Let $f(z) = \sum\_{n=0}^\infty a\_nq^n$ be the fourier expansion of a (quasi-)modular form (with $q = e^{2\pi i z}$). Consider the following related functions:


$$f\_{m,k}(z) = \sum\_{n=0}^\infty a\_{mn + k}q^n$$


$$g\_{m,k}(z) = \sum\_{n=0}^\infty a\_{mn + k}q^{mn+k}$$


for, naturally, $0 \leq k < m$.


My question: are these in any way modular? For some congruence subgroup, etc. etc.?


What if we put some restrictions on $f(z)$? Or shift the powers of $q$ somewhat?


As an example of when we *do* retain (quasi-)modularity, consider the case $f(z) = E\_2(z), m = 2$. Then it is not too hard to show that


$$
E\_2(z) = g\_{2,1}(z) + 3E\_2(2z) - 2E\_2(4z)
$$


and so it follows that $g\_{2,1}(z)$ is indeed quasi-modular for $\Gamma\_0(4)$. Since then $g\_{2,0} = E\_2(z) - g\_{2,1}(z)$, this is also quasi-modular.


So is this in general true?



Does the following operation on modular forms yield something modular?
I know nothing about quasimodular forms, so let me answer the question for modular forms in the classical sense.


First, your $f\_{m, k}$. If $k = 0$ this is essentially the image of $f$ under the Hecke operator $U\_m$, so it is always going to be modular of some level (and if f has level $\Gamma$, then $f\_{m, 0}$ will have level $ \Gamma \cap \Gamma\_0(m)$ or something like that). If $k \ne 0$ then $f\_{m, k}$ will not be modular, because $f\_{m, k}(mz) = q^{-k} g\_{m, k}(z)$, $g\_{m, k}$ is modular, and dividing by a power of $q$ destroys modularity.


Now $g\_{m, k}$. This is just a linear combination of the forms $f(z + j/m)$ for $j = 0, ..., m$ by finite Fourier theory, and all these are modular, so $g\_{m, k}$ is modular (although if $g$ is modular of level $\Gamma\_1(N)$, then $g\_{m, k}$ will have level something like $\Gamma\_1(mN) \cap \Gamma\_0(m^2 N)$ in general). 


The forms $g\_{m, k}$ are closely related to the forms
$$ g\_{\chi} = \sum\_n \chi(n) a\_n q^n $$
for Dirichlet characters $\chi$ modulo $m$, which are very interesting (and well-studied) objects; if $g$ is a Hecke eigenform then so are the forms $g\_{\chi}$ (although the $g\_{m, k}$ generally aren't).


