LECTURE 22: STANDARD K-TABLEAUX ANDGRASSMANNIAN AFFINE PERMUTATIONSWANG, QIANGThe main result of this note is the bijection between {standard k-tableaux} ↔{reduced words of Grassmannian affine permutations}.We will define these two sets and then describe the bijection.Remark 0.1. Throughout this lecture note, unless otherwise stated, n = k + 1.1. Standard k-tableauxDefinition 1.1 (Standard k-tableaux). Let γ be a (k + 1)-core, λ = b(γ). Letm = |λ|. A standard k-tableau of shape γ is a filling of γ with {1, . . . , m} suchthat the following two conditions are satisfied:(1) the filling is row strict and column strict;(2) repeated letters have the same (k + 1)-residue.Example 1.2. Let k = 2, γ = (4, 2, 2, 1, 1) is a 3-core, λ = b(γ) = (2, 1, 1, 1, 1)with m = |λ| = 6. Then the following tableau is a standard 2-tableau:1 2 3 53 54 656It is clear that each row/column is strictly increasing. The only letter ’1’ is at (1, 1)position which has 3-residue 0. The only letter ’2’ is at (1, 2) position which has3-residue 1. There are two letter ’3’s, at position (1, 3) and (2, 1), both of whichhave 3-residue 2. One letter ’4’ of residue 1. Two letter ’5’s of residue 0. Finally,two letter ’6’ of residue 2.Denote the set of all n-cores by Cn.2. Grassmannian affine permutations via an action offSnon CnIn this section we discuss a correspondence of Grassmannian affine permutationswith cores using an action offSnon Cn. Before we can do this, we need to say a bitmore about Cn.Recall that given k, each box (i, j) is assigned a k-residue given by j −i (mod k+1).Definition 2.1. Given a partition p, define Ak(p, `) to be the set of all boxes ofk-residue ` that are not already in p, but when added to p the result remains apartition. The elements in Ak(p, `) are called the addable corners of (k-)residue `.Similarly, define Rk(p, `) to be the set of all boxes of k-residue ` that are already inp and when removed from p the result remains a partition. The elements in Rk(p, `)are called the removable corners of (k-)residue `.1Theorem 2.2. For p ∈ Cn, either Ak(p, `) = ∅ or Rk(p, `) = ∅ for any `. Moreover,p ∪ Ak(p, `) ∈ Cnand p \ Rk(p, `) ∈ Cn.Proof. This can be checked explicitly. Definition 2.3 (Simple reflections offSnacting on Cn). Let si∈fSnbe a simplereflection, let γ ∈ Cn, then definesi. γ =γ ∪ Ak(γ, i) Ak(γ, i) 6= ∅γ \ Rk(γ, i) Rk(γ, i) 6= ∅γ otherwiseExample 2.4. Let k = 2 (thus n = 3), thens1.0 1 2 02 01 202=0 1 2 0 12 0 11 20 121s2.0 1 2 02 01 202=0 1 2 02 010s0.0 1 2 02 01 202=0 1 22 01 202Proposition 2.5. Definition 2.3 defines an action offSnon Cn.Proof. {si| i = 0, 1, . . . , n−1} acting on an n-core γ satisfy the braid relation, thatis(1) s2i· γ = γ for i = 0, 1, . . . , n − 1;(2) sisj· γ = sjsi· γ for |i − j| > 1;(3) sisi+1si· γ = si+1sisi+1· γ for i = 0, 1, . . . , n − 1, where addition on indicesis defined in Zn.All above can be easily verified by using the abacus representation of Cn. Theoriginal proof of above result is in the paper ”Ordering the affine symmetric group”by Lascoux (http://phalanstere.univ-mlv.fr/ al/pub engl.html). The action defined above is transitive but not simple, it is then natural toconsider the stabilizer of the ”special” element ∅ ∈ Cn. It is easily seen thatSTABfSn(∅) = Sn. Thus the map C :fSn/Sn→ Cninduced by above action isa bijection. Indeed,fSn/Snis the set of Grassmannian affine permutations.There is another point of view offSn/Sn. If we treat Snas a parabolic subgroupof (fSn, S = (s0, s1, . . . , sn−1), thenfSn/Snis in bijection tofSnS\s0, the minimalcoset representatives w.r.t Sn. In this setting,fSn/Snis equipped with the Bruhatorder inherited fromfSnS\s0.2On the Cnside, we can define the following covering relation: For p, q ∈ Cn,p  q if p = q ∪Ak(q, i) for some i, (or, equivalently, q = p \Rk(p, i)). This coveringrelation extends to a partial ordering on Cn.There is another partial ordering defined on Cn: For p, q ∈ Cn, p ⊃ q if p containsq as Young diagrams.It is clear that ⊃ is a stronger relation than , that is, p  q ⇒ p ⊃ q. In fact, ⊃is strictly stronger than  as demonstrated in the following example for n = 3: Letp = sh(0 1 2 02 0 11 2 00 12 010) and q = sh(0 1 2 02 01 202), then it is clear that p ⊃ q but p 6 q,since there can be no such a sequence (p = p0, p1, . . . , pk= q) that picovers pi+1under .The following two propositions state that, under C,  and ⊃ play exactly thesame role as the weak (left) B ruhat order >Land the Bruhat order > onfSn/Sn,respectively.Proposition 2.6. C is an isomorphism between (fSn/Sn, >L) and (Cn, ).Proof. First we note that by induction it suffices to show the correspondence be-tween the covering relations from the two sides.For the forward direction, let us assume the following inductive hypothesis: forv ∈fSn/Snand l(v) = l, if v covers w in left Bruhat order then C(v)  C(w).Now suppose u >Lv and u = siv, and let p = C(u) and q = C(v), we wantto conclude that A(q, i) 6= ∅. First we notice that it can not be the case thatA(q, i) = R(q, i) = ∅ since this would violate the bijection betweenfSn/Snand Cn.Thus, if A(q, i) = ∅ we will have R(q, i) 6= ∅.Then p = si. q = q \ R(q, i) ≺ q. Then by induction, we should have v  u, acontradiction.For the backward direction, if C(v)  C(w) then by definition C(v) = C(w) ∪A(C(w), i) for some i, and since v and w both are the minimal cose t representativeswe must have v = siw. Proposition 2.7. C is an isomorphism between (fSn/Sn, >) and (Cn, ⊃).Proof. Let us consider the following inductive hypothesis: for v, w ∈fSn/Snwherel(v) = l, v ≥ w if and only if p = C(v) ⊇ q = C(w).For the base case l = 0, above statement is clearly true.For l > 0, we know that C(v) 6= ∅, thus we can pick a corner of residue i forsome i. Then si. C(v)  C(v), and by Prop 2.6 we know siv <Lv, thus siv < v. Bylifting property we havev ≥ w ⇔ siv ≥ min(siw, w)Supp ose siw < w, then by induction, C(siv) ⊇ C(siw). So we just need to showthat A(siw, i) ⊆ C(v) (or equivalently, R(w, i) ⊆ C(v)). Pick b ∈ A(siw, i) andsuppose that b 6∈ C(siv), but then it must be the case that b ∈ A(siv, i) ⊂ C(v).Supp ose w < siw, then by induction C(siv) ⊇ C(w). Then we have C(v) ⊃C(siv) ⊇ C(w).3Conversely, if v 6≥ w then siv 6≥ siw and siv 6≥ w, …