A Beilinson-Bernstein Theorem for Analytic Quantum Groups. I

Abstract

In this two-part paper, we introduce a $p$-adic analytic analogue of Backelin and Kremnizer’s construction of the quantum flag variety of a semisimple algebraic group, when $q$ is not a root of unity and $\vert q-1\vert<1$. We then define a category of $\lambda$-twisted $D$-modules on this analytic quantum flag variety. We show that when $\lambda$ is regular and dominant and when the characteristic of the residue field does not divide the order of the Weyl group, the global section functor gives an equivalence of categories between the coherent $\lambda$-twisted $D$-modules and the category of finitely generated modules over $\widehat{U_q^\lambda}$, where the latter is a completion of the ad-finite part of the quantum group with central character corresponding to $\lambda$. Along the way, we also show that Banach comodules over the Banach completion $\widehat{ \mathcal{O}_q(B) }$ of the quantum coordinate algebra of the Borel can be naturally identified with certain topologically integrable modules.

A. Appendix A. Hopf duals of $$R$$ -Hopf algebras and their comodules

We wish to establish some duality facts to do with Hopf algebras over $R$ which are well known when working over fields but for which we couldn’t find references for Hopf algebras over more general commutative rings. Most of our proofs work using the usual arguments but one has to be a bit careful when dealing with torsion.

A.1. Hopf Duals Over $R$

For the entirety of this Section, $H$ will denote a fixed Hopf $R$-algebra. For our purposes, it will be enough to work in the case where $H$ is torsion-free. First we wish to define a notion of Hopf dual. Since $H$ has no torsion, it embeds as a sub-Hopf algebra of $H_L=H\otimes_R L$. We will define the Hopf dual to be a sub-Hopf algebra of $(H_L)^\circ$. Let $\mathcal{J}$ denote the set of ideal $I$ in $H$ such that $H/I$ is a finitely generated $R$-module. Moreover, denote by $\mathcal{J}'$ the set of ideals $I$ in $H$ such that $H/I$ is free of finite rank. Finally, write $H^*$ for $\text{Hom} _R(H, R)$. Note that $H^*$ is always torsion-free since $R$ is a domain: if $\pi f=0$ then $\pi f(u)=0$ for all $u\in H$ and so $f(u)=0$ for all $u$.

Definition A.1.

We define the Hopf dual of $H$ to be

$$H^\circ:=\{f\in H^* : f|_I=0 \text{ for some } I\in \mathcal{J}\}.$$

By the above $H^\circ$ is torsion-free.

If $n\geq 0$ and $x\in H$ we have for any $f\in H^*$ that $f(x)=0$ if and only if $f(\pi^n x)=0$. Thus if $0\neq f\in H^\circ$ then $f|_I=0$ for some $I\in\mathcal{J}$ where $H/I$ is not torsion. Moreover we then have $f|_{I_L\cap H}=0$ and so by replacing $I$ with $I_L\cap H$ we may in addition assume that $H/I$ is torsion-free. Since $R$ is a PID this shows that

$$H^\circ=\{f\in H^* : f|_I=0 \text{ for some } I\in \mathcal{J}'\}.$$

Moreover by extending scalars we may identify $H^\circ$ with an $R$-submodule of $H_L^\circ$. From this it follows by the standard arguments that $H^\circ$ is the algebra of matrix coefficients of $H$-modules which are free of finite rank over $R$. Since this collection of $H$-modules is closed under taking tensor products, direct sums and duals, and we can take dual bases, we have proved

Lemma A.2.

$H^\circ$ is an sub-Hopf $R$ -algebra of $H_L^\circ$ . In particular the algebra maps on $H^\circ$ are just the dual maps of the coalgebra maps on $H$ and vice-versa.

Remark A.3.

Some of the above arguments were implicit in Lusztig’s work, see [34, 7.1].

A.2. $H^\circ$-Comodules as $H$-Modules

We now wish to establish some correspondence between comodules over $H^\circ$ and certain $H$-modules. We call an $H$-module $M$ locally finite if for all $m\in M$, $Hm$ is finitely generated over $R$.

Proposition A.4.

Every $H^\circ$ -comodule has a canonical structure of a locally finite $H$ -module with respect to which every comodule homomorphism is an $H$ -modules homomorphism. In other words there is a canonical faithful embedding of categories between the category of $H^\circ$ -comodules and the category of locally finite $H$ -modules.

Proof.

This is just the usual argument. If $M$ is an $H^\circ$-comodule with coaction $\rho:M\to M\otimes_R H^\circ$, write $\rho(m)=\sum m_1\otimes m_2$. Then we set

$$u\cdot m=\sum m_2(u)m_1$$

for all $u\in H$. It follows from the comodule axioms that this gives a well defined module structure, i.e that $1\cdot m=m$ and that $u\cdot(u'\cdot m)=(uu')\cdot m$ for all $u,u'\in H$ and all $m\in M$. Moreover by definition of the module structure, $H\cdot m$ is finitely generated over $R$ for all $m\in M$. Finally it follows from the definition of the action that any comodule homomorphism is also a module homomorphism. $\square$

Next, we want to show that the functor we just defined is full, i.e that every $H$-module map between two $H^\circ$-comodules is a comodule homomorphism. We first need a technical result. Suppose $M$ is a locally finite $H$-module. Note that we have an $R$-module injection $\phi_M:M\to \text{Hom} _R(H,M)$ given by $\phi_M(m)(u)=um$ for all $u\in H$ and $m\in M$. Moreover we have a map

$$\theta_M: M\otimes_R H^\circ \to \text{Hom} _R(H,M)$$

given by $\theta_M(m\otimes f)(u)=f(u)m$. When the $H$-module structure on $M$ arises from an $H^\circ$-comodule structure then we have $\phi_M=\theta_M\circ \rho$. Therefore we can use this expression for $\phi_M$ as an alternative definition of the module structure on $M$. We claim that the map $\theta_M$ is injective. More generally we have the following

Lemma A.5.

Let $A$ and $B$ be $R$ -modules, $A^*= \text{Hom} _R(A,R)$ and suppose $C$ is any $R$ -submodule of $A^*$ such that $A^*/C$ has no $\pi$ -torsion. Let $M$ be any $R$ -module and set

$$\theta_{M,C}: \textit{Hom} _R(B,M)\otimes_R C \to \textit{Hom} _R(A\otimes_R B,M)$$

to be defined by $\theta_{M,C}(g\otimes f)(x\otimes y)=f(x)g(y)$ . Then the map $\theta_{M,C}$ is injective.

Proof.

Suppose that $0\neq u=\sum_{i=1}^s g_i\otimes f_i\in \text{Hom} _R(B,M)\otimes_R C$. The $R$-submodule $N$ of $\text{Hom} _R(B,M)$ generated by the $g_i$ is finitely generated, so since $R$ is a PID we can pick a generating set $n_1,\ldots, n_l, t_1,\ldots, t_m$ for $N$ such that $n_1,\ldots, n_l$ are torsion-free while $t_1,\ldots, t_m$ are $\pi$-torsion, and

$$N=\bigoplus_{i=1}^l Rn_i \oplus \bigoplus_{j=1}^m Rt_j.$$

For each $1\leq j\leq m$, let $a_j$ be the positive integer such that $Rt_j\cong R/\pi^{a_j}R$.

Now, to show that $\theta_{M,C}(u)\neq 0$ it suffices to show that the restriction of $\theta_{M,C}$ to the span of the $n_i\otimes f_j$ and $t_k\otimes f_j$ is injective. So suppose we are given

$$v=\sum r_{ij} n_i\otimes f_j +\sum r'_{kj} t_k\otimes f_j\in \ker{\theta_{M,C}}.$$

Evaluating at $x\otimes y$ we get $\sum_{i,j} r_{ij}f_j(x)n_i(y) + \sum_{k,j} r'_{kj} f_j(x)t_k(y)=0$ for all $x\in A$ and $y\in B$. In particular we have $\sum_{i,j} r_{ij}f_j(x)n_i + \sum_{k,j} r'_{kj} f_j(x)t_k=0$ for any fixed $x\in A$. Since we have a direct sum decomposition of $N$ it follows that

$$\sum_{j} r_{ij}f_j(x)=0 \quad\text{and}\quad \sum_{j} r'_{kj} f_j(x)\in \pi^{a_k}R$$

for all $x\in A$ and all $1\leq i\leq l$ and $1\leq k\leq m$. In particular, for all $k$, $\sum_{j} r'_{kj} f_j=\pi^{a_k}g_k$ for some $g_k\in C$ since $A^*/C$ has no $\pi$-torsion.

Therefore we have

$$\sum_{j} r_{ij}f_j=0 \quad\text{and}\quad \sum_{j} r'_{kj} f_j\in \pi^{a_k}C,$$

and hence

$$n_i\otimes \sum_{j} r_{ij}f_j=0=t_k\otimes \sum_{j} r'_{kj} f_j$$

for all $i,k$, and so $v=0$ as required. $\square$

Corollary A.6.

Let $M$ be an $R$ -module.

• 1. The map $\theta_M: M\otimes_R H^\circ \to \text{Hom} _R(H,M)$ is injective.

• 2. The map $M\otimes_R H^\circ\otimes_R H^\circ\to \text{Hom} _R(H\otimes_R H,M)$ sending $m\otimes f\otimes g$ to $x\otimes y\mapsto f(x)g(y)m$ is injective.

Proof.

Let $A=H$ and $C=H^\circ$. From the definition of $H^\circ$ it follows that $A^*/C$ is torsion-free. Then (i) follows immediately from the Lemma by putting $B=R$. For (ii) note that this map is simply the composite

where $\varpi(f\otimes g)(x\otimes y)=g(y)f(x)$. The map $\theta_M\otimes 1$ is injective by (i) and because $H^\circ$ is flat while the map $\varpi$ is injective by putting $B=H$ in the Lemma. $\square$

We can now deduce the result we were aiming for.

Proposition A.7.

The functor associating any $H^\circ$ -comodule to the corresponding $H$ -module is a fully faithful embedding.

Proof.

From what we have done already we just need to show that any $H$-module map $f:M\to N$ between two $H^\circ$-comodules is a comodule homomorphism. Write $\rho_M$ and $\rho_N$ for the coactions on $M$ and $N$ respectively, and pick $m\in M$ and $u\in H$. Then we know that $um=\sum m_2(u)m_1$ and we have $uf(m)=\sum m_2(u)f(m_1)$ since $f$ is a module homomorphism. On the other hand by definition of the action on $N$ we have $uf(m)=\sum f(m)_2(u)f(m)_1$. To show that $f$ is a comodule map we need to show that

$$\sum f(m_1)\otimes m_2=\sum f(m)_1\otimes f(m)_2$$

or in other words that $\rho_N\circ f=(f\otimes 1)\circ\rho_M$.

Write $\tilde{\rho}_1=\rho_N\circ f$ and $\tilde{\rho}_2=(f\otimes 1)\circ\rho_M$. Moreover recall the map $\phi: M\to \text{Hom} _R(H,M)$ given by $\phi(m)(u)=um$. Then let

$$\tilde{\phi}=\phi\circ f:M\to \text{Hom} _R(H, N)$$

so that $\tilde{\phi}(m)(u)=uf(m)$. Then by definition $\tilde{\phi}=\theta_N\circ\tilde{\rho}_1$. On the other hand by our above observation we see that $\tilde{\phi}=\theta_N\circ\tilde{\rho}_2$. Since $\theta_N$ is injective the result follows. $\square$

From now on, if $M$ is a locally finite $H$-module we will say that it is an $H^\circ$-comodule to mean that its $H$-module structure arises from an $H^\circ$-comodule structure.

In order for the above functor to be an isomorphism of categories we therefore just need to show that it is surjective. This may not be true in general, however we can write a very simple necessary and sufficient condition for an isomorphism of categories to hold. Suppose $M$ is a locally finite $H$-module and let $\phi_M: M\to \text{Hom} _R(H,M)$ be given by $\phi_M(m)(x)=x\cdot m$. We have the map $\theta_M:M\otimes_R H^\circ$ as before.

Proposition A.8.

A locally finite $H$ -module $M$ is an $H^\circ$ -comodule if and only if $\phi_M(m)$ belongs to the image of $\theta_M$ for all $m\in M$ .

Proof.

If $M$ is a comodule with coaction $\rho$, then by our observation preceding Lemma A.5 we have $\phi_M=\theta_M\circ\rho$ where $\phi_M$ comes from the induced $H$-module structure, and the result is clear. Conversely assume $\phi_M(m)$ belongs to the image of $\theta_M$ for all $m\in M$. Fix $m\in M$. Then there exists $m_1, \ldots, m_n\in M$ and $f_1, \ldots, f_n\in H^\circ$ such that for all $x\in H$, $x\cdot m=\sum_{i=1}^n f_i(x)m_i$ and we define

$$\rho(m)=\sum_{i=1}^n m_i\otimes f_i,$$

i.e $\rho(m)$ is the unique element of $M\otimes_R H^\circ$ such that $\theta_M(\rho(m))=\phi_M(m)$. We now have to check that this satisfies the comodule axioms. By definition, the counit on $H^\circ$ is defined by $\varepsilon(f)=f(1)$ and so

$$(1\otimes\varepsilon)\circ\rho(m)=\sum_{i=1}^n f_i(1)m_i=1\cdot m=m$$

as required. Finally we aim to show that the following diagram commutes:

By Corollary A.6 (2), the natural map $M\otimes_R H^\circ\otimes_R H^\circ\to \text{Hom} _R(H\otimes_R H,M)$ is injective. Hence it suffices to show that $(1\otimes\Delta)\circ\rho(m)$ and $(\rho\otimes 1)\circ\rho(m)$ act in the same way on $H\otimes_R H$ for all $m\in M$. But the former sends $x\otimes y$ to $(xy)\cdot m$ while the latter sends $x\otimes y$ to $x\cdot(y\cdot m)$ for any $x,y\in H$, which are clearly equal. $\square$

Since Lemma A.5 was quite general, the same argument as in the above proof shows the following

Lemma A.9.

Suppose $M$ is a locally finite $H$ -module and let $C$ be a subcoalgebra of $H^\circ$ such that $H^*/C$ is torsion-free. If $\phi_M(m)$ belongs to the image of $\theta_{M,C}$ for all $m\in M$ then $M$ is a $C$ -comodule.