Timeline for Any real algebraic variety is diffeomorphic to a real algebraic variety defined over $\mathbb{Q}$
Current License: CC BY-SA 4.0
26 events
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| Jun 11, 2020 at 22:09 | comment | added | Jonny Evans | Awesome, thanks @Chris! | |
| Jun 11, 2020 at 19:15 | comment | added | Chris | @JonnyEvans: the issue is settled in publications.ias.edu/sites/default/files/PUTangSp.pdf section 4.2. So there is a particular subvariety of the above $\mathbf{A}^h$, defined over $\mathbb{Q}$, such that varying the coefficients inside this subvariety defines honest deformations | |
| Jun 11, 2020 at 18:26 | comment | added | Chris | @JonnyEvans I'm starting to see your point about complete intersections and that my argument is definitely wrong: perturbing slightly the $F_i$'s is going to make the $dF_i$'s "as independent as possible" so it might cut down the dimension | |
| Jun 11, 2020 at 17:58 | comment | added | Jonny Evans | Perhaps you could fix it as follows? If you're working over a transcendental extension of Q like Q(pi) then you could basechange over the map Q(pi) --> Q(t) sending pi to another transcendental number t and get a diffeomorphic variety because the rings are basically the same (unlike if t is algebraic). Now do this for every transcendental t and you get a bunch of points in A^h which have limit points that are not transcendental because C\setminus\bar{Q} is dense in C, and at least one of these must correspond to a nonsingular variety because nonsingularity is a closed condition. | |
| Jun 11, 2020 at 17:56 | comment | added | Jonny Evans | @Chris Sorry, I misunderstood your comment, but am still confused. Imagine we're talking about twisted cubics (not complete intersections). Then generic points in A^h have empty preimage under this projection. How do you guarantee that the locus of points whose preimage is an honest deformation of X contains a \bar{Q}-point? What stops this locus being something like a line with slope pi? There seems to be something missing from the argument. | |
| Jun 11, 2020 at 9:53 | comment | added | Chris | @JonnyEvans: if I have a rigid variety then all the fibers close to $p$ are not only diffeomorphic but also biholomorphic (which implies that rigid varieties have models over $\bar{\mathbb{Q}}$). If my number $h$ is zero it means that the $F_i$'s are zero. | |
| Jun 10, 2020 at 17:45 | comment | added | Jonny Evans | @Chris But you can't necessarily perturb the coefficients of the F_i and stay in the same moduli space (it's not a complete intersection). What if you have a rigid variety defined over C: why should I be able to perturb the equations at all? (e.g. what if your number h equals zero?). This is, however, a discussion of a slightly different question, so perhaps it would be better to move the discussion to that question if you're interested. | |
| Jun 10, 2020 at 17:28 | history | edited | YCor |
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| Jun 10, 2020 at 15:34 | comment | added | Chris | @ulrich Thanks for pointing out the innacuracy. One can check the smoothness of the projection by applying the Jacobian criterion (correct me if I'm wrong) | |
| Jun 10, 2020 at 14:36 | comment | added | naf | @Chris: Your argument is not quite correct: the smoothness of a fibre does not imply that the morphism is smooth along the fibre. Of course, the statement is indeed true. | |
| Jun 10, 2020 at 13:54 | comment | added | user145520 | @Qfwfq I use the definition that it's a smooth proper scheme over $\mathrm{Spec}\:\mathbb{R}$. So the global sections of the affine line are polynomials. | |
| Jun 10, 2020 at 12:12 | comment | added | Chris | @JonnyEvans: say your smooth projective variety $X$ is given as the zero locus of $(F_1,\ldots, F_k)$ in $\mathbf{P}_{\mathbb{C}}^n$. Consider the coefficients of the $F_i$'s as indeterminates: this cuts a variety in some $\mathbf{A}^h\times\mathbf{P}^n$ and your original variety is the fiber over a point $p$ in $\mathbf{A}^h$ under projection to the first factor. Since $X$ is smooth, by openness of smoothness, the map is a topological fiber bundle over a neighborhood of $p$, which certainly contains a $\bar{\mathbb{Q}}$-point. | |
| Jun 10, 2020 at 12:00 | comment | added | Qfwfq | What's the definition of "real algebraic variety" you're using? For example, on the real line $X$, is $1/(1+x^2)$ an element of $\Gamma(X,\mathcal{O}_X)$ or not? | |
| Jun 10, 2020 at 5:53 | answer | added | David C | timeline score: 9 | |
| Jun 10, 2020 at 5:47 | comment | added | Jonny Evans | I had a related question here mathoverflow.net/questions/347739/… and I still don't understand why the answer should be positive when R is replaced by C and Q by \bar{Q} as Chris states above. | |
| Jun 10, 2020 at 5:15 | history | edited | user145520 | CC BY-SA 4.0 |
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| Jun 9, 2020 at 21:31 | comment | added | Will Sawin | That said I don't have another counterexample. | |
| Jun 9, 2020 at 21:30 | comment | added | Will Sawin | @vrz The issue is that perturbing the equations a small amount may lead to a much smaller-dimensional variety if the original equations have some redundancy, but that equations with redundancy are necessary (not everything is a complete intersection). There is no way the proof can be made to work. | |
| Jun 9, 2020 at 20:59 | comment | added | Will Sawin | That's not the issue at all, the problem is that "approximating the defining equations" is not at all useful for this question. | |
| Jun 9, 2020 at 19:27 | comment | added | Noam D. Elkies | Wait, if it's not defined over $\bf R$ how can it be a "real algebraic variety"? | |
| Jun 9, 2020 at 19:22 | history | edited | user145520 | CC BY-SA 4.0 |
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| Jun 9, 2020 at 19:20 | comment | added | Chris | I believe that Will's comment settles your question. As Moishe implicitly suggests, your question has a positive answer if you replace $\mathbb{Q}$ by $\bar{\mathbb{Q}}$. | |
| Jun 9, 2020 at 18:50 | history | edited | user145520 | CC BY-SA 4.0 |
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| Jun 9, 2020 at 18:34 | comment | added | Moishe Kohan | Do you mean defined over $\bar{{\mathbb Q}}$ rather than ${\mathbb Q}$? | |
| Jun 9, 2020 at 18:27 | comment | added | Will Sawin | Probably the fake projective planes, which have no deformations and are not even defined over $\mathbb R$, provide counterexamples. | |
| Jun 9, 2020 at 17:27 | history | asked | user145520 | CC BY-SA 4.0 |