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Consider the blowup $X$ of $\mathbb{P}^2$ at a single point $p$. Then, Orlov showed that there is a semiorthogonal decomposition $D^b(X)=\langle e,O_X,O_X(1),O_X(2)\rangle$, where $O_X(i)$ is the pullback of $O_{\mathbb{P}^2}(i)$, and $e$ can be taken to be $i_*O_E(-1)$, where $i:E\rightarrow X$ is the inclusion of the exceptional divisor. The objects $e,O_X(i)$ are in fact all exceptional, so their direct sum should be a tilting object for $D^b(X)$.

But I dont see why for example $Ext^k(e,\mathcal{O}_X(i))=0$, for $k>0$ ?

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By Grothendieck duality, $$\mathrm{R}\mathcal{H}om(i_*\mathcal{O}_E(-1),\mathcal{O}_X(i))\cong i_*\mathrm{R}\mathcal{H}om(\mathcal{O}_E(-1),i^*\mathcal{O}_X(i))\otimes \mathcal{O}_E(-1)[-1])\cong i_*\mathcal{O}_E[-1]\ ,$$thus $\mathrm{Ext}^p(i_*\mathcal{O}_E(-1),\mathcal{O}_X(i))=H^{p-1} (E,\mathcal{O}_E)=k\ $ if $\ p=1$, $0$ otherwise.

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  • $\begingroup$ Ok. You take the duality $Rf_*R\mathcal{H}om(\mathcal{F},Lf^*(\mathcal{G})\otimes \omega_f[dim(f)])\simeq R\mathcal{H}om(Rf_*\mathcal{F},\mathcal{G})$. I do not see why $\omega_f[dim(f)]$ is the same as $\mathcal{O}_E(-1)$. $\endgroup$
    – Aleksa
    Commented Mar 10, 2014 at 10:49
  • $\begingroup$ I think $\omega_i[dim(i)]$ is isomorphic to $\mathcal{O}_E(-1)[-1]$. $\endgroup$
    – Aleksa
    Commented Mar 10, 2014 at 11:29
  • $\begingroup$ You are right. So in fact $\mathrm{Ext}^1(i_*\mathcal{O}_E(-1), \mathcal{O}_X(i))=k$. Why do you believe it is 0? I didn't understand the argument. $\endgroup$
    – abx
    Commented Mar 10, 2014 at 13:10

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