Timeline for Universal homotheties for elliptic curves
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| S May 2, 2019 at 8:23 | history | bounty ended | Daniel Litt | ||
| S May 2, 2019 at 8:23 | history | notice removed | Daniel Litt | ||
| Apr 28, 2019 at 13:50 | answer | added | Chris Wuthrich | timeline score: 3 | |
| Apr 28, 2019 at 10:08 | comment | added | naf | I think the uniform boundedness statement should be pretty straightforward, at least for large primes $l$. In this case it seems very likely that the modular curve corresponding to any maximal subgroup of $PGL_2(Z/lZ)$ has genus $>1$ so by Faltings you are reduced to elliptic curves with finitely many $j$-invariants. Any twist of an elliptic curve becomes isomorphic to it over an extension of degree $2$ (for non CM curves) so the boundedness follows easily from Serre's open image theorem. For CM curves some other argument is needed... | |
| Apr 28, 2019 at 9:33 | comment | added | Daniel Litt | @ulrich: OK, I think I agree that this is enough; I'll think a bit to see if I think this uniform boundedness statement is provable. Thank you! Put another way, you argue that if $K$ is the fixed field of the kernel of the map $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \prod G_E$, then the uniform boundedness statement implies that the cyclotomic character restricted to $\text{Gal}(\overline{\mathbb{Q}}/K)$ has infinite image, which is evidently enough to conclude the desired result. | |
| Apr 27, 2019 at 5:20 | comment | added | naf | it is trivial in each $G_E$ so acts by a scalar on $T_l(E)$ and the scalar is not a root of unity because of the infinite order condition. As a final comment, I don't think you need Bogomolov's theorem for the case of finitely many curves. It should be possible to use CM theory and Serre's results for non CM curves to see this. | |
| Apr 27, 2019 at 5:13 | comment | added | naf | For the abelianisations, the idea is to look at the intersection, call it M, of all the fields inside a fixed algebraic closure of $\mathbb{Q}$ cut out by the groups $G_E$ as defined above. It seems likely that the uniform boundedness of the abelinisations should imply that the image of the $l$-adic cyclotomic character restricted to the absolute Galois group of $M$ is an open subgroup; if true, then any element $\sigma$ mapping to an element of infinite order has the desired property: | |
| Apr 27, 2019 at 5:04 | comment | added | naf | OK, here are some more details but I am not claiming that I have worked out a complete proof, and don't have time right now to think about this carefully, so make of it what you will: Firstly, I don't think the problem with CM curves is a serious issue. There are only finitely many CM elliptic curves defined over any fixed number field (of course, there are inifinitely many twists...) and one knows the Galois representations explicitly so these could probably be handled separately. | |
| Apr 26, 2019 at 14:17 | comment | added | Daniel Litt | @ulrich: This still sounds fishy to me. How can you conclude anything about the existence of a $\sigma$ trivial in $G_E$ just by knowing something about the abelianization? Could you sketch the argument you have in mind in a bit more detail? I'm also skeptical that Faltings could imply any uniform boundedness statement of this type, since it is false for CM curves, as you remark (how would an application of Faltings know which curves were CM?). | |
| Apr 26, 2019 at 4:24 | comment | added | naf | I haven't thought about this very carefully, but it seems to me that uniform boundedness of the abelianisations would imply that there exists an element $\sigma$ of the Galois group that maps trivially to each $G_E$ but maps to an element of infinite order in the maximal abelian $l$-primary quotient of the absolute Galois group of $\mathbb{Q}$. Such a $\sigma$ would satisfy the condition of the first question (again ignoring CM curves for simplicity). | |
| Apr 25, 2019 at 20:48 | history | edited | Daniel Litt | CC BY-SA 4.0 |
added 13 characters in body
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| Apr 25, 2019 at 13:55 | comment | added | Daniel Litt | @ulrich: Thanks for your comment -- can you say a bit more? In particular, I don't see why uniform boundedness of the abelianization is enough. | |
| Apr 25, 2019 at 5:37 | comment | added | naf | For an elliptic curve $E$ over $\mathbb{Q}$ let $G_E$ be the image of the Galois group in $PGL_2(\mathbb{Z}_l)$. Your first question seems to be more or less equivalent to the condition that the abelianisation of $G_E$ should have order bounded independently of $E$. For large $l$, the abelianisation is expected to be trivial (say we only consider non CM curves for simplicity) but one should be able to prove boundedness for all $l$ by looking at suitable modular curves, computing their genus and then applying Faltings's theorem. | |
| S Apr 24, 2019 at 19:39 | history | bounty started | Daniel Litt | ||
| S Apr 24, 2019 at 19:39 | history | notice added | Daniel Litt | Draw attention | |
| Apr 23, 2019 at 15:00 | history | edited | Daniel Litt | CC BY-SA 4.0 |
grammar and slight clarification
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| Apr 22, 2019 at 16:10 | history | asked | Daniel Litt | CC BY-SA 4.0 |