This question may be seen as a follow up of this one.

Let $\mathfrak{g}$ be a rank $r$ simple complex Lie algebra, and let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$. We call $\alpha_i$ the simple roots and $\varpi_i$ the fundamental weights (for $i=1,\dots ,r$). Define the set of integral weights $$\Lambda = \mathbb{Z} \varpi_1 + \dots + \mathbb{Z} \varpi_r$$ and the set of dominant integral weights $$\Lambda^+ = \mathbb{Z}^+ \varpi_1 + \dots + \mathbb{Z}^+ \varpi_r \, . $$

[EDIT: I originally defined here a partial ordering on the weight which was unfortunate and non necessary for my question. ]

Introduce finally the Weyl vector $\rho = \varpi_1 + \dots + \varpi_r $, the Weyl group $W$, and the dot action $w \cdot \lambda = w(\lambda + \rho) - \rho$ for $w \in W$. For an integral weight $\lambda$, we call $M(\lambda)$ the associated Verma module and $L(\lambda)$ the associated simple module.

Let $\lambda \in \Lambda$. If $\lambda \in -2\rho - \Lambda^+$, then the Kazhdan-Lusztig Theorem tells us that for any $w \in W$, $$\mathrm{ch}\, L (w \cdot \lambda) = \sum\limits_{x \leq w} (-1)^{\ell (x,w)} P_{x,w} (1) \, \mathrm{ch}\, M (x \cdot\lambda) \, , $$ where we use the Bruhat ordering and the length function on $W$, and the $P_{x,w} (q)$ are the Kazhdan-Lusztig polynomials.

My question is the following. Let $w \in W$ and $\lambda \in -\rho -\Lambda^+$ such that $\lambda \notin -2 \rho - \Lambda^+$. In other words, there is no element of $\Lambda^+$ in the dot orbit of $\lambda$. Is there a known expression of $\mathrm{ch}\, L (w \cdot \lambda)$ in terms of $\mathrm{ch}\, M (x \cdot\lambda)$, for $x \in W$ ?

  • $\begingroup$ In your reformulated question at the end, you start with $w \in W$ but then ignore it. Anyway, the question of having a dominant weight in the dot orbit of $\lambda$ (or not) isn't really relevant to the method I outlined. So it seems the answer is still the same, as is my comment on the next-to-last paragraph. $\endgroup$ – Jim Humphreys Aug 11 '17 at 21:22
  • $\begingroup$ I don't think I ignore $w$ in my question, it appears in $\mathrm{ch} \, L(w \cdot \lambda)$. $\endgroup$ – Antoine Aug 13 '17 at 15:02
  • $\begingroup$ I have (finally) read carefully chapter 7 and in particular section 7.10 in the book, and I think it answers all my questions after all! $\endgroup$ – Antoine Aug 13 '17 at 17:30

The answer to the question at the end is certainly yes, following the pattern in the previous question. But it's not helpful to emphasize the partial ordering of weights here. Instead, note that $\lambda$ might well be regular relative to the dot-action of the Weyl group. For example, in the case of $\mathfrak{sl}_3$ take $\lambda = -\rho -\alpha$ (where the simple roots are $\alpha, \beta$ and $\rho = \alpha + \beta$), which lies inside a shifted Weyl chamber next to the antidominant one. The KL Conjecture (soon a theorem) applies directly here, since $\lambda$ is actually in the principal block; so the character of $L(\lambda)$ is the difference of two Verma module characters. In general, you'd have to use a translation functor to move $\lambda$ around inside a Weyl chamber, or else pass to the closure (using the fact that each weight lies in the upper closure of a unique chamber); then apply the functor to those Verma modules which occur in the KL Conjecture, with the same integral coefficients.

[Note too that the preliminary remarks are somewhat out of focus, since $\lambda \leq -2\rho$ implies that $\lambda$ is a regular antidominant weight; so in this case $L(\lambda) = M(\lambda)$.]

EDIT: At first I skipped over the nonstandard definition of partial ordering of weights here (the standard definition is universally used, so a change needs some motivation and probably a different symbol). This makes my small example inapplicable, but the main point is the same: an integral weight lies inside some (shifted) Weyl chamber or else lies on one or more (shifted) reflecting hyperplanes. In the first case, just use a translation functor within that chamber to compare with the KL statement for the principal block. In the second case, use the standard geometric fact (observed by Jantzen) that every "singular" point lies in the upper closure of a unique chamber, and then translate the simple module in the principal block belonging to that chamber to the given simple module with highest weight in the upper closure.

A side remark is that most cases are uncomputable in practice, since the KL polynomials themselves typically require a long recursive computation. So most people are more interested in the theory here, which reveals the power of "geometric" thinking to resolve an algebraic question on some level. As with finite dimensional representations, there are lots of obstacles to practical computation no matter what approach is used.

  • $\begingroup$ Thanks for the answer! For the partial ordering, I didn't define it correctly in my initial post, this was a mistake. I meant the usual partial ordering that you use in your book. I'll edit my initial post accordingly. $\endgroup$ – Antoine Aug 11 '17 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.