11
$\begingroup$

Let $X$ be a projective variety over $\mathbb{C}$, denote by $D^b(X)$ the bounded derived category of coherent sheaves on $X$. Suppose we have a Fourier-Mukai functor $\Phi_{X\rightarrow X}^\mathcal{P}:D^b(X)\rightarrow D^b(X)$ being an auto-equivalence on $D^b(X)$, and further assume that $\Phi_{X\rightarrow X}^\mathcal{P}$ acts identically on objects, then is it possible that $\Phi_{X\rightarrow X}^\mathcal{P}$ fails to be the identity functor? How to find a simple example to illustrate this?

$\endgroup$
9
$\begingroup$

If $X$ is smooth and projective, then any such FM functor is in fact naturally isomorphic to the identity functor. This follows immediately from Corollary 5.23 of Huybrechts' book on Fourier-Mukai tranforms. Briefly, the idea is that the hypotheses ensure that $\mathcal{P}$ is a quasi-isomorphic to a sheaf on $X\times X$ that's supported set-theoretically on the diagonal and moreover is flat over $X$ via either projection map. One then argues that $\mathcal{P}$ is of the form $\mathcal{O}_\sigma\otimes L$, where $\mathcal{O}_\sigma$ is the structure sheaf of the graph of an automorphism of $X$ and $L$ is a line bundle pulled back from $X$.

When $X$ is not smooth, I'm not completely sure what happens. If $P$ is in fact a perfect complex on $X\times X$, then this reasoning still goes through. Otherwise, one needs to worry about the difference between $D^b(X)$ and $D_{perf}(X)$, the derived category of perfect complexes on $X$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.