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Let $f:Y\to \mathbb{A}^2$ be the blow up at $(0,0)$. $f$ is then a projective morphism hence proper. By Grothendieck duality, there exists a right adjoint of the derived image functor $Rf_*$: $f^{!}(-)=f^*(-)\otimes \omega_{Y/\mathbb{A}^2}$, where $\omega_{Y/\mathbb{A}^2}$ is the dualizing complex,.

It is very easy to describe such $\omega$ in the case of a smooth morphism, but I struggle to find what $\omega_{Y/\mathbb{A}^2}$ is concretely.

My question: How do we compute $\omega_{Y/\mathbb{A}^2}$?

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    Can you compute $f^!(\mathcal{O}_{\mathbb{A}^2})$? – 2017-01-26
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    @KReiser I think it is $O_Y(E)$ where $E$ is the exceptional divisor on $Y$. – 2017-01-26
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    If that's correct, then that's $\omega$, as the pullback of the structure sheaf is the structure sheaf. I'm having a heck of a time remembering what the definition of $f^!$ for proper maps is, so I'm not sure about checking your work on what $f^!(\mathcal{O})$ is. – 2017-01-27
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    @XuqiangQIN Your $Y$ embeds into $\mathbb{P}^2 \times \mathbb{A}^2$ pullback to relative dualizing sheaf of the latter is easy. For the former you just tensor with exterior power of the normal bundle (maybe up to shift). – 2017-04-25

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