The stabilizer of a point in the connected Lorentz group

Question

$\DeclareMathOperator\SO{SO}$For $n\geq 2$, the $n$-dimensional connected Lorentz group is defined as $$\SO(n-1,1)^{\uparrow}=\{f\in M_n (\mathbb{R}):f^T\eta f=\eta, \det (f)=1,f^0_0>0\}$$ where $$\eta=\begin{bmatrix}-1 & 0 & 0& \cdots & 0\\ 0 & 1 & 0 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}.$$

Assume that $p=(E,E,0,\dotsc,0)$. I would appreciate if someone could help me about proving this fact: The stabilizer of $p$ under $\SO(n-1,1)^{\uparrow}$ is $$\SO(n-2)\ltimes \mathbb{R}^{n-2},$$ the Euclidean group. By the stabilizer of $p$ under $\SO(n-1,1)^{\uparrow}$, I mean, $\SO(n-1,1)^{\uparrow}_p =\{ f\in \SO(n-1,1)^{\uparrow}:fp=p\}$.

Answer 1

One may prove by direct computation that in order to stabilize $p$, your matrix $f$ must be a block matrix of the form: $$ f = \begin{bmatrix} 1 & 0 & \vec{v}^T \\ 0 & 1 & -\vec{v}^T \\ && U \end{bmatrix} $$ where $\vec{v} \in \mathbb{R}^{n-2}$ and $U \in \mathrm{SO} \left( n-2 \right)$. This is because we must map the first two coordinates to themselves, but we can do any linear map on the $n-2$ remaining ones and they would still map to zero; and multiplying those coordinates by any vector $ \vec{v}^T $ still won't do anything, since they are zero. The group of all matrices of the above form is identified with the Euclidean group that you wrote down.