Does the modular form associated to cubic twist of a elliptic curve $E$ corresponds to some twist of $f_E$?

Question

Let $E$ be an elliptic curve defined over $\Bbb Q$ and $f_E$ be the modular form associated with the elliptic curve $E$. Suppose the elliptic curve $E^D$ is a quadratic twist of $E$. I understand that $f_{E^D}$ is a twist of $f_E$ by quadratic character $\mod{D}$. I would like to know when $E^d$ is a cubic twist of E, how are $f_{E^d}$, $f_E$ related?

Answer 1

I start by explaining a tiny problem that appears even for quadratic twists, the same will appear for cubic twists even if they really should be the same.

Let $E$ be an elliptic curve and $D$ a squarefree integer. On the one hand, there is a quadratic twist $E^D$. The associated Galois representations are tensored with the corresponding Dirichlet character $\chi_D$ of order $2$. The $L$-function of $E^D$ is obtained from that Galois representation and so is the modular form $f_{E^D}$ associated to the isogeny class of $E^D$.

On the other hand, starting with the modular form $f_E$ associated to the isogeny class of $E$, we obtain a quadratic twist $(f_E)^D$ obtained by multiplying the coefficients $a_n$ of $f_E$ by $\chi_D(n)$.

In most instances $f_{E^D}$ and $(f_E)^D$ are equal, but not always. Suppose that there is a prime $p$ of additive reduction for $E$, but that the elliptic curve over the field $\mathbb{Q}(\sqrt{D})$ has good reduction. (And other cases like that). Then the quadratic twist $E^D$ has good reduction over $\mathbb{Q}$. So the local factor in the Euler product of $L(E^D,s)$ at $p$ is obtained from a quadratic polynomial. It follows that $a_{p^n}(E^D)$, which is a coefficient of the Dirichlet series of $f_{E^D}$, are not all zero. Instead for $(f_E)^D$ the corresponding coefficients are all zero because the initial $a_p(f_E)$ is. Therefore in this instance the $L$-functions $L(f_{E^D},s)$ and $L((f_E)^D,s)$ differ by the local factor at $p$. Also the two modular forms won't have the same level, one has $p^2$ as a factor in the level the other is coprime to $p$. Of course, they are related by the usual map between modular forms of level dividing each other.

One easy way to see the problem is that $((f_E)^D)^D$ is not $f_E$ as the level will be larger. While $(E^D)^D$ is isomorphic to $E$

Obviously, the same problem could occur for the cubic twists. There is one more issue that may arise (and it also appears in the case that you twist with a general Dirichlet character $\chi$). Depending on your choice of normalisation for the various maps, like the one translating Dirichlet characters to Galois representations of dimension $1$ through class field theory, you might end up with the dual twist. I mean you could have $L(E,\chi,s)= L(f,\bar\chi,s)$.

Both of these are minor issues that most people are happy to ignore, because they should not interfere in principle with what you do. But we had to worry about that recently when we were looking at the integrality of special values of $L$-functions. There it made a difference in some cases.

If $D$ is coprime with the conductor $N$, then the first issue should not appear. The argument is by comparing the associated Galois representations. In the cubic case, you have to check that you chose the normalisations in such a way that they agree or that they are dual to each other.