Let $X$ be a smooth variety and $M$ a fine moduli space of certain kind of sheaves on $X$. Let $\mathcal{E}$ be the universal family on $X\times M$. Suppose there is a derived functor $F$ from $D^b(X)$ to some derived category, say $D^b(Y)$, or $D^b(A-mod)$ for some finite algebra $A$, so that $F(\mathcal{E}_t)$ is a semistable object of some kind (say semistable sheaves on $Y$ or semistable rep of $A$). This provides a map $f:M \to N$ where $N$ is the moduli of semistable objects on the target category.

My question: is this map always algebraic or is there any criterion for this map to be algebraic?

Thank you very much for your help.


If $F$ is a Fourier-Mukai functor, it can be applied not just fiberwise, but to the whole family. Namely, if $K \in D^b(X\times Y)$ is the Fourier-Mukai kernel, then by pullback it gives an object on $$ X \times Y \times M = (X \times M) \times_M (Y \times M) \subset (X \times M) \times (Y \times M). $$ The corresponding Fourier-Mukai functor $D^b(X \times M) \to D^b(Y \times M)$ applied to $\mathcal{E}$, gives an object $\mathcal{F} \in D^b(Y\times M)$ with fibers $F(\mathcal{E}_t)$, which are by assumption stable sheaves. THerefore, the object $\mathcal{F}$ gives an algebraic map $M \to N$.

  • $\begingroup$ Thank you for your answer. One follow up question: if the functor is a fully faithfull embedding of D^b(A-mod) into D^b(X) for a finite algebra $A$ and smooth $X$, is there a notion of Fourier- Mukai functor for this situation? $\endgroup$ Oct 23 '18 at 13:12
  • 1
    $\begingroup$ Yes, in this case you replace $D^b(X \times Y)$ by $D^b(X,A \otimes O_X)$, the derived category of sheaves of $A \otimes O_X$-modules on $X$. $\endgroup$
    – Sasha
    Oct 23 '18 at 14:13
  • $\begingroup$ And I suppose as in the coherent case, any fully faithful embedding of $D^b(A)$ into $D^b(X)$ would be FM in the above sense right? Thanks again! $\endgroup$ Oct 23 '18 at 14:36
  • $\begingroup$ I am not sure that this has been proved. $\endgroup$
    – Sasha
    Oct 23 '18 at 15:08

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.