Timeline for Very weak Riemann-Roch on curves (by J. Kollar)
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 10, 2020 at 21:20 | comment | added | user267839 | the way you conclude the injectivity as consequence from assumption that $C'$ is not lineary normal is not clear to me. Could you explain it more detailed? Sorry, for frequently annouying but I absolutely wants to figure out how your argument works. | |
| Apr 10, 2020 at 21:18 | comment | added | user267839 | not hit all sections in $H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'})$. So it seems that we don't have any "control" over what happens with the images of these sections under the map $H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'}) \to H^0(C,L)$. That is I not understand why $C′$ is not linearly normal implies the injectivity of this map. On the other hand if $C'$ would be linearly normal, then $H^0(\mathbb{P}^2, \mathcal{O}(1)_{\mathbb{P}^2}) \cong H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'})$ and the desired map would be injective, but | |
| Apr 10, 2020 at 21:17 | comment | added | user267839 | What do we know? Well, we assumed that $x \mapsto [\sigma_0(x):\sigma_1(x) : \sigma_2(x)]$ not factors through a line, i.e. $C'$ is not degenerated and $H^0(\mathbb{P}^2, \mathcal{O}(1)_{\mathbb{P}^2}) \to H^0(C, L)$ is injective and trivially $H^0(\mathbb{P}^2, \mathcal{O}(1)_{\mathbb{P}^2}) \to H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'})$ is injective too by the same argument with non degeneracy. But $C'$ is not linearly normal is equivalent to $H^0(\mathbb{P}^2, \mathcal{O}(1)_{\mathbb{P}^2}) \to H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'})$ not surjective, thus we | |
| Apr 10, 2020 at 21:16 | comment | added | user267839 | Nevertheless in still not understand the argument from your answer and I'm really curious to find out how it works. You say that essentially the injectivity of $H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'}) \to H^0(C,L)$ follows from that $C'$ is not linearly normal. Why? | |
| Apr 10, 2020 at 21:16 | comment | added | user267839 | Hi, meanwhile I found out that that one can also argue as follows: I think that the image $C'$ is implicitly endowed with unique reduced structure and since $C$ is irreducible, $C'$ is integral and the structure morphism $O_{C'} \to \pi_*O_C$ has a kernel is nilpotent since $\pi$ is dominant. Thus $O_{C'} \to \pi_*O_C$ is injective. And obviously the injectivity is preserved under twists by $O_{C'}(m)$ and we are done. | |
| Feb 26, 2020 at 2:42 | comment | added | user267839 | And when we know it for $m=1$, how to see that this injection also holds for every twisted version $H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(m)|_{C'}) \to H^0(C, L^m)$? | |
| Feb 26, 2020 at 2:32 | comment | added | user267839 | So we can conclude from this that asigning $X_i \to \pi^*X_i = \sigma_i$ gives an injection $kX_0 \oplus kX_1 \oplus kX_2=H^0(\mathbb{P}^2, \mathcal{O}(1)_{\mathbb{P}^2}) \to H^0(C, L)$ since $\dim_k(H^0(\mathbb{P}^2, \mathcal{O}(1)_{\mathbb{P}^2}))=3$. By construction it factorizes through $H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'})$. What I not understand is why this already imply that the induced map $H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'}) \to H^0(C, L)$ is also injective. | |
| Feb 25, 2020 at 13:31 | vote | accept | user267839 | ||
| Feb 26, 2020 at 2:32 | |||||
| Feb 25, 2020 at 7:43 | comment | added | Francesco Polizzi | Regarding the second question, you are right: also in this case, $C'$ is in general only birational to $C$. I edited the post. | |
| Feb 25, 2020 at 7:43 | comment | added | Francesco Polizzi | The projection is given by $x \mapsto [\sigma_0(x):\sigma_1(x) : \sigma_2(x)]$, where $\langle \sigma_0,\, \sigma_1, \, \sigma_2 \rangle$ span a $3$-dimensional subspace of $H^0(C, \, L)$, so a $2$-dimensional sub-system of $|L|$. If the image is $1$-dimensional, you are actually projecting on a line, not on the plane: this is a special situation, that does not happen for the general projection. | |
| Feb 25, 2020 at 7:37 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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| Feb 25, 2020 at 0:12 | comment | added | user267839 | Second question: Why assumption $\dim_k H^0(C, \, L)=3$ would imply $C \cong C'$? | |
| Feb 25, 2020 at 0:06 | comment | added | user267839 | Why it is $2$-dimensional? Indeed, it is generated by pullbacks of the $X_i \in H^0(\mathbb{P}^2, \mathcal{O}(1)_{\mathbb{P}^2})=kX_0 \oplus kX_1 \oplus kX_2$. After prozectivizing $<\pi^*X_i \vert i=0,1,2>_k-\{0\}/k^*$ is at most $3-1=2$ dimensional. So why for example it cannot be $1$ dimensional? | |
| Feb 25, 2020 at 0:06 | comment | added | user267839 | When you talking about "dimension" of sub-linear system of cls $\vert L \vert$ you mean the "projective" dimension, in sense of "dimension" of projective subspace of $H^0(C,L) - \{0\}/k^*$, not as $k$-subspace of $H^0(C,L)$, right? Question: Why the projection $\pi: C \to C'$ is induced by a $2$-dimensional sub-linear system? I assumed that $\pi$ arises from the composition $ C \subset \mathbb{P}^n \dashrightarrow \mathbb{P}^2$, right? And the space defining $\pi$ we obtain by taking the linear $k$-subspace $<\pi^*X_i \vert i=0,1,2>_k \subset H^0(C, L)$ and then projectivize it. | |
| Feb 24, 2020 at 19:13 | vote | accept | user267839 | ||
| Feb 24, 2020 at 21:26 | |||||
| Feb 24, 2020 at 18:52 | history | answered | Francesco Polizzi | CC BY-SA 4.0 |