# Is every weight of an integrable highest weight module in the Tits cone?

Let $\mathfrak{g}$ be a Kac-Moody algebra with Cartan subalgebra $\mathfrak{h}$, Weyl group $W$, and simple roots and coroots $\alpha_i, \check{\alpha_i}, i \in I$, respectively. Let $L$ be an integrable highest weight module.

Write $C$ for the dominant Weyl chamber, i.e. the locus $\{ \lambda \in \mathfrak{h}^*: (\lambda, \check\alpha_i ) \in \mathbb{R}^{\geqslant 0}, \forall i \in I \}$, and call the $W$ orbit of $C$ the Tits cone.

Does every weight of $L$ lie in the Tits cone?

If I haven't done something wrong, this is true for $\mathfrak{g}$ finite type and affine (untwisted). I am happy to restrict to the symmetrizable case, if that is easier to address.

• What do you mean by "the" Weyl chamber (or by the Tits cone)? Aside from that, it's a good idea to start with familiar examples of integrable modules such as the adjoint representation: its weights are the roots (not necessarily all real) together with 0. – Jim Humphreys Jun 29 '15 at 12:54
• @JimHumphreys Many thanks for the suggestions, I have clarified accordingly! As you say, considering the adjoint module for affine algebras shows this is false if we drop the assumption of $L$ being highest weight. – uncookedfalcon Jun 29 '15 at 23:44

Yes. The Tits cone, as the name implies, is a cone: in particular, it's convex. Any weight for a highest weight module is an affine linear combination of finitely many extremal weights (I'll leave that as an exercise; one hint is to prove it by induction on the number of simple roots you need to get to your weight from the highest weight), that is, it's the convex hull of the extremal weights for the representation. The extremal weights are obviously in the Tits cone by definition.

• Fantastic, much appreciated! – uncookedfalcon Jun 30 '15 at 3:19
• One follow up question: does every point in the intersection of the weight lattice and the convex hull of the extremal points occur with positive multiplicity (I again have arrived at a positive answer in finite and affine type, but am stumped in general)? Thanks in advance! – uncookedfalcon Jul 1 '15 at 4:57
• That is false in all types: you have to be in the same coset of the root lattice as the highest weight. With that proviso, I think it's true. Again, induct on the number of times you've had to subtract positive roots. If the weight $\mu$ isn't dominant, then its conjugate which is dominant will reduce this number, and have the same weight multiplicity. Once it's dominant, I think one should be able to find a simple root such that $\mu+\alpha_i$ is in the polytope (I don't have a slick explanation, but the picture looks right in my head). $E_i$ is injective here, so by induction, QED. – Ben Webster Jul 1 '15 at 12:40
• Ben, thank you for the reply. As you say, I forgot to mention the root lattice condition. The proof you graciously outlined is very similar to the argument I had in mind in finite and affine type. The result is indeed true, and is proposition 11.3 in Kac's Infinite Dim. Lie Algebras. Thank you again for your help! – uncookedfalcon Jul 4 '15 at 6:17