Return to Answer

This of course can be found in any reference on duality (e.g. Hartshorne "Algebraic Geometry" chap. 3) but in Hartshorne's proof it's somehow mysterious where the $\mathscr{E}xt^{n-1}_P(\mathscr{O}, \omega_{P^n})$ comes from. You can explain it by studying relative duality for morphisms (e.g. Hartshorne "Residues and Duality"), in particular for the closed immersion $i:C\to \mathbb{P}^n$. Here is another argument (in a way it's the proof from Hartshorne's AG done backwards, so that we can see how this $\mathscr{E}xt^{n-1}$ appears).

Recall that a dualizing sheaf is a sheaf $\omega$ which satisfies $H^1(C, F)^\vee = Hom_C(F, \omega)$ for every coherent $F$ on $C$.

A thing to start with is duality on $P:=\mathbb{P}^n$: $$ H^1(C, F)^\vee = H^1(P, F)^\vee = Ext^{n-1}_P(F, \omega_P). $$

We now need a natural isomorphism between $Ext^i_P(F, \omega_P)$ and $Hom_C(F, \omega)$ for some $\omega$. A good thing to do to compare $Hom$'s and $Ext$'s on $P$ and $C$ is to find a spectral sequence comparing the two. Let us look at the functors $$ Coh(P) \to Coh(C) \to Ab $$ where the first functor is $\mathscr{H}om_P(\mathscr{O}_C, -)$ and the second is $Hom_C(F, -)$, their composition $Hom_P(F, -)$. The spectral sequence of this composition is $$ E^{ij}_2 = Ext^i_C(F, \mathscr{E}xt^j_P(\mathscr{O}_X, G)) \Rightarrow Ext^{i+j}_P(F, G). $$ For $i+j=n-1$ and $G=\omega_P$ we have our group $Ext^{n-1}_P(F, \omega)$ on the right hand side, and we have $E^{0,n-1}_2 = Hom_C(F, \mathscr{E}xt^{n-1}_P(\mathscr{O}_X, \omega_P)$, which suggests we should take $\omega = \mathscr{E}xt^{n-1}_P(\mathscr{O}_X, \omega_P)$.

To prove that this works, we need to show that $E^{ij}_2 = 0$ for $j<n-1$, which is done in Hartshorne "Algebraic Geometry", chap. 3.

This of course can be found in any reference on duality (e.g. Hartshorne "Algebraic Geometry" chap. 3) but in Hartshorne's proof it's somehow mysterious where the $\mathscr{E}xt^{n-1}_P(\mathscr{O}, \omega_{P^n})$ comes from. You can explain it by studying relative duality for morphisms (e.g. Hartshorne "Residues and Duality"), in particular for the closed immersion $i:C\to \mathbb{P}^n$. Here is another argument (in a way it's the proof from Hartshorne's AG done backwards, so that we can see how this $\mathscr{E}xt^{n-1}$ appears).

Recall that a dualizing sheaf is a sheaf $\omega$ which satisfies $H^1(C, F)^\vee = Hom_C(F, \omega)$ for every coherent $F$ on $C$.

A thing to start with is duality on $P:=\mathbb{P}^n$: $$ H^1(C, F)^\vee = H^1(P, F)^\vee = Ext^{n-1}_P(F, \omega_P). $$

We now need a natural isomorphism between $Ext^i_P(F, \omega_P)$ and $Hom_C(F, \omega)$ for some $\omega$. A good thing to do to compare $Hom$'s and $Ext$'s on $P$ and $C$ is to find a spectral sequence comparing the two. Let us look at the functors $$ Coh(P) \to Coh(C) \to Ab $$ where the first functor is $\mathscr{H}om_P(\mathscr{O}_C, -)$ and the second is $Hom_C(F, -)$, their composition $Hom_P(F, -)$. The spectral sequence of this composition is $$ E^{ij}_2 = Ext^i_C(F, \mathscr{E}xt^j_P(\mathscr{O}_C, G)) \Rightarrow Ext^{i+j}_P(F, G). $$ For $i+j=n-1$ and $G=\omega_P$ we have our group $Ext^{n-1}_P(F, \omega)$ on the right hand side, and we have $E^{0,n-1}_2 = Hom_C(F, \mathscr{E}xt^{n-1}_P(\mathscr{O}_C, \omega_P)$, which suggests we should take $\omega = \mathscr{E}xt^{n-1}_P(\mathscr{O}_C, \omega_P)$.

In fact, because $C$ is a smooth curve, the above spectral sequence just boils down a to ``universal coefficients'' short exact sequence $$ 0 \to Ext^1_C(F, \mathscr{E}xt^{n-2}_P(\mathscr{O}_C, \omega_P)) \to Ext^{n-1}_P(F, \omega_P) \to Hom_C(F, \mathscr{E}xt^{n-1}_P(\mathscr{O}_C, \omega_P))\to 0. $$

To prove that the second map is an isomorphism (for any $F$), we need to show that $\mathscr{E}xt^{n-2}_P(\mathscr{O}_C, \omega_P)= 0 $, which is done in Hartshorne "Algebraic Geometry", chap. 3.

This of course can be found in any reference on duality (e.g. Hartshorne "Algebraic Geometry" chap. 3) but in Hartshorne's proof it's somehow mysterious how the $\mathscr{E}xt^{n-1}_P(\mathscr{O}, \omega_{P^n})$ appears. You can explain it by studying relative duality for morphisms (e.g. Hartshorne "Residues and Duality"), in particular for the closed immersion $i:C\to \mathbb{P}^n$. Here is another argument (in a way it's the proof from Hartshorne's AG done backwards, so that we can see how this $\mathscr{E}xt^{r-1}$ appears).

Recall that a dualizing sheaf is a sheaf $\omega$ which satisfies $H^1(C, F)^\vee = Hom_C(F, \omega)$ for every coherent $F$.

A thing to start with is duality on $P:=\mathbb{P}^n$: $$ H^1(C, F)^\vee = H^1(P, F)^\vee = Ext^{n-1}_P(F, \omega_P). $$

We now need a natural isomorphism between $Ext^i_P(F, \omega_P)$ and $Hom_C(F, \omega)$ for some $\omega$. A good thing to do to compare $Hom$'s and $Ext$'s on $P$ and $C$ is to find a spectral sequence comparing the two. Let us look at the functors $$ Coh(P) \to Coh(C) \to Ab $$ where the first functor is $\mathscr{H}om_P(\mathscr{O}_C, -)$ and the second is $Hom_C(F, -)$, their composition $Hom_P(F, -)$. The spectral sequence of this composition is $$ E^{ij}_2 = Ext^i_C(F, Ext^j_P(\mathscr{O}_X, G)) \Rightarrow Ext^{i+j}_P(F, G). $$ For $i+j=n-1$ and $G=\omega_P$ we have our group $Ext^{n-1}_P(F, \omega)$ on the right hand side, and we have $E^{0,n-1}_2 = Hom_C(F, \mathscr{E}xt^{n-1}_P(\mathscr{O}_X, \omega_P)$, which suggests we should take $\omega = \mathscr{E}xt^{n-1}_P(\mathscr{O}_X, \omega_P)$.

To prove that this works, we need to show that $E^{ij}_2 = 0$ for $j<n-1$, which is done in Hartshorne "Algebraic Geometry", chap. 3.

This of course can be found in any reference on duality (e.g. Hartshorne "Algebraic Geometry" chap. 3) but in Hartshorne's proof it's somehow mysterious where the $\mathscr{E}xt^{n-1}_P(\mathscr{O}, \omega_{P^n})$ comes from. You can explain it by studying relative duality for morphisms (e.g. Hartshorne "Residues and Duality"), in particular for the closed immersion $i:C\to \mathbb{P}^n$. Here is another argument (in a way it's the proof from Hartshorne's AG done backwards, so that we can see how this $\mathscr{E}xt^{n-1}$ appears).

Recall that a dualizing sheaf is a sheaf $\omega$ which satisfies $H^1(C, F)^\vee = Hom_C(F, \omega)$ for every coherent $F$ on $C$.

A thing to start with is duality on $P:=\mathbb{P}^n$: $$ H^1(C, F)^\vee = H^1(P, F)^\vee = Ext^{n-1}_P(F, \omega_P). $$

We now need a natural isomorphism between $Ext^i_P(F, \omega_P)$ and $Hom_C(F, \omega)$ for some $\omega$. A good thing to do to compare $Hom$'s and $Ext$'s on $P$ and $C$ is to find a spectral sequence comparing the two. Let us look at the functors $$ Coh(P) \to Coh(C) \to Ab $$ where the first functor is $\mathscr{H}om_P(\mathscr{O}_C, -)$ and the second is $Hom_C(F, -)$, their composition $Hom_P(F, -)$. The spectral sequence of this composition is $$ E^{ij}_2 = Ext^i_C(F, \mathscr{E}xt^j_P(\mathscr{O}_X, G)) \Rightarrow Ext^{i+j}_P(F, G). $$ For $i+j=n-1$ and $G=\omega_P$ we have our group $Ext^{n-1}_P(F, \omega)$ on the right hand side, and we have $E^{0,n-1}_2 = Hom_C(F, \mathscr{E}xt^{n-1}_P(\mathscr{O}_X, \omega_P)$, which suggests we should take $\omega = \mathscr{E}xt^{n-1}_P(\mathscr{O}_X, \omega_P)$.

To prove that this works, we need to show that $E^{ij}_2 = 0$ for $j<n-1$, which is done in Hartshorne "Algebraic Geometry", chap. 3.

Recall that a dualizing sheaf is a sheaf $\omega$ which satisfies $H^1(C, F)^\vee = Hom_C(F, \omega)$ for every coherent $F$.

A thing to start with is duality on $P:=\mathbb{P}^n$: $$ H^1(C, F)^\vee = H^1(P, F)^\vee = Ext^{n-1}_P(F, \omega_P). $$

We now need a natural isomorphism between $Ext^i_P(F, \omega_P)$ and $Hom_C(F, \omega)$ for some $\omega$. A good thing to do to compare $Hom$'s and $Ext$'s on $P$ and $C$ is to find a spectral sequence comparing the two. Let us look at the functors $$ Coh(P) \to Coh(C) \to Ab $$ where the first functor is $\mathscr{H}om_P(\mathscr{O}_C, -)$ and the second is $Hom_C(F, -)$, their composition $Hom_P(F, -)$. The spectral sequence of this composition is $$ E^{ij}_2 = Ext^i_C(F, Ext^j_P(\mathscr{O}_X, G)) \Rightarrow Ext^{i+j}_P(F, G). $$ For $i+j=n-1$ and $G=\omega_P$ we have our group $Ext^{n-1}_P(F, \omega)$ on the right hand side, and we have $E^{0,n-1}_2 = Hom_C(F, \mathscr{E}xt^{n-1}_P(\mathscr{O}_X, \omega_P)$, which suggests we should take $\omega = \mathscr{E}xt^{n-1}_P(\mathscr{O}_X, \omega_P)$.

To prove that this works, we need to show that $E^{ij}_2 = 0$ for $j<n-1$.

This of course can be found in any reference on duality (e.g. Hartshorne "Algebraic Geometry" chap. 3) but in Hartshorne's proof it's somehow mysterious how the $\mathscr{E}xt^{n-1}_P(\mathscr{O}, \omega_{P^n})$ appears. You can explain it by studying relative duality for morphisms (e.g. Hartshorne "Residues and Duality"), in particular for the closed immersion $i:C\to \mathbb{P}^n$. Here is another argument (in a way it's the proof from Hartshorne's AG done backwards, so that we can see how this $\mathscr{E}xt^{r-1}$ appears).

Recall that a dualizing sheaf is a sheaf $\omega$ which satisfies $H^1(C, F)^\vee = Hom_C(F, \omega)$ for every coherent $F$.

A thing to start with is duality on $P:=\mathbb{P}^n$: $$ H^1(C, F)^\vee = H^1(P, F)^\vee = Ext^{n-1}_P(F, \omega_P). $$

We now need a natural isomorphism between $Ext^i_P(F, \omega_P)$ and $Hom_C(F, \omega)$ for some $\omega$. A good thing to do to compare $Hom$'s and $Ext$'s on $P$ and $C$ is to find a spectral sequence comparing the two. Let us look at the functors $$ Coh(P) \to Coh(C) \to Ab $$ where the first functor is $\mathscr{H}om_P(\mathscr{O}_C, -)$ and the second is $Hom_C(F, -)$, their composition $Hom_P(F, -)$. The spectral sequence of this composition is $$ E^{ij}_2 = Ext^i_C(F, Ext^j_P(\mathscr{O}_X, G)) \Rightarrow Ext^{i+j}_P(F, G). $$ For $i+j=n-1$ and $G=\omega_P$ we have our group $Ext^{n-1}_P(F, \omega)$ on the right hand side, and we have $E^{0,n-1}_2 = Hom_C(F, \mathscr{E}xt^{n-1}_P(\mathscr{O}_X, \omega_P)$, which suggests we should take $\omega = \mathscr{E}xt^{n-1}_P(\mathscr{O}_X, \omega_P)$.

To prove that this works, we need to show that $E^{ij}_2 = 0$ for $j<n-1$, which is done in Hartshorne "Algebraic Geometry", chap. 3.