The Combinatorial Curve Neighborhoods of Affine Flag Manifold in

Type A(1)n−1

Songul Aslan

Dissertation submitted to the Faculty of the

Virginia Polytechnic Institute and State University

in partial fulfillment of the requirements for the degree of

Doctor of Philosophy

in

Mathematics

Constantin L. Mihalcea, Chair

William J. Floyd

Mark M. Shimozono

Daniel D. Orr

May 15, 2019

Blacksburg, Virginia

Keywords: Affine Flag Manifolds, Schubert Varieties, Curve Neighborhoods, Moment

Graph, Combinatorial Curve Neighborhoods

Copyright 2019, Songul Aslan

The Combinatorial Curve Neighborhoods of Affine Flag Manifold in Type

A(1)n−1

Songul Aslan

(ABSTRACT)

Let X be the affine flag manifold of Lie type A(1)n−1 where n ≥ 3 and let Waff be the associated

affine Weyl group. The moment graph for X encodes the torus fixed points (which are

elements of the affine Weyl group Waff) and the torus stable curves in X . Given a fixed point

u ∈ Waff and a degree d = (d0, d1, ..., dn−1) ∈ Zn≥0, the combinatorial curve neighborhood is

the set of maximal elements in the moment graph of X which can be reached from u′ ≤ u

by a chain of curves of total degree ≤ d. In this thesis we give combinatorial formulas and

algorithms for calculating these elements.

The Combinatorial Curve Neighborhoods of Affine Flag Manifold in Type

A(1)n−1

Songul Aslan

(GENERAL AUDIENCE ABSTRACT)

The study of curves on flag manifolds is motivated by questions in enumerative geometry

and physics. To a space of curves and incidence conditions one can associate a combinatorial

object called the ‘combinatorial curve neighborhood’. For a fixed degree d and a (Schubert)

cycle, the curve neighborhood consists of the set of all elements in the Weyl group which can

be reached from the given cycle by a path of fixed degree in the moment graph of the flag

manifold, and are also Bruhat maximal with respect to this property. For finite dimensional

flag manifolds, a description of the curve neighborhoods helped answer questions related to

the quantum cohomology and quantum K theory rings, and ultimately about enumerative

geometry of the flag manifolds.

In this thesis we study the situation of the affine flag manifolds, which are infinite dimen-

sional. We obtain explicit combinatorial formulas and recursions which characterize the

curve neighborhoods for flag manifolds of affine Lie type A. Among the conclusions, we men-

tion that, unlike in the finite dimensional case, the curve neighborhoods have more than one

component, although all components have the same length. In general, calculations reflect

a close connection between the curve neighborhoods and the Lie combinatorics of the affine

root system, especially the imaginary roots.

Contents

1 Introduction 1

1.1 Statement of results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Sketch of Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Preliminaries 16

2.1 Weyl Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.2 Affine Weyl Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Dynkin Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.4 Hecke Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Preliminary Lemmas and zd 23

3.1 The element zd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2 Lemma about the Hecke Product . . . . . . . . . . . . . . . . . . . . . . . . 27

iv

3.3 Lemma about Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.4 Lemmas about Lengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4 The Moment Graph and Curve Neighborhoods 45

5 Calculation of the Curve Neighborhoods 48

5.1 Curve Neighborhood Γd(id) for Finite Degrees . . . . . . . . . . . . . . . . . 48

5.2 Curve Neighborhood Γc(id) . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

5.3 Curve Neighborhood Γc+α(id) . . . . . . . . . . . . . . . . . . . . . . . . . . 51

5.4 Curve Neighborhood Γmc+α(id) . . . . . . . . . . . . . . . . . . . . . . . . . 61

5.5 Curve Neighborhood Γmc(id) . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

5.6 Curve Neighborhood Γd(id) for any d > c . . . . . . . . . . . . . . . . . . . 71

5.7 Reduction from Γd(w) to Γd(id) . . . . . . . . . . . . . . . . . . . . . . . . . 80

Biblography 82

v

List of Figures

1.1 The moment graph for the affine flag manifold of type A(1)2 . . . . . . . . . . . 4

2.1 Dynkin diagram of affine root system in type A(1)n−1. . . . . . . . . . . . . . . 20

vi

Chapter 1

Introduction

Let X = G/P be a flag manifold defined by a semisimple complex Lie group G and a

parabolic subgroup P and let Ω ⊂ X be a Schubert variety. Fix an effective degree d ∈

H2(X). The (geometric) curve neighborhood Γd(Ω) is the closure of the union of all rational

curves of degree d in X which intersect Ω. The curve neighborhood Γd(Ω) was originally

studied by Buch, Chaput, Mihalcea and Perrin; see [4]. It plays a key role in the study of

the quantum cohomology and quantum K-theory ring of X. It was proved in [4] that Γd(Ω)

is irreducible whenever Ω is irreducible. In particular, if Ω is a Schubert variety in X then

Γd(Ω) is also a Schubert variety. In [5], Buch and Mihalcea provided an explicit combinatorial

formula for the Weyl group element corresponding to Γd(Ω) when Ω is a Schubert variety in

X. It has been also shown that the calculation of the curve neighborhoods is encoded in the

moment graph of X which is a graph encoding the T -fixed points and the T -stable curves

in X where T is a maximal torus of G. In [12] Withrow has studied curve neighborhoods

1

2

to compute a presentation for the small quantum cohomology ring of a particular Bott-

Samelson variety in Type A. Moreover, Mihalcea and Shifler use the technique of curve

neighborhoods to prove a Chevalley formula in the equivariant quantum cohomology of

the odd symplectic Grassmannian, see [11]. The curve neighborhood Γd(Ω) has also been

studied in the case when X is an affine flag manifold by Mare and Mihalcea, see [9]. Mare

and Mihalcea have defined an affine version of the quantum cohomology ring and gave a

combinatorial description of the curve neighborhood for “small” degrees. Furthermore, a

combinatorial formula for the curve neighborhood of affine flag manifold of Lie type A11

has been recently given by Norton and Mihalcea; see [10]. In this dissertation we give a

combinatorial formula for the curve neighborhoods of Schubert varieties in the affine flag

manifold of Lie type A(1)n−1 for any degree where n ≥ 3.

1.1 Statement of results

Here, we will introduce some notation and recall some definitions before we give an overview

of our results. Let G be the special linear group SLn(C). Let B ⊂ G be the Borel subgroup,

the set of upper triangular matrices, and let T ⊂ B be the maximal torus, the set of diagonal

matrices. We denote by g the Lie algebra sln(C) of G. Let Π be the standard root system

associated to the triple (T,B,G). Also, let ∆ = α1, α2, ..., αn−1 ⊂ Π be the set of simple

roots. This determines a partition of Π = Π+ t Π− such that Π+ = Π ∩⊕n−1

i=1 Z≥0 αi is the

set of positive roots and Π− = Π ∩⊕n−1

i=1 Z≤0 αi is the set of negative roots (these roots are

3

described in chapter 2). Denote by W the Weyl group. The Weyl group W is generated by

the simple reflections, s1, s2, ..., sn−1, where si := sαi , i = 1, 2, ..., n−1. For w ∈ W, the length

of w is denoted by `(w). We set `(id) = 0. Now, let gaff be the affine Kac-Moody Lie algebra

associated to the Lie algebra g. Denote by Πaff the affine root system associated to gaff. Let

∆aff = α0, α1, ..., αn−1 ⊂ Πaff be the affine simple roots. This determines a partition of

Πaff into positive and negative affine roots: Πaff = Π+aff tΠ−aff where Π−aff = −Π+

aff. We denote

by Πre,+aff the set of affine positive real roots. Let Waff be the affine Weyl group of type A

(1)n−1

which is generated by the simple reflections s0, s1, .., sn−1 where si := sαi , i = 0, 1, ..., n− 1.

For w ∈ Waff, the length of w is denoted by `(w). We set `(id) = 0. Now, let X be the affine

flag manifold in type A(1)n−1. The (undirected) moment graph for X is given by the following

two conditions:

• The set of vertices is Waff.

• There is an edge between u, v ∈ Waff in the moment graph if and only if there exists

an affine positive real root α such that v = usα. This situation is denoted by

uα−→ v

We say that the degree of this edge is α.

A chain from u to v in the moment graph is a succession of adjacent edges, starting with u

and ending with v;

π : u = u0β0−→ u1

β1−→ ...βk−2−→ uk−1

βk−1−→ uk = v.

4

The degree of the chain π is deg(π) = β0 + β1 + ...+ βk−1. A chain is called increasing if the

length is increasing at each step i.e `(ui) > `(ui−1) for all i. There is a partial order on the

elements of Waff which is known as Bruhat partial order. It is defined by u < v if and only

if there exists an increasing chain starting with u and ending with v.

Example 1.1.1. Assume that X is the affine flag manifold associated to A(1)2 . Then the

moment graph for X up to elements that can be obtained by a chain of degree at most

α0 + α1 + α2 is given in the following figure with each edge labelled by its degree;

Figure 1.1: The moment graph for the affine flag manifold of type A(1)2 .

5

Remark 1.1.2. The vertices and edges of this graph correspond to the T -fixed points and

T -stable curves in the affine flag manifold, see [7, §12] for details.

A degree d is an n-tuple of nonnegative integers (d0, d1, ..., dn−1). There is a natural partial

order on degrees: If d = (d0, ..., dn−1) and d′ = (d′0, ..., d′n−1) then d ≥ d′ if and only if di ≥ d′i

for all i ∈ 0, 1, ..., n− 1.

Now, let a degree d and an element u of Waff be given. Then inspired by the geometric

definition of curve neighborhoods in [5] and [9], the (combinatorial) curve neighborhood

Γd(u) is the set of elements v in Waff such that:

1) v can be joined to u′ ≤ u (in the moment graph) by a chain of degree ≤ d;

2) The elements v are maximal among all elements satisfying (1).

In other words, to compute Γd(u) we first consider all the paths in the moment graph

that start with u′ such that u′ ≤ u and have a total degree at most d. Then the curve

neighborhood is given by the set of the elements with the maximal length that can be

reached by the paths.

Example 1.1.3. Assume that X is the affine flag manifold associated to A(1)2 . The moment

graph for X is shown in Figure 1.1. Now, let’s consider all the paths in the moment graph

that start with the identity element and have a degree at most α0+α1+α2. Then the elements

which can be reached by the paths and have the maximal length are the top elements in the

6

moment graph so

Γ(α0+α1+α2)(id) = s2s0s1s0, s1s2s1s0, s1s0s2s0, s0s2s0s1, s0s1s0s2, s0s1s2s1.

Now, we will state our first result which allows us to reduce the calculation of Γd(w) for any

given degree d and w ∈ Waff to the calculation of Γd(id).

Theorem 1.1.4. Let w ∈ Waff and d be any degree. Then

Γd(w) = max w · u : u ∈ Γd(id) .

Next, we will consider the curve neighborhood of the identity element at degree d =

(d1, d2, ..., dn−1) where di = 0 for some i = 1, 2, ..., n − 1. Here, using the automorphism

of the Dynkin diagram ϕ, d can be identified with a finite degree, see Sections 2.3 and 5.1

for details. In this case, by [5], the curve neighborhood Γd(id) consists of a unique element;

Γd(id) = zd (1.1)

where zd := sα · zd−α such that α is a maximal root which is smaller than and equal to d, we

refer to definition 3.1.1 for details. One might also see Theorem 3.1.2 where we show how zd

can be simplified.

Our next result is a corollary of Equation 1.1. Let α ∈ Πre,+aff such that α < c. Then

Γα(id) = sα. (1.2)

7

Example 1.1.5. Assume that the affine Weyl group is associated to A(1)4 and α = α0 + α4.

Then Γα(id) = sα0+α4 by Equation 1.2.

The rest of our results are about the instances where the degrees are not finite and proven

by using some techniques which are different than those one can find in [5] and [9].

Now, we will state the result for the curve neighborhood of the identity element at c which is

the degree corresponding to the imaginary coroot. In this case the neighborhood is described

in terms of translations. See Section 2.2 in chapter 2 for details.

Theorem 1.1.6. We have

1) Γc(id) = tγ : γ ∈ Π.

2) |Γc(id)| = n(n− 1).

3) For all w ∈ Γc(id), `(w) = 2(n− 1).

Example 1.1.7. Suppose that the affine Weyl group Waff is associated to A(1)2 . Then by

Theorem 1.1.6

Γc(id) = tα1 , tα2 , tα1+α2 , t−α1 , t−α2 , t−(α1+α2).

Moreover, for any w ∈ Γc(id), `(w) = 2(n− 1) = 2 · 2 = 4.

In the next theorem, we will generalize the case in Theorem 1.1.6 by considering a constant

m multiple of c where m ≥ 2 is a positive integer.

Theorem 1.1.8. Let m ≥ 2 be a positive integer. Then we have

8

1) Γmc(id) = tmγ : γ ∈ Π.

2) |Γmc(id)| = n(n− 1).

3) For all w ∈ Γmc(id), `(w) = 2m(n− 1).

Example 1.1.9. Suppose that the affine Weyl group Waff is associated to A(1)2 . Then by

Theorem 1.1.8

Γ10c(id) = t10α1 , t10α2 , t10(α1+α2), t−10α1 , t−10α2 , t−10(α1+α2).

Moreover, for any w ∈ Γ10c(id), `(w) = 2m(n− 1) = 2 · 10 · 2 = 40.

In the following theorem we will provide the result for the curve neighborhood of the identity

element at degree c+α where α ∈ Πre,+aff and α < c which plays a key role in calculating the

most general case in this thesis.

Theorem 1.1.10. Let α ∈ Πre,+aff be such that α < c. Then

1) Γc+α(id) = tβ′sα : β′ ∈ Πre,+aff (α) ∪ tβ′−csα : β′ ∈ Πre,+

aff (α)

2) |Γc+α(id)| = |β′ : β′ ∈ Πre,+aff (α)|

3) For all w ∈ Γc+α(id), `(w) = 2(n− 1) + `(sα)

where Πre,+aff (α) := β′ ∈ Πre,+

aff : β′ < c and either β′ ≤ c− α or both β′ > α and β′ ⊥ α.

Example 1.1.11. Let Waff be the Weyl group of type A(1)4 . We compute Γc+α(id) where α =

α0 +α4. Note that, we have six positive real roots which are smaller than c−α = α1 +α2 +α3;

9

β′1 = α1, β′2 = α2, β

′3 = α3, β

′4 = α1 + α2, β

′5 = α2 + α3, β

′6 = α1 + α2 + α3. Also, we have

only one positive root which is smaller than c, strictly bigger than α and perpendicular to

α; β′7 = α0 + α1 + α3 + α4. So by Theorem 1.1.10, we get

Γc+α(id) = tβ′1sα, tβ′2sα, tβ′3sα, tβ′4sα, tβ′5sα, tβ′6sα, tβ′7−csα

Moreover, for any w ∈ Γc+α(id), we have `(w) = 2(n−1)+`(sα) = 2(n−1)+2 supp(α)−1 =

2 · 4 + 2 · 2− 1 = 11.

Next, we will consider the generalization of the case in the previous theorem to the instance

where the degree is given by mc+ α such that m ∈ Z>0, α ∈ Πre,+aff and α < c.

Theorem 1.1.12. Let α < c be an affine positive real root and m be a positive integer. Then

1) Γmc+α(id) = tmβ′sα : β′ ∈ Πre,+aff (α) ∪ tm(β′−c)sα : β′ ∈ Πre,+

aff (α)

2) |Γmc+α(id)| = |β′ : β′ ∈ Πre,+aff (α)|

3) For all w ∈ Γmc+α(id), `(w) = 2m(n− 1) + `(sα)

where Πre,+aff (α) is defined in Theorem 1.1.10.

Example 1.1.13. Suppose that the affine Weyl group Waff is of type A(1)4 . We compute

Γ12c+α(id) where α = α0 + α4. Then, by Theorem 1.1.12, we get

Γ12c+α(id) = t12β′1sα, t12β′2

sα, t12β′3sα, t12β′4

sα, t12β′5sα, t12β′6

sα, t12(β′7−c)sα

10

where β′1 = α1, β′2 = α2, β

′3 = α3, β

′4 = α1 + α2, β

′5 = α2 + α3, β

′6 = α1 + α2 + α3 and

β′7 = α0 + α1 + α3 + α4, see Example 5.3.6. Moreover, for any w ∈ Γ12c+α(id), we have

`(w) = 2m(n− 1) + `(sα) = 2m(n− 1) + 2 supp(α)− 1 = 2 · 12 · 4 + 2 · 2− 1 = 99.

Last, we will state the result for the most general case: the curve neighborhood of the identity

element at any degree d which is strictly bigger than c.

Theorem 1.1.14. Let d = (d0, d1, ..., dn−1) > c be a degree and m = mind0, d1, ..., dn−1.

Also, assume that zd−mc = sγ1sγ2 ...sγk for some k, where this expression is obtained in

Theorem 3.1.2. Then

1) Γd(id) = tmβ′zd−mc : β′ ∈ Πre,+aff (d−mc) ∪ tm(β′−c)zd−mc : β′ ∈ Πre,+

aff (d−mc)

2) |Γd(id)| = |β′ : β′ ∈ Πre,+aff (d−mc)|

3) For all w ∈ Γd(id), `(w) = 2m(n− 1) + `(zd−mc)

where Πre,+aff (d−mc) is the set of β′ ∈ Πre,+

aff such that β′ < c and either β′ ∩ γi = ∅ or both

β′ > γi and β′ ⊥ γi for any i = 1, ..., k.

Example 1.1.15. LetWaff be the affine Weyl group associated to A(1)3 and d = (6, 5, 8, 5). Then

d = 5c+(1, 0, 3, 0) so m = 5 and d−5c = (1, 0, 3, 0). Also, zd−5c = sα0 ·sα2 ·sα2 ·sα2 = sα0sα2 .

Hence by Theorem 1.1.14 we get

Γd(id) = t5β′1zd−5c, t5β′2zd−5c, t5β′4zd−5c, t5(β′3−c)zd−5c

where β′1 = α1, β′2 = α3, β′3 = α0 +α1 +α3, β′4 = α1 +α2 +α3. Moreover, for all w ∈ Γd(id),

`(w) = 2m(n− 1) + `(zd−mc) = 2 · 5 · 3 + 2 = 32.

11

1.2 Sketch of Proofs

In this section, we will summarize some of the proofs of the results. In all cases below, we

first consider a path idβ1−→ sβ1

β2−→ sβ1sβ2 ...βr−→ w = sβ1sβ2 ...sβr in the moment graph such

that∑r

i=1 βi ≤ d. Note that, by Remark 4.0.2 we can assume that w = sβ1 · sβ2 · ... · sβr

where∑r

i=1 βi = d and βi < c for all i.

Proof of Γc(id): Let w = sβ1 · sβ2 · ... · sβr such that∑r

i=1 βi = c and βi < c for all i. Now,

note that∑r

i=2 βi = c− β1 is an affine positive real root. Moreover, sβ2 · sβ3 · ... · sβr ≤ sc−β1

by Corollary 3.2.2. Thus w = sβ1 · (sβ2 · ... · sβr) ≤ sβ1 · sc−β1 . But by Lemma 3.4.3, sβ1sc−β1

is reduced so sβ1 · sc−β1 = sβ1sc−β1 . We also know that sβ1sc−β1 = tγ for some γ ∈ Π, see the

proof of Lemma 3.4.1.

Proof of Γc+α(id): Let w = sβ1 · sβ2 · ... · sβr be such that∑r

i=1 βi = c + α and βi < c for

all i. By Lemma 5.3.2, we can also suppose that∑t

i=1 βi = c and∑r

i=t+1 βi = α for some t.

Then

w = (sβ1 · ... · sβt) · (sβt+1 · sβt+2 · ... · sβr) ≤ (sµ · sµ′) · sα

for some µ, µ′ ∈ Πre,+aff such that µ + µ′ = c by by Equation 5.1 and Theorem 5.1.3. Fur-

thermore, by using Lemma 5.3.3 we are able to show that sµ · sµ′ · sα ≤ sβ · sβ′ · sα where

β + β′ = c and either β′ ≤ c− α or both β′ > α and β′ ⊥ α. In addition, sβsβ′sα is reduced

so sβ · sβ′ · sα = sβsβ′sα by Lemma 3.4.6, and 3.4.8. Last, sβsβ′ = tβ′ if α0 /∈ supp(β′) and

sβsβ′ = tβ′−c if α0 ∈ supp(β′), see Section 5.3 for details.

12

Proof of Γmc(id): Let w = sβ1 · sβ2 · ... · sβr be such that∑r

i=1 βi = mc and βi < c for all i.

Now, note that∑r−1

i=1 βi = mc− βr = (m− 1)c+ c− βr. Let α := c− βr and α′ := βr. Then

by Equation 5.7 we have sβ1sβ2 ...sβr−1 ≤ (sβsβ′)m−1sα for some positive real roots, β and β′

such that β + β′ = c and either β′ ≤ c− α or both β′ > α and β′ ⊥ α. Then, we get

w = sβ1sβ2 ...sβr ≤ (sβ1sβ2 ...sβr−1) · sβr ≤ (sβsβ′)m−1sα · sα′ .

Furthermore, by Lemmas 5.4.1 and 5.4.2 we get (sβsβ′)m−1sα · sα′ ≤ (sν · sν′)m for some

ν, ν ′ ∈ Πre,+aff such that ν + ν ′ = c. Last, (sνsν′)

m = tmγ for some γ ∈ Π, see Section 5.5 for

details.

Proof of Γmc+α(id): In this case we prove the result by induction on m. Note that, if m = 1

then we have nothing but the case Γc+α(id). Now, assume that the statement is true for

m = k. We prove that it is also true for m = k + 1. Let w = sβ1 · sβ2 · ... · sβr be such that∑ri=1 βi = (k + 1)c + α and βi < c for all i. Moreover, by Lemma 5.3.2, we can suppose

that there is an integer p such that∑p

i=1 βi = c and∑r

i=p+1 βi = kc+ α. Then by Equation

5.1 we have sβ1 · sβ2 · ... · sβp ≤ sγsγ′ for some γ, γ′ ∈ Πre,+aff such that γ + γ′ = c and

sβp+1 · ... · sβr ≤ (sνsν′)ksα for some ν, ν ′ ∈ Πre,+

aff such that ν + ν ′ = c where either ν ′ ≤ c−α

or both ν ′ > α and ν ′ ⊥ α by the induction assumption. Thus,

w = (sβ1 · sβ2 · ... · sβp) · (sβp+1 · ... · sβr) ≤ (sγsγ′) · ((sνsν′)ksα) ≤ (sγ · sγ′) · ((sν · sν′)k · sα).

Then by using Lemmas 5.3.3, 5.4.1, 5.4.3 and 5.4.2 we are able to prove that

(sγ · sγ′) · ((sν · sν′)k · sα) ≤ (sβ · sβ′)k+1 · sα

13

for some β, β′ ∈ Πre,+aff such that β + β′ = c and either β′ ≤ c − α or both β′ > α and

β′ ⊥ α. Moreover, by Lemmas 3.4.6 and 3.4.8, (sβsβ′)k+1sα is reduced so (sβ · sβ′)k+1 · sα =

(sβsβ′)k+1sα. Last, the result follows by the fact that sβsβ′ = tβ′ if α0 /∈ supp(β′) and

sβsβ′ = tβ′−c if α0 ∈ supp(β′), see Section 5.4 for details.

Proof of Γd(id) for d > c: Let w = sβ1 · sβ2 · ... · sβr such that∑r

i=1 βi = d and βi < c

for all i. In addition, we can suppose that there is an integer p such that∑p

i=1 βi = c and∑ri=p+1 βi = d− c, by Lemma 5.3.2. We will prove the statement by induction on m.

First, suppose that m = 1. Then by Equation 5.1, sβ1 · sβ2 · ... · sβp ≤ sα · sα′ for some

α, α′ ∈ Πre,+aff such that α+ α′ = c. Moreover, sβp+1 · ... · sβr ≤ zd−c, by Theorem 5.1.1. Thus

w = (sβ1 · sβ2 · ... · sβp) · (sβp+1 · ... · sβr) ≤ (sα · sα′) · zd−c.

Now, by Lemma 5.6.1 we have sα · sα′ · sγ1 · sγ2 · ... · sγk ≤ sβ · sβ′ · sγ1 · sγ2 · ... · sγk for some

affine positive real roots, β, β′ ∈ Πre,+aff such that we have either β′ ∩ γi = ∅ or both β′ > γi

and β′ ⊥ γi for any i = 1, 2, ..., k.

Now, we will suppose that the statement is true for m and prove that it is also true for

m+ 1. Again, by Equation 5.1 we have sβ1 · sβ2 · ... · sβp ≤ sα · sα′ for some α, α′ ∈ Πre,+aff such

that α + α′ = c and sβp+1 · ... · sβr ≤ (sµ · sµ′)m · zd−(m+1)c for some µ, µ′ ∈ Πre,+aff such that

µ+ µ′ = c where either µ′ ∩ γi = ∅ or µ′ > γj and µ′ ⊥ γj by induction assumption. Thus

w = (sβ1 · sβ2 · ... · sβp) · (sβp+1 · ... · sβr) ≤ (sα · sα′) · (sµ · sµ′)m · zd−(m+1)c.

Now, observe that by Equation 5.2 and Lemma 5.6.2 we have (sα ·sα′) ·(sµ ·sµ′)m ·zd−(m+1)c ≤

(sβ ·sβ′)m+1 ·sγ1 ·sγ2 · ... ·sγk for some affine positive real roots, β, β′ such that β+β′ = c where

14

either β′ ∩ γi = ∅ or both β′ > γi and β′ ⊥ γi for any i = 1, 2, ..., k. Furthermore, by Lemma

3.4.9 , zd−(m+1)c = sγ1sγ2 ...sγk is reduced so sγ1sγ2 ...sγk = sγ1 · sγ2 · ... · sγk which follows by

(sβ · sβ′)m+1 · sγ1 · sγ2 · ... · sγk = (sβ · sβ′)m+1 · (sγ1sγ2 ...sγk) = (sβ · sβ′)m+1 · zd−(m+1)c. Now

by Lemma 3.4.10, the element (sβsβ′)m+1zd−(m+1)c is reduced so (sβ · sβ′)m+1 · zd−(m+1)c =

(sβsβ′)m+1zd−(m+1)c.

Here, we will provide an outline of the rest of the thesis. Chapter 2 covers the background

on the Weyl and the affine Weyl group with the associated root systems corresponding to

the affine flag manifold as well as the Hecke product that we will need throughout the thesis.

In addition, we prove several preliminary lemmas which will be used for proving the results.

In chapter 3 we recall some basic facts about the affine flag manifold of type A(1)n−1 and give

the definition of the moment graph for the affine flag manifold and the (combinatorial) curve

neighborhoods.

Chapter 4 is dedicated to the results. We discuss Γd(id) in Sections 5.1 through 5.6. In

particular, in Section 5.1 we prove Equations 1.1 and 1.2 which are the results for the finite

degrees. We provide the proof of Theorem 1.1.6 in Section 5.2 which states the result for

the degree c. The result has a central role in proving the rest of the results in this thesis.

In Section 5.3 we prove Theorem 1.1.10 which covers the result for the degree c + α where

α ∈ Πre,+aff and α < c. The result is generalized in Theorem 1.1.12 which is proved in Section

5.4. We then present the proof of Theorem 1.1.8 which generalizes the case in Theorem 1.1.6

in Section 5.5. Furthermore, we prove Theorem 1.1.14 that provides the result for the most

general case in Section 5.6. In Section 5.7 we prove Theorem 1.1.4 which implies that for any

15

given w ∈ Waff and degree d, calculation of Γd(w) can be reduced to calculation of Γd(id).

Each result is followed by an example of the corresponding case.

Chapter 2

Preliminaries

In this chapter we will set up notations and recall some basic facts about the affine root

system in type A(1)n−1. We refer to [2], [7, §1], [8, §3], and chapters 1 through 8 in [6] for

further details.

2.1 Weyl Group

Let G be the Lie group SLn(C). Let B ⊂ G be the Borel subgroup, the set of upper

triangular matrices, and T ⊂ B be the maximal torus, the set of diagonal matrices. We

denote by g the Lie algebra sln(C) of G. Let E be the subspace of Rn which consists of n-

tuples for which the coordinates sum to 0. Let ε1, ε2, ..., εn be the standard basis for Rn and

let αi := εi− εi+1 for 1 ≤ i ≤ n− 1. Let ∆ = α1, α2, ..., αn−1 ⊂ h∗ be the simple roots and

∆∨ = α∨1 , α∨2 , ..., α∨n−1 ⊂ h be the corresponding coroot simple roots, where h is the Cartan

16

17

subalgebra of g. Let Q =⊕n−1

i=1 Zαi ⊂ h∗ and Q∨ =⊕n−1

i=1 Zα∨i ⊂ h be the root and coroot

lattice. Let Π = εi−εj : i 6= j be the standard root system associated to the triple (T,B,G)

corresponding to ∆, where Π = Π+tΠ− such that Π+ = Π∩⊕n−1

i=1 Z≥0 αi = εi−εj : i < j

is the set of positive roots and Π− = Π ∩⊕n−1

i=1 Z≤0 αi = εi − εj : i > j is the set of

negative roots. Due to the fact that αi = α∨i for all i in type A, we may identify Q with Q∨,

and h∗ with h. Let 〈, 〉 : h∗ × h → C be the natural pairing. Denote by W the Weyl group.

The Weyl group W is generated by the simple reflections, s1, s2, ..., sn−1, where si := sαi ,

i = 1, 2, ..., n− 1. For w ∈ W, the length of w, denoted `(w), is the smallest positive integer

k such that w = si1si2 ...sik , 1 ≤ ij ≤ n− 1. Such an expression is called a reduced expression

of w. We set `(id) = 0. W acts on h∗ by

si(µ) = µ− 〈µ, α∨i 〉αi for µ ∈ h∗

For all w ∈ W,µ, λ ∈ h∗, we have 〈w · µ,w · λ〉 = 〈µ, λ〉. For each α ∈ Π there is a w ∈ W

and 1 ≤ i ≤ n− 1 such that α = w · αi. The reflection of α is given by sα = wsiw−1 which

is independent of the choice of w and i. Moreover, it is well-known that `(w) = |α ∈ Π+ :

w · α < 0| for all w ∈ W . We can identify W with the symmetric group Sn via the map

si 7→ (i, i+ 1).

2.2 Affine Weyl Group

Let gaff be the affine Kac-Moody Lie algebra associated to the Lie algebra g. We have

gaff = L(g) ⊕ Cc ⊕ Cd, where L(g) := g ⊗ C[t, t−1] (t ∈ C∗) is the space of all Laurent

18

polynomials in g and c is a central element with respect to the Lie bracket in gaff. We

denote by haff the Cartan subalgebra of gaff which is given by haff := h ⊕ Cc ⊕ Cd. Let

〈, 〉 : h∗aff × haff → C be the natural pairing. The affine root system Πaff associated to gaff

consists of

• mδ + α, where α ∈ Π and m ∈ Z, which are called the affine real roots.

• mδ, m ∈ Z \ 0, which are called the imaginary roots.

For any root α ∈ Π ⊂ Πaff we identify α as a linear function on h∗ whose restriction to h

is α and 〈α, c〉 = 〈α, d〉 = 0 and the imaginary root δ ∈ h∗ is defined by δ|h ⊕ Cc = 0 and

〈δ, d〉 = 1. Let ∆aff = α0 := δ − θ, α1, ..., αn−1 ⊂ Πaff be the affine simple roots where

θ = α1+...+αn−1 ∈ Π is the highest root. This determines a partition of Πaff into positive and

negative affine roots: Πaff = Π+aff t Π−aff where Π−aff = −Π+

aff and Π+aff consists of the elements

mδ+α such that either m > 0 or both α ∈ Π+ and m = 0. Denote by Πre,+aff the set of affine

positive real roots. We denote by Qaff =⊕n−1

i=0 Zαi ⊂ h∗aff and Q∨aff =⊕n−1

i=0 Zα∨i ⊂ haff the

affine root and coroot lattices. For a, b ∈⊕n−1

i=0 Zαi we say that a ≤ b if b − a is a linear

combination of non-negative coefficients of α0, α1, ..., αn−1. Denote by a < b if a ≤ b and

a 6= b. Furthermore, in type A we identify δ with c so we set c = α0 + θ =∑n−1

i=0 αi.

Let Waff = WnQ∨ be the affine Weyl group corresponding to W . We denote by tµ the image

of µ ∈ Q∨ in Waff. For all w ∈ W and µ ∈ Q∨, we have tw·µ = wtµw−1. Waff is generated by

the simple reflections, s0, s1, ..., sn−1, where si := sαi , i = 0, 1, ..., n− 1 and

si(µ) = µ− 〈µ, α∨i 〉αi for µ ∈ h∗aff and 0 ≤ i ≤ n− 1.

19

The affine root system Πaff is Waff-invariant. For an affine real root α, we have α = w · αi

for some w ∈ Waff and 0 ≤ i ≤ n− 1. Then the associated reflection sα is independent of the

choice of w and i. Also, sα = wsiw−1 and sα(µ) = µ− 〈µ, α∨〉α for µ ∈ h∗aff. For an affine

real root β = α + mδ we have sβ = sαtmα∨ = sαtmα and, in particular s0 = sθt−θ∨ = sθt−θ,

see [8, §3] for further details.

Let S = s0, s1, s2, ..., sn−1. Then (Waff, S) is a Coxeter system with the following relations;

s2i = 1 for all i

sisi+1si = si+1sisi+1 for all i

sisj = sjsi for i, j not adjacent, i 6= j

where the indices are taken modulo n; see [3, p.263]. Each w 6= e in Waff can be written

in the form w = si1si2 ...sik for some sij (not necessarily distinct) in S. If k is as small as

possible, call it the length of w, written l(w), and call any expression of w as a product of k

elements of S a reduced expression. We set `(e) = 0. For an α ∈ Πaff if α =∑n−1

i=0 aiαi then

the support of α is supp(α) = αi : ai 6= 0. We have the following equations:

• If α ∈ Πre,+aff such that α < c then `(sα) = 2|supp(α)| − 1.

• If w ∈ Waff then `(w) = |α ∈ Π+aff : w · α < 0|.

• If x = wtλ ∈ Waff then by Lemma 3.1 in [8], we have

`(x) =∑γ∈Π+

|χ(w · γ < 0) + 〈λ, γ〉| (2.1)

where χ(P ) = 1 if P is true and χ(P ) = 0 otherwise.

20

Given u, v ∈ Waff, we say that the product uv is reduced if `(uv) = `(u) + `(v).

Next, we will set up some notations for affine positive real roots which are smaller than c; let

pi,i := αi for 0 ≤ i ≤ n−1, and pi,j := αi+αi+1+...+αj−1+αj for 0 ≤ i < j ≤ n−1 such that

j− i < n−1 and pi,j := α0 +α1 + ...+αj +αi+αi+1 + ...+αn−1 where 0 ≤ j < i−1 ≤ n−2.

Note that, spi,i = si and spi,j = sisi+1...sj−1sjsj−1...si−1si = sjsj−1...si+1sisi+1...sj−1sj if

i < j, and

spi,j = sjsj−1...s1s0sisi+1...sn−2sn−1sn−2...si+1sis0s1...sj−1sj

= sisjsj−1...s1s0si+1...sn−2sn−1sn−2...si+1s0s1...sj−1sjsi

if i > j, are some of the reduced expressions for the reflections which will be used throughout

this paper.

2.3 Dynkin Diagram

In this section, we will consider the Dynkin diagram for Πaff which is given in the figure

below.

Figure 2.1: Dynkin diagram of affine root system in type A(1)n−1.

21

Let ϕ : ∆aff → ∆aff be a map given by ϕ(αi) = αi+1 if 0 ≤ i ≤ n − 2 and ϕ(αn−1) = α0.

Note that, 〈αi, α∨j 〉 = 〈ϕ(αi), ϕ(αj)∨〉 for all i, j, since the Dynkin diagram for Πaff is simply

laced and circular. Thus ϕ is an automorphism of the diagram. It is clear that the order of

ϕ is n and ϕ preserves the partial order ”≤” on the roots.

2.4 Hecke Product

We describe some of the curve neighborhoods in terms of the Hecke product. For that reason,

we recall the definition of the Hecke product and some of its properties in this section. We

refer to [5, §3] for further details. For u ∈ Waff and i ∈ 0, 1, ..., n− 1, define

u · si =

usi If `(usi) > `(u)

u otherwise.

Let u, v ∈ Waff and let v = si1si2 ...sik be any reduced expression for v. Define the Hecke

product of u and v by

u · v = u · si1 · si2 · ... · sik ,

where the simple reflections are multiplied to u in left to right order. This product is

independent of the chosen reduced expressions for v. It provides a monoid structure on the

affine Weyl group Waff. Furthermore, we have the following properties of the Hecke product:

Let u, v, v′, w ∈ Waff.

a) The Hecke product is associative, i.e. (u · v) · w = u · (v · w).

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b) If v ≤ v′ then u · v · w ≤ u · v′ · w.

c) We have u ≤ u · v, v ≤ u · v, uv ≤ u · v, and `(u · v) ≤ `(u) + `(v).

d) If uv is reduced then uv = u · v and `(u · v) = `(u) + `(v).

e) The element v = (w · u)u−1 satisfies v ≤ w and vu = v · u = w · u.

Chapter 3

Preliminary Lemmas and zd

In this chapter we will prove some lemmas which are used throughout the thesis. We will

also discuss the element zd and prove a theorem about it that is crucial for some of the

results.

3.1 The element zd

In this section we give the definition of zd which is the unique element in Γd(id) where

d = (d0, d1, ..., dn−1) ∈ Qaff such that di = 0 for some i. Furthermore, the most general case

of the curve neighborhood Γd(id) where d > c is described in terms of zd. This section also

includes a theorem where we show how one can simplify zd. We refer the reader to [5, §4]

and [1] for more details.

23

24

Definition 3.1.1. Let d = (d0, d1, ..., dn−1) ∈ Qaff such that di = 0 for some i. If d = 0 then

set zd = id. Otherwise we set zd = sα · zd−α where α is any maximal root which is smaller

than and equal to d.

zd is well-defined by induction on d.

Theorem 3.1.2. Let d = (d0, d1, ..., dn−1) be a degree such that di = 0 for some i. Then zd =

sγ1sγ2 ...sγr for some positive real roots, γ1, γ2, ..., γr such that either supp(γi) and supp(γj)

are disconnected or both γi ⊥ γj and γi, γj are comparable, for any 1 ≤ i, j ≤ r such that

i 6= j. Moreover, zd = sγ1sγ2 ...sγr is reduced and `(zd) =∑r

i=1(2 |supp(γi)| − 1).

Proof. Let zd = sβ1 · sβ2 · ... · sβr for some positive real roots, β1, β2, ..., βr. By definition,

we have either supp(βi) and supp(βj) are disconnected or βi, βj are comparable, for any

1 ≤ i, j ≤ r such that i 6= j, see [5, §4]. So sβi · sβj = sβj · sβi for any 1 ≤ i, j ≤ r, by

Lemma 3.2.1. Now, assume that βi and βj are comparable but not perpendicular for some

1 ≤ i, j ≤ r such that i 6= j. Here, we can suppose that βi ≥ βj, without loss of generality.

Now, if βi = pk,l for some 0 ≤ k, l ≤ n− 1 then we have three cases; either βj = pk,q where

q 6= l or βj = pq,l where q 6= k or βj = pk,l. First, suppose that βj = pk,q where q 6= l.

Note that, both sβi and sβj have a reduced expression which starts and ends with the simple

reflection, sk this implies that sβi · sk = sβi . We will show that sβi · sβj = sβi · sβj−αk where

αk is the simple root. Here, we will use the convention, sa := id if a = 0. Now, observe that

sβj = sk · sβj−αk · sk. Furthermore, sβi · sβj−αk = sβj−αk · sβi since βj − αk < βi by Lemma

25

3.2.1. Thus

sβi · sβj = sβi · sk · sβj−αk · sk = sβi · sβj−αk · sk = sβj−αk · sβi · sk

= sβj−αk · sβi = sβi · sβj−αk .

Here, notice that βi and βj − αk are perpendicular. Second, suppose that βj = pq,l where

q 6= k. Then, one can show that sβi · sβj = sβi · sβj−αl where αl is the simple root, by using

similar arguments in the previous case. Also, note that βi and βj − αl are perpendicular.

Last, assume that βj = pk,l. Then again by the previous two cases one can show that

sβi · sβj = sβi · sβj−αk−αl where αk and αl are the simple reflections. Moreover, βi and

βj −αk−αl are perpendicular. Hence, we can assume that either supp(βi) and supp(βj) are

disconnected or both βi, βj are comparable and βi ⊥ βj, for any 1 ≤ i, j ≤ r such that i 6= j.

Next, we will show that zd = sβ1sβ2 ...sβr is reduced. Note that βi < c for all i, by def-

inition. So `(sβi) = 2|supp(βi)| − 1 and we have `(zd) = `(sβ1sβ2 ...sβr) ≤∑r

i=1 `(sβi) =∑ri=1 2|supp(βi)| − 1. Furthermore, for a root ν, we have sβ1sβ2 ...sβr(ν) = ν −

∑ri=1 <

ν, β∨i > βi since βi ⊥ βj, for any 1 ≤ i, j ≤ r such that i 6= j. Now, assume that ν is an affine

positive real root such that ν ≤ βi for some i. Also, assume that ν and βi are not perpen-

dicular. This implies that < ν, β∨i > is either equal to 2 or 1. if < ν, β∨i >= 2 then βi = ν.

But then ν ⊥ βj for all j ∈ 1, 2, ..., r \ i, which follows by sβ1sβ2 ...sβr(ν) = −ν < 0.

Now, suppose that < ν, β∨i >= 1 then we have ν ⊥ βj for all j such that supp(βi) and

supp(βj) are disconnected since ν ≤ βi. So if ν and βj are not perpendicular for some βj

where βj 6= βi we have to have βj < βi which implies that < ν, β∨j > is either equal to 1

or −1 and there is at most one such root as βj. Now, if < ν, β∨j >= 1 then either βj < ν

26

or ν < βj but since βj < βi and βj ⊥ βi we have to have βj < ν. If < ν, β∨j >= −1 then

βj ∩ ν = ∅. Thus we have either sβ1sβ2 ...sβr(ν) = ν − βi− βj or sβ1sβ2 ...sβr(ν) = ν − βi + βj.

Here, observe that in either case sβ1sβ2 ...sβr(ν) < 0. Now, let βi = pk,l and ν = pt,q. Then

t = k or q = l. If k ≤ l then there are 2l − 2k + 1 = 2|supp(βi)| − 1 such ν and if

k > l then there are 2n + 2l − 2k + 1 = 2|supp(βi)| − 1 such ν. Hence, if A is the set

of all positive real roots, ν such that sβ1sβ2 ...sβr(ν) < 0 where ν ≤ βi and, ν and βi are

not perpendicular for some i = 1, 2, .., r then |A| =∑r

i=1 2|supp(βi)| − 1. Also, note that

A ⊆ γ ∈ Πre,+aff : sβ1sβ2 ...sβr(γ) < 0 and we have

|A| =∑r

i=1 2|supp(βi)| − 1 ≤ |γ ∈ Πre,+aff : sβ1sβ2 ...sβr(ν) < 0|

= `(sβ1sβ2 ...sβl) ≤∑r

i=1 2|supp(βi)| − 1

which follows by `(sβ1sβ2 ...sβr) =∑r

i=1 2|supp(βi)| − 1. Thus sβ1sβ2 ...sβr is reduced.

Next, we will give an example for computing zd.

Example 3.1.3. Let d = (5, 0, 2, 2, 3, 0, 4). Observe that α0 +α6 and α2 +α3 +α4 are maximal

roots of d which have disconnected supports and d = 4(α0 +α6) + 2(α2 +α3 +α4) +α0 +α4

so by definition we get

zd = sα0+α6 · sα0+α6 · sα0+α6 · sα0+α6 · sα2+α3+α4 · sα2+α3+α4 · sα0 · sα4 .

But notice that sα0+α6 · sα0+α6 = sα0+α6 and sα2+α3+α4 · sα2+α3+α4 = sα2+α3+α4 · sα3 , see

the proof of Lemma 3.2.1, so we have zd = sα0+α6 · sα2+α3+α4 · sα3 · sα0 · sα4 . Now, also

note that any two reflections that appear in zd Hecke commute by Lemma 3.2.1. Moreover,

27

sα0+α6 · sα0 = sα0+α6 and sα2+α3+α4 · sα4 = sα2+α3+α4 , again see the proof of Lemma 3.2.1.

Thus zd = sα0+α6 · sα2+α3+α4 · sα3 . But this multiplication for zd is reduced; see the proof of

Theorem 3.1.2. So zd = sα0+α6sα2+α3+α4sα3 . Furthermore,

`(zd) = 2|supp(α0 + α6)| − 1 + 2|supp(α2 + α3 + α4)| − 1 + 2|supp(α3)| − 1 = 9.

3.2 Lemma about the Hecke Product

Here, we will consider a lemma which allows us to manipulate a given Hecke multiplication

of two reflections without changing the sum of the corresponding roots.

Lemma 3.2.1. Let α, β ∈ Πre,+aff such that α, β < c. Then

1) Assume that α ≤ β. Then

sα · sβ = sβ · sα

2) Suppose that γ = α ∩ β 6= ∅ and γ 6= α, β. Then

a) If γ is a root then α− γ and β − γ are also roots. Moreover,

sα · sβ = sα · sβ−γ · sγ = sγ · sα−γ · sβ

b) If γ is not a root then γ = γ1 + γ2 for some positive real roots γ1, γ2. Also,

α− γ,α− γ1, α− γ2, β − γ, β − γ1 and β − γ2 are all roots. Moreover,

sα · sβ = sα · sβ−γ · sγ1 · sγ2 = sγ1 · sγ2 · sα−γ · sβ

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3) Assume that α ∩ β = ∅. Then

a) If α + β is a root then sα · sβ ≤ sα+β.

b) If α + β is not a root then sα · sβ = sβ · sα.

Proof. Let a simple reflection sq and a positive real root pi,j < c be given. We will show that

sq · spi,j = spi,j · sq if αq ∈ supp(pi,j). First, assume that q = i. Then we have three cases;

either i = j or i < j or i > j. If i = j then the equality is clear since pi,j = pi,i = αi and

sq = spi,j = si in this case. Now, suppose that i < j. Then since si · si = si we have

sq · spi,j = si · si · si+1 · ... · sj−1 · sj · sj−1 · ... · si−1 · si

= si · si+1 · ... · sj−1 · sj · sj−1 · ... · si−1si · si = spi,j · sq.

The case, i > j is similar. If q = j then we have the equalities by the same argument since

spi,j has a reduced expression which starts and ends with sj in all sub cases. Now assume

that q is neither i nor j. Here, we have two sub cases; i < j or i > j. First, assume that

i < j. Then i < q < j and we have

sq · spi,j = sq · si · si+1 · ... · (sq−1 · sq) · ... · sj−1 · sj · sj−1 · ... · (sq · sq−1) · ... · si−1 · si

= si · si+1 · ... · (sq · sq−1 · sq) · ... · sj−1 · sj · sj−1 · ... · (sq · sq−1) · ... · si−1 · si

= si · si+1 · ... · (sq−1 · sq · sq−1) · ... · sj−1 · sj · sj−1 · ... · (sq · sq−1) · ... · si−1 · si

= si · si+1 · ... · (sq−1 · sq) · ... · sj−1 · sj · sj−1 · ... · (sq−1 · sq · sq−1) · ... · si−1 · si

= si · si+1 · ... · (sq−1 · sq) · ... · sj−1 · sj · sj−1 · ... · (sq · sq−1 · sq) · ... · si−1 · si

= si · si+1 · ... · (sq−1 · sq) · ... · sj−1 · sj · sj−1 · ... · (sq · sq−1) · ... · si−1 · si · sq

= spi,j · sq

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The case, i > j is similar.

1) Assume that α ≤ β. Suppose that α = si1 · si2 · ... · sir is a reduced expression for α. Now,

since α ≤ β we have supp(α) = i1, i2, ..., ir ⊂ supp(β). Thus the reflection sik will Hecke

commute with β for all k by the argument above. So we will have

α · β = si1 · si2 · ... · sir · β = β · si1 · si2 · ... · sir = β · α.

2) Case a: Let γ = α∩ β be a root and γ 6= α, β. First, we will show that sα = u · sα−γ · u−1

and sβ = u−1 · sβ−γ · u for some permutation u such that γ = u · u−1. We have several cases

here;

Case a1: Assume that α = pi,j and β = pk,l where 0 ≤ i < k < j < l ≤ n− 1. Then γ = pk,j.

Furthermore, α− γ = pi,k−1 and β − γ = pj+1,l, so both are roots. Now, note that

sα = sj · sj−1 · ... · sk+1 · sk · sk−1 · ... · si+1 · si · si+1 · ... · sk−1 · sk · sk+1 · ... · sj−1 · sj,

sβ = sk · sk+1 · ... · sj−1 · sj · sj+1 · ... · sl−1 · sl · sl−1 · ...sj+1 · sj · sj−1 · ... · sk+1 · sk,

sγ = sj · sj−1 · ... · sk+1 · sk · sk+1 · ... · sj−1 · sj and sα−γ = sk−1 · ... · si+1 · si · si+1 · ... · sk−1 and

sβ−γ = sj+1 · ... · sl−1 · sl · sl−1 · ...sj+1 are some reduced expressions for the reflections. So we

can take u = sj · sj−1 · ... · sk+1 · sk which will imply that sγ = u · u−1 since sk · sk = sk.

Case a2: Suppose that α = pi,j and β = pk,l where 0 ≤ k < j < l ≤ i ≤ n−1. Then γ = pk,j.

Moreover, α− γ = pi,n−1 and β − γ = pj+1,l, hence both are roots. Again, if we consider the

reduced expressions;

sα = sj · sj−1 · ... · s1 · s0 · si · si+1 · ... · sn−1 · sn · sn−1 · ... · si+1 · si · s0 · s1... · sj−1 · sj,

30

sβ = s0 · s1... · sj−1 · sj · sj+1 · ...sl−1 · sl · sl−1... · sj+1 · sj · sj−1... · s1 · s0,

sγ = sj · sj−1 · ... · s1 · s0 · s1... · sj−1 · sj and sα−γ = si · si+1 · ... · sn−1 · sn · sn−1 · ... · si+1 · si

and sβ−γ = sj+1 · ...sl−1 · sl · sl−1... · sj+1 we can take u = sj · sj−1 · ... · s1 · s0.

Case a3: Suppose that α = pi,j and β = pk,l where 0 ≤ j ≤ k < i < l ≤ n− 1. Then γ = pi,l.

Also, α− γ = p0,j and β − γ = pk,i−1, thus both are roots. Similar to a2).

Case a4: Suppose that α = pi,j and β = pl,k where 0 ≤ k < j ≤ l < i ≤ n−1. Then γ = pi,k.

Moreover, α− γ = pk+1,j and β − γ = pl,i−1, so both are roots. Note that,

sα = sj · ... · sk · ... · s1 · s0 · si · si+1 · ... · sn−1 · ... · si−1 · si · s0 · s1 · ... · sk · ... · sj

= sj · ... · sk · ... · s1 · si · si+1 · ... · s0 · sn−1 · s0 · ... · si−1 · si · s1 · ... · sk · ... · sj

= sj · ... · sk · ... · s1 · si · si+1 · ... · sn−1 · s0 · sn−1 · ... · si−1 · si · s1 · ... · sk · ... · sj

= si · si+1 · ... · sn−1 · sj · ... · sk · ... · s1 · s0 · s1 · ... · sk · ... · sj · sn−1 · ... · si−1 · si

= si · si+1 · ... · sn−1 · s0 · s1 · ... · sk · ... · sj · ... · sk · ... · s1 · s0 · sn−1 · ... · si−1 · si,

sβ = sk · ... · s1 · s0 · sl · sl+1 · ... · si · ...sn−1 · ... · si · ... · sl+1 · sl · s0 · s1 · ... · sk

= sk · ... · s0 · sn−1 · ...si · si−1 · ... · sl+1 · sl · sl+1 · ... · si−1 · si · ... · sn−1 · s0 · ... · sk,

sγ = sk · .. · s1 · s0 · si · si+1 · ... · sn−1 · ... · si+1 · si · s0 · s1 · ... · sk

= sk · .. · s1 · si · si+1 · ... · s0 · sn−1 · s0 · ... · si+1 · si · s1 · ... · sk

= sk · .. · s1 · si · si+1 · ... · sn−1 · s0 · sn−1 · ... · si+1 · si · s1 · ... · sk

= si · si+1 · ... · sn−1 · sk · ... · s1 · s0 · s1 · ... · sk · sn−1 · ... · si+1 · si

= si · si+1 · ... · sn−1 · s0 · s1 · ... · sk · ... · s1 · s0 · sn−1 · ... · si+1 · si,

31

and sα−γ = spk+1,j= sk+1 · sk+2 · ... · sj · .. · sk+2 · sk+1, and

sβ−γ = spl,i−1= sl · sl+1 · ... · si−1 · ... · sl+1 · sl

= si−1 · si−2 · ... · sl+1 · sl · sl+1 · ... · si−2 · si−1.

Thus, we can take u = si · si+1 · ... · sn−1 · s0 · s1 · ... · sk since sk · sk = sk.

Case a5: Assume that α = pi,j and β = pl,k where 0 ≤ j < k ≤ i < l ≤ n− 1. Then γ = pl,j.

Furthermore, α− γ = pi,l−1 and β − γ = pj+1,k, hence both are roots. This case is similar to

a4).

Now, note that, in all cases above we have sα = u · sα−γ · u−1, sβ = u−1 · sβ−γ · u where

sγ = u · u−1. Also, note that u−1 · sβ = sβ · u−1 since each simple root that appears in u−1 is

in the support of β since γ < β. Hence,

sα · sβ = u · sα−γ · u−1 · sβ = u · sα−γ · sβ · u−1

= u · sα−γ · u−1 · sβ−γ · u · u−1 = sα · sβ−γ · sγ

Case b: Now, assume that γ = α ∩ β is not a root. First, we will show that γ = γ1 + γ2 for

some γ1, γ2 ∈ Πre,+aff and α − γ1, α − γ2, α − γ, β − γ1, β − γ2 and β − γ are all roots. Here,

we have two cases;

Case b1: Suppose that α = pi,j and β = pk,l where 0 ≤ k ≤ j < i − 1 ≤ l − 1 < n − 2.

Note that, γ = α ∩ β = pk,j + pi,l so we can take γ1 = pk,j and γ2 = pi,l. Also, note that

α − γ1 = pi,k−1, α − γ2 = pl+1,j, α − γ = pl+1,k−1, β − γ1 = pj+1,l, β − γ2 = pk,i−1, and

β − γ = pj+1,i−1, thus all are roots.

Case b2: Assume that α = pi,j and β = pk,l where 0 < k ≤ j < i− 1 ≤ l− 1 ≤ n− 2. Again,

32

γ = α ∩ β = pk,j + pi,l so we can take γ1 = pk,j and γ2 = pi,l. Furthermore, α − γ1 = pi,k−1,

α− γ2 = pl+1,j, α− γ = pl+1,k−1, β − γ1 = pj+1,l, β − γ2 = pk,i−1, and β − γ = pj+1,i−1, so all

are roots.

Now, note that, by the same arguments in part a) above we can get some reduced expressions

for sα and sβ such that sα = u1 · sα−γ1 · u−11 and sβ−γ1 = u−1

1 · sβ−γ1 · u1 for a permutation

u1 such that sγ1 = u1 · u−11 since both α − γ1 and β − γ1 are roots. Similarly, we can write

sα−γ1 = u2 · sα−γ1−γ2 · u−12 and sβ−γ1 = u−1

2 · sβ−γ1−γ2 · u2 for a permutation u2 such that

sγ2 = u2 ·u−12 since both α−γ1−γ2 and β−γ1−γ2 are roots. Also, note that u−1

1 ·sβ = sβ ·u−11

and u−12 · sβ = sβ · u−1

2 since each simple root that appears in u−11 and u−1

2 is in the support

of β by the fact that γ1, γ2 < β. Furthermore, supp(γ1) and supp(γ2) are disconnected so

sγ1 ·sγ2 = sγ2 ·sγ1 and more specifically, u1 ·sγ2 = sγ2 ·u1 since any simple root which appears

in u1 is in the support of γ1 by the equation sγ1 = u1 · u−11 . Thus,

sα · sβ = u1 · sα−γ1 · u−11 · sβ = u1 · u2 · sα−γ1−γ2 · u−1

2 · u−11 · sβ

= u1 · u2 · sα−γ1−γ2 · sβ · u−12 · u−1

1

= u1 · u2 · sα−γ1−γ2 · u−12 · sβ−γ2 · u2 · u−1

2 · u−11

= u1 · u2 · sα−γ1−γ2 · u−12 · u−1

1 · sβ−γ2−γ1 · u1 · u2 · u−12 · u−1

1

= u1 · sα−γ1 · u−11 · sβ−γ2−γ1 · u1 · sγ2 · u−1

1 = sα · sβ−γ · sγ2 · u1 · u−11

= sα · sβ−γ · sγ2 · sγ1 = sα · sβ−γ · sγ1 · sγ2

3) Case a: Note that, supp(α)∩ supp(β) = ∅ in this case and since both α and β are smaller

than c, the multiplication sαsβ is reduced so sαsβ = sα · sβ. Now, by Corollary 5.1.5 we get

sα · sβ = sαsβ ≤ sα+β.

33

Case b: Here, supp(α) and supp(β) are disconnected. Hence, the multiplication, sαsβ is

reduced and by the braid relations any simple reflection that appears in sα will commute

with all simple reflections which appear in sβ so sαsβ = sβsα. Thus, sα · sβ = sαsβ = sβsα =

sβ · sα.

Corollary 3.2.2. Let w = sβ1 · sβ2 · ... · sβr be such that βi ∈ Πre,+aff for all i and

∑ri=1 βi = α

where α < c. Then w ≤ sα.

Proof. The proof follows by using Lemma 3.2.1 part 3) repeatedly.

3.3 Lemma about Decomposition

Lemma 3.3.1. Let α, β ∈ Πre,+aff such that β < c and α = mc + β for an integer m ≥ 1.

Then sα = (sβsβ′)msβ where β + β′ = c.

Proof. First, we will show that sα = sβsγsβ where γ = mc− β. Note that, 〈γ, β∨〉 = 〈mc−

β, β∨〉 = −2, hence sβ(γ) = γ − 〈γ, β∨〉β = γ + 2β = α. So sα = ssβ(γ) = sβsγs−1β = sβsγsβ.

Next, we will prove the statement by induction on m. First, suppose that m = 1. Then

sα = sβsγsβ where γ = mc − β = c − β. Now, assume that the statement is true for

m = k for a positive integer k. We will prove that it is also true for m = k + 1. Let

α = (k + 1)c + β where β ∈ Πre,+aff and β < c. Again, we can write sα = sβsγsβ where

γ = (k+ 1)c− β. Note that, γ = (k+ 1)c− β = kc+ c− β and β′ := c− β is a positive root

34

which is smaller than c. So by the induction assumption we can write sγ = (sβ′sβ)ksβ′ . Thus

sα = sβsγsβ = sβ(sβ′sβ)ksβ′sβ = (sβsβ′)k+1sβ.

Corollary 3.3.2. Let α ∈ Πre,+aff such that α > c. Then sα = sβ1sβ2 ...sβk for some k where

βi ∈ Πre,+aff such that βi < c for all i and

∑ki=1 βi = α.

Proof. Now, suppose that α = mc + β for some integer m such that m ≥ 1 and an affine

positive real root β which is smaller than c. Then by Lemma 3.3.1 we have sα = (sβsβ′)msβ

where β + β′ = c.

3.4 Lemmas about Lengths

The main goal of this section is to compute the lengths of some elements of the affine Weyl

group Waff which appear in the curve neighborhoods.

Lemma 3.4.1. Let β, β′, α, α′ ∈ Πre,+aff such that α + α′ = β + β′ = c. Also, suppose that

β 6= α. Then (sβsβ′)m 6= (sαsα′)

m for any positive integer m.

Proof. First, assume that α0 /∈ supp(β). Now, sβ′ = sc−β = s−βt−β = sβt−β. Thus sβsβ′ =

sβsc−β = sβsβt−β = t−β. So (sβsβ′)m = (t−β)m = t−mβ. Second, assume that α0 ∈ supp(β).

Then sβ = sc−(c−β) = s−(c−β)t−(c−β) = s(c−β)t−(c−β). Hence

sβsβ′ = s(c−β)t−(c−β)sc−β = tsc−β(−(c−β))

= t(c−β)

35

So (sβsβ′)m = (tc−β)m = tm(c−β). Similarly, we have either (sαsα′)

m = t−mα or (sαsα′)m =

tm(c−α). So if (sβsβ′)m = (sαsα′)

m then either t−mβ = t−mα, t−mβ = tm(c−α), tm(c−β) = t−mα,

or tm(c−β) = tm(c−α) so we have either β = α, β − α = c, or α − β = c, but this is a

contradiction.

Lemma 3.4.2. Let β ∈ Πre,+aff such that β < c. Then

∑γ∈Π+ |〈β, γ〉| = 2(n− 1).

Proof. First, assume that α0 /∈ supp(β). Let β = pi,j = εi− εj+1 such that 1 ≤ i < j ≤ n− 1

and γ ∈ Π+. We have several cases here;

• If γ = β then 〈β, γ〉 = 2. If γ < β then γ = pk,l such that either i < k < l < j which

implies 〈β, γ〉 = 0, or k = i < l < j or i < k < l = j which implies 〈β, γ〉 = 1 and

there are 2(j − i) such γ.

• If supp(γ) and supp(β) are disconnected then 〈β, γ〉 = 0.

• If β+γ is a root then γ = pk,l such that either 1 ≤ k < l = i−1 or k = j+1 < l ≤ n−1

and so 〈β, γ〉 = −1. We have n− j + i− 2 such γ.

• If β ∩ γ 6= ∅ and β ∩ γ 6= β, γ then γ = pk,l such that either 1 ≤ k < i < l < j or

i < k < j < l ≤ n− 1 which implies 〈β, γ〉 = 0.

• If γ > β then γ = pk,l such that either 1 ≤ k < i < j < l ≤ n − 1 which implies

〈β, γ〉 = 0, or k = i < j < l ≤ n− 1 or 1 ≤ k < i < j = l which follows by 〈β, γ〉 = 1

and there are n− j + i− 2 such γ.

36

Thus∑

γ∈Π+ |〈β, γ〉| = 2 + (n − j + i − 2) + 2(j − i) + (n − j + i − 2) = 2(n − 1). Next, if

α0 ∈ supp(β) then the proof is similar.

Lemma 3.4.3. Let m be a positive integer and β, β′ ∈ Πre,+aff such that β + β′ = c. Then

(sβsβ′)m is reduced for any positive integer m. In particular, `((sβsβ′)

m) = 2m(n− 1).

Proof. First, note that, since both β and β′ are smaller than c we have `(β) = 2|supp(β)|−1

and `(β′) = 2|supp(β′)| − 1. Also, |supp(β)|+ |supp(β′)| = n. So

`((sβsβ′)m) ≤ m(`(β) + `(β′)) = m(2|supp(β)| − 1 + 2|supp(β′)| − 1) = 2m(n− 1).

Now, assume that α0 /∈ supp(β). Then we have (sβsβ′)m = t−mβ; see the proof of Lemma

3.4.1. Moreover,

`(t−mβ) =∑γ∈Π+

|χ(γ < 0) + 〈−mβ, γ〉|

by Equation (2.1). Here, note that, χ(γ < 0) = 0 for all γ ∈ Π+ and 〈−mβ, γ〉 = −m〈β, γ〉

so

`(t−mβ) =∑γ∈Π+

| −m〈β, γ〉| = m∑γ∈Π+

|〈β, γ〉|.

By Lemma 3.4.2,∑

γ∈Π+ |〈β, γ〉| = 2(n − 1). So `(t−mβ) = 2m(n − 1). Next, assume that

α0 ∈ supp(β). Then we have (sβsβ′)m = tm(c−β), see the proof of Lemma 3.4.1. By the similar

arguments above, one can show that `(tm(c−β)) = 2m(n− 1). Hence (sβsβ′)m is reduced and

`((sβsβ′)m) = 2m(n− 1).

37

Remark 3.4.4. Alternatively, one can prove Lemma 3.4.3 by using the following arguments:

For any β∨ ∈ Q∨ and w ∈ W we have `(tβ∨) = `(tw(β)). So one may restrict to the case

where β∨ is a dominant root. Then we have `(tβ∨) = 〈β∨, 2ρ〉 where ρ =∑n−1

i=1 ωi is the sum

of the fundamental weights, see [8]. But note that 〈α∨i , ωj〉 = δij for 1 ≤ i, j ≤ n− 1 and if

we take β = θ =∑n−1

i=1 αi we get 〈β∨, 2ρ〉 = 2(n− 1).

Lemma 3.4.5. Let β, α ∈ Πre,+aff and γ ∈ Π+ such that α ≤ β < c and sα(γ) < 0. Then

〈sα(β), γ〉 ≤ 0.

Proof. First, assume that α0 /∈ supp(β). Also, assume that β = pi,j such that 1 ≤ i < j ≤

n − 1. Now, let α = pk,l. Then β ≥ α implies that either 1 ≤ i < k < l < j ≤ n − 1,

1 ≤ i = k < l < j ≤ n − 1 or 1 ≤ i < k < l = j ≤ n − 1 and 〈β, α〉 ≥ 0, in either

case. Now, sα(γ) < 0 implies that 〈α∨, γ〉 = 〈α, γ〉 > 0 since both α and γ are positive

roots. So either 〈α, γ〉 = 2 or 〈α, γ〉 = 1. If 〈α, γ〉 = 2 then α = γ which follows by

〈sα(β), γ〉 = 〈β, sα(γ)〉 = 〈β, sα(α)〉 = 〈β,−α〉 = −〈β, α〉 ≤ 0. Now, if 〈α, γ〉 = 1 then

sα(γ) = γ − α < 0 and so γ < α. Thus, if γ = ps,q then we either have k < s < q < l or

k = s < q < l or k < s < q = l. But in either case we get 〈β, γ〉 ≤ 〈β, α〉; see the cases

in the proof of Lemma 3.4.3. So 〈sα(β), γ〉 = 〈β, sα(γ)〉 = 〈β, γ − α〉 = 〈β, γ〉 − 〈β, α〉 ≤ 0.

Assume that β is a simple root. Then β ≥ α ≥ γ implies that β = α = γ which implies

〈sα(β), γ〉 = 〈sα(α), α〉 = 〈−α, α〉 = −2.

Second, if α0 ∈ supp(β) then the statement follows by similar arguments to the previous

case.

38

Lemma 3.4.6. Let β, β′, α ∈ Πre,+aff such that β + β′ = c and β′ ≤ c−α. Then (sβsβ′)

msα is

reduced for any positive integer m. In particular, `((sβsβ′)msα) = 2m(n− 1) + `(sα).

Proof. First, observe that we can assume that α0 /∈ supp(β) by the automorphism of the

Dynkin diagram in Section 2.3. Then (sβsβ′)m = t−mβ; see the proof of Lemma 3.4.1. So

(sβsβ′)msα = t−mβsα. We also have t−mβsα = sαtsα(−mβ) since sαt−mβsα = tsα(−mβ). Now, by

Equation (2.1), we have

`(sαtsα(−mβ)) =∑

γ∈Π+ |χ(sα(γ) < 0) + 〈sα(−mβ), γ〉|

=∑

γ∈Π+ |χ(sα(γ) < 0)−m〈sα(β), γ〉|.

Now, note that, χ(sα(γ) < 0) = 0 if sα(γ) > 0 and χ(sα(γ) < 0) = 1 otherwise. Thus

`(sαtsα(−mβ)) = m∑

sα(γ)>0

|〈sα(β), γ〉|+∑

sα(γ)<0

|1−m〈sα(β), γ〉|

where γ ∈ Π+. Moreover, β ≥ α since β′ ≤ c− α. Here, notice that, for γ ∈ Π+ if sα(γ) < 0

then 〈sα(β), γ〉 ≤ 0 by Lemma 3.4.5, which implies that |1−m〈sα(β), γ〉| = 1+m|〈sα(β), γ〉|.

Thus

`(sαtsα(−mβ)) = m∑

sα(γ)>0 |〈sα(β), γ〉|+∑

sα(γ)<0 |1−m〈sα(β), γ〉|

= m∑

sα(γ)>0 |〈sα(β), γ〉|+∑

sα(γ)<0(1 +m|〈sα(β), γ〉|)

= m∑

γ∈Π+ |〈sα(β), γ〉|+ |γ ∈ Π+ : sα(γ) < 0|

Now, observe that, sα can be considered as an element of W since α0 /∈ supp(α) due to the

fact that β ≥ α, so |γ ∈ Π+ : sα(γ) < 0| = `(sα). Also, β ≥ α implies that 〈β, α∨〉 is

39

either 0, 1, or 2. Then sα(β) is either ±β or β−α and so |sα(β)| is a positive real root which

is smaller than c. Thus, by Lemma 3.4.2, we get∑

γ∈Π+ |〈sα(β), γ〉| = 2(n− 1) which follows

by `(sαtsα(−mβ)) = 2m(n− 1) + `(sα) so `((sβsβ′)msα) = 2m(n− 1) + `(sα).

On the other hand, `((sβsβ′)msα) ≤ `((sβsβ′)

m) + `(sα) = 2m(n − 1) + `(sα) by Lemma

3.4.3.

Lemma 3.4.7. Let β, α ∈ Πre,+aff and γ ∈ Π+ such that β < c − α < c and β ⊥ α. Also,

suppose that sα(γ) < 0. Then 〈sα(β), γ〉 = 0.

Proof. First, note that sα(β) = β since β ⊥ α. Moreover, 〈γ, α∨〉 > 0 since α and γ are both

positive roots and sα(γ) < 0. This implies that either γ = α or γ < α. But in either case we

have β ⊥ γ since β < c− α and β ⊥ α. So 〈sα(β), γ〉 = 〈β, γ〉 = 0.

Lemma 3.4.8. Let β, β′, α ∈ Πre,+aff such that β + β′ = c. Also, suppose that β′ > α and

β′ ⊥ α. Then (sβsβ′)msα is reduced for any positive integer m. In particular, `((sβsβ′)

msα) =

2m(n− 1) + `(sα).

Proof. First, note that we can assume that α0 /∈ supp(α) by the automorphism of the

Dynkin diagram in Section 2.3. Observe that, β′ > α and β′ ⊥ α imply that β < c − α

and β ⊥ α since β + β′ = c. Now, assume that α0 /∈ supp(β) then (sβsβ′)m = t−mβ; see

the proof of Lemma 3.4.1. So (sβsβ′)msα = t−mβsα. We also have t−mβsα = sαtsα(−mβ) since

40

sαt−mβsα = tsα(−mβ). Now, by Equation (2.1), we have

`(sαtsα(−mβ)) =∑

γ∈Π+ |χ(sα(γ) < 0) + 〈sα(−mβ), γ〉|

=∑

γ∈Π+ |χ(sα(γ) < 0)−m〈sα(β), γ〉|.

Now, note that, χ(sα(γ) < 0) = 0 if sα(γ) > 0 and χ(sα(γ) < 0) = 1 otherwise. Thus

`(sαtsα(−mβ)) = m∑

sα(γ)>0

|〈sα(β), γ〉|+∑

sα(γ)<0

|1−m〈sα(β), γ〉|

where γ ∈ Π+. Furthermore, since β < c − α and β ⊥ α we have 〈sα(β), γ〉 = 0 if γ ∈ Π+

and sα(γ) < 0 by Lemma 3.4.7, which implies that |1−m〈sα(β), γ〉| = 1. Thus

`(sαtsα(−mβ)) = m∑

sα(γ)>0 |〈sα(β), γ〉|+∑

sα(γ)<0 |1−m〈sα(β), γ〉|

= m∑

sα(γ)>0 |〈sα(β), γ〉|+∑

sα(γ)<0(1)

= m∑

γ∈Π+ |〈sα(β), γ〉|+ |γ ∈ Π+ : sα(γ) < 0|

Now, observe that, since α0 /∈ supp(α) we can write sα ∈ W so |γ ∈ Π+ : sα(γ) < 0| =

`(sα). Also, note that 〈β, α∨〉 = 0 since β ⊥ α, which follows by sα(β) = β. Thus, we get∑γ∈Π+ |〈sα(β), γ〉| =

∑γ∈Π+ |〈β, γ〉|. Now, by Lemma 3.4.2, we also have

∑γ∈Π+ |〈β, γ〉| =

2(n − 1) which follows by `(sαtsα(−mβ)) = 2m(n − 1) + `(sα) so `((sβsβ′)mssα) = 2m(n −

1) + `(sα). Next, assume that α0 ∈ supp(β) then one can prove the statement by the similar

arguments above.

On the other hand, `((sβsβ′)msα) ≤ `((sβsβ′)

m) + `(sα) = 2m(n − 1) + `(sα) by Lemma

3.4.3.

41

Lemma 3.4.9. Let β, γ1, γ2, ..., γk ∈ Πre,+aff such that β < c and for any 1 ≤ i, j ≤ k such that

i 6= j, either supp(γi) and supp(γj) are disconnected or both γi ⊥ γj and γi, γj are comparable.

Also, suppose that either γi ≤ β or both β ∩ γi = ∅ and β ⊥ γi for any 1 ≤ i ≤ k. Moreover,

let γ ∈ Π+ such that sγ1sγ2 ...sγk(γ) < 0. Then < sγk ...sγ2sγ1(β), γ >≤ 0.

Proof. First, observe that < sγk ...sγ2sγ1(β), γ >=< β, sγ1sγ2 ...sγk(γ) > . Now, note that

< β, γi > is either 2, 1 or 0 since we have either γi ≤ β or both β ∩ γi = ∅ and β ⊥ γi.

Also, note that sγ1sγ2 ...sγk(γ) = γ −∑k

i=1 < γ, γ∨i > γi since γi ⊥ γj for any 1 ≤ i, j ≤ k

such that i 6= j. Now, if < γ, γ∨i >≤ 0 for all i = 1, 2, ..., k then sγ1sγ2 ...sγk(γ) ≥ γ > 0.

So < γ, γ∨i >> 0 for some i. This implies that either < γ, γ∨i >= 2 or < γ, γ∨i >= 1. Now,

if < γ, γ∨i >= 2 then γ = γi. But then γ ⊥ γj for all j ∈ 1, 2, ..., k \ i, which follows

by sγ1sγ2 ...sγk(γ) = −γi. Then < β, sγ1sγ2 ...sγk(γ) >=< β,−γi >= − < β, γi >. Here,

note that if < β, γi >6= 0 then β is not perpendicular to γi which implies that γi ≤ β and

< β, γi > is either 2 or 1 hence we are done with this case. Now, suppose that < γ, γ∨i >= 1.

Then either γ < γi or γi < γ. Now suppose that γ < γi. Then we have three cases;

• γ ⊥ γj for all j ∈ 1, 2, ..., k \ i which follows by sγ1sγ2 ...sγk(γ) = γ − γi < 0.

Then < β, sγ1sγ2 ...sγk(γ) >=< β, γ − γi >=< β, γ > − < β, γi > . If < β, γi >= 2

then β = γi and < β, γ >= 1 in this case. If < β, γi >= 1 then either < β, γ >= 1

or < β, γ >= 0. Last, if < β, γi >= 0 then < β, γ >= 0. In all cases, we get

< β, γ > − < β, γi >≤ 0.

• For an index q, < γ, γ∨q >= −1 and γ ⊥ γj for all j ∈ 1, 2, ..., k \ i, q. Thus

42

sγ1sγ2 ...sγk(γ) = γ − γi + γq < 0. Now, we have either supp(γi) and supp(γq) are

disconnected or γq < γi or γi < γq. Now, by the facts that γ < γi and < γ, γ∨q >= −1

we have to have γq < γi. So

< β, sγ1sγ2 ...sγk(γ) >=< β, γ − γi + γq >=< β, γ > − < β, γi > + < β, γq > .

If < β, γi >= 2 then β = γi which follows by < β, γ >= 1 and < β, γq >= 0. If

< β, γi >= 1 then < β, γq >= 0 since γq is smaller than γi and γq ⊥ γi. Also, we have

either < β, γ >= 1 or < β, γ >= 0 in this case since γ < γi ≤ β. Last, if < β, γi >= 0

then < β, γ >=< β, γq >= 0 since both γq and γ are smaller than γi. Note that, in all

cases, we get < β, γ > − < β, γi > + < β, γq >≤ 0.

• For an index r 6= i, < γ, γ∨q >= 1 and γ ⊥ γj for all j ∈ 1, 2, ..., k \ i, r. Then

either γ < γr or γr < γ. But since either supp(γi) and supp(γr) are disconnected or

both γi ⊥ γr and γi and γr are comparable we have to have γr < γ which implies that

γr < γi. Thus sγ1sγ2 ...sγk(γ) = γ − γi − γr. So

< β, sγ1sγ2 ...sγk(γ) >=< β, γ − γi + γr >=< β, γ > − < β, γi > + < β, γr > .

If < β, γi >= 2 then β = γi which follows by < β, γ >= 1 and < β, γr >= 0. If

< β, γi >= 1 then < β, γr >= 0 since γr is smaller than γi and γr ⊥ γi. Also, we have

either < β, γ >= 1 or < β, γ >= 0 in this case since γ < γi ≤ β. Last, if < β, γi >= 0

then < β, γ >=< β, γr >= 0 since both γr and γ are smaller than γi. Note that, in all

cases, we get < β, γ > − < β, γi > + < β, γr >≤ 0.

The case where γi < γ is similar.

43

Lemma 3.4.10. Let β, β′ ∈ Πre,+aff such that β + β′ = c and let d = (d0, d1, ..., dn−1) be a

degree where di = 0 for some i. Also, assume that zd = sγ1sγ2 ...sγk for some k where this

expression is obtained in Theorem 3.1.2. If either β′ ∩ γi = ∅ or both β′ > γi and β′ ⊥ γi

for any 1 ≤ i ≤ k then w = (sβsβ′)mzd is reduced for any positive integer m. In particular,

`((sβsβ′)mzd) = 2m(n− 1) + `(zd).

Proof. Note that by Lemma 3.4.3 we have `((sβsβ′)m) = 2m(n − 1) so `((sβsβ′)

mzd) ≤

`((sβsβ′)m)+`(zd) = 2m(n−1)+`(zd). We will show that `((sβsβ′)

mzd) = 2m(n−1)+`(zd)

to prove that w = (sβsβ′)mzd is reduced for any positive integer m.

First, we can suppose that d0 = 0 by the automorphism of the Dynkin diagram, ϕ; see

Section (2.3) which implies that α0 /∈ supp(γi) for any i = 1, 2, ..., k so γi ∈ Π+ for all

i. Now, suppose that α0 /∈ supp(β). Then by Lemma 3.4.1, we have (sβsβ′)m = t−mβ. So

w = (sβsβ′)mzd = (sβsβ′)

msγ1sγ2 ...sγk = t−mβsγ1sγ2 ...sγk . We also have t−mβsγ1sγ2 ...sγk =

sγ1sγ2 ...sγktsγk ...sγ2sγ1 (−mβ). Now, by Lemma 3.1 in [8], we get

`(w) =∑

γ∈Π+ |χ(sγ1 ...sγk(γ) < 0)+ < sγk ...sγ1(−mβ), γ > |

=∑

γ∈Π+ |χ(sγ1 ...sγk(γ) < 0)−m < sγk ...sγ1(β), γ > |.

Now, note that χ(sγ1 ...sγk(γ) < 0) = 0 if sγ1 ...sγk(γ) > 0 and χ(sγ1 ...sγk(γ) < 0) = 1 other-

wise. Thus

`(w) = m∑

sγ1 ...sγk (γ)>0

| < sγk ...sγ1(β), γ > |+∑

sγ1 ...sγk (γ)<0

|1−m < sγk ...sγ1(β), γ > |

44

where γ ∈ Π+. Moreover, either γi ≤ β or both β ∩ γi = ∅ and β ⊥ γi for any 1 ≤ i ≤ k

since we have either β′ ∩ γi = ∅ or both β′ > γi and β′ ⊥ γi for any 1 ≤ i ≤ k. Here, notice

that, for γ ∈ Π+ if sγ1sγ2 ...sγk(γ) < 0 then < sγk ...sγ2sγ1(β), γ >≤ 0 by Lemma 3.4.9, which

implies that |1−m < sγk ...sγ1(β), γ > | = 1 +m| < sγk ...sγ1(β), γ > |. Hence,

`(w) = m∑

sγ1 ...sγk (γ)>0 | < sγk ...sγ1(β), γ > |+∑

sγ1 ...sγk (γ)<0 |1−m < sγk ...sγ1(β), γ > |

= m∑

sγ1 ...sγk (γ)>0 | < sγk ...sγ1(β), γ > |+∑

sγ1 ...sγk (γ)<0(1 +m| < sγk ...sγ1(β), γ > |)

= m∑

γ∈Π+ | < sγk ...sγ1(β), γ > |+ |γ ∈ Π+ : sγ1 ...sγk(γ) < 0|.

Here, the fact that γi ≤ β for all i and the assumptions on γi’s force us to have sγk ...sγ1(β) =

±β, sγk ...sγ1(β) = β − γi or sγk ...sγ1(β) = β − γi − γq for some 1 ≤ i, q ≤ k. In all these

cases |sγk ...sγ1(β)| is a positive real root which is smaller than c. Thus, we get∑

γ∈Π+ | <

sγk ...sγ1(β), γ > | = 2(n− 1) by Lemma 3.4.2. Furthermore, |γ ∈ Π+ : sγ1 ...sγk(γ) < 0| =

`(sγ1 ...sγk) = `(zd) since γi ∈ Π+ for all i.

If α0 ∈ supp(β) then the proof is similar.

Chapter 4

The Moment Graph and Curve

Neighborhoods

In this chapter we recall some basic facts about the affine flag manifold of type A(1)n−1 and

give the definition of the moment graph and the (combinatorial) curve neighborhoods. We

refer to[7] especially §12 and §13 for further details. Let G be the Lie group SLn(C). Let

B ⊂ G be the Borel subgroup, the set of upper triangular matrices. Now, let G(C[t, t−1])

be the group of Laurent polynomial loops from C∗ to G and G := C∗ n G(C[t, t−1]) where

C∗ acts by loop rotation. Let B be the standard Iwahori subgroup of G determined by B.

The affine flag manifold in type A(1)n−1 is given by X := G/B. The Schubert varieties and

the curve neighborhoods of Schubert varieties in this context were defined in [9, §5]. In the

following we give a different, but equivalent definition, based on the moment graph, which

generalizes the observations from [5, §5.2].

45

46

The (undirected) moment graph for X is the graph given by the following data:

• The set of V of vertices is the group Waff;

• Let u, v ∈ V be vertices. Then there is an edge from u to v iff there exists an affine

root α such that v = usα. We denote this situation by

uα−→ v

and we say that the degree of this edge is α.

A chain between u and v in the moment graph is a succession of adjacent edges starting

with u and ending with v

π : u = u0β0−→ u1

β1−→ ...βk−2−→ uk−1

βk−1−→ uk = v.

The degree of the chain π is deg(π) = β0 + β1 + ... + βk−1. A chain is called increasing if

`(ui) > `(ui−1) for all i. Define the (Bruhat) partial ordering on the elements of Waff by

u < v iff there exists an increasing chain starting with u and ending with v.

A degree is a tuple of nonnegative integers d = (d0, ..., dn−1). Notice that it has n com-

ponents, corresponding to the n affine simple roots α0, ..., αn−1. There is a natural partial

order on degrees: If d = (d0, ..., dn−1) and d′ = (d′0, ..., d′n−1) then d ≥ d′ iff di ≥ d′i for all

i ∈ 0, 1, ..., n− 1.

Definition 4.0.1. Fix a degree d and u ∈ Waff. The (combinatorial) curve neighborhood

is the set Γd(u) consisting of elements v ∈ Waff such that:

47

1) There exists a chain of some degree d′ ≤ d from u′ ≤ u to v in the moment graph of

X ;

2) The elements v are maximal among all of those satisfying the condition in 1).

Remark 4.0.2. Let idβ1−→ sβ1

β2−→ sβ1sβ2 ...βq−→ w = sβ1sβ2 ...sβq be a path in the moment

graph such that∑q

i=1 βi < d. Then there are some βq+1, βq+2, ..., βp ∈ Πre,+aff such that∑p

i=1 βi = d. Note that, by Corollary 3.3.2 we can assume that βi < c for all i. Also, observe

that

w = sβ1sβ2 ...sβq ≤ sβ1 · sβ2 · ... · sβq ≤ sβ1 · sβ2 · ... · sβq · sβq+1 · sβq+2 · ... · sβp .

Now, if one can show that sβ1 · sβ2 · ... · sβq · sβq+1 · sβq+2 · ... · sβp ≤ u for some u ∈ Waff such

that u can be reached by a path of degree at most d that starts with id then it will follow

that u ∈ Γd(id).

Chapter 5

Calculation of the Curve

Neighborhoods

In this chapter we will prove our results. In Section 5.7 we will state Theorem 5.7.1 which

implies that it is sufficient to calculate Γd(id) to compute Γd(w) for any given degree d and

w ∈ Waff. For that reason we will mainly focus on the (combinatorial) curve neighborhood

of the identity element at some degree d.

5.1 Curve Neighborhood Γd(id) for Finite Degrees

Let d = (d0, d1, ..., dn−1) ∈ Qaff. If d0 = 0 then d ∈ Q. If d0 6= 0 but di = 0 for some i 6= 0

then for an integer 1 ≤ k ≤ n− 1 we will have ϕk(d) = d′ where d′ = (d′0, d′1, ..., d

′n−1) such

that d′0 = 0 and ϕ is the automorphism of the Dynkin diagram; see Section (2.3). So we will

48

49

consider d as an element of the finite root lattice, Q, whenever di = 0 for some i.

Theorem 5.1.1. Let d = (d0, d1, ..., dn−1) be a degree such that di = 0 for some i. Then

Γd(id) = zd.

Proof. Here, we are in the finite case since di = 0 for some i. So we get the conclusion by

the finite case, see [5].

Example 5.1.2. Let Waff be the affine Weyl group associated to A(1)6 and d = (5, 0, 2, 2, 3, 0, 4).

Then by Theorem 5.1.1, we get Γd(id) = zd where zd = sα0+α6sα2+α3+α4sα3 , see Example

3.1.3.

Theorem 5.1.3. Let α ∈ Πre,+aff such that α < c. Then Γα(id) = sα.

Proof. Let α =∑n−1

i=0 aiαi. Then ai = 0 for some i since α < c and α ∈ Πre,+aff . So

Γα(id) = zα. Now, observe that zα = sα by definition.

Example 5.1.4. Assume that the affine Weyl group is associated to A(1)4 and α = α0 + α4.

Then Γα(id) = sα0+α4 by Theorem 5.1.3.

Corollary 5.1.5. Let β ∈ Πre,+aff such that β < c. If

∑ri=1 βi ≤ β for some βi ∈ Πre,+

aff then

sβ1sβ2 ...sβr ≤ sβ.

Proof. Let w = sβ1sβ2 ...sβr . Then w can be reached by a chain of degree∑r

i=1 βi = β that

starts with the identity element in the moment graph. Thus w ≤ sβ since Γβ(id) = sβ, by

Theorem 5.1.3.

50

5.2 Curve Neighborhood Γc(id)

Here we will consider one of our basic yet crucial results. The result will play a key role in

proving the rest of the results in this thesis.

Theorem 5.2.1. We have

1) Γc(id) = tγ : γ ∈ Π.

2) |Γc(id)| = n(n− 1).

3) For all w ∈ Γc(id), `(w) = 2(n− 1).

Proof. 1) Let idβ1−→ sβ1

β2−→ sβ1sβ2 ...βr−→ w = sβ1sβ2 ...sβr be a path in the moment graph

such that∑r

i=1 βi ≤ c. Note that, by Remark 4.0.2 we can assume that w = sβ1 · sβ2 · ... · sβr .

where∑r

i=1 βi = c and βi < c for all i. Now, note that∑r

i=2 βi = c− β1 is an affine positive

real root. Moreover, sβ2 ·sβ3 · ... ·sβr ≤ sc−β1 by Corollary 3.2.2. Thus w = sβ1 ·(sβ2 · ... ·sβr) ≤

sβ1 · sc−β1 . But by Lemma 3.4.3, sβ1sc−β1 is reduced so sβ1 · sc−β1 = sβ1sc−β1 . We also know

that sβ1sc−β1 = tγ for some γ ∈ Π, see the proof of Lemma 3.4.1.

2) Let β, β′, µ, µ′ ∈ Πre,+aff such that µ + µ′ = β + β′ = c. Then sβsβ′ 6= sµsµ′ if β 6= µ by

Lemma 3.4.1. Also, note that |β ∈ Πre,+aff : β < c| = n(n− 1).

3) Let w = sβsβ′ ∈ Γc(id) for some positive real roots, β, β′ such that β + β′ = c. Then

`(w) = 2(n− 1), by Lemma 3.4.3.

Remark 5.2.2. Note that, tβ′ = sβsβ′ if α0 ∈ supp(β) and tβ′−c = sβsβ′ if α0 /∈ supp(β),

51

where β, β′ ∈ Πre,+aff such that β + β′ = c; see the proof of Lemma 3.4.1. Also,

β′ : β′ ∈ Πre,+aff , β′ < c, α0 /∈ supp(β′) t β′ − c : β′ ∈ Πre,+

aff , β′ < c, α0 ∈ supp(β′) = Π.

Thus we have

Γc(id) = sβsβ′ : β, β′ ∈ Πre,+aff such that β + β′ = c. (5.1)

Example 5.2.3. Suppose that the affine Weyl group Waff is associated to A(1)2 . Then

Γc(id) = tα1 , tα2 , tα1+α2 , t−α1 , t−α2 , t−(α1+α2).

Moreover, for any w ∈ Γc(id), `(w) = 2(n− 1) = 2 · 2 = 4.

5.3 Curve Neighborhood Γc+α(id)

Here, we will state a result which inspires us to calculate the most general case. First, we

will have some preliminary lemmas.

Lemma 5.3.1. let α < c be a positive real root. Then sα · sα · sc−α ≤ sα · sc−α · sα and

sc−α · sα · sα ≤ sα · sc−α · sα.

Proof. First, assume that α = pi,i = αi for some i. Then sα = si and since si · si = si we get

sα · sα · sc−α = si · si · sc−α = si · sc−α ≤ si · sc−α · si = sα · sc−α · sα.

Now, assume that α = pi,j where i 6= j. Then sα = si·spi,j−αi ·si and spi,j−αi = sj ·spi,j−αi−αj ·sj.

Note that sα · si = sα and similarly sα · sj = sα. Also, by Lemma 3.2.1, sα Hecke commutes

52

with spi,j−αi and spi,j−αi−αj since pi,j − αi < α and pi,j − αi − αj < α. We also have

spi,j−αi−αj < sα by Corollary 5.1.5. Furthermore, supp(pi,j − αi − αj) and supp(c− pi,j) are

disconnected so spi,j−αi−αj Hecke commutes with sc−pi,j . Thus,

sα · sα · sc−α = sα · si · spi,j−αi · si · sc−α = sα · spi,j−αi · si · sc−α

= spi,j−αi · sα · si · sc−α = spi,j−αi · sα · sc−α

= sj · spi,j−αi−αj · sj · sα · sc−α = sj · spi,j−αi−αj · sα · sc−α

= sj · sα · spi,j−αi−αj · sc−α = sα · spi,j−αi−αj · sc−α

= sα · sc−α · spi,j−αi−αj ≤ sα · sc−α · sα.

The proof of the inequality sc−α · sα · sα ≤ sα · sc−α · sα is similar.

Lemma 5.3.2. Let w = sβ1 · sβ2 · ... · sβr where βi ∈ Πre,+aff and βi < c for all i. Also, assume

that d =∑r

i=1 βi = d′+d′′ where d′ is the biggest degree that 0 < d′ ≤ c and d′′ = d−d′ > 0

may or may not be the degree of a root. Then there are some positive real roots, β′1, β′2, ..., β

′r′

such that w ≤ sβ′1 · sβ′2 · ... · sβ′r′ where∑t

i=1 β′i = d′ and

∑r′

i=t+1 β′i = d′′ for some t and β′i < c

for all i.

Proof. First, assume that there is an integer b such that∑b

i=1 βi = d′. Then∑r′

i=b+1 β′i = d′′

so we can take t = b and βi = β′i for all i. Now, suppose that for any integer b such that

1 ≤ b ≤ r we have∑b

i=1 βi 6= d′. We will prove the statement by induction on d. Here,

for the base case we have d = 2βi, for some simple root βi. So d′ = d′′ = βi. Note that, if

w = sβ1 · sβ2 · ... · sβr where βi ∈ Πre,+aff and βi < c for all i such that

∑ri=1 βi = d = 2βi then

we have to have w = sβi · sβi . But, notice that, w satisfies the statement, so we are done

53

with the base case. Now assume that the statement is true for all degrees that are smaller

than d. We will prove that it is also true for d.

1) Suppose that βr < d′′. Then since d−βr < d and d−βr = d′+d′′−βr where d′′−βr > 0

by the induction assumption we have some real positive roots, β′1, β′2, ..., β

′r′ , such that

w = (sβ1 · sβ2 · ... · sβr−1) · sβr ≤ (sβ′1 · sβ′2 · ... · sβ′r′ ) · sβr

where∑t

i=1 β′i = d′ and

∑r′

i=t+1 β′i = d′′ − βr for some t and β′i < c for all i. This

implies that we are done with this case since∑t

i=1 β′i = d′ and (

∑r′

i=t+1 β′i) + βr = d′′.

2) Assume that β1 < d′′. Then again since d− β1 < d and d− β1 = d′ + d′′ − β1 where

d′′ − β1 > 0 by induction assumption we have some real positive roots, β′1, β′2, ..., β

′r′

such that

w = sβ1 · (sβ2 · ... · sβr−1 · sβr) ≤ sβ1 · (sβ′1 · sβ′2 · ... · sβ′r′ )

where∑t

i=1 β′i = d′ and

∑r′

i=t+1 β′i = d′′ − β1 for some t and β′i < c for all i. Now,

observe that β′r′ < d′′ − β1 < d′′. Thus, this case follows by the previous case.

3) Suppose that β1 ∩ d′′ = ∅. Then the statement will follow by the same arguments in

the previous cases.

4) Assume that βi∩βj 6= ∅ for some i, j such that 2 ≤ i < j ≤ r. Then∑r

i=2 βi = d−β1 <

d can be written as D + D′ where D′ is the biggest degree such that 0 < D′ ≤ d′ ≤ c

and 0 < D′′ < d′′. Thus by the induction assumption there are some real positive

54

roots, β′1, β′2, ..., β

′r′ , such that

sβ2 · sβ3 · ... · sβr ≤ sβ′1 · sβ′2 · ... · sβ′r′

where∑t

i=1 β′i = D′ and

∑r′

i=t+1 β′i = D′′ for some t and β′i < c for all i. This follows

by

w = sβ1 · (sβ2 · ... · sβr) ≤ sβ1 · (sβ′1 · sβ′2 · ... · sβ′r′ )

Note that β′r′ ≤ D′′ ≤ d′′ thus we are done with this case.

5) Suppose that βi ∩ βj = ∅ for any i, j such that 2 ≤ i < j ≤ r. This implies that∑ri=2 βi ≤ c so d′′ ≤ β1 and d < 2c. We have two main cases here:

5a) Assume that∑r

i=2 βi < c. Now, for any i such that 2 ≤ i ≤ r − 1 if βi + βi+1 is a

root then by Lemma 3.2.1 we have sβi ·sβi+1≤ sβi+βi+1

. So if we repeatedly combine

consecutive roots in sβ2 · ... ·sβr which add up to a root we can assume that no two

consecutive roots in the multiplication add up to a root. By the same lemma, this

implies that any consecutive roots of the multiplication Hecke commute. So, we

can assume that for any i, j such that 2 ≤ i, j ≤ r, sβi ·sβj = sβj ·sβi . Thus, we can

also suppose that β1∩β2 6= ∅. Now, if β1∩β2 ≤ d′′ is not a root and β1∩β2 = γ1+γ2

for some positive real roots, γ1, γ2 we will have sβ1 · sβ2 = sγ2 · sγ1 · sβ1−γ1−γ2 · sβ2

by Lemma 3.2.1. So we get

w = (sβ1 · sβ2) · sβ3 · ... · sβr ≤ (sγ2 · sγ1 · sβ1−γ1−γ2 · sβ2) · sβ3 · ... · sβr .

55

Note that γ2 < d′′ so we are done with this case by case 2) above. Now assume

that β1 ∩ β2 ≤ d′′ is a root. Let γ := β1 ∩ β2. Then by Lemma 3.2.1 we have

sβ1 · sβ2 = sγ · sβ1−γ · sβ2 . So,

w = (sβ1 · sβ2) · sβ3 · ... · sβr ≤ (sγ · sβ1−γ · sβ2) · sβ3 · ... · sβr .

Now, if γ < d′′ then again we are done by case 2) above. Assume that γ = d′′

which implies that for any i ≥ 3 we have β1 ∩ βi = ∅ and d′′ ∩ βi = ∅. Here, we

will consider the multiplication, sβ1 · sβ2 · sβ3 · ... · sβr . By Hecke commutation, we

can assume that sβ2 is the last reflection of the multiplication. Now, if the sum

of β1 and β3 is not a root then they will Hecke commute by Lemma 3.2.1 and we

will have w = sβ1 · sβ3 · sβ4 · ... · sβr · sβ2 = sβ3 · sβ1 · sβ4 · ... · sβr · sβ2 . But this implies

that we are done with this case by case 3) above since β3 ∩ d′′ = ∅. Now, if the

sum is a root then we will replace sβ1 · sβ3 with sβ1+β3 by the same lemma to get

w = sβ1 ·sβ3 ·sβ4 · ... ·sβr ·sβ2 ≤ sβ1+β3 ·sβ4 · ... ·sβr ·sβ2 . Next, we will apply the same

argument to β1 +β3 and β4 etc. So either we will be done by case 3) above that is

because w can be written in a way that the first root which appears in it is sβi such

that βi∩d′′ = ∅ or we will get w = sβ1 · sβ3 · sβ4 · ... · sβr · sβ2 ≤ sβ1+β3+β4+...+βr · sβ2 .

Last, note that, (β1 + β3 + β4 + ...+ βr) ∩ β2 = γ = d′′. By Lemma 3.2.1 we can

write sβ1+β3+β4+...+βr · sβ2 = sβ1+β3+β4+...+βr · sβ2−γ · sγ. Hence, we have done with

this case as well.

5b) Suppose that∑r

i=2 βi = c then d′ = c and β1 = d′′. Now, note that, by Equation

5.1 there is a positive real root, β < c such that sβ2 · sβ3 · ... · sβr ≤ sβ · sc−β. So

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we get

w = sβ1 · (sβ2 · ... · sβr) ≤ sβ1 · sβ · sc−β.

Here, we can assume that w = sβ1 · sβ · sc−β;

i) If β1 ∩ β = ∅ then β1 ≤ c − β. Now, first assume that β1 + β = c then

c−β = β1 = d′′ and this implies that w is already in the desired form. Second,

if β1 + β < c is not a root then by Lemma 3.2.1 we have sβ1 · sβ = sβ · sβ1 .

Also, by the same lemma we have sβ1 · sc−β = sc−β · sβ1 since β1 ≤ c − β.

Hence,

w = sβ1 · sβ · sc−β = sβ · sc−β · sβ1 .

So we are done with this case as well. Last, suppose that β1 +β < c is a root

then by Lemma 3.2.1 we have sβ1 · sβ ≤ sβ1+β. So we get

w = sβ1 · sβ · sc−β ≤ sβ1+β · sc−β.

Now, note that (β1 + β) ∩ (c − β) = β1 and by the same lemma we have

sβ1+β · sc−β = sβ1+β · sc−β−β1 · sβ1 . Thus, the statement follows.

ii) If β1 ∩ β 6= ∅ then we have several cases here; first, suppose if β1 = β then by

Lemma 5.3.1 we have

w = sβ1 · sβ · sc−β = sβ1 · sβ1 · sc−β1 ≤ sβ1 · sc−β1 · sβ1 .

Thus, we are done with case. Next, if β1 6= β then either β1 < β, β1 > β or

β1 is incomparable with β. Now, if β1 < β or β1 > β then sβ1 and sβ will

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Hecke commute by Lemma 3.2.1 so have w = sβ1 · sβ · sc−β = sβ · sβ1 · sc−β.

Now, if β1 < β then β1 + c− β < c so we are done with this case by case 5a)

above. Now, if β1 > β then β1 ∩ (c− β) 6= ∅ so the statement follows by case

4) above. Last, assume that β1 is incomparable with β. Then β1 intersects

with both β and c− β which implies that β1 ∩ β < d′′. Then by Lemma 3.2.1

we can replace the multiplication, sβ1 · sβ in w with sγ2 · sγ1 · sβ1−γ1−γ2 · sβ if

β1 ∩ β is not a root and β1 ∩ β = γ1 + γ2 for some positive real roots, γ1, γ2

and with sγ · sβ1−γ · sβ if β1 ∩ β is a root and β1 ∩ β = γ. So, the first root of

the permutation, w is either γ2 or γ, and both roots are strictly smaller than

d′′. But then the statement follows by case 2) above.

Lemma 5.3.3. Let µ, µ′ and α ∈ Πre,+aff such that µ+µ′ = c and α < c. Then there are some

positive real roots β and β′ such that sµ · sµ′ · sα ≤ sβ · sβ′ · sα where β + β′ = c and either

β′ ≤ c− α or both β′ > α and β′ ⊥ α.

Proof. We have three cases;

1) If µ′ ≤ c− α then take β′ = µ′.

2) If µ′ > c− α then µ′ ∩ α 6= ∅. Note that, µ′ ∩ α 6= µ′, α. We have two cases here;

a) If µ′ ∩α is a root then let γ := µ′ ∩α. Now, by Lemma 3.2.1 we get sµ · sµ′ · sα =

sµ ·sγ ·sµ′−γ ·sα. Here, note that µ+γ is a root since µ+γ = c−µ′+γ = c−(µ′−γ)

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and µ′ − γ is a root. Thus

sµ · sµ′ · sα = sµ · sγ · sµ′−γ · sα ≤ sµ+γ · sµ′−γ · sα.

Note that, µ′ − γ = c− α and (µ + γ) + (µ′ − γ) = c so we will take β′ = µ′ − γ

in this case.

b) If µ′ ∩ α is not a root then by Lemma 3.2.1 we have µ′ ∩ α = γ1 + γ2 for some

positive real roots γ1 and γ2 and sµ · sµ′ · sα = sµ · sγ1 · sγ2 · sµ′−γ · sα. Again, here

µ+ γ is a root since both µ+ α = c− µ′ + γ = c− (µ′ − γ) and µ′ − γ are roots.

Thus

sµ · sµ′ · sα = sµ · sγ1 · sγ2 · sµ′−γ · sα ≤ sµ+γ1+γ2 · sµ′−γ · sα.

Now, observe that, µ′ − γ = c − α and (µ + γ1 + γ2) + (µ′ − γ) = c so again we

will take β′ = µ′ − γ in this case.

3) If µ′ is incomparable with c − α then again µ′ ∩ α 6= ∅. But then we will either have

the same two cases in the second case which then we will be done by the arguments in

the case or µ′ ≤ α or µ′ > α.

a) Assume that µ′ ≤ α. If µ′ = α then sµ · sµ′ · sα = sc−α · sα · sα. Now, by Lemma

5.3.1, we have sc−α · sα · sα ≤ sα · sc−α · sα. So we can take β′ = c−α in this case.

Now, suppose that µ′ < α. Then sµ′ and sα Hecke commute by Lemma 3.2.1 part

a). Also, µ ∩ α 6= ∅. Note that the intersection of two positive roots is either a

root or the sum of two positive roots. Now, assume that µ ∩ α is a root. Let

γ := µ ∩ α. Then again by Lemma 3.2.1 we have sµ · sα = sγ · sµ−γ · sα. Thus

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sµ · sµ′ · sα = sµ · sα · sµ′ = sγ · sµ−γ · sα · sµ′ = sγ · sµ−γ · sµ′ · sα. Here, note that

µ+ µ′ − γ = c− γ so sµ−γ · sµ′ ≤ sc−γ by Corollary 5.1.5. So sγ · sµ−γ · sµ′ · sα ≤

sγ · sc−γ · sα. Note that, c− γ > c−α, hence this case falls into the case 2) above.

So we are done with this case. If µ ∩ α is not a root then the proof is similar.

b) Assume that µ′ > α. Now, if µ′ ⊥ α then we can take β′ = µ′. Now, assume

that µ′ is not perpendicular to α. Then < µ′, α >= 1 which implies sα(µ′) =

µ′− < µ′, α∨ > α = µ′ − α. So µ′ − α is a root. Also, sµ′ and sα Hecke commute

by Lemma 3.2.1. Furthermore, µ + α = c − µ′ + α = c − (µ′ − α) is also a

root and (µ + α) ∩ µ′ = α since µ and µ′ are disjoint and µ′ > α. We also have

sµ · sα ≤ sµ+α by Corollary 5.1.5. Thus sµ · sµ′ · sα = sµ · sα · sµ′ ≤ sµ+α · sµ′ . Here,

note that sµ+α · sµ′ = sµ+α · sµ′−α · sα by case 2) in Lemma 3.2.1. Now, observe

that (µ+ α) + (µ′ − α) = c and µ′ − α ≤ c− α. Thus, we can take β′ = µ′ − α in

this case.

We will state the result in the following theorem.

Theorem 5.3.4. Let α ∈ Πre,+aff be such that α < c. Then

1) Γc+α(id) = tβ′sα : β′ ∈ Πre,+aff (α) ∪ tβ′−csα : β′ ∈ Πre,+

aff (α)

2) |Γc+α(id)| = |β′ : β′ ∈ Πre,+aff (α)|

3) For all w ∈ Γc+α(id), `(w) = 2(n− 1) + `(sα)

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where Πre,+aff (α) := β′ ∈ Πre,+

aff : β′ < c and either β′ ≤ c− α or both β′ > α and β′ ⊥ α.

Proof. 1) Let idβ1−→ sβ1

β2−→ sβ1sβ2 ...βr−→ w = sβ1sβ2 ...sβr be a path in the moment graph

such that∑r

i=1 βi ≤ c+α. Note that, by Remark 4.0.2 we can assume that w = sβ1 ·sβ2 ·...·sβr .

where∑r

i=1 βi = c+ α and βi < c for all i. Now, by Lemma 5.3.2, we can also suppose that∑ti=1 βi = c and

∑ri=t+1 βi = α for some t. Then

w = (sβ1 · ... · sβt) · (sβt+1 · sβt+2 · ... · sβr) ≤ (sµ sµ′) · sα

for some µ, µ′ ∈ Πre,+aff such that µ + µ′ = c by Equation 5.1 and Corollary 3.2.2. Here, we

have (sµsµ′) · sα ≤ (sµ · sµ′) · sα. Furthermore, sµ · sµ′ · sα ≤ sβ · sβ′ · sα where β + β′ = c and

either β′ ≤ c− α or both β′ > α and β′ ⊥ α, by Lemma 5.3.3. Now, by Lemmas 3.4.6 and

3.4.8 sβsβ′sα is reduced so sβ ·sβ′ ·sα = sβsβ′sα. Hence Γc+α(id) = (sβsβ′)sα : β, β′ ∈ Πre,+aff

where β+β′ = c such that either β′ ≤ c−α or both β′ > α and β′ ⊥ α. Now, also note that

sβsβ′ = tβ′ if α0 /∈ supp(β′) and sβsβ′ = tβ′−c if α0 ∈ supp(β′), see the proof of Lemma 3.4.1.

2) Let sβsβ′sα and sνsν′sα ∈ Γc+β(id) be given. We need to show that sβsβ′sα 6= sνsν′sα if

β′ 6= ν ′. Now assume that β′ 6= ν ′ but sβsβ′sα = sνsν′sα. By Lemmas 3.4.6 and 3.4.8, both

sβsβ′sα and sνsν′sα are reduced so the equality, sβsβ′sα = sνsν′sα implies that sβsβ′ = sνsν′

but this is a contradiction by Lemma 3.4.1.

3) Let w = sβsβ′sα ∈ Γc+α(id). Then by Lemmas 3.4.6 and 3.4.8, `(w) = 2(n−1)+`(sα).

Remark 5.3.5. Let α ∈ Πre,+aff be such that α < c. Now, note that, tβ′ = sβsβ′ if α0 ∈ supp(β)

and tβ′−c = sβsβ′ if α0 /∈ supp(β) where β, β′ ∈ Πre,+aff such that β + β′ = c, see the proof of

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Lemma 3.4.1. So by Theorem 5.3.4 we get

Γc+α(id) = sβsβ′sα : β′ ∈ Πre,+aff (α) such that β + β′ = c. (5.2)

Example 5.3.6. Let Waff be the Weyl group of type A(1)4 . We compute Γc+α(id) where α =

α0 +α4. Note that, we have six positive real roots which are smaller than c−α = α1 +α2 +α3;

β′1 = α1, β′2 = α2, β

′3 = α3, β

′4 = α1 + α2, β

′5 = α2 + α3, β

′6 = α1 + α2 + α3. Also, we have

only one positive root which is smaller than c, strictly bigger than α and perpendicular to

α; β′7 = α0 + α1 + α3 + α4. So by Theorem 5.3.4, we get

Γc+α(id) = tβ′1sα, tβ′2sα, tβ′3sα, tβ′4sα, tβ′5sα, tβ′6sα, tβ′7−csα

Moreover, for any w ∈ Γc+α(id), we have `(w) = 2(n−1)+`(sα) = 2(n−1)+2 supp(α)−1 =

2 · 4 + 2 · 2− 1 = 11.

5.4 Curve Neighborhood Γmc+α(id)

This section is devoted to a generalization of the result in the previous section. We will

begin with some lemmas.

Lemma 5.4.1. Let β, β′, ν and ν ′ ∈ Πre,+aff such that ν + ν ′ = β + β′ = c and ν ′ ≤ β′. Then

sν · sν′ · sβ · sβ′ ≤ (sν · sν′)2.

Proof. Here we have two main cases;

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1) ν ′ + β is a root. Then c − (ν ′ + β) = c − ν ′ − β = ν − β = β′ − ν ′ is also a root. We

have sν′ · sβ ≤ sν′+β and (ν ′ + β) ∩ β′ = ν ′ so by Lemma 3.2.1 part 2a) we get

sν · sν′ · sβ · sβ′ ≤ sν · sν′+β · sβ′ ≤ sν · sν′+β · sβ′−ν′ · sν′ .

Also, we have ν ∩ (ν ′ + β) = β and β + β′ − ν ′ = c− ν ′ = ν which implies

sν · sν′+β · sβ′−ν′ · sν′ ≤ sν · sν′ · sβ · sβ′−ν′ · sν′ ≤ sν · sν′ · sν · sν′ .

by Lemma 3.2.1 part 2a).

2) ν ′ + β is not a root. Note that, since ν ′ + β < c and β < ν we have sν′ · sβ = sβ · sν′

and sν · sβ = sβ · sν which implies that sν · sν′ · sβ · sβ′ = sβ · sν · sν′ · sβ′ . Also note that,

ν ′ + β = c− ν + c− β′ = 2c− (ν + β′) and since ν ′ + β is not a root ν + β′ is also not

a root. Then sν · sβ′ = sβ′ · sν . Furthermore, sν′ · sβ′ = sβ′ · sν′ since ν ′ < β′. We get

sβ · sν · sν′ · sβ′ = sβ · sν · sβ′ · sν′ . Note that, ν ∩ β′ 6= ∅. Here we have two cases;

a) If γ := ν ∩ β′ is a root then ν − γ and β′ − γ are also roots. By Lemma 3.2.1

we get sβ · sν · sβ′ · sν′ = sβ · sν · sβ′−γ · sγ · sν′ . Also note that, ν ′ = β′ − γ and

β+ γ = c− β′+ γ = c− (β′− γ) = c− ν ′ = ν. Again, if we also use the facts that

sβ · sν′ = sν′ · sβ and sβ · sν = sν · sβ we will obtain

sβ · sν · sβ′−γ · sγ · sν′ = sβ · sν · sν′ · sγ · sν′ = sν · sβ · sν′ · sγ · sν′

= sν · sν′ · sβ · sγ · sν′ ≤ sν · sν′ · sν · sν′since sβ · sγ ≤ sβ+γ = sν .

b) If ν∩β′ is not a root then there are some roots γ1 and γ2 such that ν∩β′ = γ1 +γ2.

Now, by Lemma 3.2.1 part 2b) we have sν · sβ′ = sν · sβ′−γ1−γ2 · sγ1 · sγ2 . We also

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have β′ − γ1 − γ2 = ν ′ and β + γ1 + γ2 = ν which implies that

sν · sν′ · sβ · sβ′ = sβ · sν · sβ′ · sν′ = sβ · sν · sβ′−γ1−γ2 · sγ1 · sγ2 · sν′

= sβ · sν · sν′ · sγ1 · sγ2 · sν′Again, sβ commutes with sν and sν′ . Thus,

sβ · sν · sν′ · sγ1 · sγ2 · sν′ = sν · sν′ · sβ · sγ1 · sγ2 · sν′ ≤ sν · sν′ · sβ+γ1+γ2 · sν′

= sν · sν′ · sν · sν′due to the fact that sβ · sγ1 · sγ2 ≤ sβ+γ1+γ2 = sν .

Lemma 5.4.2. Let β, β′, ν and ν ′ ∈ Πre,+aff such that ν + ν ′ = β + β′ = c. Also, assume that

ν ′ > β and ν ′ ⊥ β. Then sν · sν′ · sβ · sβ′ ≤ (sβ · sβ′)2.

Proof. First, note that sν′ and sβ Hecke commute by part 1) in Lemma 3.2.1. Also, by the

assumptions on ν ′ and β, the support of ν and β are disconnected which implies that sν and

sβ Hecke commute. So we get sν · sν′ · sβ · sβ′ = sβ · sν · sν′ · sβ′ . Here, the fact that the

support of ν and β are disconnected forces us to have ν ′ ∩ β′ 6= ∅ and ν ′ ∩ β′ not to be a

root. But then ν ′ ∩ β′ = γ1 + γ2 for some γ1, γ2 ∈ Πre,+aff . By part 2) in Lemma 3.2.1, we get

sν′ · sβ′ = sγ1 · sγ2 · sν′−γ1−γ2 · sβ′ . Now, observe that ν + γ1 + γ2 = β′ and ν ′ − γ1 − γ2 = β.

Thus

sβ · sν · sν′ · sβ′ = sβ · sν · sγ1 · sγ2 · sν′−γ1−γ2 · sβ′ ≤ sβ · sν+γ1+γ2 · sβ · sβ′ = sβ · sβ′ · sβ · sβ′ .

Lemma 5.4.3. Let β, β′, ν and ν ′ ∈ Πre,+aff such that ν + ν ′ = β + β′ = c. Also, assume that

ν ′ ≤ β′, β′ > α, and β′ ⊥ α. Then, for any positive integer m, sν · sν′ · (sβ · sβ′)m · sα ≤

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(sµ · sµ′)m+1 · sα for some positive real roots µ, µ′ such that µ+ µ′ = c and either µ′ ≤ c− α

or both µ′ > α and µ′ ⊥ α.

Proof. First, observe that β′ > α and β′ ⊥ α imply that c − α > β and (c − α) ⊥ β. Also,

supp(β) and supp(α) are disconnected. So α Hecke commute with both β′ and β. We have

several cases here;

1) If ν ′ = β′ then take µ′ = β′.

2) If ν ′∩α = ∅ then ν ′ ≤ c−α. Note that, by Lemma 5.4.1, we have sν ·sν′ ·sβ ·sβ′ ≤ (sν ·sν′)2

since ν ′ ≤ β′ which follows by sν · sν′ · (sβ · sβ′)m · sα ≤ (sν · sν′)m+1 · sα. So we can take

µ′ = ν ′.

3) If ν ′ ∩ α 6= ∅ then we have several cases;

• If we have both ν ′ > α and ν ′ ⊥ α then by the same arguments in case 2) we can take

µ′ = ν ′.

• If v′ = α then sν ·sν′ ·sα = sc−α ·sα ·sα. By Lemma 5.3.1, sc−α ·sα ·sα ≤ sα ·sc−α ·sα. Thus

sν ·sν′ ·(sβ ·sβ′)m·sα = sν ·sν′ ·sα·(sβ ·sβ′)m ≤ sα·sc−α·sα·(sβ ·sβ′)m = sα·sc−α·(sβ ·sβ′)m·sα.

Here, observe that by Lemma 5.4.2, sα · sc−α · sβ · sβ′ ≤ (sβ · sβ′)2 since c− α > β and

(c−α) ⊥ β which follows by sα · sc−α · (sβ · sβ′)m · sα ≤ (sβ · sβ′)m+1 · sα. Hence we can

take µ′ = β′ in this case.

• If ν ′ < α then ν ′ and α Hecke commute by part 1) in Lemma 3.2.1. Also, note that

ν ∩ α 6= ∅. Now, assume that ν ∩ α is a root, let γ := ν ∩ α. Then, by part 2) in

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Lemma 3.2.1 we have sν · sα = sγ · sν−γ · sα. Here, note that ν − γ + ν ′ = c − γ is

a root and by the same lemma sν−γ · sν′ ≤ sc−γ. Hence sν · sν′ · sα = sν · sα · sν′ =

sγ · sν−γ · sα · sν′ = sγ · sν−γ · sν′ · sα ≤ sγ · sc−γ · sα. Moreover, c− γ > c− α > β and

(c−γ) ⊥ β since (c−α) ⊥ β. Then by Lemma 5.4.2, sγ ·sc−γ ·sβ ·sβ′ ≤ (sβ ·sβ′)2 which

follows by sν · sν′ · (sβ · sβ′)m · sα ≤ sν · sν′ · sα · (sβ · sβ′)m ≤ sγ · sc−γ · sα · (sβ · sβ′)m =

sγ · sc−γ · (sβ · sβ′)m · sα ≤ (sβ · sβ′)m+1 · sα. Hence we can take µ′ = β′ in this case.

Now, if ν ∩ α is not a root then one can show that we can take µ′ = β′, by the same

arguments in this case.

• If ν ′ > α but ν ′ is not perpendicular to α then < ν ′, α∨ >= 1 so sα(ν ′) = ν ′ − α.

So ν ′ − α is a root which implies that ν + α = c − ν ′ + α = c − (ν ′ − α) is also a

root. Thus sν · sν′ · sα = sν · sα · sν′ = sν+α · sν′ by Lemma 3.2.1. Here, note that

(ν + α) ∩ ν ′ = α. So by the same lemma we get sν+α · sν′ = sν+α · sν′−α · sα. Now,

observe that ν ′−α ≤ β′ since both ν ′ and α are smaller than β′. Then by Lemma 5.4.1,

sν+α ·sν′−α ·sβ ·sβ′ ≤ (sν+α ·sν′−α)2 which follows by sν ·sν′ · (sβ ·sβ′)m ·sα = sν ·sν′ ·sα ·

(sβ ·sβ′)m ≤ sν+α ·sν′−α ·sα ·(sβ ·sβ′)m ≤ sν+α ·sν′−α ·(sβ ·sβ′)m ·sα ≤ (sν+α ·sν′−α)m+1 ·sα.

Last, note that ν ′ − α ≤ c− α so we can take µ′ = ν ′ − α in this case.

• Assume that ν ′ ∩ α 6= ∅ and also ν ′ ∩ α 6= ν ′, α. Then ν ′ ∩ α is a root since ν ′ + α

would be bigger than c otherwise, and that would contradict the facts that β′ > α and

β′ > ν ′. Now, let γ := ν ′∩α. Then ν ′−γ is a root and sν′ ·sα = sγ ·sν′−γ ·sα by Lemma

3.2.1. Also, note that ν+γ = c−ν ′+γ = c−(ν ′−γ) is a root so by the same lemma we

have sν ·sγ ≤ sν+γ. Thus sν ·sν′ ·sα = sν ·sγ ·sν′−γ ·sα ≤ sν+γ ·sν′−γ ·sα. Here, note that

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ν ′ − γ ≤ β′ since ν ′ ≤ β′. Then by Lemma 5.4.1, sν+γ · sν′−γ · sβ · sβ′ ≤ (sν+γ · sν′−γ)2

which follows by sν ·sν′ ·(sβ ·sβ′)m ·sα = sν ·sν′ ·sα ·(sβ ·sβ′)m ≤ sν+γ ·sν′−γ ·sα ·(sβ ·sβ′)m ≤

sν+γ · sν′−γ · (sβ · sβ′)m · sα ≤ (sν+γ · sν′−γ)m+1 · sα. Last, note that ν ′− γ ≤ c−α so we

can take µ′ = ν ′ − γ in this case.

The result of this section is:

Theorem 5.4.4. Let α < c be an affine positive real root and let m be a positive integer.

Then

1) Γmc+α(id) = tmβ′sα : β′ ∈ Πre,+aff (α) ∪ tm(β′−c)sα : β′ ∈ Πre,+

aff (α)

2) |Γmc+α(id)| = |β′ : β′ ∈ Πre,+aff (α)|

3) For all w ∈ Γmc+α(id), `(w) = 2m(n− 1) + `(sα)

where Πre,+aff (α) := β′ ∈ Πre,+

aff : β′ < c and either β′ ≤ c− α or both β′ > α and β′ ⊥ α.

Proof. 1) We will prove the statement by induction on m. Assume that m = 1. Then the

statement is true by Theorem 5.3.4. Now, assume that the statement is true for m = k. We

will prove that it is also true for m = k+ 1. Let idβ1−→ sβ1

β2−→ sβ1sβ2 ...βr−→ w = sβ1sβ2 ...sβr

be a path in the moment graph such that∑r

i=1 βi ≤ (k+1)c+α. Note that, by Remark 4.0.2

we can assume that w = sβ1 · sβ2 · ... · sβr where∑r

i=1 βi = (k+ 1)c+α and βi < c for all i. It

is enough to show that w ≤ (sβsβ′)k+1sα for some β, β′ ∈ Πre,+

aff such that β + β′ = c where

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either β′ ≤ c−α or both β′ > α and β′ ⊥ α since tβ′ = sβsβ′ if α0 ∈ supp(β) and tβ′−c = sβsβ′

if α0 /∈ supp(β), see the proof of Lemma 3.4.1. Moreover, by Lemma 5.3.2 we can make the

assumption that there is an integer p such that∑p

i=1 βi = c and∑r

i=p+1 βi = kc + α. Then

by Equation 5.1 we have sβ1 · sβ2 · ... · sβp ≤ sγsγ′ for some γ, γ′ ∈ Πre,+aff such that γ + γ′ = c

and sβp+1 · ... · sβr ≤ (sβsβ′)ksα for some β, β′ ∈ Πre,+

aff such that β + β′ = c where either

β′ ≤ c−α or both β′ > α and β′ ⊥ α by the induction assumption. Here, note that, sγsγ′ is

reduced by Lemma 3.4.3. Moreover, (sβsβ′)ksα is also reduced by Lemmas 3.4.6 and 3.4.8.

Thus,

w = (sβ1 · sβ2 · ... · sβp) · (sβp+1 · ... · sβr) ≤ (sγsγ′) · ((sβsβ′)ksα) = (sγ · sγ′) · ((sβ · sβ′)k · sα).

Now note that by Lemma 5.3.3 there are ν, ν ′ ∈ Πre,+aff such that sγ · sγ′ · sβ ≤ sν · sν′ · sβ

where ν + ν ′ = c and either ν ′ ≤ c− β = β′ or both ν ′ > β and ν ′ ⊥ β.

• If ν ′ ≤ c − β = β′ then by Lemma 5.4.1, we get sν · sν′ · sβ · sβ′ ≤ (sν · sν′)2 which

follows by

w ≤ (sγ · sγ′) · (sβ · sβ′)k · sα ≤ sν · sν′ · (sβ · sβ′)k · sα ≤ (sν · sν′)k+1 · sα.

Now, if we also have β′ ≤ c − α then c − α ≥ β′ ≥ ν ′ so we are done with this case.

Now if β′ > α and β′ ⊥ α then by Lemma 5.4.3, sν ·sν′ ·(sβ ·sβ′)m ·sα ≤ (sµ ·sµ′)m+1 ·sα

for some positive real roots, µ, µ′ such that µ + µ′ = c and either µ′ ≤ c − α or both

µ′ > α and µ′ ⊥ α. Hence we are done with this case too.

• If ν ′ > β and ν ′ ⊥ β then by Lemma 5.4.2 we get sν · sν′ · sβ · sβ′ ≤ (sβ · sβ′)2 which

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follows by

w ≤ (sγ · sγ′) · (sβ · sβ′)k · sα ≤ sν · sν′ · (sβ · sβ′)k · sα ≤ (sβ · sβ′)k+1 · sα.

Thus we are done with all cases.

Moreover, by Lemmas 3.4.6 and 3.4.8, (sβsβ′)k+1sα is reduced so (sβ·sβ′)k+1·sα = (sβsβ′)

k+1sα.

2) Let (sβsβ′)msα and (sνsν′)

msα ∈ Γmc+α(id). We need to show that (sβsβ′)msα 6= (sνsν′)

msα

if β′ 6= ν ′. Now assume that β′ 6= ν ′ but

(sβsβ′)msα = (sνsν′)

msα (5.3)

. By Lemmas 3.4.6 and 3.4.8 both (sβsβ′)msα and (sνsν′)

msα are reduced, so Equation 5.3

implies that (sβsβ′)m = (sνsν′)

m but this is a contradiction by Lemma 3.4.1.

3) Let w = (sβsβ′)msα ∈ Γmc+α(id). Then `(w) = 2m(n− 1) + `(sα), by Lemmas 3.4.6 and

3.4.8.

Remark 5.4.5. a) Let α ∈ Πre,+aff be such that α < c. Now, note that, tmβ′ = (sβsβ′)

m

if α0 ∈ supp(β) and tm(β′−c) = (sβsβ′)m if α0 /∈ supp(β) where β, β′ ∈ Πre,+

aff such that

β + β′ = c and m is a positive integer, see the proof of Lemma 3.4.1. So by Theorem 5.4.4

we get

Γmc+α(id) = (sβsβ′)msα : β′ ∈ Πre,+aff (α) such that β + β′ = c. (5.4)

b) For any w = sβ1 · sβ2 · ... · sβr . where∑r

i=1 βi = mc + α we have w ≤ u such that

u ∈ Γmc+α(id), see the arguments in the proof of Theorem 5.4.4.

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Example 5.4.6. Suppose that the affine Weyl groupWaff is of typeA(1)4 .We compute Γ12c+α(id)

where α = α0 + α4. Then, by Theorem 5.4.4, we get

Γ12c+α(id) = t12β′1sα, t12β′2

sα, t12β′3sα, t12β′4

sα, t12β′5sα, t12β′6

sα, t12(β′7−c)sα

where β′1 = α1, β′2 = α2, β

′3 = α3, β

′4 = α1 + α2, β

′5 = α2 + α3, β

′6 = α1 + α2 + α3 and

β′7 = α0 + α1 + α3 + α4, see Example 5.3.6. Moreover, for any w ∈ Γ12c+α(id), we have

`(w) = 2m(n− 1) + `(sα) = 2m(n− 1) + 2 supp(α)− 1 = 2 · 12 · 4 + 2 · 2− 1 = 99.

5.5 Curve Neighborhood Γmc(id)

Theorem 5.5.1. Let m ≥ 2 be a positive integer. Then we have

1) Γmc(id) = tmγ : γ ∈ Π.

2) |Γmc(id)| = n(n− 1).

3) For all w ∈ Γmc(id), `(w) = 2m(n− 1).

Proof. 1) Let idβ1−→ sβ1

β2−→ sβ1sβ2 ...βr−→ w = sβ1sβ2 ...sβr be a path in the moment graph

such that∑r

i=1 βi ≤ mc. Note that, by Remark 4.0.2 we can assume that w = sβ1 ·sβ2 ·...·sβr .

where∑r

i=1 βi = mc and βi < c for all i. Now, note that∑r−1

i=1 βi = mc−βr = (m−1)c+c−βr.

Let α := c−βr and α′ := βr. Then by Remark 5.4.5 parts a) and b) we have sβ1 ·sβ2 ·...·sβr−1 ≤

(sβsβ′)m−1sα for some positive real roots β and β′ such that β+β′ = c and either β′ ≤ c−α

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or both β′ > α and β′ ⊥ α. Then, we get

w = (sβ1 · sβ2 · ... · sβr−1) · sβr ≤ (sβsβ′)m−1sα · sα′ .

Now, note that (sβsβ′)m−1sα = (sβ · sβ′)m−1 · sα, since = (sβsβ′)

m−1sα is reduced by Lemma

3.4.6 and Lemma 3.4.8. Furthermore, if β′ ≤ c − α then sβ · sβ′ · sα · sα′ ≤ (sβ · sβ′)2 by

Lemma 5.4.1 which follows by (sβsβ′)m−1sα · sα′ = (sβ · sβ′)m−1 · sα · sα′ ≤ (sβ · sβ′)m. If

β′ > α and β′ ⊥ α then sβ · sβ′ · sα · sα′ ≤ (sα · sα′)2 by Lemma 5.4.2 which follows by

(sβsβ′)m−1sα · sα′ = (sβ · sβ′)m−1 · sα · sα′ ≤ (sα · sα′)m. We also know that (sαsα′)

m = tmγ for

some γ ∈ Π, see the proof of Lemma 3.4.1.

2) Let β, β′, ν, ν ′ ∈ Πre,+aff such that ν + ν ′ = β + β′ = c. Then (sβsβ′)

m 6= (sνsν′)m for any

positive integer m if β 6= ν by Lemma 3.4.1. Also, note that |β ∈ Πre,+aff : β < c| =

n(n− 1).

3) Let w = (sβsβ′)m ∈ Γmc(id) for some positive real roots β, β′ such that β + β′ = c. Then

`(w) = 2m(n− 1) by Lemma 3.4.3.

Remark 5.5.2. Note that, tmβ′ = sβsβ′ if α0 ∈ supp(β) and tm(β′−c) = sβsβ′ if α0 /∈ supp(β),

where β, β′ ∈ Πre,+aff such that β+β′ = c and m is a positive integer, see the proof of Lemma

3.4.1. Furthermore,

β′ : β′ ∈ Πre,+aff , β′ < c, α0 /∈ supp(β′) t β′ − c : β′ ∈ Πre,+

aff , β′ < c, α0 ∈ supp(β′) = Π.

Thus we have

Γmc(id) = (sβsβ′)m : β, β′ ∈ Πre,+aff such that β + β′ = c. (5.5)

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Example 5.5.3. Suppose that the affine Weyl group Waff is associated to A(1)2 . Then

Γ10c(id) = t10α1 , t10α2 , t10(α1+α2), t−10α1 , t−10α2 , t−10(α1+α2).

Moreover, for any w ∈ Γ10c(id), `(w) = 2m(n− 1) = 2 · 10 · 2 = 40.

5.6 Curve Neighborhood Γd(id) for any d > c

Finally, we will discuss the most general case in this section. In order to prove the result we

will first consider two lemmas.

Lemma 5.6.1. Let α, α′ ∈ Πre,+aff such that α + α′ = c Also, let γ1, γ2, ..., γk ∈ Πre,+

aff such

that supp(γi) and supp(γj) are disconnected for any 1 ≤ i, j ≤ k such that i 6= j. Then

sα · sα′ · sγ1 · sγ2 · ... · sγk ≤ sβ · sβ′ · sγ1 · sγ2 · ... · sγk for some affine positive real roots,

β, β′ ∈ Πre,+aff such that either β′ ∩ γi = ∅ or both β′ > γi and β′ ⊥ γi for any i = 1, 2, ..., k.

Proof. We will prove the statement by the induction on k. Let k = 1. Then by Equation

5.2, (sα · sα′) · sγ1 ≤ (sβ · sβ′) · sγ1 for some β, β′ ∈ Πre,+aff such that β + β′ = c where either

β′ ≤ c− γ1 i.e β′ ∩ γ1 = ∅ or both β′ > γ1 and β′ ⊥ γ1. So we are done with this case. Now

we will assume that the statement is true for k and prove that it is also true for k + 1 i.e

we will show that (sα · sα′) · sγ1 · sγ2 · ... · sγk+1≤ (sβ · sβ′) · sγ1 · sγ2 · ... · sγk+1

for some affine

positive real roots, β, β′ ∈ Πre,+aff such that either β′ ∩ γi = ∅ or both β′ > γi and β′ ⊥ γi for

any i = 1, 2, ..., k + 1. Now note that (sα · sα′) · sγ1 · sγ2 · ... · sγk ≤ (sν · sν′) · sγ1 · sγ2 · ... · sγk

for some ν, ν ′ ∈ Πre,+aff such that ν + ν ′ = c where ν ′ ∩ γi = ∅ or both ν ′ > γi and ν ′ ⊥ γi for

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any i = 1, 2, ..., k by the induction assumption. So there is a subset I of γ1, γ2, ..., γk such

that ν ′ ∩ γi = ∅ for all γi ∈ I and ν ′ > γj and ν ′ ⊥ γj for all γj ∈ γ1, γ2, ..., γk \ I. Thus

(sα ·sα′)·sγ1 ·sγ2 ·...·sγk ·sγk+1≤ (sν ·sν′ ·sγ1 ·sγ2 ·...·sγk)·sγk+1

. Now, if we also have ν ′∩γk+1 = ∅

or both ν ′ > γk+1 and ν ′ ⊥ γk+1 then we are done. Assume not i.e assume that ν ′∩ γk+1 6= ∅

and ν ′ is not strictly bigger than or not perpendicular to γk+1. Note that sγi and sγj Hecke

commute for any 1 ≤ i, j ≤ k+1 so sν ·sν′ ·sγ1 ·sγ2 · ... ·sγk ·sγk+1= sν ·sν′ ·sγk+1

·sγ1 ·sγ2 · ... ·sγk .

Here we have several cases;

a) Assume that ν ′ ∩ γk+1 6= ν ′, γk+1 and ν ′ ∩ γk+1 is a root. Let γ := ν ′ ∩ γk+1. Then

ν ′ − γ is a root and sν′ · sγk+1= sγ · sν′−γ · sγk+1

by Lemma 3.2.1. Also, note that ν + γ =

c − ν ′ + γ = c − (ν ′ − γ) is a root so by the same lemma we have sν · sγ ≤ sν+γ. Thus

sν ·sν′ ·sγk+1= sν ·sγ ·sν′−γ ·sγk+1

≤ sν+γ ·sν′−γ ·sγk+1which follows by sν ·sν′ ·sγk+1

·sγ1 ·sγ2 ·...·sγk ≤

sν+γ ·sν′−γ ·sγk+1·sγ1 ·sγ2 · ... ·sγk . Here, note that (ν ′−γ)∩γk+1 = ∅ and (ν ′−γ)∩γi = ∅ for

all γi ∈ I and (ν ′ − γ) > γj and (ν ′ − γ) ⊥ γj for all γj ∈ γ1, γ2, ..., γk \ I since ν ′ − γ < ν ′

and supp(γ) and supp(γi) are disconnected for any i = 1, 2, ..., k. So we can take β′ = ν ′− γ

in this case. The case where ν ′ ∩ γk+1 is not a root is similar.

b) If ν ′ ∩ γk+1 = γk+1 then γk+1 ≤ ν ′.

• Assume that v′ = γk+1. Then sν · sν′ · sγk+1= sc−γk+1

· sγk+1· sγk+1

. By Lemma 5.3.1,

sc−γk+1·sγk+1

·sγk+1≤ sγk+1

·sc−γk+1·sγk+1

which follows by sν ·sν′ ·sγk+1·sγ1 ·sγ2 · ... ·sγk ≤

sγk+1· sc−γk+1

· sγk+1· sγ1 · sγ2 · ... · sγk . Here, observe that (c − γk+1) ∩ γk+1 = ∅. Also,

(c− γk+1) > γi and (c− γk+1) ⊥ γi for all i = 1, 2, ..., k since supp(γk+1) and supp(γi)

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are disconnected for any i = 1, 2, ..., k. Hence we can take β′ = c− γk+1 in this case.

• Assume that γk+1 < ν ′. Now since ν ′ is not perpendicular to γk+1 we have < ν ′, γ∨k+1 >=

1 so sγk+1(ν ′) = ν ′ − γk+1. So ν ′ − γk+1 is a root which implies that ν + γk+1 =

c−ν ′+γk+1 = c−(ν ′−γk+1) is also a root. Thus sν ·sν′ ·sγk+1= sν ·sγk+1

·sν′ = sν+γk+1·sν′

by Lemma 3.2.1. Here, note that (ν+ γk+1)∩ ν ′ = γk+1. So by the same lemma we get

sν+γk+1· sν′ = sν+γk+1

· sν′−γk+1· sγk+1

which follows by sν · sν′ · sγk+1· sγ1 · sγ2 · ... · sγk ≤

sν+γk+1· sν′−γk+1

· sγk+1· sγ1 · sγ2 · ... · sγk . Here, observe that (ν ′ − γk+1) ∩ γk+1 = ∅.

Also, (ν ′ − γk+1) ∩ γi = ∅ for all γi ∈ I and (ν ′ − γk+1) > γj and (ν ′ − γk+1) ⊥ γj

for all γj ∈ γ1, γ2, ..., γk \ I since ν ′ − γk+1 < ν ′ and supp(γk+1) and supp(γi) are

disconnected for any i = 1, 2, ..., k. So we can take β′ = ν ′ − γk+1 in this case.

c) If ν ′ ∩ γk+1 = ν ′ then ν ′ ≤ γk+1. We already consider the case where ν ′ = γk+1 in

case b) above so assume that ν ′ < γk+1. Then sν′ and sγk+1Hecke commute by part 1)

in Lemma 3.2.1. Also, note that ν ∩ γk+1 6= ∅. Now, assume that ν ∩ γk+1 is a root; let

γ := ν ∩ γk+1. Then, by part 2) in Lemma 3.2.1 we have sν · sγk+1= sγ · sν−γ · sγk+1

. Here,

note that ν − γ + ν ′ = c − γ is a root and by the same lemma sν−γ · sν′ ≤ sc−γ. Hence

sν · sν′ · sγk+1= sν · sγk+1

· sν′ = sγ · sν−γ · sγk+1· sν′ = sγ · sν−γ · sν′ · sγk+1

≤ sγ · sc−γ · sγk+1

which follows by sν · sν′ · sγk+1· sγ1 · sγ2 · ... · sγk ≤ sγ · sc−γ · sγk+1

· sγ1 · sγ2 · ... · sγk . Moreover,

(c−γ)∩γk+1 6= ∅ and (c−γ)∩γk+1 6= c−γ, γk+1 since γ < γk+1. Also, observe that (c−γ) > γi

and (c− γ) ⊥ γi for all i = 1, 2, ..., k since supp(γk+1) and supp(γi) are disconnected for any

i = 1, 2, ..., k. Thus we can consider this case under case a) above. Now, if ν ∩ γk+1 is not a

root then the proof is similar.

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Lemma 5.6.2. Let ν, ν ′, µ, µ′ ∈ Πre,+aff such that ν + ν ′ = µ + µ′ = c and either ν ′ ≤ µ′ or

both ν ′ > µ and ν ′ ⊥ µ. Also, let γ1, γ2, ..., γk ∈ Πre,+aff such that supp(γi) and supp(γj) are

disconnected for any 1 ≤ i, j ≤ k such that i 6= j. Furthermore, suppose that either µ′∩γi = ∅

or both µ′ > γi and µ′ ⊥ γi for any 1 ≤ i ≤ k. Then sν · sν′ · (sµ · sµ′)m · sγ1 · sγ2 · ... · sγk ≤

(sβ · sβ′)m+1 · sγ1 · sγ2 · ... · sγk for some affine positive real roots, β, β′ ∈ Πre,+aff such that either

β′ ∩ γi = ∅ or both β′ > γi and β′ ⊥ γi for any i = 1, 2, ..., k where m ≥ 1 is an integer.

Proof. First, if ν ′ = µ′ then we can take β′ = µ′. We will assume that ν ′ 6= µ′. If both ν ′ > µ

and ν ′ ⊥ µ are true then by Lemma 5.4.2 we get sν · sν′ · (sµ · sµ′)m ≤ (sµ · sµ′)m+1 which

follows by sν · sν′ · (sµ · sµ′)m · sγ1 · sγ2 · ... · sγk ≤ (sµ · sµ′)m+1 · sγ1 · sγ2 · ... · sγk . Thus we

can take β′ = µ′ in this case. Suppose that ν ′ < µ′. Now, observe that there is a subset,

I of γ1, γ2, ..., γk such that ν ′ ∩ γi = ∅ for all γi ∈ I and ν ′ > γj and ν ′ ⊥ γj for all

γj ∈ γ1, γ2, ..., γk \ I since we have either µ′ ∩ γi = ∅ or both µ′ > γi and µ′ ⊥ γi for any

1 ≤ i ≤ k. Then ν ′ ∩ γi = ∅ for all γi ∈ I since ν ′ < µ′. Here, we have two cases;

a) If we have either ν ′∩γj = ∅ or ν ′ > γj and ν ′ ⊥ γj for any γj such that γj ∈ γ1, γ2, ..., γk\I

then we can take β′ = ν ′ since sν ·sν′ · (sµ ·sµ′)m ≤ (sν ·sν′)m+1 by Lemma 5.4.1 which follows

by sν · sν′ · (sµ · sµ′)m · zd−(m+1)c ≤ (sν · sν′)m+1 · zd−(m+1)c.

b) Assume that we have neither ν ′ ∩ γj = ∅ nor both ν ′ > γj and ν ′ ⊥ γj for all γj ∈ B for

some B ⊆ γ1, γ2, ..., γk\I. Note that sγi and sγj Hecke commutes for any 1 ≤ i, j ≤ k so we

can assume that γ1 ∈ B. Now, note that supp(µ) and supp(γ1) are disconnected since µ′ > γ1

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and µ′ ⊥ γ1 so γ1 Hecke commute with sµ and sµ′ . Thus sν · sν′ · (sµ · sµ′)m · sγ1 · sγ2 · ... · sγk =

sν · sν′ · sγ1 · (sµ · sµ′)msγ2 · ... · sγk . Here, we have several cases;

• If ν ′ = γ1 then sν ·sν′ ·sγ1 = sc−γ1 ·sγ1 ·sγ1 . By Lemma 5.3.1, sc−γ1 ·sγ1 ·sγ1 ≤ sγ1 ·sc−γ1 ·sγ1

which follows by sν ·sν′ ·sγ1 ·(sµ ·sµ′)m ·sγ2 ·...·sγk ≤ sγ1 ·sc−γ1 ·sγ1 ·(sµ ·sµ′)m ·sγ2 ·...·sγk =

sγ1 · sc−γ1 · (sµ · sµ′)m · sγ1 · sγ2 · ... · sγk . Here, observe that (c− γ1) > µ and (c− γ1) ⊥ µ

since µ′ > γ1 and µ′ ⊥ γ1. So by Lemma 5.4.2 we get sγ1 ·sc−γ1 ·(sµ ·sµ′)m ≤ (sµ ·sµ′)m+1

which follows by sγ1 · sc−γ1 · (sµ · sµ′)m · sγ1 · sγ2 · ... · sγk . ≤ (sµ · sµ′)m+1 · sγ1 · sγ2 · ... · sγk .

Thus we can take β′ = µ′ in this case.

• If ν ′ < γ1 then ν ∩ γ1 6= ∅ since ν + ν ′ = c. Assume that ν ∩ γ1 is a root and let

γ := ν ∩ γ1. Then, by part 2) in Lemma 3.2.1 we have sν · sγ1 = sγ · sν−γ · sγ1 . Here,

note that ν − γ + ν ′ = c− γ is a root and by the same lemma sν−γ · sν′ ≤ sc−γ. Hence

sν ·sν′ ·sγ1 = sν ·sγ1 ·sν′ = sγ ·sν−γ ·sγ1 ·sν′ = sγ ·sν−γ ·sν′ ·sγ1 ≤ sγ ·sc−γ ·sγ1 which follows

by sν · sν′ · sγ1 · (sµ · sµ′)m · sγ2 · ... · sγk ≤ sγ · sc−γ · sγ1 · (sµ · sµ′)m · sγ2 · ... · sγk = sγ · sc−γ ·

(sµ · sµ′)m · sγ1 · sγ2 · ... · sγk . Here, observe that (c− γ) > µ and (c− γ) ⊥ µ since γ < γ1

where µ′ > γ1 and µ′ ⊥ γ1. So by Lemma 5.4.2 we get sγ ·sc−γ ·(sµ ·sµ′)m ≤ (sµ ·sµ′)m+1

which follows by sγ · sc−γ · (sµ · sµ′)m · sγ1 · sγ2 · ... · sγk . ≤ (sµ · sµ′)m+1 · sγ1 · sγ2 · ... · sγk .

Again we can take β′ = µ′ in this case. If ν ∩ γ1 is not a root then the proof is similar.

• If γ1 < ν ′ then since ν ′ is not perpendicular to γ1 we have < ν ′, γ∨1 >= 1 so sγ1(ν′) =

ν ′−γ1. So ν ′−γ1 is a root which implies that ν+γ1 = c−ν ′+γ1 = c− (ν ′−γ1) is also

a root. Thus sν · sν′ · sγ1 = sν · sγ1 · sν′ = sν+γ1 · sν′ by Lemma 3.2.1. Here, note that

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(ν + γ1) ∩ ν ′ = γ1. So by the same lemma we get sν+γ1 · sν′ = sν+γ1 · sν′−γ1 · sγ1 which

follows by sν ·sν′ ·sγ1 · (sµ ·sµ′)m ·sγ2 · ... ·sγk ≤ sν+γ1 ·sν′−γ1 ·sγ1 · (sµ ·sµ′)m ·sγ2 · ... ·sγk =

sν+γ1 · sν′−γ1 · (sµ · sµ′)m · sγ1 · sγ2 · ... · sγk . Here, observe that (ν ′ − γ1) ∩ γ1 = ∅.

Now, if B = γ1 then for any 1 ≤ i ≤ k we have either (ν ′ − γ1) ∩ γi = ∅ or both

(ν ′ − γ1) > γi and (ν ′ − γ1) ⊥ γi. Also, note that (ν ′ − γ1) < µ′ since ν ′ < µ′

which implies that sν+γ1 · sν′−γ1 · (sµ · sµ′)m ≤ (sν+γ1 · sν′−γ1)m+1 which follows by

sν+γ1 · sν′−γ1 · (sµ · sµ′)m · sγ1 · sγ2 · ... · sγk . ≤ (sν+γ1 · sν′−γ1)m+1 · sγ1 · sγ2 · ... · sγk . Thus

we can take β′ = ν ′− γ1 in this case. Now, suppose that |B| > 1. Then B can have at

most two elements since we have neither ν ′ ∩ γj = ∅ nor both ν ′ > γj and ν ′ ⊥ γj for

all γj ∈ B which implies that ν ′ intersect with γj and it is not strictly bigger than and

not perpendicular to γj for any γj ∈ B and also by the fact that supp(γi) and supp(γj)

are disconnected for any 1 ≤ i, j ≤ k. Now, let γq ∈ B such that γq 6= γ1. Then one

can show that sν+γ1 ·sν′−γ1 ·(sµ ·sµ′)m ·sγ1 ·sγ2 · ... ·sγk . ≤ (sβ ·sβ′)m+1 ·sγ1 ·sγ2 · ... ·sγk for

some affine positive real roots β, β′ ∈ Πre,+aff such that either β′ ∩ γi = ∅ or both β′ > γi

and β′ ⊥ γi for any i = 1, 2, ..., k by considering the relationships between ν ′ − γ1 and

γq which are identical with those between ν ′ and γ1 above under case b).

The result for the most general case is:

Theorem 5.6.3. Let d = (d0, d1, ..., dn−1) > c be a degree and m = mind0, d1, ..., dn−1.

Also, assume that zd−mc = sγ1sγ2 ...sγk for some k where this expression is obtained in Theo-

77

rem 3.1.2. Then

1) Γd(id) = tmβ′zd−mc : β′ ∈ Πre,+aff (d−mc) ∪ tm(β′−c)zd−mc : β′ ∈ Πre,+

aff (d−mc)

2) |Γd(id)| = |β′ : β′ ∈ Πre,+aff (d−mc)|

3) For all w ∈ Γd(id), `(w) = 2m(n− 1) + `(zd−mc)

where Πre,+aff (d−mc) is the set of β′ ∈ Πre,+

aff such that β′ < c and either β′ ∩ γi = ∅ or both

β′ > γi and β′ ⊥ γi for any i = 1, ..., k.

Proof. 1) First, note that we have either supp(γi) and supp(γj) are disconnected or both

γi ⊥ γj and γi, γj are comparable, for any 1 ≤ i, j ≤ k such that i 6= j; see Theorem 3.1.2.

Now, we define B := γi1 , γi2 , ..., γit ⊆ γ1, γ2, ..., γk such that for any 1 ≤ i ≤ k, we have

γi ≤ γil for some γil ∈ B. Note that, any two elements of B have disconnected supports.

Now, observe that for a root γij ∈ B and β′ ∈ Πre,+aff such that β′ ∩ γij = ∅ or both β′ > γij

and β′ ⊥ γij implies that we have either β′ ∩ γi = ∅ or both β′ > γi and β′ ⊥ γi for all

1 ≤ i ≤ k such that γi ≤ γij , respectively. Also, for any 1 ≤ i ≤ k such that β′ ∩ γi = ∅

or both β′ > γi and β′ ⊥ γi implies that we have either β′ ∩ γij = ∅ or both β′ > γij and

β′ ⊥ γij where γi ≤ γij , respectively. So we may assume that B = γ1, γ2, ..., γk i.e we will

suppose that supp(γi) and supp(γj) are disconnected for any 1 ≤ i, j ≤ k such that i 6= j.

Let idβ1−→ sβ1

β2−→ sβ1sβ2 ...βr−→ w = sβ1sβ2 ...sβr be a path in the moment graph such that∑r

i=1 βi ≤ d. Note that by Remark 4.0.2 we can assume that w = sβ1 · sβ2 · ... · sβr where∑ri=1 βi = d and βi < c for all i. Now, note that tmβ′ and tm(β′−c) are given by (sβsβ′)

m

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for some β, β′ ∈ Πre,+aff such that β + β′ = c; see the proof of Lemma 3.4.1. So it is enough

to show that w ≤ (sβsβ′)mzd−mc where we have either β′ ∩ γi = ∅ or both β′ > γi and

β′ ⊥ γi for any i = 1, 2, ..., k. Furthermore, we can assume that there is an integer p such

that∑p

i=1 βi = c and∑r

i=p+1 βi = d − c, by Lemma 5.3.2. We will prove the statement by

the induction on m.

First, suppose that m = 1. Then by Equation 5.1, sβ1 · sβ2 · ... · sβp ≤ sα · sα′ for some

α, α′ ∈ Πre,+aff such that α+ α′ = c. Moreover, sβp+1 · ... · sβr ≤ zd−c, by Theorem 5.1.1. Thus

w = (sβ1 · sβ2 · ... · sβp) · (sβp+1 · ... · sβr) ≤ (sα · sα′) · zd−c.

Now, by Lemma 5.6.1 we have sα · sα′ · sγ1 · sγ2 · ... · sγk ≤ sβ · sβ′ · sγ1 · sγ2 · ... · sγk for some

affine positive real roots, β, β′ ∈ Πre,+aff such that we have either β′ ∩ γi = ∅ or both β′ > γi

and β′ ⊥ γi for any i = 1, 2, ..., k.

Now, we will suppose that the statement is true for m and prove that it is also true for

m+ 1. Again, by Equation 5.1 we have sβ1 · sβ2 · ... · sβp ≤ sα · sα′ for some α, α′ ∈ Πre,+aff such

that α + α′ = c and sβp+1 · ... · sβr ≤ (sµ · sµ′)m · zd−(m+1)c for some µ, µ′ ∈ Πre,+aff such that

µ+ µ′ = c where either µ′ ∩ γi = ∅ or µ′ > γj and µ′ ⊥ γj by induction assumption. Thus

w = (sβ1 · sβ2 · ... · sβp) · (sβp+1 · ... · sβr) ≤ (sα · sα′) · (sµ · sµ′)m · zd−(m+1)c.

Now, observe that sα ·sα′ ·sµ ≤ sν ·sν′ ·sµ for some ν, ν ′ ∈ Πre,+aff such that ν+ν ′ = c and either

ν ′ ≤ µ′ or both ν ′ > µ and ν ′ ⊥ µ by Equation 5.2. Hence sα · sα′ · (sµ · sµ′)m · zd−(m+1)c ≤

(sν · sν′) · (sµ · sµ′)m · zd−(m+1)c. By Lemma 5.6.2 we have sν · sν′ · (sµ · sµ′)m · sγ1 · sγ2 · ... · sγk ≤

(sβ ·sβ′)m+1 ·sγ1 ·sγ2 · ... ·sγk for some affine positive real roots, β, β′ such that β+β′ = c where

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either β′ ∩ γi = ∅ or both β′ > γi and β′ ⊥ γi for any i = 1, 2, ..., k. Furthermore, by Lemma

3.4.9 , zd−(m+1)c = sγ1sγ2 ...sγk is reduced so sγ1sγ2 ...sγk = sγ1 · sγ2 · ... · sγk which follows by

(sβ · sβ′)m+1 · sγ1 · sγ2 · ... · sγk = (sβ · sβ′)m+1 · (sγ1sγ2 ...sγk) = (sβ · sβ′)m+1 · zd−(m+1)c. Now

by Lemma 3.4.10, the element (sβsβ′)m+1zd−(m+1)c is reduced so (sβ · sβ′)m+1 · zd−(m+1)c =

(sβsβ′)m+1zd−(m+1)c.

2) Let (sβsβ′)mzd−mc and (sνsν′)

mzd−mc ∈ Γd(id). We need to show that (sβsβ′)mzd−mc 6=

(sνsν′)mzd−mc if β′ 6= ν ′. Now assume that β′ 6= ν ′ but

(sβsβ′)msγ1sγ2 ...sγk = (sνsν′)

msγ1sγ2 ...sγk . (5.6)

By Lemma 3.4.10 both (sβsβ′)msγ1sγ2 ...sγk and (sνsν′)

msγ1sγ2 ...sγk are reduced, so Equation

5.6 implies that (sβsβ′)m = (sνsν′)

m but this is a contradiction by Lemma 3.4.1.

3) Let w ∈ Γd(id). Then by Lemma 3.4.10 we have `(w) = 2m(n− 1) + `(zd−mc).

Remark 5.6.4. Now, note that both tmβ′ and tm(β′−c) are given by (sβsβ′)m where β, β′ ∈ Πre,+

aff

such that β+β′ = c and m is a positive integer; see the proof of Lemma 3.4.1. So by Theorem

5.6.3 we get

Γd(id) = (sβsβ′)mzd−mc : β′ ∈ Πre,+aff (d−mc) such that β + β′ = c. (5.7)

Example 5.6.5. Let Waff be the affine Weyl group associated to A(1)3 and d = (6, 5, 8, 5). Then

d = 5c+(1, 0, 3, 0) so m = 5 and d−5c = (1, 0, 3, 0). Also, zd−5c = sα0 ·sα2 ·sα2 ·sα2 = sα0sα2 .

80

Hence by Theorem 1.1.14 we get

Γd(id) = t5β′1zd−5c, t5β′2zd−5c, t5β′4zd−5c, t5(β′3−c)zd−5c

where β′1 = α1, β′2 = α3, β′3 = α0 +α1 +α3, β′4 = α1 +α2 +α3. Moreover, for all w ∈ Γd(id),

`(w) = 2m(n− 1) + `(zd−mc) = 2 · 5 · 3 + 2 = 32.

5.7 Reduction from Γd(w) to Γd(id)

In this section we will discuss our last result which indicates that to calculate Γd(w) for any

given degree d and w ∈ Waff one only needs to calculate Γd(id).

Theorem 5.7.1. Let w ∈ Waff and d be any degree. Then

Γd(w) = max w · u : u ∈ Γd(id) .

Proof. Let

wβ1−→ wsβ1

β2−→ wsβ1sβ2 ...βk−→ wsβ1sβ2 ...sβk

be a path in the moment graph such that∑k

i=1 βi ≤ d. Then

wsβ1sβ2 ...sβk ≤ w · (sβ1sβ2 ...sβk) ≤ w · u

for some u ∈ Γd(id). To complete the proof we need to show that the element w · u can be

reached by a path which starts with an element that is smaller than and equal to w and has

a degree at most d. Now, note that w ·u = vu for some v ∈ Waff such that v ≤ w by property

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e) of the Hecke product in Section 2.4. In addition, u = sβ′1sβ′2 ...sβ′t for some positive affine

real roots β′i, i = 1, 2, ..., t such that∑t

i=1 β′i ≤ d since u ∈ Γd(id). Thus

vβ′1−→ vsβ′1

β′2−→ vsβ′1sβ′2 ...β′t−→ vsβ′1sβ′2 ...sβ′k = vu

is the path we are looking for.

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