0
$\begingroup$

Hi:

Given a finite dimensional Lie superalgebra $G$ over the field of complex numbers with decomposition $G_{-1}+G_{0}+G_{1}$, where $G_{0}$ is the even part and $G_{-1}+G_{1}$ is the odd part. Suppose $F$ is the category of all finite dimensional weight $G_{0}$-modules. Is it a general fact that all irreducible (or indecomposable) highest weight modules in $F$ are projective ? And, I'm not sure if the answer depends on the type of $G$.

Thanks!!

$\endgroup$
2
  • 1
    $\begingroup$ Do I understand correctly that $G_0$ is just a finite dimensional Lie algebra? $\endgroup$
    – David Hill
    Commented Apr 22, 2013 at 14:34
  • 3
    $\begingroup$ it is not clear what role the algebra $G$ plays in this. $\endgroup$ Commented Apr 24, 2013 at 4:31

1 Answer 1

1
$\begingroup$

Then the answer is yes, finite dimensional irreducible $G_0$-modules are projective in the category of finite dimensional modules.

Note that $G_0=G_0'\oplus Z$, where $G_0'$ is semisimple and $Z$ is the center. Let $V$ be a finite dimensional $G_0$-module. Then, by Weyl's theorem, $V$ is completely irreducible as a $G_0'$-module. The elements of $Z$ act as scalars on each irreducible summand of $V$, by Schur's lemma. This means that the decomposition of $V$ into irreducible $G_0'$-modules is a decomposition as $G_0$-modules. It now follows that the category of finite dimensional $G_0$-modules is semi-simple, and irreducible=projective.

EDIT: This answer is based on the assumption that $G$ is a simple Lie superalgebra, so $G_0$ is reductive. At one time, there was a comment/answer indicating that this was the case of interest.

$\endgroup$

You must log in to answer this question.

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