9
$\begingroup$

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

$\endgroup$
3
  • 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

13
$\begingroup$
  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 $$ A = MT = T^4ST^2ST^3ST^2ST^2ST. $$

$\endgroup$
2
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.