show · mf.gl2.history.new all knowls · up · search:

In addition to the different situations we have seen above, which involve modular functions and forms, there are also many ways to create new modular forms from given ones.

#### Vector space or graded ring structure

The most trivial way to construct a modular form from given ones is to use the vector space structure: if $f$ and $g$ have the same weight and multiplier system on some group $G$, then of course so does $\lambda f+\mu g$, for any constants $\lambda$ and $\mu$. Another way is to use the fact that the set of all modular forms with trivial multiplier for a given group has the structure of a graded ring. Even more is true: if $f$ and $g$ are modular for the same group $G$, then so is the product $fg$, whose weight will be the sum of the weights and multiplier system, the product of the multiplier systems.

#### Differentiation

If we differentiate the modular identity $f(\gamma z)=v(\gamma)(cz+d)^kf(z)$ and use $(\gamma z)'=(cz+d)^{-2}$, we obtain $f'(\gamma z)=v(\gamma)((cz+d)^{k+2}f'(z)+k(cz+d)^{k+1}f(z))\;.$ Thus $f'$ is almost a modular form (it is in fact a quasi-modular form) of weight $k+2$, and it is exactly modular if $k=0$. There are many ways to "repair" this defect of modularity: two of them involve modifying the differentiation operator by using the auxiliary functions $1/y=1/\Im(z)$ or the quasi-Eisenstein series $E_2$. Another is to take suitable combinations which remove the extra terms that prevent modularity: for instance, if $f$ is of weight $k$ and $g$ of weight $\ell$, then $f^\ell/g^k$ is of weight $0$, so its derivative is really modular of weight $2$, and expanding shows that $\ell f'g-kfg'$ is modular of weight $k+\ell+2$. This is a special case of a series of bilinear operators called the Rankin‒Cohen operators.

#### Changing the group

If we accept to modify the group on which a function is modular, there are many other ways to create new modular forms. For instance, if $f$ is modular on some group, then $f(mz)$ will also be modular of the same weight on some other group for any $m\in\Q^{\times}$. A similar construction implies that if $f(z)=\sum_{n\ge n_0}a_nq^{n/N}$ is modular, then so is $\sum_{n\equiv r\pmod{M}}a_nq^{n/N}$ and $\sum_{n\ge n_0}\psi(n)a_nq^{n/N}$ for any periodic function $\psi$. This last construction is called twisting by $\psi$.

#### Enlarging the group

Even more interesting is the possibility to enlarge the group on which a function is modular: if $f$ is modular on some subgroup $H$ of finite index of some other group $G$, say with trivial multiplier system, and if $(\gamma_i)$ is a system of left coset representatives of $H\backslash G$, so that $G=\bigsqcup_iH\gamma_i$, then it is clear that $\sum_i$ $f|_k\gamma_i$ will be modular on $G$. This is a special case of the averaging procedure mentioned at the very beginning. An important example combining the above two methods is the construction of Hecke operators: let $p$ be a prime. If, for instance, $f$ is modular of weight $k$ on the full modular group $\Gamma$, the functions $f((z+j)/p)$ and $f(pz)$ will be modular only on the subgroup $\Gamma_0(p)$ of $\Gamma$, but it is immediate to show that the linear combination $g=\sum_{0\le j\le p-1}f((z+j)/p)+p^kf(pz)$ is again modular on the full modular group $\Gamma$, and we can define the Hecke operator $T(p)$ by $T(p)(f)=g/p$.

Authors:
Knowl status:
• Review status: beta
• Last edited by Andreea Mocanu on 2016-04-01 15:21:10
Referred to by:
History: