Petersson square of twist

I should know this, but can't find a reference. Let $F$ be a cusp form, not necessarily an eigenform, on some congruence subgroup. It seems experimentally that the ratio of the Petersson square of a twist of $F$ (say by some quadratic character) to the Petersson square of $F$ belongs to $\mathbb Q(F)$. Is this true, and what is the proof ? Or are there conditions on the twist ? Is this directly related to Manin's rationality theorem, i.e., are the periods $\omega^{\pm}$ of the twist related to those of $F$ ?

• "Petersson square" = $\langle F, F \rangle$ w.r.t. Petersson inner product? For an eigenform, at least, the relation should follow from the formula where $\langle F, F \rangle$ appears in the leading coefficient of the asymptotic growth of $\sum_{n \leq x} |a_n(F)|^2$; twisting by a character of conductor $q$ only removes the terms for which $\gcd(n,q) > 1$, so we need also the growth of $\sum_{n \leq x} |a_{dn}(F)|^2$ for each $d|q$, and then relate $a_{dn}$ with $a_n$ via multiplicativity. – Noam D. Elkies Oct 28 '17 at 4:47
• Have you experimented with forms which are combinations of newfoms of different level? Do you still get rational relations if you twist by characters which are ramified at places the levels of some but not other of the newforms? – Kimball Oct 28 '17 at 15:57

For an eigenform, the product of the periods associated to $f$ ($\omega^{\pm}$ in your notation) is equal modulo $\mathbf{Q}(f)^\times$ to $i^{1-k} G(\chi) \langle f, f\rangle$, where $G(\chi)$ is the Gauss sum of the nebentypus character of $f$, and $k$ is the weight. This is Theorem 1 (iv) of Shimura's On the periods of modular forms (Math. Ann. 229, 1977).
Since the periods of the twist $f_\varepsilon$ are $G(\varepsilon)$ times the periods of $f$, and $G(\varepsilon^2 \chi) / [G(\chi) G(\varepsilon)^2]$ lies in the field of values of $\chi$ and $\varepsilon$, it follows that the ratio $\langle f_\varepsilon, f_\varepsilon\rangle / \langle f, f \rangle$ lies in $\mathbf{Q}(f, \varepsilon)^\times$ (and hence in $\mathbf{Q}(f)^\times$ if $\varepsilon$ is quadratic).