On the graded algebras associated with Hecke symmetries

TL;DR

This paper investigates the algebraic properties like Koszulness and Gorensteinness of quantum symmetric and intertwining algebras linked to Hecke symmetries, extending previous results to cases where q is a root of unity.

Contribution

It extends the analysis of graded algebra properties associated with Hecke symmetries without restricting the parameter q, especially when q is a root of unity.

Findings

Established conditions for Koszulness and Gorensteinness of these algebras.

Identified restrictions on modules for Hecke algebras at roots of unity.

Extended previous results to more general q values.

Abstract

We consider quantum symmetric algebras, FRT bialgebras and, more generally, intertwining algebras for pairs of Hecke symmetries which represent quantum hom-spaces. The paper makes an attempt to investigate Koszulness and Gorensteinness of those graded algebras without a restriction on the parameter q of the Hecke relation used earlier. When q is a root of 1, positive results require a restriction on the indecomposable modules for the Hecke algebras of type A that can occur as direct summands of representations in the tensor powers of the base space.

Equations

$$[n{]}_q=1+q+\dots +q^{n-1}\in k$$

$$A(R^{\prime },R)\text{mathchar}9219{}_{A(R)}?\mathrm{and}?\text{mathchar}9219{}_{A(R)}A(R,R^{\prime })$$

$$B_{\lambda }=\{{\tau }_j\in B_n\mid j\ne {\lambda }_1+\dots +{\lambda }_{i}for\; eachi\le k\}$$

$$\eqalign{{\cal D}({{S}}_{\nu}/{{S}}_{\lambda})&{}=\{\pi\in{{S}}_{\nu}\mid\pi\tau_{i}>\pi\hbox{ for all $\tau_{i}\in{{B}}_{\lambda}$}\}\cr&{}=\{\pi\in{{S}}_{\nu}\mid\ell(\pi\sigma)=\ell(\pi)+\ell(\sigma)\hbox{ for all $\sigma\in{{S}}_{\lambda}$}\}\cr}$$

$$\mathcal{D}(S_{\mu }\backslash S_{\nu }/S_{\lambda })=\mathcal{D}(S_{\nu }/S_{\lambda })\cap \mathcal{D}(S_{\mu }\backslash S_{\nu }).$$

$$\openup 3.0pt\halign{\hbox to\displaywidth{$\hfil\displaystyle#\hfil$}\cr 0.0pt{\hfil$\displaystyle T_{i}T_{j}T_{i}=T_{j}T_{i}T_{j}\hbox{\ \ when $|i-j|=1$},\qquad T_{i}T_{j}=T_{j}T_{i}\hbox{\ \ when $|i-j|>1$},\cr 0.0pt{\hfil$\displaystyle(T_{i}-q)(T_{i}+1)=0\quad\hbox{for $i=1,\ldots,n-1$}.\cr}}}$$

$$M={\mathcal{H}}_n{\otimes }_{{\mathcal{H}}_{\lambda }}k(\chi )$$

$$\eqalign{{\cal D}_{\lambda}&{}=A_{i}(\lambda)\cup\tau_{i}A_{i}(\lambda)\cup B_{i}(\lambda),\quad\hbox{a disjoint union, where}\cr\vskip 3.0pt plus 1.0pt minus 1.0pt\cr A_{i}(\lambda)&{}=\{\sigma\in{\cal D}_{\lambda}\mid\tau_{i}\sigma\in{\cal D}_{\lambda}{\rm\ and\ }\tau_{i}\sigma>\sigma\}\,,\cr\tau_{i}A_{i}(\lambda)&{}=\{\sigma\in{\cal D}_{\lambda}\mid\tau_{i}\sigma\in{\cal D}_{\lambda}{\rm\ and\ }\tau_{i}\sigma<\sigma\}\,,\cr B_{i}(\lambda)&{}=\{\sigma\in{\cal D}_{\lambda}\mid\tau_{i}\sigma\notin{\cal D}_{\lambda}\}=\{\sigma\in{\cal D}_{\lambda}\mid\sigma^{-1}\tau_{i}\mskip 1.0mu\sigma\in{{B}}_{\lambda}\}.}$$

$$T_{i}T_{\sigma}c=\cases{T_{\tau_{i}\sigma}c&if $\sigma\in A_{i}(\lambda)$,\cr\vskip 3.0pt plus 1.0pt minus 1.0pt\cr(q-1)\,T_{\sigma}c+q\,T_{\tau_{i}\sigma}c&if $\sigma\in\tau_{i}A_{i}(\lambda)$,\cr\vskip 3.0pt plus 1.0pt minus 1.0pt\cr\chi_{\sigma}(T_{i})\,T_{\sigma}c\quad\hbox{with $\chi_{\sigma}(T_{i})=\chi(T_{\sigma^{-1}\tau_{i}\sigma})=\chi(T_{\sigma^{-1}(i)})$}&if $\sigma\in B_{i}(\lambda)$\cr}$$

$$M=\underset{\pi \in {}_{\mu }{\mathcal{D}}_{\lambda }}{⨁}M(\pi ),M(\pi )\cong {\mathcal{H}}_{\mu }{\otimes }_{{\mathcal{H}}_{\nu (\pi )}}k({\chi }_{\pi }),$$

$${\mathrm{Hom}}_{{\mathcal{H}}_n}(N,{\mathcal{H}}_n{\otimes }_{{\mathcal{H}}_{\lambda }}U)\cong {\mathrm{Hom}}_{{\mathcal{H}}_{\lambda }}(N,U)$$

$$\dim {\mathrm{Hom}}_{{\mathcal{H}}_n}(N,M)=f(\zeta ,\chi )$$

$$\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(N,M)\cong\mathop{\rm Hom}\nolimits_{{\cal H}_{\mu}}\bigl{(}{k}(\zeta),M\bigr{)}\cong\bigoplus_{\pi\in\,{}_{\mu}\mskip-1.0mu{\cal D}_{\mskip-1.0mu\lambda}}\mathop{\rm Hom}\nolimits_{{\cal H}_{\mu}}\bigl{(}{k}(\zeta),M(\pi)\bigr{)}.$$

$$\mathop{\rm Hom}\nolimits_{{\cal H}_{\mu}}\bigl{(}{k}(\zeta),M(\pi)\bigr{)}\cong\mathop{\rm Hom}\nolimits_{{\cal H}_{\nu(\pi)}}\bigl{(}{k}(\zeta),{k}(\chi_{\pi})\bigr{)},$$

$$\dim {\mathrm{Hom}}_{{\mathcal{H}}_n}(N,M)=\dim {\mathrm{Hom}}_{{\mathcal{H}}_n}(M,N).$$

$$\dim {\mathrm{Hom}}_{{\mathcal{H}}_n}(N^{\prime },M^{\prime })=\dim {\mathrm{Hom}}_{{\mathcal{H}}_n}(M^{\prime },N^{\prime })$$

$${\mathcal{H}}_n(t{)}_K={\mathcal{H}}_n(t){\otimes }_{O}K$$

$$\chi^{t}(T_{i})=\cases{t&if $\chi(T_{i})=q$,\cr\vskip 3.0pt plus 1.0pt minus 1.0pt\cr-1&otherwise,}\qquad\quad\zeta^{t}(T_{i})=\cases{t&if $\zeta(T_{i})=q$,\cr\vskip 3.0pt plus 1.0pt minus 1.0pt\cr-1&otherwise.}$$

$$M(t)={\mathcal{H}}_n(t){\otimes }_{{\mathcal{H}}_{\lambda }(t)}O({\chi }^t),N(t)={\mathcal{H}}_n(t){\otimes }_{{\mathcal{H}}_{\mu }(t)}O({\zeta }^t).$$

$$\dim_{K}\mathop{\rm Hom}\nolimits_{{\cal H}_{n}(t)_{K}}\bigl{(}N(t)\otimes_{O}K,\,M(t)\otimes_{O}K\bigr{)}=f(\zeta^{t},\chi^{t}).$$

$$\mathop{\rm Hom}\nolimits_{{\cal H}_{n}(t)_{K}}\bigl{(}N(t)\otimes_{O}K,\,M(t)\otimes_{O}K\bigr{)}\cong\mathop{\rm Hom}\nolimits_{{\cal H}_{n}(t)}\bigl{(}N(t),\,M(t)\bigr{)}\otimes_{O}K.$$

$$\mathop{\rm Hom}\nolimits_{{\cal H}_{n}(t)}\bigl{(}N(t),M(t)\bigr{)}\otimes_{O}{k}\cong\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(N,M)$$

$$\mathop{\rm Hom}\nolimits_{{\cal H}_{n}(t)}\bigl{(}N^{\prime}(t),M^{\prime}(t)\bigr{)}\otimes_{O}{k}\cong\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(N^{\prime},M^{\prime}).$$

$$\eqalign{\dim_{\mskip 1.0mu{k}}\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(N^{\prime},M^{\prime})&{}=\dim_{K}\mathop{\rm Hom}\nolimits_{{\cal H}_{n}(t)}\bigl{(}N^{\prime}(t),M^{\prime}(t)\bigr{)}\otimes_{O}K\cr&{}=\dim_{K}\mathop{\rm Hom}\nolimits_{{\cal H}_{n}(t)_{K}}\bigl{(}N^{\prime}(t)\otimes_{O}K,\,M^{\prime}(t)\otimes_{O}K\bigr{)}.\cr}$$

$$\dim_{\mskip 1.0mu{k}}\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(M^{\prime},N^{\prime})=\dim_{K}\mathop{\rm Hom}\nolimits_{{\cal H}_{n}(t)_{K}}\bigl{(}M^{\prime}(t)\otimes_{O}K,\,N^{\prime}(t)\otimes_{O}K\bigr{)},$$

$$\dim\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}({k}_{\rm alt},M)=\cases{1&\rm if $\chi$ is the alternating representation of ${\cal H}_{\lambda}$,\cr\vskip 3.0pt plus 1.0pt minus 1.0pt\cr 0&\rm otherwise.}$$

$$\{{\tau }_i\mid d_{\lambda }\in B_i(\lambda )\}=d_{\lambda }{B}_{\lambda }{d}_{\lambda }^{-1}.$$

$$\text{mathaccent}869\chi (T_i)=\chi (\text{mathaccent}869T_i)=q-1-\chi (T_i)for\; eachi\mathrm{with}{\tau }_i\in B_{\lambda }.$$

$$\varphi T_{i}-\mathaccent 869{T}_{i}\varphi=\psi_{i}T^{2}_{i}-(T_{i}+\mathaccent 869{T}_{i})\psi_{i}T_{i}+T_{i}\mathaccent 869{T}_{i}\psi_{i}=\psi_{i}\bigl{(}T^{2}_{i}-(q-1)T_{i}-q\bigr{)}=0.$$

$${\psi }_i{T}_i-T_i{\psi }_i=(q+1{)}^{-2}(\text{mathaccent}869T_i-T_i{)}^2\phi =\phi$$

Full text

On the graded algebras associated with Hecke symmetries

Serge Skryabin

Institute of Mathematics and Mechanics, Kazan Federal University,Kremlevskaya St. 18, 420008 Kazan, Russia E-mail: Serge.Skryabin@kpfu.ru

Introduction

Let $V$ be a finite dimensional vector space over a field ${k}$. A Hecke symmetry with parameter $0\neq q\in{k}$ is any linear operator $R:V^{\otimes 2}\to V^{\otimes 2}$ which satisfies the Hecke relation $(R+{\rm id})(R-q\cdot{\rm id})=0$ and the braid relation $R_{1}R_{2}R_{1}=R_{2}R_{1}R_{2}$ where $R_{1}=R\otimes{\rm id}_{V}$ and $R_{2}={\rm id}_{V}\otimes R$ are linear operators on $V^{\otimes 3}$; we will be saying that $R$ is a Hecke symmetry on $V$. The symmetries with parameter $q=1$ were considered by Lyubashenko [15]. Many notions and results originated in his work were later generalized to the case $q\neq 1$.

The $R$-symmetric algebra ${{S}}(V,R)$ and the $R$-skewsymmetric algebra $\Lambda(V,R)$ are two factor algebras of the tensor algebra ${{T}}(V)$. They are regarded as analogs of the symmetric and the exterior algebras of $V$. Since the braid equation is just a slightly different form of the quantum Yang-Baxter equation, there is also a bialgebra $A(R)$ given by the Faddeev-Reshetikhin-Takhtajan construction [20]. This bialgebra coacts on $V$ universally with respect to the property that the induced coaction on $V^{\otimes 2}$ commutes with $R$ [14]. In particular, ${{S}}(V,R)$ and $\Lambda(V,R)$ are $A(R)$-comodule algebras. The graded algebras ${{S}}(V,R)$, $\Lambda(V,R)$, $A(R)$ are quadratic in the sense that they are generated by homogeneous elements of degree 1 and their defining relations are of degree 2.

Gurevich’s work on Hecke symmetries [7] was motivated by the construction of quantum groups not necessarily arising as deformations of the classical objects. Even disregarding the Hopf algebraic aspect, Hecke symmetries provide a large class of graded algebras with predictable properties meaningful from the viewpoint of noncommutative algebraic geometry. However, general results on these algebras have been known under the assumption that the $q$-integers

[TABLE]

are nonzero for all integers $n>0$ (this means that $q$ is not a root of 1 with the exception that $q=1$ is allowed when $\mathop{\rm char}\nolimits{k}=0$). The assumption $\mathop{\rm char}\nolimits{k}=0$ was also used, but it is less relevant. The present paper makes an attempt to investigate several questions without the aforementioned restriction on $q$. Particularly, we are interested in Koszulness and Gorensteinness of those graded algebras.

A Hecke symmetry $R$ on $V$ gives rise, for each $n>0$, to a representation of the Hecke algebra ${\cal H}_{n}$ of type $A_{n-1}$ in the vector space ${{T}}_{n}(V)=V^{\otimes n}$. If $[n]_{q}\neq 0$ for all $n$, then these Hecke algebras are semisimple, and we will refer to this case as the semisimple case. Semisimplicity was the main driving force in the earlier results on the graded algebras associated with Hecke symmetries.

For $q$ a root of 1 we cannot be too optimistic, as an example at the end of section 3 shows. This example suggests that the properties of the graded algebras depend on the kind of the Hecke algebra representations in the tensor powers of $V$. We will say that an indecomposable ${\cal H}_{n}$-module has a $1$-dimensional (respectively, trivial) source if it is a direct summand of an ${\cal H}_{n}$-module induced from a $1$-dimensional (respectively, the trivial $1$-dimensional) representation of a parabolic subalgebra. This terminology is explained by the Hecke algebra version of the Green correspondence in the modular representation theory of finite groups (see Du [5]). The following two conditions on a Hecke symmetry $R$ will emerge in the statements:

The 1-dimensional source condition. For each $n>0$ all indecomposable direct summands of $\,V^{\otimes n}$ regarded as an ${\cal H}_{n}$-module with respect to the representation arising from $R$ have $1$-dimensional sources.

The trivial source condition. For each $n>0$ all indecomposable direct summands of $\,V^{\otimes n}$ regarded as an ${\cal H}_{n}$-module with respect to the representation arising from $R$ have trivial sources.

In the semisimple case the trivial source condition is obviously satisfied since indecomposable modules are simple, and each simple ${\cal H}_{n}$-module is a direct summand of the cyclic free module. We have to consider the weaker 1-dimensional source condition in order to include the supersymmetry on a ${{Z}}/2{{Z}}$-graded vector space (in that example $q=1$, so that ${\cal H}_{n}$ is just the group algebra of the symmetric group ${{S}}_{n}$, and $V^{\otimes n}$ is a direct sum of ${\cal H}_{n}$-modules induced from not necessarily trivial $1$-dimensional representations of parabolic subalgebras, but ${\cal H}_{n}$ is not semisimple when $n\geq\mathop{\rm char}\nolimits{k}>0$). This condition is also satisfied for the Hecke symmetries related to the standard quantum supergroups. There is one Hecke symmetry on a 2-dimensional vector space for which $q$ is a 4th root of 1 and the 1-dimensional source condition fails to hold (see section 3). However, this Hecke symmetry is not closed. This raises the following

Question. Does the $1$-dimensional source condition hold for every closed Hecke symmetry?**

A closed symmetry extends to a braiding on a monoidal subcategory of the category of finite dimensional vector spaces containing $V$ and its dual objects (see [7] for the precise definition). The results we are going to present do not depend on the closedness of $R$.

Theorem 3.1. Suppose that $R$ satisfies the $1$-dimensional source condition. Then the $R$-symmetric algebra ${{S}}(V,R)$ and the $R$-skewsymmetric algebra $\Lambda(V,R)$ are Koszul. Their Hilbert series satisfy the relation  $h_{{{S}}(V,R)}(t)\,h_{\Lambda(V,R)}(-t)=1$.

In the semisimple case exactness of certain complexes proved by Gurevich amounts to the conclusion of the theorem stated above, although Koszulness of graded algebras was not mentioned in [7] explicitly. By a systematic use of various projectors all considerations in [7] were done in terms of subspaces rather than factor spaces of the tensor powers of the initial vector space. The realization of Koszul complexes based on projectors is not appropriate for arbitrary $q$, however. Koszulness in the semisimple case was also considered by Phùng Hô Hai [8]. We will discuss more general results in a moment.

Theorem 4.5. Suppose that $R$ satisfies the trivial source condition. Suppose also that $\dim\Lambda_{n}(V,R)=1$ and $\Lambda_{n+1}(V,R)=0$ for some $n>0$. Then $\Lambda(V,R)$ is a Frobenius algebra,* while ${{S}}(V,R)$ is a Gorenstein algebra of global dimension $n$.*

The subscripts here and elsewhere in the paper indicate the homogeneous components of graded spaces. The first conclusion in Theorem 4.5 is equivalent to nondegeneracy of the bilinear pairings $\Lambda_{i}(V,R)\times\Lambda_{n-i}(V,R)\to\Lambda_{n}(V,R)$ arising from the multiplication. This was proved by Gurevich in the semisimple case.

By a Gorenstein algebra we mean any positively graded algebra $A=\bigoplus_{i=0}^{\infty}A_{i}$ with $A_{0}={k}$ for which $\mathop{\rm Ext}\nolimits_{A}^{\bullet}({k},A)$ is 1-dimensional. No Noetherian conditions are requested. The two conclusions in Theorem 4.5 are closely related (see [18]), and the essential job will be to prove the first one.

Let now $R^{\prime}$ be a second Hecke symmetry on another finite dimensional vector space $V^{\prime}$. Suppose that $R$ and $R^{\prime}$ have the same parameter $q$ of the Hecke relation. Then there is a graded algebra $A(R^{\prime},R)$ whose construction generalizes that of $A(R)$. In fact $A(R^{\prime},R)=A(R)$ when $R^{\prime}=R$. The algebra $A(R^{\prime},R)$ in a different notation was introduced by Phùng Hô Hai [9] under the name “quantum hom-space”. We will consider yet another graded algebra $E(R^{\prime},R)$ whose relationship with $A(R^{\prime},R)$ is similar to that between $\Lambda(V,R)$ and ${{S}}(V,R)$. In a different notation this algebra was also introduced in [9] under the name “exterior algebra on the quantum hom-space”.

Theorem 6.2. Suppose that both $R$ and $R^{\prime}$ satisfy the $1$-dimensional source condition. Then the graded algebras $A(R^{\prime},R)$ and $E(R^{\prime},R)$ are Koszul. Their Hilbert series satisfy the relation  $h_{A(R^{\prime},R)}(t)\,h_{E(R^{\prime},R)}(-t)=1$.

In the semisimple case this was proved in [9]. The argument used by Phùng Hô Hai is based on the observation that Koszulness of $A(R^{\prime},R)$ is equivalent to a certain property of ${\cal H}_{n}^{\rm op}\otimes{\cal H}_{n}$-modules concerned with distributivity of collections of subspaces in these modules. If $R_{k,q}$ is the Hecke symmetry corresponding to the standard quantum $GL_{k}$ with parameter $q$, then $A(R_{k,q})$ is known to possess a PBW basis, and its Koszulness follows from Priddy’s theorem [19]. If $[n]_{q}\neq 0$ for all $n$, the semisimplicity of ${\cal H}_{n}^{\rm op}\otimes{\cal H}_{n}$ ensures then the desired property for all simple modules, and therefore for arbitrary ${\cal H}_{n}^{\rm op}\otimes{\cal H}_{n}$-modules. This in turn yields Koszulness of $A(R^{\prime},R)$ for arbitrary Hecke symmetries with the same parameter $q$.

In the present paper we prove directly exactness of certain complexes defined with respect to representations of ${\cal H}_{n}^{\rm op}\otimes{\cal H}_{n}$, and thus we derive Theorem 6.2 solely from Hecke algebra considerations, avoiding the use of Priddy’s theorem.

Theorem 6.6. Suppose that both $R$ and $R^{\prime}$ satisfy the trivial source condition. If $\dim E_{n}(R^{\prime},R)=1$ and $E_{n+1}(R^{\prime},R)=0$ for some integer $n>0,$ then $E(R^{\prime},R)$ is a Frobenius algebra,* while $A(R^{\prime},R)$ is a Gorenstein algebra of global dimension $n$.*

If $\dim V^{\prime}=1$ and $R^{\prime}$ is the multiplication by $q$, then $A(R^{\prime},R)={{S}}(V,R)$ and $E(R^{\prime},R)=\Lambda(V,R)$. Thus the first two results discussed in this introduction are a special case of the two subsequent ones. We nevertheless provide separate proofs in this special case. It serves as a model for the more complicated Theorems 6.2 and 6.6 where we need some lengthy verifications done in section 5 of the paper.

The quantum hom-spaces were considered in [9] in connection with the quantum version of the classical invariant theory. There is an even more obvious role played by the algebra $A(R^{\prime},R)$. Being equipped with an $A(R^{\prime})\,$-$A(R)$ bicomodule structure, it gives rise to functors between the corepresentation categories of the two bialgebras $A(R)$ and $A(R^{\prime})$. For each coalgebra $C$ let ${}^{C}\!{\cal M}$ and ${\cal M}^{C}$ stand for the categories of left and right comodules. The sign $\,\mathchar 9219\relax_{C}\,$ denotes the cotensor product of comodules.

Theorem 7.2. Suppose that for each $n>1$ the indecomposable ${\cal H}_{n}$-modules isomorphic to direct summands of ${{T}}_{n}(V^{\prime})$ are the same as those isomorphic to direct summands of ${{T}}_{n}(V)$. Then the functors

[TABLE]

are braided monoidal equivalences $\,\vphantom{{\cal M}}{}^{A(R)}{\cal M}{}\mathrel{\smash{\mathop{\longrightarrow}\limits}}\vphantom{{\cal M}}{}^{A(R^{\prime})}{\cal M}\,$ and $\,{\cal M}^{A(R)}{}\mathrel{\smash{\mathop{\longrightarrow}\limits}}{}{\cal M}^{A(R^{\prime})}$.

This should be compared with the monoidal equivalences between the corepresentation categories of two Hopf algebras. According to a result of Schauenburg [21] such equivalences are given by the cotensor product functors determined by the so-called bi-Galois algebras. In this way Bichon [2] and Mrozinski [17] showed that the categories of comodules over certain Hopf algebras associated with bilinear forms are monoidally equivalent to the respective categories defined for the standard quantizations of $SL_{2}$ and $GL_{2}$. Those Hopf algebras correspond to a special class of Hecke symmetries. Phùng Hô Hai dealt with the Hopf envelopes of the bialgebras $A(R)$ for arbitrary Hecke symmetries under the previously mentioned restriction on $q$ and the characteristic of ${k}$. By the main result of [10] the category of comodules over such a Hopf algebra is determined, up to monoidal equivalence, by the parameter $q$ and the birank of $R$.

Theorem 7.2 is a similar, to some extent, result for bialgebras, though the bicomodule algebra $A(R,R^{\prime})$ is definitely not bi-Galois (Galois algebras exist only for Hopf algebras). In fact, everything what is needed for the equivalence here is present already in the construction of the algebras involved. Note that there are no restrictions on $R$ in Theorem 7.2.

1. Hecke algebra preliminaries

We denote by ${{S}}_{n}$ the symmetric group of permutations of the set $\{1,\ldots,n\}$. Let ${{B}}_{n}=\{\tau_{1},\ldots,\tau_{n-1}\}$ be the set of basic transpositions $\tau_{i}=(i,i+1)$. The length $\ell(\sigma)$ of a permutation $\sigma\in{{S}}_{n}$ is the smallest number of factors in the expressions of $\sigma$ as product of basic transpositions. By the letter $e$ we denote the identity permutation and also the trivial subgroup of ${{S}}_{n}$.

The subgroups of ${{S}}_{n}$ generated by subsets of ${{B}}_{n}$ are called Young subgroups and are parametrized traditionally by compositions of $n$, i.e. by finite sequences of positive integers summing up to $n$. The Young subgroup ${{S}}_{\lambda}$ labelled by a composition $\lambda=(\lambda_{1},\ldots,\lambda_{k})$ is generated by the set

[TABLE]

and is isomorphic to ${{S}}_{\lambda_{1}}\times\ldots\times{{S}}_{\lambda_{k}}$. In particular, the subgroup ${{S}}_{i,\,n-i}$ corresponding to the composition $(i,n-i)$ is generated by $\{\tau_{j}\in{{B}}_{n}\mid j\neq i\}$. For each $\lambda$ the pair $({{S}}_{\lambda},{{B}}_{\lambda})$ is a Coxeter system. We will use standard facts concerning Coxeter groups and the respective Hecke algebras. For reference see, e.g., [3], [6].

Given a pair of Young subgroups ${{S}}_{\lambda}\subset{{S}}_{\nu}$, each coset of ${{S}}_{\lambda}$ in ${{S}}_{\nu}$ contains a unique element of minimal length called the distinguished coset representative. We denote by ${\cal D}({{S}}_{\nu}/{{S}}_{\lambda})$ and ${\cal D}({{S}}_{\lambda}\backslash{{S}}_{\nu})$ the sets of distinguished representatives of the respective cosets. Recall that

[TABLE]

where comparison $\pi\tau_{i}>\pi$ refers to the Bruhat order (recall that $|\ell(\pi\tau_{i})-\ell(\pi)|=1$, and $\pi\tau_{i}>\pi$ if and only if $\ell(\pi\tau_{i})>\ell(\pi)$). The set $\,{\cal D}({{S}}_{\lambda}\backslash{{S}}_{\nu})={\cal D}({{S}}_{\nu}/{{S}}_{\lambda})^{-1}\,$ admits similar characterizations. For another Young subgroup ${{S}}_{\mu}$ of ${{S}}_{\nu}$ the set of distinguished ${{S}}_{\mu}$-$\,{{S}}_{\lambda}$ double coset representatives is

[TABLE]

We will also use shorter notation: ${\cal D}_{\lambda}={\cal D}({{S}}_{n}/{{S}}_{\lambda})$, ${}_{\mu}{\cal D}={\cal D}({{S}}_{\mu}\backslash{{S}}_{n})$, and ${}_{\mu}{\cal D}_{\lambda}={\cal D}({{S}}_{\mu}\backslash{{S}}_{n}/{{S}}_{\lambda})$.

Let ${k}$ be the ground field. The Hecke algebra ${\cal H}_{n}={\cal H}_{n}(q)$ of type $A_{n-1}$ with parameter $q\in{k}$ is presented by generators $T_{1},\ldots,T_{n-1}$ and relations

[TABLE]

It has a standard basis $\{T_{\sigma}\mid\sigma\in{{S}}_{n}\}$ characterized by the properties that $T_{e}=1$ and $T_{\tau_{i}\sigma}=T_{i}T_{\sigma}$ whenever $\tau_{i}\sigma>\sigma$. Recall that $T_{\pi}T_{\sigma}=T_{\pi\sigma}$ for each pair $\pi,\sigma\in{{S}}_{n}$ such that $\ell(\pi\sigma)=\ell(\pi)+\ell(\sigma)$.

The definition of ${\cal H}_{n}(q)$ makes sense also when $q=0$. We will use this algebra called the [math]-Hecke algebra on several occasions.

The elements $\{T_{\sigma}\mid\sigma\in{{S}}_{\lambda}\}$ form a basis for the parabolic subalgebra ${\cal H}_{\lambda}$ generated by $\{T_{i}\mid\tau_{i}\in{{B}}_{\lambda}\}$. For each $k<n$ we identify the symmetric group ${{S}}_{k}$ with the subgroup of ${{S}}_{n}$ generated by $\{\tau_{i}\mid 0<i<k\}$ and the Hecke algebra ${\cal H}_{k}$ with the subalgebra of ${\cal H}_{n}$ generated by $\{T_{i}\mid 0<i<k\}$. By a convention ${{S}}_{0}={{S}}_{1}=e$ and ${\cal H}_{0}={\cal H}_{1}={k}$.

For each 1-dimensional representation of a parabolic subalgebra ${\cal H}_{\lambda}$ of ${\cal H}_{n}$ given by an algebra homomorphism $\chi:{\cal H}_{\lambda}\to{k}$ we denote by ${k}(\chi)$ the corresponding 1-dimensional ${\cal H}_{\lambda}$-module. The induced ${\cal H}_{n}$-module

[TABLE]

has a basis $\{T_{\sigma}c\mid\sigma\in{\cal D}_{\lambda}\}$ where $c=1\otimes 1$ is the canonical generator of $M$. Here and later by a basis we mean a basis over the ground field ${k}$.

By Deodhar’s lemma (see [6, 2.1.2]) for each $\tau_{i}\in{{B}}_{n}$ and $\sigma\in{\cal D}_{\lambda}$ we have either $\tau_{i}\sigma\in{\cal D}_{\lambda}$ or $\tau_{i}\in\sigma{{B}}_{\lambda}\sigma^{-1}$. Hence

[TABLE]

If $\sigma\in B_{i}(\lambda)$, then $\tau_{i}\sigma=\sigma\tau_{j}>\sigma$ for some $\tau_{j}\in{{B}}_{\lambda}$, and it follows that $T_{i}T_{\sigma}=T_{\tau_{i}\sigma}=T_{\sigma}T_{j}$. In this case $T_{\sigma}c$ is an eigenvector for the linear operator $(T_{i})_{M}$ by which $T_{i}$ acts on $M$. Let us denote by $\chi_{\sigma}(T_{i})$ the corresponding eigenvalue.

Note that $\sigma^{-1}(i)<\sigma^{-1}(i+1)$ when $\tau_{i}\sigma>\sigma$. Since $\sigma^{-1}\tau_{i}\sigma$ is the transposition of $\sigma^{-1}(i)$ and $\sigma^{-1}(i+1)$, the equality $\tau_{i}\sigma=\sigma\tau_{j}$ implies then that $j=\sigma^{-1}(i)$. Thus the action of $T_{i}$ on the basis elements of $M$ is given by the formulas

[TABLE]

The restriction of the induced ${\cal H}_{n}$-module $M$ to a second parabolic subalgebra ${\cal H}_{\mu}$ of ${\cal H}_{n}$ is given by the Mackey formula [4, Th. 2.7]:

[TABLE]

where $\nu(\pi)$ is the composition of $n$ such that ${{S}}_{\nu(\pi)}={{S}}_{\mu}\cap\pi{{S}}_{\lambda}\pi^{-1}$ and $\chi_{\pi}$ is the 1-dimensional representation of the corresponding parabolic subalgebra ${\cal H}_{\nu(\pi)}$ with the values $\chi_{\pi}(T_{i})$ on the generators $T_{i}$ of ${\cal H}_{\nu(\pi)}$ defined in the preceding paragraph. Here $M(\pi)$ is the ${\cal H}_{\mu}$-submodule of $M$ generated by $T_{\pi}c$. Its basis over ${k}$ is formed by the elements $T_{\sigma}c$ with $\sigma\in{\cal D}_{\lambda}\cap{{S}}_{\mu}\pi{{S}}_{\lambda}$.

The assignments $T_{i}\mapsto q$ (respectively, $T_{i}\mapsto-1$) for each $i$ such that $\tau_{i}\in{{B}}_{\lambda}$ define the trivial (respectively, alternating) representation ${\cal H}_{\lambda}\to{k}$. We denote by ${k}_{\rm triv}$ and ${k}_{\rm alt}$ the corresponding ${\cal H}_{\lambda}$-modules. They make sense for each parabolic subalgebra of ${\cal H}_{n}$, in particular, for ${\cal H}_{n}$ itself. If $q\neq 0$, then every 1-dimensional ${\cal H}_{n}$-module is isomorphic to either ${k}_{\rm triv}$ or ${k}_{\rm alt}$. If $q=-1$, then ${k}_{\rm triv}={k}_{\rm alt}$, and in fact all 1-dimensional representations of any parabolic subalgebra coincide.

If $q\neq 0$ then the induction functor from any parabolic subalgebra is isomorphic to the coinduction functor. In other words,

[TABLE]

for any left ${\cal H}_{n}$-module $N$ and left ${\cal H}_{\lambda}$-module $U$. This is a general property of the Hecke algebras of Coxeter groups which we view as part of the Frobenius reciprocity (see [4, Th. 2.5, 2.6] and [6, 9.1.7]). In particular, ${\cal H}_{n}$ can be replaced in the isomorphism above by any parabolic subalgebra larger than the given ${\cal H}_{\lambda}$.

Lemma 1.1. Let $M={\cal H}_{n}\otimes_{{\cal H}_{\lambda}}{k}(\chi)$ and $N={\cal H}_{n}\otimes_{{\cal H}_{\mu}}{k}(\zeta)$ be ${\cal H}_{n}$-modules induced from $1$-dimensional representations of two parabolic subalgebras of ${\cal H}_{n}$. If $q\neq 0$ then

[TABLE]

where $f(\zeta,\chi)=\#\{\pi\in{}_{\mu}{\cal D}_{\lambda}\mid\zeta(T_{i})=\chi_{\pi}(T_{i}){\rm\ for\ all\ }i{\rm\ with\ }\tau_{i}\in{{S}}_{\mu}\cap\pi{{S}}_{\lambda}\pi^{-1}\}$.

Proof. Consider the Mackey decomposition $M=\bigoplus M(\pi)$ with respect to ${\cal H}_{\mu}$. Then

[TABLE]

Recall that for each $\pi$ the ${\cal H}_{\mu}$-module $M(\pi)$ is induced from the $1$-dimensional representation $\chi_{\pi}$ of the parabolic subalgebra ${\cal H}_{\nu(\pi)}$ corresponding to the Young subgroup ${{S}}_{\mu}\cap\pi{{S}}_{\lambda}\pi^{-1}$. By the Frobenius reciprocity

[TABLE]

and this space is nonzero if and only if $\zeta$ agrees with $\chi_{\pi}$ on ${\cal H}_{\nu(\pi)}$.   $\mathchar 9219\relax$

The next lemma will be crucial for establishing the relation between the Hilbert series of the pairs of graded algebras in Theorems 3.1 and 6.2.

Lemma 1.2. Suppose that $q\neq 0$. Let $M$ and $N$ be finite dimensional ${\cal H}_{n}$-modules whose indecomposable direct summands all have $1$-dimensional sources. Then

[TABLE]

Proof. Since $\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}$ is an additive functor of both arguments, it suffices to check the desired equality when $M$ and $N$ are indecomposable modules, so that they are direct summands of ${\cal H}_{n}$-modules induced from $1$-dimensional representations of parabolic subalgebras. If ${\cal H}_{n}$ is semisimple, then the indecomposable modules are simple, and the conclusion is obviously true since for two simple modules $\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(N,M)=0$ unless $N\cong M$.

If ${\cal H}_{n}$ is not semisimple, we will apply the specialization argument. Let $M$ and $N$ be as in Lemma 1.1. In this case Lemma 1.1 gives the exact values for the dimensions of $\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(N,M)$ and $\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(M,N)$, but we have to show that

[TABLE]

whenever $N^{\prime}$ is a direct summand of $N$ and $M^{\prime}$ is a direct summand of $M$.

Let $O$ be the completion of the polynomial ring ${k}[t]$ in one indeterminate $t$ at its maximal ideal generated by $t-q$. Then $O$ is a complete discrete valuation ring with residue field isomorphic to ${k}$. The specialization homomorphism $O\to{k}$ sends $t$ to $q$. Denote by $K$ the field of fractions of $O$. Let ${\cal H}_{n}(t)$ be the Hecke algebra of type $A_{n-1}$ with parameter $t$ over the ring $O$. Then ${\cal H}_{n}(t)\otimes_{O}{k}\cong{\cal H}_{n}$, while

[TABLE]

is a semisimple Hecke algebra of type $A_{n-1}$ over the field $K$.

Let ${\cal H}_{\lambda}(t)$ and ${\cal H}_{\mu}(t)$ be the parabolic subalgebras of ${\cal H}_{n}(t)$ corresponding to the two compositions $\lambda$ and $\mu$ of $n$. Define ring homomorphisms $\chi^{t}:{\cal H}_{\lambda}(t)\to O$ and $\zeta^{t}:{\cal H}_{\mu}(t)\to O$ by the formulas

[TABLE]

Let $O(\chi^{t})=O$ with the ${\cal H}_{\lambda}(t)$-module structure given by $\chi^{t}$, and $O(\zeta^{t})=O$ the similar ${\cal H}_{\mu}(t)$-module with respect to $\zeta^{t}$. Put

[TABLE]

These are $O$-free ${\cal H}_{n}(t)$-modules such that $M(t)\otimes_{O}{k}\cong M$ and $N(t)\otimes_{O}{k}\cong N$. The ${\cal H}_{n}(t)_{K}$-modules $M(t)\otimes_{O}K$ and $N(t)\otimes_{O}K$ are induced from $1$-dimensional representations of parabolic subalgebras of ${\cal H}_{n}(t)_{K}$. So Lemma 1.1 yields

[TABLE]

By exactness of the functor $?\otimes_{O}K$, we have

[TABLE]

Since the $O$-module \mathop{\rm Hom}\nolimits_{{\cal H}_{n}(t)}\bigl{(}N(t),M(t)\bigr{)} is torsionfree, it has to be free of rank equal to $f(\zeta^{t},\chi^{t})$. For each pair of indices $i,j$ such that $\tau_{i}\in{{B}}_{\mu}$ and $\tau_{j}\in{{B}}_{\lambda}$ it is seen from the definition of $\chi^{t},\,\zeta^{t}$ that $\zeta^{t}(T_{i})=\chi^{t}(T_{j})$ if and only if $\zeta(T_{i})=\chi(T_{j})$ since $\zeta(T_{i})$ and $\chi(T_{j})$ can be equal to only $q$ or $-1$. Therefore $\zeta^{t}(T_{i})=\chi^{t}_{\pi}(T_{i})$ for some $\pi\in{}_{\mu}{\cal D}_{\lambda}$ and $i$ such that $\tau_{i}\in{{S}}_{\mu}\cap\pi{{S}}_{\lambda}\pi^{-1}$ if and only if $\zeta(T_{i})=\chi_{\pi}(T_{i})$. It follows that $f(\zeta^{t},\chi^{t})=f(\zeta,\chi)$, as defined in Lemma 1.1.

A homomorphism $\varphi:N(t)\to M(t)$ induces zero map $N\to M$ if and only if $\,\mathop{\rm Im}\varphi\subset(t-q)M(t)$, so that (t-q)^{-1}\varphi\in\mathop{\rm Hom}\nolimits_{{\cal H}_{n}(t)}\bigl{(}N(t),M(t)\bigr{)} for such a $\varphi$. This shows that \,\mathop{\rm Hom}\nolimits_{{\cal H}_{n}(t)}\bigl{(}N(t),M(t)\bigr{)}\otimes_{O}{k}\, embeds in $\,\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(N,M)$, but then

[TABLE]

by comparison of dimensions.

As a special case we get \bigl{(}\mathop{\rm End}\nolimits_{{\cal H}_{n}(t)}M(t)\bigr{)}\otimes_{O}{k}\cong\mathop{\rm End}\nolimits_{{\cal H}_{n}}M. As is well-known, in this situation each idempotent of $\mathop{\rm End}\nolimits_{{\cal H}_{n}}M$ can be lifted to an idempotent of $\mathop{\rm End}\nolimits_{{\cal H}_{n}(t)}M(t)$. Direct summands of an arbitrary module are determined by idempotents in its endomorphism ring. Thus we can find a direct summand $M^{\prime}(t)$ of the ${\cal H}_{n}(t)$-module $M(t)$ such that $M^{\prime}(t)\otimes_{O}{k}\cong M^{\prime}$. Similarly, there is a direct summand $N^{\prime}(t)$ of $N(t)$ satisfying $N^{\prime}(t)\otimes_{O}{k}\cong N^{\prime}$. Being submodules of $O$-free modules, both $M^{\prime}(t)$ and $N^{\prime}(t)$ are themselves $O$-free.

Since every ${\cal H}_{n}$-module homomorphism $N\to M$ lifts to an ${\cal H}_{n}(t)$-module homomorphism $N(t)\to M(t)$, it follows that every ${\cal H}_{n}$-module homomorphism $N^{\prime}\to M^{\prime}$ lifts to an ${\cal H}_{n}(t)$-module homomorphism $N^{\prime}(t)\to M^{\prime}(t)$. This entails

[TABLE]

Since \mathop{\rm Hom}\nolimits_{{\cal H}_{n}(t)}\bigl{(}N^{\prime}(t),M^{\prime}(t)\bigr{)} is a free $O$-module, we deduce that

[TABLE]

By symmetry

[TABLE]

and the equality $\dim\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(N^{\prime},M^{\prime})=\dim\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(M^{\prime},N^{\prime})$ follows from the already discussed semisimple case.   $\mathchar 9219\relax$

A suitable version of Lemma 1.1 is valid also when $q=0$. Later we will need only a special case of that fact:

Lemma 1.3. Let $M={\cal H}_{n}\otimes_{{\cal H}_{\lambda}}{k}(\chi)$. Then

[TABLE]

Proof. This conclusion is a consequence of the Frobenius reciprocity when $q\neq 0$. For $q=0$ it is derived as follows. Since $T_{i}^{2}=-T_{i}$, it is seen from the formulas for the action of $T_{i}$ that $T_{i}M$ is contained in the linear span of the basis elements $T_{\sigma}c$ with $\sigma\in{\cal D}_{\lambda}\mathchar 9586\relax A_{i}(\lambda)$. If $v\in M$ is such that $T_{i}v=-v$ for all $i=1,\ldots,n-1$, then $v\in\bigcap_{i=1}^{n-1}T_{i}M$.

Let $w_{n}$ and $w_{\lambda}$ be the longest elements of ${{S}}_{n}$ and ${{S}}_{\lambda}$, respectively. By [6, 2.2.1] $d_{\lambda}=w_{n}w_{\lambda}$ is the unique element of maximal length in ${\cal D}_{\lambda}$, and ${\cal D}_{\lambda}$ consists precisely of all suffixes of $d_{\lambda}$ which are elements $\sigma\in{{S}}_{n}$ satisfying $\ell(d_{\lambda})=\ell(d_{\lambda}\sigma^{-1})+\ell(\sigma)$. If $\sigma\in{\cal D}_{\lambda}$ and $\sigma\neq d_{\lambda}$, then there exists $\tau_{i}\in{{B}}_{n}$ such that $\tau_{i}\sigma>\sigma$ and $\tau_{i}\sigma$ is a suffix of $d_{\lambda}$, so that $\tau_{i}\sigma\in{\cal D}_{\lambda}$, i.e. $\sigma\in A_{i}(\lambda)$. This shows that $T_{\sigma}c$ can be involved with nonzero coefficient in the expression for $v$ only when $\sigma=d_{\lambda}$, i.e. $v$ has to be a scalar multiple of $T_{d_{\lambda}}c$.

Now $T_{d_{\lambda}}c$ is an eigenvector for all operators $(T_{i})_{M}$. Moreover, the 1-dimensional ${\cal H}_{n}$-submodule of $M$ spanned by $T_{d_{\lambda}}c$ is isomorphic to ${k}_{\rm alt}$ if and only if $\chi_{d_{\lambda}}(T_{i})$ is equal to $-1$ for each $i$ such that $d_{\lambda}\in B_{i}(\lambda)$. Since the conjugation by $w_{n}$ (respectively, by $w_{\lambda}$) map ${{B}}_{n}$ (respectively, ${{B}}_{\lambda}$) onto itself, we have $d_{\lambda}{{B}}_{\lambda}d_{\lambda}^{-1}\subset{{B}}_{n}$. This means that

[TABLE]

Since $d_{\lambda}\tau_{j}\mskip 1.0mud_{\lambda}^{-1}=\tau_{d_{\lambda}(j)}$ and $\chi_{d_{\lambda}}(T_{d_{\lambda}(j)})=\chi(T_{j})$ for each $\tau_{j}\in{{B}}_{\lambda}$, the previous condition on $\chi_{d_{\lambda}}$ is equivalent to the condition that $\chi$ is alternating.   $\mathchar 9219\relax$

Lemma 1.4. Suppose that $q\neq 0$. Let $M={k}(\chi)\otimes_{{\cal H}_{\lambda}}{\cal H}_{n}$ be the right ${\cal H}_{n}$-module induced from a $1$-dimensional representation $\chi$ of the parabolic subalgebra ${\cal H}_{\lambda}$. Considering the dual space $M^{*}$ as a left ${\cal H}_{n}$-module with respect to the natural action of ${\cal H}_{n},$ we have $\,M^{*}\cong{\cal H}_{n}\otimes_{{\cal H}_{\lambda}}{k}(\chi)$.

Proof. Since M^{*}\cong\mathop{\rm Hom}\nolimits_{{\cal H}_{\lambda}}\bigl{(}{\cal H}_{n},{k}(\chi)\bigr{)}, the conclusion is a consequence of the Frobenius reciprocity.   $\mathchar 9219\relax$

Put $\mathaccent 869{T}_{i}=q-1-T_{i}$ for each $i=1,\ldots,n-1$. The assignment $T_{i}\mapsto\mathaccent 869{T}_{i}$ extends to an involutive automorphism of ${\cal H}_{n}$. We denote by $\mathaccent 869{M}$ the ${\cal H}_{n}$-module whose underlying vector space coincides with $M$, but the new action of $T_{i}$ is given by the original action of $\mathaccent 869{T}_{i}$.

Lemma 1.5. If $M={\cal H}_{n}\otimes_{{\cal H}_{\lambda}}{k}(\chi),$ then $\mathaccent 869{M}\cong{\cal H}_{n}\otimes_{{\cal H}_{\lambda}}{k}(\mathaccent 869{\chi})$ where $\mathaccent 869{\chi}:{\cal H}_{\lambda}\to{k}$ is the $1$-dimensional representation such that

[TABLE]

Proof. Let $c$ be the canonical generator of $M$. Then $\mathaccent 869{T}_{i}c=\mathaccent 869{\chi}(T_{i})c$ for each $i$ such that $\tau_{i}\in{{B}}_{\lambda}$. Clearly $c$ generates also $\mathaccent 869{M}$. Hence there is a surjective homomorphism ${\cal H}_{n}\otimes_{{\cal H}_{\lambda}}{k}(\mathaccent 869{\chi})\to\mathaccent 869{M}$ which has to be bijective since the two modules here have equal dimensions.   $\mathchar 9219\relax$

Lemma 1.6. Suppose that $q\neq-1$. Let $M$ and $N$ be two left ${\cal H}_{n}$-modules. Given a ${k}$-linear map $\varphi:N\to M,$ the following conditions are equivalent:**

*(a) **for each $i=1,\ldots,n-1$ there exists a ${k}$-linear map $\psi_{i}:N\to M$ such that $\varphi=\psi_{i}T_{i}-T_{i}\psi_{i}\,$ *(i.e. $\varphi(x)=\psi_{i}(T_{i}x)-T_{i}\psi_{i}(x)$ for all $x\in N$),

(b) $\varphi$ is a homomorphism of ${\cal H}_{n}$-modules $N\to\mathaccent 869{M}$.

If $\,q=-1,$ then (a) $\Rightarrow$ (b).

Proof. Note that $T_{i}+\mathaccent 869{T}_{i}=q-1$ and $T_{i}\mathaccent 869{T}_{i}=-q$. If $\psi_{i}$ satisfies (a), then

[TABLE]

Hence (a) implies that $\varphi T_{i}=\mathaccent 869{T}_{i}\varphi$ for all $i$, i.e. $\varphi\in\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(N,\mathaccent 869{M})$. Conversely, if $\varphi T_{i}=\mathaccent 869{T}_{i}\varphi$, then taking $\psi_{i}=(q+1)^{-2}(\mathaccent 869{T}_{i}-T_{i})\varphi$, we get

[TABLE]

since $\,(\mathaccent 869{T}_{i}-T_{i})^{2}=(\mathaccent 869{T}_{i}+T_{i})^{2}-4T_{i}\mathaccent 869{T}_{i}=(q-1)^{2}+4q=(q+1)^{2}$.   $\mathchar 9219\relax$

Lemma 1.7. Suppose that $q=-1$. Let $M={\cal H}_{n}\otimes_{{\cal H}_{\lambda}}{k}$ and $N={\cal H}_{n}\otimes_{{\cal H}_{\mu}}{k}$ be ${\cal H}_{n}$-modules induced from the $1$-dimensional representations of two parabolic subalgebras. Denote by $c$ and $c^{\prime}$ their canonical generators. For a homomorphism $\varphi:N\to M$ the following conditions are equivalent:**

(a) $\varphi$ factors through a free ${\cal H}_{n}$-module,**

(b) $\varphi(c^{\prime})\in x_{\mu}M$ where $x_{\mu}=\sum_{\sigma\in{{S}}_{\mu}}T_{\sigma},$

(c) $\varphi(c^{\prime})\in(T_{i}+1)M$ for each $i$ such that $\tau_{i}\in{{B}}_{\mu}\,,$

*(d) *for each $i=1,\ldots,n-1$ there exists a ${k}$-linear map $\psi_{i}:N\to M$ such that $\varphi=T_{i}\psi_{i}+\psi_{i}T_{i}+2\psi_{i},$ i.e. $\varphi=\psi_{i}T_{i}-\mathaccent 869{T}_{i}\psi_{i}$ where $\,\mathaccent 869{T}_{i}=-2-T_{i}$.

The space of all homomorphisms $N\to M$ satisfying (a) – (d) has a basis indexed by the set $\{\pi\in{}_{\mu}{\cal D}_{\lambda}\mid{{S}}_{\mu}\cap\pi{{S}}_{\lambda}\pi^{-1}=e\}$ of distinguished representatives of the double cosets with the trivial intersection property.

Proof. (a)$\,\Rightarrow\,$(b) Since the algebra ${\cal H}_{\mu}$ is Frobenius, its socle contains the left module ${k}$ with multiplicity 1. The unique 1-dimensional left ideal of ${\cal H}_{\mu}$ is spanned by $x_{\mu}$. If $F$ is any free ${\cal H}_{n}$-module, then $F$ is free also as an ${\cal H}_{\mu}$-module, and therefore every homomorphism $N\to F$ sends $c^{\prime}$ into $x_{\mu}F$. From a factorization $N\to F\to M$ of $\varphi$ we deduce that $\varphi(c^{\prime})\in x_{\mu}M$.

(b)$\,\Rightarrow\,$(c) This is clear since $x_{\mu}=(T_{i}+1)\sum_{\sigma\in{\cal D}(\langle\tau_{i}\rangle\backslash{{S}}_{\mu})}T_{\sigma}$ where $\langle\tau_{i}\rangle$ is the 2-element subgroup of ${{S}}_{\mu}$ generated by $\tau_{i}$.

(c)$\,\Rightarrow\,$(d) Let us fix $i$ and construct the desired map $\psi_{i}$ by specifying its values on the basis elements $\{T_{\sigma}c^{\prime}\mid\sigma\in{\cal D}_{\mu}\}$ of $N$.

If $\sigma\in B_{i}(\mu)$, then $\tau_{i}\sigma=\sigma\tau_{j}$ for some $\tau_{j}\in{{B}}_{\mu}$. Hence $T_{i}T_{\sigma}=T_{\tau_{i}\sigma}=T_{\sigma}T_{j}$. The equality $\varphi(v)=T_{i}\psi_{i}(v)+\psi_{i}(T_{i}v)+2\psi_{i}(v)$ for the element $v=T_{\sigma}c^{\prime}$ is equivalent to $\varphi(v)=(T_{i}+1)\psi_{i}(v)$ since $T_{i}v=-v$. An element $\psi_{i}(v)\in M$ satisfying the required equality can be found. Indeed, $\varphi(v)=T_{\sigma}\varphi(c^{\prime})\in(T_{i}+1)M$ since $\varphi(c^{\prime})\in(T_{j}+1)M$.

Suppose now that $\sigma\in A_{i}(\mu)$. In this case we put

[TABLE]

Then the equality $\varphi(v)=T_{i}\psi_{i}(v)+\psi_{i}(T_{i}v)+2\psi_{i}(v)$ holds for $v=T_{\sigma}c^{\prime}$ and also for $v=T_{\tau_{i}\sigma}c^{\prime}=T_{i}T_{\sigma}c^{\prime}$. In this way the value of $\psi_{i}$ has been determined on all basis elements of $N$.

(d)$\,\Rightarrow\,$(c) If $\tau_{i}\in{{B}}_{\mu}$, then $T_{i}c^{\prime}=-c^{\prime}$, whence

[TABLE]

(c)$\,\Rightarrow\,$(b) For $\pi\in{}_{\mu}{\cal D}_{\lambda}$ let ${{S}}_{\nu(\pi)}={{S}}_{\mu}\cap\pi{{S}}_{\lambda}\pi^{-1}$. We have $M=\bigoplus_{\pi\in\,{}_{\mu}{\cal D}_{\lambda}}M(\pi)$ where $M(\pi)$ is the ${\cal H}_{\mu}$-submodule of $M$ with a basis

[TABLE]

Since \mathop{\rm Hom}\nolimits_{{\cal H}_{\mu}}\bigl{(}{k},M(\pi)\bigr{)}\cong{k} for each $\pi$, the vector space $\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}(N,M)$ has a basis $\{\varphi_{\pi}\mid\pi\in{}_{\mu}{\cal D}_{\lambda}\}$ where the homomorphism $\varphi_{\pi}:N\to M$ is defined by the rule

[TABLE]

We can write $\varphi=\sum\alpha_{\pi}\varphi_{\pi}$ with $\alpha_{\pi}\in{k}$. If $\tau_{i}\in{{B}}_{\mu}$, then $(T_{i}+1)M$ is a direct sum of its subspaces $(T_{i}+1)M(\pi)$, whence the condition $\varphi(c^{\prime})\in(T_{i}+1)M$ implies that $\varphi_{\pi}(c^{\prime})\in(T_{i}+1)M$ for each $\pi\in{}_{\mu}{\cal D}_{\lambda}$ such that $\alpha_{\pi}\neq 0$. If $\tau_{i}\in{{B}}_{\nu(\pi)}$ for some $\pi$, then $T_{i}T_{\pi}c=-T_{\pi}c$, and since the elements $T_{\sigma}c$ with $\sigma\in{\cal D}_{\lambda}$, $\sigma\neq\pi$ span a $T_{i}$-invariant subspace, we deduce that $\varphi_{\pi}(c^{\prime})\notin(T_{i}+1)M$. In this case $\alpha_{\pi}=0$. On the other hand, if ${{S}}_{\nu(\pi)}=e$, then the ${\cal H}_{\mu}$-module $M(\pi)$ is freely generated by $T_{\pi}c$, which shows that $\varphi_{\pi}(c^{\prime})=x_{\mu}T_{\pi}c\in x_{\mu}M$.

(b)$\,\Rightarrow\,$(a) Let $p\in M$ be such that $\varphi(c^{\prime})=x_{\mu}p$. Then $\varphi$ is the composite of the homomorphism $N\to{\cal H}_{n}$ sending $c^{\prime}$ to $x_{\mu}$ and the homomorphism ${\cal H}_{n}\to M$ sending $1$ to $p$.

In the course of the proof we have seen that $\varphi$ satisfies the equivalent conditions (b) and (c) if and only if $\varphi$ is a linear combination of the homomorphisms $\varphi_{\pi}$ with $\pi\in{}_{\mu}{\cal D}_{\lambda}$ and ${{S}}_{\nu(\pi)}=e$. This establishes the final conclusion.   $\mathchar 9219\relax$

2. Complexes associated with representations of ${\cal H}_{n}$

Given an ordered collection of subspaces $U_{1},\ldots,U_{n-1}$ of a vector space $M$, we denote by K_{\bullet}\bigl{(}M;(U_{i})\bigr{)} the complex of vector spaces

[TABLE]

constructed as follows. For $0\leq i\leq n$ put

[TABLE]

with the convention that $\Upsilon_{0}=\Upsilon_{1}=M$ and $\Sigma_{n-1}=\Sigma_{n}=0$. The differentials $\partial_{i}$ are induced by the inclusions $\,\Upsilon_{i}\subset\Upsilon_{i-1}$, $\,\Sigma_{i}\subset\Sigma_{i-1}$.

Exactness of this complex gives an inductive step for the verification that the lattice of subspaces of $M$ generated by $U_{1},\ldots,U_{n-1}$ is distributive [18, Ch. 1, Prop. 7.2]. Complexes of this kind are responsible for Koszulness of the graded algebras, as discussed in the next section.

We are interested in the case when $M$ is a left module over the Hecke algebra ${\cal H}_{n}={\cal H}_{n}(q)$ and the subspaces $U_{1},\ldots,U_{n-1}$ are defined by one of the two conditions below:

(a)  $U_{i}=\mathop{\rm Ker}\,(T_{i}-q)_{M}$ for each $i$,

(b)  $U_{i}=\mathop{\rm Im}\,(T_{i}+1)_{M}$ for each $i$.

where we denote by $x_{M}$ the linear operator by which an element $x\in{\cal H}_{n}$ acts on $M$. Since $(T_{i}-q)(T_{i}+1)=0$, we have

[TABLE]

both in (a) and (b). If $q\neq-1$ then (a) is equivalent to (b), and so there is a difference between the two conditions only when $q=-1$.

For each $i=0,\ldots,n$ we have identified ${{S}}_{i}$ with the subgroup of ${{S}}_{n}$ generated by the set of basic transpositions $\{\tau_{j}\mid 0<j<i\}$. Denote by ${{S}}^{\mathchar 1103\relax}_{i}$ the subgroup of ${{S}}_{n}$ generated by $\{\tau_{j}\mid i<j<n\}$. Thus ${{S}}^{\mathchar 1103\relax}_{i}\cong{{S}}_{n-i}$. In particular, ${{S}}^{\mathchar 1103\relax}_{i}$ is the trivial subgroup $e$ for $i=n$ and for $i=n-1$.

Lemma 2.1. For $0\leq i<n$ consider the following elements of ${\cal H}_{n}\,$:**

[TABLE]

where ${\cal D}_{i}={\cal D}({{S}}_{i+1}/{{S}}_{i})$ and ${\cal D}^{\mathchar 1103\relax}_{i}={\cal D}({{S}}^{\mathchar 1103\relax}_{i+1}\backslash{{S}}^{\mathchar 1103\relax}_{i})$ are the sets of distinguished coset representatives. Then

(i) $x_{i}$ maps $\Upsilon_{i}$ to $\Upsilon_{i+1}$ and $y_{i}$ maps $\Sigma_{i}$ to $\Sigma_{i+1},$

(ii) $x_{i}y_{i}$ induces a linear map $s_{i}:K_{i}\to K_{i+1},$

(iii) $\partial_{i+1}s_{i}+s_{i-1}\partial_{i}=[n]_{q}\mathop{\rm Id}\nolimits,\,$ assuming that $s_{-1}=0$.

If $[n]_{q}\neq 0$ then the complex K_{\bullet}\bigl{(}M;(U_{i})\bigr{)} is exact.

Proof. The inclusions $x_{i}\Upsilon_{i}\subset\Upsilon_{i+1}$ and $y_{i}\Sigma_{i}\subset\Sigma_{i+1}$ are special cases of Lemma 2.2 (see below) applied, respectively, to the pairs of Young subgroups ${{S}}_{i}\subset{{S}}_{i+1}$ and ${{S}}^{\mathchar 1103\relax}_{i+1}\subset{{S}}^{\mathchar 1103\relax}_{i}$. The subspace $U_{k}$ is stable under the action of any $T_{j}$ with $|j-k|>1$ since $T_{j}T_{k}=T_{k}T_{j}$. It follows that $\Upsilon_{i}$ is stable under any $T_{\sigma}$ with $\sigma\in{{S}}^{\mathchar 1103\relax}_{i}$, and therefore $y_{i}\Upsilon_{i}\subset\Upsilon_{i}$. On the other hand, $\Sigma_{i+1}$ is stable under any $T_{\sigma}$ with $\sigma\in{{S}}_{i+1}$, which yields $x_{i}\Sigma_{i+1}\subset\Sigma_{i+1}$. Hence $x_{i}y_{i}$ maps $\Upsilon_{i}$ to $\Upsilon_{i+1}$ and $\Sigma_{i}$ to $\Sigma_{i+1}$. Thus (i) and (ii) have been checked. Next, note that

[TABLE]

Suppose that $0<i<n$. Then ${\cal D}_{i}=\{e\}\cup{\cal D}_{i-1}\tau_{i}$ and ${\cal D}^{\mathchar 1103\relax}_{i-1}=\{e\}\cup\tau_{i}{\cal D}^{\mathchar 1103\relax}_{i}\,$. Since $T_{e}=1$, $\,T_{\sigma\tau_{i}}=T_{\sigma}T_{i}$ for $\sigma\in{\cal D}_{i-1}$ and $T_{\tau_{i}\sigma}=T_{i}T_{\sigma}$ for $\sigma\in{\cal D}^{\mathchar 1103\relax}_{i}$, we get

[TABLE]

If $j>i$ then $T_{j}v+v\in(T_{j}+1)M\subset U_{j}\subset\Sigma_{i}$ for all $v\in M$. Hence $T_{\sigma}v\equiv(-1)^{\ell(\sigma)}v$ modulo $\Sigma_{i}$ for $\sigma\in{\cal D}^{\mathchar 1103\relax}_{i}$, and therefore

[TABLE]

If $j<i$, then $T_{j}v=qv$ for all $v\in\Upsilon_{i}$. Hence $T_{\sigma}v=q^{\ell(\sigma)}v$ for $\sigma\in{\cal D}_{i-1}$, and therefore $\,x_{i-1}v=\sum\limits_{k=0}^{i-1}q^{k}v\,$. It follows that

[TABLE]

for all $v\in\Upsilon_{i}$. Since the map $\partial_{i+1}s_{i}+s_{i-1}\partial_{i}:K_{i}\to K_{i}$ is induced by the action of the element $x_{i}y_{i}+x_{i-1}y_{i-1}$ on $M$, this proves (iii) when $0<i<n$.

Note also that $x_{0}=1$, while $y_{0}v\equiv[n]_{q}v$ modulo $\Sigma_{0}$ for all $v\in M$. At the upper boundary $y_{n-1}=1$, while $x_{n-1}v=[n]_{q}v$ for all $v\in\Upsilon_{n}$. Hence $\partial_{1}s_{0}=[n]_{q}\mathop{\rm Id}\nolimits$ on $K_{0}$ and $s_{n-1}\partial_{n}=[n]_{q}\mathop{\rm Id}\nolimits$ on $K_{n}$, which yields (iii) for $i=0$ and $i=n$. The final conclusion is immediate from (iii).   $\mathchar 9219\relax$

Lemma 2.2. For each Young subgroup ${{S}}_{\lambda}$ of ${{S}}_{n}$ put

[TABLE]

Suppose that ${{S}}_{\lambda}$ and ${{S}}_{\mu}$ are two Young subgroups such that ${{S}}_{\lambda}\subset{{S}}_{\mu}$. Then

[TABLE]

where  $x=\sum_{\sigma\in{\cal D}({{S}}_{\mu}/{{S}}_{\lambda})}T_{\sigma}$  and  $y=\sum_{\sigma\in{\cal D}({{S}}_{\lambda}\backslash{{S}}_{\mu})}(-1)^{\ell(\sigma)}q^{m-\ell(\sigma)}T_{\sigma}$  with $m=\max\{\ell(\sigma)\mid\sigma\in{\cal D}({{S}}_{\lambda}\backslash{{S}}_{\mu})\}$.

Proof. Fix any basic transposition $\tau_{i}\in{{B}}_{\mu}$. As in section 1 we have

[TABLE]

(disjoint unions) where $A={{S}}_{\mu}\cap A_{i}(\lambda)$ and $B={{S}}_{\mu}\cap B_{i}(\lambda)$. Hence

[TABLE]

Recall that $(T_{i}+1)M\subset U_{i}$. Suppose that $\sigma\in B$. Then $\tau_{i}\sigma=\sigma\tau_{j}>\sigma$ for some $\tau_{j}\in{{B}}_{\lambda}$, and therefore $T_{i}T_{\sigma}=T_{\sigma}T_{j}$. For each $v\in\Upsilon(\lambda)$ we have $T_{i}T_{\sigma}v=qT_{\sigma}v$ since $T_{j}v=qv$. Moreover, in case (b) $v=(T_{j}+1)w$ for some $w\in M$, and then $T_{\sigma}v=(T_{i}+1)T_{\sigma}w$. It follows that $T_{\sigma}v\in U_{i}$ both in (a) and (b). Since this inclusion holds for each $\sigma\in B$, we deduce that $x\Upsilon(\lambda)\subset U_{i}$.

Similarly, if $\sigma\in B^{-1}$, then $T_{\sigma}T_{i}=T_{j}T_{\sigma}$ for some $j$ such that $\tau_{j}\in{{B}}_{\lambda}$. For each $v\in U_{i}$ we have $T_{i}v=qv$, whence $T_{j}T_{\sigma}v=qT_{\sigma}v$. Moreover, in case (b) $v=(T_{i}+1)w$ for some $w\in M$, and then $T_{\sigma}v=(T_{j}+1)T_{\sigma}w$. Hence $T_{\sigma}v\in U_{j}\subset\Sigma(\lambda)$ both in (a) and (b). Since $q-T_{i}$ annihilates $U_{i}$, it follows that $yU_{i}\subset\Sigma(\lambda)$.   $\mathchar 9219\relax$

Lemma 2.3. Let $\chi:{\cal H}_{\lambda}\to{k}$ be a $1$-dimensional representation of a parabolic subalgebra ${\cal H}_{\lambda}$ of ${\cal H}_{n}$. Consider the standard basis $\{v_{\sigma}\mid\sigma\in{\cal D}_{\lambda}\}$ of the induced ${\cal H}_{n}$-module $M={\cal H}_{n}\otimes_{{\cal H}_{\lambda}}{k}(\chi)$ where $v_{\sigma}=T_{\sigma}c$ in the notation of section $1$. Then $\mathop{\rm Ker}\,(T_{i}-q)_{M}$ is spanned by the elements

[TABLE]

and $\,\mathop{\rm Im}\,(T_{i}+1)_{M}$ is spanned by the elements

[TABLE]

Proof. The module $M$ is a direct sum of $T_{i}$-invariant subspaces $M(\sigma)$ with $\sigma$ in $A_{i}(\lambda)\cup B_{i}(\lambda)$ where $M(\sigma)$ is spanned by two elements $v_{\sigma}$, $v_{\tau_{i}\sigma}$ for $\sigma\in A_{i}(\lambda)$ and by the single element $v_{\sigma}$ for $\sigma\in B_{i}(\lambda)$. Obviously $\mathop{\rm Ker}\,(T_{i}-q)_{M}$ and $\mathop{\rm Im}\,(T_{i}+1)_{M}$ are sums of their intersections with those subspaces. From the formulas for the action of $T_{i}$ on $M$ it is clear that

[TABLE]

when $\sigma\in A_{i}(\lambda)$. If $\sigma\in B_{i}(\lambda)$, then $v_{\sigma}$ is an eigenvector for the operator $(T_{i})_{M}$ with the eigenvalue $\chi_{\sigma}(T_{i})$. In this case $v_{\sigma}\in\mathop{\rm Ker}\,(T_{i}-q)_{M}$ if and only if $\chi_{\sigma}(T_{i})=q$, and $v_{\sigma}\in\mathop{\rm Im}\,(T_{i}+1)_{M}$ if and only if $\chi_{\sigma}(T_{i})\neq-1$.   $\mathchar 9219\relax$

By Lemma 2.1 the complex K_{\bullet}\bigl{(}M;(U_{i})\bigr{)} is exact for any left ${\cal H}_{n}$-module $M$ when $[n]_{q}\neq 0$. If $q=0$, then this result does apply since $[n]_{q}=1$ in this case. For arbitrary $q$ we have to restrict the class of ${\cal H}_{n}$-modules:

Proposition 2.4. Let $M$ be a finite dimensional ${\cal H}_{n}$-module whose indecomposable direct summands all have $1$-dimensional sources. With subspaces $U_{1},\ldots,U_{n-1}$ defined by either (a) or (b) the complex K_{\bullet}\bigl{(}M;(U_{i})\bigr{)} is exact.

Proof. Note that each $U_{i}$ depends on $M$ functorially, and therefore the construction of K_{\bullet}\bigl{(}M;(U_{i})\bigr{)} gives a functor from the category of ${\cal H}_{n}$-modules to the category of complexes. Since this functor is additive, the conclusion of Proposition 2.4 holds for any given ${\cal H}_{n}$-module $M$ if and only if the conclusion holds for each indecomposable direct summand of $M$. This shows that it suffices to give the proof assuming that $M={\cal H}_{n}\otimes_{{\cal H}_{\lambda}}{k}(\chi)$ for some $1$-dimensional representation $\chi:{\cal H}_{\lambda}\to{k}$ of a parabolic subalgebra of ${\cal H}_{n}$.

By Lemma 2.1 the conclusion is true when $q=0$. Suppose that $q\neq 0$. Consider the parabolic subalgebra ${\cal H}_{\lambda}(0)$ of the 0-Hecke algebra ${\cal H}_{n}(0)$ corresponding to the same composition $\lambda$ of $n$. Let $\mathaccent 23{T}_{1},\ldots,\mathaccent 23{T}_{n-1}$ stand for the canonical generators of ${\cal H}_{n}(0)$. There is a $1$-dimensional representation $\xi:{\cal H}_{\lambda}(0)\to{k}$ defined on the generators $\{\mathaccent 23{T}_{i}\mid 0<i<n,\ \tau_{i}\in{{S}}_{\lambda}\}$ of ${\cal H}_{\lambda}(0)$ as follows. When $q\neq-1$ set

[TABLE]

When $q=-1$ set

[TABLE]

Let $\{v_{\sigma}\mid\sigma\in{\cal D}_{\lambda}\}$ be the standard basis of $M$ and $\{v^{0}_{\sigma}\mid\sigma\in{\cal D}_{\lambda}\}$ a similar basis of the induced ${\cal H}_{n}(0)$-module $M^{0}={\cal H}_{n}(0)\otimes_{{\cal H}_{\lambda}(0)}{k}(\xi)$. The assignments $v_{\sigma}\mapsto v^{0}_{\sigma}$ define a linear isomorphism $M\cong M^{0}$. In view of Lemma 2.3 the subspace $U_{i}$ of $M$ is mapped onto a similar subspace $U^{0}_{i}=\{v\in M^{0}\mid\mathaccent 23{T}_{i}v=0\}$ of $M^{0}$, for each $i=1,\ldots,n-1$. Hence the complex K_{\bullet}\bigl{(}M;(U_{i})\bigr{)} is isomorphic to K_{\bullet}\bigl{(}M^{0};(U^{0}_{i})\bigr{)}. But the latter complex is exact, as we have observed already.   $\mathchar 9219\relax$

Corollary 2.5. Under the same assumption about $M$ the complex K_{\bullet}\bigl{(}M;(U_{i})\bigr{)} is exact also when $U_{i}=(T_{i}-q)M$ for each $i=1,\ldots,n-1$.

Proof. Let $\mathaccent 869{M}$ be $M$ with the ${\cal H}_{n}$-module structure twisted by the automorphism of ${\cal H}_{n}$ sending $T_{i}$ to $q-1-T_{i}$ for each $i$. The ${\cal H}_{n}$-module $\mathaccent 869{M}$ has the same submodules as $M$, but with the twisted action of ${\cal H}_{n}$. So it follows from Lemma 1.5 that all indecomposable direct summands of $\mathaccent 869{M}$ have $1$-dimensional sources provided this holds for $M$. Since $T_{i}-q$ acts on $M$ as $-(T_{i}+1)$ acts on $\mathaccent 869{M}$, we have $U_{i}=(T_{i}+1)\mathaccent 869{M}$. Therefore Corollary 2.5 follows from Proposition 2.4 applied to $\mathaccent 869{M}$.   $\mathchar 9219\relax$

If $[n]_{q}\neq 0$ then the conclusion of Corollary 2.5 holds without any restriction on $M$ in view of Lemma 2.1. In particular, this is true for $q=0$.

3. Koszulness of the $R$-symmetric algebras

Let $A=\bigoplus_{n=0}^{\infty}A_{n}$ be a quadratic graded algebra generated by some vector space $V=A_{1}$. This means that $A\cong{{T}}(V)/I$ where $I$ is the ideal of the tensor algebra ${{T}}(V)$ generated by a vector subspace $U\subset V^{\otimes 2}$. The books [16] and [18] provide general reference on quadratic algebras. For each $n>1$ and $0<i<n$ put

[TABLE]

and $\Upsilon^{(n)}=\bigcap_{i=1}^{n-1}U_{i}^{(n)}$. Put also $\Upsilon^{(0)}={k}$ and $\Upsilon^{(1)}=V$. The right Koszul complex $K_{\bullet}(A)$ is the complex of right $A$-modules

[TABLE]

where $\partial_{i}\,$, for each $i>0$, is the restriction of the $A$-linear map

[TABLE]

for $t\in V^{\otimes(i-1)}$, $v\in V$ and $a\in A$. The grading of $A$ gives rise to a decomposition of $K_{\bullet}(A)$ into a direct sum of subcomplexes

[TABLE]

There is an isomorphism of complexes K_{\bullet}^{(n)}(A)\cong K_{\bullet}\bigl{(}V^{\otimes n};(U_{i}^{(n)})\bigr{)}, the latter having been defined in section 2. Indeed, letting  $\Upsilon_{i}^{(n)}=\bigcap_{j<i}U_{j}^{(n)}$,  $\Sigma_{i}^{(n)}=\sum_{j>i}U_{j}^{(n)}$, we have

[TABLE]

Since $\Sigma_{0}^{(k)}=\sum_{j=1}^{k-1}U_{j}^{(k)}$ coincides with the $k$th homogeneous component of the ideal $I$, it follows that $V^{\otimes k}/\Sigma_{0}^{(k)}\cong A_{k}$ for each $k$. Hence

[TABLE]

It is easy to see that these linear isomorphisms between the homogeneous components of the two complexes are compatible with the differentials.

Note that H_{0}\bigl{(}K_{\bullet}(A)\bigr{)}\cong A/A_{+}\cong{k} where $A_{+}=VA=\sum_{n>0}A_{n}$. The algebra $A$ is said to be Koszul if the complex $K_{\bullet}(A)$ is acyclic in all positive degrees, i.e. $K_{\bullet}(A)$ is a resolution of the trivial right $A$-module ${k}$. There are several equivalent characterizations of this property (see [18]). By a fundamental result of Backelin [1] $A$ is Koszul if and only if $U_{1}^{(n)},\ldots,U_{n-1}^{(n)}$ generate a distributive lattice of subspaces of $V^{\otimes n}$ for each $n>1$. From this it is easy to see that Koszulness of an algebra is a left-right symmetric property.

Assume further on that $\dim V<\infty$. Then $\dim A_{n}<\infty$ for all $n$. The Hilbert series $h_{A}(t)$ of $A$ is the formal power series in one indeterminate $t$ whose coefficients are the dimensions of the homogeneous components $A_{n}$:

[TABLE]

For each $n$ identify ${{T}}_{n}(V^{*})$ with the dual of the vector space ${{T}}_{n}(V)$ using the bilinear pairing

[TABLE]

The quadratic dual $A^{!}$ of $A$ is the factor algebra of the tensor algebra ${{T}}(V^{*})$ by the ideal generated by the subspace

[TABLE]

The $n$th homogeneous component of this ideal is then the subspace

[TABLE]

Hence $A^{!}_{n}={{T}}_{n}(V^{*})/(\Upsilon^{(n)})^{\perp}\cong(\Upsilon^{(n)})^{*}$, and it follows that

[TABLE]

If $A$ is Koszul, then $\sum_{i=0}^{n}(-1)^{i}(\dim A_{i})(\dim\Upsilon^{(n-i)})=0$ for each $n>0$, which entails the well-known relation between the Hilbert series of $A$ and $A^{!}\,$:

[TABLE]

Let $R$ be a Hecke symmetry on a vector space $V$, and let $0\neq q\in{k}$ be the parameter of the Hecke relation satisfied by $R$. For each $n\geq 0$ there is a representation of the Hecke algebra ${\cal H}_{n}={\cal H}_{n}(q)$ in ${{T}}_{n}(V)$ such that each generator $T_{i}\,$, $\,0<i<n$, acts by means of the linear operator

[TABLE]

Recall that ${\cal H}_{0}={\cal H}_{1}={k}$. We assume that $\dim V<\infty$.

The algebras ${{S}}(V,R)$ and $\Lambda(V,R)$ are defined as the factor algebras of ${{T}}(V)$ by the ideals generated, respectively, by the subspaces

[TABLE]

Theorem 3.1. Suppose that $R$ satisfies the $1$-dimensional source condition. Then the $R$-symmetric algebra ${{S}}(V,R)$ and the $R$-skewsymmetric algebra $\Lambda(V,R)$ are Koszul. Their Hilbert series satisfy the relation  $h_{{{S}}(V,R)}(t)\,h_{\Lambda(V,R)}(-t)=1$.

Proof. In terms of the ${\cal H}_{n}$-module structure arising from $R$ the previously defined subspaces of ${{T}}_{n}(V)$ are

[TABLE]

The assumption about $R$ means that each indecomposable direct summand of the ${\cal H}_{n}$-module ${{T}}_{n}(V)$ has a $1$-dimensional source. Hence the complex K_{\bullet}\bigl{(}V^{\otimes n},(U_{i}^{(n)})\bigr{)} is exact, for each $n>0$, by Proposition 2.4 and Corollary 2.5. Hence so is the isomorphic complex $K_{\bullet}^{(n)}(A)$. Since $K_{\bullet}^{(0)}(A)$ is concentrated in degree 0, it follows that H_{i}\bigl{(}K_{\bullet}(A)\bigr{)}=H_{i}\bigl{(}K_{\bullet}^{(0)}(A)\bigr{)}=0 for all $i>0$.

Let $\Upsilon^{(n)}=\bigcap_{i=1}^{n-1}U_{i}^{(n)}$ be defined with respect to $A=\Lambda(V,R)$, i.e.

[TABLE]

This is the largest subspace of ${{T}}_{n}(V)$ on which ${\cal H}_{n}$ operates trivially. On the other hand, ${{S}}_{n}(V,R)$ is the largest factor space of ${{T}}_{n}(V)$ on which ${\cal H}_{n}$ operates trivially. It follows that

[TABLE]

whence $\dim\Lambda^{!}_{n}(V,R)=\dim{{S}}_{n}(V,R)$ according to Lemma 1.2. The relation between the Hilbert series of ${{S}}(V,R)$ and $\Lambda(V,R)$ follows now from the relation between $h_{A}$ and $h_{A^{!}}$.   $\mathchar 9219\relax$

There are several transformations of the Hecke symmetry $R$. Put

[TABLE]

where $\tau$ is the flip operator $v_{1}\otimes v_{2}\mapsto v_{2}\otimes v_{1}$ on ${{T}}_{2}(V)$. Let $R^{*}$ be the linear operator on ${{T}}_{2}(V^{*})\cong{{T}}_{2}(V)^{*}$ adjoint to $R$. Given an algebra $A$, we denote by $A^{\rm op}$ the algebra with the same set of elements but with the opposite multiplication.

Lemma 3.2. The operators $\mathaccent 869{R},$ $R^{\rm op},$ $R^{*}$ are Hecke symmetries with the same parameter $q$ as $R$. We have

[TABLE]

If $q\neq-1,$ then $\,{{S}}(V,\mathaccent 869{R})=\Lambda(V,R)\,$ and $\,\Lambda(V,\mathaccent 869{R})={{S}}(V,R)\,$.

All verifications are straightforward. Note also that these Hecke symmetries $\mathaccent 869{R}$, $R^{\rm op}$, $R^{*}$ satisfy the $1$-dimensional source condition provided so does $R$. If $R$ satisfies the trivial source condition, so do $R^{\rm op}$ and $R^{*}$. The inverse operator $R^{-1}$ is a Hecke symmetry with parameter $q^{-1}$ giving rise to the same pair of quadratic graded algebras as the pair ${{S}}(V,R)$, $\Lambda(V,R)$ obtained from $R$.

We end this section with an example showing that ${{S}}(V,R)$, $\Lambda(V,R)$ are not always Koszul. Let $V$ be a 2-dimensional vector space with a basis $x,y$. Assume that $\mathop{\rm char}\nolimits{k}\neq 2$. We start with the $R$-matrix $R_{H0.2}$ in the notation of Hietarinta [12, p. 1732]:

[TABLE]

In a slightly different form an equivalent matrix appeared under the label $R_{1}$ in the list of Hlavatý [13, p. 1663]. This matrix represents an operator satisfying the quantum Yang-Baxter equation. Composing with the flip of tensorands we obtain a diagonalizable operator satisfying the braid equation whose characteristic polynomial is $(t^{2}-2t+2)^{2}$. A final scaling yields the matrix

[TABLE]

of a Hecke symmetry with eigenvalues $-1$, $q$ where $q$ is a primitive 4th root of 1. The matrix is written in the basis $x^{2},xy,yx,y^{2}$ of ${{T}}_{2}(V)$. One eigenspace of $R$ is spanned by $y^{2}-qx^{2}$, $xy-qyx$. It gives the defining relations $y^{2}=qx^{2}$, $xy=qyx$ of ${{S}}(V,R)$. The first relation shows that $y^{2}$ is central in ${{S}}(V,R)$. But $xy^{2}=-y^{2}x$ according to the second relation, whence $xy^{2}=0$. From this it is clear that ${{S}}_{3}(V,R)=0$.

Similarly, the algebra $\Lambda(V,R)$ has the defining relations $x^{2}=qy^{2}$, $yx=qxy$. It is isomorphic to ${{S}}(V,R)$. Hence

[TABLE]

Moreover, the quadratic dual algebras are isomorphic to the original ones. Thus the standard relation between their Hilbert series is not satisfied.

In this example ${\cal H}_{4}$ is the first nonsemisimple algebra in the family of Hecke algebras. Since all its proper parabolic subalgebras are semisimple, the indecomposable ${\cal H}_{4}$-modules with a 1-dimensional source are either 1-dimensional or projective. It can be checked that the ${\cal H}_{4}$-module $V^{\otimes 4}$ is a direct sum of simple 2-dimensional submodules. There are two nonisomorphic simple modules of dimension 2. One of them is projective, but the other is not. Thus $R$ does not satisfy the 1-dimensional source condition.

4. Nondegeneracy of the multiplication maps

The aim of this section is to prove that the $R$-skewsymmetric algebra $\Lambda=\Lambda(V,R)$ is Frobenius under suitable assumptions. Recall that $\Lambda={{T}}/I$ where ${{T}}={{T}}(V)$ is the tensor algebra of $V$ and $I$ is its homogeneous ideal generated by the subspace $U=\mathop{\rm Ker}\,(R-q\cdot\mathop{\rm Id}\nolimits)$ of ${{T}}_{2}$. We will be omitting the sign $\otimes$ when referring to the multiplication in ${{T}}$.

The next lemma provides the main step in tackling the problem. Recall that we denote by ${{S}}_{1,k-1}$ the subgroup of ${{S}}_{k}$ generated by $\{\tau_{i}\mid 1<i<k\}$.

Lemma 4.1. Fix some $n>1$ and put

[TABLE]

If $0<k<n,$ then $y_{k}$ maps $L_{k}$ into $VL_{k-1}$.

Proof. Note that $VL_{k-1}=\{b\in{{T}}_{k}\mid b{{T}}_{n-k+1}\subset VI_{n}\}$. Thus we have to show that $(y_{k}L_{k}){{T}}_{n-k+1}\subset VI_{n}$. Let $a\in L_{k}$. Then $a{{T}}_{n-k}\subset I_{n}$, whence

[TABLE]

We will work inside the ${\cal H}_{n+1}$-module ${{T}}_{n+1}$. In conformance with the notation of section 2 put $U_{i}=\mathop{\rm Ker}\,(T_{i}-q)_{{{T}}_{n+1}}$ for each $i=1,\ldots,n$. Note that

[TABLE]

By Lemma 2.2 applied with $\,{{S}}_{\lambda}={{S}}_{1,n}$, $\,{{S}}_{\mu}={{S}}_{n+1}\,$ and $n$ replaced by $n+1$ we get $\,y_{n+1}I_{n+1}\subset VI_{n}\,$. Therefore

[TABLE]

Since ${\cal D}({{S}}_{1,k-1}\backslash{{S}}_{k})=\{e,\tau_{1},\tau_{1}\tau_{2}\,,\ldots,\tau_{1}\tau_{2}\cdots\tau_{k-1}\}$, we have

[TABLE]

In particular, $y_{n+1}=\sum_{i=0}^{n}(-1)^{i}q^{n-i}p_{i}$. As follows immediately from the braid relations between $T_{1},\ldots,T_{n}$, the element $p_{n}$ has the property that $T_{i+1}p_{n}=p_{n}T_{i}$, and therefore

[TABLE]

for each $i=1,\ldots,n-1$. Since $p_{n}$ is invertible in ${\cal H}_{n+1}$, we deduce that $p_{n}U_{i}=U_{i+1}$ for those values of $i$. Hence $p_{n}$ maps $I_{n}V$, and in particular the subspace $a{{T}}_{n-k+1}$, into $VI_{n}$.

If $i<n$, then $p_{i}=p_{n}T_{n}^{-1}\cdots T_{i+1}^{-1}$. For each $j=k+1,\ldots,n$ the element $T_{j}$ operates on the second space in the decomposition ${{T}}_{n+1}={{T}}_{k}\otimes{{T}}_{n-k+1}$, which implies that $a{{T}}_{n-k+1}=a\otimes{{T}}_{n-k+1}$ is stable under the action of $T_{j}$ and $T_{j}^{-1}$. It follows that

[TABLE]

Now $\,y_{k}=q^{k-1-n}(y_{n+1}-\sum_{i=k}^{n}(-1)^{i}q^{n-i}p_{i})$. The previous inclusions entail

[TABLE]

Since for each $j=1,\ldots,k-1$ the element $T_{j}$ operates on the first space in the decomposition ${{T}}_{n+1}={{T}}_{k}\otimes{{T}}_{n-k+1}$, so too does $y_{k}$. Hence

[TABLE]

yielding $\,y_{k}a\in VL_{k-1}$.   $\mathchar 9219\relax$

The reader should note that the proof of Lemma 4.1 uses only the braid relations between $T_{1},\ldots,T_{n}$. Therefore Lemma 4.1 holds more generally when $R$ is a linear operator on $V^{\otimes 2}$ satisfying the braid equation but not necessarily the quadratic Hecke equation, and $q$ is any eigenvalue of $R$ used in the definition of $\Lambda$. Such an operator (called a Yang-Baxter operator in the literature) gives rise to representations of the Artin braid groups $B_{k}$. The elements $T_{\sigma}$ with $\sigma\in{{S}}_{k}$ make sense in $B_{k}$, and the element $y_{k}$ is defined for each $k$ in the group algebra of $B_{k}$. This observation will be essential later (see Lemma 6.5).

Lemma 4.2. Suppose that $\Lambda_{n}\neq 0$ for some $n>1$. If $0\neq a\in\Lambda_{k}$ where either $k=1$ or $k=2$, then $\,a\Lambda_{n-k}\neq 0\,$ and $\,\Lambda_{n-k}\,a\neq 0$.

Proof. In the notation of Lemma 4.1 $L_{0}=0$ since ${{T}}_{0}={k}$ and ${{T}}_{n}\neq I_{n}$. By Lemma 4.1 $y_{1}L_{1}\subset VL_{0}=0$ and $y_{2}L_{2}\subset VL_{1}$. Since $y_{1}=1$, we get $L_{1}=0$. Hence $y_{2}L_{2}=0$. This means that $L_{2}\subset\mathop{\rm Ker}\,(T_{1}-q)_{{{T}}_{2}}=I_{2}$ since $y_{2}=q-T_{1}$. In fact $L_{2}=I_{2}$ since the opposite inclusion is obvious. Thus $L_{k}=I_{k}$ for $k=1$ and for $k=2$. Passing to the factor algebra $\Lambda={{T}}/I$, we deduce that $\Lambda_{n-k}$ has zero left annihilator in $\Lambda_{k}$. In view of Lemma 3.2 we can apply this conclusion also to $\Lambda^{\rm op}$. Hence $\Lambda_{n-k}$ has zero right annihilator in $\Lambda_{k}$.   $\mathchar 9219\relax$

Corollary 4.3. Suppose that $\dim\Lambda_{n}=1$ and $\Lambda_{n+1}=0$. If $n=2$ or $n=3$, then $\Lambda$ is a Frobenius algebra.

Proof. By Lemma 4.2 the bilinear pairings $\Lambda_{k}\times\Lambda_{n-k}\to\Lambda_{n}$ arising from the multiplication in $\Lambda$ are nondegenerate for $k=1$ and $k=2$.   $\mathchar 9219\relax$

Lemma 4.4. Let $M$ be a finite dimensional ${\cal H}_{n}$-module whose indecomposable direct summands all have trivial sources. Let ${{S}}_{\lambda}$ be a Young subgroup of ${{S}}_{n}$ and

[TABLE]

Put $\Sigma(\lambda)=\sum_{\{i\,\mid\,\tau_{i}\in{{B}}_{\lambda}\}}U_{i}$ and $\Sigma(n)=\sum_{i=1}^{n-1}U_{i}$ where $U_{i}=\mathop{\rm Ker}\,(T_{i}-q)_{M}$ for each $i$. Then

[TABLE]

Proof. The inclusion $y\,\Sigma(n)\subset\Sigma(\lambda)$ is a special case of Lemma 2.2. So we have only to prove that $a\in\Sigma(n)$ for each $a\in M$ such that $ya\in\Sigma(\lambda)$. This assertion holds for any given ${\cal H}_{n}$-module $M$ if and only if it holds for each indecomposable direct summand of $M$. Therefore we may assume that $M={\cal H}_{n}\otimes_{{\cal H}_{\nu}}{k}_{\rm triv}$ where ${\cal H}_{\nu}$ is a parabolic subalgebra of ${\cal H}_{n}$.

If ${\cal H}_{\nu}\neq{k}$, then $T_{j}\in{\cal H}_{\nu}$ for some $j$. In this case $T_{j}c=qc$ where $c$ is the canonical generator of $M$, whence $c\in U_{j}\subset\Sigma(n)$. Observing that $\,(T_{i}+1)M\subset U_{i}\subset\Sigma(n)\,$ for each $i=1,\ldots,n-1$, we see that $\Sigma(n)$ is stable under the action of all $T_{1},\ldots,T_{n-1}$, i.e. $\Sigma(n)$ is an ${\cal H}_{n}$-submodule of $M$. But then $\Sigma(n)=M$, and the desired conclusion is obviously true.

It remains to consider the case when ${\cal H}_{\nu}={k}$, and therefore $M={\cal H}_{n}$ is a cyclic free ${\cal H}_{n}$-module. Since ${\cal H}_{n}$ is a free module over its subalgebra ${k}+{k}T_{i}$, we deduce that $U_{i}=(T_{i}+1)M$ for each $i=1,\ldots,n-1$. Hence $M/\Sigma(n)$ is the largest factor module of $M$ on which each $T_{i}$ operates as minus identity transformation. Clearly

[TABLE]

which shows that $\Sigma(n)$ is a subspace of codimension 1 in $M$.

The subspace $Y=\{a\in M\mid ya\in\Sigma(\lambda)\}$ contains $\Sigma(n)$ by Lemma 2.2. Suppose that $Y\neq\Sigma(n)$. Then we must have $Y=M$. In particular, $1\in Y$, which means that $y\in\Sigma(\lambda)$. However,

[TABLE]

where $J$ is the ideal of ${\cal H}_{\lambda}$ generated by $\{T_{i}+1\mid\tau_{i}\in{{B}}_{\lambda}\}$, i.e. $J$ is the annihilator of the alternating representation of ${\cal H}_{\lambda}$. Since ${\cal H}_{n}$ is a free left ${\cal H}_{\lambda}$-module with a basis $\{T_{\sigma}\mid\sigma\in{\cal D}({{S}}_{\lambda}\backslash{{S}}_{n})\}$, the inclusion $y\in J{\cal H}_{n}$ entails $1\in J$, which is impossible. This contradiction proves that $Y=\Sigma(n)$.   $\mathchar 9219\relax$

Theorem 4.5. Suppose that $R$ satisfies the trivial source condition. Suppose also that $\dim\Lambda_{n}(V,R)=1$ and $\Lambda_{n+1}(V,R)=0$ for some $n>0$. Then $\Lambda(V,R)$ is a Frobenius algebra,* while ${{S}}(V,R)$ is a Gorenstein algebra of global dimension $n$.*

Proof. The left kernel of the bilinear pairing $\Lambda_{k}\times\Lambda_{n-k}\to\Lambda_{n}$ arising from the multiplication in $\Lambda$ is nothing else but the image of $L_{k}$ in $\Lambda_{k}={{T}}_{k}/I_{k}$ where $L_{k}$ is the subspace of ${{T}}_{k}$ introduced in Lemma 4.1. To show that the left kernel vanishes we have to prove that $L_{k}=I_{k}$. But this can be done by induction on $k$. Indeed, if $L_{k-1}=I_{k-1}$, then $y_{k}L_{k}\subset VI_{k-1}$ by Lemma 4.1. Note that

[TABLE]

The indecomposable direct summands of the ${\cal H}_{k}$-module $M={{T}}_{k}$ all have trivial sources by the assumption about $R$, and we can apply Lemma 4.4 with $n$ replaced by $k$ and ${{S}}_{\lambda}={{S}}_{1,k-1}$. In this case $y=y_{k}$, $\Sigma(\lambda)=VI_{k-1}$, $\Sigma(k)=I_{k}$, and the conclusion of Lemma 4.4 gives the desired inclusion $L_{k}\subset I_{k}$.

Thus the pairings $\Lambda_{k}\times\Lambda_{n-k}\to\Lambda_{n}$ have trivial left kernels for all $k=0,\ldots,n$. Then $\dim\Lambda_{k}\leq\dim\Lambda_{n-k}$, and since this inequality holds also with $k$ replaced by $n-k$, we have in fact an equality. Thus all the above pairings are nondegenerate, which is a necessary and sufficient condition for $\Lambda$ to be a Frobenius algebra.

By Theorem 3.1 $h_{\Lambda}(t)=h_{{{S}}}(-t)^{-1}=h_{{{S}}^{\mskip 1.0mu!}}(t)$ where ${{S}}={{S}}(V,R)$. Thus the two graded algebras $\Lambda$ and ${{S}}^{\mskip 1.0mu!}$ have homogeneous components of equal dimensions. In particular, $\dim{{S}}^{\mskip 1.0mu!}_{n}=1$ and ${{S}}^{\mskip 1.0mu!}_{n+1}=0$. But ${{S}}^{\mskip 1.0mu!}=\Lambda(V^{*}\!,R^{*})$ by Lemma 3.2, and the Hecke symmetry $R^{*}$ satisfies the trivial source condition. We deduce that ${{S}}^{\mskip 1.0mu!}$ is a Frobenius algebra by the already established part of Theorem 4.5. Since ${{S}}$ is Koszul, its Gorensteinness follows then from [18, Remark 2 on p. 25].   $\mathchar 9219\relax$

5. Auxiliary results for the tensor product of two Hecke algebras

This section collects several results needed to deal with the algebras $A(R^{\prime},R)$ and $E(R^{\prime},R)$ in the next section. One of our goals is to investigate exactness of the complexes K_{\bullet}\bigl{(}{\cal M};(U_{i})\bigr{)} for certain collections of subspaces in a module ${\cal M}$ over the tensor product ${\cal H}^{\prime}_{n}\otimes{\cal H}_{n}$ of two Hecke algebras ${\cal H}_{n}={\cal H}_{n}(q)$ and ${\cal H}^{\prime}_{n}={\cal H}_{n}(q^{-1})$. Here $q\neq 0$.

We identify ${\cal H}_{n}$ and ${\cal H}^{\prime}_{n}$ with their canonical images in ${\cal H}^{\prime}_{n}\otimes{\cal H}_{n}$. Denote by $T_{1},\ldots,T_{n-1}$ the standard generators of ${\cal H}_{n}$ and by $T^{\prime}_{1},\ldots,T^{\prime}_{n-1}$ those of ${\cal H}^{\prime}_{n}$. Put ${\cal T}_{i}=T_{i}T^{\prime}_{i}$, i.e.

[TABLE]

for $i=1,\ldots,n-1$. Since the elements of ${\cal H}_{n}$ commute with those of ${\cal H}^{\prime}_{n}$, the elements ${\cal T}_{1},\ldots,{\cal T}_{n-1}$ satisfy the braid relations. However, in general only the cubic relations $({\cal T}_{i}-1)({\cal T}_{i}+q)({\cal T}_{i}+q^{-1})=0$ hold rather than the quadratic ones.

Proposition 5.1. Suppose that ${\cal M}=M^{\prime}\otimes M$ where $M$ is an ${\cal H}_{n}$-module and $M^{\prime}$ is an ${\cal H}^{\prime}_{n}$-module such that all indecomposable direct summands of $M$ and $M^{\prime}$ have $1$-dimensional sources. Let $U_{i}=({\cal T}_{i}-1){\cal M}$ for each $i=1,\ldots,n-1$. Then the complex K_{\bullet}\bigl{(}{\cal M};(U_{i})\bigr{)} is exact.

Since direct sum decompositions of $M$ and $M^{\prime}$ give rise to a direct sum decomposition of K_{\bullet}\bigl{(}{\cal M};(U_{i})\bigr{)}, in proving the exactness of that complex we need only to consider the case when

[TABLE]

where $\,\chi:{\cal H}_{\lambda}\to{k}$ and $\,\chi^{\prime}:{\cal H}^{\prime}_{\mu}\to{k}$ are 1-dimensional representations of parabolic subalgebras ${\cal H}_{\lambda}\,$, ${\cal H}^{\prime}_{\mu}$ of the respective Hecke algebras. By abuse of notation we will use the same letter $\chi$ for the 1-dimensional representation ${\cal H}^{\prime}_{\mu}\otimes{\cal H}_{\lambda}\to{k}$ which restricts to the given representations of ${\cal H}_{\lambda}$ and ${\cal H}^{\prime}_{\mu}$. In particular, $\chi(x)=\chi^{\prime}(x)$ for $x\in{\cal H}^{\prime}_{\mu}$. The ${\cal H}^{\prime}_{n}\otimes{\cal H}_{n}$-module

[TABLE]

has a generator $c$ such that $T_{i}c=\chi(T_{i})c$ for all $i$ with $\tau_{i}\in{{B}}_{\lambda}$ and $T^{\prime}_{i}c=\chi(T^{\prime}_{i})c$ for all $i$ with $\tau_{i}\in{{B}}_{\mu}$. Consider the standard bases $\{T_{\sigma}\mid\sigma\in{{S}}_{n}\}$, $\,\{T^{\prime}_{\sigma}\mid\sigma\in{{S}}_{n}\}$ for ${\cal H}_{n}$ and ${\cal H}^{\prime}_{n}$. Then ${\cal M}$ has a vector space basis

[TABLE]

In ${\cal M}$ we obtain a filtration of vector subspaces $0=F_{-1}{\cal M}\subset F_{0}{\cal M}\subset F_{1}{\cal M}\subset\ldots$ taking $F_{p}{\cal M}$ to be the linear span of the elements

[TABLE]

Our strategy is to relate the question we study for ${\cal M}$ to a similar question for the associated graded vector space $\mathop{\rm gr}\nolimits^{F}\!{\cal M}$ (cf. [18, Ch. 1, Cor. 7.3]):

Lemma 5.2. Put $\Upsilon_{i}=\bigcap_{j<i}U_{j}$ and $\Sigma_{i}=\sum_{j>i}U_{j}$. Assume that

[TABLE]

for all $i=0,\ldots,n$. If the complex K_{\bullet}\bigl{(}\mathop{\rm gr}\nolimits^{F}\!{\cal M};(\mathop{\rm gr}\nolimits^{F}U_{i})\bigr{)} is exact, then so too is the complex K_{\bullet}\bigl{(}{\cal M};(U_{i})\bigr{)}.

Proof. All subspaces of ${\cal M}$ are endowed with the induced filtrations. Under the assumptions stated there is an exact sequence of complexes

[TABLE]

where the last complex is exact, for each $p\geq 0$. Induction on $p$ shows that the complex K_{\bullet}\bigl{(}F_{p}{\cal M};(F_{p}U_{i})\bigr{)} is exact. But $F_{p}{\cal M}={\cal M}$, $F_{p}U_{i}=U_{i}$ for large $p$.   $\mathchar 9219\relax$

With $M$ and $M^{\prime}$ assumed to be fixed, all conditions needed for an application of Lemma 5.2 will be verified in Lemmas 5.3–5.8. This will accomplish a proof of Proposition 5.1.

In Lemma 5.3 the subspaces $\mathop{\rm gr}\nolimits^{F}U_{i}$ of $\mathop{\rm gr}\nolimits^{F}\!{\cal M}$ will be determined explicitly. A module structure over the 0-Hecke algebra ${\cal H}_{n}(0)$ will be constructed on $\mathop{\rm gr}\nolimits^{F}\!{\cal M}$ in Lemma 5.4. It will enable us to derive exactness of the complex K_{\bullet}\bigl{(}\mathop{\rm gr}\nolimits^{F}\!{\cal M};(\mathop{\rm gr}\nolimits^{F}U_{i})\bigr{)} from the results of section 2. Comparison of the subspaces $\Upsilon_{i}$, $\Sigma_{i}$ in ${\cal M}$ with their counterparts in $\mathop{\rm gr}\nolimits^{F}\!{\cal M}$ will be provided by Lemmas 5.7, 5.8.

Denote by $x_{\cal M}$ the linear operator by which an element $x\in{\cal H}^{\prime}_{n}\otimes{\cal H}_{n}$ acts on ${\cal M}$. The generators $T_{i}$, $T^{\prime}_{i}$ act on ${\cal M}$ by the formulas similar to those for the action of $T_{i}$ on $M$. Recall the subsets $A_{i}(\lambda)$, $B_{i}(\lambda)$ of ${\cal D}_{\lambda}$ defined in section 1. There are similar subsets of ${\cal D}_{\mu}$. We have

[TABLE]

If $\pi\in B_{i}(\lambda)$, then $T_{\pi}T^{\prime}_{\sigma}c$ is an eigenvector for the operator $(T_{i})_{\cal M}$ with the eigenvalue $\chi_{\pi}(T_{i})=\chi(T_{\pi^{-1}(i)})$. If $\sigma\in B_{i}(\mu)$, then $T_{\pi}T^{\prime}_{\sigma}c$ is an eigenvector for $(T^{\prime}_{i})_{\cal M}$ with the eigenvalue $\chi_{\sigma}(T^{\prime}_{i})=\chi(T^{\prime}_{\sigma^{-1}(i)})$.

Each homogeneous component $\mathop{\rm gr}\nolimits^{F}_{p}\!{\cal M}$ of $\mathop{\rm gr}\nolimits^{F}\!{\cal M}$ has a basis

[TABLE]

where $v_{\pi,\mskip 1.0mu\sigma}=T_{\pi}T^{\prime}_{\sigma}c+F_{p-1}{\cal M}$. Hence $\{v_{\pi,\mskip 1.0mu\sigma}\mid\pi\in{\cal D}_{\lambda},\ \sigma\in{\cal D}_{\mu}\}$ is a basis for $\mathop{\rm gr}\nolimits^{F}\!{\cal M}$.

Lemma 5.3. For each $i=1,\ldots,n-1$ the space $\mathop{\rm gr}\nolimits^{F}U_{i}$ has a basis consisting of the following elements:**

[TABLE]

Proof. Fixing $i$, let ${\cal M}(\pi,\sigma)$ be the $\{T_{i},T^{\prime}_{i}\}$-invariant subspace of ${\cal M}$ spanned by

[TABLE]

If $\pi\in A_{i}(\lambda)$ and $\sigma\in A_{i}(\mu)$, then these 4 elements are linearly independent. A basis for ${\cal M}(\pi,\sigma)$ is formed by 2 elements $T_{\pi}T^{\prime}_{\sigma}c$, $T_{\tau_{i}\pi}T^{\prime}_{\sigma}c$ when $\pi\in A_{i}(\lambda)$, $\sigma\in B_{i}(\mu)$, and by 2 elements $T_{\pi}T^{\prime}_{\sigma}c$, $T_{\pi}T^{\prime}_{\tau_{i}\sigma}c$ when $\pi\in B_{i}(\lambda)$, $\sigma\in A_{i}(\mu)$. Computing the action of ${\cal T}_{i}$, we deduce that $U_{i}\cap{\cal M}(\pi,\sigma)$ is spanned in the first case by

[TABLE]

in the second by $\chi_{\sigma}(T^{\prime}_{i})T_{\tau_{i}\pi}T^{\prime}_{\sigma}c-T_{\pi}T^{\prime}_{\sigma}c$, in the third by $\chi_{\pi}(T_{i})T_{\pi}T^{\prime}_{\tau_{i}\sigma}c-T_{\pi}T^{\prime}_{\sigma}c$. If $\pi\in B_{i}(\lambda)$ and $\sigma\in B_{i}(\mu)$, then ${\cal M}(\pi,\sigma)$ is spanned by a single element $T_{\pi}T^{\prime}_{\sigma}c$ which is an eigenvector for the operator by which ${\cal T}_{i}$ acts on ${\cal M}$ with the eigenvalue $\chi_{\pi}(T_{i})\chi_{\sigma}(T^{\prime}_{i})$. In this case ${\cal M}(\pi,\sigma)\subset U_{i}$ if and only if $\chi_{\pi}(T_{i})\chi_{\sigma}(T^{\prime}_{i})\neq 1$.

In each case there is a basis for $\mathop{\rm gr}\nolimits^{F}(U_{i}\cap{\cal M}(\pi,\sigma))$ given by the respective elements in the statement of Lemma 5.3.

Note that ${\cal M}$ is a direct sum of these subspaces ${\cal M}(\pi,\sigma)$ with $\pi\in A_{i}(\lambda)\cup B_{i}(\lambda)$ and $\sigma\in A_{i}(\mu)\cup B_{i}(\mu)$. This direct sum decomposition is compatible with the filtration of ${\cal M}$. Furthermore, we have $U_{i}=\bigoplus(U_{i}\cap{\cal M}(\pi,\sigma))$ since each ${\cal M}(\pi,\sigma)$ is stable under the action of ${\cal T}_{i}$, whence $\mathop{\rm gr}\nolimits^{F}U_{i}=\bigoplus\mathop{\rm gr}\nolimits^{F}(U_{i}\cap{\cal M}(\pi,\sigma))$.   $\mathchar 9219\relax$

With the next goal to describe an ${\cal H}_{n}(0)$-module structure it will be more convenient to index the basis elements of $\mathop{\rm gr}\nolimits^{F}\!{\cal M}$ by the pairs of cosets since this will allow us to exploit the natural actions of ${{S}}_{n}$ on ${{S}}_{n}/{{S}}_{\lambda}$ and on ${{S}}_{n}/{{S}}_{\mu}$.

For $x\in{{S}}_{n}/{{S}}_{\lambda}$ and $y\in{{S}}_{n}/{{S}}_{\mu}$ with their distinguished representatives $\pi\in{\cal D}_{\lambda}$ and $\sigma\in{\cal D}_{\mu}$ we put $v_{x,\mskip 1.0muy}=v_{\pi,\mskip 1.0mu\sigma}$, and we will write $\chi_{x}(T_{i})$, $\chi_{y}(T^{\prime}_{i})$, instead of $\chi_{\pi}(T_{i})$, $\chi_{\sigma}(T^{\prime}_{i})$. Note that $\pi\in B_{i}(\lambda)$ if and only if $\tau_{i}x=x$. Hence $\chi_{x}(T_{i})$ is defined if $\tau_{i}x=x$. Similarly, $\chi_{y}(T^{\prime}_{i})$ is defined if $\tau_{i}y=y$.

Consider the partial orders on ${{S}}_{n}/{{S}}_{\lambda}$ and ${{S}}_{n}/{{S}}_{\mu}$ transferred from the Bruhat orders on ${\cal D}_{\lambda}$ and ${\cal D}_{\mu}$. For $x$ and $\pi$ as in the preceding paragraph, we have $\tau_{i}x>x$ if and only if $\pi\in A_{i}(\lambda)$. Similarly, $\tau_{i}y>y$ if and only if $\sigma\in A_{i}(\mu)$. The next lemma applies to $N=\mathop{\rm gr}\nolimits^{F}\!{\cal M}$.

Lemma 5.4. Let $N$ be a vector space with a basis $\{v_{x,\mskip 1.0muy}\mid x\in{{S}}_{n}/{{S}}_{\lambda},\ y\in{{S}}_{n}/{{S}}_{\mu}\}$. Define linear operators $\mathaccent 23{T}_{1},\ldots,\mathaccent 23{T}_{n-1}$ on $N$ by the rule

[TABLE]

Then $\mathaccent 23{T}_{1},\ldots,\mathaccent 23{T}_{n-1}$ satisfy the defining relations of the [math]-Hecke algebra ${\cal H}_{n}(0)$.

Proof. It is checked immediately that $\mathaccent 23{T}_{i}^{2}=-\mathaccent 23{T}_{i}$. Also, we have to show that

[TABLE]

With the aim to do this express $\mathaccent 23{T}_{i}=\Phi_{i}+\Psi_{i}$ as the sum of two linear operators defined by the formulas

[TABLE]

In particular, each $v_{x,\mskip 1.0muy}$ is an eigenvector for $\Phi_{i}$. Note also that $\Psi_{i}v_{x,\mskip 1.0muy}$ is always a scalar multiple of $v_{\tau_{i}x,\mskip 1.0mu\tau_{i}y}$, and $\Psi_{i}v_{x,\mskip 1.0muy}\neq 0$ if and only if either $\tau_{i}y>y$ or $\tau_{i}y=y$ and $\tau_{i}x>x$.

Suppose first that $|i-j|=1$. We claim that

[TABLE]

Note that $\Phi_{i}\Phi_{j}\Phi_{i}v_{x,\mskip 1.0muy}\neq 0$ if and only if both $\Phi_{i}v_{x,\mskip 1.0muy}\neq 0$ and $\Phi_{j}v_{x,\mskip 1.0muy}\neq 0$. If these inequalities hold, then $\Phi_{i}\Phi_{j}\Phi_{i}v_{x,\mskip 1.0muy}=-v_{x,\mskip 1.0muy}$. Since this description is symmetric in $i$ and $j$, we get  $\Phi_{i}\Phi_{j}\Phi_{i}v_{x,\mskip 1.0muy}=\Phi_{j}\Phi_{i}\Phi_{j}v_{x,\mskip 1.0muy}$  in all cases.

Next, $\Phi_{j}\Psi_{i}v_{x,\mskip 1.0muy}$ is either 0 or equal to $-\Psi_{i}v_{x,\mskip 1.0muy}$. From the definition of $\Psi_{i}$ it is clear that $\Psi_{i}^{2}=0$. Hence $\Psi_{i}\Phi_{j}\Psi_{i}v_{x,\mskip 1.0muy}=0$.

We have checked the first two identities. Before we proceed with the others let us make several remarks. The two transpositions $\tau_{i}$, $\tau_{j}$ generate a subgroup $\langle\tau_{i},\tau_{j}\rangle$ of ${{S}}_{n}$ isomorphic to ${{S}}_{2}$. Each $\langle\tau_{i},\tau_{j}\rangle$-orbit in ${{S}}_{n}/{{S}}_{\lambda}$ has a smallest element. In fact, a coset $x\in{{S}}_{n}/{{S}}_{\lambda}$ is minimal in its $\langle\tau_{i},\tau_{j}\rangle$-orbit if and only if $\tau_{i}x\geq x$ and $\tau_{j}x\geq x$, if and only if the distinguished representative $\pi_{x}$ of $x$ lies in ${\cal D}(\langle\tau_{i},\tau_{j}\rangle\backslash{{S}}_{n}/{{S}}_{\lambda})$.

If $x$ is minimal in its orbit, then the stabilizer $\langle\tau_{i},\tau_{j}\rangle\cap\pi_{x}{{S}}_{\lambda}\pi_{x}^{-1}$ of $x$ in $\langle\tau_{i},\tau_{j}\rangle$ is a parabolic subgroup by the general properties of Coxeter groups. Hence there are exactly 4 possibilities for this stabilizer: the trivial subgroup, the subgroups $\langle\tau_{i}\rangle$, $\langle\tau_{j}\rangle$ generated by one of the two transpositions, and the whole $\langle\tau_{i},\tau_{j}\rangle$. In the first case $\langle\tau_{i},\tau_{j}\rangle x$ is isomorphic as a poset to ${{S}}_{2}$ with the Bruhat order. In the second case the orbit contains 3 elements forming a chain $x<\tau_{j}x<\tau_{i}\tau_{j}x$. The third case has a similar description with $i$ and $j$ interchanged. In the last case $\langle\tau_{i},\tau_{j}\rangle x$ is the single element set $\{x\}$.

The action of the longest element $w=\tau_{i}\tau_{j}\tau_{i}=\tau_{j}\tau_{i}\tau_{j}$ of the group $\langle\tau_{i},\tau_{j}\rangle$ reverses order on each $\langle\tau_{i},\tau_{j}\rangle$-orbit in ${{S}}_{n}/{{S}}_{\lambda}$. From this it is clear that each $\langle\tau_{i},\tau_{j}\rangle$-orbit has a largest element, and $x\in{{S}}_{n}/{{S}}_{\lambda}$ is maximal in its orbit if and only if $\tau_{i}x\leq x$ and $\tau_{j}x\leq x$.

Since $w\tau_{i}=\tau_{i}\tau_{j}=\tau_{j}w$, we have $\tau_{i}x=x$ if and only if $wx$ is fixed by $\tau_{j}$. We claim that in this case $\chi_{x}(T_{i})=\chi_{wx}(T_{j})=\chi_{\tau_{i}\tau_{j}x}(T_{j})$. If $\tau_{j}x\neq x$, then $x$ is either the smallest or the largest element in its $\langle\tau_{i},\tau_{j}\rangle$-orbit, and $wx=\tau_{i}\tau_{j}x$ is, respectively, the largest or the smallest element in this orbit with the distinguished representative $\pi_{wx}=\tau_{i}\tau_{j}\pi_{x}\in{\cal D}_{\lambda}$ in both cases, so that

[TABLE]

Suppose now that $\tau_{j}x=x$. Then both $\pi_{x}^{-1}\tau_{i}\pi_{x}$ and $\pi_{x}^{-1}\tau_{j}\pi_{x}$ are in ${{B}}_{\lambda}$. If these two transpositions are $\tau_{k}$ and $\tau_{l}$, then $\tau_{k}\tau_{l}\tau_{k}=\tau_{l}\tau_{k}\tau_{l}$ since $\tau_{i}\tau_{j}\tau_{i}=\tau_{j}\tau_{i}\tau_{j}$. This means that $|k-l|=1$, and so $\chi(T_{k})=\chi(T_{l})$. Hence

[TABLE]

All the previous observations apply also to the $\langle\tau_{i},\tau_{j}\rangle$-orbits in ${{S}}_{n}/{{S}}_{\mu}$.

Let us now turn to the third identity $\Psi_{i}\Psi_{j}\Psi_{i}=\Psi_{j}\Psi_{i}\Psi_{j}$.  Consider

[TABLE]

where $w=\tau_{i}\tau_{j}\tau_{i}$. Note that $\Psi_{i}\Psi_{j}\Psi_{i}v_{x,\mskip 1.0muy}\neq 0$ if and only if for each $k=1,2,3$ we have $y_{k}\geq y_{k-1}$, and if $y_{k}=y_{k-1}$ then $x_{k}>x_{k-1}$. Similarly, $\Psi_{j}\Psi_{i}\Psi_{j}v_{x,\mskip 1.0muy}\neq 0$ if and only if for each $k=1,2,3$ we have $y_{k}^{*}\geq y^{*}_{k-1}$, and if $y^{*}_{k}=y^{*}_{k-1}$ then $x^{*}_{k}>x^{*}_{k-1}$.

Since $x^{*}_{k}=wx_{3-k}$, $y^{*}_{k}=wy_{3-k}$ and $w$ reverses order on $\langle\tau_{i},\tau_{j}\rangle$-orbits in ${{S}}_{n}/{{S}}_{\lambda}$ and ${{S}}_{n}/{{S}}_{\mu}$, we see that $\Psi_{i}\Psi_{j}\Psi_{i}v_{x,\mskip 1.0muy}\neq 0$ if and only if $\Psi_{j}\Psi_{i}\Psi_{j}v_{x,\mskip 1.0muy}\neq 0$. Furthermore, if these two elements are nonzero, then

[TABLE]

where two numbers $a$, $a^{*}$ are the cardinalities of the sets of integers $1\leq k\leq 3$ such that $\,y_{k}>y_{k-1}$, $\,x_{k}<x_{k-1}\,$ in the case of $a$ and $\,y^{*}_{k}>y^{*}_{k-1}$, $\,x^{*}_{k}<x^{*}_{k-1}\,$ in the case of $a^{*}$. Since the assignment $k\mapsto 4-k$ gives a bijection between these two sets, we have $a=a^{*}$, whence  $\Psi_{i}\Psi_{j}\Psi_{i}v_{x,\mskip 1.0muy}=\Psi_{j}\Psi_{i}\Psi_{j}v_{x,\mskip 1.0muy}\,$.

The fourth identity. Recall that $\Psi_{i}\Psi_{j}v_{x,\mskip 1.0muy}$ is always a scalar multiple of $v_{x^{*}\!,\mskip 1.0muy^{*}}$ where $x^{*}=\tau_{i}\tau_{j}x$, $\,y^{*}=\tau_{i}\tau_{j}y$. For $\Psi_{i}\Psi_{j}\Phi_{i}v_{x,\mskip 1.0muy}\neq 0$ to hold, it is necessary and sufficient that $\Psi_{i}\Psi_{j}v_{x,\mskip 1.0muy}\neq 0$ and $\Phi_{i}v_{x,\mskip 1.0muy}\neq 0$, while $\Phi_{j}\Psi_{i}\Psi_{j}v_{x,\mskip 1.0muy}\neq 0$ holds if and only if $\Psi_{i}\Psi_{j}v_{x,\mskip 1.0muy}\neq 0$ and $\Phi_{j}v_{x^{*}\!,y^{*}}\neq 0$.

Since $\tau_{i}\tau_{j}\tau_{i}$ reverses the order on each $\langle\tau_{i},\tau_{j}\rangle$-orbit in ${{S}}_{n}/{{S}}_{\lambda}$, we have $\tau_{i}x<x$ if and only if $\tau_{j}x^{*}<x^{*}$. Also, $\tau_{i}x=x$ if and only if $\tau_{j}x^{*}=x^{*}$, and in this case $\chi_{x^{*}}(T_{j})=\chi_{x}(T_{i})$. Similarly, $\tau_{i}y<y$ if and only if $\tau_{j}y^{*}<y^{*}$, and $\tau_{i}y=y$ if and only if $\tau_{j}y^{*}=y^{*}$. If $\tau_{i}y=y$, then $\chi_{y^{*}}(T^{\prime}_{j})=\chi_{y}(T^{\prime}_{i})$.

Hence $\chi_{x^{*}}(T_{j})\chi_{y^{*}}(T^{\prime}_{j})=\chi_{x}(T_{i})\chi_{y}(T^{\prime}_{i})$ whenever $\tau_{i}x=x$ and $\tau_{i}y=y$ hold simultaneously. So it follows that $\Phi_{i}v_{x,\mskip 1.0muy}\neq 0$ if and only if $\Phi_{j}v_{x^{*}\!,\mskip 1.0muy^{*}}\neq 0$, in which case

[TABLE]

The fifth identity. All three terms $\Phi_{i}\Psi_{j}\Phi_{i}v_{x,\mskip 1.0muy}$, $\Psi_{j}\Phi_{i}\Phi_{j}v_{x,\mskip 1.0muy}$, $\Phi_{j}\Phi_{i}\Psi_{j}v_{x,\mskip 1.0muy}$ equal 0 when $\Psi_{j}v_{x,\mskip 1.0muy}=0$. Suppose that $\Psi_{j}v_{x,\mskip 1.0muy}\neq 0$. Then $\tau_{j}y\geq y$, and if $\tau_{j}y=y$ then $\tau_{j}x>x$. Hence $\tau_{j}y>y$ whenever $\tau_{j}x\leq x$.

If $\tau_{j}x\geq x$, we must have $\Phi_{j}v_{x,\mskip 1.0muy}=0$ and $\Phi_{j}v_{\tau_{j}x,\tau_{j}y}\neq 0$ by the definition of $\Phi_{j}$. In this case $\Psi_{j}\Phi_{i}\Phi_{j}v_{x,\mskip 1.0muy}=0$, while $\Phi_{j}\Phi_{i}\Psi_{j}v_{x,\mskip 1.0muy}=-\Phi_{i}\Psi_{j}v_{x,\mskip 1.0muy}$. Furthermore, the inequality $\Phi_{i}\Psi_{j}v_{x,\mskip 1.0muy}\neq 0$ is only possible when either $\tau_{i}\tau_{j}x<\tau_{j}x$ or $\tau_{i}\tau_{j}x=\tau_{j}x$ and $\tau_{i}\tau_{j}y\leq\tau_{j}y$. These conditions imply that $\tau_{j}x$ is the largest element of the $\langle\tau_{i},\tau_{j}\rangle$-orbit of $x$, and therefore $\tau_{i}x<x$ when $x<\tau_{j}x$. If $\tau_{j}x=x$, we have either $\tau_{i}x<x$ or $\tau_{i}x=x$ and $\tau_{i}\tau_{j}y\leq\tau_{j}y$. In particular, $\tau_{j}y$ is the largest element of the $\langle\tau_{i},\tau_{j}\rangle$-orbit of $y$ in the case when $\tau_{i}x=\tau_{j}x=x$, but then $\tau_{i}y<y$ since $y<\tau_{j}y$. In each of these cases we deduce that $\Phi_{i}v_{x,\mskip 1.0muy}\neq 0$, whence

[TABLE]

The same equalities are trivially true when $\tau_{j}x\geq x$, but $\Phi_{i}\Psi_{j}v_{x,\mskip 1.0muy}=0$.

Consider the remaining case when $\tau_{j}x<x$. Here $\Phi_{j}v_{x,\mskip 1.0muy}\neq 0$ and $\Phi_{j}v_{\tau_{j}x,\tau_{j}y}=\nobreak 0$, so that $\Phi_{j}\Phi_{i}\Psi_{j}v_{x,\mskip 1.0muy}=0$, while $\Psi_{j}\Phi_{i}\Phi_{j}v_{x,\mskip 1.0muy}=-\Psi_{j}\Phi_{i}v_{x,\mskip 1.0muy}$. If now $\Phi_{i}v_{x,\mskip 1.0muy}=0$, then all the three terms we look at vanish. If $\Phi_{i}v_{x,\mskip 1.0muy}\neq 0$, then $\tau_{i}x\leq x$, which implies that $x$ is the largest element in its $\langle\tau_{i},\tau_{j}\rangle$-orbit. But this entails $\tau_{i}\tau_{j}x<\tau_{j}x$ since $\tau_{j}x<x$, and it follows that $\Phi_{i}v_{\tau_{j}x,\tau_{j}y}\neq 0$. Hence

[TABLE]

We have checked all the required relations between $\Phi_{i},\Phi_{j},\Psi_{i},\Psi_{j}$ for any pair $i,j$ with $|i-j|=1$. By symmetry they hold also with $i$ and $j$ interchanged. The braid relation $\mathaccent 23{T}_{i}\mathaccent 23{T}_{j}\mathaccent 23{T}_{i}=\mathaccent 23{T}_{j}\mathaccent 23{T}_{i}\mathaccent 23{T}_{j}$ is now immediate.

Suppose that $|i-j|>1$. Considering the subgroup $\langle\tau_{i},\tau_{j}\rangle$ generated by $\tau_{i},\tau_{j}$, it is still true that its longest element $\tau_{i}\tau_{j}=\tau_{j}\tau_{i}$ reverses order on each $\langle\tau_{i},\tau_{j}\rangle$-orbit in ${{S}}_{n}/{{S}}_{\lambda}$ and ${{S}}_{n}/{{S}}_{\mu}$. If $\tau_{i}x=x$ for some $x\in{{S}}_{n}/{{S}}_{\lambda}$, then $\chi_{\tau_{j}x}(T_{i})=\chi_{x}(T_{i})$, and if $\tau_{i}y=y$ for some $y\in{{S}}_{n}/{{S}}_{\mu}$, then $\chi_{\tau_{j}y}(T^{\prime}_{i})=\chi_{y}(T^{\prime}_{i})$. The arguments similar to those used in the case $|i-j|=1$, but this time much shorter, show that

[TABLE]

Hence  $\mathaccent 23{T}_{i}\mathaccent 23{T}_{j}=(\Phi_{i}+\Psi_{i})(\Phi_{j}+\Psi_{j})=(\Phi_{j}+\Psi_{j})(\Phi_{i}+\Psi_{i})=\mathaccent 23{T}_{j}\mathaccent 23{T}_{i}\,$.   $\mathchar 9219\relax$

Lemma 5.5. Put $N=\mathop{\rm gr}\nolimits^{F}\!{\cal M}$ and denote by $\mathaccent 23{T}_{1},\ldots,\mathaccent 23{T}_{n-1}$ the canonical generators of the [math]-Hecke algebra ${\cal H}_{n}(0)$. With the ${\cal H}_{n}(0)$-module structure on $N$ defined by the formulas in the statement of Lemma 5.4 we have $\mathop{\rm gr}\nolimits^{F}U_{i}=\mathaccent 23{T}_{i}N$ for each $i,$ and there is an isomorphism of ${\cal H}_{n}(0)$-modules

[TABLE]

where $\nu(\pi)$ is the composition of $n$ such that ${{S}}_{\nu(\pi)}={{S}}_{\mu}\cap\pi{{S}}_{\lambda}\pi^{-1},$ $\,{\cal H}_{\nu(\pi)}(0)$ the corresponding parabolic subalgebra of ${\cal H}_{n}(0),$ and $\xi_{\pi}:{\cal H}_{\nu(\pi)}(0)\to{k}$ the representation such that

[TABLE]

Proof. In terms of the indexation of the basis elements of $N$ by the pairs $(x,y)\in{{S}}_{n}/{{S}}_{\lambda}\times{{S}}_{n}/{{S}}_{\mu}$ the image of the linear operator $\mathaccent 23{T}_{i}:N\to N$ is spanned by the elements $v_{x,\mskip 1.0muy}$ with

[TABLE]

and by the elements $v_{x,\mskip 1.0muy}-qv_{\tau_{i}x,\tau_{i}y}$ with $\tau_{i}x<x$, $\tau_{i}y>y$.

If $\pi\in{\cal D}_{\lambda}$ is the distinguished representative of $x$, then $\tau_{i}x<x$ if and only if $\pi\in\tau_{i}A_{i}(\lambda)$, and $\tau_{i}x=x$ if and only if $\pi\in B_{i}(\lambda)$. Similarly, if $\sigma\in{\cal D}_{\mu}$ is the distinguished representative of $y$, then $\tau_{i}y>y$, $\tau_{i}y<y$, or $\tau_{i}y=y$ depending on whether $\sigma$ is in $A_{i}(\mu)$, $\tau_{i}A_{i}(\mu)$, or $B_{i}(\mu)$, respectively. Comparison with the description given in Lemma 5.3 yields the desired equality $\,\mathop{\rm gr}\nolimits^{F}U_{i}=\mathaccent 23{T}_{i}N$.

We claim that the ${\cal H}_{n}(0)$-module $N$ is generated by the set

[TABLE]

where $e_{\mu}={{S}}_{\mu}$ is the coset of the identity element. If $\tau_{i}y>y$ for some $y\in{{S}}_{n}/{{S}}_{\mu}$ and $\tau_{i}\in{{B}}_{n}$, then $v_{x,\tau_{i}y}$ equals $\mathaccent 23{T}_{i}v_{\tau_{i}x,y}$ when $\tau_{i}x\leq x$ and equals $q^{-1}(1+\mathaccent 23{T}_{i})v_{\tau_{i}x,y}$ when $\tau_{i}x>x$. From this it follows by induction on $y$ that each basis element $v_{x,\mskip 1.0muy}$ lies in the submodule of $N$ generated by $\{v_{x^{\prime}\!,\mskip 1.0mue_{\mu}}\mid x^{\prime}\in{{S}}_{n}/{{S}}_{\lambda}\}$. On the other hand, $v_{\tau_{i}x^{\prime}\!,\mskip 1.0mue_{\mu}}=\mathaccent 23{T}_{i}v_{x^{\prime}\!,\mskip 1.0mue_{\mu}}$ if $\tau_{i}x^{\prime}>x^{\prime}$ for some $\tau_{i}\in{{B}}_{\mu}$. Induction on $x^{\prime}$ shows that $v_{x^{\prime}\!,\mskip 1.0mue_{\mu}}$ lies in the submodule of $N$ generated by $v_{x,\mskip 1.0mue_{\mu}}$ where $x$ is the smallest element of the ${{S}}_{\mu}$-orbit of $x^{\prime}$. This proves the claim about the generating set of $N$.

Denote by $N(x)$ the submodule of $N$ generated by $v_{x,e_{\mu}}$. From the formulas in the statement of Lemma 5.4 it is clear that $N(x)$ is contained in the subspace of $N$ spanned by $\{v_{\sigma x,\mskip 1.0mu\sigma e_{\mu}}\mid\sigma\in{{S}}_{n}\}$. If $\tau_{i}x\geq x$ for all $\tau_{i}\in{{B}}_{\mu}$, then the considerations in the preceding paragraph in fact show that $v_{\sigma x,\mskip 1.0mu\sigma e_{\mu}}\in N(x)$ for all $\sigma\in{{S}}_{n}$. In this case $N(x)$ has a basis consisting of the elements $v_{x^{\prime}\!,\mskip 1.0muy^{\prime}}$ with $(x^{\prime}\!,y^{\prime})$ in the orbit of $(x,e_{\mu})$ with respect to the diagonal action of ${{S}}_{n}$ on ${{S}}_{n}/{{S}}_{\lambda}\times{{S}}_{n}/{{S}}_{\mu}\,$, and therefore

[TABLE]

where $\mathop{\rm St}(x,e_{\mu})$ stands for the stabilizer of $(x,e_{\mu})$ in ${{S}}_{n}$ with respect to that action. The condition that $\tau_{i}x\geq x$ for all $\tau_{i}\in{{B}}_{\mu}$ means precisely that the distinguished representative of $x$ lies in ${}_{\mu}{\cal D}_{\lambda}$. If $\pi$ is this representative, then $\mathop{\rm St}(x,e_{\mu})={{S}}_{\nu(\pi)}$.

Now we put $N(\pi)=N(\pi{{S}}_{\lambda})$ for each $\pi\in{}_{\mu}{\cal D}_{\lambda}$. In other words, $N(\pi)$ is the submodule of $N$ generated by $v_{\pi,\mskip 1.0mue}$. Then $\,\dim N(\pi)=({{S}}_{n}:{{S}}_{\nu(\pi)})\,$ by the above. If $\tau_{i}\in{{B}}_{\nu(\pi)}$, then $\tau_{i}\pi{{S}}_{\lambda}=\pi{{S}}_{\lambda}$ and $\tau_{i}{{S}}_{\mu}={{S}}_{\mu}$, whence

[TABLE]

by comparison of the definition of $\xi_{\pi}$ with the last two formulas in the statement of Lemma 5.4. Hence there is a surjective homomorphism of ${\cal H}_{n}(0)$-modules

[TABLE]

Comparing the dimensions, we deduce that this map is an isomorphism.

From the preceding discussion it is also clear that each basis element of $N$ lies in exactly one submodule $N(\pi)$. Thus $N=\bigoplus_{\pi\in\,{}_{\mu}\mskip-1.0mu{\cal D}_{\mskip-1.0mu\lambda}}N(\pi)$.   $\mathchar 9219\relax$

Lemma 5.6. The complex K_{\bullet}\bigl{(}\mathop{\rm gr}\nolimits^{F}\!{\cal M};(\mathop{\rm gr}\nolimits^{F}U_{i})\bigr{)} is exact.

Proof. By Lemma 5.5 there is an ${\cal H}_{n}(0)$-module structure on $N=\mathop{\rm gr}\nolimits^{F}\!{\cal M}$ with the property that $\mathop{\rm gr}\nolimits^{F}U_{i}=\mathaccent 23{T}_{i}N$ for each $i$. Therefore Corollary 2.5 applies.   $\mathchar 9219\relax$

Lemma 5.7. Put $\,\Sigma=\sum\limits_{i=1}^{n-1}U_{i},$ $\,\Sigma^{\mathop{\rm gr}\nolimits}=\sum\limits_{i=1}^{n-1}\mathop{\rm gr}\nolimits^{F}U_{i},$ $\,\Upsilon=\bigcap\limits_{i=1}^{n-1}U_{i},$ $\Upsilon^{\mathop{\rm gr}\nolimits}=\bigcap\limits_{i=1}^{n-1}\mathop{\rm gr}\nolimits^{F}U_{i}$. We have $\,\mathop{\rm gr}\nolimits^{F}\Sigma=\Sigma^{\mathop{\rm gr}\nolimits}$ and $\,\mathop{\rm gr}\nolimits^{F}\Upsilon=\Upsilon^{\mathop{\rm gr}\nolimits}$.

Proof. The inclusions $\,\Sigma^{\mathop{\rm gr}\nolimits}\subset\mathop{\rm gr}\nolimits^{F}\Sigma\,$ and $\,\mathop{\rm gr}\nolimits^{F}\Upsilon\subset\Upsilon^{\mathop{\rm gr}\nolimits}\,$ are always true, and so we need only to compare the dimensions. By Lemma 5.5 $\,N=\mathop{\rm gr}\nolimits^{F}\!{\cal M}\,$ is a direct sum of ${\cal H}_{n}(0)$-submodules

[TABLE]

Now $N/\Sigma^{\mathop{\rm gr}\nolimits}=N/\sum\mathaccent 23{T}_{i}N$ is the largest factor module of $N$ annihilated by each $\mathaccent 23{T}_{i}$. Since

[TABLE]

[TABLE]

As $(\mathaccent 23{T}_{i}+1)\mathaccent 23{T}_{i}=0$, we can also write $\mathop{\rm gr}\nolimits^{F}U_{i}=\mathop{\rm Ker}\,(\mathaccent 23{T}_{i}+1)_{N}$. Hence $\Upsilon^{\mathop{\rm gr}\nolimits}$ is the largest submodule of $N$ on which each $\mathaccent 23{T}_{i}$ acts as $-\mathop{\rm Id}\nolimits$. Since

[TABLE]

we deduce from Lemma 1.3 that

[TABLE]

Next we are going to determine ${\cal M}/\Sigma$ and $\Upsilon$. For this we will need two different interpretations of the spaces $U_{i}$. The assignment $T_{i}\mapsto(T^{\prime}_{i})^{-1}$ extends to an algebra antiisomorphism ${\cal H}_{n}\to{\cal H}^{\prime}_{n}$ under which ${\cal H}_{\mu}$ is mapped onto ${\cal H}^{\prime}_{\mu}$. It allows us to view $M^{\prime}$ as a right ${\cal H}_{n}$-module and ${k}(\chi^{\prime})$ as a right ${\cal H}_{\mu}$-module. Clearly,

[TABLE]

The space $U_{i}$ is spanned by all elements ${\cal T}_{i}(u\otimes v)-u\otimes v=u{T_{i}}^{-1}\otimes T_{i}v-u\otimes v$ with $u\in M^{\prime}$ and $v\in M$. Replacing here $u$ with $uT_{i}$, we rewrite these elements as $u\otimes T_{i}v-uT_{i}\otimes v$. Since $T_{1},\ldots,T_{n-1}$ generate ${\cal H}_{n}$, the space $\Sigma$ is spanned by all elements $u\otimes hv-uh\otimes v$ with $u\in M^{\prime}$, $v\in M$ and $h\in{\cal H}_{n}$. It follows that

[TABLE]

where the last isomorphism is a consequence of the Mackey decomposition formula since the restriction of $M$ to ${\cal H}_{\mu}$ is a direct sum of modules ${\cal H}_{\mu}\otimes_{{\cal H}_{\nu(\pi)}}{k}(\chi_{\pi})$ with $\pi\in{}_{\mu}{\cal D}_{\lambda}$. Now ${k}(\chi^{\prime})\otimes_{{\cal H}_{\nu(\pi)}}{k}(\chi_{\pi})\neq 0$ if and only if ${\cal H}_{\nu(\pi)}$ operates in ${k}(\chi^{\prime})$ and in ${k}(\chi_{\pi})$ via the same homomorphism ${\cal H}_{\nu(\pi)}\to{k}$, i.e. $\chi(T^{\prime}_{i})^{-1}=\chi_{\pi}(T_{i})$ for all $i$ such that $\tau_{i}\in{{B}}_{\nu(\pi)}$. We get

[TABLE]

Thus $\Sigma$ has the same codimension in ${\cal M}$ as $\Sigma^{\mathop{\rm gr}\nolimits}$ in $N$. Since the filtration on ${\cal M}$ is exhaustive and separating, the dimension and the codimension of subspaces of ${\cal M}$ are preserved under passage to the associated graded spaces. The first equality in the statement of Lemma 5.7 is now clear.

We may view ${\cal M}=M^{\prime}\otimes M$ as an ${\cal H}_{n}$-bimodule. By the previous description $U_{i}$ is spanned by all elements $T_{i}\psi-\psi T_{i}$ with $\psi\in{\cal M}$. The dual vector space ${M^{\prime}}^{*}$ is a left ${\cal H}_{n}$-module in a natural way. Identifying ${\cal M}$ with $\mathop{\rm Hom}\nolimits_{\,{k}}({M^{\prime}}^{*},M)$ by means of the canonical bijection, we see that $\Upsilon$ consists precisely of those ${k}$-linear maps ${M^{\prime}}^{*}\to M$ that satisfy condition (a) of Lemma 1.6.

Suppose first that $q\neq-1$. Then $\Upsilon=\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}({M^{\prime}}^{*},\mathaccent 869{M})$. By Lemmas 1.4, 1.5

[TABLE]

For each $i$ with $\tau_{i}\in{{B}}_{\mu}$ the generator $T_{i}$ of ${\cal H}_{\mu}$ acts on ${k}(\chi^{\prime})$ as the multiplication by $\chi(T^{\prime}_{i})^{-1}$. Therefore Lemma 1.1 yields

[TABLE]

If $\tau_{i}\in{{B}}_{\nu(\pi)}$, then $\mathaccent 869{\chi}_{\pi}(T_{i})=\chi_{\pi}(\mathaccent 869{T}_{i})$. Since $\chi_{\pi}(\mathaccent 869{T}_{i})+\chi_{\pi}(T_{i})=q-1$ and $\chi_{\pi}(T_{i})$ equals either $q$ or $-1$, we always have $\chi_{\pi}(\mathaccent 869{T}_{i})\neq\chi_{\pi}(T_{i})$. Therefore $\chi(T^{\prime}_{i})^{-1}=\chi_{\pi}(\mathaccent 869{T}_{i})$ if and only if $\chi(T^{\prime}_{i})^{-1}\neq\chi_{\pi}(T_{i})$. Thus

[TABLE]

If $q=-1$, then Lemma 1.7 with $M$ replaced by $\mathaccent 869{M}$ shows that $\Upsilon$ consists precisely of those ${\cal H}_{n}$-module homomorphisms ${M^{\prime}}^{*}\to\mathaccent 869{M}$ which factor through a free module. By the last assertion in that lemma

[TABLE]

On the other hand, $\chi(T^{\prime}_{i})=\chi_{\pi}(T_{i})=-1$, and therefore $\chi_{\pi}(T_{i})\chi(T^{\prime}_{i})=1$, for each $i$ such that $\tau_{i}\in{{B}}_{\nu(\pi)}$. This means that in the case $q=-1$ the earlier formula for the dimension of $\Upsilon^{\mathop{\rm gr}\nolimits}$ counts only those permutations $\pi\in{}_{\mu}{\cal D}_{\lambda}$ for which the set ${{B}}_{\nu(\pi)}$ is empty, i.e. ${{S}}_{\nu(\pi)}$ is the trivial group.

We conclude that $\dim\Upsilon=\dim\Upsilon^{\mathop{\rm gr}\nolimits}$ both for $q\neq-1$ and for $q=-1$. This proves the second equality in the statement of Lemma 5.7.   $\mathchar 9219\relax$

Lemma 5.8. Put $\Upsilon_{i}=\bigcap_{j<i}U_{j}$ and $\Sigma_{i}=\sum_{j>i}U_{j}$. Then

[TABLE]

Proof. Let ${\cal H}_{i,n-i}\subset{\cal H}_{n}$ and ${\cal H}^{\prime}_{i,n-i}\subset{\cal H}^{\prime}_{n}$ be the parabolic subalgebras corresponding to the subgroup ${{S}}_{i,n-i}$ of ${{S}}_{n}$ generated by $\{\tau_{j}\in{{B}}_{n}\mid j\neq i\}$. Consider the Mackey decompositions

[TABLE]

where $M(\pi)$ is the ${\cal H}_{i,n-i}$-submodule of $M$ generated by $T_{\pi}\otimes 1$ and $M^{\prime}(\sigma)$ is the ${\cal H}^{\prime}_{i,n-i}$-submodule of $M^{\prime}$ generated by $T^{\prime}_{\sigma}\otimes 1$. They give rise to the decomposition of ${\cal M}$ as a direct sum of $\,{\cal H}^{\prime}_{i,n-i}\otimes{\cal H}_{i,n-i}$-submodules

[TABLE]

This decomposition is compatible with the filtration on ${\cal M}$. Also, if $j\neq i$, then U_{j}=\bigoplus\bigl{(}U_{j}\cap{\cal M}(\pi,\sigma)\bigr{)} since each summand ${\cal M}(\pi,\sigma)$ is stable under the action of ${\cal T}_{j}$. It follows that \Upsilon_{i}=\bigoplus\bigl{(}\Upsilon_{i}\cap{\cal M}(\pi,\sigma)\bigr{)} and \Sigma_{i}=\bigoplus\bigl{(}\Sigma_{i}\cap{\cal M}(\pi,\sigma)\bigr{)}.

Next, ${\cal H}_{i,n-i}\cong{\cal H}_{i}\otimes{\cal H}^{\mathchar 1103\relax}_{i}$ where ${\cal H}_{i}$ and ${\cal H}^{\mathchar 1103\relax}_{i}$ are the subalgebras of ${\cal H}_{n}$ generated, respectively, by $\{T_{j}\mid j<i\}$ and $\{T_{j}\mid j>i\}$. Since the ${\cal H}_{i,n-i}$-module $M(\pi)$ is induced from a 1-dimensional module over a parabolic subalgebra of ${\cal H}_{i,n-i}$, we have

[TABLE]

where $M(\pi)_{1}$ is an ${\cal H}_{i}$-module induced from a 1-dimensional module over a parabolic subalgebra of ${\cal H}_{i}$ and $M(\pi)_{2}$ is an ${\cal H}^{\mathchar 1103\relax}_{i}$-module induced from a 1-dimensional module over a parabolic subalgebra of ${\cal H}^{\mathchar 1103\relax}_{i}$. Let $\,{\cal H}^{\prime}_{i,n-i}\cong{\cal H}^{\prime}_{i}\otimes{{\cal H}^{\prime}}^{\mskip-2.0mu\mathchar 1103\relax}_{\mskip-5.0mui}\,$ and

[TABLE]

be similar decompositions. Then $\,{\cal M}(\pi,\sigma)\cong{\cal M}(\pi,\sigma)_{1}\otimes{\cal M}(\pi,\sigma)_{2}\,$ where

[TABLE]

Note that ${\cal H}^{\prime}_{i}\otimes{\cal H}_{i}\cong{\cal H}_{i}(q^{-1})\otimes{\cal H}_{i}(q)$ and ${{\cal H}^{\prime}}^{\mskip-2.0mu\mathchar 1103\relax}_{\mskip-5.0mui}\otimes{\cal H}^{\mathchar 1103\relax}_{i}\cong{\cal H}_{n-i}(q^{-1})\otimes{\cal H}_{n-i}(q)$. The ${\cal H}^{\prime}_{i}\otimes{\cal H}_{i}$-module ${\cal M}(\pi,\sigma)_{1}$ and the ${{\cal H}^{\prime}}^{\mskip-2.0mu\mathchar 1103\relax}_{\mskip-5.0mui}\otimes{\cal H}^{\mathchar 1103\relax}_{i}$-module ${\cal M}(\pi,\sigma)_{2}$ satisfy the same assumptions that we have imposed on the ${\cal H}^{\prime}_{n}\otimes{\cal H}_{n}$-module ${\cal M}$. In particular, we obtain filtrations $F^{1}$, $F^{2}$ on these two modules by the construction we have done for ${\cal M}$. Then the tensor product filtration $F^{t}$ on ${\cal M}(\pi,\sigma)$ differs from the filtration induced from that on ${\cal M}$ only by a shift of the filtration degrees.

Note that ${\cal T}_{j}$ lies in ${\cal H}^{\prime}_{i}\otimes{\cal H}_{i}$ when $j<i$ and in ${{\cal H}^{\prime}}^{\mskip-2.0mu\mathchar 1103\relax}_{\mskip-5.0mui}\otimes{\cal H}^{\mathchar 1103\relax}_{i}$ when $j>i$. Hence

[TABLE]

With $\,\Upsilon(\pi,\sigma)=\bigcap_{j<i}U_{j}(\pi,\sigma)\,$ and $\,\Sigma(\pi,\sigma)=\sum_{j>i}U_{j}(\pi,\sigma)\,$ we get

[TABLE]

By Lemma 5.7 applied to ${\cal M}(\pi,\sigma)_{1}$ and ${\cal M}(\pi,\sigma)_{2}$ we have

[TABLE]

and it follows that

[TABLE]

The equalities of the left and right hand sides above hold then also with $F^{t}$ replaced by the original filtration $F$ on ${\cal M}$ since this change results in the same associated graded spaces with shifted degrees of homogeneous components. Summing up over all pairs $(\pi,\sigma)\in{\cal D}({{S}}_{i,n-i}\backslash{{S}}_{n}/{{S}}_{\lambda})\times{\cal D}({{S}}_{i,n-i}\backslash{{S}}_{n}/{{S}}_{\mu})$, we arrive at the final conclusions.   $\mathchar 9219\relax$

Now the proof of Proposition 5.1 is complete. There is also a version of this result for a different collection of subspaces in ${\cal M}$:

Proposition 5.9. With the same assumptions about ${\cal M}$ as in Proposition 5.1 the complex K_{\bullet}\bigl{(}{\cal M};(U_{i})\bigr{)} is exact also in the case when $U_{i}=\mathop{\rm Ker}\,({\cal T}_{i}-1)_{\cal M}$ for each $i=1,\ldots,n-1$.

Proof. The dual space ${\cal M}^{*}\cong{M^{\prime}}^{*}\otimes M^{*}$ is a right ${\cal H}^{\prime}_{n}\otimes{\cal H}_{n}$-module in a natural way. We will view $M^{*}$ as a left ${\cal H}_{n}$-module and ${M^{\prime}}^{*}$ as a left ${\cal H}^{\prime}_{n}$-module by means of the antiautomorphisms of ${\cal H}_{n}$ and ${\cal H}^{\prime}_{n}$ such that $T_{i}\mapsto T_{n-i}$, $T^{\prime}_{i}\mapsto T^{\prime}_{n-i}$ for each $i=1,\ldots,n-1$. Since these antiautomorphisms map parabolic subalgebras onto parabolic subalgebras, the class of modules induced from 1-dimensional representations of parabolic subalgebras is preserved under passing to the duals in this way. Hence all indecomposable direct summands of $M^{*}$ and ${M^{\prime}}^{*}$ have $1$-dimensional sources, and so ${\cal M}^{*}$ satisfies the assumptions of Proposition 5.1.

For each subspace $S\subset{\cal M}$ put $S^{\perp}=\{f\in{\cal M}^{*}\mid f(S)=0\}$. Then $\Sigma_{i}^{\perp}=\bigcap_{j>i}U_{j}^{\perp}$, $\Upsilon_{i}^{\perp}=\sum_{j<i}U_{j}^{\perp}$, and

[TABLE]

which is the component of the complex K_{\bullet}\bigl{(}{\cal M}^{*};(U^{\perp}_{n-1},\ldots,U^{\perp}_{1})\bigr{)} in degree $n-i$. Thus K_{\bullet}\bigl{(}{\cal M};(U_{i})\bigr{)}\vphantom{)}^{*} is isomorphic to the complex K_{\bullet}\bigl{(}{\cal M}^{*};(U^{\perp}_{n-1},\ldots,U^{\perp}_{1})\bigr{)} with the degrees shifted by $n$. Note also that $U_{n-i}^{\perp}={\cal M}^{*}({\cal T}_{n-i}-1)=({\cal T}_{i}-1){\cal M}^{*}$ for each $i$. Therefore the latter complex is exact by Proposition 5.1.   $\mathchar 9219\relax$

The next lemma provides a key ingredient in the proof of Theorem 6.6.

Lemma 5.10. Suppose that ${\cal M}=M^{\prime}\otimes M$ where $M$ is an ${\cal H}_{n}$-module and $M^{\prime}$ is an ${\cal H}^{\prime}_{n}$-module such that all indecomposable direct summands of $M$ and $M^{\prime}$ have trivial sources. Put $\,\Sigma=\sum_{i=1}^{n-1}U_{i},$ $\,\Sigma_{1}=\sum_{i>1}U_{i}\,$ where $U_{i}=\mathop{\rm Ker}\,({\cal T}_{i}-1)_{\cal M}$ for each $i,$ and

[TABLE]

Then $\,y\Sigma\subset\Sigma_{1}\,$ and,* moreover*,* $\,y_{\cal M}^{-1}(\Sigma_{1})=\Sigma$.*

Proof. The inclusion $y\Sigma\subset\Sigma_{1}$ is proved exactly as in Lemma 2.2. For this one needs only the braid relations satisfied by ${\cal T}_{1},\ldots,{\cal T}_{n-1}$. The equality $y_{\cal M}^{-1}(\Sigma_{1})=\Sigma$ will follow from injectivity of the map $\varphi:{\cal M}/\Sigma\to{\cal M}/\Sigma_{1}$ induced by the linear operator $y_{\cal M}$. Since all verifications can be done on direct summands of ${\cal M}$ it suffices to consider the case when

[TABLE]

for some parabolic subalgebras ${\cal H}_{\lambda}\,$, ${\cal H}^{\prime}_{\mu}$. Let $c$ be the canonical generator of ${\cal M}$. By Lemma 5.11 stated below a basis for ${\cal M}/\Sigma$ is formed by the cosets of elements $T_{\pi}c$ with $\pi\in{}_{\mu}{\cal D}_{\lambda}^{e}$ where

[TABLE]

is the set of distinguished representatives of the double cosets with the trivial intersection property. There is a similar basis for ${\cal M}/\Sigma_{1}$ obtained as follows. Consider the Mackey decompositions

[TABLE]

with respect to the parabolic subalgebras ${\cal H}_{1,n-1}$ and ${\cal H}^{\prime}_{1,n-1}$. We have

[TABLE]

where $\nu(\alpha)$ and $\nu^{\prime}(\beta)$ are the compositions of $n$ such that

[TABLE]

The ${\cal H}^{\prime}_{1,n-1}\otimes{\cal H}_{1,n-1}$-submodule ${\cal M}(\alpha,\beta)$ of ${\cal M}$ generated by $T_{\alpha}T^{\prime}_{\beta}c$ is isomorphic to $M^{\prime}(\beta)\otimes M(\alpha)$. Since ${\cal M}$ is a direct sum of these submodules for different $\alpha$ and $\beta$, we get

[TABLE]

Note that ${\cal H}_{1,n-1}\cong{\cal H}_{n-1}(q)$, ${\cal H}^{\prime}_{1,n-1}\cong{\cal H}_{n-1}(q^{-1})$. Thus Lemma 5.11 applies to each ${\cal M}(\alpha,\beta)$ viewed as an ${\cal H}_{n-1}(q^{-1})\otimes{\cal H}_{n-1}(q)$-module. We will need only those summands in the decomposition of ${\cal M}/\Sigma_{1}$ which are indexed by the pairs $(\alpha,\beta)$ with $\beta=e$. Put $\mu^{1}=\nu^{\prime}(e)$, so that ${{S}}_{\mu^{1}}={{S}}_{1,n-1}\cap{{S}}_{\mu}$, and put

[TABLE]

By Lemma 5.11 ${\cal M}(\alpha,e)/\Sigma(\alpha,e)$ has a basis formed by the cosets of elements

[TABLE]

The equality $T_{\sigma}T_{\alpha}=T_{\sigma\alpha}$ here is explained by the fact that $\sigma\in{{S}}_{1,n-1}$, while $\alpha$ is the shortest element in the coset ${{S}}_{1,n-1}\alpha$, so that $\ell(\sigma\alpha)=\ell(\sigma)+\ell(\alpha)$.

We claim that the assignment $(\sigma,\alpha)\mapsto\sigma\alpha$ gives a bijection

[TABLE]

By the Mackey decomposition of coset representatives (see [6, Lemma 2.1.9]) ${\cal D}_{\lambda}$ consists precisely of those elements $\pi\in{{S}}_{n}$ which can be written as $\pi=\sigma\alpha$ for some $\alpha\in{\cal D}({{S}}_{1,n-1}\backslash{{S}}_{n}/{{S}}_{\lambda})$ and $\sigma\in{\cal D}({{S}}_{1,n-1}/{{S}}_{\nu(\alpha)})$. The pair $(\sigma,\alpha)$ is uniquely determined by $\pi$ since $\alpha$ is the shortest element in the double coset ${{S}}_{1,n-1}\pi{{S}}_{\lambda}$. Furthermore, for each $\tau_{i}\in{{B}}_{\mu^{1}}$ we have $\ell(\tau_{i}\pi)=\ell(\tau_{i}\sigma)+\ell(\alpha)$ since $\tau_{i}\sigma\in{{S}}_{1,n-1}$, whence $\tau_{i}\pi>\pi$ if and only if $\tau_{i}\sigma>\sigma$. This shows that $\pi\in{}_{\mu^{1}}{\cal D}$ if and only if $\sigma\in{}_{\mu^{1}}{\cal D}$.

Lastly, for each $\rho\in{{S}}_{\mu^{1}}$ we have $\rho\sigma\in{{S}}_{1,n-1}$, and by the Mackey decomposition the equality $\rho\pi=(\rho\sigma)\alpha$ implies that $\rho\pi\in{\cal D}_{\lambda}$ if and only if $\rho\sigma\in{\cal D}({{S}}_{1,n-1}/{{S}}_{\nu(\alpha)})$. The double coset ${{S}}_{\mu^{1}}\pi{{S}}_{\lambda}$ has the trivial intersection property if and only if $\rho\pi\in{\cal D}_{\lambda}$ for all $\rho\in{{S}}_{\mu^{1}}$, while ${{S}}_{\mu^{1}}\sigma{{S}}_{\nu(\alpha)}$ has the trivial intersection property if and only if $\rho\sigma\in{\cal D}({{S}}_{1,n-1}/{{S}}_{\nu(\alpha)})$ for all $\rho\in{{S}}_{\mu^{1}}$. We see that these properties are equivalent. Thus $\pi\in{}_{\mu^{1}}{\cal D}_{\lambda}^{e}$ if and only if $\sigma\in{\cal D}^{e}({{S}}_{\mu^{1}}\backslash{{S}}_{1,n-1}/{{S}}_{\nu(\alpha)})$, and bijectivity of the map considered above has been established.

It follows from the preceding discussion that the vector space

[TABLE]

has a basis formed by the cosets of all elements $T_{\pi}c$ with $\pi\in{}_{\mu^{1}}{\cal D}_{\lambda}^{e}$. Consider the map $\psi:{\cal M}/\Sigma\to Q$ obtained as the composite of $\varphi$ with the projection $p_{Q}$ of ${\cal M}/\Sigma_{1}$ onto $Q$. We will check that $\psi$ is injective. Once this has been done, the injectivity of $\varphi$ will be clear, and the proof of Lemma 5.10 will be complete.

We have ${\cal D}({{S}}_{1,n-1}\backslash{{S}}_{n})=\{\sigma_{j}\mid j=0,\ldots,n-1\}$ where $\sigma_{0}=e$, $\sigma_{1}=\tau_{1}$, and generally $\sigma_{j}=\tau_{1}\tau_{2}\cdots\tau_{j}$ for $j>1$. Therefore $y=\sum_{j=0}^{n-1}(-1)^{j}{\cal T}_{\sigma_{j}}$.

Let $\pi\in{}_{\mu}{\cal D}_{\lambda}^{e}$, i.e. $\pi$ is the distinguished representative of a ${{S}}_{\mu}\,$-$\,{{S}}_{\lambda}$ double coset with the trivial intersection property. Then the coset $T_{\pi}c+\Sigma$ is a basis element of ${\cal M}/\Sigma$ which is sent by $\psi$ to the element $p_{Q}(yT_{\pi}c+\Sigma_{1})\in Q$. We have

[TABLE]

Note that $\sigma_{j}\tau_{i}>\sigma_{j}$ for all $i=1,\ldots,n-1$ except for $i=j$. Hence $\sigma_{j}\in{\cal D}_{\mu}$ for $j>0$ if and only if $\tau_{j}\notin{{S}}_{\mu}$. If $\tau_{j}\in{{S}}_{\mu}$, then $T^{\prime}_{\sigma_{j}}c=T^{\prime}_{\sigma_{j-1}}T^{\prime}_{j}c=q^{-1}T^{\prime}_{\sigma_{j-1}}c$. For each $j$ it follows by induction that

[TABLE]

where $k(j)$ is the largest integer $k$ such that $0\leq k\leq j$ and $\sigma_{k}\in{\cal D}_{\mu}$. This entails

[TABLE]

Denote by $m$ the largest integer such that $0\leq m<n$ and $k(m)=0$. Then $k(j)>0$, and therefore $\sigma_{k(j)}\neq e$ for all $j>m$. In particular, $p_{Q}(T_{\sigma_{j}}T_{\pi}T^{\prime}_{\sigma_{j}}c+\Sigma_{1})=0$ for $j>m$.

If $j\leq m$, then $\sigma_{j}\in{{S}}_{\mu}$, whence $T_{\sigma_{j}}T_{\pi}=T_{\sigma_{j}\pi}$ with $\sigma_{j}\pi\in{\cal D}_{\lambda}$ by the conditions on $\pi$. Moreover, $\sigma_{j}\pi\in{}_{\mu^{1}}{\cal D}_{\lambda}^{e}$. Indeed, for each $\rho\in{{S}}_{\mu^{1}}$ we have $\rho\sigma_{j}\pi\in{\cal D}_{\lambda}$ since $\rho\sigma_{j}\in{{S}}_{\mu}$. This means that the double coset ${{S}}_{\mu^{1}}\sigma_{j}\pi{{S}}_{\lambda}$ has the trivial intersection property. But $\ell(\rho\sigma_{j})=\ell(\rho)+\ell(\sigma_{j})$ since $\rho\in{{S}}_{1,n-1}$, and therefore

[TABLE]

This shows that $\sigma_{j}\pi\in{}_{\mu^{1}}{\cal D}$.

We conclude that $\psi(T_{\pi}c+\Sigma)$ equals $\sum_{j=0}^{m}(-1)^{j}q^{-j}(T_{\sigma_{j}\pi}c+\Sigma_{1})$, which is a linear combination of distinct basis elements of $Q$ with nonzero coefficients. Note that all elements $\sigma_{j}\pi$ with $j=0,\ldots,m$ belong to the same coset ${{S}}_{\mu}\pi$ having $\pi$ as its shortest representative. Therefore the expressions for the images under $\psi$ of two different basis elements of ${\cal M}/\Sigma$ involve disjoint sets of basis elements of $Q$. Injectivity of $\psi$ and $\varphi$ is now clear.   $\mathchar 9219\relax$

Lemma 5.11. Let $M={\cal H}_{n}\otimes_{{\cal H}_{\lambda}}{k}_{\rm triv}$ and $M^{\prime}={\cal H}^{\prime}_{n}\otimes_{{\cal H}^{\prime}_{\mu}}{k}_{\rm triv}$. With the notation as in Lemma 5.10 the set $\{T_{\pi}c+\Sigma\mid\pi\in{}_{\mu}{\cal D}_{\lambda}^{e}\}$ is a basis for ${\cal M}/\Sigma$.

Proof. Consider $M^{\prime}$ as a right ${\cal H}_{n}$-module by means of the algebra antiisomorphism ${\cal H}_{n}\to{\cal H}^{\prime}_{n}$ sending $T_{i}$ to $(T^{\prime}_{i})^{-1}$. Then $M^{\prime}\cong{k}_{\rm triv}\otimes_{{\cal H}_{\mu}}{\cal H}_{n}$ and ${\cal M}=M^{\prime}\otimes M$ is an ${\cal H}_{n}$-bimodule with respect to the ${\cal H}_{n}$-module structures on $M$ and $M^{\prime}$. We have

[TABLE]

for each $i=1,\ldots,n-1$. If $u=T_{i}m-m\mathaccent 869{T}_{i}$ for some $m\in{\cal M}$ where $\mathaccent 869{T}_{i}=q-1-T_{i}$, then $u\in U_{i}$ since

[TABLE]

This shows that $U_{i}\supset\mathaccent 869{U}_{i}$ where $\mathaccent 869{U}_{i}=\{T_{i}m-m\mathaccent 869{T}_{i}\mid m\in{\cal M}\}$. If $q\neq-1$, then $U_{i}=\mathaccent 869{U}_{i}$ since each element $u\in U_{i}$ can be written as

[TABLE]

Note that $\mathaccent 869{U}_{i}$ is spanned by all elements $w\otimes T_{i}v-w\mathaccent 869{T}_{i}\otimes v$ with $v\in M$ and $w\in M^{\prime}$. Setting $\,\mathaccent 869{\Sigma}=\sum_{i=1}^{n-1}\mathaccent 869{U}_{i}\,$, we deduce that

[TABLE]

where $\mathaccent 869{M}\vphantom{M}^{\prime}$ is the right ${\cal H}_{n}$-module obtained from $M^{\prime}$ by composing the original action of ${\cal H}_{n}$ with the automorphism of ${\cal H}_{n}$ sending $T_{i}$ to $\mathaccent 869{T}_{i}$ for each $i$. By the right hand version of Lemma 1.5 we have $\mathaccent 869{M}\vphantom{M}^{\prime}\cong{k}_{\rm alt}\otimes_{{\cal H}_{\mu}}{\cal H}_{n}$. Hence

[TABLE]

since $M\cong\bigoplus\,({\cal H}_{\mu}\otimes_{{\cal H}_{\nu(\pi)}}{k}_{\rm triv})$ by the Mackey formula. Here $\nu(\pi)$ is the composition of $n$ such that ${{S}}_{\nu(\pi)}={{S}}_{\mu}\cap\pi{{S}}_{\lambda}\pi^{-1}$. For each $\pi$ the respective summand in the above decomposition of $\mathaccent 869{M}\vphantom{M}^{\prime}\otimes_{{\cal H}_{n}}M$ is spanned by the image of $T_{\pi}c$.

In view of the previous isomorphisms ${\cal M}/\mathaccent 869{\Sigma}$ has a basis consisting of the cosets of elements $T_{\pi}c$ with $\pi\in{}_{\mu}{\cal D}_{\lambda}$ such that the alternating representation of ${\cal H}_{\nu(\pi)}$ coincides with the trivial representation. If $q\neq-1$, this condition on $\pi$ means precisely that ${{S}}_{\nu(\pi)}=e$, i.e. $\pi\in{}_{\mu}{\cal D}_{\lambda}^{e}$. In this case we also have $\Sigma=\mathaccent 869{\Sigma}$, and the conclusion of Lemma 5.11 follows.

Suppose further that $q=-1$. In this case ${\cal M}/\Sigma$ is a quotient of ${\cal M}/\mathaccent 869{\Sigma}$ since $\mathaccent 869{\Sigma}\subset\nobreak\Sigma$. Hence ${\cal M}/\Sigma$ is spanned by the cosets of elements $T_{\pi}c$ with $\pi\in{}_{\mu}{\cal D}_{\lambda}$. If $\tau_{i}\in{{S}}_{\nu(\pi)}$ for some $i$, then $T_{\pi}c\in U_{i}\subset\Sigma$ since $T_{i}T_{\pi}c=-T_{\pi}c$ and $T^{\prime}_{i}T_{\pi}c=-T_{\pi}c$. It follows that ${\cal M}/\Sigma$ is spanned by the cosets of elements $T_{\pi}c$ with $\pi\in{}_{\mu}{\cal D}_{\lambda}^{e}$ only. It remains to prove that such cosets are linearly independent. But the dual space $({\cal M}/\Sigma)^{*}$ is identified with the subspace of the ${\cal H}_{n}$-bimodule

[TABLE]

consisting of all ${k}$-linear maps $f:M\to{M^{\prime}}^{*}$ with the property that for each $i$, $0<i<n$, there exists a ${k}$-linear map $f_{i}:M\to{M^{\prime}}^{*}$ such that $f=T_{i}f_{i}-f_{i}T_{i}$, i.e. $f=T_{i}f_{i}+f_{i}\mathaccent 869{T}_{i}+2f_{i}$. Here ${M^{\prime}}^{*}$ is a left ${\cal H}_{n}$-module with respect to the action of ${\cal H}_{n}$ arising naturally from the right action on $M^{\prime}$. By Lemmas 1.4, 1.5

[TABLE]

By Lemma 1.7 the ${k}$-linear maps $f$ considered above are precisely those ${\cal H}_{n}$-module homomorphisms $\mathaccent 869{M}\to{M^{\prime}}^{*}$ that factor through a free module, and the space of such homomorphisms has a basis indexed by the set ${}_{\lambda}^{\vphantom{e}}{\cal D}_{\mu}^{e}$. Since this set is in a bijection with ${}_{\mu}{\cal D}_{\lambda}^{e}$ by the map $\sigma\mapsto\sigma^{-1}$, the space ${\cal M}/\Sigma$ has dimension equal to the cardinality of ${}_{\mu}{\cal D}_{\lambda}^{e}$, and we are done.   $\mathchar 9219\relax$

6. Intertwining algebras for a pair of Hecke symmetries

Let $V$ and $V^{\prime}$ be two finite dimensional vector spaces over the field ${k}$. Let $R$ be a Hecke symmetry on $V$ and $R^{\prime}$ a Hecke symmetry on $V^{\prime}$ satisfying the Hecke relation with the same parameter $q$. For each $n\geq 0$ we will view ${{T}}_{n}(V)$ and ${{T}}_{n}(V^{\prime})$ as left modules over the Hecke algebra ${\cal H}_{n}={\cal H}_{n}(q)$ with respect to the representations arising from $R$ and $R^{\prime}$.

The tensor algebra ${{T}}({V^{\prime}}^{*}\otimes V)$ embeds canonically into ${{T}}({V^{\prime}}^{*})\otimes{{T}}(V)$. Under this embedding ${{T}}_{n}({V^{\prime}}^{*}\otimes V)$ is mapped onto ${{T}}_{n}({V^{\prime}}^{*})\otimes{{T}}_{n}(V)$. Identify ${{T}}_{n}(V^{*})$ and ${{T}}_{n}({V^{\prime}}^{*})$ with the duals of the vector spaces ${{T}}_{n}(V)$ and ${{T}}_{n}(V^{\prime})$ as in section 3.

Let ${R^{\prime}}^{*}$ be the Hecke symmetry on ${V^{\prime}}^{*}$ adjoint to $R^{\prime}$ (see Lemma 3.2). The inverse operator $({R^{\prime}}^{*})^{-1}$ is a Hecke symmetry with parameter $q^{-1}$. Denote by ${\cal R}$ the linear operator on ${{T}}_{2}({V^{\prime}}^{*}\otimes V)$ which corresponds to the operator $({R^{\prime}}^{*})^{-1}\otimes R$ acting on ${{T}}_{2}({V^{\prime}}^{*})\otimes{{T}}_{2}(V)$.

Define $A(R^{\prime},R)$ and $E(R^{\prime},R)$ as the factor algebras of ${{T}}({V^{\prime}}^{*}\otimes V)$ by the ideals generated, respectively, by $\mathop{\rm Im}\,({\cal R}-\mathop{\rm Id}\nolimits)$ and $\mathop{\rm Ker}\,({\cal R}-\mathop{\rm Id}\nolimits)$. Under the isomorphism

[TABLE]

these two subspaces of ${{T}}_{2}({V^{\prime}}^{*}\otimes V)$ are mapped, respectively, onto

[TABLE]

In the case when $R^{\prime}=R$ the first subspace gives a well-known presentation of the FRT bialgebra $A(R)=A(R,R)$ (see [11, section 4]).

Lemma 6.1. The quadratic dual algebras $A(R^{\prime},R)^{!}$ and $E(R^{\prime},R)^{!}$ are isomorphic,* respectively*,* to $E(R,R^{\prime})$ and $A(R,R^{\prime})$. If $q\neq-1,$ then $A(R^{\prime},R)=E(R^{\prime},\mathaccent 869{R})$ and $E(R^{\prime},R)=A(R^{\prime},\mathaccent 869{R})$ where $\mathaccent 869{R}=(q-1)\mathop{\rm Id}\nolimits-R$.*

Proof. Identifying the dual space of ${{T}}_{2}({V^{\prime}}^{*})\otimes{{T}}_{2}(V)$ with ${{T}}_{2}(V^{*})\otimes{{T}}_{2}(V^{\prime})$, we have

[TABLE]

The subspaces of ${{T}}_{2}(V^{*})\otimes{{T}}_{2}(V^{\prime})$ in the left hand sides of these equalities define the algebras $A(R^{\prime},R)^{!}$ and $E(R^{\prime},R)^{!}$.

If $q\neq-1$, then $\mathop{\rm Id}\nolimits\otimes R$ and ${R^{\prime}}^{*}\otimes\mathop{\rm Id}\nolimits$ are commuting diagonalizable operators with two eigenvalues $-1$ and $q$. Since ${{T}}_{2}({V^{\prime}}^{*})\otimes{{T}}_{2}(V)$ is a sum of common eigenspaces of these two operators, it is clear that

[TABLE]

We thus obtain the second conclusion.   $\mathchar 9219\relax$

Put ${\cal H}^{\prime}_{n}={\cal H}_{n}(q^{-1})$, as in section 5. Consider ${{T}}_{n}({V^{\prime}}^{*})$ as a left ${\cal H}^{\prime}_{n}$-module with respect to the representation arising from the Hecke symmetry $({R^{\prime}}^{*})^{-1}$. We thus obtain a left ${\cal H}^{\prime}_{n}\otimes{\cal H}_{n}$-module structure on ${{T}}_{n}({V^{\prime}}^{*}\otimes V)\cong{{T}}_{n}({V^{\prime}}^{*})\otimes{{T}}_{n}(V)$.

Recall that ${\cal T}_{i}=T^{\prime}_{i}\otimes T_{i}\in{\cal H}^{\prime}_{n}\otimes{\cal H}_{n}$ where $T^{\prime}_{1},\ldots,T^{\prime}_{n-1}$ are the standard generators of ${\cal H}^{\prime}_{n}$. The action of ${\cal T}_{i}$ on ${{T}}_{n}({V^{\prime}}^{*}\otimes V)$ is given by the operator

[TABLE]

Theorem 6.2. Suppose that both $R$ and $R^{\prime}$ satisfy the $1$-dimensional source condition. Then the graded algebras $A(R^{\prime},R)$ and $E(R^{\prime},R)$ are Koszul. Their Hilbert series satisfy the relation  $h_{A(R^{\prime},R)}(t)\,h_{E(R^{\prime},R)}(-t)=1$.

Proof. Put $U_{i}^{(n)}={{T}}_{i-1}({V^{\prime}}^{*}\otimes V)\otimes U\otimes{{T}}_{n-i-1}({V^{\prime}}^{*}\otimes V)\subset{{T}}_{n}({V^{\prime}}^{*}\otimes V)$ for a subspace $U\subset{{T}}_{2}({V^{\prime}}^{*}\otimes V)$ and $i=1,\ldots,n-1$. We have

[TABLE]

In both cases the complex K_{\bullet}\bigl{(}{{T}}_{n}({V^{\prime}}^{*}\otimes V),(U_{i}^{(n)})\bigr{)} is exact for each $n>0$ by Propositions 5.1 and 5.9. But these complexes are precisely the direct summands in the decomposition of the right Koszul complex $K_{\bullet}(A)$ (see section 3) where $A$ is $A(R^{\prime},R)$ in the first case and $E(R^{\prime},R)$ in the second. Hence $K_{\bullet}(A)$ is acyclic in all positive degrees.

Thus $A(R^{\prime},R)$ and $E(R^{\prime},R)$ are Koszul. The Hilbert series of the algebra $E(R^{\prime},R)^{!}$ is $\sum\,(\dim\Upsilon^{(n)})t^{n}$ where the spaces $\Upsilon^{(n)}$ are determined in Lemma 6.4 below. Making use also of Lemmas 6.3 and 1.2, we get

[TABLE]

This shows that $h_{A(R^{\prime},R)}(t)=h_{E(R^{\prime},R)^{!}}(t)$, and the final conclusion in the statement of Theorem 6.2 reduces to the standard relation between the Hilbert series of the Koszul algebra $E(R^{\prime},R)$ and its quadratic dual.   $\mathchar 9219\relax$

Lemma 6.3. \,\,A_{n}(R^{\prime},R)\cong\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}\bigl{(}{{T}}_{n}(V),{{T}}_{n}(V^{\prime})\bigr{)}^{*}.

Proof. The ideal $I$ of the algebra ${{T}}({V^{\prime}}^{*}\otimes V)$ defining its factor algebra $A(R^{\prime},R)$ has homogeneous components $I_{n}=0$ for $n\leq 1$ and $I_{n}=\sum_{i=1}^{n-1}U_{i}^{(n)}$ for $n>1$ where $U_{i}^{(n)}$ are the subspaces of ${{T}}_{n}({V^{\prime}}^{*}\otimes V)$ defined in the proof of Theorem 6.2 with $U=\mathop{\rm Im}\,({\cal R}-\mathop{\rm Id}\nolimits)$.

The right ${\cal H}_{n}$-module structure on ${{T}}_{n}(V^{\prime})^{*}$ obtained in a natural way from the left module structure on ${{T}}_{n}(V^{\prime})$ allows us to view ${{T}}_{n}({V^{\prime}}^{*}\otimes V)\cong{{T}}_{n}(V^{\prime})^{*}\otimes{{T}}_{n}(V)$ as an ${\cal H}_{n}$-bimodule. The left action of $T^{\prime}_{i}$ on ${{T}}_{n}({V^{\prime}}^{*})\cong{{T}}_{n}(V^{\prime})^{*}$ is the same as the right action of $T_{i}^{-1}$. Then $T^{\prime}_{i}a=aT_{i}^{-1}$, and so ${\cal T}_{i}a=T_{i}T^{\prime}_{i}a=T_{i}aT_{i}^{-1}$, for all $a\in{{T}}_{n}({V^{\prime}}^{*}\otimes V)$. Hence $U_{i}^{(n)}$ is spanned by all elements $T_{i}aT_{i}^{-1}-a$ or, equivalently, by all elements $T_{i}a-aT_{i}$ with $a\in{{T}}_{n}({V^{\prime}}^{*}\otimes V)$. Under the canonical isomorphisms of ${\cal H}_{n}$-bimodules

[TABLE]

the orthogonal of $U_{i}^{(n)}$ in that vector space is

[TABLE]

Hence \,A_{n}(R^{\prime},R)^{*}\cong I_{n}^{\perp}=\bigcap_{i=1}^{n-1}(U_{i}^{(n)})^{\perp}=\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}\bigl{(}{{T}}_{n}(V),{{T}}_{n}(V^{\prime})\bigr{)}.   $\mathchar 9219\relax$

Lemma 6.4. The right Koszul complex for the algebra $E(R^{\prime},R)$ has components

[TABLE]

Proof. Recall from section 3 that $\Upsilon^{(n)}=\bigcap_{i=1}^{n-1}U_{i}^{(n)}$ where $U_{i}^{(n)}$ are as defined in the proof of Theorem 6.2 with $U=\mathop{\rm Ker}\,({\cal R}-\mathop{\rm Id}\nolimits)$. In terms of the ${\cal H}_{n}$-bimodule structure on ${{T}}_{n}({V^{\prime}}^{*}\otimes V)$ we have

[TABLE]

for each $i$. It follows that $\Upsilon^{(n)}=\{a\in{{T}}_{n}({V^{\prime}}^{*}\otimes V)\mid xa=ax\hbox{ for all }x\in{\cal H}_{n}\}$. The canonical isomorphisms of ${\cal H}_{n}$-bimodules

[TABLE]

map $\Upsilon^{(n)}$ onto \mathop{\rm Hom}\nolimits_{{\cal H}_{n}}\bigl{(}{{T}}_{n}(V^{\prime}),{{T}}_{n}(V)\bigr{)}.   $\mathchar 9219\relax$

Lemma 6.5. Let $I$ be the ideal of the algebra ${{T}}={{T}}({V^{\prime}}^{*}\otimes V)$ defining its factor algebra $E(R^{\prime},R)$. Fix some $n>1$ and put

[TABLE]

where ${\cal T}_{\sigma}=T_{\sigma}T^{\prime}_{\sigma}\,$. If $0<k<n,$ then $\,y_{k}L_{k}\subset{{T}}_{1}L_{k-1}$.

This lemma is proved by exactly the same arguments as those used for Lemma 4.1 (see the remarks following the proof of that lemma).

Theorem 6.6. Suppose that both $R$ and $R^{\prime}$ satisfy the trivial source condition. If $\dim E_{n}(R^{\prime},R)=1$ and $E_{n+1}(R^{\prime},R)=0$ for some integer $n>0,$ then $E(R^{\prime},R)$ is a Frobenius algebra,* while $A(R^{\prime},R)$ is a Gorenstein algebra of global dimension $n$.*

Proof. We proceed as in the proof of Theorem 4.5. Let ${{T}}$ and $I$ be as in Lemma 6.5. By induction on $k$ we can show that $L_{k}=I_{k}$ for each $k=0,\ldots,n-1$. Indeed, if the equality $L_{k-1}=I_{k-1}$ holds for some $k$, then $y_{k}L_{k}\subset{{T}}_{1}I_{k-1}$ by Lemma 6.5. Now we apply Lemma 5.10 with $n$ replaced by $k$ and ${\cal M}={{T}}_{k}$. In the notation of that lemma we then have $y=y_{k}$, $\Sigma=I_{k}$ and $\Sigma_{1}={{T}}_{1}I_{k-1}$. The inclusion $yL_{k}\subset\Sigma_{1}$ entails $L_{k}\subset y_{\cal M}^{-1}(\Sigma_{1})=\Sigma$, i.e. $L_{k}=I_{k}$. Thus the multiplication pairing

[TABLE]

has zero left kernel. Since this holds also with $k$ replaced by $n-k$, comparison of dimensions shows that the pairing is nondegenerate. This means that the algebra $E(R^{\prime},R)$ is Frobenius. By Lemma 6.1 $A(R^{\prime},R)^{!}\cong E(R,R^{\prime})$. Since

[TABLE]

we have $\dim E_{n}(R,R^{\prime})=1$ and $E_{n+1}(R,R^{\prime})=0$. Hence $E(R,R^{\prime})$ is also Frobenius, and $A(R^{\prime},R)$ is Gorenstein by [18, Remark 2 on p. 25].   $\mathchar 9219\relax$

7. Monoidal equivalences

Let $V,V^{\prime},V^{\prime\prime}$ be three finite dimensional vector spaces over ${k}$ and $R,R^{\prime},R^{\prime\prime}$ Hecke symmetries on the respective spaces with the same parameter $q$. For each $n\geq 0$ we equip $\,{{T}}_{n}={{T}}_{n}(V)$, $\,{{T}}^{\prime}_{n}={{T}}_{n}(V^{\prime})\,$ and $\,{{T}}^{\prime\prime}_{n}={{T}}_{n}(V^{\prime\prime})\,$ with the ${\cal H}_{n}$-module structures arising from $R,R^{\prime},R^{\prime\prime}$. There is a ${k}$-linear map

[TABLE]

whose dual map $\,\Delta_{n}^{*}:A_{n}(R^{\prime},R)^{*}\otimes A_{n}(R,R^{\prime\prime})^{*}{}\mathrel{\smash{\mathop{\longrightarrow}\limits}}{}A_{n}(R^{\prime},R^{\prime\prime})^{*}\,$ is, in terms of the identifications of Lemma 6.3, the map

[TABLE]

given by the composition of homomorphisms. In particular, $A_{n}(R)$ is endowed with a comultiplication dual to the multiplication in the algebra $\,\mathop{\rm End}\nolimits_{{\cal H}_{n}}\!{{T}}_{n}$. Thus $A_{n}(R)$ is a coalgebra. Also, $A_{n}(R^{\prime},R)$ has an $A_{n}(R^{\prime}),A_{n}(R)$-bicomodule structure dual to the $\,\mathop{\rm End}\nolimits_{{\cal H}_{n}}\!{{T}}^{\prime}_{n}\,,\mathop{\rm End}\nolimits_{{\cal H}_{n}}\!{{T}}_{n}\,$-bimodule structure on $\mathop{\rm Hom}\nolimits_{{\cal H}_{n}}({{T}}_{n},{{T}}^{\prime}_{n})$.

Let ${\cal H}_{m,n}$ be the parabolic subalgebra of ${\cal H}_{m+n}$ generated by $\{T_{i}\mid i\neq m\}$. Then

[TABLE]

Noting that the multiplication maps $\,A_{m}(R^{\prime},R)\otimes A_{n}(R^{\prime},R){}\mathrel{\smash{\mathop{\longrightarrow}\limits}}{}A_{m+n}(R^{\prime},R)\,$ are dual to the inclusion maps

[TABLE]

it is easy to see that the map

[TABLE]

obtained from the previously defined maps $\Delta_{n}$ on the homogeneous components, is an algebra homomorphism. In particular, $A(R)$ is a bialgebra, while $A(R^{\prime},R)$ is an $A(R^{\prime}),A(R)$-bicomodule algebra.

If $C$ is a coalgebra, $\rho:X\to X\otimes C$ and $\lambda:Y\to C\otimes Y$ are right and left comodule structures on two vector spaces, then the cotensor product $X\mathchar 9219\relax_{C}Y$ is defined as the kernel of the ${k}$-linear map

[TABLE]

Suppose that $X$, $Y$ and $C$ are finite dimensional. Passing to the dual spaces, we get then an exact sequence

[TABLE]

which shows that $(X\mathchar 9219\relax_{C}Y)^{*}\cong X^{*}\otimes_{C^{*}}Y^{*}$. Since $\Delta_{n}^{*}$ factors through

[TABLE]

it follows that $\,\mathop{\rm Im}\Delta_{n}\subset A_{n}(R^{\prime},R)\mathchar 9219\relax_{A_{n}(R)}A_{n}(R,R^{\prime\prime})$ for each $n$. Hence $\Delta_{R^{\prime}\!,R,R^{\prime\prime}}$ is in fact an algebra homomorphism $\,A(R^{\prime},R^{\prime\prime}){}\mathrel{\smash{\mathop{\longrightarrow}\limits}}{}A(R^{\prime},R)\mathchar 9219\relax_{A(R)}A(R,R^{\prime\prime})$.

Lemma 7.1. Suppose that for each $n>1$ each indecomposable direct summand of the ${\cal H}_{n}$-module ${{T}}_{n}(V^{\prime})$ is isomorphic to a direct summand of the ${\cal H}_{n}$-module ${{T}}_{n}(V)$. Then $\Delta_{R^{\prime}\!,R,R^{\prime\prime}}$ gives an isomorphism of algebras

[TABLE]

Proof. For any ${\cal H}_{n}$-modules $X,Y,Z$ there is a canonical map

[TABLE]

which is obviously bijective when $Y=X$. Since the collection of these maps with varying $Y$ give a natural transformation of additive functors of $Y$, such a map is bijective also when $Y$ is a direct sum of modules isomorphic to direct summands of $X$. In particular, we may take $X={{T}}_{n}(V)$, $Y={{T}}_{n}(V^{\prime})$, $Z={{T}}_{n}(V^{\prime\prime})$. Passing to the dual spaces, we deduce that $\Delta_{n}$ maps $A_{n}(R^{\prime},R^{\prime\prime})$ bijectively onto $A_{n}(R^{\prime},R)\mathchar 9219\relax_{A_{n}(R)}A_{n}(R,R^{\prime\prime})$.   $\mathchar 9219\relax$

Theorem 7.2. Suppose that for each $n>1$ the indecomposable ${\cal H}_{n}$-modules isomorphic to direct summands of ${{T}}_{n}(V^{\prime})$ are the same as those isomorphic to direct summands of ${{T}}_{n}(V)$. Then the functors

[TABLE]

are braided monoidal equivalences $\,\vphantom{{\cal M}}{}^{A(R)}{\cal M}{}\mathrel{\smash{\mathop{\longrightarrow}\limits}}\vphantom{{\cal M}}{}^{A(R^{\prime})}{\cal M}\,$ and $\,{\cal M}^{A(R)}{}\mathrel{\smash{\mathop{\longrightarrow}\limits}}{}{\cal M}^{A(R^{\prime})}$.

Proof. The functor $F=A(R^{\prime},R)\mathchar 9219\relax_{A(R)}\hbox{}?$ has a quasiinverse $F^{\prime}=A(R,R^{\prime})\mathchar 9219\relax_{A(R^{\prime})}\hbox{}?$ since

[TABLE]

for left $A(R)$-comodules $X$ by the associativity of cotensor products and by Lemma 7.1, and, similarly, $FF^{\prime}\cong\mathop{\rm Id}\nolimits$. Thus $F$ is a category equivalence.

There are homomorphisms of $A(R^{\prime})$-comodules $\xi_{XY}:F(X)\otimes F(Y)\to F(X\otimes Y)$, natural in $X,Y\in\vphantom{{\cal M}}^{A(R)}{\cal M}$, obtained as restrictions of the maps

[TABLE]

We will show that $\xi_{XY}$ is an isomorphism. Since $A(R)$ is a direct sum of its subcoalgebras $A_{n}(R)$, each left $A(R)$-comodule $X$ can be written as $X=\oplus_{n=0}^{\infty}X_{n}$ where $X_{n}$ is a left $A_{n}(R)$-comodule for each $n$. Therefore it suffices to prove bijectivity of $\xi_{XY}$ assuming $X$ to be an $A_{m}(R)$-comodule, $Y$ an $A_{n}(R)$-comodule for some $m,n$. Since every comodule is a sum of finite dimensional subcomodules, we may also assume that $\dim X<\infty$ and $\dim Y<\infty$. Then $\,F(X)=A_{m}(R^{\prime},R)\mathchar 9219\relax_{A_{m}(R)}X$, and

[TABLE]

where $\,{{T}}_{m}={{T}}_{m}(V)$, $\,{{T}}^{\prime}_{m}={{T}}_{m}(V^{\prime})$.  Similarly,

[TABLE]

Identifying ${\cal H}_{m}\otimes{\cal H}_{n}$ with the subalgebra ${\cal H}_{m,n}$ of ${\cal H}_{m+n}$, we get

[TABLE]

The dual map $F(X\otimes Y)^{*}\to F(X)^{*}\otimes F(Y)^{*}$ is identified with the canonical map

[TABLE]

arising from the inclusions $\mathop{\rm Hom}\nolimits_{{\cal H}_{m+n}}({{T}}_{m+n},{{T}}^{\prime}_{m+n})\subset\mathop{\rm Hom}\nolimits_{{\cal H}_{m,n}}({{T}}_{m+n},{{T}}^{\prime}_{m+n})$ and $\mathop{\rm End}\nolimits_{{\cal H}_{m+n}}\!\!{{T}}_{m+n}\subset\mathop{\rm End}\nolimits_{{\cal H}_{m,n}}\!\!{{T}}_{m+n}$. This map is bijective by Lemma 7.3 below.

We have shown that $F(X)\otimes F(Y)\cong F(X\otimes Y)$, naturally in $X$ and $Y$. Coherence of these isomorphisms is clear from the construction. The trivial $A(R)$-comodule $A_{0}(R)={k}$ is sent by $F$ to the trivial $A(R^{\prime})$-comodule $A_{0}(R^{\prime},R)={k}$. Thus $F$ is a monoidal equivalence.

Let $b$ and $b^{\prime}$ be the braidings in the categories $\vphantom{{\cal M}}{}^{A(R)}{\cal M}$ and $\vphantom{{\cal M}}{}^{A(R^{\prime})}{\cal M}$, respectively, such that b_{\hbox{\scriptstyle V^{}V^{}}}=R^{*} and b_{\hbox{\scriptstyle{V^{\prime}}^{}{V^{\prime}}^{}}}={R^{\prime}}^{*}\!. We have to verify commutativity of the diagrams

[TABLE]

In fact, it suffices to do this only for $X=Y=V^{*}$. By the general properties of the braidings the diagram will then be commutative for $X={{T}}_{m}(V^{*})$, $Y={{T}}_{n}(V^{*})$, and therefore also when $X$ and $Y$ are subfactors of direct sums of left $A(R)$-comodules isomorphic to tensor powers of $V^{*}$. But every finite dimensional left $A(R)$-comodule is realized in this way since ${{T}}_{n}(V^{*})\cong{{T}}_{n}^{*}$ is a faithful right $\mathop{\rm End}\nolimits_{{\cal H}_{n}}\!{{T}}_{n}$-module for each $n$. For infinite dimensional comodules commutativity of the diagram will follow from the fact that $F$ commutes with inductive direct limits. Now

[TABLE]

The last isomorphism here is explained by the fact that the evaluation map

[TABLE]

is obviously bijective when $M={{T}}_{n}$, and therefore it is bijective whenever $M$ is a direct sum of ${\cal H}_{n}$-modules isomorphic to direct summands of ${{T}}_{n}$. This can be applied with $M={{T}}^{\prime}_{n}$. Note that the bijection $F({{T}}_{n}^{*})^{*}\cong{{T}}^{\prime}_{n}$ obtained in this way is an isomorphism of ${\cal H}_{n}$-modules.

In particular, we have $F(V^{*})\cong{V^{\prime}}^{*}$ and $F({{T}}_{2}^{*})\cong{{{T}}^{\prime}_{2}}^{\!*}$. The generator $T_{1}$ of ${\cal H}_{2}$ acts via $R$ on ${{T}}_{2}$ and via $R^{\prime}$ on ${{T}}^{\prime}_{2}$. This entails the commutativity of the diagram

[TABLE]

Thus we have verified all the required properties of the functor $F$. Consideration of the other case in the statement of Theorem 7.2 is quite similar.   $\mathchar 9219\relax$

Lemma 7.3. Let $A$ be a ring,* $B$ its subring*,* $X$ a left $A$-module*,* and let $M$ be a left $\mathop{\rm End}\nolimits_{B}X$-module. The canonical map*

[TABLE]

is bijective whenever $Y$ is a finite direct sum of left $A$-modules isomorphic to direct summands of $X$.

Proof. The conclusion is obvious when $Y=X$. Since we deal here with a natural transformation of two additive functors of $Y$, the conclusion then holds in full generality.   $\mathchar 9219\relax$

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