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
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).