6
$\begingroup$

Let $\mathfrak{g}$ be a semi-simple Lie algebra.

So in characteristic $0$, the Grothendieck group of a block of category $\mathcal{O}$ is given by the classes of the Verma modules. Unlike the simples (or projectives), Verma modules are very easy to understand explicitly, and one can easily perform computations with them (eg. involving translation functors).

In characteristic $p$, the analogous object is the category of modules with a fixed central character: one must specify both a Harish-Chandra character, and a character of the Frobenius (or p-) center. Now each simple is a quotient of a baby Verma module; however the classes of the baby Verma's no longer give a basis in the Grothendieck group -- in fact, they all have the same class!

I'm looking for a basis for the Grothendieck group of this category, where the corresponding objects are easy to understand (unlike the simples or projectives). In fact, I'd be happy with a collection of objects whose images span the Grothendieck group. I'm most interested in the case where the Frobenius character is 0, but the Harish-Chandra character is singular.

I've heard that the Weyl modules (coming from the char $p$ version of Borel-Weil theory) might be useful here; though I'm not entirely sure if they can be defined in this generality. This link might be relevant: Weyl modules and reduction modulo $p$.

$\endgroup$
2
$\begingroup$

I'm not sure what sources you are mainly relying on, but there are several points to be made:

1) When you say the Lie algebra is "semisimple", I suspect you mean (as people sometimes do when using shorthand) the Lie algebra of a semisimple algebraic group. There are lots of other simple Lie algebras (mostly classified for $p>3$ in the work of Premet-Strade and others) which don't come from algebraic groups, and their representations are less well studied. Anyway, it's probably best to limit to simple algebraic groups and their Lie algebras, the latter usually but not always being simple. (For example, the special linear Lie algebra of trace zero $n \times n$ matrices over an algebraically closed field of characteristic $p>0$ has a center consisting of scalar matrices when $p|n$.)

2) The algebraic group framework provides some useful tools for studying the Lie algebra representations, as seen in Jantzen's book Representations of Algebraic Groups: you can construct "mixed" group schemes, using a torus action to restore some ordering to "highest weights", but at the cost of creating an infinite rank Grothendieck group. Anyway, the simple modules for the Lie algebra always come from simple modules for the group (where the highest weights have to be "restricted"); then Steinberg's tensor product theorem produces all simple modules for the group.

3) The blocks in these various situations have been worked out by now, but for the Lie algebras alone there seems to be no realistic candidate for an easy basis of the corresponding Grothendieck group. As you say, the (baby) Verma modules for such a block all have the same composition factor multiplicities and define a single element of the Grothendieck group.

4) The Lie algebras here admit infinitely many simple modules, mostly not coming directly from the group and depending on a linear functional $\chi$ in the dual Lie algebra. For a short survey with references (not up-to-date), see my 1998 paper here. The conjecture there on blocks of the finite dimensional reduced enveloping algebra for $\chi$ was proved by Brown-Gordon here.

5) The methods of geometric representation theory have been used by Bezrukavnikov and others to get better information about the Lie algebra representations. But there still seems to be no easy way to approach this in terms of a Grothendieck group basis coming from modules that are easy to construct. The simple modules are mostly still elusive. (But those with singular "highest weights" are usually obtainable using Jantzen's translation functors from those with regular weights.)

$\endgroup$
  • $\begingroup$ Thanks for the response. I think the linear function $\chi$ that you refer to is the character of the Frobenius center that I mentioned. I don't necessarily need a "natural" basis - a collection of objects which span the Grothendieck group would be good enough for me. What if one applies translation functors to Weyl modules (so that they land inside the singular categories) -- would that be useful? $\endgroup$ – Vinoth Sep 14 '15 at 19:30
  • $\begingroup$ I meant to mention that "Weyl modules" are defined directly for the algebraic groups, where they play the role of universal highest weight modules by Kempf's theorem. (See notes posted on my homepage.) They don't play this role for the Lie algebras even when $\chi =0$. You need to be careful about which category you are working in when you talk about Grothendieck groups. $\endgroup$ – Jim Humphreys Sep 14 '15 at 20:26
  • $\begingroup$ Okay - do you mean your notes in the AMS Bulletins? Are there are Weyl modules which lie in the category where the Frobenius character $\chi = 0$, and the Harish-Chandra character is singular (perhaps $\Gamma(O(\lambda))$, where $\lambda$ is a singular weight)? I don't need them to play the role of highest weight modules; all I want is a spanning set in $K^0$. Thanks. $\endgroup$ – Vinoth Sep 15 '15 at 5:52
  • $\begingroup$ No, these are some of the unpublished notes on my homepage: people.math.umass.edu/~jeh/pub/weyl.pdf (It's important to realize that baby Verma modules don't fit into natural categories of modules for the groups, but simple modules do.) $\endgroup$ – Jim Humphreys Sep 15 '15 at 15:26
  • $\begingroup$ Thanks, I've had a look. One last question: so the block I'm interested in (with $0$ p-character, and singular HC character) contains infinitely many Weyl modules. Do the classes of these modules span the Grothendieck group? I think this should be true, since two modules with different characters have different images in $K^0$ (and one can compute the characters of these modules). $\endgroup$ – Vinoth Sep 17 '15 at 16:51

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.