3

Let $X$ be a smooth, complex projective algebraic surface. Let $C,D$ be two nonzero effective divisors on it. Then in literature one can find the following exact sequence : $0 \to \mathcal O_D(-C) \to \mathcal O_{C+D} \to \mathcal O_C \to 0$.

I'm a bit confused regarding whether it's a short exact sequence on $X$ as they are line bundles supported on different curves.

Here my question is : in the above sequence are we taking the pushforward of these line bundles (under the inclusion map) to $X$?i.e. how one appropriately interprets this sequence?

Moreover, can we twist the above sequence by line bundles on $X$ as follows : for example can we tensor the above sequence by $\mathcal O_X(C)$ to obtain : $0 \to \mathcal O_D \to \mathcal O_{C+D}(C) \to \mathcal O_C(C) \to 0$.?

Any clarification from anyone is appreciated

New
  • 347

1 Answers1

4

The original sequence can be considered either as an exact sequence on the (reducible) curve $C \cup D$, or (via pushforward) as an exact sequence on the surface $X$.

Of course, you can twist this sequence by any line bundle. If you choose to use the line bundle $\mathcal{O}_X(C)$ (or its restriction to $C \cup D$), you will get an exact sequence with $\mathcal{O}_C(C)$ on the right and $\mathcal{O}_D$ on the left, but I don't think it is a good idea to use your notation for the sheaf in the middle (because $C$ is not a Cartier divisor on $C \cup D$), it is better to denote it, say, by $\mathcal{O}_X(C)\vert_{C \cup D}$.

Sasha
  • 17,011
  • thank you very much for your answer. This notation is borrowed from Friedmans book on Algebraic surface. I have one further question: if we somehow know that $E.( C+D) <0$, for some other divisor $E$, then can we conclude that $ h^0( \mathcal O_{ C+D}( E))=0$?...I mean the standard operations regarding cohomology of canonical sequence of curves on surfaces: can they be carried out here also? – New Mar 21 '22 at 19:55
  • 1
    Not necessarily. For instance, if $\mathcal{O}D(E-C)$ has a global section then so does $\mathcal{O}_X(E)\vert{C \cup D}$, but if at the same time $E \cdot C$ is very negative, then $E \cdot (C + D)$ is negative as well. – Sasha Mar 21 '22 at 20:32
  • suppose we know that $D.( E-C) <0$ and also $ E.C<0$, then can we say that the middle term $h^0(\mathcal O_{ C+D}( E))= 0$? Or is this not necessarily the case ( perhaps by one of your previous arguments?) – New Mar 21 '22 at 20:39
  • 1
    I guess you assume $C$ and $D$ to be both reduced and irreducible. In that case both sheaves $\mathcal{O}D(E-C)$ and $\mathcal{O}_C(E)$ have no global sections, and hence the long exact sequence of cohomology implies that the sheaf $\mathcal{O}_X(E)\vert{C \cup D}$ (that stands between them) also has no golbal sections. – Sasha Mar 21 '22 at 20:53
  • in Friedmans book this sequence is introduced in the setting where $C,D$ need not be necessarily reduced and irreducible. You mean this global section vanishing phenomenon can take place only when both of them are reduced and irreducible? – New Mar 21 '22 at 20:56
  • 1
    Precisely (e.g., if $C = C_1 \cup C_2$ and $E - C_2$ has global section on $C_1$ while $E \cdot C_2$ is very negative). – Sasha Mar 21 '22 at 21:07