Let $\mathfrak{g}$ be a Kac-Moody Lie algebra (actually, I'd already be happy with an answer addressing the case where $\mathfrak{g}$ is a simple Lie algebra over $\mathbb{C}$).

1st ?: I'm wondering what results are known about the characters of $L(\lambda)$ (the irreducible module of highest weight $\lambda$) when $\lambda$ is not dominant. I know very little Lie theory (just the basics really, like in Kac's book) and am not at all up to date on the current literature.

I do know that in principle one could use KL polynomials to calculate characters, but that's not the kind of answer I'd like. Rather, I'd be interested in closed formulas like the Weyl-Kac character formula or explicit combinatorial descriptions of the weight space decomposition with respect to a Cartan. I'm sure no general results are known; I'll be happy with formulas for particular examples or special classes.

2nd ?: Is there a reason we shouldn't hope for such formulas for non-integrable modules in general? (Other than that current techniques don't provide them)

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    $\begingroup$ These questions are really broad. Already in the case of finite dimensional simple Lie algebras, there is a lot of theory for arbitrary integral highest weights (relative to BGG category $\mathcal{O}$ as in my 2008 AMS book). But here no closed character formula like Weyl's can be expected in general, given the lack of Weyl group symmetry and complexity of KL polynomial values. For Kac-Moody algebras in general there is also lots of theory, but it gets even more complicated. Many people have worked on this, such as Kashiwara, Tanisaki, Kumar, .... $\endgroup$ Dec 17, 2010 at 16:31
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    $\begingroup$ P.S. Besides the vast amount of research literature out there, a fairly recent survey paper might add some perspective: MR2057399 (2005h:17046), Tanisaki, Toshiyuki (J-OSAKC), Character formulas of Kazhdan-Lusztig type. Representations of finite dimensional algebras and related topics in Lie theory and geometry, 261–276, Fields Inst.Commun., 40, Amer. Math. Soc., Providence, RI, 2004. $\endgroup$ Dec 17, 2010 at 16:40
  • $\begingroup$ Jim, thanks for your comments and the reference. I can't seem to find the survey right now. I didn't intend for the question to be broad: by "I'll be happy with formulas for particular examples or special classes" I meant that I'd appreciate a reference to a paper where the characters of some particular non-integrable irreducibles are computed...(cont'd) $\endgroup$ Dec 19, 2010 at 23:03
  • $\begingroup$ ...I'm not convinced that it's too much to ask for combinatorial formulas for weight space dimension in special classes of examples. $\endgroup$ Dec 19, 2010 at 23:04
  • $\begingroup$ I'm only commenting here, not having been closely involved with Kac-Moody theory. But as far as I know, only for integrable representations are there rich combinatorial results. Even in the finite dimensional case, the combinatorics for non-dominant highest weights get tricky (except for antidominant weights). Asking for "combinatorial formulas" is OK but may be arbitrarily difficult to carry out in the absence of Weyl group symmetry. Even in integrable cases, concrete results require a lot of work (and motivation) for affine Lie algebras beyond the Weyl-Kac formula. $\endgroup$ Dec 20, 2010 at 23:28

1 Answer 1


As one special case you may consider the Kac-Wakimoto admissible representations for affine Kac-Moody algebras. These representations are a bit more general than the integrable ones and satisfy modular invariance properties.

Kac-Wakimoto gave a character formula for them in http://www.pnas.org/content/85/14/4956.short

A highest weight module $L(\lambda)$ for an affine Lie algebra $\mathfrak{g}$ is admissible when the weight lattice of integral weights with respect to $\lambda$ has the same dimension as the original weight lattice of $\mathfrak{g}$, and $\lambda$ satisfies a natural nonnegativity constraint with respect to the root lattice.

  • $\begingroup$ Thanks! Sorry I took so long to respond; haven't checked MO in a while. I didn't know about this paper, I'll have a look. $\endgroup$ Jan 14, 2011 at 3:04

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