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In this Mathoverflow question, Examples of Eigensheaves outside of langlands, David Ben-Zvi says

" Given a G -space X you can recover quasicoherent sheaves on X from sheaves on X/G (ie equivariant sheaves) as eigenobjects for the natural action of Rep(G)= QC(BG) on QC(X/G)."

Here QC seems to denote say the dg enhancement of the derived category of quasicoherent sheaves. I understand that there is a natural map from

$X/G \to BG$

and the monoidal action is by pulling back along this map. What I don't see is how to formulate the notion of eigensheaf in this setting. Can someone give some more detail on this statement, e.g. what is the definition of eigensheaf in this statement, what is the precise way to recover the derived category of quasi-coherent sheaves on $X$, etc.? I get the feeling that at least in spirit this is some really concrete thing, so a down-to-earth explanation would be much preferable then a reference to the Geometric Laglands literature.

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Given a group $G$ acting on a category $QC(X)$, you get a sheaf of quasicoherent categories over $BG$, whose fiber at $pt\rightarrow BG$ is $QC(X)$. Global sections of this sheaf are exactly invariants $QC(X/G)$ with an action of "global functions" $QC(BG)$. See http://arxiv.org/abs/1306.4304 for a reference on quasicoherent sheaves of categories.

For nice groups $G$ (affine algebraic groups of finite type), $BG$ is 1-affine. In other words, the functor of taking global sections is an equivalence between sheaves of categories over $BG$ and $QC(BG)$-modules.

So, to recover $QC(X)$, you have to localize $QC(X/G)$ over $BG$ and take the fiber over $pt\rightarrow BG$. This is simply $$QC(X) = Vect\otimes_{QC(BG)}QC(X/G),$$ where $QC(BG)\rightarrow Vect$ is the pullback to the basepoint $pt\rightarrow BG$. It acquires an action of $Vect\otimes_{QC(BG)} Vect\cong QC(\Omega BG)\cong QC(G)$. Of course, one can just write $X = pt\times_{BG} X/G$ and use the fact that all stacks are perfect, so $QC(pt\times_{BG} X/G) \cong QC(pt) \otimes_{QC(BG)} QC(X/G)$.

$QC(BG)$ is rigid (since $BG$ is a perfect stack), so instead of looking at an action of $QC(BG)$, by adjunction one can look at a coaction of $QC(BG)^\vee\cong QC(BG)$ and take the "cotensor" product. This is just a totalization of the cobar resolution $QC(X/G)\rightrightarrows QC(BG)\otimes QC(X/G)\rightrightarrows...$

To conclude, an object in $QC(X)$ is an object $\mathcal{F}\in QC(X/G)$ with an identification $\mathcal{O}_{triv}\boxtimes\mathcal{F} \cong (f\times id)_* \mathcal{F}$ and higher homotopies expressing associativity. Here $\mathcal{O}_{triv}$ is the skyscraper sheaf at the trivial bundle (under the identification $QC(BG)\cong\mathcal{O}(G)-comod$ it is $\mathcal{O}(G)$ as a comodule over itself) and $f\times id:X\rightarrow BG\times X$. In other words, $\mathcal{F}$ is an eigensheaf with eigenvalue $\mathcal{O}_{triv}$.

Here is one way to think of the "cobar" presentation. Quasicoherent sheaves on $X$ are the same as sheaves on $X/G$ with an action of $p_*\mathcal{O}_X$ for $p:X\rightarrow X/G$. By base change, $p_*\mathcal{O}_X\cong f^*\mathcal{O}_{triv}$. Finally, $f^*\mathcal{O}_{triv}\otimes\mathcal{F}\rightarrow\mathcal{F}$ is equivalent to a map $\mathcal{O}_{triv}\boxtimes\mathcal{F}\rightarrow (f\times id)_*\mathcal{F}$ on $BG\times X$. Its restriction to $pt\times X\rightarrow BG\times X$ is an isomorphism, hence the map itself is an isomorphism (the forgetful functor $Rep(G)\rightarrow Vect$ reflects isomorphisms).

ADDED (in response to a question in the comments): An $H$-equivariant sheaf (or a $D$-module, or a local system) on $G/N$ is a sheaf $\mathcal{F}$ on $G/N$ together with an isomorphism $p_1^*\mathcal{F}\cong a^*\mathcal{F}$ for $a: G/N\times H\rightarrow G/N$ and $p_1$ the projection to the first factor. One can also write this as $a^*\mathcal{F}\cong \mathcal{F}\boxtimes \mathcal{O}_H$, i.e. $\mathcal{F}$ is an eigensheaf with eigenvalue $\mathcal{O}_H$. Similarly, given any multiplicative sheaf $\Lambda$ on $H$, you can consider $\Lambda$-twisted $H$-equivariant sheaves on $G/N$: these are sheaves $\mathcal{F}$ together with an isomorphism $a^*\mathcal{F}\cong \mathcal{F}\boxtimes \Lambda$. (A multiplicative sheaf is a sheaf $\Lambda$ on $H$ together with an isomorphism $m^*\Lambda\cong \Lambda\boxtimes \Lambda$ for $m:H\times H\rightarrow H$ the multiplication map.) Multiplicativity is needed to make sense of the associativity conditions on the action. Remark: just as above, you can write this as a cotensor product: $QC(G/N)\otimes^{QC(H)} Vect$ for $Vect\rightarrow QC(H)$ the inclusion of $\Lambda$.

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  • $\begingroup$ So just to be clear the term eigensheaf in this context means we have another action of the monoidal category QC(BG) on Vect and it is the fact that QC(X) = $\endgroup$
    – user36931
    Aug 8 '13 at 3:19
  • $\begingroup$ sorry $ QC(X)= Vect \otimes_{QC(BG)} QC(X/G) $ ? $\endgroup$
    – user36931
    Aug 8 '13 at 3:20
  • $\begingroup$ I suppose what really peaked my fancy and why I'm trying to understand the meaning of eigensheaf is the following follow up in the above MO answer---"Another kind of example is given by monodromic sheaves --- namely, you might ask to weaken the condition of equivariance of a sheaf under a group action by twisting, so rather than getting an invariant (aka equivariant) object (eigenobject with trivial eigenvalue) you get a nontrivial eigenvalue." $\endgroup$
    – user36931
    Aug 8 '13 at 3:47
  • $\begingroup$ It seems to imply that we can consider dg-categories corresponding to a non-trivial character of a group as well, which is how I interpreted the term "eigensheaves." Though I don't know how to formulate this. $\endgroup$
    – user36931
    Aug 8 '13 at 3:49
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    $\begingroup$ Just to add to Pavel's excellent response: the result in question (recovering $QC(X)$ from $QC(X/G)$) follows as Pavel said from the identification of categories of sheaves on fiber products with tensor product categories (and can in fact be checked very concretely). Thus it is much easier than the "1-affineness" of BG, which a recent and deep theorem of Gaitsgory, though that theorem places the result in a very useful context. $\endgroup$ Aug 9 '13 at 20:50

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