My question is a lemma in the following paper.

Serre J P, Stark H M. Modular forms of weight 1/2 Modular functions of one variable VI. Springer Berlin Heidelberg, 1977: 27-67.

Let $\chi:(\mathbb{Z}/N\mathbb{Z})^*\longrightarrow\mathbb{C}^*$ be a character ($\mod N$), and $\kappa$ be a positive odd integer. A function $f$ on upper half plane $H$ is called a modular form of type $(\kappa,\chi)$ on $\Gamma_0(N)$ if

  1. $f(\gamma z) = \chi(d) j(\gamma,z)^\kappa f (z)$ for every $\gamma=\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)$ in $\Gamma_0$ this makes sense since $4\mid N$;
  2. f is holomorphic, both on H and at the cusps.

One then calls $\kappa/2$ the weight of $f$, and $\chi$ its character. The space of such forms will be denoted by $M_0(N,\kappa/2,\chi).$

Besides, operator $V(m)$ is defined by \begin{align} [f\mid V(m)](z):=f(mz). \end{align} Equivalently, if $f(z)=\sum_{n=0}^{\infty}a_nq^n$, then $f|V(m)=\sum_{n=0}^{\infty}a_nq^{mn}$.

With the notation above, they claim in their Lemma 2 in page 40 that

The operator $V(m)$ takes $M_0(N,\kappa/2,\chi)$ to $M_0(Nm,\kappa/2,\chi\chi_m)$.

I do not understand why $f|V(m)$ is in $M_0(Nm,\kappa/2,\chi\chi_m)$, where $\chi_m$ is the Kronecker character for $\mathbb{Q}(\sqrt{m})$. Any help will be appreciated. :)

  • $\begingroup$ Hi @VerMoriaty, what is the $\chi_m$ here? I can't claim to be an expert (but hope to be after I finish my PhD in a few years!). It seems that the operator will change the level quite clearly (we now have $q^{mn}$, and also leave the weight unchanged $\endgroup$ – TheMathsGeek Mar 9 '17 at 16:41
  • $\begingroup$ @TheMathsGeek I'd say$\chi_m(n) = 1_{gcd(m,n)=1}$ is the trivial character modulo $m$. The notation of Serre is unusual. Most people write that $f (z) \in M_\kappa(N,\chi)$ and $f(m z) \in M_\kappa(N,\chi \chi_m)$ and say if $f$ is not holomorphic at the cusps. $\endgroup$ – reuns Jun 6 '17 at 2:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.