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Let $\pi :\widetilde{\mathbb{P}^{3}} \longrightarrow \mathbb{P}^{3}$ be the blowup of $\mathbb{P}^{3}$ along a regular curve $\mathcal{C}$, with exceptional divisor $E$. We know the following: If $X$ is a projective nonsingular variety over an algebraically closed field $\mathcal{K}$, then dualizing sheaf $\omega^{\circ}_{X}$ is isomorphic to the canonical sheaf $\omega_{X}$.

What's the dualizing sheaf of $\widetilde{\mathbb{P}^3}$?

Any help is welcome.

red_trumpet
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Allan Ramos
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  • What sort of answer are you looking for? As you've said, on this variety the dualizing sheaf is isomorphic to the canonical sheaf - what information do you want in addition to this or instead of this statement? – KReiser Apr 10 '19 at 01:47
  • What's the dualizing sheaf of $\widetilde{\mathbb{P}}^{3}$? What is the contribution of exceptional divisor? – Allan Ramos Apr 10 '19 at 11:51
  • $\omega_{\mathbb{P}^{3}} = \mathcal{O}{\mathbb{P}^{3}}(-4)$. Then, $\omega{\widetilde{\mathbb{P}}^{3}} = $?? – Allan Ramos Apr 10 '19 at 12:05

1 Answers1

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The canonical sheaf of a blowup $f:Y\to X$ satisfies the relation $K_Y = f^*(K_X) + (c-1)E$ where $c$ is the codimension of what you're blowing up and $E$ is the exceptional divisor (see Griffiths and Harris page 187 for a proof, for example - the case where there's a nontrivial meromorphic top form is easy and you may wish to try that yourself). Since the canonical and dualizing sheaves coincide here, you have your answer.

KReiser
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