Resolutions of standard modules over KLR algebras of type $A$

TL;DR

This paper constructs explicit projective resolutions of standard modules over Khovanov-Lauda-Rouquier algebras of type A, advancing understanding of their homological properties in finite Lie types.

Contribution

It provides explicit projective resolutions of standard modules in type A KLR algebras, a novel contribution to the homological study of these algebras.

Findings

Explicit projective resolutions constructed for standard modules

Enhanced understanding of homological properties of KLR algebras in type A

Supports further algebraic and categorical investigations

Abstract

Khovanov-Lauda-Rouquier algebras $R_\theta$ of finite Lie type are affine quasihereditary with standard modules $\Delta(\pi)$ labeled by Kostant partitions $\pi$ of $\theta$. In type $A$, we construct explicit projective resolutions of standard modules $\Delta(\pi)$.

Equations

$$\Lambda :=\{\lambda =({\lambda }_a,\dots ,{\lambda }_b)\mid 0\le {\lambda }_a,\dots ,{\lambda }_b\le m\}.$$

$$P_{\mu }\mapsto P_{\lambda },x{e}_{\mu }\mapsto x{e}_{\mu }{d}_n^{\mu ,\lambda }$$

$$d_n:P_{n+1}:=\underset{\mu \in \Lambda (n+1)}{⨁}{P}_{\mu }⟶P_n:=\underset{\lambda \in \Lambda (n)}{⨁}{P}_{\lambda }.$$

$$\mathsf{p}:P_0=q^{s_0}{R}_{\theta }{e}_0⟶\Delta ({\alpha }^m),x{e}_0\mapsto x{v}_{{\alpha }^m},$$

$$0⟶P_{m(b-a+1)}⟶\cdots ⟶P_{n+1}\stackrel{d_n}{⟶}{P}_n⟶\dots \stackrel{d_0}{⟶}{P}_0\stackrel{\mathsf{p}}{⟶}\Delta ({\alpha }^m)⟶0$$

$${\tt c}_{i,j}:=\left\{\begin{array}[]{ll}2&\hbox{if $i=j$,}\\ -1&\hbox{if $|i-j|=1$,}\\ 0&\hbox{otherwise.}\end{array}\right.$$

$$\{1_{\mathbf{i}}\mid \mathbf{i}\in I^{\theta }\}\cup \{y_1,\dots ,y_d\}\cup \{{\psi }_1,\dots ,{\psi }_{d-1}\}$$

$$y_r{y}_s=y_s{y}_r;$$

$$1_{\mathbf{i}}{1}_{\mathbf{j}}={\delta }_{\mathbf{i},\mathbf{j}}{1}_{\mathbf{i}}\text{and}\sum _{\mathbf{i}\in I^{\theta }}{1}_{\mathbf{i}}=1_{\theta };$$

$$y_r{1}_{\mathbf{i}}=1_{\mathbf{i}}{y}_r\text{and}{\psi }_r{1}_{\mathbf{i}}=1_{s_r\cdot \mathbf{i}}{\psi }_r;$$

$$({\psi }_r{y}_t-y_{s_r(t)}{\psi }_r)1_{\mathbf{i}}={\delta }_{i_r,i_{r+1}}({\delta }_{t,r+1}-{\delta }_{t,r})1_{\mathbf{i}};$$

$${\psi }_r^2{1}_{\mathbf{i}}=\{\begin{array}{ll}0& \text{if }{i}_r=i_{r+1},\\ {\epsilon }_{i_r,i_{r+1}}(y_r-y_{r+1})1_{\mathbf{i}}& \text{if }|i_r-i_{r+1}|=1,\\ 1_{\mathbf{i}}& \text{otherwise};\end{array}$$

$${\psi }_r{\psi }_s={\psi }_s{\psi }_r\text{ if }|r-s|>1;$$

$$({\psi }_{r+1}{\psi }_r{\psi }_{r+1}-{\psi }_r{\psi }_{r+1}{\psi }_r)1_{\mathbf{i}}=\{\begin{array}{ll}{\epsilon }_{i_r,i_{r+1}}{1}_{\mathbf{i}}& \text{if }|i_r-i_{r+1}|=1\text{ and }{i}_r=i_{r+2},\\ 0& \text{otherwise}.\end{array}$$

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}\pgfsys@endscope}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} \par \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}$$

$$\{{\psi }_w{y}_1^{k_1}\dots y_d^{k_d}{1}_{\mathbf{i}}\mid w\in {\mathfrak{S}}_d,k_1,\dots ,k_d\in {\mathbb{Z}}_{\ge 0},\mathbf{i}\in I^{\theta }\},$$

$$\{y_1^{k_1}\dots y_d^{k_d}{\psi }_w{1}_{\mathbf{i}}\mid w\in {\mathfrak{S}}_d,k_1,\dots ,k_d\in {\mathbb{Z}}_{\ge 0},\mathbf{i}\in I^{\theta }\}.$$

$$1_{{\theta }_1,\dots ,{\theta }_t}:=\sum _{{\mathbf{i}}^1\in I^{{\theta }_1},\dots ,{\mathbf{i}}^t\in I^{{\theta }_t}}{1}_{{\mathbf{i}}^1\cdots {\mathbf{i}}^t}\in R_{\theta }.$$

$${\iota }_{{\theta }_1,\dots ,{\theta }_t}:R_{{\theta }_1}\otimes \cdots \otimes R_{{\theta }_t}\hookrightarrow 1_{{\theta }_1,\dots ,{\theta }_t}{R}_{{\theta }_1+\cdots +{\theta }_t}{1}_{{\theta }_1,\dots ,{\theta }_t}$$

$$r_1\circ \cdots \circ r_t:={\iota }_{{\theta }_1,\dots ,{\theta }_t}(r_1\otimes \cdots \otimes r_t).$$

$$1_{{\mathbf{i}}^1}\circ \cdots \circ 1_{{\mathbf{i}}^t}=1_{{\mathbf{i}}^1\cdots {\mathbf{i}}^t}({\mathbf{i}}^1\in I^{{\theta }_1},\dots ,{\mathbf{i}}^t\in I^{{\theta }_t}).$$

$$y_{0,d}:=\prod _{r=1}^d{y}_r^{r-1},1_{i^{(d)}}:={\psi }_{w_{0,d}}{y}_{0,d},\text{and}{1}_{i^{(d)}}^{\prime }:=y_{0,d}{\psi }_{w_{0,d}}.$$

$$1_{\mathbf{i}}=1_{i_1^{(d_1)}\cdots i_r^{(d_r)}}:=1_{i_1^{(d_1)}}\circ \cdots \circ 1_{i_r^{(d_r)}}\in R_{\theta }.$$

$$1_{{\mathbf{i}}^1}\circ \cdots \circ 1_{{\mathbf{i}}^t}=1_{{\mathbf{i}}^1\cdots {\mathbf{i}}^t}({\mathbf{i}}^1\in I_{\mathrm{div}}^{{\theta }_1},\dots ,{\mathbf{i}}^t\in I_{\mathrm{div}}^{{\theta }_t}).$$

$$\psi_{w_{0,d}}=:\leavevmode\hbox to22.85pt{\vbox to14.66pt{\pgfpicture\makeatletter\hbox{\hskip 2.32098pt\lower-6.87352pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{{}}{}{}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@invoke{ }\pgfsys@color@gray@fill{0}\pgfsys@invoke{ }\definecolor[named]{pgffillcolor}{rgb}{0,0,0}\pgfsys@setlinewidth{1.0pt}\pgfsys@invoke{ } {}{}{{}}{}{{}}{}\pgfsys@moveto{9.10509pt}{-6.37352pt}\pgfsys@stroke\pgfsys@invoke{ }\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{4.50238pt}{-0.91652pt}\pgfsys@invoke{ 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$${\psi }_{t\to r}:={\psi }_{(t\to r)}={\psi }_{t-1}{\psi }_{t-2}\dots {\psi }_r\in R_{\theta },$$

$$1_{\bm{i}}\psi_{t\to r}=\leavevmode\hbox to116.77pt{\vbox to35.79pt{\pgfpicture\makeatletter\hbox{\hskip 6.45557pt\lower-9.60509pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{{}}{}{}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@invoke{ }\pgfsys@color@gray@fill{0}\pgfsys@invoke{ }\definecolor[named]{pgffillcolor}{rgb}{0,0,0}\pgfsys@setlinewidth{1.0pt}\pgfsys@invoke{ } \par{}{}{{}}{}{{}} {}{}{}\pgfsys@moveto{0.0pt}{9.10509pt}\pgfsys@lineto{0.0pt}{-9.10509pt}\pgfsys@stroke\pgfsys@invoke{ }\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}}{}{}{}{}{} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{ 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$$1_{i^{(d)}}{\psi }_{d-1}{\psi }_{d-2}\dots {\psi }_1{1}_{i{i}^{(d-1)}}$$

$$=1_{i^{(d)}}{\psi }_{w_{0,d}}(1_{{\alpha }_i}\circ y_{0,d-1})$$

$$={\psi }_{w_{0,d}}(1_{{\alpha }_i}\circ y_{0,d-1})$$

$$={\psi }_{d-1}{\psi }_{d-2}\dots {\psi }_1(1_{{\alpha }_i}\circ {\psi }_{w_{0,d-1}}{y}_{0,d-1})$$

$$={\psi }_{d\to 1}{1}_{i{i}^{(d-1)}},$$

$${\mathrm{Hom}}_{R_{\theta }}(U,V{)}_d:={\mathrm{hom}}_{R_{\theta }}(q^{d}U,V)={\mathrm{hom}}_{R_{\theta }}(U,q^{-d}V).$$

$${\mathrm{Ind}}_{{\theta }_1,\dots ,{\theta }_t}=R_{\theta }{1}_{{\theta }_1,\dots ,{\theta }_t}{\otimes }_{R_{{\theta }_1}\otimes \cdots \otimes R_{{\theta }_t}}-:(R_{{\theta }_1}\otimes \cdots \otimes R_{{\theta }_t})-Mod\to R_{\theta }-Mod.$$

Full text

The second author was supported by the NSF grants DMS-1161094, DMS-1700905, the Max-Planck-Institut, the Fulbright Foundation, and the DFG Mercator program through the University of Stuttgart.

1. Introduction

Let $R_{\theta,{\mathbb{F}}}$ be a Khovanov-Lauda-Rouquier (KLR) algebra of finite Lie type over a field ${\mathbb{F}}$ corresponding to $\theta\in Q_{+}$. It is known that $R_{\theta,{\mathbb{F}}}$ is affine quasihereditary [4, 1, 7, 8]. In particular, it has finite global dimension and comes with a family of standard modules $\{\Delta(\pi)_{\mathbb{F}}\mid\pi\in\operatorname{KP}(\theta)\}$ and proper standard modules $\{\bar{\Delta}(\pi)_{\mathbb{F}}\mid\pi\in\operatorname{KP}(\theta)\}$, where $\operatorname{KP}(\theta)$ denotes the set of the Kostant partitions of $\theta$. The (proper) standard modules have well-understood formal characters. Moreover, $L(\pi)_{\mathbb{F}}:=\operatorname{head}\Delta(\pi)_{\mathbb{F}}\cong\operatorname{head}\bar{\Delta}(\pi)_{\mathbb{F}}$ is irreducible, and $\{L(\pi)_{\mathbb{F}}\mid\pi\in\operatorname{KP}(\theta)\}$ is a complete set of irreducible $R_{\theta,{\mathbb{F}}}$-modules up to isomorphism and degree shift. Finally, the projective cover $P(\pi)_{\mathbb{F}}$ of $L(\pi)_{\mathbb{F}}$ has a finite $\Delta$-filtration and the (graded) decomposition number $d_{\sigma,\pi}^{\mathbb{F}}:=[\bar{\Delta}(\sigma)_{\mathbb{F}}:L(\pi)_{\mathbb{F}}]_{q}$ equals the (well-defined) multiplicity $(P(\pi)_{\mathbb{F}}:\Delta(\sigma)_{\mathbb{F}})_{q}$, see [1, Corollary 3.14].

The KLR algebra is defined over the integers, so we have a $\mathbb{Z}$-algebra $R_{\theta}=R_{\theta,\mathbb{Z}}$ with $R_{\theta,{\mathbb{F}}}=R_{\theta}\otimes_{\mathbb{Z}}{\mathbb{F}}$. The standard modules have natural integral forms $\Delta(\pi)$ with $\Delta(\pi)_{\mathbb{F}}\cong\Delta(\pi)\otimes_{\mathbb{Z}}{\mathbb{F}}$, see [9, §4.2]. Moreover, as illustrated in [9, §4], understanding $p$-torsion in the $\mathbb{Z}$-module $\operatorname{Ext}_{R_{\theta}}^{m}(\Delta(\pi),\Delta(\sigma))$ is relevant for comparing decomposition numbers $d_{\sigma,\pi}^{\mathbb{C}}$ and $d_{\sigma,\pi}^{\mathbb{F}}$ for a field ${\mathbb{F}}$ of characteristic $p$. This motivates our interest in $\operatorname{Ext}_{R_{\theta}}^{m}(\Delta(\pi),\Delta(\sigma))$ and projective resolutions of (integral forms of) standard modules.

The problem of constructing a projective resolution of $\Delta(\pi)$ reduces easily to the semicuspidal case. To be more precise, let us fix a convex total order $\preceq$ on the set $\Phi_{+}$ of positive roots. A Kostant partition of $\theta$ is a sequence $\pi=(\beta_{1}^{m_{1}},\ldots,\beta_{t}^{m_{t}})$ such that $m_{1},\dots,m_{t}\in\mathbb{Z}_{>0}$, $\beta_{1}\succ\cdots\succ\beta_{t}$ are positive roots, and $m_{1}\beta_{1}+\cdots+m_{t}\beta_{t}=\theta$. Then $\Delta(\pi)\cong\Delta(\beta_{1}^{m_{1}})\circ\dots\circ\Delta(\beta_{t}^{m_{t}})$ where ‘$\circ$’ stands for induction product. Since induction product preserves projective modules, it is enough to resolve the standard modules of the form $\Delta(\beta_{s}^{m_{s}})$, which are exactly the semicuspidal standard modules. Moreover, the case where $\beta_{s}$ is a simple root is easy: the algebra $R_{m\alpha_{i}}$ is the nil-Hecke algebra of rank $m$ and the standard module $\Delta(\alpha_{i}^{m})$ is the projective indecomposable module $P(\alpha_{i}^{m})=R_{m\alpha_{i}}1_{i^{(m)}}$ for an explicit primitive idempotent $1_{i^{(m)}}\in R_{m\alpha_{i}}$, see [5, §2.2].

Let us now specialize to Lie type $A_{\infty}$. Then every non-simple positive root is of the form $\alpha=\alpha_{a}+\alpha_{a+1}+\dots+\alpha_{b+1}$ for integers $a\leq b$, and we will work with the lexicographic convex order on the positive roots. Let $\theta:=m\alpha$. We consider the set of compositions

[TABLE]

Let $\Lambda(n):=\{\lambda\in\Lambda\mid\lambda_{a}+\dots+\lambda_{b}=n\}$. In Section 3, for each $\lambda\in\Lambda$, we define explicit idempotents $e_{\lambda}\in R_{\theta}$ as concatenations of ‘divided power idempotents’ of the form $1_{i^{(m)}}$ mentioned above. We define projective modules $P_{\lambda}:=q^{s_{\lambda}}R_{\theta}e_{\lambda}$, where $q^{s_{\lambda}}$ stands for the grading shift by an explicitly defined integer $s_{\lambda}$. For every $\mu\in\Lambda(n+1)$ and $\mu\in\Lambda(n)$ we then define explicit elements $d_{n}^{\mu,\lambda}\in e_{\mu}R_{\theta}e_{\lambda}$ so that

[TABLE]

is an $R_{\theta}$-module homomorphism. Taking direct sum over all such gives us a homomorhism

[TABLE]

We note that $P_{n}=0$ for $n>m(b-a+1)$ and that $d_{n}^{\mu,\lambda}=0$ unless the compositions $\mu$ and $\lambda$ differ in just one part. We also have a natural map

[TABLE]

where $v_{\alpha^{m}}$ is the standard generator of $\Delta(\alpha^{m})$ of weight $a^{m}(a+1)^{m}\cdots(b+1)^{m}$, see (3.33).

Theorem A**.**

We have that

[TABLE]

is a projective resolution of the standard module $\Delta(\alpha^{m})$.

When $m=1$ this is a version of the resolution constructed in [1, Theorem 4.12], and our resolution can be thought of as the ‘thick calculus generalization’ (cf. [6]) of that resolution.

For an arbitrary $\theta$, we denote $\Delta:=\bigoplus_{\pi\in\operatorname{KP}(\theta)}\Delta(\pi)$. In [2], we will use the resolution from Theorem A to describe the algebra $\operatorname{Ext}_{R_{\theta}}^{*}(\Delta,\Delta)$ in two special cases: (1) $\theta$ is a positive root in type $A$ and (2) Lie type is $A_{2}$, i.e. $\theta$ is of the form $r\alpha_{1}+s\alpha_{2}$.

We now describe the contents of the paper and the main idea of the proof. After reviewing KLR algebras and standard modules in Section 2, we introduce the necessary combinatorial notation and define the resolution $P_{\bullet}$ in Section 3.

In order to prove that $P_{\bullet}$ is a resolution of $\Delta(\alpha^{m})$, we want to show that it is a direct summand of a resolution $Q_{\bullet}$ of $q^{m(m-1)/2}\Delta(\alpha)^{\circ m}$ introduced in §3.3. The resolution $Q_{\bullet}$ is obtained by taking the $m$th induced power of the resolution constructed in [1, Theorem 4.12].

To check that $P_{\bullet}$ is a direct summand of $Q_{\bullet}$, in §3.4 we construct what will end up being a pair of chain maps $f:P_{\bullet}\to Q_{\bullet}$ and $g:Q_{\bullet}\to P_{\bullet}$ with $g\circ f=\mathop{\mathrm{id}}\nolimits$. The main difficulty is to verify that $f$ and $g$ are indeed chain maps. This verification occupies Sections 4 and 5. Modulo the fact that $f$ and $g$ are chain maps, Theorem A is proved in §3.7.

2. Preliminaries

2.1. Basic notation

Throughout, we work over an arbitrary commutative unital ground $\Bbbk$ (since everything is defined over $\mathbb{Z}$, one could just consider the case $\Bbbk=\mathbb{Z}$).

For $r,s\in\mathbb{Z}$, we use the segment notation $[r,s]:=\{t\in\mathbb{Z}\mid r\leq t\leq s\},\ [r,s):=\{t\in\mathbb{Z}\mid r\leq t<s\},\ \text{etc.}$

Let $q$ be a variable, and $\mathbb{Z}((q))$ be the ring of Laurent series. For a non-negative integer $n$ we define $[n]:=(q^{n}-q^{-n})/(q-q^{-1})$ and $[n]^{!}:=[1][2]\cdots[n].$

We denote by $\mathfrak{S}_{d}$ the symmetric group on $d$ letters. It is a Coxeter group with generators $\{s_{r}:=(r,r+1)\mid 1\leq r<d\}$ and the corresponding length function $\ell$. The longest element of $\mathfrak{S}_{d}$ is denoted $w_{0}$ or $w_{0,d}$. An element $w\in\mathfrak{S}_{d}$ is called fully commutative if it is possible to go between any two reduced decompositions of $w$ using only the relations of the form $s_{r}s_{t}=s_{t}s_{r}$ for $|r-t|>1$. By definition, $\mathfrak{S}_{d}$ acts on $[1,d]$ on the left. For a set $I$ the $d$-tuples from $I^{d}$ are written as words ${\bm{i}}=i_{1}\cdots i_{d}$. The group $\mathfrak{S}_{d}$ acts on $I^{d}$ via place permutations as $w\cdot{\bm{i}}=i_{w^{-1}(1)}\cdots i_{w^{-1}(d)}$.

Given a composition $\lambda=(\lambda_{1},\dots,\lambda_{k})$ of $d$, we have the corresponding standard parabolic subgroup $\mathfrak{S}_{\lambda}:=\mathfrak{S}_{\lambda_{1}}\times\dots\times\mathfrak{S}_{\lambda_{k}}\leq\mathfrak{S}_{d}$. For compositions $\lambda$ and $\mu$ of $d$, we denote by ${\mathscr{D}}^{\lambda}$ (resp. ${}^{\mu}{\mathscr{D}}$, resp. ${}^{\mu}{\mathscr{D}}^{\lambda}$) the set of the shortest coset representatives for $\mathfrak{S}_{d}/\mathfrak{S}_{\lambda}$ (resp. $\mathfrak{S}_{\mu}\backslash\mathfrak{S}_{d}$, resp. $\mathfrak{S}_{\mu}\backslash\mathfrak{S}_{d}/\mathfrak{S}_{\lambda}$). The following is well-known and can be deduced for example from [3, Lemma 1.6]:

Lemma 2.1**.**

Let $\lambda,\mu$ be compositions of $d$ and $w\in{}^{\mu}{\mathscr{D}}$. Then there exist unique elements $x\in{}^{\mu}{\mathscr{D}}^{\lambda}$ and $y\in\mathfrak{S}_{\lambda}$ such that $w=xy$ and $\ell(w)=\ell(x)+\ell(y)$.

2.2. KLR Algebras

From now on, we set $I:=\mathbb{Z}$. If $i,j\in I$ with $|i-j|=1$ we set ${\varepsilon}_{i,j}:=j-i\in\{1,-1\}$. For $i,j\in I$, we set

[TABLE]

We identify $I$ with the set of vertices of the Dynkin diagram of type $A_{\infty}$ so that the numbers ${\tt c}_{i,j}$ are the entries of the corresponding Cartan matrix. The corresponding simple roots are denoted $\{\alpha_{i}\mid i\in I\}$ and set of the positive roots is $\Phi_{+}:=\{\alpha_{r}+\dots+\alpha_{s}\mid r\leq s\}$. The root lattice is $Q:=\bigoplus_{i\in I}\mathbb{Z}\cdot\alpha_{i}$, and we set Q_{+}:=\{\sum_{i}m_{i}\alpha_{i}\in Q\mid m_{i}\in\mathbb{Z}_{\geq 0}\ \text{for all i}\}. For $\theta=\sum_{i}m_{i}\alpha_{i}\in Q_{+}$, we define its height ${\operatorname{ht}}(\theta):=\sum_{i}m_{i}$. For $\theta\in Q_{+}$ of height $d$, we define $I^{\theta}:=\{{\bm{i}}=i_{1}\cdots i_{d}\in I^{d}\mid\alpha_{i_{1}}+\dots+\alpha_{i_{d}}=\theta\}$. If ${\bm{i}}\in I^{\theta}$ and ${\bm{j}}\in I^{\eta}$ then the concatenation of words ${\bm{i}}{\bm{j}}$ is an element of $I^{\theta+\eta}$.

Let $\theta\in Q_{+}$ be of height $d$. The KLR algebra [5, 10] is the unital $\Bbbk$-algebra $R_{\theta}$ (with identity denoted $1_{\theta}$) with generators

[TABLE]

and defining relations

[TABLE]

The algebra $R_{\theta}$ is graded with $\deg 1_{{\bm{i}}}=0$; $\deg(y_{s})=2$; $\deg(\psi_{r}1_{{\bm{i}}})=-{\tt c}_{i_{r},i_{r+1}}$.

We will use the Khovanov-Lauda [5] diagrammatic notation for elements of $R_{\theta}$. In particular, for ${\bm{i}}=i_{1}\cdots i_{d}\in I^{\theta}$, $1\leq r<d$ and $1\leq s\leq d$, we denote

[TABLE]

For each element $w\in\mathfrak{S}_{n}$, fix a reduced expression $w=s_{r_{1}}\cdots s_{r_{l}}$ which determines an element $\psi_{w}=\psi_{r_{1}}\cdots\psi_{r_{l}}$ (depending on the choice of a reduced expression).

Theorem 2.2**.**

[5, Theorem 2.5]**, [10, Theorem 3.7] Let $\theta\in Q_{+}$ and $d={\operatorname{ht}}(\theta)$. Then the following sets are $\Bbbk$-bases of $R_{\theta}$:

[TABLE]

2.3. Parabolic subalgebras and divided power idempotents

Let $\theta_{1},\dots,\theta_{t}\in Q_{+}$ and set $\theta:=\theta_{1}+\dots+\theta_{t}$. Set

[TABLE]

Then we have an algebra embedding

[TABLE]

obtained by horizontal concatenation of the Khovanov-Lauda diagrams. For $r_{1}\in R_{\theta_{1}},\dots,r_{t}\in R_{\theta_{t}}$, when there is no confusion, we often write

[TABLE]

For example,

[TABLE]

We fix for the moment $i\in I$, $d\in\mathbb{Z}_{>0}$ and take $\theta=d\alpha_{i}$, in which case $R_{d\alpha_{i}}$ is isomorphic to the rank $d$ nil-Hecke algebra $\mathcal{NH}_{d}$. Following [5], we consider the following elements of $R_{d\alpha_{i}}$:

[TABLE]

The following is well-known, see for example [5, §2.2]:

Lemma 2.5**.**

In $R_{d\alpha_{i}}$, the elements $1_{i^{(d)}}$ and $1_{i^{(d)}}^{\prime}$ are idempotents. Moreover,

1. (i)

$\psi_{w_{0,d}}f\psi_{w_{0,d}}=0$* for any polynomial $f$ in $y_{1},\dots,y_{d}$ of degree less than $d(d-1)/2$.* 2. (ii)

$\psi_{w_{0,d}}y_{0,d}\psi_{w_{0,d}}=\psi_{w_{0,d}}$.

Now, let $\theta\in Q_{+}$ be arbitrary. We define $I^{\theta}_{\mathrm{div}}$ to be the set of all expressions of the form $i_{1}^{(d_{1})}\cdots i_{r}^{(d_{r})}$ with $d_{1},\dots,d_{r}\in\mathbb{Z}_{\geq 0}$, $i_{1},\dots,i_{r}\in I$ and $d_{1}\alpha_{i_{1}}+\cdots+d_{r}\alpha_{i_{r}}=\theta$. We refer to such expressions as divided power words. We identify $I^{\theta}$ with the subset of $I^{\theta}_{\mathrm{div}}$ which consists of all divided power words as above with all $d_{k}=1$. We use the same notation for concatenation of divided power words as for concatenation of words. For ${\bm{i}}=i_{1}^{(d_{1})}\cdots i_{r}^{(d_{r})}\in I^{\theta}_{\mathrm{div}}$, we define the divided power idempotent

[TABLE]

Then we have the following analogue of (2.4):

[TABLE]

To be used as part of the Khovanov-Lauda diagrammatics, we denote

[TABLE]

For example, if $d=3$, we have

[TABLE]

More generally, we denote

[TABLE]

For any $1\leq r<t\leq d$, we denote the cycle $(t,t-1,\dots,r)\in\mathfrak{S}_{d}$ by $(t\to r)$. Note that $(t\to r)=s_{t-1}s_{t-2}\cdots s_{r}$ is a unique reduced decomposition. So

[TABLE]

In terms of the diagrammatic notation we have

[TABLE]

Lemma 2.8**.**

In the algebra $R_{d\alpha_{i}}$, we have

1. (i)

If $r_{1}+\dots+r_{t}=d$ then $1_{i^{(r_{1})}\cdots i^{(r_{t})}}\psi_{w_{0,d}}=\psi_{w_{0,d}}$ and $1_{i^{(r_{1})}\cdots i^{(r_{t})}}1_{i^{(d)}}=1_{i^{(d)}}$. 2. (ii)

$1_{i^{(d)}}\psi_{d\to 1}1_{i\,i^{(d-1)}}=\psi_{d\to 1}1_{i\,i^{(d-1)}}$.

Proof.

(i) Write $\psi_{w_{0,d}}=(\psi_{w_{0,r_{1}}}\circ\dots\circ\psi_{w_{0,r_{t}}})\psi_{u}$ for some $u\in\mathfrak{S}_{d}$ and use Lemma 2.5.

(ii) We have

[TABLE]

where we have used Lemma 2.5(ii) for the third equality. ∎

The following lemma easily follows from the defining relations of $R_{\theta}$:

Lemma 2.9**.**

Let $1\leq r<s\leq d$, $t\in(r,s)$, $u\in[r,s)$, and ${\bm{i}}\in I^{\theta}$. Then in $R_{\theta}$ we have:

1. (i)

$1_{\bm{i}}\psi_{s\to r}\psi_{t}=1_{\bm{i}}\psi_{t-1}\psi_{s\to r}$* unless $i_{s}=i_{t-1}=i_{t}\pm 1$.* 2. (ii)

$1_{\bm{i}}\psi_{s\to r}y_{u+1}=1_{\bm{i}}y_{u}\psi_{s\to r}$* unless $i_{s}=i_{u}$.*

2.4. Modules over $R_{\theta}$

Let $\theta\in Q_{+}$. We denote by $R_{\theta}\!\operatorname{-Mod}$ the category of graded left $R_{\theta}$-modules. The morphisms in this category are all homogeneous degree zero $R_{\theta}$-homomorphisms, which we denote $\hom_{R_{\theta}}(-,-)$. For $V\in R_{\theta}\!\operatorname{-Mod}$, 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 series $a=a(q)=\sum_{n}a_{n}q^{n}\in\mathbb{Z}[q,q^{-1}]$ with non-negative coefficients, we set $aV:=\bigoplus_{n}(q^{n}V)^{\oplus a_{n}}$. For $U,V\in R_{\theta}\!\operatorname{-Mod}$, we set $\operatorname{Hom}_{R_{\theta}}(U,V):=\bigoplus_{d\in\mathbb{Z}}\operatorname{Hom}_{R_{\theta}}(U,V)_{d}$, where

[TABLE]

We define $\operatorname{Ext}^{m}_{R_{\theta}}(U,V)$ similarly in terms of $\operatorname{ext}^{m}_{R_{\theta}}(U,V)$.

For $\theta_{1},\dots,\theta_{t}\in Q_{+}$ and $\theta:=\theta_{1}+\dots+\theta_{t}$, recalling (2.3), we have a functor

[TABLE]

For $V_{1}\in R_{\theta_{1}}\!\operatorname{-Mod},\dots,V_{t}\in R_{\theta_{t}}\!\operatorname{-Mod}$, we denote by $V_{1}\boxtimes\dots\boxtimes V_{t}$ the $\Bbbk$-module $V_{1}\otimes\dots\otimes V_{t}$, considered naturally as an $(R_{\theta_{1}}\otimes\dots\otimes R_{\theta_{t}})$-module, and set

[TABLE]

For $v_{1}\in V_{1},\dots,v_{t}\in V_{t}$, we denote

[TABLE]

Since $R_{\theta}1_{\theta_{1},\dots,\theta_{t}}$ is a free $R_{\theta_{1}}\otimes\dots\otimes R_{\theta_{t}}$-module of finite rank by Theorem 2.2, we get the following well-known properties:

Lemma 2.10**.**

The functor ${\mathrm{Ind}}_{\theta_{1},\dots,\theta_{t}}$ is exact and sends finitely generated projectives to finitely generated projectives.

The following lemma is easy to check using the fact that $1_{\bm{k}}$ is an idempotent in $R_{\chi}$ for ${\bm{k}}\in I^{\chi}_{\mathrm{div}}$:

Lemma 2.11**.**

For ${\bm{i}}\in I^{\theta}_{\mathrm{div}}$ and ${\bm{j}}\in I^{\eta}_{\mathrm{div}}$, we have $R_{\theta}1_{{\bm{i}}}\circ R_{\eta}1_{\bm{j}}\cong R_{\theta+\eta}1_{{\bm{i}}{\bm{j}}}$.

2.5. Standard modules

The algebra $R_{\theta}$ is affine quasihereditary in the sense of [7]. In particular, it comes with an important class of standard modules, which we now describe explicitly, referring to [1, §3]. We are going to work with the lexicographic convex order $\preceq$ on $\Phi_{+}$, i.e. for $\alpha=\alpha_{r}+\cdots+\alpha_{s}\in\Phi_{+}$ and $\alpha^{\prime}=\alpha_{r^{\prime}}+\cdots+\alpha_{s^{\prime}}\in\Phi_{+}$, we have $\alpha\prec\alpha^{\prime}$ if and only if either $r<r^{\prime}$ or $r=r^{\prime}$ and $s<s^{\prime}$.

Fix $\alpha=\alpha_{r}+\cdots+\alpha_{s}\in\Phi_{+}$ of height $l:=s-r+1$, and set

[TABLE]

We define the $R_{\alpha}$-module $\Delta(\alpha)$ to be the cyclic $R_{\alpha}$-module generated by a vector $v_{\alpha}$ of degree [math] with defining relations

• •

$1_{\bm{i}}v_{\alpha}=\delta_{{\bm{i}},{\bm{i}}_{\alpha}}v_{\alpha}$ for all ${\bm{i}}\in I^{\alpha}$;

• •

$\psi_{t}v_{\alpha}=0$ for all $1\leq t<l$;

• •

$y_{t}v_{\alpha}=y_{u}v_{\alpha}$ for all $1\leq t,u\leq l$.

By [1, Corollary 3.5], $\Delta(\alpha)$ is indeed the standard module corresponding to $\alpha$. In fact, if $\Bbbk=\mathbb{Z}_{p}$, by [9, (4.11)], $\Delta(\alpha)$ is a universal $\Bbbk$-form of the standard module in the sense of [9, §4.2].

The module $\Delta(\alpha)$ can be considered as an $(R_{\alpha},\Bbbk[x])$-bimodule with the right action given by $v_{\alpha}x:=y_{1}v_{\alpha}.$ Then it is easy to check that there is an isomorphism of graded $\Bbbk$-modules $\Bbbk[x]\to\Delta(\alpha),\ f\mapsto v_{\alpha}f.$

Fix an integer $m\in\mathbb{Z}_{>0}$. Let $\mathcal{NH}_{m}$ be the rank $m$ nil-Hecke algebra with standard generators $x_{1},\dots,x_{m}$, $\tau_{1},\dots,\tau_{m-1}$. For any $i\in I$, there is an isomorphism ${\varphi}:R_{m\alpha_{i}}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathcal{NH}_{m},\ y_{t}\mapsto x_{t},\ \psi_{u}\mapsto\tau_{u}$, and we have the idempotent

[TABLE]

The $R_{m\alpha}$-module $\Delta(\alpha)^{\circ m}$ is cyclicly generated by $v_{\alpha}^{\circ m}$. As explained in [1, §3.2], $\mathcal{NH}_{m}$ acts on $\Delta(\alpha)^{\circ m}$ on the right so that

[TABLE]

where $z$ is the longest element of ${\mathscr{D}}^{(l,l)}$.

Define

[TABLE]

As in [1, Lemma 3.10], we have an isomorphism of $R_{m\alpha}$-modules

[TABLE]

Given $\theta\in Q_{+}$ and a Kostant partition $\pi=(\beta_{1}^{m_{1}},\ldots,\beta_{t}^{m_{t}})\in\operatorname{KP}(\theta)$ as in the Introduction, we define the corresponding standard module

[TABLE]

If $\Bbbk$ is a field, the modules $\{\Delta(\pi)\mid\pi\in\operatorname{KP}(\theta)\}$ are the standard modules for an affine quasihereditary structure on the algebra $R_{\theta}$, see [1, 7]. If $\Bbbk=\mathbb{Z}$ or $\mathbb{Z}_{p}$, they can be thought of as integral forms of the standard modules, see [9, §4].

3. Semicuspidal resolution

Throughout the subsection, we fix $m\in\mathbb{Z}_{>0}$, $a,b\in\mathbb{Z}$ with $a\leq b$, and set

[TABLE]

We denote

[TABLE]

and $\theta:=m\alpha$. Note that $l={\operatorname{ht}}(\alpha)$ and $d={\operatorname{ht}}(\theta)$. Our goal is to construct a resolution $P_{\bullet}=P^{\alpha^{m}}_{\bullet}$ of the semicuspidal standard module $\Delta(\alpha^{m})$.

3.1. Combinatorics

We consider the set of compositions

[TABLE]

For $\lambda\in\Lambda$, we denote $|\lambda|:=\lambda_{a}+\dots+\lambda_{b}$, and for $n\in\mathbb{Z}_{\geq 0}$, we set

[TABLE]

Let $a\leq i\leq b$. We set

[TABLE]

with $1$ in the $i$th position.

Let $\lambda\in\Lambda$. Set

[TABLE]

We also associate to $\lambda$ a composition $\omega_{\lambda}$ of $d$ with $2n+1$ non-negative parts:

[TABLE]

Let $i\in[a,b]$. We denote

[TABLE]

where $r_{a-1}^{-}(\lambda)$ is interpreted as [math], and $r_{b+1}^{+}(\lambda)$ is interpreted as $d-\sum_{s=a}^{b}\lambda_{s}$. Moreover, denote

[TABLE]

Define

[TABLE]

Observe that for all $j\in[a,b+1]$, we have

[TABLE]

We also consider the sets of multicompositions

[TABLE]

Note that by definition all $\delta^{(r)}_{i}\in\{0,1\}$. For $1\leq r\leq m$ and $a\leq i\leq b$, we define ${\text{\bf e}}^{r}_{i}\in{\bm{\Lambda}}(1)$ to be the multicomposition whose $r$th component is $\mathtt{e}_{i}$ and whose other components are zero. For ${\bm{\delta}}\in{\bm{\Lambda}}$, denote

[TABLE]

Fix ${\bm{\delta}}=(\delta^{(1)},\dots,\delta^{(m)})\in{\bm{\Lambda}}$. Define

[TABLE]

For $i\in[a,b]$ we define

[TABLE]

Observe that for any $j\in[a,b+1]$, we have

[TABLE]

For $\lambda\in\Lambda$, ${\bm{\delta}}\in{\bm{\Lambda}}$, and $i\in[a,b]$, we define some signs:

[TABLE]

($\tau_{\lambda},\tau_{\bm{\delta}}$ are not to be confused with $\tau_{w}\in\mathcal{NH}_{d}$ which will not be used again).

Lemma 3.1**.**

Let ${\bm{\delta}}=(\delta^{(1)},\dots,\delta^{(m)})\in{\bm{\Lambda}}$, $i\in[a,b]$ and $r\in[1,m]$. If $\delta_{i}^{(r)}=0$, then

[TABLE]

Proof.

Let $\lambda:=\lambda^{\bm{\delta}}$ and ${\bm{\gamma}}:={\bm{\delta}}+\text{\bf e}_{i}^{r}$. Writing ‘$\equiv$’ for ‘$\equiv\text{ }(\text{\rm mod }2)\,$’, we have to prove

[TABLE]

Note that

[TABLE]

so the required comparison boils down to

[TABLE]

which is easy to see. ∎

Lemma 3.3**.**

Let ${\bm{\gamma}}=(\gamma^{(1)},\dots,\gamma^{(m)})\in{\bm{\Lambda}}$, $i\in[a,b]$ and $r\in[1,m]$. If $\gamma_{i}^{(r)}=1$, then

[TABLE]

Proof.

Let $\mu:=\lambda^{\bm{\gamma}}$, ${\bm{\delta}}:={\bm{\gamma}}-\text{\bf e}_{i}^{r}$, and $\lambda:=\lambda^{\bm{\delta}}$. Writing ‘$\equiv$’ for ‘$\equiv\text{ }(\text{\rm mod }2)\,$’, we have to prove the comparison

[TABLE]

Note that

[TABLE]

So, using also (3.2), the required comparison boils down to

[TABLE]

which is easy to see. ∎

3.2. The resolutions $P_{\bullet}^{\alpha^{m}}$ and $P_{\bullet}^{\pi}$

Let $\lambda\in\Lambda$. Recalling the divided power word ${\bm{i}}^{\lambda}\in I^{\theta}_{\mathrm{div}}$ and the integer $s_{\lambda}$ from $\S\ref{SSComb}$, we set

[TABLE]

In particular, $P_{\lambda}$ is a projective left $R_{\theta}$-module. Note that $1_{{\bm{j}}^{\lambda}}e_{\lambda}=e_{\lambda}1_{{\bm{j}}^{\lambda}}=e_{\lambda}$. Further, set for any $n\in\mathbb{Z}_{\geq 0}$:

[TABLE]

Note that $P_{n}=0$ for $n>d-m$. The projective resolution $P_{\bullet}=P^{\alpha^{m}}_{\bullet}$ of $\Delta(\alpha^{m})$ will be of the form

[TABLE]

To describe the boundary maps $d_{n}$, we first consider a more general situation. Suppose we are given two sets of idempotents $\{e_{a}\mid a\in A\}$ and $\{f_{b}\mid b\in B\}$ in an algebra $R$. An $A\times B$ matrix $D:=(d^{a,b})_{a\in A,b\in B}$ with every $d^{a,b}\in e_{a}Rf_{b}$ then yields the homomorphism between the projective $R$-modules

[TABLE]

which we refer to as the right multiplication with $D$.

We now define a $\Lambda(n+1)\times\Lambda(n)$ matrix $D_{n}$ with entries $d_{n}^{\mu,\lambda}\in e_{\mu}R_{\theta}e_{\lambda}$. Let $\lambda\in\Lambda(n)$ and $a\leq i\leq b$ be such that $\lambda_{i}<m$. Recalling (2.7), define

[TABLE]

Note that $\psi_{\lambda;i}1_{{\bm{j}}^{\lambda}}=1_{{\bm{j}}^{\lambda+\mathtt{e}_{i}}}\psi_{\lambda;i}$. Recalling the sign $\mathtt{sgn}_{\lambda;i}$ from §3.1, we now set

[TABLE]

Diagrammatically, for $\mu=\lambda+\mathtt{e}_{i}$ as above, we have

[TABLE]

We now set the boundary map $d_{n}$ to be the right multiplication with $D_{n}$:

[TABLE]

Example 3.5**.**

Let $a=b=1$ and $m=2$. Then the resolution $P_{\bullet}$ is

[TABLE]

where $d_{1}$ is a right multiplication with

[TABLE]

and $d_{0}$ is a right multiplication with

[TABLE]

It is far from clear that $\ker d_{n}={\mathrm{im}\,}d_{n+1}$ but at least the following is easy to see:

Lemma 3.6**.**

The homomorphisms $d_{n}$ are homogeneous of degree [math] for all $n$.

Proof.

Let $\lambda\in\Lambda(n)$ be such that $\lambda_{i}<m$ for some $a\leq i\leq b$, so that $\mu:=\lambda+\mathtt{e}_{i}\in\Lambda(n+1)$. The homomorphism $d_{n}=\rho_{D_{n}}$ is a right multiplication with the matrix $D_{n}$. Its $(\mu,\lambda)$-component is a homomorphism $P_{\mu}\to P_{\lambda}$ obtained by the right multiplication with $\pm e_{\mu}\psi_{\lambda;i}e_{\lambda}$. Recall that $P_{\mu}=q^{s_{\mu}}R_{\theta}e_{\mu}$ and $P_{\lambda}=q^{s_{\lambda}}R_{\theta}e_{\lambda}$. So we just need to show that $s_{\mu}=s_{\lambda}+\deg(e_{\mu}\psi_{\lambda;i}e_{\lambda})$. This is an easy computation using the fact that by definition we have $\deg(e_{\mu}\psi_{\lambda;i}e_{\lambda})=m-2\lambda_{i}$. ∎

Let $\eta\in Q_{+}$ be arbitrary and $\pi=(\beta_{1}^{m_{1}},\dots,\beta_{t}^{m_{t}})\in\operatorname{KP}(\eta)$. We will define the resolution $P_{\bullet}^{\pi}$ of $\Delta(\pi)=\Delta(\beta_{1}^{m_{1}})\circ\cdots\circ\Delta(\beta_{t}^{m_{t}})$ using the general notion of the induced product of chain complexes.

Let $(C_{\bullet},d_{C})$ and $(D_{\bullet},d_{D})$ be chain complexes of left $R_{\eta^{\prime}}$-modules and $R_{\eta^{\prime\prime}}$-modules, respectively. We define a complex of $R_{\eta^{\prime}+\eta^{\prime\prime}}$-modules

[TABLE]

with differential given by

[TABLE]

Lemma 3.8**.**

If $C_{\bullet}$ and $D_{\bullet}$ are projective resolutions of modules $M$ and $N$, respectively, then $C_{\bullet}\circ D_{\bullet}$ is a projective resolution of $M\circ N$.

Proof.

This follows from Lemma 2.10 and [11, Lemma 2.7.3]. ∎

We now define

[TABLE]

which, by Lemma 3.8, will turn out to be a projective resolution of $\Delta(\pi)$ in view of Theorem 3.34.

3.3. The resolution $Q_{\bullet}$

In order to check that $P_{\bullet}$ is a resolution of $\Delta(\alpha^{m})$, we show that it is a direct summand of a known resolution $Q_{\bullet}$ of $q^{m(m-1)/2}\Delta(\alpha)^{\circ m}$. To describe the latter resolution, let us first consider the special case $m=1$.

Lemma 3.10**.**

We have that $P^{\alpha}_{\bullet}$ is a resolution of $\Delta(\alpha)$.

Proof.

This is a special case of [1, Theorem 4.12], corresponding to the standard choice of $(\alpha_{i+1}+\dots+\alpha_{j},\alpha_{i})$ as the minimal pair for an arbitrary positive root $\alpha_{i}+\dots+\alpha_{j}$ in the definition of ${\bm{i}}_{\alpha,\sigma}$, see [1, §4.5]. ∎

Let $Q_{\bullet}$ be the resolution $q^{m(m-1)/2}(P^{\alpha}_{\bullet})^{\circ m}$, cf. (3.7). To describe $Q_{\bullet}$ more explicitly, let $n\in\mathbb{Z}_{\geq 0}$ and ${\bm{\delta}}=(\delta^{(1)},\dots,\delta^{(m)})\in{\bm{\Lambda}}(n)$. Recalling the definitions of §3.1, we set

[TABLE]

For example taking each $\delta^{(r)}$ to be $0\in\Lambda(0)$, we get

[TABLE]

and $e_{{\bm{0}}}=1_{(a\,(a+1)\cdots(b+1))^{m}}.$ Further for any $n\in\mathbb{Z}_{\geq 0}$, we set

[TABLE]

The projective resolution $Q_{\bullet}$ is

[TABLE]

with the augmentation map

[TABLE]

see §2.5, and $c_{n}$ being right multiplication with the ${\bm{\Lambda}}(n+1)\times{\bm{\Lambda}}(n)$ matrix $C=(c_{n}^{{\bm{\gamma}},{\bm{\delta}}})$ defined as follows. If ${\bm{\delta}}+\text{\bf e}^{r}_{i}\in{\bm{\Lambda}}$ for some $r\in[1,m]$ and $i\in[a,b]$, i.e. $\delta^{(r)}_{i}=0$, we set

[TABLE]

Recalling the signs defined in §3.1, for ${\bm{\delta}}\in{\bm{\Lambda}}(n)$ and ${\bm{\gamma}}\in{\bm{\Lambda}}(n+1)$, we now define

[TABLE]

The fact that $Q_{\bullet}$ is indeed isomorphic to the resolution $q^{m(m-1)/2}(P^{\alpha}_{\bullet})^{\circ m}$ is easily checked using the isomorphism $R_{\theta}e_{\bm{\delta}}\cong R_{\theta}1_{{\bm{j}}^{\delta^{(1)}}}\circ\dots\circ R_{\theta}1_{{\bm{j}}^{\delta^{(m)}}}$, which comes from Lemma 2.11.

Example 3.14**.**

Let $a=b=1$ and $m=2$. Then the resolution $Q_{\bullet}$ is

[TABLE]

where $c_{1}$ is a right multiplication with the matrix $(-1_{2121}\psi_{3}\ \ 1_{2121}\psi_{1})$, and $c_{0}$ is a right multiplication with the matrix $\left(\begin{matrix}1_{2112}\psi_{1}\\ 1_{1221}\psi_{3}\end{matrix}\right).$

3.4. Comparison maps

We now construct what will end up being a pair of chain maps $f:P_{\bullet}\to Q_{\bullet}$ and $g:Q_{\bullet}\to P_{\bullet}$ with $g\circ f=\mathop{\mathrm{id}}\nolimits$. As usual, $f_{n}$ and $g_{n}$ will be given as right multiplications with certain matrices $F_{n}$ and $G_{n}$, respectively.

Let $\lambda\in\Lambda$. Recall the definitions of §3.1. We denote by $w^{\lambda}_{0}$ the longest element of the parabolic subgroup $\mathfrak{S}_{\omega_{\lambda}}\leq\mathfrak{S}_{d}$. We also denote

[TABLE]

Let ${\bm{\delta}}=(\delta^{(1)},\dots,\delta^{(m)})\in{\bm{\Lambda}}$. We define $u({\bm{\delta}})\in\mathfrak{S}_{d}$ as follows: for all $i=a,\dots,b$, the permutation $u({\bm{\delta}})$ maps:

1. $\bullet$

the elements of $U^{\pm}_{i}(\lambda^{\bm{\delta}})$ increasingly to the elements of $U^{\pm}_{i}({\bm{\delta}})$; 2. $\bullet$

the elements of $U_{b+1}(\lambda^{\bm{\delta}})$ increasingly to the elements of $U_{b+1}({\bm{\delta}})$.

Set $w({\bm{\delta}}):=u({\bm{\delta}})^{-1}$. Then $w({\bm{\delta}})$ can also be characterized as the element of $\mathfrak{S}_{d}$ which for all $i=a,\dots,b$, maps the elements of $U^{(\pm)}_{i}({\bm{\delta}})$ increasingly to the elements of $U^{(\pm)}_{i}(\lambda^{\bm{\delta}})$ and the elements of $U_{b+1}({\bm{\delta}})$ increasingly to the elements of $U_{b+1}(\lambda^{\bm{\delta}})$.

Recall the signs $\sigma_{\bm{\delta}}$ and $\tau_{{\bm{\delta}}}$ defined in §3.1. We now define $F_{n}$ as the $\Lambda(n)\times{\bm{\Lambda}}(n)$-matrix with the entries $f_{n}^{\lambda,{\bm{\delta}}}$ defined for any $\lambda\in\Lambda(n),{\bm{\delta}}\in{\bm{\Lambda}}(n)$ as follows:

[TABLE]

We define $G_{n}$ as the ${\bm{\Lambda}}(n)\times\Lambda(n)$-matrix with the entries $g_{n}^{{\bm{\delta}},\lambda}$ defined for any ${\bm{\delta}}\in{\bm{\Lambda}}(n),\lambda\in\Lambda(n)$ as follows:

[TABLE]

Example 3.15**.**

Let $m=d=2$ as in Examples 3.5 and 3.14. Then:

[TABLE]

Lemma 3.16**.**

Let ${\bm{\delta}}\in{\bm{\Lambda}}(n)$ and $\lambda=\lambda^{\bm{\delta}}$. Then:

1. (i)

$\deg(\psi_{w({\bm{\delta}})}e_{\bm{\delta}})=\frac{m(m-1)(l-1)}{2}-mn+\sum_{i=a}^{b}\lambda_{i}^{2}$, 2. (ii)

$\deg(f_{n}^{\lambda,{\bm{\delta}}})=-\frac{m(m-1)(l+1)}{2}+mn-\sum_{i=a}^{b}\lambda_{i}^{2}$, 3. (iii)

$\deg(g_{n}^{{\bm{\delta}},\lambda})=\frac{m(m-1)(l+1)}{2}-mn+\sum_{i=a}^{b}\lambda_{i}^{2}$.

Proof.

(i) We prove this by induction on $m$. Denote the right hand side by $R(m)$ and the left hand side by $L(m)$. If $m=1$ then $w({\bm{\delta}})=1$, so $L(1)=0$. Moreover,

[TABLE]

Let $m>1$. It suffices to prove that $R(m)-R(m-1)=L(m)-L(m-1)$. Let $\delta^{(m)}=({\varepsilon}_{a},\dots,{\varepsilon}_{b})$. Then, since all ${\varepsilon}_{i}$ are [math] or $1$, we have

[TABLE]

On the other hand, consider the Khovanov-Lauda diagram of $\psi_{w({\bm{\delta}})}e_{\bm{\delta}}$. The bottom positions of the diagram correspond to to the letters if the word ${\bm{j}}^{\bm{\delta}}$, and so the rightmost $l$ bottom positions of this diagram correspond to the letters of ${\bm{j}}^{\delta^{(m)}}$. In other words, counting from the right, the sequence of colors of these positions is $a^{{\varepsilon}_{a}},\dots,b^{{\varepsilon}_{b}},b+1,b^{1-{\varepsilon}_{b}},\dots,a^{1-{\varepsilon}_{a}}$. Note that the strings which originate in these positions do not intersect each other, so $L(m)-L(m-1)$ equals the sum of the degrees of the intersections of these strings with the other strings of the diagram, i.e.

[TABLE]

which is easily seen to equal the expression for $R(m)-R(m-1)$ obtained above.

(ii) This follows from (i) since $\deg(f_{n}^{\lambda,{\bm{\delta}}})=\deg(\psi_{w({\bm{\delta}})}e_{\bm{d}})+\deg(e_{\lambda}\psi_{w_{0}^{\lambda}})$ and

[TABLE]

(iii) This follows from (i) since

[TABLE]

∎

Corollary 3.17**.**

The homomorphisms $f_{n}$ and $g_{n}$ are homogeneous of degree [math] for all $n$.

Proof.

Let ${\bm{\delta}}\in{\bm{\Lambda}}(n)$ and $\lambda=\lambda^{\bm{\delta}}$. The homomorphism $f_{n}$ is a right multiplication with the matrix $F_{n}$. Its $(\lambda,{\bm{\delta}})$-component is a homomorphism $P_{\lambda}\to Q_{\bm{\delta}}$ obtained by the right multiplication with $f_{n}^{\lambda,{\bm{\delta}}}$. Recall that $P_{\lambda}=q^{s_{\lambda}}R_{\theta}e_{\lambda}$ and $Q_{\bm{\delta}}=q^{n+m(m-1)/2}R_{\theta}e_{\bm{\delta}}$. So we just need to show that $s_{\lambda}=n+m(m-1)/2+\deg(f_{n}^{\lambda,{\bm{\delta}}})$, which easily follows from Lemma 3.16(ii).

The homomorphism $g_{n}$ is a right multiplication with the matrix $G_{n}$. Its $({\bm{\delta}},\lambda)$-component is a homomorphism $Q_{\bm{\delta}}\to P_{\lambda}$ obtained by the right multiplication with $g_{n}^{{\bm{\delta}},\lambda}$. So we just need to show that $n+m(m-1)/2=s_{\lambda}+\deg(g_{n}^{{\bm{\delta}},\lambda})$, which easily follows from Lemma 3.16(iii). ∎

Corollary 3.18**.**

Suppose ${\bm{\delta}},{\bm{\varepsilon}}\in{\bm{\Lambda}}(n)$ are such that $\lambda^{{\bm{\delta}}}=\lambda^{{\bm{\varepsilon}}}$. Then $\deg(\psi_{w({\bm{\delta}})}e_{\bm{\delta}})=\deg(\psi_{w({\bm{\varepsilon}})}e_{{\bm{\varepsilon}}})$.

3.5. Independence of reduced decompositions

Throughout this subsection we fix ${\bm{\delta}}\in{\bm{\Lambda}}(n)$ and set $\lambda:=\lambda^{\bm{\delta}}$.

Recall that in general the element $\psi_{w}\in R_{\theta}$ depends on a choice of a reduced decomposition of $w\in\mathfrak{S}_{d}$. While it is clear from the form of the braid relations in the KLR algebra that $e_{\lambda}\psi_{w_{0}^{\lambda}}$ does not depend on a choice of a reduced decomposition of $w_{0}^{\lambda}$, it is not obvious that a similar statement is true for $\psi_{w({\bm{\delta}})}e_{\bm{\delta}}$ and $e_{\bm{\delta}}\psi_{u({\bm{\delta}})}$. So a priori the elements $f_{n}^{\lambda,{\bm{\delta}}}=\pm e_{\lambda}\psi_{w_{0}^{\lambda}}\psi_{w({\bm{\delta}})}e_{\bm{\delta}}$ and $g_{n}^{{\bm{\delta}},\lambda}=\pm e_{\bm{\delta}}\psi_{u({\bm{\delta}})}y^{\lambda}e_{\lambda}$ might depend on choices of reduced decompositions of $w({\bm{\delta}})$ and $u({\bm{\delta}})$. In this subsection we will prove that this is not the case, and so in a sense the maps $f_{n}$ and $g_{n}$ are canonical.

Recall the composition $\omega_{\lambda}$ and the words ${\bm{j}}^{\lambda},{\bm{j}}^{\bm{\delta}}$ from §3.1.

Lemma 3.19**.**

The element $w({\bm{\delta}})$ is the unique element of   ${}^{\omega_{\lambda}}{\mathscr{D}}^{(l^{m})}$ with $w({\bm{\delta}})\cdot{\bm{j}}^{\bm{\delta}}={\bm{j}}^{\lambda}$.

Proof.

That $w({\bm{\delta}})\in{}^{\omega_{\lambda}}{\mathscr{D}}^{(l^{m})}$ and $w({\bm{\delta}})\cdot{\bm{j}}^{\bm{\delta}}={\bm{j}}^{\lambda}$ follows from the definitions. To prove the uniqueness statement, let $w\in{}^{\omega_{\lambda}}{\mathscr{D}}^{(l^{m})}$ and $w\cdot{\bm{j}}^{\bm{\delta}}={\bm{j}}^{\lambda}$. Since $w\in{}^{\omega_{\lambda}}{\mathscr{D}}$, it maps the elements of $U_{b+1}({\bm{\delta}})$ increasingly to the elements of $U_{b+1}(\lambda)$. By definition, ${\bm{j}}^{{\bm{\delta}}}={\bm{j}}^{\delta^{(1)}}\dots\bm{j}^{\delta^{(m)}}$. For every $r\in[1,m]$, the entries of ${\bm{j}}^{\delta^{(r)}}$ have the following properties: (1) each $i\in[a,b+1]$ appears among them exactly once; (2) the entries that precede $b+1$ appear in the increasing order; (3) the entries that succeed $b+1$ appear in the decreasing order. Since $w\in{\mathscr{D}}^{(l^{m})}$, it maps the positions corresponding to the entries in (2) to the positions which are to the left of the positions occupied with $b+1$ in ${\bm{j}}^{\lambda}$, and it maps the positions corresponding to the entries in (3) to the positions which are to the right of the positions occupied with $b+1$ in ${\bm{j}}^{\lambda}$. In other words, for all $i\in[a,b]$, the permutation $w$ maps the elements of $U_{i}^{\pm}({\bm{\delta}})$ to the elements of $U_{i}^{\pm}(\lambda)$. As $w\in{}^{\omega_{\lambda}}{\mathscr{D}}$, it now follows that for every $i\in[a,b]$, the permutation $w$ maps the elements of $U_{i}^{\pm}({\bm{\delta}})$ to the elements of $U_{i}^{\pm}(\lambda)$ increasingly. We have shown that $w=w({\bm{\delta}})$. ∎

Lemma 3.20**.**

Let ${\varepsilon},\delta\in\Lambda^{\alpha}$, and ${\bm{j}}^{\varepsilon}=w\cdot{\bm{j}}^{\delta}$ for some $w\in\mathfrak{S}_{l}$. Then either ${\varepsilon}=\delta$ and $w=1$, or $\deg(\psi_{w}1_{{\bm{j}}^{\delta}})>0$.

Proof.

Since every $i\in[a,b+1]$ appears in ${\bm{j}}^{\delta}$ exactly once, ${\varepsilon}=\delta$ implies $w=1$. On the other hand, if ${\varepsilon}\neq\delta$, let $i$ be maximal with ${\varepsilon}_{i}\neq\delta_{i}$. Then the strings colored $i$ and $i+1$ in the Khovanov-Lauda diagram $D$ for $\psi_{w}1_{{\bm{j}}^{\delta}}$ intersect (for any choice of a reduced decomposition of $w$), which contributes a degree $1$ crossing into $D$. On the other hand, since every $j\in[a,b+1]$ appears in ${\bm{j}}^{\delta}$ exactly once, $D$ has no same color crossings, which are the only possible crossings of negative degree. The lemma follows. ∎

Lemma 3.21**.**

Suppose that $w\in{}^{(l^{m})}{\mathscr{D}}^{\omega_{\lambda}}$ and $w\cdot{\bm{j}}^{\lambda}$ is of the form ${\bm{i}}^{(1)}\dots\bm{i}^{(m)}$ with ${\bm{i}}^{(1)},\dots,{\bm{i}}^{(m)}\in I^{\alpha}$. Then $w\cdot{\bm{j}}^{\lambda}={\bm{j}}^{\bm{\varepsilon}}$ for some ${\bm{\varepsilon}}\in{\bm{\Lambda}}$ with $\lambda^{\bm{\varepsilon}}=\lambda$.

Proof.

Ler $r\in[1,m]$. By assumption, the entries $i^{(r)}_{1},\dots,i^{(r)}_{l}$ of ${\bm{i}}^{(r)}$ have the following properties: (1) each $i\in[a,b+1]$ appears among them exactly once; (2) the entries that precede $b+1$ appear in the increasing order; (3) the entries that succeed $b+1$ appear in the decreasing order. The result follows. ∎

Lemma 3.22**.**

Let $P=\{w\in{}^{\omega_{\lambda}}{\mathscr{D}}\mid w\cdot{\bm{j}}^{{\bm{\delta}}}={\bm{j}}^{\lambda}\}$. Then $w({\bm{\delta}})\in P$ and $\deg(\psi_{w({\bm{\delta}})}e_{\bm{\delta}})<\deg(\psi_{w}e_{\bm{\delta}})$ for any $w\in P\setminus\{w({\bm{\delta}})\}$.

Proof.

It is clear that $w({\bm{\delta}})\in P$. On the other hand, by Lemma 2.1, an arbitrary $w\in P$ can be written uniquely in the form $w=xy$ with $x\in{}^{\omega_{\lambda}}{\mathscr{D}}^{(l^{m})}$, $y\in\mathfrak{S}_{(l^{m})}$ and $\ell(xy)=\ell(x)+\ell(y)$.

Since $xy\cdot{\bm{j}}^{\bm{\delta}}={\bm{j}}^{\lambda}$, we have $y\cdot{\bm{j}}^{\bm{\delta}}=x^{-1}\cdot{\bm{j}}^{\lambda}$. As $x^{-1}\in{}^{(l^{m})}{\mathscr{D}}^{\omega_{\lambda}}$, it follows from Lemma 3.21, that $y\cdot{\bm{j}}^{\bm{\delta}}=x^{-1}\cdot{\bm{j}}^{\lambda}$ is of the form ${\bm{j}}^{\bm{\varepsilon}}$ for some ${\bm{\varepsilon}}\in{\bm{\Lambda}}$ with $\lambda^{{\bm{\varepsilon}}}=\lambda$. By Lemma 3.19, $x=w({\bm{\varepsilon}})$. If $w\neq w({\bm{\delta}})$, then $y\neq 1$ and we have

[TABLE]

where we have used Corollary 3.18 for the second equality and Lemma 3.20 for the inequality. ∎

Lemma 3.23**.**

The element $f_{n}^{\lambda,{\bm{\delta}}}=\sigma_{\bm{\delta}}e_{\lambda}\psi_{w_{0}^{\lambda}}\psi_{w({\bm{\delta}})}e_{\bm{\delta}}$ is independent of the choice of reduced expressions for $w_{0}^{\lambda}$ and $w({\bm{\delta}})$.

Proof.

It is clear from the form of the braid relations in the KLR algebra that $e_{\lambda}\psi_{w_{0}^{\lambda}}$ is independent of the choice of a reduced expression for $w_{0}^{\lambda}$. On the other hand, if $\psi_{w({\bm{\delta}})}e_{\bm{\delta}}$ and ${\psi_{w({\bm{\delta}})}^{\prime}}e_{\bm{\delta}}$ correspond to different reduced expressions of $w({\bm{\delta}})$, it follows from the defining relations of the KLR algebra and Theorem 2.2 that $\psi_{w({\bm{\delta}})}e_{\bm{\delta}}-{\psi_{w({\bm{\delta}})}^{\prime}}e_{\bm{\delta}}$ is a linear combination of elements of the form $\psi_{u}ye_{\bm{\delta}}$ with $u\in\mathfrak{S}_{d}$, $y\in\Bbbk[y_{1},\dots,y_{d}]$ such that, $\deg(\psi_{u}ye_{\bm{\delta}})=\deg(\psi_{w({\bm{\delta}})}e_{\bm{\delta}})$ and $u{\bm{j}}^{\bm{\delta}}={\bm{j}}^{\lambda}$. We have to prove $e_{\lambda}\psi_{w_{0}^{\lambda}}\psi_{u}ye_{\bm{\delta}}=0$. Suppose otherwise.

Since we are using any preferred reduced decompositions for $u$, we may assume in addition that $u\in{}^{\omega_{\lambda}}{\mathscr{D}}$, since otherwise $e_{\lambda}\psi_{w_{0}^{\lambda}}\psi_{u}=0$. Now by Lemma 3.22, $\deg(\psi_{u}e_{\bm{\delta}})>\deg(\psi_{w({\bm{\delta}})}e_{\bm{\delta}})$, whence $\deg(\psi_{u}ye_{\bm{\delta}})>\deg(\psi_{w({\bm{\delta}})}e_{\bm{\delta}})$, giving a contradiction. ∎

Lemma 3.24**.**

The element $g_{n}^{{\bm{\delta}},\lambda}=\tau_{\bm{\delta}}e_{\bm{\delta}}\psi_{u({\bm{\delta}})}y^{\lambda}e_{\lambda}$ is independent of the choice of a reduced expression for $u({\bm{\delta}})$.

Proof.

The argument is similar to that of the previous lemma. If $e_{\bm{\delta}}\psi_{u({\bm{\delta}})}$ and $e_{\bm{\delta}}\psi_{u({\bm{\delta}})}^{\prime}$ correspond to different reduced expressions of $u({\bm{\delta}})$, then $e_{\bm{\delta}}\psi_{u({\bm{\delta}})}-e_{\bm{\delta}}\psi_{u({\bm{\delta}})}^{\prime}$ is a linear combination of elements of the form $e_{\bm{\delta}}y\psi_{w}$ with $w\in\mathfrak{S}_{d}$, $y\in\Bbbk[y_{1},\dots,y_{d}]$ such that $\deg(e_{\bm{\delta}}y\psi_{w})=\deg(e_{\bm{\delta}}\psi_{u({\bm{\delta}})})$ and $w^{-1}{\bm{j}}^{\bm{\delta}}={\bm{j}}^{\lambda}$. Moreover, in the KL dialgram of $\psi_{w}1_{{\bm{j}}^{\lambda}}$ the strings colored $b+1$ do not cross each other, since this was the case for the KL diagram of $\psi_{u({\bm{\delta}})}1_{{\bm{j}}^{\lambda}}$. Hence, if $e_{\bm{\delta}}y\psi_{w}y^{\lambda}e_{\lambda}\neq 0$, we may assume in addition that $w^{-1}\in{}^{\omega_{\lambda}}{\mathscr{D}}$. Now, using Lemma 3.22, we conclude that $\deg(e_{\bm{\delta}}y\psi_{w})>\deg(e_{\bm{\delta}}\psi_{u({\bm{\delta}})})$, getting a contradiction. ∎

3.6. Splitting

In this subsection, we aim to show that $g\circ f=\mathop{\mathrm{id}}\nolimits$. We fix $\lambda\in\Lambda(n)$ throughout the subsection. We need to prove $\sum_{{\bm{\delta}}\in{\bm{\Lambda}}(n)}f_{n}^{\lambda,{\bm{\delta}}}g_{n}^{{\bm{\delta}},\lambda}=e_{\lambda}$. Since $f_{n}^{\lambda,{\bm{\delta}}}=0$ unless $\lambda^{\bm{\delta}}=\lambda$, this is equivalent to

[TABLE]

Let ${\bm{\delta}}\in{\bm{\Lambda}}(n)$ with $\lambda^{\bm{\delta}}=\lambda$. We say that ${\bm{\delta}}$ is initial if $a$ preceeds $a+1$ in ${\bm{j}}^{\delta^{(r)}}$ for $r\in[1,m-\lambda_{a}]$ and $a$ succeeds $a+1$ in ${\bm{j}}^{\delta^{(r)}}$ for $r\in(m-\lambda_{a},m]$. In other words, ${\bm{\delta}}$ is initial if $\delta^{(r)}_{a}=0$ for $r\in[1,m-\lambda_{a}]$ and $\delta^{(r)}_{a}=1$ for $r\in(m-\lambda_{a},m]$.

Let $w\in\mathfrak{S}_{d}$ and $1\leq r,s\leq d$. We say that $(r,s)$ is an inversion pair for $w$ if $r<s$, $w(r)>w(s)$, and ${\bm{j}}^{\lambda}_{s}-{\bm{j}}^{\lambda}_{r}=\pm 1$.

Lemma 3.25**.**

Let ${\bm{\delta}}\in{\bm{\Lambda}}(n)$ be initial with $\lambda^{\bm{\delta}}=\lambda$. Set $\bar{\alpha}=\alpha_{a+1}+\dots+\alpha_{b}$, $\bar{\theta}=m\bar{\alpha}$, $\bar{\lambda}=(\lambda_{a+1},\dots,\lambda_{b})$, $\bar{n}:=\lambda_{a+1}+\dots+\lambda_{b}$, and $\bar{\bm{\delta}}=(\bar{\delta}^{(1)},\dots,\bar{\delta}^{(m)})$, where $\bar{\delta}^{(r)}=(\delta^{(r)}_{a+1},\dots,\delta^{(r)}_{b})$ for all $r\in[1,m]$. Then

[TABLE]

Proof.

By definition,

[TABLE]

Throughout the proof, ‘inversion pair’ means ‘inversion pair for $w({\bm{\delta}})$’. Recall that $w({\bm{\delta}})=u({\bm{\delta}})^{-1}$. Since ${\bm{\delta}}$ is initial, in the Khovanov-Lauda diagram for $\psi_{w({\bm{\delta}})}$ (for any choice of reduced expression) no strings of color $a$ cross each other. We want to apply quadratic relations on pairs of strings one of which has color $a$ and the other has color $a+1$. These correspond to inversion pairs $(r,s)$ with $r\in U^{-}_{a}(\lambda),s\not\in U^{-}_{a}(\lambda)$ or $s\in U^{+}_{a}(\lambda),r\not\in U^{+}_{a}(\lambda)$.

Note that there are exactly $r-1$ inversion pairs of the form $(r,s)$ when $r\in U^{-}_{a}(\lambda)$ and $d-s$ inversion pairs of the form $(r,s)$ when $s\in U^{+}_{a}(\lambda)$. Applying the corresponding quadratic relations, we see that $f_{n}^{\lambda,{\bm{\delta}}}g_{n}^{{\bm{\delta}},\lambda}$ equals

[TABLE]

where $(*)$ a sum of elements of the form

[TABLE]

with $X\in R_{\bar{\theta}}$, $Y^{\pm}$ a polynomial in the variables $y_{r}$ with $r\in U_{a}^{\pm}(\lambda)$, and $\deg Y^{-}+\deg Y^{+}<\deg y_{0,m-\lambda_{a}}+\deg y_{0,\lambda_{a}}^{\prime}$. By Lemma 2.5(i), we have $(*)=0$. So by Lemma 2.5(ii), the expression (3.27) equals the right hand side of (3.26). ∎

Define ${\bm{\delta}}_{\lambda}=(\delta_{\lambda}^{(1)},\cdots,\delta_{\lambda}^{(m)})$ to be the unique element of ${\bm{\Lambda}}(n)$ such that for each $a\leq i\leq b$ we have:

• •

$i$ precedes $b+1$ in ${\bm{j}}^{\delta_{\lambda}^{(r)}}$ for $1\leq r\leq m-\lambda_{i}$;

• •

$i$ succeeds $b+1$ in ${\bm{j}}^{\delta_{\lambda}^{(r)}}$ for $m-\lambda_{i}<r\leq m$.

Note that $\lambda^{{\bm{\delta}}_{\lambda}}=\lambda$ but ${\bm{\delta}}_{\lambda^{\bm{\delta}}}$ in general differs from ${\bm{\delta}}$.

Lemma 3.28**.**

Let ${\bm{\delta}}\in{\bm{\Lambda}}$ satisfy $\lambda^{\bm{\delta}}=\lambda$. Then

[TABLE]

Proof.

If ${\bm{\delta}}={\bm{\delta}}_{\lambda}$, the result follows by induction on ${\operatorname{ht}}(\alpha)$ using from Lemma 3.25. If ${\bm{\delta}}\neq{\bm{\delta}}_{\lambda}$, we may assume using Lemma 3.25 that ${\bm{\delta}}$ is not initial. This implies that for some $r\in[1,m)$, we have $\delta^{(r)}_{a}=1$ and $\delta^{(r+1)}_{a}=0$, i.e. the last entry of the word ${\bm{j}}^{\delta^{(r)}}$ and the first entry of the word ${\bm{j}}^{\delta^{(r+1)}}$ are both equal to $a$. It follows that a$$a

is a sub-diagram of a Khovanov-Lauda diagram for $\psi_{w({\bm{\delta}})}\psi_{u({\bm{\delta}})}y^{\lambda}e_{\lambda}$, so $f_{n}^{\lambda,{\bm{\delta}}}g_{n}^{{\bm{\delta}},\lambda}=\pm e_{\lambda}\psi_{w_{0}^{\lambda}}\psi_{w({\bm{\delta}})}\psi_{u({\bm{\delta}})}y^{\lambda}e_{\lambda}=0.$ ∎

Corollary 3.29**.**

For any $n\in\mathbb{Z}_{\geq 0}$, we have $g_{n}\circ f_{n}=\mathop{\mathrm{id}}\nolimits$.

3.7. Proof of Theorem A, assuming $f$ and $g$ are chain maps

In Sections 4 and 5, we will prove that $f$ and $g$ constructed above are chain maps. The goal of this subsection is to demonstrate that this is sufficient to establish our main result.

Let

[TABLE]

We also define $W_{0}$ to be the longest element of ${\mathscr{D}}^{(l^{m})}$. For $1\leq r<m$, we have a fully commutative element

[TABLE]

In fact, if $w_{0,m}=s_{r_{1}}\dots s_{r_{N}}$ is a reduced decomposition in $\mathfrak{S}_{m}$, then $W_{0}=z_{r_{1}}\dots z_{r_{N}}$ in $\mathfrak{S}_{d}$ with $\ell(W_{0})=\ell(z_{r_{1}})+\dots+\ell(z_{r_{N}})$. Recalling (3.11) we have

[TABLE]

Then the following is easy to check:

Lemma 3.30**.**

Let $P=\{w\in{\mathscr{D}}^{(l^{m})}\mid w\cdot{\bm{j}}^{\bm{0}}={\bm{j}}^{\bm{0}}\}$. Then $W_{0}\in P$ and $\deg(\psi_{W_{0}}e_{\bm{0}})<\deg(\psi_{w}e_{\bm{0}})$ for any $w\in P\setminus\{W_{0}\}$

Lemma 3.31**.**

Let $\psi_{W_{0}}$ and $\psi_{W_{0}}^{\prime}$ in $R_{\theta}$ correspond to different reduced expressions of $W_{0}\in\mathfrak{S}_{d}$. Then $\psi_{W_{0}}v_{\alpha}^{\circ m}=\psi_{W_{0}}^{\prime}v_{\alpha}^{\circ m}$ in $\Delta(\alpha)^{\circ m}$.

Proof.

It follows from the defining relations of the KLR algebra and Theorem 2.2 that $\psi_{W_{0}}e_{\bm{0}}-\psi_{W_{0}}^{\prime}e_{\bm{0}}$ is a linear combination of elements of the form $\psi_{w}e_{\bm{0}}$ where $w\cdot{\bm{j}}^{\bm{0}}={\bm{j}}^{\bm{0}}$, $w\neq W_{0}$, and $\deg(\psi_{w}e_{\bm{0}})=\deg(\psi_{W_{0}}e_{\bm{0}})$. Moreover, for each such $w$, by Lemma 2.1, there exist (unique) elements $x\in{\mathscr{D}}^{(l^{m})}$ and $y\in\mathfrak{S}_{(l^{m})}$ such that $w=xy$ and $\ell(w)=\ell(x)+\ell(y)$. Since $\psi_{y}v_{\alpha}^{\circ m}=0$ if $y\neq 1$, we may assume $w\in{\mathscr{D}}^{(l^{m})}$. We now have $w\in{\mathscr{D}}^{(l^{m})}$, $w\cdot{\bm{j}}^{\bm{0}}={\bm{j}}^{\bm{0}}$, and $w\neq W_{0}$. By Lemma 3.30, no such element $w$ can exist, so $(\psi_{W_{0}}e_{\bm{0}}-\psi_{W_{0}}^{\prime}e_{\bm{0}})v_{\alpha}^{\circ m}=0$. ∎

Lemma 3.32**.**

There is a choice of reduced expression of $W_{0}$ for which $g_{0}^{{\bm{0}},0}f_{0}^{0,{\bm{0}}}=e_{{\bm{0}}}Y_{0}\psi_{W_{0}}e_{{\bm{0}}}$.

Proof.

Note that $\ell(u({\bm{0}})w_{0}^{0}w({\bm{0}}))=\ell(u({\bm{0}}))+\ell(w_{0}^{0})+\ell(w({\bm{0}}))$ and $u({\bm{0}})w_{0}^{0}w({\bm{0}})=W_{0}$. So for an appropriate choice of reduced expression for $W_{0}$, we have $\psi_{u({\bm{0}})}\psi_{w_{0}^{0}}\psi_{w({\bm{0}})}=\psi_{W_{0}}$. We now compute:

[TABLE]

where the first equality is by definition of $g_{0}^{{\bm{0}},0}$ and $f_{0}^{0,{\bm{0}}}$, the second follows from Lemma 2.5(ii), the third by the defining relations of $R_{\theta}$ and the observation that $Y_{0}=u({\bm{0}})\cdot y^{0}$, and the fourth from the remark at the beginning of the proof. ∎

Note that

[TABLE]

where $g$ is the longest element of $\mathfrak{S}_{d}$ with $g\cdot({\bm{i}}_{\alpha}^{m})=a^{m}(a+1)^{m}\cdots(b+1)^{m}$. Let

[TABLE]

We refer to $v_{\alpha^{m}}$ as the standard generator of $\Delta(\alpha^{m})$. It can be checked directly but also follows from Theorem 3.34 that it does generate $\Delta(\alpha^{m})$. Note that $e_{0}v_{\alpha^{m}}=v_{\alpha^{m}}$, so there is a homomorphism

[TABLE]

where $v_{\alpha^{m}}$ is the standard generator of $\Delta(\alpha^{m})$, see (3.33).

Theorem 3.34**.**

If $f$ and $g$ are chain maps then $P_{\bullet}=P_{\bullet}^{\alpha^{m}}$ is a projective resolution of $\Delta(\alpha^{m})$, with the augmentation map ${\sf p}$.

Proof.

The modules $P_{n}$ are projective by construction. By Corollary 3.29, $P_{\bullet}$ is a complex, isomorphic to a direct summand of the complex $Q_{\bullet}$. Since $Q_{\bullet}$ is a resolution of $q^{m(m-1)/2}\Delta(\alpha)^{\circ m}$, it follows from the assumptions that $P_{\bullet}$ is exact in degrees $>0$ and its [math]th cohomology is given by ${\sf q}f_{0}g_{0}(Q_{0})$ where ${\sf q}:Q_{0}\to q^{m(m-1)/2}\Delta(\alpha)^{\circ m}$ is the augmentation map (3.12).

Recalling the $(R_{\theta},\mathcal{NH}_{m})$-bimodule structure of $q^{m(m-1)/2}\Delta(\alpha)^{\circ m}$ from §2.5 and the idempotent $e_{m}\in\mathcal{NH}_{m}$ defined in (2.12), using Lemma 3.31, we have

[TABLE]

where $w_{0,m}=s_{r_{1}}\cdots s_{r_{N}}$ is a reduced decomposition in $\mathfrak{S}_{m}$. We now have:

[TABLE]

where the first equality follows from Lemma 3.32, the second from the definition of ${\sf q}$, the third from (3.35), and the last two from the definitions.

It remains to note that ${\sf q}(f_{0}(e_{0}))=v_{\alpha^{m}}$. For, note that ${\sf q}(f_{0}(e_{0}))=f_{0}v_{\alpha}^{\circ m}$, and ${\sf q}(f_{0}(e_{0}))\in{\sf q}(f_{0}(P_{0}))={\sf q}(f_{0}(g_{0}(Q_{0})))=\Delta(\alpha^{m})$. So $f_{0}v_{\alpha}^{\circ m}=f_{0}v_{\alpha}^{\circ m}e_{m}=v_{\alpha^{m}}$. ∎

4. Verification that $f$ is a chain map

We continue with the running assumptions of the previous section. In addition, throughout the section we fix

[TABLE]

4.1. Special reduced expressions

Recall the notation of §3.1. Let $\lambda\in\Lambda^{\alpha^{m}}$ and $\delta\in\Lambda^{\alpha}$ be such that $\bar{\lambda}:=\lambda-\delta\in\Lambda^{\alpha^{m-1}}$. For $i\in[a,b+1]$, we denote by $p_{i}=p_{i}(\delta)\in[1,l]$ the position occupied by $i$ in ${\bm{j}}^{\delta}$, and set

[TABLE]

Let

[TABLE]

Note that $\{p_{a},\dots,p_{b+1}\}=[1,l]$. Define $x_{\delta}^{\lambda}\in\mathfrak{S}_{d}$ to be the permutation which maps $p_{i}$ to $q_{i}$ for all $i\in[a,b+1]$, and maps the elements of $[l+1,d]$ increasingly to the elements of $[1,d]\setminus Q$. It is easy to see that $x_{\delta}^{\lambda}$ is fully commutative, so $\psi_{x_{\delta}^{\lambda}}$ is well defined, and $1_{{\bm{j}}^{\lambda}}\psi_{x_{\delta}^{\lambda}}=\psi_{x_{\delta}^{\lambda}}1_{{\bm{j}}^{\delta}{\bm{j}}^{\bar{\lambda}}}$.

Now let $\lambda=\lambda^{\bm{\delta}}$, and $r\in[1,m]$. Define

[TABLE]

Define $x({\bm{\delta}},1):=x_{\delta^{(1)}}^{\lambda}\in\mathfrak{S}_{d}$. More generally, define for all $r=1,\dots,m-1$, the permutations $x({\bm{\delta}},r)\in\mathfrak{S}_{d}$ so that $x({\bm{\delta}},r)$ is the image of

[TABLE]

under the natural embedding $\mathfrak{S}_{(r-1)l}\times\mathfrak{S}_{(m-r+1)l}{\hookrightarrow}\mathfrak{S}_{d}$.

Recall the element $w({\bm{\delta}})\in\mathfrak{S}_{d}$ defined in §3.4. The following lemma follows from definitions:

Lemma 4.2**.**

We have $w({\bm{\delta}})=x({\bm{\delta}},1)\cdots x({\bm{\delta}},m-1)$ and $\ell(w({\bm{\delta}}))=\ell(x({\bm{\delta}},1))+\dots+\ell(x({\bm{\delta}},m-1))$.

In view of the lemma, when convenient, we will always choose reduced decompositions so that

[TABLE]

By Lemma 3.23, we then have

[TABLE]

Example 4.5**.**

Let $a=1$, $b=2$, $\lambda=(2,2)$, and {\bm{\delta}}=\big{(}(1,0),(1,1),(0,1),(0,0)\big{)}, so that $m=4$ and $n=4$. Then $f_{n}^{\lambda,{\bm{\delta}}}$ is

[TABLE]

4.2. A commutation lemma

It will be convenient to use the following notation. Let $\lambda\in\Lambda(n)$. Consider the parabolic (non-unital) subalgebra

[TABLE]

We have

[TABLE]

The natural (unital) embeddings of the algebras

[TABLE]

into $R^{\lambda}$, allow us to consider them as (non-unital) subalgebras of $R_{\theta}$. We denote the corresponding (non-unital) algebra embeddings by

[TABLE]

For example, setting

[TABLE]

for all $i\in[a,b]$, can write $\psi_{w_{0}^{\lambda}}\in R_{\theta}$ as a commuting product

[TABLE]

Lemma 4.7**.**

Let $i\in[a,b]$ with $\mu_{i}>0$, and $\lambda:=\mu-\mathtt{e}_{i}$. Then

[TABLE]

Proof.

We have

[TABLE]

where we have used (4.6) for the second equality, Lemma 2.9(i) for the third equality, and the definition of $\psi_{\lambda,i}$ as an explicit cycle element for the fourth equality. ∎

Example 4.8**.**

We illustrate Lemma 4.7 diagramatically. Let $a=1$, $b=2$, $m=4$ and $\mu=(2,3)$. Then, Lemma 4.7 claims the following equality:

[TABLE]

4.3. $f$ is a chain map

Let $\lambda=\lambda^{\bm{\delta}}$. For $i\in[a,b+1]$, let $q_{i}=q_{i}(\lambda,\delta^{(1)})$ be defined as in (4.1), and consider the cycle

[TABLE]

Let $c$ be the commuting product of cycles:

[TABLE]

Recall also the elements defined in (3.13).

Lemma 4.9**.**

Suppose $i\in[a,b]$ is such that $\mu_{i}>0$ and $\lambda:=\mu-\mathtt{e}_{i}=\lambda^{\bm{\delta}}$. Set $\bar{\theta}=(m-1)\alpha$, $\bar{\mu}:=\mu-\delta^{(1)}$, and $\bar{\lambda}:=\lambda-\delta^{(1)}$. Then $1_{{\bm{j}}^{\mu}}\psi_{\lambda;i}e_{\lambda}\psi_{w_{0}^{\lambda}}\psi_{x({\bm{\delta}},1)}$ equals

[TABLE]

Proof.

By Lemma 2.5(ii), we have $e_{\lambda}\psi_{w_{0}^{\lambda}}=1_{{\bm{j}}^{\lambda}}\psi_{w_{0}^{\lambda}}$. So using also Lemma 4.7, we get

[TABLE]

As usual, we denote by $p_{j}$ the position occupied by $j$ in $\delta^{(1)}$ and $q_{j}$ be defined as in (4.1). By Lemma 4.2, the permutation $x({\bm{\delta}},1)$ maps $p_{j}$ to $q_{j}$ for all $j\in[a,b+1]$, and the elements of $[l+1,d]$ increasingly to the elements of $[1,d]\setminus Q$.

Case 1: $\delta^{(1)}_{i}=0$. In this case we have $q_{i}=l_{i}^{-}(\lambda)$. So the KLR diagram $D$ of $1_{{\bm{j}}^{\mu}}\psi_{l_{i}^{+}(\mu)\to l_{i}^{-}(\lambda)}\psi_{x({\bm{\delta}},1)}$ has an $i$-string $S$ from the position $p_{i}$ in the bottom to the position $l_{i}^{+}(\mu)$ in the top, and the only $(i,i)$-crossings in $D$ will be the crossings of the string $S$ with the $m-\mu_{i}$ strings which originate in the positions $L_{i}^{-}(\mu)$ in the top. The $(i+1)$-string $T$ in $D$ originating in position $p_{i+1}$ in the bottom is to the right of all $(i,i)$ crossings. Pulling $T$ to the left produces error terms, which arise from opening $(i,i)$-crossings, but all of them, except the last one, amount to zero, when multiplied on the left by $1_{{\bm{j}}^{\mu}}\psi_{w_{0}^{\mu}}$. The last error term is equal to $\psi_{x_{\delta^{(1)}}^{\mu}}\iota_{\alpha,\bar{\theta}}(1_{\alpha}\otimes\psi_{l_{i}^{+}(\bar{\mu})\to l_{i}^{-}(\bar{\lambda})}))$, and the result of pulling $T$ past all $(i,i)$-crossings gives $\psi_{x({\bm{\delta}}+\text{\bf e}_{i}^{1},1)}\psi_{{\bm{\delta}};1,i}$. Multiplying on the left by $1_{{\bm{j}}^{\mu}}\psi_{w_{0}^{\mu}}$, gives

[TABLE]

and it remains to observe using Lemma 2.9 that

[TABLE]

Case 2: $\delta^{(1)}_{i}=1$. Let $D$ be the KLR diagram of $1_{{\bm{j}}^{\mu}}\psi_{l_{i}^{+}(\mu)\to l_{i}^{-}(\lambda)}\psi_{x({\bm{\delta}},1)}$. Let $S$ be the $i$-string originating in the position $l^{+}_{i}(\mu)$ in the top row of $D$, and $T$ be the $(i+1)$-string originating in the position $p_{i+1}$ in the bottom row of $D$. The quadratic relation on these strings produces a difference of two terms, one with a dot on $S$ and the other with a dot on $T$. The term with a dot on $T$ equals [math] after multiplying on the left by $1_{{\bm{j}}^{\mu}}\psi_{w_{0}^{\mu}}$. The term with a dot on $S$, when multiplied on the left by $1_{{\bm{j}}^{\mu}}\psi_{w_{0}^{\mu}}$, yields

[TABLE]

where we have used Lemma 2.9 for the last equality. ∎

Example 4.10**.**

We illustrate Lemma 4.9 diagramatically. Let $a=1$, $b=2$, $m=4$, $\mu=(2,3)$ and ${\bm{\delta}}=((1,0),(0,1),(1,1),(0,1))$.

First, take $i=2$. Then $\delta_{1}^{(2)}=0$ and Lemma 4.9 claims the following:

[TABLE]

On the other hand, if $i=1$, we have $\delta_{1}^{(1)}=1$, and Lemma 4.9 posits the following equality.

[TABLE]

Corollary 4.11**.**

If $\mu_{i}>0$ for some $i\in[a,b]$ and $\lambda:=\mu-\mathtt{e}_{i}=\lambda^{\bm{\delta}}$, then

[TABLE]

Proof.

The proof is by induction on $m$, the induction base $m=1$ being obvious. Let $\bar{\bm{\delta}}:=(\delta^{(2)},\dots,\delta^{(m)})$, $\bar{\theta}=(m-1)\alpha$, $\bar{\mu}:=\mu-\delta^{(1)}$, and $\bar{\lambda}:=\lambda-\delta^{(1)}$. By (4.3), we have

[TABLE]

Now we apply Lemma 4.9. We consider the case $\delta_{i}^{(1)}=0$, the case $\delta_{i}^{(1)}=1$ being similar. Then we get the following expression for the right hand side of (4.13):

[TABLE]

Opening parentheses, we get two summands $S_{1}+S_{2}$. Note that

[TABLE]

Moreover, using the inductive assumption for the third equality below, we get that $S_{2}$ equals

[TABLE]

As $\delta_{i}^{(1)}=0$, we have $\sum_{s=2}^{r-1}\delta_{i}^{(s)}=\sum_{s=1}^{r-1}\delta_{i}^{(s)}$. So

[TABLE]

Moreover,

[TABLE]

So $S_{2}$ equals

[TABLE]

where we have used Lemma 2.9(i) to see the first equality. Thus $S_{1}+S_{2}$ equals the right hand side of (4.12). ∎

The following statement means that $f$ is chain map:

Proposition 4.14**.**

Let $\mu\in\Lambda(n+1)$ and ${\bm{\delta}}\in{\bm{\Lambda}}(n)$. Then

[TABLE]

Proof.

By definition, $d_{n}^{\mu,\lambda}=0$ unless $\lambda=\mu-\mathtt{e}_{i}$ for some $i\in[a,b]$, and $f_{n}^{\lambda,{\bm{\delta}}}=0$ unless $\lambda=\lambda^{\bm{\delta}}$. On the other hand, $f_{n}^{\mu,{\bm{\gamma}}}=0$ unless $\mu=\lambda^{\bm{\gamma}}$, and $c_{n}^{{\bm{\gamma}},{\bm{\delta}}}=0$, unless ${\bm{\delta}}={\bm{\gamma}}-\text{\bf e}_{i}^{r}$ for some $i\in[a,b]$ and $r\in[1,m]$. So we may assume that $\mu=\lambda^{{\bm{\delta}}+\text{\bf e}_{i}^{r}}$ for some $i\in[a,b]$ and $r\in[1,m]$ such that $\delta_{i}^{(r)}=0$. In this case, letting $\lambda:=\lambda^{{\bm{\delta}}}$, we have to prove

[TABLE]

By definition of the elements involved, this means

[TABLE]

Equivalently, we need to prove

[TABLE]

which, in view of Corollary 4.11, is equivalent to the statement that

[TABLE]

for all $r\in[1,m]$ such that $\delta_{i}^{(r)}=0$. But this is Lemma 3.1. ∎

5. Verification that $g$ is a chain map

We continue with the running assumptions of Section 3. In addition, throughout the section we fix $n\in\mathbb{Z}_{\geq 0}$, $\lambda\in\Lambda(n)$ and ${\bm{\gamma}}=(\gamma^{(1)},\dots,\gamma^{(m)})\in{\bm{\Lambda}}(n+1)$.

5.1. Special reduced expressions and a commutation lemma

Recall the notation of §3.1. Let $\mu\in\Lambda^{\alpha^{m}}$ and $\gamma\in\Lambda^{\alpha}$ be such that $\bar{\mu}:=\mu-\gamma\in\Lambda^{\alpha^{m-1}}$. For $i\in[a,b+1]$, we denote

[TABLE]

Let $Q:=\{q^{a},\dots,q^{b+1}\}$. Note that $\{p^{a},\dots,p^{b+1}\}=(d-l,d]$.

Define $z_{\mu}^{\gamma}\in\mathfrak{S}_{d}$ to be the permutation which maps $q^{i}$ to $p^{i}$ for all $i\in[a,b+1]$, and maps the elements of $[1,d]\setminus Q$ increasingly to the elements of $[1,d-l]$. It is easy to see that $z_{\mu}^{\gamma}$ is fully commutative, so $\psi_{z_{\mu}^{\gamma}}$ is well defined, and $\psi_{z_{\mu}^{\gamma}}1_{{\bm{j}}^{\mu}}=1_{{\bm{j}}^{\bar{\mu}}{\bm{j}}^{\gamma}}\psi_{z_{\mu}^{\gamma}}$.

Now let $\mu:=\lambda^{\bm{\gamma}}$, and $r\in[1,m]$. Define

[TABLE]

Define $z({\bm{\gamma}},m):=z_{\mu}^{\gamma^{(m)}}\in\mathfrak{S}_{d}$. More generally, define for all $r=2,\dots,m$, the permutations

[TABLE]

Recall the element $u({\bm{\gamma}})\in\mathfrak{S}_{d}$ defined in §3.4. The following lemma follows from definitions:

Lemma 5.6**.**

We have $u({\bm{\gamma}})=z({\bm{\gamma}},2)\cdots z({\bm{\gamma}},m)$ and $\ell(u({\bm{\gamma}}))=\ell(z({\bm{\gamma}},1))+\dots+\ell(z({\bm{\gamma}},m))$.

In view of the lemma, when convenient, we will always choose reduced decompositions so that

[TABLE]

In view of Lemma 3.23, we have

[TABLE]

Example 5.9**.**

Let $a=1$, $b=2$, $\mu=(2,2)$, and {\bm{\gamma}}=\big{(}(1,0),(1,1),(0,1),(0,0)\big{)}, so that $m=4$ and $n=4$. Then $g_{n}^{\lambda,{\bm{\gamma}}}$ is

[TABLE]

Lemma 5.10**.**

Let $i\in[a,b]$ with $\lambda_{i}<m$, and $\mu:=\lambda+\mathtt{e}_{i}$. Then

[TABLE]

Proof.

Recall the notation of §4.2, we have that $y^{\mu}e_{\mu}\psi_{\lambda;i}e_{\lambda}$ equals

[TABLE]

where we have used Lemma 2.9 for the second equality and Lemma 2.8 for the last equality. ∎

Example 5.11**.**

We illustrate Lemma 5.10 diagramatically. Let $a=1$, $b=2$, $m=4$, $\mu=(2,3)$ and $\lambda=(1,3)$ so that $i=1$ using the notation of Lemma 5.10. Then, the lemma claims the following equality.

[TABLE]

5.2. $g$ is a chain map

Recall that we have fixed $\lambda\in\Lambda(n)$ and ${\bm{\gamma}}\in{\bm{\Lambda}}(n+1)$.

Lemma 5.12**.**

Suppose that $i\in[a,b]$ is such that $\lambda_{i}<m$ and $\mu:=\lambda+\mathtt{e}_{i}=\lambda^{\bm{\gamma}}$. Set $\bar{\theta}=(m-1)\alpha$, $\bar{\mu}:=\mu-\gamma^{(m)}$, and $\bar{\lambda}:=\lambda-\gamma^{(m)}$. Then $\psi_{z({\bm{\gamma}},m)}y^{\mu}e_{\mu}\psi_{\lambda;i}e_{\lambda}$ equals

[TABLE]

Proof.

Using Lemma 5.10, we get

[TABLE]

Recalling (5.1),(5.5), the permutation $z({\bm{\gamma}},m)$ maps $q^{j}$ to $p^{j}$ for all $j\in[a,b+1]$, and the elements of $[1,d]\setminus Q$ increasingly to the elements of $[1,d-l]$.

Case 1: $\gamma^{(m)}_{i}=1$. In this case we have $q^{i}=r_{i}^{+}(\lambda)$. So the KLR diagram $D$ of $\psi_{z({\bm{\gamma}},m)}\psi_{\lambda;i}1^{{\bm{j}}^{\lambda}}$ has an $i$-string $S$ from the position $r_{i}^{-}(\lambda)$ in the bottom to the position $p^{i}$ in the top, and the only $(i,i)$-crossings in $D$ will be the crossings of the string $S$ with the $\lambda_{i}$ strings which originate in the positions $R_{i}^{+}(\lambda)$ in the bottom. The $(i+1)$-string $T$ in $D$ originating in the position $p_{i+1}$ in the top is to the left of all $(i,i)$ crossings. Pulling $T$ to the right produces error terms, which arise from opening $(i,i)$-crossings, but all of them, except the last one, amount to zero, when multiplied on the right by $y^{\lambda}e_{\lambda}$. The last error term is equal to $-\iota_{\bar{\theta},\alpha}(\psi_{\bar{\lambda};i}\otimes 1_{\alpha})\psi_{z^{\gamma^{(m)}}_{\lambda}},$ and the result of pulling $T$ past all $(i,i)$-crossings gives $\psi_{{\bm{\gamma}}-\text{\bf e}_{i}^{m};m,i}\psi_{z({\bm{\gamma}}-\text{\bf e}_{i}^{m},m)}$. Multiplying on the left by $y^{\lambda}e_{\lambda}$, gives

[TABLE]

and it remains to observe using Lemma 2.9 that

[TABLE]

Case 2: $\gamma^{(m)}_{i}=0$. Let $D$ be the KLR diagram of $\psi_{z({\bm{\gamma}},m)}\psi_{\lambda;i}1_{{\bm{j}}^{\lambda}}$. Let $S$ be the $i$-string originating in the position $r^{-}_{i}(\lambda)$ in the bottom row of $D$, and $T$ be the $(i+1)$-string originating in the position $p^{i+1}$ in the top row of $D$. The quadratic relation on these strings produces a linear combination of two diagrams, one with a dot on $S$ and the other with a dot on $T$. The term containing a dot on $T$ produces a term which is zero after multiplying on the right by $y^{\lambda}e^{\lambda}$. The term containing a dot on $S$, when multiplied on the right by $y^{\lambda}e^{\lambda}$, yields

[TABLE]

where we have used Lemmas 2.9 and 2.8 to deduce the last equality. ∎

Example 5.13**.**

We illustrate Lemma 5.12 diagramatically. Using the notation of the lemma, let $a=1$, $b=2$, $\mu=(2,2)$, $\lambda=(1,2)$, $m=4$ and $i=1$. Then, if {\bm{\gamma}}=\big{(}(1,0),(1,1),(0,1),(0,0)\big{)}, we are in the $\gamma_{i}^{(m)}=0$ case and the lemma claims the following equality.

[TABLE]

If instead we let {\bm{\gamma}}=\big{(}(0,0),(1,1),(0,1),(1,0)\big{)}, we are in the $\gamma_{i}^{(m)}=1$ case and the lemma claims the following equality.

[TABLE]

Corollary 5.14**.**

Suppose that $i\in[a,b]$ is such that $\lambda_{i}<m$ and $\mu:=\lambda+\mathtt{e}_{i}=\lambda^{\bm{\gamma}}$. Then

[TABLE]

Proof.

The proof is by induction on $m$, the induction base $m=1$ being obvious. Let $\bar{\gamma}:=(\gamma^{(1)},\dots,\gamma^{(m-1)})$, $\bar{\theta}=(m-1)\alpha$, $\bar{\mu}:=\mu-\gamma^{(m)}$, and $\bar{\lambda}:=\lambda-\gamma^{(m)}$. By (5.7), we have

[TABLE]

Now we apply Lemma 5.12. We consider the case $\gamma_{i}^{(m)}=1$, the case $\gamma_{i}^{(m)}=0$ being similar. Then we get the following expression for the right hand side of (5.15):

[TABLE]

Opening parentheses, we get two summands $S_{1}+S_{2}$. Note that

[TABLE]

Moreover, using the inductive assumption, for the third equality below, $S_{2}$ equals

[TABLE]

Thus $S_{1}+S_{2}$ equals the right hand side of the expression in the corollary. ∎

The following statement means that $g$ is chain map:

Proposition 5.16**.**

Let $\lambda\in\Lambda(n)$ and ${\bm{\gamma}}\in{\bm{\Lambda}}(n+1)$. Then

[TABLE]

Proof.

By definition, $d_{n}^{\mu,\lambda}=0$ unless $\mu=\lambda+\mathtt{e}_{i}$ for some $i\in[a,b]$, and $g^{{\bm{\gamma}},\mu}_{n+1}=0$ unless $\mu=\lambda^{\bm{\gamma}}$. On the other hand, $g^{{\bm{\delta}},\lambda}_{n}=0$ unless $\lambda=\lambda^{\bm{\delta}}$, and $c_{n}^{{\bm{\gamma}},{\bm{\delta}}}=0$, unless ${\bm{\delta}}={\bm{\gamma}}-\text{\bf e}_{i}^{r}$ for some $i\in[a,b]$ and $r\in[1,m]$. So we may assume that there exists $i\in[a,b]$ such that $\lambda^{\bm{\gamma}}=\lambda+\mathtt{e}_{i}$, in which case, setting $\mu:=\lambda^{\bm{\gamma}}$, we have to prove

[TABLE]

By definition of the elements involved, this means

[TABLE]

Equivalently, we need to prove

[TABLE]

which, in view of Corollary 5.14, is equivalent to the statement that

[TABLE]

for all $r\in[1,m]$ such that $\gamma_{i}^{(r)}=1$. But this is Lemma 3.3. ∎