Orbit of $z$ under $SL_2({\mathbb Z})$ action

Question

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.

Answer 1

After spending some time on this problem, I have figured out a fairly complete answer, so here I am answering my own question. The orbit of $i$ under the action of the modular group is symmetric around the imaginary axis, and has period 1 (this is easy to see). So it is enough if we describe the orbit in the rectangle $R=\lbrace z: 0\leq\mbox{Re}(z)\leq 1/2, 0< \mbox{Im}(z)\leq 1\rbrace$. Since the imaginary part is always $1/N$ for some positive integer $N$, the orbit in this rectangle is made of finitely many points on the horizontal lines $\mbox{Im}(z)=1/N$. So in particular there are no points with $\mbox{Im}(z)>1$ or in the strip $1/2<\mbox{Im}(z) <1$. But not all integers $N$ will appear in the denominators of the imaginary parts. That is because the denominator is always of the form $N=c^2+d^2$, with $c,d$ relatively prime integers. For each pair $(c,d)$ of relatively prime integers, there will be at least one point of the orbit of $i$ with $\mbox{Im}(z)=1/(c^2+d^2)$ in the rectangle $R$. The number of points on each of these line segments will be equal to the number of solutions of the equation $x^2+y^2=N$ with $x,y$ relatively prime. This number can be determined by then prime factorization of $N$ (see the this post ) as follows. If there is a prime divisor of $N$ of form $4k+3$, then there are no solutions. If all the prime divisors of $N$ are of form $4k+1$ and there are $s$ of them, then there are $2^{s-1}$ solutions of $x^2+y^2=N$. This means that for such an $N$, there will be $2^{s-1}$ elements of the orbit of $i$ in the rectangle $R$ with imaginary part $1/N$. The real part will all be of the form $(ac+bd)/N$, where $a,b$ are such that $ad-bc=1$, with $a^2+b^2$ minimal. In other words, the $a,b$ are the unique integers obtained by the Euclidean algorithm to find the solution of $xd-yc=1$. It turns out that the first 16 integers $N$ for which $x^2+y^2=N$ has solutions in relatively prime integers are $1, 2, 5, 10, 13, 17, 25, 26, 29, 34, 37, 41, 50, 53, 58, 61$, and for each of these there is only one solution $(c,d)$. So the first 16 "floors" of the orbit of $i$ in the rectangle $R$ have only one point. The 17th floor, corresponding to $N=65$, has two solutions: $1^2+8^2=4^2+7^2$, so there will be two points in $R$ with $\mbox{Im}(z)=1/65$. Continuing this way, descending the levels, we found that the number of points in $R$ at each level is always a power of $2$.