Timeline for Nef divisors on abelian varieties are pullbacks of ample ones
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9 events
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| Feb 24, 2020 at 11:55 | comment | added | TartagliaTriangle | @PiotrAchinger, I don't understand how the divisor descends to quotient and why it should be ample. Also, why $T(L)$ has to be finite? $L$ is not supposed to be ample. | |
| Feb 23, 2020 at 17:26 | comment | added | TartagliaTriangle | Thanks to YosemiteStan and PiotrAchinger for their useful comments. Can you write them as an answer, so I can accept them? | |
| Feb 21, 2020 at 20:22 | comment | added | Piotr Achinger | Thinking more algebraically, it seems that the quotient should be taken by (the connected component of) $T(L) = \{ x\in A \,:\, T_x^* L \cong L\}$. Indeed, the restriction of $L$ to $T(L)^\circ$ lies in its $\mathrm{Pic}^0$, so by twisting by an element of $\mathrm{Pic}^0(A)$ one should be able to ensure $L$ trivial on $T(L)^\circ$. Then it descends to the quotient and is ample there (I think finiteness of $T(L)$ will ensure this). | |
| Feb 21, 2020 at 18:30 | comment | added | Yosemite Stan | ... A minor annoyance is that one might not be able to descend the $\alpha$ in the Appell-Humbert theorem (e.g. in @wnx's answer - indeed this is true for "most" strictly nef line bundles with a given $c_1$). However, if you only want up to topological equivalence, i.e. to say that there is a quotient $A\rightarrow B$ such that $D$ is the image of some ample divisor under $\mathrm{NS}(B)\rightarrow \mathrm{NS}(A)$ then you should be okay I think. | |
| Feb 21, 2020 at 18:23 | comment | added | Yosemite Stan | If you work in the complex setting I think the Appel-Humbert theorem should give you essentially what you want (en.wikipedia.org/wiki/Appell%E2%80%93Humbert_theorem). The only caveat, is that you might have to work "up to translation." Nefness I think corresponds to positive semidefiniteness of the Hermitian form H in the Appell-Humbert theorem. Then the kernel of this form should define a sublattice corresponding to a subcomplex torus. If you quotient by this subtorus, I think it should be straightforward to check that the Hermitian form "descends" and is ample on that quotient torus. | |
| Feb 21, 2020 at 17:40 | review | Close votes | |||
| Feb 28, 2020 at 18:20 | |||||
| Feb 21, 2020 at 15:19 | comment | added | Ennio Mori cone | No, take a generic degree 0 divisor on an elliptic curve. | |
| Feb 21, 2020 at 13:42 | history | edited | TartagliaTriangle | CC BY-SA 4.0 |
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| Feb 21, 2020 at 11:54 | history | asked | TartagliaTriangle | CC BY-SA 4.0 |