Let $S$ be an abelian or K3 surface, $H$ an ample class on $S$, $v\in H^{ev}(S,\mathbb{Z})$ and $M$ the moduli space of $H$-stable sheaves on $S$ with invariants fixed by $v$. Let $\mathcal{E}$ be a universal sheaf on $M\times S$. I indicate with $\pi_{ij}$ the projection from $M\times S\times M$ to the $i^{th}$ and $j^{th}$ factors. My goal is to show that the relative Ext sheaf $$ \mathcal{E}xt^2_{\pi_{13}}(\pi^*_{12}\mathcal{E},\pi^*_{23}\mathcal{E})$$ is a line bundle supported on the diagonal $\Delta\subset M\times M$. I had solved this problem in the following way. First of all let's recall the (one of the corollary) Base Change Theorem as stated in Mumford, Abelian Varieties:

let $f:X\to Y$ be a proper morphism of noetherian schemes and $\mathcal{F}$ a coherent sheaf on $X$. If $\dim H^p(X_y,\mathcal{F}_y)$ is constant for every $y\in Y$, then $R^pf_*\mathcal{F}$ is locally free and $R^pf_*\mathcal{F}\otimes_{\mathcal{O}_Y}k(y)=H^p(X_y,\mathcal{F}_y)$, where $k(y)$ is the field at $y$.

Let's apply this with $f=\pi_{13}$ and $\mathcal{F}=\mathcal{H}om(\pi^*_{12}\mathcal{E},\pi^*_{23}\mathcal{E})$. Since $M$ parametrizes stable sheaves on $S$, the latter sheaves is supported on the diagonal. On the other hand, since both $M$ and $S$ are symplectic, Serre duality assures us that also $\mathcal{E}xt^2_{\pi_{13}}(\pi^*_{12}\mathcal{E},\pi^*_{23}\mathcal{E})$ is supported on the diagonal. If $(m,m)=([F],[F])\in\Delta$, where $[F]$ is the class of a stable sheaf on $S$, then $$\dim H^2(X_{(m,m)},\mathcal{F}_{(m,m)})=\dim H^2(S,\mathcal{H}om_{\mathcal{O}_S}(F,F))=$$$$=\dim Ext^2(F,F)=\dim End(F)^{\vee}=1$$ where we used Serre duality and the fact that a stable sheaf is simple. Hence the BCT implies that $R^2\pi_{13,*}\mathcal{H}om(\pi^*_{12}\mathcal{E},\pi^*_{23}\mathcal{E})=\mathcal{E}xt^2_{\pi_{13}}(\pi^*_{12}\mathcal{E},\pi^*_{23}\mathcal{E})$ is locally free of rank $1$.

Anyway, I have recently looked at the Mukai paper, 'On the moduli space of bundle on K3 surfaces I', Proposition 4.10, where there is Mukai's proof of this statement. It turns out that his proof is quite different: he uses a fact about the diagonal, e.g. the ideal of sheaves $\mathcal{I}_{\Delta}$ annihilates $\mathcal{E}xt^2_{\pi_{13}}(\pi^*_{12}\mathcal{E},\pi^*_{23}\mathcal{E})$. Of course I trust Mukai more than myself. Nevertheless, I cannot see why and where I need that remark in my argument.

Can anybody point out some errors in my argument and suggest how to fix them?

Thank you very much.

EDIT: as Bernie noted, I cannot use the classical Base Change Theorem in this case and I need the ext-formulation (see the referement in the comments under Bernie's answer). Up to this modification, the conclusion looks 'formally' right. Anyway, why does Mukai need that $\mathcal{I}_{\Delta}$ annihilates $\mathcal{E}xt^2_{\pi_{13}}(\pi^*_{12}\mathcal{E},\pi^*_{23}\mathcal{E})$?


1 Answer 1


I cannot comment due to reputation and haven't done the computations, but I see two possible problems:

  • usually one has a spectral sequence $(R^if_{*})(\mathcal{E}xt^j(M,N)) => \mathcal{E}xt_f^{i+j}(M,N)$ because $f_{*}\mathcal{H}om(M,-)$ is a composition of functors.

  • $Ext^2(F,F)=H^2(\mathcal{H}om(F,F))$ only holds if $F$ is locally free, due to local to global spectral sequence for $Ext$.

What may be helpful: there is a base change theorem for relative Ext-sheaves, ma be this can help you

  • $\begingroup$ thank you for your comments. With your first remark, do you mean that I cannot write $R^2\pi_{13,*}\mathcal{H}om(\mathcal{E},\mathcal{E})=\mathcal{E}xt^2_{\pi_{13}}(\mathcal{E},\mathcal{E})$? This should be because the spectral sequence doesn't converge in general at page $2$..? $\endgroup$
    – User3773
    May 17, 2015 at 14:23
  • $\begingroup$ For what concern the second remark, maybe I can avoid to use it. In fact, I can pass from $H^2(\mathcal{H}om(F,F))$ to $H^0(\mathcal{H}om(F,F))^{\vee}$ by duality and get the same information. What do you think about? $\endgroup$
    – User3773
    May 17, 2015 at 14:25
  • $\begingroup$ Yes, that is what i meant with regards to the spectral sequence, we cannot write the relative Ext in general in that form. For the second question: i don't think $\mathcal{H}om(F,F)$ is selfdual if $F$ is not locally free, that is Serre duality does not work. We only have $Ext^2(F,F)\cong Hom(F,F)^{\vee}$. I recommend to look in the article "Universal family of extensions" of H.Lange. There is a base change formula which may help you. $\endgroup$
    – Bernie
    May 17, 2015 at 14:50
  • $\begingroup$ Thanks, I'm going to read at that paper and let you know! $\endgroup$
    – User3773
    May 17, 2015 at 15:20
  • $\begingroup$ Ok, I have checked. Basically, as I could image, the Base Change for ext follows from the classical one up to a locally free resolution. The remarks you have pointed out are solved if I use this 'new' formulation. Thank you, I'll edit the question to point out this. $\endgroup$
    – User3773
    May 18, 2015 at 11:22

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