Coset Representatives for $U(2,\mathbb{Z})$ in $SL(2,\mathbb{Z})$

Question

If $SL(2,\mathbb{Z})$ denotes the modular group and if $U(2,\mathbb{Z})$ denotes the subgroup of upper triangular matrices with diagonal entries equal to $1$, then is there a (nice) description of distinct coset representatives for $U(2,\mathbb{Z})$ inside $SL(2,\mathbb{Z})$?

I tried using the free product decomposition of $SL(2,\mathbb{Z})$ in terms of $S$ and $ST$ (where $S$ is the "inversion" and $T$ is the "translation"), but that is not leading me anywhere. Any help would be appreciated.

Answer 1

Write down the product

$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & \lambda \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & a\lambda +b \\ c & c\lambda+d\end{bmatrix}. $$

You can pick representatives based on $[\begin{smallmatrix}a\lambda+b\\ c\lambda+d\end{smallmatrix}]$ having minimal-size. If you need to break ties, maybe find the one that's closest to a specific ray (clockwise or counterclockwise). I wouldn't be surprised if there are better ones for depending on what your objective is. Alternatively, you could stipulate $a\lambda+b$ is within a certain set of coset representatives for integers mod $a$, like $\{0,1,\cdots,|a|-1\}$ or $[-|a|/2,|a|/2)$, or similarly for the lower right corner instead, but then you'd have to stipulate what to do when the modulus is $0$.

Answer 2

The following argument is inspired from @runway44's answer. We shall answer the question inside $PSL(2,\mathbb{Z})$.

Suppose we have two $SL(2,\mathbb{Z})$ matrices given by $$ \gamma_1 = \begin{pmatrix}a&b\\c&d\end{pmatrix}\mbox{ and }\gamma_2 = \begin{pmatrix}a'&b'\\c'&d'\end{pmatrix}. $$ Then $\gamma_1$ and $\gamma_2$ are in the same coset modulo $U(2,\mathbb{Z})$ if and only if $\gamma_1\gamma_2^{-1} \in U(2,\mathbb{Z})$, that is $$ \begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}d'&-b'\\-c'&a'\end{pmatrix} \in U(2,\mathbb{Z}). $$ In particular we see that $cd' = c'd$. If $dd'\neq 0$, then this can be rewritten as $c/d = c'/d'$. Since the pairs $(c,d)$ and $(c',d')$ are coprime, we have $c=c'$ and $d=d'$. Now suppose without loss of generality that $d=0$. This forces $c$ to be $1$. This give us that $d'=0$ and therefore $c'=1$.

I think this should solve the question.