Timeline for Is $h^1(X,O_X)$ always equal to the dimension of the Albanese?
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6 events
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| Jul 21, 2019 at 12:23 | comment | added | naf | Yes, if $X$ is normal then both questions have a positive answer. It is not true though that the Albanese is $Pic^0$ (even if $X$ is smooth); it is the dual abelian variety. | |
| Jul 21, 2019 at 8:25 | comment | added | Harry | @ulrich Thank you for your comment. Don't know how I overlooked that.... The answer to both questions is positive if $X$ is normal, right? For the first question, by Albanese I mean of $X$ I mean the data of a semi-abelian variety $G$ and a universal map $X\to G$. But since $X$ is proper, $G$ will be an abelian variety. | |
| Jul 21, 2019 at 8:21 | comment | added | Harry | @WillSawin In Corollary A.11 of the linked paper it is claimed that every (pointed) variety $X$ has an Albanese morphism. This is defined as the data of a semi-abelian variety $G$ and a universal morphism $X\to G$. However, as $X$ is proper, the Albanese variety $G$ will be (in this case) an abelian variety. In the case that $X$ is proper over $\mathbb{C}$, this coincides with $Pic^0_{red} = Pic^0$. | |
| Jul 21, 2019 at 7:28 | comment | added | naf | The answer to the first question depends on what you mean by Albanese. The second question has a negative answer: consider a nodal or cuspidal rational curve. | |
| Jul 21, 2019 at 6:04 | comment | added | Will Sawin | Are you assuming $X$ normal, or not? If not, what's the definition of the Albanese? | |
| Jul 20, 2019 at 23:46 | history | asked | Harry | CC BY-SA 4.0 |