Seminormal forms and cyclotomic quiver Hecke algebras of type A

Abstract

This paper shows that the cyclotomic quiver Hecke algebras of type A, and the gradings on these algebras, are intimately related to the classical seminormal forms. We start by classifying all seminormal bases and then give an explicit “integral” closed formula for the Gram determinants of the Specht modules in terms of the combinatorics associated with the KLR grading. We then use seminormal forms to give a deformation of the KLR algebras of type A. This makes it possible to study the cyclotomic quiver Hecke algebras in terms of the semisimple representation theory and seminormal forms. As an application we construct a new distinguished graded cellular basis of the cyclotomic KLR algebras of type A.

1 Introduction

The quiver Hecke algebras are a remarkable family of algebras that were introduced independently by Khovanov and Lauda [21, 22] and Rouquier [31]. These algebras are attached to an arbitrary oriented quiver, they are $\mathbb {Z}$-graded and they categorify the negative part of the associated quantum group. Over a field, Brundan and Kleshchev showed that the cyclotomic quiver Hecke algebras of type A, which are certain quotients of the quiver Hecke algebras of type A, are isomorphic to the cyclotomic Hecke algebras of type A.

The quiver Hecke algebras have a homogeneous presentation by generators and relations. As a consequence they have well-defined integral forms. Unlike Hecke algebras, which are generically semisimple, the cyclotomic quiver Hecke algebras are typically not semisimple even over the rational field. As a result the cyclotomic quiver Hecke algebras are rarely isomorphic to the cyclotomic Hecke algebras over an arbitrary ring.

The first main result of this paper shows that the cyclotomic quiver Hecke algebras of type A admit a one-parameter deformation. Moreover, this deformation is isomorphic to the cyclotomic Hecke algebra defined over the corresponding ring. Before we can state this result we need some notation.

Fix integers $n\ge 0$ and $e>1$ and let $\varGamma _e$ be the oriented quiver with vertex set $I=\mathbb {Z}/e\mathbb {Z}$ and edges $i\rightarrow i+1$, for $i\in I$. Given $i\in I$ let $\hat{\imath }\ge 0$ be any integer such that $i=\hat{\imath }+e\mathbb {Z}$. For each dominant weight $\varLambda$ for the corresponding Kac–Moody algebra $\mathfrak g(\varGamma _e)$, there exists a cyclotomic quiver Hecke algebra $\mathcal {R}^\varLambda _{n}$ and a cyclotomic Hecke algebra $\mathcal {H}^\varLambda _{n}$. To each tuple $\mathbf {i}\in I^n$ we associate the set of standard tableaux ${\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$ with residue sequence $\mathbf {i}$. All of these terms are defined in Sect. 3.1.

Like the cyclotomic quiver Hecke algebra, our deformation of $\mathcal {R}^\varLambda _{n}$ is adapted to the choice of e through the choice of base ring ${\mathcal {O}}$, which must be an e-idempotent subring (Definition 4.1). This definition ensures that the cyclotomic Hecke algebras are semisimple over $\fancyscript{K}$, the field of fractions of ${\mathcal {O}}$, and that $\mathcal {H}^\varLambda _n({\mathcal {O}})\otimes _{\mathcal {O}}K$ is a cyclotomic quiver Hecke algebra whenever $K={\mathcal {O}}/\mathfrak m$, for $\mathfrak m$ a maximal ideal of ${\mathcal {O}}$. For $t\in {\mathcal {O}}$ and $d\in \mathbb {Z}$ let $[d]=[d]_t$ be the corresponding quantum integer, so that $[d]=(t^d-1)/(t-1)$ if $t\ne 1$ or $[d]=d$ if $t=1$.

We can now state our first main result.

Theorem A

Suppose that $1<e<\infty$ and that $({\mathcal {O}},t)$ is an e-idempotent subring Then the algebra $\mathcal {H}^\varLambda _n({\mathcal {O}})$ is generated as an ${\mathcal {O}}$-algebra by the elements

$$\begin{aligned} \{f_{\mathbf {i}}^{\mathcal {O}}~|~\mathbf {i}\in I^n\}\cup \{\psi ^{\mathcal {O}}_r\,|\,1\le r<n\}\cup \{y^{{\mathcal {O}}}_r\,|\,1\le r\le n\} \end{aligned}$$

subject only to the following relations : 

$$\begin{aligned}&\prod _{\begin{array}{c} 1\le l\le \ell \\ \kappa _l\equiv i_1\text { }(\text {mod } e) \end{array}} (y^{{\mathcal {O}}}_1-[\kappa _l-\hat{\imath }_1])f_{\mathbf {i}}^{\mathcal {O}}=0,\\&f_{\mathbf {i}}^{\mathcal {O}}f_{\mathbf {j}}^{\mathcal {O}} = \delta _{\mathbf {i}\mathbf {j}} f_{\mathbf {i}}^{\mathcal {O}}, \quad {\textstyle \sum _{\mathbf {i}\in I^n}} f_{\mathbf {i}}^{\mathcal {O}}= 1, \quad y^{{\mathcal {O}}}_r f_{\mathbf {i}}^{\mathcal {O}}= f_{\mathbf {i}}^{\mathcal {O}}y^{{\mathcal {O}}}_r,\\&\psi ^{\mathcal {O}}_r f_{\mathbf {i}}^{\mathcal {O}}= f_{s_r{\cdot }\mathbf {i}}^{\mathcal {O}} \psi ^{\mathcal {O}}_r,\quad y^{{\mathcal {O}}}_r y^{{\mathcal {O}}}_s = y^{{\mathcal {O}}}_s y^{{\mathcal {O}}}_r,\\&\psi ^{\mathcal {O}}_r y^{{\mathcal {O}}}_{r+1} f_{\mathbf {i}}^{\mathcal {O}}=(y^{{\mathcal {O}}}_r\psi ^{\mathcal {O}}_r+\delta _{i_ri_{r+1}})f_{\mathbf {i}}^{\mathcal {O}},\quad y^{{\mathcal {O}}}_{r+1}\psi ^{\mathcal {O}}_rf_{\mathbf {i}}^{\mathcal {O}}=(\psi ^{\mathcal {O}}_r y^{{\mathcal {O}}}_r+\delta _{i_ri_{r+1}})f_{\mathbf {i}}^{\mathcal {O}},\\&\psi ^{\mathcal {O}}_r y^{{\mathcal {O}}}_s = y^{{\mathcal {O}}}_s \psi ^{\mathcal {O}}_r,\quad \text {if }s \ne r,r+1,\\&\psi ^{\mathcal {O}}_r \psi ^{\mathcal {O}}_s = \psi ^{\mathcal {O}}_s \psi ^{\mathcal {O}}_r,\quad \text {if }|r-s|>1,\\&(\psi ^{\mathcal {O}}_r)^2f_{\mathbf {i}}^{\mathcal {O}}= {\left\{ \begin{array}{ll} (y^{\langle 1+\rho _r(\mathbf {i})\rangle }_r-y^{{\mathcal {O}}}_{r+1}) (y^{\langle 1-\rho _r(\mathbf {i})\rangle }_{r+1}-y^{{\mathcal {O}}}_r)f_{\mathbf {i}}^{\mathcal {O}},&{} \text {if }i_r\leftrightarrows i_{r+1},\\ (y^{\langle 1+\rho _r(\mathbf {i})\rangle }_r-y^{{\mathcal {O}}}_{r+1})f_{\mathbf {i}}^{\mathcal {O}},&{} \text {if }i_r\rightarrow i_{r+1},\\ (y^{\langle 1-\rho _r(\mathbf {i})\rangle }_{r+1}-y^{{\mathcal {O}}}_r)f_{\mathbf {i}}^{\mathcal {O}},&{} \text {if }i_r\leftarrow i_{r+1},\\ 0,&{}\text {if }i_r=i_{r+1},\\ f_{\mathbf {i}}^{\mathcal {O}},&{}\text {otherwise}, \end{array}\right. } \end{aligned}$$

and where $\big (\psi ^{\mathcal {O}}_r\psi ^{\mathcal {O}}_{r+1}\psi ^{\mathcal {O}}_r-\psi ^{\mathcal {O}}_{r+1}\psi ^{\mathcal {O}}_r\psi ^{\mathcal {O}}_{r+1}\big )f_{\mathbf {i}}^{\mathcal {O}}$ is equal to

$$\begin{aligned} {\left\{ \begin{array}{ll} (y^{\langle 1+\rho _r(\mathbf {i})\rangle }_r+y^{\langle 1+\rho _r(\mathbf {i})\rangle }_{r+2}- y^{\langle 1+\rho _r(\mathbf {i})\rangle }_{r+1}-y^{\langle 1-\rho _r(\mathbf {i})\rangle }_{r+1})f_{\mathbf {i}}^{\mathcal {O}}, &{}\text {if }i_{r+2}=i_r\rightleftarrows i_{r+1},\\ -t^{1+\rho _r(\mathbf {i})}f_{\mathbf {i}}^{\mathcal {O}},&{}\text {if }i_{r+2}=i_r\rightarrow i_{r+1},\\ f_{\mathbf {i}}^{\mathcal {O}},&{}\text {if }i_{r+2}=i_r\leftarrow i_{r+1},\\ 0,&{}\text {otherwise}, \end{array}\right. } \end{aligned}$$

where $\rho _r(\mathbf {i})=\hat{\imath }_r-\hat{\imath }_{r+1}$ and $y^{\langle d\rangle }_r=t^dy^{{\mathcal {O}}}_r+[d],$ for $d\in \mathbb {Z}.$

As we explain in Corollary 2.10 by taking e large enough this result also applies when $e=0$. The appendix gives a direct treatment of this case.

To help the reader interpret Theorem A we include the following special case of this result that gives a new presentation of the group algebra of the symmetric group over the ring $\mathbb {Z}_{(p)}$, where p is an integer prime and $\mathbb {Z}_{(p)}$ is the localisation of $\mathbb {Z}$ at the prime ideal $p\mathbb {Z}$.

Corollary 1.1

Suppose that $e=p$ be an odd prime number and let $I=\mathbb {Z}/p\mathbb {Z}$ and $\varLambda =\varLambda _0$. Then the group algebra $\mathbb {Z}_{(p)}\mathfrak S_n$ is generated as an $\mathbb {Z}_{(p)}$-algebra by the elements

$$\begin{aligned} \{f_{\mathbf {i}}^{\mathcal {O}}~|~\mathbf {i}\in I^n\}\cup \{\psi ^{\mathcal {O}}_r~|~1\le r<n\}\cup \{y^{{\mathcal {O}}}_s~|~1\le s\le n\} \end{aligned}$$

subject only to the relations : 

$$\begin{aligned}&(y^{{\mathcal {O}}}_1)^{(\varLambda ,\alpha _{i_1})}e(\mathbf {i})=0,\quad f_{\mathbf {i}}^{\mathcal {O}}f_{\mathbf {j}}^{\mathcal {O}} = \delta _{\mathbf {i}\mathbf {j}} f_{\mathbf {i}}^{\mathcal {O}},\quad \sum _{\mathbf {i}\in I^n}f_{\mathbf {i}}^{\mathcal {O}}= 1, \quad y^{{\mathcal {O}}}_r f_{\mathbf {i}}^{\mathcal {O}}= f_{\mathbf {i}}^{\mathcal {O}}y^{{\mathcal {O}}}_r,\\&\psi ^{\mathcal {O}}_r f_{\mathbf {i}}^{\mathcal {O}}= f_{s_r{\cdot }\mathbf {i}}^{\mathcal {O}} \psi ^{\mathcal {O}}_r, \quad y^{{\mathcal {O}}}_r y^{{\mathcal {O}}}_s = y^{{\mathcal {O}}}_s y^{{\mathcal {O}}}_r,\\&\psi ^{\mathcal {O}}_r y^{{\mathcal {O}}}_{r+1} f_{\mathbf {i}}^{\mathcal {O}}=(y^{{\mathcal {O}}}_r\psi ^{\mathcal {O}}_r+ \delta _{i_ri_{r+1}})f_{\mathbf {i}}^{\mathcal {O}},\quad y^{{\mathcal {O}}}_{r+1}\psi ^{\mathcal {O}}_rf_{\mathbf {i}}^{\mathcal {O}}=(\psi ^{\mathcal {O}}_r y^{{\mathcal {O}}}_r+ \delta _{i_ri_{r+1}})f_{\mathbf {i}}^{\mathcal {O}},\\&\psi ^{\mathcal {O}}_r y^{{\mathcal {O}}}_s = y^{{\mathcal {O}}}_s \psi ^{\mathcal {O}}_r,\quad \text {if }s \ne r,r+1,\\&\psi ^{\mathcal {O}}_r \psi ^{\mathcal {O}}_s = \psi ^{\mathcal {O}}_s \psi ^{\mathcal {O}}_r,\quad \text {if }|r-s|>1,\\&(\psi ^{\mathcal {O}}_r)^2f_{\mathbf {i}}^{\mathcal {O}}= {\left\{ \begin{array}{ll} (y^{{\mathcal {O}}}_r-y^{{\mathcal {O}}}_{r+1})f_{\mathbf {i}}^{\mathcal {O}},&{} \text {if } i_r\rightarrow i_{r+1}\ne 0,\\ (y^{{\mathcal {O}}}_{r}+p-y^{{\mathcal {O}}}_{r+1})f_{\mathbf {i}}^{\mathcal {O}},&{} \text {if }i_r\rightarrow i_{r+1}=0,\\ (y^{{\mathcal {O}}}_{r+1}-y^{{\mathcal {O}}}_r)f_{\mathbf {i}}^{\mathcal {O}},&{} \text {if }0\ne i_r\leftarrow i_{r+1},\\ (y^{{\mathcal {O}}}_{r+1}+p-y^{{\mathcal {O}}}_r)f_{\mathbf {i}}^{\mathcal {O}},&{} \text {if } 0=i_r\leftarrow i_{r+1},\\ 0,&{}\text {if }i_r=i_{r+1},\\ f_{\mathbf {i}}^{\mathcal {O}},&{}\text {otherwise}, \end{array}\right. }\\&(\psi ^{\mathcal {O}}_r\psi ^{\mathcal {O}}_{r+1}\psi ^{\mathcal {O}}_r-\psi ^{\mathcal {O}}_{r+1}\psi ^{\mathcal {O}}_r\psi ^{\mathcal {O}}_{r+1})f_{\mathbf {i}}^{\mathcal {O}}= {\left\{ \begin{array}{ll} -f_{\mathbf {i}}^{\mathcal {O}},&{}\text {if }i_{r+2}=i_r\rightarrow i_{r+1},\\ f_{\mathbf {i}}^{\mathcal {O}},&{}\text {if }i_{r+2}=i_r\leftarrow i_{r+1},\\ 0,&{}\text {otherwise}, \end{array}\right. } \end{aligned}$$

for all admissible r, s and $\mathbf {i}\in I^n.$

Except for the cyclotomic relation and the last two relations (that is, the quadratic relations and the braid relations for $\psi ^{\mathcal {O}}_1,\dots \psi ^{\mathcal {O}}_{n-1}$), all of the relations in Theorem A coincide with the corresponding KLR-relations in $\mathcal {R}^\varLambda _{n}$. Interestingly, only the “Jucys–Murphy like elements” $y^{{\mathcal {O}}}_r$ need to be modified in order to define a deformation of $\mathcal {R}^\varLambda _{n}$. Over a field $K={\mathcal {O}}/\mathfrak m$, the presentation in Theorem A collapses to give the KLR algebra $\mathcal {R}^\varLambda _{n}$ because the definition of an idempotent subring ensures that $t^{1+\rho _r(\mathbf {i})}\otimes 1_K=1$ and $y_r^{\langle 1\pm \rho _r(\mathbf {i})\rangle }\otimes 1_K=y^{{\mathcal {O}}}_r\otimes 1_K$, for $1\le r\le n$.

As a first application of Theorem A, Corollary 4.31 gives what appears to be tight upper bounds on the nilpotency indices of the elements $y_1,\ldots ,y_n$ in the cyclotomic quiver Hecke algebras of type A. Previously such a result was known only in the special case of the linear quiver or, equivalently, when $e=0$.

To prove Theorem A we work almost entirely inside the semisimple representation theory of the cyclotomic Hecke algebras $\mathcal {H}^\varLambda _{n}$. We show that definition of the quiver Hecke algebra $\mathcal {R}^\varLambda _{n}$, and its grading, is implicit in Young’s seminormal form. With hindsight, using the perspective afforded by this paper, it is not too much of an exaggeration to say that Murphy could have discovered the cyclotomic quiver Hecke algebras in 1983 soon after writing his paper on the Nakayama conjecture [29].

Our proof of Theorem A gives another explanation for the KLR relations and a more conceptual proof of one direction in Brundan and Kleshchev’s isomorphism theorem [6] (see Theorem 2.9). In fact, we give a new proof of the Brundan–Kleshchev isomorphism theorem by using the Ariki–Brundan–Kleshchev categorification theorem [2, 7] to bound the dimension of the algebras defined by the presentation in Theorem A.

For the algebras of type A the authors have constructed a graded cellular basis $\{\psi _{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$ for $\mathcal {R}^\varLambda _{n}$ [14]. Here ${\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ is the set of all pairs of standard tableaux of the same shape, where the shape is a multipartition of n. The element $\psi _{{\mathfrak s}{\mathfrak t}}$ is homogeneous of degree $\deg _e{\mathfrak s}+\deg _e{\mathfrak t}$, where $\deg _e\,{:}\,{\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})\!\longrightarrow \!\mathbb {Z}$ is the combinatorial degree function introduced by Brundan et al. [8]. Li [24] has shown that $\{\psi _{{\mathfrak s}{\mathfrak t}}\}$ is a graded cellular basis of $\mathcal {R}^\varLambda _{n}$ over an arbitrary ring. In particular, the KLR algebra $\mathcal {R}^\varLambda _{n}$ is always free of rank $\dim \mathcal {H}^\varLambda _{n}(K)$, for K a field.

One of the problems with the basis $\{\psi _{{\mathfrak s}{\mathfrak t}}\}$ is that, because the KLR generators $\psi _r$, for $1\le r<n$, do not satisfy the braid relations, the basis elements $\psi _{{\mathfrak s}{\mathfrak t}}$ depend upon a choice of reduced expression for certain permutations $d({\mathfrak s}),d({\mathfrak t})\in \mathfrak S_n$ associated with the tableaux ${\mathfrak s}$ and ${\mathfrak t}$; see Sect. 2.4. As a consequence, the results of [14] constructs different $\psi$-bases for different choices of reduced expressions for the elements of $\mathfrak S_n$. The different $\psi$-bases constructed in this way are closely related and it would be advantageous to be able to make a canonical choice of basis, however, until now it has not been clear how to do this.

Fix a modular system $(\fancyscript{K},{\mathcal {O}},K)$ as in Sect. 6 and consider the corresponding cyclotomic Hecke algebras $(\mathcal {H}^\varLambda _n(\fancyscript{K}),\mathcal {H}^\varLambda _n({\mathcal {O}}),\mathcal {H}^\varLambda _{n})$, where $\mathcal {H}^\varLambda _{n}=\mathcal {H}^\varLambda _{n}(K)$. The algebra $\mathcal {H}^\varLambda _n(\fancyscript{K})$ is semisimple and has a seminormal basis $\{f_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}, \mathcal {H}^\varLambda _n({\mathcal {O}})$ is a free ${\mathcal {O}}$-subalgebra of $\mathcal {H}^\varLambda _n(\fancyscript{K})$ and $\mathcal {H}^\varLambda _{n}\cong \mathcal {H}^\varLambda _n({\mathcal {O}})\otimes _{\mathcal {O}}K$. Finally, we recall that the set ${\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ comes equipped with a natural partial order $\blacktriangleright$; see Sect. 2.4.

Theorem B

Suppose that K is a field of characteristic zero and that $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n}).$ Then there is a unique element $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\in \mathcal {H}^\varLambda _n({\mathcal {O}})$ such that

• (a) $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}=f_{{\mathfrak s}{\mathfrak t}} +\sum _{({\mathfrak u},{\mathfrak v})\blacktriangleright ({\mathfrak s},{\mathfrak t})}p^{{\mathfrak s}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x^{-1})f_{{\mathfrak u}{\mathfrak v}},$ where if $({\mathfrak u},{\mathfrak v})\blacktriangleright ({\mathfrak s},{\mathfrak t})$ then $p^{{\mathfrak s}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x)\in xK[x]$ and $\deg p^{{\mathfrak s}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x)\le \tfrac{1}{2}(\deg {\mathfrak u}-\deg {\mathfrak s}+\deg {\mathfrak v}-\deg {\mathfrak t}).$

• (b) $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\otimes _{\mathcal {O}}1_K=B_{{\mathfrak s}{\mathfrak t}}'+C_{{\mathfrak s}{\mathfrak t}},$ where $B_{{\mathfrak s}{\mathfrak t}}'$ is homogeneous of degree $\deg {\mathfrak s}+\deg {\mathfrak t}$ and $C_{{\mathfrak s}{\mathfrak t}}$ is a sum of homogeneous terms of degree strictly larger than $\deg B_{{\mathfrak s}{\mathfrak t}}.$

Moreover,  $\{B_{{\mathfrak s}{\mathfrak t}}'\,|\,({\mathfrak s},{\mathfrak t})\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})\}$ is a graded cellular basis of $\mathcal {H}^\varLambda _{n}$ and

$$\begin{aligned} B_{{\mathfrak s}{\mathfrak t}}'=\psi _{{\mathfrak s}{\mathfrak t}}+\sum _{({\mathfrak u},{\mathfrak v})\blacktriangleright ({\mathfrak s},{\mathfrak t})}r_{{\mathfrak u}{\mathfrak v}}\psi _{{\mathfrak u}{\mathfrak v}}, \end{aligned}$$

for some $r_{{\mathfrak u}{\mathfrak v}}\in K.$

There is a similar graded cellular basis of $\mathcal {H}^\varLambda _n({\mathcal {O}})$ when K is a field of positive characteristic, however, its’ description is more complicated because the corresponding polynomials $p^{{\mathfrak s}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x)$ do not necessarily satisfy the degree bound in Theorem B(a). The construction of the B-basis is reminiscent of the Kazhdan–Lusztig basis [20]. The B-basis can depend on the choice of multicharge.

As remarked above, the basis element $\psi _{{\mathfrak s}{\mathfrak t}}$ depends upon choices of reduced expression for the permutations $d({\mathfrak s}),d({\mathfrak t})\in \mathfrak S_n$, yet for any choice $B_{{\mathfrak s}{\mathfrak t}}$ is equal to $\psi _{{\mathfrak s}{\mathfrak t}}$ plus a linear combination of more dominant terms by Theorem B. The B-basis elements depend only on the indexing tableaux, and not on choices of reduced expressions. In this sense, the B-basis corrects for a deficiency in the definition of the $\psi$-bases.

To prove the two theorems above, we define a seminormal basis of a semisimple Hecke algebra to be a basis of $\mathcal {H}^\varLambda _{n}$ of simultaneous eigenvectors for the Gelfand–Zetlin subalgebra of $\mathcal {H}^\varLambda _{n}$. Seminormal bases are classical objects that are ubiquitous in the literature, having been rediscovered many times since were first introduced for the symmetric groups by Young in 1900 [37].

Seminormal basis elements are a basis of eigenvectors for the action of the Jucys–Murphy elements on the regular representation of $\mathcal {H}^\varLambda _{n}(K)$. Eigenbases are, of course, only unique up to scalar multiplication. This paper starts by introducing seminormal coefficient systems that gives a combinatorial framework for describing the structure constants of the algebra in terms of the choice of eigenvectors. The real surprise is that seminormal coefficient systems encode the KLR grading.

The close connections between the semisimple representation theory and the KLR gradings is made even more explicit in the third main result of this paper that gives a closed formula for the Gram determinants of the semisimple Specht modules of these algebras. Closed formulas for these determinants already exist in the literature [4, 16–18], however, all of these formulas describe these determinants as rational functions (or rational numbers in the degenerate case). The theorem below gives the first integral formula for these determinants.

In order to state the closed integral formulas for the Gram determinant of the Specht module $S^{\varvec{\lambda }}$, for a multipartition ${\varvec{\lambda }}$ define

$$\begin{aligned} \deg _e({\varvec{\lambda }})=\sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})}\deg _e({\mathfrak t})\in \mathbb {Z}, \end{aligned}$$

where ${\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ is the set of standard ${\varvec{\lambda }}$-tableaux. Let $\varPhi _e(t)\in \mathbb {Z}[t]$ be the eth cyclotomic polynomial for $e>1$. We prove the following (see Theorem 3.13 for a more precise statement).

Theorem C

Suppose that $\mathcal {H}^\varLambda _{n}$ is a semisimple cyclotomic Hecke algebra over $\mathbb {Q}(t),$ with Hecke parameter t. Let ${\varvec{\lambda }}$ be a multipartition of n. Then the Gram determinant of the Specht module $S^{\varvec{\lambda }}$ is equal to

$$\begin{aligned} t^N\prod _{e>1}\varPhi _e(t)^{\deg _e({\varvec{\lambda }})}, \end{aligned}$$

for a known integer N. In particular,  $\deg _e({\varvec{\lambda }})\ge 0,$ for all $e\in \{0,2,3,4,\ldots \}.$

As the integers $\deg _e({\varvec{\lambda }})$ are defined combinatorially, it should be possible to give a purely combinatorial proof that $\deg _e({\varvec{\lambda }})\ge 0$. In Sect. 3.3 we give two representation theoretic proofs of this result. The first proof is elementary but not very enlightening. The second proof uses deep positivity properties of the graded decomposition numbers of $\mathcal {H}^\varLambda _{n}(\mathbb {C})$ to show that the tableau combinatorics of $\mathcal {H}^\varLambda _{n}$ provides a framework for giving purely combinatorial formulas for the graded dimensions of the simple $\mathcal {H}^\varLambda _{n}$-modules and for the graded decomposition numbers of $\mathcal {H}^\varLambda _{n}$. Interestingly, we show that there is a close connection between the graded dimensions of the simple $\mathcal {H}^\varLambda _{n}$-modules and the graded decomposition numbers for $\mathcal {H}^\varLambda _{n}$. Note that in characteristic zero, the graded decomposition numbers of $\mathcal {H}^\varLambda _{n}$ are parabolic Kazhdan–Lusztig polynomials of type A [7], so our results show that the tableau combinatorics leads to combinatorial formulas for these polynomials. Unfortunately, we are only able to prove that such formulas exist and we are not able to make them explicit or to show that they are canonical in any way.

The outline of this paper is as follows. Section 2 defines the cyclotomic Hecke algebras of type A, giving a uniform presentation for the degenerate and non-degenerate algebras. Previously these algebras have been treated separately in the literature. We then recall the basic results about these algebras that we need from the literature, including Brundan and Kleshchev’s isomorphism theorem [6]. Section 3 develops the theory of seminormal bases for these algebras in full generality. We completely classify the seminormal bases of $\mathcal {H}^\varLambda _{n}$ and then use them to prove Theorem C, thus establishing a link between the semisimple representation theory of $\mathcal {H}^\varLambda _{n}$ and the quiver Hecke algebra $\mathcal {R}^\varLambda _{n}$. Using this we prove the existence of combinatorial formulas for the graded dimensions of the simple modules and the graded decomposition numbers of $\mathcal {H}^\varLambda _{n}$. In Sect. 4 we use the theory of seminormal forms to construct a deformation of the cyclotomic quiver Hecke algebras of type A, culminating with the proof of Theorem A. Section 5 builds on Theorem A to give a quicker construction of the graded cellular basis of $\mathcal {H}^\varLambda _{n}(K)$, over a field K, which was one of the main results of [14]. Finally, in Sect. 6 we use Theorem A to show that $\mathcal {H}^\varLambda _{n}(K)$ has the distinguished graded cellular basis described in Theorem B.

2 Cyclotomic Hecke algebras

This chapter defines the cyclotomic Hecke and quiver Hecke algebras of type A and it introduces some of the basic machinery that we need for understanding these algebras. We give a new presentation for the cyclotomic Hecke algebras of type A, which simultaneously captures the degenerate and non-degenerate cyclotomic Hecke algebras that currently appear in the literature, and then we recall the results from the literature that we need, including Brundan and Kleshchev’s isomorphism theorem [6].

2.1 Quiver combinatorics

Fix an integer $e\in \{0,2,3,4,\dots \}$ and let $\varGamma _e$ be the oriented quiver with vertex set $I=\mathbb {Z}/e\mathbb {Z}$ and edges $i\longrightarrow i+1$, for $i\in I$. If $i,j\in I$ and i and j are not connected by an edge in $\varGamma _e$ then we write

To the quiver $\varGamma _e$ we attach the Cartan matrix $(c_{ij})_{i,j\in I}$, where

$$\begin{aligned} c_{i,j}= {\left\{ \begin{array}{ll} 2,&{}\text {if } i=j,\\ -1,&{}\text {if }i\rightarrow j\text { or }i\leftarrow j,\\ -2,&{}\text {if }i\leftrightarrows j,\\ 0,&{}\text {otherwise}, \end{array}\right. } \end{aligned}$$

Let $\widehat{\mathfrak {sl}}_e$ be the corresponding Kac–Moody algebra [19] with fundamental weights $\{\varLambda _i\,|\,i\in I\}$, positive weight lattice $P^+_e=\sum _{i\in I}\mathbb {N}\varLambda _i$ and positive root lattice $Q^+=\bigoplus _{i\in I}\mathbb {N}\alpha _i$. Let $(\cdot ,\cdot )$ be the bilinear form determined by

$$\begin{aligned} (\alpha _i,\alpha _j)=c_{ij}\quad \text {and}\quad (\varLambda _i,\alpha _j)=\delta _{ij},\qquad \text {for }i,j\in I. \end{aligned}$$

More details can be found, for example, in [19, Chapter 1].

Fix, once and for all, a multicharge ${\varvec{\kappa }}=(\kappa _1,\ldots ,\kappa _\ell )\in \mathbb {Z}^\ell$ that is a sequence of integers such that if $e\ne 0$ then $\kappa _l-\kappa _{l+1}\ge n$ for $1\le l<\ell$. Define $\varLambda =\varLambda _e({\varvec{\kappa }})=\varLambda _{\bar{\kappa }_1}+\dots +\varLambda _{\bar{\kappa }_\ell }$, where $\bar{\kappa }=\kappa \text { }(\text {mod } e)$. Equivalently, $\varLambda$ is the unique element of $P^+_e$ such that

$$\begin{aligned} (\varLambda ,\alpha _i) = \#\{1\le l\le \ell ~|~ \kappa _l\equiv i\text { }(\text {mod } e)\},\quad \text {for all }i\in I. \end{aligned}$$

(2.1)

All of the bases for the modules and algebras in this paper depend implicitly on the choice of ${\varvec{\kappa }}$ even though the algebras themselves depend only on $\varLambda$.

2.2 Cyclotomic Hecke algebras

This section defines the cyclotomic Hecke algebras of type A and explains the connection between these algebras and the degenerate and non-degenerate Hecke algebras of type $G(\ell ,1,n)$.

Fix an integral domain ${\mathcal {O}}$ that contains an invertible element $\xi \in {\mathcal {O}}^\times$.

Definition 2.1

Fix integers $n\ge 0$ and $\ell \ge 1$. Then the cyclotomic Hecke algebra of type A with Hecke parameter $\xi \in {\mathcal {O}}^\times$ and cyclotomic parameters $Q_1,\ldots ,Q_\ell \in {\mathcal {O}}$ is the unital associative ${\mathcal {O}}$-algebra $\mathcal {H}_n=\mathcal {H}_n({\mathcal {O}},\xi ,Q_1,\ldots ,Q_\ell )$ with generators $L_1,\ldots ,L_n, T_1,\ldots ,T_{n-1}$ that are subject to the relations

$$\begin{aligned}&\prod _{l=1}^\ell (L_1-Q_l)=0,\quad (T_r+1)(T_r-\xi )=0, \\&L_rL_t=L_tL_r,\qquad T_rT_s=T_sT_r\quad \text {if }|r-s|>1,\\&T_sT_{s+1}T_s=T_{s+1}T_sT_{s+1},\qquad T_rL_t=L_tT_r,\quad \text {if }t\ne r,r+1,\\&L_{r+1}(T_r-\xi +1)=T_rL_r+1, \end{aligned}$$

where $1\le r<n, 1\le s<n-1$ and $1\le t\le n$.

Remark 2.2

If $\xi =1$ then, by definition, $\mathcal {H}_n$ is a degenerate cyclotomic Hecke algebra of type $G(\ell ,1,n)$. If $\xi \ne 1$ then $\mathcal {H}_n$ is (isomorphic to) an integral cyclotomic Hecke algebra of type $G(\ell ,1,n)$. To see this define $L_k'=(\xi -1)L_k+1$, for $1\le k\le n$, and observe that $\mathcal {H}_n$ is generated by $L_1',T_1,\ldots ,T_{n-1}$ subject to the usual relations for these algebras as originally defined by Ariki and Koike [3]. It is now easy to verify our claim. For each $1\le m\le n$, an eigenvector for $L_m$ of eigenvalue $[k]_{\xi }$ is the same as an eigenvector for $L'_m$ of eigenvalue $\xi ^k$. The presentation of $\mathcal {H}_n$ in Definition 2.1 unifies the definition of the ‘degenerate’ and ‘non-degenerate’ Hecke algebras, which corresponds to the cases where $\xi =1$ or $\xi \ne 1$, respectively.

Let $\mathfrak S_n$ be the symmetric group on n letters. For $1\le r<n$ let $s_r=(r,r+1)$ be the corresponding simple transposition. Then $\{s_1,\ldots ,s_{n-1}\}$ is the standard set of Coxeter generators for $\mathfrak S_n$. A reduced expression for $w\in \mathfrak S_n$ is a word $w=s_{r_1},\ldots ,s_{r_k}$ if k is minimal, for $1\le r_j<n$ and $1\le j\le k$. If $w=s_{r_1}\cdots s_{r_k}$ is reduced then set $T_w=T_{r_1}\cdots T_{r_k}$. Then $T_w$ is independent of the choice of reduced expression since the braid relations hold in $\mathcal {H}_n$. It follows, arguing as in [3, Theorem 3.3], that $\mathcal {H}_n$ is free as an ${\mathcal {O}}$-module with basis

$$\begin{aligned} \{L_1^{a_1}\cdots L_n^{a_n} T_w\,|\,0\le a_1,\ldots ,a_n<\ell \text { and }w\in \mathfrak S_n\}. \end{aligned}$$

Consequently, $\mathcal {H}_n$ is free as an ${\mathcal {O}}$-module of rank $\ell ^n n!$, which is the order of the complex reflection group of type $G(\ell ,1,n)$.

We now restrict our attention to the case of integral cyclotomic parameters. Recall that for any integer d and $t\in {\mathcal {O}}$ the quantum integer $[d]_t$ is

$$\begin{aligned}{}[d]_t={\left\{ \begin{array}{ll} 1+t+\cdots +t^{d-1},&{}\text {if }d\ge 0,\\ -(t^{-1}+t^{-2}+\cdots +t^d),&{}\text {if }d<0. \end{array}\right. } \end{aligned}$$

When t is understood we write $[d]=[d]_t$. Set $[d]_t^!=[d]^!=[1][2]\cdots [d]$ when $d>0$.

An integral cyclotomic Hecke algebra is a cyclotomic Hecke algebra $\mathcal {H}_n$ with cyclotomic parameters of the form $Q_r=[\kappa _r]_\xi$, for $\kappa _1,\ldots ,\kappa _\ell \in \mathbb {Z}$. The sequence of integers ${\varvec{\kappa }}=(\kappa _1,\ldots ,\kappa _\ell )\in \mathbb {Z}^\ell$ is the multicharge of $\mathcal {H}_n$.

Translating the Morita equivalence theorems of [11, Theorem 1.1] and [5, Theorem 5.19] into the current setting, every cyclotomic Hecke algebras of type A is Morita equivalent to a direct sum of tensor products of integral cyclotomic Hecke algebras. Therefore, there is no loss of generality in restricting our attention to the integral cyclotomic Hecke algebras of type A.

Recall that $\varLambda \in P^+_e$ and that we have fixed an integer $e\in \{0,2,3,4,\ldots \}$. We assume that ${\mathcal {O}}$ contains an invertible element $\xi \in {\mathcal {O}}^\times$ such that $[e]_\xi =0$ if $e>0$ and $[f]_\xi \ne 0$ for all $f\ge 1$ if $e=0$. Hence, either:

• (a) $\xi =1$ and e is prime and equal to the characteristic of ${\mathcal {O}}$,

• (b) $e>0$ and $\xi$ is a primitive eth root of unity, or,

• (c) $e=0$ and $\xi$ is not a root of unity.

In addition, fix a multicharge ${\varvec{\kappa }}$ so that $\varLambda =\varLambda _e({\varvec{\kappa }})$ as in (2.1).

Let $\mathcal {H}^\varLambda _{n}=\mathcal {H}^\varLambda _{n}({\mathcal {O}})$ be the integral cyclotomic Hecke algebra $\mathcal {H}_n({\mathcal {O}},\xi ,{\varvec{\kappa }})$. Using the definitions it is easy to see that, up to isomorphism, $\mathcal {H}^\varLambda _{n}$ depends only on $\xi$ and $\varLambda$. In fact, by Theorem 2.9 below, it depends only on e and $\varLambda$. Nonetheless, many of the constructions that follow, particularly the definitions of bases, depend upon the choice of ${\varvec{\kappa }}$.

2.3 Graded algebras and cellular bases

This section recalls the definitions and results from the representation theory of (graded) cellular algebras that we need.

Let A be a unital associative ${\mathcal {O}}$-algebra that is free and of finite rank as an ${\mathcal {O}}$-module. In this paper a graded module will always mean a $\mathbb {Z}$-graded module. That is, an ${\mathcal {O}}$-module M that has a decomposition $M=\bigoplus _{n\in \mathbb {Z}}M_d$ as an ${\mathcal {O}}$-module. If $m\in M_d$, for $d\in \mathbb {Z}$, then m is homogeneous of degree d and we set $\deg m=d$. If M is a graded ${\mathcal {O}}$-module and $s\in \mathbb {Z}$ let $M\langle s\rangle$ be the graded ${\mathcal {O}}$-module obtained by shifting the grading on M up by s; that is, $M\langle s\rangle _d=M_{d-s}$, for $d\in \mathbb {Z}$.

Similarly a graded algebra is a unital associative ${\mathcal {O}}$-algebra $A=\bigoplus _{d\in \mathbb {Z}}A_d$ that is a graded ${\mathcal {O}}$-module such that $A_dA_e\subseteq A_{d+e}$, for all $d,e\in \mathbb {Z}$. It follows that $1\in A_0$ and that $A_0$ is a graded subalgebra of A. A graded (right) A-module is a graded ${\mathcal {O}}$-module M such that $\underline{M}$ is an $\underline{A}$-module and $M_dA_e\subseteq M_{d+e}$, for all $d,e\in \mathbb {Z}$, where $\underline{M}$ and $\underline{A}$ mean forgetting the $\mathbb {Z}$-grading structures on M and A respectively. Graded submodules, graded left A-modules and so on are all defined in the obvious way.

The following definition extends Graham and Lehrer’s [12] definition of cellular algebras to the graded setting.

Definition 2.3

(Graded cellular algebras [12, 14]) Suppose that A is an ${\mathcal {O}}$-algebra that is free of finite rank over ${\mathcal {O}}$. A cell datum for A is an ordered triple $(\mathcal {P},T,C)$, where $(\mathcal {P},\vartriangleright )$ is the weight poset, $T(\lambda )$ is a finite set for $\lambda \in \mathcal {P}$, and

$$\begin{aligned} C\,{:}\,\coprod _{\lambda \in \mathcal {P}}T(\lambda )\times T(\lambda )\!\longrightarrow \!A; ({\mathfrak s},{\mathfrak t})\mapsto c_{{\mathfrak s}{\mathfrak t}}, \end{aligned}$$

is an injective function such that:

$(\hbox {GC}_1)$ The set $\{c_{{\mathfrak s}{\mathfrak t}}\,|\,{\mathfrak s},{\mathfrak t}\in T(\lambda )\text { for }\lambda \in \mathcal {P}\}$ is an ${\mathcal {O}}$-basis of A.

$(\hbox {GC}_2)$ If ${\mathfrak s},{\mathfrak t}\in T(\lambda )$, for some $\lambda \in \mathcal {P}$, and $a\in A$ then there exist scalars $r_{{\mathfrak t}{\mathfrak v}}(a)$, which do not depend on ${\mathfrak s}$, such that

$$\begin{aligned} c_{{\mathfrak s}{\mathfrak t}} a=\sum _{{\mathfrak v}\in T(\lambda )}r_{{\mathfrak t}{\mathfrak v}}(a)c_{{\mathfrak s}{\mathfrak v}}\quad (\hbox {mod } {A^{\vartriangleright \lambda }}), \end{aligned}$$

where $A^{\vartriangleright \lambda }$ is the ${\mathcal {O}}$-submodule of A spanned by $\{c_{{\mathfrak a}{\mathfrak b}}\,|\,\mu \vartriangleright \lambda \text { and }{\mathfrak a},{\mathfrak b}\in T(\mu )\}$.

$(\hbox {GC}_3)$ The ${\mathcal {O}}$-linear map $*\,{:}\,A\!\longrightarrow \!A$ determined by $(c_{{\mathfrak s}{\mathfrak t}})^*=c_{{\mathfrak t}{\mathfrak s}}$, for all $\lambda \in \mathcal {P}$ and all ${\mathfrak s},{\mathfrak t}\in T(\lambda )$, is an anti-isomorphism of A.

A cellular algebra is an algebra that has a cell datum. If A is a cellular algebra with cell datum $(\mathcal {P},T,C)$ then the basis $\{c_{{\mathfrak s}{\mathfrak t}}\,|\,\lambda \in \mathcal {P}\text { and } {\mathfrak s},{\mathfrak t}\in T(\lambda )\}$ is a cellular basis of A with $*$ its cellular algebra anti-automorphism.

If, in addition, A is a $\mathbb {Z}$-graded algebra then a graded cell datum for A is a cell datum $(\mathcal {P},T,C)$ together with a degree function

$$\begin{aligned} \deg \,{:}\,\coprod _{\lambda \in \mathcal {P}}T(\lambda )\!\longrightarrow \!\mathbb {Z}\end{aligned}$$

such that

$(\hbox {GC}_d)$ the element $c_{{\mathfrak s}{\mathfrak t}}$ is homogeneous of degree $\deg c_{{\mathfrak s}{\mathfrak t}}=\deg ({\mathfrak s})+\deg ({\mathfrak t})$, for all $\lambda \in \mathcal {P}$ and ${\mathfrak s},{\mathfrak t}\in T(\lambda )$.

In this case, A is a graded cellular algebra with graded cellular basis $\{c_{{\mathfrak s}{\mathfrak t}}\}$.

Fix a (graded) cellular algebra A with graded cellular basis $\{c_{{\mathfrak s}{\mathfrak t}}\}$. If $\lambda \in \mathcal {P}$ then the graded cell module is the ${\mathcal {O}}$-module $C^\lambda$ with basis $\{c_{{\mathfrak t}}\,|\,{\mathfrak t}\in T(\lambda )\}$ and with A-action

$$\begin{aligned} c_{{\mathfrak t}} a=\sum _{{\mathfrak v}\in T(\lambda )}r_{{\mathfrak t}{\mathfrak v}}(a)c_{{\mathfrak v}}, \end{aligned}$$

where the scalars $r_{{\mathfrak t}{\mathfrak v}}(a)\in {\mathcal {O}}$ are the same scalars appearing in $(\hbox {GC}_2)$. One of the key properties of the graded cell modules is that by [14, Lemma 2.7] they come equipped with a homogeneous bilinear form $\langle \ ,\ \rangle$ of degree zero that is determined by the equation

$$\begin{aligned} \langle c_{\mathfrak t}, c_{\mathfrak u}\rangle c_{{\mathfrak s}{\mathfrak v}}\equiv c_{{\mathfrak s}{\mathfrak t}}c_{{\mathfrak u}{\mathfrak v}}\quad (\hbox {mod }A^{\vartriangleright \lambda }), \end{aligned}$$

(2.2)

for ${\mathfrak s}, {\mathfrak t}, {\mathfrak u}, {\mathfrak v}\in T(\lambda )$. The radical of this form

$$\begin{aligned} {{\mathrm{rad}}}C^\lambda =\{x\in C^\lambda \,|\,\langle x,y\rangle =0 \text { for all }y\in C^\lambda \} \end{aligned}$$

is a graded A-submodule of $C^\lambda$ so that $D^\lambda =C^\lambda /{{\mathrm{rad}}}C^\lambda$ is a graded A-module. It is shown in [14, Theorem 2.10] that

$$\begin{aligned} \{D^\lambda \langle k\rangle ~|~\lambda \in \mathcal {P}, D^\lambda \ne 0 \text { and }k\in \mathbb {Z}\} \end{aligned}$$

is a complete set of pairwise non-isomorphic irreducible (graded) A-modules when ${\mathcal {O}}$ is a field.

2.4 Multipartitions and tableaux

A partition of d is a weakly decreasing sequence $\lambda =(\lambda _1,\lambda _2,\ldots )$ of non-negative integers such that $|\lambda |=\lambda _1+\lambda _2+\cdots =d$. An $\ell$-multipartition of n is an $\ell$-tuple ${\varvec{\lambda }}=(\lambda ^{(1)},\ldots ,\lambda ^{(\ell )})$ of partitions such that $|\lambda ^{(1)}|+\cdots +|\lambda ^{(\ell )}|=n$. We identify the multipartition ${\varvec{\lambda }}$ with its diagram that is the set of nodes $[\![{\varvec{\lambda }}]\!]=\{(l,r,c)\,|\,1\le c\le \lambda ^{(l)}_r\text { for }1\le l\le \ell \}$, which we think of as an ordered $\ell$-tuple of arrays of boxes in the plane. For example, if ${\varvec{\lambda }}=(3,1^2|2,1|3,2)$ then

In this way we talk of the rows, columns and components of ${\varvec{\lambda }}$.

Given two nodes $\alpha =(l,r,c)$ and $\beta =(l',r',c')$ then $\beta$ is below $\alpha$, or $\alpha$ is above $\beta$, if $(l,r,c)<(l',r',c')$ in the lexicographic order.

The set of multipartitions of n becomes a poset ordered by dominance where ${\varvec{\lambda }}$ dominates ${\varvec{\mu }}$, or ${\varvec{\lambda }}\trianglerighteq {\varvec{\mu }}$, if

$$\begin{aligned} \sum _{k=1}^{l-1}|\lambda ^{(k)}|+\sum _{j=1}^i\lambda ^{(l)}_j \ge \sum _{k=1}^{l-1}|\mu ^{(k)}|+\sum _{j=1}^i\mu ^{(l)}_j, \end{aligned}$$

for $1\le l\le \ell$ and $i\ge 1$. If ${\varvec{\lambda }}\trianglerighteq {\varvec{\mu }}$ and ${\varvec{\lambda }}\ne {\varvec{\mu }}$ then write ${\varvec{\lambda }}\vartriangleright {\varvec{\mu }}$. Let $\mathcal {P}^{\varLambda }_{n}=(\mathcal {P}^{\varLambda }_{n},\trianglerighteq )$ be the poset of multipartitions of n ordered by dominance.

Fix a multipartition ${\varvec{\lambda }}$. Then a $\varvec{{\varvec{\lambda }}}$-tableau is a bijective map ${\mathfrak t}\,{:}\,[\![{\varvec{\lambda }}]\!]\!\longrightarrow \!\{1,2,\ldots ,n\}$, which we identify with a labelling of $[\![{\varvec{\lambda }}]\!]$ by $\{1,2,\ldots ,n\}$. For example,

are both ${\varvec{\lambda }}$-tableaux when ${\varvec{\lambda }}=(3,1^2|2,1|3,2)$ as above. In this way we speak of the rows, columns and components of tableaux. If ${\mathfrak t}$ is a tableau and $1\le k\le n$ set ${{\mathrm{comp}}}_{\mathfrak t}(k)=l$ if k appears in the lth component of ${\mathfrak t}$.

A ${\varvec{\lambda }}$-tableau is standard if its entries increase along rows and columns in each component. Both of the tableaux above are standard. Let ${\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ be the set of standard ${\varvec{\lambda }}$-tableaux and let ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})=\bigcup _{{\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}}{\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$. Similarly set ${\mathrm{Std}}^2({\varvec{\lambda }})=\{({\mathfrak s},{\mathfrak t})\,|\,{\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\}$ and ${\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})=\{({\mathfrak s},{\mathfrak t})\,|\,{\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\text { for some }{\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}\}$.

If ${\mathfrak t}$ is a ${\varvec{\lambda }}$-tableau set ${{\mathrm{Shape}}}({\mathfrak t})={\varvec{\lambda }}$ and let ${\mathfrak t}_{\downarrow m}$ be the subtableau of ${\mathfrak t}$ that contains the numbers $\{1,2,\ldots ,m\}$. If ${\mathfrak t}$ is a standard ${\varvec{\lambda }}$-tableau then ${{\mathrm{Shape}}}({\mathfrak t}_{\downarrow m})$ is a multipartition for all $m\ge 0$. We extend the dominance ordering to the set of all standard tableaux by defining ${\mathfrak s}\trianglerighteq {\mathfrak t}$ if

$$\begin{aligned} {{\mathrm{Shape}}}({\mathfrak s}_{\downarrow m})\trianglerighteq {{\mathrm{Shape}}}({\mathfrak t}_{\downarrow m}), \end{aligned}$$

for $1\le m\le n$. As before, we write ${\mathfrak s}\vartriangleright {\mathfrak t}$ if ${\mathfrak s}\trianglerighteq {\mathfrak t}$ and ${\mathfrak s}\ne {\mathfrak t}$. We extend the dominance ordering to ${\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ by declaring that $({\mathfrak s},{\mathfrak t})\trianglerighteq ({\mathfrak u},{\mathfrak v})$ if ${\mathfrak s}\trianglerighteq {\mathfrak u}$ and ${\mathfrak t}\trianglerighteq {\mathfrak v}$. Similarly, $({\mathfrak s},{\mathfrak t})\vartriangleright ({\mathfrak u},{\mathfrak v})$ if $({\mathfrak s},{\mathfrak t})\trianglerighteq ({\mathfrak u},{\mathfrak v})$ and $({\mathfrak s},{\mathfrak t})\ne ({\mathfrak u},{\mathfrak v})$.

It is easy to see that there are unique standard ${\varvec{\lambda }}$-tableaux ${\mathfrak t}^{\varvec{\lambda }}$ and ${\mathfrak t}_{\varvec{\lambda }}$ such that ${\mathfrak t}^{\varvec{\lambda }}\trianglerighteq {\mathfrak t}\trianglerighteq {\mathfrak t}_{\varvec{\lambda }}$, for all ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$. The tableau ${\mathfrak t}^{\varvec{\lambda }}$ has the numbers $1,2,\ldots ,n$ entered in order from left to right along the rows of ${\mathfrak t}^{\lambda ^{(1)}}$, and then ${\mathfrak t}^{\lambda ^{(2)}},\ldots ,{\mathfrak t}^{\lambda ^{(\ell )}}$ and similarly, ${\mathfrak t}_{\varvec{\lambda }}$ is the tableau with the numbers $1,\ldots ,n$ entered in order down the columns of ${\mathfrak t}^{\lambda ^{(\ell )}},\ldots ,{\mathfrak t}^{\lambda ^{(2)}},{\mathfrak t}^{\lambda ^{(1)}}$. When ${\varvec{\lambda }}=(3,1^2|2,1|3,2)$ then the two ${\varvec{\lambda }}$-tableaux displayed above are ${\mathfrak t}^{\varvec{\lambda }}$ and ${\mathfrak t}_{\varvec{\lambda }}$.

Given a standard ${\varvec{\lambda }}$-tableau ${\mathfrak t}$ define $d({\mathfrak t})\in \mathfrak S_n$ to be the permutation such that ${\mathfrak t}={\mathfrak t}^{\varvec{\lambda }}d({\mathfrak t})$. Let $\le$ be the Bruhat order on $\mathfrak S_n$ with the convention that $1\le w$ for all $w\in \mathfrak S_n$. By a well-known result of Ehresmann and James, if ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ then ${\mathfrak s}\trianglerighteq {\mathfrak t}$ if and only if $d({\mathfrak s})\le d({\mathfrak t})$; see, for example [26, Theorem 3.8].

Recall from Sect. 2.1 that we have fixed a multicharge $\kappa \in \mathbb {Z}^\ell$. The residue of the node $A=(l,r,c)$ is ${{\mathrm{res}}}(A) = \kappa _l+c-r \text { }(\text {mod } e)$ (where we adopt the convention that $i\equiv i\text { }(\text {mod } 0)$, for $i\in \mathbb {Z}$). Thus, ${{\mathrm{res}}}(A)\in I$. A node A is an i-node if ${{\mathrm{res}}}(A)=i$. If ${\mathfrak t}$ is a ${\varvec{\mu }}$-tableaux and $1\le k\le n$ then the residue of k in ${\mathfrak t}$ is ${{\mathrm{res}}}_{\mathfrak t}(k)={{\mathrm{res}}}(A)$, where $A\in {\varvec{\mu }}$ is the unique node such that ${\mathfrak t}(A)=k$. The residue sequence of ${\mathfrak t}$ is

$$\begin{aligned} {{\mathrm{res}}}({\mathfrak t}) = ({{\mathrm{res}}}_{\mathfrak t}(1),{{\mathrm{res}}}_{\mathfrak t}(2),\ldots ,{{\mathrm{res}}}_{\mathfrak t}(n))\in I^n. \end{aligned}$$

As an important special case we set $\mathbf {i}^\mu ={{\mathrm{res}}}({\mathfrak t}^{\varvec{\mu }})$, for ${\varvec{\mu }}\in \mathcal {P}^{\varLambda }_{n}$.

Refine the dominance ordering on the set of standard tableaux by defining ${\mathfrak s}{{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}{\mathfrak t}$ if ${\mathfrak s}\trianglerighteq {\mathfrak t}$ and ${{\mathrm{res}}}({\mathfrak s})={{\mathrm{res}}}({\mathfrak t})$. Similarly, we write $({\mathfrak s},{\mathfrak t}){{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}({\mathfrak u},{\mathfrak v})$ if $({\mathfrak s},{\mathfrak t})\trianglerighteq ({\mathfrak u},{\mathfrak v}), {{\mathrm{res}}}({\mathfrak s})={{\mathrm{res}}}({\mathfrak u})$ and ${{\mathrm{res}}}({\mathfrak t})={{\mathrm{res}}}({\mathfrak v})$ and $({\mathfrak s},{\mathfrak t})\blacktriangleright ({\mathfrak u},{\mathfrak v})$ now has the obvious meaning.

Following Brundan et al. [8, Definition. 3.5] we now define the degree of a standard tableau. Suppose that ${\varvec{\mu }}\in \mathcal {P}^{\varLambda }_{n}$. A node A is an addable node of ${\varvec{\mu }}$ if $A\notin {\varvec{\mu }}$ and ${\varvec{\mu }}\cup \{A\}$ is (the diagram of) a multipartition of $n+1$. Similarly, a node B is a removable node of ${\varvec{\mu }}$ if $B\in {\varvec{\mu }}$ and ${\varvec{\mu }}{\setminus }\{B\}$ is a multipartition of $n-1$. Suppose that A is an i-node and define integers

If ${\mathfrak t}$ is a standard ${\varvec{\mu }}$-tableau define its degree inductively by setting $\deg _e({\mathfrak t})=0$, if $n=0$, and if $n>0$ then

$$\begin{aligned} \deg _e({\mathfrak t})=\deg _e({\mathfrak t}{}_{\downarrow (n-1)})+d_A({\varvec{\mu }}), \end{aligned}$$

(2.3)

where $A={\mathfrak t}^{-1}(n)$. When e is understood we write $\deg ({\mathfrak t})$.

The following result shows that the degrees of the standard tableaux are almost completely determined by the Cartan matrix $(c_{ij})$ of $\varGamma _e$.

Lemma 2.4

(Brundan et al. [8, Proposition 3.13]) Suppose that ${\mathfrak s}$ and ${\mathfrak t}$ are standard tableaux such that ${\mathfrak s}\vartriangleright {\mathfrak t}={\mathfrak s}(r,r+1),$ where $1\le r<n$ and $\mathbf {i}\in I^n.$ Let $\mathbf {i}={{\mathrm{res}}}({\mathfrak s}).$ Then $\deg _e({\mathfrak s})=\deg _e({\mathfrak t})+c_{i_ri_{r+1}}.$

2.5 The Murphy basis and cyclotomic Specht modules

The cyclotomic Hecke algebra $\mathcal {H}^\varLambda _{n}$ is a cellular algebra with several different cellular bases. This section introduces one of these bases, the Murphy basis, and uses it to define the Specht modules and simple modules of $\mathcal {H}^\varLambda _{n}$.

Fix a multipartition ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$. Following [10, Definition 3.14] and [4, §6], if ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ define $m_{{\mathfrak s}{\mathfrak t}}=T_{d({\mathfrak s})^{-1}}m_{\varvec{\lambda }}T_{d({\mathfrak t})}$, where $m_{\varvec{\lambda }}= u_{\varvec{\lambda }}x_{\varvec{\lambda }}$ where

$$\begin{aligned} u_{\varvec{\lambda }}=\prod _{1\le l<\ell }\prod _{r=1}^{|\lambda ^{(1)}|+\cdots +|\lambda ^{(l)}|} \xi ^{-\kappa _{l+1}}(L_r-[\kappa _{l+1}]) \quad \text {and}\quad x_{\varvec{\lambda }}=\sum _{w\in \mathfrak S_{\varvec{\lambda }}}T_w. \end{aligned}$$

Let $*$ be the unique anti-isomorphism of $\mathcal {H}^\varLambda _{n}$ that fixes each of the generators $T_1,\ldots ,T_{n-1},L_1,\ldots ,L_n$ of Definition 2.1.

Theorem 2.5

([10, Theorem 3.26] and [4, Theorem 6.3]) The cyclotomic Hecke algebra $\mathcal {H}^\varLambda _{n}$ is free as an ${\mathcal {O}}$-module with cellular basis

$$\begin{aligned} \{m_{{\mathfrak s}{\mathfrak t}}~|~{\mathfrak s}, {\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\text { for }{\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}\} \end{aligned}$$

with respect to the weight poset $(\mathcal {P}^{\varLambda }_{n},\trianglerighteq )$ and automorphism $*$.

Proof

This theorem can be proved uniformly in all cases by modifying the argument of [10, Theorem 3.26], however, for future reference we explain how to deduce this result from the literature for the degenerate and non-degenerate algebras.

First suppose that $\xi =1$. Then the element $m_{\varvec{\lambda }}$, for ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$, coincides exactly with the corresponding elements defined for the non-degenerate cyclotomic Hecke algebras in [4, §6]. It follows that $\{m_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in \mathcal {P}^{\varLambda }_{n}\}$ is the Murphy basis of the degenerate cyclotomic Hecke algebra $\mathcal {H}^\varLambda _{n}$ defined in [4, §6] and that the theorem is just a restatement of [4, Theorem 6.3] when $\xi =1$.

Now suppose that $\xi \ne 1$ and, as in Remark 2.2, let $L_r'=(\xi -1)L_r+1$ be the ‘non-degenerate’ Jucys–Murphy elements for $\mathcal {H}^\varLambda _{n}$, for $1\le r\le n$. An application of the definitions shows that if $\kappa \in \mathbb {Z}$ then

$$\begin{aligned} \xi ^{-\kappa }(L_r-[\kappa ])=\frac{\xi ^{-\kappa }}{\xi -1}(L_r'-\xi ^{\kappa }). \end{aligned}$$

Therefore, $u_{\varvec{\lambda }}$ is a scalar multiple of the element $u_{\varvec{\lambda }}^+$ given by [10, Definition 3.1,3.5]. Consequently, if $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ then $m_{{\mathfrak s}{\mathfrak t}}$ is a scalar multiple of the corresponding Murphy basis element from [10, Definition 3.14]. Hence, the theorem is an immediate consequence of [10, Theorem 3.26] in the non-degenerate case.

Suppose that ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$. The (cyclotomic) Specht module $\underline{S}^{\varvec{\lambda }}$ is the cell module associated to ${\varvec{\lambda }}$ using the (ungraded) cellular basis $\{m_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$. We underline $\underline{S}^{\varvec{\lambda }}$ to emphasize that $\underline{S}^{\varvec{\lambda }}$ is not graded. When ${\mathcal {O}}$ is a field let $\underline{D}^{\varvec{\lambda }}=\underline{S}^{\varvec{\lambda }}/{{\mathrm{rad}}}\underline{S}^{\varvec{\lambda }}$ and set $\mathcal {K}^{\varLambda }_{n}=\{{\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}\,|\,\underline{D}^{\varvec{\lambda }}\ne 0\}$. Ariki [2] has given a combinatorial description of the set $\mathcal {K}^{\varLambda }_{n}$. By the theory of cellular algebras [12], $\{\underline{D}^{\varvec{\mu }}\,|\,{\varvec{\mu }}\in \mathcal {K}^{\varLambda }_{n}\}$ is a complete set of pairwise non-isomorphic irreducible $\mathcal {H}^\varLambda _{n}$-modules.

The following well-known fact is fundamental to all of the results in this paper.

Lemma 2.6

Suppose that $1\le r\le n$ and that ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }}),$ for ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}.$ Then

$$\begin{aligned} m_{{\mathfrak s}{\mathfrak t}}L_r \equiv [c_r({\mathfrak t})]m_{{\mathfrak s}{\mathfrak t}} +\sum _{\begin{array}{c} {\mathfrak v}\vartriangleright {\mathfrak t}\\ {\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }}) \end{array}}r_{\mathfrak v}m_{{\mathfrak s}{\mathfrak v}}\quad (\mathrm{mod}\ \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n), \end{aligned}$$

for some $r_{\mathfrak v}\in {\mathcal {O}}$.

Proof

If $\xi =1$ then this is a restatement of [4, Lemma 6.6]. If $\xi \ne 1$ then

$$\begin{aligned} m_{{\mathfrak s}{\mathfrak t}}L_r'=\xi ^{c_r({\mathfrak t})}m_{{\mathfrak s}{\mathfrak t}} +\sum _{{\mathfrak v}\vartriangleright {\mathfrak t}}r'_{\mathfrak v}m_{{\mathfrak s}{\mathfrak t}} \quad (\mathrm{mod}\ \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n), \end{aligned}$$

for some $r'_v\in {\mathcal {O}}$, by [17, Proposition 3.7] (and the notational translations given in the proof of Theorem 2.5). As $L_r=(L'_r-1)/(\xi -1)$ the result follows.

2.6 Cyclotomic quiver Hecke algebras

Brundan and Kleshchev [6] have given a very different presentation of $\mathcal {H}^\varLambda _{n}$. This presentation is more difficult to work with but it has the advantage of showing that $\mathcal {H}^\varLambda _{n}$ is a $\mathbb {Z}$-graded algebra.

Definition 2.7

(Brundan–Kleshchev [6]) Suppose that $n\ge 0$ and $e\in \{0,2,3,4,\dots \}$. The cyclotomic Khovanov–Lauda–Rouquier algebra, or cyclotomic quiver Hecke algebra, of weight $\varLambda$ and type $\varGamma _e$ is the unital associative ${\mathcal {O}}$-algebra $\mathcal {R}^\varLambda _{n}=\mathcal {R}^\varLambda _{n}({\mathcal {O}})$ with generators

$$\begin{aligned} \{\psi _1,\ldots ,\psi _{n-1}\} \cup \{ y_1,\ldots ,y_n \} \cup \{e(\mathbf {i})~|~\mathbf {i}\in I^n\} \end{aligned}$$

and relations

$$\begin{aligned}&y_1^{(\varLambda ,\alpha _{i_1})}e(\mathbf {i})=0,\quad e(\mathbf {i}) e(\mathbf {j})= \delta _{\mathbf {i}\mathbf {j}} e(\mathbf {i}),\quad {\sum _{\mathbf {i}\in I^n}} e(\mathbf {i})= 1,\nonumber \\&y_r e(\mathbf {i}) = e(\mathbf {i}) y_r,\quad \psi _r e(\mathbf {i})= e(s_r{\cdot }\mathbf {i}) \psi _r, \quad y_r y_s=y_s y_r, \end{aligned}$$

$$\begin{aligned}&\psi _r y_{r+1} e(\mathbf {i})=(y_r\psi _r+\delta _{i_ri_{r+1}})e(\mathbf {i}),\quad y_{r+1}\psi _re(\mathbf {i})=(\psi _r y_r+\delta _{i_ri_{r+1}})e(\mathbf {i}),\nonumber \\ \end{aligned}$$

(2.4)

$$\begin{aligned}&\displaystyle \psi _r y_s=y_s \psi _r,\quad \text {if }s \ne r,r+1,\nonumber \\&\displaystyle \psi _r \psi _s=\psi _s \psi _r,\quad \text {if }|r-s|>1, \end{aligned}$$

(2.5)

for $\mathbf {i},\mathbf {j}\in I^n$ and all admissible r and s. Moreover, $\mathcal {R}^\varLambda _{n}$ is naturally $\mathbb {Z}$-graded with degree function determined by

$$\begin{aligned} \deg e(\mathbf {i})=0,\quad \deg y_r=2\quad \text {and}\quad \deg \psi _s e(\mathbf {i})=-c_{i_s,i_{s+1}}, \end{aligned}$$

for $1\le r\le n, 1\le s<n$ and $\mathbf {i}\in I^n$.

Remark 2.8

The presentation of $\mathcal {R}^\varLambda _{n}$ given in Definition 2.7 differs by a choice of signs with the definition given in [6, Theorem 1.1]. The presentation of $\mathcal {R}^\varLambda _{n}$ given above agrees with that used in [23] as the orientation of the quiver is reversed in [23].

The connection between the cyclotomic quiver Hecke algebras of type $\varGamma _e$ and the cyclotomic Hecke algebras of type $G(\ell ,1,n)$ is given by the following remarkable result of Brundan and Kleshchev.

Theorem 2.9

(Brundan–Kleshchev’s isomorphism theorem [6, Theorem 1.1]) Suppose that ${\mathcal {O}}=K$ is a field,  $\xi \in K$ as above,  and that $\varLambda =\varLambda ({\varvec{\kappa }})$. Then there is an isomorphism of algebras $\underline{\mathcal {R}}^\varLambda _{n}\cong \mathcal {H}^\varLambda _{n}$.

Rouquier [31, Corollary 3.20] has, independently, given another proof of Theorem 2.9.

We prove a stronger version of Theorem 2.9 in Theorem 4.26 below. For now we note the following simple corollary of Theorem 2.9. Recall that a choice of multicharge ${\varvec{\kappa }}$ determines a dominant weight $\varLambda _e({\varvec{\kappa }})$.

Corollary 2.10

Suppose that $n\ge 0, {\varvec{\kappa }}=(\kappa _1,\ldots ,\kappa _\ell )\in \mathbb {Z}^\ell$ and that

$$\begin{aligned} e>\max \{n+\kappa _k-\kappa _l~|~1\le k,l\le \ell \}. \end{aligned}$$

Fix invertible scalars $\xi _0\in K$ and $\xi _e\in K$ such that $\xi _0$ is not a root of unity and $\xi _e$ is a primitive eth root of unity. Then the cyclotomic Hecke algebras $\mathcal {H}^{\varLambda _0(\kappa )}_{K,\xi _0}$ and $\mathcal {H}^{\varLambda _e({\varvec{\kappa }})}_{\fancyscript{K},\xi _e}$ are isomorphic $\mathbb {Z}$-graded K-algebras.

Proof

Let $\mathcal {R}^\varLambda _{n}(0)\cong \mathcal {H}_n(K,\xi _0,{\varvec{\kappa }})$ and $\mathcal {R}^\varLambda _{n}(e)\cong \mathcal {H}_n(K,\xi _e,{\varvec{\kappa }})$ be the corresponding cyclotomic quiver Hecke algebras as in Theorem 2.9. By [14, Lemma 4.1], $e(\mathbf {i})\ne 0$ if and only if $\mathbf {i}={{\mathrm{res}}}({\mathfrak t})$, for some standard tableau ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$. The definition of e ensures that if $\mathbf {i}=\mathbf {i}^{{\mathfrak t}}$ then $i_r=i_{r+1}$ or $i_r=i_{r+1}\pm 1$ if and only if $i_r\equiv i_{r+1}\text { }(\text {mod } e)$ or $i_r\equiv i_{r+1}\pm 1\text { }(\text {mod } e)$. Therefore, $\mathcal {R}^\varLambda _{n}(0)\cong \mathcal {R}^\varLambda _{n}(e)$ arguing directly from the presentations of the cyclotomic quiver Hecke algebras given in Definition 2.7. Hence, the result follows by Theorem 2.9.

Therefore, without loss of generality, we may assume that $e>0$. In the appendix we show how to modify the results and definitions in this paper to cover the case when $e=0$ directly.

Under the assumptions of the corollary we note that the algebras $\mathcal {H}^\varLambda _{K,\xi _0}$ and $\mathcal {H}^\varLambda _{K,\xi _e}$ are Morita equivalent by the main result of [11]. That these algebras are actually isomorphic is another miracle provided by Brundan and Kleshchev’s isomorphism theorem.

3 Seminormal forms for Hecke algebras

In this chapter we develop the theory of seminormal forms in a slightly more general context than appears in the literature. In particular, in this paper a seminormal basis will be a basis for $\mathcal {H}^\varLambda _{n}$ rather than a basis of a Specht module of $\mathcal {H}^\varLambda _{n}$. We also treat all of the variations of the seminormal bases simultaneously as this will give us the flexibility to change seminormal forms when we use them in the next chapter to study the connections between $\mathcal {H}^\varLambda _{n}$ and the cyclotomic quiver Hecke algebra $\mathcal {R}^\varLambda _{n}$.

3.1 Content functions and the Gelfand–Zetlin algebra

Underpinning Brundan and Kleshchev’s isomorphism theorem (Theorem 2.9) is the decomposition of any $\mathcal {H}^\varLambda _{n}$-module into a direct sum of generalised eigenspaces for the Jucys–Murphy elements $L_1,\ldots ,L_n$. This section studies the action of the Jucys–Murphy elements on $\mathcal {H}^\varLambda _{n}$. The results in this section are well-known, at least to experts, but they are needed in the sequel.

The content of the node $\gamma =(l,r,c)$ is the integer

$$\begin{aligned} {{\mathrm{c}}}_\gamma =\kappa _l-r+c \end{aligned}$$

If ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ is a standard ${\varvec{\lambda }}$-tableau and $1\le k\le n$ then the content of k in ${\mathfrak t}$ is ${{\mathrm{c}}}_k({\mathfrak t})={{\mathrm{c}}}_\gamma$, where ${\mathfrak t}(\gamma )=k$ for $\gamma \in [\![{\varvec{\lambda }}]\!]$.

Definition 3.1

Let $\fancyscript{K}$ be a field and suppose that $t\in \fancyscript{K}^\times$ is an invertible element of $\fancyscript{K}$. The pair $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ if

$$\begin{aligned}{}[n]^!_t\prod _{1\le l<m\le \ell }\prod _{-n<d<n}[\kappa _l-\kappa _m+d]_t\in {\fancyscript{K}}^{\times }. \end{aligned}$$

Fix a multicharge ${\varvec{\kappa }}\in \mathbb {Z}^\ell$ and let $\mathcal {H}^\varLambda _n({\mathcal {O}})$ be the Hecke algebra defined over ${\mathcal {O}}$ with parameter t. In spite of our notation, note that $\mathcal {H}^\varLambda _n({\mathcal {O}})$ depends only on ${\varvec{\kappa }}$ and not directly on $\varLambda =\varLambda _e({\varvec{\kappa }})$. Let $\fancyscript{K}$ be a field that contains the field of fractions of ${\mathcal {O}}$. Then $\mathcal {H}^\varLambda _n(\fancyscript{K})=\mathcal {H}^\varLambda _n({\mathcal {O}})\otimes _{\mathcal {O}}\fancyscript{K}$.

Throughout this chapter we are going to work with the Hecke algebras $\mathcal {H}^\varLambda _n({\mathcal {O}})$ and $\mathcal {H}^\varLambda _n(\fancyscript{K})=\mathcal {H}^\varLambda _n({\mathcal {O}})\otimes _{\mathcal {O}}\fancyscript{K}$, however, we have in mind the situation of Theorem 2.9. By assumption $e>0$, so we can replace the multicharge ${\varvec{\kappa }}$ with $(\kappa _1+a_1e,\kappa _2+a_2e,\ldots ,\kappa _\ell +a_\ell e)$, for any integers $a_1,\ldots ,a_\ell \in \mathbb {Z}$, without changing the dominant weight $\varLambda =\varLambda _e({\varvec{\kappa }})$. In view of Definition 3.1 we therefore assume that

$$\begin{aligned} \kappa _l-\kappa _{l+1}\ge n,\quad \text {for }1\le l<\ell . \end{aligned}$$

(3.1)

Until further notice, we fix a multicharge ${\varvec{\kappa }}\in \mathbb {Z}^\ell$ satisfying (3.1) and consider the algebra $\mathcal {H}^\varLambda _n({\mathcal {O}})$ with parameter t.

Although we do not need this, we remark that it follows from [1] and [4, Theorem 6.11] that $\mathcal {H}^\varLambda _{n}(\fancyscript{K},t)$ is semisimple if and only if $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$. Our main use of the separation condition is the following fundamental fact that is easily proved by induction on n; see, for example [17, Lemma 3.12].

Lemma 3.2

Suppose that $\fancyscript{K}$ is an integral domain and $t\in \fancyscript{K}^\times$ is invertible. Then the following are equivalent : 

• (a) The pair $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n}),$

• (b) If ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ then ${\mathfrak s}={\mathfrak t}$ if and only if $[c_r({\mathfrak s})]=[c_r({\mathfrak t})],$ for $1\le r\le n.$

Following [30], define the Gelfand–Zetlin subalgebra of $\mathcal {H}^\varLambda _{n}$ to be the algebra $\fancyscript{L}({\mathcal {O}})=\langle L_1,\ldots ,L_n\rangle$. The aim of this section is to understand the semisimple representation theory of $\fancyscript{L}=\fancyscript{L}({\mathcal {O}})$. It follows from Definition 2.1 that $\fancyscript{L}$ is a commutative subalgebra of $\mathcal {H}^\varLambda _{n}$.

If ${\mathcal {O}}$ is an integral domain then it follows from Lemma 2.6 that, as an $(\fancyscript{L},\fancyscript{L})$-bimodule, $\mathcal {H}^\varLambda _{n}({\mathcal {O}})$ has a composition series with composition factors that are ${\mathcal {O}}$-free of rank 1 upon which $L_r$ acts as multiplication by $[c_r({\mathfrak s})]$ from the left and as multiplication by $[c_r({\mathfrak t})]$ from the right. Obtaining a better description of $\fancyscript{L}$, and of $\mathcal {H}^\varLambda _{n}$ as an $(\fancyscript{L},\fancyscript{L})$-bimodule, in the non-semisimple case is likely to be important. For example, the dimension of $\fancyscript{L}$ over a field is not known in general.

Proposition 3.3

(cf. [3, Proposition 3.17]) Suppose that $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n}),$ where $\fancyscript{K}$ is a field and $0\ne t\in \fancyscript{K}.$ Then $\mathcal {H}^\varLambda _{n}(\fancyscript{K})$ is a semisimple $(\fancyscript{L},\fancyscript{L})$-bimodule with decomposition

$$\begin{aligned} \mathcal {H}^\varLambda _{n}(\fancyscript{K})=\bigoplus _{\begin{array}{c} {\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}\\ {\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }}) \end{array}}H_{{\mathfrak s}{\mathfrak t}}, \end{aligned}$$

where $H_{{\mathfrak s}{\mathfrak t}}=\{h\in \mathcal {H}^\varLambda _{n}\,|\,L_r h=[c_r({\mathfrak s})]h\text { and }hL_r=[c_r({\mathfrak t})]h, \text { for } 1\le r\le n\}$ is one dimensional.

Proof

By Lemma 2.6, the Jucys–Murphy elements $L_1,\ldots ,L_n$ are a family of JM-elements for $\mathcal {H}^\varLambda _{n}$ in the sense of [28, Definition 2.4]. Therefore, the result is a special case of [28, Theorem 3.7].

Key to the proof of the results in [28] are the following elements that have their origins in the work of Murphy [29]. For ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ define

$$\begin{aligned} F_{\mathfrak t}=\prod _{k=1}^n\prod _{\begin{array}{c} c\in \fancyscript{C}\\ {[{{\mathrm{c}}}_k({\mathfrak t})]\ne [c]} \end{array}} \frac{L_k-[c]}{[{{\mathrm{c}}}_k({\mathfrak t})]-[c]} \end{aligned}$$

(3.2)

where $\fancyscript{C}=\{c_r({\mathfrak t})\,|\,1\le r\le n\text { and }{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})\}$ is the set of the possible contents that can appear in a standard tableau of size n. By definition, $F_{\mathfrak t}\in \fancyscript{L}(\fancyscript{K})$ and it follows directly from Proposition 3.3 that if $h_{{\mathfrak u}{\mathfrak v}}\in H_{{\mathfrak u}{\mathfrak v}}$ then

$$\begin{aligned} F_{\mathfrak s}h_{{\mathfrak u}{\mathfrak v}} F_{\mathfrak t}= \delta _{{\mathfrak s}{\mathfrak u}}\delta _{{\mathfrak v}{\mathfrak t}}h_{{\mathfrak s}{\mathfrak t}}, \end{aligned}$$

(3.3)

for all $({\mathfrak s},{\mathfrak t}),({\mathfrak u},{\mathfrak v})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$. Therefore, $H_{{\mathfrak s}{\mathfrak t}}=F_{\mathfrak s}\mathcal {H}^\varLambda _{n}F_{\mathfrak t}$.

By Proposition 3.3 we can write $1=\sum _{{\mathfrak s},{\mathfrak t}}e_{{\mathfrak s}{\mathfrak t}}$ for unique $e_{{\mathfrak s}{\mathfrak t}}\in H_{{\mathfrak s}{\mathfrak t}}$. Since $F_{\mathfrak t}=F_{\mathfrak t}^{*}$, the last displayed equation implies that $F_{\mathfrak t}=e_{{\mathfrak t}{\mathfrak t}}\in H_{{\mathfrak t}{\mathfrak t}}$ is an idempotent. Consequently,

$$\begin{aligned} \fancyscript{L}(\fancyscript{K})=\bigoplus _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})} H_{{\mathfrak t}{\mathfrak t}}=\bigoplus _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})}\fancyscript{K}F_{\mathfrak t}. \end{aligned}$$

In particular, $F_{\mathfrak t}$ is a primitive idempotent in $\fancyscript{L}(\fancyscript{K})$. If follows that $\fancyscript{L}(\fancyscript{K})$ is a split semisimple algebra of dimension $\#{\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$.

3.2 Seminormal forms

Seminormal bases for $\mathcal {H}^\varLambda _{n}$ are well-known in the literature, having their origins in the work of Young [37]. In this section we classify the seminormal bases of $\mathcal {H}^\varLambda _{n}$. This characterisation of seminormal forms appears to be new, even in the special case of the symmetric groups, although some of the details will be familiar to experts.

Throughout this section we assume that $\fancyscript{K}$ is a field, $0\ne t\in \fancyscript{K}$ and that $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$. Recall the decomposition $\mathcal {H}^\varLambda _{n}=\bigoplus _{({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})}H_{{\mathfrak s}{\mathfrak t}}$ from Proposition 3.3.

Define an anti-involution on an algebra A to be an algebra anti-automorphism of A of order 2.

Definition 3.4

Suppose that $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and let $\iota$ be an anti-involution on $\mathcal {H}^\varLambda _n(\fancyscript{K})$. An $\iota$-seminormal basis of $\mathcal {H}^\varLambda _{n}(\fancyscript{K})$ is a basis of the form $\{f_{{\mathfrak s}{\mathfrak t}}\,|\,f_{{\mathfrak s}{\mathfrak t}}=\iota (f_{{\mathfrak t}{\mathfrak s}})\in H_{{\mathfrak s}{\mathfrak t}}\text { for }({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$.

Recall that $*$ is the unique anti-involution of $\mathcal {H}^\varLambda _n(\fancyscript{K})$ that fixes each of the generators $T_1,\ldots ,T_{n-1},L_1,\ldots ,L_n$. Then $m_{{\mathfrak s}{\mathfrak t}}^{*}=m_{{\mathfrak t}{\mathfrak s}}$, for all $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$. The assumption that $f_{{\mathfrak s}{\mathfrak t}}^{*}=f_{{\mathfrak t}{\mathfrak s}}$ is not essential for what follows but it is natural because we want to work within the framework of cellular algebras.

In order to describe the action of $\mathcal {H}^\varLambda _{n}$ on its seminormal bases, if ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ then define the integers

$$\begin{aligned} \rho _r({\mathfrak t})=c_r({\mathfrak t})-c_{r+1}({\mathfrak t}),\quad \text {for }1\le r<n. \end{aligned}$$

(3.4)

Then $\rho _r({\mathfrak t})$ is the ‘axial distance’ between r and $r+1$ in the tableau ${\mathfrak t}$.

Definition 3.5

A $*$-seminormal coefficient system for $\mathcal {H}^\varLambda _n(\fancyscript{K})$ is a set of scalars ${\varvec{\alpha }}=\{\alpha _r({\mathfrak s})\,|\, 1\le r<n \text { and } {\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})\}$ in $\fancyscript{K}$ such that if ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and $1\le r<n$ then:

• (a) $\alpha _r({\mathfrak t})\alpha _{r+1}({\mathfrak t}s_r)\alpha _r({\mathfrak t}s_rs_{r+1}) =\alpha _{r+1}({\mathfrak t})\alpha _r({\mathfrak t}s_{r+1})\alpha _{r+1}({\mathfrak t}s_{r+1}s_r)$ if $r<n-1$,

• (b) $\alpha _r({\mathfrak t})\alpha _k({\mathfrak t}s_r)=\alpha _k({\mathfrak t})\alpha _r({\mathfrak t}s_k)$ if $1\le k<n$ and $|r-k|>1$,

• (c) if ${\mathfrak v}={\mathfrak t}(r,r+1)$ then $\alpha _r({\mathfrak v})=0$ if ${\mathfrak v}\notin {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and otherwise

  $$\begin{aligned} \alpha _r({\mathfrak t})\alpha _r({\mathfrak v})= \dfrac{[1+\rho _r({\mathfrak t})][1+\rho _r({\mathfrak v})]}{[\rho _r({\mathfrak t})][\rho _r({\mathfrak v})]}. \end{aligned}$$

We will see that conditions (a) and (b) correspond to the braid relations satisfied by $T_1,\ldots ,T_{n-1}$ and that (c) corresponds to the quadratic relations. Quite surprisingly, as the proof of Theorem 3.13 below shows, Definition 3.5(c) also encodes the KLR grading on $\mathcal {H}^\varLambda _{n}$.

Usually, we omit the $*$ and simply call ${\varvec{\alpha }}$ a seminormal coefficient system.

Example 3.6

A nice ‘rational’ seminormal coefficient system is given by

$$\begin{aligned} \alpha _r({\mathfrak t})={\left\{ \begin{array}{ll} \frac{[1+\rho _r({\mathfrak t})]}{[\rho _r({\mathfrak t})]}, &{} \text {if }{\mathfrak t}(r,r+1) \text { is standard},\\ 0,&{}\text {otherwise}, \end{array}\right. } \end{aligned}$$

for ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and $1\le r<n$.

Example 3.7

By Proposition 3.11 below, the following seminormal coefficient system is associated with the Murphy basis of $\mathcal {H}^\varLambda _{n}$: if ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ set ${\mathfrak v}={\mathfrak t}(r,r+1)$ and define

$$\begin{aligned} \alpha _r({\mathfrak t})={\left\{ \begin{array}{ll} 1&{} \text {if }{\mathfrak v}\text { is standard and }{\mathfrak t}\vartriangleright {\mathfrak v},\\ \frac{[1+\rho _r({\mathfrak t})][1+\rho _r({\mathfrak v})]}{[\rho _r({\mathfrak t})][\rho _r({\mathfrak v})]}, &{} \text {if }{\mathfrak v}\text { is standard and }{\mathfrak v}\vartriangleright {\mathfrak t},\\ 0,&{}\text {otherwise}, \end{array}\right. } \end{aligned}$$

for $1\le r<n$.

Another seminormal coefficient system, which is particularly well adapted to Brundan and Kleshchev’s Graded Isomorphism Theorem 2.9, is given in Sect. 5.1.

Lemma 3.8

Suppose that $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and that $\{f_{{\mathfrak s}{\mathfrak t}}\}$ is a seminormal basis of $\mathcal {H}^\varLambda _{n}$. Then there exists a unique seminormal coefficient system ${\varvec{\alpha }}$ such that if $1\le r<n$ and $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ then

$$\begin{aligned} f_{{\mathfrak s}{\mathfrak t}}T_r = \alpha _r({\mathfrak t})f_{{\mathfrak s}{\mathfrak v}}-\frac{1}{[\rho _r({\mathfrak t})]}f_{{\mathfrak s}{\mathfrak t}}, \end{aligned}$$

where ${\mathfrak v}={\mathfrak t}(r,r+1)$ and $f_{{\mathfrak s}{\mathfrak t}}=0$ if $({\mathfrak s},{\mathfrak t})\notin {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$.

Proof

The uniqueness statement is automatic, since $\{f_{{\mathfrak s}{\mathfrak t}}\}$ is a basis of $\mathcal {H}^\varLambda _n(\fancyscript{K})$, so we need to prove that such a seminormal coefficient system ${\varvec{\alpha }}$ exists.

Fix $({\mathfrak s},t)\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ and $1\le r<n$ and write

$$\begin{aligned} f_{{\mathfrak s}{\mathfrak t}}T_r = \sum _{({\mathfrak u},{\mathfrak v})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})}a_{{\mathfrak u}{\mathfrak v}}f_{{\mathfrak u}{\mathfrak v}}, \end{aligned}$$

for some $a_{{\mathfrak u}{\mathfrak v}}\in \fancyscript{K}$. Multiplying on the left by $F_{\mathfrak s}$ it follows that $a_{{\mathfrak u}{\mathfrak v}}\ne 0$ only if ${\mathfrak u}={\mathfrak s}$. If $k\ne r,r+1$ then $L_k$ commutes with $T_r$ so it follows $a_{{\mathfrak s}{\mathfrak v}}\ne 0$ only if $[c_k({\mathfrak v})]=[c_k({\mathfrak t})]$, for $k\ne r,r+1$. Using Definition 3.1, and arguing as in Lemma 3.2, this implies that $a_{{\mathfrak s}{\mathfrak v}}\ne 0$ only if ${\mathfrak v}\in \{{\mathfrak t},{\mathfrak t}(r,r+1)\}$. Therefore, we can write

$$\begin{aligned} f_{{\mathfrak s}{\mathfrak t}}T_r=\alpha _r({\mathfrak t})f_{{\mathfrak s}{\mathfrak v}}+\alpha '_r({\mathfrak t})f_{{\mathfrak s}{\mathfrak t}}, \end{aligned}$$

for some $\alpha _r({\mathfrak t}),\alpha '_r({\mathfrak t})\in \fancyscript{K}$, where ${\mathfrak v}={\mathfrak t}(r,r+1)$. (Here, and below, we adopt the convention that $f_{{\mathfrak s}{\mathfrak v}}=0$ if either of ${\mathfrak s}$ or ${\mathfrak v}$ is not standard.) By Definition 2.1, $T_rL_r=L_{r+1}(T_r-t+1)-1$, so multiplying both sides of the last displayed equation on the right by $L_r$ and comparing the coefficient of $f_{{\mathfrak s}{\mathfrak t}}$ on both sides shows that

$$\begin{aligned}{}[c_{r+1}({\mathfrak t})](\alpha '_r({\mathfrak t})-t+1)-1 = \alpha '_r({\mathfrak t})[c_r({\mathfrak t})]. \end{aligned}$$

Hence, $\alpha '_r({\mathfrak t})=-1/[\rho _r({\mathfrak t})]$ as claimed. If ${\mathfrak v}$ is not standard then we set $\alpha _r({\mathfrak t})=0$. If ${\mathfrak v}$ is standard then comparing the coefficient of $f_{{\mathfrak s}{\mathfrak t}}$ on both sides of

$$\begin{aligned} \left( \alpha _r({\mathfrak t})f_{{\mathfrak s}{\mathfrak v}}-\frac{1}{[\rho _r({\mathfrak t})]}f_{{\mathfrak s}{\mathfrak t}}\right) T_r =f_{{\mathfrak s}{\mathfrak t}}T_r^2=f_{{\mathfrak s}{\mathfrak t}}((t-1)T_r+t) \end{aligned}$$

shows that $\alpha _r({\mathfrak t})\alpha _r({\mathfrak v})=\tfrac{[1+\rho _r({\mathfrak t})][1+\rho _r({\mathfrak v})]}{[\rho _r({\mathfrak t})][\rho _r({\mathfrak v})]}$ in accordance with Definition 3.5(c).

If $1\le r<s-1<n-1$ and $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ then $(f_{{\mathfrak s}{\mathfrak t}}T_r)T_s=f_{{\mathfrak s}{\mathfrak t}}(T_rT_s)=f_{{\mathfrak s}{\mathfrak t}}(T_sT_r)=(f_{{\mathfrak s}{\mathfrak t}}T_s)T_r$. By a direct calculation, we can deduce that $\alpha _r({\mathfrak t})\alpha _s({\mathfrak t}(r,r+1))=\alpha _s({\mathfrak t})\alpha _r({\mathfrak t}(s,s+1))$.

Finally, it remains to show that Definition 3.5(a) holds. If $1\le r<n$ then $T_rT_{r+1}T_r=T_{r+1}T_rT_{r+1}$ by Definition 2.1. On the other hand, if we set ${\mathfrak t}_1={\mathfrak t}(r,r+1), {\mathfrak t}_2={\mathfrak t}(r+1,r+2), {\mathfrak t}_{12}={\mathfrak t}_1(r+1,r+2), {\mathfrak t}_{21}={\mathfrak t}_2(r,r+1)$ and ${\mathfrak t}_{121}={\mathfrak t}_{212}={\mathfrak t}(r,r+2)$ then direct calculation shows that $0=f_{{\mathfrak s}{\mathfrak t}}(T_rT_{r+1}T_r-T_{r+1}T_rT_{r+1})$ is equal to

$$\begin{aligned}&-\left( \frac{1}{[\rho _r({\mathfrak t})]^2[\rho _{r+1}({\mathfrak t})]} -\frac{1}{[\rho _r({\mathfrak t})][\rho _{r+1}({\mathfrak t})]^2} +\frac{\alpha _r({\mathfrak t})\alpha _r({\mathfrak t}_1)}{[\rho _{r+1}({\mathfrak t}_1)]} -\frac{\alpha _{r+1}({\mathfrak t})\alpha _{r+1}({\mathfrak t}_2)}{[\rho _r({\mathfrak t}_2)]}\right) f_{{\mathfrak s}{\mathfrak t}}\\&\quad +\,\alpha _r({\mathfrak t})\left( \frac{1}{[\rho _r({\mathfrak t}_1)][\rho _{r+1}({\mathfrak t}_1)]} +\frac{1}{[\rho _r({\mathfrak t})][\rho _{r+1}({\mathfrak t})]} -\frac{1}{[\rho _{r+1}({\mathfrak t})][\rho _{r+1}({\mathfrak t}_1)]}\right) f_{{\mathfrak s}{\mathfrak t}_1}\\&\quad -\,\alpha _{r+1}({\mathfrak t})\left( \frac{1}{[\rho _r({\mathfrak t}_2)][\rho _{r+1}({\mathfrak t}_2)]} +\frac{1}{[\rho _r({\mathfrak t})][\rho _{r+1}({\mathfrak t})]} -\frac{1}{[\rho _r({\mathfrak t})][\rho _r({\mathfrak t}_2)]}\right) f_{{\mathfrak s}{\mathfrak t}_2}\\&\quad -\,\alpha _r({\mathfrak t})\alpha _{r+1}({\mathfrak t}_1)\left( \frac{1}{[\rho _r({\mathfrak t}_{12})]} -\frac{1}{[\rho _{r+1}({\mathfrak t})]}\right) f_{{\mathfrak s}{\mathfrak t}_{12}}\\&\quad +\,\alpha _{r+1}({\mathfrak t})\alpha _r({\mathfrak t}_2)\left( \frac{1}{[\rho _{r+1}({\mathfrak t}_{21})]} -\frac{1}{[\rho _r({\mathfrak t})]}\right) f_{{\mathfrak s}{\mathfrak t}_{21}}\\&\quad +\,(\alpha _r({\mathfrak t})\alpha _{r+1}({\mathfrak t}_1)\alpha _r({\mathfrak t}_{12}) -\alpha _{r+1}({\mathfrak t})\alpha _r({\mathfrak t}_2)\alpha _{r+1}({\mathfrak t}_{21}))f_{{\mathfrak s}{\mathfrak t}_{121}}. \end{aligned}$$

By our conventions, if any tableau ${\mathfrak t}_?$ is not standard then $f_{{\mathfrak s}{\mathfrak t}_?}$ and the corresponding ${\varvec{\alpha }}$-coefficient are both zero. As the coefficient of $f_{{\mathfrak s}{\mathfrak t}_{121}}$ in the last displayed equation is zero it follows that Definition 3.5(a) holds. Consequently, ${\varvec{\alpha }}=\{\alpha _r({\mathfrak t})\}$ is a seminormal coefficient system, completing the proof. (It is not hard to see, using Definition 3.5 and identities like $\rho _r({\mathfrak t}_1)=-\rho _r({\mathfrak t})$ and $\rho _r({\mathfrak t}_{12})=\rho _{r+1}({\mathfrak t})$, that the remaining coefficients in the last displayed equation are automatically zero.)

Lemma 3.8 really says that acting from the right on a seminormal basis determines a seminormal coefficient system. Similarly, the left action on a seminormal basis determines a seminormal coefficient system. In general, the seminormal coefficient systems attached to the left and right actions will be different, however, because we are assuming that our seminormal bases are $*$-invariant these left and right coefficient systems coincide. Thus, for $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ and $1\le r<n$ we also have $T_rf_{{\mathfrak s}{\mathfrak t}} = \alpha _r({\mathfrak s})f_{{\mathfrak u}{\mathfrak t}}-\frac{1}{[\rho _r({\mathfrak s})]}f_{{\mathfrak s}{\mathfrak t}}$, where ${\mathfrak u}={\mathfrak s}(r,r+1)$.

Exactly as eigenvectors are not uniquely determined by their eigenvalues, seminormal bases are not uniquely determined by seminormal coefficient systems. We now fully characterize seminormal bases—and prove a converse to Lemma 3.8.

Recall that a set of idempotents in an algebra is complete if they sum to 1.

Theorem 3.9

(The seminormal basis theorem) Suppose that $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and that ${\varvec{\alpha }}$ is a seminormal coefficient system for $\mathcal {H}^\varLambda _n(\fancyscript{K})$. Then $\mathcal {H}^\varLambda _n(\fancyscript{K})$ has a $*$-seminormal basis $\{f_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$ such that if $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ then

$$\begin{aligned} f_{{\mathfrak s}{\mathfrak t}}^{*}=f_{{\mathfrak t}{\mathfrak s}},\quad f_{{\mathfrak s}{\mathfrak t}}L_k=[c_k({\mathfrak t})]f_{{\mathfrak s}{\mathfrak t}}\quad \text {and}\quad f_{{\mathfrak s}{\mathfrak t}}T_r=\alpha _r({\mathfrak t})f_{{\mathfrak s}{\mathfrak v}}-\frac{1}{[\rho _r({\mathfrak t})]}f_{{\mathfrak s}{\mathfrak t}}, \end{aligned}$$

(3.5)

where ${\mathfrak v}={\mathfrak t}(r,r+1)$ and $f_{{\mathfrak s}{\mathfrak v}}=0$ if ${\mathfrak v}$ is not standard. Moreover,  there exist non-zero scalars $\gamma _{\mathfrak t}\in \fancyscript{K},$ for ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n}),$ such that

$$\begin{aligned} F_{\mathfrak u}f_{{\mathfrak s}{\mathfrak t}} F_{\mathfrak v}= \delta _{{\mathfrak u}{\mathfrak s}}\delta _{{\mathfrak t}{\mathfrak v}}f_{{\mathfrak s}{\mathfrak t}},\quad f_{{\mathfrak s}{\mathfrak t}}f_{{\mathfrak u}{\mathfrak v}}=\delta _{{\mathfrak t}{\mathfrak u}}\gamma _{\mathfrak t}f_{{\mathfrak s}{\mathfrak v}},\quad \text {and}\quad F_{\mathfrak t}=\frac{1}{\gamma _{\mathfrak t}}f_{{\mathfrak t}{\mathfrak t}}. \end{aligned}$$

(3.6)

Furthermore,  $\{F_{\mathfrak t}\,|\,{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})\}$ is a complete set of pairwise orthogonal primitive idempotents. In particular,  every irreducible $\mathcal {H}^\varLambda _n(\fancyscript{K})$-module is isomorphic to $F_{\mathfrak s}\mathcal {H}^\varLambda _n(\fancyscript{K}),$ for some ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n}),$ and $F_{\mathfrak s}\mathcal {H}^\varLambda _n(\fancyscript{K})\cong F_{\mathfrak u}\mathcal {H}^\varLambda _n(\fancyscript{K})$ if and only if ${{\mathrm{Shape}}}({\mathfrak s})={{\mathrm{Shape}}}({\mathfrak u}).$

Finally,  the basis $\{f_{{\mathfrak s}{\mathfrak t}}\,|\,{\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\text { for }{\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}\}$ is uniquely determined by the choice of seminormal coefficient system ${\varvec{\alpha }}$ and the scalars $\{\gamma _{{\mathfrak t}^{\varvec{\lambda }}}\,|\,{\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}\}\subseteq \fancyscript{K}^\times .$

Proof

For each ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$ fix an arbitrary pair of tableaux and a non-zero element $f_{{\mathfrak s}{\mathfrak t}}\in H_{{\mathfrak s}{\mathfrak t}}$. Then $f_{{\mathfrak s}{\mathfrak t}}$ is a simultaneous eigenvector for all of the elements of $\fancyscript{L}$, where they act from the left and from the right.

Now, suppose that $1\le r<n$ and that ${\mathfrak v}={\mathfrak t}(r,r+1)$ is standard. Then $\alpha _r({\mathfrak t})\ne 0$ so we can set $f_{{\mathfrak s}{\mathfrak v}}=\frac{1}{\alpha _r({\mathfrak t})}f_{{\mathfrak s}{\mathfrak t}}(T_r+\frac{1}{[\rho _r({\mathfrak t})]})$. Equivalently, $f_{{\mathfrak s}{\mathfrak t}}T_r=\alpha _r({\mathfrak t})f_{{\mathfrak s}{\mathfrak v}}-\frac{1}{[\rho _r({\mathfrak t})]}f_{{\mathfrak s}{\mathfrak t}}$. Then using the relations in $\mathcal {H}^\varLambda _n(\fancyscript{K})$ and the defining properties of the seminormal coefficient system ${\varvec{\alpha }}$, it is straightforward to check that $f_{{\mathfrak s}{\mathfrak v}}L_k=[c_k({\mathfrak v})]f_{{\mathfrak s}{\mathfrak v}}$, so that $f_{{\mathfrak s}{\mathfrak v}}\in H_{{\mathfrak s}{\mathfrak v}}$. Moreover, $f_{{\mathfrak s}{\mathfrak v}}\ne 0$ since $f_{{\mathfrak s}{\mathfrak t}}=\frac{1}{\alpha _r({\mathfrak v})}f_{{\mathfrak s}{\mathfrak v}}(T_r+\frac{1}{[\rho _r({\mathfrak v})]})$.

More generally, it is easy to see that if ${\mathfrak v}$ is any ${\varvec{\lambda }}$-tableau then there is a sequence of standard tableaux ${\mathfrak v}_0={\mathfrak s}, {\mathfrak v}_1,\ldots ,{\mathfrak v}_z={\mathfrak v}$ such that ${\mathfrak v}_{i+1}={\mathfrak v}_i(r_i,r_i+1)$, for some integers $1\le r_i<n$. Therefore, continuing in this way it follows that given two tableaux ${\mathfrak u},{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ we can define non-zero elements $f_{{\mathfrak u}{\mathfrak v}}\in H_{{\mathfrak u}{\mathfrak v}}$ that satisfy (3.5). It follows that, once $f_{{\mathfrak s}{\mathfrak t}}$ is fixed, there is at most one choice of elements $\{f_{{\mathfrak u}{\mathfrak v}}\,|\,{\mathfrak u},{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\}$, such that (3.5) holds.

To complete the proof that the seminormal coefficient system determines a seminormal basis we need to check that the elements $f_{{\mathfrak u}{\mathfrak v}}$ from the last paragraph are well-defined. That is, we need to show that $f_{{\mathfrak u}{\mathfrak v}}$ is independent of the choice of the sequences of simple transpositions that link ${\mathfrak u}$ and ${\mathfrak v}$ to ${\mathfrak s}$ and ${\mathfrak t}$, respectively. Equivalently, we need to prove that the action of $\mathcal {H}^\varLambda _n(\fancyscript{K})$ given by (3.5) respects the relations of $\mathcal {H}^\varLambda _n(\fancyscript{K})$. Using (3.5), all of the relations in Definition 2.1 are easy to check except for the braid relations of length three that hold by virtue of the argument of Lemma 3.8. Hence, by choosing elements $f_{{\mathfrak s}{\mathfrak t}}\in H_{{\mathfrak s}{\mathfrak t}}$, for $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2({\varvec{\lambda }})$ and ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$, the seminormal coefficient system determines a unique seminormal basis.

Using (3.3) it is straightforward to prove (3.6) so we leave these details to the reader; cf. [28, Theorem 3.16]. In particular, this shows that $F_{\mathfrak s}=\frac{1}{\gamma _{\mathfrak s}}f_{{\mathfrak s}{\mathfrak s}}$ is an idempotent. To show that $F_{\mathfrak s}$ is primitive, suppose that a is a non-zero element of $F_{\mathfrak s}\mathcal {H}^\varLambda _n(\fancyscript{K})$. By (3.5), $a=\sum _{{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})}r_{\mathfrak v}f_{{\mathfrak s}{\mathfrak v}}$, for some $r_{\mathfrak v}\in \fancyscript{K}$. Fix ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ such that $r_{\mathfrak t}\ne 0$. Then $f_{{\mathfrak s}{\mathfrak t}}=1/r_{\mathfrak t}a F_{\mathfrak t}\in F_{\mathfrak s}\mathcal {H}^\varLambda _n(\fancyscript{K})$. Using (3.5) we deduce that $F_{\mathfrak s}\mathcal {H}^\varLambda _n(\fancyscript{K})$ has basis $\{f_{{\mathfrak s}{\mathfrak v}}\,|\,{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\}$. Consequently, $a\mathcal {H}^\varLambda _{n}=F_{\mathfrak s}\mathcal {H}^\varLambda _n(\fancyscript{K})$, showing that $F_{\mathfrak s}\mathcal {H}^\varLambda _n(\fancyscript{K})$ is irreducible. Therefore, $F_{\mathfrak s}$ is a primitive idempotent in $\mathcal {H}^\varLambda _n(\fancyscript{K})$.

The last paragraph, together with Definition 3.5(c), implies that if ${\mathfrak s},{\mathfrak u}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ then $F_{\mathfrak s}\mathcal {H}^\varLambda _{n}\cong F_{\mathfrak u}\mathcal {H}^\varLambda _{n}$ where an isomorphism is given by $f_{{\mathfrak s}{\mathfrak t}}\mapsto f_{{\mathfrak u}{\mathfrak t}}$, for ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$. Consequently, if ${\mathfrak s}$ and ${\mathfrak u}$ are standard tableaux of different shape then $F_{\mathfrak s}\mathcal {H}^\varLambda _{n}\not \cong F_{\mathfrak u}\mathcal {H}^\varLambda _{n}$ because the multiplicity of $S^{\varvec{\lambda }}\cong F_{\mathfrak s}\mathcal {H}^\varLambda _{n}(\fancyscript{K})$ in $\mathcal {H}^\varLambda _{n}(\fancyscript{K})$ is $\#{\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ by the Wedderburn theorem.

Finally, it remains to show that the basis $\{f_{{\mathfrak s}{\mathfrak t}}\}$ is uniquely determined by ${\varvec{\alpha }}$ and the choice of the $\gamma$-coefficients $\{\gamma _{{\mathfrak t}^{\varvec{\lambda }}}\,|\,{\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}\}$. If ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ then we have shown that, once $f_{{\mathfrak s}{\mathfrak t}}$ is fixed, there is a unique seminormal basis $\{f_{{\mathfrak u}{\mathfrak v}}\,|\,{\mathfrak u},{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\}$ satisfying (3.5). In particular, taking ${\mathfrak s}={\mathfrak t}^{\varvec{\lambda }}={\mathfrak t}$ and fixing $f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ determines these basis elements. By (3.6) the choice of $f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ also uniquely determines $\gamma _{{\mathfrak t}^{\varvec{\lambda }}}$. Conversely, by setting $f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}=\gamma _{{\mathfrak t}^{\varvec{\lambda }}}F_{{\mathfrak t}^{\varvec{\lambda }}}$ for any choice of non-zero scalars $\gamma _{{\mathfrak t}^{\varvec{\lambda }}}\in \fancyscript{K}$, for ${\varvec{\lambda }}\in \fancyscript{K}$, the seminormal coefficient system ${\varvec{\alpha }}$ determines a unique seminormal basis.

The results that follow are independent of the choice of seminormal coefficient system ${\varvec{\alpha }}$, however, the choice of $\gamma$-coefficients will be important—and in what follows it will be useful to be able to vary both the seminormal coefficient system ${\varvec{\alpha }}$ and the $\gamma$-coefficients.

The proof of Theorem 3.9 implies that the choice of $\gamma _{{\mathfrak t}^{\varvec{\lambda }}}$ determines all of the scalars $\gamma _{\mathfrak s}$, for ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$. In what follows we need the following result that makes the relationship between these coefficients more explicit.

Corollary 3.10

Suppose that ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and that ${\mathfrak v}={\mathfrak t}(r,r+1)$ is standard,  where $1\le r<n.$ Then $\alpha _r({\mathfrak v})\gamma _{\mathfrak t}=\alpha _r({\mathfrak t})\gamma _{\mathfrak v}.$

Proof

Applying (3.5) and (3.6) several times each,

$$\begin{aligned} \gamma _{\mathfrak v}f_{{\mathfrak v}{\mathfrak v}}= & {} f_{{\mathfrak v}{\mathfrak v}}f_{{\mathfrak v}{\mathfrak v}} =\frac{1}{\alpha _r({\mathfrak t})}f_{{\mathfrak v}{\mathfrak t}}\left( T_r+\frac{1}{[\rho _r({\mathfrak t})]}\right) f_{{\mathfrak v}{\mathfrak v}} =\frac{1}{\alpha _r({\mathfrak t})}f_{{\mathfrak v}{\mathfrak t}}T_rf_{{\mathfrak v}{\mathfrak v}}\\= & {} \frac{1}{\alpha _r({\mathfrak t})}f_{{\mathfrak v}{\mathfrak t}} \left( \alpha _r({\mathfrak v})f_{{\mathfrak t}{\mathfrak v}}-\frac{1}{[\rho _r({\mathfrak v})]}f_{{\mathfrak v}{\mathfrak v}}\right) =\frac{\alpha _r({\mathfrak v})}{\alpha _r({\mathfrak t})}f_{{\mathfrak v}{\mathfrak t}}f_{{\mathfrak t}{\mathfrak v}}\\= & {} \frac{\alpha _r({\mathfrak v})}{\alpha _r({\mathfrak t})}\gamma _{\mathfrak t}f_{{\mathfrak v}{\mathfrak v}}. \end{aligned}$$

Comparing coefficients, $\alpha _r({\mathfrak t})\gamma _{\mathfrak v}=\alpha _r({\mathfrak v})\gamma _{\mathfrak t}$ as required.

3.3 Seminormal bases and the Murphy basis

In this section we compute the Gram determinant of the Specht modules of $\mathcal {H}^\varLambda _{n}$, with respect to the Murphy basis, as a product of cyclotomic polynomials when $\xi \ne 1$ or as a product of primes when $\xi =1$. These determinants are already explicitly known [4, 16–18] but all existing formulas describe them as products of rational functions, or of rational numbers in the degenerate case.

By Theorem 2.5, the Murphy basis $\{m_{{\mathfrak s}{\mathfrak t}}\}$ is a cellular basis for $\mathcal {H}^\varLambda _{n}$ over an arbitrary ring. In this section we continue to work with the generic Hecke algebra $\mathcal {H}^\varLambda _{n}=\mathcal {H}^\varLambda _n({\mathcal {O}})$ with parameter t and multicharge ${\varvec{\kappa }}$ satisfying (3.1).

As $(\fancyscript{K},{\mathfrak t})$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$, for ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ we can define

$$\begin{aligned} f_{{\mathfrak s}{\mathfrak t}}=F_{\mathfrak s}m_{{\mathfrak s}{\mathfrak t}} F_{\mathfrak t}. \end{aligned}$$

By Lemma 2.6, $f_{{\mathfrak s}{\mathfrak t}}\equiv m_{{\mathfrak s}{\mathfrak t}}+\sum r_{{\mathfrak u}{\mathfrak v}}m_{{\mathfrak u}{\mathfrak v}}\text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n)$, for some $r_{{\mathfrak u}{\mathfrak v}}\in \fancyscript{K}$ where $r_{{\mathfrak u}{\mathfrak v}}\ne 0$ only if $({\mathfrak u},{\mathfrak v})\vartriangleright ({\mathfrak s},{\mathfrak t})$. It follows that $\{f_{{\mathfrak s}{\mathfrak t}}\}$ is a seminormal basis of $\mathcal {H}^\varLambda _{n}(\fancyscript{K})$ in the sense of Definition 3.4.

For ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$ set $[{\varvec{\lambda }}]_t^!=\prod _{l=1}^\ell \prod _{r\ge 1}[\lambda ^{(l)}_r]_t^!\in \mathbb {N}[t]$.

Proposition 3.11

The basis $\{f_{{\mathfrak s}{\mathfrak t}}\,|\,{\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\text { for }{\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}\}$ is the $*$-seminormal basis of $\mathcal {H}^\varLambda _n(\fancyscript{K})$ determined by the seminormal coefficient system defined in Example 3.7 and the choices

$$\begin{aligned} \gamma _{{\mathfrak t}^{\varvec{\lambda }}} = [{\varvec{\lambda }}]_t^!\prod _{1\le l<m\le \ell } \prod _{(l,r,c)\in [{\varvec{\lambda }}]}[\kappa _l-r+c-\kappa _m], \end{aligned}$$

for ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}.$

Proof

This is equivalent to [27, Theorem 2.11] in the non-degenerate case and to [4, Proposition 6.8] in the degenerate case, however, rather than translating the notation from these two papers it is easier to prove this directly.

As noted above, $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and $f_{{\mathfrak s}{\mathfrak t}}\equiv m_{{\mathfrak s}{\mathfrak t}}+\sum r_{{\mathfrak u}{\mathfrak v}}m_{{\mathfrak u}{\mathfrak v}}\text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n)$, for some $r_{{\mathfrak u}{\mathfrak v}}\in \fancyscript{K}$ where $r_{{\mathfrak u}{\mathfrak v}}\ne 0$ only if $({\mathfrak u},{\mathfrak v})\vartriangleright ({\mathfrak s},{\mathfrak t})$. Therefore, in view of (3.6), $\{f_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$ is a $*$-seminormal basis of $\mathcal {H}^\varLambda _{n}(\fancyscript{K})$. By Theorem 3.9, this basis is determined by a seminormal coefficient system ${\varvec{\alpha }}$ and by a choice of scalars $\{\gamma _{{\mathfrak t}^{\varvec{\lambda }}}\,|\,{\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}\}$. If ${\mathfrak t}\vartriangleright {\mathfrak v}={\mathfrak t}(r,r+1)$ then, by definition, $m_{{\mathfrak s}{\mathfrak t}}T_r=m_{{\mathfrak s}{\mathfrak v}}$. The transition matrix between the $\{m_{{\mathfrak s}{\mathfrak t}}\}$ and $\{f_{{\mathfrak s}{\mathfrak t}}\}$ is unitriangular so, in view of Theorem 3.9, $f_{{\mathfrak s}{\mathfrak t}}T_r=f_{{\mathfrak s}{\mathfrak v}}-\frac{1}{[\rho _r({\mathfrak t})]}f_{{\mathfrak s}{\mathfrak t}}$. Therefore, by Definition 3.5(c), the seminormal coefficient system corresponding to the basis $\{f_{{\mathfrak s}{\mathfrak t}}\}$ is the one appearing in Example 3.7.

It remains to determine the scalars $\{\gamma _{{\mathfrak t}^{\varvec{\lambda }}}\,|\,{\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}\}$ corresponding to $\{f_{{\mathfrak s}{\mathfrak t}}\}$. It is well-known, and easy to prove using the relations in $\mathcal {H}^\varLambda _{n}$, that $x_{\varvec{\lambda }}^2=[{\varvec{\lambda }}]^!_tx_{\varvec{\lambda }}$. Therefore, by Lemma 2.6,

$$\begin{aligned} f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}^2\equiv [{\varvec{\lambda }}]^!_t m_{\varvec{\lambda }}u_{\varvec{\lambda }}\equiv [{\varvec{\lambda }}]^!_t\prod _{1\le l<m\le \ell } \prod _{(l,r,c)\in [{\varvec{\lambda }}]}[\kappa _l-r+c-\kappa _m] \cdot m_{\varvec{\lambda }}\quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n). \end{aligned}$$

Hence, $\gamma _{{\mathfrak t}^{\varvec{\lambda }}}=[{\varvec{\lambda }}]^!_t\prod _{1\le l<m\le \ell } \prod _{(l,r,c)\in [{\varvec{\lambda }}]}[\kappa _l-r+c-\kappa _m]$ by (3.6).

As noted after Theorem 2.5, the Murphy basis $\{m_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$ of $\mathcal {H}^\varLambda _{n}$ gives a basis $\{m_{\mathfrak t}\,|\,{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\}$ of each Specht module $\underline{S}^{\varvec{\lambda }}$, for ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$. For example, we can set $m_{\mathfrak t}=m_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}+\mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n$, for ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$. By (2.2), the cellular basis equips the Specht module $\underline{S}^{\varvec{\lambda }}$ with an inner product $\langle \ ,\ \rangle$. The matrix

$$\begin{aligned} \underline{\mathcal {G}^{\varvec{\lambda }}}=\big (\langle m_{\mathfrak s},m_{\mathfrak t}\rangle \big )_{{\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})} \end{aligned}$$

is the Gram matrix of $\underline{S}^{{\varvec{\lambda }}}$ with respect to the Murphy basis. Similarly, the seminormal basis yields a second basis $\{f_{\mathfrak t}\,|\,{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\}$ of $\underline{S}^{\varvec{\lambda }}(\fancyscript{K})$, where $f_{\mathfrak t}=m_{\mathfrak t}F_{\mathfrak t}=f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}+\mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n$, for ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$. The transition matrix between these two bases is unitriangular, so by (3.6) we have

$$\begin{aligned} \det \underline{\mathcal {G}^{\varvec{\lambda }}}= \det \big (\langle f_{\mathfrak s},f_{\mathfrak t}\rangle \big ) = \prod _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})}\gamma _{\mathfrak t}. \end{aligned}$$

(3.7)

This ‘classical’ formula for $\det \underline{\mathcal {G}^{\varvec{\lambda }}}$ is well-known as it is the cornerstone used to prove the formula for $\det \underline{\mathcal {G}^{\varvec{\lambda }}}$ as a rational function in [17, Theorem 3.35]. The following definition will allow us to give an ‘integral’ closed formula for $\det \underline{\mathcal {G}^{\varvec{\lambda }}}$.

Definition 3.12

Suppose that $e\in \{0,2,3,4,\ldots \}, p$ is a prime integer and that ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$ is a multipartition of n. Define

$$\begin{aligned} \deg _e({\varvec{\lambda }})=\sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})}\deg _e({\mathfrak t})\quad \text {and}\quad {{\mathrm{Deg}}}_p({\varvec{\lambda }})=\sum _{k\ge 1}\deg _{p^k}({\varvec{\lambda }}). \end{aligned}$$

By definition, $\deg _e({\varvec{\lambda }})$ and ${{\mathrm{Deg}}}_p({\varvec{\lambda }})$ are integers that, a priori, could be positive, negative or zero. In fact, the next result shows that they are always non-negative integers, although we do not know of a direct combinatorial proof of this. By definition, the integers $\deg _e({\varvec{\lambda }})$ and ${{\mathrm{Deg}}}_p({\varvec{\lambda }})$ depend on ${\varvec{\kappa }}$ and e. Our definitions ensure that the tableau degrees $\deg _e({\mathfrak t})$, for ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$, coincide with (2.3) when $\varLambda =\varLambda _e({\varvec{\kappa }})$.

For $k\in \mathbb {N}$, let $\varPhi _k=\varPhi _k(t)$ be the kth cyclotomic polynomial in t. As is well-known, these polynomials are pairwise distinct irreducible polynomials in $\mathbb {Z}[t]$ and

$$\begin{aligned}{}[n] = \prod _{1<d|n}\varPhi _d(t), \end{aligned}$$

(3.8)

whenever $n\ge 1$.

Theorem 3.13

Suppose that $\kappa _l-\kappa _{l+1}>n,$ for $1\le l<\ell ,$ and that ${\mathcal {O}}=\mathbb {Z}[t,t^{-1}].$ Then

$$\begin{aligned} \det \underline{\mathcal {G}^{\varvec{\lambda }}}=t^{\ell ({\varvec{\lambda }})}\prod _{e\ge 2}\varPhi _e(t)^{\deg _e({\varvec{\lambda }})}, \end{aligned}$$

where $\ell ({\varvec{\lambda }})=\sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})}\ell (d({\mathfrak t})).$

Proof

As remarked above, $\det \underline{\mathcal {G}^{\varvec{\lambda }}}=\prod _{\mathfrak t}\gamma _{\mathfrak t}$. Therefore, to prove the theorem it is enough to show that if ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ then

$$\begin{aligned} \gamma _{\mathfrak t}= t^{\ell (d({\mathfrak t}))}\prod _{e>1}\varPhi _e(t)^{\deg _e({\mathfrak t})}. \end{aligned}$$

We prove this by induction on the dominance ordering.

Suppose first that ${\mathfrak t}={\mathfrak t}^{\varvec{\lambda }}$. Then Proposition 3.11 gives an explicit formula for $\gamma _{{\mathfrak t}^{\varvec{\lambda }}}$ and, using (2.3), it is straightforward to check by induction on n that our claim is true in this case. Suppose then that ${\mathfrak t}^{\varvec{\lambda }}\vartriangleright {\mathfrak t}$. Then we can write ${\mathfrak t}={\mathfrak s}(r,r+1)$ for some ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ such that ${\mathfrak s}\vartriangleright {\mathfrak t}$, and where $1\le r<n$. Therefore, using induction, Corollary 3.10 and the seminormal coefficient system of Proposition 3.11,

$$\begin{aligned} \gamma _{\mathfrak t}=t^{\ell (d({\mathfrak s}))} \frac{[1+\rho _r({\mathfrak s})][1+\rho _r({\mathfrak t})]}{[\rho _r({\mathfrak s})][\rho _r({\mathfrak t})]} \prod _{e>1}\varPhi _e(t)^{\deg _e({\mathfrak s})}. \end{aligned}$$

By definition, $[k]=-t^k[-k]$, for any $k\in \mathbb {Z}$. Now $\rho _r({\mathfrak s})=-\rho _r({\mathfrak t})>0$ by (3.1), so

$$\begin{aligned} \frac{[1+\rho _r({\mathfrak s})][1+\rho _r({\mathfrak t})]}{[\rho _r({\mathfrak s})][\rho _r({\mathfrak t})]} =t\,\frac{[1+\rho _r({\mathfrak s})][-\rho _r({\mathfrak t})-1]}{[\rho _r({\mathfrak s})][-\rho _r({\mathfrak t})]} =t\,\prod _{e>1}\varPhi _e(t)^{d_e}, \end{aligned}$$

where, according to (3.8), the integer $d_e$ is given in terms of the quiver $\varGamma _e$ by

$$\begin{aligned} d_e={\left\{ \begin{array}{ll} -2,&{}\text {if }i_r=i_{r+1},\\ 2,&{}\text {if }i_r\leftrightarrows i_{r+1},\\ 1,&{}\text {if }i_r\leftarrow i_{r+1}\text { or }i_r\rightarrow i_{r+1},\\ 0,&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Applying Lemma 2.4 now completes the proof of our claim—and hence proves the theorem.

Remark 3.14

We can remove the factor $t^{\ell ({\varvec{\lambda }})}$ from Theorem 3.13 by rescaling the generators $T_1,\ldots ,T_{n-1}$ so that the quadratic relations in Definition 2.1 become $(T_r-t^{\frac{1}{2}})(T_r+t^{-\frac{1}{2}})$, for $1\le r<n$. Note that the integer $d_e$ in the proof of Theorem 3.9 is equal to the degree of the homogeneous generator $\psi _r e(\mathbf {i})$ in the cyclotomic KLR algebra $\mathcal {R}^\varLambda _{n}$.

Setting $t=1$ gives the degenerate cyclotomic Hecke algebras. As a special case, the next result gives an integral closed formula for the Gram determinants of the Specht modules of the symmetric groups.

Corollary 3.15

Suppose that $\kappa _l-\kappa _{l+1}>n,$ for $1\le l<\ell ,$ and that ${\mathcal {O}}=\mathbb {Z}$ and $t=1.$ Then

$$\begin{aligned} \det \underline{\mathcal {G}^{\varvec{\lambda }}}= \prod _{\begin{array}{c} 0<p\in \mathbb {Z}\\ p\text { prime} \end{array}}p^{{{\mathrm{Deg}}}_p({\varvec{\lambda }})}, \end{aligned}$$

for ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}.$

Proof

This follows by setting $t=1$ in Theorem 3.13 and using the following well-known property of the cyclotomic polynomials:

$$\begin{aligned} \varPhi _e(1) = {\left\{ \begin{array}{ll} p,&{}\text {if }e=p^k\text { for some }k\ge 1,\\ 1,&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Corollary 3.16

Suppose that $e\in \{0,2,3,4,5,\ldots \}$ and that $p>0$ is an integer prime. Then $\deg _e({\varvec{\lambda }})\ge 0$ and ${{\mathrm{Deg}}}_p({\varvec{\lambda }})\ge 0,$ for all ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}.$

Proof

As the Murphy basis is defined over $\mathbb {Z}[t,t^{-1}]$, the Gram determinant $\det \underline{\mathcal {G}^{\varvec{\lambda }}}$ belongs to $\mathbb {Z}[t,t^{-1}]$. Therefore, $\deg _e({\varvec{\lambda }})\ge 0$ whenever $e>1$ by Theorem 3.13. Consequently, ${{\mathrm{Deg}}}_p({\varvec{\lambda }})\ge 0$. Finally, if $e\gg 0$ then $\deg _0({\mathfrak t})=\deg _e({\mathfrak t})$ for any ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$, so $\deg _e({\varvec{\lambda }})\ge 0$ for $e\in \{0,2,3,4,\ldots \}$ as claimed.

The statement of Corollary 3.16 is purely combinatorial so it should have a direct combinatorial proof. We now give a second representation theoretic proof of this result that suggests that a combinatorial proof may be difficult.

A graded set is a set D equipped with a degree function $\deg \,{:}\,D\!\longrightarrow \!\mathbb {Z}$. Let q be an indeterminate over $\mathbb {Z}$ and define the q-cardinality and degree of D to be

$$\begin{aligned} |D|_q=\sum _{d\in D}q^{\deg d}\in \mathbb {N}[q,q^{-1}] \quad \text {and}\quad \deg D=\sum _{d\in D}\deg d\in \mathbb {Z}. \end{aligned}$$

If D is a graded set and $z\in \mathbb {Z}$ let $q^zD$ be the graded set with the same elements as D but where the shifted degree function is shifted so that $d\in D$ now has degree $z+\deg d$. More generally, if $f(q)\in \mathbb {N}[q,q^{-1}]$ let f(q)D be the graded set that is the disjoint union of the appropriate number of shifted copies of D. For example $(2+q)D=D\sqcup D\sqcup qD$. By definition, $|f(q)D|_q=f(q)|D|_q$.

If $e\in \{0,2,3,4,\dots \}$ let ${\mathop {\mathrm{Std}}\nolimits }_e({\varvec{\lambda }})$ be the graded set with elements ${\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ and degree function ${\mathfrak t}\mapsto \deg _e({\mathfrak t})$, for ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }_e({\varvec{\lambda }})$.

Fix $e\in \{0,2,3,4,\dots \}$ and consider the Hecke algebra $\mathcal {H}^\varLambda _{n}(\mathbb {C})$ over $\mathbb {C}$ with Hecke parameter $\xi$, a primitive eth root of unity if $e>0$ or a non-root of unity if $e=0$. Let $S^{\varvec{\lambda }}$ be the graded Specht module introduced in [8] (see Sect. 5.2), and let $D^{\varvec{\mu }}=S^{\varvec{\mu }}/{{\mathrm{rad}}}S^{\varvec{\mu }}$ be the graded simple quotient of $S^{\varvec{\mu }}$, as in [14]. Let $\mathcal {K}^{\varLambda }_{n}$ be the set of Kleshchev multipartitions so that $\{D^{\varvec{\mu }}\langle k\rangle \,|\,{\varvec{\mu }}\in \mathcal {K}^{\varLambda }_{n}\text { and }k\in \mathbb {Z}\}$ is a complete set of non-isomorphic graded simple $\mathcal {H}^\varLambda _{n}$-modules. As recalled in Sect. 5.2, $S^{\varvec{\lambda }}$ comes equipped with a homogeneous basis $\{\psi _{\mathfrak t}\,|\,{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }_e({\varvec{\lambda }})\}$. Let $d_{{\varvec{\lambda }}{\varvec{\mu }}}(q)=[S^{\varvec{\lambda }}{:}D^{\varvec{\mu }}]_q$ be the corresponding graded decomposition number.

Fix a total ordering $\prec$ on ${\mathop {\mathrm{Std}}\nolimits }_e({\varvec{\lambda }})$ that extends the dominance ordering, such as the lexicographic ordering. Suppose that ${\varvec{\mu }}\in \mathcal {K}^{\varLambda }_{n}$. By Gaussian elimination, there exists a graded subset ${\mathop {\mathrm{DStd}}\nolimits _e}({\varvec{\mu }})$ of ${\mathop {\mathrm{Std}}\nolimits }_e({\varvec{\mu }})$ and a homogeneous basis $\{C_{\mathfrak t}\,|\,{\mathfrak t}\in {\mathop {\mathrm{DStd}}\nolimits _e}({\varvec{\mu }})\}$ of $D^{\varvec{\mu }}$ such that

$$\begin{aligned} C_{\mathfrak t}=\psi _{\mathfrak t}+\sum _{{\mathfrak v}\prec {\mathfrak t}}c_{{\mathfrak t}{\mathfrak v}}\psi _{\mathfrak v}+{{\mathrm{rad}}}S^{\varvec{\mu }}, \end{aligned}$$

for some $c_{{\mathfrak t}{\mathfrak v}}\in \mathbb {C}$ such that $c_{{\mathfrak t}{\mathfrak v}}\ne 0$ only if $\deg {\mathfrak v}=\deg {\mathfrak t}$ and ${{\mathrm{res}}}({\mathfrak v})={{\mathrm{res}}}({\mathfrak t})$. In particular, ${{\mathrm{\dim _q}}}D^{\varvec{\lambda }}=|{\mathop {\mathrm{DStd}}\nolimits _e}({\varvec{\lambda }})|_q$. Repeating this argument, with the composition factors that appear in successive layers of the radical filtration of $S^{\varvec{\lambda }}$, shows that there exists a bijection of graded sets

$$\begin{aligned} \varTheta _{\varvec{\lambda }}: {\mathop {\mathrm{Std}}\nolimits }_e({\varvec{\lambda }})\overset{\sim }{\longrightarrow }\bigsqcup _{{\varvec{\mu }}\in \mathcal {K}^{\varLambda }_{n}} d_{{\varvec{\lambda }}{\varvec{\mu }}}(q){\mathop {\mathrm{DStd}}\nolimits _e}({\varvec{\mu }}). \end{aligned}$$

Now if ${\varvec{\mu }}\in \mathcal {K}^{\varLambda }_{n}$ then $D^{\varvec{\mu }}\cong (D^{\varvec{\mu }})^\circledast$, so that $\deg {\mathop {\mathrm{DStd}}\nolimits _e}({\varvec{\mu }})=0$. It follows that $\deg q^z{\mathop {\mathrm{DStd}}\nolimits _e}({\varvec{\mu }})=z\dim \underline{D}^{\varvec{\mu }}$, for $z\in \mathbb {Z}$. Therefore, using the bijection $\varTheta _{\varvec{\lambda }}$,

$$\begin{aligned} \deg _e({\varvec{\lambda }}) = \deg {\mathop {\mathrm{Std}}\nolimits }_e({\varvec{\lambda }}) =\sum _{{\varvec{\mu }}\in \mathcal {K}^{\varLambda }_{n}}d_{{\varvec{\lambda }}{\varvec{\mu }}}(q)\deg {\mathop {\mathrm{DStd}}\nolimits _e}({\varvec{\mu }}) =\sum _{{\varvec{\mu }}\in \mathcal {K}^{\varLambda }_{n}}d_{{\varvec{\lambda }}{\varvec{\mu }}}'(1)\dim \underline{D}^{\varvec{\mu }}, \end{aligned}$$

where $d'_{{\varvec{\lambda }}{\varvec{\mu }}}(1)$ is the derivative of the graded decomposition number $d_{\lambda {\varvec{\mu }}}(q)$ evaluated at $q=1$. As we are working with the Hecke algebra $\mathcal {H}^\varLambda _{n}(\mathbb {C})$ in characteristic zero, $d_{{\varvec{\lambda }}{\varvec{\mu }}}(q)\in \mathbb {N}[q]$ by [7, Corollary 5.15]. Consequently, $\deg _e({\varvec{\lambda }})\ge 0$. Hence, the (deep) fact that $d_{{\varvec{\lambda }}{\varvec{\mu }}}(q)\in \mathbb {N}[q]$ leads to an alternative proof of Corollary 3.16.

In characteristic zero the graded cyclotomic Schur algebras is Koszul by [15, Theorem C] when $e=0$ and by [25] and [32, Proposition 7.8,7.9] in general. This implies that the Jantzen and grading filtrations of the graded Weyl modules, and hence of the graded Specht modules, coincide. Therefore, Corollary 3.16 is compatible with this Koszulity Conjecture via Ryom-Hansen’s [33, Theorem 1] description of the Jantzen sum formula; see also [38, Theorem 2.11].

The construction of the sets ${\mathop {\mathrm{DStd}}\nolimits _e}({\varvec{\mu }})$ given above is not unique because it involves many choices. It natural to ask if there is a canonical choice of basis for $S^{\varvec{\lambda }}$ that uniquely determines the sets ${\mathop {\mathrm{DStd}}\nolimits _e}({\varvec{\mu }})$ and the bijections $\varTheta _{\varvec{\lambda }}$. For level 2 such bijections are implicit in [9, §9] when $e=0$ and [15, Appendix] generalizes this to include the cases when $e>n$. It is interesting to note that the sets ${\mathop {\mathrm{DStd}}\nolimits _e}({\varvec{\mu }})$, together with the bijections $\varTheta _{\varvec{\lambda }}$, determine the graded decomposition numbers. More explicitly, if ${\mathfrak s}\in {\mathop {\mathrm{DStd}}\nolimits _e}({\varvec{\mu }})$ then

$$\begin{aligned} d_{{\varvec{\lambda }}{\varvec{\mu }}}(q)= \sum _{{\mathfrak t}\in \varTheta _{{\varvec{\lambda }}}^{-1}({\mathfrak s})}q^{\deg {\mathfrak t}-\deg {\mathfrak s}}, \end{aligned}$$

where we abuse notation and let $\varTheta _{\varvec{\lambda }}^{-1}({\mathfrak s})$ be the set of tableaux in ${\mathop {\mathrm{Std}}\nolimits }_e({\varvec{\lambda }})$ that are mapped onto a (shifted) copy of ${\mathfrak s}$ by $\varTheta _{\varvec{\lambda }}$. In particular, we can take ${\mathfrak s}={\mathfrak t}^{\varvec{\mu }}$ because it is easy to see that we must have ${\mathfrak t}^{\varvec{\mu }}\in {\mathop {\mathrm{DStd}}\nolimits _e}({\varvec{\mu }})$ whenever ${\varvec{\mu }}\in \mathcal {K}^{\varLambda }_{n}$. Hence, we have shown that the KLR tableau combinatorics leads to closed combinatorial formulas for the parabolic Kazhdan–Lusztig polynomials $d_{{\varvec{\lambda }}{\varvec{\mu }}}(q)$, and the graded simple dimensions ${{\mathrm{\dim _q}}}D^{\varvec{\mu }}$: both families of polynomials can be described as the q-cardinalities of graded sets of tableaux.

4 Integral Quiver Hecke algebras

The Seminormal Basis Theorem 3.9 compactly describes much of the semisimple representation theory of $\mathcal {H}^\varLambda _n(\fancyscript{K})$. For symmetric groups, Murphy [29] showed that seminormal bases can also be used to study the non-semisimple representation theory. Murphy’s ideas were extended to the cyclotomic Hecke algebras in [27, 28]. In this section we further extend Murphy’s ideas to connect seminormal bases and the KLR grading on $\mathcal {H}^\varLambda _{n}$.

4.1 Lifting idempotents

As Sect. 3.2, we continue to assume that ${\varvec{\kappa }}$ satisfies (3.1) and that $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$, where $\fancyscript{K}$ is a field and $0\ne t\in \fancyscript{K}$. If ${\mathcal {O}}$ is a subring of $\fancyscript{K}$ then we identify $\mathcal {H}^\varLambda _n({\mathcal {O}})$ with the obvious ${\mathcal {O}}$-subalgebra of $\mathcal {H}^\varLambda _{n}(\fancyscript{K})$ so that $\mathcal {H}^\varLambda _{n}(\fancyscript{K})\cong \mathcal {H}^\varLambda _n({\mathcal {O}})\otimes _{\mathcal {O}}\fancyscript{K}$ as $\fancyscript{K}$-algebras.

Let $J({\mathcal {O}})$ be the Jacobson radical of ${\mathcal {O}}$, the intersection of all of the maximal ideals of ${\mathcal {O}}$.

Definition 4.1

Suppose that ${\mathcal {O}}$ is a subring of $\fancyscript{K}$ and $t\in {\mathcal {O}}^{\times }$. Then $({\mathcal {O}},t)$ is an e-idempotent subring of $\fancyscript{K}$ if the following hold:

• (a) $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$;

• (b) $[k]_t$ is invertible in ${\mathcal {O}}$ whenever $k\not \equiv 0\text { }(\text {mod } e)$, for $k\in \mathbb {Z}$; and

• (c) $[k]_t\in J({\mathcal {O}})$ whenever $k\in e\mathbb {Z}$.

When e and t are understood, we simply call ${\mathcal {O}}$ an idempotent subring. Note that $\fancyscript{K}$ contains the field of fractions of ${\mathcal {O}}$, so Definition 4.1(a) ensures that $\mathcal {H}^\varLambda _{n}(\fancyscript{K})$ is semisimple and that it has a seminormal basis. Until further notice, we fix such a $*$-seminormal basis $\{f_{{\mathfrak s}{\mathfrak t}}\}$, together with the corresponding seminormal coefficient system ${\varvec{\alpha }}$ and $\gamma$-coefficients.

Let $({\mathcal {O}},t)$ be an e-idempotent subring and suppose $c\not \equiv d\text { }(\text {mod } e)$, for $c,d\in \mathbb {Z}$. Then $[c]-[d]=t^d[c-d]$ is invertible in ${\mathcal {O}}$. We use this fact below without mention.

Example 4.2

The following local rings are all examples of idempotent subrings.

• (a) Suppose that $\fancyscript{K}=\mathbb {Q}$ and $t=1$. Then $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and ${\mathcal {O}}=\mathbb {Z}_{(p)}$ is a p-idempotent subring of $\mathbb {Q}$ for any prime p.

• (b) Let K be any field and set $\fancyscript{K}=K(x)$, where x is an indeterminate over K, and $t=x+\xi$, where $\xi$ is a primitive eth root of unity in K. Then ${\mathcal {O}}=K[x]_{(x)}$ is an e-idempotent subring of $\fancyscript{K}$.

• (c) Let $\fancyscript{K}=\mathbb {Q}(x,\xi )$, where x is an indeterminate over $\mathbb {Q}$ and $\xi =\exp (2\pi i/e)$ is a primitive eth root of unity in $\mathbb {C}$. Let $t=x+\xi$. Then $(\fancyscript{K},t)$ separates ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and ${\mathcal {O}}=\mathbb {Z}[x,\xi ]_{(x)}$ is an e-idempotent subring of $\fancyscript{K}$.

• (d) Maintain the notation of the last example and let $p>1$ be a prime not dividing e. Let $\varPhi _{e,p}(x)$ be a polynomial in $\mathbb {Z}[x]$ whose reduction modulo p is the minimum polynomial of a primitive eth root of unity in an algebraically closed field of characteristic p. Then ${\mathcal {O}}=\mathbb {Z}[x,\xi ]_{(x,p,\varPhi _{e,p}(\xi ))}$ is an e-idempotent subring of $\mathbb {C}(x)$.

Suppose that $\mathbf {i}\in I^n$ and set ${\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})=\{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})\,|\,{{\mathrm{res}}}({\mathfrak t})=\mathbf {i}\}$. Define the residue idempotent $f_{\mathbf {i}}^{\mathcal {O}}$ by

$$\begin{aligned} f_{\mathbf {i}}^{\mathcal {O}}=\sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}F_{\mathfrak t}. \end{aligned}$$

(4.1)

By Theorem 3.9, $f_{\mathbf {i}}^{\mathcal {O}}$ is an idempotent in $\mathcal {H}^\varLambda _n(\fancyscript{K})$. In the rest of this section, we fix a seminormal basis $\{f_{{\mathfrak s}{\mathfrak t}}\}$ of $\mathcal {H}^\varLambda _n(\fancyscript{K})$ that is determined by a seminormal coefficient system $\{\alpha _r({\mathfrak s})\}$ and a choice of $\gamma _{{\mathfrak t}^{\varvec{\lambda }}}$. Then we have that $f_{\mathbf {i}}^{\mathcal {O}}=\sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{1}{\gamma _{\mathfrak t}}f_{{\mathfrak t}{\mathfrak t}}$.

Lemma 4.3

Suppose that ${\mathcal {O}}$ is an idempotent subring of $\fancyscript{K}$ and that $\mathbf {i}\in I^n.$ Then $f_{\mathbf {i}}^{\mathcal {O}}\in \fancyscript{L}({\mathcal {O}}).$ In particular,  $f_{\mathbf {i}}^{\mathcal {O}}$ is an idempotent in $\mathcal {H}^\varLambda _n({\mathcal {O}}).$

Proof

This result is proved when ${\mathcal {O}}$ is a discrete valuation ring in [28, Lemma 4.2], however, our weaker assumptions necessitate a different proof. Motivated, in part, by the proof of [29, Theorem 2.1], if ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$ define

$$\begin{aligned} F_{\mathfrak t}' = \prod _{k=1}^n \prod _{\begin{array}{c} c\in \fancyscript{C}\\ {{\mathrm{c}}}_k({\mathfrak t})\not \equiv c\text { }(\text {mod } e) \end{array}} \frac{L_k-[c]}{[{{\mathrm{c}}}_k({\mathfrak t})]-[c]}. \end{aligned}$$

Since ${\mathcal {O}}$ is an e-idempotent subring, $F_{\mathfrak t}'\in \fancyscript{L}({\mathcal {O}})\subset \mathcal {H}^\varLambda _n({\mathcal {O}})$. By Theorem 3.9, $\sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})}F_{\mathfrak s}$ is the identity element of $\mathcal {H}^\varLambda _n(\fancyscript{K})$ so, using (3.5), we see that

$$\begin{aligned} F_{\mathfrak t}' = \sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})}F_{\mathfrak t}' F_{\mathfrak s}= \sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})} a_{{\mathfrak s}{\mathfrak t}} F_{\mathfrak s}, \end{aligned}$$

where $a_{{\mathfrak s}{\mathfrak t}}=\prod _{k,c} ([{{\mathrm{c}}}_k({\mathfrak s})]-[c])/([{{\mathrm{c}}}_k({\mathfrak t})]-[c])\in {\mathcal {O}}$. In particular, $a_{{\mathfrak t}{\mathfrak t}}=1$. If ${\mathfrak s}\notin {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$ then there exists an integer k such that ${{\mathrm{res}}}_k({\mathfrak s})\ne {{\mathrm{res}}}_k({\mathfrak t})$, so $[{{\mathrm{c}}}_k({\mathfrak s})]-[{{\mathrm{c}}}_k({\mathfrak t})]\in {\mathcal {O}}^\times$ and $a_{{\mathfrak s}{\mathfrak t}}=0$. Therefore, $F_{\mathfrak t}'=\sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}a_{{\mathfrak s}{\mathfrak t}}F_{\mathfrak s}$. Consequently, $f_{\mathbf {i}}^{\mathcal {O}}F_{\mathfrak t}' = F_{\mathfrak t}' = F_{\mathfrak t}'f_{\mathbf {i}}^{\mathcal {O}}$ by (3.6). Notice that $F_{\mathfrak t}'F_{\mathfrak s}'=F_{\mathfrak s}'F_{\mathfrak t}'$ because $\fancyscript{L}(\fancyscript{K})$ is a commutative subalgebra of $\mathcal {H}^\varLambda _n(\fancyscript{K})$. Therefore,

$$\begin{aligned} \prod _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}(f_{\mathbf {i}}^{\mathcal {O}}-F_{\mathfrak t}')=f_{\mathbf {i}}^{\mathcal {O}}+ \sum _{\begin{array}{c} {\mathfrak t}_1,\ldots ,{\mathfrak t}_k\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})\\ \text {distinct with }k>0 \end{array}} (-1)^kF_{{\mathfrak t}_1}'F_{{\mathfrak t}_2}'\dots F_{{\mathfrak t}_k}'. \end{aligned}$$

On the other hand, since $f_{\mathbf {i}}^{\mathcal {O}}=\sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}F_{\mathfrak s}$ and $a_{{\mathfrak t}{\mathfrak t}}=1$,

$$\begin{aligned} \prod _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}(f_{\mathbf {i}}^{\mathcal {O}}-F_{\mathfrak t}') =\prod _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\sum _{\begin{array}{c} {\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})\\ {\mathfrak s}\ne {\mathfrak t} \end{array}} (1-a_{{\mathfrak s}{\mathfrak t}})F_{\mathfrak s}= 0, \end{aligned}$$

because $F_{\mathfrak s}F_{\mathfrak t}=0$ whenever ${\mathfrak s}\ne {\mathfrak t}$ by (3.6). Combining the last two equations,

$$\begin{aligned} f_{\mathbf {i}}^{\mathcal {O}}=\sum _{\begin{array}{c} {\mathfrak t}_1,\ldots ,{\mathfrak t}_k\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})\\ \text {distinct with }k>0 \end{array}} (-1)^{k+1}F_{{\mathfrak t}_1}'F_{{\mathfrak t}_2}'\cdots F_{{\mathfrak t}_k}'. \end{aligned}$$

In particular, $f_{\mathbf {i}}^{\mathcal {O}}\in \fancyscript{L}({\mathcal {O}})$ as we wanted to show.

Corollary 4.4

Suppose that ${\mathcal {O}}$ is an idempotent subring of $\fancyscript{K}$. Then $\{f_{\mathbf {i}}^{\mathcal {O}}\,|\,\mathbf {i}\in I^n\}$ is a complete set of pairwise orthogonal idempotents in $\mathcal {H}^\varLambda _n({\mathcal {O}})$.

Proof

By Theorem 3.9, $\{F_{\mathfrak t}\,|\,{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})\}$ is a complete set of pairwise orthogonal idempotents in $\mathcal {H}^\varLambda _n(\fancyscript{K})$. Hence, the result follows from Lemma 4.3.

If $\phi \in {\mathcal {O}}[X_1,\ldots ,X_n]$ is a polynomial in indeterminates $X_1,\ldots ,X_n$ over ${\mathcal {O}}$ then set $\phi (L)=\phi (L_1,\ldots ,L_n)\in \fancyscript{L}({\mathcal {O}})$. If ${\mathfrak s}$ is a tableau let $\phi ({\mathfrak s})=\phi ([c_1({\mathfrak s})],\ldots ,[c_n({\mathfrak s})])$ be the scalar in ${\mathcal {O}}$ obtained by evaluating the polynomial $\phi$ on the contents of ${\mathfrak s}$; that is, setting $X_1=[c_1({\mathfrak s})], \dots , X_n=[c_n({\mathfrak s})]$. Then, $\phi (L)f_{{\mathfrak s}{\mathfrak t}}=\phi ({\mathfrak s})f_{{\mathfrak s}{\mathfrak t}}$, for all $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$.

Ultimately, the next result will allow us to ‘renormalise’ intertwiners of the residue idempotents $f_{\mathbf {i}}^{\mathcal {O}}$, for $\mathbf {i}\in I^n$, so that they depend only on e rather than on $\xi$.

Proposition 4.5

Suppose that $\mathbf {i}\in I^n$ and $\phi \in {\mathcal {O}}[X_1,\ldots ,X_n]$ is a polynomial such that $\phi ({\mathfrak t})$ is invertible in ${\mathcal {O}},$ for all ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$. Then

$$\begin{aligned} f^\phi _\mathbf {i}=\sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{1}{\phi ({\mathfrak t})}F_{\mathfrak t}\in \fancyscript{L}({\mathcal {O}}). \end{aligned}$$

In particular,  $f^\phi _\mathbf {i}\in \mathcal {H}^\varLambda _n({\mathcal {O}}).$

Proof

By assumption, $\phi ({\mathfrak s})$ is invertible in ${\mathcal {O}}$ for all ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$. In particular, $f^\phi _\mathbf {i}$ is a well-defined element of $\fancyscript{L}(\fancyscript{K})$. It remains to show that $f^\phi _\mathbf {i}\in \fancyscript{L}({\mathcal {O}})$.

As in Lemma 4.3, for each ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$ define

$$\begin{aligned} F_{\mathfrak t}'=\prod _{\begin{array}{c} c\in \fancyscript{C}\\ {{\mathrm{c}}}_k({\mathfrak t})\not \equiv c\text { }(\text {mod } e) \end{array}} \frac{L_k-[c]}{[{{\mathrm{c}}}_k({\mathfrak t})]-[c]}\in \fancyscript{L}({\mathcal {O}}), \end{aligned}$$

and write $F_{\mathfrak t}'=\sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}a_{{\mathfrak s}{\mathfrak t}}F_{\mathfrak s}$ for some $a_{{\mathfrak s}{\mathfrak t}}\in {\mathcal {O}}$. Recall from the proof of Lemma 4.3 that $a_{{\mathfrak t}{\mathfrak t}}=1$.

Motivated by the definition of $F_{\mathfrak t}'$, set $F^\phi _{\mathfrak t}=\tfrac{\phi (L)}{\phi ({\mathfrak t})}F'_{{\mathfrak t}}$. Then $F_{\mathfrak t}^\phi \in \fancyscript{L}({\mathcal {O}})$ and

$$\begin{aligned} F_{\mathfrak t}^\phi =\sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}a_{{\mathfrak s}{\mathfrak t}}\frac{\phi (L)}{\phi ({\mathfrak t})}F_{\mathfrak s}=F_{\mathfrak t}+\sum _{\begin{array}{c} {\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})\\ {\mathfrak s}\ne {\mathfrak t} \end{array}}\frac{a_{{\mathfrak s}{\mathfrak t}}\phi ({\mathfrak s})}{\phi ({\mathfrak t})}F_{\mathfrak s}\end{aligned}$$

by (3.5). Consequently, $F^\phi _{\mathfrak t}f_{\mathbf {i}}^{\mathcal {O}}=F^\phi _{\mathfrak t}=f_{\mathbf {i}}^{\mathcal {O}}F^\phi _{\mathfrak t}$. The idempotents $\{F_{\mathfrak s}\,|\,{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})\}$ are pairwise orthogonal, so

$$\begin{aligned} f^\phi _\mathbf {i}F_{\mathfrak t}^\phi =\left( \sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{1}{\phi ({\mathfrak s})}F_{\mathfrak s}\right) \left( \sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{a_{{\mathfrak s}{\mathfrak t}}\phi ({\mathfrak s})}{\phi ({\mathfrak t})}F_{\mathfrak s}\right) =\sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{a_{{\mathfrak s}{\mathfrak t}}}{\phi ({\mathfrak t})}F_{\mathfrak s}=\frac{1}{\phi ({\mathfrak t})}F_{\mathfrak t}'. \end{aligned}$$

Therefore, $f_\mathbf {i}^\phi F_{\mathfrak t}^\phi \in \fancyscript{L}({\mathcal {O}})$, for all ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$. By (3.5), $f_\mathbf {i}^\phi f_{\mathbf {i}}^{\mathcal {O}}=f_\mathbf {i}^\phi =f_{\mathbf {i}}^{\mathcal {O}}f_\mathbf {i}^\phi$, so this implies that $f^\phi _\mathbf {i}(f_{\mathbf {i}}^{\mathcal {O}}-F^\phi _{\mathfrak t})\equiv f^\phi _\mathbf {i}\text { }(\text {mod } \fancyscript{L}({\mathcal {O}}))$. Hence, working modulo $\fancyscript{L}({\mathcal {O}})$,

$$\begin{aligned} f^\phi _\mathbf {i}\equiv f^\phi _\mathbf {i}\prod _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\ (f_{\mathbf {i}}^{\mathcal {O}}-F_{\mathfrak t}^\phi ) =f^\phi _\mathbf {i}\prod _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\sum _{\begin{array}{c} {\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})\\ {\mathfrak s}\ne {\mathfrak t} \end{array}} \frac{a_{{\mathfrak s}{\mathfrak t}}\phi ({\mathfrak s})}{\phi ({\mathfrak t})}F_{\mathfrak s}=0, \end{aligned}$$

where the last equality follows using the orthogonality of the idempotents $F_{\mathfrak s}$ once again. Therefore, $f^\phi _\mathbf {i}\in \fancyscript{L}({\mathcal {O}})$, completing the proof.

Let $\phi$ be a polynomial in ${\mathcal {O}}[X_1,\ldots ,X_n]$ satisfying the assumptions of Proposition 4.5. Then $\phi (L)f^\phi _\mathbf {i}=f_{\mathbf {i}}^{\mathcal {O}}=f^\phi _\mathbf {i}\phi (L)$ by (3.5). Abusing notation, in this situation we write

$$\begin{aligned} \frac{1}{\phi (L)}f_{\mathbf {i}}^{\mathcal {O}}= f^\phi _\mathbf {i}=\sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{1}{\phi ({\mathfrak s})}F_{\mathfrak s}=f_{\mathbf {i}}^{\mathcal {O}}\frac{1}{\phi (L)}\in \fancyscript{L}({\mathcal {O}}). \end{aligned}$$

Note that, either by direction calculation or because $\fancyscript{L}$ is commutative, we are justified in writing $f_{\mathbf {i}}^{\mathcal {O}}\tfrac{1}{\phi (L)}=\tfrac{1}{\phi (L)}f_{\mathbf {i}}^{\mathcal {O}}$.

We need the following three special cases of Proposition 4.5. For $1\le r<n$ define $M_r=1-L_r+tL_{r+1}$ and $M_r'=1+tL_r-L_{r+1}$, for $1\le r<n$. Applying the definitions, if $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ then

$$\begin{aligned} M_r f_{{\mathfrak s}{\mathfrak t}}=t^{c_r({\mathfrak s})}[1-\rho _r({\mathfrak s})]f_{{\mathfrak s}{\mathfrak t}}\quad \text {and}\quad M_r'f_{{\mathfrak s}{\mathfrak t}}= t^{c_{r+1}({\mathfrak s})}[1+\rho _r({\mathfrak s})]f_{{\mathfrak s}{\mathfrak t}}. \end{aligned}$$

(4.2)

Our main use of Proposition 4.5 is the following application that corresponds to taking $\phi (L)$ be to $L_r-L_{r+1}, M_r$ and $M_r'$, respectively.

Corollary 4.6

Suppose that ${\mathcal {O}}$ is an e-idempotent subring,  $1\le r<n$ and $\mathbf {i}\in I^n.$

• (a) If $i_r\ne i_{r+1}$ then $\dfrac{1}{L_r-L_{r+1}}f_{\mathbf {i}}^{\mathcal {O}}=\displaystyle \sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{t^{-c_{r+1}({\mathfrak t})}}{[\rho _r({\mathfrak t})]}F_{\mathfrak t}\in \fancyscript{L}({\mathcal {O}}).$

• (b) If $i_r\ne i_{r+1}+1$ then $\dfrac{1}{M_r}f_{\mathbf {i}}^{\mathcal {O}}=\displaystyle \sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{t^{-c_r({\mathfrak t})}}{[1-\rho _r({\mathfrak t})]}F_{\mathfrak t}\in \fancyscript{L}({\mathcal {O}}).$

• (c) If $i_r\ne i_{r+1}-1$ then $\dfrac{1}{M_r'}f_{\mathbf {i}}^{\mathcal {O}}=\displaystyle \sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{t^{-c_{r+1}({\mathfrak t})}}{[1+\rho _r({\mathfrak t})]}F_{\mathfrak t}\in \fancyscript{L}({\mathcal {O}}).$

4.2 Intertwiners

By Theorem 2.9, if K is a field then the KLR generators of $\mathcal {H}^\varLambda _{n}(K)$ satisfy $\psi _r e(\mathbf {i})=e(s_r\cdot \mathbf {i})\psi _r$. This section defines analogous elements in $\mathcal {H}^\varLambda _n({\mathcal {O}})$ that intertwine the residue idempotents $f_{\mathbf {i}}^{\mathcal {O}}$, for $\mathbf {i}\in I^n$.

Lemma 4.7

Suppose that $i_r=i_{r+1}$, for some $\mathbf {i}\in I^n$ and $1\le r<n$. Then $T_rf_{\mathbf {i}}^{\mathcal {O}}=f_{\mathbf {i}}^{\mathcal {O}}T_r$.

Proof

This follows directly from the Seminormal Basis Theorem 3.9. In more detail, note that if ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$ then r and $r+1$ cannot appear in the same row or in the same column of ${\mathfrak t}$. Therefore,

$$\begin{aligned} T_rf_{\mathbf {i}}^{\mathcal {O}}-f_{\mathbf {i}}^{\mathcal {O}}T_r =\sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{1}{\gamma _{\mathfrak t}}(T_rf_{{\mathfrak t}{\mathfrak t}}-f_{{\mathfrak t}{\mathfrak t}}T_r) =\sum _{\begin{array}{c} {\mathfrak t},{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})\\ {\mathfrak v}={\mathfrak t}(r,r+1) \end{array}} \left( \frac{\alpha _r({\mathfrak t})}{\gamma _{\mathfrak t}}-\frac{\alpha _r({\mathfrak v})}{\gamma _{\mathfrak v}}\right) f_{{\mathfrak v}{\mathfrak t}}, \end{aligned}$$

by (3.5). By Corollary 3.10, if ${\mathfrak v}={\mathfrak t}(r,r+1)$ then $\alpha _r({\mathfrak t})\gamma _{\mathfrak v}=\alpha _r({\mathfrak v})\gamma _{\mathfrak t}$. Hence, $T_rf_{\mathbf {i}}^{\mathcal {O}}=f_{\mathbf {i}}^{\mathcal {O}}T_r$ as claimed.

Remark 4.8

In the special case of the symmetric groups, Ryom-Hansen [34, §3] has proved an analogue of Lemma 4.7.

Using (3.5), it is easy to verify that $T_rf_{\mathbf {i}}^{\mathcal {O}}\ne f_{\mathbf {j}}^{\mathcal {O}} T_r$ if $\mathbf {j}=s_r\cdot \mathbf {i}\ne \mathbf {i}$, for $1\le r<n$ and $\mathbf {i}\in I^n$. The following elements will allow us to correct for this.

Lemma 4.9

Suppose that $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ and $1\le r<n$. Let ${\mathfrak u}={\mathfrak s}(r,r+1)$. Then $(T_rL_r-L_rT_r)f_{{\mathfrak s}{\mathfrak t}}=\alpha _r({\mathfrak s})t^{c_{r+1}({\mathfrak s})}[\rho _r({\mathfrak s})]f_{{\mathfrak u}{\mathfrak t}}.$

Proof

Using (3.5) we obtain

$$\begin{aligned} (T_rL_r-L_rT_r)f_{{\mathfrak s}{\mathfrak t}} =\alpha _r({\mathfrak s})\big ([c_r({\mathfrak s})]-[c_{r+1}({\mathfrak s})]\big )f_{{\mathfrak u}{\mathfrak t}} =\alpha _r({\mathfrak s})t^{c_{r+1}({\mathfrak s})}[\rho _r({\mathfrak s})]f_{{\mathfrak u}{\mathfrak t}}, \end{aligned}$$

where, as usual, we set $f_{{\mathfrak u}{\mathfrak t}}=0$ if ${\mathfrak u}$ is not standard.

Applying the $*$ anti-involution, $f_{{\mathfrak s}{\mathfrak t}}(T_rL_r-L_rT_r)=-\alpha _r({\mathfrak t})t^{c_{r+1}({\mathfrak t})}[\rho _r({\mathfrak t})]f_{{\mathfrak s}{\mathfrak v}}$, where ${\mathfrak v}={\mathfrak t}(r,r+1)$.

Lemma 4.10

Suppose that $i_r\ne i_{r+1}$, for some $\mathbf {i}\in I^n$ and $1\le r<n$. Set $\mathbf {j}=s_r\cdot \mathbf {i}$. Then $(T_rL_r-L_rT_r)f_{\mathbf {i}}^{\mathcal {O}}=f_{\mathbf {j}}^{\mathcal {O}}(T_rL_r-L_rT_r).$

Proof

By definition, $f_{\mathbf {i}}^{\mathcal {O}}=\sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\tfrac{1}{\gamma _{\mathfrak s}}f_{{\mathfrak s}{\mathfrak s}}$ so, by Lemma 4.9,

$$\begin{aligned} (T_rL_r-L_rT_r)f_{\mathbf {i}}^{\mathcal {O}}= & {} \sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{1}{\gamma _{\mathfrak s}}(T_rL_r-L_rT_r)f_{{\mathfrak s}{\mathfrak s}}\\= & {} \sum _{\begin{array}{c} {\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})\\ {\mathfrak u}={\mathfrak s}(r,r+1)\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n}) \end{array}} \frac{\alpha _r({\mathfrak s})t^{c_{r+1}({\mathfrak s})}[\rho _r({\mathfrak s})]}{\gamma _{\mathfrak s}}f_{{\mathfrak u}{\mathfrak s}}. \end{aligned}$$

Note that if ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$ and ${\mathfrak u}={\mathfrak s}(r,r+1)$ is standard then $\mathfrak {u}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {j})$. Similarly,

$$\begin{aligned} f_{\mathbf {j}}^{\mathcal {O}}(T_rL_r-L_rT_r)= \sum _{\begin{array}{c} {\mathfrak u}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {j})\\ {\mathfrak s}={\mathfrak u}(r,r+1)\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i}) \end{array}} -\frac{\alpha _r({\mathfrak u})t^{c_{r+1}({\mathfrak u})}[\rho _r({\mathfrak u})]}{\gamma _{\mathfrak u}}f_{{\mathfrak u}{\mathfrak s}}. \end{aligned}$$

By (3.5), the tableaux in ${\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$ and ${\mathop {\mathrm{Std}}\nolimits }(\mathbf {j})$ that have r and $r+1$ in the same row or in the same column do not contribute to the right hand sides of either of the last two equations. Moreover, the map ${\mathfrak s}\mapsto {\mathfrak u}={\mathfrak s}(r,r+1)$ defines a bijection from the set of tableaux in ${\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$ such that r and $r+1$ appear in different rows and columns to the set of tableaux in ${\mathop {\mathrm{Std}}\nolimits }(\mathbf {j})$ that have r and $r+1$ in different rows and columns. In particular, $(T_rL_r-L_rT_r)f_{\mathbf {i}}^{\mathcal {O}}=0$ if and only if $f_{\mathbf {j}}^{\mathcal {O}}(T_rL_r-L_rT_r)=0$.

To complete the proof suppose that ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$ and that ${\mathfrak u}={\mathfrak s}(r,r+1)\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {j})$. Now, $\alpha _r({\mathfrak u})\gamma _{\mathfrak s}=\alpha _r({\mathfrak s})\gamma _{\mathfrak u}$, by Corollary 3.10, and $\rho _r({\mathfrak u})=-\rho _r({\mathfrak s})$, by definition. So

$$\begin{aligned} \frac{-\alpha _r({\mathfrak u})t^{c_{r+1}({\mathfrak u})}[\rho _r({\mathfrak u})]}{\gamma _{\mathfrak u}} =\frac{-\alpha _r({\mathfrak s})t^{c_{r}({\mathfrak s})}[-\rho _r({\mathfrak s})]}{\gamma _{\mathfrak s}} =\frac{\alpha _r({\mathfrak s})t^{c_{r+1}({\mathfrak s})}[\rho _r({\mathfrak s})]}{\gamma _{\mathfrak s}}. \end{aligned}$$

Hence, comparing the equations above, $\big (T_rL_r-L_rT_r\big )f_{\mathbf {i}}^{\mathcal {O}}=f_{\mathbf {j}}^{\mathcal {O}}\big (T_rL_r-L_rT_r\big )$ as required.

Recall the definitions of $M_r$ and $M_r'$ from (4.2), for $1\le r<n$. We finish this section by giving the commutation relations for the elements $M_r, M_r', (1+T_r)$ and $(T_rL_r-L_rT_r)$. These will be important later.

Lemma 4.11

Suppose that $1\le r<n$. Then

$$\begin{aligned} (T_rL_r-L_rT_r)M_r=M_r'(T_rL_r-L_rT_r) \quad \text {and}\quad (T_r-t)M_r=M_r'(1+T_r). \end{aligned}$$

Proof

Both formulas can be proved by applying the relations in Definition 2.1. Alternatively, suppose that $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ and set ${\mathfrak v}={\mathfrak t}(r,r+1)$. Then, by (4.2) and Lemma 4.9,

$$\begin{aligned} f_{{\mathfrak s}{\mathfrak t}}(T_rL_r-L_rT_r)M_r= & {} -\alpha _r({\mathfrak t})t^{2c_r({\mathfrak v})}[\rho _r({\mathfrak t})][1+\rho _r({\mathfrak t})]f_{{\mathfrak s}{\mathfrak v}}\\= & {} f_{{\mathfrak s}{\mathfrak t}}M_r'(T_rL_r-L_rT_r), \end{aligned}$$

where the last equality follows because $c_r({\mathfrak v})=c_{r+1}({\mathfrak t})$ and $c_{r+1}({\mathfrak v})=c_r({\mathfrak t})$. As the regular representation is a faithful, this implies the first formula. The second formula can be proved similarly.

4.3 The integral KLR generators

In Lemmas 4.7 and 4.10, we have found elements in $\mathcal {H}^\varLambda _n({\mathcal {O}})$ that intertwine the residue idempotents $f_{\mathbf {i}}^{\mathcal {O}}$. These intertwiners are not quite the elements that we need, however, because they still depend on t, rather than just on e. To remove this dependence on t we will use Proposition 4.5 to renormalise these elements.

By Lemma 4.3, if $h\in \mathcal {H}^\varLambda _n({\mathcal {O}})$ then $h=\sum _{\mathbf {i}\in I^n}hf_{\mathbf {i}}^{\mathcal {O}}$, so that h is completely determined by its projections onto the spaces $\mathcal {H}^\varLambda _n({\mathcal {O}})f_{\mathbf {i}}^{\mathcal {O}}$. We use this observation to define analogues of the KLR generators in $\mathcal {H}^\varLambda _n({\mathcal {O}})$.

Recall from (4.2) that $M_r=1-L_r+tL_{r+1}$. By Corollary 4.6, if $i_r\ne i_{r+1}+1$ then $M_r$ acts invertibly on $f_{\mathbf {i}}^{\mathcal {O}}\mathcal {H}^\varLambda _n({\mathcal {O}})$ so $\tfrac{1}{M_r}f_{\mathbf {i}}^{\mathcal {O}}$ is a well-defined element of $\mathcal {H}^\varLambda _n({\mathcal {O}})$.

As in the introduction, define an embedding $I\hookrightarrow \mathbb {Z}; i\mapsto \hat{\imath }$ by defining $\hat{\imath }$ to be any integer such that $i=\hat{\imath }+e\mathbb {Z}$, for $i\in I$.

Definition 4.12

Suppose that $1\le r<n$. Define elements $\psi _r^{\mathcal {O}}=\sum _{\mathbf {i}\in I^n}\psi _r^{\mathcal {O}}f_{\mathbf {i}}^{\mathcal {O}}$ in $\mathcal {H}^\varLambda _n({\mathcal {O}})$ by

$$\begin{aligned} \psi _r^{\mathcal {O}}f_{\mathbf {i}}^{\mathcal {O}}={\left\{ \begin{array}{ll} (T_r+1)\frac{t^{\hat{\imath }_r}}{M_r}f_{\mathbf {i}}^{\mathcal {O}},&{}\text {if }i_r=i_{r+1},\\ (T_rL_r-L_rT_r)t^{-\hat{\imath }_r}f_{\mathbf {i}}^{\mathcal {O}}, &{}\text {if }i_r=i_{r+1}+1,\\ (T_rL_r-L_rT_r)\frac{1}{M_r}f_{\mathbf {i}}^{\mathcal {O}}, &{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

If $1\le r\le n$ then define $y^{{\mathcal {O}}}_r=\sum _{\mathbf {i}\in I^n}t^{-\hat{\imath }_r}(L_r-[\hat{\imath }_r])f_{\mathbf {i}}^{\mathcal {O}}$.

The order of the terms in the definition of $\psi ^{\mathcal {O}}_r$ matters because $M_r$ does not commute with $T_r+1$ or with $T_rL_r-L_rT_r$ (see Lemma 4.11), although $M_r$ does commute with $f_{\mathbf {i}}^{\mathcal {O}}$. Notice that $\psi ^{\mathcal {O}}_r$ is independent of the choice of seminormal coefficient system because the residue idempotents $f_{\mathbf {i}}^{\mathcal {O}}$ are independent of this choice.

One subtlety of Definition 4.12, which we will pay for later, is that it makes use of the embedding $I\hookrightarrow \mathbb {Z}$ in order to give meaning to expressions like $t^{\pm \hat{\imath }_r}$.

Remark 4.13

Unravelling the definitions, the element $\psi ^{\mathcal {O}}_r\otimes _{\mathcal {O}}1_K$ is a scalar multiple of the choice of KLR generators for $\mathcal {H}^\varLambda _n(\fancyscript{K})$ made by Stroppel and Webster [36, (27)]. Similarly, $y^{{\mathcal {O}}}_r\otimes _{\mathcal {O}}1_K$ is a multiple of the KLR generator $y_r$ defined by Brundan and Kleshchev [6, (4.21)].

Proposition 4.14

The algebra $\mathcal {H}^\varLambda _n({\mathcal {O}})$ is generated by the elements

$$\begin{aligned} \{f_{\mathbf {i}}^{\mathcal {O}}\,|\,\mathbf {i}\in I^n\}\cup \{\psi ^{\mathcal {O}}_r\,|\,1\le r<n\}\cup \{y^{{\mathcal {O}}}_r\,|\,1\le r\le n\}. \end{aligned}$$

Proof

Let H be the ${\mathcal {O}}$-subalgebra of $\mathcal {H}^\varLambda _n({\mathcal {O}})$ generated by the elements in the statement of the proposition. We need to show that $H=\mathcal {H}^\varLambda _n({\mathcal {O}})$. Directly from the definitions, if $1\le r\le n$ then $L_r=\sum _\mathbf {i}(t^{\hat{\imath }_r}y^{{\mathcal {O}}}_r+[i_r])f_{\mathbf {i}}^{\mathcal {O}}\in H$. Therefore, the Gelfand–Zetlin algebra $\fancyscript{L}({\mathcal {O}})$ is contained in H. Consequently, $M_r\in H$, for $1\le r<n$. By Definition 2.1, $L_rT_r-T_rL_r=T_r(L_{r+1}-L_r)-1+(1-t)L_{r+1}$. By Corollary 4.6(a), if $i_r\ne i_{r+1}$ then $\tfrac{1}{L_r-L_{r+1}}f_{\mathbf {i}}^{\mathcal {O}}\in \fancyscript{L}({\mathcal {O}})\subseteq H$. Therefore, since $M_r$ and $f_{\mathbf {i}}^{\mathcal {O}}$ commute, we can write

$$\begin{aligned} T_r f_{\mathbf {i}}^{\mathcal {O}}= {\left\{ \begin{array}{ll} \big (t^{-\hat{\imath }_r}\psi ^{\mathcal {O}}_rM_r-1\big )f_{\mathbf {i}}^{\mathcal {O}},&{}\text {if }i_r=i_{r+1},\\ \big (-t^{\hat{\imath }_r}\psi ^{\mathcal {O}}_r+1+(t-1)L_{r+1}\big )\frac{1}{L_{r+1}-L_r}f_{\mathbf {i}}^{\mathcal {O}}, &{}\text {if }i_r=i_{r+1}+1\\ \big (-\psi ^{\mathcal {O}}_rM_r+1+(t-1)L_{r+1}\big )\frac{1}{L_{r+1}-L_r}f_{\mathbf {i}}^{\mathcal {O}}, &{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

by Definition 4.12. Hence, $T_r=\sum _\mathbf {i}T_rf_{\mathbf {i}}^{\mathcal {O}}\in H$. As $T_1,\ldots ,T_{n-1},L_1,\ldots ,L_n$ generate $\mathcal {H}^\varLambda _n({\mathcal {O}})$ this implies that $H=\mathcal {H}^\varLambda _n({\mathcal {O}})$, completing the proof.

We now use the seminormal form to show that the elements in the statement of Proposition 4.14 satisfy most of the relations of Definition 2.7.

Lemma 4.15

Suppose that $1\le r<n$ and $\mathbf {i}\in I^n.$ Then $\psi ^{\mathcal {O}}_r f_{\mathbf {i}}^{\mathcal {O}}=f_{\mathbf {j}}^{\mathcal {O}}\psi ^{\mathcal {O}}_r,$ where $\mathbf {j}=s_r\cdot \mathbf {i}.$

Proof

By Lemma 4.3 and Proposition 4.5, respectively, $M_r$ and $f_{\mathbf {i}}^{\mathcal {O}}$ both belong to $\fancyscript{L}({\mathcal {O}})$, which is a commutative algebra. Therefore, $\frac{1}{M_r}f_{\mathbf {i}}^{\mathcal {O}}$ and $f_{\mathbf {i}}^{\mathcal {O}}$ commute. If $i_r=i_{r+1}$ then

$$\begin{aligned} \psi ^{\mathcal {O}}_rf_{\mathbf {i}}^{\mathcal {O}}=(T_r+1)\frac{t^{\hat{\imath }_r}}{M_r}f_{\mathbf {i}}^{\mathcal {O}}=(T_r+1)f_{\mathbf {i}}^{\mathcal {O}}\frac{t^{\hat{\imath }_r}}{M_r}f_{\mathbf {i}}^{\mathcal {O}}=f_{\mathbf {i}}^{\mathcal {O}}(T_r+1)\frac{t^{\hat{\imath }_r}}{M_r}f_{\mathbf {i}}^{\mathcal {O}}=f_{\mathbf {i}}^{\mathcal {O}}\psi ^{\mathcal {O}}_r, \end{aligned}$$

where the third equality comes from Lemma 4.7. The remaining cases follow similarly using Lemma 4.11

As we will work with right modules we need the right-handed analogue of Definition 4.12. Note that if $i_r\ne i_{r+1}+1$ then $f_{\mathbf {i}}^{\mathcal {O}}\tfrac{1}{M_r}=\tfrac{1}{M_r}f_{\mathbf {i}}^{\mathcal {O}}\in \mathcal {H}^\varLambda _n({\mathcal {O}})$ by Proposition 4.5. Similarly, if $i_r\ne i_{r+1}-1$ then $f_{\mathbf {i}}^{\mathcal {O}}\frac{1}{M_r'}=\frac{1}{M_r'}f_{\mathbf {i}}^{\mathcal {O}}\in \mathcal {H}^\varLambda _n({\mathcal {O}})$. It follows that all of the expressions in the next lemma make sense.

Lemma 4.16

Suppose $1\le r<n$ and $\mathbf {i}\in I^n.$ Then

$$\begin{aligned} f_{\mathbf {i}}^{\mathcal {O}}\psi ^{\mathcal {O}}_r = {\left\{ \begin{array}{ll} f_{\mathbf {i}}^{\mathcal {O}}\frac{t^{\hat{\imath }_{r+1}}}{M_r'}(T_r-t),&{}\text {if }i_i=i_{r+1},\\ f_{\mathbf {i}}^{\mathcal {O}}(T_rL_r-L_rT_r)t^{-\hat{\imath }_{r+1}},&{}\text {if }i_r=i_{r+1}-1,\\ f_{\mathbf {i}}^{\mathcal {O}}\frac{1}{M_r'}(T_rL_r-L_rT_r),&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

By Lemma 4.15, $f_{\mathbf {i}}^{\mathcal {O}}\psi ^{\mathcal {O}}_r=f_{\mathbf {i}}^{\mathcal {O}}\psi ^{\mathcal {O}}_rf_{\mathbf {j}}^{\mathcal {O}}$ where $\mathbf {j}=s_r\cdot \mathbf {i}$. Therefore,

$$\begin{aligned} f_{\mathbf {i}}^{\mathcal {O}}\psi ^{\mathcal {O}}_r = {\left\{ \begin{array}{ll} f_{\mathbf {i}}^{\mathcal {O}}(1+T_r)\frac{t^{\hat{\imath }_{r+1}}}{M_r}f_{\mathbf {j}}^{\mathcal {O}},&{}\text {if }i_i=i_{r+1},\\ f_{\mathbf {i}}^{\mathcal {O}}(T_rL_r-L_rT_r)t^{-\hat{\imath }_{r+1}}f_{\mathbf {j}}^{\mathcal {O}}, &{}\text {if }i_r=i_{r+1}-1,\\ f_{\mathbf {i}}^{\mathcal {O}}(T_rL_r-L_rT_r)\frac{1}{M_r}f_{\mathbf {j}}^{\mathcal {O}}, &{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

To complete the proof apply Lemma 4.11.

Lemma 4.17

Suppose that $\mathbf {i},\mathbf {j}\in I^n$ and $1\le r,s\le n.$ Then

$$\begin{aligned} \sum _{\mathbf {i}\in I^n}f_{\mathbf {i}}^{\mathcal {O}}=1, \,f_{\mathbf {i}}^{\mathcal {O}}f_{\mathbf {j}}^{\mathcal {O}}=\delta _{\mathbf {i}\mathbf {j}}f_{\mathbf {i}}^{\mathcal {O}},\,y^{{\mathcal {O}}}_r f_{\mathbf {i}}^{\mathcal {O}}=f_{\mathbf {i}}^{\mathcal {O}}y^{{\mathcal {O}}}_r\,\text {and}\,y^{{\mathcal {O}}}_ry^{{\mathcal {O}}}_s=y^{{\mathcal {O}}}_sy^{{\mathcal {O}}}_r. \end{aligned}$$

Moreover,  if $s\ne r, r+1$ then $\psi ^{\mathcal {O}}_r y^{{\mathcal {O}}}_s=y^{{\mathcal {O}}}_s\psi ^{\mathcal {O}}_r,$ for $1\le r<n$ and $1\le s\le n.$

Proof

The elements $f_{\mathbf {i}}^{\mathcal {O}}$, for $\mathbf {i}\in I^n$, form a complete set of pairwise orthogonal idempotents by Lemma 4.3, which gives the first two relations. Since $y_r, f_{\mathbf {i}}^{\mathcal {O}}\in \fancyscript{L}({\mathcal {O}})$ and $\fancyscript{L}({\mathcal {O}})$ is a commutative algebra, all of the elements $f_{\mathbf {i}}^{\mathcal {O}}, y^{{\mathcal {O}}}_r$ and $y^{{\mathcal {O}}}_s$ commute.

Now suppose that $s\ne r, r+1$. Then $y_s^{{\mathcal {O}}}$ commutes with $\frac{1}{M_r}f_{\mathbf {i}}^{{\mathcal {O}}}$ and with $T_r$. Hence, $\psi ^{\mathcal {O}}_r f_{\mathbf {i}}^{{\mathcal {O}}}y^{{\mathcal {O}}}_s=y^{{\mathcal {O}}}_s\psi ^{\mathcal {O}}_r f_{\mathbf {i}}^{{\mathcal {O}}}$, for any $\mathbf {i}\in I^n$. Therefore, $\psi ^{\mathcal {O}}_r y^{{\mathcal {O}}}_s=y^{{\mathcal {O}}}_s\psi ^{\mathcal {O}}_r$.

Lemma 4.18

Suppose that $\mathbf {i}\in I^n$. Then

$$\begin{aligned} \prod _{\begin{array}{c} 1\le l\le \ell \\ \kappa _l\equiv i_1\text { }(\text {mod } e) \end{array}} (y^{{\mathcal {O}}}_1-[\kappa _l-\hat{\imath }_1])f_{\mathbf {i}}^{\mathcal {O}}=0. \end{aligned}$$

Proof

By Definition 2.1, $\prod _{l=1}^{\ell }(L_1-[\kappa _l])=0$ so that $\prod _{l=1}^{\ell }(L_1-[\kappa _l])f_{\mathbf {i}}^{\mathcal {O}}=0$, for all $\mathbf {i}\in I$. If $\kappa _l\not \equiv i_1\text { }(\text {mod } e)$ then $[\hat{\imath }_1]\ne [\kappa _l]$ so that $(L_1-[\kappa _l])$ acts invertibly on $f_{\mathbf {i}}^{\mathcal {O}}\mathcal {H}^\varLambda _{n}$ by Proposition 4.5. Consequently, by Definition 4.12,

$$\begin{aligned} 0=\prod _{\begin{array}{c} 1\le l\le \ell \\ \kappa _l\equiv i_1\text { }(\text {mod } e) \end{array}} (t^{\hat{\imath }_1}y^{{\mathcal {O}}}_1+[\hat{\imath }_1]-[\kappa _l])f_{\mathbf {i}}^{\mathcal {O}}=t^{\hat{\imath }_1\langle \varLambda ,\alpha _{i_1}\rangle }\!\!\!\! \prod _{\begin{array}{c} 1\le l\le \ell \\ \kappa _l\equiv i_1\text { }(\text {mod } e) \end{array}} (y^{{\mathcal {O}}}_1-[\kappa _l-\hat{\imath }_1])f_{\mathbf {i}}^{\mathcal {O}}. \end{aligned}$$

As t is invertible in ${\mathcal {O}}$, the lemma follows.

Suppose that ${\mathfrak s}$ is a standard tableau, $\mathbf {i}={{\mathrm{res}}}({\mathfrak s})\in I^n$ and $1\le r<n$. Define

$$\begin{aligned} \beta _r({\mathfrak s})= {\left\{ \begin{array}{ll} \dfrac{t^{\hat{\imath }_{r}-c_{r}({\mathfrak s})}\alpha _r({\mathfrak s})}{[1-\rho _r({\mathfrak s})]}, &{}\text {if }i_r=i_{r+1},\\ t^{c_{r+1}({\mathfrak s})-\hat{\imath }_r}\alpha _r({\mathfrak s})[\rho _r({\mathfrak s})], &{}\text {if }i_r=i_{r+1}+1,\\ \dfrac{t^{-\rho _r({\mathfrak s})}\alpha _r({\mathfrak s})[\rho _r({\mathfrak s})]}{[1-\rho _r({\mathfrak s})]}, &{}\text {otherwise}, \end{array}\right. } \end{aligned}$$

(4.3)

and

$$\begin{aligned} \widehat{\beta }_r({\mathfrak s})= {\left\{ \begin{array}{ll} \dfrac{t^{\hat{\imath }_{r+1}-c_{r+1}({\mathfrak s})}\alpha _r({\mathfrak s})}{[1+\rho _{r}({\mathfrak s})]}, &{}\text {if }i_r=i_{r+1},\\ -t^{c_{r+1}({\mathfrak s})-\hat{\imath }_{r+1}}\alpha _r({\mathfrak s})[\rho _r({\mathfrak s})], &{}\text {if }i_r=i_{r+1}-1,\\ -\dfrac{\alpha _r({\mathfrak s})[\rho _r({\mathfrak s})]}{[1+\rho _r({\mathfrak s})]}, &{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

(4.4)

These scalars describe the action of $\psi ^{\mathcal {O}}_r$ upon the seminormal basis.

Lemma 4.19

Suppose that $1\le r<n$ and that $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n}).$ Set $\mathbf {i}={{\mathrm{res}}}({\mathfrak s}), \mathbf {j}={{\mathrm{res}}}({\mathfrak t}), {\mathfrak u}={\mathfrak s}(r,r+1)$ and ${\mathfrak v}={\mathfrak t}(r,r+1).$ Then

$$\begin{aligned} \psi ^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}} = \beta _r({\mathfrak s})f_{{\mathfrak u}{\mathfrak t}} -\delta _{i_ri_{r+1}}\frac{t^{\hat{\imath }_{r+1}-c_{r+1}({\mathfrak s})}}{[\rho _r({\mathfrak s})]}f_{{\mathfrak s}{\mathfrak t}},\\ \end{aligned}$$

and

$$\begin{aligned} f_{{\mathfrak s}{\mathfrak t}}\psi ^{\mathcal {O}}_r =\widehat{\beta }_r({\mathfrak t})f_{{\mathfrak s}{\mathfrak v}} -\delta _{j_rj_{r+1}}\frac{t^{\hat{\jmath }_{r+1}-c_{r+1}({\mathfrak t})}}{[\rho _r({\mathfrak t})]}f_{{\mathfrak s}{\mathfrak t}}. \end{aligned}$$

Similarly,  $y^{{\mathcal {O}}}_rf_{{\mathfrak s}{\mathfrak t}}=[c_r({\mathfrak s})-\hat{\imath }_r]f_{{\mathfrak s}{\mathfrak t}},$ and $f_{{\mathfrak s}{\mathfrak t}}y^{{\mathcal {O}}}_r=[c_r({\mathfrak t})-\hat{\jmath }_r]f_{{\mathfrak s}{\mathfrak t}},$ for $1\le r\le n.$

Proof

Applying Definition 4.12 and (3.5),

$$\begin{aligned} y^{{\mathcal {O}}}_r f_{{\mathfrak s}{\mathfrak t}}=t^{-\hat{\imath }_r}([c_r({\mathfrak s})]-[\hat{\imath }_r])f_{{\mathfrak s}{\mathfrak t}}=[c_r({\mathfrak s})-\hat{\imath }_r]f_{{\mathfrak s}{\mathfrak t}}. \end{aligned}$$

The proof that $f_{{\mathfrak s}{\mathfrak t}}y^{{\mathcal {O}}}_r=[c_r({\mathfrak t})-\hat{\jmath }_r]f_{{\mathfrak s}{\mathfrak t}}$ is similar. We now consider $\psi ^{\mathcal {O}}_r$.

By (3.6), if $\mathbf {k}\in I^n$ then $f_{\mathbf {k}}^{{\mathcal {O}}}f_{{\mathfrak s}{\mathfrak t}}=\delta _{\mathbf {i}\mathbf {k}}f_{{\mathfrak s}{\mathfrak t}}$. We use this observation below without mention. By Lemma 4.9, $(T_rL_r-L_rT_r)f_{{\mathfrak s}{\mathfrak t}}=\alpha _r({\mathfrak s})t^{c_{r+1}({\mathfrak s})}[\rho _r({\mathfrak s})]f_{{\mathfrak u}{\mathfrak t}}$. Hence, $\psi ^{\mathcal {O}}_r f_{{\mathfrak s}{\mathfrak t}}=\beta _r({\mathfrak s})f_{{\mathfrak u}{\mathfrak t}}$ when $i_r\ne i_{r+1}$ by Definition 4.12 and (4.2). Now suppose that $i_r=i_{r+1}$. Then, using (4.2) and (3.5),

$$\begin{aligned} \psi ^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}= & {} (1+T_r)\frac{t^{\hat{\imath }_r}}{M_r}f_{{\mathfrak s}{\mathfrak t}} =\frac{t^{\hat{\imath }_r-c_r({\mathfrak s})}}{[1-\rho _r({\mathfrak s})]}\left( \alpha _r({\mathfrak s})f_{{\mathfrak u}{\mathfrak t}} +\left( 1-\frac{1}{[\rho _r({\mathfrak s})]}\right) f_{{\mathfrak s}{\mathfrak t}}\right) \\= & {} \beta _r({\mathfrak s})f_{{\mathfrak u}{\mathfrak t}}-\frac{t^{\hat{\imath }_{r+1}-c_{r+1}({\mathfrak s})}}{[\rho _r({\mathfrak s})]}f_{{\mathfrak s}{\mathfrak t}}, \end{aligned}$$

as required. The formula for $f_{{\mathfrak s}{\mathfrak t}}\psi ^{\mathcal {O}}_r$ is proved similarly using Lemma 4.16 in place of Definition 4.12.

Note that, in general, $\psi ^{\mathcal {O}}_r f_{{\mathfrak s}{\mathfrak t}}\ne (f_{{\mathfrak t}{\mathfrak s}}\psi ^{\mathcal {O}}_r)^{*}$.

The next relation can also be proved using Lemmas 4.11 and 4.16.

Corollary 4.20

Suppose that $|r-t|>1,$ for $1\le r,t<n.$ Then $\psi ^{\mathcal {O}}_r\psi ^{\mathcal {O}}_t=\psi ^{\mathcal {O}}_t\psi ^{\mathcal {O}}_r.$

Proof

It follows easily from Lemma 4.19 that $\psi _r\psi _t f_{{\mathfrak s}{\mathfrak t}}=\psi _t\psi _r f_{{\mathfrak s}{\mathfrak t}}$, for all $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$. Hence, by Lemma 4.3, $\psi ^{\mathcal {O}}_r\psi ^{\mathcal {O}}_tf_{\mathbf {i}}^{\mathcal {O}}=\psi ^{\mathcal {O}}_t\psi ^{\mathcal {O}}_rf_{\mathbf {i}}^{\mathcal {O}}$, for all $\mathbf {i}\in I^n$.

Lemma 4.21

Suppose that $1\le r<n$ and $\mathbf {i}\in I^n.$ Then

$$\begin{aligned} \psi ^{\mathcal {O}}_r y^{{\mathcal {O}}}_{r+1}f_{\mathbf {i}}^{\mathcal {O}}=(y^{{\mathcal {O}}}_r\psi ^{\mathcal {O}}_r+\delta _{i_ri_{r+1}})f_{\mathbf {i}}^{\mathcal {O}}\quad \text {and}\quad y^{{\mathcal {O}}}_{r+1}\psi ^{\mathcal {O}}_rf_{\mathbf {i}}^{\mathcal {O}}=(\psi ^{\mathcal {O}}_ry^{{\mathcal {O}}}_r+\delta _{i_ri_{r+1}})f_{\mathbf {i}}^{\mathcal {O}}. \end{aligned}$$

Proof

Both formulas can be proved similarly, so we consider only the first one. We prove the stronger result that $\psi ^{\mathcal {O}}_r y^{{\mathcal {O}}}_{r+1}f_{{\mathfrak s}{\mathfrak t}}=(y^{{\mathcal {O}}}_r\psi ^{\mathcal {O}}_r+\delta _{i_ri_{r+1}})f_{{\mathfrak s}{\mathfrak t}}$, whenever $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ and ${{\mathrm{res}}}({\mathfrak s})=\mathbf {i}$. By (4.1) this implies the lemma.

Suppose first that $i_r=i_{r+1}$. Then, using Lemma 4.19,

$$\begin{aligned} \psi ^{\mathcal {O}}_ry^{{\mathcal {O}}}_{r+1}f_{{\mathfrak s}{\mathfrak t}} =[c_{r+1}({\mathfrak s})-\hat{\imath }_{r+1}]\left( \beta _r({\mathfrak s})f_{{\mathfrak u}{\mathfrak t}} -\frac{t^{\hat{\imath }_{r+1}-c_{r+1}({\mathfrak s})}}{[\rho _r({\mathfrak s})]}f_{{\mathfrak s}{\mathfrak t}}\right) . \end{aligned}$$

On the other hand, by Lemma 4.19 and (4.3),

$$\begin{aligned} (y^{{\mathcal {O}}}_r\psi ^{\mathcal {O}}_r+1)f_{{\mathfrak s}{\mathfrak t}}= & {} [c_r({\mathfrak u})-\hat{\imath }_{r+1}]\beta _r({\mathfrak s})f_{{\mathfrak u}{\mathfrak t}} +\left( 1-\frac{t^{\hat{\imath }_r-c_{r+1}({\mathfrak s})}[c_r({\mathfrak s})-\hat{\imath }_r]}{[\rho _r({\mathfrak s})]} \right) f_{{\mathfrak s}{\mathfrak t}}\\= & {} [c_r({\mathfrak u})-\hat{\imath }_{r+1}]\beta _r({\mathfrak s})f_{{\mathfrak u}{\mathfrak t}} +\frac{[\hat{\imath }_{r+1}-c_{r+1}({\mathfrak s})]}{[\rho _r({\mathfrak s})]}f_{{\mathfrak s}{\mathfrak t}}. \end{aligned}$$

Therefore, $\psi ^{\mathcal {O}}_ry^{{\mathcal {O}}}_{r+1}f_{{\mathfrak s}{\mathfrak t}}=(y^{{\mathcal {O}}}_r\psi ^{\mathcal {O}}_r+1)f_{{\mathfrak s}{\mathfrak t}}$ since $c_r({\mathfrak u})=c_{r+1}({\mathfrak s})$ and $i_r=i_{r+1}$.

If $i_r\ne i_{r+1}$ then the calculation is easier because

$$\begin{aligned} \psi ^{\mathcal {O}}_ry^{{\mathcal {O}}}_{r+1}f_{{\mathfrak s}{\mathfrak t}} =[c_{r+1}({\mathfrak s})-\hat{\imath }_{r+1}]\beta _r({\mathfrak s})f_{{\mathfrak u}{\mathfrak t}} =y^{{\mathcal {O}}}_r\psi ^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}, \end{aligned}$$

where, for the last equality, we again use the fact that $c_r({\mathfrak u})=c_{r+1}({\mathfrak s})$.

The following simple combinatorial identity largely determines both the quadratic and the (deformed) braid relations for the $\psi ^{\mathcal {O}}_r$, for $1\le r<n$. This result can be viewed as a graded analogue of the defining property Definition 3.5(c) of a seminormal coefficient system.

Lemma 4.22

Suppose that $1\le r<n$ and ${\mathfrak s},{\mathfrak u}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ with ${\mathfrak u}={\mathfrak s}(r,r+1)$ and ${{\mathrm{res}}}({\mathfrak s})=\mathbf {i}\in I^n,$ for ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}.$ Then

$$\begin{aligned} \beta _r({\mathfrak s})\beta _r({\mathfrak u})&={\left\{ \begin{array}{ll} t^{c_r({\mathfrak s})+c_{r+1}({\mathfrak s})-\hat{\imath }_r-\hat{\imath }_{r+1}}[1-\rho _r({\mathfrak s})][1+\rho _r({\mathfrak s})], &{}\text {if }i_r\leftrightarrows i_{r+1},\\ t^{c_{r+1}({\mathfrak s})-\hat{\imath }_{r+1}}[1+\rho _r({\mathfrak s})],&{}\text {if }i_r\rightarrow i_{r+1},\\ t^{c_r({\mathfrak s})-\hat{\imath }_r}[1-\rho _r({\mathfrak s})],&{}\text {if }i_r\leftarrow i_{r+1},\\ -\frac{t^{2\hat{\imath }_r-2c_{r+1}({\mathfrak s})}}{[\rho _r({\mathfrak s})]^2},&{}\text {if }i_r=i_{r+1},\\ 1,&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

The lemma follows directly from the definition of $\beta _r({\mathfrak s})$ using Definition 3.5(c).

It is time to pay the price for the failure of the embedding $I\hookrightarrow \mathbb {Z}$ to extend to an embedding of quivers. Together with the cyclotomic relation, this is place where the KLR grading fails to lift to the algebra $\mathcal {H}^\varLambda _n({\mathcal {O}})$. Recall from Definition 4.12 that $y^{{\mathcal {O}}}_r f_{\mathbf {i}}^{\mathcal {O}}=t^{-\hat{\imath }_r}(L_r-[\hat{\imath }_r])f_{\mathbf {i}}^{\mathcal {O}}$, where $1\le r\le n$ and $\mathbf {i}\in I^n$. For $d\in \mathbb {Z}$ define

$$\begin{aligned} y^{\langle d\rangle }_rf_{\mathbf {i}}^{\mathcal {O}}=t^{d-\hat{\imath }_r}(L_r-[\hat{\imath }_r-d])f_{\mathbf {i}}^{\mathcal {O}}=(t^dy^{{\mathcal {O}}}_r+[d])f_{\mathbf {i}}^{\mathcal {O}}. \end{aligned}$$

(4.5)

In particular, $y^{\langle 0\rangle }_r=y^{{\mathcal {O}}}_r$ and $y^{\langle d\rangle }_r\otimes _{\mathcal {O}}1_K=y^{{\mathcal {O}}}_r\otimes _{\mathcal {O}}1_K$ whenever e divides $d\in \mathbb {Z}$,

As a final piece of notation, set $\rho _r(\mathbf {i})=\hat{\imath }_r-\hat{\imath }_{r+1}\in \mathbb {Z}$, for $\mathbf {i}\in I^n$ and $1\le r<n$.

Proposition 4.23

Suppose that $1\le r<n$ and $\mathbf {i}\in I^n$. Then

$$\begin{aligned} (\psi ^{\mathcal {O}}_r)^2f_{\mathbf {i}}^{\mathcal {O}}= {\left\{ \begin{array}{ll} (y^{\langle 1+\rho _r(\mathbf {i})\rangle }_r-y^{{\mathcal {O}}}_{r+1})(y^{\langle 1-\rho _r(\mathbf {i})\rangle }_{r+1}-y^{{\mathcal {O}}}_r)f_{\mathbf {i}}^{\mathcal {O}},&{} \text {if }i_r\leftrightarrows i_{r+1},\\ (y^{\langle 1+\rho _r(\mathbf {i})\rangle }_r-y^{{\mathcal {O}}}_{r+1})f_{\mathbf {i}}^{\mathcal {O}},&{} \text {if }i_r\rightarrow i_{r+1},\\ (y^{\langle 1-\rho _r(\mathbf {i})\rangle }_{r+1}-y^{{\mathcal {O}}}_r)f_{\mathbf {i}}^{\mathcal {O}},&{} \text {if }i_r\leftarrow i_{r+1},\\ 0,&{}\text {if }i_r=i_{r+1},\\ f_{\mathbf {i}}^{\mathcal {O}},&{}\text {otherwise.} \end{array}\right. } \end{aligned}$$

Proof

Once again, by (4.1) it is enough to prove the corresponding formulas for $(\psi ^{\mathcal {O}}_r)^2 f_{{\mathfrak s}{\mathfrak t}}$, where $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ and $\mathbf {i}={{\mathrm{res}}}({\mathfrak s})$.

Suppose that $i_r=i_{r+1}$. Let ${\mathfrak u}={\mathfrak s}(r,r+1)$ and $\mathbf {j}={{\mathrm{res}}}({\mathfrak u})$. By Lemma 4.19,

$$\begin{aligned} (\psi ^{\mathcal {O}}_r)^2f_{{\mathfrak s}{\mathfrak t}}= & {} \left( \frac{t^{2\hat{\imath }_r-2c_{r+1}({\mathfrak s})}}{[\rho _r({\mathfrak s})]^2}+\beta _r({\mathfrak s})\beta _r({\mathfrak u})\right) f_{{\mathfrak s}{\mathfrak t}}\\&-\left( \frac{\beta _r({\mathfrak s})t^{\hat{\imath }_r-c_r({\mathfrak s})}}{[\rho _r({\mathfrak u})]} +\frac{\beta _r({\mathfrak s})t^{\hat{\jmath }_r-c_r({\mathfrak u})}}{[\rho _r({\mathfrak s})]}\right) f_{{\mathfrak u}{\mathfrak t}}. \end{aligned}$$

Note that $\rho _r({\mathfrak s})=-\rho _r({\mathfrak u})$ and $i_r=j_r$, so that $t^{\hat{\jmath }_r-c_r({\mathfrak u})}[\rho _r({\mathfrak u})]=-t^{\hat{\imath }_r-c_r({\mathfrak s})}[\rho _r({\mathfrak s})]$. Hence, using Lemma 4.22, $(\psi ^{\mathcal {O}}_r)^2f_{{\mathfrak s}{\mathfrak t}}=0$ when $i_r=i_{r+1}$ as claimed.

Now suppose that $i_r\ne i_{r+1}$. Then, by Lemmas 4.19 and 4.22,

$$\begin{aligned} (\psi ^{\mathcal {O}}_r)^2f_{{\mathfrak s}{\mathfrak t}}= & {} \beta _r({\mathfrak s})\beta _r({\mathfrak u})f_{{\mathfrak s}{\mathfrak t}}\\= & {} {\left\{ \begin{array}{ll} t^{c_r({\mathfrak s})+c_{r+1}({\mathfrak s})-\hat{\imath }_r-\hat{\imath }_{r+1}}[1-\rho _r({\mathfrak s})][1+\rho _r({\mathfrak s})]f_{{\mathfrak s}{\mathfrak t}}, &{}\text {if }i_r\leftrightarrows i_{r+1},\\ t^{c_{r+1}({\mathfrak s})-\hat{\imath }_{r+1}}[1+\rho _r({\mathfrak s})]f_{{\mathfrak s}{\mathfrak t}},&{}\text {if }i_r\rightarrow i_{r+1},\\ t^{c_{r}({\mathfrak s})-\hat{\imath }_{r}}[1-\rho _r({\mathfrak s})]f_{{\mathfrak s}{\mathfrak t}},&{}\text {if }i_r\leftarrow i_{r+1},\\ f_{{\mathfrak s}{\mathfrak t}},&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

As in Lemma 4.19, if $d\in \mathbb {Z}$ then $y^{\langle d\rangle }_rf_{{\mathfrak s}{\mathfrak t}}=[c_r({\mathfrak s})-\hat{\imath }_r+d]f_{{\mathfrak s}{\mathfrak t}}$. So, if $i_r\rightarrow i_{r+1}$ then

$$\begin{aligned} (y^{\langle 1+\rho _r(\mathbf {i})\rangle }_r-y^{{\mathcal {O}}}_{r+1})f_{{\mathfrak s}{\mathfrak t}}= & {} ([c_r({\mathfrak s})+1-\hat{\imath }_{r+1}]-[c_{r+1}({\mathfrak s})-\hat{\imath }_{r+1}])f_{{\mathfrak s}{\mathfrak t}}\\= & {} t^{c_{r+1}({\mathfrak s})-\hat{\imath }_{r+1}}[1+\rho _r({\mathfrak s})]f_{{\mathfrak s}{\mathfrak t}} =(\psi ^{\mathcal {O}}_r)^2f_{{\mathfrak s}{\mathfrak t}}. \end{aligned}$$

The cases when $i_r\rightarrow i_{r+1}$ and $i_r\leftrightarrows i_{r+1}$ are similar.

Set $\mathcal {B}^{\mathcal {O}}_r=\psi ^{\mathcal {O}}_r\psi ^{\mathcal {O}}_{r+1}\psi ^{\mathcal {O}}_r-\psi ^{\mathcal {O}}_{r+1}\psi ^{\mathcal {O}}_r\psi ^{\mathcal {O}}_{r+1}$, for $1\le r<n-1$.

Proposition 4.24

Suppose that $1\le r<n$ and ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }\ ({\varvec{\lambda }})$, with ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$ for $\mathbf {i}\in I^n$. Then

$$\begin{aligned} \mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}= {\left\{ \begin{array}{ll} (y^{\langle 1+\rho _r(\mathbf {i})\rangle }_r+y^{\langle 1+\rho _r(\mathbf {i})\rangle }_{r+2}- y^{\langle 1+\rho _r(\mathbf {i})\rangle }_{r+1}-y^{\langle 1-\rho _r(\mathbf {i})\rangle }_{r+1})f_{{\mathfrak s}{\mathfrak t}}, &{}\text {if }i_{r+2}=i_r\rightleftarrows i_{r+1},\\ -t^{1+\rho _r(\mathbf {i})}f_{{\mathfrak s}{\mathfrak t}}, &{}\text {if }i_{r+2}=i_r\rightarrow i_{r+1},\\ f_{{\mathfrak s}{\mathfrak t}}, &{}\text {if }i_{r+2}=i_r\leftarrow i_{r+1},\\ 0,&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Proof

We mimic the proof of the braid relations from Lemma 3.8.

Define (not necessarily standard) tableaux ${\mathfrak u}_1={\mathfrak s}(r,r+1), {\mathfrak u}_2={\mathfrak s}(r+1,r+2), {\mathfrak u}_{12}={\mathfrak u}_1(r+1,r+2), {\mathfrak u}_{21}={\mathfrak u}_2(r,r+1)$ and ${\mathfrak u}_{121}={\mathfrak u}_{12}(r,r+1)={\mathfrak u}_{212}$. To ease notation set $i=i_r, j=i_{r+1}$ and $k=i_{r+2}$. The relationship between these tableaux, and their residues $\{{{\mathrm{res}}}_s({\mathfrak u})\,|\,r\le s\le r+2\}=\{i,j,k\}$, is illustrated in the following diagram.

Note that if any tableau ${\mathfrak u}\in \{{\mathfrak u}_1,{\mathfrak u}_2,{\mathfrak u}_{12},{\mathfrak u}_{21},{\mathfrak u}_{121}\}$ is not standard then, by definition, $f_{{\mathfrak u}{\mathfrak t}}=0$ so this term can be ignored in all of the calculations below.

We need to compute $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}$. To start with, observe that by Lemma 4.19, the coefficient of $f_{{\mathfrak u}_{121}{\mathfrak t}}$ in $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}$ is equal to

$$\begin{aligned} \beta _r({\mathfrak s})\beta _{r+1}({\mathfrak u}_{1})\beta _r({\mathfrak u}_{12}) -\beta _{r+1}({\mathfrak s})\beta _r({\mathfrak u}_{2})\beta _{r+1}({\mathfrak u}_{21}). \end{aligned}$$

By definition, the scalars $[\rho _r({\mathfrak s})]$ and $[1-\rho _r({\mathfrak s})]$ are determined by the positions of r and $r+1$ in ${\mathfrak s}$, so it is easy to see that

$$\begin{aligned} \rho _r({\mathfrak s})= & {} \rho _{r+1}({\mathfrak u}_{21}),\quad \rho _r({\mathfrak u}_1)=\rho _{r+1}({\mathfrak u}_{121}),\quad \rho _r({\mathfrak u}_{2})=\rho _{r+1}({\mathfrak u}_{1}),\nonumber \\ \rho _r({\mathfrak u}_{12})= & {} \rho _{r+1}({\mathfrak s}),\quad \rho _r({\mathfrak u}_{21})=\rho _{r+1}({\mathfrak u}_{12}),\quad \rho _r({\mathfrak u}_{121})=\rho _{r+1}({\mathfrak u}_{2}). \end{aligned}$$

(4.6)

Observe that $\alpha _r({\mathfrak s})\alpha _{r+1}({\mathfrak u}_1)\alpha _r({\mathfrak u}_{12}) =\alpha _{r+1}({\mathfrak s})\alpha _r({\mathfrak u}_2)\alpha _{r+1}(u_{21})$ by Definition 3.5(a). Keeping track of the exponent of t, (4.3) and (4.6) now imply that $\beta _r({\mathfrak s})\beta _{r+1}({\mathfrak u}_1)\beta _r({\mathfrak u}_{12}) =\beta _{r+1}({\mathfrak s})\beta _r({\mathfrak u}_2)\beta _{r+1}({\mathfrak u}_{21})$. Note that Definition 3.5(a) is crucial here. Therefore, the coefficient of $f_{{\mathfrak u}_{121}{\mathfrak t}}$ in $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}$ is zero for any choice of i, j and k. As the coefficient of $f_{{\mathfrak u}_{121}{\mathfrak t}}$ in $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}$ is always zero we will omit $f_{{\mathfrak u}_{121}{\mathfrak t}}$ from most of the calculations that follow.

There are five cases to consider.

Case 1. i, j and k are pairwise distinct.

By Lemma 4.19 and the last paragraph,

$$\begin{aligned} \mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}} =\big (\beta _r({\mathfrak s})\beta _{r+1}({\mathfrak u}_1)\beta _r({\mathfrak u}_{12}) -\beta _{r+1}({\mathfrak s})\beta _r({\mathfrak u}_2)\beta _{r+1}({\mathfrak u}_{21})\big )f_{{\mathfrak u}_{121}{\mathfrak t}}=0, \end{aligned}$$

as required by the statement of the proposition.

Case 2. $i=j\ne k$.

In this case, using Lemma 4.19,

$$\begin{aligned} \mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}} =\beta _{r+1}({\mathfrak s})\beta _r({\mathfrak u}_2)\left( -\frac{t^{\hat{\imath }-c_{r+1}({\mathfrak s})}}{[\rho _r({\mathfrak s})]} +\frac{t^{\hat{\imath }-c_{r+2}({\mathfrak u}_{21})}}{[\rho _{r+1}({\mathfrak u}_{21})]}\right) f_{{\mathfrak u}_{21}{\mathfrak t}}. \end{aligned}$$

Now $\rho _r({\mathfrak s})=\rho _{r+1}({\mathfrak u}_{21})$ and $c_{r+1}({\mathfrak s})=c_{r+2}({\mathfrak u}_{21})$, as in (4.6). Hence, $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}=0$ when $i=j\ne k$.

Case 3. $i\ne j=k$.

This is almost identical to Case 2, so we leave the details to the reader.

Case 4. $i=k\ne j$.

Typographically, it is convenient to set $c=c_r({\mathfrak s}), c'=c_{r+1}({\mathfrak s})$ and $c''=c_{r+2}({\mathfrak s})$. According to the statement of the proposition, this is the only case where $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}\ne 0$. Using Lemma 4.19, we see that

$$\begin{aligned} \mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}= & {} \left( -\beta _r({\mathfrak s})\frac{t^{\hat{\imath }-c_{r+2}({\mathfrak u}_1)}}{[\rho _{r+1}({\mathfrak u}_1)]}\beta _r({\mathfrak u}_1) +\beta _{r+1}({\mathfrak s})\frac{t^{\hat{\imath }-c_{r+1}({\mathfrak u}_2)}}{[\rho _r({\mathfrak u}_2)]}\beta _{r+1}({\mathfrak u}_2)\right) f_{{\mathfrak s}{\mathfrak t}}.\\= & {} \frac{t^{\hat{\imath }-c''}}{[c-c'']} (-\beta _r({\mathfrak s})\beta _r({\mathfrak u}_1)+\beta _{r+1}({\mathfrak s})\beta _{r+1}({\mathfrak u}_2))f_{{\mathfrak s}{\mathfrak t}}. \end{aligned}$$

Expanding the last equation using Lemma 4.22 shows that $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}} =$

$$\begin{aligned}&\quad \begin{array}{ll} -\dfrac{t^{c}[1-\rho _r({\mathfrak s})][1+\rho _r({\mathfrak s})] -t^{c''}[1-\rho _{r+1}({\mathfrak s})][1+\rho _{r+1}({\mathfrak s})]}{t^{c''-c'+\hat{\jmath }}[c-c'']}f_{{\mathfrak s}{\mathfrak t}}, &{}\quad \text {if }i\leftrightarrows j,\\ -\dfrac{[1+\rho _r({\mathfrak s})]-[1-\rho _{r+1}({\mathfrak s})]}{t^{c''-\hat{\imath }-c'+\hat{\jmath }}[c-c'']}f_{{\mathfrak s}{\mathfrak t}}, &{}\quad \text {if }i\rightarrow j,\\ -\dfrac{t^{c}[1-\rho _r({\mathfrak s})]-t^{c''}[1+\rho _{r+1}({\mathfrak s})]}{t^{c''}[c-c'']}f_{{\mathfrak s}{\mathfrak t}}, &{}\quad \text {if }i\leftarrow j,\\ 0,&{}\text {otherwise.} \end{array} \end{aligned}$$

(Note that, by assumption, the case $i=j$ does not arise.) If $i\leftrightarrows j$ then a straightforward calculation shows that in this case

$$\begin{aligned} \mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}= & {} -([c'-\hat{\jmath }+2]+[c'-\hat{\jmath }]-[c+1-\hat{\jmath }]-[c''+1-\hat{\jmath }])f_{{\mathfrak s}{\mathfrak t}}\\= & {} -\big (y^{\langle 1+\rho _r(\mathbf {i})\rangle }_{r+1}+y^{\langle 1-\rho _r(\mathbf {i})\rangle }_{r+1}-y^{\langle 1+\rho _r(\mathbf {i})\rangle }_r-y^{\langle 1+\rho _r(\mathbf {i})\rangle }_{r+2}\big )f_{{\mathfrak s}{\mathfrak t}}, \end{aligned}$$

where the last equality uses Lemma 4.21 and the observation that, because $e=2$, we have $\{1\pm \rho _r(\mathbf {i})\}=\{0,2\}$ and $\{\hat{\imath },\hat{\jmath }\}=\{0,1\}$. A similar, but easier, calculation shows that if $i\rightarrow j$ then $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}=-t^{1+\hat{\imath }-\hat{\jmath }}f_{{\mathfrak s}{\mathfrak t}}=-t^{1+\rho _r(\mathbf {i})}f_{{\mathfrak s}{\mathfrak t}}$ and if $i\leftarrow j$ then $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}=f_{{\mathfrak s}{\mathfrak t}}$. If $i\ne j$ and $i{{\mathrm{\,{/}\text {---}\,}}}j$ then we have already seen that $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}=0$, so this completes the proof of Case 4.

Case 5. $i=j=k$.

We continue to use the notation for $c,c',c''$ from Case 4. By Lemma 4.19 (compare with the proof of Lemma 3.8), $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}$ is equal to

$$\begin{aligned}&-\left( \frac{t^{3\hat{\imath }-2c'-c''}}{[\rho _r({\mathfrak s})]^2[\rho _{r+1}({\mathfrak s})]} -\frac{t^{3\hat{\imath }-c'-2c''}}{[\rho _{r+1}({\mathfrak s})]^2[\rho _r({\mathfrak s})]} +\frac{t^{\hat{\imath }-c''}\beta _r({\mathfrak s})\beta _r({\mathfrak u}_1)}{[\rho _{r+1}({\mathfrak u}_1)]} -\frac{t^{\hat{\imath }-c''}\beta _{r+1}({\mathfrak s})\beta _{r+1}({\mathfrak u}_2)}{[\rho _r({\mathfrak u}_2)]}\right) f_{{\mathfrak s}{\mathfrak t}}\\&\quad +t^{2\hat{\imath }}\beta _r({\mathfrak s})\left( \frac{t^{-c''-c}}{[\rho _{r+1}({\mathfrak u}_1)][\rho _r({\mathfrak u}_1)]} +\frac{t^{-c'-c''}}{[\rho _r({\mathfrak s})][\rho _{r+1}({\mathfrak s})]} -\frac{t^{-2c''}}{[\rho _{r+1}({\mathfrak s})][\rho _{r+1}({\mathfrak u}_1)]}\right) f_{{\mathfrak u}_1{\mathfrak t}}\\&\quad +t^{2\hat{\imath }}\beta _{r+1}({\mathfrak s})\left( \frac{t^{-c'-c''}}{[\rho _r({\mathfrak s})][\rho _r({\mathfrak u}_2)]} -\frac{t^{-c''-c'}}{[\rho _r({\mathfrak u}_2)][\rho _{r+1}({\mathfrak u}_2)]} -\frac{t^{-c''-c'}}{[\rho _{r+1}({\mathfrak s})][\rho _r({\mathfrak s})]}\right) f_{{\mathfrak u}_2{\mathfrak t}}\\&\quad -t^{\hat{\imath }-c''}\beta _r({\mathfrak s})\beta _{r+1}({\mathfrak u}_1) \left( \frac{1}{[\rho _r({\mathfrak u}_{12})]}-\frac{1}{[\rho _{r+1}({\mathfrak s})]}\right) f_{{\mathfrak u}_{12}{\mathfrak t}}\\&\quad -t^{\hat{\imath }-c'}\beta _{r+1}({\mathfrak s})\beta _r({\mathfrak u}_2) \left( \frac{1}{[\rho _r({\mathfrak s})]}-\frac{1}{[\rho _{r+1}({\mathfrak u}_{21})]}\right) f_{{\mathfrak u}_{21}{\mathfrak t}}. \end{aligned}$$

Using (4.6) it is easy to see that the coefficients of $f_{{\mathfrak u}_{12}{\mathfrak t}}$ and$f_{{\mathfrak u}_{21}{\mathfrak t}}$ are both zero. On the other hand, if $t\ne 1$ then the coefficient of $t^{2\hat{\imath }}\beta _r({\mathfrak s})f_{{\mathfrak u}_1{\mathfrak t}}$ in $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}$ is

$$\begin{aligned} \frac{t-1}{(t^{c'}-t^c)(t^c-t^{c''})} +\frac{t-1}{(t^{c}-t^{c'})(t^{c'}-t^{c''})} -\frac{t-1}{(t^{c'}-t^{c''})(t^c-t^{c''})}=0. \end{aligned}$$

The case when $t=1$ now follows by specialisation. Similarly, the coefficient of $f_{{\mathfrak u}_2{\mathfrak t}}$ in $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}$ is also zero. Finally, using Lemma 4.22 and (4.6), the coefficient of $f_{{\mathfrak s}{\mathfrak t}}$ in $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}$ is zero as the four terms above, which give the coefficient of $f_{{\mathfrak s}{\mathfrak t}}$ in the displayed equation, cancel out in pairs. Hence, $\mathcal {B}^{\mathcal {O}}_rf_{{\mathfrak s}{\mathfrak t}}=0$ when $i=j=k$, as required.

This completes the proof.

4.4 A deformation of the quiver Hecke algebra

Using the results of the last two sections we now describe $\mathcal {H}^\varLambda _{n}({\mathcal {O}})$ by generators and relations using the ‘${\mathcal {O}}$-KLR generators’ of $\mathcal {H}^\varLambda _n({\mathcal {O}})$.

Let $R_{n}({\mathcal {O}})$ be the abstract algebra defined by the generators and relations in the statement of Theorem A. We abuse notation and use the same symbols for the generators of $R_{n}({\mathcal {O}})$ and the corresponding elements in $\mathcal {H}^\varLambda _n({\mathcal {O}})$ that we defined in Sect. 4.3. The previous section shows that there is a surjection $R_{n}({\mathcal {O}})\twoheadrightarrow \mathcal {H}^\varLambda _n({\mathcal {O}})$. We want to prove that this map is an isomorphism.

The next lemma, which is modelled on [6, Lemma 2.1], will be used to show that $R_{n}({\mathcal {O}})$ is finitely generated as an ${\mathcal {O}}$-module.

Lemma 4.25

Suppose that $1\le r\le n$ and $\mathbf {i}\in I^n.$ Then there exists a multiset $X_r(\mathbf {i})\subseteq e\mathbb {Z}$ such that

$$\begin{aligned} \prod _{c\in X_r(\mathbf {i})}(y^{{\mathcal {O}}}_r-[c])f_{\mathbf {i}}^{\mathcal {O}}=0. \end{aligned}$$

Proof

We argue by induction on r. If $r=1$ then the relations in $R_{n}({\mathcal {O}})$ say that we can take $X_1(\mathbf {i})$ to be the multiset with elements $\kappa _l-\hat{\imath }_1$, where $\kappa _l\equiv i_1\text { }(\text {mod } e)$ for $1\le l\le \ell$. By induction we assume that we have proved the result for $y^{{\mathcal {O}}}_r$ and use this to prove the result for $y^{{\mathcal {O}}}_{r+1}$. There are three cases to consider. Throughout, let $\mathbf {j}={\mathfrak s}_r\cdot \mathbf {i}$.

Case 1. $i_{r+1}{{\mathrm{\,{/}\text {---}\,}}}i_r$.

Set $X_{r+1}(\mathbf {i})=X_r(\mathbf {j})$. Then, using the relations in $R_{n}({\mathcal {O}})$,

$$\begin{aligned} \prod _{c\in X_{r+1}(\mathbf {i})}(y^{{\mathcal {O}}}_{r+1}-[c])f_{\mathbf {i}}^{\mathcal {O}}= & {} \prod _{c\in X_{r+1}(\mathbf {i})}(y^{{\mathcal {O}}}_{r+1}-[c])(\psi ^{\mathcal {O}}_r)^2f_{\mathbf {i}}^{\mathcal {O}}\\= & {} \psi ^{\mathcal {O}}_r\prod _{c\in X_r(\mathbf {j})}(y^{{\mathcal {O}}}_r-[c])f_{\mathbf {j}}^{\mathcal {O}}\psi ^{\mathcal {O}}_r =0, \end{aligned}$$

where the last equality follows by induction.

Case 2. $i_r\rightarrow i_{r+1}$ or $i_r\leftarrow i_{r+1}$

We consider only the case when $i_r\rightarrow i_{r+1}$. The case when $i_r\leftarrow i_{r+1}$ is similar. Using the quadratic relation for $\psi ^{\mathcal {O}}_r$ in $R_{n}({\mathcal {O}})$,

$$\begin{aligned} (y^{{\mathcal {O}}}_{r+1}-[c])f_{\mathbf {i}}^{\mathcal {O}}=t^{1+\rho _i(\mathbf {i})}(y^{{\mathcal {O}}}_r-[c-1-\rho _i(\mathbf {i})]) f_{\mathbf {i}}^{\mathcal {O}}-(\psi ^{\mathcal {O}}_r)^2f_{\mathbf {i}}^{\mathcal {O}}. \end{aligned}$$

Let $X_{r+1}(\mathbf {i})$ be the disjoint union $X_r(\mathbf {j})\sqcup X^+_{r+1}(\mathbf {i})$, where $X^+_{r+1}(\mathbf {i})$ is the multiset $\{c+1+\rho _r(\mathbf {i})\,|\,c\in X_r(\mathbf {i})\}$. If $d=c+1+\rho _r(\mathbf {i})\in X^+_{r+1}(\mathbf {i})$ then, by the last displayed equation,

$$\begin{aligned} \prod _{c'\in X_{r+1}(\mathbf {i})}(y^{{\mathcal {O}}}_{r+1}-[c'])f_{\mathbf {i}}^{\mathcal {O}}= & {} \prod _{{c'\in X_{r+1}(\mathbf {i}){\setminus }\{d\}}}\quad (y^{{\mathcal {O}}}_{r+1}-[c'])\cdot \big (t^{1+\rho _r(\mathbf {i})}(y^{{\mathcal {O}}}_r-[c])-(\psi ^{\mathcal {O}}_r)^2\big )f_{\mathbf {i}}^{\mathcal {O}}\\= & {} t^{1+\rho _r(\mathbf {i})}(y^{{\mathcal {O}}}_r-[c])\prod _{c'\in X_{r+1}(\mathbf {i}){\setminus }\{d\}} (y^{{\mathcal {O}}}_{r+1}-[c'])f_{\mathbf {i}}^{\mathcal {O}}\\&-\psi ^{\mathcal {O}}_r\prod _{c'\in X_{r+1}(\mathbf {i}){\setminus }\{d\}} (y^{{\mathcal {O}}}_r-[c'])f_{\mathbf {j}}^{\mathcal {O}}\psi ^{\mathcal {O}}_r. \end{aligned}$$

The second summand is zero by induction because $X_r(\mathbf {j})$ is contained in $X_{r+1}(\mathbf {i})$. Therefore, arguing this way for every $d\in X^+_{r+1}(\mathbf {i})$, there exists $N\in \mathbb {Z}$ such that

$$\begin{aligned} \prod _{c'\in X_{r+1}(\mathbf {i})}(y^{{\mathcal {O}}}_{r+1}-[c'])f_{\mathbf {i}}^{\mathcal {O}}=t^{Ne} \prod _{c'\in X_r(\mathbf {j})}(y^{{\mathcal {O}}}_{r+1}-[c'])\cdot \prod _{c\in X_r(\mathbf {i})}(y^{{\mathcal {O}}}_r-[c])f_{\mathbf {i}}^{\mathcal {O}}=0, \end{aligned}$$

where we again use induction for the last equality. This completes the proof of the inductive step when $i_r\rightarrow i_{r+1}$.

Case 3. $i_r\leftrightarrows i_{r+1}$.

This is similar to Case 2 but slightly more involved. Define $X_{r+1}(\mathbf {i})=X_r(\mathbf {j})\sqcup X^+_{r+1}(\mathbf {i})\sqcup X^-_{r+1}(\mathbf {i})$, where $X^\pm _{r+1}=\{c\pm 1+\rho _r(\mathbf {i})\,|\,c\in X_r(\mathbf {i})\}$. If $c\in X_r(\mathbf {i})$ set

$$\begin{aligned} c^{+}:=c+1+\rho _r(\mathbf {i}),\quad c^{-}:=c-1+\rho _r(\mathbf {i}). \end{aligned}$$

Using the equality

$$\begin{aligned} (\psi ^{\mathcal {O}}_r)^2f_{\mathbf {i}}^{\mathcal {O}}{=}(t^{1+\rho _r(\mathbf {i})}y^{{\mathcal {O}}}_r+[1+\rho _r(\mathbf {i})] -y^{{\mathcal {O}}}_{r+1})(t^{1-\rho _r(\mathbf {i})}y^{{\mathcal {O}}}_{r+1}+[1-\rho _r(\mathbf {i})]-y^{{\mathcal {O}}}_r)f_{\mathbf {i}}^{\mathcal {O}}, \end{aligned}$$

we see that

$$\begin{aligned} (\psi ^{\mathcal {O}}_r)^2f_{\mathbf {i}}^{\mathcal {O}}=-t^{1-\rho _r(\mathbf {i})}(y^{{\mathcal {O}}}_{r+1}-[c^{+}])(y^{{\mathcal {O}}}_{r+1}-[c^{-}])f_{\mathbf {i}}^{\mathcal {O}}+(y^{{\mathcal {O}}}_r-[c])f_{\mathbf {i}}^{\mathcal {O}}F_r(y,c), \end{aligned}$$

where $F_r(y,c)$ is a polynomial in $y^{{\mathcal {O}}}_r$ and $y^{{\mathcal {O}}}_{r+1}$ with coefficients in $\mathbb {Z}[t,t^{-1}]$. Hence,

$$\begin{aligned}&\prod _{c'\in X_{r+1}(\mathbf {i})}(y^{{\mathcal {O}}}_{r+1}-[c'])f_{\mathbf {i}}^{\mathcal {O}}\\&\quad =\prod _{c'\in X_{r+1}(\mathbf {i}){\setminus }\{c^{+},c^{-}\}} (y^{{\mathcal {O}}}_{r+1}-[c'])\cdot t^{\rho _r(\mathbf {i})-1}((y^{{\mathcal {O}}}_r-[c])f_{\mathbf {i}}^{\mathcal {O}}F_r(y,c)-(\psi ^{\mathcal {O}}_r)^2f_{\mathbf {i}}^{\mathcal {O}})\\&\quad =t^{\rho _r(\mathbf {i})-1}(y^{{\mathcal {O}}}_r-[c]) \prod _{c'\in X_{r+1}(\mathbf {i}){\setminus }\{c^{+},c^{-}\}} (y^{{\mathcal {O}}}_{r+1}-[c'])f_{\mathbf {i}}^{\mathcal {O}}F_r(y,c)\\&\qquad -t^{\rho _r(\mathbf {i})-1}\psi ^{\mathcal {O}}_r\prod _{c'\in X_{r+1}(\mathbf {i}){\setminus }\{c^{+},c^{-}\}} (y^{{\mathcal {O}}}_r-[c'])f_{\mathbf {j}}^{\mathcal {O}}\psi ^{\mathcal {O}}_r. \end{aligned}$$

The second summand is zero by induction because $X_r(\mathbf {j})$ is contained in $X_{r+1}(\mathbf {i})$. Therefore, arguing this way for every $c\in X_{r}(\mathbf {i})$, there exists $N\in \mathbb {Z}$ such that

$$\begin{aligned} \prod _{c'\in X_{r+1}(\mathbf {i})}(y^{{\mathcal {O}}}_{r+1}-[c'])f_{\mathbf {i}}^{\mathcal {O}}{=}t^{Ne} \prod _{c'\in X_r(\mathbf {j})}(y^{{\mathcal {O}}}_{r+1}-[c'])\cdot \prod _{c\in X_r(\mathbf {i})}(y^{{\mathcal {O}}}_r-[c])f_{\mathbf {i}}^{\mathcal {O}}F_r(y)=0, \end{aligned}$$

where $F_r(y)$ is a polynomial in $y^{{\mathcal {O}}}_1,\ldots ,y^{{\mathcal {O}}}_{n}$ with coefficients in $\mathbb {Z}[t,t^{-1}]$ and we again use induction for the last equality. This completes the proof of the inductive step when $i_r\leftrightarrows i_{r+1}$.

Case 4. $i_{r+1}=i_r$.

Let $\phi _r=\psi ^{\mathcal {O}}_r(y^{{\mathcal {O}}}_r-y^{{\mathcal {O}}}_{r+1})f_{\mathbf {i}}^{\mathcal {O}}$. Then $\phi _r\psi ^{\mathcal {O}}_r=-2\psi ^{\mathcal {O}}_rf_{\mathbf {i}}^{\mathcal {O}}$, so that $(1+\phi _r)^2f_{\mathbf {i}}^{\mathcal {O}}=f_{\mathbf {i}}^{\mathcal {O}}$. Moreover, an easy albeit uninspiring calculation reveals that

$$\begin{aligned} (1+\phi _r)y^{{\mathcal {O}}}_r(1+\phi _r)f_{\mathbf {i}}^{\mathcal {O}}=(y^{{\mathcal {O}}}_r+\phi _ry^{{\mathcal {O}}}_r+y^{{\mathcal {O}}}_r\phi _r+\phi _ry^{{\mathcal {O}}}_r\phi _r)f_{\mathbf {i}}^{\mathcal {O}}=y^{{\mathcal {O}}}_{r+1}f_{\mathbf {i}}^{\mathcal {O}}. \end{aligned}$$

Therefore, setting $X_{r+1}(\mathbf {i})=X_r(\mathbf {i})$,

$$\begin{aligned} \prod _{c\in X_{r+1}(\mathbf {i})}(y^{{\mathcal {O}}}_{r+1}-[c])f_{\mathbf {i}}^{\mathcal {O}}=(1+\phi _r)\prod _{c\in X_r(\mathbf {i})}(y^{{\mathcal {O}}}_r-[c])f_{\mathbf {i}}^{\mathcal {O}}(1+\phi _r)=0, \end{aligned}$$

where the last equality follows by induction. This completes the proof.

Suppose that $({\mathcal {O}},t)$ is an idempotent subring of $\fancyscript{K}$. So far we have not used the assumption that $[de]\in J({\mathcal {O}})$, for $d\in \mathbb {Z}$. This comes into play in the next theorem, which is Theorem A from the introduction.

Theorem 4.26

Suppose that $({\mathcal {O}},t)$ is an e-idempotent subring of $\fancyscript{K}$. Then $R_{n}({\mathcal {O}})\cong \mathcal {H}^\varLambda _n({\mathcal {O}})$ as ${\mathcal {O}}$-algebras.

Proof

By the results in the last two sections, the elements given in Definition 4.12 satisfy all of the relations of the corresponding generators of $R_{n}({\mathcal {O}})$. Hence, by Proposition 4.14, there is a surjective ${\mathcal {O}}$-algebra homomorphism $\theta {:}R_{n}({\mathcal {O}})\twoheadrightarrow \mathcal {H}^\varLambda _n({\mathcal {O}})$, which maps the generators of $R_{n}({\mathcal {O}})$ to the corresponding elements of $\mathcal {H}^\varLambda _n({\mathcal {O}})$.

If $w\in \mathfrak S_n$ then set $\psi ^{\mathcal {O}}_w=\psi ^{\mathcal {O}}_{r_1}\cdots \psi ^{\mathcal {O}}_{r_k}$, where $w=s_{r_1}\cdots s_{r_k}$ is a reduced expression for w. In general, $\psi ^{\mathcal {O}}_w$ will depend upon the choice of reduced expression, however, using the relations in $R_{n}({\mathcal {O}})$ it follows that every element in $R_{n}({\mathcal {O}})$ can be written as a linear combination of elements of the form $f(y)\psi _we(\mathbf {i})$, where $f(y)\in {\mathcal {O}}[y^{{\mathcal {O}}}_1,\ldots ,y^{{\mathcal {O}}}_n], w\in \mathfrak S_n$ and $\mathbf {i}\in I^n$. Therefore, $R_{n}({\mathcal {O}})$ is finitely generated as an ${\mathcal {O}}$-module by Lemma 4.25.

Now suppose that $\mathfrak m$ is a maximal ideal of ${\mathcal {O}}$ and let $K={\mathcal {O}}/\mathfrak m\cong {\mathcal {O}}_{\mathfrak m}/\mathfrak m{\mathcal {O}}_{\mathfrak m}$ and $\zeta =t+\mathfrak m$. Then $1+\zeta +\dots +\zeta ^{e-1}=0$ in K, since $[e]\in J({\mathcal {O}})\subseteq \mathfrak m$. Note also that $1+\zeta +\dots +\zeta ^{k-1}\ne 0$ if $k\notin e\mathbb {Z}$ since ${\mathcal {O}}$ is an e-idempotent subring. Consequently, $y^{\langle de\rangle }_r\otimes 1_K=y^{{\mathcal {O}}}_r\otimes 1_K$, for all $d\in \mathbb {Z}$. It is easy to see that all of the shifts $1\pm \rho _r(\mathbf {i})$ appearing in the statement of theorem are equal to either 0 or to e. Therefore, upon base change to K the relations of $R_n({\mathcal {O}}_{\mathfrak m})\otimes _{{\mathcal {O}}_{\mathfrak m}} K$ coincide with the relations of the quiver Hecke algebra $\mathcal {R}^\varLambda _{n}(K)$, see Definition 2.7 and Theorem 2.9. Consequently, $R_n({\mathcal {O}}_{\mathfrak m})\otimes _{{\mathcal {O}}_{\mathfrak m}} K\cong \mathcal {R}^\varLambda _{n}(K)$, so that $\dim R_n({\mathcal {O}}_{\mathfrak m})\otimes _{{\mathcal {O}}_{\mathfrak m}} K=\dim \mathcal {H}^\varLambda _{n}(K)$ by [7, Theorem 4.20].

By the last paragraph, if $K={\mathcal {O}}/\mathfrak m$, for any maximal ideal $\mathfrak m$ of ${\mathcal {O}}$, then $\dim R_n({\mathcal {O}}_{\mathfrak m})\otimes _{{\mathcal {O}}_{\mathfrak m}} K=\dim \mathcal {R}_{n}^{\varLambda }(K)=\ell ^nn!$. Moreover, by the second paragraph of the proof, $R_n({\mathcal {O}}_{\mathfrak m})$ is a finitely generated ${\mathcal {O}}_{\mathfrak m}$-algebra. Therefore, Nakayama’s lemma applies and it implies that $R_n({\mathcal {O}}_{\mathfrak m})$ is a free ${\mathcal {O}}_{\mathfrak m}$-module of rank $\ell ^nn!$. Hence, the map $\theta _{\mathfrak m}:R_n({\mathcal {O}}_{\mathfrak m})\overset{\sim }{\longrightarrow }\mathcal {H}^\varLambda _{n}({\mathcal {O}}_{\mathfrak m})$ is an isomorphism of ${\mathcal {O}}_{\mathfrak m}$-algebras. It follows that $\theta$ is an isomorphism of ${\mathcal {O}}$-algebras, as required.

The proof of Theorem 4.26 gives the following.

Corollary 4.27

Suppose that $K={\mathcal {O}}/\mathfrak m,$ where $\mathfrak m$ is a maximal ideal of ${\mathcal {O}}.$ Then

$$\begin{aligned} \mathcal {R}^\varLambda _{n}(K)\cong R_{n}({\mathcal {O}})\otimes _{\mathcal {O}}K\cong \mathcal {H}^\varLambda _{n}(K). \end{aligned}$$

Remark 4.28

(a) The proof of Theorem 4.26 uses [7, Theorem 4.20] to bound the rank of $R_{n}({\mathcal {O}})$. The proof of [7, Theorem 4.20] does not depend on Brundan–Kleshchev’s isomorphism Theorem 2.9 [6, Theorem 1.1]. Instead, [7, Theorem 4.20] depends on the Ariki–Brundan–Kleshchev Categorification Theorem [7, Theorem 4.18]. Consequently, Theorem 4.26 gives a new proof of Brundan and Kleshchev’s Isomorphism Theorem 2.9. It should be possible to prove Theorem 4.26 directly, without appealing to [7, Theorem 4.18], by adapting the arguments of [6, Theorem 3.3].

(b) In proving Theorem 2.9, Brundan and Kleshchev [6] construct a family of isomorphisms $\mathcal {R}^\varLambda _{n}\overset{\sim }{\longrightarrow }\mathcal {H}^\varLambda _n(\fancyscript{K})$ that depend on a choice of polynomials $Q_r(\mathbf {i})$ that can be varied subject to certain constraints. In our setting this amounts to choosing certain ‘scalars’ $q_r(\mathbf {i})$, which are rational functions in $L_r$ and $L_{r+1}$, such that $q_r(\mathbf {i})f_{\mathbf {i}}^{\mathcal {O}}\in \mathcal {H}^\varLambda _n({\mathcal {O}})$ and then defining

$$\begin{aligned} \psi _r^{\mathcal {O}}f_{\mathbf {i}}^{\mathcal {O}}={\left\{ \begin{array}{ll} (T_r+1)\frac{t^{\hat{\imath }_r}}{M_r}f_{\mathbf {i}}^{\mathcal {O}},&{}\text {if }i_r=i_{r+1},\\ (T_rL_r-L_rT_r)q_r(\mathbf {i})f_{\mathbf {i}}^{\mathcal {O}}, &{}\text {otherwise},\\ \end{array}\right. } \end{aligned}$$

such that the corresponding $\beta$-coefficients still satisfy the constraints of Lemma 4.22 and Definition 3.5(a). To make this more precise, as in Lemma 4.19 write

$$\begin{aligned} \psi ^{\mathcal {O}}_r f_{{\mathfrak s}{\mathfrak t}}=\beta '_r({\mathfrak s})f_{{\mathfrak u}{\mathfrak t}} -\delta _{i_ri_{r+1}}\frac{t^{\hat{\imath }_{r+1}-c_{r+1}({\mathfrak s})}}{[\rho _r({\mathfrak s})]}, \end{aligned}$$

where ${\mathfrak u}={\mathfrak s}(r,r+1), \beta '_r({\mathfrak s})\in \fancyscript{K}, {\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i}), \mathbf {i}\in I^n$ and $1\le r<n$. (Explicitly, $\beta '_r({\mathfrak s})=t^{c_{r+1}({\mathfrak s})}[\rho _r({\mathfrak s})]q_r({\mathfrak s})$ by Lemma 4.9, where $q_r({\mathfrak s})\in \fancyscript{K}$ is the scalar such that $q_r(\mathbf {i})f_{{\mathfrak s}{\mathfrak t}}=q_r({\mathfrak s})f_{{\mathfrak s}{\mathfrak t}}$.) Then we require that the scalars $\beta '_r({\mathfrak s})$ satisfy Lemma 4.22 and the “braid relation” of Definition 3.5(a). If the $q_r(\mathbf {i})$ are chosen so that these two identities are satisfied then it is easy to see that argument used to prove Theorem 4.26 applies, virtually without change, using these more general elements. The key point is that Lemma 4.22 still holds. The corresponding identities in Brundan and Kleshchev’s work are [6, (3.28), (3.29), (4.34) and (4.35)].

We end this section by using Theorem 4.26 to give an upper bound for the nilpotency index of the KLR generators $y_1,\ldots ,y_n$. If $1\le r\le n$ and $\mathbf {i}\in I^n$ set

$$\begin{aligned} \fancyscript{D}_{r}(\mathbf {i})=\{c_r({\mathfrak t})-\hat{\imath }_r\,|\,{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})\} \end{aligned}$$

and define $d_r(\mathbf {i})=\#\fancyscript{D}_{r}(\mathbf {i})$. For example, $\fancyscript{D}_{1}(\mathbf {i})\subseteq \{\kappa _1-\hat{\imath }_1,\ldots ,\kappa _\ell -\hat{\imath }_1\}$ so that $d_1(\mathbf {i})=(\varLambda ,\alpha _{i_1})$.

Two nodes $\gamma =(l,r,c)$ and $\gamma '=(l',r',c')$ are on the same diagonal if they have the same content. That is, $\gamma$ and $\gamma '$ are on the same diagonal if and only if $l=l'$ and $c-r=c'-r'$. The set of diagonals is indexed by pairs (l, d), with $1\le l\le \ell$ and $d\in \mathbb {Z}$, and where the corresponding diagonal is the set of nodes ${\mathbb {D}}_{l,d}=\{(l,r,c)\,|\,\kappa _l+c-r=d\}$. Hence, $d_r(\mathbf {i})=\#\fancyscript{D}_{r}(\mathbf {i})$ counts the number of different diagonals that r appears on in ${\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$. More precisely, we have:

Lemma 4.29

Suppose that $1\le r\le n$ and $\mathbf {i}\in I^n.$ Then

$$\begin{aligned} d_r(\mathbf {i})=\#\{(l,d)\,|\,d\equiv i_r\text { }(\text {mod } e) \text { and } {\mathfrak t}^{-1}(r)\in {\mathbb {D}}_{l,d} \text { for some }{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})\}. \end{aligned}$$

That is,  $d_r(\mathbf {i})$ is equal to the number of distinct diagonals that r appears on for some tableau ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i}).$

The next result is a stronger version of Lemma 4.25. We do not know how to prove this result using only the relations in $R_{n}({\mathcal {O}})$.

Proposition 4.30

Suppose that $1\le r\le n$ and $\mathbf {i}\in I^n.$ Then

$$\begin{aligned} \prod _{c\in \fancyscript{D}_{r}(\mathbf {i})}(y^{{\mathcal {O}}}_r-[c])f_{\mathbf {i}}^{\mathcal {O}}=0. \end{aligned}$$

Proof

By Lemmas 4.3 and 4.19,

$$\begin{aligned} \prod _{c\in \fancyscript{D}_{r}(\mathbf {i})}(y^{{\mathcal {O}}}_r-[c])f_{\mathbf {i}}^{\mathcal {O}}= & {} \sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\prod _{c\in \fancyscript{D}_{r}(\mathbf {i})}(y^{{\mathcal {O}}}_r-[c])\frac{1}{\gamma _{\mathfrak t}}f_{{\mathfrak t}{\mathfrak t}}\\= & {} \sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{1}{\gamma _{\mathfrak t}}\prod _{c\in \fancyscript{D}_{r}(\mathbf {i})}([c_r({\mathfrak t})-\hat{\imath }_r]-[c])f_{{\mathfrak t}{\mathfrak t}} =0, \end{aligned}$$

where the last equality follows because $c_r({\mathfrak t})-\hat{\imath }_r\in \fancyscript{D}_{r}(\mathbf {i})$, for all ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$.

Even though Proposition 4.30 is very easy to prove within our framework, it gives strong information about the nilpotency index of $y_re(\mathbf {i})$, for $\mathbf {i}\in I^n$ and $1\le e\le n$. By Proposition 4.30, and Corollary 4.27, we have the following.

Corollary 4.31

Suppose that $\mathbf {i}\in I^n$ and $1\le r\le n.$ Then $y_r^{d_r(\mathbf {i})}e(\mathbf {i})=0$ in $\mathcal {R}^\varLambda _{n}.$

When $e=0$ Brundan and Kleshchev [6, Conjecture 2.3] conjectured that $y_r^\ell =0$, for $1\le r\le n$. Hoffnung and Lauda proved this conjecture as the main result of their paper [13]. Using Corollary 4.31 we obtain a quick proof of this result and, at the same time, a generalization of it to include the cases when $e>n$.

Corollary 4.32

Suppose that $e=0$ or $e>n$. Then $y_r^{\ell }=0,$ for $1\le r\le n.$

Proof

If $e=0$ then we may assume that $e\gg 0$ by Corollary 2.10, so the results of this section apply. Hence, we may assume that $e>n$.

To prove the corollary it is enough to show that $y_r^\ell e(\mathbf {i})=0$, whenever $\mathbf {i}={{\mathrm{res}}}({\mathfrak t})$ for some standard tableau ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$. By Corollary 4.31, this will follow if we show that each component contains at most one diagonal with content congruent to $i_r$ upon which r can appear in any standard tableau ${\mathfrak s}$ with ${{\mathrm{res}}}({\mathfrak s})=\mathbf {i}$. Suppose by way of contradiction that there exists a standard tableau ${\mathfrak s}$, with ${{\mathrm{res}}}({\mathfrak s})=\mathbf {i}$, and such that r appears in the same component of ${\mathfrak s}$ and ${\mathfrak t}$ but on different diagonals. Then the axial distance between the nodes ${\mathfrak s}^{-1}(r)$ and ${\mathfrak t}^{-1}(r)$ is at least e, so every residue in I must appear in any connected path between these two nodes. As ${{\mathrm{res}}}({\mathfrak s})=\mathbf {i}={{\mathrm{res}}}({\mathfrak t})$ it follows that $\{i_1,\ldots ,i_r\}=I$. This is a contradiction, however, because $|I|=e>n\ge r$.

5 Integral bases for $\mathcal {H}^\varLambda _n({\mathcal {O}})$

Now that we have proved Theorem A, we begin to use the machinery of seminormal forms to study the cyclotomic quiver Hecke algebras $\mathcal {R}^\varLambda _{n}$. In this chapter we reconstruct the ‘natural’ homogeneous bases for the cyclotomic Hecke algebras $\mathcal {H}^\varLambda _{n}(K)$ and their Specht modules over a field.

5.1 The $\psi$-basis

Theorem 4.26 links the KLR grading on $\mathcal {H}^\varLambda _{n}\cong \mathcal {R}^\varLambda _{n}$ with the semisimple representation theory of $\mathcal {H}^\varLambda _n(\fancyscript{K})$. We next want to try and understand the graded Specht modules of $\mathcal {H}^\varLambda _{n}$ [8, 14, 23] in terms of the seminormal form. We start by lifting the homogeneous basis $\{\psi _{{\mathfrak s}{\mathfrak t}}\}$ of $\mathcal {H}^\varLambda _{n}$ to $\mathcal {H}^\varLambda _n({\mathcal {O}})$. This turns out to be easier than the approach taken in [14]. Throughout this section, ${\mathcal {O}}$ is an e-idempotent subring of $\fancyscript{K}$.

By Theorem 4.26, there is a unique anti-isomorphism $\diamond$ of $\mathcal {H}^\varLambda _n({\mathcal {O}})$ such that

$$\begin{aligned} (\psi ^{\mathcal {O}}_r)^\diamond =\psi ^{\mathcal {O}}_r,\quad (y^{{\mathcal {O}}}_s)^\diamond =y^{{\mathcal {O}}}_s\quad \text {and}\quad (f_{\mathbf {i}}^{\mathcal {O}})^\diamond =f_{\mathbf {i}}^{\mathcal {O}}, \end{aligned}$$

for $1\le r<n, 1\le s\le n$ and $\mathbf {i}\in I^n$. Lemma 4.19 shows that, in general, the automorphisms $*$ and $\diamond$ do not coincide.

Recall from Definition 3.4 that a $\diamond$-seminormal basis of $\mathcal {H}^\varLambda _n(\fancyscript{K})$ is a basis $\{f_{{\mathfrak s}{\mathfrak t}}\}$ of two-sided eigenvectors for $\fancyscript{L}$ such that $f_{{\mathfrak s}{\mathfrak t}}=f_{{\mathfrak t}{\mathfrak s}}^\diamond$, for all $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$. We define a $\diamond$-seminormal coefficient system to be a set of scalars $\{\beta _r({\mathfrak t})\}$ that satisfy the identity in Lemma 4.22 and the “braid relations” of Definition 3.5(a) (with $\alpha$ replaced by $\beta$) as well as the relation Definition 3.5(b) (with $\alpha$ replaced by $\beta$). The reader may check that the $\diamond$-seminormal coefficients correspond to the more general setup considered in Remark 4.28(b).

The main difference between a $*$-seminormal basis and a $\diamond$-seminormal basis is that $T_rf_{{\mathfrak s}{\mathfrak t}}=(f_{{\mathfrak t}{\mathfrak s}}T_r)^{*}$ for a $*$-seminormal basis whereas $\psi ^{\mathcal {O}}_r f_{{\mathfrak s}{\mathfrak t}}=(f_{{\mathfrak t}{\mathfrak s}}\psi ^{\mathcal {O}}_r)^\diamond$ for a $\diamond$-seminormal basis.

Lemma 5.1

Suppose that $\{f_{{\mathfrak s}{\mathfrak t}}\}$ is a $\diamond$-seminormal basis of $\mathcal {H}^\varLambda _n(\fancyscript{K}).$ Then there exists a unique $\diamond$-seminormal coefficient system $\{\beta _r({\mathfrak t})\}$ such that if $1\le r<n$ and $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ then

$$\begin{aligned} f_{{\mathfrak s}{\mathfrak t}}\psi ^{\mathcal {O}}_r=\beta _r({\mathfrak t})f_{{\mathfrak s}{\mathfrak v}} -\delta _{i_ri_{r+1}}\frac{t^{\hat{\imath }_{r+1}-c_{r+1}({\mathfrak t})}}{[\rho _r({\mathfrak t})]}f_{{\mathfrak s}{\mathfrak t}}, \end{aligned}$$

where ${\mathfrak v}={\mathfrak t}(r,r+1)$ and ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$, for $\mathbf {i}\in I^n$. Conversely,  as in Theorem 3.9, a $\diamond$-seminormal coefficient system,  together with a choice of scalars $\{\gamma _{{\mathfrak t}^{\varvec{\lambda }}}\,|\,{\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}\},$ determines a unique $\diamond$-seminormal basis.

Proof

By (4.3), a set of scalars $\{\beta _r({\mathfrak t})\}$ is a $\diamond$-seminormal coefficient system if and only if $\{\alpha _r({\mathfrak t})\}$ is a $*$-seminormal coefficient system, where

$$\begin{aligned} \alpha _r({\mathfrak t})={\left\{ \begin{array}{ll} \beta _r({\mathfrak t})t^{c_{r}({\mathfrak t})-\hat{\imath }_{r}}[1-\rho _r({\mathfrak t})], &{}\text {if }i_r=i_{r+1},\\ \dfrac{\beta _r({\mathfrak t})t^{\hat{\imath }_r-c_{r+1}({\mathfrak t})}}{[\rho _r({\mathfrak t})]}, &{}\text {if }i_r=i_{r+1}+1,\\ \dfrac{\beta _r({\mathfrak t})[1-\rho _r({\mathfrak t})]}{[\rho _r({\mathfrak t})]}, &{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Therefore, as seminormal coefficient systems are determined by the action of the corresponding generators of $\mathcal {H}^\varLambda _{n}$ on its right regular representation, the result follows from Theorem 3.9 and Lemma 4.19.

Henceforth, we will work with $\diamond$-seminormal bases. Lemma 5.1 also describes the left action of $\psi ^{\mathcal {O}}_r$ on the $\diamond$-seminormal basis because $\psi ^{\mathcal {O}}_r f_{{\mathfrak s}{\mathfrak t}}=(f_{{\mathfrak t}{\mathfrak s}}\psi ^{\mathcal {O}}_r)^\diamond$.

Exactly as in Theorem 3.9, if $\{f_{{\mathfrak s}{\mathfrak t}}\}$ is a $\diamond$-seminormal basis then there exists scalars $\gamma _{\mathfrak t}\in \fancyscript{K}$ such that $f_{{\mathfrak s}{\mathfrak t}}f_{{\mathfrak u}{\mathfrak v}}=\delta _{{\mathfrak u}{\mathfrak t}}\gamma _{\mathfrak t}f_{{\mathfrak s}{\mathfrak v}}$, for $({\mathfrak s},{\mathfrak t}),({\mathfrak u},{\mathfrak v})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$. Repeating the argument of Corollary 3.10, these scalars satisfy the following recurrence relation.

Corollary 5.2

Suppose that ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and that ${\mathfrak v}={\mathfrak t}(r,r+1)$ is standard,  where $1\le r<n.$ Then $\beta _r({\mathfrak v})\gamma _{\mathfrak t}=\beta _r({\mathfrak t})\gamma _{\mathfrak v}.$

Motivated by [14], we now define a new basis of $\mathcal {H}^\varLambda _n({\mathcal {O}})$ that is cellular with respect to the anti-involution $\diamond$. Fix ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$ and let $\mathbf {i}^{\varvec{\lambda }}=(i^{\varvec{\lambda }}_1,\ldots ,i^{\varvec{\lambda }}_n)$, so that $i^{\varvec{\lambda }}_r={{\mathrm{res}}}_{{\mathfrak t}^{\varvec{\lambda }}}(r)$ for $1\le r\le n$. Following [14, Definition 4.7], define

for $1\le r\le n$.

Up until now we have worked with an arbitrary seminormal basis of $\mathcal {H}^\varLambda _n(\fancyscript{K})$. In order to define a ‘nice’ basis of $\mathcal {H}^\varLambda _n({\mathcal {O}})$ that is compatible with Theorem 4.26 we now fix the choice of $\gamma$-coefficients by requiring that

$$\begin{aligned} \gamma _{{\mathfrak t}^{\varvec{\lambda }}} = \prod _{r=1}^n\prod _{\alpha \in \fancyscript{A}_{\varvec{\lambda }}(r)} [{{\mathrm{c}}}_r({\mathfrak t}^{\varvec{\lambda }})-c_\alpha ], \end{aligned}$$

(5.1)

for all ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$. Together with a choice of seminormal coefficient system, this determines $\gamma _{\mathfrak t}$ for all ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ by Corollary 5.2. By definition, $\gamma _{{\mathfrak t}^{\varvec{\lambda }}}$ is typically a non-invertible element of ${\mathcal {O}}$. Nonetheless, if $\mathbf {i}\in I^n$ then $f_{\mathbf {i}}^{\mathcal {O}}=\sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\tfrac{1}{\gamma _{\mathfrak s}}f_{{\mathfrak s}{\mathfrak s}}$ belongs to $\mathcal {H}^\varLambda _n({\mathcal {O}})$ by Lemma 4.3.

We also fix a choice of seminormal coefficient system by requiring that $\beta _r({\mathfrak s})=1$ whenever ${\mathfrak s}\vartriangleright {\mathfrak t}={\mathfrak s}(r,r+1)$, for ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and $1\le r<n$. More precisely, if $\mathbf {i}\in I$ and ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$ then we define

$$\begin{aligned} \beta _r({\mathfrak s})= {\left\{ \begin{array}{ll} 1,&{}\text {if }{\mathfrak s}\vartriangleright {\mathfrak t}\text { or }i_r{{\mathrm{\,{/}\text {---}\,}}}i_{r+1},\\ -\frac{t^{2\hat{\imath }_r-2c_{r+1}({\mathfrak s})}}{[\rho _r({\mathfrak s})]^2}, &{}\text {if }{\mathfrak t}\vartriangleright {\mathfrak s}\text { and }i_r=i_{r+1},\\ t^{c_r({\mathfrak s})+c_{r+1}({\mathfrak s})-\hat{\imath }_r-\hat{\imath }_{r+1}}[1{-}\rho _r({\mathfrak s})][1{+}\rho _r(s)], &{}\text {if }{\mathfrak t}\vartriangleright {\mathfrak s}\text { and }i_r\leftrightarrows i_{r+1},\\ t^{c_{r}({\mathfrak s})-\hat{\imath }_{r}}[1{-}\rho _r({\mathfrak s})], &{}\text {if }{\mathfrak t}\vartriangleright {\mathfrak s}\text { and }i_r\leftarrow i_{r+1},\\ t^{c_{r+1}({\mathfrak s})-\hat{\imath }_{r+1}}[1{+}\rho _r(s)], &{}\text {if }{\mathfrak t}\vartriangleright {\mathfrak s}\text { and }i_r\rightarrow i_{r+1}.\\ \end{array}\right. }\quad \qquad \end{aligned}$$

(5.2)

where ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and ${\mathfrak t}={\mathfrak s}(r,r+1)$ is standard, for $1\le r<n$. The reader is invited to check that this defines a $\diamond$-seminormal coefficient system. As the definition of $\psi ^{\mathcal {O}}_r$ is independent of the choice of seminormal coefficient system this choice is not strictly necessary for what follows but it simplifies many of the formulas.

By Lemma 5.1, this choice of $\diamond$-seminormal coefficient system and $\gamma$-coefficients determines a unique $\diamond$-seminormal basis $\{f_{{\mathfrak s}{\mathfrak t}}\}$ of $\mathcal {H}^\varLambda _n(\fancyscript{K})$. We will use this basis to define new homogeneous basis of $\mathcal {H}^\varLambda _{n}$. The first step is to define

$$\begin{aligned} y^{{\varvec{\lambda }}}_{{\mathcal {O}}}f_{\mathbf {i}^{\varvec{\lambda }}}^{\mathcal {O}}= & {} \prod _{r=1}^n\prod _{\alpha \in \fancyscript{A}_{\varvec{\lambda }}(r)} t^{-c_r({\mathfrak t}^{\varvec{\lambda }})}(L_r-[c_\alpha ])f_{\mathbf {i}^{\varvec{\lambda }}}^{\mathcal {O}}\\= & {} \prod _{r=1}^n\prod _{\alpha \in \fancyscript{A}_{\varvec{\lambda }}(r)} t^{\hat{\imath }^{\varvec{\lambda }}_r-c_r({\mathfrak t}^{\varvec{\lambda }})}(y^{{\mathcal {O}}}_r-[c_\alpha -\hat{\imath }^{\varvec{\lambda }}_r])f_{\mathbf {i}^{\varvec{\lambda }}}^{\mathcal {O}}, \end{aligned}$$

where the second equation follows by rewriting $L_kf_{\mathbf {i}}^{\mathcal {O}}$ in terms of $y_kf_{\mathbf {i}}^{\mathcal {O}}$ as in the proof of Proposition 4.14. In particular, these equations show that $y^{{\varvec{\lambda }}}_{{\mathcal {O}}}f_{\mathbf {i}^{\varvec{\lambda }}}^{\mathcal {O}}\otimes _{\mathcal {O}}1_K$ is a monomial in $y_1,\ldots ,y_n$ and, further, that it is (up to a sign) equal to the element $y^{\varvec{\lambda }}$ defined in [14, Definition 4.15].

The next result is a essentially a translation of [14, Lemma 4.13] into the current setting for the special case of the tableau ${\mathfrak t}^{\varvec{\lambda }}$.

Lemma 5.3

Suppose that ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}.$ Then there exist scalars $a_{\mathfrak s}\in \fancyscript{K}$ such that

$$\begin{aligned} y^{{\varvec{\lambda }}}_{{\mathcal {O}}}f_{\mathbf {i}^{\varvec{\lambda }}}^{\mathcal {O}}=f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}+\sum _{{\mathfrak s}\blacktriangleright {\mathfrak t}^{\varvec{\lambda }}} a_{\mathfrak s}f_{{\mathfrak s}{\mathfrak s}}. \end{aligned}$$

In particular,  $y^{{\varvec{\lambda }}}_{{\mathcal {O}}}f_{\mathbf {i}^{\varvec{\lambda }}}^{\mathcal {O}}$ is a non-zero element of $\mathcal {H}^\varLambda _n({\mathcal {O}})$.

Proof

By Lemma 4.3, $f_{\mathbf {i}^{\varvec{\lambda }}}^{\mathcal {O}}=\sum _{\mathfrak s}\frac{1}{\gamma _{\mathfrak s}}f_{{\mathfrak s}{\mathfrak s}}$, so that $y^{{\varvec{\lambda }}}_{{\mathcal {O}}}f_{\mathbf {i}^{\varvec{\lambda }}}^{\mathcal {O}}=\sum _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i}^{\varvec{\lambda }})}a_{\mathfrak s}f_{{\mathfrak s}{\mathfrak s}}$, for some $a_{\mathfrak s}\in \fancyscript{K}$, by (3.5). It remains to show that $a_{{\mathfrak t}^{\varvec{\lambda }}}=1$ and that $a_{\mathfrak s}\ne 0$ only if ${\mathfrak s}{{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}{\mathfrak t}^{\varvec{\lambda }}$. Using (3.5), and recalling the definition of $\gamma _{{\mathfrak t}^{\varvec{\lambda }}}$ from (5.1),

$$\begin{aligned} \frac{1}{\gamma _{{\mathfrak t}^{\varvec{\lambda }}}}y^{{\varvec{\lambda }}}_{{\mathcal {O}}}f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}} =\frac{1}{\gamma _{{\mathfrak t}^{\varvec{\lambda }}}}\prod _{r=1}^n\prod _{\alpha \in \fancyscript{A}_{\varvec{\lambda }}(r)} t^{-c_r({\mathfrak t}^{\varvec{\lambda }})}([c_r({\mathfrak t}^{\varvec{\lambda }})]-[c_\alpha ])\cdot f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}} =f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}. \end{aligned}$$

To complete the proof we claim that there exist scalars $a_{\mathfrak s}(k)\in \fancyscript{K}, 1\le k\le n$, such that

where $a_{{\mathfrak t}^{\varvec{\lambda }}}(k)=1$. We prove this by induction on k. If $k=1$ then the result is immediate from (3.5). Suppose that $k>1$. By induction, it is enough to show that

$$\begin{aligned} (L_k-[c_\alpha ])f_{{\mathfrak s}{\mathfrak s}} =([c_\alpha ]-[c_k({\mathfrak s})])f_{{\mathfrak s}{\mathfrak s}}=0 \end{aligned}$$

whenever and , for ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i}^{\varvec{\lambda }})$. Fix such a tableau  ${\mathfrak s}$. Since we must have $({\mathfrak s}_{\downarrow k})^{(l)}=\emptyset$ whenever $l>{{\mathrm{comp}}}_{{\mathfrak t}^{\varvec{\lambda }}}(k)$, so the node $\alpha ={\mathfrak s}^{-1}(k)$ must be below $({\mathfrak t}^{\varvec{\lambda }})^{-1}(k)$. Therefore, $\alpha \in \fancyscript{A}_{\varvec{\lambda }}(k)$, and $c_k({\mathfrak s})={{\mathrm{c}}}_\alpha$ for this $\alpha$, and forcing $a_{\mathfrak s}(k)=0$ as claimed. This completes the proof.

For each $w\in \mathfrak S_n$ we now fix a reduced expression $w=s_{r_1}\dots s_{r_k}$ for w, with $1\le r_j<n$ for $1\le j\le k$, and define $\psi ^{\mathcal {O}}_w=\psi ^{\mathcal {O}}_{r_1}\dots \psi ^{\mathcal {O}}_{r_k}$. By Theorem 4.26 the elements $\psi ^{\mathcal {O}}_r$ do not satisfy the braid relations so, in general, $\psi ^{\mathcal {O}}_w$ will depend upon this (fixed) choice of reduced expression.

Definition 5.4

Suppose that ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$. Define

$$\begin{aligned} \psi ^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}=(\psi ^{\mathcal {O}}_{d({\mathfrak s})})^\diamond y^{{\varvec{\lambda }}}_{{\mathcal {O}}}f_{\mathbf {i}^{\varvec{\lambda }}}^{\mathcal {O}}\psi ^{\mathcal {O}}_{d({\mathfrak t})}, \end{aligned}$$

for ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$.

We can now lift the graded cellular basis of [14, Definitions 5.1] to $\mathcal {H}^\varLambda _n({\mathcal {O}})$.

Theorem 5.5

Suppose that ${\mathcal {O}}$ is an idempotent subring. Then

$$\begin{aligned} \{\psi ^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\,|\,{\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\mu }})\text { for }{\varvec{\mu }}\in \mathcal {P}^{\varLambda }_{n}\} \end{aligned}$$

is a cellular basis of $\mathcal {H}^\varLambda _n({\mathcal {O}})$ with respect to the anti-involution $\diamond$.

Proof

In view of (3.5) and Lemma 4.19, Lemma 5.3 implies that

$$\begin{aligned} \psi ^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}= f_{{\mathfrak s}{\mathfrak t}} +\sum _{({\mathfrak u},{\mathfrak v})\blacktriangleright ({\mathfrak s},{\mathfrak t})}a_{{\mathfrak u}{\mathfrak v}}f_{{\mathfrak u}{\mathfrak v}}, \end{aligned}$$

(5.3)

for some $a_{{\mathfrak u}{\mathfrak v}}\in \fancyscript{K}$. Therefore, $\{\psi ^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$ is a basis of $\mathcal {H}^\varLambda _n(\fancyscript{K})$. In fact, these elements are a basis for $\mathcal {H}^\varLambda _n({\mathcal {O}})$ because if $h\in \mathcal {H}^\varLambda _n({\mathcal {O}})$ then we can write $h=\sum r_{{\mathfrak u}{\mathfrak v}}f_{{\mathfrak u}{\mathfrak v}}$, for some $r_{{\mathfrak u}{\mathfrak v}}\in \fancyscript{K}$. Pick $({\mathfrak s},{\mathfrak t})$ to be minimal with respect to dominance such that $r_{{\mathfrak s}{\mathfrak t}}\ne 0$. Then $r_{{\mathfrak s}{\mathfrak t}}\in {\mathcal {O}}$ because $h\in \mathcal {H}^\varLambda _n({\mathcal {O}})$. Consequently, $h-r_{{\mathfrak s}{\mathfrak t}}\psi ^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\in \mathcal {H}^\varLambda _n({\mathcal {O}})$ so, by continuing in this way, we can write h as a linear combination of the $\psi$-basis.

It remains to show that the $\psi$-basis is cellular with respect to the anti-involution $\diamond$. By definition, if ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$ then $y^{{\varvec{\lambda }}}_{{\mathcal {O}}}$ and $f_{\mathbf {i}^{\varvec{\lambda }}}^{\mathcal {O}}$ commute and they are fixed by the automorphism $\diamond$. Therefore, $\big (\psi ^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\big )^\diamond =\psi ^{\mathcal {O}}_{{\mathfrak t}{\mathfrak s}}$, for all ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$. By Lemma 5.1, the $\diamond$-seminormal basis $\{f_{{\mathfrak s}{\mathfrak t}}\}$ is a cellular basis with cellular anti-involution $\diamond$. It remains to verify $(\hbox {GC}_2)$ from Definition 2.3. As in Theorem 3.9, the seminormal basis $\{f_{{\mathfrak u}{\mathfrak v}}\}$ is cellular. Therefore, if $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2({\varvec{\lambda }})$ and $h\in \mathcal {H}^\varLambda _n({\mathcal {O}})$ then, using (5.3) twice,

$$\begin{aligned} \psi ^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}h= & {} (\psi ^{\mathcal {O}}_{d({\mathfrak s})})^\diamond \psi ^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}} \equiv (\psi ^{\mathcal {O}}_{d({\mathfrak s})})^\diamond \left( f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}+\sum _{{\mathfrak v}\vartriangleright {\mathfrak t}}a_{\mathfrak v}f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}\right) h\equiv (\psi ^{\mathcal {O}}_{d({\mathfrak s})})^\diamond \sum _{{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})}a'_{\mathfrak v}f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}\\\equiv & {} (\psi ^{\mathcal {O}}_{d({\mathfrak s})})^\diamond \sum _{{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }}}b_{\mathfrak v}\psi ^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}\quad \equiv \sum _{{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }}}b_{\mathfrak v}\psi ^{\mathcal {O}}_{{\mathfrak s}{\mathfrak v}}\quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n), \end{aligned}$$

where $a_{\mathfrak v},a_{\mathfrak v}'\in \fancyscript{K}$ and $b_{\mathfrak v}\in {\mathcal {O}}$ with the scalars $b_{\mathfrak v}$ being independent of ${\mathfrak s}$. Hence, $(\hbox {GC}_2)$ holds, completing the proof.

If $K={\mathcal {O}}/\mathfrak m$ for some maximal ideal $\mathfrak m$ of ${\mathcal {O}}$ then $\mathcal {H}^\varLambda _{n}(K)\cong \mathcal {H}^\varLambda _{n}({\mathcal {O}})\otimes _{\mathcal {O}}K$. Set $\psi _{{\mathfrak s}{\mathfrak t}}=\psi ^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\otimes 1_K$.

Corollary 5.6

[14, Theorem 5.8] Suppose that $K={\mathcal {O}}/\mathfrak m$ for some maximal ideal $\mathfrak m$ of ${\mathcal {O}}$. Then $\{\psi _{{\mathfrak s}{\mathfrak t}}\,|\,{\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\mu }})\text { for }{\varvec{\mu }}\in \mathcal {P}^{\varLambda }_{n}\}$ is a graded cellular basis of $\mathcal {H}^\varLambda _{n}(K)$ with $\deg \psi _{{\mathfrak s}{\mathfrak t}}=\deg {\mathfrak s}+\deg {\mathfrak t},$ for $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n}).$

5.2 Graded Specht modules and Gram determinants

By Theorem 5.5, $\{\psi ^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\}$ is a cellular basis of $\mathcal {H}^\varLambda _n({\mathcal {O}})$ so we can use it to define Specht modules for $\mathcal {H}^\varLambda _n({\mathcal {O}})$ that specialise to the graded Specht modules in characteristic zero and in positive characteristic.

Definition 5.7

Suppose that ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$. The Specht module $S^{\varvec{\lambda }}({\mathcal {O}})$ is the right $\mathcal {H}^\varLambda _n({\mathcal {O}})$-module with basis $\{\psi ^{\mathcal {O}}_{\mathfrak t}\,|\,{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\}$, where $\psi ^{\mathcal {O}}_{\mathfrak t}=\psi ^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}+\mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n({\mathcal {O}})$.

By Theorem 5.5 and [14, Corollary 5.10], ignoring the grading, $S^{\varvec{\lambda }}({\mathcal {O}})\otimes _{\mathcal {O}}K$ can be identified with the graded Specht module $S^{\varvec{\lambda }}$ of $\mathcal {H}^\varLambda _{n}$ defined by Brundan, Kleshchev and Wang [8]. The action of $\mathcal {H}^\varLambda _n(\fancyscript{K})$ on a graded Specht module is completely determined by the relations for these modules that are given in [23]. In contrast, in view of (5.3) and Theorem 4.26, the action of $\mathcal {H}^\varLambda _n({\mathcal {O}})$ on the Specht module $S^{\varvec{\lambda }}({\mathcal {O}})$ is completely determined by the (choice of) seminormal form.

We now turn to computing the determinant of the Gram matrix

$$\begin{aligned} \mathcal {G}^{{\varvec{\lambda }}}= \big (\langle \psi ^{\mathcal {O}}_{\mathfrak s},\psi ^{\mathcal {O}}_{\mathfrak t}\rangle \big )_{{\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})}. \end{aligned}$$

A priori, it is unclear how the bilinear form on $S^{\varvec{\lambda }}({\mathcal {O}})$ is related to the usual (ungraded) bilinear from on the Specht module that is defined using the Murphy basis that we considered in Theorem 3.13. The main problem in relating these two bilinear forms is that the cellular algebra anti-involutions $*$ and $\diamond$, which are used to define these bilinear forms, are different.

Note that the cellular algebra anti-involutions $*$ and $\diamond$ on $\mathcal {H}^\varLambda _n({\mathcal {O}})$ naturally extend to anti-involutions on the algebra $\mathcal {H}^\varLambda _n(\fancyscript{K})$. The key point to understanding the graded bilinear form is the following.

Lemma 5.8

Suppose that ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n}).$ Then $(F_{\mathfrak t})^\diamond =F_{\mathfrak t}.$

Proof

By definition, $F_{\mathfrak t}$ is a linear combination of products of Jucys–Murphy elements, so it can also be written as a polynomial, with coefficients in $\fancyscript{K}$, in $y^{{\mathcal {O}}}_r, f_{\mathbf {i}}^{{\mathcal {O}}}$, for $1\le r\le n$ and $\mathbf {i}\in I^n$. As $(y^{{\mathcal {O}}}_r)^\diamond =y^{{\mathcal {O}}}_r, (f_{\mathbf {i}}^{{\mathcal {O}}})^{\diamond }=f_{\mathbf {i}}^{{\mathcal {O}}}$, for $1\le r\le n$ and $\mathbf {i}\in I^n$, the result follows.

Recall that if ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ then $\psi ^{\mathcal {O}}_{\mathfrak t}=\psi ^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}+\mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n$ is a basis element of the Specht module $S^{\varvec{\lambda }}({\mathcal {O}})$. In order to compute $\det \mathcal {G}^{{\varvec{\lambda }}}$, set $f_{\mathfrak t}=\psi ^{\mathcal {O}}_{\mathfrak t}F_{\mathfrak t}$, for ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$. Recall that $S^{\varvec{\lambda }}(\fancyscript{K})=S^{\varvec{\lambda }}({\mathcal {O}})\otimes _{{\mathcal {O}}}\fancyscript{K}$.

Lemma 5.9

Suppose that ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}.$ Then $\{f_{\mathfrak t}\,|\,{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\}$ is a basis of $S^{\varvec{\lambda }}(\fancyscript{K}).$ Moreover,  $\det \mathcal {G}^{{\varvec{\lambda }}}=\det \big (\langle f_{\mathfrak s},f_{\mathfrak t}\rangle \big )=\prod _{{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})}\gamma _{{\mathfrak s}}.$

Proof

By definition, $f_{\mathfrak t}=f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}+(\mathcal {H}^\varLambda _n(\fancyscript{K}))^{\rhd {\varvec{\lambda }}}$. Therefore, $f_{\mathfrak t}\in S^{\varvec{\lambda }}(\fancyscript{K})$ and $f_{\mathfrak t}=\psi ^{\mathcal {O}}_{\mathfrak t}+\sum _{{\mathfrak v}\vartriangleright {\mathfrak t}}r_{{\mathfrak t}{\mathfrak v}} \psi ^{\mathcal {O}}_{\mathfrak v}$ by (5.3), for some scalars $r_{{\mathfrak t}{\mathfrak v}}\in \fancyscript{K}$. Set $r_{{\mathfrak t}{\mathfrak t}}=1$ and $U=\big (r_{{\mathfrak t}{\mathfrak v}}\big )$. Then $\{f_{\mathfrak t}\,|\,{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\}$ is a $\fancyscript{K}$-basis of $S^{\varvec{\lambda }}(\fancyscript{K})$ and $\mathcal {G}^{{\varvec{\lambda }}}=(U^{-1})^{tr}(\langle f_{\mathfrak s},f_{\mathfrak t}\rangle )U^{-1}$ Taking determinants shows that $\det \mathcal {G}^{{\varvec{\lambda }}}=\deg (\langle f_{\mathfrak s},f_{\mathfrak t}))$ since U is unitriangular. To complete the proof observe that $\langle f_{\mathfrak s},f_{\mathfrak t}\rangle f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}\,\, f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak s}}f_{{\mathfrak t}{\mathfrak t}^{\varvec{\lambda }}}=\delta _{{\mathfrak s}{\mathfrak t}}\gamma _{\mathfrak s}f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}\text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n)$, where we are implicitly using Lemma 5.8. The result follows.

Lemma 5.9 is subtly different from (3.7) because, in spite of our notation, the $\gamma _{\mathfrak t}$’s appearing in the two formulas satisfy different recurrence relations. It is not hard to show that the quotient of $\gamma _{\mathfrak t}$, as defined in this section, by the $\gamma _{\mathfrak t}$ defined in Sect. 3.3 in a unit in ${\mathcal {O}}$, for all ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$.

Lemma 5.10

Suppose that ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }}),$ for ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}.$ Then $\gamma _{\mathfrak t}=u_{\mathfrak t}\varPhi _e(t)^{\deg _e({\mathfrak t})},$ for some unit $u_{\mathfrak t}\in {\mathcal {O}}^\times .$

Proof

We argue by induction on the dominance order on ${\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$. If ${\mathfrak t}={\mathfrak t}^{\varvec{\lambda }}$ then (5.1) ensures that $\gamma _{{\mathfrak t}^{\varvec{\lambda }}}=u_{{\mathfrak t}^{\varvec{\lambda }}}\varPhi _e(t)^{\deg _e({\mathfrak t}^{\varvec{\lambda }})}$, for some unit $u_{{\mathfrak t}^{\varvec{\lambda }}}\in {\mathcal {O}}$. Now suppose that ${\mathfrak t}^{\varvec{\lambda }}\vartriangleright {\mathfrak t}$. Then there exists a standard tableau ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ such that ${\mathfrak s}\vartriangleright {\mathfrak t}$ and ${\mathfrak t}={\mathfrak s}(r,r+1)$, where $1\le r<n$. Arguing exactly as in Corollary 5.2 shows that $\beta _r({\mathfrak s})\gamma _{\mathfrak t}=\beta _r({\mathfrak t})\gamma _{\mathfrak s}$. Therefore, $\gamma _{\mathfrak t}=\tfrac{\beta _r({\mathfrak t})}{\beta _r({\mathfrak s})}\gamma _{\mathfrak s}=\beta _r({\mathfrak t})\gamma _{\mathfrak s}$. Hence, the lemma follows by induction exactly as in the proof of Theorem 3.13.

Remark 5.11

Looking at the definition of a $\diamond$-seminormal coefficient system shows that the quantities $\tfrac{\beta _r({\mathfrak t})}{\beta _r({\mathfrak s})}$, which are used in the proof of Lemma 5.10, are independent of the choice of $\diamond$-seminormal coefficient system. This shows that the choice of $\diamond$-seminormal coefficient system made in (5.2) really is only for convenience.

By general nonsense, the determinants of $\mathcal {G}^{{\varvec{\lambda }}}$ and $\underline{\mathcal {G}^{\varvec{\lambda }}}$ differ by a scalar in $\fancyscript{K}$. The last two results readily imply the next theorem, the real content of which is that this scalar is a unit in ${\mathcal {O}}$.

Theorem 5.12

Suppose that ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}.$ Then $\det \mathcal {G}^{{\varvec{\lambda }}}=u\varPhi _e(t)^{\deg _e({\varvec{\lambda }})},$ for some unit $u\in {\mathcal {O}}^\times$. Consequently,  $\det \mathcal {G}^{{\varvec{\lambda }}}=u'\det \underline{\mathcal {G}^{\varvec{\lambda }}},$ for some unit $u'\in {\mathcal {O}}^\times .$

If $\mathbf {i}\in I^n$ and ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$ let ${\mathop {\mathrm{Std}}\nolimits }_\mathbf {i}({\varvec{\lambda }})=\{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\,|\,{{\mathrm{res}}}({\mathfrak t})=\mathbf {i}\}$.

The Specht module $S^{\varvec{\lambda }}$ over ${\mathcal {O}}$ decomposes as a direct sum of generalised eigenspaces as an $\fancyscript{L}({\mathcal {O}})$-module: $S^{\varvec{\lambda }}=\bigoplus _{\mathbf {i}\in I^n} S_\mathbf {i}^{\varvec{\lambda }}$, where $S^{\varvec{\lambda }}_\mathbf {i}=S^{\varvec{\lambda }}f_{\mathbf {i}}^{\mathcal {O}}$. The weight space $S^{\varvec{\lambda }}_\mathbf {i}$ has basis $\{\psi ^{\mathcal {O}}_{\mathfrak t}\,|\,{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }_\mathbf {i}({\varvec{\lambda }})\}$ and the bilinear linear form $\langle \ ,\ \rangle$ on $S^{\varvec{\lambda }}$ respects the weight space decomposition of $S^{\varvec{\lambda }}$. Set

$$\begin{aligned} \deg _{e,\mathbf {i}}({\varvec{\lambda }})=\sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }_\mathbf {i}({\varvec{\lambda }})} \deg {\mathfrak t}. \end{aligned}$$

and let $\mathcal {G}^{{\varvec{\lambda }}}_\mathbf {i}$ be restriction of the Gram matrix of $S^{\varvec{\lambda }}$ to $S^{\varvec{\lambda }}_\mathbf {i}$, for $\mathbf {i}\in I^n$. Then we have the following refinement of Theorem 5.12 (and Theorem 3.13).

Corollary 5.13

Suppose that ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$ and $\mathbf {i}\in I^n$. Then $\det \mathcal {G}^{{\varvec{\lambda }}}_\mathbf {i}=u_\mathbf {i}\varPhi _e(t)^{\deg _{e,\mathbf {i}}({\varvec{\lambda }})},$ for some unit $u_\mathbf {i}\in {\mathcal {O}}^{\times }.$ Moreover,  $\deg _{e,\mathbf {i}}({\varvec{\lambda }})\ge 0.$

6 A distinguished homogeneous basis for $\mathcal {H}^\varLambda _{n}$

The $\psi$-basis of $\mathcal {H}^\varLambda _{n}({\mathcal {O}})$, the homogeneous bases of $\mathcal {H}^\varLambda _{n}$ constructed in [14], and the homogeneous basis of the graded Specht modules given by Brundan, Kleshchev and Wang [8], are all indexed by pairs of standard tableaux. Unfortunately, unlike in the ungraded case, these basis elements depend upon choices of reduced expressions for the permutations corresponding to these tableaux. In this section we construct new bases for these modules that depend only on the corresponding tableaux.

6.1 A new basis of $\mathcal {H}^\varLambda _{n}({\mathcal {O}})$

To construct our new basis for $\mathcal {H}^\varLambda _{n}$ we need to work over a complete discrete valuation ring. We start by setting up the necessary machinery.

Recall that the algebra $\mathcal {H}^\varLambda _{n}$ is defined over the field K with parameter $\xi$ and that $e>1$ is minimal such that $[e]_\xi =0$. Let x be an indeterminate over K and let ${\mathcal {O}}=K[x]_{(x)}$ and $t=x+\xi$. Then $({\mathcal {O}},t)$ is an idempotent subring by Example 4.2(b) and K(x) is the field of fractions of ${\mathcal {O}}$. Note that ${\mathcal {O}}$ is a local ring with maximal ideal $\mathfrak m=x{\mathcal {O}}$.

Let ${\widehat{\mathcal {O}}}$ be the $\mathfrak m$-adic completion of ${\mathcal {O}}$. Then ${\widehat{\mathcal {O}}}$ is a complete discrete valuation ring with field of fractions K((x)) Let $\widehat{\fancyscript{K}}=K((x))$ be the $\mathfrak m$-adic completion of K(x). Then ${\widehat{\mathcal {O}}}$ is an idempotent subring of $\widehat{\fancyscript{K}}$.

Define a valuation on $\widehat{\fancyscript{K}}^\times$ by setting $\nu _x(a)=n$ if $a=ux^n$, where $n\in \mathbb {Z}$ and $u\in {\widehat{\mathcal {O}}}^\times$ is a unit in ${\widehat{\mathcal {O}}}$. We need to work with a complete discrete valuation ring because of the following fundamental but elementary fact that is proved, for example, as [35, Proposition II.5].

Lemma 6.1

Suppose that $a\in \widehat{\fancyscript{K}}.$ Then a can be written uniquely as a convergent series

$$\begin{aligned} a=\sum _{n\in \mathbb {Z}}a_nx^n,\quad \text {with }a_n\in K, \end{aligned}$$

such that if $a\ne 0$ then $a_n\ne 0$ only if $n\ge \nu _x(a).$ Moreover,  $a\in {\widehat{\mathcal {O}}}$ if and only if $a_n=0$ for all $n<0.$

In particular, $x^{-1}K[x^{-1}]\cap {\widehat{\mathcal {O}}}=0$, where we embed $x^{-1}K[x^{-1}]$ into $\widehat{\fancyscript{K}}$ in the obvious way.

Theorem 6.2

Suppose that $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n}).$ There exists a unique element $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\in \mathcal {H}^\varLambda _{n}({\widehat{\mathcal {O}}})$ such that

$$\begin{aligned} B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}=f_{{\mathfrak s}{\mathfrak t}}+\sum _{\begin{array}{c} ({\mathfrak u},{\mathfrak v})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\\ big({\mathfrak u},{\mathfrak v})\blacktriangleright ({\mathfrak s},{\mathfrak t}) \end{array}} p_{{\mathfrak u}{\mathfrak v}}^{{\mathfrak s}{\mathfrak t}}(x^{-1})f_{{\mathfrak u}{\mathfrak v}}, \end{aligned}$$

where $p_{{\mathfrak u}{\mathfrak v}}^{{\mathfrak s}{\mathfrak t}}(x)\in xK[x].$ Moreover,  $\{B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$ is a cellular basis of $\mathcal {H}^\varLambda _{n}({\widehat{\mathcal {O}}}).$

Proof

The existence of an element $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}$ with the required properties follows directly from (5.3) and Lemma 6.1 using Gaussian elimination. (See the proof of Proposition 6.4, below, which proves a stronger result in characteristic zero.) To prove uniqueness of the element $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}$, suppose, by way of contradiction, that there exist two elements $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}$ and $B_{{\mathfrak s}{\mathfrak t}}'$ in $\mathcal {H}^\varLambda _{n}({\widehat{\mathcal {O}}})$ with the required properties. Then $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}-B'_{{\mathfrak s}{\mathfrak t}}=\sum r_{{\mathfrak u}{\mathfrak v}} f_{{\mathfrak u}{\mathfrak v}}\in \mathcal {H}^\varLambda _{n}({\widehat{\mathcal {O}}})$ and, by assumption, $r_{{\mathfrak u}{\mathfrak v}}\in x^{-1}K[x^{-1}]$ with $r_{{\mathfrak u}{\mathfrak v}}\ne 0$ only if $({\mathfrak u},{\mathfrak v})\blacktriangleright ({\mathfrak s},{\mathfrak t})$. Pick $({\mathfrak a},{\mathfrak b})$ minimal with respect to dominance such that $r_{{\mathfrak a}{\mathfrak b}}\ne 0$. Then, by Theorem 5.5, if we write $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}-B'_{{\mathfrak s}{\mathfrak t}}$ as a linear combination of $\psi$-basis elements then $\psi _{{\mathfrak a}{\mathfrak b}}^{{\widehat{\mathcal {O}}}}$ appears with coefficient $r_{{\mathfrak a}{\mathfrak b}}$. Therefore, $r_{{\mathfrak a}{\mathfrak b}}\in x^{-1}K[x^{-1}]\cap {\widehat{\mathcal {O}}}=0$, a contradiction. Hence, $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}=B'_{{\mathfrak s}{\mathfrak t}}$ as claimed.

By (5.3), the transition matrix between the B-basis and the $\psi$-basis is unitriangular, so $\{B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\}$ is a basis of $\mathcal {H}^\varLambda _{n}({\widehat{\mathcal {O}}})$. To show that the B-basis is cellular we need to check properties $(\hbox {GC}_1)$–$(\hbox {GC}_3)$ from Definition 2.3. We have already verified $(\hbox {GC}_1)$ Moreover, $(\hbox {GC}_3)$ holds because $(B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}})^\diamond =B^{\mathcal {O}}_{{\mathfrak t}{\mathfrak s}}$ by the uniqueness of $B^{\mathcal {O}}_{{\mathfrak t}{\mathfrak s}}$ since $\{f_{{\mathfrak u}{\mathfrak v}}\}$ is a $\diamond$-seminormal basis. It remains to prove $(\hbox {GC}_2)$, which we do in three steps.

Step 1. We claim that if $h\in \mathcal {H}^\varLambda _{n}({\widehat{\mathcal {O}}})$ and ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ then

$$\begin{aligned} B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}h \equiv \sum _{{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})} b_{\mathfrak v}B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}} \quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n), \end{aligned}$$

for some scalars $b_{\mathfrak v}\in {\widehat{\mathcal {O}}}$ that depend only on ${\mathfrak t}, {\mathfrak v}$ and h (and not on ${\mathfrak t}^{\varvec{\lambda }}$).

To see this first note that $\psi ^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}=f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}+\sum _{{\mathfrak v}\blacktriangleright {\mathfrak t}}a_{\mathfrak v}f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}$ by (5.3), for some $a_{\mathfrak v}\in K(x)$. Therefore, it follows by induction on the dominance order that if ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ then

$$\begin{aligned} B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}=f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}+\sum _{{\mathfrak v}\blacktriangleright {\mathfrak t}}p_{{\mathfrak t}{\mathfrak v}}f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}\quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n), \end{aligned}$$

for some $p_{{\mathfrak t}{\mathfrak v}}\in x^{-1}K[x^{-1}]$. As the seminormal basis is cellular, and the transition matrix between the seminormal basis and the B-basis is unitriangular, our claim now follows.

Step 2. As the Specht module $S^{\varvec{\lambda }}$ is cyclic there exists an element $D^{\mathcal {O}}_{\mathfrak t}\in \mathcal {H}^\varLambda _{n}({\widehat{\mathcal {O}}})$ such that $B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}\equiv B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}D^{\mathcal {O}}_{\mathfrak t}\text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n)$. We claim that

$$\begin{aligned} B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\equiv (D^{\mathcal {O}}_s)^\diamond B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}} D^{\mathcal {O}}_{\mathfrak t}\quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n), \end{aligned}$$

for all ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$.

To prove this claim, embed $\mathcal {H}^\varLambda _{n}({\widehat{\mathcal {O}}})$ in $\mathcal {H}^\varLambda _{n}(\widehat{\fancyscript{K}})$. Note that $f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}f_{{\mathfrak u}{\mathfrak v}}=0$ if ${\mathfrak u}\ne {\mathfrak t}^{\varvec{\lambda }}$, so we may assume that $D^{\mathcal {O}}_{\mathfrak t}\equiv \sum _{\mathfrak v}q_{{\mathfrak t}{\mathfrak v}}f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}\text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n)$, for some $q_{{\mathfrak t}{\mathfrak v}}\in \widehat{\fancyscript{K}}$. Then

$$\begin{aligned} B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}\equiv B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}D^{\mathcal {O}}_{\mathfrak t}=\sum _{{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})}\gamma _{{\mathfrak t}^{\varvec{\lambda }}}q_{{\mathfrak t}{\mathfrak v}}f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}\quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n). \end{aligned}$$

Therefore, $q_{{\mathfrak t}{\mathfrak v}}=\tfrac{1}{\gamma _{{\mathfrak t}^{\varvec{\lambda }}}}p_{{\mathfrak t}{\mathfrak v}}$, where $p_{{\mathfrak t}{\mathfrak v}}\in \delta _{{\mathfrak t}{\mathfrak v}}+x^{-1}K[x^{-1}]$ is as in Step 1. In particular, $q_{{\mathfrak t}{\mathfrak t}}=\tfrac{1}{\gamma _{{\mathfrak t}^{\varvec{\lambda }}}}$ and $q_{{\mathfrak t}{\mathfrak v}}\ne 0$ only if ${\mathfrak v}{{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}{\mathfrak t}$. Consequently,

$$\begin{aligned} (D^{\mathcal {O}}_s)^\diamond B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}D^{\mathcal {O}}_{\mathfrak t}\equiv & {} \sum _{\begin{array}{c} ({\mathfrak u},{\mathfrak v}){{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}({\mathfrak s},{\mathfrak t})\\ {\mathfrak u},{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }}) \end{array}} q_{{\mathfrak s}{\mathfrak u}}q_{{\mathfrak t}{\mathfrak v}}f_{{\mathfrak u}{\mathfrak t}^{\varvec{\lambda }}}f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}} =\sum _{({\mathfrak u},{\mathfrak v}){{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}({\mathfrak s},{\mathfrak t})}\gamma _{{\mathfrak t}^{\varvec{\lambda }}}^2q_{{\mathfrak s}{\mathfrak u}}q_{{\mathfrak t}{\mathfrak v}}f_{{\mathfrak u}{\mathfrak v}}\\= & {} f_{{\mathfrak s}{\mathfrak t}}+\sum _{({\mathfrak u},{\mathfrak v})\blacktriangleright ({\mathfrak s},{\mathfrak t})}p_{{\mathfrak s}{\mathfrak u}}p_{{\mathfrak t}{\mathfrak v}}f_{{\mathfrak u}{\mathfrak v}} \quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n). \end{aligned}$$

By construction, $(D^{\mathcal {O}}_s)^\diamond B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}D^{\mathcal {O}}_{\mathfrak t}\in \mathcal {H}^\varLambda _{n}({\widehat{\mathcal {O}}})$. Consequently, our claim now follows using the uniqueness property of $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}$ since $p_{{\mathfrak s}{\mathfrak u}}p_{{\mathfrak t}{\mathfrak v}}\in x^{-1}K[x^{-1}]$ when ${\mathfrak s}\ne {\mathfrak u}$ or ${\mathfrak t}\ne {\mathfrak v}$.

Step 3. We can now verify $(\hbox {GC}_2)$. If $h\in \mathcal {H}^\varLambda _{n}({\widehat{\mathcal {O}}})$ then, using steps 1 and 2,

$$\begin{aligned} B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}h\equiv (D^{\mathcal {O}}_s)^\diamond B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}h \equiv \sum _{{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})}b_{\mathfrak v}(D^{\mathcal {O}}_s)^\diamond B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}} \equiv \sum _{{\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})}b_{\mathfrak v}B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak v}} \quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n), \end{aligned}$$

where $b_{\mathfrak v}$ depends only on ${\mathfrak t}, {\mathfrak v}$ and h and not on ${\mathfrak s}$. Hence, the B-basis satisfies all of the cellular basis axioms and the theorem is proved.

By Theorem 6.2, if $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ then $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\in \mathcal {H}^\varLambda _{n}({\widehat{\mathcal {O}}})$, however, our notation suggests that $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\in \mathcal {H}^\varLambda _{n}({\mathcal {O}})$, where ${\mathcal {O}}=K[x]_{(x)}$. The next result justifies our notation and shows that we can always work over the ring ${\mathcal {O}}$.

Corollary 6.3

Let ${\mathcal {O}}=K[x]_{(x)}.$ Then $\{B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$ is a graded cellular basis of $\mathcal {H}^\varLambda _{n}({\mathcal {O}}).$

Proof

Fix $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$. Then it is enough to prove that $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\in \mathcal {H}^\varLambda _{n}({\mathcal {O}})$. First note that by construction the $\diamond$-seminormal basis is defined over the rational function field K(x), so $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}$ is defined over the ring $R=K(x)\cap {\widehat{\mathcal {O}}}$ since if $({\mathfrak u},{\mathfrak v})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ then $p^{{\mathfrak s}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x^{-1})\in K[x^{-1}]\subset K(x)$ by Theorem 6.2. Every element of K(x) can be written in the form f(x) / g(x), for $f(x),g(x)\in K[x]$ with $\gcd (f,g)=1$. Expanding f / g into a power series, as in Lemma 6.1, it is not difficult to see that if $f/g\in {\widehat{\mathcal {O}}}$ then $g(0)\ne 0$. Therefore, $R\subseteq {\mathcal {O}}$ so that $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}$ is defined over ${\mathcal {O}}$ as claimed.

By similar arguments, $D^{\mathcal {O}}_{\mathfrak t}\in \mathcal {H}^\varLambda _n({\mathcal {O}})$, for all ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$.

If K is a field of characteristic zero then we can determine the degree of the polynomials $p^{{\mathfrak s}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}\ne 0$, for $({\mathfrak u},{\mathfrak v}){{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$.

Proposition 6.4

Suppose that K is a field of characteristic zero. Suppose that $({\mathfrak u},{\mathfrak v})\blacktriangleright ({\mathfrak s},{\mathfrak t})$ for $({\mathfrak s},{\mathfrak t}),({\mathfrak u},{\mathfrak v})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n}).$ Then $p^{{\mathfrak s}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x)\in xK[x]$ and

$$\begin{aligned} \deg p^{{\mathfrak s}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x)\le \tfrac{1}{2}(\deg {\mathfrak u}-\deg {\mathfrak s}+\deg {\mathfrak v}-\deg {\mathfrak t}). \end{aligned}$$

In particular,  $p^{{\mathfrak s}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x)\ne 0$ only if $\deg {\mathfrak u}+\deg {\mathfrak v}>\deg {\mathfrak s}+\deg {\mathfrak t}.$

Proof

We argue by induction on the dominance orders on $\mathcal {P}^{\varLambda }_{n}$ and ${\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$. Note that $\deg p(x)=d$ if and only if $\nu _x\big (p(x^{-1})\big )=-d$. For convenience, throughout the proof given two tableaux ${\mathfrak s},{\mathfrak u}\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ set $\deg ({\mathfrak s},{\mathfrak u})=\deg {\mathfrak s}-\deg {\mathfrak u}$. Therefore, the proposition is equivalent to the claim that $\nu _x\big (p^{{\mathfrak s}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x^{-1})\big ) \ge \frac{1}{2}\big (\deg ({\mathfrak s},{\mathfrak u})+\deg ({\mathfrak t},{\mathfrak v})\big )$.

Suppose first that ${\varvec{\lambda }}=(n|0|\cdots |0)$. Then ${\mathfrak s}={\mathfrak t}^{\varvec{\lambda }}={\mathfrak t}$ and $\psi ^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}=f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ so there is nothing to prove. Hence, we may assume that ${\varvec{\lambda }}\ne (n|0|\cdots |0)$ and that the proposition holds for all more dominant shapes.

Next, consider the case when ${\mathfrak s}={\mathfrak t}^{\varvec{\lambda }}={\mathfrak t}$. By the proof of Lemma 5.3, if ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i}^{\varvec{\lambda }})$ and ${\mathfrak s}{{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}{\mathfrak t}^{\varvec{\lambda }}$ then $y^{{\varvec{\lambda }}}_{{\mathcal {O}}}f_{{\mathfrak s}{\mathfrak s}}=u_{\mathfrak s}'\gamma _{{\mathfrak t}^{\varvec{\lambda }}}f_{{\mathfrak s}{\mathfrak s}}$ for some unit $u'_{\mathfrak s}\in {\mathcal {O}}^\times$. Therefore, by Lemma 5.10, there exist units $u_{\mathfrak s}\in {\mathcal {O}}^\times$ so that in $\mathcal {H}^\varLambda _{n}(K(x))$

$$\begin{aligned} \psi ^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}=\sum _{{\mathfrak s}{{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}{\mathfrak t}^{\varvec{\lambda }}}\frac{u_{\mathfrak s}'\gamma _{{\mathfrak t}^{\varvec{\lambda }}}}{\gamma _{\mathfrak s}}f_{{\mathfrak s}{\mathfrak s}} =f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}+\sum _{{\mathfrak s}\blacktriangleright {\mathfrak t}^{\varvec{\lambda }}}u_{\mathfrak s}\varPhi _e(t)^{\deg ({\mathfrak t}^{\varvec{\lambda }},{\mathfrak s})}f_{{\mathfrak s}{\mathfrak s}}. \end{aligned}$$

Since $t=x+\xi$, the constant term of $\varPhi _e(t)$ is $\varPhi _e(\xi )=0$, so x divides $\varPhi _e(t)$ and $\nu _x(\varPhi _e(t)^{\deg ({\mathfrak t}^{\varvec{\lambda }},{\mathfrak s})})=\deg ({\mathfrak t}^{\varvec{\lambda }},{\mathfrak s})$ since the coefficient of x in $\varPhi _e(t)$ is non-zero. (If K is field of positive characteristic this may not be true.) Expanding each unit $u_{\mathfrak s}$ into a power series, as in Lemma 6.1, the coefficient of $f_{{\mathfrak s}{\mathfrak s}}$ can be written as $b_{\mathfrak s}+c_{\mathfrak s}$ where $b_{\mathfrak s}\in x^{-1}K[x^{-1}]$ and $c_{\mathfrak s}\in {\mathcal {O}}$. In particular, if $b_{\mathfrak s}\ne 0$ and $c_{\mathfrak s}\ne 0$ then $\nu _x(c_{\mathfrak s})\ge 0>\nu _x(b_{\mathfrak s})$ and $\nu _x(c_{\mathfrak s})>\nu _x(b_{\mathfrak s})\ge \deg ({\mathfrak t}^{\varvec{\lambda }},{\mathfrak s})$. Pick ${\mathfrak t}$ minimal with respect to dominance such that $c_{\mathfrak t}\ne 0$. Note that $\nu _x(c_{\mathfrak t})\ge \deg ({\mathfrak t}^{\varvec{\lambda }},{\mathfrak t})$, with equality only if $b_{\mathfrak t}=0$. Using induction, replace $\psi ^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ with the element $A_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}=\psi ^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}-c_{\mathfrak t}B^{\mathcal {O}}_{{\mathfrak t}{\mathfrak t}}$. By construction $A_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}\in \mathcal {H}^\varLambda _{n}({\mathcal {O}})$ and, by (5.3), the coefficient of $f_{{\mathfrak t}{\mathfrak t}}$ in $A_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ is $b_{\mathfrak t}\in x^{-1}K[x^{-1}]$. If $({\mathfrak u},{\mathfrak v}){{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}\ ({\mathfrak t},{\mathfrak t})$ then, $f_{{\mathfrak u}{\mathfrak v}}$ appears in $B^{\mathcal {O}}_{{\mathfrak t}{\mathfrak t}}$ with coefficient $p^{{\mathfrak t}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x^{-1})$ and, by induction, $\nu _x(p^{{\mathfrak t}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x^{-1}))\ge \tfrac{1}{2}(\deg ({\mathfrak t},{\mathfrak u})+\deg ({\mathfrak t},{\mathfrak v}))$. Therefore,

$$\begin{aligned} \nu _x(c_{\mathfrak t}p^{{\mathfrak t}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x^{-1}))= & {} \nu _x(c_{\mathfrak t})+ \nu _x(p^{{\mathfrak t}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x^{-1})) \ge \deg ({\mathfrak t}^{\varvec{\lambda }},{\mathfrak t})+\tfrac{1}{2}(\deg ({\mathfrak t},{\mathfrak u})+\deg ({\mathfrak t},{\mathfrak v}))\\= & {} \tfrac{1}{2}(\deg ({\mathfrak t}^{\varvec{\lambda }},{\mathfrak u})+\deg ({\mathfrak t}^{\varvec{\lambda }},{\mathfrak v})). \end{aligned}$$

It follows that if $f_{{\mathfrak u}{\mathfrak v}}$ appears in $A_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ with non-zero coefficient $a_{{\mathfrak u}{\mathfrak v}}$ then

$$\begin{aligned} \nu _x(a_{{\mathfrak u}{\mathfrak v}})\ge \tfrac{1}{2}(\deg ({\mathfrak t}^{\varvec{\lambda }},{\mathfrak u})+\deg ({\mathfrak t}^{\varvec{\lambda }},{\mathfrak v})). \end{aligned}$$

If $A_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ now has the required properties then we can set $B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}=A_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$. Otherwise, let $({\mathfrak s},{\mathfrak t})$ be a pair of tableau that is minimal with respect to dominance such that the coefficient of $f_{{\mathfrak s}{\mathfrak t}}$ in $A_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ is of the form $b_{{\mathfrak s}{\mathfrak t}}+c_{{\mathfrak s}{\mathfrak t}}$ with $c_{{\mathfrak s}{\mathfrak t}}\ne 0, \nu _x(c_{{\mathfrak s}{\mathfrak t}})\ge 0, b_{{\mathfrak s}{\mathfrak t}}\in x^{-1}K[x^{-1}]$ and $\nu _x(b_{{\mathfrak s}{\mathfrak t}})\ge \tfrac{1}{2}(\deg ({\mathfrak t}^{\varvec{\lambda }},{\mathfrak s})+\deg ({\mathfrak t}^{\varvec{\lambda }},{\mathfrak t}))$. Replacing $A_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ with $A_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}-c_{{\mathfrak s}{\mathfrak t}}B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}$ and continuing in this way we will, in a finite number of steps, construct an element $B'_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ with all of the required properties. By the uniqueness statement in Theorem 6.2, $B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}=B'_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ so this proves the proposition for the polynomials $p^{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}_{{\mathfrak u}{\mathfrak v}}(x^{-1})$.

Finally, suppose that $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2({\varvec{\lambda }})$ with $({\mathfrak t}^{\varvec{\lambda }},{\mathfrak t}^{\varvec{\lambda }})\vartriangleright ({\mathfrak s},{\mathfrak t})$. Without loss of generality, suppose that ${\mathfrak s}={\mathfrak a}(r,r+1)$ where ${\mathfrak a}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$, for $\mathbf {i}\in I^n$, and ${\mathfrak a}\vartriangleright {\mathfrak s}$. Using Lemma 4.19,

$$\begin{aligned} \psi ^{\mathcal {O}}_rB^{\mathcal {O}}_{{\mathfrak a}{\mathfrak t}}= & {} \sum _{({\mathfrak u},{\mathfrak v}) {{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}({\mathfrak a},{\mathfrak t})}p^{{\mathfrak a}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x^{-1})\psi ^{\mathcal {O}}_rf_{{\mathfrak u}{\mathfrak v}}\\= & {} \sum _{({\mathfrak u},{\mathfrak v}){{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}({\mathfrak a},{\mathfrak t})}p^{{\mathfrak a}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}}(x^{-1}) \left( \beta _r({\mathfrak u})f_{{\mathfrak u}(r,r+1),{\mathfrak v}} -\delta _{i_ri_{r+1}}\frac{t^{\hat{\imath }_{r+1}-c_{r+1}({\mathfrak u})}}{[\rho _r({\mathfrak u})]}f_{{\mathfrak u}{\mathfrak v}}\right) . \end{aligned}$$

By induction, $\nu _x(p^{{\mathfrak a}{\mathfrak t}}_{{\mathfrak u}{\mathfrak v}})\ge \frac{1}{2}(\deg ({\mathfrak a},{\mathfrak u})+\deg ({\mathfrak t},{\mathfrak v}))$. Therefore, using Lemma 5.10 (as in the proof of Theorem 3.13), it follows that if $c_{{\mathfrak u}{\mathfrak v}}\ne 0$ is the coefficient of $f_{{\mathfrak u}{\mathfrak v}}$ in the last equation then $\nu _x(c_{{\mathfrak u}{\mathfrak v}})\ge \tfrac{1}{2}\big (\deg ({\mathfrak s},{\mathfrak u})+\deg ({\mathfrak t},{\mathfrak v})\big )$. Hence, the proposition follows by repeating the argument of the last paragraph.

6.2 A distinguished homogeneous basis of$\mathcal {H}^\varLambda _{n}(K)$

This section uses Theorem 6.2 to construct a new graded cellular basis of $\mathcal {H}^\varLambda _{n}(K)$. The existence of such a basis is not automatically guaranteed by Theorem 6.2 because the elements $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\otimes 1_K$, for $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$, are not necessarily homogeneous.

The isomorphisms $K\cong {\mathcal {O}}/x{\mathcal {O}}\cong {\widehat{\mathcal {O}}}/x{\widehat{\mathcal {O}}}$ extend to K-algebra isomorphisms

$$\begin{aligned} \mathcal {H}^\varLambda _{n}(K)\cong \mathcal {H}^\varLambda _{n}({\mathcal {O}})\otimes _{\mathcal {O}}K\cong \mathcal {H}^\varLambda _{n}({\widehat{\mathcal {O}}})\otimes _{{\widehat{\mathcal {O}}}}1_K. \end{aligned}$$

We identify these three K-algebras.

Lemma 6.5

Suppose that $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n}).$ Then

$$\begin{aligned} B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\otimes 1_K=\psi _{{\mathfrak s}{\mathfrak t}}+\sum _{({\mathfrak u},{\mathfrak v})\blacktriangleright ({\mathfrak s},{\mathfrak t})}a_{{\mathfrak u}{\mathfrak v}}\psi _{{\mathfrak u}{\mathfrak v}}, \end{aligned}$$

for some $a_{{\mathfrak u}{\mathfrak b}}\in K$. In particular,  the homogeneous component of $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\otimes 1_K$ of degree $\deg {\mathfrak s}+\deg {\mathfrak t}$ is non-zero.

Proof

This is immediate from Theorem 6.2, (5.3) and Corollary 5.6.

Recall from Step 2 in the proof of Theorem 6.2 that for each ${\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ there exists an element $D^{\mathcal {O}}_{\mathfrak v}\in \mathcal {H}^\varLambda _{n}({\mathcal {O}})$ such that $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\equiv (D^{\mathcal {O}}_{\mathfrak s})^\diamond B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}D^{\mathcal {O}}_{\mathfrak t}\text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n)$.

Definition 6.6

Suppose that ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$.

• (a) If ${\mathfrak v}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ let $D_{\mathfrak v}$ be the homogeneous component of $D^{\mathcal {O}}_{\mathfrak v}\otimes 1_K$ of degree $\deg {\mathfrak v}-\deg {\mathfrak t}^{\varvec{\lambda }}$.

• (b) Define $B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ to be the homogeneous component of $B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}\otimes 1_K$ of degree $2\deg {\mathfrak t}^{\varvec{\lambda }}$. More generally, if ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ define $B_{{\mathfrak s}{\mathfrak t}} = D_{\mathfrak s}^\diamond B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}} D_{\mathfrak t}.$

By Theorem 6.2, $(B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}^{{\mathcal {O}}})^\diamond =B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}^{{\mathcal {O}}}$ which implies that $B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}^\diamond =B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$. Consequently, if ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$ then $B_{{\mathfrak s}{\mathfrak t}}^\diamond =B_{{\mathfrak t}{\mathfrak s}}$. If $B_{{\mathfrak s}{\mathfrak t}}\ne 0$ then, by construction, $B_{{\mathfrak s}{\mathfrak t}}$ is homogeneous of degree $\deg {\mathfrak s}+\deg {\mathfrak t}$. Unfortunately, it is not clear from the definitions that $B_{{\mathfrak s}{\mathfrak t}}$ is non-zero.

Proposition 6.7

Suppose that $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$. Then

$$\begin{aligned} B_{{\mathfrak s}{\mathfrak t}}\equiv \psi _{{\mathfrak s}{\mathfrak t}}+\sum _{({\mathfrak u},{\mathfrak v})\blacktriangleright ({\mathfrak s},{\mathfrak t})}b_{{\mathfrak u}{\mathfrak v}}\psi _{{\mathfrak u}{\mathfrak v}} \quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n), \end{aligned}$$

for some $b_{{\mathfrak u}{\mathfrak v}}\in K$. In particular, $B_{{\mathfrak s}{\mathfrak t}}\ne 0$.

Proof

Fix ${\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}$ and suppose that ${\mathfrak s},{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})$. If ${\mathfrak s}={\mathfrak t}={\mathfrak t}^{\varvec{\lambda }}$ then $B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ is the homogeneous component of $B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}\otimes 1_{K}$ of degree $2\deg {\mathfrak t}^{\varvec{\lambda }}$, so the result is just Lemma 6.5 in this case. Now consider the case when ${\mathfrak s}={\mathfrak t}^{\varvec{\lambda }}$ and ${\mathfrak t}$ is an arbitrary standard ${\varvec{\lambda }}$-tableau. Then, since $B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}^{{\mathcal {O}}}\equiv \psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}^{{\mathcal {O}}}\text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n)$,

$$\begin{aligned} B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}\otimes 1_{K}\equiv (\psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}^{{\mathcal {O}}}\otimes 1_{K}) (D^{\mathcal {O}}_{\mathfrak t}\otimes 1_K)\quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n). \end{aligned}$$

Looking at the homogeneous component of degree $\deg {\mathfrak t}^{\varvec{\lambda }}+\deg {\mathfrak t}$ shows that

$$\begin{aligned} B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}= B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}D_{\mathfrak t}\equiv \psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}+\sum _{{\mathfrak v}\blacktriangleright {\mathfrak t}} a_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}\psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}\quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n), \end{aligned}$$

by Lemma 6.5. Set $b_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}=a_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}$ with $b_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}=1$. Similarly,

$$\begin{aligned} D_{\mathfrak s}^\diamond \psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}\equiv D_{\mathfrak s}^\diamond B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}=B_{{\mathfrak s}{\mathfrak t}^{\varvec{\lambda }}} \equiv \sum _{{\mathfrak u}{{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}{\mathfrak s}}b_{{\mathfrak u}{\mathfrak t}^{\varvec{\lambda }}}\psi _{{\mathfrak u}{\mathfrak t}^{\varvec{\lambda }}}\quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n), \end{aligned}$$

where $b_{{\mathfrak u}{\mathfrak t}^{\varvec{\lambda }}}=a_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak u}}$ with $b_{{\mathfrak s}{\mathfrak t}^{\varvec{\lambda }}}=1$. By Corollary 5.6, $\{\psi _{{\mathfrak u}{\mathfrak v}}\}$ is a graded cellular basis of $\mathcal {H}^\varLambda _{n}(K)$ so, working modulo $\mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n$,

$$\begin{aligned} B_{{\mathfrak s}{\mathfrak t}}= & {} D_{\mathfrak s}^\diamond B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}} D_{\mathfrak t}\equiv \sum _{{\mathfrak v}{{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}{\mathfrak t}} b_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}D_{\mathfrak s}^\diamond \psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}} \equiv \sum _{{\mathfrak v}{{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}{\mathfrak t}} \sum _{{\mathfrak u}{{\mathrm{\,{\underline{{\blacktriangleright }}}\,}}}{\mathfrak s}} b_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}} b_{{\mathfrak u}{\mathfrak t}^{\varvec{\lambda }}}\psi _{{\mathfrak u}{\mathfrak v}}\\= & {} \psi _{{\mathfrak s}{\mathfrak t}} + \sum _{({\mathfrak u},{\mathfrak v})\blacktriangleright ({\mathfrak s},{\mathfrak t})} b_{{\mathfrak u}{\mathfrak t}^{\varvec{\lambda }}}b_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}\psi _{{\mathfrak u}{\mathfrak v}}\quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n). \end{aligned}$$

Setting $b_{{\mathfrak u}{\mathfrak v}}={\mathfrak b}_{{\mathfrak s}{\mathfrak t}^{\varvec{\lambda }}}b_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak v}}$ completes the proof.

Combining these results gives us a new graded cellular basis of $\mathcal {H}^\varLambda _{n}$.

Theorem 6.8

Suppose that K is a field. Then $\{B_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$ is a graded cellular basis of $\mathcal {H}^\varLambda _{n}(K)$ with cellular algebra automorphism $\diamond$.

Proof

By Proposition 6.7 and Corollary 5.6, $\{B_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t}\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$ is a basis of $\mathcal {H}^\varLambda _{n}(K)$. By definition, if $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ then $B_{{\mathfrak s}{\mathfrak t}}$ is homogeneous of degree $\deg {\mathfrak s}+\deg {\mathfrak t}$ and $B_{{\mathfrak s}{\mathfrak t}}^\diamond =B_{{\mathfrak t}{\mathfrak s}}$. Therefore, the basis $\{B_{{\mathfrak s}{\mathfrak t}}\}$ satisfies $(\hbox {GC}_1), (\hbox {GC}_3)$ and $(\hbox {GC}_d)$ from Definition 2.3. Finally, since $B_{{\mathfrak s}{\mathfrak t}}\equiv D_{\mathfrak s}^\diamond B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}} D_{\mathfrak t}\text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n), (\hbox {GC}_2)$ follows by repeating the argument from Step 3 in the proof of Theorem 6.2.

The graded cellular basis $\{B_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$ of $\mathcal {H}^\varLambda _{n}(K)$ is distinguished in the sense that, unlike $\psi _{{\mathfrak s}{\mathfrak t}}$, the element $B_{{\mathfrak s}{\mathfrak t}}$ depends only on $({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})$ and not on a choice of reduced expressions for the permutations $d({\mathfrak s})$ and $d({\mathfrak t})$.

Example 6.9

We give an example to show what B-basis elements look like. Suppose that K is a field of characteristic zero, that $e>2$, and let $\varLambda =2\varLambda _0+\varLambda _1$. Fix a multicharge ${\varvec{\kappa }}=(\kappa _1,\kappa _2,\kappa _3)$ such that ${\varvec{\kappa }}\equiv (0,1,0) \text { }(\text {mod } e)$ and ${\varvec{\kappa }}$ satisfies (3.1). We use the notation of the last two sections, so we work over the rings $(\fancyscript{K},{\widehat{\mathcal {O}}},K)$ and $t=x+\xi \in {\mathcal {O}}$.

Let ${\varvec{\lambda }}=(1|1|1)$ and set

The permutation $d({\mathfrak t})$ has two reduced expressions: $s_1s_2s_1$ and $s_2s_1s_2$. Let $\psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}$ and $\hat{\psi }_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}$, respectively, be the $\psi$-basis elements corresponding to these two reduced expressions. By Definition 2.7, $\psi _1\psi _2\psi _2e(0,1,0)=\psi _2\psi _1\psi _2e(0,1,0)-e(0,1,0),$ so $\psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}=\hat{\psi }_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}-\psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}\ne \hat{\psi }_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}$, where $0\ne \psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}=y_1e(0,1,0)$. The set of standard tableau with residue sequence (0, 1, 0) and which are dominant than or equal to ${\mathfrak t}$ is $\{{\mathfrak t}^{\varvec{\nu }}, {\mathfrak t}^{\varvec{\nu }},{\mathfrak t}^{\varvec{\lambda }},{\mathfrak t}\}$, where ${\varvec{\mu }}=(2|-|1)$ and ${\varvec{\nu }}=(1|1^2|-)$. Of these tableaux, only ${\mathfrak t}$ and ${\mathfrak t}^{\varvec{\lambda }}$ have degree 1, so it follows that $B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}=\psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$ and $B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}=\psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}+c\psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$, for some $c\in K$. To compute c it is enough to work with the seminormal basis $\{f_{\mathfrak s}\,|\,{\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }({\varvec{\lambda }})\}$ of the Specht module $S^{\varvec{\lambda }}$ over ${\mathcal {O}}$. Using Lemma 5.1 and (5.2),

$$\begin{aligned} \psi ^{\mathcal {O}}_{\mathfrak t}=f_{{\mathfrak t}^{\varvec{\lambda }}}\psi ^{\mathcal {O}}_1\psi ^{\mathcal {O}}_2\psi ^{\mathcal {O}}_1 =f_{{\mathfrak t}} -\frac{t^{\kappa _2-1-\kappa _3}[1+\kappa _1-\kappa _2]_t}{[\kappa _1-\kappa _3]_t}f_{{\mathfrak t}^{\varvec{\lambda }}}. \end{aligned}$$

Since ${\varvec{\kappa }}\equiv (0,1,0)$ we can write $1+\kappa _1-\kappa _2=ae$ and $\kappa _1-\kappa _3=be$, for some $a,b\in \mathbb {Z}$. Moreover, $a,b\ne 0$ by (3.1). It is straightforward to check that when $x=0$ the coefficient of $f_{{\mathfrak t}^{\varvec{\lambda }}}$ above is equal to $-\frac{a}{b}$, so this coefficient is invertible in ${\widehat{\mathcal {O}}}$. Hence, $B^{\mathcal {O}}_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}=f_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}+c_1f_{{\mathfrak t}^{{\varvec{\mu }}}{\mathfrak t}^{{\varvec{\mu }}}} +c_2f_{{\mathfrak t}^{{\varvec{\nu }}}{\mathfrak t}^{{\varvec{\nu }}}}\in \mathcal {H}^\varLambda _n({\mathcal {O}})$, for some $c_1,c_2\in x^{-1}K[x^{-1}]$. Since $\deg {\mathfrak t}^{{\varvec{\mu }}}=\deg {\mathfrak t}^{{\varvec{\nu }}}=2>1$, we conclude that $B_{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}=\psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}}+\frac{a}{b}\psi _{{\mathfrak t}^{\varvec{\lambda }}{\mathfrak t}^{\varvec{\lambda }}}$.

We have not yet proved Theorem B from the introduction because it is not clear that $B_{{\mathfrak s}{\mathfrak t}}$ is the homogeneous component of $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\otimes 1_K$ of degree $\deg {\mathfrak s}+\deg {\mathfrak t}$. In fact, there is no reason why this should be true.

As in Theorem B, suppose that K is a field of characteristic zero. Using Proposition 6.4, it follows by induction on the dominance ordering that the homogeneous components of $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\otimes 1_K$ have degree greater than or equal to $\deg {\mathfrak s}+\deg {\mathfrak t}$. Moreover, if $B_{{\mathfrak s}{\mathfrak t}}'$ is the homogeneous component of $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\otimes 1_K$ of degree $\deg {\mathfrak s}+\deg {\mathfrak t}$ then

$$\begin{aligned} B_{{\mathfrak s}{\mathfrak t}}'\equiv B_{{\mathfrak s}{\mathfrak t}}\quad \text { }(\text {mod } \mathcal {H}^{\vartriangleright {\varvec{\lambda }}}_n)\end{aligned}$$

by the proof of Theorem 6.2 (specifically the definition of $D^{\mathcal {O}}_{\mathfrak s}$ and $D^{\mathcal {O}}_{\mathfrak t}$). In particular, $B_{{\mathfrak s}{\mathfrak t}}'\ne 0$. As $B_{{\mathfrak s}{\mathfrak t}}'$ is the minimal homogeneous component of $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\otimes 1_K$, Theorem 6.2 readily implies that $\{B_{{\mathfrak s}{\mathfrak t}}'\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$ is a graded cellular basis of $\mathcal {H}^\varLambda _{n}$. Hence, all of the claims in Theorem B now follow.

If K is a field of positive characteristic it is not clear if $B^{\mathcal {O}}_{{\mathfrak s}{\mathfrak t}}\otimes 1_K$ has homogeneous components of degree less than $\deg {\mathfrak s}+\deg {\mathfrak t}$. It is precisely for this reason that we need the elements $D_{\mathfrak s}$ and $D_{\mathfrak t}$ in Definition 6.6.

Appendix: Seminormal forms for the linear quiver

In this appendix we show how the results in this paper work when $e=0$ so that $\xi \in K$ is either not a root of unity or $\xi =1$ and K is a field of characteristic zero. In order to define a modular system we have to leave the case where the cyclotomic parameters $Q_1,\ldots ,Q_\ell$ are integral, that is, when $Q_l=[\kappa _l]$ for $1\le l\le \ell$. This causes quite a few notational inconveniences, but otherwise the story is much the same as for the case when $e>0$. We do not develop the full theory of “0-idempotent subrings” here. Rather, we show just one way of proving the results in this paper when $e=0$.

Fix a field K and $0\ne \xi \in K$ of quantum characteristic e. That is, either $\xi =1$ and K is a field of characteristic zero or $\xi ^d\ne 1$ for $d\in \mathbb {Z}$. The multicharge ${\varvec{\kappa }}\in \mathbb {Z}^\ell$ is arbitrary.

Let ${\mathcal {O}}=\mathbb {Z}[x,\xi ]_{(x)}$ be the localisation of $\mathbb {Z}[x,\xi ]$ at the principal ideal generated by x. Let $\fancyscript{K}=\mathbb {Q}(x,\xi )$ be the field of fractions of ${\mathcal {O}}$. Define $\mathcal {H}^\varLambda _n({\mathcal {O}})$ to be the cyclotomic Hecke algebra of type A with Hecke parameter $t=\xi$, a unit in ${\mathcal {O}}$, and cyclotomic parameters

$$\begin{aligned} Q_l=x^l+[\kappa _l],\quad \text {for }1\le l\le \ell , \end{aligned}$$

where, as before, $[k]=[k]_t$ for $k\in \mathbb {Z}$. Then $\mathcal {H}^\varLambda _{n}(\fancyscript{K})=\mathcal {H}^\varLambda _n({\mathcal {O}})\otimes _{\mathcal {O}}\fancyscript{K}$ is split semisimple in view of Ariki’s semisimplicity condition [1]. Moreover, by definition, $\mathcal {H}^\varLambda _{n}(K)\cong \mathcal {H}^\varLambda _n({\mathcal {O}})\otimes _{\mathcal {O}}K$, where we consider K as an ${\mathcal {O}}$-module by setting x act on K as multiplication by zero.

Define a new content function for $\mathcal {H}^\varLambda _n({\mathcal {O}})$ by setting

$$\begin{aligned} C_\gamma =t^{c-r}x^{l}+[\kappa _l+c-r], \end{aligned}$$

for a node $\gamma =(l,r,c)$. We will also need the previous definition of contents below. If ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ is a tableau and $1\le k\le n$ then set $C_k({\mathfrak t})=C_\gamma$, where $\gamma$ is the unique node such that ${\mathfrak t}(\gamma )=k$.

As in Sect. 2.5, let $\{m_{{\mathfrak s}{\mathfrak t}}\,|\,({\mathfrak s},{\mathfrak t})\in {\mathrm{Std}}^2(\mathcal {P}^{\varLambda }_{n})\}$ be the Murphy basis of $\mathcal {H}^\varLambda _n({\mathcal {O}})$. Then the analogue of Lemma 2.6 is that if $1\le r\le n$ then

$$\begin{aligned} m_{{\mathfrak s}{\mathfrak t}}L_r=C_r({\mathfrak t})m_{{\mathfrak s}{\mathfrak t}}+\sum _{({\mathfrak u},{\mathfrak v})\vartriangleright ({\mathfrak s},{\mathfrak t})}r_{{\mathfrak u}{\mathfrak v}}m_{{\mathfrak u}{\mathfrak v}}, \end{aligned}$$

for some $r_{{\mathfrak u}{\mathfrak v}}\in {\mathcal {O}}$. As in Sect. 3.1 define a $*$-seminormal basis of $\mathcal {H}^\varLambda _{n}(\fancyscript{K})$ to be a basis $\{f_{{\mathfrak s}{\mathfrak t}}\}$ of simultaneous two-sided eigenvectors for $L_1,\ldots ,L_n$ such that $f_{{\mathfrak s}{\mathfrak t}}^{*}=f_{{\mathfrak t}{\mathfrak s}}$.

Define a seminormal coefficient system for $\mathcal {H}^\varLambda _n({\mathcal {O}})$ to be a set of scalars ${\varvec{\alpha }}=\{\alpha _r({\mathfrak s})\}$ satisfying Definition 3.5(a), (b) and such that if ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ and ${\mathfrak u}={\mathfrak s}(r,r+1)\in {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$ then

$$\begin{aligned} \alpha _r({\mathfrak s})\alpha _r({\mathfrak u}) = \frac{(1-C_r({\mathfrak s})+tC_r({\mathfrak u}))(1+tC_r({\mathfrak s})-C_r({\mathfrak u}))}{P_r({\mathfrak s})P_r({\mathfrak u})}, \end{aligned}$$

(A1)

where $P_r({\mathfrak s})=C_r({\mathfrak u})-C_r({\mathfrak s})$, and where $\alpha _r({\mathfrak s})=0$ if ${\mathfrak u}\notin {\mathop {\mathrm{Std}}\nolimits }(\mathcal {P}^{\varLambda }_{n})$.

As in Theorem 3.9, each seminormal basis of $\mathcal {H}^\varLambda _n(\fancyscript{K})$ is determined by a seminormal coefficient system ${\varvec{\alpha }}=\{\alpha _r({\mathfrak s})\}$, such that

$$\begin{aligned} T_rf_{{\mathfrak s}{\mathfrak t}}=\alpha _r({\mathfrak s})f_{{\mathfrak u}{\mathfrak t}}+\frac{1+(t-1)C_{r+1}({\mathfrak s})}{P_r({\mathfrak s})}f_{{\mathfrak s}{\mathfrak t}}, \quad \text {where }{\mathfrak u}={\mathfrak s}(r,r+1), \end{aligned}$$

together with a set of scalars $\{\gamma _{{\mathfrak t}^{\varvec{\lambda }}}\,|\,{\varvec{\lambda }}\in \mathcal {P}^{\varLambda }_{n}\}$. Notice that $I=\mathbb {Z}$, since $e=0$, so if $\mathbf {i}\in I^n$ then ${\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$ if and only if $c_r({\mathfrak t})=i_r$ and, in turn, this is equivalent to the constant term of $C_r({\mathfrak t})$ being equal to $[i_r]$, for $1\le r\le n$. Arguing as in Lemma 4.3,

$$\begin{aligned} f_{\mathbf {i}}^{\mathcal {O}}= \sum _{{\mathfrak t}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})}\frac{1}{\gamma _{\mathfrak t}}f_{{\mathfrak t}{\mathfrak t}}\in \mathcal {H}^\varLambda _n({\mathcal {O}}). \end{aligned}$$

With these definitions in place all of the arguments in Sect. 4 go through with only minor changes. In particular, if $1\le r\le n$ and $\mathbf {i}\in I^n$ then Definition 4.12 should be replaced by

$$\begin{aligned} \psi ^{\mathcal {O}}_rf_{\mathbf {i}}^{\mathcal {O}}={\left\{ \begin{array}{ll} (T_r+1)\frac{1}{M_r}f_{\mathbf {i}}^{\mathcal {O}},&{}\text {if }i_r=i_{r+1}\\ (T_rL_r-L_rT_r)f_{\mathbf {i}}^{\mathcal {O}},&{}\text {if }i_r=i_{r+1}+1,\\ (T_rL_r-L_rT_r)\frac{1}{M_r}f_{\mathbf {i}}^{\mathcal {O}},&{}\text {otherwise,}\\ \end{array}\right. } \end{aligned}$$

and $y^{{\mathcal {O}}}_rf_{\mathbf {i}}^{\mathcal {O}}=\big (L_r-C_r({\mathfrak t})\big )f_{\mathbf {i}}^{\mathcal {O}}$ where, as before, $M_r=1-L_r+tL_{r+1}$. With these new definitions, if ${\mathfrak s}\in {\mathop {\mathrm{Std}}\nolimits }(\mathbf {i})$, for $\mathbf {i}\in I^m$, and $1\le r\le n$ then Lemma 4.19 becomes

$$\begin{aligned} \psi ^{\mathcal {O}}_r f_{{\mathfrak s}{\mathfrak t}}=B_r({\mathfrak s})f_{{\mathfrak s}{\mathfrak t}}+\frac{\delta _{i_ri_{r+1}}}{P_r({\mathfrak s})}f_{{\mathfrak u}{\mathfrak t}}, \end{aligned}$$

where ${\mathfrak u}={\mathfrak s}(r,r+1)$ and

$$\begin{aligned} B_r({\mathfrak s})={\left\{ \begin{array}{ll} \frac{\alpha _r({\mathfrak s})}{1-C_r({\mathfrak s})+tC_{r+1}({\mathfrak s})},&{}\text {if }i_r=i_{r+1},\\ \alpha _r({\mathfrak s})P_r({\mathfrak s}),&{}\text {if }i_r=i_{r+1}+1,\\ \frac{\alpha _r(s)P_r({\mathfrak s})}{1-C_r({\mathfrak s})+tC_{r+1}({\mathfrak s})},&{}\text {otherwise.} \end{array}\right. } \end{aligned}$$

Observe that if ${\mathfrak u}={\mathfrak s}(r,r+1)$ is a standard tableau then, using (A1), the definitions imply that

$$\begin{aligned} B_r({\mathfrak s})B_r({\mathfrak u}) = {\left\{ \begin{array}{ll} \frac{1}{P_r({\mathfrak s})P_r({\mathfrak u})},&{}\text {if }i_r=i_{r+1},\\ (1-C_r({\mathfrak s})+tC_r({\mathfrak u}))(1+tC_r({\mathfrak s})-C_r({\mathfrak u})),&{}\text {if }i_r\leftrightarrows i_{r+1},\\ (1+tC_r({\mathfrak s})-C_r({\mathfrak u})),&{}\text {if }i_r\rightarrow i_{r+1},\\ (1-C_r({\mathfrak s})+tC_r({\mathfrak u})),&{}\text {if }i_r\leftarrow i_{r+1},\\ 1,&{}\text {otherwise.} \end{array}\right. } \end{aligned}$$

Comparing this with Lemma 4.22, it is now easy to see that analogues of Proposition 4.23 and Proposition 4.24 both hold in this situation. Hence, repeating the arguments of Sect. 4.4, a suitable modification of Theorem A also holds. Similarly, the construction of the bases in Sects. 5 and Sect. 6 now goes though largely without change.