Timeline for (1,1)-form that does not come from a divisor
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 20, 2021 at 14:43 | comment | added | Daniel Litt | Just for concreteness — in David Speyer’s example, the two-form $dz_1\wedge d\overline{z_2}$ is not in the span of the image of the cycle class map, where $dz_i$ is the pullback of any nonzero invariant differential form on $E_i$. | |
| Feb 19, 2021 at 14:31 | comment | added | David E Speyer | A simple example is the product of two non-isogenous elliptic curves, $E_1$ and $E_2$. Then $H^{(1,1)}(E_1 \times E_2)$ is $4$-dimensional. If $E_1$ and $E_2$ are not isogenous, then $\mathrm{Pic}(E_1 \times E_2)$ is only two dimensional, spanned by the classes of $E_1 \times \{ \mathrm{point} \}$ and $\{ \mathrm{point} \} \times E_2$. | |
| Feb 19, 2021 at 14:28 | comment | added | Todd Trimble | Let's please not rush to close this, before giving this relative newcomer a chance to respond. This may yet elicit a good answer. | |
| Feb 19, 2021 at 6:51 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
added 36 characters in body
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| Feb 19, 2021 at 5:34 | review | Close votes | |||
| Feb 24, 2021 at 3:03 | |||||
| Feb 18, 2021 at 23:28 | comment | added | R. van Dobben de Bruyn | As soon as $\operatorname{rk} \operatorname{NS}(M) < \dim H^{1,1}(M)$, you will have many such classes. This happens for example on a K3 surface of Picard rank less than $20$. Can you say a little more what you mean by "interesting"? | |
| Feb 18, 2021 at 23:01 | history | asked | user145752 | CC BY-SA 4.0 |