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

Question

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?

Answer 1

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.

Answer 2

In response to Jim's challenge, I think I've found the easiest approach a bit belatedly: any module in category $\mathcal{O}$ is finitely generated over $U(\mathfrak{n}_-)$, and any Verma module is free of rank 1 over it. Thus, if $M$ is infinite dimensional, and $N$ a Verma module, the tensor product $M\otimes N$ is free of infinite rank over $U(\mathfrak{n}_-)$, and thus not in $\mathcal{O}$.