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For a complex surface $X$ with a line bundle $L$, the base locus Bs$|L|$ consists of $0$-dimensional and $1$-dimensional components. The fixed part of $|L|$ is the $1$-dimensional locus and of Bs$|L|$ and we denote it by $F$.

  1. Why is $h^0(X,O(F))=1$?
  2. How should I think of the map $L(-F)\hookrightarrow L$, which induces an isomorphism $H^0(X,L(-F))\cong H^0(X,L)$?

As to the second question, I have trouble about how to understand the line bundle $L(-F)=L\otimes O(-F)$.

If I understand global sections of $L(-F)$ are global sections of $L$ which are zeros on $F$, the last isomorphism $H^0(X,L(-F))\cong H^0(X,L)$ is reasonable; since any global section of $L$ has zeros on $F$, they are global section of $L(-F)$.

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up vote 3 down vote accepted

Let's illustrate with an example. Say $X$ is the blow-up of $\mathbb P^2$ at a point, with exceptional divisor $E$. Think about the line bundle $\mathcal O_X(H+E)$, where $H$ is a line not through the blown-up point. I think it's easier to think about this in the language of divisors and linear systems, but you can translate it back to line bundles.

What does the linear system $|H+E|$ look like in our case? You can check that a divisor in this linear equivalence class consists of $E$ together with some line $L$ (or $2E+R$, where $R$ is the strict transform of a line through the blown-up point). In particular, $E$ is in the base locus, and the fixed part $F$ is just $E$.

What's $dim H^0(X,\mathcal O_X(F))$? There's no way it can be greater than 1 -- if it were, it would mean that $F$ moves in a linear system of dimension at least $1$. But that means that at least one component of $F$ isn't in the base locus of $F$ at all (it's not fixed), and there's no way such a component could be in the base locus of $L$. So it shouldn't have been included in $F$ in the first place.

At the level of linear series, the map $L(-F) \to L$ just says to take an element of $L(-F)$, add on $F$, and you get something in the linear series $|L|$. In our case, something $|L(-E)|$ will just be a line, and you get something in $L$ by throwing in $E$ too.

Your interpretation at the end seems correct to me. Note that there's always a map $H^0(X,L(-F)) \to H^0(X,L)$ whenever $F$ is effective, but it's only an isomorphism if $F$ is the fixed part.

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Thank you for the answer. –  M. K. Nov 10 '12 at 12:54
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