Timeline for Is there a geography of Hodge numbers for minimal general type algebraic surfaces?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 2, 2022 at 20:08 | comment | added | Jef | See superficie.info for a nice visual tool and many examples! | |
| Jun 1, 2022 at 21:15 | comment | added | Hacon | I think that $2h^{0,1}-4\leq h^{0,1}\leq h^{0,2}$ is optimal.... Since $\chi \geq 1$, we have $h^{0,1}\leq h^{0,2}$, and by a result of Beauville MR0688038 we have $h^{0,2}\geq 2h^{0,1}-4$ and Debarre MR0688038 shows that $K^2\geq 2h^{0,2}+2(h^{0,1}-4)$. (In the maximal Albanese case Pardini shows the optimal $c_1^2\geq 4\chi $). Clearly $h^{0,1}$ and $h^{0,2}$ are unbounded and of course if we fix $c_1^2$, we have a bounded family of surfaces and hence $h^{0,1}$ and $h^{0,2}$ are bounded. Maybe we should be then asking about bounds for $h^{0,i}/c_1^2$ (as the one given by Debarre). | |
| Jun 1, 2022 at 19:13 | comment | added | Jason Starr | You are correct. My formula is already wrong for $\mathbb{CP}^2$. Sorry about the mistake. | |
| Jun 1, 2022 at 18:57 | comment | added | Will Chen | @JasonStarr Thanks for your questions, I've clarified a bit in the OP. Also I think your formula should be $h^{1,1} - 2h^{1,0} = (5c_2 - c_1^2)/6$? | |
| Jun 1, 2022 at 18:49 | history | edited | Will Chen | CC BY-SA 4.0 |
Clarified what I meant by "relations" in the first question.
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| Jun 1, 2022 at 17:53 | comment | added | Jason Starr | Very similar to my previous comment, you also have that $h^{1,1}-2h^{1,0}$ equals $(11c_2-c_1^2)/6$. This is nonnegative by the Bogomolov-Miyaoka-Yau inequality. | |
| Jun 1, 2022 at 17:46 | comment | added | Jason Starr | "My naive question is -- Are there any known relations that must be satisfied for the Hodge number $h^{0,1}$, $h^{0,2}$, $h^{1,1}$ of MSPGT surfaces?" Just to clarify, you are asking for relations in addition to Noether's formula: $1-h^{0,1}+h^{0,2} = (c_1^2+c_2)/12$ (which in your case is nonnegative). Is it correct that you already include Noether's formula in the list of relations that you are considering? | |
| Jun 1, 2022 at 14:11 | comment | added | Will Sawin | I think the general philosophy is that $h^{0,1}$ being zero is the most common situation, which is why $h^{0,1}$ is called the "irregularity". So for any $c_1^2, c_2$ that can occur, it most likely can be satisfied with $h^{0,1}=0$, and there's perhaps a lesser chance it can be satisfied with $h^{0,1}$ larger. | |
| Jun 1, 2022 at 14:05 | history | asked | Will Chen | CC BY-SA 4.0 |