Timeline for Complex manifold defined over $\mathbb{Q}$
Current License: CC BY-SA 4.0
27 events
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| Oct 7, 2020 at 16:50 | history | edited | user164740 | CC BY-SA 4.0 |
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| Oct 6, 2020 at 20:43 | history | edited | user164740 | CC BY-SA 4.0 |
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| Oct 6, 2020 at 17:41 | comment | added | François Brunault | It seems to me that every complex torus satisfies your condition. Because you only ask for a single point in $U_i \cap U_j$, while this open set may not be connected. | |
| Oct 6, 2020 at 17:20 | history | edited | user164740 | CC BY-SA 4.0 |
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| Oct 6, 2020 at 12:19 | history | edited | user164740 | CC BY-SA 4.0 |
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| Oct 6, 2020 at 12:18 | history | undeleted | user164740 | ||
| Oct 6, 2020 at 11:30 | history | deleted | user164740 | via Vote | |
| Oct 6, 2020 at 11:29 | answer | added | Robert Bryant | timeline score: 3 | |
| Oct 3, 2020 at 6:41 | history | edited | user164740 | CC BY-SA 4.0 |
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| Oct 3, 2020 at 4:59 | comment | added | reuns | @FrançoisBrunault Otherwise we could say that $\Bbb{P^1(C)}$ and $\Bbb{C/(Z+iZ)}$ are defined over $K=\Bbb{Q}$ and $K=\Bbb{Q}(i)$ respectively because there are finitely many charts $\phi_j:U_j\to X$ such that $\phi_i^{-1}\circ \phi_j$ sends $K\cap U_j\to K\cap U_i$. Over what field is defined $y^2=x^3+5x+1$ ? (no CM) | |
| Oct 1, 2020 at 9:45 | history | edited | user164740 | CC BY-SA 4.0 |
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| Sep 29, 2020 at 21:59 | history | edited | user164740 | CC BY-SA 4.0 |
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| Sep 29, 2020 at 20:58 | history | edited | user164740 | CC BY-SA 4.0 |
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| Sep 29, 2020 at 20:48 | history | edited | user164740 | CC BY-SA 4.0 |
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| Sep 29, 2020 at 20:43 | history | edited | user164740 | CC BY-SA 4.0 |
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| Sep 29, 2020 at 20:33 | history | edited | user164740 | CC BY-SA 4.0 |
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| Sep 29, 2020 at 19:39 | history | edited | user164740 | CC BY-SA 4.0 |
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| Sep 29, 2020 at 19:21 | comment | added | François Brunault | @JoeT Right. I just don't see another sensible definition. | |
| Sep 29, 2020 at 19:20 | comment | added | user164740 | @FrançoisBrunault then a generic complex torus is defined over $\mathbb{Q}$. | |
| Sep 29, 2020 at 19:19 | comment | added | François Brunault | One could ask that the field of meromorphic functions is the base change from $\mathbb{Q}$ to $\mathbb{C}$ of a function field $F$ over $\mathbb{Q}$ (and the field of constants of $F$ is $\mathbb{Q}$). (Of course such an $F$ is not unique, even in the algebraic case we cannot expect uniqueness.) | |
| Sep 29, 2020 at 17:02 | history | edited | user164740 | CC BY-SA 4.0 |
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| Sep 29, 2020 at 16:56 | comment | added | Jason Starr | A projective manifold defined over $\mathbb{Q}$ may have no "rational point", i.e., it may have no (complex) point whose coordinates on the charts of a "rational atlas" are all rational coordinates. | |
| Sep 29, 2020 at 16:53 | comment | added | Jason Starr | Taylor series with respect to what coordinates? The Taylor expansion of the constant polynomial $z$ about the point $z=\pi$ does not have rational coefficients. | |
| Sep 29, 2020 at 16:51 | comment | added | Will Sawin | Moishezon manifolds (compact complex manifolds whose field of meromorphic functions has the expected transcendence degree) are equivalent to proper algebraic spaces, so for them one could ask that the associated algebraic space is defined over $\mathbb Q$. | |
| Sep 29, 2020 at 16:48 | history | edited | user164740 | CC BY-SA 4.0 |
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| Sep 29, 2020 at 16:46 | comment | added | Chris | Any open subset of $\mathbb{C}^n$ satisfies your condition. You may need compactness to hope for something more interesting. | |
| Sep 29, 2020 at 16:19 | history | asked | user164740 | CC BY-SA 4.0 |