My first encounter with Category $\mathcal{O}$ was (perhaps unusually) learning about Kac-Moody algebras from Kac's book. Kac takes the following definition:

The Category $\mathcal{O}$ has objects $\mathfrak{g}(A)$-modules $V$ which are $\mathfrak{h}$-diagonalizable with finite dimensional weight spaces such that there exists a finite number of elements $\lambda_1, \dots, \lambda_n \in \mathfrak{h}^*$ such that $P(V) \subset \bigcup_{i=1}^n \mathfrak{h}_{\leq \lambda_i}$, where $P(V)$ is the set of weights for $V$.

In particular, this definition allows that for any modules $V, W \in \mathcal{O}$, we have $V \otimes W \in \mathcal{O}$.

To contrast, most authors writing about Category $\mathcal{O}$ in the finite semisimple case (for example, in Humphrey's book) take as axioms

(1) $V$ is a finitely-generated $U(\mathfrak{g})$-module

(2) $V$ is $\mathfrak{h}$-semisimple

(3) $V$ is locally $\mathfrak{n}$-finite

With this choice of axioms, we no longer have that $\mathcal{O}$ is closed under tensor products.

What is the motivation/necessity for taking the differences in axioms between these two cases? Why should one not always take the formulation in Kac for "finite-type" (i.e., finite semisimple) Kac-Moody algebras?

  • $\begingroup$ Without getting into the essence of your question, I should point out that Victor Kac wrpte several versions of his book (with two publisjers from 1983 to 1990). Mpst people refer to his third edition (Cambridge, 1990). There are possibly more readable accounts of the category $\mathcal{O}$ in the Wiley texts by Moody-Pianzola (1995) and by Carter (2005). All of this started in 1976 with a paper by BGG in Ruissian on the finite dimensional simple Lie algebra case. See also the paper by Deodhar-Gabber-Kac :mathscinet.ams.org/mathscinet-getitem?mr=663417 $\endgroup$ – Jim Humphreys Apr 13 '19 at 20:49
  • $\begingroup$ Many crucial properties of the BGG category $\mathcal{O}$, for example, "every object has finite length", do not hold in the Kac setting. $\endgroup$ – Victor Protsak Apr 14 '19 at 1:49

Your Answer

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

Browse other questions tagged or ask your own question.