5
$\begingroup$

Let $G$ be a complex reductive Lie group, $B$ a Borel subgroup, with which to define "dominant weight". Let $\lambda$ be an integral weight, not necessarily dominant, but nonetheless giving a one-dimensional $B$-rep $\mathbb C_\lambda$.

Then we can define the Verma module $U{\mathfrak g} \otimes_{U\mathfrak b} \mathbb C_\lambda$, which has compatible actions of $\mathfrak g$ and $B$ (it's a "$(\mathfrak g,B)$-module"), and a unique irreducible quotient $V_\lambda$, again a $(\mathfrak g,B)$-module.

If $\lambda$ is dominant, then $V_\lambda$ is actually a (finite-dimensional) $G$-irrep, so we know how to compute its weight multiplicities in manifestly positive ways, e.g. counting Littelmann paths. For $\lambda$ not dominant, $V_\lambda$ is infinite-dimensional, but its weight multiplicities are still finite (since they're bounded by those of the Verma module).

Are there combinatorial formulae for the weight multiplicities of $V_\lambda$, when $\lambda$ is not dominant?

If so, references please!

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ Can you explain a bit more about the nature of the $B$-action? In any case, it seems that your $V_\lambda$ is the usual irreducible $\mathfrak{g}$-quotient of the Verma module (often denoted $L(\lambda)$). The Kazhdan-Lusztig algorithm gives a sort of "combinatorial formula", far from positive, for the weight multiplicities; but it's extremely hard to compute in most cases. So I'm not optimistic. (By the way, I guess $G$ here should be semisimple?) $\endgroup$
    – Jim Humphreys
    Commented Oct 3, 2014 at 20:46
  • $\begingroup$ I meant to have reductive there; I'll put it in. Yes your $L(\lambda)$ is my $V_\lambda$, insofar as they're unique. What could there be to say about "the nature of the $B$-action"? 1-d reps are of $B/B' \cong T$, so weights, like $\lambda$. $\endgroup$
    – Allen Knutson
    Commented Oct 4, 2014 at 0:53
  • $\begingroup$ I second what Jim says - I doubt that there is anything better than what is given by the Kazhdan-Lusztig conjecture. $\endgroup$
    – Alexander Braverman
    Commented Oct 4, 2014 at 2:56
  • $\begingroup$ P.S. I still don't understand how the notion of $(\mathfrak{g}, B)$-module is defined, or what extra information it might contain. Note that a "weight" for $\mathfrak{g}$ (or a Cartan subalgebra) is the differential of a weight for a torus. $\endgroup$
    – Jim Humphreys
    Commented Oct 4, 2014 at 20:34
  • 1
    $\begingroup$ (1) When $G\geq H$, a $(\mathfrak{g},H)$-module $V$ is one with actions of $\mathfrak{g}$ and of $H$, from each of which one can derive an action of $\mathfrak h$; the two actions should agree. (2) As for what extra information it should contain beyond that of a $\mathfrak{g}$-representation: a $(\mathfrak g,B)$-irrep is in category $\mathcal O$, unless I'm very confused. (3) My weight $\lambda$ was assumed integral, so better to think of it as a weight for $T$ (or $B$ since $B/[B,B] \equiv T$) than for $\mathfrak t$. $\endgroup$
    – Allen Knutson
    Commented Oct 5, 2014 at 0:12

2 Answers 2

Reset to default
3
$\begingroup$

Maybe it should be emphasized that the question here concerns only Lie algebra representations, so the initial group language (about Lie groups or perhaps algebraic groups) is unnecessary. The framework is the 1976 BGG category $\mathcal{O}$ for a semisimple Lie algebra over $\mathbb{C}$ or other algebraically closed field of characteristic 0. In general, each linear functional $\lambda \in \mathfrak{h}^*$ (for a fixed Cartan subalgebra $\mathfrak{h}$ of the given Lie algebra $\mathfrak{g}$) defines a Verma module and its unique simple quotient, while other possible highest weight modules occupy intermediate positions.

Here the weights are assumed to be integral. Such weights form an $\ell$-dimensional lattice $\Lambda \subset \mathfrak{h}^*$ (where $\ell = \dim \mathfrak{h}$), which can be viewed also as the abstract weight lattice of the underlying root system or as the character group $X(T)$ of a maximal torus $T$ having Lie algebra $\mathfrak{h}$ in a simply connected algebraic group with Lie algebra $\mathfrak{g}$. (But since $G$ usually doesn't act analytically or rationally on modules in category $\mathcal{O}$, the potential rational $T$-action doesn't add anything to the given action of $\mathfrak{h}$.)

As Victor observes, there are "trivial" cases in which there exists a combinatorial formula for the weight multiplicities of a given (integral) weight $\mu$ in a simple highest weight module of highest weight $\lambda$. A Verma module is in fact simple if and only if its highest weight is antidominant (see section 4.4 in my book on category $\mathcal{O}$). For a parabolic (= generalized) Verma module, Jantzen developed a rather complicated criterion (see sections 9.12-9.13). But even in such "trivial" cases, it's necessary to compute Kostant's partition function.

Other than this case, and the narrow case in type $A$ which Victor cites, I'm unaware of any infinite dimensional simple modules admitting such combinatorial formulas. Naturally it's impossible to prove that such formulas can't exist elsewhere. To approach this experimentally would be extremely challenging, since the combination of Kostant's partition function (for weight multiplicities in Verma modules) and the Kazhdan-Lusztig polynomial values at 1 will lead to lengthy calculations involving many cancellations even in fairly small ranks. The resulting raw data might or might not suggest any patterns, so a clever a priori theoretical approach would probably be needed to get new formulas. While the formulation and proof of the Kazhdan-Lusztig conjecture was a major breakthrough in terms of theoretical understanding of the infinite dimensional modules in question, the applications (to unitary representations of Lie groups in particular) continue to be quite indirect.

Improve this answer
$\endgroup$
2
  • $\begingroup$ The "simple iff antidominant" statement is only in the presence of the $\lambda$ integral assumption, right? $\endgroup$
    – Allen Knutson
    Commented Oct 11, 2014 at 18:12
  • $\begingroup$ @Allen: No, it's a general fact (if "antidominant" is defined in the natural way: see 3.5 for discussion of terminology). However, there seems to be no simple unified proof, so I wound up treating integral weights in 4.4 but waited until 4.8 to deal with non-integral ones. Here as elsewhere in the subject it's annoying to make detours into more elaborate arguments for arbitary $\lambda \in \mathfrak{h}^*$. And sometimes the integral case is all one wants. $\endgroup$
    – Jim Humphreys
    Commented Oct 12, 2014 at 13:33
2
$\begingroup$

The "trivial" case is when the simple highest weight module is a generalized Verma module (i.e. the module induced from a character of a parabolic). The multiplicity is given by the variant of Kostant's partition function; this can be reformulated in terms of lattice points in polyhedra, etc.

There is also Mathieu-Papadopoulo character formula (in the general linear case). The proof is somewhat involved; I worked out a simpler proof in a special case based on the theory of standard monomials.

Improve this answer
$\endgroup$
2
  • $\begingroup$ That "trivial" case is exactly when $\lambda$ (which I assumed integral) is regular antidominant, right? At least without your "generalized". $\endgroup$
    – Allen Knutson
    Commented Oct 4, 2014 at 0:56
  • $\begingroup$ That's right. In the generalized Verma module case, the highest weight $\lambda$ is antidominant with respect to a subset $S_P$ of the set $S$ simple roots and is orthogonal to its complement in $S$; the latter condition can be replaced with "dominant" if we allow finite-dimensional the inducing module to be finite-dimensional (rather than one-dimensional), which complicates the character formula somewhat. $\endgroup$
    – Victor Protsak
    Commented Oct 6, 2014 at 20:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .