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Good Morning,

These questions are a result of trying to understand the proof of Proposition III.18 in Beauville's book 'Complex Algebraic Surfaces'.

Here is the setup - everything is smooth, projective, algebraic, complex.

For a curve $C$, let $E \to C$ be a rank 2 vector bundle, and let $S=\mathbb{P}_C(E)$ be the associated geometrically ruled surface; label the surjection $S \to C$ as $p$. Let $\mathcal{O}_S(1)$ be the tautological bundle of $S$.

Let $F (\cong \mathbb{P}^1)$ be a fibre of $p$. The first equality I don't understand is $F.\mathcal{O}_S(1)=1$. The 'rough' justification I have for this is that the tautological bundle corresponds to a section of $S\to C$, and hence meets $F$ exactly once with multiplicity 1, but this doesn't totally make sense to me. Why exactly does the tautological bundle correspond to a section? (Or if this is nonsense, how else may I see the equality?).

Secondly, there is the claim that $\mathcal{O}_S(1).p^* det(E)=deg(E)$. Could someone please explain why this is true?

I suspect that both of my problems are related to not understanding what $\mathcal{O}_S(1)$ is, so here is the construction of it that I'm using. Consider the pullback bundle $p^* E \to S$. Define a sub-line bundle $N$ of $p^* E$ as follows. A point $s \in S$ is actually equal to a one dimensional subspace of $E_{p(s)}$. Choose $N_s$ to be this one dimensional subspace of $(p^*E)_s = E_{p(s)}$. Then define $\mathcal{O}_S(1)$ by

$ 0 \to N \to p^* E \to \mathcal{O}_S(1) \to 0 $.

This question is a bit vague, but could someone motivate this definition for me, or try to explain what is going on here? Is this the same as the pullback of a hyperplane in some projective embedding?

Thank you,


edit: not totally sure why the latex code isn't working at the end, apologies.

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I'm purposely going to give a hint rather than a solution: The sheaf $\mathcal{O}_{P(E)}(1)$ restricts to $\mathcal{O}(1)$ on the fibres by design. So it is a section in this case. So the rough justification of the first fact is correct. It also means that $p^*det(E)$ restricts to $det(E)$ on the section. Taking the degree, gives the second claim. – Donu Arapura Sep 10 '11 at 15:17
Thanks Donu. Not sure how to accept a comment as an answer. Cheers, Robert – Robert Garbary Sep 11 '11 at 16:01

1 Answer 1

I think Donu's explaination is clear.

(1) For $F.\mathcal{O}_S(1)=1$.

Denote Closed immersion $i:F\rightarrow S$. then restriction $i^*\mathcal{O}_S(1)=\mathcal{O}_F(1)$. Thus $F.\mathcal{O}_S(1)=deg(\mathcal{O}_F(1))=1$.

(2) For $\mathcal{O}_S(1).p^∗det(E)=deg(E)$

Denote $e$ a divisor correspond to $det(E)$ over curve $C$, then $e=n_1e_1+...+n_re_r, e_i$ are single point on $C$. $deg(E)=n_1+...+n_r$. Also, $\mathcal{O}_S(1).p^*det(E)=n_1(\mathcal{O}_S(1).p^*e_1) +...+n_r(\mathcal{O}_S(1).p^*e_r)=deg(E)$, since $\mathcal{O}_S(i).p^*e_i=1$.

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