TITLE: Confusion about Subcategories of Category $\mathcal{O}$
QUESTION [8 upvotes]: So, in learning about category $\mathcal{O}$ representations of a semisimple Lie algebra $\mathfrak{g}$, I've come across two natural kinds of subcategories, and I think I'm confused about their structure and the relationship between them. (In particular, if they are equivalent or not).
Fix $\lambda \in \mathfrak{h}^{*}$ and let $\chi_{\lambda}:Z(\mathfrak{g}) \rightarrow \mathbb{C}$ denote the corresponding central character ($Z(\mathfrak{g})$ denoting the center of $U\mathfrak{g}$)
Type 1: Modules $M$ such that $(z - \chi_{\lambda}(z))\cdot m = 0$ for all $m \in M$, $z \in Z(\mathfrak{g})$. Let's call this $\text{Mod}_{\mathcal{O}}(\mathfrak{g},\lambda)$
Type 2: Modules $M$ such that $(z - \chi_{\lambda}(z))^{n} \cdot m = 0$ for all $m \in M$, $z \in Z(\mathfrak{g})$, where $n$ is an integer possibly depending on $z$ and $m$. Let's call this $\mathcal{O}_{\lambda}$.
Now one clear reason to prefer the $\mathcal{O}_{\lambda}$ is that category $\mathcal{O}$ is a direct sum of these subcategories. Additionally, if $L \in \mathcal{O}_{\lambda}$ is simple, it's projective cover will also be in $\mathcal{O}_{\lambda}$, but not necessarily in $\text{Mod}_{\mathcal{O}}(\mathfrak{g},\lambda)$.
But it seems like, in some cases, these categories should be equivalent?
For simplicity, let $\lambda = 0$. If I'm understanding correctly:
1. Beilinson-Bernstein says that there is an equivalence of categories: $\text{Mod}_{\mathcal{O}}(\mathfrak{g},0) \cong \text{Mod}_{B}(D_{X})$, where the second thing is weakly $B$-equivariant $D$-modules on the flag variety $X$.
2. Riemann-Hilbert says that there is an equivalence $\text{Mod}_{B}(D_X) \cong \mathcal{P}_{B}(X)$, the second category being perverse sheaves on $X$ whose cohomology is constant along $B$-orbits.
3. In the BGS paper on Koszul duality, it is claimed that $D^{b}(\mathcal{O}_{0}) \cong D^{b}_{B}(X)$, the latter being the bounded derived category of constructible sheaves on $X$ whose cohomology is constant along $B$-orbits, and this congruence respects the $t-$structures giving $\mathcal{O}_{0} \cong \mathcal{P}_{B}(X)$.
Is this all correct? If it is, it would seem to establish an equivalence $\mathcal{O}_{0} \cong \text{Mod}_{\mathcal{O}}(\mathfrak{g},0)$. Which seems plausible, but also somewhat strange, since one is a strict subcategory of the other (not that that by itself would imply non-equivalence..). And even trying the simplest case, $\mathfrak{sl}_2$, I am unable to realize this equivalence, for example, the category $\text{Mod}_{\mathcal{O}}(\mathfrak{sl}_2,0)$ only seems to have four indecomposables (since it doesn't have the projective cover of $L(-2)$).

REPLY [10 votes]: The source of your confusion is the difference between B-equivariance and being smooth along B-orbits.  Since in step 3, you want only smoothness along B-orbits, you have to do the same thing in step 1.  In order to do that, you need to take some modules not in category $\mathcal{O}$, and instead consider modules where the Cartan subalgebra $\mathfrak{h}$ acts locally finitely (not necessarily semi-simply!), and the subalgebra $\mathfrak{n}$ nilpotently.  
The important point here is that now there's no containment of Type 1 or Type 2 modules: there are some modules on which the center acts semi-simply and the Cartan subalgebra does not, and some where it's vice versa.  These categories are equivalent by an old theorem of Soergel.  This theorem is secretly hiding in the result cited in BGS.