3
votes
1answer
127 views

Monstrous Moonshine for $M_{24}$?

This is connected to my MO post "Monstrous Moonshine for $M_{24}$ and K3?". In page 44 of this paper, eqn(7.16) and (7.19) yield, ...
2
votes
1answer
99 views

Calculating index of a subgroup

Compute the index $[Γ( 1 ) ′ : Γ_0 ( N ) ′ ]$ where $Γ(1)' := SL(2,\mathbb{Z})$ $Γ_0(N)':= \{ \begin{pmatrix} a & b\\ c & d\\ \end{pmatrix} \in Γ(1)' : c \equiv 0 \mod{N} \} $ I'm ...
1
vote
1answer
18 views

Degree and ramification index of a natural projection

Let $\Gamma$ a subcongruence group of $\text{SL}_2(\mathbb{Z})$, $\mathbb{H}^* = \mathbb{H} \cup \{\infty\} \cup \mathbb{Q}$ and $\overline\Gamma = \Gamma / (\Gamma\cap\{\pm 1\})$. Given the natural ...
5
votes
3answers
150 views

Purpose of cusps

In the theory of of modular forms, there is the set of of cusps defined by $\mathbb{P}^1 (\mathbb{Q})= \mathbb{Q} \cup \{\infty\}$. For an subgroup $\Gamma < \text{SL}_2(\mathbb{Z})$ of finite ...
3
votes
1answer
90 views

writing $M : \Gamma_{n,0} \backslash \Gamma_n$

Let $\Gamma_n = \operatorname{Sp}_n(\Bbb Z)$ and $\Gamma_{n,0} \subset \Gamma_n$ be a subgroup. We write $$ M = \begin{pmatrix}A & B\\ C & D \end{pmatrix} \in \operatorname{Sp}_n(\mathbb{Z})$$ ...
1
vote
2answers
71 views

High-order elements of $SL_2(\mathbb{Z})$ have no real eigenvalues

Let $\gamma=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z})$, $k$ the order of $\gamma$, i.e. $\gamma^k=1$ and $k=\min\{ l : \gamma^l = 1 \}$. I have to show that $\gamma$ ...
5
votes
2answers
201 views

How to calculate $|\operatorname {SL}_2(\mathbb Z/N\mathbb Z)|$?

the answer should be $$|\operatorname {SL}_2(\mathbb Z/N\mathbb Z)|=N^{3}\prod_{p|N}(1-{1 /p^2})$$ But first how to prove $$|\operatorname {SL}_2(\mathbb Z/p^e\mathbb Z)|=p^{3e}(1-{1 /p^2})$$
9
votes
1answer
306 views

Relation between congruence subgroups. $\Gamma(M)\Gamma(N) = \Gamma(\gcd(M,N))$

I'm hoping there's a pleasant way to solve this one. Prove that $\Gamma(M)\Gamma(N) = \Gamma(\gcd(M,N))$. Showing that $\Gamma(M)\Gamma(N) \subset \Gamma(\gcd(M,N))$ is rather straight forward, ...