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Let $G$ be a finite group, let $k$ be a large enough field of characteristic $p>0$. Let $p\mid |G|$.

Broué's abelian defect group conjecture states the following:

Let $B$ be a block of $kG$ with abelian defect group $D$ and let $b\in \text{Bl}(N_G(D))$ be its Brauer correspondent. Then the derived categories $D^b (\text{mod}(B))$ and $D^b(\text{mod}(b))$ of bounded complexes of finitely generated modules over $B$ and $b$ are equivalent as triangulated categories.

Alperin's weight conjecture (blockwise version) states the following:

Let $B$ be a block of $kG$. Then the number of irreducible Brauer characters belonging to $B$ equals the number of $G$-conjugacy classes of $p$-weights of $B$.

Is there a relationship between these two conjectures? For example: does one conjecture imply the other?

The motivation for this question is: both Alperin's weight conjecture and Broué's abelian defect group conjecture can be expressed via trivial source modules (*).

Thank you very much for the help.

(*) Small footnote: in the case of Broué's abelian defect group conjecture this means that all modules occurring in the cochain complex realising such a derived equivalence are supposed to be $p$-permutation modules. Hence, this is more restrictive than the general conjecture.

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Yes. Broué's conjecture implies the abelian defect case of Alperin's weight conjecture. This makes one think there might be a structural statement like Broué's conjecture, for non-abelian defect groups, that would imply Alperin's conjecture in general. Many attempts have been made, to little avail; Rouquier has conjectured that if the hyperfocal subgroup of the defect group is abelian then there is a derived equivalence as in Broué's conjecture, but that's about it.

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    $\begingroup$ Here, hyperfocal has to be interpreted with respect to the fusion system on the defect group determined by the block; for the principal block it is the usual hyperfocal subgroup of the Sylow subgroup. $\endgroup$ Commented Sep 2, 2023 at 13:29
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    $\begingroup$ Thank you very much for the answer and the explanations! $\endgroup$ Commented Sep 2, 2023 at 14:58

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