I need to show that $SO_{2,1}^+$ is path connected.
First of all here are a couple of definitions we are using in the lecture:
Let $n=r+s$ for $r,s \in \mathbb{N}$.
$$I_{r,s} := diag(\underbrace{ 1,\ldots,1}_{\text{$r$ times}},\underbrace{ -1,\ldots,-1}_{\text{$s$ times}})$$ $$O_{r,s} := \{M \in M_n \mid M^TI_{r,s}M = I_{r,s} \}$$ $$SO_{r,s} := \{M \in M_n \mid M^TI_{r,s}M = I_{r,s} \text{ and } det(M) = 1 \}$$ $$SO_{r,s}^+ := \{M \in M_n \mid M \in SO_{r,s} \text{ and } z > 0 \text{ for } (x,y,z)^T: = M\cdot(0,0,1)^T \}$$
Remark: Everything here is over $\mathbb{R}$.
I have already seen in the lecture that $SO_2$ is path-connected, but I do not understand how I should prove this for $SO_{r,s}^+$. Could you please give me a hint?