$F=\{z\in\Bbb{H}:\ |z|>1,\ 2|\Re(z)|<\lambda\}$ is fundamental domain for $G_\lambda$(the subgroup of $SL(2,\Bbb{R})$ generated by $S$ and $T_\lambda$

Question

Let $0<\lambda<2$ be a real number and $G_\lambda$ be the subgroup of $SL(2,\Bbb{R})$ generated by $S=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$ and $T_\lambda=\begin{pmatrix}1 & \lambda\\ 0 & 1\end{pmatrix}$. Prove that, $$F_\lambda=\{z\in\Bbb{H}:\ |z|>1,\ 2|\Re(z)|<\lambda\}$$ is a fundamental domain for group $G_\lambda$.

I can prove the statement for $\lambda=1$ using the following two lemmas-

1. For $\tau'\in\Bbb{H}$, there exists $\tau \in \Bbb{H}$ such that $\tau=A\tau'$ for some $A\in G_1$ with $|\tau|\ge 1$ and $2|\Re(\tau)|\le 1$

2. If $z\in F_1$ and $Az\in F_1$ for some $A\in G_1$, then $A= I$.

But for $\lambda=1$, $G_1=SL(2,\Bbb{Z})$ (hence the entries of the matrices are integers) and I've used the concept of Lattice to prove the above two lemmas. I cannot generalize the proof for any $\lambda\in (0,2)$.

Can anyone help me to prove the statement? Thanks for help in advance.

Answer 1

Answered in the comments, but just to close this out and give an elementary counterexample:

Let

$$\zeta = \frac{-\lambda+ \sqrt{\lambda^2 - 4}}{2}.$$ $$\zeta' = \frac{\lambda+ \sqrt{\lambda^2 - 4}}{2}.$$

Assuming $0 < \lambda < 2$ these are both in the upper half plane $\mathbb{H}$. Note that $|\zeta| = |\zeta'| = 1$ and $2 |\mathrm{Re}(\zeta)| = 2 |\mathrm{Re}(\zeta')| = 1$. These are the two "corners" of the hyperbolic triangle $F_{\lambda}$. Moreover $\zeta \zeta' = -1$. Hence $T \zeta = \zeta'$, $S \zeta' = \zeta$, and thus, if $R = ST$, that

$$R \zeta = \zeta.$$

The stabilizer in $\mathrm{PSL}_2(\mathbb{R})$ of any point in $\mathbb{H}$ is isomorphic to $\mathrm{SO}_2(\mathbb{R})$, that is, will be a hyperbolic rotation around $\zeta$. But now you see a problem; if this rotation is not given by an element of finite order, the orbit of any point different from $\zeta$ under interates of $R \in G$ will be dense in a hyperbolic circle. This is certainly incompatible with the claim. Equivalently, it must be the case that $R$ has finite order and the eigenvalues of $R$ are roots of unity. But $\zeta$ itself is an eigenvalue of $R$, so if $\zeta$ is not a root of unity then the statement is certainly wrong.

To be completely explicit, let $\lambda = 6/5$, so $\zeta = -3/5 + 4i/5$, and

$$R =TS = \left( \begin{matrix} 6/5 & -1 \\ 0 & 1 \end{matrix} \right).$$

Now for any $x \in F_{\lambda}$, $R^n x$ will be dense and hence infinitely many other iterates will be inside $F_{\lambda}$. For example, if $x = 2 i$, then

$$R^{10} x = \frac{-8648173707066 + 38146972656250 i}{25886783980445} $$

lies in $F_{\lambda}$ but is not equal to $x$.