Easy computation by Ravi Vakil (jet bundles)

Question

I landed on some short notes by Ravi Vakil from the 90s, the Beginner's Guide to Jet Bundles from the Point of View of Algebraic Geometry.

The notes are very clear but on the very first page there is an easy computation whose result seems wrong to me. We are working on $\mathbb P^2$ an Vakil says that $$ c_2(\mathcal O(d) \otimes (\mathcal O \oplus \Omega))=3(d-1)^2$$ EDIT: where $\Omega=T^*\mathbb P^2$ is the cotangent bundle.

If we call $H$ the hyperplane class in $A\mathbb P^2$, I think that the total Chern class for $\Omega$ is $$c(\Omega)=1-3H+6H^2 \ \ (\star)$$ If $(\star)$ above is correct, we should have that $$c_2(\mathcal O(d)\oplus \Omega(d))=d^2 + (2d-3)\cdot d+(d^2-3d+3),\ \ (\square)$$ which is not the above value.

I couldn't find anyone pointing this error out on the internet, therefore I'm supposing that I am the one who is wrong: I think that my mistake is somewhere in the $(\square)$ evaluation, and maybe I am missing something about the general techniques of evaluation of total Chern classes.

Answer 1

The equation $(\star)$ is incorrect.

We have $c(\Omega) = 1 - 3H + 3H^2$ because $\Omega = T^*\mathbb{CP}^2 \cong \overline{T\mathbb{CP}^2}$, $c_i(\overline{E}) = (-1)^ic_i(E)$, and $c(T\mathbb{CP}^2) = (1 + H)^3 = 1 + 3H + 3H^2$.

By using the formula for $c_2(V\otimes L)$ from this question, we obtain the correct conclusion:

\begin{align*} c_2(\mathcal{O}(d)\otimes(\mathcal{O}\oplus\Omega)) &= c_2(\mathcal{O}\oplus\Omega) + (3-1)c_1(\mathcal{O}(d))c_1(\mathcal{O}\oplus\Omega) + \binom{3}{2}c_1(\mathcal{O}(d))^2\\ &= c_2(\Omega) + 2c_1(\mathcal{O}(d))c_1(\Omega) + 3c_1(\mathcal{O}(d))^2\\ &= 3H^2 + 2dH(-3H) + 3(dH)^2\\ &= 3H^2 -6dH^2 + 3d^2H^2\\ &= (3d^2 - 6d + 3)H^2\\ &= 3(d^2 - 2d + 1)H^2\\ &= 3(d-1)^2H^2. \end{align*}