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5
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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
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4
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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
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3
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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
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2
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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 dualizing complex $ f^! \mathcal{O}Y $ is the dualizing sheaf for $X$.
Starting from
$ Hom{\mathcal_{O}_X}(F, f^! \mathcal{O}Y) \simeq R 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 $Hom(H^1(X,F)[-i], \mathbb{C}) = H^{1-i}((X,F)'$.
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
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1
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connections between Grothendieck's and Serre's duality
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 dualizing complex $ f^! \mathcal{O}Y $ is the dualizing sheaf for $X$.
Starting from
$ Hom{\mathcal_{O}_X}(F, f^! \mathcal{O}Y) \simeq R 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 $Hom(H^1(X,F)[-i], \mathbb{C}) = H^{1-i}((X,F)'$.
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
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