When is the q-multiplicity of a weight a power of q?
Pamela E. Harris
Department of Mathematics and Statistics, Williams College, United States
[email protected]
,
Margaret Rahmoeller
Department of Mathematics, Computer Science, and Physics, Roanoke College
[email protected]
,
Lisa Schneider
Department of Mathematics and Computer Science, Salisbury University
[email protected]
and
Anthony Simpson
Department of Mathematics and Statistics, Williams College, United States
[email protected]
Abstract.
Berenshtein and Zelevinskii provided an exhaustive list of pairs of weights (λ,μ) of simple Lie algebras g (up to Dynkin diagram isomorphism) for which the multiplicity of the weight μ in the representation of g with highest weight λ is equal to one.
Using Kostant’s weight multiplicity formula we describe and enumerate the contributing terms to the multiplicity for subsets of these pairs of weights and show that, in these cases, the cardinality of these contributing sets is enumerated by (multiples of) Fibonacci numbers. We conclude by using these results to compute the associated q-multiplicity for the pairs of weights considered, and conjecture that in all cases the q-multiplicity of such pairs of weights is given by a power of q.
Key words and phrases:
Kostant’s weight multiplicity formula, weight q-multiplicities, Kostant’s partition function, Weyl alternation sets, Fibonacci numbers, combinatorial representation theory
1. Introduction
Throughout this manuscript we let g be a simple complex Lie algebra and h a Cartan subalgebra with dimh=r. Following the notation of [GW], we let Φ denote the set of roots corresponding to (g,h), Φ+⊆Φ be a set of positive roots, and Δ⊆Φ+ be the set of simple roots. The set of integral and dominant integral weights are denoted by P(g) and P+(g) respectively. The Weyl group is denoted by W and recall that W is generated by the reflections s1,…,sr, where si is the root reflection corresponding to the simple root αi∈Δ. For any w∈W we let ℓ(w) denote the length of w, which represents the minimum nonnegative integer k such that σ is a product of k reflections.
The theorem of the highest weight states that every irreducible (complex) representation of g is a highest weight representation L(λ) with highest weight λ. If μ is a weight of L(λ) then one can compute the multiplicity of this weight using Kostant’s weight multiplicity formula [KMF]:
[TABLE]
where ℘ denotes Kostant’s partition function ℘:h∗→Z, which is the nonnegative integer-valued function such that for each ξ∈h∗, ℘(ξ) counts the number of ways ξ may be written as a nonnegative linear combination of positive roots.
Although Equation (1) provides a way to compute weight multiplicities, its practical use is limited. For example, since the order of the Weyl group (which indexes the sum) increases factorially in terms of the rank of the Lie algebra, the number of terms in the sum presents a complication in its practical use. The second complication arises due to the lack of general closed formulas for the value of the partition function involved. One thing that has been noted in past research, is that many of the terms contribute trivially to the sum, i.e. the value of the partition function is zero, and hence these terms do not contribute to the overall multiplicity. This latter point has motivated work in determining the Weyl alternation set (corresponding to the integral weights λ and μ), defined by
[TABLE]
Note that a Weyl group element σ is in A(λ,μ) if and only if the expression σ(λ+ρ)−(μ+ρ) can be written as a nonnegative Z-linear combination of positive roots.
Determining Weyl alternation sets provides a way to describe the complexity in computing weight multiplicities. Since its definition in 2011, there are only a few cases where a concrete description of the elements of these sets exists. This includes the case where λ denotes the highest root of the classical Lie algebras and, hence, L(λ) corresponds to the adjoint representation. In this case, Harris, Insko, and Williams established that the cardinalities of the Weyl alternation sets A(λ,0)
are given by linear recurrences with constant coefficients [PHThesis, Harris, HIW]. In 2017, Chang, Harris, and Insko described the sets A(β,μ), where β is the sum of all of the simple roots in a classical Lie algebra and μ is an integral weight, and showed that the cardinality of these sets are given by the Fibonacci numbers or multiples of the Lucas numbers [CHI].
In this work we extend this body of knowledge by considering pairs of weights (λ,μ) for which it is known that m(λ,μ)=1, but whose Weyl alternation sets are unknown.
For the simple Lie algebras, a complete list of pairs of weight (λ,μ) satisfying m(λ,μ)=1 (up to isomorphism of Dynkin diagram) were provided in the work of Berenshtein and Zelevinskii [BZ, Theorem 1.3] and is given as follows.
- (List A)
Type Ar (r≥1): λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and
(ℓ−∑1≤i≤rimi)∈(r+1)N.
2. (List B)
Type Br (r≥2): λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ is even and
(ℓ−1)=(∑1≤i<rimi)+rmr/2.
3. (List G)
Type G2:
λ=ℓϖ2, μ=m1ϖ1+m2ϖ2, where m1,m2∈Z+ and 3ℓ−1=2m1+3m2
λ=ϖ1, μ=0
where in all cases ϖ1 and ϖ2 denote fundamental weights of the respective Lie algebra.
In Sections 2 and 3 we provide a description of the elements of Weyl alternation sets A(λ,μ) in the Lie algebra of type Ar and Br, respectively, for certain pairs of weights (λ,μ) by establishing the following main results.
Theorem 1.1**.**
In type Ar with r≥1 and ℓ∈Z+:
- (1)
If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where ℓ,mi∈Z+ and (ℓ−∑imi)=0mod(r+1), then s1∈/A(λ,μ) for any choice of ℓ and r.
2. (2)
If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and (ℓ−∑1≤i≤rimi)=0, mj=0 for all 3≤j≤r, then
A(λ,μ)={1}.
3. (3)
If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and (ℓ−∑1≤i≤rimi)=0 and mj=0 for all 4≤j≤r and m3=0, then
A(λ,μ)={1,s2}.
4. (4)
If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and (ℓ−∑1≤i≤rimi)=0, mj=0 for all 5≤j≤r and m4=0, then
[TABLE]
Theorem 1.2**.**
In type Br with r≥3:
- (1)
Let ℓ be a positive odd integer. Then σ∈A(ℓϖ1,(ℓ−1−4k)ϖ1+2kϖ2) where k∈Z≥0 if and only if σ is a product of nonconsecutive si’s with 1<i≤r.
2. (2)
If ℓ,r∈2Z≥0 under the condition ℓ−1=∑1≤i<rimi+2rmr with mi∈2Z≥0, then A(ℓϖ1,μ)=∅.
3. (3)
All σ∈A(ℓϖ1,(ℓ−1−4j−6k)ϖ1+2jϖ2+2kϖ3) where j,k∈Z≥0 and k=0 are products of nonconsecutive si’s with 1<i≤r or products of the form {si1…sins2s3} with nonconsecutive sij’s for 4<ij≤r.
Note that specializing Theorem 1.2 to ℓ=1 and μ=0 recovers the result presented in [CHI, Theorem 3.1], which shows that the cardinality of the associated Weyl alternation set is given by a Fibonacci number. Extending our results to the remaining pairs of weights listed in List A and List B is much more difficult since the Weyl group action on the highest weight of these representations is not as straightforward to describe.
The complications that arise in describing these Weyl alternation sets are twofold. First, for a given λ there is a finite set of possible weights μ, which are linear combinations of fundamental weights satisfying the needed conditions. Second, whenever μ includes a fundamental weight with high index, one must consider many more conditions in order to determine whether certain Weyl group elements appear in the corresponding Weyl alternation set.
In the Lie algebra of type G2, we provide a complete description of the Weyl alternation sets A(λ,μ) for all pairs of weights (λ,μ) given in (• ‣ (List G)). In Section 4, we present the proof of the following.
Theorem 1.3**.**
Let λ=ℓϖ2 with ℓ≥1 be a weight of the Lie algebra of type G2. If μ=m1ϖ1+m2ϖ2, where m1,m2∈Z≥0 and 3ℓ−1=2m1+3m2, then
A(λ,μ)={1,s1} when λ=0 and A(0,μ)=∅.
Our last set of results is concerned with the q-multiplicity for pairs of weights (λ,μ) in the lists provided above. To make this precise, recall that for a weight ξ, the value of the q-analog of Kostant’s partition function is the polynomial ℘q(ξ)=c0+c1q+⋯+ckqk, where cj denotes number of ways to write ξ as a nonnegative integral sum of exactly j positive roots. Then the q-analog of Kostant’s weight multiplicity formula is defined, in [LL], as:
mq(λ,μ)=σ∈W∑(−1)ℓ(σ)℘q(σ(λ+ρ)−(μ+ρ)).
We provide the q-multiplicity for the pairs of weights considered in Theorems 1.1, 1.2, and 1.3. We state these results below and provide their proof in Section 5.
Theorem 1.4**.**
In type Ar with r≥1 and ℓ∈Z+:
- (1)
If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and (ℓ−∑1≤i≤rimi)=0, mj=0 for all 3≤j≤r, then
mq(λ,μ)=qm2.
2. (2)
If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and (ℓ−∑1≤i≤rimi)=0 and mj=0 for all 4≤j≤r and m3=0, then
mq(λ,μ)=qm2+3m3.
3. (3)
If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and (ℓ−∑1≤i≤rimi)=0, mj=0 for all 5≤j≤r and m4=0, then
mq(λ,μ)=qm2+3m3+6 when m4=1 and mq(λ,μ)=qm2+3m3+6m4 when m4≥2.
Theorem 1.5**.**
In type Br with r≥3:
- (1)
If ℓ is a positive odd integer and k∈Z≥0, then mq(ℓϖ1,(ℓ−1−4k)ϖ1+2kϖ2)=q2k+r.
2. (2)
If ℓ,r∈2Z≥0 under the condition ℓ−1=∑1≤i<rimi+2rmr with mi∈2Z≥0, then mq(ℓϖ1,μ)=0.
3. (3)
If ℓ is a positive odd integer and j,k∈Z≥0 with k=0, then mq(ℓϖ1,(ℓ−1−4j−6k)ϖ1+2jϖ2+2kϖ3)=qr+2j+6k−2.
Theorem 1.6**.**
Let λ=ℓϖ2 with ℓ≥1 be a weight of G2. If μ=m1ϖ1+m2ϖ2, where m1,m2∈Z≥0 and 3ℓ−1=2m1+3m2, then
mq(λ,μ)=qn+2 where m1=3n+1.
Section 6 ends the manuscript by providing ample evidence for the following.
Conjecture 1.1**.**
If λ,μ are a pair of weights for which m(λ,μ)=1, then mq(λ,μ)=qf(r), where f(r) is a function of r, the rank of the Lie algebra.
This conjecture is rather curious, as there are infinite ways that the q-multiplicity when evaluated at q=1 would yield a value of one. Yet what we see is that the only term remaining, after a tremendous amount of (polynomial) cancellation, is simply a power of q. Unfortunately, the techniques we present do not lend themselves well to provide a direct proof of the above conjecture, and we would welcome a new approach to such a problem.
Lastly, in Section 6, we present a new direction for research related to a construction of a poset we call the Weyl alternation poset. This poset is created by computing all Weyl alternation sets A(λ,μ) of a specified simple Lie algebra, and considering the set containment of these Weyl alternation sets. We present the cases of the Lie algebras of types A2, B2, and G2, in which we fix μ=0 and vary λ in the integral weight lattice.
Studying these posets would be of interest in general, but would require techniques beyond the scope presented here.
2. Type A
In this section, we consider the Lie algebra of type Ar for r≥1. The set of simple roots is given by Δ={α1,α2,⋯,αr},
and the set of positive roots is given by
[TABLE]
For 1≤i≤r, the fundamental weights are defined by
[TABLE]
The weight ρ is defined as the half sum of the positive roots, ρ=21∑α∈Φ+α, which is equivalent to ρ=ϖ1+ϖ2+⋯+ϖr. As noted previously, the Weyl group elements are generated by reflections about the hyperplanes that lie perpendicular to the simple roots. We denote these simple reflections by si, where 1≤i≤r, whose action on the simple roots is defined by
si(αj)=αj if ∣i−j∣>1, si(αj)=−αj if i=j, and si(αj)=αi+αj if ∣i−j∣=1.
We also recall that the Weyl group elements act on the fundamental weights by
[TABLE]
We provide a proof of Theorem 1.1 by establishing the following technical results.
Proposition 2.1**.**
If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where ℓ,mi∈Z+ and (ℓ−∑1≤i≤rimi)=0mod(r+1), then s1∈/A(λ,μ) for any choice of ℓ and r in Z+.
Proof.
Note that ℓ=(r+1)n+∑imi for some n∈Z≥0 and that s1(λ)=s1(ℓϖ1)=ℓ(ϖ1−α1). Hence,
[TABLE]
Note that the coefficient of α1 in the last displayed equation is
[TABLE]
Since 1≤i≤r and ℓ, r, and mi are all nonnegative integers, this coefficient is negative for all choices of r and ℓ. Thus, ℘(s1(λ+ρ)−ρ−μ)=0 and s1∈/A(λ,μ).
∎
The following result establishes part 2 of Theorem 1.1.
Proposition 2.2**.**
Consider the Lie algebra of type Ar with r≥1 and let ℓ∈Z+. If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and (ℓ−∑1≤i≤rimi)=0, mj=0 for all 3≤j≤r, then
A(λ,μ)={1}.
Proof.
We begin by establishing that under the given conditions, si∈/A(λ,μ) for any 1<i≤r. Hence, it suffices to show that if 1<i≤r, then
[TABLE]
and there exists 1≤j≤r such that cj<0.
For 1<i≤r note
[TABLE]
Hence, as desired, the coefficient ci=−1 and si∈/A(λ,μ).
To complete the proof it suffices to note that since μ=m1ϖ1+m2ϖ2, and ℓ=m1+2m2, we have
[TABLE]
Thus A(λ,μ)={1}.
∎
The following result establishes part 3 of Theorem 1.1.
Proposition 2.3**.**
Type Ar (r≥1): If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and (ℓ−∑1≤i≤rimi)=0, mj=0 for all 4≤j≤r and m3=0, then
A(λ,μ)={1,s2}.
Proof.
We begin by establishing that under the given conditions si∈/A(λ,μ) for any 2<i≤r.
Hence, it suffices to show that if 2<i≤r, then
[TABLE]
and there exists 1≤j≤r such that cj<0.
For 2<i≤r note
[TABLE]
Hence, as desired, the coefficient of αi is negative, and si∈/A(λ,μ), whenever 2<i≤r.
To complete the proof it suffices to show that 1 and s2 are both in A(λ,μ). First note that since μ=m1ϖ1+m2ϖ2+m3ϖ3, we know ℓ=m1+2m2+3m3, hence ℓ−m1=2m2+3m3. Then
[TABLE]
Since m3≥1, we have established that 1,s2∈A(λ,μ).
∎
The following result establishes part 4 of Theorem 1.1.
Proposition 2.4**.**
Type Ar (r≥1): If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and (ℓ−∑1≤i≤rimi)=0, mj=0 for all 5≤j≤r and m4=0, then
A(λ,μ)={1,s2,s3,s2s3} when m4=1 and A(λ,μ)={1,s2,s3,s2s3,s3s2,s2s3s2} when m4≥2.
Proof.
For i≥2, we compute
[TABLE]
Note that, in this case ℓ=m1+2m2+3m3+4m4; hence ℓ−m1=2m2+3m3+4m4. If i≥4, then si(λ+ρ)−ρ−μ has a negative coefficient for αi, and hence si∈/A(λ,μ). Thus, A(λ,μ) can only contain Weyl group elements formed by the simple reflections s2 and s3.
Using the similar calculations for i=j where i,j≥2 and ∣i−j∣=1, note that
[TABLE]
where the same conversion from fundamental weights to simple roots applies. Notice that if m4=1, s2s3 is the only element of this form in the Weyl alternation set. If m4≥2, then s3s2 is also in the Weyl alternation set. Also, we have
[TABLE]
so s2s3s2 is also in the Weyl alternation set when m4≥2.
∎
2.1. Low rank examples: types A2 and A3
In this section, we consider the Lie algebra of types A2 and A3. By fixing the weight μ and varying the weight λ, we can make a diagram over the fundamental weight lattice of the Lie algebra by coloring two weights with the same color when they have the same Weyl alternation set.
Figure 1 presents Weyl alternation diagrams in the Lie algebra of type A2 where we fix a weight μ in the root lattice and vary λ over the fundamental weight lattice. The case of μ=0 (Figure 1(a)) first appeared in [PHThesis] and later, Harris in collaboration with Lescinsky and Mabie, generalized these results to all integral weights μ of sl3(C) [HLM]. Figures 1(b) and 1(c) illustrate the cases where μ=ϖ1+ϖ2 and μ=2ϖ1+2ϖ2, respectively.
As depicted in Figure 1, the colored regions represent the intersection of specific hyperplanes (within the fundamental weight lattice) arising from the action of the elements of the Weyl group on the input of the partition function in (2).
That is, each element σ of the Weyl group gives rise to a region on the fundamental weight lattice arising from the intersection of some hyperplanes, from which one can create Weyl alternation diagrams.
To illustrate how to create Weyl alternation diagrams we fix r=3. For each pair (ℓϖ1,μ) for which m(ℓϖ1,μ)=1, we have that μ=m1ϖ1+m2ϖ2+m3ϖ3 and so ℓ−(m1+2m2+3m3)=4p for p∈Z≥0. We also have that
[TABLE]
Hence, if i≥2, then
[TABLE]
The above equations imply that
s2∈A(λ,μ) if and only if m3+2p≥1,
s3∈A(λ,μ) if and only if p≥1
s2s3∈A(λ,μ) if and only if m3+2p≥2 and p≥1
s3s2∈A(λ,μ) if and only if m3+2p≥1 and p≥2
s2s3s2∈A(λ,μ) if and only if m3+2p≥2 and p≥2
Hence, if p=0, then A(λ,μ)={1}; if p=1, then A(λ,μ)={1,s2,s3,s2s3}; and if p≥2, then A(λ,μ)={1,s2,s3,s2s3,s3s2,s2s3s2}.
From the inequalities above, the inclusion of σ∈W in the Weyl alternation set A(λ,μ) as prescribed relies only on the value of the nonnegative integer p.
We note that the diagrams illustrating the allowable μ is either one of the highlighted weights in Figure 2 or there is no such μ, meaning that A(λ,μ)=∅.
These low rank computations illustrate the complexity in determining whether an element σ of the Weyl group is in a specific Weyl alternation set A(λ,μ). Deciding whether or not σ is in A(λ,μ) requires determining if σ(λ+ρ)−ρ−μ lies in a specified region of the integral root lattice. Although this may sound simple, the number of such regions as we vary over all pairs of weights of a Lie algebra is currently unknown.
3. Type B
In this section, we consider the Lie algebra of type Br for r≥2. The set of simple roots is given by Δ={α1,…,αr}
and the set of positive roots is given by
Φ+={εi−εj,εi+εj:1≤i<j≤n}∪{εi:1≤i≤r}.
The fundamental weights are defined by ϖi=ε1+⋯+εi for 1≤i≤r−1 and ϖr=21(ε1+ε2+⋯+εr). The weight ρ is defined as half the sum of the positive roots, which is equivalent to ρ=ϖ1+⋯+ϖr. The Weyl group elements are generated by the simple root reflections s1,s2,…,sr which act on the simple roots and fundamental weights as follows. If 1≤i≤r−1, then si(αi)=−αi, si(αi−1)=αi−1+αi, si(αi+1)=αi+αi+1, and sr(αr)=−αr, sr(αr−1)=αr−1+2αr.
For any 1≤i,j≤r,
[TABLE]
We provide a proof of Theorem 1.2 by establishing the following technical results.
Lemma 3.1**.**
In type Br with ℓ,r∈2Z≥0 under the condition ℓ−1=∑1≤i<rimi+2rmr with mi∈2Z≥0, then A(ℓϖ1,μ)=∅.
Proof.
Suppose ℓ is even. If r is even, then 2rmr is even since mr is even. Thus, ∑1≤i<rimi+2rmr is also even. Then ℓ=1+∑1≤i<rimi+2rmr must be odd. Since ℓ is even, then there are no values of the mi’s (1≤i≤r) which satisfy the condition. Hence, A(ℓϖ1,μ)=∅.
∎
Proposition 3.1**.**
Let σ=si1si2⋯sik where the indices of the simple reflections form a collection of nonconsecutive integers 2≤i1,…,ik≤r. If λ=ℓϖ1 and μ=(ℓ−1−2b−3c)ϖ1+bϖ2+cϖ3, for ℓ,b,c nonnegative integers such that ℓ>b and ℓ>c, then we have σ(λ+ρ)−μ−ρ is a nonnegative Z-linear combination of positive roots.
Proof.
Let σ=si1si2⋯sik for some collection of nonconsecutive integers 2≤i1,…,ik≤r. Note that σ(λ)=λ and σ(ρ)=ρ−∑j=1kαij. Hence, if λ=ℓϖ1 and μ=(ℓ−1−2b−3c)ϖ1+bϖ2+cϖ3, for ℓ,b,c nonnegative integers such that ℓ>b and ℓ>c, then we have σ(λ+ρ)−μ−ρ=λ−μ−∑j=1kαij=ϖ1+(b+2c)α1+cα2−∑j=1kαij.
∎
Proposition 3.2**.**
In type Br with r≥3 and an odd, positive integer ℓ, all σ∈A(ℓϖ1,(ℓ−1−4k)ϖ1+2kϖ2) where k∈Z≥0 are products of nonconsecutive si’s with 1<i≤r.
Proof.
Let λ=ℓϖ1 and μ=(ℓ−1−4k)ϖ1+2kw2 for some k∈Z≥0. First, we note that ℓ−1=ℓ−1−4k+2(2k) so ℓ>2k and λ≻μ.
For r≥3, note that ℓϖ1=ℓ(α1+⋯+αr). We claim that the only elements σ∈W for which ℘(σ(ℓϖ1+ρ)−ρ−(ℓ−1−4k)ϖ1−2kϖ2)>0 are σ=si1si2⋯sim such that i1,i2,…,im are nonconsecutive integers between 2 and r (inclusive).
(⇐) Let σ=1, then 1(λ+ρ)−(μ+ρ)=λ−μ=ϖ1+2kα1 is a nonnegative Z-linear combination of positive roots. Thus 1∈A(λ,μ). Proposition 3.1 implies that if σ=si1si2⋯sim for some nonconsecutive integers 2≤i1,…,im≤r, then σ∈A(λ,μ).
(⇒) Suppose σ∈A(λ,μ). We proceed by induction on ℓ(σ). If ℓ(σ)=0, then σ=1, which satisfies the needed condition. If ℓ(σ)=1, then σ=si for some 1≤i≤r. If i=1, then s1(λ+ρ)−(μ+ρ)=ϖ1+2kα1−(ℓ+1)α1, which implies s1∈/A(λ,μ). Thus, σ∈A(λ,μ) cannot contain s1 in its reduced word expression.
If 1<i≤r, then si(λ+ρ)−(μ+ρ)=ϖ1+2kα1−αi, and si∈A(λ,μ) and si is of the required form.
If ℓ(σ)=2, then σ=sisj for distinct integers i,j satisfying 1<i,j≤r.
Without loss of generality, assume i<j.
If i,j are consecutive integers, then
i=j−1, with 1<i,j<r or i=r−1 and j=r. For the case 1<i,j<r, note
sj−1sj(λ+ρ)−(μ+ρ)=λ−μ−2αj−1−αj.
For the case i=r−1 and j=r, note
sr−1sr(λ+ρ)−(μ+ρ)=λ−μ−2αr−1−αr.
None of these can be written as a nonnegative Z-linear combination of positive roots. Thus, sr−1sr, srsr−1, sj−1sj, sjsj−1∈/A(λ,μ).
Moreover, any σ∈W containing sjsj−1 or sj−1sj in its reduced word expression cannot be in A(λ,μ) for all 2<j≤r. The case where i,j are nonconsecutive was already considered in Proposition 3.1.
∎
Before stating our next result we recall that the Fibonacci numbers are defined by the recurrence relation Fn=Fn−1+Fn−2 for all n≥3 and F1=F2=1.
Corollary 3.1**.**
In type Br with r≥3, if ℓ is an odd positive integer and k∈Z≥0, then ∣A(ℓϖ1,(ℓ−1−4k)ϖ1+2kϖ2)∣=Fr+1.
Proof.
We show ∣A(λ,μ)∣=Fr+1 for Br using induction. First, if r=2, then ∣A(λ,μ)∣=∣{1,s2}∣=2=F3. If r=3, then ∣A(λ,μ)∣=∣{1,s2,s3}∣=3=F4. Assume for all r, 4≤r≤m, ∣A(λ,μ)∣=∣{1,s2}∣=Fr+1, the (r+1)st Fibonacci number. We consider the case when r=m+1. All elements σ∈W consisting of nonconsecutive products of the generators s2,s3,…,sm will either contain sm+1 or not. If they don’t contain sm+1, then by our inductive hypothesis, the number of Weyl group elements consisting of nonconsecutive products of s2,s3,…,sm is given by Fm+1. If the Weyl group elements contain sm+1, then we must count the number of nonconsecutive products of s2,s3,…,sm−1, which by our inductive hypothesis is given by Fm. Therefore, ∣A(λ,μ)∣=Fm+Fm+1=Fm+2.
∎
The following result establishes part 3 of Theorem 1.2.
Proposition 3.3**.**
In type Br with r>3 and an odd, positive integer ℓ, all σ∈A(ℓϖ1,(ℓ−1−4j−6k)ϖ1+2jϖ2+2kϖ3) where j∈Z≥0, k∈Z>0, ℓ>2j, and ℓ>4k are either products of nonconsecutive si’s with 1<i≤r or products of the form si1…sims2s3 with nonconsecutive sip’s for 4<ip≤r.
Proof.
(⇐) Let σ=1. Then 1(λ+ρ)−ρ−μ=ϖ1+(2j+4k)α1+2kα2, which is a nonnegative Z-linear combination of positive roots. Thus 1∈A(λ,μ). Note that by Proposition 3.1, σ=si1si2⋯sim∈A(λ,μ). We need only show that σ=si1si2⋯sims2s3∈A(λ,μ). Note that si1⋯sims2s3(λ+ρ)=si1⋯sim(λ+ρ−2α2−α3)=λ+ρ−∑p=1kαip+si1⋯sik(−2α2−α3). Since the sip’s are a collection of nonconsecutive integers between 5 and r, inclusive, si1⋯sim(−2α2−α3)=−2α2−α3. Hence, si1⋯sims2s3(λ+ρ)−ρ−μ=ϖ1+(2j+4k)α1+(2k−2)α2−α3−∑p=1kαip∈A(λ,μ) if k=0.
(⇒) Suppose σ∈A(λ,μ). Note that λ−μ=ϖ1+(2j+4k)α1+2kα2. We proceed by induction on ℓ(σ). If ℓ(σ)=0, then σ=1, which satisfies the needed condition. If ℓ(σ)=1, then σ=si for some 1≤i≤r. If i=1, then s1(λ+ρ)−μ−ρ=ϖ1+(2j+4k)α1+2kα2−(ℓ+1)α1∈/A(λ,μ). Thus σ∈A(λ,μ) cannot contain s1 in its reduced word expression. If 1<i≤r, then si(λ+ρ)−μ−ρ=ϖ1+(2j+4k)α1+2kα2−αi and so si∈A(λ,μ) and si is of the required form.
Now suppose ℓ(σ)=2. Then σ=sism for distinct integers i,m such that 1<i,m≤r. Without loss of generality, assume i<m. If i=2 and m=3, then we have s2s3(λ+ρ)−μ−ρ=ϖ1+(2j+4k)α1+2kα2−2α2−α3, which means s2s3∈A(λ,μ) if k=0 and s2s3 is of the required form. Note that s3s2(λ+ρ)−μ−ρ=λ−μ−α2−2α3, which is not in A(λ,μ) and is not of the required form. If i,m are any other consecutive pair of integers, then i=m−1 with either 2<i,m<r or i=r−1 and m=r. Suppose 2<i,m<r. Then sm−1sm(λ+ρ)−μ−ρ=λ−μ−2αm−1−αm. If i=r−1 and m=r, then sr−1sr(λ+ρ)−μ−ρ=λ−μ−2αr−1−αr. Note that neither of these cases can be written as a nonnegative Z-linear combination of positive roots. Thus, sr−1sr,srsr−1,sm−1sm,smsm−1∈/A(λ,μ) for all cases except s2s3. Moreover, any σ∈W containing smsm−1 or sm−1sm in its reduced word expression cannot be in A(λ,μ) for all 2<m≤r, except s2s3. If i,m are nonconsecutive, then by Proposition 3.1, sism∈A(λ,μ).
∎
Corollary 3.2**.**
In type Br with r>3, if ℓ is an odd positive integer and j∈Z≥0, k∈Z>0, with ℓ>2j, and ℓ>4k , then ∣A(ℓϖ1,(ℓ−1−4j−6k)ϖ1+2jϖ2+2kϖ3)∣=2Fr.
Proof.
With λ,μ as given, we show ∣A(λ,μ)∣=2Fr for Br. As in Corollary 3.1, we know there are Fr+1 elements coming from the products of nonconsecutive si’s, 1<i≤r. By a similar argument, there are Fr−2 elements σ=si1si2⋯sims2s3 for some nonconsecutive integers 4<i1,i2,…,im≤r. Thus, ∣A(λ,μ)∣=Fr+1+Fr−2=Fr+Fr−1+Fr−2=2Fr.
∎
The condition for Br such that ℓ is even and r is odd is quite different from the three conditions we posed in Theorem 1.2. When r is odd and ℓ is even, the constraint ℓ−1=∑1≤i<rimi+2rmr forces mr=2(2k+1), since mi is even for all i. The Weyl alternation sets A(λ,μ) for B3 are easy to compute and remain quite small. But when r=5, the Weyl alternation sets grow very quickly. In the three conditions discussed in Theorem 1.2, the elements of the Weyl alternation sets had a nice pattern of being products of mostly nonconsecutive simple reflections; however, we don’t see similar patterns for the case when r is odd and ℓ is even, as seen in the tables in Appendix A. As the coefficient mr increases, the elements in the Weyl alternation set A(λ,μ) become even more complicated.
3.1. Low rank examples: B2 and B3
In this section we consider the Lie algebra of types B2 and B3. By fixing the weight μ and varying the weight λ we can make a diagram over the fundamental weight lattice of the Lie algebra by coloring two weights with the same color when they have the same Weyl alternation set.
Figure 3 presents the Weyl alternation diagram when μ=0. Observe that there are 24 distinct colored dots and the integral weights without a colored dot represent those weights that have empty Weyl alternation set. This illustrates the 25 distinct Weyl alternation types given in [PHThesis, Theorem 2.3.1].
Now we fix r=3. For each pair (ℓϖ1,μ) for which m(ℓϖ1,μ)=1, we have that μ=m1ϖ1+m2ϖ2+m3ϖ3 and m1,m2,m3∈2Z≥0 so ℓ=m1+2m2+23m3. We also have that
[TABLE]
Therefore, if m3=0, then A(λ,μ)={1,s2,s3}. If m3∈2Z>0, then A(λ,μ)={1,s2,s3,s2s3}. To display when σ∈W is an element of A(λ,μ), we provide Figure 4, which show different views of the allowed choices for μ.
4. Type G2
The positive roots of G2 are
Φ+={α1,α2,α1+α2,2α1+α2,3α1+α2,3α1+2α2} and the fundamental weights in terms of the simple roots are given by
ϖ1=2α1+α2 and
ϖ2=3α1+2α2.
Theorem 4.1**.**
Let λ=ℓϖ2 with ℓ≥1 be a weight of G2. If μ=m1ϖ1+m2ϖ2, where m1,m2∈Z≥0 and 3ℓ−1=2m1+3m2, then
A(λ,μ)={1,s1} when λ=0 and A(0,μ)=∅.
Proof.
Let λ=ℓϖ2 with ℓ≥1 be a weight of G2.
If μ=m1ϖ1+m2ϖ2, where m1,m2∈Z+ and 3ℓ−1=2m1+3m2, then we can parametrize m1=3n+1 and m2=ℓ−1−2n, with 0≤n≤⌊2ℓ−2⌋. Then
[TABLE]
and some simple computations establish
[TABLE]
Based on these computations and the fact that 0≤n≤⌊2ℓ−2⌋, we have that the coefficient of at least one of the simple roots is negative for the Weyl group elements that are not 1 or s1. This means that all Weyl group elements, aside from 1 and s1, will not contribute to the multiplicity m(λ,μ). Thereby implying that A(λ,μ)={1,s1} whenever ℓ≥1 and μ=(3n+1)ϖ1+(ℓ−1−2n)ϖ2. In the case where ℓ=0, then there are no μ satisfying the required condition, hence A(λ,μ)=∅.
∎
Figure 5 gives the Weyl alternation diagram of G2, when we take μ=0 and vary λ over the Z-linear combinations of the fundamental weights of G2. Observe that there are 60 distinct colored dots and the integral weights without a colored dot represent those weights that have empty Weyl alternation set. This illustrates the 61 distinct Weyl alternation types given in [PHThesis, Theorem 2.6.1]. Figure 6 displays the allowable choices of μ that result in the Weyl alternation set as stated in Theorem 4.1.
5. A q−analog
Recall that for a weight ξ, the value of the q-analog of Kostant’s partition function is the polynomial ℘q(ξ)=c0+c1q+⋯+ckqk, where cj denotes number of ways to write ξ as a nonnegative integral sum of exactly j positive roots. Then the q-analog of Kostant’s weight multiplicity formula is defined, in [LL], as:
mq(λ,μ)=σ∈W∑(−1)ℓ(σ)℘q(σ(λ+ρ)−(μ+ρ)).
Of course, for multiplicity one pairs of weights evaluating mq(λ,μ) at q=1 recovers the fact that m(λ,μ)=1. In what follows we provide the q-multiplicity for the pairs of weights considered in Theorems 1.1, 1.2, and 1.3.
In Appendix B we provide tables for the value of mq(ℓϖ1,μ), for multiplicity one pairs in type Ar (with 1≤r≤10) and Br (with 3≤r≤10) in the cases where 1≤ℓ≤100 and with all possible μ so that m(ℓϖ1,μ)=1. These tables provide ample evidence in support of the following.
Conjecture 1.1.
If λ,μ are a pair of weights for which m(λ,μ)=1, then mq(λ,μ)=qf(r), where f(r) is a function of r, the rank of the Lie algebra.
Unfortunately, the techniques we present, do not lend themselves well to provide a direct proof of the above conjecture and we would welcome a new approach to such a problem.
5.1. Type A
In this section we provide formulas for the q-multiplicity of the pairs of weight described in Theorem 1.1. To do so we first recall that for any nonnegative integers n,m,
[TABLE]
see [HarrisLauber, Proposition 3.1] for a proof. Our work will also require the following technical result.
Proposition 5.1**.**
If m,n,k are nonnegative integers, then
[TABLE]
Proof.
It is known that ℘(mα1+nα2+kα2)=∑f=0min(m,n,k)∑d=0min(m−f,n−f)∑e=0min(n−f−d,k−f)1, where
d accounts for the number of times the positive root α1+α2 appears
e accounts for the number of times the positive root α2+α3 appears and
f accounts for the number of times the positive root α1+α2+α3 appears
in the partition of the weight mα1+nα2+kα3, [HPPS2018, Proposition 2]. Note that it is enough to count these instances as they uniquely determine the number of times we must use the positive roots α1, α2, and α3. That is we must use m−d−f of the positive root α1, n−d−e−f of the positive root α2, and k−e−f of the positive root α3. It suffices to keep track of the total number of positive roots used in each of these partitions. This is given by d+e+f+(m−d−f)+(n−d−e−f)+(k−e−f)=m+n+k−d−e−2f from which the result follows.
∎
Theorem 1.4.
In type Ar with r≥1 and ℓ∈Z+:
- (1)
If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and (ℓ−∑1≤i≤rimi)=0, mj=0 for all 3≤j≤r, then
mq(λ,μ)=qm2.
2. (2)
If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and (ℓ−∑1≤i≤rimi)=0 and mj=0 for all 4≤j≤r and m3=0, then
mq(λ,μ)=qm2+3m3.
3. (3)
If λ=ℓϖ1, μ=∑1≤i≤rmiϖi, where mi∈Z+ and (ℓ−∑1≤i≤rimi)=0, mj=0 for all 5≤j≤r and m4=0, then
mq(λ,μ)=qm2+3m3+6 when m4=1 and mq(λ,μ)=qm2+3m3+6m4 when m4≥2.
Proof.
To establish part (1), recall from part (2) of Theorem 1.1 that A(λ,μ)={1} and 1(λ+ρ)−ρ−μ=λ−μ=m2α1 (see equation (3)). Hence mq(λ,μ)=℘q(m2α1)=qm2.
To establish part (2), recall from part (3) of Theorem 1.1 that A(λ,μ)={1,s2} and 1(λ+ρ)−ρ−μ=λ−μ=(m2+2m3)α1+m3α2 and s2(λ,μ)=(m2+2m3)α1+(m3−1)α2 (see equation (4)). Hence
[TABLE]
To establish part (3), recall from part (4) of Theorem 1.1 that A(λ,μ)={1,s2,s3,s2s3} when m4=1 and A(λ,μ)={1,s2,s3,s2s3,s3s2,s2s3s2} when m4≥2. From the proof of Proposition 2.4 note that when m4=1 we have
[TABLE]
Hence, by Proposition 5.1 we have
[TABLE]
Using the same technique as above, we find that
[TABLE]
while using (5) yields
[TABLE]
Therefore
[TABLE]
In the case where m4≥2 we know A(λ,μ)={1,s2,s3,s2s3,s3s2,s2s3s2}. From the proof of Proposition 2.4 note that when m4≥2 we have
[TABLE]
By applying Proposition 5.1 to Equation (6) we have that
[TABLE]
We now split the sum indexed by d in Equation (7) to reflect when the fact that m3+2m4−f−d≤m4−f if and only if m3+m4≤d. This implies that ℘q(1(λ+ρ)−ρ−μ) is equal to
[TABLE]
By applying the geometric sum formula to Equation (8) we find that ℘q(1(λ+ρ)−ρ−μ) is equal to
[TABLE]
Using the same technique as above we find that
[TABLE]
where
[TABLE]
By taking the sum of the above terms, with their sign corresponding to the length of the associated Weyl group element, we find that
[TABLE]
5.2. Type B
In this section we provide formulas for the q-multiplicity of the pairs of weight described in Theorem 1.2.
We begin by proving the following technical results.
Lemma 5.1**.**
If 1≤i<j≤r, then
[TABLE]
Proof.
The only positive roots that we can use are of the form αi′+⋯+αj′ where i≤i′≤j′≤j. We now proceed by induction on j−i. If j−i=0, then we are partitioning αi, which only has one partition with a single positive root. Thus ℘q(αi)=q=q(1+q)0. If j−i=1, then we are partitioning αi+αi+1. The possible partitions are the one consisting of only the single simple roots αi and αi+1, and the partition using the positive root αi+αi+1, which contributes a q2 and q to ℘q(αi+αi+1), respectively. Thus ℘q(αi+αi+1)=q2+q=q(1+q)1=q(1+q)2−1. Assume that the result holds for j−i≤k−1. Now we count the the partitions of αi+αi+1+⋯+αj+αj+1 by considering which positive root contains αj+1. These positive roots are of the form αℓ+⋯+αj+1, with i≤ℓ≤j+1. Thus
[TABLE]
By the induction hypothesis, this yields
[TABLE]
as claimed.
∎
Lemma 5.2**.**
If ℓ≥3 and x,y∈Z>0, then
[TABLE]
where ⟨m,n⟩:=℘q(mα1+nα2) for all nonnegative integers m and n.
Proof.
We will account for all partitions of xα1+yα2+α3+α4+⋯+αℓ by considering which positive root contains αℓ. These roots are of the form αi+⋯+αℓ where 1≤i≤ℓ. Thus
[TABLE]
where the factor of q is accounting for the use of the positive root αi+⋯+αℓ. We now proceed by induction on ℓ. Notice that when ℓ=3, by Equation (9) we know
[TABLE]
When ℓ=4, by Equation (9) we know
[TABLE]
Now assume that for any 3≤k≤ℓ−1 the result holds. Notice that when k=ℓ, using the recurrence in Equation (9) and the induction hypothesis we have
[TABLE]
as claimed.
∎
Theorem 1.5.
In type Br with r≥3:
- (1)
If ℓ is a positive odd integer and k∈Z≥0, then mq(ℓϖ1,(ℓ−1−4k)ϖ1+2kϖ2)=q2k+r.
2. (2)
If ℓ,r∈2Z≥0 under the condition ℓ−1=∑1≤i<rimi+2rmr with mi∈2Z≥0, then mq(ℓϖ1,μ)=0.
3. (3)
If j,k∈Z≥0 and k=0, then mq(ℓϖ1,(ℓ−1−4j−6k)ϖ1+2jϖ2+2kϖ3)=qr+2j+6k−2.
Proof.
Part (1): By Lemma 3.2 σ∈A(ℓϖ1,(ℓ−1−4k)ϖ1+2kϖ2) with k∈Z≥0 if and only if σ=si1si2⋯sim where 2≤i1,i2,…,im≤r are nonconsecutive integers. Without loss of generality, we assume that 2≤i1<i2<…<im≤r. We consider the cases where sr is a factor in σ and when it is not.
Case 1. Suppose σ=si1⋯simsr, where 2≤i1<i2<⋯<im≤r−2 are nonconsecutive integers. Then ℓ(σ)=1+m where m≥0. Notice
[TABLE]
Let τ=σ(λ+ρ)−ρ−μ=τ0+τ1+⋯+τm−1+τm, where
[TABLE]
Partitioning τ as a sum of positive roots reduces to partitioning each of the weights τ1,…,τm. Since the sets of positive roots used to partition τi and τj are disjoint for all i=j, we know that
[TABLE]
Then by Lemma 5.2 we have that
[TABLE]
By Lemma 5.1 we have that
[TABLE]
and
[TABLE]
Thus
[TABLE]
Now notice that there exist (mr−2−m) elements σ∈A(λ,μ) such that σ contains sr and ℓ(σ)=1+m≥1, while \max\{\ell(\sigma):\sigma\in\mathcal{A}(\lambda,\mu)\mbox{ containing s_{r}}\}=\lfloor\frac{r-2}{2}\rfloor. So we may compute that
[TABLE]
Proposition 3.3 in [Harris] established that if r≥0, then
[TABLE]
Applying this result to Equation (11) yields
[TABLE]
Case 2. Suppose σ=si1⋯sim, where 2≤i1,i2,…,im≤r−1 are nonconsecutive integers. Then ℓ(σ)=m where m≥0. Notice
[TABLE]
Let τ=σ(λ+ρ)−ρ−μ=τ0+τ1+⋯+τm−1+τm, where
[TABLE]
As in the previous case, partitioning τ as a sum of positive roots reduces to partitioning each of the weights τ0,τ1,…,τm. Since the sets of positive roots used to partition τi and τj are disjoint for all i=j, we know that
[TABLE]
Then by Lemma 5.2 we have that
[TABLE]
By Lemma 5.1 we have that
[TABLE]
and
[TABLE]
Thus
[TABLE]
Now notice that there exist (mr−1−m) elements σ∈A(λ,μ) such that σ does not contains sr and ℓ(σ)=m≥0, while \max\{\ell(\sigma):\sigma\in\mathcal{A}(\lambda,\mu)\mbox{ not containing s_{r}}\}=\lfloor\frac{r-1}{2}\rfloor. So we may compute that
[TABLE]
Applying (12) to Equation (11) yields
[TABLE]
The result then follows from (13) and (16) since
[TABLE]
Part (2) follows directly from the fact that in this case A(ℓϖ1,μ)=∅.
Part (3). In this case σ∈A(λ,μ) is of the form σ=si1si2⋯sim with 2≤i1,i2,…,im≤r are nonconsecutive integers or σ=si1si2⋯sims2s3 with 5≤i1,i2,…,im≤r are nonconsecutive integers. We consider cases depending on whether σ∈A(λ,μ) contains sr or not, and whether it contains one of s2,s3, or s2s3, or contains none of them. Then using the same techniques as we did in Part (1) of this proof we can compute the value of the q-multiplicity by taking the sum of the terms given in Table 1, where
[TABLE]
Thus the q-multiplicity can be computed by taking the sum of the terms in the Table 1 which establishes
[TABLE]
∎
5.3. Type G2
In this section we provide a formula for the q-multiplicity of the pairs of weight described in Theorem 1.3.
Theorem 1.6.
Let λ=ℓϖ2 with ℓ≥1 be a weight of G2. If μ=m1ϖ1+m2ϖ2, where m1,m2∈Z≥0 and 3ℓ−1=2m1+3m2, then
mq(λ,μ)=qn+2 where m1=3n+1.
Proof.
Since the positive roots of G2 are
Φ+={α1,α2,α1+α2,2α1+α2,3α1+α2,3α1+2α2}
and by Proposition 4.1 we have that if λ=ℓϖ2 with ℓ≥1 and μ=m1ϖ1+m2ϖ2, where m1,m2∈Z≥0 and 3ℓ−1=2m1+3m2, then
[TABLE]
as claimed.
∎
6. Future Work
We provide a few directions for future study. The first direction would be to provide a proof of Conjecture 1.1. Note that in Appendix B we provide ample evidence in support of this conjecture, yet the complication remaining is that the Weyl alternation sets are very complicated to compute for all remaining multiplicity one pairs of weights given in [BZ].
Another interesting direction for research is to
study the Weyl alternation poset. This is the poset arising from the set containment of the Weyl alternation sets A(λ,μ) for all integral weights of a Lie algebra. By fixing μ=0 and varying λ in the integral weight lattice, Figures 7, 8, and 9 illustrate the Weyl alternation posets of the Lie algebras of type A2, B2 and G2, respectively.
Studying these posets opens a new avenue for research in this field.
References
\enddoc@text
Appendix A Tables of Weyl alternation sets
In this section we consider the Lie algebras of type Br (with 2≤r≤6) and provide tables for the value of mq(λ,μ) for pairs of weights λ=ℓϖ1 (with 1≤ℓ≤10) and μ such that m(λ,μ)=1 along with the Weyl alternation sets. Note that for ℓ∈Z>0 the set Mℓ denotes the set of weights μ for which m(ℓϖ1,μ)=1. These tables show the rapid growth of the Weyl alternation sets and illustrate the complication in the description of its elements.
Appendix B Tables in support of Conjecture 1.1
In this section we consider the Lie algebra of type Ar (1≤r≤10) and provide tables for the value of mq(λ,μ) for pairs of weights λ=ℓϖ1 (with 1≤ℓ≤10) and μ such that m(λ,μ)=1. Note that for ℓ∈Z>0 the set Mℓ denotes the set of weights μ for which m(ℓϖ1,μ)=1. In all of these cases note that mq(λ,μ) is a power of q.
