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I know the following theorems by Serre:

1, The 2-dim l-adic representation associated to a non-CM elliptic curve is open.

2, The 2-dim l-adic representation associated the weight-12 cusp form $\Delta$ has open image (even before Deligne's construction of 2-dim l-adic representations).

So is there any general theorem about When the image of a 2-dim l-adic representation associated to a modular form is open?

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    $\begingroup$ When the weight is bigger than 1 and it's not CM. Ribet, 70s. $\endgroup$
    – eric
    Oct 14 '15 at 6:19
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    $\begingroup$ In the weight two case, and more generally for abelian varieties with real multiplications, a simple proof is also possible with Faltings's isogeny theorem. $\endgroup$ Oct 14 '15 at 6:40
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    $\begingroup$ @eric Your claim, although apparently widely believed, is false. See my answer below $\endgroup$ Oct 14 '15 at 8:35
  • $\begingroup$ Many thanks @David Loeffler! Obviously I meant "open in the Mumford-Tate group" (or something) :-) [I am assuming that the point of Mumford-Tate groups is that they dig me out of this hole?] $\endgroup$
    – eric
    Oct 15 '15 at 8:34
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This is more subtle than it looks. I asked exactly the same question some years back (see here); but I'm not going to flag this question as duplicate, because the answer that was given to my question at the time, which was identical to @eric's comment above, was wrong.

The correct answer is this. Let $f$ be a normalized non-CM newform. Associated to $f$ there is a number field $L = \mathbf{Q}(a_n(f) : n \ge 1)$, and for each prime $\mathfrak{P}$ of $L$, we get a Galois representation $$ \rho_{f, \mathfrak{P}}: G_{\mathbf{Q}} \to \operatorname{GL}_2(L_{\mathfrak{P}}). $$

It is not, however, generally the case that this has open image. There is a trivial obstruction coming from the fact that the determinant of $\rho_{f, \mathfrak{P}}$ is the product of a finite-order character and a power of the cyclotomic character, so there is a finite-index subgroup of $G_{\mathbf{Q}}$ whose determinant lands in $\mathbf{Q}_p^\times$, which is generally much smaller than $L_{\mathfrak{P}}^\times$.

One can still hope that the image contains an open subgroup of $\operatorname{SL}_2(L_{\mathfrak{P}})$, but this also turns out to be false, for a much more subtle reason. Consider the subgroup $\Gamma \subseteq \operatorname{Aut}(L / \mathbf{Q})$ consisting of those automorphisms $\sigma$ for which $f^\sigma$ is equal to the twist of $f$ by some Dirichlet character ("inner twists"). The fixed field of $\Gamma$ is a subfield $F \subseteq L$, and Momose showed that there's a finite-index subgroup of $G_{\mathbf{Q}}$ such that for any $\tau$ in this subgroup $\rho_{f, \mathfrak{P}}(\tau)$ has trace in $F_{\mathfrak{q}}$, where $\mathfrak{q}$ is the prime of $F$ below $\mathfrak{P}$. So if $F_\mathfrak{q} \ne L_{\mathfrak{P}}$, which occurs for infinitely many $\mathfrak{P}$ if $F \ne L$, then the image of $\rho_{f, \mathfrak{P}}$ cannot contain an open subgroup of $\operatorname{SL}_2(L_{\mathfrak{P}})$.

One might still hope that the image at least contains an open subgroup of $\operatorname{SL}_2(F_{\mathfrak{q}})$ but this isn't generally true either -- there is an obstruction coming from a Brauer group, which can genuinely be nontrivial! But this weaker statement does, at least, hold for all but finitely many $\mathfrak{P}$, as was shown by Momose.

If $f$ has level 1, then the group $\Gamma$ of inner twists is always trivial, so $F = L$ and the image contains an open subgroup of $\operatorname{SL}_2(L_\mathfrak{P})$ for all $\mathfrak{P}$. More generally, if $f$ has trivial character it's "usually" the case that there are no nontrivial inner twists. But every non-CM form of non-trivial character has at least one nontrivial inner twist, coming from the action of complex conjugation, so this bad behaviour is unavoidable.

All of this is very nicely explained in Ribet's paper "On l-adic representations attached to modular forms. II" (Glasgow Math. J. 27, 1985, 185--194), which you can find on Ribet's website here.

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    $\begingroup$ But if $A$ is an abelian variety whose endomorphism ring is an order in a totally real field of dimension $\dim{A}$, so in particular an abelian variety of $\mathrm{GL}_2$ type, defining a two dimensional $\mathfrak{l}$-adic representation, it does follow by Faltings that that representation has open image (Is this right?). So the statement on open image must be true for weight two non-CM normalized newforms for $\Gamma_1(N)$? $\endgroup$ Oct 14 '15 at 9:04
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    $\begingroup$ (The older question/answer should be edited with a link to this one, or some mention of the right answer) $\endgroup$
    – Myshkin
    Oct 14 '15 at 9:05
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    $\begingroup$ I see, the determinental condition must be accounted for, so we can't say open image. But then, in the case I wrote about, it seems to me that we do get an open subgroup of $\mathrm{SL}_2(L_{\mathfrak{P}})$, with equality for all but finitely many $\mathfrak{P}$. Or I may be misreading page 4 of Ribet's review of Serre's book (Abelian $l$-adic representations). $\endgroup$ Oct 14 '15 at 9:35
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    $\begingroup$ I have added a CW answer to the older question pointing out to this one. Perhaps David could accept it instead of the other, to avoid risks of confusion. $\endgroup$
    – Joël
    Oct 14 '15 at 13:38
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    $\begingroup$ @VesselinDimitrov You say "abelian varieties with real multiplications (which is what weight two non-CM eigenforms for Γ1(N) give rise to)." This is not true; there are examples of mod forms giving rise to abelian varieties with quaternion multiplication, for instance. Even when the ab var does have endomorphism algebra an order in a totally real field, this field isn't generally the same as the field of coefficients of the form, which will not be totally real when the character is nontrivial. $\endgroup$ Oct 14 '15 at 20:03

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