2.1 KLR algebras

We follow closely the set-up of [Reference Brundan, Kleshchev and McNamaraBKM14]. In particular, $\unicode[STIX]{x1D6F7}$ is an irreducible root system with simple roots $\{\unicode[STIX]{x1D6FC}_{i}\mid i\in I\}$ and $\unicode[STIX]{x1D6F7}^{+}$ is the corresponding set of positive roots. Denote by $Q$ the root lattice and by $Q^{+}\subset Q$ the set of $\mathbb{Z}_{{\geqslant}0}$ -linear combinations of simple roots, and write $\text{ht}(\unicode[STIX]{x1D6FC})=\sum _{i\in I}c_{i}$ for $\unicode[STIX]{x1D6FC}=\sum _{i\in I}c_{i}\unicode[STIX]{x1D6FC}_{i}\in Q^{+}$ . The standard symmetric bilinear form $Q\times Q\rightarrow \mathbb{Z},\;(\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD})\mapsto \unicode[STIX]{x1D6FC}\cdot \unicode[STIX]{x1D6FD}$ is normalized so that $d_{i}:=(\unicode[STIX]{x1D6FC}_{i}\cdot \unicode[STIX]{x1D6FC}_{i})/2$ is equal to $1$ for the short simple roots $\unicode[STIX]{x1D6FC}_{i}$ . We also set $d_{\unicode[STIX]{x1D6FD}}:=(\unicode[STIX]{x1D6FD}\cdot \unicode[STIX]{x1D6FD})/2$ for all $\unicode[STIX]{x1D6FD}\in \unicode[STIX]{x1D6F7}^{+}$ . The Cartan matrix is $C=(c_{i,j})_{i,j\in I}$ with $c_{i,j}:=(\unicode[STIX]{x1D6FC}_{i}\cdot \unicode[STIX]{x1D6FC}_{j})/d_{i}$ .

Fix a commutative unital ring $\Bbbk$ and an element $\unicode[STIX]{x1D6FC}\in Q^{+}$ of height $n$ . The symmetric group $S_{n}$ with simple transpositions $s_{r}:=(r~r+1)$ acts on the set

$$\begin{eqnarray}I^{\unicode[STIX]{x1D6FC}}:=\bigg\{\boldsymbol{i}=i_{1}\cdots i_{n}\in I^{n}\;\bigg|\;\mathop{\sum }_{j=1}^{n}\unicode[STIX]{x1D6FC}_{i_{j}}=\unicode[STIX]{x1D6FC}\bigg\}\end{eqnarray}$$

on the left by place permutations. Choose signs $\unicode[STIX]{x1D716}_{i,j}$ for all $i,j\in I$ with $c_{ij}<0$ so that $\unicode[STIX]{x1D716}_{i,j}\unicode[STIX]{x1D716}_{j,i}=-1$ . With this data, Khovanov and Lauda [Reference Khovanov and LaudaKL09, Reference Khovanov and LaudaKL11] and Rouquier [Reference RouquierRou08] define the $\Bbbk$ -algebra $H_{\unicode[STIX]{x1D6FC}}$ with unit $1_{\unicode[STIX]{x1D6FC}}$ , called the KLR algebra, given by generators

$$\begin{eqnarray}\{1_{\boldsymbol{i}}\mid \boldsymbol{i}\in I^{\unicode[STIX]{x1D6FC}}\}\cup \{x_{1},\ldots ,x_{n}\}\cup \{\unicode[STIX]{x1D70F}_{1},\ldots ,\unicode[STIX]{x1D70F}_{n-1}\}\end{eqnarray}$$

subject only to the following relations:

• ∙ $x_{r}x_{s}=x_{s}x_{r}$ ;

• ∙ $1_{\boldsymbol{i}}1_{\boldsymbol{j}}=\unicode[STIX]{x1D6FF}_{\boldsymbol{i},\boldsymbol{j}}1_{\boldsymbol{i}}$ and $\sum _{\boldsymbol{i}\in I^{\unicode[STIX]{x1D6FC}}}1_{\boldsymbol{i}}=1_{\unicode[STIX]{x1D6FC}}$ ;

• ∙ $x_{r}1_{\boldsymbol{i}}=1_{\boldsymbol{i}}x_{r}$ and $\unicode[STIX]{x1D70F}_{r}1_{\boldsymbol{i}}=1_{s_{r}\cdot \boldsymbol{i}}\unicode[STIX]{x1D70F}_{r}$ ;

• ∙ $(x_{t}\unicode[STIX]{x1D70F}_{r}-\unicode[STIX]{x1D70F}_{r}x_{s_{r}(t)})1_{\boldsymbol{i}}=\unicode[STIX]{x1D6FF}_{i_{r},i_{r+1}}(\unicode[STIX]{x1D6FF}_{t,r+1}-\unicode[STIX]{x1D6FF}_{t,r})1_{\boldsymbol{i}};$

• ∙ $\unicode[STIX]{x1D70F}_{r}^{2}1_{\boldsymbol{i}}=\left\{\begin{array}{@{}ll@{}}0 & \text{if }i_{r}=i_{r+1},\\ \unicode[STIX]{x1D700}_{i_{r},i_{r+1}}({x_{r}}^{-c_{i_{r},i_{r+1}}}-{x_{r+1}}^{-c_{i_{r+1},i_{r}}})1_{\boldsymbol{i}} & \text{if }c_{i_{r},i_{r+1}}<0,\\ 1_{\boldsymbol{i}} & \text{otherwise;}\end{array}\right.$

• ∙ $\unicode[STIX]{x1D70F}_{r}\unicode[STIX]{x1D70F}_{s}=\unicode[STIX]{x1D70F}_{s}\unicode[STIX]{x1D70F}_{r}$ if $|r-s|>1$ ;

• ∙ $(\unicode[STIX]{x1D70F}_{r+1}\unicode[STIX]{x1D70F}_{r}\unicode[STIX]{x1D70F}_{r+1}-\unicode[STIX]{x1D70F}_{r}\unicode[STIX]{x1D70F}_{r+1}\unicode[STIX]{x1D70F}_{r})1_{\boldsymbol{i}}=\left\{\begin{array}{@{}ll@{}}\displaystyle \sum _{a+b=-1-c_{i_{r},i_{r+1}}}\!\!\!\unicode[STIX]{x1D700}_{i_{r},i_{r+1}}x_{r}^{a}x_{r+2}^{b}1_{\boldsymbol{i}} & \text{if }c_{i_{r},i_{r+1}}<0\text{ and }i_{r}=i_{r+2},\\ 0 & \text{otherwise.}\end{array}\right.$

The KLR algebra is graded with $\deg 1_{\boldsymbol{i}}=0$ , $\deg (x_{r}1_{\boldsymbol{i}})=2d_{i_{r}}$ and $\deg (\unicode[STIX]{x1D70F}_{r}1_{\boldsymbol{i}})=-\unicode[STIX]{x1D6FC}_{i_{r}}\cdot \unicode[STIX]{x1D6FC}_{i_{r+1}}$ .

For each element $w\in S_{n}$ , fix a reduced decomposition $w=s_{r_{1}}\cdots s_{r_{l}}$ and set $\unicode[STIX]{x1D70F}_{w}=\unicode[STIX]{x1D70F}_{r_{1}}\cdots \unicode[STIX]{x1D70F}_{r_{l}}\in H_{\unicode[STIX]{x1D6FC}}$ (this element depends in general on the choice of reduced decomposition).

Theorem 2.1 (Basis theorem [Reference Khovanov and LaudaKL09, Theorem 2.5]).

The sets

(2.2) $$\begin{eqnarray}\{\unicode[STIX]{x1D70F}_{w}x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}1_{\boldsymbol{i}}\}\quad \text{and}\quad \{x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}\unicode[STIX]{x1D70F}_{w}1_{\boldsymbol{i}}\},\end{eqnarray}$$

with $w$ running over $S_{n}$ , $a_{r}$ running over $\mathbb{Z}_{{\geqslant}0}$ and $\boldsymbol{i}$ running over $I^{\unicode[STIX]{x1D6FC}}$ , are $\Bbbk$ -bases for $H_{\unicode[STIX]{x1D6FC}}$ .

It follows that $H_{\unicode[STIX]{x1D6FC}}$ is Noetherian if $\Bbbk$ is, which we shall always assume from now on. It also follows that for any $1\leqslant r\leqslant n$ , the subalgebra $\Bbbk [x_{r}]\subseteq H_{\unicode[STIX]{x1D6FC}}$ generated by $x_{r}$ is isomorphic to the polynomial algebra $\Bbbk [x]$ ; this fact will be used often without further comment. Moreover, for each $\boldsymbol{i}\in I^{\unicode[STIX]{x1D6FC}}$ , the subalgebra ${\mathcal{P}}(\boldsymbol{i})\subseteq 1_{\boldsymbol{i}}H_{\unicode[STIX]{x1D6FC}}1_{\boldsymbol{i}}$ generated by $\{x_{r}1_{\boldsymbol{i}}\mid 1\leqslant r\leqslant n\}$ is isomorphic to a polynomial algebra in $n$ variables. By defining ${\mathcal{P}}_{\unicode[STIX]{x1D6FC}}:=\bigoplus _{\boldsymbol{i}\in I^{\unicode[STIX]{x1D6FC}}}{\mathcal{P}}(\boldsymbol{i})$ , we obtain a linear action of $S_{n}$ on ${\mathcal{P}}_{\unicode[STIX]{x1D6FC}}$ given by

$$\begin{eqnarray}w\cdot x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}1_{\boldsymbol{i}}=x_{w(1)}^{a_{1}}\cdots x_{w(n)}^{a_{n}}1_{w\cdot \boldsymbol{i}}\end{eqnarray}$$

for any $w\in S_{n}$ , $\boldsymbol{i}\in I^{\unicode[STIX]{x1D6FC}}$ and $a_{1},\ldots ,a_{n}\in \mathbb{Z}_{{\geqslant}0}$ . Setting $\unicode[STIX]{x1D6EC}(\unicode[STIX]{x1D6FC}):={\mathcal{P}}_{\unicode[STIX]{x1D6FC}}^{S_{n}}$ , we have the following result.

Let $\Bbbk$ be Noetherian. If $H$ is a Noetherian graded $\Bbbk$ -algebra, we denote by $H\text{-}\text{mod}$ the category of finitely generated graded left $H$ -modules. The morphisms in this category are all homogeneous degree-zero $H$ -homomorphisms, which we denote by $\text{hom}_{H}(-,-)$ . For $V\in H\text{-}\text{mod}$ , let $q^{d}V$ denote its grading shift by $d$ , so if $V_{m}$ is the degree- $m$ component of $V$ , then $(q^{d}V)_{m}=V_{m-d}$ . More generally, for a Laurent polynomial $a=a(q)=\sum _{d}a_{d}q^{d}\in \mathbb{Z}[q,q^{-1}]$ with non-negative coefficients, we set $aV:=\bigoplus _{d}(q^{d}V)^{\oplus a_{d}}$ .

For $U,V\in H\text{-}\text{mod}$ , we set $\text{Hom}_{H}(U,V):=\bigoplus _{d\in \mathbb{Z}}\text{Hom}_{H}(U,V)_{d}$ , where

$$\begin{eqnarray}\text{Hom}_{H}(U,V)_{d}:=\text{hom}_{H}(q^{d}U,V)=\text{hom}_{H}(U,q^{-d}V).\end{eqnarray}$$

We define $\text{ext}_{H}^{m}(U,V)$ and $\text{Ext}_{H}^{m}(U,V)$ similarly. Since $U$ is finitely generated, $\text{Hom}_{H}(U,V)$ can be identified in the obvious way with the set of all $H$ -module homomorphisms ignoring the gradings. A similar result holds for $\text{Ext}_{H}^{m}(U,V)$ , since $U$ has a resolution by finitely generated projective modules. We use $\cong$ to denote an isomorphism in $H\text{-}\text{mod}$ and $\simeq$ an isomorphism up to a degree shift, i.e.  $V\simeq W$ if and only if $V\cong q^{n}W$ for some $n\in \mathbb{Z}$ .

Let $q$ be a variable, and let $\mathbb{Z}((q))$ be the ring of Laurent series. The quantum integers $[n]=(q^{n}-q^{-n})/(q-q^{-1})$ and expressions like $1/(1-q^{2})$ are always interpreted as elements of $\mathbb{Z}((q))$ .

From now on until the end of § 3, we assume that $\Bbbk$ is a field. A graded $\Bbbk$ -vector space $V=\bigoplus _{m\in \mathbb{Z}}V_{m}$ is said to be Laurentian if the graded components $V_{m}$ are finite dimensional for all $m\in \mathbb{Z}$ and $V_{m}=0$ for $m\ll 0$ . The graded dimension of a Laurentian vector space $V$ is

$$\begin{eqnarray}\text{dim}_{q}\,V:=\mathop{\sum }_{m\in \mathbb{Z}}(\dim V_{m})q^{m}\in \mathbb{Z}((q)).\end{eqnarray}$$

We always work in the category $H_{\unicode[STIX]{x1D6FC}}\text{-}\text{mod}$ . Note that $H_{\unicode[STIX]{x1D6FC}}$ is Laurentian as a vector space; therefore so is any $V\in H_{\unicode[STIX]{x1D6FC}}\text{-}\text{mod}$ , and then so are all $1_{\boldsymbol{i}}V$ for $\boldsymbol{i}\in I^{\unicode[STIX]{x1D6FC}}$ . The formal character of $V\in H_{\unicode[STIX]{x1D6FC}}\text{-}\text{mod}$ is an element of $\bigoplus _{\boldsymbol{i}\in I^{\unicode[STIX]{x1D6FC}}}\mathbb{Z}((q))\cdot \boldsymbol{i}$ defined as

$$\begin{eqnarray}\text{ch}_{q}\,V:=\mathop{\sum }_{\boldsymbol{i}\in I^{\unicode[STIX]{x1D6FC}}}(\text{dim}_{q}\,1_{\boldsymbol{i}}V)\cdot \boldsymbol{i}.\end{eqnarray}$$

Note that $\text{ch}_{q}\,(q^{d}V)=q^{d}\text{ch}_{q}\,(V)$ , where the first $q^{d}$ means the degree shift. We refer to $1_{\boldsymbol{i}}V$ as the $\boldsymbol{i}$ -weight space of $V$ and to its vectors as vectors of weight  $\boldsymbol{i}$ .

There is an anti-automorphism $\unicode[STIX]{x1D704}:H_{\unicode[STIX]{x1D6FC}}\rightarrow H_{\unicode[STIX]{x1D6FC}}$ which fixes all the generators. Given $V\in H_{\unicode[STIX]{x1D6FC}}\text{-}\text{mod}$ , we let

$$\begin{eqnarray}V^{\circledast }:=\text{Hom}_{\mathbb{ k}}(V,\Bbbk ),\end{eqnarray}$$

viewed as a left $H_{\unicode[STIX]{x1D6FC}}$ -module via $\unicode[STIX]{x1D704}$ . Note that in general $V^{\circledast }$ is not finitely generated as an $H_{\unicode[STIX]{x1D6FC}}$ -module, but we will apply $\circledast$ only to finite-dimensional modules. In that case, we have $\text{ch}_{q}\,V^{\circledast }=\overline{\text{ch}_{q}\,V}$ , where the bar means the bar-involution, i.e. the automorphism of $\mathbb{Z}[q,q^{-1}]$ that swaps $q$ and $q^{-1}$ extended to $\bigoplus _{\boldsymbol{i}\in I^{\unicode[STIX]{x1D6FC}}}\mathbb{Z}[q,q^{-1}]\cdot \boldsymbol{i}$ .

Let $\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m}\in Q^{+}$ and $\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FD}_{1}+\cdots +\unicode[STIX]{x1D6FD}_{m}$ . Consider the set of concatenations

$$\begin{eqnarray}I^{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m}}:=\{\boldsymbol{i}^{1}\cdots \boldsymbol{i}^{m}\mid \boldsymbol{i}^{1}\in I^{\unicode[STIX]{x1D6FD}_{1}},\ldots ,\boldsymbol{i}^{m}\in I^{\unicode[STIX]{x1D6FD}_{m}}\}\subseteq I^{\unicode[STIX]{x1D6FC}}.\end{eqnarray}$$

There is a natural (non-unital) algebra embedding $H_{\unicode[STIX]{x1D6FD}_{1}}\otimes \cdots \otimes H_{\unicode[STIX]{x1D6FD}_{m}}\rightarrow H_{\unicode[STIX]{x1D6FC}}$ , which sends the unit $1_{\unicode[STIX]{x1D6FD}_{1}}\otimes \cdots \otimes 1_{\unicode[STIX]{x1D6FD}_{m}}$ to the idempotent

(2.4) $$\begin{eqnarray}1_{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m}}:=\mathop{\sum }_{\boldsymbol{i}\in I^{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m}}}1_{\boldsymbol{i}}\in H_{\unicode[STIX]{x1D6FC}}.\end{eqnarray}$$

We have an exact induction functor

$$\begin{eqnarray}\text{Ind}_{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m}}^{\unicode[STIX]{x1D6FC}}=H_{\unicode[STIX]{x1D6FC}}1_{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m}}\otimes _{H_{\unicode[STIX]{x1D6FD}_{1}}\otimes \cdots \otimes H_{\unicode[STIX]{x1D6FD}_{m}}}\,-~~:~(H_{\unicode[STIX]{x1D6FD}_{1}}\otimes \cdots \otimes H_{\unicode[STIX]{x1D6FD}_{m}})\text{-}\text{mod}\rightarrow H_{\unicode[STIX]{x1D6FC}}\text{-}\text{mod}.\end{eqnarray}$$

For $V_{1}\in H_{\unicode[STIX]{x1D6FD}_{1}}\text{-}\text{mod},\ldots ,V_{m}\in H_{\unicode[STIX]{x1D6FD}_{m}}\text{-}\text{mod}$ , we denote by $V_{1}\boxtimes \cdots \boxtimes V_{m}$ the vector space $V_{1}\otimes \cdots \otimes V_{m}$ , considered naturally as an $(H_{\unicode[STIX]{x1D6FD}_{1}}\otimes \cdots \otimes H_{\unicode[STIX]{x1D6FD}_{m}})$ -module, and set

$$\begin{eqnarray}V_{1}\circ \cdots \circ V_{m}:=\text{Ind}_{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m}}^{\unicode[STIX]{x1D6FC}}V_{1}\boxtimes \cdots \boxtimes V_{m}.\end{eqnarray}$$

2.2 Standard modules

The KLR algebras $H_{\unicode[STIX]{x1D6FC}}$ are known to be affine quasihereditary in the sense of [Reference KleshchevKle15b]; see [Reference KatoKat14, Reference Brundan, Kleshchev and McNamaraBKM14, Reference Kleshchev and LoubertKL15]. Central to this theory is the notion of standard modules, whose definition depends on the choice of a certain partial order. We first fix a convex order on $\unicode[STIX]{x1D6F7}^{+}$ , i.e. a total order such that whenever $\unicode[STIX]{x1D6FE}$ , $\unicode[STIX]{x1D6FD}$ and $\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FD}$ all belong to $\unicode[STIX]{x1D6F7}^{+}$ , $\hspace{2.22198pt}\unicode[STIX]{x1D6FE}\leqslant \unicode[STIX]{x1D6FD}$ implies $\unicode[STIX]{x1D6FE}\leqslant \unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D6FD}\leqslant \unicode[STIX]{x1D6FD}$ . By [Reference PapiPap94], there is a one-to-one correspondence between convex orders on $\unicode[STIX]{x1D6F7}^{+}$ and reduced decompositions of the longest element in the corresponding Weyl group.

A Kostant partition of $\unicode[STIX]{x1D6FC}\in Q^{+}$ is a tuple $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D706}_{1},\ldots ,\unicode[STIX]{x1D706}_{r})$ of positive roots $\unicode[STIX]{x1D706}_{1}\geqslant \unicode[STIX]{x1D706}_{2}\geqslant \cdots \geqslant \unicode[STIX]{x1D706}_{r}$ such that $\unicode[STIX]{x1D706}_{1}+\cdots +\unicode[STIX]{x1D706}_{r}=\unicode[STIX]{x1D6FC}$ . Let $\text{KP}(\unicode[STIX]{x1D6FC})$ denote the set of all Kostant partitions of $\unicode[STIX]{x1D6FC}$ , and for $\unicode[STIX]{x1D706}$ as above define $\unicode[STIX]{x1D706}_{m}^{\prime }=\unicode[STIX]{x1D706}_{r-m+1}$ . Now we have a bilexicographical partial order on $\text{KP}(\unicode[STIX]{x1D6FC})$ , also denoted by ${\leqslant}$ , i.e. if $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D706}_{1},\ldots ,\unicode[STIX]{x1D706}_{r}),\unicode[STIX]{x1D707}=(\unicode[STIX]{x1D707}_{1},\ldots ,\unicode[STIX]{x1D707}_{s})\in \text{KP}(\unicode[STIX]{x1D6FC})$ then $\unicode[STIX]{x1D706}<\unicode[STIX]{x1D707}$ if and only if the following two conditions are satisfied:

• ∙ $\unicode[STIX]{x1D706}_{1}=\unicode[STIX]{x1D707}_{1},\ldots ,\unicode[STIX]{x1D706}_{l-1}=\unicode[STIX]{x1D707}_{l-1}$ and $\unicode[STIX]{x1D706}_{l}<\unicode[STIX]{x1D707}_{l}$ for some $l$ ;

• ∙ $\unicode[STIX]{x1D706}_{1}^{\prime }=\unicode[STIX]{x1D707}_{1}^{\prime },\ldots ,\unicode[STIX]{x1D706}_{m-1}^{\prime }=\unicode[STIX]{x1D707}_{m-1}^{\prime }$ and $\unicode[STIX]{x1D706}_{m}^{\prime }>\unicode[STIX]{x1D707}_{m}^{\prime }$ for some $m$ .

To every $\unicode[STIX]{x1D706}\in \text{KP}(\unicode[STIX]{x1D6FC})$ , McNamara [Reference McNamaraMcn15] (cf. [Reference Kleshchev and RamKR11, Theorem 7.2]) associates an absolutely irreducible finite-dimensional $\circledast$ -self-dual $H_{\unicode[STIX]{x1D6FC}}$ -module $L(\unicode[STIX]{x1D706})$ so that $\{L(\unicode[STIX]{x1D706})\mid \unicode[STIX]{x1D706}\in \text{KP}(\unicode[STIX]{x1D6FC})\}$ is a complete irredundant set of irreducible $H_{\unicode[STIX]{x1D6FC}}$ -modules, up to isomorphism and degree shift. Since $L(\unicode[STIX]{x1D706})$ is $\circledast$ -self-dual, its formal character is bar-invariant. The key special case is where $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D6FC})$ for $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6F7}^{+}$ , in which case $L(\unicode[STIX]{x1D706})=L(\unicode[STIX]{x1D6FC})$ is called a cuspidal irreducible module. For $m\in \mathbb{Z}_{{>}0}$ , we write $(\unicode[STIX]{x1D6FC}^{m})$ for the Kostant partition $(\unicode[STIX]{x1D6FC},\ldots ,\unicode[STIX]{x1D6FC})\in \text{KP}(m\unicode[STIX]{x1D6FC})$ , where $\unicode[STIX]{x1D6FC}$ appears $m$ times. The cuspidal modules have the following nice property.

If $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D706}_{1},\ldots ,\unicode[STIX]{x1D706}_{r})\in \text{KP}(\unicode[STIX]{x1D6FC})$ , the reduced standard module is defined to be

(2.6) $$\begin{eqnarray}\bar{\unicode[STIX]{x1D6E5}}(\unicode[STIX]{x1D706}):=q^{s(\unicode[STIX]{x1D706})}L(\unicode[STIX]{x1D706}_{1})\circ \cdots \circ L(\unicode[STIX]{x1D706}_{r})\end{eqnarray}$$

for a specific degree shift $s(\unicode[STIX]{x1D706})$ , whose description will not be important. Note that the Grothendieck group of finite-dimensional graded $H_{\unicode[STIX]{x1D6FC}}$ -modules can be considered as a $\mathbb{Z}[q,q^{-1}]$ -module with $q[V]=[qV]$ . By [Reference McNamaraMcn15, Theorem 3.1] (cf. [Reference Kleshchev and RamKR11, 7.2, 7.4]), the $H_{\unicode[STIX]{x1D6FC}}$ -module $\bar{\unicode[STIX]{x1D6E5}}(\unicode[STIX]{x1D706})$ has simple head $L(\unicode[STIX]{x1D706})$ , and in the Grothendieck group we have

(2.7) $$\begin{eqnarray}[\bar{\unicode[STIX]{x1D6E5}}(\unicode[STIX]{x1D706})]=[L(\unicode[STIX]{x1D706})]+\mathop{\sum }_{\unicode[STIX]{x1D707}<\unicode[STIX]{x1D706}}d_{\unicode[STIX]{x1D706},\unicode[STIX]{x1D707}}[L(\unicode[STIX]{x1D707})]\end{eqnarray}$$

for some coefficients $d_{\unicode[STIX]{x1D706},\unicode[STIX]{x1D707}}\in \mathbb{Z}[q,q^{-1}]$ , called the (graded) decomposition numbers. The decomposition numbers depend on the characteristic of the ground field $\Bbbk$ .

Let $P(\unicode[STIX]{x1D706})$ denote a projective cover of $L(\unicode[STIX]{x1D706})$ in $H_{\unicode[STIX]{x1D6FC}}\text{-}\text{mod}$ . For $V\in H_{\unicode[STIX]{x1D6FC}}\text{-}\text{mod}$ we define the (graded) composition multiplicity

$$\begin{eqnarray}[V:L(\unicode[STIX]{x1D706})]_{q}:=\text{dim}_{q}\,\text{Hom}(P(\unicode[STIX]{x1D706}),V)\in \mathbb{Z}((q)).\end{eqnarray}$$

The standard module $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})$ is defined as the largest quotient of $P(\unicode[STIX]{x1D706})$ all of whose composition factors are of the form $L(\unicode[STIX]{x1D707})$ with $\unicode[STIX]{x1D707}\leqslant \unicode[STIX]{x1D706}$ ; see [Reference KatoKat14, Corollary 4.13], [Reference Brundan, Kleshchev and McNamaraBKM14, Corollary 3.16] and [Reference KleshchevKle15b, (4.2)]. We note that while the irreducible modules $L(\unicode[STIX]{x1D706})$ are all finite dimensional, the standard modules $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})$ are always infinite dimensional. The standard modules have the usual nice properties.

To construct the standard modules more explicitly, let us first assume that $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6F7}^{+}$ and explain how to construct the cuspidal standard module $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ . Put

$$\begin{eqnarray}q_{\unicode[STIX]{x1D6FC}}:=q^{\unicode[STIX]{x1D6FC}\cdot \unicode[STIX]{x1D6FC}/2}.\end{eqnarray}$$

By [Reference Brundan, Kleshchev and McNamaraBKM14, Lemma 3.2], for each $n\in \mathbb{Z}_{{>}0}$ there exists a unique, up to isomorphism, indecomposable $H_{\unicode[STIX]{x1D6FC}}$ -module $\unicode[STIX]{x1D6E5}_{n}(\unicode[STIX]{x1D6FC})$ such that there are short exact sequences

$$\begin{eqnarray}\displaystyle & \displaystyle 0\rightarrow q_{\unicode[STIX]{x1D6FC}}^{2(n-1)}L(\unicode[STIX]{x1D6FC})\rightarrow \unicode[STIX]{x1D6E5}_{n}(\unicode[STIX]{x1D6FC})\rightarrow \unicode[STIX]{x1D6E5}_{n-1}(\unicode[STIX]{x1D6FC})\rightarrow 0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle 0\rightarrow q_{\unicode[STIX]{x1D6FC}}^{2}\unicode[STIX]{x1D6E5}_{n-1}(\unicode[STIX]{x1D6FC})\rightarrow \unicode[STIX]{x1D6E5}_{n}(\unicode[STIX]{x1D6FC})\rightarrow L(\unicode[STIX]{x1D6FC})\rightarrow 0, & \displaystyle \nonumber\end{eqnarray}$$

where we are using the convention that $\unicode[STIX]{x1D6E5}_{0}(\unicode[STIX]{x1D6FC})=0$ . This yields an inverse system

$$\begin{eqnarray}\cdots \rightarrow \unicode[STIX]{x1D6E5}_{2}(\unicode[STIX]{x1D6FC})\rightarrow \unicode[STIX]{x1D6E5}_{1}(\unicode[STIX]{x1D6FC})\rightarrow \unicode[STIX]{x1D6E5}_{0}(\unicode[STIX]{x1D6FC}),\end{eqnarray}$$

and we have $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})\cong \varprojlim \unicode[STIX]{x1D6E5}_{n}(\unicode[STIX]{x1D6FC})$ ; see [Reference Brundan, Kleshchev and McNamaraBKM14, Corollary 3.16].

Let $m\in \mathbb{Z}_{{>}0}$ . An explicit endomorphism $e_{m}\in \text{End}_{H_{m\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})^{\circ m})^{\text{op}}$ is defined in [Reference Brundan, Kleshchev and McNamaraBKM14, § 3.2], and then

(2.9) $$\begin{eqnarray}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC}^{m})\cong q_{\unicode[STIX]{x1D6FC}}^{m(m-1)/2}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})^{\circ m}e_{m}.\end{eqnarray}$$

Finally, for an arbitrary $\unicode[STIX]{x1D6FC}\in Q^{+}$ and $\unicode[STIX]{x1D706}\in \text{KP}(\unicode[STIX]{x1D6FC})$ , gather together the equal parts of $\unicode[STIX]{x1D706}$ to write $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D706}_{1}^{m_{1}},\ldots ,\unicode[STIX]{x1D706}_{s}^{m_{s}}),$ with $\unicode[STIX]{x1D706}_{1}>\cdots >\unicode[STIX]{x1D706}_{s}$ . Then, by [Reference Brundan, Kleshchev and McNamaraBKM14, (3.5)],

(2.10) $$\begin{eqnarray}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})\cong \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{1}^{m_{1}})\circ \cdots \circ \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{s}^{m_{s}}).\end{eqnarray}$$

Thus, cuspidal standard modules are building blocks for arbitrary standard modules. We will need some of their additional properties. Let $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6F7}^{+}$ . If $\unicode[STIX]{x1D706}\in \text{KP}(\unicode[STIX]{x1D6FC})$ is minimal such that $\unicode[STIX]{x1D706}>(\unicode[STIX]{x1D6FC})$ , then by [Reference Brundan, Kleshchev and McNamaraBKM14, Lemma 2.6], $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE})$ for positive roots $\unicode[STIX]{x1D6FD}>\unicode[STIX]{x1D6FC}>\unicode[STIX]{x1D6FE}$ . In this case, $(\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE})$ is called a minimal pair for $\unicode[STIX]{x1D6FC}$ and we write $\text{mp}(\unicode[STIX]{x1D6FC})$ for the set of all such pairs. The following result proved in [Reference Brundan, Kleshchev and McNamaraBKM14, §§ 3.1 and 4.3] describes some of the important properties of $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ .

2.3 Endomorphisms of standard modules

We shall denote by $x_{\unicode[STIX]{x1D6FC}}$ the degree- $2d_{\unicode[STIX]{x1D6FC}}$ endomorphism of $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ which corresponds to $x$ under the algebra isomorphism $\text{End}_{H_{\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC}))\cong \Bbbk [x]$ in Theorem 2.11(iii).

Proof. It follows from the construction of $x_{\unicode[STIX]{x1D6FC}}$ in [Reference Brundan, Kleshchev and McNamaraBKM14, Theorem 3.3] that $x_{\unicode[STIX]{x1D6FC}}$ is injective and $x_{\unicode[STIX]{x1D6FC}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC}))\cong q_{\unicode[STIX]{x1D6FC}}^{2}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ . This in particular implies the first statement.

Let $V\subseteq \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ be a submodule and $f:V\rightarrow \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ the natural inclusion. First, assume that $V$ is indecomposable. By Theorem 2.11(ii), up to degree shift, $V$ is isomorphic to $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ or $\unicode[STIX]{x1D6E5}_{n}(\unicode[STIX]{x1D6FC})$ for some $n\geqslant 1$ . If $V\simeq \unicode[STIX]{x1D6E5}_{n}(\unicode[STIX]{x1D6FC})$ then $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})/V$ is infinite dimensional and has simple head, so by Theorem 2.11(ii) again, $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})/V\simeq \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ . Then the short exact sequence

$$\begin{eqnarray}0\rightarrow V\rightarrow \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})\rightarrow \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})/V\rightarrow 0\end{eqnarray}$$

splits by projectivity in Theorem 2.11(ii), contradicting indecomposability of $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ . If instead $V\simeq \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ , consider the composition

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})\xrightarrow[{}]{{\sim}}V\xrightarrow[{}]{f}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC}).\end{eqnarray}$$

This produces a graded endomorphism of $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ , so that $V=x_{\unicode[STIX]{x1D6FC}}^{s}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC}))$ for some $s\geqslant 0$ . Since there are inclusions $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})\supset x_{\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})\supset x_{\unicode[STIX]{x1D6FC}}^{2}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})\supset \cdots \,,$ the general case follows from the case where $V$ is indecomposable.◻

Again let $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6F7}^{+}$ . We next consider the standard modules of the form $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC}^{m})$ . By functoriality, the endomorphism $\text{id}^{\otimes (r-1)}\otimes x_{\unicode[STIX]{x1D6FC}}\otimes \text{id}^{\otimes (m-r)}$ of the $H_{\unicode[STIX]{x1D6FC}}^{\otimes m}$ -module $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})^{\boxtimes m}$ induces an endomorphism $X_{r}$ of the $H_{m\unicode[STIX]{x1D6FC}}$ -module $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})^{\circ m}$ . The endomorphisms

$$\begin{eqnarray}X_{1},\ldots ,X_{m}\in \text{End}_{H_{m\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})^{\circ m})\end{eqnarray}$$

commute. Moreover, in [Reference Brundan, Kleshchev and McNamaraBKM14, § 3.2], some additional endomorphisms $\unicode[STIX]{x2202}_{1},\ldots ,\unicode[STIX]{x2202}_{m-1}\in \text{End}_{H_{m\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})^{\circ m})$ are constructed, and it is proved in [Reference Brundan, Kleshchev and McNamaraBKM14, Lemmas 3.7–3.9] that the algebra $\text{End}_{H_{m\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})^{\circ m})^{\text{op}}$ is isomorphic to the nil-Hecke algebra $\mathit{NH}_{m}$ , with $\unicode[STIX]{x2202}_{1},\ldots ,\unicode[STIX]{x2202}_{m-1}$ and (appropriately scaled) $X_{1},\ldots ,X_{m}$ corresponding to the standard generators of $\mathit{NH}_{m}$ . The element $e_{m}$ used in (2.9) is an explicit idempotent in $\mathit{NH}_{m}$ . Consider the algebra of symmetric functions

$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{\unicode[STIX]{x1D6FC},m}:=\Bbbk [X_{1},\ldots ,X_{m}]^{S_{m}}=Z(\mathit{NH}_{m}),\end{eqnarray}$$

with the variables $X_{r}$ in degree $2d_{\unicode[STIX]{x1D6FC}}$ . Note that $\text{dim}_{q}\,\unicode[STIX]{x1D6EC}_{\unicode[STIX]{x1D6FC},m}=1/\prod _{r=1}^{m}(1-q_{\unicode[STIX]{x1D6FC}}^{2r})$ . It is known (see, e.g., [Reference Kleshchev, Loubert and MiemietzKLM13, Theorem 4.4(iii)]) that

(2.14) $$\begin{eqnarray}e_{m}\mathit{NH}_{m}e_{m}=e_{m}\unicode[STIX]{x1D6EC}_{\unicode[STIX]{x1D6FC},m}\cong \unicode[STIX]{x1D6EC}_{\unicode[STIX]{x1D6FC},m}.\end{eqnarray}$$

Proof. Assertion (i) is [Reference Brundan, Kleshchev and McNamaraBKM14, Lemma 3.10], and (ii) follows from [Reference KleshchevKle15b, Lemma 4.11], since $(\unicode[STIX]{x1D6FC}^{m})$ is minimal in $\text{KP}(\unicode[STIX]{x1D6FC})$ by convexity. By (i) and (ii), we have that $\text{dim}_{q}\,\text{End}_{H_{m\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC}^{m}))=1/\prod _{r=1}^{m}(1-q_{\unicode[STIX]{x1D6FC}}^{2r})$ .

(iii) We have that $\mathit{NH}_{m}=\text{End}_{H_{m\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})^{\circ m})^{\text{op}}$ acts naturally on $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})^{\circ m}$ on the right, and so $\unicode[STIX]{x1D6EC}_{\unicode[STIX]{x1D6FC},m}=Z(\mathit{NH}_{m})$ acts naturally on $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC}^{m})=\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})^{\circ m}e_{m}$ . This defines an embedding $\unicode[STIX]{x1D6EC}_{\unicode[STIX]{x1D6FC},m}\rightarrow \text{End}_{H_{m\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC}^{m}))$ . This embedding must be an isomorphism because of the dimensions.

(iv) In view of Lemma 2.13, every non-zero

$$\begin{eqnarray}f\in \Bbbk [X_{1},\ldots ,X_{m}]\subseteq \mathit{NH}_{m}=\text{End}_{H_{m\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})^{\circ m})^{\text{op}}\end{eqnarray}$$

acts as an injective linear operator on $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})^{\circ m}$ . The result now follows from (2.14) and (ii).◻

Finally, we consider a general case. Let $\unicode[STIX]{x1D6FC}\in Q^{+}$ and $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D706}_{1}^{m_{1}},\ldots ,\unicode[STIX]{x1D706}_{s}^{m_{s}})\in \text{KP}(\unicode[STIX]{x1D6FC})$ with $\unicode[STIX]{x1D706}_{1}>\cdots >\unicode[STIX]{x1D706}_{s}$ . By functoriality of induction, we have a natural embedding

(2.16) $$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{\unicode[STIX]{x1D706}_{1},m_{1}}\otimes \cdots \otimes \unicode[STIX]{x1D6EC}_{\unicode[STIX]{x1D706}_{s},m_{s}}\rightarrow \text{End}_{H_{\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})),\quad f_{1}\otimes \cdots \otimes f_{s}\mapsto f_{1}\circ \cdots \circ f_{s}.\end{eqnarray}$$

Theorem 2.17. Let $\unicode[STIX]{x1D6FC}\in Q^{+}$ and $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D706}_{1}^{m_{1}},\ldots ,\unicode[STIX]{x1D706}_{s}^{m_{s}})\in \text{KP}(\unicode[STIX]{x1D6FC})$ with $\unicode[STIX]{x1D706}_{1}>\cdots >\unicode[STIX]{x1D706}_{s}$ . Then

$$\begin{eqnarray}\text{End}_{H_{\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}))\cong \unicode[STIX]{x1D6EC}_{\unicode[STIX]{x1D706}_{1},m_{1}}\otimes \cdots \otimes \unicode[STIX]{x1D6EC}_{\unicode[STIX]{x1D706}_{s},m_{s}}\end{eqnarray}$$

via (2.16), and every non-zero $H_{\unicode[STIX]{x1D6FC}}$ -endomorphism of $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})$ is injective.

Proof. It is easy to see from Theorem 2.15(iv) that every non-zero endomorphism in the image of the embedding (2.16) is injective. To see that there are no other endomorphisms, we first use adjointness of $\text{End}$ and $\text{Res}$ to show that $\text{End}_{H_{\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}))$ is isomorphic to

$$\begin{eqnarray}\text{Hom}_{H_{m_{1}\unicode[STIX]{x1D706}_{1}}\otimes \cdots \otimes H_{m_{s}\unicode[STIX]{x1D706}_{s}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{1}^{m_{1}})\boxtimes \cdots \boxtimes \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{s}^{m_{s}}),\text{Res}_{m_{1}\unicode[STIX]{x1D706}_{1},\ldots ,m_{s}\unicode[STIX]{x1D706}_{s}}^{\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})),\end{eqnarray}$$

and then note that by the Mackey theorem, as in [Reference McNamaraMcn15, Lemma 3.3] for instance, we have $\text{Res}_{m_{1}\unicode[STIX]{x1D706}_{1},\ldots ,m_{s}\unicode[STIX]{x1D706}_{s}}^{\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})\cong \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{1}^{m_{1}})\boxtimes \cdots \boxtimes \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{s}^{m_{s}})$ .◻

3.1 Proof of Theorem A modulo a hypothesis

The following hypothesis concerns a fundamental property of cuspidal standard modules and is probably true beyond finite Lie types.

The goal of this subsection is to prove Theorem A assuming the hypothesis. In § 3.2 the hypothesis will be proved using results of Kashiwara and Park, while in § 3.3 we will give a more elementary proof for simply laced types.

While Hypothesis 3.1 claims that every $\Bbbk [x_{r}]$ acts freely on $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ , no $\Bbbk [x_{r},x_{s}]$ does.

One can say more about the polynomial $f$ in the lemma; see, for example, Proposition 3.14.

Now let $\unicode[STIX]{x1D6FC}\in Q^{+}$ be arbitrary of height $n$ , and let $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D706}_{1}\geqslant \cdots \geqslant \unicode[STIX]{x1D706}_{l})\in \text{KP}(\unicode[STIX]{x1D6FC})$ . Setting

$$\begin{eqnarray}S_{\unicode[STIX]{x1D706}}:=S_{\text{ht}(\unicode[STIX]{x1D706}_{1})}\times \cdots \times S_{\text{ht}(\unicode[STIX]{x1D706}_{l})}\subset S_{n},\end{eqnarray}$$

integers $r,s\in \{1,\ldots ,n\}$ are said to be $\unicode[STIX]{x1D706}$ -equivalent, written $r{\sim}_{\unicode[STIX]{x1D706}}s$ , if they belong to the same orbit of the action of $S_{\unicode[STIX]{x1D706}}$ on $\{1,\ldots ,n\}$ . Finally, recalling the idempotents (2.4), we set

$$\begin{eqnarray}1_{\unicode[STIX]{x1D706}}:=1_{\unicode[STIX]{x1D706}_{1},\ldots ,\unicode[STIX]{x1D706}_{l}}.\end{eqnarray}$$

Proof. Write $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D706}_{1}\geqslant \cdots \geqslant \unicode[STIX]{x1D706}_{l})$ and $\unicode[STIX]{x1D707}=(\unicode[STIX]{x1D707}_{1}\geqslant \cdots \geqslant \unicode[STIX]{x1D707}_{m})$ . The assumption $1_{\unicode[STIX]{x1D706}}\unicode[STIX]{x1D70F}_{w}1_{\unicode[STIX]{x1D707}}\neq 0$ implies that $\boldsymbol{i}^{\unicode[STIX]{x1D706}}=w\cdot \boldsymbol{i}^{\unicode[STIX]{x1D707}}$ for some $\boldsymbol{i}^{\unicode[STIX]{x1D706}}\in I^{\unicode[STIX]{x1D706}_{1},\ldots ,\unicode[STIX]{x1D706}_{l}}$ and $\boldsymbol{i}^{\unicode[STIX]{x1D707}}\in I^{\unicode[STIX]{x1D707}_{1},\ldots ,\unicode[STIX]{x1D707}_{m}}$ . Write $\boldsymbol{i}^{\unicode[STIX]{x1D706}}:=\boldsymbol{i}_{1}^{\unicode[STIX]{x1D706}}\cdots \boldsymbol{i}_{l}^{\unicode[STIX]{x1D706}}$ with $\boldsymbol{i}_{a}^{\unicode[STIX]{x1D706}}\in I^{\unicode[STIX]{x1D706}_{a}}$ for all $a$ , and $\boldsymbol{i}^{\unicode[STIX]{x1D707}}:=\boldsymbol{i}_{1}^{\unicode[STIX]{x1D707}}\cdots \boldsymbol{i}_{m}^{\unicode[STIX]{x1D707}}$ with $\boldsymbol{i}_{b}^{\unicode[STIX]{x1D707}}\in I^{\unicode[STIX]{x1D707}_{b}}$ for all $b$ . Assume for a contradiction that for every $1\leqslant r<n$ we have that $r{\sim}_{\unicode[STIX]{x1D706}}r+1$ implies $w^{-1}(r){\sim}_{\unicode[STIX]{x1D707}}w^{-1}(r+1)$ . Then there is a partition $\{1,\ldots ,l\}=\bigsqcup _{b=1}^{m}A_{b}$ such that $\unicode[STIX]{x1D707}_{b}=\sum _{a\in A_{b}}\unicode[STIX]{x1D706}_{a}$ for all $b=1,\ldots ,m$ . By convexity (cf. [Reference Brundan, Kleshchev and McNamaraBKM14, Lemma 2.4]), we have

$$\begin{eqnarray}\min \{\unicode[STIX]{x1D706}_{a}\mid a\in A_{b}\}\leqslant \unicode[STIX]{x1D707}_{b}\leqslant \max \{\unicode[STIX]{x1D706}_{a}\mid a\in A_{b}\}.\end{eqnarray}$$

This implies $\unicode[STIX]{x1D706}\geqslant \unicode[STIX]{x1D707}$ .◻

Theorem 3.5. Let $\unicode[STIX]{x1D6FC}\in Q^{+}$ and $\unicode[STIX]{x1D706},\unicode[STIX]{x1D707}\in \text{KP}(\unicode[STIX]{x1D6FC})$ . If $\unicode[STIX]{x1D706}\neq \unicode[STIX]{x1D707}$ , then

$$\begin{eqnarray}\text{Hom}_{H_{\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))=0.\end{eqnarray}$$

Proof. Let $n=\text{ht}(\unicode[STIX]{x1D6FC})$ and write $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D706}_{1}\geqslant \cdots \geqslant \unicode[STIX]{x1D706}_{l})$ and $\unicode[STIX]{x1D707}=(\unicode[STIX]{x1D707}_{1}\geqslant \cdots \geqslant \unicode[STIX]{x1D707}_{m})$ . It suffices to prove that

$$\begin{eqnarray}\text{Hom}_{H_{\unicode[STIX]{x1D6FC}}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{1})\circ \cdots \circ \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{l}),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}_{1})\circ \cdots \circ \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}_{m}))=0.\end{eqnarray}$$

Suppose not, and let $\unicode[STIX]{x1D711}$ be a non-zero homomorphism. By Theorem 2.8(ii), we may assume that $\unicode[STIX]{x1D706}<\unicode[STIX]{x1D707}$ . Using Lemma 3.3, pick a generator $v\in \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{1})\circ \cdots \circ \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{l})$ such that $v=1_{\unicode[STIX]{x1D706}}v$ and, for any $r{\sim}_{\unicode[STIX]{x1D706}}r+1$ , there is a non-zero polynomial $f\in \Bbbk [x,y]$ with $f(x_{r},x_{r+1})v=0$ . Then $f(x_{r},x_{r+1})\unicode[STIX]{x1D711}(v)=0$ as well.

Denote by $S^{\unicode[STIX]{x1D707}}$ the set of shortest-length coset representatives for $S_{n}/S_{\unicode[STIX]{x1D707}}.$ Then we can write $\unicode[STIX]{x1D711}(v)=\sum _{w\in S^{\unicode[STIX]{x1D707}}}\unicode[STIX]{x1D70F}_{w}\otimes v_{w}$ for some $v_{w}\in \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}_{1})\otimes \cdots \otimes \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}_{m})$ . Since $\unicode[STIX]{x1D711}(v)=1_{\unicode[STIX]{x1D706}}\unicode[STIX]{x1D711}(v)$ and $1_{\unicode[STIX]{x1D707}}v_{w}=v_{w}$ , we have that $1_{\unicode[STIX]{x1D706}}\unicode[STIX]{x1D70F}_{w}1_{\unicode[STIX]{x1D707}}\neq 0$ whenever $v_{w}\neq 0$ . In particular, if $u\in S^{\unicode[STIX]{x1D707}}$ is an element of maximal length such that $v_{u}\neq 0$ , then by Lemma 3.4 we have $r{\sim}_{\unicode[STIX]{x1D706}}r+1$ and $u^{-1}(r)\not {\sim}_{\unicode[STIX]{x1D707}}u^{-1}(r+1)$ for some $1\leqslant r<n$ .

Now we have

$$\begin{eqnarray}\displaystyle f(x_{r},x_{r+1})\unicode[STIX]{x1D711}(v) & = & \displaystyle f(x_{r},x_{r+1})\mathop{\sum }_{w\in S^{\unicode[STIX]{x1D707}}}\unicode[STIX]{x1D70F}_{w}\otimes v_{w}\nonumber\\ \displaystyle & = & \displaystyle f(x_{r},x_{r+1})\unicode[STIX]{x1D70F}_{u}\otimes v_{u}+\mathop{\sum }_{w\neq u}f(x_{r},x_{r+1})\unicode[STIX]{x1D70F}_{w}\otimes v_{w}\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D70F}_{u}\otimes f(x_{u^{-1}(r)},x_{u^{-1}(r+1)})v_{u}+(\ast ),\nonumber\end{eqnarray}$$

where $(\ast )$ is a sum of elements of the form $\unicode[STIX]{x1D70F}_{w}\otimes v_{w}^{\prime }$ with $v_{w}^{\prime }\in \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}_{1})\otimes \cdots \otimes \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}_{m})$ and $w\in S^{\unicode[STIX]{x1D707}}\setminus \{u\}$ . The last equality holds because in $H_{\unicode[STIX]{x1D6FC}}$ , for all $1\leqslant t\leqslant n$ and $w\in S_{n}$ we have that $x_{t}\unicode[STIX]{x1D70F}_{w}=\unicode[STIX]{x1D70F}_{w}x_{w^{-1}(t)}+(\ast \ast )$ , where $(\ast \ast )$ is a linear combination of elements of the form $\unicode[STIX]{x1D70F}_{y}$ with $y\in S_{n}$ being Bruhat smaller than $w$ .

Since $u^{-1}(r)\not {\sim}_{\unicode[STIX]{x1D707}}u^{-1}(r+1)$ , there are distinct integers $a,b\in \{1,\ldots ,m\}$ and integers $1\leqslant c\leqslant \text{ht}(\unicode[STIX]{x1D707}_{a})$ and $1\leqslant d\leqslant \text{ht}(\unicode[STIX]{x1D707}_{b})$ such that for any pure tensor $v=v^{1}\otimes \cdots \otimes v^{m}\in \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}_{1})\otimes \cdots \otimes \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}_{m})$ and any $s,t\in \mathbb{Z}_{{\geqslant}0}$ , we have

$$\begin{eqnarray}x_{u^{-1}(r)}^{s}x_{u^{-1}(r+1)}^{t}v=v^{1}\otimes \cdots \otimes x_{c}^{s}v^{a}\otimes \cdots \otimes x_{d}^{t}v^{b}\otimes \cdots \otimes v^{m}.\end{eqnarray}$$

By Hypothesis 3.1, $f(x_{u^{-1}(r)},x_{u^{-1}(r+1)})v_{u}\neq 0$ . Hence $f(x_{r},x_{r+1})\unicode[STIX]{x1D711}(v)\neq 0$ , which gives a contradiction.◻

3.2 Proof of Hypothesis 3.1 using the Kashiwara–Park lemma

We begin with a key lemma which follows immediately from the results of [Reference Kashiwara and ParkKP15].

Lemma 3.6. Let $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6F7}^{+}$ , $n=\text{ht}(\unicode[STIX]{x1D6FC})$ and $i\in I$ . Define

$$\begin{eqnarray}\mathfrak{p}_{i,\unicode[STIX]{x1D6FC}}:=\mathop{\sum }_{\boldsymbol{i}\in I^{\unicode[STIX]{x1D6FC}}}\bigg(\mathop{\prod }_{r\in [1,n],i_{r}=i}x_{r}\bigg)1_{\boldsymbol{i}}.\end{eqnarray}$$

Then $\mathfrak{p}_{i,\unicode[STIX]{x1D6FC}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})\neq 0$ .

3.3 Elementary proof of Hypothesis 3.1 for simply laced types

Throughout this subsection, we assume that the root system $\unicode[STIX]{x1D6F7}$ is of (finite) $\mathit{ADE}$ type. Let $\unicode[STIX]{x1D6FC}=a_{1}\unicode[STIX]{x1D6FC}_{1}+\cdots +a_{l}\unicode[STIX]{x1D6FC}_{l}\in Q^{+}$ and $n=\text{ht}(\unicode[STIX]{x1D6FC})=a_{1}+\cdots +a_{l}$ . Pick a permutation $(i_{1},\ldots ,i_{l})$ of $(1,\ldots ,l)$ with $a_{i_{1}}>0$ , and define $\boldsymbol{i}:=i_{1}^{a_{i_{1}}}\cdots i_{l}^{a_{i_{l}}}\in I^{\unicode[STIX]{x1D6FC}}$ . Then the stabilizer of $\boldsymbol{i}$ in $S_{n}$ is the standard parabolic subgroup

$$\begin{eqnarray}S_{\boldsymbol{i}}:=S_{a_{i_{1}}}\times \cdots \times S_{a_{i_{l}}}.\end{eqnarray}$$

Let $S^{\boldsymbol{i}}$ be a set of coset representatives for $S_{n}/S_{\boldsymbol{i}}$ . Then by Theorem 2.3, the element

(3.8) $$\begin{eqnarray}z=z_{\boldsymbol{i}}:=\mathop{\sum }_{w\in S^{\boldsymbol{i}}}(x_{w(1)}+\cdots +x_{w(a_{i_{1}})})1_{w\cdot \boldsymbol{i}}\end{eqnarray}$$

is central of degree $2$ in $H_{\unicode[STIX]{x1D6FC}}$ . For any $1\leqslant r\leqslant n$ , note that

(3.9) $$\begin{eqnarray}a_{i_{1}}x_{r}=z-\mathop{\sum }_{w\in S^{\boldsymbol{i}}}((x_{w(1)}-x_{r})+\cdots +(x_{w(a_{i_{1}})}-x_{r}))1_{w\cdot \boldsymbol{i}}.\end{eqnarray}$$

Let $H_{\unicode[STIX]{x1D6FC}}^{\prime }$ be the subalgebra of $H_{\unicode[STIX]{x1D6FC}}$ generated by

$$\begin{eqnarray}\{1_{\boldsymbol{i}}\mid \boldsymbol{i}\in I^{\unicode[STIX]{x1D6FC}}\}\cup \{\unicode[STIX]{x1D70F}_{r}\mid 1\leqslant r<n\}\cup \{x_{r}-x_{r+1}\mid 1\leqslant r<n\}.\end{eqnarray}$$

For the reader’s convenience, we reprove a result from [Reference Brundan and KleshchevBK12, Lemma 3.1].

Now let $\unicode[STIX]{x1D6FC}$ be a positive root. Then one can always find an index $i_{1}$ with $a_{i_{1}}\cdot 1_{\Bbbk }\neq 0$ , so in this case we always have (3.11) for an appropriate choice of $\boldsymbol{i}$ . We always assume that this choice has been made. Following [Reference Brundan and KleshchevBK12], we can now present another useful description of the cuspidal standard module $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ . Denote by $L^{\prime }(\unicode[STIX]{x1D6FC})$ the restriction of the cuspidal irreducible module $L(\unicode[STIX]{x1D6FC})$ from $H_{\unicode[STIX]{x1D6FC}}$ to $H_{\unicode[STIX]{x1D6FC}}^{\prime }$ .

Using the description of $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ from Lemma 3.12(ii), we can now establish Hypothesis 3.1.

Proof. By (3.9), we can write $x_{r}=(1/a_{i_{1}})z+(\ast ),$ where $(\ast )$ is an element of $H_{\unicode[STIX]{x1D6FC}}^{\prime }$ . For each $1\leqslant m\leqslant N$ , we have

$$\begin{eqnarray}x_{r}^{b}(1\otimes v_{m})=\biggl(\frac{1}{a_{i_{1}}}\biggr)^{\!b}z^{b}\otimes v_{m}+(\ast \ast ),\end{eqnarray}$$

where $(\ast \ast )$ is a linear combination of terms of the form $z^{c}\otimes v_{t}$ with $c<b$ . So $\{1\otimes v_{1},\ldots ,1\otimes v_{N}\}$ is a basis of the free $\Bbbk [x_{r}]$ -module $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FC})$ .◻

The following strengthening of Lemma 3.3 is not needed for the proof of Theorem A, but we include it for completeness.

4.1 Changing scalars

In this subsection we develop a usual formalism of modular representation theory for KLR algebras. There will be nothing surprising here, but we need to exercise care since we work with infinite-dimensional algebras and often with infinite-dimensional modules.

Recall from § 2 that for a left Noetherian graded algebra $H$ , we denote by $H\text{-}\text{mod}$ the category of finitely generated graded $H$ -modules, for which we have the groups $\text{ext}_{H}^{i}(V,W)$ and $\text{Ext}_{H}^{i}(V,W)$ . To deal with change of scalars in Ext groups, we will use the following version of the universal coefficient theorem.

Theorem 4.1 (Universal coefficient theorem).

Let $V_{R}$ and $W_{R}$ be modules in $H_{\unicode[STIX]{x1D6FC},R}\text{-}\text{mod}$ , free as $R$ -modules, and let $\Bbbk$ be an $R$ -algebra. Then for every $j\in \mathbb{Z}_{{\geqslant}0}$ there is an exact sequence of (graded) $\Bbbk$ -modules

$$\begin{eqnarray}\displaystyle 0 & \rightarrow & \displaystyle \text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{j}(V_{R},W_{R})\otimes _{R}\Bbbk \rightarrow \text{Ext}_{H_{\unicode[STIX]{x1D6FC},\Bbbk }}^{j}(V_{R}\otimes _{R}\Bbbk ,W_{R}\otimes _{R}\Bbbk )\nonumber\\ \displaystyle & \rightarrow & \displaystyle \text{Tor}_{1}^{R}(\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{j+1}(V_{R},W_{R}),\Bbbk )\rightarrow 0.\nonumber\end{eqnarray}$$

In particular,

$$\begin{eqnarray}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{j}(V_{R},W_{R})\otimes _{R}K\cong \text{Ext}_{H_{\unicode[STIX]{x1D6FC},K}}^{j}(V_{R}\otimes _{R}K,W_{R}\otimes _{R}K).\end{eqnarray}$$

We need another standard result, whose proof is omitted.

Lemma 4.2. Let $\Bbbk =K$ or $F$ , let $V_{R},W_{R}\in H_{\unicode[STIX]{x1D6FC},R}\text{-}\text{mod}$ be free as $R$ -modules, and let

$$\begin{eqnarray}0\rightarrow W_{R}\stackrel{\unicode[STIX]{x1D704}}{\longrightarrow }E_{R}\stackrel{\unicode[STIX]{x1D70B}}{\longrightarrow }V_{R}\rightarrow 0\end{eqnarray}$$

be the extension corresponding to a class $\unicode[STIX]{x1D709}\in \text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(V_{R},W_{R})$ . Identifying $\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(V_{R},W_{R})\otimes _{R}\Bbbk$ with a subgroup of $\text{Ext}_{H_{\unicode[STIX]{x1D6FC},\Bbbk }}^{1}(V_{R}\otimes _{R}\Bbbk ,W_{R}\otimes _{R}\Bbbk )$ , we have that

$$\begin{eqnarray}0\rightarrow W_{R}\otimes _{R}\Bbbk \xrightarrow[{}]{\unicode[STIX]{x1D704}\otimes \text{id}_{\Bbbk }}E_{R}\otimes _{R}\Bbbk \xrightarrow[{}]{\unicode[STIX]{x1D70B}\otimes \text{id}_{\Bbbk }}V_{R}\otimes _{R}\Bbbk \rightarrow 0\end{eqnarray}$$

is the extension corresponding to a class $\unicode[STIX]{x1D709}\otimes 1_{\Bbbk }\in \text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(V_{R},W_{R})\otimes _{R}\Bbbk$ .

Let $\Bbbk =K$ or $F$ , and let $V_{\Bbbk }\in H_{\unicode[STIX]{x1D6FC},\Bbbk }\text{-}\text{mod}$ . We say that $V_{R}\in H_{\unicode[STIX]{x1D6FC},R}\text{-}\text{mod}$ is an $R$ -form of $V_{\Bbbk }$ if every graded component of $V_{R}$ is free of finite rank as an $R$ -module and, upon identifying $H_{\unicode[STIX]{x1D6FC},R}\otimes _{R}\Bbbk$ with $H_{\unicode[STIX]{x1D6FC},\Bbbk }$ , we have $V_{R}\otimes _{R}\Bbbk \cong V_{\Bbbk }$ as $H_{\unicode[STIX]{x1D6FC},\Bbbk }$ -modules. If $\Bbbk =K$ , by a full lattice in $V_{K}$ we mean a (graded) $R$ -submodule $V_{R}$ of $V_{K}$ such that every graded component $V_{d,R}$ of $V_{R}$ is a finite-rank free $R$ -module which generates the graded component $V_{d,K}$ as a $K$ -module. If $V_{R}$ is an $H_{\unicode[STIX]{x1D6FC},R}$ -invariant full lattice in $V_{K}$ , it is an $R$ -form of $V_{K}$ . Now we can see that every $V_{K}\in H_{\unicode[STIX]{x1D6FC},K}\text{-}\text{mod}$ has an $R$ -form: pick $H_{\unicode[STIX]{x1D6FC},K}$ -generators $v_{1},\ldots ,v_{r}$ and define $V_{R}:=H_{\unicode[STIX]{x1D6FC},R}\cdot v_{1}+\cdots +H_{\unicode[STIX]{x1D6FC},R}\cdot v_{1}$ .

The projective indecomposable modules over $H_{\unicode[STIX]{x1D6FC},F}$ have projective $R$ -forms. Indeed, any $P(\unicode[STIX]{x1D706})_{F}$ is of the form $H_{\unicode[STIX]{x1D6FC},F}e_{\unicode[STIX]{x1D706},F}$ for some degree-zero idempotent $e_{\unicode[STIX]{x1D706},F}$ . By the basis theorem, the degree-zero component $H_{\unicode[STIX]{x1D6FC},F,0}$ of $H_{\unicode[STIX]{x1D6FC},F}$ is defined over $R$ ; more precisely, we have $H_{\unicode[STIX]{x1D6FC},\Bbbk ,0}=H_{\unicode[STIX]{x1D6FC},R,0}\otimes _{R}\Bbbk$ for $\Bbbk =K$ or $F$ . Since $H_{\unicode[STIX]{x1D6FC},F,0}$ is finite dimensional, by the classical theorem on lifting idempotents [Reference Curtis and ReinerCR81, (6.7)], there exists an idempotent $e_{\unicode[STIX]{x1D706},R}\in H_{\unicode[STIX]{x1D6FC},R,0}$ such that $e_{\unicode[STIX]{x1D706},F}=e_{\unicode[STIX]{x1D706},R}\otimes 1_{F}$ , and

$$\begin{eqnarray}P(\unicode[STIX]{x1D706})_{R}:=H_{\unicode[STIX]{x1D6FC},R}e_{\unicode[STIX]{x1D706},R}\end{eqnarray}$$

is an $R$ -form of $P(\unicode[STIX]{x1D706})_{F}$ . The notation $P(\unicode[STIX]{x1D706})_{R}$ will be reserved for this specific $R$ -form of $P(\unicode[STIX]{x1D706})_{F}$ . Note that while the $H_{\unicode[STIX]{x1D6FC},R}$ -module $P(\unicode[STIX]{x1D706})_{R}$ is indecomposable, it is not in general true that $P(\unicode[STIX]{x1D706})_{R}\otimes _{R}K\cong P(\unicode[STIX]{x1D706})_{K}$ ; see Lemma 4.8 for more information.

Let $V_{K}\in H_{\unicode[STIX]{x1D6FC},K}\text{-}\text{mod}$ and let $V_{R}$ be an $R$ -form of $V_{K}$ . The $H_{\unicode[STIX]{x1D6FC},F}$ -module $V_{R}\otimes _{R}F$ is called a reduction modulo $p$ of $V_{K}$ . Reduction modulo $p$ in general depends on the choice of $V_{R}$ . However, as usual, we have a result of the following form.

Lemma 4.3. If $V_{K}\in H_{\unicode[STIX]{x1D6FC},K}\text{-}\text{mod}$ and $V_{R}$ is an $R$ -form of $V_{K}$ , then for any $\unicode[STIX]{x1D706}\in \text{KP}(\unicode[STIX]{x1D6FC})$ we have

$$\begin{eqnarray}[V_{R}\otimes _{R}F:L(\unicode[STIX]{x1D706})_{F}]_{q}=\dim _{q}^{K}\text{Hom}_{H_{\unicode[STIX]{x1D6FC},K}}(P(\unicode[STIX]{x1D706})_{R}\otimes _{R}K,V_{K}).\end{eqnarray}$$

In particular, the composition multiplicities $[V_{R}\otimes _{R}F:L(\unicode[STIX]{x1D706})_{F}]_{q}$ are independent of the choice of the $R$ -form $V_{R}$ .

Proof. We have

$$\begin{eqnarray}[V_{R}\otimes _{R}F:L(\unicode[STIX]{x1D706})_{F}]_{q}=\dim _{q}^{F}\text{Hom}_{H_{\unicode[STIX]{x1D6FC},F}}(P(\unicode[STIX]{x1D706})_{F},V_{R}\otimes _{R}F).\end{eqnarray}$$

By the universal coefficient theorem,

$$\begin{eqnarray}\text{Hom}_{H_{\unicode[STIX]{x1D6FC},F}}(P(\unicode[STIX]{x1D706})_{F},V_{R}\otimes _{R}F)\cong \text{Hom}_{H_{\unicode[STIX]{x1D6FC},R}}(P(\unicode[STIX]{x1D706})_{R},V_{R})\otimes _{R}F.\end{eqnarray}$$

Moreover, note that $\text{Hom}_{H_{\unicode[STIX]{x1D6FC},R}}(P(\unicode[STIX]{x1D706})_{R},V_{R})$ is $R$ -free with (graded) rank equal to

$$\begin{eqnarray}\dim _{q}^{\Bbbk }\text{Hom}_{H_{\unicode[STIX]{x1D6FC},R}}(P(\unicode[STIX]{x1D706})_{R},V_{R})\otimes _{R}\Bbbk\end{eqnarray}$$

for $\Bbbk =F$ or $K$ . Now, by the universal coefficient theorem again, we have that

$$\begin{eqnarray}\dim _{q}^{K}\text{Hom}_{H_{\unicode[STIX]{x1D6FC},R}}(P(\unicode[STIX]{x1D706})_{R},V_{R})\otimes _{R}K=\dim _{q}^{K}\text{Hom}_{H_{\unicode[STIX]{x1D6FC},K}}(P(\unicode[STIX]{x1D706})_{R}\otimes _{R}K,V_{R}\otimes _{R}K),\end{eqnarray}$$

which completes the proof, since $V_{R}\otimes _{R}K\cong V_{K}$ .◻

Our main interest is in reduction modulo $p$ of the irreducible $H_{\unicode[STIX]{x1D6FC},K}$ -modules $L(\unicode[STIX]{x1D706})_{K}$ . Pick a non-zero homogeneous vector $v\in L(\unicode[STIX]{x1D706})_{K}$ and define

$$\begin{eqnarray}L(\unicode[STIX]{x1D706})_{R}:=H_{\unicode[STIX]{x1D6FC},R}\cdot v.\end{eqnarray}$$

Then $L(\unicode[STIX]{x1D706})_{R}$ is an $H_{\unicode[STIX]{x1D6FC},R}$ -invariant full lattice in $L(\unicode[STIX]{x1D706})_{K}$ , and upon reducing modulo $p$ we get an $H_{\unicode[STIX]{x1D6FC},F}$ -module $L(\unicode[STIX]{x1D706})_{R}\otimes _{R}F$ . In general, $L(\unicode[STIX]{x1D706})_{R}\otimes _{R}F$ is not $L(\unicode[STIX]{x1D706})_{F}$ , although this happens ‘often’, for example for cuspidal modules, as stated in the following lemma.

To generalize this lemma to irreducible modules of the form $L(\unicode[STIX]{x1D6FC}^{m})$ , we need to observe that induction and restriction commute with extension of scalars. More precisely, for $\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m}\in Q^{+}$ , $\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FD}_{1}+\cdots +\unicode[STIX]{x1D6FD}_{m}$ and any ground ring $\Bbbk$ , we denote by $H_{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m};\Bbbk }$ the algebra $H_{\unicode[STIX]{x1D6FD}_{1},\Bbbk }\otimes _{\Bbbk }\cdots \otimes _{\Bbbk }H_{\unicode[STIX]{x1D6FD}_{m},\Bbbk }$ identified as usual with a (non-unital) subalgebra of $H_{\unicode[STIX]{x1D6FC},\Bbbk }$ . Now the following lemma is immediate.

Lemma 4.5. Let $V_{R}\in H_{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m};R}\text{-}\text{mod}$ and $W_{R}\in H_{\unicode[STIX]{x1D6FC},R}\text{-}\text{mod}$ . Then for any $R$ -algebra $\Bbbk$ , there are natural isomorphisms of $H_{\unicode[STIX]{x1D6FC},\Bbbk }$ -modules

$$\begin{eqnarray}(\text{Ind}_{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m}}^{\unicode[STIX]{x1D6FC}}V_{R})\otimes _{R}\Bbbk \cong \text{Ind}_{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m}}^{\unicode[STIX]{x1D6FC}}(V_{R}\otimes _{R}\Bbbk )\end{eqnarray}$$

and of $H_{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m};\Bbbk }$ -modules

$$\begin{eqnarray}(\text{Res}_{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m}}^{\unicode[STIX]{x1D6FC}}W_{R})\otimes _{R}\Bbbk \cong \text{Res}_{\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m}}^{\unicode[STIX]{x1D6FC}}(W_{R}\otimes _{R}\Bbbk ).\end{eqnarray}$$

Let $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6F7}^{+}$ and $m\in \mathbb{Z}_{{>}0}$ . If $\Bbbk$ is a field, then by Lemma 2.5 we have $L(\unicode[STIX]{x1D6FC}^{m})_{\Bbbk }\simeq L(\unicode[STIX]{x1D6FC})_{\Bbbk }^{\circ m}$ . By Lemma 4.5,

$$\begin{eqnarray}L(\unicode[STIX]{x1D6FC}^{m})_{R}:=(L(\unicode[STIX]{x1D6FC})_{R})^{\circ m}\end{eqnarray}$$

satisfies $L(\unicode[STIX]{x1D6FC}^{m})_{R}\otimes _{R}\Bbbk \simeq L(\unicode[STIX]{x1D6FC}^{m})_{\Bbbk }$ for $\Bbbk =K$ or $F$ . Taking into account Lemmas 4.3 and 4.4, we get the next result.

It was conjectured in [Reference Kleshchev and RamKR11, Conjecture 7.3] that the reduction modulo $p$ of $L(\unicode[STIX]{x1D706})_{K}$ is always $L(\unicode[STIX]{x1D706})_{F}$ , but counterexamples are given in [Reference WilliamsonWil14] (see also [Reference Brundan, Kleshchev and McNamaraBKM14, Example 2.16]). Still, it is important to understand when we have $L(\unicode[STIX]{x1D706})_{R}\otimes _{R}F\cong L(\unicode[STIX]{x1D706})_{F}$ .

At least, we always have the following property.

Lemma 4.8. Let $\unicode[STIX]{x1D6FC}\in Q^{+}$ and $\unicode[STIX]{x1D706}\in \text{KP}(\unicode[STIX]{x1D6FC})$ . Then in the Grothendieck group of finite-dimensional $H_{\unicode[STIX]{x1D6FC},F}$ -modules we have

(4.9) $$\begin{eqnarray}[L(\unicode[STIX]{x1D706})_{R}\otimes _{R}F]=[L(\unicode[STIX]{x1D706})_{F}]+\mathop{\sum }_{\unicode[STIX]{x1D707}<\unicode[STIX]{x1D706}}a_{\unicode[STIX]{x1D706},\unicode[STIX]{x1D707}}[L(\unicode[STIX]{x1D707})_{F}]\end{eqnarray}$$

for some bar-invariant Laurent polynomials $a_{\unicode[STIX]{x1D706},\unicode[STIX]{x1D707}}\in \mathbb{Z}[q,q^{-1}]$ . Moreover,

$$\begin{eqnarray}P(\unicode[STIX]{x1D706})_{R}\otimes _{R}K\cong P(\unicode[STIX]{x1D706})_{K}\oplus \bigoplus _{\unicode[STIX]{x1D707}>\unicode[STIX]{x1D706}}a_{\unicode[STIX]{x1D707},\unicode[STIX]{x1D706}}P(\unicode[STIX]{x1D707})_{K}.\end{eqnarray}$$

Proof. Let $\Bbbk =K$ or $F$ , and consider the reduced standard module $\bar{\unicode[STIX]{x1D6E5}}(\unicode[STIX]{x1D706})_{\Bbbk }$ ; see (2.6). In view of (2.7), we can write

$$\begin{eqnarray}[L(\unicode[STIX]{x1D706})_{\Bbbk }]:=[\bar{\unicode[STIX]{x1D6E5}}(\unicode[STIX]{x1D706})_{\Bbbk }]+\mathop{\sum }_{\unicode[STIX]{x1D707}<\unicode[STIX]{x1D706}}f_{\unicode[STIX]{x1D706},\unicode[STIX]{x1D707}}^{\Bbbk }[\bar{\unicode[STIX]{x1D6E5}}(\unicode[STIX]{x1D707})_{\mathbb{ k}}]\end{eqnarray}$$

for some $f_{\unicode[STIX]{x1D706},\unicode[STIX]{x1D707}}^{\Bbbk }\in \mathbb{Z}[q,q^{-1}]$ . Using Lemmas 4.4 and 4.5 induction on the bilexicographical order on $\text{KP}(\unicode[STIX]{x1D706})$ , we deduce that (4.9) holds for some, not necessarily bar-invariant, coefficients $a_{\unicode[STIX]{x1D706},\unicode[STIX]{x1D707}}\in \mathbb{Z}[q,q^{-1}]$ . Then we also have

$$\begin{eqnarray}\text{ch}_{q}\,(L(\unicode[STIX]{x1D706})_{R}\otimes _{R}F)=\text{ch}_{q}\,(L(\unicode[STIX]{x1D706})_{F})+\mathop{\sum }_{\unicode[STIX]{x1D707}<\unicode[STIX]{x1D706}}a_{\unicode[STIX]{x1D706},\unicode[STIX]{x1D707}}\text{ch}_{q}\,(L(\unicode[STIX]{x1D707})_{F}).\end{eqnarray}$$

Since reduction modulo $p$ preserves formal characters, the left-hand side is bar-invariant. Moreover, every $\text{ch}_{q}\,(L(\unicode[STIX]{x1D707})_{F})$ is bar-invariant. This implies that the coefficients $a_{\unicode[STIX]{x1D706},\unicode[STIX]{x1D707}}$ are also bar-invariant, since by [Reference Khovanov and LaudaKL09, Theorem 3.17] the formal characters $\{\text{ch}_{q}\,L(\unicode[STIX]{x1D708})_{F}\mid \unicode[STIX]{x1D708}\in \text{KP}(\unicode[STIX]{x1D6FC})\}$ are linearly independent.

Finally, for any $\unicode[STIX]{x1D707}\in \text{KP}(\unicode[STIX]{x1D706})$ we have

$$\begin{eqnarray}a_{\unicode[STIX]{x1D707},\unicode[STIX]{x1D706}}=\dim _{q}^{K}\text{Hom}_{H_{\unicode[STIX]{x1D6FC},K}}(P(\unicode[STIX]{x1D706})_{R}\otimes _{R}K,L(\unicode[STIX]{x1D707})_{K}),\end{eqnarray}$$

thanks to Lemma 4.3. This implies the second statement. ◻

4.2 Integral forms of standard modules

Our next goal is to construct some special $R$ -forms of standard modules. We call an $H_{\unicode[STIX]{x1D6FC},R}$ -module $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R}$ a universal $R$ -form of a standard module if it is an $R$ -form for both $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{K}$ and $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{F}$ . We show how to construct these for all $\unicode[STIX]{x1D706}$ .

By Theorem 2.8(i), for any field $\Bbbk$ and $\unicode[STIX]{x1D6FD}\in R^{+}$ , the standard module $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{\Bbbk }$ has simple head $L(\unicode[STIX]{x1D6FD}^{m})_{\Bbbk }$ . Pick a homogeneous generator $v\in \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{K}$ and consider the $R$ -form $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{R}:=H_{m\unicode[STIX]{x1D6FD},R}\cdot v$ of $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{K}$ . Further, for any $\unicode[STIX]{x1D6FC}\in Q^{+}$ and $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D706}_{1}^{m_{1}},\ldots ,\unicode[STIX]{x1D706}_{s}^{m_{s}})\in \text{KP}(\unicode[STIX]{x1D6FC})$ with $\unicode[STIX]{x1D706}_{1}>\cdots >\unicode[STIX]{x1D706}_{s}$ , we define the following $R$ -form of $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{K}$ (cf. Lemma 4.5):

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R}:=\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{1}^{m_{1}})_{R}\circ \cdots \circ \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{s}^{m_{s}})_{R}.\end{eqnarray}$$

Let

$$\begin{eqnarray}1_{(\unicode[STIX]{x1D706}),R}:=1_{m_{1}\unicode[STIX]{x1D706}_{1},\ldots ,m_{s}\unicode[STIX]{x1D706}_{s};R}.\end{eqnarray}$$

Then, for an appropriate set $S^{(\unicode[STIX]{x1D706})}$ of coset representatives in a symmetric group, we have that $\{\unicode[STIX]{x1D70F}_{w}1_{(\unicode[STIX]{x1D706}),R}\mid w\in S^{(\unicode[STIX]{x1D706})}\}$ is a basis of $H_{\unicode[STIX]{x1D6FC},R}1_{(\unicode[STIX]{x1D706}),R}$ considered as a right $H_{m_{1}\unicode[STIX]{x1D706}_{1},\ldots ,m_{s}\unicode[STIX]{x1D706}_{s};R}$ -module. So

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R}=\bigoplus _{w\in S^{(\unicode[STIX]{x1D706})}}\unicode[STIX]{x1D70F}_{w}1_{(\unicode[STIX]{x1D706}),R}\otimes \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{1}^{m_{1}})_{R}\otimes \cdots \otimes \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{s}^{m_{s}})_{R}.\end{eqnarray}$$

In particular, choosing $v_{t}\in \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{t}^{m_{t}})_{K}$ with $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}_{t}^{m_{t}})_{R}=H_{m_{t}\unicode[STIX]{x1D706}_{t},R}\cdot v_{t}$ for all $1\leqslant t\leqslant s$ and setting $v:=1_{(\unicode[STIX]{x1D706}),K}\otimes v_{1}\otimes \cdots \otimes v_{s}$ , we have

(4.11) $$\begin{eqnarray}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R}=H_{\unicode[STIX]{x1D6FC},R}\cdot v.\end{eqnarray}$$

Now we show that $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R}$ is a universal $R$ -form.

Proof. In view of (2.10) and Lemma 4.5, we may assume that $\unicode[STIX]{x1D706}$ is of the form $(\unicode[STIX]{x1D6FD}^{m})$ for a positive root $\unicode[STIX]{x1D6FD}$ so that $\unicode[STIX]{x1D6FC}=m\unicode[STIX]{x1D6FD}$ . By Lemma 4.3, for any $\unicode[STIX]{x1D707}\in \text{KP}(\unicode[STIX]{x1D6FC})$ we have

$$\begin{eqnarray}[\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{R}\otimes _{R}F:L(\unicode[STIX]{x1D707})_{F}]_{q}=\dim _{q}^{K}\text{Hom}_{H_{\unicode[STIX]{x1D6FC},K}}(P(\unicode[STIX]{x1D707})_{R}\otimes _{R}K,\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{K}).\end{eqnarray}$$

By convexity, $(\unicode[STIX]{x1D6FD}^{m})$ is a minimal element of $\text{KP}(\unicode[STIX]{x1D6FC})$ . So Lemma 4.8 implies that all composition factors of $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{R}\otimes _{R}F$ are $\simeq L(\unicode[STIX]{x1D6FD}^{m})_{F}$ . Moreover,

$$\begin{eqnarray}[\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{R}\otimes _{R}F:L(\unicode[STIX]{x1D6FD}^{m})_{F}]_{q}=[\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{K}:L(\unicode[STIX]{x1D6FD}^{m})_{K}]_{q}=[\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{F}:L(\unicode[STIX]{x1D6FD}^{m})_{F}]_{q}.\end{eqnarray}$$

By construction, $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{R}$ is cyclic, hence so is $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{R}\otimes _{R}F$ . Therefore $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{R}\otimes _{R}F$ is a module with simple head and belongs to the category of all modules in $H_{\unicode[STIX]{x1D6FC},F}\text{-}\text{mod}$ with composition factors $\simeq L(\unicode[STIX]{x1D6FD}^{m})_{F}$ . Since $(\unicode[STIX]{x1D6FD}^{m})$ is minimal in $\text{KP}(\unicode[STIX]{x1D6FC})$ , we have that $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{F}$ is the projective cover of $L(\unicode[STIX]{x1D6FD}^{m})_{F}$ in this category; see [Reference KleshchevKle15b, Lemma 4.11]. So there is a surjective homomorphism from $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{F}$ onto $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D6FD}^{m})_{R}\otimes _{R}F$ . This has to be an isomorphism since we have proved that the two modules have the same composition multiplicities.◻

From now on, the notation $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R}$ will be reserved for a universal $R$ -form.

Proof. By the universal coefficient theorem, for any $j\geqslant 0$ we can embed

$$\begin{eqnarray}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{j}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})\otimes _{R}F\end{eqnarray}$$

into $\text{Ext}_{H_{\unicode[STIX]{x1D6FC},F}}^{j}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{F},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{F})$ . So assertion (i) follows from Theorem A, and (iii) follows from Theorem 2.8(iii). Now statement (ii) also follows from the universal coefficient theorem, since

$$\begin{eqnarray}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R})=0\end{eqnarray}$$

by (iii), which makes the $\text{Tor}_{1}$ term trivial.◻

Given an $R$ -module $V$ , denote by $V^{\mathtt{Tors}}$ its torsion submodule. Torsion in Ext groups

$$\begin{eqnarray}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{j}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})^{\mathtt{Tors}}\end{eqnarray}$$

is of importance for Problem 4.7; see Remark 4.17. The following result was surprising to us.

Theorem 4.14. Let $\unicode[STIX]{x1D6FC}\in Q^{+}$ and $\unicode[STIX]{x1D706},\unicode[STIX]{x1D707}\in \text{KP}(\unicode[STIX]{x1D6FC})$ . Then the $R$ -module

$$\begin{eqnarray}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})\end{eqnarray}$$

is torsion-free.

Proof. By Proposition 4.13(iii), we may assume that $\unicode[STIX]{x1D706}\neq \unicode[STIX]{x1D707}$ . By the universal coefficient theorem, there is an exact sequence

$$\begin{eqnarray}\displaystyle 0 & \rightarrow & \displaystyle \text{Hom}_{H_{\unicode[STIX]{x1D6FC},R}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})\otimes _{R}F\rightarrow \text{Hom}_{H_{\unicode[STIX]{x1D6FC},F}}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{F},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{F})\nonumber\\ \displaystyle & \rightarrow & \displaystyle \text{Tor}_{1}^{R}(\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R}),F)\rightarrow 0.\nonumber\end{eqnarray}$$

By Theorem A, the middle term vanishes; hence the third term also vanishes, which implies the theorem. ◻

We will need the following generalization.

Proof. Apply induction on the length of the $\unicode[STIX]{x1D6E5}$ -filtration, the induction base coming from Theorem 4.14. If the filtration has length greater than $1$ , we have an exact sequence

$$\begin{eqnarray}0\rightarrow V_{1}\rightarrow V\rightarrow V_{2}\rightarrow 0,\end{eqnarray}$$

such that the inductive assumption applies to $V_{1}$ and $V_{2}$ . Then we get a long exact sequence

$$\begin{eqnarray}\displaystyle & & \displaystyle \text{Hom}_{H_{\unicode[STIX]{x1D6FC},R}}(V_{1},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})\rightarrow \text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(V_{2},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})\nonumber\\ \displaystyle & & \displaystyle \quad \rightarrow \text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(V,\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})\rightarrow \text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(V_{1},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R}).\nonumber\end{eqnarray}$$

By Proposition 4.13(i), the first term vanishes. By the inductive assumption, the second and fourth terms are torsion-free. Hence so is the third term. ◻

While we have just proved that there is no torsion in $\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})$ , the following result reveals the importance of torsion in $\text{Ext}^{2}$ -groups.

Corollary 4.16. Let $\unicode[STIX]{x1D6FC}\in Q^{+}$ and $\unicode[STIX]{x1D706},\unicode[STIX]{x1D707}\in \text{KP}(\unicode[STIX]{x1D6FC})$ . We have

$$\begin{eqnarray}\displaystyle & & \displaystyle \dim _{q}^{F}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},F}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{F},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{F})\nonumber\\ \displaystyle & & \displaystyle \quad =\dim _{q}^{K}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},K}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{K},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{K})+\text{d}_{q}^{R}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{2}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})^{\mathtt{Tors}}.\nonumber\end{eqnarray}$$

In particular,

$$\begin{eqnarray}\dim _{q}^{F}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},F}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{F},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{F})=\dim _{q}^{K}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},K}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{K},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{K})\end{eqnarray}$$

if and only if the $R$ -module $\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{2}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})$ is torsion-free.

Proof. By the universal coefficient theorem, there is an exact sequence

$$\begin{eqnarray}\displaystyle 0 & \rightarrow & \displaystyle \text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})\otimes _{R}F\rightarrow \text{Ext}_{H_{\unicode[STIX]{x1D6FC},F}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{F},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{F})\nonumber\\ \displaystyle & \rightarrow & \displaystyle \text{Tor}_{1}^{R}(\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{2}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R}),F)\rightarrow 0\nonumber\end{eqnarray}$$

and an isomorphism

$$\begin{eqnarray}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})\otimes _{R}K\cong \text{Ext}_{H_{\unicode[STIX]{x1D6FC},K}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{K},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{K}).\end{eqnarray}$$

The last isomorphism and Theorem 4.14 imply

$$\begin{eqnarray}\dim _{q}^{K}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},K}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{K},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{K})=\text{d}_{q}^{R}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R}).\end{eqnarray}$$

On the other hand,

$$\begin{eqnarray}\text{d}_{q}^{R}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{2}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})^{\mathtt{Tors}}=\dim _{q}^{F}\text{Tor}_{1}^{R}(\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{2}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R}),F),\end{eqnarray}$$

so the result now follows from the exactness of the first sequence. ◻

Remark 4.17. By Theorem 4.14, lack of torsion in $\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{2}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})$ is equivalent to the fact that the extension groups $\text{Ext}_{H_{\unicode[STIX]{x1D6FC}}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))$ have the same graded dimension in characteristic $0$ and characteristic $p$ . This is relevant for Problem 4.7. However, we do not understand the precise connection between Problem 4.7 and lack of torsion in the groups $\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{2}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})$ . For example, we do not know if such lack of torsion for all $\unicode[STIX]{x1D706}$ and $\unicode[STIX]{x1D707}$ implies (or is equivalent to) the James conjecture having positive solution. In the next section we establish a different statement of that nature. Set

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{\Bbbk }:=\bigoplus _{\unicode[STIX]{x1D706}\in \text{KP}(\unicode[STIX]{x1D6FC})}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{\Bbbk }.\end{eqnarray}$$

By the universal coefficient theorem, all groups $\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{j}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})$ are torsion-free if and only if the dimension of the $\Bbbk$ -algebras $\text{Ext}_{H_{\unicode[STIX]{x1D6FC},\Bbbk }}^{\bullet }(\unicode[STIX]{x1D6E5}_{\mathbb{ k}},\unicode[STIX]{x1D6E5}_{\Bbbk })$ is the same for $\Bbbk =K$ and $\Bbbk =F$ , and

$$\begin{eqnarray}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},\Bbbk }}^{\bullet }(\unicode[STIX]{x1D6E5}_{\Bbbk },\unicode[STIX]{x1D6E5}_{\Bbbk })\cong \text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{\bullet }(\unicode[STIX]{x1D6E5}_{R},\unicode[STIX]{x1D6E5}_{R})\otimes _{R}\Bbbk\end{eqnarray}$$

for $\Bbbk =K$ or $F$ . We do not know if the James conjecture has positive solution under the assumption that all groups $\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{j}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})$ are torsion-free.

4.3 Integral forms of projective modules in characteristic zero

Recall that by lifting idempotents, we have constructed projective $R$ -forms $P(\unicode[STIX]{x1D706})_{R}$ of the projective indecomposable modules $P(\unicode[STIX]{x1D706})_{F}$ . Our next goal is to construct some interesting $R$ -forms of the projective modules $P(\unicode[STIX]{x1D706})_{K}$ . As we cannot denote them by $P(\unicode[STIX]{x1D706})_{R}$ , we will have to use the notation $Q(\unicode[STIX]{x1D706})_{R}$ . We will construct $Q(\unicode[STIX]{x1D706})_{R}$ using the usual ‘universal extension procedure’ applied to universal $R$ -forms of the standard modules, but in our ‘infinite-dimensional integral’ situation we need to be rather careful. We begin with some lemmas.

For $\unicode[STIX]{x1D706}\in \text{KP}(\unicode[STIX]{x1D6FC})$ and $\Bbbk \in \{F,K,R\}$ , we consider the endomorphism algebra

$$\begin{eqnarray}B_{\unicode[STIX]{x1D706},\Bbbk }:=\text{End}_{H_{\unicode[STIX]{x1D6FC},\Bbbk }}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{\Bbbk })^{\text{op}}.\end{eqnarray}$$

By Proposition 4.13(ii), we have $B_{\unicode[STIX]{x1D706},F}\cong B_{\unicode[STIX]{x1D706},R}\otimes F$ and $B_{\unicode[STIX]{x1D706},K}\cong B_{\unicode[STIX]{x1D706},R}\otimes K$ . Note that $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{\Bbbk }$ is naturally a right $B_{\unicode[STIX]{x1D706},\Bbbk }$ -module. We need to know that this $B_{\unicode[STIX]{x1D706},\Bbbk }$ -module is finitely generated. In fact, we will prove that it is free of finite rank. First of all, this is known over a field.

The following general lemma, whose proof is omitted, will help us to transfer the result of Lemma 4.19 from $K$ and $F$ to  $R$ .

Proof. Let $\unicode[STIX]{x1D706}=(\unicode[STIX]{x1D706}_{1}^{m_{1}},\ldots ,\unicode[STIX]{x1D706}_{s}^{m_{s}})$ for positive roots $\unicode[STIX]{x1D706}_{1}>\cdots >\unicode[STIX]{x1D706}_{s}$ . Choose $v=1_{(\unicode[STIX]{x1D706}),K}\otimes v_{1}\otimes \cdots \otimes v_{s}$ as in (4.11). There is a submodule $M\subset \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{K}$ with $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{K}/M\cong \bar{\unicode[STIX]{x1D6E5}}(\unicode[STIX]{x1D706})_{K}$ . Pick $h_{1},\ldots ,h_{N}\in H_{\unicode[STIX]{x1D6FC},R}$ such that $\{h_{1}v+M,\ldots ,h_{N}v+M\}$ is an $R$ -basis of $\bar{\unicode[STIX]{x1D6E5}}(\unicode[STIX]{x1D706})_{R}=H_{\unicode[STIX]{x1D6FC},R}\cdot (v+M)$ . By Lemma 4.19,

$$\begin{eqnarray}\{h_{1}v\otimes 1_{\Bbbk },\ldots ,h_{N}v\otimes 1_{\Bbbk }\}\end{eqnarray}$$

is a $B_{\unicode[STIX]{x1D706},\Bbbk }$ -basis of $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{\!R}\,\otimes _{R}\,\Bbbk$ for $\Bbbk \,=\,\!K\!$ or $F$ . Now apply Proposition 4.13(ii) and Lemma 4.20. ◻

Lemma 4.24 (Universal extension procedure).

Let $\Bbbk \in \{F,K,R\}$ and $\unicode[STIX]{x1D707}\in \text{KP}(\unicode[STIX]{x1D6FC})$ , and let $V_{\Bbbk }$ be an indecomposable $H_{\unicode[STIX]{x1D6FC},\Bbbk }$ -module with a finite $\unicode[STIX]{x1D6E5}$ -filtration, all of whose subfactors are of the form $\simeq \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{\Bbbk }$ for $\unicode[STIX]{x1D706}\not \geqslant \unicode[STIX]{x1D707}$ . If $\Bbbk =R$ , assume in addition that $V_{R}\otimes _{R}K$ is indecomposable. Let

$$\begin{eqnarray}r(q):=\text{d}_{q}^{B_{\unicode[STIX]{x1D707},\Bbbk }}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},\Bbbk }}^{1}(V_{\mathbb{ k}},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{\Bbbk })\in \mathbb{Z}[q,q^{-1}].\end{eqnarray}$$

Then there exists an $H_{\unicode[STIX]{x1D6FC},\Bbbk }$ -module $E(V_{\Bbbk },\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{\Bbbk })$ with the following properties:

1. (i) $E(V_{\Bbbk },\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{\Bbbk })$ is indecomposable;

2. (ii) $\text{Ext}_{H_{\unicode[STIX]{x1D6FC},\Bbbk }}^{1}(E(V_{\Bbbk },\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{\Bbbk }),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{\Bbbk })=0$ ;

3. (iii) there is a short exact sequence

   $$\begin{eqnarray}0\rightarrow \overline{r(q)}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{\Bbbk }\rightarrow E(V_{\Bbbk },\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{\Bbbk })\rightarrow V_{\Bbbk }\rightarrow 0.\end{eqnarray}$$

Proof. In this proof we drop $H_{\unicode[STIX]{x1D6FC},\Bbbk }$ from the indices and write $\text{Ext}^{1}$ for $\text{Ext}_{H_{\unicode[STIX]{x1D6FC},\Bbbk }}^{1}$ etc. Also, when it is unlikely to cause confusion, we drop $\Bbbk$ from the indices. Let $\unicode[STIX]{x1D709}_{1},\ldots ,\unicode[STIX]{x1D709}_{r}$ be a minimal set of homogeneous generators of $\text{Ext}^{1}(V,\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))$ as a $B_{\unicode[STIX]{x1D707}}$ -module, and let $d_{s}:=\deg (\unicode[STIX]{x1D709}_{s})$ for $s=1,\ldots ,r$ so that $r(q)=\sum _{s}q^{d_{s}}$ . The extension

$$\begin{eqnarray}0\rightarrow q^{-d_{1}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})\rightarrow E_{1}\rightarrow V\rightarrow 0,\end{eqnarray}$$

corresponding to $\unicode[STIX]{x1D709}_{1}$ , yields the long exact sequence

$$\begin{eqnarray}\displaystyle \text{Hom}(q^{-d_{1}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))\stackrel{\unicode[STIX]{x1D711}}{\longrightarrow }\text{Ext}^{1}(V,\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))\stackrel{\unicode[STIX]{x1D713}}{\longrightarrow }\text{Ext}^{1}(E_{1},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))\rightarrow 0. & & \displaystyle \nonumber\end{eqnarray}$$

Here we have used the fact that $\text{Ext}^{1}(q^{-d_{1}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))=0$ ; see Proposition 4.13(iii). Note that $q^{-d_{1}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})=\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})$ as $H_{\unicode[STIX]{x1D6FC}}$ -modules but with degrees shifted down by $d_{1}$ . So we can consider the identity map $\text{id}:q^{-d_{1}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})\rightarrow \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})$ , which has degree $d_{1}$ . The connecting homomorphism $\unicode[STIX]{x1D711}$ maps this identity map to $\unicode[STIX]{x1D709}_{1}$ . It follows that $\text{Ext}^{1}(E_{1},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))$ is generated as a $B_{\unicode[STIX]{x1D707}}$ -module by the elements $\bar{\unicode[STIX]{x1D709}}_{2}:=\unicode[STIX]{x1D713}(\unicode[STIX]{x1D709}_{2}),\ldots ,\bar{\unicode[STIX]{x1D709}}_{r}:=\unicode[STIX]{x1D713}(\unicode[STIX]{x1D709}_{r})$ . Repeating the argument $r-1$ more times, we get an extension

$$\begin{eqnarray}0\rightarrow q^{-d_{1}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})\oplus \cdots \oplus q^{-d_{r}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})=\overline{r(q)}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})\rightarrow E\rightarrow V\rightarrow 0\end{eqnarray}$$

such that in the corresponding long exact sequence

$$\begin{eqnarray}\displaystyle & \displaystyle \text{Hom}(E,\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))\stackrel{\unicode[STIX]{x1D712}}{\longrightarrow }\text{Hom}(\overline{r(q)}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})) & \displaystyle \nonumber\\ \displaystyle & \displaystyle \stackrel{\unicode[STIX]{x1D711}}{\longrightarrow }\text{Ext}^{1}(V,\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))\rightarrow \text{Ext}^{1}(E,\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))\rightarrow 0, & \displaystyle \nonumber\end{eqnarray}$$

for all $s=1,\ldots ,r$ we have $\unicode[STIX]{x1D711}(\unicode[STIX]{x1D70B}_{s})=\unicode[STIX]{x1D709}_{s}$ , where $\unicode[STIX]{x1D70B}_{s}$ is the (degree- $d_{s}$ ) projection onto the $s$ th summand. In particular, $\unicode[STIX]{x1D711}$ is surjective and $\text{Ext}^{1}(E,\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))=0$ .

It remains to prove that $E$ is indecomposable. We first prove this when $\Bbbk$ is a field. In that case, if $E=E^{\prime }\oplus E^{\prime \prime }$ , then both $E^{\prime }$ and $E^{\prime \prime }$ have finite $\unicode[STIX]{x1D6E5}$ -filtrations; see [Reference KleshchevKle15b, Corollary 7.10]. Since $\text{Ext}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}))=0$ for $\unicode[STIX]{x1D706}\not >\unicode[STIX]{x1D707}$ , there is a partition $J^{\prime }\sqcup J^{\prime \prime }=\{1,\ldots ,r\}$ such that there are submodules

$$\begin{eqnarray}M^{\prime }\cong \bigoplus _{j\in J^{\prime }}q^{d_{j}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})\subseteq E^{\prime },\quad M^{\prime \prime }\cong \bigoplus _{j\in J^{\prime \prime }}q^{d_{j}}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})\subseteq E^{\prime \prime },\end{eqnarray}$$

and $E^{\prime }/M^{\prime }$ and $M^{\prime \prime }/E^{\prime \prime }$ have $\unicode[STIX]{x1D6E5}$ -filtrations. Since $\text{Hom}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}),V)=0$ , we deduce that $V\cong E^{\prime }/M^{\prime }\oplus E^{\prime \prime }/M^{\prime \prime }$ . As $V$ is indecomposable, we may assume that $E^{\prime }/M^{\prime }=0$ . Then some projection $\unicode[STIX]{x1D70B}_{s}$ lifts to a homomorphism $E\rightarrow \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})$ , which shows that this $\unicode[STIX]{x1D70B}_{s}$ is in the image of $\unicode[STIX]{x1D712}$ and hence in the kernel of $\unicode[STIX]{x1D711}$ , which is a contradiction.

Now let $\Bbbk =R$ . Note that $V$ and $E$ are free as $R$ -modules, since all the $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D708})_{R}$ are. If $E_{R}$ is decomposable, then so is $E_{R}\otimes K$ ; therefore it suffices to prove that $E_{R}\otimes K$ is indecomposable. In view of Corollary 4.15, the $B_{\unicode[STIX]{x1D707},K}$ -module

$$\begin{eqnarray}\text{Ext}^{1}(V_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{R})\otimes _{R}K\cong \text{Ext}^{1}(V_{R}\otimes _{R}K,\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{K})\end{eqnarray}$$

is minimally generated by $\unicode[STIX]{x1D709}_{1,R}\otimes 1_{K},\ldots ,\unicode[STIX]{x1D709}_{r,R}\otimes 1_{K}$ . It follows, using Lemma 4.2, that $E_{R}\otimes _{R}K\cong E_{K}$ , where $E_{K}$ is constructed using the universal extension procedure starting with the indecomposable module $V_{K}:=V_{R}\otimes _{R}K$ as in the first part of the proof of the lemma. By the field case established in the previous paragraph, $E_{K}$ is indecomposable.◻

Let $\unicode[STIX]{x1D706}\in \text{KP}(\unicode[STIX]{x1D6FC})$ . For $\Bbbk \in \{R,K,F\}$ , we construct a module $Q(\unicode[STIX]{x1D706})_{\Bbbk }$ by starting with $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{\Bbbk }$ and repeatedly applying the universal extension procedure. To simplify notation, we drop some of the indices $\Bbbk$ if this is unlikely to lead to confusion. Given Laurent polynomials $r_{0}(q),r_{1}(q),\ldots ,r_{m}(q)\in \mathbb{Z}[q,q^{-1}]$ with non-negative coefficients and Kostant partitions $\unicode[STIX]{x1D706}^{0},\unicode[STIX]{x1D706}^{1},\ldots ,\unicode[STIX]{x1D706}^{m}\in \text{KP}(\unicode[STIX]{x1D6FC})$ , we will use the notation

$$\begin{eqnarray}V=r_{0}(q)\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{0})\mid r_{1}(q)\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{1})\mid \cdots \mid r_{m}(q)\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{m})\end{eqnarray}$$

to indicate that the $H_{\unicode[STIX]{x1D6FC}}$ -module $V$ has a filtration $V=V_{0}\supseteq V_{1}\supseteq \cdots \supseteq V_{m+1}=(0)$ such that $V_{s}/V_{s+1}\cong r_{s}(q)\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{s})$ for $s=0,1,\ldots ,m$ .

If $\text{Ext}_{H_{\unicode[STIX]{x1D6FC}}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))=0$ for all $\unicode[STIX]{x1D707}\in \text{KP}(\unicode[STIX]{x1D6FC})$ , we set $Q(\unicode[STIX]{x1D706})_{\Bbbk }:=\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{\Bbbk }$ . Otherwise, let $\unicode[STIX]{x1D706}^{1,\Bbbk }\in \text{KP}(\unicode[STIX]{x1D6FC})$ be minimal with $\text{Ext}_{H_{\unicode[STIX]{x1D6FC}}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{1,\Bbbk }))\neq 0$ . Note that this $\unicode[STIX]{x1D706}^{1,\Bbbk }$ could indeed depend on the ground ring $\Bbbk$ , hence the notation. Also notice that $\unicode[STIX]{x1D706}^{1,\Bbbk }>\unicode[STIX]{x1D706}$ . Let

$$\begin{eqnarray}E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk })_{\mathbb{ k}}:=E(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{1,\Bbbk }));\end{eqnarray}$$

see Lemma 4.24. By construction, we have

$$\begin{eqnarray}E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk })_{\mathbb{ k}}=\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})\mid \overline{r_{1,\Bbbk }(q)}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{1,\Bbbk }),\end{eqnarray}$$

where

$$\begin{eqnarray}r_{1,\Bbbk }(q)=\text{d}_{q}^{B_{\unicode[STIX]{x1D706}^{1,\Bbbk }}}\text{Ext}_{H_{\unicode[STIX]{x1D6FC}}}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{1,\Bbbk })).\end{eqnarray}$$

This Laurent polynomial may depend on $\Bbbk$ , hence the notation. If

$$\begin{eqnarray}\text{Ext}_{H_{\unicode[STIX]{x1D6FC}}}^{1}(E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk }),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))=0\end{eqnarray}$$

for all $\unicode[STIX]{x1D707}\in \text{KP}(\unicode[STIX]{x1D6FC})$ , we set $Q(\unicode[STIX]{x1D706})_{\Bbbk }:=E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk })_{\Bbbk }$ . Otherwise, let $\unicode[STIX]{x1D706}^{2,\Bbbk }\in \text{KP}(\unicode[STIX]{x1D6FC})$ be minimal with $\text{Ext}_{H_{\unicode[STIX]{x1D6FC}}}^{1}(E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk }),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{2,\Bbbk }))\neq 0.$ Note that $\unicode[STIX]{x1D706}^{2,\Bbbk }>\unicode[STIX]{x1D706}$ and $\unicode[STIX]{x1D706}^{2,\Bbbk }\neq \unicode[STIX]{x1D706}^{1,\Bbbk }$ . Let

$$\begin{eqnarray}E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk },\unicode[STIX]{x1D706}^{2,\Bbbk })_{\mathbb{ k}}:=E(E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk }),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{2,\Bbbk })).\end{eqnarray}$$

By construction, we have

$$\begin{eqnarray}E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk },\unicode[STIX]{x1D706}^{2,\Bbbk })_{\mathbb{ k}}=\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})\mid \overline{r_{1,\Bbbk }(q)}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{1,\Bbbk })\mid \overline{r_{2,\Bbbk }(q)}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{2,\Bbbk }),\end{eqnarray}$$

where

$$\begin{eqnarray}r_{2,\Bbbk }(q)=\text{d}_{q}^{B_{\unicode[STIX]{x1D706}^{2,\Bbbk }}}\text{Ext}_{H_{\unicode[STIX]{x1D6FC}}}^{1}(E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk }),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{2,\Bbbk })).\end{eqnarray}$$

If $\text{Ext}_{H_{\unicode[STIX]{x1D6FC}}}^{1}(E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk },\unicode[STIX]{x1D706}^{2,\Bbbk }),\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707}))=0$ for all $\unicode[STIX]{x1D707}\in \text{KP}(\unicode[STIX]{x1D6FC})$ , we set

$$\begin{eqnarray}Q(\unicode[STIX]{x1D706})_{\Bbbk }:=E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk },\unicode[STIX]{x1D706}^{2,\Bbbk }).\end{eqnarray}$$

Since in each step we have to pick $\unicode[STIX]{x1D706}^{t,\Bbbk }>\unicode[STIX]{x1D706}$ , which does not belong to $\{\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk },\ldots ,\unicode[STIX]{x1D706}^{t-1,\Bbbk }\}$ , the process will stop after finitely many steps, and we will obtain an indecomposable module

$$\begin{eqnarray}E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk },\ldots ,\unicode[STIX]{x1D706}^{m_{\Bbbk },\Bbbk })_{\mathbb{ k}}=\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})\mid \overline{r_{1,\Bbbk }(q)}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{1,\Bbbk })\mid \cdots \mid \overline{r_{m_{\Bbbk },\Bbbk }(q)}\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{m_{\Bbbk },\Bbbk }),\end{eqnarray}$$

where

(4.25) $$\begin{eqnarray}r_{t,\Bbbk }(q)=\text{d}_{q}^{B_{\unicode[STIX]{x1D706}^{t,\Bbbk }}}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},\Bbbk }}^{1}(E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk },\ldots ,\unicode[STIX]{x1D706}^{t-1,\Bbbk })_{\mathbb{ k}},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{t,\Bbbk })_{\mathbb{ k}})\end{eqnarray}$$

for all $1\leqslant t\leqslant m_{\Bbbk }$ , such that

$$\begin{eqnarray}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},\Bbbk }}^{1}(E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk },\ldots ,\unicode[STIX]{x1D706}^{m_{\Bbbk },\Bbbk })_{\mathbb{ k}},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D707})_{\Bbbk })=0\end{eqnarray}$$

for all $\unicode[STIX]{x1D707}\in \text{KP}(\unicode[STIX]{x1D6FC})$ . We set

$$\begin{eqnarray}Q(\unicode[STIX]{x1D706})_{\Bbbk }:=E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,\Bbbk },\ldots ,\unicode[STIX]{x1D706}^{m_{\Bbbk },\Bbbk })_{\mathbb{ k}}.\end{eqnarray}$$

Proof. Property (i) follows from the construction and Lemma 4.18. Assertion (ii) follows from (i), the construction, and Theorem 2.8(v).

To prove (iii) and (iv), we show by induction on $t=0,1,\ldots$ , that we can choose $\unicode[STIX]{x1D706}^{t,R}=\unicode[STIX]{x1D706}^{t,K}$ , $r_{t,R}(q)=r_{t,K}(q)$ and

(4.27) $$\begin{eqnarray}E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,R},\ldots ,\unicode[STIX]{x1D706}^{t,R})_{R}\otimes _{R}K\cong E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,K},\ldots ,\unicode[STIX]{x1D706}^{t,K})_{K}.\end{eqnarray}$$

The induction base is simply the statement $\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R}\otimes _{R}K\cong \unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{K}$ . For the induction step, assume $t>0$ and that the claim has been proved for all $s<t$ .

Let $\unicode[STIX]{x1D709}_{1,R},\ldots ,\unicode[STIX]{x1D709}_{r,R}$ be a minimal set of generators of the $B_{\unicode[STIX]{x1D706}^{t,R},R}$ -module

$$\begin{eqnarray}\text{Ext}_{H_{\unicode[STIX]{x1D6FC},R}}^{1}(E(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706}^{1,R},\ldots ,\unicode[STIX]{x1D706}^{t-1,R})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{t,R})_{R}),\end{eqnarray}$$

so that

$$\begin{eqnarray}\displaystyle r_{t,R}(q)=\deg (\unicode[STIX]{x1D709}_{1,R})+\cdots +\deg (\unicode[STIX]{x1D709}_{r,R}). & & \displaystyle \nonumber\end{eqnarray}$$

Using Corollary 4.15 and the universal coefficient theorem, we deduce that $\unicode[STIX]{x1D706}^{t,K}$ can be chosen to be $\unicode[STIX]{x1D706}^{t,R}$ and the $B_{\unicode[STIX]{x1D706}^{t,R},K}$ -module

$$\begin{eqnarray}\text{Ext}^{1}(\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706})_{R},\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{t,R})_{R})\otimes _{R}K\cong \text{Ext}^{1}(V_{R}\otimes _{R}K,\unicode[STIX]{x1D6E5}(\unicode[STIX]{x1D706}^{t,R})_{K})\end{eqnarray}$$

is minimally generated by $\unicode[STIX]{x1D709}_{1,R}\otimes 1_{K},\ldots ,\unicode[STIX]{x1D709}_{r,R}\otimes 1_{K}$ , so that $r_{t,K}(q)=r_{t,R}(q)$ . Finally, (4.27) comes from Lemma 4.2.◻

In view of Theorem 4.26(i), $Q(\unicode[STIX]{x1D706})_{R}$ is not in general an $R$ -form of $Q(\unicode[STIX]{x1D706})_{F}\cong P(\unicode[STIX]{x1D706})_{F}$ . For every $\unicode[STIX]{x1D706}\in \text{KP}(\unicode[STIX]{x1D6FC})$ , define the $H_{\unicode[STIX]{x1D6FC},F}$ -module

$$\begin{eqnarray}X(\unicode[STIX]{x1D706}):=Q(\unicode[STIX]{x1D706})_{R}\otimes F.\end{eqnarray}$$