Every reference I've seen on modular forms seems to jump from the general definition of a modular form for congruence subgroups to studying modular forms just for $\Gamma_0(N)$.

Once upon a time, I saw something like

$$M_k(\Gamma(N)) \cong\bigoplus_{\chi\mod N} M_k(N^2,\chi)$$

where $M_k(\Gamma(N))$ is the vector space of all weight $k$ modular forms for $\Gamma(N)$, and on the right side $\chi$ ranges over all dirichlet characters mod $N$, and $M_k(N^2,\chi)$ is the vector space of weight $k$ modular forms $f$ satisfying $$f(\gamma z) = \chi(d)(cz+d)^kf(z)$$ for all $\gamma = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\in \Gamma_0(N^2)$.

However, I don't see how this makes sense, since $\Gamma(N)$ is not a subgroup of $\Gamma_0(N^2)$.

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    $\begingroup$ If you conjugate $\Gamma(N)$ by $\begin{bmatrix} 1 & 0 \\ 0 & N \end{bmatrix}$ you get a subgroup of $\Gamma_0(N^2)$. $\endgroup$ Commented Oct 26, 2016 at 16:54
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    $\begingroup$ See Section 3.5 of my paper: arxiv.org/pdf/1609.06740.pdf (this is for Maass forms instead of holomorphic modular forms, but the same argument applies in both cases). $\endgroup$ Commented Oct 26, 2016 at 17:07
  • $\begingroup$ @TylerLawson Is it obvious that conjugate of $\Gamma(N)$ by $\begin{bmatrix} 1 & 0 \\ 0 & N\end{bmatrix}$ should be (a) normal inside $\Gamma_0(N)$, and (b) have quotient isomorphic to $(\mathbb{Z}/N\mathbb{Z})^\times$? $\endgroup$ Commented Oct 26, 2016 at 17:26
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    $\begingroup$ This question is a bit silly. It might "suffice" to do Gamma_0(N) if you're interested in, say, elliptic curves, but it certainly does not suffice if you're interested in, say, weight 1 modular forms. Whether it not it suffices will depend on the context. A slightly more interesting question is why Gamma_1(N) suffices, but this has been answered in the comments above; any modular form for a congruence subgroup gives rise to an automorphic representation which will contain an element invariant under Gamma_1(N). $\endgroup$
    – znt
    Commented Oct 26, 2016 at 22:11
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    $\begingroup$ @znt, I think you're reading "modular forms just for $\Gamma_0(N)$" to mean modular forms of level $q$ with trivial nebentypus (often denoted by $\mathcal{M}_k\left(\Gamma_0(N)\right)$), whereas the question clearly means modular forms of level $q$ with any nebentypus $\chi$ (often written as $\mathcal{M}_k\left(\Gamma_0(N),\chi\right)$ instead of $\mathcal{M}_k(N,\chi)$ to emphasise this point). Of course, each of these spaces is a subspace of $\mathcal{M}_k\left(\Gamma_1(N)\right)$. $\endgroup$ Commented Oct 26, 2016 at 23:06

1 Answer 1


Conjugating $\Gamma(N)$ by $\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & N \end{bmatrix}$ $$\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & N \end{bmatrix} \displaystyle\Gamma(N)\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & 1/N \end{bmatrix} \scriptstyle \ \ = \ \ \underbrace{\left\{\ \scriptstyle\begin{bmatrix} aN+1 & b \\ cN^2 & dN+1\end{bmatrix} \in SL_2(\mathbb{Z})\right\}}_{\displaystyle\tilde{\Gamma}_1(N^2)}\displaystyle \ \supset \Gamma_1(N^2)$$

Define the linear operator $T : M_k(\Gamma(N)) \to M_k(\tilde{\Gamma}_1(N^2)), \ \ T f(\tau) = f (N\tau)$,

and the usual linear operators for showing $M_k(\Gamma_1(N^2)) =\displaystyle \bigoplus_{\chi \bmod N^2}M_k(\Gamma_0(N^2),\chi)$

  • For $gcd(d,N^2)=1$, let $\langle d \rangle : M_k(\Gamma_1(N^2)) \to M_k(\Gamma_1(N^2)) $, $\ \ \langle d \rangle g = g|_k\gamma, \quad \gamma \in \Gamma_0(N^2), \quad\gamma_d \equiv d \bmod N^2$ (which is well-defined, not depending on the chosen $\gamma$). Note that $\langle d d' \rangle = \langle d \rangle\langle d' \rangle$

  • And for a $\chi \bmod N^2$ : $$ \pi_\chi g= \frac{1}{\varphi(N^2)}\sum_{\begin{array}{l}d \bmod N^2\\gcd(d,N^2) =1\end{array}} \overline{\chi(d) }\langle d \rangle g$$ $\pi_\chi$ is an orthogonal projection $M_k(\Gamma_1(N^2)) \to M_k(\Gamma_0(N^2),\chi)$, and $\displaystyle\sum_{\chi \bmod N^2} \pi_\chi g=\langle 1\rangle g= g$ and for any $\chi \ne \chi'$ : $\pi_\chi \pi_{\chi'} = 0$

  • Finally, $Tf = \langle dN+1\rangle T f$, so that $\langle dN+d'\rangle T f= \langle d'\rangle T f$ and hence $\pi_\chi T f = 0$ whenever $\chi$ isn't a character $\bmod N$.

Thus $\sum_{\chi \bmod N} \pi_\chi Tf = Tf$, and together with $M_k(\Gamma_0(N^2),\chi) \subset M_k(\tilde{\Gamma}_1(N^2))$ for any $\chi \bmod N$,

it means that $$M_k(\Gamma(N)) \simeq M_k(\tilde{\Gamma}_1(N^2)) =\bigoplus_{\chi \bmod N}M_k(\Gamma_0(N^2),\chi)$$

  • $\begingroup$ (I'd like to know if you had a different idea for showing it) $\endgroup$
    – reuns
    Commented Oct 26, 2016 at 23:21
  • $\begingroup$ There's an error here. The map $T$ does not map modular forms for $\Gamma(N)$ to modular forms for $\tilde{\Gamma}_{1}(N^{2})$, rather it maps in the other direction. You can see this because if $f$ is a modular form for $\Gamma(N)$, then since $\Gamma(N)$ contains $\begin{bmatrix} 1 & N \\ 0 & 1 \end{bmatrix}$ we will have $f(\tau+N) = f(\tau)$ and so $f$ will have an expansion in terms of $q^{1/N}$. With $T$ as it is defined, $Tf$ will have an expansion in powers of $q^{1/N^{2}}$. $\endgroup$ Commented Oct 26, 2016 at 23:40
  • $\begingroup$ The answer has now been edited to fix the error. $\endgroup$ Commented Oct 27, 2016 at 21:50

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