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Sándor Kovács
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First of all, I think that all of $\mathscr F_C$'s global sections appear as global sections of $\mathscr F_{C'}$, so the answer to your second question is that it only happens in that trivial case you're mentioning.

Observe that $\nu$ is an isomorphism outside of finitely many points, so the induced morphism $\mathscr F_C\to \nu_*\mathscr F_{C'}$ is injective with a torsion cokernel. In other words, we have a short exact sequence: $$ 0 \to \mathscr F_C \to \nu_*\mathscr F_{C'} \to \mathscr Q \to 0. $$ where $\mathscr Q$ is torsion. This shows which sections of $\nu_*\mathscr F_{C'}$ do not come from $\mathscr F_C$: the ones that map to something non-zero in $\mathscr Q$. It also shows that every global section of $\mathscr F_C$ appear as a global section of $\mathscr F_{C'}$. Of course, this is not very helpful since it is hard to compute $\mathscr Q$ or what sections map to non-zero sections there.


A somewhat different approach is this: Let $\mathscr T$ denote the torsion subsheaf of $\nu^*\mathscr F_C$, so one has a short exact sequence: $$ 0\to \mathscr T \to \nu^*\mathscr F_C \to \mathscr F_{C'} \to 0. $$ Notice that the sheaf on the left hand side is torsion so it has no $H^1$. Therefore one obtains a short exact sequence of global sections: $$ 0\to H^0(C',\mathscr T) \to H^0(C',\nu^*\mathscr F_C) \to H^0(C,\mathscr F_{C'}) \to 0. $$$$ 0\to H^0(C',\mathscr T) \to H^0(C',\nu^*\mathscr F_C) \to H^0(C',\mathscr F_{C'}) \to 0. $$ In particular all the global sections of $\mathscr F_{C'}$ come from those of $\nu^*\mathscr F_C$.

Next apply $\nu_*$ to the first short exact sequence and notice that $\nu$ is finite and hence $\nu_*$ is exact: $$ 0\to \nu_*\mathscr T \to \nu_*\nu^*\mathscr F_C \to \nu_*\mathscr F_{C'} \to 0. $$ Also notice that the global sections of these sheaves are the same as the global sections of the ones they are the push forwards of.

Finally, notice that the natural morphism $\mathscr F_C\to \nu_*\nu^*\mathscr F_C$ is injective by the torsion-free assumption.

So, all the global sections of $\mathscr F_{C'}$ come from $\nu^*\mathscr F_C$. We throw out the torsion sections and then the ones you are looking for are the global sections of $\nu_*\nu^*\mathscr F_C$ that are not sections of the subsheaf generated by torsion plus the global sections of the original $\mathscr F_C$.

I guess this is still not a very good answer to your first question, but perhaps the moral is that you should look at all of these sheaves on $C$.

First of all, I think that all of $\mathscr F_C$'s global sections appear as global sections of $\mathscr F_{C'}$, so the answer to your second question is that it only happens in that trivial case you're mentioning.

Observe that $\nu$ is an isomorphism outside of finitely many points, so the induced morphism $\mathscr F_C\to \nu_*\mathscr F_{C'}$ is injective with a torsion cokernel. In other words, we have a short exact sequence: $$ 0 \to \mathscr F_C \to \nu_*\mathscr F_{C'} \to \mathscr Q \to 0. $$ where $\mathscr Q$ is torsion. This shows which sections of $\nu_*\mathscr F_{C'}$ do not come from $\mathscr F_C$: the ones that map to something non-zero in $\mathscr Q$. It also shows that every global section of $\mathscr F_C$ appear as a global section of $\mathscr F_{C'}$. Of course, this is not very helpful since it is hard to compute $\mathscr Q$ or what sections map to non-zero sections there.


A somewhat different approach is this: Let $\mathscr T$ denote the torsion subsheaf of $\nu^*\mathscr F_C$, so one has a short exact sequence: $$ 0\to \mathscr T \to \nu^*\mathscr F_C \to \mathscr F_{C'} \to 0. $$ Notice that the sheaf on the left hand side is torsion so it has no $H^1$. Therefore one obtains a short exact sequence of global sections: $$ 0\to H^0(C',\mathscr T) \to H^0(C',\nu^*\mathscr F_C) \to H^0(C,\mathscr F_{C'}) \to 0. $$ In particular all the global sections of $\mathscr F_{C'}$ come from those of $\nu^*\mathscr F_C$.

Next apply $\nu_*$ to the first short exact sequence and notice that $\nu$ is finite and hence $\nu_*$ is exact: $$ 0\to \nu_*\mathscr T \to \nu_*\nu^*\mathscr F_C \to \nu_*\mathscr F_{C'} \to 0. $$ Also notice that the global sections of these sheaves are the same as the global sections of the ones they are the push forwards of.

Finally, notice that the natural morphism $\mathscr F_C\to \nu_*\nu^*\mathscr F_C$ is injective by the torsion-free assumption.

So, all the global sections of $\mathscr F_{C'}$ come from $\nu^*\mathscr F_C$. We throw out the torsion sections and then the ones you are looking for are the global sections of $\nu_*\nu^*\mathscr F_C$ that are not sections of the subsheaf generated by torsion plus the global sections of the original $\mathscr F_C$.

I guess this is still not a very good answer to your first question, but perhaps the moral is that you should look at all of these sheaves on $C$.

First of all, I think that all of $\mathscr F_C$'s global sections appear as global sections of $\mathscr F_{C'}$, so the answer to your second question is that it only happens in that trivial case you're mentioning.

Observe that $\nu$ is an isomorphism outside of finitely many points, so the induced morphism $\mathscr F_C\to \nu_*\mathscr F_{C'}$ is injective with a torsion cokernel. In other words, we have a short exact sequence: $$ 0 \to \mathscr F_C \to \nu_*\mathscr F_{C'} \to \mathscr Q \to 0. $$ where $\mathscr Q$ is torsion. This shows which sections of $\nu_*\mathscr F_{C'}$ do not come from $\mathscr F_C$: the ones that map to something non-zero in $\mathscr Q$. It also shows that every global section of $\mathscr F_C$ appear as a global section of $\mathscr F_{C'}$. Of course, this is not very helpful since it is hard to compute $\mathscr Q$ or what sections map to non-zero sections there.


A somewhat different approach is this: Let $\mathscr T$ denote the torsion subsheaf of $\nu^*\mathscr F_C$, so one has a short exact sequence: $$ 0\to \mathscr T \to \nu^*\mathscr F_C \to \mathscr F_{C'} \to 0. $$ Notice that the sheaf on the left hand side is torsion so it has no $H^1$. Therefore one obtains a short exact sequence of global sections: $$ 0\to H^0(C',\mathscr T) \to H^0(C',\nu^*\mathscr F_C) \to H^0(C',\mathscr F_{C'}) \to 0. $$ In particular all the global sections of $\mathscr F_{C'}$ come from those of $\nu^*\mathscr F_C$.

Next apply $\nu_*$ to the first short exact sequence and notice that $\nu$ is finite and hence $\nu_*$ is exact: $$ 0\to \nu_*\mathscr T \to \nu_*\nu^*\mathscr F_C \to \nu_*\mathscr F_{C'} \to 0. $$ Also notice that the global sections of these sheaves are the same as the global sections of the ones they are the push forwards of.

Finally, notice that the natural morphism $\mathscr F_C\to \nu_*\nu^*\mathscr F_C$ is injective by the torsion-free assumption.

So, all the global sections of $\mathscr F_{C'}$ come from $\nu^*\mathscr F_C$. We throw out the torsion sections and then the ones you are looking for are the global sections of $\nu_*\nu^*\mathscr F_C$ that are not sections of the subsheaf generated by torsion plus the global sections of the original $\mathscr F_C$.

I guess this is still not a very good answer to your first question, but perhaps the moral is that you should look at all of these sheaves on $C$.

Source Link
Sándor Kovács
  • 42.9k
  • 2
  • 109
  • 155

First of all, I think that all of $\mathscr F_C$'s global sections appear as global sections of $\mathscr F_{C'}$, so the answer to your second question is that it only happens in that trivial case you're mentioning.

Observe that $\nu$ is an isomorphism outside of finitely many points, so the induced morphism $\mathscr F_C\to \nu_*\mathscr F_{C'}$ is injective with a torsion cokernel. In other words, we have a short exact sequence: $$ 0 \to \mathscr F_C \to \nu_*\mathscr F_{C'} \to \mathscr Q \to 0. $$ where $\mathscr Q$ is torsion. This shows which sections of $\nu_*\mathscr F_{C'}$ do not come from $\mathscr F_C$: the ones that map to something non-zero in $\mathscr Q$. It also shows that every global section of $\mathscr F_C$ appear as a global section of $\mathscr F_{C'}$. Of course, this is not very helpful since it is hard to compute $\mathscr Q$ or what sections map to non-zero sections there.


A somewhat different approach is this: Let $\mathscr T$ denote the torsion subsheaf of $\nu^*\mathscr F_C$, so one has a short exact sequence: $$ 0\to \mathscr T \to \nu^*\mathscr F_C \to \mathscr F_{C'} \to 0. $$ Notice that the sheaf on the left hand side is torsion so it has no $H^1$. Therefore one obtains a short exact sequence of global sections: $$ 0\to H^0(C',\mathscr T) \to H^0(C',\nu^*\mathscr F_C) \to H^0(C,\mathscr F_{C'}) \to 0. $$ In particular all the global sections of $\mathscr F_{C'}$ come from those of $\nu^*\mathscr F_C$.

Next apply $\nu_*$ to the first short exact sequence and notice that $\nu$ is finite and hence $\nu_*$ is exact: $$ 0\to \nu_*\mathscr T \to \nu_*\nu^*\mathscr F_C \to \nu_*\mathscr F_{C'} \to 0. $$ Also notice that the global sections of these sheaves are the same as the global sections of the ones they are the push forwards of.

Finally, notice that the natural morphism $\mathscr F_C\to \nu_*\nu^*\mathscr F_C$ is injective by the torsion-free assumption.

So, all the global sections of $\mathscr F_{C'}$ come from $\nu^*\mathscr F_C$. We throw out the torsion sections and then the ones you are looking for are the global sections of $\nu_*\nu^*\mathscr F_C$ that are not sections of the subsheaf generated by torsion plus the global sections of the original $\mathscr F_C$.

I guess this is still not a very good answer to your first question, but perhaps the moral is that you should look at all of these sheaves on $C$.