LECTURE 6: PROOF OF CHARACTERIZATION THEOREMBRANDON BARRETTE AND MIHAELA IFRIM1. Characterization TheoremTheorem 1.1. (Characterization Theorem) Let W be a group and S ⊂ W a gen-erating set such that s2= e ∀s ∈ S. Then the following are equivalent:(1) (W, S) is a Coxeter system(2) W has the Exchange Property(3) W has the Deletion PropertyProof. Proof will follow after some propositions and corollaries. Proposition 1.2. (Sn, S) is a Coxeter system of type An−1.Proof. By the Characterization Theorem, it suffices to show that (Sn, S) satis-fies the Exchange property. Notice the following properties show that Snis oftype An−1:sisj= sjsiif |i − j| > 1sisi+1si= si+1sisi+1s2i= eNext, suppose w = si1. . . sikis a reduced word such that`(si1. . . siksi) < `(si1. . . sik) (∗)Then, if we can prove this claim, we will have proven that (Sn, S) satisfies theExchange property.Claim. si1. . . siksi= si1. . . ˆsij. . . sikfor some 1 ≤ j ≤ k.Proof. Let a = w(i+1) and b = w(i). We proved last lecture that `(y) = inv(y) ∀y ∈Sn. Then (∗) implies that b < a and a is to the left of b in e in one line notationand a is to the right of b in w in one line notation. This implies that ∃j such that ais to the left of b in si1. . . sij−1, and a is to the right of b in si1. . . sij. This implies,in one line notation, that si1. . . sikis the same as si1. . . ˆsij. . . sikexcept that a andb are interchanged. This completes the proof of the claim. Therefore, since the claim is proven, (Sn, S) satisfies the Exchange property. Proof. (of the Characterization Theorem)(1) =⇒ (2) This is a special case of the strong exchange property.(2) =⇒ (3) This was already proved last lecture.(3) =⇒ (2) Suppose that `(ss1. . . sk) ≤ `(s1. . . sk) = k. This means thatw = s1. . . skis reduced. By the Deletion property, two letters can be deletedfrom ss1· · · skto obtain an expression for sw. We have two cases:Date: January 16, 2009.1Case 1:Suppose s is not deleted, then ss1. . . sk= ss1. . . ˆsi. . . ˆsj. . . sk. But then s1. . . sk=w = s1. . . ˆsi. . . ˆsj. . . sk. This implies `(w) < k, which is a contradiction sincew = s1. . . skis already reduced.Case 2:Suppose s is deleted, therefore sw = s1. . . ˆsi. . . skfor some 1 ≤ i ≤ kTherefore the Exchange property is satisfied.(2) =⇒ (1) Suppose (W, S) has the Exchange property. Let s1. . . sr= e be arelation in W , we need to show that this follows from a Coxeter relation, and thus(W, S) will be Coxeter system.By the Deletion property (since (2) ⇔ (3)) r = 2k which implies the relationis equivalent to s1. . . sk= s01. . . s0k.Claim. Any relations1. . . sk= s01. . . s0k(∗∗)is a consequence of pairwise relations (ss0)m(s,s0)= e.Before we begin the proof of the claim, it is important to understand someterminology. We say a relation is fine if this claim holds.Proof. (of claim) Perform induction on k.Show true for k = 1Here s = s0therefore s2= e, therefore true by assumptions.Assume true for k − 1That is, s1. . . sk−1= s01. . . s0k−1Prove true for kThere are two cases to consider here:Case 1: s1. . . skis not reduced.Then, this implies that ∃ 1 ≤ i ≤ k such that si+1. . . skis reduced, but si. . . skisnot reduced, thus by the Exchange property, we have:si+1. . . sk= sisi+1. . . ˆsj. . . skfor some i < j ≤ k. Since the length < k, thisexpression is fine. Therefore:s1. . . sk= s01. . . s0kbecomes s1. . . sisisi+1. . . ˆsj. . . sk= s01. . . s0kis also fine since length < k.Case 2: s1. . . skis reduced.Then, WLOG, we can assume that s16= s01since otherwise (∗∗) reduces to a rela-tion of length < k.By the Exchange property:s1. . . si= s01s1. . . si−1for some 1 ≤ i ≤ k (†)Which impliess1. . . ˆsi. . . sk= s02. . . s0kfor some 1 ≤ i ≤ k2Which is fine by induction.If i < k, then † is also fine since s01. . . s0k= s01s1. . . ˆsi. . . sk, which is fine, andimplies s1. . . sk= s01. . . s0kby †.If i = k, then s1. . . sk−1= s02. . . s0k, which is fine since the length < k. This alsoimplies s01s1. . . sk−1= s01. . . s0kwhich is fine by multiplying by s01.Now, we need to show s01s1. . . sk−1= s1. . . skis fine.To do this, repeat the argument with this new relation. One of two things willhappen. Either, the fineness will be settles by Case 1, or we will end up in Case 2and the relation will reduce to:s1s01s1. . . sk−2= s01s1. . . sk−1Is this fine? To answer the question, we repeat the argument yet again. If we fallin Case 1, we are done, otherwise we fall in Case 2 and again reduce the relation.Repeating this we will get the following relation:s1s01s1s01. . . = s01s1s01s1. . .which is a Coxeter relation.Therefore by mathematical induction, our claim is proven. Since our claim holds, we have that (W, S) is a Coxeter system. This completesthe proof of the Characterization Theorem.