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First Rickard (in Splendid Equivalences: Derived Categories and Permutation Modules ) and then Rouquier (Block theory via stable and Rickard equivalences, Appendix A.1) define splendid equivalences between (principal) blocks of algebras $A$ of $\mathbb{K}G$ and $B$ of $\mathbb{K}H$ with an isomorphic defect group $P$ this way:
$1)$ There exists a complex $X$ of finitely generated $(A,B)$-bimodules such that $\operatorname{Hom}_A (X,X) \simeq B$ and $\operatorname{Hom}_B(X,X) \simeq A$ in the homotopy category of complexes of $B$-modules (resp. $A$-modules), and all terms of $X$ are projective as left and right modules (this is called a split-endomorphism two sided tilting complex)
$2)$ All the terms of $X$, considered as modules of the group algebra of $G \times H$, are relatively projective with respect to the diagonal embedding of $P$ and are $p$-permutation modules (direct summands of permutation modules)

Now, reading both papers, I haven't really been able to understand the requirement that these modules have to be $p$-permutation modules, meaning I don't understand what we lose if we have a split-endomorphism two sided tilting complex made of $\operatorname{diag}(P)$-relatively projective modules that are not $p$-permutation modules.

I am kind of new to this theory, so I'm probably missing something huge. Thanks to anyone who will help me.

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  • $\begingroup$ Permutation (and $p$-permutation) modules encode combinatorial information. Perhaps that's why... $\endgroup$ – Dima Pasechnik Feb 13 '16 at 11:22
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The motivation for the definition was an attempt to explain structurally the phenomenon of an "isotypy". This makes sense for arbitrary blocks, but let's stick to principal blocks for simplicity.

Suppose $G$ is a finite group with abelian Sylow $p$-subgroup $P$, and $H=N_G(P)$ is the normalizer of $P$. Then Broué's Abelian Defect Group Conjecture predicts that the derived categories of the principal blocks of $kG$ and $kH$ (where $k$ is a sufficiently large field of characteristic $p$) should be equivalent. For any subgroup $Q\leq P$, the principal blocks of $kC_G(Q)$ and $kC_H(Q)$ should also have equivalent derived categories, and it seems reasonable to hope for some kind of compatibility between these equivalences for varying $Q$. This is what, at the level of character theory, Broué's notion of "isotypy" gave.

If you have a splendid equivalence between the principal blocks of $kG$ and $kH$, then you can obtain one between each pair $kC_G(Q)$ and $kC_H(Q)$ by applying a certain functor (the "Brauer construction") to the complex $X$. The main reason that it's important that the terms of $X$ are $p$-permutation modules is that, although it's possible to define the Brauer construction for general modules, it behaves much better for $p$-permutation modules, and this is needed to prove that it gives a tilting complex for $kC_G(Q)$ and $kC_H(Q)$.

There are also a couple of secondary reasons.

One is that it's useful to be able to pass between characteristic $p$ and characteristic zero by considering representation theory over a complete discrete valuation ring $\mathcal{O}$ of characteristic zero with residue field $k$. The fact that $p$-permutation modules over $k$ lift to ($p$-permutation) modules over $\mathcal{O}G$, which is not true for modules in general, is needed to prove that a splendid equivalence lifts to characteristic zero.

The other, rather vaguer, is that if there's some geometric origin for the tilting complexes then one might expect that the complexes should have terms related to permutation modules.

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    $\begingroup$ Thank you! I admit I skipped the isotypy notion, now everything makes a lot more sense. Just a small thing: in the second paragraph you meant characteristic $p$ or am I missing something? $\endgroup$ – AnalysisStudent0414 Feb 13 '16 at 13:08
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    $\begingroup$ @AnalysisStudent0414 Yes, sorry, I meant characteristic $p$. I've fixed it. $\endgroup$ – Jeremy Rickard Feb 14 '16 at 0:13

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