Characteristic polynomial of a matrix which stabilizes a point

Question

I'm confused on a couple of points in this paragraph in A First Course in Modular Forms, section 2.3.

Returning to elliptic points, suppose $\tau \in \mathscr H$ is fixed by a nontrivial transformation $\begin{pmatrix} a & b \\ c & d\end{pmatrix} \in \operatorname{SL}_2(\mathbb Z)$. Then $a \tau + b = c \tau^2 + d \tau$; solving for $\tau$ with the quadratic equation ($c = 0$ is impossible since $\tau \not\in \mathbb Q)$ and remembering that $\tau \in \mathscr H$ shows that $|a+d| < 2$. Thus the characteristic polynomial of $\gamma$ is $x^2 +1$ or $x^2 \pm x + 1$. Since $\gamma$ satisfies its characteristic polynomial, one of $\gamma^4 = I, \gamma^3 = I, \gamma^6 = I$ holds.

First, from where can we conclude $|a+d| < 2$? Next, how are they making conclusions about the characteristic polynomial

$$\det \begin{pmatrix} t - a & -b \\ -c & t - d\end{pmatrix} = t^2 + (-a-d)t + ad - bc = t^2 + (-a-d)t + 1$$

from the equation $a \tau + b = c \tau^2 + d \tau$?

Answer 1

For the fixed points to be in the upper half plane, the discriminant of $c\tau^2+d\tau=a\tau+b$ needs to be negative. That amounts to $(a+d)^2<4$.

Answer 2

• Since $\tau$ is complex and $a,b,c,d$ are real, the discriminant of the quadratic equation for $\tau$ must be negative: $(d-a)^2+4bc<0$. This inequality may be rewritten as $$(d+a)^2<4(ad-bc)=4\text{.}$$

• Since $a$ and $d$ are integers, $a+d$ has to be an integer of absolute value less than $2$. Therefore $a+d\in \{-1,0,1\}$.