# Category of typical representations for Lie superalgebras

I am interested in studying the category of typical representations over basic classical simple Lie superalgebras. In particular, I want to know

1) is this category semisimple (character determines a typical representation)?

2)is this category closed under tensor product?

3) are there non-trivial one-dimensional representations in this category? and

4) What is the connection between this category to the category $$\mathcal O$$ for Lie superalgebras?

In Theorem 1, Kac has given that, typical representation splits in any finite-dimensional representation (i.e. if it is a submodule or a factor-module of a finite-dimensional G-module, then it is a direct summand).

Is this means that 1) is true?

Basically, I want to know how much bad (or not nice) is this category compared to the BCG category $$\mathcal{O}$$ or $$\mathcal{O}^{int}$$ of Kac-Moody Lie algebras.

Thank you :) .

Kac (in the linked article) talks about finite dimensional representations (of basic Lie superalgebras). So lets assume that.

1) Yes, the category is semisimple. This is contained in the article by Kac (Theorem 1).

2) No. Kac shows that the typical representations are projective. Therefore a tensor product of typical representations will decompose into a direct sum of projective representations. These consists of the typical representations and the indecomposable projective covers of the irreducible atypical representations. The only exception is the category of all (finite dimensional) representations of $$\mathfrak{osp}(1|2n)$$ which is semisimple.

3) No, typical representations have superdimension 0, unless you are in the $$\mathfrak{osp}(1|2n)$$-case.

I think it is wrong to compare the category of typical representations (which is quite easy) to the BGG category $$\mathcal{O}$$. The latter is a highly non-semisimple category; and the computation of characters, decomposition numbers and so on leeds into deep mathematics like Kazhdan-Lusztig theory. The BGG category $$\mathcal{O}$$ can be defined for Lie superalgebras as well, so why not compare those two? There is a nice overview article by John Brundan about the analog of the category $$\mathcal{O}$$ in the super case, see

https://pages.uoregon.edu/brundan/papers/superofont.pdf

Surprisingly it makes a lot of sense to compare the category of finite dimensional representations to some version of classical category $$\mathcal{O}$$. Work of Brundan-Stroppel

https://arxiv.org/abs/0907.2543

shows that the blocks of the general linear superalgebra $$\mathfrak{gl}(m|n)$$ are equivalent to some limit version of blocks of some parabolic category $$\mathcal{O}$$. For more results and precise statements see the article by Brundan.

• Nice answer with excellent references. Thank you. So for $B(0,n)$ only this category will be closed under tensor product? Can you please explain your answer to the question (3) little more. I am not familiar with the term super dimension. please explain. Also, are you saying in the case of $\mathfrak g = B(0,n)$ there non-trivial one-dimensional modules? Is it mean to say, $\frac{\mathfrak g}{[g,g]}$ is non-trivial? Kindly explain to me. Thank you, – GA316 Jun 13 '19 at 4:37
• Yes, this will only happen for $\mathfrak{osp}(1|2n)$. The superdimension is the dimension of the even part minus the dimension of the odd part. Projective representations always have superdimension 0 if your category is not semisimple (so in particular the typical ones). No, $\mathfrak{osp}(1|2n)$ does not have nontrivial one-dimensional irreducible representations. This Lie superalgebra is simple and hence the kernel of a representations is either zero or the entire Lie superalgebra as for ordinary simple Lie algebras. – Thorsten Heidersdorf Jun 14 '19 at 13:25
• Thanks for your kind reply and it is very helpful. Have a nice day :) – GA316 Jun 14 '19 at 15:16