$\text{SL}_2(\mathbb{Z})$ and continued fractions?

I know the following facts: $$\text{SL}_2(\mathbb{Z})$$ is generated by everyone's favorite matrices $$\begin{equation*} S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \end{equation*}$$ and $$\begin{equation*} T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \end{equation*}$$ and $$\text{SL}_2(\mathbb{Z})$$ acts transitively on $$\mathbb{P}^1(\mathbb{Q})$$.

I have been told that the answer to the following questions have something to do with continued fraction expansions, but I would like to find a reference/pointers to the particulars.

(1) How do I write a given matrix $$A \in \text{SL}_2(\mathbb{Z})$$ in terms of $$S$$ and $$T$$?

(2) For a given $$p/q \in \mathbb{Q}$$, how do I find an element $$A \in \text{SL}_2(\mathbb{Z})$$ with $$A(\infty) = p/q$$?

• Better take $x_+=T$ and $x_-=T^\top$ as the generators and look at what a multiplication by $x_+^k$ or $x_-^k$ on the left does to the first column of the matrix. Apply Euclidean algorithm. Commented Nov 8, 2021 at 19:37
• Question (2) is trivial: you can assume $(p,q)=1$. Choose $a,b$ such that $ap+bq=1$ and take $A=\begin{pmatrix} p & -b\\ q & a \end{pmatrix}$.
– abx
Commented Nov 8, 2021 at 19:38
• You can read about it in Oleg Karpenkov: Geometry of Continued Fractions. Commented Nov 9, 2021 at 7:20

2. If you expand $$p/q$$ into a continued fraction then the successive convergents, as columns of a $$2 \times 2$$ matrix, have determinant $$\pm 1$$. Provided $$p/q$$ is in reduced form and $$q > 0$$, the last convergent $$p_n/q_n$$ in the continued fraction for $$p/q$$ will have $$p_n = p$$ and $$q_n = q$$. Let the second to last convergent be $$p_{n-1}/q_{n-1}$$. Then $$p_{n-1}q_n - q_{n-1}p_n = \pm 1$$, and it is easy to pass from such an equation to a $$2 \times 2$$ integral matrix with determinant $$1$$ and first column $$\binom{p}{q}$$.
Let's illustrate these algorithms by starting with the second question on the rational number $$p/q = 37/11$$, which is in reduced form. What is a matrix in $${\rm SL}_2(\mathbf Z)$$ with first column $$\binom{37}{11}$$? The continued fraction of $$37/11$$ is $$[3,2,1,3]$$ and the successive convergents in this continued fraction are $$3/1$$, $$7/2$$, $$10/3$$, and $$37/11$$. Using the last two convergents, we obtain $$\det(\begin{smallmatrix}10&37\\3&11\end{smallmatrix}) = -1$$, so $$11 \cdot 10 - 37 \cdot 3 = -1$$, so $$37(3) - 11(10) = 1$$. Thus the matrix $$A = (\begin{smallmatrix}37&10\\11&3\end{smallmatrix})$$ is in $${\rm SL}_2(\mathbf Z)$$ with first column $$\binom{37}{11}$$, so $$A(\infty) = 37/11$$.
Next, for the matrix $$A = (\begin{smallmatrix}37&10\\11&3\end{smallmatrix})$$ in $${\rm SL}_2(\mathbf Z)$$, how can we write $$A$$ in terms of $$S$$ and $$T$$? For this we will use a continued fraction for the first column ratio $$37/11$$ using nearest integers from above rather than from below: $$37/11 = 4 - 1/(2 - 1/(3 - 1/(2 - 1/2)))$$. Using the entries $$4, 2, 3, 2$$, and $$2$$, form the matrix product $$M = T^4ST^2ST^3ST^2ST^2S$$. Its first column will be $$\binom{37}{11}$$ but its second column might not match that of $$A$$, so $$M^{-1}A$$ will be a power of $$T$$. Indeed, $$M = (\begin{smallmatrix}37&-27\\11&-8\end{smallmatrix})$$ and $$M^{-1}A = (\begin{smallmatrix}1&1\\0&1\end{smallmatrix}) = T$$, so $$A = MT = T^4ST^2ST^3ST^2ST^2ST.$$
• And, unsurprisingly, the "word" in $S,T$ that moves a given point $z_o$ in the upper half-plane into the standard fundamental domain, (up to normalizations and inverses...) moves the copy of the fundamental domain containing $z_o$ with cusp $p/q$ that copy to the standard one, and $p/q$ to $i\infty$, etc. I still remember, ages ago, being fairly amazed, but immediately convinced, when Nick Katz off-handedly remarked that the theory of continued fractions was about $SL(2,\mathbb Z)$. :) Commented Nov 8, 2021 at 22:40