Suppose we are given two smooth projective varieties $X$ and $Y$. We consider the derived categories $D^b(X)$ and $D^b(Y)$ of coherent shaeves. Suppose furthermore we are given semiorthogonal decompositions $D^b(X)=<A,B>$ and $D^b(Y)=<C,D>$ and that we have equivalences of triangulated categories: $A\simeq D$ and $B\simeq C$. Is there a way to construct an exact additive functor $F:D^b(X)\rightarrow D^b(Y)$ from this?

$\begingroup$ Do you want the functor to be compatible with the equivalences in some way? If not then you can just take the zero functor. $\endgroup$– SashaDec 8, 2013 at 15:34

$\begingroup$ Ok. Zero functor allways works... but maybe compatible with the triangulated equivalences from above and nontrivial. I really have no idea how to construct any nontrivial functor from these data. $\endgroup$– AleksaDec 8, 2013 at 15:39
1 Answer
You can compose the projection $D^b(X) \to A$ with the equivalence $A \cong D$ and the embedding $D \to D^b(Y)$. This will be compatible with one of the equivalences. Or you can consider an analogous composition $D^b(X) \to B \to C \to D^b(Y)$. Moreover, you can take the direct sum of these two functors. This will be compatible with both equivalences. But if you want to get more interesting functor you will need some compatibility between the gluing data of the decompositions.
EDIT. Let $X = Y = P^2$ and consider the semiorthogonal decomposition $$ D^b(P^2) = \langle O,O(1),O(2) \rangle. $$ Let us construct a FMfunctor interchanging the $O(1)$ and $O(2)$ components. First, the projection to $O(2)$ is given by $RHom(O(2),)$. The FMfunctor which projects to $O(2)$ and then embbeds into $O(1)$ is thus given by the object $$ O(2)\boxtimes O(1) $$ on $P^2\times P^2$. Further, the projection to $O(1)$ by $RHom(T(1)[1],)$, so the FMfunctor which projects to $O(1)$ and then embbeds into $O(2)$ is given by the object $$ \Omega(1)\boxtimes O(2)[1]. $$ So, in the end you can take the direct sum of these two objects as a kernel. This will give you a FMfunctor compatible with your equivalences.

$\begingroup$ Maybe the question is elementary but how can a glueing compatibility be involved in constructing a functor? Under what kind of assumptions on the glueing the resulting functor is a FourierMukai transform? $\endgroup$– AleksaDec 8, 2013 at 19:19

$\begingroup$ If the gluing data is the same for both $D^b(X)$ and $D^b(Y)$ then (under appropriate technical assumptions) one can construct an equivalence of $D^b(X)$ and $D^b(Y)$. BTW, in this case it is automatically a FMfunctor. $\endgroup$– SashaDec 8, 2013 at 20:02

$\begingroup$ Ok. Take for example $X=Y=\mathbb{P}^2$. We have semiorthogonal decomposition $D^b(X)=<\mathcal{O}_X,\mathcal{O}_X(1),\mathcal{O}_X(2)>$ and for $Y$ the same. And we consider equivalences $<\mathcal{O}_X(1)>\simeq <\mathcal{O}_Y(2)>$ and $<\mathcal{O}_X(2)>\simeq <\mathcal{O}_Y(1)>$. Can one may construct a functor compatible with the equivalences that may is a FMfunctor? Under some assumptions this would give a birational map $X\rightarrow Y$ $\endgroup$– AleksaDec 9, 2013 at 2:32

$\begingroup$ Where I can find something on that topic? I am searching for literature on how one can construct functors from glueing data as you pointed out above. $\endgroup$– AleksaDec 9, 2013 at 2:39