Let $\mathfrak{g}$ be a finite dimensional complex semisimple Lie algebra (according to a comment of Victor Ostrik, we need to further require that $\mathfrak{g}$ is simple) 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 $\mathcal{O}$.

In an answer to this MO question, Jim Humphreys showed that an object $M$ in the category $\mathcal{O}$ is “tensor-closed” (meaning that $M\otimes N$ is in $\mathcal{O}$ whenever $N$ is) if and only if $M$ is finite dimensional. He also wrote a note on it.

My question is: if $M$ and $N$ in $\mathcal{O}$ are both infinite dimensional, then is it always true that $M\otimes N$ is no longer in $\mathcal{O}$? In particular let $L(\lambda)$ and $L(\mu)$ be the unique simple module with highest weights $\lambda$ and $\mu\notin \Lambda^{+}$, is $L(\lambda)\otimes L(\mu)$ in the category $\mathcal{O}$?

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    $\begingroup$ There is a silly counterexample if ${\mathfrak g}=sl(2)\oplus sl(2)$: tensor product of Verma modules for each copy of $sl(2)$ is a Verma module for ${\mathfrak g}$. So you might want ${\mathfrak g}$ to be simple in your question.. $\endgroup$ – Victor Ostrik Sep 16 '19 at 16:54
  • $\begingroup$ @VictorOstrik, I wouldn't say that's a silly counterexample at all. I did my PhD on category $\mathcal{O}$ and even wrote a paper on tensor products of infinite dimensional modules (albeit I left academia 11 years ago), and my instinct was that "of course the product has to be outside of $\mathcal{O}$". Personally, I very easily make the assumption "semi-simple" = "simple". $\endgroup$ – Johan Kåhrström Sep 16 '19 at 22:09
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    $\begingroup$ @Victor: I can't follow your brief comment which needs to be expanded. The first sentene is already unclear to me. $\endgroup$ – Jim Humphreys Sep 17 '19 at 11:31
  • $\begingroup$ @JimHumphreys, I understood it as follows: $\mathfrak{g}$ being of type $A_1\times A_1$, the Weyl group is the dihedral group $D_4$ (four elements), let $s, t\in D_4$ be the simple reflections. Then $L(s\cdot 0)$ and $L(t\cdot 0)$ are 'Verma modules for each copy of $\mathfrak{sl}_2$' in a sense. Also, $L(s\cdot 0) \otimes L(t\cdot 0)\simeq L(st\cdot 0)$, which is a Verma module for $\mathfrak{g}$. $\endgroup$ – Johan Kåhrström Sep 17 '19 at 12:31
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    $\begingroup$ @Jim: Sorry for being so sloppy. Let $V$ be a Verma module for $sl(2)$ (with arbitrary highest weight); using two projections ${\mathfrak g}\to sl(2)$ we can consider $V$ as a ${\mathfrak g}-$module in two different ways. Let us call the resulting ${\mathfrak g}-$modules $V_1$ and $V_2$. Then both $V_1$ and $V_2$ are in the category ${\mathcal O}$ for ${\mathfrak g}$, and $V_1\otimes V_2$ is Verma module over ${\mathfrak g}$. $\endgroup$ – Victor Ostrik Sep 17 '19 at 18:19

The answer to the question here is yes. (More generallly, if $M$ and $N$ lie in $\mathcal(O)$ and are both infinite dimensional, then $M \otimes N \notin \mathcal{O}$.)

The proof is eaiest to write down in case $\mathfrak{g} = \mathfrak{sl}_2$ with $M = L(\lambda)$ and $N=L(\mu)$ with $\lambda, \mu \in \mathbb{Z}$ and $\lambda, \mu \leq -1$; but the idea is similar for other pairs. The main point is that $M \otimes N$ fails to have a formal character (i.e., fails to be of finite Jordan-- Holder length). For example, when $\lambda =\mu = -1$, the weights of $L(\lambda) \otimes L(\mu)$ are $-2, -4, -4, -6,-6,-6, \dots$, so the infinitely many composition factors are those of $M(-2), M(-4), M(-6), \dots$ This is contrary to the finite length of all modules in $\mathcal{O}$.

Note here that $L(\lambda) = M(\lambda)$ is a simple Verma module and similarly $L(\mu)=M(\mu)$. $L(-2) = M(-2)$, etc.

ADDED: The point of the preceding note is to give examples where it's easy to specify the formal character of a simple module without getting into the details of the Kazhdan-Lusztig methods. But I'd also expect the rank 1 case to be sufficient for dealing with the general case: for a suitable simple root, consider how representations of $\mathfrak{g}$ restrict to the rank 1 Lie algebra corresponding to the root, etc. This approach hasn't been thoroughly worked out, however.

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  • $\begingroup$ So the key to this argument is to show that in the general case, for any pair of 'infinite' modules in $\mathcal{O}$ there is always an $\mathfrak{sl}_2$-subalgebra $\mathfrak{n}_-\oplus\mathfrak{h}\oplus\mathfrak{n}_+$ such that both $M$ an $N$ are $\mathcal{U}(\mathfrak{n}_-)$-free and hence direct sums of (infinitely many) Verma modules as $\mathfrak{sl}_2$-modules.This clearly fails in @Victor's counterexample, so relies $\endgroup$ – Johan Kåhrström Sep 18 '19 at 8:53
  • $\begingroup$ Apologies, pressed enter too early and failed to edit so here's the rest: ... so the existence of such a root clearly relies on $\mathfrak{g}$ being simple. Is $\rho$ (the half-sum of all positive roots) always a proportional to a root? I suspect that root would be the root to look at if that's the case. $\endgroup$ – Johan Kåhrström Sep 18 '19 at 9:00
  • $\begingroup$ @Johan: As you say, $\mathfrak{g}$ should be simple No, $\rho$ isn't always proportional to a root, as seen in the tables at the end of Chapter VI in Bourbaki. $\endgroup$ – Jim Humphreys Sep 18 '19 at 17:33

Edit: As @ZhaotingWei points out this is wrong. In fact, in my example we seem to have $M\simeq \bigoplus_{n=4}^\infty \Delta(-n\rho)$ which is certainly not in $\mathcal{O}$.

Inspired by @VictorOstrik's answer, I believe (but there may be holes in my logic) that this is a counter example in $\mathfrak{sl}_3$:

Let $\rho$ be the half-sum of positive roots, $s, t$ be the simple reflections in $S_3$, and for $w\in S_3$, $\lambda\in\mathfrak{h}^*$, let $w\cdot \lambda = w(\lambda + \rho) - \rho$ denote the usual 'dot action', then consider $M = L(st\cdot 0) \otimes L(ts\cdot 0)$ (for a visual see page 27 in my paper mentioned above). By weight considerations, it is clear that $M$ will have a highest weight vector $-3\rho$. This is an anti-dominant weight, so the Verma module $\Delta(-3\rho)$ must occur as a subquotient of $M$.

Now, the non-zero weight spaces of $L(st\cdot 0)$ and $L(ts\cdot 0)$ all have dimension 1 (I believe, it's been a while so I may be wrong on this). Working out the formal character of $M$ then shows that $\operatorname{ch} M = \operatorname{ch} \Delta(-3\rho)$, so we must have $M\simeq\Delta(-3\rho)$ which is in $\mathcal{O}$.

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  • $\begingroup$ It seems that $\text{M}\neq \text{ch}\Delta(-3\rho)$. For example, in the Verma module $\Delta(-3\rho)$ the multiplicity of weight $-4\rho$ is $2$. However in $M=L(st\cdot 0)\otimes L(ts\cdot 0)$ the multiplicity of weight $-4\rho$ is $3$: it contains $v_{-2\alpha-\beta}\otimes w_{-2\alpha-3\beta}$, $v_{-3\alpha-2\beta}\otimes w_{-\alpha-2\beta}$, $v_{-2\alpha-2\beta}\otimes w_{-2\alpha-2\beta}$, which are linearly independent. $\endgroup$ – Zhaoting Wei Sep 17 '19 at 14:39
  • $\begingroup$ But the weight space $-2\alpha-2\beta$ is zero-dimensional for both $L(st\cdot 0)$ and $L(ts\cdot 0)$. If you don't trust my picture that I linked, this follows from the fact that it is one-dimensional for $\Delta(st\cdot 0)$, $\Delta(ts\cdot 0)$ and $\Delta(sts\cdot 0)$ and $L(st\cdot 0) \simeq \Delta(st\cdot 0) / \Delta(sts\cdot 0)$ and $L(ts\cdot 0)\simeq \Delta(ts\cdot 0) / \Delta(sts\cdot 0)$. $\endgroup$ – Johan Kåhrström Sep 17 '19 at 15:04
  • $\begingroup$ Sorry, I mean $v_{-2\alpha-\beta}\otimes w_{-2\alpha-3\beta}$, $v_{-3\alpha-2\beta}\otimes w_{-\alpha-2\beta}$, and $v_{-3\alpha-\beta}\otimes w_{-\alpha-3\beta}$. I think $v_{-3\alpha-\beta}$ is non-zero in $L(st\cdot 0)$ and $w_{-\alpha-3\beta}$ is non-zero in $L(ts\cdot 0)$. $\endgroup$ – Zhaoting Wei Sep 17 '19 at 15:30
  • $\begingroup$ Ah yes, you’re right. $\endgroup$ – Johan Kåhrström Sep 17 '19 at 15:43

I will answer the question in the affirmative for $\mathfrak{sl}_n$, and in general provide a property of the Weyl group that would imply an affirmative answer in general. This property is true in type A, but not for the other rank 2 cases. The other rank 2 Lie algebras are small enough to be dealt with easily.

Let me assume for now that λ is regular and integral. Write $\lambda = x\cdot \lambda'$, where x is in the Weyl group and λ' is dominant. Write ch(L(λ)) as a rational function in reduced form, and let T be the set of positive roots α such that $(1-e^\alpha)$ does not appear in the denominator. Let G be the subgroup generated by all reflections in all elements of T. I claim that if g is in G, then gx≤x in Bruhat order.

Switch to indexing simples and Vermas by the Weyl group. We have $$\operatorname{ch}(L(x))=\sum_y P_{x,y}\operatorname{ch}(M(y)).$$ The claim follows from two simple observations. Firstly, for $(1-e^\alpha)$ to not appear in the denominator, we must have $P_{x,y}+P_{x,sy}=0$, where s is the reflection for the root α. Secondly, that Px,x=1 and Px,y=0 unless y≤x.

Now we turn to the question. Consider two simples L(λ) and L(μ). From the above discussion we obtain two Weyl group elements x and y, together with two sets of positve roots Tx and Ty, and groups Gx and Gy generated by the corresponding reflections, so that $G_xx\leq x$ and $G_y y\leq y$. If $L(\lambda)\otimes L(\mu)$ is in category O, then $T_x\cup T_y$ is all positive roots by looking at the character. For L(λ) and L(μ) to be infinite dimensional, we must have that x and y are not the identity.

Now we have a condition on the Weyl group which is necessary for a negative answer. In the symmetric group, subgroups generated by reflections are products of smaller symmetric groups. It is easy to see that for any two proper such subgroups, there is a reflection which is not contained in either. Thus in type A, since every positive root appears in either Tx or Ty, either Gx or Gy is the entire symmetric group, which then contradicts the Bruhat ordering condition as x and y are not the identity.

If we move beyond regular or integral weights, then it becomes even harder to cancel factors in the denominator of the character, so the argument should be easier. Beyond type A, I haven't checked any non-type A Weyl groups of rank at least three to see if my criterion is enough to answer the question.

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  • $\begingroup$ By "If $L(\lambda)\otimes L(\mu)$ is infinite dimensional, then $T_x\cup T_y$ is all positive roots by looking at the character." I think you mean "If $L(\lambda)\otimes L(\mu)$ is in $\mathcal{O}$, then $T_x\cup T_y$ is all positive roots by looking at the character." Isn't it? $\endgroup$ – Zhaoting Wei Sep 19 '19 at 18:58
  • $\begingroup$ Moreover, by "For $L(\lambda)$ and $L(\mu)$ to be finite dimensional, we must have that x and y are not the identity." I think you mean "For $L(\lambda)$ and $L(\mu)$ to be infinite dimensional, we must have that x and y are not the identity.". Isn't it? $\endgroup$ – Zhaoting Wei Sep 19 '19 at 18:59
  • $\begingroup$ Yes to both. I've edited the answer appropriately. $\endgroup$ – Peter McNamara Sep 19 '19 at 22:55
  • $\begingroup$ I'm sorry I don't get how to write Ch$(L(\lambda))$ as a rational function in reduced form. Is there a generalization of Weyl character formula to $\lambda\notin \Lambda^+$ or it is a consequence of the Kazhdan-Lusztig conjecture? $\endgroup$ – Zhaoting Wei Sep 20 '19 at 14:22
  • $\begingroup$ Since the Vermas have characters which are rational functions in the e^alphas, and the simple characters are finite combinations of the Vermas, the simple characters are also rational functions. $\endgroup$ – Peter McNamara Sep 20 '19 at 23:21

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