Whittaker Categories, Properly Stratified Categories and Fock Space Categorification for Lie Superalgebras

Abstract

We study various categories of Whittaker modules over a type I Lie superalgebra realized as cokernel categories that fit into the framework of properly stratified categories. These categories are the target of the Backelin functor $\Gamma _\zeta$. We show that these categories can be described, up to equivalence, as Serre quotients of the BGG category $\mathcal O$ and of certain singular categories of Harish-Chandra $({\mathfrak {g}},{\mathfrak {g}}_{{\bar{0}}})$-bimodules. We also show that $\Gamma _\zeta$ is a realization of the Serre quotient functor. We further investigate a q-symmetrized Fock space over a quantum group of type A and prove that, for general linear Lie superalgebras our Whittaker categories, the functor $\Gamma _\zeta$ and various realizations of Serre quotients and Serre quotient functors categorify this q-symmetrized Fock space and its q-symmetrizer. In this picture, the canonical and dual canonical bases in this q-symmetrized Fock space correspond to tilting and simple objects in these Whittaker categories, respectively.

Appendices

Appendix A: Structural Modules in $\mathcal O^{\nu \text {-pres}}$

The goal of this section is to describe the simple, standard and proper standard objects in $\mathcal O^{\nu \text {-pres}}.$

For a given module $M\in \mathcal O$, let $\text {Tr}_{\nu }(M)$ denote the sum of the images of all homomorphisms from the $\nu$-admissible projective modules to M. For a fixed $\lambda \in {\Lambda (\nu )}$, denote by $A(\lambda )$ the kernel of the canonical epimorphism $P(\lambda ) \twoheadrightarrow M(\lambda )$. Consider the following modules:

$$\begin{aligned}&S(\lambda ):= P(\lambda )/\text {Tr}_{\nu }(\textrm{rad}P(\lambda )), \end{aligned}$$

(8.1)

$$\begin{aligned}&D(\lambda ):= P(\lambda )/\text {Tr}_{\nu }(A(\lambda )), \end{aligned}$$

(8.2)

Define $Q(\lambda )$ to be the quotient of $P(\lambda )$ modulo the sum of the images of all homomorphisms $P(\mu )\rightarrow P(\lambda )$, where $\mu \in {\Lambda (\nu )}$ and $\mu >\lambda$. Denote the natural projection by

$$\begin{aligned} \varpi : P(\lambda )\rightarrow Q(\lambda ). \end{aligned}$$

(8.3)

By construction, the modules $S(\lambda ), D(\lambda )$ and $Q(\lambda )$ lie in $\mathcal O^{\nu \text {-pres}}$. We recall that $\textbf{T}_{w_0^\nu }$ denotes the twisting functor on $\mathcal O$ associated to $w^\nu _0$; see Sect. 6.2.1. The following proposition describe these modules as structural objects with respect to the stratified structure on $\mathcal O^{\nu \text {-pres}}$:

Proposition 49

Suppose that ${\mathfrak {g}}$ is one of the Lie superalgebras from the series in (2.2)–(2.4). For any $\lambda \in {\Lambda (\nu )}$, we have

$$\begin{aligned}&S(\lambda )\cong \textbf{T}_{w_0^\nu }L(\lambda )\text { and } \pi (S(\lambda ))\cong \pi (L(\lambda )), \end{aligned}$$

(8.4)

$$\begin{aligned}&D(\lambda )\cong \textbf{T}_{w_0^\nu }M(\lambda )\text { and } \pi (D(\lambda ))\cong \overline{\Delta }(\lambda ), \end{aligned}$$

(8.5)

$$\begin{aligned}&\pi (Q(\lambda ))\cong \Delta (\lambda ). \end{aligned}$$

(8.6)

We will prove Proposition 49 using the following lemmas.

Lemma 50

For any weight $\lambda \in {\mathfrak {h}}^*$, there is a direct summand P of ${\text {Res}}P(\lambda )$ such that $P\cong P_0(\lambda )$ and $U({\mathfrak {g}})P =P(\lambda )$.

Proof

We first decompose ${\text {Res}}P(\lambda ) =\bigoplus _{i=1}^{\ell }P_0(\mu _i)$. Consider the canonical quotients

$$\begin{aligned}&p: P(\lambda ) \twoheadrightarrow L(\lambda ), \end{aligned}$$

(8.7)

$$\begin{aligned}&q:K(L_0(\lambda ))\twoheadrightarrow L(\lambda ), \end{aligned}$$

(8.8)

where p factors through q. By (8.8), the ${\mathfrak {g}}_{{\bar{0}}}$-module ${\text {Res}}L(\lambda )$ has a quotient isomorphic to $L_0(\lambda )$. Therefore, ${\text {Res}}P(\lambda )$ has $L_0(\lambda )$ as a quotient as well. This implies that there is $1\le i\le \ell$ such that $\mu _i=\lambda$, and, if we let $P:=P_0(\mu _i)$, then $P\not \subseteq \ker (p)$. The fact that $U({\mathfrak {g}})P =P(\lambda )$ then follows by combining the uniqueness of the maximal submodule of $P(\lambda )$ with $p(P)\ne 0$. $\square$

For any $\lambda \in {\Lambda (\nu )}$, define

$$\begin{aligned} {\Lambda (\nu )}_{>\lambda }:=\{\mu \in {\Lambda (\nu )}\text { such that } \mu >\lambda \}. \end{aligned}$$

Lemma 51

For each $\lambda \in \Lambda (\nu )$, there is a short exact sequence of ${\mathfrak {g}}$-modules as follows:

$$\begin{aligned}&0\rightarrow X \rightarrow {\text {Ind}}P_0(\lambda ) \rightarrow K(P_0(\lambda )) \rightarrow 0, \end{aligned}$$

Here $X = U({\mathfrak {g}}_{-1})\cdot ({\mathfrak {g}}_1U({\mathfrak {g}}_1))\otimes P_0(\lambda )\subset {\text {Ind}}P_0(\lambda )$ has a Kac flag, subquotients of which are isomorphic to direct sums of the modules of the form $K(P_0(\gamma ))$, where $\gamma \in {{\Lambda (\nu )}_{>\lambda }}$.

Furthermore, X is an epimorphic image of a direct sum of modules of the form $P(\gamma )$, where $\gamma \in {{\Lambda (\nu )}_{>\lambda }}.$

Proof

For any $k>1$, we may observe that

$$\begin{aligned}&\,{\text {Hom}}_{{\mathfrak {g}}_0}(\Lambda ^k \mathfrak g_1\otimes P_0(\lambda ), L_0(\gamma )) = [ \Lambda ^k \mathfrak g_1^*\otimes L_0(\gamma ): L_0(\lambda )], \end{aligned}$$

which implies that $\Lambda ^k({\mathfrak {g}}_1)\otimes P_0(\lambda )$ decomposes into a direct sum of modules $P_0(\gamma )$, with $\gamma \in {{\Lambda (\nu )}_{>\lambda }}$. In particular, the ${\mathfrak {g}}_{\ge 0}$-module ${\text {Ind}}_{{\mathfrak {g}}_0}^{{\mathfrak {g}}_{\ge 0}}P_0(\lambda )$ has a filtration, subquotients of which are direct sums of $P_0(\gamma )$ such that ${\mathfrak {g}}_1P_0(\gamma )=0$ and $\gamma \in {{\Lambda (\nu )}_{>\lambda }}\cup \{\lambda \}$. It follows that the module

$$\begin{aligned}&{\text {Ind}}P_0(\lambda ) \cong {\text {Ind}}_{{\mathfrak {g}}_{\ge 0}}^{\mathfrak {g}}{\text {Ind}}_{{\mathfrak {g}}_0}^{{\mathfrak {g}}_{\ge 0}} P_0(\lambda ), \end{aligned}$$

admits a Kac flag, subquotients of which are of the form $K(P_0(\gamma ))$, where we have $\gamma \in {{\Lambda (\nu )}_{>\lambda }}\cup \{\lambda \}$. By [CCM, Theorem 51], each $K(P_0(\gamma ))$ has simple top, which is, automatically, isomorphic to $L(\gamma )$. Consequently, $K(P_0(\gamma ))$ is a quotient of $P(\gamma )$. This completes the proof. $\square$

Proof of Proposition 49

The isomorphisms in (8.4), (8.5) are proved in [Chi1, Theorem 12]. It remains to prove the isomorphism in (8.6). To see this, we first recall the definitions of the Kac functor $K(-)$ and the module $\Delta _0(\lambda )$ from Sects. 5 and 5.1. We shall show that $K(\Delta _0'(\lambda ))\cong Q(\lambda )$.

Let $Q_0(\lambda )$ denote the quotient of $P_0(\lambda )$ by the sum of the traces of all $P_0(\mu )$, where $\mu \in {{\Lambda (\nu )}_{>\lambda }}$, in $P_0(\lambda )$.

Let P be the ${\mathfrak {g}}_0$-submodule of ${\text {Res}}P(\lambda )$ described in Lemma 50. Recalling the map $\varpi$ from (8.3), we see that $\varpi (P)\ne 0$. In particular, the ${\mathfrak {g}}_0$-submodule $\varpi (P)$ is a quotient of $P\cong P_0(\lambda )$. We are going to show that there is an ${\mathfrak {g}}_{{\bar{0}}}$-epimorphism

$$\begin{aligned}&Q_0(\lambda ) \twoheadrightarrow \varpi (P), \end{aligned}$$

(8.9)

equivalently, we shall show that $[\varpi (P):L_0(\gamma )]=0$, for any $\gamma \in {{\Lambda (\nu )}_{>\lambda }}$.

We first claim that ${\mathfrak {g}}_1 \varpi (P)=0,$ namely, that ${\mathfrak {g}}_1 P\in \ker (\varpi )$. To see this, we consider the epimorphism ${\text {Ind}}(P) \twoheadrightarrow U({\mathfrak {g}})P$ sending $x\otimes v$ to xv, for any $x\in U({\mathfrak {g}})$ and $v\in P$. By Lemma 51, the ${\mathfrak {g}}$-module $U({\mathfrak {g}}_{-1}){\mathfrak {g}}_1U({\mathfrak {g}}_1)P$ is an epimorphic image of a direct sum of $P(\gamma )$’s, where $\gamma \in {{\Lambda (\nu )}_{>\lambda }}$. Consequently, we have ${\mathfrak {g}}_1P\subseteq \ker (\varpi )$, as desired.

Next, we suppose that $[\varpi (P):L_0(\gamma )]>0$, for some weight $\gamma$. Then there is a ${\mathfrak {g}}_0$-submodule Y of $\varpi (P)$ having a quotient $L_0(\gamma )$. Consider

$$\begin{aligned}&\,{\text {Hom}}_{\mathfrak {g}}(K(Y), Q(\lambda )) =\,{\text {Hom}}_{{\mathfrak {g}}_0}(Y, Q(\lambda )^{{\mathfrak {g}}_1}) \supseteq \,{\text {Hom}}_{{\mathfrak {g}}_0}(Y, \varpi (P)) \ne 0, \end{aligned}$$

where $Q(\lambda )^{{\mathfrak {g}}_1}$ denotes the ${\mathfrak {g}}_1$-invariants of $Q(\lambda )$. This leads to a non-zero homomorphism from $P(\gamma )$ to $Q(\lambda )$. Consequently, $\gamma$ cannot be a weight in ${{\Lambda (\nu )}_{>\lambda }}$. This proves (8.9).

Finally, we note that $U({\mathfrak {g}})\varpi (P)=Q(\lambda )$ by Lemma 50. Then we consider

$$\begin{aligned}&\,{\text {Hom}}_{\mathfrak {g}}(K(\Delta '_0(\lambda )), Q(\lambda )) =\,{\text {Hom}}_{{\mathfrak {g}}_0}(\Delta '_0(\lambda ), Q(\lambda )^{{\mathfrak {g}}_1}). \end{aligned}$$

Since $Q_0(\lambda )$ is isomorphic to the parabolically induced ${\mathfrak {g}}_{{\bar{0}}}$-module $\Delta '_0(\lambda )$ from the projective-injective $\mathfrak l_\nu$-module $P(\mathfrak l_\nu , \lambda )$, we obtain an epimorphism from $K(\Delta '_0(\lambda ))$ to $Q(\lambda )$.

Finally, it remains to show that $K(\Delta _0'(\lambda ))$ is an epimorphic image of $Q(\lambda )$. By [CCM, Theorem 51], the top of $K(\Delta _0'(\lambda ))$ is simple, and so it is isomorphic to $L(\lambda )$. Then we get an epimorphism $f: P(\lambda )\twoheadrightarrow K(\Delta _0'(\lambda ))$. It remains to show that, if $\eta \in {\Lambda (\nu )}$ is such that $[K(\Delta _0'(\lambda )): L(\eta )]\ne 0$, then $\eta \not > \lambda$. We will show the stronger statement that $\eta \le \lambda$. To see this, we observe that

$$\begin{aligned}&\textrm{ch}K(\Delta _0'(\lambda )) = \sum _{\mu \in W_\nu \cdot \lambda } \textrm{ch}M(\mu ), \end{aligned}$$

which implies that there is $\mu \in W_\nu \cdot \lambda$ such that $[M(\mu ):L(\eta )]\ne 0$. Since $\mu \in W_\nu \cdot \lambda$ and $\lambda , \eta \in {\Lambda (\nu )}$, we have $[M(\mu )/M(\lambda ):L(\eta )]=0$ and thus $[M(\mu ):L(\eta )] = [M(\lambda ):L(\eta )]$. The conclusion follows. $\square$

Example 52

Consider ${\mathfrak {g}}=\mathfrak {sl}(2)$ with $W_\nu =W$. Let $\lambda \in {\Lambda (\nu )}$. If $\lambda$ is regular, then $S(\lambda )=D(\lambda )$ is isomorphic to the the dual Verma module of highest weight $w_0^\nu \cdot \lambda$ and $Q(\lambda )= P(\lambda )$. If $\lambda$ is singular, then $S(\lambda ) =D(\lambda ) =Q(\lambda )$.

Appendix B: A Realization of ${\overline{\mathcal O}}$ and Its Graded Version

In this appendix, we realize ${\overline{\mathcal O}}$ as the category of finite-dimensional (locally unital) modules over a locally unital algebra. In the case ${\mathfrak {g}}=\mathfrak {gl}(m|n)$, we study the graded version of ${\overline{\mathcal O}}$ and show that all structural modules in ${\overline{\mathcal O}}$ have graded lifts.

1.1 Morita equivalence for $\mathcal O$

We define A to be the locally unital algebra of $\mathcal O$, that is

$$\begin{aligned}&A:=\bigoplus _{\lambda , \mu \in \Lambda } \,{\text {Hom}}_{\mathfrak g}(P(\lambda ), P(\mu )). \end{aligned}$$

For $\lambda \in {\Lambda (\nu )}$, denote by $1_\lambda$ the identity endomorphism on $P(\lambda )$. Then $\{1_\lambda |~\lambda \in \Lambda \}$ forms a complete set of mutually orthogonal idempotents and $A = \bigoplus _{\lambda ,\mu \in \Lambda }1_\mu A 1_\lambda$. Define $\text {mof}$-A to be the category of finite-dimensional locally unital right A-modules. This means that the objects of $\text {mof}$-A are finite dimensional right A-modules M such that $M= \bigoplus _{\lambda \in \Lambda } M1_\lambda$ and morphisms in $\text {mof}$-A are homomorphism of A-modules. Then we have a Morita type equivalence (c.f. [BLW, Section 2.1])

$$\begin{aligned}&T(-):\mathcal O\xrightarrow {\cong }\text {mof-}A, \end{aligned}$$

via the functor $T(M):=\bigoplus _{\lambda \in \Lambda } \,{\text {Hom}}_{\mathfrak {g}}(P(\lambda ), M)$, where $M\in \mathcal O$. We may note that $T(P(\lambda )) = 1_\lambda A$ and $\dim 1_\lambda A, \dim A1_\lambda <\infty$.

In particular, for a given $\lambda \in \Lambda$ (respectively $\mu \in \Lambda$), there are only finitely many $\mu \in \Lambda$ (respectively $\lambda \in \Lambda$) such that $\,{\text {Hom}}_{\mathfrak {g}}(P(\lambda ), P(\mu ))\ne 0$. Hence $A= \bigoplus _{\lambda ,\mu \in \Lambda }\,{\text {Hom}}_A(1_\lambda A,1_\mu A)$.

1.2 Morita equivalence for ${\overline{\mathcal O}}$

We define a subalgebra B of A as follows

$$\begin{aligned}&B:=\bigoplus _{\lambda , \mu \in {\Lambda (\nu )}} \,{\text {Hom}}_{\mathfrak g}(P(\mu ), P(\lambda ))=\bigoplus _{\lambda ,\mu \in {\Lambda (\nu )}} 1_\lambda A 1_\mu . \end{aligned}$$

(9.1)

Similarly, we define ${\text {mof-}B}$ to be the category of all finite dimensional locally unital right B-modules.

Lemma 53

There is an equivalence $\overline{\mathcal O}\cong {\text {mof-}B}$.

Proof

Let $\overline{{\text {mof-}A}}$ denote the Serre quotient of ${\text {mof-}A}$ by the Serre subcategory $I_\nu$ of all modules $M\in {\text {mof-}A}$ which satisfy $M1_\lambda =0$, for all $\lambda \in {\Lambda (\nu )}$. We observe that $I_\nu =T(\mathcal I^\nu )$. Therefore, $\overline{{\text {mof-}A}}$ is equivalent to $\overline{\mathcal O}$ through the Morita equivalence functor T. Therefore we have a commutative diagram

where $\pi ^A$ denotes the quotient functor from ${\text {mof-}A}$ to $\overline{{\text {mof-}A}}$.

In order to complete the proof we define a functor $\pi ': {\text {mof-}A}\rightarrow {\text {mof-}B}$ by letting

$$\begin{aligned}&\pi ': M\mapsto \bigoplus _{\lambda \in {\Lambda (\nu )}} M1_\lambda . \end{aligned}$$

We are now in the position to invoke [CP, Lemma A.2.1] which says that $\pi '$ induces an equivalence $\widetilde{\pi '}: \overline{{\text {mof-}A}}\rightarrow {\text {mof-}B}$ such that $\pi '= \widetilde{\pi '} \circ \pi ^A$. $\square$

By the proof of Lemma 53, there are functors $T^B: \overline{\mathcal O} \xrightarrow {\cong } {\text {mof-}B}$ and $\pi ': {\text {mof-}A}\rightarrow {\text {mof-}B}$ that make the following diagram commutative:

Corollary 54

The exact functor

$$\begin{aligned}&\pi '\circ T: \mathcal O\rightarrow {\text {mof-}B}, ~M\mapsto \bigoplus _{\lambda \in {\Lambda (\nu )}}\,{\text {Hom}}_{\mathfrak {g}}(P(\lambda ),M) \end{aligned}$$

(9.2)

satisfies the universal property for the Serre quotient ${\overline{\mathcal O}}$.

1.3 Graded category $\mathcal O$ for $\mathfrak {gl}(m|n)$

We now consider the case that ${\mathfrak {g}}=\mathfrak {gl}(m|n)$. Recall the simple-preserving contragredient duality $(-)^\vee :\mathcal O\rightarrow \mathcal O$ from Sect. 2.5. In this case, the algebra A admits a positive grading such that A is a standard Koszul algebra. We refer to [Bru4, Sections 3,4] and [BLW] for the details. With respect to this grading, the idempotents $1_\lambda$, for $\lambda \in \Lambda$, have degree zero.

Example 55

For ${\mathfrak {g}}=\mathfrak {gl}(1|1)$, the corresponding category $\mathcal O$ is equivalent to the category of finite-dimensional locally unital right modules over the path algebra of the following infinite quiver

modulo the relations $x_iy_i=y_{i+1}x_{i+1},~x_{i+1}x_{i} =y_{i}y_{i+1}=0,$ for all $i\in {\mathbb Z}$. The grading is given by $\deg (x_i)=\deg (y_i)=1.$

We denote by ${\text {gmof-}A}$ the category of graded finite-dimensional unital right A-modules with homogenous homomorphisms of degree zero. Also, we denote by $\mathbb F^A$ the functor from ${\text {gmof-}A}$ to ${\text {mof-}A}$ forgetting the grading. A object $\dot{X}\in {\text {gmof-}A}$ is called a graded lift of $X\in {\text {mof-}A}$ (respectively $X\in \mathcal O$) provided that $\mathbb F^A(\dot{X})\cong X$ (respectively $\mathbb F^A(\dot{X})\cong T(X)$). Let $\langle 1 \rangle$ denote the degree shift functor defined on graded modules, namely, for a given graded module M, $M\langle 1 \rangle$ has the same module structure but $(M\langle 1 \rangle )_{n}:=M_{n-1}$, for any homogenous subspace $M_n$. For $m\in {\mathbb Z}$, we set $\langle m \rangle := \langle 1\rangle ^m.$

Due to positivity of the grading, for $\lambda \in \Lambda$, we can choose a graded lift $\dot{L}(\lambda )$ of $L(\lambda )$, a graded lift $\dot{M}(\lambda )$ of $M(\lambda )$, a graded lift $\dot{M}(\lambda )^\vee$ of $M(\lambda )^\vee$, a graded lift $\dot{P}(\lambda )$ of $P(\lambda )$ and a graded lift $\dot{I}(\lambda )$ of $I(\lambda )$, such that the canonical maps

$$\begin{aligned}&\dot{P}(\lambda ) \twoheadrightarrow \dot{M}(\lambda ) \twoheadrightarrow \dot{L}(\lambda ),\\&\dot{L}(\lambda ) \hookrightarrow \dot{M}(\lambda )^\vee \twoheadrightarrow \dot{I}(\lambda ) \end{aligned}$$

are homogenous of degree zero.

1.4 Graded category ${\overline{\mathcal O}}$ for $\mathfrak {gl}(m|n)$

Since B is a positively graded algebra, we can denote by

$$\begin{aligned} {\dot{\overline{\mathcal O}}}:= {\text {gmof-}B}\end{aligned}$$

the category of graded finite-dimensional unital right B-modules whose morphisms are all homogenous homomorphisms of degree zero. We consider $\dot{\overline{\mathcal O}}$ as a graded version of $\overline{\mathcal O}$.

Let $\mathbb F^B: {\text {gmof-}B}\rightarrow {\text {mof-}B}$ be the functor which forgets the grading. A module X in ${\text {mof-}B}$ (respectively $X \in \overline{\mathcal O}$) is said to have a graded lift $\dot{X}\in {\dot{\overline{\mathcal O}}}$ if $\mathbb F^B(\dot{X})\cong X$ (respectively $T^B(X)\cong \mathbb F^B(\dot{X})$).

Lemma 56

Let $\dot{I_\nu }$ be the Serre subcategory of ${\text {gmof-}A}$ generated by $\{T(L(\lambda ))\langle i\rangle |~\lambda \in {\Lambda (\nu )},~i\in {\mathbb Z}\}$. Then, the functor

$$\begin{aligned} {\dot{\pi }}: {\text {gmof-}A}\rightarrow {\dot{\overline{\mathcal O}}},~ M\mapsto \bigoplus _{\lambda \in {\Lambda (\nu )}} M1_\lambda , \end{aligned}$$

gives rise to an equivalence ${\text {gmof-}A}/I_\nu \cong {\dot{\overline{\mathcal O}}}.$

Proof

The proof is similar to the proof of Lemma 53. $\square$

Let $\lambda \in {\Lambda (\nu )}$. We recall modules $S(\lambda ), D(\lambda )$ and $Q(\lambda )$ from Sect. 8; see also Proposition 49. We note that $\pi (M(\lambda )^\vee )\cong \overline{{\nabla }}(\lambda )$.

Lemma 57

Let $\lambda \in {\Lambda (\nu )}$. Then the modules $S(\lambda )$, $D(\lambda )$ and $Q(\lambda )$ from Proposition 49, and $M(\lambda )^\vee$ have graded lifts in ${\text {gmof-}A}$.

Proof

The proof is similar to the proof of [BLW, Lemma 5.5]. $\square$

For a given $\lambda \in {\Lambda (\nu )},$ we let $\dot{Q}(\lambda )$ be the quotient of $\dot{P}(\lambda )$ by the submodule generated by the image of all homomorphisms $\dot{P}(\mu )\langle m \rangle \rightarrow \dot{P}(\lambda )$, for $m\in {\mathbb Z}$ and $\mu \in {\Lambda (\nu )}$. From Lemma 57, it is easy to see that $\dot{Q}(\lambda )$ is a graded lift of $Q(\lambda )$.

Let ${\dot{X}} \in {\text {gmof-}A}$ be a graded lift of $X\in {\text {mof-}A}$. We may observe that $\dot{\pi }({\dot{X}})$ is a graded lift of $\pi '(X)\in {\text {mof-}B}$ since

$$\begin{aligned} \mathbb F^B({\dot{\pi }} (\dot{X})) \cong \mathbb F^B(\bigoplus _{\lambda \in {\Lambda (\nu )}} \dot{X}1_\lambda ) \cong \bigoplus _{\lambda \in {\Lambda (\nu )}} X1_\lambda \cong \pi '(X). \end{aligned}$$

Therefore, for any $\lambda \in {\Lambda (\nu )}$,

$$\begin{aligned}&\dot{P}^B(\lambda ):={\dot{\pi }}({\dot{P}}(\lambda )),~{\dot{\Delta }}^B(\lambda ):={\dot{\pi }}({\dot{Q}}(\lambda )), \\&\dot{\overline{{\nabla }}}^B(\lambda ):=\dot{\pi }({\dot{M}}(\lambda )^\vee ),~{\dot{L}}^B(\lambda ):=\dot{\pi }({\dot{L}}(\lambda )), \end{aligned}$$

are the graded lifts of $\pi (P(\lambda )), \Delta (\lambda ), \overline{{\nabla }}(\lambda )\text { and }\pi (L(\lambda ))$, respectively. In particular, we have the following graded BGG reciprocity.

Lemma 58

(Graded BGG reciprocity). Let $\lambda ,\mu \in {\Lambda (\nu )}$. Then ${\dot{\pi }}({\dot{P}}(\lambda ))$ has a ${\dot{\Delta }}$-flag. Furthermore, for any $m\in {\mathbb Z}$, we have

$$\begin{aligned}&(\dot{P}^B(\lambda ): \dot{\Delta }^B(\mu )\langle m \rangle ) = [ \dot{\overline{{\nabla }}}^B(\mu )\langle m \rangle : {\dot{L}}^B(\lambda )]. \end{aligned}$$

Proof

The first assertion is proved using Lemma 21 and an argument similar to the proof of [MSt1, Lemma 8.6]. For the proof of the graded BGG reciprocity, we note the following:

$$\begin{aligned}&\bigoplus _{i\in {\mathbb Z}}{\text {Ext}}^1_{{\dot{\overline{\mathcal O}}}}(\dot{\Delta }^B(\lambda )\langle i\rangle , \dot{\overline{{\nabla }}}^B(\mu )) \cong {\text {Ext}}^1_{{\text {mof-}B}}(T^B{\Delta }(\lambda ), T^B{\overline{{\nabla }}}(\mu ))\cong {\text {Ext}}^1_{\overline{\mathcal O}}({\Delta }(\lambda ), {\overline{{\nabla }}}(\mu )), \end{aligned}$$

which is zero by Theorem 24. Similarly, we calculate

$$\begin{aligned} \bigoplus _{i\in {\mathbb Z}}\,{\text {Hom}}_{{\dot{\overline{\mathcal O}}}}(\dot{\Delta }^B(\lambda )\langle i\rangle , \dot{\overline{{\nabla }}}^B(\mu ))&\cong \,{\text {Hom}}_{{\text {mof-}B}}(T^B{\Delta }(\lambda ), T^B{\overline{{\nabla }}}(\mu )) \\ {}&\cong \,{\text {Hom}}_{\overline{\mathcal O}}({\Delta }(\lambda ), {\overline{{\nabla }}}(\mu )), \end{aligned}$$

which is zero, for $\lambda \ne \mu$, and is isomorphic to $\mathbb {C}$, for $\lambda =\mu$, by Lemma 21. Since ${\dot{\Delta }}^B(\lambda )$ has a quotient isomorphic to $\dot{L}^B(\lambda )$, it follows that

$$\begin{aligned} \,{\text {Hom}}_{{\dot{\overline{\mathcal O}}}}(\dot{\Delta }^B(\lambda ), \dot{\overline{{\nabla }}}^B(\lambda )\cong \mathbb {C}. \end{aligned}$$

The claim of the lemma follows now by a standard argument; see, e.g., [Hu, Section 3.11].

$\square$

For $\lambda ,\mu \in {\Lambda (\nu )}$, we define

$$\begin{aligned} {[}\dot{\overline{{\nabla }}}^B(\lambda ): \dot{L}^B(\mu )]_q:= \sum _{m\ge 0} [\dot{\overline{{\nabla }}}^B(\lambda ): \dot{L}^B(\mu )\langle m \rangle ] q^m. \end{aligned}$$

Similarly, we define $(\dot{P}^B(\lambda ): {\dot{\Delta }}^B(\mu ))_q$. We have the following graded version of Corollary 45.

Proposition 59

For $\lambda ,\mu \in \Lambda (\nu )$, we have

$$\begin{aligned}&(\dot{P}^B(\mu ): {\dot{\Delta }}^B(\lambda ))_q=[\dot{\overline{{\nabla }}}^B(\lambda ):\dot{L}^B(\mu )]_q =\texttt {t}_{-f_\lambda \cdot w_0^\nu ,-f_\mu \cdot w_0^\nu }(q). \end{aligned}$$

Proof

We calculate

$$\begin{aligned}&\sum _{m\ge 0}[\dot{\overline{{\nabla }}}^B(\lambda )\langle m\rangle : \dot{L}^B(\mu )]q^m \\&\quad =\sum _{m\ge 0}[{\dot{\pi }}({\dot{M}}(\lambda )^\vee ): {\dot{\pi }}( \dot{L}(\mu )\langle -m\rangle )]q^m\\&\quad =\sum _{m\ge 0}[\dot{M}(\lambda )^\vee : \dot{L}(\mu )\langle -m\rangle ]q^m \\&\quad =\sum _{m\ge 0}[\dot{M}(\lambda ): \dot{L}(\mu )\langle m\rangle ]q^m, \end{aligned}$$

which is equal to the matrix coefficient of the matrix inverse to $\left( \ell _{f_{\mu },f_\lambda }(q^{-1})\right)$ by [BLW, Section 5.9]; see also [Bru4, Theorem 3.6]. Now, by [Bru2, Corollary 2.24] this inverse matrix is precisely $\left( t_{-f_{\lambda },-f_{\mu }}(q)\right)$. By Proposition 9, the polynomial $t_{-f_{\lambda },-f_{\mu }}(q)$ is equal to $\texttt {t}_{-f_\lambda \cdot w_0^\nu ,-f_\mu \cdot w_0^\nu }(q)$, which proves the claim. $\square$