Unipotent Hecke algebras: the structure, representation theory, and combinatorics

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 26 March 2015

This is an excerpt of the PhD thesis Unipotent Hecke algebras: the structure, representation theory, and combinatorics by F. Nathaniel Edgar Thiem.

Unipotent Hecke algebras

The algebras

Important subgroups of a Chevalley group

Let G=GV be a Chevalley group defined with a 𝔤-module V as in Section 2.2.2. Recall that R+ is the set of positive roots according to some fixed set of simple roots {α1,α2,,α} of 𝔤 (Section 2.2.1). The group G contains a subgroup U given by U=xα(t)|αR+,t𝔽q, which decomposes as U=αR+Uα, whereUα= xα(t)|t𝔽q, with uniqueness of expression for any fixed ordering of the positive roots [Ste1967, Lemma 18]. For each αR+, the map Uα 𝔽q+ xα(t) t is a group homomorphism.

For αR, define the maps sα: 𝔥* 𝔥* γ γ-γ(Hα)α, (3.1) sα: 𝔥 = Z(𝔤)𝔥s 𝔥 H+Hβ H+Hβ-β(Hα)Hα, forβR. (3.2)

The Weyl group of G is W=sα|αR and has a presentation given by W= s1,s2,,s |si2=1, (sisj)mij =1,1ij ,mij>0, si=sαi. If w=si1si2sir with r minimal, then the length of w is (w)=r.

Let 𝔥 be as in (2.3). If q>3, then the subgroup T= hH(t)| H𝔥,t𝔽q* has its normalizer in G given by N= wα(t),h| αR,hT,t𝔽q* ,wherewα(t)= xα(t)x-α (-t-1)xα(t). If αR, then hHα(t)=wα(t)wα(1)-1. Write hα(t)=hHα(t) and hi(t)=hαi(t).

There is a natural surjection from N onto the Weyl group W with kernel T given by π: N W wα(t) sα, forαR,t𝔽q*, h 1, forhT. (3.3) Suppose vN. Then for each minimal expression π(v)= si1si2 sir, with(π(v)) =r, there is a unique decomposition of v as v=v1v2 vrvT, wherevk= wik(1) andvTT. (3.4) Write ξi=wi(1). (3.5)

Unipotent Hecke algebras

Let G be a finite Chevalley group. Fix a nontrivial homomorphism ψ:𝔽q+*. If μ: R+ 𝔽q α μα satisfiesμα=0 for allαnot simple, (3.6) then the map ψμ: U * xα(t) ψ(μαt) (3.7) is a linear character of U. In fact, with the exception of a few degenerate special cases of G (which can be avoided if q>3), all linear characters of U are of this form [Yok1969, Theorem 1].

The unipotent Hecke algebra (G,U,ψμ) is μ=EndG (IndUG(ψμ)). (3.8) Using the anti-isomorphism (2.2), view μ as the subset of G μ=eμG eμ,where eμ=1U uUψμ (u-1)u. (3.9)

The choices μ and ψ

This section assumes all linear characters of U are of the form ψμ for some nontrivial homomorphismψ:𝔽q+* and μ as in (3.6).

The group T normalizes Uα, since hxα(t)h-1 =xα(α(h)t), forαR+,t 𝔽q,hT,α(h) 𝔽q*. Thus, if χ:U* is a linear character, then hχ: U * xα(t) χ(hxα(t)h-1) is a linear character for every hT.

Suppose χ:U* is a linear character. For every hT, IndUG(χ)= IndUG(hχ).

Proof.

Recall that IndUG(hψμ) (g)=1U xGxgx-1U hψμ(xgx-1)= 1U xGxgx-1U ψμ(hxgx-1h-1). Since T normalizes U, the sum over all xG such that xgx-1U is the same as the sum over hxG such that hxgx-1h-1U. Thus, IndUG(hψμ) (g)=1U xGxgx-1U ψμ(xgx-1) =IndUG(ψμ)(g).

The type of a linear character χ:U* is the set J{α1,α2,,α} such that χ(xαi(t)) 1for allt𝔽qif and only if αiJ. A unipotent Hecke algebra (G,U,χ) has type J if χ has type J.

(a) Fix a nontrivial homomorphism ψ:𝔽q+*. If ψ:𝔽q+* is a homomorphism, then there exists k𝔽q such that ψ(t)= ψ(kt), for allt𝔽q+. (b) The linear characters χ:U* and hχ:U* have the same type for any hT. (c) Let J{α1,α2,,α} and 𝒮J={typeJlinear characters ofU}. Then the number of T-orbits of 𝒮J is ZJ(q-1)JT, whereZJ= { hT|α(h) =1,forαJ } . (d) The number of distinct (nonisomorphic) type J unipotent Hecke algebras is at most ZJ(q-1)JT.

Proof.

(a) The map 𝔽q+ Hom(𝔽q+,*) k ψk: 𝔽q+ * t ψ(kt) is a group homomorphism, and the kernel is trivial because ψ is nontrivial. Since Hom(𝔽q+,)=q, the map is also surjective.

(b) Since Uα𝔽q+, the restriction of χ to Uα gives a linear character of 𝔽q+. Part (a) implies that for every αR+ there exists kα𝔽q such that χ(xα(t))= ψ(kαt) Furthermore, if χ has type J, then kα0 if and only if αJ. Write χ=ψ(k1,,k) whereki=kαi. Since hxα(t)h-1=xα(α(h)t) for hT, hχ= ψ(α1(h)k1,,α(h)k), and α(h)kα0 if and only if kα0 if and only if αJ. Thus, the action of T preserves the type of χ.

(c) Suppose a group K acts on a set 𝒮. Then by [Isa1994, Theorem 4.18] the number of orbits of the K action is 1K gKCard {s𝒮|g(s)=s}. In this case, K=T acts on the set 𝒮=𝒮J= { ψ(k1,k2,,k) |ki=0, unlessαiJ } . Furthermore, hψ(k1,k2,,k) =ψ(k1,k2,,k) if and only ifαi(h)=1 for allαiJ. Finally, 𝒮J=(q-1)J.

Examples. 1. If J=, then J=0 and ZJ=T, so there is a unique unipotent Hecke algebra of type J; in this case, ψμ is trivial and μ is the Yokonuma algebra. 2. A character is in general position if it has type J={α1,α2,,α}. In this case, ZJ is the center of G, and J=, so the number of type J unipotent Hecke algebras is at most Z(G)(q-1)T .

A natural basis

The group G has a double-coset decomposition G=vNUvU, [Ste1967, Theorem 4 andB=UT] (3.10) and if Nμ = {vN|eμveμ0} = { vV|u,vuv-1 Uimpliesψμ(u) =ψμ(vuv-1) } (3.11) then the set {eμveμ|vNμ} is a basis for μ [CRe1981, Prop. 11.30].

Theorem 3.7 gives a set of relations similar to those of the Yokonuma algebra (Example 1, below) for evaluating the product (eμueμ) (eμveμ), withu,vNμ in any unipotent Hecke algebra μ.

Examples.

1. The Yokonuma Hecke algebra. If μα=0 for all positive roots α, then ψμ=1 is the trivial character and N1=N. Let Tv=e1ve1 for vN, with Ti=Tξi (ξi as in (3.5)) and TH(t)=ThH(t). If vN has a decomposition v=v1v2vrVT (as in (3.4)), then Tv=Ti1Ti2 TirTvT(See Chapter 6). Thus, the Yokonuma algebra 1 has generators Ti, Th for 1i, hT (see [Yok1969-2]) with relations, Ti2 = q-1THαi (-1)+q-1 t𝔽q* THαi (t-1)Ti, 1i, TiTjTi mijterms = TjTiTj mijterms , (sish)mij=1, TiTh = TsijTi, hT, ThTk = Thk, h,kT. These relations give an “efficient” way to compute arbitrary products (e1ue1)(e1ve1) in 1. There is a surjective map from the Yokonuma algebra onto the Iwahori-Hecke algebra that sends Th1 for all hT. “Setting Th=1 in the Yokonuma algebra relations recovers relations for the Iwahori-Hecke algebra, Ti2=q-1+ q-1(q-1) Ti, TiTj mijterms = TjTi mijterms . Furthermore, there is a surjective map from the Iwahori Hecke algebra onto the Weyl group given by mapping Tisi and q1. Thus, by “setting Ti=si and q=1 we retrieve the Coxeter relations of W, si2=1, sisjsj mijterms = sjsisj mijterms . For a more detailed discussion of the Yokonuma algebra, see Chapter 6.

2. The Gelfand-Graev Hecke algebra. By definition, if μα0 for all simple roots α, then ψμ is in general position. In this case, the Gelfand-Graev character IndUG(ψμ) is multiplicity free as a G-module ([Yok1968],[Ste1967, Theorem 49]). The corresponding Hecke algebra μ is therefore commutative. However, decomposing the product (eμueμ)(eμveμ) into basis elements is more challenging than in the Yokonuma case [Cha1976, Cur1988, Rai2002].

Parabolic subalgebras of μ

Let ψμ:UG be as in (3.7). Fix a subset J{α1,α2,,α} such that ifμαi0, then αiJ. For example, if ψμ is in general position, then J={α1,α2,,α}, but if ψμ is trivial, then J could be any subset.

Let WJ=siW|αiJ, PJ=U,T,WJ andRJ=-span {αiJ}R. Then PJ has subgroups LJ=T,WJ,Uα|αRJ andUJ=Uα|αR+-RJ (3.12) (a Levi subgroup and the unipotent radical of PJ, respectively). Note that UJLJ=PJ, UJLJ=1,and, in fact, PJ=UJLJ.

Define the idempotents of U, eμJ=1LJU uLJU ψμ(u-1)u andeJ= 1UJ uUJu, (3.13) so that eμ=eμJeJ.

The group homomorphisms PJ LJ lu l and PJ G lu lu forlLJ,uUJ, induce maps InfLJPJ: {LJ-modules} {PJ-modules} M eJM, IndPJG: {PJ-modules} {G-modules} M GPJM whose composition is the map IndfLJG. Note that in the special case when LJe is an irreducible LJ-module with corresponding idempotent e, then IndfLJG: {LJ-modules} {G-modules} LJe GeeJ.

The map ψμ:U* restricts to a linear character ResULJU(ψμ):LJU*. To make the notation less heavy-handed, write ψμ:LJU*, for ResULJU(ψμ).

Let ψμ be as in (3.7). Then IndUG(ψμ) IndfLJG (IndULJLJ(ψμ)).

Proof.

Recall IndUG(ψμ)Geμ. On the other hand, IndULJLJ (ψμ) LJeμJimplies IndfLJG (IndULJLJ(ψμ)) GeμJeJ, where eμJ is as in (3.13). But eμJeJ=eμ, so IndUG(ψμ) Geμ eμJeJ IndfLJG(IndULJLJ(ψμ)).

The map θ: EndLJ(IndULJLJ(ψμ)) μ eμJveμJ eμveμ, forvLJNμ, is an injective algebra homomorphism.

Proof.

Let vNLJ. Since eμJveμJLJ, eJUJ and LJUJ=1, eμJveμJ=0 if and only ifeJ eμJveμJ=0. Because LJ normalizes UJ, both eμJ and v commute with eJ. Therefore, eμJveμJ=0 if and only ifeJeμJ veJeμJ= eμveμ=0, and {eμJveμJ0} if and only ifvNμLJ. Consequently, θ is both well-defined and injective.

Consider θ(eμJueμJ) θ(eμJveμJ)= eμueμ eμveμ= eμJeJu eμJeJv eJeμJ. Since u commutes with eJ, θ(eμJueμJ) θ(eμJveμJ) = eμJeJu eμJveJ eμJ = eμueμJv eμ = θ(eμJueμJveμJ), and so θ is an algebra homomorphism.

Write J=θ (EndLJ(IndULJLJ(ψμ))) μ (3.14) The J are “parabolic” subalgebras of μ in that they have a similar relationship to the representation theory of μ as parabolic subgroups PJ have with the representation theory of G.

Weight space decompositions for μ-modules

An important special case of Theorem 3.4 is when J=Jμ= {αi{α1,α2,,α}|μαi0}, so that ψμ has type Jμ. Write Lμ=LJμ, Wμ=WJμ, etc.

The algebra μ is a nonzero commutative subalgebra of μ.

Proof.

As a character of ULμ, ψμ is in general position, so IndLμULμ(ψμ) is a Gelfand-Graev module and μ is a Gelfand-Graev Hecke algebra (and therefore commutative).

Since μ is commutative, all the irreducible modules are one-dimensional; let ˆμ be an indexing set for the irreducible modules of μ. Suppose V is an μ-module. As an μ-module, Vγˆμ VγwhereVγ= {vV|xv=γ(x)v,xμ}. If γˆμ, then Vγ is the γ-weight space of V, and V has a weight γ if Vγ0. See Chapter 6 for an application of this decomposition.

Examples 1. In the Yokonuma algebra ψμ=1, J1= and 1=e1Te1. 2. In the Gelfand-Graev Hecke algebra case, Jμ={α1,α2,,α} and μ=μ (confirming, but not proving, that μ is commutative).

Multiplication of basis elements

This section examines the decomposition of products in terms of the natural basis (eμueμ) (eμveμ)= vNμ cuvv (eμveμ). In particular, Theorem 3.7, below, gives a set of braid-like relations (similar to those of the Yokonuma algebra) for manipulating the products, and Corollary 3.9 gives a recursive formula for computing these products.

Chevalley group relations

The relations governing the interaction between the subgroups N, U, and T will be critical in describing the Hecke algebra multiplication in the following section. They can all be found in [Ste1967].

The subgroup U=xα(t)|αR+,t𝔽q has generators {xα(t)|αR+,t𝔽q}, with relations xα(a) xβ(b) xα(a)-1 xβ(b)-1 = γ=iα+jβR+i,j>0 xγ(zγaibj), (U1) xα(a) xα(b) = xα(a+b), (U2) where zγ depends on i,j,α,β, but not on a,b𝔽q [Ste1967]. The zγ have been explicitly computed for various types in [Dem1965, Ste1967] (See also Appendix A).

The subgroup N has generators {ξi,hH(t)|i=1,2,,,H𝔥,t𝔽q*}, with relations ξi2 = hi(-1), (N1) ξiξjξiξj mijterms = ξjξiξjξi mijterms , where(sisj)mij =1inW, (N2) ξihH(t) = hsi(H) (t)ξi, (N3) hH(a) hH(b) = hH(ab), (N4) hH(a) hH(b) = hH(b) hH(a), forH,H𝔥, (N5) hH(a) hH(a) = hH+H(a), forH,H𝔥, (N6) hH1(t1) hH2(t2) hHk(tk)=1, ift1λj(H1) tkλj(Hk)=1 for all1jn, (N7) where λj:𝔥 depends on V as in (2.6).

The double-coset decomposition of G (3.10) implies G=U,N. Thus, G is generated by {xα(a),ξi,hH(b)|αR+,a𝔽q,i=1,2,,,H𝔥,b𝔽q*} with relations (U1)-(N7) and ξixα(t) xi-1 = xsi(α) (ciαt), whereciα= ±1(ciαi =-1) (UN1) hxα(b)h-1 = xα(α(h)b), forhT, (UN2) ξixi(t) ξi = xi(t-1) hi(t-1) ξixi(t-1), wherexi(t)=xαi (t)andt0, (UN3) where for αR and hH(t)T, α(hH(t))= tα(H). (3.15)

Fix a ψμ:U* as in (3.7). For k𝔽q, let eα(k)=1q t𝔽qψ (-μαkt)xα (t)with the convention eα=eα(1). (3.16) Note that for any given ordering of the positive roots, the decomposition U=αR+Uα implieseμ=αR+ eα. (3.17) In particular, given any αR+, we may choose the ordering of the positive roots to have eα appear either first or last. Therefore, since eα is an idempotent, eμeα=eμ= eαeμ. (3.18)

If w=si1si2sirW with r minimal, then let Rw = {αR+|w(α)R-} = { αir, sir (αir-1), ,sirsir-1 si2(αi1) } , (3.19) where the second equality is in [Bou2002, VI.1, Corollary 2 of Proposition 17].

Let vN, and let w=π(v) (with π:NW as in (3.3)). ξieα (k)ξi-1 = esiα (ciαk), forαR+, 1in-1, (E1) veαv-1 = ewα, forαRw, vNμ, (E2) heα(k) h-1 = eα(kα(h)-1), forhT, (E3) eμxα(t) = ψ(μαt)eμ= xα(t)eμ. (E4)

Proof.

(E1) Using relation (UN1), we have ξieα(k) ξi-1 = 1qt𝔽q ψ(-μαkt) ξixα(t) ξi-1 = 1qt𝔽q ψ(-μαkt) xsiα (ciαt) = 1qt𝔽q ψ(-μαciαkt) xsiα(t) = esiα(ciαk). (E2) Suppose αRw. Since vNμ, ψ(μαt) = ψμ(xα(t)) = ψμ(vxα(t)v-1) = ψμ(xwα(kt)) (by (UN1)) = ψ(μwαkt), for somek0. In particular, μα=kμwα and we may conclude veαv-1= 1qt𝔽q ψ(-μαt) xwα(kt)= 1qt𝔽q ψ(-μαk-1t) xwα(t)= ewα. (E3) Since hxα(t)h-1=xα(α(h)t), we have heα(k)h-1= 1qt𝔽qψ (-kt)xα(α(h)t) =t𝔽qψ (-ktα(h)-1) xα(t)=eα (kα(h)-1). (E4) Note that eαxα(t) = 1qa𝔽q ψ(-μαa) xα(a)xα(t) = 1qa𝔽q ψ(-μαa)xα (a+t) = 1qa𝔽q ψ(-μα(a-t)) xα(a) = 1qa𝔽q ψ(-μαa)ψ (μαt)xα(a) = ψ(μαt)eα. Therefore, by (3.18), eμxα(t)=eμeαxα(t)=ψ(μαt)eμ.

Local Hecke algebra relations

Let u=u1u2uruTN decompose according to si1si2sirW. For 1kr define constants ck=±1 and roots βkR+ by the equation xβk(ckt)= (uk+1ur)-1 xαik(t) (uk+1ur). (3.20) Note that Rπ(u)={β1,β2,,βr} (see (3.19)). Define fu𝔽q[y1,y2,,yr] by fu=- μβ1c1β1(uT)y1- μβ2c2β2(uT)y2-- μβrcrβr(uT)yr (3.21) and for k=1,2,,r, let uk(t)=ξik (t),where ξi(t)=ξi xi(t). (3.22)

Let u=u1u2uruT, v=v1v2vsvTNμ decompose according to si1si2sirW and sj1sj2sjsW, respectively, as in (3.4). Then (a) (eμueμ) (eμveμ)= 1qrt𝔽qr (ψfμ)(t) eμ(u1(t1)u2(t2)ur(tr)) (v1v2vs) heμ, where h=vTv-1uTvT. (b) The following local relations suffice to compute the product (eμueμ)(eμveμ). t𝔽q (ψf)(t) ξi(t)ξi = (ψf)(0) ξi2+ t𝔽q* (ψf)(t) xi(t-1) ξihi(t) xi(t-1), (1) ξixα(t) = xsi(α) (ciαt)ξi, (2) xα(t)h = hxα(α(h)-1t), (3) eμxα(t) = ψ(μαt)eμ= xα(t)eμ, (4) (ψf)(t) (ψg)(t) = (ψ(f+g))(t), forf,g𝔽q [y1±1,,yr±1] ,t𝔽qr, (5) hα(t)ξi = ξihsi(α) (t), (6) ξi(a)xα(b) = γ=mαi+nαR+m,n0 xsiγ (cisi(γ)zγambn) ξi(a),where ααi, (7) a𝔽qΦ (a)ξi(a) xi(b) = a𝔽qΦ (a-b)ξi(a), for some mapΦ: 𝔽qG, (8) hα(a) hα(b) = hα(ab), (9) ξiξjξiξj mijterms = ξjξiξjξi mijterms , wheremij is the order ofsisj inW. (10)

Proof.

(a) Order the positive roots so that by (3.17) eμueμveμ = eμu (αRweα) eβ1eβ2 eβrveμ (definitionβk) = eμ(αRwewα) ueβ1eβ2 eβrveμ (Lemma 3.6, E2) = eμueβ1 eβ2eβr veμ(Lemma 3.6, E4) = eμu1u2 uruTeβ1 eβ2eβr veμ = eμu1u2ur eβ1(μβ1β1(uT)) eβ2(μβ2β2(uT)) eβr(μβrβr(uT)) uTveμ(Lemma 3.6, E3) = eμ u1eαi1(uβ1c1β1(uT)) u2eαi2(uβ2c2β2(uT)) ureαir(uβrcrβr(uT)) uTveμ(Lemma 3.6, E1) = eμ u1eαi1(uβ1c1β1(uT)) u2eαi2(uβ2c2β2(uT)) ureαir(uβrcrβr(uT)) v1vsvT v-1uTveμ = eμqr t1,,tr𝔽q ψ(-μβ1c1t1β1(uT)) u1(t1) ψ(-μβrcrtrβr(uT)) ur(tr) v1vsheμ (definitioneα) = 1qrt𝔽qr (ψfu)(t) eμu1(t1) ur(tr)v1 vsheμ,(by (5)) where h=vTv-1uTvT, as desired.

(b) First, note that these relations are in fact correct (though not necessarily sufficient): (1) comes from (UN3); (2) comes from (UN1); (3) comes from (UN2); (4) comes from (E4); (5) comes from the multiplicativity of ψ; (6) comes from (N3); (7) comes from (U1) and (UN1); (8) comes from (U2); (9) is (N4); and (10) is (N2). It therefore remains to show sufficiency.

By (a) we may write (eμueμ) (eμveμ) =1qr t𝔽qr (ψf)(t) eμu1(t1) ur(tr) v1vsheμ for some f𝔽q[y1,,yr] and hT. Say tk is resolved if the only part of the sum depending on tk is (ψf). The product is reduced when all the tk are resolved. I will show how to resolve tr and the result will follow by induction.

Use relation (2) to define the constant d and the root γR by (v1v2vs)-1 xαir(t) (v1v2vs)= xγ(dt) (where(π(v)) =s). (3.23) There are two possible situations:

Case 1. γR+,

Case 2. γR-.

In Case 1, (eμueμ) (eμveμ) = 1qrt𝔽qr (ψf)(t) eμu1(t1) urxir(tr) v1vsheμ (by (a)) = 1qrt𝔽qr (ψf)(t) eμu1(t1) ur-1(tr-1) urv1vs xγ(dtr) heμ(by(2)) = 1qrt𝔽qr (ψf)(t) eμu1(t1) ur-1(tr-1) urv1vsh xγ(dγ(h)-1tr)eμ_ (by(3)) = 1qrt𝔽qr (ψf)(t) eμu1(t1) ur-1(tr-1) urv1vsh ψ(μγdγ(h)-1tr)_ eμ (by(4)) = 1qrt𝔽qr (ψf)(t) eμu1(t1) ur-1(tr-1) urv1vsh (ψμγdγ(h)-1yr)(tr) eμ = 1qrt𝔽qr (ψg)(t) eμu1(t1) ur-1(tr-1) urv1vsheμ, (by(5)) where g=f+μγdγ(h)-1yr. We have resolved tr in Case 1.

In Case 2, γR-, so we can no longer move xir(tr) past the vj. Instead, (eμueμ) (eμveμ) = eμqr t𝔽qr (ψf)(t)_ u1(t1) ur-1(tr-1) urxir(tr) urur-1v1 vsheμ = eμqr t𝔽qr-1 u1(t1) ur-1(tr-1) tr𝔽q (ψf)(t,tr) urxir(tr)ur_ ur-1v1 vsheμ = eμqr t𝔽qr-1 u1(t1) ur-1(tr-1) (ψf)(t,0) ur2ur-1_ v1vsheμ (by(1)) + eμqr t𝔽qr-1 u1(t1) ur-1(tr-1) tr𝔽q* (ψf)(t,tr) xir(tr-1) hir(tr-1) urxir(tr-1) ur-1 v1vsheμ = eμqr t𝔽qr-1 u1(t1) ur-1(tr-1) (ψf)(t,0) ur v1vsheμ + eμqr t𝔽qr-1 u1(t1) ur-1(tr-1) tr𝔽q* (ψg)(t,tr) xir(tr-1) hir(-tr) v1vsheμ, (by(2,3,4)) where g=f-μγdγ(h)-1yr-1 (same as in the analogous steps in Case 1). = eμqr t𝔽qr-1 (ψf)(t,0) u1(t1) ur-1(tr-1) ur v1vsheμ + eμqr t𝔽qr-1 u1(t1) ur-1(tr-1) tr𝔽q* (ψg)(t,tr)xir(tr-1) v1vshγ (-tr)heμ, (by(6)) = eμqr t𝔽qr-1 (ψf)(t,0) u1(t1) ur-1(tr-1) ur v1vsheμ + eμqr t𝔽qr-1tr𝔽q* (ψφ(g)) (t,tr) (βR+xβ(aβ(t,tr))) u1(t1) ur-1(tr-1) v1vsheμ (by(7,8)) where φ:𝔽q[y1,,yr]𝔽q[y1,,yr] catalogues the substitutions to g due to (8), the aβ(y1,y2,,yr-1)𝔽q[y1,,yr-1,yr] are determined by repeated applications of (7) and (8), and h=hγ(-tr)hT. = 1qr t𝔽qr-1 (ψf) (t,0) eμ u1(t1) ur-1(tr-1) urv1vsheμ +1qrt𝔽qr-1tr𝔽q* (ψφ(g)) (t,tr)eμ (βR+ψ(μβaβ(t,tr))) u1(t1) ur-1(tr-1) v1vsheμ (by(4)) = 1qr t𝔽qr-1 (ψf) (t,0) eμ u1(t1) ur-1(tr-1) urv1vsheμ +1qrt𝔽qr-1tr𝔽q* (ψg2) (t,tr)eμ u1(t1) ur-1(tr-1) v1vsheμ (by(5)) where g2=φ(g)+βR+μβaβ(y1,,yr-1,yr-1). Now tr is resolved for Case 2, as desired.

(Resolving tk). Let u=u1u2ukN decompose according to si1si2sik (with uT=1). Suppose vN and f𝔽q[y1,y2,,yk]. Define γR and d by the equation v-1xik(t)v=xγ(dt). Then Case 1 If (π(ukv))>(π(v)), then t𝔽qk (ψf)(t) eμu1(t1) uk(tk)veμ= t𝔽qk (ψ(f+μγdcγyk_)) (t)eμu1(t1) uk-1(tk-1) ukv_eμ. Case 2 If (π(ukv))<(π(v)), then t𝔽qk (ψf)(t) eμu1(t1) uk(tk)veμ = t𝔽qk,tk=0_ (ψf)(t)eμ u1(t1) uk-1(tk-1) ukv_eμ + t𝔽qk,tk𝔽q*_ (ψ(φk(f)-μγdyk-1_)) (t)eμ u1(t1) uk-1(tk-1) hik(tk-1)v_eμ, where φk:𝔽q[yq±1,,yk±1]𝔽q[y1±1,,yk±1] is given by t𝔽qktk𝔽q* (ψf)(t)eμ u1(t1) uk-1(tk-1) xik(tk-1)= t𝔽qktk𝔽q* (ψφk(f)_) (t)eμu1(t1) uk-1(tk-1).

Proof.

This Lemma puts v in the place of uTv in the proof of Theorem 3.7, (b), and summarizes the steps taken in Case 1 and Case 2.

Global Hecke algebra relations

Fix u=u1u2uruTNμ, decomposed according si1si2sirW (see (3.4)). Suppose vNμ and let v=uTv.

For 0kr, let τ=(τ1,τ2,,τr-k) be such that τi{+0,-0,1}, where +0, -0, and 1 are symbols. If τ has r-k elements, then the colength of τ is (τ)=k. For example, if r=10 and τ=(-0,1,+0,+0,1,1), then (τ)=4. For i{+0,-0,1}, let (i,τ)= (i,τ1,τ2,,τr-k). By convention, if (τ)=r, then τ=.

Suppose (τ)=k. Define Ξτ(u,v)= 1qrt𝔽qτ (ψfτ)(t) eμu1(t1) uk(tk)vτ (t)eμ, (3.24) where 𝔽qτ = { t𝔽qr| fork<ir, ifτi-k=+0, thenti𝔽q, ifτi-k=-0, thenti=0, ifτi-k=1, thenti𝔽q*. } ; (3.25) vτ(t) = hik+1 (tk+1)τ1 uk+11-τ1 hir (tr)τr-k ur1-τr-kv, (3.26) with +0=-0=0, 1=1 in (3.26); and fτ is defined recursively by f=fu = -μβ1c1β1(uT)y1 -μβ2c2β2(uT)y2 - -μβrcrβr(uT)yr ,(as in (3.21)), (3.27) f(i,τ) = { fτ+μγτ dτyk, ifi=±0, φk(fτ)- μγτdτ yk-1, ifi=1, (3.28) where (vτ)-1xαik(t)vτ=xγτ(dτt) and the map φk is as in Lemma 3.8, Case 2.

Remarks. 1. By (3.24) and Theorem 3.7 (a), Ξ(u,v)=(eμueμ)(eμveμ) (recall, v=uTv). 2. If (τ)=0 so that τ is a string of length r, then (a) Ξτ(u,v)=1qrt𝔽qτ(ψfτ)(t)eμvτeμ has no remaining factors of the form uk(tk), (b) Ξτ(u,v)=0 unless vτ(t)Nμ for some t𝔽q.

The following corollary gives relations for expanding Ξτ(u,v) (beginning with Ξ(u,v)) as a sum of terms of the form Ξτ with (τ)=(τ)-1. When each term has colength 0 (or length r), then we will have decomposed the product (eμueμ)(eμveμ) in terms of the basis elements of μ.

In summary, while we compute fτ recursively by removing elements from τ, we compute the product (eμueμ)(eμveμ) by progressively adding elements to τ.

(The Global Alternative). Let u,vNμ such that u=u1u2uruT decomposes according to a minimal expression in W. Let v=uTv. Then (a) (eμueμ)(eμveμ)=Ξ(u,v). (b) If (τ)=k, then Ξτ(u,v)= { Ξ(+0,τ) (u,v), if(π(ukvτ)) >(π(vτ)), Ξ(-0,τ) (u,v)+ Ξ(1,τ) (u,v), if(π(ukvτ)) <(π(vτ)).

Proof.

(a) follows from Remark 1.

(b) Suppose (τ)=k. Note that Ξτ(u,v) = 1qrt𝔽qτ (ψfτ) (t)eμu1(t1) uk(tk)vτeμ = 1qrt(𝔽qr-k)τ t𝔽qk (ψfτ) (t,t)eμu1 (t1)uk (tk)vτeμ where (𝔽qr-k)τ={(tk+1,,tr)𝔽qr-k|restrictions according toτ} (as in (3.25)). Apply Lemma 3.8 to the inside sum with ffτ, vvτ. Note that the Lemma relations imply {t𝔽qk} becomes { {t𝔽qk}, if in Case 1, {t𝔽qk|tk=0}, if in Case 2, first sum, {t𝔽qk|tk𝔽q*}, if in Case 2, second sum, fτ becomes { f(+0,τ), if in Case 1, f(-0,τ), if in Case 2, first sum, f(1,τ), if in Case 2, second sum, vτ becomes { v(+0,τ), if in Case 1, v(-0,τ), if in Case 2, first sum, v(1,τ), if in Case 2, second sum. Thus Ξτ(u,v)= { Ξ(+0,τ) (u,v), if Case 1, Ξ(-0,τ) (u,v)+ Ξ(1,τ) (u,v), if Case 2, as desired.

Notes and References

This is an excerpt of the PhD thesis Unipotent Hecke algebras: the structure, representation theory, and combinatorics by F. Nathaniel Edgar Thiem.

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