Are matrices in $\text{PSL}(2, \mathbb{Z})$ conjugate to their inverses?

Question

As I understand it, this comes down to calculating the slope of the expanding eigenvector of each matrix... but I am having trouble with the details. I feel that the fact that we have identified every matrix with its negation means that this should be true, but again, I have not been able to make anything precise. Can anyone help with this?

Answer 1

They are not. Consider $$g=\begin{pmatrix}1&1\\0&1\end{pmatrix}\in\operatorname{PSL}_2(\mathbb Z).$$ Every conjugate of $g$ is (the projection of a matrix) of the form $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}=\begin{pmatrix}1-ac&a^2\\-c^2&1+ac\end{pmatrix}$$ for some $(a,b,c,d)\in\mathbb Z^4$ with $ad-bc=1$. If this is to be $g^{-1}$, we must have $$\begin{pmatrix}1-ac&a^2\\-c^2&1+ac\end{pmatrix}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$$ (these matrices cannot be negatives of each other since they both have trace $2$), which implies $a^2=-1$. This has no solutions over $\mathbb Z$.