Combinatorics of Local Regions

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 27 April 2013

Combinatorics of Local Regions

Two conjectures are stated in [Ram1998-2, (1.3) and (1.11)] when W is a crystallographic reflection group. The first gives necessary and sufficient conditions for ℱ(γ,J) (as defined in (2.16)) to be nonempty when γ is dominant and the second determines the form of ℱ(γ,J) as an interval in the weak Bruhat order when γ is dominant and integral. Loszoncy [Los1999] proved the second conjecture (Theorem 5.2 below). His theorem implies the nonemptiness conjecture of [Ram1998-2] under the additional assumption that γ is integral. Here we review Loszoncy’s proof and prove the nonemptiness conjecture in full generality. We give an example (Example 5.4) to show that integrality is necessary in Theorem 5.2. Finally, we provide Example 5.7, which shows that one cannot expect analogous statements to hold when W is noncrystallographic.

Let R be the root system of a finite real reflection group W and fix a set of positive roots R+={α>0} in R. A set of positive roots S is closed if it satisfies the condition:

If α,β∈S and  a,b>0 are such that aα+ bβ∈R+, then aα+ bβ∈S.

The following theorem characterizes the sets which appear as inversion sets of elements of W. Recall that R(w) denotes the inversion set of w; see equation (2.4). This result is in [Bjo1984, Proposition 2], but is stated there without proof and we are not aware of a published proof. The following proof was shown to us by J. Stembridge and appears in the thesis of D. Waugh [Wau1995].

Theorem 5.1. Let W be a real reflection group. A set of positive roots S is equal to R(w) for some element w∈W if and only if S is closed and Sc=R+\S is closed.

Proof.

⟹: Let w∈W and suppose that α,β∈R(w) and aα+bβ is a positive root. Then w(aα+bβ)=a(wα)+b(wβ) is a negative root since wα and wβ are both negative roots. So R(w) is closed. Similarly, one shows that R(w)c is closed.

⟸: Assume that S is closed and that Sc is closed. We will construct w such that R(w)=S by finding a reduced word w=si1…sik for w. This is done by induction on the size of S, with the induction step being the combination of the two steps below.

Step 1: S contains a simple root.

Let α be a root of minimal height in S and assume that α=∑icαiαi, cαi∈ℝ≥0, is not simple. Then

⟨α,αi⟩>0 for some i,since 0<⟨α,α⟩ =∑i=1ncαi ⟨a,αi⟩.

Since α is not simple, α≠αi, and so both sαiα and αi are positive roots. Since sαiαi=α-⟨α,αi∨⟩αi and αi both have lower height than α, then both must be in Sc. But then the equation

α=sαiα+ ⟨α,αi∨⟩ αi

contradicts the assumption that Sc is closed. So α is simple.

Step 2: Let αi1 be a simple root in S and let S1=si1(S\{αi1}).

Claim: S1 is closed and S1c is closed.

Let α,β∈S1 and assume that aα+bβ is a positive root. Then

si1 (aα+bβ)=a si1α+b si1β∈Sand aα+bβ∈S1,

or

asi1α+b si1β=αi1 andaα+bβ=- αi1.

The second is impossible since si1αi1 is not a positive root. So aα+bβ∈S1 and S1 is closed.

Let α,β∈S1c and suppose that aα+bβ is a positive root. Since si1α and si1β are not in S, si1(aα+bβ)∉S. So aα+bβ∉S1. Thus S1c is closed.

□

An element γ∈𝔥ℂ is dominant (resp. integral) if γ(αi)∈ℝ≥0 (resp. γ(αi)∈ℤ) for all simple roots αi. The closure S‾ of a set of positive roots S is the smallest closed set of positive roots containing S.

Theorem 5.2. Let W be a crystallographic reflection group and let R be the crystallographic root system of W. Let γ∈𝔥ℂ be dominant and integral, and set

Z(γ)= {α>0 | ⟨γ,α⟩=0} andP(γ)= {α>0 | ⟨γ,α⟩=1} .

Let J⊆P(γ) be such that

if β∈J,α∈ Z(γ)and β-α∈R+, then β-α∈J,

and set

ℱ(γ,J)= { w∈W | R(w) ∩Z(γ)=∅, R (w)∩P(γ)=J } .

Then there exist elements wmin,wmax∈W such that

R(wmin)= J‾,R (wmax)= (P(γ)\J)∪Z(γ)‾c ,and ℱ(γ,J)= [wmin,wmax],

where Kc denotes the complement of K in R+ and [wmin,wmax] denotes the interval between wmin and wmax in the weak Bruhat order.

Proof.

By Theorem 5.1, the element wmin∈W will exist if J‾c is closed. Assume that β=β1+β2 where β∈J‾, β1,β2∈R+. We must show that β1∈J‾ or β2∈J‾. Since β∈J‾,

β=δ1+…δm with δi∈J.

We will decompose β=δ1+…+δm into two pieces β1=δ1+…+δk+η1 and β2=η2+δk+2+…+δm, via the following inductive procedure. Since

0< ⟨ β1+β2, β1+β2 ⟩ =∑i ⟨β1+β2,δi⟩ ,then  ⟨β1+β2,δj⟩ >0for some j.

By reindexing the δi we can assume that j=1. Thus ⟨β1,δ1⟩>0 or ⟨β2,δ1⟩>0 and we may assume that ⟨β1,δ1⟩>0. Since sδ1β1=β1-⟨β1,δ1∨⟩δ1 is a root and R is crystallographic, β1-δ1 is also a root. If β1-δ1 is a negative root, then

β1=β1andβ =(δ1-β1)+ δ2+…+δm,

gives the desired decomposition. If β1-δ1∈R+, then

β1+β2= δ1+ ((β1-δ1)+β2) and (β1-δ1)+ β2=δ2+…+ δm,

and so we may inductively apply this procedure to decompose β′= (β1-δ1)+ β2=δ2+…+ δm.

In this way we conclude that, after possible reindexing of the δi, either

β1=δ1+…+δk andβ2=δk+1 +…+δm,

or

β1=δ1+…+δk +η1andβ2= η2+δk+2+…+ δm,

where η1 and η2 are positive roots such that η1+η2=δk+1. In the first case it is immediate that β1,β2∈J‾. In the second case ⟨γ,δk+1⟩=⟨γ,η1+η2⟩=1, and so ⟨γ,η1⟩≤1 and ⟨γ,η2⟩≤1. Thus, since γ is dominant and integral, one of η1,η2 is in Z(γ) and the other is in P(γ). If η1∈Z(γ), η2=δk+1-η1 and the condition on J implies that η2∈J. Similarly, if η2∈Z(γ), then η1∈J. Thus β1∈J‾ or β2∈J‾. So J‾c is closed. Since J‾ is closed and J‾c is closed, Theorem 5.1 shows that there is an element wmin∈W such that R(wmin)=J‾.

The same method can be used to establish the existence of wmax: one must show that (P(γ)\J)∪Z(γ)‾c is closed and this is accomplished by similar arguments.

By the definition of ℱ(γ,J) an element w∈W is in ℱ(γ,J) if

J‾⊆R(w)⊆ (P(γ)\J)∪Z(γ)‾c .

Since the weak Bruhat order is the order determined by inclusions of R(w) [Bjo1984, Proposition 3] the result is a consequence of the existence of the elements wmin and wmax.

□

Remark 5.3. An alternative way to establish the existence of wmax in the proof of Theorem 5.2 is to use the conjugation involution

ℱ(γ,J) ⟷1-1 ℱ(γ,J)′ w ⟷ wu-1 where (γ,J)′= (-uγ,-u(P(γ)\J)), (5.1)

where u is the minimal length coset representative of w0Wγ and w0 is the longest element of W. That this is a well defined involution is proved in [Ram1998-2, (1.7)]. This involution takes wmax for ℱ(γ,J) to wmin for ℱ(γ,J)′. In terms of the weak Bruhat order, the structure of the interval ℱ(γ,J)′ is the same as the structure of the interval ℱ(γ,J) but with all relations reversed.

Example 5.4. The integrality of γ is necessary in Theorem 5.2. Let W=I2(4) be the dihedral group of order 8 (the Weyl group of type C2). The root system for type C2 is determined by simple roots

α1=2ε1and α2=ε2-ε1

where {ε1,ε2} is an orthonormal basis of 𝔥ℝ*=ℝ2. Let c1=c2=1 be the parameters for ℍ. If γ=(1/2)ε2 (see Figure 3), then Z(γ)={α1}, P(γ)={α1+2α2}, and γ is dominant but γ(α2) is not integral. The set J=P(γ) satisfies the condition in Theorem 5.2, but J‾=J is not an inversion set for any w∈W since J‾c is not closed.

The following method of reducing to the integral root subsystem of a weight is standard in the theory of highest weight modules for finite dimensional complex semisimple Lie algebras; see [Jan1980]. This method turns out to be an efficient tool for reducing the nonemptiness conjecture of [Ram1998-2] to the statement in Theorem 5.2.

Let R[γ]={α∈R | ⟨γ,α∨⟩∈ℤ}. For any α,β∈R[γ],

⟨γ,(sαβ)∨⟩ =⟨sαγ,β∨⟩= ⟨γ,β∨⟩- ⟨γ,α∨⟩ ⟨α,β∨⟩∈ℤ,

and so R[γ] is a root system with Weyl group W[γ]=⟨sα | α∈R[γ]⟩⊆W. If τ∈W[γ], then the R[γ]-inversion set of τ is

R[γ](τ)= { α>0 | τα< 0,α∈R[γ] } =R(τ)∩R[γ] . Hα1 Hα2 Hα1+2α2 Hα1+α2+δ Hα1+α2 Hα1+2α2+δ Hα1+2α2-δ Hα2+δ Hα2-δ Hα1-δ Hα1+δ Hα1+α2-δ γ

Theorem 5.5. Let W be a crystallographic reflection group and let R be the crystallographic root system of W. Let γ∈𝔥ℂ such that Re(γ) is dominant and set

Z(γ)= {α>0 | ⟨γ,α⟩=0} andP(γ)= {α>0 | ⟨γ,α⟩=1}.

Let J⊆P(γ) be such that

if β∈J, α∈ Z(γ) and β -α∈R+,then β-α∈J.

Then ℱ(γ,J)={w∈W | R(w)∩Z(γ), R(w)∩P(γ)=J} is nonempty.

Proof.

Since γ is dominant and integral for the root system R[γ], it follows from Theorem 5.2 that there is an element w in W[γ] such that

R[γ](w)∩ Z(γ)=∅and R[γ](w)∩ P(γ)=J,

where R[γ](w)={α∈R[γ] | α>0,wα<0}. Usually R(w) is strictly larger than R[γ](w) but it is still true that

R(w)∩Z(γ)=∅ andR(w)∩ P(γ)=J,

since all roots of P(γ) and Z(γ) are in R[γ]. So w∈ℱ(γ,J).

□

When W is crystallographic, we can use the method of the proof of Theorem 5.5 in combination with the result of Theorem 5.2 to give a precise description of the set ℱ(γ,J) for all central characters γ∈𝔥ℂ. By choosing γ appropriately in its W-orbit we may assume that Re(γ) is dominant.

Define

W[γ]= { σ∈W | R (σ)∩R[γ] =∅ } .

Each w∈W has a unique expression

w=στwithσ∈ W[γ],τ∈ W[γ],

and

R(w)∩R[γ]= R(τ)∩R[γ]= R[γ](τ).

In this way the elements of W[γ] are coset representatives of the cosets in W/W[γ].

Since P(γ)⊆R[γ] and Z(γ)⊆R[γ] it follows that

ℱ(γ,J)= { στ∈W | σ∈ W[γ],τ∈ ℱ[γ](γ,J) } , (5.2)

where

ℱ[γ](γ,J)= { τ∈W[γ] |  Rγ(τ)∩P(γ) =J,R(w)∩ Z(γ)=∅ } . (5.3)

Since ℱ(γ,J)=ℱ(Re(γ),J) and γ is dominant and integral for the root system R[γ], Theorem 5.2 has the following corollary.

Corollary 5.6. With notations and assumptions as in Theorem 5.5

ℱ(γ,J)= ℱ[γ](γ,J)= W[γ]· [τmax,τmin],

where, ℱ[γ](γ,J) is as in (5.3), τmax and τmin in W[γ] are determined by R[γ](τmax)=J‾ and R[γ](τmin)=(P(γ)\J)∪Z(γ)‾c, where the complement is taken in the set of positive roots of R[γ].

This refined version of Theorem 5.2 is reminiscent of the reduction to real central character given in [BMo1993].

The following example shows that Theorem 5.5 does not naturally extend to noncrystallographic reflection groups. Note that such a generalization necessarily involves modifying the closure condition on J to be

if β∈J,α∈Z(γ) ,a∈ℝ>0, and β-a α∈R+,then  β-aα∈J.

Example 5.7. Let W=I2(n) be the dihedral group of order 2n, n even, with root system chosen as in Section 3 (so all roots are the same length). Let γ be such that Z(γ)={β0} and P(γ)={βn/4,βn/2,β3n/4} (this γ is an example of γq in Table 1). Then the subset J={βn/4,β3n/4}⊆P(γ) satisfies the generalized closure condition above since βn/2 cannot be written as βn/4-aβ0 for any a∈ℝ>0. However, ℱ(γ,J)=∅ since there are no chambers which are on the positive side of both Hβ0 and Hβn/2 and on the negative side of both Hβn/4 and Hβ3n/4.

Notes and References

This is an excerpt of a paper entitled Representations of graded Hecke algebras, written by Cathy Kriloff (Department of Mathematics, Idaho State University, Pocatello, Idaho 83209-8085) and Arun Ram.

Research of the first author supported in part by an NSF-AWM Mentoring Travel Grant. Research of the second author supported in part by National Security Agency grant MDA904-01-1-0032 and EPSRC Grant GR K99015.

page history