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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. :)

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  • $\begingroup$ 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 $\endgroup$ Commented Mar 9, 2017 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
    Commented Jun 6, 2017 at 2:31

1 Answer 1

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Some notations first, (please skip to the answer if familiar):

It will be easier if we work with the $slash$ notation, $\big\vert_{k}$. Since the slash notation is not defined now, let us first do it. Let $$ G= \left\{ (\alpha,\phi(z)): \alpha = \left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)\in GL^+_2(\mathbb{Q}) \right\},$$ where $\phi(z)$ is a holomorphic function on $\mathbb{H}$ such that $\phi^2(z)=t\frac{cz+d}{\det(\alpha)}$, where $|t|=1$. Then, $$ f\big\vert_{k}(\alpha,\phi(z)) = (\phi(z))^{-k}f(\alpha z). $$ Also, $G$ forms a group under the group law: $$ (\alpha,\phi(z))(\beta,\psi(z)):= (\alpha\beta, \phi(\beta z)\psi(z)). $$

The $V(m)$ operator is given by $$ f|V(m) = m^{-k/4}f\big\vert_{k}\left(\begin{pmatrix}m&0\\0&1\end{pmatrix}, m^{-1/4} \right) = m^{-k/4}(m^{-1/4})^{-k}f\left(\begin{pmatrix}m&0\\0&1\end{pmatrix}z \right) = f(mz).$$

(These are given in the mentioned paper by Serre-Stark).

Answer: I will use the notation $M_{k/2}(\Gamma_{0}(N),\chi)$ for $M_0(N,k/2,\chi)$.

Let $f\in M_{k/2}(\Gamma_{0}(N),\chi)$. To prove that $$ g = f|V(m) \in M_{k/2}(\Gamma_{0}(mN),\chi\chi_m);$$ i.e., for $\gamma=\begin{pmatrix}a&b\\mNc&d\end{pmatrix}\in \Gamma_0(mN)$, we need to show that $ g|_k(\gamma,j(\gamma,z)) = \chi\chi_m(d)g. $

\begin{align} g\big\vert_k (\gamma,j(\gamma,z)) &= m^{-k/4}f\big\vert_k \left[\begin{pmatrix}m&0\\0&1\end{pmatrix}, m^{-1/4} \right]\left[ \begin{pmatrix}a&b\\mNc&d\end{pmatrix},\left(\frac{mNc}{d}\right)\left(\frac{-4}{d}\right)^{-1/2}(mNcz+d)^{1/2} \right] \\ &= m^{-k/4}f\big\vert_k \left[\begin{pmatrix}m&0\\0&1\end{pmatrix}\begin{pmatrix}a&b\\mNc&d\end{pmatrix}, m^{-1/4} \left(\frac{mNc}{d}\right)\left(\frac{-4}{d}\right)^{-1/2}(mNcz+d)^{1/2} \right] \\ &= m^{-k/4}f\big\vert_k \left[\begin{pmatrix}a&mb\\Nc&d\end{pmatrix}\begin{pmatrix}m&0\\0&1\end{pmatrix}, \left(\frac{m}{d}\right)\left(\frac{Nc}{d}\right)\left(\frac{-4}{d}\right)^{-1/2}(mNcz+d)^{1/2}m^{-1/4} \right] \\ &= m^{-k/4}f\big\vert_k \left[\begin{pmatrix}a&mb\\Nc&d\end{pmatrix}, \left(\frac{m}{d}\right)\left(\frac{Nc}{d}\right)\left(\frac{-4}{d}\right)^{-1/2}(Ncz+d)^{1/2} \right] \left[\begin{pmatrix}m&0\\0&1\end{pmatrix}, m^{-1/4} \right] \\ &= m^{-k/4}f\big\vert_k \left[\begin{pmatrix}a&mb\\Nc&d\end{pmatrix}, \left(\frac{Nc}{d}\right)\left(\frac{-4}{d}\right)^{-1/2}(Ncz+d)^{1/2} \right] \\ &\hspace{50mm}\left[\begin{pmatrix}1&0\\0&1\end{pmatrix}, \left(\frac{m}{d}\right) \right]\left[\begin{pmatrix}m&0\\0&1\end{pmatrix}, m^{-1/4} \right] \\ &= m^{-k/4}\chi(d)f\big\vert_k \left[\begin{pmatrix}1&0\\0&1\end{pmatrix}, \left(\frac{m}{d}\right) \right]\left[\begin{pmatrix}m&0\\0&1\end{pmatrix}, m^{-1/4} \right] \\ &= m^{-k/4}\chi(d) \left(\frac{m}{d}\right)^{-k}f\big\vert_k \left[\begin{pmatrix}m&0\\0&1\end{pmatrix}, m^{-1/4} \right] \\ &= \chi(d) \left(\frac{m}{d}\right) m^{-k/4} f\big\vert_k \left[\begin{pmatrix}m&0\\0&1\end{pmatrix}, m^{-1/4} \right] \\ &= \chi(d) \chi_m(d) f|V(m)\\ &= \chi\chi_m(d) g, \end{align} as required; this completes the checking of transformation law.

I am not quite clear about checking holomorphicity of $f|V(m)$; but if $f$ is holomorphic(at $\mathbb{H}$ and at cusps), I think it should be true for $f|V(m)$ also. I don't know how to prove that part.

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