Let $A$ be an abelian variety and $\hat A$ be the dual abelian variety. If $P$ is the (normalized) Poincare line bundle, then Mukai defines $R\hat S:D(A)\to D (\hat A)$ via $R\hat S(?)=Rp_{\hat A,*}(Lp_A^*(?)\otimes P)$ and $RS:D(\hat A)\to D(A)$ via $R S(?)=Rp_{ A,*}(Lp_{\hat A}^*(?)\otimes P)$. He then shows that $R\hat S\circ R\hat S =(-1_A)^*[-g]$ on $D^b_{qcoh}(A)$ and $R\hat S\circ R\hat S =(-1_{\hat A})^*[-g]$ on $D^b_{qcoh}(\hat A)$. The proof uses the projection formula which can fail for non-quasi coherent sheaves and hence for $D(A)$. My question is: Is there a counter example to $R\hat S\circ R\hat S =(-1_A)^*[-g]$ in $D(A)$?

  • 2
    $\begingroup$ What do you denote by $D(A)$? $\endgroup$
    – Sasha
    Jun 28 '16 at 8:01
  • $\begingroup$ The derived category of $\mathcal O _A$ modules. I am not so worried about bounded vs unbounded, but I am more worried / interested about the non-quasi coherent case. $\endgroup$
    – Hacon
    Jun 28 '16 at 11:50

I think it is clear that $R\hat{S}$ is not an equivalence on the categories of all $O$-modules. Indeed, if it were an equivalence, it would send product to product, and also preserve quasi-coherence. Let $M_x$ be a sky-scraper sheaf at $x\in A$; its image under $R\hat{S}$ is an invertible $O$-module on $A^\vee$ (up to a shift). However, an infinite product of copies of $M_x$ is quasi-coherent, while the infinite product of invertible $O$-modules is not quasi-coherent.


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