1. Introduction
In [Reference SoergelSoe90], Soergel gave a combinatorial description of the category 
$\mathcal {O}$ for semisimple Lie algebras. This celebrated work has many applications: another proof of the Kazhdan–Lusztig conjecture (which does not rely on the theory of 
$\mathcal {D}$-modules), Koszul duality of the category 
$\mathcal {O}$, etc. Later, Soergel [Reference SoergelSoe07] defined a certain category purely in terms of combinatorics of Coxeter systems, without any representation theory. This category describes the category 
$\mathcal {O}$. More precisely, this category is equivalent to the category of projective modules in (a deformed version of) the principal block of the category 
$\mathcal {O}$ (cf. [Reference SoergelSoe90, Reference FiebigFie06]).
Let 
$(W,S)$ be a Coxeter system and 
$V$ a reflection-faithful representation of 
$W$. Then Soergel attached the category of Soergel bimodules. When 
$W$ is the Weyl group of a semisimple Lie algebra, we may take 
$V$ to be a Cartan subalgebra, and Soergel's category is the category which describes the category 
$\mathcal {O}$.
Fiebig used this category (or, more precisely, the category of sheaves on moment graphs, which is equivalent to the category of Soergel bimodules [Reference FiebigFie08a]) to give an alternative proof of Lusztig's conjecture which says that the irreducible characters of an algebraic group over an algebraically closed field of positive characteristic are given by affine Kazhdan–Lusztig polynomials if the characteristic is large enough. He used the Soergel bimodules attached to an affine Weyl group 
$(W,S)$ and a Cartan subalgebra 
$V$ of the corresponding affine Lie algebra. If the coefficient field of 
$V$ has characteristic zero, then 
$V$ is reflection faithful. However if the characteristic is positive, then this is not reflection faithful. So he used a lifting to characteristic zero and used the theory of Soergel bimodules over a characteristic-zero field.
However, it is of course more natural to use positive-characteristic objects directly. Elias and Williamson gave an alternative category of the category of Soergel bimodules which works well even with a non-reflection-faithful representation [Reference Elias and WilliamsonEW16]. Riche and Williamson [Reference Riche and WilliamsonRW18] gave a conjecture which claims that this category describes the category of algebraic representations of an algebraic group over any field of positive characteristic. As an application, this description gives a character formula of tilting modules in terms of the 
$p$-canonical basis defined using the category of Elias–Williamson. Recently this character formula was proved by Achar, Makisumi, Riche and Williamson [Reference Achar, Makisumi, Riche and WilliamsonAMRW19] for the principal block when 
$p$ is greater than the Coxeter number.
Elias and Williamson gave the definition of their category as a certain diagrammatic category with generators and relations. Such generators and relations appeared in a study of Soergel bimodules; however, the definition seems completely different from the original definition of Soergel bimodules. The aim of this paper is to give a ‘bimodule-theoretic’ definition of this category, that is, we define a new category and prove that the new category is equivalent to that of Elias and Williamson.
We first recall the category of Soergel bimodules. Let 
$R = S(V)$ be the symmetric algebra of 
$V$ and, for 
$s\in S$, let 
$R^{s}$ be the subalgebra of 
$s$-invariants. Let 
$(n)$ be a shift of grading defined by 
$M(n)^{i} = M^{i + n}$ for 
$n\in \mathbb {Z}$, where 
$M = {\bigoplus} _i\, M^{i}$ is a graded module. Then a Soergel bimodule is a graded 
$R$-bimodule which is a direct summand of a direct sum of modules of the form
\begin{equation} R\otimes_{R^{s_1}}R\otimes_{R^{s_2}}\cdots\otimes_{R^{s_l}}R(n) \end{equation}
for 
$s_1,\ldots ,s_l\in S$ and 
$n\in \mathbb {Z}$. Let 
$\mathcal {S}\mathrm {Bimod}$ be the category of Soergel bimodules and 
$[\mathcal {S}\mathrm {Bimod}]$ the split Grothendieck group of 
$\mathcal {S}\mathrm {Bimod}$. Then Soergel [Reference SoergelSoe07] proved the following result which we call Soergel's categorification theorem.
(1) For each
$w\in W$, there exists a unique indecomposable module $B(w)\in \mathcal {S}\mathrm {Bimod}$ which satisfies the following properties.
(a) For a reduced expression
$w = s_1\cdots s_l$, $B(w)$ appears as a direct summand of $R\otimes _{R^{s_1}}\cdots \otimes _{R^{s_l}}R(l)$ with multiplicity one and the module $R\otimes _{R^{s_1}}\cdots \otimes _{R^{s_l}}R(l)$ is a direct sum of $B(v)(l)$ where $v < w$ and $l\in \mathbb {Z}$.(b) Any object in
$\mathcal {S}\mathrm {Bimod}$ is a direct sum of modules $B(w)(n)$ ($w\in W$, $n\in \mathbb {Z}$).
(2) The algebra
$[\mathcal {S}\mathrm {Bimod}]$ is isomorphic to the Hecke algebra attached to $(W,S)$.
We prove the generalization of this theorem to 
$V$ which is not reflection faithful in this paper.
1.1 A representation $V$ and our category
Let 
$V$ be a finite-dimensional representation of 
$W$ over a field 
$\mathbb {K}$. We assume that there exist 
$\alpha _s\in V$ and 
$\alpha _s^{\vee }\in V^{*}$ (where 
$V^{*}$ is the dual of 
$V$) for each 
$s\in S$ such that:
(1)
$\langle \alpha ^{\vee }_s,\alpha _s\rangle = 2$ for any $s\in S$;(2)
$s(v) = v - \langle \alpha _s^{\vee },v\rangle \alpha _s$ for any $s\in S$ and $v\in V$;(3)
$\alpha _s\ne 0$ and $\alpha _s^{\vee }\ne 0$.
The data 
$(V,\{\alpha _s\}_{s\in S},\{\alpha _s^{\vee }\}_{s\in S})$ is called a realization in [Reference Elias and WilliamsonEW16, Definition 3.1] if it satisfies a technical condition. To prove our categorification theorem, we need one more assumption. In this introduction, we make the following assumption for simplicity. See § 3.2 for details.
(4) For each
$s,t\in S$ such that the order of $st$ is finite, the representation of $V$ restricted to the group generated by $\{s,t\}$ is reflection faithful.
Under these assumptions, we consider the following category. Set 
$R = S(V)$ and let 
$Q$ be the fraction field of 
$R$. We consider the following category: the objects of the category are 
$(M,(M_Q^{w})_{w\in W},\xi _M)$ which satisfy the following condition.
$M$ is a graded 
$R$-bimodule, 
$M_Q^{w}$ is an 
$(R,Q)$-bimodule and 
$\xi _M\colon M\otimes _{R}Q\xrightarrow {\sim }{\bigoplus} _{w\in W}\, M_Q^{w}$ is an 
$(R,Q)$-isomorphism such that 
$f\in R$ and 
$m\in M_Q^{w}$ we have 
$mf = w(f)m$.
The morphisms we consider are degree-zero homomorphisms as 
$R$-bimodules which upon tensoring with 
$Q$ are compatible with isomorphisms 
$\xi _M$. It is easy to see that the 
$R$-bimodule (1.1) naturally has such a decomposition, and we say that 
$M$ is a Soergel bimodule if it appears as a direct summand of a sum of modules of this type. We now state the main theorem of this paper.
Theorem 1.1
(1) For this category we have Soergel's categorification theorem.
(2) This category is equivalent to the category of Elias and Williamson.
1.2 Sheaves on moment graphs
As mentioned above, Fiebig used sheaves on moment graphs to give an alternative proof of Lusztig's conjecture [Reference FiebigFie11] for sufficiently large primes. In this paper we prove the following result.
Theorem 1.2 Assume that the Goresky–Kottwitz–MacPherson (GKM) condition (see § 5.1) holds. Our category is equivalent to a certain full subcategory of 
$\mathcal {Z}$-modules (see § 5.1 for the precise definition) where 
$\mathcal {Z}$ is the structure algebra of the moment graph attached to 
$(W,S)$.
As a corollary, the category of Elias and Williamson is equivalent to the full subcategory of 
$\mathcal {Z}$-modules. If 
$(W,S,V)$ comes from a Kac–Moody group, this is proved using parity sheaves by combining works of Fiebig and Williamson [Reference Fiebig and WilliamsonFW14] and Riche and Williamson [Reference Riche and WilliamsonRW18].
1.3 Proof
Even though the definitions and theorems are similar to that of Soergel [Reference SoergelSoe07], it seems difficult to follow his argument in our setting. For example, he proved that after a suitable localization, the modules decompose into modules attached to rank one [Reference SoergelSoe07, Lemma 6.10]. However, this does not hold in our case. Fiebig used similar arguments in his study of moment graphs. To use this argument, he assumed the GKM condition on the moment graph. This condition does not follow from our assumptions on 
$V$.
There is another point in Soergel's argument which we cannot apply to our case. He considered the ‘standard module’ 
$\Delta _x$ for each 
$x\in W$ and proved that there exists an extension between 
$\Delta _x$ and 
$\Delta _y$ only when 
$x^{-1}y$ is a reflection (a conjugation of an element in 
$S$). Therefore the category 
$\mathcal {F}_\Delta$ consisting of the objects which admit a ‘standard flag’ behaves well. In our case, however, there are more extensions and the analogue of the category 
$\mathcal {F}_\Delta$ does not seem to behave well.
We analyze the modules in (1.1) directly using light leaves introduced by Libedinsky [Reference LibedinskyLib08]. Using Soergel's theorem, Libedinsky proved that the light leaves give a basis of a certain space of homomorphisms. In this paper, we prove Libedinsky's result directly and use it to prove Soergel's categorification theorem. The argument is new even for the original case.
1.4 Application
Let 
$G$ be a connected reductive group defined over an algebraically closed field of positive characteristic and 
$T$ a maximal torus. Then we give an action of our category (or equivalently, the category of Elias and Williamson) attached to the affine Weyl group of 
$G$ on a regular block of 
$G_1T$-modules where 
$G_1$ is the Frobenius kernel [Reference AbeAbe19]. This is a 
$G_1T$-version of a conjecture of Riche and Williamson [Reference Riche and WilliamsonRW18].
1.5 Organization of the paper
In the next section we introduce our category and give basic properties of it. We also introduce the notation on Hecke algebras. In § 3 we recall the definition of light leaves. Using the light leaves, we prove freeness of certain modules and calculate the graded ranks. In § 4 we prove the categorification theorem based on the theorems in § 3. In the final section we compare our category with the other categories.
2. The category
We construct the main category in this paper and prove some basic properties. We follow the notation in [Reference Elias and WilliamsonEW16].
2.1 A representation
Throughout this paper let 
$(W,S)$ be a Coxeter system such that 
$\#S < \infty$, and let 
$\mathbb {K}$ be a Noetherian integral domain. The length function 
$W\to \mathbb {Z}_{\ge 0}$ of 
$W$ is denoted by 
$\ell$ and the Bruhat order on 
$W$ is denoted by 
$\le$.
Let 
$(V,\{\alpha _s\}_{s\in S},\{\alpha _s^{\vee }\}_{s\in S})$ be a triple such that:
•
$V$ is a free $\mathbb {K}$-module of finite rank with a $\mathbb {K}$-linear action of $W$;•
$\alpha _s\in V$, $\alpha _s^{\vee }\in V^{*}$ where $V^{*} = \operatorname {Hom}_{\mathbb {K}}(V,\mathbb {K})$.
We assume that this satisfies:
(1)
$\langle \alpha _s^{\vee },\alpha _s\rangle = 2$ for each $s\in S$;(2)
$s(v) = v - \langle \alpha _s^{\vee },v\rangle \alpha _s$ for $s\in S$ and $v\in V$;(3)
$\alpha _s^{\vee }\colon V\to \mathbb {K}$ is surjective and $\alpha _s\ne 0$ for any $s\in S$.
Note that the third condition follows from the first condition if 
$2\in \mathbb {K}^{\times }$. Later we will add one more assumption (Assumption 3.2).
A triple which satisfies conditions (1) and (2) is called a realization if it satisfies one more technical condition in [Reference Elias and WilliamsonEW16, Definition 3.1]. Condition (3) is a part of the condition called Demazure surjectivity [Reference Elias and WilliamsonEW16, Assumption 3.9].
Set 
$R = S(V)$ and let 
$Q$ be the fraction field of 
$R$. We regard 
$R$ as a graded algebra via 
$\deg (V) = 2$. The action of the group 
$W$ on 
$V$ extends uniquely on 
$R$ by automorphisms of algebras. We remark that 
$R$ is a Noetherian integral domain.
We call 
$t\in W$ a reflection if it is conjugate to an element in 
$S$. Let 
$t = wsw^{-1}$ be a reflection where 
$s\in S$ and 
$w\in W$. Set 
$\alpha _t = w(\alpha _s)$. This depends on the choice of 
$s,w$ in general, so we fix such 
$s,w$ to define 
$\alpha _t$. By the following lemma, 
$\mathbb {K}^{\times } \alpha _t$ does not depend on the choice of 
$s,w$.
Lemma 2.1 If 
$wsw^{-1} = s'$ where 
$s,s'\in S$ and 
$w\in W$, then 
$\alpha _{s'}\in \mathbb {K}^{\times } w(\alpha _s)$.
Proof. Take 
$\delta \in V$ such that 
$\langle \alpha _s^{\vee },\delta \rangle = 1$. Then we have 
$s(\delta ) = \delta - \alpha _s$. Hence 
$s'(w(\delta )) = ws(\delta ) = w(\delta ) - w(\alpha _s)$. On the other hand, we have 
$s'(w(\delta )) = w(\delta ) - r\alpha _{s'}$ where 
$r = \langle \alpha _{s'}^{\vee },w(\delta )\rangle \in \mathbb {K}$. Hence 
$w(\alpha _s) = r\alpha _{s'}$. Replacing 
$s,w,s'$ with 
$s',w^{-1},s$ respectively, there exists 
$r'$ such that 
$w^{-1}(\alpha _{s'}) = r'\alpha _s$. Therefore 
$w(\alpha _s) = r\alpha _{s'} = rr'w(\alpha _s)$. Since 
$V$ is free, 
$\mathbb {K}$ is an integral domain and 
$w(\alpha _s)\ne 0$, we have 
$rr' = 1$. Therefore 
$r,r'\in \mathbb {K}^{\times }$.
2.2 The category
First we define the category 
$\mathcal {C}'$ as follows. The objects of 
$\mathcal {C}'$ are 
$(M,(M_Q^{w})_{w\in W},\xi _M)$ where 
$M$ is a graded 
$R$-bimodule, 
$M_Q^{w}$ is an 
$(R,Q)$-bimodule for each 
$w\in W$ and 
$\xi _M$ is an isomorphism 
$\xi _M\colon M\otimes _{R}Q\xrightarrow {\sim } {\bigoplus} _{w\in W}\, M_Q^{w}$ as 
$(R,Q)$-bimodules such that:
•
$M_Q^{w} \ne 0$ only for finite $w$;• for
$f\in R$ and $m\in M_Q^{w}$, we have $mf = w(f)m$.
A morphism 
$\varphi \colon (M,(M_Q^{w})_{w\in W},\xi _M)\to (N,(N_Q^{w})_{w\in W},\xi _N)$ in 
$\mathcal {C}'$ is a degree-zero 
$R$-bimodule homomorphism 
$\varphi \colon M\to N$ such that 
$\xi _N\circ \varphi _Q\circ \xi _M^{-1}(M_Q^{w})\subset N_Q^{w}$ for any 
$w\in W$ where 
$\varphi _Q\colon M\otimes _{R}Q\to N\otimes _{R}Q$ is the induced homomorphism. We often omit 
$M_Q^{w}$ and 
$\xi _M$ from the notation. So we just say 
$M\in \mathcal {C}'$.
Remark 2.2 By the second condition, the left action of 
$f\in R\setminus \{0\}$ on 
$M_Q^{w}$ is invertible. Therefore it is also invertible on 
$M\otimes _R Q$, and therefore 
$M\otimes _R Q$ is a 
$Q$-bimodule. The isomorphism 
$\xi _M\colon M\otimes _R Q \simeq {\bigoplus} _{w\in W}\, M_Q^{w}$ is an isomorphism as 
$Q$-bimodules.
Remark 2.3 If 
$V$ is a faithful 
$W$-representation, then, for any 
$(M,(M_Q^{w})_{w\in W},\xi _M)\in \mathcal {C}'$ and 
$(N,(N_Q^{w})_{w\in W},\xi _N)\in \mathcal {C}'$, an 
$R$-bimodule homomorphism 
$M\to N$ is automatically a morphism in 
$\mathcal {C}'$. Therefore the functor 
$(M,(M_Q^{w})_{w\in W},\xi _M)\mapsto M$ is a fully faithful functor from 
$\mathcal {C}'$ to the category of graded 
$R$-bimodules. Hence the category 
$\mathcal {C}'$ is equivalent to a full subcategory of the category of graded 
$R$-bimodules.
We also define the full subcategory 
$\mathcal {C}$ of 
$\mathcal {C}'$ as follows. For 
$M\in \mathcal {C}'$, 
$M\in \mathcal {C}$ if and only if 
$M$ is finitely generated as an 
$R$-bimodule and flat as a right 
$R$-module. Let 
$M\in \mathcal {C}$. Since 
$M$ is flat as a right 
$R$-module, it is torsion-free. Hence we have 
$M\hookrightarrow M\otimes _R Q = {\bigoplus} _{w\in W}\, M_Q^{w}$. In particular, 
$M$ is torsion-free as a left 
$R$-module. As in Remark 2.2, 
$f\notin R\setminus \{0\}$ acts invertibly from the left on 
$M\otimes _R Q$. Hence we have a homomorphism 
$Q\otimes _R M\hookrightarrow M\otimes _R Q$. It is not difficult to see that this is an isomorphism.
For 
$w\in W$, let 
$R_w$ be an object of 
$\mathcal {C}$ defined as follows. As a left 
$R$-module, 
$R_w = R$ and the bimodule structure is given by 
$mf = w(f)m$ for 
$m\in R_w$ and 
$f\in R$. The module 
$(R_w)_Q^{x}$ is given by
\[ (R_w)_Q^{x} = \begin{cases} Q & (x = w),\\ 0 & (x\ne w). \end{cases} \]
We denote 
$Q_w = R_w\otimes _R Q$ which is a 
$Q$-bimodule.
Let 
$M\in \mathcal {C}$.
• For
$m\in M$, let $m_w$ be the image of $m$ under $M\to {\bigoplus} _{w\in W}\, M_Q^{w}\twoheadrightarrow M_Q^{w}$.• We set
$M_Q = M\otimes _R Q$.• For
$I\subset W$, let $M_I$ (respectively, $M^{I}$) be the inverse image of ${\bigoplus} _{w\in I}\, M_Q^{w}$ in $M$ (respectively, the image of $M$ in ${\bigoplus} _{w\in I}\, M_Q^{w}$). We regard $M_I\subset M^{I}\subset {\bigoplus} _{w\in I}\, M_Q^{w}$.• For
$w\in W$, we write $M_w$ (respectively, $M^{w}$) for $M_{\{w\}}$ (respectively, $M^{\{w\}}$).• We write
$\operatorname {supp}_W(M) = \{w\in W\mid M_Q^{w}\ne 0\}$ and $\operatorname {supp}_W(m) = \{w\in W\mid m_w \ne 0\}$ for $m\in M$.
We have 
$M_I = \{m\in M\mid \operatorname {supp}_W(m)\subset I\}$.
Lemma 2.4 The modules 
$M_I$ and 
$M^{I}$ are both objects in 
$\mathcal {C}'$ such that 
$(M_I)_Q \simeq (M^{I})_Q \simeq {\bigoplus} _{w\in I}\, M_Q^{w}$.
Proof. Since 
$M_I\hookrightarrow M^{I}\subset {\bigoplus} _{w\in I}\, M_Q^{w}$, we have 
$M_I\otimes Q\hookrightarrow M^{I}\otimes Q\subset {\bigoplus} _{w\in I}\, M_Q^{w}$. Hence it is sufficient to prove that 
$M_I\otimes Q\hookrightarrow {\bigoplus} _{w\in I}\, M_Q^{w}$ is surjective. Let 
$m\in {\bigoplus} _{w\in I}\, M_Q^{w} \subset M_Q$ and take 
$0\ne f\in R$ such that 
$fm \in M$. Then we have 
$fm_w = (fm)_w = 0$ for any 
$w\in W\setminus I$. Since 
$M_Q^{w}$ is torsion-free, 
$m_w = 0$. Therefore 
$m\in (M_I)_Q$.
The following lemma is clear from the definitions.
Lemma 2.5 Let 
$I\subset W$ and 
$M,N\in \mathcal {C}$ such that 
$\operatorname {supp}_W(N)\subset I$. Then we have 
$\operatorname {Hom}_{\mathcal {C}'}(M,N)\simeq \operatorname {Hom}_{\mathcal {C}'}(M^{I},N)$ and 
$\operatorname {Hom}_{\mathcal {C}'}(N,M)\simeq \operatorname {Hom}_{\mathcal {C}'}(N,M_I)$.
Lemma 2.6 Any 
$M\in \mathcal {C}$ is finitely generated as a left (respectively, right) 
$R$-module.
Proof. We only prove that 
$M$ is finitely generated as a left 
$R$-module. Since 
$M^{w}$ is a quotient of 
$M$, this is also a finitely generated 
$R$-bimodule. The formula 
$mf = w(f)m$ for 
$f\in R$, 
$m\in M^{w}$ says that 
$M^{w}$ is also finitely generated as a left 
$R$-module. Since 
$M\hookrightarrow {\bigoplus} _{w\in W}\, M^{w}$ and 
$M^{w} \ne 0$ only for finite 
$w$, 
$M$ is a finitely generated left 
$R$-module.
We define a tensor product 
$M\otimes N$ of 
$M = (M,(M_Q^{w})_{w\in W},\xi _M),(N,(N_Q^{w})_{w\in W},\xi _N)\in \mathcal {C}$ as follows. As an 
$R$-bimodule, 
$M\otimes N = M\otimes _R N$; put 
$(M\otimes N)_Q^{w} = {\bigoplus} _{xy = w}\, M_Q^{x}\otimes _Q N_Q^{y}$. An isomorphism 
$\xi _{M\otimes N}\colon (M\otimes _{R} N)_Q\simeq {\bigoplus} _{w\in W}\, (M\otimes N)_Q^{w}$ is naturally defined as 
$M\otimes _{R} N\otimes _{R}Q \simeq M\otimes _{R}Q\otimes _{Q}N\otimes _{R}Q \simeq M_Q\otimes _{Q}N_Q \simeq {\bigoplus} _{x,y\in W}\, M_Q^{x}\otimes _{Q} N_Q^{y}$ 
${\bigoplus} _{w\in W}\,(M\otimes N)_Q^{w}$ where we used the fact that 
$N_Q$ is a 
$Q$-bimodule. This gives a monoidal category structure on 
$\mathcal {C}$. The unit object is 
$R_e$ where 
$e$ is the unit element of 
$W$. The following result is obvious from the definition.
Lemma 2.7 We have 
$\operatorname {supp}_W(M\otimes N) = \{xy \mid x\in \operatorname {supp}_W(M),y\in \operatorname {supp}_W(N)\}$.
2.3 Notes on gradings
Let 
$M = {\bigoplus} _{i\in \mathbb {Z}}\,M^{i}$ be a graded left 
$R$-module, right 
$R$-module or 
$R$-bimodule. The grading shift 
$M(1)$ is defined as 
$M(1)^{i} = M^{i + 1}$. Therefore, for an element 
$f\in R$ of degree 
$k$, the multiplication 
$m\mapsto fm$ gives a degree-zero homomorphism 
$M\to M(k)$ and the submodule 
$fR$ is isomorphic to 
$R(-k)$. If 
$M$ is a graded free left 
$R$-module, namely 
$M\simeq {\bigoplus} _i\, R(n_i)$ for 
$n_i\in \mathbb {Z}$, we define its graded rank 
$\operatorname {grk}(M)$ by 
$\operatorname {grk}(M) = \sum _i v^{n_i}$ where 
$v$ is an indeterminate. Obviously we have 
$\operatorname {grk}(M_1\oplus M_2) = \operatorname {grk}(M_1) + \operatorname {grk}(M_2)$ and 
$\operatorname {grk}(M(1)) = v\operatorname {grk}(M)$. If 
$M$ has a basis 
$\{m_i\}$, then 
$\operatorname {grk}(M) = \sum _i v^{-\deg (m_i)}$.
Lemma 2.8 Let 
$M\in \mathcal {C}$, 
$w\in W$ and assume that 
$M^{w}$ is a graded free left 
$R$-module. Then 
$M^{w}\simeq {\bigoplus} _i\, R_w(n_i)$ for some 
$n_i\in \mathbb {Z}$ as an object in 
$\mathcal {C}$.
Proof. The assumption says that we have an isomorphism 
$M^{w}\simeq {\bigoplus} _i\, R_w(n_i)$ as left 
$R$-modules for some 
$n_i\in \mathbb {Z}$. Since the right action of 
$f\in R$ is equal to the left action of 
$w(f)$ on both sides, this is an isomorphism as 
$R$-bimodules.
Lemma 2.9 Assume that 
$\mathbb {K}$ is a field. Let 
$M$ be a finitely generated graded free left 
$R$-module with the graded rank 
$p$ and 
$0 = M_0\subset M_1\subset M_2\subset \cdots \subset M_r = M$ be a filtration as graded left 
$R$-modules. Assume that there exists a graded free left 
$R$-submodule 
$N_i\subset M_i/M_{i - 1}$ with the graded rank 
$q^{(i)}$ such that 
$\sum _i q^{(i)} = p$. Then we have 
$N_i = M_i/M_{i - 1}$.
Proof. Let 
$N_i^{l}$ (respectively, 
$M_i^{l}$, 
$M^{l}$) be the 
$l$th graded piece of 
$N_i$. Then we have 
$\sum _i\dim N_i^{l} = \dim M^{l}$ by the assumption and we have 
$\dim M^{l} = \sum \dim (M_i/M_{i - 1})^{l}$. Hence 
$N_i^{l} = (M_i/M_{i - 1})^{l}$ for any 
$l$. Therefore we have 
$N_i = M_i/M_{i - 1}$.
For 
$M,N\in \mathcal {C}'$, we put 
$\operatorname {Hom}_{\mathcal {C}'}^{\bullet }(M,N) = {\bigoplus} _{n\in \mathbb {Z}}\,\operatorname {Hom}_{\mathcal {C}'}(M,N(n))$. This has a natural structure of a graded 
$R$-bimodule defined by 
$(afb)(m) = af(bm)$ where 
$a,b\in R$, 
$f\in \operatorname {Hom}_{\mathcal {C}'}^{\bullet }(M,N)$ and 
$m\in M$. We also use a similar notation for other situations like the category of graded left 
$R$-modules.
2.4 Soergel bimodules
Let 
$s\in S$ and set 
$R^{s} = \{f\in R\mid s(f) = f\}$. We define 
$B_s\in \mathcal {C}$ by 
$B_s = R\otimes _{R^{s}}R(1)$. A structure as an object in 
$\mathcal {C}$ (namely, 
$Q$-bimodules 
$(B_s)_Q^{w}$ and 
$\xi _{B_s}$) is determined by the condition 
$\operatorname {supp}_W(B_s) = \{e,s\}$ up to isomorphisms. Fix 
$\delta _s\in V$ such that 
$\langle \alpha _s^{\vee },\delta _s\rangle = 1$. Then an explicit description of the 
$Q$-bimodule 
$(B_s)_Q^{w}$ is
\begin{equation} \begin{aligned} (B_s)_Q^{e} & = Q(\delta_s \otimes 1 - 1 \otimes s(\delta_s))\simeq Q_e,\\ (B_s)_Q^{s} & = Q(\delta_s \otimes 1 - 1 \otimes \delta_s)\simeq Q_s. \end{aligned} \end{equation}
The projection 
$B_s\to (B_s)_Q^{e}$ (respectively, 
$B_s\to (B_s)_Q^{s}$) is given by 
$f\otimes g\mapsto fg$ (respectively, 
$f\otimes g\mapsto fs(g)$). To prove this description is straightforward using the following lemma.
Lemma 2.10 [Reference Elias and WilliamsonEW16, Claim 3.11]
We have 
$R = R^{s}\oplus \delta _s R^{s}$. Therefore 
$B_s \simeq R(1)\oplus R(-1)$ as left or right 
$R$-modules.
Let 
$M\in \mathcal {C}$. Then we have 
$B_s\otimes M\in \mathcal {C}$. We have 
$(B_s\otimes M)_Q^{w} = {\bigoplus} _{xy = w}\,(B_s)_Q^{x}\otimes M_Q^{y} = Q_e\otimes _Q M_Q^{w}\oplus Q_s\otimes _Q M_Q^{sw}$. We have 
$Q_e\otimes M_Q^{w}\simeq M_Q^{w}$ and 
$Q_s\otimes M_Q^{sw}\simeq M_Q^{sw}$ as right 
$Q$-modules. As left 
$Q$-modules, we have 
$Q_e\otimes M_Q^{w}\simeq M_Q^{w}$ but 
$Q_s\otimes M_Q^{sw}$ is not isomorphic to 
$M_Q^{sw}$. The action on the left-hand side is twisted by 
$s$, that is, the action of 
$f\in R$ on 
$Q_s\otimes M_Q^{sw}$ is equal to the action of 
$s(f)$ on 
$M_Q^{sw}$.
As 
$R$-bimodules, 
$B_s\otimes M = R\otimes _{R^{s}}M$. In terms of this description, 
$(B_s\otimes M)_Q^{w}$ is given as follows.
Lemma 2.11 Let 
$s\in S$ and 
$M\in \mathcal {C}$.
(1) The projection
$R\otimes _{R^{s}}M = B_s\otimes M\to (B_s\otimes M)_Q^{w} = M_Q^{w}\oplus M_Q^{sw}$ is given by $f\otimes m\mapsto (fm_w,s(f)m_{sw})$.(2) We have
\begin{align*} (B_s\otimes M)_Q^{w} = \{(\delta _s\otimes m - 1\otimes s(\delta _s)m) + (\delta _s\otimes m' - 1\otimes \delta _s m')\mid m\in M_Q^{w},m'\in M_Q^{sw}\}. \end{align*}
The following result is straightforward, for example from the above lemma.
Lemma 2.12 Let 
$s\in S$ and 
$M\in \mathcal {C}$. We have 
$(B_s\otimes M)_{Q}^{w} + (B_s\otimes M)_{Q}^{sw} = B_s\otimes _R M_Q^{w} + B_s\otimes _R M_Q^{sw}$.
Let 
$\mathcal {BS}$ be the full subcategory of 
$\mathcal {C}$ such that the objects are 
$\{B_{s_1}\otimes \cdots \otimes B_{s_l}(n)\mid s_1,\ldots ,s_l\in S,n\in \mathbb {Z}\}$ and the category 
$\mathcal {S}\mathrm {Bimod}$ is defined as a full subcategory of 
$\mathcal {C}$ whose objects are direct summands of objects in the category 
$\mathcal {BS}$. Obviously these categories are stable under the tensor products.
For 
$\underline {x} = (s_1,\ldots ,s_l)\in S^{l}$, we put 
$B_{\underline {x}} = B_{s_1}\otimes \cdots \otimes B_{s_l}$. By Lemma 2.7, we get the following result.
Lemma 2.13 We have 
$\operatorname {supp}_W(B_{\underline {x}}) = \{s_1^{e_1}\cdots s_l^{e_l}\mid e_i \in \{0,1\}\}$. In particular, if 
$\underline {x}$ is a reduced expression for 
$x\in W$, then 
$\operatorname {supp}_W(B_{\underline {x}}) = \{y\in W\mid y\le x\}$.
Since 
$B_s\otimes M \simeq M(-1) \oplus M(1)$ (respectively, 
$M\otimes B_s\simeq M(-1)\oplus M(1)$) as a right (respectively, left) 
$R$-module, we get the following result.
Lemma 2.14 The module 
$B_{\underline {x}}$ is graded free as a left (respectively, right) 
$R$-module and its graded rank is 
$(v + v^{-1})^{l}$.
Lemma 2.15 Let 
$M,N\in \mathcal {C}$ and 
$s\in S$. Then 
$\operatorname {Hom}_{\mathcal {C}}(B_s\otimes M,N)\simeq \operatorname {Hom}_{\mathcal {C}}(M,B_s\otimes N)$.
Proof. We regard 
$B_s\otimes M = R\otimes _{R^{s}}M(1)$ and 
$B_s\otimes N = R\otimes _{R^{s}}N(1)$. Take 
$\delta \in V$ such that 
$\langle \alpha _s^{\vee },\delta \rangle = 1$. Let 
$\varphi \colon B_s\otimes M\to N$ and define 
$\psi \colon M\to B_s\otimes N$ by 
$\psi (m) = 1\otimes \varphi (1\otimes \delta m) - s(\delta )\otimes \varphi (1\otimes m)$. Then, as in [Reference LibedinskyLib08, Lemma 3.3], 
$\psi$ is an 
$R$-bimodule homomorphism and this correspondence gives an isomorphism between the space of 
$R$-bimodule homomorphisms. We prove that this correspondence preserves homomorphisms in 
$\mathcal {C}$.
Let 
$a(m) = 1\otimes \delta m - s(\delta )\otimes m$ and 
$b(m) = 1\otimes \delta m - \delta \otimes m$. We have 
$a(m) = (\delta - (\delta + s(\delta )))\otimes m + 1\otimes \delta m $ 
$= \delta \otimes m - 1\otimes (\delta + s(\delta ))m + 1\otimes \delta m = \delta \otimes m - 1\otimes s(\delta )m$. Hence 
$(B_s \otimes M)_Q^{w} = a(M_Q^{w}) + b(M_Q^{sw})$ by Lemma 2.11. By the definition of 
$\psi$ and Lemma 2.11, the image of 
$\psi (m)$ in 
$(B_s\otimes N)_Q^{w}$ is 
$(\varphi (a(m))_w,\varphi (b(m))_{sw})$. Therefore 
$\psi (m)$ is a homomorphism in 
$\mathcal {C}$ if and only if 
$\varphi (a(M_Q^{w}))_{y} = 0$ and 
$\varphi (b(M_Q^{w}))_{sy} = 0$ for any 
$w,y\in W$ such that 
$y\ne w$, if and only if 
$\varphi (a(M_Q^{w}))\subset N_Q^{w}$ and 
$\varphi (b(M_Q^{sw}))\subset N_Q^{w}$ for any 
$w\in W$, if and only if 
$\varphi$ is a homomorphism in 
$\mathcal {C}$.
2.5 The Hecke algebra
In this paper we use the following definition of the Hecke algebra. Let 
$v$ be an indeterminate. The 
$\mathbb {Z}[v^{\pm 1}]$-algebra 
$\mathcal {H}$ is generated by 
$\{H_w\mid w\in W\}$ and defined by the following relations.
•
$(H_s - v^{-1})(H_s + v) = 0$ for any $s\in S$.• If
$\ell (w_1) + \ell (w_2) = \ell (w_1w_2)$ for $w_1,w_2\in W$, we have $H_{w_1w_2} = H_{w_1}H_{w_2}$.
For 
$s\in S$, we put 
$\underline {H}_s = H_s + v$, and for 
$\underline {x} = (s_1,\ldots ,s_l)\in S^{l}$, we put 
$\underline {H}_{\underline {x}} = \underline {H}_{s_1}\cdots \underline {H}_{s_l}$.
It is known that 
$\{H_w\mid w\in W\}$ is a basis of the 
$\mathbb {Z}[v^{\pm 1}]$-module 
$\mathcal {H}$. We define 
$p_{\underline {x}}^{w}\in \mathbb {Z}[v^{\pm 1}]$ by 
$\underline {H}_{\underline {x}} = \sum _{w\in W}p_{\underline {x}}^{w} H_w$.
Lemma 2.16 We have 
$\sum _{w\in W}v^{\ell (w)}p_{\underline {x}}^{w}(v^{-1}) = (v + v^{-1})^{l}$.
Proof. By the defining relations, 
$H_w\mapsto v^{-\ell (w)}$ gives an algebra homomorphism 
$\mathcal {H}\to \mathbb {Z}[v^{\pm 1}]$. Applying this homomorphism to 
$\underline {H}_{\underline {x}} = \sum _{w\in W}p_{\underline {x}}^{w} H_w$, we get 
$(v + v^{-1})^{l} = \sum _{w\in W}p_{\underline {x}}^{w}(v)v^{-\ell (w)}$. Replacing 
$v$ with 
$v^{-1}$, we get the lemma.
Lemma 2.17 The dimension of 
$B_{\underline {x}}^{w} \otimes Q$ as a 
$Q$-vector space is 
$p_{\underline {x}}^{w}(1)$.
Proof. After specializing 
$v$ to 
$1$, 
$\mathcal {H}$ is isomorphic to the group algebra 
$\mathbb {Z}[W]$ via 
$\mathcal {H}\ni H_w\mapsto w\in \mathbb {Z}[W]$. Therefore we have 
$(s_1 + 1)\cdots (s_l + 1) = \sum _w p_{(s_1,\ldots ,s_l)}^{w}(1)w$. Hence we have
\[ p_{(s,s_1,\ldots,s_l)}^{w}(1) = p_{(s_1,\ldots,s_l)}^{w}(1) + p_{(s_1,\ldots,s_l)}^{sw}(1). \]
On the other hand, we have 
$(B_s\otimes M)_Q^{w} = M_Q^{w}\oplus M_Q^{sw}$ for any 
$M\in \mathcal {C}$. Hence we have 
$\dim _Q(B_s\otimes M)^{w}_Q = \dim _Q(M_Q^{w}) + \dim _Q (M_Q^{sw})$. Now we get the lemma by induction on the length of 
$\underline {x}$.
2.6 Duality
Let 
$M\in \mathcal {C}$. Define a new module 
$D(M)\in \mathcal {C}'$ by
\begin{align*} D(M) & = \operatorname{Hom}^{{\bullet}}_{\text{-}R}(M,R),\\ D(M)_Q^{w} & = \operatorname{Hom}_{\text{-}Q}(M_Q^{w},Q). \end{align*}
Here 
$\operatorname {Hom}^{\bullet }_{\text{-}R}$ means the space of homomorphisms as right 
$R$-modules and the 
$R$-bimodule structure is given by 
$(afb)(m) = f(amb)$ for 
$f\in D(M)$, 
$a,b\in R$ and 
$m\in M$. Since 
$M$ is a finitely generated right 
$R$-module, we have 
$D(M)\otimes Q = \operatorname {Hom}_{\text{-}Q}(M_Q,Q) = {\bigoplus} _{w\in W}\, D(M)_Q^{w}$. Therefore 
$D(M)\in \mathcal {C}'$. For 
$\varphi \in D(M)$ and 
$w\in W$, 
$\varphi _w$ is the restriction of 
$\varphi \otimes \mathrm {Id}\colon M\otimes _{R}Q \simeq {\bigoplus} _{w\in W}\, M_Q^{w} \to Q$ to 
$M_Q^{w}$.
Lemma 2.18 We have 
$D(M^{I})\simeq D(M)_I$ for any 
$I\subset W$.
Proof. Since 
$M^{I}$ is a quotient of 
$M$, we have that 
$D(M^{I})\subset D(M)$ and 
$\psi \in D(M)$ is in 
$D(M^{I})$ if and only if 
$\psi$ is zero on 
$M_{W\setminus I}$. Since 
$R$ is an integral domain, 
$\psi$ is zero on 
$M_{W\setminus I}$ if and only if 
$\mathrm {Id}\otimes \psi$ is zero on 
$M_{W\setminus I}\otimes Q = {\bigoplus} _{w\in W\setminus I}\,M_Q^{w}$. That is, 
$\psi \in D(M)_I$.
Lemma 2.19 Let 
$M\in \mathcal {C}$ and 
$w\in W$ and assume that 
$M^{w}$ is graded free as a left 
$R$-module.
(1) The module
$D(M)_w$ is also a graded free left $R$-module and its graded rank is given by $\operatorname {grk}(D(M)_w)(v) = \operatorname {grk}(M^{w})(v^{-1})$.(2) We have
$D(D(M)_w)\simeq M^{w}$.
Proof. We may assume 
$M^{w} = R_w(n)$ for some 
$n\in \mathbb {Z}$. Then 
$D(R_w(n))\simeq R_w(-n)$. Since 
$D(M^{w})\simeq D(M)_w$, we get the first part. We also have 
$D(D(M^{w}))\simeq M^{w}$. Therefore we have 
$D(D(M)_w)\simeq D(D(M^{w}))\simeq M^{w}$.
Lemma 2.20 We have 
$D(B_s\otimes M)\simeq B_s\otimes D(M)$ for any 
$M\in \mathcal {C}$ and 
$s\in S$. In particular, we have that 
$D(B_{\underline {x}})\simeq B_{\underline {x}}$ for any 
$\underline {x}\in S^{l}$ and 
$D$ preserves 
$\mathcal {BS}$ and 
$\mathcal {S}\mathrm {Bimod}$.
Proof. We regard 
$B_s\otimes (\cdot ) = R\otimes _{R^{s}}(\cdot )$.
Take 
$\delta \in V$ such that 
$\langle \alpha _s^{\vee },\delta \rangle = 1$. For 
$\varphi \in D(B_s\otimes M)$, define 
$\varphi _1,\varphi _2\colon M\to R$ by 
$\varphi _1(m) = \varphi (1\otimes m)$ and 
$\varphi _2(m) = \varphi (\delta \otimes m)$. It is clear that these are right 
$R$-homomorphisms, hence define elements of 
$D(M)$. Set 
$\Phi (\varphi ) = 1\otimes \varphi _2 - s(\delta )\otimes \varphi _1\in B_s\otimes D(M)$.
We prove that 
$\Phi \colon D(B_s\otimes M)\to B_s\otimes D(M)$ is an 
$R$-bimodule homomorphism. Clearly we have 
$(\varphi f)_1 = \varphi _1 f$ and 
$(\varphi f)_2 = \varphi _2 f$, hence 
$\Phi$ is a right 
$R$-module homomorphism. If 
$f\in R^{s}$, then 
$(f\varphi )_1(m) = \varphi (f\otimes m) = \varphi (1\otimes fm) = (f(\varphi )_1)(m)$. Hence 
$(f\varphi )_1 = f\varphi _1$, and similarly we have 
$(f\varphi )_2 = f\varphi _2$. Therefore 
$\Phi (f\varphi ) = 1\otimes f\varphi _2 - s(\delta )\otimes f\varphi _1 = f(1\otimes \varphi _2 - s(\delta )\otimes \varphi _1) = f\Phi (\varphi )$. We prove 
$\Phi (\delta \varphi ) = \delta \Phi (\varphi )$. We have
\begin{align*} ((\delta \varphi )_2)(m) &= \varphi (\delta ^{2}\otimes m) \!=\! \varphi (\delta (\delta + s(\delta ))\otimes m \!- \delta s(\delta )\otimes m) \!=\! \varphi (\delta \otimes (\delta \!+ s(\delta ))m) - \varphi (1\otimes \delta s(\delta )m)\\ & = \varphi _2((\delta + s(\delta ))m) - \varphi _1(\delta s(\delta )m)= ((\delta + s(\delta ))\varphi _2 - \delta s(\delta )\varphi _1)(m). \end{align*}
We also have 
$(\delta \varphi )_1 = \varphi _2$. Hence
\begin{align*} \Phi (\delta \varphi ) &= 1\otimes ((\delta + s(\delta ))\varphi _2 - \delta s(\delta )\varphi _1) - s(\delta )\otimes \varphi _2 = (\delta + s(\delta ))\otimes \varphi _2 - \delta s(\delta )\otimes \varphi _1 - s(\delta )\otimes \varphi _2\\ &= \delta (1\otimes \varphi _2 - s(\delta )\otimes \varphi _1) = \delta \Phi (\varphi ). \end{align*}
By Lemma 2.10, 
$\Phi$ is a left 
$R$-module, hence an 
$R$-bimodule homomorphism.
We prove that 
$\Phi$ is a morphism in 
$\mathcal {C}'$. Let 
$\varphi \in D(B_s\otimes M)_Q^{w}$. Since 
$B_s\otimes M_Q^{y}\subset (B_s\otimes M)_Q^{y} + (B_s\otimes M)_Q^{sy}$ by Lemma 2.12, 
$\varphi _1,\varphi _2$ is zero on 
$M_Q^{y}$ if 
$y\ne w,sw$. Hence 
$\operatorname {supp}_W(\varphi _1),\operatorname {supp}_W(\varphi _2)\subset \{w,sw\}$. Therefore 
$\operatorname {supp}_W(\Phi (\varphi ))\subset \{w,sw\}$.
The homomorphism 
$\varphi$ is zero on 
$(B_s\otimes M)^{sw}_Q$. By Lemma 2.11, 
$\varphi (\delta \otimes m - 1\otimes s(\delta ) m) = 0$ for 
$m\in M_Q^{sw}$. That is, we have 
$(\varphi _2)_{sw} = s(\delta )(\varphi _1)_{sw}$. Similarly, we have 
$(\varphi _2)_w = \delta (\varphi _1)_w$. The image of 
$\Phi (\varphi )$ in 
$(B_s\otimes D(M))_Q^{sw} = D(M)_Q^{sw}\oplus D(M)_Q^{w}$ is 
$((\varphi _2)_{sw} - s(\delta )(\varphi _1)_{sw},(\varphi _2)_w - \delta (\varphi _1)_w)$ and this is zero. Therefore 
$\operatorname {supp}_W(\Phi (\varphi ))\subset \{w\}$.
Since 
$B_s = 1\otimes R\oplus \delta _s\otimes R$, we have 
$D(B_s\otimes M) = D(1\otimes M\oplus \delta _s\otimes M) = D(1\otimes M)\oplus D(\delta _s\otimes M)$. Then 
$\Phi$ sends 
$D(1\otimes M)$ (respectively, 
$D(\delta _s\otimes M)$) to 
$s(\delta )\otimes D(M)$ (respectively, 
$1\otimes D(M)$). Since 
$B_s\otimes D(M) = s(\delta )\otimes D(M)\oplus 1\otimes D(M)$, we have proved that 
$\Phi$ is an isomorphism.
3. Light leaves
In this section we construct light leaves. These are certain homomorphisms between objects of the form 
$B_{\underline {x}}$ where 
$\underline {x}$ is a sequence of elements in 
$S$. They were constructed by Libedinsky [Reference LibedinskyLib08] first in the case of the original Soergel bimodules and also constructed by Elias and Williamson [Reference Elias and WilliamsonEW16] for the diagrammatic category. We follow their constructions and prove properties of such homomorphisms. In particular, we give a basis of 
$B_{\underline {x}}^{w}$. In the next section, using this basis, we calculate the character (which will be defined in the next section) of 
$B_{\underline {x}}$. This is key to the proof of our categorification theorem.
3.1 Notation
Let 
$\underline {x} \in S^{l}$ and 
$\boldsymbol {e} = (\boldsymbol {e}_1,\ldots ,\boldsymbol {e}_l)\in \{0,1\}^{l}$. We set 
$\underline {x}^{\boldsymbol {e}} = s_1^{\boldsymbol {e}_1}\ldots s_l^{\boldsymbol {e}_l}\in W$. Let 
$x_0 = 1,x_1 = s_1^{\boldsymbol {e}_1}, x_2 = s_1^{\boldsymbol {e}_1}s_2^{\boldsymbol {e}_2},\ldots ,x_l = s_1^{\boldsymbol {e}_1}\cdots s_l^{\boldsymbol {e}_l} = \underline {x}^{\boldsymbol {e}}$. Using this sequence, we add a label to 
$\boldsymbol {e}$ at each index. We assign 
$U$ to the index 
$i$ if 
$x_{i - 1}s_i > x_{i - 1}$ and 
$D$ otherwise. The defect 
$d(\boldsymbol {e})$ of 
$\boldsymbol {e}$ is defined by
\[ d(\boldsymbol{e}) = \#\{i\mid \text{the label is }U\text{ at }i, \boldsymbol{e}_i = 0\} - \#\{i\mid \text{the label is }D\text{ at }i, \boldsymbol{e}_i = 0\}. \]
Of course, this number depends on 
$\underline {x}$, not just on 
$\boldsymbol {e}$.
Lemma 3.1 [Reference Elias and WilliamsonEW16, Lemma 2.7]
We have 
$p_{\underline {x}}^{w}(v) = \sum _{\underline {x}^{\boldsymbol {e}} = w}v^{d(\boldsymbol {e})}$.
3.2 Assumption
For 
$\underline {x}\in S^{l}$, define 
$u_{\underline {x}}\in B_{\underline {x}}$ by 
$u_{\underline {x}} = (1\otimes 1)\otimes (1\otimes 1)\otimes \cdots \otimes (1\otimes 1)$. To construct the light leaves, we need the following assumption.
Assumption 3.2 For any 
$s,t\in S$ such that 
$s\ne t$ and the order 
$m$ of 
$st$ is finite, we have the following assertion. Set 
$\underline {x} = (s,t,\ldots )\in S^{m}$ and 
$\underline {y} = (t,s,\ldots )\in S^{m}$. Then there exists a degree-zero morphism 
$B_{\underline {x}}\to B_{\underline {y}}$ in 
$\mathcal {C}$ which sends 
$u_{\underline {x}}$ to 
$u_{\underline {y}}$.
We retain this assumption throughout the rest of this paper. It seems to be difficult to check this condition directly. We have the following sufficient condition.
Lemma 3.3 If the following condition holds for any 
$s,t\in S$ such that 
$st$ is of finite order, Assumption 3.2 holds.
• Let
$T'$ be the reflections in the group $\langle s,t\rangle$ generated by $\{s,t\}$. For any $t_1,t_2\in T'$ with $t_1\ne t_2$, there exists $v\in V$ such that $\langle v,\alpha _{t_1}^{\vee } \rangle = 0$, $\langle v,\alpha _{t_2}^{\vee }\rangle = 1$.
Assume that 
$\mathbb {K}$ is a field and the representation 
$V$ restricted to 
$\langle s,t\rangle$ is reflection faithful. Then Soergel's categorification theorem holds for 
$\langle s,t\rangle$. Let 
$w_0$ be the longest element in 
$\langle s,t\rangle$. Then there exists a corresponding indecomposable object 
$B(w_0)$ which is a direct summand of 
$B_{\underline {x}}$ and 
$B_{\underline {y}}$. Therefore we have a map 
$B_{\underline {x}}\twoheadrightarrow B(w_0)\hookrightarrow B_{\underline {y}}$, and after suitable normalization this gives the desired map of the assumption. The proof of the above lemma basically follows this argument. In the proof below, we follow Soergel's argument in [Reference SoergelSoe07].
In the rest of this subsection let 
$s,t\in S$ be as in the lemma and set 
$W' = \langle s,t\rangle$. Let 
$T'$ be the set of reflections in 
$W'$.
Lemma 3.4 The representation 
$V$ restricted to 
$W'$ is faithful.
Proof. Let 
$w\in W'$ and assume that 
$w$ acts trivially on 
$V$. Assume that the length of 
$w$ is odd. If 
$\ell (w) > 1$, then, for 
$s' = s$ or 
$t$, we have 
$\ell (s'ws') < \ell (w)$ and 
$s'ws'$ is also acts trivially on 
$V$. Therefore we conclude that 
$s$ or 
$t$ acts trivially. This is a contradiction to 
$\alpha _{s'}^{\vee },\alpha _{s'}\ne 0$ for 
$s'\in \{s,t\}$. Now assume that 
$\ell (w)$ is even. Then there exist 
$t_1,t_2\in T'$ such that 
$w = t_1t_2$. If 
$w\ne 1$, then 
$t_1\ne t_2$. Take 
$v\in V$ as in the assumption of Lemma 3.3. Then 
$t_1(v) = v$ and 
$t_2(v) = v - \alpha _{t_2}\ne v$. Therefore 
$t_1(v)\ne t_2(v)$. Hence the action of 
$w$ is not trivial.
From this lemma it is sufficient to prove the existence of an 
$R$-bimodule homomorphism in the assumption.
We also note the following: if 
$t_1,t_2\in T'$ are different, then 
$\{\alpha _{t_1}^{\vee },\alpha _{t_2}^{\vee }\}$ is linearly independent. That is, 
$\{\alpha _{t_1}^{\vee },\alpha _{t_2}^{\vee }\}$ generates the two-dimensional space over 
$\operatorname {Frac}(\mathbb {K})$. Therefore 
$\operatorname {Frac}(\mathbb{K})\alpha_{t_1}^{\vee} + \operatorname {Frac}(\mathbb{K})\alpha_{t_2}^{\vee} = \operatorname {Frac}(\mathbb{K})\alpha_{s}^{\vee} + \operatorname {Frac}(\mathbb{K})\alpha_{t}^{\vee}$. That is, the two-dimensional space does not depend on 
$t_1,t_2$.
For each 
$x\in W'$, set 
$\operatorname {Gr}(x) = \{(v,x^{-1}(v))\in V^{*}\times V^{*}\mid v\in V^{*}\}$, and for each 
$A\subset W'$, let 
$R(A)$ be the space of regular functions on 
${\bigcup} _{x\in A}\operatorname {Gr}(x)$. In other words, 
$R(A)$ is the image of 
$R\otimes _{\mathbb {K}} R$ in 
$R^{A}$ under the map 
$f\otimes g\mapsto (fx(g))_{x\in A}$; here we use an identification 
$\operatorname {Gr}(x)\simeq V^{*}$ via 
$(v,x^{-1}(v))\mapsto v$. The embedding is an isomorphism after tensoring 
$Q$, that is, we have 
$R(A)\otimes Q\simeq Q^{A}$ and this decomposition gives the structure of an object of 
$\mathcal {C}'$ to 
$R(A)$. In this subsection we consider an object of 
$\mathcal {C}'$ as graded right 
$R\otimes _{\mathbb {K}}R$-modules, rather than graded 
$R$-bimodules.
The element 
$s\in S$ acts on 
$V^{*}\times V^{*}$ by 
$(v_1,v_2)\mapsto (v_1,s(v_2))$. If 
$A$ is stable under the right action of 
$s$, then 
${\bigcup} _{x\in A}\operatorname {Gr}(x)$ is stable under this action. Hence 
$s$ acts on 
$R(A)$. We also have an action of 
$s$ on 
$R\otimes _{\mathbb {K}}R$ defined by 
$f\otimes g\mapsto f\otimes s(g)$. Then the natural map 
$R\otimes _{\mathbb {K}}R\to R(A)$ is 
$s$-equivariant. Let 
$R(A)^{+}$ be the image of 
$R\otimes _{\mathbb {K}} R^{s}$ in 
$R(A)$. Then 
$R(A)^{+}$ is contained in 
$s$-invariants in 
$R(A)$.
Lemma 3.5 Assume that 
$As = A$. Then 
$R(A)^{+}\otimes _{R^{s}}R\to R(A)$ given by 
$z\otimes f\mapsto z(1\otimes f)$ is an isomorphism.
Proof. It is obvious that the map is surjective. Therefore it is sufficient to prove that the map is injective. Take 
$\delta _s\in V$ such that 
$\langle \alpha _s^{\vee },\delta _s\rangle = 1$. Then we have 
$R = R^{s}\oplus R^{s}\delta _s$. Therefore it is sufficient to prove that 
$R(A)^{+}\cap R(A)^{+}(1\otimes \delta _s) = 0$.
Let 
$z_1,z_2\in R(A)^{+}$ and assume that 
$z_1 = z_2(1\otimes \delta _s)$. Let 
$((z_i)_x)$ be the image of 
$z_i$ in 
$R^{A}$ for 
$i = 1,2$. By 
$z_1 = z_2(1\otimes \delta _s)$, we have 
$(z_1)_x = (z_2)_{x}x(\delta _s)$. By replacing 
$x$ with 
$xs$, we have 
$(z_1)_{xs} = (z_2)_{xs}xs(\delta _s)$. Since 
$z_1,z_2$ are 
$s$-invariant, we have 
$(z_1)_{xs} = (z_1)_x$ and 
$(z_2)_{xs} = (z_2)_{x}$. Therefore we get 
$(z_2)_{x}(x(\delta _s) - xs(\delta _s)) = 0$. Hence 
$(z_2)_{x}x(\alpha _s) = 0$. We get 
$(z_2)_{x} = 0$ for any 
$x\in A$ and this implies 
$z_2 = 0$.
Lemma 3.6 Let 
$x\in W'$ and assume that 
$xs > x$. Set 
$A = \{y\in W'\mid y\le x\}$. Then 
$R(A)\otimes _{R}B_s\simeq R(A\cup As)(1)\oplus R(A\cap As)(-1)$.
Proof. Since 
$xs > x$, we have 
$A\setminus As = \{x,xr\}$ for some 
$r\in T'\setminus \{s\}$. Take 
$v_0\in V$ such that 
$\langle \alpha _r^{\vee },v_0\rangle = 0$ and 
$\langle \alpha _s^{\vee },v_0\rangle = 1$ and consider 
$z_0 = x(v_0)\otimes 1 - 1\otimes v_0$. Note that we have the restriction map 
$R(A)\to R(A\cap As)$.
Claim Let 
$f\in R(A)$. Then we have 
$fz_0 = 0$ in 
$R(A)$ if and only if 
$f = 0$ in 
$R(A\cap As)$.
Recall the embedding 
$R(A)\to R^{A}$. Let 
$(f_y)_{y\in A}$ be the image of 
$f$. Then the image of 
$fz_0$ is 
$(f_y(x(v_0) - y(v_0)))_{y\in A}$. When 
$y \in \{x,xr\} = A\setminus As$, it follows that 
$x(v_0) - y(v_0) = 0$ since 
$r(v_0) = v_0$ from the assumption on 
$v_0$. Therefore if 
$f = 0$ in 
$R(A\cap As)$, then since 
$f_y = 0$ for any 
$y\in A\cap As$, we have 
$f_y(x(v_0) - y(v_0)) = 0$ for any 
$y\in A$. We get 
$fz_0 = 0$ in 
$R(A)$.
We prove 
$x(v_0) - y(v_0) \ne 0$ for any 
$y\in A\cap As$. This gives the reverse implication. We assume that 
$y^{-1}x(v_0) = v_0$. Then we also have 
$y^{-1}xr(v_0) = v_0$. Either 
$y^{-1}x$ or 
$y^{-1}xr$ is in 
$T'$. Let 
$r'$ be the element in 
$T'$. Then 
$r(v_0) = r'(v_0) = v_0$ and since 
$y\ne x,rx$, we have 
$r\ne r'$. Therefore 
$\langle \alpha _r^{\vee },v_0\rangle = \langle \alpha _{r'}^{\vee },v_0\rangle = 0$. Hence 
$v_0$ is orthogonal to 
$\operatorname {Frac}(\mathbb {K})\alpha _r + \operatorname {Frac}(\mathbb {K})\alpha _{r'}$. Recall that this space does not depend on 
$r,r'$. Hence 
$v_0$ is orthogonal to 
$\operatorname {Frac}(\mathbb {K})\alpha _r + \operatorname {Frac}(\mathbb {K})\alpha _s$. This contradicts the assumption 
$\langle \alpha _s,v_0\rangle = 1$. We get the claim.
By the claim, we have an isomorphism 
$R(A\cap As)(-2)\simeq R(A)z_0$. Let 
$M$ be the sub-
$(R,R^{s})$-module of 
$R(A)$ generated by 
$z_0$. Then the above isomorphism implies 
$R(A\cap As)^{+} (-2)\simeq M$.
Next consider the restriction map 
$R(A\cup As)\to R(A)$. The subset 
$A\cup As$ is stable under the right multiplication of 
$s$. Let 
$R(A\cup As)^{s = 1}$ be the 
$s$-fixed part. Then 
$R(A\cup As)^{+}\subset R(A\cup As)^{s = 1}$ and the above restriction map gives an injective map 
$R(A\cup As)^{s = 1}\hookrightarrow R(A)$. Let 
$N$ be the image of 
$R(A\cup As)^{+}\hookrightarrow R(A)$. We have 
$N\simeq R(A\cup As)^{+}$.
To prove 
$R(A)\otimes _{R}B_s\simeq R(A\cup As)(1)\oplus R(A\cap As)(-1)$, it is sufficient to prove 
$R(A) = M\oplus N$ by Lemma 3.5. The subspace 
$M + N$ is the image of 
$(R\otimes R^{s})z_0 + R\otimes R^{s}$. Since 
$z_0 = x(v_0)\otimes 1 + 1 \otimes v_0$, we have 
$(R\otimes R^{s})z_0 + R\otimes R^{s} = R\otimes R^{s}v_0 + R\otimes R^{s}$. We have 
$R = R^{s} + R^{s}v_0$ since 
$\langle \alpha _s^{\vee },v_0\rangle = 1$. Hence 
$(R\otimes R^{s})z_0 + R\otimes R^{s} = R\otimes R$. Therefore 
$M + N = R(A)$.
Let 
$m\in M$ and 
$n\in N$. We can write 
$m = fz_0$ for some 
$f\in \operatorname {Im}(R\otimes R^{s}\to R(A))$. We assume 
$m = n$. Consider the image of 
$f$ in 
$R(A\cap As)$ and let 
$(f_y)_{y\in A\cap As}$ be the image in 
$R^{A\cap As}$. We also denote the image of 
$n$ in 
$R^{A\cap As}$ by 
$(n_y)_{y\in A\cap As}$. Then, for each 
$y\in A\cap As$, we have 
$f_y(x(v_0) - y(v_0)) = n_{y}$. We also have 
$f_{ys}(x(v_0) - ys(v_0)) = n_{ys}$. Since 
$f$ and 
$n$ are 
$s$-invariant, 
$f_{ys} = f_{y}$, 
$n_{ys} = n_y$. Therefore 
$f_{y}(ys(v_0) - y(v_0)) = 0$. That is, we have 
$f_{y}y(\alpha _s) = 0$. We get 
$f_y = 0$ and hence 
$fz_0\in M$ is zero.
We have proved 
$R(A)\otimes _{R}B_s\simeq R(A\cup As)(1)\oplus R(A\cap As)(-1)$. For 
$A = \{y\in W'\mid y\le x\}$ with 
$xs > x$, we have 
$A\cup As = \{y\in W'\mid y\le xs\}$. We set 
$R(\le x) = R(\{y\in W'\mid y\le x\})$ for 
$x\in W'$. Then 
$R(\le xs)(1)$ is a direct summand of 
$R(\le x)\otimes _{R}B_s$ if 
$xs > x$. Replacing 
$s$ with 
$t$, 
$R(\le yt)(1)$ is a direct summand of 
$R(\le y)\otimes _{R}B_s$ if 
$yt > y$. Therefore we have the following consequence. Let 
$\underline {w} \in \{s,t\}^{\ell (w)}$ be a reduced expression for 
$w\in W'$. Then 
$B_{\underline {w}}$ has a direct summand 
$R(\le w)(\ell (w))$. Let 
$w_0\in W'$ be the longest element. Then for 
$\underline {x}$ and 
$\underline {y}$ as in Assumption 3.2, 
$R(\le w_0)(\ell (w_0))$ is a direct summand of both 
$B_{\underline {x}}$ and 
$B_{\underline {y}}$. Therefore we get a non-zero homomorphism 
$\Phi \colon B_{\underline {x}}\twoheadrightarrow R(\le w_0)(\ell (w_0))\hookrightarrow B_{\underline {y}}$.
Lemma 3.7 The homomorphism 
$\Phi$ induces an isomorphism 
$B_{\underline {x}}^{w_0}\simeq B_{\underline {y}}^{w_0}$.
Proof. By induction, we can prove that 
$B_{\underline {w}}\simeq R(\le w)(\ell (w))\oplus M$ with 
$\operatorname {supp}_{W'}(M)\subset \{y \in W'\mid y < w\}$ for any 
$w\in W'$ and a reduced expression 
$\underline {w}$ for 
$w$. Therefore 
$B_{\underline {x}} = R(\le w_0)(\ell (w_0))\oplus M$ with 
$\operatorname {supp}_{W'}(M)\subset W'\setminus \{w_0\}$. Hence 
$B_{\underline {x}}^{w_0} = R(\le w_0)(\ell (w_0))^{w_0}$ since 
$M^{w_0} = 0$. Therefore the projection 
$B_{\underline {x}}\twoheadrightarrow R(\le w_0)(\ell (w_0))$ induces an isomorphism 
$B_{\underline {x}}^{w_0}\simeq R(\le w_0)(\ell (w_0))^{w_0}$. Similarly, the embedding 
$R(\le w_0)(\ell (w_0))\hookrightarrow B_{\underline {y}}$ induces an isomorphism 
$R(\le w_0)(\ell (w_0))^{w_0}\hookrightarrow B_{\underline {y}}^{w_0}$. We get the lemma by the construction of 
$\Phi$.
Define 
$\varphi _{\underline {x}}\colon B_{\underline {x}}\simeq R\otimes _{R^{s}}R\otimes _{R^{t}}\otimes \cdots \otimes R\to R(-\ell (w_0))$ by
\begin{align*} f_0\otimes f_1\otimes \ldots \otimes f_{\ell (w_0)}\mapsto f_0(s(f_1)) (st(f_2))\cdots (st\cdots (f_l)). \end{align*}
(Recall that 
$\underline {x} = (s,t,\ldots )$.) Then this satisfies 
$\varphi _{\underline {x}}(uf) = \varphi _{\underline {x}}(w_0(f)u)$ for any 
$u\in B_{\underline {x}}$ and 
$f\in R$. Therefore it factors through 
$B_{\underline {x}}\to B_{\underline {x}}^{w_0}$. We get 
$\overline {\varphi }_{\underline {x}}\colon B_{\underline {x}}^{w_0}\to R(-\ell (w_0))$. This is surjective. On the other hand, after tensoring 
$Q$, both are one-dimensional 
$Q$-vector spaces by Lemma 2.17. Therefore it is an isomorphism. In particular, 
$\overline {\varphi }_{\underline {x}}$ is injective, hence it is an isomorphism.
Note that the 
$(-\ell (w))$th-degree part of 
$B_{\underline {x}}$ (respectively, 
$B_{\underline {y}}$) is a free 
$\mathbb {K}$-module of rank one with a basis 
$u_{\underline {x}}$ (respectively, 
$u_{\underline {y}}$). Therefore 
$\Phi (u_{\underline {x}}) = cu_{\underline {y}}$ for some 
$c\in \mathbb {K}$. Let 
$\bar {u}_{\underline {x}}\in B_{\underline {x}}^{w_0}$ (respectively, 
$\bar {u}_{\underline {y}}\in B_{\underline {y}}^{w_0}$) be the image of 
$u_{\underline {x}}$ (respectively, 
$u_{\underline {y}}$). Since 
$\varphi _{\underline {x}}(u_{\underline {x}}) = 1$, we have the following commutative diagram.
We have that 
$c\in \mathbb {K}^{\times }$ and 
$c^{-1}\Phi$ gives a desired homomorphism.
3.3 Light leaves
We recall the definition of light leaves [Reference LibedinskyLib08] following the notation of [Reference Elias and WilliamsonEW16]. We need one more item of notation.
Let 
$w\in W$ and let 
$\underline {x},\underline {y}\in S^{\ell (w)}$ be two reduced expressions for 
$w$. Then by a fundamental property of a Coxeter system, there exists a sequence 
$\underline {x}^{0} = \underline {x},\underline {x}^{1},\ldots ,\underline {x}^{r} = \underline {y}$ such that each 
$\underline {x}^{i}$ and 
$\underline {x}^{i + 1}$ only differ by a single braid relation. At each step, we can attach a homomorphism 
$B_{\underline {x}^{i}}\to B_{\underline {x}^{i + 1}}$ using the homomorphism in Assumption 3.2. We denote the composition of these maps by 
$\mathrm {rex}$. Note that 
$\mathrm {rex}$ sends 
$u_{\underline {x}}$ to 
$u_{\underline {y}}$. Of course, this homomorphism is not unique. We fix it for any such two reduced expressions. See [Reference Elias and WilliamsonEW16, § 4.2] for details.
For the definition of light leaves, we use the following maps. Let 
$s\in S$. First set 
$\partial _s(f) = (f - s(f))/\alpha _s$. Put
\[ \begin{array}{rll} m^{s}\colon B_s\to R & f_1\otimes f_2\mapsto f_1f_2 & \text{degree }1,\\ i_0^{s}\colon B_s\otimes B_s\to B_s & f_1\otimes f_2\otimes f_3 \mapsto f_1\partial_s(f_2)\otimes f_3 & \text{degree }-1,\\ i_1^{s}\colon B_s\otimes B_s \to R & f_1\otimes f_2\otimes f_3 \mapsto f_1\partial_s(f_2)f_3 & \text{degree }0. \end{array} \]
Here 
$f_1,f_2,f_3\in R$ and, in the definition of 
$i_0^{s},i_1^{s}$, we regard 
$B_s\otimes B_s = R\otimes _{R^{s}}R\otimes _{R^{s}}R$. Since 
$V$ is a faithful 
$\{e,s\}$-representation, these are homomorphisms in 
$\mathcal {C}$.
Definition 3.8 Let 
$\underline {x} = (s_1,\ldots ,s_l)\in S^{l}$, 
$\boldsymbol {e} = (\boldsymbol {e}_1,\ldots ,\boldsymbol {e}_l) \in \{0,1\}^{l}$. Then we define a light leaf 
$\mathit {LL}_{\underline {x},\boldsymbol {e}}$ which is a morphism from 
$B_{\underline {x}}$ to 
$B_{\underline {w}}$ in 
$\mathcal {C}$ where 
$\underline {w}$ is a fixed reduced expression for 
$w = \underline {x}^{\boldsymbol {e}}$. Let 
$\underline {x}_{\le k} = (s_1,\ldots ,s_k)$, 
$\boldsymbol {e}_{\le k} = (\boldsymbol {e}_1,\ldots ,\boldsymbol {e}_k)$ and 
$w_k = \underline {x}_{\le k}^{\boldsymbol {e}_{\le k}}$. We fix a reduced expression 
$\underline {w}_k$ for 
$w_k$ and define 
$\mathit {LL}_k\colon B_{\underline {x}_{\le k}}\to B_{\underline {w}_k}$ inductively as follows.
(U0) For
$\boldsymbol {e}_k = 0$ and $w_{k - 1}s_k > w_{k - 1}$, we have
\[ B_{\underline{x}_{\le k - 1}}\otimes B_{s_k}\xrightarrow{\mathit{LL}_{k - 1}\otimes \mathrm{Id}_{B_{s_k}}}B_{\underline{w}_{k - 1}}\otimes B_{s_k}\xrightarrow{\mathrm{Id}_{B_{\underline{w}_{k - 1}}}\otimes m^{s_k}} B_{\underline{w}_{k - 1}}. \](U1) For
$\boldsymbol {e}_k = 1$ and $w_{k - 1}s_k > w_{k - 1}$, we have
\[ B_{\underline{x}_{k - 1}}\otimes B_{s_k}\xrightarrow{\mathit{LL}_{k - 1}\otimes\mathrm{Id}_{B_{s_k}}} B_{\underline{w}_{k - 1}}\otimes B_{s_k} \xrightarrow{\mathrm{rex}}B_{\underline{w}_k}. \](D0) For
$\boldsymbol {e}_k = 0$ and $w_{k - 1}s_k < w_{k - 1}$, let $(t_1,\ldots ,t_{p - 1},s_k)$ be a reduced expression for $w_{k - 1}$ ending with $s_k$. We have
\begin{align*} & \hskip0.25pc B_{\underline{x}_{\le k - 1}}\otimes B_{s_k}\xrightarrow{\mathit{LL}_{k - 1}\otimes \mathrm{Id}_{B_{s_k}}}B_{\underline{w}_{k - 1}}\otimes B_{s_k}\xrightarrow{\mathrm{rex}\otimes\mathrm{Id}_{B_{s_k}}}B_{(t_1,\ldots ,t_{p - 1},s_k)}\otimes B_{s_k}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ \ \xrightarrow{\mathrm{Id}_{B_{(t_1,\ldots ,t_{p - 1})}}\otimes i_0^{s_k}}B_{(t_1,\ldots ,t_{p - 1},s_k)}\xrightarrow{\mathrm{rex}}B_{\underline{w}_k}.\end{align*}(D1) For
$\boldsymbol {e}_k = 1$ and $w_{k - 1}s_k < w_{k - 1}$, let $(t_1,\ldots ,t_{p - 1},s_k)$ be a reduced expression for $w_{k - 1}$ ending with $s_k$. We have
\begin{align*} &\hskip0.2pc B_{\underline{x}_{\le k - 1}}\otimes B_{s_k}\xrightarrow{\mathit{LL}_{k - 1}\otimes \mathrm{Id}_{B_{s_k}}}B_{\underline{w}_{k - 1}}\otimes B_{s_k}\xrightarrow{\mathrm{rex}\otimes \mathrm{Id}_{B_{s_k}}}B_{(t_1,\ldots, t_{p - 1},s_k)}\otimes B_{s_k}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ \ \xrightarrow{\mathrm{Id}_{B_{(t_1,\ldots ,t_{p - 1})}}\otimes i_1^{s_k}}B_{(t_1,\ldots ,t_{p - 1})}\xrightarrow{\mathrm{rex}}B_{\underline{w}_k}.\end{align*}
Finally, we put 
$\mathit {LL}_{\underline {x},\boldsymbol {e}} = \mathit {LL}_l$. By the construction, the degree of 
$\mathit {LL}_{\underline {x},\boldsymbol {e}}$ is 
$d(\boldsymbol {e})$.
We fix a sequence 
$\underline {x} = (s_1,\ldots ,s_l)$ in this subsection.
Lemma 3.9 Let 
$\boldsymbol {e},\boldsymbol {f}\in \{0,1\}^{l}$ such that 
$\underline {x}^{\boldsymbol {e}} = \underline {x}^{\boldsymbol {f}}$. Assume that the labels of 
$\boldsymbol {e}$ and 
$\boldsymbol {f}$ at 
$i$ are the same for all 
$i = 1,\ldots ,l$. Then we have 
$\boldsymbol {e} = \boldsymbol {f}$.
Proof. We prove 
$\boldsymbol {e}_i = \boldsymbol {f}_i$ by backward induction on 
$i$. Assume that we have 
$\boldsymbol {e}_j = \boldsymbol {f}_j$ for any 
$j > i$. Since 
$s_1^{\boldsymbol {e}_1}\cdots s_l^{\boldsymbol {e}_l} = s_1^{\boldsymbol {f}_1}\cdots s_l^{\boldsymbol {f}_l}$ by the assumption and 
$s_{i + 1}^{\boldsymbol {e}_{i + 1}}\cdots s_l^{\boldsymbol {e}_l} = s_{i + 1}^{\boldsymbol {f}_{i + 1}}\cdots s_l^{\boldsymbol {f}_l}$ by inductive hypothesis, we have 
$s_1^{\boldsymbol {e}_1}\cdots s_i^{\boldsymbol {e}_i} = s_1^{\boldsymbol {f}_1}\cdots s_i^{\boldsymbol {f}_i}$.
Assume that the label of 
$\boldsymbol {e}$ at 
$i$ (hence that of 
$\boldsymbol {f}$ at 
$i$) is 
$U$, 
$\boldsymbol {e}_i = 1$, 
$\boldsymbol {f}_i = 0$. Set 
$w = s_1^{\boldsymbol {e}_1}\cdots s_{i - 1}^{\boldsymbol {e}_{i - 1}}$. Then 
$s_1^{\boldsymbol {f}_1}\cdots s_{i - 1}^{\boldsymbol {f}_{i - 1}} = ws_i$ since 
$s_1^{\boldsymbol {e}_1}\cdots s_{i}^{\boldsymbol {e}_{i}} = s_1^{\boldsymbol {f}_1}\cdots s_{i}^{\boldsymbol {f}_{i}}$. Then since the label of 
$\boldsymbol {e}$ at 
$i$ is 
$U$, we have 
$ws_i > w$, and since the label of 
$\boldsymbol {f}$ at 
$i$ is also 
$U$, we have 
$(ws_i)s_i > ws_i$. This is a contradiction. We also have a contradiction for the other cases. Hence we have 
$\boldsymbol {e}_i = \boldsymbol {f}_i$.
Let 
$w\in W$. Using this lemma, we can define a total order 
${<_{\underline {x},w}}$ on 
$\{\boldsymbol {e}\in S^{l}\mid \underline {x}^{\boldsymbol {e}} = w\}$ as follows: 
$\boldsymbol {f} < \boldsymbol {e}$ if and only if there exists 
$i$ such that
• the labels of
$\boldsymbol {e}$ and $\boldsymbol {f}$ are the same at any $j < i$,• the label of
$\boldsymbol {e}$ is $D$ at $i$,• the label of
$\boldsymbol {f}$ is $U$ at $i$.
We just write 
$<$ for 
$<_{\underline {x},w}$ if there is no risk of confusion.
Fix 
$\delta _s\in V$ such that 
$\langle \alpha _s^{\vee },\delta _s\rangle = 1$ for each 
$s\in S$. For 
$\boldsymbol {e} \in \{0,1\}^{l}$, we define 
$b_{\underline {x},\boldsymbol {e}}\in B_{\underline {x}}$ by 
$b_{\underline {x},\boldsymbol {e}} = b_1\otimes \cdots \otimes b_l$ where 
$b_i\in B_{s_i}$ is defined by
\[ b_i = \begin{cases} 1\otimes 1 & (\text{the label of }\boldsymbol{e}\text{ at }i\text{ is }U),\\ \delta_{s_i} \otimes 1 & (\text{the label of }\boldsymbol{e}\text{ at }i\text{ is }D). \end{cases} \]Then we have the following proposition.
Proposition 3.10 Let 
$w\in W$, 
$\boldsymbol {e},\boldsymbol {f}\in \{0,1\}^{l}$ and assume that 
$\underline {x}^{\boldsymbol {e}} = \underline {x}^{\boldsymbol {f}} = w$. Fix a reduced expression 
$\underline {w}$ for 
$w$. Then
\[ \mathit{LL}_{\underline{x},\boldsymbol{e}}(b_{\underline{x},\boldsymbol{f}}) = \begin{cases} u_{\underline{w}} & (\boldsymbol{f} = \boldsymbol{e}),\\ 0 & (\boldsymbol{f} < \boldsymbol{e}). \end{cases} \]
In particular, 
$\{\mathit {LL}_{\underline {x},\boldsymbol {e}}\mid \underline {x}^{\boldsymbol {e}} = w\}$ is linearly independent.
Proof. Let 
$\underline {x}_{\le k}$, 
$\boldsymbol {e}_{\le k}$, 
$\underline {w}_k$ as in Definition 3.8 and 
$\boldsymbol {f}_{\le k}$ similarly. We prove that if the labels of 
$\boldsymbol {e}$ and 
$\boldsymbol {f}$ are the same at 
$i \le k$, then 
$\mathit {LL}_k(b_{\underline {x}_{\le k},\boldsymbol {f}_{\le k}}) = u_{\underline {w}_k}$ by induction on 
$k$.
Assume that the label of 
$\boldsymbol {e}$ at 
$k$ is 
$U$. Then we have 
$b_{\underline {x}_{\le k},\boldsymbol {f}_{\le k}} = b_{\underline {x}_{\le k - 1},\boldsymbol {f}_{\le k - 1}}\otimes (1\otimes 1)$.
If 
$\boldsymbol {e}_k = 0$, then we have
\[ \mathit{LL}_k(b_{\underline{x}_{\le k},\boldsymbol{f}_{\le k}}) = \mathit{LL}_{k - 1}(b_{\underline{x}_{\le k - 1},\boldsymbol{f}_{\le k - 1}})\otimes m^{s_k}(1\otimes 1) = u_{\underline{w}_{k - 1}}\otimes m^{s_k}(1\otimes 1) = u_{\underline{w}_{k}}. \]
If 
$\boldsymbol {e}_k = 1$, then
\[ \mathit{LL}_k(b_{\underline{x}_{\le k},\boldsymbol{f}_{\le k}}) = \mathrm{rex}(\mathit{LL}_{k - 1}(b_{\underline{x}_{\le k - 1},\boldsymbol{f}_{\le k - 1}})\otimes (1\otimes 1)) = \mathrm{rex}(u_{\underline{w}_{k - 1}}\otimes (1\otimes 1)) = u_{\underline{w}_k}. \]
Assume that the label of 
$\boldsymbol {e}$ at 
$k$ is 
$D$. Then we have 
$b_{\underline {x}_{\le k},\boldsymbol {f}_{\le k}} = b_{\underline {x}_{\le k - 1},\boldsymbol {f}_{\le k - 1}}\otimes (\delta _{s_k}\otimes 1)$. Let 
$(t_1,\ldots ,t_{p - 1},s_k)$ be a reduced expression for 
$w_{k - 1}$. If 
$\boldsymbol {e}_k = 0$, then we have
\begin{align*} \mathit{LL}_k(b_{\underline{x}_{\le k},\boldsymbol{f}_{\le k}}) & = \mathrm{rex}((\mathrm{Id}\otimes i_0^{s_k})(\mathrm{rex}(\mathit{LL}_{k - 1}(b_{\underline{x}_{\le k - 1},\boldsymbol{f}_{\le k - 1}}))\otimes (\delta_{s_k}\otimes 1)))\\ & = \mathrm{rex}((\mathrm{Id}\otimes i_0^{s_k})(u_{(t_1,\ldots,t_{p - 1},s_k)}\otimes (\delta_{s_k}\otimes 1)))\\ & = \mathrm{rex}(u_{(t_1,\ldots,t_{p - 1})}\otimes i_0^{s_k}(1\otimes \delta_{s_k}\otimes 1))\\ & = \mathrm{rex}(u_{(t_1,\ldots,t_{p - 1})}\otimes (\partial_{s_k}(\delta_{s_k})\otimes 1))\\ & = \mathrm{rex}(u_{(t_1,\ldots,t_{p - 1},s_k)})\\ & = u_{\underline{w}_k}. \end{align*} If 
$\boldsymbol {e}_k = 1$, then we have
\begin{align*} \mathit{LL}_k(b_{\underline{x}_{\le k},\boldsymbol{f}_{\le k}}) & = \mathrm{rex}((\mathrm{Id}\otimes i_1^{s_k})(\mathrm{rex}(\mathit{LL}_{k - 1}(b_{\underline{x}_{\le k - 1},\boldsymbol{f}_{\le k - 1}}))\otimes (\delta_{s_k}\otimes 1)))\\ & = \mathrm{rex}((\mathrm{Id}\otimes i_1^{s_k})(u_{(t_1,\ldots,t_{p - 1},s_k)}\otimes (\delta_{s_k}\otimes 1)))\\ & = \mathrm{rex}(u_{(t_1,\ldots,t_{p - 1})}\otimes i_1^{s_k}(1\otimes \delta_{s_k}\otimes 1))\\ & = \mathrm{rex}(u_{(t_1,\ldots,t_{p - 1})}\partial_{s_k}(\delta_{s_k}))\\ & = \mathrm{rex}(u_{(t_1,\ldots,t_{p - 1})})\\ & = u_{\underline{w}_{k}}. \end{align*}
This is the end of the induction. In particular, we have 
$\mathit {LL}_{\underline {x},\boldsymbol {e}}(b_{\underline {x},\boldsymbol {e}}) = u_{\underline {w}}$.
Assume that 
$\boldsymbol {f} < \boldsymbol {e}$ and take 
$k$ such that the labels of 
$\boldsymbol {e}$ and 
$\boldsymbol {f}$ are the same at any 
$i < k$ and the label of 
$\boldsymbol {e}$ (respectively, 
$\boldsymbol {f}$) at 
$k$ is 
$D$ (respectively, 
$U$). Then we have 
$b_{\underline {x}_{\le k},\boldsymbol {f}_{\le k}} = b_{\underline {x}_{\le k - 1},\boldsymbol {f}_{\le k - 1}}\otimes (1\otimes 1)$. Let 
$(t_1,\ldots ,t_{p - 1},s_k)$ be a reduced expression for 
$w_{k - 1}$. If 
$\boldsymbol {e}_k = 0$, then we have
\begin{align*} \mathit{LL}_k(b_{\underline{x}_{\le k},\boldsymbol{f}_{\le k}}) & = \mathrm{rex}((\mathrm{Id}\otimes i_0^{s_k})(\mathrm{rex}(\mathit{LL}_{k - 1}(b_{\underline{x}_{\le k - 1},\boldsymbol{f}_{\le k - 1}}))\otimes (1\otimes 1)))\\ & = \mathrm{rex}(u_{(t_1,\ldots,t_{p - 1})}\otimes i_0^{s_k}(1\otimes 1\otimes 1))\\ & = \mathrm{rex}(u_{(t_1,\ldots,t_{p - 1})}\otimes (\partial_{s_k}(1)\otimes 1))\\ & = 0. \end{align*}
If 
$\boldsymbol {e}_k = 1$, then we have
\begin{align*} \mathit{LL}_k(b_{\underline{x}_{\le k},\boldsymbol{f}_{\le k}}) & = \mathrm{rex}((\mathrm{Id}\otimes i_1^{s_k})(\mathrm{rex}(\mathit{LL}_{k - 1}(b_{\underline{x}_{\le k - 1},\boldsymbol{f}_{\le k - 1}}))\otimes (1\otimes 1)))\\ & = \mathrm{rex}(u_{(t_1,\ldots,t_{p - 1})}\otimes i_1^{s_k}(1\otimes 1\otimes 1))\\ & = \mathrm{rex}(u_{(t_1,\ldots,t_{p - 1})}\partial_{s_k}(1))\\ & = 0. \end{align*} These calculations imply 
$\mathit {LL}_{\underline {x},\boldsymbol {e}}(b_{\underline {x},\boldsymbol {f}}) = 0$.
Remark 3.11 Since the degree of 
$\mathit {LL}_{\underline {x},\boldsymbol {e}}$ (respectively, 
$u_{\underline {w}}$) is 
$d(\boldsymbol {e})$ (respectively, 
$-\ell (w)$), 
$\mathit {LL}_{\underline {x},\boldsymbol {e}}(b_{\underline {x},\boldsymbol {e}}) = u_{\underline {w}}$ implies 
$\deg (b_{\underline {x},\boldsymbol {e}}) = -d(\boldsymbol {e})-\ell (w)$.
3.4 A basis of $B_{\underline {x}}^{w}$
Let 
$w\in W$ with a reduced expression 
$\underline {w}$ and 
$\underline {x}\in S^{l}$, 
$\boldsymbol {e}\in \{0,1\}^{l}$ such that 
$\underline {x}^{\boldsymbol {e}} = w$. Let 
$b_{\underline {x},\boldsymbol {e}}^{w}$ be the image of 
$b_{\underline {x},\boldsymbol {e}}\in B_{\underline {x}}$ in 
$B_{\underline {x}}^{w}$.
Let 
$\underline {w} = (t_1,\ldots ,t_r)$. We define a morphism 
$\varphi _{\underline {w}}\colon B_{\underline {w}}\to R_w$ in 
$\mathcal {C}$ by 
$\varphi _{\underline {w}}(f_0\otimes \cdots \otimes f_r) = f_0(t_1(f_1))\cdots (t_1\cdots t_r(f_r))$ where 
$f_0,\ldots ,f_r\in R$ and we identify 
$B_{\underline {w}} = R\otimes _{R^{t_1}}R\otimes _{R^{t_2}}\cdots \otimes _{R^{t_l}}R$.
Theorem 3.12 Fix 
$\underline {x}$, 
$w$ and 
$\underline {w}$.
(1) The left
$R$-module $B_{\underline {x}}^{w}$ has a basis $\{b^{w}_{\underline {x},\boldsymbol {e}}\mid \underline {x}^{\boldsymbol {e}} = w\}$.(2) The left
$R$-module $B_{\underline {x}}^{w}$ is graded free and its graded rank $\operatorname {grk}(B_{\underline {x}}^{w})$ is given by $\sum _{\underline {x}^{\boldsymbol {e}} = w}v^{d(\boldsymbol {e}) + \ell (w)} = v^{\ell (w)}p_{\underline {x}}^{w}(v)$.(3) The homomorphisms
$\{\varphi _{\underline {w}}\circ \mathit {LL}_{\underline {x},\boldsymbol {e}}\mid \underline {x}^{\boldsymbol {e}} = w\}$ are a basis of $\operatorname {Hom}_{\mathcal {C}}^{\bullet }(B_{\underline {x}},R_w)$.
Proof. By Proposition 3.10, we have
\[ \varphi_{\underline{w}}(\mathit{LL}_{\underline{x},\boldsymbol{e}}(b_{\underline{x},\boldsymbol{f}})) = \begin{cases} 1 & (\boldsymbol{f} = \boldsymbol{e}),\\ 0 & (\boldsymbol{f} < \boldsymbol{e}). \end{cases} \]
Since 
$\varphi _{\underline {w}}\circ \mathit {LL}_{\underline {x},\boldsymbol {e}}\colon B_{\underline {x}}\to R_w$ is a homomorphism in 
$\mathcal {C}$, it induces 
$\psi _{\boldsymbol {e}}\colon B_{\underline {x}}^{w}\to R_w$ and we have
\[ \psi_{\boldsymbol{e}}(b_{\underline{x},\boldsymbol{f}}^{w}) = \begin{cases} 1 & (\boldsymbol{f} = \boldsymbol{e}),\\ 0 & (\boldsymbol{f} < \boldsymbol{e}). \end{cases} \]
Inductively on 
$\boldsymbol {e}$, we can take 
$\psi '_{\boldsymbol {e}} \in \psi _{\boldsymbol {e}} + \sum _{\boldsymbol {e}' > \boldsymbol {e}}R\psi _{\boldsymbol {e}'}$ such that 
$\psi '_{\boldsymbol {e}}(b_{\underline {x},\boldsymbol {f}}^{w}) = \delta _{\boldsymbol {e},\boldsymbol {f}}$ (Kronecker's delta). That is, 
$m\mapsto \sum _{\underline {x}^{\boldsymbol {e}} = w}\psi '_{\boldsymbol {e}}(m)b_{\underline {x},\boldsymbol {e}}^{w}$ gives a splitting of the embedding 
${\bigoplus} _{\underline {x}}^{\boldsymbol {e}}$ 
${= w}\,Rb_{\underline {x},\boldsymbol {e}}^{w}\hookrightarrow B_{\underline {x}}^{w}$.
Take 
$N$ such that 
$B_{\underline {x}}^{w} = ({\bigoplus} _{\underline {x}^{\boldsymbol {e}} = w}\,Rb_{\underline {x},\boldsymbol {e}}^{w})\oplus N$. Then we have 
$(B_{\underline {x}}^{w})_Q = ({\bigoplus} _{\underline {x}^{\boldsymbol {e}} = w}\,Qb_{\underline {x},\boldsymbol {e}}^{w})\oplus N_Q$. We have 
$\dim _Q (B_{\underline {x}}^{w})_Q = p_{\underline {x}}^{w}(1)$ by Lemma 2.17. By Lemma 3.1, we also have 
$\dim _Q({\bigoplus} _{\underline {x}^{\boldsymbol {e}} = w}\,Qb_{\underline {x},\boldsymbol {e}}^{w}) = p_{\underline {x}}^{w}(1)$. Hence 
$N_Q = 0$. That is, 
$N$ is a torsion module. Since 
$N\subset B_{\underline {x}}^{w}\subset (B_{\underline {x}})_Q^{w}$, 
$N$ is a torsion-free module. Hence 
$N = 0$.
(2) follows from (1).
We prove (3). Since 
$\{b_{\underline {x},\boldsymbol {e}}^{w}\}$ is a basis of 
$B_{\underline {x}}^{w}$, 
$\{\psi '_{\boldsymbol {e}}\}$ is a basis of 
$\operatorname {Hom}^{\bullet }_R(B_{\underline {x}}^{w},R)$ which is dual to 
$\{b_{\underline {x},\boldsymbol {e}}\}$. By 
$\psi '_{\boldsymbol {e}} \in \psi _{\boldsymbol {e}} + \sum _{\boldsymbol {e}' > \boldsymbol {e}}R\psi _{\boldsymbol {e}'}$, 
$\{\psi _{\boldsymbol {e}}\}$ is also a basis. Since 
$\operatorname {Hom}^{\bullet }_R(B_{\underline {x}}^{w},R)\simeq \operatorname {Hom}^{\bullet }_{\mathcal {C}}(B_{\underline {x}},R_w)$ and 
$\psi _{\boldsymbol {e}}$ corresponds to 
$\varphi _{\underline {w}}\circ \mathit {LL}_{\underline {x},\boldsymbol {e}}$ by the definition of 
$\psi _{\boldsymbol {e}}$, we get (3).
With Lemma 2.19, we have the following corollary.
Corollary 3.13 The left 
$R$-module 
$B_{\underline {x},w}$ is graded free and its graded rank is given by 
$v^{-\ell (w)}p_{\underline {x}}^{w}(v^{-1})$.
Corollary 3.14 The homomorphism 
$\varphi _{\underline {w}}\colon B_{\underline {w}}\to R_w(\ell (w))$ induces 
$B_{\underline {w}}^{w}\simeq R_w(\ell (w))$.
Proof. Since 
$\varphi _{\underline {w}}$ is a homomorphism in 
$\mathcal {C}$, this induces 
$B_{\underline {w}}^{w}\to R_w(\ell (w))$. This is obviously surjective. By the above theorem, 
$B_{\underline {w}}^{w}$ is free of rank one. Hence 
$B_{\underline {w}}^{w}\twoheadrightarrow R_w(\ell (w))$ is an isomorphism.
Corollary 3.15 Let 
$w\in W$ with a reduced expression 
$\underline {w}$ and 
$B\in \mathcal {S}\mathrm {Bimod}$. Then 
$\operatorname {Hom}^{\bullet }_{\mathcal {C}}(B,B_{\underline {w}})\to \operatorname {Hom}^{\bullet }_R(B^{w},B_{\underline {w}}^{w})$ is surjective.
3.5 Elements supported on a closed subset
We call a subset 
$I\subset W$ closed if 
$w_1\in I$, 
$w_2\in W$, 
$w_2\le w_1$ implies 
$w_2\in I$.
Lemma 3.16 Let 
$I$ be a finite closed subset and 
$w\in I$ a maximal element with respect to the Bruhat order. Then there exists a enumeration 
$w_1,w_2,\ldots$ of elements of 
$W$ such that 
$\{w_1,\ldots ,w_i\}$ is closed for any 
$i$, 
$w = w_{\#I}$ and 
$I = \{w_1,\ldots ,w_{\#I}\}$.
Proof. Set 
$k = \#I$. Let 
$w_1,\ldots ,w_{k - 1}$ be an enumeration of elements of 
$I\setminus \{w\}$ such that 
$w_i\le w_j$ implies 
$i\le j$. Let 
$w_{k + 1},\ldots$ be a similar enumeration of elements of 
$W\setminus I$ and put 
$w_k = w$. We prove that 
$\{w_1,\ldots ,w_i\}$ is always closed. If 
$i \le k$ then it is obvious. Assume that 
$i > k$ and 
$w_j\le w_i$. If 
$j\le k$, then we have 
$j\le i$. If 
$j > k$, then we have 
$j\le i$ by the assumption on the enumeration 
$w_{k + 1},\ldots .$ In any case, 
$w_j\in \{1,\ldots ,w_i\}$.
A light leaf 
$\mathit {LL}_{\underline {x},\boldsymbol {e}}\colon B_{\underline {x}}\to B_{\underline {w}}$ gives a homomorphism 
$B_{\underline {w}}\simeq D(B_{\underline {w}})\xrightarrow {D(\mathit {LL}_{\underline {x},\boldsymbol {e}})} D(B_{\underline {x}})\simeq B_{\underline {x}}$. We denote this homomorphism by 
$\mathit {LL}_{\underline {x},\boldsymbol {e}}^{*}$. Let 
$\pi _{\underline {x}}^{w}\colon B_{\underline {x}}\to B_{\underline {x}}^{w}$ be the projection. If 
$I\subset W$ and 
$w\in I$, then 
$B_{\underline {x},I\setminus \{w\}}$ is the kernel of 
$B_{\underline {x},I}\to B_{\underline {x}}^{w}$. Therefore we also write 
$\pi _{\underline {x}}^{w}$ for the projection 
$B_{\underline {x},I}\to B_{\underline {x},I}/B_{\underline {x},I\setminus \{w\}}$.
Theorem 3.17 Let 
$I$ be a closed subset and 
$w$ a maximal element in 
$I$. Set 
$I' = I\setminus \{w\}$. Let 
$\underline {w}$ be a reduced expression for 
$w$ and 
$\underline {x}\in S^{l}$. Then 
$\{\pi _{\underline {x}}^{w}(\mathit {LL}_{\underline {x},\boldsymbol {e}}^{*}(u_{\underline {w}}))\mid \underline {x}^{\boldsymbol {e}} = w\}$ is a basis of the left 
$R$-module 
$B_{\underline {x},I}/B_{\underline {x},I'}$.
Proof. Note that since 
$\{y\in W\mid y\le w\}\subset I$, we have 
$\operatorname {supp}_W(B_{\underline {w}})\subset I$. Hence the image of any homomorphism from 
$B_{\underline {w}}$ to 
$B_{\underline {x}}$ is contained in 
$B_{\underline {x},I}$. Therefore 
$\pi _{\underline {x}}^{w}(\mathit {LL}^{*}_{\underline {x},\boldsymbol {e}}(u_{\underline {w}}))\in B_{\underline {x},I}/B_{\underline {x},I'}$.
First we prove that 
$\{\pi _{\underline {x}}^{w}(\mathit {LL}_{\underline {x},\boldsymbol {e}}^{*}(u_{\underline {w}}))\mid \underline {x}^{\boldsymbol {e}} = w\}$ is linearly independent. It is sufficient to prove that this set is linearly independent over 
$Q$.
Recall that we have homomorphisms
\[ B_{\underline{x}}\xrightarrow{\mathit{LL}_{\underline{x},\boldsymbol{e}}} B_{\underline{w}}\to B_{\underline{w}}^{w}. \]
The set of these elements where 
$\boldsymbol {e}$ satisfies 
$\underline {x}^{\boldsymbol {e}} = w$ is linearly independent by Theorem 3.12. Therefore the dualized maps
\[ B_{\underline{w},w}\hookrightarrow B_{\underline{w}}\xrightarrow{\mathit{LL}^{*}_{\underline{x},\boldsymbol{e}}}B_{\underline{x}} \]
are also linearly independent. (Note that 
$B_{\underline {w},w}$ and 
$B_{\underline {x}}$ are both graded free as right 
$R$-modules, hence 
$D(D(B_{\underline {w},w}))\simeq B_{\underline {w},w}$ and 
$D(D(B_{\underline {x}}))\simeq B_{\underline {x}}$.) This map factors through 
$B_{\underline {x},w}\hookrightarrow B_{\underline {x}}$. Therefore the induced homomorphisms 
$B_{\underline {w},w}\to B_{\underline {x},w}$ are linearly independent. Since 
$B_{\underline {x},w}$ is torsion free, the maps 
$B_{\underline {w},w}\otimes _R Q\xrightarrow {\eta _{\underline {x},\boldsymbol {e},Q}}B_{\underline {x},w}\otimes _R Q$ obtained by tensoring with 
$Q$ are also linearly independent.
The left 
$R$-module 
$B_{\underline {w}}^{w}$ is free of rank one by Theorem 3.12. Since 
$D(B_{\underline {w}}^{w})\simeq B_{\underline {w},w}$, this is also true for 
$B_{\underline {w},w}$. Therefore the dimension of both 
$B_{\underline {w},w}\otimes _R Q$ and 
$B_{\underline {w}}^{w}\otimes _R Q$ is one and hence we have 
$B_{\underline {w},w}\otimes _R Q\xrightarrow {\sim }B_{\underline {w}}^{w}\otimes _R Q$. Therefore, for any 
$0\ne q\in B_{\underline {w},w}\otimes _R Q\simeq B_{\underline {w}}^{w}\otimes _R Q$, 
$\{\eta _{\underline {x},\boldsymbol {e},Q}(q)\mid \underline {x}^{\boldsymbol {e}} = w\}\subset B_{\underline {x},w}\otimes _R Q$ is linearly independent. In particular, we can take 
$q$ as the image of 
$u_{\underline {w}}\in B_{\underline {w}}$. Therefore
\[ \{\eta_{\underline{x},\boldsymbol{e},Q}(\pi_{\underline{w}}^{w}(u_{\underline{w}}))\mid \underline{x}^{\boldsymbol{e}} = w\}\subset B_{\underline{x},w}\otimes_R Q\subset (B_{\underline{x},I}/B_{\underline{x},I'})\otimes_R Q \]is linearly independent. Observing the commutative diagram
we have that 
$\{\pi _{\underline {x}}^{w}(\mathit {LL}_{\underline {x},\boldsymbol {e}}^{*}(u_{\underline {w}}))\mid \underline {x}^{\boldsymbol {e}} = w\}$ is linearly independent over 
$Q$.
Let 
$w_1,w_2,\ldots$ be an enumeration as in the previous lemma. Fix a reduced expression 
$\underline {w}_k$ for 
$w_k$. Set 
$I(k) = \{w_1,\ldots ,w_k\}$. We have a filtration 
$\{B_{\underline {x},I(k)}\}_k$ of 
$B_{\underline {x}}$ and, as we have proved above,
\begin{equation} \mathop{\bigoplus}\limits_{\underline{x}^{\boldsymbol{e}} = w_k}\,R\pi_{\underline{x}}^{w_k}(\mathit{LL}^{*}_{\underline{x},\boldsymbol{e}}(u_{\underline{w}_k}))\hookrightarrow B_{\underline{x},I(k)}/B_{\underline{x},I(k - 1)}. \end{equation} We prove that this is an isomorphism. First we prove that this map is an isomorphism when 
$\mathbb {K}$ is a field. We have 
$\deg (\mathit {LL}^{*}_{\underline {x},\boldsymbol {e}}) = d(\boldsymbol {e})$ and 
$\deg (u_{\underline {w}_k}) = -\ell (w_k)$. Therefore the graded rank of 
${\bigoplus} _{\underline {x}^{\boldsymbol {e}} = w_k}\, R\pi _{\underline {x}}^{w}(\mathit {LL}^{*}_{\underline {x},\boldsymbol {e}}(u_{\underline {w}_k}))$ is 
$\sum _{\underline {x}^{\boldsymbol {e}} = w_k}v^{-d(\boldsymbol {e}) + \ell (w_k)} = v^{\ell (w_k)}p_{\underline {x}}^{w_k}(v^{-1})$. By Lemma 2.16, the sum 
$\sum _k v^{\ell (w_k)}p_{\underline {x}}^{w_k}(v^{-1})$ is 
$(v + v^{-1})^{l}$ and this is equal to the graded rank of 
$B_{\underline {x}}$ by Lemma 2.14. Hence, if 
$\mathbb {K}$ is a field, (3.1) is an isomorphism by Lemma 2.9.
Now we prove that the map (3.1) is an isomorphism for any Noetherian domain 
$\mathbb {K}$. Let 
$\mathfrak {m}$ be a maximal ideal of 
$\mathbb {K}$. Then in the commutative diagram
the horizontal map is an isomorphism. Hence the vertical map is surjective. Therefore, for any 
$k$, 
$B_{\underline {x},I(k)}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}) \to (B_{\underline {x}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))_{I(k)}$ is surjective.
We prove that the map (3.1) is an isomorphism by backward induction on 
$k$. By inductive hypothesis, 
$B_{\underline {x},I(k')}/B_{\underline {x},I(k' - 1)}$ is graded free for any 
$k' > k$. Hence 
$B_{\underline {x}}/B_{\underline {x},I(k)}$ is graded free. Therefore 
$B_{\underline {x},I(k)}$ is a direct summand of 
$B_{\underline {x}}$. Hence 
$B_{\underline {x},I(k)}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m})\to (B_{\underline {x}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))_{I(k)}$ is injective, hence it is an isomorphism. By applying the five lemma to the commutative diagram with exact columns
the vertical map in (3.2) is an isomorphism.
Hence (3.1) is an isomorphism after tensoring 
$\mathbb {K}/\mathfrak {m}$. Note that both sides are finitely generated as 
$R$-modules. Indeed, the left-hand side is obviously finitely generated and the right-hand side is finitely generated since it is a subquotient of finitely generated 
$R$-module 
$B_{\underline {x}}$. Therefore each graded piece of both sides is a finitely generated 
$\mathbb {K}$-module. Therefore by applying Nakayama's lemma to each graded piece, (3.1) is surjective after localizing at 
$\mathfrak {m}$. Hence it is an isomorphism. Since this is true for any maximal ideal, (3.1) itself is an isomorphism.
Remark 3.18 In the proof, we proved that 
$B_{\underline {x},I(k)}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m})\simeq (B_{\underline {x}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))_{I(k)}$. Since any closed subset can be realized as 
$I(k)$ for some 
$k$ (Lemma 3.16) and a certain sequence 
$\emptyset = I(0)\subset I(1)\subset \cdots$, it is true after replacing 
$I(k)$ with any closed subset. Therefore. for any 
$B\in \mathcal {S}\mathrm {Bimod}$ and any closed subset 
$I$, the natural map 
$B_I\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m})\to (B\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))_I$ is an isomorphism.
For 
$M\in \mathcal {C}$ and 
$w\in W$, we put 
$M_{\le w} = M_{\{y\in W\mid y\le w\}}$. We define 
$M_{< w}$ in the obvious way.
Corollary 3.19 Let 
$B\in \mathcal {S}\mathrm {Bimod}$.
(1) Let
$I\subset W$ be a closed subset and $w\in I$ a maximal element. Then we have $B_{\le w}/B_{< w}\xrightarrow {\sim }B_{I}/B_{I\setminus \{w\}}$.(2) Let
$\underline {w}$ be a reduced expression for $w\in W$. Then the map $\operatorname {Hom}^{\bullet }_{\mathcal {C}}(B_{\underline {w}},B)\to \operatorname {Hom}^{\bullet }_R(B_{\underline {w}}^{w},B_{\le w}/B_{< w})$ is surjective.
Proof. We may assume 
$B = B_{\underline {x}}$ for some 
$\underline {x}\in S^{l}$. By Theorem 3.17, any element in 
$B_{\underline {x},I}/B_{\underline {x},I\setminus \{w\}}$ has a representative of the form 
$\sum c_{\boldsymbol {e}}\mathit {LL}_{\underline {x},\boldsymbol {e}}^{*}(u_{\underline {w}})$ with 
$c_{\boldsymbol {e}}\in R$. Since 
$\operatorname {supp}_W(u_{\underline {w}})\subset \{y\in W\mid y\le w\}$, 
$\operatorname {supp}_W(\mathit {LL}_{\underline {x},\boldsymbol {e}}^{*}(u_{\underline {w}}))\subset \{y\in W\mid y\le w\}$. Hence 
$\sum c_{\boldsymbol {e}}\mathit {LL}_{\underline {x},\boldsymbol {e}}^{*}(u_{\underline {w}})\in B_{\underline {x},\le w}$. We get (1).
We prove (2). The 
$R$-module 
$B_{\underline {w}}^{w}$ is free and 
$\pi _{\underline {w}}^{w}(u_{\underline {w}})$ is a basis of this module. Hence 
$\operatorname {Hom}^{\bullet }_R(B_{\underline {w}}^{w},B_{\underline {x},\le w}/B_{\underline {x},< w})\simeq B_{\underline {x},\le w}/B_{\underline {x},< w}$ by 
$\psi \mapsto \psi (\pi _{\underline {w}}^{w}(u_{\underline {w}}))$ By the functionality of 
$B\to B^{w}$ for 
$B\in \mathcal {C}$, we have 
$\psi (\pi _{\underline {w}}^{w}(u_{\underline {w}})) = \pi _{\underline {x}}^{w}(\psi (u_{\underline {w}}))$. Therefore it is sufficient to prove that 
$\operatorname {Hom}^{\bullet }_{\mathcal {C}}(B_{\underline {w}},B_{\underline {x}})\to B_{\underline {x},\le w}/B_{\underline {x},< w}$ defined by 
$\varphi \mapsto \pi _{\underline {x}}^{w}(\varphi (u_{\underline {w}}))$ is surjective. This is clear from Theorem 3.17.
The following proposition is a generalization of [Reference SoergelSoe07, Satz 6.6].
Proposition 3.20 Let 
$B\in \mathcal {S}\mathrm {Bimod}$ and 
$w\in W$. Consider the element 
$\prod _{tw < w}\alpha _t\in R$ where 
$t$ runs through reflections in 
$W$. Then we have 
$B_w\xrightarrow {\sim }(\prod _{tw < w}\alpha _t)(B_{\le w}/B_{< w})$.
Proof. We may assume 
$B = B_{\underline {x}}$. It is well known that 
$\#\{t\mid tw < w\} = \ell (w)$. Therefore 
$\deg (\prod _{tw < w}\alpha _t) = 2\ell (w)$. Hence by Corollary 3.13 and Theorem 3.17, both sides are graded free with the same graded rank.
First we assume that 
$\underline {x}$ is a reduced expression for 
$w$. We denote 
$\underline {x}$ by 
$\underline {w}$. We prove 
$B_{\underline {w},\le w}/B_{\underline {w},< w}\simeq B_{\underline {w}}^{w}$ by induction on 
$\ell (w)$.
Let 
$\underline {w} = (s_1,\ldots ,s_l)$ and set 
$s = s_1$, 
$\underline {sw} = (s_2,\ldots ,s_l)$. Then 
$\underline {sw}$ is a reduced expression for 
$sw$. Let 
$\delta _s\in V$ such that 
$\langle \alpha _s^{\vee },\delta _s\rangle = 1$. Take 
$1\otimes m + \delta _s\otimes m'\in R\otimes _{R^{s}}B_{\underline {sw}} = B_{\underline {w}}$ and assume that 
$1\otimes m + \delta _s\otimes m'\in B_{\underline {w},w}$. By Lemma 2.12, 
$\operatorname {supp}_W(m')\in \{w,sw\}$. We also have 
$m'_w = 0$ since 
$B_{\underline {sw}}^{w} = 0$. Therefore 
$m'\in B_{\underline {sw},sw}$. By inductive hypothesis, we have 
$m'\in (\prod _{tsw < sw}\alpha _t)B_{\underline {sw}}^{sw}$.
Since 
$1\otimes m + \delta _s\otimes m'\in B_{\underline {w},w}$, 
$(1\otimes m + \delta _s\otimes m')_{sw} = 0$. Hence we have 
$m_{sw} + \delta _s m'_{sw} = 0$. Therefore 
$m_{sw} + s(\delta _s) m'_{sw} = (-\delta _s + s(\delta _s))m'_{sw} = -\alpha _s m'_{sw}\in \alpha _s (\prod _{tsw < sw}\alpha _t)B_{\underline {sw}}^{sw}$. It is well known that 
$\{t\mid tw < w\} = \{s\}\cup \{sts^{-1}\mid tsw < sw\}$. Hence 
$\prod _{tw < w}\alpha _t = \alpha _s s(\prod _{tsw < sw}\alpha _t) = s(-\alpha _s\prod _{tsw < sw}\alpha _t)$. Therefore 
$m_{sw} + s(\delta _s) m'_{sw}\in s(\prod _{tw < w}\alpha _t)B_{\underline {sw}}^{sw}$. Take 
$n\in B_{\underline {sw}}$ such that 
$m_{sw} + s(\delta _s) m'_{sw} = s(\prod _{tw < w}\alpha _t)n_{sw}$. Now we have 
$(1\otimes m + \delta _s\otimes m')_w = (m_w + \delta _s m'_w,m_{sw} + s(\delta _s)m'_{sw})$ and, since 
$m_w,m'_w\in B_{\underline {sw}}^{w} = 0$, we get 
$(1\otimes m + \delta _s\otimes m')_w = (0,s(\prod _{tw < w}\alpha _t)n_{sw}) = (\prod _{tw < w}\alpha _t)(0,n_{sw})$. (Recall that the left action of 
$R$ on the second factor of 
$(B_s\otimes B_{\underline {sw}})_Q^{w} = (B_{\underline {sw}})_Q^{w}\oplus (B_{\underline {sw}})_Q^{sw}$ is twisted by 
$s$.) Again using 
$n_{w}\in B_{\underline {sw}}^{w} = 0$, we have 
$(\prod _{tw < w}\alpha _t)(0,n_{sw}) = (\prod _{tw < w}\alpha _t)(n_w,n_{sw}) = (\prod _{tw < w}\alpha _t)(1\otimes n)_w\in (\prod _{tw < w}\alpha _t)B_ {\underline {w}}^{w}$.
To finish the proof for 
$\underline {x} = \underline {w}$, we prove that 
$B_{\underline {w},w}\hookrightarrow (\prod _{tw < w}\alpha _t)B_{\underline {w}}^{w}$ is surjective using an argument in the proof of Theorem 3.17. Since both sides has the same graded rank, this is surjective if 
$\mathbb {K}$ is a field. Let 
$\mathfrak {m}\subset \mathbb {K}$ be a maximal ideal. The natural surjective map 
$B_{\underline {w}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m})\to B_{\underline {w}}^{w}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m})$ induces a surjective map 
$(B_{\underline {w}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))^{w}\to B_{\underline {w}}^{w}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m})$. Both sides are graded free with graded rank 
$v^{\ell (w)}$ (by Theorem 3.12, for instance). Therefore it is an isomorphism. Take the dual of this isomorphism. By Lemma 2.18, we have 
$D((B_{\underline {w}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))^{w})\simeq D(B_{\underline {w}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))_w$. Since 
$B_{\underline {w}}$ is free as a right 
$R$-module, we have 
$D(B_{\underline {w}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))\simeq D(B_{\underline {w}})\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m})$. By Lemma 2.20, we have 
$D(B_{\underline {w}})\simeq B_{\underline {w}}$. Since 
$B_{\underline {w}}^{w}$ is also free, with Lemmas 2.18 and 2.20, we also have 
$D(B_{\underline {w}}^{w}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))\simeq B_{\underline {w},w}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m})$. Hence 
$(B_{\underline {w}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))_w\simeq B_{\underline {w},w}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m})$. Therefore, by tensoring 
$\mathbb {K}/\mathfrak {m}$, the map 
$B_{\underline {w},w}\hookrightarrow (\prod _{tw < w}\alpha _t)B_{\underline {w}}^{w}$ becomes 
$(B_{\underline {w}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))_w\hookrightarrow (\prod _{tw < w}\alpha _t)(B_{\underline {w}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))^{w}$. Since the map 
$B_{\underline {w},w}\hookrightarrow (\prod _{tw < w}\alpha _t)B_{\underline {w}}^{w}$ is surjective if 
$\mathbb {K}$ is a field, 
$(B_{\underline {w}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))_w\hookrightarrow (\prod _{tw < w}\alpha _t)(B_{\underline {w}}\otimes _{\mathbb {K}}(\mathbb {K}/\mathfrak {m}))^{w}$ is surjective. Therefore by Nakayama's lemma, 
$B_{\underline {w},w}\hookrightarrow (\prod _{tw < w}\alpha _t)B_{\underline {w}}^{w}$ is surjective after localizing at 
$\mathfrak {m}$. Since this is true for any maximal ideal 
$\mathfrak {m}$, 
$B_{\underline {w},w}\hookrightarrow (\prod _{tw < w}\alpha _t)B_{\underline {w}}^{w}$ is surjective.
We prove the proposition for 
$B = B_{\underline {x}}$ with 
$\underline {x}\in S^{l}$. By an argument above using localization at maximal ideals and Remark 3.18, it is sufficient to prove that 
$(\prod _{tw < w}\alpha _t)B_{\underline {x},\le w}/B_{\underline {x},< w}\subset B_{\underline {x},w}$. Set 
$f = \prod _{tw < w}\alpha _t$ and let 
$fm$ be an element of the left-hand side. Then we may assume 
$m = \mathit {LL}^{*}_{\underline {x},\boldsymbol {e}}(\pi _{\underline {w}}^{w}(u_{\underline {w}}))$ by Theorem 3.17. Since we have proved the proposition for 
$B = B_{\underline {w}}$, 
$f\pi _{\underline {w}}^{w}(u_{\underline {w}})\in B_{\underline {w},w}$. Hence 
$fm = \mathit {LL}^{*}_{\underline {x},\boldsymbol {e}}(f\pi _{\underline {w}}^{w}(u_{\underline {w}}))\in B_{\underline {x},w}$.
4. The categorification theorem
In this section we assume that 
$\mathbb {K}$ is a complete local ring. Therefore a direct summand of a graded free 
$R$-module is again graded free.
4.1 The classification of indecomposable objects
Theorem 4.1
(1) For each
$w\in W$, there exists an indecomposable object $B(w)\in \mathcal {S}\mathrm {Bimod}$ such that $\operatorname {supp}_W(B(w))\subset \{x\in W\mid x\le w\}$ and $B(w)^{w}\simeq R_w(\ell (w))$. Moreover, $B(w)$ is unique up to isomorphism.(2) For any indecomposable object
$B\in \mathcal {S}\mathrm {Bimod}$, there exists a unique $(w,n)\in W\times \mathbb {Z}$ such that $B\simeq B(w)(n)$.(3) We have
$D(B(w))\simeq B(w)$.(4) For a reduced expression
$\underline {w}$ for $w\in W$, we have $B_{\underline {w}} = B(w)\oplus {\bigoplus} _{y < w,n\in \mathbb {Z}}\, B(y)(n)^{m_{n,y}}$ for some $m_{n,y}\in \mathbb {Z}_{\ge 0}$.
Proof. Fix a reduced expression 
$\underline {w}$ for 
$w$. Then we have 
$B_{\underline {w}}^{w}\simeq R_w(\ell (w))$. Therefore there exists a unique indecomposable direct summand 
$B(w)$ of 
$B_{\underline {w}}$ such that 
$B(w)^{w} = B_{\underline {w}}^{w}(\ell (w))$. This satisfies the condition of (1) and, since 
$D(B_{\underline {w}})\simeq B_{\underline {w}}$, we have 
$D(B(w))\simeq B(w)$.
It only remains to show that any object in 
$\mathcal {S}\mathrm {Bimod}$ is a direct sum of 
$B(w)(n)$. Let 
$B\in \mathcal {S}\mathrm {Bimod}$. Take 
$w\in W$ such that 
$B^{w}\ne 0$ and 
$\ell (w)$ is maximal with respect to this condition. Set 
$I = \{y\in W\mid \ell (y)\le \ell (w)\}$. This is a closed subset of 
$W$. The condition of 
$w$ and the definition of 
$I$ says that 
$B_{I} = B$. Hence 
$B_I/B_{I\setminus \{w\}} = B^{w}$.
Since 
$B^{w}$ is graded free and 
$B_{\underline {w}}^{w}\simeq R_w(\ell (w))$, there exists 
$n\in \mathbb {Z}$ such that 
$B_{\underline {w}}(n)^{w}$ is a direct summand of 
$B^{w}$. Let 
$i\colon B_{\underline {w}}(n)^{w}\hookrightarrow B^{w}$ and 
$p\colon B^{w}\twoheadrightarrow B_{\underline {w}}(n)^{w}$ be the embeddings from and the projection to the direct summand. Then by Corollaries 3.15 and 3.19(2), there exist 
$B_{\underline {w}}(n)\to B$ and 
$B\to B_{\underline {w}}(n)$ which are lifts of 
$i$ and 
$p$, respectively. Since 
$B(w)(n)$ is a direct summand such that 
$B(w)(n)^{w} = B_{\underline {w}}(n)^{w}$, composing the embedding from or the projection to 
$B(w)(n)$, we get lifts 
$\tilde {\imath }\colon B(w)(n)\to B$ and 
$\tilde {p}\colon B\to B(w)(n)$. The composition 
$\tilde {p}\circ \tilde {\imath }$ is identity on 
$B(w)(n)^{w}$, hence 
$\tilde {p}\circ \tilde {\imath }\in \operatorname {End}(B(w)(n))$ is not nilpotent. Since 
$B(w)(n)$ is indecomposable, 
$\operatorname {End}(B(w)(n))$ is local. Hence 
$\tilde {p}\circ \tilde {\imath }$ is an isomorphism. Therefore 
$B(w)(n)$ is a direct summand of 
$B$. The last statement follows from the argument with supports.
It is easy to see that 
$B(s) = B_s$ for any 
$s\in S$.
4.2 The categorification
Let 
$[\mathcal {S}\mathrm {Bimod}]$ be the split Grothendieck group of the category 
$\mathcal {S}\mathrm {Bimod}$. Then this has a structure of 
$\mathbb {Z}[v^{\pm 1}]$-algebra via 
$v[B] = [B(1)]$ and 
$[B_1][B_2] = [B_1\otimes B_2]$ where 
$[B]$ is the image of 
$B\in \mathcal {S}\mathrm {Bimod}$ in 
$[\mathcal {S}\mathrm {Bimod}]$. By Theorem 4.1, 
$\{[B(w)]\mid w\in W\}$ is a 
$\mathbb {Z}[v^{\pm 1}]$-basis of 
$[\mathcal {S}\mathrm {Bimod}]$. By Theorem 4.1(4), 
$\{[B_{\underline {w}}]\mid w\in W\}$ is also a 
$\mathbb {Z}[v^{\pm 1}]$-basis of 
$[\mathcal {S}\mathrm {Bimod}]$, here we fix a reduced expression 
$\underline {w}$ for each 
$w\in W$. In particular, 
$\{[B_{\underline {x}}]\mid \underline {x}\in S^{l},l\in \mathbb {Z}_{\ge 0}\}$ generates 
$[\mathcal {S}\mathrm {Bimod}]$
By Theorem 3.12, the 
$R$-module 
$B_{\underline {x}}^{w}$ is graded free. Therefore the 
$R$-module 
$B^{w}$ for any 
$B\in \mathcal {S}\mathrm {Bimod}$ is graded free. We define the character map 
$\operatorname {ch}\colon [\mathcal {S}\mathrm {Bimod}]\to \mathcal {H}$ by
\[ \operatorname{ch}(B) = \sum_{w\in W}v^{-\ell(w)}\operatorname{grk}(B^{w})H_w. \]Proposition 4.2 
$\operatorname {ch}(B_{\underline {x}}) = \underline {H}_{\underline {x}}$.
Proof. Clear from Theorem 3.12.
In particular, 
$\operatorname {ch}$ is an algebra homomorphism on 
$\sum _{\underline {x}\in S^{l},l\in \mathbb {Z}_{\ge 0}}\mathbb {Z}[v^{\pm 1}][B_{\underline {x}}]$. Since 
$\{[B_{\underline {x}}]\mid \underline {x}\in S^{l},l\in \mathbb {Z}_{\ge 0}\}$ generates 
$[\mathcal {S}\mathrm {Bimod}]$, 
$\operatorname {ch}$ is an algebra homomorphism from 
$[\mathcal {S}\mathrm {Bimod}]$ to 
$\mathcal {H}$.
For a reduced expression 
$\underline {w}$ for each 
$w\in W$, we have 
$\underline {H}_{\underline {w}}\in H_w + \sum _{y < w}\mathbb {Z}[v^{\pm 1}]H_y$. Hence 
$\{\underline {H}_{\underline {w}}\mid w\in W\}$ is a basis of 
$\mathcal {H}$. Since 
$\operatorname {ch}$ sends a basis 
$\{[B_{\underline {w}}]\mid w\in W\}$ to a basis 
$\{\underline {H}_{\underline {w}}\mid w\in W\}$, we get the following categorification theorem.
Theorem 4.3 The algebra homomorphism 
$\operatorname {ch}$ is an isomorphism 
$[\mathcal {S}\mathrm {Bimod}]\simeq \mathcal {H}$.
4.3 A formula for homomorphisms
We define:
• an involution
$h\mapsto \bar {h}$ on $\mathcal {H}$ by $\overline {\sum _{w\in W}a_w(v)H_w} = \sum _{w\in W}a_w(v^{-1})H_{w^{-1}}^{-1}$;• an anti-involution
$\omega \colon \mathcal {H}\to \mathcal {H}$ by $\omega (\sum _{w\in W}a_w(v)H_w) = \sum _{w\in W}a_w(v^{-1})H_w^{-1}$;• a
$\mathbb {Z}[v^{\pm 1}]$-linear map $\varepsilon \colon \mathcal {H}\to \mathbb {Z}[v^{\pm 1}]$ by $\varepsilon (\sum _{w\in W}a_wH_w) = a_e$. We also put $\overline {\varepsilon }(h) = \overline {\varepsilon (\overline {h})}$.
The following lemma follows from a straightforward calculation.
Lemma 4.4 The linear map 
$\varepsilon$ is a trace, that is, it satisfies 
$\varepsilon (hh') = \varepsilon (h'h)$ for any 
$h,h'\in \mathcal {H}$. The linear map 
$\overline {\varepsilon }$ is also a trace.
Lemma 4.5 Let 
$s_1,\ldots ,s_l\in S$.
(1) We have
$\omega (\underline {H}_{(s_1,\ldots ,s_l)}) = \underline {H}_{(s_l,\ldots ,s_1)}$ and $\overline {\underline {H}_{(s_1,\ldots ,s_l)}} = \underline {H}_{(s_1,\ldots ,s_l)}$.(2) We have
$p_{(s_l,\ldots ,s_1)}^{w^{-1}} = p_{(s_1,\ldots ,s_l)}^{w}$ for any $w\in W$.
Proof. A direct calculation shows that 
$\overline {\underline {H}_s} = \underline {H}_s$ and 
$\omega (\underline {H}_s) = \underline {H}_s$. Part (1) follows from this. For (2), we have
\begin{align*} \sum_{w\in W}p_{(s_l,\ldots,s_1)}^{w^{{-}1}}(v)H_{w} & = \sum_{w\in W}p_{(s_l,\ldots,s_1)}^{w}(v)H_{w^{{-}1}}\\ & = \overline{\omega\biggl(\sum_{w\in W}p_{(s_l,\ldots,s_1)}^{w}(v)H_w\biggr)}\\ & = \overline{\omega(H_{(s_l,\ldots,s_1)})}\\ & = H_{(s_1,\ldots,s_l)}\\ & = \sum_{w\in W}p_{(s_1,\ldots,s_l)}^{w}H_w. \end{align*}
Comparing the coefficient of 
$H_w$, we get the lemma.
Theorem 4.6 Let 
$B_1,B_2\in \mathcal {S}\mathrm {Bimod}$. Then 
$\operatorname {Hom}^{\bullet }_{\mathcal {S}\mathrm {Bimod}}(B_1,B_2)$ is graded free as a left 
$R$-module and its graded rank is given by
\[ \operatorname{grk}(\operatorname{Hom}^{{\bullet}}_{\mathcal{S}\mathrm{Bimod}}(B_1,B_2)) = \overline{\varepsilon}(\omega(\operatorname{ch}(B_1))\operatorname{ch}(B_2)) . \]Proof. For 
$B_1 = B_{(s_1,\ldots ,s_l)}$ and 
$B_2 = B_{(t_1,\ldots ,t_r)}$, by Lemma 2.15, we have
\begin{equation} \operatorname{Hom}^{{\bullet}}_{\mathcal{S}\mathrm{Bimod}}(B_1,B_2) \simeq \operatorname{Hom}^{{\bullet}}_{\mathcal{S}\mathrm{Bimod}}(R_e,B_{(s_l,\ldots,s_1)}B_{(t_1,\ldots,t_r)}) \simeq (B_{(s_l,\ldots,s_1,t_1,\ldots,t_r)})_e \end{equation}
and this is a graded free left 
$R$-module by Corollary 3.13. Therefore 
$\operatorname {Hom}^{\bullet }_{\mathcal {S}\mathrm {Bimod}}(B_1,B_2)$ is graded free for any 
$B_1,B_2\in \mathcal {S}\mathrm {Bimod}$.
The map 
$(B_1,B_2)\mapsto \operatorname {grk}(\operatorname {Hom}^{\bullet }_{\mathcal {S}\mathrm {Bimod}}(B_1,B_2)) - \overline {\varepsilon }(\omega (\operatorname {ch}(B_1))\operatorname {ch}(B_2))$ defines a bilinear form on the 
$\mathbb {Z}$-module 
$[\mathcal {S}\mathrm {Bimod}]\simeq \mathcal {H}$ which we denote by 
$f$. The following properties follow from a straightforward calculation:
\begin{gather} f(v^{{-}1}h_1,h_2) = f(h_1,vh_2) = vf(h_1,h_2), \end{gather}
\begin{gather} f(\underline{H}_sh_1,h_2) = f(h_1,\underline{H}_sh_2). \end{gather} If 
$B_1 = R_e$ and 
$B_2 = B_{\underline {x}}$, then 
$\operatorname {Hom}^{\bullet }_{\mathcal {S}\mathrm {Bimod}}(B_1,B_2) = (B_{\underline {x}})_e$ has graded rank 
$p_{\underline {x}}^{e}(v^{-1})$ by Corollary 3.13. By Proposition 4.2 and 
$\overline {\underline {H}_{\underline {x}}} = \underline {H}_{\underline {x}}$, 
$\overline {\varepsilon }(\omega (\operatorname {ch}(R_e))\operatorname {ch}(B_{\underline {x}})) = \overline {\varepsilon (\overline {\underline {H}_{\underline {x}}})} = \overline {\varepsilon (\underline {H}_{\underline {x}})} = \overline {p_{\underline {x}}^{e}(v)} = p_{\underline {x}}^{e}(v^{-1})$. Hence 
$f(1,\underline {H}_{\underline {x}}) = 0$. Therefore by (4.3), we have 
$f(\underline {H}_{(s_1,\ldots ,s_l)},\underline {H}_{(t_1,\ldots ,t_r)}) = f(1,\underline {H}_{(s_l,\ldots ,s_1,t_1,\ldots ,t_r)}) = 0$. Since 
$\{\underline {H}_{\underline {x}}\}$ spans 
$\mathcal {H}$ as a 
$\mathbb {Z}[v^{\pm 1}]$-module, by (4.2), 
$f = 0$ on 
$\mathcal {H}\times \mathcal {H}$.
Corollary 4.7 The graded rank of 
$\operatorname {Hom}^{\bullet }_{\mathcal {S}\mathrm {Bimod}}(B_{\underline {x}},B_{\underline {y}})$ is 
$\sum _{w\in W}p_{\underline {x}}^{w}(v^{-1})p_{\underline {y}}^{w}(v^{-1})$.
Proof. Let 
$\underline {x} = (s_1,\ldots ,s_l)$ and set 
$\underline {x'} = (s_l,\ldots ,s_1)$. Then we have 
$\omega (\underline {H}_{\underline {x}}) = \underline {H}_{\underline {x'}}$. Hence 
$\overline {\varepsilon }(\omega (\underline {H}_{\underline {x}})\underline {H}_{\underline {y}}) = \overline {\varepsilon }(\underline {H}_{\underline {x'}}\underline {H}_{\underline {y}})$. Since 
$\overline {\underline {H}_{\underline {x'}}} = \underline {H}_{\underline {x'}}$ and 
$\overline {\underline {H}_{\underline {y}}} = \underline {H}_{\underline {y}}$, we have 
$\overline {\varepsilon }(\underline {H}_{\underline {x'}}\underline {H}_{\underline {y}}) = \overline {\varepsilon (\underline {H}_{\underline {x'}}\underline {H}_{\underline {y}})}$. We have
\begin{align*} \varepsilon(\underline{H}_{\underline{x'}}\underline{H}_{\underline{y}}) & = \varepsilon\biggl(\biggl(\sum_{w_1\in W}p_{\underline{x'}}^{w_1}H_{w_1}\biggr)\biggl(\sum_{w_2\in W}p_{\underline{y}}^{w_2}H_{w_2}\biggr)\biggr)\\ & = \sum_{w_1,w_2\in W}p_{\underline{x'}}^{w_1}p_{\underline{y}}^{w_2}\varepsilon(H_{w_1}H_{w_2}). \end{align*}
We have 
$\varepsilon (H_{w_1}H_{w_2}) = \delta _{w_1^{-1},w_2}$ [Reference LibedinskyLib08, (4.3)]. Since 
$p_{\underline {x'}}^{w_1} = p_{\underline {x}}^{w_1^{-1}}$ by Lemma 4.5, we get the corollary.
5. Relations with other categories
5.1 Sheaves on moment graphs
Define an algebra 
$\mathcal {Z}$ by
\[ \mathcal{Z} = \biggl\{(z_w)\in \prod_{w\in W}R\,{\bigg|}\,z_{tw}\equiv z_w\!\!\!\!\pmod{\alpha_t}\text{ for any }w\in W\text{ and any reflection }t\biggr\}. \]
This is an 
$R$-algebra via 
$f(z_w) = (fz_w)$ for 
$f\in R$ and 
$(z_w)\in \mathcal {Z}$. This is the structure algebra of the moment graph attached to 
$(W,S)$. Fiebig developed the theory of sheaves on moment graphs; see [Reference FiebigFie08a, Reference FiebigFie08b]. In particular, he proved that the category of Soergel bimodules in the original sense is equivalent to a certain full subcategory of 
$\mathcal {Z}$-modules when the representation 
$V$ is reflection faithful. We generalize it.
If 
$f\in R$, then 
$(w(f))_{w\in W}\in \mathcal {Z}$. For any graded 
$\mathcal {Z}$-module 
$M$, we define a right action of 
$f$ as the action of 
$(w(f))_{w\in W}\in \mathcal {Z}$. Hence 
$M$ is an 
$R$-bimodule. To make a graded 
$\mathcal {Z}$-module 
$M$ into an object of 
$\mathcal {C}'$, we need a finiteness assumption on 
$M$.
For a subset 
$I\subset W$, set
\[ \mathcal{Z}^{I} = \biggl\{(z_w)\in \prod_{w\in I}R\,\bigg|\kern-3.5pt\begin{array}{l} z_{tw}\equiv z_w\pmod{\alpha_t}\text{ for any }w\in I\\ \text{and any reflection }t\in W\text{ such that }tw\in I \end{array}\!\!\!\!\biggr\}. \]
We have a canonical homomorphism 
$\mathcal {Z}\to \mathcal {Z}^{I}$. Let 
${\mathcal {Z}}\textrm {-Mod}^{f}$ be the full subcategory of the category of graded 
$\mathcal {Z}$-modules which consists of modules such that the action of 
$\mathcal {Z}$ factors through 
$\mathcal {Z}\to \mathcal {Z}^{I}$ for some finite 
$I\subset W$.
For a 
$\mathcal {Z}$-module 
$M$, let 
$M_Q = Q\otimes _R M$ and set
\[ M_Q^{x} = \{m\in M_Q\mid (z_w)m = z_xm \text{ for any }(z_w)\in \mathcal{Z}\}. \]
Then if 
$M\in {\mathcal {Z}}\textrm {-Mod}^{f}$, we have 
$M_Q = {\bigoplus} _{x\in W}\, M_Q^{x}$ [Reference FiebigFie08a, § 3.2]. It is easy to see that this defines an object of 
$\mathcal {C}'$. Hence this gives a functor 
$F\colon {\mathcal {Z}}\textrm {-Mod}^{f}\to \mathcal {C}'$.
Proposition 5.1 Let 
$M,N\in {\mathcal {Z}}\textrm {-Mod}^{f}$ and assume that 
$N$ is torsion-free as an 
$R$-module. Then we have 
$\operatorname {Hom}_{{\mathcal {Z}}\textrm {-Mod}^{f}}(M,N)\xrightarrow {\sim }\operatorname {Hom}_{\mathcal {C}'}(F(M),F(N))$.
Proof. Let 
$\varphi \colon F(M)\to F(N)$ be a homomorphism in 
$\mathcal {C}'$. Then 
$\varphi$ induces a map 
$\varphi _w\colon M_Q^{w} \to N_Q^{w}$. For 
$m\in M$, 
$z = (z_w)\in \mathcal {Z}$ and 
$w\in W$, we have 
$\varphi (zm)_w = \varphi _w((zm)_w) = \varphi _w(z_wm_w) = z_w\varphi _w(m_w) = z_w(\varphi (m)_w) = (z\varphi (m))_w$. Since the map 
$N\to N_Q = {\bigoplus} _{w\in W}\,N_Q^{w}$ is injective by the assumption, we have 
$\varphi (zm) = z\varphi (m)$.
Let 
$s\in S$ and define 
$\mathcal {Z}^{s} = \{(z_w)\in \mathcal {Z}\mid z_{ws} = z_w\}$. This is a subalgebra of 
$\mathcal {Z}$. We say that 
$V$ satisfies the GKM condition if 
$\{\alpha _{t_1},\alpha _{t_2}\}$ is linearly independent for any reflections 
$t_1\ne t_2$.
Lemma 5.2 Assume that 
$V$ satisfies GKM condition. Let 
$s\in S$ and 
$\delta \in V$ such that 
$\langle \alpha _s^{\vee },\delta \rangle = 1$. Then we have 
$\mathcal {Z} = \mathcal {Z}^{s}\oplus (w(\delta ))_{w\in W}\mathcal {Z}^{s}$.
Proof. Let 
$z = (z_w)_{w\in W}$. For each 
$w\in W$, there exists 
$y_w\in R$ such that 
$z_{w} - z_{ws} = w(\alpha _s)y_w$. Let 
$t\ne wsw^{-1}$ be a reflection. Then 
$tw(\alpha _s)\equiv w(\alpha _s)\pmod {\alpha _t}$. Hence 
$w(\alpha _s)(y_w - y_{tw})\equiv w(\alpha _s)y_w - tw(\alpha _s)y_{tw}$ 
$= (z_w - z_{tw}) - (z_{ws} - z_{tws})\equiv 0\pmod {\alpha _t}$. By the GKM condition, 
$\alpha _t$ and 
$w(\alpha _s)$ are linearly independent. Hence 
$y_w\equiv y_{tw}\pmod {\alpha _t}$. On the other hand, for 
$t = wsw^{-1}$, we have 
$y_{tw} = y_{ws} = y_w$ by the definition of 
$y_w$. Hence 
$y = (y_w)_{w\in W}\in \mathcal {Z}^{s}$. We also set 
$x_w = z_w - w(\delta )y_w$. Then we have 
$x_{ws} = z_{ws} - ws(\delta )y_w = z_w - w(\alpha _s)y_w - ws(\delta )y_w$ 
$= z_w - w(\alpha _s)y_w - (w(\delta ) - w(\alpha _s))y_w = z_w - w(\delta )y_w = x_w$. Hence 
$x = (x_w)_{w\in W}\in \mathcal {Z}^{s}$ and we have 
$z = x + (w(\delta ))_{w\in W}y$. Reversing this argument, we can get the uniqueness of 
$x,y$.
Proposition 5.3 Assume that 
$V$ satisfies GKM condition. Let 
$M\in {Z}\textrm {-Mod}^{f}$. Then 
$F(\mathcal {Z}\otimes _{\mathcal {Z}^{s}}M)\simeq F(M)\otimes _R B_s$.
Proof. Take 
$\delta \in V$ such that 
$\langle \alpha _s^{\vee },\delta \rangle = 1$. Define 
$F(M)\otimes _R B_s = F(M)\otimes _{R^{s}}R\to \mathcal {Z}\otimes _{\mathcal {Z}^{s}}M$ by 
$m\otimes f\mapsto (w(f))_{w\in W}\otimes m$. Since 
$F(M)\otimes _{R^{s}}R = F(M)\otimes 1\oplus F(M)\otimes \delta$ and 
$\mathcal {Z}\otimes _{\mathcal {Z}^{s}}M = 1\otimes M\oplus (w(\delta ))_{w\in W}\otimes M$, this homomorphism is an isomorphism.
Define a 
$\mathcal {Z}$-module structure on 
$R$ by 
$(z_w)_{w\in W}f = z_ef$ for 
$(z_w)_{w\in W}\in \mathcal {Z}$ and 
$f\in R$ and denote this 
$\mathcal {Z}$-module by 
$R(e)$. Then 
$F(R(e)) = R_e$. Let 
${\mathcal {Z}}\textrm {-Mod}^{\textrm {S}}$ be the full subcategory of 
${\mathcal {Z}}\textrm {-Mod}^{f}$ consisting of the direct summands of direct sums of 
$\{\mathcal {Z}\otimes _{\mathcal {Z}^{s_1}}\cdots \otimes _{\mathcal {Z}^{s_l}}R(e)(n)\mid s_1,\ldots ,s_l\in S,n\in \mathbb {Z}\}$. The following theorem follows from the above argument.
Theorem 5.4 Assume that 
$V$ satisfies GKM condition. The functor 
$F$ induces an equivalence 
${\mathcal {Z}}\textrm {-Mod}^{\mathrm {S}}\to \mathcal {S}\mathrm {Bimod}$.
5.2 Double leaves
Let 
$\underline {x}\in S^{l}$, 
$\underline {y}\in S^{l'}$, 
$\boldsymbol {e}\in \{0,1\}^{l}$ and 
$\boldsymbol {f}\in \{0,1\}^{l'}$ such that 
$\underline {x}^{\boldsymbol {e}} = \underline {y}^{\boldsymbol {f}}$. Set 
$w = \underline {x}^{\boldsymbol {e}}$ and fix its reduced expression 
$\underline {w}$. Then we have 
$\mathit {LL}_{\underline {x},\boldsymbol {e}}\colon B_{\underline {x}}\to B_{\underline {w}}$ and 
$\mathit {LL}^{*}_{\underline {y},\boldsymbol {f}}\colon B_{\underline {w}}\to B_{\underline {y}}$. We put 
$\mathbb {LL}_{\boldsymbol {e},\boldsymbol {f}} = \mathit {LL}^{*}_{\underline {y},\boldsymbol {f}}\circ \mathit {LL}_{\underline {x},\boldsymbol {e}}\colon B_{\underline {x}}\to B_{\underline {y}}$ which we call a double leaf.
Theorem 5.5 The set of double leaves 
$\{\mathbb {LL}_{\boldsymbol {e},\boldsymbol {f}}\mid \underline {x}^{\boldsymbol {e}} = \underline {y}^{\boldsymbol {f}}\}$ is a basis of 
$\operatorname {Hom}^{\bullet }_{\mathcal {BS}}(B_{\underline {x}},B_{\underline {y}})$.
Proof. The degree of 
$\mathbb {LL}_{\boldsymbol {e},\boldsymbol {f}}$ is 
$d(\boldsymbol {e}) + d(\boldsymbol {f})$. Therefore we have 
$\sum _{\underline {x}^{\boldsymbol {e}} = \underline {y}^{\boldsymbol {f}}}v^{-\deg (\mathbb {LL}_{\boldsymbol {e},\boldsymbol {f}})} = \sum _{w\in W}\sum _{\underline {x}^{\boldsymbol {e}} = w}v^{-d(\boldsymbol {e})}\sum _{\underline {x}^{\boldsymbol {f}} = w}v^{-d(\boldsymbol {f})} = \sum _{w\in W}p_{\underline {x}}^{w}(v^{-1})p_{\underline {y}}^{w}(v^{-1})$. This is equal to the graded rank of 
$\operatorname {Hom}^{\bullet }_{\mathcal {BS}}(B_{\underline {x}},B_{\underline {y}})$ by Corollary 4.7. Therefore it is sufficient to prove that the set of double leaves is linearly independent.
Assume that 
$\sum c_{\boldsymbol {e},\boldsymbol {f}}\mathbb {LL}_{\boldsymbol {e},\boldsymbol {f}} = 0$. Set 
$I = \{\underline {x}^{\boldsymbol {e}}\mid \underline {x}^{\boldsymbol {e}} = \underline {y}^{\boldsymbol {f}},c_{\boldsymbol {e},\boldsymbol {f}}\ne 0\}$ and assume that 
$I$ is not empty. Let 
$w\in I$ be an element with maximal length. Let 
$\bar {I} = \{x\in W\mid \text{ there exists } y\in I$ 
$\bar {I} = \{x\in W\mid \text{ there exists } y\in I$. Then 
$w$ is also has maximal length in 
$\bar {I}$ and 
$\bar {I}$ is closed. Set 
$\bar {I}' = \bar {I}\setminus \{w\}$. Then 
$c_{\boldsymbol {e},\boldsymbol {f}}\mathbb {LL}_{\boldsymbol {e},\boldsymbol {f}}$ factors through 
$B_{\underline {y},\bar {I}}\hookrightarrow B_{\underline {y}}$. Since 
$\bar {I}'\supset I\setminus \{w\}$, we have 
$\pi _{\underline {y}}^{w}\circ (c_{\boldsymbol {e},\boldsymbol {f}}\mathbb {LL}_{\boldsymbol {e},\boldsymbol {f}}) = 0$ unless 
$\underline {x}^{\boldsymbol {e}} = w$.
Set 
$E = \{\boldsymbol {e}\mid c_{\boldsymbol {e},\boldsymbol {f}}\ne 0, \underline {x}^{\boldsymbol {e}} = w\}$. Recall that we have a total order 
${<} = {<_{\underline {x},w}}$ on this set. Let 
$\boldsymbol {e}'\in E$ be the minimal element. Then, for 
$\boldsymbol {e}\in E$, we have 
$\mathit {LL}_{\underline {x},\boldsymbol {e}}(b_{\underline {x},\boldsymbol {e}'}) = 0$ unless 
$\boldsymbol {e} = \boldsymbol {e}'$ by Proposition 3.10. Therefore we have 
$\sum _{\underline {y}^{\boldsymbol {f}} = w}c_{\boldsymbol {e}',\boldsymbol {f}}(\pi _{\underline {y}}^{w}(\mathbb {LL}_{\boldsymbol {e}',\boldsymbol {f}}(b_{\underline {x},\boldsymbol {e}'}))) = 0$. Since 
$\mathit {LL}_{\underline {x},\boldsymbol {e}'}(b_{\underline {x},\boldsymbol {e}'}) = u_{\underline {w}}$ (Proposition 3.10), we have 
$\sum _{\underline {y}^{\boldsymbol {f}} = w}{c_{\boldsymbol {e}',\boldsymbol {f}}}\pi _{\underline {y}}^{w}(\mathit {LL}^{*}_{\underline {y},\boldsymbol {f}}(u_{\underline {w}})) = 0$. By Theorem 3.17, 
$c_{\boldsymbol {e}',\boldsymbol {f}} = 0$ for any 
$\boldsymbol {f}$ such that 
$\underline {y}^{\boldsymbol {f}} = w$. This is a contradiction.
5.3 The Elias–Williamson category
Let 
$\mathcal {D}$ be the category defined in [Reference Elias and WilliamsonEW16]. In this subsection we assume that the triple 
$(V,\{\alpha _s\}_{s\in S},\{\alpha ^{\vee }_s\}_{s\in S})$ satisfies the following conditions.
(1) The triple
$(V,\{\alpha _s\}_{s\in S},\{\alpha ^{\vee }_s\}_{s\in S})$ is a realization, that is, it satisfies the technical condition [Reference Elias and WilliamsonEW16, Definition 3.1]. The realization also satisfies the Demazure surjectivity, that is, $\alpha _s\colon V^{*}\to \mathbb {K}$ is surjective for any $s\in S$.(2) For any
$s,t\in S$ such that the order of $st$ is finite, we have:
• local non-degeneracy [Reference EliasEli16, Assumption 3.13];
• lesser invertibility [Reference EliasEli16, Assumption 3.14].
From the first assumption, the set of double leaves is a basis of the space of homomorphisms in 
$\mathcal {D}$. That is, Theorem 5.5 holds in the context of 
$\mathcal {D}$; see [Reference Elias and WilliamsonEW16, Theorem 6.12] for the detailed statement.
From the second assumption, we have a well-defined functor 
$\mathcal {F}$ from 
$\mathcal {D}$ to the category of graded 
$R$-bimodules [Reference Elias and WilliamsonEW16, Claim 5.14]. The assumptions in [Reference Elias and WilliamsonEW16] are slightly implicit. In the proof in [Reference Elias and WilliamsonEW16], to construct the functor and to prove that the functor satisfies a part of defining relations of 
$\mathcal {D}$, they use results of Elias [Reference EliasEli16]. Elias assumed the two conditions in his paper [Reference EliasEli16, Definition 6.18]. (In fact, he also assumed Demazure surjectivity, which we have already assumed.)
Remark 5.6 Assume that 
$\mathbb {K}$ is a field. Let 
$s,t\in S$ such that the order of 
$st$ is finite. If the restriction of 
$V$ to 
$\langle s,t\rangle$ is finite, then the assumptions hold [Reference EliasEli16, § 3.9].
Lemma 5.7 We make the above assumptions. Let 
$s,t\in S$ such that the order 
$m$ of 
$st$ is finite. Then the image of the 
$2m$-valent vertex by 
$\mathcal {F}$ satisfies Assumption 3.2.
Proof. We may assume 
$S = \{s,t\}$. Set 
$\underline {x} = (s,t,\ldots )\in S^{m}$ and 
$\underline {y} = (t,s,\ldots )\in S^{m}$. The graded rank of 
$\operatorname {grk} \operatorname {Hom}_{\mathcal {S}\mathrm {Bimod}}(B_{\underline {x}},B_{\underline {y}})$ is 
$\overline {\varepsilon }(\omega (\operatorname {ch}(B_{\underline {x}}))\operatorname {ch}(B_{\underline {y}})) = \overline {\varepsilon (\underline {H}_{\underline {x}}\underline {H}_{\underline {y}})}$ by Theorem 4.6. The constant term of this (that is, the dimension of the space of degree-zero homomorphisms from 
$B_{\underline {x}}$ to 
$B_{\underline {y}}$) is 
$1$ (see the proof of [Reference LibedinskyLib08, Proposition 4.3]). In particular, there exists only one morphism which satisfies Assumption 3.2. Let 
$\psi \colon B_{\underline {x}}\to B_{\underline {y}}$ be the morphism.
Let 
$\varphi \colon \underline {x}\to \underline {y}$ be the 
$2m$-valent vertex. Then 
$\varphi$ is homogeneous of degree zero and hence 
$\mathcal {F}(\varphi )$ is also homogeneous of degree zero. Therefore, as we have shown above, 
$\mathcal {F}(\varphi ) = c\psi$ for some 
$c\in \mathbb {K}$.
To prove 
$c = 1$, we use the calculation in localizations. Let 
$\mathcal {S}\mathrm {Bimod}_Q$ be the full subcategory of 
$Q$-bimodules such that 
$X\in \mathcal {S}\mathrm {Bimod}_Q$ if and only if 
$X$ is a direct sum of 
$Q_w$ with 
$w\in W$. We fix an isomorphism 
$(B_s)_Q\simeq Q_e\oplus Q_s$ as 
$f\otimes g\mapsto (fg,fs(g))$. Then, by extending this, we have an isomorphism 
$(B_{\underline {x}})_Q \simeq {\bigoplus} _{\mathbf {e}\in \{0,1\}^{m}}\,Q_{\underline {x}^{\mathbf {e}}}$. For each 
$B_{\underline {x}}\in \mathcal {BS}$, by attaching 
${\bigoplus} _{\mathbf {e}\in \{0,1\}^{m}}\,Q_{\underline {x}^{\mathbf {e}}}$, we have the functor 
$\mathcal {G}\colon \mathcal {BS}\to \mathcal {S}\mathrm {Bimod}_Q$. Define 
$\operatorname {Kar}(\mathcal {D}_Q),\mathcal {D}_Q^{\mathrm {std}},\mathrm {S}\mathit {td}$ and 
$\mathcal {F}^{\mathrm {std}}$ as in [Reference Elias and WilliamsonEW16, Definitions 5.15, 5.16, 5.22 and 4.9]. Then 
$\mathcal {D}\xrightarrow {\mathcal {F}}\mathcal {BS}\xrightarrow {\mathcal {G}} \mathcal {S}\mathrm {Bimod}_Q$ and 
$\mathcal {D}\xrightarrow {\mathcal {S}\mathit {td}} \operatorname {Kar}(\mathcal {D}_Q)\xrightarrow {\mathcal {F}^{\mathrm {std}}}\mathcal {S}\mathrm {Bimod}_Q$ are the same (see the proof of [Reference Elias and WilliamsonEW16, Proposition 5.27]). Let 
$w_0$ be the longest element in 
$\langle s,t\rangle$. Then 
$Q_{w_0}$ is a direct summand of both 
$\mathcal {G}(B_{\underline {x}})$ and 
$\mathcal {G}(B_{\underline {y}})$ with multiplicity one. We consider the composition 
$Q_{w_0}\hookrightarrow \mathcal {G}(B_{\underline {x}})\xrightarrow {\eta }\mathcal {G}(B_{\underline {y}})\to Q_{w_0}$ for 
$\eta = \mathcal {G}(\mathcal {F}(\varphi ))$ and 
$\mathcal {G}(\psi )$. To prove 
$c = 1$, it is sufficient to prove that the composition is the identity for both 
$\eta = \mathcal {G}(\mathcal {F}(\varphi ))$ and 
$\eta = \mathcal {G}(\psi )$. For 
$\eta = \mathcal {G}(\mathcal {F}(\varphi ))$, it is the identity [Reference Elias and WilliamsonEW16, § 5.5]. For 
$\eta = \mathcal {G}(\psi )$, take 
$q\in Q$ such that the composition is given by 
$a\mapsto qa$. By the definition, the projection 
$(B_{\underline {x}})_Q\to Q_{w_0}$ is given by 
$\varphi _{\underline {x}}$ which is defined in § 3.4. Hence the following diagram commutes.
Since 
$\psi$ satisfies Assumption 3.2, we have 
$\mathcal {G}(\psi )(u_{\underline {x}}) = u_{\underline {y}}$. By the definition of 
$\varphi _{\underline {x}}$, we have 
$\varphi _{\underline {x}}(u_{\underline {x}}) = 1$. Hence by considering the image of 
$u_{\underline {x}}\in B_{\underline {x}}$ in the above diagram, we conclude that 
$q = 1$.
The functor 
$\mathcal {F}$ is defined as a functor from 
$\mathcal {D}$ to the category of graded 
$R$-bimodules. We can also consider it as a functor to 
$\mathcal {BS}$ as follows. For 
$\underline {x}$, we put 
$\mathcal {F}(\underline {x}) = B_{\underline {x}}$. We have to prove that 
$\mathcal {F}(\varphi )$ is a morphism in 
$\mathcal {BS}$ for any morphism 
$\varphi$ in 
$\mathcal {D}$. The morphisms in 
$\mathcal {D}$ is generated by three types of morphisms: univalent vertices, trivalent vertices and the morphism in the previous lemma. If a morphism 
$\varphi$ in 
$\mathcal {D}$ is a univalent or trivalent vertex then it is easy to see that 
$\mathcal {F}(\varphi )$ is a morphism in 
$\mathcal {BS}$. By the previous lemma, if 
$\varphi$ is the morphism in that lemma, 
$\mathcal {F}(\varphi )$ is a morphism in 
$\mathcal {BS}$. Therefore 
$\mathcal {F}$ defines a functor 
$\mathcal {D}\to \mathcal {BS}$ which we denote by the same letter 
$\mathcal {F}$.
By flipping the diagram upside-down, we have a contravariant functor 
$D'\colon \mathcal {D}\to \mathcal {D}$. Recall that we also have the functor 
$D\colon \mathcal {BS}\to \mathcal {BS}$.
Lemma 5.8 We have 
$D\circ \mathcal {F}\simeq \mathcal {F}\circ D'$.
Proof. We fix an isomorphism 
$D(B_{\underline {x}})\simeq B_{\underline {x}}$ obtained from the proof of Lemma 2.20. Then, for any 
$B\in \mathcal {D}$, 
$D\circ \mathcal {F}(B)\simeq \mathcal {F}(D'(B))$ naturally. We prove 
$D(\mathcal {F}(\varphi )) = \mathcal {F}(D'(\varphi ))$ for any morphism 
$\varphi$ in 
$\mathcal {D}$. If 
$\varphi$ is a univalent or trivalent vertex, it is easy to see that 
$D(\mathcal {F}(\varphi )) = \mathcal {F}(D(\varphi ))$ by direct calculation. Let 
$m$ be the order of 
$st$ with 
$s,t\in S$ and assume that 
$m < \infty$. Let 
$\varphi$ be the 
$2m$-valent vertices. To prove 
$D(\mathcal {F}(\varphi )) = \mathcal {F}(D'(\varphi ))$, by the previous lemma, it is sufficient to prove that the dual of the morphism satisfying Assumption 3.2 again satisfies Assumption 3.2.
Let 
$\underline {x},\underline {y}$ be as in the previous lemma and 
$\psi \colon B_{\underline {x}}\to B_{\underline {y}}$ (respectively, 
$\psi '\colon B_{\underline {y}}\to B_{\underline {x}}$) be the morphism satisfying Assumption 3.2. Then 
$D(\psi ) \colon B_{\underline {y}}\to B_{\underline {x}}$ is homogeneous of degree zero, hence 
$D(\psi ) = c\psi '$ for some 
$c\in \mathbb {K}$ since such degree-zero morphism is unique up to constant multiple. Let 
$w_0$ be the longest element in 
$\langle s,t\rangle$. Then 
$Q_{w_0}$ is a direct summand of 
$(B_{\underline {x}})_Q$ and 
$(B_{\underline {y}})_Q$ with multiplicity one. Let 
$p_{\underline {x}}$ and 
$i_{\underline {x}}$ be the projection to and the embedding from 
$Q_{w_0}$, and the same for 
$p_{\underline {y}}$ and 
$i_{\underline {y}}$. To prove 
$c = 1$, it is sufficient to prove that 
$Q_{w_0}\xrightarrow {i_{\underline {y}}} (B_{\underline {y}})_Q\simeq D(B_{\underline {y}})_Q\xrightarrow {D(\psi )}D(B_{\underline {x}})_Q\simeq (B_{\underline {x}})_Q\xrightarrow {p_{\underline {x}}} Q_{w_0}$ is the identity since 
$\psi '$ satisfies this by the proof of the previous lemma. Obviously 
$Q_{w_0}\xrightarrow {D(p_{\underline {y}})}D(B_{\underline {y}})\xrightarrow {D(\psi )}D(B_{\underline {x}})\xrightarrow {D(i_{\underline {x}})}Q_{w_0}$ is the identity. Hence it is sufficient to prove that 
$D(p_{\underline {y}}) = i_{\underline {y}}$ under the fixed isomorphism 
$D(B_{\underline {y}})\simeq B_{\underline {y}}$. By the construction of 
$p_{\underline {y}},i_{\underline {y}}$, it is sufficient to prove the same thing for 
$B_s$, namely, the dual of 
$(B_s)_Q\to Q_s$ is 
$Q_s\hookrightarrow (B_s)_Q$. This can be checked directly.
Theorem 5.9 The category 
$\mathcal {D}$ is equivalent to 
$\mathcal {BS}$. Therefore the category 
$\operatorname {Kar}(\mathcal {D})$ is equivalent to 
$\mathcal {S}\mathrm {Bimod}$.
Proof. It is sufficient to prove that 
$\mathcal {F}$ is fully faithful. By the constructions, Lemmas 5.7 and 5.8, the functor 
$\mathcal {F}$ sends double leaves to double leaves. Since the set of double leaves gives a basis of the space of homomorphisms in both categories, the functor 
$\mathcal {F}$ is fully faithful.
Acknowledgements
The author was supported by JSPS KAKENHI Grant Number 18H01107. The author thanks Henning Haahr-Andersen and Masaharu Kaneda for reading the paper and giving helpful comments. The author also thanks the referee for a careful reading and giving many helpful comments.
