Skip to main content
10 events
when toggle format what by license comment
May 18, 2015 at 16:19 comment added User3773 Yes I forgot HuybrechtsLehn book! In fact their use of spectral sequence is just a rigorous way to use Base Change avoiding our digression on ext-Base Change. I have really appreciated your comments on the ideal $\mathcal{I}_{\Delta}$: my argument works when I only look at $\Delta$ and the restriction of my sheaf to it. Since I want to see this as a sheaf on the whole $M\times M$, I need to globalize and this is the way. Thank you very much for your suggestions, my view is pretty clearer now.
May 18, 2015 at 15:00 comment added Bernie I don't know if the two approaches are basically the same, maybe this is the case. But Mukai needs the fact that $\mathcal{I}_{\Delta}$ annihilates the sheaf, because then the relative Ext is the pushforward of a sheaf on the diagonal to the whole of $M\times M$. This sheaf is the line bundle Mukai talks about. Meanwhile I found another proof of this fact in Huybrechts and Lehn, it is in the chapter about low dimensional examples of moduli spaces on K3-surfaces, it explains what happens a little bit more.
May 18, 2015 at 11:24 comment added User3773 Anyway, I still cannot see why Mukai uses a different approach. Or if the one used by him is the same one under some remarks...
May 18, 2015 at 11:22 comment added User3773 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.
May 17, 2015 at 15:20 comment added User3773 Thanks, I'm going to read at that paper and let you know!
May 17, 2015 at 14:50 comment added Bernie 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.
May 17, 2015 at 14:25 comment added User3773 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?
May 17, 2015 at 14:23 comment added User3773 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 $2$..?
May 17, 2015 at 13:50 history edited Bernie CC BY-SA 3.0
added 105 characters in body
May 17, 2015 at 12:03 history answered Bernie CC BY-SA 3.0