- Let $f$ be modular of level $p^nN$, $(p,n) = 1$, $p > 2$ with character $\chi\psi\eta$, where $\chi$ has conductor dividing $N$, $\psi$ conductor power of $p$ and order power of $p$, and $\eta$ conductor $p$ and order dividing $p-1$. Since $p$ is odd, one can write $\psi = \xi^{-2}$. (i) Why is the character of $f \otimes \xi$ equal to $\chi\eta$; (ii) why is the reduction mod $p$ of this equal to the reduction of $f$; (iii) why is for $r \gg n$ $f \otimes \xi$ modular with respect to $\Gamma_0(p^r) \cap \Gamma_1(pN)$? Can this be proven using the converse theorem?
2.a Why is the twisted Eisenstein series $G = a_0 + \sum_{n=1}^\infty\sum_{d \mid n}\eta^{-1}(d)d^{i-1}q^n$ of Nebentypus $\eta^{-1}$ with respect to $\Gamma_0(p)$?
2.b Why is $fG$ modular with respect to $\Gamma_0(p^r) \cap \Gamma_1(N)$, wenn $f \in S_k(\Gamma_0(p^r) \cap\ \Gamma_1(pN), \chi\eta)$ ist?
(The article is here: math.berkeley.edu/~ribet/Articles/motives.pdf )
Best Answer
William Stein has answered your question (i). As for your question (ii), since $\xi$ has $p$-power order, and since any $p$-power root of unity is congruent to $1$ modulo the unique prime ideal lying over $p$ in $\mathbb Q$ adjoin the $p$-power roots of unity, we see that $f\otimes \xi$ is congruent to $f$ modulo any prime ideal lying over $p$ in the field of definition of $f$ adjoin the $p$-power roots of unity.
Finally, you have already noted in (i) that the character of $f\otimes \xi$ is just $\chi\eta$. Since the conductor of $\xi$ divides $N$ and the conductor of $\eta$ divides $p$, we see that the conductor of $\chi\eta$ divides $p N$. This gives (iii). (Note that (iii) is simply a statement about the conductor of the character of $f\otimes \xi$: In general, a modular form $g$ of level $C$ is modular on $\Gamma_0(C)\cap \Gamma_1(D)$, rather than just $\Gamma_1(C)$, for some integer $D$ dividing $C$, if the character of $g$, a priori a character of $(\mathbb Z/C)^{\times}$, actually factors through $(\mathbb Z/D)^{\times}$. In your particular case, $f$ has conductor a power of $p$ times $N$, and $\xi$ has conductor a power of $p$, so $f\otimes \xi$ has conductor a power of $p$ times $N$. The conductor of its character divide $p N$, as already noted, and so (iii) follows.)