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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.

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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 $M \to N$.

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  • $\begingroup$ 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, 2018 at 13:12
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    $\begingroup$ 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, 2018 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, 2018 at 14:36
  • $\begingroup$ I am not sure that this has been proved. $\endgroup$
    – Sasha
    Oct 23, 2018 at 15:08

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