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.


1 Answer 1


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 $\Gamma_{1}(N)$.

  • $\begingroup$ Iwaniec's in the library, so I'll grab it tomorrow. looks at clock later today. $\endgroup$ Jan 20, 2016 at 5:43
  • $\begingroup$ So, if I get this... 36n+2 is already not in Gamma_0; however, the Sturm bound for something involving this form is just 3 times the bound it would have if it were, the 3 arising from the fact that 18 is divisible by 3^2, so for various 36n+i, the factor required for any given f_i will be at worst 6 times as much? And then any product f_i f_j, or sum of several of these, at worst 36 times as much? (Thanks for your help with these, Jeremy; I'm obviously meddling with forces I don't entirely understand.) $\endgroup$ Jan 20, 2016 at 5:49
  • 1
    $\begingroup$ 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 $\Gamma_{0}(N)$. $\endgroup$ Jan 20, 2016 at 6:51
  • $\begingroup$ That's exactly what I needed, and I should be able to apply it to related problems. Thank you. $\endgroup$ Jan 20, 2016 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.