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LSpice
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The stabilizer of a point in the connected Lorentz group

$\DeclareMathOperator\SO{SO}$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 $$\eta=\begin{bmatrix}-1 & 0 & 0& \cdots & 0\\ 0 & 1 & 0 & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots &\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}.$$$$\eta=\begin{bmatrix}-1 & 0 & 0& \cdots & 0\\ 0 & 1 & 0 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}.$$

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

The stabilizer of the connected Lorentz group

$\DeclareMathOperator\SO{SO}$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 $$\eta=\begin{bmatrix}-1 & 0 & 0& \cdots & 0\\ 0 & 1 & 0 & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots &\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}.$$

Assume that $p=(E,E,0,\ldots ,0)$. 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, $\SO(n-1,1)^{\uparrow}_p =\{ f\in \SO(n-1,1)^{\uparrow}:fp=p\}$.

The stabilizer of a point in the connected Lorentz group

$\DeclareMathOperator\SO{SO}$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, $\SO(n-1,1)^{\uparrow}_p =\{ f\in \SO(n-1,1)^{\uparrow}:fp=p\}$.

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

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

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 $$\eta=\begin{bmatrix}-1 & 0 & 0& \cdots & 0\\ 0 & 1 & 0 & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots &\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}.$$

Assume that $p=(E,E,0,\ldots ,0)$. 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, $SO(n-1,1)^{\uparrow}_p =\{ fp=p:f\in SO(n-1,1)^{\uparrow}\}$.

$\DeclareMathOperator\SO{SO}$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 $$\eta=\begin{bmatrix}-1 & 0 & 0& \cdots & 0\\ 0 & 1 & 0 & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots &\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}.$$

Assume that $p=(E,E,0,\ldots ,0)$. 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, $\SO(n-1,1)^{\uparrow}_p =\{ f\in \SO(n-1,1)^{\uparrow}:fp=p\}$.

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Mahtab
Mahtab
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The stabilizer of the connected Lorentz group

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 $$\eta=\begin{bmatrix}-1 & 0 & 0& \cdots & 0\\ 0 & 1 & 0 & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots &\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}.$$

Assume that $p=(E,E,0,\ldots ,0)$. 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, $SO(n-1,1)^{\uparrow}_p =\{ fp=p:f\in SO(n-1,1)^{\uparrow}\}$.