Let $\mathbb{R}^{t,s}$ be the the vector space $\mathbb{R}^{t+s}$ equipped with the non degenerate indefinite scalar product $\eta$ that satisfies:
$$\eta(e_i,e_i)=-1\text{ for $1\leq i\leq t$}\qquad \eta(e_i,e_i)=1\text{ for $t+1\leq i\leq s+t$}$$ and: $$\eta(e_i,e_j)=0\text{ for }i\neq j$$ Then the Pseudo Orthogonal group is the Lie group defined by the set $$O(t,s)=\{A\in GL_{s+t}(\mathbb{R}):A\eta A^T=\eta\}$$ The special pseudo orthogonal group is given by: $$SO(t,s)=\{A\in O(t,s):\det(A)=1\}$$ Examine the subspaces $V_-$ and $V_+$ spanned by $\{e_1,\dots,e_t\}$ and $\{e_{t+1},e_{t+s}\}$ respectively. We say that $A\in O(t,s)$ preserves the time orientation of $\mathbb{R}^{t,s}$ if and only if $A|_{V_-}$ is and orientation preserving isomorphism onto its image. One can check that this then implies that if $A$ is written as the block matrix: $$A=\begin{pmatrix} A_{tt}&A_{ts}\\ A_{ts}&A_{ss} \end{pmatrix}$$ where $A_{ij}$ is an $i$ by $j$ matrix that $A$ preserves the time orientation if and only if $\det(A_{tt})>0$. We thus define the special orthochronus pseudo orthogonal group as: $$SO^+(t,s)=\{A\in SO(t,s):\det(A_{tt})>0\}$$ Apparently, this is the connected component of the identity. I have zero idea how to show this thought. In the Euclidean norm case I can use the face that as smooth manifolds: $$SO(n)/SO(n-1)\cong S^{n-1}$$ and proceed by induction since $SO(2)\cong S^1$ and $SO(1)=\{1\}$. I would ideally like to do something similar here, by looking at the level set $\eta^{-1}(1)$, and then somehow argue that isotropy group of any $p\in\eta^{-1}(1)$ is some other $SO^(t',s')$ such that $t'<t$ and $s'<s$, but I can't see the path forward.
I guess one could also show that as manifolds $SO^+(t,s)\cong SO(t)\times SO(s)\times\mathbb{R}^{rs}$ but this feels like a large endeavor for a small result.
Is there perhaps a way to show this directly?