6
$\begingroup$

Let $G$ be a finite group. Let $p$ be a prime dividing $|G|$. Let $K:=\overline{\mathbb{F}_p}$.

Let $b$ be a $p$-block of $G$ with abelian defect group $D$. Let $H:=N_G(D)$. Let $c$ be the Brauer correspondent of $b$.

M. Broué conjectured in the 90's that $b$ and $c$ are derived equivalent under these assumptions.

J. Rickard strengthened Broué's conjecture by conjecturing that equivalence can be given by a bounded chain complex of $p$-permutation bimodules whose indecomposable direct summands have vertices contained in $\Delta (D)$.

This means the following.

Let $Y$ be a virtual $KGb-KHc$-bimodule that is a linear combination of $p$-permutation bimodules that are finitely generated projective as left and right modules, satisfying $Y\otimes_{KHc}Y^{*}=[KGb]$ and $Y^{*}\otimes_{KGb}Y=[KHc]$ in the appropriate Grothendieck groups of bimodules.

More specifically, $Y\in T^{\Delta}(b,c)$, where $Y\in T^{\Delta}(b,c)$ denotes (in our context of Rickard's strengthening of Broué's abelian defect group conjecture) the Grothendieck group, w.r.t. $\oplus$, of the category of $p$-permutation $(b, c)$-bimodules, all of whose indecomposable direct summands have vertices contained in $\Delta(D) := \{(x, x) | x\in D\}$.

In particular, $Y$ induces a splendid virtual Morita equivalence.

Question: What is known about possible ways / strategies to (maybe computationally) lift such a splendid virtual Morita equivalence up to a splendid derived equivalence?

If not in general, is there something known, if $Y$ has very few indecomposable direct summands?

I've seen in the literature so far that there is a related article by D. Craven and R. Rouquier (see https://doi.org/10.1016/j.aim.2013.07.010) where they provide lifts of stable equivalences to perverse equivalences and they construct derived equivalences from these.

I would be particularly interested in the question of what references in the literature contain known ways or ideas on how to attack the problem to lift splendid virtual Morita equivalences up to splendid derived equivalences, but, of course, any other related literature would also be gratefully appreciated.

$\endgroup$

1 Answer 1

6
$\begingroup$

In almost all cases I know of where people have proved derived equivalences between blocks of finite groups, the proof hasn't really gone that way (i.e., finding a virtual bimodule and refining it to a splendid tilting complex). In fact, usually the virtual bimodule doesn't appear explicitly at all, although in most cases it would probably be possible to calculate it if you wanted.

There are two exceptions I can think of, where there was an "obvious" choice of a virtual bimodule, and later it was proved that there was a corresponding derived equivalence.

The first example arises for blocks of symmetric groups. About 30 years ago, I noticed a virtual bimodule that gave a virtual Morita equivalence between certain pairs of blocks of symmetric groups, such that if these could be lifted to derived equivalences then all blocks of symmetric groups with the same defect group would be derived equivalent. I found a natural way to turn them into a complex, but couldn't prove much about it. But later, Chuang and Rouquier proved that it did work in:

Chuang, Joseph; Rouquier, Raphaël, Derived equivalences for symmetric groups and $\mathfrak{sl}_2$-categorification., Ann. Math. (2) 167, No. 1, 245-298 (2008). ZBL1144.20001.

The second example is "Alvis-Curtis duality" for finite reductive groups in non-defining characteristic. Broué had conjectured that this was given by a derived equivalence, and there was an obvious choice of virtual bimodule (and a fairly obvious way to make it into a complex), which Marc Cabanes and I proved to work in:

Cabanes, Marc; Rickard, Jeremy, Alvis-Curtis duality as an equivalence of derived categories, Collins, Michael J. (ed.) et al., Modular representation theory of finite groups. Proceedings of a symposium, University of Virginia, Charlottesville, VA, USA, May 8-15, 1998. Berlin: de Gruyter. 157-174 (2001). ZBL1001.20002.

$\endgroup$
1
  • $\begingroup$ Thank you very much for the answer! $\endgroup$ Commented Jan 7, 2021 at 13:09

You must log in to answer this question.

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