Decomposing Inversion Sets of Permutations and Applications to Faces of the Littlewood-Richardson...

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DECOMPOSING INVERSION SETS OF PERMUTATIONS AND

APPLICATIONS TO FACES OF THE LITTLEWOOD-RICHARDSON

CONE

R. DEWJI, I. DIMITROV, A. MCCABE, M. ROTH, D. WEHLAU, AND J. WILSON

Abstract. If σ ∈ Sn is a permutation of {1, 2, . . . , n}, the inversion set of σ is Φ(σ) ={(i, j) | 1 ≤ i < j ≤ n, σ(i) > σ(j)}. We describe all r-tuples σ1, σ2, . . . , σr ∈ Sn such that∆+

n = {(i, j) | 1 ≤ i < j ≤ n} is the disjoint union of Φ(σ1),Φ(σ2), . . . ,Φ(σr). Using thisdescription we prove that certain faces of the Littlewood-Richardson cone are simplicialand provide an algorithm for writing down their sets of generating rays. We also considerand solve the analogous problem for the Weyl groups of root systems of type B and C

and provide some enumerative results.Keywords: Inversion set, Simple permutation, Littlewood-Richardson cone, Catalan num-bers.

1. Introduction

Let n be a positive integer and consider a root system ∆n. Fix a set of positive roots ∆+n

so that ∆n = ∆+n ⊔∆−

n where ∆−n = −∆+

n . Let W denote the corresponding Weyl group.For σ ∈ W, the inversion set of σ, Φ(σ) is defined by Φ(σ) := {v ∈ ∆+

n | σ · v ∈ ∆−n }. We

are concerned with ways to express the positive roots as a disjoint union of inversion sets:∆+

n = Φ(σ1) ⊔ Φ(σ2) ⊔ · · · ⊔ Φ(σr) where σ1, σ2, . . . , σr ∈ W.We consider here only root systems of type A, B or C. For simplicity we describe the

problem and our results first for root systems of type A. The Weyl group of the root systemAn is W = Sn+1 the symmetric group on n + 1 letters. For ease of notation, we replacen + 1 by n and work with An−1 and Sn. Thus ∆n refers to the rank n − 1 root systemof type An−1. Root systems of type B and C have isomorphic Weyl groups and so yieldidentical answers to our questions. We describe our results for type B/C root systems in§4–5.

For σ ∈ Sn we write σ = (σ(1), σ(2), . . . , σ(n)). Idn denotes the identity permutation:Idn = (1, 2, . . . , n) and wo = wo[n] = (n, n − 1, . . . , 1) ∈ Sn. Often we will write wo todenote wo[n] when the value of n is clear from the context. We use the symbol ⊔ to denotea disjoint union.

Date: October 27, 2011.2010 Mathematics Subject Classification. 05E15; 05A05, 05E10, 52B20.

1

http://arxiv.org/abs/1110.5880v1

2 DEWJI, DIMITROV, MCCABE, ROTH, WEHLAU, AND WILSON

For the root system of type An−1 we may take ∆+n := {(i, j) | 1 ≤ i < j ≤ n} for

the positive roots. The element (1, n) is the highest root. The elements (i, i + 1) with1 ≤ i ≤ n− 1 are the simple roots.

Often we will use ∆+ to denote ∆+n when the value of n is clear from the context. Given

σ ∈ Sn, the inversion set of σ is the set Φ(σ) := {(i, j) ∈ ∆+n | σ(i) > σ(j)}. Note that

Φ(Id) = ∅ and Φ(wo[n]) = ∆+n .

It is not hard to see that the element σ ∈ Sn is determined by its inversion set Φ(σ).Thus there are exactly n! inversion sets contained in ∆+

n .

Definition 1.1. A decomposition of an inversion set Φ(σ) is a set of disjoint inversion setsΦ(σ1),Φ(σ2), . . . ,Φ(σr) such that

Φ(σ) = Φ(σ1) ⊔ Φ(σ2) ⊔ · · · ⊔ Φ(σr) .

The decomposition is called trivial if Φ(σ) = Φ(σa) for some a with 1 ≤ a ≤ r.

We say that an element σ ∈ Sn (and its inversion set Φ(σ)) is reducible if there existsa non-trivial decomposition of Φ(σ). Otherwise we say that σ (and Φ(σ)) is irreducible.Note that Φ(Id) = ∅ may occur in a decomposition.

Solving the following problem was the motivation for this article.

Problem 1.2. Identify all decompositions of ∆+n :

∆+n = Φ(σ1) ⊔ Φ(σ2) ⊔ · · · ⊔Φ(σr) .

Here the ordering of the σi is not relevant.

We are interested in this problem because of its relation to two problems: (i) determiningthe regular codimension n faces of the Littlewood-Richardson cone; and (ii) studying thecup product of the cohomology of line bundles on homogeneous varieties. We brieflydescribe the relation to the Littlewood-Richardson cone in the next paragraph. Moredetails and the relation of Problem 1.2 to the cup product of line bundles appear in §7.

If A is a Hermitian matrix, denote by λ = (λ1 ≥ λ2 ≥ . . . ≥ λn) ∈ Rn its eigenvalues and

let R3n+ = {(λ, µ, ν) |λi ≥ λi+1, µi ≥ µi+1, νi ≥ νi+1 for 1 ≤ i ≤ n − 1}. In 1912 H. Weyl

posed the following question: For which triples (λ, µ, ν) ∈ R3n+ do there exist Hermitian

matrices A,B,C such that C = A + B and whose eigenvalues are λ, µ, ν respectively. In1962 A. Horn proved that the set of such triples is a polyhedral cone C′ and conjecturedinequalities determining C′. Horn’s conjecture was proved in the 1990’s by Klyachko andKnutson and Tao, see [F] for a nice exposition on Horn’s conjecture. It is worth mentioningthat the lattice points of C′ are exactly the triples (λ, µ, ν) for which the correspondingLittlewood-Richardson coefficient cνλ,µ is nonzero. N. Ressayre described all regular faces

of C′, i.e. faces that intersect the interior of R3n+ . In particular (after symmetrizing the

problem, as described in §7), the regular faces of codimension n are exactly the intersectionof R3n

+ with the subspaces defined by

w−11 λ+ w−1

2 µ+ w−13 ν = 0

DECOMPOSING INVERSION SETS AND THE LITTLEWOOD-RICHARDSON CONE 3

for w1, w2, w3 ∈ Sn with the property that ∆+n = Φ(w1) ⊔ Φ(w2) ⊔ Φ(w3).

Definition 1.3. Let σ ∈ Sn. An interval (of size t) is a set of consecutive integers {i, i +1, i + 2, . . . , i + t− 1}. A block (of size t) of the permutation σ is an interval {i, i + 1, i +2, . . . , i+ t−1} of size t such that the set {σ(i), σ(i+1), . . . , σ(i+ t−1)} is also an interval(of size t). Every permutation in Sn has n blocks of size 1 and a block of size n. If σ ∈ Sn

has no blocks of size t for all 1 < t < n then we say that σ is simple1.

Example 1.4. The permutation σ = (9, 7, 1, 5, 3, 4, 6, 8, 2) ∈ S9 has a block of size 8 cor-responding to the interval {2, 3, . . . 9} and a block size 4 corresponding to the interval{4, 5, 6, 7}. The permutation τ = (5, 2, 6, 1, 4, 7, 3) ∈ S6 has no non-trivial blocks and so issimple.

A block of size t + 1 for the permutation σ corresponds to a t × t closed square in[1, n] × [1, n] which contains t+ 1 points of the the graph of σ. Hence σ is simple if theredoes not exist a t × t closed square in containing t + 1 points of the graph of σ with2 ≤ t ≤ n− 1.

To state our results we need to introduce an inflation procedure to describe permutationsinductively. We describe this procedure heuristically as follows. We consider a permutationon n letters as a shuffling of a deck of n cards. We shuffle as follows. First cut thedeck into m piles of sizes z1, z2, . . . , zm respectively. Shuffle each of these piles accordingto a permutation σa ∈ Sza . Finally reassemble the piles in an order determined by apermutation σ0 ∈ Sm. The resulting permutation in Sn is denoted by σ0[σ1, σ2, . . . , σm]and is called an inflation of σ0.

Note that a permutation α ∈ Sn is simple if and only if α cannot be expressed as aninflation α = σ[β1, β2, . . . , βr] with 2 ≤ r ≤ n− 1.

Definition 1.5. A permutation σ ∈ Sn is called plus-decomposable if σ may be writtenin the form σ = Id2[α, β]. Otherwise σ is plus-indecomposable. Similarly, σ ∈ Sn is calledminus-decomposable if σ may be written in the form σ = wo[2][α, β]. Otherwise σ isminus-indecomposable.

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

Theorem 1.6 ([AAK][Theorem 1]). Let n ≥ 2. For every permutation α ∈ Sn there existsa simple permutation σ ∈ Sm and permutations β1, β2, . . . , βm such that α = σ[β1, β2, . . . , βm].Moreover if σ 6= Id2 and σ 6= wo[2] then β1, β2, . . . , βm and σ are unique. If σ = Id2 thenβ1, β2 and σ are unique if we add the additional condition that β1 is plus-indecomposable.Similarly, if σ = wo[2] then β1, β2 and σ are unique if we add the additional condition thatβ1 is minus-indecomposable. �

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

1We warn the reader that some authors use the terminology connected rather than simple.


Theorem 1.7. Let n ≥ 2. For every permutation α ∈ Sn there exists a permutationσ ∈ Sm and permutations β1, β2, . . . , βm such that α = σ[β1, β2, . . . , βm] where either σis simple and m ≥ 4 or σ = Idm or σ = wo[m]. Furthermore this expression for α isunique if we require that m be maximal when σ = Idm or σ = wo[m], i.e., that each βa isplus-indecomposable when σ = Id and each βa is minus-indecomposable when σ = wo. �

Definition 1.8. We say that α is expressed in simple form when we write α = σ[β1, β2, . . . , βm]in the form guaranteed by Theorem 1.7, i.e, when σ is simple with m ≥ 4 or σ = wo[m] orIdm with m maximal.

We now state our main result.

Theorem 1.9. Let ∆+n = Φ(w1)⊔Φ(w2)⊔· · ·⊔Φ(wr) be a decomposition with all Φ(wa) 6= ∅.

Then (after possibly reordering) we have

w1 = σ1[β11, β12, . . . , β1m]

w2 = σ2[β21, β22, . . . , β2m]

...

wr = σr[βr1, βr2, . . . , βrm]

where

(†) Φ(β1b) ⊔ Φ(β2b) ⊔ · · · ⊔ Φ(βrb) = ∆+zb

is a decomposition ∀b = 1, 2, . . . ,m

and σq = σq+1 = · · · = σr = Idm with{

σ1 = wo[m] and q=2, if w1 is minus-decomposable;

σ1 is simple, σ2 = woσ1 and q=3, if w1 is minus-indecomposable.

Moreover, the above decomposition of ∆+n is irreducible if and only if the decompositions

(†) are irreducible for all b, exactly one of βa1, βa2, . . . , βam is not equal to Id for eacha = q, q + 1, . . . , r, and

{

m = 2 and β11 = Idz1 and β12 = Idz2 , if w1 is minus-decomposable;

β1b = β2b = Idzb for all b = 1, 2, . . . ,m, if w1 is minus-indecomposable.

Example 1.10. Let n = 8 and let w1 = (5, 3, 4, 8, 1, 2, 6, 7), w2 = (4, 5, 6, 1, 7, 8, 3, 2), w3 =(1, 3, 2, 4, 6, 5, 7, 8). Then ∆+

8 = Φ(w1) ⊔ Φ(w2) ⊔ Φ(w3), m = 4, and

w1 = (2, 4, 1, 3)[(3, 1, 2), (1), (1, 2), (1, 2)]w2 = (3, 1, 4, 2)[(1, 2, 3), (1), (1, 2), (2, 1)]w3 = (1, 2, 3, 4)[(1, 3, 2), (1), (2, 1), (1, 2)].

Also β11 = (21)[(1), (12)] = (2, 3, 1), β21 = (12)[(1), (12)] = Id3, β31 = (12)[(1), (21)] =(1, 3, 2) and all other βij = Id. Note that ∆+

3 = Φ(β11)⊔Φ(β21)⊔Φ(β31) = Φ(β11)⊔Φ(β31).


The recursive form of this theorem allows us to inductively solve many problems concern-ing decompositions. For example, in §6 we exploit this recursiveness to obtain a number ofresults enumerating various solutions to the main problem. In §7 we use the form to provea result about the decompositions which yields an algorithm producing all generating rayson a given regular codimension n face of the Littlewood-Richardson cone.

2. Preliminaries

It is easy to see that an inversion set Φ must satisfy the following two conditions:

(i) If (i, j), (j, k) ∈ Φ then (i, k) ∈ Φ. (closed condition)(ii) If (i, j), (j, k) /∈ Φ then (i, k) /∈ Φ. (co-closed condition)

In [K, Proposition 5.10] it is shown that any subset Φ ⊆ ∆+n which satisfies the above two

conditions is an inversion set. Thus Φ ⊂ ∆+n is an inversion set if and only if both Φ and

Φc satisfy the closed condition.The graph of a permutation σ is the set of n lattice points {(i, σ(i)) | i = 1, 2, . . . , n}

considered as a subset of [1, n]× [1, n] ⊂ R2. There are

(

n2

)

line segments joining points ofthe graph of σ. Each such line segment corresponds to an element of ∆+

n and those withnegative slope correspond to the elements Φ(σ), i.e., (i, j) ∈ Φ(σ) if and only if the linesegment joining the points (i, σ(i)) and (j, σ(j)) has negative slope.

We have already noted that Φ(wo) = ∆+. The following lemma gives another indicationof the importance of wo.

Lemma 2.1. Let σ ∈ Sn. Then ∆+n = Φ(σ) ⊔ Φ(woσ).

Proof. The graph of woσ is obtained from the graph of σ by reflecting in the line y = n/2.Using the characterization of Φ(σ) as those positive roots whose corresponding line segmenthas negative slope completes the proof of the lemma. �

Proposition 2.2. (1) Suppose ∆+n = Φ(σ1)⊔Φ(σ2)⊔ . . .Φ(σr) is a decomposition and

let A be any subset of {1, 2, . . . , r}. Then ∃σ ∈ Sn such that Φ(σ) = ⊔a∈AΦ(σa).(2) Every non-empty inversion set Φ(σ) contains at least one simple root.

Proof. Clearly it suffices to prove the first assertion for doubleton sets A = {p, q}. Thusit suffices to show that Φ(σp) ⊔ Φ(σq) is both closed and co-closed. For ease of notation,we will assume A = {1, 2}. First we show that Φ(σ1) ⊔ Φ(σ2) is co-closed. Suppose that(i, j), (j, k) /∈ Φ(σ1) ⊔ Φ(σ2). Then for b = 1, 2 we have (i, k) /∈ Φ(σb) since Φ(σb) isco-closed. Thus (i, k) /∈ Φ(σ1) ⊔Φ(σ2) which shows Φ(σ1) ⊔ Φ(σ2) is co-closed.

To see that Φ(σ1)⊔Φ(σ2) is closed, suppose that (i, j), (j, k) ∈ Φ(σ1)⊔Φ(σ2). Then forb = 3, 4, . . . , r we have (i, j), (j, k) /∈ Φ(σb) and thus (i, k) /∈ Φ(σb) since Φ(σb) is co-closed.Hence (i, k) /∈ ⊔r

b=3Φ(σb) which implies that (i, k) ∈ Φ(σ1) ⊔ Φ(σ2). This shows that thatΦ(σ1) ⊔ Φ(σ2) is closed and completes the proof of the first assertion.

The second assertion follows easily from the fact that if σ(i) < σ(i + 1) for all i =1, 2, . . . , n− 1 then σ = Idn. �


Note that the hypothesis that ∆+n = Φ(σ1)⊔Φ(σ2)⊔ . . .Φ(σr) is necessary in the above

proposition; arbitrary unions of inversion sets need not be inversion sets. For example,consider n = 3, σ1 = (2, 1, 3) and σ2 = (1, 3, 2). Then Φ(σ1) = {(1, 2)}, Φ(σ2) = {(2, 3)}and Φ(σ1) ⊔ Φ(σ2) is not closed and so is not an inversion set.

The first assertion of Proposition 2.2 implies that if Φ(σ) = Φ(σ1) ⊔Φ(σ2) ⊔ · · · ⊔Φ(σr)and A is any subset of {1, 2, . . . , r} then ∃α ∈ Sn with Φ(α) = ⊔a∈AΦ(σa). This factfollows easily from the first assertion of the above proposition and the fact that ∆+

n =Φ(woσ) ⊔ Φ(σ) = Φ(woσ) ⊔ Φ(σ1) ⊔ Φ(σ2) ⊔ · · · ⊔ Φ(σr). In particular, an element σ ∈ Sn

is reducible if and only if there exist non-identity elements α1, α2 ∈ Sn with Φ(σ) =Φ(α1)⊔Φ(α2). Note that we may also write this last equation as the decomposition ∆+

n =Φ(woσ)⊔Φ(α1)⊔Φ(α2). These considerations show that in our study of decompositions of∆+ we may focus our attention on irreducible decompositions, i.e., decompositions ∆+

n =Φ(σ1) ⊔ Φ(σ2) ⊔ · · · ⊔ Φ(σr) where σa is irreducible for a = 1, 2, . . . , r.

Definition 2.3. Two sequences x1, x2, . . . , xn and y1, y2, . . . , yn each comprised of n dis-tinct real numbers are order isomorphic if xi > xj if and only if yi > yj.

Let σ ∈ Sn and suppose F is some subset of {1, 2, . . . , n} with m = |F|. WriteF = {i1, i2, . . . , im} where i1 < i2 < · · · < im. Restricting σ to F yields a sequenceσ(i1), σ(i2), . . . , σ(im) which is order isomorphic to the sequence µ(1), µ(2), . . . , µ(m) cor-responding to a unique element µ ∈ Sm. We denote this element µ by µ = θF (σ). IfF ⊂ {1, 2, . . . , n} we write ∆+

Fto denote the set ∆+

F:= {(i, j) ∈ ∆+

n | i, j ∈ F}.

3. Inflation Procedure

Now we want to describe the inflation procedure for permutations. For a history of theinflation procedure and a discussion of a number of applications we refer the reader to thesurvey article of Brignall [B].

The following definition will be useful. Write {1, 2, . . . , n} = I1⊔I2⊔· · ·⊔Im as a disjointunion of ordered intervals. A subset F ⊂ {1, 2, . . . , n} is admissible if |F ∩ Ia| = 1 for alla = 1, 2, . . . ,m.

In addition to the heuristic description given in §1, we may describe the inflation proce-dure as follows. Write {1, 2, . . . , n} as an ordered disjoint union of intervals: {1, 2, . . . , n} =I1 ⊔ I2 ⊔ · · · ⊔ Im where each Ia is an interval and Ia < Ib if a < b, i.e., if a < b andia ∈ Ia, ib ∈ Ib then ia < ib. Put za = |Ia|. Take σ0 ∈ Sm and σa ∈ Sza for a = 1, 2, . . . ,m.Then σ := σ0[σ1, σ2, . . . , σm] is characterized by the following two conditions.

(1) θF (σ) = σ0 for every admissible F .(2) θIa(σ) = σa for all a = 1, 2, . . . ,m.

The proof of the following lemma is straightforward and is left to the reader.

Lemma 3.1. Suppose α = α0[β1, β2, . . . , βm] where βa ∈ Sza for a = 1, 2, . . . ,m. DefineI1 := {1, 2, . . . , z1}, I2 := {z1 + 1, z1 + 2, . . . , z1 + z2}, . . . , Im := {z1 + z2 + · · · + zm−1 +


1, . . . , n}. Let Ψa denote the order preserving bijection Ψa : Ia → {1, 2, . . . .za}. Then

Φ(α) = {(ia, ib) | ia ∈ Ia, ib ∈ Ib, (a, b) ∈ Φ(α0)} ⊔(

⊔ma=1Ψ

−1a (Φ(βa))

)

.

Lemma 3.2. Let σ0 ∈ Sm and let z1, z2, . . . , zm be positive integers. Put n := z1 + z2 +· · ·+ zm. The permutation σ := σ0[Idz1 , Idz2 , . . . , Idzm ] ∈ Sn is irreducible if and only if σ0is irreducible.

Proof. Let {1, 2, . . . , n} = I1⊔I2⊔· · ·⊔Im be the decomposition into intervals correspondingto the the inflation σ := σ0[Idz1 , Idz2 , . . . , Idzm ]. By Lemma 3.1, we have Φ(σ) = {(i, j) ∈∆+

n | i ∈ Ia, j ∈ Ib, (a, b) ∈ Φ(σ0)}.First suppose that σ0 is reducible and write Φ(σ0) = Φ(α0) ⊔ Φ(β0). Then Φ(σ) =

Φ(α)⊔Φ(β) where α := α0[Idz1 , Idz2 , . . . , Idzn ] and β := β0[Idz1 , Idz2 , . . . , Idzn ]. Thus if σ0is irreducible then so is σ.

Next suppose that σ is reducible and write Φ(σ) = Φ(α)⊔Φ(β) where α, β ∈ Sn. Considera positive root (p, q) ∈ ∆+

n . Define a and b by p ∈ Ia and q ∈ Ib. If a = b then (p, q) /∈ Φ(σ).Suppose then that a < b. Again if (a, b) /∈ Φ(σ0) then (p, q) /∈ Φ(σ). Thus we suppose that(a, b) ∈ Φ(σ). Write Ia = {u, u + 1, . . . , u + za − 1} and Ib = {v, v + 1, . . . , v + zn − 1}.Then (u + za − 1, v) ∈ Φ(σ) and thus, without loss of generality, (u + za − 1, v) ∈ Φ(β).Then α(u+ za − 1) < α(v). This implies that (Ia, Ib) ∩ Φ(α) = ∅ where (Ia, Ib) := {(i, j) |i ∈ Ia, j ∈ Ib}. Since (Ia, Ib) ⊂ Φ(σ), this implies that (Ia, Ib) ⊂ Φ(β). This shows thatΦ(β) = {(i, j) | i ∈ Ic, j ∈ Id, (c, d) ∈ T} = ⊔(c,d)∈T (Ic, Id) for some set T ⊂ ∆+

m. Moreover,the fact that Φ(β) is both closed and co-closed implies that the set T is also both closed andco-closed and so T = Φ(β0) for some non-identity permutation β0 ∈ Sm. This shows thatβ = β0[Idz1 , Idz2 , . . . , Idzm]. Similarly α = α0[Idz1 , Idz2 , . . . , Idzm ] for some non-identityα0 ∈ Sm. Thus Φ(σ0) = Φ(α0) ⊔ Φ(β0) is reducible. �

The following proposition follows easily from the two preceding lemmas.

Proposition 3.3. Suppose α = α0[β1, β2, . . . , βm] where βa ∈ Sza for a = 1, 2, . . . ,m.

Let A ⊂ {1, 2, . . . ,m} and define γa =

{

βa, if a ∈ A;

Idza , if a /∈ Aand γ′a =

{

Idza , if a ∈ A;

βa, if a /∈ A.

Define σ0 = α0[Idz1 , Idz2 , . . . , Idzn ], σ1 = Id[γ1, γ2, . . . , γm] and σ2 = Id[γ′1, γ′2, . . . , γ

′m]

Then Φ(γ) = Φ(σ0) ⊔ Φ(σ1) ⊔ Φ(σ2). Therefore, if α 6= Id is irreducible, then exactly oneof the permutations α0, β1, β2, . . . , βm is a non-identity permutation. In particular, if α isirreducible with α0 6= Id then α = α0[Id, Id, . . . .Id] where α0 is irreducible. �

Lemma 3.4. Let σ ∈ Sn. The permutation σ is simple if and only if woσ is simple.

Proof. This result follows immediately from the fact that that the graph of woσ is obtainedfrom the graph of σ by reflecting in the line y = n/2. �

Definition 3.5. Let σ ∈ Sn, and F ,F ′ ⊂ {1, 2, . . . , n}. If Φ(σ) ∩∆+F∩∆+

F ′ 6= ∅ then wesay that F and F ′ are σ-connected. Let F and F ′ be subsets of {1, 2, . . . , n}. If there exist


subsets F1,F2, . . . ,Ft ⊂ {1, 2, . . . , n} such that F = F1, F′ = Ft and Fa is σ-connected to

Fa+1 for a = 1, 2, . . . , t− 1 then we say that F and F ′ are σ-path-connected.

We will make use of the following lemma a number of times.

Lemma 3.6. Let σ ∈ Sn and F ,F ′ ⊂ {1, 2, . . . , n}. Write µ = θF(σ) and µ′ = θF ′(θ).Suppose that Φ(σ) = Φ(σ1) ⊔ Φ(σ2) ⊔ · · · ⊔Φ(σr).

(1) If µ is irreducible then there exists δ(F) with 1 ≤ δ(F) ≤ n such that Φ(σ)∩∆+F ⊂

Φ(σδ(F)).(2) If µ and µ′ are both irreducible and F and F ′ are σ-path connected then δ(F) =

δ(F ′).

Proof. Let |F| = m and suppose that µ is irreducible. Put µa = θF (σa) for a = 1, 2, . . . , r.There exists an order preserving bijection Ψ : F → {1, 2, . . . ,m}. It is easy to see that(i, j) ∈ Φ(α) ∩∆+

Fif and only if (Ψ(i),Ψ(j)) ∈ Φ(θF (α)) for all α ∈ Sn. Thus Ψ identifies

Φ(α) ∩∆+F

with Φ(θF (α)). Intersecting Φ(σ) = Φ(σ1) ⊔ Φ(σ2) ⊔ · · · ⊔ Φ(σr) with ∆+F

andusing this identification we get Φ(µ) = Φ(µ1) ⊔Φ(µ2) ⊔ · · · ⊔Φ(µr). Since µ is irreducible,there exists δ(F) such that Φ(µ) ⊂ Φ(µδ(F)). Therefore Φ(σ) ∩∆+

F ⊂ Φ(σδ(F)).For the second assertion, suppose that µ and µ′ are irreducible. Clearly it suffices to

consider the case where F and F ′ are σ-connected. By the above, Φ(σ) ∩∆+F⊂ Φ(σδ(F))

and Φ(σ) ∩ ∆+F ′ ⊂ Φ(σδ(F ′)). But F and F ′ are σ-connected implies that there exists

(i, j) ∈ Φ(σ) ∩∆+F ∩∆+

F ′ . Thus (i, j) ∈ Φ(σδ(F)) ∩ Φ(σδ(F ′)). Hence δ(F) = δ(F ′). �

Corollary 3.7. Suppose F1,F2, . . . ,Fs ⊂ {1, 2, . . . , n} where Fi and Fj are σ-path-connectedfor all 1 ≤ i, j ≤ s. Further suppose that θFi

(σ) is irreducible for all i = 1, 2, . . . , s andΦ(σ) ⊂ ∪s

i=1∆+Fi. Then σ is irreducible.

Proof. Suppose that Φ(σ) = Φ(σ1)⊔Φ(σ2)⊔· · ·⊔Φ(σr). By the lemma, we have j = δ(F1) =δ(F2) = · · · = δ(Fr) with Φ(θ)∩∆+

Fi⊂ Φ(σj). Therefore Φ(σ) = ⊔r

i=1(Φ(σ)∩∆+Fi) ⊂ Φ(σj)

and thus the decomposition of Φ(σ) is trivial. �

Remark 3.8. Suppose σ = σ0[σ1, σ2, . . . , σm] where σ0 is irreducible. Further suppose thatthe sets F1,F2, . . . ,Ft are all admissible (with respect to σ). If the Fi are all σ-pathconnected and Φ(σ) ⊂ ∪s

i=1∆+Fi

then the above corollary applies and shows that σ isirreducible.

Definition 3.9. A permutation σ ∈ Sn is atomic if σ(i + 1) 6= σ(i) + 1 for all i =1, 2, . . . , n− 1.

Definition 3.10. Let n = 2m be even with n ≥ 4. A permutation σ is exceptional if σ isone of the following permutations

(1) σ = (2, 4, 6, . . . , n− 2, n, 1, 3, 5, . . . , n− 3, n − 1),(2) σ = (m+ 1, 1,m + 2, 2,m + 3, 3, . . . , 2m− 1,m− 1, 2m,m),(3) σ = (n− 1, n − 3, n− 5, . . . , 3, 1, n, n − 2, n − 4, . . . , 4, 2),


(4) σ = (m, 2m,m− 1, 2m − 1,m− 2, 2m − 2, . . . , 2,m+ 2, 1,m + 1).

Lemma 3.11. Let σ ∈ Sn be exceptional. Then σ is atomic and irreducible.

Proof. It is easily seen that all of these permutations are atomic.(1) Suppose σ = (2, 4, 6, . . . , n − 2, n, 1, 3, 5, . . . , n − 3, n − 1). Then Φ(σ) has only one

simple root, (m,m+ 1) and so is irreducible.For the remaining cases, we proceed by induction and use Corollary 3.7 repeatedly. If

n = 4, it is easy to check that the only two exceptional permutations (2,4,1,3) and (3,1,4,2)are irreducible. To prove the induction step, let n ≥ 6.

(2) Suppose σ = (m + 1, 1,m + 2, 2,m + 3, 3, . . . , 2m − 1,m − 1, 2m,m). Let F ′ ={1, 2, . . . , 2m − 2} and F ′ := {3, 4, . . . , 2m}. We let µ = θF (σ) = θF ′(σ). Then µ =(m + 1, 1,m + 2, 2,m + 3, 3, . . . , 2m − 1,m − 1) is again exceptional and so is irreducibleby the induction hypothesis. Furthermore F and F ′ are σ-connected since (3, 4) ∈ Φ(σ)with 3, 4 ∈ F ∩ F ′. Now Φ(σ) ∩ (∆+

F ∪ ∆+F ′) = Φ(σ) \ {(2, 2m)}. If we take F ′′ =

{2, 3, 2m − 1, 2m} then θF ′′(σ) = (3, 1, 2, 4) is irreducible with (2, 2m) ∈ Φ(σ) ∩∆+F ′′ and

(2, 3) ∈ Φ(σ) ∩∆+F ∩∆+

F ′′ . Hence Corollary 3.7 implies that σ is irreducible.(3) Suppose σ = (n − 1, n − 3, n − 5, . . . , 3, 1, n, n − 2, n − 4, . . . , 4, 2). Take F =

{1, 2, . . . , n} \ {1,m + 1} and F ′ = {1, 2, . . . , n} \ {m, 2m}. Then θF (σ) = θF ′(σ) =σ = (n− 3, n− 5, . . . , 3, 1, n− 2, n− 4, . . . , 4, 2) is again exceptional and so is irreducible bythe induction hypothesis. Also the root (2, 2m− 1) shows that F and F ′ are σ-connected.Then Φ(σ)∩(∆+

F∪∆+F ′) = Φ(σ)\{(1,m), (1, 2m), (m+1, 2m)}. For F ′′ = {1,m,m+1, 2m}

we have θF ′′(σ) = (3, 1, 4, 2) which is irreducible with Φ(σ)∩∆+F ′′ = {(1,m), (1, 2m), (m +

1, 2m)}. Finally if we take F ′′′ = {2,m,m + 1, 2m} we have θF ′′′(θ) = (3, 1, 4, 2) and F ′′′

is σ-connected to both F and F ′′ since (2,m), (m + 1, 2m) ∈ Φ(σ). Thus σ is irreducible.(4) Finally suppose that σ = (m, 2m,m−1, 2m−1,m−2, 2m−2, . . . , 2,m+2, 1,m+1).

It is straightforward to verify that σ = (3, 6, 2, 5, 1, 4) is irreducible. Thus we suppose2m ≥ 8. Let F = {1, 2, . . . , 2m} \ {2, 2m − 1}, F ′ = {1, 2, . . . , 2m} \ {m, 2m} and F ′′ ={1, 2, . . . , 2m} \ {1,m+ 1}. Then θF (σ) = θF ′(σ) = θF ′(σ) = (m− 1, 2m− 2,m− 2, 2m−3, . . . , 2,m+1, 1,m) is exceptional and so is irreducible by induction. Also F and F ′ are σ-connected by (1, 3) and F and F ′′ are σ-connected by (2m−2, 2m). Since ∆+

F∪∆+

F ′∪∆+F ′′ =

∆+2m, Corollary 3.7 implies that σ is irreducible. �

Definition 3.12. Let σ ∈ Sn. Choose k with 1 ≤ k ≤ n and put F = {1, 2, . . . , n} \ {k}.The permutation σ◦ = θF (σ) ∈ Sn−1 is called a one point deletion of σ.

The following theorem, expressed in the language of posets, was first proved by Schmerland Trotter [ST]. For a proof using permutations we refer the reader to [AA][Theorem 5].

Theorem 3.13. Let n ≥ 2 and suppose σ ∈ Sn is simple but not exceptional. Then σ hasa one point deletion σ◦ which is simple. �

Proposition 3.14. Let σ ∈ Sn with n ≥ 4. The permutation σ is simple if and only if itis atomic and irreducible.


Proof. First suppose σ is atomic and irreducible and express σ in simple form: σ =σ0[β1, β2, . . . , βm] with βa ∈ Sza for a = 1, 2, . . . ,m. By Lemma 3.2, exactly one of the per-mutations σ0, β1, β2, . . . , βm is a non-identity permutation. If σ0 = Idm then the simplicityof σ implies that β1 6= Id. Similarly, if σ0 = Idm we cannot have βm = Id. Thus σ0 6= Id.Therefore σ = σ0[Idz1 , Idz2 , . . . , Idzm ]. Since σ is atomic, this implies that za = 1 for all a.But then σ = σ0 is simple.

Next we suppose that σ is simple. It follows immediately, from the definition that σ isatomic. We prove that simple implies irreducible by induction. For the base case, we haven = 4 and this implies that σ is exceptional. Thus σ is irreducible by Lemma 3.11.

We prove the induction step by contradiction. Suppose n ≥ 5 and that σ ∈ Sn issimple. If σ is exceptional then the result follows from Lemma 3.11. Thus we supposethat σ is not exceptional and assume by way of contradiction that Φ(σ) = Φ(σ1) ⊔ Φ(σ2).Let the simple permutation σ◦ ∈ Sn−1 be a one point deletion of σ. By the inductionhypothesis, σ◦ is irreducible. Suppose that σ◦ = θF (σ) where F = {1, 2, . . . , n} \ {k}and σ(k) = ℓ. Since σ◦ is irreducible we may assume without loss of generality that{(i, j) ∈ Φ(σ) | i 6= k, j 6= k} ⊆ Φ(σ1) . Assume, by way of contradiction, that Φ(σ2) 6= ∅.Write Φ(σ2) = {(i1, k), (i2, k), . . . , (ip, k)} ⊔ {(k, i′1), (k, i

′2), . . . , (k, i

′q)} and ja = σ(ia) for

a = 1, 2, . . . , p and j′a = σ(i′a) for a = 1, 2, . . . a.Define s = max{k − ia, ja − ℓ | 1 ≤ a ≤ p}. We consider two cases. For the first

case, suppose that s = k − ib for some b with 1 ≤ b ≤ p. If s = n − 1 then we musthave k = n and ib = 1. Since σ is simple, jb 6= 1 and ℓ 6= 1. Hence, i0 := σ−1(1)satisfies 1 < i0 < k = n. Taking F = {1, i0, n}, we find θF (σ) = (312). Then θ carriesthe decomposition Φ(σ) = Φ(σ1) ⊔ Φ(σ2) to a non-trivial decomposition of (312) since(1, i0) ∈ Φ(σ1) and (1, n) ∈ Φ(σ2). This contradiction shows s 6= n− 1.

The s× s square s := [k− s, k]× [ℓ, ℓ+ s] is contained in [1, n]× [1, n]. Since σ is simpleand 1 ≤ s ≤ n − 2, the graph of σ has at most s points within s. Then the simplicityof σ implies that there is a point V = (i0, j0) of the graph of σ either strictly above s

or strictly below s. If V is above s, i.e, if k − s ≤ i0 ≤ k and ℓ + s < j0 then takingF = {ib, i0, k}, we find θF (σ) = (231). Then θ the decomposition of Φ(σ) induces a non-trivial decomposition of (231) since (ib, k) ∈ Φ(σ2) and (i0, k) ∈ Φ(σ1). This contradictionshows V must be strictly below s. Thus k− s ≤ i0 ≤ k and j0 < ℓ. For F = {ib, i0, k}, wehave θF (σ) = (312). This yields a non-trivial decomposition of (312) since (ib, i0) ∈ Φ(σ1)and (ib, k) ∈ Φ(σ2). Again we have a contradiction.

For the second case, suppose that s = jb− ℓ for some b with 1 ≤ b ≤ p. If s = n−1, thenjb = n and ℓ = 1. The simplicity of σ implies that ib 6= n and k 6= n. With F = {ib, k, n}we have θF (σ) = (312). Since (ib, k) ∈ Φ(σ2) and (ib, n) ∈ Φ(σ1), the decomposition ofΦ(σ) induces a non-trivial decomposition of (312). This contradiction shows that s 6= n−1,

Therefore 1 ≤ s ≤ n− 2, By the simplicity of σ, the square s contains at most s pointsof the graph of σ. Hence there is a point V = (i0, j0) of the graph of σ either strictly to theright of s or strictly to the left of s. If V is to the left of s, i.e, if i0 < k−s ≤ ib < k andℓ ≤ j0 ≤ ℓ+ s then we consider F = {i0, ib, k}. Then θF (σ) = (231). By the definition of


s, we have (i0, k) ∈ Φ(σ1) Also (ib, k) ∈ Φ(σ2). Hence the decomposition of Φ(σ) inducesa non-trivial decomposition of (231), a contradiction. Conversely, if V is to the right of s

then ib < k < i0 and ℓ ≤ j0 ≤ ℓ + s. Taking F = {ib, k, i0} we have θF(σ) = (312). Now(ib, k) ∈ Φ(σ2) and (ib, i0) ∈ Φ(σ1). Thus the decomposition of Φ(σ) yields a non-trivialdecomposition of (312), a contradiction.

These contradictions imply that p = 0, i.e., there are no roots of the form (ia, k) inΦ(σ2). The same proof works, mutatis mutandis, to show that there are no roots of theform (k, i′a) ∈ Φ(σ2). Thus Φ(σ2) must be empty and so Φ(σ) is irreducible as required. �

The following corollary follows immediately from Lemma 3.4 and Proposition 3.14. Thecorollary is required for our proof of Theorem 1.9 and is the motivation for proving Propo-sition 3.14.

Corollary 3.15. Suppose that σ ∈ Sn is simple. Then woσ is irreducible. �

We now give our proof of Theorem 1.9

Proof. First we suppose that ∆+n = Φ(w1)⊔Φ(w2)⊔ · · · ⊔Φ(wr) is an irreducible decompo-

sition. Without loss of generality, the highest root (1, n) is an element of Φ(w1). Expressw1 = σ1[β11, β12, . . . , β1m] in simple form. It is easy to see that (1, n) ∈ Φ(w1) implies thatσ1 6= Id.

By Lemma 3.2, σ1 is irreducible. If σ1 = wo[m] this implies that m = 2 since ∆+m =

Φ(wo[m]) = Φ(σ)⊔Φ(wo[m]σ) gives a non-trivial decomposition for any σ ∈ Sm\{wo[m], Idm}when m ≥ 3. Since w1 is irreducible, by Lemma 3.2, we must have w1 = wo[2][Id, Id] if w1

is minus-decomposable.If σ1 6= wo, then σ1 must be simple and m ≥ 4. Let {1, 2, . . . , n} = I1 ⊔ I2 ⊔ · · · ⊔ Im be

the intervals corresponding to the simple form w1 = σ1[β11, β12, . . . , β1m] with |Ia| = za.Since w1 is irreducible, we must have w1 = σ1[Idz1 , Idz2 , . . . , Idzm ]. We have Φ(wow1) =Φ(w2) ⊔ Φ(w3) ⊔ · · · ⊔ Φ(wr). Let F be an admissible set. Then θF (w1) = σ1. SinceΦ(wo[n]w1) = Φ(w1)

c, we must have wow1 = (woσ1)[wo, wo, . . . , wo]. Then since F isadmissible, θF (wow1) = woσ1. The element woσ1 is simple by Corollary 3.15. ApplyingLemma 3.6, we see that there exists δ(F) such that Φ(wow1) ⊂ Φ(wδ(F)).

Suppose F is admissible and 1 ≤ a ≤ m. We claim that there exists a root (s, t) ∈Φ(woσ1) s 6= a and t 6= a. This is clear since otherwise woσ has Idm−1 as a one pointdeletion. Since m ≥ 4, it is clear that this cannot happen since woσ is simple. If |Ia| > 1then let ia ∈ F ∩ Ia and i′a ∈ Ia \ F and construct a new admissible set by replacing ia byi′a, i.e., F

′ := (F \ {ia}) ∪ {i′a}. Write {is} = Is ∩ F = Is ∩ F ′ and {it} = It ∩ F = It ∩ F ′.Since (s, t) ∈ Φ(woσ1) it follows that (is, it) ∈ Φ(wow1) ∩ ∆+

F ∩ ∆+F ′ . Therefore F and

F ′ are wow1-connected. From this it follows that any two admissible sets are wow1-path-connected. This implies that δ(F) is constant for all admissible sets F . Without loss ofgenerality, we have δ(F) = 2 for all admissible sets F .

Hence Φ(woσ1) ⊂ Φ(θF (w2)). Assume, by way of contradiction, that this is a properinclusion. Then there exists (a, b) ∈ Φ(θF (w2))\Φ(woσ1). Then (ia, ib) ∈ Φ(w1)∩∆

+Fwhere


{ia} = F ∩ Ia and {ib} = Ib∩F . But (a, b) ∈ Φ(θF (w2)) implies that (ia, ib) ∈ Φ(w2)∩∆+F .

Thus (ia, ib) ∈ Φ(w1) ∩ Φ(w2). This contradiction shows that woσ1 = θF(w2).Since woσ1 = θF (w2) for all admissible sets F , we see that

{(ia, ib) ∈ Φ(w2) | ia ∈ Ia, ib ∈ Ib, a 6= b} = {(ia, ib) ∈ ∆+n | ia ∈ Ia, ib ∈ Ib, (a, b) ∈ Φ(woσ1)} .

Hence Φ(w2) = {(ia, ib) ∈ ∆+n | ia ∈ Ia, ib ∈ Ib, (a, b) ∈ Φ(woσ1)} ⊔

(

⊔mb=1∆

+Ib∩ Φ(w2)

)

.

From this it is clear that w2 = woσ1[β21, β22, . . . , β2m] where β2b = θIb(w2) for all b =1, 2, . . . ,m.

But since Φ(w2) is irreducible, we must have β21 = β22 = · · · = β2m = Id and thereforew2 = woσ1[Idz1 , Idz2 , . . . , Idzm ] as required.

Hence we have shown

w1 =

{

wo[2][Id, Id], if w1 is minus-decomposable;

σ1[Id, Id, . . . , Id] and w2 = woσ1[Id, Id, . . . , Id], if w1 is minus-indecomposable.

Therefore

{(i, j) ∈ ∆+n | i ∈ Ia, j ∈ Ib, a 6= b} =

{

Φ(w1), if w1 is minus-decomposable;

Φ(w1) ⊔ Φ(w2), if w1 is minus-indecomposable.

From this it follows that all other wa have the form wa = Idm[βa1, βa2, . . . , βam]. LetΨb denote the order preserving bijection Ψb : Ib → {1, 2, . . . , zb} for b = 1, 2, . . . ,m.Then Φ(wa) = ⊔m

b=1Ψ−1(Φ(βab)) for all a = 3, 4, . . . , r. Thus we have shown that the

characterization of irreducible decompositions given in the statement of the theorem iscorrect.

For general decompositions we use the fact that every decomposition may be realizedby beginning with an irreducible decomposition and then merging some collections of theirreducible inversion sets. We consider the effect of such mergers by examining effect ofmerging pairs of inversion sets.

Let w = σ[β1, β2, . . . , βm] and w′ = σ′[β′1, β

′2, . . . , β

′m] with βb, β

′b ∈ Szb for b = 1, 2, . . . ,m

and σ, σ′ ∈ Sn. It is easy to see that

Φ(w) ⊔ Φ(w′) = Φ(µ)

where µ = ν[γ1, γ2, . . . , γm] where Φ(σ) ⊔ Φ(σ′) = Φ(ν) and Φ(βb) ⊔ Φ(β′b) = Φ(γb) for

b = 1, 2, . . . ,m.Since Φ(σ) ⊔ Φ(woσ) = Φ(wo), it follows that the characterization of general decompo-

sitions given in the statement of the theorem is correct. �

4. Decompositions for Type B

Here we consider root systems of type Bn. In this section and the next, we willconsider the Weyl group W(Am) ∼= Sm+1 as the group of all permutations of the set{e1, e2, . . . , em+1}. With this notation, the positive roots are ∆+

Am= {(i, j) = ei − ej | 1 ≤

i < j ≤ m}.


We denote the positive roots for the root system of type Bn by

∆+Bn

= {εi − εj | 1 ≤ i < j ≤ n} ⊔ {εi + εj | 1 ≤ i < j ≤ n} ⊔ {εi | 1 ≤ i ≤ n}.

The set of simple roots in ∆+Bn

is {εi − εi+1 | 1 ≤ i ≤ n− 1} ⊔ {εn}.For type Bn the Weyl group W(Bn) is the set of signed permutations of the set

{ε1, ε2, . . . , εn, 0,−εn, . . . ,−ε2,−ε1}.

These are the permutations σ of this set such that σ(0) = 0 and σ(−εi) = −σ(εi) for all1 ≤ i ≤ n. Abstractly, W(Bn) ∼= Sn ⋊ (Z/2Z)n.

For 1 ≤ i ≤ 2n + 1 we define i′ := (2n + 2) − i. It is convenient to define εi = −εi′ for1 ≤ i ≤ 2n+ 1 and εn+1 = 0.

We embed W(Bn) in W(A2n) by the group homomorphism σ 7→ σ̃ where σ̃(ei) = ek ifσ(εi) = εk. The condition that σ(−εi) = −σ(εi) implies that if σ̃(ei) = ek then σ̃(ei′) = ek′ .This condition is equivalent to the condition that the graph of σ̃ is invariant under a rotationof π radians about the point (n + 1, n + 1) ∈ R

2. We say that a permutation in S2n+1 issymmetric if it satisfies these two equivalent conditions.

Consider the involution µ of ∆+A2n

induced by µ(ei) = −ei′ . Note that the elements of

∆+A2n

fixed pointwise by µ are precisely the positive roots ei − ei′ for i = 1, 2, . . . , n.

We define the map ρ : ∆A2n→ ∆Bn

by ρ(ei − ej) =

{

εi − εj , if j 6= i′, i;

εi = (εi − εj)/2 if j = i′.

Since σ̃(ei − ei′) = ek − ek′ we have σ(ρ(α)) = ρ(σ̃(α)) for all α ∈ ∆A2n. Observe that

ρ(µ(α)) = ρ(α) for all α ∈ ∆+A2n

.

Suppose α = ei − ej ∈ ∆+A2n

.

Observe that ρ(α) =

εi − εj , if 1 ≤ i < j ≤ n;

εi, if 1 ≤ i ≤ n, j = n+ 1;

εi + εj′ , if 1 ≤ i ≤ n, n+ 2 ≤ j ≤ 2n+ 1 with j 6= i′;

εi, if 1 ≤ i = j′ ≤ n;

εj′ , if i = n+ 1 < j ≤ 2n+ 1;

εi′ − εj′ , if n+ 2 ≤ i < j ≤ 2n+ 1.

.

Thus ρ(∆+A2n

) = ∆+Bn

. Clearly ρ(∆−A2n

) = ∆−Bn

also.

Proposition 4.1. Suppose σ ∈ W(Bn). Then ρ(Φ(σ̃)) = Φ(σ).

Proof. Let α ∈ ∆+A2n

and let β = σ̃(α). Then ρ(α) ∈ ∆+Bn

. Now α ∈ Φ(σ̃) if and only if

β ∈ ∆−A2n

if and only ρ(β) ∈ ∆−Bn

. Since ρ(β) = σ(ρ(α)) this implies α ∈ Φ(σ̃) if and only

if ρ(α) ∈ Φ(σ). Thus ρ(Φ(σ̃)) = Φ(σ) �

Lemma 4.2. Let σ ∈ W(Bn) and let 1 ≤ i ≤ n. Then ei − en+1 ∈ Φ(σ̃) if and only ifei − ei′ ∈ Φ(σ̃). Furthermore α ∈ Φ(σ̃) if and only if µ(α) ∈ Φ(σ̃) for all α ∈ ∆+

A2n. In

particular, if Φ(σ̃) ∩ ρ−1(γ) 6= ∅ then Φ(σ̃) ⊃ ρ−1(γ) for all γ ∈ ∆+Bn

.


Proof. First we suppose that ei − en+1 ∈ Φ(σ̃). Then µ(ei − en+1) = en+1 − ei′ ∈ Φ(σ̃).Hence by the closed property ei − ei′ = (ei − en+1) + (en+1 − ei′) ∈ Φ(σ̃).

Conversely, suppose ei − en+1 /∈ Φ(σ̃). Then µ(ei − en+1) = en+1 − ei′ /∈ Φ(σ̃) and theco-closed property shows that ei − ei′ /∈ Φ(σ̃).

For the second assertion write α = ei − ej (with i < j) and write σ̃(α) = ek − eℓ. Thenσ̃(µ(α)) = σ̃(ej′ − ei′) = eℓ′ − ek′ . Hence α ∈ Φ(σ̃) if and only if k > ℓ if and only if ℓ′ > k′

if and only if µ(α) ∈ Φ(σ̃).The final assertion follows from the above. �

The following corollary follows immediately from the preceding lemma.

Corollary 4.3. Let σ1, σ2 ∈ W(Bn) with Φ(σ̃1) ∩ Φ(σ̃2) = ∅. Then ρ(Φ(σ̃1) ⊔ Φ(σ̃2)) =ρ(Φ(σ̃1))⊔ρ(Φ(σ̃2)). In particular, suppose σ1, σ2, . . . , σr ∈ W(Bn). Then ∆+

Bn= ⊔r

i=1Φ(σr)

if and only if ∆+A2n

= ⊔ri=1Φ(σ̃r). Furthermore the element σ1 ∈ W(Bn) is indecomposable

if and only if σ̃1 ∈ W(A2n) is indecomposable. �

The embedding of W(Bn) into W(A2n) allows us to define an inflation operation forWeyl groups of type B as follows. Let m ≤ n and let {1, 2, . . . ,m} = I1 ⊔ I2 ⊔ . . . Irbe a decomposition into intervals. Put zt = |It| for t = 1, 2, . . . , r. Suppose that σ0 ∈W(Br), τr+1 ∈ W(Bn−m) and τt ∈ Szt for t = 1, 2, . . . , r. We form the inflation σ̃ =σ̃0[τ1, τ2, . . . , τr, τ̃r+1, τr+2, . . . , τ2r+1] where τ2r+2−t = woτtwo for t = 1, 2, . . . , r. Thenσ̃ ∈ W(A2n) is symmetric and so corresponds to an element σ ∈ W(Bn). We say that σ isan inflation in W(Bn).

An element σ ∈ W(Bn) which cannot be realized as such an inflation in W(Bn) exceptwith r = 0 or r = n is said to be simple in W(Bn).

The proof of the following lemma is straight forward and left to the reader.

Lemma 4.4. If α[τ1, τ2, . . . , τs] ∈ W(A2n) is symmetric then α must be symmetric, s isodd, and τs+1−t = woτtwo for all t = 1, 2, . . . , s.

Proposition 4.5. Let σ ∈ W(Bn). Then σ is simple in W(Bn) if and only if σ̃ is simple(in W(A2n)).

Proof. If σ̃ is simple in W(A2n), then σ̃ cannot be realized as a non-trivial inflation and socannot be realized as a non-trivial inflation in W(Bn).

Conversely, suppose that σ is simple in W(Bn). Assume by way of contradiction that σ̃is not simple. Then σ̃ = α[τ1, τ2, . . . , τs]. Since σ̃ ∈ S2n+1 is symmetric, Lemma 4.4 showsthat σ is an inflation in W(Bn). �


Theorem 4.6. Let ∆+Bn

= Φ(w1)⊔Φ(w2)⊔· · ·⊔Φ(wr) be a decomposition with all Φ(wa) 6=∅. Then (after possibly reordering) we have

w̃1 = σ̃1[β11, β12, . . . , β1(2s+1)]

w̃2 = σ̃2[β21, β22, . . . , β2(2s+1)]

...

w̃r = σ̃r[βr1, βr2, . . . , βr(2s+1)]

where σ1, σ2, . . . , σr ∈ W(Bs), βqt = woβq(2s+2−t)wo for all q = 1, 2, . . . , r and all t =1, 2, . . . , 2s + 1 and

(†) Φ(β1b) ⊔ Φ(β2b) ⊔ · · · ⊔ Φ(βrb) = ∆+Az

b

is a decomposition ∀b = 1, 2, . . . , 2s+ 1

and σq = σq+1 = · · · = σr = Id with{

σ̃1 = wo[2s+ 1] and q = 2, if w1 is minus-decomposable;

σ1 is W(Bs) simple, σ̃2 = woσ̃1 and q = 3, if w1 is minus-indecomposable.

Moreover, the above decomposition of ∆+Bn

is irreducible if and only if the decompositions(†) are irreducible for all b, exactly one of βa1, βa2, . . . , βa(s+1) is not equal to Id for eacha = q, q + 1, . . . , r, and

{

s = 1 and β11 = β13 = Idz1 and β12 = Idz2 , if w1 is minus-decomposable;

β1b = β2b = Idzb for all b = 1, 2, . . . , 2s+ 1, if w1 is minus-indecomposable.

5. Decompositions for Type C

In this section we consider root systems of type Cn.We denote the positive roots for the root system of type Cn by

∆+Bn

= {εi − εj | 1 ≤ i < j ≤ n} ⊔ {εi + εj | 1 ≤ i ≤ j ≤ n}.

The set of simple roots in ∆+Cn

is {εi − εi+1 | 1 ≤ i ≤ n− 1} ⊔ {2εn}.The Weyl group W(Cn) is the set of signed permutations of the set

{ε1, ε2, . . . , εn,−εn, . . . ,−ε2,−ε1}.

Abstractly, W(Cn) ∼= W(Bn) ∼= Sn ⋊ (Z/2Z)n.For 1 ≤ i ≤ 2n we define i′ := (2n+ 1)− i and put εi = −εi′ for 1 ≤ i ≤ 2n.We embed W(Cn) in W(A2n−1) by the group homomorphism σ 7→ σ̃ where σ̃(ei) = ek

if σ(εi) = εk. Again the condition that σ(−εi) = −σ(εi) implies that if σ̃(ei) = ek thenσ̃(ei′) = ek′ . This condition is equivalent to the condition that the graph of σ̃ is invariantunder a rotation of π radians about the point (n + 1/2, n + 1/2) ∈ R

2. We say that apermutation in S2n is symmetric if it satisfies these two equivalent conditions.

As in the previous section, we define µ to be the involution of ∆+A2n−1

induced by

µ(ei) = −ei′ . Then µ has no fixed points in ∆+A2n−1

.


We define the map ρ : ∆A2n−1→ ∆Cn

by ρ(ei − ej) = εi − εj. Then ρ(µ(α)) = ρ(α) for

all α ∈ ∆+A2n

.

Suppose α = ei − ej ∈ ∆+A2n−1

.

Then ρ(α) =

εi − εj , if 1 ≤ i < j ≤ n;

εi + εj′ , if 1 ≤ i ≤ n, n+ 1 ≤ j ≤ 2n with j 6= i′;

2εi, if 1 ≤ i = j′ ≤ n;

εi′ − εj′ , if n+ 1 ≤ i < j ≤ 2n.

.

Thus ρ(∆+A2n−1

) = ∆+Cn

and ρ(∆−A2n−1

) = ∆−Cn

.

As in the previous section, this implies that ρ(Φ(σ̃)) = Φ(σ) for σ ∈ W(Cn). Also ifΦ(σ̃) ∩ ρ−1(γ) 6= ∅ then Φ(σ̃) ⊃ ρ−1(γ) for all σ ∈ W(Cn) and γ ∈ ∆Cn

.The following corollary follows the above considerations.

Corollary 5.1. Let σ1, σ2 ∈ W(Cn) with Φ(σ̃1) ∩ Φ(σ̃2) = ∅. Then ρ(Φ(σ̃1) ⊔ Φ(σ̃2)) =ρ(Φ(σ̃1))⊔ρ(Φ(σ̃2)). In particular, suppose σ1, σ2, . . . , σr ∈ W(Cn). Then ∆+

Cn= ⊔r

i=1Φ(σr)

if and only if ∆+A2n−1

= ⊔ri=1Φ(σ̃r). Furthermore the element σ1 ∈ W(Cn) is indecompos-

able if and only if σ̃1 ∈ W(A2n−1) is indecomposable. �

As before, the embedding of W(Cn) into W(A2n−1) allows us to define an inflationoperation for Weyl groups of type C as follows. Let {1, 2, . . . , n} = I1 ⊔ I2 ⊔ . . . Ir be adecomposition into intervals. Put zt = |It| for t = 1, 2, . . . , r. Suppose that σ0 ∈ W(Cr)and τt ∈ Szt for t = 1, 2, . . . , r. We form the inflation σ̃ = σ̃0[τ1, τ2, . . . , τr, τ̃r+1, . . . , τ2r]where τ2r+1−t = woτtwo for t = 1, 2, . . . , r. Then σ̃ ∈ W(A2n−1) is symmetric and socorresponds to an element σ ∈ W(Cn). We say that σ is an inflation in W(Cn).

An element σ ∈ W(Cn) which cannot be realized as such an inflation in W(Cn) exceptwith r = 1 or r = n is said to be simple in W(Cn).

As before we have have the following two results.

Lemma 5.2. If α[τ1, τ2, . . . , τs] ∈ W(A2n−1) is symmetric then α must be symmetric, s iseven and τs+1−t = woτtwo for all t = 1, 2, . . . , s.

Proposition 5.3. Let σ ∈ W(Cn). Then σ is simple in W(Cn) if and only if σ̃ is simple(in W(A2n−2)).

Finally we obtain the analogue of Theorem 1.9 for type C.

Theorem 5.4. Let ∆+Cn

= Φ(w1)⊔Φ(w2)⊔· · ·⊔Φ(wr) be a decomposition with all Φ(wa) 6=∅. Then (after possibly reordering) we have

w̃1 = σ̃1[β11, β12, . . . , β1(2s)]

w̃2 = σ̃2[β21, β22, . . . , β2(2s)]

...

w̃r = σ̃r[βr1, βr2, . . . , βr(2s)]


where σ1, σ2, . . . , σr ∈ W(Cs), βqt = woβq(2s+1−t)wo for all q = 1, 2, . . . , r and all t =1, 2, . . . , 2s and

(†) Φ(β1b) ⊔ Φ(β2b) ⊔ · · · ⊔ Φ(βrb) = ∆+Azb

is a decomposition ∀b = 1, 2, . . . , 2s

and σq = σq+1 = · · · = σr = Id with{

σ̃1 = wo[2s] and q = 2, if w1 is minus-decomposable;

σ1 is W(Cs) simple, σ̃2 = woσ̃1 and q = 3, if w1 is minus-indecomposable.

Moreover, the above decomposition of ∆+Cn

is irreducible if and only if the decompositions(†) are irreducible for all b, exactly one of βa1, βa2, . . . , βas is not equal to Id for eacha = q, q + 1, . . . , r, and

{

s = 1 and β11 = β13 = Idz1 and β12 = Idz2 , if w1 is minus-decomposable;

β1b = β2b = Idzb for all b = 1, 2, . . . , 2s, if w1 is minus-indecomposable.

6. Enumerative Results

The inductive description for a decomposition provided by Theorems 1.9, 4.6, and 5.4allows us to use generating series or recursion to enumerate many different types of decom-positions. We give a few examples.

Let sn be the number of simple pairs in Sn, i.e., the number of subsets {w,wow} withw ∈ Sn and both w and wow simple (note that by Proposition 3.14, w is simple if andonly if wow is simple). Let SA(z) =

∑

n≥0 snzn = z2 + z4 +3z5 + · · · be the corresponding

generating function. By [AAK, page 5] we have the following description of S(z). LetF (z) =

∑

n≥1 n!zn and G(z) =

∑

n≥1 gnzn its functional inverse, i.e., the function defined

by the relation G(F (z)) = z. Then s1 = 0, s2 = 1, and sn = −gn/2− (−1)n for n ≥ 3.

Number of decompositions into irreducibles. Let an be the number of decomposi-tions ∆+

n = Φ(σ1) ⊔ Φ(σ2) ⊔ · · · ⊔ Φ(σr), where each σk ∈ Sn is irreducible, and where weignore the order in the decomposition. Set A(z) =

∑

n≥1 anzn be the generating series.

Theorem 1.9 leads to the relation A(z) = SA(A(z)) + z, which recursively determines thecoefficients an. Here are the low order terms of A(z):

A(z) = z + z2 + 2z3 + 6z4 + 23z5 + 114z6 + 717z7 + 5510z8 + 49570z9 + 504706z10 + · · ·

Decompositions of Maximal length. If σ 6= e then the inversion set Φ(σ) must containat least one simple root. Since there are only n − 1 simple roots, any decomposition∆+

n = Φ(σ1)⊔Φ(σ2)⊔ · · · ⊔Φ(σr), with no σa = e must satisfy r ≤ n− 1. Let CatA(n− 1)denote the number of decompositions of ∆+

n into exactly n − 1 non-empty inversion sets.(Thus each inversion set appearing in the decomposition must contain exactly one simpleroot).

Lemma 6.1. CatA(n) =1

n+1

(2nn

)

, the nth Catalan number.


Proof. We consider decompositions of the form ∆+n = Φ(σ1) ⊔ Φ(σ2) ⊔ · · · ⊔ Φ(σr) and

compute CatA(n− 1). Without loss of generality, the longest root e1−en ∈ Φ(σ1). Supposethat ek − ek+1 is the simple root in Φ(σ1). Then σ1(k + 1) < σ1(k + 2) < · · · < σ1(n) <σ1(1) < σ1(2) · · · < σ1(k) and therefore σ1 = (n − k + 1, n − k + 2, . . . , n, 1, 2 . . . , n− k) =(1, 2)[Idk, Idn−k]. Let I1 := {1, 2, . . . , k} and I2 := {k1, k + 2, . . . , n}. Then Φ(σ1) ={(ei − ej ∈ ∆+

n | i ∈ I1, j ∈ I2} = {ei − ej ∈ ∆+n | i ≤ k, j ≥ k + 1}. Therefore

∆+I1⊔∆+

I2= Φ(σ2)⊔Φ(σ3)⊔· · ·⊔Φ(σn−1). Without loss of generality, ∆+

I1= Φ(σ2)⊔Φ(σ3)⊔

· · ·⊔Φ(σk−1) and ∆+I2

= Φ(σk+1)⊔Φ(σk+2)⊔· · ·⊔Φ(σn−1). This yields the recursion relation

CatA(n − 1) =∑n−1

t=1 CatA(t− 1)CatA(n − t− 1) =∑n−2

t=0 CatA(t)CatA(n− t− 2). ThusCatA(n) =

∑nt=1 CatA(t− 1)CatA(n− t). Since CatA(1) = 1 and CatA(2) = 2 we see that

CatA(n) satisfies the usual recursion relation for the Catalan numbers. �

This incarnation of the Catalan numbers does not currently seem to appear on RichardStanley’s list [S] of 198 combinatorial interpretations of the Catalan numbers.

Type B/C results. Theorems 4.6 and 5.4 lead to similar recursions in types B/C. LetSB(z) be the generating series for the number of simple pairs in type Bn/Cn. Equivalentlythe coefficient of zn in SB(z) is the number of pairs of simple elements in S2n+1 eachof which are symmetric. The isomorphism W(Bn) ∼= W(Cn) implies that this is alsothe number of pairs of simple symmetric elements in S2n+1. One deduces the functionalequation

SB(F (z)) = 1−1

1 + F (2z)−

2F (z)

1 + F (z),

(where F (z) =∑

n≥1 n!zn as above) which determines SB(z). Here are some low order

terms:

SB(z) = 2z2 + 10z3 + 90z4 + 966z5 + 12338z6 + 181470z7 + 3018082z8 + 55995486z9 + · · ·

Decompositions into Irreducibles. Let bn be the number of decompositions of the pos-itive roots in types Bn/Cn into disjoint unions of inversion sets, and set B(z) =

∑

n≥1 bnzn

to be the generating function. Theorem 4.6 leads to the relation

B(z) =SB(A(z))

1− SB(A(z)),

which completely determines B(z). Here are the low order terms of B(z):

B(z) = z+3z2+14z3+100z4+973z5+11804z6+168809z7+2757930z8+50522912z9+ · · ·

Bn/Cn Catalan numbers. Let CatB(n) be the number of decompositions of the positiveroots of Bn/Cn into disjoint unions of inversion sets, where each inversion set contains asingle simple root. The isomorphism W(Bn) ∼= W(Cn) implies that the number of suchdecompositions is the same for types Bn and Cn. As in type A, these are the decompositions


of maximal length (subject to the restriction that each inversion set is non-empty) and thusare irreducible decompositions.

Proposition 6.2. The numbers CatB(n) satisfy the recursion CatB(n) = CatB(n− 1) +

2∑n−2

k=0 CatA(n− k − 1)CatB(k), and thus

∑

n≥1

CatB(n)zn =

1

(1− 4z)1

2 + z.

Proof. We consider theBn case. Suppose then that ∆+Bn

= Φ(α1)⊔Φ(α2)⊔· · ·⊔Φ(αn) where

each αi ∈ W(Bn) and each Φ(αi) contains a single simple root of ∆+Bn

. Without loss of gen-

erality Φ(α̃1) contains e1 − e2n+1. By Theorem 4.6, we have α̃1 = σ̃1[β11, β12, . . . , β1(2s+1)]where σ1 ∈ W(Bs) and βt = woβ2s+2−two for all t = 1, 2, . . . , s and either σ1 = wo[3] or σ1is simple and Φ(σ1) contains a single simple root. Thus σ̃1 is simple, symmetric and Φ(σ̃1)contains a pair of A2n simple roots of the form ei − ei+1, ei′−1 − ei′ . It is not hard to seethat this forces σ̃1 = wo[3], σ̃1 = (25314) or σ̃1 = (41352). The last possibility is excludedby the fact that Φ(α̃1) contains the highest root.

First suppose that σ̃1 = wo[3] and let {1, 2, . . . , 2n+1} = I1⊔I2⊔I3 be the correspondingdecomposition into intervals with |I1| = |I3| = n−k and |I2| = 2k+1 where 0 ≤ k ≤ n−1.Then α̃j = Id3[βj1, βj2, βj3] for j = 2, 3, . . . , n. Furthermore, without loss of generality,∆+

I1= Φ(β21) ⊔ Φ(β31) ⊔ · · · ⊔ Φ(β(n−k)1) is a maximal length decomposition of a root

system of type An−k−1. There are CatA(n − k − 1) such decompositions. (We also have∆+

I3= Φ(woβ23wo) ⊔ Φ(woβ33wo) ⊔ · · · ⊔ Φ(woβ(n−k)3wo).) Finally ∆+

I2= Φ(β(n−k+1)2) ⊔

Φ(β(n−k+2)2)⊔· · ·⊔Φ(βn2) is a maximal symmetric decomposition. There are CatB(k) such

decompositions. Thus there are∑n−1

k=0 CatA(n− k − 1)CatB(k) maximal decompositions of

∆+Bn

with α̃1 = wo[3].Next suppose that σ̃1 = (25314) and let {1, 2, . . . , 2n + 1} = I1 ⊔ I2 ⊔ · · · ⊔ I5 be the

corresponding decomposition. Then, as above, α̃2 ⊔ α̃3 ⊔ · · · ⊔ α̃n comprises maximal Atype decompositions of ∆+

I1and ∆+

I2and a maximal symmetric decomposition of ∆+

I3. Thus

there are

n−1∑

z1=1

n−z1∑

z2=1

CatA(z1)CatA(z2)CatB(n − z1 − z2) =n−2∑

k=0

∑

z1+z2=n−k

CatB(k)CatA(z1)CatA(z2)

=

n−2∑

k=0

CatB(k)CatA(n− k − 1)

maximal decompositions of ∆+Bn

with α̃1 = (25314).


Adding the contributions of the two cases gives

CatB(n) = CatB(n− 1) + 2

n−2∑

k=0

CatA(n− k − 1)CatB(k)

as claimed. This easily implies the stated form of the generating function. �

Remark. We have chosen to call these numbers the “type B/C Catalan numbers”, sincethey come from an enumerative problem about Coxeter groups which yields the usualCatalan numbers in the type A case. There is at least one other use of the term “Catalannumbers 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 Bn/Cn numbers are(

2nn

)

– different from the numbers given by the recursion andgenerating function above.

Number of decompositions into triples. The most important case – in any type – ofthe problems motivating these questions about decompositions is the case of decompositionsinto a disjoint union of three inversion sets. As described in §7 this corresponds to thethe case of the eigenvalues of three Hermitian matrices summing to zero (respectivelythe cup product of two cohomology groups into a third, after a similar symmetrization).The corresponding enumerative/classification problem is to write down all triples σ1, σ2,σ3 ∈ Sn (again disregarding order) with ∆+

n = Φ(σ1) ⊔ Φ(σ2) ⊔ Φ(σ3). We make thefurther restriction that no σj = e (all such triples are of the form (w,wow, e) and henceelementary to understand). Theorems 1.9, 4.6, and 5.4 provide a recursive way to generateand enumerate all such triples. Briefly, the method is a parallel recursion keeping track ofnot only the triples of the kind above, but also the subset of those triples where σ1 = wo[m]for some m. At each step, the new triples of each kind depend on the triples of both kindsfor 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 An

and Bn/Cn cases.

n An triples Bn/Cn triples1 12 1 43 3 334 17 3515 129 42106 1116 554957 10474 8004768 104604 126541649 1101012 21987018710 12153179 4206375350


11 140397525 8853945910312 1697555983 204350223836513 21516940295 5144087684339614 286680892462 140360832902047315 4028129552836 4125759267109814616 59885247963954 129904589082135016217 944511887685826 4359671883982555338118 15828354015222453 155287140302163070093619 281880601827533671 5848850283297579107742120 5327985147037232973 2322044948865982864468235

7. Motivation and an application

In this section we explain our initial motivation for this work and give an application ofTheorem 1.9. For clarity of exposition we discuss only the case of type A but everythingcarries over to the cases of types B and C. The two motivating problems both have versionsinvolving an arbitrary number of factors (which lead to Problem 1.2) but for simplicity weonly describe the case of two factors which leads to considering decompositions into threeinversion sets.

Regular faces of the Littlewood-Richardson cone. To describe how our work relatesto the Littlewood-Richardson cone we first convert the problem of eigenvalues of Hermitianmatrices to its symmetric version, i.e., instead of Hermitian matrices A,B,C satisfyingC = A + B we will consider Hermitian matrices A,B,C satisfying A + B + C = 0. Itis clear that the cone C′′, analogous to the cone C′ described in §1 is contained in thehyperplane V defined by

λ1 + . . . + λn + µ1 + . . .+ µn + ν1 + . . . + νn = 0

and contains the two-dimensional subspace W ⊂ V of (Rn)3 spanned by(1, . . . , 1, 0, . . . , 0,−1, . . . ,−1) and (0, . . . , 0, 1, . . . , 1,−1, . . . ,−1). Denote by C image ofC′′ under the projection V → V/W . We will use again (λ, µ, ν) to denote the projectionof a point in V to V/W . The natural coordinates in V/W are λ = (a1, . . . , an−1), µ =(b1, . . . , bn−1), and ν = (c1, . . . , cn−1), where ai = λi−λi+1, bi = µi−µi+1, and ci = νi−νi+1

for 1 ≤ i ≤ n−1. Clearly V/W ∼= (Rn−1)3 and Sn acts naturally on each of the componentsof (Rn−1)3: we fix the natural basis {ei− ei+1 | 1 ≤ i ≤ n− 1} of Rn−1 and the action of Sn

is by permuting the indices of this basis. The cone C is a pointed polyhedral cone of fulldimension. Each of the coordinate hyperplanes ai = 0, bi = 0, and ci = 0 for 1 ≤ i ≤ n− 1is a facet of C. Let (Rn−1)3+ denote the dominant cone defined by ai ≥ 0, bi ≥ 0, ci ≥ 0.A face of C is regular if it intersects the interior of (Rn−1)3+. N. Ressayre proved that the


regular faces of C have codimension at most n− 1. Furthermore, the faces of codimensionn− 1 are exactly the intersection of (Rn−1)3+ with the subspaces Tw1,w2,w3

defined by

w−11 λ+ w−1

2 µ+ w−13 ν = 0

for (w1, w2, w3) with the property that ∆+n = Φ(w1) ⊔ Φ(w2) ⊔ Φ(w3), see Theorem C in

[R]. Let (w1, w2, w3) be such a triple and denote by Cw1,w2,w3the corresponding face of C,

i.e. Cw1,w2,w3= Tw1,w2,w3

∩ (Rn−1)3+ = Tw1,w2,w3∩ C.

Note that Cw1,w2,w3is described by its defining hyperplanes - n − 1 from the equation

w−11 λ + w−1

2 µ + w−13 ν = 0 and 3(n − 1) from ai = 0, bi = 0, and ci = 0. It is difficult to

conclude from this description what its defining rays are. We will now show that Theorem1.9 allows us to conclude that Cw1,w2,w3

is a simplicial cone and provides an algorithm forwriting down its defining rays. (The fact that Cw1,w2,w3

is a simplicial cone also follows fromsome results in [DR].) In this section it will be convenient to identify the elements of ∆+

n

with the vectors ei−ej . Consider the inner product in Rn−1 defined by (λ, ei−ej) := λi−λj .

It is immediate that, for 1 ≤ i < j ≤ n,

(λ, ei − ej) = ai + . . .+ aj−1.

This product is Sn-invariant; in particular we have (w−1λ, ei− ej) = (λ,w(ei− ej)) for anyw ∈ Sn and ei − ej ∈ ∆+

n . To obtain defining a set of defining equations for Tw1,w2,w3it is

sufficient to chose a basis {α1, . . . , αn−1 of Rn−1 consisting of elements of ∆+n and write

(w−11 λ+ w−1

2 µ+w−13 ν, αi) = 0

for 1 ≤ i ≤ n − 1. Consider the form of the equation (w−11 λ + w−1

2 µ + w−13 ν, α) = 0

for α ∈ ∆+n . Exactly one of w1(α), w2(α), w3(α) is negative, say w1(α) = −(ei − ej),

w2(α) = ek − el, and w3(α) = ep − eq. Then (w−11 λ+w−1

2 µ+ w−13 ν, α) = 0 becomes

ai + . . .+ aj−1 = bk + . . .+ bl−1 + cp + . . . + cq−1.

This equation is especially simple when −w1(α) is simple, i.e. when j = i + 1. Then itbecomes

ai = bk + . . . + bl−1 + cp + . . .+ cq−1.

Borrowing from elementary linear algebra, we call ai an α-pivot variable and bk, . . . , bl−1,cp, . . . , cq−1 α-free variables.

Proposition 7.1. Assume that ∆+n = Φ(w1) ⊔ Φ(w2) ⊔Φ(w3). The set

Sw1,w2,w3= {α ∈ ∆+

n | −w1(α) is simple or − w2(α) is simple or − w3(α) is simple}

is a basis of Rn−1. Furthermore, this set can be labeled {α1, . . . , αn−1} so that, for i < j,the αi-pivot variable is not an αj-free variable.

Proof. Let wi = σi[βi1, βi2, . . . , βim] and let I = I1 ⊔ I2 ⊔ . . . ⊔ Im be the correspondingdecomposition into intervals of I = {1, 2, . . . , n}. Assume α = ei − ej ∈ Sw1,w2,w3

. Definethe level of α inductively as follows: if i and j belong to different parts of I, then thelevel of α is one; otherwise, i, j ∈ Ik and the level of α is one plus the level of α for the


decomposition ∆+zk

= Φ(β1k)⊔Φ(β2k)⊔Φ(β3k). Consider the projection I → {1, 2, . . . ,m}.Under this projection the level one elements of Sw1,w2,w3

are sent to the elements of Sσ1,σ2,σ3

which form a basis since either σ1 = wo or σ2 = woσ1. The elements of level greater thanone are sent to zero. On the other hand, by a simple inductive argument, the elements oflevel greater than one form bases in the subspace generated by {ei−ej | i, j in the same Ik}.Combining the above we conclude that Sw1,w2,w3

is a basis of Rn−1.To prove the second assertion, we order Sw1,w2,w3

linearly so that elements of lower levelcome before elements of higher level. Notice first that if α1 is of level one and α2 is of levelgreater than one, than no α1-pivot variable is α2-free. Now assume that both α1 and α2

are of level one. Passing to the projection as above, we conclude again that no α1-pivotvariable is α2-free. �

We call the αi-pivot variables simply pivot variables of Cw1,w2,w3and the rest of ai, bi, ci

we call free variables.

Corollary 7.2. Cw1,w2,w3is a simplicial cone.

Proof. It follows from Proposition 7.1 that there are exactly n − 1 pivot variables. Fur-thermore, by ordering them as above we can start from the bottom and replace any pivotvariable appearing in the expression of another pivot variable by its expression. When wereach the top equation, every pivot variable will have become expressed with non-negativecoefficients in terms of the free variables only. �

Example 7.3. We continue with Example 1.10. Recall that w1 = (5, 3, 4, 8, 1, 2, 6, 7), w2 =(4, 5, 6, 1, 7, 8, 3, 2), w3 = (1, 3, 2, 4, 6, 5, 7, 8) and ∆+

8 = Φ(w1) ⊔ Φ(w2) ⊔ Φ(w3). The setSw1,w2,w3

together with the corresponding equations by level is:

Level 1: e2 − e6 : a2 = b5 + b6 + b7+ c3 + c4e4 − e8 : a7 = b1+ c4 + c5 + c6 + c7e1 − e7 : b3 = a5+ c1 + c2 + c3 + c4 + c5 + c6

Level 2: e1 − e3 : a4 = b4 + b5+ c1e5 − e6 : c5 = a1+ b7e7 − e8 : b2 = a6+ c7

Level 3: e2 − e3 : c2 = a3+ b5.

The pivot variables c2 and c5 appear in the expressions for a7 and b3 and need to bereplaced. After the appropriate substitutions we obtain that the generating rays r1,. . . , r14of Cw1,w2,w3

corresponding to the free variables a1, a3, a5, a6, b1, b4, b5, b6, b7, c1, c3, c4, c6, c7


respectively are:

a1 a2 a3 a4 a5 a6 a7 b1 b2 b3 b4 b5 b6 b7 c1 c2 c3 c4 c5 c6 c7r1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0r2 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0r3 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0r4 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0r5 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0r6 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0r7 0 1 0 1 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0r8 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0r9 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0r10 0 0 0 1 0 0 0 0 0 3 0 0 0 0 1 0 0 0 0 0 0r11 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0r12 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0r13 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0r14 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1.

Cup products of line bundles on homogeneous varieties. Let G = GLn(C), let B ⊂G be a Borel subgroup, and let X = G/B. The Picard group of X is isomorphic to Z

n andhence the line bundles on X are parametrized by Z

n. We denote by Lλ the line bundle on Xwhich corresponds to the B-character −λ. We call λ ∈ Z

n dominant if λ1 ≥ λ2 ≥ . . . ≥ λn

and strictly dominant if λ1 > λ2 > . . . > λn. Let ρ = (n − 1, n − 2, . . . , 0) ∈ Zn. We call

λ ∈ Zn regular if there exists w ∈ Sn such that w·λ := w(λ+ρ)−ρ is dominant. Clearly, such

a w is unique and we denote it by wλ. The celebrated Borel-Weil-Bott theorem calculatesthe cohomology groups Hq(X,Lλ). More precisely, it states that Hq(X,Lλ) is zero unlessλ is regular and q equals the length of wλ. In the latter case, Hq(X,Lλ) ∼= V (wλ · λ)∗,where V (µ) denotes the irreducible G-module with highest weight µ. In [DR] two of usstudied the following question: For what pairs λ, µ,∈ Z

n is the cup product map

Hq1(X,Lλ)⊗Hq2(X,Lµ)∪

−→ Hq1+q2(X,Lλ+µ)

nonzero provided that the all cohomology groups above are nonzero? Theorem I in [DR]states that Φ(wλ+µ) = Φ(wλ) ⊔ Φ(wµ) is a necessary and sufficient condition for non-vanishing of the cup product map above.

8. Acknowledgement

This work was partially supported by NSERC. In particular, most of it was done withthe support of NSERC’s Undergraduate Summer Research Awards program.

References

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[AAK] M.H. Albert, M.D. Atkinson and M. Klazar, The enumeration of simple permutations, J. IntegerSeq., 6 (2003) Article 03.4.4, 18 pp. (electronic).

[B] R. Brignall, A survey of simple permutations. Permutation patterns, 41–65, London Math. Soc.Lecture Note Ser., 376, Cambridge Univ. Press, Cambridge, 2010.

[DR] I. Dimitrov and M. Roth, Cup product of line bundles on homogeneous varieties and generalized

PRV components of multiplicity one, arXiv:0909.2280v1.[F] W. Fulton, Eigenvalues, invariant factors, highest weights, and Schubert calculus, Bull. Amer. Math.

Soc. (N.S.) 37 (2000), 209249.[K] B. Kostant, Lie algebra cohomology and the generalized Borel-Weil theorem, Ann. Math. 74 (1961)

329–387.[R] N. Ressayre, Geometric invariant theory and generalized eigenvalue problem II, Annales de l Institut

Fourier, to appear.[S] R. Stanley, Catalan Addendum. http://www-math.mit.edu/∼rstan/ec/[ST] J. H. Schmerl and W. T. Trotter, Critically indecomposable partially ordered sets, graphs, tourna-

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Department of Mathematics and Statistics Queen’s University, Kingston, Ontario, Canada,

K7L 3N6

E-mail address: [email protected]


K7L 3N6


760 Lawrence Ave. West, Toronto, Ontario, Canada, M6A 1B7



K7L 3N6


Department of Mathematics and Computer Science, Royal Military College, Kingston,

Ontario, Canada, K7K 5L0


Department of Mathematics, University of Chicago, Chicago, IL, 60637 USA


http://arxiv.org/abs/0909.2280

http://www-math.mit.edu/~rstan/ec/

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