Good description of orbits of upper half plane under $SL_2 (Z)$

It’s known that $SL_2(Z)$ acts on $H=\{z\, |\, Im(z)>0\}$, is there a good description of orbits of $i$ and $w$, other than directly write down $=\{ \frac{ac|z|^2+bc\bar z+adz+bd}{c^2|z|^2+dc\bar z +dcz+d^2}; ad-bc=1 \}$?

Solutions Collecting From Web of "Good description of orbits of upper half plane under $SL_2 (Z)$"

I give some more or less useful criteria, but mostly without proofs,
so these should be regarded as exercises.

Some conditions use popular modular forms and functions.
Other conditions are elementary, using only divisibility or requiring
some expressions to evaluate to integers. You do not need modforms for those,
so if you are only interested in elementary conditions,
you can ignore the others. There are some interesting connections however.

Functions $\mathbb{H}\mapsto\mathbb{C}$ used herein:

  • Dedekind eta function
    $\eta$, which is nonzero on $\mathbb{H}$,
  • Modular Weber functions
    $$\begin{align}
    \mathfrak{f}(\tau) &= \frac{\eta^2(\tau)}{\eta(\tau/2)\,\eta(2\tau)}
    & \mathfrak{f}_1(\tau) &= \frac{\eta(\tau/2)}{\eta(\tau)}
    & \mathfrak{f}_2(\tau) &= \frac{\sqrt{2}\,\eta(2\tau)}{\eta(\tau)} \\
    \gamma_2 &= \mathfrak{f}^8\mathfrak{f}_1^8
    + \mathfrak{f}^8\mathfrak{f}_2^8 – \mathfrak{f}_1^8\mathfrak{f}_2^8
    & \gamma_3 &= \frac{1}{2}\left|\begin{matrix}
    1 & \phantom{+}\mathfrak{f}^8 & \mathfrak{f}^{16} \\
    1 & -\mathfrak{f}_1^8 & \mathfrak{f}_1^{16} \\
    1 & -\mathfrak{f}_2^8 & \mathfrak{f}_2^{16}
    \end{matrix}\right| \\
    \text{Note:}\quad
    \mathfrak{f}\,\mathfrak{f}_1\,\mathfrak{f}_2 &= \sqrt{2}
    & \mathfrak{f}^8 &= \mathfrak{f}_1^8 + \mathfrak{f}_2^8
    & \mathfrak{f}^{24} &= \mathfrak{f}_1^{24} + \mathfrak{f}_2^{24} + 48
    \end{align}$$
  • Klein’s invariant
    $$\operatorname{j} = \gamma_2^3 = \gamma_3^2+1728$$
  • Eisenstein series
    $$\begin{align}
    \operatorname{E}_4 &= \gamma_2\eta^8
    &\operatorname{E}_6 &= \gamma_3\eta^{12}
    \end{align}$$
  • Jacobi Thetanulls
    $$\begin{align}
    \varTheta_{00} &= \mathfrak{f}^2\eta
    &\varTheta_{01} &= \mathfrak{f}_1^2\eta
    &\varTheta_{10} &= \mathfrak{f}_2^2\eta
    \end{align}$$
  • Modular lambda function
    $$\lambda = \frac{\mathfrak{f}_2^8}{\mathfrak{f}^8}$$

Some modform-based conditions,
each equivalent to $\tau$ being in the orbit of
$\mathrm{i}$ under the action of $\operatorname{SL}(2,\mathbb{Z})$ are:

  • $\operatorname{E}_6(\tau) = 0$
  • $\gamma_3(\tau) = 0$
  • $\operatorname{j}(\tau)-1728 = 0$ (double zero)
  • The tuple $\left(\mathfrak{f}^{24}(\tau), -\mathfrak{f}_1^{24}(\tau),
    -\mathfrak{f}_2^{24}(\tau)\right)$ is some permutation of $(-8,-8,64)$
  • $\left\{\mathfrak{f}^{24}(\tau), -\mathfrak{f}_1^{24}(\tau),
    -\mathfrak{f}_2^{24}(\tau)\right\}\cap\{-8,64\}\neq\emptyset$
  • The tuple $\left(\left|\mathfrak{f}(\tau)\right|,
    \left|\mathfrak{f}_1(\tau)\right|, \left|\mathfrak{f}_2(\tau)\right|
    \right)$ is some permutation of $(\sqrt[8]{2},\sqrt[8]{2},\sqrt[4]{2})$
  • As before, but two matches suffice
  • The set $\left\{\mathfrak{f}^8(\tau), -\mathfrak{f}_1^8(\tau),
    -\mathfrak{f}_2^8(\tau)\right\}$ has less than three elements
  • As before, exactly two elements
  • The set $\left\{\varTheta_{00}^4(\tau), -\varTheta_{01}^4(\tau),
    -\varTheta_{10}^4(\tau)\right\}$ has less than three elements
  • As before, exactly two elements
  • $\lambda(\tau) \in \left\{-1,\frac{1}{2},2\right\}$
  • The tuple $\left(\operatorname{E}_4\!\left(\frac{\tau}{2}\right),
    \operatorname{E}_4\!\left(\frac{\tau+1}{2}\right),
    16\operatorname{E}_4(2\tau)\right)$ is some permutation of
    $\left(-4\operatorname{E}_4(\tau),
    11\operatorname{E}_4(\tau), 11\operatorname{E}_4(\tau)\right)$
  • $\left\{\operatorname{E}_4\!\left(\frac{\tau}{2}\right),
    \operatorname{E}_4\!\left(\frac{\tau+1}{2}\right),
    16\operatorname{E}_4(2\tau)\right\}\cap
    \left\{-4\operatorname{E}_4(\tau), 11\operatorname{E}_4(\tau)\right\}
    \neq\emptyset$
  • The set $\left\{\operatorname{E}_4\!\left(\frac{\tau}{2}\right),
    \operatorname{E}_4\!\left(\frac{\tau+1}{2}\right),
    16\operatorname{E}_4(2\tau)\right\}$ has less than three elements

To prove these, it is probably easiest to begin with
showing $\operatorname{E}_6(\mathrm{i}) = 0$ using modular symmetries.
Then show $\operatorname{j}(\tau) = 1728\iff\operatorname{E}_6(\tau) = 0$.
Then use the fact that $\operatorname{j}$ maps the fundamental domain,
including half its boundary with nonnegative real part,
bijectively to $\mathbb{C}$ and deduce that
$\operatorname{E}_6$ has no zeros other than $\mathrm{i}$
in the fundamental domain. Then use the fact that $\operatorname{j}$
is invariant under modular transforms, and conclude that the set of zeros of
$\operatorname{E}_6$ equals the orbit of $\mathrm{i}$.
Transforming to the remaining conditions can mostly be done with the
identities given in the introduction.
For the conditions with $\operatorname{E}_4$, the relations with Thetanulls
in another answer may be helpful.

A geometric condition:

  • $\tau$ is where a Ford circle
    touches another Ford circle or the line $\Im\tau=1$

To prove that, it may be helpful to generate the set of Ford circles,
united with the line $\Im\tau=1$, from the orbit of the line segment
$\Im\tau=1$, $\Re\tau\in[0,1)$, under the action of the modular group.

More elementary conditions, pairwise equivalent:

$$\begin{align}
&\exists\begin{pmatrix}a&b\\c&d\end{pmatrix}
\in\operatorname{SL}(2,\mathbb{Z})\colon\
\tau = \frac{{a\mathrm{i}} + b}{{c\mathrm{i}} + d}
= \frac{ac + bd + \mathrm{i}}{c^2 + d^2}
\tag{1}\\
&\exists N,u\in\mathbb{Z};\,N\mid(u^2+1)\colon\
\tau = \frac{u + \mathrm{i}}{N}
\tag{2}\\
&\frac{1}{\Im\tau}\in\mathbb{Z}\text{ and }
\frac{\Re\tau}{\Im\tau}\in\mathbb{Z}\text{ and }
\frac{1}{\Im(-\tau^{-1})}\in\mathbb{Z}
\tag{3}
\end{align}$$
Proof sketch:

  • $(1)\implies(2)$:
    Set $N=c^2+d^2$ and $u=ac+bd$, then use
    $(ac+bd)^2+(ad-bc)^2 = (a^2+b^2)(c^2+d^2)$ to deduce $N\mid(u^2+1)$.
  • $(2)\implies(1)$: From $\Im\tau>0$ deduce $N>0$.
    Use induction on $N$.

    • The base case $N=1$ can be fulfilled with
      $\begin{pmatrix}a&b\\c&d\end{pmatrix}
      = \begin{pmatrix}1&u\\0&1\end{pmatrix}$
    • For any $M\in\mathbb{N}$, assume as induction hypothesis
      that $(2)\implies(1)$ holds for all $N\leq M$ and all $u\in\mathbb{Z}$
      such that $N\mid(u^2+1)$.
      Now let $N$ be the least integer greater than $M$ for which
      $-1$ is a quadratic residue modulo $N$, so there exists $u$ such that
      $N\mid(u^2+1)$. Set $m$ to the integer obtained by rounding
      the quotient $u/N$ upwards or downwards. Then set
      $$\begin{align}
      v &= mN-u
      &\therefore\quad
      |v| &\leq N-1
      & v^2 &\equiv u^2\pmod{N}
      \\k &= \frac{v^2+1}{N}
      &\therefore\quad
      k &\in \{1,2,\ldots,N-1\}
      & k &\mid (v^2+1)
      \end{align}$$
      Since $-1$ is a quadratic residue (congruent to $v^2$) modulo $k$,
      we have not only $k<N$, but $k\leq M$.
      By the induction hypothesis,
      $$\begin{align}
      \exists\begin{pmatrix}p&q\\r&s\end{pmatrix}
      \in\operatorname{SL}(2,\mathbb{Z})\colon\
      \frac{{p\mathrm{i}} + q}{{r\mathrm{i}} + s}
      &= \frac{v + \mathrm{i}}{k}
      \\\therefore\quad
      \begin{pmatrix}a&b\\c&d\end{pmatrix}
      = \begin{pmatrix}m&-1\\1&0\end{pmatrix}
      \begin{pmatrix}p&q\\r&s\end{pmatrix}
      &\in\operatorname{SL}(2,\mathbb{Z})
      \\\implies
      \frac{{a\mathrm{i}} + b}{{c\mathrm{i}} + d}
      = m – \frac{k}{v+\mathrm{i}}
      &= m – \frac{v-\mathrm{i}}{N}
      = \frac{u + \mathrm{i}}{N}
      \end{align}$$
      So the hypothesis also holds for $N$.
  • $(2)\implies(3)$:
    We have $\frac{1}{\Im\tau} = N$,
    $\frac{\Re\tau}{\Im\tau} = u$,
    $\frac{1}{\Im(-\tau^{-1})} = \frac{u^2+1}{N}$.

  • $(3)\implies(2)$:
    Setting $N=\frac{1}{\Im\tau}$, $u = \frac{\Re\tau}{\Im\tau}$
    implies $\tau=\frac{u+\mathrm{i}}{N}$ and
    $\frac{u^2+1}{N} = \frac{1}{\Im(-\tau^{-1})}\in\mathbb{Z}$.

Example:
Consider $u=8$, $N=u^2+1=65$. This implies
$\operatorname{E}_6\!\left(\frac{8+\mathrm{i}}{65}\right)=0$.
However, we also have
$\operatorname{E}_6\!\left(\frac{18+\mathrm{i}}{65}\right)=0$,
and indeed $65\mid(18^2+1)$.

Remark:
Consequently, $65\mid(18^2-8^2)=(18-8)(18+8)$, and computing
$\operatorname{gcd}(65,18-8)=5$ reveals a factor of $65$.

In principle, one might use that to factor integers $N$ that are representable
as the sum of two coprime squares, with the help of some method for finding
zeros of $\operatorname{E}_6$ on a given line segment in $\mathbb{H}$,
e.g. $\Im\tau=\frac{1}{N}$, $\Re\tau\in\left[0,\frac{1}{2}\right]$.
In practice however, modforms oscillate so wildly at small $\Im\tau$
that the usual zero-finding methods cannot outperform any established integer
factoring method.

Therefore, while it might be tempting to say that factoring integers can be
as easy as finding zeros of modforms on a given line segment,
it is probably more adequate to state that locating zeros of modforms can be
as hard as factoring integers.
End of remark.

Let us consider orbits of $\omega = \exp\frac{2\pi\mathrm{i}}{3}$ now.
Proofs are analogous to those indicated above and left as exercise.

Some modform-based conditions,
each equivalent to $\tau$ being in the orbit of
$\omega$ under the action of $\operatorname{SL}(2,\mathbb{Z})$ are:

  • $\operatorname{E}_4(\tau) = 0$
  • $\gamma_2(\tau) = 0$
  • $\operatorname{j}(\tau) = 0$ (triple zero)
  • $\mathfrak{f}^{24}(\tau) = -\mathfrak{f}_1^{24}(\tau)
    = -\mathfrak{f}_2^{24}(\tau) = 16$
  • $16\in\{\mathfrak{f}^{24}(\tau), -\mathfrak{f}_1^{24}(\tau),
    -\mathfrak{f}_2^{24}(\tau)\}$
  • $\frac{\mathfrak{f}_1^8(\tau)}{\mathfrak{f}_2^8(\tau)}
    = -\frac{\mathfrak{f}^8(\tau)}{\mathfrak{f}_1^8(\tau)}
    = -\frac{\mathfrak{f}_2^8(\tau)}{\mathfrak{f}^8(\tau)}
    \in \{\omega^{\pm1}\}$
  • The set $\left\{\frac{\mathfrak{f}_1^8(\tau)}{\mathfrak{f}_2^8(\tau)},
    -\frac{\mathfrak{f}^8(\tau)}{\mathfrak{f}_1^8(\tau)},
    -\frac{\mathfrak{f}_2^8(\tau)}{\mathfrak{f}^8(\tau)},
    \omega,\omega^{-1}\right\}$ has less than five elements
  • $\left|\mathfrak{f}(\tau)\right|
    = \left|\mathfrak{f}_1(\tau)\right|
    = \left|\mathfrak{f}_2(\tau)\right|
    = \sqrt[6]{2}$
  • The set $\left\{\left|\mathfrak{f}(\tau)\right|,
    \left|\mathfrak{f}_1(\tau)\right|,
    \left|\mathfrak{f}_2(\tau)\right|,
    \sqrt[6]{2}\right\}$ has less than three elements
  • $\widehat\eta\!\left(\frac{\tau+1}{2}\right)
    = \widehat\eta\!\left(\frac{\tau}{2}\right)
    = \frac{\widehat\eta(\tau)}{\sqrt[12]{2}}
    = \widehat\eta(2\tau)
    = \frac{\widehat\eta(\omega)}{\sqrt[12]{2}}
    = \frac{\sqrt[6]{2}\sqrt{\pi}}
    {\sqrt{3}\,\Gamma^{3/2}\!\left(\frac{2}{3}\right)}$

    where
    $\widehat\eta(\tau) = \sqrt[4]{\Im\tau}\left|\eta(\tau)\right|$
    (invariant under modular transforms)

  • The set $\left\{\widehat\eta\!\left(\frac{\tau+1}{2}\right),
    \widehat\eta\!\left(\frac{\tau}{2}\right),
    \frac{\widehat\eta(\tau)}{\sqrt[12]{2}},
    \widehat\eta(2\tau)\right\}$ has less than three elements
  • $\frac{\varTheta_{01}^4(\tau)}{\varTheta_{10}^4(\tau)}
    = -\frac{\varTheta_{00}^4(\tau)}{\varTheta_{01}^4(\tau)}
    = -\frac{\varTheta_{10}^4(\tau)}{\varTheta_{00}^4(\tau)}
    \in \{\omega^{\pm1}\}$
  • The set $\left\{\frac{\varTheta_{01}^4(\tau)}{\varTheta_{10}^4(\tau)},
    -\frac{\varTheta_{00}^4(\tau)}{\varTheta_{01}^4(\tau)},
    -\frac{\varTheta_{10}^4(\tau)}{\varTheta_{00}^4(\tau)},
    \omega, \omega^{-1}\right\}$ has less than five elements
  • $\left|\varTheta_{00}(\tau)\right|
    = \left|\varTheta_{01}(\tau)\right|
    = \left|\varTheta_{10}(\tau)\right|
    = \sqrt[3]{2}\left|\eta(\tau)\right|$
  • The set $\left\{\left|\varTheta_{00}(\tau)\right|,
    \left|\varTheta_{01}(\tau)\right|,
    \left|\varTheta_{10}(\tau)\right|,
    \sqrt[3]{2}\left|\eta(\tau)\right|\right\}$ has less than three elements
  • $\left|\varTheta_{00}^4(\tau)+\varTheta_{01}^4(\tau)\right|
    = \left|\varTheta_{00}^4(\tau)+\varTheta_{10}^4(\tau)\right|
    = \left|\varTheta_{01}^4(\tau)-\varTheta_{10}^4(\tau)\right|$
  • $\lambda(\tau) \in \left\{-\omega^{\pm1}\right\}$
  • $\operatorname{E}_6\!\left(\frac{\tau}{2}\right)
    = \operatorname{E}_6\!\left(\frac{\tau+1}{2}\right)
    = 64\operatorname{E}_6(2\tau)
    = 22\operatorname{E}_6(\tau)$
  • The set $\left\{\operatorname{E}_6\!\left(\frac{\tau}{2}\right),
    \operatorname{E}_6\!\left(\frac{\tau+1}{2}\right),
    64\operatorname{E}_6(2\tau)\right\}$ has less than three elements

More elementary conditions, pairwise equivalent:

$$\begin{align}
&\exists\begin{pmatrix}a&b\\c&d\end{pmatrix}
\in\operatorname{SL}(2,\mathbb{Z})\colon\
\tau = \frac{{a\omega} + b}{{c\omega} + d}
= \frac{ac – bc + bd + \omega}{c^2 – cd + d^2}
\tag{1′}\\
&\exists N,u\in\mathbb{Z};\,N\mid(u^2-u+1)\colon\
\tau = \frac{u + \omega}{N}
\tag{2′}\\
&\frac{\Im\omega}{\Im\tau}\in\mathbb{Z}\text{ and }
\Re\tau\frac{\Im\omega}{\Im\tau}+\frac{1}{2}\in\mathbb{Z}\text{ and }
\frac{\Im\omega}{\Im(-\tau^{-1})}\in\mathbb{Z}
\tag{3′}
\end{align}$$

Proofs here are analoguous to those for $(1),(2),(3)$, but with slightly
more complicated expressions, e.g. $u^2-u+1$ instead of $u^2+1$ etc.
In the induction step, set $m$ to the rounded $\frac{2u-1}{2N}$
and $v=mN-u+1$.

One way to make an alternate description is: The orbit of $i$ (or $w$) is the set of all points that are fix points of some element of order $4$ (or $6$) in $SL_2(\Bbb Z)$. Then you can express the order condition as a condition about the trace.