LECTURE 10: WEAK BRUHAT ORDERSTEVEN PON1. LatticesExample 1.1. Bruhat order in B2(Coxeter diagram:•a•b)S = {a, b}, T = {a, b, bab, aba}.We can draw a graph showing the covering relations of Bruhat order on B2:abab = babaabannnnnbabPPPPPabggggggggggggggbaWWWWWWWWWWWWWWagggggggggggggggbWWWWWWWWWWWWWWWeQQQQQQQmmmmmmmDefinition 1.2. An element z in a poset is the meet (or greatest lower bound) ofa subset A if(1) z ≤ y, ∀y ∈ A(2) u ≤ y ∀y ∈ A ⇒ u ≤ zWe denote the meet of A by ∧A. If A = {x, y} we denote the meet by x ∧ y.Note: I f the meet exists, then it is unique.Definition 1.3. A poset P for which every ∅ 6= A ⊆ P has a meet is called ameet-semilattice.Definition 1.4. Similarly, we can define the join, or least uppper bound, of asubset of a poset, and a join-semilattice. A lattice is a poset which is both a meet-semilattice and a join-semilattice.Note that the Bruhat graph in Example 1.1 above is not a lattice. However,when we can obtain a lattice if instead of Bruhat order we use weak Bruhat order.2. Weak Bruhat OrderWeak Bruhat order is especially useful in studying the combinatorics of reducedwords; for example, enumerating the number of reduced words of a given Coxetergroup element. Intuitively, two elements are comparable in Bruhat order if one isa subword of the other. In weak Bruhat order, two words are comparable if oneword is a prefix (or suffix) of the other. There are two weak orders, left and rightweak Bruhat order, corresponding to if we are considering prefixes or suffixes.Definition 2.1. Let (W, S) be a Coxeter system, and let u, w ∈ W . Let ≤Rand≤Ldenote right and left (weak) Bruhat order, respectively. Then:Date: January 28, 2009.1(1) u ≤Rw if w = us1· · · sk, where si∈ S, s.t. `(us1· · · si) = `(u) + i, for1 ≤ i ≤ k.(2) u ≤Lw if w = s1· · · sku, where si∈ S, s.t. `(s1· · · siu) = `(u) + i, for1 ≤ i ≤ k.Remark 2.2. Note that left and right weak orders are distinct, but they are iso-morphic by the map w → w−1.Weak Bruhat order called “weak” because u ≤Rw ⇒ u ≤ w, and u ≤Lw ⇒u ≤ w.Example 2.3. We can draw the covering relations for weak Bruhat order:(1) Let W = S3.s1s2s1= s2s1s2s1s2kkkkkks2s1TTTTTTs1s2eTTTTTTTTTjjjjjjjjj(2) Let W = B2.abab = babaabannnnnbabPPPPPab baabeQQQQQQQmmmmmmmNote that in the examples above, we do get lattices.In the case of Sn, there is a simple test: for x, y ∈ Sn, x ≤Ry ⇔ y can beobtained from x by a sequence of adjacent transpositions that increase the inversionnumber at each step.Example 2.4. Let 263154 ∈ S6be given in 1-line notation. We can multiply bys4on the right (acting on positions) to get 263514, which increases the inversionnumber. We could then multiply by s1, then s5, then s2to get the sequence:x = 263154 →s4263514 →s1623514 →s5623541 →s2632541 = yTherefore, x ≤Ry.Proposition 2.5. Properties of Weak Order(1) There is a 1-1 correspondence between reduced words for w ∈ W and max-imal chains in [e, w]R.(2) u ≤Rw ⇔ `(u) + `(u−1w) = `(w).(3) If W is finite, then w ≤ w0for all w ∈ W .(4) Prefix property: u ≤Rw ⇔ there exist reduced expressions u = s1· · · skandw = s1· · · sksk+1· · · s0k.(5) Chain property: u <Rw ⇒ there is a chain u = u0<Ru1<R· · · <Ruk=w such that `(ui) = `(u)+ i for 0 ≤ i ≤ k.(6) Let s ∈ DL(u) ∩ DL(w). Then u ≤Rw ⇔ su ≤Rsw.2Proposition 2.6. Let u, w ∈ W . Then u ≤Rw ⇔ TL(u) ⊆ TL(w), where TL(u) ={t ∈ T | `(tu) ≤ `(u)}.Proof. (⇒) Let u = s1· · · sk, w = s1· · · sk· · · sqbe reduced words. ThenTL(u) = {s1s2· · · si· · · s2s1| 1 ≤ i ≤ k} ⊆ {s1s2· · · si· · · s2s1| 1 ≤ i ≤q} = TL(w).(⇐) Suppose u = s1· · · skis reduced. Let ti= s1s2· · · si· · · s2s1for 1 ≤ i ≤k. Assume TL(u) = {t1, · · · , tk} ⊆ TL(w). We claim there is a reducedexpression w = s1· · · sis01· · · s0q−i, for 0 ≤ i ≤ k. For i = 0, this is triviallytrue since this just means there exists a reduced word for w. Now supposethe claim is true for some i, 0 ≤ i < k. By assumption, ti+1∈ TL(w). Weknow that tj6= ti+1for j ≤ i by a lemma from a previous lecture (usingthat s1· · · si+1is reduced). Then since we can write w = s1· · · sis01· · · s0q−i,we can write ti+1= s1· · · sis01· · · s0m· · · s01si· · · s1for some 1 ≤ m ≤ q − i.Thenw = t2i+1w = (s1· · · si+1· · · s1)(s1· · · sis01· · ·ˆs0m· · · s0q−i)= s1· · · si+1s01· · ·ˆs0m· · · s0q−i.Then u ≤Rw is equivalent to the claim for i = k by the Prefix Property. Corollary 2.7. w → TL(w) provides an order and rank-preserving embeddingW ,→lattice of finite subsets of T .Proposition 2.8. If W is finite,(1) w → w0w and w → ww0are anti-automorphisms of weak order and(2) w → w0ww0is an automorphism of weak order.Proof. We will prove (2), as (1) is similar.For all s ∈ S, sw0= w0s0for some s0∈ S, since w0Sw0= S. Suppose w ≤Rws.Then w0wsw0= w0ww0s0≤Rw0ww0since `(w0wsw0) = `(ws) = `(w) + 1 =`(w0ww0) + 1 > `(w0ww0).