# Projective objects in BGG category $\mathcal{O}$ are projective $U(\mathfrak{g})$-modules?

Let $\mathfrak{g}$ be a finite dimensional semi-simple complex Lie algebra. Then, BGG category $\mathcal{O}$ is defined to be the full subcategory of finitely generated $U(\mathfrak{g})$-modules of those modules which are weight modules and locally $U(\mathfrak{n})$-finite. It is known that $\mathcal{O}$ is not extension-closed in the category of (finitely generated) $U(\mathfrak{g})$-modules, see e.g. this math.stackexchange question. In particular (since $\mathcal{O}$ is closed under factor modules), it can't be true that $\mathcal{O}$ contains all finitely generated projective $U(\mathfrak{g})$-modules. I was wondering about the following:

Is there any projective object $P(\lambda)$ in $\mathcal{O}$ which is projective as a finitely generated $U(\mathfrak{g})$-module. Or put another way: Is any finitely generated projective $U(\mathfrak{g})$-module contained in $\mathcal{O}$?

• @DietrichBurde I don't see how. Can you elaborate? Sep 2 '15 at 12:01
• @DietrichBurde That's exactly my question: It is a projective object in BGG category $\mathcal{O}$, but there could be exact sequences of $U(\mathfrak{g})$-modules with $P(\lambda)$ as an end term where the middle term is not in $\mathcal{O}$, and which do not split. Sep 2 '15 at 12:06
• $\mathcal O$ is really, really small inside the category of all $U(g)$-modules. Sep 2 '15 at 16:49
• @MarianoSuárez-Alvarez Yeah, but it's still getting a lot of attention since it was introduced. Sep 2 '15 at 16:54
• Sure. The problem is that the whole category is way too large. $\mathcal O$ is at the same time very small and large enough that it contains lots of modules that are important in nature and exhibits very rich behaviour. That's why it is so significant. Sep 2 '15 at 17:11

The restriction of a projective $U(\mathfrak{g})$-module to $U(\mathfrak{h})$ is projective, but the restriction of an object of category $\mathcal{O}$ is a direct sum of one-dimensional modules.

• So they are infinitely far away from being projective. Thanks. Sep 2 '15 at 16:56
• @JulianKuelshammer Well, I'm not sure any $U(\mathfrak{g})$-module is infinitely far away from being projective, but yes, as far as they could be. Sep 2 '15 at 17:06