# What are the “tensor-closed” object of the BGG category $\mathcal{O}$ of a semisimple Lie algebra $\mathfrak{g}$?

Let $\mathfrak{g}$ be a finite dimensional complex semisimple Lie algebra and we can consider its BGG category $\mathcal{O}$. It is well-known that $\mathcal{O}$ is not closed under tensor product, i.e. take two $\mathfrak{g}$-modules $M$ and $N$, the tensor product $M\otimes N$ is still a $\mathfrak{g}$-module but not necessarily in the category $\mathcal{O}$. However, if we take $M$ to be a finite dimensional $\mathfrak{g}$-module (which by definition is in the category $\mathcal{O}$), then for any $N\in \text{obj}\mathcal{O}$, we have $M\otimes N\in \text{obj}\mathcal{O}$.

$\textbf{My question}$ is: If $M\in \text{obj}\mathcal{O}$ has the property that for any $N\in \text{obj}\mathcal{O}$, we have $M\otimes N\in \text{obj}\mathcal{O}$, is the always true that $M$ is a finite dimensional $\mathfrak{g}$-module?

• Have you looked at sl_2 yet? I think that will give a negative answer to your question. – David Hill Jul 14 '15 at 15:34
• @DavidHill For $\mathfrak{sl}(2)$ the only thing I know is that the Verma module $M(\lambda)$, which is infinite dimensional, does not satisfy the property above. Actually I don't have many examples in head. Do you have a counter-example? – Zhaoting Wei Jul 14 '15 at 17:18
• No, I was mistaken. I think one can probably show that the tensor product of a Verma module with an infinite dimensional module is not finitely generated. – David Hill Jul 14 '15 at 20:32
• @Zhaoting: My answer is already long enough, but if you have any trouble filling in details in the general case let me know by email. The full argument gets a bit long. – Jim Humphreys Jul 16 '15 at 15:14

## 2 Answers

The question is interesting because it's a natural one which leads to an impasse for research and (probably as a result) apparently didn't make it into the literature even though people have long recognized that the answer is yes. Probably the most intriguing aspect of the question is how efficiently it can be answered: how few elementary facts about category $\mathcal{O}$ are actually needed to construct a rigorous proof. For example, does one need to know about the existence of enough projectives in $\mathcal{O}$?

Here is a fairly straightforward treatment of the rank 1 case, which makes Ben's helpful sketch more transparent, though it would require more combinatorial detail about weight spaces to make the general argument rigorous. In general it's probably natural to carry out some reductions first, including the (unnecessary) assumption that all weights are integral. The axioms imply fairly directly that all objects in $\mathcal{O}$ have finite Jordan-Holder length (finite number of composition factors). Since all simple modules are highest weight modules $L(\lambda)$ (the unique quotient of a Verma module $M(\lambda)$), and $\mathcal{O}$ is closed under subquotients, it's enough to assume that $M = L(\lambda)$ is infinite dimensional, which just means that $\lambda$ is not dominant. Now choose $N$ to be a simple Verma module $M(\mu)$ (so $\mu$ is antidominant).

For the rank 1 simple Lie algebra $\mathfrak{sl}_2$, integral weights can be treated just as integers. Here the possibilities for simple and Verma modules in $\mathcal{O}$ are very limited: the finite dimensional simples $L(\lambda)$ with $\lambda \geq 0$ and the infinite dimensional simples of negative highest weight, which are also Verma modules. Moreover, the weight spaces here all have dimension 1, with weights forming a string $\lambda, \lambda -2, \dots$ But in a tensor product $M \otimes N = M(\lambda) \otimes M(\mu)$ with $\lambda, \mu$ negative, the dimensions grow very rapidly. As Ben observes, this creates a contradiction. In particular, a weight space $(M \otimes N)_\nu$ with $\nu = \lambda+\mu - \theta$ and $\theta \in \mathbb{Z}^+$ can be arbitrarily large. On the other hand, if $M \otimes N \in \mathcal{O}$, it has only a finite number $n$ of composition factors and thus each weight space has dimension at most $n$.

ADDED: For the general case I've tried here to write down a concise proof with references, though it's unclear to me what approach will be "easiest" in terms of using only the most basic properties of $\mathcal{O}$ such as finite generation.

• Could you explain why the dimension of the weight spaces is bounded by the number of composition factors? – Mike Pierce Jul 25 '17 at 18:50
• @Mike: Note that what I've written only deals with a rank 1 Lie algebra, so you can take a shortcut here because all Verma modules or simple modules have 1-dimensional weight spaces. For the general case I tried to write up a short but somewhat more subtle proof in the linked note, which I hope is convincing though I haven't gone back to this recently. – Jim Humphreys Jul 25 '17 at 22:25

Yes. Any finitely generated module in category O is a quotient of a finite generated projective, and thus a quotient of a module with a finite Verma module filtration. Thus, any infinite dimensional module tensored with a Verma module has weight multiplicities that grow too fast (faster than a constant times the Kostant partition function), so it can't be in category O.