Group generated by commuting elements of PSL(2, $\mathbb{R})$

Question

I read a question that says given $f,g \in PSL(2,\mathbb{R})$, such that $f \circ g = g \circ f$, show $g(Fix(f)) = Fix(f)$ where $Fix(f) = \{x \in \mathbb{H} \cup \partial \mathbb{H}| f(x) =x\}$. Then we are asked to show that $\Gamma = \langle f,g \rangle$ is not Fuchsian. I believe this is not well-posed since Abelian Fuchsian groups exist, I believe. Am I correct?

Answer 1

If $f,g$ are commuting elements of $PSL(2,\mathbb R)$ then the group $\langle f,g\rangle$ they generate may or may not be discrete, and hence using the definition in your comment that group may or may not be Fuchsian. So if Fuchsian is defined in your source as you say it is, then you are correct, the problem is not well-posed.

For example, if $f = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ and $g = \begin{bmatrix}1 & \sqrt{2} \\ 0 & 1\end{bmatrix}$ then $\langle f,g\rangle$ is not discrete hence not Fuchsian.

For another example, if $f = g = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ then $\langle f,g\rangle$ is discrete, hence Fuchsian.

Or, to make it more interesting, if $f = \begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix}$ and $g = \begin{bmatrix}1 & 3 \\ 0 & 1\end{bmatrix}$, then $\langle f,g\rangle$ is discrete, hence Fuchsian.

The reason that I pressed you in the comments regarding how Fuchsian was defined in the source that you were consulting is that every abelian subgroup of $PSL(2,\mathbb R)$ is an elementary group (meaning one whose limit set in $\mathbb R \cup \{\infty\}$ is $0$, $1$ or $2$ points), and I believe that there may be some older sources where elementary subgroups are disallowed in the definition of Fuchsian. But I don't think that is standard usage now (if it ever was). Nowadays people are perfectly happy to speak about elementary Fuchsian groups.