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Let $\mathcal{E}$ be a rank two vector bundle on $\mathbb{P}^2$ fitting in the following exact sequence

$$0\rightarrow \mathcal{O}_{\mathbb{P}^2}\rightarrow \mathcal{E}\rightarrow \mathcal{I}_p(-1)\rightarrow 0$$

where $\mathcal{I}_p$ is the ideal sheaf of a point $p\in\mathbb{P}^2$. Assume that $\mathcal{E}$ is normalized, $c_{1}(\mathcal{E})\in\{-1,0\}$.

How can we compute the Chern classes of $\mathcal{E}$ ?

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1 Answer 1

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As $\mathcal{O}_{\mathbb{P}^2}$ is trivial, then multiplicativity of Chern classes in exact sequences implies: $$ c_*(\mathcal{E}) = c_*(\mathcal{I}_p(-1)). $$ We can compute $c_*(\mathcal{I}_p(-1))$ by noting that $p\in \mathbb{P}^2$ is a complete intersection of 2 lines. Therefore, the $(-1)$-twist of the Koszul complex is exact: $$ 0\rightarrow \mathcal{O}_{\mathbb{P}^2} (-3) \rightarrow \mathcal{O}_{\mathbb{P}^2}(-2)^{\oplus 2} \rightarrow \mathcal{I}_p(-1)\rightarrow 0. $$ Thus $$ c_*(I_p(-1)) = c_*(\mathcal{O}_{\mathbb{P}^2}(-2)^{\oplus 2})/c_*(\mathcal{O}_{\mathbb{P}^2}(-3)) = (1-4h+4h^2)/(1-3h)\\ =(1-4h+4h^2)(1+3h+9h^2)\\ = 1-h+(9-12+4)h^2 = 1-h+h^2, $$ where $h\in H^2(\mathbb{P}^2,\mathbb{Z})$ is the hyperplane generator. Thus: $c_1(\mathcal{E})=-1$ and $c_2(\mathcal{E}) = 1$.

I.e. $\int_{[\mathbb{P}^2]} c_2(\mathcal{E}) = 1$.

One can also use Grothendieck-Riemann-Roch to compute $c_*(\mathcal{O}_p) = 1-h^2$, and then use the ideal sequence for $\mathcal{O}_p$.

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