Timeline for Picard group of a cubic hypersurface
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 6, 2021 at 14:56 | vote | accept | CommunityBot | ||
| Jan 18, 2021 at 9:08 | history | edited | abx | CC BY-SA 4.0 |
added 343 characters in body; Post Made Community Wiki
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| Jan 17, 2021 at 14:00 | comment | added | Puzzled | One can cut out $2D$ on $X\setminus V$ intersecting $X$ with a quadric hypersurface. So it seems that there is a $2$-torsion divisor in $\text{Pic}(X\setminus V)$. | |
| Jan 17, 2021 at 13:58 | comment | added | Puzzled | @ abx. Consider the divisor $D$ in $X$ defined by $\{Z_2Z_3-Z_1Z_4 = Z_1Z_2-Z_0Z_4 = Z_1^2-Z_0Z_3 = 0\}$. This is a cone with vertex $[0:\dots:0:1]$ over a $2$-dimensional cubic scroll $S$ contained in $\{Z_5=0\}$. The restriction of $D$ to $X\setminus V$ is Cartier. If $D = X\cap H$ for some hypersurface $H\subset\mathbb{P}^5$ then $H$ must be a hyperplane. On the other hand, if $H$ is a hyperplane such that $H\cap X = D$ then $H$ must be the hyperplane generated by $S$ which does not contain the vetrtex of $D$. So, it seems that $D$ can not be cut out on $X\setminus V$ by any hypersurface. | |
| Dec 21, 2020 at 12:58 | comment | added | abx | I am talking about Proposition 3.3, p. 101, same version. | |
| Dec 21, 2020 at 9:55 | comment | added | user168611 | I am looking at this arxiv version arxiv.org/pdf/math/0511279.pdf. At page 101 they say that $Pic(X)\rightarrow Pic(X\setminus V)$ is surjective but they do not claim that it is an isomorphism. | |
| Dec 21, 2020 at 5:42 | comment | added | abx | This is Proposition 3.3 of SGA 2, Exposé XI. | |
| Dec 20, 2020 at 23:17 | comment | added | user168611 | Sorry, I found the Theorem in SGA2 saying that the local rings of $X$ are parafactorial but why does this imply that $Pic(X)\cong Pic(X\setminus V)$? | |
| Dec 14, 2020 at 17:53 | comment | added | user168611 | Ok. So $Pic(X)\cong\mathbb{Z}$ and it's generated by the hyperplane section. Thank you. | |
| Dec 14, 2020 at 17:40 | comment | added | abx | No power of $\mathscr{O}(1)$ is trivial (Lefschetz gives an isomorphism $\operatorname{Pic}(\mathbb{P}^5)\rightarrow \operatorname{Pic}(X)$). And yes, you can probably give a proof using that $X\smallsetminus V$ is homogeneous. | |
| Dec 14, 2020 at 17:15 | comment | added | user168611 | Thank you very much for the answer. Is it clear what is the smallest power of $\mathcal{O}(1)$ that is trivial? Do you know if there is an alternative argument not using the fact $X$ is a complete intersection? For instance, using the fact that $X\setminus V$ is an orbit of an algebraic group. | |
| Dec 14, 2020 at 16:39 | history | answered | abx | CC BY-SA 4.0 |