$SL(2, \Bbb Z)$ has only one cusp

Question

Let $\Gamma$ be a congruence subgroup of $SL(2, \Bbb Z)$. A cusp is an equivalence of $\Bbb Q\cup\{\infty\}$ under $\Gamma$-action.

What's the meaning of "equivalence $\Bbb Q\cup\{\infty\}$ under $\Gamma$-action?

How to use the above definition to show that $SL(2,\Bbb Z)$ has only one cusp?

Answer 1

For your first question, as you probably know the group $SL(2,\mathbb Z)$ acts on the set $\mathbb Q \cup \{\infty\}$ by the following formula, known as "fractional linear" transformation: given $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL(2,\mathbb Z)$, and given $\frac{r}{s} \in \mathbb Q \cup \{\infty\}$ (with $r,s \in \mathbb Z$), we have $$M \cdot \frac{r}{s} = \frac{a \frac{r}{s}+b}{c \frac{r}{s}+d} = \frac{ar+bs}{cr+ds} $$

The "action equation" holds, namely $$M \cdot (N \cdot \frac{r}{s}) = MN \cdot \frac{r}{s} $$ where $MN$ denotes matrix multiplication. Also $$I \cdot \frac{r}{s} = \frac{r}{s} $$ Given a subgroup $\Gamma < SL(2,\mathbb Z)$ (for example, a congruence subgroup), the orbit of $\frac{r}{s}$ under the action of $\Gamma$ is the set $$\mathcal O_\Gamma \left(\frac{r}{s}\right) = \{M \cdot \frac{r}{s} \mid M \in \Gamma\} $$ As a consequence of the action equations, the set of orbits $$\{\mathcal O_\Gamma\bigl(\frac{r}{s}\bigr) \mid \frac{r}{s} \in \mathbb Q \cup \{\infty\}\} $$ is a partition of the set $\mathbb Q \cup \{\infty\}$, meaning that if two orbits are unequal then they are disjoint.

So to answer your first question, the meaning of "equivalence of $\mathbb Q \cup \{\infty\}$ under the $\Gamma$-action" is simply equivalence with respect to the orbit partition, two elements of $\mathbb Q \cup \{\infty\}$ being equivalent if and only if their orbits are equal, if and only if each one is in the orbit of the other.

To answer your second question, what you must prove is that the action of $SL(2,\mathbb Z)$ on $\mathbb Q \cup \{\infty\}$ has just one orbit. Equivalently, you can prove that for every $\frac{r}{s}$ there exists $M \in SL(2,\mathbb Z)$ such that $M \cdot \frac{r}{s} = \frac{0}{1}$. To do this, first cancel all common factors of $r$ and $s$ so that the fraction $\frac{r}{s}$ is in lowest terms, equivalently $$gcd(r,s)=1 $$ It follows that there exist $t,u \in \mathbb Z$ such that $$tr+su=1 $$ Let $$M = \begin{pmatrix} s & -r \\ t & u \end{pmatrix}\in SL(2,\mathbb{Z}) $$

Answer 2

Pick a fundamental domain (note that there are many choices but we typically pick the one with $|x| \leq 1/2$ and above the semi-circle $|z| = 1$). Your definition defines a cusp (in the fundamental domain of your choosing) as any elements of $\mathbb{Q} \cup \{ \infty \}$ that can be mapped into the fundamental domain of your choosing by an application of $\Gamma$.

To show that $SL(2, \mathbb{Z})$ has only one cusp pick the standard fundamental domain (to make things easy) and then compute for which $x \in \mathbb{Q} \cup \{\infty\}$ there exists integers $a,b,c,d$ with $ad - b c = 1$ such that $(a x + b) / (c x + d)$ lands in this fundamental domain. Clearly $x = \infty$ is OK because we can choose $a = d = 1$ and $b = c = 0$ and then we map $\infty$ to $\infty$ which is in the fundamental domain. So it remains to check that for all other $x \in \mathbb{Q}$ there is no integers $a,b,c,d$ with $ad - b c = 1$ such that $(a x + b) / (c x + d)$ is in the fundamental domain which is clear because $x$ is a real number therefore $(a x + b) / (c x + d)$ is also a real number, but your fundamental domain doesn't touch the real line.