# How different can characters be for a sum of modular forms to still be in Gamma_0?

I have a modular form I am constructing out of sums and products of various dissected divisor-sum series, namely forms of the type $$f_i = \sum_{j=0}^\infty \sigma_1(36j+i) q^{36j+i}.$$

Each of these can be constructed from the standard forms $\sum \sigma_1(2n+1) q^{2n+1}$ and $\sum \sigma_1(3n+1) q^{3n+1}$ by various quadratic twists and magnifications (applications of $U(d)$ and $V(d)$). (Note: this increases the level by factors up to $36^2$, so if there is a better way to do that I'm all ears.)

I'd like to use Sturm's theorem on the resulting form, but if I understand correctly (I certainly might not!), summing modular forms in $M_k(\Gamma_0(N_1),\chi)$ and $M_k(\Gamma_0(N_2),\psi)$ means that I am verifiably in $M_k(\Gamma_1(lcm(N_1,N_2)),\phi)$ for some $\phi$, and not $M_k(\Gamma_0(lcm(N_1,N_2)),\phi)$. The Sturm bound for $\Gamma_1$ is substantially larger than for $\Gamma_0$, possibly infeasibly so. Arguments I have seen involving similar sums in the prior literature don't seem to worry about that, though.

Is there an argument for this restricted case that I am still in $\Gamma_0$, or am I forced to move to $\Gamma_1$?

Lacking that, is there a better bound I can state on, say, $\sum \sigma_1(36j+2) q^{36j+2}$ than $N=5832 = 9 \cdot 18^2 \cdot 2$? ($3n+1$ has original weight 9, quadratic twist mod 18 to get $18n+1$, magnify by 2 to get $36n+2$?)

I have Ono's "Web" personally and Koblitz in the library, and can try for any other citations via interlibrary loan. Many thanks.

The bad news: Yes, you are forced to go to $\Gamma_{1}(N)$, and not $\Gamma_{0}(N)$. In particular, let's say that $E = \sum \sigma_{1}(2n+1) q^{2n+1}$ is the usual weight $2$ level $4$ Eisenstein series. Denote by $E_{\chi}$, the twist of $E$ by the Dirichlet character modulo $\chi$. Then, you get that $$\sum \sigma_{1}(36n+2) q^{36n+2} = \left(\sum_{\chi \bmod 18} E_{\chi}\right) | V(2).$$ The issue here is that $E_{\chi}$ is a modular form (of some level) with character $\chi^{2}$, and not all the Dirichlet characters modulo $18$ are quadratic. The up shot is that the form above lives in $M_{2}(\Gamma_{0}(N)) \oplus M_{2}(\Gamma_{0}(N),\psi) \oplus M_{2}(\Gamma_{0}(N),\psi^{2})$, where $\psi$ is a Dirichlet character modulo $9$ with order $3$. (By the way, the best way to get a good bound on $N$ is to look at Iwaniec's Topics in Classical Automorphic Forms. He has the best theorem about levels of twists.)
The good news: What you're really getting in the end is a modular form in $M_{2}(\Gamma)$, where $\Gamma$ is an index $3$ subgroup of $\Gamma_{0}(N)$ (for some $N$). The Sturm bound for $M_{2}(\Gamma)$ is just $(k/12) [{\rm SL}_{2}(\mathbb{Z}) : \Gamma]$, and so you don't have to make your computer sweat quite as much as if you were working with $\Gamma_{1}(N)$.
• The "3 times the bound" arises because the subgroup of squares of Dirichlet characters of order $36$ has order $3$. This factor of $3$ would show up for many, but not all of the $36n+i$ (for other choices of $i$, like $18$ obviously) the form transforms for $\Gamma_{0}(N)$. The point is that for sums and products of such forms, you are still in $M_{k}(\Gamma)$, where $\Gamma$ has index $3$ in $\Gamma_{0}(N)$. Jan 20, 2016 at 6:51