# Translation functors for category $\mathcal O$

Let $\mathfrak g = \mathfrak n^- \oplus \mathfrak h \oplus \mathfrak n$ be a semisimple Lie algebra over $\mathbb C$ and let $\mathcal U$ be its enveloping algebra. Category $\mathcal O$ is the category of finitely generated $\mathcal U(\mathfrak h)$-semisimple modules such that the action of $\mathcal U(\mathfrak n)$ is locally finite. (In particular, finite dimensional modules are in category $\mathcal O$.) One can show that $\mathcal O$ is a direct sum of subcategories $\mathcal O_{\chi_\lambda}$, each of which consists of modules where the center $Z(\mathfrak g)$ of $\mathcal U$ acts by a fixed (EDIT: generalized) character $\chi_\lambda:Z(\mathfrak g) \to \mathbb C$.

If $L$ is a finite dimensional module, then we can define a shift functor $T_\lambda^\mu: \mathcal O_{\chi_\lambda} \to \mathcal O_{\chi_\mu}$ by the formula $T_\lambda^\mu(M) = pr_\mu\circ (L \otimes M)$. (Here we wrote $pr_\mu$ for the projection $\mathcal O \to \mathcal O_{\chi_\mu}$.) These translation functors have many nice properties and in certain cases are nice enough to be equivalences of categories. However, in general their behavior depends quite subtly on the geometry of the root system and on $\mu, \lambda$.

I had a reference request and a naive question:

1) Are there other definitions of translation functors, perhaps ones that behave more simply?

2) Are there simple sufficient conditions which show that two blocks $\mathcal O_{\chi_\lambda}$ and $\mathcal O_{\chi_\mu}$ are not equivalent? In particular, if $\lambda - \mu \in \Lambda_r$ (the root lattice) is it possible for $\mathcal O_{\chi_\lambda}$ and $\mathcal O_{\chi_\mu}$ to be in-equivalent?

(Retag as appropriate.)

EDIT: As Victor kindly pointed out, $\mathcal O_{\chi_\lambda}$ consists of modules where the endomorphisms $z-\chi_\lambda(z)$ are nilpotent, not necessarily 0. Also, to attempt to clarify (1), I was just wondering if there are natural functors between different blocks of $\mathcal O$ that don't come from tensoring with a finite dimensional module.

• For 1): It's not an alternative definition, but you can look at translation out of and onto the wall through Soergel's "Kombinatorikfunktor". Under this functor, translation onto the wall corresponds to restriction while translation out of the wall corresponds to induction. – Hanno Becker Aug 10 '11 at 17:12
• A technical remark: the category $\mathcal{O}_{\chi_\lambda}$ consists of the modules where $Z(\mathfrak{g})$ acts by a generalized character $\chi_\lambda$, just as in the first step of the Jordan block decomposition of a matrix. – Victor Protsak Aug 10 '11 at 22:05
• @Hanno: By the way, Soergel studied with Jantzen in Hamburg, though of course Soergel's later work developed in different directions. – Jim Humphreys Aug 14 '11 at 21:56

It's probably relevant to note that the translation functors are shadows of "obvious" functors: the categories of $\lambda$- and $\lambda+\nu$-twisted $D$-modules on the flag varieties are equivalent for any $\lambda$ and any integral $\nu$, with the functor being simply tensor product by the line bundle $L_{\nu}$. (Thus the category of twisted D-modules depends on a parameter $exp(\lambda)$ in the dual torus GROUP, not just Lie algebra). The translation functors result from the fact that the global sections from D-modules to representations does not commute with these obvious equivalences (eg global sections is sometimes an equivalence to representations of the associated central character and sometimes not, though it always has both left and right adjoints which we can use to go up and down).
For 2), the answer is yes. In $\mathfrak{sl}(2)$, most blocks containing a representation with integral highest weight have 2 simple objects: a finite dimensional rep with highest weight $n$ and an infinite dimensional Verma with highest weight $-n-2$ (the Verma of highest weight $n$ is an extension of the former by the latter, so they are in the same block). However, when $n=-1$, the finite dimensional rep goes away, and we only have one simple. In general, the blocks with highest weights $\lambda$ such that $\lambda+\rho$ lies on the wall of the Weyl chambers are always going to be weird.