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.

generalizedcharacter $\chi_\lambda$, just as in the first step of the Jordan block decomposition of a matrix. – Victor Protsak Aug 10 '11 at 22:05