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I recently asked this question on Math.StackExchange with no answer so far. So I thought maybe I can find an answer here.

Let $M(k,\Gamma_0(N))$ be a space of modular forms of weight $k$ on $\Gamma_0(N)$. Each $f \in M(k,\Gamma_0(N))$ has a Fourier expansion of the form $$ f(\tau)=\sum_{n\in \mathbb{N}}a(n)\,q^n \quad \text{where}\quad q=\mathrm{e}^{2\pi \mathrm{i}\tau} . $$

Now, let $g(\tau)$ be a function obtained from $f(\tau)$ by omitting all $a(n)$ such that $\gcd(n,N)\neq 1$, i.e. $$ g(\tau)=\sum_{\substack{n\in \mathbb{N}\\(n,N)=1}}a(n)\,q^n . $$

Question: Is $g(\tau)$ a modular form? What is its level?

Many thanks.

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    $\begingroup$ Dear @user31009, perhaps you should wait at least a few days (instead of 20 hours) before reposting your Math.StackExchange question on MathOverflow. Quick reposting leads to duplication of effort, and is frowned upon by both communities. $\endgroup$
    – Ricardo Andrade
    Commented Feb 21, 2014 at 19:44
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    $\begingroup$ Not to mention that questions can be migrated. You can flag a question for migration by a moderator, in which case people can follow the link to the other site, and there won't be duplication. $\endgroup$
    – Ben Webster ♦
    Commented Feb 21, 2014 at 20:42
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    $\begingroup$ I think this is a perfectly good question and should not be closed. $\endgroup$
    – GH from MO
    Commented Feb 21, 2014 at 23:54

3 Answers 3

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$$ g(\tau)=f(\tau)\otimes \left(\tfrac{N^2}{\cdot}\right)=\sum_{n\in \mathbb{N}}\left(\tfrac{N^2}{n}\right)a(n)\,q^n \quad \text{where}\left(\tfrac{N^2}{\cdot}\right) \text{ is the Kronecker symbol}.$$ This means that $g(\tau)$ is just the twist of $f(\tau)$ by a principle character. Indeed we have $$g(\tau)\in M(k,\Gamma_0(N^3)).$$

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  • $\begingroup$ In fact a stronger conclusion can be drawn, see my response. $\endgroup$
    – GH from MO
    Commented Feb 22, 2014 at 2:17
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We can express the form $g(\tau)$ as $$ g(\tau)=\sum_{d\mid N}\mu(d)\sum_{\substack{n\in\mathbb{N}\\d\mid n}} a(n)\,q^n.$$ In the notation of Atkin-Lehner (Hecke operators on $\Gamma_0(m)$, Math. Ann. 185 (1970), 134-160), the inner sum is $(f\mid U_d)\mid B_d$, which lies in $M(k,\Gamma_0(dN))$ by Lemmata 2, 6, 14 in the paper. This implies that $$ g\in M(k,\Gamma_0(N^2)). $$

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The answer is yes it is a modular form. For this (and the level) see Section 3.8 of Soma Purkait's thesis.

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    $\begingroup$ This seems like a bit more generality than necessary - doesn't the modularity follow straightforwardly from properties of $U_p$ and $V_p$? $\endgroup$
    – S. Carnahan ♦
    Commented Feb 21, 2014 at 21:15
  • $\begingroup$ @Carnahan: You are right, see my response. $\endgroup$
    – GH from MO
    Commented Feb 21, 2014 at 23:54

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