3
$\begingroup$

In many references about modular forms, I see that they just only focus on two congruence subgroups of $SL_2(\mathbb{Z})$, namely

$$\Gamma_1(N)=\bigg\{ \gamma \in SL_2(\mathbb{Z}) \bigg\vert \gamma \equiv \begin{pmatrix} 1 &* \\ 0 &1 \end{pmatrix} \mod N \bigg\}$$

$$\Gamma_0(N)=\bigg\{ \gamma \in SL_2(\mathbb{Z}) \bigg\vert \gamma \equiv \begin{pmatrix} * &* \\ 0 &* \end{pmatrix} \mod N \bigg\}$$

Why shouldn't we consider about other congruence subgroups? e.g., $\Gamma_2(N)=\bigg\{ \gamma \in SL_2(\mathbb{Z}) \bigg\vert \gamma \equiv \begin{pmatrix} * &1 \\ 0 &* \end{pmatrix} \mod N \bigg\}$

$\endgroup$

2 Answers 2

4
$\begingroup$

The reason why we tend to study $\Gamma_0(N)$ and $\Gamma_1(N)$ is that modular forms with respect to these subgroups are easy to work with and have many applications. In many cases we can also reduce more general problems about modular forms to problems about modular forms with respect to $\Gamma_1(N)$ or $\Gamma_0(N)$.

For instance, for any congruence subgroup $\Gamma$ of level $N$, we have $M_k(\Gamma)\subseteq M_k(\Gamma(N))$ and an injective map from $M_k(\Gamma(N))$ to $M_k(\Gamma_1(N^2))$ (let me know if you want more details about this). Hence, modular forms with respect to $\Gamma$ can be viewed as forms with respect to $\Gamma_1(N^2)$.

Moreover, we can decompose $M_k(\Gamma_1(N))$ as $M_k(\Gamma_1(N))=\bigoplus_{\chi}M_k(\Gamma_0(N),\chi)$ into the different spaces of modular forms with character. In particular, $M_k(\Gamma_0(N),\chi_0)=M_k(\Gamma_0(N))$ where $\chi_0$ is the trivial Dirichlet character mod $N$.

So in essence, studying modular forms for any congruence subgroup reduces to the case of studying modular forms for $\Gamma_1(N)$ or $\Gamma_0(N)$.

Edit (For additional requested info)

First, I'll elaborate on the injection $M_k(\Gamma(N))\to M_k(\Gamma_1(N^2))$. So, consider the matrix \begin{equation*} A_N=\begin{pmatrix}N&0\\0&1\end{pmatrix}. \end{equation*} We have \begin{equation} A_N^{-1}\Gamma(N)A_N=\Gamma^*(N):=\left\{\begin{pmatrix}a&b/N\\cN&d\end{pmatrix}\::\:\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma(N)\right\}\qquad (1) \end{equation}

Now $\Gamma_1(N^2)\subseteq\Gamma^*(N)$ so that there is a canonical injection $M_k(\Gamma^*(N))\to M_k(\Gamma_1(N^2))$. Then, $M_k(\Gamma(N))$ is isomorphic to $M_k(\Gamma^*(N))$ via the map $M_k(\Gamma(N))\to M_k(\Gamma^*(N))$, $f\mapsto f|_kA_N$. Putting these maps together gives an injection $M_k(\Gamma(N))\to M_k(\Gamma_1(N^2))$ as claimed.

Example application: In Li's paper "Newforms and Functional Equations", she focuses on defining and analysing newforms with respect to $\Gamma_0(N)$ and character $\chi$. However, by the above discussion, we can extend this theory to newforms for any congruence subgroup.

As a final remark, note that there are other congruence subgroups that are studied in literature. For instance, we have the subgroup $\Gamma^*(N)$ from Equation $(1)$ and the family of subgroups \begin{equation*} \Gamma_0(M,N)=\left\{\begin{pmatrix}a&b\\c&d\end{pmatrix}\::\:M\mid c,\:N\mid b\right\}. \end{equation*} that also appear in Li's paper. These subgroups are usually considered to simplify notation and keep track of things but generally we are interested in results regarding $\Gamma(N)$, $\Gamma_0(N)$ or $\Gamma_1(N)$.

$\endgroup$
3
  • 1
    $\begingroup$ Thanks @Daniel. The injective property $M_k(\Gamma(N)) \to M_k(\Gamma_1(N^2))$ is new for me. Does it mean $\Gamma_1(N^2) \subset \Gamma(N)$? Also, if we can reduce $M_k(\Gamma)$ to $M_k(\Gamma_1(N))$ and $M_k(\Gamma_0(N))$ why people, in many situations, consider other subgroups? $\endgroup$
    – Desunkid
    Commented May 25, 2020 at 5:48
  • 1
    $\begingroup$ These are both good questions! I will add another edit to my answer in a few hours when I'm free, answering both in detail. $\endgroup$ Commented May 25, 2020 at 6:08
  • 1
    $\begingroup$ Thanks @Daniel. It's great $\endgroup$
    – Desunkid
    Commented May 25, 2020 at 10:28
1
$\begingroup$

If $\begin{pmatrix}* & * \\ 0 & 1 \end{pmatrix}$ is in $SL_2$, then its determinant is $1$, thus the top left coefficient is $1$. Does it answer your question ?

$\endgroup$
2
  • $\begingroup$ Thanks. Maybe my previous example is so stupid. I added a new one. $\endgroup$
    – Desunkid
    Commented May 24, 2020 at 22:14
  • 1
    $\begingroup$ I think your new example is not a subgroup ! $\endgroup$
    – Didier
    Commented May 24, 2020 at 22:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .