Timeline for Reconstruct a variety from the category of locally free sheaves
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 15, 2022 at 17:02 | comment | added | SashaP | @MartinBrandenburg Sorry which exactly statement you feel is false? I think it is true that if $Vect(X)\simeq Vect(\mathbb{P^n})$ then $X\simeq\mathbb{P^n}$ as sketched in my previous comments. This should more generally work for any variety with Neron-Sever rank 1 in place of $\mathbb{P}^n$. | |
| Nov 15, 2022 at 11:25 | comment | added | Martin Brandenburg | Oh, I see. Actually my gut feeling is that the answer is No. Already for $\mathbb{P}^1$. | |
| Nov 15, 2022 at 10:51 | comment | added | SashaP | Though I realized that my general argument is not quite correct: I was too hasty to claim that we can figure out which of the briationallly equivalent varieties $R(X,L)$ is $X$ itself, I edited the answer accordingly, so the original question is still open. | |
| Nov 15, 2022 at 10:50 | comment | added | SashaP | @MartinBrandenburg If $L$ is an ample line bundle on a projective variety $X$ then $X$ is isomorphic to $Proj \oplus H^0(X,L^n)$, see e.g. stacks.math.columbia.edu/tag/01Q1 Hence if we are able to prove that an equivalence $Vect(X)\simeq Vect(Y)$ carries an ample line bundle $L$ on $X$ to an ample line bundle $M$ on $Y$ then $X\simeq Proj \oplus H^0(X,L^n)\simeq Proj \oplus H^0(Y,M^n)\simeq Y$. This is the situation in this example because an ample line bundle on $X$ is necessarily carried to an ample line bundle on $\mathbb{P}^n$. | |
| Nov 15, 2022 at 10:39 | history | edited | SashaP | CC BY-SA 4.0 |
added 292 characters in body
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| Nov 15, 2022 at 2:10 | comment | added | Martin Brandenburg | Why is X isomorphic to P^n here? Sorry for the stupid question. | |
| Nov 15, 2022 at 1:35 | comment | added | SashaP | @MartinBrandenburg Maybe it is helpful to say that for your example of a variety $X$ with $Vect(X)=Vect(\mathbb{P}^n)$ what I'm suggesting comes down to simply turning your attempted argument around. Let's choose a very ample line bundle $L$ on $X$, it is carried to some $O(k)$ under this equivalence. The graded rings corresponding to $L$ and to $O(k)$ are isomorphic so we necessarily have $k>0$, and this proves that $X\simeq \mathbb{P}^n$ because $O(k)$ is very ample. | |
| Nov 15, 2022 at 1:18 | comment | added | Martin Brandenburg | Unfortunately I lack the knowledge about algebraic geometry to understand this properly. But it seems that you overcome the problem that the property of being ample cannot be formulated within the category $\mathbf{Vect}(X)$ (again, the epimorphism issue!) by finding other conditions on the associated graded ring? | |
| Nov 15, 2022 at 0:52 | history | edited | LSpice | CC BY-SA 4.0 |
`\mathrm` -> `\operatorname`, and -> `\mathit`
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| Nov 14, 2022 at 23:36 | history | answered | SashaP | CC BY-SA 4.0 |