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.)

  • 1
    $\begingroup$ Then every finite-dimensional representation is completely reducible. So we don't need "non irreducible". $\endgroup$ Commented Apr 13, 2019 at 18:38
  • 2
    $\begingroup$ $U(\mathfrak g)$ is not finite dimensional ! $\endgroup$ Commented Apr 13, 2019 at 18:38
  • $\begingroup$ @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 $dim_C(g)=0$. $\endgroup$
    – user45765
    Commented Apr 13, 2019 at 18:49
  • $\begingroup$ @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. $\endgroup$
    – user45765
    Commented Apr 13, 2019 at 18:53
  • 2
    $\begingroup$ @user45765 “Every fg module is semisimple” is equivalent to “every module is semisimple” for rings with identity. $\endgroup$
    – rschwieb
    Commented Apr 13, 2019 at 22:00

1 Answer 1


Just to give a definite answer :

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 $\mathcal O$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .