# 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.

• @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
• 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