Specht modules for quiver Hecke algebras of type $C$

TL;DR

This paper constructs Specht modules for cyclotomic quiver Hecke algebras of type C, providing explicit bases and character formulas in infinite type, and proposing conjectures for affine type.

Contribution

It introduces Specht modules for type C quiver Hecke algebras, detailing their bases, properties, and character formulas, and extends the theory to affine type with conjectures.

Findings

Homogeneous bases for Specht modules in type C∞

Graded character formulas for these modules

Properties under exact functors $E_i^Λ$ and $F_i^Λ$

Abstract

We construct and investigate Specht modules $\mathcal{S}^\lambda$ for cyclotomic quiver Hecke algebras in type $C^{(1)}_\ell$ and $C_\infty$, which are labelled by multipartitions $\lambda$. It is shown that in type $C_\infty$, the Specht module $\mathcal{S}^\lambda$ has a homogeneous basis indexed by standard tableaux of shape $\lambda$, which yields a graded character formula and good properties with the exact functors $E_i^\Lambda$ and $F_i^\Lambda$. For type $C^{(1)}_\ell$, we propose a conjecture.

Equations

$$a_{ij}=\{\begin{array}{ll}2& \text{if }i=j,\\ -2& \text{if }(i,j)=(1,0),\\ -1& \text{if }i=j\pm 1\text{ and }(i,j)\ne (1,0),\\ 0& \text{otherwise}.\end{array}$$

$$\mathsf{A}=(a_{ij}{)}_{i,j\in I}=\left(\begin{array}{ccccccc}2& -1& 0& \cdots & 0& 0& 0\\ -2& 2& -1& \cdots & 0& 0& 0\\ 0& -1& 2& \cdots & 0& 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 0& 0& 0& \cdots & 2& -1& 0\\ 0& 0& 0& \cdots & -1& 2& -2\\ 0& 0& 0& \cdots & 0& -1& 2\end{array}\right)$$

$${\mathsf{P}}^{+}=\{\Lambda \in \mathsf{P}\mid ⟨{\alpha }_i^{\vee },\Lambda ⟩⩾0\text{ for all }i\in I\}$$

$$w[a,b](x)=\left\{\begin{array}[]{ll}x+b&\hbox{ if }1\leqslant x\leqslant a,\\ x-a&\hbox{ if }a<x\leqslant a+b,\\ x&\hbox{ if }a+b<x\leqslant n.\end{array}\right.$$

$$\bigl{(}\begin{smallmatrix}1&2&\cdots&a&a+1&a+2&\cdots&a+b\\ b+1&b+2&\cdots&b+a&1&2&\cdots&b\end{smallmatrix}\bigr{)}.$$

$$w[a,b]\nu =({\nu }_{a+1},\dots ,{\nu }_{a+b},{\nu }_1,\dots ,{\nu }_a,{\nu }_{a+b+1},\dots ,{\nu }_n).$$

$$s_{b+1}w[2,b]$$

$$=(s_b{s}_{b+1})\underline{s_b}(\underline{s_{b-1}}{s}_{b-2}\dots s_1)(\underline{s_b}{s}_{b-1}\dots s_2)$$

$$=(s_b{s}_{b+1})(s_{b-1}{s}_b)\underline{s_{b-1}}(\underline{s_{b-2}}\dots s_1)(\underline{s_{b-1}}\dots s_2)$$

$$\dots \dots \dots \dots$$

$$=(s_b{s}_{b+1})(s_{b-1}{s}_b)\dots (s_2{s}_3)\underline{s_2}(\underline{s_1})(\underline{s_2})$$

$$=(s_b{s}_{b+1})(s_{b-1}{s}_b)\dots (s_2{s}_3)(s_1{s}_2)s_1=w[2,b]s_1,$$

$${\mathsf{sh}}_k(w)=s_{i_1+k}\dots s_{i_r+k}.$$

$$|{\lambda }^{(1)}|+\cdots +|{\lambda }^{(t-1)}|+\sum _{j=1}^k{\lambda }_j^{(t)}⩾|{\mu }^{(1)}|+\cdots +|{\mu }^{(t-1)}|+\sum _{j=1}^k{\mu }_j^{(t)}$$

$$\{(r,c,t)\in {\mathbb{Z}}_{⩾1}\times {\mathbb{Z}}_{⩾1}\times \{1,\dots ,l\}\mid c⩽{\lambda }_r^{(t)}\}.$$

$$\displaystyle[\lambda]=\$$

$$\varnothing$$

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$$f_{\ell }(0+2\ell \mathbb{Z})=0,f_{\ell }(\ell +2\ell \mathbb{Z})=\ell ,$$

$$f_{\ell }(k+2\ell \mathbb{Z})=f_{\ell }(2\ell -k+2\ell \mathbb{Z})=k\text{ for }1⩽k⩽\ell -1.$$

$$\mathrm{res}(A)=\overline{{\kappa }_t+c-r}.$$

$$\mathrm{cont}(\lambda )=\sum _{A\in [\lambda ]}{\alpha }_{\mathrm{res}(A)}\in {\mathsf{Q}}^{+}.$$

$$\varnothing$$

$$d_i(\lambda )$$

$$d_A(\lambda )$$

$$d^A(\lambda )$$

$$q^{h_i}\lambda =q^{d_i(\lambda )}\lambda ,e_i\lambda =\sum _A{q}^{d_A(\lambda )}\lambda \nearrow A,f_i\lambda =\sum _A{q}^{-d^A(\lambda )}\lambda \swarrow A,$$

$$\mathcal{F}(\kappa )\simeq \mathcal{F}({\kappa }_1)\otimes \cdots \otimes \mathcal{F}({\kappa }_l).$$

$$\Delta :K_i⟼K_i\otimes K_i,e_i⟼e_i\otimes K_i+1\otimes e_i,f_i⟼f_i\otimes 1+K_i^{-1}\otimes f_i,$$

$$p_{\kappa }:\mathcal{F}(\kappa )\text{relbar}\text{joinrel}\twoheadrightarrow V_q(\Lambda ),\varnothing ⟼v_{\Lambda },$$

$$\mathrm{res}(\mathtt{T})=(\mathrm{res}({\mathtt{T}}^{-1}(1)),\mathrm{res}({\mathtt{T}}^{-1}(2)),\dots ,\mathrm{res}({\mathtt{T}}^{-1}(n))).$$

$$\displaystyle\mathtt{T}^{\lambda}=\$$

$$\varnothing$$

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$$\mathrm{deg}\mathtt{T}:=\mathrm{deg}{\mathtt{T}}_{\downarrow n-1}+d_A(\lambda ).$$

$$\mathrm{deg}\mathtt{T}=0+1+0+0+1+2+0+0+0-1=3.$$

Full text

1 S.A. is supported by the JSPS Grant-in-Aid for Scientific Research 15K04782.

2 E.P. is supported by the National Research Foundation of Korea(NRF) Grant funded by the Korean Government(MSIP)(NRF-2014R1A1A1002178).

3 L.S. was an International Research Fellow of the Japan Society for the Promotion of Science while this research was conducted.

Introduction

Representations of Hecke algebras and the symmetric group have been studied for over a century and Specht modules play important roles in the representation theory. Nowadays, we realise that on the one hand the Hecke algebras generalise to cyclotomic quiver Hecke algebras (or Khovanov–Lauda–Rouquier algebras) in the direction of categorification of quantum groups $U_{q}(\mathfrak{g})$ [17, 18, 25], and on the other hand that the Hecke algebras are cellular algebras and Specht modules are their cell modules.

In the affine type $A_{\ell}^{(1)}$ case, cellular algebras and Specht modules for cyclotomic quiver Hecke algebras were studied via the isomorphism to cyclotomic Hecke algebras given in [6]. It was shown that the cyclotomic quiver Hecke algebras of affine type $A_{\ell}^{(1)}$ have graded cellular structure [12]. Graded Specht modules in type $A_{\ell}^{(1)}$ were constructed and studied using the combinatorics of multipartitions [8, 21]. But so far little is known about Specht modules for cyclotomic quiver Hecke algebras of other types. We remark that it was proved in [20] that quiver Hecke algebras of finite type are graded affine cellular algebras.

The first and second authors [3] studied cyclotomic quiver Hecke algebras $R^{\Lambda_{0}}(n)$ for the fundamental weight $\Lambda_{0}$ in type $C^{(1)}_{\ell}$, in which a graded dimension formula for $R^{\Lambda_{0}}(n)$ is given by using the $C$-type Fock space $\mathcal{F}$ [15, 19, 24]. This Fock space $\mathcal{F}$ is constructed by folding the usual $A$-type Fock space, so the dimension formula is described in terms of combinatorics of Young diagrams. In affine type $A_{\ell}^{(1)}$, graded Specht modules are deeply related to the $A$-type Fock space. It was shown in [7] that the graded decomposition numbers of graded Specht modules can be described in terms of combinatorics of Young diagrams via the Fock space which is the $q$-version of the first author’s result [1]. One can expect that cyclotomic quiver Hecke algebras of type $C$ and the Fock space $\mathcal{F}$ of type $C$ exhibit similar properties to those of type $A$ – thus it is worth considering Specht modules for cyclotomic quiver Hecke algebras of type $C$.

In this paper, we construct Specht modules $\mathcal{S}^{\lambda}$ for cyclotomic quiver Hecke algebras of affine type $C_{\ell}^{(1)}$ and type $C_{\infty}$ which are labelled by multipartitions $\lambda$. This is inspired by [21]. Let $\mathsf{A}$ be the Cartan matrix of type $C_{\ell}^{(1)}$ or $C_{\infty}$, and $U_{q}(\mathsf{A})$ the quantum group associated with $\mathsf{A}$. We set $R(\beta)$ to be the quiver Hecke algebra associated with $\mathsf{A}$ and denote by $E_{i}^{\Lambda}$ and $F_{i}^{\Lambda}$ the functors categorifying Chevalley generators $e_{i}$ and $f_{i}$ of $U_{q}(\mathsf{A})$ on the highest weight irreducible module $V_{q}(\Lambda)$. Let $\mathscr{P}^{l}_{n}$ be the set of $l$-multipartitions of $n$ with a multicharge $\kappa=(\kappa_{1},\dots,\kappa_{l})\in\mathbb{Z}^{l}$. For $\lambda\in\mathscr{P}^{l}_{n}$, we first construct a permutation module $\mathcal{M}^{\lambda}_{\kappa}$ which has a basis indexed by row-strict tableaux of $\lambda$. These permutation modules are built from more fundamental building blocks, namely they are convolution products of the one-dimensional $R(\beta)$-modules $\mathcal{L}(k;\ell)$ defined by $\eqref{Eq: def of L}$. The modules $\mathcal{L}(k;\ell)$ take a role as the segment modules, which are given in [21], corresponding to segments in type $A_{\infty}$ and $A_{n}^{(1)}$. We also define a module $\mathcal{M}^{\lambda}_{\kappa,A}$ for each Garnir node $A\in[\lambda]$ and construct homomorphisms between $\mathcal{M}^{\lambda}_{\kappa,A}$ and $\mathcal{M}^{\lambda}_{\kappa}$, which give an interpretation of Garnir elements in terms of quiver Hecke algebras. We then define a Specht module $\mathcal{S}^{\lambda}$ using the cokernel of homomorphisms between $\mathcal{M}^{\lambda}_{\kappa,A}$ and $\mathcal{M}^{\lambda}_{\kappa}$, see Definition 3.8. The Specht module $\mathcal{S}^{\lambda}$ is spanned by homogeneous elements indexed by standard tableaux of shape $\lambda$ (Corollary 3.13). We prove in Corollary 3.21 that, in type $C_{\infty}$, this spanning set of $\mathcal{S}^{\lambda}$ is in fact a basis. Thus we have a graded character formula for $\mathcal{S}^{\lambda}$ in terms of standard tableaux and a description of $[E_{i}^{\Lambda}\mathcal{S}^{\lambda}]$ in terms of $[\mathcal{S}^{\lambda\nearrow b}]$. Here, $\lambda\nearrow b$ is the Young diagram obtained from $\lambda$ by deleting a removable node $b$. We remark that $\mathcal{S}^{\lambda}$ is not necessarily simple, even in the case of level 1 and type $C_{\infty}$. We also investigate a connection between Specht modules $\mathcal{S}^{\lambda}$ and the Fock space $\mathcal{F}$ of type $C$, which provides a description of $[F_{i}^{\Lambda}\mathcal{S}^{\lambda}]$ in terms of $[\mathcal{S}^{\lambda\swarrow b}]$, where $\lambda\swarrow b$ is the Young diagram obtained from $\lambda$ by adding an addable node $b$, see Corollary 3.23. Recently, the third author provided semisimplicity criteria for the cyclotomic quiver Hecke algebras of type $C_{\infty}$ and $C^{(1)}_{n}$ using the Specht modules [26].

The paper is organised as follows. In Section 1, we review the combinatorics of tableaux and the Fock space of type $C$, and prove lemmas on Garnir nodes. In Section 2, we recall the notion of quiver Hecke algebras, and prove several lemmas on computations of products of $\psi_{i}$ and convolution products of modules for proving our main theorem. In Section 3, we construct and investigate Specht modules $\mathcal{S}^{\lambda}$ and provide the main theorems with examples. Section 4 is devoted to proving Theorem 3.19. We may carry out the computation in a manner knot theorists do, but we have found an algebraic proof, which is easier to access for representation theorists. In Section 5, we propose a conjecture for type $C^{(1)}_{\ell}$.

1. Combinatorics of Tableaux

1.1. Lie theory notation

Let $\ell\in\{2,3,\dots\}\cup\{\infty\}$ and $I=\mathbb{Z}_{\geqslant 0}$ if $\ell=\infty$ or $I=\{0,1,2,\ldots,\ell\}$ otherwise.

For $\ell=\infty$, the corresponding Cartan matrix $\mathsf{A}=(a_{ij})_{i,j\in I}$ of type $C_{\infty}$ is given by

[TABLE]

Otherwise, the affine Cartan matrix of type $C^{(1)}_{\ell}$ is given by

[TABLE]

We adopt standard notation from [13] for the root datum; in particular we have simple roots $\{\alpha_{i}\mid i\in I\}$ and fundamental weights $\{\Lambda_{i}\mid i\in I\}$ in the weight lattice $\mathsf{P}$, and simple coroots $\{\alpha_{i}^{\vee}\mid i\in I\}$ in the dual weight lattice $\mathsf{P}^{\vee}$. There is an invariant symmetric bilinear form $(-,-)$ on $\mathsf{P}$ satisfying $(\Lambda_{i},\alpha_{j})=d_{j}\delta_{ij}$ and $(\alpha_{i},\alpha_{j})=d_{i}a_{ij}$ where $d=(2,1,1,\dots)$ if $\ell=\infty$ or $d=(2,1\dots,1,2)$ if $\ell<\infty$. Let

[TABLE]

be the set of dominant integral weights, where $\langle$ , $\rangle$ is the natural pairing. We denote by $\mathsf{Q}:=\bigoplus_{i\in I}\mathbb{Z}\alpha_{i}$ the root lattice and $\mathsf{Q}^{+}=\bigoplus_{i\in I}\mathbb{Z}_{\geqslant 0}\alpha_{i}$ is the positive cone of the root lattice. Note that the null root in type $C^{(1)}_{\ell}$ is given by $\delta=\alpha_{0}+2\alpha_{1}+\dots+2\alpha_{\ell-1}+\alpha_{\ell}.$

1.2. The symmetric group and multipartitions

Denote by $\mathfrak{S}_{n}$ the symmetric group on $n$ letters, with Coxeter generators $s_{1},\dots,s_{n-1}$. For a permutation $w\in\mathfrak{S}_{n}$, a reduced expression for $w$ is an expression $w=s_{i_{1}}\dots s_{i_{r}}$ of minimal length; $r=\ell(w)$ is the length of $w$.

We denote by $\mathfrak{S}_{m+n}/\mathfrak{S}_{m}{\times}\mathfrak{S}_{n}$ the set of distinguished left coset representatives of $\mathfrak{S}_{m}{\times}\mathfrak{S}_{n}$ in $\mathfrak{S}_{m+n}$, i.e. $\ell(ws_{i})=\ell(w)+1$ for $w\in\mathfrak{S}_{m+n}/\mathfrak{S}_{m}{\times}\mathfrak{S}_{n}$ and $i\neq m$.

For $a,b\in\mathbb{Z}_{\geqslant 0}$ with $a+b\leqslant n$, we define $w[a,b]\in\mathfrak{S}_{n}$ by

[TABLE]

In two-line notation, $w[a,b]$ is

[TABLE]

Throughout the paper, $w\in\mathfrak{S}_{n}$ permutes letters of a tableau, but permutes places of $\nu=(\nu_{1},\ldots,\nu_{n})\in I^{n}$ as $w\nu=(\nu_{w^{-1}(1)},\dots,\nu_{w^{-1}(n)})$. In particular,

[TABLE]

Lemma 1.1**.**

Let $a,b\geqslant 1$.

• (1)

$w[a,b]=(s_{b}\dots s_{a+b-1})\dots(s_{1}\dots s_{a})=(s_{b}\dots s_{1})\dots(s_{a+b-1}\dots s_{a})$.

• (2)

$s_{b+1}w[2,b]=w[2,b]s_{1}$.

Proof.

It is easy to see (1). Using the braid relations, we have

[TABLE]

which complete the proof of (2). Here, the underlines indicate generators to which we can apply the braid relation. ∎

The following easy lemma will be useful to us later. Note that the equality $s_{b+1}w[2,b]=w[2,b]s_{1}$ in Lemma 1.1(2) is a special case of this, but the importance of Lemma 1.1(2) lies in the ‘long-hand derivation’ of this equality, which we will utilise later in Lemma 2.14.

Lemma 1.2**.**

Let $w\in\mathfrak{S}_{n}$ and $1\leqslant i\leqslant n-1$. If $w(i+1)=w(i)+1$, then $s_{w(i)}w=ws_{i}$.

For a reduced expression $w=s_{i_{1}}\dots s_{i_{r}}$ and $k\in\mathbb{Z}_{\geqslant 0}$ with $i_{j}<n-k$ for $1\leqslant j\leqslant r$, we set

[TABLE]

Note that $\mathsf{sh}_{k}(w)$ does not depend on the choice of reduced expressions. For $a,b,c\in\mathbb{Z}_{\geqslant 0}$, we define the block permutation $\mathrm{S}_{2}(c,a,b)$ to be $\mathsf{sh}_{c}(w[a,b])$.

Definition 1.3**.**

For $v,w\in\mathfrak{S}_{n}$, we write $v\succcurlyeq w$ if there is a reduced expression for $v$ which has an expression for $w$ as a subsequence. We write $v\succ w$ if $v\succcurlyeq w$ and $v\neq w$. This partial order is called the Bruhat order.

The left order (sometimes called the weak Bruhat order) is given by $v\geqslant_{L}w$ if there is a reduced expression for $v$ which has a reduced expression for $w$ as a suffix – that is, $v=s_{i_{1}}\dots s_{i_{r}}w$ for some $i_{1},\dots,i_{r}$ with $r=\ell(v)-\ell(w)$.

We fix an integer $l\geqslant 1$ throughout, which we refer to as the level.

Definition 1.4**.**

For $n\geqslant 0$, a partition of $n$ is a weakly decreasing sequence of non-negative integers $\lambda=(\lambda_{1},\lambda_{2},\dots)$ such that the sum $|\lambda|=\lambda_{1}+\lambda_{2}+\cdots$ is equal to $n$. If $\lambda$ is a partition of $n$ we write $\lambda\vdash n$. We write $\varnothing$ for the unique partition of 0. Note that we will in general omit trailing zeros for partitions.

An $l$-multipartition of $n$ is an $l$-tuple of partitions $\lambda=(\lambda^{(1)},\dots,\lambda^{(l)})$ such that the total size $\sum_{i=1}^{l}|\lambda^{(i)}|=n$. We denote the set of $l$-multipartitions of $n$ by $\mathscr{P}^{l}_{n}$ and set $\mathscr{P}^{l}:=\cup_{n\geqslant 0}\mathscr{P}^{l}_{n}$.

Similarly, a composition is a sequence $\mu=(\mu_{1},\mu_{1},\dots)$ of non-negative integers, and an $l$-multicomposition is an $l$-tuple of compositions.

If $\lambda$ and $\mu$ are $l$-multicompositions of $n$, we say that $\lambda$ dominates $\mu$, and write $\lambda\trianglerighteqslant\mu$ if

[TABLE]

for all $1\leqslant t\leqslant l$ and $k\geqslant 0$.

For any $\lambda\in\mathscr{P}^{l}_{n}$, we define its Young diagram $[\lambda]$ to be the set

[TABLE]

We will depict a Young diagram for a partition using the English convention, and for a multipartition $\lambda$ as a column vector of Young diagrams for the components $\lambda^{(1)},\dots,\lambda^{(l)}$. If $l=1$, then we write simply $(r,c)$ for $(r,c,t)$.

Example 1.5**.**

Let $\lambda=((4,3,1,1),\varnothing,(3,2,1))\in\mathscr{P}^{3}_{15}$. Then we write

[TABLE]

With this convention, we say that for nodes $A=(r,c,t),A^{\prime}=(r^{\prime},c^{\prime},t^{\prime})\in[\lambda]$, $A$ is below $A^{\prime}$ if $t>t^{\prime}$ or if $t=t^{\prime}$ with $r>r^{\prime}$, and $A$ is above $A^{\prime}$ if $A^{\prime}$ is below $A$.

We define $f_{\ell}:\mathbb{Z}\rightarrow I$ by $k\mapsto|k|$ if $\ell=\infty$ and, if $\ell\neq\infty$, $f_{\ell}:\mathbb{Z}/2\ell\mathbb{Z}\rightarrow I$ by

[TABLE]

Let $p$ be the natural projection $\mathbb{Z}\rightarrow\mathbb{Z}/2\ell\mathbb{Z}$ if $\ell\neq\infty$ and $p=\textrm{id}$ if $\ell=\infty$. Then we define $\pi_{\ell}=f_{\ell}\circ p:\mathbb{Z}\rightarrow I$. We denote $\pi_{\ell}(k)$ by $\overline{k}$, for $k\in\mathbb{Z}$, if there is no confusion.

Now we fix a multicharge $\kappa=(\kappa_{1},\ldots,\kappa_{l})\in\mathbb{Z}^{l}$ and define $\Lambda\in\mathsf{P}^{+}$ by $\Lambda=\sum_{i=1}^{l}\Lambda_{\overline{\kappa_{i}}}.$ Let $\lambda$ be an $l$-multipartition. Then, to any node $A=(r,c,t)\in[\lambda]$ we may associate its residue by

[TABLE]

If $\mathrm{res}(A)=i$, we call $A$ an $i$-node. Thus, $l$-multipartitions may be coloured by $I$. We define the content of $\lambda$ to be

[TABLE]

Example 1.6**.**

For $\lambda=((4,3,1,1),\varnothing,(3,2,1))$ as above, and $\kappa=(2,0,-1)$, the residues of $[\lambda]$ are given as follows.

[TABLE]

We also have $\operatorname{cont}(\lambda)=2\alpha_{0}+5\alpha_{1}+3\alpha_{2}+3\alpha_{3}+\alpha_{4}+\alpha_{5}$.

We say that a node $A$ is removable (resp. addable) if $[\lambda]\setminus A$ (resp. $[\lambda]\cup A$) is a valid Young diagram for a multipartition of $n-1$ (resp. $n+1$). We write $\lambda\nearrow A$ (resp. $\lambda\swarrow A$) as shorthand for the multipartition whose Young diagram is $[\lambda]\setminus A$ (resp. $[\lambda]\cup A$). For an $i$-node $A\in[\lambda]$, we set

[TABLE]

We define $\mathcal{F}(\kappa)$ to be a $\mathbb{Q}(q)$-vector space with basis consisting of the coloured $l$-multipartitions. Then $\mathcal{F}(\kappa)$ has a $U_{q}(\mathfrak{g}(\mathsf{A}))$-module structure defined by

[TABLE]

where $A$ runs over all removable $i$-nodes and all addable $i$-nodes respectively. The above description of $\mathcal{F}(\kappa)$ matches with that of the type A Fock space given in [7, Section 3.6], which is slightly different from [19, 24]. We call $\mathcal{F}(\kappa)$ the level $l$ Fock space with multicharge $\kappa$. Note that the weight of a coloured $l$-multipartition $\lambda$ is $\Lambda-\operatorname{cont}(\lambda)$, and there is a $U_{q}(\mathfrak{g}(\mathsf{A}))$-module isomorphism

[TABLE]

Here, the $U_{q}(\mathfrak{g}(\mathsf{A}))$-module structure of the tensor product comes from the comultiplication of $U_{q}(\mathfrak{g}(\mathsf{A}))$ given by, for $i\in I$,

[TABLE]

where $K_{i}=q^{\frac{(\alpha_{i},\alpha_{i})}{2}h_{i}}$. (cf. [7, Section 3.1])

Let $\Lambda=\sum_{t=1}^{l}\Lambda_{\overline{\kappa_{t}}}$ and $V_{q}(\Lambda)$ the irreducible highest weight $U_{q}(\mathfrak{g}(\mathsf{A}))$-module with highest weight $\Lambda$. As $\varnothing$ is a highest weight vector of $\mathcal{F}(\kappa)$ with highest weight $\Lambda$, we have a canonical $U_{q}(\mathfrak{g}(\mathsf{A}))$-module epimorphism

[TABLE]

where $v_{\Lambda}$ is a highest weight vector of $V_{q}(\Lambda)$.

1.3. Tableaux

We will mostly adopt the notation of [8, 21] for tableaux.

Let $\lambda\in\mathscr{P}^{l}_{n}$. A $\lambda$-tableau is a bijection $\mathtt{T}:[\lambda]\rightarrow\{1,\dots,n\}$. We depict $\mathtt{T}$ by filling each node $(r,c,t)\in[\lambda]$ with $\mathtt{T}(r,c,t)$. We say that a tableau $\mathtt{T}$ is row-strict if the entries increase along the rows of each component of $\mathtt{T}$, and column-strict if the entries increase down the columns of each component of $\mathtt{T}$. If $\mathtt{T}$ is both row- and column-strict, we call it standard. We denote the set of standard tableaux by $\mathsf{Std}(\lambda)$, the set of row-strict tableaux by $\mathsf{RowStd}(\lambda)$ and the set of row-strict tableaux which are not standard by $\mathsf{Row}(\lambda)$ = $\mathsf{RowStd}(\lambda)\setminus\mathsf{Std}(\lambda)$. Note that the symmetric group $\mathfrak{S}_{n}$ acts naturally on the left on the set of tableaux.

For each $\lambda$-tableau $\mathtt{T}$, we have the associated residue sequence

[TABLE]

Let $\mathtt{T}^{\lambda}$ be the initial tableau, which is the distinguished tableau where we fill the nodes with $1,\dots,n$ first along successive rows in $\lambda^{(1)}$, then $\lambda^{(2)}$, and so on. Then for each $\lambda$-tableau, $\mathtt{T}$, we may define the permutation $w^{\mathtt{T}}\in\mathfrak{S}_{n}$ by $w^{\mathtt{T}}\mathtt{T}^{\lambda}=\mathtt{T}$ and the length $\ell(\mathtt{T})\in\mathbb{Z}_{\geqslant 0}$ by $\ell(\mathtt{T})=\ell(w^{\mathtt{T}})$. We simply write $T\geqslant_{L}T^{\prime}$ when $w^{T}\geqslant_{L}w^{T^{\prime}}$ for tableaux $T$ and $T^{\prime}$.

Example 1.7**.**

Continuing our previous example,

[TABLE]

We define the dominance order on $\lambda$-tableaux by setting $\mathtt{T}\trianglerighteqslant\mathtt{S}$ if and only if $w^{\mathtt{T}}\preccurlyeq w^{\mathtt{S}}$. The matchup of terminology and notation with the dominance order on partitions is justified by Lemma 1.8. Note in particular that $\mathtt{T}^{\lambda}\trianglerighteqslant\mathtt{T}$ for all $\lambda$-tableaux $\mathtt{T}$.

First, we introduce one more concept. Let $\mathtt{T}$ be a $\lambda$-tableau and $0\leqslant m\leqslant n$. We denote by $\mathtt{T}_{\downarrow m}$ the set of nodes of $[\lambda]$ whose entries are less than or equal to $m$. If $\mathtt{T}\in\mathsf{Std}(\lambda)$, then $\mathtt{T}_{\downarrow m}$ is a tableau for some multipartition, which we call $\operatorname{Shp}(\mathtt{T}_{\downarrow m})$. If $\mathtt{T}\in\mathsf{Row}(\lambda)$, then $\mathtt{T}_{\downarrow m}$ is a tableau for some multicomposition, which we also call $\operatorname{Shp}(\mathtt{T}_{\downarrow m})$.

Lemma 1.8**.**

[23, Theorem 3.8]** Suppose $\lambda\in\mathscr{P}^{l}_{n}$ and $\mathtt{T},\mathtt{S}\in\mathsf{RowStd}(\lambda)$. Then $\mathtt{T}\trianglerighteqslant\mathtt{S}$ if and only if $\operatorname{Shp}(\mathtt{T}_{\downarrow m})\trianglerighteqslant\operatorname{Shp}(\mathtt{S}_{\downarrow m})$ for all $1\leqslant m\leqslant n$.

For any $\lambda\in\mathscr{P}^{l}_{n}$ and $\mathtt{T}\in\mathsf{Std}(\lambda)$ we define the degree $\deg\mathtt{T}$ of $\mathtt{T}$ as follows. If $n=0$ then $\mathtt{T}$ is the unique $\varnothing$-tableau and we set $\deg\mathtt{T}:=0$. Otherwise, let $A=\mathtt{T}^{-1}(n)\in[\lambda]$ and suppose $A$ is an $i$-node. We set inductively

[TABLE]

Example 1.9**.**

Let $\ell=\infty$, $\kappa=(2,-1)$ and $\lambda=((2,2,1),(3,2))$. Then the residue pattern of $\lambda$ is

[TABLE]

and if $\mathtt{T}$ is the tableau

[TABLE]

we have

[TABLE]

The nodes contributing to the degree are those containing the entries 2 (a 1-node), 5 (a 2-node), 6 (a 0-node) and 10 (a 1-node).

1.4. Garnir tableaux

Definition 1.10**.**

Let $\lambda\in\mathscr{P}^{l}_{n}$ and $A=(r,c,t)\in[\lambda]$. We call $A$ a Garnir node if $(r+1,c,t)\in[\lambda]$. For a Garnir node $A\in[\lambda]$, the Garnir belt $\mathbf{B}^{A}$ is the set of nodes

[TABLE]

Finally, for a Garnir node $A\in[\lambda]$, the Garnir tableau $\mathtt{G}^{A}$ is the $\lambda$-tableau which agrees with the initial tableau $\mathtt{T}^{\lambda}$ outside of $\mathbf{B}^{A}$ and has the entries $u,u+1,\dots,v$ from the bottom left to the top right of $\mathbf{B}^{A}$, where $u=\mathtt{T}^{\lambda}(r,c,t)$ and $v=\mathtt{T}^{\lambda}(r+1,c,t)$. Then

[TABLE]

where $a=\sum_{i=1}^{t-1}|\lambda^{(i)}|+\sum_{j=1}^{r-1}\lambda^{(t)}_{j}+c-1$. Note that $S_{2}(a,\lambda^{(t)}_{r}-c+1,c)$ is $321$-avoiding so that $w^{\mathtt{G}^{A}}$ is fully commutative. See [4, Lemma 2.1] for example.

Example 1.11**.**

Let $\lambda=((4,3,1,1),\varnothing,(3,2,1))$ and $A=(1,3,1)$. Then the Garnir tableau $\mathtt{G}^{A}$, with the Garnir belt $\mathbf{B}^{A}$ shaded, is as follows.

[TABLE]

The following lemma is an easy generalisation of [23, Lemma 3.14] but we include a proof for the reader’s convenience. This lemma and Lemma 1.13 will be used in the proof of Theorem 3.19 in Section 4.

Lemma 1.12**.**

Let $\lambda\in\mathscr{P}^{l}_{n}$ and $\mathtt{T}\in\mathsf{Row}(\lambda)$. If $A=(r,c,t)\in[\lambda]$ with $\mathtt{T}(r,c,t)>\mathtt{T}(r+1,c,t)$, then there is an element $w\in\mathfrak{S}_{n}$ such that $\mathtt{T}=w\mathtt{G}^{A}$ and $\ell(\mathtt{T})=\ell(w)+\ell(\mathtt{G}^{A})$. That is, $w^{\mathtt{T}}\geqslant_{L}w^{\mathtt{G}^{A}}$. Conversely, if $\mathtt{T}=w\mathtt{G}^{A}$ with $\ell(\mathtt{T})=\ell(w)+\ell(\mathtt{G}^{A})$ then $\mathtt{T}\in\mathsf{Row}(\lambda)$.

Proof.

Let $u:=\mathtt{T}^{\lambda}(r,c,t)$, $v:=\mathtt{T}^{\lambda}(r+1,c,t)$, $a:=\mathtt{T}(r,c,t)$ and $b:=\mathtt{T}(r+1,c,t)$. If $\mathtt{T}=\mathtt{G}^{A}$, the result is clear. So we suppose that $\mathtt{T}\neq\mathtt{G}^{A}$, and we will choose a basic transposition $s_{i}$ such that $s_{i}\mathtt{T}\in\mathsf{Row}(\lambda)$ and $s_{i}\mathtt{T}\vartriangleright\mathtt{T}$, from which the result follows by (reverse) induction on the dominance order $\vartriangleright$.

If $\mathtt{T}$ coincides with $\mathtt{T}^{\lambda}$ outside of $\mathbf{B}^{A}$, there is a gap in the reading word of $\mathtt{T}$ in either the first or the second row of $\mathbf{B}^{A}$ – otherwise $u,u+1,\dots,v$ are split into two sets of consecutive numbers and as $\mathtt{T}(r,c,t)>\mathtt{T}(r+1,c,t)$ the only way to fill in the numbers is $\mathtt{T}=\mathtt{G}^{A}$. Thus, we may choose $i+1$ in the first row and $i$ in the second row for some $i$ with $(i,i+1)\neq(b,a)$, so that $s_{i}\mathtt{T}(r,c,t)>s_{i}\mathtt{T}(r+1,c,t)$.

Otherwise, we may choose $s_{i}$ so that $s_{i}\mathtt{T}(r,c,t)>s_{i}\mathtt{T}(r+1,c,t)$ as follows. First suppose that the reading word of $\mathtt{T}$ begins $1,2,\dots,m,m^{\prime}$ for some $m^{\prime}>m+1$ and $m<u$. Then setting $i=m^{\prime}-1$ suffices. Next suppose that the reading word of $\mathtt{T}$ ends $m^{\prime},m,m+1,\dots,n$ for some $v<m^{\prime}<m-1$. Then, setting $i=m^{\prime}$ suffices.

For the converse statement, we argue by induction on $\ell(\mathtt{T})$. Suppose $\mathtt{T}=w\mathtt{G}^{A}$ with $\ell(\mathtt{T})=\ell(w)+\ell(\mathtt{G}^{A})$ and $s_{i}\mathtt{T}\vartriangleleft\mathtt{T}$. Then if $s_{i}\mathtt{T}$ is standard, so is $\mathtt{T}$, by [9, Lemma 1.5]. But this contradicts the induction hypothesis. ∎

Lemma 1.13**.**

Let $\lambda\in\mathscr{P}^{l}_{n}$ and $\mathtt{T}\in\mathsf{Row}(\lambda)$.

1. (1)

If $\mathtt{T}(r,c,t)=\mathtt{T}(r+1,c,t)+1$, then there is an element $w\in\mathfrak{S}_{n}$ such that, for $A:=(r,c,t)\in[\lambda]$,

1. (i)

$\mathtt{T}=w\mathtt{G}^{A}$, 2. (ii)

$s_{p}w=ws_{q}$, where $p=\mathtt{T}(r+1,c,t)$ and $q=\mathtt{G}^{A}(r+1,c,t)$. 2. (2)

If $\mathtt{T}(r,c+1,t)=\mathtt{T}(r,c,t)+1$, then there is a Garnir node $A\in[\lambda]$ and $w\in\mathfrak{S}_{n}$ such that

1. (i)

$\mathtt{G}^{A}(r,c+1,t)=\mathtt{G}^{A}(r,c,t)+1$, 2. (ii)

$\mathtt{T}=w\mathtt{G}^{A}$, 3. (iii)

$s_{p}w=ws_{q}$, where $p=\mathtt{T}(r,c,t)$ and $q=\mathtt{G}^{A}(r,c,t)$.

Proof.

1. (1)

Part (i) follows from Lemma 1.12. For part (ii), note that $\mathtt{G}^{A}(r,c,t)=q+1$ by definition, so that

[TABLE]

It follows from Lemma 1.2 that $ws_{q}=s_{w(q)}w=s_{p}w$. 2. (2)

We begin by choosing a node $A=(r^{\prime},c^{\prime},t^{\prime})\in[\lambda]$ such that $\mathtt{T}(r^{\prime},c^{\prime},t^{\prime})>\mathtt{T}(r^{\prime}+1,c^{\prime},t^{\prime})$ and $\mathtt{G}^{A}(r,c+1,t)=\mathtt{G}^{A}(r,c,t)+1$ as follows.

If $\mathtt{T}(r,c+1,t)>\mathtt{T}(r+1,c+1,t)$ then we know that $\mathtt{T}(r,c,t)>\mathtt{T}(r+1,c,t)$, so we may choose $A=(r,c,t)$.

Next, suppose $\mathtt{T}(r-1,c,t)>\mathtt{T}(r,c,t)$. Then we have $\mathtt{T}(r-1,c+1,t)>\mathtt{T}(r,c+1,t)$, and we may choose $A=(r-1,c+1,t)$.

Otherwise, if $\mathtt{T}(r,c+1,t)<\mathtt{T}(r+1,c+1,t)$ and $\mathtt{T}(r-1,c,t)<\mathtt{T}(r,c,t)$, then, as $\mathtt{T}\in\mathsf{Row}(\lambda)$, there is some node $A=(r^{\prime},c^{\prime},t^{\prime})$ such that $\mathtt{T}(r^{\prime},c^{\prime},t^{\prime})>\mathtt{T}(r^{\prime}+1,c^{\prime},t^{\prime})$ and $(r^{\prime},c^{\prime},t^{\prime})\neq(r-1,c,t),(r,c+1,t)$. Since $\mathtt{G}^{A}(x,y+1,z)=\mathtt{G}^{A}(x,y,z)+1$ holds unless $(x,y,z)=(r^{\prime},c^{\prime}-1,t^{\prime}),(r^{\prime}+1,c^{\prime},t^{\prime})$, $(r,c,t)\neq(r^{\prime},c^{\prime}-1,t^{\prime}),(r^{\prime}+1,c^{\prime},t^{\prime})$ implies $\mathtt{G}^{A}(r,c+1,t)=\mathtt{G}^{A}(r,c,t)+1$. Hence (i) is proved.

Now, by Lemma 1.12 and the fact that $\mathtt{T}(r^{\prime},c^{\prime},t^{\prime})>\mathtt{T}(r^{\prime}+1,c^{\prime},t^{\prime})$, there is some $w\in\mathfrak{S}_{n}$ such that $\mathtt{T}=w\mathtt{G}^{A}$ and $\ell(\mathtt{T})=\ell(w)+\ell(\mathtt{G}^{A})$. We have proved (ii). Moreover,

[TABLE]

Thus, (iii) follows from Lemma 1.2.∎

Lemma 1.14**.**

Let $A$ and $B$ be distinct Garnir nodes of $\lambda\in\mathscr{P}^{l}_{n}$. Then there is a unique tableau $\mathtt{G}^{A,B}\in\mathsf{Row}(\lambda)$ such that

1. (1)

$\mathtt{G}^{A,B}\geqslant_{L}\mathtt{G}^{A}$* and $\mathtt{G}^{A,B}\geqslant_{L}\mathtt{G}^{B}$,* 2. (2)

$\mathtt{T}\geqslant_{L}\mathtt{G}^{A,B}$* for any $\mathtt{T}\in\mathsf{RowStd}(\lambda)$ with $\mathtt{T}\geqslant_{L}\mathtt{G}^{A}$ and $\mathtt{T}\geqslant_{L}\mathtt{G}^{B}$.*

Proof.

It is known that $\mathsf{RowStd}(\lambda)$ is a lattice with respect to the left order. See for example [5, Theorem 7.1] (with some slight modification to generalise to $\mathsf{RowStd}(\lambda)$). Thus $\mathtt{G}^{A,B}=\mathtt{G}^{A}\vee\mathtt{G}^{B}$. ∎

We redefine $\mathtt{G}^{A,B}$ in Definition 1.15 below in a more concrete manner and show in LABEL:construction_of_G^{A and LABEL:B} that it coincides with $\mathtt{G}^{A,B}$ in Lemma 1.14.

Definition 1.15**.**

Suppose $A,B\in[\lambda]$ are distinct Garnir nodes. We define the sets $\mathbf{B}^{A}(2)$ and $\mathbf{B}^{B}(1)$ to be the second row of $\mathbf{B}^{A}$ and the first row of $\mathbf{B}^{B}$, respectively.

We define the generalised Garnir belt $\mathbf{B}^{A,B}$ of $[\lambda]$ to be the following set of nodes.

1. (1)

If $\mathbf{B}^{A}\cap\mathbf{B}^{B}=\emptyset$, then $\mathbf{B}^{A,B}:=\mathbf{B}^{A}\cup\mathbf{B}^{B}$. 2. (2)

If $A=(r,c,t)$ and $B=(r,c^{\prime},t)$ for some $c^{\prime}>c$, then

[TABLE]

In this case, we set $\mathbf{B}^{A,B}(1)=\mathbf{B}^{A}\setminus\mathbf{B}^{B}(1)$ and $\mathbf{B}^{A,B}(2)=\mathbf{B}^{B}\setminus\mathbf{B}^{A}(2)$. 3. (3)

If $A=(r,c,t)$ and $B=(r-1,c^{\prime},t)$ for some $c^{\prime}\geqslant c$, then

[TABLE]

Finally, we define the generalised Garnir tableau in the first two cases above to be the $\lambda$-tableau $\mathtt{G}^{A,B}$ which agrees with $\mathtt{T}^{\lambda}$ outside of $\mathbf{B}^{A,B}$ and has the entries of $\mathbf{B}^{A,B}$ as follows:

1. (1)

If $\mathbf{B}^{A}\cap\mathbf{B}^{B}=\emptyset$, then we fill each of $\mathbf{B}^{A}$ and $\mathbf{B}^{B}$ as in $\mathtt{G}^{A}$ and $\mathtt{G}^{B}$, respectively. 2. (2)

If $A=(r,c,t)$ and $B=(r,c^{\prime},t)$ for some $c^{\prime}\geqslant c$, then we first fill the entries of $\mathbf{B}^{A,B}(1)$, from bottom left to top right, and then we fill the entries of $\mathbf{B}^{A,B}(2)$, from bottom left to top right.

In the third case above, $\mathtt{G}^{A,B}$ is defined as follows.

1. (3)

If $A=(r,c,t)$ and $B=(r-1,c^{\prime},t)$ for some $c^{\prime}\geqslant c$, $\mathtt{G}^{A,B}$ is defined to be the $\lambda$-tableau which agrees with $\mathtt{T}^{\lambda}$ outside of the three rows of $[\lambda]$ which contain elements of $\mathbf{B}^{A,B}$, and we fill the entries of these three rows first in order along rows above $\mathbf{B}^{A,B}$, then from bottom left to top right in $\mathbf{B}^{A,B}$, and finally in order along rows below $\mathbf{B}^{A,B}$.

Example 1.16**.**

Let $\lambda=((1),(10,9,6,2))$ and $A=(2,3,2)\in[\lambda]$. Then we have the following tableaux $\mathtt{G}^{A,B}$ in cases corresponding to Definition 1.15, where we have shaded the generalised Garnir belts $\mathbf{B}^{A,B}$ in each case.

1. (1)

Let $B=(1,1,2)$. Then

[TABLE] 2. (2)

Let $B=(2,6,2)$. Then

[TABLE] 3. (3)

Let $B=(1,6,2)$. Then

[TABLE]

Lemma 1.17**.**

The construction of $\mathtt{G}^{A,B}$ in Definition 1.15 satisfies $\mathtt{G}^{A,B}=\mathtt{G}^{A}\vee\mathtt{G}^{B}$ and thus coincides with the tableau $\mathtt{G}^{A,B}$ defined in Lemma 1.14.

Proof.

It is easy to see that $\mathtt{G}^{A,B}\geqslant_{L}\mathtt{G}^{A}$ and $\mathtt{G}^{A,B}\geqslant_{L}\mathtt{G}^{B}$, so that we have $\mathtt{G}^{A,B}\geqslant_{L}\mathtt{G}^{A}\vee\mathtt{G}^{B}$. If the inequality were strict, then there exists a basic transposition $s$ such that

[TABLE]

However, the explicit construction of $\mathtt{G}^{A,B}$ shows that either $s\mathtt{G}^{A,B}\not\geqslant_{L}\mathtt{G}^{A}$ or $s\mathtt{G}^{A,B}\not\geqslant_{L}\mathtt{G}^{B}$ occurs for any basic transposition $s$ with $\mathtt{G}^{A,B}>_{L}s\mathtt{G}^{A,B}$. Hence, we must have equality. ∎

Lemma 1.18**.**

Let $A=(r,c,t)$ and $B=(r^{\prime},c^{\prime},t^{\prime})$ be Garnir nodes of $\lambda\in\mathscr{P}^{l}_{n}$.

1. (1)

If $\mathbf{B}^{A}\cap\mathbf{B}^{B}=\emptyset$ then $w^{\mathtt{G}^{A,B}}$ is fully commutative. 2. (2)

If $r=r^{\prime}$ and $t=t^{\prime}$ then $w^{\mathtt{G}^{A,B}}$ is fully commutative. 3. (3)

If $\mathtt{G}^{A,B}=w^{A}\mathtt{G}^{A}=w^{B}\mathtt{G}^{B}$, then $w^{A}$ and $w^{B}$ are fully commutative.

Proof.

Take $w^{A},w^{B}\in\mathsf{W}$ such that $\mathtt{G}^{A,B}=w^{A}\mathtt{G}^{A}=w^{B}\mathtt{G}^{B}$. We consider the three cases (1), (2) and (3) in Definition 1.15.

• •

In the first case, $\mathtt{G}^{A,B}=w^{\mathtt{G}^{A}}\mathtt{G}^{B}=w^{\mathtt{G}^{B}}\mathtt{G}^{A}$ and it is clear that each of $w^{\mathtt{G}^{A}}$ and $w^{\mathtt{G}^{B}}$ are of the form $S_{2}(c,a,b)=\mathsf{sh}_{c}w[a,b]$ for some $a,b,c$. Further, $w^{\mathtt{G}^{A,B}}$ has a unique descent pattern of $2143$. Thus, $w^{A}=w^{\mathtt{G}^{B}}$, $w^{B}=w^{\mathtt{G}^{A}}$ and $w^{\mathtt{G}^{A,B}}$ are $321$-avoiding. This implies that (1) holds, and (3) holds when $\mathbf{B}^{A}\cap\mathbf{B}^{B}=\emptyset$.

• •

In the second case, $w^{A}$ is a shift of $w[\lambda^{(t)}_{r}-c^{\prime}+1,c^{\prime}-c]$ and $w^{B}$ is a shift of $w[c^{\prime}-c,c]$. Thus $w^{A}$ and $w^{B}$ are $321$-avoiding. Further, the two-line notation for $w^{\mathtt{G}^{A,B}}$ is

[TABLE]

up to shift. Hence, $w^{\mathtt{G}^{A,B}}$ is $321$-avoiding, which yields that (2) holds, and (3) holds when $r=r^{\prime}$ and $t=t^{\prime}$.

• •

In the third case, $w^{A}$ and $w^{B}$ are $w[\lambda^{(t)}_{r-1}-c^{\prime}+1,c+c^{\prime}]$ and $w[\lambda^{(t)}_{r-1}+\lambda^{(t)}_{r}-c-c^{\prime}+2,c]$ up to shift, respectively. Thus, they are also $321$-avoiding, which completes the proof of $(3)$.

∎

2. Quiver Hecke algebras

2.1. Affine and cyclotomic quiver Hecke algebras

In this subsection, $\mathsf{A}$ is an arbitrary symmetrisable Cartan matrix.

Let $\mathcal{O}$ be a unital commutative ring and we fix a system of polynomials $Q_{i,j}(u,v)\in\mathcal{O}[u,v]$ for $i,j\in I$ of the form

[TABLE]

where $t_{i,j;p,q}\in\mathcal{O}$ are such that $t_{i,j;-a_{ij},0}\in\mathcal{O}^{\times}$ and $Q_{i,j}(u,v)=Q_{j,i}(v,u)$.

For $\nu\in I^{n}$ and $\nu^{\prime}\in I^{n^{\prime}}$, we denote the concatenation of $\nu$ and $\nu^{\prime}$ by $\nu*\nu^{\prime}\in I^{n+n^{\prime}}$. Here, we understand that $I^{0}:=\{\emptyset\}$ and $\emptyset*\nu=\nu*\emptyset=\nu$.

Definition 2.1**.**

The cyclotomic quiver Hecke algebra $R^{\Lambda}(n)$ associated with polynomials $(Q_{i,j}(u,v))_{i,j\in I}$ and $\Lambda\in\mathsf{P}^{+}$ is the $\mathbb{Z}$-graded unital $\mathcal{O}$-algebra generated by

[TABLE]

subject to the following relations.

[TABLE]

for all admissible $r,s,\nu,\nu^{\prime}$, and $x_{1}^{\langle\alpha^{\vee}_{\nu_{1}},\Lambda\rangle}e(\nu)=0$ for $\nu\in I^{n}$.

The algebra $R^{\Lambda}(n)$ is given a $\mathbb{Z}$-grading by setting

[TABLE]

for all admissible $r,s$ and $\nu$.

For $\beta\in\mathsf{Q}^{+}$ with ${\rm ht}(\beta)=n$, we set

[TABLE]

Then $e(\beta):=\sum_{\nu\in I^{\beta}}e(\nu)$ is a central idempotent. We define $R^{\Lambda}(\beta):=R^{\Lambda}(n)e(\beta)$, which is also an $\mathcal{O}$-algebra. It is clear that $R^{\Lambda}(\beta)$ may be defined by the same set of relations if we replace $I^{n}$ with $I^{\beta}$. We have the following decomposition of $R^{\Lambda}(n)$ into a direct sum of $\mathcal{O}$-algebras.

[TABLE]

When we drop the relation $x_{1}^{\langle\alpha^{\vee}_{\nu_{1}},\Lambda\rangle}e(\nu)=0$ for $\nu\in I^{\beta}$, we obtain the quiver Hecke algebra $R(\beta)$.

For each element $w\in\mathfrak{S}_{n}$, we fix a preferred reduced expression $w=s_{i_{1}}\dots s_{i_{t}}$ and define

[TABLE]

Note that $\psi_{w}$ depends on the choice of reduced expressions of $w$ unless $w$ is fully commutative. The following comes from the defining relations.

Proposition 2.2** ([8, Proof of Proposition 2.5]).**

For two reduced expressions $s_{i_{1}}\dots s_{i_{t}}=s_{j_{1}}\dots s_{j_{t}}$ for an element $w\in\mathfrak{S}_{n}$, $(\psi_{i_{1}}\dots\psi_{i_{t}}-\psi_{j_{1}}\dots\psi_{j_{t}})e(\nu)$ can be written as a linear combination of elements of the form $\psi_{u}f(x)e(\nu)$, where $u\prec w$ with $\ell(u)\leqslant\ell(w)-3$, and $f(x)$ is a polynomial in the generators $x_{1},\dots,x_{n}$.

Theorem 2.3** ([17, 18, 25]).**

Let $\beta\in\mathsf{Q}^{+}$ with ${\rm ht}(\beta)=n$. Then the set

[TABLE]

is an $\mathcal{O}$-basis of $R(\beta)$.

Proposition 2.4**.**

Suppose that $Q_{ij}(u,v)$ have integral coefficients. We denote the cyclotomic quiver Hecke algebra defined over $\mathbb{Z}$ by $R_{\mathbb{Z}}^{\Lambda}(n)$. Then $R_{\mathbb{Z}}^{\Lambda}(n)$ is free of finite rank over $\mathbb{Z}$. Further, $R^{\Lambda}(n)\simeq R_{\mathbb{Z}}^{\Lambda}(n)\otimes_{\mathbb{Z}}\mathcal{O}$ as $\mathcal{O}$-algebras.

Proof.

We prove by induction on $n$ that $R^{\Lambda}(n)$ is a projective $\mathcal{O}$-module. It is clear that $R^{\Lambda}(0)=\mathcal{O}$ is a projective $\mathcal{O}$-module. Suppose that $R^{\Lambda}(n-1)$ is a projective $\mathcal{O}$-module. By [14, Thm 4.5] , $R^{\Lambda}(n)$ is a projective $R^{\Lambda}(n-1)$-module. Thus, the induction hypothesis implies that $R^{\Lambda}(n)$ is a projective $\mathcal{O}$-module. Applying the argument to $\mathcal{O}=\mathbb{Z}$ and noting that $\mathbb{Z}$ is a principal ideal domain, we deduce that $R_{\mathbb{Z}}^{\Lambda}(n)$ is a free $\mathbb{Z}$-module of finite rank.

As the defining relations of $R_{\mathbb{Z}}^{\Lambda}(n)$ hold in $R^{\Lambda}(n)$, the $\mathbb{Z}$-algebra homomorphism

[TABLE]

given by mapping the generators $\psi_{i},x_{j},e(\nu)$ to the corresponding generators is well-defined. Hence we have a surjective $\mathcal{O}$-algebra homomorphism

[TABLE]

On the other hand, as the defining relations of $R^{\Lambda}(n)$ hold in $R_{\mathbb{Z}}^{\Lambda}(n)\otimes_{\mathbb{Z}}\mathcal{O}$, we have a surjective $\mathcal{O}$-algebra homomorphism

[TABLE]

Thus, $R^{\Lambda}(n)\simeq R_{\mathbb{Z}}^{\Lambda}(n)\otimes_{\mathbb{Z}}\mathcal{O}$. ∎

Note that our choices $\eqref{Qij for affine C}$ and $\eqref{Qij for C_infty}$ of $Q_{ij}(u,v)$ being integral coefficients allow us to define the cyclotomic Hecke algebra over $\mathbb{Z}$.

2.2. The $C_{\infty}$ case

In this subsection, we carry out some computations in type $C_{\infty}$. We choose the following system of polynomials $Q_{i,j}(u,v)$ as our preferred choice: if the Cartan matrix $\mathsf{A}$ is of type $C^{(1)}_{\ell}$ then, for $i<j$,

[TABLE]

and if the Cartan matrix $\mathsf{A}$ is of type $C_{\infty}$ then, for $i<j$,

[TABLE]

Note that if we assume that $\mathcal{O}$ is a field and that any element of $\mathcal{O}$ has a square root, then other choices of the polynomials $Q_{i,j}(u,v)$ yield isomorphic algebras [2, Lemma 3.2]. Further we have the following graded dimension formulas. For $\nu\in I^{n}$, let

[TABLE]

Theorem 2.5**.**

For $\nu,\nu^{\prime}\in I^{\beta}$, we have

[TABLE]

where $\operatorname{rank}_{q}M:=\sum_{k\in\mathbb{Z}}\operatorname{rank}_{\mathcal{O}}(M_{k})q^{k}$ for a free graded $\mathcal{O}$-module $M=\bigoplus_{k\in\mathbb{Z}}M_{k}$.

Proof.

By virtue of Proposition 2.4, it suffices to prove the result when $\mathcal{O}$ is a field. The irreducible highest weight $U_{q}(\mathfrak{g}(\mathsf{A}))$-module with highest weight $\sum_{i=1}^{l}\Lambda_{\overline{\kappa_{i}}}\in\mathsf{P}^{+}$ is realised as the submodule $U_{q}(\mathfrak{g}(\mathsf{A}))\varnothing\subseteq\mathcal{F}(\kappa)$. Thus, the proof is entirely similar to [3]. The only difference is that we use the tensor product Fock space $\mathcal{F}(\kappa)$. ∎

Now we assume that the Cartan matrix $\mathsf{A}$ is of type $C_{\infty}$ and prepare some technical results. We consider fully commutative elements $S_{2}(c,a,b)=\mathsf{sh}_{c}w[a,b]$. Then $\psi_{w}$ for $w=S_{2}(c,a,b)$ does not depend on the choice of a preferred reduced expression. We denote it by $\Psi_{2}(c,a,b)$. If $c=0$ we denote it by $\Psi[a,b]$ instead. We have

[TABLE]

For $1\leqslant x\leqslant y$, we define

[TABLE]

Then Lemma 1.1(1) implies

[TABLE]

In particular, $\psi$-generators that appear in $\Psi_{2}(c,a,b)$ are $\psi_{c+1},\dots,\psi_{c+a+b-1}$. We also have the following formulae.

[TABLE]

Remark 2.6*.*

The algebra $R(\beta)$ admits an anti-involution which fixes the generators. Then it sends $\Psi_{2}(c,a,b)$ to $\Psi_{2}(c,b,a)$ because

[TABLE]

Definition 2.7**.**

For $a_{1},\dots,a_{t}\in\mathbb{Z}_{\geqslant 0}$, we define a block transposition $S_{i}(a_{1},\dots,a_{t})$ by

[TABLE]

Then it is fully commutative and we may define $\Psi_{i}(a_{1},\dots,a_{t})$ by

[TABLE]

More generally, we define block permutations $S_{i_{1}}\dots S_{i_{p}}(a_{1},\dots,a_{t})$ by

[TABLE]

and the corresponding $\Psi_{i_{1}}\dots\Psi_{i_{p}}(a_{1},\dots,a_{t})$ by

[TABLE]

Observing that $s_{i_{1}},\dots,s_{i_{p}}$ permute places, the following is clear.

Lemma 2.8**.**

Let $w=s_{i_{1}}\dots s_{i_{p}}\in\mathfrak{S}_{t}$ and $a_{1},\dots,a_{t}\in\mathbb{Z}_{\geqslant 0}$. If we define

[TABLE]

then the two-line notation of $S_{i_{1}}\dots S_{i_{p}}(a_{1},\dots,a_{t})$ is given as follows.

[TABLE]

Corollary 2.9**.**

Suppose that each $S_{i}$ is given by the reduced expressions in Lemma 1.1(1). Then $S_{i_{1}}\dots S_{i_{p}}(a_{1},\dots,a_{t})$ is a reduced expression if and only if $S_{i_{1}}\dots S_{i_{p}}(1,\dots,1)$ is.

The two-line notation may be used to represent $\Psi_{i_{1}}\dots\Psi_{i_{p}}(a_{1},\dots,a_{t})e(\nu)$ by diagrams.

Example 2.10**.**

Let $\nu^{1}\in I^{a}$, $\nu^{2}\in I^{b}$, $\nu^{3}\in I^{c}$, for $a,b,c\geqslant 1$, and $\nu=\nu^{1}*\nu^{2}*\nu^{3}$. Then $\Psi[a+b,c]e(\nu)$ is represented by

[TABLE]

and it follows that $\Psi[a+b,c]e(\nu)=\Psi_{1}\Psi_{2}(a,b,c)e(\nu)$.

Corollary 2.11**.**

Let $\underline{a}=(a_{1},\dots,a_{t})$. If $j\neq i\pm 1$ then $\Psi_{i}\Psi_{j}(\underline{a})=\Psi_{j}\Psi_{i}(\underline{a})$.

Lemma 2.12**.**

Suppose that the Cartan matrix $\mathsf{A}$ is of type $C_{\infty}$. Let $\nu=(\nu_{1},\nu_{2},\dots,\nu_{n})\in I^{n}$ and $a,b\in\mathbb{Z}_{>0}$ with $a<n$ and $a+b\leqslant n$.

1. (1)

If $|\nu_{i}-\nu_{a+1}|\geqslant 2$, for $1\leqslant i\leqslant a$, then $\Psi[1,a]\Psi[a,1]e(\nu)=e(\nu)$. 2. (2)

If $|\nu_{1}-\nu_{i}|\geqslant 2$, for $2\leqslant i\leqslant a+1$, then $\Psi[a,1]\Psi[1,a]e(\nu)=e(\nu)$. 3. (3)

If $\nu_{1}=\nu_{a+1}\neq\nu_{2},\dots,\nu_{a}$, then $x_{a+1}\Psi[1,a]e(\nu)=(\Psi[1,a]x_{1}+\Psi[1,a-1])e(\nu)$. 4. (4)

If $\nu_{1}=\nu_{a+1}\neq\nu_{2},\dots,\nu_{a}$, then $x_{1}\Psi[a,1]e(\nu)=(\Psi[a,1]x_{a+1}-\Psi_{2}(1,a-1,1))e(\nu)$. 5. (5)

If $|\nu_{i}-\nu_{j}|\geqslant 2$, for $1\leqslant i\leqslant a$, $a+1\leqslant j\leqslant a+b$, then $\Psi[b,a]\Psi[a,b]e(\nu)=e(\nu)$.

Proof.

(1) As $\Psi[1,a]=\psi_{a}\dots\psi_{1}$ and $\Psi[a,1]=\psi_{1}\dots\psi_{a}$, we have

[TABLE]

where $\mu^{1}=(\nu_{1},\nu_{a+1},\nu_{2},\dots,\nu_{a},\nu_{a+2},\dots,\nu_{n})$, then

[TABLE]

where $\mu^{2}=(\nu_{1},\nu_{2},\nu_{a+1},\nu_{3},\dots,\nu_{a},\nu_{a+2},\dots,\nu_{n})$, and so on. We continue the computation in this way and we reach $\Psi[1,a]\Psi[a,1]e(\nu)=e(\nu)$.

(2) The proof is similar to (1) and left to the reader.

(3) By the assumption, we have

[TABLE]

(4) By a similar computation to (3), we have

[TABLE]

(5) We recall the formulas

[TABLE]

Then repeated use of (2) proves the result. ∎

Lemma 2.13**.**

Suppose that the Cartan matrix $\mathsf{A}$ is of type $C_{\infty}$. Then

[TABLE]

where $\epsilon(r,\nu)$ is given as follows.

[TABLE]

Lemma 2.14**.**

Let $\nu=(\mu_{1}\mu_{2}\nu_{1}\dots\nu_{b})\in I^{b+2}$ for $b\geqslant 1$. Then

[TABLE]

where $c_{k}$ is given by

[TABLE]

Proof.

We follow Lemma 1.1(2). Then,

[TABLE]

where the error terms are computed by using Lemma 2.13. Here, the underlines indicate generators to which we can apply the braid-type relation. ∎

Lemma 2.15**.**

Let $a,b\geqslant 1$ and $1\leqslant i\leqslant a-1$. Then

[TABLE]

where $c_{k}=x_{k+i-1}+x_{k+i+1}$ if $(\nu_{i},\nu_{i+1},\nu_{a+k})=(1,0,1)$, $c_{k}=1$ if $(\nu_{i},\nu_{i+1},\nu_{a+k})=(j,j\pm 1,j)$ for some $j\geqslant 0$ such that $\nu_{i+1}\neq 0$, and $c_{k}=0$ otherwise.

Proof.

As the $\psi$-generators that appear in $\Psi_{2}(c,a,b)$ are $\psi_{c+1},\dots,\psi_{c+a+b-1}$, we have

[TABLE]

where $\mu=(\nu_{1}\dots\nu_{i+1}\nu_{a+1}\dots\nu_{a+b}\nu_{i+2}\dots\nu_{a}\nu_{a+b+1}\dots\nu_{n})$. We apply Lemma 2.14 to substitute

[TABLE]

Then we have the desired formula. ∎

We may also compute $(\psi_{b+i_{1}}\dots\psi_{b+i_{r}}\Psi[a,b]-\Psi[a,b]\psi_{i_{1}}\dots\psi_{i_{r}})e(\nu)$ by applying Lemma 2.15 to $(\psi_{b+i_{k}}\Psi[a,b]-\Psi[a,b]\psi_{i_{k}})e(s_{i_{k+1}}\dots s_{i_{r}}\nu)$ in the expression

[TABLE]

In particular, we obtain the following.

Lemma 2.16**.**

Let $a,b,c\geqslant 1$ and $1\leqslant m\leqslant b$. Then,

[TABLE]

where $c_{st}=x_{m+s+t-2}+x_{m+s+t}$ if $(\nu_{m+s-1},\nu_{m+a},\nu_{a+b+t})=(1,0,1)$, $c_{st}=1$ if $(\nu_{m+s-1},\nu_{m+a},$ $\nu_{a+b+t})=(j,j\pm 1,j)$ for some $j$ such that $\nu_{m+a}\neq 0$, $c_{st}=0$ otherwise.

Proof.

As $\Psi_{2}(c+m-1,a,1)=\psi_{c+m}\dots\psi_{c+m+a-1}$, the left-hand side is

[TABLE]

where $\mu^{s}=s_{m+s}\dots s_{m+a-1}\nu=(\nu_{1}\dots\nu_{m+s-1}\nu_{m+a}\nu_{m+s}\dots\widehat{\nu_{m+a}}\dots\nu_{n})$. Thus, we apply Lemma 2.15 with $i=m+s-1$. ∎

Remark 2.17*.*

We have prepared Lemma 2.16 for computing

[TABLE]

in later sections. If $a=0$ or $b=0$ or $c=0$ then it is zero. Thus we assume $a,b,c\geqslant 1$. First we observe that

[TABLE]

Hence we compute $\Psi_{2}(c,a,b)\Psi[a+b,c]e(\nu)-\Psi[a+b,c]\Psi_{1}(a,b,c)e(\nu)$, which is equal to

[TABLE]

since $\Psi_{2}(c,a,b)=\Psi_{2}(c+b-1,a,1)\dots\Psi_{2}(c,a,1)$. Then

[TABLE]

where $w[a,m-1]\nu=(\nu_{a+1}\dots\nu_{a+m-1}\nu_{1}\dots\nu_{a}\nu_{a+m}\dots\nu_{n})$. To compute

[TABLE]

using Lemma 2.16, we check whether $(\nu_{s},\nu_{a+m},\nu_{a+b+t})$ is $(1,0,1)$ or $(j,j+1,j)$ for $j\geqslant 0$, or $(j,j-1,j)$ for $j\geqslant 2$, for any $1\leqslant s\leqslant a$ and $1\leqslant t\leqslant c$.

2.3. Module categories

In the subsequent Subsections 2.3, 2.4 and 2.5, we keep the assumption that $\mathsf{A}$ is an arbitrary symmetrisable Cartan matrix but assume that $\mathcal{O}$ is a field.

We denote by $R(\beta)\text{-}\mathrm{proj}$ and $R(\beta)\text{-}\mathrm{gmod}$ the full subcategories in the category $R(\beta)\text{-}\mathrm{Mod}$ of graded $R(\beta)$-modules which consist of finitely generated projective graded $R(\beta)$-modules or finite dimensional graded $R(\beta)$-modules, respectively. We set

[TABLE]

Similarly, $R^{\Lambda}(\beta)\text{-}\mathrm{proj}$ and $R^{\Lambda}(\beta)\text{-}\mathrm{gmod}$ are the full subcategories in the category $R^{\Lambda}(\beta)\text{-}\mathrm{Mod}$ of graded $R^{\Lambda}(\beta)$-modules which consist of finitely generated projective graded $R^{\Lambda}(\beta)$-modules or finite dimensional graded $R^{\Lambda}(\beta)$-modules, respectively. We set

[TABLE]

Let us denote by $q$ the grading shift functor, i.e. $(qM)_{k}=M_{k-1}$ for a graded module $M=\bigoplus_{k\in\mathbb{Z}}M_{k}$. For $M\in R(\beta)\text{-}\mathrm{gmod}$, the $q$-character of $M$ is defined by

[TABLE]

where $\dim_{q}V:=\sum_{k\in\mathbb{Z}}\dim(V_{k})q^{k}$ for a graded vector space $V=\bigoplus_{k\in\mathbb{Z}}V_{k}$.

For graded $R(\beta)$-modules $M$ and $N$, we denote by $\mathrm{Hom}_{R(\beta)}(M,N)$ the space of degree preserving module homomorphisms. If $f\in\mathrm{Hom}_{R(\beta)}(q^{k}M,N)$, we set $\deg(f):=k.$ Then we define the following graded vector space:

[TABLE]

We write $\mathrm{Hom}(M,N)$ and $\mathrm{HOM}(M,N)$ if there is no confusion. For $\beta,\beta^{\prime}\in\mathsf{Q}^{+}$, we set

[TABLE]

Definition 2.18**.**

Let $M$ be a graded $R(\beta)$-module, $N$ a graded $R(\beta^{\prime})$-module. Then the convolution product $M\circ N$ is the graded $R(\beta+\beta^{\prime})$-module defined by

[TABLE]

Let $\mathbb{A}=\mathbb{Z}[q,q^{-1}]$. We denote by $[R\text{-}\mathrm{proj}]$ and $[R\text{-}\mathrm{gmod}]$ the Grothendieck groups of $R\text{-}\mathrm{proj}$ and $R\text{-}\mathrm{gmod}$ respectively. The convolution product makes $[R\text{-}\mathrm{proj}]$ and $[R\text{-}\mathrm{gmod}]$ into $\mathbb{A}$-algebras, and we have the following theorem.

Theorem 2.19** ([17, 18, 25]).**

There exist isomorphisms of $\mathbb{A}$-algebras

[TABLE]

Now we explain the cyclotomic categorification theorem proved by Kang and Kashiwara. For this, we introduce the induction and restriction functors $F_{i}^{\Lambda}$ and $E_{i}^{\Lambda}$, for $i\in I$, as follows.

• •

The induction functors $F_{i}^{\Lambda}:R^{\Lambda}(\beta)\text{-}\mathrm{Mod}\to R^{\Lambda}(\beta+\alpha_{i})\text{-}\mathrm{Mod}$ are defined by

[TABLE]

• •

The restriction functors $E_{i}^{\Lambda}:R^{\Lambda}(\beta)\text{-}\mathrm{Mod}\to R^{\Lambda}(\beta-\alpha_{i})\text{-}\mathrm{Mod}$ are defined by

[TABLE]

The following theorem is proved by showing that $F_{i}^{\Lambda}$ and $E_{i}^{\Lambda}$ are biadjoint functors. The action of the Chevalley generators on the left-hand side of each of the isomorphisms in the theorem is given by the linear operators induced by the functors: for $\beta\in\mathsf{Q}^{+}$,

[TABLE]

Theorem 2.20** ([14, Theorem 6.2]).**

For $\Lambda\in\mathsf{P}^{+}$, there exist $U_{\mathbb{A}}(\mathfrak{g})$-module isomorphisms

[TABLE]

2.4. Convolution product for cyclotomic quiver Hecke algebras

The aim of this subsection is to prove the following.

Proposition 2.21**.**

If $M\in R^{\Lambda}(\beta)\text{-}\mathrm{gmod}$ for $\Lambda\in\mathsf{P}^{+}$ and $N\in R^{\Lambda^{\prime}}(\beta^{\prime})\text{-}\mathrm{gmod}$ for $\Lambda^{\prime}\in\mathsf{P}^{+}$, then $M\circ N\in R^{\Lambda+\Lambda^{\prime}}(\beta+\beta^{\prime})\text{-}\mathrm{gmod}.$

Proof.

Let $m={\rm ht}(\beta)$ and $n={\rm ht}(\beta^{\prime})$. We may assume that $M$ and $N$ are non-zero modules and $m,n>0$, so that we may take non-zero elements $a\in e(\nu_{1}\dots\nu_{m})M$ and $b\in e(\nu_{m+1}\dots\nu_{m+n})N$, for some $\nu=(\nu_{1},\nu_{2},\dots,\nu_{m+n})\in I^{m+n}$. As $e(\nu_{1}\dots\nu_{m})\neq 0$ and $e(\nu_{m+1}\dots\nu_{m+n})\neq 0$, the defining relations

[TABLE]

imply $\langle\alpha^{\vee}_{\nu_{1}},\Lambda\rangle>0$ and $\langle\alpha^{\vee}_{\nu_{m+1}},\Lambda^{\prime}\rangle>0$. We take $w\in\mathfrak{S}_{m+n}/\mathfrak{S}_{m}{\times}\mathfrak{S}_{n}$. Note that

[TABLE]

so that $w$ is fully commutative and

[TABLE]

To prove the assertion, it suffices to show that

[TABLE]

where $l=\langle\alpha^{\vee}_{\nu_{w^{-1}(1)}},\Lambda\rangle$ and $l^{\prime}=\langle\alpha^{\vee}_{\nu_{w^{-1}(1)}},\Lambda^{\prime}\rangle$.

If $w^{-1}(1)=1$, then $x_{1}\psi_{w}=\psi_{w}x_{1}$. Thus, we have

[TABLE]

Suppose that $w^{-1}(1)=m+1$. We set $u=ws_{m}s_{m-1}\dots s_{1}$, whose two-line notation is

[TABLE]

so that $u(1)=1$ and $\ell(us_{1}s_{2}\dots s_{m})=\ell(u)+\ell(s_{1}s_{2}\dots s_{m})=\ell(u)+m$. As $w$ is fully commutative, $\psi_{w}=\psi_{u}\psi_{1}\psi_{2}\dots\psi_{m}$. It follows from

[TABLE]

and $s_{1}\dots s_{t-1}(\nu_{1}\dots\nu_{m})=(\nu_{t}\nu_{1}\dots\widehat{\nu_{t}}\dots\nu_{m})$ that

[TABLE]

Continuing this process, we have

[TABLE]

which completes the proof. ∎

2.5. Dual space for the convolution product

Let $\tau:R(\beta)\rightarrow R(\beta)$ be the graded anti-involution which is the identity on generators. For M\in\text{R(\beta)\text{-}\mathrm{gmod}}, we define $M^{\circledast}:=\mathrm{HOM}_{\mathcal{O}}(M,\mathcal{O})$ to be the dual of $M$ whose $R(\beta)$-action is given by $(xf)(v)=f(\tau(x)v)$ for $x\in R(\beta)$, $f\in M^{\circledast}$ and $v\in M$.

We take self-dual simple modules M\in\text{R(\beta)\text{-}\mathrm{gmod}} and N\in\text{R(\gamma)\text{-}\mathrm{gmod}} with $m={\rm ht}(\beta)$ and $n={\rm ht}(\gamma)$. Let $\mathfrak{b}_{M}$ and $\mathfrak{b}_{N}$ be bases of $M$ and $N$ over $\mathcal{O}$ respectively. Then

[TABLE]

is a basis of $M\circ N$. We define

[TABLE]

to be the dual basis of $\mathfrak{b}_{M\circ N}$, i.e. $\xi_{w}^{x,y}(\psi_{w^{\prime}}\otimes x^{\prime}\otimes y^{\prime})=\delta_{(w,x,y),(w^{\prime},x^{\prime},y^{\prime})}$. It is known that there is an $R(\beta+\gamma)$-module isomorphism

[TABLE]

which sends $1\otimes y\otimes x$ to $\xi_{w[m,n]}^{x,y}$ for $y\in N$ and $x\in M$. See [22, Theorem 2.2(2)].

Lemma 2.22**.**

The isomorphism $N\circ M\simeq q^{-(\beta,\gamma)}(M\circ N)^{\circledast}$ sends $\psi_{w}\otimes y\otimes x\in N\circ M$ to

[TABLE]

for some $a_{w^{\prime},x^{\prime},y^{\prime}}\in\mathcal{O}$, and $\xi_{w^{-1}w[m,n]}^{x,y}\in q^{-(\beta,\gamma)}(M\circ N)^{\circledast}$ to

[TABLE]

for some $b_{w^{\prime},x^{\prime},y^{\prime}}\in\mathcal{O}$.

Proof.

The first assertion is clear because $\psi_{w}\xi_{w[m,n]}^{x,y}$ has the desired form. The second assertion follows from the first. ∎

3. Specht modules in affine and infinite type C

In this section, we introduce Specht modules for cyclotomic quiver Hecke algebras in type $C^{(1)}_{\ell}$ or $C_{\infty}$ and provide a basis theorem for Specht modules in type $C_{\infty}$. From now until Definition 3.11, we assume that $\mathcal{O}$ is a field.

3.1. The modules $\mathcal{L}(k;\ell)$

For $k\in\mathbb{Z}$ and $\ell\in\mathbb{Z}_{>0}$, let

[TABLE]

Then $\mathcal{L}(k;\ell)=\mathcal{O}\mathfrak{l}_{(k;\ell)}$ is the one-dimensional graded $R(\beta_{(k;\ell)})$-module defined by $\deg(\mathfrak{l}_{(k;\ell)})=0$ and

[TABLE]

for $1\leqslant i\leqslant\ell$, $1\leqslant j\leqslant\ell-1$, $\nu\in I^{\beta_{(k;\ell)}}$. If there is no confusion, we write $\mathfrak{l}$ for $\mathfrak{l}_{(k;\ell)}$ and we sometimes write $\mathcal{L}(\overline{k},\overline{k+1},\ldots,\overline{k+\ell-1})$ instead of $\mathcal{L}(k;\ell)$.

Let $k\in\mathbb{Z}$ and $\ell_{1},\ell_{2}\in\mathbb{Z}_{\geqslant 0}$. As

[TABLE]

as an $R(\beta_{(k;\ell_{1})})\otimes R(\beta_{(k+\ell_{1};\ell_{2})})$-module by construction, we have

[TABLE]

by Frobenius reciprocity so that there exists a surjective $R(\beta_{(k;\ell_{1}+\ell_{2})})$-module homomorphism

[TABLE]

sending $\mathfrak{l}\otimes\mathfrak{l}$ to $\mathfrak{l}$. Taking the dual of $p$, we have the graded monomorphism

[TABLE]

Then, noting that $p(\psi_{w}\otimes\mathfrak{l}\otimes\mathfrak{l})=\psi_{w}\mathfrak{l}$ implies $\iota(\mathfrak{l})=\xi_{1}^{\mathfrak{l},\mathfrak{l}}$, $\eqref{dual}$ from Lemma 2.22 shows that

[TABLE]

with $\psi_{w}(\mathfrak{l}\otimes\mathfrak{l})\in e(\nu_{(k;\ell_{1}+\ell_{2})})\mathcal{L}(k+\ell_{1};\ell_{2})\circ\mathcal{L}(k;\ell_{1})$ and $\deg(\psi_{w}(\mathfrak{l}\otimes\mathfrak{l}))=0$ whenever $a_{w}\neq 0$. Here, $\deg(\mathfrak{l}\otimes\mathfrak{l})=(\beta_{(k;\ell_{1})},\beta_{(k+\ell_{1};\ell_{2})})$ because of the shift. The following lemma is easy to see by construction.

Lemma 3.1**.**

Define

[TABLE]

1. (1)

Let $\mathfrak{l}_{1}=\mathfrak{l}_{(k;\ell_{1})}$ and $\mathfrak{l}_{2}=\mathfrak{l}_{(k+\ell_{1};\ell_{2})}$. Then

[TABLE]

with $\psi_{w}(\mathfrak{l}_{2}\otimes\mathfrak{l}_{1})\in e(\nu_{(k;\ell_{1}+\ell_{2})})\mathcal{L}(k+\ell_{1};\ell_{2})\circ\mathcal{L}(k;\ell_{1})$ and $\deg(\psi_{w}(\mathfrak{l}_{2}\otimes\mathfrak{l}_{1}))=0$ whenever $a_{w}\neq 0$. 2. (2)

${\rm im}(r)$* is isomorphic to $\mathcal{L}(k;\ell_{1}+\ell_{2})$.*

Corollary 3.2**.**

If the Cartan matrix is of type $C_{\infty}$, then

[TABLE]

Proof.

In type $C_{\infty}$, we know by examining residues that $e(\nu)\mathcal{L}(k+\ell_{1};\ell_{2})\circ\mathcal{L}(k;\ell_{1})\neq 0$ if and only if $\nu$ is a shuffle of $\nu_{(k+\ell_{1};\ell_{2})}$ and $\nu_{(k;\ell_{1})}$. Thus it is straightforward to check that

[TABLE]

which completes the proof by Lemma 3.1. ∎

Remark 3.3*.*

It is easy to show that $\mathcal{L}(k;\ell)$ admits an affinization for any $k$ and $\ell$. If $\mathsf{A}$ is of type $C_{\infty}$, then $\mathcal{L}(k;\ell)$ is real and $r$ in Lemma 3.1 is the $R$-matrix [16]. Note that, if $\mathsf{A}$ is affine, $\mathcal{L}(k;\ell)$ is not real in general.

Proposition 3.4**.**

Let $k\in\mathbb{Z}$ and $a,b,c\in\mathbb{Z}_{\geqslant 0}$ with $b\geqslant c>0$. Then, there is a non-zero $R(\beta_{(k;a)}+\beta_{(k-1;a+b+1)}+\beta_{(k+a;c-1)})$-module homomorphism

[TABLE]

such that

[TABLE]

If the Cartan matrix $\mathsf{A}$ is of type $C_{\infty}$ then $a_{w}=0$ for all $w\prec S_{2}(a,b,a+1)$.

Proof.

Combining Lemma 3.1 with the surjectivity of $p$, we have a non-zero homomorphism

[TABLE]

Lemma 3.1 (2) tells us that the image of the first homomorphism is isomorphic to

[TABLE]

which is generated by

[TABLE]

by Lemma 3.1 (1). Thus it gives a non-zero homomorphism

[TABLE]

such that $\mathsf{g}(\mathfrak{l}\otimes\mathfrak{l}\otimes\mathfrak{l})$ has the desired form. ∎

3.2. The modules $\mathcal{S}^{\lambda}$

Let $\lambda=(\lambda_{1},\lambda_{2},\ldots,\lambda_{t})\in\mathscr{P}^{1}_{n}$ with a charge $\kappa\in\mathbb{Z}$. Note that the level of $\lambda$ is 1. Let $\beta:=\operatorname{cont}(\lambda)$ and define

[TABLE]

For a Garnir node $A=(r,c)\in[\lambda]$, let

[TABLE]

where $\lambda_{<r}=(\lambda_{1},\ldots,\lambda_{r-1})$ and $\lambda_{>r+1}=(\lambda_{r+2},\ldots,\lambda_{t})$. We denote by $m^{\lambda}_{\kappa}$ (resp. $m^{\lambda}_{\kappa,A}$) the distinguished generator $\mathfrak{l}\otimes\dots\otimes\mathfrak{l}$ of $\mathcal{M}^{\lambda}_{\kappa}$. (resp. $\mathcal{M}^{\lambda}_{\kappa,A}$.) By Proposition 3.4, we have the non-zero homomorphism

[TABLE]

which gives the induced homomorphism

[TABLE]

Definition 3.5**.**

Let $\lambda\in\mathscr{P}^{1}_{n}$ with a charge $\kappa\in\mathbb{Z}$. Then we define, for a Garnir node $A$,

[TABLE]

By Proposition 3.4 and $\eqref{exp of GA}$, we have

[TABLE]

Remark 3.6*.*

If the Cartan matrix $\mathsf{A}$ is of type $C_{\infty}$, then $g^{\lambda}_{\kappa,A}=\psi_{w^{\mathtt{G}^{A}}}m^{\lambda}_{\kappa}$. This is reminiscent of the Garnir element defined in [21] for type $A_{\infty}$.

Lemma 3.7**.**

We have

• (i)

$x_{i}g^{\lambda}_{\kappa,A}=0$, for $1\leqslant i\leqslant n$,

• (ii)

$\psi_{j}g^{\lambda}_{\kappa,A}=0$* unless $s_{j}\mathtt{G}^{A}\in\mathsf{Row}(\lambda)$.*

Proof.

For $1\leqslant i\leqslant n$, we have

[TABLE]

Let $A=(r,c)\in\lambda$ and $l_{p}=\sum_{k=1}^{p}\lambda_{p}$. Considering the definition of $\mathtt{G}^{A}$, we know that

[TABLE]

Thus, by the construction of $m^{\lambda}_{\kappa,A}$,

[TABLE]

unless $s_{j}\mathtt{G}^{A}\in\mathsf{Row}(\lambda)$. ∎

We define $\mathcal{H}^{\lambda}_{\kappa}:\oplus_{A}q^{-p_{\kappa,A}^{\lambda}}\mathcal{M}^{\lambda}_{\kappa,A}\rightarrow\mathcal{M}^{\lambda}_{\kappa}$ as the sum of $\mathcal{H}^{\lambda}_{\kappa,A}$ over Garnir nodes $A$ of $\lambda$ and set

[TABLE]

If there is no possibility of confusion, we will drop the subscript $\kappa$ from our notation, i.e. we will simply write $\mathcal{M}^{\lambda}$, $m^{\lambda}$, $g^{\lambda}_{A}$, $\mathcal{S}^{\lambda}$, etc.

Definition 3.8**.**

For $\lambda=(\lambda^{(1)},\dots,\lambda^{(l)})\in\mathscr{P}^{l}_{n}$ and $\kappa=(\kappa_{1},\dots,\kappa_{l})\in\mathbb{Z}^{l}$, we define

[TABLE]

We write $\mathcal{S}^{\lambda}(\mathcal{O})$ when we need to emphasise the field.

Remark 3.9*.*

By Theorem 2.3, the set

[TABLE]

is an $\mathcal{O}$-basis of $\mathcal{M}^{\lambda}$.

Note that $\deg\psi_{w^{\mathtt{T}}}m^{\lambda}=\deg\psi_{w^{\mathtt{T}}}e(\mathrm{res}(\mathtt{T}))$ by definition. Then we have the following result.

Proposition 3.10**.**

If $\mathtt{T}\in\mathsf{Std}(\lambda)$, then $\deg\psi_{w^{\mathtt{T}}}m^{\lambda}=\deg\mathtt{T}-\deg\mathtt{T}^{\lambda}$.

Proof.

We closely follow the proof of [8, Proposition 3.13]. If $\mathtt{T}=\mathtt{T}^{\lambda}$, then we have

[TABLE]

Thus, it suffices to prove our statement in the case that $\mathtt{S},\mathtt{T}\in\mathsf{Std}(\lambda)$ are such that $\ell(\mathtt{S})=\ell(\mathtt{T})+1$ and $\mathtt{S}=s_{r}\mathtt{T}$. Let $\mathrm{res}(\mathtt{T})=(\nu_{1},\nu_{2},\dots,\nu_{n})$. We may assume that $r=n-1$. We want to show that $\deg\mathtt{T}-\deg\mathtt{S}=(\alpha_{\nu_{n-1}},\alpha_{\nu_{n}})$. Let $A=\mathtt{T}^{-1}(n)$ and $B=\mathtt{T}^{-1}(n-1)$. By assumption, $B$ is above $A$ in $[\lambda]$. Now,

[TABLE]

Note that $\mathtt{T}_{\downarrow n-2}=\mathtt{S}_{\downarrow n-2}$, and since $B$ is above $A$, $d_{A}(\lambda)=d_{A}(\lambda\nearrow B)$. So we must show that $d_{B}(\lambda\nearrow A)-d_{B}(\lambda)=(\alpha_{\nu_{n-1}},\alpha_{\nu_{n}})$.

If $\mathrm{res}(A)=\mathrm{res}(B)=i$, then removing $A$ leads to the disappearance of a removable $i$-node and the appearance of a new addable $i$-node below $B$, so that $d_{B}(\lambda\nearrow A)-d_{B}(\lambda)=4$ if $i=0$, or $2$ otherwise.

If $\mathrm{res}(A)=0$ and $\mathrm{res}(B)=1$, removing $A$ leaves either one fewer addable $1$-node and one extra removable $1$-node, or two extra removable $1$-nodes, or two fewer addable $1$-nodes, so that $d_{B}(\lambda\nearrow A)-d_{B}(\lambda)=-2$.

If $\mathrm{res}(A)=1$ and $\mathrm{res}(B)=0$, removing $A$ leaves either one extra removable [math]-node or one fewer addable [math]-node, so that $d_{B}(\lambda\nearrow A)-d_{B}(\lambda)=-2$.

If $\mathrm{res}(A)=\ell-1$ and $\mathrm{res}(B)=\ell$ or $\mathrm{res}(A)=\ell$ and $\mathrm{res}(B)=\ell-1$ in type $C^{(1)}_{\ell}$, similar arguments show that $d_{B}(\lambda\nearrow A)-d_{B}(\lambda)=-2$.

If $\mathrm{res}(A)=i\pm 1$ and $\mathrm{res}(B)=i$, with neither residue equal to [math] or $\ell$, then removing $A$ leaves either one extra removable $i$-node or one fewer addable $i$-node, so $d_{B}(\lambda\nearrow A)-d_{B}(\lambda)=-1$.

In all other cases, removing $A$ does not change the degree, so $d_{B}(\lambda\nearrow A)-d_{B}(\lambda)=0$. ∎

We denote by $\overline{m}^{\lambda}$ the image of $m^{\lambda}$ under the projection $q^{\deg(\mathtt{T}^{\lambda})}\mathcal{M}^{\lambda}\rightarrow\mathcal{S}^{\lambda}$.

Definition 3.11**.**

Let $\mathcal{O}$ be an integral domain. Then for $\lambda\in\mathscr{P}^{l}_{n}$ and $\kappa\in\mathbb{Z}^{l}$, we define $\mathcal{S}^{\lambda}_{\kappa}(\mathcal{O})$ over $\mathcal{O}$ to be the lattice $R_{\mathcal{O}}(\operatorname{cont}(\lambda))\overline{m}^{\lambda}$ generated by $\overline{m}^{\lambda}$ in $\mathcal{S}^{\lambda}_{\kappa}(\mathbb{F})$, where $\mathbb{F}=\operatorname{Frac}(\mathcal{O})$ and $R_{\mathcal{O}}(\operatorname{cont}(\lambda))$ is the quiver Hecke algebra over $\mathcal{O}$.

From now on, let $\mathcal{O}$ denote an arbitrary integral domain.

Theorem 3.12**.**

Let $\lambda\in\mathscr{P}^{1}_{n}$ with a charge $\kappa\in\mathbb{Z}$, and let $\beta=\operatorname{cont}(\lambda)$.

1. (1)

$\mathcal{S}^{\lambda}$* is generated by $\{\psi_{w^{\mathtt{T}}}\overline{m}^{\lambda}\mid\mathtt{T}\in\mathsf{Std}(\lambda)\}$ as an $\mathcal{O}$-module.* 2. (2)

$\mathcal{S}^{\lambda}$* is a graded $R^{\Lambda_{\overline{\kappa}}}(\beta)$-module.*

Proof.

(1) For $\ell=0,1,\dots$, we define

[TABLE]

Then $\eqref{basis of M}$ implies that $\mathcal{S}^{\lambda}$ is generated by $\bigcup_{\ell\geqslant 0}A_{\ell}$ as an $\mathcal{O}$-module, so it suffices to show that

[TABLE]

for all $\ell\geqslant 0$ by induction on $\ell$. If $\ell=0$, there is nothing to prove. Suppose that $\ell>0$ and take $\mathtt{T}\in\mathsf{RowStd}(\lambda)$ with $\ell=\ell(\mathtt{T})$. We will show that $\psi_{w^{\mathtt{T}}}\overline{m}^{\lambda}\in{\rm Span}_{\mathcal{O}}B_{\ell}$. Since it is trivial when $\mathtt{T}\in\mathsf{Std}(\lambda)$, we assume that $\mathtt{T}\in\mathsf{Row}(\lambda)$ and prove $\psi_{w^{\mathtt{T}}}\overline{m}^{\lambda}\in{\rm Span}_{\mathcal{O}}B_{\ell-1}$. We set

[TABLE]

By Lemma 1.12, there are a Garnir node $A\in[\lambda]$ and an element $w\in\mathfrak{S}_{n}$ such that $\mathtt{T}=w\mathtt{G}^{A}$ and $\ell(\mathtt{T})=\ell(w)+\ell(\mathtt{G}^{A})$. It follows from Proposition 2.2 and $\eqref{basis of M}$ that

[TABLE]

By $\eqref{g_A and Garnir elt}$, we have

[TABLE]

which implies that

[TABLE]

proving $\psi_{w^{\mathtt{T}}}\overline{m}^{\lambda}\in{\rm Span}_{\mathcal{O}}B_{\ell-1}$ by the induction hypothesis ${\rm Span}_{\mathcal{O}}A_{\ell-1}={\rm Span}_{\mathcal{O}}B_{\ell-1}$.

(2) It follows from (1) that it suffices to prove $x_{1}^{\langle\alpha_{\nu_{1}}^{\vee},\Lambda_{\overline{\kappa}}\rangle}e(\nu)\psi_{w^{\mathtt{T}}}m^{\lambda}=0$, for $\mathtt{T}\in\mathsf{Std}(\lambda)$. But if $\mathtt{T}\in\mathsf{Std}(\lambda)$ then $w^{\mathtt{T}}(1)=1$, so that $\psi_{w^{\mathtt{T}}}$ is a product of $\psi_{2},\dots,\psi_{n-1}$ and $x_{1}\psi_{w^{\mathtt{T}}}=\psi_{w^{\mathtt{T}}}x_{1}$ holds. Then, since $e(\nu)\psi_{w^{\mathtt{T}}}m^{\lambda}\neq 0$ implies $\nu_{1}=\overline{\kappa}$,

[TABLE]

Corollary 3.13**.**

Let $l\in\mathbb{Z}_{>0}$, $\lambda\in\mathscr{P}^{l}_{n}$, $\kappa=(\kappa_{1},\dots,\kappa_{l})\in\mathbb{Z}^{l}$, and let $\beta=\operatorname{cont}(\lambda)$.

1. (1)

$\mathcal{S}^{\lambda}$* is generated by $\{\psi_{w^{\mathtt{T}}}\overline{m}^{\lambda}\mid\mathtt{T}\in\mathsf{Std}(\lambda)\}$ as an $\mathcal{O}$-module.* 2. (2)

Let $\Lambda=\Lambda_{\overline{\kappa_{1}}}+\dots+\Lambda_{\overline{\kappa_{l}}}$. Then $\mathcal{S}^{\lambda}$ is a graded $R^{\Lambda}(\beta)$-module.

Proof.

This follows from Theorem 3.12 and Proposition 2.21. ∎

Definition 3.14**.**

Let $l\in\mathbb{Z}_{>0}$, $\kappa=(\kappa_{1},\dots,\kappa_{l})\in\mathbb{Z}^{l}$ and $\Lambda=\Lambda_{\overline{\kappa_{1}}}+\dots+\Lambda_{\overline{\kappa_{l}}}$. Then we call the graded $R^{\Lambda}(\beta)$-modules $\mathcal{S}^{\lambda}$, for $\lambda\in\mathscr{P}^{l}_{n}$, Specht modules.

Remark 3.15*.*

One can easily construct a ‘column version’ of the Specht modules by the same argument. (cf. [21, Section 7]).

Example 3.16**.**

Let $\kappa\in\mathbb{Z}$ and $\lambda=(n),\lambda^{\prime}=(1^{n})\in\mathscr{P}^{1}_{n}$. It is straightforward to prove that

[TABLE]

In particular, $\mathcal{S}^{\lambda}_{0}\simeq\mathcal{S}^{\lambda^{\prime}}_{0}$.

Example 3.17**.**

Suppose that $\mathsf{A}$ is of type $C_{\infty}$ or $C_{\ell}^{(1)}$ with $\ell>2$. Let $\kappa=-1$ and $\lambda=(4),\mu=(3,1)\in\mathscr{P}^{1}_{4}$. As $\lambda$ has no Garnir nodes, we have

[TABLE]

Since $\mu$ has only one Garnir node $(1,1)$, we have $p_{(1,1)}^{\mu}=(\alpha_{2},\alpha_{1}+\alpha_{0}+\alpha_{1})=-2$ and

[TABLE]

Thus, we have $\mathcal{G}^{\lambda}\simeq q^{2}\mathcal{L}(2101)$ and

[TABLE]

The epimorphism $\mathcal{L}(101)\circ\mathcal{L}(2)\twoheadrightarrow\mathcal{L}(1012)$ gives the epimorphism

[TABLE]

which tells us that $\mathcal{S}^{\mu}$ is not simple and the head of $\mathcal{S}^{\mu}$ is isomorphic to $\mathcal{S}^{\lambda}$.

Example 3.18**.**

Suppose that $\mathsf{A}$ is of type $C_{\infty}$ or $C_{\ell}^{(1)}$ with $\ell>2$. Let $\kappa=0$ and $\lambda=(3,2,1)\in\mathscr{P}^{1}_{6}$. Then $\deg(\mathtt{T}^{\lambda})=1$, $\mathcal{M}^{\lambda}=\mathcal{L}(012)\circ\mathcal{L}(10)\circ\mathcal{L}(2)$ and the Garnir nodes of $\lambda$ are $A_{1}:=(1,1)$, $A_{2}:=(1,2)$ and $A_{3}:=(2,1)$. Since

[TABLE]

we have

[TABLE]

Thus, $\mathcal{G}^{\lambda}=\langle\psi_{1}\psi_{2}\psi_{3}m^{\lambda},\psi_{3}\psi_{2}\psi_{4}\psi_{3}m^{\lambda},\psi_{4}\psi_{5}m^{\lambda}\rangle\subset\mathcal{M}^{\lambda}$ and

[TABLE]

3.3. Basis theorem for type $C_{\infty}$

Suppose that the Cartan matrix is of type $C_{\infty}$. Then we have the following basis theorem for Specht modules, whose proof is postponed to Section 4.

Theorem 3.19**.**

Let $\lambda\in\mathscr{P}^{1}_{n}$ with a charge $\kappa\in\mathbb{Z}$. Then the set $\{\psi_{w^{\mathtt{T}}}\overline{m}^{\lambda}\mid\mathtt{T}\in\mathsf{Std}(\lambda)\}$ is an $\mathcal{O}$-basis of $\mathcal{S}^{\lambda}$. Moreover, we have the following graded character formula.

[TABLE]

Corollary 3.20**.**

In the Grothendieck group of $R^{\Lambda_{\overline{\kappa}}}(n-1)\text{-}\mathrm{gmod}$, we have

[TABLE]

where $b$ runs over all removable $i$-nodes.

Proof.

We rewrite the graded character formula from Theorem 3.19 as follows.

[TABLE]

where $b$ runs over all removable nodes. Thus,

[TABLE]

where $b$ runs over all removable $i$-nodes. ∎

Let $l\in\mathbb{Z}_{>0}$, $\kappa=(\kappa_{1},\ldots,\kappa_{l})\in\mathbb{Z}^{l}$ and $\Lambda=\Lambda_{\overline{\kappa}_{1}}+\dots+\Lambda_{\overline{\kappa}_{l}}$. One can easily prove Corollary 3.21 from Theorem 3.19 and Corollary 3.20.

Corollary 3.21**.**

Let $\lambda\in\mathscr{P}^{l}_{n}$.

1. (1)

The set $\{\psi_{w^{\mathtt{T}}}\overline{m}^{\lambda}\mid\mathtt{T}\in\mathsf{Std}(\lambda)\}$ is an $\mathcal{O}$-basis of $\mathcal{S}^{\lambda}$. Moreover,

[TABLE] 2. (2)

In the Grothendieck group of $R^{\Lambda}(n-1)\text{-}\mathrm{gmod}$, we have

[TABLE]

where $b$ runs over all removable $i$-nodes.

We revisit the Fock space $\mathcal{F}(\kappa)$. As $V_{q}(\Lambda)\simeq V_{q}(\Lambda)^{\vee}$, by Theorem 2.20, we can identify $V_{q}(\Lambda)\simeq V_{q}(\Lambda)^{\vee}\simeq\mathbb{Q}(q)\otimes_{\mathbb{A}}[R^{\Lambda}\text{-}\mathrm{gmod}]$. Thus, from $\eqref{Eq: pk}$, we have the $U_{q}(\mathfrak{g}(\mathsf{A}))$-module epimorphism

[TABLE]

Proposition 3.22**.**

For $\lambda\in\mathscr{P}^{l}_{n}$, we have

[TABLE]

Proof.

It is obvious that $p_{\kappa}(\varnothing)=[\mathcal{S}^{\varnothing}]$ and ${\rm wt}(\varnothing)={\rm wt}([\mathcal{S}^{\varnothing}])=\Lambda$. Since both of $\mathcal{F}(\kappa)$ and $\mathbb{Q}(q)\otimes_{\mathbb{A}}[R^{\Lambda}\text{-}\mathrm{gmod}]$ are integrable $U_{q}(\mathfrak{g}(\mathsf{A}))$-modules and $\mathbb{Q}(q)\otimes_{\mathbb{A}}[R^{\Lambda}\text{-}\mathrm{gmod}]$ is simple, it suffices to show that $e_{i}(p_{\kappa}(\lambda)-[\mathcal{S}^{\lambda}])=0$ for all $\lambda$. By $\eqref{Eq: def of Fock sp}$, Corollary 3.21 and the induction hypothesis, we have

[TABLE]

which completes the proof. ∎

Corollary 3.23 follows from $\eqref{Eq: def of Fock sp}$, $\eqref{Eq: categorification}$ and Proposition 3.22.

Corollary 3.23**.**

Let $\beta=\operatorname{cont}(\lambda)$. In the Grothendieck group of $R^{\Lambda}(n+1)\text{-}\mathrm{gmod}$, we have

[TABLE]

where $b$ runs over all addable $i$-nodes.

Example 3.24**.**

We use the same notation as in Example 3.18. Let $\mu=(2,2)$, $\mu_{1}=(3,2)$ and $\mu_{2}=(2,2,1)$. By Theorem 3.19, we have

[TABLE]

where $[n]_{q}:=\frac{q^{n}-q^{-n}}{q-q^{-1}}$ for $n\in\mathbb{Z}_{\geqslant 0}$. Let $B_{1}=(1,3)$ and $B_{2}=(3,1)$. Then

[TABLE]

By Corollary 3.20 and Corollary 3.23, we have

[TABLE]

Remark 3.25*.*

It looks like we need a modified version of the upper global basis in the Fock space to describe the simple modules. It is an interesting problem to characterise the elements in the Fock space which correspond to the simple modules.

4. Proof of Theorem 3.19

We assume that the Cartan matrix $\mathsf{A}$ is of type $C_{\infty}$ and take the parameters $\eqref{Qij for C_infty}$ for the quiver Hecke algebra $R(\beta)$. Let us fix $\lambda\in\mathscr{P}^{1}_{n}$, $\kappa\in\mathbb{Z}$ and $\beta=\operatorname{cont}(\lambda)$.

Definition 4.1**.**

For $t\in\mathbb{Z}_{>0}$, we define

[TABLE]

Remark 4.2*.*

Note that we require $w\mathtt{G}^{A}\in\mathsf{Row}(\lambda)$ in the definition of $\mathcal{G}^{\lambda}_{<t}$. In Theorem 4.15 below, we will eliminate the possibility that $\mathcal{G}^{\lambda}$ is strictly larger than $\sum_{t\in\mathbb{Z}_{>0}}\mathcal{G}^{\lambda}_{<t}$.

Lemmas 4.3 and 4.4 are needed for proving Lemma 4.5.

Lemma 4.3**.**

Let $\mathtt{T}\in\mathsf{RowStd}(\lambda)$.

1. (1)

If

1. (i)

$\lambda=(\lambda_{1},\lambda_{2})$* with $\lambda_{2}>0$,* 2. (ii)

$\mathrm{res}(\mathtt{T}^{-1}(n))=\overline{\kappa+\lambda_{1}-1}$* and $\mathrm{res}(\mathtt{T}^{-1}(n-1))=\overline{\kappa+\lambda_{1}-2}$,*

then $\mathtt{T}(1,\lambda_{1})=n$. 2. (2)

If

1. (i)

$\lambda=(\lambda_{1},\lambda_{2},\lambda_{3})$* with $\lambda_{3}>0$,* 2. (ii)

$\mathrm{res}(\mathtt{T}^{-1}(n))=\overline{\kappa+\lambda_{1}-1}$, $\mathrm{res}(\mathtt{T}^{-1}(n-1))=\overline{\kappa+\lambda_{1}-2}$ and $\mathrm{res}(\mathtt{T}^{-1}(n-2))=\overline{\kappa+\lambda_{1}-3}$, 3. (iii)

$\mathrm{res}(3,\lambda_{3})\neq\mathrm{res}(1,\lambda_{2})$,

then $\mathtt{T}(1,\lambda_{1})=n$.

Proof.

(1) As $\mathtt{T}\in\mathsf{RowStd}(\lambda)$, $\mathtt{T}^{-1}(n)$ = $(1,\lambda_{1})$ or $(2,\lambda_{2})$. If $\kappa+\lambda_{1}-1\leqslant 0$ then $(1,\lambda_{1})$ is the only node of residue $\overline{\kappa+\lambda_{1}-1}$, so that $\mathtt{T}(1,\lambda_{1})=n$. Suppose that $\kappa+\lambda_{1}-1>0$ and $\mathtt{T}^{-1}(n)$ = $(2,\lambda_{2})$. Then the assumption (ii) implies that

[TABLE]

and $\mathtt{T}^{-1}(n-1)$ = $(1,\lambda_{1})$ or $(2,\lambda_{2}-1)$. Thus, $\mathrm{res}(\mathtt{T}^{-1}(n-1))$ is either $\kappa+\lambda_{1}-1$ or $\kappa+\lambda_{1}$, which are not equal to $\overline{\kappa+\lambda_{1}-2}=\kappa+\lambda_{1}-2$.

(2) As $\mathtt{T}\in\mathsf{RowStd}(\lambda)$, we know that $\mathtt{T}^{-1}(n)$ = $(1,\lambda_{1})$, $(2,\lambda_{2})$ or $(3,\lambda_{3})$. Further, by the same reasoning as in (1), $\kappa+\lambda_{1}-1>0$ and $\mathtt{T}^{-1}(n-1)\neq(1,\lambda_{1})$ hold if $\mathtt{T}^{-1}(n)$ = $(2,\lambda_{2})$ or $(3,\lambda_{3})$.

Suppose that $\mathtt{T}^{-1}(n)=(2,\lambda_{2})$. Then

[TABLE]

and $\mathtt{T}^{-1}(n-1)$ = $(2,\lambda_{2}-1)$ or $(3,\lambda_{3})$. Thus, $\mathrm{res}(\mathtt{T}^{-1}(n-1))\geqslant\kappa+\lambda_{1}$, which is not equal to $\overline{\kappa+\lambda_{1}-2}=\kappa+\lambda_{1}-2$.

Now suppose that $\mathtt{T}^{-1}(n)=(3,\lambda_{3})$. Then

[TABLE]

and $\mathtt{T}^{-1}(n-1)$ = $(2,\lambda_{2})$ or $(3,\lambda_{3}-1)$. Then $\mathrm{res}(\mathtt{T}^{-1}(n-1))$ = $\kappa+\lambda_{1}-2<\mathrm{res}(\mathtt{T}^{-1}(n))$ implies that $\mathtt{T}^{-1}(n-1)$ = $(2,\lambda_{2})$. In particular, $\mathtt{T}^{-1}(n-2)$ = $(1,\lambda_{1})$, $(2,\lambda_{2}-1)$ or $(3,\lambda_{3}-1)$.

If $\kappa+\lambda_{1}-1=1$, then $\mathrm{res}(\mathtt{T}^{-1}(n-1))=0$ by condition (ii). It follows that $\kappa+\lambda_{2}-2=0$ and $\kappa+\lambda_{3}-3=-1$, so that $\lambda_{1}=\lambda_{2}=\lambda_{3}$. But this contradicts condition (iii).

If $\kappa+\lambda_{1}-1\geqslant 2$, then, since $\mathrm{res}(\mathtt{T}^{-1}(n-1))=\kappa+\lambda_{1}-2$, we have one of the following.

• (a)

$\kappa+\lambda_{2}-2=\kappa+\lambda_{1}-2>0$ and $\lambda_{1}=\lambda_{2}$.

• (b)

$\kappa+\lambda_{2}-2=-(\kappa+\lambda_{1}-2)<0$ and $\lambda_{3}=\lambda_{2}$.

Suppose that we are in case (a). Then

[TABLE]

which contradicts condition (iii). Suppose that we are in case (b). Then none of $(1,\lambda_{1})$, $(2,\lambda_{2}-1)$ or $(3,\lambda_{3}-1)$ has residue $\kappa+\lambda_{1}-3$.∎

Lemma 4.4**.**

Let $A=(r,c)$ be a Garnir node of $[\lambda]$. Then there is no tableau $\mathtt{T}\in\mathsf{RowStd}(\lambda)$ such that

[TABLE]

Proof.

We take $\mathtt{T}\in\mathsf{RowStd}(\lambda)$ such that

[TABLE]

Note that Lemma 1.8 implies that

[TABLE]

Thus, we may assume that $\lambda=(\lambda_{1},c)$ and $r=1$. As $\mathrm{res}(\mathtt{T})=\mathrm{res}(\mathtt{G}^{A})$, we have $\mathtt{T}(1,\lambda_{1})=n$ by Lemma 4.3(1).

If $\lambda_{1}=c$, then we have $\mathtt{T}=\mathtt{G}^{A}$ by $\eqref{eq: seq1}$.

If $\lambda_{1}>c$, then we have

[TABLE]

which, by induction on $\lambda_{1}-c$, implies that $\mathtt{T}_{\downarrow n-1}=\mathtt{G}^{A}_{\downarrow n-1}$. Therefore, we have $\mathtt{T}=\mathtt{G}^{A}$. ∎

Lemma 4.5**.**

Let $A=(r,c)$ and $B=(r^{\prime},c^{\prime})$ be Garnir nodes of $[\lambda]$ with $c\leqslant c^{\prime}$. Suppose that either $(r\neq r^{\prime}+1)$ or $(r=r^{\prime}+1$ and $\mathrm{res}(B)\neq\mathrm{res}(r+1,c))$. Then there is no tableau $\mathtt{T}\in\mathsf{RowStd}(\lambda)$ such that

[TABLE]

Proof.

Let $\mathtt{T}\in\mathsf{RowStd}(\lambda)$ such that

[TABLE]

and $\mathbf{B}^{A}$ and $\mathbf{B}^{B}$ the Garnir belts corresponding to $A$ and $B$ respectively. If $\mathbf{B}^{A}\cap\mathbf{B}^{B}=\emptyset$, then we may argue as in the proof of Lemma 4.4 to see that $\mathtt{T}=\mathtt{G}^{A,B}$. So we assume that $\mathbf{B}^{A}\cap\mathbf{B}^{B}\neq\emptyset$. Then, we have two cases – either $r=r^{\prime}$, or $r=r^{\prime}+1$.

First suppose that $r=r^{\prime}$. By Lemma 1.8, we may assume that $\lambda=(\lambda_{1},c^{\prime})$, $c\neq c^{\prime}$ and $r=r^{\prime}=1$. Since $\mathrm{res}(\mathtt{T})=\mathrm{res}(\mathtt{G}^{A,B})$, we have $\mathtt{T}(1,\lambda_{1})=n$ by Lemma 4.3(1).

Suppose that $\lambda_{1}=c^{\prime}$ and consider the condition

[TABLE]

where $p=c^{\prime}-c+1$. As $\mathtt{G}^{A,B}_{\downarrow n-p}$ is a Garnir tableau with Garnir node $A$, we conclude that $\mathtt{T}_{\downarrow n-p}=\mathtt{G}^{A,B}_{\downarrow n-p}$ by Lemma 4.4. Since $\mathtt{T}\in\mathsf{RowStd}(\lambda)$ and $\mathtt{T}^{-1}(n)=(1,\lambda_{1})$, it follows that the entries $n-p+1,\dots,n-1$ must appear in the nodes $(2,c+1),\dots,(2,c^{\prime})$ respectively, and thus $\mathtt{T}=\mathtt{G}^{A,B}$.

If $\lambda_{1}>c^{\prime}$, then

[TABLE]

which, by induction on $\lambda_{1}-c^{\prime}$, implies that $\mathtt{T}_{\downarrow n-1}=\mathtt{G}^{A,B}_{\downarrow n-1}$. Thus we have $\mathtt{T}=\mathtt{G}^{A,B}$.

Next, suppose that $r=r^{\prime}+1$. By Lemma 1.8, we may assume that $\lambda=(\lambda_{1},\lambda_{2},c)$ and $r^{\prime}=1$. Note that $\mathrm{res}(1,c^{\prime})\neq\mathrm{res}(3,c)$ by our assumption and $\mathtt{T}(1,k)=k$ for $k=1,2,\ldots,c^{\prime}-1$ by Lemma 1.8. We now proceed by induction on $\lambda_{2}-c^{\prime}$.

First, suppose that $\lambda_{2}=c^{\prime}$. We must have $\mathtt{T}(1,\lambda_{1})=n$, by Lemma 4.3(2).

If $\lambda_{1}=c^{\prime}$, then we define row-strict tableaux $\mathtt{T}^{\prime}$ and $\mathtt{G}^{\prime}$ of shape $(\lambda_{2},c)$ by

[TABLE]

Then $\mathtt{G}^{\prime}$ becomes a Garnir tableau with Garnir node $A$ and

[TABLE]

which implies that $\mathtt{T}^{\prime}=\mathtt{G}^{\prime}$ by Lemma 4.4. Thus we have $\mathtt{T}=\mathtt{G}^{A,B}$.

If $\lambda_{1}>c^{\prime}$, then

[TABLE]

which implies that $\mathtt{T}_{\downarrow n-1}=\mathtt{G}^{A,B}_{\downarrow n-1}$ by induction on $\lambda_{1}-c^{\prime}$. Thus we have $\mathtt{T}=\mathtt{G}^{A,B}$.

Now suppose that $\lambda_{2}>c^{\prime}$. Then $n$ is located in the node $(2,\lambda_{2})$ in $\mathtt{G}^{A,B}$. Thus $\mathtt{T}^{-1}(n)$ = $(2,\lambda_{2})$ or $(3,c)$ since $\mathtt{T}\trianglerighteqslant\mathtt{G}^{A,B}$. Suppose that $\mathtt{T}^{-1}(n)=(3,c)$. Since $\mathrm{res}(\mathtt{T})=\mathrm{res}(\mathtt{G}^{A,B})$, we have $\mathrm{res}(3,c)=\mathrm{res}(2,\lambda_{2})$ and therefore $\kappa+\lambda_{2}-2>0$ and $\kappa+c-3<0$.

If $\lambda_{2}=c^{\prime}+1$, then $\mathrm{res}(3,c)=\mathrm{res}(2,\lambda_{2})=\mathrm{res}(1,c^{\prime})$ which is a contradiction.

If $\lambda_{2}>c^{\prime}+1$, then we have $\mathrm{res}(\mathtt{T}^{-1}(n-1))=\kappa+\lambda_{2}-3=\mathrm{res}(\mathtt{T}^{-1}(n))-1$. Then $\kappa+c-3<0$ implies that $\mathtt{T}^{-1}(n-1)$ cannot be in the third row, so that $\mathtt{T}^{-1}(n-1)$ = $(1,\lambda_{1})$ or $(2,\lambda_{2})$. But then $\mathrm{res}(\mathtt{T}^{-1}(n-1))\neq\kappa+\lambda_{2}-3$, another contradiction. Therefore we must have $\mathtt{T}^{-1}(n)=(2,\lambda_{2})$.

Thus we have

[TABLE]

which implies, by induction on $\lambda_{2}-c^{\prime}$, that $\mathtt{T}_{\downarrow n-1}=\mathtt{G}^{A,B}_{\downarrow n-1}$. We conclude that $\mathtt{T}=\mathtt{G}^{A,B}$. ∎

4.1. A lemma for block braid relations

Lemma 4.6**.**

Let $\nu=\nu^{1}*\nu^{2}*\nu^{3}$ where $a_{1},a_{3}\geqslant 1$ and

[TABLE]

for some $i\geqslant 0$. We set $a_{2}=2i+1$ and $\underline{a}=(1,a_{1}-1,a_{2},a_{3}-1,1)$. Then

[TABLE]

is given as follows.

1. (1)

Suppose $i\neq 0$. Then it is equal to

[TABLE] 2. (2)

Suppose $i=0$. Then it is equal to $\Psi_{1}\Psi_{4}\Psi_{2}\Psi_{3}\Psi_{2}(\underline{a})(x_{1}+x_{a_{1}+a_{3}+1})e(\nu)$.

Proof.

Following Remark 2.17, we compute

[TABLE]

where $X_{k}=\Psi_{2}(a_{3}+k-1,a_{1},1)\Psi[a_{1}+a_{2},a_{3}]-\Psi[a_{1}+a_{2},a_{3}]\Psi_{2}(k-1,a_{1},1)$.

Then, Lemma 2.16 tells that the term $\Psi_{2}(a_{3}+k,a_{1},a_{2}-k)X_{k}\Psi[a_{1},k-1]e(\nu)$ survives only if for some $1\leqslant s\leqslant a_{1}$ and $1\leqslant t\leqslant a_{3}$,

• •

the $s$th entry of $\nu^{1}=(i+1,i+2,\dots,i+a_{1})$,

• •

the $k$th entry of $\nu^{2}=(i,\dots,0\dots,i)$,

• •

the $t$th entry of $\nu^{3}=(i+a_{3},\dots,i+1)$

form a triple of the form $(1,0,1)$, $(j,j+1,j)$ for $j\geqslant 0$, or $(j,j-1,j)$ for $j\geqslant 2$. Hence, either $(s,k,t)=(1,1,a_{3})$ or $(1,a_{2},a_{3})$ is possible. Thus, if $i\neq 0$ we insert

[TABLE]

and $X_{k}=0$ for $k\neq 1,a_{2}$, to obtain

[TABLE]

On the other hand, if $i=0$ we obtain

[TABLE]

(1) Suppose that $i\neq 0$. We write

[TABLE]

and let $\nu=(i+1)*\nu^{a}*(i)*\nu^{b}*(i)*\nu^{c}*(i+1)$. Then the first term is

[TABLE]

where $\underline{a}=(1,a_{1}-1,1,a_{2}-2,1,a_{3}-1,1)$. Then, following the recipe in Remark 2.17, we know that there is no error term in $\Psi_{5}\Psi_{4}\Psi_{5}(\underline{b})e(\mu)-\Psi_{4}\Psi_{5}\Psi_{4}(\underline{b})e(\mu)$, so that

[TABLE]

where $\mu=(i+1)*(i)*\nu^{c}*\nu^{a}*\nu^{b}*(i+1)*(i)$ and $\underline{b}=(1,1,a_{3}-1,a_{1}-1,a_{2}-2,1,1)$. Then, Lemma 2.12(1) implies

[TABLE]

where $\mu^{\prime}=(i+1)*(i)*\nu^{c}*\nu^{b}*(i+1)*(i)*\nu^{a}$ and $\underline{b}^{\prime}=(1,1,a_{3}-1,a_{2}-2,1,1,a_{1}-1)$. We continue with similar arguments:

[TABLE]

where $\mu^{\prime\prime}=(i+1)*(i)*\nu^{c}*\nu^{b}*\nu^{a}*(i+1)*(i)$ and $\underline{b}^{\prime\prime}=(1,1,a_{3}-1,a_{2}-2,a_{1}-1,1,1)$,

[TABLE]

where $\mu^{\prime\prime\prime}=(i+1)*(i)*\nu^{a}*\nu^{b}*\nu^{c}*(i)*(i+1)$ and $\underline{b}^{\prime\prime\prime}=(1,1,a_{1}-1,a_{2}-2,a_{3}-1,1,1)$, and after one more step, we obtain

[TABLE]

Then, we can check that this is equal to $\Psi_{1}\Psi_{4}\Psi_{2}\Psi_{3}\Psi_{2}(\underline{a})(x_{a_{1}+a_{2}}+x_{a_{1}+a_{2}+a_{3}})e(\nu)$ if we change $\underline{a}$ to $\underline{a}=(1,a_{1}-1,a_{2},a_{3}-1,1)$.

Next, we consider the second term

[TABLE]

where $\underline{a}=(1,a_{1}-1,1,a_{2}-2,1,a_{3}-1,1)$. Then

[TABLE]

where $\mu=(i)*\nu^{b}*(i+1)*\nu^{c}*(i)*(i+1)*\nu^{a}$ and $\underline{b}=(1,a_{2}-2,1,a_{3}-1,1,1,a_{1}-1)$,

[TABLE]

where $\mu^{\prime}=(i)*(i+1)*\nu^{b}*\nu^{a}*(i)*\nu^{c}*(i+1)$ and $\underline{b}^{\prime}=(1,1,a_{2}-2,a_{1}-1,1,a_{3}-1,1)$,

[TABLE]

where $\mu^{\prime\prime}=(i)*(i+1)*\nu^{c}*\nu^{b}*(i)*(i+1)*\nu^{a}$ and $\underline{b}^{\prime\prime}=(1,1,a_{3}-1,a_{2}-2,1,1,a_{1}-1)$,

[TABLE]

where $\mu^{\prime\prime\prime}=(i+1)*(i)*\nu^{b}*\nu^{a}*(i)*\nu^{c}*(i+1)$ and $\underline{b}^{\prime\prime\prime}=(1,1,a_{2}-2,a_{1}-1,1,a_{3}-1,1)$. Then this is equal to $\Psi_{1}\Psi_{4}\Psi_{2}\Psi_{3}\Psi_{2}(\underline{a})(x_{1}+x_{a_{1}+1})e(\nu)$ if we change $\underline{a}$ to $\underline{a}=(1,a_{1}-1,a_{2},a_{3}-1,1)$.

(2) Suppose that $i=0$. We write $\nu^{1}=(1)*\nu^{a},\;\nu^{2}=(0),\;\nu^{3}=\nu^{c}*(1)$ as before, and let $\nu=(1)*\nu^{a}*(0)*\nu^{c}*(1)$ and $\underline{a}=(1,a_{1}-1,a_{3}-1,1)$. Then

[TABLE]

which is the desired result. ∎

4.2. A special three row case

To handle the case that the Garnir belt of $\mathtt{G}^{A,B}$ has three rows, we may assume that $\lambda=(\lambda_{1},\lambda_{2},\lambda_{3})$ with $\lambda_{3}>0$ and that Garnir nodes are $A=(2,c)$, $B=(1,c^{\prime})$ with $c\leqslant c^{\prime}$. In this subsection, we consider a special case, that is, we assume

• (i)

The first row of the Garnir belt has residues $\nu^{1}_{r}=(i+1,i+2,\dots,\lambda_{1}-c^{\prime}+i+1)$ from left to right.

• (ii)

The second row of the Garnir belt has residues $\nu^{2}_{m}=(i,i-1,\dots,1,0,1,\dots,i-1,i)$ from left to right.

• (iii)

The third row of the Garnir belt has residues $\nu^{3}_{l}=(c+i,c+i-1,\dots,i+1)$ from left to right.

In particular, $\mathrm{res}(A)=i$ and $\mathrm{res}(B)=i+1$. We denote

• (i)

the residues of the first row of $\lambda$ by $\nu^{1}_{l}*\nu^{1}_{r}$,

• (ii)

the residues of the second row of $\lambda$ by $\nu^{2}_{l}*\nu^{2}_{m}*\nu^{2}_{r}$,

• (iii)

the residues of the third row of $\lambda$ by $\nu^{3}_{l}*\nu^{3}_{r}$,

respectively. Pictorially, if $c\neq 1$ and $c^{\prime}\neq\lambda_{2}$ then

[TABLE]

Recall from Lemma 1.18 that there are fully commutative elements $w^{A}$ and $w^{B}$ such that

[TABLE]

We will show that $\psi_{w^{A}}g_{A}^{\lambda}\equiv\psi_{w^{B}}g_{B}^{\lambda}\pmod{\mathcal{G}^{\lambda}_{<\ell(\mathtt{G}^{A,B})}}$ in Lemma 4.11. We build up to this with several smaller lemmas, as the calculation is quite lengthy.

We denote the length of $\nu^{1}_{l},\nu^{1}_{r},\nu^{2}_{l},\nu^{2}_{m},\nu^{2}_{r},\nu^{3}_{l},\nu^{3}_{r}$ by $a^{1}_{l},a^{1}_{r},a^{2}_{l},a^{2}_{m},a^{2}_{r},a^{3}_{l},a^{3}_{r}$, respectively, and define

[TABLE]

Lemma 4.7**.**

We have

[TABLE]

where

[TABLE]

Proof.

Note that $a^{1}_{r}\geqslant 1$, $a^{2}_{m}\geqslant 1$, $a^{3}_{l}\geqslant 1$ by definition. Then, by considering the two-line notation for $w^{\mathtt{G}^{A}}$ and $w^{\mathtt{G}^{B}}$ as in Lemma 2.8, we know that

[TABLE]

Similarly, we have $w^{A}=S_{4}S_{3}S_{2}(\underline{a}^{\prime})$ and $w^{B}=S_{3}S_{4}S_{5}(\underline{a}^{\prime\prime})$. Thus,

[TABLE]

where $\mu=\nu^{1}_{l}*\nu^{2}_{l}*\nu^{1}_{r}*\nu^{2}_{m}*\nu^{3}_{l}*\nu^{2}_{r}*\nu^{3}_{r}$. We apply Lemma 4.6 to compute

[TABLE]

Then the result is as follows.

• (i)

If $i\neq 0$ then

[TABLE]

where $X_{1}=x_{a^{1}_{l}+a^{2}_{l}+1}+x_{a^{1}_{l}+a^{1}_{r}+a^{2}_{l}+a^{2}_{m}+a^{3}_{l}}$ and $X_{2}=x_{a^{1}_{l}+a^{1}_{r}+a^{2}_{l}+1}+x_{a^{1}_{l}+a^{1}_{r}+a^{2}_{l}+a^{2}_{m}}$.

• (ii)

If $i=0$ then

[TABLE]

where $X_{1}=x_{a^{1}_{l}+a^{2}_{l}+1}+x_{a^{1}_{l}+a^{1}_{r}+a^{2}_{l}+a^{3}_{l}+1}$.

As $\Psi_{7}\Psi_{6}\Psi_{2}\Psi_{3}(\underline{b})$ does not touch the fifth block of $\underline{b}$, if $i\neq 0$ then

[TABLE]

Lemma 4.8**.**

Let

[TABLE]

1. (1)

If $a^{2}_{l}=0$, then

[TABLE]

Otherwise,

[TABLE] 2. (2)

If $a^{2}_{r}=0$, then

[TABLE]

Otherwise,

[TABLE]

Proof.

First, note that $\Psi_{7}\Psi_{6}\Psi_{2}\Psi_{3}(\underline{b})$ is the product of

[TABLE]

in this order, where $\nu^{1}_{r}=(i+1)*\dot{\nu}^{1}_{r}$, $\nu^{3}_{l}=\dot{\nu}^{3}_{l}*(i+1)$.

1. (1)

Moving $x_{a^{1}_{l}+a^{2}_{l}+1}$ to the right,

[TABLE]

We apply Lemma 2.12(3) to $(x_{a^{1}_{l}+a^{2}_{l}+1}\Psi_{2}-\Psi_{2}x_{a^{1}_{l}+1})e(\mu)$, where

[TABLE]

If $a^{2}_{l}=0$ then $\Psi_{2}$ and $\Psi_{3}$ are the identity and

[TABLE]

Now suppose that $a^{2}_{l}\geqslant 1$. Note that $a^{3}_{l}=a^{2}_{l}+1\geqslant 2$. We write

[TABLE]

Then, after applying Lemma 2.12(3), we obtain

[TABLE]

Then we can continue as follows.

[TABLE]

where $\nu^{\prime}=\nu^{1}_{l}*\dot{\nu}^{2}_{l}*(i+1)*(i+1)*\dot{\nu}^{1}_{r}*\nu^{2}_{m}*\dot{\nu}^{3}_{l}*\nu^{2}_{r}*(i+1)*\nu^{3}_{r}$ and $\underline{c}^{\prime}=(a^{1}_{l},a^{2}_{l}-1,1,1,a^{1}_{r}-1,a^{2}_{m},a^{3}_{l}-1,a^{2}_{r},1,a^{3}_{r}),$ so that $\Psi_{5}\Psi_{6}\Psi_{5}(\underline{c}^{\prime})e(\nu^{\prime})=\Psi_{6}\Psi_{5}\Psi_{6}(\underline{c}^{\prime})e(\nu^{\prime})$ and we have

[TABLE]

Hence, $\Psi_{4}\Psi_{7}\Psi_{8}\Psi_{5}\Psi_{6}\Psi_{5}\Psi_{2}\Psi_{4}\Psi_{3}\Psi_{7}(\underline{c})m^{\lambda}$ is equal to

[TABLE] 2. (2)

Similarly, if we move $x_{a^{1}_{l}+a^{1}_{r}+a^{2}_{l}+a^{2}_{m}+a^{3}_{l}}$ to the right,

[TABLE]

and we apply Lemma 2.12(4) to $(x_{a^{1}_{l}+a^{1}_{r}+a^{2}_{l}+a^{2}_{m}+a^{3}_{l}}\Psi_{7}-\Psi_{7}x_{a^{1}_{l}+a^{1}_{r}+a^{2}_{l}+a^{2}_{m}+a^{2}_{r}+a^{3}_{l}})e(\mu)$, where

[TABLE]

If $a^{2}_{r}=0$ then $\Psi_{6}$ and $\Psi_{7}$ are the identity and

[TABLE]

Now suppose that $a^{2}_{r}\geqslant 1$. We write $\nu=\nu^{1}_{l}*(i+1)*\dot{\nu}^{1}_{r}*\nu^{2}_{l}*\nu^{2}_{m}*(i+1)*\dot{\nu}^{2}_{r}*\dot{\nu}^{3}_{l}*(i+1)*\nu^{3}_{r}$. Then, after applying Lemma 2.12(4), we obtain

[TABLE]

Hence, $-(\Psi_{3}\Psi_{2}\Psi_{8}\Psi_{4})(\Psi_{6}\Psi_{5}\Psi_{6})\Psi_{7}\Psi_{4}\Psi_{3}(\underline{d})m^{\lambda}$ is equal to

[TABLE]

Lemma 4.9**.**

Let $\underline{c}$ and $\underline{d}$ be as in Lemma 4.8. Then

[TABLE]

are reduced.

Proof.

By Corollary 2.9, it suffices to check that they have length 10 if $\underline{c}=\underline{d}=(1,\dots,1)$, which is easy to check. ∎

Lemma 4.10**.**

Let $A^{\prime}=(2,c-1)$, $B^{\prime}=(1,c^{\prime}+1)$. Then we have the following.

1. (1)

Let $a^{2}_{l}\geqslant 1$ and $a^{2}_{r}=0$. Then

[TABLE]

for some reduced expression of $w\in\mathfrak{S}_{n}$ with $\ell(w\mathtt{G}^{A^{\prime}})=\ell(w)+\ell(\mathtt{G}^{A^{\prime}})<\ell(\mathtt{G}^{A,B})$. 2. (2)

Let $a^{2}_{r}\geqslant 1$ and $a^{2}_{l}=0$. Then

[TABLE]

for some reduced expression of $w\in\mathfrak{S}_{n}$ with $\ell(w\mathtt{G}^{B^{\prime}})=\ell(w)+\ell(\mathtt{G}^{B^{\prime}})<\ell(\mathtt{G}^{A,B})$.

Proof.

The result follows readily if we can show that the second term in each of Lemma 4.8(1) and (2) are zero under the corresponding conditions, since $g^{\lambda}_{A^{\prime}}=\Psi_{5}\Psi_{6}\Psi_{7}(\underline{c})m^{\lambda}$ and $g^{\lambda}_{B^{\prime}}=\Psi_{5}\Psi_{4}\Psi_{3}(\underline{d})m^{\lambda}$. (We note that under the corresponding hypotheses, $\Psi_{7}(\underline{c})$ in (1) and $\Psi_{3}(\underline{d})$ in (2) are the identity.)

1. (1)

For the second term from Lemma 4.8(1), it suffices to consider

[TABLE]

where $\nu^{\prime}=\nu^{1}_{l}*(i+1)*\dot{\nu}^{2}_{l}*\dot{\nu}^{1}_{r}*(i+1)*\dot{\nu}^{3}_{l}*\nu^{2}_{m}*\nu^{2}_{r}*(i+1)*\nu^{3}_{r}$ and

[TABLE]

For this, we need to compute

[TABLE]

When $a^{1}_{r}=1$, this error term is zero, and we are done. So we may assume that $a^{1}_{r}\geqslant 2$ and continue as follows. The above expression is equal to

[TABLE]

by Lemma 2.16, but we need a different formula here: first, we apply Lemma 2.16 to obtain

[TABLE]

Then, we apply the anti-involution of $R(\beta)$ to obtain

[TABLE]

We use this formula to compute the second term. To state the result in this case, we change $\underline{c}$ to

[TABLE]

The second term is then

[TABLE]

We continue as follows.

[TABLE]

where the third equality above follows from Lemma 2.12(1) and

[TABLE]

Thus, if we write

[TABLE]

by abuse of notation, we have the result. 2. (2)

For the second term from Lemma 4.8(2), it suffices to consider

[TABLE]

where $\nu^{\prime}=\nu^{1}_{l}*(i+1)*\nu^{2}_{l}*\nu^{2}_{m}*\dot{\nu}^{1}_{r}*(i+1)*\dot{\nu}^{3}_{l}*\dot{\nu}^{2}_{r}*(i+1)*\nu^{3}_{r}$ and

[TABLE]

For this, we need to compute

[TABLE]

again. But, our assumption that $a^{2}_{l}=0$ is equivalent to $a^{3}_{l}=1$, which implies that this error term is zero. Thus if we write

[TABLE]

by abuse of notation, the result follows. ∎

Lemma 4.11**.**

We have

[TABLE]

Proof.

Lemma 4.8 implies that

[TABLE]

for some reduced expressions of $w,w^{\prime}\in\mathfrak{S}_{n}$ such that $\ell(w)+\ell(\mathtt{G}^{A^{\prime}})<\ell(\mathtt{G}^{A,B})$ and $\ell(w^{\prime})+\ell(\mathtt{G}^{B^{\prime}})<\ell(\mathtt{G}^{A,B})$. Here, $\gamma$ and $\delta$ are the second terms appearing in Lemma 4.8(1) and (2) respectively, and are both zero by Lemma 4.10 unless $a^{2}_{l}\geqslant 1$ and $a^{2}_{r}\geqslant 1$. Lemma 4.9 implies that $\psi_{w}g_{A^{\prime}}^{\lambda}$ and $\psi_{w^{\prime}}g_{B^{\prime}}^{\lambda}$ belong to $\mathcal{G}^{\lambda}_{<\ell(\mathtt{G}^{A,B})}$; we may also have that $\psi_{w}g_{A^{\prime}}^{\lambda}$ or $\psi_{w^{\prime}}g_{B^{\prime}}^{\lambda}$ are zero, in the degenerate cases that $a^{2}_{l}=0$ or $a^{2}_{r}=0$, respectively, by Lemma 4.8.

So it remains to show that $\gamma+\delta\equiv 0\pmod{\mathcal{G}^{\lambda}_{<\ell(\mathtt{G}^{A,B})}}$ when $a^{2}_{l}\geqslant 1$ and $a^{2}_{r}\geqslant 1$. In fact, we will show that $\gamma+\delta=0$. We continue by further calculation with $\gamma$ and $\delta$.

As in Lemma 4.8(1), we compute $\gamma$ by using

[TABLE]

To state the result, we change $\underline{c}$ to

[TABLE]

and write $\nu=\nu^{1}_{l}*(i+1)*(i+2)*\ddot{\nu}^{1}_{r}*\dot{\nu}^{2}_{l}*(i+1)*\nu^{2}_{m}*(i+1)*\dot{\nu}^{2}_{r}*\ddot{\nu}^{3}_{l}*(i+2)*(i+1)*\nu^{3}_{r}$. Then

[TABLE]

where

[TABLE]

We use

[TABLE]

to continue as follows.

[TABLE]

where, in the final equality, we have used that $\Psi_{11}(\underline{c})m^{\lambda}=\psi_{a^{1}_{l}+a^{1}_{r}+a^{2}_{l}+a^{2}_{m}+a^{2}_{r}+a^{3}_{l}-1}m^{\lambda}=0$ since $a^{3}_{l}\geqslant 2$, and $\underline{e}=(a^{1}_{l},a^{2}_{l}-1,1,a^{3}_{l}-2,1,1,a^{2}_{m},a^{1}_{r}-2,1,a^{2}_{r}-1,1,1,a^{3}_{r})$,

[TABLE]

Similarly, we compute $\delta$ as in the proof of Lemma 4.8(2) by using the same equality as above. We replace $\underline{d}$ with $\underline{c}$:

[TABLE]

and write $\nu=\nu^{1}_{l}*(i+1)*(i+2)*\ddot{\nu}^{1}_{r}*\dot{\nu}^{2}_{l}*(i+1)*\nu^{2}_{m}*(i+1)*\dot{\nu}^{2}_{r}*\ddot{\nu}^{3}_{l}*(i+2)*(i+1)*\nu^{3}_{r}$. Then

[TABLE]

where we have used Lemma 2.12(2) for the fourth equality, $\Psi_{2}(\underline{c})m^{\lambda}=\psi_{a^{1}_{l}+1}m^{\lambda}=0$ (since $a^{1}_{r}\geqslant a^{2}_{r}+1\geqslant 2$) in the final equality, and

[TABLE]

It’s easy to see, by applying three further braid relations which don’t yield error terms, that (†) and (‡) are negations of each other, so that $\gamma+\delta=0$, and the proof is complete. ∎

4.3. Proof of Theorem 3.19

Lemma 4.12**.**

Let $A$ and $B$ be Garnir nodes of $[\lambda]$. Let $w^{A}$ and $w^{B}$ be the fully commutative elements given in Lemma 1.18. Then we have

[TABLE]

Proof.

We may assume that $A=(r,c)$ is to the left of $B=(r^{\prime},c^{\prime})$ in $[\lambda]$.

Suppose $r=r^{\prime}+1$ and $\mathrm{res}(B)=\mathrm{res}(r+1,c)$. Without loss of generality, we may assume that $[\lambda]$ has 3 rows and $r^{\prime}=1$. Then the assertion holds by Lemma 4.11.

Otherwise, it follows from Proposition 2.2 and Corollary 3.2 that

[TABLE]

where $w$ runs over all elements such that (i) $w\prec w^{\mathtt{G}^{A,B}}$ and (ii) $e(\mathrm{res}(\mathtt{G}^{A,B}))\psi_{w}m^{\lambda}=\psi_{w}m^{\lambda}$. By Lemma 4.5, we conclude that

[TABLE]

Lemma 4.13**.**

Let $\mathtt{T}\in\mathsf{Row}(\lambda)$. Suppose that $\mathtt{T}=w\mathtt{G}^{A}$ for $w\in\mathfrak{S}_{n}$ and a Garnir node $A$ of $[\lambda]$ with $\ell(\mathtt{T})=\ell(w)+\ell(\mathtt{G}^{A})$.

1. (1)

If $\mathtt{T}=u\mathtt{G}^{B}$ for $u\in\mathfrak{S}_{n}$ and a Garnir node $B$ of $[\lambda]$ with $\ell(\mathtt{T})=\ell(u)+\ell(\mathtt{G}^{B})$, then

[TABLE] 2. (2)

We have

[TABLE] 3. (3)

For $\sigma\in\mathfrak{S}_{n}$ with $\ell(\sigma)+\ell(\mathtt{G}^{A})<\ell(\mathtt{T})$,

[TABLE]

Proof.

First, we prove (1), (2), and (3) for Garnir tableaux. Let $\mathtt{T}=\mathtt{G}^{A}$. If $\mathtt{G}^{A}=u\mathtt{G}^{B}$ for some $u\in\mathfrak{S}_{n}$ and some Garnir node $B$ of $[\lambda]$ with $\ell(\mathtt{T})=\ell(u)+\ell(\mathtt{G}^{B})$, then it implies $\mathtt{G}^{A}\geqslant_{L}\mathtt{G}^{B}$, so that $\mathtt{G}^{A,B}=\mathtt{G}^{A}$ follows. Thus $B=A$ and $u=\mathrm{id}$ by Lemma 1.17, proving (1). Assertion (3) also holds obviously since there is no $\sigma\in\mathfrak{S}_{n}$ such that $\ell(\sigma)+\ell(\mathtt{G}^{A})<\ell(\mathtt{G}^{A})$. Assertion (2) follows from Lemma 3.7.

Now we prove (1), (2), and (3) by induction on $l:=\ell(\mathtt{T})$. If $l=\min\{\ell(\mathtt{T})\mid\mathtt{T}\in\mathsf{Row}(\lambda)\}$, then $\mathtt{T}$ is a Garnir tableau and there is nothing to prove. We assume that (1), (2), and (3) hold for all $\mathtt{T}^{\prime}=w^{\prime}\mathtt{G}^{A^{\prime}}\in\mathsf{Row}(\lambda)$ with

[TABLE]

(1) We consider $\mathtt{T}=w\mathtt{G}^{A}=u\mathtt{G}^{B}$ for some $w,u\in\mathfrak{S}_{n}$ and Garnir nodes $A,B$ of $[\lambda]$ with $\ell(\mathtt{T})=l=\ell(w)+\ell(\mathtt{G}^{A})=\ell(u)+\ell(\mathtt{G}^{B})$. By Lemma 1.14, there is $v\in\mathfrak{S}_{n}$ such that

[TABLE]

which tells us that

[TABLE]

where $w^{A}$ and $w^{B}$ are given in Lemma 1.18. Then, by Proposition 2.2 and the induction hypothesis on (3), we know that if $\psi_{\sigma}$ appears as an error term in $(\psi_{w}-\psi_{v}\psi_{w^{A}})e(\mathrm{res}(\mathtt{G}^{A}))$ then $\psi_{\sigma}g^{\lambda}_{A}\in\mathfrak{\mathtt{G}}^{\lambda}_{<\ell(\mathtt{T})}$. Similarly, if $\psi_{\sigma}$ appears as an error term in $(\psi_{u}-\psi_{v}\psi_{w^{B}})e(\mathrm{res}(\mathtt{G}^{B}))$ then $\psi_{\sigma}g^{\lambda}_{B}\in\mathfrak{\mathtt{G}}^{\lambda}_{<\ell(\mathtt{T})}$.

By Lemma 4.12 and the induction hypothesis, we have

[TABLE]

Thus, (1) holds for $\ell(\mathtt{T})=l$.

(2) For $i=1,\ldots,n$, it follows from

[TABLE]

that

[TABLE]

by the induction hypothesis. It remains to prove that $\psi_{j}\psi_{w}g_{A}^{\lambda}\equiv 0$ unless $s_{j}\mathtt{T}\in\mathsf{Row}(\lambda)$ and $s_{j}\mathtt{T}\vartriangleleft\mathtt{T}$. There are two cases:

• (i)

$s_{j}\mathtt{T}\notin\mathsf{Row}(\lambda)$ (i.e. $s_{j}\mathtt{T}\notin\mathsf{RowStd}(\lambda)$ or $s_{j}\mathtt{T}\in\mathsf{Std}(\lambda)$),

• (ii)

$s_{j}\mathtt{T}\in\mathsf{Row}(\lambda)$ and $s_{j}\mathtt{T}\vartriangleright\mathtt{T}$.

1. (i)

If $s_{j}\mathtt{T}\notin\mathsf{RowStd}(\lambda)$, then there is a node $(r,c)\in[\lambda]$ such that

[TABLE]

By Lemma 1.13 (2), we can take $B\in[\lambda]$ and $u\in\mathfrak{S}_{n}$ such that

1. (a)

$\mathtt{G}^{B}(r,c+1)=\mathtt{G}^{B}(r,c)+1$, 2. (b)

$\mathtt{T}=u\mathtt{G}^{B}$, 3. (c)

$s_{j}u=us_{p}$ where $p=\mathtt{G}^{B}(r,c)$.

Thus, we have

[TABLE]

where $\nu=e(\mathrm{res}(\mathtt{G}^{B}))$. Since (1) holds for the length $l$, Proposition 2.2 implies that $\psi_{w}g_{A}^{\lambda}-\psi_{u}g_{B}^{\lambda}\in\mathcal{G}^{\lambda}_{\leqslant l-3}$, so that $\psi_{j}\psi_{w}g_{A}^{\lambda}\equiv\psi_{j}\psi_{u}g_{B}^{\lambda}\pmod{\mathcal{G}^{\lambda}_{<\ell(\mathtt{T})}}$ by the induction hypothesis on (3). Similarly, Propositions 2.2 and 3.7 imply that $\psi_{j}\psi_{u}g_{B}^{\lambda}\equiv\psi_{u}\psi_{p}g_{B}^{\lambda}\pmod{\mathcal{G}^{\lambda}_{<\ell(\mathtt{T})}}$, and $\psi_{p}g_{B}^{\lambda}=0$, so that

[TABLE]

Suppose that $s_{j}\mathtt{T}\in\mathsf{Std}(\lambda)$. Then there is a node $C=(r,c)\in[\lambda]$ such that

[TABLE]

By Lemma 1.13(1), there is a permutation $v\in\mathfrak{S}_{n}$ such that

1. (a)

$\mathtt{T}=v\mathtt{G}^{C}$, 2. (b)

$s_{j}v=vs_{q}$ where $q=\mathtt{G}^{C}(r+1,c)$.

In a similar manner as above, we have

[TABLE] 2. (ii)

We assume that $\mathtt{S}:=s_{j}\mathtt{T}\in\mathsf{Row}(\lambda)$ and $\mathtt{S}\vartriangleright\mathtt{T}$. Note that $\ell(\mathtt{T})=\ell(\mathtt{S})+1$. Then

[TABLE]

for some Garnir node $B\in[\lambda]$ and $u\in\mathfrak{S}_{n}$. As (1) holds for the length $l$, we have

[TABLE]

where $\nu=(\nu_{1},\dots,\nu_{n})\in I^{n}$ is the residue sequence of $u\mathtt{G}^{B}$.

By (i) and (ii) the assertion (2) holds for $\ell(\mathtt{T})=l$.

(3) Suppose that $\mathtt{T}=w\mathtt{G}^{A}$ for $w\in\mathfrak{S}_{n}$ and a Garnir node $A$ of $[\lambda]$ with $\ell(\mathtt{T})=\ell(w)+\ell(\mathtt{G}^{A})$. Let $\sigma=s_{i_{1}}\dots s_{i_{r}}$ be a reduced expression for $\sigma$, so that $\psi_{\sigma}=\psi_{i_{1}}\dots\psi_{i_{r}}$ and $\ell(\sigma)+\ell(\mathtt{G}^{A})<l$. Note that we do not assume that $\ell(\sigma\mathtt{G}^{A})=\ell(\sigma)+\ell(\mathtt{G}^{A})$.

If $\ell(\sigma\mathtt{G}^{A})=\ell(\sigma)+\ell(\mathtt{G}^{A})$, then $\sigma\mathtt{G}^{A}\in\mathsf{Row}(\lambda)$ by Lemma 1.12. Thus, we are done.

If $\ell(\sigma\mathtt{G}^{A})<\ell(\sigma)+\ell(\mathtt{G}^{A})$, then there is some $k$ such that $\ell(s_{i_{k}}\dots s_{i_{r}}\mathtt{G}^{A})=\ell(s_{i_{k}}\dots s_{i_{r}})+\ell(\mathtt{G}^{A})$ and $\ell(s_{i_{k-1}}\dots s_{i_{r}}\mathtt{G}^{A})<\ell(s_{i_{k}}\dots s_{i_{r}}\mathtt{G}^{A})$. Once again, by the induction hypothesis on (2), we have that $\psi_{i_{k-1}}\dots\psi_{i_{r}}g_{A}^{\lambda}\in\mathcal{G}^{\lambda}_{<\ell(s_{i_{k}}\dots s_{i_{r}}\mathtt{G}^{A})}$ and may conclude that $\psi_{i_{1}}\dots\psi_{i_{r}}g_{A}^{\lambda}\in\mathcal{G}^{\lambda}_{<l}$ by induction.

Thus, assertion (3) holds for $\ell(\mathtt{T})=l$, which completes the proof. ∎

Corollary 4.14**.**

For each $\mathtt{T}\in\mathsf{Row}(\lambda)$, we fix a Garnir node $A$ and $w\in\mathfrak{S}_{n}$ such that $\mathtt{T}=w\mathtt{G}^{A}$, and set $g_{\mathtt{T}}^{\lambda}=\psi_{w}g_{A}^{\lambda}+\mathcal{G}^{\lambda}_{<\ell(\mathtt{T})}$. Then

1. (1)

The element $g_{\mathtt{T}}^{\lambda}\in\mathcal{G}^{\lambda}/\mathcal{G}^{\lambda}_{<\ell(\mathtt{T})}$ does not depend on the choice of $A$ or the choice of reduced expression for $w$. 2. (2)

The $\mathcal{O}$-submodule $\sum_{t>0}\mathcal{G}^{\lambda}_{<t}$ is an $R(\beta)$-submodule of $\mathcal{M}^{\lambda}$.

Proof.

Parts (1) and (2) follow from Lemma 4.13 (1) and Lemma 4.13 (2), respectively. ∎

Recall that $\mathcal{G}^{\lambda}={\rm im}\ \mathcal{H}^{\lambda}$ and $\mathcal{S}^{\lambda}=q^{\deg(\mathtt{T}^{\lambda})}{\rm coker}\ \mathcal{H}^{\lambda}$ for $\lambda\vdash n$.

Theorem 4.15**.**

1. (1)

The $\mathcal{O}$-submodules $\{\mathcal{G}^{\lambda}_{<t}\}_{t\in\mathbb{Z}_{>0}}$ give a filtration of $\mathcal{G}^{\lambda}$. 2. (2)

For $t\in\mathbb{Z}_{>0}$, $\mathcal{G}^{\lambda}_{<t+1}/\mathcal{G}^{\lambda}_{<t}$ is a free $\mathcal{O}$-module with basis

[TABLE]

Proof.

Corollary 4.14(2) implies that $\mathcal{G}^{\lambda}=\sum_{t>0}\mathcal{G}^{\lambda}_{<t}$, which is (1). Then (2) is clear. ∎

We are now ready to prove Theorem 3.19.

Proof of Theorem 3.19.

Let us consider

[TABLE]

for some $a_{\mathtt{T}^{1}},\dots,a_{\mathtt{T}^{t}}\in\mathcal{O}$ and $\mathtt{T}_{1},\dots,\mathtt{T}_{t}\in\mathsf{Std}(\lambda)$. By $\eqref{basis of M}$ and Theorem 4.15, we have

[TABLE]

which implies that $\{\psi_{w^{\mathtt{T}}}\overline{m}^{\lambda}\mid\mathtt{T}\in\mathsf{Std}(\lambda)\}$ is linearly independent in $\mathcal{S}^{\lambda}$. Thus the assertion follows from Theorem 3.12(1). ∎

5. A conjecture in type $C^{(1)}_{\ell}$

We end with a conjecture giving the elements $g^{\lambda}_{A}$ explicitly in type $C^{(1)}_{\ell}$, as well as a basis of $\mathcal{S}^{\lambda}$ in this type. In Remark 3.6, we noted the similarity between the Garnir elements in type $A_{\infty}$ and $C_{\infty}$. Similarly, we expect that the affine type $C$ case resembles that of affine type $A$. Our constructions in this section closely follow [21, Section 5]. We fix $\lambda\in\mathscr{P}^{l}_{n}$ and $\kappa\in\mathbb{Z}^{l}$ throughout, as well as a Garnir node $A=(r,c,t)\in[\lambda]$. Recall the definition of the Garnir tableau $\mathtt{G}^{A}$ from earlier, as well as the residue pattern in type $C^{(1)}_{\ell}$ – in particular the natural projection $p:\mathbb{Z}\rightarrow\mathbb{Z}/2\ell\mathbb{Z}$.

Definition 5.1**.**

A brick is a set of $2\ell$ adjacent nodes in the same row of the Garnir belt $\mathbf{B}^{A}$, $\{(a,b,t),(a,b+1,t),\dots,(a,b+2\ell-1,t)\}$, such that $p(\kappa_{t}+b-a)=p(\kappa_{t}+c-r)$.

We denote by $k$ the number of bricks contained in $\mathbf{B}^{A}$, and label the bricks $B_{1},B_{2},\dots,B_{k}$ from left-to-right along row $r+1$ and then from left-to-right along row $r$. We now introduce permutations which transpose adjacent bricks. Let $d$ be the smallest entry of $B_{1}$ in $\mathtt{G}^{A}$. Then for $1\leqslant r<k$, the permutation

[TABLE]

may be thought of as transposing $B_{r}$ and $B_{r+1}$. We have the corresponding elements $\sigma_{r}=\sigma_{r}^{A}:=(-1)^{\ell}\psi_{w_{r}}\in R(\beta)$. We further define $\tau_{r}=\tau_{r}^{A}:=\sigma_{r}+1$. We should emphasise here that we have a $(-1)^{\ell}$ in our definition of $\sigma_{r}$, which differs from the definition of row Specht modules in [21, Section 5.4]. However a similar minus sign occurs in their definition of corresponding elements in column Specht modules [21, Section 7.1]. We suspect that this minus sign is merely an artefact of our choice of the polynomials $Q_{i,j}(u,v)$.

Note that any permutation $u$ of bricks may be written as a reduced expression $u=w_{r_{1}}\dots w_{r_{i}}$, and if $u$ is fully commutative we have a well-defined element $\tau_{u}:=\tau_{r_{1}}\dots\tau_{r_{i}}$.

We define $\mathtt{T}^{A}$ to be the tableau obtained from $\mathtt{G}^{A}$ by rearranging the entries in the bricks $B_{1},\dots,B_{k}$ so that they are in order along row $r$ and then row $r+1$. This is the most dominant row-strict tableau which may be obtained from $\mathtt{G}^{A}$ by acting by our brick permutations $w_{r}$.

Example 5.2**.**

Let $\ell=2,\lambda=(15,7,3)\in\mathscr{P}^{1}_{25}$, and $A=(1,5)$. We depict the Garnir tableau $\mathtt{G}^{A}$ below with Garnir belt shaded and bricks $B_{1},B_{2},B_{3}$ labelled, as well as the tableau $\mathtt{T}^{A}$.

[TABLE]

Note that $\mathtt{G}^{A}=w_{1}w_{2}\mathtt{T}^{A}$.

Conjecture 5.3**.**

Let $\lambda\in\mathscr{P}^{l}_{n}$, and let $A\in[\lambda]$ be a Garnir node. Suppose $\mathbf{B}^{A}$ contains $a$ bricks in the first row, and $b$ in the second. Then in type $C^{(1)}_{\ell}$, the Garnir element $g^{\lambda}_{A}$ is

[TABLE]

where the sum is over all $u\in\mathfrak{S}_{a+b}/\mathfrak{S}_{a}\times\mathfrak{S}_{b}$. Furthermore, Theorem 3.19 and Corollary 3.21 hold in type $C^{(1)}_{\ell}$, giving a homogeneous basis of $\mathcal{S}^{\lambda}$ indexed by standard $\lambda$-tableaux.

Example 5.4**.**

Continuing from the previous example,

[TABLE]

Evidence for our conjecture is provided by many examples we have computed in GAP [11].

Finally, we note that the above form for $g^{\lambda}_{A}$ does not instantly yield a clean expression in terms of basis elements of $\mathcal{M}^{\lambda}$ – this can be seen in Example 5.4 where $w_{1}\mathtt{T}^{A}m^{\lambda}$ is not row-strict. Fayers [10] has addressed this problem in type $A$, and in fact if our conjecture holds, then his work automatically applies to our $g^{\lambda}_{A}$ too.