Timeline for Are most Kähler manifolds non-projective?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Dec 15, 2016 at 13:22 | comment | added | Donu Arapura | OK, I'll take a look when I get a chance. | |
| Dec 15, 2016 at 12:54 | comment | added | Jorge Vitório Pereira | See mathoverflow.net/questions/231144/… | |
| Dec 14, 2016 at 23:38 | comment | added | Jorge Vitório Pereira | I don't see why the algebraic dimension of $S$ is one. As a matter of fact, i think it can be equal to two in some examples. There are foliations on the product $C\times E$ transverse to $\pi$ and without algebraic leaves. This leads to a presentation of the product as a suspension with infinite $h$. | |
| Dec 14, 2016 at 17:58 | comment | added | ACL | Addendum for my own edification: since $E$ is commutative, morphisms $h\colon \Gamma\to E$ factor through the abelianization $\Gamma^{\rm ab}$ of $\Gamma$. The presentation of $\Gamma$ or the Hurewicz theorem show that the group $\Gamma^{\rm ab}$ is isomorphic to $\mathbf Z^{2g}$, hence is free abelian, and non trivial. Consequently, there are plenty of morphisms $h\colon\Gamma \to E$ with infinite image. | |
| Dec 14, 2016 at 17:21 | history | edited | Donu Arapura | CC BY-SA 3.0 |
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| Dec 14, 2016 at 14:55 | history | edited | Donu Arapura | CC BY-SA 3.0 |
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| Dec 13, 2016 at 21:08 | history | answered | Donu Arapura | CC BY-SA 3.0 |