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Let $S$ be a projective smooth curve over an arbitrary field. Let $F$ and $E$ be locally free sheaves on $S$ of rank 1 and 2 making the following sequence exact: $$ 0\rightarrow \mathcal{O}_S\rightarrow E \rightarrow F \rightarrow 0$$
If we denote $P(E)$ and $P(F)$ the projective bundles over $S$ associated with $E$ and $F$, we know that $P(F)$ is a divisor in $P(E)$ and that $P(F)$ belongs to the linear system of $\mathcal{O}_{P(E)}(1)$. I would like to compute the intersection number $$(\mathcal{O}_{P(E)}(1) \cdot \mathcal{O}_{P(E)}(1))_{P(E)}$$ or equivalently $$(P(F) \cdot P(F))_{P(E)}.$$ It should be well-known that it coincides with $\text{deg}(F)$, but I can't see why and I can't find any reference. Any help would be much appreciated, thank you very much.

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1 Answer 1

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First, let us gather some facts:

Let $\pi: P(E) \to S$ be the projective bundle map.

$1)$ We have a graded isomorphism (Hartshorne, II.7.11)

$$\operatorname{Sym}(E) \cong \bigoplus_{d \in \mathbb Z} \pi_* \mathcal O_{P(E)}(d)$$

$2)$ We have $R^i \pi_*\mathcal O_{P(E)}(d)=0$ for all $d>0,i>0$ (Hartshorne, Exercise III.8.4), hence we can compute the cohomology groups $H^i(P(E),\mathcal O_{P(E)}(d))$ after push-forward.

$3)$ Given an exact sequence $0 \to A \to B \to C \to 0$ of sheaves of modules, we have an exact sequence (see http://stacks.math.columbia.edu/tag/01CJ) $$A \otimes \operatorname{Sym}^{n-1} B \to \operatorname{Sym}^{n} B \to \operatorname{Sym}^{n} C \to 0. ~~~~(\star)$$ If $A$ is locally free of rank $1$ and $B,C$ are locally free, we have exactness at the left side, i.e. $$0 \to A \otimes \operatorname{Sym}^{n-1} B \to \operatorname{Sym}^{n} B \to \operatorname{Sym}^{n} C \to 0$$ is exact. This is easy : The ranks in $(\star)$ are additive in this case, hence the kernel of the first map must be torsion, but it is torsion free as a submodule of a free module.

In our case, we get an exact sequence

$$0 \to \operatorname{Sym}^{n-1}(E) \to \operatorname{Sym}^{n}(E) \to F^n \to 0.$$

$4)$ We have the asymptotic Riemann-Roch formula: If $\mathcal L$ is a line bundle on a $d$-dimensional projective variety, we have

$\chi(\mathcal L^n) = \frac{(\mathcal L^d)}{d!}n^d + \text{lower degree terms...}$

Here $(\mathcal L^d)$ denotes the self-intersection number of $\mathcal L$. You can find this in the book of Lazarsfeld 'Positivity ...' as theorem 1.1.24.


Let us put these together: Let $L=\mathcal O_{P(E)}(1)$.

We have $H^i(P(E),L^n)=H^i(S,\pi_*L^n)=H^i(S,\operatorname{Sym}^n(E))$ , hence $\chi(L^n)=\chi(\operatorname{Sym}^n(E))$.

The exact sequence from $3)$ thus gives us $$\chi(L^n)-\chi(L^{n-1})=\chi(F^n) ~~~ (\bullet)$$ by the additivity of the euler characteristic.

Together with the asymptotic Riemann-Roch formulas for $L$ and $F$ $$\chi(L^n)=\frac{(L.L)}{2}n^2 + c_1n+c_0,$$ $$\chi(F^n)=\deg(F)n + d_0,$$ this yields $(L.L)=\deg(F)$, since these are the linear coefficients of the left-hand side and the right-hand side of $(\bullet)$ respectively.

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