Equivalence of reduced expressions

Symmetric Groups and Iwahori–Hecke Algebras

4.1 The symmetric groups

4.1.5 Equivalence of reduced expressions

We now state the so-calledexchange theorem.

Theorem 4.8.Let si1· · ·sir be a reduced expression for w ∈ Sn, where r = λ(w). If λ(wsj) < λ(w) for some j ∈ {1, . . . , n−1}, then there is k ∈ {1, . . . , r} such that ws_j =s_i1· · ·s!_ik· · ·s_ir. If λ(s_jw) < λ(w) for some j∈ {1, . . . , n−1}, then there isk∈ {1, . . . , r}such thats_jw=s_i1· · ·s!_ik· · ·s_ir. Proof. We saw in the proof of Lemma 4.7 (a) that if t1, . . . , tr are the trans- positions defined by (4.7), then I(w) = {t1, . . . , tr}. If λ(wsj)< λ(w), then s_j ∈ I(w) by Lemma 4.7 (c). Therefore, s_j = t_k for some k ∈ {1, . . . , r}. By (4.8),

wsj=wtk =si₁· · ·s!i_k· · ·si_r.

The second claim is deduced from the first one by replacingw withw⁻¹. Corollary 4.9.Let w∈Sn. If λ(ws_j)< λ(w) for some j ∈ {1, . . . , n−1}, then there is a reduced expression forwending withs_j. Ifλ(s_jw)< λ(w)for some j ∈ {1, . . . , n−1}, then there is a reduced expression for w beginning withsj.

This is a direct corollary of the previous theorem: ifλ(wsj)< λ(w), then wsj = si₁· · ·s!i_k· · ·si_r and w = si₁· · ·s!i_k· · ·si_rsj is a reduced expression for w, since its length is equal to r = λ(w). The second claim is proven similarly.

We conclude with a lemma needed in the proof of Lemma 4.18 below.

Lemma 4.10.If λ(siwsj) = λ(w) and λ(siw) = λ(wsj) for w ∈ Sn and somei, j∈ {1, . . . , n−1}, thensiw=wsj andsiwsj =w.

Proof. (a) Suppose first that λ(siw) = λ(wsj) > λ(siwsj) = λ(w). By Lemma 4.6,

I(s_iw) =w⁻¹I(s_i)wΔI(w) ={w⁻¹s_iw}ΔI(w).

Sinceλ(siwsj) < λ(siw) and λ(wsj)> λ(w), Lemma 4.7(c) implies that sj

belongs toI(siw), but not toI(w). Thereforesj=w⁻¹siw; hencesiw=wsj

andsiwsj=ws²_j =w.

(b) If λ(siw) = λ(wsj) < λ(siwsj) = λ(w), then we apply a similar argument, using thatI(w) =I(si(siw)) ={w⁻¹siw}ΔI(siw).

S1(i, j)S2∼S1(j, i)S2 (4.9) for allS1, S2∈Mn and alli, j∈ {1, . . . , n−1} such that|i−j| ≥2, and

S1(i, j, i)S2∼S1(j, i, j)S2 (4.10) for allS₁, S₂∈M_n and alli, j∈ {1, . . . , n−1} such that|i−j|= 1. Observe that equivalent sequences have the same length. The equivalence relation∼ has been devised so that

(i₁, . . . , i_k)∼(j₁, . . . , j_k)∈M_n =⇒ s_i1· · ·s_ik=s_j1· · ·s_jk∈Sn. Lemma 4.11.Ifsi₁· · ·si_k andsj₁· · ·sj_k are reduced expressions for the same permutationw∈Sn, then(i1, . . . , ik)∼(j1, . . . , jk)inMn.

This lemma shows that for any w ∈ Sn, we can pass from one reduced expression forwto any other reduced expression forwusing only the relations

sisj =sjsi if|i−j| ≥2, sisjsi=sjsisj if|i−j|= 1.

Proof. We prove the lemma by induction onk. Ifk= 0, thenw= 1 has only one reduced expression. Ifk= 1, thenw=s_i for someiandw has only one reduced expression.

Assume that k ≥2. From the equality si₁· · ·si_k = sj₁· · ·sj_k we deduce thatsi₂· · ·si_k =si₁sj₁· · ·sj_k. Sincesi₂· · ·si_k is reduced,

λ(si₁sj₁· · ·sj_k) =λ(si₂· · ·si_k) =k−1< k=λ(sj₁· · ·sj_k). Therefore, by Theorem 4.8, there is an integerpwith 1≤p≤ksuch that

si₂· · ·si_k=si₁sj₁· · ·sj_k=sj₁· · ·s!j_p· · ·sj_k. (4.11) Sincesi₂· · ·si_k andsj₁· · ·s!j_p· · ·sj_krepresent the same permutation and have the same lengthk−1 and sincesi₂· · ·si_kis reduced,sj₁· · ·s!j_p· · ·sj_k is also re- duced. By the induction hypothesis, (i2, . . . , ik)∼(j1, . . . ,jp, . . . , jk). Hence,

(i1, i2, . . . , ik)∼(i1, j1, . . . ,jp, . . . , jk). (4.12) The wordsj₁· · ·sj_p, being a part of the reduced word sj₁· · ·sj_k, is reduced.

The second equality in (4.11) implies that si1sj1· · ·sj_p−1 and sj1· · ·sjp are equal in Sn. Since these words have the same length pand one of them is reduced, so is the other one. Ifp < k, then we apply the induction hypothesis to these words, obtaining (i₁, j₁, . . . , j_p−1)∼(j₁, . . . , j_p). From this and (4.12), we obtain

(i₁, i₂, . . . , i_k)∼(i₁, j₁, . . . ,j_p, . . . , j_k)

= (i1, j1, . . . , jp−1)(jp+1, . . . , jk)

∼(j₁, . . . , j_p)(j_p+1, . . . , j_k)

= (j1, . . . , jp, jp+1, . . . , jk), which was to be proven.

Ifp=k, then (4.12) becomes (i1, i2, . . . , ik)∼(i1, j1, . . . , jk−1). This equi- valence implies that

s_i1s_j1· · ·s_jk−1 =s_i1s_i2· · ·s_ik=s_j1s_j2· · ·s_jk. Summarizing our argument, we see that to prove the implication

si₁· · ·si_k=sj₁· · ·sj_k =⇒ (i1, . . . , ik)∼(j1, . . . , jk), it is enough to prove the implication

si₁sj₁· · ·sj_k−1 =sj₁· · ·sj_k =⇒ (i1, j1, . . . , jk−1)∼(j1, . . . , jk). (4.13) We now start the argument all over again with the reduced expressions sj1· · ·sj_k = si1sj1· · ·sj_k−1. Proceeding as above, we show that in order to prove (4.13), it is enough to prove the implication

sj₁si₁sj₁· · ·sj_k−2=si₁sj₁sj₂· · ·sj_k−1

=⇒ (j₁, i₁, j₁, . . . , j_k−2)∼(i₁, j₁, j₂, . . . , j_k−1). (4.14) We first prove (4.14) when|i1−j1| ≥2. Thensi₁sj₁ =sj₁si₁ and

si₁sj₁sj₁· · ·sj_k−2 =sj₁si₁sj₁· · ·sj_k−2 =si₁sj₁sj₂· · ·sj_k−1. Multiplying on the left bysj₁si₁ inSn, we obtain

sj₁· · ·sj_k−2=sj₂· · ·sj_k−1.

Both sides are reduced expressions of lengthk−2. By the induction assump- tion, (j1, . . . , jk−2)∼(j2, . . . , jk−1). From this and (4.9), we obtain

(j1, i1, j1, . . . , jk−2) = (j1, i1)(j1, . . . , jk−2)

∼(i1, j1)(j2, . . . , jk−1)

= (i₁, j₁, j₂, . . . , j_k−1), which was to be proven.

If |i₁−j₁| = 1, then we proceed again as above and reduce the proof of (4.14) to showing that the equality

si₁sj₁si₁sj₁· · ·sj_k−3 =sj₁si₁sj₁sj₂· · ·sj_k−2

implies that (i1, j1, i1, j1, . . . , jk−3) ∼ (j1, i1, j1, j2, . . . , jk−2). This and the equalitysj₁si₁sj₁=si₁sj₁si₁ imply that

sj₁si₁sj₁sj₁· · ·sj_k−3 =si₁sj₁si₁sj₁· · ·sj_k−3 =sj₁si₁sj₁sj₂· · ·sj_k−2, which, after left multiplication bysj1si1sj1, gives

sj1· · ·sj_k−3=sj2· · ·sj_k−2.

Since both sides of this equality are reduced expressions of lengthk−3, we can apply the induction hypothesis and obtain (j1, . . . , jk−3)∼(j2, . . . , jk−2).

From this and (4.10),

(i1, j1, i1, j1, . . . , jk−3) = (i1, j1, i1)(j1, . . . , jk−3)

∼(j1, i1, j1)(j2, . . . , jk−2)

= (j₁, i₁, j₁, j₂, . . . , j_k−2),

which was to be proven.

The following theorem is useful for defining maps from the symmetric groups to monoids.

Theorem 4.12.For any monoidM and anyx1, . . . , xn−1∈M satisfying the relations

xixj=xjxi if |i−j| ≥2, x_ix_jx_i=x_jx_ix_j if |i−j|= 1, there is a set-theoretic mapρ:Sn→M defined by

ρ(w) =xi₁· · ·xi_k,

for anyw∈Sn and any reduced expression w=si1· · ·si_k. Proof. Define a monoid homomorphismρ:Mn→M by

ρ(i₁, . . . , i_k) =x_i1· · ·x_ik

for all (i1, . . . , ik)∈Mn. We claim thatρ(S) =ρ(S) for allS, S ∈Mn such that S ∼S. Indeed, by definition of the equivalence ∼, it suffices to prove the claim whenS =S1(i, j)S2 (resp.S=S1(i, j, i)S2) andS =S1(j, i)S2

(resp.S=S1(j, i, j)S2) forS1,S2∈Mn andi, j∈ {1, . . . , n−1}such that

|i−j| ≥2 (resp.|i−j|= 1). By the assumptions of the theorem, ρ(S1(i, j)S2) =ρ(S1)xixjρ(S2)

=ρ(S1)xjxiρ(S2)

=ρ(S1(j, i)S2) if|i−j| ≥2, and

ρ(S1(i, j, i)S2) =ρ(S1)xixjxiρ(S2)

=ρ(S1)xjxixjρ(S2)

=ρ(S1(j, i, j)S2) if|i−j|= 1.

To prove the theorem, we need only check that ρ is well defined, i.e., if si1· · ·si_k andsj1· · ·sj_k are reduced expressions forw∈Sn, then

xi₁· · ·xi_k =xj₁· · ·xj_k.

By Lemma 4.11, (i1, . . . , ik)∼(j1, . . . , jk) inMn. By the claim above, xi1· · ·xi_k =ρ(i1, . . . , ik) =ρ(j1, . . . , jk) =xj1· · ·xj_k. 4.1.6 The longest element ofSn

Letw0∈Sn be the permutationi→n+ 1−ifor alli∈ {1, . . . , n−1}: w0=

1 2 . . . n−1 n n n−1 . . . 2 1

. (4.15)

It is clear thatw0is the only permutationw∈Sn such thatw(i)> w(j) for alli,j∈ {1, . . . , n−1}withi < j. In other words,w=w0 if and only if the setI(w) consists of all transpositions. By Lemma 4.7 (a),λ(w0) =n(n−1)/2 andλ(w)< n(n−1)/2 for w=w0. Because of this, w0 is called thelongest element ofSn. We record two other properties ofw0 (the second one will be used in Section 6.5.2).

Lemma 4.13.If w∈Sn satisfiesλ(wsi)< λ(w) for all i ∈ {1, . . . , n−1}, thenw=w0.

Proof. By Lemma 4.7 (c), si ∈I(w) for alli. Thenw(i)> w(i+ 1) for alli.

The only permutation satisfying these inequalities isw0. Lemma 4.14.For any u, v∈Sn such thatuv=w0,

λ(u) +λ(v) =λ(w0). Proof. The lemma trivially holds for u=w₀ andv= 1.

We claim that for any u∈Sn,u=w₀, there is a sequences_i1, . . . , s_ir of simple transpositions such thatus_i1· · ·s_ir =w₀andλ(us_i1· · ·s_ir) =λ(u)+r.

Before we prove the claim, let us show that it implies the lemma foru and v=u⁻¹w₀=s_i1· · ·s_ir. Clearly,λ(v)≤r and

λ(w0) =λ(uv)≤λ(u) +λ(v)≤λ(u) +r=λ(usi₁· · ·si_r) =λ(w0). Therefore,λ(u) +λ(v) =λ(w0).

Let us now establish the claim. Since u = w0, by Lemma 4.13, there issi1 such thatλ(usi1)≥λ(u). By Lemma 4.5, we haveλ(usi1) =λ(u) + 1. If λ(us_i1) =λ(w₀), thenus_i1=w₀, sincew₀is the unique element ofSnof max- imal length, and we are done. Ifλ(us_i1)< λ(w₀), then again by Lemma 4.13, we can finds_i2 such thatλ(us_i1s_i2)≥λ(us_i1). Then

λ(usi₁si₂) =λ(usi₁) + 1 =λ(u) + 2.

Ifλ(usi1si2) =λ(w0), thenusi1si2=w0and we are done. If not, we continue as above until we find the required sequencesi1, . . . , sir.

Exercise 4.1.1.Using Theorem 4.8, prove that ifλ(si1· · ·sir)< r, then there arep, q∈ {1, . . . , r} such thatp < q and

s_i1· · ·s_ir =s_i1· · ·s!_ip· · ·s!_iq· · ·s_ir, wheres!i_p ands!i_q are removed on the right-hand side.

Exercise 4.1.2.Deduce Theorem 4.1 from Theorem 4.12, using the latter to construct a left inverseSn→Gn ofϕ:Gn→Sn.

Exercise 4.1.3.(a) Show thatwk,=sksk−1· · ·sis a reduced word for each pair (k, ) such that 1≤≤k≤n−1.

(b) Prove that the wordwk₁,₁wk₂,₂· · ·wk_r,_r obtained by concatenating words as in (a) is reduced fork1< k2<· · ·< kr.

Exercise 4.1.4.For any integerk≥1, set [k]q = 1 +q+· · ·+q^k−1 ∈Z[q].

Show that

w∈Sn

q^λ(w)= [1]q[2]q[3]q· · ·[n]q. (Hint: Use Exercise 4.1.3 and Corollary 4.4.)

Exercise 4.1.5.Prove that for any w ∈ Sn and any i = 1, . . . , n−1, the equalityλ(s_iw) =λ(w) + 1 holds if and only ifw⁻¹(i)< w⁻¹(i+ 1).