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 $A(\infty) = p/q$?

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Nov 8, 2021 at 19:37
  • 1
    $\begingroup$ Question (2) is trivial: you can assume $(p,q)=1$. Choose $a,b$ such that $ap+bq=1$ and take $A=\begin{pmatrix} p & -b\\ q & a \end{pmatrix}$. $\endgroup$
    – abx
    Commented Nov 8, 2021 at 19:38
  • $\begingroup$ You can read about it in Oleg Karpenkov: Geometry of Continued Fractions. $\endgroup$ Commented Nov 9, 2021 at 7:20

1 Answer 1

  1. See Remark 2.2 here.

  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 $$ A = MT = T^4ST^2ST^3ST^2ST^2ST. $$

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    $\begingroup$ 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 $SL(2,\mathbb Z)$. :) $\endgroup$ Commented Nov 8, 2021 at 22:40
  • $\begingroup$ Thank you very much - this is exactly what I was looking for and your explanation was very helpful! $\endgroup$
    – Sprotte
    Commented Nov 9, 2021 at 18:55

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