Timeline for Examples of degree zero, rank one reflexive sheaves without r-th roots
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Oct 14, 2018 at 15:08 | comment | added | Chen | @JasonStarr Thank you, now I understand. By degree, I still mean the definition I wrote. Is there some general strategy for producing such examples. I understand (like in the case of quadrics), the Picard rank greater than $1$ gives rise to such examples. Could you also give an example of a reflexive sheaf of rank one, degree zero but not invertible and does not have an r-th root. | |
Oct 14, 2018 at 15:02 | comment | added | Jason Starr | "I am missing something." With the definition that you wrote above, certainly every $k$-point of $\text{Pic}^0(X)$ does give a degree zero invertible sheaf. However, with the definition that you wrote above, there are also degree zero invertible sheaves that are not algebraically equivalent to the structure sheaf. For instance, the invertible sheaf that I have twice listed in the comments above has degree zero with respect to the fixed polarization, yet it is not algebraically equivalent to zero. | |
Oct 14, 2018 at 14:58 | comment | added | Chen | @JasonStarr I am missing something. I want $k$-points of $Pic^0(X)$ to be "degree zero" invertible sheaves. | |
Oct 14, 2018 at 14:52 | comment | added | Jason Starr | "Moreover, if $X$ is smooth, then the Picard variety is an abelian variety ..." This is not correct. According to your definition, the group of isomorphism classes of degree zero, rank one reflexive sheaves on a quadric hypersurface $X$ in $\mathbb{P}^3$ is a free cyclic group generated by the class of $\text{pr}_1^*\mathcal{O}(1)\otimes \text{pr}_2^*\mathcal{O}(-1)$. It is not the set of $k$-points of an Abelian variety. It is not divisible. The example I wrote in my comment above is a counterexample for every integer $r>1$. | |
Oct 14, 2018 at 14:44 | comment | added | Chen | @JasonStarr My definition of degree is the same as the one in Huybrects-Lehn page 13. I have edited the question to fix a polarisation at the beginning. If I understand correctly, your example is of degree zero. Moreover, if $X$ is smooth, then the Picard variety is an abelian variety so, any invertible sheaf has an $r$-th root. So, my question is not interesting if $X$ is smooth. | |
Oct 14, 2018 at 14:34 | history | edited | Chen | CC BY-SA 4.0 |
added 65 characters in body
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Oct 14, 2018 at 14:33 | comment | added | Jason Starr | What is your definition of "degree zero sheaf"? For a smooth quadric hypersurface $X$ in $\mathbb{P}^3$ that is isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$, do you consider $\text{pr}_1^*\mathcal{O}(1)\otimes \text{pr}_2^*\mathcal{O}(-1)$ to have degree zero? | |
Oct 14, 2018 at 14:30 | history | asked | Chen | CC BY-SA 4.0 |