Simple quotient of Verma Module $ M_\mathfrak{b}(\lambda) $

Question

I'm starting to look into Lie Superalgebras and understand a bit more about simple f.d. modules. Right now I'm studying fairly simple examples to try and get a feel for it. I'm trying to find the simple modules of $ \mathfrak{gl}(1|1) $ and I thought of doing this through the Verma modules for a fixed Borel $ \mathfrak{b} $ (As far as I can tell there are only 2 distinct borels).

My issue is that I'm uncertain of how to find the simple quotient of a given Verma module.

I would appreciate help with either the non-super case (I.e. given a Lie algebra and a Verma module over how would I compute it's simple quotient) or the case I am actually looking at now.

Answer 1

I am not sure what you mean by computing the simple quotient. There is a general theorem stating that any Verma module has a maximal proper submodule (which then is automatically uniquely determined). The quotient by this maximal proper submodule is (almost tautologically) simple. What this maximal proper submodule looks like depends a lot on the Verma module in question. (I am referring here to the non-super case, I don't know the details about the super case.) There are Verma modules which themselves are simple, and others for which the simple quotient is finite dimensional. The main deep result here is that for dominant integral highest weight the quotient indeed is finite dimensional. I don't think that there is much more that you can say in general.