Timeline for Are there any quadratic functions on an abelian variety not from the height machine?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 17, 2022 at 19:14 | vote | accept | user unknown | ||
| Oct 17, 2022 at 19:14 | comment | added | user unknown | I see! So the point is, a linear function on $\mathbb{R}^\mathbb{N}$ is basically an infinite real vector with no restriction on the coordinates. The space of linear functions is not a direct sum but a direct product. That's very helpful. Thanks a lot! | |
| Oct 17, 2022 at 18:28 | comment | added | Will Sawin | @userunknown If it had a countable basis, it could have only countably many linearly independent vectors (at most $n$ using each set of $n$ basis vectors, and there are countably many sets of $n$ basis vectors for all $n$) but in fact the vectors $(1,t,t^2,\dots)$ are linearly independent for all real $t$. | |
| Oct 17, 2022 at 18:25 | comment | added | Will Sawin | @userunknown Yes, it's also true for linear functions. Exactly for the reason you say: A linear function on a countable-dimensional vector space is determined by countably many values, i.e. it's the product $\mathbb R^{\mathbb N}$ of countably many copies of $\mathbb R$. But that product does not have countable dimension, i.e. it lacks a countable basis. | |
| Oct 17, 2022 at 17:49 | comment | added | user unknown | Actually I'm not quite sure why this second point is correct: if we take a (countable) real basis of $X(\bar K)\otimes_\mathbb{Z} \mathbb{R}$, the quadratic function as a sum of linear form and quadratic form, with the quadratic form determined by the induced inner product, seems that it can be determined by countably many values. Why is it uncountable? | |
| Oct 17, 2022 at 17:17 | comment | added | user unknown | Thanks a lot! I know very little about functional analysis; is there an easy way to think about this fact: "the space of quadratic functions on a countably-infinite-dimensional vector space has uncountable dimension"? Is it also true for linear functions? | |
| Oct 17, 2022 at 12:59 | comment | added | Will Sawin | @userunknown See edited version. | |
| Oct 17, 2022 at 12:58 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| Oct 17, 2022 at 2:43 | comment | added | user unknown | Sorry I should have been more clear on it. I really mean the map from the infinite dimensional space to the infinite dimensional space. The quadratic function here could contain a linear part, so for example, an algebraically trivial line bundle can give a nonzero linear function. The reference I'm looking at is Lang's Fundamentals of Diophantine geometry, chapter 5. | |
| Oct 17, 2022 at 2:28 | history | answered | Will Sawin | CC BY-SA 4.0 |