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Let $E$ be an elliptic curve over a number field.

When $E$ has no CM and is a $\mathbb Q$-curve (i.e. it is $\overline{\mathbb Q}$-isogenous to all of its conjugates), it is nowadays known that $E$ is a factor up to isogeny of $J_1(N)$. In a nice paper X. Guitart and J. Quer define a subclass of $\mathbb Q$-curves, called strongly modular curves, which enjoy the property that their $L$-function is a product of $L$-functions of newforms. Thus, the $L$-function of these curves can be analitically continued to $\mathbb C$.

If $E$ has CM, the $L$-function of $E$ is known to admit an analytic continuation to $\mathbb C$ because it coincides with a product of $L$-functions of Hecke characters.

My first question is: are $L$-functions of elliptic curves with CM also products of $L$-functions of (classical) newforms? It is not clear at all to me what is the connection between $L$-functions of cuspforms and $L$-functions of Hecke characters.

My second question is: what is the state of art of the problem? Namely, is there any other class of elliptic curves whose $L$-function is known to have an analytic continuation?

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  • $\begingroup$ As it was already pointed out, the general answer to your first question is no. But I believe you yourself already know it. I guess you want to know the relations between 2-dim cuspidal reps and Hecke characters, like when they have the same L function. And it will be related to automorphic induction. $\endgroup$
    – user42690
    Commented Nov 25, 2015 at 2:39

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As said in the comment above, the answer to your first question is no. Hecke L-functions are expected to factor as a product of irreducible cuspidal automorphic representations of $\mathrm{GL}_n(\mathbb{A})$ (they are known to be a product of Artin L-functions, but we don't know those are entire in general). In the case of an elliptic curve with CM (using Deuring's theorem) this means

$$L_E(s)=L(s,\chi)L(s,\bar\chi)=\prod_{i=1}^r L(s,\pi_i)^{e_i}$$

But in general those representations $\pi$ are $n$-dimensional.

To see this in the simplest case, consider a Hecke L-function with trivial character, that is, $L(s,1)=\zeta_K(s)$, the Dedekind zeta function of the number field. We know from Aramata-Brauer (in black) and conditionally on the strong Artin conjecture (in red) that

$$\zeta_K(s)=\zeta(s)\prod_{\rho \neq 1} L(s,\rho)=\color{red}{\zeta(s)\prod_{\pi \neq 1} L(s,\pi)}$$

where the product goes over all the nontrivial irreducible representations of $\mathrm{Gal}(K/\mathbb{Q})$.

Now, in order for $\zeta_K$ to be expressible in terms of classical newforms, all (or some, depending on how one interprets your question) those representations of that Galois groups have to be of degree $2$. But that's definitely not true. Number fields with $\mathrm{Gal}(K/\mathbb{Q}) \cong S_n$, $n>4$ are alredy a counterexample.

Of course classical newforms also show up, but I'm not sure that there's a nice characterization of when that happens, depending on $E$, $K$ and $\mathrm{End}_K(E)\otimes\mathbb{Q}$.

Regarding the second question, analytic continuation (a special case of the Hasse-Weil conjecture) is only known in general when $E$ has CM (Deuring), when $E$ is defined over $\mathbb{Q}$ (Wiles-Breuil-Conrad-Diamond-Taylor) and when $K$ is real quadratic (Freitas-Hung-Siksek).

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    $\begingroup$ It is not true that $L(s,\chi)$ is a product of Dirichlet $L$-functions when $K$ is abelian over $\mathbb{Q}$. For example, let $\chi$ be a quadratic Hecke character of $K$, and let $L/K$ be the quadratic extension attached to $\chi$. Then $\zeta_L(s)=\zeta_K(s)L(s,\chi)$, and the condition is equivalent to $L/\mathbb{Q}$ being abelian, which is stronger than $K/\mathbb{Q}$ being abelian. Another example is provided by any classical CM elliptic curve $E/\mathbb{Q}$. Indeed, in this case $L(s,E)=L(s,\chi)$ with $K/\mathbb{Q}$ quadratic, but $L(s,E)$ does not factor into Dirichlet $L$-functions. $\endgroup$
    – GH from MO
    Commented Nov 23, 2015 at 0:01
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    $\begingroup$ @GHfromMO Thanks, you are right, I'll correct it. I was thinking of something else. $\endgroup$
    – Myshkin
    Commented Nov 23, 2015 at 0:14
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    $\begingroup$ I think when $E$ is defined over real quadratic fields, modularity is also known now. (Freitas - Le Hung - Siksek 15) $\endgroup$
    – Pig
    Commented Nov 23, 2015 at 0:30
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    $\begingroup$ @user31814 Thanks! I mentioned that result myself one month ago here, but it completely slipped my mind. $\endgroup$
    – Myshkin
    Commented Nov 23, 2015 at 0:43
  • $\begingroup$ @Myshkin: thanks for your answer! Yes, that is the paper I was referring to. $\endgroup$
    – Ferra
    Commented Nov 23, 2015 at 8:39

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