# field of definition of Fricke involution applied to a modular form

Let $F\in M_k(\Gamma_0(N),\chi)$, not necessarily an eigenform nor cuspidal, but assume that $\Bbb Q(F)$ is a number field $K$. What can one say of $\Bbb Q(F|_kW_N)$, where $W_N$ is the Fricke involution ? Experiments seem to show that the Fourier coefficients of $F|_kW_N$ divided by $\sqrt{Q}$ for some positive or negative divisor of $N$ (probably linked to the conductor of $\chi$) also lie in $K$.

More generally same question for a general Atkin--Lehner involution $W_Q$, and also for $1/2$-integral weight (in which case even a fourth root may be necessary).

It may be possible to start with a corresponding result for newforms, but I have not seen how to complete the argument.

P.S. Since I tested mainly with real characters, $\sqrt{Q}$ may of course be the Gauss sum associated to $\chi$. But the questions remain.

• To clarify, are you saying there is a choice of $Q|N$ such that all Fourier coefficients divided by $\sqrt Q$ lie in $K$, or can $Q$ vary with the coefficients? Sep 9 '17 at 23:21
• Can you say that $F$ is a $K$-linear combination of eigenforms with coefficient fields in $K$? Sep 9 '17 at 23:24
• Yes, $Q$ is a fixed positive or negative divisor of $N$. But $\sqrt{Q}$ could instead be the Gauss sum for $\chi$ or something similar. Sep 10 '17 at 8:27

Your guess that this is related to the Gauss sum is correct. If $f$ has integer weight, level $N$ and coefficients in $K$, then $W_N(f) / \tau(\chi)$ has coefficients in $K$ also.

This follows from a lovely theorem of Shimura, which states that $M_k(\Gamma(N), \mathbf{Q}[\zeta_N])$ is preserved by the action of $SL_2(\mathbf{Z} / N)$, and that the actions of $SL_2(\mathbf{Z} / N)$ and $\operatorname{Gal}(\mathbf{Q}[\zeta_N] / \mathbf{Q})$ piece together into an action of $GL_2(\mathbf{Z} / N)$ if you identify the Galois group with the matrices of the form $\begin{pmatrix} * & 0 \\ 0 & 1\end{pmatrix}$ (or it might be $\begin{pmatrix} 1 & 0 \\ 0 & *\end{pmatrix}$, I can't remember).

To use this here, consider $f$ as an element of $M_k(\Gamma(N), K)$, and look at the action of $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$; the result is $W_N(f)(z/N)$ up to a power of $N$, which has the same coefficient field as $W_N(f)$ [edit: possibly up to a factor of $\sqrt N$ in odd weights, depending on you conventions], and Shimura's result now tells you that this has coefficients in $K(\zeta_N)$ and you can read off exactly how $Gal(K(\zeta_N) / K)$ acts.

The same argument will tell you how partial Atkin--Lehner operators $W_Q$ for $Q \| N$ act, with a little more bookkeeping -- if I remember correctly, you get the Gauss sum of the Q-primary part of $\chi$ coming out.

A reference for this is: Ohta, "P-adic Eichler--Shimura isomorphisms" (Crelle #463, 1995), sections 3.5 and 3.6. Lemma 3.5.2 on page 83 is a reciprocity law describing how Galois acts on $W_Q(f)$, which is (I hope!) equivalent to the statement I gave above; and section 3.6 (pages 86-9) gives a detailed proof of the lemma using Katz's algebraic description of modular forms.

(I have no idea about the half-integer weight case.)

• Thanks! perfect explanation, I will look at the paper you mention. Sep 10 '17 at 17:09
• Unfortunately, the result you mention seems to be correct (experimentally) only for even weights. In odd weights, you clearly must also divide by a Gauss sum, but also by (something like) the square root of the level. Simplest (counter)example: the Eisenstein series of weight 3 and character -3, level 3. Sep 11 '17 at 20:04
• @HenriCohen How are you normalising the action of $W_N$? There are various conventions as to whether you have $N^k$ or $N^{k/2}$ or $N^{k-2}$ etc, which of course can throw off the normalisations when $k$ is odd. I think Ohta defines $W_N(f)(z) = N^{-1} z^k f(-1/Nz)$, which means that the operator is not an involution in weights $\ne 2$. If you are normalising so that $W_N^2 = 1$ then that probably accounts for the discrepancy. Sep 11 '17 at 20:12
• Yes, I do normalize so that $W_N^2=1$, so the discrepancy is explained. Sep 12 '17 at 8:30