cat}_\text {A}(n-1)}\). Without loss of generality, the highest root \(e_1-e_n \in {\Phi ({\alpha _1})}\). Suppose that \(e_k-e_{k+1}\) is the simple root in \({\Phi ({\alpha _1})}\). Then \(\alpha _1(k+1)< \alpha _1(k+2)< \dots< \alpha _1(n)< \alpha _1(1)< \alpha _1(2) \dots < \alpha _1(k)\) and therefore \(\alpha _1 = (n-k+1, n-k+2, \dots , n, 1 ,2 \dots , n-k) = (1,2)[{ I}_k,{ I}_{n-k}]\). Let \(U_1 := \{1,2,\dots ,k\}\) and \(U_2 := \{k+1,k+2,\dots ,n\}\). Then \({\Phi ({\alpha _1})} = \{ (e_i-e_j \in \Delta _n^+ \mid i \in U_1, j \in U_2\} = \{ e_i-e_j \in \Delta _n^+ \mid i \leqslant k, j \geqslant k+1\}\). Therefore \(\Delta _{U_1}^+ \sqcup \Delta _{U_2}^+ = {\Phi ({\alpha _2})} \sqcup {\Phi ({\alpha _3})} \sqcup \dots \sqcup {\Phi ({\alpha _{n-1}})}\). Without loss of generality, \(\Delta _{U_1}^+ = {\Phi ({\alpha _2})} \sqcup {\Phi ({\alpha _3})} \sqcup \dots \sqcup {\Phi ({\alpha _{k-1}})}\) and \(\Delta _{U_2}^+ = {\Phi ({\alpha _{k+1}})} \sqcup {\Phi ({\alpha _{k+2}})} \sqcup \dots \sqcup {\Phi ({\alpha _{n-1}})}\). This yields the recursion relation \({\text {Cat}_\text {A}(n-1)} = \sum _{t=1}^{n-1} {\text {Cat}_\text {A}(t-1)}{\text {Cat}_\text {A}(n-t-1)} = \sum _{t=0}^{n-2} {\text {Cat}_\text {A}(t)}{\text {Cat}_\text {A}(n-t-2)}\). Thus \({\text {Cat}_\text {A}(n)} = \sum _{t=1}^{n} {\text {Cat}_\text {A}(t-1)}{\text {Cat}_\text {A}(n-t)}\). Since \({\text {Cat}_\text {A}(1)}=1\) and \({\text {Cat}_\text {A}(2)}=2\) we see that \({\text {Cat}_\text {A}(n)}\) satisfies the usual recursion relation for the Catalan numbers. \(\square \)

This incarnation of the Catalan numbers is number 186 in Richard Stanley’s monograph [12].

*Type B/C results*Theorem 5.4 leads to similar recursions in types

*B*/

*C*. Let \(S_{B}(z)\) be the generating series for the number of simple pairs in type \(B_n/C_n\). Equivalently the coefficient of \(z^n\) in \(S_{B}(z)\) is the number of pairs of simple elements in \(S_{2n+1}\) each of which are symmetric. The isomorphism \({\mathcal W}(B_n) \cong {\mathcal W}(C_n)\) implies that this is also the number of pairs of simple symmetric elements in \(S_{2n}\). One deduces the functional equation

*Decompositions into irreducibles*Let \(b_n\) be the number of decompositions of the positive roots in types \(B_n/C_n\) into disjoint unions of irreducible inversion sets, and let \(B(z) = \sum _{n\geqslant 1} b_n z^n\) to be the generating function. Theorem 5.4 leads to the relation

*B*(

*z*). Here are the low order terms of

*B*(

*z*):

*Catalan numbers*Let \({\text {Cat}_\text {B}(n)}\) be the number of decompositions of the positive roots of \(B_n/C_n\) into disjoint unions of inversion sets, where each inversion set contains a single simple root. The isomorphism \({\mathcal W}(B_n) \cong {\mathcal W}(C_n)\) implies that the number of such decompositions is the same for types \(B_n\) and \(C_n\). As in type

*A*, these are the decompositions of maximal length (subject to the restriction that each inversion set is non-empty) and thus are irreducible decompositions.

### Proposition 6.2

### Proof

We consider the \(B_n\) case. Suppose then that \(\Delta _{B_n}^+ = {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup \dots \sqcup {\Phi ({\alpha _n})}\) where each \(\alpha _i \in {\mathcal W}(B_n)\) and each \({\Phi ({\alpha _i})}\) contains a single simple root of \(\Delta _{B_n}^+\). Without loss of generality \({\Phi ({\iota (\alpha _1)})}\) contains \(e_1 - e_{2n+1}\). By Theorem 5.4, we have \( \alpha _1 = \sigma _1[[\beta _{11},\beta _{12},\dots ,\beta _{1s};\beta _{1(s+1)}]]\) where \(\sigma _1 \in {\mathcal W}(B_s)\) and either Open image in new window or \(\sigma _1\) is simple and \({\Phi ({\sigma _1})}\) contains a single simple root. Thus if Open image in new window then \(\iota (\sigma _1)\) is simple, symmetric and \({\Phi ({\iota (\sigma _1)})}\) contains a pair of \(A_{2n}\) simple roots of the form \(e_i-e_{i+1}, e_{i'-1}-e_{i'}\), where \(i' = 2n+2 - i\). It is not hard to see that this forces Open image in new window, \(\iota (\sigma _1) = (41{,}352)\) or \(\iota (\sigma _1) = (25{,}314)\). The last possibility is excluded by the fact that \({\Phi ({\iota (\sigma _1)})}\) contains the highest root \(e_1-e_5\).

First suppose that Open image in new window and let \(\{1,2,\dots ,2n+1\}=U_1 \sqcup U_2 \sqcup U_3\) be the corresponding decomposition into intervals with \(|U_1|=|U_3|=n-k\) and \(|U_2|=2k+1\) where \(0\leqslant k \leqslant n-1\). Then \(\iota (\alpha _j) = { I}_3[\beta _{j1},\beta _{j2},\beta _{j3}]\) for \(j=2,3,\dots ,n\). Furthermore, without loss of generality, \(\Delta _{U_1}^+ = {\Phi ({\beta _{21}})} \sqcup {\Phi ({\beta _{31}})} \sqcup \dots \sqcup {\Phi ({\beta _{(n-k)1}})}\) is a maximal length decomposition of a root system of type \(A_{n-k-1}\). There are \({\text {Cat}_\text {A}(n-k-1)}\) such decompositions. (We also have Open image in new window.) Finally \(\Delta _{U_2}^+ = {\Phi ({{\beta }_{(n-k+1)2}})} \sqcup {\Phi ({{\beta }_{(n-k+2)2}})} \sqcup \dots \sqcup {\Phi ({{\beta }_{n2}})}\) is a maximal symmetric decomposition. There are \({\text {Cat}_\text {B}(k)}\) such decompositions. Thus there are \(\sum _{k=0}^{n-1} {\text {Cat}_\text {A}(n-k-1)}{\text {Cat}_\text {B}(k)}\) maximal decompositions of \(\Delta _{B_n}^+\) with Open image in new window.

*A*type decompositions of \(\Delta _{U_1}^+\) and \(\Delta _{U_2}^+\) and a maximal symmetric decomposition of \(\Delta _{U_3}^+\). Thus there are

These numbers appear in other combinatorial settings, see A081696 in the Sloane on-line encyclopedia of integer sequences.

*Remark * We have chosen to call these numbers the “type *B* / *C* Catalan numbers”, since they come from an enumerative problem about Coxeter groups which yields the usual Catalan numbers in the type *A* case. There is at least one other use of the term “Catalan numbers for other types” in the literature, again stemming from an enumerative problem (generalizing non-crossing partitions) valid for all Coxeter groups. In this second problem, the type \(B_n/C_n\) numbers are \(\left( {\begin{array}{c}2n\\ n\end{array}}\right) \) (see [3, pg. 39])—different from the numbers given by the recursion and generating function above.

*Number of decompositions into triples*The most important case—in any type—of the problems motivating these questions about decompositions is the case of decompositions into a disjoint union of three inversion sets. As described in Sect. 1.2 this corresponds to the case of the eigenvalues of three Hermitian matrices summing to zero (respectively the cup product of two cohomology groups into a third, after a similar symmetrization). The corresponding enumerative/classification problem is to write down all triples \(\alpha _1\), \(\alpha _2\), \(\alpha _3\in S_n\) (again disregarding order) with \(\Delta _n^{+}= {\Phi ({\alpha _1})} \sqcup {\Phi ({\alpha _2})} \sqcup {\Phi ({\alpha _3})}\). We make the further restriction that no \(\alpha _j={ I}\) (all such triples are of the form Open image in new window and hence elementary to understand). Theorems 1.11 and 5.4 provide a recursive way to generate and enumerate all such triples. Briefly, the method is a parallel recursion keeping track of not only the triples of the kind above, but also the subset of those triples where Open image in new window for some

*m*. At each step, the new triples of each kind depend on the triples of both kinds for smaller

*n*. (We omit the exact description of the recursion since, although elementary, it is slightly messy.) Here is a small table of the number of such triples, and both the \(A_n\) and \(B_n/C_n\) cases.

## 7 Decomposing a single inversion set

^{3}

Let \(\alpha = \sigma [\beta _1, \ldots , \beta _m]\) be the simple form of \(\alpha \). To list all ordered decompositions^{4} of \({\Phi ({\alpha })}\) we proceed as follows:

*Step 1*Write all decompositions

*Step 2*For every decomposition of Step 1 write the decompositions

*Step 3*Write all partitions \(\mathcal {U}\) of the set \(\{1, 2, \ldots , m\}\) into \(l \geqslant 4\) intervals \(U_1, U_2, \ldots , U_l\) of lengths \(z_1, z_2, \ldots , z_l\) and for each such partition construct the following elements:

*Step 4*Write all decompositions

*Step 5*For every decomposition of Step 1 and every simple \(\sigma \in S_l\) write the decompositionsThese complete the list of all decompositions of \({\Phi ({\alpha })}\).

*l*intervals.

The problem of decomposing a single inversion set can be solved algorithmically for types *B* and *C* as well and, furthermore, one can also discuss the decomposition of a given inversion set into the disjoint union of a fixed number of inversion sets. These descriptions are analogous to the one given above and we omit them here.

## 8 Parameterizing regular codimension *n* faces of the Littlewood–Richardson cone

The Horn conjecture (now a theorem due to Klyachko and Knutson and Tao) describes the Littlewood–Richardson cone in terms of its supporting hyperplanes. The article [6] is a good reference for the history and solution of this conjecture. One may also seek to describe the cone in terms of its generating rays. In this section we explain how our work allows us to find the extremal rays lying on the smallest regular faces of the cone. In addition, this method proves that these faces are simplicial. For clarity of exposition we discuss only the case of type *A* but everything carries over to the cases of types *B* and *C*.

*Regular faces of the Littlewood–Richardson cone*To describe how our results relate to the Littlewood–Richardson cone we first convert the problem of eigenvalues of Hermitian matrices to its symmetric version as in Sect. 1.2. I.e., instead of Hermitian matrices

*A*,

*B*,

*C*satisfying \(C = A+B\) we will consider Hermitian matrices

*A*,

*B*,

*C*satisfying \(A+B+C = 0\). It is clear that the cone \(\mathcal {C''}\) of such triples is contained in the hyperplane

*V*defined by the trace condition

*V*to

*V*/

*W*. The natural coordinates in

*V*/

*W*are \(\lambda = (a_1, \ldots , a_{n-1})\), \(\mu = (b_1, \ldots , b_{n-1})\), and \(\nu = (c_1, \ldots , c_{n-1})\), where \(a_i = \lambda _i - \lambda _{i+1}\), \(b_i = \mu _i - \mu _{i+1}\), and \(c_i = \nu _i - \nu _{i+1}\) for \(1 \leqslant i \leqslant n-1\). Clearly \(V/W \cong ({\mathbb R}^{n-1})^3\) and \(S_n\) acts naturally on each of the components of \(({\mathbb R}^{n-1})^3\): we fix the natural basis \(\{{e}_i - {e}_{i+1} \, | \, 1 \leqslant i \leqslant n-1\}\) of \({\mathbb R}^{n-1}\) and the action of \(S_n\) is by permuting the indices of this basis. The cone \(\mathcal {C}\) is a pointed polyhedral cone of full dimension. Each of the coordinate hyperplanes \(a_i = 0\), \(b_i=0\), and \(c_i = 0\) for a fixed

*i*with \(1 \leqslant i \leqslant n-1\) is a facet of \(\mathcal {C}\). Let \({\mathbb R}^{3n-3}_{\geqslant 0}\subset ({\mathbb R}^{n-1})^3\) denote the dominant cone defined by \(a_i \geqslant 0\), \(b_i \geqslant 0\), \(c_i \geqslant 0\) for all \(1 \leqslant i \leqslant n-1\). A face of \(\mathcal {C}\) is called

*regular*if it intersects the interior of \({\mathbb R}^{3n-3}_{\geqslant 0}\). Ressayre proved that the regular faces of \(\mathcal {C}\) have codimension at most \(n-1\). Furthermore, the faces of codimension \(n-1\) are exactly the intersection of \({\mathbb R}^{3n-3}_{\geqslant 0}\) with the codimension \(n-1\) subspaces \(T_{\alpha _1, \alpha _2, \alpha _3}\) defined by

*v*-

*pivot variable*and \(b_k, \ldots , b_{l-1}\), \(c_p, \ldots , c_{q-1}\)

*v*-

*free variables*in this case.

### Proposition 8.1

### Proof

Let \(\alpha _i = \sigma _i[\beta _{i1}, \beta _{i2}, \ldots , \beta _{im}]\), \(i=1,2,3\), be the expression of the \(\alpha _i\) in terms of inflations guaranteed by Theorem 1.11, and let \(\{1,2, \ldots , n\} = U_1 \sqcup U_2 \sqcup \ldots \sqcup U_m\) be the corresponding decomposition into intervals. Assume \(v = {e}_i - {e}_j \in S_{\alpha _1,\alpha _2, \alpha _3}\). Define the *level of**v* inductively as follows: if *i* and *j* belong to different parts of \(\{1,\ldots , n\}\), then the level of *v* is one; otherwise, \(i, j \in U_k\) and the level of *v* is one plus the level of *v* for the decomposition \(\Delta _{z_k}^+ = \Phi (\beta _{1k}) \sqcup \Phi (\beta _{2k}) \sqcup \Phi (\beta _{3k})\). Consider the projection \(\{1,2,\ldots , n\} \rightarrow \{1,2, \ldots , m\}\) induced by the decomposition into intervals. Under this projection the level one elements of \(S_{\alpha _1, \alpha _2, \alpha _3}\) are sent bijectively to the elements of \(S_{\sigma _1, \sigma _2, \sigma _3}\), and these latter elements form a basis of \({\mathbb R}^{m-1}\) since, by Theorem 1.11, \((\sigma _1,\sigma _2,\sigma _3)=(\sigma _1,J\sigma _1,I)\), and for such a triple the statement about being a basis is easily checked. The elements of level greater than one are sent to zero under this map. On the other hand, by induction, the elements of level greater than one form a basis of the subspace generated by \(\{{e}_i - {e}_j\, | \, i, j { \text{ in } \text{ the } \text{ same } } U_k\}\). Combining the above we conclude that \(S_{\alpha _1, \alpha _2, \alpha _3}\) is a basis of \({\mathbb R}^{n-1}\).

To prove the second assertion, we order \(S_{\alpha _1, \alpha _2, \alpha _3}\) linearly so that elements of lower level come before elements of higher level. Notice first that if \(v_1\) is of level one and \(v_2\) is of level greater than one, than no \(v_1\)-pivot variable is \(v_2\)-free. Now assume that both \(v_1\) and \(v_2\) are of level one. Passing to the projection as above, we conclude again that no \(v_1\)-pivot variable is \(v_2\)-free. \(\square \)

We call the \(v_i\)-pivot variables simply *pivot variables* of \(C_{\alpha _1,\alpha _2,\alpha _3}\) and the rest of \(a_i, b_i, c_i\) we call *free variables*.

### Corollary 8.2

\(C_{\alpha _1,\alpha _2,\alpha _3}\) is a simplicial cone.

### Proof

It follows from Proposition 8.1 that there are exactly \(n-1\) pivot variables. Furthermore, by ordering them as above we can start from the bottom and replace any pivot variable appearing in the expression of another pivot variable by its expression. When we reach the top equation, every pivot variable will have become expressed with non-negative coefficients in terms of the free variables only. \(\square \)

### Example 8.3

## Footnotes

- 1.
In keeping with standard usage in algebra, a better term for these kinds of permutations would be

*indecomposable*. However, the term “indecomposable permutation” is already established in the literature for a different class of permutations. Moreover, following [2], we use the terms “plus- or minus-indecomposibles” in relation to the inflation procedure in Sect. 1.3, and wish to avoid a conflict of terminology. - 2.
We warn the reader that some authors use the terminology

*connected*rather than simple. - 3.
We thank Lukas Katthän for asking us this question after a previous version of this paper appeared on ArXiv, see [9].

- 4.
We choose to list the ordered decompositions of \({\Phi ({\alpha })}\) to simplify the formula for counting them.

## Notes

### Acknowledgements

The authors thank the referee who suggested a number of improvements in the exposition. This work was partially supported by NSERC. In particular, most of it was done with the support of NSERC’s Undergraduate Summer Research Awards program.

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