3
$\begingroup$

Let $X$ be a smooth projective variety over the complex numbers, and let $\text{Coh}(X)$ be the category of coherent sheaves on $X$. Consider the dg-category $\text{Perf}(X)$ of perfect complexes on $X$, which is a full dg-subcategory of the dg-category of complexes of objects in $\text{Coh}(X)$. Let $\text{H}_{\text{qe}}$ be the homotopy category of $\text{Perf}(X)$.

Let $Y$ be another smooth projective variety over the complex numbers, and similarly define $\text{Coh}(Y)$, $\text{Perf}(Y)$, and its homotopy category $\text{H}_{\text{qe}}(Y)$.

Suppose we have a fully faithful dg-functor $F: \text{Perf}(X) \to \text{Perf}(Y)$ that induces an equivalence of triangulated categories $\text{H}_{\text{qe}}(X) \to \text{H}_{\text{qe}}(Y)$.

Does there exist an algebraic morphism $f: Y \to X$, such that the derived pushforward functor $Rf_*: \text{Perf}(X) \to \text{Perf}(Y)$ is isomorphic to $F$ in the homotopy category of dg-functors between $\text{Perf}(X)$ and $\text{Perf}(Y)$? Furthermore, if such an $f$ exists, is it unique up to isomorphism?

Any relevant papers would be appreciated.

$\endgroup$
2
  • 2
    $\begingroup$ Wouldn't you expect it to be induced by some cohomological correspondence instead of an actual morphism? Or is there reason to believe the situation is different than for the full derived categories? $\endgroup$ Commented Oct 6 at 16:04
  • 1
    $\begingroup$ @R.vanDobbendeBruyn: It's possible that the dg-enhancement of the derived category provides more rigidity, which might allow for the existence of a morphism where only a cohomological correspondence exists in the case of the derived category alone. $\endgroup$
    – Nhan Le
    Commented Oct 6 at 16:13

1 Answer 1

4
$\begingroup$

The answer to your first question is no. Indeed let $X$ and $Y$ be smooth projective varieties and let $F: \operatorname{Perf}_{dg}X \to \operatorname{Perf}_{dg}Y$ be a dg-functor inducing an equivalence on the homotopy category (=the triangulated category of perfect complexes). Then the induced functor (which I will also call $F$) is of Fourier-Mukai type. There is a priori no reason for this functor to be induced from an actual isomorphism $f:X \to Y$. For an example, there exist K3 surfaces $X$ and $Y$ which are derived equivalent but not isomorphic ($Y$ is a fine moduli space of sheaves on $X$) see the book of Huybrechts on K3 surfaces for a proof and further references.

If we make the further assumption that the functor $F$ sends the skyscraper sheaves of closed points on $X$ to shifts of the skyscraper sheaves of closed points on $Y$, then in fact $F$ is isomorphic to $Rf_*(- \otimes L)[k]$ where $L$ is a line bundle and $f:X \to Y$ is an isomorphism of varieties. In general both $L$ and $[k]$ are nontrivial so this functor is not isomorphic to $Rf_*$, however I am not sure if the isomorphism $f$ appearing above is unique.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .