About stabilisers under Möbius transformations
问题内容
The action of $\mathrm{SL}_2(\mathbb{Z})$ on $\mathbb{H}$ (the upper-half plane) is defined by the Möbius transformation $$\gamma\cdot z\longrightarrow \frac{az+b}{cz+d}\qquad\text{where}\qquad\gamma=\begin{pmatrix} a & b \\ c& d \end{pmatrix}\in\mathrm{SL}_2(\mathbb{Z}).$$
I found that $\gamma \cdot i=i$ implies that $\gamma\in \left\langle S \right\rangle$ where $S=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$, and that $\gamma \cdot \omega=\omega$ implies $\gamma\in\left\langle W\right\rangle$ where $W=\begin{bmatrix} 0 & 1 \\ -1 & -1 \end{bmatrix}$. Note that $\omega^3=1$ and $W^3=I$, and $i^2=-1$ and $S^4=I$.
Now if $\gamma\cdot z=z$ and $z^n=1$, does that imply that $\gamma^n=\pm \ I$? Is this true or false? The reason I wrote $\pm\ I$ is that the Möbius transformation with respect to $\gamma$ and $-\gamma$ are the same.
回答 (1)
Write $$ \gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathrm{SL}(2, \mathbb{Z}). $$ Then if $\gamma z = z$, we have $$ \frac{az + b}{cz + d} = z \implies az + b = cz^2 + d, $$ or equivalently $$ cz^2 + (d - a) z - b = 0. \tag{1}$$
Thus $z$ is a root of a quadratic polynomial with real coefficients.
If $c = 0$, then either $a = d = 1$ or $a = d = -1$, and ultimately we have $\gamma = \pm I$.
If $c \neq 0$, then we have a bit more work to do.
Recall from the quadratic formula that a quadratic $Ax^2 + Bx + C$ has two real roots, a repeated root, or two complex roots depending on its discriminant $D := B^2 - 4AC$.
The discriminant of the quadratic equation in (1) is $D := (d - a)^2 + 4bc$. Using that $ad - bc = 1$, this can be rewritten as $D = (a + d)^2 - 4$. As $z$ is a complex root, we know that $D < 0$, which implies that $$ (a + d)^2 - 4 < 0 \iff \lvert a + d \rvert < 2. $$
(Aside: this fact is sometimes stated in a terse form, that $\gamma$ is either the $\pm I$ or "elliptic").
This now reduces one to checking what happens in three cases. I'll do one and just say what happens in the other two.
The case when $a + d = 0$
Going back to $(1)$, we have the equation $$ cz^2 - 2az - b = 0. $$ The quadratic formula shows the solutions are $$ z = \frac{a \pm \sqrt{a^2 + bc}}{c}. $$ And $ad - bc = 1$ combined with $a = -d$ implies that $a^2 + bc = -1$. Thus the solutions are $$ z = \frac{a \pm i}{c}. $$ And for our case, the fixed point is the one where $z \in \mathbb{H}$.
Now impose the condition that $\lvert z \rvert = 1$. This shows that $$ \lvert z \rvert^2 = \frac{a^2 + 1}{c^2} = 1. $$ Rearranging, this shows that $c^2 - a^2 = 1$, or that $(c + a)(c - a) = 1$. As $a$ and $c$ are integers, this means that both factors are $1$ or both factors are $-1$. These end up giving solutions $(a, c) = (0, \pm 1)$ and fixed points $z = \pm i$. The one in the upper halfplane is $i$.
Other cases: $a + d = \pm 1$
The two other cases are very similar. In both, the fixed point takes the shape $$ z = \frac{a - d + i \sqrt{3}}{2c}. $$ Similar analysis shows that this reduces to $\omega$ with $\omega^3 = 1$.
And in total, the claim is right and these are the only such points.