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Apr 8, 2014 at 14:35 comment added user39380 Ah..you are right, I am sorry..Thanks for your patience!
Apr 8, 2014 at 14:33 comment added abx I don't understand the notation $O_{-F}$. Aren't you confusing $\mathcal{O}_X(F)$, which is an invertible line bundle on $X$, with $\mathcal{O}_F$, the structure sheaf of $F$, supported on $F\subset X$?
Apr 8, 2014 at 14:24 comment added user39380 Then $O_{-F}$ and $O_{-2E}$ are not isomorphic? Since they are the dual line bundle of $O_{F}$ and $O_{2E}$..$O_F\otimes_{O_X} O_{-F}=O_X$...?
Apr 8, 2014 at 14:10 comment added abx $\mathcal{O}_F$ and $\mathcal{O}_{2E}$ are certainly not isomorphic, they have different support (namely $F$ and $E$).
Apr 8, 2014 at 14:06 comment added user39380 Since we used $H^1(O_{-F})=H^1(O_{-2E})$ in the prove above, I was thinking there is no difference if we apply it to the 0th cohomology group. Maybe I had faute with understanding something?
Apr 8, 2014 at 14:02 comment added user39380 Since we have $F$ is linear equivalent to $2E$, $O_F$ and $O_{2E}$ define the same invertible sheaf, can we argue $H^0(O_{2E})=k$ directly by this?
Apr 8, 2014 at 12:53 comment added abx Yes and yes ...
Apr 8, 2014 at 12:31 comment added user39380 Is that because $H^0(O_X)=H^0(O_F)=k$? Also is pencil here just mean the existence of an elliptic fibration?
Apr 8, 2014 at 12:18 comment added abx 1) Every Enriques surface has such a pencil, look at any book on surfaces (e.g. Barth etc.). 2) The linear system $|2E|$ is a pencil of elliptic curves; take a smooth $F\in |2E|$. Since $H^1(X,\mathcal{O}_X)=0$, the exact sequence $0\rightarrow \mathcal{O}_X(-F)\rightarrow \mathcal{O}_X\rightarrow \mathcal{O}_F\rightarrow 0$ gives $H^1(X,\mathcal{O}_X(-F))=0$.
Apr 8, 2014 at 12:13 comment added user39380 Thanks for your answer! But how do you get the example? Also I find it hard to check the $H^1(X,O_X(-2E))=0$, is there an easy way to show that?
Apr 2, 2014 at 10:04 vote accept CommunityBot
Apr 2, 2014 at 6:17 history answered abx CC BY-SA 3.0