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We have the sequence $0 \rightarrow \Omega^1_{\mathbb{P}^2}\rightarrow3\mathcal{O}_{\mathbb{P}^2}(-1)\rightarrow \mathcal{O}_{\mathbb{P}^2}\rightarrow 0$.

Can we write a exact sequece such that $\Omega^1_{\mathbb{P}^2}$ is on the right?

Sorry if the question was not properly written, I'm looking for a exact sequece of the form $0 \leftarrow \Omega^1_{\mathbb{P}^2}\leftarrow \bigoplus\mathcal{O}_{\mathbb{P}^2}(a_{1i})\leftarrow \bigoplus \mathcal{O}_{\mathbb{P}^2}(a_{2i})\leftarrow \cdots$ (with all the terms given by sums of line bundles)

If it exists how can I contruct it?

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  • $\begingroup$ I think you'll need to give more details to get a good answer; as it stands, the answer is obviously (but unhelpfully) "yes". What kind of terms do you want to allow in the exact sequence? $\endgroup$
    – user5117
    Commented Oct 3, 2013 at 11:48
  • $\begingroup$ Search for "Euler sequence". It is covered, for example, in Hartshorne's book. $\endgroup$
    – Martin Brandenburg
    Commented Oct 4, 2013 at 10:20

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$$ 0 \to O(-3) \to O(-2)^{\oplus 3} \to \Omega^1 \to 0. $$

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  • $\begingroup$ Thanks. By computing the global sections I guessed it could be that but it could only work if we already knew that there exists a resolution. Why does it exists? Are we using the relation between the tangent and the cotangent? I thought it only worked for cohomologies and not for the sheaves themselves... $\endgroup$
    – Bajouca
    Commented Oct 3, 2013 at 12:46
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    $\begingroup$ $\Omega^1 = T(-3)$. $\endgroup$
    – Sasha
    Commented Oct 3, 2013 at 12:52

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