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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 $\mathcal{O}$?

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  • $\begingroup$ @DietrichBurde I don't see how. Can you elaborate? $\endgroup$ Sep 2 '15 at 12:01
  • $\begingroup$ @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. $\endgroup$ Sep 2 '15 at 12:06
  • $\begingroup$ $\mathcal O$ is really, really small inside the category of all $U(g)$-modules. $\endgroup$ Sep 2 '15 at 16:49
  • $\begingroup$ @MarianoSuárez-Alvarez Yeah, but it's still getting a lot of attention since it was introduced. $\endgroup$ Sep 2 '15 at 16:54
  • $\begingroup$ 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. $\endgroup$ Sep 2 '15 at 17:11
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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.

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  • $\begingroup$ So they are infinitely far away from being projective. Thanks. $\endgroup$ Sep 2 '15 at 16:56
  • $\begingroup$ @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. $\endgroup$ Sep 2 '15 at 17:06

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