Is there a category of representations of a simple Lie algebra on which its Weyl group naturally acts?

For any simple Lie algebra $\mathfrak{g}$, is there a category $C$ of (possibly infinite-dimensional) representations of $\mathfrak{g}$ such the Weyl group $W$ of the corresponding root system acts in a nontrivial way on $C$? In other words, can we find such a category $C$ and for each $g \in W$, an endofunctor

$$A(g) : C \to C$$

together with natural isomorphisms

$$\alpha_{g,h} : A(g) A(h) \stackrel{\sim}{\rightarrow} A(g h)$$

perhaps obeying the obvious coherence laws?

I'd be even happier if certain weights $\lambda \in \mathfrak{h}^*$ (where $\mathfrak{h}$ is the Cartan) somehow gave rise to simple objects $R_\lambda \in C$ in such a way that

$$A(g) R_\lambda \cong R_{g(\lambda)}$$

At one point I hoped the Bernstein-Gelfand-Gelfand category $\mathcal{O}$ would do the job here, because for each weight $\lambda \in \mathfrak{h}^*$ I believe there's a simple object $L_\lambda \in \mathcal{O}$. But someone more knowledgeable than me persuaded me that no endofunctors $A(g)$ sending $L_\lambda$ to $L_{g(\lambda)}$, or perhaps just no exact such functors, exist on category $\mathcal{O}$. I would love to be wrong here, or at least to learn how close (or far) the Weyl group comes to being able to act on category $\mathcal{O}$ in such a way that

$$A(g) L_\lambda \cong L_{g(\lambda)}$$

Maybe we can't find such an action with exact or even right exact functors, but we can still do it with functors that preserve direct sums.

• Do you want isomorphism classes of representations? Are you aware that the intertwiner of parabolic induced representations of reductive groups over local fields are families of operators indexed by the Weyl group associated to the corresponding Levi-subgroup. – Marc Palm Sep 12 '13 at 10:44
• So first of all, one should probably consider just the principal block of $\mathcal{O}$, so the irreducibles are indexed by the Weyl group. The twisting functors come close, but they don't satisfy quite the correct relations when applied to the irreducibles. – Tobias Kildetoft Sep 12 '13 at 10:56
• I don't know what "do I want isomorphism classes of representations?" means. I want a category whose objects are some of the representations of $\mathfrak{g}$, on which the Weyl group has a natural (and nontrivial) action. – John Baez Sep 12 '13 at 12:53
• I recall the section on twisting functors in Humphrey's book to be a bit on the short side. There is a paper by Andersen and Stroppel, "Twisting functors on $\mathcal{O}$" which I recall has a quite good introduction to the subject. – Tobias Kildetoft Sep 12 '13 at 13:07
• There are a number of endofunctors on $\mathcal O$ that you may want to look at in addition to the twisting functors such as Arkhipov's, Enright's and Irving's functors (caveat: I am far from an expert on this!). Humphreys' book has some material on these functors; also see the papers "On Arkhipov's and Enright's functors" by Khomenko and Mazorchuk and "On functors associated to a simple root" by Mazorchuk and Stroppel. I think, though, that these functors only satisfy braid relations and not Weyl group relations, so this may not be quite what you're looking for. – Chuck Hague Sep 12 '13 at 14:43

To follow up on what Tobias and Chuck wrote, I think you should reconsider category $\mathcal{O}$ or better yet, its principal block $\mathcal{O}_0$, but with the proviso that you have to make some sacrifices. One sacrifice is that you have to work in the derived category, not the abelian, and the other sacrifice is that you must accept a braid group action, not a Weyl group one.

Once you do this, there are a beautiful pair of commuting braid group actions on $D^b(\mathcal{O}_0)$ given by shuffling and twisting functors. If you identify the Grothendieck group of $D^b(\mathcal{O}_0)$ with $\mathbb{Z}[W]$ by sending the class of the Verma module $M_{w\cdot \rho-\rho}$ to $w\in \mathbb{Z}[W]$, then these categorify the left and right actions of right and left actions of $W$ on itself, except that the functors satisfy the braid relations, not the Weyl group relations. One secret reason for this is that they really categorify the right and left action of the Hecke algebra on itself, so you can't really expect more than the braid relations.

This action also really thinks that it is a Weyl group action in the following sense: the most natural way of defining them actually assigns a functor in a each element of the Weyl group, and the proof of the braid relations is really that $T_wT_{w'}=T_{ww'}$ if $\ell(w)+\ell(w')=\ell(ww')$. Perhaps this is slicing it a little finely, but to me that says you should really think of it as a Weyl group action that got upgraded a bit.

EDIT: Yes, I mean the Artin group, which you can define as the group freely generated by the elements of the Weyl group, modulo the relation $T_wT_{w'}=T_{ww'}$ if $\ell(w)+\ell(w')=\ell(ww')$.

If you want a monoidal category, then maybe you should think about the category of Harish-Chandra bimodules, discussed in work of Bernstein and Gelfand (that's BGG with lower Gelfand multiplicity). Let $U(\mathfrak{g})_0=U(\mathfrak{g})/I_0$ denote $U(\mathfrak{g})$ modulo the ideal $I_0$ generated by central elements that act trivially on the trivial representation. The category I have is essentially the category of bimodules you get over $U(\mathfrak{g})_0$ by looking taking all sums, quotients and extensions of the bimodules $B_V$ gotten from $U(\mathfrak{g})\otimes V$ for all finite dimensional reps $V$ by killing the left and right action of $I_0$. The right action on $U(\mathfrak{g})\otimes V$ is the obvious one ignoring $V$, and the left action is by the coproduct (this is the bimodule where tensor product of it with a left module over $U(\mathfrak{g})$ is the same as usual Hopf algebra tensoring).

The derived category of HC-bimodules is monoidal under the usual tensor product, and there is a functor from the braid group (thought of as a monoidal category) to this category, which corresponds to the shuffling functors.

• By the the 'braid group', do you mean the Artin group which has the same standard generators $s_i$ as the Weyl group (one per dot in the Dynkin diagram), and the same relations $(s_i s_j)^{m_{ij}} = 1$ for $i \ne j$, but lacking the relations $s_i^2 = 1$? I'm hoping you'll answer yes. The other 'sacrifice' you demand, working with the derived category of the principal block of category $\mathcal{O}$, is no problem for me, except someday I'd like a category with a monoidal structure. – John Baez Sep 13 '13 at 5:18
• @JohnBaez Yes, that is the group you get an action of (I had forgotten that the twisting functors only had the braid relations when I wrote my original comment). – Tobias Kildetoft Sep 13 '13 at 6:58
• +1 for Gelfand multiplicity :) $[BGG:G] = (1+q)$? – Geordie Williamson Sep 14 '13 at 6:39
• @GeordieWilliamson Depends on how you decide to grade your Gelfands. Is multiplication by q a generation earlier, or a generation later? – Ben Webster Sep 18 '13 at 13:18

I can cook up such category on my Foreman grill. Let $\tilde{U}(g)=S(h)\otimes_{S(h)^W}U(g)$ be the extended enveloping algebra of $g$. $W$ acts on $\tilde{U}(g)$, hence, it acts on the category of its representations, all of whom are $g$-modules.

This action is rather silly on blocks with generic central character: $W$ just moves the blocks around. However, if the central character has a non-trivial stabilizer in $W$, the action has non-trivial content.