The modular group $\mbox{SL}_2({\mathbb Z})=\left\lbrace\begin{pmatrix} a & b\\ c & d\end{pmatrix}: a,b,c,d \in {\mathbb Z}, ad-bc=1 \right\rbrace$ acts on the upper half plane ${\mathbb H}=\lbrace z\in {\mathbb C}: \mbox{Im}(z)>0\rbrace $ by $\begin{pmatrix} a & b\\ c & d\end{pmatrix}z=\dfrac{aż+b}{cz+d}$. For a fixed $\gamma \in \mbox{SL}_2({\mathbb Z})$, the geometry of this action is easily described and there are plenty of pictures everywhere, for example modular group action. But I cannot find a geometric picture of the orbit of a fixed point of $\mathbb H$. In other words, if $z\in {\mathbb H}$, what does the set $\lbrace \gamma(z): \gamma \in \mbox{SL}_2({\mathbb Z})\rbrace$ look like? I imagine that is because it is a very complicated set, as can be seen for example by this StackExchange post describing conditions in terms of intricate algebraic conditions just for the orbit of $i$. But still, I would imagine that it should be possible to produce some sort of computer plot to get an idea of what the orbit of $i$ or other special points are. I searched the web but could not find anything.
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2Some vertices of a tree, a discrete subset. If you take $i$ and $e^{2\pi i / 3}$ and the part of circle connecting them and take the orbit of that connecting arc under the group action you get a tree. – psl2Z Mar 22 '24 at 00:19
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@psi2Z would you have a reference for that result? – Math101 Mar 22 '24 at 23:36