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Let $f$ be a modular form of weight 2 for the group $\Gamma_0(N)$, with Fourier expansion $$ \sum_{n\geq 0} a(n)\ q^n. $$

Let $d$ be an integer dividing $N$, and consider the Fourier series $\sum b(n)q^n$, where $$ b(n) = a \left( n / gcd(n,d) \right). $$

I want to believe that this latter series is the Fourier expansion of another modular form, perhaps it's some combination of Hecke operators acting on $f$? In my example, $N$ and $d$ are squarefree, if that helps.

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1 Answer 1

In general, its can't be possible. The L-function associated to $b(n)$ is $$L(g,s)=\sum_{n\ge 1}\frac{b(n)}{n^s}=M(s)\sum_{(n,d)=1}\zeta(s)\sum_{n| d^\infty} \frac{a(n)}{n^s},$$where $M(s)$ is holomorphic and convergent for $s>1$.

we note the symmetry fact $\zeta(s) \sim \zeta(1-s) $, and $$\sum_{n} \frac{a(n)}{n^s} \sim \sum_{n} \frac{a(n)}{n^{2-s}}.$$ There does not exist a weight corresponding to $b(n )$, such that $L(g,k-s)=L(g,s)$

by converse theorem in book topics[Henryk Iwaniec chap 6], $L(g,s)$ does not come from a modular form.

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