# Relation between semisimple Lie algebra completely reducibility and semisimple ring

Let $$\frak g$$ be a semi-simple finite dimensional Lie algebra over the complex numbers $$\mathbb C$$. Then every non irreducible representation of $$\frak g$$ is completely reducible.

Q1: Is category f.g. left $$U(\mathfrak{g})$$-Mod is equivalent to the category finite dimensional $$\operatorname{Rep}(\mathfrak{g})$$?

Q2: If every f.g $$U(\mathfrak{g})$$ module is completely reducible (in other words, they can be written as direct sums of simples) and above category equivalence holds, then from completely reducibility, I deduce for semi-simple Lie algebra, $$U(\mathfrak{g})$$ is semisimple as a ring. Hence, I would expect all $$U(\mathfrak{g})$$ modules to be projective and injective at the same time. Is this correct?

Q3: If Q1, Q2 are wrong, what is the relationship between semi-simple Lie algebras and semi-simple rings? (I would not expect for infinite dimensional representation, above categorical equivalence holds. The "semi-simple" ring should be interpreted as every f.g. module is decomposed into simple modules only.)

• Then every finite-dimensional representation is completely reducible. So we don't need "non irreducible". – Dietrich Burde Apr 13 at 18:38
• $U(\mathfrak g)$ is not finite dimensional ! – Nicolas Hemelsoet Apr 13 at 18:38
• @NicolasHemelsoet Ah, my dumbness. I mistaken the product with lie bracket which is not associative. Thanks for the correction. So $U(g)$ is never finite dimensional unless $dim_C(g)=0$. – user45765 Apr 13 at 18:49
• @DietrichBurde Is semi-simpleness in representation same as semi-simpleness as ring sense? I mean semi simple here for f.g. objects only. The ring semi-simpleness demands much stronger statement for any modules, they decompose into simples. – user45765 Apr 13 at 18:53
• @user45765 “Every fg module is semisimple” is equivalent to “every module is semisimple” for rings with identity. – rschwieb Apr 13 at 22:00

Q1 : No. For example, any simple modules can be expressed as some quotient of $$U(\mathfrak g)$$ (look for "Verma modules"), and there is so such module in $$Rep(\mathfrak g)$$.
Q2 : No. For example, the natural sequence $$0 \to M(-2) \to M(0) \to L(0) \to 0$$ does not split (here $$M(-2), M(0)$$ are two Verma modules for $$\mathfrak{sl}_2$$ and $$L(0)$$ is the trivial representation). Note that $$M(0), M(-2)$$ are infinite-dimensional.
Q3 : The only relation is that $$Rep(\mathfrak g)$$ is a semisimple category when $$\mathfrak g$$ is semisimple. So your last sentence is the correct interpretation.
Remark : $$U(\mathfrak g)$$ is a rather complicated ring. For more information you can look at Dixmier's book on Universal Envelopping algebra. Also you can look at Humphrey's book about BGG category $$\mathcal O$$.