<s>[INST] Fixed points of group action Let $G=SL_2(Z)$ work on $\{z \in C | Im(z) \gt 0\}$ by $\begin{pmatrix}a&b \\ c&d\end{pmatrix}(z)=\frac{az+b}{cz+d}$.
First of all I want to find which elements of G leave $\zeta_3$ fixed.
Writing it out gives me $c\zeta_3^2+(d-a)\zeta_3-b=0$. If I now use $\zeta_3=e^{\frac{2\pi i}{3}}$ i find $a+c-d=0, -a+2b+c+d=0$ and together with $ad-bc=1$ (det=1) and the fact that a,b,c,d are from Z I can calculate the solutions with a computer. However there must be a much more elegant solution to this problem. Help would be much appreciated
Next I want to show that the action has no fixed points if |a+d|>2. Thus $cz^2+(d-a)z-b=0$ has no solutions (with Im(z)>0) when |a+d|>2
Solutions are $z=\frac{a-d}{2c}+\frac{\sqrt{(d-a)^2+4bc}}{2c}$, which has imaginary part $\sqrt{|(d-a)^2+4bc|}/2c$. But nowhere I can see the |a+d| coming in, so I'm stuck here?
Thanks in advance
 [/INST] You don't need a computer. From $a+c-d=0$, you can substitute $d = a+c$ into $-a+2b+c+d = 0$ to get $2(b+c) = 0$. Thus you are at
$$\begin{pmatrix}a&b\\-b&a-b \end{pmatrix}.$$
Now the determinant gives you $a(a-b) + b^2 = 1$, or $(2a-b)^2 + 3b^2 = 4$, so you have


*

*$b = 0$, and $a = \pm 1$, which gives the identity, or

*$b = 1$, and $a \in \{0,1\}$, or

*$b = -1$, and $a\in \{0,-1\}$ which gives the same maps as the previous.


So, up to multiplication by $-1$:
$$\begin{pmatrix} 1&0\\0&1\end{pmatrix}: z\mapsto z;\quad \begin{pmatrix} 0 & -1\\1&1\end{pmatrix}: z \mapsto -\frac{1}{z+1};\quad \begin{pmatrix}-1&-1\\1&0 \end{pmatrix}: z \mapsto -\left(1+\frac1z\right).$$
We have the fixed points
$$\frac{a-d \pm \sqrt{(a-d)^2+4bc}}{2c} = \frac{a-d \pm\sqrt{a^2+d^2 -4(ad-bc)}}{2c} = \frac{a-d\pm\sqrt{a^2+d^2-4}}{2c},$$
and for $c \neq 0$, they are real if $a^2+d^2 \geqslant 4$, which is the case unless $\lvert a\rvert \leqslant 1$ and $\lvert d\rvert \leqslant 1$.
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