LECTURE 5: CHARACTERIZATION THEOREM ANDEXAMPLESMIHAELA IFRIM AND BRANDON BARRETTE1. Strong Exchange PropertyWe now review the Strong Exchange Property Theorem from last lecture.Theorem 1.1. Let w = s1s2. . . skbe a reduced expression for w ∈ W with si∈ Sand let t ∈ T . Then `(tw) < `(w) implies that:(1.1) tw = s1· · · ˆsi· · · skfor some 1 ≤ i ≤ k.Corollary 1.2. Let w = s1· · · skbe a reduced word and let t ∈ T . Then the fol-lowing are equivalent:(1) `(tw ) < `(w);(2) tw = s1· · · ˆsi· · · skfor some i;(3) t = s1s2· · · si· · · s2s1.Proposition 1.3. Deletion PropertyLet w = s1· · · skbe such that `(w) < k.Then w = s1· · · ˆsi· · · ˆsj· · · skfor some 1 ≤ i < j ≤ k.Proof. We choose i maximal such that sisi+1· · · skis not reduced and therefore`(si· · · sk) < `(si+1· · · sk). By the Strong Exchange Property we obtain:si· · · sk= si+1· · · ˆsj· · · skfor some 1 < j ≤ k.Using the equality above we obtain:w = s1· · · sk= s1· · · ˆsi· · · ˆsj· · · skwhich ends our proof. Corollary 1.4. The following properties hold:(1) Any word w = s1· · · skcontains a reduced word as a subword by deleting aneven number of letters.(2) Suppose that s1· · · sk= s01· · · s0kand also suppose that both are reduced. Then⇒ {s1, . . . , sk} = {s01, . . . , s0k}.(3) S is a minimal generating set for W .Proof. (1) follows from Deletion Property.(2) Suppose ∃ sjwhich is not included in the set I := {s01, . . . , s0k}. Here we choosej minimal with the property just mentioned. By Corollary 1.2, if t = s1· · · sj· · · s1then there must exists an i such thats1· · · sj· · · s1= s01· · · s0i· · · s01Date: January 14, 2009.1for some i.Therefore sj= (sj−1· · · s1)(s01· · · s0i· · · s01)(s1· · · sj−1) - where all are letters in I.Using the Deletion Property we can find a reduced subword of the right-hand side,but this will give us:sj= s0a∈ Iwhich is a contradiction with the assumption that sjis not in I.(3) Follows from (2) since no element s ∈ S can be written as a product of otherelements in S. 2. Characterization of Coxeter groupsWe will assume that W is an arbitrary group. Let S ⊆ W be a generating setsuch that s2= e, ∀s ∈ S. Therefore the concept of length, `(w), where w ∈ Wstill makes sense and the concept of reduced expressions also still makes sense.In this new context, we say that the system (W, S) has the Exchange or Deletionprop e rty if the following hold:The Exchange PropertyLet w = s1· · · skbe reduced, and let s ∈ S. Then `(sw) < `(w) ⇒ sw =si· · · bsi· · · skfor some i, 1 ≤ i < j < k.The Del etion PropertyIf w = s1· · · sk, then `(w) < k ⇒ w = si· · · bsi· · · bsj· · · skfor 1 ≤ i < j < k.Theorem 2. 1. Characterization TheoremLet W be a group group and let S ⊆ W be a generating set with s2= e ∀s ∈ S.Then the following are equivalent:(1) (W, S) is a Coxeter system.(2) W satisfies the Exchange Property.(3) W satisfies the Deletion Property.Proof. The proof will be presented in the next lecture. Now let’s look at the following example:Example 2.2. Snis the well known group of permutations of [n]. Snis gener-ated by S = {s1, · · · , sn−1} where si= (i, i + 1). In one line notation, we havesi= [1, · · · , i − 1, i + 1, i, · · · , n]. Recall that s2i= e.2Fixing x ∈ Snwe recall that:Right action by si:Then xsiis obtained from x by interchanging the positions of x(i) and x(i + 1).For example we have [31524]s3= [31254].Left action by si:Then six is contained from x by interchanging the values i and i + 1.A numerical example is the following s3[31524] = [41523]and this shows that Sgenerates Sn.Definition 2.3. The inversion number of x ∈ Snis given by the followingexpression:inv(x) =| {(i, j) | i < j, x(i) > x(j)} | .Looking at the definition it is easy to see that the following lemma holds:Lemma 2.4. The following equality holds:inv(xsi) = inv(x) +(1 if x(i) < x(i + 1)−1 if x(i) > x(i + 1)The property that we will prove now shows a very useful relation between thelength of a word and the number of inversions of the word.Proposition 2.5. We have the following relation:`(x) = inv(x), ∀x ∈ Sn.Proof. We know that we have `(e) = inv(e). Then by the Lemma 2.4 we obtainthat inv(x) ≤ `(x).Claim. `(x) ≤ inv(x).Proof. (of the claim) Since inv(x) = 0 ⇒ x = e ⇒ `(e) = 0. Hence the claim istrue for inv(x) = 0. We proceed by induction on inv(x). Let x ∈ Snbe such thatinv(x) = k + 1. Then x 6= e ⇒ ∃s ∈ S such that inv(xs) = k. By the inductionhyphothesis `(xs) ≤ k ⇒ `(x) ≤ k + 1 = inv(x).This finishes the proof. We recall from our previous lectures that the descent set DR(x) = {s ∈ S |`(xs) < `(x)}.Proposition 2.6. For Snwe have DR(x) = {si∈ S | such that x(i) > x(i + 1)}.This implies that the definition of DR(x) that we wrote above is the same with thenotion we just stated in the statement of the proposition.Proof. By the Proposition 2.5 we have:DR(x) = {s ∈ S | inv(xs) < inv(x)} = {si∈ S | such that x(i) > x(i + 1)}. Proposition 2.7. Using the Characterization Theorem we can prove that (Sn, S)is a Coxeter system of type An−1.The proof will be given in the next