Veronese embeddings and locally free resolutions

Question

Let $i : \mathbf P^1 \to \mathbf P^2$ be the second Veronese embedding. Clearly, $i_\star \mathcal O_{\mathbf P^1}$ has a locally free resolution of the form

$$ 0 \to \mathcal O_{\mathbf P^2} (-2) \to \mathcal O_{\mathbf P^2} \to i_\star \mathcal O_{\mathbf P^1} \to 0 $$

More generally, for any integer $n$, we have $\mathcal O_{\mathbf P^1} (2n) \cong \mathcal O_{ \mathbf P^1} \otimes i^\star \mathcal O_{\mathbf P^2} (n)$, so $i_\star \mathcal O_{\mathbf P^1} (2n) $ has a resolution

$$ 0 \to \mathcal O_{\mathbf P^2} (n-2) \to \mathcal O_{\mathbf P^2} (n) \to i_\star \mathcal O_{\mathbf P^1}(2n) \to 0 $$

My question is: how can I construct a locally free resolution for $i_\star O_{\mathbf P^1} (2n+1)$?

(I should add that it would make my life easier if the sheaves in this resolution are direct sums of invertible sheaves on $\mathbf P^2$.)

Answer 1

The resolution looks like, $$0\to\mathcal{O}_{\mathbb{P}^2}(n-1)^{\oplus 2}\to \mathcal{O}_{\mathbb{P}^2}(n)^{\oplus 2}\to i_*\mathcal{O}_{\mathbb{P}^1}(2n+1)\to 0,$$ with the determinant of the $2\times 2$ matrix appearing in the left most map is just the equation of the quadric defining $i(\mathbb{P}^1)$.