Exercise 2.9.1

Question

Prove that for each $U \in \mathrm{SL}(2,\mathbb{Z})$ there exists $k \in \mathbb{Z}_{\geq 0}$ and exponents $s_i, t_i \in \Z,\ 1 \leq i \leq k$ such that
$$U = T^{s_1} S ^{t_1} T^{s_2} S^{t_2}T^{s_3} \cdots S^{t_k} T ^{s_k}.$$

Richard Ganaye · Accepted Answer

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\begin{proof}
We take a reduced positive definite form $f$ whose automorphism group is $\mathrm{Aut}^+(f) = \{-I, I\}$, for instance $f = (1,0,5)$ (see Theorem 2.5.10).

Define $g = f\cdot U^{-1}$. Then the positive definite form $g$ is properly equivalent to $f$, thus the reduced form in the proper class of $g$ is $f$. The reduction uses only matrices $S,T$ so that there is a sequence $A_1,\ldots,A_r$, where $A_i \in \{S,T, S^{-1}, T^{-1}\}$ such that
$g\cdot (A_1A_2\cdots A_r) = f$
(See chapter 5).
This gives
$$f \cdot(U^{-1} A_1\cdots A_r) = f,$$
so that 
$$U^{-1} A_1\cdots A_r \in \mathrm{Aut}^+(f) = \{-I, I\}.$$
If $U^{-1} A_1\cdots A_r = I$, then $U = A_1\cdots A_r \in \langle S, T angle$. Otherwise, $U^{-1} A_1\cdots A_r = -I$, and
$$U = -A_1\cdots A_r= S^2 A_1\cdots A_r \in \langle S, T angle.$$
We have proved that $S,T$ are generators of $\mathrm{SL}(2,\Z)$:
$$\mathrm{SL}(2,\Z) = \langle S, T angle.$$
In other words, for each $U \in \mathrm{SL}(2,\Z)$ there exists $k \in \N$ and exponents $s_i, t_i \in \Z,\ 1 \leq i \leq k$ such that
$$U = T^{s_1} S ^{t_1} T^{s_2} S^{t_2}T^{s_3} \cdots S^{t_k} T ^{s_k}.$$
(See a numerical example in Ex. 2.9.2.)
\end{proof}

Exercise 2.9.1

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Exercise 2.9.1

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