# $\text{SL}_2(\mathbb Z)$ acts on upper plane $\mathbb H$. What kind of points have non-trivial stabilizer? And how many orbits are there?

$$\text{SL}_2(\mathbb Z)$$ acts on upper plane $$\mathbb H= \{z \in \mathbb{C} | \Im(z) > 0 \}$$ via Mobius transformation. $$\text{ For } \gamma =\begin{bmatrix} a &b \\c&d \end{bmatrix} \in\text{SL}_2(\mathbb Z), \ \gamma z =\begin{bmatrix} a &b \\c&d \end{bmatrix}\cdot z = \frac{az +b}{cz+d}$$

Stabilizer of $$z$$ means set $$\{\gamma \in \text{SL}_2(\Bbb Z), \gamma z=z\}$$.

I want to know what kind of points have non-trivial stabilizer and the number of orbits.

My effort: For $$z \in \Bbb H$$, suppose $$z = x + i y$$.

$$\text{For } \gamma =\begin{bmatrix} a &b \\c&d \end{bmatrix}, \gamma z =\frac{az+b}{cz+d}=z\iff az+b=cz^2+dz$$

$$\iff ax+ayi+b = cx^2-cy^2+2cxyi + dx +dyi$$ $$\iff ay=2cxy+dy\ \&\ ax+b=cx^2-cy^2+dx$$ $$\iff a=2cx+d \ \&\ ax+b=cx^2-cy^2+dx$$ $$\implies b=-c(x^2+y^2),\ \gamma =\begin{bmatrix} 2cx+d &-c(x^2+y^2) \\c&d \end{bmatrix}.$$

$$\gamma \in \text{SL}_2(\Bbb Z),\ (2cx+d)\times d-(-c(x^2+y^2))\times c=1$$ $$\implies (cx+d)^2+(cy)^2=1.$$

Then how to proceed? Thanks in advance.

I haven't learnt modular form yet, and I don't know if these help:

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

Orbit of complex unit $$i$$ under moebius tranformation in $$SL_2(\mathbb{Z})$$

Edit:

GTM$$105$$, Serge Lang, SL$$_2(\mathbb R)$$ might help.

Comment:

It's an exercise of section about group action on set, and before this section the book just introduces definition and basic property of group, so this problem is a bit more difficult than I thought.

• I think it's a mistake to write $z$ as $x+iy$. Instead the quadratic formula works fine. Using $ad-bc=1$ you get the discriminant of your quadratic is $(a+d)^2-4$. You need this negative to have a point in the upper half plane, so $|a+d|<2$. Since $a$ and $d$ are integers, this gives finitely many cases to check. – Hempelicious Dec 10 '18 at 13:27

I just realized that the points with stabilizers of order $$2$$ are in the same orbit, and the points with stabilizers of order $$3$$ are in the same orbit. There are therefore exactly two orbits with non-trivial stabilizers. The previous answer is too long and the browser is becoming slow, so I have to add another answer.

In the case of stabilizers of order $$2$$, recall that $$c\mid a^2+1$$. Using the knowledge about Gaussian integers (in particular, the unique factorization property), $$c$$ can be written as $$r^2+s^2=(r+si)(r-si)$$ for some $$r,s\in\Bbb{Z}$$ such that $$r-si$$ divides $$a+i$$. That is, $$\frac{a+i}{r-si}=p+qi$$ for some $$p,q\in\Bbb{Z}$$. Then observe that $$\begin{pmatrix}q&p\\s&r\end{pmatrix}\cdot i=\frac{p+qi}{r+si}=\frac{(p+qi)(r-si)}{r^2+s^2}=\frac{a+i}{r^2+s^2}=\frac{a}{c}+\frac{i}{c}.$$ Thus, $$\frac{a}{c}+\frac{i}{c}$$ is in the orbit of $$i$$.

In the case of stabilizers of order $$3$$, we can assume WLOG that we are in the case $$d=-a+1$$, where $$c\mid a^2-a+1$$. Let $$\omega$$ denote $$-\frac12+\frac{\sqrt{3}i}{2}$$. Using the knowledge about Eisenstein integers (in particular, the unique factorization property), $$c$$ can be written as $$r^2-rs+s^2=(r+s\omega)(r+s\overline{\omega})$$ for some $$r,s\in \Bbb Z$$ such that $$r+s\overline{\omega}$$ divides $$a+\omega$$. That is, $$\frac{a+\omega}{r+s\overline{\omega}}=p+q\omega$$ for some $$p,q\in \Bbb Z$$. Finally, observe that $$\begin{pmatrix}q&p\\s&r\end{pmatrix}\cdot\omega=\frac{p+q\omega}{r+s\omega}=\frac{(p+q\omega)(r+s\overline{\omega})}{r^2-rs+s^2}=\frac{a+\omega}{c}=\frac{2a-1}{2c}+\frac{\sqrt{3}i}{2c}.$$ Thus, $$\frac{2a-1}{2c}+\frac{\sqrt{3}i}{2c}$$ is in the orbit of $$\omega$$.

The group $$\mathrm{PSL}_2(\mathbf{Z})$$ is a free product $$C_2\ast C_3$$, and hence its nontrivial torsion elements have order 2 or 3, and has 1 conjugacy class of cyclic subgroup of order 2 (a representative is $$z\mapsto -1/z$$ with fixed point $$i$$), 1 conjugacy class of cyclic subgroup of order 3 (a representative being generated by $$z\mapsto -1/(1+z)$$, fixing $$j$$, the unique root of $$1+z+z^2$$ in the upper half-plane). The elements of infinite order have no fixed point on $$\mathbf{H}$$.

So the points with nontrivial stabilizer are those in the orbit of $$i$$ and the orbit of $$j$$. They are not in the same orbit since otherwise the stabilizer should contain an element of order 6.

• Thanks. And could you please explain why consider $\text{PSL}_2(\Bbb Z)$ and the relation to points with nontrivial stabilizer in $\Bbb H$? – Andrews Dec 10 '18 at 0:30
• @Andrews: the center of $SL(2,\mathbb{Z})$ acts trivially on the half-plane, so you usually mod it out. – Hempelicious Dec 10 '18 at 4:43

At some point you have $$b=-c(x^2+y^2)$$ and $$a=2cx+d$$. If $$c=0$$, then $$b=0$$, in which case the element $$\begin{pmatrix} a&b\\c&d\end{pmatrix}$$ of the projective special linear group $$\operatorname{PSL}_2(\Bbb Z)$$ of this form is the identity element (recalling that $$\operatorname{PSL}_2(\Bbb Z)=\operatorname{SL}_2(\Bbb{Z})/\{\pm I\}$$), so we are not interested in this case. So, $$c$$ is assumed to be non-zero, and can without loss of generality assumed to be positive.

That is, $$x=\frac{a-d}{2c}$$, so that $$-\frac{b}{c}=x^2+y^2=\frac{(a-d)^2}{4c^2}+y^2.$$ Since $$y>0$$, we have $$y=\frac{\sqrt{4-(a+d)^2}}{2c},$$ which means $$d=-a$$ or $$d=-a\pm1$$.

In the case $$d=-a$$, we must have $$x=\frac{a}{c}$$ and $$y=\frac{1}{c}$$. For a given point $$z=x+yi$$ of this form that is fixed by a non-trivial $$\gamma\in\operatorname{PSL}_2(\Bbb{Z})$$, we can assume that $$\gamma=\begin{pmatrix}a&b\\c&-a\end{pmatrix}\wedge bc=-(a^2+1).\tag{1}$$ That is, $$\gamma^2$$ is the identity of $$\operatorname{PSL}_2(\Bbb{Z})$$. The points in $$\Bbb H$$ with a stabilizer of order $$2$$ are of the form $$z=x+yi$$, where $$x=\frac{a}{c}\wedge y=\frac{1}{2c}$$ for some integer $$a$$ and for some integer $$c>0$$, such that $$a^2+1$$ is divisible by $$c$$ (so that there exists $$b\in \Bbb Z$$ such that $$bc=-(a^2+1)$$). The number of orbits for a given $$a$$ is $$\sigma_0(a^2+1)$$, where $$\sigma_0$$ is the divisor counting function.

For the case $$d=-a+1$$, we see that $$x=\frac{2a-1}{2c}$$ and $$y=\frac{\sqrt{3}}{2c}$$. Then, $$z=x+yi$$ is fixed by $$\gamma=\begin{pmatrix}a&b\\c&-a+1\end{pmatrix}\wedge bc=-(a^2-a+1).\tag{2}$$ Note that $$\gamma^3$$ is the identity of $$\operatorname{PSL}_2(\Bbb{Z})$$. Thus, for a given $$a$$, there are corresponding $$\sigma_0(a^2-a+1)$$ points $$z\in \Bbb H$$.

For the case $$d=-a-1$$, we see that $$x=\frac{2a+1}{2c}$$ and $$y=\frac{\sqrt{3}}{2c}$$. Then, $$z=x+yi$$ is fixed by $$\gamma=\begin{pmatrix}a&b\\c&-a-1\end{pmatrix}\wedge bc=-(a^2+a+1).\tag{3}$$ Note that $$\gamma^3$$ is the identity of $$\operatorname{PSL}_2(\Bbb{Z})$$. Thus, for a given $$a$$, there are corresponding $$\sigma_0(a^2+a+1)$$ points $$z\in \Bbb H$$. (It can be seen that this case is identical to the previous case via the transformation $$a\mapsto a-1$$.)

For example, $$x=\frac{3}{5}$$ and $$y=\frac{1}{5}$$ fit the bill (with $$a=3$$, $$b=-2$$, $$c=5$$, and $$d=-a=-3$$) for a point with stabilizer of order $$2$$, with $$\gamma=\begin{pmatrix}3&-2\\5&-3\end{pmatrix}.$$ For the same $$a=3$$, there are three more points with stablizers of order $$2$$, i.e., with $$c=1$$, $$c=2$$, and $$c=10$$. That is, for $$a=3$$, we have in total four points with non-trivial stabilizers $$\gamma$$ of order $$2$$: $$z=3+i$$ with $$\gamma=\begin{pmatrix}3&-10\\1&-3\end{pmatrix}$$, $$z=\frac{3}{2}+\frac{i}{2}$$ with $$\gamma=\begin{pmatrix}3&-5\\2&-3\end{pmatrix}$$, $$z=\frac{3}{5}+\frac{i}{5}$$ with $$\gamma=\begin{pmatrix}3&-2\\5&-3\end{pmatrix}$$, and $$z=\frac{3}{10}+\frac{i}{10}$$ with $$\gamma=\begin{pmatrix}3&-1\\10&-3\end{pmatrix}$$.

For the same $$a=3$$, there are also four points with stabilizers of order $$3$$. That is, for $$a=3$$, we have four points with non-trivial stabilizers $$\gamma$$ of order $$3$$: $$z=\frac52+\frac{\sqrt{3}i}{2}$$ with $$\gamma=\begin{pmatrix}3&-7\\1&-2\end{pmatrix}$$, $$z=\frac{5}{14}+\frac{\sqrt{13}i}{14}$$ with $$\gamma=\begin{pmatrix}3&-1\\7&-2\end{pmatrix}$$, $$z=\frac{7}{2}+\frac{\sqrt{3}i}{2}$$ with $$\gamma=\begin{pmatrix}3&-13\\1&-4\end{pmatrix}$$, and $$z=\frac{7}{26}+\frac{\sqrt{3}i}{26}$$ with $$\gamma=\begin{pmatrix}3&-1\\13&-4\end{pmatrix}$$.

In conclusion, there are three kinds of points $$z\in\Bbb H$$---those with trivial stabilizers, those with stabilizers of order $$2$$, and those with stabilizers of order $$3$$. The stablizers in non-trivial cases are generated by $$\gamma$$ given in (1) and (2).