Proving if $\gamma\in SL(2,\mathbb{Z})$ has order $6$, then $\gamma$ is conjugate to $\begin{bmatrix}0 & -1\\ 1 & 1\end{bmatrix}^{\pm1}$

Question

I am reading Diamond's book on modular forms, and I came across the proof that the stabilizers of elliptic points on the upper half-plane are cyclic. To prove that, Diamond starts to show that a transformation that fixes a point must have order $1, 2, 3, 4$ or $6$. For $\operatorname{ord}(\gamma)=6$, the author proceeds to argue that $\gamma$ is conjugate to matrices $\begin{bmatrix}0 & -1\\ 1 & 1\end{bmatrix}^{\pm1}$.

However, I found the language he uses a little bit confusing.

Since $\gamma^6=I$, the lattice $L=\mathbb{Z}^2$ of integral column vectors is a module over $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$, defining the scalar product $(a+b\mu_6)\cdot v$ for $a,b\in\mathbb{Z}$ and $v\in L$ by the matrix-by-vector product $(aI+b\gamma)v$

In this case,

• What does the sentence "the lattice $L=\mathbb{Z}^2$ of integral column vectors" mean? Is $L$ just the set $\left\{\begin{bmatrix}a\\ b\end{bmatrix}:a,b\in\mathbb{Z}\right\}$?
• Also, is defining $(a+b\mu_6)\cdot v$ by $(aI+b\gamma)v$ an action to ensure that $(sr)\cdot v=s\cdot(r\cdot v)$ as$$(aI+b\gamma)(cI+d\gamma)=acI+bd\gamma^2+(ad+bc)\gamma=acI+bd(\gamma-I)+(ad+bc)\gamma=(ac-bd)I+(bc+ad+bd)I$$which is consistent with multiplications in $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$?
• And finally, how can we show that $L$ is finitely generated over $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$? Can this be done by simply saying that it can be generated by $\begin{bmatrix}1\\ 0\end{bmatrix}$ and $\begin{bmatrix}0\\ 1\end{bmatrix}$ with coefficients $a, b\in\mathbb{Z}\subseteq\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$?

Thanks in advance. Any help will be appreciated.

Answer 1

• Completely right.
• You simply have to show that the given identification $\mathbb Z[\mu_6] \to M_{2\times 2}(\mathbb Z)$ is a ring homomorphism. The only unclear part is that the identification preserves multiplication. Instead of doing it by hand what you tried to do you could realize that $\mu_6$ and $\gamma$ are both zeros of the polynomial $x^2-x+1$. So the square part breaks down on both sides the same way.
• Right, $L$ is generated by those two vectors as an abstract group so in particular as $\mathbb Z[\mu_6]$ module.

By the way $\mathbb Z[\mu_6]$ is a principal ideal domain like I am (see my user name ).