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Let $X$ be a K3 surface and fix an ample line bundle on $X$. Let $v\in \widetilde{H}(X,\mathbb{Z})$ be a Mukai vector and $M(v)$ be the moduli space of semi-stable coherent sheaves on $X$ with Mukai vector $v$. Under certain conditions on $v$, we could show that $M(v)$ is a fine moduli space and parametrizes only stable sheaves and in addition dim$M(v)=2$. As a result there exists a universal sheaf $\mathcal{E}$ on $X\times M(v)$. Let $M\subset M(v)$ be a connected component. Then we have the following theorem of Mukai.

Under above assumptions, the integral transform $$ \Phi_{\mathcal{E}}:D^b(X)\to D^b(M) $$ gives an equivalence of trangulated categories, where $D^b(-)$ denotes the bounded derived category of coherent sheaves. See Huybrechts Section 10.3 for details.

The proof of above theorem in Huybrechts depends on the general criterion of equivalence of integral transforms rather than giving an inverse functor. Nevertheless, in the analogue theorem for abelian varieties, an inverse functor could be given.

My question is: for a K3 surface $X$, could we construct an inverse functor for $\Phi_{\mathcal{E}}$ and realize it as a integral transform?

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    $\begingroup$ Orlov's theorem states that any exact fully faithful functor between $D^b Coh$ of smooth projective varieties can be realised as an integral transform, at least if you allow arbitrary objects of $D^b(X × Y)$ as integral kernels. That certainly applies to your case, so yes, the inverse functor is an integral transform. An inverse functor is in particular right adjoint, and a right adjoint to an integral transform can be realised as an integral transform by the dual sheaf, considered as an object of $D^b(M × X)$ and dualized in that category. $\endgroup$
    – Anton Fetisov
    Commented May 11, 2017 at 0:43
  • $\begingroup$ @AntonFetisov I see, it's so obvious. $\endgroup$
    – Zhaoting Wei
    Commented May 11, 2017 at 1:15

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