LECTURE 20: THE AFFINE SYMMETRIC GROUPBRANDON BARRETTE1. Recap from Last LectureRecall:˜Snis the affine symmetric group. Elements ω ∈˜Snare bijections from Zto itself satisfying:(1) ω(i + n) = ω (i) + n ∀i ∈ Z(2)Pni=1ω(i) =n+12Remark 1.1. For all ω ∈˜Snand i, j ∈ Z, ω(i) 6≡ ω(j) mod n ⇐⇒ i 6≡ j mod n.This will be useful later.2. Affine InversionDefinition 2.1. The affine inversion number of v ∈˜Snisfinv(v) = |{(i, j) ∈ [n] × Z | i < j, v(i) > v(j)}|.Proposition 2.2. `(v) =finv(v) ∀v ∈˜SnProof. Before we begin the proof of the proposition, we first give a claim:Claim:finv(v) ≤ `(v).Proof of claim.It can be checked directly from the definition that for 1 ≤ i ≤ n − 1(∗)finv(vsi) =(finv(v) + 1 if v(i) < v(i + 1)finv(v) − 1 if v(i) > v(i + 1).The same is also true for i = 0 however, it is not obvious.It is clear from the definition that (i, j) with 2 ≤ i ≤ n − 1 is an inversion of v ifand only if (i, s0(j)) is an inversion of vs0. Hence it remains to consider the casesi = 1 and i = n.Assume v(n) < v(n + 1). If j > n + 1 and (n, j) is an inversion of v, then(n, s0(j)) is an inversion of vs0. Also, (n, s0(j)) is an inversion of vs0and (n, j) isnot an inversion of v ⇐⇒ v(n + 1) ≥ v(j) ≥ v(n). (†)Similarly, if j > 1 and (1, j) is an inversion of vs0, then (1, s0(j)) is an inver-sion of v. Also, (1, s0(j)) is an inversion of vs0and (1, j) is not and inversion ofvs0⇐⇒ v(1) ≥ v(s0(j)) ≥ v(0). (‡)Since v(i + n) = v(i) + n, the cardinality of (†) and (‡) are equal.Date: February 23, 2009.1Note: (n, n + 1) is an inversion of vs0but not of v, which means thatfinv(vs0) =finv(v) + 1.Now, assume v(n) > v(n + 1).By similar arguments as above,finv(vs0) =finv(v) − 1. Therefore the recursionformula (*) for inversion has been proven.To finish the proof of the claim notice thatfinv(e) = 0 = `(e). By the recursionformula we just proved, we have thefinv(v) ≤ `(v), thus proving the claim.To prove the proposition, thatfinv(v) = `(v) we use induction onfinv.Iffinv(v) = 0, then we must havev(1) < v(2) < · · · < v(n) < v(n + 1) = v(1) + nwhich imples that v = e.Next, supposefinv(v) = t + 1 > 0 and assume by induction thatfinv(u) ≤ t =⇒finv(u) = `(u). Sincefinv(v) > 0, we have v 6= e =⇒ ∃s ∈ S such thatfinv(vs) = t.Then, by the induction hypothesis, we have thatfinv(vs) = `(vs) = t =⇒ `(v) ≤ t + 1 =⇒ `(v) ≤finv(v).Therefore we have shown thatfinv(v) = `(v). A consequence of the previous result is a simple description of the descent set ofaffine permutations.Proposition 2.3. If v ∈˜Snthen DR(v) = {si∈ S | v(i) > v(i + 1)}Proof. By the previous proposition, we have thatDR(v) = {si∈ S |finv(vsi) <finv(v)}.The rest follows from (∗), the recursion formula for inversion. Proposition 2.4. (˜Sn, S) with S = {s0, . . . , sn−1} is a Coxeter system.Proof. This is very similar to the case of the symmetric group. Proposition 2.5. For 0 ≤ i ≤ n − 1, let J = S\{si}. Then:(1) (˜Sn)J= Stab([i + 1, n + i])(2) (˜Sn)J= {v ∈˜Sn| v(1) < v(2) < · · · < v(i), v(i + 1) < · · · < v(n + 1)}Proof. (1) Obvious(2) Recall (˜Sn)J= {v ∈˜Sn| vs > v ∀s ∈ J} by definition. Then, by applyingthe recursion formula for inversion ( ∗), we have our result. 23. Minimal representatives uJin (˜Sn)J, J = S\siLet uJbe the minimal coset represe ntative of (˜Sn)Jfor J = S\si. By Proposi-tion 2.5, in complete notation uJis obtained from u by rearranging the entries{u(i + 1 + kn), . . . , u(i + n + kn)} in increasing order ∀k ∈ Z.Example 3.1. Let n = 5.For u = [−3, 6, 3, −5, 14], notice first that this satisfies the two c onditions foraffine permutations stated in the beginning of these notes.Define our set J = {s0, s1, s2, s4} (s3removed), then we can write u as:u = . . . | − 3 6 3 − 5 14 | 2 11 8 0 19 | . . .We choose −5 2 8 11 14 in increasing order. Then we can write:uJ= [3, 6, 9, −5, 2]Where we obtain the first 3 entries in window notation by subtracting −5 of thelast 3 elements in −5 2 8 11 14.Definition 3.2. The elements in (˜Sn)Jfor J = {s1, . . . , sn−1} are called the Grass-mannian elements.Remark 3.3. By a lemma we proved previously (in the section about parabolicsubgroups), u ∈˜Snis Grassmannian:⇐⇒ no reduced expression for u ends in letters in J⇐⇒ every reduced expression for u ends in s0⇐⇒ in window notation, [u(1), . . . , u(n)], all entries are increasing.4. Reflections for˜SnFor a, b ∈ Z, with a 6≡ b mod n, then defineta,b:=Yr∈Z(a + rn, b + rn)Note: si= ti,i+1for 0 ≤ i < n.Proposition 4.1. The set of reflection of˜Snis:{ti,j+kn| 1 ≤ i < j ≤ n, k ∈ Z}.Proof. Let u ∈˜Sn, 0 ≤ i < n, then we have that:usiu−1=Yr∈Z(u(i) + rn, u(i + 1) + rn)Since u is any element in˜Sn, u(i) and u(i + 1) can be any two elements of Z notcongruent mod n.