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I am reading chapter 3 of A First Course in Modular Forms but have troubles in Exercise 3.7.1 (c) and (d).

(c) Show that the $\Gamma_0^{\pm}(N)$-conjugacy class of $\gamma \in \Gamma_0(N)$ is the union of the $\Gamma_0(N)$ conjugacy classes of $\gamma$ and $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \gamma \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$. Show that if $\gamma$ has order $4$ or $6$ then this union is disjoint.

(d) Let $\gamma = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}^{-1} = \begin{bmatrix} 3 & -2 \\ 5 & -3 \end{bmatrix}$, an order-$4$ element of $\Gamma_0(5)$. Show that $\gamma$ is not conjugate to its inverse in $\Gamma_0(5)$.

Here let $\mathrm{GL}_2(\mathbb{Z})$ be the group of invertible $2 \times 2$ matrices with integer entries, $\mathrm{SL}_2(\mathbb{Z})$ the group of $2 \times 2$ matrices with integer entries and determinant $1$, and \begin{align*} \Gamma_0^{\pm}(N) &= \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \mathrm{GL}_2(\mathbb{Z}) : \begin{bmatrix} a & b \\ c & d \end{bmatrix} \equiv \begin{bmatrix} * & * \\ 0 & * \end{bmatrix} \pmod{N} \right\} \\ \Gamma_0(N) &= \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \mathrm{SL}_2(\mathbb{Z}) : \begin{bmatrix} a & b \\ c & d \end{bmatrix} \equiv \begin{bmatrix} * & * \\ 0 & * \end{bmatrix} \pmod{N} \right\}. \end{align*}

I have proved the first part of (c): the map \begin{align*} \left(\Gamma_0^{\pm}(N) \setminus \Gamma_0(N)\right) &\longrightarrow \Gamma_0(N) \\ \alpha &\longmapsto \alpha \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \end{align*} is a bijection, so \begin{align*} \left\{ \beta \gamma \beta^{-1}: \beta \in \Gamma_0^{\pm}(N) \setminus \Gamma_0(N) \right\} = \left\{ \alpha \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \gamma \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \alpha^{-1}: \alpha \in \Gamma_0(N) \right\}. \end{align*}

How to show the remaining parts? For part (d), I think one cannot apply (c) directly because (d) serves as an example of the following statement in P.93

The extended conjugacy class of $\gamma$ under $\Gamma_0^{\pm}(N)$ is not in general the union of the conjugacy class of $\gamma$ and $\gamma^{-1}$ under $\Gamma_0(N)$.

fyhung88
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1 Answers1

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Let

$$\Gamma_0^-(N)=\left\{ \begin{bmatrix} a & b\\ c & d \end{bmatrix}\in\Gamma_0^\pm(N):ad-bc=-1 \right\},$$

so the $\Gamma_0^\pm(N)$-conjugacy classes of $\gamma$ is the union of $\Gamma_0(N)$-conjugacy classes of $\gamma$ and $\Gamma_0^-(N)$-conjugacy classes of $\gamma$. Let $\alpha=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\in\Gamma_0^-(N)$, then

$$\alpha\gamma\alpha^{-1}=\begin{bmatrix} a & -b\\ c & -d \end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}\gamma\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}\begin{bmatrix} -d & b\\ -c & a \end{bmatrix},$$

where $\begin{bmatrix} a & -b\\ c & -d \end{bmatrix}\in\Gamma_0(N)$ and $\begin{bmatrix} -d & b\\ -c & a \end{bmatrix}=\begin{bmatrix} a & -b\\ c & -d \end{bmatrix}^{-1}$. Let $\alpha=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\in\Gamma_0(N)$ this time, then

$$\alpha\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}\gamma\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}\alpha^{-1}=\begin{bmatrix} a & -b\\ c & -d \end{bmatrix}\gamma\begin{bmatrix} d & -b\\ c & -a \end{bmatrix},$$

where $\begin{bmatrix} a & -b\\ c & -d \end{bmatrix}\in\Gamma_0^-(N)$ and $\begin{bmatrix} d & -b\\ c & -a \end{bmatrix}=\begin{bmatrix} a & -b\\ c & -d \end{bmatrix}^{-1}$.

Thus the $\Gamma_0^-(N)$-conjugacy classes of $\gamma$ is equal to the $\Gamma_0(N)$-conjugacy classes of $\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}\gamma\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}$. The first statement of (c) is true.

By Proposition 2.3.3, if $\gamma$ has order $4$ then $\gamma$ is conjugate to $\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}^{\pm1}=\pm\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}$ in $\mathrm{SL}_2(\mathbb Z)$. If this two conjugacy classes intersects, i.e. $\gamma$ is conjugate to $\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}\gamma\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}$ in $\Gamma_0(N)$, then $\gamma$ is conjugate to $\gamma$ in $\Gamma_0^-(N)$, then $\pm\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}$ is conjugate to $\pm\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}$ in $\mathrm{GL}_2^-(\mathbb Z)$, where

$$\mathrm{GL}_2^-(\mathbb Z)=\left\{ \begin{bmatrix} a & b\\ c & d \end{bmatrix}\in\mathrm{GL}_2(\mathbb Z):ad-bc=-1 \right\}.$$

Assuming $\pm\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}=\pm\alpha\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}\alpha^{-1}$ and $\alpha=\begin{bmatrix} a & b\\ c & d \end{bmatrix}\in\mathrm{GL}_2^-(\mathbb Z)$, then

$$\pm\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}=\pm\begin{bmatrix} -(ac+bd) & a^2+b^2\\ -(c^2+d^2) & ac+bd \end{bmatrix}.$$

Thus $a^2+b^2=-1$, a contradiction.

If $\gamma$ has order $6$, we deduce that $\begin{bmatrix} 0 & -1\\ 1 & 1 \end{bmatrix}$ and $\begin{bmatrix} 1 & 1\\ -1 & 0 \end{bmatrix}$ are conjugate to themselves in $\mathrm{GL}_2^-(\mathbb Z)$ by similar way, and which yields

$$\begin{bmatrix} 0 & -1\\ 1 & 1 \end{bmatrix}=\begin{bmatrix} bc-ac-bd & a^2+b^2-ab\\ cd-c^2-d^2 & ac+bd-ad \end{bmatrix},$$

$$\begin{bmatrix} 1 & 1\\ -1 & 0 \end{bmatrix}=\begin{bmatrix} ac+bd-ad & ab-a^2-b^2\\ c^2+d^2-cd & bc-ac-bd \end{bmatrix}.$$

Thus we have $\left( a-\frac12b \right)^2+\frac34b^2=-1$ and $\left( c-\frac12d \right)^2+\frac34d^2=-1$, contradictions.

Kevin
  • 994