Labeling points of the flag variety <math> <mi>G</mi><mo>/</mo><mi>B</mi> </math>

Labeling points of the flag variety G/B

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

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 2 February 2009

Labeling points of the flag variety G/B

In this section we follow [St, Ch 8] to show the points of the flag variety are naturally indexed by labeled walks. This sis the first step in making a precise connection between the points of the flag variety and the alcove walk theory in [Ra].

Let G be a Tits group as in (3.5) over the field 𝔽=ℂ. The root subgroups

𝒳 α = x α c | c∈ℂ   for  α∈ R re ,  satisfy   w 𝒳 β w -1 = 𝒳 wβ ,
for w∈W and β ∈ R re , since h α ∨ c 𝒳 β h α ∨ c -1 = 𝒳 β and n α 𝒳 β n α -1 = 𝒳 s α ,β . As a group 𝒳 α is isomorphic to ℂ under addition.

The flag variety is G/B, where the subgroup

B  is generated by  T  and   x α f  for  α∈ R re + , f∈ ℂ.

Let w∈W. The inversion set of w is

R w = α∈ R re + | w -1 α ∉ R re + andl w =Card R w

is the length of w. View a reduced expression w → = s i 1 … s i l in the generators in (2.16) as a walk in W starting at 1 and ending at w,
1 → s i 1 → s i 1 s i 2 → … → s i 1 … s i l

Letting x i c = x α i c and n i = n α i 1 , the following theorem shows that

BwB= x i 1 c 1 n i 1 -1 x i 2 c 2 n i 2 -1 … x i l c l n i l -1 B| c 1 ,…, c l ∈ℂ
so that yhe G/B points of BwB are in bijection with labelings of the edges of the walk by complex numbers c 1 ,…, c l . The elements of R w are
β 1 = α i 1 , β 2 = s i 1 α i 2 ,…, β l = s i 1 … s i l-1 α i l ,
and the first relation in (1.6) gives
x i 1 c 1 n i 1 -1 x i 2 c 2 n i 2 -1 … x i l c l n i l -1 = x β 1 ± c 1 … x β l ± c l n w ,
where n w = n i 1 -1 … n i l -1 .

[St, Thm. 15 and Lemma 43] Let w∈W and let n w be a representative of w in N. If R w = β 1 … β l then x β 1 c 1 … x β l c l n w | c 1 ,…… c l ∈ℂ is a set of representatives of the B cosets is BwB.

Proof. The conceptual reason for this is that BwB = ( ∏ α∈ R re + 𝒳 α ) n w B= n w ( ∏ w -1 α ∉ R re + 𝒳 w -1 α )( ∏ w -1 α ∈ R re + 𝒳 w -1 α )B = n w ( ∏ w -1 α ∉ R re + 𝒳 w -1 α )B=( ∏ α∈R w 𝒳 α ) n w B = x β 1 c 1 … x β l c l n w B| c 1 ,…, c l ∈𝔽 .

Since R re + may be infinite there is a subtlety in the decomposition and ordering of the product of 𝒳 α in the second "equality" and it is necessary to proceed more carefully. Choose a reduced decomposition w= s i 1 … s i l and let β 1 ,…, β l be the ordering of R w from (4.6).

Step 1: Since R w ⊆ R re + there is an inclusion x β 1 c 1 … x β l c l n w B| c 1 ,…, c l ∈𝔽 ⊆BwB. To prove equality proceed by induction on l.

Base case: Suppose that w= s j . Let α∈ R re + and c,d ∈ℂ. If c=0 or α, α j is a prenilpotent pair then, by relation (3.2),

x α d x α j c n j -1 B= x α j c' n j -1 B, for some  c'∈ℂ.

If α i , α j is not a prenilpotent pair and c≠0 then α, α j is a prenilpotent pair and by (3.2), x α d x α j c n j -1 B= x α d x - α j c -1 n j -1 B= x - α j c -1 n j -1 B= x α j c n j -1 B.

Thus x α j c n j -1 B| c∈ℂ is B invariant and so B s j B= x α j c n j -1 B| c∈ℂ .

Induction step: If w= s i 1 … s i l is reduced and if l w s j >l w then, by induction, Bw s j B⊆BwB.B s j B= s β 1 c 1 … x β l c l x w α j n w n j -1 B| c 1,…, c l ,c∈ 𝔽 , so that Bw s j B= x β 1 c 1 … x β l+1 c l+1 n w s j B| c 1 ,…, c l+1 ∈ ℂ   with   β l+1 =w α j .

Step 2: Prove that BwB=BvB if and only if w=v bu induction on l w .

Base case: Suppose that l w =0. Then BwB=BvB implies that v ∈B so there is a representative n v of v such that n v ∈ B ∩ N. Then v R re + ⊆ R re + since n v 𝒳 α n v -1 = 𝒳 vα ∈ B for α∈ R re + . So l v =0. Thus, by (2.16), v=1.

Induction step: Assume BwB=BvB and s j is such that l w s j <l w . Since BvB.B s j B ⊆ BvB ∪ Bv s j B (see [St, Lemma 25], Bw s j B⊆BwB.B s j B=BvB.B s j B ⊆ BvB ∪ Bv s j B=BwB∪Bv s j B

Thus, by induction, w s j =w or w s j =v s j . Since w s j ≠w, it follows that w=v.

Step 3: Let us show that if x α i 1 c 1 n i 1 -1 … x α i l c l n i l -1 B= n i 1 x i 1 c 1 ' - c 1 n i 1 -1 … x i l c l ' n il -1 B is in B s i 2 … s i l B. If c 1 ' ≠ c 1 then n i 1 -1 x i 1 c 1 ' - c 1 n i 1 ∈B s i 1 B and the right hand side is contained in n i 1 -1 x i 1 c 1 ' - c 1 n i 1 B s i 1 … s i l B⊆B s i 1 B.B s i 2 … s i l B.

By Step 2 this is impossible and so c 1 ' = c 1 . Then by induction, c i ' = c ' for i=1,2, …,l .

Step 4: From the definition of R w it follows that if α,β∈R w and α+β∈ R re then α+β∈R w and if α,β ∈ R w then α,β form a prenilpotent pair. Thus by [St, Lemma 17], any total order on the set R w can be taken in the statement of the theorem.

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Suppose that λ ∈ 𝔥 * is dominant integral and M λ is an integrable highest weight representation of G generated by a highest weight vector v λ + . Then the set BwB v λ + contains the vector w v λ + and is contained in the sum ⊕ ν ≥ wλ M λ ν of the weight spaces with weights ≥wλ. This is another way to show that if w≠v then BvB≠BwB and accomplish Step 2 in the proof of the above theorem.

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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