I am working on the following exercise:
Let
$$r_t := \begin{pmatrix} 1 &0 &0 \\ 0 &\cosh t & \sinh t \\ 0 &\sinh t & \cosh t \end{pmatrix}. $$ Show that for $M \in SO_{2,1}^+$ there is a rotation $\rho$ (in the common sense) around the $z$-axis and some $t \in \mathbb{R}$ such that $r_t \circ \rho \circ M \cdot (0,0,1)^T = (0,0,1)^T$.
Here are the definitions we are using in the lecture:
$$I_{r,s} := diag(\underbrace{ 1,\ldots,1}_{\text{$r$ times}},\underbrace{ -1,\ldots,-1}_{\text{$s$ times}})$$ $$SO_{r,s} := \{M \in M_n \mid M^TI_{r,s}M = I_{r,s} \text{ and } det(M) = 1 \}$$ $$SO_{r,s}^+ := \{M \in M_n \mid M \in SO_{r,s} \text{ and } z > 0 \text{ for } (x,y,z)^T: = M\cdot(0,0,1)^T \}$$
I do not see how I could show that. Could you please give me a hint?
based on pages 23 and 124 in Magnus, Noneuclidean Tesselations and Their Groups.
You asked for $M^T D M = D,$ where $D$ is the three by three diagonal matrix with diagonal elements $1,1,-1$ in that order. For some real $a,b,c,d$ with $ad-bc=1,$ we get
$$ M = \left( \begin{array}{ccc} \frac{1}{2} \left( a^2 - b^2 - c^2 + d^2 \right)&ab+cd&\frac{1}{2} \left( a^2 + b^2 - c^2 - d^2 \right) \\ ac+bd&ad+bc&ac-bd \\ \frac{1}{2} \left( a^2 - b^2 + c^2 - d^2 \right)&ab-cd&\frac{1}{2} \left( a^2 + b^2 + c^2 + d^2 \right) \\ \end{array} \right). $$
Furthermore, all $M$ arise this way. The reduction they are asking amounts to taking the two by two matrix $ \left( \begin{array}{rr} a&b \\ c & d \end{array} \right) $ in $SL_2 \mathbb R$ back to the identity, in three steps.