# Do “most” modular forms have no extra twists?

Let $f$ be a modular form -- more specifically, a normalized new eigenform which is not of CM type.

We say $f$ has extra twists if there exists some $\sigma \in \operatorname{Aut}( \mathbf{C})$ such that the Galois conjugate $f^\sigma$ is equal to the twist of $f$ by some non-trivial Dirichlet character, so $a_n(f)^\sigma = \chi(n) a_n(f)$ for all $n$ coprime to the level.

Any newform of nontrivial character has at least one extra twist (because the complex conjugate $\bar{f}$ is a twist of $f$.) But browsing some tables suggests that extra twists are pretty unusual for newforms of trivial character.

• Do "most" non-CM newforms of trivial character (in some precisely quantifiable sense) have no extra twists?
• Do "most" newforms of non-trivial character have no extra twists other than the obvious one?

(This question is prompted by this earlier question, which is subtle precisely when extra twists exist.)

If a newform $f \in \mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N),\varepsilon)$ has an inner twist by some $\sigma \in \operatorname{Aut}(\mathbb{C})$, then $f^{\sigma}$ is a newform of the same level as $f$. Moreover, if $\varepsilon$ is trivial, then so is the nebentypus of $f^{\sigma}$ (see (3.8) of Ribet's paper), and so any inner twist must arise from a quadratic Dirichlet character.

Ribet's paper actually shows (see (3.9)) that if $N$ is squarefree, then there are no nontrivial inner twists. More precisely, he shows that if $N$ is squarefree and $\chi$ is a quadratic Dirichlet character, then the twist $f \otimes \chi$ cannot have level $N$ and trivial nebentypus.

This is not hard to see by a local argument: if $N$ is squarefree, then for any $p \mid N$, the local component $\pi_p$ of the associated automorphic representation must be isomorphic to $\mathrm{St}$, the Steinberg representation of conductor $p$ associated to the trivial character. Any nontrivial twist of $f$ leaving the nebentypus unchanged must necessarily be quadratic, but any twist of $\mathrm{St}$ by a ramified quadratic character has conductor at least $p^2$, not $p$ (and in fact exactly $p^2$ if $p$ is odd). Similarly, at any unramified place, the twist of an unramified principal series representation by a ramified quadratic character has conductor at least $p^2$. So the level of $f \otimes \chi$ is strictly greater than $N$.

But if $N$ is not squarefree, then this is no longer the case! Indeed, in a recent paper (Theorem 6.4), I proved the following result (well, technically I proved it for Maaß cusp forms, but it generalises in an obvious way to holomorphic cusp forms).

Fix a nonsquarefree odd integer $N$, and let $N' > 1$ be squarefree and such that $N'^2 \mid N$. Let $\mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N))_{\mathrm{nonmon}(\varepsilon_{\mathrm{quad}(N')})}$ denote the vector space spanned by newforms $f$ of weight $k$, level $N$, and trivial nebentypus such that $f$ does not have CM by the quadratic Dirichlet character $\varepsilon_{\mathrm{quad}(N')}$ modulo $N'$, but that the twist $f \otimes \varepsilon_{\mathrm{quad}(N')}$ is also a newform of level $N$ and trivial nebentypus. Then $\frac{\dim \mathcal{S}_{k}^{\mathrm{new}} (\Gamma_0(N))_{\mathrm{nonmon} (\varepsilon _{\mathrm{quad}(N')} )} }{\dim \mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N))} \sim \prod_{\substack{p \mid N' \\ p^2 \parallel N}} \left(1 - \frac{p}{p^2 - p - 1}\right)$ as $k$ tends to infinity over the even integers.

Furthermore, this still holds if we replace $\mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N))_{\mathrm{nonmon}(\varepsilon_{\mathrm{quad}(N')})}$ with $\bigcap_{\substack{N^* \mid N' \\ N^* > 1}} \mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N))_{\mathrm{nonmon}(\varepsilon _{\mathrm{quad}(N^*)})}.$

A similar result holds even when $N$ is squarefree if $\varepsilon$ is nontrivial; see Proposition 6.5 of my paper.

That inner twists occur in abundance when $N$ is not squarefree was observed as far back as 1977 by Ribet; at the bottom of page 48 of this paper, Ribet writes

It would be of interest to give an a priori construction of forms with extra twists. If the level $N$ is divisible by a high power of a prime, these forms seem to be more the rule than the exception.

• I was just re-visiting this (5 years later!) and it struck me that this doesn't answer the question. This argument shows that $f \otimes \varepsilon$ often has the same weight and level as $f$; but that doesn't imply $f$ and $f \otimes \varepsilon$ are often Galois conjugate which is a much stronger condition. In other words, the argument doesn't prove that forms with extra twists exist -- it just shows that one can't use the argument from Ribet's paper to prove that they don't. – David Loeffler May 13 '20 at 6:31

This could be hard (though I admit all analytic number theory questions look hard to me).

Indeed, we know quite a lot about the endomorphism algebra $X$ (built from cocycles attached to extra twists) of newforms thanks to the results of K.Ribet, F.Momose, J.Quer (in weight 2) and Brown-Gahte (in weight $k>2$); see Endomorphism algebras of motives attached to elliptic modular forms (J. Number Theory 95 (2002), no. 1, 14-37) and references therein.

Section 4 of the above article gives criteria for the set of quadratic extra twists to be non-trivial in terms of the $p$-adic valuation of $a^2_p$ (and more precisely about the parity of its valuation) so the question of whether most eigencuspforms have no extra twists might be closely related to the question whether most eigencuspforms are $p$-ordinary. Certainly, this is expected to be true but I don't think much is known in weight $k>2$ about this.

Let me addd an elementary analytic remark. The asymptotic natural density of square-free integers is $1/\zeta(2)\simeq 0,60$ so most integers are square-free. Hence (?), most eigencuspforms have square-free conductor. As an eigencuspform with square-free conductor and trivial character has no extra twist (a theorem of Ribet), perhaps the answer to question 1 is indeed yes for some sense of most.