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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.

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    $\begingroup$ 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. $\endgroup$
    – naf
    Apr 25 '19 at 5:37
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    $\begingroup$ 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). $\endgroup$
    – naf
    Apr 26 '19 at 4:24
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    $\begingroup$ 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. $\endgroup$
    – naf
    Apr 27 '19 at 5:04
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    $\begingroup$ 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: $\endgroup$
    – naf
    Apr 27 '19 at 5:13
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    $\begingroup$ 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. $\endgroup$
    – naf
    Apr 27 '19 at 5:20
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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.

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  • $\begingroup$ 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... $\endgroup$ Apr 28 '19 at 15:30

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