# Universal homotheties for elliptic curves

Let $$K$$ be a number field and $$E_1, \cdots, E_n$$ elliptic curves over $$K$$. Let $$\ell$$ be a prime. Then there exists an element $$\sigma \in \text{Gal}(\overline{K}/K)$$ such that $$\sigma$$ acts on $$T_\ell(E_i)$$ via a non-root-of-unity scalar for all $$i$$. (Here $$T_\ell(E_i)$$ denotes the $$\ell$$-adic Tate module of $$E_i$$.)

This is a result of Bogomolov, who shows that for any Abelian variety $$A/K$$, there exists an element of $$\text{Gal}(\overline{K}/K)$$ which acts on $$T_\ell(A)$$ via a non-root-of-unity scalar; applying this result to $$\prod_i E_i$$ gives the claim.

I would like to know if an analogous statement is true for infinite sets of elliptic curves. In particular:

Fix a prime $$\ell$$. Does there exist an element $$\sigma\in \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$$ such that for all elliptic curves $$E/\mathbb{Q}$$, $$\sigma$$ acts on $$T_\ell(E)$$ via a non-root-of-unity scalar?

I am also interested in the following stronger variant:

Fix a prime $$\ell$$. Does there exist an element $$\sigma\in \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$$ such that for all number fields $$K\subset \overline{\mathbb{Q}}$$ and all elliptic curves $$E/K$$, there exists an integer $$N$$ such that $$\sigma^N$$ acts via a non-root-of-unity scalar on $$T_\ell(E)$$?

Note that the above makes sense because for $$N$$ sufficiently divisible, $$\sigma^N\in \text{Gal}(\overline{\mathbb{Q}}/K)$$.

Of course this question is related to uniform boundedness conjectures, but my hope is that it can be answered independently from them.

• 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. – ulrich Apr 25 '19 at 5:37
• @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. – Daniel Litt Apr 25 '19 at 13:55
• 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). – ulrich Apr 26 '19 at 4:24
• @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?). – Daniel Litt Apr 26 '19 at 14:17
• 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. – ulrich Apr 27 '19 at 5:04

Fix the prime $$\ell$$. I hope: There exists a constant $$C$$ such that for any elliptic curve $$E/\mathbb{Q}$$ the image of the Galois representation $$\rho\colon \operatorname{Gal} ( \bar{\mathbb{Q}}/\mathbb{Q}) \to \operatorname{GL}_2(\mathbb{Z}_{\ell})$$ contains $$\bigl\{x^C \bigm\vert x\in \mathbb{Z}_{\ell}^{\times}\bigr\}$$.

Carola Eckstein in her thesis http://cds.cern.ch/record/897530/files/cer-002567978.pdf answers this partially. If $$E$$ has complex multiplication then $$C=54$$ will do. If $$\ell>163$$ then $$C=12$$ is ok. Agnès David https://arxiv.org/abs/1007.4725 makes further progress and shows that for $$\ell>23$$ and different from $$37$$, $$43$$, $$67$$ and $$163$$ then $$C=2$$ works. I don't know more about the question for smaller $$\ell$$, but as ulrich hints in his comments above, one should be able to prove the existence of $$C$$.

Suppose there is such a $$C$$. Given an elliptic curve $$E/\mathbb{Q}$$, let $$K_E$$ be the field fixed by the image of $$\operatorname{im}(\rho)\to \operatorname{PGL}_2(\mathbb{Z}_\ell)$$ (as in ulrich's comment). Set $$K_\infty = \mathbb{Q}(\mu_{\ell^{\infty}})$$. The image of the homotheties under $$\det:\operatorname{im}(\rho) \to \operatorname{Gal}(K_\infty/\mathbb{Q})\cong \mathbb{Z}_{\ell}^{\times}$$ will contain all $$2C$$-th powers. Therefore there is a finite extension $$F/\mathbb{Q}$$ inside $$K_\infty$$ such that $$K_E\cap K_\infty\subset F$$ for all elliptic curves $$E$$.

In particular the compositum $$\mathcal{K}$$ of any (or all) $$K_E$$ intersects $$K_\infty$$ in $$F$$ and hence any element of $$\operatorname{Gal}(\bar{\mathbb{Q}}/\mathcal{K})$$ that maps to an element $$\sigma$$ of infinite order in $$\operatorname{Gal}(K_\infty/F)$$ will work.

The reference above will also show that this works for many $$\ell$$ over any fixed number field instead of $$\mathbb{Q}$$.

I have no answer for the second question, yet.

• I think the existence of such a C should follow from work of Cadoret-Tamagawa, in a fashion more or less analogous to what ulrich sketches. I’ll try to check the details, and see if this can be used to do my second question... – Daniel Litt Apr 28 '19 at 15:30