High-order elements of $SL_2(\mathbb{Z})$ have no real eigenvalues

Question

Let $\gamma=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb{Z})$, $k$ the order of $\gamma$, i.e. $\gamma^k=1$ and $k=\min\{ l : \gamma^l = 1 \}$. I have to show that $\gamma$ has no real eigenvalues if $k>2$.

The eigenvalues of $\gamma$ are $\gamma_{1,2} = \frac{1}{2} (a+d \pm \sqrt{(a+d)^2-4})$, i.e. I have to show that $(a+d)^2<4$ for $k>2$.

How can I prove this? I have determined the first powers of $\gamma$ to get the condition directly from $\gamma^k = 1$ but I failed. Probably, there is an easier way?

Answer 1

Assume there is a real eigenvalue. Then the minimal polynomial of $\gamma$ is a divisor of $X^k-1$ and has degree at most $2$ and has at least one real root. If its degree is $2$, the other root must also be real. The only real roots of unity are $\pm1$, so the minimal polynomial os one of $X-1$, $X+1$ or $(X-1)(X+1)=X^2-1$. All three are divisors of $X^2-1$, i.e. we find $\gamma^2=1$.

Answer 2

In dimension $2$, if there is one real eigenvalue the characteristic polynomial splits (two real eigenvalues, possibly equal). If $\lambda$ is eigenvalue of $\gamma$ satisfying $\gamma^k=1$ then $\lambda^k=1$, and with $\lambda\in\Bbb R$ this means $\lambda\in\{-1,+1\}$. Also the minimal polynomial of $\gamma$ divides $X^k-1$ so has no multiple roots; therefore $\gamma$ is diagonalisable and $\gamma^2=1$, so $k\leq2$. That $\gamma\in SL_2(\Bbb Z)$ has nothing to do with it (apart for the dimension $2$).