1
$\begingroup$

Consider a moduli scheme $P$ of sheaves $F$ on a Calabi-Yau n-fold $X$ with a section $s\in H^0(F)$. Alternatively, such objects can be described as maps $\mathcal{O}_X \xrightarrow{s} F$ called pairs. Take the universal pair $\mathcal{O}_{X\times P}\xrightarrow{\mathfrak{s}}\mathcal{F}$ on $X\times P$, and construct its push forward $$ H^\bullet(\mathcal{O}_X)\otimes \mathcal{O}_P\xrightarrow{R\pi_*(\mathfrak{s})}R\pi_*(\mathcal{F}) $$ along the projection $\pi: X\times P\to P$. There is then a composition $\kappa$ of morphisms $$ \mathcal{O}_P[-n]\cong H^n(\mathcal{O}_X)[-n]\otimes \mathcal{O}_P\to H^\bullet(\mathcal{O}_X)\otimes \mathcal{O}_P\xrightarrow{R\pi_*(\mathfrak{s})}R\pi_*(\mathcal{F})\,, $$ where the first map is induced by the inclusion of the fourth cohomology group of $\mathcal{O}_X$.

Assuming that the family $\mathcal{F}$ satisfies $R^i\pi_*(\mathcal{F}) = 0$ for $i>0$ (i.e., $H^i(F)=0$ for $i>0$ for all such pairs $\mathcal{O}_X\to F$), is it true that $\kappa =0$ in the derived category of sheaves on $P$?

I tried taking all sorts of $\pi_*$-acyclic resolutions of $\mathcal{O}_{X\times P}$ and $\mathcal{F}$ and finding an actual morphism of complexes between them, but to no avail. At the same time, I did not see why this statement would be false.

$\endgroup$

1 Answer 1

4
$\begingroup$

This is not true in general. Consider, for instance, the moduli space of skyscraper sheaves on an elliptic curve $X$. Thus, $P = X$, $\mathcal{F} = \Delta_*\mathcal{O}_X$ is the structure sheaf of the diagonal, and $$ s \colon \mathcal{O}_{X \times X} \to \Delta_*\mathcal{O}_X $$ is the natural morphism. Note that the induced morphism $$ H^1(X, \mathcal{O}_X) \oplus H^1(X, \mathcal{O}_X) = H^1(X \times X, \mathcal{O}_{X \times X}) \stackrel{s}\to H^1(X \times X, \Delta_*\mathcal{O}_X) = H^1(X, \mathcal{O}_X) $$ is the summation; in particular, its restriction to each summand in the left-hand side is the identity. It remains to note that your composition morphism $$ \mathcal{O}_X[-1] \to R\pi_*\mathcal{O}_{X \times X} \stackrel{R\pi_*s}\to R\pi_*\mathcal{F} $$ corresponds by adjucntion to the composition $$ \mathcal{O}_{X \times X} \stackrel{\pi^*\epsilon}\to \mathcal{O}_{X \times X} \stackrel{s}\to \Delta_*\mathcal{O}_X, $$ where $\epsilon$ is the generator of $H^1(X, \mathcal{O}_X)$.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you! Perhaps let me comment that this argument works very strict Calabi-Yau n-fold. Also, you don't need to use the adjunction at the end. The statement follows from unwrapping your $H^1(X, \mathcal{O}_X) \oplus H^1(X, \mathcal{O}_X) = H^1(X \times X, \mathcal{O}_{X \times X}) \stackrel{s}\to H^1(X \times X, \Delta_*\mathcal{O}_X) = H^1(X, \mathcal{O}_X)$. $\endgroup$
    – Arkadij
    Commented Sep 10 at 7:31

You must log in to answer this question.

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