Hi, I would like to show that if $f: X \rightarrow Y=Spec \, \mathbb{C}$, where $X$ is a nonsingular complex projective variety, is the projection to a point, then the complex $f^! \mathcal{O}_Y$, appearing in Grothendieck's duality, is the dualizing sheaf for $X$. Let $F$ be a coherent sheaf on $X$. Starting from $Hom_{\mathcal{O}_X}(F, f^! \mathcal{O}_Y) \simeq Hom_{{O}_Y}(Rf_* F, \mathcal{O}_Y)$ applying the cohomology functor $H^i$ we obtain $Ext^i(F, f^! \mathcal{O}_Y) \simeq Ext^i(Rf_* F, \mathcal{O}_Y).$ Using Yoneda's Formula, the right term becomes $Hom^i_{D(Y)}(Rf_* F, \mathcal{O}_Y) \cong Hom_{D(Y)}(Rf_* F, \mathcal{O}_Y[i]) \cong Hom(\widetilde{H^1(X,F)}[-i], \mathcal{O}_Y),$ where, for the last isomorphism, I use Theorem 8.5 from Hartshorne'a Algebraic Geometry, p.251. Now, by Corollary 5.5 pag.151 from Hartshorne, and remembering that $\Gamma(Y, \mathcal{O}_Y)= \mathbb{C}$, the last term is equal to $Hom(H^1(X,F)[-i], \mathbb{C}) = H^{1-i}((X,F)'$H^{1-i}(X,F)'$. Now, we have $Hom^i_{D(X)}(F, f^! \mathcal{O}_Y) \cong H^{1-i}(X,F)'$ and, shifting by $(-n+1-i)$, $Hom(F, f^! \mathcal{O_Y}[-n+1]) \cong H^n(X,F)'$. So, $f^!(\mathcal{O}_Y)[-n+1]$ is a dualizing sheaf for $X$. But we know that, for a nonsingular projective variety the (unique) dualizing sheaf is the canonical sheaf $\omega$. Thus, we must have $f^!\mathcal{O}_Y[-n+1] \cong \omega$, then $f^! \mathcal{O}_Y = \omega[n-1]$. I should obtain$f^! \mathcal{O}_Y = \omega[n]$... what's wrong? Thank you 4 added 23 characters in body; added 39 characters in body; added 2 characters in body Hi, I would like to show that if$f: X \rightarrow Y=Spec \, \mathbb{C}$, where$X$is a nonsingular complex projective variety, is the projection to a point, then the complex $f^! \mathcal{O}_Y$, appearing in Grothendieck's duality, is the dualizing sheaf for$X$. Let $F$ be a coherent sheaf on $X$. Starting from $ Hom_{\mathcal_{O}_X}(FHom_{\mathcal{O}_X}(F, f^! \mathcal{O}_Y) \simeq R Hom_{O}_Y}(Rf_Hom_{{O}_Y}(Rf_* F, \mathcal{O}_Y) $ applying the cohomology functor$ H^i $we obtain $Ext^i(F, f^! \mathcal{O}Y) mathcal{O}_Y) \simeq Ext^i(RfExt^i(Rf_* F, \mathcal{O}Y).$mathcal{O}_Y).$ Using Yoneda's Formula, the right term becomes $Hom^i{D(Y)}(Rf* Hom^i_{D(Y)}(Rf_* F, \mathcal{O}Y) mathcal{O}_Y) \cong Hom{D(Y)}(RfHom_{D(Y)}(Rf_* F, \mathcal{O}_Y[i]) \cong Hom(\widetilde{H^1(X,F)}[-i], \mathcal{O}_Y),$ where, for the last isomorphism, I use Theorem 8.5 from Hartshorne'a Algebraic Geometry, p.251. Now, by Corollary 5.5 pag.151 from Hartshorne, and remembering that $\Gamma(Y, mathcal{O}Y)= \mathbb{C}$, mathcal{O}_Y)= \mathbb{C}$, the last term is equal to $Hom(H^1(X,F)[-i], \mathbb{C}) = H^{1-i}((X,F)'$. Now, we have $Hom^i{D(X)}(F, Hom^i_{D(X)}(F, f^! \mathcal{O}_Y) \cong H^{1-i}(X,F)'$ and, shifting by $(-n+1-i)$, (-n+1-i)$, $Hom(F, f^! \mathcal{O_Y}[-n+1]) \cong H^n(X,F)'$. So, $f^!(\mathcal{O}_Y)[-n+1]$ is a dualizing sheaf for $X$. X$. But we know that, for a nonsingular projective variety the (unique) dualizing sheaf is the canonical sheaf $\omega$. \omega$. Thus, we must have $f^!\mathcal{O}_Y[-n+1] \cong \omega$, omega$, then $f^! \mathcal{O}_Y = \omega[n-1]$.omega[n-1]$.
I should obtain $f^! \mathcal{O}_Y = \omega[n]$... what's wrong? Thank you
Hi, I would like to show , using Grothendieck duality, that if $f: X \rightarrow Y=Spec \, \mathbb{C}$, where $X$ is a nonsingular complex projective variety, is the projection to a point, then the complex $f^! f^! \mathcal{O}Y$ mathcal{O}_Y$, appearing in Grothendieck's duality, is the dualizing sheaf for$X$. Starting from $ Hom{\mathcal_{O}_X}(F, Hom_{\mathcal_{O}_X}(F, f^! \mathcal{O}Y) mathcal{O}_Y) \simeq R Hom{O}Y}(RfHom_{O}_Y}(Rf_* F, \mathcal{O}_Y) $ applying the cohomology functor$ H^i $we obtain$Ext^i(F, f^! \mathcal{O}Y) \simeq Ext^i(Rf* F, \mathcal{O}Y).$Using Yoneda's Formula, the right term becomes$Hom^i{D(Y)}(Rf* F, \mathcal{O}Y) \cong Hom{D(Y)}(Rf* F, \mathcal{O}_Y[i]) \cong Hom(\widetilde{H^1(X,F)}[-i], \mathcal{O}_Y),$where, for the last isomorphism, I use Theorem 8.5 from Hartshorne'a Algebraic Geometry, p.251. Now, by Corollary 5.5 pag.151 from Hartshorne, and remembering that$\Gamma(Y, mathcal{O}Y)= \mathbb{C}$, the last term is equal to$Hom(H^1(X,F)[-i], \mathbb{C}) = H^{1-i}((X,F)'$. Now, we have$Hom^i{D(X)}(F, f^! \mathcal{O}_Y) \cong H^{1-i}(X,F)'$and, shifting by$(-n+1-i)$,$Hom(F, f^! \mathcal{O_Y}[-n+1]) \cong H^n(X,F)'$. So,$f^!(\mathcal{O}_Y)[-n+1]$is a dualizing sheaf for$X$. But we know that, for a nonsingular projective variety the (unique) dualizing sheaf is the canonical sheaf$\omega$. Thus, we must have$f^!\mathcal{O}_Y[-n+1] \cong \omega$, then$f^! \mathcal{O}_Y = \omega[n-1]$. I should obtain$f^! \mathcal{O}_Y = \omega[n]\$... what's wrong? Thank you