The Euclidean descent word multiplies back to the matrix

The Euclidean descent of a unimodular $2\times2$ matrix, written as a word in the generators. Let $S=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$ and $T^n=\left(\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\right)$, and for $M\in\mathrm{SL}(2,\mathbb Z)$ iterate the Euclidean step $M\mapsto (M\cdot T^{-n})\cdot S$, $n=M_{01}/M_{00}$, recording at each step the factor $S^{-1}T^{n}$; when $M_{00}$ reaches $0$ the two-element terminal word $S^{\pm1}T^{k}$ closes the expansion. The node records that word, BurauDescent.word M, and the statement proved here is

This is the combinatorial engine behind the descent section $\rho$ of the reduced braid quotient: it turns an arbitrary element of $\mathrm{SL}(2,\mathbb Z)$ into an explicit product of the two generators, on which the multiplication rules for $\rho$ are checked generator by generator.